begin








definition
  mode set;
end;

definition let x,y be set;
  pred x = y;
  reflexivity;
  symmetry;
end;

notation let x,y be set;
  antonym x <> y for x = y;
end;

definition let x,X be set;
  pred x in X;
  asymmetry;
end;















begin

 reserve x,y,z,u,N,M,X,Y,Z for set;

theorem 
 (for x holds x in X iff x in Y) implies X = Y;

 definition let y; func { y } -> set means
  x in it iff x = y;
  correctness;
  let z; func { y, z } -> set means
   x in it iff x = y or x = z;
   correctness;
   commutativity;
 end;

 definition let X,Y;
  pred X c= Y means
   x in X implies x in Y;
  reflexivity;
 end;

 definition let X;
   func union X -> set means
    x in it iff ex Y st x in Y & Y in X;
   correctness;
 end;

theorem 
 x in X implies ex Y st Y in X & not ex x st x in X & x in Y;

 scheme Fraenkel { A()-> set, P[set, set] }:
  ex X st for x holds x in X iff ex y st y in A() & P[y,x]
  provided for x,y,z st P[x,y] & P[x,z] holds y = z
  proof
    thus thesis;
  end;

  definition let x,y;
  func [x,y] equals
   { { x,y }, { x } };
   correctness;
 end;

 definition let X,Y;
  pred X,Y are_equipotent means
    ex Z st
       (for x st x in X ex y st y in Y & [x,y] in Z) &
       (for y st y in Y ex x st x in X & [x,y] in Z) &
        for x,y,z,u st [x,y] in Z & [z,u] in Z holds x = z iff y = u;
 end;

























theorem
  ex M st N in M &
     (for X,Y holds X in M & Y c= X implies Y in M) &
     (for X st X in M ex Z st Z in M & for Y st Y c= X holds Y in Z) &
     (for X holds X c= M implies X,M are_equipotent or X in M);














begin

reserve X, Y, Z, x, y, z for set;

scheme
  Separation { A()-> set, P[set] } :
  ex X being set st for x being set holds x in X iff x in A() & P[x]
  proof
  defpred Q[set,set] means $1 = $2 & P[$2];
A1: for x,y,z st Q[x,y] & Q[x,z] holds y = z;
  consider X such that
A2: for x holds x in X iff ex y st y in A() & Q[y,x] from TARSKI:sch 1(A1);
  take X;
  let x;
  x in X iff ex y st y in A() & y = x & P[x] by A2;
  hence thesis;
end;

definition
  let X be set;
  attr X is empty means
  :Def1:
  not ex x being set st x in X;
end;

registration
  cluster empty for set;
  existence
  proof
    defpred P[set] means contradiction;
    consider Y such that
A1: x in Y iff x in the set & P[x] from Separation;
    take Y;
    thus not ex x being set st x in Y by A1;
  end;
end;

definition
  func {} -> set equals
  the empty set;
  coherence;
  let X,Y be set;
  func X \/ Y -> set means
  :Def3:
  x in it iff x in X or x in Y;
  existence
  proof
    take union {X,Y};
    let x;
    thus x in union {X,Y} implies x in X or x in Y
    proof
      assume x in union {X,Y};
      then ex Z being set st x in Z & Z in {X,Y} by TARSKI:def 4;
      hence thesis by TARSKI:def 2;
    end;
    X in {X,Y} & Y in {X,Y} by TARSKI:def 2;
    hence thesis by TARSKI:def 4;
  end;
  uniqueness
  proof
    let A1, A2 be set such that
A3: x in A1 iff x in X or x in Y and
A4: x in A2 iff x in X or x in Y;
    now
      let x;
      x in A1 iff x in X or x in Y by A3;
      hence x in A1 iff x in A2 by A4;
    end;
    hence thesis by TARSKI:1;
  end;
  commutativity;
  idempotence;
  func X /\ Y -> set means
  :Def4:
  x in it iff x in X & x in Y;
  existence
  proof
    defpred P[set] means $1 in Y;
    thus ex Z being set st for x holds x in Z iff x in X & P[x] from
    Separation;
  end;
  uniqueness
  proof
    let A1, A2 be set such that
A5: x in A1 iff x in X & x in Y and
A6: x in A2 iff x in X & x in Y;
    now
      let x;
      x in A1 iff x in X & x in Y by A5;
      hence x in A1 iff x in A2 by A6;
    end;
    hence thesis by TARSKI:1;
  end;
  commutativity;
  idempotence;
  func X \ Y -> set means
  :Def5:
  x in it iff x in X & not x in Y;
  existence
  proof
    defpred P[set] means not $1 in Y;
    thus ex Z being set st for x holds x in Z iff x in X & P[x] from
    Separation;
  end;
  uniqueness
  proof
    let A1, A2 be set such that
A7: x in A1 iff x in X & not x in Y and
A8: x in A2 iff x in X & not x in Y;
    now
      let x;
      x in A1 iff x in X & not x in Y by A7;
      hence x in A1 iff x in A2 by A8;
    end;
    hence thesis by TARSKI:1;
  end;
end;

definition
  let X, Y be set;
  func X \+\ Y -> set equals
  (X \ Y) \/ (Y \ X);
  correctness;
  commutativity;
  pred X misses Y means
  :Def7:
  X /\ Y = {};
  symmetry;
  pred X c< Y means
  :Def8:
  X c= Y & X <> Y;
  irreflexivity;
  asymmetry
  proof
    let X,Y;
    assume
A1: X c= Y;
    assume X <> Y;
    assume Y c= X;
    then for x holds x in X iff x in Y by A1,TARSKI:def 3;
    hence thesis by TARSKI:1;
  end;
  pred X,Y are_c=-comparable means
  X c= Y or Y c= X;
  reflexivity;
  symmetry;
  redefine pred X = Y means
  X c= Y & Y c= X;
  compatibility
  proof
    thus X = Y implies X c= Y & Y c= X;
    assume X c= Y & Y c= X;
    then for x holds x in X iff x in Y by TARSKI:def 3;
    hence thesis by TARSKI:1;
  end;
end;

notation
  let X, Y be set;
  antonym X meets Y for X misses Y;
end;

theorem
  x in X \+\ Y iff not (x in X iff x in Y)
proof
  x in X \+\ Y iff x in X \ Y or x in Y \ X by Def3;
  hence thesis by Def5;
end;

theorem
  (for x holds not x in X iff (x in Y iff x in Z)) implies X = Y \+\ Z
proof
  assume
A1: not x in X iff (x in Y iff x in Z);
  now
    let x;
    x in X iff x in Y & not x in Z or x in Z & not x in Y by A1;
    then x in X iff x in Y \ Z or x in Z \ Y by Def5;
    hence x in X iff x in Y \+\ Z by Def3;
  end;
  hence thesis by TARSKI:1;
end;

registration
  cluster {} -> empty;
  coherence;
end;

registration
  let x1 be set;
  cluster { x1 } -> non empty;
  coherence
  proof
    x1 in {x1} by TARSKI:def 1;
    hence ex x being set st x in {x1};
  end;
  let x2 be set;
  cluster { x1, x2 } -> non empty;
  coherence
  proof
    x1 in {x1,x2} by TARSKI:def 2;
    hence ex x being set st x in {x1,x2};
  end;
end;

registration
  cluster non empty for set;
  existence
  proof
    take { the set };
    thus thesis;
  end;
end;

registration
  let D be non empty set, X be set;
  cluster D \/ X -> non empty;
  coherence
  proof
    consider x being set such that
A1: x in D by Def1;
    x in D \/ X by A1,Def3;
    hence ex x being set st x in D \/ X;
  end;
  cluster X \/ D -> non empty;
  coherence;
end;

Lm1: X is empty implies X = {}
proof
  assume not ex x st x in X;
  then x in {} iff x in X by Def1;
  hence thesis by TARSKI:1;
end;

theorem Th3:
  X meets Y iff ex x st x in X & x in Y
proof
  hereby
    assume X meets Y;
    then X /\ Y <> {} by Def7;
    then X /\ Y is not empty by Lm1;
    then consider x such that
A1: x in X /\ Y by Def1;
    take x;
    thus x in X & x in Y by A1,Def4;
  end;
  given x such that
A2: x in X & x in Y;
  x in X /\ Y by A2,Def4;
  then X /\ Y <> {} by Def1;
  hence thesis by Def7;
end;

theorem
  X meets Y iff ex x st x in X /\ Y
proof
  hereby
    assume X meets Y;
    then X /\ Y <> {} by Def7;
    then X /\ Y is not empty by Lm1;
    then consider x such that
A1: x in X /\ Y by Def1;
    take x;
    thus x in X /\ Y by A1;
  end;
  assume ex x st x in X /\ Y;
  then X /\ Y <> {} by Def1;
  hence thesis by Def7;
end;

theorem
  X misses Y & x in X \/ Y implies x in X & not x in Y or x in Y & not x in X
  by Def3,Th3;

scheme
  Extensionality { X,Y() -> set, P[set] } : X() = Y()
provided
A1: for x holds x in X() iff P[x] and
A2: for x holds x in Y() iff P[x]
proof
A3: x in Y() implies x in X()
  proof
    assume x in Y();
    then P[x] by A2;
    hence thesis by A1;
  end;
  x in X() implies x in Y()
  proof
    assume x in X();
    then P[x] by A1;
    hence thesis by A2;
  end;
  hence thesis by A3,TARSKI:1;
end;

scheme
  SetEq { P[set] } :
  for X1,X2 being set st
    (for x being set holds x in X1 iff P[x]) &
    (for x being set holds x in X2 iff P[x]) holds X1 = X2
  proof
  let X1,X2 be set such that
A1: for x being set holds x in X1 iff P[x] and
A2: for x being set holds x in X2 iff P[x];
  thus thesis from Extensionality(A1,A2);
end;

begin 


theorem
  X c< Y implies ex x st x in Y & not x in X
proof
  assume X c< Y;
  then (not for x holds x in X iff x in Y) & X c= Y by Def8,TARSKI:1;
  hence thesis by TARSKI:def 3;
end;



theorem
  X <> {} implies ex x st x in X
proof
  assume
A1: X <> {};
  assume not ex x st x in X;
  then X is empty by Def1;
  hence thesis by A1,Lm1;
end;














begin








theorem
  for X being set holds X \/ {} = X
proof
  let X be set;
  thus X \/ {} c= X
  proof
    let x be set;
    assume x in X \/ {};
    then x in X or x in {} by XBOOLE_0:def 3;
    hence thesis by XBOOLE_0:def 1;
  end;
  let x be set;
  assume x in X;
  hence thesis by XBOOLE_0:def 3;
end;

theorem
  for X being set holds X /\ {} = {}
proof
  let X be set;
  thus X /\ {} c= {}
  proof
    let x be set;
    assume x in X /\ {};
    hence thesis by XBOOLE_0:def 4;
  end;
  let x be set;
  assume x in {};
  hence thesis by XBOOLE_0:def 1;
end;

theorem
  for X being set holds X \ {} = X
proof
  let X be set;
  thus X \ {} c= X
  proof
    let x be set;
    assume x in X \ {};
    hence thesis by XBOOLE_0:def 5;
  end;
  let x be set;
A1: not x in {} by XBOOLE_0:def 1;
  assume x in X;
  hence thesis by A1,XBOOLE_0:def 5;
end;

theorem
  for X being set holds {} \ X = {}
proof
  let X be set;
  thus {} \ X c= {}
  proof
    let x be set;
    assume x in {} \ X;
    hence thesis by XBOOLE_0:def 5;
  end;
  let x be set;
  assume x in {};
  hence thesis by XBOOLE_0:def 1;
end;

theorem
  for X being set holds X \+\ {} = X
proof
  let X be set;
  thus X \+\ {} c= X
  proof
    let x be set;
    assume x in X \+\ {};
    then
A1: x in X \ {} or x in {} \ X by XBOOLE_0:def 3;
    per cases by A1,XBOOLE_0:def 5;
    suppose
      x in X & not x in {};
      hence thesis;
    end;
    suppose
      x in {} & not x in X;
      hence thesis by XBOOLE_0:def 1;
    end;
  end;
  let x be set;
A2: not x in {} by XBOOLE_0:def 1;
  assume x in X;
  then x in X \ {} by A2,XBOOLE_0:def 5;
  hence thesis by XBOOLE_0:def 3;
end;

reserve x,X for set;

Lm1: X is empty implies X = {}
proof
  assume not ex x st x in X;
  then x in {} iff x in X by XBOOLE_0:def 1;
  hence thesis by TARSKI:1;
end;

theorem
  for X being set st X is empty holds X = {} by Lm1;

theorem
  for x, X being set st x in X holds X is non empty by XBOOLE_0:def 1;

theorem
  for X, Y being set st X is empty & X <> Y holds Y is non empty
proof
  let X, Y be set;
  assume that
A1: X is empty and
A2: X <> Y;
  X = {} by A1,Lm1;
  hence thesis by A2,Lm1;
end;
 



















begin

reserve x,A,B,X,X9,Y,Y9,Z,V for set;


theorem Th1:
  X c= Y & Y c= Z implies X c= Z
proof
  assume that
A1: X c= Y and
A2: Y c= Z;
  let x;
  assume x in X;
  then x in Y by A1,TARSKI:def 3;
  hence thesis by A2,TARSKI:def 3;
end;

theorem Th2:
  {} c= X
proof
  let x;
  thus thesis;
end;

theorem Th3:
  X c= {} implies X = {}
proof
  assume X c= {};
  hence X c= {} & {} c= X by Th2;
end;

theorem Th4:
  (X \/ Y) \/ Z = X \/ (Y \/ Z)
proof
  thus (X \/ Y) \/ Z c= X \/ (Y \/ Z)
  proof
    let x;
    assume x in (X \/ Y) \/ Z;
    then x in X \/ Y or x in Z by XBOOLE_0:def 3;
    then x in X or x in Y or x in Z by XBOOLE_0:def 3;
    then x in X or x in Y \/ Z by XBOOLE_0:def 3;
    hence thesis by XBOOLE_0:def 3;
  end;
  let x;
  assume x in X \/ (Y \/ Z);
  then x in X or x in Y \/ Z by XBOOLE_0:def 3;
  then x in X or x in Y or x in Z by XBOOLE_0:def 3;
  then x in X \/ Y or x in Z by XBOOLE_0:def 3;
  hence thesis by XBOOLE_0:def 3;
end;

theorem
  (X \/ Y) \/ Z = (X \/ Z) \/ (Y \/ Z)
proof
  (X \/ Y) \/ Z = X \/ ((Z \/ Z) \/ Y) by Th4
    .= X \/ (Z \/ (Z \/ Y)) by Th4
    .= (X \/ Z) \/ (Y \/ Z) by Th4;
  hence thesis;
end;

theorem
  X \/ (X \/ Y) = X \/ Y
proof
  X \/ (X \/ Y) = (X \/ X) \/ Y by Th4
    .= X \/ Y;
  hence thesis;
end;

theorem Th7:
  X c= X \/ Y
proof
  let x;
  thus thesis by XBOOLE_0:def 3;
end;

theorem Th8:
  X c= Z & Y c= Z implies X \/ Y c= Z
proof
  assume
A1: X c= Z & Y c= Z;
  let x;
  assume x in X \/ Y;
  then x in X or x in Y by XBOOLE_0:def 3;
  hence thesis by A1,TARSKI:def 3;
end;

theorem
  X c= Y implies X \/ Z c= Y \/ Z
proof
  assume
A1: X c= Y;
  let x;
  assume x in X \/ Z;
  then x in X or x in Z by XBOOLE_0:def 3;
  then x in Y or x in Z by A1,TARSKI:def 3;
  hence thesis by XBOOLE_0:def 3;
end;

theorem
  X c= Y implies X c= Z \/ Y
proof
  Y c= Z \/ Y by Th7;
  hence thesis by Th1;
end;

theorem
  X \/ Y c= Z implies X c= Z
proof
  X c= X \/ Y by Th7;
  hence thesis by Th1;
end;

theorem Th12:
  X c= Y implies X \/ Y = Y
proof
  assume
A1: X c= Y;
  thus X \/ Y c= Y
  proof
    let x;
    assume x in X \/ Y;
    then x in X or x in Y by XBOOLE_0:def 3;
    hence thesis by A1,TARSKI:def 3;
  end;
  let x;
  thus thesis by XBOOLE_0:def 3;
end;

theorem
  X c= Y & Z c= V implies X \/ Z c= Y \/ V
proof
  assume
A1: X c= Y;
  assume
A2: Z c= V;
  let x;
  assume x in X \/ Z;
  then x in X or x in Z by XBOOLE_0:def 3;
  then x in Y or x in V by A1,A2,TARSKI:def 3;
  hence thesis by XBOOLE_0:def 3;
end;

theorem
  (Y c= X & Z c= X & for V st Y c= V & Z c= V holds X c= V) implies X = Y \/ Z
proof
  assume that
A1: Y c= X & Z c= X and
A2: Y c= V & Z c= V implies X c= V;
  Y c= Y \/ Z & Z c= Y \/ Z by Th7;
  hence X c= Y \/ Z by A2;
  thus thesis by A1,Th8;
end;

theorem
  X \/ Y = {} implies X = {};

theorem Th16:
  (X /\ Y) /\ Z = X /\ (Y /\ Z)
proof
  thus (X /\ Y) /\ Z c= X /\ (Y /\ Z)
  proof
    let x;
    assume
A1: x in (X /\ Y) /\ Z;
    then
A2: x in Z by XBOOLE_0:def 4;
A3: x in X /\ Y by A1,XBOOLE_0:def 4;
    then
A4: x in X by XBOOLE_0:def 4;
    x in Y by A3,XBOOLE_0:def 4;
    then x in Y /\ Z by A2,XBOOLE_0:def 4;
    hence thesis by A4,XBOOLE_0:def 4;
  end;
  let x;
  assume
A5: x in X /\ (Y /\ Z);
  then
A6: x in Y /\ Z by XBOOLE_0:def 4;
  then
A7: x in Y by XBOOLE_0:def 4;
A8: x in Z by A6,XBOOLE_0:def 4;
  x in X by A5,XBOOLE_0:def 4;
  then x in X /\ Y by A7,XBOOLE_0:def 4;
  hence thesis by A8,XBOOLE_0:def 4;
end;

theorem Th17:
  X /\ Y c= X
proof
  let x;
  thus thesis by XBOOLE_0:def 4;
end;

theorem
  X c= Y /\ Z implies X c= Y
proof
  Y /\ Z c= Y by Th17;
  hence thesis by Th1;
end;

theorem Th19:
  Z c= X & Z c= Y implies Z c= X /\ Y
proof
  assume
A1: Z c= X & Z c= Y;
  let x;
  assume x in Z;
  then x in X & x in Y by A1,TARSKI:def 3;
  hence thesis by XBOOLE_0:def 4;
end;

theorem
  (X c= Y & X c= Z & for V st V c= Y & V c= Z holds V c= X) implies X = Y /\ Z
proof
  assume that
A1: X c= Y & X c= Z and
A2: V c= Y & V c= Z implies V c= X;
  thus X c= Y /\ Z by A1,Th19;
  Y /\ Z c= Y & Y /\ Z c= Z implies Y /\ Z c= X by A2;
  hence thesis by Th17;
end;

theorem
  X /\ (X \/ Y) = X
proof
  thus X /\ (X \/ Y) c= X
  proof
    let x;
    thus thesis by XBOOLE_0:def 4;
  end;
  let x;
  assume
A1: x in X;
  then x in X \/ Y by XBOOLE_0:def 3;
  hence thesis by A1,XBOOLE_0:def 4;
end;

theorem Th22:
  X \/ (X /\ Y) = X
proof
  thus X \/ (X /\ Y) c= X
  proof
    let x;
    assume x in X \/ (X /\ Y);
    then x in X or x in X /\ Y by XBOOLE_0:def 3;
    hence thesis by XBOOLE_0:def 4;
  end;
  let x;
  thus thesis by XBOOLE_0:def 3;
end;

theorem Th23:
  X /\ (Y \/ Z) = X /\ Y \/ X /\ Z
proof
  thus X /\ (Y \/ Z) c= X /\ Y \/ X /\ Z
  proof
    let x;
    assume
A1: x in X /\ (Y \/ Z);
    then x in Y \/ Z by XBOOLE_0:def 4;
    then
A2: x in Y or x in Z by XBOOLE_0:def 3;
    x in X by A1,XBOOLE_0:def 4;
    then x in X /\ Y or x in X /\ Z by A2,XBOOLE_0:def 4;
    hence thesis by XBOOLE_0:def 3;
  end;
  let x;
  assume x in X /\ Y \/ X /\ Z;
  then x in X /\ Y or x in X /\ Z by XBOOLE_0:def 3;
  then
A3: x in X & x in Y or x in X & x in Z by XBOOLE_0:def 4;
  then x in Y \/ Z by XBOOLE_0:def 3;
  hence thesis by A3,XBOOLE_0:def 4;
end;

theorem Th24:
  X \/ Y /\ Z = (X \/ Y) /\ (X \/ Z)
proof
  thus X \/ Y /\ Z c= (X \/ Y) /\ (X \/ Z)
  proof
    let x;
    assume x in X \/ Y /\ Z;
    then x in X or x in Y /\ Z by XBOOLE_0:def 3;
    then x in X or x in Y & x in Z by XBOOLE_0:def 4;
    then x in X \/ Y & x in X \/ Z by XBOOLE_0:def 3;
    hence thesis by XBOOLE_0:def 4;
  end;
  let x;
  assume
A1: x in (X \/ Y) /\ (X \/ Z);
  then x in X \/ Z by XBOOLE_0:def 4;
  then
A2: x in X or x in Z by XBOOLE_0:def 3;
  x in X \/ Y by A1,XBOOLE_0:def 4;
  then x in X or x in Y by XBOOLE_0:def 3;
  then x in X or x in Y /\ Z by A2,XBOOLE_0:def 4;
  hence thesis by XBOOLE_0:def 3;
end;

theorem
  (X /\ Y) \/ (Y /\ Z) \/ (Z /\ X) = (X \/ Y) /\ (Y \/ Z) /\ (Z \/ X)
proof
  thus X /\ Y \/ Y /\ Z \/ Z /\ X = (X /\ Y \/ Y /\ Z \/ Z) /\ (X /\ Y \/ Y /\
  Z \/ X) by Th24
    .= (X /\ Y \/ (Y /\ Z \/ Z)) /\ (X /\ Y \/ Y /\ Z \/ X) by Th4
    .= (X /\ Y \/ Z) /\ (X /\ Y \/ Y /\ Z \/ X) by Th22
    .= (X /\ Y \/ Z) /\ (X /\ Y \/ X \/ Y /\ Z) by Th4
    .= (X /\ Y \/ Z) /\ (X \/ Y /\ Z) by Th22
    .= (X \/ Z) /\ (Y \/ Z) /\ (X \/ Y /\ Z) by Th24
    .= (X \/ Z) /\ (Y \/ Z) /\ ((X \/ Y) /\ (X \/ Z)) by Th24
    .= (X \/ Y) /\ ((Y \/ Z) /\ (X \/ Z) /\ (X \/ Z)) by Th16
    .= (X \/ Y) /\ ((Y \/ Z) /\ ((X \/ Z) /\ (X \/ Z))) by Th16
    .= (X \/ Y) /\ (Y \/ Z) /\ (Z \/ X) by Th16;
end;

theorem Th26:
  X c= Y implies X /\ Z c= Y /\ Z
proof
  assume
A1: X c= Y;
  let x;
  assume
A2: x in X /\ Z;
  then x in X by XBOOLE_0:def 4;
  then
A3: x in Y by A1,TARSKI:def 3;
  x in Z by A2,XBOOLE_0:def 4;
  hence thesis by A3,XBOOLE_0:def 4;
end;

theorem
  X c= Y & Z c= V implies X /\ Z c= Y /\ V
proof
  assume that
A1: X c= Y and
A2: Z c= V;
  let x;
  assume
A3: x in X /\ Z;
  then x in Z by XBOOLE_0:def 4;
  then
A4: x in V by A2,TARSKI:def 3;
  x in X by A3,XBOOLE_0:def 4;
  then x in Y by A1,TARSKI:def 3;
  hence thesis by A4,XBOOLE_0:def 4;
end;

theorem Th28:
  X c= Y implies X /\ Y = X
proof
  assume
A1: X c= Y;
  thus X /\ Y c= X by Th17;
  let x;
  assume
A2: x in X;
  then x in Y by A1,TARSKI:def 3;
  hence thesis by A2,XBOOLE_0:def 4;
end;

theorem
  X /\ Y c= X \/ Z
proof
  X /\ Y c= X & X c= X \/ Z by Th7,Th17;
  hence thesis by Th1;
end;

theorem
  X c= Z implies X \/ Y /\ Z = (X \/ Y) /\ Z
proof
  assume
A1: X c= Z;
  thus X \/ Y /\ Z c= (X \/ Y) /\ Z
  proof
    let x;
    assume x in X \/ Y /\ Z;
    then
A2: x in X or x in Y /\ Z by XBOOLE_0:def 3;
    then x in X or x in Y & x in Z by XBOOLE_0:def 4;
    then
A3: x in (X \/ Y) by XBOOLE_0:def 3;
    x in Z by A1,A2,TARSKI:def 3,XBOOLE_0:def 4;
    hence thesis by A3,XBOOLE_0:def 4;
  end;
  let x;
  assume
A4: x in (X \/ Y) /\ Z;
  then x in X \/ Y by XBOOLE_0:def 4;
  then
A5: x in X or x in Y by XBOOLE_0:def 3;
  x in Z by A4,XBOOLE_0:def 4;
  then x in X & x in Z or x in Y /\ Z by A5,XBOOLE_0:def 4;
  hence thesis by XBOOLE_0:def 3;
end;

theorem
  (X /\ Y) \/ (X /\ Z) c= Y \/ Z
proof
  now
    let x;
    assume x in (X /\ Y) \/ (X /\ Z);
    then x in (X /\ Y) or x in (X /\ Z) by XBOOLE_0:def 3;
    then x in X & x in Y or x in X & x in Z by XBOOLE_0:def 4;
    hence x in Y \/ Z by XBOOLE_0:def 3;
  end;
  hence thesis by TARSKI:def 3;
end;

Lm1: X \ Y = {} iff X c= Y
proof
  thus X \ Y = {} implies X c= Y
  proof
    assume
A1: X \ Y = {};
    let x;
    assume x in X & not x in Y;
    hence contradiction by A1,XBOOLE_0:def 5;
  end;
  assume
A2: X c= Y;
  now
    let x;
    x in X & not x in Y iff contradiction by A2,TARSKI:def 3;
    hence x in X \ Y iff x in {} by XBOOLE_0:def 5;
  end;
  hence thesis by TARSKI:1;
end;

theorem
  X \ Y = Y \ X implies X = Y
proof
  assume
A1: X \ Y = Y \ X;
  now
    let x;
    x in X & not x in Y iff x in Y \ X by A1,XBOOLE_0:def 5;
    hence x in X iff x in Y by XBOOLE_0:def 5;
  end;
  hence thesis by TARSKI:1;
end;

theorem Th33:
  X c= Y implies X \ Z c= Y \ Z
proof
  assume
A1: X c= Y;
  let x;
  assume
A2: x in X \ Z;
  then x in X by XBOOLE_0:def 5;
  then
A3: x in Y by A1,TARSKI:def 3;
  not x in Z by A2,XBOOLE_0:def 5;
  hence thesis by A3,XBOOLE_0:def 5;
end;

theorem Th34:
  X c= Y implies Z \ Y c= Z \ X
proof
  assume
A1: X c= Y;
  let x;
  assume
A2: x in Z \ Y;
  then not x in Y by XBOOLE_0:def 5;
  then
A3: not x in X by A1,TARSKI:def 3;
  x in Z by A2,XBOOLE_0:def 5;
  hence thesis by A3,XBOOLE_0:def 5;
end;

Lm2: X \ (Y /\ Z) = (X \ Y) \/ (X \ Z)
proof
  thus X \ (Y /\ Z) c= (X \ Y) \/ (X \ Z)
  proof
    let x;
    assume
A1: x in X \ (Y /\ Z);
    then not x in (Y /\ Z) by XBOOLE_0:def 5;
    then
A2: not x in Y or not x in Z by XBOOLE_0:def 4;
    x in X by A1,XBOOLE_0:def 5;
    then x in (X \ Y) or x in (X \ Z) by A2,XBOOLE_0:def 5;
    hence thesis by XBOOLE_0:def 3;
  end;
  (X \ Y) c= X \ (Y /\ Z) & X \ Z c= X \ (Y /\ Z) by Th17,Th34;
  hence thesis by Th8;
end;

theorem
  X c= Y & Z c= V implies X \ V c= Y \ Z
proof
  assume X c= Y & Z c= V;
  then X \ V c= Y \ V & Y \ V c= Y \ Z by Th33,Th34;
  hence thesis by Th1;
end;

theorem Th36:
  X \ Y c= X
proof
  let x;
  thus thesis by XBOOLE_0:def 5;
end;

theorem
  X \ Y = {} iff X c= Y by Lm1;

theorem
  X c= Y \ X implies X = {}
proof
  assume
A1: X c= Y \ X;
  thus X c= {}
  proof
    let x;
    assume
A2: x in X;
    then x in Y \ X by A1,TARSKI:def 3;
    hence thesis by A2,XBOOLE_0:def 5;
  end;
  thus thesis by Th2;
end;

theorem Th39:
  X \/ (Y \ X) = X \/ Y
proof
  thus X \/ (Y \ X) c= X \/ Y
  proof
    let x;
    assume x in X \/ (Y \ X);
    then x in X or x in Y \ X by XBOOLE_0:def 3;
    then x in X or x in Y by XBOOLE_0:def 5;
    hence thesis by XBOOLE_0:def 3;
  end;
  let x;
  assume x in X \/ Y;
  then x in X or x in Y & not x in X by XBOOLE_0:def 3;
  then x in X or x in Y \ X by XBOOLE_0:def 5;
  hence thesis by XBOOLE_0:def 3;
end;

theorem
  (X \/ Y) \ Y = X \ Y
proof
  thus x in (X \/ Y) \ Y implies x in X \ Y
  proof
    assume
A1: x in (X \/ Y) \ Y;
    then x in (X \/ Y) by XBOOLE_0:def 5;
    then
A2: x in X or x in Y by XBOOLE_0:def 3;
    not x in Y by A1,XBOOLE_0:def 5;
    hence thesis by A2,XBOOLE_0:def 5;
  end;
  thus x in X \ Y implies x in (X \/ Y) \ Y
  proof
    assume
A3: x in X \ Y;
    then x in X or x in Y by XBOOLE_0:def 5;
    then
A4: x in (X \/ Y) by XBOOLE_0:def 3;
    not x in Y by A3,XBOOLE_0:def 5;
    hence thesis by A4,XBOOLE_0:def 5;
  end;
end;

theorem Th41:
  (X \ Y) \ Z = X \ (Y \/ Z)
proof
  thus x in (X \ Y) \ Z implies x in X \ (Y \/ Z)
  proof
    assume
A1: x in (X \ Y) \ Z;
    then
A2: not x in Z by XBOOLE_0:def 5;
A3: x in (X \ Y) by A1,XBOOLE_0:def 5;
    then
A4: x in X by XBOOLE_0:def 5;
    not x in Y by A3,XBOOLE_0:def 5;
    then not x in (Y \/ Z) by A2,XBOOLE_0:def 3;
    hence thesis by A4,XBOOLE_0:def 5;
  end;
  thus x in X \ (Y \/ Z) implies x in (X \ Y) \ Z
  proof
    assume
A5: x in X \ (Y \/ Z);
    then
A6: not x in (Y \/ Z) by XBOOLE_0:def 5;
    then
A7: not x in Y by XBOOLE_0:def 3;
A8: not x in Z by A6,XBOOLE_0:def 3;
    x in X by A5,XBOOLE_0:def 5;
    then x in (X \ Y) by A7,XBOOLE_0:def 5;
    hence thesis by A8,XBOOLE_0:def 5;
  end;
end;

theorem Th42:
  (X \/ Y) \ Z = (X \ Z) \/ (Y \ Z)
proof
  thus (X \/ Y) \ Z c= (X \ Z) \/ (Y \ Z)
  proof
    let x;
    assume
A1: x in (X \/ Y) \ Z;
    then x in (X \/ Y) by XBOOLE_0:def 5;
    then x in X & not x in Z or x in Y & not x in Z by A1,XBOOLE_0:def 3,def 5;
    then x in (X \ Z) or x in (Y \ Z) by XBOOLE_0:def 5;
    hence thesis by XBOOLE_0:def 3;
  end;
  let x;
  assume x in (( X \ Z) \/ (Y \ Z));
  then x in (X \ Z) or x in (Y \ Z) by XBOOLE_0:def 3;
  then
A2: x in X & not x in Z or x in Y & not x in Z by XBOOLE_0:def 5;
  then x in (X \/ Y) by XBOOLE_0:def 3;
  hence thesis by A2,XBOOLE_0:def 5;
end;

theorem
  X c= Y \/ Z implies X \ Y c= Z
proof
  assume
A1: X c= Y \/ Z;
  let x;
  assume
A2: x in X \ Y;
  then x in X by XBOOLE_0:def 5;
  then
A3: x in Y \/ Z by A1,TARSKI:def 3;
  not x in Y by A2,XBOOLE_0:def 5;
  hence thesis by A3,XBOOLE_0:def 3;
end;

theorem
  X \ Y c= Z implies X c= Y \/ Z
proof
  assume
A1: x in X \ Y implies x in Z;
  let x;
  assume x in X;
  then x in X \ Y or x in Y by XBOOLE_0:def 5;
  then x in Z or x in Y by A1;
  hence thesis by XBOOLE_0:def 3;
end;

theorem
  X c= Y implies Y = X \/ (Y \ X)
proof
  assume
A1: X c= Y;
  now
    let x;
    x in Y iff x in X or x in (Y \ X) by A1,TARSKI:def 3,XBOOLE_0:def 5;
    hence x in Y iff x in X \/ (Y \ X) by XBOOLE_0:def 3;
  end;
  hence thesis by TARSKI:1;
end;

theorem
  X \ (X \/ Y) = {}
proof
  X c= X \/ Y by Th7;
  hence thesis by Lm1;
end;

theorem Th47:
  X \ X /\ Y = X \ Y
proof
  now
    let x;
    x in X & not x in X /\ Y iff x in X & not x in Y by XBOOLE_0:def 4;
    hence x in X \ X /\ Y iff x in X \ Y by XBOOLE_0:def 5;
  end;
  hence thesis by TARSKI:1;
end;

theorem
  X \ (X \ Y) = X /\ Y
proof
  thus x in X \ (X \ Y) implies x in X /\ Y
  proof
    assume
A1: x in X \ (X \ Y);
    then not x in (X \ Y) by XBOOLE_0:def 5;
    then
A2: not x in X or x in Y by XBOOLE_0:def 5;
    x in X by A1,XBOOLE_0:def 5;
    hence thesis by A2,XBOOLE_0:def 4;
  end;
  thus x in X /\ Y implies x in X \ (X \ Y)
  proof
    assume
A3: x in X /\ Y;
    then not x in X or x in Y by XBOOLE_0:def 4;
    then
A4: not x in (X \ Y) by XBOOLE_0:def 5;
    x in X by A3,XBOOLE_0:def 4;
    hence thesis by A4,XBOOLE_0:def 5;
  end;
end;

theorem Th49:
  X /\ (Y \ Z) = (X /\ Y) \ Z
proof
  now
    let x;
    x in X & x in Y & not x in Z iff x in X & x in Y & not x in Z;
    then x in X & x in (Y \ Z) iff x in (X /\ Y) & not x in Z by XBOOLE_0:def 4
,def 5;
    hence x in X /\ (Y \ Z) iff x in (X /\ Y) \ Z by XBOOLE_0:def 4,def 5;
  end;
  hence thesis by TARSKI:1;
end;

theorem Th50:
  X /\ (Y \ Z) = X /\ Y \ X /\ Z
proof
A1: X /\ Y c= X by Th17;
  X /\ Y \ X /\ Z = ((X /\ Y) \ X) \/ ((X /\ Y) \ Z) by Lm2
    .= {} \/ ((X /\ Y) \ Z) by A1,Lm1
    .= (X /\ Y) \ Z;
  hence thesis by Th49;
end;

theorem Th51:
  X /\ Y \/ (X \ Y) = X
proof
  thus X /\ Y \/ (X \ Y) c= X
  proof
    let x;
    assume x in X /\ Y \/ (X \ Y);
    then x in X /\ Y or x in (X \ Y) by XBOOLE_0:def 3;
    hence thesis by XBOOLE_0:def 4,def 5;
  end;
  let x;
  assume x in X;
  then x in X & x in Y or x in (X\Y) by XBOOLE_0:def 5;
  then x in X /\ Y or x in (X \ Y) by XBOOLE_0:def 4;
  hence thesis by XBOOLE_0:def 3;
end;

theorem Th52:
  X \ (Y \ Z) = (X \ Y) \/ X /\ Z
proof
  thus x in X \ (Y \ Z) implies x in (X \ Y) \/ X /\ Z
  proof
    assume
A1: x in X \ (Y \ Z);
    then not x in (Y \ Z) by XBOOLE_0:def 5;
    then x in X & not x in Y or x in X & x in Z by A1,XBOOLE_0:def 5;
    then x in (X \ Y) or x in X /\ Z by XBOOLE_0:def 4,def 5;
    hence thesis by XBOOLE_0:def 3;
  end;
  thus x in (X \ Y) \/ X /\ Z implies x in X \ (Y \ Z)
  proof
    assume x in (X \ Y) \/ X /\ Z;
    then x in (X \ Y) or x in X /\ Z by XBOOLE_0:def 3;
    then
A2: x in X & not x in Y or x in X & x in Z by XBOOLE_0:def 4,def 5;
    then not x in (Y \ Z) by XBOOLE_0:def 5;
    hence thesis by A2,XBOOLE_0:def 5;
  end;
end;

theorem
  X \ (Y \/ Z) = (X \ Y) /\ (X \ Z)
proof
  X \(Y \/ Z) c= X \ Y & X \ (Y \/ Z) c= X \ Z by Th7,Th34;
  hence X \ (Y \/ Z) c= (X \ Y) /\ (X \ Z) by Th19;
  let x;
  assume
A1: x in (X \ Y) /\ (X \ Z);
  then
A2: x in (X \ Y) by XBOOLE_0:def 4;
  then
A3: x in X by XBOOLE_0:def 5;
  x in (X \ Z) by A1,XBOOLE_0:def 4;
  then
A4: not x in Z by XBOOLE_0:def 5;
  not x in Y by A2,XBOOLE_0:def 5;
  then not x in (Y \/ Z) by A4,XBOOLE_0:def 3;
  hence thesis by A3,XBOOLE_0:def 5;
end;

theorem
  X \ (Y /\ Z) = (X \ Y) \/ (X \ Z) by Lm2;

theorem Th55:
  (X \/ Y) \ (X /\ Y) = (X \ Y) \/ (Y \ X)
proof
  x in (X \/ Y) \ (X /\ Y) iff x in (X \ Y) \/ (Y \ X)
  proof
    thus x in (X \/ Y) \ (X /\ Y) implies x in (X \ Y) \/ (Y \ X)
    proof
      assume
A1:   x in (X \/ Y) \ (X /\ Y);
      then not x in (X /\ Y) by XBOOLE_0:def 5;
      then
A2:   not x in X or not x in Y by XBOOLE_0:def 4;
      x in (X \/ Y) by A1,XBOOLE_0:def 5;
      then x in X or x in Y by XBOOLE_0:def 3;
      then x in (X \ Y) or x in( Y \ X) by A2,XBOOLE_0:def 5;
      hence thesis by XBOOLE_0:def 3;
    end;
    assume x in (X \ Y) \/ (Y \ X);
    then x in (X \ Y) or x in (Y \ X) by XBOOLE_0:def 3;
    then x in X & not x in Y or x in Y & not x in X by XBOOLE_0:def 5;
    then ( not x in (X /\ Y))& x in (X \/ Y) by XBOOLE_0:def 3,def 4;
    hence thesis by XBOOLE_0:def 5;
  end;
  hence thesis by TARSKI:1;
end;

Lm3: X c= Y & Y c< Z implies X c< Z
proof
  assume that
A1: X c= Y and
A2: Y c< Z;
  Y c= Z by A2,XBOOLE_0:def 8;
  hence X c= Z & X <> Z by A1,A2,Th1,XBOOLE_0:def 10;
end;

theorem
  X c< Y & Y c< Z implies X c< Z
proof
  assume that
A1: X c< Y and
A2: Y c< Z;
  X c= Y by A1,XBOOLE_0:def 8;
  hence thesis by A2,Lm3;
end;

theorem
  not (X c< Y & Y c< X);

theorem
  X c< Y & Y c= Z implies X c< Z
proof
  assume that
A1: X c< Y and
A2: Y c= Z;
  X c= Y by A1,XBOOLE_0:def 8;
  hence X c= Z & X <> Z by A1,A2,Th1,XBOOLE_0:def 10;
end;

theorem
  X c= Y & Y c< Z implies X c< Z by Lm3;

theorem Th60:
  X c= Y implies not Y c< X
proof
  assume X c= Y & Y c= X & X <> Y;
  hence contradiction by XBOOLE_0:def 10;
end;

theorem
  X <> {} implies {} c< X
proof
  assume
A1: X <> {};
  thus {} c= X by Th2;
  thus thesis by A1;
end;

theorem
  not X c< {}
proof
  assume
A1: X c< {};
  then X c= {} by XBOOLE_0:def 8;
  hence contradiction by A1,Th3;
end;



theorem Th63:
  X c= Y & Y misses Z implies X misses Z
proof
  assume X c= Y & Y /\ Z = {};
  then X /\ Z = {} by Th3,Th26;
  hence thesis by XBOOLE_0:def 7;
end;

theorem
  A c= X & B c= Y & X misses Y implies A misses B
proof
  assume that
A1: A c= X and
A2: B c= Y and
A3: X misses Y;
  A misses Y by A1,A3,Th63;
  hence thesis by A2,Th63;
end;

theorem
  X misses {}
proof
  assume X meets {};
  then ex x st x in X & x in {} by XBOOLE_0:3;
  hence contradiction;
end;

theorem
  X meets X iff X <> {}
proof
  hereby
    assume X meets X;
    then ex x st x in X & x in X by XBOOLE_0:3;
    hence X <> {};
  end;
  assume X <> {};
  then X /\ X <> {};
  hence thesis by XBOOLE_0:def 7;
end;

theorem
  X c= Y & X c= Z & Y misses Z implies X = {}
proof
  assume X c= Y & X c= Z & Y /\ Z = {};
  hence thesis by Th3,Th19;
end;


theorem Th68:
  for A being non empty set st A c= Y & A c= Z holds Y meets Z
proof
  let A be non empty set;
  consider x being set such that
A1: x in A by XBOOLE_0:def 1;
  assume A c= Y & A c= Z;
  then x in Y & x in Z by A1,TARSKI:def 3;
  hence thesis by XBOOLE_0:3;
end;

theorem
  for A being non empty set st A c= Y holds A meets Y by Th68;

theorem Th70:
  X meets Y \/ Z iff X meets Y or X meets Z
proof
  thus X meets Y \/ Z implies X meets Y or X meets Z
  proof
    assume X meets Y \/ Z;
    then consider x such that
A1: x in X & x in Y \/ Z by XBOOLE_0:3;
    x in X & x in Y or x in X & x in Z by A1,XBOOLE_0:def 3;
    hence thesis by XBOOLE_0:3;
  end;
A2: X meets Z implies X meets Y \/ Z
  proof
    assume X meets Z;
    then consider x such that
A3: x in X and
A4: x in Z by XBOOLE_0:3;
    x in Y \/ Z by A4,XBOOLE_0:def 3;
    hence thesis by A3,XBOOLE_0:3;
  end;
  X meets Y implies X meets Y \/ Z
  proof
    assume X meets Y;
    then consider x such that
A5: x in X and
A6: x in Y by XBOOLE_0:3;
    x in Y \/ Z by A6,XBOOLE_0:def 3;
    hence thesis by A5,XBOOLE_0:3;
  end;
  hence thesis by A2;
end;

theorem
  X \/ Y = Z \/ Y & X misses Y & Z misses Y implies X = Z
proof
  assume that
A1: X \/ Y = Z \/ Y and
A2: X /\ Y = {} and
A3: Z /\ Y = {};
  thus X c= Z
  proof
    let x be set such that
A4: x in X;
    X c= Z \/ Y by A1,Th7;
    then
A5: x in Z \/ Y by A4,TARSKI:def 3;
    not x in Y by A2,A4,XBOOLE_0:def 4;
    hence thesis by A5,XBOOLE_0:def 3;
  end;
  let x be set such that
A6: x in Z;
  Z c= X \/ Y by A1,Th7;
  then
A7: x in X \/ Y by A6,TARSKI:def 3;
  not x in Y by A3,A6,XBOOLE_0:def 4;
  hence thesis by A7,XBOOLE_0:def 3;
end;

theorem
  X9 \/ Y9 = X \/ Y & X misses X9 & Y misses Y9 implies X = Y9
proof
  assume
A1: X9 \/ Y9 = X \/ Y;
  assume X misses X9 & Y misses Y9;
  then
A2: X /\ X9 = {} & Y /\ Y9 = {} by XBOOLE_0:def 7;
  thus X = X /\ (X9 \/ Y9) by A1,Th7,Th28
    .= X /\ X9 \/ X /\ Y9 by Th23
    .= (X \/ Y) /\ Y9 by A2,Th23
    .= Y9 by A1,Th7,Th28;
end;

theorem
  X c= Y \/ Z & X misses Z implies X c= Y
proof
  assume that
A1: X c= Y \/ Z and
A2: X /\ Z = {};
  X /\ (Y \/ Z)= X by A1,Th28;
  then Y /\ X \/ {} = X by A2,Th23;
  hence thesis by Th17;
end;

theorem Th74:
  X meets Y /\ Z implies X meets Y
proof
  assume X meets Y /\ Z;
  then consider x such that
A1: x in X and
A2: x in Y /\ Z by XBOOLE_0:3;
  x in Y by A2,XBOOLE_0:def 4;
  hence thesis by A1,XBOOLE_0:3;
end;

theorem
  X meets Y implies X /\ Y meets Y
proof
  assume X meets Y;
  then consider x being set such that
A1: x in X and
A2: x in Y by XBOOLE_0:3;
  x in X /\ Y by A1,A2,XBOOLE_0:def 4;
  hence thesis by A2,XBOOLE_0:3;
end;

theorem
  Y misses Z implies X /\ Y misses X /\ Z
proof
  assume Y misses Z;
  then (X /\ Z) misses Y by Th74;
  hence thesis by Th74;
end;

theorem
  X meets Y & X c= Z implies X meets Y /\ Z
proof
  assume that
A1: X meets Y and
A2: X c= Z;
  now
    assume
A3: X /\ (Y /\ Z) = {};
    X /\ Y = (X /\ Z) /\ Y by A2,Th28
      .= {} by A3,Th16;
    hence contradiction by A1,XBOOLE_0:def 7;
  end;
  hence thesis by XBOOLE_0:def 7;
end;

theorem
  X misses Y implies X /\ (Y \/ Z) = X /\ Z
proof
  assume X misses Y;
  then X /\ Y = {} by XBOOLE_0:def 7;
  hence X /\ (Y \/ Z) = {} \/ X /\ Z by Th23
    .= X /\ Z;
end;

theorem Th79:
  X \ Y misses Y
proof
  not ex x st x in (X \ Y) /\ Y
  proof
    given x such that
A1: x in (X \ Y) /\ Y;
    x in X \ Y & x in Y by A1,XBOOLE_0:def 4;
    hence contradiction by XBOOLE_0:def 5;
  end;
  hence (X \ Y) /\ Y = {} by XBOOLE_0:def 1;
end;

theorem
  X misses Y implies X misses Y \ Z
proof
  assume
A1: X misses Y;
  assume X meets Y \ Z;
  then consider x such that
A2: x in X and
A3: x in Y \ Z by XBOOLE_0:3;
  x in Y by A3,XBOOLE_0:def 5;
  hence thesis by A1,A2,XBOOLE_0:3;
end;

theorem
  X misses Y \ Z implies Y misses X \ Z
proof
A1: X misses Y \ Z iff X /\ (Y \ Z) = {} by XBOOLE_0:def 7;
  X /\ (Y \ Z) = Y /\ X \ Z by Th49
    .= Y /\ (X \ Z) by Th49;
  hence thesis by A1,XBOOLE_0:def 7;
end;

theorem
  X \ Y misses Y \ X
proof
  assume X \ Y meets Y \ X;
  then consider x such that
A1: x in X \ Y and
A2: x in Y \ X by XBOOLE_0:3;
  x in X by A1,XBOOLE_0:def 5;
  hence thesis by A2,XBOOLE_0:def 5;
end;

theorem Th83:
  X misses Y iff X \ Y = X
proof
  thus X misses Y implies X \ Y = X
  proof
    assume
A1: X /\ Y = {};
    thus for x holds x in X \ Y implies x in X by XBOOLE_0:def 5;
    let x;
    not x in X /\ Y implies not x in X or not x in Y by XBOOLE_0:def 4;
    hence thesis by A1,XBOOLE_0:def 5;
  end;
  assume
A2: X \ Y = X;
  not ex x st x in X /\ Y
  proof
    given x such that
A3: x in X /\ Y;
    x in X & x in Y by A3,XBOOLE_0:def 4;
    hence contradiction by A2,XBOOLE_0:def 5;
  end;
  hence thesis by XBOOLE_0:4;
end;

theorem
  X meets Y & X misses Z implies X meets Y \ Z
proof
  assume that
A1: X meets Y and
A2: X misses Z;
  X /\ (Y \ Z) = X /\ Y \ X /\ Z by Th50
    .= X /\ Y \ {} by A2,XBOOLE_0:def 7;
  hence X /\ (Y \ Z) <> {} by A1,XBOOLE_0:def 7;
end;

theorem
  X c= Y implies X misses Z \ Y
proof
  assume
A1: X c= Y;
  thus X /\ (Z \ Y) = Z /\ X \ Y by Th49
    .= Z /\ (X \ Y) by Th49
    .= Z /\ {} by A1,Lm1
    .= {};
end;

theorem Th86:
  X c= Y & X misses Z implies X c= Y \ Z
proof
  assume
A1: X c= Y & X /\ Z = {};
  let x;
  assume x in X;
  then x in Y & not x in Z by A1,TARSKI:def 3,XBOOLE_0:def 4;
  hence thesis by XBOOLE_0:def 5;
end;

theorem
  Y misses Z implies (X \ Y) \/ Z = (X \/ Z) \ Y
proof
  assume
A1: Y misses Z;
  thus (X \/ Z) \ Y = (X \ Y) \/ (Z \ Y) by Th42
    .= (X \ Y) \/ Z by A1,Th83;
end;

theorem Th88:
  X misses Y implies (X \/ Y) \ Y = X
proof
  assume
A1: X misses Y;
  thus (X \/ Y) \ Y = (X \ Y) \/ (Y \ Y) by Th42
    .= (X \ Y) \/ {} by Lm1
    .= X by A1,Th83;
end;

theorem Th89:
  X /\ Y misses X \ Y
proof
  now
    let x;
    not (x in X & x in Y & not x in Y);
    hence not (x in X /\ Y & x in X \ Y) by XBOOLE_0:def 4,def 5;
  end;
  hence thesis by XBOOLE_0:3;
end;

theorem
  X \ (X /\ Y) misses Y
proof
  X \ (X /\ Y) = X \ Y by Th47;
  hence thesis by Th79;
end;

theorem
  (X \+\ Y) \+\ Z = X \+\ (Y \+\ Z)
proof
  set S1 = X \ (Y \/ Z), S2 = Y \ (X \/ Z), S3 = Z \ (X \/ Y), S4 = X /\ Y /\
  Z;
  thus (X \+\ Y) \+\ Z = (((X \ Y) \ Z) \/ ((Y \ X) \ Z)) \/ (Z \ ((X \ Y) \/
  (Y \ X))) by Th42
    .= (S1 \/ ((Y \ X) \ Z)) \/ (Z \ ((X \ Y) \/ (Y \ X))) by Th41
    .= (S1 \/ S2) \/ (Z \ ((X \ Y) \/ (Y \ X))) by Th41
    .= (S1 \/ S2) \/ (Z \ ((X \/ Y) \ (X /\ Y))) by Th55
    .= (S1 \/ S2) \/ (S4 \/ S3) by Th52
    .= (S1 \/ S2 \/ S4) \/ S3 by Th4
    .= (S1 \/ S4 \/ S2) \/ S3 by Th4
    .= (S1 \/ S4) \/ (S2 \/ S3) by Th4
    .= (S1 \/ X /\ (Y /\ Z)) \/ (S2 \/ S3) by Th16
    .= X \ ((Y \/ Z) \ (Y /\ Z)) \/ (S2 \/ S3) by Th52
    .= X \ ((Y \ Z) \/ (Z \ Y)) \/ (S2 \/ (Z \ (Y \/ X))) by Th55
    .= X \ ((Y \ Z) \/ (Z \ Y)) \/ ((Y \ (Z \/ X)) \/ (Z \ Y \ X)) by Th41
    .= X \ ((Y \ Z) \/ (Z \ Y)) \/ ((Y \ Z \ X) \/ (Z \ Y \ X)) by Th41
    .= X \+\ (Y \+\ Z) by Th42;
end;

theorem
  X \+\ X = {} by Lm1;

theorem Th93:
  X \/ Y = (X \+\ Y) \/ X /\ Y
proof
  thus X \/ Y = ((X \ Y) \/ X /\ Y) \/ Y by Th51
    .= (X \ Y) \/ (X /\ Y \/ Y) by Th4
    .= (X \ Y) \/ Y by Th22
    .= (X \ Y) \/ ((Y \ X) \/ (Y /\ X)) by Th51
    .= (X \+\ Y) \/ X /\ Y by Th4;
end;

Lm4: X /\ Y misses X \+\ Y
proof
  X /\ Y misses X \ Y & X /\ Y misses Y \ X by Th89;
  hence thesis by Th70;
end;

Lm5: X \+\ Y = (X \/ Y) \ X /\ Y
proof
  thus X \+\ Y = (X \ X /\ Y) \/ (Y \ X) by Th47
    .= (X \ X /\ Y) \/ (Y \ X /\ Y) by Th47
    .= (X \/ Y) \ X /\ Y by Th42;
end;

theorem
  X \/ Y = X \+\ Y \+\ X /\ Y
proof
  X /\ Y misses X \+\ Y by Lm4;
  then (X \+\ Y) \ X /\ Y = X \+\ Y & X /\ Y \ (X \+\ Y) = X /\ Y by Th83;
  hence thesis by Th93;
end;

theorem
  X /\ Y = X \+\ Y \+\ (X \/ Y)
proof
  X \+\ Y = (X \/ Y) \ X /\ Y by Lm5;
  then X \+\ Y c= X \/ Y by Th36;
  then
A1: (X \+\ Y) \ (X \/ Y) = {} by Lm1;
  X \/ Y = (X \+\ Y) \/ X /\ Y by Th93;
  hence thesis by A1,Lm4,Th88;
end;

theorem
  X \ Y c= X \+\ Y by Th7;

theorem
  X \ Y c= Z & Y \ X c= Z implies X \+\ Y c= Z by Th8;

theorem
  X \/ Y = X \+\ (Y \ X)
proof
A1: Y \ X \ X = Y \ (X \/ X) by Th41
    .= Y \ X;
  X \ (Y \ X) = (X \ Y) \/ X /\ X by Th52
    .= X by Th12,Th36;
  hence thesis by A1,Th39;
end;

theorem
  (X \+\ Y) \ Z = (X \ (Y \/ Z)) \/ (Y \ (X \/ Z))
proof
  thus (X \+\ Y) \ Z = (X \ Y \ Z) \/ (Y \ X \ Z) by Th42
    .= (X \ (Y \/ Z)) \/ (Y \ X \ Z) by Th41
    .= (X \ (Y \/ Z)) \/ (Y \ (X \/ Z)) by Th41;
end;

theorem
  X \ Y = X \+\ (X /\ Y)
proof
  X /\ Y c= X by Th17;
  then X /\ Y \ X = {} by Lm1;
  hence thesis by Th47;
end;

theorem
  X \+\ Y = (X \/ Y) \ X /\ Y by Lm5;

theorem
  X \ (Y \+\ Z) = X \ (Y \/ Z) \/ X /\ Y /\ Z
proof
  thus X \ (Y \+\ Z) = X \ ((Y \/ Z) \ Y /\ Z) by Lm5
    .= X \ (Y \/ Z) \/ X /\ (Y /\ Z) by Th52
    .= X \ (Y \/ Z) \/ X /\ Y /\ Z by Th16;
end;

theorem
  X /\ Y misses X \+\ Y by Lm4;

theorem
  X c< Y or X = Y or Y c< X iff X,Y are_c=-comparable
proof
  thus X c< Y or X = Y or Y c< X implies X,Y are_c=-comparable
  proof
    assume X c< Y or X = Y or Y c< X;
    hence X c= Y or Y c= X by XBOOLE_0:def 8;
  end;
  assume X c= Y or Y c= X;
  hence thesis by XBOOLE_0:def 8;
end;

begin 

theorem
  for X, Y being set st X c< Y holds Y \ X <> {}
proof
  let X, Y be set;
  assume
A1: X c< Y;
  assume Y \ X = {};
  then Y c= X by Lm1;
  hence thesis by A1,Th60;
end;

theorem Th106:
  X c= A \ B implies X c= A & X misses B
proof
  assume
A1: X c= A \ B;
  A \ B c= A by Th36;
  hence X c= A by A1,Th1;
  now
    let x;
    assume x in X;
    then x in A \ B by A1,TARSKI:def 3;
    hence not x in B by XBOOLE_0:def 5;
  end;
  hence thesis by XBOOLE_0:3;
end;

theorem
  X c= A \+\ B iff X c= A \/ B & X misses A /\ B
proof
  A \+\ B = (A \/ B) \ A /\ B by Lm5;
  hence thesis by Th86,Th106;
end;

theorem
  X c= A implies X /\ Y c= A
proof
  X /\ Y c= X by Th17;
  hence thesis by Th1;
end;

theorem Th109:
  X c= A implies X \ Y c= A
proof
  X \ Y c= X by Th36;
  hence thesis by Th1;
end;

theorem
  X c= A & Y c= A implies X \+\ Y c= A
proof
  assume X c= A & Y c= A;
  then X \ Y c= A & Y \ X c= A by Th109;
  hence thesis by Th8;
end;

theorem Th111:
  (X /\ Z) \ (Y /\ Z) = (X \ Y) /\ Z
proof
  thus (X /\ Z) \ (Y /\ Z) = ((X /\ Z) \ Y) \/ ((X /\ Z) \ Z) by Lm2
    .= ((X /\ Z) \ Y) \/ (X /\ (Z \ Z)) by Th49
    .= ((X /\ Z) \ Y) \/ (X /\ {}) by Lm1
    .= (X \ Y) /\ Z by Th49;
end;

theorem
  (X /\ Z) \+\ (Y /\ Z) = (X \+\ Y) /\ Z
proof
  thus (X /\ Z) \+\ (Y /\ Z) = ((X \ Y) /\ Z) \/ ((Y /\ Z) \ (X /\ Z)) by Th111
    .= ((X \ Y) /\ Z) \/ ((Y \ X) /\ Z) by Th111
    .= (X \+\ Y) /\ Z by Th23;
end;

theorem
  X \/ Y \/ Z \/ V = X \/ (Y \/ Z \/ V)
proof
  X \/ Y \/ Z \/ V = X \/ Y \/ (Z \/ V) by Th4
    .= X \/ (Y \/ (Z \/ V)) by Th4
    .= X \/ (Y \/ Z \/ V) by Th4;
  hence thesis;
end;

theorem
  for A,B,C,D being set st A misses D & B misses D & C misses D holds A
  \/ B \/ C misses D
proof
  let A,B,C,D be set;
  assume A misses D & B misses D;
  then
A1: A \/ B misses D by Th70;
  assume C misses D;
  hence thesis by A1,Th70;
end;

theorem
  not A c< {}
proof
  assume
A1: A c< {};
  then A c= {} by XBOOLE_0:def 8;
  hence contradiction by A1,Th3;
end;

theorem
  X /\ (Y /\ Z) = (X /\ Y) /\ (X /\ Z)
proof
  thus X /\ (Y /\ Z) = X /\ X /\ Y /\ Z by Th16
    .= X /\ (X /\ Y) /\ Z by Th16
    .= (X /\ Y) /\ (X /\ Z) by Th16;
end;

theorem
  for P,G,C being set st C c= G holds P \ C = (P \ G) \/ (P /\ (G \ C))
proof
  let P,G,C be set;
  assume C c= G;
  then
A1: P \ G c= P \ C by Th34;
  thus P \ C c= (P \ G) \/ (P /\ (G \ C))
  proof
    let x be set;
    assume x in P \ C;
    then x in P & not x in G or x in P & x in G & not x in C by XBOOLE_0:def 5;
    then x in P \ G or x in P & x in G \ C by XBOOLE_0:def 5;
    then x in P \ G or x in P /\ (G \ C) by XBOOLE_0:def 4;
    hence thesis by XBOOLE_0:def 3;
  end;
  P /\ (G \ C) = (P /\ G) \ C & (P /\ G) \ C c= P \ C by Th17,Th33,Th49;
  hence thesis by A1,Th8;
end;














begin

reserve x,x1,x2,x3,x4,x5,x6,x7,x8,x9,x10,y,X,Z for set;

Lm1: x in union({X,{y}}) iff x in X or x=y
proof
A1: x in union({X,{y}}) implies x in X or x in {y}
  proof
    assume x in union({X,{y}});
    then ex Z st x in Z & Z in {X,{y}} by TARSKI:def 4;
    hence thesis by TARSKI:def 2;
  end;
A2: x in {y} iff x=y by TARSKI:def 1;
  X in {X,{y}} & {y} in {X,{y}} by TARSKI:def 2;
  hence thesis by A1,A2,TARSKI:def 4;
end;

definition
  let x1,x2,x3;
  func { x1,x2,x3 } -> set means
  :Def1:
  x in it iff x=x1 or x=x2 or x=x3;
  existence
  proof
    take union({{x1,x2},{x3}});
    let x;
    x in { x1,x2 } iff x = x1 or x = x2 by TARSKI:def 2;
    hence thesis by Lm1;
  end;
  uniqueness
  proof
    defpred P[set] means $1=x1 or $1=x2 or $1=x3;
    for X1,X2 being set st (for x being set holds x in X1 iff P[x]) & (for
    x being set holds x in X2 iff P[x]) holds X1 = X2 from XBOOLE_0:sch 3;
    hence thesis;
  end;
end;

definition
  let x1,x2,x3,x4;
  func { x1,x2,x3,x4 } -> set means
  :Def2:
  x in it iff x=x1 or x=x2 or x=x3 or x=x4;
  existence
  proof
    take union({{x1,x2,x3},{x4}});
    let x;
    x in { x1,x2,x3 } iff x = x1 or x = x2 or x = x3 by Def1;
    hence thesis by Lm1;
  end;
  uniqueness
  proof
    defpred P[set] means $1=x1 or $1=x2 or $1=x3 or $1=x4;
    for X1,X2 being set st (for x being set holds x in X1 iff P[x]) & (for
    x being set holds x in X2 iff P[x]) holds X1 = X2 from XBOOLE_0:sch 3;
    hence thesis;
  end;
end;

definition
  let x1,x2,x3,x4,x5;
  func { x1,x2,x3,x4,x5 } -> set means
  :Def3:
  x in it iff x=x1 or x=x2 or x=x3 or x=x4 or x=x5;
  existence
  proof
    take union({{x1,x2,x3,x4},{x5}});
    let x;
    x in { x1,x2,x3,x4 } iff x = x1 or x = x2 or x = x3 or x = x4 by Def2;
    hence thesis by Lm1;
  end;
  uniqueness
  proof
    defpred P[set] means $1=x1 or $1=x2 or $1=x3 or $1=x4 or $1=x5;
    for X1,X2 being set st (for x being set holds x in X1 iff P[x]) & (for
    x being set holds x in X2 iff P[x]) holds X1 = X2 from XBOOLE_0:sch 3;
    hence thesis;
  end;
end;

definition
  let x1,x2,x3,x4,x5,x6;
  func { x1,x2,x3,x4,x5,x6 } -> set means
  :Def4:
  x in it iff x=x1 or x=x2 or x =x3 or x=x4 or x=x5 or x=x6;
  existence
  proof
    take union({{x1,x2,x3,x4,x5},{x6}});
    let x;
    x in {x1,x2,x3,x4,x5} iff x=x1 or x=x2 or x=x3 or x=x4 or x=x5 by Def3;
    hence thesis by Lm1;
  end;
  uniqueness
  proof
    defpred P[set] means $1=x1 or $1=x2 or $1=x3 or $1=x4 or $1=x5 or $1=x6;
    for X1,X2 being set st (for x being set holds x in X1 iff P[x]) & (for
    x being set holds x in X2 iff P[x]) holds X1 = X2 from XBOOLE_0:sch 3;
    hence thesis;
  end;
end;

definition
  let x1,x2,x3,x4,x5,x6,x7;
  func { x1,x2,x3,x4,x5,x6,x7 } -> set means
  :Def5:
  x in it iff x=x1 or x=x2 or x=x3 or x=x4 or x=x5 or x=x6 or x=x7;
  existence
  proof
    take union({{x1,x2,x3,x4,x5,x6},{x7}});
    let x;
    x in { x1,x2,x3,x4,x5,x6 } iff x=x1 or x=x2 or x=x3 or x=x4 or x=x5 or
    x=x6 by Def4;
    hence thesis by Lm1;
  end;
  uniqueness
  proof
    defpred P[set] means $1=x1 or $1=x2 or $1=x3 or $1=x4 or $1=x5 or $1=x6 or
    $1=x7;
    for X1,X2 being set st (for x being set holds x in X1 iff P[x]) & (for
    x being set holds x in X2 iff P[x]) holds X1 = X2 from XBOOLE_0:sch 3;
    hence thesis;
  end;
end;

definition
  let x1,x2,x3,x4,x5,x6,x7,x8;
  func { x1,x2,x3,x4,x5,x6,x7,x8 } -> set means
  :Def6:
  x in it iff x=x1 or x= x2 or x=x3 or x=x4 or x=x5 or x=x6 or x=x7 or x=x8;
  existence
  proof
    take union({{x1,x2,x3,x4,x5,x6,x7},{x8}});
    let x;
    x in { x1,x2,x3,x4,x5,x6,x7 } iff x=x1 or x=x2 or x=x3 or x=x4 or x=x5
    or x = x6 or x=x7 by Def5;
    hence thesis by Lm1;
  end;
  uniqueness
  proof
    defpred P[set] means $1=x1 or $1=x2 or $1=x3 or $1=x4 or $1=x5 or $1=x6 or
    $1=x7 or $1=x8;
    for X1,X2 being set st (for x being set holds x in X1 iff P[x]) & (for
    x being set holds x in X2 iff P[x]) holds X1 = X2 from XBOOLE_0:sch 3;
    hence thesis;
  end;
end;

definition
  let x1,x2,x3,x4,x5,x6,x7,x8,x9;
  func { x1,x2,x3,x4,x5,x6,x7,x8,x9 } -> set means
  :Def7:
  x in it iff x=x1 or
  x=x2 or x=x3 or x=x4 or x=x5 or x=x6 or x=x7 or x=x8 or x=x9;
  existence
  proof
    take union({{x1,x2,x3,x4,x5,x6,x7,x8},{x9}});
    let x;
    x in { x1,x2,x3,x4,x5,x6,x7,x8 } iff x=x1 or x=x2 or x=x3 or x=x4 or x
    =x5 or x = x6 or x=x7 or x=x8 by Def6;
    hence thesis by Lm1;
  end;
  uniqueness
  proof
    defpred P[set] means $1=x1 or $1=x2 or $1=x3 or $1=x4 or $1=x5 or $1=x6 or
    $1=x7 or $1=x8 or $1=x9;
    thus for X, Y being set st (for x being set holds x in X iff P[x]) & (for
    x being set holds x in Y iff P[x]) holds X = Y from XBOOLE_0:sch 3;
  end;
end;

definition
  let x1,x2,x3,x4,x5,x6,x7,x8,x9,x10;
  func { x1,x2,x3,x4,x5,x6,x7,x8,x9,x10 } -> set means
  :Def8:
  x in it iff x=x1
  or x=x2 or x=x3 or x=x4 or x=x5 or x=x6 or x=x7 or x=x8 or x=x9 or x=x10;
  existence
  proof
    take union({{x1,x2,x3,x4,x5,x6,x7,x8,x9},{x10}});
    let x;
    x in { x1,x2,x3,x4,x5,x6,x7,x8,x9 } iff x=x1 or x=x2 or x=x3 or x=x4
    or x=x5 or x = x6 or x=x7 or x=x8 or x=x9 by Def7;
    hence thesis by Lm1;
  end;
  uniqueness
  proof
    defpred P[set] means $1=x1 or $1=x2 or $1=x3 or $1=x4 or $1=x5 or $1=x6 or
    $1=x7 or $1=x8 or $1=x9 or $1=x10;
    thus for X, Y being set st (for x being set holds x in X iff P[x]) & (for
    x being set holds x in Y iff P[x]) holds X = Y from XBOOLE_0:sch 3;
  end;
end;

theorem Th1:
  { x1,x2 } = { x1 } \/ { x2 }
proof
  now
    let x;
    x in { x1,x2 } iff x=x1 or x=x2 by TARSKI:def 2;
    then x in { x1,x2 } iff x in { x1 } or x in { x2 } by TARSKI:def 1;
    hence x in { x1,x2 } iff x in { x1 } \/ { x2 } by XBOOLE_0:def 3;
  end;
  hence thesis by TARSKI:1;
end;

theorem Th2:
  { x1,x2,x3 } = { x1 } \/ { x2,x3 }
proof
  now
    let x;
    x in { x1,x2,x3 } iff x=x1 or x=x2 or x=x3 by Def1;
    then x in { x1,x2,x3 } iff x in { x1 } or x in { x2,x3 } by TARSKI:def 1
,def 2;
    hence x in { x1,x2,x3 } iff x in { x1 } \/ { x2,x3 } by XBOOLE_0:def 3;
  end;
  hence thesis by TARSKI:1;
end;

theorem Th3:
  { x1,x2,x3 } = { x1,x2 } \/ { x3 }
proof
  thus { x1,x2,x3 } = { x1 } \/ { x2,x3 } by Th2
    .= { x1 } \/ ({ x2 } \/ { x3 }) by Th1
    .= { x1 } \/ { x2 } \/ { x3 } by XBOOLE_1:4
    .= { x1,x2 } \/ { x3 } by Th1;
end;

Lm2: { x1,x2,x3,x4 } = { x1,x2 } \/ { x3,x4 }
proof
  now
    let x;
    x in { x1,x2,x3,x4 } iff x=x1 or x=x2 or x=x3 or x=x4 by Def2;
    then x in { x1,x2,x3,x4 } iff x in { x1,x2 } or x in { x3,x4 } by
TARSKI:def 2;
    hence x in { x1,x2,x3,x4 } iff x in { x1,x2 } \/ { x3,x4 } by
XBOOLE_0:def 3;
  end;
  hence thesis by TARSKI:1;
end;

theorem Th4:
  { x1,x2,x3,x4 } = { x1 } \/ { x2,x3,x4 }
proof
  thus { x1,x2,x3,x4 } = { x1,x2 } \/ { x3,x4 } by Lm2
    .= { x1 } \/ { x2 } \/ { x3,x4 } by Th1
    .= { x1 } \/ ({ x2 } \/ { x3,x4 }) by XBOOLE_1:4
    .= { x1 } \/ { x2,x3,x4 } by Th2;
end;

theorem
  { x1,x2,x3,x4 } = { x1,x2 } \/ { x3,x4 } by Lm2;

theorem Th6:
  { x1,x2,x3,x4 } = { x1,x2,x3 } \/ { x4 }
proof
  thus { x1,x2,x3,x4 } = { x1,x2 } \/ { x3,x4 } by Lm2
    .= { x1,x2 } \/ ({ x3 } \/ { x4 }) by Th1
    .= { x1,x2 } \/ { x3 } \/ { x4 } by XBOOLE_1:4
    .= { x1,x2,x3 } \/ { x4 } by Th3;
end;

Lm3: { x1,x2,x3,x4,x5 } = { x1,x2,x3 } \/ { x4,x5 }
proof
  now
    let x;
    x in { x1,x2,x3,x4,x5 } iff x=x1 or x=x2 or x=x3 or x=x4 or x=x5 by Def3;
    then x in { x1,x2,x3,x4,x5 } iff x in { x1,x2,x3 } or x in { x4,x5 } by
Def1,TARSKI:def 2;
    hence x in { x1,x2,x3,x4,x5 } iff x in { x1,x2,x3 } \/ { x4,x5 } by
XBOOLE_0:def 3;
  end;
  hence thesis by TARSKI:1;
end;

theorem Th7:
  { x1,x2,x3,x4,x5 } = { x1 } \/ { x2,x3,x4,x5 }
proof
  thus { x1,x2,x3,x4,x5 } = { x1,x2,x3 } \/ { x4,x5 } by Lm3
    .= { x1 } \/ { x2,x3 } \/ { x4,x5 } by Th2
    .= { x1 } \/ ({ x2,x3 } \/ { x4,x5 }) by XBOOLE_1:4
    .= { x1 } \/ { x2,x3,x4,x5 } by Lm2;
end;

theorem Th8:
  { x1,x2,x3,x4,x5 } = { x1,x2 } \/ { x3,x4,x5 }
proof
  thus { x1,x2,x3,x4,x5 } = { x1,x2,x3 } \/ { x4,x5 } by Lm3
    .= { x1,x2 } \/ { x3 } \/ { x4,x5 } by Th3
    .= { x1,x2 } \/ ({ x3 } \/ { x4,x5 }) by XBOOLE_1:4
    .= { x1,x2 } \/ { x3,x4,x5 } by Th2;
end;

theorem
  { x1,x2,x3,x4,x5 } = { x1,x2,x3 } \/ { x4,x5 } by Lm3;

theorem Th10:
  { x1,x2,x3,x4,x5 } = { x1,x2,x3,x4 } \/ { x5 }
proof
  thus { x1,x2,x3,x4,x5 } = { x1,x2,x3 } \/ { x4,x5 } by Lm3
    .= { x1,x2,x3 } \/ ({ x4 } \/ { x5 }) by Th1
    .= { x1,x2,x3 } \/ { x4 } \/ { x5 } by XBOOLE_1:4
    .= { x1,x2,x3,x4 } \/ { x5 } by Th6;
end;

Lm4: { x1,x2,x3,x4,x5,x6 } = { x1,x2,x3 } \/ { x4,x5,x6 }
proof
  now
    let x;
    x in { x1,x2,x3,x4,x5,x6 } iff x=x1 or x=x2 or x=x3 or x=x4 or x=x5 or
    x=x6 by Def4;
    then
    x in { x1,x2,x3,x4,x5,x6 } iff x in { x1,x2,x3 } or x in { x4,x5,x6 }
    by Def1;
    hence x in { x1,x2,x3,x4,x5,x6 } iff x in { x1,x2,x3 } \/ { x4,x5,x6 } by
XBOOLE_0:def 3;
  end;
  hence thesis by TARSKI:1;
end;

theorem Th11:
  { x1,x2,x3,x4,x5,x6 } = { x1 } \/ { x2,x3,x4,x5,x6 }
proof
  thus { x1,x2,x3,x4,x5,x6 } = { x1,x2,x3 } \/ { x4,x5,x6 } by Lm4
    .= { x1 } \/ { x2,x3 } \/ { x4,x5,x6 } by Th2
    .= { x1 } \/ ({ x2,x3 } \/ { x4,x5,x6 }) by XBOOLE_1:4
    .= { x1 } \/ { x2,x3,x4,x5,x6 } by Th8;
end;

theorem Th12:
  { x1,x2,x3,x4,x5,x6 } = { x1,x2 } \/ { x3,x4,x5,x6 }
proof
  thus { x1,x2,x3,x4,x5,x6 } = { x1,x2,x3 } \/ { x4,x5,x6 } by Lm4
    .= { x1,x2 } \/ { x3 } \/ { x4,x5,x6 } by Th3
    .= { x1,x2 } \/ ({ x3 } \/ { x4,x5,x6 }) by XBOOLE_1:4
    .= { x1,x2 } \/ { x3,x4,x5,x6 } by Th4;
end;

theorem
  { x1,x2,x3,x4,x5,x6 } = { x1,x2,x3 } \/ { x4,x5,x6 } by Lm4;

theorem Th14:
  { x1,x2,x3,x4,x5,x6 } = { x1,x2,x3,x4 } \/ { x5,x6 }
proof
  thus { x1,x2,x3,x4,x5,x6 } = { x1,x2,x3 } \/ { x4,x5,x6 } by Lm4
    .= { x1,x2,x3 } \/ ({ x4 } \/ { x5,x6 }) by Th2
    .= { x1,x2,x3 } \/ { x4 } \/ { x5,x6 } by XBOOLE_1:4
    .= { x1,x2,x3,x4 } \/ { x5,x6 } by Th6;
end;

theorem
  { x1,x2,x3,x4,x5,x6 } = { x1,x2,x3,x4,x5 } \/ { x6 }
proof
  thus { x1,x2,x3,x4,x5,x6 } = { x1,x2,x3 } \/ { x4,x5,x6 } by Lm4
    .= { x1,x2,x3 } \/ ({ x4,x5 } \/ { x6 }) by Th3
    .= { x1,x2,x3 } \/ { x4,x5 } \/ { x6 } by XBOOLE_1:4
    .= { x1,x2,x3,x4,x5 } \/ { x6 } by Lm3;
end;

Lm5: { x1,x2,x3,x4,x5,x6,x7 } = { x1,x2,x3,x4 } \/ { x5,x6,x7 }
proof
  now
    let x;
A1: x in { x5,x6,x7 } iff x=x5 or x=x6 or x=x7 by Def1;
    x in { x1,x2,x3,x4 } iff x=x1 or x=x2 or x=x3 or x=x4 by Def2;
    hence x in { x1,x2,x3,x4,x5,x6,x7 } iff x in { x1,x2,x3,x4 } \/ { x5,x6,x7
    } by A1,Def5,XBOOLE_0:def 3;
  end;
  hence thesis by TARSKI:1;
end;

theorem Th16:
  { x1,x2,x3,x4,x5,x6,x7 } = { x1 } \/ { x2,x3,x4,x5,x6,x7 }
proof
  thus { x1,x2,x3,x4,x5,x6,x7 } = { x1,x2,x3,x4 } \/ { x5,x6,x7 } by Lm5
    .= { x1 } \/ { x2,x3,x4 } \/ { x5,x6,x7 } by Th4
    .= { x1 } \/ ({ x2,x3,x4 } \/ { x5,x6,x7 }) by XBOOLE_1:4
    .= { x1 } \/ { x2,x3,x4,x5,x6,x7 } by Lm4;
end;

theorem Th17:
  { x1,x2,x3,x4,x5,x6,x7 } = { x1,x2 } \/ { x3,x4,x5,x6,x7 }
proof
  thus { x1,x2,x3,x4,x5,x6,x7 } = { x1,x2,x3,x4 } \/ { x5,x6,x7 } by Lm5
    .= { x1,x2 } \/ { x3,x4 } \/ { x5,x6,x7 } by Lm2
    .= { x1,x2 } \/ ({ x3,x4 } \/ { x5,x6,x7 }) by XBOOLE_1:4
    .= { x1,x2 } \/ { x3,x4,x5,x6,x7 } by Th8;
end;

theorem Th18:
  { x1,x2,x3,x4,x5,x6,x7 } = { x1,x2,x3 } \/ { x4,x5,x6,x7 }
proof
  thus { x1,x2,x3,x4,x5,x6,x7 } = { x1,x2,x3,x4 } \/ { x5,x6,x7 } by Lm5
    .= { x1,x2,x3 } \/ { x4 } \/ { x5,x6,x7 } by Th6
    .= { x1,x2,x3 } \/ ({ x4 } \/ { x5,x6,x7 }) by XBOOLE_1:4
    .= { x1,x2,x3 } \/ { x4,x5,x6,x7 } by Th4;
end;

theorem
  { x1,x2,x3,x4,x5,x6,x7 } = { x1,x2,x3,x4 } \/ { x5,x6,x7 } by Lm5;

theorem
  { x1,x2,x3,x4,x5,x6,x7 } = { x1,x2,x3,x4,x5 } \/ { x6,x7 }
proof
  thus { x1,x2,x3,x4,x5,x6,x7 } = { x1,x2,x3,x4 } \/ { x5,x6,x7 } by Lm5
    .= { x1,x2,x3,x4 } \/ ({ x5 } \/ { x6,x7 }) by Th2
    .= { x1,x2,x3,x4 } \/ { x5 } \/ { x6,x7 } by XBOOLE_1:4
    .= { x1,x2,x3,x4,x5 } \/ { x6,x7 } by Th10;
end;

theorem
  { x1,x2,x3,x4,x5,x6,x7 } = { x1,x2,x3,x4,x5,x6 } \/ { x7 }
proof
  thus { x1,x2,x3,x4,x5,x6,x7 } = { x1,x2,x3,x4 } \/ { x5,x6,x7 } by Lm5
    .= { x1,x2,x3,x4 } \/ ({ x5,x6 } \/ { x7 }) by Th3
    .= { x1,x2,x3,x4 } \/ { x5,x6 } \/ { x7 } by XBOOLE_1:4
    .= { x1,x2,x3,x4,x5,x6 } \/ { x7 } by Th14;
end;

Lm6: { x1,x2,x3,x4,x5,x6,x7,x8 } = { x1,x2,x3,x4 } \/ { x5,x6,x7,x8 }
proof
  now
    let x;
A1: x in { x5,x6,x7,x8 } iff x=x5 or x=x6 or x=x7 or x=x8 by Def2;
    x in { x1,x2,x3,x4 } iff x=x1 or x=x2 or x=x3 or x=x4 by Def2;
    hence
    x in { x1,x2,x3,x4,x5,x6,x7,x8 } iff x in { x1,x2,x3,x4 } \/ { x5,x6,
    x7,x8 } by A1,Def6,XBOOLE_0:def 3;
  end;
  hence thesis by TARSKI:1;
end;

theorem
  { x1,x2,x3,x4,x5,x6,x7,x8 } = { x1 } \/ { x2,x3,x4,x5,x6,x7,x8 }
proof
  thus { x1,x2,x3,x4,x5,x6,x7,x8 } = { x1,x2,x3,x4 } \/ { x5,x6,x7,x8 } by Lm6
    .= { x1 } \/ { x2,x3,x4 } \/ { x5,x6,x7,x8 } by Th4
    .= { x1 } \/ ({ x2,x3,x4 } \/ { x5,x6,x7,x8 }) by XBOOLE_1:4
    .= { x1 } \/ { x2,x3,x4,x5,x6,x7,x8 } by Th18;
end;

theorem Th23:
  { x1,x2,x3,x4,x5,x6,x7,x8 } = { x1,x2 } \/ { x3,x4,x5,x6,x7,x8 }
proof
  thus { x1,x2,x3,x4,x5,x6,x7,x8 } = { x1,x2,x3,x4 } \/ { x5,x6,x7,x8 } by Lm6
    .= { x1,x2 } \/ { x3,x4 } \/ { x5,x6,x7,x8 } by Lm2
    .= { x1,x2 } \/ ({ x3,x4 } \/ { x5,x6,x7,x8 }) by XBOOLE_1:4
    .= { x1,x2 } \/ { x3,x4,x5,x6,x7,x8 } by Th12;
end;

theorem Th24:
  { x1,x2,x3,x4,x5,x6,x7,x8 } = { x1,x2,x3 } \/ { x4,x5,x6,x7,x8 }
proof
  thus { x1,x2,x3,x4,x5,x6,x7,x8 } = { x1,x2,x3,x4 } \/ { x5,x6,x7,x8 } by Lm6
    .= { x1,x2,x3 } \/ { x4 } \/ { x5,x6,x7,x8 } by Th6
    .= { x1,x2,x3 } \/ ({ x4 } \/ { x5,x6,x7,x8 }) by XBOOLE_1:4
    .= { x1,x2,x3 } \/ { x4,x5,x6,x7,x8 } by Th7;
end;

theorem
  { x1,x2,x3,x4,x5,x6,x7,x8 } = { x1,x2,x3,x4 } \/ { x5,x6,x7,x8 } by Lm6;

theorem
  { x1,x2,x3,x4,x5,x6,x7,x8 } = { x1,x2,x3,x4,x5 } \/ { x6,x7,x8 }
proof
  thus { x1,x2,x3,x4,x5,x6,x7,x8 } = { x1,x2,x3,x4 } \/ { x5,x6,x7,x8 } by Lm6
    .= { x1,x2,x3,x4 } \/ ({x5 } \/ { x6,x7,x8 }) by Th4
    .= { x1,x2,x3,x4 } \/ {x5 } \/ { x6,x7,x8 } by XBOOLE_1:4
    .= { x1,x2,x3,x4,x5 } \/ { x6,x7,x8 } by Th10;
end;

theorem
  { x1,x2,x3,x4,x5,x6,x7,x8 } = { x1,x2,x3,x4,x5,x6 } \/ { x7,x8 }
proof
  thus { x1,x2,x3,x4,x5,x6,x7,x8 } = { x1,x2,x3,x4 } \/ { x5,x6,x7,x8 } by Lm6
    .= { x1,x2,x3,x4 } \/ ({ x5,x6 } \/ { x7,x8 }) by Lm2
    .= { x1,x2,x3,x4 } \/ { x5,x6 } \/ { x7,x8 } by XBOOLE_1:4
    .= { x1,x2,x3,x4,x5,x6 } \/ { x7,x8 } by Th14;
end;

theorem
  { x1,x2,x3,x4,x5,x6,x7,x8 } = { x1,x2,x3,x4,x5,x6,x7 } \/ { x8 }
proof
  thus { x1,x2,x3,x4,x5,x6,x7,x8 } = { x1,x2,x3,x4 } \/ { x5,x6,x7,x8 } by Lm6
    .= { x1,x2,x3,x4 } \/ ({ x5,x6,x7 } \/ { x8 }) by Th6
    .= { x1,x2,x3,x4 } \/ { x5,x6,x7 } \/ { x8 } by XBOOLE_1:4
    .= { x1,x2,x3,x4,x5,x6,x7 } \/ { x8 } by Lm5;
end;

theorem Th29:
  { x1,x1 } = { x1 }
proof
  now
    let x;
    x in { x1,x1 } iff x = x1 by TARSKI:def 2;
    hence x in { x1,x1 } iff x in { x1 } by TARSKI:def 1;
  end;
  hence thesis by TARSKI:1;
end;

theorem Th30:
  { x1,x1,x2 } = { x1,x2 }
proof
  thus { x1,x1,x2 } = { x1,x1 } \/ { x2 } by Th3
    .= { x1 } \/ { x2 } by Th29
    .= { x1,x2 } by Th1;
end;

theorem Th31:
  { x1,x1,x2,x3 } = { x1,x2,x3 }
proof
  thus { x1,x1,x2,x3 } = { x1,x1 } \/ { x2,x3 } by Lm2
    .= { x1 } \/ { x2,x3 } by Th29
    .= { x1,x2,x3 } by Th2;
end;

theorem Th32:
  { x1,x1,x2,x3,x4 } = { x1,x2,x3,x4 }
proof
  thus { x1,x1,x2,x3,x4 } = { x1,x1 } \/ { x2,x3,x4 } by Th8
    .= { x1 } \/ { x2,x3,x4 } by Th29
    .= { x1,x2,x3,x4 } by Th4;
end;

theorem Th33:
  { x1,x1,x2,x3,x4,x5 } = { x1,x2,x3,x4,x5 }
proof
  thus { x1,x1,x2,x3,x4,x5 } = { x1,x1 } \/ { x2,x3,x4,x5 } by Th12
    .= { x1 } \/ { x2,x3,x4,x5 } by Th29
    .= { x1,x2,x3,x4,x5 } by Th7;
end;

theorem Th34:
  { x1,x1,x2,x3,x4,x5,x6 } = { x1,x2,x3,x4,x5,x6 }
proof
  thus { x1,x1,x2,x3,x4,x5,x6 } = { x1,x1 } \/ { x2,x3,x4,x5,x6 } by Th17
    .= { x1 } \/ { x2,x3,x4,x5,x6 } by Th29
    .= { x1,x2,x3,x4,x5,x6 } by Th11;
end;

theorem Th35:
  { x1,x1,x2,x3,x4,x5,x6,x7 } = { x1,x2,x3,x4,x5,x6,x7 }
proof
  thus { x1,x1,x2,x3,x4,x5,x6,x7 } = { x1,x1 } \/ { x2,x3,x4,x5,x6,x7 } by Th23
    .= { x1 } \/ { x2,x3,x4,x5,x6,x7 } by Th29
    .= { x1,x2,x3,x4,x5,x6,x7 } by Th16;
end;

theorem
  { x1,x1,x1 } = { x1 }
proof
  thus { x1,x1,x1 } = { x1,x1 } by Th30
    .= { x1 } by Th29;
end;

theorem Th37:
  { x1,x1,x1,x2 } = { x1,x2 }
proof
  thus { x1,x1,x1,x2 } = { x1,x1,x2 } by Th31
    .= { x1,x2 } by Th30;
end;

theorem Th38:
  { x1,x1,x1,x2,x3 } = { x1,x2,x3 }
proof
  thus { x1,x1,x1,x2,x3 } = { x1,x1,x2,x3 } by Th32
    .= { x1,x2,x3 } by Th31;
end;

theorem Th39:
  { x1,x1,x1,x2,x3,x4 } = { x1,x2,x3,x4 }
proof
  thus { x1,x1,x1,x2,x3,x4 } = { x1,x1,x2,x3,x4 } by Th33
    .= { x1,x2,x3,x4 } by Th32;
end;

theorem Th40:
  { x1,x1,x1,x2,x3,x4,x5 } = { x1,x2,x3,x4,x5 }
proof
  thus { x1,x1,x1,x2,x3,x4,x5 } = { x1,x1,x2,x3,x4,x5 } by Th34
    .= { x1,x2,x3,x4,x5 } by Th33;
end;

theorem Th41:
  { x1,x1,x1,x2,x3,x4,x5,x6 } = { x1,x2,x3,x4,x5,x6 }
proof
  thus { x1,x1,x1,x2,x3,x4,x5,x6 } = { x1,x1,x2,x3,x4,x5,x6 } by Th35
    .= { x1,x2,x3,x4,x5,x6 } by Th34;
end;

theorem
  { x1,x1,x1,x1 } = { x1 }
proof
  thus { x1,x1,x1,x1 } = { x1,x1 } by Th37
    .= { x1 } by Th29;
end;

theorem Th43:
  { x1,x1,x1,x1,x2 } = { x1,x2 }
proof
  thus { x1,x1,x1,x1,x2 } = { x1,x1,x2 } by Th38
    .= { x1,x2 } by Th30;
end;

theorem Th44:
  { x1,x1,x1,x1,x2,x3 } = { x1,x2,x3 }
proof
  thus { x1,x1,x1,x1,x2,x3 } = { x1,x1,x2,x3 } by Th39
    .= { x1,x2,x3 } by Th31;
end;

theorem Th45:
  { x1,x1,x1,x1,x2,x3,x4 } = { x1,x2,x3,x4 }
proof
  thus { x1,x1,x1,x1,x2,x3,x4 } = { x1,x1,x2,x3,x4 } by Th40
    .= { x1,x2,x3,x4 } by Th32;
end;

theorem Th46:
  { x1,x1,x1,x1,x2,x3,x4,x5 } = { x1,x2,x3,x4,x5 }
proof
  thus { x1,x1,x1,x1,x2,x3,x4,x5 } = { x1,x1,x2,x3,x4,x5 } by Th41
    .= { x1,x2,x3,x4,x5 } by Th33;
end;

theorem
  { x1,x1,x1,x1,x1 } = { x1 }
proof
  thus { x1,x1,x1,x1,x1 } = { x1,x1 } by Th43
    .= { x1 } by Th29;
end;

theorem Th48:
  { x1,x1,x1,x1,x1,x2 } = { x1,x2 }
proof
  thus { x1,x1,x1,x1,x1,x2 } = { x1,x1,x2 } by Th44
    .= { x1,x2 } by Th30;
end;

theorem Th49:
  { x1,x1,x1,x1,x1,x2,x3 } = { x1,x2,x3 }
proof
  thus { x1,x1,x1,x1,x1,x2,x3 } = { x1,x1,x2,x3 } by Th45
    .= { x1,x2,x3 } by Th31;
end;

theorem Th50:
  { x1,x1,x1,x1,x1,x2,x3,x4 } = { x1,x2,x3,x4 }
proof
  thus { x1,x1,x1,x1,x1,x2,x3,x4 } = { x1,x1,x2,x3,x4 } by Th46
    .= { x1,x2,x3,x4 } by Th32;
end;

theorem
  { x1,x1,x1,x1,x1,x1 } = { x1 }
proof
  thus { x1,x1,x1,x1,x1,x1 } = { x1,x1 } by Th48
    .= { x1 } by Th29;
end;

theorem Th52:
  { x1,x1,x1,x1,x1,x1,x2 } = { x1,x2 }
proof
  thus { x1,x1,x1,x1,x1,x1,x2 } = { x1,x1,x2 } by Th49
    .= { x1,x2 } by Th30;
end;

theorem Th53:
  { x1,x1,x1,x1,x1,x1,x2,x3 } = { x1,x2,x3 }
proof
  thus { x1,x1,x1,x1,x1,x1,x2,x3 } = { x1,x1,x2,x3 } by Th50
    .= { x1,x2,x3 } by Th31;
end;

theorem
  { x1,x1,x1,x1,x1,x1,x1 } = { x1 }
proof
  thus { x1,x1,x1,x1,x1,x1,x1 } = { x1,x1 } by Th52
    .= { x1 } by Th29;
end;

theorem Th55:
  { x1,x1,x1,x1,x1,x1,x1,x2 } = { x1,x2 }
proof
  thus { x1,x1,x1,x1,x1,x1,x1,x2 } = { x1,x1,x2 } by Th53
    .= { x1,x2 } by Th30;
end;

theorem
  { x1,x1,x1,x1,x1,x1,x1,x1 } = { x1 }
proof
  thus { x1,x1,x1,x1,x1,x1,x1,x1 } = { x1,x1 } by Th55
    .= { x1 } by Th29;
end;

theorem Th57:
  { x1,x2,x3 } = { x1,x3,x2 }
proof
  thus { x1,x2,x3 } = { x1 } \/ { x2,x3 } by Th2
    .= { x1,x3,x2 } by Th2;
end;

theorem Th58:
  { x1,x2,x3 } = { x2,x1,x3 }
proof
  thus { x1,x2,x3 } = { x1,x2 } \/ { x3 } by Th3
    .= { x2,x1,x3 } by Th3;
end;

theorem Th59:
  { x1,x2,x3 } = { x2,x3,x1 }
proof
  thus { x1,x2,x3 } = { x2,x3 } \/ { x1 } by Th2
    .= { x2,x3,x1 } by Th3;
end;

theorem Th60:
  { x1,x2,x3 } = { x3,x2,x1 }
proof
  thus { x1,x2,x3 } = { x3,x1,x2 } by Th59
    .= { x3,x2,x1 } by Th57;
end;

theorem Th61:
  { x1,x2,x3,x4 } = { x1,x2,x4,x3 }
proof
  thus { x1,x2,x3,x4 } = { x1 } \/ { x2,x3,x4 } by Th4
    .= { x1 } \/ { x2,x4,x3 } by Th57
    .= { x1,x2,x4,x3 } by Th4;
end;

theorem
  { x1,x2,x3,x4 } = { x1,x3,x2,x4 }
proof
  thus { x1,x2,x3,x4 } = { x1 } \/ { x2,x3,x4 } by Th4
    .= { x1 } \/ { x3,x2,x4 } by Th58
    .= { x1,x3,x2,x4 } by Th4;
end;

theorem Th63:
  { x1,x2,x3,x4 } = { x1,x3,x4,x2 }
proof
  thus { x1,x2,x3,x4 } = { x1 } \/ { x2,x3,x4 } by Th4
    .= { x1 } \/ { x3,x4,x2 } by Th59
    .= { x1,x3,x4,x2 } by Th4;
end;

theorem Th64:
  { x1,x2,x3,x4 } = { x1,x4,x3,x2 }
proof
  thus { x1,x2,x3,x4 } = { x1 } \/ { x2,x3,x4 } by Th4
    .= { x1 } \/ { x4,x3,x2 } by Th60
    .= { x1,x4,x3,x2 } by Th4;
end;

theorem Th65:
  { x1,x2,x3,x4 } = { x2,x1,x3,x4 }
proof
  thus { x1,x2,x3,x4 } = { x1,x2,x3 } \/ { x4 } by Th6
    .= { x2,x1,x3 } \/ { x4 } by Th58
    .= { x2,x1,x3,x4 } by Th6;
end;

Lm7: { x1,x2,x3,x4 } = { x2,x3,x1,x4 }
proof
  thus { x1,x2,x3,x4 } = { x1,x2,x3 } \/ { x4 } by Th6
    .= { x2,x3,x1 } \/ { x4 } by Th59
    .= { x2,x3,x1,x4 } by Th6;
end;

theorem
  { x1,x2,x3,x4 } = { x2,x1,x4,x3 }
proof
  thus { x1,x2,x3,x4 } = { x2,x3,x1,x4 } by Lm7
    .= { x2,x1,x4,x3 } by Th63;
end;

theorem
  { x1,x2,x3,x4 } = { x2,x3,x1,x4 } by Lm7;

theorem
  { x1,x2,x3,x4 } = { x2,x3,x4,x1 }
proof
  thus { x1,x2,x3,x4 } = { x2,x3,x1,x4 } by Lm7
    .= { x2,x3,x4,x1 } by Th61;
end;

theorem Th69:
  { x1,x2,x3,x4 } = { x2,x4,x1,x3 }
proof
  thus { x1,x2,x3,x4 } = { x2,x3,x1,x4 } by Lm7
    .= { x2,x4,x1,x3 } by Th64;
end;

theorem
  { x1,x2,x3,x4 } = { x2,x4,x3,x1 }
proof
  thus { x1,x2,x3,x4 } = { x2,x3,x1,x4 } by Lm7
    .= { x2,x4,x3,x1 } by Th63;
end;

Lm8: { x1,x2,x3,x4 } = { x3,x2,x1,x4 }
proof
  thus { x1,x2,x3,x4 } = { x1,x2,x3 } \/ { x4 } by Th6
    .= { x3,x2,x1 } \/ { x4 } by Th60
    .= { x3,x2,x1,x4 } by Th6;
end;

theorem
  { x1,x2,x3,x4 } = { x3,x2,x1,x4 } by Lm8;

theorem
  { x1,x2,x3,x4 } = { x3,x2,x4,x1 }
proof
  thus { x1,x2,x3,x4 } = { x3,x2,x1,x4 } by Lm8
    .= { x3,x2,x4,x1 } by Th61;
end;

theorem
  { x1,x2,x3,x4 } = { x3,x4,x1,x2 }
proof
  thus { x1,x2,x3,x4 } = { x3,x2,x1,x4 } by Lm8
    .= { x3,x4,x1,x2 } by Th64;
end;

theorem Th74:
  { x1,x2,x3,x4 } = { x3,x4,x2,x1 }
proof
  thus { x1,x2,x3,x4 } = { x3,x2,x1,x4 } by Lm8
    .= { x3,x4,x2,x1 } by Th63;
end;

theorem
  { x1,x2,x3,x4 } = { x4,x2,x3,x1 }
proof
  thus { x1,x2,x3,x4 } = { x3,x4,x2,x1 } by Th74
    .= { x4,x2,x3,x1 } by Lm7;
end;

theorem
  { x1,x2,x3,x4 } = { x4,x3,x2,x1 }
proof
  thus { x1,x2,x3,x4 } = { x3,x4,x2,x1 } by Th74
    .= { x4,x3,x2,x1 } by Th65;
end;

Lm9: { x1,x2,x3,x4,x5,x6,x7,x8,x9 } = { x1,x2,x3,x4 } \/ { x5,x6,x7,x8,x9 }
proof
  now
    let x;
A1: x in { x5,x6,x7,x8,x9 } iff x=x5 or x=x6 or x=x7 or x=x8 or x=x9 by Def3;
    x in { x1,x2,x3,x4 } iff x=x1 or x=x2 or x=x3 or x=x4 by Def2;
    hence
    x in { x1,x2,x3,x4,x5,x6,x7,x8,x9 } iff x in { x1,x2,x3,x4 } \/ { x5,
    x6,x7,x8,x9 } by A1,Def7,XBOOLE_0:def 3;
  end;
  hence thesis by TARSKI:1;
end;

theorem
  { x1,x2,x3,x4,x5,x6,x7,x8,x9 } = { x1 } \/ { x2,x3,x4,x5,x6,x7,x8,x9 }
proof
  thus { x1,x2,x3,x4,x5,x6,x7,x8,x9 } = { x1,x2,x3,x4 } \/ { x5,x6,x7,x8,x9 }
  by Lm9
    .= ({ x1 } \/ { x2,x3,x4 }) \/ { x5,x6,x7,x8,x9 } by Th4
    .= { x1 } \/ ({ x2,x3,x4 } \/ { x5,x6,x7,x8,x9 }) by XBOOLE_1:4
    .= { x1 } \/ { x2,x3,x4,x5,x6,x7,x8,x9 } by Th24;
end;

theorem
  { x1,x2,x3,x4,x5,x6,x7,x8,x9 } = { x1,x2 } \/ { x3,x4,x5,x6,x7,x8,x9 }
proof
  thus { x1,x2,x3,x4,x5,x6,x7,x8,x9 } = { x1,x2,x3,x4 } \/ { x5,x6,x7,x8,x9 }
  by Lm9
    .= { x1,x2 } \/ { x3,x4 } \/ { x5,x6,x7,x8,x9 } by Lm2
    .= { x1,x2 } \/ ({ x3,x4 } \/ { x5,x6,x7,x8,x9 }) by XBOOLE_1:4
    .= { x1,x2 } \/ { x3,x4,x5,x6,x7,x8,x9 } by Th17;
end;

theorem
  { x1,x2,x3,x4,x5,x6,x7,x8,x9 } = { x1,x2,x3 } \/ { x4,x5,x6,x7,x8,x9 }
proof
  thus { x1,x2,x3,x4,x5,x6,x7,x8,x9 } = { x1,x2,x3,x4 } \/ { x5,x6,x7,x8,x9 }
  by Lm9
    .= { x1,x2,x3 } \/ { x4 } \/ { x5,x6,x7,x8,x9 } by Th6
    .= { x1,x2,x3 } \/ ({ x4 } \/ { x5,x6,x7,x8,x9 }) by XBOOLE_1:4
    .= { x1,x2,x3 } \/ { x4,x5,x6,x7,x8,x9 } by Th11;
end;

theorem
  { x1,x2,x3,x4,x5,x6,x7,x8,x9 } = { x1,x2,x3,x4 } \/ { x5,x6,x7,x8,x9 }
  by Lm9;

theorem
  { x1,x2,x3,x4,x5,x6,x7,x8,x9 } = { x1,x2,x3,x4,x5 } \/ { x6,x7,x8,x9 }
proof
  thus { x1,x2,x3,x4,x5,x6,x7,x8,x9 } = { x1,x2,x3,x4 } \/ { x5,x6,x7,x8,x9 }
  by Lm9
    .= { x1,x2,x3,x4 } \/ ({x5 } \/ { x6,x7,x8,x9 }) by Th7
    .= { x1,x2,x3,x4 } \/ {x5 } \/ { x6,x7,x8,x9 } by XBOOLE_1:4
    .= { x1,x2,x3,x4,x5 } \/ { x6,x7,x8,x9 } by Th10;
end;

theorem
  { x1,x2,x3,x4,x5,x6,x7,x8,x9 } = { x1,x2,x3,x4,x5,x6 } \/ { x7,x8,x9 }
proof
  thus { x1,x2,x3,x4,x5,x6,x7,x8,x9 } = { x1,x2,x3,x4 } \/ { x5,x6,x7,x8,x9 }
  by Lm9
    .= { x1,x2,x3,x4 } \/ ({ x5,x6 } \/ { x7,x8,x9 }) by Th8
    .= { x1,x2,x3,x4 } \/ { x5,x6 } \/ { x7,x8,x9 } by XBOOLE_1:4
    .= { x1,x2,x3,x4,x5,x6 } \/ { x7,x8,x9 } by Th14;
end;

theorem
  { x1,x2,x3,x4,x5,x6,x7,x8,x9 } = { x1,x2,x3,x4,x5,x6,x7 } \/ { x8,x9 }
proof
  thus { x1,x2,x3,x4,x5,x6,x7,x8,x9 } = { x1,x2,x3,x4 } \/ { x5,x6,x7,x8,x9 }
  by Lm9
    .= { x1,x2,x3,x4 } \/ ({ x5,x6,x7 } \/ { x8,x9 }) by Lm3
    .= { x1,x2,x3,x4 } \/ { x5,x6,x7 } \/ { x8,x9 } by XBOOLE_1:4
    .= { x1,x2,x3,x4,x5,x6,x7 } \/ { x8,x9 } by Lm5;
end;

theorem
  { x1,x2,x3,x4,x5,x6,x7,x8,x9 } = { x1,x2,x3,x4,x5,x6,x7,x8 } \/ { x9 }
proof
  thus { x1,x2,x3,x4,x5,x6,x7,x8,x9 } = { x1,x2,x3,x4 } \/ { x5,x6,x7,x8,x9 }
  by Lm9
    .= { x1,x2,x3,x4 } \/ ({ x5,x6,x7,x8 } \/ { x9 }) by Th10
    .= { x1,x2,x3,x4 } \/ { x5,x6,x7,x8 } \/ { x9 } by XBOOLE_1:4
    .= { x1,x2,x3,x4,x5,x6,x7,x8 } \/ { x9 } by Lm6;
end;

theorem
  { x1,x2,x3,x4,x5,x6,x7,x8,x9,x10 } = { x1,x2,x3,x4,x5,x6,x7,x8,x9 } \/
  { x10 }
proof
  now
    let x;
A1: x in {x10} iff x=x10 by TARSKI:def 1;
    x=x1 or x=x2 or x=x3 or x=x4 or x=x5 or x=x6 or x=x7 or x=x8 or x=x9
    or x=x10 iff x in { x1,x2,x3,x4,x5,x6,x7,x8,x9 } or x = x10 by Def7;
    hence
    x in { x1,x2,x3,x4,x5,x6,x7,x8,x9,x10 } iff x in { x1,x2,x3,x4,x5,x6,
    x7,x8,x9 } \/ {x10} by A1,Def8,XBOOLE_0:def 3;
  end;
  hence thesis by TARSKI:1;
end;

begin 

theorem 
  for x, y, z being set st x <> y & x <> z holds {x, y, z} \ {x} = {y, z }
proof
  let x, y, z be set such that
A1: x <> y & x <> z;
  hereby
    let a be set;
    assume
A2: a in {x, y, z} \ {x};
    then a in {x, y, z} by XBOOLE_0:def 5;
    then
A3: a = x or a = y or a = z by Def1;
    not a in {x} by A2,XBOOLE_0:def 5;
    hence a in {y, z} by A3,TARSKI:def 1,def 2;
  end;
  let a be set;
  assume a in {y, z};
  then
A4: a = y or a = z by TARSKI:def 2;
  then
A5: not a in {x} by A1,TARSKI:def 1;
  a in {x, y, z} by A4,Def1;
  hence thesis by A5,XBOOLE_0:def 5;
end;



theorem
  for x1,x2,x3 being set holds {x2,x1} \/ {x3,x1} = {x1,x2,x3}
proof
  let x1,x2,x3 be set;
  thus {x2,x1} \/ {x3,x1} = {x2,x1,x3,x1} by Lm2
    .= {x1,x1,x2,x3} by Th69
    .= {x1,x2,x3} by Th31;
end;














begin 

reserve x,x1,x2,x3,x4,y,y1,y2,y3,y4,z,z1,z2,z2,z4 for set;

definition let x;
 attr x is pair means
:Dfx: ex x1,x2 st x = [x1,x2];
end;

registration let x1,x2;
 cluster [x1,x2] -> pair;
 coherence by Dfx;
end;

reserve X,Y for set;

Lx4: {x} = {y1,y2} implies x = y1
proof
  assume { x } = { y1,y2 };
  then y1 in { x } by TARSKI:def 2;
  hence thesis by TARSKI:def 1;
end;

Lx5:  {x} = {y1,y2} implies y1 = y2
proof
  assume
A1: { x } = { y1,y2 };
  then x = y1 by Lx4;
  hence thesis by A1,Lx4;
end;

Lx6: { x1,x2 } = { y1,y2 } implies x1 = y1 or x1 = y2
proof
  x1 in { x1,x2 } by TARSKI:def 2;
  hence thesis by TARSKI:def 2;
end;

theorem Th1:
  [x1,x2] = [y1,y2] implies x1 = y1 & x2 = y2
proof
  assume
A1: [x1,x2] = [y1,y2];
  per cases;
  suppose
A3: y1 <> y2;
    then
A4: {x1} <> {y1,y2} by Lx5;
    then {x1} = {y1} by A1,Lx6;
   then x1 in {y1} by TARSKI:def 1;
   hence
A5: x1 = y1 by TARSKI:def 1;
    {y1,y2} = {x1,x2} by A1,A4,Lx6;
   hence thesis by A3,A5,Lx6;
  end;
  suppose
A6: y1 = y2;
    then {{x1,x2},{x1}} = {{y1},{y1}} by A1,ENUMSET1:29
        .= {{y1}} by ENUMSET1:29;
    then { y1 } = { x1,x2 } by Lx4;
   hence thesis by A6,Lx4;
  end;
end;

definition
  let x;
  assume x is pair;
   then consider x1,x2 such that
A1: x = [x1,x2] by Dfx;
  func x`1 means
:Def1: x = [y1,y2] implies it = y1;
  existence
  proof
    take x1;
    thus thesis by A1,Th1;
  end;
  uniqueness
  proof
    let z1,z2;
    assume that
A2: x = [y1,y2] implies z1 = y1 and
A3: x = [y1,y2] implies z2 = y1;
   thus z1 = x1 by A1,A2
     .= z2 by A1,A3;
  end;
  func x`2 means
:Def2: x = [y1,y2] implies it = y2;
  existence
  proof
    take x2;
    thus thesis by A1,Th1;
  end;
  uniqueness
  proof
    let z1,z2;
    assume that
A4:  x = [y1,y2] implies z1 = y2 and
A5:  x = [y1,y2] implies z2 = y2;
   thus z1 = x2 by A1,A4
    .= z2 by A1,A5;
  end;
end;

registration let x1,x2;
 reduce [x1,x2]`1 to x1;
 reducibility by Def1;
 reduce [x1,x2]`2 to x2;
 reducibility by Def2;
end;

registration
 cluster pair for set;
 existence
 proof
  take [the set, the set], the set, the set;
  thus thesis;
 end;
end;

registration let x be pair set;
 reduce [x`1,x`2] to x;
 reducibility
  proof
   consider x1,x2 such that
W:  x = [x1,x2] by Dfx;
   thus [x`1,x`2] = [x1,x`2] by W,Def1
     .= x by W,Def2;
  end;
end;

theorem
 for a,b being pair set st a`1 = b`1 & a`2 = b`2
  holds a = b
proof let a,b be pair set such that
Z: a`1 = b`1 & a`2 = b`2;
  a = [a`1,a`2] & b = [b`1,b`2];
 hence a = b by Z;
end;

begin 

definition
  let x1,x2,x3;
  func [x1,x2,x3] equals
  [[x1,x2],x3];
  coherence;
end;

definition let x;
 attr x is triple means
:Dfy: ex x1,x2,x3 st x = [x1,x2,x3];
end;

registration let x1,x2,x3;
 cluster [x1,x2,x3] -> triple;
 coherence by Dfy;
end;

theorem Th3:
  [x1,x2,x3] = [y1,y2,y3] implies x1 = y1 & x2 = y2 & x3 = y3
proof
  assume
A1: [x1,x2,x3] = [y1,y2,y3];
  then [x1,x2] = [y1,y2] by Th1;
  hence thesis by A1,Th1;
end;

registration
 cluster triple for set;
 existence
 proof
  take [the set, the set, the set], the set, the set, the set;
  thus thesis;
 end;
 cluster triple -> pair for set;
 coherence
  proof let x be set;
   assume ex x1,x2,x3 st x = [x1,x2,x3];
   hence thesis;
  end;
end;

definition
  let x be set;
  func x`1_3 equals x`1`1;
   coherence;
  func x`2_3 equals x`1`2;
   coherence;
end;

notation
 let x be set;
 synonym x`3_3 for x`2;
end;

registration let x1,x2,x3;
 reduce [x1,x2,x3]`1_3 to x1;
 reducibility
  proof
   thus [x1,x2,x3]`1_3 = [[x1,x2],x3]`1`1 .= x1;
  end;
 reduce [x1,x2,x3]`2_3 to x2;
 reducibility
  proof
   thus [x1,x2,x3]`2_3 = [[x1,x2],x3]`1`2 .= x2;
  end;
 reduce [x1,x2,x3]`3_3 to x3;
 reducibility
  proof
   thus [x1,x2,x3]`3_3 = [[x1,x2],x3]`2 .= x3;
  end;
end;

registration let x be triple set;
 reduce [x`1_3,x`2_3,x`3_3] to x;
 reducibility
  proof
   consider x1,x2,x3 such that
W:  x = [x1,x2,x3] by Dfy;
    [x1,x2,x3]`1_3 = x1 & [x1,x2,x3]`2_3 = x2 & [x1,x2,x3]`3_3 = x3;
   hence thesis by W;
  end;
end;

theorem
 for a,b being triple set st a`1_3 = b`1_3 & a`2_3 = b`2_3 & a`3_3 = b`3_3
  holds a = b
proof let a,b be triple set such that
Z: a`1_3 = b`1_3 & a`2_3 = b`2_3 & a`3_3 = b`3_3;
  a = [a`1_3,a`2_3,a`3_3] & b = [b`1_3,b`2_3,b`3_3];
 hence a = b by Z;
end;

begin 

definition
  let x1,x2,x3,x4;
  func [x1,x2,x3,x4] equals
  [[x1,x2,x3],x4];
  coherence;
end;

definition let x;
 attr x is quadruple means
:Dfz: ex x1,x2,x3,x4 st x = [x1,x2,x3,x4];
end;

registration let x1,x2,x3,x4;
 cluster [x1,x2,x3,x4] -> quadruple;
 coherence by Dfz;
end;

theorem
 [x1,x2,x3,x4] = [y1,y2,y3,y4]
   implies x1 = y1 & x2 = y2 & x3 = y3 & x4 = y4
proof
  assume
A1: [x1,x2,x3,x4] = [y1,y2,y3,y4];
  then [x1,x2,x3] = [y1,y2,y3] by Th1;
  hence thesis by A1,Th3,Th1;
end;

registration
 cluster quadruple for set;
 existence
 proof
  take [the set, the set, the set, the set],
    the set, the set, the set, the set;
  thus thesis;
 end;
 cluster quadruple -> triple for set;
 coherence
  proof let x;
   given x1,x2,x3,x4 such that
G:  x = [x1,x2,x3,x4];
    x = [[x1,x2],x3,x4] by G;
   hence thesis;
  end;
end;

definition
 let x be set;
  func x`1_4 equals x`1`1`1;
   coherence;
  func x`2_4 equals x`1`1`2;
  coherence;
end;

notation let x be set;
 synonym x`3_4 for x`2_3;
 synonym x`4_4 for x`2;
end;

registration let x1,x2,x3,x4;
 reduce [x1,x2,x3,x4]`1_4 to x1;
 reducibility
  proof
   thus [x1,x2,x3,x4]`1_4 = [[x1,x2,x3],x4]`1`1`1
          .= [x1,x2,x3]`1_3
          .= x1;
  end;
 reduce [x1,x2,x3,x4]`2_4 to x2;
 reducibility
  proof
   thus [x1,x2,x3,x4]`2_4 = [[x1,x2,x3],x4]`1`1`2
          .= [x1,x2,x3]`2_3
          .= x2;
  end;
 reduce [x1,x2,x3,x4]`3_4 to x3;
 reducibility
  proof
   thus [x1,x2,x3,x4]`3_4 = [[x1,x2,x3],x4]`1`2
          .= x3;
  end;
 reduce [x1,x2,x3,x4]`4_4 to x4;
 reducibility
  proof
   thus [x1,x2,x3,x4]`4_4 = [[x1,x2,x3],x4]`2
          .= x4;
  end;
end;

registration let x be quadruple set;
 reduce [x`1_4,x`2_4,x`3_4,x`4_4] to x;
 reducibility
  proof
   consider x1,x2,x3,x4 such that
W:  x = [x1,x2,x3,x4] by Dfz;
    [x1,x2,x3,x4]`1_4 = x1 & [x1,x2,x3,x4]`2_4 = x2 &
    [x1,x2,x3,x4]`3_4 = x3 & [x1,x2,x3,x4]`4_4 = x4;
   hence thesis by W;
  end;
end;

reserve X,X1,X2,X3,X4,Y for set;



theorem Pre1:
 [x,y] in X implies x in union union X
  proof assume
A1: [x,y] in X;
   {x} in [x,y] by TARSKI:def 2;
   then
A2: {x} in union X by A1,TARSKI:def 4;
    x in {x} by TARSKI:def 1;
   hence x in union union X by A2,TARSKI:def 4;
  end;

theorem Pre2:
 [x,y] in X implies y in union union X
  proof assume
A1: [x,y] in X;
   {x,y} in [x,y] by TARSKI:def 2;
   then
A2: {x,y} in union X by A1,TARSKI:def 4;
    y in {x,y} by TARSKI:def 2;
   hence y in union union X by A2,TARSKI:def 4;
  end;



definition
  let X be set;
  func proj1 X -> set means
:Def4:
  x in it iff ex y st [x,y] in X;
  existence
  proof
    defpred Q[set] means ex y st [$1,y] in X;
    consider Y such that
A0: for x holds x in Y iff x in union union X & Q[x] from XBOOLE_0:sch 1;
    take Y;
    let x;
    thus x in Y implies ex y st [x,y] in X by A0;
    given y such that
A1:  [x,y] in X;
    x in union union X by A1,Pre1;
   hence thesis by A1,A0;
  end;
  uniqueness
  proof
    let X1,X2;
    assume that
A2: for x holds x in X1 iff ex y st [x,y] in X and
A3: for x holds x in X2 iff ex y st [x,y] in X;
    now
      let x;
      x in X1 iff ex y st [x,y] in X by A2;
      hence x in X1 iff x in X2 by A3;
    end;
    hence thesis by TARSKI:1;
  end;
  func proj2 X -> set means
:Def5:
  x in it iff ex y st [y,x] in X;
  existence
  proof
    defpred Q[set] means ex y st [y,$1] in X;
    consider Y such that
A0: for x holds x in Y iff x in union union X & Q[x] from XBOOLE_0:sch 1;
    take Y;
    let x;
    thus x in Y implies ex y st [y,x] in X by A0;
    given y such that
A1:  [y,x] in X;
    x in union union X by A1,Pre2;
   hence thesis by A1,A0;
  end;
  uniqueness
  proof
    let X1,X2;
    assume that
A2: for x holds x in X1 iff ex y st [y,x] in X and
A3: for x holds x in X2 iff ex y st [y,x] in X;
    now
      let x;
      x in X1 iff ex y st [y,x] in X by A2;
      hence x in X1 iff x in X2 by A3;
    end;
    hence thesis by TARSKI:1;
  end;
end;

theorem Th10:
 X c= Y implies proj1 X c= proj1 Y
proof assume
A1: X c= Y;
 let x;
 assume x in proj1 X;
  then consider y such that
A2: [x,y] in X by Def4;
  [x,y] in Y by A1,A2,TARSKI:def 3;
 hence thesis by Def4;
end;

theorem Th11:
 X c= Y implies proj2 X c= proj2 Y
proof assume
A1: X c= Y;
 let x;
 assume x in proj2 X;
  then consider y such that
A2: [y,x] in X by Def5;
  [y,x] in Y by A1,A2,TARSKI:def 3;
 hence thesis by Def5;
end;

definition
  let X be set;
  func proj1_3 X -> set equals
   proj1 proj1 X;
  coherence;
  func proj2_3 X -> set equals
   proj2 proj1 X;
  coherence;
end;

notation let X be set;
 synonym proj3_3 X for proj2 X;
end;

theorem Th12:
 X c= Y implies proj1_3 X c= proj1_3 Y
proof assume X c= Y;
  then proj1 X c= proj1 Y by Th10;
 hence thesis by Th10;
end;

theorem Th13:
 X c= Y implies proj2_3 X c= proj2_3 Y
proof assume X c= Y;
  then proj1 X c= proj1 Y by Th10;
 hence thesis by Th11;
end;

theorem Th14:
 x in proj1_3 X implies ex y,z st [x,y,z] in X
proof assume x in proj1_3 X;
  then consider y such that
W1: [x,y] in proj1 X by Def4;
  consider z such that
W2: [[x,y],z] in X by W1,Def4;
 take y,z;
 thus thesis by W2;
end;

theorem Th14a:
 [x,y,z] in X implies x in proj1_3 X
 proof
  assume [x,y,z] in X;
   then [x,y] in proj1 X by Def4;
  hence thesis by Def4;
 end;

theorem Th15:
 x in proj2_3 X implies ex y,z st [y,x,z] in X
proof assume x in proj2_3 X;
  then consider y such that
W1: [y,x] in proj1 X by Def5;
  consider z such that
W2: [[y,x],z] in X by W1,Def4;
 take y,z;
 thus thesis by W2;
end;

theorem Th15a:
 [x,y,z] in X implies y in proj2_3 X
 proof
  assume [x,y,z] in X;
   then [x,y] in proj1 X by Def4;
  hence thesis by Def5;
 end;

definition
  let X be set;
func proj1_4 X -> set equals
   proj1 proj1_3 X;
  coherence;
  func proj2_4 X -> set equals
   proj2 proj1_3 X;
  coherence;
end;

notation let X be set;
  synonym proj3_4 X for proj2_3 X;
  synonym proj4_4 X for proj2 X;
end;

theorem Th17:
 X c= Y implies proj1_4 X c= proj1_4 Y
proof assume X c= Y;
  then proj1_3 X c= proj1_3 Y by Th12;
 hence thesis by Th10;
end;

theorem Th18:
 X c= Y implies proj2_4 X c= proj2_4 Y
proof assume X c= Y;
  then proj1_3 X c= proj1_3 Y by Th12;
 hence thesis by Th11;
end;

theorem Th19:
 x in proj1_4 X implies ex x1,x2,x3 st [x,x1,x2,x3] in X
proof assume x in proj1_4 X;
  then consider x1 such that
W1: [x,x1] in proj1_3 X by Def4;
  consider x2 such that
W2: [[x,x1],x2] in proj1 X by W1,Def4;
  consider x3 such that
W3: [[[x,x1],x2],x3] in X by W2,Def4;
 take x1,x2,x3;
 thus thesis by W3;
end;

theorem Th19a:
 [x,x1,x2,x3] in X implies x in proj1_4 X
proof
 assume [x,x1,x2,x3] in X;
  then [[x,x1],x2,x3] in X;
  then [x,x1] in proj1_3 X by Th14a;
 hence thesis by Def4;
end;

theorem Th20:
 x in proj2_4 X implies ex x1,x2,x3 st [x1,x,x2,x3] in X
proof assume x in proj2_4 X;
  then consider x1 such that
W1: [x1,x] in proj1_3 X by Def5;
  consider x2 such that
W2: [[x1,x],x2] in proj1 X by W1,Def4;
  consider x3 such that
W3: [[[x1,x],x2],x3] in X by W2,Def4;
 take x1,x2,x3;
 thus thesis by W3;
end;

theorem Th20a:
 [x1,x,x2,x3] in X implies x in proj2_4 X
proof
 assume [x1,x,x2,x3] in X;
  then [[x1,x],x2,x3] in X;
  then [x1,x] in proj1_3 X by Th14a;
 hence thesis by Def5;
end;

theorem
 for a,b being quadruple set
  st a`1_4 = b`1_4 & a`2_4 = b`2_4 & a`3_4 = b`3_4 & a`4_4 = b`4_4
  holds a = b
proof let a,b be quadruple set such that
Z: a`1_4 = b`1_4 & a`2_4 = b`2_4 & a`3_4 = b`3_4 & a`4_4 = b`4_4;
  a = [a`1_4,a`2_4,a`3_4,a`4_4] & b = [b`1_4,b`2_4,b`3_4,b`4_4];
 hence a = b by Z;
end;

begin 

registration let X be empty set;
 cluster proj1 X -> empty;
 coherence
  proof
   assume proj1 X is non empty;
    then consider x such that
W:   x in proj1 X by XBOOLE_0:def 1;
    ex y st [x,y] in X by W,Def4;
   hence contradiction;
  end;
end;

registration let X be empty set;
 cluster proj2 X -> empty;
 coherence
  proof
   assume proj2 X is non empty;
    then consider x such that
W:   x in proj2 X by XBOOLE_0:def 1;
    ex y st [y,x] in X by W,Def5;
   hence contradiction;
  end;
end;

registration let X be empty set;
 cluster proj1_3 X -> empty;
 coherence;
end;

registration let X be empty set;
 cluster proj2_3 X -> empty;
 coherence;
end;

registration let X be empty set;
 cluster proj1_4 X -> empty;
 coherence;
end;

registration let X be empty set;
 cluster proj2_4 X -> empty;
 coherence;
end;

theorem Th22:
 proj1(X \/ Y) = proj1 X \/ proj1 Y
proof
  thus proj1(X \/ Y) c= proj1 X \/ proj1 Y
  proof
   let x;
   assume x in proj1(X \/ Y);
    then consider y such that
A1: [x,y] in X \/ Y by Def4;
    [x,y] in X or [x,y] in Y by A1,XBOOLE_0:def 3;
    then x in proj1 X or x in proj1 Y by Def4;
   hence thesis by XBOOLE_0:def 3;
  end;
A3: proj1 Y c= proj1(X \/ Y) by Th10,XBOOLE_1:7;
   proj1 X c= proj1(X \/ Y) by Th10,XBOOLE_1:7;
  hence proj1 X \/ proj1 Y c= proj1 (X \/ Y) by A3,XBOOLE_1:8;
end;

theorem
  proj1(X /\ Y) c= proj1 X /\ proj1 Y
proof
  proj1(X /\ Y) c= proj1 X & proj1(X /\ Y) c= proj1 Y by Th10,XBOOLE_1:17;
 hence thesis by XBOOLE_1:19;
end;

theorem Th24:
  proj1 X \ proj1 Y c= proj1(X \ Y)
proof
 let x;
 assume
A1: x in proj1 X \ proj1 Y;
  then x in proj1 X by XBOOLE_0:def 5;
  then consider y such that
A2: [x,y] in X by Def4;
  not x in proj1 Y by A1,XBOOLE_0:def 5;
  then not [x,y] in Y by Def4;
  then [x,y] in X \ Y by A2,XBOOLE_0:def 5;
 hence thesis by Def4;
end;

theorem
  proj1 X \+\ proj1 Y c= proj1(X \+\ Y)
proof
  proj1 X \ proj1 Y c= proj1(X \ Y) & proj1 Y \ proj1 X c= proj1(Y \ X)
     by Th24;
  then proj1 X \+\ proj1 Y c= proj1(X\Y) \/ proj1(Y\X) by XBOOLE_1:13;
 hence thesis by Th22;
end;

theorem Th26:
 proj2(X \/ Y) = proj2 X \/ proj2 Y
proof
 thus proj2(X \/ Y) c= proj2 X \/ proj2 Y
  proof
   let y;
   assume y in proj2(X \/ Y);
    then consider x such that
A1: [x,y] in X \/ Y by Def5;
    [x,y] in X or [x,y] in Y by A1,XBOOLE_0:def 3;
    then y in proj2 X or y in proj2 Y by Def5;
   hence thesis by XBOOLE_0:def 3;
  end;
A3: proj2 Y c= proj2(X \/ Y) by Th11,XBOOLE_1:7;
  proj2 X c= proj2(X \/ Y) by Th11,XBOOLE_1:7;
 hence proj2 X \/ proj2 Y c= proj2(X \/ Y) by A3,XBOOLE_1:8;
end;

theorem
  proj2(X /\ Y) c= proj2 X /\ proj2 Y
proof
  let y;
  assume y in proj2(X /\ Y);
  then consider x such that
A3: [x,y] in X /\ Y by Def5;
  [x,y] in Y by A3,XBOOLE_0:def 4;
  then
A4: y in proj2 Y by Def5;
  [x,y] in X by A3,XBOOLE_0:def 4;
  then y in proj2 X by Def5;
  hence thesis by A4,XBOOLE_0:def 4;
end;

theorem Th28:
  proj2 X \ proj2 Y c= proj2(X \ Y)
proof
  let y;
  assume
A3: y in proj2 X \ proj2 Y;
  then y in proj2 X by XBOOLE_0:def 5;
  then consider x such that
A4: [x,y] in X by Def5;
  not y in proj2 Y by A3,XBOOLE_0:def 5;
  then not [x,y] in Y by Def5;
  then [x,y] in X \ Y by A4,XBOOLE_0:def 5;
  hence thesis by Def5;
end;

theorem
   proj2 X \+\ proj2 Y c= proj2(X \+\ Y)
proof
  proj2 X \ proj2 Y c= proj2(X \ Y) & proj2 Y \ proj2 X c= proj2(Y \ X)
      by Th28;
  then proj2 X \+\ proj2 Y c= proj2(X\Y) \/ proj2(Y\X) by XBOOLE_1:13;
  hence thesis by Th26;
end;

theorem Th30:
 proj1_3(X \/ Y) = proj1_3 X \/ proj1_3 Y
proof
 thus proj1_3(X \/ Y) = proj1(proj1 X \/ proj1 Y) by Th22
     .= proj1_3 X \/ proj1_3 Y by Th22;
end;

theorem
  proj1_3(X /\ Y) c= proj1_3 X /\ proj1_3 Y
proof
  proj1_3(X /\ Y) c= proj1_3 X & proj1_3(X /\ Y) c= proj1_3 Y
       by Th12,XBOOLE_1:17;
 hence thesis by XBOOLE_1:19;
end;

theorem Th32:
  proj1_3 X \ proj1_3 Y c= proj1_3(X \ Y)
proof
 let x;
 assume
A1: x in proj1_3 X \ proj1_3 Y;
  then x in proj1_3 X by XBOOLE_0:def 5;
  then consider y,z such that
A2: [x,y,z] in X by Th14;
  not x in proj1_3 Y by A1,XBOOLE_0:def 5;
  then not [x,y,z] in Y by Th14a;
  then [x,y,z] in X \ Y by A2,XBOOLE_0:def 5;
 hence thesis by Th14a;
end;

theorem
  proj1_3 X \+\ proj1_3 Y c= proj1_3(X \+\ Y)
proof
  proj1_3 X \ proj1_3 Y c= proj1_3(X \ Y) &
  proj1_3 Y \ proj1_3 X c= proj1_3(Y \ X) by Th32;
  then proj1_3 X \+\ proj1_3 Y c= proj1_3(X\Y) \/ proj1_3(Y\X) by XBOOLE_1:13;
 hence thesis by Th30;
end;

theorem Th34:
 proj2_3(X \/ Y) = proj2_3 X \/ proj2_3 Y
proof
 thus proj2_3(X \/ Y) = proj2(proj1 X \/ proj1 Y) by Th22
     .= proj2_3 X \/ proj2_3 Y by Th26;
end;

theorem
  proj2_3(X /\ Y) c= proj2_3 X /\ proj2_3 Y
proof
  proj2_3(X /\ Y) c= proj2_3 X & proj2_3(X /\ Y) c= proj2_3 Y
   by Th13,XBOOLE_1:17;
 hence thesis by XBOOLE_1:19;
end;

theorem Th36:
  proj2_3 X \ proj2_3 Y c= proj2_3(X \ Y)
proof
 let x;
 assume
A1: x in proj2_3 X \ proj2_3 Y;
  then x in proj2_3 X by XBOOLE_0:def 5;
  then consider y,z such that
A2: [y,x,z] in X by Th15;
  not x in proj2_3 Y by A1,XBOOLE_0:def 5;
  then not [y,x,z] in Y by Th15a;
  then [y,x,z] in X \ Y by A2,XBOOLE_0:def 5;
 hence thesis by Th15a;
end;

theorem
  proj2_3 X \+\ proj2_3 Y c= proj2_3(X \+\ Y)
proof
  proj2_3 X \ proj2_3 Y c= proj2_3(X \ Y) &
  proj2_3 Y \ proj2_3 X c= proj2_3(Y \ X) by Th36;
  then proj2_3 X \+\ proj2_3 Y c= proj2_3(X\Y) \/ proj2_3(Y\X) by XBOOLE_1:13;
 hence thesis by Th34;
end;

theorem Th38:
 proj1_4(X \/ Y) = proj1_4 X \/ proj1_4 Y
proof
 thus proj1_4(X \/ Y) = proj1(proj1_3 X \/ proj1_3 Y) by Th30
     .= proj1_4 X \/ proj1_4 Y by Th22;
end;

theorem
  proj1_4(X /\ Y) c= proj1_4 X /\ proj1_4 Y
proof
  proj1_4(X /\ Y) c= proj1_4 X & proj1_4(X /\ Y) c= proj1_4 Y
   by Th17,XBOOLE_1:17;
 hence thesis by XBOOLE_1:19;
end;

theorem Th32:
  proj1_4 X \ proj1_4 Y c= proj1_4(X \ Y)
proof
 let x;
 assume
A1: x in proj1_4 X \ proj1_4 Y;
  then x in proj1_4 X by XBOOLE_0:def 5;
  then consider x1,x2,x3 such that
A2: [x,x1,x2,x3] in X by Th19;
  not x in proj1_4 Y by A1,XBOOLE_0:def 5;
  then not [x,x1,x2,x3] in Y by Th19a;
  then [x,x1,x2,x3] in X \ Y by A2,XBOOLE_0:def 5;
 hence thesis by Th19a;
end;

theorem
  proj1_4 X \+\ proj1_4 Y c= proj1_4(X \+\ Y)
proof
  proj1_4 X \ proj1_4 Y c= proj1_4(X \ Y) &
  proj1_4 Y \ proj1_4 X c= proj1_4(Y \ X) by Th32;
  then proj1_4 X \+\ proj1_4 Y c= proj1_4(X\Y) \/ proj1_4(Y\X) by XBOOLE_1:13;
 hence thesis by Th38;
end;

theorem Th34:
 proj2_4(X \/ Y) = proj2_4 X \/ proj2_4 Y
proof
 thus proj2_4(X \/ Y) = proj2(proj1_3 X \/ proj1_3 Y) by Th30
     .= proj2_4 X \/ proj2_4 Y by Th26;
end;

theorem
  proj2_4(X /\ Y) c= proj2_4 X /\ proj2_4 Y
proof
  proj2_4(X /\ Y) c= proj2_4 X & proj2_4(X /\ Y) c= proj2_4 Y
    by Th18,XBOOLE_1:17;
 hence thesis by XBOOLE_1:19;
end;

theorem Th36:
  proj2_4 X \ proj2_4 Y c= proj2_4(X \ Y)
proof
 let x;
 assume
A1: x in proj2_4 X \ proj2_4 Y;
  then x in proj2_4 X by XBOOLE_0:def 5;
  then consider x1,x2,x3 such that
A2: [x1,x,x2,x3] in X by Th20;
  not x in proj2_4 Y by A1,XBOOLE_0:def 5;
  then not [x1,x,x2,x3] in Y by Th20a;
  then [x1,x,x2,x3] in X \ Y by A2,XBOOLE_0:def 5;
 hence thesis by Th20a;
end;

theorem
  proj2_4 X \+\ proj2_4 Y c= proj2_4(X \+\ Y)
proof
  proj2_4 X \ proj2_4 Y c= proj2_4(X \ Y) &
  proj2_4 Y \ proj2_4 X c= proj2_4(Y \ X) by Th36;
  then proj2_4 X \+\ proj2_4 Y c= proj2_4(X\Y) \/ proj2_4(Y\X) by XBOOLE_1:13;
 hence thesis by Th34;
end;














begin

 reserve x,y,X1,X2,X3,X4,X5,X6,Y,Y1,Y2,Y3,Y4,Y5,Z,Z1,Z2,Z3,Z4,Z5 for set;
 reserve X for non empty set;




theorem Th1:
  ex Y st Y in X & Y misses X
proof
  consider x such that
W:  x in X by XBOOLE_0:def 1;
   consider Y such that
A1: Y in X & not ex x st x in X & x in Y by TARSKI:2,W;
  take Y;
  thus thesis by A1,XBOOLE_0:3;
end;

theorem
  ex Y st Y in X & for Y1 st Y1 in Y holds Y1 misses X
proof
  defpred P[set] means $1 meets X;
  consider Z such that
A1: for Y holds Y in Z iff Y in union X & P[Y] from XBOOLE_0:sch 1;
  consider Y such that
A2: Y in X \/ Z and
A3: Y misses X \/ Z by Th1;
  assume
A4: not thesis;
  now
    assume
A5: Y in X;
    then consider Y1 such that
A6: Y1 in Y and
A7: not Y1 misses X by A4;
    Y1 in union X by A5,A6,TARSKI:def 4;
    then Y1 in Z by A1,A7;
    then Y1 in X \/ Z by XBOOLE_0:def 3;
    hence contradiction by A3,A6,XBOOLE_0:3;
  end;
  then Y in Z by A2,XBOOLE_0:def 3;
  then Y meets X by A1;
  hence contradiction by A3,XBOOLE_1:70;
end;

theorem
  ex Y st Y in X & for Y1,Y2 st Y1 in Y2 & Y2 in Y holds Y1 misses X
proof
  defpred P[set] means ex Y1 st Y1 in $1 & Y1 meets X;
  consider Z1 such that
A1: for Y holds Y in Z1 iff Y in union X & P[Y] from XBOOLE_0:sch 1;
  defpred Q[set] means $1 meets X;
  consider Z2 such that
A2: for Y holds Y in Z2 iff Y in union union X & Q[Y] from XBOOLE_0:sch 1;
  consider Y such that
A3: Y in X \/ Z1 \/ Z2 and
A4: Y misses X \/ Z1 \/ Z2 by Th1;
A5: now
    assume
A6: Y in Z1;
    then consider Y1 such that
A7: Y1 in Y and
A8: Y1 meets X by A1;
    Y in union X by A1,A6;
    then Y1 in union union X by A7,TARSKI:def 4;
    then Y1 in Z2 by A2,A8;
    then Y1 in X \/ Z1 \/ Z2 by XBOOLE_0:def 3;
    hence contradiction by A4,A7,XBOOLE_0:3;
  end;
  assume
A9: not thesis;
A10: now
    assume
A11: Y in X;
    then consider Y1,Y2 such that
A12: Y1 in Y2 and
A13: Y2 in Y and
A14: not Y1 misses X by A9;
    Y2 in union X by A11,A13,TARSKI:def 4;
    then Y2 in Z1 by A1,A12,A14;
    then Y2 in X \/ Z1 by XBOOLE_0:def 3;
    then Y2 in X \/ Z1 \/ Z2 by XBOOLE_0:def 3;
    hence contradiction by A4,A13,XBOOLE_0:3;
  end;
  Y in X \/ (Z1 \/ Z2) by A3,XBOOLE_1:4;
  then Y in Z1 \/ Z2 by A10,XBOOLE_0:def 3;
  then Y in Z2 by A5,XBOOLE_0:def 3;
  then Y meets X by A2;
  then Y meets X \/ Z1 by XBOOLE_1:70;
  hence contradiction by A4,XBOOLE_1:70;
end;

theorem
  ex Y st Y in X &
   for Y1,Y2,Y3 st Y1 in Y2 & Y2 in Y3 & Y3 in Y holds Y1 misses X
proof
  defpred P[set] means ex Y1,Y2 st Y1 in Y2 & Y2 in $1 & Y1 meets X;
  consider Z1 such that
A1: for Y holds Y in Z1 iff Y in union X & P[Y] from XBOOLE_0:sch 1;
  defpred R[set] means $1 meets X;
  defpred Q[set] means ex Y1 st Y1 in $1 & Y1 meets X;
  consider Z2 such that
A2: for Y holds Y in Z2 iff Y in union union X & Q[Y] from XBOOLE_0:sch 1;
  consider Z3 such that
A3: for Y holds Y in Z3 iff Y in union union union X & R[Y]
    from XBOOLE_0:sch 1;
  consider Y such that
A4: Y in X \/ Z1 \/ Z2 \/ Z3 and
A5: Y misses X \/ Z1 \/ Z2 \/ Z3 by Th1;
A6: now
    assume
A7: Y in Z2;
    then consider Y1 such that
A8: Y1 in Y and
A9: Y1 meets X by A2;
    Y in union union X by A2,A7;
    then Y1 in union union union X by A8,TARSKI:def 4;
    then Y1 in Z3 by A3,A9;
    then Y1 in X \/ Z1 \/ Z2 \/ Z3 by XBOOLE_0:def 3;
    hence contradiction by A5,A8,XBOOLE_0:3;
  end;
A10: now
    assume
A11: Y in Z1;
    then consider Y1,Y2 such that
A12: Y1 in Y2 and
A13: Y2 in Y and
A14: Y1 meets X by A1;
    Y in union X by A1,A11;
    then Y2 in union union X by A13,TARSKI:def 4;
    then Y2 in Z2 by A2,A12,A14;
    then Y2 in X \/ Z1 \/ Z2 by XBOOLE_0:def 3;
    then Y meets X \/ Z1 \/ Z2 by A13,XBOOLE_0:3;
    hence contradiction by A5,XBOOLE_1:70;
  end;
  set V = X \/ Z1 \/ Z2 \/ Z3;
A15: V = X \/ (Z1 \/ Z2) \/ Z3 by XBOOLE_1:4
    .= X \/ (Z1 \/ Z2 \/ Z3) by XBOOLE_1:4;
  assume
A16: not thesis;
  now
    assume
A17: Y in X;
    then consider Y1,Y2,Y3 such that
A18: Y1 in Y2 & Y2 in Y3 and
A19: Y3 in Y and
A20: not Y1 misses X by A16;
    Y3 in union X by A17,A19,TARSKI:def 4;
    then Y3 in Z1 by A1,A18,A20;
    then Y3 in X \/ Z1 by XBOOLE_0:def 3;
    then Y3 in X \/ Z1 \/ Z2 by XBOOLE_0:def 3;
    then Y3 in X \/ Z1 \/ Z2 \/ Z3 by XBOOLE_0:def 3;
    hence contradiction by A5,A19,XBOOLE_0:3;
  end;
  then Y in Z1 \/ Z2 \/ Z3 by A15,A4,XBOOLE_0:def 3;
  then Y in Z1 \/ (Z2 \/ Z3) by XBOOLE_1:4;
  then Y in Z2 \/ Z3 by A10,XBOOLE_0:def 3;
  then Y in Z3 by A6,XBOOLE_0:def 3;
  then Y meets X by A3;
  hence contradiction by A15,A5,XBOOLE_1:70;
end;

theorem
  ex Y st Y in X &
   for Y1,Y2,Y3,Y4 st Y1 in Y2 & Y2 in  Y3 & Y3 in Y4 & Y4 in Y
    holds Y1 misses X
proof
  defpred P[set] means ex Y1,Y2,Y3 st Y1 in Y2 & Y2 in Y3 & Y3 in $1 & Y1
  meets X;
  consider Z1 such that
A1: for Y holds Y in Z1 iff Y in union X & P[Y] from XBOOLE_0:sch 1;
  defpred S[set] means $1 meets X;
  defpred R[set] means ex Y1 st Y1 in $1 & Y1 meets X;
  defpred Q[set] means ex Y1,Y2 st Y1 in Y2 & Y2 in $1 & Y1 meets X;
  consider Z2 such that
A2: for Y holds Y in Z2 iff Y in union union X & Q[Y] from XBOOLE_0:sch 1;
  consider Z4 such that
A3: for Y holds Y in Z4 iff Y in union union union union X & S[Y] from
  XBOOLE_0:sch 1;
  consider Z3 such that
A4: for Y holds Y in Z3 iff Y in union union union X & R[Y] from
  XBOOLE_0:sch 1;
  consider Y such that
A5: Y in X \/ Z1 \/ Z2 \/ Z3 \/ Z4 and
A6: Y misses X \/ Z1 \/ Z2 \/ Z3 \/ Z4 by Th1;
A7: now
    assume
A8: Y in Z3;
    then consider Y1 such that
A9: Y1 in Y and
A10: Y1 meets X by A4;
    Y in union union union X by A4,A8;
    then Y1 in union union union union X by A9,TARSKI:def 4;
    then Y1 in Z4 by A3,A10;
    then Y1 in X \/ Z1 \/ Z2 \/ Z3 \/ Z4 by XBOOLE_0:def 3;
    hence contradiction by A6,A9,XBOOLE_0:3;
  end;
A11: now
    assume
A12: Y in Z1;
    then consider Y1,Y2,Y3 such that
A13: Y1 in Y2 & Y2 in Y3 and
A14: Y3 in Y and
A15: Y1 meets X by A1;
    Y in union X by A1,A12;
    then Y3 in union union X by A14,TARSKI:def 4;
    then Y3 in Z2 by A2,A13,A15;
    then Y3 in X \/ Z1 \/ Z2 by XBOOLE_0:def 3;
    then Y meets X \/ Z1 \/ Z2 by A14,XBOOLE_0:3;
    then Y meets X \/ Z1 \/ Z2 \/ Z3 by XBOOLE_1:70;
    hence contradiction by A6,XBOOLE_1:70;
  end;
A16: X \/ Z1 \/ Z2 \/ Z3 \/ Z4 = X \/ (Z1 \/ Z2) \/ Z3 \/ Z4 by XBOOLE_1:4
    .= X \/ (Z1 \/ Z2 \/ Z3) \/ Z4 by XBOOLE_1:4
    .= X \/ (Z1 \/ Z2 \/ Z3 \/ Z4) by XBOOLE_1:4;
A17: now
    assume
A18: Y in Z2;
    then consider Y1,Y2 such that
A19: Y1 in Y2 and
A20: Y2 in Y and
A21: Y1 meets X by A2;
    Y in union union X by A2,A18;
    then Y2 in union union union X by A20,TARSKI:def 4;
    then Y2 in Z3 by A4,A19,A21;
    then Y2 in X \/ Z1 \/ Z2 \/ Z3 by XBOOLE_0:def 3;
    then Y meets X \/ Z1 \/ Z2 \/ Z3 by A20,XBOOLE_0:3;
    hence contradiction by A6,XBOOLE_1:70;
  end;
  assume
A22: not thesis;
  now
    assume
A23: Y in X;
    then consider Y1,Y2,Y3,Y4 such that
A24: Y1 in Y2 & Y2 in Y3 & Y3 in Y4 and
A25: Y4 in Y and
A26: not Y1 misses X by A22;
    Y4 in union X by A23,A25,TARSKI:def 4;
    then Y4 in Z1 by A1,A24,A26;
    then Y4 in X \/ Z1 by XBOOLE_0:def 3;
    then Y4 in X \/ Z1 \/ Z2 by XBOOLE_0:def 3;
    then Y4 in X \/ Z1 \/ Z2 \/ Z3 by XBOOLE_0:def 3;
    then Y4 in X \/ Z1 \/ Z2 \/ Z3 \/ Z4 by XBOOLE_0:def 3;
    hence contradiction by A6,A25,XBOOLE_0:3;
  end;
  then Y in Z1 \/ Z2 \/ Z3 \/ Z4 by A16,A5,XBOOLE_0:def 3;
  then Y in Z1 \/ (Z2 \/ Z3) \/ Z4 by XBOOLE_1:4;
  then Y in Z1 \/ (Z2 \/ Z3 \/ Z4) by XBOOLE_1:4;
  then Y in Z2 \/ Z3 \/ Z4 by A11,XBOOLE_0:def 3;
  then Y in Z2 \/ (Z3 \/ Z4) by XBOOLE_1:4;
  then Y in Z3 \/ Z4 by A17,XBOOLE_0:def 3;
  then Y in Z4 by A7,XBOOLE_0:def 3;
  then Y meets X by A3;
  hence contradiction by A16,A6,XBOOLE_1:70;
end;

theorem
  ex Y st Y in X &
   for Y1,Y2,Y3,Y4,Y5 st Y1 in Y2 &  Y2 in Y3 & Y3 in Y4 & Y4 in Y5 & Y5 in Y
    holds Y1 misses X
proof
  defpred P[set] means ex Y1,Y2,Y3,Y4 st Y1 in Y2 & Y2 in Y3 & Y3 in Y4 & Y4
  in $1 & Y1 meets X;
  consider Z1 such that
A1: for Y holds Y in Z1 iff Y in union X & P[Y] from XBOOLE_0:sch 1;
  defpred T[set] means $1 meets X;
  defpred S[set] means ex Y1 st Y1 in $1 & Y1 meets X;
  defpred R[set] means ex Y1,Y2 st Y1 in Y2 & Y2 in $1 & Y1 meets X;
  defpred Q[set] means ex Y1,Y2,Y3 st Y1 in Y2 & Y2 in Y3 & Y3 in $1 & Y1
  meets X;
  consider Z2 such that
A2: for Y holds Y in Z2 iff Y in union union X & Q[Y] from XBOOLE_0:sch
  1;
  consider Z5 such that
A3: for Y holds Y in Z5 iff Y in union union union union union X & T[Y]
  from XBOOLE_0:sch 1;
  consider Z3 such that
A4: for Y holds Y in Z3 iff Y in union union union X & R[Y] from
  XBOOLE_0:sch 1;
  consider Z4 such that
A5: for Y holds Y in Z4 iff Y in union union union union X & S[Y] from
  XBOOLE_0:sch 1;
  set V = X \/ Z1 \/ Z2 \/ Z3 \/ Z4 \/ Z5;
  consider Y such that
A6: Y in V and
A7: Y misses V by Th1;
A8: V = X \/ (Z1 \/ Z2) \/ Z3 \/ Z4 \/ Z5 by XBOOLE_1:4
    .= X \/ (Z1 \/ Z2 \/ Z3) \/ Z4 \/ Z5 by XBOOLE_1:4
    .= X \/ (Z1 \/ Z2 \/ Z3 \/ Z4) \/ Z5 by XBOOLE_1:4
    .= X \/ (Z1 \/ Z2 \/ Z3 \/ Z4 \/ Z5) by XBOOLE_1:4;
A9: now
    assume
A10: Y in Z1;
    then consider Y1,Y2,Y3,Y4 such that
A11: Y1 in Y2 & Y2 in Y3 & Y3 in Y4 and
A12: Y4 in Y and
A13: Y1 meets X by A1;
    Y in union X by A1,A10;
    then Y4 in union union X by A12,TARSKI:def 4;
    then Y4 in Z2 by A2,A11,A13;
    then Y4 in X \/ Z1 \/ Z2 by XBOOLE_0:def 3;
    then Y meets X \/ Z1 \/ Z2 by A12,XBOOLE_0:3;
    then Y meets X \/ Z1 \/ Z2 \/ Z3 by XBOOLE_1:70;
    then Y meets X \/ Z1 \/ Z2 \/ Z3 \/ Z4 by XBOOLE_1:70;
    hence contradiction by A7,XBOOLE_1:70;
  end;
A14: now
    assume
A15: Y in Z2;
    then consider Y1,Y2,Y3 such that
A16: Y1 in Y2 & Y2 in Y3 and
A17: Y3 in Y and
A18: Y1 meets X by A2;
    Y in union union X by A2,A15;
    then Y3 in union union union X by A17,TARSKI:def 4;
    then Y3 in Z3 by A4,A16,A18;
    then Y3 in X \/ Z1 \/ Z2 \/ Z3 by XBOOLE_0:def 3;
    then Y3 in X \/ Z1 \/ Z2 \/ Z3 \/ Z4 by XBOOLE_0:def 3;
    then Y3 in V by XBOOLE_0:def 3;
    hence contradiction by A7,A17,XBOOLE_0:3;
  end;
A19: now
    assume
A20: Y in Z3;
    then consider Y1,Y2 such that
A21: Y1 in Y2 and
A22: Y2 in Y and
A23: Y1 meets X by A4;
    Y in union union union X by A4,A20;
    then Y2 in union union union union X by A22,TARSKI:def 4;
    then Y2 in Z4 by A5,A21,A23;
    then Y2 in X \/ Z1 \/ Z2 \/ Z3 \/ Z4 by XBOOLE_0:def 3;
    then Y2 in V by XBOOLE_0:def 3;
    hence contradiction by A7,A22,XBOOLE_0:3;
  end;
A24: now
    assume
A25: Y in Z4;
    then consider Y1 such that
A26: Y1 in Y and
A27: Y1 meets X by A5;
    Y in union union union union X by A5,A25;
    then Y1 in union union union union union X by A26,TARSKI:def 4;
    then Y1 in Z5 by A3,A27;
    then Y1 in V by XBOOLE_0:def 3;
    hence contradiction by A7,A26,XBOOLE_0:3;
  end;
  assume
A28: not thesis;
  now
    assume
A29: Y in X;
    then consider Y1,Y2,Y3,Y4,Y5 such that
A30: Y1 in Y2 & Y2 in Y3 & Y3 in Y4 & Y4 in Y5 and
A31: Y5 in Y and
A32: not Y1 misses X by A28;
    Y5 in union X by A29,A31,TARSKI:def 4;
    then Y5 in Z1 by A1,A30,A32;
    then Y5 in X \/ Z1 by XBOOLE_0:def 3;
    then Y5 in X \/ Z1 \/ Z2 by XBOOLE_0:def 3;
    then Y5 in X \/ Z1 \/ Z2 \/ Z3 by XBOOLE_0:def 3;
    then Y meets X \/ Z1 \/ Z2 \/ Z3 by A31,XBOOLE_0:3;
    then Y meets X \/ Z1 \/ Z2 \/ Z3 \/ Z4 by XBOOLE_1:70;
    hence contradiction by A7,XBOOLE_1:70;
  end;
  then Y in Z1 \/ Z2 \/ Z3 \/ Z4 \/ Z5 by A8,A6,XBOOLE_0:def 3;
  then Y in Z1 \/ (Z2 \/ Z3) \/ Z4 \/ Z5 by XBOOLE_1:4;
  then Y in Z1 \/ (Z2 \/ Z3 \/ Z4) \/ Z5 by XBOOLE_1:4;
  then Y in Z1 \/ (Z2 \/ Z3 \/ Z4 \/ Z5) by XBOOLE_1:4;
  then Y in Z2 \/ Z3 \/ Z4 \/ Z5 by A9,XBOOLE_0:def 3;
  then Y in Z2 \/ (Z3 \/ Z4) \/ Z5 by XBOOLE_1:4;
  then Y in Z2 \/ (Z3 \/ Z4 \/ Z5) by XBOOLE_1:4;
  then Y in Z3 \/ Z4 \/ Z5 by A14,XBOOLE_0:def 3;
  then Y in Z3 \/ (Z4 \/ Z5) by XBOOLE_1:4;
  then Y in Z4 \/ Z5 by A19,XBOOLE_0:def 3;
  then Y in Z5 by A24,XBOOLE_0:def 3;
  then Y meets X by A3;
  hence contradiction by A8,A7,XBOOLE_1:70;
end;




theorem
  not ( X1 in X2 & X2 in X3 & X3 in X1)
proof
A1: X2 in { X1,X2,X3 } & X3 in { X1,X2,X3 } by ENUMSET1:def 1;
A2: X1 in { X1,X2,X3 } by ENUMSET1:def 1;
  then consider T being set such that
A3: T in { X1,X2,X3 } and
A4: { X1,X2,X3 } misses T by Th1;
  T = X1 or T = X2 or T = X3 by A3,ENUMSET1:def 1;
  hence thesis by A2,A1,A4,XBOOLE_0:3;
end;

theorem
  not ( X1 in X2 & X2 in X3 & X3 in X4 & X4 in X1)
proof
A1: X2 in { X1,X2,X3,X4 } & X3 in { X1,X2,X3,X4 } by ENUMSET1:def 2;
A2: X4 in { X1,X2,X3,X4 } by ENUMSET1:def 2;
A3: X1 in { X1,X2,X3,X4 } by ENUMSET1:def 2;
  then consider T being set such that
A4: T in { X1,X2,X3,X4 } and
A5: { X1,X2,X3,X4 } misses T by Th1;
  T = X1 or T = X2 or T = X3 or T = X4 by A4,ENUMSET1:def 2;
  hence thesis by A3,A1,A2,A5,XBOOLE_0:3;
end;

theorem
  not ( X1 in X2 & X2 in X3 & X3 in X4 & X4 in X5 & X5 in X1)
proof
  assume that
A1: X1 in X2 and
A2: X2 in X3 and
A3: X3 in X4 and
A4: X4 in X5 and
A5: X5 in X1;
  set Z = { X1,X2,X3,X4,X5 };
A6: for Y st Y in Z holds Z meets Y
  proof
    let Y such that
A7: Y in Z;
    now
      per cases by A7,ENUMSET1:def 3;
      suppose
A8:     Y = X1;
        take y = X5;
        thus y in Z & y in Y by A5,A8,ENUMSET1:def 3;
      end;
      suppose
A9:     Y = X2;
        take y = X1;
        thus y in Z & y in Y by A1,A9,ENUMSET1:def 3;
      end;
      suppose
A10:    Y = X3;
        take y = X2;
        thus y in Z & y in Y by A2,A10,ENUMSET1:def 3;
      end;
      suppose
A11:    Y = X4;
        take y = X3;
        thus y in Z & y in Y by A3,A11,ENUMSET1:def 3;
      end;
      suppose
A12:    Y = X5;
        take y = X4;
        thus y in Z & y in Y by A4,A12,ENUMSET1:def 3;
      end;
    end;
    hence thesis by XBOOLE_0:3;
  end;
  X1 in { X1,X2,X3,X4,X5 } by ENUMSET1:def 3;
  hence contradiction by A6,Th1;
end;

theorem
  not ( X1 in X2 & X2 in X3 & X3 in X4 & X4 in X5 & X5 in X6 & X6 in X1)
proof
  assume that
A1: X1 in X2 and
A2: X2 in X3 and
A3: X3 in X4 and
A4: X4 in X5 and
A5: X5 in X6 and
A6: X6 in X1;
  set Z = { X1,X2,X3,X4,X5,X6 };
A7: for Y st Y in Z holds Z meets Y
  proof
    let Y such that
A8: Y in Z;
    now
      per cases by A8,ENUMSET1:def 4;
      suppose
A9:     Y = X1;
        take y = X6;
        thus y in Z & y in Y by A6,A9,ENUMSET1:def 4;
      end;
      suppose
A10:    Y = X2;
        take y = X1;
        thus y in Z & y in Y by A1,A10,ENUMSET1:def 4;
      end;
      suppose
A11:    Y = X3;
        take y = X2;
        thus y in Z & y in Y by A2,A11,ENUMSET1:def 4;
      end;
      suppose
A12:    Y = X4;
        take y = X3;
        thus y in Z & y in Y by A3,A12,ENUMSET1:def 4;
      end;
      suppose
A13:    Y = X5;
        take y = X4;
        thus y in Z & y in Y by A4,A13,ENUMSET1:def 4;
      end;
      suppose
A14:    Y = X6;
        take y = X5;
        thus y in Z & y in Y by A5,A14,ENUMSET1:def 4;
      end;
    end;
    hence thesis by XBOOLE_0:3;
  end;
  X1 in { X1,X2,X3,X4,X5,X6 } by ENUMSET1:def 4;
  hence contradiction by A7,Th1;
end;














begin

reserve v,x,x1,x2,y,y1,y2,z,A,B,X,X1,X2,X3,X4,Y,Y1,Y2,Z,N,M for set;

registration
  let x, y be set;
  cluster [x,y] -> non empty;
  coherence;
end;

Lm1: {x} c= X iff x in X
proof
  x in {x} by TARSKI:def 1;
  hence {x} c= X implies x in X by TARSKI:def 3;
  assume
A1: x in X;
  let y;
  thus thesis by A1,TARSKI:def 1;
end;

Lm2: Y c= X & not x in Y implies Y c= X \ { x }
proof
  assume
A1: Y c= X & not x in Y;
  let y;
  assume y in Y;
  then y in X & not y in { x } by A1,TARSKI:def 1,def 3;
  hence thesis by XBOOLE_0:def 5;
end;

Lm3: Y c= { x } iff Y = {} or Y = { x }
proof
  thus Y c= { x } implies Y = {} or Y = { x }
  proof
    assume
A1: Y c= { x };
    x in Y or not x in Y;
    then { x } c= Y or Y c= { x } \ { x } by A1,Lm1,Lm2;
    then { x } = Y or Y c= {} by A1,XBOOLE_0:def 10,XBOOLE_1:37;
    hence thesis by XBOOLE_1:3;
  end;
  thus thesis by XBOOLE_1:2;
end;

definition
  let X;
  defpred IT[set] means $1 c= X;
  func bool X means
  :Def1:
  Z in it iff Z c= X;
  existence
  proof
    consider M such that
A1: X in M & for X,Y holds X in M & Y c= X implies Y in M and
    for X st X in M ex Z st Z in M & for Y st Y c= X holds Y in Z and
    for X holds X c= M implies X,M are_equipotent or X in M by TARSKI:3;
    consider IT being set such that
A2: Y in IT iff Y in M & IT[Y] from XBOOLE_0:sch 1;
    take IT;
    thus thesis by A2,A1;
  end;
  uniqueness
  proof
    thus for X1,X2 being set st (for x being set holds x in X1 iff IT[x]) & (
    for x being set holds x in X2 iff IT[x]) holds X1 = X2 from XBOOLE_0:sch 3;
  end;
end;

definition
  let X1,X2;
  defpred X[set] means ex x,y st x in X1 & y in X2 & $1 = [x,y];
  func [: X1,X2 :] means
  :Def2:
  z in it iff ex x,y st x in X1 & y in X2 & z = [x,y];
  existence
  proof
A1: X1 c= X1 \/ X2 by XBOOLE_1:7;
    consider X such that
A2: for z holds z in X iff z in bool(bool(X1 \/ X2)) & X[z] from
    XBOOLE_0:sch 1;
    take X;
    let z;
    thus z in X implies ex x,y st x in X1 & y in X2 & z = [x,y] by A2;
    given x,y such that
A3: x in X1 and
A4: y in X2 and
A5: z = [x,y];
    {x,y} c= X1 \/ X2
    proof
      let z;
      assume z in {x,y};
      then z = x or z = y by TARSKI:def 2;
      hence thesis by A3,A4,XBOOLE_0:def 3;
    end;
    then
A6: {x,y} in bool(X1 \/ X2) by Def1;
    {x} c= X1 by A3,Lm1;
    then {x} c= X1 \/ X2 by A1,XBOOLE_1:1;
    then
A7: {x} in bool(X1 \/ X2) by Def1;
    {{x,y},{x}} c= bool(X1 \/ X2)
    proof
      let z;
      assume z in {{x,y},{x}};
      hence thesis by A7,A6,TARSKI:def 2;
    end;
    then {{x,y},{x}} in bool(bool(X1 \/ X2)) by Def1;
    hence thesis by A2,A3,A4,A5;
  end;
  uniqueness
  proof
    thus for X1,X2 being set st (for x being set holds x in X1 iff X[x]) & (
    for x being set holds x in X2 iff X[x]) holds X1 = X2 from XBOOLE_0:sch 3;
  end;
end;

definition
  let X1,X2,X3;
  func [: X1,X2,X3 :] equals
  [:[:X1,X2:],X3:];
  coherence;
end;

definition
  let X1,X2,X3,X4;
  func [: X1,X2,X3,X4 :] equals
  [:[:X1,X2,X3:],X4:];
  coherence;
end;

begin





theorem
  bool {} = { {} }
proof
  now
    let x be set;
    x c= {} iff x = {} by XBOOLE_1:3;
    hence x in bool {} iff x in { {} } by Def1,TARSKI:def 1;
  end;
  hence thesis by TARSKI:1;
end;

theorem
  union {} = {}
proof
  now
    given x such that
A1: x in union {};
    ex X being set st x in X & X in {} by A1,TARSKI:def 4;
    hence contradiction;
  end;
  hence thesis by XBOOLE_0:7;
end;





theorem Th3:
  {x} c= {y} implies x = y
proof
  x in { x } by TARSKI:def 1;
  then {x} = {y} implies x = y by TARSKI:def 1;
  hence thesis by Lm3;
end;

theorem Th4:
  {x} = {y1,y2} implies x = y1
proof
  assume { x } = { y1,y2 };
  then y1 in { x } by TARSKI:def 2;
  hence thesis by TARSKI:def 1;
end;

theorem
  {x} = {y1,y2} implies y1 = y2
proof
  assume
A1: { x } = { y1,y2 };
  then x = y1 by Th4;
  hence thesis by A1,Th4;
end;

theorem Th6:
  { x1,x2 } = { y1,y2 } implies x1 = y1 or x1 = y2
proof
  x1 in { x1,x2 } by TARSKI:def 2;
  hence thesis by TARSKI:def 2;
end;

theorem Th7:
  {x} c= {x,y}
proof
  let z;
  assume z in {x};
  then z=x by TARSKI:def 1;
  hence thesis by TARSKI:def 2;
end;

Lm4: {x} \/ X c= X implies x in X
proof
  assume
A1: {x} \/ X c= X;
  assume not x in X;
  then not x in {x} \/ X by A1,TARSKI:def 3;
  then not x in {x} by XBOOLE_0:def 3;
  hence thesis by TARSKI:def 1;
end;

theorem
  {x} \/ {y} = {x} implies x = y
proof
  assume {x} \/ {y} = {x};
  then y in {x} or x in {y} by Lm4;
  hence thesis by TARSKI:def 1;
end;

theorem
  {x} \/ {x,y} = {x,y}
proof
  x in {x,y} by TARSKI:def 2;
  hence thesis by XBOOLE_1:12,Lm1;
end;

Lm6: {x} misses X implies not x in X
proof
A1: x in {x} by TARSKI:def 1;
  assume {x} /\ X = {} & x in X;
  hence contradiction by A1,XBOOLE_0:def 4;
end;

theorem
  {x} misses {y} implies x <> y
proof
  assume {x} misses {y};
  then not x in {y} by Lm6;
  hence thesis by TARSKI:def 1;
end;

Lm7: not x in X implies {x} misses X
proof
  assume
A1: not x in X;
  thus {x} /\ X c= {}
  proof
    let y;
    assume y in {x} /\ X;
    then y in {x} & y in X by XBOOLE_0:def 4;
    hence thesis by A1,TARSKI:def 1;
  end;
  thus thesis by XBOOLE_1:2;
end;

theorem Th11:
  x <> y implies {x} misses {y}
proof
  assume x <> y;
  then not x in {y} by TARSKI:def 1;
  hence thesis by Lm7;
end;

Lm8: X /\ {x} = {x} implies x in X
proof
  assume X /\ {x} = {x};
  then x in X /\ {x} by TARSKI:def 1;
  hence thesis by XBOOLE_0:def 4;
end;

theorem
  {x} /\ {y} = {x} implies x = y
proof
  assume {x} /\ {y} = {x};
  then x in {y} or y in {x} by Lm8;
  hence thesis by TARSKI:def 1;
end;

Lm9: x in X implies X /\ {x} = {x} by XBOOLE_1:28,Lm1;

theorem
  {x} /\ {x,y} = {x}
proof
  x in {x,y} by TARSKI:def 2;
  hence thesis by XBOOLE_1:28,Lm1;
end;

Lm10: {x} \ X = {x} iff not x in X by Lm6,Lm7,XBOOLE_1:83;

theorem
  {x} \ {y} = {x} iff x <> y
proof
  {x} \ {y} = {x} iff not x in {y} by Lm6,Lm7,XBOOLE_1:83;
  hence thesis by TARSKI:def 1;
end;

Lm11: {x} \ X = {} iff x in X by Lm1,XBOOLE_1:37;

theorem
  {x} \ {y} = {} implies x = y
proof
  assume {x} \ {y} = {};
  then x in {y} by Lm1,XBOOLE_1:37;
  hence thesis by TARSKI:def 1;
end;

theorem
  {x} \ {x,y} = {}
proof
  x in {x,y} by TARSKI:def 2;
  hence thesis by Lm1,XBOOLE_1:37;
end;

Lm12: {x,y} \ X = {x} iff not x in X & (y in X or x = y)
proof
  thus {x,y} \ X = {x} implies not x in X & (y in X or x = y)
  proof
    assume
A1: {x,y} \ X = {x};
    then x in {x,y} \ X by TARSKI:def 1;
    hence not x in X by XBOOLE_0:def 5;
A2: y in {x,y} by TARSKI:def 2;
    assume not y in X;
    then y in {x} by A1,A2,XBOOLE_0:def 5;
    hence thesis by TARSKI:def 1;
  end;
  assume
A3: ( not x in X)&( y in X or x=y);
  z in {x,y} \ X iff z=x
  proof
    z in {x,y} \ X iff z in {x,y} & not z in X by XBOOLE_0:def 5;
    hence thesis by A3,TARSKI:def 2;
  end;
  hence thesis by TARSKI:def 1;
end;

theorem
  x <> y implies { x,y } \ { y } = { x }
proof
  assume x <> y;
  then
A1: not x in {y} by TARSKI:def 1;
  y in {y} by TARSKI:def 1;
  hence thesis by A1,Lm12;
end;

theorem
  {x} c= {y} implies x = y by Th3;

theorem
  {z} c= {x,y} implies z = x or z = y
proof
A1: z in {z} by TARSKI:def 1;
  assume {z} c= {x,y};
  then z in {x,y} by A1,TARSKI:def 3;
  hence thesis by TARSKI:def 2;
end;

theorem Th20:
  {x,y} c= {z} implies x = z
proof
A1: x in {x,y} by TARSKI:def 2;
  assume {x,y} c= {z};
  then x in {z} by A1,TARSKI:def 3;
  hence thesis by TARSKI:def 1;
end;

theorem
  {x,y} c= {z} implies {x,y} = {z}
proof
  assume {x,y} c= {z};
  then x=z & y=z by Th20;
  hence thesis by ENUMSET1:29;
end;

Lm13: X <> {x} & X <> {} implies ex y st y in X & y <> x
proof
  assume that
A1: X <> {x} and
A2: X <> {};
  per cases;
  suppose
A3: not x in X;
    consider y being set such that
A4: y in X by A2,XBOOLE_0:7;
    take y;
    thus thesis by A3,A4;
  end;
  suppose
A5: x in X;
    consider y such that
A6: y in X & not y in {x} or y in {x} & not y in X by A1,TARSKI:1;
    take y;
    thus thesis by A5,A6,TARSKI:def 1;
  end;
end;

Lm14: Z c= {x1,x2} iff Z = {} or Z = {x1} or Z = {x2} or Z = {x1,x2}
proof
  thus Z c= {x1,x2} implies Z={} or Z={x1} or Z={x2} or Z={x1,x2}
  proof
    assume that
A1: Z c= {x1,x2} and
A2: Z <> {} and
A3: Z <> {x1} and
A4: Z <> {x2};
    now
      let z;
      thus z in Z implies z in {x1,x2} by A1,TARSKI:def 3;
A5:   now
        assume
A6:     not x1 in Z;
        consider y such that
A7:     y in Z and
A8:     y<>x2 by A2,A4,Lm13;
        y in {x1,x2} by A1,A7,TARSKI:def 3;
        hence contradiction by A6,A7,A8,TARSKI:def 2;
      end;
A9:   now
        assume
A10:    not x2 in Z;
        consider y such that
A11:    y in Z and
A12:    y<>x1 by A2,A3,Lm13;
        y in {x1,x2} by A1,A11,TARSKI:def 3;
        hence contradiction by A10,A11,A12,TARSKI:def 2;
      end;
      assume z in {x1,x2};
      hence z in Z by A5,A9,TARSKI:def 2;
    end;
    hence thesis by TARSKI:1;
  end;
  thus thesis by Th7,XBOOLE_1:2;
end;

theorem
  {x1,x2} c= {y1,y2} implies x1 = y1 or x1 = y2
proof
  assume {x1,x2} c= {y1,y2};
  then {x1,x2}={} or {x1,x2}={y1} or {x1,x2}={y2} or {x1,x2}={y1,y2} by Lm14;
  hence thesis by Th4,Th6;
end;

theorem
  x <> y implies { x } \+\ { y } = { x,y }
proof
  assume
A1: x <> y;
  z in { x } \+\ { y } iff z=x or z=y
  proof
A2: z in {y} iff z=y by TARSKI:def 1;
    z in {x} iff z=x by TARSKI:def 1;
    hence thesis by A1,A2,XBOOLE_0:1;
  end;
  hence thesis by TARSKI:def 2;
end;

theorem
  bool { x } = { {} , { x }}
proof
  now
    let y;
    y c= { x } iff y = {} or y = { x } by Lm3;
    hence y in bool { x } iff y in { {}, { x }} by Def1,TARSKI:def 2;
  end;
  hence thesis by TARSKI:1;
end;

Lm15: X in A implies X c= union A
proof
  assume
A1: X in A;
  let x;
  thus thesis by A1,TARSKI:def 4;
end;

theorem
  union { x } = x
proof
A1: union { x } c= x
  proof
    let y;
    assume y in union { x };
    then ex Y being set st y in Y & Y in { x } by TARSKI:def 4;
    hence thesis by TARSKI:def 1;
  end;
  x in { x } by TARSKI:def 1;
  then x c= union { x } by Lm15;
  hence thesis by A1,XBOOLE_0:def 10;
end;

Lm16: union { X,Y } = X \/ Y
proof
  x in union {X,Y} iff x in X or x in Y
  proof
    thus x in union {X,Y} implies x in X or x in Y
    proof
      assume x in union {X,Y};
      then ex Z st x in Z & Z in {X,Y} by TARSKI:def 4;
      hence thesis by TARSKI:def 2;
    end;
    X in {X,Y} & Y in {X,Y} by TARSKI:def 2;
    hence thesis by TARSKI:def 4;
  end;
  hence thesis by XBOOLE_0:def 3;
end;

theorem
  union {{x},{y}} = {x,y}
proof
  thus union {{x},{y}} = {x} \/ {y} by Lm16
    .= {x,y} by ENUMSET1:1;
end;





Lm17: [x,y] in [:X,Y:] iff x in X & y in Y
proof
  thus [x,y] in [:X,Y:] implies x in X & y in Y
  proof
    assume [x,y] in [:X,Y:];
    then ex x1,y1 st x1 in X & y1 in Y & [x,y]=[x1,y1] by Def2;
    hence thesis by XTUPLE_0:1;
  end;
  thus thesis by Def2;
end;

canceled;

theorem
  [x,y] in [:{x1},{y1}:] iff x = x1 & y = y1
proof
  x=x1 & y=y1 iff x in {x1} & y in {y1} by TARSKI:def 1;
  hence thesis by Lm17;
end;

theorem
  [:{x},{y}:] = {[x,y]}
proof
  now
    let z;
    thus z in [:{x},{y}:] implies z in {[x,y]}
    proof
      assume z in [:{x},{y}:];
      then consider x1,y1 such that
A1:   x1 in {x} & y1 in {y} and
A2:   z=[x1,y1] by Def2;
      x1=x & y1=y by A1,TARSKI:def 1;
      hence thesis by A2,TARSKI:def 1;
    end;
    assume z in {[x,y]};
    then
A3: z=[x,y] by TARSKI:def 1;
    x in {x} & y in {y} by TARSKI:def 1;
    hence z in [:{x},{y}:] by A3,Lm17;
  end;
  hence thesis by TARSKI:1;
end;

theorem Th30:
  [:{x},{y,z}:] = {[x,y],[x,z]} & [:{x,y},{z}:] = {[x,z],[y,z]}
proof
  now
    let v;
A1: z in{y,z} by TARSKI:def 2;
    thus v in [:{x},{y,z}:] implies v in {[x,y],[x,z]}
    proof
      assume v in [:{x},{y,z}:];
      then consider x1,y1 such that
A2:   x1 in {x} and
A3:   y1 in {y,z} and
A4:   v=[x1,y1] by Def2;
A5:   y1=y or y1=z by A3,TARSKI:def 2;
      x1=x by A2,TARSKI:def 1;
      hence thesis by A4,A5,TARSKI:def 2;
    end;
    assume v in {[x,y],[x,z]};
    then
A6: v=[x,y] or v=[x,z] by TARSKI:def 2;
    x in{x} & y in{y,z} by TARSKI:def 1,def 2;
    hence v in [:{x},{y,z}:] by A6,A1,Lm17;
  end;
  hence [:{x},{y,z}:] = {[x,y],[x,z]} by TARSKI:1;
  now
    let v;
A7: z in{z} by TARSKI:def 1;
    thus v in [:{x,y},{z}:] implies v in {[x,z],[y,z]}
    proof
      assume v in [:{x,y},{z}:];
      then consider x1,y1 such that
A8:   x1 in {x,y} and
A9:   y1 in {z} and
A10:  v=[x1,y1] by Def2;
A11:  x1=x or x1=y by A8,TARSKI:def 2;
      y1=z by A9,TARSKI:def 1;
      hence thesis by A10,A11,TARSKI:def 2;
    end;
    assume v in {[x,z],[y,z]};
    then
A12: v=[x,z] or v=[y,z] by TARSKI:def 2;
    x in{x,y} & y in{x,y} by TARSKI:def 2;
    hence v in [:{x,y},{z}:] by A12,A7,Lm17;
  end;
  hence thesis by TARSKI:1;
end;





theorem
  {x} c= X iff x in X by Lm1;

theorem Th32:
  {x1,x2} c= Z iff x1 in Z & x2 in Z
proof
  x1 in {x1,x2} & x2 in {x1,x2} by TARSKI:def 2;
  hence {x1,x2} c= Z implies x1 in Z & x2 in Z by TARSKI:def 3;
  assume
A1: x1 in Z & x2 in Z;
  let z;
  assume z in {x1,x2};
  hence thesis by A1,TARSKI:def 2;
end;





theorem
  Y c= { x } iff Y = {} or Y = { x } by Lm3;

theorem
  Y c= X & not x in Y implies Y c= X \ { x } by Lm2;

theorem
  X <> {x} & X <> {} implies ex y st y in X & y <> x by Lm13;

theorem
  Z c= {x1,x2} iff Z = {} or Z = {x1} or Z = {x2} or Z = {x1,x2} by Lm14;





theorem Th37:
  {z} = X \/ Y implies X = {z} & Y = {z} or X = {} & Y = {z} or X
  = {z} & Y = {}
proof
  assume
A1: {z} = X \/ Y;
  X <> {} or Y <> {} by A1;
  hence thesis by A1,Lm3,XBOOLE_1:7;
end;

theorem
  {z} = X \/ Y & X <> Y implies X = {} or Y = {}
proof
  assume {z} = X \/ Y;
  then X={z} & Y={z} or X={} & Y={z} or X={z} & Y={} by Th37;
  hence thesis;
end;

theorem
  {x} \/ X c= X implies x in X by Lm4;

theorem
  x in X implies {x} \/ X = X by XBOOLE_1:12,Lm1;

theorem
  {x,y} \/ Z c= Z implies x in Z
proof
  assume
A1: {x,y} \/ Z c= Z;
  assume not x in Z;
  then not x in {x,y} \/ Z by A1,TARSKI:def 3;
  then not x in {x,y} by XBOOLE_0:def 3;
  hence thesis by TARSKI:def 2;
end;

theorem
  x in Z & y in Z implies {x,y} \/ Z = Z by XBOOLE_1:12,Th32;

theorem
  {x} \/ X <> {};

theorem
  {x,y} \/ X <> {};





theorem
  X /\ {x} = {x} implies x in X by Lm8;

theorem
  x in X implies X /\ {x} = {x} by XBOOLE_1:28,Lm1;

theorem
  x in Z & y in Z implies {x,y} /\ Z = {x,y} by XBOOLE_1:28,Th32;

theorem
  {x} misses X implies not x in X by Lm6;

theorem Th49:
  {x,y} misses Z implies not x in Z
proof
A1: x in {x,y} by TARSKI:def 2;
  assume {x,y} /\ Z = {} & x in Z;
  hence contradiction by A1,XBOOLE_0:def 4;
end;

theorem
  not x in X implies {x} misses X by Lm7;

theorem Th51:
  not x in Z & not y in Z implies {x,y} misses Z
proof
  assume
A1: ( not x in Z)& not y in Z;
  assume {x,y} meets Z;
  then consider z such that
A2: z in {x,y} /\ Z by XBOOLE_0:4;
  z in {x,y} & z in Z by A2,XBOOLE_0:def 4;
  hence contradiction by A1,TARSKI:def 2;
end;

theorem
  {x} misses X or {x} /\ X = {x} by Lm7,Lm9;

theorem
  {x,y} /\ X = {x} implies not y in X or x = y
proof
A1: y in {x,y} by TARSKI:def 2;
  assume {x,y} /\ X = {x} & y in X;
  then y in {x} by A1,XBOOLE_0:def 4;
  hence thesis by TARSKI:def 1;
end;

theorem
  x in X & (not y in X or x = y) implies {x,y} /\ X = {x}
proof
  assume
A1: x in X &( not y in X or x=y);
  z in {x,y} /\ X iff z=x
  proof
    z in {x,y} /\ X iff z in {x,y} & z in X by XBOOLE_0:def 4;
    hence thesis by A1,TARSKI:def 2;
  end;
  hence thesis by TARSKI:def 1;
end;

theorem
  {x,y} /\ X = {x,y} implies x in X
proof
  assume {x,y} /\ X = {x,y};
  then x in {x,y} /\ X & y in {x,y} /\ X or x in X /\ {x,y} & y in X /\ {x,y}
  by TARSKI:def 2;
  hence thesis by XBOOLE_0:def 4;
end;





theorem Th56:
  z in X \ {x} iff z in X & z <> x
proof
  z in X \ {x} iff z in X & not z in {x} by XBOOLE_0:def 5;
  hence thesis by TARSKI:def 1;
end;

theorem Th57:
  X \ {x} = X iff not x in X
proof
  X \ {x} = X iff X misses {x} by XBOOLE_1:83;
  hence thesis by Lm6,Lm7;
end;

theorem
  X \ {x} = {} implies X = {} or X = {x}
proof
  assume X \ {x} = {};
  then X c= {x} by XBOOLE_1:37;
  hence thesis by Lm3;
end;

theorem
  {x} \ X = {x} iff not x in X by Lm6,Lm7,XBOOLE_1:83;

theorem
  {x} \ X = {} iff x in X by Lm1,XBOOLE_1:37;

theorem
  {x} \ X = {} or {x} \ X = {x} by Lm10,Lm11;

theorem
  {x,y} \ X = {x} iff not x in X & (y in X or x = y) by Lm12;

theorem
  {x,y} \ X = {x,y} iff not x in X & not y in X by Th49,Th51,XBOOLE_1:83;

theorem Th64:
  {x,y} \ X = {} iff x in X & y in X
proof
  {x,y} \ X = {} iff {x,y} c= X by XBOOLE_1:37;
  hence thesis by Th32;
end;

theorem
  {x,y} \ X = {} or {x,y} \ X = {x} or {x,y} \ X = {y} or {x,y} \ X = {x ,y}
proof
  assume that
A1: {x,y} \ X <> {} and
A2: {x,y} \ X <> {x} and
A3: {x,y} \ X <> {y};
A4: not x in X & x<>y or y in X by A3,Lm12;
  x in X or not y in X & x<>y by A2,Lm12;
  hence thesis by A1,A4,Th51,XBOOLE_1:83,Th64;
end;

theorem
  X \ {x,y} = {} iff X = {} or X = {x} or X = {y} or X = {x,y}
proof
  X \ {x,y} = {} iff X c= {x,y} by XBOOLE_1:37;
  hence thesis by Lm14;
end;





theorem
  A c= B implies bool A c= bool B
proof
  assume
A1: A c= B;
  let x;
  assume x in bool A;
  then x c= A by Def1;
  then x c= B by A1,XBOOLE_1:1;
  hence thesis by Def1;
end;

theorem
  { A } c= bool A
proof
  A in bool A by Def1;
  hence thesis by Lm1;
end;

theorem
  bool A \/ bool B c= bool (A \/ B)
proof
  let x;
  assume x in bool A \/ bool B;
  then x in bool A or x in bool B by XBOOLE_0:def 3;
  then
A1: x c= A or x c= B by Def1;
  A c= A \/ B & B c= A \/ B by XBOOLE_1:7;
  then x c= A \/ B by A1,XBOOLE_1:1;
  hence thesis by Def1;
end;

theorem
  bool A \/ bool B = bool (A \/ B) implies A,B are_c=-comparable
proof
  assume
A1: bool A \/ bool B = bool (A \/ B);
  A \/ B in bool (A \/ B) by Def1;
  then A \/ B in bool A or A \/ B in bool B by A1,XBOOLE_0:def 3;
  then
A2: A \/ B c= A or A \/ B c= B by Def1;
  A c= A \/ B & B c= A \/ B by XBOOLE_1:7;
  hence A c= B or B c= A by A2,XBOOLE_0:def 10;
end;

theorem
  bool (A /\ B) = bool A /\ bool B
proof
  now
    let x;
    A /\ B c= A & A /\ B c= B by XBOOLE_1:17;
    then x c= A & x c= B iff x c= A /\ B by XBOOLE_1:1,19;
    then x in bool A & x in bool B iff x in bool (A /\ B) by Def1;
    hence x in bool (A /\ B) iff x in bool A /\ bool B by XBOOLE_0:def 4;
  end;
  hence thesis by TARSKI:1;
end;

theorem
  bool (A \ B) c= { {} } \/ (bool A \ bool B)
proof
  let x;
  assume x in bool (A \ B);
  then
A1: x c= A \ B by Def1;
  then x misses B by XBOOLE_1:63,79;
  then A \ B c= A & x /\ B = {} by XBOOLE_0:def 7,XBOOLE_1:36;
  then x = {} or x c= A & not x c= B by A1,XBOOLE_1:1,28;
  then x in { {} } or x in bool A & not x in bool B by Def1,TARSKI:def 1;
  then x in { {} } or x in bool A \ bool B by XBOOLE_0:def 5;
  hence thesis by XBOOLE_0:def 3;
end;

theorem
  bool (A \ B) \/ bool (B \ A) c= bool (A \+\ B)
proof
  let x;
  assume x in bool (A \ B) \/ bool (B \ A);
  then x in bool (A \ B) or x in bool (B \ A) by XBOOLE_0:def 3;
  then
A1: x c= A \ B or x c= B \ A by Def1;
  x c= (A \ B) \/ (B \ A)
  proof
    let z;
    assume z in x;
    then z in A \ B or z in B \ A by A1,TARSKI:def 3;
    hence thesis by XBOOLE_0:def 3;
  end;
  hence thesis by Def1;
end;





theorem
  X in A implies X c= union A by Lm15;

theorem
  union { X,Y } = X \/ Y by Lm16;

theorem
  (for X st X in A holds X c= Z) implies union A c= Z
proof
  assume
A1: for X st X in A holds X c= Z;
  let y;
  assume y in union A;
  then consider Y such that
A2: y in Y and
A3: Y in A by TARSKI:def 4;
  Y c= Z by A1,A3;
  hence thesis by A2,TARSKI:def 3;
end;

theorem Th77:
  A c= B implies union A c= union B
proof
  assume
A1: A c= B;
  let x;
  assume x in union A;
  then consider Y such that
A2: x in Y and
A3: Y in A by TARSKI:def 4;
  Y in B by A1,A3,TARSKI:def 3;
  hence thesis by A2,TARSKI:def 4;
end;

theorem
  union (A \/ B) = union A \/ union B
proof
A1: union (A \/ B) c= union A \/ union B
  proof
    let x;
    assume x in union (A \/ B);
    then consider Y such that
A2: x in Y and
A3: Y in A \/ B by TARSKI:def 4;
    Y in A or Y in B by A3,XBOOLE_0:def 3;
    then x in union A or x in union B by A2,TARSKI:def 4;
    hence thesis by XBOOLE_0:def 3;
  end;
  union A c= union (A \/ B) & union B c= union (A \/ B) by Th77,XBOOLE_1:7;
  then union A \/ union B c= union (A \/ B) by XBOOLE_1:8;
  hence thesis by A1,XBOOLE_0:def 10;
end;

theorem Th79:
  union (A /\ B) c= union A /\ union B
proof
  let x;
  assume x in union (A /\ B);
  then consider Y such that
A1: x in Y and
A2: Y in A /\ B by TARSKI:def 4;
  Y in B by A2,XBOOLE_0:def 4;
  then
A3: x in union B by A1,TARSKI:def 4;
  Y in A by A2,XBOOLE_0:def 4;
  then x in union A by A1,TARSKI:def 4;
  hence thesis by A3,XBOOLE_0:def 4;
end;

theorem Th80:
  (for X st X in A holds X misses B) implies union A misses B
proof
  assume
A1: for X st X in A holds X misses B;
  assume union(A) meets B;
  then consider z being set such that
A2: z in union(A) /\ B by XBOOLE_0:4;
  z in union(A) by A2,XBOOLE_0:def 4;
  then consider X such that
A3: z in X and
A4: X in A by TARSKI:def 4;
  z in B by A2,XBOOLE_0:def 4;
  then z in X /\ B by A3,XBOOLE_0:def 4;
  hence contradiction by A1,A4,XBOOLE_0:4;
end;

theorem
  union bool A = A
proof
  now
    let x;
    thus x in union bool A implies x in A
    proof
      assume x in union bool A;
      then consider X such that
A1:   x in X and
A2:   X in bool A by TARSKI:def 4;
      X c= A by A2,Def1;
      hence thesis by A1,TARSKI:def 3;
    end;
    assume x in A;
    then { x } c= A by Lm1;
    then
A3: { x } in bool A by Def1;
    x in { x } by TARSKI:def 1;
    hence x in union bool A by A3,TARSKI:def 4;
  end;
  hence thesis by TARSKI:1;
end;

theorem
  A c= bool union A
proof
  let x;
  assume x in A;
  then x c= union A by Lm15;
  hence thesis by Def1;
end;

theorem
  (for X,Y st X<>Y & X in A \/ B & Y in A \/ B holds X misses Y) implies
  union(A /\ B) = union A /\ union B
proof
  assume
A1: for X,Y st X<>Y & X in A \/ B & Y in A \/ B holds X misses Y;
A2: union A /\ union B c= union (A /\ B)
  proof
    let x;
    assume
A3: x in union A /\ union B;
    then x in union A by XBOOLE_0:def 4;
    then consider X such that
A4: x in X and
A5: X in A by TARSKI:def 4;
    x in union B by A3,XBOOLE_0:def 4;
    then consider Y such that
A6: x in Y and
A7: Y in B by TARSKI:def 4;
    now
B8:   x in X /\ Y by A4,A6,XBOOLE_0:def 4;
      assume
A9:   X<>Y;
      X in A \/ B & Y in A \/ B by A5,A7,XBOOLE_0:def 3;
      hence contradiction by A1,A9,B8,XBOOLE_0:4;
    end;
    then Y in A /\ B by A5,A7,XBOOLE_0:def 4;
    hence thesis by A6,TARSKI:def 4;
  end;
  union (A /\ B) c= union A /\ union B by Th79;
  hence thesis by A2,XBOOLE_0:def 10;
end;





theorem Th84:
  A c= [:X,Y:] & z in A implies ex x,y st x in X & y in Y & z = [ x,y]
proof
  assume A c= [:X,Y:] & z in A;
  then z in [:X,Y:] by TARSKI:def 3;
  hence thesis by Def2;
end;

theorem Th85:
  z in [:X1, Y1:] /\ [:X2, Y2:] implies ex x,y st z = [x,y] & x
  in X1 /\ X2 & y in Y1 /\ Y2
proof
  assume
A1: z in [:X1, Y1:] /\ [:X2, Y2:];
  then z in [:X1, Y1:] by XBOOLE_0:def 4;
  then consider x1,y1 such that
A2: x1 in X1 and
A3: y1 in Y1 and
A4: z=[x1,y1] by Def2;
  z in [:X2, Y2:] by A1,XBOOLE_0:def 4;
  then consider x2,y2 such that
A5: x2 in X2 and
A6: y2 in Y2 and
A7: z=[x2,y2] by Def2;
  y1=y2 by A4,A7,XTUPLE_0:1;
  then
A8: y1 in Y1 /\ Y2 by A3,A6,XBOOLE_0:def 4;
  x1=x2 by A4,A7,XTUPLE_0:1;
  then x1 in X1 /\ X2 by A2,A5,XBOOLE_0:def 4;
  hence thesis by A4,A8;
end;

theorem
  [:X,Y:] c= bool bool (X \/ Y)
proof
  let z;
  assume z in [:X,Y:];
  then consider x,y such that
A1: x in X and
A2: y in Y and
A3: z = [x,y] by Def2;
  z c= bool (X \/ Y)
  proof
    let u be set;
    assume u in z;
    then
A4: u = {x,y} or u = {x} by A3,TARSKI:def 2;
    X c= X \/ Y & {x} c= X by A1,Lm1,XBOOLE_1:7;
    then
A5: {x} c= X \/ Y by XBOOLE_1:1;
    x in X \/ Y & y in X \/ Y by A1,A2,XBOOLE_0:def 3;
    then {x,y} c= X \/ Y by Th32;
    hence thesis by A5,A4,Def1;
  end;
  hence thesis by Def1;
end;

theorem
  [x,y] in [:X,Y:] iff x in X & y in Y by Lm17;

theorem Th88:
  [x,y] in [:X,Y:] implies [y,x] in [:Y,X:]
proof
  [x,y] in [:X,Y:] implies x in X & y in Y by Lm17;
  hence thesis by Lm17;
end;

theorem
  (for x,y holds [x,y] in [:X1,Y1:] iff [x,y] in [:X2,Y2:]) implies [:X1
  ,Y1:]=[:X2,Y2:]
proof
  assume
A1: for x,y holds [x,y] in [:X1,Y1:] iff [x,y] in [:X2,Y2:];
  now
    let z;
    thus z in [:X1,Y1:] implies z in [:X2,Y2:]
    proof
      assume
A2:   z in [:X1,Y1:];
      then ex x,y st x in X1 & y in Y1 & [x,y]=z by Def2;
      hence thesis by A1,A2;
    end;
    assume
A3: z in [:X2,Y2:];
    then ex x,y st x in X2 & y in Y2 & [x,y]=z by Def2;
    hence z in [:X1,Y1:] by A1,A3;
  end;
  hence thesis by TARSKI:1;
end;

Lm18: A c= [:X1,Y1:] & B c= [:X2,Y2:] & (for x,y holds [x,y] in A iff [x,y] in
B) implies A = B
proof
  assume that
A1: A c= [:X1,Y1:] and
A2: B c= [:X2,Y2:] and
A3: for x,y holds [x,y] in A iff [x,y] in B;
  now
    let z;
A4: z in B implies ex x,y st x in X2 & y in Y2 & z=[x,y] by A2,Th84;
    z in A implies ex x,y st x in X1 & y in Y1 & z=[x,y] by A1,Th84;
    hence z in A iff z in B by A3,A4;
  end;
  hence thesis by TARSKI:1;
end;

Lm19: ( for z st z in A ex x,y st z = [x,y]) & (for z st z in B ex x,y st z=[x
,y]) & (for x,y holds [x,y] in A iff [x,y] in B) implies A = B
proof
  assume that
A1: for z st z in A ex x,y st z=[x,y] and
A2: for z st z in B ex x,y st z=[x,y] and
A3: for x,y holds [x,y] in A iff [x,y] in B;
  now
    let z;
A4: z in B implies ex x,y st z=[x, y ] by A2;
    z in A implies ex x,y st z=[x,y] by A1;
    hence z in A iff z in B by A3,A4;
  end;
  hence thesis by TARSKI:1;
end;

theorem Th90:
  [:X,Y:] = {} iff X = {} or Y = {}
proof
  thus [:X,Y:] = {} implies X={} or Y={}
  proof
    assume
A1: [:X,Y:] = {};
    assume X<>{};
    then consider x being set such that
A2: x in X by XBOOLE_0:7;
    assume Y<>{};
    then consider y being set such that
A3: y in Y by XBOOLE_0:7;
    [x,y] in [:X,Y:] by A2,A3,Def2;
    hence contradiction by A1;
  end;
  assume
A4: not thesis;
  then consider z being set such that
A5: z in [:X,Y:] by XBOOLE_0:7;
A6: now
    assume
A7: Y={};
    ex x,y st x in X & y in Y & [x,y]=z by A5,Def2;
    hence contradiction by A7;
  end;
  now
    assume
A8: X={};
    ex x,y st x in X & y in Y & [x,y]=z by A5,Def2;
    hence contradiction by A8;
  end;
  hence contradiction by A4,A6;
end;

theorem
  X <> {} & Y <> {} & [:X,Y:] = [:Y,X:] implies X = Y
proof
  assume X<>{};
  then consider x being set such that
A1: x in X by XBOOLE_0:7;
  assume Y<>{};
  then consider y being set such that
A2: y in Y by XBOOLE_0:7;
  assume
A3: [:X,Y:]=[:Y,X:];
  z in X iff z in Y
  proof
    thus z in X implies z in Y
    proof
      assume z in X;
      then [z,y] in [:Y,X:] by A2,A3,Lm17;
      hence thesis by Lm17;
    end;
    assume z in Y;
    then [z,x] in [:X,Y:] by A1,A3,Lm17;
    hence thesis by Lm17;
  end;
  hence thesis by TARSKI:1;
end;

theorem
  [:X,X:] = [:Y,Y:] implies X = Y
proof
  assume
A1: [:X,X:] = [:Y,Y:];
  x in X iff x in Y
  proof
    x in X iff [x,x] in [:X,X:] by Lm17;
    hence thesis by A1,Lm17;
  end;
  hence thesis by TARSKI:1;
end;

Lm20: z in [:X,Y:] implies ex x,y st [x,y] = z
proof
  assume z in [:X,Y:];
  then ex x,y st x in X & y in Y & [x,y]=z by Def2;
  hence thesis;
end;

theorem
  X c= [:X,X:] implies X = {}
proof
  assume
A1: X c= [:X,X:];
  assume X <> {};
  then consider z being set such that
A2: z in X by XBOOLE_0:7;
  consider Y such that
A3: Y in X \/ union X and
A4: X \/ union X misses Y by XREGULAR:1,A2;
  now
    assume
A5: Y in X;
    then consider x,y such that
A6: Y=[x,y] by Lm20,A1,TARSKI:def 3;
A7: {x} in Y by A6,TARSKI:def 2;
    Y c= union X by A5,Lm15;
    then {x} in union X by A7,TARSKI:def 3;
    then {x} in X \/ union X by XBOOLE_0:def 3;
    hence contradiction by A4,A7,XBOOLE_0:3;
  end;
  then Y in union X by A3,XBOOLE_0:def 3;
  then consider Z such that
A8: Y in Z and
A9: Z in X by TARSKI:def 4;
  Z in [:X,X:] by A1,A9,TARSKI:def 3;
  then consider x,y such that
A10: x in X and
  y in X and
A11: Z=[x,y] by Def2;
  Y={x} or Y={x,y} by A8,A11,TARSKI:def 2;
  then
A12: x in Y by TARSKI:def 1,def 2;
  x in X \/ union X by A10,XBOOLE_0:def 3;
  hence contradiction by A4,A12,XBOOLE_0:3;
end;

theorem
  Z <> {} & ([:X,Z:] c= [:Y,Z:] or [:Z,X:] c= [:Z,Y:]) implies X c= Y
proof
  assume Z<>{};
  then consider z being set such that
A1: z in Z by XBOOLE_0:7;
  assume
A2: [:X,Z:] c= [:Y,Z:] or [:Z,X:] c= [:Z,Y:];
  let x;
  assume x in X;
  then [x,z] in [:X,Z:] & [z,x] in [:Z,X:] by A1,Def2;
  then [x,z] in [:Y,Z:] or [z,x] in [:Z,Y:] by A2,TARSKI:def 3;
  hence thesis by Lm17;
end;

theorem Th95:
  X c= Y implies [:X,Z:] c= [:Y,Z:] & [:Z,X:] c= [:Z,Y:]
proof
  assume
A1: X c= Y;
  thus [:X,Z:] c= [:Y,Z:]
  proof
    let z;
    assume z in [:X,Z:];
    then consider x,y such that
A2: x in X and
A3: y in Z & z=[x,y] by Def2;
    x in Y by A1,A2,TARSKI:def 3;
    hence thesis by A3,Def2;
  end;
  let z;
  assume z in [:Z,X:];
  then consider y,x such that
A4: y in Z and
A5: x in X and
A6: z=[y,x] by Def2;
  x in Y by A1,A5,TARSKI:def 3;
  hence thesis by A4,A6,Def2;
end;

theorem Th96:
  X1 c= Y1 & X2 c= Y2 implies [:X1,X2:] c= [:Y1,Y2:]
proof
  assume X1 c= Y1 & X2 c= Y2;
  then [:X1,X2:] c= [:Y1,X2:] & [:Y1,X2:] c= [:Y1,Y2:] by Th95;
  hence thesis by XBOOLE_1:1;
end;

theorem Th97:
  [:X \/ Y, Z:] = [:X, Z:] \/ [:Y, Z:] & [:Z, X \/ Y:] = [:Z, X:] \/ [:Z, Y:]
proof
A1: for z st z in [:X \/ Y, Z:] ex x,y st z=[x,y] by Lm20;
A2: for x,y holds [x,y] in [:X \/ Y, Z:] iff [x,y] in [:X, Z:] \/ [:Y, Z:]
  proof
    let x,y;
    thus [x,y] in [:X \/ Y, Z:] implies [x,y] in [:X, Z:] \/ [:Y, Z:]
    proof
      assume
A3:   [x,y] in [:X \/ Y, Z:];
      then x in X \/ Y by Lm17;
      then
A4:   x in X or x in Y by XBOOLE_0:def 3;
      y in Z by A3,Lm17;
      then [x,y] in [:X,Z:] or [x,y] in [:Y,Z:] by A4,Lm17;
      hence thesis by XBOOLE_0:def 3;
    end;
    assume [x,y] in [:X, Z:] \/ [:Y, Z:];
    then [x,y] in [:X, Z:] or [x,y] in [:Y, Z:] by XBOOLE_0:def 3;
    then
A5: x in X & y in Z or x in Y & y in Z by Lm17;
    then x in X \/ Y by XBOOLE_0:def 3;
    hence thesis by A5,Lm17;
  end;
A6: z in [:X1,X2:] \/ [:Y1,Y2:] implies ex x,y st z=[x,y]
  proof
    assume z in [:X1,X2:] \/ [:Y1,Y2:];
    then z in [:X1,X2:] or z in [:Y1,Y2:] by XBOOLE_0:def 3;
    hence thesis by Lm20;
  end;
  then for z st z in [:X,Z:] \/ [:Y,Z:] ex x,y st z=[x,y];
  hence
A7: [:X \/ Y, Z:] = [:X, Z:] \/ [:Y, Z:] by A1,A2,Lm19;
A8: for y,x holds [y,x] in [:Z, X \/ Y:] iff [y,x] in [:Z, X:] \/ [:Z, Y:]
  proof
    let y,x;
A9: [x,y]in[:X, Z:] or [x,y]in[:Y,Z:] iff [y,x]in[:Z,X:] or [y,x]in[:Z,Y
    :] by Th88;
    [y,x] in [:Z, X \/ Y:] iff [x,y] in [:X \/ Y, Z:] by Th88;
    hence thesis by A7,A9,XBOOLE_0:def 3;
  end;
A10: for z st z in [:Z,X \/ Y:] ex x,y st z=[x,y] by Lm20;
  for z st z in [:Z,X:] \/ [:Z,Y:] ex x,y st z=[x,y] by A6;
  hence thesis by A10,A8,Lm19;
end;

theorem
  [:X1 \/ X2, Y1 \/ Y2:] = [:X1,Y1:] \/ [:X1,Y2:] \/ [:X2,Y1:] \/ [:X2, Y2:]
proof
  thus [:X1 \/ X2, Y1 \/ Y2:] = [:X1, Y1 \/ Y2:] \/ [:X2, Y1 \/ Y2:] by Th97
    .= [:X1,Y1:] \/ [:X1,Y2:] \/ [:X2,Y1 \/ Y2:] by Th97
    .= [:X1,Y1:] \/ [:X1,Y2:] \/ ([:X2,Y1:] \/ [:X2,Y2:]) by Th97
    .= [:X1,Y1:] \/ [:X1,Y2:] \/ [:X2,Y1:] \/ [:X2,Y2:] by XBOOLE_1:4;
end;

theorem
  [:X /\ Y, Z:] = [:X, Z:] /\ [:Y, Z:] & [:Z, X /\ Y:] = [:Z, X:] /\ [:Z , Y :]
proof
A1: for x,y holds [x,y] in [:X /\ Y, Z:] iff [x,y] in [:X, Z:] /\ [:Y, Z :]
  proof
    let x,y;
    thus [x,y] in [:X /\ Y, Z:] implies [x,y] in [:X, Z:] /\ [:Y, Z:]
    proof
      assume
A2:   [x,y] in [:X /\ Y, Z:];
      then
A3:   x in X /\ Y by Lm17;
A4:   y in Z by A2,Lm17;
      x in Y by A3,XBOOLE_0:def 4;
      then
A5:   [x,y] in [:Y,Z:] by A4,Lm17;
      x in X by A3,XBOOLE_0:def 4;
      then [x,y] in [:X,Z:] by A4,Lm17;
      hence thesis by A5,XBOOLE_0:def 4;
    end;
    assume
A6: [x,y] in [:X, Z:] /\ [:Y, Z:];
    then [x,y] in [:Y, Z:] by XBOOLE_0:def 4;
    then
A7: x in Y by Lm17;
A8: [x,y] in [:X, Z:] by A6,XBOOLE_0:def 4;
    then x in X by Lm17;
    then
A9: x in X /\ Y by A7,XBOOLE_0:def 4;
    y in Z by A8,Lm17;
    hence thesis by A9,Lm17;
  end;
  [:X, Z:] /\ [:Y, Z:] c= [:X, Z:] by XBOOLE_1:17;
  hence
A10: [:X /\ Y, Z:] = [:X, Z:] /\ [:Y, Z:] by A1,Lm18;
A11: for y,x holds [y,x] in [:Z, X /\ Y:] iff [y,x] in [:Z, X:] /\ [:Z, Y:]
  proof
    let y,x;
A12: [x,y]in[:X, Z:] & [x,y]in[:Y,Z:] iff [y,x]in[:Z,X:] & [y,x]in[:Z,Y:]
    by Th88;
    [y,x] in [:Z, X /\ Y:] iff [x,y] in [:X /\ Y, Z:] by Th88;
    hence thesis by A10,A12,XBOOLE_0:def 4;
  end;
  [:Z, X:] /\ [:Z, Y:] c= [:Z, X:] by XBOOLE_1:17;
  hence thesis by A11,Lm18;
end;

theorem Th100:
  [:X1 /\ X2, Y1 /\ Y2:] = [:X1,Y1:] /\ [:X2, Y2:]
proof
  Y1 /\ Y2 c= Y2 & X1 /\ X2 c= X2 by XBOOLE_1:17;
  then
A1: [:X1 /\ X2, Y1 /\ Y2:] c= [:X2, Y2:] by Th96;
A2: [:X1, Y1:] /\ [:X2, Y2:] c= [:X1 /\ X2, Y1 /\ Y2:]
  proof
    let z;
    assume z in [:X1, Y1:] /\ [:X2, Y2:];
    then ex x,y st z=[x,y] & x in X1 /\ X2 & y in Y1 /\ Y2 by Th85;
    hence thesis by Def2;
  end;
  Y1 /\ Y2 c= Y1 & X1 /\ X2 c= X1 by XBOOLE_1:17;
  then [:X1 /\ X2, Y1 /\ Y2:] c= [:X1, Y1:] by Th96;
  then [:X1 /\ X2, Y1 /\ Y2:] c= [:X1, Y1:] /\ [:X2, Y2:] by A1,XBOOLE_1:19;
  hence thesis by A2,XBOOLE_0:def 10;
end;

theorem
  A c= X & B c= Y implies [:A,Y:] /\ [:X,B:] = [:A,B:]
proof
  assume that
A1: A c= X and
A2: B c= Y;
A3: [:A,Y:] /\ [:X,B:] = [:A /\ X, Y /\ B:] by Th100;
  A /\ X = A by A1,XBOOLE_1:28;
  hence thesis by A2,A3,XBOOLE_1:28;
end;

theorem Th102:
  [:X \ Y, Z:] = [:X, Z:] \ [:Y, Z:] & [:Z, X \ Y:] = [:Z, X:] \ [:Z, Y:]
proof
A1: for x,y holds [x,y] in [:X \ Y, Z:] iff [x,y] in [:X, Z:] \ [:Y, Z:]
  proof
    let x,y;
    thus [x,y] in [:X \ Y, Z:] implies [x,y] in [:X, Z:] \ [:Y, Z:]
    proof
      assume
A2:   [x,y] in [:X \ Y, Z:];
      then
A3:   x in X \ Y by Lm17;
      then
A4:   x in X by XBOOLE_0:def 5;
      not x in Y by A3,XBOOLE_0:def 5;
      then
A5:   not [x,y] in [:Y,Z:] by Lm17;
      y in Z by A2,Lm17;
      then [x,y] in [:X,Z:] by A4,Lm17;
      hence thesis by A5,XBOOLE_0:def 5;
    end;
    assume
A6: [x,y] in [:X, Z:] \ [:Y, Z:];
    then
A7: [x,y] in [:X, Z:] by XBOOLE_0:def 5;
    then
A8: y in Z by Lm17;
    not [x,y] in [:Y, Z:] by A6,XBOOLE_0:def 5;
    then
A9: not (x in Y & y in Z) by Lm17;
    x in X by A7,Lm17;
    then x in X \ Y by A7,A9,Lm17,XBOOLE_0:def 5;
    hence thesis by A8,Lm17;
  end;
  [:X, Z:] \ [:Y, Z:] c= [:X, Z:] by XBOOLE_1:36;
  hence
A10: [:X \ Y, Z:] = [:X, Z:] \ [:Y, Z:] by A1,Lm18;
A11: for y,x holds [y,x] in [:Z, X \ Y:] iff [y,x] in [:Z, X:] \ [:Z, Y:]
  proof
    let y,x;
A12: [x,y]in[:X, Z:]& not[x,y]in[:Y,Z:] iff [y,x]in[:Z,X:]& not[y,x]in[:Z,
    Y :] by Th88;
    [y,x] in [:Z, X \ Y:] iff [x,y] in [:X \ Y, Z:] by Th88;
    hence thesis by A10,A12,XBOOLE_0:def 5;
  end;
  [:Z, X:] \ [:Z, Y:] c= [:Z, X:] by XBOOLE_1:36;
  hence thesis by A11,Lm18;
end;

theorem
  [:X1,X2:] \ [:Y1,Y2:] = [:X1\Y1,X2:] \/ [:X1,X2\Y2:]
proof
A1: [:Y1,X2:] /\ [:X1,Y2:] = [:Y1 /\ X1, X2 /\ Y2:] by Th100;
  Y1 /\ X1 c= Y1 & X2 /\ Y2 c= Y2 by XBOOLE_1:17;
  then
A2: [:Y1 /\ X1, X2 /\ Y2:] c= [:Y1,Y2:] by Th96;
A3: [:X1\Y1,X2:] \/ [:X1,X2\Y2:] c= [:X1,X2:] \ [:Y1, Y2:]
  proof
    let z;
A4: now
      assume z in [:X1\Y1,X2:];
      then consider x,y such that
A5:   x in X1\Y1 and
A6:   y in X2 and
A7:   z=[x,y] by Def2;
      not x in Y1 by A5,XBOOLE_0:def 5;
      then
A8:   not z in [:Y1,Y2:] by A7,Lm17;
      x in X1 by A5,XBOOLE_0:def 5;
      then z in [:X1,X2:] by A6,A7,Lm17;
      hence thesis by A8,XBOOLE_0:def 5;
    end;
A9: now
      assume z in [:X1,X2\Y2:];
      then consider x,y such that
A10:  x in X1 and
A11:  y in X2\Y2 and
A12:  z=[x,y] by Def2;
      not y in Y2 by A11,XBOOLE_0:def 5;
      then
A13:  not z in [:Y1,Y2:] by A12,Lm17;
      y in X2 by A11,XBOOLE_0:def 5;
      then z in [:X1,X2:] by A10,A12,Lm17;
      hence thesis by A13,XBOOLE_0:def 5;
    end;
    assume z in [:X1\Y1,X2:] \/ [:X1,X2\Y2:];
    hence thesis by A4,A9,XBOOLE_0:def 3;
  end;
  [:X1\Y1,X2:] = [:X1,X2:] \ [:Y1,X2:] & [:X1,X2\Y2:] = [:X1,X2:] \ [:X1,
  Y2 :] by Th102;
  then [:X1\Y1,X2:] \/ [:X1,X2\Y2:] = [:X1,X2:] \ [:Y1 /\ X1, X2 /\ Y2:] by A1,
XBOOLE_1:54;
  then [:X1,X2:] \ [:Y1,Y2:] c= [:X1\Y1,X2:] \/ [:X1,X2\Y2:] by A2,XBOOLE_1:34;
  hence thesis by A3,XBOOLE_0:def 10;
end;

theorem Th104:
  X1 misses X2 or Y1 misses Y2 implies [:X1,Y1:] misses [:X2,Y2:]
proof
  assume
A1: X1 misses X2 or Y1 misses Y2;
  assume not thesis;
  then consider z being set such that
A2: z in [:X1,Y1:] /\ [:X2,Y2:] by XBOOLE_0:4;
  ex x,y st z=[x,y] & x in X1 /\ X2 & y in Y1 /\ Y2 by A2,Th85;
  hence contradiction by A1,XBOOLE_0:4;
end;

theorem
  [x,y] in [:{z},Y:] iff x = z & y in Y
proof
A1: x in {z} iff x=z by TARSKI:def 1;
  hence [x,y] in [:{z},Y:] implies x=z & y in Y by Lm17;
  thus thesis by A1,Lm17;
end;

theorem
  [x,y] in [:X,{z}:] iff x in X & y = z
proof
A1: y in {z} iff y=z by TARSKI:def 1;
  hence [x,y] in [:X,{z}:] implies x in X & y=z by Lm17;
  thus thesis by A1,Lm17;
end;

theorem
  X <> {} implies [:{x},X:] <> {} & [:X,{x}:] <> {} by Th90;

theorem
  x <> y implies [:{x},X:] misses [:{y},Y:] & [:X,{x}:] misses [:Y,{y}:]
proof
  assume x<>y;
  then {x} misses {y} by Th11;
  hence thesis by Th104;
end;

theorem
  [:{x,y},X:] = [:{x},X:] \/ [:{y},X:] & [:X,{x,y}:] = [:X,{x}:] \/ [:X, {y} :]
proof
  {x,y}={x} \/ {y} by ENUMSET1:1;
  hence thesis by Th97;
end;

theorem Th110:
  X1 <> {} & Y1 <> {} & [:X1,Y1:] = [:X2,Y2:] implies X1 = X2 & Y1 = Y2
proof
  assume
A1: X1<>{};
  then consider x being set such that
A2: x in X1 by XBOOLE_0:7;
  assume
A3: Y1<>{};
  then consider y being set such that
A4: y in Y1 by XBOOLE_0:7;
  assume
A5: [:X1,Y1:] = [:X2,Y2:];
  then
A6: [:X2,Y2:] <> {} by A1,A3,Th90;
  then
A7: Y2 <> {} by Th90;
  z in X1 iff z in X2
  proof
    consider y2 being set such that
A8: y2 in Y2 by A7,XBOOLE_0:7;
    thus z in X1 implies z in X2
    proof
      assume z in X1;
      then [z,y] in [:X2,Y2:] by A4,A5,Lm17;
      hence thesis by Lm17;
    end;
    assume z in X2;
    then [z,y2] in [:X2,Y2:] by A8,Lm17;
    hence thesis by A5,Lm17;
  end;
  hence X1 = X2 by TARSKI:1;
A9: X2 <> {} by A6,Th90;
  z in Y1 iff z in Y2
  proof
    consider x2 being set such that
A10: x2 in X2 by A9,XBOOLE_0:7;
    thus z in Y1 implies z in Y2
    proof
      assume z in Y1;
      then [x,z] in [:X2,Y2:] by A2,A5,Lm17;
      hence thesis by Lm17;
    end;
    assume z in Y2;
    then [x2,z] in [:X2,Y2:] by A10,Lm17;
    hence thesis by A5,Lm17;
  end;
  hence thesis by TARSKI:1;
end;

theorem
  X c= [:X,Y:] or X c= [:Y,X:] implies X = {}
proof
  assume
A1: X c= [:X,Y:] or X c= [:Y,X:];
  assume
A2: X <> {};
A3: now
    defpred P[set] means ex Y st $1=Y & ex z st z in Y & z in X;
    consider Z such that
A4: for y holds y in Z iff y in union X & P[y] from XBOOLE_0:sch 1;
    consider Y2 such that
A5: Y2 in X \/ Z and
A6: X \/ Z misses Y2 by XREGULAR:1,A2;
    now
      assume
A7:   not ex Y2 st Y2 in X & for Y1 st Y1 in Y2 holds for z holds not
      z in Y1 or not z in X;
      now
        assume
A8:     Y2 in X;
        then consider Y1 such that
A9:     Y1 in Y2 and
A10:    ex z st z in Y1 & z in X by A7;
        Y1 in union X by A8,A9,TARSKI:def 4;
        then Y1 in Z by A4,A10;
        then Y1 in X \/ Z by XBOOLE_0:def 3;
        hence contradiction by A6,A9,XBOOLE_0:3;
      end;
      then Y2 in Z by A5,XBOOLE_0:def 3;
      then ex X2 st Y2=X2 & ex z st z in X2 & z in X by A4;
      then consider z such that
A11:  z in Y2 and
A12:  z in X;
      z in X \/ Z by A12,XBOOLE_0:def 3;
      hence contradiction by A6,A11,XBOOLE_0:3;
    end;
    then consider Y1 such that
A13: Y1 in X and
A14: not ex Y2 st Y2 in Y1 & ex z st z in Y2 & z in X;
A15: now
      given y,x such that
A16:  x in X and
A17:  Y1 = [y,x];
A18:  x in {y,x} by TARSKI:def 2;
      {y,x} in Y1 by A17,TARSKI:def 2;
      hence contradiction by A14,A16,A18;
    end;
    assume X c= [:Y,X:];
    then Y1 in [:Y,X:] by A13,TARSKI:def 3;
    then ex y,x st y in Y & x in X & Y1 = [y,x] by Def2;
    hence contradiction by A15;
  end;
  now
    consider z being set such that
A19: z in X by A2,XBOOLE_0:7;
    consider Y1 such that
A20: Y1 in X \/ union X and
A21: X \/ union X misses Y1 by XREGULAR:1,A19;
    assume
A22: X c= [:X,Y:];
    now
      assume
A23:  Y1 in X;
      then consider x,y such that
A24:  Y1 = [x,y] by Lm20,A22,TARSKI:def 3;
A25:  {x} in Y1 by A24,TARSKI:def 2;
      Y1 c= union X by A23,Lm15;
      then {x} in union X by A25,TARSKI:def 3;
      then {x} in X \/ union X by XBOOLE_0:def 3;
      hence contradiction by A21,A25,XBOOLE_0:3;
    end;
    then Y1 in union X by A20,XBOOLE_0:def 3;
    then consider Y2 such that
A26: Y1 in Y2 and
A27: Y2 in X by TARSKI:def 4;
    Y2 in [:X,Y:] by A22,A27,TARSKI:def 3;
    then consider x,y such that
A28: x in X and
    y in Y and
A29: Y2=[x,y] by Def2;
    Y1={x} or Y1={x,y} by A26,A29,TARSKI:def 2;
    then
A30: x in Y1 by TARSKI:def 1,def 2;
    x in X \/ union X by A28,XBOOLE_0:def 3;
    hence contradiction by A21,A30,XBOOLE_0:3;
  end;
  hence thesis by A1,A3;
end;

theorem
  ex M st N in M & (for X,Y holds X in M & Y c= X implies Y in M) & (for
  X holds X in M implies bool X in M) & for X holds X c= M implies X,M
  are_equipotent or X in M
proof
  consider M such that
A1: N in M and
A2: for X,Y holds X in M & Y c= X implies Y in M and
A3: for X st X in M ex Z st Z in M & for Y st Y c= X holds Y in Z and
A4: for X holds X c= M implies X,M are_equipotent or X in M by TARSKI:3;
  take M;
  thus N in M by A1;
  thus for X,Y holds X in M & Y c= X implies Y in M by A2;
  thus for X holds X in M implies bool X in M
  proof
    let X;
    assume X in M;
    then consider Z such that
A5: Z in M and
A6: for Y st Y c= X holds Y in Z by A3;
    now
      let Y;
      assume Y in bool X;
      then Y c= X by Def1;
      hence Y in Z by A6;
    end;
    hence thesis by A2,A5,TARSKI:def 3;
  end;
  thus thesis by A4;
end;

reserve e,X,X1,X2,Y1,Y2 for set;

theorem
  e in [:X1,Y1:] & e in [:X2,Y2:] implies e in [:X1 /\ X2, Y1 /\ Y2:]
proof
  assume e in [:X1,Y1:] & e in [:X2,Y2:];
  then e in [:X1,Y1:] /\ [:X2,Y2:] by XBOOLE_0:def 4;
  hence thesis by Th100;
end;

begin 


theorem Th114:
  [:X1,X2:] c= [:Y1,Y2:] & [:X1,X2:] <> {} implies X1 c= Y1 & X2 c= Y2
proof
  assume that
A1: [:X1,X2:] c= [:Y1,Y2:] and
A2: [:X1,X2:] <> {};
A3: [:X1,X2:] = [:X1,X2:] /\ [:Y1,Y2:] by A1,XBOOLE_1:28
    .= [:X1/\Y1,X2/\Y2:] by Th100;
  X1 <> {} & X2 <> {} by A2,Th90;
  then X1 = X1/\Y1 & X2 = X2/\Y2 by A3,Th110;
  hence thesis by XBOOLE_1:17;
end;



theorem
  for A being non empty set, B,C,D being set st [:A,B:] c= [:C,D:] or [:
  B,A:] c= [:D,C:] holds B c= D
proof
  let A be non empty set, B,C,D be set such that
A1: [:A,B:] c= [:C,D:] or [:B,A:] c= [:D,C:];
  per cases;
  suppose
    B = {};
    hence thesis by XBOOLE_1:2;
  end;
  suppose
    B <> {};
    then [:A,B:] <> {} & [:B,A:] <> {} by Th90;
    hence thesis by A1,Th114;
  end;
end;

theorem
  x in X implies (X\{x})\/{x}=X
proof
  assume
A1: x in X;
  thus (X\{x})\/{x}c=X
  proof
    let y;
    assume y in (X\{x})\/{x};
    then y in X\{x} or y in {x} by XBOOLE_0:def 3;
    hence thesis by A1,Th56,TARSKI:def 1;
  end;
  thus X c= (X\{x})\/{x}
  proof
    let y;
    assume y in X;
    then y in {x} or y in X\{x} by XBOOLE_0:def 5;
    hence thesis by XBOOLE_0:def 3;
  end;
end;

theorem
  not x in X implies (X\/{x})\{x}=X
proof
A1: (X \/ {x}) \ {x} = X \ {x} by XBOOLE_1:40;
  assume not x in X;
  hence thesis by A1,Th57;
end;



theorem
  for x,y,z,Z being set holds Z c= {x,y,z} iff Z = {} or Z = {x} or Z =
  {y} or Z = {z} or Z = {x,y} or Z = {y,z} or Z = {x,z} or Z = {x,y,z}
proof
  let x,y,z,Z be set;
  hereby
    assume that
A1: Z c= {x,y,z} and
A2: Z <> {} and
A3: Z <> {x} and
A4: Z <> {y} and
A5: Z <> {z} and
A6: Z <> {x,y} and
A7: Z <> {y,z} and
A8: Z <> {x,z};
    {x,y,z} c= Z
    proof
      let a be set;
A9:   now
        {x,y,z} \ {x} = ({x} \/ {y,z}) \ {x} by ENUMSET1:2
          .= {y,z} \ {x} by XBOOLE_1:40;
        then
A10:    {x,y,z} \ {x} c= {y,z} by XBOOLE_1:36;
        assume not x in Z;
        then Z c= {x,y,z} \ {x} by A1,Lm2;
        hence contradiction by A2,A4,A5,A7,Lm14,A10,XBOOLE_1:1;
      end;
A11:  now
        {x,y,z} \ {y} = ({x,y} \/ {z}) \ {y} by ENUMSET1:3
          .= ({x} \/ {y} \/ {z}) \ {y} by ENUMSET1:1
          .= ({x} \/ {z} \/ {y}) \ {y} by XBOOLE_1:4
          .= ({x,z} \/ {y}) \ {y} by ENUMSET1:1
          .= {x,z} \ {y} by XBOOLE_1:40;
        then
A12:    {x,y,z} \ {y} c= {x,z} by XBOOLE_1:36;
        assume not y in Z;
        then Z c= {x,y,z} \ {y} by A1,Lm2;
        hence contradiction by A2,A3,A5,A8,Lm14,A12,XBOOLE_1:1;
      end;
A13:  now
        {x,y,z} \ {z} = ({x,y} \/ {z}) \ {z} by ENUMSET1:3
          .= {x,y} \ {z} by XBOOLE_1:40;
        then
A14:    {x,y,z} \ {z} c= {x,y} by XBOOLE_1:36;
        assume not z in Z;
        then Z c= {x,y,z} \ {z} by A1,Lm2;
        hence contradiction by A2,A3,A4,A6,Lm14,A14,XBOOLE_1:1;
      end;
      assume a in {x,y,z};
      hence thesis by A9,A11,A13,ENUMSET1:def 1;
    end;
    hence Z = {x,y,z} by A1,XBOOLE_0:def 10;
  end;
  assume
A15: Z = {} or Z = {x} or Z = {y} or Z = {z} or Z = {x,y} or Z = {y,z}
  or Z = {x,z} or Z = {x,y,z};
  per cases by A15;
  suppose
    Z = {};
    hence thesis by XBOOLE_1:2;
  end;
  suppose
    Z = {x};
    then Z c= {x} \/ {y,z} by XBOOLE_1:7;
    hence thesis by ENUMSET1:2;
  end;
  suppose
A16: Z = {y};
    {x,y} c= {x,y} \/ {z} by XBOOLE_1:7;
    then
A17: {x,y} c= {x,y,z} by ENUMSET1:3;
    Z c= {x,y} by A16,Th7;
    hence thesis by A17,XBOOLE_1:1;
  end;
  suppose
    Z = {z};
    then Z c= {x,y} \/ {z} by XBOOLE_1:7;
    hence thesis by ENUMSET1:3;
  end;
  suppose
    Z = {x,y};
    then Z c= {x,y} \/ {z} by XBOOLE_1:7;
    hence thesis by ENUMSET1:3;
  end;
  suppose
    Z = {y,z};
    then Z c= {x} \/ {y,z} by XBOOLE_1:7;
    hence thesis by ENUMSET1:2;
  end;
  suppose
A18: Z = {x,z};
A19: {x,z} \/ {y} = {x} \/ {z} \/ {y} by ENUMSET1:1
      .= {x} \/ ({y} \/ {z}) by XBOOLE_1:4
      .= {x} \/ {y,z} by ENUMSET1:1;
    Z c= {x,z} \/ {y} by A18,XBOOLE_1:7;
    hence thesis by A19,ENUMSET1:2;
  end;
  suppose
    Z = {x,y,z};
    hence thesis;
  end;
end;



theorem
  N c= [:X1,Y1:] & M c= [:X2,Y2:] implies N \/ M c= [:X1 \/ X2,Y1 \/ Y2 :]
proof
  assume N c= [:X1,Y1:] & M c= [:X2,Y2:];
  then
A1: N \/ M c= [:X1,Y1:] \/ [:X2,Y2:] by XBOOLE_1:13;
  X1 c= X1 \/ X2 & Y1 c= Y1 \/ Y2 by XBOOLE_1:7;
  then
A2: [:X1,Y1:] c= [:X1 \/ X2,Y1 \/ Y2:] by Th96;
  Y2 c= Y1 \/ Y2 & X2 c= X1 \/ X2 by XBOOLE_1:7;
  then [:X2,Y2:] c= [:X1 \/ X2,Y1 \/ Y2:] by Th96;
  then [:X1,Y1:] \/ [:X2,Y2:] c= [:X1 \/ X2,Y1 \/ Y2:] by A2,XBOOLE_1:8;
  hence thesis by A1,XBOOLE_1:1;
end;

Lm21: not x in X & not y in X implies {x,y} misses X
proof
  assume
A1: ( not x in X)& not y in X;
  thus {x,y} /\ X c= {}
  proof
    let z;
    assume z in {x,y} /\ X;
    then z in {x,y} & z in X by XBOOLE_0:def 4;
    hence thesis by A1,TARSKI:def 2;
  end;
  thus thesis by XBOOLE_1:2;
end;

theorem Th120:
  not x in X & not y in X implies X = X \ {x,y}
proof
  X \ {x,y} = X iff X misses {x,y} by XBOOLE_1:83;
  hence thesis by Lm21;
end;

theorem
  not x in X & not y in X implies X = X \/ {x,y} \ {x,y}
proof
A1: (X \/ {x,y}) \ {x,y} = X \ {x,y} by XBOOLE_1:40;
  assume ( not x in X)& not y in X;
  hence thesis by A1,Th120;
end;



definition
  let x1, x2, x3 be set;
  pred x1, x2, x3 are_mutually_different means
  x1 <> x2 & x1 <> x3 & x2 <> x3;
end;

definition
  let x1, x2, x3, x4 be set;
  pred x1, x2, x3, x4 are_mutually_different means
  x1 <> x2 & x1 <> x3 & x1 <> x4 & x2 <> x3 & x2 <> x4 & x3 <> x4;
end;



definition
  let x1, x2, x3, x4, x5 be set;
  pred x1, x2, x3, x4, x5 are_mutually_different means
  x1 <> x2 & x1 <> x3 &
x1 <> x4 & x1 <> x5 & x2 <> x3 & x2 <> x4 & x2 <> x5 & x3 <> x4 & x3 <> x5 & x4
  <> x5;
end;



definition
  let x1, x2, x3, x4, x5, x6 be set;
  pred x1, x2, x3, x4, x5, x6 are_mutually_different means
  x1 <> x2 & x1 <> x3
& x1 <> x4 & x1 <> x5 & x1 <> x6 & x2 <> x3 & x2 <> x4 & x2 <> x5 & x2 <> x6 &
  x3 <> x4 & x3 <> x5 & x3 <> x6 & x4 <> x5 & x4 <> x6 & x5 <> x6;
end;

definition
  let x1, x2, x3, x4, x5, x6, x7 be set;
  pred x1, x2, x3, x4, x5, x6, x7 are_mutually_different means
  x1 <> x2 & x1
<> x3 & x1 <> x4 & x1 <> x5 & x1 <> x6 & x1 <> x7 & x2 <> x3 & x2 <> x4 & x2 <>
x5 & x2 <> x6 & x2 <> x7 & x3 <> x4 & x3 <> x5 & x3 <> x6 & x3 <> x7 & x4 <> x5
  & x4 <> x6 & x4 <> x7 & x5 <> x6 & x5 <> x7 & x6 <> x7;
end;



theorem
  [:{x1,x2},{y1,y2}:] = {[x1,y1],[x1,y2],[x2,y1],[x2,y2]}
proof
  thus [:{x1,x2},{y1,y2}:] = [:{x1}\/{x2},{y1,y2}:] by ENUMSET1:1
    .= [:{x1},{y1,y2}:] \/ [:{x2},{y1,y2}:] by Th97
    .= {[x1,y1],[x1,y2]} \/ [:{x2},{y1,y2}:] by Th30
    .= {[x1,y1],[x1,y2]} \/ {[x2,y1],[x2,y2]} by Th30
    .= {[x1,y1],[x1,y2],[x2,y1],[x2,y2]} by ENUMSET1:5;
end;



theorem
  x <> y implies A \/ {x} \ {y} = A \ {y} \/ {x} by Th11,XBOOLE_1:87;



definition
  let X;
  attr X is trivial means
:Def10: x in X & y in X implies x = y;
end;

registration
  cluster empty -> trivial for set;
  coherence
  proof
    let X;
    assume
A1: X is empty;
    let x,y;
    thus thesis by A1;
  end;
end;

registration
  cluster non trivial -> non empty for set;
  coherence;
end;

registration
  let x be set;
  cluster {x} -> trivial;
  coherence
  proof
    let y,z;
    assume that
A1: y in {x} and
A2: z in {x};
    y =x by A1,TARSKI:def 1;
    hence thesis by A2,TARSKI:def 1;
  end;
end;

registration
 cluster trivial non empty for set;
 existence
  proof
   take {{}};
   thus thesis;
  end;
end;



theorem
  for A,B,C,p being set st A c= B & B /\ C = {p} & p in A holds A /\ C = {p}
proof
  let A,B,C,p be set such that
A1: A c= B;
  assume
A2: B /\ C = {p};
  p in B /\ C by A2,TARSKI:def 1;
  then
A4: p in C by XBOOLE_0:def 4;
  assume p in A;
  then p in A /\ C by A4,XBOOLE_0:def 4;
  hence thesis by A2,Lm3,A1,XBOOLE_1:26;
end;



theorem
  for A,B,C,p being set st A /\ B c= {p} & p in C & C misses B holds A
  \/ C misses B
proof
  let A,B,C,p be set such that
A1: A /\ B c= {p} and
A2: p in C and
A3: C misses B;
  {p} c= C by A2,Lm1;
  then A /\ B c= C by A1,XBOOLE_1:1;
  then A /\ B /\ B c= C /\ B by XBOOLE_1:27;
  then
A4: A /\ (B /\ B) c= C /\ B by XBOOLE_1:16;
  (A \/ C) /\ B = A /\ B \/ C /\ B by XBOOLE_1:23
    .= C /\ B by A4,XBOOLE_1:12
    .= {} by A3,XBOOLE_0:def 7;
  hence thesis by XBOOLE_0:def 7;
end;

theorem
  for A,B being set st for x,y being set st x in A & y in B holds x
  misses y holds union A misses union B
proof
  let A,B be set such that
A1: for x,y being set st x in A & y in B holds x misses y;
  for y being set st y in B holds union A misses y
  proof
    let y be set;
    assume y in B;
    then for x being set st x in A holds x misses y by A1;
    hence thesis by Th80;
  end;
  hence thesis by Th80;
end;



registration
  let X be set, Y be empty set;
  cluster [:X, Y:] -> empty;
  coherence by Th90;
end;

registration
  let X be empty set, Y be set;
  cluster [:X, Y:] -> empty;
  coherence by Th90;
end;



theorem
 not A in [:A,B:]
  proof
   assume A in [:A,B:];
    then consider x,y such that
A1:   x in A & y in B & A = [x,y] by Th84;
    x = {x} or x = {x,y} by A1,TARSKI:def 2;
    then x in x by TARSKI:def 1,def 2;
   hence contradiction;
  end;

theorem
 [x,{x}] in [:{x},[x,{x}]:]
  proof
    [:{x},[x,{x}]:] = { [x,{x}], [x,{x,{x}}] } by Th30;
   hence thesis by TARSKI:def 2;
  end;

theorem
 B in [:A,B:] implies ex x st x in A & B = [x,{x}]
  proof
   assume B in [:A,B:];
    then consider x,y such that
A1:   x in A & y in B & B = [x,y] by Th84;
   take x;
   thus x in A by A1;
   per cases by A1,TARSKI:def 2;
   suppose y = {x};
    hence thesis by A1;
   end;
   suppose y = {x,y};
     then y in y by TARSKI:def 2;
    hence thesis;
   end;
  end;

theorem
 B c= A & A is trivial implies B is trivial
proof assume that
A1: B c= A and
A2: A is trivial;
  let x,y;
 assume x in B & y in B;
  then x in A & y in A by A1,TARSKI:def 3;
 hence thesis by A2,Def10;
end;

registration
  cluster non trivial for set;
  existence
  proof
   take {{},{{}}}, {},{{}};
   thus {} in {{},{{}}} & {{}} in  {{},{{}}} by TARSKI:def 2;
   thus {} <> {{}};
  end;
end;

theorem Th131:
 X is non empty trivial implies ex x st X = {x}
 proof
  assume X is non empty;
   then consider x such that
A1:  x in X by XBOOLE_0:def 1;
  assume
A2:  X is trivial;
  take x;
  for y holds y in X iff x = y by A2,A1,Def10;
  hence thesis by TARSKI:def 1;
 end;

theorem
  for x being set, X being trivial set st x in X holds X = {x}
proof
  let x be set, X be trivial set;
  assume
A1: x in X;
  then ex x being set st X = {x} by Th131;
  hence thesis by A1,TARSKI:def 1;
end;



theorem
  for a,b,c,X being set st a in X & b in X & c in X holds {a,b,c} c= X
proof
  let a,b,c,X be set;
  assume a in X & b in X & c in X;
  then {a,b} c= X & {c} c= X by Lm1,Th32;
  then {a,b} \/ {c} c= X by XBOOLE_1:8;
  hence thesis by ENUMSET1:3;
end;



theorem
 [x,y] in X implies x in union union X & y in union union X
proof
  assume
A1: [x,y] in X;
  { x } in [x,y] by TARSKI:def 2;
  then
A2: {x} in union X by A1,TARSKI:def 4;
  {x,y} in [x,y] by TARSKI:def 2;
  then
A3: {x,y} in union X by A1,TARSKI:def 4;
  y in {x,y} & x in {x} by TARSKI:def 1,def 2;
  hence thesis by A3,A2,TARSKI:def 4;
end;

theorem Th135:
  X c= Y \/ {x} implies x in X or X c= Y
proof
  assume that
A1: X c= Y \/ {x} and
A2: not x in X;
  X = X /\ (Y \/ {x}) by A1,XBOOLE_1:28
    .= X /\ Y \/ X /\ {x} by XBOOLE_1:23
    .= X /\ Y \/ {} by A2,Lm7,XBOOLE_0:def 7
    .= X /\ Y;
  hence thesis by XBOOLE_1:17;
end;

theorem
  x in X \/ {y} iff x in X or x = y
proof
  x in X \/ {y} iff x in X or x in {y} by XBOOLE_0:def 3;
  hence thesis by TARSKI:def 1;
end;

theorem
  X \/ {x} c= Y iff x in Y & X c= Y
proof
 X c= Y & {x} c= Y implies X \/ {x} c= Y by XBOOLE_1:8;
  hence thesis by Lm1,XBOOLE_1:11;
end;

theorem
  for A, B, a being set st A c= B & B c= A \/ {a} holds A \/ {a} = B or A = B
proof
  let A, B, a be set;
  assume that
A1: A c= B and
A2: B c= A \/ {a};
  assume that
A3: A \/ {a} <> B and
A4: A <> B;
A5: not A \/ {a} c= B by A2,A3,XBOOLE_0:def 10;
A6: not a in B
  proof
    assume a in B;
    then {a} c= B by Lm1;
    hence thesis by A1,A5,XBOOLE_1:8;
  end;
  not B c= A by A1,A4,XBOOLE_0:def 10;
  hence thesis by A2,A6,Th135;
end;















begin

reserve E,X,x,x1,x2,x3,x4,x5,x6,x7,x8,x9,x10 for set;

registration
  let X be set;
  cluster bool X -> non empty;
  coherence by ZFMISC_1:def 1;
end;

registration
  let x1, x2, x3;
  cluster { x1, x2, x3 } -> non empty;
  coherence by ENUMSET1:def 1;
  let x4;
  cluster { x1, x2, x3, x4 } -> non empty;
  coherence by ENUMSET1:def 2;
  let x5;
  cluster { x1, x2, x3, x4, x5 } -> non empty;
  coherence by ENUMSET1:def 3;
  let x6;
  cluster { x1, x2, x3, x4, x5, x6 } -> non empty;
  coherence by ENUMSET1:def 4;
  let x7;
  cluster { x1, x2, x3, x4, x5, x6, x7 } -> non empty;
  coherence by ENUMSET1:def 5;
  let x8;
  cluster { x1, x2, x3, x4, x5, x6, x7, x8 } -> non empty;
  coherence by ENUMSET1:def 6;
  let x9;
  cluster { x1, x2, x3, x4, x5, x6, x7, x8, x9 } -> non empty;
  coherence by ENUMSET1:def 7;
  let x10;
  cluster { x1, x2, x3, x4, x5, x6, x7, x8, x9, x10 } -> non empty;
  coherence by ENUMSET1:def 8;
end;

definition
  let X;
  mode Element of X means :Def1:
  it in X if X is non empty otherwise it is empty;
  existence by XBOOLE_0:def 1;
  consistency;
  sethood
  proof
   per cases;
   case X is non empty;
    take X;
    thus thesis;
   end;
   case X is empty;
    take {{}};
    let y be set;
    thus thesis by TARSKI:def 1;
   end;
  end;
end;

definition
  let X;
  mode Subset of X is Element of bool X;
end;

registration
  let X be non empty set;
  cluster non empty for Subset of X;
  existence
  proof
    X in bool X by ZFMISC_1:def 1;
    then X is Subset of X by Def1;
    hence thesis;
  end;
end;

registration
  let X1,X2 be non empty set;
  cluster [: X1,X2 :] -> non empty;
  coherence
  proof
    consider x2 being set such that
A1: x2 in X2 by XBOOLE_0:def 1;
    consider x1 being set such that
A2: x1 in X1 by XBOOLE_0:def 1;
    [x1,x2] in [:X1,X2:] by A2,A1,ZFMISC_1:def 2;
    hence thesis;
  end;
end;

registration
  let X1,X2,X3 be non empty set;
  cluster [: X1,X2,X3 :] -> non empty;
  coherence;
end;

registration
  let X1,X2,X3,X4 be non empty set;
  cluster [: X1,X2,X3,X4 :] -> non empty;
  coherence;
end;

definition
  let D be non empty set, X be non empty Subset of D;
  redefine mode Element of X -> Element of D;
  coherence
  proof
    let x be Element of X;
    X in bool D by Def1;
    then
A1: X c= D by ZFMISC_1:def 1;
    x in X by Def1;
    then x in D by A1,TARSKI:def 3;
    hence thesis by Def1;
  end;
end;

Lm1: for X being Subset of E, x being set st x in X holds x in E
proof
  let X be Subset of E, x be set such that
A1: x in X;
  X in bool E by Def1;
  then X c= E by ZFMISC_1:def 1;
  hence thesis by A1,TARSKI:def 3;
end;

registration
  let E;
  cluster empty for Subset of E;
  existence
  proof
    {} c= E by XBOOLE_1:2;
    then {} in bool E by ZFMISC_1:def 1;
    then {} is Subset of E by Def1;
    hence thesis;
  end;
end;

definition
  let E;
  func {}E -> Subset of E equals
  {};
  coherence
  proof
    {} c= E by XBOOLE_1:2;
    then {} in bool E by ZFMISC_1:def 1;
    hence thesis by Def1;
  end;
  correctness;
  func [#] E -> Subset of E equals
  E;
  coherence
  proof
    E in bool E by ZFMISC_1:def 1;
    hence thesis by Def1;
  end;
  correctness;
end;

registration
  let E;
  cluster {}E -> empty;
  coherence;
end;

theorem
  {} is Subset of X
proof
  {}X = {};
  hence thesis;
end;

reserve A,B,C for Subset of E;

theorem Th2:
  (for x being Element of E holds x in A implies x in B) implies A c= B
proof
  assume
A1: for x being Element of E holds x in A implies x in B;
  let x be set;
  assume
A2: x in A;
  then x in E by Lm1;
  then x is Element of E by Def1;
  hence thesis by A1,A2;
end;

theorem Th3:
  (for x being Element of E holds x in A iff x in B) implies A = B
proof
  assume for x being Element of E holds x in A iff x in B;
  hence A c= B & B c= A by Th2;
end;

theorem Th4:
  A <> {} implies ex x being Element of E st x in A
proof
  assume A <> {};
  then consider x being set such that
A1: x in A by XBOOLE_0:def 1;
  x in E by A1,Lm1;
  then x is Element of E by Def1;
  hence thesis by A1;
end;

definition
  let E,A;
  func A` -> Subset of E equals
  E \ A;
  coherence
  proof
    E \ A c= E by XBOOLE_1:36;
    then E \ A in bool E by ZFMISC_1:def 1;
    hence thesis by Def1;
  end;
  correctness;
  involutiveness
  proof
    let S,T be Subset of E;
    assume
A1: S = E \ T;
    T in bool E by Def1;
    then T c= E by ZFMISC_1:def 1;
    hence T = {} \/ E /\ T by XBOOLE_1:28
      .= (E \ E) \/ E /\ T by XBOOLE_1:37
      .= E \ S by A1,XBOOLE_1:52;
  end;
  let B;
  redefine func A \/ B -> Subset of E;
  coherence
  proof
    B in bool E by Def1;
    then
A2: B c= E by ZFMISC_1:def 1;
    A in bool E by Def1;
    then A c= E by ZFMISC_1:def 1;
    then A \/ B c= E by A2,XBOOLE_1:8;
    then A \/ B in bool E by ZFMISC_1:def 1;
    hence thesis by Def1;
  end;
  redefine func A \+\ B -> Subset of E;
  coherence
  proof
    B in bool E by Def1;
    then B \ A c= B & B c= E by XBOOLE_1:36,ZFMISC_1:def 1;
    then
A3: B \ A c= E by XBOOLE_1:1;
    A in bool E by Def1;
    then A \ B c= A & A c= E by XBOOLE_1:36,ZFMISC_1:def 1;
    then A \ B c= E by XBOOLE_1:1;
    then (A \ B) \/ (B \ A) c= E by A3,XBOOLE_1:8;
    then A \+\ B in bool E by ZFMISC_1:def 1;
    hence thesis by Def1;
  end;
end;

definition
  let X,Y be set;
  redefine func X \ Y -> Subset of X;
  coherence
  proof
    X \ Y c= X by XBOOLE_1:36;
    then X \ Y in bool X by ZFMISC_1:def 1;
    hence thesis by Def1;
  end;
end;

definition
  let E,A,X;
  redefine func A \ X -> Subset of E;
  coherence
  proof
    A in bool E by Def1;
    then A \ X c= A & A c= E by XBOOLE_1:36,ZFMISC_1:def 1;
    then A \ X c= E by XBOOLE_1:1;
    then A \ X in bool E by ZFMISC_1:def 1;
    hence A \ X is Subset of E by Def1;
  end;
end;

definition
  let E,A,X;
  redefine func A /\ X -> Subset of E;
  coherence
  proof
    A in bool E by Def1;
    then A /\ X c= A & A c= E by XBOOLE_1:17,ZFMISC_1:def 1;
    then A /\ X c= E by XBOOLE_1:1;
    then A /\ X in bool E by ZFMISC_1:def 1;
    hence thesis by Def1;
  end;
end;

definition
  let E,X,A;
  redefine func X /\ A -> Subset of E;
  coherence
  proof
    A /\ X is Subset of E;
    hence thesis;
  end;
end;

theorem
  (for x being Element of E holds x in A iff x in B or x in C) implies A
  = B \/ C
proof
  assume
A1: for x being Element of E holds x in A iff x in B or x in C;
  now
    let x be Element of E;
    assume x in A;
    then x in B or x in C by A1;
    hence x in B \/ C by XBOOLE_0:def 3;
  end;
  hence A c= B \/ C by Th2;
  now
    let x be Element of E;
    assume x in B \/ C;
    then x in B or x in C by XBOOLE_0:def 3;
    hence x in A by A1;
  end;
  hence thesis by Th2;
end;

theorem
  (for x being Element of E holds x in A iff x in B & x in C) implies A
  = B /\ C
proof
  assume
A1: for x being Element of E holds x in A iff x in B & x in C;
  now
    let x be Element of E;
    assume x in A;
    then x in B & x in C by A1;
    hence x in B /\ C by XBOOLE_0:def 4;
  end;
  hence A c= B /\ C by Th2;
  now
    let x be Element of E;
    assume x in B /\ C;
    then x in B & x in C by XBOOLE_0:def 4;
    hence x in A by A1;
  end;
  hence thesis by Th2;
end;

theorem
  (for x being Element of E holds x in A iff x in B & not x in C)
  implies A = B \ C
proof
  assume
A1: for x being Element of E holds x in A iff x in B & not x in C;
  now
    let x be Element of E;
    assume x in A;
    then x in B & not x in C by A1;
    hence x in B \ C by XBOOLE_0:def 5;
  end;
  hence A c= B \ C by Th2;
  now
    let x be Element of E;
    assume x in B \ C;
    then x in B & not x in C by XBOOLE_0:def 5;
    hence x in A by A1;
  end;
  hence thesis by Th2;
end;

theorem
  (for x being Element of E holds x in A iff not(x in B iff x in C))
  implies A = B \+\ C
proof
  assume
A1: for x being Element of E holds x in A iff not(x in B iff x in C);
  now
    let x be Element of E;
    assume x in A;
    then x in B & not x in C or x in C & not x in B by A1;
    then x in B \ C or x in C \ B by XBOOLE_0:def 5;
    hence x in B \+\ C by XBOOLE_0:def 3;
  end;
  hence A c= B \+\ C by Th2;
  now
    let x be Element of E;
    assume x in B \+\ C;
    then x in B \ C or x in C \ B by XBOOLE_0:def 3;
    then x in B & not x in C or x in C & not x in B by XBOOLE_0:def 5;
    hence x in A by A1;
  end;
  hence thesis by Th2;
end;

theorem
  [#] E = ({} E)`;

theorem Th10:
  A \/ A` = [#]E
proof
  A in bool E by Def1;
  then A c= E by ZFMISC_1:def 1;
  hence thesis by XBOOLE_1:45;
end;

theorem
  A \/ [#]E = [#]E
proof
  A in bool E by Def1;
  then A c= E by ZFMISC_1:def 1;
  hence thesis by XBOOLE_1:12;
end;

theorem Th12:
  A c= B iff B` c= A`
proof
  thus A c= B implies B` c= A` by XBOOLE_1:34;
A1: E \ B` = B`` .= B;
  E \ A` = A`` .= A;
  hence thesis by A1,XBOOLE_1:34;
end;

theorem
  A \ B = A /\ B`
proof
  A in bool E by Def1;
  then
A1: A c= E by ZFMISC_1:def 1;
  thus A /\ B` = (A /\ E) \ B by XBOOLE_1:49
    .= A \ B by A1,XBOOLE_1:28;
end;

theorem
  (A \ B)` = A` \/ B
proof
  B in bool E by Def1;
  then
A1: B c= E by ZFMISC_1:def 1;
  thus (A \ B)` = (E \ A) \/ E /\ B by XBOOLE_1:52
    .= A` \/ B by A1,XBOOLE_1:28;
end;

theorem
  (A \+\ B)` = A /\ B \/ A` /\ B`
proof
  A in bool E by Def1;
  then
A1: A c= E by ZFMISC_1:def 1;
  thus (A \+\ B)` = E \ (A \/ B) \/ E /\ A /\ B by XBOOLE_1:102
    .= A /\ B \/ (E \ (A \/ B)) by A1,XBOOLE_1:28
    .= A /\ B \/ (A`) /\ (B`) by XBOOLE_1:53;
end;

theorem
  A c= B` implies B c= A`
proof
  assume A c= B`;
  then B``c= A` by Th12;
  hence thesis;
end;

theorem
  A` c= B implies B` c= A
proof
  assume A` c= B;
  then B` c= A`` by Th12;
  hence thesis;
end;

theorem
  A c= A` iff A = {}E
proof
  thus A c= A` implies A = {}E by XBOOLE_1:38;
  A in bool E by Def1;
  hence thesis by ZFMISC_1:def 1;
end;

theorem
  A` c= A iff A = [#]E
proof
  thus A` c= A implies A = [#]E
  proof
    assume A` c= A;
    hence A = A \/ A` by XBOOLE_1:12
      .= [#]E by Th10;
  end;
  assume A = [#]E;
  then A` = {} by XBOOLE_1:37;
  hence thesis by XBOOLE_1:2;
end;

theorem
  X c= A & X c= A` implies X = {} by XBOOLE_1:67,79;

theorem
  (A \/ B)` c= A`
proof
  A c= A \/ B by XBOOLE_1:7;
  hence thesis by Th12;
end;

theorem
  A` c= (A /\ B)`
proof
  A /\ B c= A by XBOOLE_1:17;
  hence thesis by Th12;
end;

theorem Th23:
  A misses B iff A c= B`
proof
  thus A misses B implies A c= B`
  proof
    assume
A1: A misses B;
    let x;
    assume
A2: x in A;
    then
A3: not x in B by A1,XBOOLE_0:3;
    x in E by A2,Lm1;
    hence thesis by A3,XBOOLE_0:def 5;
  end;
  assume
A4: A c= B`;
  assume A meets B;
  then consider x such that
A5: x in A and
A6: x in B by XBOOLE_0:3;
  x in E \ B by A4,A5,TARSKI:def 3;
  hence thesis by A6,XBOOLE_0:def 5;
end;

theorem
  A misses B` iff A c= B
proof
  B``= B;
  hence thesis by Th23;
end;

theorem
  A misses B & A` misses B` implies A = B`
proof
  assume that
A1: A misses B and
A2: A` misses B`;
A3: A in bool E by Def1;
  thus A c= B`
  proof
    let x;
    assume
A4: x in A;
    then
A5: not x in B by A1,XBOOLE_0:3;
    A c= E by A3,ZFMISC_1:def 1;
    then x in E by A4,TARSKI:def 3;
    hence thesis by A5,XBOOLE_0:def 5;
  end;
  let x;
A6: x in A` implies not x in B` by A2,XBOOLE_0:3;
  assume
A7: x in B`;
  then x in E by XBOOLE_0:def 5;
  hence thesis by A7,A6,XBOOLE_0:def 5;
end;

theorem
  A c= B & C misses B implies A c= C`
proof
  assume A c= B & C misses B;
  then A misses C by XBOOLE_1:63;
  hence thesis by Th23;
end;





theorem
  (for a being Element of A holds a in B) implies A c= B
proof
  assume
A1: for a being Element of A holds a in B;
  let a be set;
  assume a in A;
  then a is Element of A by Def1;
  hence thesis by A1;
end;

theorem
  (for x being Element of E holds x in A) implies E = A
proof
  assume
A1: for x being Element of E holds x in A;
  thus E c= A
  proof
    let a be set;
    assume a in E;
    then a is Element of E by Def1;
    hence thesis by A1;
  end;
  A in bool E by Def1;
  hence thesis by ZFMISC_1:def 1;
end;

theorem
  E <> {} implies for B for x being Element of E st not x in B holds x in B`
proof
  assume
A1: E <> {};
  let B be Subset of E;
  let x be Element of E;
  assume
A2: not x in B;
  x in E by A1,Def1;
  hence thesis by A2,XBOOLE_0:def 5;
end;

theorem Th30:
  for A,B st for x being Element of E holds x in A iff not x in B holds A = B`
proof
  let A,B be Subset of E;
  assume
A1: for x being Element of E holds x in A iff not x in B;
  thus A c= B`
  proof
    let x be set;
    assume
A2: x in A;
    A in bool E by Def1;
    then A c= E by ZFMISC_1:def 1;
    then x in E by A2,TARSKI:def 3;
    then x is Element of E by Def1;
    then
A3: not x in B by A1,A2;
    x in E by A2,Lm1;
    hence thesis by A3,XBOOLE_0:def 5;
  end;
  let x be set;
  assume
A4: x in B`;
  B` in bool E by Def1;
  then B` c= E by ZFMISC_1:def 1;
  then x in E by A4,TARSKI:def 3;
  then reconsider x as Element of E by Def1;
  not x in B by A4,XBOOLE_0:def 5;
  hence thesis by A1;
end;

theorem
  for A,B st for x being Element of E holds not x in A iff x in B holds A = B`
proof
  let A,B;
  assume for x being Element of E holds not x in A iff x in B;
  then for x being Element of E holds x in A iff not x in B;
  hence thesis by Th30;
end;

theorem
  for A,B st for x being Element of E holds not(x in A iff x in B) holds A = B`
proof
  let A,B;
  assume for x being Element of E holds not(x in A iff x in B);
  then for x being Element of E holds x in A iff not x in B;
  hence thesis by Th30;
end;

reserve x1,x2,x3,x4,x5,x6,x7,x8 for Element of X;

theorem
  X <> {} implies {x1} is Subset of X
proof
  assume X <> {};
  then
A1: x1 in X by Def1;
  {x1} c= X
  proof
    let x;
    assume x in {x1};
    hence thesis by A1,TARSKI:def 1;
  end;
  then {x1} in bool X by ZFMISC_1:def 1;
  hence thesis by Def1;
end;

theorem
  X <> {} implies {x1,x2} is Subset of X
proof
  assume X <> {};
  then
A1: x1 in X & x2 in X by Def1;
  {x1,x2} c= X
  proof
    let x;
    assume x in {x1,x2};
    hence thesis by A1,TARSKI:def 2;
  end;
  then {x1,x2} in bool X by ZFMISC_1:def 1;
  hence thesis by Def1;
end;

theorem
  X <> {} implies {x1,x2,x3} is Subset of X
proof
  assume
A1: X <> {};
  then
A2: x3 in X by Def1;
A3: x1 in X & x2 in X by A1,Def1;
  {x1,x2,x3} c= X
  proof
    let x;
    assume x in {x1,x2,x3};
    hence thesis by A3,A2,ENUMSET1:def 1;
  end;
  then {x1,x2,x3} in bool X by ZFMISC_1:def 1;
  hence thesis by Def1;
end;

theorem
  X <> {} implies {x1,x2,x3,x4} is Subset of X
proof
  assume
A1: X <> {};
  then
A2: x3 in X & x4 in X by Def1;
A3: x1 in X & x2 in X by A1,Def1;
  {x1,x2,x3,x4} c= X
  proof
    let x;
    assume x in {x1,x2,x3,x4};
    hence thesis by A3,A2,ENUMSET1:def 2;
  end;
  then {x1,x2,x3,x4} in bool X by ZFMISC_1:def 1;
  hence thesis by Def1;
end;

theorem
  X <> {} implies {x1,x2,x3,x4,x5} is Subset of X
proof
  assume
A1: X <> {};
  {x1,x2,x3,x4,x5} c= X
  proof
    let x;
A2: x in { x1,x2,x3,x4,x5 } implies x=x1 or x=x2 or x=x3 or x=x4 or x=x5
    by ENUMSET1:def 3;
    assume x in {x1,x2,x3,x4,x5};
    hence thesis by A1,A2,Def1;
  end;
  then {x1,x2,x3,x4,x5} in bool X by ZFMISC_1:def 1;
  hence thesis by Def1;
end;

theorem
  X <> {} implies {x1,x2,x3,x4,x5,x6} is Subset of X
proof
  assume
A1: X <> {};
  {x1,x2,x3,x4,x5,x6} c= X
  proof
    let x;
A2: x in { x1,x2,x3,x4,x5,x6} implies x=x1 or x=x2 or x=x3 or x=x4 or x=x5
    or x=x6 by ENUMSET1:def 4;
    assume x in {x1,x2,x3,x4,x5,x6};
    hence thesis by A1,A2,Def1;
  end;
  then {x1,x2,x3,x4,x5,x6} in bool X by ZFMISC_1:def 1;
  hence thesis by Def1;
end;

theorem
  X <> {} implies {x1,x2,x3,x4,x5,x6,x7} is Subset of X
proof
  assume
A1: X <> {};
  {x1,x2,x3,x4,x5,x6,x7} c= X
  proof
    let x;
A2: x in { x1,x2,x3,x4,x5,x6,x7} implies x=x1 or x=x2 or x=x3 or x=x4 or x
    =x5 or x=x6 or x=x7 by ENUMSET1:def 5;
    assume x in {x1,x2,x3,x4,x5,x6,x7};
    hence thesis by A1,A2,Def1;
  end;
  then {x1,x2,x3,x4,x5,x6,x7} in bool X by ZFMISC_1:def 1;
  hence thesis by Def1;
end;

theorem
  X <> {} implies {x1,x2,x3,x4,x5,x6,x7,x8} is Subset of X
proof
  assume
A1: X <> {};
  {x1,x2,x3,x4,x5,x6,x7,x8} c= X
  proof
    let x;
A2: x in { x1,x2,x3,x4,x5,x6,x7,x8} implies x=x1 or x=x2 or x=x3 or x=x4
    or x=x5 or x=x6 or x=x7 or x=x8 by ENUMSET1:def 6;
    assume x in {x1,x2,x3,x4,x5,x6,x7,x8};
    hence thesis by A1,A2,Def1;
  end;
  then {x1,x2,x3,x4,x5,x6,x7,x8} in bool X by ZFMISC_1:def 1;
  hence thesis by Def1;
end;

theorem
  x in X implies {x} is Subset of X
proof
  assume x in X;
  then {x} c= X by ZFMISC_1:31;
  then {x} in bool X by ZFMISC_1:def 1;
  hence thesis by Def1;
end;

scheme
  SubsetEx { A()-> set, P[set] } : ex X being Subset of A() st for x holds x
  in X iff x in A() & P[x] proof
  consider X being set such that
A1: for x holds x in X iff x in A() & P[x] from XBOOLE_0:sch 1;
  X c= A()
  proof
    let x;
    thus thesis by A1;
  end;
  then X in bool A() by ZFMISC_1:def 1;
  then reconsider X as Subset of A() by Def1;
  take X;
  thus thesis by A1;
end;

scheme
  SubsetEq {X() -> set, X1,X2()-> Subset of X(), P[set]}: X1() = X2()
provided
A1: for y being Element of X() holds y in X1() iff P[y] and
A2: for y being Element of X() holds y in X2() iff P[y]
proof
  for x being Element of X() holds x in X1() iff x in X2()
  proof
    let x be Element of X();
    hereby
      assume x in X1();
      then P[x] by A1;
      hence x in X2() by A2;
    end;
    assume x in X2();
    then P[x] by A2;
    hence thesis by A1;
  end;
  hence thesis by Th3;
end;

definition
  let X, Y be non empty set;
  redefine pred X misses Y;
  irreflexivity
  proof
    let A be non empty set;
    ex x being set st x in A by XBOOLE_0:def 1;
    hence thesis by XBOOLE_0:3;
  end;
end;

definition
  let X, Y be non empty set;
  redefine pred X meets Y;
  reflexivity
  proof
    let A be non empty set;
    thus not A misses A;
  end;
end;

definition
  let S be set;
  func choose S -> Element of S equals
  the Element of S;
  correctness;
end;

begin 

reserve Y for set;

Lm2: (for x st x in X holds x in Y) implies X is Subset of Y
proof
  assume for x st x in X holds x in Y;
  then X c= Y by TARSKI:def 3;
  then X in bool Y by ZFMISC_1:def 1;
  hence thesis by Def1;
end;

Lm3: x in A implies x is Element of E
proof
  assume x in A;
  then x in E by Lm1;
  hence thesis by Def1;
end;



scheme
  SubsetEx { A() -> non empty set, P[set] } : ex B being Subset of A() st for
  x being Element of A() holds x in B iff P[x] proof
  consider B being set such that
A1: for x being set holds x in B iff x in A() & P[x] from XBOOLE_0:sch 1;
  for x being set holds x in B implies x in A() by A1;
  then reconsider B as Subset of A() by Lm2;
  take B;
  let x be Element of A();
  x in A() by Def1;
  hence thesis by A1;
end;



scheme
  SubComp{A() -> set, F1,F2() -> Subset of A(), P[set]}: F1() = F2()
provided
A1: for X being Element of A() holds X in F1() iff P[X] and
A2: for X being Element of A() holds X in F2() iff P[X]
proof
  thus F1() c= F2()
  proof
    let x;
    assume
A3: x in F1();
    then reconsider X = x as Element of A() by Lm3;
    P[X] by A1,A3;
    hence thesis by A2;
  end;
  let x;
  assume
A4: x in F2();
  then reconsider X = x as Element of A() by Lm3;
  P[X] by A2,A4;
  hence thesis by A1;
end;



theorem
  A` = B` implies A = B
proof
  assume A` = B`;
  hence A = B`` .= B;
end;



registration
  let X be empty set;
  cluster -> empty for Subset of X;
  coherence
  proof
    let Y be Subset of X;
    Y in bool X by Def1;
    then Y c= X by ZFMISC_1:def 1;
    hence thesis by XBOOLE_1:3;
  end;
end;



definition
  let E be set;
  let A be Subset of E;
  attr A is proper means
  A <> E;
end;

registration
  let E be set;
  cluster [#]E -> non proper;
  coherence
  proof
    thus [#]E = E;
  end;
end;

registration
  let E be set;
  cluster non proper for Subset of E;
  existence
  proof
    take [#]E;
    thus thesis;
  end;
end;

registration
  let E be non empty set;
  cluster non proper -> non empty for Subset of E;
  coherence
  proof
    let A be Subset of E;
    assume A = E;
    hence thesis;
  end;
  cluster empty -> proper for Subset of E;
  coherence;
end;

registration
  let E be non empty set;
  cluster proper for Subset of E;
  existence
  proof
    take {} E;
    thus {} E <> E;
  end;
end;

registration
  let E be empty set;
  cluster -> non proper for Subset of E;
  coherence
  proof
    let A be Subset of E;
    thus A = E;
  end;
end;



theorem
  for X,Y,A being set, z being set st z in A & A c= [:X,Y:] ex x
  being Element of X, y being Element of Y st z = [x,y]
proof
  let X,Y,A be set, z be set;
  assume z in A & A c= [:X,Y:];
  then consider x,y being set such that
A1: x in X and
A2: y in Y and
A3: z = [x,y] by ZFMISC_1:84;
  reconsider y as Element of Y by A2,Def1;
  reconsider x as Element of X by A1,Def1;
  take x,y;
  thus thesis by A3;
end;



theorem
  for X being non empty set, A, B being non empty Subset of X st A c< B
   ex p being Element of X st p in B & A c= B \ {p}
proof
  let X be non empty set, A, B be non empty Subset of X;
  assume
A1: A c< B;
  then consider p being Element of X such that
A2: p in B \ A by Th4,XBOOLE_1:105;
A3: not p in A by A2,XBOOLE_0:def 5;
  take p;
  A c= B by A1,XBOOLE_0:def 8;
  hence thesis by A2,A3,XBOOLE_0:def 5,ZFMISC_1:34;
end;

definition
  let X be non empty set;
  redefine attr X is trivial means
  for x,y being Element of X holds x = y;
 compatibility
  proof
   thus X is trivial implies for x,y being Element of X holds x = y
    proof assume
A1:    X is trivial;
     let x,y be Element of X;
      x in X & y in X by Def1;
     hence x = y by A1,ZFMISC_1:def 10;
    end;
   assume
A2:  for x,y being Element of X holds x = y;
   let x,y be set;
    assume x in X & y in X;
     then x is Element of X & y is Element of X by Def1;
    hence thesis by A2;
  end;
end;

registration let X be non empty set;
 cluster non empty trivial for Subset of X;
 existence
  proof
    the Element of X in X by Def1;
    then {the Element of X} c= X by ZFMISC_1:31;
    then {the Element of X} in bool X by ZFMISC_1:def 1;
    then reconsider A = {the Element of X} as Subset of X by Def1;
   take A;
   thus A is non empty;
   let x,y be set;
   assume x in A & y in A;
    then x = the Element of X & y = the Element of X by TARSKI:def 1;
   hence thesis;
  end;
end;

registration
  let X be trivial set;
  cluster -> trivial for Subset of X;
  coherence
  proof
    let Y be Subset of X;
    let x,y be set;
    assume x in Y & y in Y;
     then x in X & y in X by Lm1;
    hence thesis by ZFMISC_1:def 10;
  end;
end;

registration
  let X be non trivial set;
  cluster non trivial for Subset of X;
  existence
  proof
    take [#]X;
    thus thesis;
  end;
end;

theorem
  for D being set, A being Subset of D st A is non trivial
    ex d1,d2 being Element of D st d1 in A & d2 in A & d1 <> d2
proof
  let D be set, A be Subset of D;
   assume A is non trivial;
    then consider d1,d2 being set such that
A1: d1 in A & d2 in A and
A2: d1 <> d2 by ZFMISC_1:def 10;
    d1 in D & d2 in D by A1,Lm1;
    then reconsider d1,d2 as Element of D by Def1;
   take d1,d2;
   thus d1 in A & d2 in A & d1 <> d2 by A1,A2;
end;

theorem Th46:
 for X being trivial non empty set
  ex x being Element of X st X = {x}
proof let X be trivial non empty set;
  consider x being set such that
A1:  X = {x} by ZFMISC_1:131;
  x in X by A1,TARSKI:def 1;
  then reconsider x as Element of X by Def1;
 take x;
 thus X = {x} by A1;
end;



theorem
  for X being non empty set, A being non empty Subset of X holds A
  is trivial implies ex x being Element of X st A = {x}
proof
  let X be non empty set, A be non empty Subset of X;
  assume A is trivial;
  then ex s being Element of A st A = {s} by Th46;
  hence thesis;
end;

theorem
  for X be non trivial set, x being Element of X
   ex y be set st y in X & x <> y
proof
  let X be non trivial set;
  let x be Element of X;
   consider d1,d2 being set such that
A1: d1 in X & d2 in X and
A2: d1 <> d2 by ZFMISC_1:def 10;
   x <> d1 or x <> d2 by A2;
  hence thesis by A1;
end;















begin








theorem
  for a, b being set st a in b holds a is Element of b by SUBSET_1:def 1;

theorem
  for a, b being set st a is Element of b & b is non empty holds a in b
  by SUBSET_1:def 1;

theorem Th3:
  for a, b being set holds a is Subset of b iff a c= b
proof
  let a, b be set;
  hereby
    assume a is Subset of b;
    then a in bool b by SUBSET_1:def 1;
    hence a c= b by ZFMISC_1:def 1;
  end;
  assume a c= b;
  then a in bool b by ZFMISC_1:def 1;
  hence thesis by SUBSET_1:def 1;
end;

theorem
  for a, b, c being set st a in b & b is Subset of c holds a is Element of c
proof
  let a, b, c be set;
  assume that
A1: a in b and
A2: b is Subset of c;
  b c= c by A2,Th3;
  then a in c by A1,TARSKI:def 3;
  hence thesis by SUBSET_1:def 1;
end;

theorem
  for a, b, c being set st a in b & b is Subset of c holds c is non empty;














begin

reserve X,Y,Z,Z1,Z2,D,x,y for set;

definition
  let X;
  func meet X means
  :Def1:
  for x holds x in it iff for Y holds Y in X implies
  x in Y if X <> {} otherwise it = {};
  existence
  proof
    thus X <> {} implies ex Z1 st for x holds x in Z1 iff for Z st Z in X
    holds x in Z
    proof
      defpred X[set] means for Z st Z in X holds $1 in Z;
      assume
A1:   X <> {};
      consider Y such that
A2:   for x holds x in Y iff x in union X & X[x] from XBOOLE_0:sch 1;
      take Y;
      for x holds (for Z st Z in X holds x in Z) implies x in Y
      proof
        set y = the Element of X;
        let x such that
A3:     for Z st Z in X holds x in Z;
        x in y by A1,A3;
        then x in union X by A1,TARSKI:def 4;
        hence thesis by A2,A3;
      end;
      hence thesis by A2;
    end;
    thus thesis;
  end;
  uniqueness
  proof
    let Z1,Z2;
    now
      assume that
      X <> {} and
A4:   for x holds x in Z1 iff for Y holds Y in X implies x in Y and
A5:   for x holds x in Z2 iff for Y holds Y in X implies x in Y;
      now
        let x;
        x in Z1 iff for Y holds Y in X implies x in Y by A4;
        hence x in Z1 iff x in Z2 by A5;
      end;
      hence Z1 = Z2 by TARSKI:1;
    end;
    hence thesis;
  end;
  consistency;
end;



theorem
  meet {} = {} by Def1;

theorem Th2:
  meet X c= union X
proof
A1: now
    assume
A2: X <> {};
    now
      set y = the Element of X;
      let x;
      assume x in meet X;
      then x in y by A2,Def1;
      hence x in union X by A2,TARSKI:def 4;
    end;
    hence thesis by TARSKI:def 3;
  end;
  now
    assume X = {};
    then meet X = {} by Def1;
    hence thesis by XBOOLE_1:2;
  end;
  hence thesis by A1;
end;

theorem Th3:
  Z in X implies meet X c= Z
proof
  assume
A1: Z in X;
  meet X c= Z
  proof
    let x;
    assume x in meet X;
    hence thesis by A1,Def1;
  end;
  hence thesis;
end;

theorem
  {} in X implies meet X = {} by Th3,XBOOLE_1:3;

theorem
  X <> {} & (for Z1 st Z1 in X holds Z c= Z1) implies Z c= meet X
proof
  assume that
A1: X <> {} and
A2: for Z1 st Z1 in X holds Z c= Z1;
  thus Z c= meet X
  proof
    let x such that
A3: x in Z;
    for Y st Y in X holds x in Y
    proof
      let Y;
      assume Y in X;
      then Z c= Y by A2;
      hence thesis by A3;
    end;
    hence thesis by A1,Def1;
  end;
end;

theorem Th6:
  X <> {} & X c= Y implies meet Y c= meet X
proof
  assume that
A1: X <> {} and
A2: X c= Y;
  let x;
  assume x in meet Y;
  then for Z st Z in X holds x in Z by A2,Def1;
  hence thesis by A1,Def1;
end;

theorem
  X in Y & X c= Z implies meet Y c= Z
proof
  assume that
A1: X in Y and
A2: X c= Z;
  meet Y c= X by A1,Th3;
  hence thesis by A2,XBOOLE_1:1;
end;

theorem
  X in Y & X misses Z implies meet Y misses Z by Th3,XBOOLE_1:63;

theorem
  X <> {} & Y <> {} implies meet (X \/ Y) = meet X /\ meet Y
proof
  assume that
A1: X <> {} and
A2: Y <> {};
A3: meet X /\ meet Y c= meet (X \/ Y)
  proof
    let x;
    assume x in meet X /\ meet Y;
    then
A4: x in meet X & x in meet Y by XBOOLE_0:def 4;
    now
      let Z;
      assume Z in X \/ Y;
      then Z in X or Z in Y by XBOOLE_0:def 3;
      hence x in Z by A4,Def1;
    end;
    hence thesis by A1,Def1;
  end;
  meet (X \/ Y) c= meet X & meet (X \/ Y) c= meet Y by A1,A2,Th6,XBOOLE_1:7;
  then meet (X \/ Y) c= meet X /\ meet Y by XBOOLE_1:19;
  hence thesis by A3,XBOOLE_0:def 10;
end;

theorem
  meet {x} = x
proof
A1: x c= meet {x}
  proof
    let y;
    assume y in x;
    then for Z st Z in {x} holds y in Z by TARSKI:def 1;
    hence thesis by Def1;
  end;
  x in {x} by TARSKI:def 1;
  then meet {x} c= x by Th3;
  hence thesis by A1,XBOOLE_0:def 10;
end;

theorem
  meet {X,Y} = X /\ Y
proof
A1: X in {X,Y} & Y in {X,Y} by TARSKI:def 2;
  x in meet {X,Y} iff x in X & x in Y
  proof
    thus x in meet {X,Y} implies x in X & x in Y by A1,Def1;
    assume x in X & x in Y;
    then for Z holds Z in {X,Y} implies x in Z by TARSKI:def 2;
    hence thesis by Def1;
  end;
  hence thesis by XBOOLE_0:def 4;
end;

reserve SFX,SFY,SFZ for set;

definition
  let SFX,SFY;
  pred SFX is_finer_than SFY means
  for X st X in SFX ex Y st Y in SFY & X c= Y;
  reflexivity;
  pred SFY is_coarser_than SFX means
  for Y st Y in SFY ex X st X in SFX & X c= Y;
  reflexivity;
end;

theorem
  SFX c= SFY implies SFX is_finer_than SFY
proof
  assume
A1: SFX c= SFY;
  let X such that
A2: X in SFX;
  take X;
  thus thesis by A1,A2;
end;

theorem
  SFX is_finer_than SFY implies union SFX c= union SFY
proof
  assume
A1: for X st X in SFX ex Y st Y in SFY & X c= Y;
  thus union SFX c= union SFY
  proof
    let x;
    assume x in union SFX;
    then consider Y such that
A2: x in Y and
A3: Y in SFX by TARSKI:def 4;
    ex Z st Z in SFY & Y c= Z by A1,A3;
    hence thesis by A2,TARSKI:def 4;
  end;
end;

theorem
  SFY <> {} & SFY is_coarser_than SFX implies meet SFX c= meet SFY
proof
  assume that
A1: SFY <> {} and
A2: for Z2 st Z2 in SFY ex Z1 st Z1 in SFX & Z1 c= Z2;
  meet SFX c= meet SFY
  proof
    let x such that
A3: x in meet SFX;
    for Z st Z in SFY holds x in Z
    proof
      let Z;
      assume Z in SFY;
      then consider Z1 such that
A4:   Z1 in SFX and
A5:   Z1 c= Z by A2;
      x in Z1 by A3,A4,Def1;
      hence thesis by A5;
    end;
    hence thesis by A1,Def1;
  end;
  hence thesis;
end;

theorem
  {} is_finer_than SFX
proof
  let X;
  assume X in {};
  hence thesis;
end;

theorem
  SFX is_finer_than {} implies SFX = {}
proof
  assume
A1: for X st X in SFX ex Y st Y in {} & X c= Y;
  set x = the Element of SFX;
  assume not thesis;
  then ex Y st Y in {} & x c= Y by A1;
  hence contradiction;
end;

theorem
  SFX is_finer_than SFY & SFY is_finer_than SFZ implies SFX is_finer_than SFZ
proof
  assume that
A1: for X st X in SFX ex Y st Y in SFY & X c= Y and
A2: for X st X in SFY ex Y st Y in SFZ & X c= Y;
  let X;
  assume X in SFX;
  then consider Y such that
A3: Y in SFY and
A4: X c= Y by A1;
  consider Z such that
A5: Z in SFZ & Y c= Z by A2,A3;
  take Z;
  thus thesis by A4,A5,XBOOLE_1:1;
end;

theorem
  SFX is_finer_than {Y} implies for X st X in SFX holds X c= Y
proof
  assume
A1: for X st X in SFX ex Z st Z in {Y} & X c= Z;
  let X;
  assume X in SFX;
  then ex Z st Z in {Y} & X c= Z by A1;
  hence thesis by TARSKI:def 1;
end;

theorem
  SFX is_finer_than {X,Y} implies for Z st Z in SFX holds Z c= X or Z c= Y
proof
  assume
A1: for Z1 st Z1 in SFX ex Z2 st Z2 in {X,Y} & Z1 c= Z2;
  let Z;
  assume Z in SFX;
  then consider Z2 such that
A2: Z2 in {X,Y} and
A3: Z c= Z2 by A1;
  {X,Y} = {X} \/ {Y} by ENUMSET1:1;
  then Z2 in {X} or Z2 in {Y} by A2,XBOOLE_0:def 3;
  hence thesis by A3,TARSKI:def 1;
end;

definition
  let SFX,SFY;
  func UNION (SFX,SFY) means
  :Def4:
  Z in it iff ex X,Y st X in SFX & Y in SFY & Z = X \/ Y;
  existence
  proof
    defpred X[set] means ex Z being set st $1 = Z & ex X,Y st X in SFX & Y in
    SFY & Z = X \/ Y;
    consider XX being set such that
A1: for x holds x in XX iff x in bool(union SFX \/ union SFY) & X[x]
    from XBOOLE_0:sch 1;
    take XX;
    for Z holds Z in XX iff ex X,Y st X in SFX & Y in SFY & Z = X \/ Y
    proof
      let Z;
A2:   now
        given X,Y such that
A3:     X in SFX & Y in SFY and
A4:     Z = X \/ Y;
        X c= union SFX & Y c= union SFY by A3,ZFMISC_1:74;
        then Z c= union SFX \/ union SFY by A4,XBOOLE_1:13;
        hence Z in XX by A1,A3,A4;
      end;
      now
        assume Z in XX;
        then ex Z1 st Z = Z1 & ex X,Y st X in SFX & Y in SFY & Z1 = X \/ Y by
A1;
        hence ex X,Y st X in SFX & Y in SFY & Z = X \/ Y;
      end;
      hence thesis by A2;
    end;
    hence thesis;
  end;
  uniqueness
  proof
    let Z1,Z2 be set;
    assume that
A5: for Z holds Z in Z1 iff ex X,Y st X in SFX & Y in SFY & Z = X \/ Y and
A6: for Z holds Z in Z2 iff ex X,Y st X in SFX & Y in SFY & Z = X \/ Y;
    now
      let Z;
      Z in Z1 iff ex X,Y st X in SFX & Y in SFY & Z = X \/ Y by A5;
      hence Z in Z1 iff Z in Z2 by A6;
    end;
    hence thesis by TARSKI:1;
  end;
  commutativity
  proof
    let SFZ,SFX,SFY;
    assume
A7: for Z holds Z in SFZ iff ex X,Y st X in SFX & Y in SFY & Z = X \/ Y;
    let Z;
    hereby
      assume Z in SFZ;
      then ex X,Y st X in SFX & Y in SFY & Z = X \/ Y by A7;
      hence ex Y,X st Y in SFY & X in SFX & Z = Y \/ X;
    end;
    thus thesis by A7;
  end;
  func INTERSECTION (SFX,SFY) means
  :Def5:
  Z in it iff ex X,Y st X in SFX & Y in SFY & Z = X /\ Y;
  existence
  proof
    defpred X[set] means ex Z being set st $1 = Z & ex X,Y st X in SFX & Y in
    SFY & Z = X /\ Y;
    consider XX being set such that
A8: for x holds x in XX iff x in bool(union SFX /\ union SFY) & X[x]
    from XBOOLE_0:sch 1;
    take XX;
    for Z holds Z in XX iff ex X,Y st X in SFX & Y in SFY & Z = X /\ Y
    proof
      let Z;
A9:   now
        given X,Y such that
A10:    X in SFX & Y in SFY and
A11:    Z = X /\ Y;
        X c= union SFX & Y c= union SFY by A10,ZFMISC_1:74;
        then Z c= union SFX /\ union SFY by A11,XBOOLE_1:27;
        hence Z in XX by A8,A10,A11;
      end;
      now
        assume Z in XX;
        then ex Z1 st Z = Z1 & ex X,Y st X in SFX & Y in SFY & Z1 = X /\ Y by
A8;
        hence ex X,Y st X in SFX & Y in SFY & Z = X /\ Y;
      end;
      hence thesis by A9;
    end;
    hence thesis;
  end;
  uniqueness
  proof
    let Z1,Z2 be set;
    assume that
A12: for Z holds Z in Z1 iff ex X,Y st X in SFX & Y in SFY & Z = X /\ Y and
A13: for Z holds Z in Z2 iff ex X,Y st X in SFX & Y in SFY & Z = X /\ Y;
    now
      let Z;
      Z in Z1 iff ex X,Y st X in SFX & Y in SFY & Z = X /\ Y by A12;
      hence Z in Z1 iff Z in Z2 by A13;
    end;
    hence thesis by TARSKI:1;
  end;
  commutativity
  proof
    let SFZ,SFX,SFY;
    assume
A14: for Z holds Z in SFZ iff ex X,Y st X in SFX & Y in SFY & Z = X /\ Y;
    let Z;
    hereby
      assume Z in SFZ;
      then ex X,Y st X in SFX & Y in SFY & Z = X /\ Y by A14;
      hence ex Y,X st Y in SFY & X in SFX & Z = Y /\ X;
    end;
    thus thesis by A14;
  end;
  func DIFFERENCE (SFX,SFY) means
  :Def6:
  Z in it iff ex X,Y st X in SFX & Y in SFY & Z = X \ Y;
  existence
  proof
    defpred X[set] means ex Z being set st $1 = Z & ex X,Y st X in SFX & Y in
    SFY & Z = X \ Y;
    consider XX being set such that
A15: for x holds x in XX iff x in bool union SFX & X[x] from XBOOLE_0:
    sch 1;
    take XX;
    for Z holds Z in XX iff ex X,Y st X in SFX & Y in SFY & Z = X \ Y
    proof
      let Z;
A16:  now
        given X,Y such that
A17:    X in SFX and
A18:    Y in SFY and
A19:    Z = X \ Y;
        X c= union SFX by A17,ZFMISC_1:74;
        then Z c= union SFX by A19,XBOOLE_1:1;
        hence Z in XX by A15,A17,A18,A19;
      end;
      now
        assume Z in XX;
        then ex Z1 st Z = Z1 & ex X,Y st X in SFX & Y in SFY & Z1 = X \ Y by
A15;
        hence ex X,Y st X in SFX & Y in SFY & Z = X \ Y;
      end;
      hence thesis by A16;
    end;
    hence thesis;
  end;
  uniqueness
  proof
    let Z1,Z2 be set;
    assume that
A20: for Z holds Z in Z1 iff ex X,Y st X in SFX & Y in SFY & Z = X \ Y and
A21: for Z holds Z in Z2 iff ex X,Y st X in SFX & Y in SFY & Z = X \ Y;
    now
      let Z;
      Z in Z1 iff ex X,Y st X in SFX & Y in SFY & Z = X \ Y by A20;
      hence Z in Z1 iff Z in Z2 by A21;
    end;
    hence thesis by TARSKI:1;
  end;
end;

theorem
  SFX is_finer_than UNION(SFX,SFX)
proof
  let X such that
A1: X in SFX;
  take X;
  X = X \/ X;
  hence thesis by A1,Def4;
end;

theorem
  INTERSECTION(SFX,SFX) is_finer_than SFX
proof
  let X;
  assume X in INTERSECTION(SFX,SFX);
  then consider Z1,Z2 such that
A1: Z1 in SFX and
  Z2 in SFX and
A2: X = Z1 /\ Z2 by Def5;
  take Z1;
  thus thesis by A1,A2,XBOOLE_1:17;
end;

theorem
  DIFFERENCE(SFX,SFX) is_finer_than SFX
proof
  let X;
  assume X in DIFFERENCE(SFX,SFX);
  then consider Z1,Z2 such that
A1: Z1 in SFX and
  Z2 in SFX and
A2: X = Z1 \ Z2 by Def6;
  take Z1;
  thus thesis by A1,A2;
end;

theorem
  SFX meets SFY implies meet SFX /\ meet SFY = meet INTERSECTION(SFX,SFY )
proof
  set y = the Element of SFX /\ SFY;
  assume
A1: SFX /\ SFY <> {};
  then
A2: SFY <> {};
A3: y in SFX by A1,XBOOLE_0:def 4;
A4: y in SFY by A1,XBOOLE_0:def 4;
A5: SFX <> {} by A1;
A6: meet INTERSECTION(SFX,SFY) c= meet SFX /\ meet SFY
  proof
    let x such that
A7: x in meet INTERSECTION(SFX,SFY);
    for Z st Z in SFY holds x in Z
    proof
      let Z;
      assume Z in SFY;
      then y /\ Z in INTERSECTION(SFX,SFY) by A3,Def5;
      then x in y /\ Z by A7,Def1;
      hence thesis by XBOOLE_0:def 4;
    end;
    then
A8: x in meet SFY by A2,Def1;
    for Z st Z in SFX holds x in Z
    proof
      let Z;
      assume Z in SFX;
      then Z /\ y in INTERSECTION(SFX,SFY) by A4,Def5;
      then x in Z /\ y by A7,Def1;
      hence thesis by XBOOLE_0:def 4;
    end;
    then x in meet SFX by A5,Def1;
    hence thesis by A8,XBOOLE_0:def 4;
  end;
A9: y /\ y in INTERSECTION(SFX,SFY) by A3,A4,Def5;
  meet SFX /\ meet SFY c= meet INTERSECTION(SFX,SFY)
  proof
    let x;
    assume x in meet SFX /\ meet SFY;
    then
A10: x in meet SFX & x in meet SFY by XBOOLE_0:def 4;
    for Z st Z in INTERSECTION(SFX,SFY) holds x in Z
    proof
      let Z;
      assume Z in INTERSECTION(SFX,SFY);
      then consider Z1,Z2 such that
A11:  Z1 in SFX & Z2 in SFY and
A12:  Z = Z1 /\ Z2 by Def5;
      x in Z1 & x in Z2 by A10,A11,Def1;
      hence thesis by A12,XBOOLE_0:def 4;
    end;
    hence thesis by A9,Def1;
  end;
  hence thesis by A6,XBOOLE_0:def 10;
end;

theorem
  SFY <> {} implies X \/ meet SFY = meet UNION({X},SFY)
proof
  assume
A1: SFY <> {};
  set y = the Element of SFY;
  X in {X} by TARSKI:def 1;
  then
A2: X \/ y in UNION({X},SFY) by A1,Def4;
A3: X \/ meet SFY c= meet UNION({X},SFY)
  proof
    let x;
    assume x in X \/ meet SFY;
    then
A4: x in X or x in meet SFY by XBOOLE_0:def 3;
    for Z st Z in UNION({X},SFY) holds x in Z
    proof
      let Z;
      assume Z in UNION({X},SFY);
      then consider Z1,Z2 such that
A5:   Z1 in {X} & Z2 in SFY and
A6:   Z = Z1 \/ Z2 by Def4;
      x in Z1 or x in Z2 by A4,A5,Def1,TARSKI:def 1;
      hence thesis by A6,XBOOLE_0:def 3;
    end;
    hence thesis by A2,Def1;
  end;
  meet UNION({X},SFY) c= X \/ meet SFY
  proof
    let x;
    assume
A7: x in meet UNION({X},SFY);
    assume
A8: not x in X \/ meet SFY;
    then
A9: not x in X by XBOOLE_0:def 3;
    not x in meet SFY by A8,XBOOLE_0:def 3;
    then consider Z such that
A10: Z in SFY and
A11: not x in Z by A1,Def1;
    X in {X} by TARSKI:def 1;
    then X \/ Z in UNION({X},SFY) by A10,Def4;
    then x in X \/ Z by A7,Def1;
    hence contradiction by A9,A11,XBOOLE_0:def 3;
  end;
  hence thesis by A3,XBOOLE_0:def 10;
end;

theorem
  X /\ union SFY = union INTERSECTION({X},SFY)
proof
A1: union INTERSECTION({X},SFY) c= X /\ union SFY
  proof
    let x;
    assume x in union INTERSECTION({X},SFY);
    then consider Z such that
A2: x in Z and
A3: Z in INTERSECTION({X},SFY) by TARSKI:def 4;
    consider X1,X2 be set such that
A4: X1 in {X} and
A5: X2 in SFY and
A6: Z = X1 /\ X2 by A3,Def5;
    x in X2 by A2,A6,XBOOLE_0:def 4;
    then
A7: x in union SFY by A5,TARSKI:def 4;
    X1 = X by A4,TARSKI:def 1;
    then x in X by A2,A6,XBOOLE_0:def 4;
    hence thesis by A7,XBOOLE_0:def 4;
  end;
  X /\ union SFY c= union INTERSECTION({X},SFY)
  proof
    let x;
    assume
A8: x in X /\ union SFY;
    then x in union SFY by XBOOLE_0:def 4;
    then consider Y such that
A9: x in Y and
A10: Y in SFY by TARSKI:def 4;
    x in X by A8,XBOOLE_0:def 4;
    then
A11: x in X /\ Y by A9,XBOOLE_0:def 4;
    X in {X} by TARSKI:def 1;
    then X /\ Y in INTERSECTION({X},SFY) by A10,Def5;
    hence thesis by A11,TARSKI:def 4;
  end;
  hence thesis by A1,XBOOLE_0:def 10;
end;

theorem
  SFY <> {} implies X \ union SFY = meet DIFFERENCE({X},SFY)
proof
  set y = the Element of SFY;
A1: X in {X} by TARSKI:def 1;
  assume SFY <> {};
  then
A2: X \ y in DIFFERENCE({X},SFY) by A1,Def6;
A3: meet DIFFERENCE({X},SFY) c= X \ union SFY
  proof
    let x;
    assume
A4: x in meet DIFFERENCE({X},SFY);
    for Z st Z in SFY holds not x in Z
    proof
      let Z;
      assume Z in SFY;
      then X \ Z in DIFFERENCE({X},SFY) by A1,Def6;
      then x in X \ Z by A4,Def1;
      hence thesis by XBOOLE_0:def 5;
    end;
    then for Z st x in Z holds not Z in SFY;
    then
A5: not x in union SFY by TARSKI:def 4;
    x in X \ y by A2,A4,Def1;
    hence thesis by A5,XBOOLE_0:def 5;
  end;
  X \ union SFY c= meet DIFFERENCE({X},SFY)
  proof
    let x;
    assume x in X \ union SFY;
    then
A6: x in X & not x in union SFY by XBOOLE_0:def 5;
    for Z st Z in DIFFERENCE({X},SFY) holds x in Z
    proof
      let Z;
      assume Z in DIFFERENCE({X},SFY);
      then consider Z1,Z2 such that
A7:   Z1 in {X} & Z2 in SFY and
A8:   Z = Z1 \ Z2 by Def6;
      x in Z1 & not x in Z2 by A6,A7,TARSKI:def 1,def 4;
      hence thesis by A8,XBOOLE_0:def 5;
    end;
    hence thesis by A2,Def1;
  end;
  hence thesis by A3,XBOOLE_0:def 10;
end;

theorem
  SFY <> {} implies X \ meet SFY = union DIFFERENCE ({X},SFY)
proof
A1: X in {X} by TARSKI:def 1;
A2: union DIFFERENCE({X},SFY) c= X \ meet SFY
  proof
    let x;
    assume x in union DIFFERENCE({X},SFY);
    then consider Z such that
A3: x in Z and
A4: Z in DIFFERENCE({X},SFY) by TARSKI:def 4;
    consider Z1,Z2 such that
A5: Z1 in {X} and
A6: Z2 in SFY and
A7: Z = Z1 \ Z2 by A4,Def6;
    not x in Z2 by A3,A7,XBOOLE_0:def 5;
    then
A8: not x in meet SFY by A6,Def1;
    Z1 = X by A5,TARSKI:def 1;
    hence thesis by A3,A7,A8,XBOOLE_0:def 5;
  end;
  assume
A9: SFY <> {};
  X \ meet SFY c= union DIFFERENCE({X},SFY)
  proof
    let x;
    assume
A10: x in X \ meet SFY;
    then not x in meet SFY by XBOOLE_0:def 5;
    then consider Z such that
A11: Z in SFY and
A12: not x in Z by A9,Def1;
A13: x in X \ Z by A10,A12,XBOOLE_0:def 5;
    X \ Z in DIFFERENCE({X},SFY) by A1,A11,Def6;
    hence thesis by A13,TARSKI:def 4;
  end;
  hence thesis by A2,XBOOLE_0:def 10;
end;

theorem
  union INTERSECTION (SFX,SFY) = union SFX /\ union SFY
proof
  thus union INTERSECTION(SFX,SFY) c= union SFX /\ union SFY
  proof
    let x;
    assume x in union INTERSECTION(SFX,SFY);
    then consider Z such that
A1: x in Z and
A2: Z in INTERSECTION(SFX,SFY) by TARSKI:def 4;
    consider X,Y such that
A3: X in SFX and
A4: Y in SFY and
A5: Z = X /\ Y by A2,Def5;
    x in Y by A1,A5,XBOOLE_0:def 4;
    then
A6: x in union SFY by A4,TARSKI:def 4;
    x in X by A1,A5,XBOOLE_0:def 4;
    then x in union SFX by A3,TARSKI:def 4;
    hence thesis by A6,XBOOLE_0:def 4;
  end;
  let x be set;
  assume
A7: x in union SFX /\ union SFY;
  then x in union SFX by XBOOLE_0:def 4;
  then consider X0 being set such that
A8: x in X0 & X0 in SFX by TARSKI:def 4;
  x in union SFY by A7,XBOOLE_0:def 4;
  then consider Y0 being set such that
A9: x in Y0 & Y0 in SFY by TARSKI:def 4;
  X0 /\ Y0 in INTERSECTION(SFX,SFY) & x in X0 /\ Y0 by A8,A9,Def5,
XBOOLE_0:def 4;
  hence thesis by TARSKI:def 4;
end;

theorem
  SFX <> {} & SFY <> {} implies meet SFX \/ meet SFY c= meet UNION (SFX, SFY)
proof
  set y = the Element of SFX;
  set z = the Element of SFY;
  assume SFX <> {} & SFY <> {};
  then
A1: y \/ z in UNION(SFX,SFY) by Def4;
  let x;
  assume x in meet SFX \/ meet SFY;
  then
A2: x in meet SFX or x in meet SFY by XBOOLE_0:def 3;
  for Z st Z in UNION(SFX,SFY) holds x in Z
  proof
    let Z;
    assume Z in UNION(SFX,SFY);
    then consider X,Y such that
A3: X in SFX & Y in SFY and
A4: Z = X \/ Y by Def4;
    x in X or x in Y by A2,A3,Def1;
    hence thesis by A4,XBOOLE_0:def 3;
  end;
  hence thesis by A1,Def1;
end;

theorem
  meet DIFFERENCE (SFX,SFY) c= meet SFX \ meet SFY
proof
  per cases;
  suppose
A1: SFX = {} or SFY = {};
    now
      assume DIFFERENCE (SFX,SFY) <> {};
      then consider e being set such that
A2:   e in DIFFERENCE (SFX,SFY) by XBOOLE_0:def 1;
      ex X,Y st X in SFX & Y in SFY & e = X \ Y by A2,Def6;
      hence contradiction by A1;
    end;
    then meet DIFFERENCE (SFX,SFY) = {} by Def1;
    hence thesis by XBOOLE_1:2;
  end;
  suppose
A3: SFX <> {} & SFY <> {};
    set z = the Element of SFX;
    set y = the Element of SFY;
    let x such that
A4: x in meet DIFFERENCE (SFX,SFY);
    for Z st Z in SFX holds x in Z
    proof
      let Z;
      assume Z in SFX;
      then Z \ y in DIFFERENCE (SFX,SFY) by A3,Def6;
      then x in Z \ y by A4,Def1;
      hence thesis;
    end;
    then
A5: x in meet SFX by A3,Def1;
    z \ y in DIFFERENCE(SFX,SFY) by A3,Def6;
    then x in z \ y by A4,Def1;
    then not x in y by XBOOLE_0:def 5;
    then not x in meet SFY by A3,Def1;
    hence thesis by A5,XBOOLE_0:def 5;
  end;
end;



definition
  let D be set;
  mode Subset-Family of D is Subset of bool D;
end;

reserve F,G for Subset-Family of D;
reserve P for Subset of D;

definition
  let D,F;
  redefine func union F -> Subset of D;
  coherence
  proof
    union F c= D
    proof
      let x;
      assume x in union F;
      then ex Z st x in Z & Z in F by TARSKI:def 4;
      hence thesis;
    end;
    hence thesis;
  end;
end;

definition
  let D,F;
  redefine func meet F -> Subset of D;
  coherence
  proof
    meet F c= D
    proof
A1:   meet F c= union F by Th2;
      let x;
      assume x in meet F;
      then x in union F by A1;
      hence thesis;
    end;
    hence thesis;
  end;
end;

theorem Th31:
  (for P holds P in F iff P in G) implies F=G
proof
  assume
A1: for P holds P in F iff P in G;
  thus F c= G
  proof
    let x;
    assume x in F;
    hence thesis by A1;
  end;
  let x;
  assume x in G;
  hence thesis by A1;
end;

definition
  let D,F;

  func COMPLEMENT(F) -> Subset-Family of D means
  :Def7:
  for P being Subset of D holds P in it iff P` in F;
  existence
  proof
    defpred X[Subset of D] means $1` in F;
    thus ex G being Subset-Family of D st for P being Subset of D holds P in G
    iff X[P] from SUBSET_1:sch 3;
  end;
  uniqueness
  proof
    let G,H be Subset-Family of D;
    assume that
A1: for P holds P in G iff P` in F and
A2: for P holds P in H iff P` in F;
    now
      let P;
      P in G iff P` in F by A1;
      hence P in G iff P in H by A2;
    end;
    hence thesis by Th31;
  end;
  involutiveness
  proof
    let F,G such that
A3: for P being Subset of D holds P in F iff P` in G;
    let P be Subset of D;
    P`` = P;
    hence thesis by A3;
  end;
end;

theorem Th32:
  F <> {} implies COMPLEMENT(F) <> {}
proof
  set X = the Element of F;
  assume
A1: F <> {};
  then reconsider X as Subset of D by TARSKI:def 3;
  X`` = X;
  hence thesis by A1,Def7;
end;

theorem
  F <> {} implies [#] D \ union F = meet (COMPLEMENT(F))
proof
A1: for x st x in D holds (for X st X in F holds not x in X) iff for Y st Y
  in COMPLEMENT(F) holds x in Y
  proof
    let x such that
A2: x in D;
    thus (for X st X in F holds not x in X) implies for Y st Y in COMPLEMENT(F
    ) holds x in Y
    proof
      assume
A3:   for X st X in F holds not x in X;
      let Y;
      assume
A4:   Y in COMPLEMENT(F);
      then reconsider Y as Subset of D;
      Y` in F by A4,Def7;
      then not x in Y` by A3;
      hence thesis by A2,XBOOLE_0:def 5;
    end;
    assume
A5: for Y st Y in COMPLEMENT(F) holds x in Y;
    let X;
    assume
A6: X in F;
    then reconsider X as Subset of D;
    X`` = X;
    then X` in COMPLEMENT(F) by A6,Def7;
    then x in X` by A5;
    hence thesis by XBOOLE_0:def 5;
  end;
  assume F <> {};
  then
A7: COMPLEMENT(F) <> {} by Th32;
A8: [#] D \ union F c= meet COMPLEMENT(F)
  proof
    let x;
    assume
A9: x in [#] D \ union F;
    then not x in union F by XBOOLE_0:def 5;
    then for X st X in F holds not x in X by TARSKI:def 4;
    then for Y st Y in COMPLEMENT(F) holds x in Y by A1,A9;
    hence thesis by A7,Def1;
  end;
  meet COMPLEMENT(F) c= [#] D \ union F
  proof
    let x;
    assume
A10: x in meet COMPLEMENT(F);
    then for X holds X in COMPLEMENT(F) implies x in X by Def1;
    then for Y st x in Y holds not Y in F by A1;
    then not x in union F by TARSKI:def 4;
    hence thesis by A10,XBOOLE_0:def 5;
  end;
  hence thesis by A8,XBOOLE_0:def 10;
end;

theorem
  F <> {} implies union COMPLEMENT(F) = [#] D \ meet F
proof
  assume
A1: F <> {};
A2: [#] D \ meet F c= union COMPLEMENT(F)
  proof
    let x;
    assume
A3: x in [#] D \ meet F;
    then not x in meet F by XBOOLE_0:def 5;
    then consider X such that
A4: X in F and
A5: not x in X by A1,Def1;
    reconsider X as Subset of D by A4;
    reconsider XX=X` as set;
A6: X`` = X;
    ex Y st x in Y & Y in COMPLEMENT(F)
    proof
      take XX;
      thus thesis by A3,A4,A5,A6,Def7,XBOOLE_0:def 5;
    end;
    hence thesis by TARSKI:def 4;
  end;
  union COMPLEMENT(F) c= [#] D \ meet F
  proof
    let x;
    assume
A7: x in union COMPLEMENT(F);
    then consider X such that
A8: x in X and
A9: X in COMPLEMENT(F) by TARSKI:def 4;
    reconsider X as Subset of D by A9;
    reconsider XX=X` as set;
    ex Y st Y in F & not x in Y
    proof
      take Y = XX;
      thus Y in F by A9,Def7;
      thus thesis by A8,XBOOLE_0:def 5;
    end;
    then not x in meet F by Def1;
    hence thesis by A7,XBOOLE_0:def 5;
  end;
  hence thesis by A2,XBOOLE_0:def 10;
end;

begin 


theorem
  for X being set, F being Subset-Family of X for P being Subset of X
  holds P` in COMPLEMENT F iff P in F
proof
  let X be set, F be Subset-Family of X;
  let P be Subset of X;
  P = P``;
  hence thesis by Def7;
end;

theorem Th36:
  for X being set, F,G being Subset-Family of X st COMPLEMENT F c=
  COMPLEMENT G holds F c= G
proof
  let X be set, F,G be Subset-Family of X such that
A1: COMPLEMENT F c= COMPLEMENT G;
  let x be set;
  assume
A2: x in F;
  then reconsider A = x as Subset of X;
  A in COMPLEMENT COMPLEMENT F by A2;
  then A` in COMPLEMENT F by Def7;
  then A`` in G by A1,Def7;
  hence thesis;
end;

theorem
  for X being set, F,G being Subset-Family of X holds COMPLEMENT F c= G
  iff F c= COMPLEMENT G
proof
  let X be set, F,G be Subset-Family of X;
  hereby
    assume COMPLEMENT F c= G;
    then COMPLEMENT F c= COMPLEMENT COMPLEMENT G;
    hence F c= COMPLEMENT G by Th36;
  end;
  assume F c= COMPLEMENT G;
  then COMPLEMENT COMPLEMENT F c= COMPLEMENT G;
  hence thesis by Th36;
end;

theorem
  for X being set, F,G being Subset-Family of X st COMPLEMENT F =
  COMPLEMENT G holds F = G
proof
  let X be set, F,G be Subset-Family of X;
  assume COMPLEMENT F = COMPLEMENT G;
  hence F = COMPLEMENT COMPLEMENT G .= G;
end;

theorem
  for X being set, F,G being Subset-Family of X holds COMPLEMENT(F \/ G)
  = COMPLEMENT F \/ COMPLEMENT G
proof
  let X be set, F,G be Subset-Family of X;
  for P being Subset of X holds P in COMPLEMENT F \/ COMPLEMENT G iff P`
  in F \/ G
  proof
    let P be Subset of X;
    P in COMPLEMENT F or P in COMPLEMENT G iff P` in F or P` in G by Def7;
    hence thesis by XBOOLE_0:def 3;
  end;
  hence thesis by Def7;
end;

theorem
  for X being set, F being Subset-Family of X st F = {X} holds
  COMPLEMENT F = {{}}
proof
  let X be set, F be Subset-Family of X such that
A1: F = {X};
  {} c= X by XBOOLE_1:2;
  then reconsider G = {{}} as Subset-Family of X by ZFMISC_1:31;
  reconsider G as Subset-Family of X;
  for P being Subset of X holds P in G iff P` in F
  proof
    let P be Subset of X;
    hereby
      assume P in G;
      then P = {}X by TARSKI:def 1;
      hence P` in F by A1,TARSKI:def 1;
    end;
    assume P` in F;
    then
A2: P` = [#]X by A1,TARSKI:def 1;
    P = P`` .= {} by A2,XBOOLE_1:37;
    hence thesis by TARSKI:def 1;
  end;
  hence thesis by Def7;
end;

registration
  let X be set, F be empty Subset-Family of X;
  cluster COMPLEMENT F -> empty;
  coherence
  proof
    assume COMPLEMENT F is not empty;
    then ex x being Subset of X st x in COMPLEMENT F by SUBSET_1:4;
    hence contradiction by Def7;
  end;
end;



definition
  let IT be set;
  attr IT is with_non-empty_elements means
  :Def8:
  not {} in IT;
end;

registration
  cluster non empty with_non-empty_elements for set;
  existence
  proof
    take {{{}}};
    thus {{{}}} is non empty;
    assume {} in {{{}}};
    hence contradiction by TARSKI:def 1;
  end;
end;

registration
  let A be non empty set;
  cluster { A } -> with_non-empty_elements;
  coherence
  proof
    assume {} in { A };
    hence contradiction by TARSKI:def 1;
  end;
  let B be non empty set;
  cluster { A, B } -> with_non-empty_elements;
  coherence
  proof
    assume {} in { A,B };
    hence contradiction by TARSKI:def 2;
  end;
  let C be non empty set;
  cluster { A, B, C } -> with_non-empty_elements;
  coherence
  proof
    assume {} in { A, B, C };
    hence contradiction by ENUMSET1:def 1;
  end;
  let D be non empty set;
  cluster { A, B, C, D } -> with_non-empty_elements;
  coherence
  proof
    assume {} in { A, B, C, D };
    hence contradiction by ENUMSET1:def 2;
  end;
  let E be non empty set;
  cluster { A, B, C, D, E } -> with_non-empty_elements;
  coherence
  proof
    assume {} in { A, B, C, D, E };
    hence contradiction by ENUMSET1:def 3;
  end;
  let F be non empty set;
  cluster { A, B, C, D, E, F } -> with_non-empty_elements;
  coherence
  proof
    assume {} in { A, B, C, D, E, F };
    hence contradiction by ENUMSET1:def 4;
  end;
  let G be non empty set;
  cluster { A, B, C, D, E, F, G } -> with_non-empty_elements;
  coherence
  proof
    assume {} in { A, B, C, D, E, F, G };
    hence contradiction by ENUMSET1:def 5;
  end;
  let H be non empty set;
  cluster { A, B, C, D, E, F, G, H } -> with_non-empty_elements;
  coherence
  proof
    assume {} in { A, B, C, D, E, F, G, H };
    hence contradiction by ENUMSET1:def 6;
  end;
  let I be non empty set;
  cluster { A, B, C, D, E, F, G, H, I } -> with_non-empty_elements;
  coherence
  proof
    assume {} in { A, B, C, D, E, F, G, H, I };
    hence contradiction by ENUMSET1:def 7;
  end;
  let J be non empty set;
  cluster { A, B, C, D, E, F, G, H, I, J } -> with_non-empty_elements;
  coherence
  proof
    assume {} in { A, B, C, D, E, F, G, H, I, J };
    hence contradiction by ENUMSET1:def 8;
  end;
end;

registration
  let A,B be with_non-empty_elements set;
  cluster A \/ B -> with_non-empty_elements;
  coherence
  proof
    ( not {} in A)& not {} in B by Def8;
    then not {} in A \/ B by XBOOLE_0:def 3;
    hence thesis by Def8;
  end;
end;



theorem
  union Y c= Z & X in Y implies X c= Z
proof
  assume that
A1: union Y c= Z and
A2: X in Y;
  thus X c= Z
  proof
    let x be set;
    assume x in X;
    then x in union Y by A2,TARSKI:def 4;
    hence thesis by A1;
  end;
end;



theorem
  for A,B,X being set holds ( X c= union (A \/ B) & for Y being set st Y
  in B holds Y misses X ) implies X c= union A
proof
  let A,B,X be set;
  assume X c= union (A \/ B);
  then X c= union (A \/ B) /\ X by XBOOLE_1:19;
  then X c= (union A \/ union B) /\ X by ZFMISC_1:78;
  then
A1: X c= (union A /\ X) \/ (union B /\ X) by XBOOLE_1:23;
  assume for Y st Y in B holds Y misses X;
  then union B misses X by ZFMISC_1:80;
  then
A2: union B /\ X = {} by XBOOLE_0:def 7;
  union A /\ X c= union A by XBOOLE_1:17;
  hence thesis by A2,A1,XBOOLE_1:1;
end;



definition
  let M be set;
  let B be Subset-Family of M;
  func Intersect B -> Subset of M equals
  :Def9:
  meet B if B <> {} otherwise M;
  coherence
  proof
    M c= M;
    hence thesis;
  end;
  consistency;
end;

theorem Th43:
  for X, x being set, R being Subset-Family of X st x in X holds x
  in Intersect R iff for Y being set st Y in R holds x in Y
proof
  let X, x be set, R be Subset-Family of X;
  assume
A1: x in X;
  hereby
    assume
A2: x in Intersect R;
    let Y be set;
    assume
A3: Y in R;
    then Intersect R = meet R by Def9;
    hence x in Y by A2,A3,Def1;
  end;
  assume
A4: for Y being set st Y in R holds x in Y;
  per cases;
  suppose
A5: R <> {};
    then x in meet R by A4,Def1;
    hence thesis by A5,Def9;
  end;
  suppose
    R = {};
    hence thesis by A1,Def9;
  end;
end;

theorem
  for X being set for H, J being Subset-Family of X st H c= J holds
  Intersect J c= Intersect H
proof
  let X be set;
  let H, J be Subset-Family of X such that
A1: H c= J;
  let x be set;
  assume
A2: x in Intersect J;
  then for Y be set st Y in H holds x in Y by A1,Th43;
  hence thesis by A2,Th43;
end;



registration
  let X be non empty with_non-empty_elements set;
  cluster -> non empty for Element of X;
  coherence by Def8;
end;



reserve E for set;

definition
  let E;
  attr E is empty-membered means
  :Def10:
  not ex x being non empty set st x in E;
end;

notation
  let E;
  antonym E is with_non-empty_element for E is empty-membered;
end;

registration
  cluster with_non-empty_element for set;
  existence
  proof
    take {{{}}}, {{}};
    thus thesis by TARSKI:def 1;
  end;
end;

registration
  cluster with_non-empty_element -> non empty for set;
  coherence
  proof
    let X be set;
    assume X is with_non-empty_element;
    then ex x being non empty set st x in X by Def10;
    hence thesis;
  end;
end;



registration
  let X be with_non-empty_element set;
  cluster non empty for Element of X;
  existence
  proof
    ex x being non empty set st x in X by Def10;
    hence thesis;
  end;
end;



registration
  let D be non empty with_non-empty_elements set;
  cluster union D -> non empty;
  coherence
  proof
    set d = the Element of D;
    d c= union D by ZFMISC_1:74;
    hence thesis;
  end;
end;



registration
  cluster non empty with_non-empty_elements -> with_non-empty_element for set;
  coherence
  proof
    let X be set;
    assume that
A1: X is non empty and
A2: X is with_non-empty_elements;
    consider x being Element of X such that
A3: x in [#]X by A1,SUBSET_1:4;
    reconsider x as non empty set by A2,A3;
    take x;
    thus thesis by A3;
  end;
end;



definition
  let X be set;
  mode Cover of X means
    :Def11:
    X c= union it;
  existence
  proof
    take {X};
    thus thesis by ZFMISC_1:25;
  end;
end;

theorem
  for X being set, F being Subset-Family of X holds F is Cover of X iff
  union F = X
proof
  let X be set, F be Subset-Family of X;
  thus F is Cover of X implies union F = X
  proof
    assume
A1: F is Cover of X;
    thus union F c= X;
    thus thesis by A1,Def11;
  end;
  thus thesis by Def11;
end;



theorem Th46:
  {{}} is Subset-Family of X
proof
  {} c= X by XBOOLE_1:2;
  hence thesis by ZFMISC_1:31;
end;



definition
  let X be set;
  let F be Subset-Family of X;
  attr F is with_proper_subsets means
  :Def12:
  not X in F;
end;

theorem
  for TS being set, F, G being Subset-Family of TS st F is
  with_proper_subsets & G c= F holds G is with_proper_subsets
proof
  let TS be set, F, G being Subset-Family of TS;
  assume
A1: F is with_proper_subsets & G c= F;
  assume not G is with_proper_subsets;
  then TS in G by Def12;
  hence thesis by A1,Def12;
end;

registration
  let TS be non empty set;
  cluster with_proper_subsets for Subset-Family of TS;
  existence
  proof
    reconsider F = { {} } as Subset-Family of TS by Th46;
    take F;
    assume TS in F;
    hence thesis by TARSKI:def 1;
  end;
end;

theorem
  for TS being non empty set, A, B being with_proper_subsets
  Subset-Family of TS holds A \/ B is with_proper_subsets
proof
  let TS be non empty set, A, B be with_proper_subsets Subset-Family of TS;
  assume TS in A \/ B;
  then TS in A or TS in B by XBOOLE_0:def 3;
  hence thesis by Def12;
end;



registration
  cluster non trivial -> with_non-empty_element for set;
  coherence
  proof
    let W be set;
    assume W is non trivial;
    then consider w1,w2 being set such that
A1: w1 in W & w2 in W and
A2: w1 <> w2 by ZFMISC_1:def 10;
    w1 <> {} or w2 <> {} by A2;
    hence thesis by A1,Def10;
  end;
end;



definition
  let X be set;
  redefine func bool X -> Subset-Family of X;
  coherence by ZFMISC_1:def 1;
end;



theorem
  for A being non empty set, b being set st A <> {b} ex a being
  Element of A st a <> b
proof
  let A be non empty set, b be set such that
A1: A <> {b};
  assume
A2: for a being Element of A holds a = b;
  now
    set a0 = the Element of A;
    let a be set;
    thus a in A implies a = b by A2;
    assume
A3: a = b;
    a0 = b by A2;
    hence a in A by A3;
  end;
  hence contradiction by A1,TARSKI:def 1;
end;















begin

reserve A,X,X1,X2,Y,Y1,Y2,a,b,c,d,x,y,z for set;

definition
  let IT be set;
  attr IT is Relation-like means
  :Def1:
  x in IT implies ex y,z st x = [y,z];
end;

registration
  cluster empty -> Relation-like for set;
  coherence
  proof
    let X be set;
    assume X is empty;
    hence for p being set st p in X ex x,y st [x,y] = p;
  end;
end;

definition
  mode Relation is Relation-like set;
end;

reserve P,P1,P2,Q,R,S for Relation;

registration
  let R be Relation;
  cluster -> Relation-like for Subset of R;
  coherence
  proof
    let A be Subset of R;
    let x;
    thus thesis by Def1;
  end;
end;

scheme
  RelExistence{A,B() -> set, P[set,set]}: ex R being Relation st for x,y holds
  [x,y] in R iff x in A() & y in B() & P[x,y]
  proof
  ex R being Relation st for p being set holds p in R iff p in [:A(),B():]
  & ex x,y st p=[x,y] & P[x,y]
  proof
    defpred Q[set] means ex x,y st $1=[x,y] & P[x,y];
    consider B being set such that
A1: for p being set holds p in B iff p in [:A(),B():] & Q[p] from
    XBOOLE_0:sch 1;
    for p being set st p in B holds ex x,y st p = [x,y]
    proof
      let p be set;
      assume p in B;
      then ex x,y st p=[x,y] & P[x,y] by A1;
      hence thesis;
    end;
    then B is Relation by Def1;
    hence thesis by A1;
  end;
  then consider R being Relation such that
A2: for p being set holds p in R iff p in [:A(),B():] & ex x,y st p=[x,y] &
    P[x,y];
  take R;
  let x,y;
  thus [x,y] in R implies x in A() & y in B() & P[x,y]
  proof
    assume
A3: [x,y] in R;
    then consider xx,yy being set such that
A4: [x,y]=[xx,yy] and
A5: P[xx,yy] by A2;
A6: [x,y] in [:A(),B():] by A2,A3;
    x=xx by A4,XTUPLE_0:1;
    hence thesis by A4,A5,A6,XTUPLE_0:1,ZFMISC_1:87;
  end;
  thus x in A() & y in B() & P[x,y] implies [x,y] in R
  proof
    assume that
A7: x in A() & y in B() and
A8: P[x,y];
    [x,y] in [:A(),B():] by A7,ZFMISC_1:87;
    hence thesis by A2,A8;
  end;
end;

definition
  let P,R;
  redefine pred P = R means
  for a,b holds [a,b] in P iff [a,b] in R;
  compatibility
  proof
    thus P = R implies for a,b holds [a,b] in P iff [a,b] in R;
    assume
A1: for a,b holds [a,b] in P iff [a,b] in R;
    thus P c= R
    proof
      let x;
      assume
A2:   x in P;
      then ex y,z st x = [y,z] by Def1;
      hence x in R by A1,A2;
    end;
    let x;
    assume
A3: x in R;
    then ex y,z st x = [y,z] by Def1;
    hence x in P by A1,A3;
  end;
end;

registration
  let P,X;
  cluster P /\ X -> Relation-like;
  coherence
  proof
    P /\ X c= P by XBOOLE_1:17;
    hence thesis;
  end;
  cluster P \ X -> Relation-like;
  coherence;
end;

registration
  let P,R;
  cluster P \/ R -> Relation-like;
  coherence
  proof
    let x;
A1: x in R implies ex y,z st x = [y,z] by Def1;
    x in P implies ex y,z st x = [y,z] by Def1;
    hence x in P \/ R implies ex y,z st x = [y,z] by A1,XBOOLE_0:def 3;
  end;
end;

registration
  let R,S be Relation;
  cluster R \+\ S -> Relation-like;
  coherence;
end;

registration
  let a, b be set;
  cluster {[a,b]} -> Relation-like;
  coherence
  proof
    let x;
    assume x in {[a,b]};
    then x = [a,b] by TARSKI:def 1;
    hence thesis;
  end;
  cluster [:a,b:] -> Relation-like;
  coherence
  proof
    let z;
    assume z in [:a,b:];
    then ex x,y st x in a & y in b & [x,y]=z by ZFMISC_1:def 2;
    hence thesis;
  end;
end;

registration
  let a, b, c, d be set;
  cluster {[a,b],[c,d]} -> Relation-like;
  coherence
  proof
    {[a,b],[c,d]} = {[a,b]} \/ {[c,d]} by ENUMSET1:1;
    hence thesis;
  end;
end;

definition
  let P,A;
  redefine pred P c= A means
  for a,b holds [a,b] in P implies [a,b] in A;
  compatibility
  proof
    thus P c= A implies for a,b holds [a,b] in P implies [a,b] in A;
    assume
A1: for a,b holds [a,b] in P implies [a,b] in A;
    let x;
    assume
A2: x in P;
    then ex y,z st x = [y,z] by Def1;
    hence thesis by A1,A2;
  end;
end;



notation
  let R be Relation;
  synonym dom R for proj1 R;
end;

theorem Th1:
  dom(P \/ R) = dom P \/ dom R by XTUPLE_0:23;

theorem Th2:
  dom(P /\ R) c= dom P /\ dom R by XTUPLE_0:24;

theorem
  dom P \ dom R c= dom(P \ R) by XTUPLE_0:25;



notation
  let R be Relation;
  synonym rng R for proj2 R;
end;

canceled 3;

theorem Th7:
  R c= [:dom R, rng R:]
proof
  let y,z;
  assume [y,z] in R;
  then y in dom R & z in rng R by XTUPLE_0:def 12,def 13;
  hence thesis by ZFMISC_1:87;
end;

theorem
  R /\ [:dom R, rng R:] = R by Th7,XBOOLE_1:28;

theorem Th9:
  R = {[x,y]} implies dom R = {x} & rng R = {y}
proof
  assume
A1: R = {[x,y]};
  z in dom R iff z in {x}
  proof
    thus z in dom R implies z in {x}
    proof
      assume z in dom R;
      then consider b such that
A2:   [z,b] in R by XTUPLE_0:def 12;
      [z,b] = [x,y] by A1,A2,TARSKI:def 1;
      then z=x by XTUPLE_0:1;
      hence thesis by TARSKI:def 1;
    end;
    assume z in {x};
    then z=x by TARSKI:def 1;
    then [z,y] in R by A1,TARSKI:def 1;
    hence thesis by XTUPLE_0:def 12;
  end;
  hence dom R = {x} by TARSKI:1;
  z in rng R iff z in {y}
  proof
    thus z in rng R implies z in {y}
    proof
      assume z in rng R;
      then consider a such that
A3:   [a,z] in R by XTUPLE_0:def 13;
      [a,z] = [x,y] by A1,A3,TARSKI:def 1;
      then z = y by XTUPLE_0:1;
      hence thesis by TARSKI:def 1;
    end;
    assume z in {y};
    then z = y by TARSKI:def 1;
    then [x,z] in R by A1,TARSKI:def 1;
    hence thesis by XTUPLE_0:def 13;
  end;
  hence thesis by TARSKI:1;
end;

theorem
  R = {[a,b],[x,y]} implies dom R = {a,x} & rng R = {b,y}
proof
  assume
A1: R = {[a,b],[x,y]};
  thus dom R = {a,x}
  proof
    thus dom R c= {a,x}
    proof
      let z;
      assume z in dom R;
      then consider c such that
A2:   [z,c] in R by XTUPLE_0:def 12;
      [z,c] = [a,b] or [z,c] = [x,y] by A1,A2,TARSKI:def 2;
      then z = a or z = x by XTUPLE_0:1;
      hence thesis by TARSKI:def 2;
    end;
    let z;
    assume z in {a,x};
    then z = a or z = x by TARSKI:def 2;
    then [z,b] in R or [z,y] in R by A1,TARSKI:def 2;
    hence thesis by XTUPLE_0:def 12;
  end;
  thus rng R c= {b,y}
  proof
    let z;
    assume z in rng R;
    then consider d such that
A3: [d,z] in R by XTUPLE_0:def 13;
    [d,z] = [a,b] or [d,z] = [x,y] by A1,A3,TARSKI:def 2;
    then z = b or z = y by XTUPLE_0:1;
    hence thesis by TARSKI:def 2;
  end;
  let z;
  assume z in {b,y};
  then z = b or z = y by TARSKI:def 2;
  then [a,z] in R or [x,z] in R by A1,TARSKI:def 2;
  hence thesis by XTUPLE_0:def 13;
end;

theorem Th11:
  P c= R implies dom P c= dom R & rng P c= rng R by XTUPLE_0:8,9;

theorem Th12:
  rng(P \/ R) = rng P \/ rng R by XTUPLE_0:27;

theorem Th13:
  rng(P /\ R) c= rng P /\ rng R by XTUPLE_0:28;

theorem
  rng P \ rng R c= rng(P \ R) by XTUPLE_0:29;



definition
 canceled 2;
  let R;
  func field R -> set equals
  dom R \/ rng R;
  coherence;
end;

theorem
  [a,b] in R implies a in field R & b in field R
proof
  assume
A1: [a,b] in R;
  then a in dom R by XTUPLE_0:def 12;
  hence a in field R by XBOOLE_0:def 3;
  b in rng R by A1,XTUPLE_0:def 13;
  hence thesis by XBOOLE_0:def 3;
end;

theorem
  P c= R implies field P c= field R
proof
  assume P c= R;
  then dom P c= dom R & rng P c= rng R by Th11;
  hence thesis by XBOOLE_1:13;
end;

theorem Th17:
  field {[x,y]} = {x,y}
proof
  set R = {[x,y]};
  dom R = {x} & rng R = {y} by Th9;
  hence thesis by ENUMSET1:1;
end;

theorem
  field(P \/ R) = field P \/ field R
proof
  thus field(P \/ R) = dom P \/ dom R \/ rng(P \/ R) by Th1
    .= dom P \/ dom R \/ (rng P \/ rng R) by Th12
    .= dom P \/ dom R \/ rng P \/ rng R by XBOOLE_1:4
    .= field P \/ dom R \/ rng R by XBOOLE_1:4
    .= field P \/ field R by XBOOLE_1:4;
end;

theorem
  field(P /\ R) c= field P /\ field R
proof
  let x;
  assume x in field(P /\ R);
  then
A1: x in dom(P /\ R) or x in rng(P /\ R) by XBOOLE_0:def 3;
  x in dom P /\ dom R or x in rng P /\ rng R implies (x in dom P or x in
  rng P) & (x in dom R or x in rng R) by XBOOLE_0:def 4;
  then
A2: x in dom P /\ dom R or x in rng P /\ rng R implies x in dom P \/ rng P &
  x in dom R \/ rng R by XBOOLE_0:def 3;
  dom(P /\ R) c= dom P /\ dom R & rng(P /\ R) c= rng P /\ rng R by Th2,Th13;
  hence thesis by A1,A2,XBOOLE_0:def 4;
end;



definition
  let R;
  func R~ -> Relation means
  :Def7:
  [x,y] in it iff [y,x] in R;
  existence
  proof
    defpred Q[set,set] means [$2,$1] in R;
    consider P such that
A1: for x,y holds [x,y] in P iff x in rng R & y in dom R & Q[x,y] from
    RelExistence;
    take P;
    let x,y;
    [y,x] in R implies y in dom R & x in rng R by XTUPLE_0:def 12,def 13;
    hence thesis by A1;
  end;
  uniqueness
  proof
    let P1,P2;
    assume that
A2: [x,y] in P1 iff [y,x] in R and
A3: [x,y] in P2 iff [y,x] in R;
    let x,y;
    [x,y] in P1 iff [y,x] in R by A2;
    hence [x,y] in P1 iff [x,y] in P2 by A3;
  end;
  involutiveness;
end;

theorem Th20:
  rng R = dom(R~) & dom R = rng(R~)
proof
  thus rng R c= dom(R~)
  proof
    let u be set;
    assume u in rng R;
    then consider x such that
A1: [x,u] in R by XTUPLE_0:def 13;
    [u,x] in R~ by A1,Def7;
    hence thesis by XTUPLE_0:def 12;
  end;
  thus dom(R~) c= rng R
  proof
    let u be set;
    assume u in dom(R~);
    then consider x such that
A2: [u,x] in R~ by XTUPLE_0:def 12;
    [x,u] in R by A2,Def7;
    hence thesis by XTUPLE_0:def 13;
  end;
  thus dom R c= rng(R~)
  proof
    let u be set;
    assume u in dom R;
    then consider x such that
A3: [u,x] in R by XTUPLE_0:def 12;
    [x,u] in R~ by A3,Def7;
    hence thesis by XTUPLE_0:def 13;
  end;
  let u be set;
  assume u in rng(R~);
  then consider x such that
A4: [x,u] in R~ by XTUPLE_0:def 13;
  [u,x] in R by A4,Def7;
  hence thesis by XTUPLE_0:def 12;
end;

theorem
  field R = field(R~)
proof
  thus field R = rng(R~) \/ rng R by Th20
    .= field(R~) by Th20;
end;

theorem
  (P /\ R)~ = P~ /\ R~
proof
  let x,y;
  [x,y] in (P /\ R)~ iff [y,x] in P /\ R by Def7;
  then [x,y] in (P /\ R)~ iff [y,x] in P & [y,x] in R by XBOOLE_0:def 4;
  then [x,y] in (P /\ R)~ iff [x,y] in P~ & [x,y] in R~ by Def7;
  hence thesis by XBOOLE_0:def 4;
end;

theorem
  (P \/ R)~ = P~ \/ R~
proof
  let x,y;
  [x,y] in (P \/ R)~ iff [y,x] in P \/ R by Def7;
  then [x,y] in (P \/ R)~ iff [y,x] in P or [y,x] in R by XBOOLE_0:def 3;
  then [x,y] in (P \/ R)~ iff [x,y] in P~ or [x,y] in R~ by Def7;
  hence thesis by XBOOLE_0:def 3;
end;

theorem
  (P \ R)~ = P~ \ R~
proof
  let x,y;
  [x,y] in (P \ R)~ iff [y,x] in P \ R by Def7;
  then [x,y] in (P \ R)~ iff [y,x] in P & not [y,x] in R by XBOOLE_0:def 5;
  then [x,y] in (P \ R)~ iff [x,y] in P~ & not [x,y] in R~ by Def7;
  hence thesis by XBOOLE_0:def 5;
end;



definition
  let P,R be set;
  func P(#)R -> Relation means
  :Def8:
  [x,y] in it iff ex z st [x,z] in P & [z,y] in R;
  existence
  proof
    defpred Z[set,set] means ex z st [$1,z] in P & [z,$2] in R;
    consider Q such that
A1: for x,y holds [x,y] in Q iff x in proj1 P & y in proj2 R & Z[x,y] from
    RelExistence;
    take Q;
    let x,y;
    thus [x,y] in Q implies ex z st [x,z] in P & [z,y] in R by A1;
    given z such that
A2: [x,z] in P & [z,y] in R;
    x in proj1 P & y in proj2 R by A2,XTUPLE_0:def 12,def 13;
    hence thesis by A1,A2;
  end;
  uniqueness
  proof
    let P1,P2;
    assume that
A3: [x,y] in P1 iff ex z st [x,z] in P & [z,y] in R and
A4: [x,y] in P2 iff ex z st [x,z] in P & [z,y] in R;
    let x,y;
    [x,y] in P1 iff ex z st [x,z] in P & [z,y] in R by A3;
    hence [x,y] in P1 iff [x,y] in P2 by A4;
  end;
end;

notation
  let P,R;
  synonym P*R for P(#)R;
end;

theorem Th25:
  dom(P*R) c= dom P
proof
  let x;
  assume x in dom(P*R);
  then consider y such that
A1: [x,y] in P*R by XTUPLE_0:def 12;
  ex z st [x,z] in P & [z,y] in R by A1,Def8;
  hence thesis by XTUPLE_0:def 12;
end;

theorem Th26:
  rng(P*R) c= rng R
proof
  let y;
  assume y in rng(P*R);
  then consider x such that
A1: [x,y] in P*R by XTUPLE_0:def 13;
  ex z st [x,z] in P & [z,y] in R by A1,Def8;
  hence thesis by XTUPLE_0:def 13;
end;

theorem
  rng R c= dom P implies dom(R*P) = dom R
proof
  assume
A1: y in rng R implies y in dom P;
  thus dom(R*P) c= dom R by Th25;
  let x;
  assume x in dom R;
  then consider y such that
A2: [x,y] in R by XTUPLE_0:def 12;
  y in rng R by A2,XTUPLE_0:def 13;
  then y in dom P by A1;
  then consider z such that
A3: [y,z] in P by XTUPLE_0:def 12;
  [x,z] in R*P by A2,A3,Def8;
  hence thesis by XTUPLE_0:def 12;
end;

theorem
  dom P c= rng R implies rng(R*P) = rng P
proof
  assume
A1: for y st y in dom P holds y in rng R;
  thus rng(R*P) c= rng P by Th26;
  let z;
  assume z in rng P;
  then consider y such that
A2: [y,z] in P by XTUPLE_0:def 13;
  y in dom P by A2,XTUPLE_0:def 12;
  then y in rng R by A1;
  then consider x such that
A3: [x,y] in R by XTUPLE_0:def 13;
  [x,z] in R*P by A2,A3,Def8;
  hence thesis by XTUPLE_0:def 13;
end;

theorem Th29:
  P c= R implies Q*P c= Q*R
proof
  assume
A1: P c= R;
  let x,y;
  assume [x,y] in Q*P;
  then ex z st [x,z] in Q & [z,y] in P by Def8;
  hence thesis by A1,Def8;
end;

theorem Th30:
  P c= Q implies P*R c= Q*R
proof
  assume
A1: P c= Q;
  let x,y;
  assume [x,y] in P*R;
  then ex z st [x,z] in P & [z,y] in R by Def8;
  hence thesis by A1,Def8;
end;

theorem
  P c= R & Q c= S implies P*Q c= R*S
proof
  assume
A1: P c= R & Q c= S;
  let x,y;
  assume [x,y] in P*Q;
  then ex z st [x,z] in P & [z,y] in Q by Def8;
  hence [x,y] in R*S by A1,Def8;
end;

theorem
  P*(R \/ Q) = (P*R) \/ (P*Q)
  proof
    let x,y;
    hereby
      assume [x,y] in P*(R \/ Q);
      then consider z such that
A1:   [x,z] in P and
A2:   [z,y] in R \/ Q by Def8;
      [z,y] in R or [z,y] in Q by A2,XBOOLE_0:def 3;
      then [x,y] in P*R or [x,y] in P*Q by A1,Def8;
      hence [x,y] in (P*R) \/ (P*Q) by XBOOLE_0:def 3;
    end;
    assume
A3: [x,y] in (P*R) \/ (P*Q);
    per cases by A3,XBOOLE_0:def 3;
    suppose [x,y] in P*Q;
      then consider z such that
A4:   [x,z] in P and
A5:   [z,y] in Q by Def8;
      [z,y] in R \/ Q by A5,XBOOLE_0:def 3;
      hence [x,y] in P*(R \/ Q) by A4,Def8;
    end;
    suppose [x,y] in P*R;
      then consider z such that
A6:   [x,z] in P and
A7:   [z,y] in R by Def8;
      [z,y] in R \/ Q by A7,XBOOLE_0:def 3;
      hence [x,y] in P*(R \/ Q) by A6,Def8;
    end;
  end;

theorem
  P*(R /\ Q) c= (P*R) /\ (P*Q)
proof
  let x,y;
  assume [x,y] in P*(R /\ Q);
  then consider z such that
A1: [x,z] in P and
A2: [z,y] in R /\ Q by Def8;
  [z,y] in Q by A2,XBOOLE_0:def 4;
  then
A3: [x,y] in P*Q by A1,Def8;
  [z,y] in R by A2,XBOOLE_0:def 4;
  then [x,y] in P*R by A1,Def8;
  hence [x,y] in (P*R) /\ (P*Q) by A3,XBOOLE_0:def 4;
end;

theorem
  (P*R) \ (P*Q) c= P*(R \ Q)
proof
  let a,b;
  assume
A1: [a,b] in (P*R) \ (P*Q);
  then consider y such that
A2: [a,y] in P and
A3: [y,b] in R by Def8;
  not [a,b] in P*Q by A1,XBOOLE_0:def 5;
  then not [a,y] in P or not [y,b] in Q by Def8;
  then [y,b] in R \ Q by A2,A3,XBOOLE_0:def 5;
  hence [a,b] in P*(R \ Q) by A2,Def8;
end;

theorem
  (P*R)~ = R~*P~
proof
  let a,b;
  hereby
    assume [a,b] in (P*R)~;
    then [b,a] in P*R by Def7;
    then consider y such that
A1: [b,y] in P & [y,a] in R by Def8;
    [y,b] in P~ & [a,y] in R~ by A1,Def7;
    hence [a,b] in R~*P~ by Def8;
  end;
  assume [a,b] in R~*P~;
  then consider y such that
A2: [a,y] in R~ & [y,b] in P~ by Def8;
  [y,a] in R & [b,y] in P by A2,Def7;
  then [b,a] in P*R by Def8;
  hence [a,b] in (P*R)~ by Def7;
end;

theorem Th36:
  (P*R)*Q = P*(R*Q)
proof
  let a,b;
  hereby
    assume [a,b] in (P*R)*Q;
    then consider y such that
A1: [a,y] in P*R and
A2: [y,b] in Q by Def8;
    consider x such that
A3: [a,x] in P and
A4: [x,y] in R by A1,Def8;
    [x,b] in R*Q by A2,A4,Def8;
    hence [a,b] in P*(R*Q) by A3,Def8;
  end;
  assume [a,b] in P*(R*Q);
  then consider y such that
A5: [a,y] in P and
A6: [y,b] in R*Q by Def8;
    consider x such that
A7: [y,x] in R and
A8: [x,b] in Q by A6,Def8;
    [a,x] in P*R by A5,A7,Def8;
    hence [a,b] in (P*R)*Q by A8,Def8;
  end;



registration
  cluster non empty for Relation;
  existence
  proof
    {[{},{}]} is non empty;
    hence thesis;
  end;
end;

registration
  let f be non empty Relation;
  cluster dom f -> non empty;
  coherence
  proof
    ex x1,x2 being set st the  Element of f = [x1,x2] by Def1;
    hence thesis by XTUPLE_0:def 12;
  end;
  cluster rng f -> non empty;
  coherence
  proof
    ex x1,x2 being set st the Element of f = [x1,x2] by Def1;
    hence thesis by XTUPLE_0:def 13;
  end;
end;

theorem Th37:
  (for x,y holds not [x,y] in R) implies R = {}
proof
  assume
A1: for x,y holds not [x,y] in R;
  assume
A2: not thesis;
  then ex x,y st the Element of R = [x,y] by Def1;
  hence contradiction by A1,A2;
end;

theorem Th38:
  dom {} = {} & rng {} = {}
proof
  thus dom {} = {}
  proof let x,y be set;
   thus [x,y] in dom {} implies [x,y] in {}
    proof
      assume [x,y] in dom {};
      then ex z st [[x,y],z] in {};
      hence thesis;
    end;
    thus thesis;
  end;
  let x,y be set;
  thus [x,y] in rng{} implies [x,y] in {}
   proof assume [x,y] in rng{};
    then ex z st [z,[x,y]] in {};
    hence thesis;
   end;
  thus thesis;
end;

theorem Th39:
  {}*R = {} & R*{} = {}
proof
  thus {}*R = {}
  proof
    let x,y;
    hereby
      assume [x,y] in {}*R;
      then ex z st [x,z] in {} & [z,y] in R by Def8;
      hence [x,y] in {};
    end;
    thus thesis;
  end;
  let x,y;
  hereby
    assume [x,y] in R*{};
    then ex z st [x,z] in R & [z,y] in {} by Def8;
    hence [x,y] in {};
  end;
  thus thesis;
end;

registration
  let X be empty set;
  cluster dom X -> empty;
  coherence;
  cluster rng X -> empty;
  coherence;
  let R;
  cluster X*R -> empty;
  coherence by Th39;
  cluster R*X -> empty;
  coherence by Th39;
end;

theorem
  field {} = {};

theorem Th41:
  dom R = {} or rng R = {} implies R = {};

theorem
  dom R = {} iff rng R = {} by Th38,Th41;

registration
  let X be empty set;
  cluster X~ -> empty;
  coherence
  proof
    for x,y st [x,y] in {}~ holds contradiction by Def7;
    hence thesis by Th37;
  end;
end;

theorem
  {}~ = {};

theorem
  rng R misses dom P implies R*P = {}
proof
  assume
A1: rng R /\ dom P = {};
  assume R*P <> {};
  then consider x,z such that
A2: [x,z] in R*P by Th37;
  consider y such that
A3: [x,y] in R & [y,z] in P by A2,Def8;
  y in rng R & y in dom P by A3,XTUPLE_0:def 12,def 13;
  hence contradiction by A1,XBOOLE_0:def 4;
end;

definition
  let R be Relation;
  attr R is non-empty means
  :Def9:
  not {} in rng R;
end;

registration
  cluster non-empty for Relation;
  existence
  proof
    take {};
    thus not {} in rng {};
  end;
end;

registration
  let R be Relation, S be non-empty Relation;
  cluster R*S -> non-empty;
  coherence
  proof
    rng(R*S) c= rng S by Th26;
    hence not {} in rng(R*S) by Def9;
  end;
end;



definition
  let X;
  func id X -> Relation means
  :Def10:
  [x,y] in it iff x in X & x = y;
  existence
  proof
    defpred Z[set,set] means $1=$2;
    consider R such that
A1: for x,y holds [x,y] in R iff x in X & y in X & Z[x,y] from
    RelExistence;
    take R;
    let x,y;
    thus thesis by A1;
  end;
  uniqueness
  proof
    let P1,P2;
    assume that
A2: [x,y] in P1 iff x in X & x=y and
A3: [x,y] in P2 iff x in X & x=y;
    let x,y;
    [x,y] in P1 iff x in X & x=y by A2;
    hence [x,y] in P1 iff [x,y] in P2 by A3;
  end;
end;

registration let X;
 reduce dom id X to X;
 reducibility
  proof
    thus dom id X c= X
    proof
      let x;
      assume x in dom(id X);
      then ex y st [x,y] in id X by XTUPLE_0:def 12;
      hence x in X by Def10;
    end;
    let x;
    assume x in X;
    then [x,x] in id X by Def10;
    hence x in dom(id X) by XTUPLE_0:def 12;
  end;
 reduce rng id X to X;
 reducibility
  proof
  thus rng id X c= X
  proof
    let x;
    assume x in rng id X;
    then consider y such that
A1: [y,x] in id X by XTUPLE_0:def 13;
    x = y by A1,Def10;
    hence thesis by A1,Def10;
  end;
  let x;
  [x,x] in id X iff x in X by Def10;
  hence thesis by XTUPLE_0:def 13;
  end;
end;

theorem
  dom id X = X & rng id X = X;

registration let X;
 reduce (id X)~ to id X;
 reducibility
  proof
  let x,y;
  thus [x,y] in (id X)~ implies [x,y] in id X
  proof
    assume [x,y] in (id X)~;
    then [y,x] in id X by Def7;
    hence thesis by Def10;
  end;
  assume
A1: [x,y] in id X;
  then x = y by Def10;
  hence thesis by A1,Def7;
  end;
end;

theorem
  (id X)~ = id X;

theorem
  (for x st x in X holds [x,x] in R) implies id X c= R
proof
  assume
A1: for x st x in X holds [x,x] in R;
  let x,y;
  assume [x,y] in id X;
  then x in X & x=y by Def10;
  hence thesis by A1;
end;

theorem Th48:
  [x,y] in (id X)*R iff x in X & [x,y] in R
proof
  thus [x,y] in (id X)*R implies x in X & [x,y] in R
  proof
    assume [x,y] in (id X)*R;
    then ex z st [x,z] in id X & [z,y] in R by Def8;
    hence thesis by Def10;
  end;
  assume x in X;
  then [x,x] in id X by Def10;
  hence thesis by Def8;
end;

theorem Th49:
  [x,y] in R*id Y iff y in Y & [x,y] in R
proof
  thus [x,y] in R*id Y implies y in Y & [x,y] in R
  proof
    assume [x,y] in R*id Y;
    then consider z such that
A1: [x,z] in R and
A2: [z,y] in id Y by Def8;
    z = y by A2,Def10;
    hence thesis by A1,A2,Def10;
  end;
  y in Y implies [y,y] in id Y by Def10;
  hence thesis by Def8;
end;

theorem Th50:
  R*(id X) c= R & (id X)*R c= R
proof
  thus [x,y] in R*id X implies [x,y] in R
  proof
    assume [x,y] in R*id X;
    then ex z st [x,z] in R & [z,y] in id X by Def8;
    hence thesis by Def10;
  end;
  thus [x,y] in (id X)*R implies [x,y] in R
  proof
    assume [x,y] in (id X)*R;
    then ex z st [x,z] in id X & [z,y] in R by Def8;
    hence thesis by Def10;
  end;
end;

theorem Th51:
  dom R c= X implies (id X)*R = R
proof
  assume
A1: dom R c= X;
A2: R c= (id X)*R
  proof
    let x,y;
    assume
A3: [x,y] in R;
    then x in dom R by XTUPLE_0:def 12;
    then [x,x] in id X by A1,Def10;
    hence thesis by A3,Def8;
  end;
  (id X)*R c= R by Th50;
  hence thesis by A2,XBOOLE_0:def 10;
end;

theorem
  (id dom R)*R = R by Th51;

theorem Th53:
  rng R c= Y implies R*(id Y) = R
proof
  assume
A1: rng R c= Y;
A2: R c= R*(id Y)
  proof
    let x,y;
    assume
A3: [x,y] in R;
    then y in rng R by XTUPLE_0:def 13;
    then [y,y] in (id Y) by A1,Def10;
    hence thesis by A3,Def8;
  end;
  R*(id Y) c= R by Th50;
  hence thesis by A2,XBOOLE_0:def 10;
end;

theorem
  R*(id rng R) = R by Th53;

theorem
  id {} = {}
proof
  dom(id {}) = {};
  hence thesis;
end;

theorem
  rng P2 c= X & P2*R = id dom P1 & R*P1 = id X implies P1 = P2
proof
  (P2*R)*P1 = P2*(R*P1) & id(dom P1)*P1 = P1 by Th36,Th51;
  hence thesis by Th53;
end;

definition
  let R,X;
  func R|X -> Relation means
  :Def11:
  [x,y] in it iff x in X & [x,y] in R;
  existence
  proof
    defpred Z[set,set] means ($1 in X & [$1,$2] in R);
    consider P such that
A1: for x,y holds [x,y] in P iff x in dom R & y in rng R & Z[x,y] from
    RelExistence;
    take P;
    let x,y;
    x in X & [x,y] in R implies x in dom R & y in rng R
      by XTUPLE_0:def 12,def 13;
    hence thesis by A1;
  end;
  uniqueness
  proof
    let P1,P2;
    assume that
A2: [x,y] in P1 iff x in X & [x,y] in R and
A3: [x,y] in P2 iff x in X & [x,y] in R;
    let x,y;
    [x,y] in P1 iff x in X & [x,y] in R by A2;
    hence [x,y] in P1 iff [x,y] in P2 by A3;
  end;
end;

registration
  let R be Relation, X be empty set;
  cluster R|X -> empty;
  coherence
  proof
    not [x,y] in R|{} by Def11;
    hence thesis by Th37;
  end;
end;

theorem Th57:
  x in dom(R|X) iff x in X & x in dom R
proof
  thus x in dom(R|X) implies x in X & x in dom R
  proof
    assume x in dom(R|X);
    then consider y such that
A1: [x,y] in R|X by XTUPLE_0:def 12;
    [x,y] in R by A1,Def11;
    hence thesis by A1,Def11,XTUPLE_0:def 12;
  end;
  assume
A2: x in X;
  assume x in dom R;
  then consider y such that
A3: [x,y] in R by XTUPLE_0:def 12;
  [x,y] in R|X by A2,A3,Def11;
  hence thesis by XTUPLE_0:def 12;
end;

theorem Th58:
  dom(R|X) c= X
proof
  let x;
  thus thesis by Th57;
end;

theorem Th59:
  R|X c= R
proof
  thus [x,y] in R|X implies [x,y] in R by Def11;
end;

theorem Th60:
  dom(R|X) c= dom R
proof
  let x;
  thus thesis by Th57;
end;

theorem Th61:
  dom(R|X) = dom R /\ X
proof
  x in dom(R|X) iff x in dom R /\ X
  proof
    x in dom(R|X) iff x in dom R & x in X by Th57;
    hence thesis by XBOOLE_0:def 4;
  end;
  hence thesis by TARSKI:1;
end;

theorem
  X c= dom R implies dom(R|X) = X
proof
  dom(R|X) = dom R /\ X by Th61;
  hence thesis by XBOOLE_1:28;
end;

theorem
  (R|X)*P c= R*P by Th30,Th59;

theorem
  P*(R|X) c= P*R by Th29,Th59;

theorem
  R|X = (id X)*R
proof
  let x,y;
  [x,y] in R|X iff [x,y] in R & x in X by Def11;
  hence thesis by Th48;
end;

theorem
  R|X = {} iff dom R misses X
proof
  thus R|X = {} implies dom R misses X
  proof
    assume
A1: R|X = {};
    thus dom R /\ X = {}
    proof
      thus dom R /\ X c= {}
      proof
        let x be set;
        assume
A2:     x in (dom R) /\ X;
        then x in dom R by XBOOLE_0:def 4;
        then
A3:     ex y st [x,y] in R by XTUPLE_0:def 12;
        x in X by A2,XBOOLE_0:def 4;
        hence thesis by A1,A3,Def11;
      end;
      thus thesis by XBOOLE_1:2;
    end;
  end;
  assume
A4: (dom R) /\ X = {};
  let x,y;
  hereby
    assume
A5: [x,y] in R|X;
    then x in X by Def11;
    then
A6: not x in dom R by A4,XBOOLE_0:def 4;
    [x,y] in R by A5,Def11;
    hence [x,y] in {} by A6,XTUPLE_0:def 12;
  end;
  thus thesis;
end;

theorem Th67:
  R|X = R /\ [:X,rng R:]
proof
  set P = R /\ [:X,rng R:];
  let x,y;
  thus [x,y] in R|X implies [x,y] in P
  proof
    assume
A1: [x,y] in R|X;
    then
A2: x in X by Def11;
A3: [x,y] in R by A1,Def11;
    then y in rng R by XTUPLE_0:def 13;
    then [x,y] in [:X,rng R:] by A2,ZFMISC_1:def 2;
    hence thesis by A3,XBOOLE_0:def 4;
  end;
  assume
A4: [x,y] in P;
  then [x,y] in [:X,rng R:] by XBOOLE_0:def 4;
  then
A5: x in X by ZFMISC_1:87;
  [x,y] in R by A4,XBOOLE_0:def 4;
  hence thesis by A5,Def11;
end;

theorem Th68:
  dom R c= X implies R|X = R
proof
  assume dom R c= X; then
A1: [:dom R,rng R:] c= [:X,rng R:] by ZFMISC_1:95;
  R c= [:dom R,rng R:] & R|X = R /\ [:X,rng R:] by Th7,Th67;
  hence thesis by A1,XBOOLE_1:1,28;
end;

registration let R;
 reduce R|dom R to R;
 reducibility by Th68;
end;

theorem
  R|dom R = R;

theorem Th70:
  rng(R|X) c= rng R
proof
  (R|X) c= R by Th59;
  hence thesis by Th11;
end;

theorem Th71:
  (R|X)|Y = R|(X /\ Y)
proof
  let x,y;
A1: [x,y] in R|X iff [x,y] in R & x in X by Def11;
A2: [x,y] in R|(X /\ Y) iff [x,y] in R & x in X /\ Y by Def11;
  [x,y] in (R|X)|Y iff [x,y] in R|X & x in Y by Def11;
  hence thesis by A1,A2,XBOOLE_0:def 4;
end;

registration let R,X;
 reduce R|X|X to R|X;
 reducibility
  proof
  X /\ X = X;
  hence thesis by Th71;
  end;
end;

theorem
  (R|X)|X = R|X;

theorem
  X c= Y implies (R|X)|Y = R|X
proof
  X c= Y implies X /\ Y = X by XBOOLE_1:28;
  hence thesis by Th71;
end;

theorem
  Y c= X implies (R|X)|Y = R|Y
proof
  Y c= X implies X /\ Y = Y by XBOOLE_1:28;
  hence thesis by Th71;
end;

theorem Th75:
  X c= Y implies R|X c= R|Y
proof
  assume
A1: X c= Y;
  let x,y;
  assume [x,y] in R|X;
  then x in X & [x,y] in R by Def11;
  hence [x,y] in R|Y by A1,Def11;
end;

theorem Th76:
  P c= R implies P|X c= R|X
proof
  assume
A1: P c= R;
  let x,y;
  assume [x,y] in P|X;
  then [x,y] in P & x in X by Def11;
  hence thesis by A1,Def11;
end;

theorem Th77:
  P c= R & X c= Y implies P|X c= R|Y
proof
  assume P c= R & X c= Y;
  then P|X c= R|X & R|X c= R|Y by Th75,Th76;
  hence thesis by XBOOLE_1:1;
end;

theorem Th78:
  R|(X \/ Y) = (R|X) \/ (R|Y)
proof
  let x,y;
  hereby
    assume
A1: [x,y] in R|(X \/ Y);
    then x in X \/ Y by Def11; then
A2: x in X or x in Y by XBOOLE_0:def 3;
    [x,y] in R by A1,Def11;
    then [x,y] in R|X or [x,y] in R|Y by A2,Def11;
    hence [x,y] in (R|X) \/ (R|Y) by XBOOLE_0:def 3;
  end;
  assume
A3: [x,y] in (R|X) \/ (R|Y);
  per cases by A3,XBOOLE_0:def 3;
  suppose
A4: [x,y] in R|Y;
    then x in Y by Def11;
    then
A5: x in X \/ Y by XBOOLE_0:def 3;
    [x,y] in R by A4,Def11;
    hence [x,y] in R|(X \/ Y) by A5,Def11;
  end;
  suppose
A6: [x,y] in R|X;
    then x in X by Def11;
    then
A7: x in X \/ Y by XBOOLE_0:def 3;
    [x,y] in R by A6,Def11;
    hence [x,y] in R|(X \/ Y) by A7,Def11;
  end;
end;

theorem
  R|(X /\ Y) = (R|X) /\ (R|Y)
  proof
    let x,y;
    hereby
      assume
A1:   [x,y] in R|(X /\ Y); then
A2:   x in X /\ Y by Def11;
A3:   [x,y] in R by A1,Def11;
      x in Y by A2,XBOOLE_0:def 4; then
A4:   [x,y] in R|Y by A3,Def11;
      x in X by A2,XBOOLE_0:def 4;
      then [x,y] in R|X by A3,Def11;
      hence [x,y] in (R|X) /\ (R|Y) by A4,XBOOLE_0:def 4;
    end;
    assume
A5: [x,y] in (R|X) /\ (R|Y);
    then [x,y] in R|Y by XBOOLE_0:def 4; then
A6: x in Y by Def11;
A7: [x,y] in R|X by A5,XBOOLE_0:def 4;
    then x in X by Def11; then
A8: x in X /\ Y by A6,XBOOLE_0:def 4;
    [x,y] in R by A7,Def11;
    hence [x,y] in R|(X /\ Y) by A8,Def11;
  end;

theorem Th80:
  R|(X \ Y) = R|X \ R|Y
proof
  let x,y;
  hereby
    assume
A1: [x,y] in R|(X \ Y); then
A2: x in X \ Y by Def11;
    then not x in Y by XBOOLE_0:def 5;
    then
A3: not [x,y] in R|Y by Def11;
    [x,y] in R by A1,Def11;
    then [x,y] in R|X by A2,Def11;
    hence [x,y] in R|X \ R|Y by A3,XBOOLE_0:def 5;
  end;
  assume
A4: [x,y] in R|X \ R|Y; then
A5: [x,y] in R by Def11;
    not [x,y] in R|Y by A4,XBOOLE_0:def 5; then
A6: not x in Y or not [x,y] in R by Def11;
    x in X by A4,Def11;
    then x in X \ Y by A4,A6,Def11,XBOOLE_0:def 5;
    hence [x,y] in R|(X \ Y) by A5,Def11;
end;

registration
  let R be empty Relation, X be set;
  cluster R|X -> empty;
  coherence
  proof
    for x,y st [x,y] in {}|X holds contradiction by Def11;
    hence thesis by Th37;
  end;
end;

theorem
  R|{} = {};

theorem
  {}|X = {};

theorem
  (P*R)|X = (P|X)*R
proof
  let x,y;
  hereby
    assume
A1: [x,y] in (P*R)|X;
    then [x,y] in P*R by Def11;
    then consider a such that
A2: [x,a] in P and
A3: [a,y] in R by Def8;
    x in X by A1,Def11;
    then [x,a] in P|X by A2,Def11;
    hence [x,y] in (P|X)*R by A3,Def8;
  end;
  assume [x,y] in (P|X)*R;
  then consider a such that
A4: [x,a] in P|X and
A5: [a,y] in R by Def8;
  [x,a] in P by A4,Def11;
  then
A6: [x,y] in P*R by A5,Def8;
  x in X by A4,Def11;
  hence [x,y] in (P*R)|X by A6,Def11;
end;

definition
  let Y,R;
  func Y|`R -> Relation means
  :Def12:
  [x,y] in it iff y in Y & [x,y] in R;
  existence
  proof
    defpred Z[set,set] means $2 in Y & [$1,$2] in R;
    consider P such that
A1: for x,y holds [x,y] in P iff x in dom R & y in rng R & Z[x,y] from
    RelExistence;
    take P;
    let x,y;
    y in Y & [x,y] in R implies x in dom R & y in rng R
       by XTUPLE_0:def 12,def 13;
    hence thesis by A1;
  end;
  uniqueness
  proof
    let P1,P2;
    assume that
A2: [x,y] in P1 iff y in Y & [x,y] in R and
A3: [x,y] in P2 iff y in Y & [x,y] in R;
    let x,y;
    [x,y] in P1 iff y in Y & [x,y] in R by A2;
    hence [x,y] in P1 iff [x,y] in P2 by A3;
  end;
end;

registration
  let R be Relation, X be empty set;
  cluster X|`R -> empty;
  coherence
  proof
    not [x,y] in X|`R by Def12;
    hence thesis by Th37;
  end;
end;

theorem Th84:
  y in rng(Y|`R) iff y in Y & y in rng R
proof
  thus y in rng(Y|`R) implies y in Y & y in rng R
  proof
    assume y in rng(Y|`R);
    then consider x such that
A1: [x,y] in Y|`R by XTUPLE_0:def 13;
    [x,y] in R by A1,Def12;
    hence thesis by A1,Def12,XTUPLE_0:def 13;
  end;
  assume
A2: y in Y;
  assume y in rng R;
  then consider x such that
A3: [x,y] in R by XTUPLE_0:def 13;
  [x,y] in Y|`R by A2,A3,Def12;
  hence thesis by XTUPLE_0:def 13;
end;

theorem Th85:
  rng(Y|`R) c= Y
proof
  let y;
  thus thesis by Th84;
end;

theorem Th86:
  Y|`R c= R
proof
  thus [x,y] in Y|`R implies [x,y] in R by Def12;
end;

theorem Th87:
  rng(Y|`R) c= rng R
proof
  Y|`R c= R by Th86;
  hence thesis by Th11;
end;

theorem Th88:
  rng(Y|`R) = rng R /\ Y
proof
  rng(Y|`R) c= Y & rng(Y|`R) c= rng R by Th85,Th87;
  hence rng(Y|`R) c= rng R /\ Y by XBOOLE_1:19;
  let y;
  assume
A1: y in rng R /\ Y;
  then y in rng R by XBOOLE_0:def 4;
  then consider x such that
A2: [x,y] in R by XTUPLE_0:def 13;
  y in Y by A1,XBOOLE_0:def 4;
  then [x,y] in Y|`R by A2,Def12;
  hence thesis by XTUPLE_0:def 13;
end;

theorem
  Y c= rng R implies rng(Y|`R) = Y
proof
  assume Y c= rng R;
  then rng R /\ Y = Y by XBOOLE_1:28;
  hence thesis by Th88;
end;

theorem
  (Y|`R)*P c= R*P by Th30,Th86;

theorem
  P*(Y|`R) c= P*R by Th29,Th86;

theorem
  Y|`R = R*(id Y)
proof
  let x,y;
  [x,y] in Y|`R iff y in Y & [x,y] in R by Def12;
  hence thesis by Th49;
end;

theorem Th93:
  Y|`R = R /\ [:dom R,Y:]
proof
  set P = R /\ [:dom R,Y:];
  let x,y;
  thus [x,y] in Y|`R implies [x,y] in P
  proof
    assume
A1: [x,y] in Y|`R;
    then
A2: y in Y by Def12;
A3: [x,y] in R by A1,Def12;
    then x in dom R by XTUPLE_0:def 12;
    then [x,y] in [:dom R,Y:] by A2,ZFMISC_1:def 2;
    hence thesis by A3,XBOOLE_0:def 4;
  end;
  assume
A4: [x,y] in P;
  then [x,y] in [:dom R,Y:] by XBOOLE_0:def 4; then
A5: y in Y by ZFMISC_1:87;
  [x,y] in R by A4,XBOOLE_0:def 4;
  hence thesis by A5,Def12;
end;

theorem Th94:
  rng R c= Y implies Y|`R = R
proof
  assume rng R c= Y; then
A1: [:dom R,rng R:] c= [:dom R,Y:] by ZFMISC_1:95;
  R c= [:dom R,rng R:] & Y|`R = R /\ [:dom R,Y:] by Th7,Th93;
  hence thesis by A1,XBOOLE_1:1,28;
end;

registration let R;
 reduce rng R|`R to R;
 reducibility by Th94;
end;

theorem
  rng R|`R=R;

theorem Th96:
  Y|`(X|`R) = (Y /\ X)|`R
proof
  let x,y;
A1: [x,y] in X|`R iff [x,y] in R & y in X by Def12;
A2: [x,y] in (Y /\ X)|`R iff [x,y] in R & y in Y /\ X by Def12;
  [x,y] in Y|`(X|`R) iff [x,y] in X|`R & y in Y by Def12;
  hence thesis by A1,A2,XBOOLE_0:def 4;
end;

registration let Y,R;
 reduce Y|`(Y|`R) to Y|`R;
 reducibility
proof
  Y /\ Y = Y;
  hence thesis by Th96;
end;
end;

theorem
  Y|`(Y|`R) = Y|`R;

theorem
  X c= Y implies Y|`(X|`R) = X|`R
proof
  X c= Y implies Y /\ X = X by XBOOLE_1:28;
  hence thesis by Th96;
end;

theorem
  Y c= X implies Y|`(X|`R) = Y|`R
proof
  Y c= X implies Y /\ X = Y by XBOOLE_1:28;
  hence thesis by Th96;
end;

theorem Th100:
  X c= Y implies X|`R c= Y|`R
proof
  assume
A1: X c= Y;
  let x,y;  [x,y] in X|`R iff [x,y] in R & y in X by Def12;
  hence thesis by A1,Def12;
end;

theorem Th101:
  P1 c= P2 implies Y|`P1 c= Y|`P2
proof
  assume
A1: P1 c= P2;
  let x,y;
  assume [x,y] in Y|`P1;
  then [x,y] in P1 & y in Y by Def12;
  hence thesis by A1,Def12;
end;

theorem
  P1 c= P2 & Y1 c= Y2 implies Y1|`P1 c= Y2|`P2
proof
  assume P1 c= P2 & Y1 c= Y2;
  then Y1|`P1 c= Y1|`P2 & Y1|`P2 c= Y2|`P2 by Th100,Th101;
  hence thesis by XBOOLE_1:1;
end;

theorem
  (X \/ Y)|`R = (X|`R) \/ (Y|`R)
proof
  let x,y;
A1: y in X \/ Y iff y in X or y in Y by XBOOLE_0:def 3;
A2: [x,y] in (X|`R) \/ (Y|`R) iff [x,y] in X|`R or [x,y] in Y|`R
                 by XBOOLE_0:def 3;
  [x,y] in (X \/ Y)|`R iff y in X \/ Y & [x,y] in R by Def12;
  hence thesis by A1,A2,Def12;
end;

theorem
  (X /\ Y)|`R = X|`R /\ Y|`R
proof
  let x,y;
A1: y in X /\ Y iff y in X & y in Y by XBOOLE_0:def 4;
A2: [x,y] in X|`R /\ Y|`R iff [x,y] in X|`R & [x,y] in Y|`R by XBOOLE_0:def 4;
  [x,y] in (X /\ Y)|`R iff y in X /\ Y & [x,y] in R by Def12;
  hence thesis by A1,A2,Def12;
end;

theorem
  (X \ Y)|`R = X|`R \ Y|`R
proof
  let x,y;
A1: y in X \ Y iff y in X & not y in Y by XBOOLE_0:def 5;
A2: [x,y] in X|`R \ Y|`R iff [x,y] in X|`R & not [x,y] in Y|`R
          by XBOOLE_0:def 5;
  [x,y] in X|`R iff y in X & [x,y] in R by Def12;
  hence thesis by A1,A2,Def12;
end;

theorem
  {}|`R = {};

theorem
  Y|`{} = {}
proof
  for x,y st [x,y] in Y|`{} holds contradiction by Def12;
  hence thesis by Th37;
end;

theorem
  Y|`(P*R) = P*(Y|`R)
proof
  let x,y;
  hereby
    assume
A1: [x,y] in Y|`(P*R);
    then [x,y] in P*R by Def12;
    then consider a such that
A2: [x,a] in P and
A3: [a,y] in R by Def8;
    y in Y by A1,Def12;
    then [a,y] in Y|`R by A3,Def12;
    hence [x,y] in P*(Y|`R) by A2,Def8;
  end;
  assume [x,y] in P*(Y|`R);
  then consider a such that
A4: [x,a] in P and
A5: [a,y] in Y|`R by Def8;
  [a,y] in R by A5,Def12;
  then
A6: [x,y] in P*R by A4,Def8;
  y in Y by A5,Def12;
  hence [x,y] in Y|`(P*R) by A6,Def12;
end;

theorem
  (Y|`R)|X = Y|`(R|X)
proof
  let x,y;
A1: [x,y] in R & x in X iff [x,y] in R|X by Def11;
  [x,y] in (Y|`R) iff [x,y] in R & y in Y by Def12;
  hence thesis by A1,Def11,Def12;
end;



definition
  let R,X;
  func R.:X -> set means
  :Def13:
  y in it iff ex x st [x,y] in R & x in X;
  existence
  proof
    defpred Z[set] means ex x st [x,$1] in R & x in X;
    consider Y such that
A1: for y holds y in Y iff y in rng R & Z[y] from XBOOLE_0:sch 1;
    take Y;
    let y;
    thus y in Y implies ex x st [x,y] in R & x in X by A1;
    given x such that
A2: [x,y] in R and
A3: x in X;
    y in rng R by A2,XTUPLE_0:def 13;
    hence thesis by A1,A2,A3;
  end;
  uniqueness
  proof
    let Y1,Y2;
    assume that
A4: y in Y1 iff ex x st [x,y] in R & x in X and
A5: y in Y2 iff ex x st [x,y] in R & x in X;
    now
      let y;
      y in Y1 iff ex x st [x,y] in R & x in X by A4;
      hence y in Y1 iff y in Y2 by A5;
    end;
    hence thesis by TARSKI:1;
  end;
end;

theorem Th110:
  y in R.:X iff ex x st x in dom R & [x,y] in R & x in X
proof
  thus y in R.:X implies ex x st x in dom R & [x,y] in R & x in X
  proof
    assume y in R.:X;
    then consider x such that
A1: [x,y] in R & x in X by Def13;
    take x;
    thus thesis by A1,XTUPLE_0:def 12;
  end;
  thus thesis by Def13;
end;

theorem Th111:
  R.:X c= rng R
proof
  let y;
  assume y in R.:X;
  then ex x st [x,y] in R & x in X by Def13;
  hence thesis by XTUPLE_0:def 13;
end;

theorem
  R.:X = R.:(dom R /\ X)
proof
  y in R.:X iff y in R.:(dom R /\ X)
  proof
    thus y in R.:(X) implies y in R.:(dom R /\ X)
    proof
      assume y in R.:(X);
      then consider x such that
A1:   x in dom R and
A2:   [x,y] in R and
A3:   x in X by Th110;
      x in dom R /\ X by A1,A3,XBOOLE_0:def 4;
      hence thesis by A2,Def13;
    end;
    assume y in R.:(dom R /\ X);
    then consider x such that
    x in dom R and
A4: [x,y] in R and
A5: x in dom R /\ X by Th110;
    x in X by A5,XBOOLE_0:def 4;
    hence thesis by A4,Def13;
  end;
  hence thesis by TARSKI:1;
end;

theorem Th113:
  R.:dom R = rng R
proof
  thus R.:dom R c= rng R by Th111;
  let y;
  assume y in rng R;
  then consider x such that
A1: [x,y] in R by XTUPLE_0:def 13;
  x in dom R by A1,XTUPLE_0:def 12;
  hence thesis by A1,Def13;
end;

theorem
  R.:X c= R.:(dom R)
proof
  R.:X c= rng R by Th111;
  hence thesis by Th113;
end;

theorem
  rng(R|X) = R.:X
proof
  y in rng(R|X) iff y in R.:X
  proof
    thus y in rng(R|X) implies y in R.:X
    proof
      assume y in rng(R|X);
      then consider x such that
A1:   [x,y] in R|X by XTUPLE_0:def 13;
      x in X & [x,y] in R by A1,Def11;
      hence thesis by Def13;
    end;
    assume y in R.:X;
    then consider x such that
A2: [x,y] in R & x in X by Def13;
    [x,y] in R|X by A2,Def11;
    hence thesis by XTUPLE_0:def 13;
  end;
  hence thesis by TARSKI:1;
end;

registration let R; let X be empty set;
 cluster R.:X -> empty;
 coherence
 proof
  set y = the Element of R.:{};
  y in R.:{} implies ex x st [x,y] in R & x in {} by Def13;
  hence thesis;
 end;
end;

registration let R be empty Relation; let X;
 cluster R.:X -> empty;
 coherence
 proof
  assume not thesis;
  then ex x st [x,the Element of {}.:X] in {} & x in X by Def13;
  hence contradiction;
 end;
end;

canceled 2;

theorem
  R.:X = {} iff dom R misses X
proof
  set y = the Element of R.:X;
  thus R.:X = {} implies dom R misses X
  proof
    assume
A1: R.:X = {};
    assume not thesis;
    then consider x be set such that
A2: x in dom R and
A3: x in X by XBOOLE_0:3;
    ex y st [x,y] in R by A2,XTUPLE_0:def 12;
    hence contradiction by A1,A2,A3,Th110;
  end;
  assume
A4: dom R /\ X ={};
  assume not thesis;
  then ex x st x in dom R & [x,y] in R & x in X by Th110;
  hence contradiction by A4,XBOOLE_0:def 4;
end;

theorem
  X <> {} & X c= dom R implies R.:X <> {}
proof
  assume that
A1: X <> {} and
A2: X c= dom R;
  set x = the Element of X;
A3: x in dom R by A1,A2,TARSKI:def 3;
  then ex y st [x,y] in R by XTUPLE_0:def 12;
  hence thesis by A1,A3,Th110;
end;

theorem
  R.:(X \/ Y) = R.:X \/ R.:Y
  proof
    thus R.:(X \/ Y) c= R.:X \/ R.:Y
    proof
      let y;
      assume y in R.:(X \/ Y);
      then consider x such that
A1:   [x,y] in R and
A2:   x in X \/ Y by Def13;
      x in X or x in Y by A2,XBOOLE_0:def 3;
      then y in R.:X or y in R.:Y by A1,Def13;
      hence y in R.:X \/ R.:Y by XBOOLE_0:def 3;
    end;
    let y;
    assume
A3: y in R.:X \/ R.:Y;
    per cases by A3,XBOOLE_0:def 3;
    suppose y in R.:Y;
      then consider x such that
A4:   [x,y] in R and
A5:   x in Y by Def13;
      x in X \/ Y by A5,XBOOLE_0:def 3;
      hence y in R.:(X \/ Y) by A4,Def13;
    end;
    suppose y in R.:X;
      then consider x such that
A6:   [x,y] in R and
A7:   x in X by Def13;
      x in X \/ Y by A7,XBOOLE_0:def 3;
      hence y in R.:(X \/ Y) by A6,Def13;
    end;
  end;

theorem
  R.:(X /\ Y) c= R.:X /\ R.:Y
proof
  let y;
  assume y in R.:(X /\ Y);
  then consider x such that
A1: [x,y] in R and
A2: x in X /\ Y by Def13;
  x in Y by A2,XBOOLE_0:def 4;
  then
A3: y in R.:Y by A1,Def13;
  x in X by A2,XBOOLE_0:def 4;
  then y in R.:X by A1,Def13;
  hence thesis by A3,XBOOLE_0:def 4;
end;

theorem
  R.:X \ R.:Y c= R.:(X \ Y)
proof
  let y;
  assume
A1: y in R.:X \ R.:Y;
  then consider x such that
A2: [x,y] in R and
A3: x in X by Def13;
  not y in R.:Y by A1,XBOOLE_0:def 5;
  then not x in Y or not [x,y] in R by Def13;
  then x in X \ Y by A2,A3,XBOOLE_0:def 5;
  hence thesis by A2,Def13;
end;

theorem Th123:
  X c= Y implies R.:X c= R.:Y
proof
  assume
A1: X c= Y;
  let y;
  assume y in R.:X;
  then ex x st [x,y] in R & x in X by Def13;
  hence thesis by A1,Def13;
end;

theorem Th124:
  P c= R implies P.:X c= R.:X
proof
  assume
A1: P c= R;
  let y;
  assume y in P.:X;
  then ex x st [x,y] in P & x in X by Def13;
  hence thesis by A1,Def13;
end;

theorem
  P c= R & X c= Y implies P.:X c= R.:Y
proof
  assume P c= R & X c= Y;
  then P.:X c= R.:X & R.:X c= R.:Y by Th123,Th124;
  hence thesis by XBOOLE_1:1;
end;

theorem
  (P*R).:X = R.:(P.:X)
proof
  y in (P*R).:X iff y in R.:(P.:X)
  proof
    thus y in (P*R).:X implies y in R.:(P.:X)
    proof
      assume y in (P*R).:X;
      then consider x such that
A1:   [x,y] in P*R and
A2:   x in X by Def13;
      consider z such that
A3:   [x,z] in P and
A4:   [z,y] in R by A1,Def8;
      z in P.:X by A2,A3,Def13;
      hence thesis by A4,Def13;
    end;
    assume y in R.:(P.:X);
    then consider x such that
A5: [x,y] in R and
A6: x in P.:X by Def13;
    consider z such that
A7: [z,x] in P and
A8: z in X by A6,Def13;
    [z,y] in P*R by A5,A7,Def8;
    hence thesis by A8,Def13;
  end;
  hence thesis by TARSKI:1;
end;

theorem Th127:
  rng(P*R) = R.:(rng P)
proof
  z in rng(P*R) iff z in R.:(rng P)
  proof
    thus z in rng(P*R) implies z in R.:(rng P)
    proof
      assume z in rng(P*R);
      then consider x such that
A1:   [x,z] in P*R by XTUPLE_0:def 13;
      consider y such that
A2:   [x,y] in P and
A3:   [y,z] in R by A1,Def8;
      y in rng P by A2,XTUPLE_0:def 13;
      hence thesis by A3,Def13;
    end;
    assume z in R.:(rng P);
    then consider y such that
A4: [y,z] in R and
A5: y in rng P by Def13;
    consider x such that
A6: [x,y] in P by A5,XTUPLE_0:def 13;
    [x,z] in P*R by A4,A6,Def8;
    hence thesis by XTUPLE_0:def 13;
  end;
  hence thesis by TARSKI:1;
end;

theorem
  (R|X).:Y c= R.:Y by Th59,Th124;

theorem Th129:
  for R be Relation, X, Y be set st X c= Y holds (R|Y).:X = R.:X
proof
  let R be Relation, X, Y be set such that
A1: X c= Y;
  thus (R|Y).:X c= R.:X by Th59,Th124;
  let y be set;
  assume y in R.:X;
  then consider x1 be set such that
A2: [x1,y] in R and
A3: x1 in X by Def13;
  ex x be set st [x,y] in R|Y & x in X
  proof
    take x1;
    thus [x1,y] in R|Y by A1,A2,A3,Def11;
    thus thesis by A3;
  end;
  hence thesis by Def13;
end;

theorem
  (dom R) /\ X c= (R~).:(R.:X)
proof
  let x;
  assume
A1: x in (dom R) /\ X;
  then x in dom R by XBOOLE_0:def 4;
  then consider y such that
A2: [x,y] in R by XTUPLE_0:def 12;
A3: [y,x] in R~ by A2,Def7;
  x in X by A1,XBOOLE_0:def 4;
  then y in R.:X by A2,Def13;
  hence thesis by A3,Def13;
end;



definition
  let R,Y;
  func R"Y -> set means
  :Def14:
  x in it iff ex y st [x,y] in R & y in Y;
  existence
  proof
    defpred Z[set] means ex y st [$1,y] in R & y in Y;
    consider X such that
A1: for x holds x in X iff x in dom R & Z[x] from XBOOLE_0:sch 1;
    take X;
    let x;
    thus x in X implies ex y st [x,y] in R & y in Y by A1;
    given y such that
A2: [x,y] in R and
A3: y in Y;
    x in dom R by A2,XTUPLE_0:def 12;
    hence thesis by A1,A2,A3;
  end;
  uniqueness
  proof
    let X1,X2;
    assume that
A4: x in X1 iff ex y st [x,y] in R & y in Y and
A5: x in X2 iff ex y st [x,y] in R & y in Y;
    now
      let x;
      x in X1 iff ex y st [x,y] in R & y in Y by A4;
      hence x in X1 iff x in X2 by A5;
    end;
    hence thesis by TARSKI:1;
  end;
end;

theorem Th131:
  x in R"Y iff ex y st y in rng R & [x,y] in R & y in Y
proof
  thus x in R"Y implies ex y st y in rng R & [x,y] in R & y in Y
  proof
    assume x in R"Y;
    then consider y such that
A1: [x,y] in R & y in Y by Def14;
    take y;
    thus thesis by A1,XTUPLE_0:def 13;
  end;
  thus thesis by Def14;
end;

theorem Th132:
  R"Y c= dom R
proof
  let x;
  assume x in R"Y;
  then ex y st [x,y] in R & y in Y by Def14;
  hence thesis by XTUPLE_0:def 12;
end;

theorem
  R"Y = R"(rng R /\ Y)
proof
  x in R"Y iff x in R"(rng R /\ Y)
  proof
    thus x in R"Y implies x in R"(rng R /\ Y)
    proof
      assume x in R"Y;
      then consider y such that
A1:   y in rng R and
A2:   [x,y] in R and
A3:   y in Y by Th131;
      y in rng R /\ Y by A1,A3,XBOOLE_0:def 4;
      hence thesis by A2,Def14;
    end;
    assume x in R"(rng R /\ Y);
    then consider y such that
    y in rng R and
A4: [x,y] in R and
A5: y in rng R /\ Y by Th131;
    y in Y by A5,XBOOLE_0:def 4;
    hence thesis by A4,Def14;
  end;
  hence thesis by TARSKI:1;
end;

theorem Th134:
  R"rng R = dom R
proof
  thus R"rng R c= dom R by Th132;
  let x;
  assume x in dom R;
  then consider y such that
A1: [x,y] in R by XTUPLE_0:def 12;
  y in rng R by A1,XTUPLE_0:def 13;
  hence thesis by A1,Def14;
end;

theorem
  R"Y c= R"rng R
proof
  R"Y c= dom R by Th132;
  hence thesis by Th134;
end;

registration let R; let Y be empty set;
 cluster R"Y -> empty;
 coherence
 proof
  set x = the Element of R"{};
  x in R"{} implies ex y st [x,y] in R & y in {} by Def14;
  hence thesis;
 end;
end;

registration let R be empty Relation; let Y;
 cluster R"Y -> empty;
 coherence
 proof
  assume not thesis;
  then ex y st [the Element of {}"Y,y] in {} & y in Y by Def14;
  hence contradiction;
 end;
end;

canceled 2;

theorem
  R"Y = {} iff rng R misses Y
proof
  set x = the Element of R"Y;
  thus R"Y = {} implies rng R misses Y
  proof
    assume
A1: R"Y = {};
    assume not thesis;
    then consider y being set such that
A2: y in rng R and
A3: y in Y by XBOOLE_0:3;
    ex x st [x,y] in R by A2,XTUPLE_0:def 13;
    hence contradiction by A1,A2,A3,Th131;
  end;
  assume
A4: rng R /\ Y = {};
  assume not thesis;
  then ex y st y in rng R & [x,y] in R & y in Y by Th131;
  hence contradiction by A4,XBOOLE_0:def 4;
end;

theorem
  Y <> {} & Y c= rng R implies R"Y <> {}
proof
  assume that
A1: Y <> {} and
A2: Y c= rng R;
  set y = the Element of Y;
A3: y in rng R by A1,A2,TARSKI:def 3;
  then ex x st [x,y] in R by XTUPLE_0:def 13;
  hence thesis by A1,A3,Th131;
end;

theorem
  R"(X \/ Y) = R"X \/ R"Y
  proof
    thus R"(X \/ Y) c= R"X \/ R"Y
    proof
      let x;
      assume x in R"(X \/ Y);
      then consider y such that
A1:   [x,y] in R and
A2:   y in X \/ Y by Def14;
      y in X or y in Y by A2,XBOOLE_0:def 3;
      then x in R"X or x in R"Y by A1,Def14;
      hence x in R"X \/ R"Y by XBOOLE_0:def 3;
    end;
    let x;
    assume
A3: x in R"X \/ R"Y;
    per cases by A3,XBOOLE_0:def 3;
    suppose x in R"Y;
      then consider y such that
A4:   [x,y] in R and
A5:   y in Y by Def14;
      y in X \/ Y by A5,XBOOLE_0:def 3;
      hence x in R"(X \/ Y) by A4,Def14;
    end;
    suppose x in R"X;
      then consider y such that
A6:   [x,y] in R and
A7:   y in X by Def14;
      y in X \/ Y by A7,XBOOLE_0:def 3;
      hence x in R"(X \/ Y) by A6,Def14;
    end;
  end;

theorem
  R"(X /\ Y) c= R"X /\ R"Y
proof
  let x;
  assume x in R"(X /\ Y);
  then consider y such that
A1: [x,y] in R and
A2: y in X /\ Y by Def14;
  y in Y by A2,XBOOLE_0:def 4;
  then
A3: x in R"Y by A1,Def14;
  y in X by A2,XBOOLE_0:def 4;
  then x in R"X by A1,Def14;
  hence thesis by A3,XBOOLE_0:def 4;
end;

theorem
  R"X \ R"Y c= R"(X \ Y)
proof
  let x;
  assume
A1: x in R"X \ R"Y;
  then consider y such that
A2: [x,y] in R and
A3: y in X by Def14;
  not x in R"Y by A1,XBOOLE_0:def 5;
  then not y in Y or not [x,y] in R by Def14;
  then y in X \ Y by A2,A3,XBOOLE_0:def 5;
  hence thesis by A2,Def14;
end;

theorem Th143:
  X c= Y implies R"X c= R"Y
proof
  assume
A1: X c= Y;
  let x;
  assume x in R"X;
  then ex y st [x,y] in R & y in X by Def14;
  hence thesis by A1,Def14;
end;

theorem Th144:
  P c= R implies P"Y c= R"Y
proof
  assume
A1: P c= R;
  let x;
  assume x in P"Y;
  then ex y st [x,y] in P & y in Y by Def14;
  hence thesis by A1,Def14;
end;

theorem
  P c= R & X c= Y implies P"X c= R"Y
proof
  assume P c= R & X c= Y;
  then P"X c= R"X & R"X c= R"Y by Th143,Th144;
  hence thesis by XBOOLE_1:1;
end;

theorem
  (P*R)"Y = P"(R"Y)
proof
  x in (P*R)"Y iff x in P"(R"Y)
  proof
    thus x in (P*R)"Y implies x in P"(R"Y)
    proof
      assume x in (P*R)"Y;
      then consider y such that
A1:   [x,y] in P*R and
A2:   y in Y by Def14;
      consider z such that
A3:   [x,z] in P and
A4:   [z,y] in R by A1,Def8;
      z in R"Y by A2,A4,Def14;
      hence thesis by A3,Def14;
    end;
    assume x in P"(R"Y);
    then consider y such that
A5: [x,y] in P and
A6: y in R"Y by Def14;
    consider z such that
A7: [y,z] in R and
A8: z in Y by A6,Def14;
    [x,z] in P*R by A5,A7,Def8;
    hence thesis by A8,Def14;
  end;
  hence thesis by TARSKI:1;
end;

theorem Th147:
  dom(P*R) = P"(dom R)
proof
  z in dom(P*R) iff z in P"(dom R)
  proof
    thus z in dom(P*R) implies z in P"(dom R)
    proof
      assume z in dom(P*R);
      then consider x such that
A1:   [z,x] in P*R by XTUPLE_0:def 12;
      consider y such that
A2:   [z,y] in P and
A3:   [y,x] in R by A1,Def8;
      y in dom R by A3,XTUPLE_0:def 12;
      hence thesis by A2,Def14;
    end;
    assume z in P"(dom R);
    then consider y such that
A4: [z,y] in P and
A5: y in dom R by Def14;
    consider x such that
A6: [y,x] in R by A5,XTUPLE_0:def 12;
    [z,x] in P*R by A4,A6,Def8;
    hence thesis by XTUPLE_0:def 12;
  end;
  hence thesis by TARSKI:1;
end;

theorem
  (rng R) /\ Y c= (R~)"(R"Y)
proof
  let y;
  assume
A1: y in (rng R) /\ Y;
  then y in rng R by XBOOLE_0:def 4;
  then consider x such that
A2: [x,y] in R by XTUPLE_0:def 13;
A3: [y,x] in R~ by A2,Def7;
  y in Y by A1,XBOOLE_0:def 4;
  then x in R"Y by A2,Def14;
  hence thesis by A3,Def14;
end;

begin 

definition
  let R;
  attr R is empty-yielding means
  :Def15:
  rng R c= {{}};
end;

theorem
  R is empty-yielding iff for X st X in rng R holds X = {}
proof
  hereby
    assume R is empty-yielding;
    then rng R c= {{}} by Def15;
    hence for X st X in rng R holds X = {} by TARSKI:def 1;
  end;
  assume
A1: for X st X in rng R holds X = {};
  let Y;
  assume Y in rng R;
  then Y = {} by A1;
  hence thesis by TARSKI:def 1;
end;



theorem
  for f,g being Relation, A,B being set st f|A = g|A & f|B = g|B holds
  f|(A \/ B) = g|(A \/ B)
proof
  let f,g be Relation, A,B be set;
  assume f|A = g|A & f|B = g|B;
  hence f|(A \/ B) = g|A \/ g|B by Th78
    .= g|(A \/ B) by Th78;
end;

theorem
  for X being set, f,g being Relation st dom g c= X & g c= f holds g c= f|X
proof
  let X be set, f,g be Relation;
  assume dom g c= X & g c= f;
  then g|dom g c= f|X by Th77;
  hence thesis;
end;

theorem
  for f being Relation, X being set st X misses dom f holds f|X = {}
proof
  let f be Relation, X be set such that
A1: X /\ dom f = {};
  thus f|X = (f|dom f)|X
    .= f|({} qua set) by A1,Th71
    .= {};
end;



theorem
  for f,g being Relation, A,B being set st A c= B & f|B = g|B holds f|A = g|A
proof
  let f,g be Relation, A,B be set;
  assume that
A1: A c= B and
A2: f|B = g|B;
A3: A = B /\ A by A1,XBOOLE_1:28;
  hence f|A = (f|B)|A by Th71
    .= g|A by A2,A3,Th71;
end;



theorem
  R|dom S = R|dom(S|dom R)
proof
  thus R|dom S = R|dom R|dom S
    .= R|(dom S /\ dom R) by Th71
    .= R|dom(S|dom R) by Th61;
end;



registration
  cluster empty -> empty-yielding for Relation;
  coherence
  proof
    let R be Relation;
    assume R is empty;
    then rng R = {};
    hence rng R c= {{}} by XBOOLE_1:2;
  end;
end;

registration
  let R be empty-yielding Relation;
  let X be set;
  cluster R|X -> empty-yielding;
  coherence
  proof
    rng R c= {{}} & rng(R|X) c= rng R by Def15,Th70;
    hence rng(R|X) c= {{}} by XBOOLE_1:1;
  end;
end;

theorem
  R|X is non empty-yielding implies R is non empty-yielding;



definition
  let R be Relation, x be set;
  func Im(R,x) equals
  R.:{x};
  correctness;
end;





scheme
  ExtensionalityR { A, B() -> Relation, P[set,set] }: A() = B()
provided
A1: for a, b being set holds [a,b] in A() iff P[a,b] and
A2: for a, b being set holds [a,b] in B() iff P[a,b]
proof
  let y,z be set;
  hereby
    assume [y,z] in A();
    then P[y,z] by A1;
    hence [y,z] in B() by A2;
  end;
  assume [y,z] in B();
  then P[y,z] by A2;
  hence thesis by A1;
end;



theorem
  dom (R | (dom R \ X)) = dom R \ X
proof
  thus dom (R | (dom R \ X)) = dom R /\ (dom R \ X) by Th61
    .= (dom R /\ dom R) \ X by XBOOLE_1:49
    .= dom R \ X;
end;

theorem
  R | X = R | (dom R /\ X)
proof
  thus R | X = (R | dom R) | X
    .= R | (dom R /\ X) by Th71;
end;



theorem
  dom [:X,Y:] c= X
proof
  let x;
  assume x in dom [:X,Y:];
  then ex y st [x,y] in [:X,Y:] by XTUPLE_0:def 12;
  hence thesis by ZFMISC_1:87;
end;

theorem
  rng [:X,Y:] c= Y
proof
  let x;
  assume x in rng [:X,Y:];
  then ex y st [y,x] in [:X,Y:] by XTUPLE_0:def 13;
  hence thesis by ZFMISC_1:87;
end;



theorem
  X <> {} & Y <> {} implies dom [:X,Y:] = X & rng [:X,Y:] = Y
proof
  assume X <> {};
  then consider a such that
A1: a in X by XBOOLE_0:def 1;
  assume Y <> {};
  then consider b such that
A2: b in Y by XBOOLE_0:def 1;
A3: now
    let x;
    assume x in X;
    then [x,b] in [:X,Y:] by A2,ZFMISC_1:87;
    hence ex y st [x,y] in [:X,Y:];
  end;
  (ex y st [x,y] in [:X,Y:]) implies x in X by ZFMISC_1:87;
  hence dom[:X,Y:] = X by A3,XTUPLE_0:def 12;
A4: now
    let y;
    assume y in Y;
    then [a,y] in [:X,Y:] by A1,ZFMISC_1:87;
    hence ex x st [x,y] in [:X,Y:];
  end;
  (ex x st [x,y] in [:X,Y:]) implies y in Y by ZFMISC_1:87;
  hence thesis by A4,XTUPLE_0:def 13;
end;

theorem
  dom R = {} & dom Q = {} implies R = Q
proof
  assume that
A1: dom R = {} and
A2: dom Q = {};
  R = {} by A1;
  hence thesis by A2;
end;

theorem
  rng R = {} & rng Q = {} implies R = Q
proof
  assume that
A1: rng R = {} and
A2: rng Q = {};
  R = {} by A1;
  hence thesis by A2;
end;

theorem
  dom R = dom Q implies dom(S*R) = dom (S*Q)
proof
  assume
A1: dom R = dom Q;
  thus dom(S*R) = S"dom R by Th147
    .= dom (S*Q) by A1,Th147;
end;

theorem
  rng R = rng Q implies rng(R*S) = rng (Q*S)
proof
  assume
A1: rng R = rng Q;
  thus rng(R*S) = S.:rng R by Th127
    .= rng (Q*S) by A1,Th127;
end;



definition
  let R be Relation, x be set;
  func Coim(R,x) equals
  R"{x};
  coherence;
end;



registration
  let R be trivial Relation;
  cluster dom R -> trivial;
  coherence
  proof
    let x,y;
    assume x in dom R;
    then consider a such that
A1: [x,a] in R by XTUPLE_0:def 12;
    assume y in dom R;
    then consider b such that
A2: [y,b] in R by XTUPLE_0:def 12;
    [x,a] = [y,b] by A1,A2,ZFMISC_1:def 10;
    hence thesis by XTUPLE_0:1;
  end;
end;

registration
  let R be trivial Relation;
  cluster rng R -> trivial;
  coherence
  proof
    let x,y;
    assume x in rng R;
    then consider a such that
A1: [a,x] in R by XTUPLE_0:def 13;
    assume y in rng R;
    then consider b such that
A2: [b,y] in R by XTUPLE_0:def 13;
    [a,x] = [b,y] by A1,A2,ZFMISC_1:def 10;
    hence thesis by XTUPLE_0:1;
  end;
end;



theorem
  rng R c= dom (S|X) implies R*(S|X) = R*S
proof
  assume
A1: rng R c= dom (S|X);
  let a,b;
  R*(S|X) c= R*S by Th29,Th59;
  hence [a,b] in R*(S|X) implies [a,b] in R*S;
  assume [a,b] in R*S;
  then consider c such that
A2: [a,c] in R and
A3: [c,b] in S by Def8;
  c in rng R by A2,XTUPLE_0:def 13;
  then c in X by A1,Th57;
  then [c,b] in S|X by A3,Def11;
  hence thesis by A2,Def8;
end;



theorem
  Q|A = R|A implies Q.:A = R.:A
proof
  assume Q|A = R|A;
  hence Q.:A = (R|A).:A by Th129
  .= R.:A by Th129;
end;



definition
  let X,R;
  attr R is X-defined means
  :Def18:
  dom R c= X;
  attr R is X-valued means
  :Def19:
  rng R c= X;
end;

Lm1:
 {} is X-defined Y-valued
 proof
  thus dom{} c= X & rng{} c= Y by XBOOLE_1:2;
 end;

registration
  let X,Y;
  cluster X-defined Y-valued for Relation;
  existence
   proof
    take {};
    thus thesis by Lm1;
   end;
end;



theorem
  for D being set, R being D-valued Relation for y being set st y in rng R
  holds y in D
proof
  let D be set, R be D-valued Relation;
  rng R c= D by Def19;
  hence thesis;
end;



registration
  let X,A;
  let R be A-valued Relation;
  cluster R|X -> A-valued;
  coherence
  proof
    rng(R|X) c= rng R & rng R c= A by Def19,Th70;
    hence rng(R|X) c= A by XBOOLE_1:1;
  end;
end;

registration
  let X,A;
  let R be A-defined Relation;
  cluster R|X -> A-defined X-defined;
  coherence
  proof
    dom(R|X) c= dom R & dom R c= A by Def18,Th60;
    hence dom(R|X) c= A by XBOOLE_1:1;
    thus dom(R|X) c= X by Th58;
  end;
end;

registration
  let X;
  cluster id X -> X-defined X-valued;
  coherence
  proof
    thus dom id X c= X & rng id X c= X;
  end;
end;



registration
  let A be set;
  let R be A-valued Relation, S be Relation;
  cluster S*R -> A-valued;
  coherence
  proof
    rng R c= A & rng(S*R) c= rng R by Def19,Th26;
    hence rng(S*R) c= A by XBOOLE_1:1;
  end;
end;

registration
  let A be set;
  let R be A-defined Relation, S be Relation;
  cluster R*S -> A-defined;
  coherence
  proof
    dom R c= A & dom(R*S) c= dom R by Def18,Th25;
    hence dom(R*S) c= A by XBOOLE_1:1;
  end;
end;



theorem
 x in X implies Im([:X,Y:],x) = Y proof assume
A1:  x in X;
   thus Im([:X,Y:],x) c= Y proof let y;
     assume y in Im([:X,Y:],x);
      then ex z st [z,y] in [:X,Y:] & z in {x} by Def13;
      hence y in Y by ZFMISC_1:87;
    end;
   let y;
   assume y in Y;
    then
A2:   [x,y] in [:X,Y:] by A1,ZFMISC_1:87;
     x in {x} by TARSKI:def 1;
   hence y in Im([:X,Y:],x) by A2,Def13;
  end;

theorem Th169:
 [x,y] in R iff y in Im(R,x)
  proof
    thus [x,y] in R implies y in Im(R,x)
    proof
      x in {x} by TARSKI:def 1;
      hence thesis by Def13;
    end;
    assume y in Im(R,x);
    then ex z st [z,y] in R & z in {x} by Def13;
    hence [x,y] in R by TARSKI:def 1;
  end;

theorem
  x in dom R iff Im(R,x) <> {}
proof
 thus x in dom R implies Im(R,x) <> {}
  proof
   assume x in dom R;
    then ex y st [x,y] in R by XTUPLE_0:def 12;
   hence Im(R,x) <> {} by Th169;
  end;
 assume Im(R,x) <> {};
  then consider y such that
A1: y in Im(R,x) by XBOOLE_0:def 1;
  [x,y] in R by A1,Th169;
 hence x in dom R by XTUPLE_0:def 12;
end;

theorem
 {} is X-defined Y-valued by Lm1;



registration
  cluster empty -> non-empty for Relation;
  coherence
  proof let R be Relation;
   assume R is empty;
   hence not {} in rng R;
  end;
end;

registration let X be set, R be X-defined Relation;
  cluster -> X-defined for Subset of R;
  coherence
  proof let S be Subset of R;
A1:  dom R c= X by Def18;
    dom S c= dom R by Th11;
   hence dom S c= X by A1,XBOOLE_1:1;
  end;
end;

registration let X be set, R be X-valued Relation;
  cluster -> X-valued for Subset of R;
  coherence
  proof let S be Subset of R;
A1:  rng R c= X by Def19;
    rng S c= rng R by Th11;
    hence rng S c= X by A1,XBOOLE_1:1;
  end;
end;



theorem
  X misses Y implies R|X|Y = {}
proof
  assume X misses Y;
  then X /\ Y = {} by XBOOLE_0:def 7;
  hence R|X|Y = R|({} qua set) by Th71
     .= {};
end;



theorem
  field {[x,x]} = {x}
proof
  thus field {[x,x]} = {x,x} by Th17
    .= {x} by ENUMSET1:29;
end;

registration let X; let R be X-defined Relation;
 reduce R|X to R;
 reducibility
proof
  dom R c= X by Def18;
  hence R|X = R by Th68;
end;
end;

registration let Y; let R be Y-valued Relation;
 reduce Y|`R to R;
 reducibility
proof
  rng R c= Y by Def19;
  hence Y|`R = R by Th94;
end;
end;

theorem
  for R being X-defined Relation holds R = R|X;

theorem
  for S being Relation, R being X-defined Relation st R c= S holds R c= S|X
proof let S be Relation, R be X-defined Relation;
  R = R|X;
  hence thesis by Th76;
end;



theorem Th176:
  dom R c= X implies R \ (R|A) = R|(X\A)
proof
  assume dom R c= X;
  hence R \ (R|A) =  R|X \ R|A by Th68
     .= R|(X\A) by Th80;
end;

theorem Th177:
  dom(R \ (R|A)) = dom R \ A
proof
  R \ (R|A) = R|(dom R \ A) by Th176;
  hence dom(R \ (R|A)) = dom R /\ (dom R \ A) by Th61
      .= (dom R /\ dom R) \ A by XBOOLE_1:49
      .= dom R \ A;
end;

theorem
  dom R \ dom(R|A) = dom(R \ (R|A))
proof
  thus dom R \ dom(R|A) = dom R \ A /\ dom R by Th61
     .= dom R \ A by XBOOLE_1:47
     .= dom(R \ (R|A)) by Th177;
end;

theorem
  dom R misses dom S implies R misses S
proof
   assume
A1: dom R misses dom S;
   assume R meets S;
   then consider x such that
A2:x in R and
A3:x in S by XBOOLE_0:3;
   consider y,z such that
A4:x = [y,z] by A2,Def1;
   y in dom R & y in dom S by A2,A3,A4,XTUPLE_0:def 12;
   hence contradiction by A1,XBOOLE_0:3;
end;

theorem
  rng R misses rng S implies R misses S
proof
  assume
A1:rng R misses rng S;
  assume R meets S;
  then consider x such that
A2: x in R and
A3: x in S by XBOOLE_0:3;
   consider y,z such that
A4: x = [y,z] by A2,Def1;
   z in rng R & z in rng S by A2,A3,A4,XTUPLE_0:def 13;
   hence contradiction by A1,XBOOLE_0:3;
end;

theorem
  X c= Y implies (R \ R|Y)|X = {}
 proof
   assume
A1: X c= Y;
   dom R /\ X c= X by XBOOLE_1:17; then
A2:  dom R /\ X c= Y by A1,XBOOLE_1:1;
    dom(R \ R|Y) = dom R \ Y by Th177;
    then dom((R \ R|Y)|X) =(dom R \ Y) /\ X by Th61
       .= dom R /\ X \ Y by XBOOLE_1:49
       .= {} by A2,XBOOLE_1:37;
   hence (R \ R|Y)|X = {};
 end;

theorem
 X c= Y implies
 for R being X-defined Relation holds R is Y-defined
 proof assume
A1: X c= Y;
  let R be X-defined Relation;
   dom R c= X by Def18;
  hence dom R c= Y by A1,XBOOLE_1:1;
 end;

theorem
 X c= Y implies
 for R being X-valued Relation holds R is Y-valued
 proof assume
A1: X c= Y;
  let R be X-valued Relation;
   rng R c= X by Def19;
  hence rng R c= Y by A1,XBOOLE_1:1;
 end;

theorem
  R c= S iff R c= S|dom R
proof
 thus R c= S implies R c= S|dom R
  proof
   assume R c= S;
    then R|dom R c= S|dom R by Th76;
   hence thesis;
  end;
  S|dom R c= S by Th59;
 hence thesis by XBOOLE_1:1;
end;

theorem
  for R being X-defined Y-valued Relation holds R c= [:X,Y:]
  proof
    let R be X-defined Y-valued Relation;
A1:  R c= [:dom R,rng R:] by Th7;
    dom R c= X & rng R c= Y by Def18,Def19;
    then [:dom R,rng R:] c= [:X,Y:] by ZFMISC_1:96;
    hence thesis by A1,XBOOLE_1:1;
  end;



theorem Th186:
  dom(X|`R) c= dom R
proof
  X|`R c= R by Th86;
  hence thesis by Th11;
end;

registration
  let A,X;
  let R be A-defined Relation;
  cluster X|`R -> A-defined;
  coherence
  proof
    dom(X|`R) c= dom R & dom R c= A by Def18,Th186;
    hence dom(X|`R) c= A by XBOOLE_1:1;
  end;
end;

registration
  let X,A;
  let R be A-valued Relation;
  cluster X|`R -> A-valued X-valued;
  coherence
  proof
    rng(X|`R) c= rng R & rng R c= A by Def19,Th87;
    hence rng(X|`R) c= A by XBOOLE_1:1;
    thus rng(X|`R) c= X by Th85;
  end;
end;

registration
 let X be empty set;
 cluster -> empty for X-defined Relation;
 coherence
  proof let R be X-defined Relation;
    dom R c= X by Def18;
   hence thesis by XBOOLE_1:3;
  end;
 cluster -> empty for X-valued Relation;
 coherence
  proof let R be X-valued Relation;
    rng R c= X by Def19;
   hence thesis by XBOOLE_1:3;
  end;
end;
















begin

reserve X,X1,X2,Y,Y1,Y2,p,x,x1,x2,y,y1,y2,z,z1,z2 for set;

definition
  let X be set;
  attr X is Function-like means
  :Def1:
  for x,y1,y2 st [x,y1] in X & [x,y2] in X holds y1 = y2;
end;

registration
  cluster empty -> Function-like for set;
  coherence
  proof
    let f be set;
    assume f is empty;
    hence for x,y1,y2 st [x,y1] in f & [x,y2] in f holds y1 = y2;
  end;
end;

registration
  cluster Function-like for Relation;
  existence
  proof
    take {};
    thus thesis;
  end;
end;

definition
  mode Function is Function-like Relation;
end;

registration
  let a, b be set;
  cluster {[a,b]} -> Function-like;
  coherence
  proof
    set X = {[a, b]};
A1: [:{a},{b}:] = X by ZFMISC_1:29;
    for x,y1,y2 be set st [x,y1] in X & [x,y2] in X holds y1 = y2
    proof
      let x,y1,y2 be set such that
A2:   [x,y1] in X and
A3:   [x,y2] in X;
      y1 = b by A1,A2,ZFMISC_1:28;
      hence thesis by A1,A3,ZFMISC_1:28;
    end;
    hence thesis by Def1;
  end;
end;

reserve f,g,g1,g2,h for Function,
  R,S for Relation;

scheme
  GraphFunc { A()->set,P[set,set] } :
  ex f st for x,y holds [x,y] in f iff x in A() & P[x,y]
provided
A1: for x,y1,y2 st P[x,y1] & P[x,y2] holds y1 = y2
proof
  consider Y such that
A2: for y holds y in Y iff ex x st x in A() & P[x,y] from TARSKI:sch 1(A1);
  defpred R[set] means ex x,y st [x,y] = $1 & P[x,y];
  consider F being set such that
A3: p in F iff p in [:A(),Y:] & R[p] from XBOOLE_0:sch 1;
  now
    thus p in F implies ex x,y st [x,y] = p
    proof
      p in F implies ex x,y st [x,y] = p & P[x,y] by A3;
      hence thesis;
    end;
    let x,y1,y2;
    assume [x,y1] in F;
    then consider x1,z1 such that
A4: [x1,z1] = [x,y1] and
A5: P[x1,z1] by A3;
A6: x = x1 & z1 = y1 by A4,XTUPLE_0:1;
    assume [x,y2] in F;
    then consider x2,z2 such that
A7: [x2,z2] = [x,y2] and
A8: P[x2,z2] by A3;
    x = x2 & z2 = y2 by A7,XTUPLE_0:1;
    hence y1 = y2 by A1,A5,A8,A6;
  end;
  then reconsider f = F as Function by Def1,RELAT_1:def 1;
  take f;
  let x,y;
  thus [x,y] in f implies x in A() & P[x,y]
  proof
    assume
A9: [x,y] in f;
    then consider x1,y1 such that
A10: [x1,y1] = [x,y] and
A11: P[x1,y1] by A3;
    [x,y] in [:A(),Y:] by A3,A9;
    hence x in A() by ZFMISC_1:87;
    x1 = x by A10,XTUPLE_0:1;
    hence thesis by A10,A11,XTUPLE_0:1;
  end;
  assume that
A12: x in A() and
A13: P[x,y];
  y in Y by A2,A12,A13;
  then [x,y] in [:A(),Y:] by A12,ZFMISC_1:87;
  hence thesis by A3,A13;
end;

definition
  let f,x;
  func f.x -> set means
  :Def2:
  [x,it] in f if x in dom f otherwise it = {};
  existence by XTUPLE_0:def 12;
  uniqueness by Def1;
  consistency;
end;

theorem Th1:
  [x,y] in f iff x in dom f & y = f.x
proof
  thus [x,y] in f implies x in dom f & y = f.x
  proof
    assume
A1: [x,y] in f;
    hence x in dom f by XTUPLE_0:def 12;
    hence thesis by A1,Def2;
  end;
  thus thesis by Def2;
end;

theorem Th2:
  dom f = dom g & (for x st x in dom f holds f.x = g.x) implies f = g
proof
  assume that
A1: dom f = dom g and
A2: for x st x in dom f holds f.x = g.x;
  let x,y;
  thus [x,y] in f implies [x,y] in g
  proof
    assume
A3: [x,y] in f;
    then
A4: x in dom f by XTUPLE_0:def 12;
    then f.x = y by A3,Def2;
    then g.x = y by A2,A4;
    hence thesis by A1,A4,Def2;
  end;
  assume
A5: [x,y] in g;
  then
A6: x in dom g by XTUPLE_0:def 12;
  then g.x = y by A5,Def2;
  then f.x = y by A1,A2,A6;
  hence thesis by A1,A6,Def2;
end;

definition
  let f;
  redefine func rng f means
  :Def3:
  for y holds y in it iff ex x st x in dom f & y = f.x;
  compatibility
  proof
    let Y;
    hereby
      assume
A1:   Y = rng f;
      let y;
      hereby
        assume y in Y;
        then consider x such that
A2:     [x,y] in f by A1,XTUPLE_0:def 13;
        take x;
        thus x in dom f & y = f.x by A2,Th1;
      end;
      given x such that
A3:   x in dom f & y = f.x;
      [x,y] in f by A3,Def2;
      hence y in Y by A1,XTUPLE_0:def 13;
    end;
    assume
A4: for y holds y in Y iff ex x st x in dom f & y = f.x;
    hereby
      let y;
      assume y in Y;
      then consider x such that
A5:   x in dom f & y = f.x by A4;
      [x,y] in f by A5,Def2;
      hence y in rng f by XTUPLE_0:def 13;
    end;
    let y;
    assume y in rng f;
    then consider x such that
A6: [x,y] in f by XTUPLE_0:def 13;
    x in dom f & y = f.x by A6,Th1;
    hence thesis by A4;
  end;
end;

theorem
  x in dom f implies f.x in rng f by Def3;

theorem Th4:
  dom f = {x} implies rng f = {f.x}
proof
  assume
A1: dom f = {x};
  y in rng f iff y in {f.x}
  proof
    thus y in rng f implies y in {f.x}
    proof
      assume y in rng f;
      then consider z such that
A2:   z in dom f and
A3:   y = f.z by Def3;
      z = x by A1,A2,TARSKI:def 1;
      hence thesis by A3,TARSKI:def 1;
    end;
    assume y in {f.x};
    then
A4: y = f.x by TARSKI:def 1;
    x in dom f by A1,TARSKI:def 1;
    hence thesis by A4,Def3;
  end;
  hence thesis by TARSKI:1;
end;

scheme
  FuncEx { A()->set,P[set,set] } :
  ex f st dom f = A() & for x st x in A() holds P[x,f.x]
provided
A1: for x,y1,y2 st x in A() & P[x,y1] & P[x,y2] holds y1 = y2 and
A2: for x st x in A() ex y st P[x,y]
proof
  defpred R[set,set] means $1 in A() & P[$1,$2];
A3: for x,y1,y2 st R[x,y1] & R[x,y2] holds y1 = y2 by A1;
  consider f being Function such that
A4: for x,y holds [x,y] in f iff x in A() & R[x,y] from GraphFunc(A3);
  take f;
  x in dom f iff x in A()
  proof
    thus x in dom f implies x in A()
    proof
      assume x in dom f;
      then ex y st [x,y] in f by XTUPLE_0:def 12;
      hence thesis by A4;
    end;
    assume
A5: x in A();
    then consider y such that
A6: P[x,y] by A2;
    [x,y] in f by A4,A5,A6;
    hence thesis by XTUPLE_0:def 12;
  end;
  hence
A7: dom f = A() by TARSKI:1;
  let x;
  assume
A8: x in A();
  then consider y such that
A9: P[x,y] by A2;
  [x,y] in f by A4,A8,A9;
  hence thesis by A7,A8,A9,Def2;
end;

scheme
  Lambda { A() -> set,F(set) -> set } :
  ex f being Function st dom f = A() & for x st x in A() holds f.x = F(x)
proof
  defpred P[set,set] means $2 = F($1);
A1: for x st x in A() ex y st P[x,y];
A2: for x,y1,y2 st x in A() & P[x,y1] & P[x,y2] holds y1 = y2;
  thus ex f being Function st dom f = A() & for x st x in A() holds P[x,f.x]
  from FuncEx(A2,A1);
end;

theorem Th5:
  X <> {} implies for y ex f st dom f = X & rng f = {y}
proof
  assume
A1: X <> {};
  let y;
  deffunc F(set) = y;
  consider f such that
A2: dom f = X and
A3: for x st x in X holds f.x = F(x) from Lambda;
  take f;
  thus dom f = X by A2;
  y1 in rng f iff y1 = y
  proof
A4: now
      set x = the Element of X;
      assume
A5:   y1 = y;
      f.x = y by A1,A3;
      hence y1 in rng f by A1,A2,A5,Def3;
    end;
    now
      assume y1 in rng f;
      then ex x st x in dom f & y1 = f.x by Def3;
      hence y1 = y by A2,A3;
    end;
    hence thesis by A4;
  end;
  hence thesis by TARSKI:def 1;
end;

theorem
  (for f,g st dom f = X & dom g = X holds f = g) implies X = {}
proof
  deffunc F(set) = {};
  assume
A1: for f,g st dom f = X & dom g = X holds f = g;
  set x = the Element of X;
  consider f being Function such that
A2: dom f = X and
A3: for x st x in X holds f.x = F(x) from Lambda;
  assume
A4: not thesis;
  then
A5: f.x = {} by A3;
  deffunc F(set) = 1;
  consider g being Function such that
A6: dom g = X and
A7: for x st x in X holds g.x = F(x) from Lambda;
  g.x = 1 by A4,A7;
  hence contradiction by A1,A2,A6,A5;
end;

theorem
  dom f = dom g & rng f = {y} & rng g = {y} implies f = g
proof
  assume that
A1: dom f = dom g and
A2: rng f = {y} and
A3: rng g = {y};
  x in dom f implies f.x = g.x
  proof
    assume
A4: x in dom f;
    then f.x in rng f by Def3;
    then
A5: f.x = y by A2,TARSKI:def 1;
    g.x in rng g by A1,A4,Def3;
    hence thesis by A3,A5,TARSKI:def 1;
  end;
  hence thesis by A1,Th2;
end;

theorem
  Y <> {} or X = {} implies ex f st X = dom f & rng f c= Y
proof
  assume
A1: Y <> {} or X = {};
A2: now
    set y = the Element of Y;
    deffunc F(set) = y;
    consider f such that
A3: dom f = X and
A4: for x st x in X holds f.x = F(x) from Lambda;
    assume X <> {};
    then
A5: y in Y by A1;
    take f;
    thus dom f = X by A3;
    z in rng f implies z in Y
    proof
      assume z in rng f;
      then ex x st x in dom f & z = f.x by Def3;
      hence thesis by A5,A3,A4;
    end;
    hence rng f c= Y by TARSKI:def 3;
  end;
  now
    assume
A6: X = {};
    take f = {};
    thus dom f = X by A6;
    thus rng f c= Y by XBOOLE_1:2;
  end;
  hence thesis by A2;
end;

theorem
  (for y st y in Y ex x st x in dom f & y = f.x) implies Y c= rng f
proof
  assume
A1: for y st y in Y ex x st x in dom f & y = f.x;
  let y;
  assume y in Y;
  then ex x st x in dom f & y = f.x by A1;
  hence thesis by Def3;
end;

notation
  let f,g;
  synonym g*f for f*g;
end;

registration
  let f,g;
  cluster g*f -> Function-like;
  coherence
  proof
    let x,y1,y2;
    assume [x,y1] in g*f;
    then consider z1 such that
A1: [x,z1] in f and
A2: [z1,y1] in g by RELAT_1:def 8;
    assume [x,y2] in g*f;
    then consider z2 such that
A3: [x,z2] in f and
A4: [z2,y2] in g by RELAT_1:def 8;
    z1 = z2 by A1,A3,Def1;
    hence thesis by A2,A4,Def1;
  end;
end;

theorem
  for h st (for x holds x in dom h iff x in dom f & f.x in dom g) & (for
  x st x in dom h holds h.x = g.(f.x)) holds h = g*f
proof
  let h;
  assume that
A1: for x holds x in dom h iff x in dom f & f.x in dom g and
A2: for x st x in dom h holds h.x = g.(f.x);
  now
    let x,y;
    hereby
      assume
A3:   [x,y] in h;
      then
A4:   x in dom h by XTUPLE_0:def 12;
      then
A5:   f.x in dom g by A1;
      take y1 = f.x;
      x in dom f by A1,A4;
      hence [x,y1] in f by Def2;
      y = h.x by A3,A4,Def2
        .= g.(f.x) by A2,A4;
      hence [y1,y] in g by A5,Def2;
    end;
    given z such that
A6: [x,z] in f and
A7: [z,y] in g;
A8: x in dom f by A6,XTUPLE_0:def 12;
    then
A9: z = f.x by A6,Def2;
A10: z in dom g by A7,XTUPLE_0:def 12;
    then
A11: x in dom h by A1,A8,A9;
    y = g.z by A7,A10,Def2;
    then y = h.x by A2,A9,A11;
    hence [x,y] in h by A11,Def2;
  end;
  hence thesis by RELAT_1:def 8;
end;

theorem Th11:
  x in dom(g*f) iff x in dom f & f.x in dom g
proof
  set h = g*f;
  hereby
    assume x in dom h;
    then consider y such that
A1: [x,y] in h by XTUPLE_0:def 12;
    consider z such that
A2: [x,z] in f and
A3: [z,y] in g by A1,RELAT_1:def 8;
    thus x in dom f by A2,XTUPLE_0:def 12;
    then z = f.x by A2,Def2;
    hence f.x in dom g by A3,XTUPLE_0:def 12;
  end;
  assume
A4: x in dom f;
  then consider z such that
A5: [x,z] in f by XTUPLE_0:def 12;
  assume f.x in dom g;
  then consider y such that
A6: [f.x,y] in g by XTUPLE_0:def 12;
  z = f.x by A4,A5,Def2;
  then [x,y] in h by A5,A6,RELAT_1:def 8;
  hence thesis by XTUPLE_0:def 12;
end;

theorem Th12:
  x in dom(g*f) implies (g*f).x = g.(f.x)
proof
  set h = g*f;
  assume
A1: x in dom h;
  then consider y such that
A2: [x,y] in h by XTUPLE_0:def 12;
  consider z such that
A3: [x,z] in f and
A4: [z,y] in g by A2,RELAT_1:def 8;
  x in dom f by A3,XTUPLE_0:def 12;
  then
A5: z = f.x by A3,Def2;
  then f.x in dom g by A4,XTUPLE_0:def 12;
  then y = g.(f.x) by A4,A5,Def2;
  hence thesis by A1,A2,Def2;
end;

theorem Th13:
  x in dom f implies (g*f).x = g.(f.x)
proof
  assume
A1: x in dom f;
  per cases;
  suppose
    f.x in dom g;
    then x in dom(g*f) by A1,Th11;
    hence thesis by Th12;
  end;
  suppose
A2: not f.x in dom g;
    then not x in dom(g*f) by Th11;
    hence (g*f).x = {} by Def2
      .= g.(f.x) by A2,Def2;
  end;
end;

theorem
  z in rng(g*f) implies z in rng g
proof
  assume z in rng(g*f);
  then consider x such that
A1: x in dom(g*f) and
A2: z = (g*f).x by Def3;
  f.x in dom g & (g*f).x = g.(f.x) by A1,Th11,Th12;
  hence thesis by A2,Def3;
end;

theorem Th15:
  dom(g*f) = dom f implies rng f c= dom g
proof
  assume
A1: dom(g*f) = dom f;
  let y;
  assume y in rng f;
  then ex x st x in dom f & y = f.x by Def3;
  hence thesis by A1,Th11;
end;

theorem
  rng f c= Y & (for g,h st dom g = Y & dom h = Y & g*f = h*f holds g = h
  ) implies Y = rng f
proof
  assume that
A1: rng f c= Y and
A2: for g,h st dom g = Y & dom h = Y & g*f = h*f holds g = h;
  Y c= rng f
  proof
    deffunc F(set) = {};
    let y;
    assume that
A3: y in Y and
A4: not y in rng f;
    defpred P[set,set] means ($1 = y implies $2 = 1) & ($1 <> y implies $2 =
    {});
A5: x in Y implies ex y1 st P[x,y1]
    proof
      assume x in Y;
      x = y implies thesis;
      hence thesis;
    end;
A6: for x,y1,y2 st x in Y & P[x,y1] & P[x,y2] holds y1 = y2;
    consider h being Function such that
A7: dom h = Y and
A8: for x st x in Y holds P[x,h.x] from FuncEx(A6,A5);
A9: dom(h*f) = dom f by A1,A7,RELAT_1:27;
    consider g being Function such that
A10: dom g = Y and
A11: x in Y implies g.x = F(x) from Lambda;
A12: dom(g*f) = dom f by A1,A10,RELAT_1:27;
    x in dom f implies (g*f).x = (h*f).x
    proof
      assume
A13:  x in dom f;
      then f.x in rng f by Def3;
      then
A14:  g.(f.x) = {} & h.(f.x) = {} by A1,A4,A11,A8;
      (g*f).x = g.(f.x) by A12,A13,Th12;
      hence thesis by A9,A13,A14,Th12;
    end;
    then
A15: g = h by A2,A10,A7,A12,A9,Th2;
    g.y = {} by A3,A11;
    hence contradiction by A3,A8,A15;
  end;
  hence thesis by A1,XBOOLE_0:def 10;
end;

registration
  let X;
  cluster id X -> Function-like;
  coherence
  proof
    let x,y1,y2;
    assume that
A1: [x,y1] in id X and
A2: [x,y2] in id X;
    x = y1 by A1,RELAT_1:def 10;
    hence thesis by A2,RELAT_1:def 10;
  end;
end;

theorem Th17:
  f = id X iff dom f = X & for x st x in X holds f.x = x
proof
  hereby
    assume
A1: f = id X;
    hence
A2: dom f = X by RELAT_1:45;
    let x;
    assume
A3: x in X;
    then [x,x] in f by A1,RELAT_1:def 10;
    hence f.x = x by A2,A3,Def2;
  end;
  assume that
A4: dom f = X and
A5: for x st x in X holds f.x = x;
  now
    let x,y;
    hereby
      assume
A6:   [x,y] in f;
      hence
A7:   x in X by A4,Th1;
      y = f.x by A6,Th1;
      hence x = y by A5,A7;
    end;
    assume
A8: x in X;
    then f.x = x by A5;
    hence x = y implies [x,y] in f by A4,A8,Th1;
  end;
  hence thesis by RELAT_1:def 10;
end;

theorem
  x in X implies (id X).x = x by Th17;

theorem Th19:
  dom(f*(id X)) = dom f /\ X
proof
  x in dom(f*(id X)) iff x in dom f /\ X
  proof
    x in dom(f*(id X)) iff x in dom f & x in X
    proof
      thus x in dom(f*(id X)) implies x in dom f & x in X
      proof
        assume x in dom(f*(id X));
        then
A1:     x in dom((id X)) & (id X).x in dom f by Th11;
        thus thesis by A1,Th17;
      end;
      assume
A2:   x in dom f;
A3:   dom((id X)) = X;
      assume
A4:   x in X;
      then (id X).x in dom f by A2,Th17;
      hence thesis by A4,A3,Th11;
    end;
    hence thesis by XBOOLE_0:def 4;
  end;
  hence thesis by TARSKI:1;
end;

theorem
  x in dom f /\ X implies f.x = (f*(id X)).x
proof
  assume x in dom f /\ X;
  then x in X by XBOOLE_0:def 4;
  then (id X).x = x & x in dom id X by Th17;
  hence thesis by Th13;
end;

theorem
  x in dom((id Y)*f) iff x in dom f & f.x in Y
proof
  dom((id Y)) = Y;
  hence thesis by Th11;
end;

theorem
  (id X)*(id Y) = id(X /\ Y)
proof
A1: dom((id X)*(id Y)) = dom((id X)) /\ Y by Th19
    .= X /\ Y;
A2: z in X /\ Y implies ((id X)*(id Y)).z = (id(X /\ Y)).z
  proof
    assume
A3: z in X /\ Y;
    then
A4: z in X by XBOOLE_0:def 4;
A5: z in Y by A3,XBOOLE_0:def 4;
    thus ((id X)*(id Y)).z = (id X).((id Y).z) by A1,A3,Th12
      .= (id X).z by A5,Th17
      .= z by A4,Th17
      .= (id(X /\ Y)).z by A3,Th17;
  end;
  X /\ Y = dom id(X /\ Y);
  hence thesis by A1,A2,Th2;
end;

theorem Th23:
  rng f = dom g & g*f = f implies g = id dom g
proof
  assume that
A1: rng f = dom g and
A2: g*f = f;
  set X = dom g;
  x in X implies g.x = x
  proof
    assume x in X;
    then ex y st y in dom f & f.y = x by A1,Def3;
    hence thesis by A2,Th13;
  end;
  hence thesis by Th17;
end;

definition
  let f;

  attr f is one-to-one means
  :Def4:
  for x1,x2 st x1 in dom f & x2 in dom f & f.x1 = f.x2 holds x1 = x2;
end;

theorem Th24:
  f is one-to-one & g is one-to-one implies g*f is one-to-one
proof
  assume that
A1: f is one-to-one and
A2: g is one-to-one;
  now
    let x1,x2;
    assume
A3: x1 in dom(g*f) & x2 in dom(g*f);
    then
A4: (g*f).x1 = g.(f.x1) & (g*f).x2 = g .(f.x2) by Th12;
A5: x1 in dom f & x2 in dom f by A3,Th11;
    assume
A6: (g*f).x1 = (g*f).x2;
    f.x1 in dom g & f.x2 in dom g by A3,Th11;
    then f.x1 = f.x2 by A2,A4,A6,Def4;
    hence x1 = x2 by A1,A5,Def4;
  end;
  hence thesis by Def4;
end;

theorem Th25:
  g*f is one-to-one & rng f c= dom g implies f is one-to-one
proof
  assume that
A1: g*f is one-to-one and
A2: rng f c= dom g;
  now
    let x1,x2;
    assume that
A3: x1 in dom f & x2 in dom f and
A4: f.x1 =f.x2;
A5: x1 in dom(g*f) & x2 in dom(g*f) by A2,A3,RELAT_1:27;
    (g*f).x1 = g.(f.x1) & (g*f).x2 = g.(f.x2) by A3,Th13;
    hence x1 = x2 by A1,A4,A5,Def4;
  end;
  hence thesis by Def4;
end;

theorem
  g*f is one-to-one & rng f = dom g implies f is one-to-one & g is one-to-one
proof
  assume that
A1: g*f is one-to-one and
A2: rng f = dom g;
A3: dom(g*f) = dom f by A2,RELAT_1:27;
  thus f is one-to-one by A1,A2,Th25;
  assume not g is one-to-one;
  then consider y1,y2 such that
A4: y1 in dom g and
A5: y2 in dom g and
A6: g.y1 = g.y2 & y1 <> y2 by Def4;
  consider x2 such that
A7: x2 in dom f and
A8: f.x2 = y2 by A2,A5,Def3;
A9: (g*f).x2 = g.(f.x2) by A7,Th13;
  consider x1 such that
A10: x1 in dom f and
A11: f.x1 = y1 by A2,A4,Def3;
  (g*f).x1 = g.(f.x1) by A10,Th13;
  hence contradiction by A1,A6,A10,A11,A7,A8,A3,A9,Def4;
end;

theorem
  f is one-to-one iff for g,h st rng g c= dom f & rng h c= dom f & dom g
  = dom h & f*g = f*h holds g = h
proof
  thus f is one-to-one implies for g,h st rng g c=dom f & rng h c=dom f & dom
  g = dom h & f*g = f*h holds g = h
  proof
    assume
A1: f is one-to-one;
    let g,h such that
A2: rng g c= dom f & rng h c= dom f and
A3: dom g = dom h and
A4: f*g = f*h;
    x in dom g implies g.x = h.x
    proof
      assume
A5:   x in dom g;
      then
A6:   g.x in rng g & h.x in rng h by A3,Def3;
      (f*g).x = f.(g.x) & (f*h).x = f.(h.x) by A3,A5,Th13;
      hence thesis by A1,A2,A4,A6,Def4;
    end;
    hence thesis by A3,Th2;
  end;
  assume
A7: for g,h st rng g c=dom f & rng h c=dom f & dom g = dom h & f*g = f*
  h holds g = h;
  x1 in dom f & x2 in dom f & f.x1 = f.x2 implies x1 = x2
  proof
    assume that
A8: x1 in dom f and
A9: x2 in dom f and
A10: f.x1 = f.x2;
    deffunc F(set) = x1;
    consider g being Function such that
A11: dom g = {{}} and
A12: for x st x in {{}} holds g.x = F(x) from Lambda;
A13: {} in {{}} by TARSKI:def 1;
    then
A14: g.{} = x1 by A12;
    then rng g = {x1} by A11,Th4;
    then
A15: rng g c= dom f by A8,ZFMISC_1:31;
    then
A16: dom(f*g) = dom g by RELAT_1:27;
    deffunc F(set) = x2;
    consider h being Function such that
A17: dom h = {{}} and
A18: for x st x in {{}} holds h.x = F(x) from Lambda;
A19: h.{} = x2 by A18,A13;
    then rng h = {x2} by A17,Th4;
    then
A20: rng h c= dom f by A9,ZFMISC_1:31;
    then
A21: dom(f*h) = dom h by RELAT_1:27;
    x in dom(f*g) implies (f*g).x = (f*h).x
    proof
      assume
A22:  x in dom(f*g);
      then
A23:  g.x = x1 by A11,A12,A16;
      (f*g).x = f.(g.x) & (f*h).x = f.(h.x) by A11,A17,A16,A21,A22,Th12;
      hence thesis by A10,A11,A18,A16,A22,A23;
    end;
    hence thesis by A7,A11,A17,A14,A19,A15,A20,A16,A21,Th2;
  end;
  hence thesis by Def4;
end;

theorem
  dom f = X & dom g = X & rng g c= X & f is one-to-one & f*g = f implies
  g = id X
proof
  assume that
A1: dom f = X and
A2: dom g = X and
A3: rng g c= X & f is one-to-one and
A4: f*g = f;
  x in X implies g.x = x
  proof
    assume
A5: x in X;
    then g.x in rng g & f.x = f.(g.x) by A2,A4,Def3,Th13;
    hence thesis by A1,A3,A5,Def4;
  end;
  hence thesis by A2,Th17;
end;

theorem
  rng(g*f) = rng g & g is one-to-one implies dom g c= rng f
proof
  assume that
A1: rng(g*f) = rng g and
A2: g is one-to-one;
  let y;
  assume
A3: y in dom g;
  then g.y in rng(g*f) by A1,Def3;
  then consider x such that
A4: x in dom(g*f) and
A5: g.y = (g*f).x by Def3;
  (g*f).x = g.(f.x) & f.x in dom g by A4,Th11,Th12;
  then
A6: y = f.x by A2,A3,A5,Def4;
  x in dom f by A4,Th11;
  hence thesis by A6,Def3;
end;

registration
  let X be set;
  cluster id X -> one-to-one;
  coherence
proof
  let x1,x2;
  assume that
A1: x1 in dom id X and
A2: x2 in dom id X;
  x1 in X by A1;
  then
A3: (id X).x1 = x1 by Th17;
  x2 in X by A2;
  hence thesis by A3,Th17;
end;
end;

canceled;

theorem
  (ex g st g*f = id dom f) implies f is one-to-one
proof
  given g such that
A1: g*f = id dom f;
  dom(g*f) = dom f by A1,RELAT_1:45;
  then rng f c= dom g by Th15;
  hence thesis by A1,Th25;
end;

registration
  cluster empty -> one-to-one for Function;
  coherence
  proof
    let f be Function;
    assume
A1: f is empty;
    let x1,x2;
    thus thesis by A1;
  end;
end;

registration
  cluster one-to-one for Function;
  existence
  proof
    take {};
    thus thesis;
  end;
end;

registration
  let f be one-to-one Function;
  cluster f~ -> Function-like;
  coherence
  proof
    let x,y1,y2;
    assume that
A1: [x,y1] in f~ and
A2: [x,y2] in f~;
A3: [y2,x] in f by A2,RELAT_1:def 7;
    then
A4: y2 in dom f by XTUPLE_0:def 12;
    then
A5: x = f.y2 by A3,Def2;
A6: [y1,x] in f by A1,RELAT_1:def 7;
    then
A7: y1 in dom f by XTUPLE_0:def 12;
    then x = f.y1 by A6,Def2;
    hence thesis by A7,A4,A5,Def4;
  end;
end;

definition
  let f;
  assume
A1: f is one-to-one;
  func f" -> Function equals
  :Def5:
  f~;
  coherence by A1;
end;

theorem Th32:
  f is one-to-one implies for g being Function holds g=f" iff dom
  g = rng f & for y,x holds y in rng f & x = g.y iff x in dom f & y = f.x
proof
  assume
A1: f is one-to-one;
  let g be Function;
  thus g = f" implies dom g = rng f & for y,x holds y in rng f & x = g.y iff x
  in dom f & y = f.x
  proof
    assume g = f";
    then
A2: g = f~ by A1,Def5;
    hence dom g = rng f by RELAT_1:20;
    let y,x;
    thus y in rng f & x = g.y implies x in dom f & y = f.x
    proof
      assume that
A3:   y in rng f and
A4:   x = g.y;
      y in dom g by A2,A3,RELAT_1:20;
      then [y,x] in g by A4,Def2;
      then
A5:   [x,y] in f by A2,RELAT_1:def 7;
      hence x in dom f by XTUPLE_0:def 12;
      hence thesis by A5,Def2;
    end;
    assume x in dom f & y = f.x;
    then
A6: [x,y] in f by Def2;
    hence y in rng f by XTUPLE_0:def 13;
    then
A7: y in dom g by A2,RELAT_1:20;
    [y,x] in g by A2,A6,RELAT_1:def 7;
    hence thesis by A7,Def2;
  end;
  assume that
A8: dom g = rng f and
A9: for y,x holds y in rng f & x = g.y iff x in dom f & y = f.x;
  let a,b be set;
  thus [a,b] in g implies [a,b] in f"
  proof
    assume
A10: [a,b] in g;
    then
A11: a in dom g by XTUPLE_0:def 12;
    then b = g.a by A10,Def2;
    then b in dom f & a = f.b by A8,A9,A11;
    then [b,a] in f by Def2;
    then [a,b] in f~ by RELAT_1:def 7;
    hence thesis by A1,Def5;
  end;
  assume [a,b] in f";
  then [a,b] in f~ by A1,Def5;
  then
A12: [b,a] in f by RELAT_1:def 7;
  then
A13: b in dom f by XTUPLE_0:def 12;
  then a = f.b by A12,Def2;
  then a in rng f & b = g.a by A9,A13;
  hence thesis by A8,Def2;
end;

theorem Th33:
  f is one-to-one implies rng f = dom(f") & dom f = rng(f")
proof
  assume f is one-to-one;
  then f" = f~ by Def5;
  hence thesis by RELAT_1:20;
end;

theorem Th34:
  f is one-to-one & x in dom f implies x = (f").(f.x) & x = (f"*f) .x
proof
  assume
A1: f is one-to-one;
  assume
A2: x in dom f;
  hence x = (f").(f.x) by A1,Th32;
  hence thesis by A2,Th13;
end;

theorem Th35:
  f is one-to-one & y in rng f implies y = f.((f").y) & y = (f*f") .y
proof
  assume
A1: f is one-to-one;
  assume
A2: y in rng f;
  hence
A3: y = f.((f").y) by A1,Th32;
  rng f = dom(f") by A1,Th33;
  hence thesis by A2,A3,Th13;
end;

theorem Th36:
  f is one-to-one implies dom(f"*f) = dom f & rng(f"*f) = dom f
proof
  assume
A1: f is one-to-one;
  then
A2: rng f = dom(f") by Th33;
  then rng(f"*f) = rng(f") by RELAT_1:28;
  hence thesis by A1,A2,Th33,RELAT_1:27;
end;

theorem Th37:
  f is one-to-one implies dom(f*f") = rng f & rng(f*f") = rng f
proof
  assume
A1: f is one-to-one;
  then
A2: rng(f") = dom f by Th33;
  then dom(f*f") = dom(f") by RELAT_1:27;
  hence thesis by A1,A2,Th33,RELAT_1:28;
end;

theorem
  f is one-to-one & dom f = rng g & rng f = dom g & (for x,y st x in dom
  f & y in dom g holds f.x = y iff g.y = x) implies g = f"
proof
  assume that
A1: f is one-to-one and
A2: dom f = rng g and
A3: rng f = dom g and
A4: for x,y st x in dom f & y in dom g holds f.x = y iff g.y = x;
A5: y in dom g implies g.y = (f").y
  proof
    assume
A6: y in dom g;
    then
A7: g.y in dom f by A2,Def3;
    then f.(g.y) = y by A4,A6;
    hence thesis by A1,A7,Th32;
  end;
  rng f = dom(f") by A1,Th32;
  hence thesis by A3,A5,Th2;
end;

theorem Th39:
  f is one-to-one implies f"*f = id dom f & f*f" = id rng f
proof
  assume
A1: f is one-to-one;
A2: x in dom(f"*f) implies (f"*f).x = x
  proof
    assume x in dom(f"*f);
    then x in dom f by A1,Th36;
    hence thesis by A1,Th34;
  end;
A3: x in dom(f*f") implies (f*f").x = x
  proof
    assume x in dom(f*f");
    then x in rng f by A1,Th37;
    hence thesis by A1,Th35;
  end;
  dom(f"*f) = dom f by A1,Th36;
  hence f"*f = id dom f by A2,Th17;
  dom(f*f") = rng f by A1,Th37;
  hence thesis by A3,Th17;
end;

theorem Th40:
  f is one-to-one implies f" is one-to-one
proof
  assume
A1: f is one-to-one;
  let y1,y2;
  assume that
A2: y1 in dom(f") and
A3: y2 in dom(f");
  y1 in rng f by A1,A2,Th32;
  then
A4: y1 = f.((f").y1) by A1,Th35;
  y2 in rng f by A1,A3,Th32;
  hence thesis by A1,A4,Th35;
end;

registration
  let f be one-to-one Function;
  cluster f" -> one-to-one;
  coherence by Th40;
  let g be one-to-one Function;
  cluster g*f -> one-to-one;
  coherence by Th24;
end;

Lm1: rng(g2) = X & f*g2 = id dom g1 & g1*f = id X implies g1 = g2
proof
A1: g1*(f*g2) = (g1*f)*g2 & g1*(id dom g1) = g1 by RELAT_1:36,51;
  assume rng(g2) = X & f*g2 = id dom g1 & g1*f = id X;
  hence thesis by A1,RELAT_1:53;
end;

theorem Th41:
  f is one-to-one & rng f = dom g & g*f = id dom f implies g = f"
proof
  assume that
A1: f is one-to-one and
A2: rng f = dom g & g*f = id dom f;
  f*f" = id rng f & rng(f") = dom f by A1,Th33,Th39;
  hence thesis by A2,Lm1;
end;

theorem
  f is one-to-one & rng g = dom f & f*g = id rng f implies g = f"
proof
  assume that
A1: f is one-to-one and
A2: rng g = dom f & f*g = id rng f;
  f"*f = id dom f & dom(f") = rng f by A1,Th33,Th39;
  hence thesis by A2,Lm1;
end;

theorem
  f is one-to-one implies (f")" = f
proof
  assume
A1: f is one-to-one;
  then rng f = dom(f") by Th33;
  then
A2: f*f" = id dom(f") by A1,Th39;
  dom f = rng(f") by A1,Th33;
  hence thesis by A1,A2,Th41;
end;

theorem
  f is one-to-one & g is one-to-one implies (g*f)" = f"*g"
proof
  assume that
A1: f is one-to-one and
A2: g is one-to-one;
  y in rng(g*f) iff y in dom(f"*g")
  proof
    thus y in rng(g*f) implies y in dom(f"*g")
    proof
      assume y in rng(g*f);
      then consider x such that
A3:   x in dom(g*f) and
A4:   y = (g*f).x by Def3;
A5:   f.x in dom g by A3,Th11;
A6:   y = g.(f.x) by A3,A4,Th12;
      then y in rng g by A5,Def3;
      then
A7:   y in dom(g") by A2,Th32;
A8:   x in dom f by A3,Th11;
      (g").(g.(f.x)) = (g"*g).(f.x) by A5,Th13
        .= (id dom g).(f.x) by A2,Th39
        .= f.x by A5,Th17;
      then (g").y in rng f by A8,A6,Def3;
      then (g").y in dom(f") by A1,Th32;
      hence thesis by A7,Th11;
    end;
    assume
A9: y in dom(f"*g");
    then y in dom(g") by Th11;
    then y in rng g by A2,Th32;
    then consider z such that
A10: z in dom g and
A11: y = g.z by Def3;
    (g").y in dom(f") by A9,Th11;
    then (g").(g.z) in rng f by A1,A11,Th32;
    then (g"*g).z in rng f by A10,Th13;
    then (id dom g).z in rng f by A2,Th39;
    then z in rng f by A10,Th17;
    then consider x such that
A12: x in dom f & z = f.x by Def3;
    x in dom(g*f) & y = (g*f).x by A10,A11,A12,Th11,Th13;
    hence thesis by Def3;
  end;
  then
A13: rng(g*f) = dom(f"*g") by TARSKI:1;
  x in dom((f"*g")*(g*f)) iff x in dom(g*f)
  proof
    thus x in dom((f"*g")*(g*f)) implies x in dom(g*f) by Th11;
    assume
A14: x in dom(g*f);
    then (g*f).x in rng(g*f) by Def3;
    hence thesis by A13,A14,Th11;
  end;
  then
A15: dom((f"*g")*(g*f)) = dom(g*f) by TARSKI:1;
  x in dom(g*f) implies ((f"*g")*(g*f)).x = x
  proof
    assume
A16: x in dom(g*f);
    then
A17: f.x in dom g by Th11;
    (g*f).x in rng(g*f) by A16,Def3;
    then
A18: g.(f.x) in dom(f"*g") by A13,A16,Th12;
A19: x in dom f by A16,Th11;
    thus ((f"*g")*(g*f)).x = (f"*g").((g*f).x) by A15,A16,Th12
      .= (f"*g").(g.(f.x)) by A16,Th12
      .= (f").((g").(g.(f.x))) by A18,Th12
      .= (f").((g"*g).(f.x)) by A17,Th13
      .= (f").((id dom g).(f.x)) by A2,Th39
      .= (f").(f.x) by A17,Th17
      .= x by A1,A19,Th34;
  end;
  then (f"*g")*(g*f) = id dom(g*f) by A15,Th17;
  hence thesis by A1,A2,A13,Th41;
end;

theorem
  (id X)" = id X
proof
  dom id X = X;
  then
A1: (id X)"*(id X) = id X by Th39;
  dom((id X)") = rng id X & rng id X = X by Th33;
  hence thesis by A1,Th23;
end;

registration
  let f,X;
  cluster f|X -> Function-like;
  coherence
  proof
    let x,y1,y2;
    assume [x,y1] in f|X & [x,y2] in f|X;
    then [x,y1] in f & [x,y2] in f by RELAT_1:def 11;
    hence thesis by Def1;
  end;
end;

theorem
  dom g = dom f /\ X & (for x st x in dom g holds g.x = f.x) implies g = f|X
proof
  assume that
A1: dom g = dom f /\ X and
A2: for x st x in dom g holds g.x = f.x;
  now
    let x,y;
    hereby
      assume
A3:   [x,y] in g;
      then
A4:   x in dom g by XTUPLE_0:def 12;
      hence x in X by A1,XBOOLE_0:def 4;
A5:   x in dom f by A1,A4,XBOOLE_0:def 4;
      y = g.x by A3,A4,Def2
        .= f.x by A2,A4;
      hence [x,y] in f by A5,Def2;
    end;
    assume
A6: x in X;
    assume
A7: [x,y] in f;
    then
A8: x in dom f by XTUPLE_0:def 12;
    then
A9: x in dom g by A1,A6,XBOOLE_0:def 4;
    y = f.x by A7,A8,Def2
      .= g.x by A2,A9;
    hence [x,y] in g by A9,Def2;
  end;
  hence thesis by RELAT_1:def 11;
end;

theorem Th47:
  x in dom(f|X) implies (f|X).x = f.x
proof
  set g = f|X;
  assume
A1: x in dom g;
  dom g = dom f /\ X by RELAT_1:61;
  then
A2: x in dom f by A1,XBOOLE_0:def 4;
  g c= f & [x,g.x] in g by A1,Def2,RELAT_1:59;
  hence g.x = f.x by A2,Def2;
end;

theorem Th48:
  x in dom f /\ X implies (f|X).x = f.x
proof
  assume x in dom f /\ X;
  then x in dom(f|X) by RELAT_1:61;
  hence thesis by Th47;
end;

theorem Th49:
  x in X implies (f|X).x = f.x
proof
  assume
A1: x in X;
  per cases;
  suppose
    x in dom f;
    then x in dom(f|X) by A1,RELAT_1:57;
    hence thesis by Th47;
  end;
  suppose
A2: not x in dom f;
    then not x in dom(f|X) by RELAT_1:57;
    hence (f|X).x = {} by Def2
      .= f.x by A2,Def2;
  end;
end;

theorem
  x in dom f & x in X implies f.x in rng(f|X)
proof
  assume that
A1: x in dom f and
A2: x in X;
  x in dom f /\ X by A1,A2,XBOOLE_0:def 4;
  then
A3: x in dom(f|X) by RELAT_1:61;
  (f|X).x = f.x by A2,Th49;
  hence thesis by A3,Def3;
end;

theorem
  X c= Y implies (f|X)|Y = f|X & (f|Y)|X = f|X by RELAT_1:73,74;

theorem
  f is one-to-one implies f|X is one-to-one
proof
  assume
A1: f is one-to-one;
  let x1,x2;
  assume that
A2: x1 in dom(f|X) and
A3: x2 in dom(f|X);
  x1 in dom f /\ X by A2,RELAT_1:61;
  then
A4: x1 in dom f by XBOOLE_0:def 4;
  x2 in dom f /\ X by A3,RELAT_1:61;
  then
A5: x2 in dom f by XBOOLE_0:def 4;
  (f|X).x1 = f.x1 & (f|X).x2 = f.x2 by A2,A3,Th47;
  hence thesis by A1,A4,A5,Def4;
end;

registration
  let Y,f;
  cluster Y|`f -> Function-like;
  coherence
  proof
    let x,y1,y2;
    assume [x,y1] in Y|`f & [x,y2] in Y|`f;
    then [x,y1] in f & [x,y2] in f by RELAT_1:def 12;
    hence thesis by Def1;
  end;
end;

theorem Th53:
  g = Y|`f iff (for x holds x in dom g iff x in dom f & f.x in Y) &
  for x st x in dom g holds g.x = f.x
proof
  hereby
    assume
A1: g = Y|`f;
    hereby
      let x;
      hereby
        assume x in dom g;
        then
A2:     [x,g.x] in g by Def2;
        then
A3:     [x,g.x] in f by A1,RELAT_1:def 12;
        hence x in dom f by XTUPLE_0:def 12;
        then f.x = g.x by A3,Def2;
        hence f.x in Y by A1,A2,RELAT_1:def 12;
      end;
      assume x in dom f;
      then
A4:   [x,f.x] in f by Def2;
      assume f.x in Y;
      then [x,f.x] in g by A1,A4,RELAT_1:def 12;
      hence x in dom g by XTUPLE_0:def 12;
    end;
    let x;
    assume x in dom g;
    then [x,g.x] in g by Def2;
    then
A5: [x,g.x] in f by A1,RELAT_1:def 12;
    then x in dom f by XTUPLE_0:def 12;
    hence f.x = g.x by A5,Def2;
  end;
  assume that
A6: for x holds x in dom g iff x in dom f & f.x in Y and
A7: for x st x in dom g holds g.x = f.x;
  now
    let x,y;
    hereby
      assume
A8:   [x,y] in g;
      then
A9:   x in dom g by XTUPLE_0:def 12;
      then
A10:  y = g.x by A8,Def2
        .= f.x by A7,A9;
      hence y in Y by A6,A9;
      x in dom f by A6,A9;
      hence [x,y] in f by A10,Def2;
    end;
    assume
A11: y in Y;
    assume
A12: [x,y] in f;
    then
A13: y = f.x by Th1;
    x in dom f by A12,XTUPLE_0:def 12;
    then
A14: x in dom g by A6,A11,A13;
    then y = g.x by A7,A13;
    hence [x,y] in g by A14,Def2;
  end;
  hence thesis by RELAT_1:def 12;
end;

theorem
  x in dom(Y|`f) iff x in dom f & f.x in Y by Th53;

theorem
  x in dom(Y|`f) implies (Y|`f).x = f.x by Th53;

theorem
  dom(Y|`f) c= dom f
proof
  let x;
  thus thesis by Th53;
end;

theorem
  X c= Y implies Y|`(X|`f) = X|`f & X|`(Y|`f) = X|`f by RELAT_1:98,99;

theorem
  f is one-to-one implies Y|`f is one-to-one
proof
  assume
A1: f is one-to-one;
  let x1,x2 such that
A2: x1 in dom(Y|`f) & x2 in dom(Y|`f) and
A3: (Y|`f).x1 = (Y|`f).x2;
A4: x1 in dom f & x2 in dom f by A2,Th53;
  (Y|`f).x1 = f.x1 & (Y|`f).x2 = f.x2 by A2,Th53;
  hence thesis by A1,A3,A4,Def4;
end;

definition
  let f,X;

  redefine func f.:X means
  :Def6:
  for y holds y in it iff ex x st x in dom f & x in X & y = f.x;
  compatibility
  proof
    let Y;
    hereby
      assume
A1:   Y = f.:X;
      let y;
      hereby
        assume y in Y;
        then consider x such that
A2:     [x,y] in f and
A3:     x in X by A1,RELAT_1:def 13;
        take x;
        thus
A4:     x in dom f by A2,XTUPLE_0:def 12;
        thus x in X by A3;
        thus y = f.x by A2,A4,Def2;
      end;
      given x such that
A5:   x in dom f and
A6:   x in X and
A7:   y = f.x;
      [x,y] in f by A5,A7,Def2;
      hence y in Y by A1,A6,RELAT_1:def 13;
    end;
    assume
A8: for y holds y in Y iff ex x st x in dom f & x in X & y = f.x;
    now
      let y;
      hereby
        assume y in Y;
        then consider x such that
A9:     x in dom f and
A10:    x in X and
A11:    y = f.x by A8;
        take x;
        thus [x,y] in f by A9,A11,Def2;
        thus x in X by A10;
      end;
      given x such that
A12:  [x,y] in f and
A13:  x in X;
      x in dom f & y = f.x by A12,Th1;
      hence y in Y by A8,A13;
    end;
    hence thesis by RELAT_1:def 13;
  end;
end;

theorem Th59:
  x in dom f implies Im(f,x) = {f.x}
proof
  assume
A1: x in dom f;
  y in f.:{x} iff y in {f.x}
  proof
    thus y in f.:{x} implies y in {f.x}
    proof
      assume y in f.:{x};
      then consider z such that
      z in dom f and
A2:   z in {x} and
A3:   y = f.z by Def6;
      z = x by A2,TARSKI:def 1;
      hence thesis by A3,TARSKI:def 1;
    end;
    assume y in {f.x};
    then
A4: y = f.x by TARSKI:def 1;
    x in {x} by TARSKI:def 1;
    hence thesis by A1,A4,Def6;
  end;
  hence thesis by TARSKI:1;
end;

theorem
  x1 in dom f & x2 in dom f implies f.:{x1,x2} = {f.x1,f.x2}
proof
  assume
A1: x1 in dom f & x2 in dom f;
  y in f.:{x1,x2} iff y = f.x1 or y = f.x2
  proof
A2: x1 in {x1,x2} & x2 in {x1,x2} by TARSKI:def 2;
    thus y in f.:{x1,x2} implies y = f.x1 or y = f.x2
    proof
      assume y in f.:{x1,x2};
      then ex x st x in dom f & x in {x1,x2} & y = f.x by Def6;
      hence thesis by TARSKI:def 2;
    end;
    assume y = f.x1 or y = f.x2;
    hence thesis by A1,A2,Def6;
  end;
  hence thesis by TARSKI:def 2;
end;

theorem
  (Y|`f).:X c= f.:X
proof
  let y;
  assume y in (Y|`f).:X;
  then consider x such that
A1: x in dom(Y|`f) and
A2: x in X and
A3: y = (Y|`f).x by Def6;
  y = f.x & x in dom f by A1,A3,Th53;
  hence thesis by A2,Def6;
end;

theorem Th62:
  f is one-to-one implies f.:(X1 /\ X2) = f.:X1 /\ f.:X2
proof
  assume
A1: f is one-to-one;
A2: f.:X1 /\ f.:X2 c= f.:(X1 /\ X2)
  proof
    let y;
    assume
A3: y in f.:X1 /\ f.:X2;
    then y in f.:X1 by XBOOLE_0:def 4;
    then consider x1 such that
A4: x1 in dom f and
A5: x1 in X1 and
A6: y = f.x1 by Def6;
    y in f.:X2 by A3,XBOOLE_0:def 4;
    then consider x2 such that
A7: x2 in dom f and
A8: x2 in X2 and
A9: y = f.x2 by Def6;
    x1 = x2 by A1,A4,A6,A7,A9,Def4;
    then x1 in X1 /\ X2 by A5,A8,XBOOLE_0:def 4;
    hence thesis by A4,A6,Def6;
  end;
  f.:(X1 /\ X2)c=f.:X1 /\ f.:X2 by RELAT_1:121;
  hence thesis by A2,XBOOLE_0:def 10;
end;

theorem
  (for X1,X2 holds f.:(X1 /\ X2) = f.:X1 /\ f.:X2) implies f is one-to-one
proof
  assume
A1: for X1,X2 holds f.:(X1 /\ X2) = f.:X1 /\ f.:X2;
  given x1,x2 such that
A2: x1 in dom f & x2 in dom f and
A3: f.x1 = f.x2 and
A4: x1 <> x2;
A5: f.:({x1}/\{x2}) = f.:{x1}/\f.: {x2} by A1;
  {x1} misses {x2} by A4,ZFMISC_1:11;
  then
A6: {x1} /\ {x2} = {} by XBOOLE_0:def 7;
  Im(f,x1) = {f.x1} & Im(f,x2) = {f.x2} by A2,Th59;
  hence contradiction by A3,A6,A5;
end;

theorem
  f is one-to-one implies f.:(X1 \ X2) = f.:X1 \ f.:X2
proof
  assume
A1: f is one-to-one;
A2: f.:(X1 \ X2) c= f.:X1 \ f.:X2
  proof
    let y;
    assume y in f.:(X1\X2);
    then consider x such that
A3: x in dom f and
A4: x in X1\X2 and
A5: y = f.x by Def6;
A6: not x in X2 by A4,XBOOLE_0:def 5;
A7: now
      assume y in f.:X2;
      then ex z st z in dom f & z in X2 & y = f.z by Def6;
      hence contradiction by A1,A3,A5,A6,Def4;
    end;
    y in f.:X1 by A3,A4,A5,Def6;
    hence thesis by A7,XBOOLE_0:def 5;
  end;
  f.:X1 \ f.: X2 c= f.:(X1 \ X2) by RELAT_1:122;
  hence thesis by A2,XBOOLE_0:def 10;
end;

theorem
  (for X1,X2 holds f.:(X1 \ X2) = f.:X1 \ f.:X2) implies f is one-to-one
proof
  assume
A1: for X1,X2 holds f.:(X1 \ X2) = f.:X1 \ f.:X2;
  given x1,x2 such that
A2: x1 in dom f & x2 in dom f and
A3: f.x1 = f.x2 and
A4: x1 <> x2;
A5: f.:({x1}\{x2}) = f.:{x1} by A4,ZFMISC_1:14;
A6: f.:({x1}\{x2}) = f.:{x1}\f.:{x2} by A1;
  Im(f,x1) = {f.x1} & Im(f,x2) = {f.x2} by A2,Th59;
  hence contradiction by A3,A5,A6,XBOOLE_1:37;
end;

theorem
  X misses Y & f is one-to-one implies f.:X misses f.:Y
proof
  assume X /\ Y = {} & f is one-to-one;
  then f.:(X /\ Y) = {} & f.:(X /\ Y) = f.:X /\ f.:Y by Th62;
  hence thesis by XBOOLE_0:def 7;
end;

theorem
  (Y|`f).:X = Y /\ f.:X
proof
  y in (Y|`f).:X iff y in Y /\ f.:X
  proof
    thus y in (Y|`f).:X implies y in Y /\ f.:X
    proof
      assume y in (Y|`f).:X;
      then consider x such that
A1:   x in dom(Y|`f) and
A2:   x in X and
A3:   y = (Y|`f).x by Def6;
A4:   y = f.x by A1,A3,Th53;
      then
A5:   y in Y by A1,Th53;
      x in dom f by A1,Th53;
      then y in f.:X by A2,A4,Def6;
      hence thesis by A5,XBOOLE_0:def 4;
    end;
    assume
A6: y in Y /\ f.:X;
    then y in f.:X by XBOOLE_0:def 4;
    then consider x such that
A7: x in dom f and
A8: x in X and
A9: y = f.x by Def6;
    y in Y by A6,XBOOLE_0:def 4;
    then
A10: x in dom(Y|`f) by A7,A9,Th53;
    then (Y|`f).x = f.x by Th53;
    hence thesis by A8,A9,A10,Def6;
  end;
  hence thesis by TARSKI:1;
end;

definition
  let f,Y;
  redefine func f"Y means
  :Def7:
  for x holds x in it iff x in dom f & f.x in Y;
  compatibility
  proof
    let X;
    hereby
      assume
A1:   X = f"Y;
      let x;
      hereby
        assume x in X;
        then
A2:     ex y st [x,y] in f & y in Y by A1,RELAT_1:def 14;
        hence x in dom f by XTUPLE_0:def 12;
        thus f.x in Y by A2,Th1;
      end;
      assume that
A3:   x in dom f and
A4:   f.x in Y;
      [x,f.x] in f by A3,Th1;
      hence x in X by A1,A4,RELAT_1:def 14;
    end;
    assume
A5: for x holds x in X iff x in dom f & f.x in Y;
    now
      let x;
      hereby
        assume
A6:     x in X;
        take y = f.x;
        x in dom f by A5,A6;
        hence [x,y] in f by Def2;
        thus y in Y by A5,A6;
      end;
      given y such that
A7:   [x,y] in f and
A8:   y in Y;
      x in dom f & y = f.x by A7,Th1;
      hence x in X by A5,A8;
    end;
    hence thesis by RELAT_1:def 14;
  end;
end;

theorem Th68:
  f"(Y1 /\ Y2) = f"Y1 /\ f"Y2
proof
  x in f"(Y1 /\ Y2) iff x in f"Y1 /\ f"Y2
  proof
A1: x in f"Y2 iff f.x in Y2 & x in dom f by Def7;
A2: x in f"(Y1 /\ Y2) iff f.x in Y1 /\ Y2 & x in dom f by Def7;
    x in f"Y1 iff f.x in Y1 & x in dom f by Def7;
    hence thesis by A1,A2,XBOOLE_0:def 4;
  end;
  hence thesis by TARSKI:1;
end;

theorem
  f"(Y1 \ Y2) = f"Y1 \ f"Y2
proof
  x in f"(Y1 \ Y2) iff x in f"Y1 \ f"Y2
  proof
A1: x in f"Y2 iff f.x in Y2 & x in dom f by Def7;
A2: x in f"(Y1 \ Y2) iff f.x in Y1 \ Y2 & x in dom f by Def7;
    x in f"Y1 iff f.x in Y1 & x in dom f by Def7;
    hence thesis by A1,A2,XBOOLE_0:def 5;
  end;
  hence thesis by TARSKI:1;
end;

theorem
  (R|X)"Y = X /\ (R"Y)
proof
  hereby
    let x;
    assume x in (R|X)"Y;
    then
A1: ex y st [x,y] in R|X & y in Y by RELAT_1:def 14;
    then
A2: x in X by RELAT_1:def 11;
    R|X c= R by RELAT_1:59;
    then x in R"Y by A1,RELAT_1:def 14;
    hence x in X /\ (R"Y) by A2,XBOOLE_0:def 4;
  end;
  let x;
  assume
A3: x in X /\ (R"Y);
  then x in R"Y by XBOOLE_0:def 4;
  then consider y such that
A4: [x,y] in R and
A5: y in Y by RELAT_1:def 14;
  x in X by A3,XBOOLE_0:def 4;
  then [x,y] in R|X by A4,RELAT_1:def 11;
  hence thesis by A5,RELAT_1:def 14;
end;

theorem
  for f being Function, A,B being set st A misses B holds f"A misses f"B
proof
  let f be Function, A,B be set;
  assume A misses B;
  then A /\ B = {} by XBOOLE_0:def 7;
  then {} = f"(A /\ B)
    .= f"A /\ f"B by Th68;
  hence thesis by XBOOLE_0:def 7;
end;

theorem Th72:
  y in rng R iff R"{y} <> {}
proof
  thus y in rng R implies R"{y} <> {}
  proof
    assume y in rng R;
    then
A1: ex x st [x,y] in R by XTUPLE_0:def 13;
    y in {y} by TARSKI:def 1;
    hence thesis by A1,RELAT_1:def 14;
  end;
  assume R"{y} <> {};
  then consider x such that
A2: x in R"{y} by XBOOLE_0:def 1;
  consider z such that
A3: [x,z] in R and
A4: z in {y} by A2,RELAT_1:def 14;
  z = y by A4,TARSKI:def 1;
  hence thesis by A3,XTUPLE_0:def 13;
end;

theorem
  (for y st y in Y holds R"{y} <> {}) implies Y c= rng R
proof
  assume
A1: for y st y in Y holds R"{y} <> {};
  let y;
  assume y in Y;
  then R"{y} <> {} by A1;
  hence thesis by Th72;
end;

theorem Th74:
  (for y st y in rng f ex x st f"{y} = {x}) iff f is one-to-one
proof
  thus (for y st y in rng f ex x st f"{y} = {x}) implies f is one-to-one
  proof
    assume
A1: for y st y in rng f ex x st f"{y} = {x};
    let x1,x2;
    assume that
A2: x1 in dom f and
A3: x2 in dom f;
    f.x1 in rng f by A2,Def3;
    then consider y1 such that
A4: f"{f.x1} = {y1} by A1;
    f.x2 in rng f by A3,Def3;
    then consider y2 such that
A5: f"{f.x2} = {y2} by A1;
    f.x1 in {f.x1} by TARSKI:def 1;
    then x1 in {y1} by A2,A4,Def7;
    then
A6: y1 = x1 by TARSKI:def 1;
    f.x2 in {f.x2} by TARSKI:def 1;
    then x2 in {y2} by A3,A5,Def7;
    hence thesis by A4,A5,A6,TARSKI:def 1;
  end;
  assume
A7: f is one-to-one;
  let y;
  assume y in rng f;
  then consider x such that
A8: x in dom f & y = f.x by Def3;
  take x;
  z in f"{y} iff z = x
  proof
    thus z in f"{y} implies z = x
    proof
      assume
A9:   z in f"{y};
      then f.z in {y} by Def7;
      then
A10:  f.z = y by TARSKI:def 1;
      z in dom f by A9,Def7;
      hence thesis by A7,A8,A10,Def4;
    end;
    y in {y} by TARSKI:def 1;
    hence thesis by A8,Def7;
  end;
  hence thesis by TARSKI:def 1;
end;

theorem Th75:
  f.:(f"Y) c= Y
proof
  let y;
  assume y in f.:(f"Y);
  then ex x st x in dom f & x in f"Y & y = f.x by Def6;
  hence thesis by Def7;
end;

theorem Th76:
  X c= dom R implies X c= R"(R.:X)
proof
  assume
A1: X c= dom R;
  let x;
  assume
A2: x in X;
  then consider Rx being set such that
A3: [x,Rx] in R by A1,XTUPLE_0:def 12;
  Rx in R.:X by A2,A3,RELAT_1:def 13;
  hence thesis by A3,RELAT_1:def 14;
end;

theorem
  Y c= rng f implies f.:(f"Y) = Y
proof
  assume
A1: Y c= rng f;
  thus f.:(f"Y) c= Y by Th75;
  let y;
  assume
A2: y in Y;
  then consider x such that
A3: x in dom f & y = f.x by A1,Def3;
  x in f"Y by A2,A3,Def7;
  hence thesis by A3,Def6;
end;

theorem
  f.:(f"Y) = Y /\ f.:(dom f)
proof
  f.:(f"Y) c= Y & f.:(f"(Y)) c= f.:(dom f) by Th75,RELAT_1:114;
  hence f.:(f"Y) c= Y /\ f.:(dom f) by XBOOLE_1:19;
  let y;
  assume
A1: y in Y /\ f.:(dom f);
  then y in f.:(dom f) by XBOOLE_0:def 4;
  then consider x such that
A2: x in dom f and
  x in dom f and
A3: y = f.x by Def6;
  y in Y by A1,XBOOLE_0:def 4;
  then x in f"Y by A2,A3,Def7;
  hence thesis by A2,A3,Def6;
end;

theorem Th79:
  f.:(X /\ f"Y) c= (f.:X) /\ Y
proof
  let y;
  assume y in f.:(X /\ f"Y);
  then consider x such that
A1: x in dom f and
A2: x in X /\ f"Y and
A3: y = f.x by Def6;
  x in f"Y by A2,XBOOLE_0:def 4;
  then
A4: y in Y by A3,Def7;
  x in X by A2,XBOOLE_0:def 4;
  then y in f.:X by A1,A3,Def6;
  hence thesis by A4,XBOOLE_0:def 4;
end;

theorem
  f.:(X /\ f"Y) = (f.:X) /\ Y
proof
  thus f.:(X /\ f"Y)c=(f.:X) /\ Y by Th79;
  let y;
  assume
A1: y in (f.:X) /\ Y;
  then y in f.:X by XBOOLE_0:def 4;
  then consider x such that
A2: x in dom f and
A3: x in X and
A4: y = f.x by Def6;
  y in Y by A1,XBOOLE_0:def 4;
  then x in f"Y by A2,A4,Def7;
  then x in X /\ f"Y by A3,XBOOLE_0:def 4;
  hence thesis by A2,A4,Def6;
end;

theorem
  X /\ R"Y c= R"(R.:X /\ Y)
proof
  let x;
  assume
A1: x in X /\ R"Y;
  then x in R"Y by XBOOLE_0:def 4;
  then consider Rx being set such that
A2: [x,Rx] in R and
A3: Rx in Y by RELAT_1:def 14;
  x in X by A1,XBOOLE_0:def 4;
  then Rx in R.:X by A2,RELAT_1:def 13;
  then Rx in R.:X /\ Y by A3,XBOOLE_0:def 4;
  hence thesis by A2,RELAT_1:def 14;
end;

theorem Th82:
  f is one-to-one implies f"(f.:X) c= X
proof
  assume
A1: f is one-to-one;
  let x;
  assume
A2: x in f"(f.:X);
  then f.x in f.:X by Def7;
  then
A3: ex z st z in dom f & z in X & f.x = f.z by Def6;
  x in dom f by A2,Def7;
  hence thesis by A1,A3,Def4;
end;

theorem
  (for X holds f"(f.:X) c= X) implies f is one-to-one
proof
  assume
A1: for X holds f"(f.:X) c= X;
  given x1,x2 such that
A2: x1 in dom f and
A3: x2 in dom f and
A4: f.x1 = f.x2 & x1 <> x2;
A5: f"(f.:{x1}) c= {x1} by A1;
A6: Im(f,x2) = {f.x2} by A3,Th59;
A7: Im(f,x1) = {f.x1} by A2,Th59;
  f.x1 in rng f by A2,Def3;
  then f"(f.:{x1}) <> {} by A7,Th72;
  then f"(f.:{x1}) = {x1} by A5,ZFMISC_1:33;
  hence contradiction by A1,A4,A7,A6,ZFMISC_1:3;
end;

theorem
  f is one-to-one implies f.:X = (f")"X
proof
  assume
A1: f is one-to-one;
  y in f.:X iff y in (f")"X
  proof
    thus y in f.:X implies y in (f")"X
    proof
      assume y in f.:X;
      then consider x such that
A2:   x in dom f and
A3:   x in X and
A4:   y = f.x by Def6;
      y in rng f by A2,A4,Def3;
      then
A5:   y in dom(f") by A1,Th32;
      (f").(f.x) = x by A1,A2,Th32;
      hence thesis by A3,A4,A5,Def7;
    end;
    assume
A6: y in (f")"X;
    then
A7: (f").y in X by Def7;
    y in dom(f") by A6,Def7;
    then y in rng(f) by A1,Th32;
    then consider x such that
A8: x in dom(f) & y = f.x by Def3;
    (f").y = x by A1,A8,Th34;
    hence thesis by A7,A8,Def6;
  end;
  hence thesis by TARSKI:1;
end;

theorem
  f is one-to-one implies f"Y = (f").:Y
proof
  assume
A1: f is one-to-one;
  x in f"Y iff x in (f").:Y
  proof
    thus x in f"Y implies x in (f").:Y
    proof
      assume
A2:   x in f"Y;
      then
A3:   f.x in Y by Def7;
A4:   x in dom f by A2,Def7;
      then f.x in rng(f) by Def3;
      then
A5:   f.x in dom(f") by A1,Th32;
      (f").(f.x) = x by A1,A4,Th32;
      hence thesis by A3,A5,Def6;
    end;
    assume x in (f").:Y;
    then consider y such that
A6: y in dom(f") and
A7: y in Y and
A8: x = (f").y by Def6;
    dom(f") = rng f by A1,Th32;
    then y = f.x & x in dom f by A1,A6,A8,Th32;
    hence thesis by A7,Def7;
  end;
  hence thesis by TARSKI:1;
end;



theorem
  Y = rng f & dom g = Y & dom h = Y & g*f = h*f implies g = h
proof
  assume that
A1: Y = rng f and
A2: dom g = Y & dom h = Y and
A3: g*f = h*f;
  y in Y implies g.y = h.y
  proof
    assume y in Y;
    then consider x such that
A4: x in dom f & y = f.x by A1,Def3;
    (g*f).x = g.y by A4,Th13;
    hence thesis by A3,A4,Th13;
  end;
  hence thesis by A2,Th2;
end;

theorem
  f.:X1 c= f.:X2 & X1 c= dom f & f is one-to-one implies X1 c= X2
proof
  assume that
A1: f.:X1 c= f.:X2 and
A2: X1 c= dom f and
A3: f is one-to-one;
  let x;
  assume
A4: x in X1;
  then f.x in f.:X1 by A2,Def6;
  then ex x2 st x2 in dom f & x2 in X2 & f.x = f.x2 by A1,Def6;
  hence thesis by A2,A3,A4,Def4;
end;

theorem Th88:
  f"Y1 c= f"Y2 & Y1 c= rng f implies Y1 c= Y2
proof
  assume that
A1: f"Y1 c= f"Y2 and
A2: Y1 c= rng f;
  let y;
  assume
A3: y in Y1;
  then consider x such that
A4: x in dom f and
A5: y = f.x by A2,Def3;
  x in f"Y1 by A3,A4,A5,Def7;
  hence thesis by A1,A5,Def7;
end;

theorem
  f is one-to-one iff for y ex x st f"{y} c= {x}
proof
  (for y ex x st f"{y} c= {x}) iff for y st y in rng f ex x st f"{y} = {x}
  proof
    thus (for y ex x st f"{y} c= {x}) implies for y st y in rng f ex x st f"{y
    } = {x}
    proof
      assume
A1:   for y ex x st f"{y} c= {x};
      let y;
      consider x such that
A2:   f"{y} c= {x} by A1;
      assume y in rng f;
      then consider x1 such that
A3:   x1 in dom f and
A4:   y = f.x1 by Def3;
      take x;
      f.x1 in {y} by A4,TARSKI:def 1;
      then f"{y} <> {} by A3,Def7;
      hence thesis by A2,ZFMISC_1:33;
    end;
    assume
A5: for y st y in rng f ex x st f"{y} = {x};
    let y;
A6: now
      set x = the set;
      assume
A7:   not y in rng f;
      take x;
      rng f misses {y} by A7,ZFMISC_1:50;
      then f"{y} = {} by RELAT_1:138;
      hence f"{y} c= {x} by XBOOLE_1:2;
    end;
    now
      assume y in rng f;
      then consider x such that
A8:   f"{y} = {x} by A5;
      take x;
      thus f"{y} c= {x} by A8;
    end;
    hence thesis by A6;
  end;
  hence thesis by Th74;
end;

theorem
  rng R c= dom S implies R"X c= (R*S)"(S.:X)
proof
  assume
A1: rng R c= dom S;
  let x;
  assume x in R"X;
  then consider Rx being set such that
A2: [x,Rx] in R and
A3: Rx in X by RELAT_1:def 14;
  Rx in rng R by A2,XTUPLE_0:def 13;
  then consider SRx being set such that
A4: [Rx,SRx] in S by A1,XTUPLE_0:def 12;
  SRx in S.:X & [x,SRx] in R*S by A2,A3,A4,RELAT_1:def 8,def 13;
  hence thesis by RELAT_1:def 14;
end;

theorem
  for f being Function st f " X = f " Y & X c= rng f & Y c= rng f holds X = Y
proof
  let f be Function;
  assume f " X = f " Y & X c= rng f & Y c= rng f;
  then X c= Y & Y c= X by Th88;
  hence thesis by XBOOLE_0:def 10;
end;

begin 


reserve e,u for set,
  A for Subset of X;

theorem
  (id X).:A = A
proof
  now
    let e;
    thus e in A implies ex u st u in dom id X & u in A & e = (id X).u
    proof
      assume
A1:   e in A;
      take e;
      thus e in dom id X by A1;
      thus e in A by A1;
      thus thesis by A1,Th17;
    end;
    assume ex u st u in dom id X & u in A & e = (id X).u;
    hence e in A by Th17;
  end;
  hence thesis by Def6;
end;



definition
  let f be Function;
  redefine attr f is empty-yielding means
:Def8:  for x st x in dom f holds f.x is empty;
  compatibility
  proof
    hereby
      assume
A1:   f is empty-yielding;
      let x;
      assume x in dom f;
      then f.x in rng f by Def3;
      hence f.x is empty by A1,RELAT_1:149;
    end;
    assume
A2: for x st x in dom f holds f.x is empty;
    let s be set;
    assume s in rng f;
    then ex e being set st e in dom f & s = f.e by Def3;
    then s = {} by A2;
    hence thesis by TARSKI:def 1;
  end;
end;



definition
  let F be Function;
  redefine attr F is non-empty means
  :Def9:
  for n being set st n in dom F holds F.n is non empty;
  compatibility
  proof
    hereby
      assume F is non-empty;
      then
A1:   not {} in rng F by RELAT_1:def 9;
      let i be set;
      assume i in dom F;
      hence F.i is non empty by A1,Def3;
    end;
    assume
A2: for n being set st n in dom F holds F.n is non empty;
    assume {} in rng F;
    then ex i being set st i in dom F & F.i = {} by Def3;
    hence contradiction by A2;
  end;
end;



registration
  cluster non-empty for Function;
  existence
  proof
    take {};
    let x be set;
    thus thesis;
  end;
end;



scheme
  LambdaB { D()->non empty set, F(set)->set } :
  ex f be Function st dom f = D() & for d be Element of D() holds f.d = F(d)
proof
  consider f be Function such that
A1: dom f = D() & for d be set st d in D() holds f.d = F(d) from Lambda;
  take f;
  thus thesis by A1;
end;



registration
  let f be non-empty Function;
  cluster rng f -> with_non-empty_elements;
  coherence
  proof
    assume {} in rng f;
    then ex x being set st x in dom f & {} = f.x by Def3;
    hence thesis by Def9;
  end;
end;



definition
  let f be Function;
  attr f is constant means
  :Def10:
  x in dom f & y in dom f implies f.x = f.y;
end;

theorem
  for A,B being set, f being Function st A c= dom f & f.:A c= B holds A c= f"B
proof
  let A,B be set, f being Function;
  assume A c= dom f;
  then
A1: A c= f"(f.:A) by Th76;
  assume f.:A c= B;
  then f"(f.:A) c= f"B by RELAT_1:143;
  hence thesis by A1,XBOOLE_1:1;
end;



theorem
  for f being Function st X c= dom f & f is one-to-one holds f"(f.:X) = X
proof
  let f be Function such that
A1: X c= dom f and
A2: f is one-to-one;
  thus f"(f.:X) c= X by A2,Th82;
  let x;
  assume
A3: x in X;
  then f.x in f.:X by A1,Def6;
  hence thesis by A1,A3,Def7;
end;



definition
  let f,g;
  redefine pred f = g means
  dom f = dom g & for x st x in dom f holds f.x = g.x;
  compatibility by Th2;
end;



registration
  cluster non-empty non empty for Function;
  existence
  proof
    consider f such that
A1: dom f = {{}} and
A2: rng f = {{{}}} by Th5;
    take f;
    not {} in rng f by A2,TARSKI:def 1;
    hence f is non-empty by RELAT_1:def 9;
    thus thesis by A1;
  end;
end;



registration
  let a be non-empty non empty Function;
  let i be Element of dom a;
  cluster a.i -> non empty;
  coherence
  proof
    a.i in rng a by Def3;
    hence thesis by RELAT_1:def 9;
  end;
end;



registration
  let f be Function;
  cluster -> Function-like for Subset of f;
  coherence
  proof
    let g be Subset of f;
    let x,y1,y2;
    assume [x,y1] in g & [x,y2] in g;
    hence thesis by Def1;
  end;
end;



theorem
  for f,g being Function, D being set st D c= dom f & D c= dom g holds f
  | D = g | D iff for x being set st x in D holds f.x = g.x
proof
  let f,g be Function;
  let D be set;
  assume that
A1: D c= dom f and
A2: D c= dom g;
A3: dom (g | D) = dom g /\ D by RELAT_1:61
    .= D by A2,XBOOLE_1:28;
  hereby
    assume
A4: f | D = g | D;
    hereby
      let x be set;
      assume
A5:   x in D;
      hence f.x = (g | D).x by A4,Th49
        .= g.x by A5,Th49;
    end;
  end;
  assume
A6: for x being set st x in D holds f.x = g.x;
A7: now
    let x be set;
    assume
A8: x in D;
    hence (f | D).x = f.x by Th49
      .= g.x by A6,A8
      .= (g | D).x by A8,Th49;
  end;
  dom (f | D) = dom f /\ D by RELAT_1:61
    .= D by A1,XBOOLE_1:28;
  hence thesis by A3,A7,Th2;
end;



theorem
  for f,g being Function, X being set st dom f = dom g & (for x being
  set st x in X holds f.x = g.x) holds f|X = g|X
proof
  let f,g be Function, X be set such that
A1: dom f = dom g and
A2: for x being set st x in X holds f.x = g.x;
A3: dom (f|X) =dom f /\ X by RELAT_1:61;
  then
A4: dom (f|X) = dom (g|X) by A1,RELAT_1:61;
  now
    let x be set;
    assume
A5: x in dom (f|X);
    then
A6: x in X by A3,XBOOLE_0:def 4;
    (f|X).x = f.x & (g|X).x = g.x by A4,A5,Th47;
    hence (f|X).x = (g|X).x by A2,A6;
  end;
  hence thesis by A4,Th2;
end;



theorem Th97:
  rng(f|{X}) c= {f.X}
proof
  let x;
  assume x in rng(f|{X});
  then consider y such that
A1: y in dom(f|{X}) and
A2: x = (f|{X}).y by Def3;
  dom(f|{X}) c= {X} by RELAT_1:58;
  then y = X by A1,TARSKI:def 1;
  then x = f.X by A1,A2,Th47;
  hence thesis by TARSKI:def 1;
end;

theorem
  X in dom f implies rng(f|{X}) ={f.X}
proof
A1: X in {X} by TARSKI:def 1;
  assume X in dom f;
  then
A2: X in dom(f|{X}) by A1,RELAT_1:57;
  thus rng(f|{X}) c= {f.X} by Th97;
  let x;
  assume x in {f.X};
  then x = f.X by TARSKI:def 1;
  then x = (f|{X}).X by A2,Th47;
  hence thesis by A2,Def3;
end;



registration
  cluster empty -> constant for Function;
  coherence
  proof
    let f be Function;
    assume
A1: f is empty;
    let x be set;
    thus thesis by A1;
  end;
end;



registration
  let f be constant Function;
  cluster rng f -> trivial;
  coherence
  proof
    per cases;
    suppose
      f is empty;
      then reconsider g = f as empty Function;
      rng g is empty;
      hence thesis;
    end;
    suppose
      f <> {};
      then consider x being set such that
A1:   x in dom f by XBOOLE_0:def 1;
      for y being set holds y in {f.x} iff ex z being set st z in dom f &
      y = f.z
      proof
        let y be set;
        hereby
          assume
A2:       y in {f.x};
          take x;
          thus x in dom f & y = f.x by A1,A2,TARSKI:def 1;
        end;
        given z being set such that
A3:     z in dom f & y = f.z;
        y = f.x by A1,A3,Def10;
        hence thesis by TARSKI:def 1;
      end;
      hence thesis by Def3;
    end;
  end;
end;

registration
  cluster non constant for Function;
  existence
  proof
    set f = {[1,1],[2,2]};
    f is Function-like
    proof
      let x,y,z be set;
      assume that
A1:   [x,y] in f and
A2:   [x,z] in f;
      [x,y] = [1,1] or [x,y] =[2,2] by A1,TARSKI:def 2;
      then
A3:   x = 1 & y = 1 or x = 2 & y = 2 by XTUPLE_0:1;
      [x,z] = [1,1] or [x,z] =[2,2] by A2,TARSKI:def 2;
      hence thesis by A3,XTUPLE_0:1;
    end;
    then reconsider f as Function;
    take f, 1, 2;
A4: [2,2] in f by TARSKI:def 2;
A5: [1,1] in f by TARSKI:def 2;
    hence
A6: 1 in dom f & 2 in dom f by A4,XTUPLE_0:def 12;
    then f.1 = 1 by A5,Def2;
    hence thesis by A4,A6,Def2;
  end;
end;

registration
  let f be non constant Function;
  cluster rng f -> non trivial;
  coherence
  proof
    assume
A1: rng f is trivial;
    per cases;
    suppose
      rng f is empty;
      then reconsider f as empty Function;
      f is trivial;
      hence thesis;
    end;
    suppose
      rng f is non empty;
      then consider x being set such that
A2:   x in rng f by XBOOLE_0:def 1;
      f is constant
      proof
        let y,z be set;
        assume that
A3:     y in dom f and
A4:     z in dom f;
A5:     f.z in rng f by A4,Def3;
        f.y in rng f by A3,Def3;
        hence f.y = x by A1,A2,ZFMISC_1:def 10
          .= f.z by A1,A2,A5,ZFMISC_1:def 10;
      end;
      hence thesis;
    end;
  end;
end;

registration
  cluster non constant -> non trivial for Function;
  coherence
  proof
    let f be Function;
    assume f is non constant;
    then consider n1,n2 being set such that
A1: n1 in dom f and
A2: n2 in dom f and
A3: f.n1 <> f.n2 by Def10;
    reconsider f as non empty Function by A1;
    f is non trivial
    proof
      reconsider x = [n1,f.n1], y = [n2,f.n2] as Element of f by A1,A2,Th1;
      take x,y;
      thus x in f & y in f;
      thus thesis by A3,XTUPLE_0:1;
    end;
    hence thesis;
  end;
end;

registration
  cluster trivial -> constant for Function;
  coherence;
end;



theorem
  for F,G be Function, X holds (G|(F.:X))*(F|X) = (G*F)|X
proof
  let F,G be Function,X;
  set Y = dom ((G*F)|X);
  now
    let x;
    thus x in dom ((G|(F.:X))*(F|X)) implies x in Y
    proof
      assume
A1:   x in dom ((G|(F.:X))*(F|X));
      then
A2:   x in dom (F|X) by Th11;
      then
A3:   x in dom F /\ X by RELAT_1:61;
      then
A4:   x in X by XBOOLE_0:def 4;
      (F|X).x in dom (G|(F.:X)) by A1,Th11;
      then F.x in dom (G|(F.:X)) by A2,Th47;
      then F.x in dom G /\ (F.:X) by RELAT_1:61;
      then
A5:   F.x in dom G by XBOOLE_0:def 4;
      x in dom F by A3,XBOOLE_0:def 4;
      then x in dom (G*F) by A5,Th11;
      then x in dom (G*F)/\ X by A4,XBOOLE_0:def 4;
      hence thesis by RELAT_1:61;
    end;
    assume x in Y;
    then
A6: x in dom (G*F) /\ X by RELAT_1:61;
    then
A7: x in dom (G*F) by XBOOLE_0:def 4;
    then
A8: F.x in dom G by Th11;
A9: x in X by A6,XBOOLE_0:def 4;
    x in dom F by A7,Th11;
    then x in dom F /\ X by A9,XBOOLE_0:def 4;
    then
A10: x in dom (F|X) by RELAT_1:61;
    x in dom F by A7,Th11;
    then F.x in F.:X by A9,Def6;
    then F.x in dom G /\ (F.:X) by A8,XBOOLE_0:def 4;
    then F.x in dom (G|(F.:X)) by RELAT_1:61;
    then (F|X).x in dom (G|(F.:X)) by A10,Th47;
    hence x in dom ((G|(F.:X))*(F|X)) by A10,Th11;
  end;
  then
A11: Y = dom ((G|(F.:X))*(F|X)) by TARSKI:1;
  now
    let x;
    assume
A12: x in Y;
    then
A13: x in dom (G*F) /\ X by RELAT_1:61;
    then x in dom (G*F) by XBOOLE_0:def 4;
    then
A14: x in dom F by Th11;
A15: x in X by A13,XBOOLE_0:def 4;
    then
A16: F.x in F.:X by A14,Def6;
    thus ((G|(F.:X))*(F|X)).x =(G|(F.:X)).((F|X).x) by A11,A12,Th12
      .= (G|(F.:X)).(F.x) by A15,Th49
      .= G.(F.x) by A16,Th49
      .= (G*F).x by A14,Th13
      .= ((G*F)|X).x by A13,Th48;
  end;
  hence thesis by A11,Th2;
end;

theorem
  for F,G be Function, X,X1 holds (G|X1)*(F|X) = (G*F)|(X /\ (F"X1))
proof
  let F,G be Function,X,X1;
  set Y = dom ((G|X1)*(F|X));
  now
    let x;
    thus x in dom ((G*F)|(X /\ (F"X1))) implies x in Y
    proof
      assume x in dom ((G*F)|(X /\ (F"X1)));
      then
A1:   x in dom (G*F) /\ (X /\ (F"X1)) by RELAT_1:61;
      then
A2:   x in dom (G*F) by XBOOLE_0:def 4;
A3:   x in X /\ (F"X1) by A1,XBOOLE_0:def 4;
      then
A4:   x in X by XBOOLE_0:def 4;
      x in dom F by A2,Th11;
      then x in dom F /\ X by A4,XBOOLE_0:def 4;
      then
A5:   x in dom (F|X) by RELAT_1:61;
      x in (F"X1) by A3,XBOOLE_0:def 4;
      then
A6:   F.x in X1 by Def7;
      F.x in dom G by A2,Th11;
      then F.x in dom G /\ X1 by A6,XBOOLE_0:def 4;
      then F.x in dom (G|X1) by RELAT_1:61;
      then (F|X).x in dom (G|X1) by A5,Th47;
      hence thesis by A5,Th11;
    end;
    assume
A7: x in Y;
    then
A8: x in dom(F|X) by Th11;
    then
A9: x in dom F /\ X by RELAT_1:61;
    then
A10: x in dom F by XBOOLE_0:def 4;
A11: x in X by A9,XBOOLE_0:def 4;
    (F|X).x in dom(G|X1) by A7,Th11;
    then F.x in dom (G|X1) by A8,Th47;
    then
A12: F.x in dom G /\ X1 by RELAT_1:61;
    then F.x in X1 by XBOOLE_0:def 4;
    then x in F"X1 by A10,Def7;
    then
A13: x in X /\ F"X1 by A11,XBOOLE_0:def 4;
    F.x in dom G by A12,XBOOLE_0:def 4;
    then x in dom (G*F) by A10,Th11;
    then x in dom (G*F) /\ (X/\(F"X1)) by A13,XBOOLE_0:def 4;
    hence x in dom ((G*F)|(X /\ (F"X1))) by RELAT_1:61;
  end;
  then
A14: Y = dom ((G*F)|(X /\ (F"X1))) by TARSKI:1;
  now
    let x;
    assume
A15: x in Y;
    then
A16: x in dom (F|X) by Th11;
    then
A17: x in dom F /\ X by RELAT_1:61;
    then
A18: x in dom F by XBOOLE_0:def 4;
A19: (F|X).x in dom (G|X1) by A15,Th11;
    then
A20: F.x in dom (G|X1) by A16,Th47;
A21: x in X by A17,XBOOLE_0:def 4;
    F.x in dom (G|X1) by A16,A19,Th47;
    then F.x in dom G /\ X1 by RELAT_1:61;
    then F.x in X1 by XBOOLE_0:def 4;
    then x in F"X1 by A18,Def7;
    then
A22: x in X /\ F"X1 by A21,XBOOLE_0:def 4;
    thus ((G|X1)*(F|X)).x =(G|X1).((F|X).x) by A15,Th12
      .= (G|X1).(F.x) by A16,Th47
      .= G.(F.x) by A20,Th47
      .= (G*F).x by A18,Th13
      .= ((G*F)|(X/\(F"X1))).x by A22,Th49;
  end;
  hence thesis by A14,Th2;
end;

theorem
  for F,G be Function,X holds X c= dom (G*F) iff X c= dom F & F.:X c= dom G
proof
  let F,G be Function,X;
  thus X c= dom (G*F) implies X c= dom F & F.:X c= dom G
  proof
    assume
A1: X c= dom (G*F);
    then for x st x in X holds x in dom F by Th11;
    hence X c= dom F by TARSKI:def 3;
    let x;
    assume x in F.:X;
    then ex y st y in dom F & y in X & x=F.y by Def6;
    hence thesis by A1,Th11;
  end;
  assume that
A2: X c= dom F and
A3: F.:X c= dom G;
  let x;
  assume
A4: x in X;
  then F.x in F.:X by A2,Def6;
  hence thesis by A2,A3,A4,Th11;
end;



definition
  let f be Function;
  assume
A1: f is non empty constant;
  func the_value_of f means
  ex x being set st x in dom f & it = f.x;
  existence
  proof
    consider x being set such that
A2: x in dom f by A1,XBOOLE_0:def 1;
    take f.x;
    thus thesis by A2;
  end;
  uniqueness by A1,Def10;
end;



registration
  let X,Y;
  cluster X-defined Y-valued for Function;
  existence
  proof
    take {};
    thus dom{} c= X & rng{} c= Y by XBOOLE_1:2;
  end;
end;

theorem
  for X being set, f being X-valued Function for x being set st x in dom
  f holds f.x in X
proof
  let X be set, f be X-valued Function;
  let x be set;
  assume x in dom f;
  then
A1: f.x in rng f by Def3;
  rng f c= X by RELAT_1:def 19;
  hence thesis by A1;
end;



definition
  let IT be set;
  attr IT is functional means
  :Def13:
  for x being set st x in IT holds x is Function;
end;

registration
  cluster empty -> functional for set;
  coherence
  proof
    let A be set;
    assume
A1: A is empty;
    let x be set;
    thus thesis by A1;
  end;
  let f be Function;
  cluster { f } -> functional;
  coherence
  proof
    let x be set;
    thus thesis by TARSKI:def 1;
  end;
  let g be Function;
  cluster { f,g } -> functional;
  coherence
  proof
    let x be set;
    thus thesis by TARSKI:def 2;
  end;
end;

registration
  cluster non empty functional for set;
  existence
  proof
    take { {} };
    thus thesis;
  end;
end;

registration
  let P be functional set;
  cluster -> Function-like Relation-like for Element of P;
  coherence
  proof
    let x be Element of P;
    per cases;
    suppose
      P is empty;
      hence thesis by SUBSET_1:def 1;
    end;
    suppose
      P is non empty;
      hence thesis by Def13;
    end;
  end;
end;

registration
  let A be functional set;
  cluster -> functional for Subset of A;
  coherence
proof
  let B be Subset of A;
  let x be set;
  thus thesis;
end;
end;



definition let g,f be Function;
 attr f is g-compatible means
:Def14: x in dom f implies f.x in g.x;
end;

theorem
  f is g-compatible & dom f = dom g implies g is non-empty proof assume
A1: f is g-compatible & dom f = dom g;
  let x;
  assume x in dom g;
  hence g.x is non empty by A1,Def14;
 end;

theorem Th104:
 {} is f-compatible
 proof
  let x;
  thus thesis;
 end;

registration
 let I be set, f be Function;
 cluster empty I-defined f-compatible for Function;
 existence
  proof
   take {};
   thus thesis by Th104,RELAT_1:171;
  end;
end;

registration let X be set;
 let f be Function, g be f-compatible Function;
 cluster g|X -> f-compatible;
 coherence
  proof let x;
A1:  dom(g|X) c= dom g by RELAT_1:60;
   assume
A2:   x in dom(g|X);
    then g.x in f.x by A1,Def14;
   hence (g|X).x in f.x by A2,Th47;
  end;
end;

registration let I be set;
 cluster non-empty I-defined for Function;
 existence
  proof
   take {};
   thus {} is non-empty;
   thus dom {} c= I by XBOOLE_1:2;
  end;
end;

theorem Th105:
 for g being f-compatible Function holds dom g c= dom f
 proof
  let g be f-compatible Function;
  let x be set;
  assume x in dom g;
   then g.x in f.x by Def14;
  hence x in dom f by Def2;
 end;

registration let X;
 let f be X-defined Function;
 cluster f-compatible -> X-defined for Function;
 coherence
  proof let g be Function;
   assume g is f-compatible;
    then
A1:   dom g c= dom f by Th105;
    dom f c= X by RELAT_1:def 18;
   hence dom g c= X by A1,XBOOLE_1:1;
  end;
end;

theorem
  for f being X-valued Function st x in dom f holds f.x is Element of X
 proof let f be X-valued Function;
  assume x in dom f;
   then
A1: f.x in rng f by Def3;
   rng f c= X by RELAT_1:def 19;
  hence f.x is Element of X by A1;
 end;



theorem
  for f being Function,A being set st f is one-to-one & A c= dom f
  holds f".:(f.:A)=A
proof
  let f be Function,A be set;
  set B = f.:A;
  assume that
A1: f is one-to-one and
A2: A c= dom f;
A3: f".:B c= A
  proof
    let y be set;
    assume y in f".:B;
    then consider x being set such that
    x in dom (f") and
A4: x in B and
A5: y=f".x by Def6;
    ex y2 being set st ( y2 in dom f)&( y2 in A)&( x=f.y2) by A4,Def6
;
    hence thesis by A1,A5,Th32;
  end;
  A c= f".:B
  proof
    let x be set;
    assume
A6: x in A;
    set y0=f.x;
A7: f".y0=x by A1,A2,A6,Th34;
    y0 in rng f by A2,A6,Def3;
    then
A8: y0 in dom (f") by A1,Th33;
    y0 in B by A2,A6,Def6;
    hence thesis by A7,A8,Def6;
  end;
  hence thesis by A3,XBOOLE_0:def 10;
end;

registration
  let A be functional set, x be set;
  let F be A-valued Function;
  cluster F.x -> Function-like Relation-like;
  coherence
  proof
    per cases;
    suppose x in dom F;
      then
A1:   F.x in rng F by Def3;
      rng F c= A by RELAT_1:def 19;
      hence thesis by A1;
    end;
    suppose not x in dom F;
      hence thesis by Def2;
    end;
  end;
end;



theorem Th108:
 x in X & x in dom f implies f.x in f.:X
  proof assume that
A1: x in X and
A2: x in dom f;
   x in X /\ dom f by A1,A2,XBOOLE_0:def 4;
   then x in dom(f|X) by RELAT_1:61;
   then
A3:  (f|X).x in rng(f|X) by Def3;
    (f|X).x = f.x by A1,Th49;
   hence f.x in f.:X by A3,RELAT_1:115;
  end;

theorem
 X <> {} & X c= dom f implies f.:X <> {}
 proof assume X <> {};
   then ex x st x in X by XBOOLE_0:def 1;
  hence thesis by Th108;
 end;

registration
  let f be non trivial Function;
  cluster dom f -> non trivial;
  coherence
  proof
   consider u,w being set such that
A1: u in f and
A2: w in f and
A3: u <> w by ZFMISC_1:def 10;
   consider u1,u2 being set such that
A4: u = [u1,u2] by A1,RELAT_1:def 1;
   consider w1,w2 being set such that
A5: w = [w1,w2] by A2,RELAT_1:def 1;
   take u1,w1;
   thus u1 in dom f & w1 in dom f by A4,A5,A1,A2,XTUPLE_0:def 12;
   thus u1 <> w1 by A1,A2,A3,A4,A5,Def1;
  end;
end;



theorem
  for B being non empty functional set, f being Function
   st f = union B
  holds dom f = union { dom g where g is Element of B: not contradiction}
      & rng f = union { rng g where g is Element of B: not contradiction }
proof
  let B be non empty functional set, f be Function such that
A1: f = union B;
  set X = { dom g where g is Element of B: not contradiction };
  now
    let x be set;
    hereby
      assume x in dom f;
      then [x,f.x] in f by Th1;
      then consider g being set such that
A2:   [x,f.x] in g and
A3:   g in B by A1,TARSKI:def 4;
      reconsider g as Function by A3;
      take Z = dom g;
      thus x in Z & Z in X by A2,A3,Th1;
    end;
    given Z being set such that
A4: x in Z and
A5: Z in X;
    consider g being Element of B such that
A6: Z = dom g by A5;
    [x,g.x] in g by A4,A6,Th1;
    then [x,g.x] in f by A1,TARSKI:def 4;
    hence x in dom f by Th1;
  end;
  hence dom f = union X by TARSKI:def 4;
  set X = { rng g where g is Element of B: not contradiction };
  now
    let y be set;
    hereby
      assume y in rng f;
      then consider x being set such that
A7:   x in dom f & y = f.x by Def3;
      [x,y] in f by A7,Th1;
      then consider g being set such that
A8:   [x,y] in g and
A9:   g in B by A1,TARSKI:def 4;
      reconsider g as Function by A9;
      take Z = rng g;
      x in dom g & y = g.x by A8,Th1;
      hence y in Z & Z in X by A9,Def3;
    end;
    given Z being set such that
A10: y in Z and
A11: Z in X;
    consider g being Element of B such that
A12: Z = rng g by A11;
    consider x being set such that
A13: x in dom g & y = g.x by A10,A12,Def3;
    [x,y] in g by A13,Th1;
    then [x,y] in f by A1,TARSKI:def 4;
    hence y in rng f by XTUPLE_0:def 13;
  end;
  hence thesis by TARSKI:def 4;
end;

theorem Th111: 
 for M being set
   st for X st X in M holds X <> {}
  ex f being Function
  st dom f = M & for X st X in M holds f.X in X
proof let M be set;
 assume
A1: for X st X in M holds X <> {};
  deffunc F(set) = the Element of $1;
  consider f being Function such that
A2: dom f = M and
A3: for x st x in M holds f.x = F(x) from Lambda;
 take f;
 thus dom f = M by A2;
 let X;
 assume
A4: X in M;
  then
A5:  f.X = the Element of X by A3;
  X <> {} by A1,A4;
 hence f.X in X by A5;
end;

scheme
  NonUniqBoundFuncEx { X() -> set, Y() -> set, P[set,set] }:
  ex f being Function st dom f = X() & rng f c= Y()
   & for x st x in X() holds P[x,f.x]
provided
A1: for x st x in X() ex y st y in Y() & P[x,y]
proof
 per cases;
 suppose
A2: X() = {};
  take {};
  thus thesis by A2,XBOOLE_1:2;
 end;
 suppose
A3: X() <> {};
  defpred Q[set,set] means for y holds y in $2 iff y in Y() & P[$1,y];
A4: for e,u1,u2 being set st e in X() & Q[e,u1] & Q[e,u2] holds u1 = u2
  proof
    let e,u1,u2 be set such that
    e in X();
    assume
A5: for y holds y in u1 iff y in Y() & P[e,y];
    defpred A[set] means $1 in Y() & P[e,$1];
A6: for x be set holds x in u1 iff A[x] by A5;
    assume
A7: for y holds y in u2 iff y in Y() & P[e,y];
A8: for x be set holds x in u2 iff A[x] by A7;
    u1 = u2 from XBOOLE_0:sch 2(A6,A8);
    hence thesis;
  end;
A9: for x st x in X() ex y st Q[x,y]
  proof
    let x such that
    x in X();
    defpred R[set] means P[x,$1];
    consider X such that
A10: y in X iff y in Y() & R[y] from XBOOLE_0:sch 1;
    take X;
    thus thesis by A10;
  end;
   consider G being Function such that
A11: dom G = X() & for x st x in X() holds Q[x,G.x] from FuncEx(A4,A9);
   reconsider D = rng G as non empty set by A11,A3,RELAT_1:42;
    now
      let X;
      assume X in D;
      then consider x such that
A12:  x in dom G & X = G.x by Def3;
      (for y holds y in X iff y in Y() & P[x,y]) &
      ex y st y in Y() & P[x,y] by A1,A11,A12;
      hence X <> {};
    end;
    then consider F be Function such that
A13: dom F = D and
A14: for X st X in D holds F.X in X by Th111;
A15: dom (F*G) = X() by A11,A13,RELAT_1:27;
    take f = F*G;
    thus dom f = X() by A11,A13,RELAT_1:27;
      rng F c= Y()
      proof
        let x;
        assume x in rng F;
        then consider y such that
A16:    y in dom F and
A17:    x = F.y by Def3;
A18:    ex z st z in dom G & y = G.z by A13,A16,Def3;
        x in y by A13,A14,A16,A17;
        hence thesis by A11,A18;
      end;
      hence rng f c= Y() by A13,RELAT_1:28;
      let x;
      assume
A19:  x in X();
      then f.x = F.(G.x) & G.x in D by A11,A15,Th12,Def3;
      then f.x in G.x by A14;
  hence thesis by A11,A19;
 end;
end;

registration
  let f be empty-yielding Function;
  let x;
  cluster f.x -> empty;
  coherence
  proof
    x in dom f or not x in dom f;
    hence thesis by Def2,Def8;
  end;
end;















begin

reserve X,Y,p,x,x1,x2,y,y1,y2,z,z1,z2 for set;
reserve f,g,h for Function;

theorem Th1:
  for G being set st G c= f holds G is Function;

theorem Th2:
  f c= g iff dom f c= dom g & for x st x in dom f holds f.x = g.x
proof
  thus f c= g implies dom f c= dom g & for x st x in dom f holds f.x = g.x
  proof
    assume
A1: f c= g;
    hence dom f c= dom g by RELAT_1:11;
    let x;
    assume x in dom f;
    then [x,f.x] in f by FUNCT_1:def 2;
    hence thesis by A1,FUNCT_1:1;
  end;
  assume that
A2: dom f c= dom g and
A3: for x st x in dom f holds f.x = g.x;
  let x,y;
  assume
A4: [x,y] in f;
  then
A5: x in dom f by FUNCT_1:1;
  y = f.x by A4,FUNCT_1:1;
  then y = g.x by A3,A5;
  hence thesis by A2,A5,FUNCT_1:def 2;
end;

theorem
  dom f = dom g & f c= g implies f = g
proof
  assume that
A1: dom f = dom g and
A2: f c= g;
  [x,y] in f iff [x,y] in g
  proof
    thus [x,y] in f implies [x,y] in g by A2;
    assume
A3: [x,y] in g;
    then x in dom f by A1,XTUPLE_0:def 12;
    then [x,f.x] in f by FUNCT_1:1;
    hence thesis by A2,A3,FUNCT_1:def 1;
  end;
  hence thesis by RELAT_1:def 2;
end;

Lm1: rng f /\ rng h = {} & x in dom f & y in dom h implies f.x <> h.y
proof
  assume
A1: rng f /\ rng h = {};
  assume x in dom f & y in dom h;
  then f.x in rng f & h.y in rng h by FUNCT_1:def 3;
  hence thesis by A1,XBOOLE_0:def 4;
end;

theorem
  [x,z] in (g*f) implies [x,f.x] in f & [f.x,z] in g
proof
  assume [x,z] in (g*f);
  then ex y st [x,y] in f & [y,z] in g by RELAT_1:def 8;
  hence thesis by FUNCT_1:1;
end;

theorem
  {[x,y]} is Function;

Lm2: [x,y] in {[x1,y1]} implies x = x1 & y = y1
proof
  assume [x,y] in {[x1,y1]};
  then [x,y] = [x1,y1] by TARSKI:def 1;
  hence thesis by XTUPLE_0:1;
end;

theorem
  f = {[x,y]} implies f.x = y
proof
  assume f = {[x,y]};
  then [x,y] in f by TARSKI:def 1;
  hence thesis by FUNCT_1:1;
end;

theorem Th7:
  dom f = {x} implies f = {[x,f.x]}
proof
  reconsider g = {[x,f.x]} as Function;
  assume
A1: dom f = {x};
  [y,z] in f iff [y,z] in g
  proof
    thus [y,z] in f implies [y,z] in g
    proof
      assume
A2:   [y,z] in f;
      then y in {x} by A1,XTUPLE_0:def 12;
      then
A3:   y = x by TARSKI:def 1;
A4:   rng f = {f.x} by A1,FUNCT_1:4;
      z in rng f by A2,XTUPLE_0:def 13;
      then z = f.x by A4,TARSKI:def 1;
      hence thesis by A3,TARSKI:def 1;
    end;
    assume [y,z] in g;
    then
A5: y = x & z = f.x by Lm2;
    x in dom f by A1,TARSKI:def 1;
    hence thesis by A5,FUNCT_1:def 2;
  end;
  hence thesis by RELAT_1:def 2;
end;

theorem
  {[x1,y1],[x2,y2]} is Function iff (x1 = x2 implies y1 = y2)
proof
  thus {[x1,y1],[x2,y2]} is Function & x1 = x2 implies y1 = y2
  proof
A1: [x1,y1] in {[x1,y1],[x2,y2]} & [x2,y2] in {[x1,y1],[x2,y2]} by TARSKI:def 2
;
    assume {[x1,y1],[x2,y2]} is Function;
    hence thesis by A1,FUNCT_1:def 1;
  end;
  assume
A2: x1 = x2 implies y1 = y2;
  now
    thus p in {[x1,y1],[x2,y2]} implies ex x,y st [x,y] = p
    proof
      assume p in {[x1,y1],[x2,y2]};
      then p = [x1,y1] or p = [x2,y2] by TARSKI:def 2;
      hence thesis;
    end;
    let x,z1,z2;
A3: [x,z1] = [x1,y1] & [x,z2] = [x1,y1] or [x,z1] = [x2,y2] & [x,z2] = [
    x2,y2] implies z1 = y1 & z2 = y1 or z1 = y2 & z2 = y2 by XTUPLE_0:1;
A4: now
      assume
A5:   [x,z1] = [x1,y1] & [x,z2] = [x2,y2] or [x,z1] = [x2,y2] & [x,z2
      ] = [x1,y1];
      then x = x1 & x = x2 by XTUPLE_0:1;
      hence z1 = z2 by A2,A5,XTUPLE_0:1;
    end;
    assume [x,z1] in {[x1,y1],[x2,y2]} & [x,z2] in {[x1,y1],[x2,y2]};
    hence z1 = z2 by A3,A4,TARSKI:def 2;
  end;
  hence thesis by FUNCT_1:def 1;
end;

theorem Th9:
  f is one-to-one iff for x1,x2,y st [x1,y] in f & [x2,y] in f holds x1 = x2
proof
  thus f is one-to-one implies for x1,x2,y st [x1,y] in f & [x2,y] in f holds
  x1 = x2
  proof
    assume
A1: f is one-to-one;
    let x1,x2,y;
    assume
A2: [x1,y] in f & [x2,y] in f;
    then
A3: f.x1 = y & f.x2 = y by FUNCT_1:1;
    x1 in dom f & x2 in dom f by A2,FUNCT_1:1;
    hence thesis by A1,A3,FUNCT_1:def 4;
  end;
  assume
A4: for x1,x2,y st [x1,y] in f & [x2,y] in f holds x1 = x2;
  let x1,x2;
  assume x1 in dom f & x2 in dom f & f.x1 = f.x2;
  then [x1,f.x1] in f & [x2,f.x1] in f by FUNCT_1:1;
  hence thesis by A4;
end;

theorem Th10:
  g c= f & f is one-to-one implies g is one-to-one
proof
  assume g c= f & f is one-to-one;
  then for x1,x2,y st [x1,y] in g & [x2,y] in g holds x1 = x2 by Th9;
  hence thesis by Th9;
end;

registration
  let f,X;
  cluster f /\ X -> Function-like;
  coherence by Th1,XBOOLE_1:17;
end;

theorem
  x in dom(f /\ g) implies (f /\ g).x = f.x
proof
  set y = (f /\ g).x;
  assume x in dom(f /\ g);
  then [x,y] in (f /\ g) by FUNCT_1:def 2;
  then [x,y] in f by XBOOLE_0:def 4;
  hence thesis by FUNCT_1:1;
end;

theorem
  f is one-to-one implies f /\ g is one-to-one by Th10,XBOOLE_1:17;

theorem
  dom f misses dom g implies f \/ g is Function
proof
  assume
A1: dom f /\ dom g = {};
  now
    thus p in f \/ g implies ex x,y st [x,y] = p by RELAT_1:def 1;
    let x,y1,y2;
    assume [x,y1] in f \/ g;
    then
A2: [x,y1] in f or [x,y1] in g by XBOOLE_0:def 3;
    assume [x,y2] in f \/ g;
    then
A3: [x,y2] in f or [x,y2] in g by XBOOLE_0:def 3;
    not (x in dom f & x in dom g) by A1,XBOOLE_0:def 4;
    hence y1 = y2 by A2,A3,FUNCT_1:def 1,XTUPLE_0:def 12;
  end;
  hence thesis by FUNCT_1:def 1;
end;

theorem
  f c= h & g c= h implies f \/ g is Function by Th1,XBOOLE_1:8;

Lm3: h = f \/ g implies (x in dom h iff x in dom f or x in dom g)
proof
  assume
A1: h = f \/ g;
  thus x in dom h implies x in dom f or x in dom g
  proof
    assume x in dom h;
    then [x,h.x] in h by FUNCT_1:def 2;
    then [x,h.x] in f or [x,h.x] in g by A1,XBOOLE_0:def 3;
    hence thesis by XTUPLE_0:def 12;
  end;
  assume x in dom f or x in dom g;
  then [x,f.x] in f or [x,g.x] in g by FUNCT_1:def 2;
  then [x,f.x] in h or [x,g.x] in h by A1,XBOOLE_0:def 3;
  hence thesis by XTUPLE_0:def 12;
end;

theorem Th15:
  x in dom g & h = f \/ g implies h.x = g.x
proof
  assume x in dom g;
  then [x,g.x] in g by FUNCT_1:def 2;
  then h = f \/ g implies [x,g.x] in h by XBOOLE_0:def 3;
  hence thesis by FUNCT_1:1;
end;

theorem
  x in dom h & h = f \/ g implies h.x = f.x or h.x = g.x
proof
  assume x in dom h;
  then [x,h.x] in h by FUNCT_1:def 2;
  then h = f \/ g implies [x,h.x] in f or [x,h.x] in g by XBOOLE_0:def 3;
  hence thesis by FUNCT_1:1;
end;

theorem
  f is one-to-one & g is one-to-one & h = f \/ g & rng f misses rng g
  implies h is one-to-one
proof
  assume that
A1: f is one-to-one & g is one-to-one and
A2: h = f \/ g and
A3: rng f /\ rng g = {};
  now
    let x1,x2;
    assume that
A4: x1 in dom h & x2 in dom h and
A5: h.x1 = h.x2;
A6: now
      assume x1 in dom f & x2 in dom g or x1 in dom g & x2 in dom f;
      then
      h.x1 = f.x1 & h.x2 = g.x2 & f.x1 <> g.x2 or h.x1 = g.x1 & h.x2 = f.
      x2 & f.x2 <> g.x1 by A2,A3,Lm1,Th15;
      hence x1 = x2 by A5;
    end;
A7: x1 in dom g & x2 in dom g implies h.x1 = g.x1 & h.x2 = g.x2 by A2,Th15;
    x1 in dom f & x2 in dom f implies h.x1 = f.x1 & h.x2 = f.x2 by A2,Th15;
    then x1 in dom f & x2 in dom f or x1 in dom g & x2 in dom g implies x1 =
    x2 by A1,A5,A7,FUNCT_1:def 4;
    hence x1 = x2 by A2,A4,A6,Lm3;
  end;
  hence thesis by FUNCT_1:def 4;
end;

canceled 2;

theorem
  f is one-to-one implies for x,y holds [y,x] in f" iff [x,y] in f
proof
  assume
A1: f is one-to-one;
  let x,y;
  dom(f") = rng(f) by A1,FUNCT_1:33;
  then y in dom(f") & x = (f").y iff x in dom f & y = f.x by A1,FUNCT_1:32;
  hence thesis by FUNCT_1:1;
end;

theorem
  f = {} implies f" = {}
proof
  assume f = {};
  then dom(f") = {} by FUNCT_1:33,RELAT_1:38;
  hence thesis;
end;

theorem
  x in dom f & x in X iff [x,f.x] in f|X
proof
  x in dom f iff [x,f.x] in f by FUNCT_1:def 2,XTUPLE_0:def 12;
  hence thesis by RELAT_1:def 11;
end;

theorem Th23:
  g c= f implies f|dom g = g
proof
  assume
A1: g c= f;
  [x,y] in (f|dom g) implies [x,y] in g
  proof
    assume
A2: [x,y] in (f|dom g);
    then x in dom g by RELAT_1:def 11;
    then
A3: [x,g.x] in g by FUNCT_1:def 2;
    [x,y] in f by A2,RELAT_1:def 11;
    hence thesis by A1,A3,FUNCT_1:def 1;
  end;
  then
A4: (f|dom g) c= g by RELAT_1:def 3;
  (g|dom g) c= (f|dom g) by A1,RELAT_1:76;
  then g c= (f|dom g);
  hence thesis by A4,XBOOLE_0:def 10;
end;

theorem
  x in dom f & f.x in Y iff [x,f.x] in Y|`f
proof
  x in dom f iff [x,f.x] in f by FUNCT_1:def 2,XTUPLE_0:def 12;
  hence thesis by RELAT_1:def 12;
end;

theorem
  g c= f & f is one-to-one implies rng g|`f = g
proof
  assume
A1: g c= f;
  assume
A2: f is one-to-one;
  [x,y] in (rng g|`f) implies [x,y] in g
  proof
    assume
A3: [x,y] in (rng g|`f);
    then y in rng g by RELAT_1:def 12;
    then
A4: ex x1 st [x1,y] in g by XTUPLE_0:def 13;
    [x,y] in f by A3,RELAT_1:def 12;
    hence thesis by A1,A2,A4,Th9;
  end;
  then
A5: (rng g|`f) c= g by RELAT_1:def 3;
  (rng g|`g) c= (rng g|`f) by A1,RELAT_1:101;
  then g c= (rng g|`f);
  hence thesis by A5,XBOOLE_0:def 10;
end;

theorem
  x in f"Y iff [x,f.x] in f & f.x in Y
proof
  thus x in f"Y implies [x,f.x] in f & f.x in Y
  proof
    assume x in f"Y;
    then ex y st [x,y] in f & y in Y by RELAT_1:def 14;
    hence thesis by FUNCT_1:1;
  end;
  thus thesis by RELAT_1:def 14;
end;

begin 


theorem
  for X being set, f,g being Function st X c= dom f & f c= g holds f|X = g|X
proof
  let X be set, f,g be Function such that
A1: X c= dom f;
  assume f c= g;
  hence f|X = g|(dom f)|X by Th23
    .= g|((dom f) /\ X) by RELAT_1:71
    .= g|X by A1,XBOOLE_1:28;
end;



theorem Th28:
  for f being Function, x being set st x in dom f holds f|{x} = {[ x,f.x]}
proof
  let f be Function, x be set such that
A1: x in dom f;
A2: x in {x} by TARSKI:def 1;
  dom(f|{x} qua Function) = dom f /\ {x} by RELAT_1:61
    .= {x} by A1,ZFMISC_1:46;
  hence f|{x} = {[x,(f|{x}).x]} by Th7
    .= {[x,f.x]} by A2,FUNCT_1:49;
end;



theorem Th29:
  for f,g being Function, x being set st dom f = dom g & f.x = g.x
  holds f|{x} = g|{x}
proof
  let f,g be Function, x be set such that
A1: dom f = dom g and
A2: f.x = g.x;
  per cases;
  suppose
A3: x in dom f;
    hence f|{x} = {[x,g.x]} by A2,Th28
      .= g|{x} by A1,A3,Th28;
  end;
  suppose
    not x in dom f;
    then
A4: {x} misses dom f by ZFMISC_1:50;
    hence f|{x} = {} by RELAT_1:66
      .= g|{x} by A1,A4,RELAT_1:66;
  end;
end;

theorem Th30:
  for f,g being Function, x,y being set st dom f = dom g & f.x = g
  .x & f.y = g.y holds f|{x,y} = g|{x,y}
proof
  let f,g be Function, x,y be set;
  assume dom f = dom g & f.x = g.x & f.y = g.y;
  then
A1: f|{x} = g|{x} & f|{y} = g|{y} by Th29;
  {x,y} = {x} \/ {y} by ENUMSET1:1;
  hence thesis by A1,RELAT_1:150;
end;

theorem
  for f,g being Function, x,y,z being set st dom f = dom g & f.x = g.x &
  f.y = g.y & f.z = g.z holds f|{x,y,z} = g|{x,y,z}
proof
  let f,g be Function, x,y,z be set;
  assume dom f = dom g & f.x = g.x & f.y = g.y & f.z = g.z;
  then
A1: f|{x,y} = g|{x,y} & f|{z} = g|{z} by Th29,Th30;
  {x,y,z} = {x,y} \/ {z} by ENUMSET1:3;
  hence thesis by A1,RELAT_1:150;
end;



registration
  let f be Function, A be set;
  cluster f \ A -> Function-like;
  coherence;
end;

theorem
  for f, g being Function st x in dom f \ dom g holds (f \ g).x = f.x
proof
  let f, g be Function such that
A1: x in dom f \ dom g;
  f \ g c= f & dom f \ dom g c= dom (f \ g) by RELAT_1:3;
  hence thesis by A1,Th2;
end;



theorem
  f c= g & f c= h implies g|dom f = h|dom f
proof
  assume that
A1: f c= g and
A2: f c= h;
  thus g|dom f = f by A1,Th23
    .= h|dom f by A2,Th23;
end;



registration let f be Function, g be Subset of f;
 cluster g-compatible -> f-compatible for Function;
 coherence
  proof let F be Function such that
A1: F is g-compatible;
   let x;
   assume x in dom F;
    then
A2:   F.x in g.x by A1,FUNCT_1:def 14;
    then x in dom g by FUNCT_1:def 2;
   hence F.x in f.x by A2,Th2;
  end;
end;



theorem Th34:
 g c= f implies g = f|dom g
 proof
  assume
A1:  g c= f;
   then dom g c= dom f by RELAT_1:11;
  hence dom g = dom(f|dom g) by RELAT_1:62;
  let x;
  assume
A2: x in dom g;
  hence g.x = f.x by A1,Th2
      .= (f|dom g).x by A2,FUNCT_1:49;
 end;

registration
 let f be Function, g be f-compatible Function;
 cluster -> f-compatible for Subset of g;
 coherence
  proof let h be Subset of g;
   h = g|dom h by Th34;
   hence thesis;
  end;
end;

theorem
 g c= f & x in X & X /\ dom f c= dom g
  implies f.x = g.x
proof assume that
A1: g c= f and
A2: x in X and
A3: X /\ dom f c= dom g;
 per cases;
 suppose x in dom g;
 hence f.x = g.x by A1,Th2;
 end;
 suppose
A4:  not x in dom g;
  then not x in X /\ dom f by A3;
  then not x in dom f by A2,XBOOLE_0:def 4;
 hence f.x = {} by FUNCT_1:def 2
   .= g.x by A4,FUNCT_1:def 2;
 end;
end;














begin

reserve X,a,b,c,x,y,z for set;
reserve P,R for Relation;

definition
  let R,X;
  pred R is_reflexive_in X means
  :Def1:
  x in X implies [x,x] in R;
  pred R is_irreflexive_in X means
  :Def2:
  x in X implies not [x,x] in R;
  pred R is_symmetric_in X means
  :Def3:
  x in X & y in X & [x,y] in R implies [ y,x] in R;
  pred R is_antisymmetric_in X means
  :Def4:
  x in X & y in X & [x,y] in R & [y, x] in R implies x = y;
  pred R is_asymmetric_in X means
  :Def5:
  x in X & y in X & [x,y] in R implies not [y,x] in R;
  pred R is_connected_in X means
  :Def6:
  x in X & y in X & x <>y implies [x,y] in R or [y,x] in R;
  pred R is_strongly_connected_in X means
  :Def7:
  x in X & y in X implies [x,y] in R or [y,x] in R;
  pred R is_transitive_in X means
  :Def8:
  x in X & y in X & z in X & [x,y] in R & [y,z] in R implies [x,z] in R;
end;

definition
  let R;
  attr R is reflexive means
  :Def9:
  R is_reflexive_in field R;
  attr R is irreflexive means
  :Def10:
  R is_irreflexive_in field R;
  attr R is symmetric means
  :Def11:
  R is_symmetric_in field R;
  attr R is antisymmetric means
  :Def12:
  R is_antisymmetric_in field R;
  attr R is asymmetric means
  :Def13:
  R is_asymmetric_in field R;
  attr R is connected means
  :Def14:
  R is_connected_in field R;
  attr R is strongly_connected means
  :Def15:
  R is_strongly_connected_in field R;
  attr R is transitive means
  :Def16:
  R is_transitive_in field R;
end;

registration
  cluster empty -> reflexive irreflexive symmetric antisymmetric asymmetric
          connected strongly_connected transitive for Relation;
  coherence
  proof
    let R be Relation;
    assume
A1: R is empty;
    thus R is reflexive
    proof
      let x;
      thus thesis by A1,RELAT_1:40;
    end;
    thus R is irreflexive
    proof
      let x;
      thus thesis by A1;
    end;
    thus R is symmetric
    proof
      let x;
      thus thesis by A1;
    end;
    thus R is antisymmetric
    proof
      let x;
      thus thesis by A1;
    end;
    thus R is asymmetric
    proof
      let x;
      thus thesis by A1;
    end;
    thus R is connected
    proof
      let x;
      thus thesis by A1,RELAT_1:40;
    end;
    thus R is strongly_connected
    proof
      let x;
      thus thesis by A1,RELAT_1:40;
    end;
    let x;
    thus thesis by A1;
  end;
end;

theorem
  R is reflexive iff id field R c= R
proof
  hereby
    assume
A3: R is reflexive;
    thus id field R c= R
    proof
      let a,b;
      assume [a,b] in id field R;
      then a in field R & a = b by RELAT_1:def 10;
      hence [a,b] in R by A3,Def1,Def9;
    end;
  end;
  assume
A2: id field R c= R;
  let a;
  assume a in field R;
  then [a,a] in id field R by RELAT_1:def 10;
  hence [a,a] in R by A2;
end;

theorem
  R is irreflexive iff id field R misses R
proof
  hereby
    assume R is irreflexive;
    then
A2: R is_irreflexive_in field R by Def10;
    now
      let a,b;
      assume
A3:   [a,b] in id (field R) /\ R;
      then [a,b] in id (field R) by XBOOLE_0:def 4;
      then
A4:   a in field R & a = b by RELAT_1:def 10;
      [a,b] in R by A3,XBOOLE_0:def 4;
      hence contradiction by A2,A4,Def2;
    end;
    hence id (field R) misses R by RELAT_1:37,XBOOLE_0:def 7;
  end;
  assume
A5: id (field R) misses R;
  let a;
  assume a in field R;
  then [a,a] in id field R by RELAT_1:def 10;
  hence thesis by A5,XBOOLE_0:3;
end;

theorem
  R is_antisymmetric_in X iff R \ id X is_asymmetric_in X
proof
  hereby
    assume
A1: R is_antisymmetric_in X;
    thus R \ id X is_asymmetric_in X
    proof
      let x,y;
      assume that
A2:   x in X and
A3:   y in X and
A4:   [x,y] in R \ id X;
      not [x,y] in id X by A4,XBOOLE_0:def 5;
      then
A5:   x <> y by A2,RELAT_1:def 10;
      [x,y] in R by A4,XBOOLE_0:def 5;
      then not [y,x] in R by A1,A2,A3,A5,Def4;
      hence not [y,x] in R \ id X by XBOOLE_0:def 5;
    end;
  end;
  assume
A6: R \ id X is_asymmetric_in X;
  let x,y;
  assume that
A7: x in X & y in X and
A8: [x,y] in R and
A9: [y,x] in R;
  assume
A10: x <> y;
  then not [y,x] in id X by RELAT_1:def 10;
  then
A11: [y,x] in R \ id X by A9,XBOOLE_0:def 5;
  not [x,y] in id X by A10,RELAT_1:def 10;
  then [x,y] in R \ id X by A8,XBOOLE_0:def 5;
  hence contradiction by A6,A7,A11,Def5;
end;

theorem
  R is_asymmetric_in X implies R \/ id X is_antisymmetric_in X
proof
  assume
A1: R is_asymmetric_in X;
  let x,y;
  assume that
A2: x in X & y in X and
A3: [x,y] in R \/ id X and
A4: [y,x] in R \/ id X;
  assume
A5: x <> y;
  then not [y,x] in id X by RELAT_1:def 10;
  then
A6: [y,x] in R by A4,XBOOLE_0:def 3;
  not [x,y] in id X by A5,RELAT_1:def 10;
  then [x,y] in R by A3,XBOOLE_0:def 3;
  hence contradiction by A1,A2,A6,Def5;
end;

canceled 7;

registration
  cluster symmetric transitive -> reflexive for Relation;
coherence
proof
  let R be Relation;
  assume that
A1: R is symmetric and
A2: R is transitive;
A3: R is_transitive_in field R by A2,Def16;
A4: R is_symmetric_in field R by A1,Def11;
    let a;
A5: now
      assume a in dom R;
      then consider b such that
A6:   [a,b] in R by XTUPLE_0:def 12;
A7:   a in field R & b in field R by A6,RELAT_1:15;
      then [b,a] in R by A4,A6,Def3;
      hence [a,a] in R by A3,A6,A7,Def8;
    end;
    now
      assume a in rng R;
      then consider b such that
A9:   [b,a] in R by XTUPLE_0:def 13;
A10:  a in field R & b in field R by A9,RELAT_1:15;
      then [a,b] in R by A4,A9,Def3;
      hence [a,a] in R by A3,A9,A10,Def8;
    end;
    hence thesis by A5,XBOOLE_0:def 3;
  end;
end;

registration
  let X;
  cluster id X -> symmetric transitive antisymmetric;
coherence
proof
  thus id X is symmetric
  proof
    let a,b;
    assume that
    a in field(id X) and
    b in field(id X) and
A1: [a,b] in id X;
    a = b by A1,RELAT_1:def 10;
    hence thesis by A1;
  end;
  thus id X is transitive
  proof
    let a,b,c;
    thus thesis by RELAT_1:def 10;
  end;
  thus id X is antisymmetric
  proof
    let a,b;
    thus thesis by RELAT_1:def 10;
  end;
end;
end;

registration
  cluster irreflexive transitive -> asymmetric for Relation;
coherence
proof
  let R be Relation;
  assume that
A4: R is_irreflexive_in field R and
A3: R is_transitive_in field R;
  let a,b;
  assume that
A5: a in field R and
A6: b in field R;
  not [a,a] in R by A4,A5,Def2;
  hence thesis by A3,A5,A6,Def8;
end;
  cluster asymmetric -> irreflexive antisymmetric for Relation;
coherence
proof
  let R be Relation;
  assume
A1: R is_asymmetric_in field R;
  then for x holds x in field R implies not [x,x] in R by Def5;
  hence R is irreflexive by Def2,Def10;
  for x,y holds x in field R & y in field R & [x,y] in R & [y,x] in R
  implies x = y by A1,Def5;
  hence thesis by Def4,Def12;
end;
end;

registration
  let R be reflexive Relation;
  cluster R~ -> reflexive;
coherence
proof
A1: R is_reflexive_in field R by Def9;
  let x;
  assume x in field(R~);
  then x in field R by RELAT_1:21;
  then [x,x] in R by A1,Def1;
  hence [x,x] in R~ by RELAT_1:def 7;
end;
end;

registration
  let R be irreflexive Relation;
  cluster R~ -> irreflexive;
coherence
proof
A1: R is_irreflexive_in field R by Def10;
  let x;
  assume x in field(R~);
  then x in field R by RELAT_1:21;
  then not [x,x] in R by A1,Def2;
  hence thesis by RELAT_1:def 7;
end;
end;

theorem
  R is reflexive implies dom R = dom(R~) & rng R = rng(R~)
proof
  assume
A1: R is reflexive; then
A2: R is_reflexive_in field R by Def9;
A3: R~ is_reflexive_in field(R~) by A1,Def9;
  now
    let x;
A4: now
      assume x in dom(R~);
      then x in field(R~) by XBOOLE_0:def 3;
      then [x,x] in R~ by A1,Def1,Def9;
      then [x,x] in R by RELAT_1:def 7;
      hence x in dom R by XTUPLE_0:def 12;
    end;
    now
      assume x in dom R;
      then x in field R by XBOOLE_0:def 3;
      then [x,x] in R by A1,Def1,Def9;
      then [x,x] in R~ by RELAT_1:def 7;
      hence x in dom(R~) by XTUPLE_0:def 12;
    end;
    hence x in dom R iff x in dom(R~) by A4;
  end;
  hence dom R = dom(R~) by TARSKI:1;
  now
    let x;
A5: now
      assume x in rng(R~);
      then x in field(R~) by XBOOLE_0:def 3;
      then [x,x] in R~ by A3,Def1;
      then [x,x] in R by RELAT_1:def 7;
      hence x in rng R by XTUPLE_0:def 13;
    end;
    now
      assume x in rng R;
      then x in field R by XBOOLE_0:def 3;
      then [x,x] in R by A2,Def1;
      then [x,x] in R~ by RELAT_1:def 7;
      hence x in rng(R~) by XTUPLE_0:def 13;
    end;
    hence x in rng R iff x in rng(R~) by A5;
  end;
  hence thesis by TARSKI:1;
end;

theorem Th13:
  R is symmetric iff R = R~
proof
  hereby
    assume R is symmetric;
    then
A2: R is_symmetric_in field R by Def11;
    now
      let a,b;
A3:   now
        assume [a,b] in R~;
        then
A4:     [b,a] in R by RELAT_1:def 7;
        then a in field R & b in field R by RELAT_1:15;
        hence [a,b] in R by A2,A4,Def3;
      end;
      now
        assume
A5:     [a,b] in R;
        then a in field R & b in field R by RELAT_1:15;
        then [b,a] in R by A2,A5,Def3;
        hence [a,b] in R~ by RELAT_1:def 7;
      end;
      hence [a,b] in R iff [a,b] in R~ by A3;
    end;
    hence R = R~ by RELAT_1:def 2;
  end;
  assume R = R~;
  then for a,b holds a in field R & b in field R & [a,b] in R implies [b,a]
  in R by RELAT_1:def 7;
  hence thesis by Def3,Def11;
end;

registration
  let P,R be reflexive Relation;
  cluster P \/ R -> reflexive;
coherence
proof
A3: R is_reflexive_in field R by Def9;
A4: P is_reflexive_in field P by Def9;
  now
    let a;
A5: now
      assume a in field P;
      then [a,a] in P by A4,Def1;
      hence [a,a] in P \/ R by XBOOLE_0:def 3;
    end;
A6: now
      assume a in field R;
      then [a,a] in R by A3,Def1;
      hence [a,a] in P \/ R by XBOOLE_0:def 3;
    end;
    assume a in field(P \/ R);
    then a in field P \/ field R by RELAT_1:18;
    hence [a,a] in P \/ R by A5,A6,XBOOLE_0:def 3;
  end;
  hence P \/ R is reflexive by Def1,Def9;
end;
  cluster P /\ R -> reflexive;
coherence
proof
A3: R is_reflexive_in field R by Def9;
A4: P is_reflexive_in field P by Def9;
  now
    let a;
    assume
A7: a in field(P /\ R);
A8: field(P /\ R) c= field P /\ field R by RELAT_1:19;
    then a in field R by A7,XBOOLE_0:def 4;
    then
A9: [a,a] in R by A3,Def1;
    a in field P by A8,A7,XBOOLE_0:def 4;
    then [a,a] in P by A4,Def1;
    hence [a,a] in P /\ R by A9,XBOOLE_0:def 4;
  end;
  hence thesis by Def1,Def9;
end;
end;

registration
  let P,R be irreflexive Relation;
  cluster P \/ R -> irreflexive;
coherence
proof
A3: P is_irreflexive_in field P by Def10;
A4: R is_irreflexive_in field R by Def10;
    let a;
A5: now
      assume a in field P;
      then
A6:   not [a,a] in P by A3,Def2;
A7:   not a in field R implies not [a,a] in R by RELAT_1:15;
      a in field R implies not [a,a] in R by A4,Def2;
      hence not [a,a] in P \/ R by A6,A7,XBOOLE_0:def 3;
    end;
A8: now
      assume a in field R;
      then
A9:   not [a,a] in R by A4,Def2;
A10:  not a in field P implies not [a,a] in P by RELAT_1:15;
      a in field P implies not [a,a] in P by A3,Def2;
      hence not [a,a] in P \/ R by A9,A10,XBOOLE_0:def 3;
    end;
    assume a in field(P \/ R);
    then a in field P \/ field R by RELAT_1:18;
    hence not [a,a] in P \/ R by A5,A8,XBOOLE_0:def 3;
  end;
  cluster P /\ R -> irreflexive;
coherence
proof
  let a;
  assume
A11: a in field(P /\ R);
  field(P /\ R) c= (field P) /\ (field R) by RELAT_1:19;
  then a in field P by A11,XBOOLE_0:def 4;
  then not [a,a] in P by Def10,Def2;
  hence thesis by XBOOLE_0:def 4;
end;
end;

registration
  let P be irreflexive Relation;
  let R be Relation;
  cluster P \ R -> irreflexive;
coherence
proof
A1: P is_irreflexive_in field P by Def10;
    let a;
A2: now
      assume a in dom(P \ R);
      then consider b such that
A3:   [a,b] in P \ R by XTUPLE_0:def 12;
      [a,b] in P by A3,XBOOLE_0:def 5;
      then a in field P by RELAT_1:15;
      hence not [a,a] in P by A1,Def2;
    end;
    now
      assume a in rng(P \ R);
      then consider b such that
A5:   [b,a] in P \ R by XTUPLE_0:def 13;
      [b,a] in P by A5,XBOOLE_0:def 5;
      then a in field P by RELAT_1:15;
      hence not [a,a] in P by A1,Def2;
    end;
    hence thesis by A2,XBOOLE_0:def 3,def 5;
end;
end;

registration
  let R be symmetric Relation;
  cluster R~ -> symmetric;
  coherence by Th13;
end;

registration
  let P,R be symmetric Relation;
  cluster P \/ R -> symmetric;
coherence
proof
A3: R is_symmetric_in field R by Def11;
A4: P is_symmetric_in field P by Def11;
  now
    let a,b;
    assume that
    a in field(P \/ R) and
    b in field(P \/ R) and
A5: [a,b] in P \/ R;
A6: now
      assume
A7:   [a,b] in R;
      then a in field R & b in field R by RELAT_1:15;
      then [b,a] in R by A3,A7,Def3;
      hence [b,a] in P \/ R by XBOOLE_0:def 3;
    end;
    now
      assume
A8:   [a,b] in P;
      then a in field P & b in field P by RELAT_1:15;
      then [b,a] in P by A4,A8,Def3;
      hence [b,a] in P \/ R by XBOOLE_0:def 3;
    end;
    hence [b,a] in P \/ R by A5,A6,XBOOLE_0:def 3;
  end;
  hence thesis by Def3,Def11;
end;
  cluster P /\ R -> symmetric;
coherence
proof
A3: R is_symmetric_in field R by Def11;
A4: P is_symmetric_in field P by Def11;
  now
    let a,b;
    assume that
A9: a in field(P /\ R) & b in field(P /\ R) and
A10: [a,b] in P /\ R;
A11: [a,b] in R by A10,XBOOLE_0:def 4;
A12: field(P /\ R) c= field P /\ field R by RELAT_1:19;
    then a in field R & b in field R by A9,XBOOLE_0:def 4;
    then
A13: [b,a] in R by A3,A11,Def3;
A14: [a,b] in P by A10,XBOOLE_0:def 4;
    a in field P & b in field P by A12,A9,XBOOLE_0:def 4;
    then [b,a] in P by A4,A14,Def3;
    hence [b,a] in P /\ R by A13,XBOOLE_0:def 4;
  end;
  hence thesis by Def3,Def11;
end;
  cluster P \ R -> symmetric;
coherence
proof
A3: R is_symmetric_in field R by Def11;
A4: P is_symmetric_in field P by Def11;
  now
    let a,b;
    assume that
    a in field(P \ R) and
    b in field(P \ R) and
A15: [a,b] in P \ R;
    not [a,b] in R by A15,XBOOLE_0:def 5;
    then
A16: not b in field R or not a in field R or not [b,a] in R by A3,Def3;
A17: [a,b] in P by A15,XBOOLE_0:def 5;
    then a in field P & b in field P by RELAT_1:15;
    then
A18: [b,a] in P by A4,A17,Def3;
    ( not b in field R or not a in field R ) implies not [b,a] in R
    by RELAT_1:15;
    hence [b,a] in P \ R by A18,A16,XBOOLE_0:def 5;
  end;
  hence thesis by Def3,Def11;
end;
end;

registration
  let R be asymmetric Relation;
  cluster R~ -> asymmetric;
coherence
proof
A1: R is_asymmetric_in field R by Def13;
  let x,y;
  assume that
A2: x in field(R~) & y in field(R~) and
A3: [x,y] in R~;
A4: [y,x] in R by A3,RELAT_1:def 7;
  x in field R & y in field R by A2,RELAT_1:21;
  then not [x,y] in R by A1,A4,Def5;
  hence thesis by RELAT_1:def 7;
end;
end;

registration
  let P be Relation;
  let R be asymmetric Relation;
  cluster P /\ R -> asymmetric;
coherence
proof
A1: R is_asymmetric_in field R by Def13;
A2: field(P /\ R) c= (field P) /\ (field R) by RELAT_1:19;
  let a,b;
  assume that
A3: a in field(P /\ R) & b in field(P /\ R) and
A4: [a,b] in P /\ R;
A5: [a,b] in R by A4,XBOOLE_0:def 4;
  a in field R & b in field R by A2,A3,XBOOLE_0:def 4;
  then not [b,a] in R by A1,A5,Def5;
  hence thesis by XBOOLE_0:def 4;
end;
  cluster R /\ P -> asymmetric;
coherence;
end;

registration
  let P be asymmetric Relation;
  let R be Relation;
  cluster P \ R -> asymmetric;
coherence
proof
A1: P is_asymmetric_in field P by Def13;
  let a,b;
  assume that
  a in field(P \ R) and
  b in field(P \ R) and
A2: [a,b] in P \ R;
A3: [a,b] in P by A2,XBOOLE_0:def 5;
  then a in field P & b in field P by RELAT_1:15;
  then not [b,a] in P by A1,A3,Def5;
  hence thesis by XBOOLE_0:def 5;
end;
end;

canceled 8;

theorem
  R is antisymmetric iff R /\ (R~) c= id (dom R)
proof
A1: now
    assume R is antisymmetric;
    then
A2: R is_antisymmetric_in field R by Def12;
    now
      let a,b;
      assume
A3:   [a,b] in R /\ (R~);
      then [a,b] in R~ by XBOOLE_0:def 4;
      then
A4:   [b,a] in R by RELAT_1:def 7;
      then
A5:   b in dom R by XTUPLE_0:def 12;
A6:   [a,b] in R by A3,XBOOLE_0:def 4;
      then a in field R & b in field R by RELAT_1:15;
      then a = b by A2,A6,A4,Def4;
      hence [a,b] in id (dom R) by A5,RELAT_1:def 10;
    end;
    hence R /\ (R~) c= id (dom R) by RELAT_1:def 3;
  end;
  now
    assume
A7: R /\ (R~) c= id (dom R);
    now
      let a,b;
      assume that
      a in field R and
      b in field R and
A8:   [a,b] in R and
A9:   [b,a] in R;
      [a,b] in R~ by A9,RELAT_1:def 7;
      then [a,b] in R /\ (R~) by A8,XBOOLE_0:def 4;
      hence a = b by A7,RELAT_1:def 10;
    end;
    hence R is antisymmetric by Def4,Def12;
  end;
  hence thesis by A1;
end;

registration
  let R be antisymmetric Relation;
  cluster R~ -> antisymmetric;
coherence
proof
  let x,y;
  assume that
A2: x in field(R~) & y in field(R~) and
A3: [x,y] in R~ & [y,x] in R~;
A4: [y,x] in R & [x,y] in R by A3,RELAT_1:def 7;
  x in field R & y in field R by A2,RELAT_1:21;
  hence x = y by A4,Def4,Def12;
end;
end;

registration
  let P be antisymmetric Relation;
  let R be Relation;
  cluster P /\ R -> antisymmetric;
coherence
proof
A1: P is_antisymmetric_in field P by Def12;
  let a,b;
  assume that
  a in field(P /\ R) & b in field(P /\ R) and
A2: [a,b] in (P /\ R) and
A3: [b,a] in (P /\ R);
A4: [b,a] in P by A3,XBOOLE_0:def 4;
A5: [a,b] in P by A2,XBOOLE_0:def 4;
  then a in field P & b in field P by RELAT_1:15;
  hence a = b by A1,A5,A4,Def4;
end;
  cluster R /\ P -> antisymmetric;
coherence;
  cluster P \ R -> antisymmetric;
coherence
proof
A1: P is_antisymmetric_in field P by Def12;
  let a,b;
  assume that
  a in field(P \ R) & b in field(P \ R) and
A6: [a,b] in P \ R and
A7: [b,a] in P \ R;
A8: [b,a] in P by A7,XBOOLE_0:def 5;
A9: [a,b] in P by A6,XBOOLE_0:def 5;
  then a in field P & b in field P by RELAT_1:15;
  hence thesis by A1,A9,A8,Def4;
end;
end;

registration
  let R be transitive Relation;
  cluster R~ -> transitive;
coherence
proof
A1: R is_transitive_in field R by Def16;
  let x,y,z;
  assume that
A2: x in field(R~) & y in field(R~) and
A3: z in field(R~) and
A4: [x,y] in R~ and
A5: [y,z] in R~;
A6: x in field R & y in field R by A2,RELAT_1:21;
A7: [y,x] in R by A4,RELAT_1:def 7;
  z in field R & [z,y] in R by A3,A5,RELAT_1:21,def 7;
  then [z,x] in R by A1,A6,A7,Def8;
  hence thesis by RELAT_1:def 7;
end;
end;

registration
  let P,R be transitive Relation;
  cluster P /\ R -> transitive;
coherence
proof
A3: R is_transitive_in field R by Def16;
A4: P is_transitive_in field P by Def16;
  let a,b,c;
  assume that
  a in field(P /\ R) & b in field(P /\ R) & c in field(P /\ R) and
A5: [a,b] in P /\ R and
A6: [b,c] in P /\ R;
A7: [b,c] in R by A6,XBOOLE_0:def 4;
  then
A8: c in field R by RELAT_1:15;
A9: [a,b] in R by A5,XBOOLE_0:def 4;
  then a in field R & b in field R by RELAT_1:15;
  then
A10: [a,c] in R by A3,A9,A7,A8,Def8;
A11: [b,c] in P by A6,XBOOLE_0:def 4;
  then
A12: c in field P by RELAT_1:15;
A13: [a,b] in P by A5,XBOOLE_0:def 4;
  then a in field P & b in field P by RELAT_1:15;
  then [a,c] in P by A4,A13,A11,A12,Def8;
  hence thesis by A10,XBOOLE_0:def 4;
end;
end;

canceled 4;

theorem
  R is transitive iff R*R c= R
proof
  hereby
    assume R is transitive;
    then
A2: R is_transitive_in field R by Def16;
    now
      let a,b;
      assume [a,b] in R*R;
      then consider c such that
A3:   [a,c] in R and
A4:   [c,b] in R by RELAT_1:def 8;
A5:   c in field R by A3,RELAT_1:15;
      a in field R & b in field R by A3,A4,RELAT_1:15;
      hence [a,b] in R by A2,A3,A4,A5,Def8;
    end;
    hence R*R c= R by RELAT_1:def 3;
  end;
  assume
A6: R*R c= R;
  let a,b,c;
  assume a in field R & b in field R & c in field R;
  assume [a,b] in R & [b,c] in R;
  then [a,c] in R*R by RELAT_1:def 8;
  hence thesis by A6;
end;

theorem
  R is connected iff [:field R,field R:] \ id (field R) c= R \/ R~
proof
  hereby
    assume R is connected;
    then
A2: R is_connected_in field R by Def14;
    now
      let x;
      now
        assume
A3:     x in [:field R, field R:] \ id (field R);
        then x in [:field R, field R:] by XBOOLE_0:def 5;
        then consider y,z such that
A4:     y in field R and
A5:     z in field R and
A6:     x = [y,z] by ZFMISC_1:def 2;
        not x in id (field R) by A3,XBOOLE_0:def 5;
        then y <> z by A4,A6,RELAT_1:def 10;
        then [y,z] in R or [z,y] in R by A2,A4,A5,Def6;
        then [y,z] in R or [y,z] in R~ by RELAT_1:def 7;
        hence x in R \/ R~ by A6,XBOOLE_0:def 3;
      end;
      hence x in [:field R, field R:] \ id (field R) implies x in R \/ R~;
    end;
    hence [:field R, field R:] \ id (field R) c= R \/ R~ by TARSKI:def 3;
  end;
  assume
A7: [:field R, field R:] \ id (field R) c= R \/ R~;
  let a,b;
  [a,b] in [:field R, field R:] \ id (field R) implies [a,b] in R \/ R~ by A7;
  then
  [a,b] in [:field R, field R:] & not [a,b] in id (field R) implies
  [a,b] in R \/ R~ by XBOOLE_0:def 5;
  then a in field R & b in field R & a <> b implies
  [a,b] in R or [a,b] in R~ by RELAT_1:def 10,XBOOLE_0:def 3,ZFMISC_1:87;
  hence thesis by RELAT_1:def 7;
end;

registration
  cluster strongly_connected -> connected reflexive for Relation;
coherence
proof
  let R be Relation;
  assume
A1: R is_strongly_connected_in field R;
  then
  for x,y holds x in field R & y in field R & x <> y implies ( [x,y] in R
  or [y,x] in R) by Def7;
  hence R is connected by Def6,Def14;
  let x;
  thus thesis by A1,Def7;
end;
end;

canceled;

theorem
  R is strongly_connected iff [:field R, field R:] = R \/ R~
proof
  hereby
    assume
A2: R is strongly_connected;
    now
      let x;
A3:   now
        assume
A4:     x in R \/ R~;
        then consider y,z such that
A5:     x = [y,z] by RELAT_1:def 1;
        [y,z] in R or [y,z] in R~ by A4,A5,XBOOLE_0:def 3;
        then [y,z] in R or [z,y] in R by RELAT_1:def 7;
        then y in field R & z in field R by RELAT_1:15;
        hence x in [:field R, field R:] by A5,ZFMISC_1:87;
      end;
      now
        assume x in [:field R, field R:];
        then consider y,z such that
A6:     y in field R & z in field R and
A7:     x = [y,z] by ZFMISC_1:def 2;
        [y,z] in R or [z,y] in R by A2,A6,Def7,Def15;
        then [y,z] in R or [y,z] in R~ by RELAT_1:def 7;
        hence x in R \/ R~ by A7,XBOOLE_0:def 3;
      end;
      hence x in [:field R, field R:] iff x in R \/ R~ by A3;
    end;
    hence [:field R, field R:] = R \/ R~ by TARSKI:1;
  end;
  assume
A8: [:field R, field R:] = R \/ R~;
  let a,b;
  a in field R & b in field R implies [a,b] in R \/ R~ by A8,ZFMISC_1:87;
  then a in field R & b in field R implies [a,b] in R or [a,b] in R~
  by XBOOLE_0:def 3;
  hence thesis by RELAT_1:def 7;
end;

theorem
  R is transitive iff for x,y,z st [x,y] in R & [y,z] in R holds [x,z] in R
proof
  hereby
    assume
A1: R is transitive;
    let x,y,z;
    assume that
A2: [x,y] in R and
A3: [y,z] in R;
A4: z in field R by A3,RELAT_1:15;
    x in field R & y in field R by A2,RELAT_1:15;
    hence [x,z] in R by A1,A2,A3,A4,Def8,Def16;
  end;
  assume for x,y,z st [x,y] in R & [y,z] in R holds [x,z] in R;
  then x in field R & y in field R & z in field R & [x,y] in R & [y,z] in R
  implies [x,z] in R;
  hence R is_transitive_in field R by Def8;
end;















begin

reserve X,Y,Z,X1,X2,X3,X4,X5,X6,x,y for set;

canceled 4;

theorem Th5:
  Y in X implies not X c= Y
proof
  assume
A1: Y in X;
  assume X c= Y;
  then Y in Y by A1;
  hence contradiction;
end;

definition
  let X;
  func succ X -> set equals
  X \/ { X };
  coherence;
end;

registration
  let X;
  cluster succ X -> non empty;
  coherence;
end;

theorem Th6:
  X in succ X
proof
  X in { X } by TARSKI:def 1;
  hence thesis by XBOOLE_0:def 3;
end;

theorem
  succ X = succ Y implies X = Y
proof
  assume that
A1: succ X = succ Y and
A2: X <> Y;
  Y in X \/ { X } by A1,Th6;
  then
A3: Y in X or Y in { X } by XBOOLE_0:def 3;
  X in Y \/ { Y } by A1,Th6;
  then X in Y or X in { Y } by XBOOLE_0:def 3;
  hence contradiction by A2,A3,TARSKI:def 1;
end;

theorem Th8:
  x in succ X iff x in X or x = X
proof
  thus x in succ X implies x in X or x = X
  proof
    assume x in succ X;
    then x in X or x in { X } by XBOOLE_0:def 3;
    hence thesis by TARSKI:def 1;
  end;
  assume x in X or x = X;
  then x in X or x in { X } by TARSKI:def 1;
  hence thesis by XBOOLE_0:def 3;
end;

theorem Th9:
  X <> succ X
proof
  assume
A1: X = succ X;
  X in succ X by Th6;
  hence contradiction by A1;
end;





reserve a,b,c,X,Y,Z,x,y,z for set;

definition
  let X;
  attr X is epsilon-transitive means
  :Def2:
  for x st x in X holds x c= X;
  attr X is epsilon-connected means
  :Def3:
  for x,y st x in X & y in X holds x in y or x = y or y in x;
end;

Lm1: {} is epsilon-transitive & {} is epsilon-connected
proof
  thus x in {} implies x c= {};
  thus x in {} & y in {} implies x in y or x = y or y in x;
end;





definition
  let IT be set;
  attr IT is ordinal means
  :Def4:
  IT is epsilon-transitive epsilon-connected;
end;

registration
  cluster ordinal -> epsilon-transitive epsilon-connected for set;
  coherence by Def4;
  cluster epsilon-transitive epsilon-connected -> ordinal for set;
  coherence by Def4;
end;

notation
  synonym number for set;
end;

registration
  cluster ordinal for number;
  existence by Lm1;
end;

definition
  mode Ordinal is ordinal number;
end;

reserve A,B,C,D for Ordinal;

theorem Th10:
  for A,B being set, C being epsilon-transitive set st A in B & B in C
   holds A in C
proof
  let A,B be set, C be epsilon-transitive set;
  assume that
A1: A in B and
A2: B in C;
  B c= C by A2,Def2;
  hence thesis by A1;
end;

theorem Th11:
  for x being epsilon-transitive set, A being Ordinal st x c< A holds x in A
proof
  let x be epsilon-transitive set, A be Ordinal;
  set a = the Element of A \ x;
  assume
A1: x c< A;
  then
A2: x c= A by XBOOLE_0:def 8;
  A \ x <> {}
  proof
    assume A \ x = {};
    then A c= x by XBOOLE_1:37;
    hence contradiction by A1,XBOOLE_1:60;
  end;
  then a in A \ x;
  then consider y such that
A3: y in A \ x and
A4: not ex a st a in A \ x & a in y by TARSKI:2;
A5: not y in x by A3,XBOOLE_0:def 5;
  now
    let a;
    assume a in x;
    then consider z such that
A6: z = a and
A7: z in x;
    z c= x by A7,Def2;
    then not y in z by A3,XBOOLE_0:def 5;
    hence a in y by A2,A3,A5,A6,A7,Def3;
  end;
  then
A8: x c= y by TARSKI:def 3;
A9: y c= A by A3,Def2;
  now
    let a;
    assume
A10: a in y;
    then not a in A \ x by A4;
    hence a in x by A9,A10,XBOOLE_0:def 5;
  end;
  then
A11: y c= x by TARSKI:def 3;
  y in A by A3;
  hence thesis by A11,A8,XBOOLE_0:def 10;
end;

theorem
  for A being epsilon-transitive set, B, C being Ordinal st A c= B & B
  in C holds A in C
proof
  let A be epsilon-transitive set, B, C be Ordinal;
  assume that
A1: A c= B and
A2: B in C;
  B c= C by A2,Def2;
  then
A3: A c= C by A1,XBOOLE_1:1;
  A <> C by A1,A2,Th5;
  then A c< C by A3,XBOOLE_0:def 8;
  hence thesis by Th11;
end;

theorem Th13:
  a in A implies a is Ordinal
proof
  assume
A1: a in A;
  then
A2: a c= A by Def2;
  now
    let y;
    assume
A3: y in a;
    then
A4: y c= A by A2,Def2;
    assume not y c= a;
    then consider b such that
A5: b in y & not b in a by TARSKI:def 3;
    b in y \ a by A5,XBOOLE_0:def 5;
    then consider z such that
A6: z in y \ a and
    not ex c st c in y \ a & c in z by TARSKI:2;
    z in y by A6;
    then z in a or z = a or a in z by A1,A4,Def3;
    hence contradiction by A3,A6,XREGULAR:7,XBOOLE_0:def 5;
  end;
  then
A7: a is epsilon-transitive by Def2;
  for y,z st y in a & z in a holds y in z or y = z or z in y by A2,Def3;
  then a is epsilon-connected by Def3;
  hence thesis by A7;
end;

theorem Th14:
  A in B or A = B or B in A
proof
  assume that
A1: not A in B and
A2: A <> B;
  not A c< B by A1,Th11;
  then not A c= B by A2,XBOOLE_0:def 8;
  then consider a such that
A3: a in A & not a in B by TARSKI:def 3;
  a in A \ B by A3,XBOOLE_0:def 5;
  then consider X such that
A4: X in A \ B and
A5: not ex a st a in A \ B & a in X by TARSKI:2;
A6: X c= A by A4,Def2;
  now
    let b;
    assume
A7: b in X;
    then not b in A \ B by A5;
    hence b in B by A6,A7,XBOOLE_0:def 5;
  end;
  then X c= B by TARSKI:def 3;
  then
A8: X c< B or X = B by XBOOLE_0:def 8;
  ( not X in B)& X is Ordinal by A4,Th13,XBOOLE_0:def 5;
  hence thesis by A4,A8,Th11;
end;

definition
  let A,B;
  redefine pred A c= B means
  for C st C in A holds C in B;
  compatibility
  proof
    thus A c= B implies for C st C in A holds C in B;
    assume
A1: for C st C in A holds C in B;
    let x;
    assume
A2: x in A;
    then reconsider C = x as Ordinal by Th13;
    C in B by A1,A2;
    hence thesis;
  end;
  connectedness
  proof
    let A,B;
    given C such that
A3: C in A & not C in B;
    let D;
    A in B or A = B or B in A by Th14;
    hence thesis by A3,Th10;
  end;
end;

theorem
  A,B are_c=-comparable
proof
  A c= B or B c= A;
  hence thesis by XBOOLE_0:def 9;
end;

theorem Th16:
  A c= B or B in A
proof
  A in B or A = B or B in A by Th14;
  hence thesis by Def2;
end;

registration
  cluster empty -> ordinal for number;
  coherence by Lm1;
end;

theorem Th17:
  x is Ordinal implies succ x is Ordinal
proof
A1: now
    let y;
A2: y in { x } implies y = x by TARSKI:def 1;
    assume y in succ x;
    hence y in x or y = x by A2,XBOOLE_0:def 3;
  end;
  assume
A3: x is Ordinal;
  now
    let y;
A4: y = x implies y c= succ x by XBOOLE_1:7;
A5: now
      assume y in x;
      then
A6:   y c= x by A3,Def2;
      x c= x \/ { x } by XBOOLE_1:7;
      hence y c= succ x by A6,XBOOLE_1:1;
    end;
    assume y in succ x;
    hence y c= succ x by A1,A5,A4;
  end;
  then
A7: succ x is epsilon-transitive by Def2;
  now
    let y,z;
    assume that
A8: y in succ x and
A9: z in succ x;
A10: z in x or z = x by A1,A9;
    y in x or y = x by A1,A8;
    hence y in z or y = z or z in y by A3,A10,Def3;
  end;
  then succ x is epsilon-connected by Def3;
  hence thesis by A7;
end;

theorem Th18:
  x is ordinal implies union x is ordinal
proof
  assume x is epsilon-transitive & x is epsilon-connected;
  then reconsider A = x as Ordinal;
  thus y in union x implies y c= union x
  proof
    assume y in union x;
    then consider z such that
A1: y in z and
A2: z in x by TARSKI:def 4;
    z in A by A2;
    then reconsider z as Ordinal by Th13;
    z c= A by A2,Def2;
    hence thesis by A1,ZFMISC_1:74;
  end;
  let y,z;
  assume that
A3: y in union x and
A4: z in union x;
  consider X such that
A5: y in X and
A6: X in x by A3,TARSKI:def 4;
A7: X in A by A6;
  consider Y such that
A8: z in Y and
A9: Y in x by A4,TARSKI:def 4;
  reconsider X,Y as Ordinal by A9,A7,Th13;
  z in Y by A8;
  then
A10: z is Ordinal by Th13;
  y in X by A5;
  then y is Ordinal by Th13;
  hence thesis by A10,Th14;
end;

registration
  cluster non empty for Ordinal;
  existence
  proof
    reconsider A = succ {} as Ordinal by Th17;
    take A;
    thus thesis;
  end;
end;

registration
  let A;
  cluster succ A -> non empty ordinal;
  coherence by Th17;
  cluster union A -> ordinal;
  coherence by Th18;
end;

theorem Th19:
  (for x st x in X holds x is Ordinal & x c= X) implies X is ordinal
proof
  assume
A1: for x st x in X holds x is Ordinal & x c= X;
  thus X is epsilon-transitive
  proof
    let x;
    assume x in X;
    hence thesis by A1;
  end;
  let x,y;
  assume x in X & y in X;
  then x is Ordinal & y is Ordinal by A1;
  hence thesis by Th14;
end;

theorem Th20:
  X c= A & X <> {} implies ex C st C in X & for B st B in X holds C c= B
proof
  set a = the Element of X;
  assume that
A1: X c= A and
A2: X <> {};
  a in X by A2;
  then consider Y such that
A3: Y in X and
A4: not ex a st a in X & a in Y by TARSKI:2;
  Y is Ordinal by A1,A3,Th13;
  then consider C such that
A5: C = Y;
  take C;
  thus C in X by A3,A5;
  let B;
  assume B in X;
  then not B in C by A4,A5;
  then B = C or C in B by Th14;
  hence thesis by Def2;
end;

theorem Th21:
  A in B iff succ A c= B
proof
  thus A in B implies succ A c= B
  proof
    assume
A1: A in B;
    then a in { A } implies a in B by TARSKI:def 1;
    then
A2: { A } c= B by TARSKI:def 3;
    A c= B by A1,Def2;
    hence thesis by A2,XBOOLE_1:8;
  end;
  assume
A3: succ A c= B;
  A in succ A by Th6;
  hence thesis by A3;
end;

theorem Th22:
  A in succ C iff A c= C
proof
  thus A in succ C implies A c= C
  proof
    assume A in succ C;
    then A in C or A in { C } by XBOOLE_0:def 3;
    hence thesis by Def2,TARSKI:def 1;
  end;
  assume
A1: A c= C;
  assume not A in succ C;
  then A = succ C or succ C in A by Th14;
  then
A2: succ C c= C by A1,Def2;
  C in succ C by Th6;
  then C c= succ C by Def2;
  then succ C = C by A2,XBOOLE_0:def 10;
  hence contradiction by Th9;
end;





scheme
  OrdinalMin { P[Ordinal] } : ex A st P[A] & for B st P[B] holds A c= B
provided
A1: ex A st P[A]
proof
  consider A such that
A2: P[A] by A1;
  defpred R[set] means ex B st $1 = B & P[B];
  consider X such that
A3: x in X iff x in succ A & R[x] from XBOOLE_0:sch 1;
  for a holds a in X implies a in succ A by A3;
  then
A4: X c= succ A by TARSKI:def 3;
  A in succ A by Th6;
  then X <> {} by A2,A3;
  then consider C such that
A5: C in X and
A6: for B st B in X holds C c= B by A4,Th20;
  C in succ A by A3,A5;
  then
A7: C c= succ A by Def2;
  take C;
  ex B st C = B & P[B] by A3,A5;
  hence P[C];
  let B;
  assume
A8: P[B];
  assume
A9: not C c= B;
  then B c< C by XBOOLE_0:def 8;
  then B in C by Th11;
  then B in X by A3,A8,A7;
  hence contradiction by A6,A9;
end;


scheme
  TransfiniteInd { P[Ordinal] } : for A holds P[A]
provided
A1: for A st for C st C in A holds P[C] holds P[A]
proof
  defpred R[set] means ex B st $1 = B & P[B];
  let A;
  set Y = succ A;
  consider Z such that
A2: x in Z iff x in Y & R[x] from XBOOLE_0:sch 1;
  now
    assume Y \ Z <> {};
    then consider C such that
A3: C in Y \ Z and
A4: for B st B in Y \ Z holds C c= B by Th20;
    now
      let B such that
A5:   B in C;
A6:   C c= Y by A3,Def2;
      now
        assume not B in Z;
        then B in Y \ Z by A5,A6,XBOOLE_0:def 5;
        then
A7:     C c= B by A4;
        C <> B by A5;
        then C c< B by A7,XBOOLE_0:def 8;
        hence contradiction by A5,Th11;
      end;
      then ex B9 being Ordinal st B = B9 & P[B9] by A2;
      hence P[B];
    end;
    then
A8: P[C] by A1;
    not C in Z by A3,XBOOLE_0:def 5;
    hence contradiction by A2,A3,A8;
  end;
  then not A in Y or A in Z by XBOOLE_0:def 5;
  then ex B st A = B & P[B] by A2,Th6;
  hence thesis;
end;





theorem Th23:
  for X st for a st a in X holds a is Ordinal holds union X is ordinal
proof
  let X such that
A1: for a st a in X holds a is Ordinal;
  thus union X is epsilon-transitive
  proof
    let x;
    assume x in union X;
    then consider Y such that
A2: x in Y and
A3: Y in X by TARSKI:def 4;
    Y is Ordinal by A1,A3;
    then
A4: x c= Y by A2,Def2;
    let a;
    assume a in x;
    hence thesis by A3,A4,TARSKI:def 4;
  end;
  let x,y;
  assume that
A5: x in union X and
A6: y in union X;
  consider Z such that
A7: y in Z and
A8: Z in X by A6,TARSKI:def 4;
  Z is Ordinal by A1,A8;
  then
A9: y is Ordinal by A7,Th13;
  consider Y such that
A10: x in Y and
A11: Y in X by A5,TARSKI:def 4;
  Y is Ordinal by A1,A11;
  then x is Ordinal by A10,Th13;
  hence thesis by A9,Th14;
end;

theorem Th24:
  for X st for a st a in X holds a is Ordinal ex A st X c= A
proof
  let X;
  assume
A1: for a st a in X holds a is Ordinal;
  then
A2: union X is Ordinal by Th23;
  then reconsider A = succ(union X) as Ordinal;
  take A;
  let a;
  assume
A3: a in X;
  then reconsider B = a as Ordinal by A1;
  b in B implies b in union X by A3,TARSKI:def 4;
  then B c= union X by TARSKI:def 3;
  hence thesis by A2,Th22;
end;

theorem Th25:
  not ex X st for x holds x in X iff x is Ordinal
proof
  given X such that
A1: for x holds x in X iff x is Ordinal;
A2: X is epsilon-transitive
  proof
    let x;
    assume x in X;
    then
A3: x is Ordinal by A1;
    thus thesis
    proof
      let a;
      assume a in x;
      then a is Ordinal by A3,Th13;
      hence thesis by A1;
    end;
  end;
  X is epsilon-connected
  proof
    let x,y;
    assume x in X & y in X;
    then x is Ordinal & y is Ordinal by A1;
    hence thesis by Th14;
  end;
  then X in X by A1,A2;
  hence contradiction;
end;

theorem Th26:
  not ex X st for A holds A in X
proof
  defpred P[set] means $1 is Ordinal;
  given X such that
A1: A in X;
  consider Y such that
A2: a in Y iff a in X & P[a] from XBOOLE_0:sch 1;
  now
    let x;
    assume
A3: x is Ordinal;
    then x in X by A1;
    hence x in Y by A2,A3;
  end;
  then x in Y iff x is Ordinal by A2;
  hence contradiction by Th25;
end;

theorem
  for X ex A st not A in X & for B st not B in X holds A c= B
proof
  let X;
  defpred P[set] means not $1 in X;
  consider B such that
A1: not B in X by Th26;
  consider Y such that
A2: a in Y iff a in succ B & P[a] from XBOOLE_0:sch 1;
  for a holds a in Y implies a in succ B by A2;
  then
A3: Y c= succ B by TARSKI:def 3;
  B in succ B by Th6;
  then Y <> {} by A1,A2;
  then consider A such that
A4: A in Y and
A5: for B st B in Y holds A c= B by A3,Th20;
  A in succ B by A2,A4;
  then
A6: A c= succ B by Def2;
  take A;
  thus not A in X by A2,A4;
  let C;
  assume
A7: not C in X;
  assume
A8: not A c= C;
  then not A in C by Def2;
  then C in A by A8,Th14;
  then C in Y by A2,A7,A6;
  hence contradiction by A5,A8;
end;





definition
  let A be set;
  attr A is limit_ordinal means
  :Def6:
  A = union A;
end;

theorem Th28:
  for A holds A is limit_ordinal iff for C st C in A holds succ C in A
proof
  let A;
  thus A is limit_ordinal implies for C st C in A holds succ C in A
  proof
    assume A is limit_ordinal;
    then
A1: A = union A by Def6;
    let C;
    assume C in A;
    then consider z such that
A2: C in z and
A3: z in A by A1,TARSKI:def 4;
    for b holds b in { C } implies b in z by A2,TARSKI:def 1;
    then
A4: { C } c= z by TARSKI:def 3;
A5: z is Ordinal by A3,Th13;
    then C c= z by A2,Def2;
    then succ C c= z by A4,XBOOLE_1:8;
    then succ C = z or succ C c< z by XBOOLE_0:def 8;
    then
A6: succ C = z or succ C in z by A5,Th11;
    z c= A by A3,Def2;
    hence thesis by A3,A6;
  end;
  assume
A7: for C st C in A holds succ C in A;
  now
    let a;
    assume
A8: a in A;
    then a is Ordinal by Th13;
    then
A9: succ a in A by A7,A8;
    a in succ a by Th6;
    hence a in union A by A9,TARSKI:def 4;
  end;
  then
A10: A c= union A by TARSKI:def 3;
  now
    let a;
    assume a in union A;
    then consider z such that
A11: a in z and
A12: z in A by TARSKI:def 4;
    z c= A by A12,Def2;
    hence a in A by A11;
  end;
  then union A c= A by TARSKI:def 3;
  then A = union A by A10,XBOOLE_0:def 10;
  hence thesis by Def6;
end;

theorem
  not A is limit_ordinal iff ex B st A = succ B
proof
  thus not A is limit_ordinal implies ex B st A = succ B
  proof
    assume not A is limit_ordinal;
    then consider B such that
A1: B in A and
A2: not succ B in A by Th28;
    take B;
    assume
A3: A <> succ B;
    succ B c= A by A1,Th21;
    then succ B c< A by A3,XBOOLE_0:def 8;
    hence contradiction by A2,Th11;
  end;
  given B such that
A4: A = succ B;
  B in A & not succ B in A by A4,Th6;
  hence thesis by Th28;
end;

reserve F,G for Function;





definition
  let IT be set;
  attr IT is T-Sequence-like means
  :Def7:
  proj1 IT is ordinal;
end;

registration
  cluster empty -> T-Sequence-like for set;
  coherence
  proof
    let E be set;
    assume E is empty;
    then proj1 E = {};
    hence thesis by Def7;
  end;
end;

definition
  mode T-Sequence is T-Sequence-like Function;
end;

registration
  let Z;
  cluster Z-valued for T-Sequence;
  existence
  proof
    reconsider L = {} as T-Sequence;
    take L;
    rng {} = {};
    hence rng L c= Z by XBOOLE_1:2;
  end;
end;

definition
  let Z;
  mode T-Sequence of Z is Z-valued T-Sequence;
end;

theorem
  {} is T-Sequence of Z
proof
  reconsider L = {} as T-Sequence;
  rng L = {};
  then rng L c= Z by XBOOLE_1:2;
  hence thesis by RELAT_1:def 19;
end;

reserve L,L1 for T-Sequence;

theorem
  dom F is Ordinal implies F is T-Sequence of rng F by Def7,RELAT_1:def 19;

registration
  let L;
  cluster dom L -> ordinal;
  coherence by Def7;
end;

theorem
  X c= Y implies for L being T-Sequence of X holds L is T-Sequence of Y
proof
  assume
A1: X c= Y;
  let L be T-Sequence of X;
  rng L c= X by RELAT_1:def 19;
  then rng L c= Y by A1,XBOOLE_1:1;
  hence thesis by RELAT_1:def 19;
end;

registration
  let L,A;
  cluster L|A -> rng L-valued T-Sequence-like;
  coherence
  proof
A1: rng(L|A) c= rng L by RELAT_1:70;
    A c= dom L implies dom(L|A) = A by RELAT_1:62;
    hence thesis by A1,Def7,RELAT_1:68,def 19;
  end;
end;

theorem
  for L being T-Sequence of X for A holds L|A is T-Sequence of X;

definition

  let IT be set;
  attr IT is c=-linear means
  :Def8:
  for x,y being set st x in IT & y in IT holds x,y are_c=-comparable;
end;

theorem
  (for a st a in X holds a is T-Sequence) & X is c=-linear implies
  union X is T-Sequence
proof
  assume that
A1: for a st a in X holds a is T-Sequence and
A2: X is c=-linear;
  union X is Relation-like Function-like
  proof
    thus a in union X implies ex b,c st [b,c] = a
    proof
      assume a in union X;
      then consider x such that
A3:   a in x and
A4:   x in X by TARSKI:def 4;
      reconsider x as T-Sequence by A1,A4;
      for a st a in x holds ex b,c st [b,c] = a by RELAT_1:def 1;
      hence thesis by A3;
    end;
    let a,b,c;
    assume that
A5: [a,b] in union X and
A6: [a,c] in union X;
    consider y such that
A7: [a,c] in y and
A8: y in X by A6,TARSKI:def 4;
    reconsider y as T-Sequence by A1,A8;
    consider x such that
A9: [a,b] in x and
A10: x in X by A5,TARSKI:def 4;
    reconsider x as T-Sequence by A1,A10;
    x,y are_c=-comparable by A2,A10,A8,Def8;
    then x c= y or y c= x by XBOOLE_0:def 9;
    hence thesis by A9,A7,FUNCT_1:def 1;
  end;
  then reconsider F = union X as Function;
  defpred P[set,set] means $1 in X & for L st L = $1 holds dom L = $2;
A11: for a,b,c st P[a,b] & P[a,c] holds b = c
  proof
    let a,b,c;
    assume that
A12: a in X and
A13: for L st L = a holds dom L = b and
    a in X and
A14: for L st L = a holds dom L = c;
    reconsider a as T-Sequence by A1,A12;
    dom a = b by A13;
    hence thesis by A14;
  end;
  consider G such that
A15: [a,b] in G iff a in X & P[a,b] from FUNCT_1:sch 1(A11);
A16: [a,b] in G implies b is Ordinal
  proof
    assume
A17: [a,b] in G;
    then a in X by A15;
    then reconsider a as T-Sequence by A1;
    dom a = b by A15,A17;
    hence thesis;
  end;
  a in rng G implies a is Ordinal
  proof
    assume a in rng G;
    then consider b such that
A18: b in dom G & a = G.b by FUNCT_1:def 3;
    [b,a] in G by A18,FUNCT_1:1;
    hence thesis by A16;
  end;
  then consider A such that
A19: rng G c= A by Th24;
  defpred P[Ordinal] means for B st B in rng G holds B c= $1;
  for B st B in rng G holds B c= A by A19,Def2;
  then
A20: ex A st P[A];
  consider A such that
A21: P[A] & for C st P[C] holds A c= C from OrdinalMin(A20);
  dom F = A
  proof
    thus a in dom F implies a in A
    proof
      assume a in dom F;
      then consider b such that
A22:  [a,b] in F by XTUPLE_0:def 12;
      consider x such that
A23:  [a,b] in x and
A24:  x in X by A22,TARSKI:def 4;
      reconsider x as T-Sequence by A1,A24;
      for L st L = x holds dom L = dom x;
      then [x,dom x] in G by A15,A24;
      then x in dom G & dom x = G.x by FUNCT_1:1;
      then dom x in rng G by FUNCT_1:def 3;
      then
A25:  dom x c= A by A21;
      a in dom x by A23,XTUPLE_0:def 12;
      hence thesis by A25;
    end;
    let a;
    assume
A26: a in A;
    then reconsider a9 = a as Ordinal by Th13;
    now
      assume
A27:  for L st L in X holds not a9 in dom L;
      now
        let B;
        assume B in rng G;
        then consider c such that
A28:    c in dom G & B = G.c by FUNCT_1:def 3;
A29:    [c,B] in G by A28,FUNCT_1:1;
        then c in X by A15;
        then reconsider c as T-Sequence by A1;
        c in X & dom c = B by A15,A29;
        hence B c= a9 by A27,Th16;
      end;
      then
A30:  A c= a9 by A21;
      a9 c= A by A26,Def2;
      then A = a by A30,XBOOLE_0:def 10;
      hence contradiction by A26;
    end;
    then consider L such that
A31: L in X & a in dom L;
    L c= F & ex b st [a,b] in L by A31,XTUPLE_0:def 12,ZFMISC_1:74;
    hence thesis by XTUPLE_0:def 12;
  end;
  hence thesis by Def7;
end;





scheme
  TSUniq { A()->Ordinal, H(T-Sequence)->set, L1, L2() -> T-Sequence } : L1() =
  L2()
provided
A1: dom L1() = A() & for B,L st B in A() & L = L1()|B holds L1().B = H(L ) and
A2: dom L2() = A() & for B,L st B in A() & L = L2()|B holds L2().B = H(L )
proof
  defpred P[set] means L1().$1 <> L2().$1;
  consider X such that
A3: x in X iff x in A() & P[x] from XBOOLE_0:sch 1;
  for b holds b in X implies b in A() by A3;
  then
A4: X c= A() by TARSKI:def 3;
  assume L1() <> L2();
  then ex a st a in A() & L1().a <> L2().a by A1,A2,FUNCT_1:2;
  then X <> {} by A3;
  then consider B such that
A5: B in X and
A6: for C st C in X holds B c= C by A4,Th20;
A7: B in A() by A3,A5;
  then
A8: B c= A() by Def2;
  then
A9: dom(L1()|B) = B by A1,RELAT_1:62;
A10: dom(L2()|B) = B by A2,A8,RELAT_1:62;
A11: now
    let C;
    assume
A12: C in B;
    then not C in X by A6,Th5;
    hence L1().C = L2().C by A3,A8,A12;
  end;
  now
    let a;
    assume
A13: a in B;
    then reconsider a9 = a as Ordinal by Th13;
    L1().a9 = L2().a9 & L1()|B.a = L1().a by A11,A9,A13,FUNCT_1:47;
    hence L1()|B.a = L2()|B.a by A10,A13,FUNCT_1:47;
  end;
  then
A14: L1()|B = L2()|B by A9,A10,FUNCT_1:2;
  L1().B = H(L1()|B) & L2().B = H(L2()|B) by A1,A2,A7;
  hence contradiction by A3,A5,A14;
end;

scheme
  TSExist { A()->Ordinal,H(T-Sequence)->set } : ex L st dom L = A() & for B,L1
  st B in A() & L1 = L|B holds L.B = H(L1) proof
  defpred S[Ordinal] means ex L st dom L = $1 & for B st B in $1 holds L.B = H
  (L|B);
A1: for B st for C st C in B holds S[C] holds S[B]
  proof
    defpred P[set,set] means $1 is Ordinal & $2 is T-Sequence & for A,L st A =
    $1 & L = $2 holds dom L = A & for B st B in A holds L.B = H(L|B);
    let B such that
A2: for C st C in B ex L st dom L = C & for D st D in C holds L.D = H( L|D);
A3: for a,b,c st P[a,b] & P[a,c] holds b = c
    proof
      let a,b,c;
      assume that
A4:   a is Ordinal and
A5:   b is T-Sequence and
A6:   for A,L st A = a & L = b holds dom L = A & for B st B in A holds
      L.B = H(L|B) and
      a is Ordinal and
A7:   c is T-Sequence and
A8:   for A,L st A = a & L = c holds dom L = A & for B st B in A holds
      L.B = H(L|B);
      reconsider c as T-Sequence by A7;
      reconsider a as Ordinal by A4;
A9:   dom c = a & for B,L st B in a & L=c|B holds c.B = H(L) by A8;
      reconsider b as T-Sequence by A5;
A10:  dom b = a & for B,L st B in a & L=b|B holds b.B = H(L) by A6;
      b = c from TSUniq(A10,A9);
      hence thesis;
    end;
    consider G such that
A11: [a,b] in G iff a in B & P[a,b] from FUNCT_1:sch 1(A3);
    defpred R[set,set] means ex L st L = G.$1 & $2 = H(L);
A12: dom G = B
    proof
      thus a in dom G implies a in B
      proof
        assume a in dom G;
        then ex b st [a,b] in G by XTUPLE_0:def 12;
        hence thesis by A11;
      end;
      let a;
      assume
A13:  a in B;
      then reconsider a9 = a as Ordinal by Th13;
      consider L such that
A14:  dom L = a9 & for D st D in a9 holds L.D = H(L|D) by A2,A13;
      for A holds for K be T-Sequence holds A = a9 & K = L implies dom K
      = A & for B holds B in A implies K.B = H(K|B) by A14;
      then [a9,L] in G by A11,A13;
      hence thesis by XTUPLE_0:def 12;
    end;
A15: for a st a in B ex b st R[a,b]
    proof
      let a;
      assume a in B;
      then consider c such that
A16:  [a,c] in G by A12,XTUPLE_0:def 12;
      reconsider L = c as T-Sequence by A11,A16;
      take H(L), L;
      thus L = G.a by A16,FUNCT_1:1;
      thus thesis;
    end;
A17: for a,b,c st a in B & R[a,b] & R[a,c] holds b = c;
    consider F such that
A18: dom F = B & for a being set st a in B holds R[a,F.a] from FUNCT_1
    :sch 2(A17,A15);
    reconsider L = F as T-Sequence by A18,Def7;
    take L;
    thus dom L = B by A18;
    let D;
    assume
A19: D in B;
    then consider K being T-Sequence such that
A20: K = G.D and
A21: F.D = H(K) by A18;
A22: [D,K] in G by A12,A19,A20,FUNCT_1:1;
    then
A23: dom K = D by A11;
A24: now
      let C;
      assume
A25:  C in D;
A26:  now
        let A,L such that
A27:    A = C and
A28:    L = K|C;
A29:    C c= D by A25,Def2;
        hence
A30:    dom L = A by A23,A27,A28,RELAT_1:62;
        let B;
        assume
A31:    B in A;
        then B c= A by Def2;
        then
A32:    K|B = L|B by A27,A28,FUNCT_1:51;
        K.B = H(K|B) by A11,A22,A27,A29,A31;
        hence L.B = H(L|B) by A28,A30,A31,A32,FUNCT_1:47;
      end;
      C in B by A19,A25,Th10;
      then [C,K|C] in G by A11,A26;
      hence G.C = K|C by FUNCT_1:1;
    end;
A33: now
      let a;
      assume
A34:  a in D;
      then reconsider A = a as Ordinal by Th13;
A35:  G.A = K|A & L|D.A = L.A by A24,A34,FUNCT_1:49;
      ex J being T-Sequence st J = G.A & F.A = H(J) by A18,A19,A34,Th10;
      hence L|D.a = K.a by A11,A22,A34,A35;
    end;
    D c= dom L by A18,A19,Def2;
    then dom(L|D) = D by RELAT_1:62;
    hence thesis by A21,A23,A33,FUNCT_1:2;
  end;
  for A holds S[A] from TransfiniteInd(A1);
  then consider L such that
A36: dom L = A() and
A37: for B st B in A() holds L.B = H(L|B);
  take L;
  thus dom L = A() by A36;
  let B,L1;
  assume B in A() & L1 = L|B;
  hence thesis by A37;
end;

scheme
  FuncTS { L() -> T-Sequence, F(Ordinal)->set, H(T-Sequence)->set } : for B st
  B in dom L() holds L().B = H(L()|B)
provided
A1: for A,a holds a = F(A) iff ex L st a = H(L) & dom L = A & for B st B
in A holds L.B = H(L|B) and
A2: for A st A in dom L() holds L().A = F(A)
proof
  consider L such that
  F(dom L()) = H(L) and
A3: dom L = dom L() and
A4: for B st B in dom L() holds L.B = H(L|B) by A1;
  now
    let b;
    assume
A5: b in dom L;
    then reconsider B = b as Ordinal by Th13;
    now
      take K = L|B;
      thus H(L|B) = H(K);
A6:   B c= dom L by A5,Def2;
      hence dom K = B by RELAT_1:62;
      let C;
      assume
A7:   C in B;
      then C c= B by Def2;
      then
A8:   L|C = K|C by FUNCT_1:51;
      K.C = L.C by A7,FUNCT_1:49;
      hence K.C = H(K|C) by A3,A4,A6,A7,A8;
    end;
    then F(B) = H(L|B) by A1
      .= L.B by A3,A4,A5;
    hence L().b = L.b by A2,A3,A5;
  end;
  then L() = L by A3,FUNCT_1:2;
  hence thesis by A4;
end;

theorem
  A c< B or A = B or B c< A
proof
  assume that
A1: ( not A c< B)& not A = B and
A2: not B c< A;
  not A c= B by A1,XBOOLE_0:def 8;
  hence contradiction by A2,XBOOLE_0:def 8;
end;

begin 


definition
  let X;
  func On X -> set means
  :Def9:
  x in it iff x in X & x is Ordinal;
  existence
  proof
    defpred P[set] means $1 is Ordinal;
    thus ex Y being set st for x being set holds x in Y iff x in X & P[x] from
    XBOOLE_0:sch 1;
  end;
  uniqueness
  proof
    defpred P[set] means $1 in X & $1 is Ordinal;
    let Y,Z such that
A1: x in Y iff P[x] and
A2: x in Z iff P[x];
    thus Y = Z from XBOOLE_0:sch 2(A1,A2);
  end;
  func Lim X -> set means
  x in it iff x in X & ex A st x = A & A is limit_ordinal;
  existence
  proof
    defpred P[set] means ex A st $1 = A & A is limit_ordinal;
    thus ex Y being set st for x being set holds x in Y iff x in X & P[x] from
    XBOOLE_0:sch 1;
  end;
  uniqueness
  proof
    defpred P[set] means $1 in X & ex A st $1 = A & A is limit_ordinal;
    let Y,Z such that
A3: x in Y iff P[x] and
A4: x in Z iff P[x];
    thus Y = Z from XBOOLE_0:sch 2(A3,A4);
  end;
end;


theorem Th36:
  for D ex A st D in A & A is limit_ordinal
proof
  let D;
  consider Field being set such that
A1: D in Field and
A2: for X,Y holds X in Field & Y c= X implies Y in Field and
A3: for X holds X in Field implies bool X in Field and
  for X holds X c= Field implies X,Field are_equipotent or X in Field by
ZFMISC_1:112;
  for X st X in On Field holds X is Ordinal & X c= On Field
  proof
    let X;
    assume
A4: X in On Field;
    then reconsider A = X as Ordinal by Def9;
A5: A in Field by A4,Def9;
    thus X is Ordinal by A4,Def9;
    let y;
    assume
A6: y in X;
    then y in A;
    then reconsider B = y as Ordinal by Th13;
    B c= A by A6,Def2;
    then B in Field by A2,A5;
    hence thesis by Def9;
  end;
  then reconsider ON = On Field as Ordinal by Th19;
  take ON;
  thus D in ON by A1,Def9;
  A in ON implies succ A in ON
  proof
A7: succ A c= bool A
    proof
      let x;
      assume x in succ A;
      then x in A or x = A by Th8;
      then x c= A by Def2;
      hence thesis;
    end;
    assume A in ON;
    then A in Field by Def9;
    then bool A in Field by A3;
    then succ A in Field by A2,A7;
    hence thesis by Def9;
  end;
  hence thesis by Th28;
end;

definition
  func omega -> set means
  :Def11:
  {} in it & it is limit_ordinal & it is
  ordinal & for A st {} in A & A is limit_ordinal holds it c= A;
  existence
  proof
    defpred P[Ordinal] means {} in $1 & $1 is limit_ordinal;
A1: ex A st P[A] by Th36;
    ex C st P[C] & for A st P[A] holds C c= A from OrdinalMin(A1);
    hence thesis;
  end;
  uniqueness
  proof
    let B,C be set;
    assume {} in B & B is limit_ordinal &( B is ordinal & for A st {} in A &
    A is limit_ordinal holds B c= A ) & {} in C & C is limit_ordinal &( C is
    ordinal & for A st {} in A & A is limit_ordinal holds C c= A );
    hence B c= C & C c= B;
  end;
end;

registration
  cluster omega -> non empty ordinal;
  coherence by Def11;
end;

definition
  let A be set;
  attr A is natural means
  :Def12:
  A in omega;
end;

registration
  cluster natural for number;
  existence
  proof
    take n = the Element of omega;
    thus n in omega;
  end;
end;

definition
  mode Nat is natural number;
end;

registration
 sethood of natural number
  proof
   take omega;
   let y be Nat;
   thus y in omega by Def12;
  end;
end;



registration
  let A be Ordinal;
  cluster -> ordinal for Element of A;
  coherence
  proof
    let x be Element of A;
    A is empty or A is non empty;
    hence thesis by Th13,SUBSET_1:def 1;
  end;
end;



registration
  cluster natural -> ordinal for number;
  coherence
  proof
    let n be number;
    assume n is natural;
    then reconsider n as Element of omega by Def12;
    n is ordinal;
    hence thesis;
  end;
end;



scheme
  ALFA { D() -> non empty set, P[set,set] }: ex F st dom F = D() & for d being
  Element of D() ex A st A = F.d & P[d,A] & for B st P[d,B] holds A c= B
provided
A1: for d being Element of D() ex A st P[d,A]
proof
  defpred Q[set,set] means ex A st A = $2 & P[$1,A] & for B st P[$1,B] holds A
  c= B;
A2: for x st x in D() ex y st Q[x,y]
  proof
    let x;
    assume x in D();
    then reconsider d = x as Element of D();
    defpred Q[Ordinal] means P[d,$1];
A3: ex A st Q[A] by A1;
    consider A such that
A4: Q[A] & for B st Q[B] holds A c= B from OrdinalMin(A3);
    reconsider y = A as set;
    take y,A;
    thus thesis by A4;
  end;
A5: for x,y,z st x in D() & Q[x,y] & Q[x,z] holds y = z
  proof
    let x,y,z such that
    x in D();
    given A1 being Ordinal such that
A6: A1 = y and
A7: ( P[x,A1])& for B st P[x,B] holds A1 c= B;
    given A2 being Ordinal such that
A8: A2 = z and
A9: ( P[x,A2])& for B st P[x,B] holds A2 c= B;
    A1 c= A2 & A2 c= A1 by A7,A9;
    hence thesis by A6,A8,XBOOLE_0:def 10;
  end;
  consider F such that
A10: dom F = D() & for x st x in D() holds Q[x,F.x] from FUNCT_1:sch 2(
  A5, A2);
  take F;
  thus dom F = D() by A10;
  let d be Element of D();
  thus thesis by A10;
end;



theorem
  succ X \ {X} = X
proof
  thus succ X \ {X} c= X
  proof
    let x;
    assume
A1: x in succ X \ {X};
    then
A2: not x in {X} by XBOOLE_0:def 5;
    x in X or x = X by A1,Th8;
    hence thesis by A2,TARSKI:def 1;
  end;
  let x;
  assume
A3: x in X;
  then x <> X;
  then
A4: not x in {X} by TARSKI:def 1;
  x in succ X by A3,Th8;
  hence thesis by A4,XBOOLE_0:def 5;
end;



registration
  cluster empty -> natural for number;
  coherence
  proof
    {} in omega by Def11;
    hence thesis by Def12;
  end;
  cluster -> natural for Element of omega;
  coherence by Def12;
end;

registration
  cluster non empty natural for number;
  existence
  proof
    take succ{};
    thus succ{} is non empty;
    omega is limit_ordinal & {} in omega by Def11;
    hence succ{} in omega by Th28;
  end;
end;



registration
  let a be natural Ordinal;
  cluster succ a -> natural;
  coherence
  proof
    omega is limit_ordinal & a in omega by Def11,Def12;
    hence succ a in omega by Th28;
  end;
end;



registration
 cluster empty -> c=-linear for set;
 coherence
  proof let X be set such that
A1: X is empty;
   let x be set;
   thus thesis by A1;
  end;
end;

registration let X be c=-linear set;
 cluster -> c=-linear for Subset of X;
 coherence
proof
 let Y be Subset of X;
   let x, y be set;
   assume x in Y & y in Y;
   hence x,y are_c=-comparable by Def8;
end;
end;















begin

reserve a,b,c,d,x,y,z,X,Y,Z for set;
reserve R,S,T for Relation;

Lm1: R is reflexive iff for x st x in field R holds [x,x] in R
proof
  R is reflexive iff R is_reflexive_in field R by RELAT_2:def 9;
  hence thesis by RELAT_2:def 1;
end;

Lm2: R is transitive iff for x,y,z st [x,y] in R & [y,z] in R holds [x,z] in R
proof
  thus R is transitive implies for x,y,z st [x,y] in R & [y,z] in R holds [x,z
  ] in R
  proof
    assume R is transitive;
    then
A1: R is_transitive_in field R by RELAT_2:def 16;
    let x,y,z;
    assume that
A2: [x,y] in R and
A3: [y,z] in R;
A4: z in field R by A3,RELAT_1:15;
    x in field R & y in field R by A2,RELAT_1:15;
    hence thesis by A1,A2,A3,A4,RELAT_2:def 8;
  end;
  assume for x,y,z st [x,y] in R & [y,z] in R holds [x,z] in R;
  then
  for x,y,z st x in field R & y in field R & z in field R & [x,y] in R & [
  y,z] in R holds [x,z] in R;
  then R is_transitive_in field R by RELAT_2:def 8;
  hence thesis by RELAT_2:def 16;
end;

Lm3: R is antisymmetric iff for x,y st [x,y] in R & [y,x] in R holds x = y
proof
  thus R is antisymmetric implies for x,y st [x,y] in R & [y,x] in R holds x =
  y
  proof
    assume R is antisymmetric;
    then
A1: R is_antisymmetric_in field R by RELAT_2:def 12;
    let x,y;
    assume that
A2: [x,y] in R and
A3: [y,x] in R;
    x in field R & y in field R by A2,RELAT_1:15;
    hence thesis by A1,A2,A3,RELAT_2:def 4;
  end;
  assume for x,y st [x,y] in R & [y,x] in R holds x = y;
  then
  for x,y st x in field R & y in field R & [x,y] in R & [y,x] in R holds x = y;
  then R is_antisymmetric_in field R by RELAT_2:def 4;
  hence thesis by RELAT_2:def 12;
end;

Lm4: R is connected iff for x,y st x in field R & y in field R & x <> y holds
[x,y] in R or [y,x] in R
proof
  R is connected iff R is_connected_in field R by RELAT_2:def 14;
  hence thesis by RELAT_2:def 6;
end;

definition
  let R,a;
  func R-Seg(a) equals
  Coim(R,a) \ {a};
  coherence;
end;

theorem Th1:
  x in R-Seg a iff x <> a & [x,a] in R
proof
  hereby
    assume
A1: x in R-Seg a;
    hence x <> a by ZFMISC_1:56;
    ex y st [x,y] in R & y in {a} by A1,RELAT_1:def 14;
    hence [x,a] in R by TARSKI:def 1;
  end;
  assume that
A2: x <> a and
A3: [x,a] in R;
  a in {a} by TARSKI:def 1;
  then x in Coim(R,a) by A3,RELAT_1:def 14;
  hence thesis by A2,ZFMISC_1:56;
end;

theorem Th2:
  x in field R or R-Seg(x) = {}
proof
  assume
A1: not x in field R;
  set y = the Element of R-Seg(x);
  assume R-Seg(x) <> {};
  then [y,x] in R by Th1;
  hence contradiction by A1,RELAT_1:15;
end;

definition
  let R;
  attr R is well_founded means
  :Def2:
  for Y st Y c= field R & Y <> {} ex a st a in Y & R-Seg a misses Y;
  let X;
  pred R is_well_founded_in X means
  :Def3:
  for Y st Y c= X & Y <> {} ex a st a in Y & R-Seg a misses Y;
end;

theorem Th3:
  R is well_founded iff R is_well_founded_in field R
proof
  thus R is well_founded implies R is_well_founded_in field R
  proof
    assume for Y st Y c= field R & Y <> {} ex a st a in Y & R-Seg(a) misses Y;
    hence for Y st Y c= field R & Y <> {} ex a st a in Y & R-Seg(a) misses Y;
  end;
  assume for Y st Y c= field R & Y <> {} ex a st a in Y & R-Seg(a) misses Y;
  hence for Y st Y c= field R & Y <> {} ex a st a in Y & R-Seg(a) misses Y;
end;

definition
  let R;
  attr R is well-ordering means
  :Def4:
  R is reflexive & R is transitive & R is
  antisymmetric & R is connected & R is well_founded;
  let X;
  pred R well_orders X means
  :Def5:
  R is_reflexive_in X & R is_transitive_in X
  & R is_antisymmetric_in X & R is_connected_in X & R is_well_founded_in X;
end;

registration
  cluster well-ordering ->
          reflexive transitive antisymmetric connected well_founded
  for Relation;
  coherence by Def4;
  cluster reflexive transitive antisymmetric connected well_founded
          -> well-ordering for Relation;
  coherence by Def4;
end;

theorem
  R well_orders field R iff R is well-ordering
proof
  thus R well_orders field R implies R is well-ordering
  proof
    assume R is_reflexive_in field R & R is_transitive_in field R & R
is_antisymmetric_in field R & R is_connected_in field R & R is_well_founded_in
    field R;
    hence R is reflexive & R is transitive & R is antisymmetric & R is
    connected & R is well_founded by Th3,RELAT_2:def 9,def 12,def 14,def 16;
  end;
  assume R is reflexive & R is transitive & R is antisymmetric & R is
  connected & R is well_founded;
  hence R is_reflexive_in field R & R is_transitive_in field R & R
is_antisymmetric_in field R & R is_connected_in field R & R is_well_founded_in
  field R by Th3,RELAT_2:def 9,def 12,def 14,def 16;
end;

theorem
  R well_orders X implies for Y st Y c= X & Y <> {} ex a st a in Y & for
  b st b in Y holds [a,b] in R
proof
  assume
A1: R well_orders X;
  then
A2: R is_reflexive_in X by Def5;
A3: R is_connected_in X by A1,Def5;
  let Y;
  assume that
A4: Y c= X and
A5: Y <> {};
  R is_well_founded_in X by A1,Def5;
  then consider a such that
A6: a in Y and
A7: R-Seg(a) misses Y by A4,A5,Def3;
  take a;
  thus a in Y by A6;
  let b;
  assume
A8: b in Y;
  then not b in R-Seg(a) by A7,XBOOLE_0:3;
  then a = b or not [b,a] in R by Th1;
  then a <> b implies [a,b] in R by A3,A4,A6,A8,RELAT_2:def 6;
  hence thesis by A2,A4,A6,RELAT_2:def 1;
end;

theorem Th6:
  R is well-ordering implies for Y st Y c= field R & Y <> {} ex a
  st a in Y & for b st b in Y holds [a,b] in R
proof
  assume
A1: R is well-ordering;
  let Y;
  assume that
A2: Y c= field R and
A3: Y <> {};
  consider a such that
A4: a in Y and
A5: R-Seg(a) misses Y by A1,A2,A3,Def2;
  take a;
  thus a in Y by A4;
  let b;
  assume
A6: b in Y;
  then not b in R-Seg(a) by A5,XBOOLE_0:3;
  then a = b or not [b,a] in R by Th1;
  then a <> b implies [a,b] in R by A1,A2,A4,A6,Lm4;
  hence thesis by A1,A2,A4,Lm1;
end;

theorem
  for R st R is well-ordering & field R <> {} ex a st a in field R & for
  b st b in field R holds [a,b] in R by Th6;

theorem
  for R st R is well-ordering for a st a in field R holds (for b st b in
field R holds [b,a] in R) or ex b st b in field R & [a,b] in R & for c st c in
  field R & [a,c] in R holds c = a or [b,c] in R
proof
  let R;
  assume
A1: R is well-ordering;
  let a such that
A2: a in field R;
  defpred P[set] means not [$1,a] in R;
  given b such that
A3: b in field R & not [b,a] in R;
  consider Z such that
A4: c in Z iff c in field R & P[c] from XBOOLE_0:sch 1;
  for b holds b in Z implies b in field R by A4;
  then
A5: Z c= field R by TARSKI:def 3;
  Z <> {} by A3,A4;
  then consider d such that
A6: d in Z and
A7: for c st c in Z holds [d,c] in R by A1,A5,Th6;
  take d;
  thus
A8: d in field R by A4,A6;
A9: not [d,a] in R by A4,A6;
  then a <> d by A6,A7;
  hence [a,d] in R by A1,A2,A8,A9,Lm4;
  let c;
  assume that
A10: c in field R and
A11: [a,c] in R;
  assume c <> a;
  then not [c,a] in R by A1,A11,Lm3;
  then c in Z by A4,A10;
  hence thesis by A7;
end;

reserve F,G for Function;

theorem Th9:
  R-Seg(a) c= field R
proof
  let b;
  assume b in R-Seg(a);
  then [b,a] in R by Th1;
  hence thesis by RELAT_1:15;
end;

definition
  let R,Y;
  func R |_2 Y -> Relation equals
  R /\ [:Y,Y:];
  coherence;
end;

theorem Th10:
  R |_2 X = X|`R|X
proof
  let x,y;
  thus [x,y] in R |_2 X implies [x,y] in X|`R|X
  proof
    assume
A1: [x,y] in R |_2 X;
    then
A2: [x,y] in [:X,X:] by XBOOLE_0:def 4;
    then
A3: x in X by ZFMISC_1:87;
A4: y in X by A2,ZFMISC_1:87;
    [x,y] in R by A1,XBOOLE_0:def 4;
    then [x,y] in X|`R by A4,RELAT_1:def 12;
    hence thesis by A3,RELAT_1:def 11;
  end;
  assume
A5: [x,y] in X|`R|X;
  then
A6: [x,y] in X|`R by RELAT_1:def 11;
  then
A7: [x,y] in R by RELAT_1:def 12;
A8: y in X by A6,RELAT_1:def 12;
  x in X by A5,RELAT_1:def 11;
  then [x,y] in [:X,X:] by A8,ZFMISC_1:87;
  hence thesis by A7,XBOOLE_0:def 4;
end;

theorem Th11:
  R |_2 X = X|`(R|X)
proof
  thus R |_2 X = X|`R|X by Th10
    .= X|`(R|X) by RELAT_1:109;
end;

Lm5: dom(X|`R) c= dom R
proof
  let x;
  assume x in dom(X|`R);
  then consider y such that
A1: [x,y] in X|`R by XTUPLE_0:def 12;
  [x,y] in R by A1,RELAT_1:def 12;
  hence thesis by XTUPLE_0:def 12;
end;

theorem Th12:
  x in field(R |_2 X) implies x in field R & x in X
proof
A1: dom(X|`R|X) = dom(X|`R) /\ X & rng(X|`(R|X)) = rng(R|X) /\ X
             by RELAT_1:61,88;
  assume x in field(R |_2 X);
  then
A2: x in dom(R |_2 X) or x in rng(R |_2 X) by XBOOLE_0:def 3;
A3: dom(X|`R) c= dom R & rng(R|X) c= rng R by Lm5,RELAT_1:70;
  R |_2 X = X|`R|X & R |_2 X = X|`(R|X) by Th10,Th11;
  then x in dom(X|`R) & x in X or x in rng(R|X) & x in X
    by A2,A1,XBOOLE_0:def 4;
  hence thesis by A3,XBOOLE_0:def 3;
end;

theorem Th13:
  field(R |_2 X) c= field R & field(R |_2 X) c= X
proof
  ( for x st x in field(R |_2 X) holds x in field R)& for x st x in field(
  R |_2 X) holds x in X by Th12;
  hence thesis by TARSKI:def 3;
end;

theorem Th14:
  (R |_2 X)-Seg(a) c= R-Seg(a)
proof
  let x;
  assume
A1: x in (R |_2 X)-Seg(a);
  then [x,a] in R |_2 X by Th1;
  then
A2: [x,a] in R by XBOOLE_0:def 4;
  x <> a by A1,Th1;
  hence thesis by A2,Th1;
end;

theorem Th15:
  R is reflexive implies R |_2 X is reflexive
proof
  assume
A1: R is reflexive;
  now
    let a;
    assume
A2: a in field(R |_2 X);
    then a in X by Th12;
    then
A3: [a,a] in [:X,X:] by ZFMISC_1:87;
    a in field R by A2,Th12;
    then [a,a] in R by A1,Lm1;
    hence [a,a] in R |_2 X by A3,XBOOLE_0:def 4;
  end;
  hence thesis by Lm1;
end;

theorem Th16:
  R is connected implies R |_2 Y is connected
proof
  assume
A1: R is connected;
  now
    let a,b;
    assume that
A2: a in field(R |_2 Y) & b in field(R |_2 Y) and
A3: a <> b;
    a in Y & b in Y by A2,Th12;
    then
A4: [a,b] in [:Y,Y:] & [b,a] in [:Y,Y:] by ZFMISC_1:87;
    a in field R & b in field R by A2,Th12;
    then [a,b] in R or [b,a] in R by A1,A3,Lm4;
    hence [a,b] in R |_2 Y or [b,a] in R |_2 Y by A4,XBOOLE_0:def 4;
  end;
  hence thesis by Lm4;
end;

theorem Th17:
  R is transitive implies R |_2 Y is transitive
proof
  assume
A1: R is transitive;
  now
    let a,b,c;
    assume that
A2: [a,b] in R |_2 Y and
A3: [b,c] in R |_2 Y;
    [a,b] in R & [b,c] in R by A2,A3,XBOOLE_0:def 4;
    then
A4: [a,c] in R by A1,Lm2;
    [b,c] in [:Y,Y:] by A3,XBOOLE_0:def 4;
    then
A5: c in Y by ZFMISC_1:87;
    [a,b] in [:Y,Y:] by A2,XBOOLE_0:def 4;
    then a in Y by ZFMISC_1:87;
    then [a,c] in [:Y,Y:] by A5,ZFMISC_1:87;
    hence [a,c] in R |_2 Y by A4,XBOOLE_0:def 4;
  end;
  hence thesis by Lm2;
end;

theorem Th18:
  R is antisymmetric implies R |_2 Y is antisymmetric
proof
  assume
A1: R is antisymmetric;
  now
    let a,b;
    assume [a,b] in R |_2 Y & [b,a] in R |_2 Y;
    then [a,b] in R & [b,a] in R by XBOOLE_0:def 4;
    hence a = b by A1,Lm3;
  end;
  hence thesis by Lm3;
end;

theorem Th19:
  (R |_2 X) |_2 Y = R |_2 (X /\ Y)
proof
  thus (R |_2 X) |_2 Y = R /\ ([:X,X:] /\ [:Y,Y:]) by XBOOLE_1:16
    .= R |_2 (X /\ Y) by ZFMISC_1:100;
end;

theorem
  (R |_2 X) |_2 Y = (R |_2 Y) |_2 X
proof
  thus (R |_2 X) |_2 Y = R |_2 (Y /\ X) by Th19
    .= (R |_2 Y) |_2 X by Th19;
end;

theorem
  (R |_2 Y) |_2 Y = R |_2 Y
proof
  let a,b;
  thus [a,b] in (R |_2 Y) |_2 Y implies [a,b] in R |_2 Y by XBOOLE_0:def 4;
  assume
A1: [a,b] in R |_2 Y;
  then [a,b] in [:Y,Y:] by XBOOLE_0:def 4;
  hence thesis by A1,XBOOLE_0:def 4;
end;

theorem Th22:
  Z c= Y implies (R |_2 Y) |_2 Z = R |_2 Z
proof
  assume
A1: Z c= Y;
  let a,b;
  thus [a,b] in (R |_2 Y) |_2 Z implies [a,b] in R |_2 Z
  proof
    assume
A2: [a,b] in (R |_2 Y) |_2 Z;
    then [a,b] in R |_2 Y by XBOOLE_0:def 4;
    then
A3: [a,b] in R by XBOOLE_0:def 4;
    [a,b] in [:Z,Z:] by A2,XBOOLE_0:def 4;
    hence thesis by A3,XBOOLE_0:def 4;
  end;
  assume
A4: [a,b] in R |_2 Z;
  then
A5: [a,b] in R by XBOOLE_0:def 4;
A6: [a,b] in [:Z,Z:] by A4,XBOOLE_0:def 4;
  then a in Z & b in Z by ZFMISC_1:87;
  then [a,b] in [:Y,Y:] by A1,ZFMISC_1:87;
  then [a,b] in R |_2 Y by A5,XBOOLE_0:def 4;
  hence thesis by A6,XBOOLE_0:def 4;
end;

theorem Th23:
  R |_2 field R = R
proof
  let x,y;
  thus [x,y] in R |_2 field R implies [x,y] in R by XBOOLE_0:def 4;
  assume
A1: [x,y] in R;
  then x in field R & y in field R by RELAT_1:15;
  then [x,y] in [:field R,field R:] by ZFMISC_1:87;
  hence thesis by A1,XBOOLE_0:def 4;
end;

theorem Th24:
  R is well_founded implies R |_2 X is well_founded
proof
  assume
A1: for Y st Y c= field R & Y <> {} ex a st a in Y & R-Seg(a) misses Y;
A2: field(R |_2 X) c= field R by Th13;
  let Y;
  assume Y c= field(R |_2 X) & Y <> {};
  then consider a such that
A3: a in Y and
A4: R-Seg(a) misses Y by A1,A2,XBOOLE_1:1;
  take a;
  thus a in Y by A3;
  assume not thesis;
  then
A5: ex b being set st b in (R |_2 X)-Seg(a) & b in Y by XBOOLE_0:3;
  (R |_2 X)-Seg(a) c= R-Seg(a) by Th14;
  hence contradiction by A4,A5,XBOOLE_0:3;
end;

theorem Th25:
  R is well-ordering implies R |_2 Y is well-ordering
proof
  assume R is well-ordering;
  hence
  R |_2 Y is reflexive transitive antisymmetric connected well_founded by Th15
,Th16,Th17,Th18,Th24;
end;

theorem Th26:
  R is well-ordering implies R-Seg(a),R-Seg(b) are_c=-comparable
proof
  assume
A1: R is well-ordering;
A2: now
    assume
A3: a in field R & b in field R;
    now
      assume a <> b;
A4:   now
        assume
A5:     [b,a] in R;
        now
          let c;
          assume
A6:       c in R-Seg(b);
          then
A7:       [c,b] in R by Th1;
          then
A8:       [c,a] in R by A1,A5,Lm2;
          c <> b by A6,Th1;
          then c <> a by A1,A5,A7,Lm3;
          hence c in R-Seg(a) by A8,Th1;
        end;
        hence R-Seg(b) c= R-Seg(a) by TARSKI:def 3;
      end;
      now
        assume
A9:     [a,b] in R;
        now
          let c;
          assume
A10:      c in R-Seg(a);
          then
A11:      [c,a] in R by Th1;
          then
A12:      [c,b] in R by A1,A9,Lm2;
          c <> a by A10,Th1;
          then c <> b by A1,A9,A11,Lm3;
          hence c in R-Seg(b) by A12,Th1;
        end;
        hence R-Seg(a) c= R-Seg(b) by TARSKI:def 3;
      end;
      hence thesis by A1,A3,A4,Lm4,XBOOLE_0:def 9;
    end;
    hence thesis;
  end;
  now
    assume R-Seg(a) = {} or R-Seg(b) = {};
    then R-Seg(a) c= R-Seg(b) or R-Seg(b) c= R-Seg(a) by XBOOLE_1:2;
    hence thesis by XBOOLE_0:def 9;
  end;
  hence thesis by A2,Th2;
end;

theorem Th27:
  R is well-ordering & b in R-Seg(a) implies (R |_2 (R-Seg(a)))
  -Seg(b) = R-Seg(b)
proof
  assume that
A1: R is well-ordering and
A2: b in R-Seg(a);
  set S = R |_2 (R-Seg(a));
  now
    let c;
    assume
A3: c in R-Seg(b);
    then
A4: [c,b] in R by Th1;
A5: [b,a] in R by A2,Th1;
    then
A6: [c,a] in R by A1,A4,Lm2;
A7: c <> b by A3,Th1;
    then c <> a by A1,A4,A5,Lm3;
    then c in R-Seg(a) by A6,Th1;
    then [c,b] in [:R-Seg(a),R-Seg(a):] by A2,ZFMISC_1:87;
    then [c,b] in S by A4,XBOOLE_0:def 4;
    hence c in S-Seg(b) by A7,Th1;
  end;
  then
A8: R-Seg(b) c= S-Seg(b) by TARSKI:def 3;
  now
    let c;
    assume
A9: c in S-Seg(b);
    then [c,b] in S by Th1;
    then
A10: [c,b] in R by XBOOLE_0:def 4;
    c <> b by A9,Th1;
    hence c in R-Seg(b) by A10,Th1;
  end;
  then S-Seg(b) c= R-Seg(b) by TARSKI:def 3;
  hence thesis by A8,XBOOLE_0:def 10;
end;

theorem Th28:
  R is well-ordering & Y c= field R implies (Y = field R or (ex a
st a in field R & Y = R-Seg(a) ) iff for a st a in Y for b st [b,a] in R holds
  b in Y )
proof
  assume that
A1: R is well-ordering and
A2: Y c= field R;
  now
    given a such that
    a in field R and
A3: Y = R-Seg(a);
    let b such that
A4: b in Y;
A5: [b,a] in R by A3,A4,Th1;
    let c such that
A6: [c,b] in R;
A7: [c,a] in R by A1,A6,A5,Lm2;
    b <> a by A3,A4,Th1;
    then c <> a by A1,A6,A5,Lm3;
    hence c in Y by A3,A7,Th1;
  end;
  hence Y = field R or (ex a st a in field R & Y = R-Seg(a) ) implies for a st
  a in Y for b st [b,a] in R holds b in Y by RELAT_1:15;
  assume
A8: for a st a in Y for b st [b,a] in R holds b in Y;
  assume Y <> field R;
  then ex d st not ( d in field R iff d in Y ) by TARSKI:1;
  then field R \ Y <> {} by A2,XBOOLE_0:def 5;
  then consider a such that
A9: a in field R \ Y and
A10: for b st b in field R \ Y holds [a,b] in R by A1,Th6;
A11: now
    let b;
    assume
A12: b in R-Seg(a);
    then
A13: [b,a] in R by Th1;
    assume
A14: not b in Y;
    b in field R by A13,RELAT_1:15;
    then b in field R \ Y by A14,XBOOLE_0:def 5;
    then
A15: [a,b] in R by A10;
    b <> a by A12,Th1;
    hence contradiction by A1,A13,A15,Lm3;
  end;
  take a;
  thus a in field R by A9;
  now
A16: not a in Y by A9,XBOOLE_0:def 5;
    let b;
    assume
A17: b in Y;
    assume not b in R-Seg(a);
    then
A18: not [b,a] in R or a = b by Th1;
    a <> b by A9,A17,XBOOLE_0:def 5;
    then [a,b] in R by A2,A1,A9,A17,A18,Lm4;
    hence contradiction by A8,A17,A16;
  end;
  hence Y = R-Seg(a) by A11,TARSKI:1;
end;

theorem Th29:
  R is well-ordering & a in field R & b in field R implies ( [a,b]
  in R iff R-Seg(a) c= R-Seg(b) )
proof
  assume that
A1: R is well-ordering and
A2: a in field R and
A3: b in field R;
  thus [a,b] in R implies R-Seg(a) c= R-Seg(b)
  proof
    assume
A4: [a,b] in R;
    let c;
    assume
A5: c in R-Seg(a);
    then
A6: [c,a] in R by Th1;
    then
A7: [c,b] in R by A1,A4,Lm2;
    c <> a by A5,Th1;
    then c <> b by A1,A4,A6,Lm3;
    hence thesis by A7,Th1;
  end;
  assume
A8: R-Seg(a) c= R-Seg(b);
   now
    assume
A9: a <> b;
    assume not [a,b] in R;
    then [b,a] in R by A2,A3,A1,A9,Lm4;
    then b in R-Seg(a) by A9,Th1;
    hence contradiction by A8,Th1;
  end;
  hence thesis by A1,A2,Lm1;
end;

theorem Th30:
  R is well-ordering & a in field R & b in field R implies ( R-Seg
  (a) c= R-Seg(b) iff a = b or a in R-Seg(b) )
proof
  assume
A1: R is well-ordering & a in field R & b in field R;
  thus R-Seg(a) c= R-Seg(b) implies a = b or a in R-Seg(b)
  proof
    assume R-Seg(a) c= R-Seg(b);
    then [a,b] in R by A1,Th29;
    hence thesis by Th1;
  end;
  now
    assume a in R-Seg(b);
    then [a,b] in R by Th1;
    hence R-Seg(a) c= R-Seg(b) by A1,Th29;
  end;
  hence thesis;
end;

theorem Th31:
  R is well-ordering & X c= field R implies field(R |_2 X) = X
proof
  assume that
A1: R is well-ordering and
A2: X c= field R;
  thus field(R |_2 X) c= X by Th13;
  let x;
  assume
A3: x in X;
  then
A4: [x,x] in [:X,X:] by ZFMISC_1:87;
  [x,x] in R by A1,A2,A3,Lm1;
  then [x,x] in R |_2 X by A4,XBOOLE_0:def 4;
  hence thesis by RELAT_1:15;
end;

theorem Th32:
  R is well-ordering implies field(R |_2 R-Seg(a)) = R-Seg(a)
proof
  R-Seg(a) c= field R by Th9;
  hence thesis by Th31;
end;

theorem Th33:
  R is well-ordering implies for Z st for a st a in field R & R
  -Seg(a) c= Z holds a in Z holds field R c= Z
proof
  assume
A1: R is well-ordering;
  let Z such that
A2: for a st a in field R & R-Seg(a) c= Z holds a in Z;
A3: now
    let a such that
A4: a in field R and
A5: for b st [b,a] in R & a <> b holds b in Z;
    now
      let b;
      assume b in R-Seg(a);
      then [b,a] in R & b <> a by Th1;
      hence b in Z by A5;
    end;
    then R-Seg(a) c= Z by TARSKI:def 3;
    hence a in Z by A2,A4;
  end;
  given a such that
A6: a in field R & not a in Z;
  field R \ Z <> {} by A6,XBOOLE_0:def 5;
  then consider a such that
A7: a in field R \ Z and
A8: for b st b in field R \ Z holds [a,b] in R by A1,Th6;
  not a in Z by A7,XBOOLE_0:def 5;
  then consider b such that
A9: [b,a] in R and
A10: b <> a and
A11: not b in Z by A3,A7;
  b in field R by A9,RELAT_1:15;
  then b in field R \ Z by A11,XBOOLE_0:def 5;
  then [a,b] in R by A8;
  hence contradiction by A1,A9,A10,Lm3;
end;

theorem Th34:
  R is well-ordering & a in field R & b in field R & (for c st c
  in R-Seg(a) holds [c,b] in R & c <> b) implies [a,b] in R
proof
  assume that
A1: R is well-ordering & a in field R & b in field R and
A2: c in R-Seg(a) implies [c,b] in R & c <> b;
  assume
A3: not [a,b] in R;
  a <> b by A1,A3,Lm1;
  then [b,a] in R by A1,A3,Lm4;
  then b in R-Seg(a) by A3,Th1;
  hence contradiction by A2;
end;

theorem Th35:
  R is well-ordering & dom F = field R & rng F c= field R & (for a
,b st [a,b] in R & a <> b holds [F.a,F.b] in R & F.a <> F.b) implies for a st a
  in field R holds [a,F.a] in R
proof
  assume that
A1: R is well-ordering & dom F = field R & rng F c= field R and
A2: [a,b] in R & a <> b implies [F.a,F.b] in R & F.a <> F.b;
  defpred P[set] means [$1,F.$1] in R;
  consider Z such that
A3: a in Z iff a in field R & P[a] from XBOOLE_0:sch 1;
  now
    let a;
    assume
A4: a in field R;
    assume
A5: R-Seg(a) c= Z;
A6: now
      let b;
      assume
A7:   b in R-Seg(a);
      then
A8:  [b,F.b] in R by A3,A5;
A9:  [b,a] in R & b <> a by A7,Th1;
      then
A10:  [F.b,F.a] in R by A2;
      hence [b,F.a] in R by A1,A8,Lm2;
      F.b <> F.a by A2,A9;
      hence b <> F.a by A1,A8,A10,Lm3;
    end;
    F.a in rng F by A1,A4,FUNCT_1:def 3;
    then [a,F.a] in R by A1,A4,A6,Th34;
    hence a in Z by A3,A4;
  end;
  then
A11: field R c= Z by A1,Th33;
  let a;
  assume a in field R;
  hence thesis by A3,A11;
end;

definition
  let R,S,F;
  pred F is_isomorphism_of R,S means
  :Def7:
  dom F = field R & rng F = field S
& F is one-to-one & for a,b holds [a,b] in R iff a in field R & b in field R &
  [F.a,F.b] in S;
end;

theorem Th36:
  F is_isomorphism_of R,S implies for a,b st [a,b] in R & a <> b
  holds [F.a,F.b] in S & F.a <> F.b
proof
  assume
A1: F is_isomorphism_of R,S;
  then
A2: dom F = field R & F is one-to-one by Def7;
  let a,b;
  assume that
A3: [a,b] in R and
A4: a <> b;
  a in field R & b in field R by A1,A3,Def7;
  hence thesis by A1,A2,A3,A4,Def7,FUNCT_1:def 4;
end;

definition
  let R,S;
  pred R,S are_isomorphic means
  :Def8:
  ex F st F is_isomorphism_of R,S;
end;

theorem Th37:
  id(field R) is_isomorphism_of R,R
proof
A1: now
    let a,b;
    thus [a,b] in R implies a in field R & b in field R & [id(field R).a,id(
    field R).b] in R
    proof
      assume
A2:   [a,b] in R;
      hence
A3:   a in field R & b in field R by RELAT_1:15;
      then id(field R).a = a by FUNCT_1:18;
      hence thesis by A2,A3,FUNCT_1:18;
    end;
    assume that
A4: a in field R and
A5: b in field R & [id(field R).a,id(field R).b] in R;
    id(field R).a = a by A4,FUNCT_1:18;
    hence [a,b] in R by A5,FUNCT_1:18;
  end;
  dom(id(field R)) = field R & rng(id(field R)) = field R;
  hence thesis by A1,Def7;
end;

theorem
  R,R are_isomorphic
proof
  take id(field R);
  thus thesis by Th37;
end;

theorem Th39:
  F is_isomorphism_of R,S implies F" is_isomorphism_of S,R
proof
  assume
A1: F is_isomorphism_of R,S;
  then
A2: F is one-to-one by Def7;
A3: rng F = field S by A1,Def7;
  hence
A4: dom(F") = field S by A2,FUNCT_1:33;
A5: dom F = field R by A1,Def7;
  hence rng(F") = field R by A2,FUNCT_1:33;
  thus F" is one-to-one by A2;
  let a,b;
  thus [a,b] in S implies a in field S & b in field S & [F".a,F".b] in R
  proof
A6: dom F = rng(F") by A2,FUNCT_1:33;
    assume
A7: [a,b] in S;
    hence
A8: a in field S & b in field S by RELAT_1:15;
    then
A9: F".a in rng(F") & F".b in rng(F") by A4,FUNCT_1:def 3;
    a = F.(F".a) & b = F.(F".b) by A3,A2,A8,FUNCT_1:35;
    hence thesis by A1,A5,A7,A6,A9,Def7;
  end;
  assume that
A10: a in field S & b in field S and
A11: [F".a,F".b] in R;
  F.(F".a) = a & F.(F".b) = b by A3,A2,A10,FUNCT_1:35;
  hence thesis by A1,A11,Def7;
end;

theorem Th40:
  R,S are_isomorphic implies S,R are_isomorphic
proof
  given F such that
A1: F is_isomorphism_of R,S;
  take F";
  thus thesis by A1,Th39;
end;

theorem Th41:
  F is_isomorphism_of R,S & G is_isomorphism_of S,T implies G*F
  is_isomorphism_of R,T
proof
  assume that
A1: dom F = field R and
A2: rng F = field S and
A3: F is one-to-one and
A4: for a,b holds [a,b] in R iff a in field R & b in field R & [F.a,F.b] in S;
  assume that
A5: dom G = field S and
A6: rng G = field T and
A7: G is one-to-one and
A8: for a,b holds [a,b] in S iff a in field S & b in field S & [G.a,G.b] in T;
  thus dom(G*F) = field R by A1,A2,A5,RELAT_1:27;
  thus rng(G*F) = field T by A2,A5,A6,RELAT_1:28;
  thus G*F is one-to-one by A3,A7;
  let a,b;
  thus [a,b] in R implies a in field R & b in field R & [(G*F).a,(G*F).b] in T
  proof
    assume
A9: [a,b] in R;
    hence a in field R & b in field R by RELAT_1:15;
    then
A10: (G*F).a = G.(F.a) & (G*F).b = G.(F.b) by A1,FUNCT_1:13;
    [F.a,F.b] in S by A4,A9;
    hence thesis by A8,A10;
  end;
  assume that
A11: a in field R & b in field R and
A12: [(G*F).a,(G*F).b] in T;
A13: (G*F).a = G.(F.a) & (G*F).b = G.(F.b) by A1,A11,FUNCT_1:13;
  F.a in field S & F.b in field S by A1,A2,A11,FUNCT_1:def 3;
  then [F.a,F.b] in S by A8,A12,A13;
  hence thesis by A4,A11;
end;

theorem Th42:
  R,S are_isomorphic & S,T are_isomorphic implies R,T are_isomorphic
proof
  given F such that
A1: F is_isomorphism_of R,S;
  given G such that
A2: G is_isomorphism_of S,T;
  take G*F;
  thus thesis by A1,A2,Th41;
end;

theorem Th43:
  F is_isomorphism_of R,S implies ( R is reflexive implies S is
  reflexive ) & ( R is transitive implies S is transitive ) & ( R is connected
implies S is connected ) & ( R is antisymmetric implies S is antisymmetric ) &
  ( R is well_founded implies S is well_founded )
proof
  assume
A1: F is_isomorphism_of R,S;
  then
A2: dom F = field R by Def7;
A3: rng F = field S by A1,Def7;
A4: F is one-to-one by A1,Def7;
  then
A5: rng(F") = dom F & dom(F") = rng F by FUNCT_1:33;
  thus R is reflexive implies S is reflexive
  proof
    assume
A6: R is reflexive;
    now
      let a;
      assume
A7:   a in field S;
      then F".a in field R by A2,A3,A5,FUNCT_1:def 3;
      then
A8:   [F".a,F".a] in R by A6,Lm1;
      a = F.(F".a) by A3,A4,A7,FUNCT_1:35;
      hence [a,a] in S by A1,A8,Def7;
    end;
    hence thesis by Lm1;
  end;
  thus R is transitive implies S is transitive
  proof
    assume
A9: R is transitive;
    now
      let a,b,c;
      assume that
A10:  [a,b] in S and
A11:  [b,c] in S;
A12:  c in field S by A11,RELAT_1:15;
      then
A13:  c = F.(F".c) by A3,A4,FUNCT_1:35;
      b in field S by A10,RELAT_1:15;
      then
A14:  b = F.(F".b) & F".b in field R by A2,A3,A4,A5,FUNCT_1:35,def 3;
      F".c in field R by A2,A3,A5,A12,FUNCT_1:def 3;
      then
A15:  [F".b,F".c] in R by A1,A11,A13,A14,Def7;
A16:  a in field S by A10,RELAT_1:15;
      then
A17:  a = F.(F".a) by A3,A4,FUNCT_1:35;
      F".a in field R by A2,A3,A5,A16,FUNCT_1:def 3;
      then [F".a,F".b] in R by A1,A10,A17,A14,Def7;
      then [F".a,F".c] in R by A9,A15,Lm2;
      hence [a,c] in S by A1,A17,A13,Def7;
    end;
    hence thesis by Lm2;
  end;
  thus R is connected implies S is connected
  proof
    assume
A18: R is connected;
    now
      let a,b;
      assume that
A19:  a in field S & b in field S and
A20:  a <> b;
A21:  a = F.(F".a) & b = F.(F".b) by A3,A4,A19,FUNCT_1:35;
      F".a in field R & F".b in field R by A2,A3,A5,A19,FUNCT_1:def 3;
      then [F".a,F".b] in R or [F".b,F".a] in R by A18,A20,A21,Lm4;
      hence [a,b] in S or [b,a] in S by A1,A21,Def7;
    end;
    hence thesis by Lm4;
  end;
  thus R is antisymmetric implies S is antisymmetric
  proof
    assume
A22: R is antisymmetric;
    now
      let a,b;
      assume that
A23:  [a,b] in S and
A24:  [b,a] in S;
A25:  a in field S by A23,RELAT_1:15;
      then
A26:  a = F.(F".a) by A3,A4,FUNCT_1:35;
A27:  b in field S by A23,RELAT_1:15;
      then
A28:  b = F.(F".b) by A3,A4,FUNCT_1:35;
A29:  F".b in field R by A2,A3,A5,A27,FUNCT_1:def 3;
      F".a in field R by A2,A3,A5,A25,FUNCT_1:def 3;
      then [F".a,F".b] in R & [F".b,F".a] in R by A1,A23,A24,A26,A28,A29,Def7;
      hence a = b by A22,A26,A28,Lm3;
    end;
    hence thesis by Lm3;
  end;
  assume
A30: for Y st Y c= field R & Y <> {} ex x st x in Y & R-Seg(x) misses Y;
  let Z;
  assume that
A31: Z c= field S and
A32: Z <> {};
A33: F"Z c= dom F by RELAT_1:132;
  then consider x such that
A34: x in F"Z and
A35: R-Seg(x) misses F"Z by A2,A3,A30,A31,A32,RELAT_1:139;
  take F.x;
  thus F.x in Z by A34,FUNCT_1:def 7;
  assume not thesis;
  then consider y being set such that
A36: y in S-Seg(F.x) and
A37: y in Z by XBOOLE_0:3;
A38: F".y in dom F by A3,A5,A31,A37,FUNCT_1:def 3;
A39: [y,F.x] in S by A36,Th1;
A40: y = F.(F".y) by A3,A4,A31,A37,FUNCT_1:35;
  then F".y in F"Z by A37,A38,FUNCT_1:def 7;
  then not F".y in R-Seg(x) by A35,XBOOLE_0:3;
  then not [F".y,x] in R or F".y = x by Th1;
  hence contradiction by A1,A2,A33,A34,A36,A40,A38,A39,Th1,Def7;
end;

theorem Th44:
  R is well-ordering & F is_isomorphism_of R,S implies S is well-ordering
proof
  assume R is reflexive transitive antisymmetric connected well_founded & F
  is_isomorphism_of R,S;
  hence S is reflexive transitive antisymmetric connected well_founded by Th43;
end;

theorem Th45:
  R is well-ordering implies for F,G st F is_isomorphism_of R,S &
  G is_isomorphism_of R,S holds F = G
proof
  assume
A1: R is well-ordering;
  let F,G;
  assume that
A2: F is_isomorphism_of R,S and
A3: G is_isomorphism_of R,S;
A4: dom F = field R by A2,Def7;
A5: S is well-ordering by A1,A2,Th44;
A6: rng F = field S by A2,Def7;
A7: G is one-to-one by A3,Def7;
A8: dom G = field R by A3,Def7;
A9: G" is_isomorphism_of S,R by A3,Th39;
  then
A10: G" is one-to-one by Def7;
A11: F is one-to-one by A2,Def7;
A12: rng G = field S by A3,Def7;
A13: F" is_isomorphism_of S,R by A2,Th39;
  then
A14: F" is one-to-one by Def7;
  for a st a in field R holds F.a = G.a
  proof
A15: dom(F") = field S by A6,A11,FUNCT_1:33;
    then
A16: dom(F"*G) = field R by A8,A12,RELAT_1:27;
A17: now
      let a,b;
      assume that
A18:  [a,b] in R and
A19:  a <> b;
A20:  [G.a,G.b] in S by A3,A18,Def7;
A21:  b in field R by A18,RELAT_1:15;
      then
A22:  F".(G.b) = (F"*G).b by A8,FUNCT_1:13;
A23:  a in field R by A18,RELAT_1:15;
      then F".(G.a) = (F"*G).a by A8,FUNCT_1:13;
      hence [(F"*G).a,(F"*G).b] in R by A13,A20,A22,Def7;
      thus (F"*G).a <> (F"*G).b by A14,A7,A16,A19,A23,A21,FUNCT_1:def 4;
    end;
A24: dom(G") = field S by A12,A7,FUNCT_1:33;
    then
A25: dom(G"*F) = field R by A4,A6,RELAT_1:27;
A26: now
      let a,b;
      assume that
A27:  [a,b] in R and
A28:  a <> b;
A29:  [F.a,F.b] in S by A2,A27,Def7;
A30:  b in field R by A27,RELAT_1:15;
      then
A31:  G".(F.b) = (G"*F).b by A4,FUNCT_1:13;
A32:  a in field R by A27,RELAT_1:15;
      then G".(F.a) = (G"*F).a by A4,FUNCT_1:13;
      hence [(G"*F).a,(G"*F).b] in R by A9,A29,A31,Def7;
      thus (G"*F).a <> (G"*F).b by A11,A10,A25,A28,A32,A30,FUNCT_1:def 4;
    end;
    let a such that
A33: a in field R;
A34: F".(G.a) = (F"*G).a by A8,A33,FUNCT_1:13;
    G.a in rng F by A6,A8,A12,A33,FUNCT_1:def 3;
    then
A35: F.(F".(G.a)) = G.a by A11,FUNCT_1:35;
    rng(F") = field R by A4,A11,FUNCT_1:33;
    then rng(F"*G) = field R by A12,A15,RELAT_1:28;
    then [a,(F"*G).a] in R by A1,A33,A16,A17,Th35;
    then
A36: [F.a,G.a] in S by A2,A34,A35,Def7;
    F.a in rng G by A4,A6,A12,A33,FUNCT_1:def 3;
    then
A37: G.(G".(F.a)) = F.a by A7,FUNCT_1:35;
A38: G".(F.a) = (G"*F).a by A4,A33,FUNCT_1:13;
    rng(G") = field R by A8,A7,FUNCT_1:33;
    then rng(G"*F) = field R by A6,A24,RELAT_1:28;
    then [a,(G"*F).a] in R by A1,A33,A25,A26,Th35;
    then [G.a,F.a] in S by A3,A38,A37,Def7;
    hence thesis by A5,A36,Lm3;
  end;
  hence thesis by A4,A8,FUNCT_1:2;
end;

definition
  let R,S;
  assume that
A1: R is well-ordering and
A2: R,S are_isomorphic;
  func canonical_isomorphism_of(R,S) -> Function means
  :Def9:
  it is_isomorphism_of R,S;
  existence by A2,Def8;
  uniqueness by A1,Th45;
end;

theorem Th46:
  R is well-ordering implies for a st a in field R holds
  not R,R |_2 (R-Seg(a)) are_isomorphic
proof
  assume
A1: R is well-ordering;
  let a such that
A2: a in field R;
  set S = R |_2 (R-Seg(a));
  set F = canonical_isomorphism_of(R,S);
  assume R,R |_2 (R-Seg(a)) are_isomorphic;
  then
A3: F is_isomorphism_of R,S by A1,Def9;
  then
A4: dom F = field R by Def7;
A5: F is one-to-one by A3,Def7;
A6: now
    let b,c;
    assume that
A7: [b,c] in R and
A8: b <> c;
    [F.b,F.c] in R |_2 (R-Seg(a)) by A3,A7,Def7;
    hence [F.b,F.c] in R by XBOOLE_0:def 4;
    b in field R & c in field R by A7,RELAT_1:15;
    hence F.b <> F.c by A4,A5,A8,FUNCT_1:def 4;
  end;
A9: rng F = field S by A3,Def7;
  field S = R-Seg(a) by A1,Th32;
  then F.a in R-Seg(a) by A2,A4,A9,FUNCT_1:def 3;
  then
A10: [F.a,a] in R & F.a <> a by Th1;
  rng F c= field R by A9,Th13;
  then [a,F.a] in R by A1,A2,A4,A6,Th35;
  hence contradiction by A1,A10,Lm3;
end;

theorem Th47:
  R is well-ordering & a in field R & b in field R & a <> b
  implies not R |_2 (R-Seg(a)),R |_2 (R-Seg(b)) are_isomorphic
proof
  assume that
A1: R is well-ordering and
A2: a in field R & b in field R and
A3: a <> b;
A4: now
    set S = R |_2 (R-Seg(a));
    assume
A5: R-Seg(b) c= R-Seg(a);
    then
A6: S |_2 (R-Seg(b)) = R |_2 (R-Seg(b)) by Th22;
A7: field S = R-Seg(a) by A1,Th32;
A8: b in R-Seg(a)
    proof
      assume not b in R-Seg(a);
      then not [b,a] in R by A3,Th1;
      then [a,b] in R by A2,A3,A1,Lm4;
      then a in R-Seg(b) by A3,Th1;
      hence contradiction by A5,Th1;
    end;
    then R-Seg(b) = S-Seg(b) by A1,Th27;
    hence thesis by A1,A7,A8,A6,Th25,Th46;
  end;
A9: now
    set S = R |_2 (R-Seg(b));
    assume
A10: R-Seg(a) c= R-Seg(b);
    then
A11: S |_2 (R-Seg(a)) = R |_2 (R-Seg(a)) by Th22;
A12: field S = R-Seg(b) & S is well-ordering by A1,Th25,Th32;
A13: a in R-Seg(b)
    proof
      assume not a in R-Seg(b);
      then not [a,b] in R by A3,Th1;
      then [b,a] in R by A2,A3,A1,Lm4;
      then b in R-Seg(a) by A3,Th1;
      hence contradiction by A10,Th1;
    end;
    then R-Seg(a) = S-Seg(a) by A1,Th27;
    hence thesis by A13,A11,A12,Th40,Th46;
  end;
  R-Seg(a),R-Seg(b) are_c=-comparable by A1,Th26;
  hence thesis by A9,A4,XBOOLE_0:def 9;
end;

theorem Th48:
  R is well-ordering & Z c= field R & F is_isomorphism_of R,S
  implies F|Z is_isomorphism_of R |_2 Z,S |_2 (F.:Z) & R |_2 Z,S |_2 (F.:Z)
  are_isomorphic
proof
  assume that
A1: R is well-ordering and
A2: Z c= field R and
A3: F is_isomorphism_of R,S;
A4: F.:Z c= rng F by RELAT_1:111;
  rng F = field S by A3,Def7;
  then
A5: F.:Z = field(S |_2 (F.:Z)) by A1,A3,A4,Th31,Th44;
A6: F is one-to-one by A3,Def7;
A7: Z = field(R |_2 Z) by A1,A2,Th31;
A8: dom F = field R by A3,Def7;
  thus F|Z is_isomorphism_of R |_2 Z,S |_2 (F.:Z)
  proof
    thus
A9: dom(F|Z) = field(R |_2 Z) by A2,A8,A7,RELAT_1:62;
    thus
A10: rng(F|Z) = field(S |_2 (F.:Z)) by A5,RELAT_1:115;
    thus F|Z is one-to-one by A6,FUNCT_1:52;
    let a,b;
    thus [a,b] in R |_2 Z implies a in field(R |_2 Z) & b in field(R |_2 Z) &
    [F|Z.a,F|Z.b] in S |_2 (F.:Z)
    proof
      assume
A11:  [a,b] in R |_2 Z;
      then [a,b] in R by XBOOLE_0:def 4;
      then
A12:  [F.a,F.b] in S by A3,Def7;
      thus
A13:  a in field(R |_2 Z) & b in field(R |_2 Z) by A11,RELAT_1:15;
      then F|Z.a in rng(F|Z) & F|Z.b in rng(F|Z) by A9,FUNCT_1:def 3;
      then
A14:  [F|Z.a,F|Z.b] in [:F.:Z,F.:Z:] by A5,A10,ZFMISC_1:87;
      F.a = F|Z.a & F.b = F|Z.b by A9,A13,FUNCT_1:47;
      hence thesis by A12,A14,XBOOLE_0:def 4;
    end;
    assume that
A15: a in field(R |_2 Z) & b in field(R |_2 Z) and
A16: [F|Z.a,F|Z.b] in S |_2 (F.:Z);
    F.a = F|Z.a & F.b = F|Z.b by A9,A15,FUNCT_1:47;
    then
A17: [F.a,F.b] in S by A16,XBOOLE_0:def 4;
A18: [a,b] in [:Z,Z:] by A7,A15,ZFMISC_1:87;
    a in field R & b in field R by A15,Th12;
    then [a,b] in R by A3,A17,Def7;
    hence thesis by A18,XBOOLE_0:def 4;
  end;
  hence thesis by Def8;
end;

theorem Th49:
  F is_isomorphism_of R,S implies for a st a in field R ex b st b
  in field S & F.:(R-Seg(a)) = S-Seg(b)
proof
  assume
A1: F is_isomorphism_of R,S;
  then
A2: dom F = field R by Def7;
  let a;
  assume
A3: a in field R;
  take b = F.a;
A4: rng F = field S by A1,Def7;
  hence b in field S by A3,A2,FUNCT_1:def 3;
A5: F is one-to-one by A1,Def7;
A6: c in S-Seg(b) implies c in F.:(R-Seg(a))
  proof
    assume
A7: c in S-Seg(b);
    then
A8: c <> b by Th1;
A9: [c,b] in S by A7,Th1;
    then
A10: c in field S by RELAT_1:15;
    then
A11: c = F.(F".c) by A4,A5,FUNCT_1:35;
    rng(F") = dom F & dom(F") = rng F by A5,FUNCT_1:33;
    then
A12: F".c in field R by A2,A4,A10,FUNCT_1:def 3;
    then [F".c,a] in R by A1,A3,A9,A11,Def7;
    then F".c in R-Seg(a) by A8,A11,Th1;
    hence thesis by A2,A11,A12,FUNCT_1:def 6;
  end;
  c in F.:(R-Seg(a)) implies c in S-Seg(b)
  proof
    assume c in F.:(R-Seg(a));
    then consider d such that
A13: d in dom F and
A14: d in R-Seg(a) and
A15: c = F.d by FUNCT_1:def 6;
    [d,a] in R by A14,Th1;
    then
A16: [c,b] in S by A1,A15,Def7;
    d <> a by A14,Th1;
    then c <> b by A3,A2,A5,A13,A15,FUNCT_1:def 4;
    hence thesis by A16,Th1;
  end;
  hence thesis by A6,TARSKI:1;
end;

theorem Th50:
  R is well-ordering & F is_isomorphism_of R,S implies for a st a
  in field R ex b st b in field S & R |_2 (R-Seg(a)),S |_2 (S-Seg(b))
  are_isomorphic
proof
  assume that
A1: R is well-ordering and
A2: F is_isomorphism_of R,S;
  let a;
  assume a in field R;
  then consider b such that
A3: b in field S & F.:(R-Seg(a)) = S-Seg(b) by A2,Th49;
  take b;
  R-Seg(a) c= field R by Th9;
  hence thesis by A1,A2,A3,Th48;
end;

theorem Th51:
  R is well-ordering & S is well-ordering & a in field R & b in
field S & c in field S & R,S |_2 (S-Seg(b)) are_isomorphic & R |_2 (R-Seg(a)),S
  |_2 (S-Seg(c)) are_isomorphic implies S-Seg(c) c= S-Seg(b) & [c,b] in S
proof
  assume that
A1: R is well-ordering and
A2: S is well-ordering and
A3: a in field R and
A4: b in field S and
A5: c in field S and
A6: R,S |_2 (S-Seg(b)) are_isomorphic and
A7: R |_2 (R-Seg(a)),S |_2 (S-Seg(c)) are_isomorphic;
  set Q = S |_2 (S-Seg(b));
  set F1 = canonical_isomorphism_of(R,Q);
A8: F1 is_isomorphism_of R,Q by A1,A6,Def9;
  then consider d such that
A9: d in field Q and
A10: F1.:(R-Seg(a)) = Q-Seg(d) by A3,Th49;
A11: S-Seg(b) = field Q by A2,Th32;
  then
A12: Q-Seg(d) = S-Seg(d) by A2,A9,Th27;
A13: rng F1 = S-Seg(b) by A8,A11,Def7;
  then
A14: Q-Seg(d) c= S-Seg(b) by A10,RELAT_1:111;
  set T = S |_2 (S-Seg(c));
  set P = R |_2 (R-Seg(a));
A15: T,P are_isomorphic by A7,Th40;
A16: d in field S by A9,Th12;
  R-Seg(a) c= field R by Th9;
  then P,Q |_2 (F1.:(R-Seg(a))) are_isomorphic by A1,A8,Th48;
  then T,Q |_2 (Q-Seg(d)) are_isomorphic by A10,A15,Th42;
  then T,S |_2 (S-Seg(d)) are_isomorphic by A10,A12,A13,Th22,RELAT_1:111;
  hence S-Seg(c) c= S-Seg(b) by A2,A5,A12,A14,A16,Th47;
  hence thesis by A2,A4,A5,Th29;
end;

theorem Th52:
  R is well-ordering & S is well-ordering implies R,S
  are_isomorphic or (ex a st a in field R & R |_2 (R-Seg(a)),S are_isomorphic )
  or ex a st a in field S & R,S |_2 (S-Seg(a)) are_isomorphic
proof
  assume that
A1: R is well-ordering and
A2: S is well-ordering;
  defpred P[set] means ex b st b in field S & R |_2 (R-Seg($1)),S |_2 (S-Seg(b
  )) are_isomorphic;
  consider Z such that
A3: a in Z iff a in field R & P[a] from XBOOLE_0:sch 1;
A4: Z c= field R
  proof
    let x;
    thus thesis by A3;
  end;
  defpred P[set,set] means $2 in field S & R |_2 (R-Seg $1),S |_2 (S-Seg $2)
  are_isomorphic;
A5: for a,b,c st P[a, b] & P[a, c] holds b = c
  proof
    let a,b,c;
    assume that
A6: b in field S and
A7: R |_2 (R-Seg(a)),S |_2 (S-Seg(b)) are_isomorphic and
A8: c in field S & R |_2 (R-Seg(a)),S |_2 (S-Seg(c)) are_isomorphic;
    S |_2 (S-Seg(b)),R |_2 (R-Seg(a)) are_isomorphic by A7,Th40;
    hence thesis by A2,A6,A8,Th42,Th47;
  end;
  consider F such that
A9: [a,b] in F iff a in field R & P[a,b] from FUNCT_1:sch 1(A5);
A10: Z = dom F
  proof
    thus a in Z implies a in dom F
    proof
      assume
A11:  a in Z;
      then consider b such that
A12:  b in field S & R |_2 (R-Seg(a)),S |_2 (S-Seg(b)) are_isomorphic by A3;
      a in field R by A3,A11;
      then [a,b] in F by A9,A12;
      hence thesis by XTUPLE_0:def 12;
    end;
    let a;
    assume a in dom F;
    then consider b such that
A13: [a,b] in F by XTUPLE_0:def 12;
A14: R |_2 (R-Seg(a)),S |_2 (S-Seg(b)) are_isomorphic by A9,A13;
    a in field R & b in field S by A9,A13;
    hence thesis by A3,A14;
  end;
A15: rng F c= field S
  proof
    let a;
    assume a in rng F;
    then consider b such that
A16: b in dom F & a = F.b by FUNCT_1:def 3;
    [b,a] in F by A16,FUNCT_1:1;
    hence thesis by A9;
  end;
A17: F is_isomorphism_of R |_2 (dom F),S |_2 (rng F)
  proof
    thus dom F = field(R |_2 (dom F)) & rng F = field(S |_2 (rng F)) by A1,A2
,A4,A15,A10,Th31;
    thus
A18: F is one-to-one
    proof
      let a,b;
      assume that
A19:  a in dom F and
A20:  b in dom F and
A21:  F.a = F.b;
A22:  [b,F.b] in F by A20,FUNCT_1:1;
      then R |_2 (R-Seg(b)),S |_2 (S-Seg(F.a)) are_isomorphic by A9,A21;
      then
A23:  S |_2 (S-Seg(F.a)),R |_2 (R-Seg(b)) are_isomorphic by Th40;
      [a,F.a] in F by A19,FUNCT_1:1;
      then
A24:  a in field R & R |_2 (R-Seg(a)),S |_2 (S-Seg(F.a)) are_isomorphic by A9;
      b in field R by A9,A22;
      hence thesis by A1,A24,A23,Th42,Th47;
    end;
    let a,b;
    set P = R |_2 (R-Seg(a));
A25: field P = R-Seg(a) & P is well-ordering by A1,Th25,Th32;
    thus [a,b] in R |_2 (dom F) implies a in field(R |_2 (dom F)) & b in field
    (R |_2 (dom F)) & [F.a,F.b] in S |_2 (rng F)
    proof
      assume
A26:  [a,b] in R |_2 (dom F);
      hence
A27:  a in field(R |_2 (dom F)) & b in field(R |_2 (dom F)) by RELAT_1:15;
      then
A28:  a in dom F by Th12;
      then
A29:  [a,F.a] in F by FUNCT_1:1;
      then
A30:  F.a in field S by A9;
A31:  b in dom F by A27,Th12;
      then
A32:  [b,F.b] in F by FUNCT_1:1;
      then
A33:  b in field R by A9;
A34:  F.b in field S & R |_2 (R-Seg(b)),S |_2 (S-Seg(F.b))
      are_isomorphic by A9,A32;
A35:  [a,b] in R by A26,XBOOLE_0:def 4;
A36:  F.b in rng F by A31,FUNCT_1:def 3;
      F.a in rng F by A28,FUNCT_1:def 3;
      then
A37:  [F.a,F.b] in [:rng F,rng F:] by A36,ZFMISC_1:87;
      a in field R by A9,A29;
      then
A38:  R-Seg(a) c= R-Seg(b) by A1,A33,A35,Th29;
A39:  R |_2 (R-Seg(a)),S |_2 (S-Seg(F.a)) are_isomorphic by A9,A29;
A40:  now
        set P = R |_2 (R-Seg(b));
A41:    field P = R-Seg(b) & P is well-ordering by A1,Th25,Th32;
        assume a <> b;
        then
A42:    a in R-Seg(b) by A35,Th1;
        then P-Seg(a) = R-Seg(a) by A1,Th27;
        then P |_2(P-Seg(a)),S |_2(S-Seg(F.a)) are_isomorphic by A39,A38,Th22;
        then [F.a,F.b] in S by A2,A30,A34,A42,A41,Th51;
        hence thesis by A37,XBOOLE_0:def 4;
      end;
      a = b implies thesis
      proof
        assume a = b;
        then [F.a,F.b] in S by A2,A30,Lm1;
        hence thesis by A37,XBOOLE_0:def 4;
      end;
      hence thesis by A40;
    end;
    assume that
A43: a in field(R |_2 (dom F)) and
A44: b in field(R |_2 (dom F)) and
A45: [F.a,F.b] in S |_2 (rng F);
A46: [F.a,F.b] in S by A45,XBOOLE_0:def 4;
A47: a in dom F by A43,Th12;
    then
A48: [a,F.a] in F by FUNCT_1:1;
    then
A49: a in field R by A9;
    assume not [a,b] in R |_2 (dom F);
    then
A50: not [a,b] in R or not [a,b] in [:dom F,dom F:] by XBOOLE_0:def 4;
    then
A51: a <> b by A1,A47,A49,Lm1,ZFMISC_1:87;
A52: b in dom F by A44,Th12;
    then
A53: [b,F.b] in F by FUNCT_1:1;
    then
A54: R |_2 (R-Seg(b)),S |_2 (S-Seg(F.b)) are_isomorphic by A9;
A55: b in field R by A9,A53;
    then
A56: [b,a] in R by A1,A47,A52,A50,A49,A51,Lm4,ZFMISC_1:87;
    then
A57: R-Seg(b) c= R-Seg(a) by A1,A49,A55,Th29;
A58: b in R-Seg(a) by A47,A52,A50,A56,Th1,ZFMISC_1:87;
    then P-Seg(b) = R-Seg(b) by A1,Th27;
    then
A59: P |_2 (P-Seg(b)),S |_2 (S-Seg(F.b)) are_isomorphic by A54,A57,Th22;
A60: F.b in field S by A9,A53;
    F.a in field S & R |_2 (R-Seg(a)),S |_2 (S-Seg(F.a)) are_isomorphic
    by A9,A48;
    then [F.b,F.a] in S by A2,A60,A58,A25,A59,Th51;
    then F.a = F.b by A2,A46,Lm3;
    hence contradiction by A18,A47,A52,A51,FUNCT_1:def 4;
  end;
A61: now
    given a such that
A62: a in field R and
A63: Z = R-Seg(a);
    given b such that
A64: b in field S and
A65: rng F = S-Seg(b);
    R |_2 (R-Seg(a)),S |_2 (S-Seg(b)) are_isomorphic by A10,A17,A63,A65,Def8;
    then a in Z by A3,A62,A64;
    hence contradiction by A63,Th1;
  end;
A66: now
    let a such that
A67: a in Z;
    consider c such that
A68: c in field S and
A69: R |_2 (R-Seg(a)),S |_2 (S-Seg(c)) are_isomorphic by A3,A67;
    let b such that
A70: [b,a] in R;
A71: a in field R by A3,A67;
    now
      set Q = S |_2 (S-Seg(c));
      set P = R |_2 (R-Seg(a));
      P is well-ordering by A1,Th25;
      then
A72:  canonical_isomorphism_of(P,Q) is_isomorphism_of P,Q by A69,Def9;
      assume a <> b;
      then
A73:  b in R-Seg(a) by A70,Th1;
      then
A74:  P-Seg(b) = R-Seg(b) by A1,Th27;
A75:  b in field R by A70,RELAT_1:15;
      then R-Seg(b) c= R-Seg(a) by A1,A71,A73,Th30;
      then
A76:  P |_2 (R-Seg(b)) = R |_2 (R-Seg(b)) by Th22;
      field P = R-Seg(a) by A1,Th32;
      then consider d such that
A77:  d in field Q and
A78:  P |_2 (P-Seg(b)),Q |_2 (Q-Seg(d)) are_isomorphic by A1,A72,A73,Th25,Th50;
A79:  S-Seg(c) = field Q by A2,Th32;
      then
A80:  Q-Seg(d) = S-Seg(d) by A2,A77,Th27;
      [d,c] in S by A77,A79,Th1;
      then
A81:  d in field S by RELAT_1:15;
      then S-Seg(d) c= S-Seg(c) by A2,A68,A77,A79,Th30;
      then
      R |_2 (R-Seg(b)),S |_2 (S-Seg(d)) are_isomorphic by A78,A74,A80,A76,Th22;
      hence b in Z by A3,A75,A81;
    end;
    hence b in Z by A67;
  end;
A82: R |_2 Z,S |_2 (rng F) are_isomorphic by A10,A17,Def8;
A83: now
    assume
A84: Z = field R;
    given a such that
A85: a in field S and
A86: rng F = S-Seg(a);
    take a;
    thus a in field S by A85;
    thus R,S |_2 (S-Seg(a)) are_isomorphic by A82,A84,A86,Th23;
  end;
A87: now
    let a such that
A88: a in rng F;
    consider c such that
A89: c in dom F & a = F.c by A88,FUNCT_1:def 3;
A90: [c,a] in F by A89,FUNCT_1:1;
    then
A91: a in field S by A9;
    let b such that
A92: [b,a] in S;
A93: R |_2 (R-Seg(c)),S |_2 (S-Seg(a)) are_isomorphic by A9,A90;
A94: c in field R by A9,A90;
    now
      set Q = S |_2 (S-Seg(a));
      set P = R |_2 (R-Seg(c));
      assume a <> b;
      then
A95:  b in S-Seg(a) by A92,Th1;
      then
A96: Q-Seg(b) = S-Seg(b) by A2,Th27;
A97: b in field S by A92,RELAT_1:15;
      then S-Seg(b) c= S-Seg(a) by A2,A91,A95,Th30;
      then
A98: Q |_2 (S-Seg(b)) = S |_2 (S-Seg(b)) by Th22;
      Q,P are_isomorphic & Q is well-ordering by A2,A93,Th25,Th40;
      then
A99: canonical_isomorphism_of(Q,P) is_isomorphism_of Q,P by Def9;
      field Q = S-Seg(a) by A2,Th32;
      then consider d such that
A100: d in field P and
A101: Q |_2 (Q-Seg(b)),P |_2 (P-Seg(d)) are_isomorphic by A2,A99,A95,Th25,Th50
;
A102: R-Seg(c) = field P by A1,Th32;
      then
A103: P-Seg(d) = R-Seg(d) by A1,A100,Th27;
      [d,c] in R by A100,A102,Th1;
      then
A104: d in field R by RELAT_1:15;
      then R-Seg(d) c= R-Seg(c) by A1,A94,A100,A102,Th30;
      then
S |_2 (S-Seg(b)),R |_2 (R-Seg(d)) are_isomorphic by A101,A96,A103,A98,Th22;
      then R |_2 (R-Seg(d)),S |_2 (S-Seg(b)) are_isomorphic by Th40;
      then [d,b] in F by A9,A97,A104;
      then d in dom F & b = F.d by FUNCT_1:1;
      hence b in rng F by FUNCT_1:def 3;
    end;
    hence b in rng F by A88;
  end;
A105: now
    assume
A106: rng F = field S;
    given a such that
A107: a in field R and
A108: Z = R-Seg(a);
    take a;
    thus a in field R by A107;
    thus R |_2 (R-Seg(a)),S are_isomorphic by A82,A106,A108,Th23;
  end;
  now
    assume
A109: Z = field R & rng F = field S;
    R |_2 field R = R & S |_2 field S = S by Th23;
    hence R,S are_isomorphic by A10,A17,A109,Def8;
  end;
  hence thesis by A1,A2,A4,A15,A66,A87,A61,A83,A105,Th28;
end;

theorem
  Y c= field R & R is well-ordering implies R,R |_2 Y are_isomorphic or
  ex a st a in field R & R |_2 (R-Seg(a)),R |_2 Y are_isomorphic
proof
  assume that
A1: Y c= field R and
A2: R is well-ordering;
A3: now
    given a such that
A4: a in field(R |_2 Y) and
A5: R,(R |_2 Y) |_2 ((R |_2 Y)-Seg(a)) are_isomorphic;
    consider F such that
A6: F is_isomorphism_of R,(R |_2 Y) |_2 ((R |_2 Y)-Seg(a)) by A5,Def8;
A7: now
      let c,b;
      assume
A8:   [c,b] in R & c <> b;
      then [F.c,F.b] in (R |_2 Y) |_2 ((R |_2 Y)-Seg(a)) by A6,Th36;
      then [F.c,F.b] in R |_2 Y by XBOOLE_0:def 4;
      hence [F.c,F.b] in R & F.c <> F.b by A6,A8,Th36,XBOOLE_0:def 4;
    end;
A9: field(R |_2 Y) = Y by A1,A2,Th31;
    field((R |_2 Y) |_2 ((R |_2 Y)-Seg(a))) = (R |_2 Y)-Seg(a) by A2,Th25,Th32;
    then
A10: rng F = (R |_2 Y)-Seg(a) by A6,Def7;
A11: dom F = field R by A6,Def7;
    then
A12: F.a in rng F by A1,A4,A9,FUNCT_1:def 3;
    then
A13: F.a <> a by A10,Th1;
    [F.a,a] in R |_2 Y by A10,A12,Th1;
    then
A14: [F.a,a] in R by XBOOLE_0:def 4;
    (R |_2 Y)-Seg(a) c= Y by A9,Th9;
    then rng F c= field R by A1,A10,XBOOLE_1:1;
    then [a,F.a] in R by A1,A2,A4,A9,A11,A7,Th35;
    hence contradiction by A13,A14,A2,Lm3;
  end;
  R |_2 Y is well-ordering by A2,Th25;
  hence thesis by A2,A3,Th52;
end;



theorem
  R,S are_isomorphic & R is well-ordering implies S is well-ordering
proof
  assume R,S are_isomorphic;
  then ex F being Function st F is_isomorphism_of R,S by Def8;
  hence thesis by Th44;
end;














begin

reserve A,B,X,X1,Y,Y1,Y2,Z,a,x,y,z for set;





definition
  let X,Y;
  mode Relation of X,Y is Subset of [:X,Y:];
end;

registration
  let X,Y;
  cluster -> Relation-like for Subset of [:X,Y:];
  coherence;
end;

registration
  let X,Y;
  cluster -> X-defined Y-valued for Relation of X,Y;
  coherence
  proof
    let R be Relation of X,Y;
    thus dom R c= X
    proof
      let x;
      assume x in dom R;
      then ex y st [x,y] in R by XTUPLE_0:def 12;
      hence thesis by ZFMISC_1:87;
    end;
    let y;
    assume y in rng R;
    then ex x st [x,y] in R by XTUPLE_0:def 13;
    hence thesis by ZFMISC_1:87;
  end;
end;

reserve P,R for Relation of X,Y;

definition
  let X,Y,R,Z;
  redefine pred R c= Z means
  for x being Element of X, y being Element of Y
  holds [x,y] in R implies [x,y] in Z;
  compatibility
  proof
    thus R c= Z implies for x being Element of X, y being Element of Y holds [
    x,y] in R implies [x,y] in Z;
    assume
A1: for x being Element of X, y being Element of Y holds [x,y] in R
    implies [x,y] in Z;
    let a,b be set;
    assume
A2: [a,b] in R;
    then reconsider a9=a as Element of X by ZFMISC_1:87;
    reconsider b9=b as Element of Y by A2,ZFMISC_1:87;
    [a9,b9] in Z by A1,A2;
    hence thesis;
  end;
end;

definition
  let X,Y,P,R;
  redefine pred P = R means
  for x being Element of X, y being Element of Y
  holds [x,y] in P iff [x,y] in R;
  compatibility
  proof
    thus P = R implies for x being Element of X, y being Element of Y holds [x
    ,y] in P iff [x,y] in R;
    assume
A1: for x being Element of X, y being Element of Y holds [x,y] in P
    iff [x,y] in R;
    let a,b be set;
    hereby
      assume
A2:   [a,b] in P;
      then reconsider a9=a as Element of X by ZFMISC_1:87;
      reconsider b9=b as Element of Y by A2,ZFMISC_1:87;
      [a9,b9] in R by A1,A2;
      hence [a,b] in R;
    end;
    assume
A3: [a,b] in R;
    then reconsider a9=a as Element of X by ZFMISC_1:87;
    reconsider b9=b as Element of Y by A3,ZFMISC_1:87;
    [a9,b9] in P by A1,A3;
    hence thesis;
  end;
end;

theorem
  A c= R implies A is Relation of X,Y by XBOOLE_1:1;

theorem
  a in R implies ex x,y st a = [x,y] & x in X & y in Y
proof
  assume
A1: a in R;
  then consider x,y such that
A2: a = [x,y] by RELAT_1:def 1;
  x in X & y in Y by A1,A2,ZFMISC_1:87;
  hence thesis by A2;
end;

theorem
  x in X & y in Y implies {[x,y]} is Relation of X,Y
proof
  assume x in X & y in Y;
  then [x,y] in [:X,Y:] by ZFMISC_1:87;
  hence thesis by ZFMISC_1:31;
end;

theorem
  for R being Relation st dom R c= X & rng R c= Y holds R is Relation of X,Y
proof
  let R be Relation;
  assume dom R c= X & rng R c= Y;
  then R c= [:dom R, rng R:] & [:dom R, rng R:] c= [:X,Y:] by RELAT_1:7
,ZFMISC_1:96;
  hence thesis by XBOOLE_1:1;
end;

theorem
  dom R c= X1 implies R is Relation of X1,Y
proof
A1: rng R c= Y by RELAT_1:def 19;
  assume dom R c= X1;
  then R c= [:dom R, rng R:] & [:dom R, rng R:] c= [:X1,Y:] by A1,RELAT_1:7
,ZFMISC_1:96;
  hence thesis by XBOOLE_1:1;
end;

theorem
  rng R c= Y1 implies R is Relation of X,Y1
proof
A1: dom R c= X by RELAT_1:def 18;
  assume rng R c= Y1;
  then R c= [:dom R, rng R:] & [:dom R, rng R:] c= [:X,Y1:] by A1,RELAT_1:7
,ZFMISC_1:96;
  hence thesis by XBOOLE_1:1;
end;

theorem
  X c= X1 & Y c= Y1 implies R is Relation of X1,Y1
proof
  assume X c= X1 & Y c= Y1;
  then [:X,Y:] c= [:X1,Y1:] by ZFMISC_1:96;
  hence thesis by XBOOLE_1:1;
end;

registration
  let X;
  let R,S be X-defined Relation;
  cluster R \/ S -> X-defined;
  coherence
  proof
A1: dom(R \/ S) = dom R \/ dom S by RELAT_1:1;
    dom R c= X & dom S c= X by RELAT_1:def 18;
    hence dom(R \/ S) c= X by A1,XBOOLE_1:8;
  end;
end;

registration
  let X;
  let R be X-defined Relation, S be Relation;
  cluster R /\ S -> X-defined;
  coherence
  proof
    R /\ S c= R by XBOOLE_1:17;
    then dom R c= X & dom(R /\ S) c= dom R by RELAT_1:11,def 18;
    hence dom(R /\ S) c= X by XBOOLE_1:1;
  end;
  cluster R \ S -> X-defined;
  coherence;
end;

registration
  let X;
  let R,S be X-valued Relation;
  cluster R \/ S -> X-valued;
  coherence
  proof
A1: rng(R \/ S) = rng R \/ rng S by RELAT_1:12;
    rng R c= X & rng S c= X by RELAT_1:def 19;
    hence rng(R \/ S) c= X by A1,XBOOLE_1:8;
  end;
end;

registration
  let X;
  let R be X-valued Relation, S be Relation;
  cluster R /\ S -> X-valued;
  coherence
  proof
    R /\ S c= R by XBOOLE_1:17;
    then rng R c= X & rng(R /\ S) c= rng R by RELAT_1:11,def 19;
    hence rng(R /\ S) c= X by XBOOLE_1:1;
  end;
  cluster R \ S -> X-valued;
  coherence;
end;

definition let X; let R be X-defined Relation;
 redefine func dom R -> Subset of X;
  coherence by RELAT_1:def 18;
end;

definition let X; let R be X-valued Relation;
 redefine func rng R -> Subset of X;
  coherence by RELAT_1:def 19;
end;

theorem
  field R c= X \/ Y
proof
  dom R \/ rng R c= X \/ Y by XBOOLE_1:13;
  hence thesis;
end;

theorem
  (for x st x in X ex y st [x,y] in R) iff dom R = X
proof
  thus (for x st x in X ex y st [x,y] in R) implies dom R = X
  proof
    assume
A1: for x st x in X ex y st [x,y] in R;
    now
      let x;
      now
        assume x in X;
        then ex y st [x,y] in R by A1;
        hence x in dom R by XTUPLE_0:def 12;
      end;
      hence x in dom R iff x in X;
    end;
    hence dom R = X by TARSKI:1;
  end;
  thus thesis by XTUPLE_0:def 12;
end;

theorem
  (for y st y in Y ex x st [x,y] in R) iff rng R = Y
proof
  thus (for y st y in Y ex x st [x,y] in R) implies rng R = Y
  proof
    assume
A1: for y st y in Y ex x st [x,y] in R;
    now
      let y;
      now
        assume y in Y;
        then ex x st [x,y] in R by A1;
        hence y in rng R by XTUPLE_0:def 13;
      end;
      hence y in rng R iff y in Y;
    end;
    hence rng R = Y by TARSKI:1;
  end;
  thus thesis by XTUPLE_0:def 13;
end;

definition
  let X,Y,R;
  redefine func R~ -> Relation of Y,X;
  coherence
  proof
    now
      let x,y;
      assume [x,y] in R~;
      then [y,x] in R by RELAT_1:def 7;
      hence [x,y] in [:Y,X:] by ZFMISC_1:88;
    end;
    hence thesis by RELAT_1:def 3;
  end;
end;

definition
  let X,Y1,Y2,Z;
  let P be Relation of X,Y1;
  let R be Relation of Y2,Z;
  redefine func P*R -> Relation of X,Z;
  coherence
  proof
    now
      let x,z;
      assume [x,z] in P*R;
      then ex y st [x,y] in P & [y,z] in R by RELAT_1:def 8;
      then x in X & z in Z by ZFMISC_1:87;
      hence [x,z] in [:X,Z:] by ZFMISC_1:87;
    end;
    hence thesis by RELAT_1:def 3;
  end;
end;

theorem
  dom (R~) = rng R & rng (R~) = dom R
proof
  now
    let x;
A1: now
      assume x in rng R;
      then consider y such that
A2:   [y,x] in R by XTUPLE_0:def 13;
      [x,y] in R~ by A2,RELAT_1:def 7;
      hence x in dom (R~) by XTUPLE_0:def 12;
    end;
    now
      assume x in dom (R~);
      then consider y such that
A3:   [x,y] in R~ by XTUPLE_0:def 12;
      [y,x] in R by A3,RELAT_1:def 7;
      hence x in rng R by XTUPLE_0:def 13;
    end;
    hence x in dom (R~) iff x in rng R by A1;
  end;
  hence dom (R~) = rng R by TARSKI:1;
  now
    let x;
A4: now
      assume x in dom R;
      then consider y such that
A5:   [x,y] in R by XTUPLE_0:def 12;
      [y,x] in R~ by A5,RELAT_1:def 7;
      hence x in rng (R~) by XTUPLE_0:def 13;
    end;
    now
      assume x in rng (R~);
      then consider y such that
A6:   [y,x] in R~ by XTUPLE_0:def 13;
      [x,y] in R by A6,RELAT_1:def 7;
      hence x in dom R by XTUPLE_0:def 12;
    end;
    hence x in rng (R~) iff x in dom R by A4;
  end;
  hence thesis by TARSKI:1;
end;

theorem
  {} is Relation of X,Y by XBOOLE_1:2;

registration
  let A be empty set, B be set;
  cluster -> empty for Relation of A,B;
  coherence;
  cluster -> empty for Relation of B,A;
  coherence;
end;

theorem Th13:
  id X c= [:X,X:]
proof
  [:X,X:] c= [:X,X:];
  then reconsider R = [:X,X:] as Relation of X,X;
  [x,y] in id X implies [x,y] in R
  proof
    assume [x,y] in id X;
    then x in X & x = y by RELAT_1:def 10;
    hence thesis by ZFMISC_1:87;
  end;
  hence thesis by RELAT_1:def 3;
end;

theorem
  id X is Relation of X,X by Th13;

theorem Th15:
  id A c= R implies A c= dom R & A c= rng R
proof
  assume
A1: id A c= R;
  thus A c= dom R
  proof
    let x;
    assume x in A;
    then [x,x] in id A by RELAT_1:def 10;
    hence thesis by A1,XTUPLE_0:def 12;
  end;
  thus A c= rng R
  proof
    let x;
    assume x in A;
    then [x,x] in id A by RELAT_1:def 10;
    hence thesis by A1,XTUPLE_0:def 13;
  end;
end;

theorem
  id X c= R implies X = dom R & X c= rng R
proof
  assume
A1: id X c= R;
  then X c= dom R by Th15;
  hence X = dom R by XBOOLE_0:def 10;
  thus thesis by A1,Th15;
end;

theorem
  id Y c= R implies Y c= dom R & Y = rng R
proof
  assume
A1: id Y c= R;
  hence Y c= dom R by Th15;
  Y c= rng R by A1,Th15;
  hence Y = rng R by XBOOLE_0:def 10;
end;

definition
  let X,Y,R,A;
  redefine func R|A -> Relation of X,Y;
  coherence
  proof
    now
      let x,y;
      assume [x,y] in R|A;
      then [x,y] in R by RELAT_1:def 11;
      hence [x,y] in [:X,Y:];
    end;
    hence thesis by RELAT_1:def 3;
  end;
end;

definition
  let X,Y,B,R;
  redefine func B|`R -> Relation of X,Y;
  coherence
  proof
    now
      let x,y;
      assume [x,y] in B|`R;
      then [x,y] in R by RELAT_1:def 12;
      hence [x,y] in [:X,Y:];
    end;
    hence thesis by RELAT_1:def 3;
  end;
end;

theorem
  R|X1 is Relation of X1,Y
proof
  now
    let x,y;
    assume [x,y] in R|X1;
    then x in X1 & y in Y by RELAT_1:def 11,ZFMISC_1:87;
    hence [x,y] in [:X1,Y:] by ZFMISC_1:87;
  end;
  hence thesis by RELAT_1:def 3;
end;

theorem
  X c= X1 implies R|X1 = R
proof
  assume
A1: X c= X1;
  now
    let x,y;
    now
      assume
A2:   [x,y] in R;
      then x in X by ZFMISC_1:87;
      hence [x,y] in R|X1 by A1,A2,RELAT_1:def 11;
    end;
    hence [x,y] in R|X1 iff [x,y] in R by RELAT_1:def 11;
  end;
  hence thesis by RELAT_1:def 2;
end;

theorem
  Y1|`R is Relation of X,Y1
proof
  now
    let x,y;
    assume [x,y] in Y1|`R;
    then y in Y1 & x in X by RELAT_1:def 12,ZFMISC_1:87;
    hence [x,y] in [:X,Y1:] by ZFMISC_1:87;
  end;
  hence thesis by RELAT_1:def 3;
end;

theorem
  Y c= Y1 implies Y1|`R = R
proof
  assume
A1: Y c= Y1;
  now
    let x,y;
    now
      assume
A2:   [x,y] in R;
      then y in Y by ZFMISC_1:87;
      hence [x,y] in Y1|`R by A1,A2,RELAT_1:def 12;
    end;
    hence [x,y] in Y1|`R iff [x,y] in R by RELAT_1:def 12;
  end;
  hence thesis by RELAT_1:def 2;
end;

definition
  let X,Y,R,A;
  redefine func R.:A -> Subset of Y;
  coherence
  proof
    R.:A c= rng R by RELAT_1:111;
    hence thesis by XBOOLE_1:1;
  end;
  redefine func R"A -> Subset of X;
  coherence
  proof
    R"A c= dom R by RELAT_1:132;
    hence thesis by XBOOLE_1:1;
  end;
end;

theorem Th22:
  R.:X = rng R & R"Y = dom R
proof
  now
    let y;
A1: now
      assume y in rng R;
      then consider x such that
A2:   [x,y] in R by XTUPLE_0:def 13;
      x in X by A2,ZFMISC_1:87;
      hence y in R.:X by A2,RELAT_1:def 13;
    end;
    now
      assume y in R.:X;
      then ex x st [x,y] in R & x in X by RELAT_1:def 13;
      hence y in rng R by XTUPLE_0:def 13;
    end;
    hence y in R.:X iff y in rng R by A1;
  end;
  hence R.:X = rng R by TARSKI:1;
  now
    let x;
A3: now
      assume x in dom R;
      then consider y such that
A4:   [x,y] in R by XTUPLE_0:def 12;
      y in Y by A4,ZFMISC_1:87;
      hence x in R"Y by A4,RELAT_1:def 14;
    end;
    now
      assume x in R"Y;
      then ex y st [x,y] in R & y in Y by RELAT_1:def 14;
      hence x in dom R by XTUPLE_0:def 12;
    end;
    hence x in R"Y iff x in dom R by A3;
  end;
  hence thesis by TARSKI:1;
end;

theorem
  R.:(R"Y) = rng R & R"(R.:X) = dom R
proof
  R"Y = dom R & R.:X = rng R by Th22;
  hence thesis by RELAT_1:113,134;
end;

scheme
  RelOnSetEx{A() -> set,B() -> set,P[set,set]}: ex R being Relation of A(),B()
  st for x,y holds [x,y] in R iff x in A() & y in B() & P[x,y] proof
  consider R being Relation such that
A1: for x,y holds [x,y] in R iff x in A() & y in B() & P[x,y] from
  RELAT_1:sch 1;
  R c= [:A(),B():]
  proof
    let x1,x2 be set;
    assume [x1,x2] in R;
    then x1 in A() & x2 in B() by A1;
    hence thesis by ZFMISC_1:87;
  end;
  then reconsider R as Relation of A(),B();
  take R;
  thus thesis by A1;
end;



definition
  let X;
  mode Relation of X is Relation of X,X;
end;

reserve D,D1,D2,E,F for non empty set;
reserve R for Relation of D,E;
reserve x for Element of D;
reserve y for Element of E;

registration
  let D be non empty set;
  cluster id D -> non empty;
  coherence
  proof
    now
      set y = the Element of D;
A1:   [y,y] in id D by RELAT_1:def 10;
      assume id D = {};
      hence contradiction by A1;
    end;
    hence thesis;
  end;
end;

theorem
  for x being Element of D holds x in dom R iff ex y being Element of E
  st [x,y] in R
proof
  let x be Element of D;
  thus x in dom R implies ex y being Element of E st [x,y] in R
  proof
    assume x in dom R;
    then consider y being set such that
A1: [x,y] in R by XTUPLE_0:def 12;
    reconsider b = y as Element of E by A1,ZFMISC_1:87;
    take b;
    thus thesis by A1;
  end;
  given y being Element of E such that
A2: [x,y] in R;
  thus thesis by A2,XTUPLE_0:def 12;
end;

theorem
  for y being set holds y in rng R iff ex x being Element of D st [x,y] in R
proof
  let y be set;
  thus y in rng R implies ex x being Element of D st [x,y] in R
  proof
    assume y in rng R;
    then consider x being set such that
A1: [x,y] in R by XTUPLE_0:def 13;
    reconsider a = x as Element of D by A1,ZFMISC_1:87;
    take a;
    thus thesis by A1;
  end;
  given x being Element of D such that
A2: [x,y] in R;
  thus thesis by A2,XTUPLE_0:def 13;
end;

theorem
  dom R <> {} implies ex y being Element of E st y in rng R
proof
  assume dom R <> {};
  then rng R <> {} by RELAT_1:42;
  then ex y being set st y in rng R by XBOOLE_0:def 1;
  hence thesis;
end;

theorem
  rng R <> {} implies ex x being Element of D st x in dom R
proof
  assume rng R <> {};
  then dom R <> {} by RELAT_1:42;
  then ex x being set st x in dom R by XBOOLE_0:def 1;
  hence thesis;
end;

theorem
  for P being (Relation of D,E), R being Relation of E,F for x, z being
  set holds [x,z] in P*R iff ex y being Element of E st [x,y] in P & [y,z] in R
proof
  let P be (Relation of D,E), R be Relation of E,F;
  let x, z be set;
  thus [x,z] in P*R implies ex y being Element of E st [x,y] in P & [y,z] in R
  proof
    assume [x,z] in P*R;
    then consider y being set such that
A1: [x,y] in P and
A2: [y,z] in R by RELAT_1:def 8;
    reconsider a = y as Element of E by A1,ZFMISC_1:87;
    take a;
    thus thesis by A1,A2;
  end;
  given y such that
A3: [x,y] in P & [y,z] in R;
  thus thesis by A3,RELAT_1:def 8;
end;

theorem
  y in R.:D1 iff ex x being Element of D st [x,y] in R & x in D1
proof
  thus y in R.:D1 implies ex x being Element of D st [x,y] in R & x in D1
  proof
    assume y in R.:D1;
    then consider x being set such that
A1: [x,y] in R and
A2: x in D1 by RELAT_1:def 13;
    reconsider a = x as Element of D by A1,ZFMISC_1:87;
    take a;
    thus thesis by A1,A2;
  end;
  given x such that
A3: [x,y] in R & x in D1;
  thus thesis by A3,RELAT_1:def 13;
end;

theorem
  x in R"D2 iff ex y being Element of E st [x,y] in R & y in D2
proof
  thus x in R"D2 implies ex y being Element of E st [x,y] in R & y in D2
  proof
    assume x in R"D2;
    then consider y being set such that
A1: [x,y] in R and
A2: y in D2 by RELAT_1:def 14;
    reconsider b = y as Element of E by A1,ZFMISC_1:87;
    take b;
    thus thesis by A1,A2;
  end;
  given y being Element of E such that
A3: [x,y] in R & y in D2;
  thus thesis by A3,RELAT_1:def 14;
end;

scheme
  RelOnDomEx{A,B() -> non empty set, P[set,set]}: ex R being Relation of A(),B
() st for x being Element of A(), y being Element of B() holds [x,y] in R iff P
  [x,y] proof
  consider R being Relation of A(),B() qua set such that
A1: for x,y being set holds [x,y] in R iff x in A() & y in B() & P[x,y]
  from RelOnSetEx;
  take R;
  thus thesis by A1;
end;

begin 


scheme
  { N()-> set, M() -> Subset of N(), F(set)->set }: ex R being Relation of M()
  st for i being Element of N() st i in M() holds Im(R,i) = F(i)
provided
A1: for i being Element of N() st i in M() holds F(i) c= M()
proof
  defpred P[set,set] means $2 in F($1);
  consider R being Relation of M() such that
A2: for x,y being set holds [x,y] in R iff x in M() & y in M() & P[x,y]
  from RelOnSetEx;
  take R;
  let i be Element of N();
  assume
A3: i in M();
  thus Im(R,i) c= F(i)
  proof
    let e be set;
    assume e in Im(R,i);
    then consider u being set such that
A4: [u,e] in R and
A5: u in {i} by RELAT_1:def 13;
    u = i by A5,TARSKI:def 1;
    hence thesis by A2,A4;
  end;
  let e be set;
  assume
A6: e in F(i);
  F(i) c= M() by A1,A3;
  then i in {i} & [i,e] in R by A2,A3,A6,TARSKI:def 1;
  hence thesis by RELAT_1:def 13;
end;

theorem
  for N being set, R,S being Relation of N st for i being set st i in N
  holds Im(R,i) = Im(S,i) holds R = S
proof
  let N be set, R,S be Relation of N such that
A1: for i being set st i in N holds Im(R,i) = Im(S,i);
  let a,b be Element of N;
  thus [a,b] in R implies [a,b] in S
  proof
    assume
A2: [a,b] in R;
    then
A3: a in dom R by XTUPLE_0:def 12;
    a in {a} by TARSKI:def 1;
    then b in Im(R,a) by A2,RELAT_1:def 13;
    then b in Im(S,a) by A1,A3;
    then ex e being set st [e,b] in S & e in {a} by RELAT_1:def 13;
    hence thesis by TARSKI:def 1;
  end;
  assume
A4: [a,b] in S;
  then
A5: a in dom S by XTUPLE_0:def 12;
  a in {a} by TARSKI:def 1;
  then b in Im(S,a) by A4,RELAT_1:def 13;
  then b in Im(R,a) by A1,A5;
  then ex e being set st [e,b] in R & e in {a} by RELAT_1:def 13;
  hence thesis by TARSKI:def 1;
end;



scheme
  {A,B() -> set, P[set,set], P,R()->Relation of A(), B()}: P() = R()
provided
A1: for p being Element of A(), q being Element of B() holds [p,q] in P(
) iff P[p,q] and
A2: for p being Element of A(), q being Element of B() holds [p,q] in R(
) iff P[p,q]
proof
  let y be Element of A(), z be Element of B();
  [y,z] in P() iff P[y,z] by A1;
  hence thesis by A2;
end;



registration
  let X,Y,Z;
  let f be Relation of [:X,Y:], Z;
  cluster dom f -> Relation-like;
  coherence;
end;

registration
  let X,Y,Z;
  let f be Relation of X, [:Y, Z:];
  cluster rng f -> Relation-like;
  coherence;
end;



theorem
  A misses X implies P|A = {}
proof
  assume A misses X;
  then A misses dom P by XBOOLE_1:63;
  hence thesis by RELAT_1:152;
end;



registration
  let R be non empty Relation, Y be non empty Subset of dom R;
  cluster R|Y -> non empty;
  coherence
  proof
    dom(R|Y) = Y by RELAT_1:62;
    hence thesis;
  end;
end;

registration
  let R be non empty Relation;
  let Y be non empty Subset of dom R;
  cluster R.:Y -> non empty;
  coherence
  proof
    R.:Y = rng(R|Y) by RELAT_1:115;
    hence thesis;
  end;
end;



registration
  let X,Y be set;
  cluster empty for Relation of X,Y;
  existence
  proof
    {} is Relation of X,Y by XBOOLE_1:2;
    hence thesis;
  end;
end;














begin

reserve x,x1,x2,y,y9,y1,y2,z,z1,z2,P,X,X1,X2,Y,Y1,Y2,V,Z for set;



theorem Th1:
  for f,g being Function st for x st x in dom f /\ dom g holds f.x
  = g.x ex h being Function st f \/ g = h
proof
  let f,g be Function such that
A1: for x st x in dom f /\ dom g holds f.x = g.x;
  defpred P[set,set] means [$1,$2] in f \/ g;
A2: for x,y1,y2 st P[x,y1] & P[x,y2] holds y1 = y2
  proof
    let x,y1,y2 such that
A3: [x,y1] in f \/ g and
A4: [x,y2] in f \/ g;
    now
      [x,y1] in f or [x,y1] in g by A3,XBOOLE_0:def 3;
      then
A5:   x in dom f & f.x = y1 or x in dom g & g.x = y1 by FUNCT_1:1;
A6:   [x,y2] in f or [x,y2] in g by A4,XBOOLE_0:def 3;
      then
A7:   x in dom f & f.x = y2 or x in dom g & g.x = y2 by FUNCT_1:1;
      per cases by A6,XTUPLE_0:def 12;
      suppose
        x in dom f & x in dom g;
        then x in dom f /\ dom g by XBOOLE_0:def 4;
        hence thesis by A1,A5,A7;
      end;
      suppose
        x in dom f & not x in dom g;
        hence thesis by A6,A5,FUNCT_1:1;
      end;
      suppose
        not x in dom f & x in dom g;
        hence thesis by A6,A5,FUNCT_1:1;
      end;
    end;
    hence thesis;
  end;
  consider h being Function such that
A8: for x,y holds [x,y] in h iff x in dom f \/ dom g & P[x,y] from
  FUNCT_1:sch 1(A2);
  take h;
  let x,y;
  thus [x,y] in f \/ g implies [x,y] in h
  proof
    assume
A9: [x,y] in f \/ g;
    then [x,y] in f or [x,y] in g by XBOOLE_0:def 3;
    then x in dom f or x in dom g by XTUPLE_0:def 12;
    then x in dom f \/ dom g by XBOOLE_0:def 3;
    hence thesis by A8,A9;
  end;
  thus thesis by A8;
end;

theorem Th2:
  for f,g,h being Function st f \/ g = h for x st x in dom f /\ dom
  g holds f.x = g.x
proof
  let f,g,h be Function such that
A1: f \/ g = h;
  let x;
  assume
A2: x in dom f /\ dom g;
  then x in dom f by XBOOLE_0:def 4;
  then
A3: h.x = f.x by A1,GRFUNC_1:15;
  x in dom g by A2,XBOOLE_0:def 4;
  hence thesis by A1,A3,GRFUNC_1:15;
end;

scheme
  LambdaC{A()->set,C[set],F(set)->set,G(set)->set}:
   ex f being Function st dom f = A() &
    for x st x in A() holds (C[x] implies f.x = F(x)) &
    (not C[x] implies f.x = G(x));
  defpred P[set,set] means (C[$1] implies $2 = F($1)) & (not C[$1] implies $2
  = G($1));
A1: for x st x in A() ex y st P[x,y]
  proof
    let x;
    not C[x] implies (C[x] implies G(x) = F(x)) & (not C[x] implies G(x) =
    G(x));
    hence thesis;
  end;
A2: for x,y1,y2 st x in A() & P[x,y1] & P[x,y2] holds y1 = y2;
  thus ex f being Function st dom f = A() & for x st x in A() holds P[x,f.x]
  from FUNCT_1:sch 2(A2,A1);
end;

Lm1: now
  let X,Y;
  take E = {};
  thus dom E c= X & rng E c= Y by XBOOLE_1:2;
end;

registration
  let X,Y;
  cluster Function-like for Relation of X,Y;
  existence
  proof
    consider E being Function such that
A1: dom E c= X & rng E c= Y by Lm1;
    reconsider E as Relation of X,Y by A1,RELSET_1:4;
    take E;
    thus thesis;
  end;
end;

definition
  let X,Y;
  mode PartFunc of X,Y is Function-like Relation of X,Y;
end;

theorem
  for f being PartFunc of X,Y st y in rng f ex x being Element of X st x
  in dom f & y = f.x
proof
  let f be PartFunc of X,Y;
  assume y in rng f;
  then ex x st x in dom f & y = f.x by FUNCT_1:def 3;
  hence thesis;
end;

theorem Th4:
  for f being Y-valued Function st x in dom f holds f.x in Y
proof
  let f be Y-valued Function;
  assume x in dom f;
  then
A1: f.x in rng f by FUNCT_1:def 3;
  rng f c= Y by RELAT_1:def 19;
  hence thesis by A1;
end;

theorem
  for f1,f2 being PartFunc of X,Y st dom f1 = dom f2 & for x being
  Element of X st x in dom f1 holds f1.x = f2.x holds f1 = f2
proof
  let f1,f2 be PartFunc of X,Y such that
A1: dom f1 = dom f2 and
A2: for x being Element of X st x in dom f1 holds f1.x = f2.x;
  for x st x in dom f1 holds f1.x = f2.x by A2;
  hence thesis by A1,FUNCT_1:2;
end;

scheme
  PartFuncEx{X,Y()->set,P[set,set]}: ex f being PartFunc of X(),Y() st (for x
holds x in dom f iff x in X() & ex y st P[x,y]) & for x st x in dom f holds P[x
  ,f.x]
provided
A1: for x,y st x in X() & P[x,y] holds y in Y() and
A2: for x,y1,y2 st x in X() & P[x,y1] & P[x,y2] holds y1 = y2
proof
A3: now
    defpred R[set] means ex y st P[$1,y];
    set y1 = the set;
    assume Y() <> {};
    defpred Q[set,set] means ((ex y st P[$1,y]) implies P[$1,$2]) & ((for y
    holds not P[$1,y]) implies $2=y1);
A4: for x st x in X() ex z st Q[x,z]
    proof
      let x such that
      x in X();
      (for y holds not P[x,y]) implies ((ex y st P[x,y]) implies P[x,y1])
      & ((for y holds not P[x,y]) implies y1=y1);
      hence thesis;
    end;
A5: for x,z1,z2 st x in X() & Q[x,z1] & Q[x,z2] holds z1 = z2 by A2;
    consider g being Function such that
A6: dom g = X() and
A7: for x st x in X() holds Q[x,g.x] from FUNCT_1:sch 2(A5,A4);
    consider X being set such that
A8: for x holds x in X iff x in X() & R[x] from XBOOLE_0:sch 1;
    set f=g|X;
A9: dom f c= X() by A6,RELAT_1:60;
    rng f c= Y()
    proof
      let y;
      assume y in rng f;
      then consider x such that
A10:  x in dom f and
A11:  y = f.x by FUNCT_1:def 3;
A12:  dom f c= X by RELAT_1:58;
      then x in X() & ex y st P[x,y] by A8,A10;
      then P[x,g.x] by A7;
      then
A13:  P[x,y] by A10,A11,FUNCT_1:47;
      x in X() by A8,A10,A12;
      hence thesis by A1,A13;
    end;
    then reconsider f as PartFunc of X(),Y() by A9,RELSET_1:4;
    take f;
    thus for x holds x in dom f iff x in X() & ex y st P[x,y]
    proof
      let x;
      dom f c= X by RELAT_1:58;
      hence x in dom f implies x in X() & ex y st P[x,y] by A8;
      assume that
A14:  x in X() and
A15:  ex y st P[x,y];
      x in X by A8,A14,A15;
      then x in dom g /\ X by A6,A14,XBOOLE_0:def 4;
      hence thesis by RELAT_1:61;
    end;
    let x;
    assume
A16: x in dom f;
    dom f c= X by RELAT_1:58;
    then ex y st P[x,y] by A8,A16;
    then P[x,g.x] by A7,A16;
    hence P[x,f.x] by A16,FUNCT_1:47;
  end;
  now
    consider f being Function such that
A17: dom f c= X() & rng f c= Y() by Lm1;
    reconsider f as PartFunc of X(),Y() by A17,RELSET_1:4;
    assume
A18: Y() = {};
    take f;
    thus for x holds x in dom f iff x in X() & ex y st P[x,y] by A1,A18;
    thus for x st x in dom f holds P[x,f.x] by A18;
  end;
  hence thesis by A3;
end;

scheme
  LambdaR{X,Y()->set,F(set)->set,P[set]}: ex f being PartFunc of X(),Y() st (
for x holds x in dom f iff x in X() & P[x]) & for x st x in dom f holds f.x = F
  (x)
provided
A1: for x st P[x] holds F(x) in Y()
proof
  defpred Q[set,set] means P[$1] & $2 = F($1);
A2: for x,y1,y2 st x in X() & Q[x,y1] & Q[x,y2] holds y1 = y2;
A3: for x,y st x in X() & Q[x,y] holds y in Y() by A1;
  consider f being PartFunc of X(),Y() such that
A4: for x holds x in dom f iff x in X() & ex y st Q[x,y] and
A5: for x st x in dom f holds Q[x,f.x] from PartFuncEx(A3,A2);
  take f;
  thus for x holds x in dom f iff x in X() & P[x]
  proof
    let x;
    thus x in dom f implies x in X() & P[x]
    proof
      assume
A6:   x in dom f;
      then ex y st P[x] & y = F(x) by A4;
      hence thesis by A6;
    end;
    assume that
A7: x in X() and
A8: P[x];
    ex y st P[x] & y = F(x) by A8;
    hence thesis by A4,A7;
  end;
  thus thesis by A5;
end;

definition
  let X,Y,V,Z;
  let f be PartFunc of X,Y;
  let g be PartFunc of V,Z;
  redefine func g*f -> PartFunc of X,Z;
  coherence
  proof
A1: dom(g*f) c= X;
    rng(g*f) c= Z by RELAT_1:def 19;
    hence thesis by A1,RELSET_1:4;
  end;
end;

theorem
  for f being Relation of X,Y holds (id X)*f = f
proof
  let f be Relation of X,Y;
  dom f c= X;
  hence thesis by RELAT_1:51;
end;

theorem
  for f being Relation of X,Y holds f*(id Y) = f
proof
  let f be Relation of X,Y;
  rng f c= Y;
  hence thesis by RELAT_1:53;
end;

theorem
  for f being PartFunc of X,Y st (for x1,x2 being Element of X st x1 in
  dom f & x2 in dom f & f.x1 = f.x2 holds x1 = x2) holds f is one-to-one
proof
  let f be PartFunc of X,Y;
  assume for x1,x2 being Element of X st x1 in dom f & x2 in dom f & f.x1 = f
  .x2 holds x1 = x2;
  then for x1,x2 st x1 in dom f & x2 in dom f & f.x1 = f.x2 holds x1 = x2;
  hence thesis by FUNCT_1:def 4;
end;

theorem
  for f being PartFunc of X,Y st f is one-to-one holds f" is PartFunc of Y,X
proof
  let f be PartFunc of X,Y such that
A1: f is one-to-one;
  rng f c= Y by RELAT_1:def 19;
  then
A2: dom(f") c= Y by A1,FUNCT_1:33;
  dom f c= X;
  then rng(f") c= X by A1,FUNCT_1:33;
  hence thesis by A2,RELSET_1:4;
end;

theorem
  for f being PartFunc of X,Y holds f|Z is PartFunc of Z,Y
proof
  let f be PartFunc of X,Y;
  dom(f|Z) c= Z & rng(f|Z) c= Y by RELAT_1:58,def 19;
  hence thesis by RELSET_1:4;
end;

theorem Th11:
  for f being PartFunc of X,Y holds f|Z is PartFunc of X,Y;

definition
  let X,Y;
  let f be PartFunc of X,Y;
  let Z be set;
  redefine func f|Z -> PartFunc of X,Y;
  coherence by Th11;
end;

theorem
  for f being PartFunc of X,Y holds Z|`f is PartFunc of X,Z
proof
  let f be PartFunc of X,Y;
  dom(Z|`f) c= X & rng(Z|`f) c= Z by RELAT_1:85;
  hence thesis by RELSET_1:4;
end;

theorem
  for f being PartFunc of X,Y holds Z|`f is PartFunc of X,Y;

theorem Th14:
  for f being Function holds Y|`f|X is PartFunc of X,Y
proof
  let f be Function;
  Y|`f|X = Y|`(f|X) by RELAT_1:109;
  then dom(Y|`f|X) c= X & rng(Y|`f|X) c= Y by RELAT_1:58,85;
  hence thesis by RELSET_1:4;
end;

theorem
  for f being PartFunc of X,Y st y in f.:X ex x being Element of X st x
  in dom f & y = f.x
proof
  let f be PartFunc of X,Y;
  assume y in f.:X;
  then ex x st x in dom f & x in X & y = f.x by FUNCT_1:def 6;
  hence thesis;
end;



theorem Th16:
  for f being PartFunc of {x},Y holds rng f c= {f.x}
proof
  let f be PartFunc of {x},Y;
  dom f = {} or dom f = {x} by ZFMISC_1:33;
  then rng f = {} or rng f = {f.x} by FUNCT_1:4,RELAT_1:42;
  hence thesis by ZFMISC_1:33;
end;

theorem
  for f being PartFunc of {x},Y holds f is one-to-one
proof
  let f be PartFunc of {x},Y;
  let x1,x2;
  assume that
A1: x1 in dom f and
A2: x2 in dom f;
  dom f<>{} implies x1 = x & x2 = x by A1,A2,TARSKI:def 1;
  hence thesis by A1;
end;

theorem
  for f being PartFunc of {x},Y holds f.:P c= {f.x}
proof
  let f be PartFunc of {x},Y;
  f.:P c= rng f & rng f c= {f.x} by Th16,RELAT_1:111;
  hence thesis by XBOOLE_1:1;
end;

theorem
  for f being Function st dom f = {x} & x in X & f.x in Y holds f is
  PartFunc of X,Y
proof
  let f be Function;
  assume that
A1: dom f = {x} and
A2: x in X and
A3: f.x in Y;
  rng f = {f.x} by A1,FUNCT_1:4;
  then
A4: rng f c= Y by A3,ZFMISC_1:31;
  dom f c= X by A1,A2,ZFMISC_1:31;
  hence thesis by A4,RELSET_1:4;
end;



theorem Th20:
  for f being PartFunc of X,{y} st x in dom f holds f.x = y
proof
  let f be PartFunc of X,{y};
  x in dom f implies f.x in {y} by Th4;
  hence thesis by TARSKI:def 1;
end;

theorem
  for f1,f2 being PartFunc of X,{y} st dom f1 = dom f2 holds f1 = f2
proof
  let f1,f2 be PartFunc of X,{y} such that
A1: dom f1 = dom f2;
  x in dom f1 implies f1.x = f2.x
  proof
    assume
A2: x in dom f1;
    then f1.x = y by Th20;
    hence thesis by A1,A2,Th20;
  end;
  hence thesis by A1,FUNCT_1:2;
end;



definition
  let f be Function;
  let X,Y be set;

  func <:f,X,Y:> -> PartFunc of X,Y equals
  Y|`f|X;
  coherence by Th14;
end;

theorem Th22:
  for f being Function holds <:f,X,Y:> c= f
proof
  let f be Function;
  (Y|`f|X) c= (Y|`f) & (Y|`f) c= f by RELAT_1:59,86;
  hence thesis by XBOOLE_1:1;
end;

theorem Th23:
  for f being Function holds dom <:f,X,Y:> c= dom f & rng <:f,X,Y :> c= rng f
proof
  let f be Function;
  <:f,X,Y:> c= f by Th22;
  hence thesis by RELAT_1:11;
end;

theorem Th24:
  for f being Function holds x in dom <:f,X,Y:> iff x in dom f & x
  in X & f.x in Y
proof
  let f be Function;
  thus x in dom <:f,X,Y:> implies x in dom f & x in X & f.x in Y
  proof
    assume
A1: x in dom <:f,X,Y:>;
    then x in dom(Y|`f) /\ X by RELAT_1:61;
    then x in dom(Y|`f) by XBOOLE_0:def 4;
    hence thesis by A1,FUNCT_1:54;
  end;
  assume that
A2: x in dom f and
A3: x in X and
A4: f.x in Y;
  x in dom(Y|`f) by A2,A4,FUNCT_1:54;
  then x in dom(Y|`f) /\ X by A3,XBOOLE_0:def 4;
  hence thesis by RELAT_1:61;
end;

theorem Th25:
  for f being Function st x in dom f & x in X & f.x in Y holds <:f
  ,X,Y:>.x = f.x
proof
  let f be Function such that
A1: x in dom f and
A2: x in X and
A3: f.x in Y;
  x in dom(Y|`f) by A1,A3,FUNCT_1:54;
  then f.x = (Y|`f).x by FUNCT_1:55
    .= (Y|`f|X).x by A2,FUNCT_1:49;
  hence thesis;
end;

theorem Th26:
  for f being Function st x in dom <:f,X,Y:> holds <:f,X,Y:>.x = f .x
proof
  let f be Function;
  assume
A1: x in dom <:f,X,Y:>;
  then x in dom f & f.x in Y by Th24;
  hence thesis by A1,Th25;
end;

theorem
  for f,g being Function st f c= g holds <:f,X,Y:> c= <:g,X,Y:>
proof
  let f,g be Function such that
A1: f c= g;
A2: dom <:f,X,Y:> c= dom f by Th23;
  now
    thus
A3: dom <:f,X,Y:> c= dom <:g,X,Y:>
    proof
      let x;
A4:   dom f c= dom g by A1,RELAT_1:11;
      assume
A5:   x in dom <:f,X,Y:>;
      then
A6:   f.x = g.x by A1,A2,GRFUNC_1:2;
      x in dom f & f.x in Y by A5,Th24;
      hence thesis by A5,A4,A6,Th24;
    end;
    let x;
    assume
A7: x in dom <:f,X,Y:>;
    then
A8: <:f,X,Y:>.x = f.x by Th26;
    <:g,X,Y:>.x = g.x by A3,A7,Th26;
    hence <:f,X,Y:>.x = <:g,X,Y:>.x by A1,A2,A7,A8,GRFUNC_1:2;
  end;
  hence thesis by GRFUNC_1:2;
end;

theorem Th28:
  for f being Function st Z c= X holds <:f,Z,Y:> c= <:f,X,Y:>
proof
  let f be Function such that
A1: Z c= X;
A2: dom <:f,Z,Y:> c= dom <:f,X,Y:>
  proof
    let x;
    assume
A3: x in dom <:f,Z,Y:>;
    then
A4: f.x in Y by Th24;
    x in Z & x in dom f by A3,Th24;
    hence thesis by A1,A4,Th24;
  end;
  now
    let x;
    assume
A5: x in dom <:f,Z,Y:>;
    then <:f,Z,Y:>.x = f.x by Th26;
    hence <:f,Z,Y:>.x = <:f,X,Y:>.x by A2,A5,Th26;
  end;
  hence thesis by A2,GRFUNC_1:2;
end;

theorem Th29:
  for f being Function st Z c= Y holds <:f,X,Z:> c= <:f,X,Y:>
proof
  let f be Function such that
A1: Z c= Y;
A2: dom <:f,X,Z:> c= dom <:f,X,Y:>
  proof
    let x;
    assume
A3: x in dom <:f,X,Z:>;
    then f.x in Z & x in dom f by Th24;
    hence thesis by A1,A3,Th24;
  end;
  now
    let x;
    assume
A4: x in dom <:f,X,Z:>;
    then <:f,X,Z:>.x = f.x by Th26;
    hence <:f,X,Z:>.x = <:f,X,Y:>.x by A2,A4,Th26;
  end;
  hence thesis by A2,GRFUNC_1:2;
end;

theorem
  for f being Function st X1 c= X2 & Y1 c= Y2 holds <:f,X1,Y1:> c= <:f, X2,Y2:>
proof
  let f be Function;
  assume X1 c= X2 & Y1 c= Y2;
  then <:f,X1,Y1:> c= <:f,X2,Y1:> & <:f,X2,Y1:> c= <:f,X2,Y2:> by Th28,Th29;
  hence thesis by XBOOLE_1:1;
end;

theorem Th31:
  for f being Function st dom f c= X & rng f c= Y holds f = <:f,X, Y:>
proof
  let f be Function such that
A1: dom f c= X & rng f c= Y;
A2: dom f c= dom <:f,X,Y:>
  proof
    let x;
    assume
A3: x in dom f;
    then f.x in rng f by FUNCT_1:def 3;
    hence thesis by A1,A3,Th24;
  end;
  dom <:f,X,Y:> c= dom f by Th23;
  then
A4: dom f = dom <:f,X,Y:> by A2,XBOOLE_0:def 10;
  for x st x in dom f holds f.x = <:f,X,Y:>.x by A2,Th26;
  hence thesis by A4,FUNCT_1:2;
end;

theorem
  for f being Function holds f = <:f,dom f,rng f:>;

theorem
  for f being PartFunc of X,Y holds <:f,X,Y:> = f
proof
  let f be PartFunc of X,Y;
  thus thesis;
end;

theorem Th34:
  <:{},X,Y:> = {}
proof
  dom {} c= X & rng {} c= Y by XBOOLE_1:2;
  hence thesis by Th31;
end;

theorem Th35:
  for f,g being Function holds (<:g,Y,Z:>*<:f,X,Y:>) c= <:g*f,X,Z :>
proof
  let f,g be Function;
A1: for x st x in dom (<:g,Y,Z:>*<:f,X,Y:>) holds (<:g,Y,Z:>*<:f,X,Y:>).x =
  <:g*f,X,Z:>.x
  proof
    let x;
    assume
A2: x in dom (<:g,Y,Z:>*<:f,X,Y:>);
    then
A3: x in dom<:f,X,Y:> by FUNCT_1:11;
    then
A4: x in dom f by Th24;
    <:f,X,Y:>.x in dom<:g,Y,Z:> by A2,FUNCT_1:11;
    then
A5: f.x in dom<:g,Y,Z:> by A3,Th26;
    then g.(f.x) in Z by Th24;
    then
A6: (g*f).x in Z by A4,FUNCT_1:13;
    f.x in dom g by A5,Th24;
    then x in dom (g*f) by A4,FUNCT_1:11;
    then
A7: x in dom <:g*f,X,Z:> by A2,A6,Th24;
    thus (<:g,Y,Z:>*<:f,X,Y:>).x = <:g,Y,Z:>.(<:f,X,Y:>.x) by A2,FUNCT_1:12
      .= <:g,Y,Z:>.(f.x) by A3,Th26
      .= g.(f.x) by A5,Th26
      .= (g*f).x by A4,FUNCT_1:13
      .= <:g*f,X,Z:>.x by A7,Th26;
  end;
  dom (<:g,Y,Z:>*<:f,X,Y:>) c= dom <:g*f,X,Z:>
  proof
    let x;
    assume
A8: x in dom (<:g,Y,Z:>*<:f,X,Y:>);
    then
A9: x in dom<:f,X,Y:> by FUNCT_1:11;
    then
A10: x in dom f by Th24;
    <:f,X,Y:>.x in dom<:g,Y,Z:> by A8,FUNCT_1:11;
    then
A11: f.x in dom<:g,Y,Z:> by A9,Th26;
    then g.(f.x) in Z by Th24;
    then
A12: (g*f).x in Z by A10,FUNCT_1:13;
    f.x in dom g by A11,Th24;
    then x in dom (g*f) by A10,FUNCT_1:11;
    hence thesis by A8,A12,Th24;
  end;
  hence thesis by A1,GRFUNC_1:2;
end;

theorem
  for f,g being Function st rng f /\ dom g c= Y holds <:g,Y,Z:>*<:f,X,Y
  :> = <:g*f,X,Z:>
proof
  let f,g be Function such that
A1: rng f /\ dom g c= Y;
A2: dom <:g*f,X,Z:> c= dom (<:g,Y,Z:>*<:f,X,Y:>)
  proof
    let x;
    assume
A3: x in dom <:g*f,X,Z:>;
    then
A4: x in dom (g*f) by Th24;
    then
A5: f.x in dom g by FUNCT_1:11;
A6: x in dom f by A4,FUNCT_1:11;
    then f.x in rng f by FUNCT_1:def 3;
    then
A7: f.x in rng f /\ dom g by A5,XBOOLE_0:def 4;
    (g*f).x in Z by A3,Th24;
    then g.(f.x) in Z by A4,FUNCT_1:12;
    then
A8: f.x in dom <:g,Y,Z:> by A1,A5,A7,Th24;
    x in dom <:f,X,Y:> & <:f,X,Y:>.x = f.x by A1,A3,A6,A7,Th24,Th25;
    hence thesis by A8,FUNCT_1:11;
  end;
  for x st x in dom <:g*f,X,Z:> holds <:g*f,X,Z:>.x = (<:g,Y,Z:>*<:f,X,Y :>).x
  proof
    let x;
    assume
A9: x in dom <:g*f,X,Z:>;
    then
A10: x in dom (g*f) by Th24;
    then
A11: f.x in dom g by FUNCT_1:11;
    x in dom f by A10,FUNCT_1:11;
    then f.x in rng f by FUNCT_1:def 3;
    then
A12: f.x in rng f /\ dom g by A11,XBOOLE_0:def 4;
    (g*f).x in Z by A9,Th24;
    then g.(f.x) in Z by A10,FUNCT_1:12;
    then
A13: f.x in dom <:g,Y,Z:> by A1,A11,A12,Th24;
    x in dom f by A10,FUNCT_1:11;
    then
A14: x in dom <:f,X,Y:> by A1,A9,A12,Th24;
    thus <:g*f,X,Z:>.x = (g*f).x by A9,Th26
      .= g.(f.x) by A10,FUNCT_1:12
      .= <:g,Y,Z:>.(f.x) by A13,Th26
      .= <:g,Y,Z:>.(<:f,X,Y:>.x) by A14,Th26
      .= (<:g,Y,Z:>*<:f,X,Y:>).x by A2,A9,FUNCT_1:12;
  end;
  then
A15: <:g*f,X,Z:> c= (<:g,Y,Z:>*<:f,X,Y:>) by A2,GRFUNC_1:2;
  (<:g,Y,Z:>*<:f,X,Y:>) c= <:g*f,X,Z:> by Th35;
  hence thesis by A15,XBOOLE_0:def 10;
end;

theorem Th37:
  for f being Function st f is one-to-one holds <:f,X,Y:> is one-to-one
proof
  let f be Function;
  assume f is one-to-one;
  then Y|`f is one-to-one by FUNCT_1:58;
  hence thesis by FUNCT_1:52;
end;

theorem
  for f being Function st f is one-to-one holds <:f,X,Y:>" = <:f",Y,X:>
proof
  let f be Function;
  assume
A1: f is one-to-one;
  then
A2: <:f,X,Y:> is one-to-one by Th37;
  y in dom (<:f,X,Y:>") iff y in dom <:f",Y,X:>
  proof
    thus y in dom (<:f,X,Y:>") implies y in dom <:f",Y,X:>
    proof
      assume y in dom (<:f,X,Y:>");
      then
A3:   y in rng <:f,X,Y:> by A2,FUNCT_1:33;
      then consider x such that
A4:   x in dom <:f,X,Y:> and
A5:   y = <:f,X,Y:>.x by FUNCT_1:def 3;
A6:   f.x = y by A4,A5,Th26;
      then
A7:   y in Y by A4,Th24;
      rng <:f,X,Y:> c= rng f by Th23;
      then y in rng f by A3;
      then
A8:   y in dom(f") by A1,FUNCT_1:32;
      dom <:f,X,Y:> c= dom f by Th23;
      then (f").y = x by A1,A4,A6,FUNCT_1:32;
      hence thesis by A4,A8,A7,Th24;
    end;
    assume
A9: y in dom <:f",Y,X:>;
    dom <:f",Y,X:> c= dom (f") by Th23;
    then y in dom(f") by A9;
    then y in rng f by A1,FUNCT_1:33;
    then consider x such that
A10: x in dom f and
A11: y = f.x by FUNCT_1:def 3;
    x =(f").(f.x) by A1,A10,FUNCT_1:34;
    then x in X by A9,A11,Th24;
    then x in dom <:f,X,Y:> by A9,A10,A11,Th24;
    then <:f,X,Y:>.x in rng <:f,X,Y:> & <:f,X,Y:>.x = f.x by Th26,FUNCT_1:def 3
;
    hence thesis by A2,A11,FUNCT_1:33;
  end;
  then
A12: dom (<:f,X,Y:>") = dom <:f",Y,X:> by TARSKI:1;
  for y st y in dom <:f",Y,X:> holds <:f",Y,X:>.y = (<:f,X,Y:>").y
  proof
    let y;
A13: rng <:f,X,Y:> c= rng f by Th23;
    assume
A14: y in dom <:f",Y,X:>;
    then y in rng <:f,X,Y:> by A2,A12,FUNCT_1:33;
    then consider x such that
A15: x in dom f and
A16: y = f.x by A13,FUNCT_1:def 3;
A17: x =(f").(f.x) by A1,A15,FUNCT_1:34;
    then x in X by A14,A16,Th24;
    then x in dom<:f,X,Y:> by A14,A15,A16,Th24;
    then (<:f,X,Y:>").(<:f,X,Y:>.x) = x & <:f,X,Y:>.x = f.x by A2,Th26,
FUNCT_1:34;
    hence thesis by A14,A16,A17,Th26;
  end;
  hence thesis by A12,FUNCT_1:2;
end;

theorem
  for f being Function holds Z|`<:f,X,Y:> = <:f,X,Z /\ Y:>
proof
  let f be Function;
  thus Z|`<:f,X,Y:> = Z|`(Y|`(f|X)) by RELAT_1:109
    .= (Z /\ Y)|`(f|X) by RELAT_1:96
    .= <:f,X,Z /\ Y:> by RELAT_1:109;
end;



definition
  let X;
  let f be X-defined Relation;
  attr f is total means
:Def2:
  dom f = X;
end;

registration
  let X be empty set, Y be set;
  cluster -> total for Relation of X,Y;
  coherence
  proof
    let R be Relation of X,Y;
    thus dom R = X;
  end;
end;

registration
  let X be non empty set, Y be empty set;
  cluster -> non total for Relation of X,Y;
  coherence
  proof
    let f be Relation of X,Y;
    assume f is total;
    then dom f = X by Def2;
    hence thesis;
  end;
end;

theorem Th40:
  for f being Function st <:f,X,Y:> is total holds X c= dom f
proof
  let f be Function such that
A1: dom <:f,X,Y:> = X;
A2: dom <:f,X,Y:> c= dom f by Th23;
  let x;
  assume x in X;
  hence thesis by A1,A2;
end;

theorem
  <:{},X,Y:> is total implies X = {}
 proof
   dom {} = {};
  hence thesis by Th40,XBOOLE_1:3;
 end;

theorem
  for f being Function st X c= dom f & rng f c= Y holds <:f,X,Y:> is total
proof
  let f be Function such that
A1: X c= dom f and
A2: rng f c= Y;
  X c= dom <:f,X,Y:>
  proof
    let x;
    assume
A3: x in X;
    then f.x in rng f by A1,FUNCT_1:def 3;
    hence thesis by A1,A2,A3,Th24;
  end;
  hence dom <:f,X,Y:> = X by XBOOLE_0:def 10;
end;

theorem
  for f being Function st <:f,X,Y:> is total holds f.:X c= Y
proof
  let f be Function such that
A1: dom <:f,X,Y:> = X;
  let y;
A2: rng <:f,X,Y:> c= Y by RELAT_1:def 19;
  assume y in f.:X;
  then consider x such that
  x in dom f and
A3: x in X & y = f.x by FUNCT_1:def 6;
  <:f,X,Y:>.x = y & <:f,X,Y:>.x in rng <:f,X,Y:> by A1,A3,Th26,FUNCT_1:def 3;
  hence thesis by A2;
end;

theorem
  for f being Function st X c= dom f & f.:X c= Y holds <:f,X,Y:> is total
proof
  let f be Function such that
A1: X c= dom f and
A2: f.:X c= Y;
  X c= dom <:f,X,Y:>
  proof
    let x;
    assume
A3: x in X;
    then f.x in f.:X by A1,FUNCT_1:def 6;
    hence thesis by A1,A2,A3,Th24;
  end;
  hence dom <:f,X,Y:> = X by XBOOLE_0:def 10;
end;



definition
  let X,Y;
  func PFuncs(X,Y) -> set means
  :Def3:
  x in it iff ex f being Function st x = f & dom f c= X & rng f c= Y;
  existence
  proof
    defpred P[set] means ex f be Function st $1 = f & dom f c= X & rng f c= Y;
    consider F being set such that
A1: z in F iff z in bool [:X,Y:] & P[z] from XBOOLE_0:sch 1;
    take F;
    let z;
    thus z in F implies ex f being Function st z = f & dom f c= X & rng f c= Y
    by A1;
    given f being Function such that
A2: z = f and
A3: dom f c= X & rng f c= Y;
    f c= [:X,Y:]
    proof
      let x,y;
      assume
A4:   [x,y] in f;
      then
A5:   x in dom f by XTUPLE_0:def 12;
      then y = f.x by A4,FUNCT_1:def 2;
      then y in rng f by A5,FUNCT_1:def 3;
      hence thesis by A3,A5,ZFMISC_1:def 2;
    end;
    hence thesis by A1,A2,A3;
  end;
  uniqueness
  proof
    let F1,F2 be set such that
A6: x in F1 iff ex f being Function st x = f & dom f c= X & rng f c= Y and
A7: x in F2 iff ex f being Function st x = f & dom f c= X & rng f c= Y;
    x in F1 iff x in F2
    proof
      x in F1 iff ex f being Function st x = f & dom f c= X & rng f c= Y by A6;
      hence thesis by A7;
    end;
    hence thesis by TARSKI:1;
  end;
end;

registration
  let X,Y;
  cluster PFuncs(X,Y) -> non empty;
  coherence
  proof
    ex f being Function st dom f c= X & rng f c= Y by Lm1;
    hence thesis by Def3;
  end;
end;

theorem Th45:
  for f being PartFunc of X,Y holds f in PFuncs(X,Y)
proof
  let f be PartFunc of X,Y;
  dom f c= X & rng f c= Y by RELAT_1:def 19;
  hence thesis by Def3;
end;

theorem Th46:
  for f being set st f in PFuncs(X,Y) holds f is PartFunc of X,Y
proof
  let f be set;
  assume f in PFuncs(X,Y);
  then ex F being Function st f = F & dom F c= X & rng F c= Y by Def3;
  hence thesis by RELSET_1:4;
end;

theorem
  for f being Element of PFuncs(X,Y) holds f is PartFunc of X,Y by Th46;

theorem
  PFuncs({},Y) = { {} }
proof
  x in PFuncs({},Y) iff x = {}
  proof
    thus x in PFuncs({},Y) implies x = {}
    proof
      assume x in PFuncs({},Y);
      then x is PartFunc of {},Y by Th46;
      hence thesis;
    end;
    {} is PartFunc of {},Y by RELSET_1:12;
    hence thesis by Th45;
  end;
  hence thesis by TARSKI:def 1;
end;

theorem
  PFuncs(X,{}) = { {} }
proof
  x in PFuncs(X,{}) iff x = {}
  proof
    thus x in PFuncs(X,{}) implies x = {}
    proof
      assume x in PFuncs(X,{});
      then x is PartFunc of X,{} by Th46;
      hence thesis;
    end;
    {} is PartFunc of X,{} by RELSET_1:12;
    hence thesis by Th45;
  end;
  hence thesis by TARSKI:def 1;
end;

theorem
  X1 c= X2 & Y1 c= Y2 implies PFuncs(X1,Y1) c= PFuncs(X2,Y2)
proof
  assume
A1: X1 c= X2 & Y1 c= Y2;
  let f be set;
  assume f in PFuncs(X1,Y1);
  then f is PartFunc of X1,Y1 by Th46;
  then f is PartFunc of X2,Y2 by A1,RELSET_1:7;
  hence thesis by Th45;
end;



definition
  let f,g be Function;
  pred f tolerates g means
  :Def4:
  for x st x in dom f /\ dom g holds f.x = g.x;
  reflexivity;
  symmetry;
end;

theorem Th51:
  for f,g being Function holds f tolerates g iff ex h being
  Function st f \/ g = h
proof
  let f,g be Function;
  (for x st x in dom f /\ dom g holds f.x = g.x) iff ex h being Function
  st f \/ g = h by Th1,Th2;
  hence thesis by Def4;
end;

theorem Th52:
  for f,g being Function holds f tolerates g iff ex h being
  Function st f c= h & g c= h
proof
  let f,g be Function;
  now
    thus (ex h being Function st f c= h & g c= h) implies ex h being Function
    st f \/ g = h
    proof
      given h being Function such that
A1:   f c= h & g c= h;
      f \/ g is Function by A1,GRFUNC_1:1,XBOOLE_1:8;
      hence thesis;
    end;
    given h being Function such that
A2: f \/ g = h;
    f c= h & g c= h by A2,XBOOLE_1:7;
    hence ex h being Function st f c= h & g c= h;
  end;
  hence thesis by Th51;
end;

theorem Th53:
  for f,g being Function st dom f c= dom g holds f tolerates g
  iff for x st x in dom f holds f.x = g.x
proof
  let f,g be Function;
  assume dom f c= dom g;
  then dom f /\ dom g = dom f by XBOOLE_1:28;
  hence thesis by Def4;
end;

theorem
  for f,g being Function st f c= g holds f tolerates g by Th52;

theorem Th55:
  for f,g being Function st dom f = dom g & f tolerates g holds f = g
proof
  let f,g be Function;
  assume that
A1: dom f = dom g and
A2: f tolerates g;
  for x st x in dom f holds f.x = g.x by A1,A2,Th53;
  hence thesis by A1,FUNCT_1:2;
end;

theorem
  for f,g being Function st dom f misses dom g holds f tolerates g
proof
  let f,g be Function;
  assume dom f misses dom g;
  then f \/ g is Function by GRFUNC_1:13;
  hence thesis by Th51;
end;

theorem
  for f,g,h being Function st f c= h & g c= h holds f tolerates g by Th52;

theorem
  for f,g being PartFunc of X,Y for h being Function st f tolerates h &
  g c= f holds g tolerates h
proof
  let f,g be (PartFunc of X,Y),h be Function such that
A1: f tolerates h and
A2: g c= f;
A3: dom g c= dom f by A2,RELAT_1:11;
  let x;
  assume
A4: x in dom g /\ dom h;
  then
A5: x in dom g by XBOOLE_0:def 4;
  then
A6: f.x = g.x by A2,GRFUNC_1:2;
  x in dom h by A4,XBOOLE_0:def 4;
  then x in dom f /\ dom h by A5,A3,XBOOLE_0:def 4;
  hence thesis by A1,A6,Def4;
end;

theorem Th59:
  for f being Function holds {} tolerates f
proof
  let f be Function;
  {} \/ f = f;
  hence thesis by Th51;
end;

theorem
  for f being Function holds <:{},X,Y:> tolerates f
proof
  let f be Function;
  <:{},X,Y:> = {} by Th34;
  hence thesis by Th59;
end;

theorem
  for f,g being PartFunc of X,{y} holds f tolerates g
proof
  let f,g be PartFunc of X,{y};
  let x;
  assume
A1: x in dom f /\ dom g;
  then x in dom f by XBOOLE_0:def 4;
  then
A2: f.x = y by Th20;
  x in dom g by A1,XBOOLE_0:def 4;
  hence thesis by A2,Th20;
end;

theorem
  for f being Function holds f|X tolerates f
proof
  let f be Function;
  (f|X) c= f by RELAT_1:59;
  hence thesis by Th52;
end;

theorem
  for f being Function holds Y|`f tolerates f
proof
  let f be Function;
  (Y|`f) c= f by RELAT_1:86;
  hence thesis by Th52;
end;

theorem Th64:
  for f being Function holds Y|`f|X tolerates f
proof
  let f be Function;
  (Y|`f|X) c= (Y|`f) & (Y|`f) c= f by RELAT_1:59,86;
  then (Y|`f|X) c= f by XBOOLE_1:1;
  hence thesis by Th52;
end;

theorem
  for f being Function holds <:f,X,Y:> tolerates f by Th64;

theorem Th66:
  for f,g being PartFunc of X,Y st f is total & g is total & f
  tolerates g holds f = g
proof
  let f,g be PartFunc of X,Y;
  assume dom f = X & dom g = X;
  hence thesis by Th55;
end;

theorem Th67:
  for f,g,h being PartFunc of X,Y st f tolerates h & g tolerates
  h & h is total holds f tolerates g
proof
  let f,g,h be PartFunc of X,Y such that
A1: f tolerates h and
A2: g tolerates h and
A3: dom h = X;
  let x;
  assume
A4: x in dom f /\ dom g;
  then x in dom f by XBOOLE_0:def 4;
  then x in dom f /\ dom h by A3,XBOOLE_0:def 4;
  then
A5: f.x = h.x by A1,Def4;
  x in dom g by A4,XBOOLE_0:def 4;
  then x in dom g /\ dom h by A3,XBOOLE_0:def 4;
  hence thesis by A2,A5,Def4;
end;

theorem Th68:
  for f,g being PartFunc of X,Y st (Y = {} implies X = {}) & f
  tolerates g ex h being PartFunc of X,Y st h is total & f tolerates h & g
  tolerates h
proof
  let f,g be PartFunc of X,Y such that
A1: Y = {} implies X = {} and
A2: f tolerates g;
  now
    per cases;
    suppose
A3:   Y = {};
      then f tolerates <:{},X,Y:> & g tolerates <:{},X,Y:>;
      hence thesis by A1,A3;
    end;
    suppose
A4:   Y <> {};
      set y = the Element of Y;
      defpred P[set,set] means ($1 in dom f implies $2 = f.$1) & ($1 in dom g
      implies $2 = g.$1) & (not $1 in dom f & not $1 in dom g implies $2 = y);
A5:   for x st x in X ex y9 st P[x,y9]
      proof
        let x such that
        x in X;
        now
          per cases;
          suppose
A6:         x in dom f & x in dom g;
            take y9 = f.x;
            thus x in dom f implies y9 = f.x;
            x in dom f /\ dom g by A6,XBOOLE_0:def 4;
            hence x in dom g implies y9 = g.x by A2,Def4;
            thus not x in dom f & not x in dom g implies y9 = y by A6;
          end;
          suppose
A7:         x in dom f & not x in dom g;
            take y9 = f.x;
            thus (x in dom f implies y9 = f.x) & (x in dom g implies y9 = g.x)
            & (not x in dom f & not x in dom g implies y9 = y) by A7;
          end;
          suppose
A8:         not x in dom f & x in dom g;
            take y9 = g.x;
            thus (x in dom f implies y9 = f.x) & (x in dom g implies y9 = g.x)
            & (not x in dom f & not x in dom g implies y9 = y) by A8;
          end;
          suppose
            not x in dom f & not x in dom g;
            hence thesis;
          end;
        end;
        hence thesis;
      end;
A9:   for x,y1,y2 st x in X & P[x,y1] & P[x,y2] holds y1 = y2;
      consider h being Function such that
A10:  dom h = X and
A11:  for x st x in X holds P[x,h.x] from FUNCT_1:sch 2(A9,A5);
      rng h c= Y
      proof
        let z;
        assume z in rng h;
        then consider x such that
A12:    x in dom h and
A13:    z = h.x by FUNCT_1:def 3;
        per cases;
        suppose
A14:      x in dom f & x in dom g;
          then z = f.x by A11,A13;
          hence thesis by A14,Th4;
        end;
        suppose
A15:      x in dom f & not x in dom g;
          then z = f.x by A11,A13;
          hence thesis by A15,Th4;
        end;
        suppose
A16:      not x in dom f & x in dom g;
          then z = g.x by A11,A13;
          hence thesis by A16,Th4;
        end;
        suppose
          not x in dom f & not x in dom g;
          then z = y by A10,A11,A12,A13;
          hence thesis by A4;
        end;
      end;
      then reconsider h9 = h as PartFunc of X,Y by A10,RELSET_1:4;
A17:  f tolerates h
      proof
        let x;
        assume x in dom f /\ dom h;
        then x in dom f by XBOOLE_0:def 4;
        hence thesis by A11;
      end;
A18:  g tolerates h
      proof
        let x;
        assume x in dom g /\ dom h;
        then x in dom g by XBOOLE_0:def 4;
        hence thesis by A11;
      end;
      h9 is total by A10,Def2;
      hence thesis by A17,A18;
    end;
  end;
  hence thesis;
end;

definition
  let X,Y;
  let f be PartFunc of X,Y;
  func TotFuncs f -> set means
  :Def5:
  x in it iff ex g being PartFunc of X,Y st g = x & g is total & f tolerates g;
  existence
  proof
    defpred P[set] means ex g being PartFunc of X,Y st g = $1 & g is total & f
    tolerates g;
    now
      consider F being set such that
A1:   for x holds x in F iff x in PFuncs(X,Y) & P[x] from XBOOLE_0:sch
      1;
      take F;
      let x;
      thus x in F implies ex g being PartFunc of X,Y st g = x & g is total & f
      tolerates g by A1;
      given g being PartFunc of X,Y such that
A2:   g = x & g is total & f tolerates g;
      g in PFuncs(X,Y) by Th45;
      hence x in F by A1,A2;
    end;
    hence thesis;
  end;
  uniqueness
  proof
    defpred P[set] means ex g being PartFunc of X,Y st g = $1 & g is total & f
    tolerates g;
    let F1,F2 be set such that
A3: for x holds x in F1 iff P[x] and
A4: for x holds x in F2 iff P[x];
    thus thesis from XBOOLE_0:sch 2(A3,A4);
  end;
end;

theorem Th69:
  for f being PartFunc of X,Y for g being set st g in TotFuncs(f)
  holds g is PartFunc of X,Y
proof
  let f be PartFunc of X,Y;
  let g be set;
  assume g in TotFuncs(f);
  then
  ex g9 being PartFunc of X,Y st g9 = g & g9 is total & f tolerates g9 by Def5;
  hence thesis;
end;

theorem Th70:
  for f,g being PartFunc of X,Y st g in TotFuncs(f) holds g is total
proof
  let f,g be PartFunc of X,Y;
  assume g in TotFuncs(f);
  then
  ex g9 being PartFunc of X,Y st g9 = g & g9 is total & f tolerates g9 by Def5;
  hence thesis;
end;

theorem Th71:
  for f being PartFunc of X,Y for g being Function st g in
  TotFuncs(f) holds f tolerates g
proof
  let f be PartFunc of X,Y;
  let g be Function;
  assume g in TotFuncs(f);
  then
  ex g9 being PartFunc of X,Y st g9 = g & g9 is total & f tolerates g9 by Def5;
  hence thesis;
end;

registration
  let X be non empty set, Y be empty set;
  let f be PartFunc of X,Y;
  cluster TotFuncs f -> empty;
  coherence
  proof
    set g = the Element of TotFuncs(f);
    assume TotFuncs(f) is not empty;
    then
    ex g9 being PartFunc of X,{} st g9 = g & g9 is total & f tolerates g9
    by Def5;
    hence contradiction;
  end;
end;

theorem Th72:
  for f being PartFunc of X,Y holds f is total iff TotFuncs f = { f}
proof
  let f be PartFunc of X,Y;
  thus f is total implies TotFuncs f = {f}
  proof
    assume
A1: f is total;
    for g being set holds g in TotFuncs f iff f = g
    proof
      let g be set;
      thus g in TotFuncs f implies f = g
      proof
        assume g in TotFuncs f;
        then
        ex g9 being PartFunc of X,Y st g9 = g & g9 is total & f tolerates
        g9 by Def5;
        hence thesis by A1,Th66;
      end;
      thus thesis by A1,Def5;
    end;
    hence thesis by TARSKI:def 1;
  end;
  assume TotFuncs f = {f};
  then f in TotFuncs f by TARSKI:def 1;
  hence thesis by Th70;
end;

theorem
  for f being PartFunc of {},Y holds TotFuncs f = {f} by Th72;

theorem
  for f being PartFunc of {},Y holds TotFuncs f = {{}} by Th72;

theorem
  for f,g being PartFunc of X,Y st TotFuncs f meets TotFuncs g holds f
  tolerates g
proof
  let f,g be PartFunc of X,Y;
  set h = the Element of TotFuncs f /\ TotFuncs g;
  assume
A1: TotFuncs f /\ TotFuncs g <> {};
  then
A2: h in TotFuncs f by XBOOLE_0:def 4;
A3: h in TotFuncs g by A1,XBOOLE_0:def 4;
  reconsider h as PartFunc of X,Y by A2,Th69;
A4: g tolerates h by A3,Th71;
  f tolerates h by A2,Th71;
  hence thesis by A2,A4,Th67,Th70;
end;

theorem
  for f,g being PartFunc of X,Y st (Y = {} implies X = {}) & f tolerates
  g holds TotFuncs f meets TotFuncs g
proof
  let f,g be PartFunc of X,Y;
  assume ( Y = {} implies X = {})& f tolerates g;
  then consider h being PartFunc of X,Y such that
A1: h is total & f tolerates h & g tolerates h by Th68;
  h in TotFuncs f & h in TotFuncs g by A1,Def5;
  hence TotFuncs f /\ TotFuncs g <> {} by XBOOLE_0:def 4;
end;

begin 

Lm2: for R being Relation of X st R = id X holds R is total
proof
  let R be Relation of X;
  assume R = id X;
  hence dom R = X by RELAT_1:45;
end;

Lm3: for R being Relation st R = id X holds R is reflexive symmetric
antisymmetric transitive
proof
  let R be Relation;
  assume
A1: R = id X;
  then
A2: dom R = X & rng R = X by RELAT_1:45;
  thus R is_reflexive_in field R
  proof
    let x;
    thus thesis by A1,A2,RELAT_1:def 10;
  end;
  thus R is_symmetric_in field R
  proof
    let x,y;
    assume that
    x in field R and
    y in field R;
    assume [x,y] in R;
    hence thesis by A1,RELAT_1:def 10;
  end;
  thus R is_antisymmetric_in field R
  proof
    let x;
    thus thesis by A1,RELAT_1:def 10;
  end;
  thus R is_transitive_in field R
  proof
    let x;
    thus thesis by A1,RELAT_1:def 10;
  end;
end;

Lm4: id X is Relation of X
proof
  dom id X c= X & rng id X c= X;
  hence thesis by RELSET_1:4;
end;

registration
  let X;
  cluster total reflexive symmetric antisymmetric transitive for Relation of X;
  existence
  proof
    reconsider R = id X as Relation of X by Lm4;
    take R;
    thus thesis by Lm2,Lm3;
  end;
end;

registration
  cluster symmetric transitive -> reflexive for Relation;
  coherence
  proof
    let R be Relation;
    assume that
A1: R is_symmetric_in field R and
A2: R is_transitive_in field R;
    let x;
    assume
A3: x in field R;
    then x in dom R or x in rng R by XBOOLE_0:def 3;
    then consider y such that
A4: [x,y] in R or [y,x] in R by XTUPLE_0:def 12,def 13;
    y in rng R or y in dom R by A4,XTUPLE_0:def 12,def 13;
    then
A5: y in field R by XBOOLE_0:def 3;
    then [x,y] in R & [y,x] in R by A1,A3,A4,RELAT_2:def 3;
    hence thesis by A2,A3,A5,RELAT_2:def 8;
  end;
end;

registration
  let X;
  cluster id X -> symmetric antisymmetric transitive;
  coherence by Lm3;
end;

definition
  let X;
  redefine func id X -> total Relation of X;
  coherence by Lm2,Lm4;
end;

scheme
  LambdaC9{ A() -> non empty set, C[set], F,G(set) -> set } :
   ex f being Function st dom f = A() &
    for x be Element of A() holds
     (C[x] implies f.x = F(x)) & (not C[x] implies f.x = G(x));
  consider f being Function such that
A1: dom f = A() & for x be set st x in A() holds (C[x] implies f.x = F(x
  )) & (not C[x] implies f.x = G(x)) from LambdaC;
  take f;
  thus thesis by A1;
end;

begin 


reserve A for set,
  f,g,h for Function;

theorem Th77:
  f tolerates g & [x,y] in f & [x,z] in g implies y = z
proof
  assume f tolerates g;
  then consider h such that
A1: h = f \/ g by Th51;
  assume [x,y] in f & [x,z] in g;
  then [x,y] in h & [x,z] in h by A1,XBOOLE_0:def 3;
  hence thesis by FUNCT_1:def 1;
end;

theorem
  A is functional & (for f,g being Function st f in A & g in A holds f
  tolerates g) implies union A is Function
proof
  assume that
A1: for x st x in A holds x is Function and
A2: for f,g being Function st f in A & g in A holds f tolerates g;
A3: now
    let x,y,z;
    assume that
A4: [x,y] in union A and
A5: [x,z] in union A;
    consider p being set such that
A6: [x,y] in p and
A7: p in A by A4,TARSKI:def 4;
    consider q being set such that
A8: [x,z] in q and
A9: q in A by A5,TARSKI:def 4;
    reconsider p,q as Function by A1,A7,A9;
    p tolerates q by A2,A7,A9;
    hence y = z by A6,A8,Th77;
  end;
  now
    let z;
    assume z in union A;
    then consider p being set such that
A10: z in p and
A11: p in A by TARSKI:def 4;
    reconsider p as Function by A1,A11;
    p = p;
    hence ex x,y st [x,y] = z by A10,RELAT_1:def 1;
  end;
  hence thesis by A3,FUNCT_1:def 1,RELAT_1:def 1;
end;



definition
  let D be set, p be D-valued Function, i be set;
  assume
A1: i in dom p;
  func p/.i -> Element of D equals
:Def6: p.i;
  coherence by A1,Th4;
end;



registration
  let X,Y be non empty set;
  cluster non empty for PartFunc of X,Y;
  existence
  proof
    reconsider p = {[choose X,choose Y]} as PartFunc of X,Y by RELSET_1:3;
    take p;
    thus thesis;
  end;
end;



registration
  let A, B be set;
  cluster PFuncs(A,B) -> functional;
  coherence
  proof
    let x be set;
    assume x in PFuncs(A,B);
    then ex f being Function st x = f & dom f c= A & rng f c= B by Def3;
    hence thesis;
  end;
end;



theorem
  for f1,f2, g being Function st rng g c= dom f1 & rng g c= dom f2
  & f1 tolerates f2 holds f1*g = f2*g
proof
  let f1,f2, g be Function;
  assume that
A1: rng g c= dom f1 and
A2: rng g c= dom f2 and
A3: f1 tolerates f2;
A4: dom (f2*g) = dom g by A2,RELAT_1:27;
A5: dom (f1*g) = dom g by A1,RELAT_1:27;
  now
    let x be set;
    assume
A6: x in dom g;
    then
A7: (f2*g).x = f2.(g.x) by A4,FUNCT_1:12;
    g.x in rng g by A6,FUNCT_1:def 3;
    then
A8: g.x in (dom f1) /\ (dom f2) by A1,A2,XBOOLE_0:def 4;
    (f1*g).x = f1.(g.x) by A5,A6,FUNCT_1:12;
    hence (f1*g).x = (f2*g).x by A3,A7,A8,Def4;
  end;
  hence thesis by A5,A4,FUNCT_1:2;
end;



theorem
 for f being Y-valued Function st x in dom(f|X)
  holds  (f|X)/.x = f/.x
proof let f be Y-valued Function;
 assume
A1: x in dom(f|X);
  then
A2: x in dom f by RELAT_1:57;
 thus (f|X)/.x = (f|X).x by A1,Def6
      .= f.x by A1,FUNCT_1:47
      .= f/.x by A2,Def6;
end;














begin

reserve v,x,x1,x2,x3,x4,y,y1,y2,y3,y4,z,z1,z2, X,X1,X2,X3,X4,Y,Y1,Y2,Y3,Y4,Y5,
  Z,Z1,Z2,Z3,Z4,Z5 for set;

reserve p for pair set;









canceled 6;

theorem Th7:
  [x,y]`1=x & [x,y]`2=y;

theorem Th8:
  [p`1,p`2] = p;

theorem
  X <> {} implies ex v st v in X & not ex x,y st (x in X or y in X) & v
  = [x,y]
proof
  assume X <> {};
  then consider Y such that
A1: Y in X and
A2: not ex Y1 st Y1 in Y & not Y1 misses X by XREGULAR:2;
  take v = Y;
  thus v in X by A1;
  given x,y such that
A3: x in X or y in X and
A4: v = [x,y];
A5: { x,y } in Y by A4,TARSKI:def 2;
  x in { x,y } & y in { x,y } by TARSKI:def 2;
  hence contradiction by A2,A5,A3,XBOOLE_0:3;
end;




theorem Th10:
  z in [:X,Y:] implies z`1 in X & z`2 in Y
proof
  assume z in [:X,Y:];
  then consider x,y such that
W: x in X & y in Y & z=[x,y] by ZFMISC_1:def 2;
   [x,y]`1 = x & [x,y]`2 = y;
  hence thesis by W;
end;

theorem
  (ex x,y st z=[x,y]) & z`1 in X & z`2 in Y implies z in [:X,Y:]
proof
  assume ex x,y st z=[x,y];
  then [z`1,z`2]=z by Th8;
  hence thesis by ZFMISC_1:def 2;
end;

theorem
  z in [:{x},Y:] implies z`1=x & z`2 in Y
proof
  assume
A1: z in [:{x},Y:];
  then z`1 in {x} by Th10;
  hence thesis by A1,Th10,TARSKI:def 1;
end;

theorem
  z in [:X,{y}:] implies z`1 in X & z`2 = y
proof
  assume
A1: z in [:X,{y}:];
  then z`2 in {y} by Th10;
  hence thesis by A1,Th10,TARSKI:def 1;
end;

theorem
  z in [:{x},{y}:] implies z`1 = x & z`2 = y
proof
  assume z in [:{x},{y}:];
  then z`1 in {x} & z`2 in {y} by Th10;
  hence thesis by TARSKI:def 1;
end;

theorem
  z in [:{x1,x2},Y:] implies (z`1=x1 or z`1=x2) & z`2 in Y
proof
  assume
A1: z in [:{x1,x2},Y:];
  then z`1 in {x1,x2} by Th10;
  hence thesis by A1,Th10,TARSKI:def 2;
end;

theorem
  z in [:X,{y1,y2}:] implies z`1 in X & (z`2 = y1 or z`2 = y2)
proof
  assume
A1: z in [:X,{y1,y2}:];
  then z`2 in {y1,y2} by Th10;
  hence thesis by A1,Th10,TARSKI:def 2;
end;

theorem
  z in [:{x1,x2},{y}:] implies (z`1=x1 or z`1=x2) & z`2 = y
proof
  assume z in [:{x1,x2},{y}:];
  then z`1 in {x1,x2} & z`2 in {y} by Th10;
  hence thesis by TARSKI:def 1,def 2;
end;

theorem
  z in [:{x},{y1,y2}:] implies z`1 = x & (z`2 = y1 or z`2 = y2)
proof
  assume z in [:{x},{y1,y2}:];
  then z`1 in {x} & z`2 in {y1,y2} by Th10;
  hence thesis by TARSKI:def 1,def 2;
end;

theorem
  z in [:{x1,x2},{y1,y2}:] implies (z`1 = x1 or z`1 = x2) & (z`2 = y1 or
  z`2 = y2)
proof
  assume z in [:{x1,x2},{y1,y2}:];
  then z`1 in {x1,x2} & z`2 in {y1,y2} by Th10;
  hence thesis by TARSKI:def 2;
end;

theorem Th20:
  (ex y,z st x = [y,z]) implies x <> x`1 & x <> x`2
proof
  given y,z such that
A1: x = [y,z];
X: [y,z]`1 = y & [y,z]`2 = z;
  now
    assume y = x;
    then {{y,z},{y}} in {y} by A1,TARSKI:def 1;
    hence contradiction by TARSKI:def 2;
  end;
  hence x <> x`1 by A1,X;
  now
    assume z = x;
    then {{y,z},{y}} in {y,z} by A1,TARSKI:def 2;
    hence contradiction by TARSKI:def 2;
  end;
  hence thesis by A1,X;
end;




reserve R for Relation;

theorem Th21:
  x in R implies x = [x`1,x`2]
proof
  assume x in R;
  then ex x1,x2 st x=[x1,x2] by RELAT_1:def 1;
  hence thesis by Th8;
end;

theorem
  X <> {} & Y <> {} implies for x being Element of [:X,Y:] holds x
  = [x`1,x`2] by Th21;

Lm1: X1 <> {} & X2 <> {} implies for x being Element of [:X1,X2:] ex xx1 being
(Element of X1), xx2 being Element of X2 st x = [xx1,xx2]

proof
  assume
A1: X1 <> {} & X2 <> {};
  let x be Element of [:X1,X2:];
  reconsider xx2 = x`2 as Element of X2 by A1,Th10;
  reconsider xx1 = x`1 as Element of X1 by A1,Th10;
  take xx1,xx2;
  thus thesis by A1,Th21;
end;

theorem Th23:
  [:{x1,x2},{y1,y2}:] = {[x1,y1],[x1,y2],[x2,y1],[x2,y2]}
proof
  thus [:{x1,x2},{y1,y2}:] = [:{x1},{y1,y2}:] \/ [:{x2},{y1,y2}:] by
ZFMISC_1:109
    .= {[x1,y1],[x1,y2]} \/ [:{x2},{y1,y2}:] by ZFMISC_1:30
    .= {[x1,y1],[x1,y2]} \/ {[x2,y1],[x2,y2]} by ZFMISC_1:30
    .= {[x1,y1],[x1,y2],[x2,y1],[x2,y2]} by ENUMSET1:5;
end;

theorem Th24:
  X <> {} & Y <> {} implies for x being Element of [:X,Y:] holds x
  <> x`1 & x <> x`2
proof
  assume
A1: X <> {} & Y <> {};
  let x be Element of [:X,Y:];
  x = [x`1,x`2] by A1,Th21;
  hence thesis by Th20;
end;





canceled;

theorem Th26:
  X <> {} implies ex v st v in X & not ex x,y,z st (x in X or y in
  X) & v = [x,y,z]
proof
  assume X <> {};
  then consider Y such that
A1: Y in X and
A2: not ex Y1,Y2,Y3 st Y1 in Y2 & Y2 in Y3 & Y3 in Y & not Y1 misses X
  by XREGULAR:4;
  take v = Y;
  thus v in X by A1;
  given x,y,z such that
A3: x in X or y in X and
A4: v = [x,y,z];
  set Y1 = { x,y }, Y2 = { Y1,{x} }, Y3 = { Y2,z };
B5: x in Y1 & y in Y1 by TARSKI:def 2;
A6: Y3 in Y by A4,TARSKI:def 2;
  Y1 in Y2 & Y2 in Y3 by TARSKI:def 2;
  hence contradiction by A2,B5,A6,A3,XBOOLE_0:3;
end;





canceled 3;

theorem Th30:
  X <> {} implies ex v st v in X & not ex x1,x2,x3,x4 st (x1 in X
  or x2 in X) & v = [x1,x2,x3,x4]
proof
  assume X <> {};
  then consider Y such that
A1: Y in X and
A2: for Y1,Y2,Y3,Y4,Y5 st Y1 in Y2 & Y2 in Y3 & Y3 in Y4 & Y4 in Y5 & Y5
  in Y holds Y1 misses X by XREGULAR:6;
  take v = Y;
  thus v in X by A1;
  given x1,x2,x3,x4 such that
A3: x1 in X or x2 in X and
A4: v = [x1,x2,x3,x4];
  set Y1 = { x1,x2 }, Y2 = { Y1,{x1} }, Y3 = { Y2,x3 }, Y4 = { Y3, {Y2}}, Y5 =
  { Y4,x4 };
A5: Y3 in Y4 & Y4 in Y5 by TARSKI:def 2;
A6: Y5 in Y by A4,TARSKI:def 2;
B7: x1 in Y1 & x2 in Y1 by TARSKI:def 2;
  Y1 in Y2 & Y2 in Y3 by TARSKI:def 2;
  hence contradiction by A2,B7,A5,A6,A3,XBOOLE_0:3;
end;





theorem Th31:
  X1 <> {} & X2 <> {} & X3 <> {} iff [:X1,X2,X3:] <> {}
proof
A1: X1 <> {} & X2 <> {} iff [:X1,X2:] <> {} by ZFMISC_1:90;
  [:[:X1,X2:],X3:] = [:X1,X2,X3:] by ZFMISC_1:def 3;
  hence thesis by A1,ZFMISC_1:90;
end;

reserve xx1 for Element of X1,
  xx2 for Element of X2,
  xx3 for Element of X3;

theorem Th32:
  X1<>{} & X2<>{} & X3<>{} implies ( [:X1,X2,X3:] = [:Y1,Y2,Y3:]
  implies X1=Y1 & X2=Y2 & X3=Y3)
proof
A1: [:X1,X2,X3:] = [:[:X1,X2:],X3:] by ZFMISC_1:def 3;
  assume
A2: X1<>{} & X2<>{};
  assume
A4: X3<>{};
  assume [:X1,X2,X3:] = [:Y1,Y2,Y3:];
  then
A5: [:[:X1,X2:],X3:] = [:[:Y1,Y2:],Y3:] by A1,ZFMISC_1:def 3;
  then [:X1,X2:] = [:Y1,Y2:] by A2,A4,ZFMISC_1:110;
  hence thesis by A2,A4,A5,ZFMISC_1:110;
end;

theorem
  [:X1,X2,X3:]<>{} & [:X1,X2,X3:] = [:Y1,Y2,Y3:] implies X1=Y1 & X2=Y2 &
  X3=Y3
proof
  assume
A1: [:X1,X2,X3:]<>{};
  then
A2: X3<>{} by Th31;
  X1<>{} & X2<>{} by A1,Th31;
  hence thesis by A2,Th32;
end;

theorem
  [:X,X,X:] = [:Y,Y,Y:] implies X = Y
proof
  assume [:X,X,X:] = [:Y,Y,Y:];
  then X<>{} or Y<>{} implies thesis by Th32;
  hence thesis;
end;

Lm2: X1 <> {} & X2 <> {} & X3 <> {} implies
 for x being Element of [:X1,X2,X3:]
  ex xx1,xx2,xx3 st x = [xx1,xx2,xx3]

proof
  assume that
A1: X1 <> {} & X2 <> {} and
A2: X3 <> {};
  let x be Element of [:X1,X2,X3:];
  reconsider x9=x as Element of [:[:X1,X2:],X3:] by ZFMISC_1:def 3;
  consider x12 being (Element of [:X1,X2:]), xx3 such that
A3: x9 = [x12,xx3] by A1,A2,Lm1;
  consider xx1,xx2 such that
A4: x12 = [xx1,xx2] by A1,Lm1;
  take xx1,xx2,xx3;
  thus thesis by A3,A4;
end;

theorem Th35:
  [:{x1},{x2},{x3}:] = { [x1,x2,x3] }
proof
  thus [:{x1},{x2},{x3}:] = [:[:{x1},{x2}:],{x3}:] by ZFMISC_1:def 3
    .= [:{[x1,x2]},{x3}:] by ZFMISC_1:29
    .= { [x1,x2,x3] } by ZFMISC_1:29;
end;

theorem
  [:{x1,y1},{x2},{x3}:] = { [x1,x2,x3],[y1,x2,x3] }
proof
  thus [:{x1,y1},{x2},{x3}:] = [:[:{x1,y1},{x2}:],{x3}:] by ZFMISC_1:def 3
    .= [:{[x1,x2],[y1,x2]},{x3}:] by ZFMISC_1:30
    .= { [x1,x2,x3],[y1,x2,x3] } by ZFMISC_1:30;
end;

theorem
  [:{x1},{x2,y2},{x3}:] = { [x1,x2,x3],[x1,y2,x3] }
proof
  thus [:{x1},{x2,y2},{x3}:] = [:[:{x1},{x2,y2}:],{x3}:] by ZFMISC_1:def 3
    .= [:{[x1,x2],[x1,y2]},{x3}:] by ZFMISC_1:30
    .= { [x1,x2,x3],[x1,y2,x3] } by ZFMISC_1:30;
end;

theorem
  [:{x1},{x2},{x3,y3}:] = { [x1,x2,x3],[x1,x2,y3] }
proof
  thus [:{x1},{x2},{x3,y3}:] = [:[:{x1},{x2}:],{x3,y3}:] by ZFMISC_1:def 3
    .= [:{[x1,x2]},{x3,y3}:] by ZFMISC_1:29
    .= { [x1,x2,x3],[x1,x2,y3] } by ZFMISC_1:30;
end;

theorem
  [:{x1,y1},{x2,y2},{x3}:] = { [x1,x2,x3],[y1,x2,x3],[x1,y2,x3],[y1,y2,
  x3] }
proof
  thus [:{x1,y1},{x2,y2},{x3}:] = [:[:{x1,y1},{x2,y2}:],{x3}:] by
ZFMISC_1:def 3
    .= [:{[x1,x2],[x1,y2],[y1,x2],[y1,y2]},{x3}:] by Th23
    .= [:{[x1,x2],[x1,y2]} \/ {[y1,x2],[y1,y2]},{x3}:] by ENUMSET1:5
    .= [:{[x1,x2],[x1,y2]},{x3}:] \/ [:{[y1,x2],[y1,y2]},{x3}:] by ZFMISC_1:97
    .= { [[x1,x2],x3],[[x1,y2],x3]} \/ [:{[y1,x2],[y1,y2]},{x3}:] by
ZFMISC_1:30
    .= { [[x1,x2],x3],[[x1,y2],x3]} \/ {[[y1,x2],x3],[[y1,y2],x3] } by
ZFMISC_1:30
    .= { [x1,x2,x3],[x1,y2,x3],[y1,x2,x3],[[y1,y2],x3] } by ENUMSET1:5
    .= { [x1,x2,x3],[y1,x2,x3],[x1,y2,x3],[y1,y2,x3] } by ENUMSET1:62;
end;

theorem
  [:{x1,y1},{x2},{x3,y3}:] = { [x1,x2,x3],[y1,x2,x3],[x1,x2,y3],[y1,x2,
  y3] }
proof
  thus [:{x1,y1},{x2},{x3,y3}:] = [:[:{x1,y1},{x2}:],{x3,y3}:] by
ZFMISC_1:def 3
    .= [:{[x1,x2],[y1,x2]},{x3,y3}:] by ZFMISC_1:30
    .= { [[x1,x2],x3],[[x1,x2],y3],[[y1,x2],x3],[[y1,x2],y3] } by Th23
    .= { [x1,x2,x3],[y1,x2,x3],[x1,x2,y3],[y1,x2,y3] } by ENUMSET1:62;
end;

theorem
  [:{x1},{x2,y2},{x3,y3}:] = { [x1,x2,x3],[x1,y2,x3],[x1,x2,y3],[x1,y2,
  y3] }
proof
  thus [:{x1},{x2,y2},{x3,y3}:] = [:[:{x1},{x2,y2}:],{x3,y3}:] by
ZFMISC_1:def 3
    .= [:{[x1,x2],[x1,y2]},{x3,y3}:] by ZFMISC_1:30
    .= { [[x1,x2],x3],[[x1,x2],y3],[[x1,y2],x3],[[x1,y2],y3] } by Th23
    .= { [x1,x2,x3],[x1,y2,x3],[x1,x2,y3],[x1,y2,y3] } by ENUMSET1:62;
end;

theorem
  [:{x1,y1},{x2,y2},{x3,y3}:] = { [x1,x2,x3],[x1,y2,x3],[x1,x2,y3],[x1,
  y2,y3], [y1,x2,x3],[y1,y2,x3],[y1,x2,y3],[y1,y2,y3] }
proof
A1: [:{[x1,x2],[x1,y2]},{x3,y3}:] = { [[x1,x2],x3],[[x1,x2],y3],[[x1,y2],x3]
  ,[[x1,y2],y3]} by Th23
    .= { [x1,x2,x3],[x1,y2,x3],[x1,x2,y3],[x1,y2,y3]} by ENUMSET1:62;
A2: [:{[y1,x2],[y1,y2]},{x3,y3}:] = { [[y1,x2],x3],[[y1,x2],y3],[[y1,y2],x3]
  ,[[y1,y2],y3]} by Th23
    .= { [y1,x2,x3],[y1,y2,x3],[y1,x2,y3],[y1,y2,y3]} by ENUMSET1:62;
  thus [:{x1,y1},{x2,y2},{x3,y3}:] = [:[:{x1,y1},{x2,y2}:],{x3,y3}:] by
ZFMISC_1:def 3
    .= [:{[x1,x2],[x1,y2],[y1,x2],[y1,y2]},{x3,y3}:] by Th23
    .= [:{[x1,x2],[x1,y2]} \/ {[y1,x2],[y1,y2]},{x3,y3}:] by ENUMSET1:5
    .= { [x1,x2,x3],[x1,y2,x3],[x1,x2,y3],[x1,y2,y3]} \/ {[y1,x2,x3],[y1,y2,
  x3],[y1,x2,y3],[y1,y2,y3] } by A1,A2,ZFMISC_1:97
    .= { [x1,x2,x3],[x1,y2,x3],[x1,x2,y3],[x1,y2,y3], [y1,x2,x3],[y1,y2,x3],
  [y1,x2,y3],[y1,y2,y3] } by ENUMSET1:25;
end;

registration
 let X1,X2,X3 be non empty set;
 cluster -> triple for Element of [:X1,X2,X3:];
 coherence
  proof let x be Element of [:X1,X2,X3:];
   ex xx1 being Element of X1, xx2 being Element of X2,
           xx3 being Element of X3 st x = [xx1,xx2,xx3] by Lm2;
   hence thesis;
  end;
end;

definition
 canceled 4;
  let X1,X2,X3 be non empty set;
  let x be Element of [:X1,X2,X3:];
  redefine func x`1_3 -> Element of X1 means
  :Def5:
  x = [x1,x2,x3] implies it = x1;
  coherence
  proof
    consider xx1 being Element of X1,xx2 being Element of X2,
             xx3 being Element of X3 such that
A6: x = [xx1,xx2,xx3] by Lm2;
    [xx1,xx2,xx3]`1_3 = xx1;
   hence x`1_3 is Element of X1 by A6;
  end;
  compatibility
   proof let y be Element of X1;
    thus y = x`1_3 implies for x1,x2,x3 holds x = [x1,x2,x3] implies y = x1
     proof assume
Z:     y = x`1_3;
      let x1,x2,x3 such that
W:     x = [x1,x2,x3];
       [x1,x2,x3]`1_3 = x1;
      hence y = x1 by W,Z;
     end;
    assume
Z:   for x1,x2,x3 holds x = [x1,x2,x3] implies y = x1;
    consider xx1 being Element of X1,xx2 being Element of X2,
             xx3 being Element of X3 such that
A6: x = [xx1,xx2,xx3] by Lm2;
     [xx1,xx2,xx3]`1_3 = xx1;
    hence y = x`1_3 by Z,A6;
   end;
  redefine func x`2_3 -> Element of X2 means
  :Def6:
  x = [x1,x2,x3] implies it = x2;
  coherence
  proof
    consider xx1 being Element of X1,xx2 being Element of X2,
             xx3 being Element of X3 such that
A6: x = [xx1,xx2,xx3] by Lm2;
    [xx1,xx2,xx3]`2_3 = xx2;
   hence x`2_3 is Element of X2 by A6;
  end;
  compatibility
   proof let y be Element of X2;
    thus y = x`2_3 implies for x1,x2,x3 holds x = [x1,x2,x3] implies y = x2
     proof assume
Z:     y = x`2_3;
      let x1,x2,x3 such that
W:     x = [x1,x2,x3];
       [x1,x2,x3]`2_3 = x2;
      hence y = x2 by W,Z;
     end;
    assume
Z:   for x1,x2,x3 holds x = [x1,x2,x3] implies y = x2;
    consider xx1 being Element of X1,xx2 being Element of X2,
             xx3 being Element of X3 such that
A6: x = [xx1,xx2,xx3] by Lm2;
     [xx1,xx2,xx3]`2_3 = xx2;
    hence y = x`2_3 by Z,A6;
   end;
  redefine func x`3_3 -> Element of X3 means
  :Def7:
  x = [x1,x2,x3] implies it = x3;
  coherence
  proof
    consider xx1 being Element of X1,xx2 being Element of X2,
             xx3 being Element of X3 such that
A6: x = [xx1,xx2,xx3] by Lm2;
    [xx1,xx2,xx3]`3_3 = xx3;
   hence x`3_3 is Element of X3 by A6;
  end;
  compatibility
   proof let y be Element of X3;
    thus y = x`3_3 implies for x1,x2,x3 holds x = [x1,x2,x3] implies y = x3
     proof assume
Z:     y = x`3_3;
      let x1,x2,x3 such that
W:     x = [x1,x2,x3];
       [x1,x2,x3]`3_3 = x3;
      hence y = x3 by W,Z;
     end;
    assume
Z:   for x1,x2,x3 holds x = [x1,x2,x3] implies y = x3;
    consider xx1 being Element of X1,xx2 being Element of X2,
             xx3 being Element of X3 such that
A6: x = [xx1,xx2,xx3] by Lm2;
     [xx1,xx2,xx3]`3_3 = xx3;
    hence y = x`3_3 by Z,A6;
   end;
end;

canceled;

theorem
 for X1,X2,X3 being non empty set
  for x being Element of [:X1,X2,
  X3:] holds x = [x`1_3,x`2_3,x`3_3];

theorem Th45:
  X c= [:X,Y,Z:] or X c= [:Y,Z,X:] or X c= [:Z,X,Y:] implies X =
  {}
proof
  assume that
A1: X c= [:X,Y,Z:] or X c= [:Y,Z,X:] or X c= [:Z,X,Y:] and
A2: X <> {};
  [:X,Y,Z:]<>{} or [:Y,Z,X:]<>{} or [:Z,X,Y:]<>{} by A1,A2;
  then reconsider X,Y,Z as non empty set by Th31;
  per cases by A1;
  suppose
A4: X c= [:X,Y,Z:];
    consider v such that
A5: v in X and
A6: for x,y,z st x in X or y in X holds v <> [x,y,z] by Th26;
    reconsider v as Element of [:X,Y,Z:] by A4,A5;
    v = [v`1_3,v`2_3,v`3_3];
    hence contradiction by A6;
  end;
  suppose
    X c= [:Y,Z,X:];
    then X c= [:[:Y,Z:],X:] by ZFMISC_1:def 3;
    hence thesis by ZFMISC_1:111;
  end;
  suppose
A7: X c= [:Z,X,Y:];
    consider v such that
A8: v in X and
A9: for z,x,y st z in X or x in X holds v <> [z,x,y] by Th26;
    reconsider v as Element of [:Z,X,Y:] by A7,A8;
    v = [v`1_3,v`2_3,v`3_3];
    hence thesis by A9;
  end;
end;

canceled;

theorem Th47:
 for X1,X2,X3 being non empty set
 for x being Element of [:X1,X2,X3:]
   holds x <> x`1_3 & x <> x`2_3 & x <> x`3_3
proof let X1,X2,X3 be non empty set;
  let x be Element of [:X1,X2,X3:];
  set Y9 = { x`1_3,x`2_3 }, Y = { Y9,{x`1_3}},
      X9 = { Y,x`3_3 }, X = { X9,{Y} };
A2: x = [x`1_3,x`2_3,x`3_3]
    .= [[x`1_3,x`2_3],x`3_3];
  then x = x`1_3 or x = x`2_3 implies X in Y9 & Y9 in Y & Y in X9 & X9 in X
    by TARSKI:def 2;
  hence x <> x`1_3 & x <> x`2_3 by XREGULAR:8;
  thus thesis by A2,Th20;
end;

theorem
  [:X1,X2,X3:] meets [:Y1,Y2,Y3:] implies X1 meets Y1 & X2 meets Y2 & X3
  meets Y3
proof
  assume
A1: [:X1,X2,X3:] meets [:Y1,Y2,Y3:];
A2: [:[:X1,X2:],X3:] = [:X1,X2,X3:] & [:[:Y1,Y2:],Y3:] = [:Y1,Y2,Y3:] by
ZFMISC_1:def 3;
  then [:X1,X2:] meets [:Y1,Y2:] by A1,ZFMISC_1:104;
  hence thesis by A2,A1,ZFMISC_1:104;
end;





theorem Th49:
  [:X1,X2,X3,X4:] = [:[:[:X1,X2:],X3:],X4:]
proof
  thus [:X1,X2,X3,X4:] = [:[:X1,X2,X3:],X4:] by ZFMISC_1:def 4
    .= [:[:[:X1,X2:],X3:],X4:] by ZFMISC_1:def 3;
end;

theorem Th50:
  [:[:X1,X2:],X3,X4:] = [:X1,X2,X3,X4:]
proof
  thus [:[:X1,X2:],X3,X4:] = [:[:[:X1,X2:],X3:],X4:] by ZFMISC_1:def 3
    .= [:X1,X2,X3,X4:] by Th49;
end;

theorem Th51:
  X1 <> {} & X2 <> {} & X3 <> {} & X4 <> {} iff [:X1,X2,X3,X4:] <>
  {}
proof
A1: [:[:X1,X2,X3:],X4:] = [:X1,X2,X3,X4:] by ZFMISC_1:def 4;
  X1 <> {} & X2 <> {} & X3 <> {} iff [:X1,X2,X3:] <> {} by Th31;
  hence thesis by A1,ZFMISC_1:90;
end;

theorem Th52:
  X1<>{} & X2<>{} & X3<>{} & X4<>{} & [:X1,X2,X3,X4:] = [:Y1,Y2,Y3
  ,Y4:] implies X1=Y1 & X2=Y2 & X3=Y3 & X4=Y4
proof
A1: [:X1,X2,X3,X4:] = [:[:X1,X2,X3:],X4:] by ZFMISC_1:def 4;
  assume
A2: X1<>{} & X2<>{} & X3<>{};
  assume
A3: X4<>{};
  assume [:X1,X2,X3,X4:] = [:Y1,Y2,Y3,Y4:];
  then
A4: [:[:X1,X2,X3:],X4:] = [:[:Y1,Y2,Y3:],Y4:] by A1,ZFMISC_1:def 4;
   [:X1,X2,X3:] = [:Y1,Y2,Y3:] by A3,A4,A2,ZFMISC_1:110;
  hence thesis by A2,A3,A4,Th32,ZFMISC_1:110;
end;

theorem
  [:X1,X2,X3,X4:]<>{} & [:X1,X2,X3,X4:] = [:Y1,Y2,Y3,Y4:] implies X1=Y1
  & X2=Y2 & X3=Y3 & X4=Y4
proof
  assume
A1: [:X1,X2,X3,X4:]<>{};
  then
A2: X3<>{} & X4<>{} by Th51;
  X1<>{} & X2<>{} by A1,Th51;
  hence thesis by A2,Th52;
end;

theorem
  [:X,X,X,X:] = [:Y,Y,Y,Y:] implies X = Y
proof
  assume [:X,X,X,X:] = [:Y,Y,Y,Y:];
  then X<>{} or Y<>{} implies thesis by Th52;
  hence thesis;
end;

reserve xx4 for Element of X4;

Lm3: X1 <> {} & X2 <> {} & X3 <> {} & X4 <> {} implies for x being Element of
[:X1,X2,X3,X4:] ex xx1,xx2,xx3,xx4 st x = [xx1,xx2,xx3,xx4]

proof
  assume that
A1: X1 <> {} & X2 <> {} & X3 <> {} and
A2: X4 <> {};
  let x be Element of [:X1,X2,X3,X4:];
  reconsider x9=x as Element of [:[:X1,X2,X3:],X4:] by ZFMISC_1:def 4;
  consider x123 being (Element of [:X1,X2,X3:]), xx4 such that
A3: x9 = [x123,xx4] by A2,Lm1,A1;
  consider xx1,xx2,xx3 such that
A4: x123 = [xx1,xx2,xx3] by A1,Lm2;
  take xx1,xx2,xx3,xx4;
  thus thesis by A3,A4;
end;

registration
 let X1,X2,X3,X4 be non empty set;
 cluster -> quadruple for Element of [:X1,X2,X3,X4:];
 coherence
  proof let x be Element of [:X1,X2,X3,X4:];
   ex xx1 being Element of X1, xx2 being Element of X2,
      xx3 being Element of X3, xx4 being Element of X4
     st x = [xx1,xx2,xx3,xx4] by Lm3;
   hence thesis;
  end;
end;

definition
  let X1,X2,X3,X4 be non empty set;
  let x be Element of [:X1,X2,X3,X4:];
  redefine func x`1_4 -> Element of X1 means
  :Def8:
  x = [x1,x2,x3,x4] implies it = x1;
  coherence
  proof
    consider xx1 being Element of X1, xx2 being Element of X2,
             xx3 being Element of X3, xx4 being Element of X4 such that
A6: x = [xx1,xx2,xx3,xx4] by Lm3;
    [xx1,xx2,xx3,xx4]`1_4 = xx1;
   hence x`1_4 is Element of X1 by A6;
  end;
  compatibility
   proof let y be Element of X1;
    thus y = x`1_4 implies for x1,x2,x3,x4 holds x = [x1,x2,x3,x4]
      implies y = x1
     proof assume
Z:     y = x`1_4;
      let x1,x2,x3,x4 such that
W:     x = [x1,x2,x3,x4];
       [x1,x2,x3,x4]`1_4 = x1;
      hence y = x1 by W,Z;
     end;
    assume
Z:   for x1,x2,x3,x4 holds x = [x1,x2,x3,x4] implies y = x1;
    consider xx1 being Element of X1, xx2 being Element of X2,
             xx3 being Element of X3, xx4 being Element of X4 such that
A6: x = [xx1,xx2,xx3,xx4] by Lm3;
     [xx1,xx2,xx3,xx4]`1_4 = xx1;
    hence y = x`1_4 by Z,A6;
   end;
  redefine func x`2_4 -> Element of X2 means
  :Def9:
  x = [x1,x2,x3,x4] implies it = x2;
  coherence
  proof
   consider xx1 being Element of X1, xx2 being Element of X2,
             xx3 being Element of X3, xx4 being Element of X4 such that
A6: x = [xx1,xx2,xx3,xx4] by Lm3;
    [xx1,xx2,xx3,xx4]`2_4 = xx2;
   hence x`2_4 is Element of X2 by A6;
  end;
  compatibility
   proof let y be Element of X2;
    thus y = x`2_4 implies for x1,x2,x3,x4 holds x = [x1,x2,x3,x4]
     implies y = x2
     proof assume
Z:     y = x`2_4;
      let x1,x2,x3,x4 such that
W:     x = [x1,x2,x3,x4];
       [x1,x2,x3,x4]`2_4 = x2;
      hence y = x2 by W,Z;
     end;
    assume
Z:   for x1,x2,x3,x4 holds x = [x1,x2,x3,x4] implies y = x2;
   consider xx1 being Element of X1, xx2 being Element of X2,
             xx3 being Element of X3, xx4 being Element of X4 such that
A6: x = [xx1,xx2,xx3,xx4] by Lm3;
     [xx1,xx2,xx3,xx4]`2_4 = xx2;
    hence y = x`2_4 by Z,A6;
   end;
  redefine func x`3_4 -> Element of X3 means
  :Def10:
  x = [x1,x2,x3,x4] implies it = x3;
  coherence
  proof
   consider xx1 being Element of X1, xx2 being Element of X2,
             xx3 being Element of X3, xx4 being Element of X4 such that
A6: x = [xx1,xx2,xx3,xx4] by Lm3;
    [xx1,xx2,xx3,xx4]`3_4 = xx3;
   hence x`3_4 is Element of X3 by A6;
  end;
  compatibility
   proof let y be Element of X3;
    thus y = x`3_4 implies for x1,x2,x3,x4 holds x = [x1,x2,x3,x4]
       implies y = x3
     proof assume
Z:     y = x`3_4;
      let x1,x2,x3,x4 such that
W:     x = [x1,x2,x3,x4];
       [x1,x2,x3,x4]`3_4 = x3;
      hence y = x3 by W,Z;
     end;
    assume
Z:   for x1,x2,x3,x4 holds x = [x1,x2,x3,x4] implies y = x3;
   consider xx1 being Element of X1, xx2 being Element of X2,
             xx3 being Element of X3, xx4 being Element of X4 such that
A6: x = [xx1,xx2,xx3,xx4] by Lm3;
     [xx1,xx2,xx3,xx4]`3_4 = xx3;
    hence y = x`3_4 by Z,A6;
   end;
  redefine func x`4_4 -> Element of X4 means
  :Def11:
  x = [x1,x2,x3,x4] implies it = x4;
  coherence
  proof
   consider xx1 being Element of X1, xx2 being Element of X2,
             xx3 being Element of X3, xx4 being Element of X4 such that
A6: x = [xx1,xx2,xx3,xx4] by Lm3;
    [xx1,xx2,xx3,xx4]`4_4 = xx4;
   hence x`4_4 is Element of X4 by A6;
  end;
  compatibility
   proof let y be Element of X4;
    thus y = x`4_4 implies for x1,x2,x3,x4 holds x = [x1,x2,x3,x4]
       implies y = x4
     proof assume
Z:     y = x`4_4;
      let x1,x2,x3,x4 such that
W:     x = [x1,x2,x3,x4];
       [x1,x2,x3,x4]`4_4 = x4;
      hence y = x4 by W,Z;
     end;
    assume
Z:   for x1,x2,x3,x4 holds x = [x1,x2,x3,x4] implies y = x4;
   consider xx1 being Element of X1, xx2 being Element of X2,
             xx3 being Element of X3, xx4 being Element of X4 such that
A6: x = [xx1,xx2,xx3,xx4] by Lm3;
     [xx1,xx2,xx3,xx4]`4_4 = xx4;
    hence y = x`4_4 by Z,A6;
   end;
end;

canceled;

theorem
 for X1,X2,X3,X4 being non empty set
 for x being Element of
  [:X1,X2,X3,X4:] holds x = [x`1_4,x`2_4,x`3_4,x`4_4];

canceled;

theorem
 for X1,X2,X3,X4 being non empty set
 for x being Element of [:X1,X2,X3,X4:]
  holds x <> x`1_4 & x <> x`2_4 & x <> x`3_4 & x <> x`4_4
proof let X1,X2,X3,X4 be non empty set;
  let x be Element of [:X1,X2,X3,X4:];
   reconsider Y = [:X1,X2:], X3, X4 as non empty set;
  reconsider x9=x as Element of [:Y,X3,X4:] by Th50;
  set Z9 = { x`1_4,x`2_4 }, Z = { Z9,{x`1_4}},
      Y9 = { Z,x`3_4 }, Y = { Y9,{Z} }, X9 =
  { Y,x`4_4 }, X = { X9,{Y} };
  x = [x`1_4,x`2_4,x`3_4,x`4_4]
    .= X;
  then
  x = x`1_4 or x = x`2_4
   implies X in Z9 & Z9 in Z & Z in Y9 & Y9 in Y & Y in X9 & X9 in X
     by TARSKI:def 2;
  hence x <> x`1_4 & x <> x`2_4 by XREGULAR:10;
  x`3_4 = (x qua set)`1`2
    .= x9`2_3;
  hence thesis by Th47;
end;

theorem
  X1 c= [:X1,X2,X3,X4:] or X1 c= [:X2,X3,X4,X1:] or X1 c= [:X3,X4,X1,X2
  :] or X1 c= [:X4,X1,X2,X3:] implies X1 = {}
proof
  assume that
A1: X1 c= [:X1,X2,X3,X4:] or X1 c= [:X2,X3,X4,X1:] or X1 c= [:X3,X4,X1,
  X2:] or X1 c= [:X4,X1,X2,X3:] and
A2: X1 <> {};
A3: [:X1,X2,X3,X4:]<>{} or [:X2,X3,X4,X1:]<>{} or [:X3,X4,X1,X2:]<>{} or [:
  X4,X1,X2,X3:]<>{} by A1,A2;
   reconsider X1,X2,X3,X4 as non empty set by A3,Th51;
  per cases by A1;
  suppose
A6: X1 c= [:X1,X2,X3,X4:];
    consider v such that
A7: v in X1 and
A8: for x1,x2,x3,x4 st x1 in X1 or x2 in X1 holds v <> [x1,x2,x3,x4]
    by Th30;
    reconsider v as Element of [:X1,X2,X3,X4:] by A6,A7;
    v = [v`1_4,v`2_4,v`3_4,v`4_4];
    hence contradiction by A8;
  end;
  suppose
    X1 c= [:X2,X3,X4,X1:];
    then X1 c= [:[:X2,X3:],X4,X1:] by Th50;
    hence thesis by Th45;
  end;
  suppose
    X1 c= [:X3,X4,X1,X2:];
    then X1 c= [:[:X3,X4:],X1,X2:] by Th50;
    hence thesis by Th45;
  end;
  suppose
A9: X1 c= [:X4,X1,X2,X3:];
    consider v such that
A10: v in X1 and
A11: for x1,x2,x3,x4 st x1 in X1 or x2 in X1 holds v <> [x1,x2,x3,x4]
    by Th30;
    reconsider v as Element of [:X4,X1,X2,X3:] by A9,A10;
    v = [v`1_4,v`2_4,v`3_4,v`4_4];
    hence thesis by A11;
  end;
end;

theorem
  [:X1,X2,X3,X4:] meets [:Y1,Y2,Y3,Y4:] implies X1 meets Y1 & X2 meets
  Y2 & X3 meets Y3 & X4 meets Y4
proof
  assume
A1: [:X1,X2,X3,X4:] meets [:Y1,Y2,Y3,Y4:];
A2: [:[:[:X1,X2:],X3:],X4:] = [:X1,X2,X3,X4:] & [:[:[:Y1,Y2:],Y3:],Y4:] = [:
  Y1, Y2,Y3,Y4:] by Th49;
  then
A3: [:[:X1,X2:],X3:] meets [:[:Y1,Y2:],Y3:] by A1,ZFMISC_1:104;
  then [:X1,X2:] meets [:Y1,Y2:] by ZFMISC_1:104;
  hence thesis by A2,A1,A3,ZFMISC_1:104;
end;

theorem
  [:{x1},{x2},{x3},{x4}:] = { [x1,x2,x3,x4] }
proof
  thus [:{x1},{x2},{x3},{x4}:] = [:[:{x1},{x2}:],{x3},{x4}:] by Th50
    .= [:{[x1,x2]},{x3},{x4}:] by ZFMISC_1:29
    .= {[[x1,x2],x3,x4]} by Th35
    .= { [x1,x2,x3,x4] };
end;



theorem Th62:
  [:X,Y:] <> {} implies for x being Element of [:X,Y:] holds x <>
  x`1 & x <> x`2
proof
  assume [:X,Y:] <> {};
  then X <> {} & Y <> {} by ZFMISC_1:90;
  hence thesis by Th24;
end;

theorem
  x in [:X,Y:] implies x <> x`1 & x <> x`2 by Th62;



reserve A1 for Subset of X1,
  A2 for Subset of X2,
  A3 for Subset of X3,
  A4 for Subset of X4;

reserve x for Element of [:X1,X2,X3:];

theorem
 for X1,X2,X3 being non empty set
 for x being Element of [:X1,X2,X3:]
 for x1,x2,x3 st x = [x1,x2,x3] holds
  x`1_3 = x1 & x`2_3 = x2 & x`3_3 = x3 by Def5,Def6,Def7;

theorem
 for X1,X2,X3 being non empty set
 for x being Element of [:X1,X2,X3:]
  st for xx1 being Element of X1,xx2 being Element of X2,
             xx3 being Element of X3 st x = [xx1,xx2,xx3] holds y1 = xx1
 holds y1 =x`1_3
proof let X1,X2,X3 be non empty set;
  let x be Element of [:X1,X2,X3:];
  assume that
A2: for xx1 being Element of X1,xx2 being Element of X2,
             xx3 being Element of X3 st x = [xx1,xx2,xx3] holds y1 = xx1;
  x = [x`1_3,x`2_3,x`3_3];
  hence thesis by A2;
end;

theorem
 for X1,X2,X3 being non empty set
 for x being Element of [:X1,X2,X3:]
  st for xx1 being Element of X1,xx2 being Element of X2,
             xx3 being Element of X3 st x = [xx1,xx2,xx3] holds  y2 = xx2
 holds y2 =x`2_3
proof let X1,X2,X3 be non empty set;
  let x be Element of [:X1,X2,X3:];
  assume that
A2: for xx1 being Element of X1,xx2 being Element of X2,
             xx3 being Element of X3 st x = [xx1,xx2,xx3] holds y2 = xx2;
  x = [x`1_3,x`2_3,x`3_3];
  hence thesis by A2;
end;

theorem
 for X1,X2,X3 being non empty set
 for x being Element of [:X1,X2,X3:]
  st for xx1 being Element of X1,xx2 being Element of X2,
             xx3 being Element of X3 st x = [xx1,xx2,xx3] holds y3 = xx3
 holds y3 =x`3_3
proof
 let X1,X2,X3 be non empty set;
  let x be Element of [:X1,X2,X3:];
  assume that
A2: for xx1 being Element of X1,xx2 being Element of X2,
             xx3 being Element of X3 st x = [xx1,xx2,xx3] holds y3 = xx3;
  x = [x`1_3,x`2_3,x`3_3];
  hence thesis by A2;
end;

theorem Th68:
  z in [: X1,X2,X3 :] implies ex x1,x2,x3 st x1 in X1 & x2 in X2 &
  x3 in X3 & z = [x1,x2,x3]
proof
  assume z in [: X1,X2,X3 :];
  then z in [:[:X1,X2:],X3:] by ZFMISC_1:def 3;
  then consider x12, x3 being set such that
A1: x12 in [:X1,X2:] and
A2: x3 in X3 and
A3: z = [x12,x3] by ZFMISC_1:def 2;
  consider x1,x2 being set such that
A4: x1 in X1 & x2 in X2 and
A5: x12 = [x1,x2] by A1,ZFMISC_1:def 2;
  z = [x1,x2,x3] by A3,A5;
  hence thesis by A2,A4;
end;

theorem Th69:
  [x1,x2,x3] in [: X1,X2,X3 :] iff x1 in X1 & x2 in X2 & x3 in X3
proof
A1: [x1,x2] in [:X1,X2:] iff x1 in X1 & x2 in X2 by ZFMISC_1:87;
  [:X1,X2,X3:] = [:[:X1,X2:],X3:] by ZFMISC_1:def 3;
  hence thesis by A1,ZFMISC_1:87;
end;

theorem
  (for z holds z in Z iff ex x1,x2,x3 st x1 in X1 & x2 in X2 & x3 in X3
  & z = [x1,x2,x3]) implies Z = [: X1,X2,X3 :]
proof
  assume
A1: for z holds z in Z iff ex x1,x2,x3 st x1 in X1 & x2 in X2 & x3 in X3
  & z = [x1,x2,x3];
  now
    let z;
    thus z in Z implies z in [:[:X1,X2:],X3:]
    proof
      assume z in Z;
      then consider x1,x2,x3 such that
A2:   x1 in X1 & x2 in X2 and
A3:   x3 in X3 & z = [x1,x2,x3] by A1;
      [x1,x2] in [:X1,X2:] by A2,ZFMISC_1:def 2;
      hence thesis by A3,ZFMISC_1:def 2;
    end;
    assume z in [:[:X1,X2:],X3:];
    then consider x12,x3 being set such that
A4: x12 in [:X1,X2:] and
A5: x3 in X3 and
A6: z = [x12,x3] by ZFMISC_1:def 2;
    consider x1,x2 being set such that
A7: x1 in X1 & x2 in X2 and
A8: x12 = [x1,x2] by A4,ZFMISC_1:def 2;
    z = [x1,x2,x3] by A6,A8;
    hence z in Z by A1,A5,A7;
  end;
  then Z = [:[:X1,X2:],X3:] by TARSKI:1;
  hence thesis by ZFMISC_1:def 3;
end;

canceled;

theorem
 for X1,X2,X3 being non empty set
 for A1 being non empty Subset of X1, A2 being non empty Subset of X2,
     A3 being non empty Subset of X3
  for x being Element of [:X1,X2,X3:] st x in [:A1,A2,A3:]
   holds x`1_3 in A1 & x`2_3 in A2 & x`3_3 in A3
proof
 let X1,X2,X3 be non empty set;
 let A1 be non empty Subset of X1, A2 be non empty Subset of X2,
     A3 be non empty Subset of X3;
  let x be Element of [:X1,X2,X3:];
  assume
 x in [:A1,A2,A3:];
  then reconsider y = x as Element of [:A1,A2,A3:];
A2: y`2_3 in A2;
A3: y`3_3 in A3;
  y`1_3 in A1;
  hence thesis by A2,A3;
end;

theorem Th73:
  X1 c= Y1 & X2 c= Y2 & X3 c= Y3 implies [:X1,X2,X3:] c= [:Y1,Y2,
  Y3:]
proof
A1: [:X1,X2,X3:] = [:[:X1,X2:],X3:] & [:Y1,Y2,Y3:] = [:[:Y1,Y2:],Y3:] by
ZFMISC_1:def 3;
  assume X1 c= Y1 & X2 c= Y2;
  then
A2: [:X1,X2:] c= [:Y1,Y2:] by ZFMISC_1:96;
  assume X3 c= Y3;
  hence thesis by A2,A1,ZFMISC_1:96;
end;



reserve x for Element of [:X1,X2,X3,X4:];

theorem
 for X1,X2,X3,X4 being non empty set
 for x being Element of [:X1,X2,X3,X4:]
 for x1,x2,x3,x4 being set st x = [x1,
  x2,x3,x4] holds x`1_4 = x1 & x`2_4 = x2 & x`3_4 = x3 & x`4_4 = x4
   by Def8,Def9,Def10,Def11;

theorem
 for X1,X2,X3,X4 being non empty set
 for x being Element of [:X1,X2,X3,X4:]
 st for xx1 being Element of X1,
             xx2 being Element of X2,
             xx3 being Element of X3,
             xx4 being Element of X4 st x = [xx1,xx2,xx3,xx4] holds y1 = xx1
 holds y1 =x`1_4
proof
 let X1,X2,X3,X4 be non empty set;
 let x be Element of [:X1,X2,X3,X4:];
  assume that
A2: for xx1 being Element of X1,
             xx2 being Element of X2,
             xx3 being Element of X3,
             xx4 being Element of X4 st x = [xx1,xx2,xx3,xx4] holds y1 = xx1;
  x = [x`1_4,x`2_4,x`3_4,x`4_4];
  hence thesis by A2;
end;

theorem
 for X1,X2,X3,X4 being non empty set
 for x being Element of [:X1,X2,X3,X4:]
 st for xx1 being Element of X1,
             xx2 being Element of X2,
             xx3 being Element of X3,
             xx4 being Element of X4 st x = [xx1,
  xx2,xx3,xx4] holds y2 = xx2
 holds y2 =x`2_4
proof
 let X1,X2,X3,X4 be non empty set;
 let x be Element of [:X1,X2,X3,X4:];
  assume that
A2: for xx1 being Element of X1,
             xx2 being Element of X2,
             xx3 being Element of X3,
             xx4 being Element of X4 st x = [xx1,xx2,xx3,xx4] holds y2 = xx2;
  x = [x`1_4,x`2_4,x`3_4,x`4_4];
  hence thesis by A2;
end;

theorem
 for X1,X2,X3,X4 being non empty set
 for x being Element of [:X1,X2,X3,X4:]
 st for xx1 being Element of X1,
             xx2 being Element of X2,
             xx3 being Element of X3,
             xx4 being Element of X4 st x = [xx1,
  xx2,xx3,xx4] holds y3 = xx3
 holds y3 =x`3_4
proof
 let X1,X2,X3,X4 be non empty set;
 let x be Element of [:X1,X2,X3,X4:];
  assume that
A2: for xx1 being Element of X1,
             xx2 being Element of X2,
             xx3 being Element of X3,
             xx4 being Element of X4 st x = [xx1,xx2,xx3,xx4] holds y3 = xx3;
  x = [x`1_4,x`2_4,x`3_4,x`4_4];
  hence thesis by A2;
end;

theorem
 for X1,X2,X3,X4 being non empty set
 for x being Element of [:X1,X2,X3,X4:]
  st for xx1 being Element of X1,
             xx2 being Element of X2,
             xx3 being Element of X3,
             xx4 being Element of X4 st x = [xx1,
  xx2,xx3,xx4] holds y4 = xx4
 holds y4 =x`4_4
proof
 let X1,X2,X3,X4 be non empty set;
 let x be Element of [:X1,X2,X3,X4:];
  assume that
A2: for xx1 being Element of X1,
             xx2 being Element of X2,
             xx3 being Element of X3,
             xx4 being Element of X4 st x = [xx1,xx2,xx3,xx4] holds y4 = xx4;
  x = [x`1_4,x`2_4,x`3_4,x`4_4];
  hence thesis by A2;
end;

theorem
  z in [: X1,X2,X3,X4 :] implies ex x1,x2,x3,x4 st x1 in X1 & x2 in X2 &
  x3 in X3 & x4 in X4 & z = [x1,x2,x3,x4]
proof
  assume z in [: X1,X2,X3,X4 :];
  then z in [:[:X1,X2,X3:],X4:] by ZFMISC_1:def 4;
  then consider x123, x4 being set such that
A1: x123 in [:X1,X2,X3:] and
A2: x4 in X4 and
A3: z = [x123,x4] by ZFMISC_1:def 2;
  consider x1, x2, x3 such that
A4: x1 in X1 & x2 in X2 & x3 in X3 and
A5: x123 = [x1,x2,x3] by A1,Th68;
  z = [x1,x2,x3,x4] by A3,A5;
  hence thesis by A2,A4;
end;

theorem
  [x1,x2,x3,x4] in [: X1,X2,X3,X4 :] iff x1 in X1 & x2 in X2 & x3 in X3
  & x4 in X4
proof
A1: [x1,x2] in [:X1,X2:] iff x1 in X1 & x2 in X2 by ZFMISC_1:87;
  [:X1,X2,X3,X4:] = [:[:X1,X2:],X3,X4:] & [x1,x2,x3,x4] = [[x1,x2],x3,x4]
  by Th50;
  hence thesis by A1,Th69;
end;

theorem
  (for z holds z in Z iff ex x1,x2,x3,x4 st x1 in X1 & x2 in X2 & x3 in
  X3 & x4 in X4 & z = [x1,x2,x3,x4]) implies Z = [: X1,X2,X3,X4 :]
proof
  assume
A1: for z holds z in Z iff ex x1,x2,x3,x4 st x1 in X1 & x2 in X2 & x3 in
  X3 & x4 in X4 & z = [x1,x2,x3,x4];
  now
    let z;
    thus z in Z implies z in [:[:X1,X2,X3:],X4:]
    proof
      assume z in Z;
      then consider x1,x2,x3,x4 such that
A2:   x1 in X1 & x2 in X2 & x3 in X3 and
A3:   x4 in X4 & z = [x1,x2,x3,x4] by A1;
      [x1,x2,x3] in [:X1,X2,X3:] by A2,Th69;
      hence thesis by A3,ZFMISC_1:def 2;
    end;
    assume z in [:[:X1,X2,X3:],X4:];
    then consider x123,x4 being set such that
A4: x123 in [:X1,X2,X3:] and
A5: x4 in X4 and
A6: z = [x123,x4] by ZFMISC_1:def 2;
    consider x1,x2,x3 such that
A7: x1 in X1 & x2 in X2 & x3 in X3 and
A8: x123 = [x1,x2,x3] by A4,Th68;
    z = [x1,x2,x3,x4] by A6,A8;
    hence z in Z by A1,A5,A7;
  end;
  then Z = [:[:X1,X2,X3:],X4:] by TARSKI:1;
  hence thesis by ZFMISC_1:def 4;
end;

canceled;

theorem
 for X1,X2,X3,X4 being non empty set,
     A1 being non empty Subset of X1, A2 being non empty Subset of X2,
     A3 being non empty Subset of X3, A4 being non empty Subset of X4
  for x being Element of [:X1,X2,X3,X4:] st x in [:A1,A2,A3,A4:]
  holds x`1_4 in A1 & x`2_4 in A2 & x`3_4 in A3 & x`4_4 in A4
proof
 let X1,X2,X3,X4 be non empty set,
     A1 be non empty Subset of X1, A2 be non empty Subset of X2,
     A3 be non empty Subset of X3, A4 be non empty Subset of X4;
  let x be Element of [:X1,X2,X3,X4:];
  assume
 x in [:A1,A2,A3,A4:];
  then reconsider y = x as Element of [:A1,A2,A3,A4:];
A2: y`2_4 in A2;
A3: y`4_4 in A4;
A4: y`3_4 in A3;
  y`1_4 in A1;
  hence thesis by A2,A4,A3;
end;

theorem Th84:
  X1 c= Y1 & X2 c= Y2 & X3 c= Y3 & X4 c= Y4
   implies [:X1,X2,X3,X4:] c= [:Y1,Y2,Y3,Y4:]
proof
A1: [:X1,X2,X3,X4:] = [:[:X1,X2,X3:],X4:] & [:Y1,Y2,Y3,Y4:] = [:[:Y1,Y2,Y3:]
  ,Y4 :] by ZFMISC_1:def 4;
  assume X1 c= Y1 & X2 c= Y2 & X3 c= Y3;
  then
A2: [:X1,X2,X3:] c= [:Y1,Y2,Y3:] by Th73;
  assume X4 c= Y4;
  hence thesis by A2,A1,ZFMISC_1:96;
end;

definition
  let X1,X2,A1,A2;
  redefine func [:A1,A2:] -> Subset of [:X1,X2:];
  coherence by ZFMISC_1:96;
end;

definition
  let X1,X2,X3,A1,A2,A3;
  redefine func [:A1,A2,A3:] -> Subset of [:X1,X2,X3:];
  coherence by Th73;
end;

definition
  let X1,X2,X3,X4,A1,A2,A3,A4;
  redefine func [:A1,A2,A3,A4:] -> Subset of [:X1,X2,X3,X4:];
  coherence by Th84;
end;

begin 



definition
  let f be Function;
  func pr1 f -> Function means
  dom it = dom f & for x being set st x in dom f
  holds it.x = (f.x)`1;
  existence
  proof
    deffunc F(set) = (f.$1)`1;
    ex g being Function st dom g = dom f & for x being set st x in dom f
    holds g.x = F(x) from FUNCT_1:sch 3;
    hence thesis;
  end;
  uniqueness
  proof
    let p1, p2 be Function such that
A1: dom p1 = dom f and
A2: for x being set st x in dom f holds p1.x = (f.x)`1 and
A3: dom p2 = dom f and
A4: for x being set st x in dom f holds p2.x = (f.x)`1;
    now
      let x be set;
      assume
A5:   x in dom f;
      then p1.x = (f.x)`1 by A2;
      hence p1.x = p2.x by A4,A5;
    end;
    hence thesis by A1,A3,FUNCT_1:2;
  end;
  func pr2 f -> Function means
  dom it = dom f & for x being set st x in dom f
  holds it.x = (f.x)`2;
  existence
  proof
    deffunc F(set) = (f.$1)`2;
    ex g being Function st dom g = dom f & for x being set st x in dom f
    holds g.x = F(x) from FUNCT_1:sch 3;
    hence thesis;
  end;
  uniqueness
  proof
    let p1, p2 be Function such that
A6: dom p1 = dom f and
A7: for x being set st x in dom f holds p1.x = (f.x)`2 and
A8: dom p2 = dom f and
A9: for x being set st x in dom f holds p2.x = (f.x)`2;
    now
      let x be set;
      assume
A10:  x in dom f;
      then p1.x = (f.x)`2 by A7;
      hence p1.x = p2.x by A9,A10;
    end;
    hence thesis by A6,A8,FUNCT_1:2;
  end;
end;

definition
  let x be set;
  func x`11 equals
  x`1`1;
  coherence;
  func x`12 equals
  x`1`2;
  coherence;
  func x`21 equals
  x`2`1;
  coherence;
  func x`22 equals
  x`2`2;
  coherence;
end;

reserve x for set;

theorem
  [[x1,x2],y]`11 = x1 & [[x1,x2],y]`12 = x2 & [x,[y1,y2]]`21 = y1 & [x,[
  y1,y2]]`22 = y2;



theorem
  x in R implies x`1 in dom R & x`2 in rng R
proof
  assume
A1: x in R;
  then x = [x`1,x`2] by Th21;
  hence thesis by A1,XTUPLE_0:def 12,def 13;
end;



theorem Th87:
 for R being non empty Relation, x being set
  holds Im(R,x) = { I`2 where I is Element of R: I`1 = x }
 proof let R be non empty Relation, x being set;
   set X = { I`2 where I is Element of R: I`1 = x };
  thus Im(R,x) c= X
   proof let z;
    assume z in Im(R,x);
     then consider y such that
A1:   [y,z] in R and
A2:   y in {x} by RELAT_1:def 13;
A3:   y = x by A2,TARSKI:def 1;
     y = [y,z]`1 & z = [y,z]`2;
    hence z in X by A1,A3;
   end;
  let z;
  assume z in X;
   then consider I being Element of R such that
A4: z= I`2 and
A5: I`1 = x;
A6: I = [I`1,I`2] by Th21;
   x in {x} by TARSKI:def 1;
  hence z in Im(R,x) by A4,A5,A6,RELAT_1:def 13;
 end;

theorem
 x in R implies x`2 in Im(R,x`1)
proof
 assume
A1: x in R;
  then x`2 in { I`2 where I is Element of R: I`1 = x`1 };
 hence thesis by A1,Th87;
end;

theorem Th89:
 x in R & y in R & x`1 = y`1 & x`2 = y`2 implies x = y
 proof assume x in R & y in R;
   then x = [x`1,x`2] & y = [y`1,y`2] by Th21;
  hence thesis;
 end;

theorem
 for R being non empty Relation, x,y being Element of R
   st x`1 = y`1 & x`2 = y`2
 holds x = y by Th89;



theorem
   proj1 proj1 {[x1,x2,x3],[y1,y2,y3]} = {x1,y1}
  proof
   thus proj1 proj1 {[x1,x2,x3],[y1,y2,y3]}
       = proj1 {[x1,x2],[y1,y2]} by RELAT_1:10
      .= {x1,y1} by RELAT_1:10;
  end;

theorem
   proj1 proj1 {[x1,x2,x3]} = {x1}
   proof
   thus proj1 proj1 {[x1,x2,x3]}
       = proj1 {[x1,x2]} by RELAT_1:9
      .= {x1} by RELAT_1:9;
   end;

scheme
  BiFuncEx{A()->set,B()->set,C()->set,P[set,set,set]}: ex f,g being Function
  st dom f = A() & dom g = A() & for x st x in A() holds P[x,f.x,g.x]
provided
A1: x in A() implies ex y,z st y in B() & z in C() & P[x,y,z]
proof
  defpred H[set,set] means for y,z st $2`1 = y & $2`2 = z holds P[$1,y,z];
A2: x in A() implies ex p being set st p in [:B(),C():] & H[x,p]
  proof
    assume x in A();
    then consider y,z such that
A3: y in B() & z in C() and
A4: P[x,y,z] by A1;
    take p=[y,z];
    thus p in [:B(),C():] by A3,ZFMISC_1:87;
    thus for y,z st p`1 = y & p`2 = z holds P[x,y,z]
    proof
      let x1,x2 be set;
      assume that
A5:   p`1 = x1 and
A6:   p`2 = x2;
      x1=y by A5,Th7;
      hence thesis by A4,A6,Th7;
    end;
  end;
  consider h being Function such that
  dom h = A() & rng h c= [:B(),C():] and
A7: for x st x in A() holds H[x,h.x] from FUNCT_1:sch 5(A2);
  deffunc g(set) = (h.$1)`2;
  deffunc f(set) = (h.$1)`1;
  consider f being Function such that
A8: dom f = A() and
A9: for x st x in A() holds f.x = f(x) from FUNCT_1:sch 3;
  consider g being Function such that
A10: dom g = A() and
A11: for x st x in A() holds g.x = g(x) from FUNCT_1:sch 3;
  take f,g;
  thus dom f = A() & dom g = A() by A8,A10;
  thus for x st x in A() holds P[x,f.x,g.x]
  proof
    let x;
    assume
A12: x in A();
    then f.x = (h.x)`1 & g.x = (h.x)`2 by A9,A11;
    hence thesis by A7,A12;
  end;
end;















begin

reserve X,Y,Z for set,
  a,b,c,d,x,y,z,u for set,
  R for Relation,
  A,B,C for Ordinal;

definition
  let X;
  func RelIncl X -> Relation means
  :Def1:
  field it = X & for Y,Z st Y in X & Z in X holds [Y,Z] in it iff Y c= Z;
  existence
  proof
    defpred P[set,set] means $1 c= $2;
    consider R such that
A1: [x,y] in R iff x in X & y in X & P[x,y] from RELAT_1:sch 1;
    take R;
    thus field R = X
    proof
      thus x in field R implies x in X
      proof
A2:     now
          assume x in dom R;
          then ex y st [x,y] in R by XTUPLE_0:def 12;
          hence thesis by A1;
        end;
A3:     now
          assume x in rng R;
          then ex y st [y,x] in R by XTUPLE_0:def 13;
          hence thesis by A1;
        end;
        assume x in field R;
        hence thesis by A2,A3,XBOOLE_0:def 3;
      end;
      let x;
      assume x in X;
      then [x,x] in R by A1;
      hence thesis by RELAT_1:15;
    end;
    let Y,Z such that
A4: Y in X & Z in X;
    thus [Y,Z] in R implies Y c= Z by A1;
    thus thesis by A1,A4;
  end;
  uniqueness
  proof
    let R1,R2 be Relation such that
A5: field R1 = X and
A6: for Y,Z st Y in X & Z in X holds [Y,Z] in R1 iff Y c= Z and
A7: field R2 = X and
A8: for Y,Z st Y in X & Z in X holds [Y,Z] in R2 iff Y c= Z;
    let x,y;
    thus [x,y] in R1 implies [x,y] in R2
    proof
      assume
A9:   [x,y] in R1;
      then
A10:  x in field R1 & y in field R1 by RELAT_1:15;
      then x c= y by A5,A6,A9;
      hence thesis by A5,A8,A10;
    end;
    assume
A11: [x,y] in R2;
    then
A12: x in field R2 & y in field R2 by RELAT_1:15;
    then x c= y by A7,A8,A11;
    hence thesis by A6,A7,A12;
  end;
end;

canceled 6;

registration
  let X;
  cluster RelIncl X -> reflexive;
  coherence
proof
A1: field RelIncl X = X by Def1;
  let a;
  assume a in field RelIncl X;
  hence thesis by A1,Def1;
end;
  cluster RelIncl X -> transitive;
  coherence
proof
  let a,b,c such that
A2: a in field RelIncl X and
A3: b in field RelIncl X and
A4: c in field RelIncl X and
A5: [a,b] in RelIncl X & [b,c] in RelIncl X;
  field RelIncl X = X by Def1;
  then a c= b & b c= c by A2,A3,A4,A5,Def1;
  then
A6: a c= c by XBOOLE_1:1;
  a in X & c in X by A2,A4,Def1;
  hence thesis by A6,Def1;
end;
  cluster RelIncl X -> antisymmetric;
  coherence
proof
A7: field RelIncl X = X by Def1;
  let a,b;
  assume a in field RelIncl X & b in field RelIncl X & [a,b] in RelIncl X & [
  b,a ] in RelIncl X;
  then a c= b & b c= a by A7,Def1;
  hence thesis by XBOOLE_0:def 10;
end;
end;

registration
  let A;
  cluster RelIncl A -> connected;
coherence
proof
  let a,b such that
A1: a in field RelIncl A & b in field RelIncl A and
  a <> b;
A2: field RelIncl A = A by Def1;
  then reconsider Y = a, Z = b as Ordinal by A1;
  Y c= Z or Z c= Y;
  hence thesis by A1,A2,Def1;
end;
  cluster RelIncl A -> well_founded;
coherence
proof
  let Y;
  assume that
A3: Y c= field RelIncl A and
A4: Y <> {};
  defpred P[set] means $1 in Y;
  set x = the Element of Y;
A5: field RelIncl A = A by Def1;
  then x in A by A3,A4,TARSKI:def 3;
  then reconsider x as Ordinal;
  x in Y by A4;
  then
A6: ex B st P[B];
  consider B such that
A7: P[B] & for C st P[C] holds B c= C from ORDINAL1:sch 1(A6);
  reconsider x = B as set;
  take x;
  thus x in Y by A7;
  set y = the Element of (RelIncl A)-Seg(x) /\ Y;
  assume
A8: (RelIncl A)-Seg(x) /\ Y <> {};
  then
A9: y in Y by XBOOLE_0:def 4;
  then reconsider C = y as Ordinal by A3,A5;
A10: y in (RelIncl A)-Seg(x) by A8,XBOOLE_0:def 4;
  then [y,x] in RelIncl A by WELLORD1:1;
  then
A11: C c= B by A3,A5,A7,A9,Def1;
A12: y <> x by A10,WELLORD1:1;
  B c= C by A7,A9;
  hence contradiction by A12,A11,XBOOLE_0:def 10;
end;
end;

theorem Th7:
  Y c= X implies (RelIncl X) |_2 Y = RelIncl Y
proof
  assume
A1: Y c= X;
  let a,b;
  thus [a,b] in (RelIncl X) |_2 Y implies [a,b] in RelIncl Y
  proof
    assume
A2: [a,b] in (RelIncl X) |_2 Y;
    then [a,b] in [:Y,Y:] by XBOOLE_0:def 4;
    then
A3: a in Y & b in Y by ZFMISC_1:87;
    [a,b] in RelIncl X by A2,XBOOLE_0:def 4;
    then a c= b by A1,A3,Def1;
    hence thesis by A3,Def1;
  end;
  assume
A4: [a,b] in RelIncl Y;
  then
A5: a in field RelIncl Y & b in field RelIncl Y by RELAT_1:15;
A6: field RelIncl Y = Y by Def1;
  then a c= b by A4,A5,Def1;
  then
A7: [a,b] in RelIncl X by A1,A5,A6,Def1;
  [a,b] in [:Y,Y:] by A5,A6,ZFMISC_1:87;
  hence thesis by A7,XBOOLE_0:def 4;
end;

theorem Th8:
  for A,X st X c= A holds RelIncl X is well-ordering
proof
  let A,X;
  (RelIncl A) |_2 X is well-ordering by WELLORD1:25;
  hence thesis by Th7;
end;

reserve H for Function;

theorem Th9:
  A in B implies A = (RelIncl B)-Seg(A)
proof
  assume
A1: A in B;
  thus a in A implies a in (RelIncl B)-Seg(A)
  proof
    assume
A2: a in A;
    then reconsider C = a as Ordinal;
A3: a <> A by A2;
A4: A c= B by A1,ORDINAL1:def 2;
    C c= A by A2,ORDINAL1:def 2;
    then [C,A] in RelIncl B by A1,A2,A4,Def1;
    hence thesis by A3,WELLORD1:1;
  end;
  let a;
  assume
A5: a in (RelIncl B)-Seg(A);
  then
A6: a <> A by WELLORD1:1;
A7: [a,A] in RelIncl B by A5,WELLORD1:1;
  then
A8: a in field RelIncl B by RELAT_1:15;
A9: field RelIncl B = B by Def1;
  then reconsider C = a as Ordinal by A8;
  C c= A by A1,A7,A8,A9,Def1;
  then C c< A by A6,XBOOLE_0:def 8;
  hence thesis by ORDINAL1:11;
end;

theorem Th10:
  RelIncl A,RelIncl B are_isomorphic implies A = B
proof
A1: field RelIncl A = A by Def1;
  assume
A2: RelIncl A,RelIncl B are_isomorphic;
A3: now
A4: field RelIncl B = B by Def1;
    assume
A5: A in B;
    then A = (RelIncl B)-Seg(A) by Th9;
    then RelIncl A = (RelIncl B) |_2 ((RelIncl B)-Seg(A)) by A4,Th7,WELLORD1:9;
    hence contradiction by A2,A5,A4,WELLORD1:40,46;
  end;
  assume A <> B;
  then
A6: A in B or B in A by ORDINAL1:14;
  then B = (RelIncl A)-Seg(B) by A3,Th9;
  then RelIncl B = (RelIncl A) |_2 ((RelIncl A)-Seg(B)) by A1,Th7,WELLORD1:9;
  hence contradiction by A2,A6,A3,A1,WELLORD1:46;
end;

theorem Th11:
  R,RelIncl A are_isomorphic & R,RelIncl B are_isomorphic implies A = B
proof
  assume that
A1: R,RelIncl A are_isomorphic and
A2: R,RelIncl B are_isomorphic;
  RelIncl A,R are_isomorphic by A1,WELLORD1:40;
  hence thesis by A2,Th10,WELLORD1:42;
end;

theorem Th12:
  for R st R is well-ordering & for a st a in field R ex A st R
  |_2 (R-Seg(a)),RelIncl A are_isomorphic ex A st R,RelIncl A are_isomorphic
proof
  let R such that
A1: R is well-ordering;
  defpred P[set,set] means for A holds A = $2 iff R |_2 (R-Seg $1),RelIncl A
  are_isomorphic;
  assume
A2: for a st a in field R ex A st R |_2 (R-Seg(a)),RelIncl A are_isomorphic;
A3: for a st a in field R ex b st P[a,b]
  proof
    let a;
    assume a in field R;
    then consider A such that
A4: R |_2 (R-Seg(a)),RelIncl A are_isomorphic by A2;
    reconsider b = A as set;
    take b;
    let B;
    thus B = b implies R |_2 (R-Seg(a)),RelIncl B are_isomorphic by A4;
    assume R |_2 (R-Seg(a)),RelIncl B are_isomorphic;
    hence thesis by A4,Th11;
  end;
A5: for b,c,d st b in field R & P[b,c] & P[b,d] holds c = d
  proof
    let b,c,d such that
A6: b in field R and
A7: A = c iff R |_2 (R-Seg(b)),RelIncl A are_isomorphic and
A8: B = d iff R |_2 (R-Seg(b)),RelIncl B are_isomorphic;
    consider A such that
A9: R |_2 (R-Seg(b)),RelIncl A are_isomorphic by A2,A6;
    A = c by A7,A9;
    hence thesis by A8,A9;
  end;
  consider H such that
A10: dom H = field R & for b st b in field R holds P[b, H.b] from
  FUNCT_1:sch 2(A5,A3);
  for a st a in rng H holds a is Ordinal
  proof
    let b;
    assume b in rng H;
    then consider c such that
A11: c in dom H and
A12: b = H.c by FUNCT_1:def 3;
    ex A st R |_2 (R-Seg(c)),RelIncl A are_isomorphic by A2,A10,A11;
    hence thesis by A10,A11,A12;
  end;
  then consider C such that
A13: rng H c= C by ORDINAL1:24;
A14: now
    let b;
    assume
A15: b in rng H;
    then consider b9 being set such that
A16: b9 in dom H and
A17: b = H.b9 by FUNCT_1:def 3;
    set V = R-Seg(b9);
    set P = R |_2 V;
    consider A such that
A18: P,RelIncl A are_isomorphic by A2,A10,A16;
    let c such that
A19: [c,b] in RelIncl C;
A20: A = b by A10,A16,A17,A18;
    now
A21:  C = field RelIncl C by Def1;
      then
A22:  c in C by A19,RELAT_1:15;
      then reconsider B = c as Ordinal;
      b in C by A19,A21,RELAT_1:15;
      then
A23:  B c= A by A20,A19,A22,Def1;
      then
A24:  (RelIncl A) |_2 B = RelIncl B by Th7;
      assume c <> b;
      then
A25:  B c< A by A20,A23,XBOOLE_0:def 8;
      then
A26:  B = (RelIncl A)-Seg(B) by Th9,ORDINAL1:11;
A27:  A = field RelIncl A by Def1;
      RelIncl A,P are_isomorphic by A18,WELLORD1:40;
      then
      canonical_isomorphism_of(RelIncl A,P) is_isomorphism_of RelIncl A,P
      by WELLORD1:def 9;
      then consider d such that
A28:  d in field P and
A29:  (RelIncl A) |_2 ((RelIncl A)-Seg(B)),P |_2 (P-Seg(d))
      are_isomorphic by A25,A27,ORDINAL1:11,WELLORD1:50;
A30:  d in field R by A28,WELLORD1:12;
A31:  P-Seg(d) = R-Seg(d) by A1,A28,WELLORD1:12,27;
      d in V by A28,WELLORD1:12;
      then [d,b9] in R by WELLORD1:1;
      then R-Seg(d) c= R-Seg(b9) by A1,A10,A16,A30,WELLORD1:29;
      then RelIncl B,R |_2 (R-Seg(d)) are_isomorphic by A29,A26,A24,A31,
WELLORD1:22;
      then R |_2 (R-Seg(d)),RelIncl B are_isomorphic by WELLORD1:40;
      then B = H.d by A10,A30;
      hence c in rng H by A10,A30,FUNCT_1:def 3;
    end;
    hence c in rng H by A15;
  end;
A32: (ex a st a in C & rng H = (RelIncl C)-Seg(a)) implies rng H is Ordinal
  by Th9;
  C = field RelIncl C & RelIncl C is well-ordering by Def1;
  then reconsider A = rng H as Ordinal by A13,A14,A32,WELLORD1:28;
  take A;
  take H;
  thus dom H = field R by A10;
  thus rng H = field RelIncl A by Def1;
A33: a in dom H implies H.a is Ordinal
  proof
    assume a in dom H;
    then H.a in A by FUNCT_1:def 3;
    hence thesis;
  end;
  thus
A34: H is one-to-one
  proof
    let a,b;
    assume that
A35: a in dom H and
A36: b in dom H and
A37: H.a = H.b;
    reconsider B = H.a as Ordinal by A33,A35;
    R |_2 (R-Seg(b)),RelIncl B are_isomorphic by A10,A36,A37;
    then
A38: RelIncl B,R |_2 (R-Seg(b)) are_isomorphic by WELLORD1:40;
    R |_2 (R-Seg(a)),RelIncl B are_isomorphic by A10,A35;
    then R |_2 (R-Seg(a)),R |_2 (R-Seg(b)) are_isomorphic by A38,WELLORD1:42;
    hence thesis by A1,A10,A35,A36,WELLORD1:47;
  end;
  let a,b;
  thus [a,b] in R implies a in field R & b in field R & [H.a,H.b] in RelIncl A
  proof
    set Z = R-Seg(b);
    set P = R |_2 Z;
A39: A = field RelIncl A & P is well-ordering by A1,Def1,WELLORD1:25;
    assume
A40: [a,b] in R;
    hence
A41: a in field R & b in field R by RELAT_1:15;
    then reconsider A9 = H.a, B9 = H.b as Ordinal by A10,A33;
A42: R |_2 (R-Seg(b)),RelIncl B9 are_isomorphic by A10,A41;
A43: A9 in A by A10,A41,FUNCT_1:def 3;
A44: B9 in A by A10,A41,FUNCT_1:def 3;
A45: R |_2 (R-Seg(a)),RelIncl A9 are_isomorphic by A10,A41;
    now
      assume a <> b;
      then
A46:  a in Z by A40,WELLORD1:1;
      then
A47:  P-Seg(a) = R-Seg(a) by A1,WELLORD1:27;
      Z c= field R by WELLORD1:9;
      then
A48:  a in field P by A1,A46,WELLORD1:31;
      A9 c= A by A43,ORDINAL1:def 2;
      then
A49:  (RelIncl A) |_2 A9 = RelIncl A9 by Th7;
      A9 = (RelIncl A)-Seg(A9) & R-Seg(a) c= R-Seg(b) by A1,A40,A41,A43,Th9,
WELLORD1:29;
      then
A50:  P |_2 (P-Seg(a)),(RelIncl A) |_2 ((RelIncl A)-Seg(A9))
      are_isomorphic by A45,A49,A47,WELLORD1:22;
      B9 = (RelIncl A)-Seg(B9) & B9 c= A by A44,Th9,ORDINAL1:def 2;
      then P,(RelIncl A) |_2 ((RelIncl A)-Seg(B9)) are_isomorphic by A42,Th7;
      hence [A9,B9] in RelIncl A by A43,A44,A39,A48,A50,WELLORD1:51;
    end;
    hence thesis by A43,Def1;
  end;
  assume that
A51: a in field R and
A52: b in field R and
A53: [H.a,H.b] in RelIncl A;
  assume
A54: not [a,b] in R;
  R is_reflexive_in field R by A1,RELAT_2:def 9;
  then
A55: a <> b by A51,A54,RELAT_2:def 1;
  R is_connected_in field R by A1,RELAT_2:def 14;
  then
A56: [b,a] in R by A51,A52,A54,A55,RELAT_2:def 6;
  then
A57: R-Seg(b) c= R-Seg(a) by A1,A51,A52,WELLORD1:29;
A58: RelIncl A is_antisymmetric_in field RelIncl A by RELAT_2:def 12;
A59: A = field RelIncl A by Def1;
  reconsider A9 = H.a, B9 = H.b as Ordinal by A10,A33,A51,A52;
A60: R |_2 (R-Seg(a)),RelIncl A9 are_isomorphic by A10,A51;
A61: R |_2 (R-Seg(b)),RelIncl B9 are_isomorphic by A10,A52;
A62: B9 in A by A10,A52,FUNCT_1:def 3;
  then B9 c= A by ORDINAL1:def 2;
  then
A63: (RelIncl A) |_2 B9 = RelIncl B9 by Th7;
  set Z = R-Seg(a);
  set P = R |_2 Z;
A64: A9 in A by A10,A51,FUNCT_1:def 3;
  then A9 = (RelIncl A)-Seg(A9) & A9 c= A by Th9,ORDINAL1:def 2;
  then
A65: P,(RelIncl A) |_2 ((RelIncl A)-Seg(A9)) are_isomorphic by A60,Th7;
A66: b in Z by A54,A56,WELLORD1:1;
  then
A67: P-Seg(b) = R-Seg(b) by A1,WELLORD1:27;
  B9 = (RelIncl A)-Seg(B9) by A62,Th9;
  then
A68: P |_2 (P-Seg(b)),(RelIncl A) |_2 ((RelIncl A)-Seg(B9)) are_isomorphic
  by A61,A63,A67,A57,WELLORD1:22;
  Z c= field R by WELLORD1:9;
  then
A69: b in field P by A1,A66,WELLORD1:31;
  P is well-ordering by A1,WELLORD1:25;
  then [B9,A9] in RelIncl A by A69,A64,A62,A59,A65,A68,WELLORD1:51;
  then H.a = H.b by A53,A58,A64,A62,A59,RELAT_2:def 4;
  hence contradiction by A10,A34,A51,A52,A55,FUNCT_1:def 4;
end;

theorem Th13:
  for R st R is well-ordering ex A st R,RelIncl A are_isomorphic
proof
  let R such that
A1: R is well-ordering;
  defpred P[set] means ex A st R |_2 (R-Seg($1)),RelIncl A are_isomorphic;
  consider Z such that
A2: a in Z iff a in field R & P[a] from XBOOLE_0:sch 1;
  now
    let a such that
A3: a in field R and
A4: R-Seg(a) c= Z;
    set P = R |_2 (R-Seg(a));
    now
      let b;
      assume
A5:   b in field P;
      then
A6:   b in R-Seg(a) by WELLORD1:12;
      then
A7:   [b,a] in R by WELLORD1:1;
      b in field R by A5,WELLORD1:12;
      then
A8:   R-Seg(b) c= R-Seg(a) by A1,A3,A7,WELLORD1:29;
      consider A such that
A9:   R |_2 (R-Seg(b)),RelIncl A are_isomorphic by A2,A4,A6;
      take A;
      P-Seg(b) = R-Seg(b) by A1,A5,WELLORD1:12,27;
      hence P |_2 (P-Seg(b)),RelIncl A are_isomorphic by A9,A8,WELLORD1:22;
    end;
    then ex A st P,RelIncl A are_isomorphic by A1,Th12,WELLORD1:25;
    hence a in Z by A2,A3;
  end;
  then field R c= Z by A1,WELLORD1:33;
  then
  for a st a in field R ex A st R |_2 (R-Seg(a)),RelIncl A are_isomorphic
  by A2;
  hence thesis by A1,Th12;
end;

definition
  let R;
  assume
A1: R is well-ordering;
  func order_type_of R -> Ordinal means
  :Def2:
  R,RelIncl it are_isomorphic;
  existence by A1,Th13;
  uniqueness by Th11;
end;

definition
  let A,R;
  pred A is_order_type_of R means
  A = order_type_of R;
end;

theorem
  X c= A implies order_type_of RelIncl X c= A
proof
  assume
A1: X c= A;
  then
A2: (RelIncl A) |_2 X = RelIncl X by Th7;
A3: RelIncl X is well-ordering by A1,Th8;
A4: now
    assume RelIncl A,(RelIncl A) |_2 X are_isomorphic;
    then RelIncl X,RelIncl A are_isomorphic by A2,WELLORD1:40;
    hence thesis by A3,Def2;
  end;
A5: now
    given a such that
A6: a in A and
A7: (RelIncl A) |_2 ((RelIncl A)-Seg(a)),(RelIncl A) |_2 X are_isomorphic;
    reconsider a as Ordinal by A6;
A8: (RelIncl A)-Seg(a) = a by A6,Th9;
A9: a c= A by A6,ORDINAL1:def 2;
    then (RelIncl A) |_2 a = RelIncl a by Th7;
    then RelIncl X,RelIncl a are_isomorphic by A2,A7,A8,WELLORD1:40;
    hence thesis by A3,A9,Def2;
  end;
  field RelIncl A = A by Def1;
  hence thesis by A1,A4,A5,WELLORD1:53;
end;

reserve f,g for Function;

definition
  let X,Y;
  redefine pred X,Y are_equipotent means
  ex f st f is one-to-one & dom f = X & rng f = Y;
  compatibility
  proof
    thus X,Y are_equipotent implies ex f st f is one-to-one & dom f = X & rng
    f = Y
    proof
      assume X,Y are_equipotent;
      then consider Z such that
A1:   for x st x in X ex y st y in Y & [x,y] in Z and
A2:   for y st y in Y ex x st x in X & [x,y] in Z and
A3:   for x,y,z,u st [x,y] in Z & [z,u] in Z holds x = z iff y = u by
TARSKI:def 6;
      set F = Z /\ [:X,Y:];
      for x,y,z st [x,y] in F & [x,z] in F holds y = z
      proof
        let x,y,z;
        assume [x,y] in F & [x,z] in F;
        then [x,y] in Z & [x,z] in Z by XBOOLE_0:def 4;
        hence thesis by A3;
      end;
      then reconsider F as Function by FUNCT_1:def 1;
      take f = F;
      thus f is one-to-one
      proof
        let x,y;
        assume that
A4:     x in dom f and
A5:     y in dom f and
A6:     f.x = f.y;
        [y,f.y] in f by A5,FUNCT_1:1;
        then
A7:     [y,f.y] in Z by XBOOLE_0:def 4;
        [x,f.x] in f by A4,FUNCT_1:1;
        then [x,f.x] in Z by XBOOLE_0:def 4;
        hence thesis by A3,A6,A7;
      end;
      thus dom f c= X
      proof
        let x;
        assume x in dom f;
        then [x,f.x] in f by FUNCT_1:1;
        then [x,f.x] in [:X,Y:] by XBOOLE_0:def 4;
        hence thesis by ZFMISC_1:87;
      end;
      thus X c= dom f
      proof
        let x;
        assume
A8:     x in X;
        then consider y such that
A9:     y in Y and
A10:    [x,y] in Z by A1;
        [x,y] in [:X,Y:] by A8,A9,ZFMISC_1:87;
        then [x,y] in f by A10,XBOOLE_0:def 4;
        hence thesis by FUNCT_1:1;
      end;
      thus rng f c= Y
      proof
        let y;
        assume y in rng f;
        then consider x such that
A11:    x in dom f & y = f.x by FUNCT_1:def 3;
        [x,y] in f by A11,FUNCT_1:1;
        then [x,y] in [:X,Y:] by XBOOLE_0:def 4;
        hence thesis by ZFMISC_1:87;
      end;
      thus Y c= rng f
      proof
        let y;
        assume
A12:    y in Y;
        then consider x such that
A13:    x in X and
A14:    [x,y] in Z by A2;
        [x,y] in [:X,Y:] by A12,A13,ZFMISC_1:87;
        then [x,y] in f by A14,XBOOLE_0:def 4;
        then x in dom f & y = f.x by FUNCT_1:1;
        hence thesis by FUNCT_1:def 3;
      end;
    end;
    (ex f st f is one-to-one & dom f = X & rng f = Y) implies ex Z st (
for x st x in X ex y st y in Y & [x,y] in Z) & (for y st y in Y ex x st x in X
& [x,y] in Z) & for x,y,z,u st [x,y] in Z & [z,u] in Z holds x = z iff y = u
    proof
      given f such that
A15:  f is one-to-one and
A16:  dom f = X and
A17:  rng f = Y;
      take Z = f;
      thus x in X implies ex y st y in Y & [x,y] in Z
      proof
        assume
A18:    x in X;
        take f.x;
        thus f.x in Y by A16,A17,A18,FUNCT_1:def 3;
        thus thesis by A16,A18,FUNCT_1:1;
      end;
      thus for y st y in Y ex x st x in X & [x,y] in Z
      proof
        let y such that
A19:    y in Y;
        take f".y;
A20:    dom(f") = rng f by A15,FUNCT_1:33;
        then
A21:    f".y in rng(f") by A17,A19,FUNCT_1:def 3;
A22:    rng(f") = dom f by A15,FUNCT_1:33;
        hence f".y in X by A16,A17,A19,A20,FUNCT_1:def 3;
        y = f.(f".y) by A15,A17,A19,FUNCT_1:35;
        hence thesis by A22,A21,FUNCT_1:1;
      end;
      let x,y,z,u;
      assume
A23:  [x,y] in Z & [z,u] in Z;
      then
A24:  x in dom f & z in dom f by FUNCT_1:1;
      y = f.x & u = f.z by A23,FUNCT_1:1;
      hence thesis by A15,A24,FUNCT_1:def 4;
    end;
    hence thesis by TARSKI:def 6;
  end;
  reflexivity
  proof
    let X;
    take id X;
    thus thesis;
  end;
  symmetry
  proof
    let X,Y;
    given f such that
A25: f is one-to-one & dom f = X & rng f = Y;
    take f";
    thus thesis by A25,FUNCT_1:33;
  end;
end;

theorem
  X,Y are_equipotent & Y,Z are_equipotent implies X,Z are_equipotent
proof
  given f such that
A1: f is one-to-one & dom f = X & rng f = Y;
  given g such that
A2: g is one-to-one & dom g = Y & rng g = Z;
  take g*f;
  thus thesis by A1,A2,RELAT_1:27,28;
end;

theorem Th16:
  R well_orders X implies field(R|_2 X) = X & R|_2 X is well-ordering
proof
  assume that
A1: R is_reflexive_in X and
A2: R is_transitive_in X and
A3: R is_antisymmetric_in X and
A4: R is_connected_in X and
A5: R is_well_founded_in X;
A6: R|_2 X is_antisymmetric_in X
  proof
    let x,y;
    assume that
A7: x in X & y in X and
A8: [x,y] in R|_2 X & [y,x] in R|_2 X;
    [x,y] in R & [y,x] in R by A8,XBOOLE_0:def 4;
    hence thesis by A3,A7,RELAT_2:def 4;
  end;
A9: R|_2 X is_well_founded_in X
  proof
    let Y;
    assume Y c= X & Y <> {};
    then consider a such that
A10: a in Y and
A11: R-Seg(a) misses Y by A5,WELLORD1:def 3;
    take a;
    thus a in Y by A10;
    assume not thesis;
    then consider x being set such that
A12: x in (R|_2 X)-Seg(a) and
A13: x in Y by XBOOLE_0:3;
    [x,a] in R|_2 X by A12,WELLORD1:1;
    then
A14: [x,a] in R by XBOOLE_0:def 4;
    x <> a by A12,WELLORD1:1;
    then x in R-Seg(a) by A14,WELLORD1:1;
    hence contradiction by A11,A13,XBOOLE_0:3;
  end;
A15: R|_2 X is_transitive_in X
  proof
    let x,y,z;
    assume that
A16: x in X and
A17: y in X and
A18: z in X and
A19: [x,y] in R|_2 X & [y,z] in R|_2 X;
    [x,y] in R & [y,z] in R by A19,XBOOLE_0:def 4;
    then
A20: [x,z] in R by A2,A16,A17,A18,RELAT_2:def 8;
    [x,z] in [:X,X:] by A16,A18,ZFMISC_1:87;
    hence thesis by A20,XBOOLE_0:def 4;
  end;
A21: R|_2 X is_connected_in X
  proof
    let x,y;
    assume that
A22: x in X & y in X and
A23: x <> y;
A24: [x,y] in [:X,X:] & [y,x] in [:X,X :] by A22,ZFMISC_1:87;
    [x,y] in R or [y,x] in R by A4,A22,A23,RELAT_2:def 6;
    hence thesis by A24,XBOOLE_0:def 4;
  end;
  thus
A25: field(R|_2 X) = X
  proof
    thus field(R|_2 X) c= X by WELLORD1:13;
    let x;
    assume x in X;
    then [x,x] in R & [x,x] in [:X,X:] by A1,RELAT_2:def 1,ZFMISC_1:87;
    then [x,x] in R|_2 X by XBOOLE_0:def 4;
    hence thesis by RELAT_1:15;
  end;
  R|_2 X is_reflexive_in X
  proof
    let x;
    assume x in X;
    then [x,x] in R & [x,x] in [:X,X:] by A1,RELAT_2:def 1,ZFMISC_1:87;
    hence thesis by XBOOLE_0:def 4;
  end;
  then R|_2 X well_orders X by A15,A6,A21,A9,WELLORD1:def 5;
  hence thesis by A25,WELLORD1:4;
end;





Lm1: R is well-ordering & X,field R are_equipotent implies ex R st R
well_orders X
proof
  assume
A1: R is well-ordering;
  given f such that
A2: f is one-to-one and
A3: dom f = X and
A4: rng f = field R;
  defpred P[set,set] means [f.$1,f.$2] in R;
  consider Q being Relation such that
A5: [x,y] in Q iff x in X & y in X & P[x,y] from RELAT_1:sch 1;
  take Q;
A6: R is_reflexive_in field R by A1,RELAT_2:def 9;
A7: field Q = X
  proof
    thus field Q c= X
    proof
      let x;
      assume that
A8:   x in field Q and
A9:   not x in X;
      for y holds not [y,x] in Q by A5,A9;
      then
A10:  not x in rng Q by XTUPLE_0:def 13;
      for y holds not [x,y] in Q by A5,A9;
      then not x in dom Q by XTUPLE_0:def 12;
      hence contradiction by A8,A10,XBOOLE_0:def 3;
    end;
    let x;
    assume
A11: x in X;
    then f.x in rng f by A3,FUNCT_1:def 3;
    then [f.x,f.x] in R by A6,A4,RELAT_2:def 1;
    then [x,x] in Q by A5,A11;
    hence thesis by RELAT_1:15;
  end;
  f is_isomorphism_of Q,R
  proof
    thus dom f = field Q & rng f = field R & f is one-to-one by A2,A3,A4,A7;
    let x,y;
    thus [x,y] in Q implies x in field Q & y in field Q & [f.x,f.y] in R by A5
,A7;
    assume x in field Q & y in field Q & [f.x,f.y] in R;
    hence thesis by A5,A7;
  end;
  then f" is_isomorphism_of R,Q by WELLORD1:39;
  then Q is well-ordering by A1,WELLORD1:44;
  hence thesis by A7,WELLORD1:4;
end;


theorem Th17:
  for X ex R st R well_orders X
proof
  deffunc F(set) = {$1};
  defpred P[set] means $1 is Ordinal;
  let X;
  consider Class being set such that
A1: X in Class and
A2: Y in Class & Z c= Y implies Z in Class and
  Y in Class implies bool Y in Class and
A3: Y c= Class implies Y,Class are_equipotent or Y in Class by ZFMISC_1:112;
  consider ON being set such that
A4: x in ON iff x in Class & P[x] from XBOOLE_0:sch 1;
  for Y st Y in ON holds Y is Ordinal & Y c= ON
  proof
    let Y;
    assume
A5: Y in ON;
    hence Y is Ordinal by A4;
    reconsider A = Y as Ordinal by A4,A5;
    let x;
    assume
A6: x in Y;
    then x in A;
    then reconsider B = x as Ordinal;
A7: B c= A by A6,ORDINAL1:def 2;
    A in Class by A4,A5;
    then B in Class by A2,A7;
    hence thesis by A4;
  end;
  then reconsider ON as Ordinal by ORDINAL1:19;
A8: ON c= Class
  proof
    let x;
    thus thesis by A4;
  end;
A9: ON,Class are_equipotent
  proof
    assume not thesis;
    then ON in Class by A3,A8;
    then ON in ON by A4;
    hence contradiction;
  end;
  field RelIncl ON = ON by Def1;
  then consider R such that
A10: R well_orders Class by A9,Lm1;
  consider f such that
A11: dom f = X & for x st x in X holds f.x = F(x) from FUNCT_1:sch 3;
A12: rng f c= Class
  proof
    let x;
    assume x in rng f;
    then consider y such that
A13: y in dom f and
A14: x = f.y by FUNCT_1:def 3;
A15: { y } c= X
    proof
      let z;
      assume z in { y };
      hence thesis by A11,A13,TARSKI:def 1;
    end;
    f.y = { y } by A11,A13;
    hence thesis by A1,A2,A14,A15;
  end;
A16: X,rng f are_equipotent
  proof
    take f;
    thus f is one-to-one
    proof
      let x,y;
      assume that
A17:  x in dom f & y in dom f and
A18:  f.x = f.y;
      f.x = { x } & f.y = { y } by A11,A17;
      hence thesis by A18,ZFMISC_1:3;
    end;
    thus thesis by A11;
  end;
  set Q = R|_2 Class;
  field Q = Class by A10,Th16;
  then
A19: field(Q|_2 rng f) = rng f by A10,A12,Th16,WELLORD1:31;
  Q is well-ordering by A10,Th16;
  hence thesis by A16,A19,Lm1,WELLORD1:25;
end;

reserve M for non empty set;


theorem
  (for X st X in M holds X <> {}) & (for X,Y st X in M & Y in M & X <> Y
holds X misses Y) implies ex Choice being set st for X st X in M ex x st Choice
  /\ X = { x }
proof
  assume that
A1: for X st X in M holds X <> {} and
A2: for X,Y st X in M & Y in M & X <> Y holds X misses Y;
  consider R such that
A3: R well_orders union M by Th17;
A4: R is_reflexive_in union M by A3,WELLORD1:def 5;
A5: R is_connected_in union M by A3,WELLORD1:def 5;
  defpred Ch[set,set] means $2 in $1 & for z st z in $1 holds [$2,z] in R;
A6: R is_antisymmetric_in union M by A3,WELLORD1:def 5;
A7: for x,y,z st x in M & Ch[x,y] & Ch[x,z] holds y = z
  proof
    let x,y,z such that
A8: x in M;
    assume that
A9: y in x and
A10: for u st u in x holds [y,u] in R;
A11: y in union M by A8,A9,TARSKI:def 4;
    assume that
A12: z in x and
A13: for u st u in x holds [z,u] in R;
A14: z in union M by A8,A12,TARSKI:def 4;
    [y,z] in R & [z,y] in R by A9,A10,A12,A13;
    hence thesis by A6,A11,A14,RELAT_2:def 4;
  end;
A15: R is_well_founded_in union M by A3,WELLORD1:def 5;
A16: for x st x in M ex y st Ch[x,y]
  proof
    let x;
    assume
A17: x in M;
    then
A18: x c= union M by ZFMISC_1:74;
    x <> {} by A1,A17;
    then consider y such that
A19: y in x and
A20: R-Seg(y) misses x by A15,A18,WELLORD1:def 3;
    take y;
    thus y in x by A19;
    let z;
    assume
A21: z in x;
    then
A22: not z in R-Seg(y) by A20,XBOOLE_0:3;
    y <> z implies [y,z] in R or [z,y] in R by A5,A18,A19,A21,RELAT_2:def 6;
    hence thesis by A4,A18,A21,A22,RELAT_2:def 1,WELLORD1:1;
  end;
  consider f such that
A23: dom f = M & for x st x in M holds Ch[x,f.x] from FUNCT_1:sch 2(A7,
  A16 );
  take Choice = rng f;
  let X such that
A24: X in M;
  take x = f.X;
  thus Choice /\ X c= { x }
  proof
    let y;
    assume
A25: y in Choice /\ X;
    then
A26: y in X by XBOOLE_0:def 4;
    y in Choice by A25,XBOOLE_0:def 4;
    then consider z such that
A27: z in dom f and
A28: y = f.z by FUNCT_1:def 3;
    assume not y in { x };
    then X <> z by A28,TARSKI:def 1;
    then X misses z by A2,A23,A24,A27;
    then
A29: X /\ z = {} by XBOOLE_0:def 7;
    f.z in z by A23,A27;
    hence contradiction by A26,A28,A29,XBOOLE_0:def 4;
  end;
  let y;
  assume y in { x };
  then
A30: y = x by TARSKI:def 1;
  then
A31: y in X by A24,A23;
  y in Choice by A23,A24,A30,FUNCT_1:def 3;
  hence thesis by A31,XBOOLE_0:def 4;
end;

begin 

theorem 
  for X being set holds RelIncl X is_reflexive_in X
proof
  let X be set;
  RelIncl X is reflexive & field RelIncl X = X by Def1;
  hence thesis by RELAT_2:def 9;
end;

theorem 
  for X being set holds RelIncl X is_transitive_in X
proof
  let X be set;
  RelIncl X is transitive & field RelIncl X = X by Def1;
  hence thesis by RELAT_2:def 16;
end;

theorem 
  for X being set holds RelIncl X is_antisymmetric_in X
proof
  let X be set;
  RelIncl X is antisymmetric & field RelIncl X = X by Def1;
  hence thesis by RELAT_2:def 12;
end;



registration
  cluster RelIncl {} -> empty;
  coherence
  proof
    for Y,Z being set st Y in {} & Z in {} holds [Y,Z] in {} iff Y c= Z;
    hence thesis by Def1,RELAT_1:40;
  end;
end;

registration
  let X be non empty set;
  cluster RelIncl X -> non empty;
  coherence
  proof
    set a = the Element of X;
    [a,a] in RelIncl X by Def1;
    hence thesis;
  end;
end;

theorem
  RelIncl {x} = {[x,x]}
proof
A1: for Y,Z being set st Y in {x} & Z in {x} holds [Y,Z] in {[x,x]} iff Y c= Z
  proof
    let Y,Z be set;
    assume that
A2: Y in {x} and
A3: Z in {x};
A4: Y = x by A2,TARSKI:def 1;
    hence [Y,Z] in {[x,x]} implies Y c= Z by A3,TARSKI:def 1;
    Z = x by A3,TARSKI:def 1;
    hence thesis by A4,TARSKI:def 1;
  end;
  field {[x,x]} = {x} by RELAT_1:173;
  hence thesis by A1,Def1;
end;

theorem
  RelIncl X c= [:X,X:]
proof
  set R = RelIncl X;
  let a,b be set;
  assume
A1: [a,b] in R;
  then b in field R by RELAT_1:15;
  then
A2: b in X by Def1;
  a in field R by A1,RELAT_1:15;
  then a in X by Def1;
  hence thesis by A2,ZFMISC_1:87;
end;














begin 

reserve P,Q,X,Y,Z,p,x,x9,x1,x2,y,z for set;

definition
  let X,Y;
  let R be Relation of X,Y;
  attr R is quasi_total means
  :Def1:
  X = dom R if Y <> {} otherwise R = {};
  consistency;
end;

registration
  let X,Y;
  cluster quasi_total for PartFunc of X,Y;
  existence
  proof
    per cases;
    suppose
A1:   Y = {};
      reconsider R = {} as PartFunc of X,Y by RELSET_1:12;
      take R;
      thus thesis by A1,Def1;
    end;
    suppose
A2:   Y <> {};
      then consider f being Function such that
A3:   X = dom f and
A4:   rng f c= Y by FUNCT_1:8;
      reconsider R = f as PartFunc of X,Y by A3,A4,RELSET_1:4;
      take R;
      thus thesis by A2,A3,Def1;
    end;
  end;
end;

registration
  let X,Y;
  cluster total -> quasi_total for Relation of X,Y;
  coherence
  proof
    let f be Relation of X,Y;
    assume
A1: dom f = X;
    per cases;
    case Y <> {};
      thus thesis by A1;
    end;
    case Y = {};
      hence thesis;
    end;
  end;
end;

definition
  let X,Y;
  mode Function of X,Y is quasi_total PartFunc of X,Y;
end;

registration let X be empty set, Y be set;
 cluster quasi_total -> total for Relation of X,Y;
 coherence;
end;

registration let X be set, Y be non empty set;
 cluster quasi_total -> total for Relation of X,Y;
 coherence
  proof let F be Relation of X,Y;
   assume F is quasi_total;
   hence X = dom F by Def1;
  end;
end;

registration let X be set;
 cluster quasi_total -> total for Relation of X,X;
 coherence
  proof
   per cases;
   suppose X = {};
    hence thesis;
   end;
   suppose X <> {};
    hence thesis;
   end;
  end;
end;

theorem
  for f being Function holds f is Function of dom f, rng f
proof
  let f be Function;
  reconsider R = f as Relation of dom f, rng f by RELSET_1:4;
  rng R <> {} or rng R = {};
  hence thesis by Def1;
end;

theorem Th2:
  for f being Function st rng f c= Y holds f is Function of dom f, Y
proof
  let f be Function;
  assume rng f c= Y;
  then reconsider R = f as Relation of dom f,Y by RELSET_1:4;
  Y = {} or Y <> {};
  then R is quasi_total by Def1;
  hence thesis;
end;

theorem
  for f being Function st dom f = X & for x st x in X holds f.x in Y
  holds f is Function of X,Y
proof
  let f be Function such that
A1: dom f = X and
A2: for x st x in X holds f.x in Y;
  rng f c= Y
  proof
    let y;
    assume y in rng f;
    then ex x st x in X & y = f.x by A1,FUNCT_1:def 3;
    hence thesis by A2;
  end;
  then reconsider R = f as Relation of dom f,Y by RELSET_1:4;
  Y = {} or Y <> {};
  then R is quasi_total by Def1;
  hence thesis by A1;
end;

theorem Th4:
  for f being Function of X,Y st Y <> {} & x in X holds f.x in rng f
proof
  let f be Function of X,Y;
  assume Y <> {};
  then dom f = X by Def1;
  hence thesis by FUNCT_1:def 3;
end;

theorem Th5:
  for f being Function of X,Y st Y <> {} & x in X holds f.x in Y
proof
  let f be Function of X,Y;
  assume Y <> {} & x in X;
  then f.x in rng f by Th4;
  hence thesis;
end;

theorem
  for f being Function of X,Y st (Y = {} implies X = {}) & rng f c= Z
  holds f is Function of X,Z
proof
  let f be Function of X,Y;
  assume Y <> {} or X = {};
  then
A1: dom f = X by Def1;
  assume
A2: rng f c= Z;
  now
    assume Z = {};
    then rng f = {} by A2;
    hence X = {} by A1,RELAT_1:42;
  end;
  hence thesis by A1,A2,Def1,RELSET_1:4;
end;

theorem
  for f being Function of X,Y st (Y = {} implies X = {}) & Y c= Z holds
  f is Function of X,Z by RELSET_1:7;

scheme
  FuncEx1{X, Y() -> set, P[set,set]}: ex f being Function of X(),Y() st for x
  st x in X() holds P[x,f.x]
provided
A1: for x st x in X() ex y st y in Y() & P[x,y]
proof
  consider f being Function such that
A2: dom f = X() & rng f c= Y() and
A3: for x st x in X() holds P[x,f.x] from FUNCT_1:sch 5(A1);
  reconsider f as Function of X(),Y() by A2,Th2;
 take f;
 thus thesis by A3;
end;

scheme
  Lambda1{X, Y() -> set, F(set)->set}: ex f being Function of X(),Y() st for x
  st x in X() holds f.x = F(x)
provided
A1: for x st x in X() holds F(x) in Y()
proof
  defpred P[set,set] means $2 = F($1);
A2: for x st x in X() ex y st y in Y() & P[x,y] by A1;
  thus ex f being Function of X(),Y() st for x st x in X() holds P[x,f.x] from
  FuncEx1(A2);
end;

definition
  let X,Y;
  func Funcs(X,Y) -> set means
  :Def2:
  x in it iff ex f being Function st x = f & dom f = X & rng f c= Y;
  existence
  proof
    defpred P[set] means ex f being Function st $1 = f & dom f = X & rng f c=
    Y;
    consider F being set such that
A1: z in F iff z in bool [:X,Y:] & P[z] from XBOOLE_0:sch 1;
    take F;
    let z;
    thus z in F implies ex f being Function st z = f & dom f = X & rng f c= Y
    by A1;
    given f being Function such that
A2: z = f and
A3: dom f = X & rng f c= Y;
    f c= [:X,Y:]
    proof
      let p;
      assume
A4:   p in f;
      then consider x,y such that
A5:   p = [x,y] by RELAT_1:def 1;
A6:   x in dom f by A4,A5,XTUPLE_0:def 12;
      then y = f.x by A4,A5,FUNCT_1:def 2;
      then y in rng f by A6,FUNCT_1:def 3;
      hence thesis by A3,A5,A6,ZFMISC_1:def 2;
    end;
    hence thesis by A1,A2,A3;
  end;
  uniqueness
  proof
    let F1,F2 be set such that
A7: x in F1 iff ex f being Function st x = f & dom f = X & rng f c= Y and
A8: x in F2 iff ex f being Function st x = f & dom f = X & rng f c= Y;
    x in F1 iff x in F2
    proof
      x in F1 iff ex f being Function st x = f & dom f = X & rng f c= Y by A7;
      hence thesis by A8;
    end;
    hence thesis by TARSKI:1;
  end;
end;

theorem Th8:
  for f being Function of X,Y st Y = {} implies X = {} holds f in Funcs(X,Y)
proof
  let f be Function of X,Y;
  assume Y = {} implies X = {};
  then
A1: dom f = X by Def1;
  rng f c= Y;
  hence thesis by A1,Def2;
end;

theorem
  for f being Function of X,X holds f in Funcs(X,X) by Th8;

registration
  let X be set, Y be non empty set;
  cluster Funcs(X,Y) -> non empty;
  coherence by Th8;
end;

registration
  let X be set;
  cluster Funcs(X,X) -> non empty;
  coherence by Th8;
end;

registration
  let X be non empty set, Y be empty set;
  cluster Funcs(X,Y) -> empty;
  coherence
  proof
    assume Funcs(X,Y) is non empty;
    then consider f being Function such that
    the Element of Funcs(X,Y) = f and
A1: dom f = X and
A2: rng f c= {} by Def2;
    rng f = {} by A2;
    hence contradiction by A1,RELAT_1:42;
  end;
end;

theorem
  for f being Function of X,Y st for y st y in Y ex x st x in X & y = f.x
  holds rng f = Y
proof
  let f be Function of X,Y such that
A1: for y st y in Y ex x st x in X & y = f.x;
  per cases;
  suppose Y = {};
   hence thesis;
  end;
  suppose A2: Y <> {};
  y in rng f iff y in Y
  proof
    dom f = X by A2,Def1;
    then y in rng f iff ex x st x in X & y = f.x by FUNCT_1:def 3;
    hence thesis by A1;
  end;
  hence thesis by TARSKI:1;
  end;
end;

theorem Th11:
  for f being Function of X,Y st y in rng f ex x st x in X & f.x = y
proof
  let f be Function of X,Y;
  assume
A1: y in rng f;
  then dom f = X by Def1;
  hence thesis by A1,FUNCT_1:def 3;
end;

theorem Th12:
  for f1,f2 being Function of X,Y st for x st x in X holds f1.x = f2.x
  holds f1 = f2
proof
  let f1,f2 be Function of X,Y;
  per cases;
  suppose Y = {};
    hence thesis;
  end;
  suppose Y <> {};
    then dom f1 = X & dom f2 = X by Def1;
    hence thesis by FUNCT_1:2;
  end;
end;

theorem Th13:
  for f being quasi_total Relation of X,Y for g being quasi_total
  Relation of Y,Z st Y = {} implies Z = {} or X = {} holds f*g is quasi_total
proof
  let f be quasi_total Relation of X,Y;
  let g be quasi_total Relation of Y,Z such that
A1: Y = {} implies Z = {} or X = {};
  per cases;
  case
A2: Z <> {};
    then
A3: dom g = Y by Def1;
    dom f = X & rng f c= Y by A1,A2,Def1;
    hence thesis by A3,RELAT_1:27;
  end;
  case Z = {};
    hence thesis;
  end;
end;

theorem
  for f being Function of X,Y for g being Function of Y,Z st Z <> {} &
  rng f = Y & rng g = Z holds rng(g*f) = Z
proof
  let f be Function of X,Y;
  let g be Function of Y,Z;
  assume Z <> {};
  then dom g = Y by Def1;
  hence thesis by RELAT_1:28;
end;

theorem Th15:
  for f being Function of X,Y, g being Function st Y <> {} & x in
  X holds (g*f).x = g.(f.x)
proof
  let f be Function of X,Y, g be Function;
  assume Y <> {};
  then X = dom f by Def1;
  hence thesis by FUNCT_1:13;
end;

theorem
  for f being Function of X,Y st Y <> {} holds rng f = Y iff for Z st Z
  <> {} for g,h being Function of Y,Z st g*f = h*f holds g = h
proof
  let f be Function of X,Y;
  assume
A1: Y <> {};
  thus rng f = Y implies for Z st Z <> {} for g,h being Function of Y,Z st g*f
  = h*f holds g = h
  proof
    assume
A2: rng f = Y;
    let Z such that
A3: Z <> {};
    let g,h be Function of Y,Z;
    dom g = Y & dom h = Y by A3,Def1;
    hence thesis by A2,FUNCT_1:86;
  end;
  assume
A4: for Z st Z <> {} for g,h being Function of Y,Z st g*f = h*f holds g = h;
  for g,h being Function st dom g = Y & dom h = Y & g*f = h*f holds g = h
  proof
    let g,h be Function;
    assume that
A5: dom g = Y and
A6: dom h = Y;
A7: rng g <> {} by A1,A5,RELAT_1:42;
A8: g is Function of Y,rng g \/ rng h by A5,Th2,XBOOLE_1:7;
    h is Function of Y,rng g \/ rng h by A6,Th2,XBOOLE_1:7;
    hence thesis by A4,A7,A8;
  end;
  hence thesis by FUNCT_1:16;
end;

theorem
  for f being Relation of X,Y holds (id X)*f = f & f*(id Y) = f
proof
  let f be Relation of X,Y;
  dom f c= X;
  hence (id X)*f = f by RELAT_1:51;
  rng f c= Y;
  hence thesis by RELAT_1:53;
end;

theorem
  for f being Function of X,Y for g being Function of Y,X st f*g = id Y
  holds rng f = Y
proof
  let f be (Function of X,Y),g be Function of Y,X;
  assume f*g = id Y;
  then rng(f*g) = Y by RELAT_1:45;
  then Y c= rng f by RELAT_1:26;
  hence thesis by XBOOLE_0:def 10;
end;

theorem
  for f being Function of X,Y st Y = {} implies X = {} holds f is
  one-to-one iff for x1,x2 st x1 in X & x2 in X & f.x1 = f.x2 holds x1 = x2
proof
  let f be Function of X,Y;
  assume Y = {} implies X = {};
  then dom f = X by Def1;
  hence thesis by FUNCT_1:def 4;
end;

theorem
  for f being Function of X,Y for g being Function of Y,Z st (Z = {}
  implies Y = {}) & g*f is one-to-one holds f is one-to-one
proof
  let f be Function of X,Y;
  let g be Function of Y,Z;
  assume Z <> {} or Y = {};
  then
A1: Y = dom g by Def1;
  rng f c= Y;
  hence thesis by A1,FUNCT_1:25;
end;

theorem
  for f being Function of X,Y st X <> {} & Y <> {} holds f is one-to-one
  iff for Z for g,h being Function of Z,X st f*g = f*h holds g = h
proof
  let f be Function of X,Y;
  assume that
A1: X <> {} and
A2: Y <> {};
A3: dom f = X by A2,Def1;
  thus f is one-to-one implies for Z for g,h being Function of Z,X st f*g = f*
  h holds g = h
  proof
    assume
A4: f is one-to-one;
    let Z;
    let g,h be Function of Z,X;
A5: rng g c= X & rng h c= X;
    dom g = Z & dom h = Z by A1,Def1;
    hence thesis by A3,A4,A5,FUNCT_1:27;
  end;
  assume
A6: for Z for g,h being Function of Z,X st f*g = f*h holds g = h;
  now
    let g,h be Function;
    assume rng g c= dom f & rng h c= dom f & dom g = dom h;
    then g is Function of dom g,X & h is Function of dom g,X by A3,Th2;
    hence f*g = f*h implies g = h by A6;
  end;
  hence thesis by FUNCT_1:27;
end;

theorem
  for f being Function of X,Y for g being Function of Y,Z st Z <> {} &
  rng(g*f) = Z & g is one-to-one holds rng f = Y
proof
  let f be Function of X,Y;
  let g be Function of Y,Z;
  assume that
A1: Z <> {} and
A2: rng(g*f) = Z and
A3: g is one-to-one;
A4: dom g = Y by A1,Def1;
  rng(g*f) c= rng g by RELAT_1:26;
  then rng g = rng(g*f) by A2,XBOOLE_0:def 10;
  then Y c= rng f by A3,A4,FUNCT_1:29;
  hence thesis by XBOOLE_0:def 10;
end;

definition
  let Y be set;
  let f be Y-valued Relation;
  attr f is onto means
  :Def3:
  rng f = Y;
end;

theorem
  for f being Function of X,Y for g being Function of Y,X st g*f = id X
   holds f is one-to-one & g is onto
proof
  let f be Function of X,Y;
  let g be Function of Y,X;
  assume that
A1: g*f = id X;
  thus f is one-to-one
   proof
    per cases;
    suppose Y = {};
     hence thesis;
    end;
    suppose Y <> {};
     then dom f = X by Def1;
    hence thesis by A1,FUNCT_1:31;
    end;
   end;
  rng(g*f) = X by A1,RELAT_1:45;
  then X c= rng g by RELAT_1:26;
  hence rng g = X by XBOOLE_0:def 10;
end;

theorem
  for f being Function of X,Y for g being Function of Y,Z st (Z = {}
  implies Y = {}) & g*f is one-to-one & rng f = Y holds f is one-to-one & g is
  one-to-one
proof
  let f be Function of X,Y;
  let g be Function of Y,Z;
  assume Z <> {} or Y = {};
  then Y = dom g by Def1;
  hence thesis by FUNCT_1:26;
end;

theorem Th25:
  for f being Function of X,Y st f is one-to-one & rng f = Y holds
  f" is Function of Y,X
proof
  let f be Function of X,Y;
  assume that
A1: f is one-to-one and
A2: rng f = Y;
A3: rng(f") c= X
  proof
    let x;
    assume x in rng(f");
    then x in dom f by A1,FUNCT_1:33;
    hence thesis;
  end;
  dom(f") = Y by A1,A2,FUNCT_1:33;
  then reconsider R = f" as Relation of Y,X by A3,RELSET_1:4;
  R is quasi_total
  proof
    per cases;
    case X <> {};
      thus thesis by A1,A2,FUNCT_1:33;
    end;
    case X = {};
      then rng f = {};
      then dom(f") = {} by A1,FUNCT_1:32;
      hence thesis;
    end;
  end;
  hence thesis;
end;

theorem
  for f being Function of X,Y st Y <> {} & f is one-to-one & x in X
  holds (f").(f.x) = x
proof
  let f be Function of X,Y;
  assume Y <> {};
  then dom f = X by Def1;
  hence thesis by FUNCT_1:34;
end;

theorem
  for X be set, Y,Z be non empty set for f be Function of X,Y
  for g be Function of Y,Z holds f is onto & g is onto implies g*f is onto
proof
  let X be set, Y,Z be non empty set;
  let f be Function of X,Y;
  let g be Function of Y,Z;
  assume that
A1: f is onto and
A2: g is onto;
  rng f = Y by A1,Def3
    .= dom g by Def1;
  then rng(g*f) = rng g by RELAT_1:28
    .= Z by A2,Def3;
  hence thesis by Def3;
end;

theorem
  for f being Function of X,Y for g being Function of Y,X st X <> {} & Y
<> {} & rng f = Y & f is one-to-one & for y,x holds y in Y & g.y = x iff x in X
  & f.x = y holds g = f"
proof
  let f be Function of X,Y;
  let g be Function of Y,X;
  assume X <> {} & Y <> {};
  then dom f = X & dom g = Y by Def1;
  hence thesis by FUNCT_1:32;
end;

theorem
  for f being Function of X,Y st Y <> {} & rng f = Y & f is one-to-one
  holds f"*f = id X & f*f" = id Y
proof
  let f be Function of X,Y;
  assume Y <> {};
  then dom f = X by Def1;
  hence thesis by FUNCT_1:39;
end;

theorem
  for f being Function of X,Y for g being Function of Y,X st X <> {} & Y
  <> {} & rng f = Y & g*f = id X & f is one-to-one holds g = f"
proof
  let f be Function of X,Y;
  let g be Function of Y,X;
  assume X <> {} & Y <> {};
  then dom f = X & dom g = Y by Def1;
  hence thesis by FUNCT_1:41;
end;

theorem
  for f being Function of X,Y st Y <> {} & ex g being Function of Y,X st
  g*f = id X holds f is one-to-one
proof
  let f be Function of X,Y;
  assume Y <> {};
  then
A1: dom f = X by Def1;
  given g being Function of Y,X such that
A2: g*f = id X;
  thus thesis by A2,A1,FUNCT_1:31;
end;

theorem Th32:
  for f being Function of X,Y st (Y = {} implies X = {}) & Z c= X
  holds f|Z is Function of Z,Y
proof
  let f be Function of X,Y such that
A1: Y = {} implies X = {} and
A2: Z c= X;
  dom f = X by A1,Def1;
  then
A3: Z = dom(f|Z) by A2,RELAT_1:62;
  rng(f|Z) c= Y;
  then reconsider R = f|Z as Relation of Z,Y by A3,RELSET_1:4;
  R is quasi_total
  proof
    per cases;
    case Y <> {};
      dom f = X by A1,Def1;
      hence thesis by A2,RELAT_1:62;
    end;
    case Y = {};
      hence thesis;
    end;
  end;
  hence thesis;
end;

theorem
  for f being Function of X,Y st X c= Z holds f|Z = f by RELSET_1:19;

theorem
  for f being Function of X,Y st Y <> {} & x in X & f.x in Z holds (Z|`f)
  .x = f.x
proof
  let f be Function of X,Y;
  assume that
A1: Y <> {} & x in X and
A2: f.x in Z;
  x in dom f by A1,Def1;
  then x in dom(Z|`f) by A2,FUNCT_1:54;
  hence thesis by FUNCT_1:55;
end;

theorem
  for f being Function of X,Y st Y <> {} for y holds (ex x st x in X & x
  in P & y = f.x) implies y in f.:P
proof
  let f be Function of X,Y;
  assume Y <> {};
  then
A1: dom f = X by Def1;
  let y;
  given x such that
A2: x in X & x in P & y = f.x;
  thus thesis by A1,A2,FUNCT_1:def 6;
end;

theorem
  for f being Function of X,Y holds f.:P c= Y;

canceled;

theorem
  for f being Function of X,Y st Y <> {} for x holds x in f"Q iff x in X
  & f.x in Q
proof
  let f be Function of X,Y;
  assume Y <> {};
  then dom f = X by Def1;
  hence thesis by FUNCT_1:def 7;
end;

theorem
  for f being PartFunc of X,Y holds f"Q c= X;

theorem Th40:
  for f being Function of X,Y st Y = {} implies X = {} holds f"Y = X
proof
  let f be Function of X,Y;
  rng f /\ Y = rng f by XBOOLE_1:28;
  then
A1: f"Y = f"(rng f) by RELAT_1:133;
  assume Y <> {} or X = {};
  then dom f = X by Def1;
  hence thesis by A1,RELAT_1:134;
end;

theorem
  for f being Function of X,Y holds (for y st y in Y holds f"{y} <> {})
  iff rng f = Y
proof
  let f be Function of X,Y;
  thus (for y st y in Y holds f"{y} <> {}) implies rng f = Y
  proof
    assume for y st y in Y holds f"{y} <> {};
    then Y c= rng f by FUNCT_1:73;
    hence thesis by XBOOLE_0:def 10;
  end;
  thus thesis by FUNCT_1:72;
end;

theorem Th42:
  for f being Function of X,Y st (Y = {} implies X = {}) & P c= X
  holds P c= f"(f.:P)
proof
  let f be Function of X,Y;
  assume Y <> {} or X = {};
  then dom f = X by Def1;
  hence thesis by FUNCT_1:76;
end;

theorem
  for f being Function of X,Y st Y = {} implies X = {} holds f"(f.:X) = X
proof
  let f be Function of X,Y;
  assume Y <> {} or X = {};
  then
A1: dom f = X by Def1;
  then f"(rng f) = X by RELAT_1:134;
  hence thesis by A1,RELAT_1:113;
end;

theorem
  for f being Function of X,Y for g being Function of Y,Z st (Z = {}
  implies Y = {}) holds f"Q c= (g*f)"(g.:Q)
proof
  let f be Function of X,Y;
  let g be Function of Y,Z;
  assume Z <> {} or Y = {};
  then
A1: dom g = Y by Def1;
  rng f c= Y;
  hence thesis by A1,FUNCT_1:90;
end;

theorem
  for f being Function of {},Y holds f.:P = {};

theorem
  for f being Function of {},Y holds f"Q = {};

theorem
  for f being Function of {x},Y st Y <> {} holds f.x in Y
proof
  let f be Function of {x},Y;
  assume Y <> {};
  then
A1: dom f = {x} by Def1;
  rng f c= Y;
  then {f.x} c= Y by A1,FUNCT_1:4;
  hence thesis by ZFMISC_1:31;
end;

theorem Th48:
  for f being Function of {x},Y st Y <> {} holds rng f = {f.x}
proof
  let f be Function of {x},Y;
  assume Y <> {};
  then dom f = {x} by Def1;
  hence thesis by FUNCT_1:4;
end;

theorem
  for f being Function of {x},Y st Y <> {} holds f.:P c= {f.x}
proof
  let f be Function of {x},Y;
  f.:P c= rng f by RELAT_1:111;
  hence thesis by Th48;
end;

theorem Th50:
  for f being Function of X,{y} st x in X holds f.x = y
proof
  let f be Function of X,{y};
  x in X implies f.x in {y} by Th5;
  hence thesis by TARSKI:def 1;
end;

theorem Th51:
  for f1,f2 being Function of X,{y} holds f1 = f2
proof
  let f1,f2 be Function of X,{y};
  x in X implies f1.x = f2.x
  proof
    assume
A1: x in X;
    then f1.x = y by Th50;
    hence thesis by A1,Th50;
  end;
  hence thesis by Th12;
end;

theorem Th52:
  for f being Function of X,X holds dom f = X
proof
  X = {} implies X = {};
  hence thesis by Def1;
end;

registration
  let X,Y be set;
  let f be quasi_total PartFunc of X,Y;
  let g be quasi_total PartFunc of X,X;
  cluster f*g -> quasi_total for PartFunc of X,Y;
  coherence
  proof
    per cases;
    suppose
      Y = {};
      hence thesis;
    end;
    suppose
A1:   Y <> {};
      then dom f = X by Def1;
      then dom(f*g) = g"X by RELAT_1:147
        .= dom g by RELSET_1:22
        .= X by Th52;
      hence thesis by A1,Def1;
    end;
  end;
end;

registration
  let X,Y be set;
  let f be quasi_total PartFunc of Y,Y;
  let g be quasi_total PartFunc of X,Y;
  cluster f*g -> quasi_total for PartFunc of X,Y;
  coherence
  proof
    per cases;
    suppose
      Y = {};
      hence thesis;
    end;
    suppose
A1:   Y <> {};
      dom f = Y by Th52;
      then dom(f*g) = g"Y by RELAT_1:147
        .= dom g by RELSET_1:22
        .= X by A1,Def1;
      hence thesis by A1,Def1;
    end;
  end;
end;

theorem Th53:
  for f,g being Relation of X,X st rng f = X & rng g = X holds rng (g*f) = X
proof
  let f,g be Relation of X,X;
  assume that
A1: rng f = X and
A2: rng g = X;
  thus rng(g*f) = f.:X by A2,RELAT_1:127
    .= X by A1,RELSET_1:22;
end;

theorem
  for f,g being Function of X,X st g*f = f & rng f = X holds g = id X
proof
  let f,g be Function of X,X;
  dom g = X by Th52;
  hence thesis by FUNCT_1:23;
end;

theorem
  for f,g being Function of X,X st f*g = f & f is one-to-one holds g = id X
proof
  let f,g be Function of X,X;
A1: rng g c= X;
  dom f = X & dom g = X by Th52;
  hence thesis by A1,FUNCT_1:28;
end;

theorem Th56:
  for f being Function of X,X holds f is one-to-one iff for x1,x2
  st x1 in X & x2 in X & f.x1 = f.x2 holds x1 = x2
proof
  let f be Function of X,X;
  dom f = X by Th52;
  hence thesis by FUNCT_1:def 4;
end;

definition
  let X, Y;
  let f be X-defined Y-valued Function;
  attr f is bijective means
  :Def4:
  f is one-to-one onto;
end;

registration
  let X, Y be set;
  cluster bijective -> one-to-one onto for PartFunc of X,Y;
  coherence by Def4;
  cluster one-to-one onto -> bijective for PartFunc of X,Y;
  coherence by Def4;
end;

registration
  let X;
  cluster bijective for Function of X,X;
  existence
  proof
    take id X;
    thus id X is one-to-one & rng id X = X;
  end;
end;

definition
  let X;
  mode Permutation of X is bijective Function of X,X;
end;

theorem Th57:
  for f being Function of X, X st f is one-to-one & rng f = X
  holds f is Permutation of X
proof
  let f be Function of X, X;
  assume that
A1: f is one-to-one and
A2: rng f = X;
  f is onto by A2,Def3;
  hence thesis by A1;
end;

theorem
  for f being Function of X,X st f is one-to-one holds for x1,x2 st x1
  in X & x2 in X & f.x1 = f.x2 holds x1 = x2 by Th56;

registration
  let X;
  let f,g be onto PartFunc of X,X;
  cluster f*g -> onto for PartFunc of X,X;
  coherence
  proof
    rng f = X & rng g = X by Def3;
    then rng(f*g) = X by Th53;
    hence thesis by Def3;
  end;
end;

registration
  let X;
  let f,g be bijective Function of X,X;
  cluster g*f -> bijective for Function of X,X;
  coherence;
end;

registration
  let X;
  cluster reflexive total -> bijective for Function of X,X;
  coherence
  proof
    let f be Function of X,X;
    assume
A1: f is reflexive total;
A3: field f = dom f \/ rng f by RELAT_1:def 6;
    thus f is one-to-one
    proof
      let x1,x2 such that
A4:   x1 in dom f and
A5:   x2 in dom f and
A6:   f.x1 = f.x2;
      x1 in field f by A3,A4,XBOOLE_0:def 3;
      then [x1,x1] in f by A1,RELAT_2:def 1,def 9;
      then
A7:   x1 = f.x1 by A4,FUNCT_1:def 2;
      x2 in field f by A3,A5,XBOOLE_0:def 3;
      then [x2,x2] in f by A1,RELAT_2:def 1,def 9;
      hence thesis by A5,A6,A7,FUNCT_1:def 2;
    end;
    thus rng f c= X;
    let x;
    assume x in X;
    then x in dom f by PARTFUN1:def 2;
    then x in field f by A3,XBOOLE_0:def 3;
    then [x,x] in f by A1,RELAT_2:def 1,def 9;
    hence thesis by XTUPLE_0:def 13;
  end;
end;

definition
  let X;
  let f be Permutation of X;
  redefine func f" -> Permutation of X;
  coherence
  proof
    dom f = X by Th52;
    then
A1: rng(f") = X by FUNCT_1:33;
    rng f = X by Def3;
    then f" is Function of X,X by Th25;
    hence thesis by A1,Th57;
  end;
end;

theorem
  for f,g being Permutation of X st g*f = g holds f = id X
proof
  let f,g be Permutation of X;
A1: rng f c= X;
  dom f = X & dom g = X by Th52;
  hence thesis by A1,FUNCT_1:28;
end;

theorem
  for f,g being Permutation of X st g*f = id X holds g = f"
proof
  let f,g be Permutation of X;
A1: dom f = X by Th52;
  rng f= X & dom g = X by Def3,Th52;
  hence thesis by A1,FUNCT_1:41;
end;

theorem
  for f being Permutation of X holds (f")*f =id X & f*(f") = id X
proof
  let f be Permutation of X;
  dom f = X & rng f = X by Def3,Th52;
  hence thesis by FUNCT_1:39;
end;

theorem Th62:
  for f being Permutation of X st P c= X holds f.:(f"P) = P & f"(f .:P) = P
proof
  let f be Permutation of X such that
A1: P c= X;
  dom f = X by Th52;
  then
A2: P c= f"(f.:P) by A1,FUNCT_1:76;
  f"(f.:P) c= P & rng f = X by Def3,FUNCT_1:82;
  hence thesis by A1,A2,FUNCT_1:77,XBOOLE_0:def 10;
end;

reserve D for non empty set;

registration
  let X,D,Z;
  let f be Function of X,D;
  let g be Function of D,Z;
  cluster g*f -> quasi_total for PartFunc of X,Z;
  coherence by Th13;
end;

definition
  let C be non empty set, D be set;
  let f be Function of C,D;
  let c be Element of C;
  redefine func f.c -> Element of D;
  coherence
  proof
    D is non empty or D is empty;
    hence thesis by Th5,SUBSET_1:def 1;
  end;
end;

scheme
  FuncExD{C, D() -> non empty set, P[set,set]}: ex f being Function of C(),D()
  st for x being Element of C() holds P[x,f.x]
provided
A1: for x being Element of C() ex y being Element of D() st P[x,y]
proof
  defpred R[set,set] means $1 in C() & $2 in D() & P[ $1,$2];
A2: for x being set st x in C() ex y being set st y in D() & R[x,y]
  proof
    let x be set;
    assume
A3: x in C();
    then ex y being Element of D() st P[x,y] by A1;
    hence thesis by A3;
  end;
  consider f being Function of C(),D() such that
A4: for x being set st x in C() holds R[x,f.x] from FuncEx1(A2);
  take f;
  let x be Element of C();
  thus thesis by A4;
end;

scheme
  LambdaD{C, D() -> non empty set, F(Element of C()) -> Element of D()}: ex f
being Function of C(),D() st for x being Element of C() holds f.x = F(x) proof
  defpred P[Element of C(),set] means $2 = F($1);
A1: for x being Element of C() ex y being Element of D() st P[x,y];
  thus ex f being Function of C(),D() st for x being Element of C() holds P[x,
  f.x] from FuncExD(A1);
end;

theorem Th63:
  for f1,f2 being Function of X,Y st for x being Element of X
  holds f1.x = f2.x holds f1 = f2
proof
  let f1,f2 be Function of X,Y;
  assume for x being Element of X holds f1.x = f2.x;
  then for x st x in X holds f1.x = f2.x;
  hence thesis by Th12;
end;

theorem Th64:
  for P being set for f being Function of X,Y for y holds y in f
  .:P implies ex x st x in X & x in P & y = f.x
proof
  let P be set;
  let f be Function of X,Y;
  let y;
  assume y in f.:P;
  then consider x such that
A1: x in dom f and
A2: x in P & y = f.x by FUNCT_1:def 6;
  take x;
  thus x in X by A1;
  thus thesis by A2;
end;

theorem
  for f being Function of X,Y for y st y in f.:P ex c being Element of X
  st c in P & y = f.c
proof
  let f be Function of X,Y;
  let y;
  assume y in f.:P;
  then consider x such that
A1: x in X and
A2: x in P & y = f.x by Th64;
  reconsider c = x as Element of X by A1;
  take c;
  thus thesis by A2;
end;

begin 

theorem Th66:
  for f being set st f in Funcs(X,Y) holds f is Function of X,Y
proof
  let f be set;
  assume f in Funcs(X,Y);
  then
  ( not(Y = {} & X <> {}))& ex f9 being Function st f9 = f & dom f9 = X &
  rng f9 c= Y by Def2;
  hence thesis by Def1,RELSET_1:4;
end;

scheme
  Lambda1C{A, B() -> set, C[set], F(set)->set, G(set)->set}: ex f being
  Function of A(),B() st for x st x in A() holds (C[ x] implies f.x = F(x)) & (
  not C[ x] implies f.x = G(x))
provided
A1: for x st x in A() holds (C[ x] implies F(x) in B()) & (not C[ x]
implies G(x) in B())
proof
A2: now
    set x = the Element of A();
    assume
A3: B() = {};
    assume
A4: A() <> {};
    then C[ x] implies F(x) in B() by A1;
    hence contradiction by A1,A3,A4;
  end;
  consider f being Function such that
A5: dom f = A() and
A6: for x st x in A() holds (C[ x] implies f.x = F(x)) & (not C[ x]
  implies f.x = G(x)) from PARTFUN1:sch 1;
  rng f c= B()
  proof
    let y;
    assume y in rng f;
    then consider x such that
A7: x in dom f and
A8: y = f.x by FUNCT_1:def 3;
A9: not C[ x] implies f.x = G(x) by A5,A6,A7;
    C[ x] implies f.x = F(x) by A5,A6,A7;
    hence thesis by A1,A5,A7,A8,A9;
  end;
  then f is Function of A(),B() by A5,A2,Def1,RELSET_1:4;
  hence thesis by A6;
end;

theorem
  for f being PartFunc of X,Y st dom f = X holds f is Function of X,Y
proof
  let f be PartFunc of X,Y;
  rng f c= Y;
  hence thesis by Th2;
end;

theorem
  for f being PartFunc of X,Y st f is total holds f is Function of X,Y;

theorem
  for f being PartFunc of X,Y st (Y = {} implies X = {}) & f is Function
  of X,Y holds f is total;

theorem
  for f being Function of X,Y st (Y = {} implies X = {}) holds
  <:f,X,Y:> is total;

registration
  let X; let f be Function of X,X;
  cluster <:f,X,X:> -> total;
  coherence;
end;

theorem Th71:
  for f being PartFunc of X,Y st Y = {} implies X = {} ex g being
  Function of X,Y st for x st x in dom f holds g.x = f.x
proof
  let f be PartFunc of X,Y such that
A1: Y = {} implies X = {};
  per cases;
  suppose Y = {};
    then reconsider g = f as Function of X,Y by A1;
    take g;
    thus thesis;
  end;
  suppose
A2: Y <> {};
    deffunc F(set) = f.$1;
    defpred P[set] means $1 in dom f;
    set y = the Element of Y;
    deffunc G(set) = y;
A3: for x st x in X holds (P[ x] implies F(x) in Y) & (not P[ x] implies G
    (x) in Y) by A2,PARTFUN1:4;
    consider g being Function of X,Y such that
A4: for x st x in X holds (P[ x] implies g.x = F(x)) & (not P[ x]
    implies g.x = G(x)) from Lambda1C(A3);
    take g;
    thus thesis by A4;
  end;
end;

theorem
  Funcs(X,Y) c= PFuncs(X,Y)
proof
  let x;
  assume x in Funcs(X,Y);
  then ex f being Function st x = f & dom f = X & rng f c= Y by Def2;
  hence thesis by PARTFUN1:def 3;
end;

theorem
  for f,g being Function of X,Y st (Y = {} implies X = {}) & f
  tolerates g holds f = g by PARTFUN1:66;

theorem
  for f,g being Function of X,X st f tolerates g holds f = g by PARTFUN1:66;

theorem Th75:
  for f being PartFunc of X,Y for g being Function of X,Y st Y =
  {} implies X = {} holds f tolerates g iff for x st x in dom f holds f.x = g.x
proof
  let f be PartFunc of X,Y;
  let g be Function of X,Y;
  assume Y = {} implies X = {};
  then dom g = X by Def1;
  hence thesis by PARTFUN1:53;
end;

theorem
  for f being PartFunc of X,X for g being Function of X,X holds f
  tolerates g iff for x st x in dom f holds f.x = g.x
proof
  let f be PartFunc of X,X;
  let g be Function of X,X;
  X = {} implies X = {};
  hence thesis by Th75;
end;

theorem Th77:
  for f being PartFunc of X,Y st Y = {} implies X = {} ex g being
  Function of X,Y st f tolerates g
proof
  let f be PartFunc of X,Y;
  assume
A1: Y = {} implies X = {};
  then consider g being Function of X,Y such that
A2: for x st x in dom f holds g.x = f.x by Th71;
  take g;
  thus thesis by A1,A2,Th75;
end;

theorem
  for f,g being PartFunc of X,X for h being Function of X,X st f
  tolerates h & g tolerates h holds f tolerates g by PARTFUN1:67;

theorem
  for f,g being PartFunc of X,Y st (Y = {} implies X = {}) & f tolerates
  g ex h being Function of X,Y st f tolerates h & g tolerates h
proof
  let f,g be PartFunc of X,Y;
  assume ( Y = {} implies X = {})& f tolerates g;
  then
  ex h being PartFunc of X,Y st h is total & f tolerates h & g tolerates h
  by PARTFUN1:68;
  hence thesis;
end;

theorem
  for f being PartFunc of X,Y for g being Function of X,Y st (Y =
  {} implies X = {}) & f tolerates g holds g in TotFuncs f by PARTFUN1:def 5;

theorem
  for f being PartFunc of X,X for g being Function of X,X st f tolerates
  g holds g in TotFuncs f by PARTFUN1:def 5;

theorem Th82:
  for f being PartFunc of X,Y for g being set st g in TotFuncs(f)
  holds g is Function of X,Y
proof
  let f be PartFunc of X,Y;
  let g be set;
  assume g in TotFuncs(f);
  then
  ex g9 being PartFunc of X,Y st g9 = g & g9 is total & f tolerates g9 by
PARTFUN1:def 5;
  hence thesis;
end;

theorem
  for f being PartFunc of X,Y holds TotFuncs f c= Funcs(X,Y)
proof
  let f be PartFunc of X,Y;
  per cases;
  suppose Y = {} & X <> {};
    hence thesis;
  end;
  suppose
A1: Y <> {} or X = {};
    let g be set;
    assume g in TotFuncs f;
    then g is Function of X,Y by Th82;
    hence thesis by A1,Th8;
  end;
end;

theorem
  TotFuncs <:{},X,Y:> = Funcs(X,Y)
proof
  per cases;
  suppose
A1: Y = {} & X <> {};
    then TotFuncs <:{},X,Y:> = {};
    hence thesis by A1;
  end;
  suppose
A2: Y = {} implies X = {};
    for g being set holds g in TotFuncs <:{},X,Y:> iff g in Funcs(X,Y)
    proof
      let g be set;
      thus g in TotFuncs <:{},X,Y:> implies g in Funcs(X,Y)
      proof
        assume g in TotFuncs <:{},X,Y:>;
        then g is Function of X,Y by Th82;
        hence thesis by A2,Th8;
      end;
      assume
A3:   g in Funcs(X,Y);
      then reconsider g9 = g as PartFunc of X,Y by Th66;
A4:   <:{},X,Y:> tolerates g9 by PARTFUN1:60;
      g is Function of X,Y by A3,Th66;
      hence thesis by A2,A4,PARTFUN1:def 5;
    end;
    hence thesis by TARSKI:1;
  end;
end;

theorem
  for f being Function of X,Y st Y = {} implies X = {} holds
  TotFuncs <:f,X,Y:> = {f} by PARTFUN1:72;

theorem
  for f being Function of X,X holds TotFuncs <:f,X,X:> = {f} by PARTFUN1:72;

theorem
  for f being PartFunc of X,{y} for g being Function of X,{y} holds
  TotFuncs f = {g}
proof
  let f be PartFunc of X,{y};
  let g be Function of X,{y};
  for h being set holds h in TotFuncs f iff h = g
  proof
    let h be set;
    thus h in TotFuncs f implies h = g
    proof
      assume h in TotFuncs f;
      then h is Function of X,{y} by Th82;
      hence thesis by Th51;
    end;
    f tolerates g by PARTFUN1:61;
    hence thesis by PARTFUN1:def 5;
  end;
  hence thesis by TARSKI:def 1;
end;

theorem
  for f,g being PartFunc of X,Y st g c= f holds TotFuncs f c= TotFuncs g
proof
  let f,g be PartFunc of X,Y such that
A1: g c= f;
  let h be set;
  assume
A2: h in TotFuncs f;
  then reconsider h9=h as PartFunc of X,Y by PARTFUN1:69;
A3: h9 is total by A2,PARTFUN1:70;
  g tolerates h9 by A1,A2,PARTFUN1:58,71;
  hence thesis by A3,PARTFUN1:def 5;
end;

theorem Th89:
  for f,g being PartFunc of X,Y st dom g c= dom f & TotFuncs f c=
  TotFuncs g holds g c= f
proof
  let f,g be PartFunc of X,Y such that
A1: dom g c= dom f;
  now
    per cases;
    suppose Y = {} & X <> {};
      hence thesis;
    end;
    suppose
A2:   Y = {} implies X = {};
      thus TotFuncs f c= TotFuncs g implies g c= f
      proof
        assume
A3:     TotFuncs f c= TotFuncs g;
        for x st x in dom g holds g.x = f.x
        proof
          let x;
          consider h being Function of X,Y such that
A4:       f tolerates h by A2,Th77;
          h in TotFuncs f by A2,A4,PARTFUN1:def 5;
          then
B5:       g tolerates h by A3,PARTFUN1:71;
          assume x in dom g;
          then x in dom f /\ dom g by A1,XBOOLE_0:def 4;
          hence thesis by B5,PARTFUN1:def 4,A2,A4,PARTFUN1:67;
        end;
        hence thesis by A1,GRFUNC_1:2;
      end;
    end;
  end;
  hence thesis;
end;

theorem Th90:
  for f,g being PartFunc of X,Y st TotFuncs f c= TotFuncs g & (
  for y holds Y <> {y}) holds g c= f
proof
  let f,g be PartFunc of X,Y such that
A1: TotFuncs f c= TotFuncs g and
A2: for y holds Y <> {y};
  now
    per cases;
    suppose Y = {};
      hence dom g c= dom f;
    end;
    suppose
A3:   Y <> {};
      thus dom g c= dom f
      proof
        deffunc F(set) = f.$1;
        defpred P[set] means $1 in dom f;
        let x such that
A4:     x in dom g and
A5:     not x in dom f;
A6:     Y <> {g.x} by A2;
        g.x in Y by A4,PARTFUN1:4;
        then consider y such that
A7:     y in Y and
A8:     y <> g.x by A6,ZFMISC_1:35;
        deffunc G(set) = y;
A9:     for x9 st x9 in X holds (P[ x9] implies F(x9) in Y) & (not P[ x9]
        implies G(x9) in Y) by A7,PARTFUN1:4;
        consider h being Function of X,Y such that
A10:    for x9 st x9 in X holds (P[ x9] implies h.x9 = F(x9)) & (not
        P[ x9] implies h.x9 = G(x9)) from Lambda1C(A9);
        f tolerates h
        proof
          let x9;
          assume x9 in dom f /\ dom h;
          then x9 in dom f by XBOOLE_0:def 4;
          hence thesis by A10;
        end;
        then
B11:    h in TotFuncs f by A3,PARTFUN1:def 5;
        x in X by A4;
        then x in dom h by A3,Def1;
        then
A12:    x in dom g /\ dom h by A4,XBOOLE_0:def 4;
        h.x = y by A4,A5,A10;
        hence contradiction by A8,B11,A12,PARTFUN1:def 4,A1,PARTFUN1:71;
      end;
    end;
  end;
  hence thesis by A1,Th89;
end;

theorem
  for f,g being PartFunc of X,Y st (for y holds Y <> {y}) & TotFuncs f =
  TotFuncs g holds f = g
proof
  let f,g be PartFunc of X,Y;
  assume
A1: for y holds Y <> {y};
  assume TotFuncs f = TotFuncs g;
  then g c= f & f c= g by A1,Th90;
  hence thesis by XBOOLE_0:def 10;
end;



registration
  let A,B be non empty set;
  cluster -> non empty for Function of A,B;
  coherence by Def1,RELAT_1:38;
end;

begin 


scheme
  LambdaSep1{D, R() -> non empty set, A() -> Element of D(), B() -> Element of
R(), F(set) -> Element of R()}: ex f being Function of D(),R() st f.A() = B() &
  for x being Element of D() st x <> A() holds f.x = F(x) proof
  defpred P[set,set] means ($1 = A() implies $2 = B()) & ($1 <> A() implies $2
  = F($1));
A1: for x being Element of D() ex y being Element of R() st P[x,y]
  proof
    let x be Element of D();
    x = A() implies thesis;
    hence thesis;
  end;
  consider f being Function of D(),R() such that
A2: for x being Element of D() holds P[x,f.x] from FuncExD(A1);
  take f;
  thus thesis by A2;
end;

scheme
  LambdaSep2{D, R() -> non empty set, A1, A2() -> Element of D(), B1, B2() ->
Element of R(), F(set) -> Element of R()}: ex f being Function of D(),R() st f.
A1() = B1() & f.A2() = B2() & for x being Element of D() st x <> A1() & x <> A2
  () holds f.x = F(x)
provided
A1: A1() <> A2()
proof
  defpred P[set,set] means ($1 = A1() implies $2 = B1()) & ($1 = A2() implies
  $2 = B2()) & ($1 <> A1() & $1 <> A2() implies $2 = F($1));
A2: for x being Element of D() ex y being Element of R() st P[x,y]
  proof
    let x be Element of D();
A3: x = A2() implies thesis by A1;
    x = A1() implies thesis by A1;
    hence thesis by A3;
  end;
  consider f being Function of D(),R() such that
A4: for x being Element of D() holds P[x,f.x] from FuncExD(A2);
  take f;
  thus thesis by A4;
end;



theorem
  for A,B being set for f being Function st f in Funcs(A,B) holds dom f
  = A & rng f c= B
proof
  let A,B be set;
  let f be Function;
  assume f in Funcs(A,B);
  then ex g being Function st f = g & dom g = A & rng g c= B by Def2;
  hence thesis;
end;



scheme
  FunctRealEx{X()->non empty set,Y()->set,F(set)->set}: ex f being Function of
  X(),Y() st for x being Element of X() holds f.x = F(x)
provided
A1: for x being Element of X() holds F(x) in Y()
proof
  defpred P[set,set] means $2 = F($1);
A2: for x being set st x in X() ex y being set st y in Y() & P[x,y] by A1;
  ex f being Function of X(),Y() st for x being set st x in X() holds P[x,
  f.x] from FuncEx1(A2);
  then consider f being Function of X(),Y() such that
A3: for x being set st x in X() holds f.x = F(x);
  for x being Element of X() holds f.x = F (x) by A3;
  hence thesis;
end;



scheme
  KappaMD{X, Y() -> non empty set, F(set) -> set}: ex f being Function of X(),
  Y() st for x being Element of X() holds f.x = F(x)
provided
A1: for x being Element of X() holds F(x) is Element of Y()
proof
A2: now
    let x be Element of X();
    F(x) is Element of Y() by A1;
    hence F(x) in Y();
  end;
  consider f being Function of X(), Y() such that
A3: for x being Element of X() holds f.x = F(x) from FunctRealEx(A2);
  take f;
  thus thesis by A3;
end;

definition
  let A,B,C be non empty set;
  let f be Function of A, [:B,C:];
  redefine func pr1 f -> Function of A,B means
  :Def5:
  for x being Element of A holds it.x = (f.x)`1;
  coherence
  proof
A1: dom pr1 f = dom f by MCART_1:def 12;
A2: rng pr1 f c= B
    proof
      let x be set;
      assume x in rng pr1 f;
      then consider y being set such that
A3:   y in dom pr1 f & x = (pr1 f).y by FUNCT_1:def 3;
      x = (f.y)`1 & f.y in [:B,C:] by A1,A3,Th5,MCART_1:def 12;
      hence thesis by MCART_1:10;
    end;
    dom pr1 f = A by A1,Def1;
    hence thesis by A2,Th2;
  end;
  compatibility
  proof
    let IT be Function of A,B;
A4: dom pr1 f = dom f by MCART_1:def 12;
    then
A5: dom pr1 f = A by Def1;
    hence IT = pr1 f implies for x being Element of A holds IT.x = (f.x)`1 by
A4,MCART_1:def 12;
    assume for x being Element of A holds IT.x = (f.x)`1;
    then
A6: for x being set st x in dom f holds IT.x = (f.x)`1;
    dom IT = dom f by A4,A5,Def1;
    hence thesis by A6,MCART_1:def 12;
  end;
  redefine func pr2 f -> Function of A,C means
  :Def6:
  for x being Element of A holds it .x = (f.x)`2;
  coherence
  proof
A7: dom pr2 f = dom f by MCART_1:def 13;
A8: rng pr2 f c= C
    proof
      let x be set;
      assume x in rng pr2 f;
      then consider y being set such that
A9:   y in dom pr2 f & x = (pr2 f).y by FUNCT_1:def 3;
      x = (f.y)`2 & f.y in [:B,C:] by A7,A9,Th5,MCART_1:def 13;
      hence thesis by MCART_1:10;
    end;
    dom pr2 f = A by A7,Def1;
    hence thesis by A8,Th2;
  end;
  compatibility
  proof
    let IT be Function of A,C;
A10: dom pr2 f = dom f by MCART_1:def 13;
    then
A11: dom pr2 f = A by Def1;
    hence IT = pr2 f implies for x being Element of A holds IT.x = (f.x)`2 by
A10,MCART_1:def 13;
    assume for x being Element of A holds IT.x = (f.x)`2;
    then
A12: for x being set st x in dom f holds IT.x = (f.x)`2;
    dom IT = dom f by A10,A11,Def1;
    hence thesis by A12,MCART_1:def 13;
  end;
end;

definition
  let A1 be set, B1 be non empty set, A2 be set, B2 be non empty set, f1 be
  Function of A1,B1, f2 be Function of A2,B2;
  redefine pred f1 = f2 means
  A1 = A2 & for a being Element of A1 holds f1.a = f2.a;
  compatibility
  proof
A1: dom f1 = A1 by Def1;
    hence
    f1 = f2 implies A1 = A2 & for a being Element of A1 holds f1.a = f2.a
    by Def1;
    assume that
A2: A1 = A2 and
A3: for a being Element of A1 holds f1.a = f2.a;
A4: dom f2 = A2 by Def1;
    for a being set st a in A1 holds f1.a = f2.a by A3;
    hence thesis by A1,A4,A2,FUNCT_1:2;
  end;
end;

definition
  let A,B be set, f1,f2 be Function of A,B;
  redefine pred f1 = f2 means
  for a being Element of A holds f1.a = f2.a;
  compatibility by Th63;
end;



theorem
  for N being set, f being Function of N, bool N ex R being Relation of
  N st for i being set st i in N holds Im(R,i) = f.i
proof
  let N be set, f be Function of N, bool N;
  deffunc F(set) = f.$1;
A1: for i being Element of N st i in [#]N holds F(i) c= [#]N
  proof
    let i be Element of N;
    assume i in [#]N;
    then f.i in bool N by Th5;
    hence thesis;
  end;
  consider R being Relation of [#]N such that
A2: for i being Element of N st i in [#]N holds Im(R,i) = F(i) from
  RELSET_1:sch 3(A1);
  reconsider R as Relation of N;
  take R;
  thus thesis by A2;
end;



theorem Th94:
  for A being Subset of X holds (id X)"A = A
proof
  let A be Subset of X;
  thus A = (id X)"((id X).:A) by Th62
    .= (id X)"A by FUNCT_1:92;
end;



reserve A,B for non empty set;

theorem
  for f being Function of A,B, A0 being Subset of A, B0 being Subset of
  B holds f.:A0 c= B0 iff A0 c= f"B0
proof
  let f be Function of A,B, A0 be Subset of A, B0 be Subset of B;
  thus f.:A0 c= B0 implies A0 c= f"B0
  proof
    assume f.:A0 c= B0;
    then
A1: f"(f.:A0) c= f"B0 by RELAT_1:143;
    A0 c= f"(f.:A0) by Th42;
    hence thesis by A1,XBOOLE_1:1;
  end;
  thus A0 c= f"B0 implies f.:A0 c= B0
  proof
    assume A0 c= f"B0;
    then
A2: f.:A0 c= f.:f"B0 by RELAT_1:123;
    f.:f"B0 c= B0 by FUNCT_1:75;
    hence thesis by A2,XBOOLE_1:1;
  end;
end;

theorem
  for f being Function of A,B, A0 being non empty Subset of A, f0 being
Function of A0,B st for c being Element of A st c in A0 holds f.c = f0.c holds
  f|A0 = f0
proof
  let f be Function of A,B, A0 be non empty Subset of A, f0 be Function of A0,
  B such that
A1: for c being Element of A st c in A0 holds f.c = f0.c;
  reconsider g = f|A0 as Function of A0,B by Th32;
  for c being Element of A0 holds g.c = f0.c
  proof
    let c be Element of A0;
    thus g.c = f.c by FUNCT_1:49
      .= f0.c by A1;
  end;
  hence thesis by Th63;
end;

theorem
  for f being Function, A0, C being set st C c= A0 holds f.:C = (f|A0).: C
proof
  let f be Function, A0, C be set;
  assume
A1: C c= A0;
  thus (f|A0).:C = (f*(id A0)).:C by RELAT_1:65
    .= f.:((id A0).:C) by RELAT_1:126
    .= f.:C by A1,FUNCT_1:92;
end;

theorem
  for f being Function, A0, D being set st f"D c= A0 holds f"D = (f|A0)" D
proof
  let f be Function, A0, D be set;
  assume
A1: f"D c= A0;
  thus (f|A0)"D = (f*(id A0))"D by RELAT_1:65
    .= (id A0)"(f"D) by RELAT_1:146
    .= f"D by A1,Th94;
end;

scheme
  MChoice{A()-> non empty set, B()-> non empty set, F(set) -> set}: ex t being
  Function of A(),B() st for a being Element of A() holds t.a in F(a)
provided
A1: for a being Element of A() holds B() meets F(a)
proof
  defpred P[set,set] means $2 in F($1);
A2: for e being set st e in A() ex u being set st u in B() & P[e,u]
                          by A1,XBOOLE_0:3;
  consider t being Function such that
A3: dom t = A() & rng t c= B() and
A4: for e being set st e in A() holds P[e,t.e] from  FUNCT_1:sch 5(A2);
  reconsider t as Function of A(),B() by A3,Def1,RELSET_1:4;
  take t;
  let a be Element of A();
  thus thesis by A4;
end;



theorem Th99:
  for X, D be non empty set, p be Function of X,D, i be Element
  of X holds p/.i = p.i
proof
  let X, D be non empty set, p be Function of X,D, i be Element of X;
  i in X;
  then i in dom p by Def1;
  hence thesis by PARTFUN1:def 6;
end;

registration
  let X, D be non empty set, p be Function of X,D, i be Element of X;
  identify p/.i with p.i;
  correctness by Th99;
end;



theorem
  for S,X being set, f being Function of S,X, A being Subset of X st X =
  {} implies S = {} holds (f"A)` = f"(A`)
proof
  let S,X be set, f be Function of S,X, A be Subset of X such that
A1: X = {} implies S = {};

  A /\ A` = {} by XBOOLE_0:def 7,XBOOLE_1:79;
  then f"A /\ f"(A`) = f"({}X) by FUNCT_1:68
    .= {};
  then
A2: f"A misses f"(A`) by XBOOLE_0:def 7;
  f"A \/ f"(A`) = f"(A \/ A`) by RELAT_1:140
    .= f"[#]X by SUBSET_1:10
    .= [#]S by A1,Th40;
  then (f"A)` /\ (f"(A`))` = ([#]S)` by XBOOLE_1:53
    .= {}S by XBOOLE_1:37;
  then (f"A)` misses (f"(A`))` by XBOOLE_0:def 7;
  hence thesis by A2,SUBSET_1:25;
end;



theorem
  for X,Y,Z being set, D being non empty set, f being Function of X,D st
  Y c= X & f.:Y c= Z holds f|Y is Function of Y,Z
proof
  let X,Y,Z be set, D be non empty set, f be Function of X,D;
  assume that
A1: Y c= X and
A2: f.:Y c= Z;
  dom f = X by Def1;
  then
A3: dom(f|Y) = Y by A1,RELAT_1:62;
A4: now
    assume Z = {};
    then rng(f|Y) = {} by A2,RELAT_1:115;
    hence Y = {} by A3,RELAT_1:42;
  end;
  rng(f|Y) c= Z by A2,RELAT_1:115;
  hence thesis by A3,A4,Def1,RELSET_1:4;
end;



definition
  let T,S be non empty set;
  let f be Function of T,S;
  let G be Subset-Family of S;
  func f"G -> Subset-Family of T means
  :Def9:
  for A being Subset of T holds A
  in it iff ex B being Subset of S st B in G & A = f"B;
  existence
  proof
    defpred P[Subset of T] means ex B being Subset of S st B in G & $1 = f"B;
    ex R being Subset-Family of T st for A being Subset of T holds A in R
    iff P[A] from SUBSET_1:sch 3;
    hence thesis;
  end;
  uniqueness
  proof
    let R1,R2 be Subset-Family of T such that
A1: for A being Subset of T holds A in R1 iff ex B being Subset of S
    st B in G & A = f"B and
A2: for A being Subset of T holds A in R2 iff ex B being Subset of S
    st B in G & A = f"B;
    for x being set holds (x in R1 iff x in R2)
    proof
      let x be set;
      thus x in R1 implies x in R2
      proof
        assume
A3:     x in R1;
        then reconsider x as Subset of T;
        ex B being Subset of S st B in G & x = f"B by A1,A3;
        hence thesis by A2;
      end;
      assume
A4:   x in R2;
      then reconsider x as Subset of T;
      ex B being Subset of S st B in G & x = f"B by A2,A4;
      hence thesis by A1;
    end;
    hence thesis by TARSKI:1;
  end;
end;

theorem
  for T,S being non empty set, f being Function of T,S, A,B being
  Subset-Family of S st A c= B holds f"A c= f"B
proof
  let T,S be non empty set;
  let f be Function of T,S;
  let A,B be Subset-Family of S;
  assume
A1: A c= B;
  let x be set;
  assume
A2: x in f"A;
  then reconsider x as Subset of T;
  ex C being Subset of S st C in B & x = f"C
  proof
    consider C being Subset of S such that
A3: C in A & x = f"C by A2,Def9;
    take C;
    thus thesis by A1,A3;
  end;
  hence thesis by Def9;
end;

definition
  let T,S be non empty set;
  let f be Function of T,S;
  let G be Subset-Family of T;
  func f.:G -> Subset-Family of S means
  :Def10:
  for A being Subset of S holds
  A in it iff ex B being Subset of T st B in G & A = f.:B;
  existence
  proof
    thus ex R being Subset-Family of S st for A being Subset of S holds A in R
    iff ex B being Subset of T st B in G & A = f.:B
    proof
      defpred P[Subset of S] means ex B being Subset of T st B in G & $1 = f.:
      B;
      ex R being Subset-Family of S st for A being Subset of S holds A in
      R iff P[A] from SUBSET_1:sch 3;
      hence thesis;
    end;
  end;
  uniqueness
  proof
    let R1,R2 be Subset-Family of S such that
A1: for A being Subset of S holds A in R1 iff ex B being Subset of T
    st B in G & A = f.:B and
A2: for A being Subset of S holds A in R2 iff ex B being Subset of T
    st B in G & A = f.:B;
    for x being set holds (x in R1 iff x in R2)
    proof
      let x be set;
      thus x in R1 implies x in R2
      proof
        assume
A3:     x in R1;
        then reconsider x as Subset of S;
        ex B being Subset of T st B in G & x = f.:B by A1,A3;
        hence thesis by A2;
      end;
      assume
A4:   x in R2;
      then reconsider x as Subset of S;
      ex B being Subset of T st B in G & x = f.:B by A2,A4;
      hence thesis by A1;
    end;
    hence thesis by TARSKI:1;
  end;
end;

theorem
  for T,S being non empty set, f being Function of T,S, A,B being
  Subset-Family of T holds A c= B implies f.:A c= f.:B
proof
  let T,S be non empty set;
  let f be Function of T,S;
  let A,B be Subset-Family of T;
  assume
A1: A c= B;
  let x be set;
  assume
A2: x in f.:A;
  then reconsider x as Subset of S;
  ex C being Subset of T st C in B & x = f.:C
  proof
    consider C being Subset of T such that
A3: C in A & x = f.:C by A2,Def10;
    take C;
    thus thesis by A1,A3;
  end;
  hence thesis by Def10;
end;

theorem
  for T,S being non empty set, f being Function of T,S, B being
  Subset-Family of S, P being Subset of S st f.:(f"B) is Cover of P holds B is
  Cover of P
proof
  let T,S be non empty set;
  let f be Function of T,S;
  let B be Subset-Family of S;
  let P be Subset of S;
  assume f.:(f"B) is Cover of P;
  then
A1: P c= union (f.:(f"B)) by SETFAM_1:def 11;
  P c= union B
  proof
    let x be set;
    assume x in P;
    then consider Y being set such that
A2: x in Y and
A3: Y in f.:(f"(B)) by A1,TARSKI:def 4;
    ex Z being set st x in Z & Z in B
    proof
      reconsider Y as Subset of S by A3;
      consider Y1 being Subset of T such that
A4:   Y1 in f"(B) and
A5:   Y = f.:Y1 by A3,Def10;
      consider Y2 being Subset of S such that
A6:   Y2 in B & Y1 = f"(Y2) by A4,Def9;
A7:   f.:(f"Y2) c= Y2 by FUNCT_1:75;
      reconsider Y2 as set;
      take Y2;
      thus thesis by A2,A5,A6,A7;
    end;
    hence thesis by TARSKI:def 4;
  end;
  hence thesis by SETFAM_1:def 11;
end;

theorem
  for T,S being non empty set, f being Function of T,S, B being
  Subset-Family of T, P being Subset of T st B is Cover of P holds f"(f.:B) is
  Cover of P
proof
  let T,S be non empty set;
  let f be Function of T,S;
  let B be Subset-Family of T;
  let P be Subset of T;
  assume B is Cover of P;
  then
A1: P c= union B by SETFAM_1:def 11;
  P c= union(f"(f.:B))
  proof
    let x be set;
    assume x in P;
    then consider Y being set such that
A2: x in Y and
A3: Y in B by A1,TARSKI:def 4;
    ex Z being set st x in Z & Z in f"(f.:B)
    proof
      reconsider Y as Subset of T by A3;
      set Z = f"(f.:Y);
      take Z;
      dom f = T by Def1;
      then
A4:   Y c= f"(f.:Y) by FUNCT_1:76;
      f.:Y in f.:B by A3,Def10;
      hence thesis by A2,A4,Def9;
    end;
    hence thesis by TARSKI:def 4;
  end;
  hence thesis by SETFAM_1:def 11;
end;

theorem
  for T,S being non empty set, f being Function of T,S, Q being
  Subset-Family of S holds union(f.:(f"(Q))) c= union Q
proof
  let T,S be non empty set;
  let f be Function of T,S;
  let Q be Subset-Family of S;
  let x be set;
  thus x in union(f.:(f"(Q))) implies x in union Q
  proof
    assume x in union(f.:(f"Q));
    then consider A being set such that
A1: x in A and
A2: A in f.:(f"Q) by TARSKI:def 4;
    reconsider A as Subset of S by A2;
    consider A1 being Subset of T such that
A3: A1 in f"Q and
A4: A = f.:A1 by A2,Def10;
    consider A2 being Subset of S such that
A5: A2 in Q & A1 = f"A2 by A3,Def9;
    f.:(f"A2) c= A2 by FUNCT_1:75;
    hence thesis by A1,A4,A5,TARSKI:def 4;
  end;
end;

theorem
  for T,S being non empty set, f being Function of T,S, P being
  Subset-Family of T holds union P c= union(f"(f.:P))
proof
  let T,S be non empty set;
  let f be Function of T,S;
  let P be Subset-Family of T;
  let x be set;
  assume x in union P;
  then consider A being set such that
A1: x in A and
A2: A in P by TARSKI:def 4;
A3: A c= T by A2;
  reconsider A as Subset of T by A2;
  reconsider A1 = f.:A as Subset of S;
  reconsider A2 = f"A1 as Subset of T;
  A c= dom f by A3,Def1;
  then
A4: A c= A2 by FUNCT_1:76;
  A1 in f.:P by A2,Def10;
  then A2 in f"(f.:P) by Def9;
  hence thesis by A1,A4,TARSKI:def 4;
end;





definition
  let X,Z be set, Y be non empty set;
  let f be Function of X,Y;
  let p be Z-valued Function;
  assume
A1: rng f c= dom p;
  func p/*f -> Function of X,Z equals
  :Def11:
  p*f;
  coherence
  proof
    dom f = X by Def1;
    then
A2:   dom(p*f) = X by A1,RELAT_1:27;
    then
A3:   p*f is total by PARTFUN1:def 2;
     rng(p*f) c= Z;
    hence thesis by A3,A2,RELSET_1:4;
  end;
end;

reserve Y for non empty set,
  f for Function of X,Y,
  p for PartFunc of Y,Z,
  x for Element of X;

theorem Th108:
  X <> {} & rng f c= dom p implies (p/*f).x = p.(f.x)
proof
  assume that
A1: X <> {} and
A2: rng f c= dom p;
  p/*f = p*f by A2,Def11;
  hence thesis by A1,Th15;
end;

theorem Th109:
  X <> {} & rng f c= dom p implies (p/*f).x = p/.(f.x)
proof
  assume that
A1: X <> {} and
A2: rng f c= dom p;
A3: f.x in rng f by A1,Th4;
  thus (p/*f).x = p.(f.x) by A1,A2,Th108
    .= p/.(f.x) by A2,A3,PARTFUN1:def 6;
end;

reserve g for Function of X,X;

theorem
  rng f c= dom p implies (p/*f)*g = p/*(f*g)
proof
A1: rng(f*g) c= rng f by RELAT_1:26;
  assume
A2: rng f c= dom p;
  hence (p/*f)*g = p*f*g by Def11
    .= p*(f*g) by RELAT_1:36
    .= p /* (f*g) by A2,A1,Def11,XBOOLE_1:1;
end;



theorem
  for X,Y being non empty set, f being Function of X,Y holds f is
  constant iff ex y being Element of Y st rng f = {y}
proof
  let X,Y be non empty set;
  let f be Function of X,Y;
  hereby
    consider x be set such that
A1: x in dom f by XBOOLE_0:def 1;
    set y = f.x;
    reconsider y as Element of Y by A1,Th5;
    assume
A2: f is constant;
    take y;
    for y9 being set holds y9 in rng f iff y9 = y
    proof
      let y9 be set;
      hereby
        assume y9 in rng f;
        then ex x9 be set st x9 in dom f & y9 = f.x9 by FUNCT_1:def 3;
        hence y9 = y by A2,A1,FUNCT_1:def 10;
      end;
      assume y9 = y;
      hence thesis by A1,Th4;
    end;
    hence rng f = {y} by TARSKI:def 1;
  end;
  given y be Element of Y such that
A3: rng f = {y};
  let x,x9 be set;
  assume x in dom f;
  then
A4: f.x in rng f by Th4;
  assume x9 in dom f;
  then
A5: f.x9 in rng f by Th4;
  thus f.x = y by A3,A4,TARSKI:def 1
    .= f.x9 by A3,A5,TARSKI:def 1;
end;



theorem
  for A,B being non empty set, x being Element of A, f being Function of
  A,B holds f.x in rng f
proof
  let A,B be non empty set, x be Element of A, f be Function of A,B;
  dom f = A by Def1;
  hence thesis by FUNCT_1:def 3;
end;



theorem Th113:
  for A,B being set, f being Function of A,B st y in rng f ex x
  being Element of A st y = f.x
proof
  let A,B be set, f be Function of A,B;
  assume y in rng f;
  then consider x being set such that
A1: x in A and
A2: f.x = y by Th11;
  reconsider x as Element of A by A1;
  take x;
  thus thesis by A2;
end;



theorem
  for A,B being non empty set, f being Function of A,B st for x being
  Element of A holds f.x in Z holds rng f c= Z
proof
  let A,B be non empty set, f be Function of A,B such that
A1: for x being Element of A holds f.x in Z;
  let y;
  assume y in rng f;
  then ex x being Element of A st f.x=y by Th113;
  hence thesis by A1;
end;

reserve X,Y for non empty set,
  Z,S,T for set,
  f for Function of X,Y,
  g for PartFunc of Y,Z,
  x for Element of X;

theorem
  g is total implies (g/*f).x = g.(f.x)
proof
  assume g is total;
  then dom g = Y by PARTFUN1:def 2;
  then rng f c= dom g;
  hence thesis by Th108;
end;

theorem
  g is total implies (g/*f).x = g/.(f.x)
proof
  assume g is total;
  then dom g = Y by PARTFUN1:def 2;
  then rng f c= dom g;
  hence thesis by Th109;
end;

theorem Th117:
  rng f c= dom (g|S) implies (g|S)/*f = g/*f
proof
  assume
A1: rng f c= dom (g|S);
  let x be Element of X;
A2: dom (g|S) c= dom g by RELAT_1:60;
A3: f.x in rng f by Th4;
  thus ((g|S)/*f).x = (g|S).(f.x) by A1,Th108
    .= g.(f.x) by A1,A3,FUNCT_1:47
    .= (g/*f).x by A1,A2,Th108,XBOOLE_1:1;
end;

theorem
  rng f c= dom (g|S) & S c= T implies (g|S)/*f = (g|T)/*f
proof
  assume
A1: rng f c= dom (g|S);
  assume S c= T;
  then g|S c= g|T by RELAT_1:75;
  then
A2: dom (g|S) c= dom (g|T) by RELAT_1:11;
  thus (g|S)/*f = g/*f by A1,Th117
    .= (g|T)/*f by A1,A2,Th117,XBOOLE_1:1;
end;



theorem
  for H being Function of D, [:A,B:], d being Element of D holds H.d = [
  pr1 H.d,pr2 H.d]
proof
  let H be Function of D, [:A,B:], d be Element of D;
  thus H.d = [(H.d)`1,(H.d)`2] by MCART_1:21
    .= [(H.d)`1,pr2 H.d] by Def6
    .= [pr1 H.d,pr2 H.d] by Def5;
end;



theorem
  for A1,A2, B1,B2 being set for f being Function of A1,A2, g being
Function of B1,B2 st f tolerates g holds f /\ g is Function of A1 /\ B1, A2 /\
  B2
proof
  let A1,A2, B1,B2 be set;
  let f be Function of A1,A2, g be Function of B1,B2 such that
A1: f tolerates g;
A2: dom (f/\g) c= dom f /\ dom g & dom f /\ dom g c= A1/\B1 by RELAT_1:2
,XBOOLE_1:27;
A3: now
    set a = the Element of A1 /\ B1;
    assume that
A4: dom f = A1 and
    A1 <> {} and
A5: dom g = B1 and
    B1 <> {};
    hereby
      assume
A6:   A1 /\ B1 <> {};
      then a in A1 by XBOOLE_0:def 4;
      then
A7:   f.a in rng f by A4,FUNCT_1:def 3;
      f.a = g.a & a in B1 by A1,A4,A5,A6,PARTFUN1:def 4,XBOOLE_0:def 4;
      then f.a in rng g by A5,FUNCT_1:def 3;
      hence A2 /\ B2 <> {} by A7,XBOOLE_0:def 4;
    end;
    thus dom (f /\ g) = A1 /\ B1
    proof
      thus dom (f /\ g) c= A1 /\ B1 by A2,XBOOLE_1:1;
      let a be set;
      assume
A8:   a in A1 /\ B1;
      then a in A1 by XBOOLE_0:def 4;
      then
A9:   [a,f.a] in f by A4,FUNCT_1:def 2;
      f.a = g.a & a in B1 by A1,A4,A5,A8,PARTFUN1:def 4,XBOOLE_0:def 4;
      then [a,f.a] in g by A5,FUNCT_1:def 2;
      then [a,f.a] in f /\ g by A9,XBOOLE_0:def 4;
      hence thesis by XTUPLE_0:def 12;
    end;
  end;
  rng (f/\g) c= rng f /\ rng g & rng f /\ rng g c= A2/\B2 by RELAT_1:13
,XBOOLE_1:27;
  then
A10: rng (f/\g) c= A2/\B2 by XBOOLE_1:1;
A11: A2 = {} or A2 <> {};
A12: B2 = {} or B2 <> {};
A13: now
    per cases;
    case A2 /\ B2 <> {};
      hence dom (f /\ g) = A1 /\ B1 by A12,A3,Def1,A11;
    end;
    case A1 /\ B1 = {};
      hence dom (f /\ g) = A1 /\ B1 by A2;
    end;
    case A2 /\ B2 = {} & A1 /\ B1 <> {};
      hence f /\ g = {} by A12,A3,Def1,A11;
    end;
  end;
  dom (f/\g) c= A1/\B1 by A2,XBOOLE_1:1;
  hence thesis by A10,A13,Def1,RELSET_1:4;
end;



registration
  let A, B be set;
  cluster Funcs(A,B) -> functional;
  coherence
  proof
    let x be set;
    assume x in Funcs(A,B);
    then ex f being Function st x = f & dom f = A & rng f c= B by Def2;
    hence thesis;
  end;
end;

definition
  let A, B be set;
  mode FUNCTION_DOMAIN of A,B -> non empty set means
    :Def12:
    for x being Element of it holds x is Function of A,B;
  existence
  proof
    take IT = {the Function of A,B};
    let g be Element of IT;
    thus thesis by TARSKI:def 1;
  end;
end;

registration
  let A, B be set;
  cluster -> functional for FUNCTION_DOMAIN of A,B;
  coherence
  proof
    let F be FUNCTION_DOMAIN of A,B;
    thus for x being set st x in F holds x is Function by Def12;
  end;
end;

theorem
  for f being Function of P,Q holds { f } is FUNCTION_DOMAIN of P,Q
proof
  let f be Function of P,Q;
  for g be Element of { f } holds g is Function of P,Q by TARSKI:def 1;
  hence thesis by Def12;
end;

theorem Th122:
  Funcs(P,B) is FUNCTION_DOMAIN of P,B
proof
  for f be Element of Funcs(P,B) holds f is Function of P,B by Th66;
  hence thesis by Def12;
end;

definition
  let A be set, B be non empty set;
  redefine func Funcs(A,B) -> FUNCTION_DOMAIN of A,B;
  coherence by Th122;
  let F be FUNCTION_DOMAIN of A,B;
  redefine mode Element of F -> Function of A,B;
  coherence by Def12;
end;

registration
  let I be set;
  cluster id I -> total for I-defined Function;
  coherence;
end;



definition
  let X,A;
  let F be Function of X,A;
  let x be set;
  assume
A1: x in X;
  redefine func F/.x equals
  F.x;
  compatibility
  proof
    let a be Element of A;
    x in dom F by A1,Def1;
    hence thesis by PARTFUN1:def 6;
  end;
end;

theorem
  for f being Function of X, X, g being X-valued Function
   holds dom(f*g) = dom g
proof
  let f be Function of X, X;
  let g be X-valued Function;
  dom f = X by Def1;
  then rng g c= dom f;
  hence thesis by RELAT_1:27;
end;

theorem
  for X being non empty set, f being Function of X,X st for x being
  Element of X holds f.x = x holds f = id X
proof
  let X be non empty set, f be Function of X,X;
  assume
A1: for x being Element of X holds f.x = x;
    let x be Element of X;
    x = (id X).x by FUNCT_1:18;
    hence thesis by A1;
end;



definition
  let O,E be set;
  mode Action of O,E is Function of O, Funcs(E,E);
end;

theorem
  for x being set, A being set
   for f,g being Function of {x}, A st f.x = g.x
   holds f = g
proof
  let x be set, A be set;
  let f,g be Function of {x}, A such that
A1: f.x = g.x;
  now
    let y be set;
    assume y in {x};
    then y = x by TARSKI:def 1;
    hence f.y = g.y by A1;
  end;
  hence thesis by Th12;
end;















begin

definition
  let f be Function;
  let a,b be set;
  func f.(a,b) -> set equals
  f.[a,b];
  correctness;
end;

reserve A for set;

definition
  let A,B be non empty set, C be set, f be Function of [:A,B:],C;
  let a be Element of A;
  let b be Element of B;
  redefine func f.(a,b) -> Element of C;
  coherence
  proof
    reconsider ab = [a,b] as Element of [:A,B:] by ZFMISC_1:def 2;
    f.ab is Element of C;
    hence thesis;
  end;
end;

reserve X,Y,Z,x,x1,x2,y,y1,y2,z,z1,z2 for set;

theorem Th1:
  for X,Y,Z being set for f1,f2 being Function of [:X,Y:],Z st for
  x,y being set st x in X & y in Y holds f1.(x,y) = f2.(x,y) holds f1 = f2
proof
  let X,Y,Z be set;
  let f1,f2 be Function of [:X,Y:],Z such that
A1: for x,y being set st x in X & y in Y holds f1.(x,y) = f2.(x,y);
  for z being set st z in [:X,Y:] holds f1.z = f2.z
  proof
    let z be set;
    assume z in [:X,Y:];
    then consider x,y being set such that
A2: x in X & y in Y and
A3: z = [x,y] by ZFMISC_1:def 2;
    f1.(x,y) = f1.z & f2.(x,y) = f2.z by A3;
    hence thesis by A1,A2;
  end;
  hence thesis by FUNCT_2:12;
end;

theorem
  for f1,f2 being Function of [:X,Y:],Z st for a being Element of X for
  b being Element of Y holds f1.(a,b) = f2.(a,b) holds f1 = f2
proof
  let f1,f2 be Function of [:X,Y:],Z;
  assume
  for a being Element of X for b being Element of Y holds f1.(a,b) = f2.(a,b);
  then for x,y being set st x in X & y in Y holds f1.(x,y) = f2.(x,y);
  hence thesis by Th1;
end;

definition
  let A be set;
  mode UnOp of A is Function of A,A;
  mode BinOp of A is Function of [:A,A:],A;
end;

definition
  let A be set, f be BinOp of A;
  let a,b be Element of A;
  redefine func f.(a,b) -> Element of A;
  coherence
  proof
   per cases;
   suppose A <> {};
    then reconsider A as non empty set;
    reconsider f as BinOp of A;
    reconsider ab = [a,b] as Element of [:A,A:] by ZFMISC_1:def 2;
    f.ab is Element of A;
    hence thesis;
   end;
   suppose
A1:  A = {};
     then not [a,b] in dom f;
     then f.(a,b) = {} by FUNCT_1:def 2;
    hence thesis by A1,SUBSET_1:def 1;
   end;
  end;
end;

reserve u for UnOp of A,
  o,o9 for BinOp of A,
  a,b,c,e,e1,e2 for Element of A;

scheme
  FuncEx2{X, Y, Z() -> set, P[set,set,set]}: ex f being Function of [:X(),Y()
  :],Z() st for x,y st x in X() & y in Y() holds P[x,y,f.(x,y)]
provided
A1: for x,y st x in X() & y in Y() ex z st z in Z() & P[x,y,z]
proof
  defpred R[set,set] means for x1,y1 st $1 = [x1,y1] holds P[x1,y1,$2];
A2: for x st x in [:X(),Y():] ex z st z in Z() & R[x,z]
  proof
    let x;
    assume x in [:X(),Y():];
    then consider x1,y1 such that
A3: x1 in X() & y1 in Y() and
A4: x = [x1,y1] by ZFMISC_1:def 2;
    consider z such that
A5: z in Z() and
A6: P[x1,y1,z] by A1,A3;
    take z;
    thus z in Z() by A5;
    let x2,y2;
    assume
A7: x = [x2,y2];
    then x1 = x2 by A4,XTUPLE_0:1;
    hence thesis by A4,A6,A7,XTUPLE_0:1;
  end;
  consider f being Function of [:X(),Y():],Z() such that
A8: for x st x in [:X(),Y():] holds R[x,f.x] from FUNCT_2:sch 1(A2);
  take f;
  let x,y;
  assume x in X() & y in Y();
  then [x,y] in [:X(),Y():] by ZFMISC_1:def 2;
  hence thesis by A8;
end;

scheme
  Lambda2{X, Y, Z() -> set, F(set,set)->set}: ex f being Function of [:X(),Y()
  :],Z() st for x,y st x in X() & y in Y() holds f.(x,y) = F(x,y)
provided
A1: for x,y st x in X() & y in Y() holds F(x,y) in Z()
proof
  defpred P[set,set,set] means $3 = F($1,$2);
A2: for x,y st x in X() & y in Y() ex z st z in Z() & P[x,y,z] by A1;
  thus ex f being Function of [:X(),Y():],Z() st for x,y st x in X() & y in Y(
  ) holds P[x,y,f.(x,y)] from FuncEx2(A2);
end;

scheme
  FuncEx2D{X, Y, Z() -> non empty set, P[set,set,set]}: ex f being Function of
[:X(),Y():],Z() st for x being Element of X() for y being Element of Y() holds
  P[x,y,f.(x,y)]
provided
A1: for x being Element of X() for y being Element of Y() ex z being
Element of Z() st P[x,y,z]
proof
  defpred R[set,set] means for x being (Element of X()),y being Element of Y()
  st $1 = [x,y] holds P[x,y,$2];
A2: for p being Element of [:X(),Y():] ex z being Element of Z() st R[p,z]
  proof
    let p be Element of [:X(),Y():];
    consider x1,y1 such that
A3: x1 in X() and
A4: y1 in Y() and
A5: p = [x1,y1] by ZFMISC_1:def 2;
    reconsider y1 as Element of Y() by A4;
    reconsider x1 as Element of X() by A3;
    consider z being Element of Z() such that
A6: P[x1,y1,z] by A1;
    take z;
    let x be Element of X(),y be Element of Y();
    assume
A7: p = [x,y];
    then x1 = x by A5,XTUPLE_0:1;
    hence thesis by A5,A6,A7,XTUPLE_0:1;
  end;
  consider f being Function of [:X(),Y():],Z() such that
A8: for p being Element of [:X(),Y():] holds R[p,f.p] from FUNCT_2:sch 3
  (A2);
  take f;
  let x be Element of X();
  let y be Element of Y();
  reconsider xy = [x,y] as Element of [:X(),Y():] by ZFMISC_1:def 2;
  P[x,y,f.xy] by A8;
  hence thesis;
end;

scheme
  Lambda2D{X, Y, Z() -> non empty set, F(Element of X(),Element of Y()) ->
Element of Z()}: ex f being Function of [:X(),Y():],Z() st for x being Element
  of X() for y being Element of Y() holds f.(x,y)=F(x,y) proof
  defpred P[Element of X(),Element of Y(),set] means $3 = F($1,$2);
A1: for x being Element of X() for y being Element of Y() ex z being Element
  of Z() st P[x,y,z];
  thus ex f being Function of [:X(),Y():],Z() st for x being Element of X()
  for y being Element of Y() holds P[x,y,f.(x,y)] from FuncEx2D(A1);
end;

definition
  let A,o;
  attr o is commutative means
  :Def2:
  for a,b holds o.(a,b) = o.(b,a);
  attr o is associative means

  for a,b,c holds o.(a,o.(b,c)) = o.(o.(a,b ),c);
  attr o is idempotent means

  for a holds o.(a,a) = a;
end;

registration
  cluster -> empty associative commutative for BinOp of {};
  coherence
  proof
    let f be BinOp of {};
    thus f is empty;
A1: f c= [:dom f, rng f:];
    hereby
      let a,b,c be Element of {};
      thus f.(a,f.(b,c)) = {} by A1,FUNCT_1:def 2
        .= f.(f.(a,b),c) by A1,FUNCT_1:def 2;
    end;
    let a,b be Element of {};
    thus f.(a,b) = {} by A1,FUNCT_1:def 2
      .= f.(b,a) by A1,FUNCT_1:def 2;
  end;
end;

definition
  let A,e,o;
  pred e is_a_left_unity_wrt o means
  :Def5:
  for a holds o.(e,a) = a;
  pred e is_a_right_unity_wrt o means
  :Def6:
  for a holds o.(a,e) = a;
end;

definition
  let A,e,o;
  pred e is_a_unity_wrt o means
  e is_a_left_unity_wrt o & e is_a_right_unity_wrt o;
end;

theorem Th3:
  e is_a_unity_wrt o iff for a holds o.(e,a) = a & o.(a,e) = a
proof
  thus e is_a_unity_wrt o implies for a holds o.(e,a) = a & o.(a,e) = a
  proof
    assume e is_a_left_unity_wrt o & e is_a_right_unity_wrt o;
    hence thesis by Def5,Def6;
  end;
  assume for a holds o.(e,a) = a & o.(a,e) = a;
  hence (for a holds o.(e,a) = a) & for a holds o.(a,e) = a;
end;

theorem Th4:
  o is commutative implies (e is_a_unity_wrt o iff for a holds o.( e,a) = a)
proof
  assume
A1: o is commutative;
  now
    thus (for a holds o.(e,a) = a & o.(a,e) = a) implies for a holds o.(e,a) =
    a;
    assume
A2: for a holds o.(e,a) = a;
    let a;
    thus o.(e,a) = a by A2;
    thus o.(a,e) = o.(e,a) by A1,Def2
      .= a by A2;
  end;
  hence thesis by Th3;
end;

theorem Th5:
  o is commutative implies (e is_a_unity_wrt o iff for a holds o.( a,e) = a)
proof
  assume
A1: o is commutative;
  now
    thus (for a holds o.(e,a) = a & o.(a,e) = a) implies for a holds o.(a,e) =
    a;
    assume
A2: for a holds o.(a,e) = a;
    let a;
    thus o.(e,a) = o.(a,e) by A1,Def2
      .= a by A2;
    thus o.(a,e) = a by A2;
  end;
  hence thesis by Th3;
end;

theorem Th6:
  o is commutative implies (e is_a_unity_wrt o iff e is_a_left_unity_wrt o)
proof
  e is_a_left_unity_wrt o iff for a holds o.(e,a) = a by Def5;
  hence thesis by Th4;
end;

theorem Th7:
  o is commutative implies (e is_a_unity_wrt o iff e is_a_right_unity_wrt o)
proof
  e is_a_right_unity_wrt o iff for a holds o.(a,e) = a by Def6;
  hence thesis by Th5;
end;

theorem
  o is commutative implies (e is_a_left_unity_wrt o iff e
  is_a_right_unity_wrt o)
proof
  assume
A1: o is commutative;
  then e is_a_unity_wrt o iff e is_a_left_unity_wrt o by Th6;
  hence thesis by A1,Th7;
end;

theorem Th9:
  e1 is_a_left_unity_wrt o & e2 is_a_right_unity_wrt o implies e1 = e2
proof
  assume that
A1: e1 is_a_left_unity_wrt o and
A2: e2 is_a_right_unity_wrt o;
  thus e1 = o.(e1,e2) by A2,Def6
    .= e2 by A1,Def5;
end;

theorem Th10:
  e1 is_a_unity_wrt o & e2 is_a_unity_wrt o implies e1 = e2
proof
  assume that
  e1 is_a_left_unity_wrt o and
A1: e1 is_a_right_unity_wrt o & e2 is_a_left_unity_wrt o and
  e2 is_a_right_unity_wrt o;
  thus thesis by A1,Th9;
end;

definition
  let A,o;
  assume
A1: ex e st e is_a_unity_wrt o;
  func the_unity_wrt o -> Element of A means
  it is_a_unity_wrt o;
  existence by A1;
  uniqueness by Th10;
end;

definition
  let A,o9,o;
  pred o9 is_left_distributive_wrt o means
  :Def9:
  for a,b,c holds o9.(a,o.(b,c )) = o.(o9.(a,b),o9.(a,c));
  pred o9 is_right_distributive_wrt o means
  :Def10:
  for a,b,c holds o9.(o.(a,b ),c) = o.(o9.(a,c),o9.(b,c));
end;

definition
  let A,o9,o;
  pred o9 is_distributive_wrt o means
  o9 is_left_distributive_wrt o & o9 is_right_distributive_wrt o;
end;

theorem Th11:
  o9 is_distributive_wrt o iff for a,b,c holds o9.(a,o.(b,c)) = o.
  (o9.(a,b),o9.(a,c)) & o9.(o.(a,b),c) = o.(o9.(a,c),o9.(b,c))
proof
  thus o9 is_distributive_wrt o implies for a,b,c holds o9.(a,o.(b,c)) = o.(o9
  .(a,b),o9.(a,c)) & o9.(o.(a,b),c) = o.(o9.(a,c),o9.(b,c))
  proof
    assume o9 is_left_distributive_wrt o & o9 is_right_distributive_wrt o;
    hence thesis by Def9,Def10;
  end;
  assume for a,b,c holds o9.(a,o.(b,c)) = o.(o9.(a,b),o9.(a,c)) & o9.(o.(a,b)
  ,c) = o.(o9.(a,c),o9.(b,c));
  hence (for a,b,c holds o9.(a,o.(b,c)) = o.(o9.(a,b),o9.(a,c))) & for a,b,c
  holds o9.(o.(a,b),c) = o.(o9.(a,c),o9.(b,c));
end;

theorem Th12:
  for A being non empty set, o,o9 being BinOp of A holds o9 is
commutative implies (o9 is_distributive_wrt o iff for a,b,c being Element of A
  holds o9.(a,o.(b,c)) = o.(o9.(a,b),o9.(a,c)))
proof
  let A be non empty set, o,o9 be BinOp of A;
  assume
A1: o9 is commutative;
  (for a,b,c being Element of A holds o9.(a,o.(b,c)) = o.(o9.(a,b),o9.(a,c
  )) & o9.(o.(a,b),c) = o.(o9.(a,c),o9.(b,c))) iff for a,b,c being Element of A
  holds o9.(a,o.(b,c)) = o.(o9.(a,b),o9.(a,c))
  proof
    thus (for a,b,c being Element of A holds o9.(a,o.(b,c)) = o.(o9.(a,b),o9.(
a,c)) & o9.(o.(a,b),c) = o.(o9.(a,c),o9.(b,c))) implies for a,b,c being Element
    of A holds o9.(a,o.(b,c)) = o.(o9.(a,b),o9.(a,c));
    assume
A2: for a,b,c being Element of A holds o9.(a,o.(b,c)) = o.(o9.(a,b),o9 .(a,c));
    let a,b,c be Element of A;
    thus o9.(a,o.(b,c)) = o.(o9.(a,b),o9.(a,c)) by A2;
    thus o9.(o.(a,b),c) = o9.(c,o.(a,b)) by A1,Def2
      .= o.(o9.(c,a),o9.(c,b)) by A2
      .= o.(o9.(a,c),o9.(c,b)) by A1,Def2
      .= o.(o9.(a,c),o9.(b,c)) by A1,Def2;
  end;
  hence thesis by Th11;
end;

theorem Th13:
  for A being non empty set, o,o9 being BinOp of A holds o9 is
commutative implies (o9 is_distributive_wrt o iff for a,b,c being Element of A
  holds o9.(o.(a,b),c) = o.(o9.(a,c),o9.(b,c)))
proof
  let A be non empty set, o,o9 be BinOp of A;
  assume
A1: o9 is commutative;
  (for a,b,c being Element of A holds o9.(a,o.(b,c)) = o.(o9.(a,b),o9.(a,c
  )) & o9.(o.(a,b),c) = o.(o9.(a,c),o9.(b,c))) iff for a,b,c being Element of A
  holds o9.(o.(a,b),c) = o.(o9.(a,c),o9.(b,c))
  proof
    thus (for a,b,c being Element of A holds o9.(a,o.(b,c)) = o.(o9.(a,b),o9.(
a,c)) & o9.(o.(a,b),c) = o.(o9.(a,c),o9.(b,c))) implies for a,b,c being Element
    of A holds o9.(o.(a,b),c) = o.(o9.(a,c),o9.(b,c));
    assume
A2: for a,b,c being Element of A holds o9.(o.(a,b),c) = o.(o9.(a,c),o9 .(b,c));
    let a,b,c be Element of A;
    thus o9.(a,o.(b,c)) = o9.(o.(b,c),a) by A1,Def2
      .= o.(o9.(b,a),o9.(c,a)) by A2
      .= o.(o9.(a,b),o9.(c,a)) by A1,Def2
      .= o.(o9.(a,b),o9.(a,c)) by A1,Def2;
    thus thesis by A2;
  end;
  hence thesis by Th11;
end;

theorem Th14:
  for A being non empty set, o,o9 being BinOp of A holds o9 is
commutative implies (o9 is_distributive_wrt o iff o9 is_left_distributive_wrt o
  )
proof
  let A be non empty set, o,o9 be BinOp of A;
  o9 is_left_distributive_wrt o iff for a,b,c being Element of A holds o9.
  (a,o.(b,c)) = o.(o9.(a,b),o9.(a,c)) by Def9;
  hence thesis by Th12;
end;

theorem Th15:
  for A being non empty set, o,o9 being BinOp of A holds o9 is
commutative implies (o9 is_distributive_wrt o iff o9 is_right_distributive_wrt
  o)
proof
  let A be non empty set, o,o9 be BinOp of A;
  o9 is_right_distributive_wrt o iff for a,b,c being Element of A holds o9
  .(o.(a,b),c) = o.(o9.(a,c),o9.(b,c)) by Def10;
  hence thesis by Th13;
end;

theorem
  for A being non empty set, o,o9 being BinOp of A holds o9 is
  commutative implies (o9 is_right_distributive_wrt o iff o9
  is_left_distributive_wrt o)
proof
  let A be non empty set, o,o9 be BinOp of A;
  assume
A1: o9 is commutative;
  then o9 is_distributive_wrt o iff o9 is_left_distributive_wrt o by Th14;
  hence thesis by A1,Th15;
end;

definition
  let A,u,o;
  pred u is_distributive_wrt o means
  :Def12:
  for a,b holds u.(o.(a,b)) = o.((u .a),(u.b));
end;














definition
 canceled 3;
  let A be non empty set, e be Element of A, o be BinOp of A;
  redefine pred e is_a_left_unity_wrt o means
  for a being Element of A holds o .(e,a) = a;
  correctness by Def5;
  redefine pred e is_a_right_unity_wrt o means
  for a being Element of A holds o.(a,e) = a;
  correctness by Def6;
end;

definition
  let A be non empty set, o9,o be BinOp of A;
  redefine pred o9 is_left_distributive_wrt o means
  for a,b,c being Element of A holds o9.(a,o.(b,c)) = o.(o9.(a,b),o9.(a,c));
  correctness by Def9;
  redefine pred o9 is_right_distributive_wrt o means
  for a,b,c being Element of A holds o9.(o.(a,b),c) = o.(o9.(a,c),o9.(b,c));
  correctness by Def10;
end;

definition
  let A be non empty set, u be UnOp of A, o be BinOp of A;
  redefine pred u is_distributive_wrt o means
  for a,b being Element of A holds u.(o.(a,b)) = o.((u.a),(u.b));
  correctness by Def12;
end;



theorem
  for f being Function of [:X,Y:],Z st x in X & y in Y & Z <> {} holds f
  .(x,y) in Z
proof
  let f be Function of [:X,Y:],Z;
  assume x in X & y in Y;
  then [x,y] in [:X,Y:] by ZFMISC_1:87;
  hence thesis by FUNCT_2:5;
end;



theorem
  for x, y, X, Y, Z being set, f being Function of [:X,Y:],Z, g being
  Function st Z <> {} & x in X & y in Y holds (g*f).(x,y) = g.(f.(x,y))
proof
  let x, y, X, Y, Z be set, f be Function of [:X,Y:],Z, g be Function such
  that
A1: Z <> {};
  assume x in X & y in Y;
  then [x,y] in [:X,Y:] by ZFMISC_1:87;
  hence thesis by A1,FUNCT_2:15;
end;



theorem
  for f being Function st dom f = [:X,Y:] holds f is constant iff for x1
  ,x2,y1,y2 st x1 in X & x2 in X & y1 in Y & y2 in Y holds f.(x1,y1)=f.(x2,y2)
proof
  let f be Function such that
A1: dom f = [:X,Y:];
  hereby
    assume
A2: f is constant;
    let x1,x2,y1,y2;
    assume x1 in X & x2 in X & y1 in Y & y2 in Y;
    then [x1,y1] in [:X,Y:] & [x2,y2] in [:X,Y:] by ZFMISC_1:87;
    hence f.(x1,y1)=f.(x2,y2) by A1,A2,FUNCT_1:def 10;
  end;
  assume
A3: for x1,x2,y1,y2 st x1 in X & x2 in X & y1 in Y & y2 in Y holds f.(x1
  ,y1)=f.(x2,y2);
  let x,y;
  assume x in dom f;
  then consider x1,y1 such that
A4: x1 in X & y1 in Y and
A5: x = [x1,y1] by A1,ZFMISC_1:84;
  assume y in dom f;
  then consider x2,y2 such that
A6: x2 in X & y2 in Y and
A7: y = [x2,y2] by A1,ZFMISC_1:84;
  thus f.x = f.(x1,y1) by A5
    .= f.(x2,y2) by A3,A4,A6
    .= f.y by A7;
end;



theorem
  for f1,f2 being PartFunc of [:X,Y:],Z st dom f1 = dom f2 & for x,y st
  [x,y] in dom f1 holds f1.(x,y)=f2.(x,y) holds f1 = f2
proof
  let f1,f2 be PartFunc of [:X,Y:],Z such that
A1: dom f1 = dom f2 and
A2: for x,y st [x,y] in dom f1 holds f1.(x,y) = f2.(x,y);
  z in dom f1 implies f1.z = f2.z
  proof
    assume
A3: z in dom f1;
    then consider x,y such that
A4: z = [x,y] by RELAT_1:def 1;
    f1.(x,y) = f2.(x,y) by A2,A3,A4;
    hence thesis by A4;
  end;
  hence thesis by A1,FUNCT_1:2;
end;

scheme
  PartFuncEx2{X,Y,Z()->set,P[set,set,set]}: ex f being PartFunc of [:X(),Y():]
,Z() st (for x,y holds [x,y] in dom f iff x in X() & y in Y() & ex z st P[x,y,z
  ]) & for x,y st [x,y] in dom f holds P[x,y,f.(x,y)]
provided
A1: for x,y,z st x in X() & y in Y() & P[x,y,z] holds z in Z() and
A2: for x,y,z1,z2 st x in X() & y in Y() & P[x,y,z1] & P[x,y,z2] holds z1 = z2
proof
  defpred Q[set,set] means for x1,y1 st $1 = [x1,y1] holds P[x1,y1,$2];
A3: for x,z st x in [:X(),Y():] & Q[x,z] holds z in Z()
  proof
    let x,z;
    assume x in [:X(),Y():];
    then
A4: ex x1,y1 st x1 in X() & y1 in Y() & x = [x1,y1] by ZFMISC_1:def 2;
    assume for x1,y1 st x = [x1,y1] holds P[x1,y1,z];
    hence thesis by A1,A4;
  end;
A5: for x,z1,z2 st x in [:X(),Y():] & Q[x,z1] & Q[x,z2] holds z1 = z2
  proof
    let x,z1,z2 such that
A6: x in [:X(),Y():] and
A7: ( for x1,y1 st x = [x1,y1] holds P[x1,y1,z1])& for x1,y1 st x = [
    x1,y1] holds P[x1,y1,z2];
    consider x1,y1 such that
A8: x1 in X() & y1 in Y() and
A9: x = [x1,y1] by A6,ZFMISC_1:def 2;
    ( P[x1,y1,z1])& P[x1,y1,z2] by A7,A9;
    hence thesis by A2,A8;
  end;
  consider f being PartFunc of [:X(),Y():],Z() such that
A10: for x holds x in dom f iff x in [:X(),Y():] & ex z st Q[x,z] and
A11: for x st x in dom f holds Q[x,f.x] from PARTFUN1:sch 2(A3,A5);
  take f;
  thus for x,y holds [x,y] in dom f iff x in X() & y in Y() & ex z st P[x,y,z ]
  proof
    let x,y;
    thus [x,y] in dom f implies x in X() & y in Y() & ex z st P[x,y,z]
    proof
      assume
A12:  [x,y] in dom f;
      hence x in X() & y in Y() by ZFMISC_1:87;
      consider z such that
A13:  for x1,y1 st [x,y] = [x1,y1] holds P[x1,y1,z] by A10,A12;
      take z;
      thus thesis by A13;
    end;
    assume x in X() & y in Y();
    then
A14: [x,y] in [:X(),Y():] by ZFMISC_1:def 2;
    given z such that
A15: P[x,y,z];
    now
      take z;
      let x1,y1;
      assume
A16:  [x,y] = [x1,y1];
      then x=x1 by XTUPLE_0:1;
      hence P[x1,y1,z] by A15,A16,XTUPLE_0:1;
    end;
    hence thesis by A10,A14;
  end;
  thus thesis by A11;
end;

scheme
  LambdaR2{X,Y,Z()->set,F(set,set)->set,P[set,set]}: ex f being PartFunc of [:
X(),Y():],Z() st (for x,y holds [x,y] in dom f iff x in X() & y in Y() & P[x,y]
  ) & for x,y st [x,y] in dom f holds f.(x,y) = F(x,y)
provided
A1: for x,y st P[x,y] holds F(x,y) in Z()
proof
  defpred Q[set,set,set] means P[$1,$2] & $3 = F($1,$2);
A2: for x,y,z1,z2 st x in X() & y in Y() & Q[x,y,z1] & Q[x,y,z2] holds z1 = z2;
A3: for x,y,z st x in X() & y in Y() & Q[x,y,z] holds z in Z() by A1;
  consider f being PartFunc of [:X(),Y():],Z() such that
A4: for x,y holds [x,y] in dom f iff x in X() & y in Y() & ex z st Q[x,y
  ,z] and
A5: for x,y st [x,y] in dom f holds Q[x,y,f.(x,y)] from PartFuncEx2(A3,
  A2 );
  take f;
  thus for x,y holds [x,y] in dom f iff x in X() & y in Y() & P[x,y]
  proof
    let x,y;
    thus [x,y] in dom f implies x in X() & y in Y() & P[x,y]
    proof
      assume
A6:   [x,y] in dom f;
      then ex z st P[x,y] & z = F(x,y) by A4;
      hence thesis by A4,A6;
    end;
    assume that
A7: x in X() & y in Y() and
A8: P[x,y];
    ex z st P[x,y] & z = F(x,y) by A8;
    hence thesis by A4,A7;
  end;
  thus thesis by A5;
end;



scheme
  PartLambda2{X,Y,Z()->set,F(set,set)->set,P[set,set]}: ex f being PartFunc of
[:X(),Y():],Z() st (for x,y holds [x,y] in dom f iff x in X() & y in Y() & P[x,
  y]) & for x,y st [x,y] in dom f holds f.(x,y) = F(x,y)
provided
A1: for x,y st x in X() & y in Y() & P[x,y] holds F(x,y) in Z()
proof
  defpred R[set,set] means $1 in X() & $2 in Y() & P[$1,$2];
A2: for x,y st R[x,y] holds F(x,y) in Z() by A1;
  consider f being PartFunc of [:X(),Y():],Z() such that
A3: ( for x,y holds [x,y] in dom f iff x in X() & y in Y() & R[x,y])&
  for x,y st [ x,y] in dom f holds f.(x,y) = F(x,y) from LambdaR2(A2);
  take f;
  thus thesis by A3;
end;

scheme
  {X,Y()->non empty set,Z()->set,F(set,set)->set, P[set,set]}: ex f being
PartFunc of [:X(),Y():],Z() st (for x being Element of X(), y being Element of
  Y() holds [x,y] in dom f iff P[x,y]) & for x being Element of X(), y being
  Element of Y() st [x,y] in dom f holds f.(x,y) = F(x,y)
provided
A1: for x being Element of X(), y being Element of Y() st P[x,y] holds F
(x,y) in Z()
proof
A2: for x,y st x in X() & y in Y() & P[x,y] holds F(x,y) in Z() by A1;
  consider f being PartFunc of [:X(),Y():],Z() such that
A3: ( for x,y holds [x,y] in dom f iff x in X() & y in Y() & P[x,y])&
  for x,y st [ x,y] in dom f holds f.(x,y) = F(x,y) from PartLambda2(A2);
  take f;
  thus thesis by A3;
end;



definition
  let A be set, f be BinOp of A, x,y be Element of A;
  redefine func f.(x,y) -> Element of A;
  coherence
  proof
    per cases;
    suppose
   A = {};
      then dom f = {};
      then f.(x,y) = {} by FUNCT_1:def 2;
      hence thesis;
    end;
    suppose
      A <> {};
      then reconsider A as non empty set;
      reconsider f as BinOp of A;
      reconsider x,y as Element of A;
      f.(x,y) is Element of A;
      hence thesis;
    end;
  end;
end;

definition
 let X,Y,Z be set; let f1,f2 being Function of [:X,Y:],Z;
 redefine pred f1 = f2 means
  for x,y being set st x in X & y in Y holds f1.(x,y) = f2.(x,y);
 compatibility by Th1;
end;














begin

reserve a,b,c,d for set,
  D,X1,X2,X3,X4 for non empty set,
  x1,y1,z1 for Element of X1,
  x2 for Element of X2,
  x3 for Element of X3,
  x4 for Element of X4,
  A1,B1 for Subset of X1;



theorem
  a in [:X1,X2:] implies ex x1,x2 st a=[x1,x2]
proof
  assume a in [:X1,X2:];
  then consider x1,x2 being set such that
A1: x1 in X1 and
A2: x2 in X2 and
A3: a=[x1,x2] by ZFMISC_1:def 2;
  reconsider x2 as Element of X2 by A2;
  reconsider x1 as Element of X1 by A1;
  take x1,x2;
  thus thesis by A3;
end;

theorem
  for x,y being Element of [:X1,X2:] st x`1=y`1 & x`2=y`2 holds x=y
proof
  let x,y be Element of [:X1,X2:];
  [x`1,x`2]=x by MCART_1:21;
  hence thesis by MCART_1:21;
end;

definition
  let X1,X2,x1,x2;
  redefine func [x1,x2] -> Element of [:X1,X2:];
  coherence by ZFMISC_1:87;
end;

definition
  let X1,X2;
  let x be Element of [:X1,X2:];
  redefine func x`1 -> Element of X1;
  coherence by MCART_1:10;
  redefine func x`2 -> Element of X2;
  coherence by MCART_1:10;
end;



theorem Th3:
  a in [: X1,X2,X3 :] iff ex x1,x2,x3 st a = [x1,x2,x3]
proof
  thus a in [: X1,X2,X3 :] implies ex x1,x2,x3 st a = [x1,x2,x3]
  proof
    assume a in [: X1,X2,X3 :];
    then a in [:[:X1,X2:],X3:] by ZFMISC_1:def 3;
    then consider x12, x3 being set such that
A1: x12 in [:X1,X2:] and
A2: x3 in X3 and
A3: a = [x12,x3] by ZFMISC_1:def 2;
    reconsider x3 as Element of X3 by A2;
    consider x1,x2 being set such that
A4: x1 in X1 and
A5: x2 in X2 and
A6: x12 = [x1,x2] by A1,ZFMISC_1:def 2;
    reconsider x2 as Element of X2 by A5;
    reconsider x1 as Element of X1 by A4;
    a = [x1,x2,x3] by A3,A6;
    hence thesis;
  end;
  given x1,x2,x3 such that
A7: a = [x1,x2,x3];
  a = [[x1,x2],x3] by A7;
  then a in [:[:X1,X2:],X3:];
  hence thesis by ZFMISC_1:def 3;
end;

theorem Th4:
  (for a holds a in D iff ex x1,x2,x3 st a = [x1,x2,x3]) implies D
  = [: X1,X2,X3 :]
proof
  assume
A1: for a holds a in D iff ex x1,x2,x3 st a = [x1,x2,x3];
  now
    let a;
    thus a in D implies a in [:[:X1,X2:],X3:]
    proof
      assume a in D;
      then consider x1,x2,x3 such that
A2:   a = [x1,x2,x3] by A1;
      a = [[x1,x2],x3] by A2;
      hence thesis;
    end;
    assume a in [:[:X1,X2:],X3:];
    then consider x12,x3 being set such that
A3: x12 in [:X1,X2:] and
A4: x3 in X3 and
A5: a = [x12,x3] by ZFMISC_1:def 2;
    reconsider x3 as Element of X3 by A4;
    consider x1,x2 being set such that
A6: x1 in X1 and
A7: x2 in X2 and
A8: x12 = [x1,x2] by A3,ZFMISC_1:def 2;
    reconsider x2 as Element of X2 by A7;
    reconsider x1 as Element of X1 by A6;
    a = [x1,x2,x3] by A5,A8;
    hence a in D by A1;
  end;
  then D = [:[:X1,X2:],X3:] by TARSKI:1;
  hence thesis by ZFMISC_1:def 3;
end;

theorem
  D = [: X1,X2,X3 :] iff for a holds a in D iff ex x1,x2,x3 st a = [x1,
  x2,x3] by Th3,Th4;

reserve x,y for Element of [:X1,X2,X3:];

definition
  let X1,X2,X3,x1,x2,x3;
  redefine func [x1,x2,x3] -> Element of [:X1,X2,X3:];
  coherence by MCART_1:69;
end;

theorem
  a =x`1_3 iff for x1,x2,x3 st x = [x1,x2,x3] holds a = x1
proof
  thus a =x`1_3 implies for x1,x2,x3 st x = [x1,x2,x3] holds a = x1
  proof
A1: x = [x`1_3,x`2_3,x`3_3];
    assume
A2: a = x`1_3;
    let x1,x2,x3;
    assume x = [x1,x2,x3];
    hence thesis by A2,A1,XTUPLE_0:3;
  end;
  thus thesis by MCART_1:65;
end;

theorem
  b =x`2_3 iff for x1,x2,x3 st x = [x1,x2,x3] holds b = x2
proof
  thus b =x`2_3 implies for x1,x2,x3 st x = [x1,x2,x3] holds b = x2
  proof
A1: x = [x`1_3,x`2_3,x`3_3];
    assume
A2: b = x`2_3;
    let x1,x2,x3;
    assume x = [x1,x2,x3];
    hence thesis by A2,A1,XTUPLE_0:3;
  end;
  thus thesis by MCART_1:66;
end;

theorem
  c =x`3_3 iff for x1,x2,x3 st x = [x1,x2,x3] holds c = x3
proof
  thus c =x`3_3 implies for x1,x2,x3 st x = [x1,x2,x3] holds c = x3
  proof
A1: x = [x`1_3,x`2_3,x`3_3];
    assume
A2: c = x`3_3;
    let x1,x2,x3;
    assume x = [x1,x2,x3];
    hence thesis by A2,A1,XTUPLE_0:3;
  end;
  thus thesis by MCART_1:67;
end;

theorem
  x`1_3=y`1_3 & x`2_3=y`2_3 & x`3_3=y`3_3 implies x=y
proof
  [x`1_3,x`2_3,x`3_3]=x;
  hence thesis by MCART_1:44;
end;

theorem Th10:
  a in [: X1,X2,X3,X4 :] iff ex x1,x2,x3,x4 st a = [x1,x2,x3,x4]
proof
  thus a in [: X1,X2,X3,X4 :] implies ex x1,x2,x3,x4 st a = [x1,x2,x3,x4]
  proof
    assume a in [: X1,X2,X3,X4 :];
    then a in [:[:X1,X2,X3:],X4:] by ZFMISC_1:def 4;
    then consider x123, x4 being set such that
A1: x123 in [:X1,X2,X3:] and
A2: x4 in X4 and
A3: a = [x123,x4] by ZFMISC_1:def 2;
    reconsider x4 as Element of X4 by A2;
    consider x1 being (Element of X1), x2 being (Element of X2), x3 being
    Element of X3 such that
A4: x123 = [x1,x2,x3] by A1,Th3;
    a = [x1,x2,x3,x4] by A3,A4;
    hence thesis;
  end;
  given x1,x2,x3,x4 such that
A5: a = [x1,x2,x3,x4];
  a = [[x1,x2,x3],x4] by A5;
  then a in [:[:X1,X2,X3:],X4:];
  hence thesis by ZFMISC_1:def 4;
end;

theorem Th11:
  (for a holds a in D iff ex x1,x2,x3,x4 st a = [x1,x2,x3,x4])
  implies D = [: X1,X2,X3,X4 :]
proof
  assume
A1: for a holds a in D iff ex x1,x2,x3,x4 st a = [x1,x2,x3,x4];
  now
    let a;
    thus a in D implies a in [:[:X1,X2,X3:],X4:]
    proof
      assume a in D;
      then consider x1,x2,x3,x4 such that
A2:   a = [x1,x2,x3,x4] by A1;
      a = [[x1,x2,x3],x4] by A2;
      hence thesis;
    end;
    assume a in [:[:X1,X2,X3:],X4:];
    then consider x123,x4 being set such that
A3: x123 in [:X1,X2,X3:] and
A4: x4 in X4 and
A5: a = [x123,x4] by ZFMISC_1:def 2;
    reconsider x4 as Element of X4 by A4;
    consider x1,x2,x3 such that
A6: x123 = [x1,x2,x3] by A3,Th3;
    a = [x1,x2,x3,x4] by A5,A6;
    hence a in D by A1;
  end;
  then D = [:[:X1,X2,X3:],X4:] by TARSKI:1;
  hence thesis by ZFMISC_1:def 4;
end;

theorem
  D = [: X1,X2,X3,X4 :] iff for a holds a in D iff ex x1,x2,x3,x4 st a =
  [x1,x2,x3,x4] by Th10,Th11;

reserve x for Element of [:X1,X2,X3,X4:];

definition
  let X1,X2,X3,X4,x1,x2,x3,x4;
  redefine func [x1,x2,x3,x4] -> Element of [:X1,X2,X3,X4:];
  coherence by MCART_1:80;
end;

theorem
  a=x`1_4 iff for x1,x2,x3,x4 st x = [x1,x2,x3,x4] holds a = x1
proof
  thus a =x`1_4 implies for x1,x2,x3,x4 st x = [x1,x2,x3,x4] holds a = x1
  proof
A1: x = [x`1_4,x`2_4,x`3_4,x`4_4];
    assume
A2: a = x`1_4;
    let x1,x2,x3,x4;
    assume x = [x1,x2,x3,x4];
    hence thesis by A2,A1,XTUPLE_0:5;
  end;
  thus thesis by MCART_1:75;
end;

theorem
  b=x`2_4 iff for x1,x2,x3,x4 st x = [x1,x2,x3,x4] holds b = x2
proof
  thus b =x`2_4 implies for x1,x2,x3,x4 st x = [x1,x2,x3,x4] holds b = x2
  proof
A1: x = [x`1_4,x`2_4,x`3_4,x`4_4];
    assume
A2: b = x`2_4;
    let x1,x2,x3,x4;
    assume x = [x1,x2,x3,x4];
    hence thesis by A2,A1,XTUPLE_0:5;
  end;
  thus thesis by MCART_1:76;
end;

theorem
  c = x`3_4 iff for x1,x2,x3,x4 st x = [x1,x2,x3,x4] holds c = x3
proof
  thus c =x`3_4 implies for x1,x2,x3,x4 st x = [x1,x2,x3,x4] holds c = x3
  proof
A1: x = [x`1_4,x`2_4,x`3_4,x`4_4];
    assume
A2: c = x`3_4;
    let x1,x2,x3,x4;
    assume x = [x1,x2,x3,x4];
    hence thesis by A2,A1,XTUPLE_0:5;
  end;
  thus thesis by MCART_1:77;
end;

theorem
  d=x`4_4 iff for x1,x2,x3,x4 st x = [x1,x2,x3,x4] holds d = x4
proof
  thus d =x`4_4 implies for x1,x2,x3,x4 st x = [x1,x2,x3,x4] holds d = x4
  proof
A1: x = [x`1_4,x`2_4,x`3_4,x`4_4];
    assume
A2: d = x`4_4;
    let x1,x2,x3,x4;
    assume x = [x1,x2,x3,x4];
    hence thesis by A2,A1,XTUPLE_0:5;
  end;
  thus thesis by MCART_1:78;
end;

theorem
  for x,y being Element of [:X1,X2,X3,X4:]
   st x`1_4=y`1_4 & x`2_4=y`2_4 & x`3_4=y
  `3_4 & x`4_4=y`4_4 holds x=y
proof
  let x,y be Element of [:X1,X2,X3,X4:];
  [x`1_4,x`2_4,x`3_4,x`4_4]=x;
  hence thesis by MCART_1:56;
end;

reserve A2 for Subset of X2,
  A3 for Subset of X3,
  A4 for Subset of X4;

scheme
  Fraenkel1 {P[set]}: for X1 holds { x1 : P[x1] } is Subset of X1 proof
  let X1;
  { x1 : P[x1] } c= X1
  proof
    let a;
    assume a in { x1 : P[x1] };
    then ex x1 st a = x1 & P[x1];
    hence thesis;
  end;
  hence thesis;
end;

scheme
  Fraenkel2 {P[set,set]}: for X1,X2 holds { [x1,x2] : P[x1,x2] } is Subset of
  [:X1,X2:] proof
  let X1,X2;
  { [x1,x2] : P[x1,x2] } c= [:X1,X2:]
  proof
    let a;
    assume a in { [x1,x2] : P[x1,x2] };
    then ex x1,x2 st a = [x1,x2] & P[x1,x2];
    hence thesis;
  end;
  hence thesis;
end;

scheme
  Fraenkel3 {P[set,set,set]}: for X1,X2,X3 holds { [x1,x2,x3] : P[x1,x2,x3] }
  is Subset of [:X1,X2,X3:] proof
  let X1,X2,X3;
  { [x1,x2,x3] : P[x1,x2,x3] } c= [:X1,X2,X3:]
  proof
    let a;
    assume a in { [x1,x2,x3] : P[x1,x2,x3] };
    then ex x1,x2,x3 st a = [x1,x2,x3] & P[x1,x2,x3];
    hence thesis;
  end;
  hence thesis;
end;

scheme
  Fraenkel4 {P[set,set,set,set]}: for X1,X2,X3,X4 holds { [x1,x2,x3,x4] : P[x1
  ,x2,x3,x4] } is Subset of [:X1,X2,X3,X4:] proof
  let X1,X2,X3,X4;
  { [x1,x2,x3,x4] : P[x1,x2,x3,x4] } c= [:X1,X2,X3,X4:]
  proof
    let a;
    assume a in { [x1,x2,x3,x4] : P[x1,x2,x3,x4] };
    then ex x1,x2,x3,x4 st a = [x1,x2,x3,x4] & P[x1,x2,x3,x4];
    hence thesis;
  end;
  hence thesis;
end;

scheme
  Fraenkel5 {P[set],Q[set]}: for X1 st for x1 holds P[x1] implies Q[x1] holds
  { y1 : P[y1] } c= { z1 : Q[z1] } proof
  let X1 such that
A1: P[x1] implies Q[x1];
  let a;
  assume a in { x1 : P[x1] };
  then consider x1 such that
A2: a = x1 and
A3: P[x1];
  Q[x1] by A1,A3;
  hence thesis by A2;
end;

scheme
  Fraenkel6 {P[set],Q[set]}: for X1 st for x1 holds P[x1] iff Q[x1] holds { y1
  : P[y1] } = { z1 : Q[z1] } proof
  let X1;
  assume
A1: P[x1] iff Q[x1];
  for X1 st for x1 holds P[x1] implies Q[x1] holds { y1 : P[y1] } c= { z1
  : Q[z1] } from Fraenkel5;
  hence { y1 : P[y1] } c= { z1 : Q[z1] } by A1;
  for X1 st for x1 holds Q[x1] implies P[x1] holds { y1 : Q[y1] } c= { z1
  : P[z1] } from Fraenkel5;
  hence thesis by A1;
end;

scheme
  SubsetD{D() -> non empty set,P[set]}: {d where d is Element of D() : P[d]}
  is Subset of D() proof
  for D being non empty set holds {d where d is Element of D : P[d]} is
  Subset of D from Fraenkel1;
  hence thesis;
end;

theorem
  X1 = { x1 : not contradiction }
proof
  defpred P[set] means not contradiction;
A1: y1 in { x1 : not contradiction };
  { x1 : P[x1] } is Subset of X1 from SubsetD;
  hence thesis by A1,SUBSET_1:28;
end;

theorem
  [:X1,X2:] = { [x1,x2] : not contradiction }
proof
  defpred P[set,set] means not contradiction;
A1: for x being Element of [:X1,X2:] holds x in { [x1,x2] : not contradiction }
  proof
    let x be Element of [:X1,X2:];
    x = [x`1,x`2] by MCART_1:21;
    hence thesis;
  end;
  for X1,X2 holds { [x1,x2] : P[x1,x2] } is Subset of [:X1,X2:] from
  Fraenkel2;
  then { [x1,x2] : not contradiction } is Subset of [:X1,X2:];
  hence thesis by A1,SUBSET_1:28;
end;

theorem
  [:X1,X2,X3:] = { [x1,x2,x3] : not contradiction }
proof
  defpred P[set,set,set] means not contradiction;
A1: for x being Element of [:X1,X2,X3:] holds x in { [x1,x2,x3] : not
  contradiction }
  proof
    let x be Element of [:X1,X2,X3:];
    x = [x`1_3,x`2_3,x`3_3];
    hence thesis;
  end;
  for X1,X2,X3 holds { [x1,x2,x3] : P[x1,x2,x3] } is Subset of [:X1,X2,X3
  :] from Fraenkel3;
  then { [x1,x2,x3] : not contradiction } is Subset of [:X1,X2,X3:];
  hence thesis by A1,SUBSET_1:28;
end;

theorem
  [:X1,X2,X3,X4:] = { [x1,x2,x3,x4] : not contradiction }
proof
  defpred P[set,set,set,set] means not contradiction;
A1: for x being Element of [:X1,X2,X3,X4:] holds x in { [x1,x2,x3,x4] : not
  contradiction }
  proof
    let x be Element of [:X1,X2,X3,X4:];
    x = [x`1_4,x`2_4,x`3_4,x`4_4];
    hence thesis;
  end;
  for X1,X2,X3,X4 holds { [x1,x2,x3,x4] : P[x1,x2,x3,x4] } is Subset of [:
  X1,X2,X3,X4:] from Fraenkel4;
  then { [x1,x2,x3,x4] : not contradiction } is Subset of [:X1,X2,X3,X4:];
  hence thesis by A1,SUBSET_1:28;
end;

theorem
  A1 = { x1 : x1 in A1 }
proof
  thus A1 c= { x1 : x1 in A1 }
  proof
    let a;
    assume a in A1;
    hence thesis;
  end;
  let a;
  assume a in { x1 : x1 in A1 };
  then ex x1 st a = x1 & x1 in A1;
  hence thesis;
end;

theorem
  [:A1,A2:] = { [x1,x2] : x1 in A1 & x2 in A2 }
proof
  thus [:A1,A2:] c= { [x1,x2] : x1 in A1 & x2 in A2 }
  proof
    let a;
    assume
A1: a in [:A1,A2:];
    then reconsider x = a as Element of [:X1,X2:];
A2: x = [x`1,x`2] by MCART_1:21;
    x`1 in A1 & x`2 in A2 by A1,MCART_1:10;
    hence thesis by A2;
  end;
  let a;
  assume a in { [x1,x2] : x1 in A1 & x2 in A2 };
  then ex x1,x2 st a = [x1,x2] & x1 in A1 & x2 in A2;
  hence thesis by ZFMISC_1:87;
end;

theorem
  [:A1,A2,A3:] = { [x1,x2,x3] : x1 in A1 & x2 in A2 & x3 in A3 }
proof
  thus [:A1,A2,A3:] c= { [x1,x2,x3] : x1 in A1 & x2 in A2 & x3 in A3 }
  proof
    let a;
    assume
B1: a in [:A1,A2,A3:];
    reconsider A1 as non empty Subset of X1 by B1,MCART_1:31;
    reconsider A2 as non empty Subset of X2 by B1,MCART_1:31;
    reconsider A3 as non empty Subset of X3 by B1,MCART_1:31;
A1: a in [:A1,A2,A3:] by B1;
    reconsider x = a as Element of [:X1,X2,X3:] by A1;
B2: x = [x`1_3,x`2_3,x`3_3];
A2:  x`3_3 in A3 by A1,MCART_1:72;
    x`1_3 in A1 & x`2_3 in A2 by A1,MCART_1:72;
    hence thesis by A2,B2;
  end;
  let a;
  assume a in { [x1,x2,x3] : x1 in A1 & x2 in A2 & x3 in A3 };
  then ex x1,x2,x3 st a= [x1,x2,x3] & x1 in A1 & x2 in A2 & x3 in A3;
  hence thesis by MCART_1:69;
end;

theorem
  [:A1,A2,A3,A4:] = { [x1,x2,x3,x4] : x1 in A1 & x2 in A2 & x3 in A3 &
  x4 in A4 }
proof
  thus [:A1,A2,A3,A4:] c= { [x1,x2,x3,x4] : x1 in A1 & x2 in A2 & x3 in A3 &
  x4 in A4 }
  proof
    let a;
    assume
B1: a in [:A1,A2,A3,A4:];
    reconsider A1 as non empty Subset of X1 by B1,MCART_1:51;
    reconsider A2 as non empty Subset of X2 by B1,MCART_1:51;
    reconsider A3 as non empty Subset of X3 by B1,MCART_1:51;
    reconsider A4 as non empty Subset of X4 by B1,MCART_1:51;
A1: a in [:A1,A2,A3,A4:] by B1;
    then reconsider x = a as Element of [:X1,X2,X3,X4:];
A2: x`3_4 in A3 & x`4_4 in A4 by A1,MCART_1:83;
A3: x = [x`1_4,x`2_4,x`3_4,x`4_4];
    x`1_4 in A1 & x`2_4 in A2 by A1,MCART_1:83;
    hence thesis by A3,A2;
  end;
  let a;
  assume a in { [x1,x2,x3,x4] : x1 in A1 & x2 in A2 & x3 in A3 & x4 in A4 };
  then ex x1,x2,x3,x4 st a = [x1,x2,x3,x4] & x1 in A1 & x2 in A2 & x3 in A3 &
  x4 in A4;
  hence thesis by MCART_1:80;
end;

theorem
  {} X1 = { x1 : contradiction }
proof
  thus {} X1 c= { x1 : contradiction }
  proof
    let a;
    thus thesis;
  end;
  let a;
  assume a in { x1 : contradiction };
  then ex x1 st a = x1 & contradiction;
  hence thesis;
end;

theorem
  A1` = { x1 : not x1 in A1 }
proof
  thus A1` c= { x1 : not x1 in A1 }
  proof
    let a;
    assume
A1: a in A1`;
    then not a in A1 by XBOOLE_0:def 5;
    hence thesis by A1;
  end;
  let a;
  assume a in { x1 : not x1 in A1 };
  then ex x1 st a = x1 & not x1 in A1;
  hence thesis by SUBSET_1:29;
end;

theorem
  A1 /\ B1 = { x1 : x1 in A1 & x1 in B1 }
proof
  thus A1 /\ B1 c= { x1 : x1 in A1 & x1 in B1 }
  proof
    let a;
    assume
A1: a in A1 /\ B1;
    then reconsider x = a as Element of X1;
    x in A1 & x in B1 by A1,XBOOLE_0:def 4;
    hence thesis;
  end;
  let a;
  assume a in { x1 : x1 in A1 & x1 in B1 };
  then ex x1 st a = x1 & x1 in A1 & x1 in B1;
  hence thesis by XBOOLE_0:def 4;
end;

theorem
  A1 \/ B1 = { x1 : x1 in A1 or x1 in B1 }
proof
  thus A1 \/ B1 c= { x1 : x1 in A1 or x1 in B1 }
  proof
    let a;
    assume
A1: a in A1 \/ B1;
    then reconsider x = a as Element of X1;
    x in A1 or x in B1 by A1,XBOOLE_0:def 3;
    hence thesis;
  end;
  let a;
  assume a in { x1 : x1 in A1 or x1 in B1 };
  then ex x1 st a = x1 & (x1 in A1 or x1 in B1);
  hence thesis by XBOOLE_0:def 3;
end;

theorem
  A1 \ B1 = { x1 : x1 in A1 & not x1 in B1 }
proof
  thus A1 \ B1 c= { x1 : x1 in A1 & not x1 in B1 }
  proof
    let a;
    assume
A1: a in A1 \ B1;
    then reconsider x = a as Element of X1;
    x in A1 & not x in B1 by A1,XBOOLE_0:def 5;
    hence thesis;
  end;
  let a;
  assume a in { x1 : x1 in A1 & not x1 in B1 };
  then ex x1 st a = x1 & x1 in A1 & not x1 in B1;
  hence thesis by XBOOLE_0:def 5;
end;

theorem Th31:
  A1 \+\ B1 = { x1 : x1 in A1 & not x1 in B1 or not x1 in A1 & x1 in B1 }
proof
  thus A1 \+\ B1 c= { x1 : x1 in A1 & not x1 in B1 or not x1 in A1 & x1 in B1}
  proof
    let a;
    assume
A1: a in A1 \+\ B1;
    then reconsider x = a as Element of X1;
    x in A1 & not x in B1 or not x in A1 & x in B1 by A1,XBOOLE_0:1;
    hence thesis;
  end;
  let a;
  assume a in { x1 : x1 in A1 & not x1 in B1 or not x1 in A1 & x1 in B1 };
  then ex x1 st a = x1 & (x1 in A1 & not x1 in B1 or not x1 in A1 & x1 in B1);
  hence thesis by XBOOLE_0:1;
end;

theorem Th32:
  A1 \+\ B1 = { x1 : not x1 in A1 iff x1 in B1 }
proof
A1: for x1 holds (x1 in A1 & not x1 in B1 or not x1 in A1 & x1 in B1) iff (
  not x1 in A1 iff x1 in B1);
  defpred Q[set] means not $1 in A1 iff $1 in B1;
  defpred P[set] means $1 in A1 & not $1 in B1 or not $1 in A1 & $1 in B1;
A2: A1 \+\ B1 = { x1 : x1 in A1 & not x1 in B1 or not x1 in A1 & x1 in B1 }
  by Th31;
  for X1 st for x1 holds P[x1] iff Q[x1] holds { y1 : P[y1] } = { z1 : Q[
  z1] } from Fraenkel6;
  hence thesis by A1,A2;
end;

theorem
  A1 \+\ B1 = { x1 : x1 in A1 iff not x1 in B1 }
proof
A1: for x1 holds (not x1 in A1 iff x1 in B1) iff (x1 in A1 iff not x1 in B1);
  defpred Q[set] means $1 in A1 iff not $1 in B1;
  defpred P[set] means not $1 in A1 iff $1 in B1;
A2: A1 \+\ B1 = { x1 : not x1 in A1 iff x1 in B1 } by Th32;
  for X1 st for x1 holds P[x1] iff Q[x1] holds { y1 : P[y1] } = { z1 : Q[
  z1] } from Fraenkel6;
  hence thesis by A1,A2;
end;

theorem
  A1 \+\ B1 = { x1 : not(x1 in A1 iff x1 in B1) }
proof
A1: for x1 holds (not x1 in A1 iff x1 in B1) iff not(x1 in A1 iff x1 in B1);
  defpred Q[set] means not($1 in A1 iff $1 in B1);
  defpred P[set] means not $1 in A1 iff $1 in B1;
A2: A1 \+\ B1 = { x1 : not x1 in A1 iff x1 in B1 } by Th32;
  for X1 st for x1 holds P[x1] iff Q[x1] holds { y1 : P[y1] } = { z1 : Q[
  z1] } from Fraenkel6;
  hence thesis by A1,A2;
end;

definition
  let D be non empty set;
  let x1 be Element of D;
  redefine func {x1} -> Subset of D;
  coherence by SUBSET_1:33;
  let x2 be Element of D;
  redefine func {x1,x2} -> Subset of D;
  coherence by SUBSET_1:34;
  let x3 be Element of D;
  redefine func {x1,x2,x3} -> Subset of D;
  coherence by SUBSET_1:35;
  let x4 be Element of D;
  redefine func {x1,x2,x3,x4} -> Subset of D;
  coherence by SUBSET_1:36;
  let x5 be Element of D;
  redefine func {x1,x2,x3,x4,x5} -> Subset of D;
  coherence by SUBSET_1:37;
  let x6 be Element of D;
  redefine func {x1,x2,x3,x4,x5,x6} -> Subset of D;
  coherence by SUBSET_1:38;
  let x7 be Element of D;
  redefine func {x1,x2,x3,x4,x5,x6,x7} -> Subset of D;
  coherence by SUBSET_1:39;
  let x8 be Element of D;
  redefine func {x1,x2,x3,x4,x5,x6,x7,x8} -> Subset of D;
  coherence by SUBSET_1:40;
end;

begin 


scheme
  SubsetFD { A, D() -> non empty set, F(set) -> Element of D(), P[set] }: { F(
  x) where x is Element of A(): P[x]} is Subset of D() proof
  set X = { F(x) where x is Element of A(): P[x]};
  X c= D()
  proof
    let y be set;
    assume y in X;
    then ex z being Element of A() st y = F(z) & P[z];
    hence thesis;
  end;
  hence thesis;
end;

scheme
  SubsetFD2 { A, B, D() -> non empty set, F(set,set) -> Element of D(), P[set,
  set] }: { F(x,y) where x is Element of A(), y is Element of B(): P[x,y]} is
  Subset of D() proof
  set X = { F(x,y) where x is Element of A(), y is Element of B(): P[x,y]};
  X c= D()
  proof
    let w be set;
    assume w in X;
    then
    ex x being Element of A(), y being Element of B() st w = F(x,y) & P[ x ,y];
    hence thesis;
  end;
  hence thesis;
end;

definition
  let D1,D2,D3 be non empty set, x be Element of [:[:D1,D2:],D3:];
  redefine func x`11 -> Element of D1;
  coherence
  proof
    x`11 = x`1`1 by MCART_1:def 14;
    hence thesis;
  end;
  redefine func x`12 -> Element of D2;
  coherence
  proof
    x`12 = x`1`2 by MCART_1:def 15;
    hence thesis;
  end;
end;

definition
  let D1,D2,D3 be non empty set, x be Element of [:D1,[:D2,D3:]:];
  redefine func x`21 -> Element of D2;
  coherence
  proof
    x`21 = x`2`1 by MCART_1:def 16;
    hence thesis;
  end;
  redefine func x`22 -> Element of D3;
  coherence
  proof
    x`22 = x`2`2 by MCART_1:def 17;
    hence thesis;
  end;
end;



scheme
  AndScheme{A()-> non empty set, P,Q[set]}: { a where a is Element of A(): P[a
  ] & Q[a] } = { a1 where a1 is Element of A(): P[a1] } /\ { a2 where a2 is
  Element of A(): Q[a2] } proof
  set A = { a where a is Element of A(): P[a] & Q[a] }, A1 = { a1 where a1 is
  Element of A(): P[a1] }, A2 = { a2 where a2 is Element of A(): Q[a2] };
  thus A c= A1 /\ A2
  proof
    let x be set;
    assume x in A;
    then consider a being Element of A() such that
A1: x = a and
A2: ( P[a])& Q[a];
    a in A1 & a in A2 by A2;
    hence thesis by A1,XBOOLE_0:def 4;
  end;
  let x be set;
  assume
A3: x in A1 /\ A2;
  then x in A2 by XBOOLE_0:def 4;
  then
A4: ex a being Element of A() st x = a & Q[a];
  x in A1 by A3,XBOOLE_0:def 4;
  then ex a being Element of A() st x = a & P[a];
  hence thesis by A4;
end;



registration
  let A be non empty set;
  cluster c=-linear non empty for Subset of A;
  existence
  proof
    set a = the Element of A;
    reconsider B = { a } as non empty Subset of A;
    take B;
    B is c=-linear
    proof
      let x,y be set;
      assume that
A1:   x in B and
A2:   y in B;
      x = a by A1,TARSKI:def 1;
      hence thesis by A2,TARSKI:def 1;
    end;
    hence thesis;
  end;
end;















begin

reserve p,q,x,x1,x2,y,y1,y2,z,z1,z2 for set;
reserve A,B,V,X,X1,X2,Y,Y1,Y2,Z for set;
reserve C,C1,C2,D,D1,D2 for non empty set;



theorem Th1:
  A c= Y implies id A = (id Y)|A
proof
  assume A c= Y;
  hence id A = id(Y /\ A) by XBOOLE_1:28
    .= (id Y)*(id A) by FUNCT_1:22
    .= (id Y)|A by RELAT_1:65;
end;

theorem Th2:
  for f,g being Function st X c= dom(g*f) holds f.:X c= dom g
proof
  let f,g be Function such that
A1: X c= dom(g*f);
  let y;
  assume y in f.:X;
  then ex x st x in dom f & x in X & y = f.x by FUNCT_1:def 6;
  hence thesis by A1,FUNCT_1:11;
end;

theorem Th3:
  for f,g being Function st X c= dom f & f.:X c= dom g holds X c= dom(g*f)
proof
  let f,g be Function such that
A1: X c= dom f and
A2: f.:X c= dom g;
  let x;
  assume
A3: x in X;
  then f.x in f.:X by A1,FUNCT_1:def 6;
  hence thesis by A1,A2,A3,FUNCT_1:11;
end;

theorem Th4:
  for f,g being Function st Y c= rng(g*f) & g is one-to-one holds g "Y c= rng f
proof
  let f,g be Function such that
A1: Y c= rng(g*f) and
A2: g is one-to-one;
  let y;
  assume
A3: y in g"Y;
  then
A4: y in dom g by FUNCT_1:def 7;
  g.y in Y by A3,FUNCT_1:def 7;
  then consider x such that
A5: x in dom(g*f) and
A6: g.y = (g*f).x by A1,FUNCT_1:def 3;
A7: x in dom f by A5,FUNCT_1:11;
  g.y = g.(f.x) & f.x in dom g by A5,A6,FUNCT_1:11,12;
  then y = f.x by A2,A4,FUNCT_1:def 4;
  hence thesis by A7,FUNCT_1:def 3;
end;

theorem Th5:
  for f,g being Function st Y c= rng g & g"Y c= rng f holds Y c= rng(g*f)
proof
  let f,g be Function such that
A1: Y c= rng g and
A2: g"Y c= rng f;
  let y;
  assume
A3: y in Y;
  then consider z such that
A4: z in dom g & y = g.z by A1,FUNCT_1:def 3;
  z in g"Y by A3,A4,FUNCT_1:def 7;
  then consider x such that
A5: x in dom f & z = f.x by A2,FUNCT_1:def 3;
  x in dom(g*f) & (g*f).x = y by A4,A5,FUNCT_1:11,13;
  hence thesis by FUNCT_1:def 3;
end;

scheme
  FuncEx3{A()->set,B()-> set,P[set,set,set]}: ex f being Function st dom f =
  [:A(),B():] & for x,y st x in A() & y in B() holds P[x,y,f.(x,y)]
provided
A1: for x,y,z1,z2 st x in A() & y in B() & P[x,y,z1] & P[x,y,z2] holds
z1 = z2 and
A2: for x,y st x in A() & y in B() ex z st P[x,y,z]
proof
  defpred R[set,set] means for x,y st $1=[x,y] holds P[x,y,$2];
  set D = [:A(), B():];
A3: for p st p in D ex z st R[p,z]
  proof
    let p;
    assume p in D;
    then consider x1,y1 such that
A4: x1 in A() & y1 in B() and
A5: p = [x1,y1] by ZFMISC_1:def 2;
    consider z such that
A6: P[x1,y1,z] by A2,A4;
    take z;
    let x,y;
    assume
A7: p=[x,y];
    then x=x1 by A5,XTUPLE_0:1;
    hence thesis by A5,A6,A7,XTUPLE_0:1;
  end;
A8: for p,y1,y2 st p in D & R[p,y1] & R[p,y2] holds y1 = y2
  proof
    let p,y1,y2;
    assume p in D;
    then consider x1,x2 such that
A9: x1 in A() & x2 in B() and
A10: p = [x1,x2] by ZFMISC_1:def 2;
    assume
    ( for x,y st p=[x,y] holds P[x,y,y1])& for x,y st p=[x,y] holds P[ x,y,y2];
    then ( P[x1,x2,y1])& P[x1,x2,y2] by A10;
    hence thesis by A1,A9;
  end;
  consider f being Function such that
A11: dom f = D and
A12: for p st p in D holds R[p,f.p] from FUNCT_1:sch 2(A8,A3);
  take f;
  thus dom f = [:A(),B():] by A11;
  let x,y;
  assume x in A() & y in B();
  then [x,y] in [:A(),B():] by ZFMISC_1:def 2;
  hence thesis by A12;
end;

scheme
  Lambda3{A()->set,B()->set,F(set,set)->set}: ex f being Function st dom f =
  [:A(),B():] & for x,y st x in A() & y in B() holds f.(x,y) = F(x,y) proof
  defpred P[set,set,set] means $3 = F($1,$2);
A1: for x,y st x in A() & y in B() ex z st P[x,y,z];
A2: for x,y,z1,z2 st x in A() & y in B() & P[x,y,z1] & P[x,y,z2] holds z1=z2;
  ex f being Function st dom f = [:A(),B():] & for x,y st x in A() & y in
  B() holds P[x,y,f.(x,y)] from FuncEx3(A2,A1);
  hence thesis;
end;

theorem Th6:
  for f,g being Function st dom f = [:X,Y:] & dom g = [:X,Y:] & for
  x,y st x in X & y in Y holds f.(x,y) = g.(x,y) holds f = g
proof
  let f,g be Function such that
A1: dom f = [:X,Y:] & dom g = [:X,Y:] and
A2: for x,y st x in X & y in Y holds f.(x,y) = g.(x,y);
  p in [:X,Y:] implies f.p = g.p
  proof
    assume p in [:X,Y:];
    then consider x,y such that
A3: x in X & y in Y and
A4: p = [x,y] by ZFMISC_1:def 2;
    f.(x,y) = g.(x,y) by A2,A3;
    hence thesis by A4;
  end;
  hence thesis by A1,FUNCT_1:2;
end;



definition
  let f be Function;
  func .:f -> Function means
  :Def1:
  dom it = bool dom f & for X st X c= dom f holds it.X = f.:X;
  existence
  proof
    defpred P[set,set] means for X st $1=X holds $2 = f.:X;
A1: for x st x in bool dom f ex y st P[x,y]
    proof
      let x such that
      x in bool dom f;
      take f.:x;
      thus thesis;
    end;
A2: for x,y1,y2 st x in bool dom f & P[x,y1] & P[x,y2] holds y1=y2
    proof
      let x,y1,y2 such that
      x in bool dom f and
A3:   for X st x=X holds y1 = f.:X and
A4:   for X st x=X holds y2 = f.:X;
      thus y1 = f.:x by A3
        .= y2 by A4;
    end;
    consider g being Function such that
A5: dom g = bool dom f & for x st x in bool dom f holds P[x,g.x] from
    FUNCT_1:sch 2(A2,A1);
    take g;
    thus thesis by A5;
  end;
  uniqueness
  proof
    let f1,f2 be Function such that
A6: dom f1 = bool dom f and
A7: for X st X c= dom f holds f1.X = f.:X and
A8: dom f2 = bool dom f and
A9: for X st X c= dom f holds f2.X = f.:X;
    for x st x in bool dom f holds f1.x = f2.x
    proof
      let x;
      assume
A10:  x in bool dom f;
      then f1.x = f.:x by A7;
      hence thesis by A9,A10;
    end;
    hence thesis by A6,A8,FUNCT_1:2;
  end;
end;

theorem Th7:
  for f being Function st X in dom(.:f) holds (.:f).X = f.:X
proof
  let f be Function;
  assume X in dom(.:f);
  then X in bool dom f by Def1;
  hence thesis by Def1;
end;

theorem
  for f being Function holds (.:f).{} = {}
proof
  let f be Function;
  {} c= dom f by XBOOLE_1:2;
  then (.:f).{} = f.:{} by Def1;
  hence thesis;
end;

theorem Th9:
  for f being Function holds rng(.:f) c= bool rng f
proof
  let f be Function;
  let y;
  assume y in rng(.:f);
  then consider x such that
A1: x in dom(.:f) & y = (.:f).x by FUNCT_1:def 3;
  y = f.:x by A1,Th7;
  then y c= rng f by RELAT_1:111;
  hence thesis;
end;

theorem
  for f being Function holds (.:f).:A c= bool rng f
proof
  let f be Function;
  (.:f).:A c= rng(.:f) & rng(.:f) c= bool rng f by Th9,RELAT_1:111;
  hence thesis by XBOOLE_1:1;
end;

theorem Th11:
  for f being Function holds (.:f)"B c= bool dom f
proof
  let f be Function;
  dom(.:f) = bool dom f by Def1;
  hence thesis by RELAT_1:132;
end;

theorem
  for f being Function of X,D holds (.:f)"B c= bool X
proof
  let f be Function of X,D;
  dom f = X by FUNCT_2:def 1;
  hence thesis by Th11;
end;

theorem Th13:
  for f being Function holds union((.:f).:A) c= f.:(union A)
proof
  let f be Function;
  let y;
  assume y in union((.:f).:A);
  then consider Z such that
A1: y in Z and
A2: Z in (.:f).:A by TARSKI:def 4;
  consider X such that
A3: X in dom(.:f) and
A4: X in A and
A5: Z = (.:f).X by A2,FUNCT_1:def 6;
  y in f.:X by A1,A3,A5,Th7;
  then consider x such that
A6: x in dom f and
A7: x in X and
A8: y = f.x by FUNCT_1:def 6;
  x in union A by A4,A7,TARSKI:def 4;
  hence thesis by A6,A8,FUNCT_1:def 6;
end;

theorem Th14:
  for f being Function st A c= bool dom f holds f.:(union A) = union((.:f).:A)
proof
  let f be Function such that
A1: A c= bool dom f;
A2: f.:(union A) c= union((.:f).:A)
  proof
    let y;
    assume y in f.:(union A);
    then consider x such that
    x in dom f and
A3: x in union A and
A4: y = f.x by FUNCT_1:def 6;
    consider X such that
A5: x in X and
A6: X in A by A3,TARSKI:def 4;
    X in bool dom f by A1,A6;
    then X in dom(.:f) by Def1;
    then
A7: (.:f).X in (.:f).:A by A6,FUNCT_1:def 6;
    y in f.:X by A1,A4,A5,A6,FUNCT_1:def 6;
    then y in (.:f).X by A1,A6,Def1;
    hence thesis by A7,TARSKI:def 4;
  end;
  union((.:f).:A) c= f.:(union A) by Th13;
  hence thesis by A2,XBOOLE_0:def 10;
end;

theorem
  for f being Function of X,D st A c= bool X holds f.:(union A) = union(
  (.:f).:A)
proof
  let f be Function of X,D;
  dom f = X by FUNCT_2:def 1;
  hence thesis by Th14;
end;

theorem Th16:
  for f being Function holds union((.:f)"B) c= f"(union B)
proof
  let f be Function;
  let x;
  assume x in union((.:f)"B);
  then consider X such that
A1: x in X and
A2: X in (.:f)"B by TARSKI:def 4;
  dom(.:f) = bool dom f by Def1;
  then
A3: X in bool dom f by A2,FUNCT_1:def 7;
  then
A4: f.x in f.:X by A1,FUNCT_1:def 6;
  (.:f).X in B by A2,FUNCT_1:def 7;
  then f.:X in B by A3,Def1;
  then f.x in union B by A4,TARSKI:def 4;
  hence thesis by A1,A3,FUNCT_1:def 7;
end;

theorem
  for f being Function st B c= bool rng f holds f"(union B) = union((.:f
  ) " B )
proof
  let f be Function such that
A1: B c= bool rng f;
A2: f"(union B) c= union((.:f)"B)
  proof
    let x;
    assume
A3: x in f"(union B);
    then f.x in union B by FUNCT_1:def 7;
    then consider Y such that
A4: f.x in Y and
A5: Y in B by TARSKI:def 4;
A6: f"Y c= dom f by RELAT_1:132;
    then f"Y in bool dom f;
    then
A7: f"Y in dom(.:f) by Def1;
    f.:(f"Y) = Y by A1,A5,FUNCT_1:77;
    then (.:f).(f"Y) in B by A5,A6,Def1;
    then
A8: f"Y in (.:f)"B by A7,FUNCT_1:def 7;
    x in dom f by A3,FUNCT_1:def 7;
    then x in f"Y by A4,FUNCT_1:def 7;
    hence thesis by A8,TARSKI:def 4;
  end;
  union((.:f)"B) c= f"(union B) by Th16;
  hence thesis by A2,XBOOLE_0:def 10;
end;

theorem
  for f,g being Function holds .:(g*f) = .:g*.:f
proof
  let f,g be Function;
  for x holds x in dom .:(g*f) iff x in dom(.:g*.:f)
  proof
    let x;
    thus x in dom .:(g*f) implies x in dom(.:g*.:f)
    proof
      assume x in dom .:(g*f);
      then
A1:   x in bool dom(g*f) by Def1;
      dom(g*f) c= dom f by RELAT_1:25;
      then
A2:   x c= dom f by A1,XBOOLE_1:1;
      then x in bool dom f;
      then
A3:   x in dom .:f by Def1;
      f.:x c= dom g by A1,Th2;
      then f.:x in bool dom g;
      then f.:x in dom .:g by Def1;
      then .:f.x in dom .:g by A2,Def1;
      hence thesis by A3,FUNCT_1:11;
    end;
    assume
A4: x in dom(.:g*.:f);
    then
A5: x in dom .:f by FUNCT_1:11;
    .:f.x in dom .:g by A4,FUNCT_1:11;
    then f.:x in dom .:g by A5,Th7;
    then
A6: f.:x in bool dom g by Def1;
    x in bool dom f by A5,Def1;
    then x c= dom(g*f) by A6,Th3;
    then x in bool dom(g*f);
    hence thesis by Def1;
  end;
  then
A7: dom .:(g*f) = dom(.:g*.:f) by TARSKI:1;
  for x st x in dom .:(g*f) holds (.:(g*f)).x = (.:g*.:f).x
  proof
    let x;
    assume
A8: x in dom .:(g*f);
    then
A9: x in bool dom(g*f) by Def1;
    then
A10: f.:x c= dom g by Th2;
    dom(g*f) c= dom f by RELAT_1:25;
    then
A11: x c= dom f by A9,XBOOLE_1:1;
    then x in bool dom f;
    then
A12: x in dom .:f by Def1;
    thus (.:(g*f)).x = (g*f).:x by A8,Th7
      .= g.:(f.:x) by RELAT_1:126
      .= .:g.(f.:x) by A10,Def1
      .= .:g.(.:f.x) by A11,Def1
      .= (.:g*.:f).x by A12,FUNCT_1:13;
  end;
  hence thesis by A7,FUNCT_1:2;
end;

theorem Th19:
  for f being Function holds .:f is Function of bool dom f, bool rng f
proof
  let f be Function;
  dom .:f = bool dom f & rng.:f c= bool rng f by Def1,Th9;
  hence thesis by FUNCT_2:def 1,RELSET_1:4;
end;

theorem Th20:
  for f being Function of X,Y st Y = {} implies X = {} holds .:f
  is Function of bool X, bool Y
proof
  let f be Function of X,Y;
  assume Y = {} implies X = {};
  then
A1: dom f = X by FUNCT_2:def 1;
  rng f c= Y by RELAT_1:def 19;
  then
A2: bool rng f c= bool Y by ZFMISC_1:67;
A3: .:f is Function of bool dom f, bool rng f by Th19;
  then rng.:f c= bool rng f by RELAT_1:def 19;
  then
A4: rng.:f c= bool Y by A2,XBOOLE_1:1;
  dom.:f = bool dom f by A3,FUNCT_2:def 1;
  hence thesis by A1,A4,FUNCT_2:def 1,RELSET_1:4;
end;

definition
  let X,D;
  let f be Function of X,D;
  redefine func .:f -> Function of bool X, bool D;
  coherence by Th20;
end;



definition
  let f be Function;
  func "f -> Function means
  :Def2:
  dom it = bool rng f & for Y st Y c= rng f holds it.Y = f"Y;
  existence
  proof
    defpred P[set,set] means for Y st $1=Y holds $2 = f"Y;
A1: for y st y in bool rng f ex x st P[y,x]
    proof
      let y such that
      y in bool rng f;
      take f"y;
      thus thesis;
    end;
A2: for y,x1,x2 st y in bool rng f & P[y,x1] & P[y,x2] holds x1=x2
    proof
      let y,x1,x2 such that
      y in bool rng f and
A3:   for Y st y=Y holds x1 = f"Y and
A4:   for Y st y=Y holds x2 = f"Y;
      thus x1 = f"y by A3
        .= x2 by A4;
    end;
    consider g being Function such that
A5: dom g = bool rng f & for y st y in bool rng f holds P[y,g.y] from
    FUNCT_1:sch 2(A2,A1);
    take g;
    thus thesis by A5;
  end;
  uniqueness
  proof
    let f1,f2 be Function such that
A6: dom f1 = bool rng f and
A7: for Y st Y c= rng f holds f1.Y = f"Y and
A8: dom f2 = bool rng f and
A9: for Y st Y c= rng f holds f2.Y = f"Y;
    for y st y in bool rng f holds f1.y = f2.y
    proof
      let y;
      assume
A10:  y in bool rng f;
      then f1.y = f"y by A7;
      hence thesis by A9,A10;
    end;
    hence thesis by A6,A8,FUNCT_1:2;
  end;
end;

theorem Th21:
  for f being Function st Y in dom("f) holds ("f).Y = f"Y
proof
  let f be Function;
  dom("f) = bool rng f by Def2;
  hence thesis by Def2;
end;

theorem Th22:
  for f being Function holds rng("f) c= bool dom f
proof
  let f be Function;
  let x;
  assume x in rng("f);
  then consider y such that
A1: y in dom("f) & x = "f.y by FUNCT_1:def 3;
  x = f"y by A1,Th21;
  then x c= dom f by RELAT_1:132;
  hence thesis;
end;

theorem
  for f being Function holds ("f).:B c= bool dom f
proof
  let f be Function;
  rng("f) c= bool dom f & ("f).:B c= rng("f) by Th22,RELAT_1:111;
  hence thesis by XBOOLE_1:1;
end;

theorem
  for f being Function holds ("f)"A c= bool rng f
proof
  let f be Function;
  dom("f) = bool rng f by Def2;
  hence thesis by RELAT_1:132;
end;

theorem Th25:
  for f being Function holds union(("f).:B) c= f"(union B)
proof
  let f be Function;
  let x;
  assume x in union(("f).:B);
  then consider X such that
A1: x in X and
A2: X in ("f).:B by TARSKI:def 4;
  consider Y such that
A3: Y in dom("f) and
A4: Y in B and
A5: X = "f.Y by A2,FUNCT_1:def 6;
A6: ("f).Y = f"Y by A3,Th21;
  Y in bool rng f by A3,Def2;
  then
A7: f.:X in B by A4,A5,A6,FUNCT_1:77;
A8: f"Y c= dom f by RELAT_1:132;
  then f.x in f.:X by A1,A5,A6,FUNCT_1:def 6;
  then f.x in union B by A7,TARSKI:def 4;
  hence thesis by A1,A5,A6,A8,FUNCT_1:def 7;
end;

theorem
  for f being Function st B c= bool rng f holds union(("f).:B) = f"( union B )
proof
  let f be Function such that
A1: B c= bool rng f;
A2: f"(union B) c= union(("f).:B)
  proof
    let x;
    assume
A3: x in f"(union B);
    then f.x in union B by FUNCT_1:def 7;
    then consider Y such that
A4: f.x in Y and
A5: Y in B by TARSKI:def 4;
    x in dom f by A3,FUNCT_1:def 7;
    then
A6: x in f"Y by A4,FUNCT_1:def 7;
    Y in bool rng f by A1,A5;
    then
A7: Y in dom("f) by Def2;
    ("f).Y = f"Y by A1,A5,Def2;
    then f"Y in ("f).:B by A5,A7,FUNCT_1:def 6;
    hence thesis by A6,TARSKI:def 4;
  end;
  union(("f).:B) c= f"(union B) by Th25;
  hence thesis by A2,XBOOLE_0:def 10;
end;

theorem Th27:
  for f being Function holds union(("f)"A) c= f.:(union A)
proof
  let f be Function;
  let y;
  assume y in union(("f)"A);
  then consider Y such that
A1: y in Y and
A2: Y in ("f)"A by TARSKI:def 4;
  dom("f) = bool rng f by Def2;
  then
A3: Y in bool rng f by A2,FUNCT_1:def 7;
  then consider x such that
A4: x in dom f & y = f.x by A1,FUNCT_1:def 3;
  "f.Y in A by A2,FUNCT_1:def 7;
  then
A5: f"Y in A by A3,Def2;
  x in f"Y by A1,A4,FUNCT_1:def 7;
  then x in union A by A5,TARSKI:def 4;
  hence thesis by A4,FUNCT_1:def 6;
end;

theorem
  for f being Function st A c= bool dom f & f is one-to-one holds union(
  ("f)"A) = f.:(union A)
proof
  let f be Function such that
A1: A c= bool dom f and
A2: f is one-to-one;
A3: f.:(union A) c= union(("f)"A)
  proof
    let y;
    assume y in f.:(union A);
    then consider x such that
A4: x in dom f and
A5: x in union A and
A6: y = f.x by FUNCT_1:def 6;
    consider X such that
A7: x in X and
A8: X in A by A5,TARSKI:def 4;
A9: f"(f.:X) c= X by A2,FUNCT_1:82;
A10: f.:X c= rng f by RELAT_1:111;
    then f.:X in bool rng f;
    then
A11: (f.:X) in dom("f) by Def2;
    X c= f"(f.:X) by A1,A8,FUNCT_1:76;
    then f"(f.:X) = X by A9,XBOOLE_0:def 10;
    then ("f).(f.:X) in A by A8,A10,Def2;
    then
A12: (f.:X) in ("f)"A by A11,FUNCT_1:def 7;
    y in (f.:X) by A4,A6,A7,FUNCT_1:def 6;
    hence thesis by A12,TARSKI:def 4;
  end;
  union(("f)"A) c= f.:(union A) by Th27;
  hence thesis by A3,XBOOLE_0:def 10;
end;

theorem Th29:
  for f being Function holds ("f).:B c= (.:f)"B
proof
  let f be Function;
  let x;
  assume x in ("f).:B;
  then consider Y such that
A1: Y in dom ("f) and
A2: Y in B and
A3: x = "f.Y by FUNCT_1:def 6;
A4: "f.Y = f"Y by A1,Th21;
  then
A5: x c= dom f by A3,RELAT_1:132;
  then x in bool dom f;
  then
A6: x in dom(.:f) by Def1;
  Y in bool rng f by A1,Def2;
  then f.:x in B by A2,A3,A4,FUNCT_1:77;
  then (.:f).x in B by A5,Def1;
  hence thesis by A6,FUNCT_1:def 7;
end;

theorem
  for f being Function st f is one-to-one holds ("f).:B = (.:f)"B
proof
  let f be Function such that
A1: f is one-to-one;
A2: (.:f)"B c= ("f).:B
  proof
    let x;
A3: f.:x c= rng f by RELAT_1:111;
    then f.:x in bool rng f;
    then
A4: f.:x in dom("f) by Def2;
    assume
A5: x in (.:f)"B;
    then
A6: x in dom(.:f) by FUNCT_1:def 7;
    then x in bool dom f by Def1;
    then
A7: x c= f"(f.:x) by FUNCT_1:76;
    f"(f.:x) c= x by A1,FUNCT_1:82;
    then x = f"(f.:x) by A7,XBOOLE_0:def 10;
    then
A8: x =("f).(f.:x) by A3,Def2;
    (.:f).x in B by A5,FUNCT_1:def 7;
    then f.:x in B by A6,Th7;
    hence thesis by A8,A4,FUNCT_1:def 6;
  end;
  ("f).:B c= (.:f)"B by Th29;
  hence thesis by A2,XBOOLE_0:def 10;
end;

theorem Th31:
  for f being Function,A be set st A c= bool dom f holds ("f)"A c= (.:f).:A
proof
  let f be Function,A be set such that
A1: A c= bool dom f;
  let y;
  assume
A2: y in ("f)"A;
  then
A3: "f.y in A by FUNCT_1:def 7;
  y in dom("f) by A2,FUNCT_1:def 7;
  then
A4: y in bool rng f by Def2;
  then
A5: f"y in A by A3,Def2;
  then f"y in bool dom f by A1;
  then
A6: f"y in dom .:f by Def1;
  f.:(f"y) = y by A4,FUNCT_1:77;
  then
A7: .:f.(f"y) = y by A1,A5,Def1;
  f"y in A by A3,A4,Def2;
  hence thesis by A7,A6,FUNCT_1:def 6;
end;

theorem Th32:
  for f being Function,A be set st f is one-to-one holds (.:f).:A c= ("f)"A
proof
  let f be Function,A be set such that
A1: f is one-to-one;
  let y;
  assume y in (.:f).:A;
  then consider x such that
A2: x in dom(.:f) and
A3: x in A and
A4: y = .:f.x by FUNCT_1:def 6;
A5: x in bool dom f by A2,Def1;
  then
A6: y = f.:x by A4,Def1;
  then
A7: x c= f"y by A5,FUNCT_1:76;
A8: y c= rng f by A6,RELAT_1:111;
  then y in bool rng f;
  then
A9: y in dom("f) by Def2;
  f"y c= x by A1,A6,FUNCT_1:82;
  then f"y in A by A3,A7,XBOOLE_0:def 10;
  then "f.y in A by A8,Def2;
  hence thesis by A9,FUNCT_1:def 7;
end;

theorem
  for f being Function,A be set st f is one-to-one & A c= bool dom f
  holds ("f)"A = (.:f).:A
proof
  let f be Function,A be set;
  assume f is one-to-one & A c= bool dom f;
  then ("f)"A c= (.:f).:A & (.:f).:A c= ("f)"A by Th31,Th32;
  hence thesis by XBOOLE_0:def 10;
end;

theorem
  for f,g being Function st g is one-to-one holds "(g*f) = "f*"g
proof
  let f,g be Function such that
A1: g is one-to-one;
  for y holds y in dom "(g*f) iff y in dom("f*"g)
  proof
    let y;
    thus y in dom "(g*f) implies y in dom("f*"g)
    proof
      assume y in dom "(g*f);
      then
A2:   y in bool rng(g*f) by Def2;
      rng(g*f) c= rng g by RELAT_1:26;
      then
A3:   y c= rng g by A2,XBOOLE_1:1;
      then y in bool rng g;
      then
A4:   y in dom "g by Def2;
      g"y c= rng f by A1,A2,Th4;
      then g"y in bool rng f;
      then g"y in dom "f by Def2;
      then "g.y in dom "f by A3,Def2;
      hence thesis by A4,FUNCT_1:11;
    end;
    assume
A5: y in dom("f*"g);
    then
A6: y in dom "g by FUNCT_1:11;
    "g.y in dom "f by A5,FUNCT_1:11;
    then g"y in dom "f by A6,Th21;
    then
A7: g"y in bool rng f by Def2;
    y in bool rng g by A6,Def2;
    then y c= rng(g*f) by A7,Th5;
    then y in bool rng(g*f);
    hence thesis by Def2;
  end;
  then
A8: dom "(g*f) = dom("f*"g) by TARSKI:1;
  for y st y in dom "(g*f) holds "(g*f).y = ("f*"g).y
  proof
    let y;
    assume
A9: y in dom "(g*f);
    then
A10: y in bool rng(g*f) by Def2;
    then
A11: g"y c= rng f by A1,Th4;
    rng(g*f) c= rng g by RELAT_1:26;
    then
A12: y c= rng g by A10,XBOOLE_1:1;
    then y in bool rng g;
    then
A13: y in dom "g by Def2;
    thus "(g*f).y = (g*f)"y by A9,Th21
      .= f"(g"y) by RELAT_1:146
      .= "f.(g"y) by A11,Def2
      .= "f.("g.y) by A12,Def2
      .= ("f*"g).y by A13,FUNCT_1:13;
  end;
  hence thesis by A8,FUNCT_1:2;
end;

theorem
  for f being Function holds "f is Function of bool rng f, bool dom f
proof
  let f be Function;
  dom"f = bool rng f & rng"f c= bool dom f by Def2,Th22;
  hence thesis by FUNCT_2:def 1,RELSET_1:4;
end;



definition
  let A,X;
  func chi(A,X) -> Function means
  :Def3:
  dom it = X & for x st x in X holds (x
  in A implies it.x = 1) & (not x in A implies it.x = {});
  existence
  proof
    defpred P[set,set] means ($1 in A implies $2 = 1) & (not $1 in A implies
    $2 = {});
A1: for x st x in X ex y st P[x,y]
    proof
      let x;
      assume x in X;
      not x in A implies ex y st y = {} & (x in A implies y = 1) & (not x
      in A implies y = {});
      hence thesis;
    end;
A2: for x,y1,y2 st x in X & P[x,y1] & P[x,y2] holds y1 = y2;
    ex f being Function st dom f = X & for x st x in X holds P[x,f.x] from
    FUNCT_1:sch 2(A2,A1);
    hence thesis;
  end;
  uniqueness
  proof
    let f1,f2 be Function such that
A3: dom f1 = X and
A4: for x st x in X holds (x in A implies f1.x = 1) & (not x in A
    implies f1.x = {}) and
A5: dom f2 = X and
A6: for x st x in X holds (x in A implies f2.x = 1) & (not x in A
    implies f2.x = {});
    for x st x in X holds f1.x=f2.x
    proof
      let x;
      assume
A7:   x in X;
      then
A8:   not x in A implies f1.x = {} & f2.x = {} by A4,A6;
      x in A implies f1.x = 1 & f2.x = 1 by A4,A6,A7;
      hence thesis by A8;
    end;
    hence thesis by A3,A5,FUNCT_1:2;
  end;
end;

theorem Th36:
  chi(A,X).x = 1 implies x in A
proof
  assume
A1: chi(A,X).x = 1;
  per cases;
  suppose
    x in X;
    hence thesis by A1,Def3;
  end;
  suppose
    not x in X;
    then not x in dom chi(A,X) by Def3;
    hence thesis by A1,FUNCT_1:def 2;
  end;
end;

theorem
  x in X \ A implies chi(A,X).x = {}
proof
  assume
A1: x in X\A;
  then not x in A by XBOOLE_0:def 5;
  hence thesis by A1,Def3;
end;

theorem
  A c= X & B c= X & chi(A,X) = chi(B,X) implies A = B
proof
  assume that
A1: A c= X and
A2: B c= X and
A3: chi(A,X) = chi(B,X);
  x in A iff x in B
  proof
    thus x in A implies x in B
    proof
      assume x in A;
      then chi(A,X).x = 1 by A1,Def3;
      hence thesis by A3,Th36;
    end;
    assume x in B;
    then chi(B,X).x = 1 by A2,Def3;
    hence thesis by A3,Th36;
  end;
  hence thesis by TARSKI:1;
end;

theorem Th39:
  rng chi(A,X) c= {{},1}
proof
  let y;
  assume y in rng chi(A,X);
  then consider x such that
A1: x in dom chi(A,X) and
A2: y = chi(A,X).x by FUNCT_1:def 3;
A3: x in A or not x in A;
  x in X by A1,Def3;
  then y = {} or y = 1 by A2,A3,Def3;
  hence thesis by TARSKI:def 2;
end;

theorem
  for f being Function of X,{{},1} holds f = chi(f"{1},X)
proof
  let f be Function of X,{{},1};
  now
    thus
A1: dom f = X by FUNCT_2:def 1;
    let x such that
A2: x in X;
    thus x in f"{1} implies f.x = 1
    proof
      assume x in f"{1};
      then f.x in {1} by FUNCT_1:def 7;
      hence thesis by TARSKI:def 1;
    end;
    assume not x in f"{1};
    then not f.x in {1} by A1,A2,FUNCT_1:def 7;
    then
A3: rng f c= {{},1} & f.x <> 1 by RELAT_1:def 19,TARSKI:def 1;
    f.x in rng f by A1,A2,FUNCT_1:def 3;
    hence f.x = {} by A3,TARSKI:def 2;
  end;
  hence thesis by Def3;
end;

definition
  let A,X;
  redefine func chi(A,X) -> Function of X,{{},1};
  coherence
  proof
    dom chi(A,X) = X & rng chi(A,X) c= {{},1} by Def3,Th39;
    hence thesis by FUNCT_2:def 1,RELSET_1:4;
  end;
end;

notation
  let Y;
  let A be Subset of Y;
  synonym incl A for id A;
end;

definition
  let Y;
  let A be Subset of Y;
  redefine func incl A -> Function of A,Y;
  coherence
  proof
A1: rng id A = A & dom id A = A;
    thus thesis by A1,FUNCT_2:2;
  end;
end;

theorem
  for A being Subset of Y holds incl A = (id Y)|A by Th1;

theorem
  for A being Subset of Y st x in A holds incl(A).x in Y
proof
  let A be Subset of Y such that
A1: x in A;
  dom incl A = A & rng incl A = A;
  then incl(A).x in A by A1,FUNCT_1:def 3;
  hence thesis;
end;



definition
  let X,Y;

  func pr1(X,Y) -> Function means
  :Def4:
  dom it = [:X,Y:] & for x,y st x in X & y in Y holds it.(x,y) = x;
  existence
  proof
    deffunc F(set,set) = $1;
    ex f being Function st dom f = [:X,Y:] & for x,y st x in X & y in Y
    holds f.(x,y) = F(x,y) from Lambda3;
    hence thesis;
  end;
  uniqueness
  proof
    let f1,f2 be Function such that
A1: dom f1 = [:X,Y:] and
A2: for x,y st x in X & y in Y holds f1.(x,y) = x and
A3: dom f2 = [:X,Y:] and
A4: for x,y st x in X & y in Y holds f2.(x,y) = x;
    for x,y st x in X & y in Y holds f1.(x,y) = f2.(x,y)
    proof
      let x,y;
      assume
A5:   x in X & y in Y;
      then f1.(x,y) = x by A2;
      hence thesis by A4,A5;
    end;
    hence thesis by A1,A3,Th6;
  end;
  func pr2(X,Y) -> Function means
  :Def5:
  dom it = [:X,Y:] & for x,y st x in X & y in Y holds it.(x,y) = y;
  existence
  proof
    deffunc F(set,set) = $2;
    ex f being Function st dom f = [:X,Y:] & for x,y st x in X & y in Y
    holds f.(x,y) = F(x,y) from Lambda3;
    hence thesis;
  end;
  uniqueness
  proof
    let f1,f2 be Function such that
A6: dom f1 = [:X,Y:] and
A7: for x,y st x in X & y in Y holds f1.(x,y) = y and
A8: dom f2 = [:X,Y:] and
A9: for x,y st x in X & y in Y holds f2.(x,y) = y;
    for x,y st x in X & y in Y holds f1.(x,y) = f2.(x,y)
    proof
      let x,y;
      assume
A10:  x in X & y in Y;
      then f1.(x,y) = y by A7;
      hence thesis by A9,A10;
    end;
    hence thesis by A6,A8,Th6;
  end;
end;

theorem Th43:
  rng pr1(X,Y) c= X
proof
  let x;
  assume x in rng pr1(X,Y);
  then consider p such that
A1: p in dom pr1(X,Y) and
A2: x = pr1(X,Y).p by FUNCT_1:def 3;
  p in [:X,Y:] by A1,Def4;
  then consider x1,y1 such that
A3: x1 in X & y1 in Y and
A4: p = [x1,y1] by ZFMISC_1:def 2;
  x =pr1(X,Y).(x1,y1) by A2,A4;
  hence thesis by A3,Def4;
end;

theorem
  Y <> {} implies rng pr1(X,Y) = X
proof
  set y = the Element of Y;
  assume
A1: Y <> {};
A2: X c= rng pr1(X,Y)
  proof
    let x;
    assume
A3: x in X;
    then [x,y] in [:X,Y:] by A1,ZFMISC_1:87;
    then
A4: [x,y] in dom pr1(X,Y) by Def4;
    pr1(X,Y).(x,y) = x by A1,A3,Def4;
    hence thesis by A4,FUNCT_1:def 3;
  end;
  rng pr1(X,Y) c= X by Th43;
  hence thesis by A2,XBOOLE_0:def 10;
end;

theorem Th45:
  rng pr2(X,Y) c= Y
proof
  let y;
  assume y in rng pr2(X,Y);
  then consider p such that
A1: p in dom pr2(X,Y) and
A2: y = pr2(X,Y).p by FUNCT_1:def 3;
  p in [:X,Y:] by A1,Def5;
  then consider x1,y1 such that
A3: x1 in X & y1 in Y and
A4: p = [x1,y1] by ZFMISC_1:def 2;
  y = pr2(X,Y).(x1,y1) by A2,A4;
  hence thesis by A3,Def5;
end;

theorem
  X <> {} implies rng pr2(X,Y) = Y
proof
  set x = the Element of X;
  assume
A1: X <> {};
A2: Y c= rng pr2(X,Y)
  proof
    let y;
    assume
A3: y in Y;
    then [x,y] in [:X,Y:] by A1,ZFMISC_1:87;
    then
A4: [x,y] in dom pr2(X,Y) by Def5;
    pr2(X,Y).(x,y) = y by A1,A3,Def5;
    hence thesis by A4,FUNCT_1:def 3;
  end;
  rng pr2(X,Y) c= Y by Th45;
  hence thesis by A2,XBOOLE_0:def 10;
end;

definition
  let X,Y;
  redefine func pr1(X,Y) -> Function of [:X,Y:],X;
  coherence
  proof
    per cases;
    suppose
   X = {} implies [:X,Y:] = {};
      dom pr1(X,Y) = [:X,Y:] & rng pr1(X,Y) c= X by Def4,Th43;
      hence thesis by FUNCT_2:2;
    end;
    suppose
      X = {} & [:X,Y:] <> {};
      hence thesis by ZFMISC_1:90;
    end;
  end;
  redefine func pr2(X,Y) -> Function of [:X,Y:],Y;
  coherence
  proof
    per cases;
    suppose
   Y = {} implies [:X,Y:] = {};
      dom pr2(X,Y) = [:X,Y:] & rng pr2(X,Y) c= Y by Def5,Th45;
      hence thesis by FUNCT_2:2;
    end;
    suppose
      Y = {} & [:X,Y:] <> {};
      hence thesis by ZFMISC_1:90;
    end;
  end;
end;

definition
  let X;
  func delta(X) -> Function means
  :Def6:
  dom it = X & for x st x in X holds it .x = [x,x];
  existence
  proof
    deffunc F(set) = [$1,$1];
    ex f being Function st dom f = X & for x st x in X holds f.x = F(x)
    from FUNCT_1:sch 3;
    hence thesis;
  end;
  uniqueness
  proof
    let f1,f2 be Function such that
A1: dom f1 = X and
A2: for x st x in X holds f1.x = [x,x] and
A3: dom f2 = X and
A4: for x st x in X holds f2.x = [x,x];
    for x st x in X holds f1.x = f2.x
    proof
      let x;
      assume
A5:   x in X;
      then f1.x = [x,x] by A2;
      hence thesis by A4,A5;
    end;
    hence thesis by A1,A3,FUNCT_1:2;
  end;
end;

theorem Th47:
  rng delta X c= [:X,X:]
proof
  let y;
  assume y in rng delta X;
  then consider x such that
A1: x in dom delta X and
A2: y = (delta X).x by FUNCT_1:def 3;
A3: x in X by A1,Def6;
  then y = [x,x] by A2,Def6;
  hence thesis by A3,ZFMISC_1:87;
end;

definition
  let X;
  redefine func delta(X) -> Function of X,[:X,X:];
  coherence
  proof
    dom delta X = X & rng delta X c= [:X,X:] by Def6,Th47;
    hence thesis by FUNCT_2:2;
  end;
end;



definition
  let f,g be Function;
  func <:f,g:> -> Function means
  :Def7:
  dom it = dom f /\ dom g & for x st x in dom it holds it.x = [f.x,g.x];
  existence
  proof
    deffunc F(set) = [f.$1,g.$1];
    ex fg being Function st dom fg = dom f /\ dom g & for x st x in dom f
    /\ dom g holds fg.x = F(x) from FUNCT_1:sch 3;
    hence thesis;
  end;
  uniqueness
  proof
    let f1,f2 be Function such that
A1: dom f1 = dom f /\ dom g and
A2: for x st x in dom f1 holds f1.x = [f.x,g.x] and
A3: dom f2 = dom f /\ dom g and
A4: for x st x in dom f2 holds f2.x = [f.x,g.x];
    for x st x in dom f /\ dom g holds f1.x=f2.x
    proof
      let x;
      assume
A5:   x in dom f /\ dom g;
      then f1.x=[f.x,g.x] by A1,A2;
      hence thesis by A3,A4,A5;
    end;
    hence thesis by A1,A3,FUNCT_1:2;
  end;
end;

registration
  let f be empty Function, g be Function;
  cluster <:f,g:> -> empty;
  coherence
  proof
    dom f = {};
    then dom<:f,g:> = {} /\ dom g by Def7;
    hence thesis;
  end;
  cluster <:g,f:> -> empty;
  coherence
  proof
    dom f = {};
    then dom<:g,f:> = {} /\ dom g by Def7;
    hence thesis;
  end;
end;

theorem Th48:
  for f,g being Function st x in dom f /\ dom g holds <:f,g:>.x = [f.x,g.x]
proof
  let f,g be Function;
  assume x in dom f /\ dom g;
  then x in dom <:f,g:> by Def7;
  hence thesis by Def7;
end;

theorem Th49:
  for f,g being Function st dom f = X & dom g = X & x in X holds
  <:f,g:>.x = [f.x,g.x]
proof
  let f,g be Function;
  assume dom f = X & dom g = X & x in X;
  then x in dom f /\ dom g;
  then x in dom <:f,g:> by Def7;
  hence thesis by Def7;
end;

theorem Th50:
  for f,g being Function st dom f = X & dom g = X holds dom <:f,g :> = X
proof
  let f,g be Function;
  dom <:f,g:> = dom f /\ dom g by Def7;
  hence thesis;
end;

theorem Th51:
  for f,g being Function holds rng <:f,g:> c= [:rng f,rng g:]
proof
  let f,g be Function;
  let q;
  assume q in rng <:f,g:>;
  then consider x such that
A1: x in dom <:f,g:> and
A2: q = <:f,g:>.x by FUNCT_1:def 3;
A3: x in dom f /\ dom g by A1,Def7;
  then x in dom f by XBOOLE_0:def 4;
  then
A4: f.x in rng f by FUNCT_1:def 3;
  x in dom g by A3,XBOOLE_0:def 4;
  then
A5: g.x in rng g by FUNCT_1:def 3;
  q = [f.x,g.x] by A1,A2,Def7;
  hence thesis by A4,A5,ZFMISC_1:87;
end;

theorem Th52:
  for f,g being Function st dom f = dom g & rng f c= Y & rng g c=
  Z holds pr1(Y,Z)*<:f,g:> = f & pr2(Y,Z)*<:f,g:> = g
proof
  let f,g be Function such that
A1: dom f = dom g and
A2: rng f c= Y & rng g c= Z;
A3: [:rng f, rng g:] c= [:Y,Z:] by A2,ZFMISC_1:96;
A4: rng <:f,g:> c= [:rng f, rng g:] by Th51;
  dom pr1(Y,Z) = [:Y, Z:] by Def4;
  then
A5: dom(pr1(Y,Z)*<:f,g:>) = dom <:f,g:> by A3,A4,RELAT_1:27,XBOOLE_1:1;
  then
A6: dom(pr1(Y,Z)*<:f,g:>) = dom f by A1,Th50;
  x in dom f implies (pr1(Y,Z)*<:f,g:>).x = f.x
  proof
    assume
A7: x in dom f;
    then
A8: f.x in rng f & g.x in rng g by A1,FUNCT_1:def 3;
    thus (pr1(Y,Z)*<:f,g:>).x = pr1(Y,Z).(<:f,g:>.x) by A6,A7,FUNCT_1:12
      .= pr1(Y,Z).(f.x,g.x) by A5,A6,A7,Def7
      .= f.x by A2,A8,Def4;
  end;
  hence pr1(Y,Z)*<:f,g:> = f by A6,FUNCT_1:2;
  dom pr2(Y,Z) = [:Y, Z:] by Def5;
  then
A9: dom(pr2(Y,Z)*<:f,g:>) = dom <:f,g:> by A3,A4,RELAT_1:27,XBOOLE_1:1;
  then
A10: dom(pr2(Y,Z)*<:f,g:>) = dom g by A1,Th50;
  x in dom g implies (pr2(Y,Z)*<:f,g:>).x = g.x
  proof
    assume
A11: x in dom g;
    then
A12: f.x in rng f & g.x in rng g by A1,FUNCT_1:def 3;
    thus (pr2(Y,Z)*<:f,g:>).x = pr2(Y,Z).(<:f,g:>.x) by A10,A11,FUNCT_1:12
      .= pr2(Y,Z).(f.x,g.x) by A9,A10,A11,Def7
      .= g.x by A2,A12,Def5;
  end;
  hence thesis by A10,FUNCT_1:2;
end;

theorem Th53:
  <:pr1(X,Y),pr2(X,Y):> = id [:X,Y:]
proof
  dom pr1(X,Y) = [:X,Y:] & dom pr2(X,Y) = [:X,Y:] by Def4,Def5;
  then
A1: dom <:pr1(X,Y),pr2(X,Y):> = [:X,Y:] by Th50;
A2: for x,y st x in X & y in Y holds <:pr1(X,Y),pr2(X,Y):>.(x,y) = (id [:X,Y
  :]).(x,y)
  proof
    let x,y;
    assume
A3: x in X & y in Y;
    then
A4: [x,y] in [:X,Y:] by ZFMISC_1:87;
    hence <:pr1(X,Y),pr2(X,Y):>.(x,y) = [pr1(X,Y).(x,y),pr2(X,Y).(x,y)] by A1
,Def7
      .= [x,pr2(X,Y).(x,y)] by A3,Def4
      .= [x,y] by A3,Def5
      .= (id [:X,Y:]).(x,y) by A4,FUNCT_1:18;
  end;
  dom id [:X,Y:] = [:X,Y:];
  hence thesis by A1,A2,Th6;
end;

theorem Th54:
  for f,g,h,k being Function st dom f = dom g & dom k = dom h & <:
  f,g:> = <:k,h:> holds f = k & g = h
proof
  let f,g,h,k be Function such that
A1: dom f = dom g and
A2: dom k = dom h and
A3: <:f,g:> = <:k,h:>;
A4: dom <:f,g:> = dom f by A1,Th50;
  x in dom f implies f.x = k.x
  proof
    assume x in dom f;
    then <:f,g:>.x = [f.x,g.x] & <:k,h:>.x = [k.x,h.x] by A3,A4,Def7;
    hence thesis by A3,XTUPLE_0:1;
  end;
  hence f = k by A2,A3,A4,Th50,FUNCT_1:2;
A5: dom <:f,g:> = dom g by A1,Th50;
  x in dom g implies g.x = h.x
  proof
    assume x in dom g;
    then <:f,g:>.x = [f.x,g.x] & <:k,h:>.x = [k.x,h.x] by A3,A5,Def7;
    hence thesis by A3,XTUPLE_0:1;
  end;
  hence thesis by A2,A3,A5,Th50,FUNCT_1:2;
end;

theorem
  for f,g,h being Function holds <:f*h,g*h:> = <:f,g:>*h
proof
  let f,g,h be Function;
  x in dom <:f*h,g*h:> iff x in dom(<:f,g:>*h)
  proof
    thus x in dom <:f*h,g*h:> implies x in dom(<:f,g:>*h)
    proof
      assume x in dom <:f*h,g*h:>;
      then
A1:   x in dom(f*h) /\ dom(g*h) by Def7;
      then
A2:   x in dom(f*h) by XBOOLE_0:def 4;
      x in dom(g*h) by A1,XBOOLE_0:def 4;
      then
A3:   h.x in dom g by FUNCT_1:11;
      h.x in dom f by A2,FUNCT_1:11;
      then h.x in dom f /\ dom g by A3,XBOOLE_0:def 4;
      then
A4:   h.x in dom <:f,g:> by Def7;
      x in dom h by A2,FUNCT_1:11;
      hence thesis by A4,FUNCT_1:11;
    end;
    assume
A5: x in dom(<:f,g:>*h);
    then
A6: x in dom h by FUNCT_1:11;
    h.x in dom <:f,g:> by A5,FUNCT_1:11;
    then
A7: h.x in dom f /\ dom g by Def7;
    then h.x in dom g by XBOOLE_0:def 4;
    then
A8: x in dom(g*h) by A6,FUNCT_1:11;
    h.x in dom f by A7,XBOOLE_0:def 4;
    then x in dom(f*h) by A6,FUNCT_1:11;
    then x in dom(f*h) /\ dom(g*h) by A8,XBOOLE_0:def 4;
    hence thesis by Def7;
  end;
  then
A9: dom <:f*h,g*h:> = dom(<:f,g:>*h) by TARSKI:1;
  for x st x in dom <:f*h,g*h:> holds <:f*h,g*h:>.x = (<:f,g:>*h).x
  proof
    let x;
    assume
A10: x in dom <:f*h,g*h:>;
    then
A11: x in dom(f*h) /\ dom(g*h) by Def7;
    then
A12: x in dom(f*h) by XBOOLE_0:def 4;
    then
A13: x in dom h by FUNCT_1:11;
A14: x in dom(g*h) by A11,XBOOLE_0:def 4;
    then
A15: h.x in dom g by FUNCT_1:11;
    h.x in dom f by A12,FUNCT_1:11;
    then
A16: h.x in dom f /\ dom g by A15,XBOOLE_0:def 4;
    thus <:f*h,g*h:>.x = [(f*h).x,(g*h).x] by A10,Def7
      .= [f.(h.x),(g*h).x] by A12,FUNCT_1:12
      .= [f.(h.x),g.(h.x)] by A14,FUNCT_1:12
      .= <:f,g:>.(h.x) by A16,Th48
      .= (<:f,g:>*h).x by A13,FUNCT_1:13;
  end;
  hence thesis by A9,FUNCT_1:2;
end;

theorem
  for f,g being Function holds <:f,g:>.:A c= [:f.:A,g.:A:]
proof
  let f,g be Function;
  let y;
  assume y in <:f,g:>.:A;
  then consider x such that
A1: x in dom <:f,g:> and
A2: x in A and
A3: y = <:f,g:>.x by FUNCT_1:def 6;
A4: x in dom f /\ dom g by A1,Def7;
  then x in dom f by XBOOLE_0:def 4;
  then
A5: f.x in f.:A by A2,FUNCT_1:def 6;
  x in dom g by A4,XBOOLE_0:def 4;
  then
A6: g.x in g.: A by A2,FUNCT_1:def 6;
  y = [f.x,g.x] by A1,A3,Def7;
  hence thesis by A5,A6,ZFMISC_1:def 2;
end;

theorem
  for f,g being Function holds <:f,g:>"[:B,C:] = f"B /\ g"C
proof
  let f,g be Function;
  x in <:f,g:>"[:B,C:] iff x in f"B /\ g"C
  proof
    thus x in <:f,g:>"[:B,C:] implies x in f"B /\ g"C
    proof
      assume
A1:   x in <:f,g:>"[:B,C:];
      then <:f,g:>.x in [:B,C:] by FUNCT_1:def 7;
      then consider y1,y2 such that
A2:   y1 in B and
A3:   y2 in C and
A4:   <:f,g:>.x = [y1,y2] by ZFMISC_1:def 2;
A5:   x in dom <:f,g:> by A1,FUNCT_1:def 7;
      then
A6:   x in dom f /\ dom g by Def7;
      then
A7:   x in dom g by XBOOLE_0:def 4;
A8:   [y1,y2] = [f.x,g.x] by A4,A5,Def7;
      then y2 = g.x by XTUPLE_0:1;
      then
A9:   x in g"C by A3,A7,FUNCT_1:def 7;
A10:  x in dom f by A6,XBOOLE_0:def 4;
      y1 =f.x by A8,XTUPLE_0:1;
      then x in f"B by A2,A10,FUNCT_1:def 7;
      hence thesis by A9,XBOOLE_0:def 4;
    end;
    assume
A11: x in f"B /\ g"C;
    then
A12: x in g"C by XBOOLE_0:def 4;
    then
A13: x in dom g by FUNCT_1:def 7;
A14: x in f"B by A11,XBOOLE_0:def 4;
    then x in dom f by FUNCT_1:def 7;
    then
A15: x in dom f /\ dom g by A13,XBOOLE_0:def 4;
    then
A16: x in dom <:f,g:> by Def7;
A17: g.x in C by A12,FUNCT_1:def 7;
    f.x in B by A14,FUNCT_1:def 7;
    then [f.x,g.x] in [:B,C:] by A17,ZFMISC_1:def 2;
    then <:f,g:>.x in [:B,C:] by A15,Th48;
    hence thesis by A16,FUNCT_1:def 7;
  end;
  hence thesis by TARSKI:1;
end;

theorem Th58:
  for f being Function of X,Y for g being Function of X,Z st (Y =
{} implies X = {}) & (Z = {} implies X = {}) holds <:f,g:> is Function of X,[:Y
  ,Z:]
proof
  let f be Function of X,Y;
  let g be Function of X,Z;
  assume
A1: ( Y = {} implies X = {})&( Z = {} implies X = {});
  per cases;
  suppose
 [:Y,Z:] = {} implies X = {};
    rng f c= Y & rng g c= Z by RELAT_1:def 19;
    then
A2: [:rng f,rng g:] c= [:Y,Z:] by ZFMISC_1:96;
    dom f = X & dom g = X by A1,FUNCT_2:def 1;
    then
A3: dom<:f,g:> = X by Th50;
    rng <:f,g:> c= [:rng f,rng g:] by Th51;
    then rng<:f,g:> c= [:Y,Z:] by A2,XBOOLE_1:1;
    hence thesis by A3,FUNCT_2:2;
  end;
  suppose
    [:Y,Z:] = {} & X <> {};
    hence thesis by A1;
  end;
end;

definition
  let X,D1,D2;
  let f1 be Function of X,D1;
  let f2 be Function of X,D2;
  redefine func <:f1,f2:> -> Function of X,[:D1,D2:];
  coherence by Th58;
end;

theorem
  for f1 being Function of C,D1 for f2 being Function of C,D2 for c
  being Element of C holds <:f1,f2:>.c = [f1.c,f2.c]
proof
  let f1 be Function of C,D1;
  let f2 be Function of C,D2;
  let c be Element of C;
  dom f1 = C & dom f2 = C by FUNCT_2:def 1;
  hence thesis by Th49;
end;

theorem
  for f being Function of X,Y for g being Function of X,Z holds rng <:f,
  g:> c= [:Y,Z:]
proof
  let f be Function of X,Y;
  let g be Function of X,Z;
  rng f c= Y & rng g c= Z by RELAT_1:def 19;
  then
A1: [:rng f,rng g:] c= [:Y,Z:] by ZFMISC_1:96;
  rng <:f,g:> c= [:rng f,rng g:] by Th51;
  hence thesis by A1,XBOOLE_1:1;
end;

theorem Th61:
  for f being Function of X,Y for g being Function of X,Z st (Y =
{} implies X = {}) & (Z = {} implies X = {}) holds pr1(Y,Z)*<:f,g:> = f & pr2(Y
  ,Z)*<:f,g:> = g
proof
  let f be Function of X,Y;
  let g be Function of X,Z;
  assume ( Y = {} implies X = {})&( Z = {} implies X = {});
  then
A1: dom f = X & dom g = X by FUNCT_2:def 1;
  rng f c= Y & rng g c= Z by RELAT_1:def 19;
  hence thesis by A1,Th52;
end;

theorem
  for f being Function of X,D1 for g being Function of X,D2 holds pr1(D1
  ,D2)*<:f,g:> = f & pr2(D1,D2)*<:f,g:> = g by Th61;

theorem
  for f1,f2 being Function of X,Y for g1,g2 being Function of X,Z st (Y
= {} implies X = {}) & (Z = {} implies X = {}) & <:f1,g1:> = <:f2,g2:> holds f1
  = f2 & g1 = g2
proof
  let f1,f2 be Function of X,Y;
  let g1,g2 be Function of X,Z;
  assume that
A1: Y = {} implies X = {} and
A2: Z = {} implies X = {};
A3: dom g1 = X & dom g2 = X by A2,FUNCT_2:def 1;
  dom f1 = X & dom f2 = X by A1,FUNCT_2:def 1;
  hence thesis by A3,Th54;
end;

theorem
  for f1,f2 being Function of X,D1 for g1,g2 being Function of X,D2 st
  <:f1,g1:> = <:f2,g2:> holds f1 = f2 & g1 = g2
proof
  let f1,f2 be Function of X,D1;
  let g1,g2 be Function of X,D2;
A1: dom g1 = X & dom g2 = X by FUNCT_2:def 1;
  dom f1 = X & dom f2 = X by FUNCT_2:def 1;
  hence thesis by A1,Th54;
end;



definition
  let f,g be Function;
  func [:f,g:] -> Function means
  :Def8:
  dom it = [:dom f, dom g:] & for x,y st
  x in dom f & y in dom g holds it.(x,y) = [f.x,g.y];
  existence
  proof
    deffunc F(set,set) = [f.$1,g.$2];
    ex F being Function st dom F = [:dom f,dom g:] & for x,y st x in dom f
    & y in dom g holds F.(x,y) = F(x,y) from Lambda3;
    hence thesis;
  end;
  uniqueness
  proof
    let fg1, fg2 be Function such that
A1: dom fg1 = [:dom f, dom g:] and
A2: for x,y st x in dom f & y in dom g holds fg1.(x,y) = [f.x,g.y] and
A3: dom fg2 = [:dom f, dom g:] and
A4: for x,y st x in dom f & y in dom g holds fg2.(x,y) = [f.x,g.y];
    for p st p in [:dom f,dom g:] holds fg1.p=fg2.p
    proof
      let p;
      assume p in [:dom f,dom g:];
      then consider x,y such that
A5:   x in dom f & y in dom g and
A6:   p = [x,y] by ZFMISC_1:def 2;
      fg1.(x,y) = [f.x,g.y] & fg2.(x,y) = [f.x,g.y] by A2,A4,A5;
      hence thesis by A6;
    end;
    hence thesis by A1,A3,FUNCT_1:2;
  end;
end;

theorem Th65:
  for f,g being Function, x,y st [x,y] in [:dom f,dom g:] holds [:
  f,g:].(x,y) = [f.x,g.y]
proof
  let f,g be Function, x,y;
  assume [x,y] in [:dom f,dom g:];
  then x in dom f & y in dom g by ZFMISC_1:87;
  hence thesis by Def8;
end;

theorem Th66:
  for f,g being Function holds [:f,g:] = <:f*pr1(dom f,dom g),g*
  pr2(dom f,dom g):>
proof
  let f,g be Function;
A1: dom pr1(dom f,dom g) = [:dom f,dom g:] by Def4;
A2: dom pr2(dom f,dom g) = [:dom f,dom g:] by Def5;
  rng pr2(dom f,dom g) c= dom g by Th45;
  then
A3: dom(g*pr2(dom f,dom g)) = [:dom f,dom g:] by A2,RELAT_1:27;
  rng pr1(dom f,dom g) c= dom f by Th43;
  then dom(f*pr1(dom f,dom g)) = [:dom f,dom g:] by A1,RELAT_1:27;
  then
A4: dom <:f*pr1(dom f,dom g),g*pr2(dom f,dom g):> = [:dom f,dom g:] by A3,Th50;
A5: for x,y st x in dom f & y in dom g holds [:f,g:].(x,y) = <:f*pr1(dom f,
  dom g),g*pr2(dom f,dom g):>.(x,y)
  proof
    let x,y;
    assume
A6: x in dom f & y in dom g;
    then
A7: [x,y] in [:dom f,dom g:] by ZFMISC_1:87;
    thus [:f,g:].(x,y) = [f.x,g.y] by A6,Def8
      .= [f.(pr1(dom f,dom g).(x,y)),g.y] by A6,Def4
      .= [f.(pr1(dom f,dom g).(x,y)),g.(pr2(dom f,dom g).(x,y))] by A6,Def5
      .= [(f*pr1(dom f,dom g)).(x,y),g.(pr2(dom f,dom g).(x,y))] by A1,A7,
FUNCT_1:13
      .= [(f*pr1(dom f,dom g)).(x,y),(g*pr2(dom f,dom g)).(x,y)] by A2,A7,
FUNCT_1:13
      .= <:f*pr1(dom f,dom g),g*pr2(dom f,dom g):>.(x,y) by A4,A7,Def7;
  end;
  dom [:f,g:] = [:dom f,dom g:] by Def8;
  hence thesis by A4,A5,Th6;
end;

theorem Th67:
  for f,g being Function holds rng [:f,g:] = [:rng f,rng g:]
proof
  let f,g be Function;
  q in rng [:f,g:] iff q in [:rng f,rng g:]
  proof
A1: dom [:f,g:]=[:dom f,dom g:] by Def8;
    thus q in rng [:f,g:] implies q in [:rng f,rng g:]
    proof
      assume q in rng [:f,g:];
      then consider p such that
A2:   p in dom [:f,g:] and
A3:   q = [:f,g:].p by FUNCT_1:def 3;
      p in [:dom f, dom g:] by A2,Def8;
      then consider x,y such that
A4:   x in dom f & y in dom g and
A5:   p = [x,y] by ZFMISC_1:def 2;
A6:   f.x in rng f & g.y in rng g by A4,FUNCT_1:def 3;
      q = [:f,g:].(x,y) by A3,A5
        .= [f.x,g.y] by A4,Def8;
      hence thesis by A6,ZFMISC_1:87;
    end;
    assume q in [:rng f,rng g:];
    then consider y1,y2 such that
A7: y1 in rng f and
A8: y2 in rng g and
A9: q = [y1,y2] by ZFMISC_1:def 2;
    consider x2 such that
A10: x2 in dom g and
A11: y2 = g.x2 by A8,FUNCT_1:def 3;
    consider x1 such that
A12: x1 in dom f and
A13: y1 = f.x1 by A7,FUNCT_1:def 3;
A14: [x1,x2] in [:dom f,dom g:] by A12,A10,ZFMISC_1:87;
    [:f,g:].(x1,x2)=q by A9,A12,A13,A10,A11,Def8;
    hence thesis by A14,A1,FUNCT_1:def 3;
  end;
  hence thesis by TARSKI:1;
end;

theorem Th68:
  for f,g being Function st dom f = X & dom g = X holds <:f,g:> =
  [:f,g:]*(delta X)
proof
  let f,g be Function such that
A1: dom f = X & dom g = X;
A2: dom delta X = X by Def6;
  rng delta X c= [:X,X:] by Th47;
  then rng delta X c= dom [:f,g:] by A1,Def8;
  then
A3: dom([:f,g:]*(delta X)) = X by A2,RELAT_1:27;
  dom f /\ dom g = X by A1;
  then
A4: dom <:f,g:> = X by Def7;
  for x st x in X holds <:f,g:>.x = ([:f,g:]*(delta X)).x
  proof
    let x;
    assume
A5: x in X;
    hence <:f,g:>.x = [f.x,g.x] by A4,Def7
      .= [:f,g:].(x,x) by A1,A5,Def8
      .= [:f,g:].((delta X).x) by A5,Def6
      .= ([:f,g:]*(delta X)).x by A3,A5,FUNCT_1:12;
  end;
  hence thesis by A4,A3,FUNCT_1:2;
end;

theorem
  [:id X, id Y:] = id [:X,Y:]
proof
  rng pr1(X,Y) c= X by Th43;
  then
A1: (id X)*pr1(X,Y) = pr1(X,Y) by RELAT_1:53;
  rng pr2(X,Y) c= Y by Th45;
  then
A2: (id Y)*pr2(X,Y) = pr2(X,Y) by RELAT_1:53;
  dom id X = X & dom id Y = Y;
  hence [:id X, id Y:] = <:pr1(X,Y),pr2(X,Y):> by A1,A2,Th66
    .= id [:X,Y:] by Th53;
end;

theorem
  for f,g,h,k being Function holds [:f,h:]*<:g,k:> = <:f*g,h*k:>
proof
  let f,g,h,k be Function;
  x in dom([:f,h:]*<:g,k:>) iff x in dom <:f*g,h*k:>
  proof
    thus x in dom([:f,h:]*<:g,k:>) implies x in dom <:f*g,h*k:>
    proof
      assume
A1:   x in dom([:f,h:]*<:g,k:>);
      then
A2:   x in dom <:g,k:> by FUNCT_1:11;
      then
A3:   x in dom g /\ dom k by Def7;
      then
A4:   x in dom g by XBOOLE_0:def 4;
A5:   x in dom k by A3,XBOOLE_0:def 4;
      <:g,k:>.x in dom [:f,h:] by A1,FUNCT_1:11;
      then [g.x,k.x] in dom [:f,h:] by A2,Def7;
      then
A6:   [g.x,k.x] in [:dom f,dom h:] by Def8;
      then k.x in dom h by ZFMISC_1:87;
      then
A7:   x in dom(h*k) by A5,FUNCT_1:11;
      g.x in dom f by A6,ZFMISC_1:87;
      then x in dom(f*g) by A4,FUNCT_1:11;
      then x in dom(f*g) /\ dom(h*k) by A7,XBOOLE_0:def 4;
      hence thesis by Def7;
    end;
    assume x in dom <:f*g,h*k:>;
    then
A8: x in dom(f*g) /\ dom(h*k) by Def7;
    then
A9: x in dom(f*g) by XBOOLE_0:def 4;
A10: x in dom(h*k) by A8,XBOOLE_0:def 4;
    then
A11: x in dom k by FUNCT_1:11;
    x in dom g by A9,FUNCT_1:11;
    then x in dom g /\ dom k by A11,XBOOLE_0:def 4;
    then
A12: x in dom <:g,k:> by Def7;
A13: k.x in dom h by A10,FUNCT_1:11;
    g.x in dom f by A9,FUNCT_1:11;
    then [g.x,k.x] in [:dom f,dom h:] by A13,ZFMISC_1:87;
    then [g.x,k.x] in dom [:f,h:] by Def8;
    then <:g,k:>.x in dom [:f,h:] by A12,Def7;
    hence thesis by A12,FUNCT_1:11;
  end;
  then
A14: dom([:f,h:]*<:g,k:>) = dom <:f*g,h*k:> by TARSKI:1;
  for x st x in dom([:f,h:]*<:g,k:>) holds ([:f,h:]*<:g,k:>).x = <:f*g,h*
  k :> . x
  proof
    let x;
    assume
A15: x in dom([:f,h:]*<:g,k:>);
    then
A16: x in dom <:g,k:> by FUNCT_1:11;
    then
A17: x in dom g /\ dom k by Def7;
    then
A18: x in dom g by XBOOLE_0:def 4;
    <:g,k:>.x in dom [:f,h:] by A15,FUNCT_1:11;
    then [g.x,k.x] in dom [:f,h:] by A16,Def7;
    then
A19: [g.x,k.x] in [:dom f,dom h:] by Def8;
    then
A20: g.x in dom f by ZFMISC_1:87;
A21: x in dom k by A17,XBOOLE_0:def 4;
A22: k.x in dom h by A19,ZFMISC_1:87;
    then
A23: x in dom(h*k) by A21,FUNCT_1:11;
    x in dom(f*g) by A20,A18,FUNCT_1:11;
    then
A24: x in dom(f*g) /\ dom(h*k) by A23,XBOOLE_0:def 4;
    thus ([:f,h:]*<:g,k:>).x = [:f,h:].(<:g,k:>.x) by A15,FUNCT_1:12
      .= [:f,h:].(g.x,k.x) by A16,Def7
      .= [f.(g.x),h.(k.x)] by A20,A22,Def8
      .= [(f*g).x,h.(k.x)] by A18,FUNCT_1:13
      .= [(f*g).x,(h*k).x] by A21,FUNCT_1:13
      .= <:f*g,h*k:>.x by A24,Th48;
  end;
  hence thesis by A14,FUNCT_1:2;
end;

theorem
  for f,g,h,k being Function holds [:f,h:]*[:g,k:] = [:f*g,h*k:]
proof
  let f,g,h,k be Function;
A1: for x,y st x in dom(f*g) & y in dom(h*k) holds ([:f,h:]*[:g,k:]).(x,y)
  = [:f*g,h*k:].(x,y)
  proof
    let x,y such that
A2: x in dom(f*g) and
A3: y in dom(h*k);
A4: g.x in dom f & k.y in dom h by A2,A3,FUNCT_1:11;
A5: x in dom g & y in dom k by A2,A3,FUNCT_1:11;
    then [x,y] in [:dom g, dom k:] by ZFMISC_1:87;
    then [x,y] in dom [:g,k:] by Def8;
    hence ([:f,h:]*[:g,k:]).(x,y) = [:f,h:].([:g,k:].(x,y)) by FUNCT_1:13
      .= [:f,h:].(g.x,k.y) by A5,Def8
      .= [f.(g.x),h.(k.y)] by A4,Def8
      .= [(f*g).x,h.(k.y)] by A2,FUNCT_1:12
      .= [(f*g).x,(h*k).y] by A3,FUNCT_1:12
      .= [:f*g,h*k:].(x,y) by A2,A3,Def8;
  end;
  p in dom([:f,h:]*[:g,k:]) iff p in [:dom(f*g),dom(h*k):]
  proof
A6: dom [:g,k:] = [:dom g, dom k:] by Def8;
A7: dom [:f,h:] = [:dom f, dom h:] by Def8;
    thus p in dom([:f,h:]*[:g,k:]) implies p in [:dom(f*g),dom(h*k):]
    proof
      assume
A8:   p in dom([:f,h:]*[:g,k:]);
      then
A9:   [:g,k:].p in dom [:f,h:] by FUNCT_1:11;
A10:  p in dom [:g,k:] by A8,FUNCT_1:11;
      then consider x1,x2 such that
A11:  x1 in dom g and
A12:  x2 in dom k and
A13:  p = [x1,x2] by A6,ZFMISC_1:def 2;
      [:g,k:].(x1,x2) =[:g,k:].p by A13;
      then
A14:  [g.x1,k.x2] in dom [:f,h:] by A6,A10,A9,A13,Th65;
      then k.x2 in dom h by A7,ZFMISC_1:87;
      then
A15:  x2 in dom(h*k) by A12,FUNCT_1:11;
      g.x1 in dom f by A7,A14,ZFMISC_1:87;
      then x1 in dom(f*g) by A11,FUNCT_1:11;
      hence thesis by A13,A15,ZFMISC_1:87;
    end;
    assume p in [:dom(f*g),dom(h*k):];
    then consider x1,x2 such that
A16: x1 in dom(f*g) & x2 in dom(h*k) and
A17: p = [x1,x2] by ZFMISC_1:def 2;
    x1 in dom g & x2 in dom k by A16,FUNCT_1:11;
    then
A18: [x1,x2] in dom [:g,k:] by A6,ZFMISC_1:87;
    g.x1 in dom f & k.x2 in dom h by A16,FUNCT_1:11;
    then [g.x1,k.x2] in dom [:f,h:] by A7,ZFMISC_1:87;
    then [:g,k:].(x1,x2) in dom [:f,h:] by A6,A18,Th65;
    hence thesis by A17,A18,FUNCT_1:11;
  end;
  then
A19: dom([:f,h:]*[:g,k:]) = [:dom(f*g),dom(h*k):] by TARSKI:1;
  [:dom(f*g),dom(h*k):] = dom [:f*g,h*k:] by Def8;
  hence thesis by A19,A1,Th6;
end;

theorem
  for f,g being Function holds [:f,g:].:[:B,A:] = [:f.:B,g.:A:]
proof
  let f,g be Function;
  q in [:f,g:].:[:B,A:] iff q in [:f.:B,g.:A:]
  proof
A1: [:dom f,dom g:] = dom [:f,g:] by Def8;
    thus q in [:f,g:].:[:B,A:] implies q in [:f.:B,g.:A:]
    proof
      assume q in [:f,g:].:[:B,A:];
      then consider p such that
A2:   p in dom [:f,g:] and
A3:   p in [:B,A:] and
A4:   q = [:f,g:].p by FUNCT_1:def 6;
      dom [:f,g:] = [:dom f,dom g:] by Def8;
      then consider x,y such that
A5:   x in dom f and
A6:   y in dom g and
A7:   p =[x,y] by A2,ZFMISC_1:def 2;
      x in B by A3,A7,ZFMISC_1:87;
      then
A8:   f.x in f.:B by A5,FUNCT_1:def 6;
      y in A by A3,A7,ZFMISC_1:87;
      then
A9:   g.y in g.:A by A6,FUNCT_1:def 6;
      q = [:f,g:].(x,y) by A4,A7;
      then q=[f.x,g.y] by A5,A6,Def8;
      hence thesis by A8,A9,ZFMISC_1:87;
    end;
    assume q in [:f.:B,g.:A:];
    then consider y1,y2 such that
A10: y1 in f.:B and
A11: y2 in g.:A and
A12: q = [y1,y2] by ZFMISC_1:def 2;
    consider x1 such that
A13: x1 in dom f and
A14: x1 in B and
A15: y1 = f.x1 by A10,FUNCT_1:def 6;
    consider x2 such that
A16: x2 in dom g and
A17: x2 in A and
A18: y2 = g.x2 by A11,FUNCT_1:def 6;
A19: [:f,g:].(x1,x2) = [f.x1,g.x2] by A13,A16,Def8;
    [x1,x2] in [:dom f,dom g:] & [x1,x2] in [:B,A:] by A13,A14,A16,A17,
ZFMISC_1:87;
    hence thesis by A12,A15,A18,A1,A19,FUNCT_1:def 6;
  end;
  hence thesis by TARSKI:1;
end;

theorem
  for f,g being Function holds [:f,g:]"[:B,A:] = [:f"B,g"A:]
proof
  let f,g be Function;
  q in [:f,g:]"[:B,A:] iff q in [:f"B,g"A:]
  proof
    thus q in [:f,g:]"[:B,A:] implies q in [:f"B,g"A:]
    proof
      assume
A1:   q in [:f,g:]"[:B,A:];
      then
A2:   [:f,g:].q in [:B,A:] by FUNCT_1:def 7;
      q in dom [:f,g:] by A1,FUNCT_1:def 7;
      then q in [:dom f,dom g:] by Def8;
      then consider x1,x2 such that
A3:   x1 in dom f and
A4:   x2 in dom g and
A5:   q = [x1,x2] by ZFMISC_1:def 2;
      [:f,g:].q = [:f,g:].(x1,x2) by A5;
      then
A6:   [f.x1,g.x2] in [:B,A:] by A3,A4,A2,Def8;
      then g.x2 in A by ZFMISC_1:87;
      then
A7:   x2 in g"A by A4,FUNCT_1:def 7;
      f.x1 in B by A6,ZFMISC_1:87;
      then x1 in f"B by A3,FUNCT_1:def 7;
      hence thesis by A5,A7,ZFMISC_1:87;
    end;
    assume q in [:f"B,g"A:];
    then consider x1,x2 such that
A8: x1 in f"B & x2 in g"A and
A9: q = [x1,x2] by ZFMISC_1:def 2;
    f.x1 in B & g.x2 in A by A8,FUNCT_1:def 7;
    then
A10: [f.x1,g.x2] in [:B,A:] by ZFMISC_1:87;
    x1 in dom f & x2 in dom g by A8,FUNCT_1:def 7;
    then
A11: [x1,x2] in [:dom f,dom g:] & [:f,g:].(x1,x2) = [f.x1,g.x2] by Def8,
ZFMISC_1:87;
    [:dom f,dom g:] = dom [:f,g:] by Def8;
    hence thesis by A9,A11,A10,FUNCT_1:def 7;
  end;
  hence thesis by TARSKI:1;
end;

theorem Th74:
  for f being Function of X,Y for g being Function of V,Z holds [:
  f,g:] is Function of [:X,V:],[:Y,Z:]
proof
  let f be Function of X,Y;
  let g be Function of V,Z;
  per cases;
  suppose
A1: [:Y,Z:] = {} implies [:X,V:] = {};
    now
      per cases by A1;
      suppose
A2:     [:Y,Z:] <> {};
        rng f c= Y & rng g c= Z by RELAT_1:def 19;
        then [:rng f,rng g:] c= [:Y,Z:] by ZFMISC_1:96;
        then
A3:     rng [:f,g:] c= [:Y,Z:] by Th67;
        Z = {} implies V = {} by A2,ZFMISC_1:90;
        then
A4:     dom g = V by FUNCT_2:def 1;
        Y = {} implies X = {} by A2,ZFMISC_1:90;
        then dom f = X by FUNCT_2:def 1;
        then dom[:f,g:] = [:X,V:] by A4,Def8;
        hence thesis by A3,FUNCT_2:2;
      end;
      suppose
A5:     [:X,V:] = {};
        then X = {} or V = {};
        then dom f = {} or dom g = {};
        then [:dom f,dom g:] = {} by ZFMISC_1:90;
        then
A6:     dom[:f,g:] = [:X,V:] by A5,Def8;
        rng f c= Y & rng g c= Z by RELAT_1:def 19;
        then [:rng f,rng g:] c= [:Y,Z:] by ZFMISC_1:96;
        then rng[:f,g:] c= [:Y,Z:] by Th67;
        hence thesis by A6,FUNCT_2:2;
      end;
    end;
    hence thesis;
  end;
  suppose
A7: [:Y,Z:] = {} & [:X,V:] <> {};
    then Y = {} or Z = {};
    then f = {} or g = {};
    then [:dom f,dom g:] = {} by RELAT_1:38,ZFMISC_1:90;
    then
A8: dom [:f,g:] = {} by Def8;
    then rng[:f,g:] = {} & dom [:f,g:] c= [:X,V:] by RELAT_1:42,XBOOLE_1:2;
    then reconsider R = [:f,g:] as Relation of [:X,V:],[:Y,Z:] by RELSET_1:4
,XBOOLE_1:2;
    [:f,g:] = {} by A8;
    then R is quasi_total by A7,FUNCT_2:def 1;
    hence thesis;
  end;
end;

definition
  let X1,X2,Y1,Y2;
  let f1 be Function of X1,Y1;
  let f2 be Function of X2,Y2;
  redefine func [:f1,f2:] -> Function of [:X1,X2:],[:Y1,Y2:];
  coherence by Th74;
end;

theorem
  for f1 being Function of C1,D1 for f2 being Function of C2,D2 for c1
being Element of C1 for c2 being Element of C2 holds [:f1,f2:].(c1,c2) = [f1.c1
  ,f2.c2]
proof
  let f1 be Function of C1,D1;
  let f2 be Function of C2,D2;
  let c1 be Element of C1;
  let c2 be Element of C2;
  dom f1 = C1 & dom f2 = C2 by FUNCT_2:def 1;
  hence thesis by Def8;
end;

theorem
  for f1 being Function of X1,Y1 for f2 being Function of X2,Y2 st (Y1 =
{} implies X1 = {}) & (Y2 = {} implies X2 = {}) holds [:f1,f2:] = <:f1*pr1(X1,
  X2),f2*pr2(X1,X2):>
proof
  let f1 be Function of X1,Y1;
  let f2 be Function of X2,Y2;
  assume ( Y1 = {} implies X1 = {})&( Y2 = {} implies X2 = {});
  then dom f1 = X1 & dom f2 = X2 by FUNCT_2:def 1;
  hence thesis by Th66;
end;

theorem
  for f1 being Function of X1,D1 for f2 being Function of X2,D2 holds [:
  f1,f2:] = <:f1*pr1(X1,X2),f2*pr2(X1,X2):>
proof
  let f1 be Function of X1,D1;
  let f2 be Function of X2,D2;
  dom f1 = X1 & dom f2 = X2 by FUNCT_2:def 1;
  hence thesis by Th66;
end;

theorem
  for f1 being Function of X,Y1 for f2 being Function of X,Y2 holds <:f1
  ,f2:> = [:f1,f2:]*(delta X)
proof
  let f1 be Function of X,Y1;
  let f2 be Function of X,Y2;
  per cases;
  suppose
    Y1 = {} or Y2 = {};
    then
A1: dom f1 = {} or dom f2 = {};
A2: dom[:f1,f2:] = [:dom f1, dom f2:] by Def8
      .= {} by A1,ZFMISC_1:90;
    dom<:f1,f2:> = dom f1 /\ dom f2 by Def7
      .= {} by A1;
    hence <:f1,f2:> = {}*delta X .= [:f1,f2:]*delta X by A2,RELAT_1:41;
  end;
  suppose
    Y1 <> {} & Y2 <> {};
    then dom f1 = X & dom f2 = X by FUNCT_2:def 1;
    hence thesis by Th68;
  end;
end;

begin 


theorem
  for f being Function holds pr1(dom f,rng f).:f = dom f
proof
  let f be Function;
  now
    let y be set;
    thus y in dom f implies ex x being set st x in dom pr1(dom f,rng f) & x in
    f & y = pr1(dom f,rng f).x
    proof
      assume
A1:   y in dom f;
      take [y,f.y];
A2:   f.y in rng f by A1,FUNCT_1:def 3;
      then [y,f.y] in [:dom f,rng f:] by A1,ZFMISC_1:87;
      hence [y,f.y] in dom pr1(dom f,rng f) by Def4;
      thus [y,f.y] in f by A1,FUNCT_1:def 2;
      thus y = pr1(dom f,rng f).(y,f.y) by A1,A2,Def4
        .= pr1(dom f,rng f).[y,f.y];
    end;
    given x being set such that
A3: x in dom pr1(dom f,rng f) and
    x in f and
A4: y = pr1(dom f,rng f).x;
    consider x1,x2 being set such that
A5: x1 in dom f & x2 in rng f and
A6: x = [x1,x2] by A3,ZFMISC_1:84;
    y = pr1(dom f,rng f).(x1,x2) by A4,A6;
    hence y in dom f by A5,Def4;
  end;
  hence thesis by FUNCT_1:def 6;
end;

theorem
  for A,B,C being non empty set, f,g being Function of A,[:B,C:] st
  pr1(B,C)*f = pr1(B,C)*g & pr2(B,C)*f = pr2(B,C)*g holds f = g
proof
  let A,B,C be non empty set, f,g be Function of A,[:B,C:] such that
A1: pr1(B,C)*f = pr1(B,C)*g & pr2(B,C)*f = pr2(B,C)*g;
  now
    let a be Element of A;
    consider b1 being Element of B, c1 being Element of C such that
A2: f.a = [b1,c1] by DOMAIN_1:1;
    consider b2 being Element of B, c2 being Element of C such that
A3: g.a = [b2,c2] by DOMAIN_1:1;
A4: pr1(B,C).(g.a) = (pr1(B,C)*g).a by FUNCT_2:15;
A5: pr1(B,C).(f.a) = (pr1(B,C)*f).a & pr2(B,C).(f.a) = (pr2(B,C)*f).a by
FUNCT_2:15;
A6: pr2(B,C).(b1,c1) = c1 & pr2(B,C).(b2,c2) = c2 by Def5;
    pr1(B,C).(b1,c1) = b1 & pr1(B,C).(b2,c2) = b2 by Def4;
    hence f.a = g.a by A1,A2,A3,A6,A5,A4,FUNCT_2:15;
  end;
  hence thesis by FUNCT_2:63;
end;

registration
  let F,G be one-to-one Function;
  cluster [:F,G:] -> one-to-one;
  coherence
proof
  let z1,z2 be set such that
A1: z1 in dom [:F,G:] and
A2: z2 in dom [:F,G:] and
A3: [:F,G:].z1 = [:F,G:].z2;
A4: dom[:F,G:] = [:dom F, dom G:] by Def8;
  then consider x1,y1 being set such that
A5: x1 in dom F and
A6: y1 in dom G and
A7: z1 = [x1,y1] by A1,ZFMISC_1:84;
A8: [:F,G:].(x1,y1) = [F.x1,G.y1] by A5,A6,Def8;
  consider x2,y2 being set such that
A9: x2 in dom F and
A10: y2 in dom G and
A11: z2 = [x2,y2] by A2,A4,ZFMISC_1:84;
A12: [:F,G:].(x2,y2) = [F.x2,G.y2] by A9,A10,Def8;
  then F.x1 = F.x2 by A3,A7,A11,A8,XTUPLE_0:1;
  then
A13: x1 = x2 by A5,A9,FUNCT_1:def 4;
  G.y1 = G.y2 by A3,A7,A11,A8,A12,XTUPLE_0:1;
  hence thesis by A6,A7,A10,A11,A13,FUNCT_1:def 4;
end;
end;

registration let A be set;
 cluster idempotent for BinOp of A;
 existence
  proof
   take pr1(A,A);
   per cases;
   suppose
A1:  A <> {};
   let a be Element of A;
    a in A by A1;
   hence pr1(A,A).(a,a) = a by Def4;
   end;
   suppose
A2:  A = {};
   let a be Element of A;
    not [a,a] in dom pr1(A,A) by A2;
   hence pr1(A,A).(a,a) = {} by FUNCT_1:def 2
       .= a by A2,SUBSET_1:def 1;
   end;
  end;
end;

registration let A be set, b be idempotent BinOp of A;
 let a be Element of A;
 reduce b.(a,a) to a;
 reducibility by BINOP_1:def 4;
end;














begin

reserve f,g,h for Function,
  A for set;

theorem Th1:
  delta A = <:id A, id A:>
proof
  thus delta A = id [:A,A:]*delta A by FUNCT_2:17
    .= [:id A, id A:]*delta A by FUNCT_3:69
    .= <:id A, id A:> by FUNCT_3:78;
end;

reserve F for Function,
  B,x,y,y1,y2,z for set;

definition
  let f;
  func f~ -> Function means
  :Def1:
  dom it = dom f & for x st x in dom f holds
(for y,z st f.x = [y,z] holds it.x = [z,y]) & (f.x = it.x or ex y,z st f.x =[y,
  z]);
  existence
  proof
    defpred P[set,set] means (for y,z st f.$1 = [y,z] holds $2 = [z,y]) & (f.
    $1 = $2 or ex y,z st f.$1 = [y,z]);
A1: now
      let x such that
      x in dom f;
      now
        per cases;
        suppose
A2:       ex y,z st f.x = [y,z];
          then consider y,z such that
A3:       f.x = [y,z];
          take y1 = [z,y];
          thus for y,z st f.x = [y,z] holds y1 = [z,y]
          proof
            let y9,z9 be set;
            assume
A4:         f.x = [y9,z9];
            then z = z9 by A3,XTUPLE_0:1;
            hence thesis by A3,A4,XTUPLE_0:1;
          end;
          thus f.x = y1 or ex y,z st f.x = [y,z] by A2;
        end;
        suppose
A5:       not ex y,z st f.x = [y,z];
          take y1 = f.x;
          thus (for y,z st f.x = [y,z] holds y1 = [z,y]) & (f.x = y1 or ex y,z
          st f.x = [y,z]) by A5;
        end;
      end;
      hence ex y st P[x,y];
    end;
A6: now
      let x,y1,y2 such that
      x in dom f;
      assume that
A7:   P[x,y1] and
A8:   P[x,y2];
      now
        per cases;
        suppose
          ex y,z st f.x = [y,z];
          then consider y,z such that
A9:       f.x = [y,z];
          y1 = [z,y] by A7,A9;
          hence y1 = y2 by A8,A9;
        end;
        suppose
          not ex y,z st f.x = [y,z];
          hence y1 = y2 by A7,A8;
        end;
      end;
      hence y1 = y2;
    end;
    thus ex g st dom g = dom f & for x st x in dom f holds P[x,g.x] from
    FUNCT_1:sch 2(A6,A1);
  end;
  uniqueness
  proof
    defpred P[Function] means for x st x in dom f holds (for y,z st f.x = [y,z
    ] holds $1.x = [z,y]) & (f.x = $1.x or ex y,z st f.x =[y,z]);
    let g1,g2 be Function such that
A10: dom g1 = dom f and
A11: P[g1] and
A12: dom g2 = dom f and
A13: P[g2];
    now
      let x;
      assume
A14:  x in dom f;
      now
        per cases;
        suppose
          ex y,z st f.x = [y,z];
          then consider y,z such that
A15:      f.x = [y,z];
          g1.x = [z,y] by A11,A14,A15;
          hence g1.x = g2.x by A13,A14,A15;
        end;
        suppose
A16:      not ex y,z st f.x = [y,z];
          then g1.x = f.x by A11,A14;
          hence g1.x = g2.x by A13,A14,A16;
        end;
      end;
      hence g1.x = g2.x;
    end;
    hence thesis by A10,A12,FUNCT_1:2;
  end;
  involutiveness
  proof
    let g,f be Function;
    assume that
A17: dom g = dom f and
A18: for x st x in dom f holds (for y,z st f.x = [y,z] holds g.x = [z,
    y]) & (f.x = g.x or ex y,z st f.x =[y,z]);
    thus dom f = dom g by A17;
    let x;
    assume
A19: x in dom g;
    thus for y,z st g.x = [y,z] holds f.x = [z,y]
    proof
      let y,z such that
A20:  g.x = [y,z];
      per cases;
      suppose
        ex z,y st f.x =[z,y];
        then consider y1,y2 such that
A21:    f.x = [y1,y2];
A22:    g.x = [y2,y1] by A17,A18,A19,A21;
        then y = y2 by A20,XTUPLE_0:1;
        hence thesis by A20,A21,A22,XTUPLE_0:1;
      end;
      suppose
        not ex z,y st f.x =[z,y];
        then f.x = g.x by A17,A18,A19;
        hence thesis by A17,A18,A19,A20;
      end;
    end;
    assume g.x <> f.x;
    then consider y,z such that
A23: f.x = [y,z] by A17,A18,A19;
    take z,y;
    thus thesis by A17,A18,A19,A23;
  end;
end;

theorem Th2:
  <:f,g:> = <:g,f:>~
proof
A1: dom <:f,g:> = dom g /\ dom f by FUNCT_3:def 7
    .= dom <:g,f:> by FUNCT_3:def 7;
A2: now
    let x;
    assume
A3: x in dom <:f,g:>;
    then
A4: <:g,f:>.x = [g.x, f.x] by A1,FUNCT_3:def 7;
    thus <:f,g:>.x = [f.x, g.x] by A3,FUNCT_3:def 7
      .= <:g,f:>~.x by A1,A3,A4,Def1;
  end;
  dom <:f,g:> = dom (<:g,f:>~) by A1,Def1;
  hence thesis by A2,FUNCT_1:2;
end;

theorem Th3:
  (f|A)~ = f~|A
proof
A1: dom (f|A) = dom f /\ A by RELAT_1:61
    .= dom (f~) /\ A by Def1
    .= dom (f~|A) by RELAT_1:61;
A2: now
    let x such that
A3: x in dom(f~|A);
A4: dom (f|A) c= dom f by RELAT_1:60;
    now
      per cases;
      suppose
        ex y,z st (f|A).x = [y,z];
        then consider y,z such that
A5:     (f|A).x = [y,z];
A6:     f.x = [y,z] by A1,A3,A5,FUNCT_1:47;
        thus (f|A)~.x = [z,y] by A1,A3,A5,Def1
          .= f~.x by A1,A3,A4,A6,Def1
          .= (f~|A).x by A3,FUNCT_1:47;
      end;
      suppose
A7:     not ex y,z st (f|A).x = [y,z];
A8:     (f|A).x = f.x by A1,A3,FUNCT_1:47;
        (f|A)~.x = (f|A).x by A1,A3,A7,Def1;
        hence (f|A)~.x = f~.x by A1,A3,A4,A7,A8,Def1
          .= (f~|A).x by A3,FUNCT_1:47;
      end;
    end;
    hence (f|A)~.x = (f~|A).x;
  end;
  dom ((f|A)~) = dom (f|A) by Def1;
  hence thesis by A1,A2,FUNCT_1:2;
end;

theorem
  (delta A)~ = delta A
proof
  thus (delta A)~ = <:id A, id A:>~ by Th1
    .= <:id A, id A:> by Th2
    .= delta A by Th1;
end;

theorem Th5:
  <:f,g:>|A = <:f|A,g:>
proof
A1: dom (<:f,g:>|A) = dom <:f,g:> /\ A by RELAT_1:61
    .= dom f /\ dom g /\ A by FUNCT_3:def 7
    .= dom f /\ A /\ dom g by XBOOLE_1:16
    .= dom (f|A) /\ dom g by RELAT_1:61;
  now
A2: dom (<:f,g:>|A) c= dom <:f,g:> by RELAT_1:60;
    let x such that
A3: x in dom (<:f,g:>|A);
A4: x in dom (f|A) by A1,A3,XBOOLE_0:def 4;
    thus (<:f,g:>|A).x = <:f,g:>.x by A3,FUNCT_1:47
      .= [f.x, g.x] by A3,A2,FUNCT_3:def 7
      .= [(f|A).x, g.x] by A4,FUNCT_1:47;
  end;
  hence thesis by A1,FUNCT_3:def 7;
end;

theorem Th6:
  <:f,g:>|A = <:f,g|A:>
proof
  thus <:f,g:>|A = <:g,f:>~|A by Th2
    .= (<:g,f:>|A)~ by Th3
    .= <:g|A,f:>~ by Th5
    .= <:f,g|A:> by Th2;
end;

definition
  let A, z be set;
  func A --> z -> set equals
  [:A, {z}:];
  coherence;
end;

registration
  let A, z be set;
  cluster A --> z -> Function-like Relation-like;
  coherence
  proof
    thus A --> z is Function-like
    proof
      let x,y1,y2;
      assume that
A1:   [x,y1] in A --> z and
A2:   [x,y2] in A --> z;
      y1 in {z} by A1,ZFMISC_1:87;
      then
A3:   y1 = z by TARSKI:def 1;
      y2 in {z} by A2,ZFMISC_1:87;
      hence thesis by A3,TARSKI:def 1;
    end;
    thus for x st x in A --> z ex y1,y2 st [y1,y2] =x by RELAT_1:def 1;
  end;
end;

theorem Th7:
  x in A implies (A --> z).x = z
proof
  assume
A1: x in A;
  z in {z} by TARSKI:def 1;
  then [x,z] in (A --> z) by A1,ZFMISC_1:87;
  hence thesis by FUNCT_1:1;
end;

theorem
  A <> {} implies rng (A --> x) = {x} by RELAT_1:160;

theorem Th9:
  rng f = {x} implies f = (dom f) --> x
proof
  assume
A1: rng f = {x};
  then dom f <> {} by RELAT_1:42;
  then dom((dom f) --> x) = dom f & rng((dom f) -->x) = {x} by RELAT_1:160;
  hence thesis by A1,FUNCT_1:7;
end;

registration
  let x;
  cluster {} --> x -> empty;
  coherence by ZFMISC_1:90;
end;

registration
  let x;
  let A be empty set;
  cluster A --> x -> empty;
  coherence;
end;

registration
  let x;
  let A be non empty set;
  cluster A --> x -> non empty;
  coherence;
end;

theorem
  dom ({} --> x) = {} & rng ({} --> x) = {};

theorem Th11:
  (for z st z in dom f holds f.z = x) implies f = dom f --> x
proof
  assume
A1: for z st z in dom f holds f.z = x;
  now
    per cases;
    suppose
A2:   dom f = {};
      dom ({} --> x) = {};
      hence thesis by A2;
    end;
    suppose
A3:   dom f <> {};
      set z = the Element of dom f;
      now
        let y;
        thus y in {x} implies ex y1 st y1 in dom f & y = f.y1
        proof
          assume y in {x};
          then y = x by TARSKI:def 1;
          then f.z = y by A1,A3;
          hence thesis by A3;
        end;
        assume ex y1 st y1 in dom f & y = f.y1;
        then y = x by A1;
        hence y in {x} by TARSKI:def 1;
      end;
      then rng f = {x} by FUNCT_1:def 3;
      hence thesis by Th9;
    end;
  end;
  hence thesis;
end;

theorem Th12:
  (A --> x)|B = A /\ B --> x
proof
A1: A = {} or A <> {};
A2: A /\ B = {} or A /\ B <> {};
A3: dom ((A --> x)|B) = dom (A --> x) /\ B by RELAT_1:61
    .= A /\ B by A1,RELAT_1:160
    .= dom (A /\ B --> x) by A2,RELAT_1:160;
  now
    let z such that
A4: z in dom (A /\ B --> x);
    A /\ B = {} or A /\ B <> {};
    then
A5: z in A /\ B by A4,RELAT_1:160;
    then
A6: z in A by XBOOLE_0:def 4;
    thus ((A --> x)|B).z = (A --> x).z by A3,A4,FUNCT_1:47
      .= x by A6,Th7
      .= (A /\ B --> x).z by A5,Th7;
  end;
  hence thesis by A3,FUNCT_1:2;
end;

theorem Th13:
  dom (A --> x) = A & rng (A --> x) c= {x}
proof
  now
    per cases;
    suppose
A1:   A = {};
      rng ({} --> x) = {};
      hence thesis by A1,XBOOLE_1:2;
    end;
    suppose
      A <> {};
      hence thesis by RELAT_1:160;
    end;
  end;
  hence thesis;
end;

theorem Th14:
  x in B implies (A --> x)"B = A
proof
  assume
A1: x in B;
  now
    per cases;
    suppose
A2:   A = {};
      thus thesis by A2;
    end;
    suppose
      A <> {};
      then
A3:   rng (A --> x) = {x} by RELAT_1:160;
      {x} c= B by A1,ZFMISC_1:31;
      then {x} /\ B = {x} by XBOOLE_1:28;
      hence (A --> x)"B = (A --> x)"{x} by A3,RELAT_1:133
        .= dom (A -->x) by A3,RELAT_1:134
        .= A by Th13;
    end;
  end;
  hence thesis;
end;

theorem
  (A --> x)"{x} = A
proof
  x in {x} by TARSKI:def 1;
  hence thesis by Th14;
end;

theorem
  not x in B implies (A --> x)"B = {}
proof
  assume
A1: not x in B;
  rng (A --> x) c= {x} by Th13;
  then rng (A --> x) misses B by A1,XBOOLE_1:63,ZFMISC_1:50;
  hence thesis by RELAT_1:138;
end;

theorem
  x in dom h implies h*(A --> x) = A --> h.x
proof
  assume
A1: x in dom h;
A2: now
    let z;
    assume
A3: z in dom (h*(A --> x));
    then z in dom (A --> x) by FUNCT_1:11;
    then
A4: z in A by Th13;
    thus (h*(A --> x)).z = h.((A --> x).z) by A3,FUNCT_1:12
      .= h.x by A4,Th7
      .= (A --> h.x).z by A4,Th7;
  end;
  dom (h*(A --> x)) = (A --> x)"dom h by RELAT_1:147
    .= A by A1,Th14
    .= dom (A --> h.x) by Th13;
  hence thesis by A2,FUNCT_1:2;
end;

theorem
  A <> {} & x in dom h implies dom(h*(A --> x)) <> {}
proof
  assume that
A1: A <> {} and
A2: x in dom h;
  set y = the Element of A;
  y in A by A1;
  then
A3: y in dom (A -->x) by Th13;
  (A --> x).y = x by A1,Th7;
  hence thesis by A2,A3,FUNCT_1:11;
end;

theorem
  (A --> x)*h = h"A --> x
proof
A1: dom ((A --> x)*h) = h"dom(A --> x) by RELAT_1:147
    .= h"A by Th13;
A2: now
    let z;
    assume
A3: z in dom ((A --> x)*h);
    then h.z in dom (A --> x) by FUNCT_1:11;
    then
A4: h.z in A by Th13;
    thus ((A --> x)*h).z = (A --> x).(h.z) by A3,FUNCT_1:12
      .= x by A4,Th7
      .= (h"A --> x).z by A1,A3,Th7;
  end;
  dom ((A --> x)*h) = dom (h"A --> x) by A1,Th13;
  hence thesis by A2,FUNCT_1:2;
end;

theorem
  (A --> [x,y])~ = A --> [y,x]
proof
A1: dom ((A --> [x,y])~) = dom (A --> [x,y]) by Def1;
  then
A2: dom ((A --> [x,y])~) = A by Th13;
A3: now
    let z;
    assume
A4: z in dom ((A --> [x,y])~);
    then (A --> [x,y]).z = [x,y] by A2,Th7;
    hence ((A --> [x,y])~).z = [y,x] by A1,A4,Def1
      .= (A --> [y,x]).z by A2,A4,Th7;
  end;
  dom ((A --> [x,y])~) = dom (A --> [y,x]) by A2,Th13;
  hence thesis by A3,FUNCT_1:2;
end;

definition
  let F,f,g;
  func F.:(f,g) -> set equals
  F * <:f,g:>;
  correctness;
end;

registration
  let F,f,g;
  cluster F.:(f,g) -> Function-like Relation-like;
  coherence;
end;

Lm1: x in dom (F*<:f,g:>) implies (F*<:f,g:>).x = F.(f.x,g.x)
proof
  assume
A1: x in dom (F*<:f,g:>);
  then
A2: x in dom <:f,g:> by FUNCT_1:11;
  thus (F*<:f,g:>).x = F.(<:f,g:>.x) by A1,FUNCT_1:12
    .= F.(f.x,g.x) by A2,FUNCT_3:def 7;
end;

theorem
  for h st dom h = dom(F.:(f,g)) & for z being set st z in dom (F.:(f,g)
  ) holds h.z = F.(f.z,g.z) holds h = F.:(f,g)
proof
  let h;
  assume that
A1: dom h = dom(F.:(f,g)) and
A2: for z being set st z in dom (F.:(f,g)) holds h.z = F.(f.z,g.z);
  now
    let z be set;
    assume
A3: z in dom (F.:(f,g));
    hence h.z = F.(f.z,g.z) by A2
      .= (F.:(f,g)).z by A3,Lm1;
  end;
  hence thesis by A1,FUNCT_1:2;
end;

theorem
  x in dom (F.:(f,g)) implies (F.:(f,g)).x = F.(f.x,g.x) by Lm1;

theorem Th23:
  f|A = g|A implies (F.:(f,h))|A = (F.:(g,h))|A
proof
  assume
A1: f|A = g|A;
  thus (F.:(f,h))|A = F*<:f,h:>|A by RELAT_1:83
    .= F*<:f|A,h:> by Th5
    .= F*<:g,h:>|A by A1,Th5
    .= (F.:(g,h))|A by RELAT_1:83;
end;

theorem Th24:
  f|A = g|A implies (F.:(h,f))|A = (F.:(h,g))|A
proof
  assume
A1: f|A = g|A;
  thus (F.:(h,f))|A = F*<:h,f:>|A by RELAT_1:83
    .= F*<:h,f|A:> by Th6
    .= F*<:h,g:>|A by A1,Th6
    .= (F.:(h,g))|A by RELAT_1:83;
end;

theorem Th25:
  F.:(f,g)*h = F.:(f*h, g*h)
proof
  thus F.:(f,g)*h = F*(<:f,g:>*h) by RELAT_1:36
    .= F.:(f*h, g*h) by FUNCT_3:55;
end;

definition
  let F,f,x;
  func F[:](f,x) -> set equals
  F * <:f, dom f --> x:>;
  correctness;
end;

registration
  let F,f,x;
  cluster F[:](f,x) -> Function-like Relation-like;
  coherence;
end;

theorem
  F[:](f,x) = F.:(f, dom f --> x);

theorem Th27:
  x in dom (F[:](f,z)) implies (F[:](f,z)).x = F.(f.x,z)
proof
A1: dom <:f, dom f --> z:> = dom f /\ dom (dom f --> z) by FUNCT_3:def 7;
  assume
A2: x in dom (F[:](f,z));
  then x in dom <:f, dom f --> z:> by FUNCT_1:11;
  then
A3: x in dom f by A1,XBOOLE_0:def 4;
  thus (F[:](f,z)).x = F.(f.x,(dom f --> z).x) by A2,Lm1
    .= F.(f.x,z) by A3,Th7;
end;

theorem
  f|A = g|A implies (F[:](f,x))|A = (F[:](g,x))|A
proof
  assume
A1: f|A = g|A;
  dom f /\ A = dom (f|A) by RELAT_1:61
    .= dom g /\ A by A1,RELAT_1:61;
  then
A2: (dom f --> x)|A = dom g /\ A --> x by Th12
    .= (dom g -->x)|A by Th12;
  thus (F[:](f,x))|A = (F.:(f, dom f --> x))|A
    .= (F.:(g, dom f --> x))|A by A1,Th23
    .= (F.:(g, dom g --> x))|A by A2,Th24
    .= (F[:](g,x))|A;
end;

theorem Th29:
  F[:](f,x)*h = F[:](f*h,x)
proof
A1: dom (dom f -->x) = dom f by Th13;
  then
A2: dom ((dom f --> x)*h) = dom (f*h) by RELAT_1:163;
A3: now
    let z;
    assume
A4: z in dom ((dom f --> x)*h);
    then
A5: h.z in dom(dom f -->x) by FUNCT_1:11;
    thus ((dom f --> x)*h).z = (dom f --> x).(h.z) by A4,FUNCT_1:12
      .= x by A1,A5,Th7;
  end;
  thus F[:](f,x)*h = F.:(f, dom f --> x)*h
    .= F.:(f*h, (dom f --> x)*h) by Th25
    .= F[:](f*h,x) by A2,A3,Th11;
end;

theorem
  F[:](f,x)*id A = F[:](f|A,x)
proof
  thus F[:](f,x)*id A = F[:](f*id A, x) by Th29
    .= F[:](f|A, x) by RELAT_1:65;
end;

definition
  let F,x,g;
  func F[;](x,g) -> set equals
  F * <:dom g --> x, g:>;
  correctness;
end;

registration
  let F,x,g;
  cluster F[;](x,g) -> Function-like Relation-like;
  coherence;
end;

theorem
  F[;](x,g) = F.:(dom g --> x, g);

theorem Th32:
  x in dom (F[;](z,f)) implies (F[;](z,f)).x = F.(z,f.x)
proof
A1: dom <:dom f --> z, f:> = dom (dom f --> z) /\ dom f by FUNCT_3:def 7;
  assume
A2: x in dom (F[;](z,f));
  then x in dom <:dom f --> z, f:> by FUNCT_1:11;
  then
A3: x in dom f by A1,XBOOLE_0:def 4;
  thus (F[;](z,f)).x = F.((dom f --> z).x, f.x) by A2,Lm1
    .= F.(z, f.x) by A3,Th7;
end;

theorem
  f|A = g|A implies (F[;](x,f))|A = (F[;](x,g))|A
proof
  assume
A1: f|A = g|A;
  dom f /\ A = dom (f|A) by RELAT_1:61
    .= dom g /\ A by A1,RELAT_1:61;
  then
A2: (dom f --> x)|A = dom g /\ A --> x by Th12
    .= (dom g -->x)|A by Th12;
  thus (F[;](x,f))|A = (F.:(dom f --> x, f))|A
    .= (F.:(dom f --> x, g))|A by A1,Th24
    .= (F.:(dom g --> x, g))|A by A2,Th23
    .= (F[;](x,g))|A;
end;

theorem Th34:
  F[;](x,f)*h = F[;](x,f*h)
proof
A1: dom (dom f -->x) = dom f by Th13;
  then
A2: dom ((dom f --> x)*h) = dom (f*h) by RELAT_1:163;
A3: now
    let z;
    assume
A4: z in dom ((dom f --> x)*h);
    then
A5: h.z in dom(dom f -->x) by FUNCT_1:11;
    thus ((dom f --> x)*h).z = (dom f --> x).(h.z) by A4,FUNCT_1:12
      .= x by A1,A5,Th7;
  end;
  thus F[;](x,f)*h = F.:(dom f --> x, f)*h
    .= F.:((dom f --> x)*h, f*h) by Th25
    .= F[;](x,f*h) by A2,A3,Th11;
end;

theorem
  F[;](x,f)*id A = F[;](x,f|A)
proof
  thus F[;](x,f)*id A = F[;](x,f*id A) by Th34
    .= F[;](x,f|A) by RELAT_1:65;
end;

reserve X for non empty set,
  Y for set,
  F for BinOp of X,
  f,g,h for Function
  of Y,X,
  x,x1,x2 for Element of X;

theorem Th36:
  F.:(f,g) is Function of Y,X
proof
  F*<:f,g:> is Function of Y,X;
  hence thesis;
end;

definition
  let X be non empty set, Z be set;
  let F be BinOp of X, f,g be Function of Z,X;
  redefine func F.:(f,g) -> Function of Z,X;
  coherence by Th36;
end;

reserve Y for non empty set,
  F for BinOp of X,
  f,g,h for Function of Y,X,
  x,
  x1,x2 for Element of X;

theorem Th37:
  for z being Element of Y holds (F.:(f,g)).z = F.(f.z,g.z)
proof
  let z be Element of Y;
  dom (F.:(f,g)) = Y by FUNCT_2:def 1;
  hence thesis by Lm1;
end;

theorem Th38:
  for h being Function of Y,X holds (for z being Element of Y
  holds h.z = F.(f.z,g.z)) implies h = F.:(f,g)
proof
  let h be Function of Y,X;
  assume
A1: for z being Element of Y holds h.z = F.(f.z,g.z);
  now
    let z be Element of Y;
    thus h.z = F.(f.z,g.z) by A1
      .= (F.:(f,g)).z by Th37;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  for g being Function of X,X holds F.:(id X, g)*f = F.:(f,g*f)
proof
  let g be Function of X,X;
  thus F.:(id X, g)*f = F.:(id X*f, g*f) by Th25
    .= F.:(f,g*f) by FUNCT_2:17;
end;

theorem
  for g being Function of X,X holds F.:(g, id X)*f = F.:(g*f,f)
proof
  let g be Function of X,X;
  thus F.:(g, id X)*f = F.:(g*f, id X*f) by Th25
    .= F.:(g*f, f) by FUNCT_2:17;
end;

theorem
  F.:(id X, id X)*f = F.:(f,f)
proof
  thus F.:(id X, id X)*f = F.:(id X*f, id X*f) by Th25
    .= F.:(id X*f, f) by FUNCT_2:17
    .= F.:(f,f) by FUNCT_2:17;
end;

theorem
  for g being Function of X,X holds F.:(id X, g).x = F.(x,g.x)
proof
  let g be Function of X,X;
  thus F.:(id X, g).x = F.(id X.x, g.x) by Th37
    .= F.(x,g.x) by FUNCT_1:18;
end;

theorem
  for g being Function of X,X holds F.:(g, id X).x = F.(g.x,x)
proof
  let g be Function of X,X;
  thus F.:(g, id X).x = F.(g.x, id X.x) by Th37
    .= F.(g.x, x) by FUNCT_1:18;
end;

theorem
  F.:(id X, id X).x = F.(x,x)
proof
  thus F.:(id X, id X).x = F.(id X.x, id X.x) by Th37
    .= F.(id X.x, x) by FUNCT_1:18
    .= F.(x,x) by FUNCT_1:18;
end;

theorem Th45:
  for A,B for x being set st x in B holds A --> x is Function of A
  , B
proof
  let A,B;
  let x be set;
A1: rng (A --> x) c= {x} by Th13;
A2: dom (A --> x) = A by Th13;
  assume
A3: x in B;
  then {x} c= B by ZFMISC_1:31;
  then rng (A --> x) c= B by A1,XBOOLE_1:1;
  hence thesis by A3,A2,FUNCT_2:def 1,RELSET_1:4;
end;

definition
  let I,i be set;
  redefine func I --> i -> Function of I,{i};
  coherence
  proof
    dom (I --> i) = I & rng (I --> i) c= {i} by Th13;
    hence thesis by FUNCT_2:def 1,RELSET_1:4;
  end;
end;

definition
  let B be non empty set, A be set, b be Element of B;
  redefine func A --> b -> Function of A,B;
  coherence by Th45;
end;

theorem
  for A,X,x holds A --> x is Function of A, X;

reserve Y for set,
  F for BinOp of X,
  f,g,h for Function
  of Y,X,
  x,x1,x2 for Element of X;

theorem Th47:
  F[:](f,x) is Function of Y,X
proof
  dom f = Y by FUNCT_2:def 1;
  then reconsider g = dom f --> x as Function of Y,X;
  F*<:f,g:> is Function of Y,X;
  hence thesis;
end;

definition
  let X be non empty set, Z be set;
  let F be BinOp of X, f be Function of Z,X, x be Element of X;
  redefine func F[:](f,x) -> Function of Z,X;
  coherence by Th47;
end;

reserve Y for non empty set,
  F for BinOp of X,
  f,g,h for Function of Y,X,
  x,
  x1,x2 for Element of X;

theorem Th48:
  for y being Element of Y holds (F[:](f,x)).y = F.(f.y,x)
proof
  let y be Element of Y;
  dom (F[:](f,x)) = Y by FUNCT_2:def 1;
  hence thesis by Th27;
end;

theorem Th49:
  (for y being Element of Y holds g.y = F.(f.y,x)) implies g = F
  [:](f,x)
proof
  assume
A1: for y being Element of Y holds g.y = F.(f.y,x);
  now
    let y be Element of Y;
    thus g.y = F.(f.y,x) by A1
      .= (F[:](f,x)).y by Th48;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  F[:](id X, x)*f = F[:](f,x)
proof
  thus F[:](id X, x)*f = F[:](id X*f, x) by Th29
    .= F[:](f,x) by FUNCT_2:17;
end;

theorem
  F[:](id X, x).x = F.(x,x)
proof
  thus F[:](id X, x).x = F.(id X.x, x) by Th48
    .= F.(x,x) by FUNCT_1:18;
end;

reserve Y for set,
  F for BinOp of X,
  f,g,h for Function
  of Y,X,
  x,x1,x2 for Element of X;

theorem Th52:
  F[;](x,g) is Function of Y,X
proof
  dom g = Y by FUNCT_2:def 1;
  then reconsider f = dom g --> x as Function of Y,X;
  F*<:f,g:> is Function of Y,X;
  hence thesis;
end;

definition
  let X be non empty set, Z be set;
  let F be BinOp of X, x be Element of X;
  let g be Function of Z,X;
  redefine func F[;](x,g) -> Function of Z,X;
  coherence by Th52;
end;

reserve Y for non empty set,
  F for BinOp of X,
  f,g,h for Function of Y,X,
  x,
  x1,x2 for Element of X;

theorem Th53:
  for y being Element of Y holds (F[;](x,f)).y = F.(x,f.y)
proof
  let y be Element of Y;
  dom (F[;](x,f)) = Y by FUNCT_2:def 1;
  hence thesis by Th32;
end;

theorem Th54:
  (for y being Element of Y holds g.y = F.(x,f.y)) implies g = F
  [;](x,f)
proof
  assume
A1: for y being Element of Y holds g.y = F.(x,f.y);
  now
    let y be Element of Y;
    thus g.y = F.(x,f.y) by A1
      .= (F[;](x,f)).y by Th53;
  end;
  hence thesis by FUNCT_2:63;
end;

reserve Y for set,
  F for BinOp of X,
  f,g,h for Function
  of Y,X,
  x,x1,x2 for Element of X;

theorem
  F[;](x, id X)*f = F[;](x,f)
proof
  thus F[;](x, id X)*f = F[;](x, id X*f) by Th34
    .= F[;](x,f) by FUNCT_2:17;
end;

theorem
  F[;](x, id X).x = F.(x,x)
proof
  thus F[;](x, id X).x = F.(x, id X.x) by Th53
    .= F.(x,x) by FUNCT_1:18;
end;

theorem
  for X,Y,Z being non empty set for f being Function of X, [:Y,Z:] for x
  being Element of X holds f~.x =[(f.x)`2,(f.x)`1]
proof
  let X,Y,Z be non empty set;
  let f be Function of X, [:Y,Z:];
  let x be Element of X;
  x in X;
  then
A1: x in dom f by FUNCT_2:def 1;
  f.x = [(f.x)`1, (f.x)`2] by MCART_1:22;
  hence thesis by A1,Def1;
end;

definition
  let X,Y,Z be non empty set;
  let f be Function of X, [:Y,Z:];
  redefine func rng f -> Relation of Y,Z;
  coherence by RELAT_1:def 19;
end;

definition
  let X,Y,Z be non empty set;
  let f be Function of X, [:Y,Z:];
  redefine func f~ -> Function of X, [:Z,Y:];
  coherence
  proof
A1: rng (f~) c= [:Z,Y:]
    proof
      let x be set;
      assume x in rng (f~);
      then consider y being set such that
A2:   y in dom (f~) and
A3:   x = f~.y by FUNCT_1:def 3;
A4:   y in dom f by A2,Def1;
      then reconsider y as Element of X;
A5:   f.y = [(f.y)`1,(f.y)`2] by MCART_1:21;
      then f~.y = [(f.y)`2,(f.y)`1] by A4,Def1;
      hence thesis by A3,A5,ZFMISC_1:88;
    end;
    X = dom f by FUNCT_2:def 1
      .= dom (f~) by Def1;
    hence thesis by A1,FUNCT_2:def 1,RELSET_1:4;
  end;
end;

theorem
  for X,Y,Z being non empty set for f being Function of X, [:Y,Z:] holds
  rng (f~) = (rng f)~
proof
  let X,Y,Z be non empty set;
  let f be Function of X, [:Y,Z:];
  let x,y be set;
  thus [x,y] in rng (f~) implies [x,y] in (rng f)~
  proof
    assume [x,y] in rng (f~);
    then consider z being set such that
A1: z in dom (f~) and
A2: [x,y] = f~.z by FUNCT_1:def 3;
A3: z in dom f by A1,Def1;
    f.z = f~~.z
      .= [y,x] by A1,A2,Def1;
    then [y,x] in rng f by A3,FUNCT_1:def 3;
    hence thesis by RELAT_1:def 7;
  end;
  assume [x,y] in (rng f)~;
  then [y,x] in rng f by RELAT_1:def 7;
  then consider z being set such that
A4: z in dom f & [y,x] = f.z by FUNCT_1:def 3;
  z in dom (f~) & f~.z = [x,y] by A4,Def1;
  hence thesis by FUNCT_1:def 3;
end;

reserve y for Element of Y;

theorem
  F is associative implies F[:](F[;](x1,f),x2) = F[;](x1,F[:](f,x2))
proof
  assume
A1: F is associative;
  per cases;
  suppose
    Y = {};
    hence thesis;
  end;
  suppose
A2: Y <> {};
    now
      let y;
      reconsider x3 = f.y as Element of X by A2,FUNCT_2:5;
      thus (F[:](F[;](x1,f),x2)).y = F.((F[;](x1,f)).y,x2) by A2,Th48
        .= F.(F.(x1,x3),x2) by A2,Th53
        .= F.(x1,F.(x3,x2)) by A1,BINOP_1:def 3
        .= F.(x1,(F[:](f,x2)).y) by A2,Th48;
    end;
    hence thesis by A2,Th54;
  end;
end;

theorem
  F is associative implies F.:(F[:](f,x),g) = F.:(f,F[;](x,g))
proof
  assume
A1: F is associative;
  per cases;
  suppose
    Y = {};
    hence thesis;
  end;
  suppose
A2: Y <> {};
    now
      let y;
      reconsider x1 = f.y, x2 = g.y as Element of X by A2,FUNCT_2:5;
      thus (F.:(F[:](f,x),g)).y = F.((F[:](f,x)).y,g.y) by A2,Th37
        .= F.(F.(x1,x),x2) by A2,Th48
        .= F.(x1,F.(x,x2)) by A1,BINOP_1:def 3
        .= F.(f.y,(F[;](x,g)).y) by A2,Th53;
    end;
    hence thesis by A2,Th38;
  end;
end;

theorem
  F is associative implies F.:(F.:(f,g),h) = F.:(f,F.:(g,h))
proof
  assume
A1: F is associative;
  per cases;
  suppose
    Y = {};
    hence thesis;
  end;
  suppose
A2: Y <> {};
    now
      let y;
      reconsider x1 = f.y, x2 = g.y, x3 = h.y as Element of X by A2,FUNCT_2:5;
      thus (F.:(F.:(f,g),h)).y = F.((F.:(f,g)).y,h.y) by A2,Th37
        .= F.(F.(f.y,g.y),h.y) by A2,Th37
        .= F.(x1,F.(x2,x3)) by A1,BINOP_1:def 3
        .= F.(f.y,(F.:(g,h)).y) by A2,Th37;
    end;
    hence thesis by A2,Th38;
  end;
end;

theorem
  F is associative implies F[;](F.(x1,x2),f) = F[;](x1,F[;](x2,f))
proof
  assume
A1: F is associative;
  per cases;
  suppose
    Y = {};
    hence thesis;
  end;
  suppose
A2: Y <> {};
    now
      let y;
      reconsider x3 = f.y as Element of X by A2,FUNCT_2:5;
      thus (F[;](F.(x1,x2),f)).y = F.(F.(x1,x2),f.y) by A2,Th53
        .= F.(x1,F.(x2,x3)) by A1,BINOP_1:def 3
        .= F.(x1,(F[;](x2,f)).y) by A2,Th53;
    end;
    hence thesis by A2,Th54;
  end;
end;

theorem
  F is associative implies F[:](f, F.(x1,x2)) = F[:](F[:](f,x1),x2)
proof
  assume
A1: F is associative;
  per cases;
  suppose
    Y = {};
    hence thesis;
  end;
  suppose
A2: Y <> {};
    now
      let y;
      reconsider x3 = f.y as Element of X by A2,FUNCT_2:5;
      thus (F[:](f, F.(x1,x2))).y = F.(f.y, F.(x1,x2)) by A2,Th48
        .= F.(F.(x3,x1),x2) by A1,BINOP_1:def 3
        .= F.(F[:](f,x1).y,x2) by A2,Th48;
    end;
    hence thesis by A2,Th49;
  end;
end;

theorem
  F is commutative implies F[;](x,f) = F[:](f,x)
proof
  assume
A1: F is commutative;
  per cases;
  suppose
    Y = {};
    hence thesis;
  end;
  suppose
A2: Y <> {};
    now
      let y;
      reconsider x1 = f.y as Element of X by A2,FUNCT_2:5;
      thus (F[;](x,f)).y = F.(x,x1) by A2,Th53
        .= F.(f.y,x) by A1,BINOP_1:def 2;
    end;
    hence thesis by A2,Th49;
  end;
end;

theorem
  F is commutative implies F.:(f,g) = F.:(g,f)
proof
  assume
A1: F is commutative;
  per cases;
  suppose
    Y = {};
    hence thesis;
  end;
  suppose
A2: Y <> {};
    now
      let y;
      reconsider x1 = f.y, x2 = g.y as Element of X by A2,FUNCT_2:5;
      thus (F.:(f,g)).y = F.(x1,x2) by A2,Th37
        .= F.(g.y,f.y) by A1,BINOP_1:def 2;
    end;
    hence thesis by A2,Th38;
  end;
end;

theorem
  F is idempotent implies F.:(f,f) = f
proof
  assume
A1: F is idempotent;
  per cases;
  suppose
A2: Y = {};
    hence F.:(f,f) = {}
      .= f by A2;
  end;
  suppose
A3: Y <> {};
    now
      let y;
      reconsider x = f.y as Element of X by A3,FUNCT_2:5;
      thus f.y = F.(x,x) by A1
        .= F.(f.y,f.y);
    end;
    hence thesis by A3,Th38;
  end;
end;

reserve Y for non empty set,
  F for BinOp of X,
  f for Function of Y,X,
  x for
  Element of X,
  y for Element of Y;

theorem
  F is idempotent implies F[;](f.y,f).y = f.y
proof
  assume
A1: F is idempotent;
  thus F[;](f.y,f).y = F.(f.y,f.y) by Th53
    .= f.y by A1;
end;

theorem
  F is idempotent implies F[:](f,f.y).y = f.y
proof
  assume
A1: F is idempotent;
  thus F[:](f,f.y).y = F.(f.y,f.y) by Th48
    .= f.y by A1;
end;



theorem
  for F,f,g being Function st [:rng f, rng g:] c= dom F holds dom(F.:(f,
  g)) = dom f /\ dom g
proof
  let F,f,g be Function such that
A1: [:rng f, rng g:] c= dom F;
  rng<:f,g:> c= [:rng f, rng g:] by FUNCT_3:51;
  hence dom(F.:(f,g)) = dom<:f,g:> by A1,RELAT_1:27,XBOOLE_1:1
    .= dom f /\ dom g by FUNCT_3:def 7;
end;



definition
  let IT be Function;
  attr IT is Function-yielding means
  :Def6:
  for x being set st x in dom IT holds IT.x is Function;
end;

registration
  cluster Function-yielding for Function;
  existence
  proof
    take the set --> the Function;
    let x be set;
    thus thesis;
  end;
end;

registration
  let B be Function-yielding Function, j be set;
  cluster B.j -> Function-like Relation-like;
  coherence
  proof
    per cases;
    suppose
      j in dom B;
      hence thesis by Def6;
    end;
    suppose
      not j in dom B;
      hence thesis by FUNCT_1:def 2;
    end;
  end;
end;

registration
  let F be Function-yielding Function, f be Function;
  cluster F * f -> Function-yielding;
  coherence
  proof
    thus F * f is Function-yielding
    proof
      let x be set;
      assume x in dom (F*f);
      then (F*f).x = F.(f.x) by FUNCT_1:12;
      hence thesis;
    end;
  end;
end;



registration
  let B;
  let c be non empty set;
  cluster B --> c -> non-empty;
  coherence
  proof
    not {} in rng(B --> c) by TARSKI:def 1;
    hence thesis by RELAT_1:def 9;
  end;
end;



theorem
  ([:X,Y:] --> z).(x,y) = z
proof
  [x,y] in [:X,Y:] by ZFMISC_1:87;
  hence thesis by Th7;
end;



reserve a,b,c for set;

definition
  let a,b,c;
  func (a,b).-->c -> Function equals
  {[a,b]} --> c;
  coherence;
end;

theorem
  ((a,b).-->c).(a,b) = c
proof
  [a,b] in {[a,b]} by TARSKI:def 1;
  hence thesis by Th7;
end;



definition
  let x,y,a,b be set;
  func IFEQ(x,y,a,b) -> set equals
  :Def8:
  a if x = y otherwise b;
  correctness;
end;

definition
  let D be set;
  let x,y be set, a,b be Element of D;
  redefine func IFEQ(x,y,a,b) -> Element of D;
  coherence
  proof
    x = y or x <> y;
    hence thesis by Def8;
  end;
end;

definition
  let x,y be set;
  func x .--> y -> set equals
  {x} --> y;
  coherence;
end;

registration
  let x,y be set;
  cluster x .--> y -> Function-like Relation-like;
  coherence;
end;

registration
  let x,y be set;
  cluster x .--> y -> one-to-one;
  coherence
  proof
    let x1,x2 be set;
    set f = x .--> y;
    assume that
A1: x1 in dom f and
A2: x2 in dom f;
A3: dom f = {x} by Th13;
    then x1 =x by A1,TARSKI:def 1;
    hence thesis by A3,A2,TARSKI:def 1;
  end;
end;

theorem Th72:
  for x,y be set holds (x .--> y).x = y
proof
  let x,y be set;
  x in {x} by TARSKI:def 1;
  hence thesis by Th7;
end;



theorem
  for a,b being set, f being Function holds a.-->b c= f iff a in dom f &
  f.a = b
proof
  let a,b be set, f be Function;
A1: dom(a.-->b) = {a} by Th13;
  then
A2: a in dom(a.-->b) by TARSKI:def 1;
  hereby
    assume
A3: a.-->b c= f;
    then {a} c= dom f by A1,GRFUNC_1:2;
    hence a in dom f by ZFMISC_1:31;
    thus f.a = (a.-->b).a by A2,A3,GRFUNC_1:2
      .= b by Th72;
  end;
  assume that
A4: a in dom f and
A5: f.a = b;
A6: now
    let x be set;
    assume x in dom(a.-->b);
    then x = a by A1,TARSKI:def 1;
    hence (a.-->b).x = f.x by A5,Th72;
  end;
  dom(a.-->b) c= dom f by A1,A4,ZFMISC_1:31;
  hence thesis by A6,GRFUNC_1:2;
end;



notation
  let o,m,r be set;
  synonym (o,m) :-> r for (o,m) .--> r;
end;

Lm2: for o,m,r be set holds (o,m) :-> r is Function of [:{o},{m}:],{r}
proof
  let o,m,r be set;
  dom((o,m) :-> r) = {[o,m]} by Th13
    .= [:{o},{m}:] by ZFMISC_1:29;
  hence thesis by FUNCT_2:def 1;
end;

definition
  let o,m,r be set;
  redefine func (o,m) :-> r -> Function of [:{o},{m}:],{r} means
  not contradiction;
  coherence by Lm2;
  compatibility
  proof
    let f be Function of [:{o},{m}:],{r};
    (o,m) .--> r is Function of [:{o},{m}:],{r} by Lm2;
    hence thesis by FUNCT_2:51;
  end;
end;



reserve x,y,z for set;

theorem
  x in dom(x .--> y)
proof
  dom(x .--> y) = {x} by Th13;
  hence thesis by TARSKI:def 1;
end;

theorem
  z in dom(x .--> y) implies z = x
proof
  dom(x .--> y) = {x} by Th13;
  hence thesis by TARSKI:def 1;
end;



theorem
  not x in A implies (x .--> y)|A = {}
proof
  assume not x in A;
  then {x} misses A by ZFMISC_1:50;
  then dom(x .--> y) misses A by Th13;
  hence thesis by RELAT_1:66;
end;



notation
  let x,y;
  synonym x :-> y for x .--> y;
end;

definition
  let m,o be set;
  redefine func m :-> o -> Function of {m}, {o};
  coherence;
end;

theorem
  for x being Element of {a} for y being Element of {b} holds ((a,b):->c
  ).(x,y) = c by TARSKI:def 1;



registration
  let f be Function-yielding Function, C be set;
  cluster f|C -> Function-yielding;
  coherence
  proof
    let i be set;
    f.i is Function;
    hence thesis by FUNCT_1:47;
  end;
end;



registration
  let A be set;
  let f be Function;
  cluster A --> f -> Function-yielding;
  coherence
  proof
    let a be set;
    thus thesis;
  end;
end;



registration
  let X,a be set;
  cluster X --> a -> constant;
  coherence
  proof
    let x,y be set;
    assume that
A1: x in dom(X --> a) and
A2: y in dom(X --> a);
    thus (X --> a).x = a by A1,Th7
      .= (X --> a).y by A2,Th7;
  end;
end;


registration
  cluster non empty constant for Function;
  existence
  proof
    take {{}} --> {};
    thus thesis;
  end;
end;


registration
  let f be constant Function, X be set;
  cluster f|X -> constant;
  coherence
  proof
    let x,y be set;
A1: dom(f|X) c= dom f by RELAT_1:60;
    assume that
A2: x in dom(f|X) and
A3: y in dom (f|X);
    thus (f|X).x = f.x by A2,FUNCT_1:47
      .= f.y by A1,A2,A3,FUNCT_1:def 10
      .= (f|X).y by A3,FUNCT_1:47;
  end;
end;


theorem
  for f being non empty constant Function ex y st for x st x in dom f
  holds f.x = y
proof
  let f be non empty constant Function;
  consider y such that
A1: y in rng f by XBOOLE_0:def 1;
  take y;
  ex x0 being set st x0 in dom f & f.x0 = y by A1,FUNCT_1:def 3;
  hence thesis by FUNCT_1:def 10;
end;


theorem
  for X being non empty set, x being set holds the_value_of (X --> x) =
  x
proof
  let X be non empty set, x be set;
  set f = X --> x;
  ex i being set st i in dom f & the_value_of f = f.i by FUNCT_1:def 12;
  hence thesis by Th7;
end;



theorem
  for f being constant Function holds f = (dom f) --> the_value_of f
proof
  let f be constant Function;
  thus dom ((dom f) --> the_value_of f) = dom f by Th13;
  let x be set;
  assume
A1: x in dom f;
  then f <> {} & ((dom f) --> the_value_of f).x = the_value_of f by Th7;
  hence thesis by A1,FUNCT_1:def 12;
end;



registration
  let X be set, Y be non empty set;
  cluster total for PartFunc of X,Y;
  existence
  proof
    consider y being set such that
A1: y in Y by XBOOLE_0:def 1;
    reconsider y as Element of Y by A1;
    take X --> y;
    thus thesis;
  end;
end;



registration
  let I, A be set;
  cluster I --> A -> I-defined;
  coherence;
end;

registration
  let I, A be set;
  cluster I .--> A -> {I}-defined;
  coherence;
end;



theorem
  (A --> x).:B c= {x};

registration
  let I be set, f be Function;
  cluster I .--> f -> Function-yielding;
  coherence;
end;



registration let I be set;
 cluster total for I-defined non-empty Function;
 existence
  proof
   take I --> 1;
    thus thesis;
  end;
end;

theorem Th82:
  x .--> y is_isomorphism_of {[x,x]},{[y,y]}
proof
  set F = x .--> y;
  set R = {[x,x]};
  set S = {[y,y]};
A1: field R = {x} by RELAT_1:173;
  hence dom F = field R by Th13;
  field S = {y} by RELAT_1:173;
  hence rng F = field S by RELAT_1:160;
  thus F is one-to-one;
  let a,b be set;
  hereby
    assume [a,b] in R;
    then [a,b] = [x,x] by TARSKI:def 1;
    then
A2: a = x & b = x by XTUPLE_0:1;
    hence a in field R & b in field R by A1,TARSKI:def 1;
    F.x = y by Th72;
    hence [F.a,F.b] in S by A2,TARSKI:def 1;
  end;
  assume a in field R & b in field R;
  then a = x & b = x by A1,TARSKI:def 1;
  hence thesis by TARSKI:def 1;
end;

theorem
  {[x,x]}, {[y,y]} are_isomorphic
proof
  take x .--> y;
  thus thesis by Th82;
end;

registration
  let I, A be set;
  cluster I --> A -> total for I-defined Function;
  coherence;
end;

theorem
  for f being Function st x in dom f
   holds x .--> f.x c= f
proof
  let f be Function;
  assume x in dom f;
  then [x, f.x] in f by FUNCT_1:1;
  then {[x, f.x]} c= f by ZFMISC_1:31;
  hence thesis by ZFMISC_1:29;
end;

registration let A be non empty set;
 let x be set, i be Element of A;
 cluster x .--> i -> A-valued;
 coherence
  proof
    rng(x .--> i) = {i} by RELAT_1:160;
   hence rng(x .--> i) c= A by ZFMISC_1:31;
  end;
end;

reserve Y for set,
        f,g for Function of Y,X,
        x for Element of X,
        y for Element of Y;



theorem
  F is associative implies F.:(F[;](x,f),g) = F[;](x,F.:(f,g))
proof
  assume
A1: F is associative;
  per cases;
  suppose
    Y = {};
    hence thesis;
  end;
  suppose
A2: Y <> {};
    now
      let y;
      reconsider x1 = f.y, x2 = g.y as Element of X by A2,FUNCT_2:5;
      thus (F[;](x,F.:(f,g))).y = F.(x,F.:(f,g).y) by A2,Th53
        .= F.(x,F.(x1,x2)) by A2,Th37
        .= F.(F.(x,x1),x2) by A1,BINOP_1:def 3
        .= F.(F[;](x,f).y,g.y) by A2,Th53;
    end;
    hence thesis by A2,Th38;
  end;
end;

registration let A be set, B be non empty set, x be Element of B;
 cluster A --> x -> B-valued;
 coherence;
end;

registration let A be non empty set, x be Element of A, y be set;
 cluster x .--> y -> A-defined;
 coherence
  proof
     dom(x .--> y) = {x} by Th13;
   hence dom(x .--> y) c= A by ZFMISC_1:31;
  end;
end;

theorem
 for x,y,A being set st x in A
 holds (x .--> y)|A = x .--> y
proof let x,y,A be set;
  assume x in A;
  then {x} c= A by ZFMISC_1:31;
  then dom(x .--> y) c= A by Th13;
  hence thesis by RELAT_1:68;
end;



registration
  let Y be functional set;
  cluster Y-valued -> Function-yielding for Function;
  coherence
  proof
    let f be Function;
    assume
A1: f is Y-valued;
    let x be set;
    thus thesis by A1;
  end;
end;



definition
  let IT be Function;
  attr IT is Relation-yielding means
  for x be set st x in dom IT holds IT.x is Relation;
end;

registration
 cluster Function-yielding -> Relation-yielding for Function;
 coherence
  proof let f be Function;
   assume
A1:  f is Function-yielding;
   let x be set;
   thus thesis by A1;
  end;
end;

registration
 cluster empty -> Function-yielding for Function;
 coherence
  proof let f be Function;
   assume
A1:  f is empty;
   let x be set; thus thesis by A1;
  end;
end;



theorem
  for X,Y,x,y being set holds X --> x tolerates Y --> y iff x = y or X misses Y
proof
  let X,Y,x,y be set;
  set f = X --> x, g = Y --> y;
A1: dom g = Y by Th13;
A2: dom f = X by Th13;
  thus f tolerates g implies x = y or X misses Y
  proof
    assume that
A3: for z being set st z in dom f /\ dom g holds f.z = g.z and
A4: x <> y;
    assume X meets Y;
    then consider z be set such that
A5: z in X and
A6: z in Y by XBOOLE_0:3;
A7: f.z = x by A5,Th7;
A8: g.z = y by A6,Th7;
    z in X /\ Y by A5,A6,XBOOLE_0:def 4;
    hence thesis by A2,A1,A3,A4,A7,A8;
  end;
  assume
A9: x = y or X misses Y;
  let z be set;
  assume
A10: z in dom f /\ dom g;
  then
A11: z in Y;
A12: z in X by A2,A10,XBOOLE_0:def 4;
  then f.z = x by Th7;
  hence thesis by A9,A12,A11,Th7,XBOOLE_0:3;
end;

reserve x,y,z,A for set;

theorem Th88:
 rng(x .--> y) = {y}
 proof
   dom(x .--> y) = {x} by Th13;
  hence rng(x .--> y) = {(x .--> y).x} by FUNCT_1:4
      .= {y} by Th72;
 end;

theorem
 z in A implies (A --> x)*(y .--> z) = y .--> x
 proof assume
A1: z in A;
A2:  dom(y .--> z) = {y} by Th13
       .= dom(y .--> x) by Th13;
   rng(y .--> z) = {z} by Th88;
   then rng(y .--> z) c= A by A1,ZFMISC_1:31;
   then rng(y .--> z) c= dom(A --> x) by Th13;
  hence dom((A --> x)*(y .--> z)) = dom(y .--> x) by A2,RELAT_1:27;
  let e be set;
  assume
A3:  e in dom((A --> x)*(y .--> z));
  thus ((A --> x)*(y .--> z)).e = (A --> x).((y .--> z).e) by A3,FUNCT_1:12
      .= (A --> x).z by A3,Th7
      .= x by A1,Th7
      .= (y .--> x).e by A3,Th7;
 end;















begin 

reserve a,b,p,x,x9,x1,x19,x2,y,y9,y1,y19,y2,z,z9,z1,z2,X,X9,Y,Y9,Z,Z9 for set;
reserve A,D,D9 for non empty set;
reserve f,g,h for Function;

Lm1: [[x,x9],[y,y9]] = [[x1,x19],[y1,y19]] implies x = x1 & y = y1 & x9= x19 &
y9 = y19
proof
  assume [[x,x9],[y,y9]] = [[x1,x19],[y1,y19]];
  then [x,x9] = [x1,x19] & [y,y9] = [y1,y19] by XTUPLE_0:1;
  hence thesis by XTUPLE_0:1;
end;

theorem Th1:
  (for z st z in Z holds ex x,y st z = [x,y]) implies ex X,Y st Z
  c= [:X,Y:]
proof
  assume
A1: for z st z in Z holds ex x,y st z = [x,y];
  defpred P[set] means ex y st [ $1,y] in Z;
  consider X such that
A2: x in X iff x in union union Z & P[ x] from XBOOLE_0:sch 1;
  defpred P[set] means ex x st [x,$1] in Z;
  consider Y such that
A3: y in Y iff y in union union Z & P[y] from XBOOLE_0:sch 1;
  take X,Y;
  let z;
  assume
A4: z in Z;
  then consider x,y such that
A5: z = [x,y] by A1;
  x in union union Z by A4,A5,ZFMISC_1:134;
  then
A6: x in X by A2,A4,A5;
  y in union union Z by A4,A5,ZFMISC_1:134;
  then y in Y by A3,A4,A5;
  hence thesis by A5,A6,ZFMISC_1:87;
end;

theorem
  g*f = (g|rng f)*f
proof
  x in dom(g*f) iff x in dom((g|rng f)*f)
  proof
A1: dom(g|rng f) = dom g /\ rng f by RELAT_1:61;
    thus x in dom(g*f) implies x in dom((g|rng f)*f)
    proof
      assume
A2:   x in dom(g*f);
      then
A3:   x in dom f by FUNCT_1:11;
      x in dom f by A2,FUNCT_1:11;
      then
A4:   f.x in rng f by FUNCT_1:def 3;
      f.x in dom g by A2,FUNCT_1:11;
      then f.x in dom(g|rng f) by A1,A4,XBOOLE_0:def 4;
      hence thesis by A3,FUNCT_1:11;
    end;
    assume
A5: x in dom((g|rng f)*f);
    then f.x in dom(g|rng f) by FUNCT_1:11;
    then
A6: f.x in dom g by A1,XBOOLE_0:def 4;
    x in dom f by A5,FUNCT_1:11;
    hence thesis by A6,FUNCT_1:11;
  end;
  then
A7: dom(g*f) = dom((g|rng f)*f) by TARSKI:1;
  x in dom(g*f) implies (g*f).x = ((g|rng f)*f).x
  proof
    assume
A8: x in dom(g*f);
    then
A9: x in dom f by FUNCT_1:11;
    then
A10: f.x in rng f by FUNCT_1:def 3;
    thus (g*f).x = g.(f.x) by A8,FUNCT_1:12
      .= (g|rng f).(f.x) by A10,FUNCT_1:49
      .= ((g|rng f)*f).x by A9,FUNCT_1:13;
  end;
  hence thesis by A7,FUNCT_1:2;
end;

theorem
  id X c= id Y iff X c= Y
proof
  thus id X c= id Y implies X c= Y
  proof
    assume
A1: id X c= id Y;
    let x;
    assume x in X;
    then [x,x] in id X by RELAT_1:def 10;
    hence thesis by A1,RELAT_1:def 10;
  end;
  assume
A2: X c= Y;
  let z;
  assume
A3: z in id X;
  then consider x,x9 such that
A4: x in X and
  x9 in X and
A5: z = [x,x9] by ZFMISC_1:84;
  x = x9 by A3,A5,RELAT_1:def 10;
  hence thesis by A2,A4,A5,RELAT_1:def 10;
end;

theorem
  X c= Y implies X --> a c= Y --> a
proof
A1: dom(X --> a) = X by FUNCOP_1:13;
  assume
A2: X c= Y;
A3: now
    let x;
    assume
A4: x in dom(X --> a);
    then (X --> a).x = a by FUNCOP_1:7;
    hence (X --> a).x = (Y --> a).x by A2,A1,A4,FUNCOP_1:7;
  end;
  dom(Y --> a) = Y by FUNCOP_1:13;
  hence thesis by A2,A1,A3,GRFUNC_1:2;
end;

theorem Th5:
  X --> a c= Y --> b implies X c= Y
proof
  assume X --> a c= Y --> b;
  then
A1: dom(X --> a) c= dom(Y --> b) by RELAT_1:11;
  dom(X --> a) = X by FUNCOP_1:13;
  hence thesis by A1,FUNCOP_1:13;
end;

theorem
  X <> {} & X --> a c= Y --> b implies a = b
proof
  assume
A1: X <> {};
  set x = the Element of X;
  assume
A2: X --> a c= Y --> b;
  then X c= Y by Th5;
  then x in Y by A1,TARSKI:def 3;
  then
A3: (Y --> b).x = b by FUNCOP_1:7;
  dom(X --> a) = X & (X --> a).x = a by A1,FUNCOP_1:7,13;
  hence thesis by A1,A2,A3,GRFUNC_1:2;
end;

theorem
  x in dom f implies x .--> f.x c= f
proof
A1: now
    let y;
    assume y in dom (x .--> f.x);
    then x = y by FUNCOP_1:75;
    hence (x .--> f.x).y = f.y by FUNCOP_1:72;
  end;
  assume
A2: x in dom f;
  dom (x .--> f.x) c= dom f
  proof
    let y;
    thus thesis by A2,FUNCOP_1:75;
  end;
  hence thesis by A1,GRFUNC_1:2;
end;



theorem
  Y|`f|X c= f
proof
  Y|`f|X c= Y|`f & Y|`f c= f by RELAT_1:59,86;
  hence thesis by XBOOLE_1:1;
end;

theorem
  f c= g implies Y|`f|X c= Y|`g|X
proof
  assume f c= g;
  then Y|`f c= Y|`g by RELAT_1:101;
  hence thesis by RELAT_1:76;
end;

definition
  let f,g;
  func f +* g -> Function means
  :Def1:
  dom it = dom f \/ dom g &
  for x st x in dom f \/ dom g holds
  (x in dom g implies it.x = g.x) &
  (not x in dom g implies it.x = f.x);
  existence
  proof
    deffunc G(set) = f.$1;
    deffunc F(set) = g.$1;
    defpred P[set] means $1 in dom g;
    thus ex F being Function st dom F = dom f \/ dom g &
    for x st x in dom f \/ dom g holds
    (P[x] implies F.x = F(x)) & (not P[x] implies F.x = G(x))
      from PARTFUN1:sch 1;
  end;
  uniqueness
  proof
    let h1,h2 be Function such that
A1: dom h1 = dom f \/ dom g and
A2: for x st x in dom f \/ dom g holds (x in dom g implies h1.x = g.x)
    & (not x in dom g implies h1.x = f.x) and
A3: dom h2 = dom f \/ dom g and
A4: for x st x in dom f \/ dom g holds (x in dom g implies h2.x = g.x)
    & (not x in dom g implies h2.x = f.x);
    for x st x in dom f \/ dom g holds h1.x = h2.x
    proof
      let x;
      assume
A5:   x in dom f \/ dom g;
      then
A6:   not x in dom g implies h1.x = f.x & h2.x = f.x by A2,A4;
      x in dom g implies h1.x = g.x & h2.x = g.x by A2,A4,A5;
      hence thesis by A6;
    end;
    hence thesis by A1,A3,FUNCT_1:2;
  end;
  idempotence;
end;

theorem Th10:
  dom f c= dom(f+*g) & dom g c= dom(f+*g)
proof
  dom(f+*g) = dom f \/ dom g by Def1;
  hence thesis by XBOOLE_1:7;
end;

theorem Th11:
  not x in dom g implies (f +* g).x = f.x
proof
  assume
A1: not x in dom g;
  per cases;
  suppose
    x in dom f;
    then x in dom f \/ dom g by XBOOLE_0:def 3;
    hence thesis by A1,Def1;
  end;
  suppose
A2: not x in dom f;
    then not x in dom f \/ dom g by A1,XBOOLE_0:def 3;
    then not x in dom(f+*g) by Def1;
    hence (f+*g).x = {} by FUNCT_1:def 2
      .= f.x by A2,FUNCT_1:def 2;
  end;
end;

theorem Th12:
  x in dom(f +* g) iff x in dom f or x in dom g
proof
  x in dom(f +* g) iff x in dom f \/ dom g by Def1;
  hence thesis by XBOOLE_0:def 3;
end;

theorem Th13:
  x in dom g implies (f+*g).x = g.x
proof
  x in dom g implies x in dom f \/ dom g by XBOOLE_0:def 3;
  hence thesis by Def1;
end;

theorem Th14:
  f +* g +* h = f +* (g +* h)
proof
A1: now
    let x such that
    x in dom f \/ dom(g +* h);
    hereby
      assume
A2:   x in dom(g +* h);
      per cases by A2,Th12;
      suppose
A3:     x in dom g & not x in dom h;
        hence (f +* g +* h).x = (f +* g).x by Th11
          .= g.x by A3,Th13
          .= (g +* h).x by A3,Th11;
      end;
      suppose
A4:     x in dom h;
        hence (f +* g +* h).x = h.x by Th13
          .= (g +* h).x by A4,Th13;
      end;
    end;
    assume
A5: not x in dom(g +* h);
    then
A6: not x in dom g by Th12;
    not x in dom h by A5,Th12;
    hence (f +* g +* h).x = (f +* g).x by Th11
      .= f.x by A6,Th11;
  end;
  dom(f +* g +* h) = dom(f +* g) \/ dom h by Def1
    .= dom f \/ dom g \/ dom h by Def1
    .= dom f \/ (dom g \/ dom h) by XBOOLE_1:4
    .= dom f \/ dom(g +* h) by Def1;
  hence thesis by A1,Def1;
end;

theorem Th15:
  f tolerates g & x in dom f implies (f+*g).x = f.x
proof
  assume that
A1: f tolerates g and
A2: x in dom f;
  now
    per cases;
    suppose
      x in dom g;
      then x in dom f /\ dom g & (f+*g).x = g.x by A2,Th13,XBOOLE_0:def 4;
      hence thesis by A1,PARTFUN1:def 4;
    end;
    suppose
      not x in dom g;
      hence thesis by Th11;
    end;
  end;
  hence thesis;
end;

theorem
  dom f misses dom g & x in dom f implies (f +* g).x = f.x
proof
  assume dom f /\ dom g = {} & x in dom f;
  then not x in dom g by XBOOLE_0:def 4;
  hence thesis by Th11;
end;

theorem Th17:
  rng(f +* g) c= rng f \/ rng g
proof
  let y;
  assume y in rng(f +* g);
  then consider x such that
A1: x in dom (f +* g) and
A2: y = (f +* g).x by FUNCT_1:def 3;
  x in dom f & not x in dom g or x in dom g by A1,Th12;
  then x in dom f & (f +* g).x = f.x or x in dom g & (f +* g).x = g.x by Th11
,Th13;
  then y in rng f or y in rng g by A2,FUNCT_1:def 3;
  hence thesis by XBOOLE_0:def 3;
end;

theorem
  rng g c= rng(f +* g)
proof
  let y;
  assume y in rng g;
  then consider x such that
A1: x in dom g & y = g.x by FUNCT_1:def 3;
  x in dom(f +* g) & (f +* g).x = y by A1,Th12,Th13;
  hence thesis by FUNCT_1:def 3;
end;

theorem Th19:
  dom f c= dom g implies f +* g = g
proof
  assume dom f c= dom g;
  then dom f \/ dom g = dom g by XBOOLE_1:12;
  then dom(f +* g) = dom g & for x st x in dom g holds (f +* g).x = g.x by Def1
;
  hence thesis by FUNCT_1:2;
end;


registration let f; let g be empty Function;
 reduce g +* f to f;
 reducibility
 proof
  dom g c= dom f by XBOOLE_1:2;
  hence thesis by Th19;
 end;
 reduce f +* g to f;
 reducibility
 proof
A1: for x st x in dom f holds (f +* g).x = f.x
  proof
    let x;
    assume x in dom f;
    then x in dom f \ dom g;
    hence thesis by Th11;
  end;
  dom f \/ dom g = dom f;
  then dom(f +* g) = dom f by Def1;
  hence thesis by A1,FUNCT_1:2;
 end;
end;

theorem Th20:
  {} +* f = f;

theorem Th21:
  f +* {} = f;

theorem
  id(X) +* id(Y) = id(X \/ Y)
proof
A1: x in dom id(X \/ Y) implies (id(X) +* id(Y)).x = id(X \/ Y).x
  proof
    assume
A2: x in dom id(X \/ Y);
    now
      per cases;
      suppose
A3:     x in Y;
        dom id Y = Y;
        hence (id(X) +* id(Y)).x = (id Y).x by A3,Th13
          .= x by A3,FUNCT_1:18
          .= id(X \/ Y).x by A2,FUNCT_1:18;
      end;
      suppose
A4:     not x in Y;
        then
A5:     x in X by A2,XBOOLE_0:def 3;
        not x in dom id Y by A4;
        hence (id(X) +* id(Y)).x = (id X).x by Th11
          .= x by A5,FUNCT_1:18
          .= id(X \/ Y).x by A2,FUNCT_1:18;
      end;
    end;
    hence thesis;
  end;
  dom(id(X) +* id(Y)) = dom id X \/ dom id Y by Def1
    .= X \/ dom id Y
    .= X \/ Y
    .= dom id(X \/ Y);
  hence thesis by A1,FUNCT_1:2;
end;

theorem
  (f +* g)|(dom g) = g
proof
  dom f \/ dom g = dom(f +* g) by Def1;
  then
A1: dom((f +* g)|(dom g)) = dom g by RELAT_1:62,XBOOLE_1:7;
  for x st x in dom g holds ((f +* g)|(dom g)).x = g.x
  proof
    let x;
    x in dom g implies (f +* g).x = g.x by Th13;
    hence thesis by A1,FUNCT_1:47;
  end;
  hence thesis by A1,FUNCT_1:2;
end;

theorem Th24:
  ((f +* g)|(dom f \ dom g)) c= f
proof
A1: for x st x in dom((f +* g)|(dom f \ dom g)) holds ((f +* g)|(dom f \ dom
  g)).x = f.x
  proof
    let x such that
A2: x in dom((f +* g)|(dom f \ dom g));
    dom((f +* g)|(dom f \ dom g)) c= dom f \ dom g by RELAT_1:58;
    then not x in dom g by A2,XBOOLE_0:def 5;
    then (f +* g).x = f.x by Th11;
    hence thesis by A2,FUNCT_1:47;
  end;
  dom((f +* g)|(dom f \ dom g)) c= dom f \ dom g by RELAT_1:58;
  then dom((f +* g)|(dom f \ dom g)) c= dom f by XBOOLE_1:1;
  hence thesis by A1,GRFUNC_1:2;
end;

theorem Th25:
  g c= f +* g
proof
  dom(f +* g) = dom f \/ dom g by Def1;
  then
A1: dom g c= dom(f +* g) by XBOOLE_1:7;
  for x st x in dom g holds (f+*g).x = g.x by Th13;
  hence thesis by A1,GRFUNC_1:2;
end;

theorem
  f tolerates g +* h implies f|(dom f \ dom h) tolerates g
proof
  assume
A1: f tolerates g +* h;
  let x;
  assume
A2: x in dom(f|(dom f \ dom h)) /\ dom g;
  then
A3: x in dom(f|(dom f \ dom h)) by XBOOLE_0:def 4;
  x in dom g by A2,XBOOLE_0:def 4;
  then
A4: x in dom(g +* h) by Th12;
A5: dom(f|(dom f \ dom h)) c= dom f \ dom h by RELAT_1:58;
  then x in dom f by A3,XBOOLE_0:def 5;
  then
A6: x in dom f /\ dom(g +* h) by A4,XBOOLE_0:def 4;
  not x in dom h by A3,A5,XBOOLE_0:def 5;
  then (g +* h).x = g.x by Th11;
  then f.x = g.x by A1,A6,PARTFUN1:def 4;
  hence thesis by A3,A5,FUNCT_1:49;
end;

theorem Th27:
  f tolerates g +* h implies f tolerates h
proof
  assume
A1: f tolerates g +* h;
  let x;
  assume
A2: x in dom f /\ dom h;
  then
A3: x in dom f by XBOOLE_0:def 4;
A4: x in dom h by A2,XBOOLE_0:def 4;
  then x in dom(g +* h) by Th12;
  then
A5: x in dom f /\ dom(g +* h) by A3,XBOOLE_0:def 4;
  (g +* h).x = h.x by A4,Th13;
  hence thesis by A1,A5,PARTFUN1:def 4;
end;

theorem Th28:
  f tolerates g iff f c= f +* g
proof
  thus f tolerates g implies f c= (f +* g)
  proof
    dom f \/ dom g = dom(f +* g) by Def1;
    then
A1: dom f c= dom(f +* g) by XBOOLE_1:7;
    assume f tolerates g;
    then for x st x in dom f holds (f +* g).x = f.x by Th15;
    hence thesis by A1,GRFUNC_1:2;
  end;
  thus thesis by Th27,PARTFUN1:54;
end;

theorem Th29:
  f +* g c= f \/ g
proof
  let p;
  assume
A1: p in f +* g;
  then consider x,y such that
A2: p = [x,y] by RELAT_1:def 1;
  x in dom(f +* g) by A1,A2,FUNCT_1:1;
  then x in dom f & not x in dom g or x in dom g by Th12;
  then
A3: x in dom f & (f +* g).x = f.x or x in dom g & (f +* g).x = g.x by Th11,Th13
;
  y = (f +* g).x by A1,A2,FUNCT_1:1;
  then p in f or p in g by A2,A3,FUNCT_1:1;
  hence thesis by XBOOLE_0:def 3;
end;

theorem Th30:
  f tolerates g iff f \/ g = f +* g
proof
  thus f tolerates g implies f \/ g = f +* g
  proof
    assume f tolerates g;
    then
A1: f c= f +* g by Th28;
A2: f +* g c= f \/ g by Th29;
    g c= f +* g by Th25;
    then f \/ g c= f +* g by A1,XBOOLE_1:8;
    hence thesis by A2,XBOOLE_0:def 10;
  end;
  thus thesis by PARTFUN1:51;
end;

theorem Th31:
  dom f misses dom g implies f \/ g = f +* g
proof
  assume dom f misses dom g;
  then f tolerates g by PARTFUN1:56;
  hence thesis by Th30;
end;

theorem Th32:
  dom f misses dom g implies f c= f +* g
proof
  assume dom f misses dom g;
  then f \/ g = f +* g by Th31;
  hence thesis by XBOOLE_1:7;
end;

theorem
  dom f misses dom g implies (f +* g)|(dom f) = f
proof
  assume dom f misses dom g;
  then
A1: dom f \ dom g = dom f by XBOOLE_1:83;
  dom((f +* g)|(dom f)) = dom(f +* g) /\ dom f by RELAT_1:61
    .= (dom f \/ dom g) /\ dom f by Def1
    .= dom f by XBOOLE_1:21;
  hence thesis by A1,Th24,GRFUNC_1:3;
end;

theorem Th34:
  f tolerates g iff f +* g = g +* f
proof
  thus f tolerates g implies f +* g = g +* f
  proof
    assume
A1: f tolerates g;
A2: for x st x in dom(f +* g) holds (f +* g).x = (g +* f).x
    proof
      let x such that
A3:   x in dom(f +* g);
      now
A4:     dom(f +* g) = dom f \/ dom g by Def1;
        per cases by A3,A4,XBOOLE_0:def 3;
        suppose
A5:       x in dom f & x in dom g;
          then
A6:       (g +* f).x = f.x by Th13;
          x in dom f /\ dom g & (f +* g).x = g.x by A5,Th13,XBOOLE_0:def 4;
          hence thesis by A1,A6,PARTFUN1:def 4;
        end;
        suppose
A7:       x in dom f & not x in dom g;
          then (f +* g).x = f.x by Th11;
          hence thesis by A7,Th13;
        end;
        suppose
A8:       not x in dom f & x in dom g;
          then (f +* g).x = g.x by Th13;
          hence thesis by A8,Th11;
        end;
      end;
      hence thesis;
    end;
    dom(f +* g) = dom g \/ dom f by Def1
      .= dom(g +* f) by Def1;
    hence thesis by A2,FUNCT_1:2;
  end;
  assume
A9: f +* g = g +* f;
  let x;
  assume
A10: x in dom f /\ dom g;
  then x in dom g by XBOOLE_0:def 4;
  then
A11: (f +* g).x = g.x by Th13;
  x in dom f by A10,XBOOLE_0:def 4;
  hence thesis by A9,A11,Th13;
end;

theorem Th35:
  dom f misses dom g implies f +* g = g +* f
proof
  assume dom f misses dom g;
  then f tolerates g by PARTFUN1:56;
  hence thesis by Th34;
end;

theorem
  for f,g being PartFunc of X,Y st g is total holds f +* g = g
proof
  let f,g be PartFunc of X,Y;
  assume dom g = X;
  then dom f c= dom g;
  hence thesis by Th19;
end;

theorem Th37:
  for f,g being Function of X,Y st Y = {} implies X = {} holds f +* g = g
proof
  let f,g be Function of X,Y;
  assume Y = {} implies X = {};
  then dom f = X & dom g = X by FUNCT_2:def 1;
  hence thesis by Th19;
end;

theorem
  for f,g being Function of X,X holds f +* g = g by Th37;

theorem
  for f,g being Function of X,D holds f +* g = g by Th37;

theorem
  for f,g being PartFunc of X,Y holds f +* g is PartFunc of X,Y
proof
  let f,g be PartFunc of X,Y;
  rng (f +* g) c= rng f \/ rng g by Th17;
  then
A1: rng(f +* g) c= Y by XBOOLE_1:1;
  dom (f +* g) = dom f \/ dom g by Def1;
  hence thesis by A1,RELSET_1:4;
end;



definition
  let f;
  func ~f -> Function means
  :Def2:
  (for x holds x in dom it iff ex y,z st x =
  [z,y] & [y,z] in dom f) & for y,z st [y,z] in dom f holds it.(z,y) = f.(y,z);
  existence
  proof
    defpred P[set,set] means ex y,z st $1 = [z,y] & $2 = f.[y,z];
    defpred P[set] means ex y st [ $1,y] in dom f;
    consider D1 being set such that
A1: x in D1 iff x in union union dom f & P[ x] from XBOOLE_0:sch 1;
    defpred P[set] means ex y st [y,$1] in dom f;
    consider D2 being set such that
A2: y in D2 iff y in union union dom f & P[y] from XBOOLE_0:sch 1;
    defpred P[set] means ex y,z st $1 = [z,y] & [y,z] in dom f;
    consider D being set such that
A3: x in D iff x in [:D2,D1:] & P[ x] from XBOOLE_0:sch 1;
A4: for x,y1,y2 st x in D & P[x,y1] & P[x,y2] holds y1 = y2
    proof
      let x,y1,y2 such that x in D;
      given y,z such that
A5:   x = [z,y] and
A6:   y1 = f.[y,z];
      given y9,z9 such that
A7:   x = [z9,y9] and
A8:   y2 = f.[y9,z9];
      z = z9 by A5,A7,XTUPLE_0:1;
      hence thesis by A5,A6,A7,A8,XTUPLE_0:1;
    end;
A9: for x st x in D ex y1 st P[x,y1]
    proof
      let x;
      assume x in D;
      then consider y,z such that
A10:  x = [z,y] and
      [y,z] in dom f by A3;
      take f.[y,z];
      thus thesis by A10;
    end;
    consider h being Function such that
A11: dom h = D and
A12: for x st x in D holds P[x,h.x] from FUNCT_1:sch 2(A4,A9);
    take h;
    thus
A13: for x holds x in dom h iff ex y,z st x = [z,y] & [y,z] in dom f
    proof
      let x;
      thus x in dom h implies ex y,z st x = [z,y] & [y,z] in dom f by A3,A11;
      given y,z such that
A14:  x = [z,y] and
A15:  [y,z] in dom f;
      y in union union dom f by A15,ZFMISC_1:134; then
A16:  y in D1 by A1,A15;
      z in union union dom f by A15,ZFMISC_1:134;
      then z in D2 by A2,A15;
      then [z,y] in [:D2,D1:] by A16,ZFMISC_1:87;
      hence thesis by A3,A11,A14,A15;
    end;
    let y,z;
    assume [y,z] in dom f;
    then [z,y] in D by A11,A13;
    then consider y9,z9 such that
A17: [z,y] = [z9,y9] and
A18: h.(z,y) = f.[y9,z9] by A12;
    z = z9 by A17,XTUPLE_0:1;
    hence thesis by A17,A18,XTUPLE_0:1;
  end;
  uniqueness
  proof
    let h1,h2 be Function such that
A19: for x holds x in dom h1 iff ex y,z st x = [z,y] & [y,z] in dom f and
A20: for y,z st [y,z] in dom f holds h1.(z,y) = f.(y,z) and
A21: for x holds x in dom h2 iff ex y,z st x = [z,y] & [y,z] in dom f and
A22: for y,z st [y,z] in dom f holds h2.(z,y) = f.(y,z);
A23: x in dom h1 implies h1.x = h2.x
    proof
      assume x in dom h1;
      then consider x1,x2 such that
A24:  x = [x2,x1] and
A25:  [x1,x2] in dom f by A19;
      h1.(x2,x1) = f.(x1,x2) by A20,A25
        .= h2.(x2,x1) by A22,A25;
      hence thesis by A24;
    end;
    x in dom h1 iff x in dom h2
    proof
      x in dom h1 iff ex y,z st x = [z,y] & [y,z] in dom f by A19;
      hence thesis by A21;
    end;
    then dom h1 = dom h2 by TARSKI:1;
    hence thesis by A23,FUNCT_1:2;
  end;
end;

theorem Th41:
  rng ~f c= rng f
proof
  let y;
  assume y in rng ~f;
  then consider x such that
A1: x in dom ~f and
A2: y = (~f).x by FUNCT_1:def 3;
  consider x1,x2 such that
A3: x = [x2,x1] and
A4: [x1,x2] in dom f by A1,Def2;
  y = (~f).(x2,x1) by A2,A3
    .= f.(x1,x2) by A4,Def2;
  hence thesis by A4,FUNCT_1:def 3;
end;

theorem Th42:
  [x,y] in dom f iff [y,x] in dom ~f
proof
  thus [x,y] in dom f implies [y,x] in dom ~f by Def2;
  assume [y,x] in dom ~f;
  then consider x1,y1 such that
A1: [y,x] = [y1,x1] and
A2: [x1,y1] in dom f by Def2;
  x1 = x by A1,XTUPLE_0:1;
  hence thesis by A1,A2,XTUPLE_0:1;
end;

theorem
  [y,x] in dom ~f implies (~f).(y,x) = f.(x,y)
proof
  assume [y,x] in dom ~f;
  then [x,y] in dom f by Th42;
  hence thesis by Def2;
end;

theorem
  ex X,Y st dom ~f c= [:X,Y:]
proof
  now
    let z;
    assume z in dom ~f;
    then ex x,y st z = [y,x] & [x,y] in dom f by Def2;
    hence ex x,y st z = [x,y];
  end;
  hence thesis by Th1;
end;

theorem Th45:
  dom f c= [:X,Y:] implies dom ~f c= [:Y,X:]
proof
  assume
A1: dom f c= [:X,Y:];
  let z;
  assume z in dom ~f;
  then ex x,y st z = [y,x] & [x,y] in dom f by Def2;
  hence thesis by A1,ZFMISC_1:88;
end;

theorem Th46:
  dom f = [:X,Y:] implies dom ~f = [:Y,X:]
proof
  assume
A1: dom f = [:X,Y:];
  hence dom ~f c= [:Y,X:] by Th45;
  let z;
  assume z in [:Y,X:];
  then consider y,x such that
A2: y in Y & x in X and
A3: z = [y,x] by ZFMISC_1:def 2;
  [x,y] in dom f by A1,A2,ZFMISC_1:def 2;
  hence thesis by A3,Def2;
end;

theorem Th47:
  dom f c= [:X,Y:] implies rng ~f = rng f
proof
  assume
A1: dom f c= [:X,Y:];
  thus rng ~f c= rng f by Th41;
  let y;
  assume y in rng f;
  then consider x such that
A2: x in dom f and
A3: y = f.x by FUNCT_1:def 3;
  consider x1,y1 such that
  x1 in X and
  y1 in Y and
A4: x =[x1,y1] by A1,A2,ZFMISC_1:84;
A5: [y1,x1] in dom ~f by A2,A4,Th42;
  y = f.(x1,y1) by A3,A4
    .= (~f).(y1,x1) by A2,A4,Def2;
  hence thesis by A5,FUNCT_1:def 3;
end;

theorem
  for f being PartFunc of [:X,Y:],Z holds ~f is PartFunc of [:Y,X:],Z
proof
  let f be PartFunc of [:X,Y:],Z;
A1: dom f c= [:X,Y:];
  then
A2: dom ~f c= [:Y,X:] by Th45;
  rng f c= Z;
  then rng ~f c= Z by A1,Th47;
  hence thesis by A2,RELSET_1:4;
end;

theorem Th49:
  for f being Function of [:X,Y:],Z st Z<>{} holds ~f is Function
  of [:Y,X:],Z
proof
  let f be Function of [:X,Y:],Z;
  assume
A1: Z <> {};
  then
A2: dom f = [:X,Y:] by FUNCT_2:def 1;
  then
A3: [:Y,X:] = dom ~f by Th46;
  rng f c= Z;
  then rng ~f c= Z by A2,Th47;
  then reconsider R = ~f as Relation of [:Y,X:],Z by A3,RELSET_1:4;
  R is quasi_total
  proof
    per cases;
    case Z <> {};
      dom f = [:X,Y:] by A1,FUNCT_2:def 1;
      hence thesis by Th46;
    end;
    case Z = {};
      hence thesis;
    end;
  end;
  hence thesis;
end;

theorem
  for f being Function of [:X,Y:],D holds ~f is Function of [:Y,X:],D by Th49;

theorem Th51:
  ~~f c= f
proof
A1: now
    let x;
    assume x in dom ~~f;
    then consider y2,z2 such that
A2: x = [z2,y2] & [y2,z2] in dom ~f by Def2;
    take y2,z2;
    thus x = [z2,y2] & [y2,z2] in dom ~ f & [z2,y2] in dom f by A2,Th42;
  end;
A3: x in dom ~~f implies (~~f).x = f.x
  proof
    assume x in dom ~~f;
    then consider y2,z2 such that
A4: x = [z2,y2] and
A5: [y2,z2] in dom ~f and
A6: [z2,y2] in dom f by A1;
    (~~f).(z2,y2) = (~f).(y2,z2) by A5,Def2
      .= f.(z2,y2) by A6,Def2;
    hence thesis by A4;
  end;
  dom ~~f c= dom f
  proof
    let x;
    assume x in dom ~~f;
    then
    ex y2,z2 st x = [z2,y2] & [y2,z2] in dom ~ f & [z2,y2] in dom f by A1;
    hence thesis;
  end;
  hence thesis by A3,GRFUNC_1:2;
end;

theorem Th52:
  dom f c= [:X,Y:] implies ~~f = f
proof
  assume
A1: dom f c= [:X,Y:];
A2: ~~f c= f by Th51;
  dom ~~ f = dom f
  proof
    thus dom ~~f c= dom f by A2,RELAT_1:11;
    let z;
    assume
A3: z in dom f;
    then consider x,y such that
    x in X and
    y in Y and
A4: z = [x,y] by A1,ZFMISC_1:84;
    [y,x] in dom ~f by A3,A4,Th42;
    hence thesis by A4,Th42;
  end;
  hence thesis by Th51,GRFUNC_1:3;
end;

theorem
  for f being PartFunc of [:X,Y:],Z holds ~~f = f
proof
  let f be PartFunc of [:X,Y:],Z;
  dom f c= [:X,Y:];
  hence thesis by Th52;
end;



definition
  let f,g;
  func |:f,g:| -> Function means
  :Def3:
  (for z holds z in dom it iff ex x,y,x9
,y9 st z = [[x,x9],[y,y9]] & [x,y] in dom f & [x9,y9] in dom g) & for x,y,x9,y9
st [x,y] in dom f & [x9,y9] in dom g holds it.([x,x9],[y,y9]) = [f.(x,y),g.(x9,
  y9)];
  existence
  proof
    defpred P[set,set] means ex x,y,x9,y9 st $1 = [[x,x9],[y,y9]] & $2 = [f.[x
    ,y],g.[x9,y9]];
    defpred P[set] means ex y st [ $1,y] in dom f;
    consider D1 being set such that
A1: x in D1 iff x in union union dom f & P[ x] from XBOOLE_0:sch 1;
    defpred P[set] means ex x st [x,$1] in dom f;
    consider D2 being set such that
A2: y in D2 iff y in union union dom f & P[y] from XBOOLE_0:sch 1;
    defpred P[set] means ex y st [ $1,y] in dom g;
    consider D19 being set such that
A3: x in D19 iff x in union union dom g & P[ x] from XBOOLE_0:sch 1;
    defpred P[set] means ex x st [x,$1] in dom g;
    consider D29 being set such that
A4: y in D29 iff y in union union dom g & P[y] from XBOOLE_0:sch 1;
    defpred P[set] means ex x,y,x9,y9 st $1 = [[x,x9],[y,y9]] & [x,y] in dom f
    & [x9,y9] in dom g;
    consider D being set such that
A5: z in D iff z in [:[:D1,D19:],[:D2,D29:]:] & P[z] from XBOOLE_0:sch
    1;
A6: for z,z1,z2 st z in D & P[z,z1] & P[z,z2] holds z1 = z2
    proof
      let z,z1,z2 such that
      z in D;
      given x,y,x9,y9 such that
A7:   z = [[x,x9],[y,y9]] and
A8:   z1 = [f.[x,y],g.[x9,y9]];
      given x1,y1,x19,y19 such that
A9:   z = [[x1,x19],[y1,y19]] and
A10:  z2 = [f.[x1,y1],g.[x19,y19]];
A11:  x9 = x19 by A7,A9,Lm1;
      x = x1 & y = y1 by A7,A9,Lm1;
      hence thesis by A7,A8,A9,A10,A11,Lm1;
    end;
A12: for z st z in D holds ex z1 st P[z,z1]
    proof
      let z;
      assume z in D;
      then consider x,y,x9,y9 such that
A13:  z = [[x,x9],[y,y9]] and
      [x,y] in dom f and
      [x9,y9] in dom g by A5;
      take [f.[x,y],g.[x9,y9]];
      thus thesis by A13;
    end;
    consider h being Function such that
A14: dom h = D and
A15: for z st z in D holds P[z,h.z] from FUNCT_1:sch 2(A6,A12);
    take h;
    thus
A16: for z holds z in dom h iff ex x,y,x9,y9 st z = [[x,x9],[y,y9]] &
    [x,y] in dom f & [x9,y9] in dom g
    proof
      let z;
      thus z in dom h implies ex x,y,x9,y9 st z = [[x,x9],[y,y9]] & [x,y] in
      dom f & [x9,y9] in dom g by A5,A14;
      given x,y,x9,y9 such that
A17:  z = [[x,x9],[y,y9]] and
A18:  [x,y] in dom f and
A19:  [x9,y9] in dom g;
      y9 in union union dom g by A19,ZFMISC_1:134;
      then
A20:  y9 in D29 by A4,A19;
      y in union union dom f by A18,ZFMISC_1:134;
      then y in D2 by A2,A18;
      then
A21:  [y,y9] in [:D2,D29:] by A20,ZFMISC_1:87;
      x9 in union union dom g by A19,ZFMISC_1:134;
      then
A22:  x9 in D19 by A3,A19;
      x in union union dom f by A18,ZFMISC_1:134;
      then x in D1 by A1,A18;
      then [x,x9] in [:D1,D19:] by A22,ZFMISC_1:87;
      then z in [:[:D1,D19:],[:D2,D29:]:] by A17,A21,ZFMISC_1:87;
      hence thesis by A5,A14,A17,A18,A19;
    end;
    let x,y,x9,y9;
    assume [x,y] in dom f & [x9,y9] in dom g;
    then [[x,x9],[y,y9]] in D by A14,A16;
    then consider x1,y1,x19,y19 such that
A23: [[x,x9],[y,y9]] = [[x1,x19],[y1,y19]] and
A24: h.[[x,x9],[y,y9]] = [f.[x1,y1],g.[x19,y19]] by A15;
A25: x9= x19 by A23,Lm1;
    x = x1 & y = y1 by A23,Lm1;
    hence thesis by A23,A24,A25,Lm1;
  end;
  uniqueness
  proof
    let h1,h2 be Function such that
A26: for z holds z in dom h1 iff ex x,y,x9,y9 st z = [[x,x9],[y,y9]] &
    [x,y] in dom f & [x9,y9] in dom g and
A27: for x,y,x9,y9 st [x,y] in dom f & [x9,y9] in dom g holds h1.([x,
    x9],[y,y9]) = [f.(x,y),g.(x9,y9)] and
A28: for z holds z in dom h2 iff ex x,y,x9,y9 st z = [[x,x9],[y,y9]] &
    [x,y] in dom f & [x9,y9] in dom g and
A29: for x,y,x9,y9 st [x,y] in dom f & [x9,y9] in dom g holds h2.([x,
    x9],[y,y9]) = [f.(x,y),g.(x9,y9)];
A30: z in dom h1 implies h1.z = h2.z
    proof
      assume z in dom h1;
      then consider x,y,x9,y9 such that
A31:  z = [[x,x9],[y,y9]] and
A32:  [x,y] in dom f & [x9,y9] in dom g by A26;
      h1.([x,x9],[y,y9]) = [f.(x,y),g.(x9,y9)] by A27,A32
        .= h2.([x,x9],[y,y9]) by A29,A32;
      hence thesis by A31;
    end;
    z in dom h1 iff z in dom h2
    proof
      z in dom h1 iff ex x,y,x9,y9 st z = [[x,x9],[y,y9]] & [x,y] in dom
      f & [x9,y9] in dom g by A26;
      hence thesis by A28;
    end;
    then dom h1 = dom h2 by TARSKI:1;
    hence thesis by A30,FUNCT_1:2;
  end;
end;

theorem Th54:
  [[x,x9],[y,y9]] in dom |:f,g:| iff [x,y] in dom f & [x9,y9] in dom g
proof
  thus [[x,x9],[y,y9]] in dom |:f,g:| implies [x,y] in dom f & [x9,y9] in dom
  g
  proof
    assume [[x,x9],[y,y9]] in dom |:f,g:|;
    then consider x1,y1,x19,y19 such that
A1: [[x,x9],[y,y9]] = [[x1,x19],[y1,y19]] and
A2: [x1,y1] in dom f & [x19,y19] in dom g by Def3;
    x = x1 & x9= x19 by A1,Lm1;
    hence thesis by A1,A2,Lm1;
  end;
  thus thesis by Def3;
end;

theorem
  [[x,x9],[y,y9]] in dom |:f,g:| implies |:f,g:|.([x,x9],[y,y9]) = [f.(x
  ,y),g.(x9,y9)]
proof
  assume [[x,x9],[y,y9]] in dom |:f,g:|;
  then [x,y] in dom f & [x9,y9] in dom g by Th54;
  hence thesis by Def3;
end;

theorem Th56:
  rng |:f,g:| c= [:rng f,rng g:]
proof
  let z;
  assume z in rng |:f,g:|;
  then consider p such that
A1: p in dom |:f,g:| and
A2: z = |:f,g:|.p by FUNCT_1:def 3;
  consider x,y,x9,y9 such that
A3: p = [[x,x9],[y,y9]] and
A4: [x,y] in dom f & [x9,y9] in dom g by A1,Def3;
A5: f.[x,y] in rng f & g.[x9,y9] in rng g by A4,FUNCT_1:def 3;
  z = |:f,g:|.([x,x9],[y,y9]) by A2,A3
    .= [f.(x,y),g.(x9,y9)] by A4,Def3;
  hence thesis by A5,ZFMISC_1:87;
end;

theorem Th57:
  dom f c= [:X,Y:] & dom g c= [:X9,Y9:] implies dom|:f,g:| c= [:[:
  X,X9:],[:Y,Y9:]:]
proof
  assume
A1: dom f c= [:X,Y:] & dom g c= [:X9,Y9:];
  let xy be set;
  assume xy in dom|:f,g:|;
  then consider x,y,x9,y9 such that
A2: xy = [[x,x9],[y,y9]] and
A3: [x,y] in dom f & [x9,y9] in dom g by Def3;
  y in Y & y9 in Y9 by A1,A3,ZFMISC_1:87;
  then
A4: [y,y9] in [:Y,Y9:] by ZFMISC_1:87;
  x in X & x9 in X9 by A1,A3,ZFMISC_1:87;
  then [x,x9] in [:X,X9:] by ZFMISC_1:87;
  hence thesis by A2,A4,ZFMISC_1:87;
end;

theorem Th58:
  dom f = [:X,Y:] & dom g = [:X9,Y9:] implies
  dom|:f,g:| = [:[:X,X9:],[:Y,Y9:]:]
proof
  assume
A1: dom f = [:X,Y:] & dom g = [:X9,Y9:];
  hence dom|:f,g:| c= [:[:X,X9:],[:Y,Y9:]:] by Th57;
  let z;
  assume z in [:[:X,X9:],[:Y,Y9:]:];
  then consider xx,yy being set such that
A2: xx in [:X,X9:] and
A3: yy in [:Y,Y9:] and
A4: z = [xx,yy] by ZFMISC_1:def 2;
  consider y,y9 such that
A5: y in Y & y9 in Y9 and
A6: yy = [y,y9] by A3,ZFMISC_1:def 2;
  consider x,x9 such that
A7: x in X & x9 in X9 and
A8: xx = [x,x9] by A2,ZFMISC_1:def 2;
  [x,y] in dom f & [x9,y9] in dom g by A1,A7,A5,ZFMISC_1:87;
  hence thesis by A4,A8,A6,Def3;
end;

theorem
  for f being PartFunc of [:X,Y:],Z for g being PartFunc of [:X9,Y9:],Z9
  holds |:f,g:| is PartFunc of [:[:X,X9:],[:Y,Y9:]:],[:Z,Z9:]
proof
  let f be PartFunc of [:X,Y:],Z;
  let g be PartFunc of [:X9,Y9:],Z9;
  rng |:f,g:| c= [:rng f,rng g:] & [:rng f,rng g:] c= [:Z,Z9:] by Th56,
ZFMISC_1:96;
  then
A1: rng|:f,g:| c= [:Z,Z9:] by XBOOLE_1:1;
  dom f c= [:X,Y:] & dom g c= [:X9,Y9:];
  then dom|:f,g:| c= [:[:X,X9:],[:Y,Y9:]:] by Th57;
  hence thesis by A1,RELSET_1:4;
end;

theorem Th60:
  for f being Function of [:X,Y:],Z for g being Function of [:X9,
Y9:],Z9 st Z <> {} & Z9 <> {} holds |:f,g:| is Function of [:[:X,X9:],[:Y,Y9:]
  :],[:Z,Z9:]
proof
  let f be Function of [:X,Y:],Z;
  let g be Function of [:X9,Y9:],Z9;
  rng |:f,g:| c= [:rng f,rng g:] & [:rng f,rng g:] c= [:Z,Z9:] by Th56,
ZFMISC_1:96;
  then
A1: rng|:f,g:| c= [:Z,Z9:] by XBOOLE_1:1;
  assume
A2: Z <> {} & Z9 <> {};
  then dom f = [:X,Y:] & dom g = [:X9,Y9:] by FUNCT_2:def 1;
  then [:[:X,X9:],[:Y,Y9:]:] = dom|:f,g:| by Th58;
  then reconsider
  R = |:f,g:| as Relation of [:[:X,X9:],[:Y,Y9:]:],[:Z,Z9:] by A1,RELSET_1:4;
  R is quasi_total
  proof
    per cases;
    case [:Z,Z9:] <> {};
      dom f = [:X,Y:] & dom g = [:X9,Y9:] by A2,FUNCT_2:def 1;
      hence thesis by Th58;
    end;
    case [:Z,Z9:] = {};
      hence thesis;
    end;
  end;
  hence thesis;
end;

theorem
  for f being Function of [:X,Y:],D for g being Function of [:X9,Y9:],D9
  holds |:f,g:| is Function of [:[:X,X9:],[:Y,Y9:]:],[:D,D9:] by Th60;

definition
  let x,y,a,b be set;
  func (x,y) --> (a,b) -> set equals
  (x .--> a) +* (y .--> b);
  correctness;
end;

registration
  let x,y,a,b be set;
  cluster (x,y) --> (a,b) -> Function-like Relation-like;
  coherence;
end;

theorem Th62:
  dom((x1,x2) --> (y1,y2)) = {x1,x2} & rng((x1,x2) --> (y1,y2)) c=
  {y1,y2}
proof
  set f = {x1} --> y1, g = {x2} -->y2, h = (x1,x2) --> (y1,y2);
  rng f \/ rng g c= {y1} \/ {y2} by XBOOLE_1:13;
  then
A1: rng f \/ rng g c= {y1,y2} by ENUMSET1:1;
  dom f = {x1} & dom g = {x2} by FUNCOP_1:13;
  then dom f \/ dom g = {x1,x2} by ENUMSET1:1;
  hence dom h = {x1,x2} by Def1;
  rng h c= rng f \/ rng g by Th17;
  hence thesis by A1,XBOOLE_1:1;
end;

theorem Th63:
  (x1 <> x2 implies ((x1,x2) --> (y1,y2)).x1 = y1) & ((x1,x2) -->
  (y1,y2)).x2 = y2
proof
  set f = {x1} --> y1, g = {x2} -->y2, h = (x1,x2) --> (y1,y2);
A1: x2 in {x2} by TARSKI:def 1;
A2: x1 in {x1} by TARSKI:def 1;
  hereby
    assume x1 <> x2;
    then not x1 in dom g by TARSKI:def 1;
    then h.x1 = f.x1 by Th11;
    hence h.x1 = y1 by A2,FUNCOP_1:7;
  end;
  {x2} = dom g by FUNCOP_1:13;
  then h.x2 = g.x2 by A1,Th13;
  hence thesis by A1,FUNCOP_1:7;
end;

theorem
  x1 <> x2 implies rng((x1,x2) --> (y1,y2)) = {y1,y2}
proof
  set h = (x1,x2) --> (y1,y2);
  assume
A1: x1 <> x2;
  thus rng h c= {y1,y2} by Th62;
  let y;
  assume y in {y1,y2};
  then y = y1 or y = y2 by TARSKI:def 2;
  then
A2: h.x1 = y or h.x2 = y by A1,Th63;
  dom h = {x1,x2} by Th62;
  then x1 in dom h & x2 in dom h by TARSKI:def 2;
  hence thesis by A2,FUNCT_1:def 3;
end;

theorem
  (x1,x2) --> (y,y) = {x1,x2} --> y
proof
  set F = (x1,x2)-->(y,y), f = {x1}-->y, g = {x2}-->y, F9 = {x1,x2}-->y;
  now
    thus
A1: dom F = {x1,x2} & dom F9 = {x1,x2} by Th62,FUNCOP_1:13;
    let x such that
A2: x in {x1,x2};
    now
      per cases by A1,A2,Th12;
      suppose
A3:     x in dom f & not x in dom g;
        then F.x = f.x by Th11;
        hence F.x = y & F9.x = y by A2,A3,FUNCOP_1:7;
      end;
      suppose
A4:     x in dom g;
        then F.x = g.x by Th13;
        hence F.x = y & F9.x = y by A2,A4,FUNCOP_1:7;
      end;
    end;
    hence F.x = F9.x;
  end;
  hence thesis by FUNCT_1:2;
end;

definition
  let A,x1,x2;
  let y1,y2 be Element of A;
  redefine func (x1,x2) --> (y1,y2) -> Function of {x1,x2},A;
  coherence
  proof
    set f = {x1} --> y1, g = {x2} -->y2, h = (x1,x2) --> (y1,y2);
    rng h c= rng f \/ rng g by Th17;
    then
A1: rng h c= A by XBOOLE_1:1;
    dom h = {x1,x2} by Th62;
    hence thesis by A1,FUNCT_2:def 1,RELSET_1:4;
  end;
end;

theorem
  for a,b,c,d being set, g being Function st dom g = {a,b} & g.a = c & g
  .b = d holds g = (a,b) --> (c,d)
proof
  let a,b,c,d be set;
  let h be Function such that
A1: dom h = {a,b} and
A2: h.a = c and
A3: h.b = d;
  set f = {a} --> c, g = {b} -->d;
A4: b in {b} by TARSKI:def 1;
A5: a in {a} by TARSKI:def 1;
A6: now
    let x such that
A7: x in dom f \/ dom g;
    thus x in dom g implies h.x = g.x
    proof
      assume x in dom g;
      then x = b by TARSKI:def 1;
      hence thesis by A3,A4,FUNCOP_1:7;
    end;
    assume not x in dom g;
    then x in dom f by A7,XBOOLE_0:def 3;
    then x = a by TARSKI:def 1;
    hence h.x = f.x by A2,A5,FUNCOP_1:7;
  end;
  dom f = {a} & dom g = {b} by FUNCOP_1:13;
  then dom h = dom f \/ dom g by A1,ENUMSET1:1;
  hence thesis by A6,Def1;
end;

theorem Th67:
  for a,b,c,d being set st a <> c holds (a,c) --> (b,d) = { [a,b],
  [c,d] }
proof
  let a,b,c,d be set such that
A1: a <> c;
  set f = {a} --> b, g = {c} --> d;
A2: dom f = {a} & dom g = {c} by FUNCOP_1:13;
  {a} --> b = {[a,b]} & {c} --> d = {[c,d]} by ZFMISC_1:29;
  hence (a,c) --> (b,d) = {[a,b]} \/ {[c,d]} by A1,A2,Th31,ZFMISC_1:11
    .= { [a,b], [c,d] } by ENUMSET1:1;
end;

theorem
  for a,b,x,y,x9,y9 being set st a <> b & (a,b) --> (x,y) = (a,b) --> (
  x9,y9) holds x = x9 & y = y9
proof
  let a,b,x,y,x9,y9 be set such that
A1: a <> b and
A2: (a,b) --> (x,y) = (a,b) --> (x9,y9);
  thus x = ((a,b) --> (x,y)).a by A1,Th63
    .= x9 by A1,A2,Th63;
  thus y = ((a,b) --> (x,y)).b by Th63
    .= y9 by A2,Th63;
end;

begin 



theorem
  for f1,f2, g1,g2 being Function st rng g1 c= dom f1 & rng g2 c= dom f2
  & f1 tolerates f2 holds (f1+*f2)*(g1+*g2) = (f1*g1)+*(f2*g2)
proof
  let f1,f2, g1,g2 be Function;
  assume that
A1: rng g1 c= dom f1 & rng g2 c= dom f2 and
A2: f1 tolerates f2;
A3: rng (g1+*g2) c= (rng g1) \/ rng g2 & dom (f1+*f2) = (dom f1) \/ dom f2
  by Def1,Th17;
  (rng g1) \/ rng g2 c= (dom f1) \/ dom f2 by A1,XBOOLE_1:13;
  then
A4: dom ((f1+*f2)*(g1+*g2)) = dom (g1+*g2) by A3,RELAT_1:27,XBOOLE_1:1
    .= (dom g1) \/ dom g2 by Def1;
A5: dom (f1*g1) = dom g1 & dom (f2*g2) = dom g2 by A1,RELAT_1:27;
A6: now
    let x be set;
A7: not x in dom g2 or x in dom g2;
    assume
A8: x in (dom g1) \/ dom g2;
    then
A9: ((f1+*f2)*(g1+*g2)).x = (f1+*f2).((g1+*g2).x) by A4,FUNCT_1:12;
    x in dom g1 or x in dom g2 by A8,XBOOLE_0:def 3;
    then
    (g1+*g2).x = g1.x & ((f1*g1)+*(f2*g2)).x = (f1*g1).x & (f1*g1).x = f1
.(g1.x) & g1.x in rng g1 & (g1.x in rng g1 implies g1.x in dom f1) or (g1+*g2).
x = g2.x & ((f1*g1)+*(f2*g2)).x = (f2*g2).x & (f2*g2).x = f2.(g2.x) & g2.x in
    rng g2 & (g2.x in rng g2 implies g2.x in dom f2) by A1,A5,A8,A7,Def1,
FUNCT_1:12,def 3;
    hence ((f1+*f2)*(g1+*g2)).x = ((f1*g1)+*(f2*g2)).x by A2,A9,Th13,Th15;
  end;
  dom ((f1*g1)+*(f2*g2)) = (dom g1) \/ dom g2 by A5,Def1;
  hence thesis by A4,A6,FUNCT_1:2;
end;



reserve A,B for set;

theorem Th70:
  dom f c= A \/ B implies f|A +* f|B = f
proof
A1: dom(f|A) = dom f /\ A & dom(f|B) = dom f /\ B by RELAT_1:61;
A2: x in dom(f|A) \/ dom(f|B) implies (x in dom(f|B) implies f.x = f|B.x) &
  (not x in dom(f|B) implies f.x = f|A.x)
  proof
    assume
A3: x in dom(f|A) \/ dom(f|B);
    thus x in dom(f|B) implies f.x = f|B.x by FUNCT_1:47;
    assume not x in dom(f|B);
    then x in dom(f|A) by A3,XBOOLE_0:def 3;
    hence thesis by FUNCT_1:47;
  end;
  assume dom f c= A \/ B;
  then dom f = dom f /\ (A \/ B) by XBOOLE_1:28
    .= dom(f|A) \/ dom(f|B) by A1,XBOOLE_1:23;
  hence thesis by A2,Def1;
end;



theorem Th71:
  for p,q being Function , A being set holds (p +* q)|A = p|A +* q|A
proof
  let p,q be Function , A be set;
A1: dom ((p +* q)|A) = dom (p +* q) /\ A by RELAT_1:61
    .= (dom p \/ dom q) /\ A by Def1
    .= (dom p /\ A) \/ (dom q /\ A) by XBOOLE_1:23
    .= dom (p|A) \/ (dom q /\ A) by RELAT_1:61
    .= dom (p|A) \/ dom (q|A) by RELAT_1:61;
  for x being set st x in dom (p|A) \/ dom (q|A) holds (x in dom (q|A)
implies ((p +* q)|A).x = (q|A).x) & (not x in dom (q|A) implies ((p +* q)|A).x
  = (p|A).x)
  proof
    let x be set;
    assume
A2: x in dom (p|A) \/ dom (q|A);
    then x in dom (p|A) or x in dom (q|A) by XBOOLE_0:def 3;
    then x in (dom p /\ A) or x in dom q /\ A by RELAT_1:61;
    then
A3: x in A by XBOOLE_0:def 4;
    hereby
      assume
A4:   x in dom (q|A);
      then x in (dom q /\ A) by RELAT_1:61;
      then
A5:   x in dom q by XBOOLE_0:def 4;
      thus ((p +* q)|A).x = (p +* q).x by A1,A2,FUNCT_1:47
        .= q.x by A5,Th13
        .= (q|A).x by A4,FUNCT_1:47;
    end;
    assume
A6: not x in dom (q|A);
    then not x in (dom q /\ A) by RELAT_1:61;
    then
A7: not x in dom q by A3,XBOOLE_0:def 4;
A8: x in dom (p|A) by A2,A6,XBOOLE_0:def 3;
    then x in dom p /\ A by RELAT_1:61;
    then x in dom p by XBOOLE_0:def 4;
    then x in dom (p +* q) by Th12;
    then x in dom (p +* q) /\ A by A3,XBOOLE_0:def 4;
    then x in dom ((p +* q)|A) by RELAT_1:61;
    hence ((p +* q)|A).x = (p +* q).x by FUNCT_1:47
      .= p.x by A7,Th11
      .= (p|A).x by A8,FUNCT_1:47;
  end;
  hence thesis by A1,Def1;
end;

theorem Th72:
  for f,g being Function, A being set st A misses dom g holds
  (f +* g)|A = f|A
proof
  let f,g be Function, A be set;
  assume A misses dom g;
  then dom g /\ A = {} by XBOOLE_0:def 7;
  then dom (g|A) = {} by RELAT_1:61;
  then g|A = {};
  hence (f +* g)|A = f|A +* {} by Th71
    .= f|A;
end;

theorem
  for f,g being Function , A being set holds dom f misses A implies
  (f +* g)|A = g|A
proof
  let f,g be Function , A be set;
  assume dom f misses A;
  then dom f /\ A = {} by XBOOLE_0:def 7;
  then dom (f|A) = {} by RELAT_1:61;
  then f|A = {};
  hence (f +* g)|A = {} +* g|A by Th71
    .= g|A;
end;

theorem
  for f,g,h being Function st dom g = dom h holds f +* g +* h = f +* h
proof
  let f,g,h be Function;
  assume
A1: dom g = dom h;
  thus f +* g +* h = f +* (g +* h) by Th14
    .= f +* h by A1,Th19;
end;

Lm2:
  for f,g being Function holds f c= g implies g +* f = g
proof
  let f,g be Function;
  assume
A1: f c= g;
  then f tolerates g by PARTFUN1:54;
  hence g+*f = f\/ g by Th30
    .= g by A1,XBOOLE_1:12;
end;

theorem
  for f being Function, A being set holds f +* f|A = f by Lm2,RELAT_1:59;

theorem
  for f,g being Function, B,C being set st dom f c= B & dom g c= C & B
  misses C holds (f +* g)|B = f & (f +* g)|C = g
proof
  let f,g be Function, B,C be set;
  assume that
A1: dom f c= B and
A2: dom g c= C and
A3: B misses C;
  dom f misses C by A1,A3,XBOOLE_1:63;
  then dom f /\ C = {} by XBOOLE_0:def 7;
  then dom (f|C) = {} by RELAT_1:61;
  then
A4: f|C = {};
  dom g misses B by A2,A3,XBOOLE_1:63;
  then dom g /\ B = {} by XBOOLE_0:def 7;
  then dom (g|B) = {} by RELAT_1:61;
  then
A5: g|B = {};
  thus (f +* g)|B = f|B +* g|B by Th71
    .= f|B by A5,Th21
    .= f by A1,RELAT_1:68;
  thus (f +* g)|C = f|C +* g|C by Th71
    .= g|C by A4,Th20
    .= g by A2,RELAT_1:68;
end;

theorem
  for p,q being Function, A being set holds dom p c= A & dom q misses A
  implies (p +* q)|A = p
proof
  let p,q be Function, A be set;
  assume that
A1: dom p c= A and
A2: dom q misses A;
  thus (p +* q )|A = p|A by A2,Th72
    .= p by A1,RELAT_1:68;
end;

theorem
  for f being Function, A,B being set holds f|(A \/ B) = f|A +* f|B
proof
  let f be Function, A,B be set;
A1: f|(A \/ B)|B = f|((A \/ B) /\ B) by RELAT_1:71
    .= f|B by XBOOLE_1:21;
A2: dom (f|(A \/ B)) c= A \/ B by RELAT_1:58;
  f|(A \/ B)|A = f|((A \/ B) /\ A) by RELAT_1:71
    .= f|A by XBOOLE_1:21;
  hence thesis by A1,A2,Th70;
end;



reserve x,y,i,j,k for set;

theorem
  (i,j):->k = [i,j].-->k;

theorem
  ((i,j):->k).(i,j) = k by FUNCOP_1:71;



theorem
  for a,b,c being set holds (a,a) --> (b,c) = a .--> c
proof
  let a,b,c be set;
  dom(a .-->c) = {a} by FUNCOP_1:13
    .= dom({a} -->b) by FUNCOP_1:13;
  hence thesis by Th19;
end;

theorem
  for x,y holds x .--> y = {[x,y]} by ZFMISC_1:29;



theorem
  for f being Function, a,b,c being set st a <> c holds (f +* (a .-->b))
  .c = f.c
proof
  let f be Function, a,b,c be set such that
A1: a <> c;
  set g = a .-->b;
  not c in dom g by A1,TARSKI:def 1;
  hence thesis by Th11;
end;

theorem
  for f being Function, a,b,c,d being set st a <> b holds
  (f +* ((a,b)-->(c,d))) .a = c & (f +* ((a,b)-->(c,d))) .b = d
proof
  let f be Function, a,b,c,d be set such that
A1: a <> b;
  set g = (a,b)-->(c,d);
A2: dom g = {a,b} by Th62;
  then a in dom g by TARSKI:def 2;
  hence (f +* g).a = g.a by Th13
    .= c by A1,Th63;
  b in dom g by A2,TARSKI:def 2;
  hence (f +* g).b = g.b by Th13
    .= d by Th63;
end;



theorem Th85:
  for a,b being set, f being Function st a in dom f & f.a = b
  holds a .--> b c= f
proof
  let a,b be set, f be Function;
  assume a in dom f & f.a = b;
  then [a,b] in f by FUNCT_1:1;
  then {[a,b]} c= f by ZFMISC_1:31;
  hence thesis by ZFMISC_1:29;
end;

theorem
  for a,b,c,d being set, f being Function st
  a in dom f & c in dom f & f.a = b & f.c = d holds (a,c) --> (b,d) c= f
proof
  let a,b,c,d be set, f be Function;
  assume that
A1: a in dom f and
A2: c in dom f and
A3: f.a = b & f.c = d;
  per cases;
  suppose
A4: a <> c;
    [a,b] in f & [c,d] in f by A1,A2,A3,FUNCT_1:1;
    then { [a,b], [c,d]} c= f by ZFMISC_1:32;
    hence thesis by A4,Th67;
  end;
  suppose
    a = c;
    hence thesis by A1,A3,Th85;
  end;
end;



theorem Th87:
  for f,g,h being Function st f c= h & g c= h holds f +* g c= h
proof
  let f,g,h be Function;
  assume f c= h & g c= h;
  then
A1: f \/ g c= h by XBOOLE_1:8;
  f +* g c= f \/ g by Th29;
  hence thesis by A1,XBOOLE_1:1;
end;



theorem
  for f, g being Function, A being set st A /\ dom f c= A /\ dom g holds
  (f+*g|A)|A = g|A
proof
  let f, g be Function, A be set;
  assume
A1: A /\ dom f c= A /\ dom g;
A2: dom (f|A) = A /\ dom f & dom (g|A) = A /\ dom g by RELAT_1:61;
  thus (f+*g|A)|A = (f|A)+*(g|A)|A by Th71
    .= (f|A)+*g|A
    .= g|A by A1,A2,Th19;
end;



theorem
  for f be Function, a,b,n,m be set holds
    (f +* (a .--> b) +* (m .--> n)).m = n
proof
  let f be Function, a,b,n,m be set;
  set mn=m .--> n;
  dom mn ={ m } by FUNCOP_1:13;
  then m in dom mn by TARSKI:def 1;
  hence (f +* (a .--> b) +* mn).m=mn.m by Th13
    .=n by FUNCOP_1:72;
end;

theorem
  for f be Function, n,m be set holds (f +* (n .--> m) +* (m .--> n)).n=
  m
proof
  let f be Function,n,m be set;
  set mn=m .--> n, nm=n .--> m;
 dom mn ={ m } by FUNCOP_1:13;
  then
A1: m in dom mn by TARSKI:def 1;
  per cases;
  suppose
A2: n=m;
    hence (f +* nm +* mn).n=mn.m by A1,Th13
      .=m by A2,FUNCOP_1:72;
  end;
  suppose
A3: n<>m;
    dom nm ={ n } by FUNCOP_1:13;
    then
A4: n in dom nm by TARSKI:def 1;
    not n in dom mn by A3,TARSKI:def 1;
    hence (f +* nm +* mn).n=(f +* nm).n by Th11
      .=nm.n by A4,Th13
      .=m by FUNCOP_1:72;
  end;
end;

theorem
  for f be Function, a,b,n,m,x be set st x <> m & x <> a holds (f +* (a
  .--> b) +* (m .--> n)).x=f.x
proof
  let f be Function, a,b,n,m,x be set;
  assume that
A1: x<>m and
A2: x<>a;
  set mn=m .--> n, nm=a .--> b;
A3: not x in dom nm by A2,TARSKI:def 1;
  not x in dom mn by A1,TARSKI:def 1;
  hence (f +* nm +* mn).x=(f +* nm).x by Th11
    .=f.x by A3,Th11;
end;



theorem
  f is one-to-one & g is one-to-one & rng f misses rng g implies
  f+*g is one-to-one
proof
  assume that
A1: f is one-to-one and
A2: g is one-to-one and
A3: rng f misses rng g;
  let x1,x2 be set;
  set fg = f+*g;
  assume that
A4: x1 in dom fg and
A5: x2 in dom fg and
A6: fg.x1 = fg.x2;
A7: x1 in dom f or x1 in dom g by A4,Th12;
A8: x2 in dom f or x2 in dom g by A5,Th12;
  per cases;
  suppose
A9: x1 in dom g & x2 in dom g;
    then fg.x1 = g.x1 & fg.x2 = g.x2 by Th13;
    hence thesis by A2,A6,A9,FUNCT_1:def 4;
  end;
  suppose
A10: x1 in dom g & not x2 in dom g;
    then x2 in dom f by A5,Th12;
    then
A11: f.x2 in rng f by FUNCT_1:def 3;
A12: g.x1 in rng g by A10,FUNCT_1:def 3;
    fg.x1 = g.x1 & fg.x2 = f.x2 by A10,Th11,Th13;
    hence thesis by A3,A6,A12,A11,XBOOLE_0:3;
  end;
  suppose
A13: not x1 in dom g & x2 in dom g;
    then x1 in dom f by A4,Th12;
    then
A14: f.x1 in rng f by FUNCT_1:def 3;
A15: g.x2 in rng g by A13,FUNCT_1:def 3;
    fg.x1 = f.x1 & fg.x2 = g.x2 by A13,Th11,Th13;
    hence thesis by A3,A6,A15,A14,XBOOLE_0:3;
  end;
  suppose
A16: not x1 in dom g & not x2 in dom g;
    then fg.x1 = f.x1 & fg.x2 = f.x2 by Th11;
    hence thesis by A1,A6,A7,A8,A16,FUNCT_1:def 4;
  end;
end;

registration let f,g be Function;
 reduce f +* g +* g to f +* g;
 reducibility
 proof
  thus f +* g +* g = f +*( g +* g ) by Th14
    .= f +* g;
 end;
end;



theorem
  for f,g being Function holds f +* g +* g = f +* g;

theorem
  for f,g,h being Function, D being set holds (f +* g)|D =h | D implies
  (h +* g) | D = (f +* g) | D
proof
  let f,g,h be Function, D be set;
  assume
A1: (f +* g)|D =h | D;
A2: dom ((f +* g) | D) = dom (f +* g) /\ D by RELAT_1:61
    .= (dom f \/ dom g) /\ D by Def1;
A3: dom ((h +* g) | D) = dom (h +* g) /\ D by RELAT_1:61
    .= (dom h \/ dom g) /\ D by Def1;
  then
A4: dom ((h +* g) | D) = (dom h /\ D) \/ (dom g /\ D) by XBOOLE_1:23
    .= ((dom f \/ dom g) /\ D) \/ (dom g /\ D) by A1,A2,RELAT_1:61
    .= ((dom f \/ dom g) \/ dom g) /\ D by XBOOLE_1:23
    .= (dom f \/ (dom g \/ dom g)) /\ D by XBOOLE_1:4
    .= (dom f \/ dom g ) /\ D;
  now
    let x be set;
    assume
A5: x in dom ((f +* g) | D);
    then
A6: x in dom f \/ dom g by A2,XBOOLE_0:def 4;
A7: x in D by A2,A5,XBOOLE_0:def 4;
A8: x in dom h \/ dom g & ((h +* g) | D).x = (h +* g).x by A2,A3,A4,A5,
FUNCT_1:47,XBOOLE_0:def 4;
    per cases;
    suppose
A9:   x in dom g;
      ((f +* g) | D).x = (f +* g).x by A5,FUNCT_1:47
        .=g.x by A6,A9,Def1;
      hence ((h +* g) | D).x =((f +* g) | D).x by A8,A9,Def1;
    end;
    suppose
      not x in dom g;
      hence ((h +* g) | D).x = h.x by A8,Def1
        .=((f +* g) | D).x by A1,A7,FUNCT_1:49;
    end;
  end;
  hence thesis by A2,A4,FUNCT_1:2;
end;

theorem
  for f,g,h being Function, D being set holds f | D =h | D implies (h +*
  g) | D = (f +* g) | D
proof
  let f,g,h be Function, D be set;
  assume
A1: f |D = h | D;
A2: dom ((f +* g) | D) = dom (f +* g) /\ D by RELAT_1:61
    .= (dom f \/ dom g) /\ D by Def1;
A3: dom ((h +* g) | D) = dom (h +* g) /\ D by RELAT_1:61
    .= (dom h \/ dom g) /\ D by Def1;
  then
A4: dom ((h +* g) | D) = (dom h /\ D) \/ (dom g /\ D) by XBOOLE_1:23
    .= dom (f| D) \/ (dom g /\ D) by A1,RELAT_1:61
    .= (dom f /\ D) \/ (dom g /\ D) by RELAT_1:61
    .= (dom f \/ dom g ) /\ D by XBOOLE_1:23;
  now
    let x be set;
    assume
A5: x in dom ((f +* g) | D);
    then
A6: x in dom f \/ dom g by A2,XBOOLE_0:def 4;
A7: x in D by A2,A5,XBOOLE_0:def 4;
A8: x in dom h \/ dom g & ((h +* g) | D).x = (h +* g).x by A2,A3,A4,A5,
FUNCT_1:47,XBOOLE_0:def 4;
    per cases;
    suppose
A9:   x in dom g;
      ((f +* g) | D).x = (f +* g).x by A5,FUNCT_1:47
        .=g.x by A6,A9,Def1;
      hence ((h +* g) | D).x =((f +* g) | D).x by A8,A9,Def1;
    end;
    suppose
A10:  not x in dom g;
      then
A11:  ((h +* g) | D).x = h.x by A8,Def1
        .=(h | D ).x by A7,FUNCT_1:49;
      thus ((f +* g) | D).x = (f +* g).x by A5,FUNCT_1:47
        .=f.x by A6,A10,Def1
        .=((h +* g) | D).x by A1,A7,A11,FUNCT_1:49;
    end;
  end;
  hence thesis by A2,A4,FUNCT_1:2;
end;



theorem Th96:
  x .--> x = id{x}
proof
  for y,z holds [y,z] in x .--> x iff y in {x} & y = z
  proof
    let y,z;
A1: x .--> x = {[x,x]} by ZFMISC_1:29;
    thus [y,z] in x .--> x implies y in {x} & y = z
    proof
      assume [y,z] in x .--> x;
      then
A2:   [y,z] = [x,x] by A1,TARSKI:def 1;
      then y = x by XTUPLE_0:1;
      hence thesis by A2,TARSKI:def 1,XTUPLE_0:1;
    end;
    assume y in {x};
    then y = x by TARSKI:def 1;
    hence thesis by A1,TARSKI:def 1;
  end;
  hence thesis by RELAT_1:def 10;
end;

theorem Th97:
  f c= g implies f+*g = g
proof
  assume
A1: f c= g;
  then f tolerates g by PARTFUN1:54;
  hence f+*g = f\/ g by Th30
    .= g by A1,XBOOLE_1:12;
end;

theorem Th98:
  f c= g implies g+*f = g
proof
  assume
A1: f c= g;
  then f tolerates g by PARTFUN1:54;
  hence g+*f = f\/ g by Th30
    .= g by A1,XBOOLE_1:12;
end;

begin 

definition
  let f,x,y;
  func f+~(x,y) equals
  f+*((x .--> y)*f);
  coherence;
end;

registration
  let f,x,y;
  cluster f+~(x,y) -> Relation-like Function-like;
  coherence;
end;

theorem Th99:
  dom(f+~(x,y)) = dom f
proof
  thus dom(f+~(x,y)) = dom f \/ dom((x .--> y)*f) by Def1
    .= dom f by RELAT_1:25,XBOOLE_1:12;
end;

theorem Th100:
  x <> y implies not x in rng(f+~(x,y))
proof
  assume
A1: x <> y;
  assume x in rng(f+~(x,y));
  then consider z such that
A2: z in dom(f+~(x,y)) and
A3: (f+~(x,y)).z = x by FUNCT_1:def 3;
A4: z in dom f by A2,Th99;
A5: now
    assume
A6: not z in dom((x .--> y)*f);
    then f.z = x by A3,Th11;
    then f.z in dom(x .--> y) by FUNCOP_1:74;
    hence contradiction by A4,A6,FUNCT_1:11;
  end;
  (x .--> y).(f.z) = ((x .--> y)*f).z by A4,FUNCT_1:13
    .= x by A3,A5,Th13;
  then f.z <> x by A1,FUNCOP_1:72;
  then not f.z in dom(x .--> y) by FUNCOP_1:75;
  hence thesis by A5,FUNCT_1:11;
end;

theorem
  x in rng f implies y in rng(f+~(x,y))
proof
  assume x in rng f;
  then consider z such that
A1: z in dom f and
A2: f.z = x by FUNCT_1:def 3;
A3: dom((x .--> y)*f) c= dom(f+~(x,y)) by Th10;
  x in dom(x.-->y) by FUNCOP_1:74;
  then
A4: z in dom((x .--> y)*f) by A1,A2,FUNCT_1:11;
  then (f+~(x,y)).z = ((x .--> y)*f).z by Th13
    .= (x .--> y).(f.z) by A1,FUNCT_1:13
    .= y by A2,FUNCOP_1:72;
  hence thesis by A4,A3,FUNCT_1:3;
end;

theorem Th102:
  f+~(x,x) = f
proof
  thus f+~(x,x) = f+*((id{x})*f) by Th96
    .= f by Th98,RELAT_1:50;
end;

theorem Th103:
  not x in rng f implies f+~(x,y) = f
proof
A1: dom (x .--> y) = {x} by FUNCOP_1:13;
  assume not x in rng f;
  then dom (x .--> y) misses rng f by A1,ZFMISC_1:50;
  then (x .--> y)*f = {} by RELAT_1:44;
  hence thesis by Th21;
end;

theorem
  rng(f+~(x,y)) c= rng f \ {x} \/ {y}
proof
  per cases;
  suppose
A1: not x in rng f;
    then f+~(x,y) = f by Th103;
    then rng(f+~(x,y)) = rng f \ {x} by A1,ZFMISC_1:57;
    hence thesis by XBOOLE_1:7;
  end;
  suppose that
A2: x in rng f and
A3: x = y;
    f+~(x,y) = f by A3,Th102;
    hence thesis by A2,A3,ZFMISC_1:116;
  end;
  suppose that
A4: x <> y;
    not x in rng(f+~(x,y)) by A4,Th100;
    then
A5: rng(f+~(x,y)) \ {x} = rng(f+~(x,y)) by ZFMISC_1:57;
    rng(x .--> y) = {y} by FUNCOP_1:8;
    then
A6: rng f \/ rng((x .--> y)*f) c= rng f \/ {y} by RELAT_1:26,XBOOLE_1:9;
    rng(f+~(x,y)) c= rng f \/ rng((x .--> y)*f) by Th17;
    then rng(f+~(x,y)) c= rng f \/ {y} by A6,XBOOLE_1:1;
    then rng(f+~(x,y)) c= rng f \/ {y} \ {x} by A5,XBOOLE_1:33;
    hence thesis by A4,ZFMISC_1:123;
  end;
end;

theorem
  f.z <> x implies (f+~(x,y)).z = f.z
proof
  assume f.z <> x;
  then not f.z in dom(x.-->y) by FUNCOP_1:75;
  then not z in dom((x.-->y)*f) by FUNCT_1:11;
  hence thesis by Th11;
end;

theorem
  z in dom f & f.z = x implies (f+~(x,y)).z = y
proof
  assume that
A1: z in dom f and
A2: f.z = x;
  f.z in dom(x.-->y) by A2,FUNCOP_1:74;
  then
A3: z in dom((x.-->y)*f) by A1,FUNCT_1:11;
  hence (f+~(x,y)).z = ((x.-->y)*f).z by Th13
    .= (x.-->y).x by A2,A3,FUNCT_1:12
    .= y by FUNCOP_1:72;
end;



theorem
  not x in dom f implies f c= f +*(x .--> y)
proof
  assume not x in dom f;
  then dom f misses {x} by ZFMISC_1:50;
  then dom f misses dom(x .--> y) by FUNCOP_1:13;
  hence thesis by Th32;
end;

theorem
  for f being PartFunc of X,Y, x,y st x in X & y in Y holds f+*(x .--> y
  ) is PartFunc of X,Y
proof
  let f be PartFunc of X,Y, x,y;
  assume that
A1: x in X and
A2: y in Y;
A3: {x} c= X by A1,ZFMISC_1:31;
  {y} c= Y by A2,ZFMISC_1:31;
  then rng(x .--> y) c= Y by FUNCOP_1:8;
  then
A4: rng f \/ rng(x .--> y) c= Y by XBOOLE_1:8;
  rng(f+*(x .--> y)) c= rng f \/ rng(x .--> y) by Th17;
  then
A5: rng(f+*(x .--> y)) c= Y by A4,XBOOLE_1:1;
  dom(f+*(x .--> y)) = dom f \/ dom(x .--> y) by Def1
    .= dom f \/ {x} by FUNCOP_1:13;
  hence thesis by A3,A5,RELSET_1:4,XBOOLE_1:8;
end;



registration
  let f be Function, g be non empty Function;
  cluster f +* g -> non empty;
  coherence
  proof
    dom (f+*g) = dom f \/ dom g by Def1;
    hence thesis;
  end;
  cluster g +* f -> non empty;
  coherence
  proof
    dom (g+*f) = dom g \/ dom f by Def1;
    hence thesis;
  end;
end;



registration
  let f,g be non-empty Function;
  cluster f+*g -> non-empty;
  coherence
  proof
    set h = f+*g;
A1: dom (f+*g) = dom f \/ dom g by Def1;
    not {} in rng h
    proof
      assume {} in rng h;
      then consider x being set such that
A2:   x in dom (f+*g) & {} = h.x by FUNCT_1:def 3;
      not x in dom g or x in dom g;
      then {} = f.x & x in dom f or {} = g.x & x in dom g by A1,A2,Def1,
XBOOLE_0:def 3;
      then {} in rng f or {} in rng g by FUNCT_1:def 3;
      hence thesis by RELAT_1:def 9;
    end;
    hence thesis by RELAT_1:def 9;
  end;
end;

definition let X,Y be set;
  let f,g be PartFunc of X,Y;
  redefine func f+* g ->  PartFunc of X,Y;
 coherence
  proof
A1:  dom(f+* g) c= dom f \/ dom g by Def1;
A2:  dom(f+* g) c= X by A1,XBOOLE_1:1;
A3:  rng(f+* g) c= rng f \/ rng g by Th17;
    rng(f+* g) c= Y by A3,XBOOLE_1:1;
    then f+* g is Relation of X,Y by A2,RELSET_1:4;
   hence thesis;
  end;
end;



theorem
  dom ((x --> y)+*(x .-->z)) = succ x
proof
  thus dom ((x --> y)+*(x .-->z))
      = dom (x --> y) \/ dom (x .-->z) by Def1
    .= x \/ dom (x .-->z) by FUNCOP_1:13
    .= x \/ {x} by FUNCOP_1:13
    .= succ x by ORDINAL1:def 1;
end;

theorem
  dom ((x --> y)+*(x .-->z)+*(succ x .-->z)) = succ succ x
proof
  thus dom ((x --> y)+*(x .-->z)+*(succ x .-->z))
      = dom ((x --> y)+*(x .-->z)) \/ dom(succ x .-->z) by Def1
     .= dom (x --> y) \/ dom (x .-->z) \/ dom(succ x .-->z) by Def1

    .= x \/ dom (x .-->z) \/ dom(succ x .-->z) by FUNCOP_1:13
    .= x \/ {x} \/ dom(succ x .-->z) by FUNCOP_1:13
    .= x \/ {x} \/ {succ x} by FUNCOP_1:13
    .= succ x \/ {succ x} by ORDINAL1:def 1
    .= succ succ x by ORDINAL1:def 1;
end;



registration let f,g be Function-yielding Function;
  cluster f+*g -> Function-yielding;
  coherence
  proof
   let x be set;
   assume x in dom(f+*g);
    then
A1:   x in dom f \/ dom g by Def1;
   per cases by A1,XBOOLE_0:def 3;
   suppose
   x in dom g;
    then (f+*g).x = g.x by Th13;
   hence (f+*g).x is Function;
   end;
   suppose
   x in dom f & not x in dom g;
    then (f+*g).x = f.x by Th11;
   hence (f+*g).x is Function;
   end;
  end;
end;



registration
  let I be set;
  let f,g be I-defined Function;
  cluster f+*g -> I-defined;
  coherence
  proof
    dom (f+*g) = dom f \/ dom g by Def1;
    then dom (f+*g) c= I;
    hence thesis by RELAT_1:def 18;
  end;
end;

registration
  let I be set;
  let f be total I-defined Function;
  let g be I-defined Function;
  cluster f+*g -> total for I-defined Function;
  coherence
  proof
    dom f = I by PARTFUN1:def 2;
    then dom (f+*g) = I \/ dom g by Def1
      .= I by XBOOLE_1:12;
    hence thesis by PARTFUN1:def 2;
  end;
  cluster g+*f -> total for I-defined Function;
  coherence
  proof
    dom f = I by PARTFUN1:def 2;
    then dom (g+*f) = I \/ dom g by Def1
      .= I by XBOOLE_1:12;
    hence thesis by PARTFUN1:def 2;
  end;
end;

registration let I be set;
  let g,h be I-valued Function;
  cluster g+*h -> I-valued;
 coherence
  proof
A1:  rng(g+*h) c= rng g \/ rng h by Th17;
   rng(g+*h) c= I by A1,XBOOLE_1:1;
   hence thesis by RELAT_1:def 19;
  end;
end;

registration let f be Function;
  let g,h be f-compatible Function;
  cluster g+*h -> f-compatible;
 coherence
  proof
   let x;
   assume
A1:  x in dom(g+*h);
A2: dom(g+*h) = dom g \/ dom h by Def1;
   per cases by A1,A2,XBOOLE_0:def 3;
   suppose
A3:   x in dom g & not x in dom h;
    then g.x in f.x by FUNCT_1:def 14;
   hence (g+*h).x in f.x by A3,Th11;
   end;
   suppose
A4:  x in dom h;
    then h.x in f.x by FUNCT_1:def 14;
   hence (g+*h).x in f.x by A4,Th13;
   end;

  end;
end;



theorem
  f|A +* f = f by Th97,RELAT_1:59;



theorem
  for R being Relation st dom R = {x} & rng R = {y}
    holds R = x .--> y
proof let R be Relation;
  assume that
A1: dom R = {x} and
A2: rng R = {y};
  set g = x .--> y;
A3: g = {[x,y]} by ZFMISC_1:29;
  for a, b being set holds [a,b] in R iff [a,b] in g
  proof
    let a, b be set;
    hereby
      assume
A4:   [a,b] in R;
      then b in rng R by XTUPLE_0:def 13;
      then
A5:   b = y by A2,TARSKI:def 1;
      a in dom R by A4,XTUPLE_0:def 12;
      then a = x by A1,TARSKI:def 1;
      hence [a,b] in g by A3,A5,TARSKI:def 1;
    end;
    assume [a,b] in g;
    then
A6: [a,b] = [x,y] by A3,TARSKI:def 1;
    then
A7: b = y by XTUPLE_0:1;
    a = x by A6,XTUPLE_0:1;
    then a in dom R by A1,TARSKI:def 1;
    then consider z being set such that
A8: [a,z] in R by XTUPLE_0:def 12;
    z in rng R by A8,XTUPLE_0:def 13;
    hence thesis by A2,A7,A8,TARSKI:def 1;
  end;
  hence thesis by RELAT_1:def 2;
end;

theorem
  (f +* (x .-->y)).x = y
proof
A1: x in {x} by TARSKI:def 1;
  then dom(x .--> y) = {x} & x in dom f \/ {x} by FUNCOP_1:13,XBOOLE_0:def 3;
 hence (f +* (x .-->y) ).x = (x .-->y).x by A1,Def1
    .= y by FUNCOP_1:72;
end;

theorem
 f +* (x .--> z1) +* (x .--> z2) = f +* (x .--> z2)
proof
A1: dom (x .--> z1) = {x} by FUNCOP_1:13
    .= dom (x .--> z2) by FUNCOP_1:13;
  thus f +* (x .--> z1) +* (x .--> z2)
   = f +* ((x .--> z1) +* (x .--> z2)) by Th14
    .= f +* (x .--> z2) by A1,Th19;
end;

registration let A be non empty set, a,b be Element of A, x,y be set;
  cluster (a,b) --> (x,y) -> A-defined;
  coherence;
end;

theorem
  dom g misses dom h implies f +* g +* h +* g = f +* g +* h
 proof assume
A1: dom g misses dom h;
  thus f +* g +* h +* g
     = f +* (g +* h) +* g by Th14
    .= f +* (g +* h +* g) by Th14
    .= f +* (h +* g +* g) by A1,Th35
    .= f +* (h +* g)
    .= f +* (g +* h) by A1,Th35
    .= f +* g +* h by Th14;
 end;

theorem
  dom f misses dom h & f c= g +* h implies f c= g
proof assume that
A1: dom f misses dom h and
A2: f c= g +* h;
A3: (g +* h)|dom f = g|dom f +* h|dom f by Th71
      .= g|dom f +* {} by A1,RELAT_1:66
      .= g|dom f;
  f|dom f = f;
  then f c= g|dom f by A2,A3,RELAT_1:76;
 hence f c= g by RELAT_1:184;
end;

theorem Th117:
  dom f misses dom h & f c= g implies f c= g +* h
proof assume that
A1: dom f misses dom h and
A2: f c= g;
A3: (g +* h)|dom f = g|dom f +* h|dom f by Th71
      .= g|dom f +* {} by A1,RELAT_1:66
      .= g|dom f;
  f|dom f = f;
  then f c= (g +* h)|dom f by A2,A3,RELAT_1:76;
 hence f c= g +* h by RELAT_1:184;
end;

theorem
 dom g misses dom h implies f +* g +* h = f +* h +* g
proof assume
A1: dom g misses dom h;
 thus f +* g +* h = f +* (g +* h) by Th14
  .= f +* (h +* g) by A1,Th35
  .= f +* h +* g by Th14;
end;

theorem
 dom f misses dom g implies (f +* g) \ g = f
proof assume
A1: dom f misses dom g;
  then
A2: f misses g by RELAT_1:179;
 thus (f +* g) \ g
     = (f \/ g) \ g by A1,Th31
    .= f by A2,XBOOLE_1:88;
end;

theorem
  dom f misses dom g implies f \ g = f
  proof
    assume dom f misses dom g;
    then f misses g by RELAT_1:179;
    hence thesis by XBOOLE_1:83;
  end;

theorem
  dom g misses dom h implies (f \ g) +* h = (f +* h) \ g
  proof
    assume
A1: dom g misses dom h;
A2: dom(f+*h) = dom f \/ dom h by Def1;
A3: dom((f\g)+*h) = dom(f\g) \/ dom h by Def1;
A4: dom((f\g)+*h) = dom((f+*h)\g)
    proof
      thus dom((f\g)+*h) c= dom((f+*h)\g)
      proof
        let x be set;
        assume
A5:     x in dom((f\g)+*h);
        per cases by A3,A5,XBOOLE_0:def 3;
        suppose that
A6:       x in dom(f\g) and
A7:       not x in dom h;
          consider y being set such that
A8:       [x,y] in f\g by A6,XTUPLE_0:def 12;
A9:       x in dom f by A8,XTUPLE_0:def 12;
          then
A10:       x in dom(f+*h) by A2,XBOOLE_0:def 3;
A11:       not [x,y] in g by A8,XBOOLE_0:def 5;
A12:       (f+*h).x = f.x by A7,Th11;
          f.x = y by A8,A9,FUNCT_1:def 2;
          then [x,y] in f+*h by A12,A10,FUNCT_1:def 2;
          then [x,y] in (f+*h)\g by A11,XBOOLE_0:def 5;
          hence thesis by XTUPLE_0:def 12;
        end;
        suppose
A13:       x in dom h;
          then
A14:       not x in dom g by A1,XBOOLE_0:3;
          x in dom(f+*h) by A2,A13,XBOOLE_0:def 3;
          then
A15:       x in dom(f+*h) \ dom g by A14,XBOOLE_0:def 5;
          dom(f+*h) \ dom g c= dom((f+*h)\g) by RELAT_1:3;
          hence thesis by A15;
        end;
      end;
      let x be set;
      assume x in dom((f+*h)\g);
      then consider y being set such that
A16:   [x,y] in (f+*h)\g by XTUPLE_0:def 12;
A17:   x in dom(f+*h) by A16,XTUPLE_0:def 12;
      then
A18:   y = (f+*h).x by A16,FUNCT_1:def 2;
      per cases by A2,A17,XBOOLE_0:def 3;
      suppose that
A19:     x in dom f and
A20:     not x in dom h;
A21:     not [x,y] in g by A16,XBOOLE_0:def 5;
        (f+*h).x = f.x by A20,Th11;
        then [x,y] in f by A19,A18,FUNCT_1:def 2;
        then [x,y] in f\g by A21,XBOOLE_0:def 5;
        then x in dom(f\g) by XTUPLE_0:def 12;
        hence thesis by A3,XBOOLE_0:def 3;
      end;
      suppose x in dom h;
        hence thesis by A3,XBOOLE_0:def 3;
      end;
    end;
    now
      let x be set;
      assume
A22:   x in dom((f\g)+*h);
      per cases by A3,A22,XBOOLE_0:def 3;
      suppose that
A23:     x in dom(f\g) and
A24:     not x in dom h;
        thus ((f\g)+*h).x = (f\g).x by A24,Th11
        .= f.x by A23,GRFUNC_1:2
        .= (f+*h).x by A24,Th11
        .= ((f+*h)\g).x by A4,A22,GRFUNC_1:2;
      end;
      suppose
A25:     x in dom h;
        then
A26:     not x in dom g by A1,XBOOLE_0:3;
        x in dom(f+*h) by A25,A2,XBOOLE_0:def 3;
        then x in dom(f+*h) \ dom g by A26,XBOOLE_0:def 5;
        hence ((f+*h)\g).x = (f+*h).x by GRFUNC_1:32
        .= h.x by A25,Th13
        .= ((f\g)+*h).x by A25,Th13;
      end;
    end;
    hence thesis by A4,FUNCT_1:2;
  end;

theorem
 for f1,f2,g1,g2 being Function
  st f1 c= g1 & f2 c= g2 & dom f1 misses dom g2
 holds f1 +* f2 c= g1 +* g2
 proof
  let f1,f2,g1,g2 be Function such that
A1: f1 c= g1 and
A2: f2 c= g2 and
A3: dom f1 misses dom g2;
A4: f1 c= g1 +* g2 by A1,A3,Th117;
   g2 c= g1 +* g2 by Th25;
   then f2 c= g1 +* g2 by A2,XBOOLE_1:1;
  hence f1 +* f2 c= g1 +* g2 by A4,Th87;
 end;

theorem Th123:
  for f, g, h being Function st f c= g holds f +* h c= g +* h
proof
  let f, g, h be Function;
  assume
A1: f c= g;
  now
    dom (f +* h) = dom f \/ dom h & dom (g +* h) = dom g \/ dom h by Def1;
    hence dom (f +* h) c= dom (g +* h) by A1,RELAT_1:11,XBOOLE_1:9;
    let x be set;
    assume x in dom (f +* h);
    then
A2: x in dom f or x in dom h by Th12;
    per cases;
    suppose
A3:   x in dom h;
      hence (f +* h).x = h.x by Th13
        .= (g +* h).x by A3,Th13;
    end;
    suppose
A4:   not x in dom h;
      then (f +* h).x = f.x & (g +* h).x = g.x by Th11;
      hence (f +* h).x = (g +* h).x by A1,A2,A4,GRFUNC_1:2;
    end;
  end;
  hence thesis by GRFUNC_1:2;
end;

theorem
  for f, g, h being Function st f c= g & dom f misses dom h holds f c= g +* h
proof
  let f, g, h be Function;
  assume f c= g;
  then
A1: f +* h c= g +* h by Th123;
  assume dom f misses dom h;
  then f c= f +* h by Th32;
  hence thesis by A1,XBOOLE_1:1;
end;

registration
  let x, y be set;
  cluster x .--> y -> trivial;
  coherence
  proof
    x .--> y = {[x,y]} by ZFMISC_1:29;
    hence thesis;
  end;
end;



theorem
  for f,g,h being Function st f tolerates g & g tolerates h & h
  tolerates f holds f+*g tolerates h
proof
  let f,g,h be Function such that
A1: f tolerates g and
A2: g tolerates h and
A3: h tolerates f;
  let x be set;
  assume
A4: x in (dom (f+*g)) /\ dom h;
  then x in dom (f+*g) by XBOOLE_0:def 4;
  then
A5: x in dom f or x in dom g by Th12;
  x in dom h by A4,XBOOLE_0:def 4;
  then
  x in (dom h) /\ dom f & (f+*g).x = f.x or x in (dom g) /\ dom h & (f+*g)
  .x = g.x by A1,A5,Th13,Th15,XBOOLE_0:def 4;
  hence thesis by A2,A3,PARTFUN1:def 4;
end;















begin









theorem
  {} is Element of omega
proof
  {} in omega by ORDINAL1:def 11;
  hence thesis by SUBSET_1:def 1;
end;
 


















begin

reserve A,A1,A2,B,C,D for Ordinal,
  X,Y for set,
  x,y,a,b,c for set,
  L,L1,L2,L3 for T-Sequence,
  f for Function;

scheme
  OrdinalInd { P[Ordinal] } : for A holds P[A]
provided
A1: P[{}] and
A2: for A st P[A] holds P[succ A] and
A3: for A st A <> {} & A is limit_ordinal & for B st B in A holds P[B]
holds P[A]
proof
A4: for A st for B st B in A holds P[B] holds P[A]
  proof
    let A such that
A5: for B st B in A holds P[B];
A6: now
      assume that
A7:   A <> {} and
A8:   for B holds A <> succ B;
      A is limit_ordinal by A8,ORDINAL1:29;
      hence thesis by A3,A5,A7;
    end;
    now
      given B such that
A9:   A = succ B;
      B in A by A9,ORDINAL1:6;
      hence thesis by A2,A5,A9;
    end;
    hence thesis by A1,A6;
  end;
  thus for A holds P[A] from ORDINAL1:sch 2(A4);
end;

theorem Th1:
  A c= B iff succ A c= succ B
proof
  A c= B iff A in succ B by ORDINAL1:22;
  hence thesis by ORDINAL1:21;
end;

theorem Th2:
  union succ A = A
proof
  thus union succ A c= A
  proof
    let x;
    assume x in union succ A;
    then consider X such that
A1: x in X and
A2: X in succ A by TARSKI:def 4;
    reconsider X as Ordinal by A2;
    X c= A by A2,ORDINAL1:22;
    hence thesis by A1;
  end;
  thus thesis by ORDINAL1:6,ZFMISC_1:74;
end;

theorem
  succ A c= bool A
proof
  let x;
  assume
A1: x in succ A;
  then reconsider B = x as Ordinal;
  x in A or x = A by A1,ORDINAL1:8;
  then B c= A by ORDINAL1:def 2;
  hence thesis;
end;

theorem
  {} is limit_ordinal
proof
  thus {} = union {} by ZFMISC_1:2;
end;

theorem Th5:
  union A c= A
proof
  let x;
  assume x in union A;
  then consider Y such that
A1: x in Y and
A2: Y in A by TARSKI:def 4;
  Y c= A by A2,ORDINAL1:def 2;
  hence thesis by A1;
end;

definition
  let L;
  func last L -> set equals
  L.(union dom L);
  coherence;
end;

theorem
  dom L = succ A implies last L = L.A by Th2;

theorem
  On X c= X
proof
  thus x in On X implies x in X by ORDINAL1:def 9;
end;

theorem Th8:
  On A = A
proof
  x in A iff x in A & x is Ordinal;
  hence thesis by ORDINAL1:def 9;
end;

theorem Th9:
  X c= Y implies On X c= On Y
proof
  assume
A1: X c= Y;
  let x;
  assume x in On X;
  then x in X & x is Ordinal by ORDINAL1:def 9;
  hence thesis by A1,ORDINAL1:def 9;
end;

theorem
  Lim X c= X
proof
  thus x in Lim X implies x in X by ORDINAL1:def 10;
end;

theorem
  X c= Y implies Lim X c= Lim Y
proof
  assume
A1: X c= Y;
  let x;
  assume x in Lim X;
  then x in X & ex A st x = A & A is limit_ordinal by ORDINAL1:def 10;
  hence thesis by A1,ORDINAL1:def 10;
end;

theorem
  Lim X c= On X
proof
  let x;
  assume x in Lim X;
  then x in X & ex A st x = A & A is limit_ordinal by ORDINAL1:def 10;
  hence thesis by ORDINAL1:def 9;
end;

theorem Th13:
  (for x st x in X holds x is Ordinal) implies meet X is Ordinal
proof
  assume
A1: for x st x in X holds x is Ordinal;
  now
    defpred P[Ordinal] means $1 in X;
    set x = the Element of X;
    assume
A2: X <> {};
    then x is Ordinal by A1;
    then
A3: ex A st P[A] by A2;
    consider A such that
A4: P[A] & for B st P[B] holds A c= B from ORDINAL1:sch 1(A3);
    for x holds x in A iff for Y st Y in X holds x in Y
    proof
      let x;
      thus x in A implies for Y st Y in X holds x in Y
      proof
        assume
A5:     x in A;
        let Y;
        assume
A6:     Y in X;
        then reconsider B = Y as Ordinal by A1;
        A c= B by A4,A6;
        hence thesis by A5;
      end;
      assume for Y st Y in X holds x in Y;
      hence thesis by A4;
    end;
    hence thesis by A2,SETFAM_1:def 1;
  end;
  hence thesis by SETFAM_1:def 1;
end;

registration
  cluster limit_ordinal for Ordinal;
  existence
  proof
    take omega;
    thus thesis by ORDINAL1:def 11;
  end;
end;

definition
  let X;

  func inf X -> Ordinal equals
  meet On X;
  coherence
  proof
    x in On X implies x is Ordinal by ORDINAL1:def 9;
    hence thesis by Th13;
  end;
  func sup X -> Ordinal means
  :Def3:
  On X c= it & for A st On X c= A holds it c= A;
  existence
  proof
    defpred P[Ordinal] means On X c= $1;
    x in On X implies x is Ordinal by ORDINAL1:def 9;
    then reconsider A = union On X as Ordinal by ORDINAL1:23;
    On X c= succ A
    proof
      let x;
      assume
A1:   x in On X;
      then reconsider B = x as Ordinal by ORDINAL1:def 9;
      B c= A by A1,ZFMISC_1:74;
      hence thesis by ORDINAL1:22;
    end;
    then
A2: ex A st P[A];
    thus ex F being Ordinal st P[F] & for A st P[A] holds F c= A from ORDINAL1
    :sch 1(A2);
  end;
  uniqueness
  proof
    let B,C;
    assume ( On X c= B & for A st On X c= A holds B c= A )&( On X c= C & for
    A st On X c= A holds C c= A );
    hence B c= C & C c= B;
  end;
end;

theorem
  A in X implies inf X c= A
proof
  assume A in X;
  then A in On X by ORDINAL1:def 9;
  hence thesis by SETFAM_1:3;
end;

theorem
  On X <> {} & (for A st A in X holds D c= A) implies D c= inf X
proof
  assume that
A1: On X <> {} and
A2: for A st A in X holds D c= A;
  let x such that
A3: x in D;
  for Y st Y in On X holds x in Y
  proof
    let Y;
    assume
A4: Y in On X;
    then reconsider A = Y as Ordinal by ORDINAL1:def 9;
    A in X by A4,ORDINAL1:def 9;
    then D c= A by A2;
    hence thesis by A3;
  end;
  hence thesis by A1,SETFAM_1:def 1;
end;

theorem
  A in X & X c= Y implies inf Y c= inf X
proof
  assume A in X;
  then
A1: On X <> {} by ORDINAL1:def 9;
  assume X c= Y;
  then On X c= On Y by Th9;
  hence thesis by A1,SETFAM_1:6;
end;

theorem
  A in X implies inf X in X
proof
  defpred P[Ordinal] means $1 in X;
  assume A in X;
  then
A1: ex A st P[A];
  consider A such that
A2: P[A] & for B st P[B] holds A c= B from ORDINAL1:sch 1(A1);
A3: A in On X by A2,ORDINAL1:def 9;
A4: now
    let x;
    thus x in A implies for Y st Y in On X holds x in Y
    proof
      assume
A5:   x in A;
      let Y;
      assume
A6:   Y in On X;
      then reconsider B = Y as Ordinal by ORDINAL1:def 9;
      Y in X by A6,ORDINAL1:def 9;
      then A c= B by A2;
      hence thesis by A5;
    end;
    assume for Y st Y in On X holds x in Y;
    hence x in A by A3;
  end;
  On X <> {} by A2,ORDINAL1:def 9;
  hence thesis by A2,A4,SETFAM_1:def 1;
end;

theorem Th18:
  sup A = A
proof
  On A = A & for B st On A c= B holds A c= B by Th8;
  hence thesis by Def3;
end;

theorem Th19:
  A in X implies A in sup X
proof
  assume A in X;
  then
A1: A in On X by ORDINAL1:def 9;
  On X c= sup X by Def3;
  hence thesis by A1;
end;

theorem Th20:
  (for A st A in X holds A in D) implies sup X c= D
proof
  assume
A1: for A st A in X holds A in D;
  On X c= D
  proof
    let x;
    assume
A2: x in On X;
    then reconsider A = x as Ordinal by ORDINAL1:def 9;
    A in X by A2,ORDINAL1:def 9;
    hence thesis by A1;
  end;
  hence thesis by Def3;
end;

theorem
  A in sup X implies ex B st B in X & A c= B
proof
  assume that
A1: A in sup X and
A2: for B st B in X holds not A c= B;
  for B st B in X holds B in A by A2,ORDINAL1:16;
  then sup X c= A by Th20;
  hence contradiction by A1,ORDINAL1:5;
end;

theorem
  X c= Y implies sup X c= sup Y
proof
  assume X c= Y;
  then
A1: On X c= On Y by Th9;
  On Y c= sup Y by Def3;
  then On X c= sup Y by A1,XBOOLE_1:1;
  hence thesis by Def3;
end;

theorem
  sup { A } = succ A
proof
A1: On { A } c= succ A
  proof
    let x;
    assume x in On { A };
    then x in { A } by ORDINAL1:def 9;
    then x = A by TARSKI:def 1;
    hence thesis by ORDINAL1:6;
  end;
  now
    A in { A } by TARSKI:def 1;
    then
A2: A in On { A } by ORDINAL1:def 9;
    let B;
    assume On { A } c= B;
    hence succ A c= B by A2,ORDINAL1:21;
  end;
  hence thesis by A1,Def3;
end;

theorem
  inf X c= sup X
proof
  let x;
  set y = the Element of On X;
  assume
A1: x in inf X;
  then reconsider y as Ordinal by ORDINAL1:def 9,SETFAM_1:1;
  On X c= sup X by Def3;
  then y in sup X by A1,SETFAM_1:1,TARSKI:def 3;
  then
A2: y c= sup X by ORDINAL1:def 2;
  x in y by A1,SETFAM_1:1,def 1;
  hence thesis by A2;
end;

scheme
  TSLambda { A()->Ordinal, F(Ordinal)->set } : ex L st dom L = A() & for A st
  A in A() holds L.A = F(A) proof
  deffunc G(set) = F(sup union { $1 });
  consider f such that
A1: dom f = A() & for x st x in A() holds f.x = G(x) from FUNCT_1:sch 3;
  reconsider f as T-Sequence by A1,ORDINAL1:def 7;
  take L = f;
  thus dom L = A() by A1;
  let A;
  assume A in A();
  hence L.A = F(sup union { A }) by A1
    .= F(sup A) by ZFMISC_1:25
    .= F(A) by Th18;
end;

definition
  let f;
  attr f is Ordinal-yielding means
  :Def4:
  ex A st rng f c= A;
end;

registration
  cluster Ordinal-yielding for T-Sequence;
  existence
  proof
    set A = the Ordinal;
    set L = the T-Sequence of A;
    take L,A;
    thus thesis by RELAT_1:def 19;
  end;
end;

definition
  mode Ordinal-Sequence is Ordinal-yielding T-Sequence;
end;

registration
  let A;
  cluster -> Ordinal-yielding for T-Sequence of A;
  coherence
  proof
    let L be T-Sequence of A;
    take A;
    thus thesis by RELAT_1:def 19;
  end;
end;

registration
  let L be Ordinal-Sequence;
  let A;
  cluster L|A -> Ordinal-yielding;
  coherence
  proof
    consider B such that
A1: rng L c= B by Def4;
    L|A is Ordinal-yielding
    proof
      take B;
      rng(L|A) c= rng L by RELAT_1:70;
      hence thesis by A1,XBOOLE_1:1;
    end;
    hence thesis;
  end;
end;

reserve fi,psi for Ordinal-Sequence;

theorem Th25:
  A in dom fi implies fi.A is Ordinal
proof
  assume A in dom fi;
  then
A1: fi.A in rng fi by FUNCT_1:def 3;
  ex B st rng fi c= B by Def4;
  hence thesis by A1;
end;

registration
  let f be Ordinal-Sequence, a be Ordinal;
  cluster f.a -> ordinal;
  coherence
  proof
    a in dom f or not a in dom f;
    hence thesis by Th25,FUNCT_1:def 2;
  end;
end;

scheme
  OSLambda { A()->Ordinal, F(Ordinal)->Ordinal } : ex fi st dom fi = A() & for
  A st A in A() holds fi.A = F(A) proof
  consider L such that
A1: dom L = A() & for A st A in A() holds L.A = F(A) from TSLambda;
  L is Ordinal-yielding
  proof
    take sup rng L;
    let x;
    assume
A2: x in rng L;
    then consider y such that
A3: y in dom L and
A4: x = L.y by FUNCT_1:def 3;
    reconsider y as Ordinal by A3;
    L.y = F(y) by A1,A3;
    then
A5: x in On rng L by A2,A4,ORDINAL1:def 9;
    On rng L c= sup rng L by Def3;
    hence thesis by A5;
  end;
  then reconsider L as Ordinal-Sequence;
  take fi = L;
  thus dom fi = A() by A1;
  let A;
  assume A in A();
  hence thesis by A1;
end;

scheme
  TSUniq1 { A()->Ordinal, B()->set, C(Ordinal,set)->set, D(Ordinal,T-Sequence)
  ->set, L1()->T-Sequence, L2()->T-Sequence } : L1() = L2()
provided
A1: dom L1() = A() and
A2: {} in A() implies L1().{} = B() and
A3: for A st succ A in A() holds L1().(succ A) = C(A,L1().A) and
A4: for A st A in A() & A <> {} & A is limit_ordinal holds L1().A = D(A,
L1()|A) and
A5: dom L2() = A() and
A6: {} in A() implies L2().{} = B() and
A7: for A st succ A in A() holds L2().(succ A) = C(A,L2().A) and
A8: for A st A in A() & A <> {} & A is limit_ordinal holds L2().A = D(A,
L2()|A)
proof
  defpred P[set] means L1().$1 <> L2().$1;
  consider X such that
A9: Y in X iff Y in A() & P[Y] from XBOOLE_0:sch 1;
  for b holds b in X implies b in A() by A9;
  then
A10: X c= A() by TARSKI:def 3;
  assume L1() <> L2();
  then ex a st a in A() & L1().a <> L2().a by A1,A5,FUNCT_1:2;
  then X <> {} by A9;
  then consider B such that
A11: B in X and
A12: for C st C in X holds B c= C by A10,ORDINAL1:20;
A13: B in A() by A9,A11;
  then
A14: B c= A() by ORDINAL1:def 2;
  then
A15: dom(L1()|B) = B & dom(L2()|B) = B by A1,A5,RELAT_1:62;
A16: now
    let C;
    assume
A17: C in B;
    then not C in X by A12,ORDINAL1:5;
    hence L1().C = L2().C by A9,A14,A17;
  end;
A18: now
    let a;
    assume
A19: a in B;
    L1()|B.a = L1().a & L2()|B.a = L2().a by A15,A19,FUNCT_1:47;
    hence L1()|B.a = L2()|B.a by A16,A19;
  end;
A20: now
    given C such that
A21: B = succ C;
A22: L1().C = L1()|B.C & L2().C = L2()|B.C by A21,FUNCT_1:49,ORDINAL1:6;
    L1().B = C(C,L1().C) & L2().B = C(C,L2().C) by A3,A7,A13,A21;
    hence L1().B = L2().B by A18,A21,A22,ORDINAL1:6;
  end;
  now
    assume that
A23: B <> {} and
A24: for C holds B <> succ C;
    B is limit_ordinal by A24,ORDINAL1:29;
    then L1().B = D(B,L1()|B) & L2().B = D(B,L2()|B) by A4,A8,A13,A23;
    hence L1().B = L2().B by A15,A18,FUNCT_1:2;
  end;
  hence contradiction by A2,A6,A9,A11,A20;
end;

scheme
  TSExist1 { A()->Ordinal, B()->set, C(Ordinal,set)->set, D(Ordinal,T-Sequence
  )->set } : ex L st dom L = A() & ({} in A() implies L.{} = B() ) & (for A st
  succ A in A() holds L.(succ A) = C(A,L.A) ) & for A st A in A() & A <> {} & A
  is limit_ordinal holds L.A = D(A,L|A) proof
  defpred P[Ordinal,T-Sequence] means dom $2 = $1 & ({} in $1 implies $2.{} =
B() ) & (for A st succ A in $1 holds $2.(succ A) = C(A,$2.A) ) & for A st A in
  $1 & A <> {} & A is limit_ordinal holds $2.A = D(A,$2|A);
  defpred R[Ordinal] means ex L st P[$1,L];
A1: for B st for C st C in B holds R[C] holds R[B]
  proof
    defpred R[set,set] means $1 is Ordinal & $2 is T-Sequence & for A,L st A =
    $1 & L = $2 holds P[A,L];
    let B such that
A2: for C st C in B ex L st P[C,L];
A3: for a,b,c st R[a,b] & R[a,c] holds b = c
    proof
      let a,b,c;
      assume that
A4:   a is Ordinal and
A5:   b is T-Sequence and
A6:   for A,L st A = a & L = b holds P[A,L] and
      a is Ordinal and
A7:   c is T-Sequence and
A8:   for A,L st A = a & L = c holds P[ A,L];
      reconsider a as Ordinal by A4;
      reconsider c as T-Sequence by A7;
A9:   dom c = a by A8;
A10:  for A st A in a & A <> {} & A is limit_ordinal holds c.A = D(A,c|A)
      by A8;
A11:  for A st succ A in a holds c.(succ A) = C(A,c.A) by A8;
A12:  {} in a implies c.{} = B() by A8;
      reconsider b as T-Sequence by A5;
A13:  {} in a implies b.{} = B() by A6;
A14:  for A st succ A in a holds b.(succ A) = C(A,b.A) by A6;
A15:  for A st A in a & A <> {} & A is limit_ordinal holds b.A = D(A,b|A)
      by A6;
A16:  dom b = a by A6;
      b = c from TSUniq1(A16,A13,A14,A15,A9,A12,A11,A10);
      hence thesis;
    end;
    consider G being Function such that
A17: [a,b] in G iff a in B & R[a,b] from FUNCT_1:sch 1(A3);
    defpred Q[set,set] means ex A,L st A = $1 & L = G.$1 & (A = {} & $2 = B()
or (ex B st A = succ B & $2 = C(B,L.B)) or A <> {} & A is limit_ordinal & $2 =
    D(A,L));
A18: dom G = B
    proof
      thus a in dom G implies a in B
      proof
        assume a in dom G;
        then ex b st [a,b] in G by XTUPLE_0:def 12;
        hence thesis by A17;
      end;
      let a;
      assume
A19:  a in B;
      then reconsider a9 = a as Ordinal;
      consider L such that
A20:  P[a9,L] by A2,A19;
      for A holds for K be T-Sequence holds A = a9 & K = L implies P[A,K]
      by A20;
      then [a9,L] in G by A17,A19;
      hence thesis by XTUPLE_0:def 12;
    end;
A21: for a st a in B ex b st Q[a,b]
    proof
      let a;
      assume
A22:  a in B;
      then reconsider A = a as Ordinal;
      consider c such that
A23:  [a,c] in G by A18,A22,XTUPLE_0:def 12;
      reconsider L = c as T-Sequence by A17,A23;
A24:  now
        given C such that
A25:    A = succ C;
        thus ex b st Q[a,b]
        proof
          take C(C,L.C), A, L;
          thus A = a & L = G.a by A23,FUNCT_1:1;
          thus thesis by A25;
        end;
      end;
A26:  now
        assume
A27:    A <> {} & for C holds A <> succ C;
        thus Q[a,D(A,L)]
        proof
          take A,L;
          thus A = a & L = G.a by A23,FUNCT_1:1;
          thus thesis by A27,ORDINAL1:29;
        end;
      end;
      now
        assume
A28:    A = {};
        thus Q[a,B()]
        proof
          take A,L;
          thus A = a & L = G.a by A23,FUNCT_1:1;
          thus thesis by A28;
        end;
      end;
      hence thesis by A24,A26;
    end;
A29: for a,b,c st a in B & Q[a,b] & Q[a,c] holds b = c
    proof
      let a,b,c such that
      a in B;
      given Ab being Ordinal,Lb being T-Sequence such that
A30:  Ab = a and
A31:  Lb = G.a and
A32:  Ab = {} & b = B() or (ex B st Ab = succ B & b = C(B,Lb.B)) or
      Ab <> {} & Ab is limit_ordinal & b = D(Ab,Lb);
      given Ac being Ordinal,Lc being T-Sequence such that
A33:  Ac = a and
A34:  Lc = G.a and
A35:  Ac = {} & c = B() or (ex B st Ac = succ B & c = C(B,Lc.B)) or
      Ac <> {} & Ac is limit_ordinal & c = D(Ac,Lc);
      now
        given C such that
A36:    Ab = succ C;
        consider A such that
A37:    Ab = succ A and
A38:    b = C(A,Lb.A) by A32,A36,ORDINAL1:29;
        consider D such that
A39:    Ac = succ D and
A40:    c = C(D,Lc.D) by A30,A33,A35,A36,ORDINAL1:29;
        A = D by A30,A33,A37,A39,ORDINAL1:7;
        hence thesis by A31,A34,A38,A40;
      end;
      hence thesis by A30,A31,A32,A33,A34,A35;
    end;
    consider F being Function such that
A41: dom F = B & for a st a in B holds Q[a,F.a] from FUNCT_1:sch 2(A29
    ,A21);
    reconsider L = F as T-Sequence by A41,ORDINAL1:def 7;
    take L;
    thus dom L = B by A41;
    thus {} in B implies L.{} = B()
    proof
      assume {} in B;
      then ex A being Ordinal, K being T-Sequence st A = {} & K = G.{} & (A =
{} & F.{} = B() or (ex B st A = succ B & F.{} = C(B,K.B)) or A <> {} & A is
      limit_ordinal & F.{} = D(A,K)) by A41;
      hence thesis;
    end;
A42: for A,L1 st A in B & L1 = G.A holds L|A = L1
    proof
      defpred P[Ordinal] means for L1 st $1 in B & L1 = G.$1 holds L|$1 = L1;
A43:  for A st for C st C in A holds P[C] holds P[A]
      proof
        let A such that
        for C st C in A for L1 st C in B & L1 = G.C holds L|C = L1;
        let L1;
        assume that
A44:    A in B and
A45:    L1 = G.A;
A46:    [A,L1] in G by A18,A44,A45,FUNCT_1:1;
        then
A47:    P[A,L1] by A17;
A48:    now
          let x;
          assume
A49:      x in A;
          then reconsider x9 = x as Ordinal;
A50:      x9 in B by A44,A49,ORDINAL1:10;
          then consider A1,L2 such that
A51:      A1 = x9 and
A52:      L2 = G.x9 and
A53:      A1 = {} & L.x9 = B() or (ex B st A1 = succ B & L.x9 = C(B,
          L2.B)) or A1 <> {} & A1 is limit_ordinal & L.x9 = D(A1,L2) by A41;
          for D,L3 st D = x9 & L3 = L1|x9 holds P[D,L3]
          proof
            let D,L3 such that
A54:        D = x9 and
A55:        L3 = L1|x9;
            x9 c= A by A49,ORDINAL1:def 2;
            hence dom L3 = D by A47,A54,A55,RELAT_1:62;
            thus {} in D implies L3.{} = B() by A47,A49,A54,A55,FUNCT_1:49
,ORDINAL1:10;
            thus succ C in D implies L3.(succ C) = C(C,L3.C)
            proof
              assume
A56:          succ C in D;
              C in succ C by ORDINAL1:6;
              then
A57:          L1|x9.C = L1.C by A54,A56,FUNCT_1:49,ORDINAL1:10;
              succ C in A & L1|x9.succ C = L1.succ C by A49,A54,A56,FUNCT_1:49
,ORDINAL1:10;
              hence thesis by A17,A46,A55,A57;
            end;
            let C;
            assume that
A58:        C in D and
A59:        C <> {} & C is limit_ordinal;
            C c= x9 by A54,A58,ORDINAL1:def 2;
            then
A60:        L1|C = L3|C by A55,FUNCT_1:51;
            C in A by A49,A54,A58,ORDINAL1:10;
            then L1.C = D(C,L3|C) by A17,A46,A59,A60;
            hence thesis by A54,A55,A58,FUNCT_1:49;
          end;
          then [x9,L1|x9] in G by A17,A50;
          then
A61:      L1|x9 = L2 by A52,FUNCT_1:1;
A62:      L|A.x = L.x by A49,FUNCT_1:49;
          now
            given D such that
A63:        x9 = succ D;
A64:        L1.x = C(D,L1.D) by A17,A46,A49,A63;
            consider C such that
A65:        A1 = succ C and
A66:        L.x9 = C(C,L2.C) by A51,A53,A63,ORDINAL1:29;
            C = D by A51,A63,A65,ORDINAL1:7;
            hence L1.x = L|A.x by A62,A61,A63,A66,A64,FUNCT_1:49,ORDINAL1:6;
          end;
          hence L1.x = L|A.x by A17,A46,A49,A51,A53,A62,A61;
        end;
        A c= dom L by A41,A44,ORDINAL1:def 2;
        then dom(L|A) = A by RELAT_1:62;
        hence thesis by A47,A48,FUNCT_1:2;
      end;
      thus for A holds P[A] from ORDINAL1:sch 2(A43);
    end;
    thus succ A in B implies L.(succ A) = C(A,L.A)
    proof
      assume
A67:  succ A in B;
      then consider C being Ordinal, K being T-Sequence such that
A68:  C = succ A and
A69:  K = G.succ A and
A70:  C = {} & F.succ A = B() or (ex B st C = succ B & F.succ A = C(B
      ,K.B)) or C <> {} & C is limit_ordinal & F.succ A = D(C,K) by A41;
A71:  K = L|succ A by A42,A67,A69;
      consider D such that
A72:  C = succ D and
A73:  F.succ A = C(D,K.D) by A68,A70,ORDINAL1:29;
      A = D by A68,A72,ORDINAL1:7;
      hence thesis by A73,A71,FUNCT_1:49,ORDINAL1:6;
    end;
    let D;
    assume that
A74: D in B and
A75: D <> {} & D is limit_ordinal;
    ex A being Ordinal, K being T-Sequence st A = D & K = G.D & (A = {}
    & F.D = B() or (ex B st A = succ B & F.D = C(B,K.B)) or A <> {} & A is
    limit_ordinal & F.D = D(A,K)) by A41,A74;
    hence thesis by A42,A74,A75,ORDINAL1:29;
  end;
  for A holds R[A] from ORDINAL1:sch 2(A1);
  hence thesis;
end;

scheme
  TSResult { L()->T-Sequence, F(Ordinal)->set, A()->Ordinal, B()->set, C(
Ordinal,set)->set, D(Ordinal,T-Sequence)->set } : for A st A in dom L() holds L
  ().A = F(A)
provided
A1: for A,x holds x = F(A) iff ex L st x = last L & dom L = succ A & L.
{} = B() & (for C st succ C in succ A holds L.succ C = C(C,L.C)) & for C st C
in succ A & C <> {} & C is limit_ordinal holds L.C = D(C,L|C) and
A2: dom L() = A() and
A3: {} in A() implies L().{} = B() and
A4: for A st succ A in A() holds L().(succ A) = C(A,L().A) and
A5: for A st A in A() & A <> {} & A is limit_ordinal holds L().A = D(A,L ()|A)
proof
  let A;
  set L = L()|succ A;
  assume A in dom L();
  then
A6: succ A c= dom L() by ORDINAL1:21;
A7: for C st succ C in succ A holds L.succ C = C(C,L.C)
  proof
    let C such that
A8: succ C in succ A;
    C in succ C by ORDINAL1:6;
    then
A9: L.C = L().C by A8,FUNCT_1:49,ORDINAL1:10;
    L.succ C = L().succ C by A8,FUNCT_1:49;
    hence thesis by A2,A4,A6,A8,A9;
  end;
A10: for C st C in succ A & C <> {} & C is limit_ordinal holds L.C = D(C,L|C )
  proof
    let C;
    assume that
A11: C in succ A and
A12: C <> {} & C is limit_ordinal;
    C c= succ A by A11,ORDINAL1:def 2;
    then
A13: L|C = L()|C by FUNCT_1:51;
    L.C = L().C by A11,FUNCT_1:49;
    hence thesis by A2,A5,A6,A11,A12,A13;
  end;
  {} c= succ A;
  then {} c< succ A by XBOOLE_0:def 8;
  then
A14: {} in succ A & L.{} = L().{} by FUNCT_1:49,ORDINAL1:11;
A15: dom L = succ A by A6,RELAT_1:62;
  then A in succ A & last L = L.A by Th2,ORDINAL1:21;
  then last L = L().A by FUNCT_1:49;
  hence thesis by A1,A2,A3,A6,A15,A14,A7,A10;
end;

scheme
  TSDef { A()->Ordinal, B()->set, C(Ordinal,set)->set, D(Ordinal,T-Sequence)->
set } : (ex x,L st x = last L & dom L = succ A() & L.{} = B() & (for C st succ
C in succ A() holds L.succ C = C(C,L.C)) & for C st C in succ A() & C <> {} & C
is limit_ordinal holds L.C = D(C,L|C) ) & for x1,x2 being set st (ex L st x1 =
last L & dom L = succ A() & L.{} = B() & (for C st succ C in succ A() holds L.
  succ C = C(C,L.C)) & for C st C in succ A() & C <> {} & C is limit_ordinal
holds L.C = D(C,L|C) ) & (ex L st x2 = last L & dom L = succ A() & L.{} = B() &
(for C st succ C in succ A() holds L.succ C = C(C,L.C)) & for C st C in succ A(
  ) & C <> {} & C is limit_ordinal holds L.C = D(C,L|C) ) holds x1 = x2 proof
  consider L such that
A1: dom L = succ A() & ({} in succ A() implies L.{} = B() ) & (for C st
succ C in succ A() holds L.(succ C) = C(C,L.C)) & for C st C in succ A() & C <>
  {} & C is limit_ordinal holds L.C = D(C,L|C) from TSExist1;
  thus ex x,L st x = last L & dom L = succ A() & L.{} = B() & (for C st succ C
  in succ A() holds L.succ C = C(C,L.C)) & for C st C in succ A() & C <> {} & C
  is limit_ordinal holds L.C = D(C,L|C)
  proof
    take x = last L, L;
    thus x = last L & dom L = succ A() by A1;
    {} c= succ A();
    then {} c< succ A() by XBOOLE_0:def 8;
    hence thesis by A1,ORDINAL1:11;
  end;
  let x1,x2 be set;
  given L1 such that
A2: x1 = last L1 and
A3: dom L1 = succ A() and
A4: L1.{} = B() and
A5: for C st succ C in succ A() holds L1.succ C = C(C,L1.C) and
A6: for C st C in succ A() & C <> {} & C is limit_ordinal holds L1.C = D
  (C,L1|C);
A7: {} in succ A() implies L1.{} = B() by A4;
  given L2 such that
A8: x2 = last L2 and
A9: dom L2 = succ A() and
A10: L2.{} = B() and
A11: for C st succ C in succ A() holds L2.succ C = C(C,L2.C) and
A12: for C st C in succ A() & C <> {} & C is limit_ordinal holds L2.C =
  D(C,L2|C);
A13: {} in succ A() implies L2.{} = B() by A10;
  L1 = L2 from TSUniq1(A3,A7,A5,A6,A9,A13,A11,A12);
  hence thesis by A2,A8;
end;

scheme
  TSResult0 { F(Ordinal)->set, B()->set, C(Ordinal,set)->set, D(Ordinal,
  T-Sequence)->set } : F({}) = B()
provided
A1: for A,x holds x = F(A) iff ex L st x = last L & dom L = succ A & L.
{} = B() & (for C st succ C in succ A holds L.succ C = C(C,L.C)) & for C st C
in succ A & C <> {} & C is limit_ordinal holds L.C = D(C,L|C)
proof
  consider L such that
A2: dom L = succ {} & ({} in succ {} implies L.{} = B() ) & (for A st
succ A in succ {} holds L.(succ A) = C(A,L.A) ) & for A st A in succ {} & A <>
  {} & A is limit_ordinal holds L.A = D(A,L|A) from TSExist1;
  B() = last L by A2,Th2,ORDINAL1:6;
  hence thesis by A1,A2,ORDINAL1:6;
end;

scheme
  TSResultS { B()->set, C(Ordinal,set)->set, D(Ordinal,T-Sequence)->set, F(
  Ordinal)->set } : for A holds F(succ A) = C(A,F(A))
provided
A1: for A,x holds x = F(A) iff ex L st x = last L & dom L = succ A & L.
{} = B() & (for C st succ C in succ A holds L.succ C = C(C,L.C)) & for C st C
in succ A & C <> {} & C is limit_ordinal holds L.C = D(C,L|C)
proof
  let A;
  consider L such that
A2: dom L = succ succ A and
A3: {} in succ succ A implies L.{} = B() and
A4: for C st succ C in succ succ A holds L.(succ C) = C(C,L.C) and
A5: for C st C in succ succ A & C <> {} & C is limit_ordinal holds L.C =
  D(C,L|C) from TSExist1;
A6: for A,x holds x = F(A) iff ex L st x = last L & dom L = succ A & L.{} =
  B() & (for C st succ C in succ A holds L.succ C = C(C,L.C)) & for C st C in
  succ A & C <> {} & C is limit_ordinal holds L.C = D(C,L|C) by A1;
A7: for B st B in dom L holds L.B = F(B) from TSResult(A6,A2,A3,A4, A5);
  then
A8: L.succ A = F(succ A) by A2,ORDINAL1:6;
  A in succ A & succ A in succ succ A by ORDINAL1:6;
  then L.A = F(A) by A2,A7,ORDINAL1:10;
  hence thesis by A4,A8,ORDINAL1:6;
end;

scheme
  TSResultL { L()->T-Sequence, A()->Ordinal, F(Ordinal)->set, B()->set, C(
  Ordinal,set)->set, D(Ordinal,T-Sequence)->set } : F(A()) = D(A(),L())
provided
A1: for A,x holds x = F(A) iff ex L st x = last L & dom L = succ A & L.
{} = B() & (for C st succ C in succ A holds L.succ C = C(C,L.C)) & for C st C
in succ A & C <> {} & C is limit_ordinal holds L.C = D(C,L|C) and
A2: A() <> {} & A() is limit_ordinal and
A3: dom L() = A() and
A4: for A st A in A() holds L().A = F(A)
proof
A5: A() in succ A() by ORDINAL1:6;
  consider L such that
A6: dom L = succ A() and
A7: {} in succ A() implies L.{} = B() and
A8: for C st succ C in succ A() holds L.(succ C) = C(C,L.C) and
A9: for C st C in succ A() & C <> {} & C is limit_ordinal holds L.C = D(
  C,L|C) from TSExist1;
  set L1 = L|A();
A10: for A,x holds x = F(A) iff ex L st x = last L & dom L = succ A & L.{} =
  B() & (for C st succ C in succ A holds L.succ C = C(C,L.C)) & for C st C in
  succ A & C <> {} & C is limit_ordinal holds L.C = D(C,L|C) by A1;
A11: for B st B in dom L holds L.B = F(B) from TSResult(A10,A6,A7,A8, A9);
A12: now
    let x;
    assume
A13: x in A();
    then reconsider x9 = x as Ordinal;
    thus L1.x = L.x9 by A13,FUNCT_1:49
      .= F(x9) by A6,A11,A5,A13,ORDINAL1:10
      .= L().x by A4,A13;
  end;
  A() c= dom L by A6,A5,ORDINAL1:def 2;
  then dom L1 = A() by RELAT_1:62;
  then L1 = L() by A3,A12,FUNCT_1:2;
  then L.A() = D(A(),L()) by A2,A9,ORDINAL1:6;
  hence thesis by A6,A11,ORDINAL1:6;
end;

scheme
  OSExist { A()->Ordinal, B()->Ordinal, C(Ordinal,Ordinal)->Ordinal, D(Ordinal
,T-Sequence)->Ordinal } : ex fi st dom fi = A() & ({} in A() implies fi.{} = B(
) ) & (for A st succ A in A() holds fi.(succ A) = C(A,fi.A) ) & for A st A in A
  () & A <> {} & A is limit_ordinal holds fi.A = D(A,fi|A) proof
  deffunc CC(Ordinal,set) = C($1,sup union {$2});
  consider L such that
A1: dom L = A() and
A2: {} in A() implies L.{} = B() and
A3: for A st succ A in A() holds L.(succ A) = CC(A,L.A) and
A4: for A st A in A() & A <> {} & A is limit_ordinal holds L.A = D(A,L|A
  ) from TSExist1;
  L is Ordinal-yielding
  proof
    take sup rng L;
    let x;
    assume
A5: x in rng L;
    then consider y such that
A6: y in dom L and
A7: x = L.y by FUNCT_1:def 3;
    reconsider y as Ordinal by A6;
A8: now
      assume that
A9:   y <> {} and
A10:  for B holds y <> succ B;
      y is limit_ordinal by A10,ORDINAL1:29;
      then L.y = D(y,L|y) by A1,A4,A6,A9;
      hence x is Ordinal by A7;
    end;
A11: On rng L c= sup rng L by Def3;
    now
      given B such that
A12:  y = succ B;
      L.y = C(B,sup union {L.B}) by A1,A3,A6,A12;
      hence x is Ordinal by A7;
    end;
    then x in On rng L by A1,A2,A5,A6,A7,A8,ORDINAL1:def 9;
    hence thesis by A11;
  end;
  then reconsider L as Ordinal-Sequence;
  take fi = L;
  thus dom fi = A() & ({} in A() implies fi.{} = B() ) by A1,A2;
  thus for A st succ A in A() holds fi.(succ A) = C(A,fi.A)
  proof
    let A;
    reconsider B = fi.A as Ordinal;
    sup union {B} = sup B by ZFMISC_1:25
      .= B by Th18;
    hence thesis by A3;
  end;
  thus thesis by A4;
end;

scheme
  OSResult { fi()->Ordinal-Sequence, F(Ordinal)->Ordinal, A()->Ordinal, B()->
Ordinal, C(Ordinal,Ordinal)->Ordinal, D(Ordinal,T-Sequence)->Ordinal } : for A
  st A in dom fi() holds fi().A = F(A)
provided
A1: for A,B holds B = F(A) iff ex fi st B = last fi & dom fi = succ A &
fi.{} = B() & (for C st succ C in succ A holds fi.succ C = C(C,fi.C)) & for C
st C in succ A & C <> {} & C is limit_ordinal holds fi.C = D(C,fi|C) and
A2: dom fi() = A() and
A3: {} in A() implies fi().{} = B() and
A4: for A st succ A in A() holds fi().(succ A) = C(A,fi().A) and
A5: for A st A in A() & A <> {} & A is limit_ordinal holds fi().A = D(A,
fi()|A)
proof
  let A;
  set fi = fi()|succ A;
  assume A in dom fi();
  then
A6: succ A c= dom fi() by ORDINAL1:21;
A7: for C st succ C in succ A holds fi.succ C = C(C,fi.C)
  proof
    let C such that
A8: succ C in succ A;
    C in succ C by ORDINAL1:6;
    then
A9: fi.C = fi().C by A8,FUNCT_1:49,ORDINAL1:10;
    fi.succ C = fi().succ C by A8,FUNCT_1:49;
    hence thesis by A2,A4,A6,A8,A9;
  end;
A10: for C st C in succ A & C <> {} & C is limit_ordinal holds fi.C = D(C,fi
  |C)
  proof
    let C;
    assume that
A11: C in succ A and
A12: C <> {} & C is limit_ordinal;
    C c= succ A by A11,ORDINAL1:def 2;
    then
A13: fi|C = fi()|C by FUNCT_1:51;
    fi.C = fi().C by A11,FUNCT_1:49;
    hence thesis by A2,A5,A6,A11,A12,A13;
  end;
  {} c= succ A;
  then {} c< succ A by XBOOLE_0:def 8;
  then
A14: {} in succ A & fi.{} = fi().{} by FUNCT_1:49,ORDINAL1:11;
A15: dom fi = succ A by A6,RELAT_1:62;
  then A in succ A & last fi = fi.A by Th2,ORDINAL1:21;
  then last fi = fi().A by FUNCT_1:49;
  hence thesis by A1,A2,A3,A6,A15,A14,A7,A10;
end;

scheme
  OSDef { A()->Ordinal, B()->Ordinal, C(Ordinal,Ordinal)->Ordinal, D(Ordinal,
T-Sequence)->Ordinal } : (ex A,fi st A = last fi & dom fi = succ A() & fi.{} =
B() & (for C st succ C in succ A() holds fi.succ C = C(C,fi.C)) & for C st C in
succ A() & C <> {} & C is limit_ordinal holds fi.C = D(C,fi|C) ) & for A1,A2 st
(ex fi st A1 = last fi & dom fi = succ A() & fi.{} = B() & (for C st succ C in
succ A() holds fi.succ C = C(C,fi.C)) & for C st C in succ A() & C <> {} & C is
limit_ordinal holds fi.C = D(C,fi|C) ) & (ex fi st A2 = last fi & dom fi = succ
A() & fi.{} = B() & (for C st succ C in succ A() holds fi.succ C = C(C,fi.C)) &
for C st C in succ A() & C <> {} & C is limit_ordinal holds fi.C = D(C,fi|C) )
  holds A1 = A2 proof
  consider fi such that
A1: dom fi = succ A() & ({} in succ A() implies fi.{} = B() ) & (for C
st succ C in succ A() holds fi.(succ C) = C(C,fi.C)) & for C st C in succ A() &
  C <> {} & C is limit_ordinal holds fi.C = D(C,fi|C) from OSExist;
  reconsider A = last fi as Ordinal;
  thus ex A,fi st A = last fi & dom fi = succ A() & fi.{} = B() & (for C st
succ C in succ A() holds fi.succ C = C(C,fi.C)) & for C st C in succ A() & C <>
  {} & C is limit_ordinal holds fi.C = D(C,fi|C)
  proof
    take A, fi;
    thus A = last fi & dom fi = succ A() by A1;
    {} c= succ A();
    then {} c< succ A() by XBOOLE_0:def 8;
    hence thesis by A1,ORDINAL1:11;
  end;
  deffunc CD(Ordinal,Ordinal) = C($1,sup union { $2 });
  let A1,A2 be Ordinal;
  given L1 being Ordinal-Sequence such that
A2: A1 = last L1 and
A3: dom L1 = succ A() and
A4: L1.{} = B() and
A5: for C st succ C in succ A() holds L1.succ C = C(C,L1.C) and
A6: for C st C in succ A() & C <> {} & C is limit_ordinal holds L1.C = D
  (C,L1|C);
A7: {} in succ A() implies L1.{} = B() by A4;
A8: for C st succ C in succ A() holds L1.(succ C) = CD(C,L1.C)
  proof
    let C such that
A9: succ C in succ A();
    reconsider x9 = L1.C as Ordinal;
    sup union { L1.C } = sup x9 by ZFMISC_1:25
      .= x9 by Th18;
    hence thesis by A5,A9;
  end;
  given L2 being Ordinal-Sequence such that
A10: A2 = last L2 and
A11: dom L2 = succ A() and
A12: L2.{} = B() and
A13: for C st succ C in succ A() holds L2.succ C = C(C,L2.C) and
A14: for C st C in succ A() & C <> {} & C is limit_ordinal holds L2.C =
  D(C,L2|C);
A15: for C st C in succ A() & C <> {} & C is limit_ordinal holds L2.C = D(C,
  L2|C) by A14;
A16: for C st C in succ A() & C <> {} & C is limit_ordinal holds L1.C = D(C,
  L1|C) by A6;
A17: for C st succ C in succ A() holds L2.(succ C) = CD(C,L2.C)
  proof
    let C such that
A18: succ C in succ A();
    reconsider x9 = L2.C as Ordinal;
    sup union { L2.C } = sup x9 by ZFMISC_1:25
      .= x9 by Th18;
    hence thesis by A13,A18;
  end;
A19: {} in succ A() implies L2.{} = B() by A12;
  L1 = L2 from TSUniq1(A3,A7,A8,A16,A11,A19,A17,A15);
  hence thesis by A2,A10;
end;

scheme
  OSResult0 { F(Ordinal)->Ordinal, B()->Ordinal, C(Ordinal,Ordinal)->Ordinal,
  D(Ordinal,T-Sequence)->Ordinal } : F({}) = B()
provided
A1: for A,B holds B = F(A) iff ex fi st B = last fi & dom fi = succ A &
fi.{} = B() & (for C st succ C in succ A holds fi.succ C = C(C,fi.C)) & for C
st C in succ A & C <> {} & C is limit_ordinal holds fi.C = D(C,fi|C)
proof
  consider fi such that
A2: dom fi = succ {} & ({} in succ {} implies fi.{} = B() ) & (for A st
  succ A in succ {} holds fi.(succ A) = C(A,fi.A) ) & for A st A in succ {} & A
  <> {} & A is limit_ordinal holds fi.A = D(A,fi|A) from OSExist;
  B() = last fi by A2,Th2,ORDINAL1:6;
  hence thesis by A1,A2,ORDINAL1:6;
end;

scheme
  OSResultS { B()->Ordinal, C(Ordinal,Ordinal)->Ordinal, D(Ordinal,T-Sequence)
  ->Ordinal, F(Ordinal)->Ordinal } : for A holds F(succ A) = C(A,F(A))
provided
A1: for A,B holds B = F(A) iff ex fi st B = last fi & dom fi = succ A &
fi.{} = B() & (for C st succ C in succ A holds fi.succ C = C(C,fi.C)) & for C
st C in succ A & C <> {} & C is limit_ordinal holds fi.C = D(C,fi|C)
proof
  let A;
  consider fi such that
A2: dom fi = succ succ A and
A3: {} in succ succ A implies fi.{} = B() and
A4: for C st succ C in succ succ A holds fi.(succ C) = C(C,fi.C) and
A5: for C st C in succ succ A & C <> {} & C is limit_ordinal holds fi.C
  = D(C,fi|C) from OSExist;
A6: for A,B holds B = F(A) iff ex fi st B = last fi & dom fi = succ A & fi.
{} = B() & (for C st succ C in succ A holds fi.succ C = C(C,fi.C)) & for C st C
  in succ A & C <> {} & C is limit_ordinal holds fi.C = D(C,fi|C) by A1;
A7: for B st B in dom fi holds fi.B = F(B) from OSResult(A6,A2,A3,A4, A5);
  then
A8: fi.succ A = F(succ A) by A2,ORDINAL1:6;
  A in succ A & succ A in succ succ A by ORDINAL1:6;
  then fi.A = F(A) by A2,A7,ORDINAL1:10;
  hence thesis by A4,A8,ORDINAL1:6;
end;

scheme
  OSResultL { fi()->Ordinal-Sequence, A()->Ordinal, F(Ordinal)->Ordinal, B()->
Ordinal, C(Ordinal,Ordinal)->Ordinal, D(Ordinal,T-Sequence)->Ordinal } : F(A())
  = D(A(),fi())
provided
A1: for A,B holds B = F(A) iff ex fi st B = last fi & dom fi = succ A &
fi.{} = B() & (for C st succ C in succ A holds fi.succ C = C(C,fi.C)) & for C
st C in succ A & C <> {} & C is limit_ordinal holds fi.C = D(C,fi|C) and
A2: A() <> {} & A() is limit_ordinal and
A3: dom fi() = A() and
A4: for A st A in A() holds fi().A = F(A)
proof
A5: A() in succ A() by ORDINAL1:6;
  consider fi such that
A6: dom fi = succ A() and
A7: {} in succ A() implies fi.{} = B() and
A8: for C st succ C in succ A() holds fi.(succ C) = C(C,fi.C) and
A9: for C st C in succ A() & C <> {} & C is limit_ordinal holds fi.C = D
  (C,fi|C) from OSExist;
  set psi = fi|A();
A10: for A,B holds B = F(A) iff ex fi st B = last fi & dom fi = succ A & fi.
{} = B() & (for C st succ C in succ A holds fi.succ C = C(C,fi.C)) & for C st C
  in succ A & C <> {} & C is limit_ordinal holds fi.C = D(C,fi|C) by A1;
A11: for B st B in dom fi holds fi.B = F(B) from OSResult(A10,A6,A7,A8, A9);
A12: now
    let x;
    assume
A13: x in A();
    then reconsider x9 = x as Ordinal;
    thus psi.x = fi.x9 by A13,FUNCT_1:49
      .= F(x9) by A6,A11,A5,A13,ORDINAL1:10
      .= fi().x by A4,A13;
  end;
  A() c= dom fi by A6,A5,ORDINAL1:def 2;
  then dom psi = A() by RELAT_1:62;
  then psi = fi() by A3,A12,FUNCT_1:2;
  then fi.A() = D(A(),fi()) by A2,A9,ORDINAL1:6;
  hence thesis by A6,A11,ORDINAL1:6;
end;

definition
  let L;
  func sup L -> Ordinal equals
  sup rng L;
  correctness;
  func inf L -> Ordinal equals
  inf rng L;
  correctness;
end;

theorem
  sup L = sup rng L & inf L = inf rng L;

definition
  let L;
  func lim_sup L -> Ordinal means
  ex fi st it = inf fi & dom fi = dom L & for
  A st A in dom L holds fi.A = sup rng (L|(dom L \ A));
  existence
  proof
    deffunc F(Ordinal) = sup rng (L|(dom L \ $1));
    consider fi such that
A1: dom fi = dom L & for A st A in dom L holds fi.A = F(A) from
    OSLambda;
    take inf fi, fi;
    thus thesis by A1;
  end;
  uniqueness
  proof
    let A1,A2 be Ordinal;
    given fi such that
A2: A1 = inf fi & dom fi = dom L and
A3: for A st A in dom L holds fi.A = sup rng (L|(dom L \ A));
    given psi such that
A4: A2 = inf psi & dom psi = dom L and
A5: for A st A in dom L holds psi.A = sup rng (L|(dom L \ A));
    now
      let x;
      assume
A6:   x in dom L;
      then reconsider A = x as Ordinal;
      fi.A = sup rng (L|(dom L \ A)) by A3,A6;
      hence fi.x = psi.x by A5,A6;
    end;
    hence thesis by A2,A4,FUNCT_1:2;
  end;
  func lim_inf L -> Ordinal means
  ex fi st it = sup fi & dom fi = dom L & for
  A st A in dom L holds fi.A = inf rng (L|(dom L \ A));
  existence
  proof
    deffunc F(Ordinal) = inf rng (L|(dom L \ $1));
    consider fi such that
A7: dom fi = dom L & for A st A in dom L holds fi.A = F(A) from
    OSLambda;
    take sup fi, fi;
    thus thesis by A7;
  end;
  uniqueness
  proof
    let A1,A2 be Ordinal;
    given fi such that
A8: A1 = sup fi & dom fi = dom L and
A9: for A st A in dom L holds fi.A = inf rng (L|(dom L \ A));
    given psi such that
A10: A2 = sup psi & dom psi = dom L and
A11: for A st A in dom L holds psi.A = inf rng (L|(dom L \ A));
    now
      let x;
      assume
A12:  x in dom L;
      then reconsider A = x as Ordinal;
      fi.A = inf rng (L|(dom L \ A)) by A9,A12;
      hence fi.x = psi.x by A11,A12;
    end;
    hence thesis by A8,A10,FUNCT_1:2;
  end;
end;

definition
  let A,fi;
  pred A is_limes_of fi means
  :Def9:
  ex B st B in dom fi & for C st B c= C &
C in dom fi holds fi.C = {} if A = {} otherwise for B,C st B in A & A in C ex D
st D in dom fi & for E being Ordinal st D c= E & E in dom fi holds B in fi.E &
  fi.E in C;
  consistency;
end;

definition
  let fi;
  given A such that
A1: A is_limes_of fi;
  func lim fi -> Ordinal means
  :Def10:
  it is_limes_of fi;
  existence by A1;
  uniqueness
  proof
    let A1,A2 be Ordinal such that
A2: (A1 = {} & ex B st B in dom fi & for C st B c= C & C in dom fi
holds fi.C = {}) or (A1 <> {} & for B,C st B in A1 & A1 in C ex D st D in dom
fi & for E being Ordinal st D c= E & E in dom fi holds B in fi.E & fi.E in C)
    and
A3: (A2 = {} & ex B st B in dom fi & for C st B c= C & C in dom fi
holds fi.C = {}) or (A2 <> {} & for B,C st B in A2 & A2 in C ex D st D in dom
fi & for E being Ordinal st D c= E & E in dom fi holds B in fi.E & fi.E in C);
A4: now
      set x = the Element of A1;
      reconsider x as Ordinal;
      assume A1 in A2;
      then consider D such that
A5:   D in dom fi and
A6:   for A st D c= A & A in dom fi holds A1 in fi.A & fi.A in succ A2
      by A3,ORDINAL1:6;
      now
        assume A1 = {};
        then consider B such that
A7:     B in dom fi and
A8:     for C st B c= C & C in dom fi holds fi.C = {} by A2;
        B c= D or D c= B;
        then consider E being Ordinal such that
A9:     E = D & B c= D or E = B & D c= B;
        fi.E = {} by A5,A7,A8,A9;
        hence contradiction by A5,A6,A7,A9;
      end;
      then consider C such that
A10:  C in dom fi and
A11:  for A st C c= A & A in dom fi holds x in fi.A & fi.A in succ A1
      by A2,ORDINAL1:6;
      C c= D or D c= C;
      then consider E being Ordinal such that
A12:  E = D & C c= D or E = C & D c= C;
      fi.E in succ A1 by A5,A10,A11,A12;
      then
A13:  fi.E in A1 or fi.E = A1 by ORDINAL1:8;
      A1 in fi.E by A5,A6,A10,A12;
      hence contradiction by A13;
    end;
    set x = the Element of A2;
    reconsider x as Ordinal;
    assume A1 <> A2;
    then A1 in A2 or A2 in A1 by ORDINAL1:14;
    then consider D such that
A14: D in dom fi and
A15: for A st D c= A & A in dom fi holds A2 in fi.A & fi.A in succ A1
    by A2,A4,ORDINAL1:6;
    now
      assume A2 = {};
      then consider B such that
A16:  B in dom fi and
A17:  for C st B c= C & C in dom fi holds fi.C = {} by A3;
      B c= D or D c= B;
      then consider E being Ordinal such that
A18:  E = D & B c= D or E = B & D c= B;
      fi.E = {} by A14,A16,A17,A18;
      hence contradiction by A14,A15,A16,A18;
    end;
    then consider C such that
A19: C in dom fi and
A20: for A st C c= A & A in dom fi holds x in fi.A & fi.A in succ A2
    by A3,ORDINAL1:6;
    C c= D or D c= C;
    then consider E being Ordinal such that
A21: E = D & C c= D or E = C & D c= C;
    fi.E in succ A2 by A14,A19,A20,A21;
    then
A22: fi.E in A2 or fi.E = A2 by ORDINAL1:8;
    A2 in fi.E by A14,A15,A19,A21;
    hence contradiction by A22;
  end;
end;

definition
  let A,fi;
  func lim(A,fi) -> Ordinal equals
  lim(fi|A);
  correctness;
end;

definition
  let L be Ordinal-Sequence;
  attr L is increasing means
  for A,B st A in B & B in dom L holds L.A in L.B;
  attr L is continuous means
  for A,B st A in dom L & A <> {} & A is
  limit_ordinal & B = L.A holds B is_limes_of L|A;
end;

definition
  let A,B;
  func A +^ B -> Ordinal means
  :Def14:
  ex fi st it = last fi & dom fi = succ B
& fi.{} = A & (for C st succ C in succ B holds fi.succ C = succ(fi.C)) & for C
  st C in succ B & C <> {} & C is limit_ordinal holds fi.C = sup(fi|C);
  correctness
  proof
    deffunc D(Ordinal,T-Sequence) = sup $2;
    deffunc C(Ordinal,Ordinal) = succ $2;
    (ex F being Ordinal,fi st F = last fi & dom fi = succ B & fi.{} = A &
(for C st succ C in succ B holds fi.succ C = C(C,fi.C)) & for C st C in succ B
& C <> {} & C is limit_ordinal holds fi.C = D(C,fi|C) ) & for A1,A2 st (ex fi
    st A1 = last fi & dom fi = succ B & fi.{} = A & (for C st succ C in succ B
    holds fi.succ C = C(C,fi.C)) & for C st C in succ B & C <> {} & C is
limit_ordinal holds fi.C = D(C,fi|C) ) & (ex fi st A2 = last fi & dom fi = succ
B & fi.{} = A & (for C st succ C in succ B holds fi.succ C = C(C,fi.C)) & for C
st C in succ B & C <> {} & C is limit_ordinal holds fi.C = D(C,fi|C) ) holds A1
    = A2 from OSDef;
    hence thesis;
  end;
end;

definition
  let A,B;
  func A *^ B -> Ordinal means
  :Def15:
  ex fi st it = last fi & dom fi = succ A
& fi.{} = {} & (for C st succ C in succ A holds fi.succ C = (fi.C)+^B) & for C
  st C in succ A & C <> {} & C is limit_ordinal holds fi.C = union sup(fi|C);
  correctness
  proof
    deffunc D(Ordinal,Ordinal-Sequence) = union sup $2;
    deffunc C(Ordinal,Ordinal) = $2+^B;
    (ex F being Ordinal,fi st F = last fi & dom fi = succ A & fi.{} = {} &
(for C st succ C in succ A holds fi.succ C = C(C,fi.C)) & for C st C in succ A
& C <> {} & C is limit_ordinal holds fi.C = D(C,fi|C) ) & for A1,A2 st (ex fi
    st A1 = last fi & dom fi = succ A & fi.{} = {} & (for C st succ C in succ A
    holds fi.succ C = C(C,fi.C)) & for C st C in succ A & C <> {} & C is
limit_ordinal holds fi.C = D(C,fi|C) ) & (ex fi st A2 = last fi & dom fi = succ
A & fi.{} = {} & (for C st succ C in succ A holds fi.succ C = C(C,fi.C)) & for
C st C in succ A & C <> {} & C is limit_ordinal holds fi.C = D(C,fi|C) ) holds
    A1 = A2 from OSDef;
    hence thesis;
  end;
end;

registration
  let O be Ordinal;
  cluster -> ordinal for Element of O;
  coherence;
end;

definition
  let A,B;
  func exp(A,B) -> Ordinal means
  :Def16:
  ex fi st it = last fi & dom fi = succ
B & fi.{} = 1 & (for C st succ C in succ B holds fi.succ C = A*^(fi.C)) & for C
  st C in succ B & C <> {} & C is limit_ordinal holds fi.C = lim(fi|C);
  correctness
  proof
    deffunc D(Ordinal,Ordinal-Sequence) = lim $2;
    deffunc C(Ordinal,Ordinal) = A*^$2;
    (ex F being Ordinal,fi st F = last fi & dom fi = succ B & fi.{} = 1 &
(for C st succ C in succ B holds fi.succ C = C(C,fi.C)) & for C st C in succ B
& C <> {} & C is limit_ordinal holds fi.C = D(C,fi|C) ) & for A1,A2 st (ex fi
    st A1 = last fi & dom fi = succ B & fi.{} = 1 & (for C st succ C in succ B
    holds fi.succ C = C(C,fi.C)) & for C st C in succ B & C <> {} & C is
limit_ordinal holds fi.C = D(C,fi|C) ) & (ex fi st A2 = last fi & dom fi = succ
B & fi.{} = 1 & (for C st succ C in succ B holds fi.succ C = C(C,fi.C)) & for C
st C in succ B & C <> {} & C is limit_ordinal holds fi.C = D(C,fi|C) ) holds A1
    = A2 from OSDef;
    hence thesis;
  end;
end;

theorem Th27:
  A+^{} = A
proof
  deffunc C(Ordinal,Ordinal) = succ $2;
  deffunc D(Ordinal,T-Sequence) = sup $2;
  deffunc F(Ordinal) = A+^$1;
A1: for B,C holds C = F(B) iff ex fi st C = last fi & dom fi = succ B & fi.
  {} = A & (for C st succ C in succ B holds fi.succ C = C(C,fi.C)) & for C st C
  in succ B & C <> {} & C is limit_ordinal holds fi.C = D(C,fi|C) by Def14;
  thus F({}) = A from OSResult0(A1);
end;

theorem Th28:
  A+^succ B = succ(A+^B)
proof
  deffunc C(Ordinal,Ordinal) = succ $2;
  deffunc D(Ordinal,T-Sequence) = sup $2;
  deffunc F(Ordinal) = A+^$1;
A1: for B,C holds C = F(B) iff ex fi st C = last fi & dom fi = succ B & fi.
  {} = A & (for C st succ C in succ B holds fi.succ C = C(C,fi.C)) & for C st C
  in succ B & C <> {} & C is limit_ordinal holds fi.C = D(C,fi|C) by Def14;
  for B holds F(succ B) = C(B,F(B)) from OSResultS(A1);
  hence thesis;
end;

theorem Th29:
  B <> {} & B is limit_ordinal implies for fi st dom fi = B & for
  C st C in B holds fi.C = A+^C holds A+^B = sup fi
proof
  deffunc C(Ordinal,Ordinal) = succ $2;
  deffunc D(Ordinal,Ordinal-Sequence) = sup $2;
  assume
A1: B <> {} & B is limit_ordinal;
  deffunc F(Ordinal) = A+^$1;
  let fi such that
A2: dom fi = B and
A3: for C st C in B holds fi.C = F(C);
A4: for B,C holds C = F(B) iff ex fi st C = last fi & dom fi = succ B & fi.
  {} = A & (for C st succ C in succ B holds fi.succ C = C(C,fi.C)) & for C st C
  in succ B & C <> {} & C is limit_ordinal holds fi.C = D(C,fi|C) by Def14;
  thus F(B) = D(B,fi) from OSResultL(A4,A1,A2,A3);
end;

theorem Th30:
  {}+^A = A
proof
  defpred P[Ordinal] means {}+^$1 = $1;
A1: for A holds P[A] implies P[succ A] by Th28;
A2: for A st A <> {} & A is limit_ordinal & for B st B in A holds P[B] holds
  P[A]
  proof
    deffunc F(Ordinal) = {}+^$1;
    let A;
    assume that
A3: A <> {} & A is limit_ordinal and
A4: for B st B in A holds {}+^B = B;
    consider fi such that
A5: dom fi = A & for B st B in A holds fi.B = F(B) from OSLambda;
A6: rng fi = A
    proof
      thus x in rng fi implies x in A
      proof
        assume x in rng fi;
        then consider y such that
A7:     y in dom fi and
A8:     x = fi.y by FUNCT_1:def 3;
        reconsider y as Ordinal by A7;
        x = {}+^y by A5,A7,A8
          .= y by A4,A5,A7;
        hence thesis by A5,A7;
      end;
      let x;
      assume
A9:   x in A;
      then reconsider B = x as Ordinal;
      {}+^B = B & fi.B = {}+^B by A4,A5,A9;
      hence thesis by A5,A9,FUNCT_1:def 3;
    end;
    {}+^A = sup fi by A3,A5,Th29
      .= sup rng fi;
    hence thesis by A6,Th18;
  end;
A10: P[{}] by Th27;
  for A holds P[A] from OrdinalInd(A10,A1,A2);
  hence thesis;
end;

Lm1: 1 = succ {};

theorem
  A+^1 = succ A
proof
  thus A+^1 = succ(A+^{}) by Lm1,Th28
    .= succ A by Th27;
end;

theorem Th32:
  A in B implies C +^ A in C +^ B
proof
  defpred P[Ordinal] means A in $1 implies C +^ A in C +^ $1;
A1: for B st for D st D in B holds P[D] holds P[B]
  proof
    let B such that
A2: for D st D in B holds A in D implies C +^ A in C +^ D and
A3: A in B;
A4: now
      given D such that
A5:   B = succ D;
A6:   now
        assume A in D;
        then
A7:     C +^ A in C +^ D by A2,A5,ORDINAL1:6;
        succ(C +^ D) = C +^ succ D & C +^ D in succ(C +^ D) by Th28,ORDINAL1:6
;
        hence thesis by A5,A7,ORDINAL1:10;
      end;
      now
        assume
A8:     not A in D;
        A c< D iff A c= D & A <> D by XBOOLE_0:def 8;
        then C +^ A in succ(C +^ D) by A3,A5,A8,ORDINAL1:11,22;
        hence thesis by A5,Th28;
      end;
      hence thesis by A6;
    end;
    now
      deffunc F(Ordinal) = C+^$1;
      consider fi such that
A9:   dom fi = B & for D st D in B holds fi.D = F(D) from OSLambda;
      fi.A = C+^A by A3,A9;
      then
A10:  C+^A in rng fi by A3,A9,FUNCT_1:def 3;
      assume for D holds B <> succ D;
      then B is limit_ordinal by ORDINAL1:29;
      then C+^B = sup fi by A3,A9,Th29
        .= sup rng fi;
      hence thesis by A10,Th19;
    end;
    hence thesis by A4;
  end;
  for B holds P[B] from ORDINAL1:sch 2(A1);
  hence thesis;
end;

theorem Th33:
  A c= B implies C +^ A c= C +^ B
proof
  assume
A1: A c= B;
  now
    assume A <> B;
    then A c< B by A1,XBOOLE_0:def 8;
    then C +^ A in C +^ B by Th32,ORDINAL1:11;
    hence thesis by ORDINAL1:def 2;
  end;
  hence thesis;
end;

theorem Th34:
  A c= B implies A +^ C c= B +^ C
proof
  defpred P[Ordinal] means A +^ $1 c= B +^ $1;
  assume
A1: A c= B;
A2: for C st for D st D in C holds P[D] holds P[C]
  proof
    let C such that
A3: for D st D in C holds A +^ D c= B +^ D;
A4: now
      given D such that
A5:   C = succ D;
A6:   B +^ C = succ(B +^ D) by A5,Th28;
      A +^ D c= B +^ D & A +^ C = succ(A +^ D) by A3,A5,Th28,ORDINAL1:6;
      hence thesis by A6,Th1;
    end;
A7: now
      deffunc F(Ordinal) = A+^$1;
      assume that
A8:   C <> {} and
A9:   for D holds C <> succ D;
      consider fi such that
A10:  dom fi = C & for D st D in C holds fi.D = F(D) from OSLambda;
A11:  now
        let D;
        assume D in rng fi;
        then consider x such that
A12:    x in dom fi and
A13:    D = fi.x by FUNCT_1:def 3;
        reconsider x as Ordinal by A12;
A14:    B +^ x in B +^ C by A10,A12,Th32;
        D = A+^x & A +^ x c= B +^ x by A3,A10,A12,A13;
        hence D in B +^ C by A14,ORDINAL1:12;
      end;
      C is limit_ordinal by A9,ORDINAL1:29;
      then A+^C = sup fi by A8,A10,Th29
        .= sup rng fi;
      hence thesis by A11,Th20;
    end;
    now
      assume
A15:  C = {};
      then A +^ C = A by Th27;
      hence thesis by A1,A15,Th27;
    end;
    hence thesis by A4,A7;
  end;
  for C holds P[C] from ORDINAL1:sch 2(A2);
  hence thesis;
end;

theorem Th35:
  {}*^A = {}
proof
  deffunc D(Ordinal,T-Sequence) = union sup $2;
  deffunc F(Ordinal) = $1*^A;
  deffunc C(Ordinal,Ordinal) = $2+^A;
A1: for B,C holds C = F(B) iff ex fi st C = last fi & dom fi = succ B & fi.
{} = {} & (for C st succ C in succ B holds fi.succ C = C(C,fi.C)) & for C st C
  in succ B & C <> {} & C is limit_ordinal holds fi.C = D(C,fi|C) by Def15;
  thus F({}) = {} from OSResult0(A1);
end;

theorem Th36:
  (succ B)*^A = B*^A +^ A
proof
  deffunc D(Ordinal,T-Sequence) = union sup $2;
  deffunc F(Ordinal) = $1*^A;
  deffunc C(Ordinal,Ordinal) = $2 +^ A;
A1: for B,C holds C = F(B) iff ex fi st C = last fi & dom fi = succ B & fi.
{} = {} & (for C st succ C in succ B holds fi.succ C = C(C,fi.C)) & for C st C
  in succ B & C <> {} & C is limit_ordinal holds fi.C = D(C,fi|C) by Def15;
  for B holds F(succ B) = C(B,F(B)) from OSResultS(A1);
  hence thesis;
end;

theorem Th37:
  B <> {} & B is limit_ordinal implies for fi st dom fi = B & for
  C st C in B holds fi.C = C*^A holds B*^A = union sup fi
proof
  deffunc D(Ordinal,Ordinal-Sequence) = union sup $2;
  assume
A1: B <> {} & B is limit_ordinal;
  deffunc C(Ordinal,Ordinal) = $2+^A;
  deffunc F(Ordinal) = $1*^A;
  let fi such that
A2: dom fi = B and
A3: for C st C in B holds fi.C = F(C);
A4: for B,C holds C = F(B) iff ex fi st C = last fi & dom fi = succ B & fi.
{} = {} & (for C st succ C in succ B holds fi.succ C = C(C,fi.C)) & for C st C
  in succ B & C <> {} & C is limit_ordinal holds fi.C = D(C,fi|C) by Def15;
  thus F(B) = D(B,fi) from OSResultL(A4,A1,A2,A3);
end;

theorem Th38:
  A*^{} = {}
proof
  defpred P[Ordinal] means $1*^{} = {};
A1: for A st P[A] holds P[succ A]
  proof
    let A;
    assume A*^{} = {};
    hence (succ A)*^{} = {}+^{} by Th36
      .= {} by Th27;
  end;
A2: for A st A <> {} & A is limit_ordinal & for B st B in A holds P[B] holds
  P[A]
  proof
    deffunc F(Ordinal) = $1*^{};
    let A;
    assume that
A3: A <> {} and
A4: A is limit_ordinal and
A5: for B st B in A holds B*^{} = {};
    consider fi such that
A6: dom fi = A & for B st B in A holds fi.B = F(B) from OSLambda;
    rng fi = succ {}
    proof
      thus x in rng fi implies x in succ {}
      proof
        assume x in rng fi;
        then consider y such that
A7:     y in dom fi and
A8:     x = fi.y by FUNCT_1:def 3;
        reconsider y as Ordinal by A7;
        x = y*^{} by A6,A7,A8
          .= {} by A5,A6,A7;
        hence thesis by TARSKI:def 1;
      end;
      let x;
      assume x in succ {};
      then
A9:   x = {} by TARSKI:def 1;
      {} c= A;
      then
A10:  {} c< A by A3,XBOOLE_0:def 8;
      then
A11:  {} in A by ORDINAL1:11;
      {} *^ {} = {} by Th35;
      then fi.x = x by A6,A10,A9,ORDINAL1:11;
      hence thesis by A6,A11,A9,FUNCT_1:def 3;
    end;
    then
A12: sup rng fi = succ {} by Th18;
    A*^{} = union sup fi by A3,A4,A6,Th37
      .= union sup rng fi;
    hence thesis by A12,Th2;
  end;
A13: P[{}] by Th35;
  for A holds P[A] from OrdinalInd(A13,A1,A2);
  hence thesis;
end;

theorem Th39:
  1*^A = A & A*^1 = A
proof
  defpred P[Ordinal] means $1*^succ {} = $1;
  thus 1*^A = {}*^A +^ A by Lm1,Th36
    .= {} +^ A by Th35
    .= A by Th30;
A1: for A st for B st B in A holds P[B] holds P[A]
  proof
    let A such that
A2: for B st B in A holds B*^(succ {}) = B;
A3: now
      deffunc F(Ordinal) = $1*^succ {};
      assume that
A4:   A <> {} and
A5:   for B holds A <> succ B;
      consider fi such that
A6:   dom fi = A & for D st D in A holds fi.D = F(D) from OSLambda;
A7:   A = rng fi
      proof
        thus A c= rng fi
        proof
          let x;
          assume
A8:       x in A;
          then reconsider B = x as Ordinal;
          x = B*^succ {} by A2,A8
            .= fi.x by A6,A8;
          hence thesis by A6,A8,FUNCT_1:def 3;
        end;
        let x;
        assume x in rng fi;
        then consider y such that
A9:     y in dom fi and
A10:    x = fi.y by FUNCT_1:def 3;
        reconsider y as Ordinal by A9;
        fi.y = y*^succ {} by A6,A9
          .= y by A2,A6,A9;
        hence thesis by A6,A9,A10;
      end;
A11:  A is limit_ordinal by A5,ORDINAL1:29;
      then A*^succ {} = union sup fi by A4,A6,Th37
        .= union sup rng fi;
      hence A*^succ {} = union A by A7,Th18
        .= A by A11,ORDINAL1:def 6;
    end;
    now
      given B such that
A12:  A = succ B;
      thus A*^(succ {}) = B*^(succ {}) +^ succ {} by A12,Th36
        .= B +^ succ {} by A2,A12,ORDINAL1:6
        .= succ(B +^ {}) by Th28
        .= A by A12,Th27;
    end;
    hence thesis by A3,Th35;
  end;
  for A holds P[A] from ORDINAL1:sch 2(A1);
  hence thesis;
end;

theorem Th40:
  C <> {} & A in B implies A*^C in B*^C
proof
A1: {} c= C;
  defpred P[Ordinal] means A in $1 implies A*^C in $1*^C;
  assume C <> {};
  then
A2: {} c< C by A1,XBOOLE_0:def 8;
  then
A3: {} in C by ORDINAL1:11;
A4: for B st for D st D in B holds P[D] holds P[B]
  proof
    let B such that
A5: for D st D in B holds A in D implies A *^ C in D *^ C and
A6: A in B;
A7: now
      given D such that
A8:   B = succ D;
A9:   now
        assume A in D;
        then
A10:    A*^C in D*^C by A5,A8,ORDINAL1:6;
A11:    D*^C +^ {} in D*^C +^ C by A2,Th32,ORDINAL1:11;
        D*^C +^ C = (succ D)*^C & D*^C +^ {} = D*^C by Th27,Th36;
        hence thesis by A8,A10,A11,ORDINAL1:10;
      end;
      now
A12:    A*^C +^ {} = A*^C by Th27;
        assume
A13:    not A in D;
        A c< D iff A c= D & A <> D by XBOOLE_0:def 8;
        then A*^C +^ {} in D*^C +^ C by A3,A6,A8,A13,Th32,ORDINAL1:11,22;
        hence thesis by A8,A12,Th36;
      end;
      hence thesis by A9;
    end;
    now
A14:  A*^C +^ {} = A*^C & A*^C +^ {} in A*^C +^ C by A2,Th27,Th32,ORDINAL1:11;
A15:  (succ A)*^C = A*^C +^ C by Th36;
      deffunc F(Ordinal) = $1*^C;
      consider fi such that
A16:  dom fi = B & for D st D in B holds fi.D = F(D) from OSLambda;
      assume for D holds B <> succ D;
      then
A17:  B is limit_ordinal by ORDINAL1:29;
      then
A18:  succ A in B by A6,ORDINAL1:28;
      then fi.(succ A) = (succ A)*^C by A16;
      then (succ A)*^C in rng fi by A16,A18,FUNCT_1:def 3;
      then
A19:  (succ A)*^C c= union sup rng fi by Th19,ZFMISC_1:74;
      B*^C = union sup fi by A6,A17,A16,Th37
        .= union sup rng fi;
      hence thesis by A19,A14,A15;
    end;
    hence thesis by A7;
  end;
  for B holds P[B] from ORDINAL1:sch 2(A4);
  hence thesis;
end;

theorem
  A c= B implies A*^C c= B*^C
proof
  assume
A1: A c= B;
A2: now
    assume
A3: C <> {};
    now
      assume A <> B;
      then A c< B by A1,XBOOLE_0:def 8;
      then A*^C in B*^C by A3,Th40,ORDINAL1:11;
      hence thesis by ORDINAL1:def 2;
    end;
    hence thesis;
  end;
  now
    assume C = {};
    then A*^C = {} by Th38;
    hence thesis;
  end;
  hence thesis by A2;
end;

theorem
  A c= B implies C*^A c= C*^B
proof
  assume
A1: A c= B;
  now
    defpred P[Ordinal] means $1*^A c= $1*^B;
    assume
A2: B <> {};
A3: for C st for D st D in C holds P[D] holds P[C]
    proof
      let C such that
A4:   for D st D in C holds D*^A c= D*^B;
A5:   now
        given D such that
A6:     C = succ D;
        D*^A c= D*^B by A4,A6,ORDINAL1:6;
        then
A7:     D*^A +^ A c= D*^B +^ A by Th34;
A8:     C*^A = D*^A +^ A & C*^B = D*^B +^ B by A6,Th36;
        D*^B +^ A c= D*^B +^ B by A1,Th33;
        hence thesis by A7,A8,XBOOLE_1:1;
      end;
A9:   now
        deffunc F(Ordinal) = $1*^A;
        assume that
A10:    C <> {} and
A11:    for D holds C <> succ D;
        consider fi such that
A12:    dom fi = C & for D st D in C holds fi.D = F(D) from OSLambda;
        now
          let D;
          assume D in rng fi;
          then consider x such that
A13:      x in dom fi and
A14:      D = fi.x by FUNCT_1:def 3;
          reconsider x as Ordinal by A13;
A15:      x*^B in C*^B by A2,A12,A13,Th40;
          D = x*^A & x*^A c= x*^B by A4,A12,A13,A14;
          hence D in C*^B by A15,ORDINAL1:12;
        end;
        then
A16:    sup rng fi c= C*^B by Th20;
        C is limit_ordinal by A11,ORDINAL1:29;
        then
A17:    C*^A = union sup fi by A10,A12,Th37
          .= union sup rng fi;
        union sup rng fi c= sup rng fi by Th5;
        hence thesis by A17,A16,XBOOLE_1:1;
      end;
      now
        assume C = {};
        then C*^A = {} by Th35;
        hence thesis;
      end;
      hence thesis by A5,A9;
    end;
    for C holds P[C] from ORDINAL1:sch 2(A3);
    hence thesis;
  end;
  hence thesis by A1;
end;

theorem Th43:
  exp(A,{}) = 1
proof
  deffunc D(Ordinal,Ordinal-Sequence) = lim $2;
  deffunc F(Ordinal) = exp(A,$1);
  deffunc C(Ordinal,Ordinal) = A*^$2;
A1: for B,C holds C = F(B) iff ex fi st C = last fi & dom fi = succ B & fi.
  {} = 1 & (for C st succ C in succ B holds fi.succ C = C(C,fi.C)) & for C st C
  in succ B & C <> {} & C is limit_ordinal holds fi.C = D(C,fi|C) by Def16;
  thus F({}) = 1 from OSResult0(A1);
end;

theorem Th44:
  exp(A,succ B) = A*^exp(A,B)
proof
  deffunc D(Ordinal,Ordinal-Sequence) = lim $2;
  deffunc F(Ordinal) = exp(A,$1);
  deffunc C(Ordinal,Ordinal) = A *^ $2;
A1: for B,C holds C = F(B) iff ex fi st C = last fi & dom fi = succ B & fi.
  {} = 1 & (for C st succ C in succ B holds fi.succ C = C(C,fi.C)) & for C st C
  in succ B & C <> {} & C is limit_ordinal holds fi.C = D(C,fi|C) by Def16;
  for B holds F(succ B) = C(B,F(B)) from OSResultS(A1);
  hence thesis;
end;

theorem Th45:
  B <> {} & B is limit_ordinal implies for fi st dom fi = B & for
  C st C in B holds fi.C = exp(A,C) holds exp(A,B) = lim fi
proof
  deffunc D(Ordinal,Ordinal-Sequence) = lim $2;
  assume
A1: B <> {} & B is limit_ordinal;
  deffunc C(Ordinal,Ordinal) = A*^$2;
  deffunc F(Ordinal) = exp(A,$1);
  let fi such that
A2: dom fi = B and
A3: for C st C in B holds fi.C = F(C);
A4: for B,C holds C = F(B) iff ex fi st C = last fi & dom fi = succ B & fi.
  {} = 1 & (for C st succ C in succ B holds fi.succ C = C(C,fi.C)) & for C st C
  in succ B & C <> {} & C is limit_ordinal holds fi.C = D(C,fi|C) by Def16;
  thus F(B) = D(B,fi) from OSResultL(A4,A1,A2,A3);
end;

theorem
  exp(A,1) = A & exp(1,A) = 1
proof
  defpred P[Ordinal] means exp(1,$1) = 1;
A1: for A st P[A] holds P[succ A]
  proof
    let A;
    assume exp(1,A) = 1;
    hence exp(1,succ A) = 1*^1 by Th44
      .= 1 by Th39;
  end;
  thus exp(A,1) = A*^exp(A,{}) by Lm1,Th44
    .= A*^1 by Th43
    .= A by Th39;
A2: for A st A <> {} & A is limit_ordinal & for B st B in A holds P[B] holds
  P[A]
  proof
    deffunc F(Ordinal) = exp(1,$1);
    let A such that
A3: A <> {} and
A4: A is limit_ordinal and
A5: for B st B in A holds exp(1,B) = 1;
    consider fi such that
A6: dom fi = A & for B st B in A holds fi.B = F(B) from OSLambda;
A7: rng fi c= { 1 }
    proof
      let x;
      assume x in rng fi;
      then consider y such that
A8:   y in dom fi and
A9:   x = fi.y by FUNCT_1:def 3;
      reconsider y as Ordinal by A8;
      x = exp(1,y) by A6,A8,A9
        .= 1 by A5,A6,A8;
      hence thesis by TARSKI:def 1;
    end;
    now
      set x = the Element of A;
      thus {} <> 1;
      let B,C such that
A10:  B in 1 & 1 in C;
      reconsider x as Ordinal;
      take D = x;
      thus D in dom fi by A3,A6;
      let E be Ordinal;
      assume that
      D c= E and
A11:  E in dom fi;
      fi.E in rng fi by A11,FUNCT_1:def 3;
      hence B in fi.E & fi.E in C by A7,A10,TARSKI:def 1;
    end;
    then
A12: 1 is_limes_of fi by Def9;
    exp(1,A) = lim fi by A3,A4,A6,Th45;
    hence thesis by A12,Def10;
  end;
A13: P[{}] by Th43;
  for A holds P[A] from OrdinalInd(A13,A1,A2);
  hence thesis;
end;

theorem
  for A ex B,C st B is limit_ordinal & C is natural & A = B +^ C
proof
  defpred Th[Ordinal] means ex B,C st B is limit_ordinal & C is natural & $1 =
  B +^ C;
A1: for A st for B st B in A holds Th[B] holds Th[A]
  proof
    let A such that
A2: for B st B in A holds Th[B];
A3: (ex B st A = succ B) implies Th[A]
    proof
      given B such that
A4:   A = succ B;
      consider C,D such that
A5:   C is limit_ordinal and
A6:   D is natural and
A7:   B = C +^ D by A2,A4,ORDINAL1:6;
      take C, E = succ D;
      thus C is limit_ordinal by A5;
      thus E in omega by A6,ORDINAL1:def 12;
      thus thesis by A4,A7,Th28;
    end;
    (for B holds A <> succ B) implies Th[A]
    proof
      assume
A8:   for D holds A <> succ D;
      take B = A, C = {};
      thus B is limit_ordinal by A8,ORDINAL1:29;
      thus C in omega by ORDINAL1:def 11;
      thus thesis by Th27;
    end;
    hence thesis by A3;
  end;
  thus for A holds Th[A] from ORDINAL1:sch 2(A1);
end;



registration
  let X be set, o be Ordinal;
  cluster X --> o -> Ordinal-yielding;
  coherence
  proof
    rng(X --> o) c= {o} & {o} c= succ o by FUNCOP_1:13,XBOOLE_1:7;
    then rng(X --> o) c= succ o by XBOOLE_1:1;
    hence thesis by Def4;
  end;
end;

registration
  let O be Ordinal, x be set;
  cluster O --> x -> T-Sequence-like;
  coherence
  proof
    dom(O --> x) = O by FUNCOP_1:13;
    hence thesis by ORDINAL1:def 7;
  end;
end;



definition
  let A,B be Ordinal;
  pred A is_cofinal_with B means
  ex xi being Ordinal-Sequence st dom xi = B &
  rng xi c= A & xi is increasing & A = sup xi;
  reflexivity
  proof
    let A;
A1: dom id A = A by RELAT_1:45;
    then reconsider xi = id A as T-Sequence by ORDINAL1:def 7;
A2: rng id A = A by RELAT_1:45;
    then reconsider xi as Ordinal-Sequence by Def4;
    take xi;
    thus dom xi = A & rng xi c= A by RELAT_1:45;
    thus xi is increasing
    proof
      let B,C;
      assume that
A3:   B in C and
A4:   C in dom xi;
      xi.C = C by A1,A4,FUNCT_1:18;
      hence thesis by A1,A3,A4,FUNCT_1:18,ORDINAL1:10;
    end;
    thus thesis by A2,Th18;
  end;
end;

reserve e,u for set;

theorem Th48:
  e in rng psi implies e is Ordinal
proof
  assume e in rng psi;
  then ex u st u in dom psi & e = psi.u by FUNCT_1:def 3;
  hence thesis;
end;

theorem
  rng psi c= sup psi
proof
  let e;
  assume
A1: e in rng psi;
  then e is Ordinal by Th48;
  hence thesis by A1,Th19;
end;

theorem
  A is_cofinal_with {} implies A = {}
proof
  given psi such that
A1: dom psi = {} and
  rng psi c= A and
  psi is increasing and
A2: A = sup psi;
  thus A = sup {} by A1,A2,RELAT_1:42
    .= {} by Th18;
end;



scheme
  OmegaInd {a()-> Nat, P[set]}:
   P[a()]
provided
A1: P[{}] and
A2: for a being Nat st P[a] holds P[succ a]
proof
  defpred P[Ordinal] means $1 is natural implies P[$1];
A3: now
    let A;
    assume
A4: P[A];
    now
      assume succ A is natural;
      then
A5:   succ A in omega by ORDINAL1:def 12;
      A in succ A by ORDINAL1:6;
      then A is Element of omega by A5,ORDINAL1:10;
      hence P[succ A] by A2,A4;
    end;
    hence P[succ A];
  end;
A6: now
    let A;
    assume
A7: A <> {};
    {} c= A;
    then {} c< A by A7,XBOOLE_0:def 8;
    then
A8: {} in A by ORDINAL1:11;
A9: not A in A;
    assume A is limit_ordinal;
    then omega c= A by A8,ORDINAL1:def 11;
    then not A in omega by A9;
    hence (for B st B in A holds P[B]) implies P[A] by ORDINAL1:def 12;
  end;
A10: P[{}] by A1;
  for A holds P[A] from OrdinalInd(A10,A3,A6);
  hence thesis;
end;

registration
  let a, b be Nat;
  cluster a +^ b -> natural;
  coherence
  proof
    defpred P[Nat] means a+^$1 is natural;
A1: now
      let b be Nat;
      assume P[b];
      then reconsider c = a+^b as Nat;
      a+^succ b = succ c by Th28;
      hence P[succ b];
    end;
A2: P[{}] by Th27;
    thus P[b] from OmegaInd(A2,A1);
  end;
end;



registration
  let x,y be set, a,b be Nat;
  cluster IFEQ(x,y,a,b) -> natural;
  coherence
  proof
    per cases;
    suppose
      x = y;
      hence thesis by FUNCOP_1:def 8;
    end;
    suppose
      x <> y;
      hence thesis by FUNCOP_1:def 8;
    end;
  end;
end;



scheme LambdaRecEx{A() -> set,G(set,set) -> set}:
   ex f being Function st dom f = omega & f.{} = A() &
    for n being Nat holds f.succ n = G(n,f.n)
 proof
  deffunc D(set,set) = {};
  consider L being T-Sequence such that
A1: dom L = omega and
A2: {} in omega implies L.{} = A() and
A3: for A being Ordinal st succ A in omega holds L.(succ A) = G(A,L.A) and
  for A being Ordinal st A in omega & A <> {} & A is limit_ordinal holds L.A
  = D(A,L|A) from TSExist1;
  take L;
  thus dom L = omega by A1;
   {} in omega by ORDINAL1:def 12;
  hence L.{} = A() by A2;
  let n be Nat;
  succ n in omega by ORDINAL1:def 12;
  hence thesis by A3;
end;

reserve n for Nat;

scheme RecUn{A() -> set, F, G() -> Function, P[set,set,set]}:
 F() = G()
provided
A1: dom F() = omega and
A2: F().{} = A() and
A3: for n holds P[n,F().n,F().(succ n)] and
A4: dom G() = omega and
A5: G().{} = A() and
A6: for n holds P[n,G().n,G().(succ n)] and
A7: for n for x,y1,y2 being set st P[n,x,y1] & P[n,x,y2] holds y1 = y2
proof
  defpred P[Nat] means F().$1 = G().$1;
A8: for n st P[n] holds P[succ n]
  proof
    let n;
    assume F().n = G().n;
    then
A9: P[n,F().n,G().(succ n)] by A6;
    P[n,F().n,F().(succ n)] by A3;
    hence thesis by A7,A9;
  end;
A10: P[{}] by A2,A5;
  for n holds P[n] proof let n;
    thus thesis from OmegaInd(A10,A8);
   end;
  then for x st x in omega holds F().x = G().x;
  hence thesis by A1,A4,FUNCT_1:2;
end;

scheme LambdaRecUn{A() -> set, F(set,set) -> set, F, G() -> Function}:
   F() = G()
provided
A1: dom F() = omega and
A2: F().{} = A() and
A3: for n holds F().(succ n) = F(n,F().n) and
A4: dom G() = omega and
A5: G().{} = A() and
A6: for n holds G().(succ n) = F(n,G().n)
proof
  defpred P[Nat,set,set] means $3=F($1,$2);
A7: for n holds P[n,G().n,G().(succ n)] by A6;
A8: for n being Nat,x,y1,y2 being set st P[n,x,y1] & P[n,x,y2]
    holds y1=y2;
A9: for n holds P[n,F().n,F().(succ n)] by A3;
  thus F() = G() from RecUn(A1,A2,A9,A4,A5,A7,A8);
end;














begin

reserve fi,psi for Ordinal-Sequence,
  A,B,C,D for Ordinal,
  X,Y for set,
  x,y for set;

theorem
  X c= succ X by XBOOLE_1:7;

theorem
  succ X c= Y implies X c= Y
proof
  X c= succ X by XBOOLE_1:7;
  hence thesis by XBOOLE_1:1;
end;

theorem
  A in B iff succ A in succ B
proof
  A in B iff succ A c= B by ORDINAL1:21;
  hence thesis by ORDINAL1:22;
end;

theorem
  X c= A implies union X is Ordinal
proof
  assume X c= A;
  then for x st x in X holds x is Ordinal;
  hence thesis by ORDINAL1:23;
end;

theorem Th5:
  union On X is Ordinal
proof
  x in On X implies x is Ordinal by ORDINAL1:def 9;
  hence thesis by ORDINAL1:23;
end;

theorem Th6:
  X c= A implies On X = X
proof
  defpred P[set] means $1 in X & $1 is Ordinal;
  assume X c= A;
  then
A1: x in X iff P[x];
A2: x in On X iff P[x] by ORDINAL1:def 9;
  thus thesis from XBOOLE_0:sch 2(A2,A1);
end;

theorem Th7:
  On {A} = {A}
proof
  {A} c= succ A by XBOOLE_1:7;
  hence thesis by Th6;
end;

theorem Th8:
  A <> {} implies {} in A
proof
A1: {} c= A by XBOOLE_1:2;
  assume A <> {};
  then {} c< A by A1,XBOOLE_0:def 8;
  hence thesis by ORDINAL1:11;
end;

theorem
  inf A = {}
proof
A1: inf A = meet A by ORDINAL2:8;
  then A <> {} implies thesis by Th8,SETFAM_1:4;
  hence thesis by A1,SETFAM_1:def 1;
end;

theorem
  inf {A} = A
proof
  thus inf {A} = meet {A} by Th7
    .= A by SETFAM_1:10;
end;

theorem
  X c= A implies meet X is Ordinal
proof
  assume X c= A;
  then inf X = meet X by Th6;
  hence thesis;
end;

registration
  let A,B;
  cluster A \/ B -> ordinal;
  coherence
  proof
    A c= B or B c= A;
    hence thesis by XBOOLE_1:12;
  end;
  cluster A /\ B -> ordinal;
  coherence
  proof
    A c= B or B c= A;
    hence thesis by XBOOLE_1:28;
  end;
end;

theorem
  A \/ B = A or A \/ B = B
proof
  A c= B or B c= A;
  hence thesis by XBOOLE_1:12;
end;

theorem
  A /\ B = A or A /\ B = B
proof
  A c= B or B c= A;
  hence thesis by XBOOLE_1:28;
end;

Lm1: 1 = succ {};

theorem Th14:
  A in 1 implies A = {}
proof
  assume A in 1;
  hence A c= {} & {} c= A by Lm1,ORDINAL1:22,XBOOLE_1:2;
end;

theorem
  1 = {{}} by Lm1;

theorem Th16:
  A c= 1 implies A = {} or A = 1
proof
  assume that
A1: A c= 1 and
A2: A <> {} and
A3: A <> 1;
  A c< 1 by A1,A3,XBOOLE_0:def 8;
  hence contradiction by A2,Th14,ORDINAL1:11;
end;

theorem
  (A c= B or A in B) & C in D implies A+^C in B+^D
proof
  assume that
A1: A c= B or A in B and
A2: C in D;
A3: B+^C in B+^D by A2,ORDINAL2:32;
  A c= B by A1,ORDINAL1:def 2;
  hence thesis by A3,ORDINAL1:12,ORDINAL2:34;
end;

theorem
  A c= B & C c= D implies A+^C c= B+^D
proof
  assume that
A1: A c= B and
A2: C c= D;
A3: B+^C c= B+^D by A2,ORDINAL2:33;
  A+^C c= B+^C by A1,ORDINAL2:34;
  hence thesis by A3,XBOOLE_1:1;
end;

theorem Th19:
  A in B & (C c= D & D <> {} or C in D) implies A*^C in B*^D
proof
  assume that
A1: A in B and
A2: C c= D & D <> {} or C in D;
A3: C c= D by A2,ORDINAL1:def 2;
  A*^D in B*^D by A1,A2,ORDINAL2:40;
  hence thesis by A3,ORDINAL1:12,ORDINAL2:42;
end;

theorem
  A c= B & C c= D implies A*^C c= B*^D
proof
  assume that
A1: A c= B and
A2: C c= D;
A3: B*^C c= B*^D by A2,ORDINAL2:42;
  A*^C c= B*^C by A1,ORDINAL2:41;
  hence thesis by A3,XBOOLE_1:1;
end;

theorem Th21:
  B+^C = B+^D implies C = D
proof
  assume that
A1: B+^C = B+^D and
A2: C <> D;
  C in D or D in C by A2,ORDINAL1:14;
  then B+^C in B+^C by A1,ORDINAL2:32;
  hence contradiction;
end;

theorem Th22:
  B+^C in B+^D implies C in D
proof
  assume that
A1: B+^C in B+^D and
A2: not C in D;
  D c= C by A2,ORDINAL1:16;
  hence contradiction by A1,ORDINAL1:5,ORDINAL2:33;
end;

theorem Th23:
  B+^C c= B+^D implies C c= D
proof
  assume
A1: B+^C c= B+^D;
  B+^C c= B+^D & B+^C <> B+^D iff B+^C c< B+^D by XBOOLE_0:def 8;
  then C = D or C in D by A1,Th21,Th22,ORDINAL1:11;
  hence thesis by ORDINAL1:def 2;
end;

theorem Th24:
  A c= A+^B & B c= A+^B
proof
A1: {}+^B c= A+^B by ORDINAL2:34,XBOOLE_1:2;
  A+^{} c= A+^B by ORDINAL2:33,XBOOLE_1:2;
  hence thesis by A1,ORDINAL2:27,30;
end;

theorem
  A in B implies A in B+^C & A in C+^B
proof
  assume
A1: A in B;
A2: B c= C+^B by Th24;
  B c= B+^C by Th24;
  hence thesis by A1,A2;
end;

theorem Th26:
  A+^B = {} implies A = {} & B = {}
proof
  assume
A1: A+^B = {};
  then
A2: B c= {} by Th24;
  A c= {} by A1,Th24;
  hence thesis by A2,XBOOLE_1:3;
end;

theorem Th27:
  A c= B implies ex C st B = A+^C
proof
  defpred P[Ordinal] means A c= $1 implies ex C st $1 = A+^C;
A1: for B st P[B] holds P[succ B]
  proof
    let B such that
A2: A c= B implies ex C st B = A+^C and
A3: A c= succ B;
A4: now
      assume A <> succ B;
      then A c< succ B by A3,XBOOLE_0:def 8;
      then A in succ B by ORDINAL1:11;
      then consider C such that
A5:   B = A+^C by A2,ORDINAL1:22;
      succ B = A+^succ C by A5,ORDINAL2:28;
      hence thesis;
    end;
    A = succ B implies succ B = A+^{} by ORDINAL2:27;
    hence thesis by A4;
  end;
A6: for B st B <> {} & B is limit_ordinal & for C st C in B holds P[C]
  holds P[B]
  proof
    deffunc F(Ordinal) = A +^ $1;
    let B such that
    B <> {} and
A7: B is limit_ordinal and
    for C st C in B holds A c= C implies ex D st C = A+^D and
A8: A c= B;
    defpred P[Ordinal] means B c= A+^$1;
A9: B = {}+^B by ORDINAL2:30;
    {}+^B c= A+^B by ORDINAL2:34,XBOOLE_1:2;
    then
A10: ex C st P[C] by A9;
    consider C such that
A11: P[C] & for D st P[D] holds C c= D from ORDINAL1:sch 1(A10);
    consider L being Ordinal-Sequence such that
A12: dom L = C & for D st D in C holds L.D = F(D) from ORDINAL2:sch 3;
A13: now
      for D st D in rng L holds D in B
      proof
        let D;
        assume D in rng L;
        then consider y such that
A14:    y in dom L and
A15:    D = L.y by FUNCT_1:def 3;
        reconsider y as Ordinal by A14;
A16:    not C c= y by A12,A14,ORDINAL1:5;
        L.y = A+^y by A12,A14;
        hence thesis by A11,A15,A16,ORDINAL1:16;
      end;
      then
A17:  sup rng L c= B by ORDINAL2:20;
A18:  C is limit_ordinal
      proof
        assume not thesis;
        then consider D such that
A19:    C = succ D by ORDINAL1:29;
        not C c= D by A19,ORDINAL1:5,6;
        then A+^D in B by A11,ORDINAL1:16;
        then
A20:    succ(A+^D) c= B by ORDINAL1:21;
        succ(A+^D) = A+^succ D by ORDINAL2:28;
        then B = succ(A+^D) by A11,A19,A20,XBOOLE_0:def 10;
        hence contradiction by A7,ORDINAL1:29;
      end;
      assume C <> {};
      then A+^C = sup L by A12,A18,ORDINAL2:29
        .= sup rng L;
      then B = A+^C by A11,A17,XBOOLE_0:def 10;
      hence thesis;
    end;
    now
      assume C = {};
      then A+^C = A by ORDINAL2:27;
      hence B = A by A8,A11,XBOOLE_0:def 10
        .= A+^{} by ORDINAL2:27;
    end;
    hence thesis by A13;
  end;
A21: P[{}]
  proof
    assume A c= {};
    then A = {} by XBOOLE_1:3;
    then {} = A+^{} by ORDINAL2:30;
    hence thesis;
  end;
  for B holds P[B] from ORDINAL2:sch 1(A21,A1,A6);
  hence thesis;
end;

theorem Th28:
  A in B implies ex C st B = A+^C & C <> {}
proof
  assume
A1: A in B;
  then A c= B by ORDINAL1:def 2;
  then consider C such that
A2: B = A+^C by Th27;
  take C;
  thus B = A+^C by A2;
  assume C = {};
  then B = A by A2,ORDINAL2:27;
  hence contradiction by A1;
end;

theorem Th29:
  A <> {} & A is limit_ordinal implies B+^A is limit_ordinal
proof
  assume that
A1: A <> {} and
A2: A is limit_ordinal;
  {} c= A by XBOOLE_1:2;
  then
A3: {} c< A by A1,XBOOLE_0:def 8;
  deffunc F(Ordinal) = B +^ $1;
  consider L being Ordinal-Sequence such that
A4: dom L = A & for C st C in A holds L.C = F(C) from ORDINAL2:sch 3;
A5: B+^A = sup L by A1,A2,A4,ORDINAL2:29
    .= sup rng L;
  for C st C in B+^A holds succ C in B+^A
  proof
    let C such that
A6: C in B+^A;
A7: now
      assume not C in B;
      then consider D such that
A8:   C = B+^D by Th27,ORDINAL1:16;
      now
        assume not D in A;
        then B+^A c= B+^D by ORDINAL1:16,ORDINAL2:33;
        then C in C by A6,A8;
        hence contradiction;
      end;
      then
A9:   succ D in A by A2,ORDINAL1:28;
      then L.(succ D) = B+^succ D by A4
        .= succ(B+^D) by ORDINAL2:28;
      then succ C in rng L by A4,A8,A9,FUNCT_1:def 3;
      hence thesis by A5,ORDINAL2:19;
    end;
    now
      assume C in B;
      then
A10:  succ C c= B by ORDINAL1:21;
A11:  (succ C)+^{} = succ C by ORDINAL2:27;
      B+^{} in B+^A by A3,ORDINAL1:11,ORDINAL2:32;
      hence thesis by A10,A11,ORDINAL1:12,ORDINAL2:34;
    end;
    hence thesis by A7;
  end;
  hence thesis by ORDINAL1:28;
end;

theorem Th30:
  A+^B+^C = A+^(B+^C)
proof
  defpred Sigma[Ordinal] means A+^B+^$1 = A+^(B+^$1);
A1: for C st Sigma[C] holds Sigma[succ C]
  proof
    let C such that
A2: A+^B+^C = A+^(B+^C);
    thus A+^B+^succ C = succ(A+^B+^C) by ORDINAL2:28
      .= A+^succ(B+^C) by A2,ORDINAL2:28
      .= A+^(B+^succ C) by ORDINAL2:28;
  end;
A3: for C st C <> {} & C is limit_ordinal & for D st D in C holds Sigma[D]
  holds Sigma[C]
  proof
    deffunc F(Ordinal) = A +^ B +^ $1;
    let C such that
A4: C <> {} and
A5: C is limit_ordinal and
A6: for D st D in C holds Sigma[D];
    consider L being Ordinal-Sequence such that
A7: dom L = C & for D st D in C holds L.D = F(D) from ORDINAL2:sch 3;
    deffunc F(Ordinal) = A +^ $1;
    consider L1 being Ordinal-Sequence such that
A8: dom L1 = B+^C & for D st D in B+^C holds L1.D = F(D) from
    ORDINAL2:sch 3;
A9: rng L c= rng L1
    proof
      let x;
      assume x in rng L;
      then consider y such that
A10:  y in dom L and
A11:  x = L.y by FUNCT_1:def 3;
      reconsider y as Ordinal by A10;
A12:  B+^y in B+^C by A7,A10,ORDINAL2:32;
      L.y = A+^B+^y by A7,A10;
      then
A13:  L.y = A+^(B+^y) by A6,A7,A10;
      L1.(B+^y) = A+^(B+^y) by A7,A8,A10,ORDINAL2:32;
      hence thesis by A8,A11,A13,A12,FUNCT_1:def 3;
    end;
A14: B+^C <> {} by A4,Th26;
    B+^C is limit_ordinal by A4,A5,Th29;
    then
A15: A+^(B+^C) = sup L1 by A8,A14,ORDINAL2:29
      .= sup rng L1;
    A+^B+^C = sup L by A4,A5,A7,ORDINAL2:29
      .= sup rng L;
    hence A+^B+^C c= A+^(B+^C) by A15,A9,ORDINAL2:22;
    let x;
    assume
A16: x in A+^(B+^C);
    then reconsider x9 = x as Ordinal;
A17: now
A18:  A c= A+^B by Th24;
      assume
A19:  not x in A+^B;
      then A+^B c= x9 by ORDINAL1:16;
      then consider E being Ordinal such that
A20:  x9 = A+^E by A18,Th27,XBOOLE_1:1;
      B c= E by A19,A20,ORDINAL1:16,ORDINAL2:32;
      then consider F being Ordinal such that
A21:  E = B+^F by Th27;
A22:  now
        assume not F in C;
        then B+^C c= B+^F by ORDINAL1:16,ORDINAL2:33;
        then A+^(B+^C) c= A+^(B+^F) by ORDINAL2:33;
        then x9 in x9 by A16,A20,A21;
        hence contradiction;
      end;
      then x = A+^B+^F by A6,A20,A21;
      hence thesis by A22,ORDINAL2:32;
    end;
A23: A+^B+^{} = A+^B by ORDINAL2:27;
    A+^B+^{} c= A+^B+^C by ORDINAL2:33,XBOOLE_1:2;
    hence thesis by A23,A17;
  end;
  A+^B+^{} = A+^B by ORDINAL2:27
    .= A+^(B+^{}) by ORDINAL2:27;
  then
A24: Sigma[{}];
  for C holds Sigma[C] from ORDINAL2:sch 1(A24,A1,A3);
  hence thesis;
end;

theorem
  A*^B = {} implies A = {} or B = {}
proof
  assume that
A1: A*^B = {} and
A2: A <> {} and
A3: B <> {};
  {} c= A by XBOOLE_1:2;
  then {} c< A by A2,XBOOLE_0:def 8;
  hence contradiction by A1,A3,ORDINAL1:11,ORDINAL2:40;
end;

theorem
  A in B & C <> {} implies A in B*^C & A in C*^B
proof
  assume that
A1: A in B and
A2: C <> {};
  {} c= C by XBOOLE_1:2;
  then {} c< C by A2,XBOOLE_0:def 8;
  then {} in C by ORDINAL1:11;
  then
A3: 1 c= C by Lm1,ORDINAL1:21;
  then
A4: 1*^B c= C*^B by ORDINAL2:41;
A5: 1*^B = B by ORDINAL2:39;
A6: B*^1 = B by ORDINAL2:39;
  B*^1 c= B*^C by A3,ORDINAL2:42;
  hence thesis by A1,A4,A6,A5;
end;

theorem Th33:
  B*^A = C*^A & A <> {} implies B = C
proof
  assume that
A1: B*^A = C*^A and
A2: A <> {} and
A3: B <> C;
  B in C or C in B by A3,ORDINAL1:14;
  then B*^A in B*^A by A1,A2,ORDINAL2:40;
  hence contradiction;
end;

theorem Th34:
  B*^A in C*^A implies B in C
proof
  assume that
A1: B*^A in C*^A and
A2: not B in C;
  C c= B by A2,ORDINAL1:16;
  hence contradiction by A1,ORDINAL1:5,ORDINAL2:41;
end;

theorem Th35:
  B*^A c= C*^A & A <> {} implies B c= C
proof
  assume
A1: B*^A c= C*^A;
  B*^A c= C*^A & B*^A <> C*^A iff B*^A c< C*^A by XBOOLE_0:def 8;
  then (A <> {} implies B = C) or B in C by A1,Th33,Th34,ORDINAL1:11;
  hence thesis by ORDINAL1:def 2;
end;

theorem Th36:
  B <> {} implies A c= A*^B & A c= B*^A
proof
  assume B <> {};
  then {} in B by Th8;
  then
A1: 1 c= B by Lm1,ORDINAL1:21;
  then
A2: A*^1 c= A*^B by ORDINAL2:42;
  1*^A c= B*^A by A1,ORDINAL2:41;
  hence thesis by A2,ORDINAL2:39;
end;

theorem
  A*^B = 1 implies A = 1 & B = 1
proof
  assume
A1: A*^B = 1;
  then
A2: B <> {} by ORDINAL2:38;
  {} c= B by XBOOLE_1:2;
  then {} c< B by A2,XBOOLE_0:def 8;
  then {} in B by ORDINAL1:11;
  then
A3: 1 c= B by Lm1,ORDINAL1:21;
A4: now
A5: B = 1*^B by ORDINAL2:39;
    assume 1 in A;
    hence contradiction by A1,A2,A3,A5,ORDINAL1:5,ORDINAL2:40;
  end;
  now
    assume A in 1;
    then A c= {} by Lm1,ORDINAL1:22;
    then A = {} by XBOOLE_1:3;
    hence contradiction by A1,ORDINAL2:35;
  end;
  hence A = 1 by A4,ORDINAL1:14;
  hence thesis by A1,ORDINAL2:39;
end;

theorem Th38:
  A in B+^C implies A in B or ex D st D in C & A = B+^D
proof
  assume that
A1: A in B+^C and
A2: not A in B;
  consider D such that
A3: A = B+^D by A2,Th27,ORDINAL1:16;
  take D;
  assume not thesis;
  then C c= D by A3,ORDINAL1:16;
  hence contradiction by A1,A3,ORDINAL1:5,ORDINAL2:33;
end;

definition
  let C,fi;

  func C+^fi -> Ordinal-Sequence means
  : Def1:
  dom it = dom fi & for A st A in dom fi holds it.A = C+^(fi.A);
  existence
  proof
    deffunc F(Ordinal) = C+^(fi.$1);
    thus ex F being Ordinal-Sequence st dom F = dom fi & for A st A in dom fi
    holds F.A = F(A) from ORDINAL2:sch 3;
  end;
  uniqueness
  proof
    let f1,f2 be Ordinal-Sequence such that
A1: dom f1 = dom fi and
A2: for A st A in dom fi holds f1.A = C+^(fi.A) and
A3: dom f2 = dom fi and
A4: for A st A in dom fi holds f2.A = C+^(fi.A);
    now
      let x;
      assume
A5:   x in dom fi;
      then reconsider A = x as Ordinal;
      thus f1.x = C+^(fi.A) by A2,A5
        .= f2.x by A4,A5;
    end;
    hence thesis by A1,A3,FUNCT_1:2;
  end;
  func fi+^C -> Ordinal-Sequence means
  dom it = dom fi & for A st A in dom fi holds it.A = (fi.A)+^C;
  existence
  proof
    deffunc F(Ordinal) = (fi.$1)+^C;
    thus ex F being Ordinal-Sequence st dom F = dom fi & for A st A in dom fi
    holds F.A = F(A) from ORDINAL2:sch 3;
  end;
  uniqueness
  proof
    let f1,f2 be Ordinal-Sequence such that
A6: dom f1 = dom fi and
A7: for A st A in dom fi holds f1.A = (fi.A)+^C and
A8: dom f2 = dom fi and
A9: for A st A in dom fi holds f2.A = (fi.A)+^C;
    now
      let x;
      assume
A10:  x in dom fi;
      then reconsider A = x as Ordinal;
      thus f1.x = (fi.A)+^C by A7,A10
        .= f2.x by A9,A10;
    end;
    hence thesis by A6,A8,FUNCT_1:2;
  end;
  func C*^fi -> Ordinal-Sequence means
  dom it = dom fi & for A st A in dom fi holds it.A = C*^(fi.A);
  existence
  proof
    deffunc F(Ordinal) = C*^(fi.$1);
    thus ex F being Ordinal-Sequence st dom F = dom fi & for A st A in dom fi
    holds F.A = F(A) from ORDINAL2:sch 3;
  end;
  uniqueness
  proof
    let f1,f2 be Ordinal-Sequence such that
A11: dom f1 = dom fi and
A12: for A st A in dom fi holds f1.A = C*^(fi.A) and
A13: dom f2 = dom fi and
A14: for A st A in dom fi holds f2.A = C*^(fi.A);
    now
      let x;
      assume
A15:  x in dom fi;
      then reconsider A = x as Ordinal;
      thus f1.x = C*^(fi.A) by A12,A15
        .= f2.x by A14,A15;
    end;
    hence thesis by A11,A13,FUNCT_1:2;
  end;
  func fi*^C -> Ordinal-Sequence means
  : Def4:
  dom it = dom fi & for A st A in dom fi holds it.A = (fi.A)*^C;
  existence
  proof
    deffunc F(Ordinal) = (fi.$1)*^C;
    thus ex F being Ordinal-Sequence st dom F = dom fi & for A st A in dom fi
    holds F.A = F(A) from ORDINAL2:sch 3;
  end;
  uniqueness
  proof
    let f1,f2 be Ordinal-Sequence such that
A16: dom f1 = dom fi and
A17: for A st A in dom fi holds f1.A = (fi.A)*^C and
A18: dom f2 = dom fi and
A19: for A st A in dom fi holds f2.A = (fi.A)*^C;
    now
      let x;
      assume
A20:  x in dom fi;
      then reconsider A = x as Ordinal;
      thus f1.x = (fi.A)*^C by A17,A20
        .= f2.x by A19,A20;
    end;
    hence thesis by A16,A18,FUNCT_1:2;
  end;
end;

theorem Th39:
  {} <> dom fi & dom fi = dom psi & (for A,B st A in dom fi & B =
  fi.A holds psi.A = C+^B) implies sup psi = C+^sup fi
proof
  assume that
A1: {} <> dom fi and
A2: dom fi = dom psi and
A3: for A,B st A in dom fi & B = fi.A holds psi.A = C+^B;
  set z = the Element of dom fi;
  reconsider z9 = fi.z as Ordinal;
A4: C+^sup rng fi c= sup rng psi
  proof
    let x;
    assume
A5: x in C+^sup rng fi;
    then reconsider A = x as Ordinal;
A6: now
      given B such that
A7:   B in sup rng fi and
A8:   A = C+^B;
      consider D such that
A9:   D in rng fi and
A10:  B c= D by A7,ORDINAL2:21;
      consider x such that
A11:  x in dom fi and
A12:  D = fi.x by A9,FUNCT_1:def 3;
      reconsider x as Ordinal by A11;
      psi.x = C+^D by A3,A11,A12;
      then C+^D in rng psi by A2,A11,FUNCT_1:def 3;
      then C+^D in sup rng psi by ORDINAL2:19;
      hence A in sup rng psi by A8,A10,ORDINAL1:12,ORDINAL2:33;
    end;
    now
      C+^z9 = psi.z by A1,A3;
      then C+^z9 in rng psi by A1,A2,FUNCT_1:def 3;
      then
A13:  C+^z9 in sup rng psi by ORDINAL2:19;
      assume
A14:  A in C;
      C c= C+^z9 by Th24;
      then A c= C+^z9 by A14,ORDINAL1:def 2;
      hence A in sup rng psi by A13,ORDINAL1:12;
    end;
    hence thesis by A5,A6,Th38;
  end;
  sup rng psi c= C+^sup rng fi
  proof
    let x;
    assume
A15: x in sup rng psi;
    then reconsider A = x as Ordinal;
    consider B such that
A16: B in rng psi and
A17: A c= B by A15,ORDINAL2:21;
    consider y such that
A18: y in dom psi and
A19: B = psi.y by A16,FUNCT_1:def 3;
    reconsider y as Ordinal by A18;
    reconsider y9 = fi.y as Ordinal;
    y9 in rng fi by A2,A18,FUNCT_1:def 3;
    then
A20: y9 in sup rng fi by ORDINAL2:19;
    B = C+^y9 by A2,A3,A18,A19;
    then B in C+^sup rng fi by A20,ORDINAL2:32;
    hence thesis by A17,ORDINAL1:12;
  end;
  hence thesis by A4,XBOOLE_0:def 10;
end;

theorem Th40:
  A is limit_ordinal implies A*^B is limit_ordinal
proof
A1: now
    deffunc F(Ordinal) = $1 *^ B;
    assume that
A2: A <> {} and
A3: A is limit_ordinal;
    consider fi such that
A4: dom fi = A & for C st C in A holds fi.C = F(C) from ORDINAL2:sch 3;
A5: A*^B = union sup fi by A2,A3,A4,ORDINAL2:37
      .= union sup rng fi;
    for C st C in A*^B holds succ C in A*^B
    proof
      let C;
      assume
A6:   C in A*^B;
      then consider X such that
A7:   C in X and
A8:   X in sup rng fi by A5,TARSKI:def 4;
      reconsider X as Ordinal by A8;
      consider D such that
A9:   D in rng fi and
A10:  X c= D by A8,ORDINAL2:21;
      consider x such that
A11:  x in dom fi and
A12:  D = fi.x by A9,FUNCT_1:def 3;
      succ C c= D by A7,A10,ORDINAL1:21;
      then
A13:  succ C in succ D by ORDINAL1:22;
      reconsider x as Ordinal by A11;
A14:  succ x in dom fi by A3,A4,A11,ORDINAL1:28;
      then fi.succ x = (succ x)*^B by A4
        .= x*^B+^B by ORDINAL2:36;
      then x*^B+^B in rng fi by A14,FUNCT_1:def 3;
      then
A15:  x*^B+^B in sup rng fi by ORDINAL2:19;
A16:  x*^B+^succ {} = succ(x*^B+^{}) by ORDINAL2:28;
      B<>{} by A6,ORDINAL2:38;
      then {} in B by Th8;
      then
A17:  succ {} c= B by ORDINAL1:21;
A18:  x*^B+^{} = x*^B by ORDINAL2:27;
      x*^B = fi.x by A4,A11;
      then succ D in sup rng fi by A12,A17,A16,A18,A15,ORDINAL1:12,ORDINAL2:33;
      hence thesis by A5,A13,TARSKI:def 4;
    end;
    hence thesis by ORDINAL1:28;
  end;
  assume A is limit_ordinal;
  hence thesis by A1,ORDINAL2:35;
end;

theorem Th41:
  A in B*^C & B is limit_ordinal implies ex D st D in B & A in D*^ C
proof
  assume that
A1: A in B*^C and
A2: B is limit_ordinal;
  deffunc F(Ordinal) = $1 *^ C;
  consider fi such that
A3: dom fi = B & for D st D in B holds fi.D = F(D) from ORDINAL2:sch 3;
  B <> {} by A1,ORDINAL2:35;
  then B*^C = union sup fi by A2,A3,ORDINAL2:37
    .= union sup rng fi;
  then consider X such that
A4: A in X and
A5: X in sup rng fi by A1,TARSKI:def 4;
  reconsider X as Ordinal by A5;
  consider D such that
A6: D in rng fi and
A7: X c= D by A5,ORDINAL2:21;
  consider x such that
A8: x in dom fi and
A9: D = fi.x by A6,FUNCT_1:def 3;
  reconsider x as Ordinal by A8;
  take E = succ x;
  thus E in B by A2,A3,A8,ORDINAL1:28;
A10: D+^{} = D by ORDINAL2:27;
A11: C <> {} by A1,ORDINAL2:38;
  E*^C = x*^C+^C by ORDINAL2:36
    .= D+^C by A3,A8,A9;
  then D in E*^C by A11,A10,Th8,ORDINAL2:32;
  hence thesis by A4,A7,ORDINAL1:10;
end;

theorem Th42:
  dom fi = dom psi & C <> {} & sup fi is limit_ordinal & (for A,B
  st A in dom fi & B = fi.A holds psi.A = B*^C) implies sup psi = (sup fi)*^C
proof
  assume that
A1: dom fi = dom psi and
A2: C <> {} and
A3: sup fi is limit_ordinal and
A4: for A,B st A in dom fi & B = fi.A holds psi.A = B*^C;
A5: (sup rng fi)*^C c= sup rng psi
  proof
    let x;
    assume
A6: x in (sup rng fi)*^C;
    then reconsider A = x as Ordinal;
    consider B such that
A7: B in sup rng fi and
A8: A in B*^C by A3,A6,Th41;
    consider D such that
A9: D in rng fi and
A10: B c= D by A7,ORDINAL2:21;
    consider y such that
A11: y in dom fi and
A12: D = fi.y by A9,FUNCT_1:def 3;
    reconsider y as Ordinal by A11;
    reconsider y9 = psi.y as Ordinal;
A13: y9 in rng psi by A1,A11,FUNCT_1:def 3;
    y9 = D*^C by A4,A11,A12;
    then
A14: D*^C in sup rng psi by A13,ORDINAL2:19;
    B*^C c= D*^C by A10,ORDINAL2:41;
    hence thesis by A8,A14,ORDINAL1:10;
  end;
  sup rng psi c= (sup rng fi)*^C
  proof
    let x;
    assume
A15: x in sup rng psi;
    then reconsider A = x as Ordinal;
    consider B such that
A16: B in rng psi and
A17: A c= B by A15,ORDINAL2:21;
    consider y such that
A18: y in dom psi and
A19: B = psi.y by A16,FUNCT_1:def 3;
    reconsider y as Ordinal by A18;
    reconsider y9 = fi.y as Ordinal;
    y9 in rng fi by A1,A18,FUNCT_1:def 3;
    then
A20: y9 in sup rng fi by ORDINAL2:19;
    B = y9*^C by A1,A4,A18,A19;
    then B in (sup rng fi)*^C by A2,A20,ORDINAL2:40;
    hence thesis by A17,ORDINAL1:12;
  end;
  hence thesis by A5,XBOOLE_0:def 10;
end;

theorem Th43:
  {} <> dom fi implies sup (C+^fi) = C+^sup fi
proof
A1: for A,B st A in dom fi & B = fi.A holds (C+^fi).A = C+^B by Def1;
  dom (C+^fi) = dom fi by Def1;
  hence thesis by A1,Th39;
end;

theorem Th44:
  {} <> dom fi & C <> {} & sup fi is limit_ordinal implies sup (fi
  *^C) = (sup fi)*^C
proof
A1: for A,B st A in dom fi & B = fi.A holds (fi*^C).A = B*^C by Def4;
  dom (fi*^C) = dom fi by Def4;
  hence thesis by A1,Th42;
end;

theorem Th45:
  B <> {} implies union(A+^B) = A+^union B
proof
  assume
A1: B <> {};
A2: now
    assume not ex C st B = succ C;
    then
A3: B is limit_ordinal by ORDINAL1:29;
    then A+^B is limit_ordinal by A1,Th29;
    then union(A+^B) = A+^B by ORDINAL1:def 6;
    hence thesis by A3,ORDINAL1:def 6;
  end;
  now
    given C such that
A4: B = succ C;
    thus union(A+^B) = union succ (A+^C) by A4,ORDINAL2:28
      .= A+^C by ORDINAL2:2
      .= A+^union B by A4,ORDINAL2:2;
  end;
  hence thesis by A2;
end;

theorem Th46:
  (A+^B)*^C = A*^C +^ B*^C
proof
  defpred S[Ordinal] means (A+^$1)*^C = A*^C +^ $1*^C;
A1: for B st S[B] holds S[succ B]
  proof
    let B such that
A2: (A+^B)*^C = A*^C +^ B*^C;
    thus (A+^succ B)*^C = (succ(A+^B))*^C by ORDINAL2:28
      .= A*^C +^ B*^C +^ C by A2,ORDINAL2:36
      .= A*^C +^ (B*^C +^ C) by Th30
      .= A*^C +^ (succ B)*^C by ORDINAL2:36;
  end;
A3: for B st B <> {} & B is limit_ordinal & for D st D in B holds S[D] holds
  S[B]
  proof
    deffunc F(Ordinal) = A +^ $1;
    let B such that
A4: B <> {} and
A5: B is limit_ordinal and
A6: for D st D in B holds S[D];
    consider fi such that
A7: dom fi = B & for D st D in B holds fi.D = F(D) from ORDINAL2:sch
    3;
    A+^B is limit_ordinal by A4,A5,Th29;
    then
A8: (A+^B)*^C is limit_ordinal by Th40;
A9: dom (fi*^C) = dom fi by Def4;
A10: now
      assume
A11:  C = {};
      then
A12:  A*^C = {} by ORDINAL2:38;
A13:  B*^C = {} by A11,ORDINAL2:38;
      (A+^B)*^C = {} by A11,ORDINAL2:38;
      hence thesis by A12,A13,ORDINAL2:27;
    end;
    deffunc F(Ordinal) = $1 *^ C;
    consider psi such that
A14: dom psi = B & for D st D in B holds psi.D = F(D) from ORDINAL2:
    sch 3;
A15: now
      let x;
      assume
A16:  x in B;
      then reconsider k = x as Ordinal;
      reconsider m = fi.k, n = psi.k as Ordinal;
      thus (fi*^C).x = m*^C by A7,A16,Def4
        .= (A+^k)*^C by A7,A16
        .= A*^C+^k*^C by A6,A16
        .= A*^C+^n by A14,A16
        .= (A*^C+^psi).x by A14,A16,Def1;
    end;
    reconsider k = psi.{} as Ordinal;
    {} in B by A4,Th8;
    then k in rng psi by A14,FUNCT_1:def 3;
    then
A17: k in sup rng psi by ORDINAL2:19;
    dom (A*^C+^psi) = dom psi by Def1;
    then
A18: fi*^C = A*^C+^psi by A7,A14,A9,A15,FUNCT_1:2;
A19: A+^B = sup fi by A4,A5,A7,ORDINAL2:29;
    now
      assume C <> {};
      then (A+^B)*^C = sup(fi*^C) by A4,A5,A7,A19,Th29,Th44
        .= A*^C+^sup psi by A4,A14,A18,Th43;
      hence (A+^B)*^C = union(A*^C+^sup psi) by A8,ORDINAL1:def 6
        .= A*^C+^union sup psi by A17,Th45
        .= A*^C+^B*^C by A4,A5,A14,ORDINAL2:37;
    end;
    hence thesis by A10;
  end;
  (A+^{})*^C = A*^C by ORDINAL2:27
    .= A*^C +^ {} by ORDINAL2:27
    .= A*^C +^ {}*^C by ORDINAL2:35;
  then
A20: S[{}];
  for B holds S[B] from ORDINAL2:sch 1(A20,A1,A3);
  hence thesis;
end;

theorem Th47:
  A <> {} implies ex C,D st B = C*^A+^D & D in A
proof
  defpred I[Ordinal] means ex C,D st $1 = C*^A+^D & D in A;
  assume
A1: A <> {};
A2: for B st B <> {} & B is limit_ordinal & for A st A in B holds I[A]
  holds I[B]
  proof
    {} in A by A1,Th8;
    then
A3: succ {} c= A by ORDINAL1:21;
    let B such that
    B <> {} and
A4: B is limit_ordinal and
    for A st A in B holds I[A];
    defpred P[Ordinal] means $1 in B & B in $1*^A;
    B*^1 = B by ORDINAL2:39;
    then
A5: B c= B*^A by A3,ORDINAL2:42;
A6: now
      assume B <> B*^A;
      then B c< B*^A by A5,XBOOLE_0:def 8;
      then B in B*^A by ORDINAL1:11;
      then
A7:   ex C st P[C] by A4,Th41;
      consider C such that
A8:   P[C] and
A9:   for C1 being Ordinal st P[C1] holds C c= C1 from ORDINAL1:sch 1
      (A7);
      now
        assume C is limit_ordinal;
        then consider C1 being Ordinal such that
A10:    C1 in C and
A11:    B in C1*^A by A8,Th41;
        C1 in B by A8,A10,ORDINAL1:10;
        hence contradiction by A9,A10,A11,ORDINAL1:5;
      end;
      then consider C1 being Ordinal such that
A12:  C = succ C1 by ORDINAL1:29;
A13:  C1 in C by A12,ORDINAL1:6;
      then C1 in B by A8,ORDINAL1:10;
      then not B in C1*^A by A9,A13,ORDINAL1:5;
      then consider D such that
A14:  B = C1*^A+^D by Th27,ORDINAL1:16;
      thus I[B]
      proof
        take C1,D;
        thus B = C1*^A+^D by A14;
        C1*^A+^D in C1*^A+^A by A8,A12,A14,ORDINAL2:36;
        hence thesis by Th22;
      end;
    end;
    B = B*^A implies B = B*^A+^{} & {} in A by A1,Th8,ORDINAL2:27;
    hence thesis by A6;
  end;
A15: for B st I[B] holds I[succ B]
  proof
    let B;
    given C,D such that
A16: B = C*^A+^D and
A17: D in A;
A18: now
      assume not succ D in A;
      then
A19:  A c= succ D by ORDINAL1:16;
      take C1 = succ C, D1 = {};
      succ D c= A by A17,ORDINAL1:21;
      then
A20:  A = succ D by A19,XBOOLE_0:def 10;
      thus C1*^A+^D1 = C1*^A by ORDINAL2:27
        .= C*^A+^A by ORDINAL2:36
        .= succ B by A16,A20,ORDINAL2:28;
      thus D1 in A by A1,Th8;
    end;
    now
      assume
A21:  succ D in A;
      take C1 = C, D1 = succ D;
      thus C1*^A+^D1 = succ B by A16,ORDINAL2:28;
      thus D1 in A by A21;
    end;
    hence thesis by A18;
  end;
A22: I[{}]
  proof
    take C = {}, D = {};
    thus {} = {}+^{} by ORDINAL2:27
      .= C*^A+^D by ORDINAL2:35;
    thus thesis by A1,Th8;
  end;
  for B holds I[B] from ORDINAL2:sch 1(A22,A15,A2);
  hence thesis;
end;

theorem Th48:
  for C1,D1,C2,D2 being Ordinal st C1*^A+^D1 = C2*^A+^D2 & D1 in A
  & D2 in A holds C1 = C2 & D1 = D2
proof
  let C1,D1,C2,D2 be Ordinal such that
A1: C1*^A+^D1 = C2*^A+^D2 and
A2: D1 in A and
A3: D2 in A;
  set B = C1*^A+^D1;
A4: now
    assume C2 in C1;
    then consider C such that
A5: C1 = C2+^C and
A6: C <> {} by Th28;
    B = C2*^A+^C*^A+^D1 by A5,Th46
      .= C2*^A+^(C*^A+^D1) by Th30;
    then
A7: D2 = C*^A+^D1 by A1,Th21;
A8: C*^A c= C*^A+^D1 by Th24;
    A c= C*^A by A6,Th36;
    hence contradiction by A3,A7,A8,ORDINAL1:5;
  end;
  now
    assume C1 in C2;
    then consider C such that
A9: C2 = C1+^C and
A10: C <> {} by Th28;
    B = C1*^A+^C*^A+^D2 by A1,A9,Th46
      .= C1*^A+^(C*^A+^D2) by Th30;
    then
A11: D1 = C*^A+^D2 by Th21;
A12: C*^A c= C*^A+^D2 by Th24;
    A c= C*^A by A10,Th36;
    hence contradiction by A2,A11,A12,ORDINAL1:5;
  end;
  hence C1 = C2 by A4,ORDINAL1:14;
  hence thesis by A1,Th21;
end;

theorem Th49:
  1 in B & A <> {} & A is limit_ordinal implies for fi st dom fi =
  A & for C st C in A holds fi.C = C*^B holds A*^B = sup fi
proof
  assume that
A1: 1 in B and
A2: A <> {} and
A3: A is limit_ordinal;
  let fi;
  assume that
A4: dom fi = A and
A5: for C st C in A holds fi.C = C*^B;
  now
    given C such that
A6: sup fi = succ C;
    consider D such that
A7: D in rng fi and
A8: C c= D by A6,ORDINAL1:6,ORDINAL2:21;
    D in sup fi by A7,ORDINAL2:19;
    then
A9: succ D c= succ C by A6,ORDINAL1:21;
    succ C c= succ D by A8,ORDINAL2:1;
    then succ C = succ D by A9,XBOOLE_0:def 10;
    then C = D by ORDINAL1:7;
    then consider x such that
A10: x in dom fi and
A11: C = fi.x by A7,FUNCT_1:def 3;
    reconsider x as Ordinal by A10;
A12: C = x*^B by A4,A5,A10,A11;
    C+^1 in C+^B by A1,ORDINAL2:32;
    then
A13: sup fi in C+^B by A6,ORDINAL2:31;
A14: (succ x)*^B = x*^B+^B by ORDINAL2:36;
A15: succ x in dom fi by A3,A4,A10,ORDINAL1:28;
    then fi.succ x = (succ x)*^B by A4,A5;
    then C+^B in rng fi by A15,A12,A14,FUNCT_1:def 3;
    hence contradiction by A13,ORDINAL2:19;
  end;
  then
A16: sup fi is limit_ordinal by ORDINAL1:29;
  A*^B = union sup fi by A2,A3,A4,A5,ORDINAL2:37;
  hence thesis by A16,ORDINAL1:def 6;
end;

theorem
  (A*^B)*^C = A*^(B*^C)
proof
  defpred P[Ordinal] means ($1*^B)*^C = $1*^(B*^C);
A1: {}*^C = {} by ORDINAL2:35;
A2: for A st P[A] holds P[succ A]
  proof
    let A such that
A3: (A*^B)*^C = A*^(B*^C);
    thus ((succ A)*^B)*^C = (A*^B+^B)*^C by ORDINAL2:36
      .= A*^(B*^C)+^B*^C by A3,Th46
      .= A*^(B*^C)+^1*^(B*^C) by ORDINAL2:39
      .= (A+^1)*^(B*^C) by Th46
      .= (succ A)*^(B*^C) by ORDINAL2:31;
  end;
A4: for A st A <> {} & A is limit_ordinal & for D st D in A holds P[D] holds
  P[A]
  proof
    let A such that
A5: A <> {} and
A6: A is limit_ordinal and
A7: for D st D in A holds (D*^B)*^C = D*^(B*^C);
A8: now
      deffunc F(Ordinal) = $1 *^ B;
      assume that
A9:   1 in B and
A10:  1 in C;
      consider fi such that
A11:  dom fi = A & for D st D in A holds fi.D = F(D) from ORDINAL2:
      sch 3;
A12:  dom(fi*^C) = A & for D st D in A holds (fi*^C).D = D*^(B*^C)
      proof
        thus dom(fi*^C) = A by A11,Def4;
        let D;
        assume
A13:    D in A;
        then
A14:    fi.D = D*^B by A11;
        (fi*^C).D = (fi.D)*^C by A11,A13,Def4;
        hence thesis by A7,A13,A14;
      end;
      1 = 1*^1 by ORDINAL2:39;
      then 1 in B*^C by A9,A10,Th19;
      then
A15:  A*^(B*^C) = sup(fi*^C) by A5,A6,A12,Th49;
      A*^B = sup fi by A5,A6,A9,A11,Th49;
      hence thesis by A5,A6,A10,A11,A15,Th40,Th44;
    end;
    now
      assume not (1 in B & 1 in C);
      then
A16:  B = {} or B = 1 or C = {} or C = 1 by Th16,ORDINAL1:16;
A17:  {}*^C = {} by ORDINAL2:35;
A18:  A*^B*^1 = A*^B by ORDINAL2:39;
A19:  (A*^B)*^{} = {} by ORDINAL2:38;
A20:  A*^1 = A by ORDINAL2:39;
      A*^{} = {} by ORDINAL2:38;
      hence thesis by A16,A17,A20,A19,A18,ORDINAL2:38,39;
    end;
    hence thesis by A8;
  end;
  {}*^B = {} by ORDINAL2:35;
  then
A21: P[{}] by A1,ORDINAL2:35;
  for A holds P[A] from ORDINAL2:sch 1(A21,A2,A4);
  hence thesis;
end;

definition
  let A,B;
  func A -^ B -> Ordinal means
  : Def5:
  A = B+^it if B c= A otherwise it = {};
  existence by Th27;
  uniqueness by Th21;
  consistency;
  func A div^ B -> Ordinal means
  : Def6:
  ex C st A = it*^B+^C & C in B if B <> {} otherwise it = {};
  consistency;
  existence by Th47;
  uniqueness by Th48;
end;

definition
  let A,B;
  func A mod^ B -> Ordinal equals
  A-^(A div^ B)*^B;
  correctness;
end;

theorem
  A in B implies B = A+^(B-^A)
proof
  assume A in B;
  then A c= B by ORDINAL1:def 2;
  hence thesis by Def5;
end;

theorem Th52:
  A+^B-^A = B
proof
  A c= A+^B by Th24;
  hence thesis by Def5;
end;

theorem Th53:
  A in B & (C c= A or C in A) implies A-^C in B-^C
proof
  assume that
A1: A in B and
A2: C c= A or C in A;
  A c= B by A1,ORDINAL1:def 2;
  then C c= B by A2,ORDINAL1:def 2,XBOOLE_1:1;
  then
A3: B = C+^(B-^C) by Def5;
  C c= A by A2,ORDINAL1:def 2;
  then A = C+^(A-^C) by Def5;
  hence thesis by A1,A3,Th22;
end;

theorem Th54:
  A-^A = {}
proof
  A+^{} = A by ORDINAL2:27;
  hence thesis by Def5;
end;

theorem
  A in B implies B-^A <> {} & {} in B-^A
proof
  assume A in B;
  then A-^A in B-^A by Th53;
  hence thesis by Th54;
end;

theorem Th56:
  A-^{} = A & {}-^A = {}
proof
A1: {}+^A = A by ORDINAL2:30;
  {} c= A by XBOOLE_1:2;
  hence A-^{} = A by A1,Def5;
  not A c= {} or A c= {};
  then thesis or A = {} by Def5,XBOOLE_1:3;
  hence thesis by A1,Def5;
end;

theorem
  A-^(B+^C) = (A-^B)-^C
proof
  now
    per cases;
    suppose
      B+^C c= A;
      then A = B+^C+^(A-^(B+^C)) by Def5;
      then A = B+^(C+^(A-^(B+^C))) by Th30;
      then C+^(A-^(B+^C)) = A-^B by Th52;
      hence thesis by Th52;
    end;
    suppose
A1:   not B+^C c= A;
A2:   now
        assume A = B+^(A-^B);
        then not C c= A-^B by A1,ORDINAL2:33;
        hence A-^B-^C = {} by Def5;
      end;
      B c= A or not B c= A;
      then
A3:   A = B+^(A-^B) or A-^B = {} by Def5;
      A-^(B+^C) = {} by A1,Def5;
      hence thesis by A3,A2,Th56;
    end;
  end;
  hence thesis;
end;

theorem
  A c= B implies C-^B c= C-^A
proof
  assume
A1: A c= B;
  then
A2: B = A+^(B-^A) by Def5;
A3: now
    assume
A4: B c= C;
    then
A5: C = B+^(C-^B) by Def5;
    A c= C by A1,A4,XBOOLE_1:1;
    then B+^(C-^B) = A+^(C-^A) by A5,Def5;
    then A+^((B-^A)+^(C-^B)) = A+^(C-^A) by A2,Th30;
    then (B-^A)+^(C-^B) = C-^A by Th21;
    hence thesis by Th24;
  end;
  now
    assume not B c= C;
    then C-^B = {} by Def5;
    hence thesis by XBOOLE_1:2;
  end;
  hence thesis by A3;
end;

theorem
  A c= B implies A-^C c= B-^C
proof
  assume
A1: A c= B;
A2: now
    assume
A3: C c= A;
    then
A4: A = C+^(A-^C) by Def5;
    C c= B by A1,A3,XBOOLE_1:1;
    then C+^(A-^C) c= C+^(B-^C) by A1,A4,Def5;
    hence thesis by Th23;
  end;
  now
    assume not C c= A;
    then A-^C = {} by Def5;
    hence thesis by XBOOLE_1:2;
  end;
  hence thesis by A2;
end;

theorem
  C <> {} & A in B+^C implies A-^B in C
proof
  assume
A1: C <> {};
A2: B+^C-^B = C by Th52;
  not B c= A implies A-^B = {} by Def5;
  hence thesis by A1,A2,Th8,Th53;
end;

theorem
  A+^B in C implies B in C-^A
proof
A1: A+^B-^A = B by Th52;
  A c= A+^B by Th24;
  hence thesis by A1,Th53;
end;

theorem
  A c= B+^(A-^B)
proof
  now
    per cases;
    suppose
      B c= A;
      hence thesis by Def5;
    end;
    suppose
A1:   not B c= A;
      then A-^B = {} by Def5;
      hence thesis by A1,ORDINAL2:27;
    end;
  end;
  hence thesis;
end;

theorem
  A*^C -^ B*^C = (A-^B)*^C
proof
A1: now
    assume
A2: not B c= A;
    then
A3: not B*^C c= A*^C or C = {} by Th35;
A4: {}*^C = {} by ORDINAL2:35;
A5: A*^{} = {} by ORDINAL2:38;
    A-^B = {} by A2,Def5;
    hence thesis by A3,A5,A4,Def5,Th56;
  end;
  now
    assume B c= A;
    then A = B+^(A-^B) by Def5;
    then A*^C = B*^C+^(A-^B)*^C by Th46;
    hence thesis by Th52;
  end;
  hence thesis by A1;
end;

theorem Th64:
  (A div^ B)*^B c= A
proof
  now
    per cases;
    suppose
      B <> {};
      then ex C st A = (A div^ B)*^B+^C & C in B by Def6;
      hence thesis by Th24;
    end;
    suppose
      B = {};
      then A div^ B = {} by Def6;
      then (A div^ B)*^B = {} by ORDINAL2:35;
      hence thesis by XBOOLE_1:2;
    end;
  end;
  hence thesis;
end;

theorem Th65:
  A = (A div^ B)*^B+^(A mod^ B)
proof
  (A div^ B)*^B c= A by Th64;
  hence thesis by Def5;
end;

theorem
  A = B*^C+^D & D in C implies B = A div^ C & D = A mod^ C
proof
  assume that
A1: A = B*^C+^D and
A2: D in C;
  thus B = A div^ C by A1,A2,Def6;
  hence thesis by A1,Th52;
end;

theorem
  A in B*^C implies A div^ C in B & A mod^ C in C
proof
A1: A = (A div^ C)*^C+^(A mod^ C) by Th65;
  assume
A2: A in B*^C;
  then C <> {} by ORDINAL2:38;
  then
A3: ex D st A = (A div^ C)*^C+^D & D in C by Def6;
  then
A4: (A div^ C)*^C c= A by Th24;
  assume not thesis;
  then B*^C c= (A div^ C)*^C by A3,A1,Th21,ORDINAL1:16,ORDINAL2:41;
  hence contradiction by A2,A4,ORDINAL1:5;
end;

theorem Th68:
  B <> {} implies A*^B div^ B = A
proof
  assume B <> {};
  then
A1: {} in B by Th8;
  A*^B = A*^B+^{} by ORDINAL2:27;
  hence thesis by A1,Def6;
end;

theorem
  A*^B mod^ B = {}
proof
A1: A*^{} = {} by ORDINAL2:38;
A2: A*^B-^A*^B = {} by Th54;
  {}-^(A*^B div^ B)*^B = {} by Th56;
  hence thesis by A1,A2,Th68;
end;

theorem
  {} div^ A = {} & {} mod^ A = {} & A mod^ {} = A
proof
A1: A = {} or A <> {};
  {} = {}*^A by ORDINAL2:35;
  hence {} div^ A = {} by A1,Def6,Th68;
  thus {} mod^ A = {} by Th56;
  thus A mod^ {} = A-^{} by ORDINAL2:38
    .= A by Th56;
end;

theorem
  A div^ 1 = A & A mod^ 1 = {}
proof
A1: A = A*^1 by ORDINAL2:39;
A2: A = A+^{} by ORDINAL2:27;
A3: {} in 1 by Th8;
  hence A div^ 1 = A by A1,A2,Def6;
  thus A mod^ 1 = A-^A by A1,A2,A3,Def6
    .= {} by Th54;
end;

begin 


theorem
  sup X c= succ union On X
proof
  reconsider A = union On X as Ordinal by Th5;
  On X c= succ A
  proof
    let x;
    assume
A1: x in On X;
    then reconsider a = x as Ordinal by ORDINAL1:def 9;
    a c= A by A1,ZFMISC_1:74;
    hence thesis by ORDINAL1:22;
  end;
  hence thesis by ORDINAL2:def 3;
end;



reserve e,u for set;

theorem
  succ A is_cofinal_with 1
proof
  deffunc F(set) = A;
  consider psi such that
A1: dom psi = 1 & for B st B in 1 holds psi.B = F(B) from ORDINAL2:sch 3;
  take psi;
  thus dom psi = 1 by A1;
  thus rng psi c= succ A
  proof
    let e;
    assume e in rng psi;
    then consider u such that
A2: u in 1 and
A3: e = psi.u by A1,FUNCT_1:def 3;
    reconsider u as Ordinal by A2;
    psi.u = A by A1,A2;
    hence thesis by A3,ORDINAL1:6;
  end;
  thus psi is increasing
  proof
    let B,C;
    assume that
A4: B in C and
A5: C in dom psi;
    thus thesis by A1,A4,A5,Th14;
  end;
A6: psi.{} = A by A1,Lm1,ORDINAL1:6;
  rng psi = {psi.{}} by A1,Lm1,FUNCT_1:4;
  hence thesis by A6,ORDINAL2:23;
end;



theorem Th74:
  for a,b being Ordinal st a+^b is natural holds a in omega & b in omega
proof
  let x,y be Ordinal such that
A1: x+^y in omega;
A2: y c= x+^y by Th24;
  x c= x+^y by Th24;
  hence thesis by A1,A2,ORDINAL1:12;
end;

registration
  let a, b be natural Ordinal;
  cluster a -^ b -> natural;
  coherence
  proof
    not b c= a or b c= a;
    then a -^ b = {} or a = b+^(a-^b) by Def5;
    hence a-^b in omega by Th74,ORDINAL1:def 11;
  end;
  cluster a *^ b -> natural;
  coherence
  proof
    defpred P[natural Ordinal] means $1*^b is natural;
A1: now
      let a be natural Ordinal;
      assume P[a];
      then reconsider c = a*^b as natural Ordinal;
      (succ a)*^b = c+^b by ORDINAL2:36;
      hence P[succ a];
    end;
A2: P[{}] by ORDINAL2:35;
    P[a] from ORDINAL2:sch 17(A2,A1);
    hence thesis;
  end;
end;

theorem
  for a,b being Ordinal st a*^b is natural non empty holds a in omega &
  b in omega
proof
  let x,y be Ordinal such that
A1: x*^y in omega;
  assume
A2: x*^y is non empty;
  then y <> {} by ORDINAL2:38;
  then
A3: x c= x*^y by Th36;
  x <> {} by A2,ORDINAL2:35;
  then y c= x*^y by Th36;
  hence thesis by A1,A3,ORDINAL1:12;
end;

definition
  let a,b be natural Ordinal;
  redefine func a+^b;
  commutativity
  proof
    let a,b be natural Ordinal;
    defpred R[natural Ordinal] means a+^$1 = $1+^a;
A1: now
      let b be natural Ordinal;
      assume
A2:   R[b];
      defpred P[natural Ordinal] means (succ b)+^$1 = succ (b+^$1);
A3:   now
        let a be natural Ordinal;
        assume
A4:     P[a];
        (succ b)+^succ a = succ ((succ b)+^a) by ORDINAL2:28
          .= succ (b+^succ a) by A4,ORDINAL2:28;
        hence P[succ a];
      end;
      (succ b)+^{} = succ b by ORDINAL2:27
        .= succ (b+^{}) by ORDINAL2:27;
      then
A5:   P[{}];
      P[a] from ORDINAL2:sch 17(A5,A3);
      hence R[succ b] by A2,ORDINAL2:28;
    end;
    a+^{} = a by ORDINAL2:27
      .= {}+^a by ORDINAL2:30;
    then
A6: R[{}];
    thus R[b] from ORDINAL2:sch 17(A6,A1);
  end;
end;

definition
  let a,b be natural Ordinal;
  redefine func a*^b;
  commutativity
  proof
    let a,b be natural Ordinal;
    defpred R[natural Ordinal] means a*^$1 = $1*^a;
A1: now
      let b be natural Ordinal;
      defpred P[natural Ordinal] means $1*^succ b = $1*^b+^$1;
      assume
A2:   R[b];
A3:   now
        let a be natural Ordinal;
        assume
A4:     P[a];
        (succ a)*^succ b = a*^(succ b)+^succ b by ORDINAL2:36
          .= a*^b+^(a+^succ b) by A4,Th30
          .= a*^b+^succ (a+^b) by ORDINAL2:28
          .= succ (a*^b+^(a+^b)) by ORDINAL2:28
          .= succ (a*^b+^b+^a) by Th30
          .= succ ((succ a)*^b+^a) by ORDINAL2:36
          .= (succ a)*^b+^succ a by ORDINAL2:28;
        hence P[succ a];
      end;
      {}*^succ b = {} by ORDINAL2:35
        .= {}+^{} by ORDINAL2:27
        .= {}*^b+^{} by ORDINAL2:35;
      then
A5:   P[{}];
      P[a] from ORDINAL2:sch 17(A5,A3);
      hence R[succ b] by A2,ORDINAL2:36;
    end;
    a*^{} = {} by ORDINAL2:38
      .= {}*^a by ORDINAL2:35;
    then
A6: R[{}];
    thus R[b] from ORDINAL2:sch 17(A6,A1);
  end;
end;














begin

reserve a,b,x,y,z,z1,z2,z3,y1,y3,y4,A,B,C,D,G,M,N,X,Y,Z,W0,W00 for set,
  R,S,T, W,W1,W2 for Relation,
  F,H,H1 for Function;

theorem Th1:
  x in field R iff ex y st ([x,y] in R or [y,x] in R)
proof
  x in (dom R \/ rng R) iff x in dom R or x in rng R by XBOOLE_0:def 3;
  hence thesis by RELAT_1:def 6,XTUPLE_0:def 12,def 13;
end;

theorem Th2:
  X <> {} & Y <> {} & W = [: X,Y :] implies field W = X \/ Y
proof
  set a = the Element of X,b = the Element of Y;
  assume that
A1: X <> {} and
A2: Y <> {} and
A3: W = [: X,Y :];
A4: x in field W implies x in X \/ Y
  proof
    assume x in field W;
    then consider y such that
A5: [x,y] in W or [y,x] in W by Th1;
A6: [y,x] in W implies x in X \/ Y
    proof
      assume [y,x] in W;
      then x in Y by A3,ZFMISC_1:87;
      hence thesis by XBOOLE_0:def 3;
    end;
    [x,y] in W implies x in X \/ Y
    proof
      assume [x,y] in W;
      then x in X by A3,ZFMISC_1:87;
      hence thesis by XBOOLE_0:def 3;
    end;
    hence thesis by A5,A6;
  end;
  x in X \/ Y implies x in field W
  proof
A7: x in X implies x in field W
    proof
      assume x in X;
      then [x,b] in W by A2,A3,ZFMISC_1:87;
      hence thesis by Th1;
    end;
A8: x in Y implies x in field W
    proof
      assume x in Y;
      then [a,x] in W by A1,A3,ZFMISC_1:87;
      hence thesis by Th1;
    end;
    assume x in X \/ Y;
    hence thesis by A7,A8,XBOOLE_0:def 3;
  end;
  hence thesis by A4,TARSKI:1;
end;

scheme
  RSeparation { A()-> set, P[Relation] } : ex B st for R being Relation holds
  R in B iff R in A() & P[R] proof
  defpred p[set,set] means $1 = $2 & ex S st S = $2 & P[S];
A1: for y,t,v being set st p[y,t] & p[y,v] holds t = v;
  consider B such that
A2: for t being set holds t in B iff ex y being set st y in A() & p[y,t]
  from TARSKI:sch 1(A1);
  take B;
  let R;
  R in B implies ex T st T in A() & T = R & P[R]
  proof
    assume R in B;
    then consider y being set such that
A3: y in A() and
A4: y = R and
A5: ex S st S = R & P[S] by A2;
    reconsider y as Relation by A4;
    take y;
    thus thesis by A3,A4,A5;
  end;
  hence R in B implies R in A() & P[R];
  thus thesis by A2;
end;

theorem Th3:
  for x,y,W st x in field W & y in field W & W is well-ordering
  holds not x in W-Seg(y) implies [y,x] in W
proof
  let x,y,W;
  assume that
A1: x in field W and
A2: y in field W and
A3: W is well-ordering;
  W is connected by A3,WELLORD1:def 4;
  then W is_connected_in field W by RELAT_2:def 14;
  then
A4: x<>y implies [x,y] in W or [y,x] in W by A1,A2,RELAT_2:def 6;
  W is reflexive by A3,WELLORD1:def 4;
  then
A5: W is_reflexive_in field W by RELAT_2:def 9;
  assume not x in W-Seg(y);
  hence thesis by A1,A5,A4,RELAT_2:def 1,WELLORD1:1;
end;

theorem Th4:
  for x,y,W st x in field W & y in field W & W is well-ordering
  holds x in W-Seg(y) implies not [y,x] in W
proof
  let x,y,W;
  assume that
A1: x in field W & y in field W and
A2: W is well-ordering;
  W is antisymmetric by A2,WELLORD1:def 4;
  then
A3: W is_antisymmetric_in field W by RELAT_2:def 12;
  assume x in W-Seg(y);
  then
A4: x<>y & [x,y] in W by WELLORD1:1;
  assume [y,x] in W;
  hence contradiction by A1,A3,A4,RELAT_2:def 4;
end;

theorem Th5:
  for F,D st (for X st X in D holds not F.X in X & F.X in union D)
ex R st field R c= union D & R is well-ordering & not field R in D & for y st y
  in field R holds R-Seg(y) in D & F.(R-Seg(y)) = y
proof
  let F,D;
  assume
A1: for X st X in D holds not F.X in X & F.X in union D;
  defpred P[Relation] means $1 is well-ordering & for y st y in field $1 holds
  $1-Seg(y) in D & F.($1-Seg(y)) = y;
  set W0=bool [: union D, union D :];
  consider G such that
A2: W in G iff W in W0 & P[W] from RSeparation;
  defpred P[set,set] means ex W st [$1,$2] in W & W in G;
  consider S such that
A3: [x,y] in S iff x in union D & y in union D & P[x,y] from RELAT_1:
  sch 1;
  take R = S;
A4: x in field R implies x in union D & ex W st x in field W & W in G
  proof
    assume x in field R;
    then consider y such that
A5: [x,y] in R or [y,x] in R by Th1;
    (x in union D & y in union D & ex S st [x,y] in S & S in G) or (y in
    union D & x in union D & ex S st [y,x] in S & S in G) by A3,A5;
    then consider S such that
A6: ( [x,y] in S or [y,x] in S)& S in G;
    thus x in union D by A3,A5;
    take S;
    thus thesis by A6,Th1;
  end;
  then x in field R implies x in union D;
  hence field R c= union D by TARSKI:def 3;
A7: for W1,W2 holds W1 in G & W2 in G implies ((W1 c= W2 & for x st x in
  field W1 holds W1-Seg(x) = W2-Seg(x) ) or (W2 c= W1 & for x st x in field W2
  holds W2-Seg(x) = W1-Seg(x) ))
  proof
    let W1,W2;
    assume that
A8: W1 in G and
A9: W2 in G;
A10: W2 is well-ordering by A2,A9;
    defpred P[set] means $1 in field W2 & W1 |_2 (W1-Seg($1)) = W2 |_2 (W2-Seg
    ($1));
    consider C such that
A11: x in C iff x in field W1 & P[x] from XBOOLE_0:sch 1;
A12: W1 is well-ordering by A2,A8;
A13: x in C implies W1-Seg(x) = W2-Seg(x)
    proof
      assume
A14:  x in C;
      for y holds y in W1-Seg(x) iff y in W2-Seg(x)
      proof
        let y;
        field (W1 |_2(W1-Seg(x))) = W1-Seg(x) & field (W2 |_2(W2-Seg(x)))
        = W2-Seg(x ) by A12,A10,WELLORD1:32;
        hence thesis by A11,A14;
      end;
      hence thesis by TARSKI:1;
    end;
A15: x in C implies W1-Seg(x) c= C
    proof
      assume
A16:  x in C;
      y in W1-Seg(x) implies y in C
      proof
        assume
A17:    y in W1-Seg(x);
        then
A18:    y in W2-Seg(x) by A13,A16;
        then
A19:    [y,x] in W2 by WELLORD1:1;
        then
A20:    y in field W2 by RELAT_1:15;
A21:    W1-Seg(y)=(W1 |_2 (W1-Seg(x)))-Seg(y) by A12,A17,WELLORD1:27
          .=(W2 |_2 (W2-Seg(x)))-Seg(y) by A11,A16
          .=W2-Seg(y) by A10,A18,WELLORD1:27;
A22:    [y,x] in W1 by A17,WELLORD1:1;
        then
A23:    y in field W1 by RELAT_1:15;
        x in field W2 by A19,RELAT_1:15;
        then
A24:    W2-Seg(y) c= W2-Seg(x) by A10,A18,A20,WELLORD1:30;
        x in field W1 by A22,RELAT_1:15;
        then W1-Seg(y) c= W1-Seg(x) by A12,A17,A23,WELLORD1:30;
        then W1 |_2 (W1-Seg(y)) = (W1 |_2 (W1-Seg(x))) |_2 (W1-Seg(y)) by
WELLORD1:22
          .= (W2 |_2 (W2-Seg(x))) |_2 (W2-Seg(y)) by A11,A16,A21
          .= W2 |_2 (W2-Seg(y)) by A24,WELLORD1:22;
        hence thesis by A11,A23,A20;
      end;
      hence thesis by TARSKI:def 3;
    end;
A25: y1 in field W1 & not y1 in C implies ex y3 st y3 in field W1 & C=W1
    -Seg(y3) & not y3 in C
    proof
      set Y = field W1 \ C;
      assume y1 in field W1 & not y1 in C;
      then Y <> {} by XBOOLE_0:def 5;
      then consider a such that
A26:  a in Y and
A27:  for b st b in Y holds [a,b] in W1 by A12,WELLORD1:6;
      take y3=a;
      x in C iff x in W1-Seg(y3)
      proof
        thus x in C implies x in W1-Seg(y3)
        proof
          assume that
A28:      x in C and
A29:      not x in W1-Seg(y3);
          x in field W1 by A11,A28;
          then
A30:      [y3,x] in W1 by A12,A26,A29,Th3;
A31:      W1-Seg(x) c= C by A15,A28;
          y3 <> x implies y3 in C
          proof
            assume y3 <> x;
            then y3 in W1-Seg(x) by A30,WELLORD1:1;
            hence thesis by A31;
          end;
          hence contradiction by A26,A28,XBOOLE_0:def 5;
        end;
        thus x in W1-Seg(y3) implies x in C
        proof
          assume that
A32:      x in W1-Seg(y3) and
A33:      not x in C;
          [x,y3] in W1 by A32,WELLORD1:1;
          then
A34:      x in field W1 by RELAT_1:15;
          then x in Y by A33,XBOOLE_0:def 5;
          then [y3,x] in W1 by A27;
          hence contradiction by A12,A26,A32,A34,Th4;
        end;
      end;
      hence thesis by A26,TARSKI:1,XBOOLE_0:def 5;
    end;
A35: x in C implies W2-Seg(x) c= C
    proof
      assume
A36:  x in C;
      let y;
      assume
A37:  y in W2-Seg(x);
      then
A38:  y in W1-Seg(x) by A13,A36;
      then
A39:  [y,x] in W1 by WELLORD1:1;
      then
A40:  y in field W1 by RELAT_1:15;
A41:  W2-Seg(y)=(W2 |_2 (W2-Seg(x)))-Seg(y) by A10,A37,WELLORD1:27
        .=(W1 |_2 (W1-Seg(x)))-Seg(y) by A11,A36
        .=W1-Seg(y) by A12,A38,WELLORD1:27;
A42:  [y,x] in W2 by A37,WELLORD1:1;
      then
A43:  y in field W2 by RELAT_1:15;
      x in field W1 by A39,RELAT_1:15;
      then
A44:  W1-Seg(y) c= W1-Seg(x) by A12,A38,A40,WELLORD1:30;
      x in field W2 by A42,RELAT_1:15;
      then W2-Seg(y) c= W2-Seg(x) by A10,A37,A43,WELLORD1:30;
      then W2 |_2 (W2-Seg(y)) = (W2 |_2 (W2-Seg(x))) |_2 (W2-Seg(y)) by
WELLORD1:22
        .= (W1 |_2 (W1-Seg(x))) |_2 (W1-Seg(y)) by A11,A36,A41
        .= W1 |_2 (W1-Seg(y)) by A44,WELLORD1:22;
      hence thesis by A11,A43,A40;
    end;
A45: y1 in field W2 & not y1 in C implies ex y3 st y3 in field W2 & C=W2
    -Seg(y3) & not y3 in C
    proof
      set Y = field W2 \ C;
      assume y1 in field W2 & not y1 in C;
      then Y <> {} by XBOOLE_0:def 5;
      then consider a such that
A46:  a in Y and
A47:  for b st b in Y holds [a,b] in W2 by A10,WELLORD1:6;
      take y3=a;
      x in C iff x in W2-Seg(y3)
      proof
        thus x in C implies x in W2-Seg(y3)
        proof
          assume that
A48:      x in C and
A49:      not x in W2-Seg(y3);
          x in field W2 by A11,A48;
          then
A50:      [y3,x] in W2 by A10,A46,A49,Th3;
A51:      W2-Seg(x) c= C by A35,A48;
          y3 <> x implies y3 in C
          proof
            assume y3 <> x;
            then y3 in W2-Seg(x) by A50,WELLORD1:1;
            hence thesis by A51;
          end;
          hence contradiction by A46,A48,XBOOLE_0:def 5;
        end;
        thus x in W2-Seg(y3) implies x in C
        proof
          assume that
A52:      x in W2-Seg(y3) and
A53:      not x in C;
          [x,y3] in W2 by A52,WELLORD1:1;
          then
A54:      x in field W2 by RELAT_1:15;
          then x in Y by A53,XBOOLE_0:def 5;
          then [y3,x] in W2 by A47;
          hence contradiction by A10,A46,A52,A54,Th4;
        end;
      end;
      hence thesis by A46,TARSKI:1,XBOOLE_0:def 5;
    end;
A55: C = field W1 or C = field W2
    proof
      assume not C = field W1;
      then ex x st not (x in C implies x in field W1) or not (x in field W1
      implies x in C) by TARSKI:1;
      then consider y3 such that
A56:  y3 in field W1 and
A57:  C=W1-Seg(y3) and
A58:  not y3 in C by A11,A25;
      assume not C = field W2;
      then ex x st not (x in C implies x in field W2) or not (x in field W2
      implies x in C) by TARSKI:1;
      then consider y4 such that
A59:  y4 in field W2 and
A60:  C=W2-Seg(y4) and
      not y4 in C by A11,A45;
A61:  y3 = F.(W2-Seg(y4)) by A2,A8,A56,A57,A60
        .= y4 by A2,A9,A59;
      z in W1 |_2 (W1-Seg(y3)) iff z in W2 |_2 (W2-Seg(y3))
      proof
A62:    z in W1 & z in [: W1-Seg(y3),W1-Seg(y3) :] implies z in W2 & z in
        [: W2-Seg(y3),W2-Seg(y3) :]
        proof
          assume that
A63:      z in W1 and
A64:      z in [: W1-Seg(y3),W1-Seg(y3) :];
          consider z1,z2 such that
A65:      z1 in W1-Seg(y3) and
A66:      z2 in W1-Seg(y3) and
A67:      z=[z1,z2] by A64,ZFMISC_1:def 2;
          z1 in W1-Seg(z2) or z1=z2 & not z1 in W1-Seg(z2) by A63,A67,
WELLORD1:1;
          then
A68:      z1 in W2-Seg(z2) or z1=z2 & not z1 in W2-Seg(z2) by A13,A57,A66;
          z1 in field W2 by A11,A57,A65;
          hence thesis by A10,A57,A60,A61,A64,A67,A68,Th3,WELLORD1:1;
        end;
        z in W2 & z in [: W2-Seg(y3),W2-Seg(y3) :] implies z in W1 & z
        in [: W1-Seg(y3),W1-Seg(y3) :]
        proof
          assume that
A69:      z in W2 and
A70:      z in [: W2-Seg(y3),W2-Seg(y3) :];
          consider z1,z2 such that
A71:      z1 in W2-Seg(y3) and
A72:      z2 in W2-Seg(y3) and
A73:      z=[z1,z2] by A70,ZFMISC_1:def 2;
          z1 in W2-Seg(z2) or z1=z2 & not z1 in W2-Seg(z2) by A69,A73,
WELLORD1:1;
          then
A74:      z1 in W1-Seg(z2) or z1=z2 & not z1 in W1-Seg(z2) by A13,A60,A61,A72;
          z1 in field W1 by A11,A60,A61,A71;
          hence thesis by A12,A57,A60,A61,A70,A73,A74,Th3,WELLORD1:1;
        end;
        hence thesis by A62,XBOOLE_0:def 4;
      end;
      then W1 |_2 (W1-Seg(y3)) = W2 |_2 (W2-Seg(y3)) by TARSKI:1;
      hence contradiction by A11,A56,A58,A59,A61;
    end;
A75: C = field W2 implies (W2 c= W1 & for x st x in field W2 holds W2-Seg
    (x) = W1-Seg(x) )
    proof
      assume
A76:  C = field W2;
      [z1,z2] in W2 implies [z1,z2] in W1
      proof
        assume
A77:    [z1,z2] in W2;
        then
A78:    z1 in W2-Seg(z2) or z1=z2 & not z1 in W2-Seg(z2) by WELLORD1:1;
        z1 in C by A76,A77,RELAT_1:15;
        then
A79:    z1 in field W1 by A11;
        z2 in C by A76,A77,RELAT_1:15;
        then z1 in W1-Seg(z2) or z1=z2 & not z1 in W1-Seg(z2) by A13,A78;
        hence thesis by A12,A79,Th3,WELLORD1:1;
      end;
      hence thesis by A13,A76,RELAT_1:def 3;
    end;
    C = field W1 implies (W1 c= W2 & for x st x in field W1 holds W1-Seg
    (x) = W2-Seg(x) )
    proof
      assume
A80:  C = field W1;
      [z1,z2] in W1 implies [z1,z2] in W2
      proof
        assume
A81:    [z1,z2] in W1;
        then
A82:    z1 in W1-Seg(z2) or z1=z2 & not z1 in W1-Seg(z2) by WELLORD1:1;
        z1 in C by A80,A81,RELAT_1:15;
        then
A83:    z1 in field W2 by A11;
        z2 in C by A80,A81,RELAT_1:15;
        then z1 in W2-Seg(z2) or z1=z2 & not z1 in W2-Seg(z2) by A13,A82;
        hence thesis by A10,A83,Th3,WELLORD1:1;
      end;
      hence thesis by A13,A80,RELAT_1:def 3;
    end;
    hence thesis by A55,A75;
  end;
A84: x in field R & y in field R & [x,y] in R & [y,x] in R implies x=y
  proof
    assume that
    x in field R and
    y in field R and
A85: [x,y] in R and
A86: [y,x] in R;
    consider W1 such that
A87: [x,y] in W1 and
A88: W1 in G by A3,A85;
    consider W2 such that
A89: [y,x] in W2 and
A90: W2 in G by A3,A86;
A91: W2 c= W1 implies x=y
    proof
      W1 is well-ordering by A2,A88;
      then W1 well_orders field W1 by WELLORD1:4;
      then
A92:  W1 is_antisymmetric_in field W1 by WELLORD1:def 5;
      assume
A93:  W2 c= W1;
      then x in field W1 & y in field W1 by A89,RELAT_1:15;
      hence thesis by A87,A89,A93,A92,RELAT_2:def 4;
    end;
    W1 c= W2 implies x=y
    proof
      W2 is well-ordering by A2,A90;
      then W2 well_orders field W2 by WELLORD1:4;
      then
A94:  W2 is_antisymmetric_in field W2 by WELLORD1:def 5;
      assume
A95:  W1 c= W2;
      then x in field W2 & y in field W2 by A87,RELAT_1:15;
      hence thesis by A87,A89,A95,A94,RELAT_2:def 4;
    end;
    hence thesis by A7,A88,A90,A91;
  end;
  then
A96: R is_antisymmetric_in field R by RELAT_2:def 4;
A97: W in G implies field W c= field R
  proof
    assume
A98: W in G;
    let x;
    assume x in field W;
    then consider y such that
A99: [x,y] in W or [y,x] in W by Th1;
A100: [y,x] in W implies [y,x] in R
    proof
      assume
A101: [y,x] in W;
      W in W0 by A2,A98;
      then ex z1,z2 st z1 in union D & z2 in union D & [y,x]=[z1,z2] by A101,
ZFMISC_1:84;
      hence thesis by A3,A98,A101;
    end;
    [x,y] in W implies [x,y] in R
    proof
      assume
A102: [x,y] in W;
      W in W0 by A2,A98;
      then ex z1,z2 st z1 in union D & z2 in union D & [x,y]=[z1,z2] by A102,
ZFMISC_1:84;
      hence thesis by A3,A98,A102;
    end;
    hence thesis by A99,A100,Th1;
  end;
A103: for y st y in field R holds R-Seg(y) in D & F.(R-Seg(y)) = y
  proof
    let y;
    assume
A104: y in field R;
    then consider W such that
A105: y in field W and
A106: W in G by A4;
A107: y in union D by A4,A104;
A108: field W c= field R by A97,A106;
A109: x in W-Seg(y) implies x in R-Seg(y)
    proof
      assume
A110: x in W-Seg(y);
      then
A111: [x,y] in W by WELLORD1:1;
      then x in field W by RELAT_1:15;
      then x in union D by A4,A108;
      then
A112: [x,y] in R by A3,A106,A107,A111;
      not x =y by A110,WELLORD1:1;
      hence thesis by A112,WELLORD1:1;
    end;
    x in R-Seg(y) implies x in W-Seg(y)
    proof
      assume
A113: x in R-Seg(y);
      then [x,y] in R by WELLORD1:1;
      then consider W1 such that
A114: [x,y] in W1 and
A115: W1 in G by A3;
A116: y in field W1 by A114,RELAT_1:15;
      not x =y by A113,WELLORD1:1;
      then x in W1-Seg(y) by A114,WELLORD1:1;
      hence thesis by A7,A105,A106,A115,A116;
    end;
    then W-Seg(y) = R-Seg(y) by A109,TARSKI:1;
    hence thesis by A2,A105,A106;
  end;
A117: x in field R & y in field R & x <>y implies [x,y] in R or [y,x] in R
  proof
    assume that
A118: x in field R and
A119: y in field R and
A120: x <>y;
    consider W2 such that
A121: y in field W2 and
A122: W2 in G by A4,A119;
    consider W1 such that
A123: x in field W1 and
A124: W1 in G by A4,A118;
A125: x in union D & y in union D by A4,A118,A119;
A126: W2 c= W1 implies [x,y] in R or [y,x] in R
    proof
      W1 is well-ordering by A2,A124;
      then W1 well_orders field W1 by WELLORD1:4;
      then
A127: W1 is_connected_in field W1 by WELLORD1:def 5;
      assume W2 c= W1;
      then field W2 c= field W1 by RELAT_1:16;
      then [x,y] in W1 or [y,x] in W1 by A120,A123,A121,A127,RELAT_2:def 6;
      hence thesis by A3,A124,A125;
    end;
    W1 c= W2 implies [x,y] in R or [y,x] in R
    proof
      W2 is well-ordering by A2,A122;
      then W2 well_orders field W2 by WELLORD1:4;
      then
A128: W2 is_connected_in field W2 by WELLORD1:def 5;
      assume W1 c= W2;
      then field W1 c= field W2 by RELAT_1:16;
      then [x,y] in W2 or [y,x] in W2 by A120,A123,A121,A128,RELAT_2:def 6;
      hence thesis by A3,A125,A122;
    end;
    hence thesis by A7,A124,A122,A126;
  end;
  then
A129: R is_connected_in field R by RELAT_2:def 6;
A130: R is_well_founded_in field R
  proof
    let Y;
    assume that
A131: Y c= field R and
A132: Y <> {};
    set y = the Element of Y;
    y in field R by A131,A132,TARSKI:def 3;
    then consider W such that
A133: y in field W and
A134: W in G by A4;
    W is well-ordering by A2,A134;
    then W well_orders field W by WELLORD1:4;
    then
A135: W is_well_founded_in field W by WELLORD1:def 5;
    set A = Y /\ field W;
A136: A c= field W by XBOOLE_1:17;
    A <> {} by A132,A133,XBOOLE_0:def 4;
    then consider a such that
A137: a in A and
A138: W-Seg(a) misses A by A135,A136,WELLORD1:def 3;
    ex b st b in Y & R-Seg(b) misses Y
    proof
      take b= a;
      thus b in Y by A137,XBOOLE_0:def 4;
      assume not thesis;
      then consider x being set such that
A139: x in R-Seg(b) and
A140: x in Y by XBOOLE_0:3;
      [x,b] in R by A139,WELLORD1:1;
      then consider W1 such that
A141: [x,b] in W1 and
A142: W1 in G by A3;
A143: b in field W1 by A141,RELAT_1:15;
      x<>b by A139,WELLORD1:1;
      then x in W1-Seg(b) by A141,WELLORD1:1;
      then
A144: x in W-Seg(a) by A7,A134,A136,A137,A142,A143;
      then [x,a] in W by WELLORD1:1;
      then x in field W by RELAT_1:15;
      then x in A by A140,XBOOLE_0:def 4;
      hence contradiction by A138,A144,XBOOLE_0:3;
    end;
    hence thesis;
  end;
A145: x in field R & y in field R & z in field R & [x,y] in R & [y,z] in R
  implies [x,z] in R
  proof
    assume that
    x in field R and
    y in field R and
    z in field R and
A146: [x,y] in R and
A147: [y,z] in R;
A148: x in union D & z in union D by A3,A146,A147;
    consider W1 such that
A149: [x,y] in W1 and
A150: W1 in G by A3,A146;
    consider W2 such that
A151: [y,z] in W2 and
A152: W2 in G by A3,A147;
    ex W st [x,y] in W & [y,z] in W & W in G
    proof
      take W = W2;
A153: not x in W1-Seg(y) implies [x,y] in W
      proof
A154:   W1 is well-ordering by A2,A150;
        then W1 well_orders field W1 by WELLORD1:4;
        then
A155:   W1 is_antisymmetric_in field W1 by WELLORD1:def 5;
        W is well-ordering by A2,A152;
        then W well_orders field W by WELLORD1:4;
        then
A156:   W is_reflexive_in field W by WELLORD1:def 5;
A157:   x in field W1 & y in field W1 by A149,RELAT_1:15;
        assume not x in W1-Seg(y);
        then [y,x] in W1 by A157,A154,Th3;
        then
A158:   x=y by A149,A157,A155,RELAT_2:def 4;
        y in field W by A151,RELAT_1:15;
        hence thesis by A158,A156,RELAT_2:def 1;
      end;
      y in field W1 & y in field W by A149,A151,RELAT_1:15;
      then W1-Seg(y) = W-Seg(y) by A7,A150,A152;
      hence thesis by A151,A152,A153,WELLORD1:1;
    end;
    then consider W such that
A159: [x,y] in W and
A160: [y,z] in W and
A161: W in G;
A162: z in field W by A160,RELAT_1:15;
    W is well-ordering by A2,A161;
    then W well_orders field W by WELLORD1:4;
    then
A163: W is_transitive_in field W by WELLORD1:def 5;
    x in field W & y in field W by A159,RELAT_1:15;
    then [x,z] in W by A159,A160,A163,A162,RELAT_2:def 8;
    hence thesis by A3,A148,A161;
  end;
  then
A164: R is_transitive_in field R by RELAT_2:def 8;
A165: x in field R implies [x,x] in R
  proof
    assume
A166: x in field R;
    then consider W such that
A167: x in field W and
A168: W in G by A4;
    W is well-ordering by A2,A168;
    then W well_orders field W by WELLORD1:4;
    then W is_reflexive_in field W by WELLORD1:def 5;
    then
A169: [x,x] in W by A167,RELAT_2:def 1;
    x in union D by A4,A166;
    hence thesis by A3,A168,A169;
  end;
A170: not field R in D
  proof
    set a0=F.(field R);
    reconsider W3 = [: field R,{ a0 } :] as Relation;
    reconsider W4 = { [a0,a0]} as Relation;
    reconsider W1=R \/ [: field R,{ a0 }:] \/ {[ a0,a0 ]} as Relation;
    {[ a0,a0 ]} c= W1 & [ a0,a0 ] in {[ a0,a0 ]} by TARSKI:def 1,XBOOLE_1:7;
    then
A171: a0 in field W1 by RELAT_1:15;
    field W4 = {a0,a0} by RELAT_1:17;
    then
A172: field W4 = {a0}\/{a0} by ENUMSET1:1;
A173: field R = {} implies field W1 =field R \/ { a0 }
    proof
      assume
A174: field R = {};
A175: field W3 = {}
      proof
        set z3 = the Element of field W3;
        assume field W3 <> {};
        then ex z2 st [z3,z2] in W3 or [z2,z3] in W3 by Th1;
        hence contradiction by A174,ZFMISC_1:90;
      end;
      field W1 =field (R \/ W3) \/ field W4 by RELAT_1:18;
      then field W1 =field R \/ {} \/ {a0} by A172,A175,RELAT_1:18;
      hence thesis;
    end;
A176: field R <> {} implies field W1 =field R \/ { a0 }
    proof
      assume field R <> {};
      then
A177: field W3 = field R \/ { a0 } by Th2;
      field W1 =field (R \/ W3) \/ field W4 by RELAT_1:18;
      then
      field W1 =field R \/ (field R \/ { a0 }) \/ {a0} by A172,A177,RELAT_1:18;
      then field W1 =(field R \/ field R) \/ { a0 } \/ {a0} by XBOOLE_1:4;
      then field W1 =(field R \/ field R) \/ ({ a0 } \/ {a0}) by XBOOLE_1:4;
      hence thesis;
    end;
A178: x in field W1 implies x in field R or x = a0
    proof
      assume x in field W1;
      then x in field R or x in {a0} by A176,A173,XBOOLE_0:def 3;
      hence thesis by TARSKI:def 1;
    end;
A179: [x,y] in W1 iff [x,y] in R or [x,y] in W3 or [x,y] in W4
    proof
      [x,y] in W1 iff ([x,y] in (R \/ W3) or [x,y] in W4) by XBOOLE_0:def 3;
      hence thesis by XBOOLE_0:def 3;
    end;
    x in field W1 & y in field W1 & x <>y implies [x,y] in W1 or [y,x] in W1
    proof
      assume that
A180: x in field W1 and
A181: y in field W1 and
A182: x <>y;
A183: not x in field R implies [x,y] in W1 or [y,x] in W1
      proof
        assume not x in field R;
        then
A184:   x = a0 by A178,A180;
A185:   y in field R implies [x,y] in W1 or [y,x] in W1
        proof
          assume y in field R;
          then [y,x] in W3 by A184,ZFMISC_1:106;
          hence thesis by A179;
        end;
        y = a0 implies [x,y] in W1 or [y,x] in W1
        proof
          assume y = a0;
          then [x,y] in W4 by A184,TARSKI:def 1;
          hence thesis by A179;
        end;
        hence thesis by A178,A181,A185;
      end;
A186: not y in field R implies [x,y] in W1 or [y,x] in W1
      proof
        assume not y in field R;
        then
A187:   y = a0 by A178,A181;
A188:   x in field R implies [y,x] in W1 or [x,y] in W1
        proof
          assume x in field R;
          then [x,y] in W3 by A187,ZFMISC_1:106;
          hence thesis by A179;
        end;
        x = a0 implies [y,x] in W1 or [x,y] in W1
        proof
          assume x = a0;
          then [y,x] in W4 by A187,TARSKI:def 1;
          hence thesis by A179;
        end;
        hence thesis by A178,A180,A188;
      end;
      x in field R & y in field R implies [x,y] in W1 or [y,x] in W1
      proof
        assume x in field R & y in field R;
        then [x,y] in R or [y,x] in R by A117,A182;
        hence thesis by A179;
      end;
      hence thesis by A183,A186;
    end;
    then
A189: W1 is_connected_in field W1 by RELAT_2:def 6;
    assume
A190: field R in D;
    for x,y holds [x,y] in W1 implies [x,y] in [: union D, union D :]
    proof
      let x,y;
      assume
A191: [x,y] in W1;
      then y in field W1 by RELAT_1:15;
      then y in field R or y=a0 by A178;
      then
A192: y in union D by A1,A4,A190;
      x in field W1 by A191,RELAT_1:15;
      then x in field R or x=a0 by A178;
      then x in union D by A1,A4,A190;
      hence thesis by A192,ZFMISC_1:def 2;
    end;
    then
A193: W1 c= [: union D,union D :] by RELAT_1:def 3;
A194: not a0 in field R by A1,A190;
A195: [x,y] in W1 & y in field R implies [x,y] in R & x in field R
    proof
      assume that
A196: [x,y] in W1 and
A197: y in field R;
A198: not [x,y] in W4
      proof
        assume [x,y] in W4;
        then [x,y] = [a0,a0] by TARSKI:def 1;
        hence contradiction by A194,A197,XTUPLE_0:1;
      end;
      not [x,y] in W3 by A194,A197,ZFMISC_1:106;
      hence [x,y] in R by A179,A196,A198;
      [x,y] in R or [x,y] in W3 or [x,y] in W4 by A179,A196;
      hence thesis by A198,RELAT_1:15,ZFMISC_1:106;
    end;
    x in field W1 & y in field W1 & [x,y] in W1 & [y,x] in W1 implies x= y
    proof
      assume that
A199: x in field W1 and
A200: y in field W1 and
A201: [x,y] in W1 and
A202: [y,x] in W1;
A203: x in field R implies x=y
      proof
        assume
A204:   x in field R;
        then
A205:   [y,x] in R by A195,A202;
A206:   y in field R by A195,A202,A204;
        then [x,y] in R by A195,A201;
        hence thesis by A84,A204,A205,A206;
      end;
A207: y in field R implies x=y
      proof
        assume
A208:   y in field R;
        then
A209:   [x,y] in R by A195,A201;
A210:   x in field R by A195,A201,A208;
        then [y,x] in R by A195,A202;
        hence thesis by A84,A208,A209,A210;
      end;
      y in field R or y =a0 by A178,A200;
      hence thesis by A178,A199,A203,A207;
    end;
    then
A211: W1 is_antisymmetric_in field W1 by RELAT_2:def 4;
A212: y in field R implies W1-Seg(y) = R-Seg(y)
    proof
      assume
A213: y in field R;
A214: x in W1-Seg(y) implies x in R-Seg(y)
      proof
        assume
A215:   x in W1-Seg(y);
        then [x,y] in W1 by WELLORD1:1;
        then
A216:   [x,y] in R by A195,A213;
        x<>y by A215,WELLORD1:1;
        hence thesis by A216,WELLORD1:1;
      end;
      x in R-Seg(y) implies x in W1-Seg(y)
      proof
        assume
A217:   x in R-Seg(y);
        then [x,y] in R by WELLORD1:1;
        then
A218:   [x,y] in W1 by A179;
        x<>y by A217,WELLORD1:1;
        hence thesis by A218,WELLORD1:1;
      end;
      hence thesis by A214,TARSKI:1;
    end;
A219: W1 is_well_founded_in field W1
    proof
      let Y;
      assume that
A220: Y c= field W1 and
A221: Y <> {};
A222: not Y c=field R implies ex a st a in Y & W1-Seg(a) misses Y
      proof
        assume not Y c= field R;
A223:   not (field R) /\ Y = {} implies ex a st a in Y & W1-Seg(a) misses Y
        proof
          set X = (field R) /\ Y;
A224:     X c= field R by XBOOLE_1:17;
          assume not (field R) /\ Y = {};
          then consider y such that
A225:     y in X and
A226:     R-Seg(y) misses X by A130,A224,WELLORD1:def 3;
A227:     R-Seg(y) /\ Y c= R-Seg(y) /\ X
          proof
            let x;
            assume
A228:       x in R-Seg(y) /\ Y;
            then
A229:       x in Y by XBOOLE_0:def 4;
A230:       x in R-Seg(y) by A228,XBOOLE_0:def 4;
            then [x,y] in R by WELLORD1:1;
            then x in field R by RELAT_1:15;
            then x in X by A229,XBOOLE_0:def 4;
            hence thesis by A230,XBOOLE_0:def 4;
          end;
          R-Seg(y) /\ X = {} by A226,XBOOLE_0:def 7;
          then W1-Seg(y) /\ Y = {} by A212,A224,A225,A227;
          then
A231:     W1-Seg(y) misses Y by XBOOLE_0:def 7;
          y in Y by A225,XBOOLE_0:def 4;
          hence thesis by A231;
        end;
        (field R) /\ Y = {} implies ex a st a in Y & W1-Seg(a) misses Y
        proof
          set y = the Element of Y;
A232:     W1-Seg(a0) c= field R
          proof
            let z be set;
            assume
A233:       z in W1-Seg(a0);
            then [z,a0] in W1 by WELLORD1:1;
            then
A234:       z in field W1 by RELAT_1:15;
            z <> a0 by A233,WELLORD1:1;
            hence thesis by A178,A234;
          end;
A235:     y in field W1 by A220,A221,TARSKI:def 3;
          assume
A236:     (field R) /\ Y = {};
          then not y in field R by A221,XBOOLE_0:def 4;
          then y = a0 by A178,A235;
          then W1-Seg(y) /\ Y = {} by A236,A232,XBOOLE_1:3,26;
          then W1-Seg(y) misses Y by XBOOLE_0:def 7;
          hence thesis by A221;
        end;
        hence thesis by A223;
      end;
      Y c= field R implies ex a st a in Y & W1-Seg(a) misses Y
      proof
        assume
A237:   Y c= field R;
        then consider b such that
A238:   b in Y & R-Seg(b) misses Y by A130,A221,WELLORD1:def 3;
        take b;
        thus thesis by A212,A237,A238;
      end;
      hence thesis by A222;
    end;
A239: for y st y in field W1 holds W1-Seg(y) in D & F.(W1-Seg(y)) = y
    proof
      let y;
A240: y in field R implies W1-Seg(y) = R-Seg(y)
      proof
        assume
A241:   y in field R;
A242:   x in W1-Seg(y) implies x in R-Seg(y)
        proof
A243:     [x,y] in W4 implies [x,y] = [a0,a0] by TARSKI:def 1;
          assume
A244:     x in W1-Seg(y);
          then [x,y] in W1 by WELLORD1:1;
          then [x,y] in (R \/ W3) or [x,y] in W4 by XBOOLE_0:def 3;
          then
A245:     [x,y] in R or [x,y] in W3 or [x,y] in W4 by XBOOLE_0:def 3;
          not x=y by A244,WELLORD1:1;
          hence thesis by A194,A241,A245,A243,WELLORD1:1,XTUPLE_0:1
,ZFMISC_1:106;
        end;
        x in R-Seg(y) implies x in W1-Seg(y)
        proof
          assume
A246:     x in R-Seg(y);
          then [x,y] in R by WELLORD1:1;
          then [x,y] in R \/ W3 by XBOOLE_0:def 3;
          then
A247:     [x,y] in W1 by XBOOLE_0:def 3;
          not x=y by A246,WELLORD1:1;
          hence thesis by A247,WELLORD1:1;
        end;
        hence thesis by A242,TARSKI:1;
      end;
A248: x in W1-Seg(a0) implies x in field R
      proof
        assume
A249:   x in W1-Seg(a0);
        then [x,a0] in W1 by WELLORD1:1;
        then
A250:   x in field W1 by RELAT_1:15;
        not x=a0 by A249,WELLORD1:1;
        hence thesis by A178,A250;
      end;
A251: x in field R implies x in W1-Seg(a0)
      proof
        assume
A252:   x in field R;
        then [x,a0] in W3 by ZFMISC_1:106;
        then [x,a0] in R \/ W3 by XBOOLE_0:def 3;
        then [x,a0] in W1 by XBOOLE_0:def 3;
        hence thesis by A194,A252,WELLORD1:1;
      end;
      assume y in field W1;
      then y in field R or y=a0 by A178;
      hence thesis by A103,A190,A240,A248,A251,TARSKI:1;
    end;
    x in field W1 & y in field W1 & z in field W1 & [x,y] in W1 & [y,z]
    in W1 implies [x,z] in W1
    proof
      assume that
A253: x in field W1 and
      y in field W1 and
A254: z in field W1 and
A255: [x,y] in W1 and
A256: [y,z] in W1;
A257: z = a0 implies [x,z] in W1
      proof
        assume
A258:   z = a0;
A259:   x = a0 implies [x,z] in W1
        proof
          assume x = a0;
          then [x,z] in W4 by A258,TARSKI:def 1;
          hence thesis by A179;
        end;
        x in field R implies [x,z] in W1
        proof
          assume x in field R;
          then [x,z] in W3 by A258,ZFMISC_1:106;
          hence thesis by A179;
        end;
        hence thesis by A178,A253,A259;
      end;
      z in field R implies [x,z] in W1
      proof
        assume
A260:   z in field R;
        then
A261:   [y,z] in R by A195,A256;
A262:   y in field R by A195,A256,A260;
        then [x,y] in R & x in field R by A195,A255;
        then [x,z] in R by A145,A260,A261,A262;
        hence thesis by A179;
      end;
      hence thesis by A178,A254,A257;
    end;
    then
A263: W1 is_transitive_in field W1 by RELAT_2:def 8;
    x in field W1 implies [x,x] in W1
    proof
A264: x = a0 implies [x,x] in W1
      proof
A265:   [a0,a0] in W4 by TARSKI:def 1;
        assume x=a0;
        hence thesis by A179,A265;
      end;
A266: x in field R implies [x,x] in W1
      proof
        assume x in field R;
        then [x,x] in R by A165;
        hence thesis by A179;
      end;
      assume x in field W1;
      hence thesis by A178,A266,A264;
    end;
    then W1 is_reflexive_in field W1 by RELAT_2:def 1;
    then W1 well_orders field W1 by A263,A211,A189,A219,WELLORD1:def 5;
    then W1 is well-ordering by WELLORD1:4;
    then W1 in G by A2,A193,A239;
    then field W1 c= field R by A97;
    hence contradiction by A1,A190,A171;
  end;
  R is_reflexive_in field R by A165,RELAT_2:def 1;
  then R well_orders field R by A164,A96,A129,A130,WELLORD1:def 5;
  hence thesis by A103,A170,WELLORD1:4;
end;

Lm1: X,M are_equipotent iff ex Z st (for x st x in X ex y st y in M & [x,y] in
Z) & (for y st y in M ex x st x in X & [x,y] in Z) & for x,z1,y,z2 st [x,z1] in
Z & [y,z2] in Z holds x = y iff z1 = z2
proof
A1: ( ex Z st (for x st x in X ex y st y in M & [x,y] in Z) & (for y st y in
  M ex x st x in X & [x,y] in Z) & for x,z1,y,z2 st [x,z1] in Z & [y,z2] in Z
  holds x = y iff z1 = z2 ) implies X,M are_equipotent
  proof
    assume ex Z st (for x st x in X ex y st y in M & [x,y] in Z) & (for y st
y in M ex x st x in X & [x,y] in Z) & for x,z1,y,z2 st [x,z1] in Z & [y,z2] in
    Z holds x = y iff z1 = z2;
    hence ex Z st (for x st x in X ex y st y in M & [x,y] in Z) & (for y st y
in M ex x st x in X & [x,y] in Z) & for x,z1,y,z2 st [x,z1] in Z & [y,z2] in Z
    holds x = y iff z1 = z2;
  end;
  X,M are_equipotent implies ex Z st (for x st x in X ex y st y in M & [x,
y] in Z) & (for y st y in M ex x st x in X & [x,y] in Z) & for x,z1,y,z2 st [x,
  z1] in Z & [y,z2] in Z holds x = y iff z1 = z2
  proof
    assume ex Z st (for x st x in X ex y st y in M & [x,y] in Z) & (for y st
y in M ex x st x in X & [x,y] in Z) & for x,z1,y,z2 st [x,z1] in Z & [y,z2] in
    Z holds x = y iff z1 = z2;
    hence thesis;
  end;
  hence thesis by A1;
end;

theorem
  for N ex R st R is well-ordering & field R = N
proof
  let N;
  consider M such that
A1: N in M & for X,Y holds X in M & Y c= X implies Y in M and
A2: for X holds X in M implies bool X in M and
A3: for X holds X c= M implies X,M are_equipotent or X in M by ZFMISC_1:112;
  defpred P[set] means not $1,M are_equipotent;
  consider D such that
A4: A in D iff A in bool M & P[A] from XBOOLE_0:sch 1;
A5: union D c= M
  proof
    let x;
    assume x in union D;
    then consider A such that
A6: x in A and
A7: A in D by TARSKI:def 4;
    A in bool M by A4,A7;
    hence thesis by A6;
  end;
  set F = id D;
  for Z st Z in D holds not F.Z in Z & F.Z in union D
  proof
    let Z;
    assume
A8: Z in D;
    not Z in Z;
    hence not F.Z in Z by A8,FUNCT_1:18;
    X in D implies X in union D
    proof
A9:   X in { X } by TARSKI:def 1;
      assume X in D;
      then
A10:  X in bool M & not X,M are_equipotent by A4;
A11:  not { X },M are_equipotent
      proof
A12:    X <> bool X
        proof
          assume X = bool X;
          then not X in bool X;
          hence contradiction by ZFMISC_1:def 1;
        end;
        assume { X },M are_equipotent;
        then consider Z such that
        for x st x in { X } ex y st y in M & [x,y] in Z and
A13:    for y st y in M ex x st x in { X } & [x,y] in Z and
A14:    for x,z1,y,z2 st [x,z1] in Z & [y,z2] in Z holds x = y iff z1
        = z2 by Lm1;
        bool X in M by A2,A3,A10;
        then consider y such that
A15:    y in {X} and
A16:    [y,bool X] in Z by A13;
        consider x such that
A17:    x in {X} and
A18:    [x,X] in Z by A3,A10,A13;
        x=X by A17,TARSKI:def 1;
        then y=x by A15,TARSKI:def 1;
        hence contradiction by A14,A12,A18,A16;
      end;
      X in M by A3,A10;
      then x in {X} implies x in M by TARSKI:def 1;
      then { X } c= M by TARSKI:def 3;
      then { X } in D by A4,A11;
      hence thesis by A9,TARSKI:def 4;
    end;
    then Z in union D by A8;
    hence thesis by A8,FUNCT_1:18;
  end;
  then consider S such that
A19: field S c= union D and
A20: S is well-ordering and
A21: not field S in D and
  for y st y in field S holds S-Seg(y) in D & F.(S-Seg(y)) = y by Th5;
  not field S c= M or field S,M are_equipotent by A4,A21;
  then consider Z such that
A22: for x st x in field S ex y st y in M & [x,y] in Z and
A23: for y st y in M ex x st x in field S & [x,y] in Z and
A24: for x,z1,y,z2 st [x,z1] in Z & [y,z2] in Z holds x = y iff z1 = z2
  by A5,A19,Lm1,XBOOLE_1:1;
  defpred P[set,set] means [$2,$1] in Z;
A25: for x st x in M ex y st P[x,y]
  proof
    let x;
    assume x in M;
    then ex y st y in field S & [y,x] in Z by A23;
    hence thesis;
  end;
A26: for x,z1,z2 st x in M & P[x,z1] & P[x,z2] holds z1 = z2 by A24;
  consider H such that
A27: dom H = M and
A28: for x st x in M holds P[x, H.x] from FUNCT_1:sch 2(A26,A25);
  defpred P[set,set] means $2={$1};
  defpred P[set] means ex x,y st $1=[x,y] & [H.x,H.y] in S;
  consider W0 such that
A29: z in W0 iff z in [: M,M :] & P[z] from XBOOLE_0:sch 1;
A30: x in field S implies x in rng H
  proof
    assume x in field S;
    then consider y such that
A31: y in M and
A32: [x,y] in Z by A22;
    set z1= H.y;
    z1 in rng H & [z1,y] in Z by A27,A28,A31,FUNCT_1:def 3;
    hence thesis by A24,A32;
  end;
  for z1,z2 st z1 in dom H & z2 in dom H & H.z1 = H.z2 holds z1 = z2
  proof
    let z1,z2;
    assume that
A33: z1 in dom H & z2 in dom H and
A34: H.z1 = H.z2;
    [H.z1,z1] in Z & [H.z2,z2] in Z by A27,A28,A33;
    hence thesis by A24,A34;
  end;
  then
A35: H is one-to-one by FUNCT_1:def 4;
  z in W0 implies ex x,y st z=[x,y]
  proof
    assume z in W0;
    then ex z1,z2 st z=[z1,z2] & [H.z1,H.z2] in S by A29;
    hence thesis;
  end;
  then reconsider W0 as Relation by RELAT_1:def 1;
A36: z in field W0 implies z in M
  proof
    assume z in field W0;
    then consider z1 such that
A37: [z,z1] in W0 or [z1,z] in W0 by Th1;
A38: [z1,z] in W0 implies z in M
    proof
      assume [z1,z] in W0;
      then [z1,z] in [: M,M :] by A29;
      hence thesis by ZFMISC_1:87;
    end;
    [z,z1] in W0 implies z in M
    proof
      assume [z,z1] in W0;
      then [z,z1] in [: M,M :] by A29;
      hence thesis by ZFMISC_1:87;
    end;
    hence thesis by A37,A38;
  end;
A39: for x st x in N ex y st P[x,y];
A40: for x,z1,z2 st x in N & P[x,z1] & P[x,z2] holds z1 = z2;
  consider H1 such that
A41: dom H1 = N and
A42: for x st x in N holds P[x, H1.x] from FUNCT_1:sch 2(A40,A39);
  for z1,z2 st z1 in dom H1 & z2 in dom H1 & H1.z1 = H1.z2 holds z1 = z2
  proof
    let z1,z2;
    assume that
A43: z1 in dom H1 and
A44: z2 in dom H1 and
A45: H1.z1 = H1.z2;
    H1.z1={z1} by A41,A42,A43;
    then
A46: z1 in H1.z2 by A45,TARSKI:def 1;
    H1.z2={z2} by A41,A42,A44;
    hence thesis by A46,TARSKI:def 1;
  end;
  then
A47: H1 is one-to-one by FUNCT_1:def 4;
  set S1=W0 |_2 rng H1;
  x in rng H implies x in field S
  proof
    assume x in rng H;
    then consider z1 such that
A48: z1 in dom H & x = H.z1 by FUNCT_1:def 3;
    ( ex z2 st z2 in field S & [z2,z1] in Z)& [x,z1] in Z by A23,A27,A28,A48;
    hence thesis by A24;
  end;
  then
A49: rng H = field S by A30,TARSKI:1;
  z in M implies z in field W0
  proof
    assume
A50: z in M;
    ex z1 st [z,z1] in W0 or [z1,z] in W0
    proof
      H.z in field S by A27,A49,A50,FUNCT_1:def 3;
      then consider z2 such that
A51:  [H.z,z2] in S or [z2,H.z] in S by Th1;
A52:  [H.z,z2] in S implies ex z1 st [z,z1] in W0 or [z1,z] in W0
      proof
        assume
A53:    [H.z,z2] in S;
        then z2 in rng H by A49,RELAT_1:15;
        then consider z3 such that
A54:    z3 in dom H and
A55:    z2=H.z3 by FUNCT_1:def 3;
        take z3;
        [z,z3] in [: M,M :] by A27,A50,A54,ZFMISC_1:87;
        hence thesis by A29,A53,A55;
      end;
      [z2,H.z] in S implies ex z1 st [z,z1] in W0 or [z1,z] in W0
      proof
        assume
A56:    [z2,H.z] in S;
        then z2 in rng H by A49,RELAT_1:15;
        then consider z3 such that
A57:    z3 in dom H and
A58:    z2=H.z3 by FUNCT_1:def 3;
        take z3;
        [z3,z] in [: M,M :] by A27,A50,A57,ZFMISC_1:87;
        hence thesis by A29,A56,A58;
      end;
      hence thesis by A51,A52;
    end;
    hence thesis by Th1;
  end;
  then
A59: field W0 = M by A36,TARSKI:1;
  for a,b holds [a,b] in W0 iff a in field W0 & b in field W0 & [H.a,H.b ] in S
  proof
    let a,b;
A60: [a,b] in W0 implies a in field W0 & b in field W0 & [H.a,H.b] in S
    proof
      assume
A61:  [a,b] in W0;
      then
A62:  [a,b] in [: M,M :] by A29;
      consider x,y such that
A63:  [a,b] = [x,y] and
A64:  [H.x,H.y] in S by A29,A61;
      a=x by A63,XTUPLE_0:1;
      hence thesis by A59,A62,A63,A64,XTUPLE_0:1,ZFMISC_1:87;
    end;
    a in field W0 & b in field W0 & [H.a,H.b] in S implies [a,b] in W0
    proof
      assume that
A65:  a in field W0 & b in field W0 and
A66:  [H.a,H.b] in S;
      [a,b] in [: M,M :] by A59,A65,ZFMISC_1:87;
      hence thesis by A29,A66;
    end;
    hence thesis by A60;
  end;
  then H is_isomorphism_of W0,S by A27,A35,A49,A59,WELLORD1:def 7;
  then
A67: (H") is_isomorphism_of S, W0 by WELLORD1:39;
  then W0 is well-ordering by A20,WELLORD1:44;
  then
A68: S1 is well-ordering by WELLORD1:25;
  defpred P[set] means ex x,y st $1=[x,y] & [H1.x,H1.y] in S1;
  consider W00 such that
A69: z in W00 iff z in [: N,N :] & P[z] from XBOOLE_0:sch 1;
  z in W00 implies ex x,y st z=[x,y]
  proof
    assume z in W00;
    then ex z1,z2 st z=[z1,z2] & [H1.z1,H1.z2] in S1 by A69;
    hence thesis;
  end;
  then reconsider W00 as Relation by RELAT_1:def 1;
A70: z in field W00 implies z in N
  proof
    assume z in field W00;
    then consider z1 such that
A71: [z,z1] in W00 or [z1,z] in W00 by Th1;
A72: [z1,z] in W00 implies z in N
    proof
      assume [z1,z] in W00;
      then [z1,z] in [: N,N :] by A69;
      hence thesis by ZFMISC_1:87;
    end;
    [z,z1] in W00 implies z in N
    proof
      assume [z,z1] in W00;
      then [z,z1] in [: N,N :] by A69;
      hence thesis by ZFMISC_1:87;
    end;
    hence thesis by A71,A72;
  end;
  rng H1 c= M
  proof
    let x;
    assume x in rng H1;
    then consider y such that
A73: y in dom H1 and
A74: x = H1.y by FUNCT_1:def 3;
    z1 in { y } implies z1 in N by A41,A73,TARSKI:def 1;
    then
A75: { y } c= N by TARSKI:def 3;
    x = { y } by A41,A42,A73,A74;
    hence thesis by A1,A75;
  end;
  then
A76: field S1 = rng H1 by A20,A59,A67,WELLORD1:31,44;
  z in N implies z in field W00
  proof
    assume
A77: z in N;
    ex z1 st [z,z1] in W00 or [z1,z] in W00
    proof
      H1.z in field S1 by A41,A76,A77,FUNCT_1:def 3;
      then consider z2 such that
A78:  [H1.z,z2] in S1 or [z2,H1.z] in S1 by Th1;
A79:  [H1.z,z2] in S1 implies ex z1 st [z,z1] in W00 or [z1,z] in W00
      proof
        assume
A80:    [H1.z,z2] in S1;
        then z2 in rng H1 by A76,RELAT_1:15;
        then consider z3 such that
A81:    z3 in dom H1 and
A82:    z2=H1.z3 by FUNCT_1:def 3;
        take z3;
        [z,z3] in [: N,N :] by A41,A77,A81,ZFMISC_1:87;
        hence thesis by A69,A80,A82;
      end;
      [z2,H1.z] in S1 implies ex z1 st [z,z1] in W00 or [z1,z] in W00
      proof
        assume
A83:    [z2,H1.z] in S1;
        then z2 in rng H1 by A76,RELAT_1:15;
        then consider z3 such that
A84:    z3 in dom H1 and
A85:    z2=H1.z3 by FUNCT_1:def 3;
        take z3;
        [z3,z] in [: N,N :] by A41,A77,A84,ZFMISC_1:87;
        hence thesis by A69,A83,A85;
      end;
      hence thesis by A78,A79;
    end;
    hence thesis by Th1;
  end;
  then
A86: field W00 = N by A70,TARSKI:1;
  for a,b holds [a,b] in W00 iff a in field W00 & b in field W00 & [H1.a
  ,H1.b] in S1
  proof
    let a,b;
A87: [a,b] in W00 implies a in field W00 & b in field W00 & [H1.a,H1.b] in S1
    proof
      assume
A88:  [a,b] in W00;
      then
A89:  [a,b] in [: N,N :] by A69;
      consider x,y such that
A90:  [a,b] = [x,y] and
A91:  [H1.x,H1.y] in S1 by A69,A88;
      a=x by A90,XTUPLE_0:1;
      hence thesis by A86,A89,A90,A91,XTUPLE_0:1,ZFMISC_1:87;
    end;
    a in field W00 & b in field W00 & [H1.a,H1.b] in S1 implies [a,b] in W00
    proof
      assume that
A92:  a in field W00 & b in field W00 and
A93:  [H1.a,H1.b] in S1;
      [a,b] in [: N,N :] by A86,A92,ZFMISC_1:87;
      hence thesis by A69,A93;
    end;
    hence thesis by A87;
  end;
  then H1 is_isomorphism_of W00,S1 by A41,A47,A76,A86,WELLORD1:def 7;
  then (H1") is_isomorphism_of S1, W00 by WELLORD1:39;
  hence thesis by A68,A86,WELLORD1:44;
end;














begin 

definition
  let f be Function;
  let a,b,c be set;
  func f.(a,b,c) -> set equals
  f.[a,b,c];
  correctness;
end;

reserve A,B,C,D,E for non empty set,
  a for Element of A,
  b for Element of B,
  c for Element of C,
  d for Element of D,
  X,Y,Z,S,x,y,z,s,t for set;

definition
  let A,B,C,D;
  let f be Function of [:A,B,C:],D;
  let a,b,c;
  redefine func f.(a,b,c) -> Element of D;
  coherence
  proof
    f.([a,b,c]) is Element of D;
    hence thesis;
  end;
end;

theorem Th1:
  for f1,f2 being Function of [:X,Y,Z:],D st for x,y,z st x in X &
  y in Y & z in Z holds f1.[x,y,z] = f2.[x,y,z] holds f1 = f2
proof
  let f1,f2 be Function of [:X,Y,Z:],D such that
A1: for x,y,z st x in X & y in Y & z in Z holds f1.[x,y,z] = f2.[x,y,z];
  for t st t in [:X,Y,Z:] holds f1.t = f2.t
  proof
    let t;
    assume t in [:X,Y,Z:];
    then ex x,y,z st x in X & y in Y & z in Z & t = [x,y,z] by MCART_1:68;
    hence thesis by A1;
  end;
  hence thesis by FUNCT_2:12;
end;

theorem Th2:
  for f1,f2 being Function of [:A,B,C:],D st for a,b,c holds f1.[a,
  b,c] = f2.[a,b,c] holds f1 = f2
proof
  let f1,f2 be Function of [:A,B,C:],D;
  assume for a,b,c holds f1.[a,b,c] = f2.[a,b,c];
  then for x,y,z st x in A & y in B & z in C holds f1.[x,y,z] = f2.[x,y,z];
  hence thesis by Th1;
end;

theorem
  for f1,f2 being Function of [:A,B,C:],D st for a being Element of A
for b being Element of B for c being Element of C holds f1.(a,b,c) = f2.(a,b,c)
  holds f1 = f2
proof
  let f1,f2 be Function of [:A,B,C:],D such that
A1: for a being Element of A for b being Element of B for c being
  Element of C holds f1.(a,b,c) = f2.(a,b,c);
  for a being Element of A for b being Element of B for c being Element of
  C holds f1.[a,b,c] = f2.[a,b,c]
  proof
    let a be Element of A;
    let b be Element of B;
    let c be Element of C;
    f1.(a,b,c) = f1.[a,b,c] & f2.(a,b,c) = f2.[a,b,c];
    hence thesis by A1;
  end;
  hence thesis by Th2;
end;

definition
  let A be set;
  mode TriOp of A is Function of [:A,A,A:],A;
end;

scheme
  FuncEx3D { X,Y,Z,T() -> non empty set, P[set,set,set,set] } : ex f being
  Function of [:X(),Y(),Z():],T() st for x being Element of X() for y being
  Element of Y() for z being Element of Z() holds P[x,y,z,f.[x,y,z]]
provided
A1: for x being Element of X() for y being Element of Y() for z being
Element of Z() ex t being Element of T() st P[x,y,z,t]
proof
  defpred Q[set,set] means ( for x being (Element of X()), y being (Element of
  Y()), z being Element of Z() st $1 = [x,y,z] holds P[x,y,z,$2]);
A2: for p being Element of [:X(),Y(),Z():] ex t being Element of T() st Q[p, t]
  proof
    let p be Element of [:X(),Y(),Z():];
    consider x1, y1, z1 be set such that
A3: x1 in X() and
A4: y1 in Y() and
A5: z1 in Z() and
A6: p = [x1,y1,z1] by MCART_1:68;
    reconsider z1 as Element of Z() by A5;
    reconsider y1 as Element of Y() by A4;
    reconsider x1 as Element of X() by A3;
    consider t being Element of T() such that
A7: P[x1,y1,z1,t] by A1;
    take t;
    let x be (Element of X()), y be (Element of Y()), z be Element of Z();
    assume
A8: p = [x,y,z];
    then x1 = x & y1 = y by A6,XTUPLE_0:3;
    hence thesis by A6,A7,A8,XTUPLE_0:3;
  end;
  consider f being Function of [:X(),Y(),Z():],T() such that
A9: for p being Element of [:X(),Y(),Z():] holds Q[p,f.p] from FUNCT_2:
  sch 3 (A2);
  take f;
  let x be Element of X();
  let y be Element of Y();
  let z be Element of Z();
  thus thesis by A9;
end;

scheme
  TriOpEx { A()->non empty set, P[ Element of A(), Element of A(), Element of
A(), Element of A()] }: ex o being TriOp of A() st for a,b,c being Element of A
  () holds P[a,b,c,o.(a,b,c)]
provided
A1: for x,y,z being Element of A() ex t being Element of A() st P[x,y,z, t]
proof
  defpred Q[Element of A(),Element of A(),Element of A(),Element of A()] means
  for r being Element of A() st r = $4 holds P[$1,$2,$3,r];
A2: for x,y,z being Element of A() ex t being Element of A() st Q[x,y,z,t]
  proof
    let x,y,z be Element of A();
    consider t being Element of A() such that
A3: P[x,y,z,t] by A1;
    take t;
    thus thesis by A3;
  end;
  consider f being Function of [:A(),A(),A():],A() such that
A4: for a,b,c being Element of A() holds Q[a,b,c,f.[a,b,c]] from
  FuncEx3D(A2 );
  take f;
  let a,b,c be Element of A();
  thus thesis by A4;
end;

scheme
  Lambda3D { X, Y, Z, T()->non empty set, F( Element of X(), Element of Y(),
Element of Z()) -> Element of T() }: ex f being Function of [:X(),Y(),Z():],T()
st for x being Element of X() for y being Element of Y() for z being Element of
  Z() holds f.[x,y,z]=F(x,y,z) proof
  defpred Q[Element of X(),Element of Y(),Element of Z(),Element of T()] means
  $4 = F($1,$2,$3);
A1: for x being Element of X() for y being Element of Y() for z being
  Element of Z() ex t being Element of T() st Q[x,y,z,t];
  ex f being Function of [:X(),Y(),Z():],T() st for x being Element of X()
for y being Element of Y() for z being Element of Z() holds Q[x,y,z,f.[x,y,z]]
  from FuncEx3D(A1);
  hence thesis;
end;

scheme
  TriOpLambda { A,B,C,D()->non empty set, O( Element of A(), Element of B(),
Element of C()) -> Element of D() }: ex o being Function of [:A(),B(),C():],D()
  st for a being Element of A(), b being Element of B(), c being Element of C()
  holds o.(a,b,c) = O(a,b,c) proof
  consider f being Function of [:A(),B(),C():],D() such that
A1: for a being Element of A(), b being Element of B(), c being Element
  of C() holds f.[a,b,c] = O(a,b,c) from Lambda3D;
  take f;
  let a be Element of A(), b be Element of B(), c be Element of C();
  thus thesis by A1;
end;



definition
  let f be Function;
  let a,b,c,d be set;
  func f.(a,b,c,d) -> set equals
  f.[a,b,c,d];
  correctness;
end;

definition
  let A,B,C,D,E;
  let f be Function of [:A,B,C,D:],E;
  let a,b,c,d;
  redefine func f.(a,b,c,d) -> Element of E;
  coherence
  proof
    f.([a,b,c,d]) is Element of E;
    hence thesis;
  end;
end;

theorem Th4:
  for f1,f2 being Function of [:X,Y,Z,S:],D st for x,y,z,s st x in
  X & y in Y & z in Z & s in S holds f1.[x,y,z,s] = f2.[x,y,z,s] holds f1 = f2
proof
  let f1,f2 be Function of [:X,Y,Z,S:],D such that
A1: for x,y,z,s st x in X & y in Y & z in Z & s in S holds f1.[x,y,z,s]
  = f2.[x,y,z,s];
  for t st t in [:X,Y,Z,S:] holds f1.t = f2.t
  proof
    let t;
    assume t in [:X,Y,Z,S:];
    then ex x,y,z,s st x in X & y in Y & z in Z & s in S & t = [x,y,z,s] by
MCART_1:79;
    hence thesis by A1;
  end;
  hence thesis by FUNCT_2:12;
end;

theorem Th5:
  for f1,f2 being Function of [:A,B,C,D:],E st for a,b,c,d holds f1
  .[a,b,c,d] = f2.[a,b,c,d] holds f1 = f2
proof
  let f1,f2 be Function of [:A,B,C,D:],E;
  assume for a,b,c,d holds f1.[a,b,c,d] = f2.[a,b,c,d];
  then
  for x,y,z,s st x in A & y in B & z in C & s in D holds f1.[x,y,z,s] = f2
  .[x,y,z,s];
  hence thesis by Th4;
end;

theorem
  for f1,f2 being Function of [:A,B,C,D:],E st for a,b,c,d holds f1.(a,b
  ,c,d) = f2.(a,b,c,d) holds f1 = f2
proof
  let f1,f2 be Function of [:A,B,C,D:],E such that
A1: for a,b,c,d holds f1.(a,b,c,d) = f2.(a,b,c,d);
  for a,b,c,d holds f1.[a,b,c,d] = f2.[a,b,c,d]
  proof
    let a,b,c,d;
    f1.(a,b,c,d) = f1.[a,b,c,d] & f2.(a,b,c,d) = f2.[a,b,c,d];
    hence thesis by A1;
  end;
  hence thesis by Th5;
end;

definition
  let A;
  mode QuaOp of A is Function of [:A,A,A,A:],A;
end;

scheme
  FuncEx4D { X, Y, Z, S, T() -> non empty set, P[set,set,set,set,set] }: ex f
  being Function of [:X(),Y(),Z(),S():],T() st for x being Element of X() for y
  being Element of Y() for z being Element of Z() for s being Element of S()
  holds P[x,y,z,s,f.[x,y,z,s]]
provided
A1: for x being Element of X() for y being Element of Y() for z being
Element of Z() for s being Element of S() ex t being Element of T() st P[x,y,z,
s,t]
proof
  defpred Q[set,set] means for x being (Element of X()), y being (Element of Y
()), z being (Element of Z()), s being Element of S() st $1 = [x,y,z,s] holds P
  [x,y,z,s,$2];
A2: for p being Element of [:X(),Y(),Z(),S():] ex t being Element of T() st
  Q[p,t]
  proof
    let p be Element of [:X(),Y(),Z(),S():];
    consider x1, y1, z1, s1 be set such that
A3: x1 in X() and
A4: y1 in Y() and
A5: z1 in Z() and
A6: s1 in S() and
A7: p = [x1,y1,z1,s1] by MCART_1:79;
    reconsider s1 as Element of S() by A6;
    reconsider z1 as Element of Z() by A5;
    reconsider y1 as Element of Y() by A4;
    reconsider x1 as Element of X() by A3;
    consider t being Element of T() such that
A8: P[x1,y1,z1,s1,t] by A1;
    take t;
    let x be Element of X(), y be Element of Y(), z be Element of Z(), s be
    Element of S();
    assume
A9: p = [x,y,z,s];
    then
A10: z1 = z by A7,XTUPLE_0:5;
    x1 = x & y1 = y by A7,A9,XTUPLE_0:5;
    hence thesis by A7,A8,A9,A10,XTUPLE_0:5;
  end;
  consider f being Function of [:X(),Y(),Z(),S():],T() such that
A11: for p being Element of [:X(),Y(),Z(),S():] holds Q[p,f.p] from
  FUNCT_2:sch 3(A2);
  take f;
  let x be Element of X();
  let y be Element of Y();
  let z be Element of Z();
  let s be Element of S();
  thus thesis by A11;
end;

scheme
  QuaOpEx { A()->non empty set, P[ Element of A(), Element of A(), Element of
A(), Element of A(), Element of A()] }: ex o being QuaOp of A() st for a,b,c,d
  being Element of A() holds P[a,b,c,d,o.(a,b,c,d)]
provided
A1: for x,y,z,s being Element of A() ex t being Element of A() st P[x,y, z,s,t]
proof
  defpred Q[Element of A(), Element of A(),Element of A(), Element of A(),
Element of A()] means for r being Element of A() st r = $5 holds P[$1,$2,$3,$4,
  $5];
A2: for x,y,z,s being Element of A() ex t being Element of A() st Q[x,y,z,s, t]
  proof
    let x,y,z,s be Element of A();
    consider t being Element of A() such that
A3: P[x,y,z,s,t] by A1;
    take t;
    thus thesis by A3;
  end;
  consider f being Function of [:A(),A(),A(),A():],A() such that
A4: for a,b,c,d being Element of A() holds Q[a,b,c,d,f.[a,b,c,d]] from
  FuncEx4D(A2);
  take f;
  let a,b,c,d be Element of A();
  thus thesis by A4;
end;

scheme
  Lambda4D { X, Y, Z, S, T() -> non empty set, F( Element of X(), Element of Y
(), Element of Z(), Element of S()) -> Element of T() }: ex f being Function of
[:X(),Y(),Z(),S():],T() st for x being Element of X() for y being Element of Y(
) for z being Element of Z() for s being Element of S() holds f.[x,y,z,s]=F(x,y
  ,z,s) proof
  defpred P[Element of X(),Element of Y(),Element of Z(),Element of S(),
  Element of T()] means $5 = F($1,$2,$3,$4);
A1: for x being Element of X() for y being Element of Y() for z being
Element of Z() for s being Element of S() ex t being Element of T() st P[x,y,z,
  s,t];
  ex f being Function of [:X(),Y(),Z(),S():],T() st for x being Element of
  X() for y being Element of Y() for z being Element of Z() for s being Element
  of S() holds P[x,y,z,s,f.[x,y,z,s]] from FuncEx4D(A1);
  hence thesis;
end;

scheme
  QuaOpLambda { A()->non empty set, O( Element of A(), Element of A(), Element
of A(), Element of A()) -> Element of A() }: ex o being QuaOp of A() st for a,b
  ,c,d being Element of A() holds o.(a,b,c,d) = O(a,b,c,d) proof
  deffunc F(Element of A(),Element of A(),Element of A(),Element of A()) = O(
  $1,$2,$3,$4);
  consider f being Function of [:A(),A(),A(),A():],A() such that
A1: for a,b,c,d being Element of A() holds f.[a,b,c,d] = F(a,b,c,d) from
  Lambda4D;
  take f;
  let a,b,c,d be Element of A();
  thus thesis by A1;
end;














begin

reserve a,b,d for set;

scheme
  Schemat0 {P[set]} : ex a st P[a]
provided
A1: for a holds P[a]
proof
  set a = the set;
  P[a] by A1;
  hence thesis;
end;

scheme
  Schemat3 {S[set,set]} : for b ex a st S[a,b]
provided
A1: ex a st for b holds S[a,b]
proof
  thus thesis by A1;
end;

scheme
  Schemat8 {P[set],Q[set]} : (for a holds P[a]) implies for a holds Q[a]
provided
A1: for a holds P[a] implies Q[a]
proof
  thus thesis by A1;
end;

scheme
  Schemat9 {P[set],Q[set]} : (for a holds P[a]) iff for a holds Q[a]
provided
A1: for a holds P[a] iff Q[a]
proof
  thus (for a holds P[a]) implies for a holds Q[a]
  proof
    assume
A2: for a holds P[a];
    now
      let a;
      P[a] iff Q[a] by A1;
      hence Q[a] by A2;
    end;
    hence thesis;
  end;
  assume
A3: for a holds Q[a];
  now
    let a;
    P[a] iff Q[a] by A1;
    hence P[a] by A3;
  end;
  hence thesis;
end;

scheme
  Schemat17 {P[set],T[]} : (for a holds P[a]) implies T[]
provided
A1: for a holds P[a] implies T[]
proof
  assume
A2: for a holds P[a];
  now
    let a;
    P[a] by A2;
    hence thesis by A1;
  end;
  hence thesis;
end;

scheme
  Schemat18a {P[set],Q[set]} : ex a st for b holds P[a] or Q[b]
provided
A1: (ex a st P[a]) or for b holds Q[b]
proof
  now
A2: now
      set a = the set;
      assume for b holds Q[b];
      then for b holds P[a] or Q[b];
      hence thesis;
    end;
    now
      given a such that
A3:   P[a];
      for b holds P[a] or Q[b] by A3;
      hence thesis;
    end;
    hence thesis by A1,A2;
  end;
  hence thesis;
end;

scheme
  Schemat18b {P[set],Q[set]} : (ex a st P[a]) or for b holds Q[b]
provided
A1: ex a st for b holds P[a] or Q[b]
proof
  thus thesis by A1;
end;

scheme
  Schemat20b {P[set],Q[set]} : ex a st for b holds P[a] or Q[b]
provided
A1: for b ex a st P[a] or Q[b]
proof
A2: (ex a st P[a]) or for b holds Q[b] by A1;
  ex a st for b holds P[a] or Q[b] from Schemat18a(A2);
  hence thesis;
end;

scheme
  Schemat22a {P[set],Q[set]} : for b ex a st P[a] & Q[b]
provided
A1: (ex a st P[a]) & for b holds Q[b]
proof
  thus thesis by A1;
end;

scheme
  Schemat22b {P[set],Q[set]} : (ex a st P[a]) & for b holds Q[b]
provided
A1: for b ex a st P[a] & Q[b]
proof
  assume
A2: (for a holds not P[a]) or ex b st not Q[b];
  per cases by A2;
  suppose
    ex b st not Q[b];
    then consider b such that
A3: not Q[b];
    now
      let d;
      assume d = b;
      ex a st P[a] & Q[b] by A1;
      hence contradiction by A3;
    end;
    hence thesis;
  end;
  suppose
A4: for a holds not P[a];
    now
      let b;
      ex a st ( P[a])& Q[b] by A1;
      hence thesis by A4;
    end;
    hence thesis;
  end;
end;

scheme
  Schemat23b {P[set],Q[set]} : ex a st for b holds P[a] & Q[b]
provided
A1: for b ex a st P[a] & Q[b]
proof
A2: for b ex a st P[a] & Q[b] by A1;
  (ex a st P[a]) & for b holds Q[b] from Schemat22b(A2);
  hence thesis;
end;

scheme
  Schemat28 {S[set,set]} : ex b st for a holds S[a,b]
provided
A1: for a,b holds S[a,b]
proof
  now
    let b;
    for a holds S[a,b] by A1;
    hence thesis;
  end;
  hence thesis;
end;

scheme
  Schemat30 {S[set,set]} : ex a st S[a,a]
provided
A1: ex a st for b holds S[a,b]
proof
  thus thesis by A1;
end;

scheme
  Schemat31 {S[set,set]} : for a ex b st S[b,a]
provided
A1: for a holds S[a,a]
proof
  thus thesis by A1;
end;

scheme
  Schemat33 {S[set,set]} : for a ex b st S[a,b]
provided
A1: for a holds S[a,a]
proof
  thus thesis by A1;
end;

scheme
  Schemat36 {S[set,set]} : ex a,b st S[a,b]
provided
A1: for b ex a st S[a,b]
proof
  set b = the set;
  ex a st S[a,b] by A1;
  hence thesis;
end;














begin

 reserve x,y,z,t,X,Y,Z,W for set;
 reserve R,S,T for Relation;

Lm1: X <> {} & Y <> {} implies dom [:X,Y:] = X & rng [:X,Y:] = Y
    by RELAT_1:160;

theorem Th1:
  dom (R /\ [:X,Y:]) c= X & rng (R /\ [:X,Y:]) c= Y
proof
    per cases;
    suppose
      X = {} or Y = {};
      then R /\ [:X,Y:] = R /\ {} by ZFMISC_1:90
        .= {};
      hence thesis by RELAT_1:38,XBOOLE_1:2;
    end;
    suppose
A1:   X <> {} & Y <> {};
      rng (R /\ [:X,Y:]) c= rng R /\ rng [:X,Y:] by RELAT_1:13; then
A2:   rng (R /\ [:X,Y:]) c= rng R /\ Y by A1,Lm1;
      dom (R /\ [:X,Y:]) c= dom R /\ dom [:X,Y:] by RELAT_1:2; then
A3:   dom (R /\ [:X,Y:]) c= dom R /\ X by A1,Lm1;
      dom R /\ X c= X by XBOOLE_1:17;
      hence dom (R /\ [:X,Y:]) c= X by A3,XBOOLE_1:1;
      rng R /\ Y c= Y by XBOOLE_1:17;
      hence rng (R /\ [:X,Y:]) c= Y by A2,XBOOLE_1:1;
    end;
end;

theorem
  X misses Y implies dom (R /\ [:X,Y:]) misses rng (R /\ [:X,Y:])
proof
  assume
A1: X /\ Y = {};
  dom (R /\ [:X,Y:]) c= X by Th1; then
A2: dom (R /\ [:X,Y:]) /\ rng (R /\ [:X,Y:]) c= X /\ rng (R /\ [:X,Y:]) by
XBOOLE_1:26;
  X /\ rng (R /\ [:X,Y:]) c= X /\ Y by Th1,XBOOLE_1:26;
  hence dom (R /\ [:X,Y:]) /\ rng (R /\ [:X,Y:]) = {} by A1,A2,XBOOLE_1:1,3;
end;

theorem Th3:
  R c= [:X,Y:] implies dom R c= X & rng R c= Y
proof
  assume R c= [:X,Y:];
  then R /\ [:X,Y:] = R by XBOOLE_1:28;
  hence thesis by Th1;
end;

theorem
  R c= [:X,Y:] implies R~ c= [:Y,X:]
proof
  assume
A1: R c= [:X,Y:];
  let z,t;
  assume [z,t] in R~;
  then [t,z] in R by RELAT_1:def 7;
  then t in X & z in Y by A1,ZFMISC_1:87;
  hence thesis by ZFMISC_1:87;
end;

theorem
  [:X,Y:]~ = [:Y,X:]
proof
  let x,y;
  thus [x,y] in [:X,Y:]~ implies [x,y] in [:Y,X:]
  proof
    assume [x,y] in [:X,Y:]~;
    then [y,x] in [:X,Y:] by RELAT_1:def 7;
    then y in X & x in Y by ZFMISC_1:87;
    hence thesis by ZFMISC_1:87;
  end;
  assume [x,y] in [:Y,X:];
  then y in X & x in Y by ZFMISC_1:87;
  then [y,x] in [:X,Y:] by ZFMISC_1:87;
  hence thesis by RELAT_1:def 7;
end;

theorem Th6:
  (R \/ S) * T = (R * T) \/ (S * T)
proof
  thus (R \/ S) * T = (R * T) \/ (S * T)
  proof
    let x,y;
    thus [x,y] in (R \/ S) * T implies [x,y] in (R * T) \/ (S * T)
    proof
      assume [x,y] in (R \/ S) * T;
      then consider z such that
A1:   [x,z] in R \/ S & [z,y] in T by RELAT_1:def 8;
      [x,z] in R & [z,y] in T or [x,z] in S & [z,y] in T by A1,XBOOLE_0:def 3;
      then [x,y] in R * T or [x,y] in S *T by RELAT_1:def 8;
      hence thesis by XBOOLE_0:def 3;
    end;
    assume A2: [x,y] in (R * T) \/ (S * T);
    per cases by A2,XBOOLE_0:def 3;
    suppose
      [x,y] in S * T;
      then consider z such that
A3:   [x,z] in S and
A4:   [z,y] in T by RELAT_1:def 8;
      [x,z] in R \/ S by A3,XBOOLE_0:def 3;
      hence thesis by A4,RELAT_1:def 8;
    end;
    suppose [x,y] in R * T;
      then consider z such that
A5:   [x,z] in R and
A6:   [z,y] in T by RELAT_1:def 8;
      [x,z] in R \/ S by A5,XBOOLE_0:def 3;
      hence thesis by A6,RELAT_1:def 8;
    end;
  end;
end;

theorem
  (X misses Y & R c= [:X,Y:] \/ [:Y,X:] & [x,y] in R & x in X implies
    not x in Y & not y in X & y in Y) &
  (X misses Y & R c= [:X,Y:] \/ [:Y,X:] & [x,y] in R & y in Y implies
    not y in X & not x in Y & x in X) &
  (X misses Y & R c= [:X,Y:] \/ [:Y,X:] & [x,y] in R & x in Y implies
    not x in X & not y in Y & y in X) &
  (X misses Y & R c= [:X,Y:] \/ [:Y,X:] & [x,y] in R & y in X implies
    not x in X & not y in Y & x in Y)
proof
  thus X misses Y & R c= [:X,Y:] \/ [:Y,X:] & [x,y] in R & x in X implies not
  x in Y & not y in X & y in Y
  proof
    assume that
A1: X misses Y and
A2: R c= [:X,Y:] \/ [:Y,X:] & [x,y] in R and
A3: x in X;
A4: not [x,y] in [:Y,X:]
    proof
      assume
A5:   [x,y] in [:Y,X:];
      not x in Y by A1,A3,XBOOLE_0:3;
      hence thesis by A5,ZFMISC_1:87;
    end;
A6: [x,y] in [:X,Y:] implies thesis
    proof
      assume [x,y] in [:X,Y:];
      then x in X & y in Y by ZFMISC_1:87;
      hence thesis by A1,XBOOLE_0:3;
    end;
    [:X,Y:] misses [:Y,X:] by A1,ZFMISC_1:104;
    hence thesis by A2,A6,A4,XBOOLE_0:5;
  end;
  thus X misses Y & R c= [:X,Y:] \/ [:Y,X:] & [x,y] in R & y in Y implies
  not y in X & not x in Y & x in X
  proof
    assume that
A7: X misses Y and
A8: R c= [:X,Y:] \/ [:Y,X:] & [x,y] in R and
A9: y in Y;
A10: not [x,y] in [:Y,X:]
    proof
      assume
A11:  [x,y] in [:Y,X:];
      not y in X by A7,A9,XBOOLE_0:3;
      hence thesis by A11,ZFMISC_1:87;
    end;
    [x,y] in [:X,Y:] implies thesis
    proof
      assume [x,y] in [:X,Y:]; then
      x in X & y in Y by ZFMISC_1:87;
      hence thesis by A7,XBOOLE_0:3;
    end;
    hence thesis by A8,A10,XBOOLE_0:def 3;
  end;
  thus X misses Y & R c= [:X,Y:] \/ [:Y,X:] & [x,y] in R & x in Y implies not
  x in X & not y in Y & y in X
  proof
    assume that
A12: X misses Y and
A13: R c= [:X,Y:] \/ [:Y,X:] & [x,y] in R and
A14: x in Y;
A15: not [x,y] in [:X,Y:]
    proof
      assume
A16:  [x,y] in [:X,Y:];
      not x in X by A12,A14,XBOOLE_0:3;
      hence thesis by A16,ZFMISC_1:87;
    end;
    [x,y] in [:Y,X:] & not [x,y] in [:X,Y:] implies thesis
    proof
      assume [x,y] in [:Y,X:] & not [x,y] in [:X,Y:]; then
      x in Y & y in X & not x in X or x in Y & y in X & not y in Y
        by ZFMISC_1:87;
      hence thesis by A12,XBOOLE_0:3;
    end;
    hence thesis by A13,A15,XBOOLE_0:def 3;
  end;
    assume that
A17: X misses Y and
A18: R c= [:X,Y:] \/ [:Y,X:] & [x,y] in R and
A19: y in X;
A20: not [x,y] in [:X,Y:]
    proof
      assume
A21:  [x,y] in [:X,Y:];
      not y in Y by A17,A19,XBOOLE_0:3;
      hence thesis by A21,ZFMISC_1:87;
    end;
    [x,y] in [:Y,X:] implies thesis
    proof
      assume [x,y] in [:Y,X:]; then
      x in Y & y in X by ZFMISC_1:87;
      hence thesis by A17,XBOOLE_0:3;
    end;
    hence thesis by A18,A20,XBOOLE_0:def 3;
end;

theorem Th8:
  R c= [:X,Y:] implies R|Z = R /\ [:Z,Y:] & (Z|`R) = R /\ [:X,Z:]
proof
  assume
A1: R c= [:X,Y:];
  thus (R|Z) = R /\ [:Z,Y:]
  proof
    let x,y;
    thus [x,y] in (R|Z) implies [x,y] in R /\ [:Z,Y:]
    proof
      assume
A2:   [x,y] in (R|Z); then
A3:   x in Z by RELAT_1:def 11;
A4:   [x,y] in R by A2,RELAT_1:def 11;
      then y in Y by A1,ZFMISC_1:87;
      then [x,y] in [:Z,Y:] by A3,ZFMISC_1:87;
      hence thesis by A4,XBOOLE_0:def 4;
    end;
    thus [x,y] in R /\ [:Z,Y:] implies [x,y] in (R|Z)
    proof
      assume
A5:   [x,y] in R /\ [:Z,Y:];
      then [x,y] in [:Z,Y:] by XBOOLE_0:def 4; then
A6:   x in Z by ZFMISC_1:87;
      [x,y] in R by A5,XBOOLE_0:def 4;
      hence thesis by A6,RELAT_1:def 11;
    end;
  end;
  let x,y;
  thus [x,y] in (Z|`R) implies [x,y] in R /\ [:X,Z:]
  proof
    assume
A7: [x,y] in (Z|`R); then
A8: y in Z by RELAT_1:def 12;
A9: [x,y] in R by A7,RELAT_1:def 12;
    then x in X by A1,ZFMISC_1:87;
    then [x,y] in [:X,Z:] by A8,ZFMISC_1:87;
    hence thesis by A9,XBOOLE_0:def 4;
  end;
    assume
A10: [x,y] in R /\ [:X,Z:];
    then [x,y] in [:X,Z:] by XBOOLE_0:def 4; then
A11: y in Z by ZFMISC_1:87;
    [x,y] in R by A10,XBOOLE_0:def 4;
    hence thesis by A11,RELAT_1:def 12;
end;

theorem
  R c= [:X,Y:] & X = Z \/ W implies R = (R|Z) \/ (R|W)
proof
  assume that
A1: R c= [:X,Y:] and
A2: X = Z \/ W;
  thus R = R /\ [:X,Y:] by A1,XBOOLE_1:28
    .= R /\ ([:Z,Y:] \/ [:W,Y:]) by A2,ZFMISC_1:97
    .= (R /\ [:Z,Y:]) \/ (R /\ [:W,Y:]) by XBOOLE_1:23
    .= (R|Z) \/ (R /\ [:W,Y:]) by A1,Th8
    .= (R|Z) \/ (R|W) by A1,Th8;
end;

theorem
  X misses Y & R c= [:X,Y:] \/ [:Y,X:] implies R|X c= [:X,Y:]
proof
  assume that
A1: X /\ Y = {} and
A2: R c= [:X,Y:] \/ [:Y,X:];
  R|X = R /\ [:X,rng R:] by RELAT_1:67;
  then R|X c= ([:X,Y:] \/ [:Y,X:]) /\ [:X,rng R:] by A2,XBOOLE_1:26;
  then
A3: R|X c= [:X,Y:] /\ [:X,rng R:] \/ [:Y,X:] /\ [:X,rng R:] by XBOOLE_1:23;
  [:Y,X:] /\ [:X,rng R:] = [:X /\ Y, X /\ rng R:] by ZFMISC_1:100
    .= {} by A1,ZFMISC_1:90;
  hence thesis by A3,XBOOLE_1:18;
end;

theorem
  R c= S implies R~ c= S~
proof
  assume R c= S;
  then R \/ S = S by XBOOLE_1:12;
  then (R~) \/ (S~) = S~ by RELAT_1:23;
  hence thesis by XBOOLE_1:7;
end;



Lm2: id X c= [:X,X:]
proof
  let x,y;
  assume [x,y] in id X;
  then x in X & x = y by RELAT_1:def 10;
  hence thesis by ZFMISC_1:87;
end;

theorem Th12:
  id(X) * id(X) = id X
proof
  dom id(X) = X;
  hence thesis by RELAT_1:51;
end;

theorem
  id({x}) = {[x,x]}
proof
  x in {x} by TARSKI:def 1;
  then [x,x] in id({x}) by RELAT_1:def 10; then
A1: {[x,x]} c= id {x} by ZFMISC_1:31;
  [:{x},{x}:] = {[x,x]} by ZFMISC_1:29;
  then id({x}) c= {[x,x]} by Lm2;
  hence thesis by A1,XBOOLE_0:def 10;
end;

theorem Th14:
  id(X \/ Y) = id(X) \/ id(Y) & id(X /\ Y) = id(X) /\ id(Y) &
  id(X \ Y) = id(X) \ id(Y)
proof
  thus id(X \/ Y) = id(X) \/ id(Y)
  proof
    let x,y;
    thus [x,y] in id(X \/ Y) implies [x,y] in id(X) \/ id(Y)
    proof
      assume
A1:   [x,y] in id(X \/ Y);
      then x in X \/ Y by RELAT_1:def 10;
      then
A2:   x in X or x in Y by XBOOLE_0:def 3;
      x = y by A1,RELAT_1:def 10;
      then [x,y] in id(X) or [x,y] in id(Y) by A2,RELAT_1:def 10;
      hence thesis by XBOOLE_0:def 3;
    end;
    assume [x,y] in id(X) \/ id(Y); then
A3: [x,y] in id(X) or [x,y] in id(Y) by XBOOLE_0:def 3;
    then x in X or x in Y by RELAT_1:def 10;
    then
A4: x in X \/ Y by XBOOLE_0:def 3;
    x = y by A3,RELAT_1:def 10;
    hence thesis by A4,RELAT_1:def 10;
  end;
  thus id(X /\ Y) = id(X) /\ id(Y)
  proof
    let x,y;
    thus [x,y] in id(X /\ Y) implies [x,y] in id(X) /\ id(Y)
    proof
      assume
A5:   [x,y] in id(X /\ Y); then
A6:   x in X /\ Y by RELAT_1:def 10;
A7:   x = y by A5,RELAT_1:def 10;
      x in Y by A6,XBOOLE_0:def 4; then
A8:   [x,y] in id(Y) by A7,RELAT_1:def 10;
      x in X by A6,XBOOLE_0:def 4;
      then [x,y] in id(X) by A7,RELAT_1:def 10;
      hence thesis by A8,XBOOLE_0:def 4;
    end;
    assume
A9: [x,y] in id(X) /\ id Y; then
A10: [x,y] in id(X) by XBOOLE_0:def 4; then
A11: x = y by RELAT_1:def 10;
    [x,y] in id(Y) by A9,XBOOLE_0:def 4; then
A12: x in Y by RELAT_1:def 10;
    x in X by A10,RELAT_1:def 10;
    then x in X /\ Y by A12,XBOOLE_0:def 4;
    hence thesis by A11,RELAT_1:def 10;
  end;
    let x,y;
    thus [x,y] in id(X \ Y) implies [x,y] in id(X) \ id(Y)
    proof
      assume
A13:  [x,y] in id(X \ Y); then
A14:  x in X \ Y by RELAT_1:def 10;
      then not x in Y by XBOOLE_0:def 5; then
A15:  not [x,y] in id(Y) by RELAT_1:def 10;
      x = y by A13,RELAT_1:def 10;
      then [x,y] in id(X) by A14,RELAT_1:def 10;
      hence thesis by A15,XBOOLE_0:def 5;
    end;
    assume
A16: [x,y] in id(X) \ id(Y); then
A17: x = y by RELAT_1:def 10;
    not [x,y] in id(Y) by A16,XBOOLE_0:def 5; then
A18: not (x in Y & x = y) by RELAT_1:def 10;
    x in X by A16,RELAT_1:def 10;
    then x in X \ Y by A16,A18,RELAT_1:def 10,XBOOLE_0:def 5;
    hence thesis by A17,RELAT_1:def 10;
end;

theorem Th15:
  X c= Y implies id X c= id Y
proof
  assume X c= Y;
  then X \/ Y = Y by XBOOLE_1:12;
  then id(X) \/ id Y = id Y by Th14;
  hence thesis by XBOOLE_1:7;
end;

theorem
  id(X \ Y) \ id X = {}
proof
  id(X \ Y) c= id(X) by Th15;
  hence thesis by XBOOLE_1:37;
end;

theorem
  R c= id dom R implies R = id dom R
proof
  assume
A1: R c= id dom R;
  let x,y;
  thus [x,y] in R implies [x,y] in id dom R by A1;
  assume
A2: [x,y] in id dom R;
  then x in dom R by RELAT_1:def 10; then
A3: ex z st [x,z] in R by XTUPLE_0:def 12;
  x = y by A2,RELAT_1:def 10;
  hence thesis by A1,A3,RELAT_1:def 10;
end;

theorem
  id X c= R \/ R~ implies id X c= R & id X c= R~
proof
  assume
A1: id X c= R \/ R~;
  for x holds x in X implies [x,x] in R & [x,x] in R~
  proof
    let x;
    assume x in X;
    then [x,x] in id(X) by RELAT_1:def 10; then
A2: [x,x] in R or [x,x] in R~ by A1,XBOOLE_0:def 3;
    hence [x,x] in R by RELAT_1:def 7;
    thus thesis by A2,RELAT_1:def 7;
  end; then
  (for x holds x in X implies [x,x] in R) &
  for x holds x in X implies [x,x] in R~;
  hence thesis by RELAT_1:47;
end;

theorem
  id X c= R implies id X c= R~
proof
  assume
A1: id X c= R;
  for x holds x in X implies [x,x] in R~
  proof
    let x;
    assume x in X;
    then [x,x] in id X by RELAT_1:def 10;
    hence thesis by A1,RELAT_1:def 7;
  end;
  hence thesis by RELAT_1:47;
end;



theorem
  R c= [:X,X:] implies R \ id dom R = R \ id X & R \ id rng R = R \ id X
proof
A1: R \ id dom R c= R \ id X
  proof
    let x,y;
    assume
A2: [x,y] in R \ id dom R; then
A3: not [x,y] in id dom R by XBOOLE_0:def 5;
    not [x,y] in id X
    proof
      assume [x,y] in id X; then
A4:   x = y by RELAT_1:def 10;
      x in dom R by A2,XTUPLE_0:def 12;
      hence contradiction by A3,A4,RELAT_1:def 10;
    end;
    hence thesis by A2,XBOOLE_0:def 5;
  end;
A5: R \ id rng R c= R \ id X
  proof
    let x,y;
    assume
A6: [x,y] in R \ id rng R; then
A7: not [x,y] in id rng R by XBOOLE_0:def 5;
    not [x,y] in id X
    proof
      assume [x,y] in id X; then
A8:   x = y by RELAT_1:def 10;
      y in rng R by A6,XTUPLE_0:def 13;
      hence contradiction by A7,A8,RELAT_1:def 10;
    end;
    hence thesis by A6,XBOOLE_0:def 5;
  end;
  assume
A9: R c= [:X,X:];
  then id dom R c= id X by Th3,Th15;
  then R \ id X c= R \ id dom R by XBOOLE_1:34;
  hence R \ id dom R = R \ id X by A1,XBOOLE_0:def 10;
  id(rng R) c= id X by A9,Th3,Th15;
  then R \ id(X) c= R \ id rng R by XBOOLE_1:34;
  hence thesis by A5,XBOOLE_0:def 10;
end;

theorem
  (id(X) * (R \ id X) = {} implies dom (R \ id X) = dom R \ X) &
  ((R \ id X) * id X = {} implies rng (R \ id X) = rng R \ X)
proof
  thus id(X) * (R \ id X) = {} implies dom (R \ id X) = dom R \ X
  proof
    assume
A1: id(X) * (R \ id X) = {};
A2: dom (R \ id X) c= dom R \ X
    proof
      let x;
A3:   x in dom R & not x in X implies thesis by XBOOLE_0:def 5;
      assume x in dom (R \ id X); then
A6:   ex y st [x,y] in (R \ id X) by XTUPLE_0:def 12;
      not x in X
      proof
        assume x in X;
        then [x,x] in id X by RELAT_1:def 10;
        hence thesis by A1,A6,RELAT_1:def 8;
      end;
      hence thesis by A6,A3,XTUPLE_0:def 12;
    end;
    dom R \ dom id X c= dom (R \ id X) by RELAT_1:3;
    then dom R \ X c= dom (R \ id X);
    hence thesis by A2,XBOOLE_0:def 10;
  end;
  thus (R \ id X) * id(X) = {} implies rng (R \ id X) = rng R \ X
  proof
    assume
A7: (R \ id X) * id X = {};
A8: rng (R \ id X) c= rng R \ X
    proof
      let y;
A9:   y in rng R & not y in X implies thesis by XBOOLE_0:def 5;
      assume y in rng(R \ id X); then
A12:  ex x st [x,y] in R \ id X by XTUPLE_0:def 13;
      not y in X
      proof
        assume y in X;
        then [y,y] in id X by RELAT_1:def 10;
        hence thesis by A7,A12,RELAT_1:def 8;
      end;
      hence thesis by A12,A9,XTUPLE_0:def 13;
    end;
    rng R \ rng id X c= rng (R \ id X) by RELAT_1:14;
    then rng R \ X c= rng (R \ id X);
    hence thesis by A8,XBOOLE_0:def 10;
  end;
end;

theorem Th22:
  (R c= R * R & R * (R \ id rng R) = {} implies id rng R c= R) &
  (R c= R * R & (R \ id dom R) * R = {} implies id dom R c= R)
proof
  thus R c= R * R & R * (R \ id rng R) = {} implies id rng R c= R
  proof
    assume that
A1: R c= R * R and
A2: R * (R \ id rng R) = {};
    for y holds y in rng R implies [y,y] in R
    proof
      let y;
      assume y in rng R;
      then consider x such that
A3:   [x,y] in R by XTUPLE_0:def 13;
      consider z such that
A4:   [x,z] in R and
A5:   [z,y] in R by A1,A3,RELAT_1:def 8;
      z = y
      proof
        assume z <> y;
        then not [z,y] in id rng R by RELAT_1:def 10;
        then [z,y] in (R \ id rng R) by A5,XBOOLE_0:def 5;
        hence contradiction by A2,A4,RELAT_1:def 8;
      end;
      hence thesis by A5;
    end;
    hence thesis by RELAT_1:47;
  end;
    assume that
A6: R c= R * R and
A7: (R \ id dom R) * R = {};
    for x holds x in dom R implies [x,x] in R
    proof
      let x;
      assume x in dom R;
      then consider y such that
A8:   [x,y] in R by XTUPLE_0:def 12;
      consider z such that
A9:   [x,z] in R and
A10:  [z,y] in R by A6,A8,RELAT_1:def 8;
      z = x
      proof
        assume z <> x;
        then not [x,z] in id dom R by RELAT_1:def 10;
        then [x,z] in R \ id dom R by A9,XBOOLE_0:def 5;
        hence contradiction by A7,A10,RELAT_1:def 8;
      end;
      hence thesis by A9;
    end;
    hence thesis by RELAT_1:47;
end;

theorem
  (R c= R * R & R * (R \ id rng R) = {} implies R /\ (id rng R) = id rng R) &
  (R c= R * R & (R \ id dom R) * R = {} implies
    R /\ (id dom R) = id dom R) by Th22,XBOOLE_1:28;

theorem
  (R * (R \ id X) = {} implies R * (R \ id rng R) = {}) &
  ((R \ id X) * R = {} implies (R \ id dom R) * R = {})
proof
  thus R * (R \ id X) = {} implies R * (R \ id rng R) = {}
  proof
    assume that
A1: R * (R \ id X) = {} and
A2: R * (R \ id rng R) <> {};
    consider x,y such that
A3: [x,y] in R * (R \ id rng R) by A2,RELAT_1:37;
    consider z such that
A4: [x,z] in R and
A5: [z,y] in R \ id rng R by A3,RELAT_1:def 8;
    z in rng R & not [z,y] in id rng R
       by A4,A5,XBOOLE_0:def 5,XTUPLE_0:def 13;
    then z <> y by RELAT_1:def 10;
    then not [z,y] in id X by RELAT_1:def 10;
    then [z,y] in R \ id X by A5,XBOOLE_0:def 5;
    hence contradiction by A1,A4,RELAT_1:def 8;
  end;
    assume that
A6: (R \ id X) * R = {} and
A7: (R \ id dom R) * R <> {};
    consider x,y such that
A8: [x,y] in (R \ id dom R) * R by A7,RELAT_1:37;
    consider z such that
A9: [x,z] in (R \ id dom R) and
A10: [z,y] in R by A8,RELAT_1:def 8;
    not [x,z] in id dom R by A9,XBOOLE_0:def 5;
    then not z in dom R or x <> z by RELAT_1:def 10;
    then not [x,z] in id X by A10,RELAT_1:def 10,XTUPLE_0:def 12;
    then [x,z] in R \ id X by A9,XBOOLE_0:def 5;
    hence contradiction by A6,A10,RELAT_1:def 8;
end;



definition
  let R;
  func CL R -> Relation equals
  R /\ id dom R;
  correctness;
end;

theorem Th25:
  [x,y] in CL R implies x in dom CL R & x = y
proof
  assume
A1: [x,y] in CL R;
  then [x,y] in id dom R by XBOOLE_0:def 4;
  hence thesis by A1,RELAT_1:def 10,XTUPLE_0:def 12;
end;

theorem Th26:
  dom CL R = rng CL R
proof
  thus dom CL R c= rng CL R
  proof
    let x;
    assume x in dom CL R;
    then consider y such that
A1: [x,y] in CL R by XTUPLE_0:def 12;
    [x,y] in id dom R by A1,XBOOLE_0:def 4;
    then [x,x] in CL R by A1,RELAT_1:def 10;
    hence thesis by XTUPLE_0:def 13;
  end;
  let x;
  assume x in rng CL R;
  then consider y such that
A2: [y,x] in CL R by XTUPLE_0:def 13;
  [y,x] in id dom R by A2,XBOOLE_0:def 4;
  then [x,x] in CL R by A2,RELAT_1:def 10;
  hence thesis by XTUPLE_0:def 12;
end;

theorem Th27:
  (x in dom CL R iff x in dom R & [x,x] in R) &
  (x in rng CL R iff x in dom R & [x,x] in R) &
  (x in rng CL R iff x in rng R & [x,x] in R) &
  (x in dom CL R iff x in rng R & [x,x] in R)
proof
A1: x in dom R & [x,x] in R implies x in dom CL R
  proof
    assume that
A2: x in dom R and
A3: [x,x] in R;
    [x,x] in id dom R by A2,RELAT_1:def 10;
    then [x,x] in (R /\ id dom R) by A3,XBOOLE_0:def 4;
    hence thesis by XTUPLE_0:def 12;
  end;
A4: x in dom CL R implies x in dom R & [x,x] in R
  proof
    assume x in dom CL R;
    then consider y such that
A5: [x,y] in CL R by XTUPLE_0:def 12;
    [x,y] in R & [x,y] in id dom R by A5,XBOOLE_0:def 4;
    hence thesis by RELAT_1:def 10;
  end;
  hence x in dom CL R iff x in dom R & [x,x] in R by A1;
  thus x in rng CL R iff x in dom R & [x,x] in R by A4,A1,Th26;
  thus x in rng CL R iff x in rng R & [x,x] in R
           by A4,A1,Th26,XTUPLE_0:def 12,def 13;
  thus thesis by A4,A1,XTUPLE_0:def 12,def 13;
end;

theorem Th28:
  CL R = id dom CL R
proof
  let x,y;
  thus [x,y] in CL R implies [x,y] in id dom CL R
  proof
    assume
A1: [x,y] in CL R;
    then [x,y] in id dom R by XBOOLE_0:def 4; then
A2: x = y by RELAT_1:def 10;
    x in dom CL R by A1,XTUPLE_0:def 12;
    hence thesis by A2,RELAT_1:def 10;
  end;
  assume
A3: [x,y] in id dom CL R;
  then x in dom CL R by RELAT_1:def 10; then
A4: ex z st [x,z] in CL R by XTUPLE_0:def 12;
  x = y by A3,RELAT_1:def 10;
  hence thesis by A4,Th25;
end;

theorem Th29:
  (R * R = R & R * (R \ CL R) = {} & [x,y] in R & x <> y implies
    x in (dom R \ dom CL R) & y in dom CL R) &
  (R * R = R & (R \ CL R) * R = {} & [x,y] in R & x <> y implies
    y in (rng R \ dom CL R) & x in dom CL R)
proof
  thus R * R = R & R * (R \ CL R) = {} & [x,y] in R & x <> y implies
    x in (dom R \ dom CL R) & y in dom CL R
  proof
    assume that
A1: R * R = R and
A2: R * (R \ CL R) = {} and
A3: [x,y] in R and
A4: x <> y;
A5: y in rng R by A3,XTUPLE_0:def 13;
    consider z such that
A6: [x,z] in R and
A7: [z,y] in R by A1,A3,RELAT_1:def 8;
A8: z = y
    proof
      assume z <> y;
      then not [z,y] in id dom R by RELAT_1:def 10;
      then not [z,y] in R /\ id dom R by XBOOLE_0:def 4;
      then [z,y] in R \ CL R by A7,XBOOLE_0:def 5;
      hence thesis by A2,A6,RELAT_1:def 8;
    end;
    not [x,y] in id dom R by A4,RELAT_1:def 10;
    then not [x,y] in R /\ id dom R by XBOOLE_0:def 4; then
A9: [x,y] in R \ CL(R) by A3,XBOOLE_0:def 5;
A10: not x in dom CL R
    proof
      assume x in dom CL R;
      then [x,x] in R by Th27;
      hence thesis by A2,A9,RELAT_1:def 8;
    end;
    x in dom R by A6,XTUPLE_0:def 12;
    hence thesis by A7,A8,A5,A10,Th27,XBOOLE_0:def 5;
  end;
  thus R * R = R & (R \ CL R) * R = {} & [x,y] in R & x <> y implies
  y in (rng R \ dom CL R) & x in dom CL R
  proof
    assume that
A11: R * R = R and
A12: (R \ CL R) * R = {} and
A13: [x,y] in R and
A14: x <> y;
A15: x in dom R by A13,XTUPLE_0:def 12;
    consider z such that
A16: [x,z] in R and
A17: [z,y] in R by A11,A13,RELAT_1:def 8;
A18: z = x
    proof
      assume z <> x;
      then not [x,z] in id dom R by RELAT_1:def 10;
      then not [x,z] in R /\ id dom R by XBOOLE_0:def 4;
      then [x,z] in R \ CL R by A16,XBOOLE_0:def 5;
      hence thesis by A12,A17,RELAT_1:def 8;
    end;
    not [x,y] in id dom R by A14,RELAT_1:def 10;
    then not [x,y] in R /\ id dom R by XBOOLE_0:def 4;
    then
A19: [x,y] in R \ CL R by A13,XBOOLE_0:def 5;
A20: not y in dom CL R
    proof
      assume y in dom CL R;
      then [y,y] in R by Th27;
      hence thesis by A12,A19,RELAT_1:def 8;
    end;
    y in rng R by A17,XTUPLE_0:def 13;
    hence thesis by A16,A18,A15,A20,Th27,XBOOLE_0:def 5;
  end;
end;

theorem
  (R * R = R & R * (R \ id dom R) = {} & [x,y] in R & x <> y implies
    x in ((dom R) \ dom CL R) & y in dom CL R) &
  (R * R = R & (R \ id dom R) * R = {} & [x,y] in R & x <> y implies
    y in ((rng R) \ dom CL R) & x in dom CL R)
proof
  R \ CL R = R \ id dom R by XBOOLE_1:47;
  hence thesis by Th29;
end;

theorem
  (R * R = R & R * (R \ id dom R) = {} implies dom CL R = rng R &
    rng CL R = rng R) &
  (R * R = R & (R \ id dom R) * R = {} implies dom CL R = dom R &
    rng CL R = dom R)
proof
  thus R * R = R & R * (R \ id dom R) = {} implies
  dom CL R = rng R & rng CL R = rng R
  proof
    assume that
A1: R * R = R and
A2: R * (R \ id dom R) = {};
    for y holds y in rng R implies y in dom CL R
    proof
      let y;
      assume y in rng R;
      then consider x such that
A3:   [x,y] in R by XTUPLE_0:def 13;
      consider z such that
A4:   [x,z] in R and
A5:   [z,y] in R by A1,A3,RELAT_1:def 8;
A6:   z = y
      proof
        assume z <> y;
        then not [z,y] in id dom R by RELAT_1:def 10;
        then [z,y] in R \ id dom R by A5,XBOOLE_0:def 5;
        hence thesis by A2,A4,RELAT_1:def 8;
      end;
      z in dom R by A5,XTUPLE_0:def 12;
      then [z,y] in id dom R by A6,RELAT_1:def 10;
      then [z,y] in R /\ id dom R by A5,XBOOLE_0:def 4;
      hence thesis by A6,XTUPLE_0:def 12;
    end;
    then
A7: rng R c= dom CL R by TARSKI:def 3;
    CL(R) c= R by XBOOLE_1:17;
    then rng CL R c= rng R by RELAT_1:11;
    then dom CL R c= rng R by Th26;
    then dom CL R = rng R by A7,XBOOLE_0:def 10;
    hence thesis by Th26;
  end;
  thus R * R = R & (R \ id dom R) * R = {} implies
  dom CL R = dom R & rng CL R = dom R
  proof
    assume that
A8: R * R = R and
A9: (R \ id dom R) * R = {};
    for x holds x in dom R implies x in dom CL R
    proof
      let x;
      assume
A10:  x in dom R;
      then consider y such that
A11:  [x,y] in R by XTUPLE_0:def 12;
      consider z such that
A12:  [x,z] in R and
A13:  [z,y] in R by A8,A11,RELAT_1:def 8;
A14:  z = x
      proof
        assume z <> x;
        then not [x,z] in id dom R by RELAT_1:def 10;
        then [x,z] in R \ id dom R by A12,XBOOLE_0:def 5;
        hence thesis by A9,A13,RELAT_1:def 8;
      end;
      then [x,z] in id dom R by A10,RELAT_1:def 10;
      then [x,z] in R /\ id dom R by A12,XBOOLE_0:def 4;
      then z in rng CL R by XTUPLE_0:def 13;
      hence thesis by A14,Th26;
    end;
    then
A15: dom R c= dom CL R by TARSKI:def 3;
    CL R c= R by XBOOLE_1:17;
    then dom CL R c= dom R by RELAT_1:11;
    then dom CL R = dom R by A15,XBOOLE_0:def 10;
    hence thesis by Th26;
  end;
end;

theorem Th32:
  dom CL R c= dom R & rng CL R c= rng R & rng CL R c= dom R & dom CL R c= rng R
proof
  CL R c= R by XBOOLE_1:17;
  hence dom CL R c= dom R & rng CL R c= rng R by RELAT_1:11;
  hence thesis by Th26;
end;

theorem
  id dom CL R c= id dom R & id rng CL R c= id dom R
proof
  dom CL R c= dom R by Th32;
  then id dom CL R c= id dom R by Th15;
  hence thesis by Th26;
end;

theorem Th34:
  id dom CL R c= R & id rng CL R c= R
proof
  thus id dom CL R c= R
  proof
    let x,y;
    assume [x,y] in id dom CL R;
    then x in dom CL R & x = y by RELAT_1:def 10;
    hence thesis by Th27;
  end;
  hence thesis by Th26;
end;

theorem Th35:
  (id X c= R & id(X) * (R \ id X) = {} implies R|X = id X) &
  (id X c= R & (R \ id X) * id X = {} implies X|`R = id X)
proof
  thus id X c= R & id(X) * (R \ id X) = {} implies R|X = id X
  proof
    assume that
A1: id(X) c= R and
A2: id(X) * (R \ id(X)) = {};
    R|X = id(X) * R by RELAT_1:65
      .= id(X) * (R \/ id(X)) by A1,XBOOLE_1:12
      .= id(X) * ((R \ id(X)) \/ id(X)) by XBOOLE_1:39
      .= {} \/ id(X) * id(X) by A2,RELAT_1:32
      .= id(X) by Th12;
    hence thesis;
  end;
    assume that
A3: id(X) c= R and
A4: (R \ id X) * id X = {};
    X|`R = R * id X by RELAT_1:92
      .= (R \/ id X) * id X by A3,XBOOLE_1:12
      .= ((R \ id X) \/ id X) * id X by XBOOLE_1:39
      .= {} \/ id(X) * id X by A4,Th6
      .= id X by Th12;
    hence thesis;
end;

theorem
  (id(dom CL R) * (R \ id(dom CL R)) = {} implies R|(dom CL R) = id dom CL R &
    R|(rng CL R) = id dom CL R) &
  ((R \ id rng CL R) * id(rng CL R) = {} implies
    (dom CL R)|`R = id dom CL R &
    (rng CL R)|`R = id rng CL R)
proof
  thus id(dom CL R) * (R \ id dom CL R) = {} implies
  R|(dom CL R) = id(dom CL R) & R|(rng CL R) = id dom CL R
  proof
    assume
A1: id(dom CL R) * (R \ id dom CL R) = {};
    id(dom CL R) c= R by Th34;
    then R|(dom CL R) = id dom CL R by A1,Th35;
    hence thesis by Th26;
  end;
    assume
A2: (R \ id rng CL R) * id rng CL R = {};
    id rng CL R c= R by Th34;
    then (rng CL R)|`R = id rng CL R by A2,Th35;
    then (dom CL R)|`R = id rng CL R by Th26;
    hence thesis by Th26;
end;

theorem
  (R * (R \ id dom R) = {} implies id(dom CL R) * (R \ id dom CL R) = {}) &
  ((R \ id dom R) * R = {} implies
    (R \ id dom CL R) * id dom CL R = {})
proof
  thus R * (R \ id dom R) = {} implies id (dom CL R) * (R \ id dom CL R) = {}
  proof
A1: id dom CL R c= R by Th34;
    assume
A2: R * (R \ id dom R) = {};
    R \ id dom R = R \ CL R by XBOOLE_1:47
      .= R \ id dom CL R by Th28;
    hence thesis by A2,A1,RELAT_1:30,XBOOLE_1:3;
  end;
A3: id dom CL R c= R by Th34;
    assume
A4: (R \ id dom R) * R = {};
    R \ id dom R = R \ CL R by XBOOLE_1:47
      .= R \ id dom CL R by Th28;
    hence thesis by A4,A3,RELAT_1:29,XBOOLE_1:3;
end;

theorem Th38:
  (S * R = S & R * (R \ id dom R) = {} implies S * (R \ id dom R) = {}) &
  (R * S = S & (R \ id dom R) * R = {} implies (R \ id dom R) * S = {})
proof
  thus S * R = S & R * (R \ id dom R) = {} implies S * (R \ id dom R) = {}
  proof
    assume S * R = S & R * (R \ id dom R) = {};
    then S * (R \ id dom R) = S * {} by RELAT_1:36
      .= {};
    hence thesis;
  end;
  assume R * S = S & (R \ id dom R) * R = {};
  then (R \ id dom R) * S = {} * S by RELAT_1:36
    .= {};
  hence thesis;
end;

theorem Th39:
  (S * R = S & R * (R \ id dom R) = {} implies CL(S) c= CL(R)) &
  (R * S = S & (R \ id dom R) * R = {} implies CL(S) c= CL(R))
proof
  thus S * R = S & R * (R \ id dom R) = {} implies CL S c= CL R
  proof
    assume that
A1: S * R = S and
A2: R * (R \ id dom R) = {};
A3: S * (R \ id dom R) = {} by A1,A2,Th38;
    for x,y holds [x,y] in CL S implies [x,y] in CL R
    proof
      let x,y;
      assume
A4:   [x,y] in CL S; then
A5:   [x,y] in id dom S by XBOOLE_0:def 4;
      [x,y] in S by A4,XBOOLE_0:def 4;
      then consider z such that
A6:   [x,z] in S and
A7:   [z,y] in R by A1,RELAT_1:def 8;
A8:   z = y
      proof
        assume z <> y;
        then not [z,y] in id dom R by RELAT_1:def 10;
        then [z,y] in R \ id dom R by A7,XBOOLE_0:def 5;
        hence contradiction by A3,A6,RELAT_1:def 8;
      end;
A9:   x = y by A5,RELAT_1:def 10;
      then x in dom R by A7,A8,XTUPLE_0:def 12;
      then
A10:  [x,y] in id dom R by A9,RELAT_1:def 10;
      [x,y] in R by A5,A7,A8,RELAT_1:def 10;
      hence thesis by A10,XBOOLE_0:def 4;
    end;
    hence thesis by RELAT_1:def 3;
  end;
  assume that
A11: R * S = S and
A12: (R \ id dom R) * R = {};
A13: (R \ id dom R) * S = {} by A11,A12,Th38;
  for x,y holds [x,y] in CL S implies [x,y] in CL R
  proof
    let x,y;
    assume
A14: [x,y] in CL S; then
A15: [x,y] in id dom S by XBOOLE_0:def 4; then
A16: x = y by RELAT_1:def 10;
    [x,y] in S by A14,XBOOLE_0:def 4;
    then consider z such that
A17: [x,z] in R and
A18: [z,y] in S by A11,RELAT_1:def 8;
    x in dom R by A17,XTUPLE_0:def 12; then
A19: [x,y] in id dom R by A16,RELAT_1:def 10;
    z = x
    proof
      assume z <> x;
      then not [x,z] in id dom R by RELAT_1:def 10;
      then [x,z] in R \ id dom R by A17,XBOOLE_0:def 5;
      hence contradiction by A13,A18,RELAT_1:def 8;
    end;
    then [x,y] in R by A15,A17,RELAT_1:def 10;
    hence thesis by A19,XBOOLE_0:def 4;
  end;
  hence thesis by RELAT_1:def 3;
end;

theorem
  (S * R = S & R * (R \ id dom R) = {} & R * S = R &
    S * (S \ id dom S) = {} implies CL S = CL R) &
  (R * S = S & (R \ id dom R) * R = {} & S * R = R &
    (S \ id dom S) * S = {} implies CL S = CL R)
proof
  thus S * R = S & R * (R \ id dom R) = {} &
  R * S = R & S * (S \ id dom S) = {} implies CL S = CL R
  proof
    assume S * R = S & R * (R \ id dom R) = {} &
      R * S = R & S * (S \ id dom S) = {};
    then CL S c= CL R & CL R c= CL S by Th39;
    hence thesis by XBOOLE_0:def 10;
  end;
  assume R * S = S & (R \ id dom R) * R = {} & S * R = R &
    (S \ id dom S) * S = {};
  then CL S c= CL R & CL R c= CL S by Th39;
  hence thesis by XBOOLE_0:def 10;
end;














begin 

definition
  let a be set;
  func NOT1 a equals
  :Def1:
  {} if a is non empty otherwise {{}};
  correctness;
end;

registration
  let a be empty set;
  cluster NOT1 a -> non empty;
  coherence by Def1;
end;

registration
  let a be non empty set;
  cluster NOT1 a -> empty;
  coherence by Def1;
end;

theorem
  NOT1 {{}} = {} & NOT1 {} = {{}} by Def1;

theorem
  for a being set holds NOT1 a is non empty iff not a is non empty;

reserve a,b,c,d,e,f,g,h for set;

theorem
  NOT1 {} is non empty;

definition
  let a,b be set;
  func AND2(a,b) equals
  :Def2:
  NOT1 {} if a is non empty & b is non empty
  otherwise {};
  correctness;
  commutativity;
end;

registration
  let a,b be non empty set;
  cluster AND2(a,b) -> non empty;
  coherence by Def2;
end;

registration
  let a be empty set, b be set;
  cluster AND2(a,b) -> empty;
  coherence by Def2;
end;

theorem
  AND2(a,b) is non empty iff a is non empty & b is non empty;

definition
  let a,b be set;
  func OR2(a, b) equals
  :Def3:
  NOT1 {} if a is non empty or b is non empty
  otherwise {};
  correctness;
  commutativity;
end;

registration
  let a be set,b be non empty set;
  cluster OR2(a,b) -> non empty;
  coherence by Def3;
end;

registration
  let a,b be empty set;
  cluster OR2(a,b) -> empty;
  coherence by Def3;
end;

theorem
  OR2(a,b) is non empty iff (a is non empty or b is non empty);

definition
  let a,b be set;
  func XOR2(a,b) equals
  :Def4:
  NOT1 {} if a is non empty & not b is non empty
  or not a is non empty & b is non empty otherwise {};
  correctness;
  commutativity;
end;

registration
  let a be empty set,b be non empty set;
  cluster XOR2(a,b) -> non empty;
  coherence by Def4;
end;

registration
  let a,b be empty set;
  cluster XOR2(a,b) -> empty;
  coherence by Def4;
end;

registration
  let a,b be non empty set;
  cluster XOR2(a,b) -> empty;
  coherence by Def4;
end;

theorem
  XOR2(a, b) is non empty iff a is non empty & not b is non empty or not
  a is non empty & b is non empty;

theorem
  not XOR2(a,a) is non empty
proof
  XOR2(a,a) is non empty iff a is non empty & not a is non empty or not a
  is non empty & a is non empty;
  hence thesis;
end;

theorem
  XOR2(a,{}) is non empty iff a is non empty;

theorem
  XOR2(a,b) is non empty iff XOR2(b,a) is non empty;

definition
  let a,b be set;
  func EQV2(a, b) equals
  :Def5:
  NOT1 {} if a is non empty iff b is non empty
  otherwise {};
  correctness;
  commutativity;
end;

registration
  let a be empty set,b be non empty set;
  cluster EQV2(a,b) -> empty;
  coherence by Def5;
end;

registration
  let a,b be empty set;
  cluster EQV2(a,b) -> non empty;
  coherence by Def5;
end;

registration
  let a,b be non empty set;
  cluster EQV2(a,b) -> non empty;
  coherence by Def5;
end;

theorem
  EQV2(a, b) is non empty iff (a is non empty iff b is non empty);

theorem
  EQV2(a,b) is non empty iff not XOR2(a,b) is non empty
proof
  EQV2(a,b) is non empty iff (a is non empty iff b is non empty);
  hence thesis;
end;

definition
  let a,b be set;
  func NAND2(a, b) equals
  :Def6:
  NOT1 {} if not (a is non empty & b is non
  empty) otherwise {};
  correctness;
  commutativity;
end;

registration
  let a be empty set, b be set;
  cluster NAND2(a,b) -> non empty;
  coherence by Def6;
end;

registration
  let a,b be non empty set;
  cluster NAND2(a,b) -> empty;
  coherence by Def6;
end;

theorem
  NAND2(a, b) is non empty iff not (a is non empty & b is non empty);

definition
  let a,b be set;
  func NOR2(a, b) equals
  :Def7:
  NOT1 {} if not (a is non empty or b is non
  empty) otherwise {};
  correctness;
  commutativity;
end;

registration
  let a,b be empty set;
  cluster NOR2(a,b) -> non empty;
  coherence by Def7;
end;

registration
  let a be non empty set, b be set;
  cluster NOR2(a,b) -> empty;
  coherence by Def7;
end;

theorem
  NOR2(a,b) is non empty iff not (a is non empty or b is non empty);

definition
  let a,b,c be set;
  func AND3(a,b,c) equals
  :Def8:
  NOT1 {} if a is non empty & b is non empty &
  c is non empty otherwise {};
  correctness;
end;

registration
  let a,b,c be non empty set;
  cluster AND3(a,b,c) -> non empty;
  coherence by Def8;
end;

registration
  let a be empty set, b,c be set;
  cluster AND3(a,b,c) -> empty;
  coherence by Def8;
  cluster AND3(b,a,c) -> empty;
  coherence by Def8;
  cluster AND3(b,c,a) -> empty;
  coherence by Def8;
end;

theorem
  AND3(a,b,c) is non empty iff a is non empty & b is non empty & c is
  non empty;

definition
  let a,b,c be set;
  func OR3(a,b,c) equals
  :Def9:
  NOT1 {} if a is non empty or b is non empty or
  c is non empty otherwise {};
  correctness;
end;

registration
  let a,b,c be empty set;
  cluster OR3(a,b,c) -> empty;
  coherence by Def9;
end;

registration
  let a be non empty set, b,c be set;
  cluster OR3(a,b,c) -> non empty;
  coherence by Def9;
  cluster OR3(b,a,c) -> non empty;
  coherence by Def9;
  cluster OR3(b,c,a) -> non empty;
  coherence by Def9;
end;

theorem
  OR3(a,b,c) is non empty iff a is non empty or b is non empty or c is
  non empty;

definition
  let a,b,c be set;
  func XOR3(a,b,c) equals
  :Def10:
  NOT1 {} if ( a is non empty & not b is non
empty or not a is non empty & b is non empty ) & not c is non empty or not ( a
is non empty & not b is non empty or not a is non empty & b is non empty ) & c
  is non empty otherwise {};
  correctness;
end;

registration
  let a,b,c be empty set;
  cluster XOR3(a,b,c) -> empty;
  coherence by Def10;
end;

registration
  let a,b be empty set, c be non empty set;
  cluster XOR3(a,b,c) -> non empty;
  coherence by Def10;
  cluster XOR3(a,c,b) -> non empty;
  coherence by Def10;
  cluster XOR3(c,a,b) -> non empty;
  coherence by Def10;
end;

registration
  let a,b be non empty set, c be empty set;
  cluster XOR3(a,b,c) -> empty;
  coherence by Def10;
  cluster XOR3(a,c,b) -> empty;
  coherence by Def10;
  cluster XOR3(c,a,b) -> empty;
  coherence by Def10;
end;

registration
  let a,b,c be non empty set;
  cluster XOR3(a,b,c) -> non empty;
  coherence by Def10;
end;

theorem
  XOR3(a,b,c) is non empty iff ( a is non empty & not b is non empty or
  not a is non empty & b is non empty ) & not c is non empty or not ( a is non
empty & not b is non empty or not a is non empty & b is non empty ) & c is non
  empty;

definition
  let a,b,c be set;
  func MAJ3(a,b,c) equals
  :Def11:
  NOT1 {} if a is non empty & b is non empty
or b is non empty & c is non empty or c is non empty & a is non empty otherwise
  {};
  correctness;
end;

registration
  let a,b be non empty set, c be set;
  cluster MAJ3(a,b,c) -> non empty;
  coherence by Def11;
  cluster MAJ3(a,c,b) -> non empty;
  coherence by Def11;
  cluster MAJ3(c,a,b) -> non empty;
  coherence by Def11;
end;

registration
  let a,b be empty set, c be set;
  cluster MAJ3(a,b,c) -> empty;
  coherence by Def11;
  cluster MAJ3(a,c,b) -> empty;
  coherence by Def11;
  cluster MAJ3(c,a,b) -> empty;
  coherence by Def11;
end;

theorem
  MAJ3(a,b,c) is non empty iff a is non empty & b is non empty or b is
  non empty & c is non empty or c is non empty & a is non empty;

definition
  let a,b,c be set;
  func NAND3(a,b,c) equals
  :Def12:
  NOT1 {} if not (a is non empty & b is non
  empty & c is non empty) otherwise {};
  correctness;
end;

theorem
  NAND3(a,b,c) is non empty iff not (a is non empty & b is non empty & c
  is non empty) by Def12;

definition
  let a,b,c be set;
  func NOR3(a,b,c) equals
  :Def13:
  NOT1 {} if not (a is non empty or b is non
  empty or c is non empty) otherwise {};
  correctness;
end;

theorem
  NOR3(a,b,c) is non empty iff not (a is non empty or b is non empty or
  c is non empty) by Def13;

definition
  let a,b,c,d be set;
  func AND4(a,b,c,d) equals
  :Def14:
  NOT1 {} if a is non empty & b is non empty
  & c is non empty & d is non empty otherwise {};
  correctness;
end;

theorem
  AND4(a,b,c,d) is non empty iff a is non empty & b is non empty & c is
  non empty & d is non empty by Def14;

definition
  let a,b,c,d be set;
  func OR4(a,b,c,d) equals
  :Def15:
  NOT1 {} if a is non empty or b is non empty
  or c is non empty or d is non empty otherwise {};
  correctness;
end;

theorem
  OR4(a,b,c,d) is non empty iff a is non empty or b is non empty or c is
  non empty or d is non empty by Def15;

definition
  let a,b,c,d be set;
  func NAND4(a,b,c,d) equals
  :Def16:
  NOT1 {} if not (a is non empty & b is non
  empty & c is non empty & d is non empty) otherwise {};
  correctness;
end;

theorem
  NAND4(a,b,c,d) is non empty iff not (a is non empty & b is non empty &
  c is non empty & d is non empty) by Def16;

definition
  let a,b,c,d be set;
  func NOR4(a,b,c,d) equals
  :Def17:
  NOT1 {} if not (a is non empty or b is non
  empty or c is non empty or d is non empty) otherwise {};
  correctness;
end;

theorem
  NOR4(a,b,c,d) is non empty iff not (a is non empty or b is non empty
  or c is non empty or d is non empty) by Def17;

definition
  let a,b,c,d,e be set;
  func AND5(a,b,c,d,e) equals
  :Def18:
  NOT1 {} if a is non empty & b is non
  empty & c is non empty & d is non empty & e is non empty otherwise {};
  correctness;
end;

theorem
  AND5(a,b,c,d,e) is non empty iff a is non empty & b is non empty & c
  is non empty & d is non empty & e is non empty by Def18;

definition
  let a,b,c,d,e be set;
  func OR5(a,b,c,d,e) equals
  :Def19:
  NOT1 {} if a is non empty or b is non
  empty or c is non empty or d is non empty or e is non empty otherwise {};
  correctness;
end;

theorem
  OR5(a,b,c,d,e) is non empty iff a is non empty or b is non empty or c
  is non empty or d is non empty or e is non empty by Def19;

definition
  let a,b,c,d,e be set;
  func NAND5(a,b,c,d,e) equals
  :Def20:
  NOT1 {} if not (a is non empty & b is
  non empty & c is non empty & d is non empty & e is non empty) otherwise {};
  correctness;
end;

theorem
  NAND5(a,b,c,d,e) is non empty iff not (a is non empty & b is non empty
  & c is non empty & d is non empty & e is non empty) by Def20;

definition
  let a,b,c,d,e be set;
  func NOR5(a,b,c,d,e) equals
  :Def21:
  NOT1 {} if not (a is non empty or b is
non empty or c is non empty or d is non empty or e is non empty) otherwise {};
  correctness;
end;

theorem
  NOR5(a,b,c,d,e) is non empty iff not (a is non empty or b is non empty
  or c is non empty or d is non empty or e is non empty) by Def21;

definition
  let a,b,c,d,e,f be set;
  func AND6(a,b,c,d,e,f) equals
  :Def22:
  NOT1 {} if a is non empty & b is non
  empty & c is non empty & d is non empty & e is non empty &f is non empty
  otherwise {};
  correctness;
end;

theorem
  AND6(a,b,c,d,e,f) is non empty iff a is non empty & b is non empty & c
  is non empty & d is non empty & e is non empty &f is non empty by Def22;

definition
  let a,b,c,d,e,f be set;
  func OR6(a,b,c,d,e,f) equals
  :Def23:
  NOT1 {} if a is non empty or b is non
  empty or c is non empty or d is non empty or e is non empty or f is non empty
  otherwise {};
  correctness;
end;

theorem
  OR6(a,b,c,d,e,f) is non empty iff a is non empty or b is non empty or
c is non empty or d is non empty or e is non empty or f is non empty by Def23;

definition
  let a,b,c,d,e,f be set;
  func NAND6(a,b,c,d,e,f) equals
  :Def24:
  NOT1 {} if not (a is non empty & b is
  non empty & c is non empty & d is non empty & e is non empty &f is non empty)
  otherwise {};
  correctness;
end;

theorem
  NAND6(a,b,c,d,e,f) is non empty iff not (a is non empty & b is non
  empty & c is non empty & d is non empty & e is non empty &f is non empty) by
Def24;

definition
  let a,b,c,d,e,f be set;
  func NOR6(a,b,c,d,e,f) equals
  :Def25:
  NOT1 {} if not (a is non empty or b is
  non empty or c is non empty or d is non empty or e is non empty or f is non
  empty) otherwise {};
  correctness;
end;

theorem
  NOR6(a,b,c,d,e,f) is non empty iff not (a is non empty or b is non
empty or c is non empty or d is non empty or e is non empty or f is non empty)
  by Def25;

definition
  let a,b,c,d,e,f,g be set;
  func AND7(a,b,c,d,e,f,g) equals
  :Def26:
  NOT1 {} if a is non empty & b is non
  empty & c is non empty & d is non empty & e is non empty & f is non empty & g
  is non empty otherwise {};
  correctness;
end;

theorem
  AND7(a,b,c,d,e,f,g) is non empty iff a is non empty & b is non empty &
  c is non empty & d is non empty & e is non empty & f is non empty & g is non
  empty by Def26;

definition
  let a,b,c,d,e,f,g be set;
  func OR7(a,b,c,d,e,f,g) equals
  :Def27:
  NOT1 {} if a is non empty or b is non
  empty or c is non empty or d is non empty or e is non empty or f is non empty
  or g is non empty otherwise {};
  correctness;
end;

theorem
  OR7(a,b,c,d,e,f,g) is non empty iff a is non empty or b is non empty
or c is non empty or d is non empty or e is non empty or f is non empty or g is
  non empty by Def27;

definition
  let a,b,c,d,e,f,g be set;
  func NAND7(a,b,c,d,e,f,g) equals
  :Def28:
  NOT1 {} if not (a is non empty & b
  is non empty & c is non empty & d is non empty & e is non empty & f is non
  empty & g is non empty) otherwise {};
  correctness;
end;

theorem
  NAND7(a,b,c,d,e,f,g) is non empty iff not (a is non empty & b is non
  empty & c is non empty & d is non empty & e is non empty & f is non empty & g
  is non empty) by Def28;

definition
  let a,b,c,d,e,f,g be set;
  func NOR7(a,b,c,d,e,f,g) equals
  :Def29:
  NOT1 {} if not (a is non empty or b
is non empty or c is non empty or d is non empty or e is non empty or f is non
  empty or g is non empty) otherwise {};
  correctness;
end;

theorem
  NOR7(a,b,c,d,e,f,g) is non empty iff not (a is non empty or b is non
  empty or c is non empty or d is non empty or e is non empty or f is non empty
  or g is non empty) by Def29;

definition
  let a,b,c,d,e,f,g,h be set;
  func AND8(a,b,c,d,e,f,g,h) equals
  :Def30:
  NOT1 {} if a is non empty & b is
non empty & c is non empty & d is non empty & e is non empty & f is non empty &
  g is non empty & h is non empty otherwise {};
  correctness;
end;

theorem
  AND8(a,b,c,d,e,f,g,h) is non empty iff a is non empty & b is non empty
& c is non empty & d is non empty & e is non empty & f is non empty & g is non
  empty & h is non empty by Def30;

definition
  let a,b,c,d,e,f,g,h be set;
  func OR8(a,b,c,d,e,f,g,h) equals
  :Def31:
  NOT1 {} if a is non empty or b is
  non empty or c is non empty or d is non empty or e is non empty or f is non
  empty or g is non empty or h is non empty otherwise {};
  correctness;
end;

theorem
  OR8(a,b,c,d,e,f,g,h) is non empty iff a is non empty or b is non empty
or c is non empty or d is non empty or e is non empty or f is non empty or g is
  non empty or h is non empty by Def31;

definition
  let a,b,c,d,e,f,g,h be set;
  func NAND8(a,b,c,d,e,f,g,h) equals
  :Def32:
  NOT1 {} if not (a is non empty &
  b is non empty & c is non empty & d is non empty & e is non empty & f is non
  empty & g is non empty & h is non empty) otherwise {};
  correctness;
end;

theorem
  NAND8(a,b,c,d,e,f,g,h) is non empty iff not (a is non empty & b is non
  empty & c is non empty & d is non empty & e is non empty & f is non empty & g
  is non empty & h is non empty) by Def32;

definition
  let a,b,c,d,e,f,g,h be set;
  func NOR8(a,b,c,d,e,f,g,h) equals
  :Def33:
  NOT1 {} if not (a is non empty or
  b is non empty or c is non empty or d is non empty or e is non empty or f is
  non empty or g is non empty or h is non empty) otherwise {};
  correctness;
end;

theorem
  NOR8(a,b,c,d,e,f,g,h) is non empty iff not (a is non empty or b is non
  empty or c is non empty or d is non empty or e is non empty or f is non empty
  or g is non empty or h is non empty) by Def33;

begin 




theorem
  for c1,x1,x2,x3,x4,y1,y2,y3,y4,c2,c3,c4,c5,n1,n2,n3,n4,n,c5b being set
  holds (MAJ3(x1,y1,c1) is non empty implies c2 is non empty) & (MAJ3(x2,y2,c2)
is non empty implies c3 is non empty) & (MAJ3(x3,y3,c3) is non empty implies c4
is non empty) & (MAJ3(x4,y4,c4) is non empty implies c5 is non empty) & (n1 is
non empty implies OR2(x1,y1) is non empty)& (n2 is non empty implies OR2(x2,y2)
  is non empty)& (n3 is non empty implies OR2(x3,y3) is non empty)& (n4 is non
empty implies OR2(x4,y4) is non empty)& (n is non empty implies AND5(c1,n1,n2,
n3,n4) is non empty)& (c5b is non empty iff OR2(c5,n) is non empty) implies (c5
  is non empty iff c5b is non empty)
proof
  let c1,x1,x2,x3,x4,y1,y2,y3,y4,c2,c3,c4,c5,n1,n2,n3,n4,n,c5b be set;
  assume that
A1: ( MAJ3(x1,y1,c1) is non empty implies c2 is non empty)&( MAJ3(x2,y2,
  c2) is non empty implies c3 is non empty) &( MAJ3(x3,y3,c3) is non empty
  implies c4 is non empty) and
A2: MAJ3(x4,y4,c4) is non empty implies c5 is non empty and
A3: n1 is non empty implies OR2(x1,y1) is non empty and
A4: n2 is non empty implies OR2(x2,y2) is non empty and
A5: n3 is non empty implies OR2(x3,y3) is non empty and
A6: n4 is non empty implies OR2(x4,y4) is non empty and
A7: n is non empty implies AND5(c1,n1,n2,n3,n4) is non empty and
A8: c5b is non empty implies OR2(c5,n) is non empty and
A9: OR2(c5,n) is non empty implies c5b is non empty;
A10: now
    assume
A11: n is non empty;
    then
A12: c1 is non empty by A7,Def18;
A13: x3 is non empty or y3 is non empty by A5,A7,A11,Def18;
A14: x2 is non empty or y2 is non empty by A4,A7,A11,Def18;
A15: x4 is non empty or y4 is non empty by A6,A7,A11,Def18;
    x1 is non empty or y1 is non empty by A3,A7,A11,Def18;
    hence MAJ3(x4,y4,c4) is non empty by A1,A12,A14,A13,A15;
  end;
  thus c5 is non empty implies c5b is non empty by A9;
  assume c5b is non empty;
  hence thesis by A2,A8,A10;
end;



definition
  let a,b be set;
  func MODADD2(a,b) equals
  :Def34:
  NOT1 {} if (a is non empty or b is non
  empty) & not(a is non empty & b is non empty) otherwise {};
  correctness;
  commutativity;
end;

theorem
  MODADD2(a,b) is non empty iff (a is non empty or b is non empty) & not
  (a is non empty & b is non empty) by Def34;



notation
  let a,b,c be set;
  synonym ADD1(a,b,c) for XOR3(a,b,c);
  synonym CARR1(a,b,c) for MAJ3(a,b,c);
end;



definition
  let a1,b1,a2,b2,c be set;

  func ADD2(a2,b2,a1,b1,c) equals
  XOR3(a2,b2,CARR1(a1,b1,c));
  coherence;
end;

definition
  let a1,b1,a2,b2,c be set;
  func CARR2(a2,b2,a1,b1,c) equals
  MAJ3(a2,b2,CARR1(a1,b1,c));
  coherence;
end;



definition
  let a1,b1,a2,b2,a3,b3,c be set;
  func ADD3(a3,b3,a2,b2,a1,b1,c) equals
  XOR3(a3,b3,CARR2(a2,b2,a1,b1,c));
  coherence;
end;

definition
  let a1,b1,a2,b2,a3,b3,c be set;
  func CARR3(a3,b3,a2,b2,a1,b1,c) equals
  MAJ3(a3,b3,CARR2(a2,b2,a1,b1,c));
  coherence;
end;



definition
  let a1,b1,a2,b2,a3,b3,a4,b4,c be set;
  func ADD4(a4,b4,a3,b3,a2,b2,a1,b1,c) equals
  XOR3(a4,b4,CARR3(a3,b3,a2,b2,a1,
  b1,c));
  coherence;
end;

definition
  let a1,b1,a2,b2,a3,b3,a4,b4,c be set;
  func CARR4(a4,b4,a3,b3,a2,b2,a1,b1,c) equals
  MAJ3(a4,b4,CARR3(a3,b3,a2,b2,a1
  ,b1,c));
  coherence;
end;






theorem
  for c1,x1,y1,x2,y2,x3,y3,x4,y4,c4, q1,p1,sd1,q2,p2,sd2,q3,p3,sd3,q4,p4
  ,sd4,cb1,cb2,l2,t2,l3,m3, t3,l4,m4,n4,t4,l5,m5,n5,o5,s1,s2,s3,s4 being set
  holds (q1 is non empty iff NOR2(x1,y1) is non empty)& (p1 is non empty iff
NAND2(x1,y1) is non empty)& (sd1 is non empty iff MODADD2(x1,y1) is non empty)&
(q2 is non empty iff NOR2(x2,y2) is non empty)& (p2 is non empty iff NAND2(x2,
  y2) is non empty)& (sd2 is non empty iff MODADD2(x2,y2) is non empty)& (q3 is
  non empty iff NOR2(x3,y3) is non empty)& (p3 is non empty iff NAND2(x3,y3) is
  non empty)& (sd3 is non empty iff MODADD2(x3,y3) is non empty)& (q4 is non
  empty iff NOR2(x4,y4) is non empty)& (p4 is non empty iff NAND2(x4,y4) is non
  empty)& (sd4 is non empty iff MODADD2(x4,y4) is non empty)& (cb1 is non empty
iff NOT1 c1 is non empty)& (cb2 is non empty iff NOT1 cb1 is non empty)& (s1 is
non empty iff XOR2(cb2,sd1) is non empty)& (l2 is non empty iff AND2(cb1,p1) is
  non empty)& (t2 is non empty iff NOR2(l2,q1) is non empty)& (s2 is non empty
iff XOR2(t2,sd2) is non empty)& (l3 is non empty iff AND2(q1,p2) is non empty)&
(m3 is non empty iff AND3(p2,p1,cb1) is non empty)& (t3 is non empty iff NOR3(
  l3,m3,q2) is non empty)& (s3 is non empty iff XOR2(t3,sd3) is non empty)& (l4
is non empty iff AND2(q2,p3) is non empty)& (m4 is non empty iff AND3(q1,p3,p2)
  is non empty)& (n4 is non empty iff AND4(p3,p2,p1,cb1) is non empty)& (t4 is
non empty iff NOR4(l4,m4,n4,q3) is non empty)& (s4 is non empty iff XOR2(t4,sd4
  ) is non empty)& (l5 is non empty iff AND2(q3,p4) is non empty)& (m5 is non
empty iff AND3(q2,p4,p3) is non empty)& (n5 is non empty iff AND4(q1,p4,p3,p2)
is non empty)& (o5 is non empty iff AND5(p4,p3,p2,p1,cb1) is non empty)& (c4 is
  non empty iff NOR5(q4,l5,m5,n5,o5) is non empty) implies (s1 is non empty iff
ADD1(x1,y1,c1) is non empty)& (s2 is non empty iff ADD2(x2,y2,x1,y1,c1) is non
  empty)& (s3 is non empty iff ADD3(x3,y3,x2,y2,x1,y1,c1) is non empty)& (s4 is
non empty iff ADD4(x4,y4,x3,y3,x2,y2,x1,y1,c1) is non empty)& (c4 is non empty
  iff CARR4(x4,y4,x3,y3,x2,y2,x1,y1,c1) is non empty)
proof
  let c1,x1,y1,x2,y2,x3,y3,x4,y4,c4, q1,p1,sd1,q2,p2,sd2,q3,p3,sd3,q4,p4,sd4,
  cb1,cb2,l2,t2,l3,m3, t3,l4,m4,n4,t4,l5,m5,n5,o5,s1,s2,s3,s4 be set;
  assume that
A1: q1 is non empty iff NOR2(x1,y1) is non empty and
A2: p1 is non empty iff NAND2(x1,y1) is non empty and
A3: sd1 is non empty iff MODADD2(x1,y1) is non empty and
A4: q2 is non empty iff NOR2(x2,y2) is non empty and
A5: p2 is non empty iff NAND2(x2,y2) is non empty and
A6: sd2 is non empty iff MODADD2(x2,y2) is non empty and
A7: q3 is non empty iff NOR2(x3,y3) is non empty and
A8: p3 is non empty iff NAND2(x3,y3) is non empty and
A9: sd3 is non empty iff MODADD2(x3,y3) is non empty and
A10: q4 is non empty iff NOR2(x4,y4) is non empty and
A11: p4 is non empty iff NAND2(x4,y4) is non empty and
A12: sd4 is non empty iff MODADD2(x4,y4) is non empty and
A13: cb1 is non empty iff NOT1 c1 is non empty and
A14: ( cb2 is non empty iff NOT1 cb1 is non empty)&( s1 is non empty iff
  XOR2(cb2, sd1) is non empty) and
A15: l2 is non empty iff AND2(cb1,p1) is non empty and
A16: ( t2 is non empty iff NOR2(l2,q1) is non empty)&( s2 is non empty
  iff XOR2(t2, sd2) is non empty) and
A17: l3 is non empty iff AND2(q1,p2) is non empty and
A18: m3 is non empty iff AND3(p2,p1,cb1) is non empty and
A19: ( t3 is non empty iff NOR3(l3,m3,q2) is non empty)&( s3 is non
  empty iff XOR2( t3,sd3) is non empty) and
A20: l4 is non empty iff AND2(q2,p3) is non empty and
A21: ( m4 is non empty iff AND3(q1,p3,p2) is non empty)&( n4 is non
empty iff AND4 ( p3,p2,p1,cb1) is non empty) & ( t4 is non empty iff NOR4(l4,m4
  ,n4,q3) is non empty)&( s4 is non empty iff XOR2(t4,sd4) is non empty) and
A22: l5 is non empty iff AND2(q3,p4) is non empty and
A23: ( m5 is non empty iff AND3(q2,p4,p3) is non empty)&( n5 is non
empty iff AND4 ( q1,p4,p3,p2) is non empty) &( o5 is non empty iff AND5( p4,p3,
  p2,p1,cb1) is non empty) and
A24: c4 is non empty iff NOR5( q4,l5,m5,n5,o5) is non empty;
A25: p1 is non empty iff not (x1 is non empty & y1 is non empty) by A2;
A26: sd1 is non empty iff (x1 is non empty or y1 is non empty) & not(x1 is
  non empty & y1 is non empty) by A3,Def34;
  hereby
    assume
A27: s1 is non empty;
    per cases by A13,A14,A27;
    suppose
      c1 is non empty & sd1 is empty;
      hence ADD1(x1,y1,c1) is non empty by A26;
    end;
    suppose
      c1 is empty & sd1 is non empty;
      hence ADD1(x1,y1,c1) is non empty by A26;
    end;
  end;
  hereby
    assume
A28: ADD1(x1,y1,c1) is non empty;
    per cases by A26,A28;
    suppose
      c1 is non empty & sd1 is empty;
      hence s1 is non empty by A13,A14;
    end;
    suppose
      c1 is empty & sd1 is non empty;
      hence s1 is non empty by A13,A14;
    end;
  end;
A29: q1 is non empty iff not (x1 is non empty or y1 is non empty) by A1;
A30: sd2 is non empty iff (x2 is non empty or y2 is non empty) & not(x2 is
  non empty & y2 is non empty) by A6,Def34;
  hereby
    assume
A31: s2 is non empty;
    per cases by A13,A15,A16,A31;
    suppose
      (c1 is non empty or p1 is empty) & q1 is empty & sd2 is empty;
      hence ADD2(x2,y2,x1,y1,c1) is non empty by A29,A25,A30;
    end;
    suppose
      (c1 is empty & p1 is non empty or q1 is non empty) & sd2 is non empty;
      hence ADD2(x2,y2,x1,y1,c1) is non empty by A29,A25,A30;
    end;
  end;
A32: cb1 is non empty iff not c1 is non empty by A13;
  hereby
    assume
A33: ADD2(x2,y2,x1,y1,c1) is non empty;
    per cases by A15,A29,A25,A30,A32,A33;
    suppose
      l2 is empty & q1 is empty & sd2 is empty;
      hence s2 is non empty by A16;
    end;
    suppose
      (l2 is non empty or q1 is non empty) & sd2 is non empty;
      hence s2 is non empty by A16;
    end;
  end;
A34: q2 is non empty iff not(x2 is non empty or y2 is non empty) by A4;
A35: p2 is non empty iff not(x2 is non empty & y2 is non empty) by A5;
A36: sd3 is non empty iff (x3 is non empty or y3 is non empty) & not(x3 is
  non empty & y3 is non empty) by A9,Def34;
  hereby
    assume
A37: s3 is non empty;
    per cases by A9,A19,A37,Def13,Def34;
    suppose
      l3 is empty & m3 is empty & q2 is empty & x3 is empty & y3 is empty;
      hence
      ADD3(x3,y3,x2,y2,x1,y1,c1) is non empty by A17,A18,A29,A25,A34,A35,A32;
    end;
    suppose
      l3 is empty & m3 is empty & q2 is empty & x3 is non empty & y3
      is non empty;
      hence
      ADD3(x3,y3,x2,y2,x1,y1,c1) is non empty by A17,A18,A29,A25,A34,A35,A32;
    end;
    suppose
A38:  l3 is non empty & sd3 is non empty;
      then
A39:  x2 is empty or y2 is empty by A5,A17;
      x1 is empty & y1 is empty by A1,A17,A38;
      hence ADD3(x3,y3,x2,y2,x1,y1,c1) is non empty by A36,A38,A39;
    end;
    suppose
A40:  m3 is non empty & sd3 is non empty;
      then c1 is empty by A13,A18;
      hence ADD3(x3,y3,x2,y2,x1,y1,c1) is non empty by A18,A25,A35,A36,A40;
    end;
    suppose
      q2 is non empty & sd3 is non empty;
      hence ADD3(x3,y3,x2,y2,x1,y1,c1) is non empty by A34,A36;
    end;
  end;
  hereby
    assume
A41: ADD3(x3,y3,x2,y2,x1,y1,c1) is non empty;
    per cases by A17,A18,A29,A25,A34,A35,A36,A32,A41;
    suppose
      l3 is empty & m3 is empty & q2 is empty & x3 is empty & y3 is empty;
      hence s3 is non empty by A9,A19,Def13,Def34;
    end;
    suppose
      l3 is empty & m3 is empty & q2 is empty & x3 is non empty & y3
      is non empty;
      hence s3 is non empty by A9,A19,Def13,Def34;
    end;
    suppose
      l3 is non empty & sd3 is non empty;
      hence s3 is non empty by A19,Def13;
    end;
    suppose
      m3 is non empty & sd3 is non empty;
      hence s3 is non empty by A19,Def13;
    end;
    suppose
      q2 is non empty & sd3 is non empty;
      hence s3 is non empty by A19,Def13;
    end;
  end;
A42: p3 is non empty iff not(x3 is non empty & y3 is non empty) by A8;
A43: sd4 is non empty iff (x4 is non empty or y4 is non empty) & not(x4 is
  non empty & y4 is non empty) by A12,Def34;
A44: q3 is non empty iff not(x3 is non empty or y3 is non empty) by A7;
  hereby
    assume
A45: s4 is non empty;
    per cases by A13,A21,A45,Def14,Def17;
    suppose
      l4 is empty & (q1 is empty or p3 is empty or p2 is empty) & (p3
is empty or p2 is empty or p1 is empty or c1 is non empty) & q3 is empty & sd4
      is empty;
      hence ADD4(x4,y4,x3,y3,x2,y2,x1,y1,c1) is non empty by A20,A29,A25,A34
,A35,A44,A42,A43;
    end;
    suppose
      l4 is non empty & sd4 is non empty;
      hence ADD4(x4,y4,x3,y3,x2,y2,x1,y1,c1) is non empty by A20,A34,A42,A43;
    end;
    suppose
      q1 is non empty & p3 is non empty & p2 is non empty & sd4 is non empty;
      hence ADD4(x4,y4,x3,y3,x2,y2,x1,y1,c1) is non empty by A29,A35,A42,A43;
    end;
    suppose
      p3 is non empty & p2 is non empty & p1 is non empty & c1 is
      empty & sd4 is non empty;
      hence ADD4(x4,y4,x3,y3,x2,y2,x1,y1,c1) is non empty by A25,A35,A42,A43;
    end;
    suppose
      q3 is non empty &sd4 is non empty;
      hence ADD4(x4,y4,x3,y3,x2,y2,x1,y1,c1) is non empty by A44,A43;
    end;
  end;
  hereby
    assume
A46: ADD4(x4,y4,x3,y3,x2,y2,x1,y1,c1) is non empty;
    thus s4 is non empty
    proof
      assume
A47:  s4 is empty;
      per cases by A13,A21,A47,Def14,Def17;
      suppose
        l4 is empty & (q1 is empty or p3 is empty or p2 is empty) & (
p3 is empty or p2 is empty or p1 is empty or c1 is non empty) & q3 is empty &
        sd4 is non empty;
        hence thesis by A20,A29,A25,A34,A35,A44,A42,A43,A46;
      end;
      suppose
        l4 is non empty & sd4 is empty;
        hence thesis by A20,A34,A42,A43,A46;
      end;
      suppose
        q1 is non empty & p3 is non empty & p2 is non empty & sd4 is empty;
        hence thesis by A29,A35,A42,A43,A46;
      end;
      suppose
        p3 is non empty & p2 is non empty & p1 is non empty & c1 is
        empty & sd4 is empty;
        hence thesis by A25,A35,A42,A43,A46;
      end;
      suppose
        q3 is non empty &sd4 is empty;
        hence thesis by A44,A43,A46;
      end;
    end;
  end;
A48: p4 is non empty iff not(x4 is non empty & y4 is non empty) by A11;
A49: q4 is non empty iff not(x4 is non empty or y4 is non empty) by A10;
  now
    assume that
A50: q4 is empty and
A51: l5 is empty and
A52: m5 is empty & n5 is empty & o5 is empty;
    per cases by A13,A23,A52,Def14,Def18;
    suppose
      p4 is empty;
      hence CARR4(x4,y4,x3,y3,x2,y2,x1,y1,c1) is non empty by A48;
    end;
    suppose
      p3 is empty;
      hence CARR4(x4,y4,x3,y3,x2,y2,x1,y1,c1) is non empty by A42,A49,A50;
    end;
    suppose
      p2 is empty & q2 is empty;
      hence CARR4(x4,y4,x3,y3,x2,y2,x1,y1,c1) is non empty by A22,A35,A44,A49
,A48,A50,A51;
    end;
    suppose
      p1 is empty & q2 is empty;
      hence CARR4(x4,y4,x3,y3,x2,y2,x1,y1,c1) is non empty by A22,A25,A34,A44
,A49,A48,A50,A51;
    end;
    suppose
      c1 is non empty & q1 is empty& q2 is empty;
      hence CARR4(x4,y4,x3,y3,x2,y2,x1,y1,c1) is non empty by A22,A29,A34,A44
,A49,A48,A50,A51;
    end;
  end;
  hence c4 is non empty implies CARR4(x4,y4,x3,y3,x2,y2,x1,y1,c1) is non empty
  by A24,Def21;
  assume
A53: CARR4(x4,y4,x3,y3,x2,y2,x1,y1,c1) is non empty;
  assume
A54: c4 is empty;
  per cases by A13,A22,A23,A24,A54,Def14,Def18,Def21;
  suppose
    q4 is non empty;
    hence thesis by A49,A53;
  end;
  suppose
    q3 is non empty & p4 is non empty;
    hence thesis by A44,A48,A53;
  end;
  suppose
    q2 is non empty & p4 is non empty & p3 is non empty;
    hence thesis by A34,A42,A48,A53;
  end;
  suppose
    q1 is non empty & p4 is non empty & p3 is non empty & p2 is non empty;
    hence thesis by A29,A35,A42,A48,A53;
  end;
  suppose
    p4 is non empty & p3 is non empty & p2 is non empty & p1 is non
    empty & c1 is empty;
    hence thesis by A25,A35,A42,A48,A53;
  end;
end;














begin 

reserve a,b,c,d for set;




theorem
  for s0,s1,s2,s3,s4,s5,s6,s7,ns0,ns1,ns2,ns3,ns4,ns5,ns6,ns7,q1,q2,q3,
nq1, nq2,nq3 being set holds (s0 is non empty iff AND3(NOT1 q3,NOT1 q2,NOT1 q1)
is not empty)& (s1 is non empty iff AND3(NOT1 q3,NOT1 q2,q1) is not empty)& (s2
  is non empty iff AND3(NOT1 q3,q2,NOT1 q1) is not empty)& (s3 is non empty iff
AND3(NOT1 q3,q2,q1) is not empty)& (s4 is non empty iff AND3(q3,NOT1 q2,NOT1 q1
) is not empty)& (s5 is non empty iff AND3(q3,NOT1 q2,q1) is not empty)& (s6 is
non empty iff AND3(q3,q2,NOT1 q1) is not empty)& (s7 is non empty iff AND3(q3,
q2,q1) is not empty)& (ns0 is non empty iff AND3(NOT1 nq3,NOT1 nq2,NOT1 nq1) is
not empty)& (ns1 is non empty iff AND3(NOT1 nq3,NOT1 nq2,nq1) is not empty)& (
  ns2 is non empty iff AND3(NOT1 nq3,nq2,NOT1 nq1) is not empty)& (ns3 is non
empty iff AND3(NOT1 nq3,nq2,nq1) is not empty)& (ns4 is non empty iff AND3(nq3,
NOT1 nq2,NOT1 nq1) is not empty)& (ns5 is non empty iff AND3(nq3,NOT1 nq2,nq1)
is not empty)& (ns6 is non empty iff AND3(nq3,nq2,NOT1 nq1) is not empty)& (ns7
  is non empty iff AND3(nq3,nq2,nq1) is not empty)& (nq1 is non empty iff (NOT1
q1) is not empty)& (nq2 is non empty iff XOR2(q1,q2) is not empty)& (nq3 is non
empty iff OR2(AND2(q3, NOT1 q1), AND2(q1, XOR2(q2,q3))) is not empty) implies (
ns1 is non empty iff s0 is non empty)& (ns2 is non empty iff s1 is non empty)&
(ns3 is non empty iff s2 is non empty)& (ns4 is non empty iff s3 is non empty)&
(ns5 is non empty iff s4 is non empty)& (ns6 is non empty iff s5 is non empty)&
(ns7 is non empty iff s6 is non empty)& (ns0 is non empty iff s7 is non empty)
proof
  let s0,s1,s2,s3,s4,s5,s6,s7,ns0,ns1,ns2,ns3,ns4,ns5,ns6,ns7,q1,q2,q3,nq1,
  nq2,nq3 be set;
  assume that
A1: s0 is non empty iff AND3(NOT1 q3,NOT1 q2,NOT1 q1) is not empty and
A2: s1 is non empty iff AND3(NOT1 q3,NOT1 q2,q1) is not empty and
A3: s2 is non empty iff AND3(NOT1 q3,q2,NOT1 q1) is not empty and
A4: s3 is non empty iff AND3(NOT1 q3,q2,q1) is not empty and
A5: s4 is non empty iff AND3(q3,NOT1 q2,NOT1 q1) is not empty and
A6: s5 is non empty iff AND3(q3,NOT1 q2,q1) is not empty and
A7: s6 is non empty iff AND3(q3,q2,NOT1 q1) is not empty and
A8: s7 is non empty iff AND3(q3,q2,q1) is not empty and
A9: ns0 is non empty iff AND3(NOT1 nq3,NOT1 nq2,NOT1 nq1) is not empty and
A10: ns1 is non empty iff AND3(NOT1 nq3,NOT1 nq2,nq1) is not empty and
A11: ns2 is non empty iff AND3(NOT1 nq3,nq2,NOT1 nq1) is not empty and
A12: ns3 is non empty iff AND3(NOT1 nq3,nq2,nq1) is not empty and
A13: ns4 is non empty iff AND3(nq3,NOT1 nq2,NOT1 nq1) is not empty and
A14: ns5 is non empty iff AND3(nq3,NOT1 nq2,nq1) is not empty and
A15: ns6 is non empty iff AND3(nq3,nq2,NOT1 nq1) is not empty and
A16: ns7 is non empty iff AND3(nq3,nq2,nq1) is not empty and
A17: ( nq1 is non empty iff NOT1 q1 is not empty)&( nq2 is non empty iff
  XOR2(q1,q2) is not empty) and
A18: nq3 is non empty iff OR2(AND2(q3, NOT1 q1), AND2(q1, XOR2(q2,q3)))
  is not empty;
  ns1 is non empty iff NOT1 nq3 is non empty & not XOR2(q1,q2) is not
  empty & not q1 is non empty by A10,A17;
  then ns1 is non empty iff (not q3 is not empty or not NOT1 q1 is not empty)
& (not q1 is not empty or not XOR2(q2,q3) is not empty) & not q2 is non empty &
  not q1 is non empty by A18;
  hence ns1 is non empty iff s0 is non empty by A1;
  ns2 is non empty iff not AND2(q3, NOT1 q1) is non empty & not AND2(q1,
  XOR2(q2,q3)) is non empty & not q2 is non empty & q1 is non empty by A11,A17
,A18,GATE_1:6;
  then ns2 is non empty iff (not q3 is non empty or q1 is non empty) & not q3
  is non empty & not q2 is non empty & q1 is non empty;
  hence ns2 is non empty iff s1 is non empty by A2;
  ns3 is non empty iff NOT1 nq3 is non empty & XOR2(q1,q2) is non empty &
  not q1 is non empty by A12,A17;
  then ns3 is non empty iff (not q3 is non empty or not NOT1 q1 is non empty)
& (not q1 is non empty or not XOR2(q2,q3) is non empty) & q2 is non empty & not
  q1 is non empty by A18;
  hence ns3 is non empty iff s2 is non empty by A3;
  ns4 is non empty iff (AND2(q3, NOT1 q1) is non empty or AND2(q1, XOR2(
q2,q3)) is non empty) & q2 is non empty & q1 is non empty by A13,A17,A18,
GATE_1:6;
  then ns4 is non empty iff not q3 is non empty & q2 is non empty & q1 is non
  empty;
  hence ns4 is non empty iff s3 is non empty by A4;
  ns5 is non empty iff nq3 is non empty & not XOR2(q1,q2) is non empty &
  not q1 is non empty by A14,A17;
  then ns5 is non empty iff (AND2(q3, NOT1 q1) is non empty or AND2(q1, XOR2(
  q2,q3)) is non empty) & not q2 is non empty & not q1 is non empty by A18;
  then ns5 is non empty iff q3 is non empty & NOT1 q2 is non empty & NOT1 q1
  is non empty;
  hence ns5 is non empty iff s4 is non empty by A5;
  ns6 is non empty iff (AND2(q3, NOT1 q1) is non empty or AND2(q1, XOR2(
  q2,q3)) is non empty) & not q2 is non empty & q1 is non empty by A15,A17,A18,
GATE_1:6;
  then ns6 is non empty iff q3 is non empty & not q2 is non empty & q1 is non
  empty;
  hence ns6 is non empty iff s5 is non empty by A6;
  ns7 is non empty iff nq3 is non empty & XOR2(q1,q2) is non empty & not
  q1 is non empty by A16,A17;
  then ns7 is non empty iff (AND2(q3, NOT1 q1) is non empty or AND2(q1, XOR2(
  q2,q3)) is non empty) & q2 is non empty & not q1 is non empty by A18;
  then
  ns7 is non empty iff q3 is non empty & q2 is non empty & NOT1 q1 is non
  empty;
  hence ns7 is non empty iff s6 is non empty by A7;
  ns0 is non empty iff not AND2(q3, NOT1 q1) is non empty & not AND2(q1,
  XOR2(q2,q3)) is non empty & q2 is non empty & q1 is non empty by A9,A17,A18,
GATE_1:6;
  then ns0 is non empty iff (not q3 is non empty or q1 is non empty) & q3 is
  non empty & q2 is non empty & q1 is non empty;
  hence thesis by A8;
end;

theorem
  AND3(AND2(a,d),AND2(b,d),AND2(c,d)) is non empty iff AND2(AND3(a,b,c),
  d) is non empty
proof
  a is non empty & b is non empty & c is non empty & d is non empty iff
  AND3(a,b,c) is non empty & d is non empty;
  hence thesis;
end;

theorem
  (not AND2(a,b) is non empty iff OR2(NOT1 a, NOT1 b) is non empty)& (
OR2(a,b) is non empty & OR2(c,b) is non empty iff OR2(AND2(a,c),b) is non empty
) & (OR2(a,b) is non empty & OR2(c,b) is non empty & OR2(d,b) is non empty iff
OR2(AND3(a,c,d),b) is non empty)& (OR2(a,b) is non empty & (a is non empty iff
  c is non empty) implies OR2(c,b) is non empty )
proof
A1: OR2(a,b) is non empty & OR2(c,b) is non empty iff (a is non empty or b
  is non empty) & (c is non empty or b is non empty);
A2: OR2(a,b) is non empty & OR2(c,b) is non empty & OR2(d,b) is non empty
iff (a is non empty or b is non empty ) & (c is non empty or b is non empty ) &
  (d is non empty or b is non empty );
  not AND2(a,b) is non empty iff not (a is non empty & b is non empty);
  hence thesis by A1,A2;
end;





theorem
  for s0,s1,s2,s3,s4,s5,s6,s7,ns0,ns1,ns2,ns3,ns4,ns5,ns6,ns7,q1,q2,q3,
  nq1, nq2,nq3,R being set holds (s0 is non empty iff AND3(NOT1 q3,NOT1 q2,NOT1
q1) is non empty)& (s1 is non empty iff AND3(NOT1 q3,NOT1 q2,q1) is non empty)&
  (s2 is non empty iff AND3(NOT1 q3,q2,NOT1 q1) is non empty)& (s3 is non empty
  iff AND3(NOT1 q3,q2,q1) is non empty)& (s4 is non empty iff AND3(q3,NOT1 q2,
NOT1 q1) is non empty)& (s5 is non empty iff AND3(q3,NOT1 q2,q1) is non empty)&
  (s6 is non empty iff AND3(q3,q2,NOT1 q1) is non empty)& (s7 is non empty iff
AND3(q3,q2,q1) is non empty)& (ns0 is non empty iff AND3(NOT1 nq3,NOT1 nq2,NOT1
  nq1) is non empty)& (ns1 is non empty iff AND3(NOT1 nq3,NOT1 nq2,nq1) is non
  empty)& (ns2 is non empty iff AND3(NOT1 nq3,nq2,NOT1 nq1) is non empty)& (ns3
  is non empty iff AND3(NOT1 nq3,nq2,nq1) is non empty)& (ns4 is non empty iff
AND3(nq3,NOT1 nq2,NOT1 nq1) is non empty)& (ns5 is non empty iff AND3(nq3,NOT1
  nq2,nq1) is non empty)& (ns6 is non empty iff AND3(nq3,nq2,NOT1 nq1) is non
  empty)& (ns7 is non empty iff AND3(nq3,nq2,nq1) is non empty)& (nq1 is non
empty iff AND2(NOT1 q1,R) is non empty)& (nq2 is non empty iff AND2(XOR2(q1,q2)
  ,R) is non empty)& (nq3 is non empty iff AND2(OR2(AND2(q3, NOT1 q1), AND2(q1,
XOR2(q2,q3))),R) is non empty) implies (ns1 is non empty iff AND2(s0,R) is non
  empty)& (ns2 is non empty iff AND2(s1,R) is non empty)& (ns3 is non empty iff
AND2(s2,R) is non empty)& (ns4 is non empty iff AND2(s3,R) is non empty)& (ns5
is non empty iff AND2(s4,R) is non empty)& (ns6 is non empty iff AND2(s5,R) is
  non empty)& (ns7 is non empty iff AND2(s6,R) is non empty)& (ns0 is non empty
  iff OR2(s7,NOT1 R) is non empty)
proof
  let s0,s1,s2,s3,s4,s5,s6,s7,ns0,ns1,ns2,ns3,ns4,ns5,ns6,ns7,q1,q2,q3,nq1,
  nq2,nq3,R be set;
  assume that
A1: s0 is non empty iff AND3(NOT1 q3,NOT1 q2,NOT1 q1) is non empty and
A2: s1 is non empty iff AND3(NOT1 q3,NOT1 q2,q1) is non empty and
A3: s2 is non empty iff AND3(NOT1 q3,q2,NOT1 q1) is non empty and
A4: s3 is non empty iff AND3(NOT1 q3,q2,q1) is non empty and
A5: s4 is non empty iff AND3(q3,NOT1 q2,NOT1 q1) is non empty and
A6: s5 is non empty iff AND3(q3,NOT1 q2,q1) is non empty and
A7: s6 is non empty iff AND3(q3,q2,NOT1 q1) is non empty and
A8: s7 is non empty iff AND3(q3,q2,q1) is non empty and
A9: ns0 is non empty iff AND3(NOT1 nq3,NOT1 nq2,NOT1 nq1) is non empty and
A10: ns1 is non empty iff AND3(NOT1 nq3,NOT1 nq2,nq1) is non empty and
A11: ns2 is non empty iff AND3(NOT1 nq3,nq2,NOT1 nq1) is non empty and
A12: ns3 is non empty iff AND3(NOT1 nq3,nq2,nq1) is non empty and
A13: ns4 is non empty iff AND3(nq3,NOT1 nq2,NOT1 nq1) is non empty and
A14: ns5 is non empty iff AND3(nq3,NOT1 nq2,nq1) is non empty and
A15: ns6 is non empty iff AND3(nq3,nq2,NOT1 nq1) is non empty and
A16: ns7 is non empty iff AND3(nq3,nq2,nq1) is non empty and
A17: nq1 is non empty iff AND2(NOT1 q1,R) is non empty and
A18: nq2 is non empty iff AND2(XOR2(q1,q2),R) is non empty and
A19: nq3 is non empty iff AND2(OR2(AND2(q3, NOT1 q1), AND2(q1, XOR2(q2,
  q3))),R) is non empty;
  ns1 is non empty iff not nq3 is non empty & ( not XOR2(q1,q2) is non
empty or not R is non empty) & not q1 is non empty & R is non empty by A10,A17
,A18;
  then ns1 is non empty iff (not q3 is non empty or not NOT1 q1 is non empty)
& (not q1 is non empty or not XOR2(q2,q3) is non empty) & not q2 is non empty &
  not q1 is non empty & R is non empty by A19;
  hence ns1 is non empty iff AND2(s0,R) is non empty by A1;
  ns2 is non empty iff not nq3 is non empty & XOR2(q1,q2) is non empty &
  R is non empty & (not NOT1 q1 is non empty or not R is non empty) by A11,A17
,A18;
  then ns2 is non empty iff not nq3 is non empty & XOR2(q1,q2) is non empty &
  R is non empty & q1 is non empty;
  then ns2 is non empty iff not AND2(q3, NOT1 q1) is non empty & not AND2(q1,
  XOR2(q2,q3)) is non empty & not q2 is non empty & q1 is non empty & R is non
  empty by A19;
  then ns2 is non empty iff (not q3 is non empty or q1 is non empty) & not q3
  is non empty & not q2 is non empty & q1 is non empty & R is non empty;
  hence ns2 is non empty iff AND2(s1,R) is non empty by A2;
  ns3 is non empty iff not nq3 is non empty & XOR2(q1,q2) is non empty &
  R is non empty & not q1 is non empty & R is non empty by A12,A17,A18;
  then ns3 is non empty iff (not q3 is non empty or not NOT1 q1 is non empty)
& (not q1 is non empty or not XOR2(q2,q3) is non empty) & q2 is non empty & not
  q1 is non empty & R is non empty by A19;
  hence ns3 is non empty iff AND2(s2,R) is non empty by A3;
  ns4 is non empty iff OR2(AND2(q3, NOT1 q1), AND2(q1, XOR2(q2,q3))) is
non empty & R is non empty & not nq2 is non empty & not AND2(NOT1 q1,R) is non
  empty by A13,A17,A19;
  then ns4 is non empty iff OR2(AND2(q3, NOT1 q1), AND2(q1, XOR2(q2,q3))) is
non empty & R is non empty & not XOR2(q1,q2) is non empty & q1 is non empty by
A18;
  then ns4 is non empty iff ( q3 is non empty & NOT1 q1 is non empty or q1 is
non empty & XOR2(q2,q3) is non empty ) & R is non empty & q2 is non empty & q1
  is non empty;
  then ns4 is non empty iff not q3 is non empty & R is non empty & q2 is non
  empty & q1 is non empty;
  hence ns4 is non empty iff AND2(s3,R) is non empty by A4;
  ns5 is non empty iff nq3 is non empty & (not XOR2(q1,q2) is non empty
  or not R is non empty) & not q1 is non empty & R is non empty by A14,A17,A18;
  then ns5 is non empty iff OR2(AND2(q3, NOT1 q1), AND2(q1, XOR2(q2,q3))) is
non empty & not q2 is non empty & not q1 is non empty & R is non empty by A19;
  then ns5 is non empty iff q3 is non empty & NOT1 q2 is non empty & NOT1 q1
  is non empty & R is non empty;
  hence ns5 is non empty iff AND2(s4,R) is non empty by A5;
  ns6 is non empty iff nq3 is non empty & XOR2(q1,q2) is non empty & R is
  non empty & (not NOT1 q1 is non empty or not R is non empty) by A15,A17,A18;
  then
  ns6 is non empty iff nq3 is non empty & XOR2(q1,q2) is non empty & R is
  non empty & q1 is non empty;
  then ns6 is non empty iff ( q3 is non empty & NOT1 q1 is non empty or q1 is
non empty & XOR2(q2,q3) is non empty ) & R is non empty & not q2 is non empty &
  q1 is non empty by A19;
  then ns6 is non empty iff q3 is non empty & R is non empty & not q2 is non
  empty & q1 is non empty;
  hence ns6 is non empty iff AND2(s5,R) is non empty by A6;
  ns7 is non empty iff nq3 is non empty & XOR2(q1,q2) is non empty & not
  q1 is non empty & R is non empty by A16,A17,A18;
  then ns7 is non empty iff OR2(AND2(q3, NOT1 q1), AND2(q1, XOR2(q2,q3))) is
  non empty & q2 is non empty & not q1 is non empty & R is non empty by A19;
  then
  ns7 is non empty iff q3 is non empty & q2 is non empty & NOT1 q1 is non
  empty & R is non empty;
  hence ns7 is non empty iff AND2(s6,R) is non empty by A7;
  ns0 is non empty iff q1 is non empty & not XOR2(q1,q2) is non empty &
not OR2(AND2(q3, NOT1 q1), AND2(q1, XOR2(q2,q3))) is non empty or not R is non
  empty by A9,A17,A18,A19;
  then ns0 is non empty iff q1 is non empty & q2 is non empty & not AND2(q3,
  NOT1 q1) is non empty & not XOR2(q2,q3) is non empty or not R is non empty;
  then ns0 is non empty iff q1 is non empty & q2 is non empty & q1 is non
  empty & q3 is non empty or not R is non empty;
  hence thesis by A8;
end;














begin 




theorem 
  for s0,s1,s2,s3,ns0,ns1,ns2,ns3,q1,q2,nq1,nq2 being set holds (s0 is
not empty iff AND2(NOT1 q2, NOT1 q1) is not empty)& (s1 is not empty iff AND2(
  NOT1 q2, q1) is not empty)& (s2 is not empty iff AND2( q2, NOT1 q1) is not
  empty)& (s3 is not empty iff AND2( q2, q1) is not empty) & (ns0 is not empty
iff AND2(NOT1 nq2,NOT1 nq1) is not empty)& (ns1 is not empty iff AND2(NOT1 nq2,
nq1) is not empty)& (ns2 is not empty iff AND2( nq2,NOT1 nq1) is not empty)& (
ns3 is not empty iff AND2( nq2, nq1) is not empty) & (nq1 is not empty iff NOT1
  q2 is not empty)& (nq2 is not empty iff q1 is not empty) implies (ns1 is not
empty iff s0 is not empty)& (ns3 is not empty iff s1 is not empty)& (ns2 is not
  empty iff s3 is not empty)& (ns0 is not empty iff s2 is not empty)
proof
  let s0,s1,s2,s3,ns0,ns1,ns2,ns3,q1,q2,nq1,nq2 be set;
  assume that
A1: s0 is not empty iff AND2(NOT1 q2,NOT1 q1) is not empty and
A2: s1 is not empty iff AND2(NOT1 q2,q1) is not empty and
A3: s2 is not empty iff AND2(q2,NOT1 q1) is not empty and
A4: s3 is not empty iff AND2(q2,q1) is not empty and
A5: ns0 is not empty iff AND2(NOT1 nq2,NOT1 nq1) is not empty and
A6: ns1 is not empty iff AND2(NOT1 nq2,nq1) is not empty and
A7: ns2 is not empty iff AND2(nq2,NOT1 nq1) is not empty and
A8: ns3 is not empty iff AND2(nq2,nq1) is not empty and
A9: ( nq1 is not empty iff NOT1 q2 is not empty)&( nq2 is not empty iff
  q1 is not empty);
  thus ns1 is not empty iff s0 is not empty by A1,A6,A9;
  thus ns3 is not empty iff s1 is not empty by A2,A8,A9;
  ns2 is not empty iff q2 is not empty & q1 is not empty by A7,A9;
  hence ns2 is not empty iff s3 is not empty by A4;
  ns0 is not empty iff q2 is not empty & NOT1 q1 is not empty by A5,A9;
  hence thesis by A3;
end;






theorem 
  for s0,s1,s2,s3,ns0,ns1,ns2,ns3,q1,q2,nq1,nq2,R being set holds (s0 is
not empty iff AND2(NOT1 q2, NOT1 q1) is not empty)& (s1 is not empty iff AND2(
  NOT1 q2, q1) is not empty)& (s2 is not empty iff AND2( q2, NOT1 q1) is not
  empty)& (s3 is not empty iff AND2( q2, q1) is not empty) & (ns0 is not empty
iff AND2(NOT1 nq2,NOT1 nq1) is not empty)& (ns1 is not empty iff AND2(NOT1 nq2,
nq1) is not empty)& (ns2 is not empty iff AND2( nq2,NOT1 nq1) is not empty)& (
ns3 is not empty iff AND2( nq2, nq1) is not empty) & (nq1 is not empty iff AND2
  (NOT1 q2, R) is not empty)& (nq2 is not empty iff AND2( q1, R) is not empty)
implies (ns1 is not empty iff AND2(s0, R) is not empty)& (ns3 is not empty iff
  AND2(s1, R) is not empty)& (ns2 is not empty iff AND2(s3, R) is not empty)& (
  ns0 is not empty iff OR2(AND2(s2, R),NOT1 R) is not empty)
proof
  let s0,s1,s2,s3,ns0,ns1,ns2,ns3,q1,q2,nq1,nq2,R be set;
  assume that
A1: s0 is not empty iff AND2(NOT1 q2,NOT1 q1) is not empty and
A2: s1 is not empty iff AND2(NOT1 q2,q1) is not empty and
A3: s2 is not empty iff AND2(q2,NOT1 q1) is not empty and
A4: s3 is not empty iff AND2(q2,q1) is not empty and
A5: ns0 is not empty iff AND2(NOT1 nq2,NOT1 nq1) is not empty and
A6: ns1 is not empty iff AND2(NOT1 nq2,nq1) is not empty and
A7: ns2 is not empty iff AND2(nq2,NOT1 nq1) is not empty and
A8: ns3 is not empty iff AND2(nq2,nq1) is not empty and
A9: ( nq1 is not empty iff AND2(NOT1 q2,R) is not empty)&( nq2 is not
  empty iff AND2( q1,R) is not empty);
  ns1 is not empty iff NOT1 q2 is not empty & R is not empty & not(q1 is
  not empty & R is not empty) by A6,A9;
  hence ns1 is not empty iff AND2(s0,R) is not empty by A1;
  ns3 is not empty iff NOT1 q2 is not empty & q1 is not empty & R is not
  empty by A8,A9;
  hence ns3 is not empty iff AND2(s1,R) is not empty by A2;
  ns2 is not empty iff not(NOT1 q2 is not empty & R is not empty) & q1 is
  not empty & R is not empty by A7,A9;
  then
  ns2 is not empty iff q2 is not empty & q1 is not empty & R is not empty;
  hence ns2 is not empty iff AND2(s3,R) is not empty by A4;
  ns0 is not empty iff not(NOT1 q2 is not empty & R is not empty) & not(
  q1 is not empty & R is not empty) by A5,A9;
  then ns0 is not empty iff q2 is not empty & NOT1 q1 is not empty & R is not
  empty or not R is not empty;
  hence thesis by A3;
end;






theorem 
  for s0,s1,s2,s3,s4,s5,s6,s7,ns0,ns1,ns2,ns3,ns4,ns5,ns6,ns7, q1,q2,q3,
nq1,nq2,nq3 being set holds (s0 is not empty iff AND3(NOT1 q3, NOT1 q2, NOT1 q1
) is not empty)& (s1 is not empty iff AND3(NOT1 q3, NOT1 q2, q1) is not empty)&
(s2 is not empty iff AND3(NOT1 q3, q2, NOT1 q1) is not empty)& (s3 is not empty
iff AND3(NOT1 q3, q2, q1) is not empty)& (s4 is not empty iff AND3( q3, NOT1 q2
  , NOT1 q1) is not empty)& (s5 is not empty iff AND3( q3, NOT1 q2, q1) is not
  empty)& (s6 is not empty iff AND3( q3, q2, NOT1 q1) is not empty)& (s7 is not
empty iff AND3( q3, q2, q1) is not empty) & (ns0 is not empty iff AND3(NOT1 nq3
,NOT1 nq2,NOT1 nq1) is not empty)& (ns1 is not empty iff AND3(NOT1 nq3,NOT1 nq2
, nq1) is not empty)& (ns2 is not empty iff AND3(NOT1 nq3, nq2,NOT1 nq1) is not
  empty)& (ns3 is not empty iff AND3(NOT1 nq3, nq2, nq1) is not empty)& (ns4 is
not empty iff AND3( nq3,NOT1 nq2,NOT1 nq1) is not empty)& (ns5 is not empty iff
  AND3( nq3,NOT1 nq2, nq1) is not empty)& (ns6 is not empty iff AND3( nq3, nq2,
  NOT1 nq1) is not empty)& (ns7 is not empty iff AND3( nq3, nq2, nq1) is not
empty) & (nq1 is not empty iff NOT1 q3 is not empty)& (nq2 is not empty iff q1
is not empty)& (nq3 is not empty iff q2 is not empty) implies (ns1 is not empty
iff s0 is not empty)& (ns3 is not empty iff s1 is not empty)& (ns7 is not empty
iff s3 is not empty)& (ns6 is not empty iff s7 is not empty)& (ns4 is not empty
  iff s6 is not empty)& (ns0 is not empty iff s4 is not empty) & (ns2 is not
  empty iff s5 is not empty)& (ns5 is not empty iff s2 is not empty)
proof
  let s0,s1,s2,s3,s4,s5,s6,s7,ns0,ns1,ns2,ns3,ns4,ns5,ns6,ns7, q1,q2,q3,nq1,
  nq2,nq3 be set;
  assume that
A1: s0 is not empty iff AND3(NOT1 q3,NOT1 q2,NOT1 q1) is not empty and
A2: s1 is not empty iff AND3(NOT1 q3,NOT1 q2,q1) is not empty and
A3: s2 is not empty iff AND3(NOT1 q3,q2,NOT1 q1) is not empty and
A4: s3 is not empty iff AND3(NOT1 q3,q2,q1) is not empty and
A5: s4 is not empty iff AND3(q3,NOT1 q2,NOT1 q1) is not empty and
A6: s5 is not empty iff AND3(q3,NOT1 q2,q1) is not empty and
A7: s6 is not empty iff AND3(q3,q2,NOT1 q1) is not empty and
A8: s7 is not empty iff AND3(q3,q2,q1) is not empty and
A9: ns0 is not empty iff AND3(NOT1 nq3,NOT1 nq2,NOT1 nq1) is not empty and
A10: ns1 is not empty iff AND3(NOT1 nq3,NOT1 nq2,nq1) is not empty and
A11: ns2 is not empty iff AND3(NOT1 nq3,nq2,NOT1 nq1) is not empty and
A12: ns3 is not empty iff AND3(NOT1 nq3,nq2,nq1) is not empty and
A13: ns4 is not empty iff AND3(nq3,NOT1 nq2,NOT1 nq1) is not empty and
A14: ns5 is not empty iff AND3(nq3,NOT1 nq2,nq1) is not empty and
A15: ns6 is not empty iff AND3(nq3,nq2,NOT1 nq1) is not empty and
A16: ns7 is not empty iff AND3(nq3,nq2,nq1) is not empty and
A17: ( nq1 is not empty iff NOT1 q3 is not empty)&( nq2 is not empty iff
  q1 is not empty) &( nq3 is not empty iff q2 is not empty);
  thus ns1 is not empty iff s0 is not empty by A1,A10,A17;
  thus ns3 is not empty iff s1 is not empty by A2,A12,A17;
  thus ns7 is not empty iff s3 is not empty by A4,A16,A17;
  ns6 is not empty iff q3 is not empty & q2 is not empty & q1 is not
  empty by A15,A17;
  hence ns6 is not empty iff s7 is not empty by A8;
  ns4 is not empty iff q3 is not empty & q2 is not empty & NOT1 q1 is not
  empty by A13,A17;
  hence ns4 is not empty iff s6 is not empty by A7;
  ns0 is not empty iff q3 is not empty & NOT1 q2 is not empty & NOT1 q1
  is not empty by A9,A17;
  hence ns0 is not empty iff s4 is not empty by A5;
  ns2 is not empty iff q3 is not empty & NOT1 q2 is not empty & q1 is not
  empty by A11,A17;
  hence ns2 is not empty iff s5 is not empty by A6;
  thus thesis by A3,A14,A17;
end;







theorem 
  for s0,s1,s2,s3,s4,s5,s6,s7,ns0,ns1,ns2,ns3,ns4,ns5,ns6,ns7, q1,q2,q3,
nq1,nq2,nq3,R being set holds (s0 is not empty iff AND3(NOT1 q3, NOT1 q2, NOT1
q1) is not empty)& (s1 is not empty iff AND3(NOT1 q3, NOT1 q2, q1) is not empty
  )& (s2 is not empty iff AND3(NOT1 q3, q2, NOT1 q1) is not empty)& (s3 is not
  empty iff AND3(NOT1 q3, q2, q1) is not empty)& (s4 is not empty iff AND3( q3,
NOT1 q2, NOT1 q1) is not empty)& (s5 is not empty iff AND3( q3, NOT1 q2, q1) is
  not empty)& (s6 is not empty iff AND3( q3, q2, NOT1 q1) is not empty)& (s7 is
not empty iff AND3( q3, q2, q1) is not empty) & (ns0 is not empty iff AND3(NOT1
nq3,NOT1 nq2,NOT1 nq1) is not empty)& (ns1 is not empty iff AND3(NOT1 nq3,NOT1
nq2, nq1) is not empty)& (ns2 is not empty iff AND3(NOT1 nq3, nq2,NOT1 nq1) is
not empty)& (ns3 is not empty iff AND3(NOT1 nq3, nq2, nq1) is not empty)& (ns4
is not empty iff AND3( nq3,NOT1 nq2,NOT1 nq1) is not empty)& (ns5 is not empty
  iff AND3( nq3,NOT1 nq2, nq1) is not empty)& (ns6 is not empty iff AND3( nq3,
nq2,NOT1 nq1) is not empty)& (ns7 is not empty iff AND3( nq3, nq2, nq1) is not
  empty) & (nq1 is not empty iff AND2(NOT1 q3, R) is not empty)& (nq2 is not
empty iff AND2( q1, R) is not empty)& (nq3 is not empty iff AND2( q2, R) is not
  empty) implies (ns1 is not empty iff AND2(s0, R) is not empty)& (ns3 is not
  empty iff AND2(s1, R) is not empty)& (ns7 is not empty iff AND2(s3, R) is not
empty)& (ns6 is not empty iff AND2(s7, R) is not empty)& (ns4 is not empty iff
AND2(s6, R) is not empty)& (ns0 is not empty iff OR2(AND2(s4, R),NOT1 R) is not
empty) & (ns2 is not empty iff AND2(s5, R) is not empty)& (ns5 is not empty iff
  AND2(s2, R) is not empty)
proof
  let s0,s1,s2,s3,s4,s5,s6,s7,ns0,ns1,ns2,ns3,ns4,ns5,ns6,ns7, q1,q2,q3,nq1,
  nq2,nq3,R be set;
  assume that
A1: s0 is not empty iff AND3(NOT1 q3,NOT1 q2,NOT1 q1) is not empty and
A2: s1 is not empty iff AND3(NOT1 q3,NOT1 q2,q1) is not empty and
A3: s2 is not empty iff AND3(NOT1 q3,q2,NOT1 q1) is not empty and
A4: s3 is not empty iff AND3(NOT1 q3,q2,q1) is not empty and
A5: s4 is not empty iff AND3(q3,NOT1 q2,NOT1 q1) is not empty and
A6: s5 is not empty iff AND3(q3,NOT1 q2,q1) is not empty and
A7: s6 is not empty iff AND3(q3,q2,NOT1 q1) is not empty and
A8: s7 is not empty iff AND3(q3,q2,q1) is not empty and
A9: ns0 is not empty iff AND3(NOT1 nq3,NOT1 nq2,NOT1 nq1) is not empty and
A10: ns1 is not empty iff AND3(NOT1 nq3,NOT1 nq2,nq1) is not empty and
A11: ns2 is not empty iff AND3(NOT1 nq3,nq2,NOT1 nq1) is not empty and
A12: ns3 is not empty iff AND3(NOT1 nq3,nq2,nq1) is not empty and
A13: ns4 is not empty iff AND3(nq3,NOT1 nq2,NOT1 nq1) is not empty and
A14: ns5 is not empty iff AND3(nq3,NOT1 nq2,nq1) is not empty and
A15: ns6 is not empty iff AND3(nq3,nq2,NOT1 nq1) is not empty and
A16: ns7 is not empty iff AND3(nq3,nq2,nq1) is not empty and
A17: ( nq1 is not empty iff AND2(NOT1 q3,R) is not empty)&( nq2 is not
empty iff AND2( q1,R) is not empty) &( nq3 is not empty iff AND2( q2,R) is not
  empty);
  ns1 is not empty iff NOT1 q3 is not empty & R is not empty & not(q2 is
not empty & R is not empty) & not(q1 is not empty & R is not empty) by A10,A17;
  hence ns1 is not empty iff AND2(s0,R) is not empty by A1;
  ns3 is not empty iff NOT1 q3 is not empty & R is not empty & not(q2 is
  not empty & R is not empty) & q1 is not empty & R is not empty by A12,A17;
  hence ns3 is not empty iff AND2(s1,R) is not empty by A2;
  ns7 is not empty iff NOT1 q3 is not empty & R is not empty & q2 is not
  empty & R is not empty & q1 is not empty & R is not empty by A16,A17;
  hence ns7 is not empty iff AND2(s3,R) is not empty by A4;
  ns6 is not empty iff not(NOT1 q3 is not empty & R is not empty) & q2 is
  not empty & R is not empty & q1 is not empty & R is not empty by A15,A17;
  then ns6 is not empty iff q3 is not empty & q2 is not empty & q1 is not
  empty & R is not empty;
  hence ns6 is not empty iff AND2(s7,R) is not empty by A8;
  ns4 is not empty iff not(NOT1 q3 is not empty & R is not empty) & q2 is
not empty & R is not empty & not(q1 is not empty & R is not empty) by A13,A17;
  then
  ns4 is not empty iff q3 is not empty & q2 is not empty & NOT1 q1 is not
  empty & R is not empty;
  hence ns4 is not empty iff AND2(s6,R) is not empty by A7;
  ns0 is not empty iff not(NOT1 q3 is not empty & R is not empty) & not(
q2 is not empty & R is not empty) & not(q1 is not empty & R is not empty) by A9
,A17;
  then ns0 is not empty iff q3 is not empty & NOT1 q2 is not empty & NOT1 q1
  is not empty & R is not empty or not(R) is not empty;
  hence ns0 is not empty iff OR2(AND2(s4, R),NOT1 R) is not empty by A5;
  ns2 is not empty iff not(NOT1 q3 is not empty & R is not empty) & not(
q2 is not empty & R is not empty) & q1 is not empty & R is not empty by A11,A17
  ;
  then
  ns2 is not empty iff q3 is not empty & NOT1 q2 is not empty & q1 is not
  empty & R is not empty;
  hence ns2 is not empty iff AND2(s5,R) is not empty by A6;
  ns5 is not empty iff NOT1 q3 is not empty & R is not empty & q2 is not
  empty & R is not empty & not(q1 is not empty & R is not empty) by A14,A17;
  hence thesis by A3;
end;







theorem 
  for s0,s1,s2,s3,s4,s5,s6,s7,s8,s9,s10,s11,s12,s13,s14,s15, ns0,ns1,ns2
,ns3,ns4,ns5,ns6,ns7,ns8,ns9,ns10,ns11,ns12,ns13,ns14,ns15, q1,q2,q3,q4,nq1,nq2
  ,nq3,nq4 being set holds (s0 is not empty iff AND4(NOT1 q4, NOT1 q3, NOT1 q2,
NOT1 q1) is not empty)& (s1 is not empty iff AND4(NOT1 q4, NOT1 q3, NOT1 q2, q1
  ) is not empty)& (s2 is not empty iff AND4(NOT1 q4, NOT1 q3, q2, NOT1 q1) is
not empty)& (s3 is not empty iff AND4(NOT1 q4, NOT1 q3, q2, q1) is not empty)&
(s4 is not empty iff AND4(NOT1 q4, q3, NOT1 q2, NOT1 q1) is not empty)& (s5 is
  not empty iff AND4(NOT1 q4, q3, NOT1 q2, q1) is not empty)& (s6 is not empty
  iff AND4(NOT1 q4, q3, q2, NOT1 q1) is not empty)& (s7 is not empty iff AND4(
  NOT1 q4, q3, q2, q1) is not empty)& (s8 is not empty iff AND4( q4, NOT1 q3,
  NOT1 q2, NOT1 q1) is not empty)& (s9 is not empty iff AND4( q4, NOT1 q3, NOT1
q2, q1) is not empty)& (s10 is not empty iff AND4( q4, NOT1 q3, q2, NOT1 q1) is
  not empty)& (s11 is not empty iff AND4( q4, NOT1 q3, q2, q1) is not empty)& (
s12 is not empty iff AND4( q4, q3, NOT1 q2, NOT1 q1) is not empty)& (s13 is not
empty iff AND4( q4, q3, NOT1 q2, q1) is not empty)& (s14 is not empty iff AND4(
q4, q3, q2, NOT1 q1) is not empty)& (s15 is not empty iff AND4( q4, q3, q2, q1)
  is not empty) & (ns0 is not empty iff AND4(NOT1 nq4, NOT1 nq3, NOT1 nq2, NOT1
  nq1) is not empty)& (ns1 is not empty iff AND4(NOT1 nq4, NOT1 nq3, NOT1 nq2,
  nq1) is not empty)& (ns2 is not empty iff AND4(NOT1 nq4, NOT1 nq3, nq2, NOT1
nq1) is not empty)& (ns3 is not empty iff AND4(NOT1 nq4, NOT1 nq3, nq2, nq1) is
  not empty)& (ns4 is not empty iff AND4(NOT1 nq4, nq3, NOT1 nq2, NOT1 nq1) is
  not empty)& (ns5 is not empty iff AND4(NOT1 nq4, nq3, NOT1 nq2, nq1) is not
empty)& (ns6 is not empty iff AND4(NOT1 nq4, nq3, nq2, NOT1 nq1) is not empty)&
(ns7 is not empty iff AND4(NOT1 nq4, nq3, nq2, nq1) is not empty)& (ns8 is not
  empty iff AND4( nq4, NOT1 nq3, NOT1 nq2, NOT1 nq1) is not empty)& (ns9 is not
empty iff AND4( nq4, NOT1 nq3, NOT1 nq2, nq1) is not empty)& (ns10 is not empty
  iff AND4( nq4, NOT1 nq3, nq2, NOT1 nq1) is not empty)& (ns11 is not empty iff
AND4( nq4, NOT1 nq3, nq2, nq1) is not empty)& (ns12 is not empty iff AND4( nq4,
nq3, NOT1 nq2, NOT1 nq1) is not empty)& (ns13 is not empty iff AND4( nq4, nq3,
NOT1 nq2, nq1) is not empty)& (ns14 is not empty iff AND4( nq4, nq3, nq2, NOT1
  nq1) is not empty)& (ns15 is not empty iff AND4( nq4, nq3, nq2, nq1) is not
empty) & (nq1 is not empty iff NOT1 q4 is not empty)& (nq2 is not empty iff q1
is not empty)& (nq3 is not empty iff q2 is not empty)& (nq4 is not empty iff q3
  is not empty) implies (ns1 is not empty iff s0 is not empty )& (ns3 is not
empty iff s1 is not empty )& (ns7 is not empty iff s3 is not empty )& (ns15 is
  not empty iff s7 is not empty )& (ns14 is not empty iff s15 is not empty)& (
ns12 is not empty iff s14 is not empty)& (ns8 is not empty iff s12 is not empty
  )& (ns0 is not empty iff s8 is not empty ) & (ns5 is not empty iff s2 is not
empty )& (ns11 is not empty iff s5 is not empty )& (ns6 is not empty iff s11 is
  not empty)& (ns13 is not empty iff s6 is not empty )& (ns10 is not empty iff
  s13 is not empty)& (ns4 is not empty iff s10 is not empty)& (ns9 is not empty
  iff s4 is not empty)& (ns2 is not empty iff s9 is not empty)
proof
  let s0,s1,s2,s3,s4,s5,s6,s7,s8,s9,s10,s11,s12,s13,s14,s15, ns0,ns1,ns2,ns3,
ns4,ns5,ns6,ns7,ns8,ns9,ns10,ns11,ns12,ns13,ns14,ns15, q1,q2,q3,q4,nq1,nq2,nq3,
  nq4 be set;
  assume that
A1: s0 is not empty iff AND4(NOT1 q4, NOT1 q3, NOT1 q2, NOT1 q1) is not
  empty and
A2: s1 is not empty iff AND4(NOT1 q4, NOT1 q3, NOT1 q2, q1) is not empty and
A3: s2 is not empty iff AND4(NOT1 q4, NOT1 q3, q2, NOT1 q1) is not empty and
A4: s3 is not empty iff AND4(NOT1 q4, NOT1 q3, q2, q1) is not empty and
A5: s4 is not empty iff AND4(NOT1 q4, q3, NOT1 q2, NOT1 q1) is not empty and
A6: s5 is not empty iff AND4(NOT1 q4, q3, NOT1 q2, q1) is not empty and
A7: s6 is not empty iff AND4(NOT1 q4, q3, q2, NOT1 q1) is not empty and
A8: s7 is not empty iff AND4(NOT1 q4, q3, q2, q1) is not empty and
A9: s8 is not empty iff AND4( q4, NOT1 q3, NOT1 q2, NOT1 q1) is not empty and
A10: s9 is not empty iff AND4( q4, NOT1 q3, NOT1 q2, q1) is not empty and
A11: s10 is not empty iff AND4( q4, NOT1 q3, q2, NOT1 q1) is not empty and
A12: s11 is not empty iff AND4( q4, NOT1 q3, q2, q1) is not empty and
A13: s12 is not empty iff AND4( q4, q3, NOT1 q2, NOT1 q1) is not empty and
A14: s13 is not empty iff AND4( q4, q3, NOT1 q2, q1) is not empty and
A15: s14 is not empty iff AND4( q4, q3, q2, NOT1 q1) is not empty and
A16: s15 is not empty iff AND4( q4, q3, q2, q1) is not empty and
A17: ns0 is not empty iff AND4(NOT1 nq4, NOT1 nq3, NOT1 nq2, NOT1 nq1)
  is not empty and
A18: ns1 is not empty iff AND4(NOT1 nq4, NOT1 nq3, NOT1 nq2, nq1) is not
  empty and
A19: ns2 is not empty iff AND4(NOT1 nq4, NOT1 nq3, nq2, NOT1 nq1) is not
  empty and
A20: ns3 is not empty iff AND4(NOT1 nq4, NOT1 nq3, nq2, nq1) is not empty and
A21: ns4 is not empty iff AND4(NOT1 nq4, nq3, NOT1 nq2, NOT1 nq1) is not
  empty and
A22: ns5 is not empty iff AND4(NOT1 nq4, nq3, NOT1 nq2, nq1) is not empty and
A23: ns6 is not empty iff AND4(NOT1 nq4, nq3, nq2, NOT1 nq1) is not empty and
A24: ns7 is not empty iff AND4(NOT1 nq4, nq3, nq2, nq1) is not empty and
A25: ns8 is not empty iff AND4( nq4, NOT1 nq3, NOT1 nq2, NOT1 nq1) is
  not empty and
A26: ns9 is not empty iff AND4( nq4, NOT1 nq3, NOT1 nq2, nq1) is not empty and
A27: ns10 is not empty iff AND4( nq4, NOT1 nq3, nq2, NOT1 nq1) is not empty and
A28: ns11 is not empty iff AND4( nq4, NOT1 nq3, nq2, nq1) is not empty and
A29: ns12 is not empty iff AND4( nq4, nq3, NOT1 nq2, NOT1 nq1) is not empty and
A30: ns13 is not empty iff AND4( nq4, nq3, NOT1 nq2, nq1) is not empty and
A31: ns14 is not empty iff AND4( nq4, nq3, nq2, NOT1 nq1) is not empty and
A32: ns15 is not empty iff AND4( nq4, nq3, nq2, nq1) is not empty and
A33: ( nq1 is not empty iff NOT1 q4 is not empty)&( nq2 is not empty iff
  q1 is not empty) & ( nq3 is not empty iff q2 is not empty)&( nq4 is not empty
  iff q3 is not empty);
  thus ns1 is not empty iff s0 is not empty by A1,A18,A33,GATE_1:20;
  thus ns3 is not empty iff s1 is not empty by A2,A20,A33,GATE_1:20;
  thus ns7 is not empty iff s3 is not empty by A4,A24,A33,GATE_1:20;
  thus ns15 is not empty iff s7 is not empty by A8,A32,A33,GATE_1:20;
  ns14 is not empty iff q4 is not empty & q3 is not empty & q2 is not
  empty & q1 is not empty by A31,A33,GATE_1:20;
  hence ns14 is not empty iff s15 is not empty by A16,GATE_1:20;
  ns12 is not empty iff q4 is not empty & q3 is not empty & q2 is not
  empty & NOT1 q1 is not empty by A29,A33,GATE_1:20;
  hence ns12 is not empty iff s14 is not empty by A15,GATE_1:20;
  ns8 is not empty iff q4 is not empty & q3 is not empty & NOT1 q2 is not
  empty & NOT1 q1 is not empty by A25,A33,GATE_1:20;
  hence ns8 is not empty iff s12 is not empty by A13,GATE_1:20;
  ns0 is not empty iff q4 is not empty & NOT1 q3 is not empty & NOT1 q2
  is not empty & NOT1 q1 is not empty by A17,A33,GATE_1:20;
  hence ns0 is not empty iff s8 is not empty by A9,GATE_1:20;
  thus ns5 is not empty iff s2 is not empty by A3,A22,A33,GATE_1:20;
  thus ns11 is not empty iff s5 is not empty by A6,A28,A33,GATE_1:20;
  ns6 is not empty iff q4 is not empty & NOT1 q3 is not empty & q2 is not
  empty & q1 is not empty by A23,A33,GATE_1:20;
  hence ns6 is not empty iff s11 is not empty by A12,GATE_1:20;
  thus ns13 is not empty iff s6 is not empty by A7,A30,A33,GATE_1:20;
  ns10 is not empty iff q4 is not empty & q3 is not empty & NOT1 q2 is
  not empty & q1 is not empty by A27,A33,GATE_1:20;
  hence ns10 is not empty iff s13 is not empty by A14,GATE_1:20;
  ns4 is not empty iff q4 is not empty & NOT1 q3 is not empty & q2 is not
  empty & NOT1 q1 is not empty by A21,A33,GATE_1:20;
  hence ns4 is not empty iff s10 is not empty by A11,GATE_1:20;
  thus ns9 is not empty iff s4 is not empty by A5,A26,A33,GATE_1:20;
  ns2 is not empty iff q4 is not empty & NOT1 q3 is not empty & NOT1 q2
  is not empty & q1 is not empty by A19,A33,GATE_1:20;
  hence thesis by A10,GATE_1:20;
end;








theorem 
  for s0,s1,s2,s3,s4,s5,s6,s7,s8,s9,s10,s11,s12,s13,s14,s15, ns0,ns1,ns2
,ns3,ns4,ns5,ns6,ns7,ns8,ns9,ns10,ns11,ns12,ns13,ns14,ns15, q1,q2,q3,q4,nq1,nq2
,nq3,nq4,R being set holds (s0 is not empty iff AND4(NOT1 q4, NOT1 q3, NOT1 q2,
NOT1 q1) is not empty)& (s1 is not empty iff AND4(NOT1 q4, NOT1 q3, NOT1 q2, q1
  ) is not empty)& (s2 is not empty iff AND4(NOT1 q4, NOT1 q3, q2, NOT1 q1) is
not empty)& (s3 is not empty iff AND4(NOT1 q4, NOT1 q3, q2, q1) is not empty)&
(s4 is not empty iff AND4(NOT1 q4, q3, NOT1 q2, NOT1 q1) is not empty)& (s5 is
  not empty iff AND4(NOT1 q4, q3, NOT1 q2, q1) is not empty)& (s6 is not empty
  iff AND4(NOT1 q4, q3, q2, NOT1 q1) is not empty)& (s7 is not empty iff AND4(
  NOT1 q4, q3, q2, q1) is not empty)& (s8 is not empty iff AND4( q4, NOT1 q3,
  NOT1 q2, NOT1 q1) is not empty)& (s9 is not empty iff AND4( q4, NOT1 q3, NOT1
q2, q1) is not empty)& (s10 is not empty iff AND4( q4, NOT1 q3, q2, NOT1 q1) is
  not empty)& (s11 is not empty iff AND4( q4, NOT1 q3, q2, q1) is not empty)& (
s12 is not empty iff AND4( q4, q3, NOT1 q2, NOT1 q1) is not empty)& (s13 is not
empty iff AND4( q4, q3, NOT1 q2, q1) is not empty)& (s14 is not empty iff AND4(
q4, q3, q2, NOT1 q1) is not empty)& (s15 is not empty iff AND4( q4, q3, q2, q1)
  is not empty) & (ns0 is not empty iff AND4(NOT1 nq4, NOT1 nq3, NOT1 nq2, NOT1
  nq1) is not empty)& (ns1 is not empty iff AND4(NOT1 nq4, NOT1 nq3, NOT1 nq2,
  nq1) is not empty)& (ns2 is not empty iff AND4(NOT1 nq4, NOT1 nq3, nq2, NOT1
nq1) is not empty)& (ns3 is not empty iff AND4(NOT1 nq4, NOT1 nq3, nq2, nq1) is
  not empty)& (ns4 is not empty iff AND4(NOT1 nq4, nq3, NOT1 nq2, NOT1 nq1) is
  not empty)& (ns5 is not empty iff AND4(NOT1 nq4, nq3, NOT1 nq2, nq1) is not
empty)& (ns6 is not empty iff AND4(NOT1 nq4, nq3, nq2, NOT1 nq1) is not empty)&
(ns7 is not empty iff AND4(NOT1 nq4, nq3, nq2, nq1) is not empty)& (ns8 is not
  empty iff AND4( nq4, NOT1 nq3, NOT1 nq2, NOT1 nq1) is not empty)& (ns9 is not
empty iff AND4( nq4, NOT1 nq3, NOT1 nq2, nq1) is not empty)& (ns10 is not empty
  iff AND4( nq4, NOT1 nq3, nq2, NOT1 nq1) is not empty)& (ns11 is not empty iff
AND4( nq4, NOT1 nq3, nq2, nq1) is not empty)& (ns12 is not empty iff AND4( nq4,
nq3, NOT1 nq2, NOT1 nq1) is not empty)& (ns13 is not empty iff AND4( nq4, nq3,
NOT1 nq2, nq1) is not empty)& (ns14 is not empty iff AND4( nq4, nq3, nq2, NOT1
  nq1) is not empty)& (ns15 is not empty iff AND4( nq4, nq3, nq2, nq1) is not
empty) & (nq1 is not empty iff AND2(NOT1 q4,R) is not empty)& (nq2 is not empty
iff AND2( q1,R) is not empty)& (nq3 is not empty iff AND2( q2,R) is not empty)&
  (nq4 is not empty iff AND2( q3,R) is not empty) implies (ns1 is not empty iff
AND2(s0,R) is not empty)& (ns3 is not empty iff AND2(s1,R) is not empty)& (ns7
is not empty iff AND2(s3,R) is not empty)& (ns15 is not empty iff AND2(s7,R) is
  not empty)& (ns14 is not empty iff AND2(s15,R) is not empty)& (ns12 is not
  empty iff AND2(s14,R) is not empty)& (ns8 is not empty iff AND2(s12,R) is not
  empty)& (ns0 is not empty iff OR2(AND2(s8,R),NOT1 R) is not empty) & (ns5 is
  not empty iff AND2(s2,R) is not empty)& (ns11 is not empty iff AND2(s5,R) is
not empty)& (ns6 is not empty iff AND2(s11,R) is not empty)& (ns13 is not empty
iff AND2(s6,R) is not empty)& (ns10 is not empty iff AND2(s13,R) is not empty)&
(ns4 is not empty iff AND2(s10,R) is not empty)& (ns9 is not empty iff AND2(s4,
  R) is not empty)& (ns2 is not empty iff AND2(s9,R) is not empty)
proof
  let s0,s1,s2,s3,s4,s5,s6,s7,s8,s9,s10,s11,s12,s13,s14,s15, ns0,ns1,ns2,ns3,
ns4,ns5,ns6,ns7,ns8,ns9,ns10,ns11,ns12,ns13,ns14,ns15, q1,q2,q3,q4,nq1,nq2,nq3,
  nq4,R be set;
  assume that
A1: s0 is not empty iff AND4(NOT1 q4, NOT1 q3, NOT1 q2, NOT1 q1) is not
  empty and
A2: s1 is not empty iff AND4(NOT1 q4, NOT1 q3, NOT1 q2, q1) is not empty and
A3: s2 is not empty iff AND4(NOT1 q4, NOT1 q3, q2, NOT1 q1) is not empty and
A4: s3 is not empty iff AND4(NOT1 q4, NOT1 q3, q2, q1) is not empty and
A5: s4 is not empty iff AND4(NOT1 q4, q3, NOT1 q2, NOT1 q1) is not empty and
A6: s5 is not empty iff AND4(NOT1 q4, q3, NOT1 q2, q1) is not empty and
A7: s6 is not empty iff AND4(NOT1 q4, q3, q2, NOT1 q1) is not empty and
A8: s7 is not empty iff AND4(NOT1 q4, q3, q2, q1) is not empty and
A9: s8 is not empty iff AND4( q4, NOT1 q3, NOT1 q2, NOT1 q1) is not empty and
A10: s9 is not empty iff AND4( q4, NOT1 q3, NOT1 q2, q1) is not empty and
A11: s10 is not empty iff AND4( q4, NOT1 q3, q2, NOT1 q1) is not empty and
A12: s11 is not empty iff AND4( q4, NOT1 q3, q2, q1) is not empty and
A13: s12 is not empty iff AND4( q4, q3, NOT1 q2, NOT1 q1) is not empty and
A14: s13 is not empty iff AND4( q4, q3, NOT1 q2, q1) is not empty and
A15: s14 is not empty iff AND4( q4, q3, q2, NOT1 q1) is not empty and
A16: s15 is not empty iff AND4( q4, q3, q2, q1) is not empty and
A17: ns0 is not empty iff AND4(NOT1 nq4, NOT1 nq3, NOT1 nq2, NOT1 nq1)
  is not empty and
A18: ns1 is not empty iff AND4(NOT1 nq4, NOT1 nq3, NOT1 nq2, nq1) is not
  empty and
A19: ns2 is not empty iff AND4(NOT1 nq4, NOT1 nq3, nq2, NOT1 nq1) is not
  empty and
A20: ns3 is not empty iff AND4(NOT1 nq4, NOT1 nq3, nq2, nq1) is not empty and
A21: ns4 is not empty iff AND4(NOT1 nq4, nq3, NOT1 nq2, NOT1 nq1) is not
  empty and
A22: ns5 is not empty iff AND4(NOT1 nq4, nq3, NOT1 nq2, nq1) is not empty and
A23: ns6 is not empty iff AND4(NOT1 nq4, nq3, nq2, NOT1 nq1) is not empty and
A24: ns7 is not empty iff AND4(NOT1 nq4, nq3, nq2, nq1) is not empty and
A25: ns8 is not empty iff AND4( nq4, NOT1 nq3, NOT1 nq2, NOT1 nq1) is
  not empty and
A26: ns9 is not empty iff AND4( nq4, NOT1 nq3, NOT1 nq2, nq1) is not empty and
A27: ns10 is not empty iff AND4( nq4, NOT1 nq3, nq2, NOT1 nq1) is not empty and
A28: ns11 is not empty iff AND4( nq4, NOT1 nq3, nq2, nq1) is not empty and
A29: ns12 is not empty iff AND4( nq4, nq3, NOT1 nq2, NOT1 nq1) is not empty and
A30: ns13 is not empty iff AND4( nq4, nq3, NOT1 nq2, nq1) is not empty and
A31: ns14 is not empty iff AND4( nq4, nq3, nq2, NOT1 nq1) is not empty and
A32: ns15 is not empty iff AND4( nq4, nq3, nq2, nq1) is not empty and
A33: ( nq1 is not empty iff AND2(NOT1 q4,R) is not empty)&( nq2 is not
empty iff AND2( q1,R) is not empty) & ( nq3 is not empty iff AND2( q2,R) is not
  empty)&( nq4 is not empty iff AND2( q3,R) is not empty);
  ns1 is not empty iff NOT1 q4 is not empty & R is not empty & not(q3 is
not empty & R is not empty) & not(q2 is not empty & R is not empty) & not(q1 is
  not empty & R is not empty) by A18,A33,GATE_1:20;
  hence ns1 is not empty iff AND2(s0,R) is not empty by A1,GATE_1:20;
  ns3 is not empty iff NOT1 q4 is not empty & R is not empty & not(q3 is
not empty & R is not empty) & not(q2 is not empty & R is not empty) & q1 is not
  empty & R is not empty by A20,A33,GATE_1:20;
  hence ns3 is not empty iff AND2(s1,R) is not empty by A2,GATE_1:20;
  ns7 is not empty iff NOT1 q4 is not empty & R is not empty & not(q3 is
  not empty & R is not empty) & q2 is not empty & R is not empty & q1 is not
  empty & R is not empty by A24,A33,GATE_1:20;
  hence ns7 is not empty iff AND2(s3,R) is not empty by A4,GATE_1:20;
  ns15 is not empty iff NOT1 q4 is not empty & R is not empty & q3 is not
empty & R is not empty & q2 is not empty & R is not empty & q1 is not empty & R
  is not empty by A32,A33,GATE_1:20;
  hence ns15 is not empty iff AND2(s7,R) is not empty by A8,GATE_1:20;
  ns14 is not empty iff not(NOT1 q4 is not empty & R is not empty) & q3
  is not empty & R is not empty & q2 is not empty & R is not empty & q1 is not
  empty & R is not empty by A31,A33,GATE_1:20;
  then ns14 is not empty iff q4 is not empty & q3 is not empty & q2 is not
  empty & q1 is not empty & R is not empty;
  hence ns14 is not empty iff AND2(s15,R) is not empty by A16,GATE_1:20;
  ns12 is not empty iff not(NOT1 q4 is not empty & R is not empty) & q3
  is not empty & R is not empty & q2 is not empty & R is not empty & not(q1 is
  not empty & R is not empty) by A29,A33,GATE_1:20;
  then ns12 is not empty iff q4 is not empty & q3 is not empty & q2 is not
  empty & NOT1 q1 is not empty & R is not empty;
  hence ns12 is not empty iff AND2(s14,R) is not empty by A15,GATE_1:20;
  ns8 is not empty iff not(NOT1 q4 is not empty & R is not empty) & q3 is
not empty & R is not empty & not(q2 is not empty & R is not empty) & not(q1 is
  not empty & R is not empty) by A25,A33,GATE_1:20;
  then
  ns8 is not empty iff q4 is not empty & q3 is not empty & NOT1 q2 is not
  empty & NOT1 q1 is not empty & R is not empty;
  hence ns8 is not empty iff AND2(s12,R) is not empty by A13,GATE_1:20;
  ns0 is not empty iff not(NOT1 q4 is not empty & R is not empty) & not(
q3 is not empty & R is not empty) & not(q2 is not empty & R is not empty) & not
  (q1 is not empty & R is not empty) by A17,A33,GATE_1:20;
  then ns0 is not empty iff q4 is not empty & NOT1 q3 is not empty & NOT1 q2
  is not empty & NOT1 q1 is not empty & R is not empty or not(R) is not empty;
  hence ns0 is not empty iff OR2(AND2(s8, R),NOT1 R) is not empty by A9,
GATE_1:20;
  ns5 is not empty iff NOT1 q4 is not empty & R is not empty & not(q3 is
not empty & R is not empty) & q2 is not empty & R is not empty & not(q1 is not
  empty & R is not empty) by A22,A33,GATE_1:20;
  hence ns5 is not empty iff AND2(s2,R) is not empty by A3,GATE_1:20;
  ns11 is not empty iff NOT1 q4 is not empty & R is not empty & q3 is not
  empty & R is not empty & not(q2 is not empty & R is not empty) & q1 is not
  empty & R is not empty by A28,A33,GATE_1:20;
  hence ns11 is not empty iff AND2(s5,R) is not empty by A6,GATE_1:20;
  ns6 is not empty iff not(NOT1 q4 is not empty & R is not empty) & not(
  q3 is not empty & R is not empty) & q2 is not empty & R is not empty & q1 is
  not empty & R is not empty by A23,A33,GATE_1:20;
  then
  ns6 is not empty iff q4 is not empty & NOT1 q3 is not empty & q2 is not
  empty & q1 is not empty & R is not empty;
  hence ns6 is not empty iff AND2(s11,R) is not empty by A12,GATE_1:20;
  ns13 is not empty iff NOT1 q4 is not empty & R is not empty & q3 is not
empty & R is not empty & q2 is not empty & R is not empty & not(q1 is not empty
  & R is not empty) by A30,A33,GATE_1:20;
  hence ns13 is not empty iff AND2(s6,R) is not empty by A7,GATE_1:20;
  ns10 is not empty iff not(NOT1 q4 is not empty & R is not empty) & q3
  is not empty & R is not empty & not(q2 is not empty & R is not empty) & q1 is
  not empty & R is not empty by A27,A33,GATE_1:20;
  then ns10 is not empty iff q4 is not empty & q3 is not empty & NOT1 q2 is
  not empty & q1 is not empty & R is not empty;
  hence ns10 is not empty iff AND2(s13,R) is not empty by A14,GATE_1:20;
  ns4 is not empty iff not(NOT1 q4 is not empty & R is not empty) & not(
  q3 is not empty & R is not empty) & q2 is not empty & R is not empty & not(q1
  is not empty & R is not empty) by A21,A33,GATE_1:20;
  then
  ns4 is not empty iff q4 is not empty & NOT1 q3 is not empty & q2 is not
  empty & NOT1 q1 is not empty & R is not empty;
  hence ns4 is not empty iff AND2(s10,R) is not empty by A11,GATE_1:20;
  ns9 is not empty iff NOT1 q4 is not empty & R is not empty & q3 is not
empty & R is not empty & not(q2 is not empty & R is not empty) & not(q1 is not
  empty & R is not empty) by A26,A33,GATE_1:20;
  hence ns9 is not empty iff AND2(s4,R) is not empty by A5,GATE_1:20;
  ns2 is not empty iff not(NOT1 q4 is not empty & R is not empty) & not(
q3 is not empty & R is not empty) & not(q2 is not empty & R is not empty) & q1
  is not empty & R is not empty by A19,A33,GATE_1:20;
  then ns2 is not empty iff q4 is not empty & NOT1 q3 is not empty & NOT1 q2
  is not empty & q1 is not empty & R is not empty;
  hence thesis by A10,GATE_1:20;
end;














begin



















theorem
  for g0,g1,g2,g3,g4,g5,g6,g7,g8,g9,g10,g11,g12, a0,a1,a2,a3,a4,a5,a6,a7
,a8,a9,a10,a11, b0,b1,b2,b3,b4,b5,b6,b7,b8,b9,b10,b11, p being set holds g12 is
not empty & (b0 is not empty iff XOR2(p,AND2(g0,a11)) is not empty)& (b1 is not
  empty iff XOR2(a0,AND2(g1,a11)) is not empty)& (b2 is not empty iff XOR2(a1,
AND2(g2,a11)) is not empty)& (b3 is not empty iff XOR2(a2,AND2(g3,a11)) is not
  empty)& (b4 is not empty iff XOR2(a3,AND2(g4,a11)) is not empty)& (b5 is not
  empty iff XOR2(a4,AND2(g5,a11)) is not empty)& (b6 is not empty iff XOR2(a5,
AND2(g6,a11)) is not empty)& (b7 is not empty iff XOR2(a6,AND2(g7,a11)) is not
  empty)& (b8 is not empty iff XOR2(a7,AND2(g8,a11)) is not empty)& (b9 is not
  empty iff XOR2(a8,AND2(g9,a11)) is not empty)& (b10 is not empty iff XOR2(a9,
AND2(g10,a11)) is not empty)& (b11 is not empty iff XOR2(a10,AND2(g11,a11)) is
not empty) implies (a11 is not empty iff AND2(g12,a11) is not empty) & (a10 is
not empty iff XOR2(b11,AND2(g11,a11)) is not empty)& (a9 is not empty iff XOR2(
b10,AND2(g10,a11)) is not empty)& (a8 is not empty iff XOR2(b9,AND2(g9,a11)) is
  not empty)& (a7 is not empty iff XOR2(b8,AND2(g8,a11)) is not empty)& (a6 is
not empty iff XOR2(b7,AND2(g7,a11)) is not empty)& (a5 is not empty iff XOR2(b6
,AND2(g6,a11)) is not empty)& (a4 is not empty iff XOR2(b5,AND2(g5,a11)) is not
  empty)& (a3 is not empty iff XOR2(b4,AND2(g4,a11)) is not empty)& (a2 is not
  empty iff XOR2(b3,AND2(g3,a11)) is not empty)& (a1 is not empty iff XOR2(b2,
AND2(g2,a11)) is not empty)& (a0 is not empty iff XOR2(b1,AND2(g1,a11)) is not
  empty)& (p is not empty iff XOR2(b0,AND2(g0,a11)) is not empty)
proof
  let g0,g1,g2,g3,g4,g5,g6,g7,g8,g9,g10,g11,g12, a0,a1,a2,a3,a4,a5,a6,a7,a8,a9
  ,a10,a11, b0,b1,b2,b3,b4,b5,b6,b7,b8,b9,b10,b11, p be set;
  assume that
A1: g12 is not empty and
A2: b0 is not empty iff XOR2(p,AND2(g0,a11)) is not empty and
A3: b1 is not empty iff XOR2(a0,AND2(g1,a11)) is not empty and
A4: b2 is not empty iff XOR2(a1,AND2(g2,a11)) is not empty and
A5: b3 is not empty iff XOR2(a2,AND2(g3,a11)) is not empty and
A6: b4 is not empty iff XOR2(a3,AND2(g4,a11)) is not empty and
A7: b5 is not empty iff XOR2(a4,AND2(g5,a11)) is not empty and
A8: b6 is not empty iff XOR2(a5,AND2(g6,a11)) is not empty and
A9: b7 is not empty iff XOR2(a6,AND2(g7,a11)) is not empty and
A10: b8 is not empty iff XOR2(a7,AND2(g8,a11)) is not empty and
A11: b9 is not empty iff XOR2(a8,AND2(g9,a11)) is not empty and
A12: b10 is not empty iff XOR2(a9,AND2(g10,a11)) is not empty and
A13: b11 is not empty iff XOR2(a10,AND2(g11,a11)) is not empty;
  thus a11 is not empty iff AND2(g12,a11) is not empty by A1;
  XOR2(b11,AND2(g11,a11)) is not empty iff b11 is not empty & not AND2(
  g11,a11) is not empty or not b11 is not empty & AND2(g11,a11) is not empty;
  hence a10 is not empty iff XOR2(b11,AND2(g11,a11)) is not empty by A13;
  XOR2(b10,AND2(g10,a11)) is not empty iff b10 is not empty & not AND2(
  g10,a11) is not empty or not b10 is not empty & AND2(g10,a11) is not empty;
  hence a9 is not empty iff XOR2(b10,AND2(g10,a11)) is not empty by A12;
  XOR2(b9,AND2(g9,a11)) is not empty iff b9 is not empty & not AND2(g9,
  a11) is not empty or not b9 is not empty & AND2(g9,a11) is not empty;
  hence a8 is not empty iff XOR2(b9,AND2(g9,a11)) is not empty by A11;
  XOR2(b8,AND2(g8,a11)) is not empty iff b8 is not empty & not AND2(g8,
  a11) is not empty or not b8 is not empty & AND2(g8,a11) is not empty;
  hence a7 is not empty iff XOR2(b8,AND2(g8,a11)) is not empty by A10;
  XOR2(b7,AND2(g7,a11)) is not empty iff b7 is not empty & not AND2(g7,
  a11) is not empty or not b7 is not empty & AND2(g7,a11) is not empty;
  hence a6 is not empty iff XOR2(b7,AND2(g7,a11)) is not empty by A9;
  XOR2(b6,AND2(g6,a11)) is not empty iff b6 is not empty & not AND2(g6,
  a11) is not empty or not b6 is not empty & AND2(g6,a11) is not empty;
  hence a5 is not empty iff XOR2(b6,AND2(g6,a11)) is not empty by A8;
  XOR2(b5,AND2(g5,a11)) is not empty iff b5 is not empty & not AND2(g5,
  a11) is not empty or not b5 is not empty & AND2(g5,a11) is not empty;
  hence a4 is not empty iff XOR2(b5,AND2(g5,a11)) is not empty by A7;
  XOR2(b4,AND2(g4,a11)) is not empty iff b4 is not empty & not AND2(g4,
  a11) is not empty or not b4 is not empty & AND2(g4,a11) is not empty;
  hence a3 is not empty iff XOR2(b4,AND2(g4,a11)) is not empty by A6;
  XOR2(b3,AND2(g3,a11)) is not empty iff b3 is not empty & not AND2(g3,
  a11) is not empty or not b3 is not empty & AND2(g3,a11) is not empty;
  hence a2 is not empty iff XOR2(b3,AND2(g3,a11)) is not empty by A5;
  XOR2(b2,AND2(g2,a11)) is not empty iff b2 is not empty & not AND2(g2,
  a11) is not empty or not b2 is not empty & AND2(g2,a11) is not empty;
  hence a1 is not empty iff XOR2(b2,AND2(g2,a11)) is not empty by A4;
  XOR2(b1,AND2(g1,a11)) is not empty iff b1 is not empty & not AND2(g1,
  a11) is not empty or not b1 is not empty & AND2(g1,a11) is not empty;
  hence a0 is not empty iff XOR2(b1,AND2(g1,a11)) is not empty by A3;
  XOR2(b0,AND2(g0,a11)) is not empty iff b0 is not empty & not AND2(g0,
  a11) is not empty or not b0 is not empty & AND2(g0,a11) is not empty;
  hence thesis by A2;
end;

















theorem
  for g0,g1,g2,g3,g4,g5,g6,g7,g8,g9,g10,g11,g12,g13,g14,g15,g16, a0,a1,
a2,a3,a4,a5,a6,a7,a8,a9,a10,a11,a12,a13,a14,a15, b0,b1,b2,b3,b4,b5,b6,b7,b8,b9,
b10,b11,b12,b13,b14,b15, p being set holds g16 is not empty & (b0 is not empty
  iff XOR2(p,AND2(g0,a15)) is not empty)& (b1 is not empty iff XOR2(a0,AND2(g1,
a15)) is not empty)& (b2 is not empty iff XOR2(a1,AND2(g2,a15)) is not empty)&
(b3 is not empty iff XOR2(a2,AND2(g3,a15)) is not empty)& (b4 is not empty iff
XOR2(a3,AND2(g4,a15)) is not empty)& (b5 is not empty iff XOR2(a4,AND2(g5,a15))
is not empty)& (b6 is not empty iff XOR2(a5,AND2(g6,a15)) is not empty)& (b7 is
not empty iff XOR2(a6,AND2(g7,a15)) is not empty)& (b8 is not empty iff XOR2(a7
,AND2(g8,a15)) is not empty)& (b9 is not empty iff XOR2(a8,AND2(g9,a15)) is not
empty)& (b10 is not empty iff XOR2(a9,AND2(g10,a15)) is not empty)& (b11 is not
empty iff XOR2(a10,AND2(g11,a15)) is not empty)& (b12 is not empty iff XOR2(a11
,AND2(g12,a15)) is not empty)& (b13 is not empty iff XOR2(a12,AND2(g13,a15)) is
  not empty)& (b14 is not empty iff XOR2(a13,AND2(g14,a15)) is not empty)& (b15
  is not empty iff XOR2(a14,AND2(g15,a15)) is not empty) implies (a15 is not
empty iff AND2(g16,a15) is not empty) & (a14 is not empty iff XOR2(b15,AND2(g15
  ,a15)) is not empty)& (a13 is not empty iff XOR2(b14,AND2(g14,a15)) is not
  empty)& (a12 is not empty iff XOR2(b13,AND2(g13,a15)) is not empty)& (a11 is
not empty iff XOR2(b12,AND2(g12,a15)) is not empty)& (a10 is not empty iff XOR2
(b11,AND2(g11,a15)) is not empty)& (a9 is not empty iff XOR2(b10,AND2(g10,a15))
is not empty)& (a8 is not empty iff XOR2(b9,AND2(g9,a15)) is not empty)& (a7 is
not empty iff XOR2(b8,AND2(g8,a15)) is not empty)& (a6 is not empty iff XOR2(b7
,AND2(g7,a15)) is not empty)& (a5 is not empty iff XOR2(b6,AND2(g6,a15)) is not
  empty)& (a4 is not empty iff XOR2(b5,AND2(g5,a15)) is not empty)& (a3 is not
  empty iff XOR2(b4,AND2(g4,a15)) is not empty)& (a2 is not empty iff XOR2(b3,
AND2(g3,a15)) is not empty)& (a1 is not empty iff XOR2(b2,AND2(g2,a15)) is not
  empty)& (a0 is not empty iff XOR2(b1,AND2(g1,a15)) is not empty)& (p is not
  empty iff XOR2(b0,AND2(g0,a15)) is not empty)
proof
  let g0,g1,g2,g3,g4,g5,g6,g7,g8,g9,g10,g11,g12,g13,g14,g15,g16, a0,a1,a2,a3,
  a4,a5,a6,a7,a8,a9,a10,a11,a12,a13,a14,a15, b0,b1,b2,b3,b4,b5,b6,b7,b8,b9,b10,
  b11,b12,b13,b14,b15, p be set;
  assume that
A1: g16 is not empty and
A2: b0 is not empty iff XOR2(p,AND2(g0,a15)) is not empty and
A3: b1 is not empty iff XOR2(a0,AND2(g1,a15)) is not empty and
A4: b2 is not empty iff XOR2(a1,AND2(g2,a15)) is not empty and
A5: b3 is not empty iff XOR2(a2,AND2(g3,a15)) is not empty and
A6: b4 is not empty iff XOR2(a3,AND2(g4,a15)) is not empty and
A7: b5 is not empty iff XOR2(a4,AND2(g5,a15)) is not empty and
A8: b6 is not empty iff XOR2(a5,AND2(g6,a15)) is not empty and
A9: b7 is not empty iff XOR2(a6,AND2(g7,a15)) is not empty and
A10: b8 is not empty iff XOR2(a7,AND2(g8,a15)) is not empty and
A11: b9 is not empty iff XOR2(a8,AND2(g9,a15)) is not empty and
A12: b10 is not empty iff XOR2(a9,AND2(g10,a15)) is not empty and
A13: b11 is not empty iff XOR2(a10,AND2(g11,a15)) is not empty and
A14: b12 is not empty iff XOR2(a11,AND2(g12,a15)) is not empty and
A15: b13 is not empty iff XOR2(a12,AND2(g13,a15)) is not empty and
A16: b14 is not empty iff XOR2(a13,AND2(g14,a15)) is not empty and
A17: b15 is not empty iff XOR2(a14,AND2(g15,a15)) is not empty;
  thus a15 is not empty iff AND2(g16,a15) is not empty by A1;
  XOR2(b15,AND2(g15,a15)) is not empty iff b15 is not empty & not AND2(
  g15,a15) is not empty or not b15 is not empty & AND2(g15,a15) is not empty;
  hence a14 is not empty iff XOR2(b15,AND2(g15,a15)) is not empty by A17;
  XOR2(b14,AND2(g14,a15)) is not empty iff b14 is not empty & not AND2(
  g14,a15) is not empty or not b14 is not empty & AND2(g14,a15) is not empty;
  hence a13 is not empty iff XOR2(b14,AND2(g14,a15)) is not empty by A16;
  XOR2(b13,AND2(g13,a15)) is not empty iff b13 is not empty & not AND2(
  g13,a15) is not empty or not b13 is not empty & AND2(g13,a15) is not empty;
  hence a12 is not empty iff XOR2(b13,AND2(g13,a15)) is not empty by A15;
  XOR2(b12,AND2(g12,a15)) is not empty iff b12 is not empty & not AND2(
  g12,a15) is not empty or not b12 is not empty & AND2(g12,a15) is not empty;
  hence a11 is not empty iff XOR2(b12,AND2(g12,a15)) is not empty by A14;
  XOR2(b11,AND2(g11,a15)) is not empty iff b11 is not empty & not AND2(
  g11,a15) is not empty or not b11 is not empty & AND2(g11,a15) is not empty;
  hence a10 is not empty iff XOR2(b11,AND2(g11,a15)) is not empty by A13;
  XOR2(b10,AND2(g10,a15)) is not empty iff b10 is not empty & not AND2(
  g10,a15) is not empty or not b10 is not empty & AND2(g10,a15) is not empty;
  hence a9 is not empty iff XOR2(b10,AND2(g10,a15)) is not empty by A12;
  XOR2(b9,AND2(g9,a15)) is not empty iff b9 is not empty & not AND2(g9,
  a15) is not empty or not b9 is not empty & AND2(g9,a15) is not empty;
  hence a8 is not empty iff XOR2(b9,AND2(g9,a15)) is not empty by A11;
  XOR2(b8,AND2(g8,a15)) is not empty iff b8 is not empty & not AND2(g8,
  a15) is not empty or not b8 is not empty & AND2(g8,a15) is not empty;
  hence a7 is not empty iff XOR2(b8,AND2(g8,a15)) is not empty by A10;
  XOR2(b7,AND2(g7,a15)) is not empty iff b7 is not empty & not AND2(g7,
  a15) is not empty or not b7 is not empty & AND2(g7,a15) is not empty;
  hence a6 is not empty iff XOR2(b7,AND2(g7,a15)) is not empty by A9;
  XOR2(b6,AND2(g6,a15)) is not empty iff b6 is not empty & not AND2(g6,
  a15) is not empty or not b6 is not empty & AND2(g6,a15) is not empty;
  hence a5 is not empty iff XOR2(b6,AND2(g6,a15)) is not empty by A8;
  XOR2(b5,AND2(g5,a15)) is not empty iff b5 is not empty & not AND2(g5,
  a15) is not empty or not b5 is not empty & AND2(g5,a15) is not empty;
  hence a4 is not empty iff XOR2(b5,AND2(g5,a15)) is not empty by A7;
  XOR2(b4,AND2(g4,a15)) is not empty iff b4 is not empty & not AND2(g4,
  a15) is not empty or not b4 is not empty & AND2(g4,a15) is not empty;
  hence a3 is not empty iff XOR2(b4,AND2(g4,a15)) is not empty by A6;
  XOR2(b3,AND2(g3,a15)) is not empty iff b3 is not empty & not AND2(g3,
  a15) is not empty or not b3 is not empty & AND2(g3,a15) is not empty;
  hence a2 is not empty iff XOR2(b3,AND2(g3,a15)) is not empty by A5;
  XOR2(b2,AND2(g2,a15)) is not empty iff b2 is not empty & not AND2(g2,
  a15) is not empty or not b2 is not empty & AND2(g2,a15) is not empty;
  hence a1 is not empty iff XOR2(b2,AND2(g2,a15)) is not empty by A4;
  XOR2(b1,AND2(g1,a15)) is not empty iff b1 is not empty & not AND2(g1,
  a15) is not empty or not b1 is not empty & AND2(g1,a15) is not empty;
  hence a0 is not empty iff XOR2(b1,AND2(g1,a15)) is not empty by A3;
  XOR2(b0,AND2(g0,a15)) is not empty iff b0 is not empty & not AND2(g0,
  a15) is not empty or not b0 is not empty & AND2(g0,a15) is not empty;
  hence thesis by A2;
end;

begin


























theorem
  for g0,g1,g2,g3,g4,g5,g6,g7,g8,g9,g10,g11,g12, a0,a1,a2,a3,a4,a5,a6,a7
  ,a8,a9,a10,a11, b0,b1,b2,b3,b4,b5,b6,b7,b8,b9,b10,b11, z,p being set holds g0
  is not empty & not z is not empty & (b0 is not empty iff XOR2(p,a11) is not
  empty)& (b1 is not empty iff XOR2(a0,AND2(g1,b0)) is not empty)& (b2 is not
empty iff XOR2(a1,AND2(g2,b0)) is not empty)& (b3 is not empty iff XOR2(a2,AND2
(g3,b0)) is not empty)& (b4 is not empty iff XOR2(a3,AND2(g4,b0)) is not empty)
& (b5 is not empty iff XOR2(a4,AND2(g5,b0)) is not empty)& (b6 is not empty iff
  XOR2(a5,AND2(g6,b0)) is not empty)& (b7 is not empty iff XOR2(a6,AND2(g7,b0))
is not empty)& (b8 is not empty iff XOR2(a7,AND2(g8,b0)) is not empty)& (b9 is
not empty iff XOR2(a8,AND2(g9,b0)) is not empty)& (b10 is not empty iff XOR2(a9
  ,AND2(g10,b0)) is not empty)& (b11 is not empty iff XOR2(a10,AND2(g11,b0)) is
  not empty) implies (b11 is not empty iff XOR2(XOR2(a10,AND2(g11,a11)),XOR2(z,
AND2(g11,p))) is not empty) & (b10 is not empty iff XOR2(XOR2(a9,AND2(g10,a11))
,XOR2(z,AND2(g10,p))) is not empty) & (b9 is not empty iff XOR2(XOR2(a8,AND2(g9
  ,a11)),XOR2(z,AND2(g9,p))) is not empty) & (b8 is not empty iff XOR2(XOR2(a7,
  AND2(g8,a11)),XOR2(z,AND2(g8,p))) is not empty) & (b7 is not empty iff XOR2(
XOR2(a6,AND2(g7,a11)),XOR2(z,AND2(g7,p))) is not empty) & (b6 is not empty iff
XOR2(XOR2(a5,AND2(g6,a11)),XOR2(z,AND2(g6,p))) is not empty) & (b5 is not empty
  iff XOR2(XOR2(a4,AND2(g5,a11)),XOR2(z,AND2(g5,p))) is not empty) & (b4 is not
empty iff XOR2(XOR2(a3,AND2(g4,a11)),XOR2(z,AND2(g4,p))) is not empty) & (b3 is
not empty iff XOR2(XOR2(a2,AND2(g3,a11)),XOR2(z,AND2(g3,p))) is not empty) & (
b2 is not empty iff XOR2(XOR2(a1,AND2(g2,a11)),XOR2(z,AND2(g2,p))) is not empty
) & (b1 is not empty iff XOR2(XOR2(a0,AND2(g1,a11)),XOR2(z,AND2(g1,p))) is not
empty) & (b0 is not empty iff XOR2(XOR2(z, AND2(g0,a11)),XOR2(z,AND2(g0,p))) is
  not empty)
proof
  let g0,g1,g2,g3,g4,g5,g6,g7,g8,g9,g10,g11,g12, a0,a1,a2,a3,a4,a5,a6,a7,a8,a9
  ,a10,a11, b0,b1,b2,b3,b4,b5,b6,b7,b8,b9,b10,b11, z,p be set;
  assume that
A1: g0 is not empty and
A2: not z is not empty and
A3: b0 is not empty iff XOR2(p,a11) is not empty and
A4: b1 is not empty iff XOR2(a0,AND2(g1,b0)) is not empty and
A5: b2 is not empty iff XOR2(a1,AND2(g2,b0)) is not empty and
A6: b3 is not empty iff XOR2(a2,AND2(g3,b0)) is not empty and
A7: b4 is not empty iff XOR2(a3,AND2(g4,b0)) is not empty and
A8: b5 is not empty iff XOR2(a4,AND2(g5,b0)) is not empty and
A9: b6 is not empty iff XOR2(a5,AND2(g6,b0)) is not empty and
A10: b7 is not empty iff XOR2(a6,AND2(g7,b0)) is not empty and
A11: b8 is not empty iff XOR2(a7,AND2(g8,b0)) is not empty and
A12: b9 is not empty iff XOR2(a8,AND2(g9,b0)) is not empty and
A13: b10 is not empty iff XOR2(a9,AND2(g10,b0)) is not empty and
A14: b11 is not empty iff XOR2(a10,AND2(g11,b0)) is not empty;
  XOR2(XOR2(a10,AND2(g11,a11)),XOR2(z,AND2(g11,p))) is not empty iff (
a10 is not empty & (not g11 is not empty or not a11 is not empty) & (not g11 is
  not empty or not p is not empty) or not a10 is not empty & g11 is not empty &
  a11 is not empty & (not g11 is not empty or not p is not empty) ) or (not a10
  is not empty or g11 is not empty & a11 is not empty ) & (a10 is not empty or
not (g11 is not empty & a11 is not empty)) & g11 is not empty & p is not empty
  by A2;
  hence b11 is not empty iff XOR2(XOR2(a10,AND2(g11,a11)),XOR2(z,AND2(g11,p)))
  is not empty by A3,A14;
  XOR2(XOR2(a9,AND2(g10,a11)),XOR2(z,AND2(g10,p))) is not empty iff ( a9
is not empty & (not g10 is not empty or not a11 is not empty) & (not g10 is not
empty or not p is not empty) or not a9 is not empty & g10 is not empty & a11 is
  not empty & (not g10 is not empty or not p is not empty) ) or (not a9 is not
  empty or g10 is not empty & a11 is not empty ) & (a9 is not empty or not (g10
  is not empty & a11 is not empty)) & g10 is not empty & p is not empty by A2;
  hence b10 is not empty iff XOR2(XOR2(a9,AND2(g10,a11)),XOR2(z,AND2(g10,p)))
  is not empty by A3,A13;
  XOR2(XOR2(a8,AND2(g9,a11)),XOR2(z,AND2(g9,p))) is not empty iff ( a8 is
  not empty & (not g9 is not empty or not a11 is not empty) & (not g9 is not
empty or not p is not empty) or not a8 is not empty & g9 is not empty & a11 is
  not empty & (not g9 is not empty or not p is not empty) ) or (not a8 is not
empty or g9 is not empty & a11 is not empty ) & (a8 is not empty or not (g9 is
  not empty & a11 is not empty)) & g9 is not empty & p is not empty by A2;
  hence b9 is not empty iff XOR2(XOR2(a8,AND2(g9,a11)),XOR2(z,AND2(g9,p))) is
  not empty by A3,A12;
  XOR2(XOR2(a7,AND2(g8,a11)),XOR2(z,AND2(g8,p))) is not empty iff ( a7 is
  not empty & (not g8 is not empty or not a11 is not empty) & (not g8 is not
empty or not p is not empty) or not a7 is not empty & g8 is not empty & a11 is
  not empty & (not g8 is not empty or not p is not empty) ) or (not a7 is not
empty or g8 is not empty & a11 is not empty ) & (a7 is not empty or not (g8 is
  not empty & a11 is not empty)) & g8 is not empty & p is not empty by A2;
  hence b8 is not empty iff XOR2(XOR2(a7,AND2(g8,a11)),XOR2(z,AND2(g8,p))) is
  not empty by A3,A11;
  XOR2(XOR2(a6,AND2(g7,a11)),XOR2(z,AND2(g7,p))) is not empty iff ( a6 is
  not empty & (not g7 is not empty or not a11 is not empty) & (not g7 is not
empty or not p is not empty) or not a6 is not empty & g7 is not empty & a11 is
  not empty & (not g7 is not empty or not p is not empty) ) or (not a6 is not
empty or g7 is not empty & a11 is not empty ) & (a6 is not empty or not (g7 is
  not empty & a11 is not empty)) & g7 is not empty & p is not empty by A2;
  hence b7 is not empty iff XOR2(XOR2(a6,AND2(g7,a11)),XOR2(z,AND2(g7,p))) is
  not empty by A3,A10;
  XOR2(XOR2(a5,AND2(g6,a11)),XOR2(z,AND2(g6,p))) is not empty iff ( a5 is
  not empty & (not g6 is not empty or not a11 is not empty) & (not g6 is not
empty or not p is not empty) or not a5 is not empty & g6 is not empty & a11 is
  not empty & (not g6 is not empty or not p is not empty) ) or (not a5 is not
empty or g6 is not empty & a11 is not empty ) & (a5 is not empty or not (g6 is
  not empty & a11 is not empty)) & g6 is not empty & p is not empty by A2;
  hence b6 is not empty iff XOR2(XOR2(a5,AND2(g6,a11)),XOR2(z,AND2(g6,p))) is
  not empty by A3,A9;
  XOR2(XOR2(a4,AND2(g5,a11)),XOR2(z,AND2(g5,p))) is not empty iff ( a4 is
  not empty & (not g5 is not empty or not a11 is not empty) & (not g5 is not
empty or not p is not empty) or not a4 is not empty & g5 is not empty & a11 is
  not empty & (not g5 is not empty or not p is not empty) ) or (not a4 is not
empty or g5 is not empty & a11 is not empty ) & (a4 is not empty or not (g5 is
  not empty & a11 is not empty)) & g5 is not empty & p is not empty by A2;
  hence b5 is not empty iff XOR2(XOR2(a4,AND2(g5,a11)),XOR2(z,AND2(g5,p))) is
  not empty by A3,A8;
  XOR2(XOR2(a3,AND2(g4,a11)),XOR2(z,AND2(g4,p))) is not empty iff ( a3 is
  not empty & (not g4 is not empty or not a11 is not empty) & (not g4 is not
empty or not p is not empty) or not a3 is not empty & g4 is not empty & a11 is
  not empty & (not g4 is not empty or not p is not empty) ) or (not a3 is not
empty or g4 is not empty & a11 is not empty ) & (a3 is not empty or not (g4 is
  not empty & a11 is not empty)) & g4 is not empty & p is not empty by A2;
  hence b4 is not empty iff XOR2(XOR2(a3,AND2(g4,a11)),XOR2(z,AND2(g4,p))) is
  not empty by A3,A7;
  XOR2(XOR2(a2,AND2(g3,a11)),XOR2(z,AND2(g3,p))) is not empty iff ( a2 is
  not empty & (not g3 is not empty or not a11 is not empty) & (not g3 is not
empty or not p is not empty) or not a2 is not empty & g3 is not empty & a11 is
  not empty & (not g3 is not empty or not p is not empty) ) or (not a2 is not
empty or g3 is not empty & a11 is not empty ) & (a2 is not empty or not (g3 is
  not empty & a11 is not empty)) & g3 is not empty & p is not empty by A2;
  hence b3 is not empty iff XOR2(XOR2(a2,AND2(g3,a11)),XOR2(z,AND2(g3,p))) is
  not empty by A3,A6;
  XOR2(XOR2(a1,AND2(g2,a11)),XOR2(z,AND2(g2,p))) is not empty iff ( a1 is
  not empty & (not g2 is not empty or not a11 is not empty) & (not g2 is not
empty or not p is not empty) or not a1 is not empty & g2 is not empty & a11 is
  not empty & (not g2 is not empty or not p is not empty) ) or (not a1 is not
empty or g2 is not empty & a11 is not empty ) & (a1 is not empty or not (g2 is
  not empty & a11 is not empty)) & g2 is not empty & p is not empty by A2;
  hence b2 is not empty iff XOR2(XOR2(a1,AND2(g2,a11)),XOR2(z,AND2(g2,p))) is
  not empty by A3,A5;
  XOR2(XOR2(a0,AND2(g1,a11)),XOR2(z,AND2(g1,p))) is not empty iff ( a0 is
  not empty & (not g1 is not empty or not a11 is not empty) & (not g1 is not
empty or not p is not empty) or not a0 is not empty & g1 is not empty & a11 is
  not empty & (not g1 is not empty or not p is not empty) ) or (not a0 is not
empty or g1 is not empty & a11 is not empty ) & (a0 is not empty or not (g1 is
  not empty & a11 is not empty)) & g1 is not empty & p is not empty by A2;
  hence b1 is not empty iff XOR2(XOR2(a0,AND2(g1,a11)),XOR2(z,AND2(g1,p))) is
  not empty by A3,A4;
  XOR2(XOR2(z,AND2(g0,a11)),XOR2(z,AND2(g0,p))) is not empty iff ( z is
  not empty & (not g0 is not empty or not a11 is not empty) & (not g0 is not
  empty or not p is not empty) or not z is not empty & g0 is not empty & a11 is
  not empty & (not g0 is not empty or not p is not empty) ) or (not z is not
  empty or g0 is not empty & a11 is not empty ) & (z is not empty or not (g0 is
  not empty & a11 is not empty)) & g0 is not empty & p is not empty by A2;
  hence thesis by A1,A3;
end;

























theorem
  for g0,g1,g2,g3,g4,g5,g6,g7,g8,g9,g10,g11,g12,g13,g14,g15,g16, a0,a1,
a2,a3,a4,a5,a6,a7,a8,a9,a10,a11,a12,a13,a14,a15, b0,b1,b2,b3,b4,b5,b6,b7,b8,b9,
  b10,b11,b12,b13,b14,b15, z,p being set holds g0 is not empty & not z is not
  empty & (b0 is not empty iff XOR2(p,a15) is not empty)& (b1 is not empty iff
  XOR2(a0,AND2(g1,b0)) is not empty)& (b2 is not empty iff XOR2(a1,AND2(g2,b0))
is not empty)& (b3 is not empty iff XOR2(a2,AND2(g3,b0)) is not empty)& (b4 is
not empty iff XOR2(a3,AND2(g4,b0)) is not empty)& (b5 is not empty iff XOR2(a4,
  AND2(g5,b0)) is not empty)& (b6 is not empty iff XOR2(a5,AND2(g6,b0)) is not
  empty)& (b7 is not empty iff XOR2(a6,AND2(g7,b0)) is not empty)& (b8 is not
empty iff XOR2(a7,AND2(g8,b0)) is not empty)& (b9 is not empty iff XOR2(a8,AND2
  (g9,b0)) is not empty)& (b10 is not empty iff XOR2(a9,AND2(g10,b0)) is not
empty)& (b11 is not empty iff XOR2(a10,AND2(g11,b0)) is not empty)& (b12 is not
empty iff XOR2(a11,AND2(g12,b0)) is not empty)& (b13 is not empty iff XOR2(a12,
  AND2(g13,b0)) is not empty)& (b14 is not empty iff XOR2(a13,AND2(g14,b0)) is
not empty)& (b15 is not empty iff XOR2(a14,AND2(g15,b0)) is not empty) implies
(b15 is not empty iff XOR2(XOR2(a14,AND2(g15,a15)),XOR2(z,AND2(g15,p))) is not
empty) & (b14 is not empty iff XOR2(XOR2(a13,AND2(g14,a15)),XOR2(z,AND2(g14,p))
  ) is not empty) & (b13 is not empty iff XOR2(XOR2(a12,AND2(g13,a15)),XOR2(z,
AND2(g13,p))) is not empty) & (b12 is not empty iff XOR2(XOR2(a11,AND2(g12,a15)
),XOR2(z,AND2(g12,p))) is not empty) & (b11 is not empty iff XOR2(XOR2(a10,AND2
(g11,a15)),XOR2(z,AND2(g11,p))) is not empty) & (b10 is not empty iff XOR2(XOR2
  (a9,AND2(g10,a15)),XOR2(z,AND2(g10,p))) is not empty) & (b9 is not empty iff
XOR2(XOR2(a8,AND2(g9,a15)),XOR2(z,AND2(g9,p))) is not empty) & (b8 is not empty
  iff XOR2(XOR2(a7,AND2(g8,a15)),XOR2(z,AND2(g8,p))) is not empty) & (b7 is not
empty iff XOR2(XOR2(a6,AND2(g7,a15)),XOR2(z,AND2(g7,p))) is not empty) & (b6 is
not empty iff XOR2(XOR2(a5,AND2(g6,a15)),XOR2(z,AND2(g6,p))) is not empty) & (
b5 is not empty iff XOR2(XOR2(a4,AND2(g5,a15)),XOR2(z,AND2(g5,p))) is not empty
) & (b4 is not empty iff XOR2(XOR2(a3,AND2(g4,a15)),XOR2(z,AND2(g4,p))) is not
empty) & (b3 is not empty iff XOR2(XOR2(a2,AND2(g3,a15)),XOR2(z,AND2(g3,p))) is
not empty) & (b2 is not empty iff XOR2(XOR2(a1,AND2(g2,a15)),XOR2(z,AND2(g2,p))
) is not empty) & (b1 is not empty iff XOR2(XOR2(a0,AND2(g1,a15)),XOR2(z,AND2(
g1,p))) is not empty) & (b0 is not empty iff XOR2(XOR2(z, AND2(g0,a15)),XOR2(z,
  AND2(g0,p))) is not empty)
proof
  let g0,g1,g2,g3,g4,g5,g6,g7,g8,g9,g10,g11,g12,g13,g14,g15,g16, a0,a1,a2,a3,
  a4,a5,a6,a7,a8,a9,a10,a11,a12,a13,a14,a15, b0,b1,b2,b3,b4,b5,b6,b7,b8,b9,b10,
  b11,b12,b13,b14,b15, z,p be set;
  assume that
A1: g0 is not empty and
A2: not z is not empty and
A3: b0 is not empty iff XOR2(p,a15) is not empty and
A4: b1 is not empty iff XOR2(a0,AND2(g1,b0)) is not empty and
A5: b2 is not empty iff XOR2(a1,AND2(g2,b0)) is not empty and
A6: b3 is not empty iff XOR2(a2,AND2(g3,b0)) is not empty and
A7: b4 is not empty iff XOR2(a3,AND2(g4,b0)) is not empty and
A8: b5 is not empty iff XOR2(a4,AND2(g5,b0)) is not empty and
A9: b6 is not empty iff XOR2(a5,AND2(g6,b0)) is not empty and
A10: b7 is not empty iff XOR2(a6,AND2(g7,b0)) is not empty and
A11: b8 is not empty iff XOR2(a7,AND2(g8,b0)) is not empty and
A12: b9 is not empty iff XOR2(a8,AND2(g9,b0)) is not empty and
A13: b10 is not empty iff XOR2(a9,AND2(g10,b0)) is not empty and
A14: b11 is not empty iff XOR2(a10,AND2(g11,b0)) is not empty and
A15: b12 is not empty iff XOR2(a11,AND2(g12,b0)) is not empty and
A16: b13 is not empty iff XOR2(a12,AND2(g13,b0)) is not empty and
A17: b14 is not empty iff XOR2(a13,AND2(g14,b0)) is not empty and
A18: b15 is not empty iff XOR2(a14,AND2(g15,b0)) is not empty;
  XOR2(XOR2(a14,AND2(g15,a15)),XOR2(z,AND2(g15,p))) is not empty iff (
a14 is not empty & (not g15 is not empty or not a15 is not empty) & (not g15 is
  not empty or not p is not empty) or not a14 is not empty & g15 is not empty &
  a15 is not empty & (not g15 is not empty or not p is not empty) ) or (not a14
  is not empty or g15 is not empty & a15 is not empty ) & (a14 is not empty or
not (g15 is not empty & a15 is not empty)) & g15 is not empty & p is not empty
  by A2;
  hence b15 is not empty iff XOR2(XOR2(a14,AND2(g15,a15)),XOR2(z,AND2(g15,p)))
  is not empty by A3,A18;
  XOR2(XOR2(a13,AND2(g14,a15)),XOR2(z,AND2(g14,p))) is not empty iff (
a13 is not empty & (not g14 is not empty or not a15 is not empty) & (not g14 is
  not empty or not p is not empty) or not a13 is not empty & g14 is not empty &
  a15 is not empty & (not g14 is not empty or not p is not empty) ) or (not a13
  is not empty or g14 is not empty & a15 is not empty ) & (a13 is not empty or
not (g14 is not empty & a15 is not empty)) & g14 is not empty & p is not empty
  by A2;
  hence b14 is not empty iff XOR2(XOR2(a13,AND2(g14,a15)),XOR2(z,AND2(g14,p)))
  is not empty by A3,A17;
  XOR2(XOR2(a12,AND2(g13,a15)),XOR2(z,AND2(g13,p))) is not empty iff (
a12 is not empty & (not g13 is not empty or not a15 is not empty) & (not g13 is
  not empty or not p is not empty) or not a12 is not empty & g13 is not empty &
  a15 is not empty & (not g13 is not empty or not p is not empty) ) or (not a12
  is not empty or g13 is not empty & a15 is not empty ) & (a12 is not empty or
not (g13 is not empty & a15 is not empty)) & g13 is not empty & p is not empty
  by A2;
  hence b13 is not empty iff XOR2(XOR2(a12,AND2(g13,a15)),XOR2(z,AND2(g13,p)))
  is not empty by A3,A16;
  XOR2(XOR2(a11,AND2(g12,a15)),XOR2(z,AND2(g12,p))) is not empty iff (
a11 is not empty & (not g12 is not empty or not a15 is not empty) & (not g12 is
  not empty or not p is not empty) or not a11 is not empty & g12 is not empty &
  a15 is not empty & (not g12 is not empty or not p is not empty) ) or (not a11
  is not empty or g12 is not empty & a15 is not empty ) & (a11 is not empty or
not (g12 is not empty & a15 is not empty)) & g12 is not empty & p is not empty
  by A2;
  hence b12 is not empty iff XOR2(XOR2(a11,AND2(g12,a15)),XOR2(z,AND2(g12,p)))
  is not empty by A3,A15;
  XOR2(XOR2(a10,AND2(g11,a15)),XOR2(z,AND2(g11,p))) is not empty iff (
a10 is not empty & (not g11 is not empty or not a15 is not empty) & (not g11 is
  not empty or not p is not empty) or not a10 is not empty & g11 is not empty &
  a15 is not empty & (not g11 is not empty or not p is not empty) ) or (not a10
  is not empty or g11 is not empty & a15 is not empty ) & (a10 is not empty or
not (g11 is not empty & a15 is not empty)) & g11 is not empty & p is not empty
  by A2;
  hence b11 is not empty iff XOR2(XOR2(a10,AND2(g11,a15)),XOR2(z,AND2(g11,p)))
  is not empty by A3,A14;
  XOR2(XOR2(a9,AND2(g10,a15)),XOR2(z,AND2(g10,p))) is not empty iff ( a9
is not empty & (not g10 is not empty or not a15 is not empty) & (not g10 is not
empty or not p is not empty) or not a9 is not empty & g10 is not empty & a15 is
  not empty & (not g10 is not empty or not p is not empty) ) or (not a9 is not
  empty or g10 is not empty & a15 is not empty ) & (a9 is not empty or not (g10
  is not empty & a15 is not empty)) & g10 is not empty & p is not empty by A2;
  hence b10 is not empty iff XOR2(XOR2(a9,AND2(g10,a15)),XOR2(z,AND2(g10,p)))
  is not empty by A3,A13;
  XOR2(XOR2(a8,AND2(g9,a15)),XOR2(z,AND2(g9,p))) is not empty iff ( a8 is
  not empty & (not g9 is not empty or not a15 is not empty) & (not g9 is not
empty or not p is not empty) or not a8 is not empty & g9 is not empty & a15 is
  not empty & (not g9 is not empty or not p is not empty) ) or (not a8 is not
empty or g9 is not empty & a15 is not empty ) & (a8 is not empty or not (g9 is
  not empty & a15 is not empty)) & g9 is not empty & p is not empty by A2;
  hence b9 is not empty iff XOR2(XOR2(a8,AND2(g9,a15)),XOR2(z,AND2(g9,p))) is
  not empty by A3,A12;
  XOR2(XOR2(a7,AND2(g8,a15)),XOR2(z,AND2(g8,p))) is not empty iff ( a7 is
  not empty & (not g8 is not empty or not a15 is not empty) & (not g8 is not
empty or not p is not empty) or not a7 is not empty & g8 is not empty & a15 is
  not empty & (not g8 is not empty or not p is not empty) ) or (not a7 is not
empty or g8 is not empty & a15 is not empty ) & (a7 is not empty or not (g8 is
  not empty & a15 is not empty)) & g8 is not empty & p is not empty by A2;
  hence b8 is not empty iff XOR2(XOR2(a7,AND2(g8,a15)),XOR2(z,AND2(g8,p))) is
  not empty by A3,A11;
  XOR2(XOR2(a6,AND2(g7,a15)),XOR2(z,AND2(g7,p))) is not empty iff ( a6 is
  not empty & (not g7 is not empty or not a15 is not empty) & (not g7 is not
empty or not p is not empty) or not a6 is not empty & g7 is not empty & a15 is
  not empty & (not g7 is not empty or not p is not empty) ) or (not a6 is not
empty or g7 is not empty & a15 is not empty ) & (a6 is not empty or not (g7 is
  not empty & a15 is not empty)) & g7 is not empty & p is not empty by A2;
  hence b7 is not empty iff XOR2(XOR2(a6,AND2(g7,a15)),XOR2(z,AND2(g7,p))) is
  not empty by A3,A10;
  XOR2(XOR2(a5,AND2(g6,a15)),XOR2(z,AND2(g6,p))) is not empty iff ( a5 is
  not empty & (not g6 is not empty or not a15 is not empty) & (not g6 is not
empty or not p is not empty) or not a5 is not empty & g6 is not empty & a15 is
  not empty & (not g6 is not empty or not p is not empty) ) or (not a5 is not
empty or g6 is not empty & a15 is not empty ) & (a5 is not empty or not (g6 is
  not empty & a15 is not empty)) & g6 is not empty & p is not empty by A2;
  hence b6 is not empty iff XOR2(XOR2(a5,AND2(g6,a15)),XOR2(z,AND2(g6,p))) is
  not empty by A3,A9;
  XOR2(XOR2(a4,AND2(g5,a15)),XOR2(z,AND2(g5,p))) is not empty iff ( a4 is
  not empty & (not g5 is not empty or not a15 is not empty) & (not g5 is not
empty or not p is not empty) or not a4 is not empty & g5 is not empty & a15 is
  not empty & (not g5 is not empty or not p is not empty) ) or (not a4 is not
empty or g5 is not empty & a15 is not empty ) & (a4 is not empty or not (g5 is
  not empty & a15 is not empty)) & g5 is not empty & p is not empty by A2;
  hence b5 is not empty iff XOR2(XOR2(a4,AND2(g5,a15)),XOR2(z,AND2(g5,p))) is
  not empty by A3,A8;
  XOR2(XOR2(a3,AND2(g4,a15)),XOR2(z,AND2(g4,p))) is not empty iff ( a3 is
  not empty & (not g4 is not empty or not a15 is not empty) & (not g4 is not
empty or not p is not empty) or not a3 is not empty & g4 is not empty & a15 is
  not empty & (not g4 is not empty or not p is not empty) ) or (not a3 is not
empty or g4 is not empty & a15 is not empty ) & (a3 is not empty or not (g4 is
  not empty & a15 is not empty)) & g4 is not empty & p is not empty by A2;
  hence b4 is not empty iff XOR2(XOR2(a3,AND2(g4,a15)),XOR2(z,AND2(g4,p))) is
  not empty by A3,A7;
  XOR2(XOR2(a2,AND2(g3,a15)),XOR2(z,AND2(g3,p))) is not empty iff ( a2 is
  not empty & (not g3 is not empty or not a15 is not empty) & (not g3 is not
empty or not p is not empty) or not a2 is not empty & g3 is not empty & a15 is
  not empty & (not g3 is not empty or not p is not empty) ) or (not a2 is not
empty or g3 is not empty & a15 is not empty ) & (a2 is not empty or not (g3 is
  not empty & a15 is not empty)) & g3 is not empty & p is not empty by A2;
  hence b3 is not empty iff XOR2(XOR2(a2,AND2(g3,a15)),XOR2(z,AND2(g3,p))) is
  not empty by A3,A6;
  XOR2(XOR2(a1,AND2(g2,a15)),XOR2(z,AND2(g2,p))) is not empty iff ( a1 is
  not empty & (not g2 is not empty or not a15 is not empty) & (not g2 is not
empty or not p is not empty) or not a1 is not empty & g2 is not empty & a15 is
  not empty & (not g2 is not empty or not p is not empty) ) or (not a1 is not
empty or g2 is not empty & a15 is not empty ) & (a1 is not empty or not (g2 is
  not empty & a15 is not empty)) & g2 is not empty & p is not empty by A2;
  hence b2 is not empty iff XOR2(XOR2(a1,AND2(g2,a15)),XOR2(z,AND2(g2,p))) is
  not empty by A3,A5;
  XOR2(XOR2(a0,AND2(g1,a15)),XOR2(z,AND2(g1,p))) is not empty iff ( a0 is
  not empty & (not g1 is not empty or not a15 is not empty) & (not g1 is not
empty or not p is not empty) or not a0 is not empty & g1 is not empty & a15 is
  not empty & (not g1 is not empty or not p is not empty) ) or (not a0 is not
empty or g1 is not empty & a15 is not empty ) & (a0 is not empty or not (g1 is
  not empty & a15 is not empty)) & g1 is not empty & p is not empty by A2;
  hence b1 is not empty iff XOR2(XOR2(a0,AND2(g1,a15)),XOR2(z,AND2(g1,p))) is
  not empty by A3,A4;
  XOR2(XOR2(z,AND2(g0,a15)),XOR2(z,AND2(g0,p))) is not empty iff ( z is
  not empty & (not g0 is not empty or not a15 is not empty) & (not g0 is not
  empty or not p is not empty) or not z is not empty & g0 is not empty & a15 is
  not empty & (not g0 is not empty or not p is not empty) ) or (not z is not
  empty or g0 is not empty & a15 is not empty ) & (z is not empty or not (g0 is
  not empty & a15 is not empty)) & g0 is not empty & p is not empty by A2;
  hence thesis by A1,A3;
end;














begin



definition
  let x0,x1,y0,y1 be set;
  func MULT210(x1,y1,x0,y0) -> set equals
  AND2(x0,y0);
  correctness;
  func MULT211(x1,y1,x0,y0) -> set equals
  ADD1(AND2(x1,y0),AND2(x0,y1),{});
  correctness;
  func MULT212(x1,y1,x0,y0) -> set equals
  ADD2({},AND2(x1,y1),AND2(x1,y0),AND2
  (x0,y1),{});
  correctness;
  func MULT213(x1,y1,x0,y0) -> set equals
  CARR2({},AND2(x1,y1),AND2(x1,y0),
  AND2(x0,y1),{});
  correctness;
end;









theorem
  for x0,x1,y0,y1,z0,z1,z2,z3,q00,q01,c01,q11,c11 being set holds (q00
is not empty iff AND2(x0,y0) is not empty)& (q01 is not empty iff XOR3(AND2(x1,
y0),AND2(x0,y1),{} ) is not empty)& (c01 is not empty iff MAJ3(AND2(x1,y0),AND2
(x0,y1),{} ) is not empty)& (q11 is not empty iff XOR3(AND2(x1,y1),{} ,c01) is
not empty)& (c11 is not empty iff MAJ3(AND2(x1,y1),{} ,c01) is not empty)& (z0
is not empty iff q00 is not empty)& (z1 is not empty iff q01 is not empty)& (z2
  is not empty iff q11 is not empty)& (z3 is not empty iff c11 is not empty)
  implies (z0 is not empty iff MULT210(x1,y1,x0,y0) is not empty)& (z1 is not
empty iff MULT211(x1,y1,x0,y0) is not empty)& (z2 is not empty iff MULT212(x1,
y1,x0,y0) is not empty)& (z3 is not empty iff MULT213(x1,y1,x0,y0) is not empty
  )
proof
  let x0,x1,y0,y1,z0,z1,z2,z3,q00,q01,c01,q11,c11 be set;
  assume that
A1: q00 is not empty iff AND2(x0,y0) is not empty and
A2: q01 is not empty iff XOR3(AND2(x1,y0),AND2(x0,y1),{} ) is not empty and
A3: c01 is not empty iff MAJ3(AND2(x1,y0),AND2(x0,y1),{} ) is not empty and
A4: q11 is not empty iff XOR3(AND2(x1,y1),{} ,c01) is not empty and
A5: c11 is not empty iff MAJ3(AND2(x1,y1),{} ,c01) is not empty and
A6: z0 is not empty iff q00 is not empty and
A7: z1 is not empty iff q01 is not empty and
A8: z2 is not empty iff q11 is not empty and
A9: z3 is not empty iff c11 is not empty;
  thus z0 is not empty iff MULT210(x1,y1,x0,y0) is not empty by A1,A6;
  thus z1 is not empty iff MULT211(x1,y1,x0,y0) is not empty by A2,A7;
  set m212 = MULT212(x1,y1,x0,y0);
  set x1y1 = AND2(x1,y1);
  set x0y1 = AND2(x0,y1);
  set x1y0 = AND2(x1,y0);
  m212 = XOR3({},x1y1,MAJ3(x1y0,x0y1,{})) by GATE_1:def 35;
  then
  m212 is not empty iff x1y1 is not empty & not MAJ3(x1y0,x0y1,{}) is not
  empty or not x1y1 is not empty & MAJ3(x1y0,x0y1,{}) is not empty;
  hence z2 is not empty iff MULT212(x1,y1,x0,y0) is not empty by A3,A4,A8;
  set m213 = MULT213(x1,y1,x0,y0);
  m213 = MAJ3({},x1y1,MAJ3(x1y0,x0y1,{})) by GATE_1:def 36;
  then m213 is not empty iff x1y1 is not empty & MAJ3(x1y0,x0y1,{}) is not
  empty;
  hence thesis by A3,A5,A9;
end;







definition
  let x0,x1,x2,y0,y1 be set;
  func MULT310(x2,x1,y1,x0,y0) -> set equals
  AND2(x0,y0);
  correctness;
  func MULT311(x2,x1,y1,x0,y0) -> set equals
  ADD1(AND2(x1,y0),AND2(x0,y1),{});
  correctness;
  func MULT312(x2,x1,y1,x0,y0) -> set equals
  ADD2(AND2(x2,y0),AND2(x1,y1),
  AND2(x1,y0),AND2(x0,y1),{});
  correctness;
  func MULT313(x2,x1,y1,x0,y0) -> set equals
  ADD3({} ,AND2(x2,y1), AND2(x2,y0)
  ,AND2(x1,y1), AND2(x1,y0),AND2(x0,y1),{});
  correctness;
  func MULT314(x2,x1,y1,x0,y0) -> set equals
  CARR3({} ,AND2(x2,y1), AND2(x2,y0
  ),AND2(x1,y1), AND2(x1,y0),AND2(x0,y1),{});
  correctness;
end;




definition
  let x0,x1,x2,y0,y1,y2 be set;
  func MULT321(x2,y2,x1,y1,x0,y0) -> set equals
  ADD1(MULT312(x2,x1,y1,x0,y0),
  AND2(x0,y2),{});
  correctness;
  func MULT322(x2,y2,x1,y1,x0,y0) -> set equals
  ADD2(MULT313(x2,x1,y1,x0,y0),
  AND2(x1,y2), MULT312(x2,x1,y1,x0,y0),AND2(x0,y2),{});
  correctness;
  func MULT323(x2,y2,x1,y1,x0,y0) -> set equals
  ADD3(MULT314(x2,x1,y1,x0,y0),
AND2(x2,y2), MULT313(x2,x1,y1,x0,y0),AND2(x1,y2), MULT312(x2,x1,y1,x0,y0),AND2(
  x0,y2),{});
  correctness;
  func MULT324(x2,y2,x1,y1,x0,y0) -> set equals
  CARR3(MULT314(x2,x1,y1,x0,y0),
AND2(x2,y2), MULT313(x2,x1,y1,x0,y0),AND2(x1,y2), MULT312(x2,x1,y1,x0,y0),AND2(
  x0,y2),{});
  correctness;
end;






theorem 
  for x0,x1,x2,y0,y1,y2,z0,z1,z2,z3,z4,z5, q00,q01,q02,c01,c02,q11,q12,
  c11,c12,q21,q22,c21,c22 being set holds (q00 is not empty iff AND2(x0,y0) is
  not empty)& (q01 is not empty iff XOR3(AND2(x1,y0),AND2(x0,y1),{} ) is not
empty)& (c01 is not empty iff MAJ3(AND2(x1,y0),AND2(x0,y1),{} ) is not empty)&
(q02 is not empty iff XOR3(AND2(x2,y0),AND2(x1,y1),{} ) is not empty)& (c02 is
  not empty iff MAJ3(AND2(x2,y0),AND2(x1,y1),{} ) is not empty)& (q11 is not
empty iff XOR3(q02,AND2(x0,y2),c01) is not empty)& (c11 is not empty iff MAJ3(
q02,AND2(x0,y2),c01) is not empty)& (q12 is not empty iff XOR3(AND2(x2,y1),AND2
(x1,y2),c02) is not empty)& (c12 is not empty iff MAJ3(AND2(x2,y1),AND2(x1,y2),
c02) is not empty)& (q21 is not empty iff XOR3(q12,{} ,c11) is not empty)& (c21
  is not empty iff MAJ3(q12,{} ,c11) is not empty)& (q22 is not empty iff XOR3(
AND2(x2,y2),c21,c12) is not empty)& (c22 is not empty iff MAJ3(AND2(x2,y2),c21,
  c12) is not empty)& (z0 is not empty iff q00 is not empty)& (z1 is not empty
iff q01 is not empty)& (z2 is not empty iff q11 is not empty)& (z3 is not empty
iff q21 is not empty)& (z4 is not empty iff q22 is not empty)& (z5 is not empty
iff c22 is not empty) implies (z0 is not empty iff MULT310(x2, x1,y1,x0,y0) is
not empty)& (z1 is not empty iff MULT311(x2, x1,y1,x0,y0) is not empty)& (z2 is
  not empty iff MULT321(x2,y2,x1,y1,x0,y0) is not empty)& (z3 is not empty iff
MULT322(x2,y2,x1,y1,x0,y0) is not empty)& (z4 is not empty iff MULT323(x2,y2,x1
  ,y1,x0,y0) is not empty)& (z5 is not empty iff MULT324(x2,y2,x1,y1,x0,y0) is
  not empty)
proof
  let x0,x1,x2,y0,y1,y2,z0,z1,z2,z3,z4,z5, q00,q01,q02,c01,c02,q11,q12,c11,c12
  ,q21,q22,c21,c22 be set;
  assume that
A1: q00 is not empty iff AND2(x0,y0) is not empty and
A2: q01 is not empty iff XOR3(AND2(x1,y0),AND2(x0,y1),{} ) is not empty and
A3: ( c01 is not empty iff MAJ3(AND2(x1,y0),AND2(x0,y1),{} ) is not
  empty)&( q02 is not empty iff XOR3(AND2(x2,y0),AND2(x1,y1),{} ) is not empty)
  and
A4: c02 is not empty iff MAJ3(AND2(x2,y0),AND2(x1,y1),{} ) is not empty and
A5: q11 is not empty iff XOR3(q02,AND2(x0,y2),c01) is not empty and
A6: c11 is not empty iff MAJ3(q02,AND2(x0,y2),c01) is not empty and
A7: q12 is not empty iff XOR3(AND2(x2,y1),AND2(x1,y2),c02) is not empty and
A8: c12 is not empty iff MAJ3(AND2(x2,y1),AND2(x1,y2),c02) is not empty and
A9: q21 is not empty iff XOR3(q12,{} ,c11) is not empty and
A10: c21 is not empty iff MAJ3(q12,{} ,c11) is not empty and
A11: q22 is not empty iff XOR3(AND2(x2,y2),c21,c12) is not empty and
A12: c22 is not empty iff MAJ3(AND2(x2,y2),c21,c12) is not empty and
A13: z0 is not empty iff q00 is not empty and
A14: z1 is not empty iff q01 is not empty and
A15: z2 is not empty iff q11 is not empty and
A16: z3 is not empty iff q21 is not empty and
A17: z4 is not empty iff q22 is not empty and
A18: z5 is not empty iff c22 is not empty;
  set x0y2 = AND2(x0,y2);
A19: c11 is not empty iff q02 is not empty & x0y2 is not empty or x0y2 is
  not empty & c01 is not empty or c01 is not empty & q02 is not empty by A6;
  thus z0 is not empty iff MULT310(x2,x1,y1,x0,y0) is not empty by A1,A13;
  thus z1 is not empty iff MULT311(x2,x1,y1,x0,y0) is not empty by A2,A14;
  set m312 = MULT312(x2,x1,y1,x0,y0);
  set x1y1 = AND2(x1,y1);
  set x0y1 = AND2(x0,y1);
  set x2y0 = AND2(x2,y0);
  set x1y0 = AND2(x1,y0);
A20: m312 = XOR3(x2y0,x1y1,MAJ3(x1y0,x0y1,{})) by GATE_1:def 35;
  then
A21: m312 is not empty iff ( x2y0 is not empty & not x1y1 is not empty or
  not x2y0 is not empty & x1y1 is not empty ) & not MAJ3(x1y0,x0y1,{}) is not
  empty or not ( x2y0 is not empty & not x1y1 is not empty or not x2y0 is not
  empty & x1y1 is not empty ) & MAJ3(x1y0,x0y1,{}) is not empty;
A22: m312 is not empty iff ( x2y0 is not empty & not x1y1 is not empty or
not x2y0 is not empty & x1y1 is not empty ) & not (x1y0 is not empty & x0y1 is
  not empty) or not ( x2y0 is not empty & not x1y1 is not empty or not x2y0 is
not empty & x1y1 is not empty ) & x1y0 is not empty & x0y1 is not empty by A20;
  q11 is not empty iff ( q02 is not empty & not x0y2 is not empty or not
  q02 is not empty & x0y2 is not empty ) & not c01 is not empty or not ( q02 is
not empty & not x0y2 is not empty or not q02 is not empty & x0y2 is not empty )
  & c01 is not empty by A5;
  hence z2 is not empty iff MULT321(x2,y2,x1,y1,x0,y0) is not empty by A3,A15
,A21;
  set x1y2 = AND2(x1,y2);
  set x2y1 = AND2(x2,y1);
A23: q12 is not empty iff ( x2y1 is not empty & not x1y2 is not empty or not
x2y1 is not empty & x1y2 is not empty ) & not c02 is not empty or not ( x2y1 is
not empty & not x1y2 is not empty or not x2y1 is not empty & x1y2 is not empty
  ) & c02 is not empty by A7;
  set m324 = MULT324(x2,y2,x1,y1,x0,y0);
  set m323 = MULT323(x2,y2,x1,y1,x0,y0);
  set m314 = MULT314(x2,x1,y1,x0,y0);
  set x2y2 = AND2(x2,y2);
A24: m314 = MAJ3({},x2y1,CARR2(x2y0,x1y1,x1y0,x0y1,{})) by GATE_1:def 38
    .= MAJ3({},x2y1,MAJ3(x2y0,x1y1,MAJ3(x1y0,x0y1,{}))) by GATE_1:def 36;
  set m322 = MULT322(x2,y2,x1,y1,x0,y0);
  set m313 = MULT313(x2,x1,y1,x0,y0);
A25: m313 = XOR3({},x2y1,CARR2(x2y0,x1y1,x1y0,x0y1,{})) by GATE_1:def 37
    .= XOR3({},x2y1,MAJ3(x2y0,x1y1,MAJ3(x1y0,x0y1,{}))) by GATE_1:def 36;
  m322 = XOR3(m313,x1y2,MAJ3(m312,x0y2,{})) by GATE_1:def 35;
  hence z3 is not empty iff MULT322(x2,y2,x1,y1,x0,y0) is not empty by A3,A4,A9
,A16,A19,A23,A22,A25;
A26: m323 = XOR3(m314,x2y2,CARR2(m313,x1y2,m312,x0y2,{})) by GATE_1:def 37
    .= XOR3(m314,x2y2,MAJ3(m313,x1y2,MAJ3(m312,x0y2,{}))) by GATE_1:def 36;
  q22 is not empty iff ( x2y2 is not empty & not c21 is not empty or not
x2y2 is not empty & c21 is not empty ) & not c12 is not empty or not ( x2y2 is
not empty & not c21 is not empty or not x2y2 is not empty & c21 is not empty )
  & c12 is not empty by A11;
  hence
  z4 is not empty iff MULT323(x2,y2,x1,y1,x0,y0) is not empty by A3,A4,A8,A10
,A17,A19,A23,A22,A25,A24,A26;
A27: m324 = MAJ3(m314,x2y2,CARR2(m313,x1y2,m312,x0y2,{})) by GATE_1:def 38
    .= MAJ3(m314,x2y2,MAJ3(m313,x1y2,MAJ3(m312,x0y2,{}))) by GATE_1:def 36;
  c22 is not empty iff x2y2 is not empty & c21 is not empty or c21 is not
  empty & c12 is not empty or c12 is not empty & x2y2 is not empty by A12;
  hence thesis by A3,A4,A8,A10,A18,A19,A23,A22,A25,A24,A27;
end;

begin 








theorem
  for x0,x1,x2,y0,y1,y2,z0,z1,z2,z3,z4,z5, q00,q01,q02,q03,c01,c02,c03,
  q11,q12,q13,c11,c12,c13 being set holds (q00 is not empty iff AND2(x0,y0) is
not empty)& (q01 is not empty iff XOR3(AND2(x1,y0),AND2(x0,y1),{}) is not empty
  )& (c01 is not empty iff MAJ3(AND2(x1,y0),AND2(x0,y1),{}) is not empty)& (q02
is not empty iff XOR3(AND2(x2,y0),AND2(x1,y1),AND2(x0,y2)) is not empty)& (c02
is not empty iff MAJ3(AND2(x2,y0),AND2(x1,y1),AND2(x0,y2)) is not empty)& (q03
  is not empty iff XOR3(AND2(x2,y1),AND2(x1,y2),{}) is not empty)& (c03 is not
empty iff MAJ3(AND2(x2,y1),AND2(x1,y2),{}) is not empty)& (q11 is not empty iff
  XOR3(q02,c01,{}) is not empty)& (c11 is not empty iff MAJ3(q02,c01,{}) is not
  empty)& (q12 is not empty iff XOR3(q03,c02,c11) is not empty)& (c12 is not
empty iff MAJ3(q03,c02,c11) is not empty)& (q13 is not empty iff XOR3(AND2(x2,
y2),c03,c12) is not empty)& (c13 is not empty iff MAJ3(AND2(x2,y2),c03,c12) is
not empty)& (z0 is not empty iff q00 is not empty)& (z1 is not empty iff q01 is
not empty)& (z2 is not empty iff q11 is not empty)& (z3 is not empty iff q12 is
not empty)& (z4 is not empty iff q13 is not empty)& (z5 is not empty iff c13 is
not empty) implies (z0 is not empty iff MULT310(x2, x1,y1,x0,y0) is not empty)&
  (z1 is not empty iff MULT311(x2, x1,y1,x0,y0) is not empty)& (z2 is not empty
iff MULT321(x2,y2,x1,y1,x0,y0) is not empty)& (z3 is not empty iff MULT322(x2,
y2,x1,y1,x0,y0) is not empty)& (z4 is not empty iff MULT323(x2,y2,x1,y1,x0,y0)
  is not empty)& (z5 is not empty iff MULT324(x2,y2,x1,y1,x0,y0) is not empty)
proof
  let x0,x1,x2,y0,y1,y2,z0,z1,z2,z3,z4,z5, q00,q01,q02,q03,c01,c02,c03,q11,q12
  ,q13,c11,c12,c13 be set;
  assume that
A1: q00 is not empty iff AND2(x0,y0) is not empty and
A2: q01 is not empty iff XOR3(AND2(x1,y0),AND2(x0,y1),{}) is not empty and
A3: c01 is not empty iff MAJ3(AND2(x1,y0),AND2(x0,y1),{}) is not empty and
A4: q02 is not empty iff XOR3(AND2(x2,y0),AND2(x1,y1),AND2(x0,y2)) is
  not empty and
A5: c02 is not empty iff MAJ3(AND2(x2,y0),AND2(x1,y1),AND2(x0,y2)) is
  not empty and
A6: q03 is not empty iff XOR3(AND2(x2,y1),AND2(x1,y2),{}) is not empty and
A7: c03 is not empty iff MAJ3(AND2(x2,y1),AND2(x1,y2),{}) is not empty and
A8: q11 is not empty iff XOR3(q02,c01,{}) is not empty and
A9: c11 is not empty iff MAJ3(q02,c01,{}) is not empty and
A10: q12 is not empty iff XOR3(q03,c02,c11) is not empty and
A11: c12 is not empty iff MAJ3(q03,c02,c11) is not empty and
A12: q13 is not empty iff XOR3(AND2(x2,y2),c03,c12) is not empty and
A13: c13 is not empty iff MAJ3(AND2(x2,y2),c03,c12) is not empty and
A14: z0 is not empty iff q00 is not empty and
A15: z1 is not empty iff q01 is not empty and
A16: z2 is not empty iff q11 is not empty and
A17: z3 is not empty iff q12 is not empty and
A18: z4 is not empty iff q13 is not empty and
A19: z5 is not empty iff c13 is not empty;
  thus z0 is not empty iff MULT310(x2,x1,y1,x0,y0) is not empty by A1,A14;
  thus z1 is not empty iff MULT311(x2,x1,y1,x0,y0) is not empty by A2,A15;
  set m312 = MULT312(x2,x1,y1,x0,y0);
  set x0y2 = AND2(x0,y2);
  set x1y1 = AND2(x1,y1);
  set x0y1 = AND2(x0,y1);
  set x2y0 = AND2(x2,y0);
  set x1y0 = AND2(x1,y0);
A20: m312 = XOR3(x2y0,x1y1,MAJ3(x1y0,x0y1,{})) by GATE_1:def 35;
  set m323 = MULT323(x2,y2,x1,y1,x0,y0);
  set m314 = MULT314(x2,x1,y1,x0,y0);
  set x2y2 = AND2(x2,y2);
  set m322 = MULT322(x2,y2,x1,y1,x0,y0);
  set m313 = MULT313(x2,x1,y1,x0,y0);
  set x1y2 = AND2(x1,y2);
  set x2y1 = AND2(x2,y1);
A21: q03 is not empty iff x2y1 is not empty & not x1y2 is not empty or not
  x2y1 is not empty & x1y2 is not empty by A6;
A22: q02 is not empty iff ( x2y0 is not empty & not x1y1 is not empty or not
  x2y0 is not empty & x1y1 is not empty ) & not x0y2 is not empty or not ( x2y0
  is not empty & not x1y1 is not empty or not x2y0 is not empty & x1y1 is not
  empty ) & x0y2 is not empty by A4;
  hence z2 is not empty iff MULT321(x2,y2,x1,y1,x0,y0) is not empty by A3,A8
,A16,A20;
A23: m313 = XOR3({},x2y1,CARR2(x2y0,x1y1,x1y0,x0y1,{})) by GATE_1:def 37
    .= XOR3({},x2y1,MAJ3(x2y0,x1y1,MAJ3(x1y0,x0y1,{}))) by GATE_1:def 36;
  m322 = XOR3(m313,x1y2,MAJ3(m312,x0y2,{})) by GATE_1:def 35;
  hence
  z3 is not empty iff MULT322(x2,y2,x1,y1,x0,y0) is not empty by A3,A5,A9,A10
,A17,A22,A21,A20,A23;
A24: m314 = MAJ3({},x2y1,CARR2(x2y0,x1y1,x1y0,x0y1,{})) by GATE_1:def 38
    .= MAJ3({},x2y1,MAJ3(x2y0,x1y1,MAJ3(x1y0,x0y1,{}))) by GATE_1:def 36;
  set m324 = MULT324(x2,y2,x1,y1,x0,y0);
A25: m324 = MAJ3(m314,x2y2,CARR2(m313,x1y2,m312,x0y2,{})) by GATE_1:def 38
    .= MAJ3(m314,x2y2,MAJ3(m313,x1y2,MAJ3(m312,x0y2,{}))) by GATE_1:def 36;
A26: m323 = XOR3(m314,x2y2,CARR2(m313,x1y2,m312,x0y2,{})) by GATE_1:def 37
    .= XOR3(m314,x2y2,MAJ3(m313,x1y2,MAJ3(m312,x0y2,{}))) by GATE_1:def 36;
  q13 is not empty iff ( x2y2 is not empty & not c03 is not empty or not
x2y2 is not empty & c03 is not empty ) & not c12 is not empty or not ( x2y2 is
not empty & not c03 is not empty or not x2y2 is not empty & c03 is not empty )
  & c12 is not empty by A12;
  hence
  z4 is not empty iff MULT323(x2,y2,x1,y1,x0,y0) is not empty by A3,A5,A7,A9
,A11,A18,A22,A21,A20,A23,A24,A26;
  c13 is not empty iff x2y2 is not empty & c03 is not empty or c03 is not
  empty & c12 is not empty or c12 is not empty & x2y2 is not empty by A13;
  hence thesis by A3,A5,A7,A9,A11,A19,A22,A21,A20,A23,A24,A25;
end;







notation
  let a1,b1,c be set;
  synonym CLAADD1(a1,b1,c) for XOR3(a1,b1,c);
  synonym CLACARR1(a1,b1,c) for MAJ3(a1,b1,c);
end;



definition
  let a1,b1,a2,b2,c be set;

  func CLAADD2(a2,b2,a1,b1,c) -> set equals
  XOR3(a2,b2,MAJ3(a1,b1,c));
  correctness;
  func CLACARR2(a2,b2,a1,b1,c) -> set equals
  OR2(AND2(a2,b2),AND2(OR2(a2,b2),
  MAJ3(a1,b1,c)));
  correctness;
end;



definition
  let a1,b1,a2,b2,a3,b3,c be set;
  func CLAADD3(a3,b3,a2,b2,a1,b1,c) -> set equals
  XOR3(a3,b3,CLACARR2(a2,b2,a1
  ,b1,c));
  correctness;
  func CLACARR3(a3,b3,a2,b2,a1,b1,c) -> set equals
  OR3(AND2(a3,b3),AND2(OR2(a3
  ,b3),AND2(a2,b2)), AND3(OR2(a3,b3),OR2(a2,b2),MAJ3(a1,b1,c)));
  correctness;
end;



definition
  let a1,b1,a2,b2,a3,b3,a4,b4,c be set;
  func CLAADD4(a4,b4,a3,b3,a2,b2,a1,b1,c) -> set equals
  XOR3(a4,b4,CLACARR3(a3
  ,b3,a2,b2,a1,b1,c));
  correctness;
  func CLACARR4(a4,b4,a3,b3,a2,b2,a1,b1,c) -> set equals
  OR4(AND2(a4,b4),AND2(
OR2(a4,b4),AND2(a3,b3)), AND3(OR2(a4,b4),OR2(a3,b3),AND2(a2,b2)), AND4(OR2(a4,
  b4),OR2(a3,b3),OR2(a2,b2),MAJ3(a1,b1,c)));
  correctness;
end;







theorem 
  for x0,x1,x2,y0,y1,y2,z0,z1,z2,z3,z4,z5, q00,q01,q02,q03,c01,c02,c03
  being set holds (q00 is not empty iff AND2(x0,y0) is not empty)& (q01 is not
empty iff XOR3(AND2(x1,y0),AND2(x0,y1),{}) is not empty)& (c01 is not empty iff
MAJ3(AND2(x1,y0),AND2(x0,y1),{}) is not empty)& (q02 is not empty iff XOR3(AND2
(x2,y0),AND2(x1,y1),AND2(x0,y2)) is not empty)& (c02 is not empty iff MAJ3(AND2
(x2,y0),AND2(x1,y1),AND2(x0,y2)) is not empty)& (q03 is not empty iff XOR3(AND2
(x2,y1),AND2(x1,y2),{}) is not empty)& (c03 is not empty iff MAJ3(AND2(x2,y1),
  AND2(x1,y2),{}) is not empty)& (z0 is not empty iff q00 is not empty)& (z1 is
  not empty iff q01 is not empty)& (z2 is not empty iff CLAADD1( q02,c01,{}) is
not empty)& (z3 is not empty iff CLAADD2( q03,c02,q02,c01,{}) is not empty)& (
z4 is not empty iff CLAADD3(AND2(x2,y2),c03,q03,c02,q02,c01,{}) is not empty)&
(z5 is not empty iff CLACARR3(AND2(x2,y2),c03,q03,c02,q02,c01,{}) is not empty)
implies (z0 is not empty iff MULT310(x2, x1,y1,x0,y0) is not empty)& (z1 is not
empty iff MULT311(x2, x1,y1,x0,y0) is not empty)& (z2 is not empty iff MULT321(
x2,y2,x1,y1,x0,y0) is not empty)& (z3 is not empty iff MULT322(x2,y2,x1,y1,x0,
y0) is not empty)& (z4 is not empty iff MULT323(x2,y2,x1,y1,x0,y0) is not empty
  )& (z5 is not empty iff MULT324(x2,y2,x1,y1,x0,y0) is not empty)
proof
  let x0,x1,x2,y0,y1,y2,z0,z1,z2,z3,z4,z5, q00,q01,q02,q03,c01,c02,c03 be set;
  assume that
A1: q00 is not empty iff AND2(x0,y0) is not empty and
A2: q01 is not empty iff XOR3(AND2(x1,y0),AND2(x0,y1),{}) is not empty and
A3: c01 is not empty iff MAJ3(AND2(x1,y0),AND2(x0,y1),{}) is not empty and
A4: q02 is not empty iff XOR3(AND2(x2,y0),AND2(x1,y1),AND2(x0,y2)) is
  not empty and
A5: c02 is not empty iff MAJ3(AND2(x2,y0),AND2(x1,y1),AND2(x0,y2)) is
  not empty and
A6: q03 is not empty iff XOR3(AND2(x2,y1),AND2(x1,y2),{}) is not empty and
A7: c03 is not empty iff MAJ3(AND2(x2,y1),AND2(x1,y2),{}) is not empty and
A8: z0 is not empty iff q00 is not empty and
A9: z1 is not empty iff q01 is not empty and
A10: z2 is not empty iff CLAADD1( q02,c01,{}) is not empty and
A11: z3 is not empty iff CLAADD2( q03,c02,q02,c01,{}) is not empty and
A12: z4 is not empty iff CLAADD3(AND2(x2,y2),c03,q03,c02,q02,c01,{}) is
  not empty and
A13: z5 is not empty iff CLACARR3(AND2(x2,y2),c03,q03,c02,q02,c01,{}) is
  not empty;
  set x0y2 = AND2(x0,y2);
  set x1y1 = AND2(x1,y1);
  set x2y0 = AND2(x2,y0);
  thus z0 is not empty iff MULT310(x2,x1,y1,x0,y0) is not empty by A1,A8;
  thus z1 is not empty iff MULT311(x2,x1,y1,x0,y0) is not empty by A2,A9;
  set m312 = MULT312(x2,x1,y1,x0,y0);
  set x0y1 = AND2(x0,y1);
  set x1y0 = AND2(x1,y0);
A14: m312 = XOR3(x2y0,x1y1,MAJ3(x1y0,x0y1,{})) by GATE_1:def 35;
  then
A15: m312 is not empty iff ( x2y0 is not empty & not x1y1 is not empty or
not x2y0 is not empty & x1y1 is not empty ) & not (x1y0 is not empty & x0y1 is
  not empty) or not ( x2y0 is not empty & not x1y1 is not empty or not x2y0 is
  not empty & x1y1 is not empty ) & x1y0 is not empty & x0y1 is not empty;
  set x1y2 = AND2(x1,y2);
  set x2y1 = AND2(x2,y1);
A16: q03 is not empty iff x2y1 is not empty & not x1y2 is not empty or not
  x2y1 is not empty & x1y2 is not empty by A6;
  set m324 = MULT324(x2,y2,x1,y1,x0,y0);
  set m323 = MULT323(x2,y2,x1,y1,x0,y0);
  set m314 = MULT314(x2,x1,y1,x0,y0);
  set x2y2 = AND2(x2,y2);
A17: m314 = MAJ3({},x2y1,CARR2(x2y0,x1y1,x1y0,x0y1,{})) by GATE_1:def 38
    .= MAJ3({},x2y1,MAJ3(x2y0,x1y1,MAJ3(x1y0,x0y1,{}))) by GATE_1:def 36;
  set m322 = MULT322(x2,y2,x1,y1,x0,y0);
  set m313 = MULT313(x2,x1,y1,x0,y0);
A18: m313 = XOR3({},x2y1,CARR2(x2y0,x1y1,x1y0,x0y1,{})) by GATE_1:def 37
    .= XOR3({},x2y1,MAJ3(x2y0,x1y1,MAJ3(x1y0,x0y1,{}))) by GATE_1:def 36;
  q02 is not empty iff ( x2y0 is not empty & not x1y1 is not empty or not
  x2y0 is not empty & x1y1 is not empty ) & not x0y2 is not empty or not ( x2y0
  is not empty & not x1y1 is not empty or not x2y0 is not empty & x1y1 is not
  empty ) & x0y2 is not empty by A4;
  hence z2 is not empty iff MULT321(x2,y2,x1,y1,x0,y0) is not empty by A3,A10
,A14;
A19: c02 is not empty iff x2y0 is not empty & x1y1 is not empty or x1y1 is
  not empty & x0y2 is not empty or x0y2 is not empty & x2y0 is not empty by A5;
  m322 = XOR3(m313,x1y2,MAJ3(m312,x0y2,{})) by GATE_1:def 35;
  then m322 is not empty iff ( m313 is not empty & not x1y2 is not empty or
not m313 is not empty & x1y2 is not empty ) & not (m312 is not empty & x0y2 is
  not empty) or not ( m313 is not empty & not x1y2 is not empty or not m313 is
  not empty & x1y2 is not empty ) & m312 is not empty & x0y2 is not empty;
  hence z3 is not empty iff MULT322(x2,y2,x1,y1,x0,y0) is not empty by A3,A4
,A11,A19,A16,A15,A18;
  m323 = XOR3(m314,x2y2,CARR2(m313,x1y2,m312,x0y2,{})) by GATE_1:def 37
    .= XOR3(m314,x2y2,MAJ3(m313,x1y2,MAJ3(m312,x0y2,{}))) by GATE_1:def 36;
  then m323 is not empty iff ( m314 is not empty & not x2y2 is not empty or
not m314 is not empty & x2y2 is not empty ) & not ( m313 is not empty & x1y2 is
not empty or x1y2 is not empty & m312 is not empty & x0y2 is not empty or m312
  is not empty & x0y2 is not empty & m313 is not empty ) or not ( m314 is not
empty & not x2y2 is not empty or not m314 is not empty & x2y2 is not empty ) &
  ( m313 is not empty & x1y2 is not empty or x1y2 is not empty & m312 is not
  empty & x0y2 is not empty or m312 is not empty & x0y2 is not empty & m313 is
  not empty );
  hence
  z4 is not empty iff MULT323(x2,y2,x1,y1,x0,y0) is not empty by A3,A4,A7,A12
,A19,A16,A15,A18,A17;
  m324 = MAJ3(m314,x2y2,CARR2(m313,x1y2,m312,x0y2,{})) by GATE_1:def 38
    .= MAJ3(m314,x2y2,MAJ3(m313,x1y2,MAJ3(m312,x0y2,{}))) by GATE_1:def 36;
  then m324 is not empty iff m314 is not empty & x2y2 is not empty or x2y2 is
not empty & ( m313 is not empty & x1y2 is not empty or x1y2 is not empty & m312
  is not empty & x0y2 is not empty or m312 is not empty & x0y2 is not empty &
  m313 is not empty ) or ( m313 is not empty & x1y2 is not empty or x1y2 is not
  empty & m312 is not empty & x0y2 is not empty or m312 is not empty & x0y2 is
  not empty & m313 is not empty ) & m314 is not empty;
  hence thesis by A3,A4,A7,A13,A19,A16,A15,A18,A17;
end;














begin

definition
  let IT be set;
  attr IT is finite means
  :Def1:
  ex p being Function st rng p = IT & dom p in omega;
end;

notation
  let IT be set;
  antonym IT is infinite for IT is finite;
end;

reserve A, B, X, Y, Z, x, y for set;
reserve f for Function;

Lm1: {x} is finite
proof
  set p = {{}} --> x;
A1: dom p = {{}} by FUNCOP_1:13;
  take p;
  for y holds y in {x} iff ex x st x in dom p & y = p.x
  proof
    let y;
    thus y in {x} implies ex x st x in dom p & y = p.x
    proof
      assume y in {x};
      then
A2:   y = x by TARSKI:def 1;
      take {};
      thus {} in dom p by A1,TARSKI:def 1;
      {} in {{}} by TARSKI:def 1;
      hence thesis by A2,FUNCOP_1:7;
    end;
    assume ex z be set st z in dom p & y = p.z;
    then y = x by FUNCOP_1:7;
    hence thesis by TARSKI:def 1;
  end;
  hence rng p = {x} by FUNCT_1:def 3;
  thus thesis by A1,ORDINAL3:15;
end;

registration
  cluster non empty finite for set;
  existence
  proof
    {the set} is finite by Lm1;
    hence thesis;
  end;
end;

registration
  cluster empty -> finite for set;
  coherence
  proof
    let A be set;
    assume
A1: A is empty;
    take {};
    thus rng {} = A by A1;
    thus thesis by ORDINAL1:def 11;
  end;
end;

scheme OLambdaC{A()->set,C[set],F(set)->set,G(set)->set}:
  ex f being Function st dom f = A() & for x being Ordinal st x in A() holds
  (C[x] implies f.x = F(x)) & (not C[x] implies f.x = G(x))
proof
  consider f being Function such that
A1: dom f = A() & for x being set st x in A() holds
  (C[x] implies f.x = F(x)) & (not C[x] implies f.x = G(x))
  from PARTFUN1:sch 1;
  take f;
  thus thesis by A1;
end;

Lm2: A is finite & B is finite implies A \/ B is finite
proof
  given p be Function such that
A1: rng p = A and
A2: dom p in omega;
  given q be Function such that
A3: rng q = B and
A4: dom q in omega;
  reconsider domp = dom p, domq = dom q as Ordinal by A2,A4;
  deffunc F(Ordinal) = p.$1;
  deffunc G(Ordinal) = q.($1-^domp);
  defpred P[set] means $1 in domp;
  consider f such that
A5: dom f = domp+^domq and
A6: for x being Ordinal st x in domp+^domq holds
  (P[x] implies f.x = F(x)) & (not P[x] implies f.x = G(x)) from OLambdaC;
  take f;
  reconsider domp, domq as natural Ordinal by A2,A4;
  thus rng f = A \/ B
  proof
    thus rng f c= A \/ B
    proof
      let y;
      assume y in rng f;
      then consider x such that
A7:   x in dom f and
A8:   y = f.x by FUNCT_1:def 3;
      reconsider x as Ordinal by A5,A7;
      per cases;
      suppose
A9:     x in domp;
        then
A10:    y = p.x by A5,A6,A7,A8;
        p.x in rng p by A9,FUNCT_1:def 3;
        hence thesis by A1,A10,XBOOLE_0:def 3;
      end;
      suppose
A11:    not x in domp;
        then
A12:    domp c= x by ORDINAL1:16;
A13:    y = q.(x-^domp) by A5,A6,A7,A8,A11;
        (domp)+^(x-^domp) = x by A12,ORDINAL3:def 5;
        then x-^domp in dom q by A5,A7,ORDINAL3:22;
        then y in B by A3,A13,FUNCT_1:def 3;
        hence thesis by XBOOLE_0:def 3;
      end;
    end;
    let x;
    assume
A14: x in A \/ B;
    per cases by A1,A3,A14,XBOOLE_0:def 3;
    suppose x in rng p;
      then consider y such that
A15:  y in dom p and
A16:  x = p.y by FUNCT_1:def 3;
A17:  dom p c= dom f by A5,ORDINAL3:24;
      then x = f.y by A5,A6,A15,A16;
      hence thesis by A15,A17,FUNCT_1:def 3;
    end;
    suppose x in rng q;
      then consider y such that
A18:  y in domq and
A19:  x = q.y by FUNCT_1:def 3;
      reconsider y as Ordinal by A18;
      set z = domp +^ y;
      domp c= z by ORDINAL3:24;
      then
A20:  not z in domp by ORDINAL1:5;
A21:  z in dom f by A5,A18,ORDINAL3:17;
      x = q.(z-^domp) by A19,ORDINAL3:52
        .= f.z by A5,A6,A20,A21;
      hence thesis by A21,FUNCT_1:def 3;
    end;
  end;
  domp+^domq is natural;
  hence thesis by A5,ORDINAL1:def 12;
end;

registration
  let x be set;
  cluster {x} -> finite;
  coherence by Lm1;
end;

registration
  let x,y be set;
  cluster {x,y} -> finite;
  coherence
  proof
    {x,y} = {x} \/ {y} by ENUMSET1:1;
    hence thesis by Lm2;
  end;
end;

registration
  let x,y,z be set;
  cluster {x,y,z} -> finite;
  coherence
  proof
    {x,y,z} = {x} \/ {y,z} by ENUMSET1:2;
    hence thesis by Lm2;
  end;
end;

registration
  let x1,x2,x3,x4 be set;
  cluster {x1,x2,x3,x4} -> finite;
  coherence
  proof
    {x1,x2,x3,x4} = {x1} \/ {x2,x3,x4} by ENUMSET1:4;
    hence thesis by Lm2;
  end;
end;

registration
  let x1,x2,x3,x4,x5 be set;
  cluster {x1,x2,x3,x4,x5} -> finite;
  coherence
  proof
    {x1,x2,x3,x4,x5} = {x1} \/ {x2,x3,x4,x5} by ENUMSET1:7;
    hence thesis by Lm2;
  end;
end;

registration
  let x1,x2,x3,x4,x5,x6 be set;
  cluster {x1,x2,x3,x4,x5,x6} -> finite;
  coherence
  proof
    {x1,x2,x3,x4,x5,x6} = {x1} \/ {x2,x3,x4,x5,x6} by ENUMSET1:11;
    hence thesis by Lm2;
  end;
end;

registration
  let x1,x2,x3,x4,x5,x6,x7 be set;
  cluster {x1,x2,x3,x4,x5,x6,x7} -> finite;
  coherence
  proof
    {x1,x2,x3,x4,x5,x6,x7} = {x1} \/ {x2,x3,x4,x5,x6,x7} by ENUMSET1:16;
    hence thesis by Lm2;
  end;
end;

registration
  let x1,x2,x3,x4,x5,x6,x7,x8 be set;
  cluster {x1,x2,x3,x4,x5,x6,x7,x8} -> finite;
  coherence
  proof
    {x1,x2,x3,x4,x5,x6,x7,x8} = {x1} \/ {x2,x3,x4,x5,x6,x7,x8} by ENUMSET1:22;
    hence thesis by Lm2;
  end;
end;

registration
  let B be finite set;
  cluster -> finite for Subset of B;
  coherence
  proof
    let A be Subset of B;
    per cases;
    suppose A is empty;
      hence thesis;
    end;
    suppose A is non empty;
      then consider x1 be set such that
A1:   x1 in A by XBOOLE_0:def 1;
      consider p be Function such that
A2:   rng p = B and
A3:   dom p in omega by Def1;
      deffunc F(set) = p.$1;
      deffunc G(set) = x1;
      defpred P[set] means p.$1 in A;
      consider q be Function such that
A4:   dom q = dom p and
A5:   for x st x in dom p holds (P[x] implies q.x = F(x)) &
      (not P[x] implies q.x = G(x)) from PARTFUN1:sch 1;
      now
        let y;
        thus y in A implies ex x st x in dom q & y = q.x
        proof
          assume
A6:       y in A;
          then consider x such that
A7:       x in dom p and
A8:       p.x = y by A2,FUNCT_1:def 3;
          take x;
          thus x in dom q by A4,A7;
          thus thesis by A5,A6,A7,A8;
        end;
        given x such that
A9:     x in dom q and
A10:    y = q.x;
        per cases;
        suppose p.x in A;
          hence y in A by A4,A5,A9,A10;
        end;
        suppose not p.x in A;
          hence y in A by A1,A4,A5,A9,A10;
        end;
      end;
      then rng q = A by FUNCT_1:def 3;
      hence thesis by A3,A4,Def1;
    end;
  end;
end;

registration
  let X,Y be finite set;
  cluster X \/ Y -> finite;
  coherence by Lm2;
end;

theorem
  A c= B & B is finite implies A is finite;

theorem
  A is finite & B is finite implies A \/ B is finite;

registration
  let A be set, B be finite set;
  cluster A /\ B -> finite;
  coherence
  proof
    A /\ B c= B by XBOOLE_1:17;
    hence thesis;
  end;
end;

registration
  let A be finite set, B be set;
  cluster A /\ B -> finite;
  coherence;
  cluster A \ B -> finite;
  coherence;
end;

theorem
  A is finite implies A /\ B is finite;

theorem
  A is finite implies A \ B is finite;

registration
  let f be Function, A be finite set;
  cluster f.:A -> finite;
  coherence
proof
  set B = dom f /\ A;
  consider p be Function such that
A1: rng p=B and
A2: dom p in omega by Def1;
  take f*p;
  rng (f*p) = f.:B by A1,RELAT_1:127;
  hence rng (f*p) = f.:A by RELAT_1:112;
  thus thesis by A1,A2,RELAT_1:27,XBOOLE_1:17;
end;
end;

theorem
  A is finite implies f.:A is finite;

reserve O for Ordinal;

theorem Th6:
  A is finite implies for X being Subset-Family of A st X <> {}
  ex x being set st x in X &
    for B being set st B in X holds x c= B implies B = x
proof
  assume
A1: A is finite;
  let X be Subset-Family of A such that
A2: X <> {};
  consider p be Function such that
A3: rng p = A and
A4: dom p in omega by A1,Def1;
  defpred P[set] means p.:$1 in X;
  consider G being set such that
A5: for x holds x in G iff x in bool dom p & P[x] from XBOOLE_0:sch 1;
  G c= bool dom p
  proof
    let x;
    assume x in G;
    hence thesis by A5;
  end;
  then reconsider GX=G as Subset-Family of dom p;
  set x = the Element of X;
  x is Subset of A by A2,TARSKI:def 3;
  then
A6: p.:(p"x) = x by A3,FUNCT_1:77;
  p"x c= dom p by RELAT_1:132;
  then
A7: GX <> {} by A2,A5,A6;
  defpred P[set] means
  $1 in omega implies for X being Subset-Family of $1 st X <> {}
  ex x being set st
  x in X & for B being set st B in X holds x c= B implies B = x;
A8: P[{}]
  proof
    assume {} in omega;
    let X be Subset-Family of {};
    assume X <> {};
    then
A9: X = {{}} by ZFMISC_1:1,33;
    take {};
    thus thesis by A9,TARSKI:def 1;
  end;
A10: P[O] implies P[succ O]
  proof
    assume
A11: O in omega implies for X being Subset-Family of O st X <> {}
    ex x being set st
    x in X & for B being set st B in X holds x c= B implies B = x;
    thus succ O in omega implies
    for X being Subset-Family of succ O st X <> {} ex x being set st
    x in X & for B being set st B in X holds x c= B implies B = x
    proof
      assume succ O in omega;
      then
A12:  succ O c= omega by ORDINAL1:def 2;
      let X be Subset-Family of succ O such that
A13:  X <> {};
      defpred P[set] means ex A st A in X & $1 = A \ {O};
      consider X1 being set such that
A14:  for x holds x in X1 iff x in bool O & P[x] from XBOOLE_0:sch 1;
      X1 c= bool O
      proof
        let x;
        assume x in X1;
        hence thesis by A14;
      end;
      then reconsider X2=X1 as Subset-Family of O;
      set y = the Element of X;
A15:  succ O = O \/ {O} by ORDINAL1:def 1;
      y is Subset of succ O by A13,TARSKI:def 3;
      then y \ {O} c= (O \/ {O}) \ {O} by A15,XBOOLE_1:33;
      then
A16:  y \ {O} c= O \ {O} by XBOOLE_1:40;
A17:  O in succ O by ORDINAL1:6;
A18:  not O in O;
      then y \ {O} c= O by A16,ZFMISC_1:57;
      then X2 <> {} by A13,A14;
      then consider Z such that
A19:  Z in X2 and
A20:  for B being set st B in X2 holds Z c= B implies B = Z by A11,A12,A17;
      consider X1 being set such that
A21:  X1 in X and
A22:  Z=X1 \ {O} by A14,A19;
A23:  O in {O} by TARSKI:def 1;
      then
A24:  not O in Z by A22,XBOOLE_0:def 5;
A25:  now
        assume
A26:    Z \/ {O} in X;
        take A=Z \/ {O};
        for B being set st B in X holds A c= B implies B = A
        proof
          let B such that
A27:      B in X;
          assume
A28:      A c= B;
A29:      now
            assume
A30:        O in B;
            set Y = B \ {O};
            {O} c= B by A30,ZFMISC_1:31;
            then
A31:        B = Y \/ {O} by XBOOLE_1:45;
            A \ {O} c= Y by A28,XBOOLE_1:33;
            then
A32:        Z \ {O} c= Y by XBOOLE_1:40;
            not O in Z by A22,A23,XBOOLE_0:def 5;
            then
A33:        Z c= Y by A32,ZFMISC_1:57;
            Y c= (O \/ {O}) \ {O} by A15,A27,XBOOLE_1:33;
            then Y c= O \ {O} by XBOOLE_1:40;
            then Y c= O by A18,ZFMISC_1:57;
            then Y in X2 by A14,A27;
            hence thesis by A20,A31,A33;
          end;
          now
            assume
A34:        not O in B;
            O in {O} by TARSKI:def 1;
            then O in A by XBOOLE_0:def 3;
            hence contradiction by A28,A34;
          end;
          hence thesis by A29;
        end;
        hence thesis by A26;
      end;
      now
        assume
A35:    not Z \/ {O} in X;
        take A=Z;
        now
          assume O in X1;
          then X1 = X1 \/ {O} by ZFMISC_1:40
            .= Z \/ {O} by A22,XBOOLE_1:39;
          hence contradiction by A21,A35;
        end;
        then
A36:    A in X by A21,A22,ZFMISC_1:57;
        for B being set st B in X holds A c= B implies B = A
        proof
          let B;
          assume
A37:      B in X;
          then B \ {O} c= (O \/ {O}) \ {O} by A15,XBOOLE_1:33;
          then B \ {O} c= O \ {O} by XBOOLE_1:40;
          then B \ {O} c= O by A18,ZFMISC_1:57;
          then
A38:      B \ {O} in X2 by A14,A37;
          assume
A39:      A c= B;
          then A \ {O} c= B \ {O} by XBOOLE_1:33;
          then
A40:      A c= B \ {O} by A24,ZFMISC_1:57;
A41:      now
            assume
A42:        O in B;
            A \/ {O} = (B \ {O}) \/ {O} by A20,A38,A40
              .= B \/ {O} by XBOOLE_1:39
              .= B by A42,ZFMISC_1:40;
            hence contradiction by A35,A37;
          end;
A43:      B c= O
          proof
            let x such that
A44:        x in B;
            x in O or x in {O} by A15,A37,A44,XBOOLE_0:def 3;
            hence thesis by A41,A44,TARSKI:def 1;
          end;
          B \ {O} = B by A41,ZFMISC_1:57;
          then B in X2 by A14,A37,A43;
          hence thesis by A20,A39;
        end;
        hence thesis by A36;
      end;
      hence thesis by A25;
    end;
  end;
A45: O <> {} & O is limit_ordinal & (for B being Ordinal st B in O holds P[B])
  implies P[O]
  proof
    assume that
A46: O <> {} and
A47: O is limit_ordinal and for B being Ordinal st B in O holds P[B] and
A48: O in omega;
    {} in O by A46,ORDINAL1:14;
    then omega c= O by A47,ORDINAL1:def 11;
    then O in O by A48;
    hence thesis;
  end;
  P[O] from ORDINAL2:sch 1(A8,A10,A45);
  then consider g being set such that
A49: g in GX and
A50: for B being set st B in GX holds g c= B implies B=g by A4,A7;
  take p.:g;
  for B st B in X holds p.:g c= B implies B = p.:g
  proof
    let B;
    assume
A51: B in X;
    assume p.:g c= B;
    then
A52: p"(p.:g) c= p"B by RELAT_1:143;
A53: g c= p"(p.:g) by A49,FUNCT_1:76;
A54: p.:(p"B) = B by A3,A51,FUNCT_1:77;
    p"B c= dom p by RELAT_1:132;
    then p"B in GX by A5,A51,A54;
    hence thesis by A50,A52,A53,A54,XBOOLE_1:1;
  end;
  hence thesis by A5,A49;
end;

scheme Finite{A()->set,P[set]} : P[A()]
provided
A1: A() is finite and
A2: P[{}] and
A3: for x,B being set st x in A() & B c= A() & P[B] holds P[B \/ {x}]
proof
  now
    assume A() <> {};
    defpred R[set] means ex B st B=$1 & P[B];
    consider G being set such that
A4: for x holds x in G iff x in bool A() & R[x] from XBOOLE_0:sch 1;
    G c= bool A()
    proof
      let x;
      assume x in G;
      hence thesis by A4;
    end;
    then reconsider GA=G as Subset-Family of A();
    {} c= A() by XBOOLE_1:2;
    then GA <> {} by A2,A4;
    then consider B such that
A5: B in GA and
A6: for X being set st X in GA holds B c= X implies B=X by A1,Th6;
A7: ex A st A = B & P[A] by A4,A5;
    now set x = the Element of A() \ B;
      assume B <> A();
      then not A() c= B by A5,XBOOLE_0:def 10;
      then
A8:   A() \ B <> {} by XBOOLE_1:37;
      then
A9:   x in A() by XBOOLE_0:def 5;
      then
A10:  P[B \/ {x}] by A3,A5,A7;
      {x} c= A() by A9,ZFMISC_1:31;
      then B \/ {x} c= A() by A5,XBOOLE_1:8;
      then
A11:  B \/ {x} in GA by A4,A10;
      not x in B by A8,XBOOLE_0:def 5;
      then not {x} c= B by ZFMISC_1:31;
      then B \/ {x} <> B by XBOOLE_1:7;
      hence contradiction by A6,A11,XBOOLE_1:7;
    end;
    hence thesis by A7;
  end;
  hence thesis by A2;
end;

Lm3: A
is finite & (for X st X in A holds X is finite) implies union A is finite
proof
  assume that
A1: A is finite and
A2: for X st X in A holds X is finite;
  defpred P[set] means ex B st B=$1 & union B is finite;
  consider G being set such that
A3: for x holds x in G iff x in bool A & P[x] from XBOOLE_0:sch 1;
  defpred P[set] means $1 in G;
  {} c= A by XBOOLE_1:2;
  then
A4: P[{}] by A3,ZFMISC_1:2;
A5: for x,B being set st x in A & B c= A & P[B] holds P[B \/ {x}]
  proof
    let x,B be set;
    assume that
A6: x in A and B c= A and
A7: B in G;
A8: ex X st X=B & union X is finite by A3,A7;
A9: x is finite by A2,A6;
A10: union (B \/ {x}) = union B \/ union {x} by ZFMISC_1:78
      .= union B \/ x by ZFMISC_1:25;
A11: {x} c= A by A6,ZFMISC_1:31;
    B in bool A by A3,A7;
    then B \/ {x} c= A by A11,XBOOLE_1:8;
    hence thesis by A3,A8,A9,A10;
  end;
  P[A] from Finite(A1,A4,A5);
  then ex X st X=A & union X is finite by A3;
  hence thesis;
end;

registration
  let A,B be finite set;
  cluster [:A,B:] -> finite;
  coherence
  proof
A1: for y holds [:A,{y}:] is finite
    proof
      let y;
      deffunc F(set) = [$1,y];
      consider f such that
A2:   dom f=A & for x st x in A holds f.x=F(x) from FUNCT_1:sch 3;
      for x holds x in rng f iff x in [:A,{y}:]
      proof
        let x;
        thus x in rng f implies x in [:A,{y}:]
        proof
          assume x in rng f;
          then consider z be set such that
A3:       z in dom f and
A4:       f.z = x by FUNCT_1:def 3;
          x = [z,y] by A2,A3,A4;
          hence thesis by A2,A3,ZFMISC_1:106;
        end;
        thus x in [:A,{y}:] implies x in rng f
        proof
          assume x in [:A,{y}:];
          then consider x1,x2 be set such that
A5:       x1 in A and
A6:       x2 in {y} and
A7:       x=[x1,x2] by ZFMISC_1:def 2;
          x2=y by A6,TARSKI:def 1;
          then x = f.x1 by A2,A5,A7;
          hence thesis by A2,A5,FUNCT_1:def 3;
        end;
      end;
      then rng f = [:A,{y}:] by TARSKI:1;
      then f.:A=[:A,{y}:] by A2,RELAT_1:113;
      hence thesis;
    end;
    defpred P[set] means ex y st y in B & $1 = [:A,{y}:];
    consider G being set such that
A8: for x holds x in G iff x in bool [:A,B:] & P[x] from XBOOLE_0:sch 1;
    for x holds x in union G iff x in [:A,B:]
    proof
      let x;
      thus x in union G implies x in [:A,B:]
      proof
        assume x in union G;
        then consider X such that
A9:     x in X and
A10:    X in G by TARSKI:def 4;
        X in bool [:A,B:]by A8,A10;
        hence thesis by A9;
      end;
      assume x in [:A,B:];
      then consider y,z be set such that
A11:  y in A and
A12:  z in B and
A13:  x=[y,z] by ZFMISC_1:def 2;
A14:  [y,z] in [:A,{z}:] by A11,ZFMISC_1:106;
      {z} c= B by A12,ZFMISC_1:31;
      then [:A,{z}:] c= [:A,B:] by ZFMISC_1:95;
      then [:A,{z}:] in G by A8,A12;
      hence thesis by A13,A14,TARSKI:def 4;
    end;
    then
A15: union G = [:A,B:] by TARSKI:1;
    deffunc F(set) = [:A,{$1}:];
    consider g being Function such that
A16: dom g = B & for x st x in B holds g.x = F(x) from FUNCT_1:sch 3;
    for x holds x in rng g iff x in G
    proof
      let x;
      thus x in rng g implies x in G
      proof
        assume x in rng g;
        then consider y such that
A17:    y in dom g and
A18:    g.y = x by FUNCT_1:def 3;
A19:    x = [:A,{y}:] by A16,A17,A18;
        {y} c= B by A16,A17,ZFMISC_1:31;
        then x c= [:A,B:] by A19,ZFMISC_1:95;
        hence thesis by A8,A16,A17,A19;
      end;
      assume x in G;
      then consider y such that
A20:  y in B and
A21:  x = [:A,{y}:] by A8;
      x = g.y by A16,A20,A21;
      hence thesis by A16,A20,FUNCT_1:def 3;
    end;
    then rng g = G by TARSKI:1;
    then
A22: g.:B = G by A16,RELAT_1:113;
    for X st X in G holds X is finite
    proof
      let X;
      assume X in G;
      then ex y st y in B & X=[:A,{y}:] by A8;
      hence thesis by A1;
    end;
    hence thesis by A15,A22,Lm3;
  end;
end;

registration
  let A,B,C be finite set;
  cluster [:A,B,C:] -> finite;
  coherence
  proof
    [:[:A,B:],C:] is finite;
    hence thesis by ZFMISC_1:def 3;
  end;
end;

registration
  let A,B,C,D be finite set;
  cluster [:A,B,C,D:] -> finite;
  coherence
  proof
    [:[:A,B,C:],D:] is finite;
    hence thesis by ZFMISC_1:def 4;
  end;
end;

registration
  let A be finite set;
  cluster bool A -> finite;
  coherence
  proof
A1: A is finite;
    defpred P[set] means bool $1 is finite;
    consider G being set such that
A2: for x holds x in G iff x in bool A & P[x] from XBOOLE_0:sch 1;
    defpred P[set] means $1 in G;
A3:  bool{} is finite by ZFMISC_1:1;
    {} c= A by XBOOLE_1:2;
    then
A4: P[{}] by A3,A2;
A5: for x,B being set st x in A & B c= A & P[B] holds P[B \/ {x}]
    proof
      let x,B be set;
      assume that
A6:   x in A and B c= A and
A7:   B in G;
A8:   now
        assume x in B;
        then {x} c= B by ZFMISC_1:31;
        hence thesis by A7,XBOOLE_1:12;
      end;
      now
        assume
A9:     not x in B;
        defpred P[set,set] means ex Y st Y=$1 & $2=Y \/ {x};
A10:     for y,y1,y2 be set st y in bool B & P[y,y1] & P[y,y2] holds y1 = y2;
A11:    for y st y in bool B ex z be set st P[y,z]
        proof
          let y such that y in bool B;
          ex Y st Y=y & y \/ {x} = Y \/ {x};
          hence thesis;
        end;
        consider f such that
A12:    dom f = bool B and
A13:    for y st y in bool B holds P[y,f.y] from FUNCT_1:sch 2(A10,A11);
A14:    bool B is finite by A2,A7;
        y in rng f iff y in bool(B \/ {x}) \ bool B
        proof
          thus y in rng f implies y in bool(B \/ {x}) \ bool B
          proof
            assume y in rng f;
            then consider x1 be set such that
A15:        x1 in dom f and
A16:        f.x1=y by FUNCT_1:def 3;
            consider X1 being set such that
A17:        X1=x1 and
A18:        f.x1= X1 \/ {x} by A12,A13,A15;
A19:        X1 \/ {x} c= B \/ {x} by A12,A15,A17,XBOOLE_1:13;
            x in {x} by TARSKI:def 1;
            then x in y by A16,A18,XBOOLE_0:def 3;
            then not y in bool B by A9;
            hence thesis by A16,A18,A19,XBOOLE_0:def 5;
          end;
          assume
A20:      y in bool(B \/ {x}) \ bool B;
          set Z=y \ {x};
A21:      now
            assume
A22:        not x in y;
            y c= B
            proof
              let z be set;
              assume
A23:          z in y;
              then not z in {x} by A22,TARSKI:def 1;
              hence thesis by A20,A23,XBOOLE_0:def 3;
            end;
            hence contradiction by A20,XBOOLE_0:def 5;
          end;
A24:      Z c= B by A20,XBOOLE_1:43;
          then consider X such that
A25:      X=Z and
A26:      f.Z = X \/ {x} by A13;
          X \/ {x} = y \/ {x} by A25,XBOOLE_1:39
            .= y by A21,ZFMISC_1:40;
          hence thesis by A12,A24,A26,FUNCT_1:def 3;
        end;
        then rng f = bool(B \/ {x}) \ bool B by TARSKI:1;
        then
A27:    f.:(bool B) = bool(B \/ {x}) \ bool B by A12,RELAT_1:113;
A28:    bool B c= bool(B \/ {x}) by XBOOLE_1:7,ZFMISC_1:67;
A29:    (bool(B \/ {x}) \ bool B) \/ bool B = bool(B \/ {x}) \/
        bool B by XBOOLE_1:39
          .= bool(B \/ {x}) by A28,XBOOLE_1:12;
A30:    {x} c= A by A6,ZFMISC_1:31;
        B in bool A by A2,A7;
        then B \/ {x} c= A by A30,XBOOLE_1:8;
        hence thesis by A2,A14,A27,A29;
      end;
      hence thesis by A8;
    end;
    P[A] from Finite(A1,A4,A5);
    hence thesis by A2;
  end;
end;

theorem Th7:
  A is finite & (for X st X in A holds X is finite) iff union A is finite
proof
  thus A is finite & (for X st X in A holds X is finite) implies
  union A is finite by Lm3;
  thus union A is finite implies
  A is finite & for X st X in A holds X is finite
  proof
    assume
A1: union A is finite;
    A c= bool union A by ZFMISC_1:82;
    hence A is finite by A1;
    let X;
    assume X in A;
    then X c= union A by ZFMISC_1:74;
    hence thesis by A1;
  end;
end;

theorem Th8:
  dom f is finite implies rng f is finite
proof
  assume dom f is finite;
  then f.:(dom f) is finite;
  hence thesis by RELAT_1:113;
end;

theorem
  Y c= rng f & f"Y is finite implies Y is finite
proof
  assume Y c= rng f;
  then f.:(f"Y) = Y by FUNCT_1:77;
  hence thesis;
end;

registration
  let X, Y be finite set;
  cluster X \+\ Y -> finite;
  coherence;
end;

registration
  let X be non empty set;
  cluster finite non empty for Subset of X;
  existence
  proof
    take {the Element of X};
    thus thesis;
  end;
end;

begin 


theorem Th10:
  for f being Function holds dom f is finite iff f is finite
proof
  let f be Function;
  thus dom f is finite implies f is finite
  proof
    assume
A1: dom f is finite;
    then
A2: rng f is finite by Th8;
    f c= [:dom f, rng f:] by RELAT_1:7;
    hence thesis by A1,A2;
  end;
  pr1(dom f,rng f).:f = dom f by FUNCT_3:79;
  hence thesis;
end;



theorem
  for F being set st F is finite & F <> {} & F is c=-linear
  ex m being set st m in F & for C being set st C in F holds m c= C
proof
  defpred P[set] means $1 <> {} implies
  ex m being set st m in $1 & for C being set st C in $1 holds m c= C;
  let F be set such that
A1: F is finite and
A2: F <> {} and
A3: F is c=-linear;
A4: P[{}];
A5: now
    let x,B be set such that
A6: x in F and
A7: B c= F and
A8: P[B];
    now per cases;
      suppose
A9:     not ex y being set st y in B & y c=x;
        assume B \/ {x} <> {};
        take m = x;
        x in {x} by TARSKI:def 1;
        hence m in B \/ {x} by XBOOLE_0:def 3;
        let C be set;
        assume C in B \/ {x};
        then
A10:    C in B or C in {x} by XBOOLE_0:def 3;
        then C,x are_c=-comparable by A3,A6,A7,ORDINAL1:def 8,TARSKI:def 1;
        then C c= x or x c= C by XBOOLE_0:def 9;
        hence m c= C by A9,A10,TARSKI:def 1;
      end;
      suppose ex y being set st y in B & y c=x;
        then consider y being set such that
A11:    y in B and
A12:    y c=x;
        assume B \/ {x} <> {};
        consider m being set such that
A13:    m in B and
A14:    for C being set st C in B holds m c= C by A8,A11;
        m c= y by A11,A14;
        then
A15:    m c= x by A12,XBOOLE_1:1;
        take m;
        thus m in B \/ {x} by A13,XBOOLE_0:def 3;
        let C be set;
        assume C in B \/ {x};
        then C in B or C in {x} by XBOOLE_0:def 3;
        hence m c= C by A14,A15,TARSKI:def 1;
      end;
    end;
    hence P[B \/ {x}];
  end;
  P[F] from Finite(A1,A4,A5);
  hence thesis by A2;
end;



theorem
  for F being set st F is finite & F <> {} & F is c=-linear
  ex m being set st m in F & for C being set st C in F holds C c= m
proof
  defpred P[set] means $1 <> {} implies
  ex m being set st m in $1 & for C being set st C in $1 holds C c= m;
  let F be set such that
A1: F is finite and
A2: F <> {} and
A3: F is c=-linear;
A4: P[{}];
A5: now
    let x,B be set such that
A6: x in F and
A7: B c= F and
A8: P[B];
    now per cases;
      suppose
A9:     not ex y being set st y in B & x c= y;
        assume B \/ {x} <> {};
        take m = x;
        x in {x} by TARSKI:def 1;
        hence m in B \/ {x} by XBOOLE_0:def 3;
        let C be set;
        assume C in B \/ {x};
        then
A10:    C in B or C in {x} by XBOOLE_0:def 3;
        then C,x are_c=-comparable by A3,A6,A7,ORDINAL1:def 8,TARSKI:def 1;
        then C c= x or x c= C by XBOOLE_0:def 9;
        hence C c= m by A9,A10,TARSKI:def 1;
      end;
      suppose ex y being set st y in B & x c= y;
        then consider y being set such that
A11:    y in B and
A12:    x c= y;
        assume B \/ {x} <> {};
        consider m being set such that
A13:    m in B and
A14:    for C being set st C in B holds C c= m by A8,A11;
        y c= m by A11,A14;
        then
A15:    x c= m by A12,XBOOLE_1:1;
        take m;
        thus m in B \/ {x} by A13,XBOOLE_0:def 3;
        let C be set;
        assume C in B \/ {x};
        then C in B or C in {x} by XBOOLE_0:def 3;
        hence C c= m by A14,A15,TARSKI:def 1;
      end;
    end;
    hence P[B \/ {x}];
  end;
  P[F] from Finite(A1,A4,A5);
  hence thesis by A2;
end;



definition
  let R be Relation;
  attr R is finite-yielding means
  :Def2:
  for x st x in rng R holds x is finite;
end;



reserve a for set;

deffunc F(set) = $1`1;

theorem
  X is finite & X c= [:Y,Z:] implies
  ex A,B being set st A is finite & A c= Y & B is finite & B c= Z &
  X c= [:A,B:]
proof
  deffunc G(set) = $1`2;
  assume that
A1: X is finite and
A2: X c= [:Y,Z:];
  consider f being Function such that
A3: dom f = X and
A4: for a st a in X holds f.a = F(a) from FUNCT_1:sch 3;
  consider g being Function such that
A5: dom g = X and
A6: for a st a in X holds g.a = G(a) from FUNCT_1:sch 3;
  take A = rng f, B = rng g;
  thus A is finite by A1,A3,Th8;
  thus A c= Y
  proof
    let a;
    assume a in A;
    then consider x being set such that
A7: x in dom f and
A8: f.x = a by FUNCT_1:def 3;
    consider y,z being set such that
A9: y in Y and z in Z and
A10: x = [y,z] by A2,A3,A7,ZFMISC_1:def 2;
    f.x = x`1 by A3,A4,A7
      .= y by A10,MCART_1:7;
    hence thesis by A8,A9;
  end;
  thus B is finite by A1,A5,Th8;
  thus B c= Z
  proof
    let a;
    assume a in B;
    then consider x being set such that
A11: x in dom g and
A12: g.x = a by FUNCT_1:def 3;
    consider y,z being set such that
    y in Y and
A13: z in Z and
A14: x = [y,z] by A2,A5,A11,ZFMISC_1:def 2;
    g.x = x`2 by A5,A6,A11
      .= z by A14,MCART_1:7;
    hence thesis by A12,A13;
  end;
  thus X c= [:A,B:]
  proof
    let a;
    assume
A15: a in X;
    then consider x,y being set such that
    x in Y and y in Z and
A16: a=[x,y] by A2,ZFMISC_1:def 2;
A17: g.a = a`2 by A6,A15
      .= y by A16,MCART_1:7;
    f.a = a`1 by A4,A15
      .= x by A16,MCART_1:7;
    then
A18: x in A by A3,A15,FUNCT_1:def 3;
    y in B by A5,A15,A17,FUNCT_1:def 3;
    hence thesis by A16,A18,ZFMISC_1:87;
  end;
end;

theorem
  X is finite & X c= [:Y,Z:] implies
  ex A being set st A is finite & A c= Y & X c= [:A,Z:]
proof
  assume that
A1: X is finite and
A2: X c= [:Y,Z:];
  consider f being Function such that
A3: dom f = X and
A4: for a st a in X holds f.a = F(a) from FUNCT_1:sch 3;
  take A = rng f;
  thus A is finite by A1,A3,Th8;
  thus A c= Y
  proof
    let a;
    assume a in A;
    then consider x being set such that
A5: x in dom f and
A6: f.x = a by FUNCT_1:def 3;
    consider y,z being set such that
A7: y in Y and z in Z and
A8: x = [y,z] by A2,A3,A5,ZFMISC_1:def 2;
    f.x = x`1 by A3,A4,A5
      .= y by A8,MCART_1:7;
    hence thesis by A6,A7;
  end;
  thus X c= [:A,Z:]
  proof
    let a;
    assume
A9: a in X;
    then consider x,y being set such that
    x in Y and
A10: y in Z and
A11: a=[x,y] by A2,ZFMISC_1:def 2;
    f.a = a`1 by A4,A9
      .= x by A11,MCART_1:7;
    then x in A by A3,A9,FUNCT_1:def 3;
    hence thesis by A10,A11,ZFMISC_1:87;
  end;
end;



registration
  cluster finite non empty for Function;
  existence
  proof
    {[{},{}]} is Function;
    hence thesis;
  end;
end;

registration
  let R be finite Relation;
  cluster dom R -> finite;
  coherence
  proof
    consider f being Function such that
A1: dom f = R and
A2: for x st x in R holds f.x = x`1 from FUNCT_1:sch 3;
    now
      let x;
      thus x in rng f implies ex y st [x,y] in R
      proof
        assume x in rng f;
        then consider a such that
A3:     a in dom f and
A4:     f.a = x by FUNCT_1:def 3;
        take a`2;
A5:     ex x,y st a = [x,y] by A1,A3,RELAT_1:def 1;
        x = a`1 by A1,A2,A3,A4;
        hence thesis by A1,A3,A5,MCART_1:8;
      end;
      given y such that
A6:   [x,y] in R;
      f.[x,y] = [x,y]`1 by A2,A6
        .= x;
      hence x in rng f by A1,A6,FUNCT_1:3;
    end;
    then rng f = dom R by XTUPLE_0:def 12;
    hence thesis by A1,Th8;
  end;
end;



registration
  let f be Function, g be finite Function;
  cluster f*g -> finite;
  coherence
  proof dom(f*g) c= dom g by RELAT_1:25;
    hence thesis by Th10;
  end;
end;



registration
  let A be finite set, B be set;
  cluster -> finite for Function of A, B;
  coherence
  proof
    let f be Function of A, B;
    dom f is finite;
    hence thesis by Th10;
  end;
end;



registration
  let x,y be set;
  cluster x .--> y -> finite;
  coherence;
end;



registration
  let R be finite Relation;
  cluster rng R -> finite;
  coherence
  proof
    consider f being Function such that
A1: dom f = R and
A2: for x st x in R holds f.x = x`2 from FUNCT_1:sch 3;
    now
      let y;
      thus y in rng f implies ex x st [x,y] in R
      proof
        assume y in rng f;
        then consider a such that
A3:     a in dom f and
A4:     f.a = y by FUNCT_1:def 3;
        take a`1;
A5:     ex x,y st a = [x,y] by A1,A3,RELAT_1:def 1;
        y = a`2 by A1,A2,A3,A4;
        hence thesis by A1,A3,A5,MCART_1:8;
      end;
      given x such that
A6:   [x,y] in R;
      f.[x,y] = [x,y]`2 by A2,A6
        .= y;
      hence y in rng f by A1,A6,FUNCT_1:3;
    end;
    then rng f = rng R by XTUPLE_0:def 13;
    hence thesis by A1,Th8;
  end;
end;



registration
  let f be finite Function, x be set;
  cluster f"x -> finite;
  coherence
  proof
    f"x c= dom f by RELAT_1:132;
    hence thesis;
  end;
end;

registration
  let f, g be finite Function;
  cluster f +* g -> finite;
  coherence
  proof
    dom (f+*g) = dom f \/ dom g by FUNCT_4:def 1;
    hence thesis by Th10;
  end;
end;



definition
  let F be set;
  attr F is centered means
  F <> {} &
  for G being set st G <> {} & G c= F & G is finite holds meet G <> {};
end;

definition
 let f be Function;
  redefine attr f is finite-yielding means
:Def4: for i being set st i in dom f holds f.i is finite;
  compatibility
  proof
    hereby
      assume
A1:   f is finite-yielding;
      let i be set such that i in dom f;
      per cases;
      suppose i in dom f;
        then f.i in rng f by FUNCT_1:3;
        hence f.i is finite by A1,Def2;
      end;
      suppose not i in dom f;
        hence f.i is finite by FUNCT_1:def 2;
      end;
    end;
    assume
A2: for i being set st i in dom f holds f.i is finite;
    let i be set;
    assume i in rng f;
    then ex x st x in dom f & i = f.x by FUNCT_1:def 3;
    hence thesis by A2;
  end;
end;



definition
  let I be set;
  let IT be I-defined Function;
  redefine attr IT is finite-yielding means
  for i being set st i in I holds IT.i is finite;
  compatibility
  proof
    hereby
      assume
A1:   IT is finite-yielding;
      let i be set such that i in I;
      per cases;
      suppose i in dom IT;
        hence IT.i is finite by A1,Def4;
      end;
      suppose not i in dom IT;
        hence IT.i is finite by FUNCT_1:def 2;
      end;
    end;
    assume
A2: for i being set st i in I holds IT.i is finite;
    let i be set;
     dom IT c= I;
    hence thesis by A2;
  end;
end;



theorem
  B is infinite implies not B in [:A,B:]
  proof assume that
A1: B is infinite and
A2: B in [:A,B:];
  ex x st x in A & B = [x,{x}] by A2,ZFMISC_1:129;
 hence thesis by A1;
end;



registration
 let I be set, f be I-defined Function;
 cluster finite I-defined f-compatible for Function;
 existence
  proof
   take {};
   thus thesis by FUNCT_1:104,RELAT_1:171;
  end;
end;

registration
 let X,Y be set;
 cluster finite X-defined Y-valued for Function;
 existence
  proof
   take {};
   thus thesis by RELAT_1:171;
  end;
end;

registration
 let X,Y be non empty set;
 cluster X-defined Y-valued non empty finite for Function;
 existence
  proof set x = the Element of X, y = the Element of Y;
   take F = x .--> y;
   thus F is X-defined;
   thus F is Y-valued;
   thus thesis;
  end;
end;

registration
  let A be set, F be finite Relation;
  cluster A|`F -> finite;
  coherence
  proof
    A|`F c= F by RELAT_1:86;
    hence thesis;
  end;
end;

registration
  let A be set, F be finite Relation;
  cluster F|A -> finite;
  coherence
  proof
    F|A c= F by RELAT_1:59;
    hence thesis;
  end;
end;

registration
  let A be finite set, F be Function;
  cluster F|A -> finite;
  coherence
  proof
    dom(F|A) c= A by RELAT_1:58;
    hence thesis by Th10;
  end;
end;

registration
  let R be finite Relation;
  cluster field R -> finite;
  coherence;
end;

registration
  cluster trivial -> finite for set;
  coherence
  proof
    let S be set such that
A1: S is trivial;
    per cases by A1,ZFMISC_1:131;
    suppose S is empty;
      hence thesis;
    end;
    suppose
      ex x being set st S = {x};
      hence thesis;
    end;
  end;
end;

registration
  cluster infinite -> non trivial for set;
  coherence;
end;

registration
  let X be non trivial set;
  cluster finite non trivial for Subset of X;
  existence
  proof
   consider a1,a2 being set such that
A1:  a1 in X & a2 in X & a1 <> a2 by ZFMISC_1:def 10;
    reconsider A = {a1,a2} as Subset of X by A1,ZFMISC_1:32;
   take A;
   thus A is finite;
    a1 in A & a2 in A by TARSKI:def 2;
   hence thesis by A1,ZFMISC_1:def 10;
  end;
end;



registration
  let x,y,a,b be set;
  cluster (x,y) --> (a,b) -> finite;
  coherence;
end;



definition
  let A be set;
  attr A is finite-membered means
:Def6: for B being set st B in A holds B is finite;
end;

registration
 cluster empty -> finite-membered for set;
 coherence
  proof let A be set;
   assume
A1:  A is empty;
   let B be set;
   thus thesis by A1;
  end;
end;

registration
  let A be finite-membered set;
  cluster -> finite for Element of A;
 coherence
  proof let B be Element of A;
   per cases;
   suppose A is empty;
   hence thesis by SUBSET_1:def 1;
   end;
   suppose A is not empty;
   hence thesis by Def6;
   end;
  end;
end;

registration
  cluster non empty finite finite-membered for set;
  existence
  proof
    take x = {{{}}};
    thus x is non empty finite;
    let y be set;
    thus thesis by TARSKI:def 1;
  end;
end;



registration
  let X be finite set;
  cluster {X} -> finite-membered;
  coherence
  proof
  let x;
  thus thesis by TARSKI:def 1;
 end;
  cluster bool X -> finite-membered;
  coherence
  proof
  let x;
  thus thesis;
 end;
  let Y be finite set;
  cluster {X,Y} -> finite-membered;
  coherence
  proof
  let x;
  thus thesis by TARSKI:def 2;
 end;
end;

registration
  let X be finite-membered set;
  cluster -> finite-membered for Subset of X;
  coherence
  proof
  let S be Subset of X;
  let x be set;
  thus thesis;
 end;
  let Y be finite-membered set;
  cluster X \/ Y -> finite-membered;
  coherence
  proof
  let x be set;
  assume x in X\/Y;
  then x in X or x in Y by XBOOLE_0:def 3;
  hence thesis;
 end;
end;

registration
  let X be finite finite-membered set;
  cluster union X -> finite;
  coherence
  proof
  for x st x in X holds x is finite;
  hence thesis by Th7;
 end;
end;

registration
  cluster non empty finite-yielding for Function;
  existence
  proof
    take F = {{}} .--> {{}};
    thus F is non empty;
    let x;
    assume x in {{{}}};
     then x = {{}} by TARSKI:def 1;
    hence F.x is finite by FUNCOP_1:72;
  end;
  cluster empty -> finite-yielding for Relation;
  coherence
  proof
    let F be Relation;
    assume
A1: F is empty;
    let x;
    thus thesis by A1;
  end;
end;

registration
  let F be finite-yielding Function, x be set;
  cluster F.x -> finite;
  coherence
   proof
    per cases;
    suppose x in dom F;
     hence thesis by Def4;
    end;
    suppose not x in dom F;
     hence thesis by FUNCT_1:def 2;
    end;
   end;
end;

registration let F be finite-yielding Relation;
 cluster rng F -> finite-membered;
 coherence
  proof let e be set;
   thus thesis by Def2;
  end;
end;














begin 

reserve X,Y,x for set;

definition
  let IT be set;
  attr IT is cup-closed means
  :Def1:
  for X,Y being set st X in IT & Y in IT holds X \/ Y in IT;
  attr IT is cap-closed means
  for X,Y being set st X in IT & Y in IT holds X /\ Y in IT;
  attr IT is diff-closed means
  :Def3:
  for X,Y being set st X in IT & Y in IT holds X \ Y in IT;
end;

definition
  let IT be set;
  attr IT is preBoolean means
  :Def4:
  IT is cup-closed diff-closed;
end;

registration
  cluster preBoolean -> cup-closed diff-closed for set;
  coherence by Def4;
  cluster cup-closed diff-closed -> preBoolean for set;
  coherence by Def4;
end;

registration
  cluster non empty cup-closed cap-closed diff-closed for set;
  existence
  proof
    take {{}};
    thus {{}} is non empty;
    thus {{}} is cup-closed
    proof
      let X,Y be set;
      assume that
A1:   X in {{}} and
A2:   Y in {{}};
      X = {} by A1,TARSKI:def 1;
      hence thesis by A2;
    end;
    thus {{}} is cap-closed
    proof
      let X,Y be set;
      assume that
A3:   X in {{}} and
      Y in {{}};
      X = {} by A3,TARSKI:def 1;
      hence thesis by TARSKI:def 1;
    end;
    thus {{}} is diff-closed
    proof
      let X,Y be set;
      assume that
A4:   X in {{}} and
      Y in {{}};
      X = {} by A4,TARSKI:def 1;
      hence thesis by TARSKI:def 1;
    end;
  end;
end;

theorem Th1:
  for A being set holds A is preBoolean iff for X,Y being set st X
  in A & Y in A holds X \/ Y in A & X \ Y in A
proof
  let A be set;
  thus A is preBoolean implies for X,Y being set st X in A & Y in A holds X \/
  Y in A & X \ Y in A by Def1,Def3;
  assume
A1: for X,Y being set st X in A & Y in A holds X \/ Y in A & X \ Y in A;
A2: A is diff-closed
  proof
    let X,Y be set;
    assume X in A & Y in A;
    hence thesis by A1;
  end;
  A is cup-closed
  proof
    let X,Y be set;
    assume X in A & Y in A;
    hence thesis by A1;
  end;
  hence thesis by A2;
end;

reserve A for non empty preBoolean set;

definition
  let A;
  let X,Y be Element of A;
  redefine func X \/ Y -> Element of A;
  correctness by Th1;
  redefine func X \ Y -> Element of A;
  correctness by Th1;
end;

theorem Th2:
  X is Element of A & Y is Element of A implies X /\ Y is Element of A
proof
  assume X is Element of A & Y is Element of A;
  then reconsider X,Y as Element of A;
  X /\ Y = X \ (X \ Y) by XBOOLE_1:48;
  hence thesis;
end;

theorem Th3:
  X is Element of A & Y is Element of A implies X \+\ Y is Element of A
proof
  assume X is Element of A & Y is Element of A;
  then reconsider X,Y as Element of A;
  X \+\ Y = (X \ Y) \/ (Y \ X);
  hence thesis;
end;

theorem
  for A being non empty set st for X,Y being Element of A holds X \+\ Y
  in A & X \ Y in A holds A is preBoolean
proof
  let A be non empty set such that
A1: for X,Y being Element of A holds X \+\ Y in A & X \ Y in A;
  now
    let X,Y be set;
    assume that
A2: X in A and
A3: Y in A;
    reconsider Z = Y \ X as Element of A by A1,A2,A3;
    X \/ Y = X \+\ Z by XBOOLE_1:98;
    hence X \/ Y in A by A1,A2;
    thus X \ Y in A by A1,A2,A3;
  end;
  hence thesis by Th1;
end;

theorem
  for A being non empty set st for X,Y being Element of A holds X \+\ Y
  in A & X /\ Y in A holds A is preBoolean
proof
  let A be non empty set such that
A1: for X,Y being Element of A holds X \+\ Y in A & X /\ Y in A;
  now
    let X,Y be set;
    assume that
A2: X in A and
A3: Y in A;
    reconsider Z1 = X \+\ Y, Z2 = X /\ Y as Element of A by A1,A2,A3;
    X \/ Y = Z1 \+\ Z2 by XBOOLE_1:94;
    hence X \/ Y in A by A1;
    X \ Y = X \+\ Z2 by XBOOLE_1:100;
    hence X \ Y in A by A1,A2;
  end;
  hence thesis by Th1;
end;

theorem
  for A being non empty set st for X,Y being Element of A holds X \+\ Y
  in A & X \/ Y in A holds A is preBoolean
proof
  let A be non empty set such that
A1: for X,Y being Element of A holds X \+\ Y in A & X \/ Y in A;
  now
    let X,Y be set;
    assume that
A2: X in A and
A3: Y in A;
    thus X \/ Y in A by A1,A2,A3;
    reconsider Z1 = X \+\ Y, Z2 = X \/ Y as Element of A by A1,A2,A3;
    X /\ Y = Z1 \+\ Z2 by XBOOLE_1:95;
    then reconsider Z = X /\ Y as Element of A by A1;
    X \ Y = X \+\ Z by XBOOLE_1:100;
    hence X \ Y in A by A1,A2;
  end;
  hence thesis by Th1;
end;

definition
  let A;
  let X,Y be Element of A;
  redefine func X /\ Y -> Element of A;
  coherence by Th2;
  redefine func X \+\ Y -> Element of A;
  coherence by Th3;
end;

theorem Th7:
  {} in A
proof
  set x = the Element of A;
  x \ x = {} by XBOOLE_1:37;
  hence thesis;
end;

theorem Th8:
  for A being set holds bool A is preBoolean
proof
  let A be set;
  now
    let X,Y be set;
    assume X in bool A & Y in bool A;
    then reconsider X9=X,Y9=Y as Subset of A;
    X9 \/ Y9 in bool A & X9 \ Y9 in bool A;
    hence X \/ Y in bool A & X \ Y in bool A;
  end;
  hence thesis by Th1;
end;

registration
  let A be set;
  cluster bool A -> preBoolean;
  coherence by Th8;
end;

theorem
  for A,B being non empty preBoolean set holds A /\ B is non empty preBoolean
proof
  let A,B be non empty preBoolean set;
  {} in A & {} in B by Th7;
  then reconsider C = A /\ B as non empty set by XBOOLE_0:def 4;
  now
    let X,Y be set;
    assume
A1: X in C & Y in C;
    then
A2: X in B & Y in B by XBOOLE_0:def 4;
    then
A3: X \/ Y in B by Th1;
A4: X \ Y in B by A2,Th1;
A5: X in A & Y in A by A1,XBOOLE_0:def 4;
    then X \/ Y in A by Th1;
    hence X \/ Y in C by A3,XBOOLE_0:def 4;
    X \ Y in A by A5,Th1;
    hence X \ Y in C by A4,XBOOLE_0:def 4;
  end;
  hence thesis by Th1;
end;



definition
  let A be set;
  func Fin A -> preBoolean set means
  :Def5:
  for X being set holds X in it iff X c= A & X is finite;
  existence
  proof
    defpred P[set] means ex y being set st y=$1 & y c= A & y is finite;
    consider P being set such that
A1: for x holds x in P iff x in bool A & P[x] from XBOOLE_0:sch 1;
    {} c= A by XBOOLE_1:2;
    then reconsider Q=P as non empty set by A1;
    for X,Y being set st X in Q & Y in Q holds X \/ Y in Q & X \ Y in Q
    proof
      let X,Y be set;
      assume that
A2:   X in Q and
A3:   Y in Q;
      consider Z1 being set such that
A4:   Z1=X and
A5:   Z1 c=A and
A6:   Z1 is finite by A1,A2;
      consider Z2 being set such that
A7:   Z2=Y and
A8:   Z2 c= A and
A9:   Z2 is finite by A1,A3;
A10:  Z1 \ Z2 c= A by A5,XBOOLE_1:1;
      Z1 \/ Z2 c= A by A5,A8,XBOOLE_1:8;
      hence thesis by A1,A4,A6,A7,A9,A10;
    end;
    then reconsider R=Q as non empty preBoolean set by Th1;
    for X being set holds X in R iff X c= A & X is finite
    proof
      let X be set;
      thus X in R implies X c= A & X is finite
      proof
        assume X in R;
        then ex Y being set st Y=X & Y c= A & Y is finite by A1;
        hence thesis;
      end;
      thus thesis by A1;
    end;
    hence thesis;
  end;
  uniqueness
  proof
    let P,Q be preBoolean set;
    assume that
A11: for X being set holds X in P iff X c= A & X is finite and
A12: for X being set holds X in Q iff X c= A & X is finite;
    for x being set holds x in P iff x in Q
    proof
      let x;
      thus x in P implies x in Q
      proof
        assume x in P;
        then x c= A & x is finite by A11;
        hence thesis by A12;
      end;
      thus x in Q implies x in P
      proof
        assume x in Q;
        then x c= A & x is finite by A12;
        hence thesis by A11;
      end;
    end;
    hence thesis by TARSKI:1;
  end;
end;

registration
  let A be set;
  cluster Fin A -> non empty;
  coherence
  proof
    {} c= A by XBOOLE_1:2;
    hence thesis by Def5;
  end;
end;

registration
  let A be set;
  cluster -> finite for Element of Fin A;
  coherence by Def5;
end;

theorem Th10:
  for A, B being set st A c= B holds Fin A c= Fin B
proof
  let A,B be set;
  assume
A1: A c= B;
  now
    let X;
    assume
A2: X in Fin A;
    then X c= A by Def5;
    then X c= B by A1,XBOOLE_1:1;
    hence X in Fin B by A2,Def5;
  end;
  hence thesis by TARSKI:def 3;
end;

theorem
  for A, B being set holds Fin (A /\ B) = Fin A /\ Fin B
proof
  let A, B be set;
  Fin (A /\ B) c= Fin A & Fin (A /\ B) c= Fin B by Th10,XBOOLE_1:17;
  hence Fin (A /\ B) c= Fin A /\ Fin B by XBOOLE_1:19;
  now
    let X;
    assume
A1: X in Fin A /\ Fin B;
    then X in Fin B by XBOOLE_0:def 4;
    then
A2: X c= B by Def5;
A3: X in Fin A by A1,XBOOLE_0:def 4;
    then X c= A by Def5;
    then X c= A /\ B by A2,XBOOLE_1:19;
    hence X in Fin (A /\ B) by A3,Def5;
  end;
  hence thesis by TARSKI:def 3;
end;

theorem
  for A, B being set holds Fin A \/ Fin B c= Fin (A \/ B)
proof
  let A, B be set;
  Fin A c= Fin (A \/ B) & Fin B c= Fin(A \/ B) by Th10,XBOOLE_1:7;
  hence thesis by XBOOLE_1:8;
end;

theorem Th13:
  for A being set holds Fin A c= bool A
proof
  let A be set;
  let x;
  assume x in Fin A;
  then x c= A by Def5;
  hence thesis;
end;

theorem Th14:
  for A being set st A is finite holds Fin A = bool A
proof
  let A be set;
  assume
A1: A is finite;
A2: bool A c= Fin A
  proof
    let x;
    assume x in bool A;
    hence thesis by A1,Def5;
  end;
  Fin A c= bool A by Th13;
  hence thesis by A2,XBOOLE_0:def 10;
end;

theorem
  Fin {} = { {} } by Th14,ZFMISC_1:1;



definition
  let A be set;
  mode Finite_Subset of A is Element of Fin A;
end;

theorem
  for A being set, X being Finite_Subset of A holds X is finite;

theorem
  for A being set, X being Finite_Subset of A holds X is Subset of A by Def5;

theorem
  for A being set, X being Subset of A st A is finite holds X is
  Finite_Subset of A by Def5;














begin

reserve X,Y for set,
  x,x1,x2,y,y1,y2,z for set,
  f,g,h for Function;

Lm1: (ex X st X <> {} & X in Y) iff union Y <> {}
proof
  thus (ex X st X <> {} & X in Y) implies union Y <> {}
  proof
    given X such that
A1: X <> {} and
A2: X in Y;
    set x = the Element of X;
    x in X by A1;
    hence thesis by A2,TARSKI:def 4;
  end;
  set x = the Element of union Y;
  assume union Y <> {};
  then consider X such that
A3: x in X and
A4: X in Y by TARSKI:def 4;
  take X;
  thus thesis by A3,A4;
end;





reserve M for non empty set;

definition
  let M;
  assume
A1: not {} in M;
  mode Choice_Function of M -> Function of M, union M means
    :Def1:
    for X st X in M holds it.X in X;
  existence
  proof
    deffunc FF(set)=$1`2;
    set x = the Element of M;
    deffunc F(set)=[:{$1},$1:];
    consider f such that
A2: dom f = M and
A3: for X st X in M holds f.X = F(X) from FUNCT_1:sch 3;
A4: x in M;
    then reconsider N = rng f as non empty set by A2,FUNCT_1:3;
A5: now
      let X,Y;
      assume that
A6:   X in N and
A7:   Y in N and
A8:   X <> Y;
      consider y such that
A9:   y in dom f and
A10:  f.y = Y by A7,FUNCT_1:def 3;
A11:  Y = [:{y},y:] by A2,A3,A9,A10;
      set z = the Element of X /\ Y;
      assume X meets Y;
      then
A12:  X /\ Y <> {} by XBOOLE_0:def 7;
      consider x such that
A13:  x in dom f and
A14:  f.x = X by A6,FUNCT_1:def 3;
      X = [:{x},x:] by A2,A3,A13,A14;
      then consider x1,x2 such that
      z = [x1,x2] and
A15:  x1 in {x} /\ {y} and
      x2 in x /\ y by A11,A12,ZFMISC_1:85;
      x1 in {x} by A15,XBOOLE_0:def 4;
      then
A16:  x1 = x by TARSKI:def 1;
      x1 in{y} by A15,XBOOLE_0:def 4;
      hence contradiction by A8,A14,A10,A16,TARSKI:def 1;
    end;
    now
      let X;
      assume X in N;
      then consider y such that
A17:  y in dom f and
A18:  f.y = X by FUNCT_1:def 3;
A19:  y <> {} by A1,A2,A17;
      X = [:{y},y:] by A2,A3,A17,A18;
      hence X <> {} by A19;
    end;
    then consider Choice being set such that
A20: for X st X in N ex x st Choice /\ X = {x} by A5,WELLORD2:18;
    defpred P[set,set] means $2 in $1 /\ Choice;
A21: for X st X in N ex y st P[X,y]
    proof
      let X;
      assume X in N;
      then consider x such that
A22:  Choice /\ X = {x} by A20;
      take x;
      thus thesis by A22,TARSKI:def 1;
    end;
A23: for X,y1,y2 st X in N & P[X,y1] & P[X,y2] holds y1 = y2
    proof
      let X,y1,y2;
      assume that
A24:  X in N and
A25:  y1 in X /\ Choice and
A26:  y2 in X /\ Choice;
      consider x such that
A27:  {x} = Choice /\ X by A20,A24;
      x = y1 by A25,A27,TARSKI:def 1;
      hence thesis by A26,A27,TARSKI:def 1;
    end;
    consider g such that
A28: dom g = N and
A29: for X st X in N holds P[X,g.X] from FUNCT_1:sch 2 (A23,A21);
A30: rng g c= union N
    proof
      let x;
      assume x in rng g;
      then consider y such that
A31:  y in dom g and
A32:  g.y = x by FUNCT_1:def 3;
      x in y /\ Choice by A28,A29,A31,A32;
      then x in y by XBOOLE_0:def 4;
      hence thesis by A28,A31,TARSKI:def 4;
    end;
    f.x in rng f by A2,FUNCT_1:def 3;
    then consider x1 such that
A33: x1 in N;
    consider x2 such that
A34: x2 in dom f and
A35: f.x2 = x1 by A33,FUNCT_1:def 3;
A36: x2 <> {} by A1,A2,A34;
    reconsider f as Function of M,N by A2,FUNCT_2:def 1,RELSET_1:4;
    consider h such that
A37: dom h = union N and
A38: for x st x in union N holds h.x = FF(x) from FUNCT_1:sch 3;
A39: rng h c= union M
    proof
      let x;
      assume x in rng h;
      then consider y such that
A40:  y in dom h and
A41:  h.y = x by FUNCT_1:def 3;
      consider Y such that
A42:  y in Y and
A43:  Y in N by A37,A40,TARSKI:def 4;
      consider z such that
A44:  z in dom f and
A45:  f.z = Y by A43,FUNCT_1:def 3;
A46:  Y = [:{z},z:] by A3,A44,A45;
      then consider x1,x2 such that
A47:  y = [x1,x2] by A42,RELAT_1:def 1;
A48:  x2 in z by A42,A46,A47,ZFMISC_1:87;
      x = [x1,x2]`2 by A37,A38,A40,A41,A47;
      then x in z by A48;
      hence thesis by A44,TARSKI:def 4;
    end;
    x1 = [:{x2},x2:] by A2,A3,A34,A35;
    then
A49: union N <> {} by A33,A36,Lm1;
    then reconsider g as Function of N, union N by A28,A30,FUNCT_2:def 1
,RELSET_1:4;
    union M <> {} by A1,A4,Lm1;
    then reconsider h as Function of union N, union M by A37,A39,FUNCT_2:def 1
,RELSET_1:4;
A50: dom(g * f) = M by A49,FUNCT_2:def 1;
    reconsider F = h * (g * f) as Function of M, union M by A49;
    take F;
    let X;
    assume
A51: X in M;
    then
A52: f.X = [:{X},X:] by A3;
    set fX = f.X;
A53: f.X in N by A2,A51,FUNCT_1:def 3;
    then g.(fX) in rng g by A28,FUNCT_1:def 3;
    then
A54: h.(g.fX) = (g.fX)`2 by A38;
A55: g.(fX) in fX /\ Choice by A29,A53;
    then consider x1,x2 such that
A56: [x1,x2] = g.(fX) by A52,RELAT_1:def 1;
    g.(fX) in fX by A55,XBOOLE_0:def 4;
    then x2 in X by A52,A56,ZFMISC_1:87;
    then
A57: h.(g.fX) in X by A56,A54,MCART_1:7;
    (g * f).X = g.fX by A2,A51,FUNCT_1:13;
    hence thesis by A50,A51,A57,FUNCT_1:13;
  end;
end;

reserve D for non empty set;

definition
  let D be set;
  func BOOL D -> set equals
  bool D \ {{}};
  coherence;
end;

registration
  let D;
  cluster BOOL D -> non empty;
  coherence
  proof
A1: not D in {{}} by TARSKI:def 1;
    D in bool D by ZFMISC_1:def 1;
    hence thesis by A1,XBOOLE_0:def 5;
  end;
end;

theorem
  not {} in BOOL D
proof
  assume not thesis;
  then not {} in {{}} by XBOOLE_0:def 5;
  hence thesis by TARSKI:def 1;
end;

theorem
  D c= X iff D in BOOL X
proof
  thus D c= X implies D in BOOL X
  proof
A1: not D in {{}} by TARSKI:def 1;
    assume D c= X;
    hence thesis by A1,XBOOLE_0:def 5;
  end;
  assume D in BOOL X;
  hence thesis;
end;

reserve P for Relation;





definition
  let X;
  mode Order of X is total reflexive antisymmetric transitive Relation of X;
end;

reserve O for Order of X;

Lm2: for R being total Relation of X holds field R = X
proof
  let R be total Relation of X;
  thus field R = X \/ rng R by PARTFUN1:def 2
    .= X by XBOOLE_1:12;
end;

theorem Th3:
  x in X implies [x,x] in O
proof
  field O = X by Lm2;
  then O is_reflexive_in X by RELAT_2:def 9;
  hence thesis by RELAT_2:def 1;
end;

theorem
  x in X & y in X & [x,y] in O & [y,x] in O implies x = y
proof
  field O = X by Lm2;
  then O is_antisymmetric_in X by RELAT_2:def 12;
  hence thesis by RELAT_2:def 4;
end;

theorem
  x in X & y in X & z in X & [x,y] in O & [y,z] in O implies [x,z] in O
proof
  field O = X by Lm2;
  then O is_transitive_in X by RELAT_2:def 16;
  hence thesis by RELAT_2:def 8;
end;

theorem
  (ex X st X <> {} & X in Y) iff union Y <> {} by Lm1;

theorem
  P is_strongly_connected_in X iff P is_reflexive_in X & P is_connected_in X
proof
  thus P is_strongly_connected_in X implies P is_reflexive_in X & P
  is_connected_in X
  proof
    assume
A1: P is_strongly_connected_in X;
    thus P is_reflexive_in X
    proof
      let x;
      assume x in X;
      hence thesis by A1,RELAT_2:def 7;
    end;
    let x,y;
    assume that
A2: x in X and
A3: y in X;
    thus thesis by A1,A2,A3,RELAT_2:def 7;
  end;
  assume that
A4: P is_reflexive_in X and
A5: P is_connected_in X;
  let x,y;
  assume that
A6: x in X and
A7: y in X;
  x = y implies thesis by A4,A6,RELAT_2:def 1;
  hence thesis by A5,A6,A7,RELAT_2:def 6;
end;

theorem Th8:
  P is_reflexive_in X & Y c= X implies P is_reflexive_in Y
proof
  assume that
A1: P is_reflexive_in X and
A2: Y c= X;
  let x;
  assume x in Y;
  hence thesis by A1,A2,RELAT_2:def 1;
end;

theorem Th9:
  P is_antisymmetric_in X & Y c= X implies P is_antisymmetric_in Y
proof
  assume that
A1: P is_antisymmetric_in X and
A2: Y c= X;
  let x,y;
  assume that
A3: x in Y and
A4: y in Y;
  thus thesis by A1,A2,A3,A4,RELAT_2:def 4;
end;

theorem Th10:
  P is_transitive_in X & Y c= X implies P is_transitive_in Y
proof
  assume that
A1: P is_transitive_in X and
A2: Y c= X;
  let x,y,z;
  assume that
A3: x in Y and
A4: y in Y and
A5: z in Y;
  thus thesis by A1,A2,A3,A4,A5,RELAT_2:def 8;
end;

theorem
  P is_strongly_connected_in X & Y c= X implies P is_strongly_connected_in Y
proof
  assume that
A1: P is_strongly_connected_in X and
A2: Y c= X;
  let x,y;
  assume that
A3: x in Y and
A4: y in Y;
  thus thesis by A1,A2,A3,A4,RELAT_2:def 7;
end;

theorem
  for R being total Relation of X holds field R = X by Lm2;

theorem Th13:
  for A being set, R being Relation of A st R is_reflexive_in A
  holds dom R = A & field R = A
proof
  let A be set, R be Relation of A such that
A1: R is_reflexive_in A;
A2: A c= dom R
  proof
    let x;
    assume x in A;
    then [x,x] in R by A1,RELAT_2:def 1;
    hence thesis by XTUPLE_0:def 12;
  end;
  hence dom R = A by XBOOLE_0:def 10;
  thus field R = A \/ rng R by A2,XBOOLE_0:def 10
    .= A by XBOOLE_1:12;
end;

begin 

reserve R,P for Relation,
  X,X1,X2,Y,Z,x,y,z,u for set,
  g,h for Function,
  O for Order of X,
  D for non empty set,
  d,d1,d2 for Element of D,
  A1,A2,B for Ordinal,
  L,L1,L2 for T-Sequence;





theorem Th14:
  dom O = X & rng O = X
proof
  thus dom O = X
  proof
    thus dom O c= X;
    let x;
    assume x in X;
    then [x,x] in O by Th3;
    hence thesis by XTUPLE_0:def 12;
  end;
  thus rng O c= X;
  let x;
  assume x in X;
  then [x,x] in O by Th3;
  hence thesis by XTUPLE_0:def 13;
end;

theorem
  field O = X
proof
  thus X = X \/ X .= dom O \/ X by Th14
    .= field O by Th14;
end;

definition
  let R;
  attr R is being_quasi-order means
  R is reflexive transitive;
  attr R is being_partial-order means
  :Def4:
  R is reflexive transitive antisymmetric;
  attr R is being_linear-order means
  :Def5:
  R is reflexive transitive antisymmetric connected;
end;

theorem
  R is being_quasi-order implies R~ is being_quasi-order
proof
  assume that
A1: R is reflexive and
A2: R is transitive;
  thus R~ is reflexive & R~ is transitive by A1,A2;
end;

theorem
  R is being_partial-order implies R~ is being_partial-order
proof
  assume that
A1: R is reflexive and
A2: R is transitive and
A3: R is antisymmetric;
  thus R~ is reflexive & R~ is transitive & R~ is antisymmetric by A1,A2,A3;
end;

Lm3: R is connected implies R~ is connected
proof
  assume
A1: x in field R & y in field R & x <> y implies [x,y] in R or [y,x] in R;
  let x,y;
  assume that
A2: x in field(R~) and
A3: y in field(R~) and
A4: x <> y;
  field R = field(R~) by RELAT_1:21;
  then [x,y] in R or [y,x] in R by A1,A2,A3,A4;
  hence thesis by RELAT_1:def 7;
end;

theorem Th18:
  R is being_linear-order implies R~ is being_linear-order
proof
  assume that
A1: R is reflexive and
A2: R is transitive and
A3: R is antisymmetric and
A4: R is connected;
  thus R~ is reflexive & R~ is transitive & R~ is antisymmetric & R~ is
  connected by A1,A2,A3,A4,Lm3;
end;

theorem
  R is well-ordering implies R is being_quasi-order & R is
  being_partial-order & R is being_linear-order
proof
  assume that
A1: R is reflexive and
A2: R is transitive and
A3: R is antisymmetric and
A4: R is connected and
  R is well_founded;
  thus R is reflexive & R is transitive & R is reflexive & R is transitive & R
is antisymmetric & R is reflexive & R is transitive & R is antisymmetric & R is
  connected by A1,A2,A3,A4;
end;

theorem
  R is being_linear-order implies R is being_quasi-order & R is
  being_partial-order
proof
  assume that
A1: R is reflexive and
A2: R is transitive and
A3: R is antisymmetric and
  R is connected;
  thus R is reflexive & R is transitive & R is reflexive & R is transitive & R
  is antisymmetric by A1,A2,A3;
end;

theorem Th21:
  R is being_partial-order implies R is being_quasi-order
proof
  assume that
A1: R is reflexive and
A2: R is transitive and
  R is antisymmetric;
  thus R is reflexive & R is transitive by A1,A2;
end;

theorem
  O is being_partial-order by Def4;

theorem
  O is being_quasi-order
proof
  O is being_partial-order by Def4;
  hence thesis by Th21;
end;

theorem
  O is connected implies O is being_linear-order by Def5;

Lm4: R c= [:field R,field R:]
proof
  let y,z;
  assume
A1: [y,z] in R;
  then
A2: z in field R by RELAT_1:15;
  y in field R by A1,RELAT_1:15;
  hence thesis by A2,ZFMISC_1:87;
end;

Lm5: R is reflexive & X c= field R implies field(R|_2 X) = X
proof
  assume that
A1: x in field R implies [x,x] in R and
A2: X c= field R;
  thus field(R|_2 X) c= X
  proof
    let y;
    thus thesis by WELLORD1:12;
  end;
  let y;
  assume
A3: y in X;
  then
A4: [y,y] in [:X,X:] by ZFMISC_1:87;
  [y,y] in R by A1,A2,A3;
  then [y,y] in R|_2 X by A4,XBOOLE_0:def 4;
  hence thesis by RELAT_1:15;
end;

theorem
  R is being_quasi-order implies R|_2 X is being_quasi-order
proof
  assume that
A1: R is reflexive and
A2: R is transitive;
  thus R|_2 X is reflexive & R|_2 X is transitive by A1,A2,WELLORD1:15,17;
end;

theorem
  R is being_partial-order implies R|_2 X is being_partial-order
proof
  assume that
A1: R is reflexive and
A2: R is transitive and
A3: R is antisymmetric;
  thus R|_2 X is reflexive & R|_2 X is transitive & R|_2 X is antisymmetric by
A1,A2,A3,WELLORD1:15,17;
end;

theorem
  R is being_linear-order implies R|_2 X is being_linear-order
proof
  assume that
A1: R is reflexive and
A2: R is transitive and
A3: R is antisymmetric and
A4: R is connected;
  thus R|_2 X is reflexive & R|_2 X is transitive & R|_2 X is antisymmetric &
  R|_2 X is connected by A1,A2,A3,A4,WELLORD1:15,16,17;
end;

registration let R be empty Relation;
 cluster field R -> empty;
 coherence;
end;

registration
 cluster empty ->  being_quasi-order being_partial-order
         being_linear-order well-ordering for Relation;
 coherence
proof
 let R be Relation such that
A1: R is empty;
  thus
A2: R is reflexive
  proof
    let x;
    assume x in field R;
    hence thesis by A1;
  end;
  thus
A3: R is transitive
  proof
    let x,y,z such that
    x in field R;
    thus thesis by A1;
  end;
  hence R is reflexive & R is transitive by A2;
  thus
A4: R is antisymmetric
  proof
    let x,y such that
    x in field R;
    thus thesis by A1;
  end;
  hence R is reflexive & R is transitive & R is antisymmetric by A2,A3;
  thus R is connected
  proof
    let x,y such that
    x in field R;
    thus thesis by A1;
  end;
  hence R is reflexive & R is transitive & R is antisymmetric & R is
  connected by A2,A3,A4;
  let Y;
  assume that
A5: Y c= field R and
A6: Y <> {};
  thus thesis by A5,A6,A1;
end;
end;

registration let X;
 cluster id X -> being_quasi-order being_partial-order;
 coherence
proof
  thus id X is reflexive & id X is transitive & id X is reflexive & id X is
  transitive & id X is antisymmetric;
end;
end;

definition
  let R,X;
  pred R quasi_orders X means
  :Def6:
  R is_reflexive_in X & R is_transitive_in X;
  pred R partially_orders X means
  :Def7:
  R is_reflexive_in X & R is_transitive_in X & R is_antisymmetric_in X;
  pred R linearly_orders X means
  R is_reflexive_in X & R is_transitive_in X &
  R is_antisymmetric_in X & R is_connected_in X;
end;

theorem Th28:
  R well_orders X implies R quasi_orders X & R partially_orders X
  & R linearly_orders X
proof
  assume that
A1: R is_reflexive_in X and
A2: R is_transitive_in X and
A3: R is_antisymmetric_in X and
A4: R is_connected_in X and
  R is_well_founded_in X;
  thus R is_reflexive_in X & R is_transitive_in X & R is_reflexive_in X & R
  is_transitive_in X & R is_antisymmetric_in X & R is_reflexive_in X & R
is_transitive_in X & R is_antisymmetric_in X & R is_connected_in X by A1,A2,A3
,A4;
end;

theorem Th29:
  R linearly_orders X implies R quasi_orders X & R partially_orders X
proof
  assume that
A1: R is_reflexive_in X and
A2: R is_transitive_in X and
A3: R is_antisymmetric_in X and
  R is_connected_in X;
  thus R is_reflexive_in X & R is_transitive_in X & R is_reflexive_in X & R
  is_transitive_in X & R is_antisymmetric_in X by A1,A2,A3;
end;

theorem
  R partially_orders X implies R quasi_orders X
proof
  assume that
A1: R is_reflexive_in X and
A2: R is_transitive_in X and
  R is_antisymmetric_in X;
  thus R is_reflexive_in X & R is_transitive_in X by A1,A2;
end;

theorem Th31:
  R is being_quasi-order implies R quasi_orders field R
proof
  assume that
A1: R is_reflexive_in field R and
A2: R is_transitive_in field R;
  thus R is_reflexive_in field R & R is_transitive_in field R by A1,A2;
end;

theorem
  R quasi_orders Y & X c= Y implies R quasi_orders X
proof
  assume that
A1: R is_reflexive_in Y and
A2: R is_transitive_in Y and
A3: X c= Y;
  thus R is_reflexive_in X & R is_transitive_in X by A1,A2,A3,Th8,Th10;
end;

Lm6: R is_reflexive_in X implies R|_2 X is reflexive
proof
  assume
A1: x in X implies [x,x] in R;
  let x;
  assume
A2: x in field(R|_2 X);
  then x in X by WELLORD1:12;
  then
A3: [x,x] in [:X,X:] by ZFMISC_1:87;
  [x,x] in R by A1,A2,WELLORD1:12;
  hence thesis by A3,XBOOLE_0:def 4;
end;

Lm7: R is_transitive_in X implies R|_2 X is transitive
proof
  assume
A1: x in X & y in X & z in X & [x,y] in R & [y,z] in R implies [x,z] in R;
  let x,y,z;
  assume that
A2: x in field(R|_2 X) and
A3: y in field(R|_2 X) and
A4: z in field(R|_2 X);
A5: z in X by A4,WELLORD1:12;
A6: x in X by A2,WELLORD1:12;
  then
A7: [x,z] in [:X,X:] by A5,ZFMISC_1:87;
  assume that
A8: [x,y] in R|_2 X and
A9: [y,z] in R|_2 X;
A10: [x,y] in R by A8,XBOOLE_0:def 4;
A11: [y,z] in R by A9,XBOOLE_0:def 4;
  y in X by A3,WELLORD1:12;
  then [x,z] in R by A1,A6,A5,A10,A11;
  hence thesis by A7,XBOOLE_0:def 4;
end;

Lm8: R is_antisymmetric_in X implies R|_2 X is antisymmetric
proof
  assume
A1: x in X & y in X & [x,y] in R & [y,x] in R implies x = y;
  let x,y;
  assume that
A2: x in field(R|_2 X) and
A3: y in field(R|_2 X);
A4: y in X by A3,WELLORD1:12;
  assume that
A5: [x,y] in R|_2 X and
A6: [y,x] in R|_2 X;
A7: [x,y] in R by A5,XBOOLE_0:def 4;
A8: [y,x] in R by A6,XBOOLE_0:def 4;
  x in X by A2,WELLORD1:12;
  hence thesis by A1,A4,A7,A8;
end;

Lm9: R is_connected_in X implies R|_2 X is connected
proof
  assume
A1: x in X & y in X & x <> y implies [x,y] in R or [y,x] in R;
  let x,y;
  assume that
A2: x in field(R|_2 X) and
A3: y in field(R|_2 X) and
A4: x <> y;
A5: y in X by A3,WELLORD1:12;
A6: x in X by A2,WELLORD1:12;
  then
A7: [x,y] in [:X,X:] by A5,ZFMISC_1:87;
A8: [y,x] in [:X,X:] by A6,A5,ZFMISC_1:87;
  [x,y] in R or [y,x] in R by A1,A4,A6,A5;
  hence thesis by A7,A8,XBOOLE_0:def 4;
end;

theorem
  R quasi_orders X implies R|_2 X is being_quasi-order
proof
  assume that
A1: R is_reflexive_in X and
A2: R is_transitive_in X;
  thus R|_2 X is reflexive & R|_2 X is transitive by A1,A2,Lm6,Lm7;
end;

theorem Th34:
  R is being_partial-order implies R partially_orders field R
proof
  assume that
A1: R is_reflexive_in field R and
A2: R is_transitive_in field R and
A3: R is_antisymmetric_in field R;
  thus R is_reflexive_in field R & R is_transitive_in field R & R
  is_antisymmetric_in field R by A1,A2,A3;
end;

theorem
  R partially_orders Y & X c= Y implies R partially_orders X
proof
  assume that
A1: R is_reflexive_in Y and
A2: R is_transitive_in Y and
A3: R is_antisymmetric_in Y and
A4: X c= Y;
  thus R is_reflexive_in X & R is_transitive_in X & R is_antisymmetric_in X by
A1,A2,A3,A4,Th8,Th9,Th10;
end;

theorem
  R partially_orders X implies R|_2 X is being_partial-order
proof
  assume that
A1: R is_reflexive_in X and
A2: R is_transitive_in X and
A3: R is_antisymmetric_in X;
  thus R|_2 X is reflexive & R|_2 X is transitive & R|_2 X is antisymmetric by
A1,A2,A3,Lm6,Lm7,Lm8;
end;

Lm10: R is_connected_in X & Y c= X implies R is_connected_in Y
proof
  assume that
A1: R is_connected_in X and
A2: Y c= X;
  let x,y;
  assume that
A3: x in Y and
A4: y in Y;
  thus thesis by A1,A2,A3,A4,RELAT_2:def 6;
end;

theorem
  R is being_linear-order implies R linearly_orders field R
proof
  assume that
A1: R is_reflexive_in field R and
A2: R is_transitive_in field R and
A3: R is_antisymmetric_in field R and
A4: R is_connected_in field R;
  thus R is_reflexive_in field R & R is_transitive_in field R & R
  is_antisymmetric_in field R & R is_connected_in field R by A1,A2,A3,A4;
end;

theorem
  R linearly_orders Y & X c= Y implies R linearly_orders X
proof
  assume that
A1: R is_reflexive_in Y and
A2: R is_transitive_in Y and
A3: R is_antisymmetric_in Y and
A4: R is_connected_in Y and
A5: X c= Y;
  thus R is_reflexive_in X & R is_transitive_in X & R is_antisymmetric_in X &
  R is_connected_in X by A1,A2,A3,A4,A5,Lm10,Th8,Th9,Th10;
end;

theorem
  R linearly_orders X implies R|_2 X is being_linear-order
proof
  assume that
A1: R is_reflexive_in X and
A2: R is_transitive_in X and
A3: R is_antisymmetric_in X and
A4: R is_connected_in X;
  thus R|_2 X is reflexive & R|_2 X is transitive & R|_2 X is antisymmetric &
  R|_2 X is connected by A1,A2,A3,A4,Lm6,Lm7,Lm8,Lm9;
end;

Lm11: R is_reflexive_in X implies R~ is_reflexive_in X
proof
  assume
A1: x in X implies [x,x] in R;
  let x;
  assume x in X;
  then [x,x] in R by A1;
  hence thesis by RELAT_1:def 7;
end;

Lm12: R is_transitive_in X implies R~ is_transitive_in X
proof
  assume
A1: x in X & y in X & z in X & [x,y] in R & [y,z] in R implies [x,z] in R;
  let x,y,z;
  assume that
A2: x in X and
A3: y in X and
A4: z in X and
A5: [x,y] in R~ and
A6: [y,z] in R~;
A7: [z,y] in R by A6,RELAT_1:def 7;
  [y,x] in R by A5,RELAT_1:def 7;
  then [z,x] in R by A1,A2,A3,A4,A7;
  hence thesis by RELAT_1:def 7;
end;

Lm13: R is_antisymmetric_in X implies R~ is_antisymmetric_in X
proof
  assume
A1: x in X & y in X & [x,y] in R & [y,x] in R implies x = y;
  let x,y;
  assume that
A2: x in X and
A3: y in X and
A4: [x,y] in R~ and
A5: [y,x] in R~;
A6: [y,x] in R by A4,RELAT_1:def 7;
  [x,y] in R by A5,RELAT_1:def 7;
  hence thesis by A1,A2,A3,A6;
end;

Lm14: R is_connected_in X implies R~ is_connected_in X
proof
  assume
A1: x in X & y in X & x <> y implies [x,y] in R or [y,x] in R;
  let x,y;
  assume that
A2: x in X and
A3: y in X and
A4: x <> y;
  [x,y] in R or [y,x] in R by A1,A2,A3,A4;
  hence thesis by RELAT_1:def 7;
end;

theorem
  R quasi_orders X implies R~ quasi_orders X
proof
  assume that
A1: R is_reflexive_in X and
A2: R is_transitive_in X;
  thus R~ is_reflexive_in X & R~ is_transitive_in X by A1,A2,Lm11,Lm12;
end;

theorem Th41:
  R partially_orders X implies R~ partially_orders X
proof
  assume that
A1: R is_reflexive_in X and
A2: R is_transitive_in X and
A3: R is_antisymmetric_in X;
  thus R~ is_reflexive_in X & R~ is_transitive_in X & R~ is_antisymmetric_in X
  by A1,A2,A3,Lm11,Lm12,Lm13;
end;

theorem
  R linearly_orders X implies R~ linearly_orders X
proof
  assume that
A1: R is_reflexive_in X and
A2: R is_transitive_in X and
A3: R is_antisymmetric_in X and
A4: R is_connected_in X;
  thus R~ is_reflexive_in X & R~ is_transitive_in X & R~ is_antisymmetric_in X
  & R~ is_connected_in X by A1,A2,A3,A4,Lm11,Lm12,Lm13,Lm14;
end;

theorem
  O quasi_orders X
proof
A1: field O = X by Lm2;
  then
A2: O is_transitive_in X by RELAT_2:def 16;
  O is_reflexive_in X by A1,RELAT_2:def 9;
  hence thesis by A2,Def6;
end;

theorem
  O partially_orders X
proof
A1: field O = X by Lm2;
  then
A2: O is_antisymmetric_in X by RELAT_2:def 12;
A3: O is_transitive_in X by A1,RELAT_2:def 16;
  O is_reflexive_in X by A1,RELAT_2:def 9;
  hence thesis by A2,A3,Def7;
end;

theorem Th45:
  R partially_orders X implies R |_2 X is Order of X
proof
  set S = R |_2 X;
A1: field S c= X by WELLORD1:13;
  rng S c= field S by XBOOLE_1:7;
  then
A2: rng S c= X by A1,XBOOLE_1:1;
  dom S c= field S by XBOOLE_1:7;
  then dom S c= X by A1,XBOOLE_1:1;
  then reconsider S as Relation of X by A2,RELSET_1:4;
  assume
A3: R partially_orders X;
A4: R |_2 X is_antisymmetric_in X
  proof
A5: R is_antisymmetric_in X by A3,Def7;
    let x,y;
    assume that
A6: x in X and
A7: y in X and
A8: [x,y] in R |_2 X and
A9: [y,x] in R |_2 X;
A10: [y,x] in R by A9,XBOOLE_0:def 4;
    [x,y] in R by A8,XBOOLE_0:def 4;
    hence thesis by A6,A7,A10,A5,RELAT_2:def 4;
  end;
A11: R |_2 X is_transitive_in X
  proof
A12: R is_transitive_in X by A3,Def7;
    let x,y,z;
    assume that
A13: x in X and
A14: y in X and
A15: z in X and
A16: [x,y] in R |_2 X and
A17: [y,z] in R |_2 X;
A18: [x,z] in [:X,X:] by A13,A15,ZFMISC_1:87;
A19: [y,z] in R by A17,XBOOLE_0:def 4;
    [x,y] in R by A16,XBOOLE_0:def 4;
    then [x,z] in R by A13,A14,A15,A19,A12,RELAT_2:def 8;
    hence thesis by A18,XBOOLE_0:def 4;
  end;
A20: R is_reflexive_in X by A3,Def7;
A21: R |_2 X is_reflexive_in X
  proof
    let x;
    assume
A22: x in X;
    then
A23: [x,x] in [:X,X:] by ZFMISC_1:87;
    [x,x] in R by A20,A22,RELAT_2:def 1;
    hence thesis by A23,XBOOLE_0:def 4;
  end;
  then
A24: field S = X by Th13;
  dom S = X by A21,Th13;
  hence thesis by A21,A24,A4,A11,PARTFUN1:def 2,RELAT_2:def 9,def 12,def 16;
end;

theorem
  R linearly_orders X implies R |_2 X is Order of X by Th29,Th45;

theorem
  R well_orders X implies R |_2 X is Order of X by Th28,Th45;

theorem
  id X quasi_orders X & id X partially_orders X
proof
    field id X = X \/ rng id X
    .= X \/ X
    .= X;
 hence thesis by Th31,Th34;
end;

definition
  let R,X;
  pred X has_upper_Zorn_property_wrt R means
  :Def9:
  for Y st Y c= X & R|_2 Y
  is being_linear-order ex x st x in X & for y st y in Y holds [y,x] in R;
  pred X has_lower_Zorn_property_wrt R means
  for Y st Y c= X & R|_2 Y is
  being_linear-order ex x st x in X & for y st y in Y holds [x,y] in R;
end;

Lm15: (R|_2 X)~ = R~|_2 X
proof
  now
    let x,y;
    thus [x,y] in R~|_2 X implies [y,x] in R|_2 X
    proof
      assume
A1:   [x,y] in R~|_2 X;
      then [x,y] in [:X,X:] by XBOOLE_0:def 4;
      then
A2:   [y,x] in [:X,X:] by ZFMISC_1:88;
      [x,y] in R~ by A1,XBOOLE_0:def 4;
      then [y,x] in R by RELAT_1:def 7;
      hence thesis by A2,XBOOLE_0:def 4;
    end;
    assume
A3: [y,x] in R|_2 X;
    then [y,x] in [:X,X:] by XBOOLE_0:def 4;
    then
A4: [x,y] in [:X,X:] by ZFMISC_1:88;
    [y,x] in R by A3,XBOOLE_0:def 4;
    then [x,y] in R~ by RELAT_1:def 7;
    hence [x,y] in R~|_2 X by A4,XBOOLE_0:def 4;
  end;
  hence thesis by RELAT_1:def 7;
end;

Lm16: now
  let R;
  thus R|_2 {} = {}|`R|{} by WELLORD1:10
    .= {};
end;

theorem Th49:
  X has_upper_Zorn_property_wrt R implies X <> {}
proof
  assume
A1: for Y st Y c= X & R|_2 Y is being_linear-order ex x st x in X & for
  y st y in Y holds [y,x] in R;
  R|_2 {} is being_linear-order by Lm16;
  then ex x st x in X & for y st y in {} holds [y,x] in R by A1,XBOOLE_1:2;
  hence thesis;
end;

theorem
  X has_lower_Zorn_property_wrt R implies X <> {}
proof
  assume
A1: for Y st Y c= X & R|_2 Y is being_linear-order ex x st x in X & for
  y st y in Y holds [x,y] in R;
  R|_2 {} is being_linear-order by Lm16;
  then ex x st x in X & for y st y in {} holds [x,y] in R by A1,XBOOLE_1:2;
  hence thesis;
end;

theorem Th51:
  X has_upper_Zorn_property_wrt R iff X has_lower_Zorn_property_wrt R~
proof
  thus X has_upper_Zorn_property_wrt R implies X has_lower_Zorn_property_wrt R
  ~
  proof
    assume
A1: for Y st Y c= X & R|_2 Y is being_linear-order ex x st x in X &
    for y st y in Y holds [y,x] in R;
    let Y;
A2: (R|_2 Y)~~ = R|_2 Y;
    assume that
A3: Y c= X and
A4: R~|_2 Y is being_linear-order;
    R~|_2 Y = (R|_2 Y)~ by Lm15;
    then consider x such that
A5: x in X and
A6: for y st y in Y holds [y,x] in R by A1,A2,A3,A4,Th18;
    take x;
    thus x in X by A5;
    let y;
    assume y in Y;
    then [y,x] in R by A6;
    hence thesis by RELAT_1:def 7;
  end;
  assume
A7: for Y st Y c= X & R~|_2 Y is being_linear-order ex x st x in X &
  for y st y in Y holds [x,y] in R~;
  let Y;
  assume that
A8: Y c= X and
A9: R|_2 Y is being_linear-order;
  R~|_2 Y = (R|_2 Y)~ by Lm15;
  then consider x such that
A10: x in X and
A11: for y st y in Y holds [x,y] in R~ by A7,A8,A9,Th18;
  take x;
  thus x in X by A10;
  let y;
  assume y in Y;
  then [x,y] in R~ by A11;
  hence thesis by RELAT_1:def 7;
end;

theorem
  X has_upper_Zorn_property_wrt R~ iff X has_lower_Zorn_property_wrt R
proof
  R~~ = R;
  hence thesis by Th51;
end;

definition
  let R,x;
  pred x is_maximal_in R means
  :Def11:
  x in field R & not ex y st y in field R & y <> x & [x,y] in R;
  pred x is_minimal_in R means
  :Def12:
  x in field R & not ex y st y in field R & y <> x & [y,x] in R;
  pred x is_superior_of R means
  :Def13:
  x in field R & for y st y in field R & y <> x holds [y,x] in R;
  pred x is_inferior_of R means
  :Def14:
  x in field R & for y st y in field R & y <> x holds [x,y] in R;
end;

theorem
  x is_inferior_of R & R is antisymmetric implies x is_minimal_in R
proof
  assume that
A1: x is_inferior_of R and
A2: R is antisymmetric;
A3: R is_antisymmetric_in field R by A2,RELAT_2:def 12;
  thus
A4: x in field R by A1,Def14;
  let y;
  assume that
A5: y in field R and
A6: y <> x and
A7: [y,x] in R;
  [x,y] in R by A1,A5,A6,Def14;
  hence thesis by A4,A5,A6,A7,A3,RELAT_2:def 4;
end;

theorem
  x is_superior_of R & R is antisymmetric implies x is_maximal_in R
proof
  assume that
A1: x is_superior_of R and
A2: R is antisymmetric;
A3: R is_antisymmetric_in field R by A2,RELAT_2:def 12;
  thus
A4: x in field R by A1,Def13;
  let y;
  assume that
A5: y in field R and
A6: y <> x and
A7: [x,y] in R;
  [y,x] in R by A1,A5,A6,Def13;
  hence thesis by A4,A5,A6,A7,A3,RELAT_2:def 4;
end;

theorem
  x is_minimal_in R & R is connected implies x is_inferior_of R
proof
  assume that
A1: x is_minimal_in R and
A2: R is connected;
  thus
A3: x in field R by A1,Def12;
  let y;
  assume that
A4: y in field R and
A5: y <> x;
  R is_connected_in field R by A2,RELAT_2:def 14;
  then [x,y] in R or [y,x] in R by A3,A4,A5,RELAT_2:def 6;
  hence thesis by A1,A4,Def12;
end;

theorem
  x is_maximal_in R & R is connected implies x is_superior_of R
proof
  assume that
A1: x is_maximal_in R and
A2: R is connected;
  thus
A3: x in field R by A1,Def11;
  let y;
  assume that
A4: y in field R and
A5: y <> x;
  R is_connected_in field R by A2,RELAT_2:def 14;
  then [x,y] in R or [y,x] in R by A3,A4,A5,RELAT_2:def 6;
  hence thesis by A1,A4,Def11;
end;

theorem
  x in X & x is_superior_of R & X c= field R & R is reflexive implies X
  has_upper_Zorn_property_wrt R
proof
  assume that
A1: x in X and
A2: x is_superior_of R and
A3: X c= field R and
A4: R is_reflexive_in field R;
  let Y such that
A5: Y c= X and
  R|_2 Y is being_linear-order;
  take x;
  thus x in X by A1;
  let y;
  assume y in Y;
  then
A6: y in X by A5;
  y = x or y <> x;
  hence thesis by A2,A3,A4,A6,Def13,RELAT_2:def 1;
end;

theorem
  x in X & x is_inferior_of R & X c= field R & R is reflexive implies X
  has_lower_Zorn_property_wrt R
proof
  assume that
A1: x in X and
A2: x is_inferior_of R and
A3: X c= field R and
A4: R is_reflexive_in field R;
  let Y such that
A5: Y c= X and
  R|_2 Y is being_linear-order;
  take x;
  thus x in X by A1;
  let y;
  assume y in Y;
  then
A6: y in X by A5;
  y = x or y <> x;
  hence thesis by A2,A3,A4,A6,Def14,RELAT_2:def 1;
end;

theorem Th59:
  x is_minimal_in R iff x is_maximal_in R~
proof
A1: field R = field(R~) by RELAT_1:21;
  thus x is_minimal_in R implies x is_maximal_in R~
  proof
    assume that
A2: x in field R and
A3: not ex y st y in field R & y <> x & [y,x] in R;
    thus x in field(R~) by A2,RELAT_1:21;
    let y;
    assume that
A4: y in field(R~) and
A5: y <> x;
    not [y,x] in R by A1,A3,A4,A5;
    hence thesis by RELAT_1:def 7;
  end;
  assume that
A6: x in field(R~) and
A7: not ex y st y in field(R~) & y <> x & [x,y] in R~;
  thus x in field R by A6,RELAT_1:21;
  let y;
  assume that
A8: y in field R and
A9: y <> x;
  not [x,y] in R~ by A1,A7,A8,A9;
  hence thesis by RELAT_1:def 7;
end;

theorem
  x is_minimal_in R~ iff x is_maximal_in R
proof
A1: field R = field(R~) by RELAT_1:21;
  thus x is_minimal_in R~ implies x is_maximal_in R
  proof
    assume that
A2: x in field(R~) and
A3: not ex y st y in field(R~) & y <> x & [y,x] in R~;
    thus x in field R by A2,RELAT_1:21;
    let y;
    assume that
A4: y in field R and
A5: y <> x;
    not [y,x] in R~ by A1,A3,A4,A5;
    hence thesis by RELAT_1:def 7;
  end;
  assume that
A6: x in field R and
A7: not ex y st y in field R & y <> x & [x,y] in R;
  thus x in field(R~) by A6,RELAT_1:21;
  let y;
  assume that
A8: y in field(R~) and
A9: y <> x;
  not [x,y] in R by A1,A7,A8,A9;
  hence thesis by RELAT_1:def 7;
end;

theorem
  x is_inferior_of R iff x is_superior_of R~
proof
A1: field R = field(R~) by RELAT_1:21;
  thus x is_inferior_of R implies x is_superior_of R~
  proof
    assume that
A2: x in field R and
A3: for y st y in field R & y <> x holds [x,y] in R;
    thus x in field(R~) by A2,RELAT_1:21;
    let y;
    assume that
A4: y in field(R~) and
A5: y <> x;
    [x,y] in R by A1,A3,A4,A5;
    hence thesis by RELAT_1:def 7;
  end;
  assume that
A6: x in field(R~) and
A7: for y st y in field(R~) & y <> x holds [y,x] in R~;
  thus x in field R by A6,RELAT_1:21;
  let y;
  assume that
A8: y in field R and
A9: y <> x;
  [y,x] in R~ by A1,A7,A8,A9;
  hence thesis by RELAT_1:def 7;
end;

theorem
  x is_inferior_of R~ iff x is_superior_of R
proof
A1: field R = field(R~) by RELAT_1:21;
  thus x is_inferior_of R~ implies x is_superior_of R
  proof
    assume that
A2: x in field(R~) and
A3: for y st y in field(R~) & y <> x holds [x,y] in R~;
    thus x in field R by A2,RELAT_1:21;
    let y;
    assume that
A4: y in field R and
A5: y <> x;
    [x,y] in R~ by A1,A3,A4,A5;
    hence thesis by RELAT_1:def 7;
  end;
  assume that
A6: x in field R and
A7: for y st y in field R & y <> x holds [y,x] in R;
  thus x in field(R~) by A6,RELAT_1:21;
  let y;
  assume that
A8: y in field(R~) and
A9: y <> x;
  [y,x] in R by A1,A7,A8,A9;
  hence thesis by RELAT_1:def 7;
end;

Lm17: R well_orders X & Y c= X implies R well_orders Y
proof
  assume that
A1: R well_orders X and
A2: Y c= X;
A3: R is_transitive_in X by A1,WELLORD1:def 5;
A4: R is_connected_in X by A1,WELLORD1:def 5;
A5: R is_antisymmetric_in X by A1,WELLORD1:def 5;
A6: R is_well_founded_in X by A1,WELLORD1:def 5;
  R is_reflexive_in X by A1,WELLORD1:def 5;
  hence R is_reflexive_in Y & R is_transitive_in Y & R is_antisymmetric_in Y &
  R is_connected_in Y by A2,A3,A5,A4,Lm10,Th8,Th9,Th10;
  let Z;
  assume that
A7: Z c= Y and
A8: Z <> {};
  Z c= X by A2,A7,XBOOLE_1:1;
  hence thesis by A8,A6,WELLORD1:def 3;
end;





reserve A,C for Ordinal;

theorem Th63:
  for R,X st R partially_orders X & field R = X & X
  has_upper_Zorn_property_wrt R ex x st x is_maximal_in R
proof
  let R;
  let FIELD be set;
  assume that
A1: R is_reflexive_in FIELD and
A2: R is_transitive_in FIELD and
A3: R is_antisymmetric_in FIELD and
A4: field R = FIELD and
A5: FIELD has_upper_Zorn_property_wrt R;
  reconsider XD = FIELD as non empty set by A5,Th49;
  consider D such that
A6: D = XD;
A7: D in bool D by ZFMISC_1:def 1;
  not D in {{}} by TARSKI:def 1;
  then reconsider M = (bool D) \ {{}} as non empty set by A7,XBOOLE_0:def 5;
  set f = the Choice_Function of M;
  defpred P[set,set] means $1 <> {} & $2 = f.$1 or $1 = {} & $2 = D;
A8: for x st x in bool D ex y st P[x,y]
  proof
    let x such that
    x in bool D;
    x = {} implies thesis;
    hence thesis;
  end;
A9: for x,y,z st x in bool D & P[x,y] & P[x,z] holds y = z;
  consider g such that
A10: dom g = bool D & for x st x in bool D holds P[x,g.x] from FUNCT_1:
  sch 2 (A9,A8);
  defpred ON[Ordinal,set] means ex L st $2 = g.{ d : for x st x in rng L holds
x <> d & [x,d] in R } & dom L = $1 & for A,L1 st A in $1 & L1 = L|A holds L.A =
  g.{ d1 : for x st x in rng L1 holds x <> d1 & [x,d1] in R };
A11: ON[A,x] & ON[A,y] implies x = y
  proof
    given L1 such that
A12: x = g.{ d1 : for x st x in rng L1 holds x <> d1 & [x,d1] in R } and
A13: dom L1 = A & for C,L st C in A & L = L1|C holds L1.C = g.{ d2 :
    for x st x in rng L holds x <> d2 & [x,d2] in R };
A14: L1 = L1|A by A13,RELAT_1:68;
    given L2 such that
A15: y = g.{ d1 : for x st x in rng L2 holds x <> d1 & [x,d1] in R } and
A16: dom L2 = A & for C,L st C in A & L = L2|C holds L2.C = g.{ d2 :
    for x st x in rng L holds x <> d2 & [x,d2] in R };
    defpred P[Ordinal] means $1 c= A implies L1|$1 = L2|$1;
A17: for A1 st for A2 st A2 in A1 holds P[A2] holds P[A1]
    proof
      let A1 such that
A18:  for A2 st A2 in A1 holds A2 c= A implies L1|A2 = L2|A2 and
A19:  A1 c= A;
A20:  dom(L2|A1) = A1 by A16,A19,RELAT_1:62;
A21:  now
        let x;
        assume
A22:    x in A1;
        then reconsider A3 = x as Ordinal;
A23:    L1.x = L1|A1.x by A22,FUNCT_1:49;
        A3 c= A by A19,A22,ORDINAL1:def 2;
        then
A24:    L1|A3 = L2|A3 by A18,A22;
A25:    L2.A3 = g.{ d2 : for x st x in rng(L2|A3) holds x <> d2 & [x,d2]
        in R } by A16,A19,A22;
        L1.A3 = g.{ d1 : for x st x in rng(L1|A3) holds x <> d1 & [x,d1]
        in R } by A13,A19,A22;
        hence L1|A1.x = L2|A1.x by A22,A24,A23,A25,FUNCT_1:49;
      end;
      dom(L1|A1) = A1 by A13,A19,RELAT_1:62;
      hence thesis by A20,A21,FUNCT_1:2;
    end;
    for A1 holds P[A1] from ORDINAL1:sch 2(A17);
    then
A26: L1|A = L2|A;
    L2 = L2|A by A16,RELAT_1:68;
    hence thesis by A12,A15,A26,A14;
  end;
  {} in {{}} by TARSKI:def 1;
  then
A27: not {} in M by XBOOLE_0:def 5;
A28: X in bool D & X <> {} implies g.X in X
  proof
    assume that
A29: X in bool D and
A30: X <> {};
    not X in {{}} by A30,TARSKI:def 1;
    then
A31: X in M by A29,XBOOLE_0:def 5;
    f.X = g.X by A10,A29,A30;
    hence thesis by A27,A31,Def1;
  end;
  defpred OM[Ordinal] means ON[$1,D];
  {} c= D by XBOOLE_1:2;
  then
A32: g.{} = D by A10;
A33: now
    let A,B,L1,L2;
    assume that
A34: dom L1 = A & for C,L st C in A & L = L1|C holds L1.C = g.{ d2 :
    for x st x in rng L holds x <> d2 & [x,d2] in R } and
A35: dom L2 = B & for C,L st C in B & L = L2|C holds L2.C = g.{ d2 :
    for x st x in rng L holds x <> d2 & [x,d2] in R };
    let C such that
A36: C c= A and
A37: C c= B;
    defpred P[Ordinal] means $1 c= C implies L1|$1 = L2|$1;
A38: for A1 st for A2 st A2 in A1 holds P[A2] holds P[A1]
    proof
      let A1 such that
A39:  for A2 st A2 in A1 holds A2 c= C implies L1|A2 = L2|A2 and
A40:  A1 c= C;
A41:  dom(L2|A1) = A1 by A35,A37,A40,RELAT_1:62,XBOOLE_1:1;
A42:  now
        let x;
        assume
A43:    x in A1;
        then reconsider A3 = x as Ordinal;
A44:    L1.x = L1|A1.x by A43,FUNCT_1:49;
        A3 c= C by A40,A43,ORDINAL1:def 2;
        then
A45:    L1|A3 = L2|A3 by A39,A43;
A46:    A3 in C by A40,A43;
        then
A47:    L2.A3 = g.{ d2 : for x st x in rng(L2|A3) holds x <> d2 & [x,d2]
        in R } by A35,A37;
        L1.A3 = g.{ d1 : for x st x in rng(L1|A3) holds x <> d1 & [x,d1]
        in R } by A34,A36,A46;
        hence L1|A1.x = L2|A1.x by A43,A45,A44,A47,FUNCT_1:49;
      end;
      dom(L1|A1) = A1 by A34,A36,A40,RELAT_1:62,XBOOLE_1:1;
      hence thesis by A41,A42,FUNCT_1:2;
    end;
    for A1 holds P[A1] from ORDINAL1:sch 2(A38);
    hence L1|C = L2|C;
  end;
A48: for d,A,B st ON[A,d] & ON[B,d] holds A = B
  proof
    let d,A,B;
    given L1 such that
A49: d = g.{ d1 : for x st x in rng L1 holds x <> d1 & [x,d1] in R } and
A50: dom L1 = A & for C,L st C in A & L = L1|C holds L1.C = g.{ d2 :
    for x st x in rng L holds x <> d2 & [x,d2] in R };
    given L2 such that
A51: d = g.{ d1 : for x st x in rng L2 holds x <> d1 & [x,d1] in R } and
A52: dom L2 = B & for C,L st C in B & L = L2|C holds L2.C = g.{ d2 :
    for x st x in rng L holds x <> d2 & [x,d2] in R };
A53: now
      set Y = { d1 : for x st x in rng L1 holds x <> d1 & [x,d1] in R };
      set X = { d1 : for x st x in rng(L1|B) holds x <> d1 & [x,d1] in R };
A54:  Y c= D
      proof
        let x;
        assume x in Y;
        then ex d1 st x = d1 & for x st x in rng L1 holds x <> d1 & [x,d1] in
        R;
        hence thesis;
      end;
      assume
A55:  B in A;
      then B c= A by ORDINAL1:def 2;
      then
A56:  L1|B = L2|B by A33,A50,A52
        .= L2 by A52,RELAT_1:68;
      L1.B = g.X by A50,A55;
      then
A57:  d in rng L1 by A50,A51,A55,A56,FUNCT_1:def 3;
A58:  now
        assume
A59:    Y <> {};
        then not Y in {{}} by TARSKI:def 1;
        then
A60:    Y in M by A54,XBOOLE_0:def 5;
        g.Y = f.Y by A10,A54,A59;
        then d in Y by A27,A49,A60,Def1;
        then ex d1 st d = d1 & for x st x in rng L1 holds x <> d1 & [x,d1] in
        R;
        hence contradiction by A57;
      end;
A61:  not d in d;
      P[Y,g.Y] by A10,A54;
      hence contradiction by A49,A61,A58;
    end;
    now
      set Y = { d1 : for x st x in rng L2 holds x <> d1 & [x,d1] in R };
      set X = { d1 : for x st x in rng(L2|A) holds x <> d1 & [x,d1] in R };
A62:  Y c= D
      proof
        let x;
        assume x in Y;
        then ex d1 st x = d1 & for x st x in rng L2 holds x <> d1 & [x,d1] in
        R;
        hence thesis;
      end;
      assume
A63:  A in B;
      then A c= B by ORDINAL1:def 2;
      then
A64:  L2|A = L1|A by A33,A50,A52
        .= L1 by A50,RELAT_1:68;
      L2.A = g.X by A52,A63;
      then
A65:  d in rng L2 by A49,A52,A63,A64,FUNCT_1:def 3;
A66:  now
        assume
A67:    Y <> {};
        then not Y in {{}} by TARSKI:def 1;
        then
A68:    Y in M by A62,XBOOLE_0:def 5;
        g.Y = f.Y by A10,A62,A67;
        then d in Y by A27,A51,A68,Def1;
        then ex d1 st d = d1 & for x st x in rng L2 holds x <> d1 & [x,d1] in
        R;
        hence contradiction by A65;
      end;
A69:  not d in d;
      P[Y,g.Y] by A10,A62;
      hence contradiction by A51,A69,A66;
    end;
    hence thesis by A53,ORDINAL1:14;
  end;
A70: ex A st OM[A]
  proof
    defpred P[set,set] means ex A st $2 = A & ON[A,$1];
    defpred OO[Ordinal] means ex d st ON[$1,d];
    assume
A71: for A holds not ON[A,D];
A72: for A st for B st B in A holds OO[B] holds OO[A]
    proof
      defpred P[set,set] means ex C st $1 = C & ON[C,$2];
      let A such that
A73:  for B st B in A ex d st ON[B,d];
A74:  for x st x in A ex y st P[x,y]
      proof
        let x;
        assume
A75:    x in A;
        then reconsider C = x as Ordinal;
        consider d such that
A76:    ON[C,d] by A73,A75;
        reconsider y = d as set;
        take y,C;
        thus thesis by A76;
      end;
A77:  for x,y,z st x in A & P[x,y] & P[x,z] holds y = z by A11;
      consider h such that
A78:  dom h = A & for x st x in A holds P[x,h.x] from FUNCT_1:sch 2(
      A77,A74 );
      reconsider h as T-Sequence by A78,ORDINAL1:def 7;
      set X = { d1 : for x st x in rng h holds x <> d1 & [x,d1] in R };
A79:  X c= D
      proof
        let x;
        assume x in X;
        then ex d1 st x = d1 & for x st x in rng h holds x <> d1 & [x,d1] in R;
        hence thesis;
      end;
A80:  ON[A,g.X]
      proof
        take h;
        thus g.X = g.{ d1 : for x st x in rng h holds x <> d1 & [x,d1] in R }
        & dom h = A by A78;
        let B,L;
        assume that
A81:    B in A and
A82:    L = h|B;
        ex C st B = C & ON[C,h.B] by A78,A81;
        then consider L1 such that
A83:    h.B = g.{d1: for x st x in rng L1 holds x <> d1 & [x,d1] in R } and
A84:    dom L1 = B & for C,L st C in B & L = L1|C holds L1.C = g.{
d1:     for x st x in rng L holds x <> d1 & [x,d1] in R };
A85:    for x st x in B holds L1.x = h|B.x
        proof
          let x;
          assume
A86:      x in B;
          then reconsider A1 = x as Ordinal;
A87:      h|B.x = h.x by A86,FUNCT_1:49;
A88:      ON[A1,L1.x]
          proof
            reconsider K = L1|A1 as T-Sequence;
            take K;
            thus L1.x = g.{d1: for x st x in rng K holds x <> d1 & [x,d1] in R
            } by A84,A86;
            A1 c= B by A86,ORDINAL1:def 2;
            hence dom K = A1 by A84,RELAT_1:62;
            let A2,L2;
            assume that
A89:        A2 in A1 and
A90:        L2 = K|A2;
            A2 c= A1 by A89,ORDINAL1:def 2;
            then
A91:        L2 = L1|A2 by A90,FUNCT_1:51;
            K.A2 = L1.A2 by A89,FUNCT_1:49;
            hence thesis by A84,A86,A89,A91,ORDINAL1:10;
          end;
          ex A2 st x = A2 & ON[A2,h.x] by A78,A81,A86,ORDINAL1:10;
          hence thesis by A11,A88,A87;
        end;
        B c= A by A81,ORDINAL1:def 2;
        then dom(h|B) = B by A78,RELAT_1:62;
        then h|B = L1 by A84,A85,FUNCT_1:2;
        hence thesis by A82,A83;
      end;
      then X <> {} by A32,A71;
      then g.X in X by A28,A79;
      then reconsider d = g.X as Element of D by A79;
      take d;
      thus thesis by A80;
    end;
A92: for A holds OO[A] from ORDINAL1:sch 2(A72);
A93: for x,y,z st P[x,y] & P[x,z] holds y = z
    proof
      let x,y,z;
      given A1 such that
A94:  y = A1 and
A95:  ON[A1,x];
      consider d1 such that
A96:  ON[A1,d1] by A92;
      given A2 such that
A97:  z = A2 and
A98:  ON[A2,x];
      d1 = x by A11,A95,A96;
      hence thesis by A48,A94,A95,A97,A98;
    end;
    consider X such that
A99: x in X iff ex y st y in D & P[y,x] from TARSKI:sch 1(A93);
    for A holds A in X
    proof
      let A;
      ex d st ON[A,d] by A92;
      hence thesis by A99;
    end;
    hence contradiction by ORDINAL1:26;
  end;
  consider A such that
A100: OM[A] & for B st OM[B] holds A c= B from ORDINAL1:sch 1(A70);
A101: for L holds { d : for x st x in rng L holds x <> d & [x,d] in R } c= D
  proof
    let L,x;
    assume x in { d : for x st x in rng L holds x <> d & [x,d] in R };
    then ex d1 st x = d1 & for x st x in rng L holds x <> d1 & [x,d1] in R;
    hence thesis;
  end;
  D in bool D by ZFMISC_1:def 1;
  then reconsider d = g.D as Element of D by A28;
A102: d in D & ON[{},d]
  proof
    deffunc H(set) = {};
    thus d in D;
    consider L such that
A103: dom L = {} & for B,L1 st B in {} & L1 = L|B holds L.B = H(L1)
    from ORDINAL1:sch 4;
    take L;
A104: D c= { d1 : for x st x in rng L holds x <> d1 & [x,d1] in R }
    proof
      let x;
      assume x in D;
      then reconsider d = x as Element of D;
      for x st x in rng L holds x <> d & [x,d] in R by A103,RELAT_1:42;
      hence thesis;
    end;
    { d1 : for x st x in rng L holds x <> d1 & [x,d1] in R } c= D by A101;
    hence d = g.{ d1 : for x st x in rng L holds x <> d1 & [x,d1] in R } by
A104,XBOOLE_0:def 10;
    thus dom L = {} by A103;
    thus thesis;
  end;
A105: {} c= A;
  defpred P[set] means ex B st B in A & ON[B,$1];
  consider X such that
A106: x in X iff x in D & P[x] from XBOOLE_0:sch 1;
  not Y in Y;
  then d <> D;
  then {} <> A by A11,A100,A102;
  then {} c< A by A105,XBOOLE_0:def 8;
  then {} in A by ORDINAL1:11;
  then reconsider X as non empty set by A106,A102;
  consider L such that
A107: D = g.{ d1 : for x st x in rng L holds x <> d1 & [x,d1] in R } and
A108: dom L = A & for B,L1 st B in A & L1 = L|B holds L.B = g.{ d2 : for
  x st x in rng L1 holds x <> d2 & [x,d2] in R } by A100;
  R is transitive by A2,A4,RELAT_2:def 16;
  then
A109: R|_2 X is transitive by WELLORD1:17;
A110: rng L c= X
  proof
    let z;
    assume z in rng L;
    then consider y such that
A111: y in dom L and
A112: z = L.y by FUNCT_1:def 3;
    reconsider y as Ordinal by A111;
    set Z = { d2 : for x st x in rng(L|y) holds x <> d2 & [x,d2] in R };
A113: Z c= D by A101;
A114: ON[y,z]
    proof
      reconsider K = L|y as T-Sequence;
      take K;
      thus z = g.{ d2 : for x st x in rng K holds x <> d2 & [x,d2] in R } by
A108,A111,A112;
      y c= dom L by A111,ORDINAL1:def 2;
      hence
A115: dom K = y by RELAT_1:62;
      let B,L1;
      assume that
A116: B in y and
A117: L1 = K|B;
      B c= y by A116,ORDINAL1:def 2;
      then
A118: L1 = L|B by A117,FUNCT_1:51;
      K.B = L.B by A115,A116,FUNCT_1:47;
      hence thesis by A108,A111,A116,A118,ORDINAL1:10;
    end;
    now
      assume Z = {};
      then z = D by A32,A108,A111,A112;
      then dom L c= y by A100,A108,A114;
      hence contradiction by A111,ORDINAL1:5;
    end;
    then g.Z in Z by A28,A113;
    then z in Z by A108,A111,A112;
    hence thesis by A106,A108,A111,A114,A113;
  end;
A119: R|_2 X is connected
  proof
    let x,y;
    assume that
A120: x in field(R|_2 X) and
A121: y in field(R|_2 X);
A122: x in X by A120,WELLORD1:12;
    then consider A1 such that
    A1 in A and
A123: ON[A1,x] by A106;
A124: y in X by A121,WELLORD1:12;
    then consider A2 such that
    A2 in A and
A125: ON[A2,y] by A106;
    reconsider x9 = x, y9 = y as Element of D by A106,A122,A124;
    consider L1 such that
A126: x9 = g.{ d1 : for x st x in rng L1 holds x <> d1 & [x,d1] in R } and
A127: dom L1 = A1 & for C,L st C in A1 & L = L1|C holds L1.C = g.{ d2
    : for x st x in rng L holds x <> d2 & [x,d2] in R } by A123;
    consider L2 such that
A128: y9 = g.{ d1 : for x st x in rng L2 holds x <> d1 & [x,d1] in R } and
A129: dom L2 = A2 & for C,L st C in A2 & L = L2|C holds L2.C = g.{ d2
    : for x st x in rng L holds x <> d2 & [x,d2] in R } by A125;
A130: [x,y] in [:X,X:] by A122,A124,ZFMISC_1:87;
A131: now
      set Y = { d1 : for x st x in rng L2 holds x <> d1 & [x,d1] in R };
      set Z = { d1 : for x st x in rng(L2|A1) holds x <> d1 & [x,d1] in R };
      not y9 in y9;
      then
A132: Y <> {} by A32,A128;
      Y c= D by A101;
      then y9 in Y by A28,A128,A132;
      then
A133: ex d1 st y9 = d1 & for x st x in rng L2 holds x <> d1 & [x,d1] in R;
      assume
A134: A1 in A2;
      then A1 c= A2 by ORDINAL1:def 2;
      then
A135: L2|A1 = L1|A1 by A33,A127,A129
        .= L1 by A127,RELAT_1:68;
      L2.A1 = g.Z by A129,A134;
      then x9 in rng L2 by A126,A129,A134,A135,FUNCT_1:def 3;
      then [x,y] in R by A133;
      hence thesis by A130,XBOOLE_0:def 4;
    end;
A136: [y,x] in [:X,X:] by A122,A124,ZFMISC_1:87;
A137: now
      set Y = { d1 : for x st x in rng L1 holds x <> d1 & [x,d1] in R };
      set Z = { d1 : for x st x in rng(L1|A2) holds x <> d1 & [x,d1] in R };
      not d1 in d1;
      then
A138: Y <> {} by A32,A126;
      Y c= D by A101;
      then x9 in Y by A28,A126,A138;
      then
A139: ex d1 st x9 = d1 & for x st x in rng L1 holds x <> d1 & [x,d1] in R;
      assume
A140: A2 in A1;
      then A2 c= A1 by ORDINAL1:def 2;
      then
A141: L1|A2 = L2|A2 by A33,A127,A129
        .= L2 by A129,RELAT_1:68;
      L1.A2 = g.Z by A127,A140;
      then y9 in rng L1 by A127,A128,A140,A141,FUNCT_1:def 3;
      then [y,x] in R by A139;
      hence thesis by A136,XBOOLE_0:def 4;
    end;
    A1 = A2 implies thesis by A11,A123,A125;
    hence thesis by A131,A137,ORDINAL1:14;
  end;
  R is antisymmetric by A3,A4,RELAT_2:def 12;
  then
A142: R|_2 X is antisymmetric;
  set Z = { d1 : for x st x in rng L holds x <> d1 & [x,d1] in R };
A143: X c= D
  proof
    let x;
    thus thesis by A106;
  end;
  R is reflexive by A1,A4,RELAT_2:def 9;
  then R|_2 X is reflexive by WELLORD1:15;
  then R|_2 X is being_linear-order by A109,A142,A119,Def5;
  then consider x such that
A144: x in D and
A145: for y st y in X holds [y,x] in R by A5,A6,A143,Def9;
  take x;
  thus x in field R by A4,A6,A144;
  let y such that
A146: y in field R and
A147: y <> x and
A148: [x,y] in R;
  reconsider y9 = y as Element of D by A4,A6,A146;
A149: now
    assume y in X;
    then [y,x] in R by A145;
    hence contradiction by A3,A4,A6,A144,A146,A147,A148,RELAT_2:def 4;
  end;
  now
    let z;
    assume
A150: z in rng L;
    then reconsider z9 = z as Element of X by A110;
    thus z <> y9 by A110,A149,A150;
A151: [z9,x] in R by A145;
    z in D by A106,A110,A150;
    hence [z,y] in R by A2,A4,A6,A144,A146,A148,A151,RELAT_2:def 8;
  end;
  then
A152: y in Z;
  Z c= D by A101;
  hence contradiction by A28,A107,A152,ORDINAL1:5;
end;

theorem Th64:
  for R,X st R partially_orders X & field R = X & X
  has_lower_Zorn_property_wrt R ex x st x is_minimal_in R
proof
  let R,X such that
A1: R partially_orders X and
A2: field R = X and
A3: X has_lower_Zorn_property_wrt R;
  R = R~~;
  then
A4: X has_upper_Zorn_property_wrt R~ by A3,Th51;
  field(R~) = X by A2,RELAT_1:21;
  then consider x such that
A5: x is_maximal_in R~ by A1,A4,Th41,Th63;
  take x;
  thus thesis by A5,Th59;
end;



theorem Th65:
  for X st X <> {} & for Z st Z c= X & Z is c=-linear ex Y st Y
in X & for X1 st X1 in Z holds X1 c= Y ex Y st Y in X & for Z st Z in X & Z <>
  Y holds not Y c= Z
proof
  let X;
  assume that
A1: X <> {} and
A2: for Z st Z c= X & Z is c=-linear ex Y st Y in X & for X1 st X1 in Z
  holds X1 c= Y;
  reconsider D = X as non empty set by A1;
  set R = RelIncl D;
A3: D has_upper_Zorn_property_wrt R
  proof
    let Z;
    assume that
A4: Z c= D and
A5: R|_2 Z is being_linear-order;
    set Q = R|_2 Z;
A6: Z c= field(R|_2 Z)
    proof
      let x;
      assume
A7:   x in Z;
      then
A8:   [x,x] in [:Z,Z:] by ZFMISC_1:87;
      [x,x] in R by A4,A7,WELLORD2:def 1;
      then [x,x] in R|_2 Z by A8,XBOOLE_0:def 4;
      hence thesis by RELAT_1:15;
    end;
    R|_2 Z is connected by A5,Def5;
    then
A9: R|_2 Z is_connected_in field(R|_2 Z) by RELAT_2:def 14;
    Z is c=-linear
    proof
      let X1,X2;
      assume that
A10:  X1 in Z and
A11:  X2 in Z;
      X1 <> X2 implies [X1,X2] in Q or [X2,X1] in Q by A9,A6,A10,A11,
RELAT_2:def 6;
      then X1 <> X2 implies [X1,X2] in R or [X2,X1] in R by XBOOLE_0:def 4;
      hence X1 c= X2 or X2 c= X1 by A4,A10,A11,WELLORD2:def 1;
    end;
    then consider Y such that
A12: Y in X and
A13: for X1 st X1 in Z holds X1 c= Y by A2,A4;
    take x = Y;
    thus x in D by A12;
    let y;
    assume
A14: y in Z;
    then y c= Y by A13;
    hence thesis by A4,A12,A14,WELLORD2:def 1;
  end;
A15: field R = D by WELLORD2:def 1;
A16: R is_antisymmetric_in D by A15,RELAT_2:def 12;
A17: R is_transitive_in D by A15,RELAT_2:def 16;
  R is_reflexive_in D by A15,RELAT_2:def 9;
  then R partially_orders D by A17,A16,Def7;
  then consider x such that
A18: x is_maximal_in R by A15,A3,Th63;
  take Y = x;
A19: Y in field R by A18,Def11;
  thus Y in X by A15,A18,Def11;
  let Z;
  assume that
A20: Z in X and
A21: Z <> Y;
  not [Y,Z] in R by A15,A18,A20,A21,Def11;
  hence thesis by A15,A19,A20,WELLORD2:def 1;
end;

theorem Th66:
  for X st X <> {} & for Z st Z c= X & Z is c=-linear ex Y st Y
in X & for X1 st X1 in Z holds Y c= X1 ex Y st Y in X & for Z st Z in X & Z <>
  Y holds not Z c= Y
proof
  let X;
  assume that
A1: X <> {} and
A2: for Z st Z c= X & Z is c=-linear ex Y st Y in X & for X1 st X1 in Z
  holds Y c= X1;
  reconsider D = X as non empty set by A1;
  set R = (RelIncl D)~;
A3: x in D implies [x,x] in R
  proof
    x in D implies [x,x] in RelIncl D by WELLORD2:def 1;
    hence thesis by RELAT_1:def 7;
  end;
A4: D has_upper_Zorn_property_wrt R
  proof
    let Z;
    assume that
A5: Z c= D and
A6: R|_2 Z is being_linear-order;
    set Q = R|_2 Z;
A7: Z c= field(R|_2 Z)
    proof
      let x;
      assume
A8:   x in Z;
      then
A9:   [x,x] in [:Z,Z:] by ZFMISC_1:87;
      [x,x] in R by A3,A5,A8;
      then [x,x] in R|_2 Z by A9,XBOOLE_0:def 4;
      hence thesis by RELAT_1:15;
    end;
    R|_2 Z is connected by A6,Def5;
    then
A10: R|_2 Z is_connected_in field(R|_2 Z) by RELAT_2:def 14;
    Z is c=-linear
    proof
      let X1,X2;
      assume that
A11:  X1 in Z and
A12:  X2 in Z;
      X1 <> X2 implies [X1,X2] in Q or [X2,X1] in Q by A10,A7,A11,A12,
RELAT_2:def 6;
      then X1 <> X2 implies [X1,X2] in R or [X2,X1] in R by XBOOLE_0:def 4;
      then X1 <> X2 implies [X1,X2] in RelIncl D or [X2,X1] in RelIncl D by
RELAT_1:def 7;
      hence X1 c= X2 or X2 c= X1 by A5,A11,A12,WELLORD2:def 1;
    end;
    then consider Y such that
A13: Y in X and
A14: for X1 st X1 in Z holds Y c= X1 by A2,A5;
    take x = Y;
    thus x in D by A13;
    let y;
    assume
A15: y in Z;
    then Y c= y by A14;
    then [Y,y] in RelIncl D by A5,A13,A15,WELLORD2:def 1;
    hence thesis by RELAT_1:def 7;
  end;
A16: field R = field RelIncl D by RELAT_1:21
    .= D by WELLORD2:def 1;
A17: field RelIncl D = D by WELLORD2:def 1;
  RelIncl D is_antisymmetric_in D by A17,RELAT_2:def 12;
  then
A18: R is_antisymmetric_in D by Lm13;
  RelIncl D is_transitive_in D by A17,RELAT_2:def 16;
  then
A19: R is_transitive_in D by Lm12;
  RelIncl D is_reflexive_in D by A17,RELAT_2:def 9;
  then R is_reflexive_in D by Lm11;
  then R partially_orders D by A19,A18,Def7;
  then consider x such that
A20: x is_maximal_in R by A16,A4,Th63;
  take Y = x;
A21: Y in field R by A20,Def11;
  thus Y in X by A16,A20,Def11;
  let Z;
  assume that
A22: Z in X and
A23: Z <> Y;
  not [Y,Z] in R by A16,A20,A22,A23,Def11;
  then not [Z,Y] in RelIncl D by RELAT_1:def 7;
  hence thesis by A16,A21,A22,WELLORD2:def 1;
end;

theorem Th67:
  for X st X <> {} & for Z st Z <> {} & Z c= X & Z is c=-linear
  holds union Z in X ex Y st Y in X & for Z st Z in X & Z <> Y holds not Y c= Z
proof
  let X such that
A1: X <> {} and
A2: for Z st Z <> {} & Z c= X & Z is c=-linear holds union Z in X;
  for Z st Z c= X & Z is c=-linear ex Y st Y in X & for X1 st X1 in Z
  holds X1 c= Y
  proof
    set x = the Element of X;
    let Z such that
A3: Z c= X and
A4: Z is c=-linear;
    Z <> {} or Z = {};
    then consider Y such that
A5: Y = union Z & Z <> {} or Y = x & Z = {};
    take Y;
    thus thesis by A1,A2,A3,A4,A5,ZFMISC_1:74;
  end;
  hence thesis by A1,Th65;
end;

theorem
  for X st X <> {} & for Z st Z <> {} & Z c= X & Z is c=-linear holds
  meet Z in X ex Y st Y in X & for Z st Z in X & Z <> Y holds not Z c= Y
proof
  let X such that
A1: X <> {} and
A2: for Z st Z <> {} & Z c= X & Z is c=-linear holds meet Z in X;
  for Z st Z c= X & Z is c=-linear ex Y st Y in X & for X1 st X1 in Z
  holds Y c= X1
  proof
    set x = the Element of X;
    let Z such that
A3: Z c= X and
A4: Z is c=-linear;
    Z <> {} or Z = {};
    then consider Y such that
A5: Y = meet Z & Z <> {} or Y = x & Z = {};
    take Y;
    thus thesis by A1,A2,A3,A4,A5,SETFAM_1:3;
  end;
  hence thesis by A1,Th66;
end;

scheme
  ZornMax{A() -> non empty set, P[set,set]}: ex x being Element of A() st for
  y being Element of A() st x <> y holds not P[x,y]
provided
A1: for x being Element of A() holds P[x,x] and
A2: for x,y being Element of A() st P[x,y] & P[y,x] holds x = y and
A3: for x,y,z being Element of A() st P[x,y] & P[y,z] holds P[x,z] and
A4: for X st X c= A() & (for x,y being Element of A() st x in X & y in X
holds P[x,y] or P[y,x]) holds ex y being Element of A() st for x being Element
of A() st x in X holds P[x,y]
proof
  consider R being Relation of A() such that
A5: for x,y being Element of A() holds [x,y] in R iff P[x,y] from
  RELSET_1:sch 2;
A6: R is_transitive_in A()
  proof
    let x,y,z;
    assume that
A7: x in A() and
A8: y in A() and
A9: z in A();
    reconsider x9 = x, y9 = y, z9 = z as Element of A() by A7,A8,A9;
    assume that
A10: [x,y] in R and
A11: [y,z] in R;
A12: P[y9,z9] by A5,A11;
    P[x9,y9] by A5,A10;
    then P[x9,z9] by A3,A12;
    hence thesis by A5;
  end;
  A() c= dom R
  proof
    let x;
    assume x in A();
    then reconsider x9 = x as Element of A();
    P[x9,x9] by A1;
    then [x,x] in R by A5;
    hence thesis by XTUPLE_0:def 12;
  end;
  then dom R = A() by XBOOLE_0:def 10;
  then
A13: A() c= field R by XBOOLE_1:7;
  field R = dom R \/ rng R;
  then
A14: field R = A() by A13,XBOOLE_0:def 10;
A15: R is_reflexive_in A()
  proof
    let x;
    assume x in A();
    then reconsider x9 = x as Element of A();
    P[x9,x9] by A1;
    hence thesis by A5;
  end;
A16: A() has_upper_Zorn_property_wrt R
  proof
    let Y such that
A17: Y c= A() and
A18: R|_2 Y is being_linear-order;
    for x,y being Element of A() st x in Y & y in Y holds P[x,y] or P[y,x ]
    proof
      let x,y be Element of A() such that
A19:  x in Y and
A20:  y in Y;
A21:  R|_2 Y is reflexive connected by A18,Def5;
      Y c= field(R|_2 Y)
      proof
        let x;
        assume
A22:    x in Y;
        then
A23:    [x,x] in [:Y,Y:] by ZFMISC_1:87;
        [x,x] in R by A15,A17,A22,RELAT_2:def 1;
        then [x,x] in R|_2 Y by A23,XBOOLE_0:def 4;
        hence thesis by RELAT_1:15;
      end;
      then
A24:  Y = field(R|_2 Y|_2 Y) by A21,Lm5
        .= field(R|_2 Y) by WELLORD1:21;
      then
A25:  R|_2 Y is_connected_in Y by A21,RELAT_2:def 14;
A26:  R|_2 Y is_reflexive_in Y by A21,A24,RELAT_2:def 9;
      x = y or x <> y;
      then [x,y] in R|_2 Y or [y,x] in R|_2 Y by A19,A20,A25,A26,RELAT_2:def 1
,def 6;
      then [x,y] in R or [y,x] in R by XBOOLE_0:def 4;
      hence thesis by A5;
    end;
    then consider y being Element of A() such that
A27: for x being Element of A() st x in Y holds P[x,y] by A4,A17;
    take y;
    thus y in A();
    let x;
    assume
A28: x in Y;
    then P[x,y] by A17,A27;
    hence thesis by A5,A17,A28;
  end;
  R is_antisymmetric_in A()
  proof
    let x,y;
    assume that
A29: x in A() and
A30: y in A();
    reconsider x9 = x, y9 = y as Element of A() by A29,A30;
    assume that
A31: [x,y] in R and
A32: [y,x] in R;
A33: P[y9,x9] by A5,A32;
    P[x9,y9] by A5,A31;
    hence thesis by A2,A33;
  end;
  then R partially_orders A() by A15,A6,Def7;
  then consider x such that
A34: x is_maximal_in R by A14,A16,Th63;
  take x;
  thus x is Element of A() by A14,A34,Def11;
  let y being Element of A();
  assume x <> y;
  then
A35: not [x,y] in R by A14,A34,Def11;
  x in A() by A14,A34,Def11;
  hence thesis by A5,A35;
end;

scheme
  ZornMin{A() -> non empty set, P[set,set]}: ex x being Element of A() st for
  y being Element of A() st x <> y holds not P[y,x]
provided
A1: for x being Element of A() holds P[x,x] and
A2: for x,y being Element of A() st P[x,y] & P[y,x] holds x = y and
A3: for x,y,z being Element of A() st P[x,y] & P[y,z] holds P[x,z] and
A4: for X st X c= A() & (for x,y being Element of A() st x in X & y in X
holds P[x,y] or P[y,x]) holds ex y being Element of A() st for x being Element
of A() st x in X holds P[y,x]
proof
  consider R being Relation of A() such that
A5: for x,y being Element of A() holds [x,y] in R iff P[x,y] from
  RELSET_1:sch 2;
A6: R is_transitive_in A()
  proof
    let x,y,z;
    assume that
A7: x in A() and
A8: y in A() and
A9: z in A();
    reconsider x9 = x, y9 = y, z9 = z as Element of A() by A7,A8,A9;
    assume that
A10: [x,y] in R and
A11: [y,z] in R;
A12: P[y9,z9] by A5,A11;
    P[x9,y9] by A5,A10;
    then P[x9,z9] by A3,A12;
    hence thesis by A5;
  end;
  A() c= dom R
  proof
    let x;
    assume x in A();
    then reconsider x9 = x as Element of A();
    P[x9,x9] by A1;
    then [x,x] in R by A5;
    hence thesis by XTUPLE_0:def 12;
  end;
  then dom R = A() by XBOOLE_0:def 10;
  then
A13: A() c= field R by XBOOLE_1:7;
  field R = dom R \/ rng R;
  then
A14: field R = A() by A13,XBOOLE_0:def 10;
A15: R is_reflexive_in A()
  proof
    let x;
    assume x in A();
    then reconsider x9 = x as Element of A();
    P[x9,x9] by A1;
    hence thesis by A5;
  end;
A16: A() has_lower_Zorn_property_wrt R
  proof
    let Y such that
A17: Y c= A() and
A18: R|_2 Y is being_linear-order;
    for x,y being Element of A() st x in Y & y in Y holds P[x,y] or P[y,x ]
    proof
      let x,y be Element of A() such that
A19:  x in Y and
A20:  y in Y;
A21:  R|_2 Y is reflexive connected by A18,Def5;
      Y c= field(R|_2 Y)
      proof
        let x;
        assume
A22:    x in Y;
        then
A23:    [x,x] in [:Y,Y:] by ZFMISC_1:87;
        [x,x] in R by A15,A17,A22,RELAT_2:def 1;
        then [x,x] in R|_2 Y by A23,XBOOLE_0:def 4;
        hence thesis by RELAT_1:15;
      end;
      then
A24:  Y = field(R|_2 Y|_2 Y) by A21,Lm5
        .= field(R|_2 Y) by WELLORD1:21;
      then
A25:  R|_2 Y is_connected_in Y by A21,RELAT_2:def 14;
A26:  R|_2 Y is_reflexive_in Y by A21,A24,RELAT_2:def 9;
      x = y or x <> y;
      then [x,y] in R|_2 Y or [y,x] in R|_2 Y by A19,A20,A25,A26,RELAT_2:def 1
,def 6;
      then [x,y] in R or [y,x] in R by XBOOLE_0:def 4;
      hence thesis by A5;
    end;
    then consider y being Element of A() such that
A27: for x being Element of A() st x in Y holds P[y,x] by A4,A17;
    take y;
    thus y in A();
    let x;
    assume
A28: x in Y;
    then P[y,x] by A17,A27;
    hence thesis by A5,A17,A28;
  end;
  R is_antisymmetric_in A()
  proof
    let x,y;
    assume that
A29: x in A() and
A30: y in A();
    reconsider x9 = x, y9 = y as Element of A() by A29,A30;
    assume that
A31: [x,y] in R and
A32: [y,x] in R;
A33: P[y9,x9] by A5,A32;
    P[x9,y9] by A5,A31;
    hence thesis by A2,A33;
  end;
  then R partially_orders A() by A15,A6,Def7;
  then consider x such that
A34: x is_minimal_in R by A14,A16,Th64;
  take x;
  thus x is Element of A() by A14,A34,Def12;
  let y being Element of A();
  assume x <> y;
  then
A35: not [y,x] in R by A14,A34,Def12;
  x in A() by A14,A34,Def12;
  hence thesis by A5,A35;
end;





theorem
  R partially_orders X & field R = X implies ex P st R c= P & P
  linearly_orders X & field P = X
proof
  assume that
A1: R partially_orders X and
A2: field R = X;
  defpred P[set] means ex P st $1 = P & R c= P & P partially_orders X & field
  P = X;
  consider Rel being set such that
A3: x in Rel iff x in bool [:X,X:] & P[x] from XBOOLE_0:sch 1;
A4: for Z st Z <> {} & Z c= Rel & Z is c=-linear holds union Z in Rel
  proof
    reconsider T = [:X,X:] as Relation;
    let Z;
    assume that
A5: Z <> {} and
A6: Z c= Rel and
A7: Z is c=-linear;
A8: union bool [:X,X:] = T by ZFMISC_1:81;
    Z c= bool [:X,X:]
    proof
      let x;
      assume x in Z;
      hence thesis by A3,A6;
    end;
    then
A9: union Z c= union bool [:X,X:] by ZFMISC_1:77;
    then reconsider S = union Z as Relation;
    set y = the Element of Z;
    y in Rel by A5,A6,TARSKI:def 3;
    then consider P such that
A10: y = P and
A11: R c= P and
A12: P partially_orders X and
    field P = X by A3;
A13: S is_antisymmetric_in X
    proof
      let x,y;
      assume that
A14:  x in X and
A15:  y in X and
A16:  [x,y] in S and
A17:  [y,x] in S;
      consider X1 such that
A18:  [x,y] in X1 and
A19:  X1 in Z by A16,TARSKI:def 4;
      consider P1 being Relation such that
A20:  X1 = P1 and
      R c= P1 and
A21:  P1 partially_orders X and
      field P1 = X by A3,A6,A19;
      consider X2 such that
A22:  [y,x] in X2 and
A23:  X2 in Z by A17,TARSKI:def 4;
      X1,X2 are_c=-comparable by A7,A19,A23,ORDINAL1:def 8;
      then
A24:  X1 c= X2 or X2 c= X1 by XBOOLE_0:def 9;
      consider P2 being Relation such that
A25:  X2 = P2 and
      R c= P2 and
A26:  P2 partially_orders X and
      field P2 = X by A3,A6,A23;
A27:  P2 is_antisymmetric_in X by A26,Def7;
      P1 is_antisymmetric_in X by A21,Def7;
      hence thesis by A14,A15,A18,A22,A24,A20,A25,A27,RELAT_2:def 4;
    end;
A28: S is_transitive_in X
    proof
      let x,y,z;
      assume that
A29:  x in X and
A30:  y in X and
A31:  z in X and
A32:  [x,y] in S and
A33:  [y,z] in S;
      consider X1 such that
A34:  [x,y] in X1 and
A35:  X1 in Z by A32,TARSKI:def 4;
      consider P1 being Relation such that
A36:  X1 = P1 and
      R c= P1 and
A37:  P1 partially_orders X and
      field P1 = X by A3,A6,A35;
      consider X2 such that
A38:  [y,z] in X2 and
A39:  X2 in Z by A33,TARSKI:def 4;
      X1,X2 are_c=-comparable by A7,A35,A39,ORDINAL1:def 8;
      then
A40:  X1 c= X2 or X2 c= X1 by XBOOLE_0:def 9;
      consider P2 being Relation such that
A41:  X2 = P2 and
      R c= P2 and
A42:  P2 partially_orders X and
      field P2 = X by A3,A6,A39;
A43:  P2 is_transitive_in X by A42,Def7;
      P1 is_transitive_in X by A37,Def7;
      then [x,z] in P1 or [x,z] in P2 by A29,A30,A31,A34,A38,A40,A36,A41,A43,
RELAT_2:def 8;
      hence thesis by A35,A39,A36,A41,TARSKI:def 4;
    end;
A44: P is_reflexive_in X by A12,Def7;
    S is_reflexive_in X
    proof
      let x;
      assume x in X;
      then [x,x] in P by A44,RELAT_2:def 1;
      hence thesis by A5,A10,TARSKI:def 4;
    end;
    then
A45: S partially_orders X by A28,A13,Def7;
A46: field S c= X
    proof
      let x;
A47:  now
        assume x in dom S;
        then consider y such that
A48:    [x,y] in S by XTUPLE_0:def 12;
        consider Y such that
A49:    [x,y] in Y and
A50:    Y in Z by A48,TARSKI:def 4;
        ex P st Y = P & R c= P & P partially_orders X & field P = X by A3,A6
,A50;
        hence thesis by A49,RELAT_1:15;
      end;
A51:  now
        assume x in rng S;
        then consider y such that
A52:    [y,x] in S by XTUPLE_0:def 13;
        consider Y such that
A53:    [y,x] in Y and
A54:    Y in Z by A52,TARSKI:def 4;
        ex P st Y = P & R c= P & P partially_orders X & field P = X by A3,A6
,A54;
        hence thesis by A53,RELAT_1:15;
      end;
      assume x in field S;
      hence thesis by A47,A51,XBOOLE_0:def 3;
    end;
A55: R c= S
    proof
      let x,y;
      assume [x,y] in R;
      hence thesis by A5,A10,A11,TARSKI:def 4;
    end;
    then X c= field S by A2,RELAT_1:16;
    then field S = X by A46,XBOOLE_0:def 10;
    hence thesis by A3,A9,A8,A55,A45;
  end;
  R c= [:X,X:] by A2,Lm4;
  then Rel <> {} by A1,A2,A3;
  then consider Y such that
A56: Y in Rel and
A57: for Z st Z in Rel & Z <> Y holds not Y c= Z by A4,Th67;
  consider P such that
A58: Y = P and
A59: R c= P and
A60: P partially_orders X and
A61: field P = X by A3,A56;
  take P;
  thus R c= P by A59;
  thus
A62: P is_reflexive_in X & P is_transitive_in X & P is_antisymmetric_in
  X by A60,Def7;
  thus P is_connected_in X
  proof
    let x,y such that
A63: x in X and
A64: y in X and
    x <> y and
A65: not [x,y] in P and
A66: not [y,x] in P;
    defpred Q[set,set] means [$1,$2] in P or [$1,x] in P & [y,$2] in P;
    consider Q being Relation such that
A67: [z,u] in Q iff z in X & u in X & Q[z,u] from RELAT_1:sch 1;
A68: field Q c= X
    proof
      let z;
A69:  now
        assume z in dom Q;
        then ex u st [z,u] in Q by XTUPLE_0:def 12;
        hence thesis by A67;
      end;
A70:  now
        assume z in rng Q;
        then ex u st [u,z] in Q by XTUPLE_0:def 13;
        hence thesis by A67;
      end;
      assume z in field Q;
      hence thesis by A69,A70,XBOOLE_0:def 3;
    end;
A71: P c= Q
    proof
      let z,u;
      assume
A72:  [z,u] in P;
      then
A73:  u in field P by RELAT_1:15;
      z in field P by A72,RELAT_1:15;
      hence thesis by A61,A67,A72,A73;
    end;
    then X c= field Q by A61,RELAT_1:16;
    then
A74: field Q = X by A68,XBOOLE_0:def 10;
A75: Q is_transitive_in X
    proof
      let a,b,c be set;
      assume that
A76:  a in X and
A77:  b in X and
A78:  c in X and
A79:  [a,b] in Q and
A80:  [b,c] in Q;
A81:  [b,c] in P or [b,x] in P & [y,c] in P by A67,A80;
      [a,b] in P or [a,x] in P & [y,b] in P by A67,A79;
      then [a,c] in P or [a,x] in P & [y,c] in P or [a,x] in P & [y,c] in P
      or [y,x] in P by A62,A63,A64,A76,A77,A78,A81,RELAT_2:def 8;
      hence thesis by A66,A67,A76,A78;
    end;
A82: Q is_antisymmetric_in X
    proof
      let a,b be set;
      assume that
A83:  a in X and
A84:  b in X and
A85:  [a,b] in Q and
A86:  [b,a] in Q;
A87:  [b,a] in P or [b,x] in P & [y,a] in P by A67,A86;
      [a,b] in P or [a,x] in P & [y,b] in P by A67,A85;
      then a = b or [a,x] in P & [y,a] in P or [y,x] in P by A62,A63,A64,A83
,A84,A87,RELAT_2:def 4,def 8;
      hence thesis by A62,A63,A64,A66,A83,RELAT_2:def 8;
    end;
    Q is_reflexive_in X
    proof
      let z;
      assume
A88:  z in X;
      then [z,z] in P by A62,RELAT_2:def 1;
      hence thesis by A67,A88;
    end;
    then
A89: Q partially_orders X by A75,A82,Def7;
A90: Q c= [:X,X:]
    proof
      let y,z;
      assume
A91:  [y,z] in Q;
      then
A92:  z in X by A67;
      y in X by A67,A91;
      hence thesis by A92,ZFMISC_1:87;
    end;
    R c= Q by A59,A71,XBOOLE_1:1;
    then Q in Rel by A3,A90,A74,A89;
    then
A93: Q = P by A57,A58,A71;
A94: [y,y] in P by A62,A64,RELAT_2:def 1;
    [x,x] in P by A62,A63,RELAT_2:def 1;
    hence contradiction by A63,A64,A65,A67,A93,A94;
  end;
  thus thesis by A61;
end;





theorem
  R c= [:field R,field R:] by Lm4;

theorem
  R is reflexive & X c= field R implies field(R|_2 X) = X by Lm5;

theorem
  R is_reflexive_in X implies R|_2 X is reflexive by Lm6;

theorem
  R is_transitive_in X implies R|_2 X is transitive by Lm7;

theorem
  R is_antisymmetric_in X implies R|_2 X is antisymmetric by Lm8;

theorem
  R is_connected_in X implies R|_2 X is connected by Lm9;

theorem
  R is_connected_in X & Y c= X implies R is_connected_in Y by Lm10;

theorem
  R well_orders X & Y c= X implies R well_orders Y by Lm17;

theorem
  R is connected implies R~ is connected by Lm3;

theorem
  R is_reflexive_in X implies R~ is_reflexive_in X by Lm11;

theorem
  R is_transitive_in X implies R~ is_transitive_in X by Lm12;

theorem
  R is_antisymmetric_in X implies R~ is_antisymmetric_in X by Lm13;

theorem
  R is_connected_in X implies R~ is_connected_in X by Lm14;

theorem
  (R|_2 X)~ = R~|_2 X by Lm15;

theorem
  R|_2 {} = {} by Lm16;

begin 


theorem
  Z is finite & Z c= rng f implies ex Y st Y c= dom f & Y is finite & f .:Y = Z
proof
  assume that
A1: Z is finite and
A2: Z c= rng f;
  per cases;
  suppose
A3: Z = {};
    take Y = {};
    thus Y c= dom f & Y is finite by XBOOLE_1:2;
    thus thesis by A3;
  end;
  suppose
A4: Z <> {};
    deffunc F(set) = f"{$1};
    consider g being Function such that
A5: dom g = Z and
A6: for x st x in Z holds g.x = F(x) from FUNCT_1:sch 3;
    reconsider AA = rng g as non empty set by A4,A5,RELAT_1:42;
    set ff = the Choice_Function of AA;
    take Y = ff.:AA;
A7: for X being set st X in AA holds X <> {}
    proof
      let X be set;
      assume X in AA;
      then consider x such that
A8:   x in dom g and
A9:   g.x=X by FUNCT_1:def 3;
      g.x = f"{x} by A5,A6,A8;
      hence thesis by A2,A5,A8,A9,FUNCT_1:72;
    end;
    then
A10: not {} in AA;
    thus
A11: Y c= dom f
    proof
      let x;
      assume x in Y;
      then consider y such that
      y in dom ff and
A12:  y in AA and
A13:  ff.y=x by FUNCT_1:def 6;
      y in g.:Z by A5,A12,RELAT_1:113;
      then consider z such that
      z in dom g and
A14:  z in Z and
A15:  g.z=y by FUNCT_1:def 6;
A16:  g.z = f"{z} by A6,A14;
      x in g.z by A10,A12,A13,A15,Def1;
      hence thesis by A16,FUNCT_1:def 7;
    end;
    AA = g.:Z by A5,RELAT_1:113;
    hence Y is finite by A1;
    set z = the Element of AA;
    set y = the Element of z;
    z <> {} by A7;
    then y in z;
    then reconsider AA9=union AA as non empty set by TARSKI:def 4;
    reconsider f9= ff as Function of AA,AA9;
A17: dom f9 = AA by FUNCT_2:def 1;
    x in f.:Y iff x in Z
    proof
      thus x in f.:Y implies x in Z
      proof
        assume x in f.:(Y);
        then consider y such that
        y in dom f and
A18:    y in Y and
A19:    f.y = x by FUNCT_1:def 6;
        consider z such that
A20:    z in dom ff and
        z in AA and
A21:    ff.z=y by A18,FUNCT_1:def 6;
        consider a being set such that
A22:    a in dom g and
A23:    g.a=z by A20,FUNCT_1:def 3;
        g.a = f"{a} by A5,A6,A22;
        then y in f"{a} by A10,A20,A21,A23,Def1;
        then f.y in {a} by FUNCT_1:def 7;
        hence thesis by A5,A19,A22,TARSKI:def 1;
      end;
      set y=ff.(g.x);
      assume
A24:  x in Z;
      then
A25:  g.x in AA by A5,FUNCT_1:def 3;
      g.x = f"{x} by A6,A24;
      then y in f"{x} by A10,A25,Def1;
      then f.y in {x} by FUNCT_1:def 7;
      then
A26:  f.y = x by TARSKI:def 1;
      ff.(g.x) in rng ff by A17,A25,FUNCT_1:def 3;
      then y in Y by A17,RELAT_1:113;
      hence thesis by A11,A26,FUNCT_1:def 6;
    end;
    hence thesis by TARSKI:1;
  end;
end;



theorem Th86:
  field R is finite implies R is finite
proof
  assume field R is finite;
  then reconsider A = field R as finite set;
  R c= [:A,A:] by Lm4;
  hence thesis;
end;

theorem
  dom R is finite & rng R is finite implies R is finite
proof
  assume that
A1: dom R is finite and
A2: rng R is finite;
  field R is finite by A1,A2;
  hence thesis by Th86;
end;



registration
  cluster order_type_of {} -> empty;
  coherence
  proof
    {},RelIncl {} are_isomorphic by WELLORD1:38;
    hence thesis by WELLORD2:def 2;
  end;
end;

theorem
 for O being Ordinal holds order_type_of RelIncl O = O proof let O be Ordinal;
  RelIncl O is well-ordering & RelIncl O,RelIncl O are_isomorphic
  by WELLORD1:38;
  hence thesis by WELLORD2:def 2;
end;














begin 

reserve x,y,z,X,Y for set;

theorem Th1:
  {x} c= {x,y,z}
proof
  {x,y,z} = {x} \/ {y,z} by ENUMSET1:2;
  hence thesis by XBOOLE_1:7;
end;

theorem Th2:
  {x,y} c= {x,y,z}
proof
  {x,y,z} = {x,y} \/ {z} by ENUMSET1:3;
  hence thesis by XBOOLE_1:7;
end;

canceled 3;

theorem Th6:
  for X,Y for f being Function holds f.:(Y \ f"X) = f.:Y \ X
proof
  let X,Y;
  let f be Function;
  now
    let x;
    thus x in f.:(Y \ f"X) implies x in f.:Y \ X
    proof
      assume x in f.:(Y \ f"X);
      then consider z such that
A1:   z in dom f and
A2:   z in Y \ f"X and
A3:   f.z = x by FUNCT_1:def 6;
      not z in f"X by A2,XBOOLE_0:def 5;
      then
A4:   not x in X by A1,A3,FUNCT_1:def 7;
      f.z in f.:Y by A1,A2,FUNCT_1:def 6;
      hence thesis by A3,A4,XBOOLE_0:def 5;
    end;
    assume
A5: x in f.:Y \ X;
    then consider z such that
A6: z in dom f and
A7: z in Y and
A8: f.z = x by FUNCT_1:def 6;
    not x in X by A5,XBOOLE_0:def 5;
    then not z in f"X by A8,FUNCT_1:def 7;
    then z in Y \ f"X by A7,XBOOLE_0:def 5;
    hence x in f.:(Y \ f"X) by A6,A8,FUNCT_1:def 6;
  end;
  hence thesis by TARSKI:1;
end;

reserve X,Y for non empty set,
  f for Function of X,Y;

theorem Th7:
  for x being Element of X holds x in f"{f.x}
proof
  let x be Element of X;
  f.x in {f.x} by TARSKI:def 1;
  hence thesis by FUNCT_2:38;
end;

theorem Th8:
  for x being Element of X holds Im(f,x) = {f.x}
proof
  let x be Element of X;
  for y holds y in f.:{x} iff y = f.x
  proof
    let y;
    thus y in f.:{x} implies y = f.x
    proof
      assume y in f.:{x};
      then ex z st z in dom f & z in {x} & f.z = y by FUNCT_1:def 6;
      hence thesis by TARSKI:def 1;
    end;
    x in {x} by TARSKI:def 1;
    hence thesis by FUNCT_2:35;
  end;
  hence thesis by TARSKI:def 1;
end;



theorem Th9:
  for B being Element of Fin X for x st x in B holds x is Element of X
proof
  let B be Element of Fin X;
A1: B c= X by FINSUB_1:def 5;
  let x;
  assume x in B;
  hence thesis by A1;
end;

theorem
  for A being (Element of Fin X), B being set for f being Function of X,
  Y st for x being Element of X holds x in A implies f.x in B holds f.:A c= B
proof
  let A be (Element of Fin X), B be set;
  let f be Function of X,Y such that
A1: for x being Element of X holds x in A implies f.x in B;
  let x be set;
  assume x in f.:A;
  then consider y being set such that
  y in dom f and
A2: y in A and
A3: x = f.y by FUNCT_1:def 6;
  reconsider y as Element of X by A2,Th9;
  x = f.y by A3;
  hence thesis by A1,A2;
end;

theorem Th11:
  for X being set for B being Element of Fin X for A being set st
  A c= B holds A is Element of Fin X
proof
  let X be set, B be Element of Fin X;
  let A be set such that
A1: A c= B;
  B c= X by FINSUB_1:def 5;
  then A c= X by A1,XBOOLE_1:1;
  hence thesis by A1,FINSUB_1:def 5;
end;

Lm1: for A being Element of Fin X holds f.: A is Element of Fin Y by
FINSUB_1:def 5;

theorem Th12:
  for B being Element of Fin X st B <> {} ex x being Element of X st x in B
proof
  let B be Element of Fin X;
  set x = the Element of B;
  assume
A1: B <> {};
  then reconsider x as Element of X by Th9;
  take x;
  thus thesis by A1;
end;

theorem Th13:
  for A being Element of Fin X holds f.:A = {} implies A = {}
proof
  let A be Element of Fin X;
  assume
A1: f.:A = {};
  assume A <> {};
  then consider x being Element of X such that
A2: x in A by Th12;
  f.x in f.:A by A2,FUNCT_2:35;
  hence contradiction by A1;
end;

registration
  let X be set;
  cluster empty for Element of Fin X;
  existence
  proof
    {} is Element of Fin X by FINSUB_1:7;
    hence thesis;
  end;
end;

definition
  let X be set;
  func {}.X -> empty Element of Fin X equals
  {};
  coherence by FINSUB_1:7;
end;

scheme
  FinSubFuncEx{ A()->non empty set, B()->(Element of Fin A()), P[set,set] } :
ex f being Function of A(), Fin A() st for b,a being Element of A() holds a in
  f.b iff a in B() & P[a,b] proof
  defpred X[set,Element of Fin A()] means for a being Element of A() holds a
  in $2 iff a in B() & P[a,$1];
A1: now
    reconsider B = B() as Subset of A() by FINSUB_1:def 5;
    let x be Element of A();
    defpred X[set] means $1 in B() & P[$1,x];
    consider y being Subset of A() such that
A2: for a being Element of A() holds a in y iff X[a] from SUBSET_1:sch
    3;
    for x being Element of A() holds x in y implies x in B by A2;
    then y c= B() by SUBSET_1:2;
    then reconsider y as Element of Fin A() by FINSUB_1:def 5;
    take y9 = y;
    thus X[x,y9] by A2;
  end;
  thus ex f being Function of A(), Fin A() st for x being Element of A() holds
  X[x,f.x] from FUNCT_2:sch 3(A1);
end;



definition
  let X be non empty set, F be BinOp of X;
  attr F is having_a_unity means
  :Def2:
  ex x being Element of X st x is_a_unity_wrt F;
end;

theorem Th14:
  for X being non empty set, F being BinOp of X holds F is
  having_a_unity iff the_unity_wrt F is_a_unity_wrt F
proof
  let X be non empty set, F be BinOp of X;
  thus F is having_a_unity implies the_unity_wrt F is_a_unity_wrt F
  proof
    assume F is having_a_unity;
    then ex x being Element of X st x is_a_unity_wrt F by Def2;
    hence thesis by BINOP_1:def 8;
  end;
  thus thesis by Def2;
end;

theorem Th15:
  for X being non empty set, F being BinOp of X st F is
having_a_unity for x being Element of X holds F.(the_unity_wrt F, x) = x & F.(x
  , the_unity_wrt F) = x
proof
  let X be non empty set, F be BinOp of X;
  assume F is having_a_unity;
  then
A1: the_unity_wrt F is_a_unity_wrt F by Th14;
  let x be Element of X;
  thus thesis by A1,BINOP_1:3;
end;



registration
  let X be non empty set;
  cluster non empty for Element of Fin X;
  existence
  proof
    set x = the Element of X;
    {x} is Subset of X by SUBSET_1:41;
    then {x} is Element of Fin X by FINSUB_1:def 5;
    hence thesis;
  end;
end;

notation
  let X be non empty set, x be Element of X;
  synonym {.x.} for {x};
  let y be Element of X;
  synonym {.x,y.} for {x,y};
  let z be Element of X;
  synonym {.x,y,z.} for {x,y,z};
end;

definition
  let X be non empty set, x be Element of X;
  redefine func {.x.} -> Element of Fin X;
  coherence
  proof
    {x} is Subset of X by SUBSET_1:41;
    hence thesis by FINSUB_1:def 5;
  end;
  let y be Element of X;
  redefine func {.x,y.} -> Element of Fin X;
  coherence
  proof
    {x,y} is Subset of X by SUBSET_1:34;
    hence thesis by FINSUB_1:def 5;
  end;
  let z be Element of X;
  redefine func {.x,y,z.} -> Element of Fin X;
  coherence
  proof
    {x,y,z} is Subset of X by SUBSET_1:35;
    hence thesis by FINSUB_1:def 5;
  end;
end;

definition
  let X be set;
  let A,B be Element of Fin X;
  redefine func A \/ B -> Element of Fin X;
  coherence by FINSUB_1:1;
end;

definition
  let X be set;
  let A,B be Element of Fin X;
  redefine func A \ B -> Element of Fin X;
  coherence by FINSUB_1:1;
end;

scheme
  FinSubInd1{ X() -> non empty set, P[set] } : for B being Element of Fin X()
  holds P[B]
provided
A1: P[{}.X()] and
A2: for B9 being (Element of Fin X()), b being Element of X() holds P[B9
] & not b in B9 implies P[B9 \/ {b}]
proof
  defpred X[set] means ex B9 being Element of Fin X() st B9 = $1 & P[B9];
  let B be Element of Fin X();
A3: for x,A being set st x in B & A c= B & X[A] holds X[A \/ {x}]
  proof
    let x,A be set such that
A4: x in B and
A5: A c= B and
A6: ex B9 being Element of Fin X() st B9 = A & P[B9];
    reconsider x9 = x as Element of X() by A4,Th9;
    reconsider A9 = A as Element of Fin X() by A5,Th11;
    take A9 \/ {.x9.};
    thus A9 \/ {.x9.} = A \/ {x};
    thus P[A9 \/ {x9}] by A2,A6,ZFMISC_1:40;
  end;
A7: B is finite;
A8: X[{}] by A1;
  X[B] from FINSET_1:sch 2(A7,A8,A3);
  hence thesis;
end;

scheme
  FinSubInd2{ X() -> non empty set, P[Element of Fin X()] } : for B being non
  empty Element of Fin X() holds P[B]
provided
A1: for x being Element of X() holds P[{.x.}] and
A2: for B1,B2 being non empty Element of Fin X() st P[B1] & P[B2] holds
P[B1 \/ B2]
proof
  defpred X[set] means $1 <> {} implies ex B9 being Element of Fin X() st B9 =
  $1 & P[B9];
  let B be non empty Element of Fin X();
A3: for x,A being set st x in B & A c= B & X[A] holds X[A \/ {x}]
  proof
    let x,A be set such that
A4: x in B and
A5: A c= B and
A6: A <> {} implies ex B9 being Element of Fin X() st B9 = A & P[B9] and
    A \/ {x} <> {};
    reconsider x9 = x as Element of X() by A4,Th9;
    reconsider A9 = A as Element of Fin X() by A5,Th11;
    take A9 \/ {.x9.};
    thus A9 \/ {.x9.} =A \/ {x};
A7: P[{.x9.}] by A1;
    per cases;
    suppose
      A = {};
      hence thesis by A1;
    end;
    suppose
      A <> {};
      hence thesis by A2,A6,A7;
    end;
  end;
A8: X[{}];
A9: B is finite;
  X[B] from FINSET_1:sch 2(A9,A8,A3);
  hence thesis;
end;

scheme
  FinSubInd3{ X() -> non empty set, P[set] } : for B being Element of Fin X()
  holds P[B]
provided
A1: P[{}.X()] and
A2: for B9 being (Element of Fin X()), b being Element of X() holds P[B9
] implies P[B9 \/ {b}]
proof
A3: for B9 being (Element of Fin X()), b being Element of X() holds P[B9] &
  not b in B9 implies P[B9 \/ {b}] by A2;
A4: P[{}.X()] by A1;
  thus for B being Element of Fin X() holds P[B] from FinSubInd1(A4, A3);
end;

definition
  let X, Y be non empty set;
  let F be BinOp of Y;
  let B be Element of Fin X;
  let f be Function of X,Y;
  assume that
A1: B <> {} or F is having_a_unity and
A2: F is commutative and
A3: F is associative;
  func F $$ (B,f) -> Element of Y means
  :Def3:
  ex G being Function of Fin X, Y
st it = G.B & (for e being Element of Y st e is_a_unity_wrt F holds G.{} = e) &
(for x being Element of X holds G.{x} = f.x) & for B9 being Element of Fin X st
B9 c= B & B9 <> {} for x being Element of X st x in B \ B9 holds G.(B9 \/ {x})
  = F.(G.B9,f.x);
  existence
  proof
    defpred X[set] means ex G being Function of Fin X, Y st (for e being
Element of Y st e is_a_unity_wrt F holds G.{} = e) & (for x being Element of X
holds G.{x} = f.x) & for B9 being Element of Fin X st B9 c= $1 & B9 <> {} for x
    being Element of X st x in $1 \ B9 holds G.(B9 \/ {x}) = F.(G.B9, f.x);
A4: for B9 being Element of Fin X, b being Element of X holds X[B9] & not
    b in B9 implies X[B9 \/ {b}]
    proof
      let B be (Element of Fin X), x be Element of X;
      given G being Function of Fin X, Y such that
A5:   for e being Element of Y st e is_a_unity_wrt F holds G.{} = e and
A6:   for x being Element of X holds G.{x} = f.x and
A7:   for B9 being Element of Fin X st B9 c= B & B9 <> {} for x being
      Element of X st x in B \ B9 holds G.(B9 \/ {x}) = F.(G.B9, f.x);
      assume
A8:   not x in B;
      thus ex G being Function of Fin X, Y st (for e being Element of Y st e
is_a_unity_wrt F holds G.{} = e) & (for x being Element of X holds G.{x} = f.x)
      & for B9 being Element of Fin X st B9 c= B \/ {x} & B9 <> {} for x9 being
Element of X st x9 in (B \/ {x}) \ B9 holds G.(B9 \/ {x9}) = F.(G.B9, f.x9)
      proof
        defpred X[set,set] means (for C being Element of Fin X st C <> {} & C
c= B & $1 = C \/ {.x.} holds $2 = F.(G.C, f.x)) & ((not ex C being Element of
        Fin X st C <> {} & C c= B & $1 = C \/ {.x.}) implies $2 = G.$1);
A9:     now
          let B9 be Element of Fin X;
          thus ex y being Element of Y st X[B9,y]
          proof
A10:        now
              given C being Element of Fin X such that
A11:          C <> {} and
A12:          C c= B and
A13:          B9 = C \/ {x};
              take y = F.(G.C, f.x);
              for C being Element of Fin X st C <> {} & C c= B & B9 = C
              \/ {x} holds y = F.(G.C, f.x)
              proof
                not x in C by A8,A12;
                then
A14:            C c= B9 \ {x} by A13,XBOOLE_1:7,ZFMISC_1:34;
                B9 \ {x} = C \ {x} by A13,XBOOLE_1:40;
                then
A15:            B9 \ {x} = C by A14,XBOOLE_0:def 10;
                let C1 be Element of Fin X such that
                C1 <> {} and
A16:            C1 c= B and
A17:            B9 = C1 \/ {x};
                not x in C1 by A8,A16;
                then
A18:            C1 c= B9 \ {x} by A17,XBOOLE_1:7,ZFMISC_1:34;
                B9 \{x} = C1 \ {x} by A17,XBOOLE_1:40;
                hence thesis by A18,A15,XBOOLE_0:def 10;
              end;
              hence thesis by A11,A12,A13;
            end;
            now
              assume
A19:          not ex C being Element of Fin X st C <> {} & C c= B &
              B9 = C \/ {x};
              take y = G.B9;
              thus for C being Element of Fin X st C <> {} & C c= B & B9 = C
              \/ {x} holds y = F.(G.C, f.x) by A19;
              thus (not ex C being Element of Fin X st C <> {} & C c= B & B9 =
              C \/ {x}) implies y = G.B9;
            end;
            hence thesis by A10;
          end;
        end;
        consider H being Function of Fin X, Y such that
A20:    for B9 being Element of Fin X holds X[B9,H.B9] from FUNCT_2:
        sch 3 (A9);
        take J = H;
        now
          given C being Element of Fin X such that
A21:      C <> {} and
A22:      C c= B and
A23:      {x} = C \/ {x};
          C = {x} by A21,ZFMISC_1:33,A23,XBOOLE_1:7;
          then x in C by TARSKI:def 1;
          hence contradiction by A8,A22;
        end;
        then
A24:    J.{x} = G.{.x.} by A20;
        thus for e being Element of Y st e is_a_unity_wrt F holds J.{} = e
        proof
          reconsider E = {} as Element of Fin X by FINSUB_1:7;
          not ex C being Element of Fin X st C <> {} & C c= B & E = C \/ { x};
          then J.E = G.E by A20;
          hence thesis by A5;
        end;
        thus for x being Element of X holds J.{x} = f.x
        proof
          let x9 be Element of X;
          now
            given C being Element of Fin X such that
A25:        C <> {} and
A26:        C c= B and
A27:        {x9} = C \/ {x};
A28:        C = {x9} by A25,ZFMISC_1:33,A27,XBOOLE_1:7;
            x = x9 by A27,XBOOLE_1:7,ZFMISC_1:18;
            then x in C by A28,TARSKI:def 1;
            hence contradiction by A8,A26;
          end;
          hence J.{x9} = G.{.x9.} by A20
            .= f.x9 by A6;
        end;
        let B9 be Element of Fin X such that
A29:    B9 c= B \/ {x} and
A30:    B9 <> {};
        let x9 be Element of X;
        assume
A31:    x9 in (B \/ {x}) \ B9;
        then
A32:    not x9 in B9 by XBOOLE_0:def 5;
        per cases;
        suppose
A33:      x in B9;
          then
A34:      x9 in B by A31,A32,ZFMISC_1:136;
          per cases;
          suppose
A36:        B9 = {x};
            hence J.(B9 \/ {.x9.}) = F.(G.{.x9.}, f.x)
                 by A20,ZFMISC_1:31,A34
              .= F.(f.x9, f.x) by A6
              .= F.(f.x, f.x9) by A2,BINOP_1:def 2
              .= F.(J.B9, f.x9) by A6,A24,A36;
          end;
          suppose
            B9 <> {x};
            then not B9 c= {x} by A30,ZFMISC_1:33;
            then
A37:        B9 \ {x} <> {} by XBOOLE_1:37;
            set C = (B9 \ {.x.}) \/ {.x9.};
            not x9 in B9 \ {x} by A31,XBOOLE_0:def 5;
            then
A38:        x9 in B \(B9 \ {x}) by A34,XBOOLE_0:def 5;
            B9 \ {x} c= B by A29,XBOOLE_1:43;
            then
A39:        C c= B by A33,ZFMISC_1:137,A31,A32,ZFMISC_1:136;
            B9 \/ {.x9.} = B9 \/ {x} \/ {x9} by A33,ZFMISC_1:40
              .= (B9 \ {x}) \/ {x} \/ {x9} by XBOOLE_1:39
              .= C \/ {.x.} by XBOOLE_1:4;
            hence J.(B9 \/ {.x9.}) = F.(G.C, f.x) by A20,A39
              .= F.(F.(G.(B9\{.x.}), f.x9), f.x) by A7,A29,A37,A38,XBOOLE_1:43
              .= F.(G.(B9\{.x.}),F.(f.x9, f.x)) by A3,BINOP_1:def 3
              .= F.(G.(B9\{x}),F.(f.x, f.x9)) by A2,BINOP_1:def 2
              .= F.(F.(G.(B9\{.x.}), f.x), f.x9) by A3,BINOP_1:def 3
              .= F.(J.((B9\{.x.})\/{.x.}), f.x9) by A20,A29,A37,XBOOLE_1:43
              .= F.(J.(B9 \/ {x}), f.x9) by XBOOLE_1:39
              .= F.(J.B9, f.x9) by A33,ZFMISC_1:40;
          end;
        end;
        suppose
A40:      not x in B9;
          then
A41:      B9 c= B by A29,ZFMISC_1:135;
A42:      not ex C being Element of Fin X st C <> {} & C c= B & B9 = C
          \/ {x} by A40,ZFMISC_1:136;
          per cases;
          suppose
A43:        x <> x9;
            then x9 in B by A31,ZFMISC_1:136;
            then
A44:        x9 in B \ B9 by A32,XBOOLE_0:def 5;
            not x in B9 \/ {x9} by A40,A43,ZFMISC_1:136;
            then not ex C being Element of Fin X st C <> {} & C c= B & B9 \/
            {x9} = C \/ {x} by ZFMISC_1:136;
            hence J.(B9 \/ {.x9.}) = G.(B9 \/ {.x9.}) by A20
              .= F.(G.B9, f.x9) by A7,A30,A41,A44
              .= F.(J.B9, f.x9) by A20,A42;
          end;
          suppose
            x = x9;
            hence J.(B9 \/ {.x9.}) = F.(G.B9, f.x9)
             by A20,A29,A30,A40,ZFMISC_1:135
              .= F.(J.B9, f.x9) by A20,A42;
          end;
        end;
      end;
    end;
A45: X[{}.X]
    proof
      consider e being Element of Y such that
A46:  (ex e being Element of Y st e is_a_unity_wrt F) implies e =
      the_unity_wrt F;
      defpred X[set,set] means (for x9 being Element of X st $1 = {x9} holds
$2 = f.x9) & (not (ex x9 being Element of X st $1 = {x9}) implies $2 = e);
A47:  now
        let x be Element of Fin X;
A48:    now
          given x9 being Element of X such that
A49:      x = {x9};
          take y = f.x9;
          thus for x9 being Element of X st x = {x9} holds y = f.x9 by A49,
ZFMISC_1:3;
          thus not (ex x9 being Element of X st x = {x9}) implies y = e by A49;
        end;
        now
          assume
A50:      not ex x9 being Element of X st x = {x9};
          take y = e;
          thus (for x9 being Element of X st x = {x9} holds y = f.x9) & (not (
          ex x9 being Element of X st x = {x9}) implies y = e) by A50;
        end;
        hence ex y being Element of Y st X[x,y] by A48;
      end;
      consider G9 being Function of Fin X, Y such that
A51:  for B9 being Element of Fin X holds X[B9,G9.B9] from FUNCT_2:
      sch 3(A47 );
      take G = G9;
      thus for e being Element of Y st e is_a_unity_wrt F holds G.{} = e
      proof
        reconsider E = {} as Element of Fin X by FINSUB_1:7;
        let e9 be Element of Y such that
A52:    e9 is_a_unity_wrt F;
        not ex x9 being Element of X st E = {x9};
        hence G.{} = e by A51
          .= e9 by A46,A52,BINOP_1:def 8;
      end;
      thus for x being Element of X holds G.{.x.} = f.x by A51;
      thus for B9 being Element of Fin X st B9 c= {}.X & B9 <> {} for x being
      Element of X st x in {}.X \ B9 holds G.(B9 \/ {x}) = F.(G.B9, f.x);
    end;
    for B being Element of Fin X holds X[B] from FinSubInd1(A45,A4);
    then consider G being Function of Fin X, Y such that
A53: ( ( for e being Element of Y st e is_a_unity_wrt F holds G.{} =
e)& for x being Element of X holds G.{x} = f.x )& for B9 being Element of Fin X
st B9 c= B & B9 <> {} for x being Element of X st x in B \ B9 holds G.(B9 \/ {x
    }) = F.(G .B9, f.x);
    take G.B, G;
    thus thesis by A53;
  end;
  uniqueness
  proof
    let x,y be Element of Y;
    given G1 being Function of Fin X, Y such that
A54: x = G1.B and
A55: for e being Element of Y st e is_a_unity_wrt F holds G1.{} = e and
A56: for x being Element of X holds G1.{x} = f.x and
A57: for B9 being Element of Fin X st B9 c= B & B9 <> {} for x being
    Element of X st x in B \ B9 holds G1.(B9 \/ {x}) = F.(G1.B9,f.x);
    given G2 being Function of Fin X, Y such that
A58: y = G2.B and
A59: for e being Element of Y st e is_a_unity_wrt F holds G2.{} = e and
A60: for x being Element of X holds G2.{x} = f.x and
A61: for B9 being Element of Fin X st B9 c= B & B9 <> {} for x being
    Element of X st x in B \ B9 holds G2.(B9 \/ {x}) = F.(G2.B9,f.x);
    per cases;
    suppose
A62:  B = {};
      thus x = the_unity_wrt F by A54,A55,A62,A1,Th14
        .= y by A58,A59,A62,A1,Th14;
    end;
    suppose
A64:  B <> {};
      defpred X[set] means $1 c= B & $1 <> {} implies G1.$1 = G2.$1;
A65:  for B9 being Element of Fin X, b being Element of X holds X[B9] &
      not b in B9 implies X[B9 \/ {b}]
      proof
        let B9 be (Element of Fin X), x be Element of X;
        assume
A66:    B9 c= B & B9 <> {} implies G1.B9 = G2.B9;
        assume
A67:    not x in B9;
        assume
A68:    B9 \/ {x} c= B;
        then
A69:    B9 c= B by ZFMISC_1:137;
        assume B9 \/ {x} <> {};
        x in B by A68,ZFMISC_1:137;
        then
A70:    x in B \ B9 by A67,XBOOLE_0:def 5;
        per cases;
        suppose
A71:      B9 = {};
          hence G1.(B9 \/ {x}) = f.x by A56
            .= G2.(B9 \/ {x}) by A60,A71;
        end;
        suppose
A72:      B9 <> {};
          hence G1.(B9 \/ {x}) = F.(G1.B9, f.x) by A57,A69,A70
            .= G2.(B9 \/ {x}) by A61,A66,A69,A70,A72;
        end;
      end;
A73:  X[{}.X];
      for B9 being Element of Fin X holds X[B9] from FinSubInd1(A73,
      A65 );
      hence thesis by A54,A58,A64;
    end;
  end;
end;

theorem Th16:
  for X, Y being non empty set for F being BinOp of Y for B being
Element of Fin X for f being Function of X,Y st (B <> {} or F is having_a_unity
) & F is idempotent & F is commutative & F is associative for IT being Element
of Y holds IT = F $$ (B,f) iff ex G being Function of Fin X, Y st IT = G.B & (
  for e being Element of Y st e is_a_unity_wrt F holds G.{} = e) & (for x being
Element of X holds G.{x} = f.x) & for B9 being Element of Fin X st B9 c= B & B9
  <> {} for x being Element of X st x in B holds G.(B9 \/ {x}) = F.(G.B9,f.x)
proof
  let X, Y be non empty set;
  let F be BinOp of Y;
  let B be Element of Fin X;
  let f be Function of X,Y;
  assume that
A1: B <> {} or F is having_a_unity and
A2: F is idempotent and
A3: F is commutative and
A4: F is associative;
  let IT be Element of Y;
  thus IT = F $$ (B,f) implies ex G being Function of Fin X, Y st IT = G.B & (
  for e being Element of Y st e is_a_unity_wrt F holds G.{} = e) & (for x being
Element of X holds G.{x} = f.x) & for B9 being Element of Fin X st B9 c= B & B9
  <> {} for x being Element of X st x in B holds G.(B9 \/ {x}) = F.(G.B9,f.x)
  proof
    assume IT = F $$ (B,f);
    then consider G being Function of Fin X, Y such that
A5: IT = G.B & for e being Element of Y st e is_a_unity_wrt F holds G.
    {} = e and
A6: for x being Element of X holds G.{x} = f.x and
A7: for B9 being Element of Fin X st B9 c= B & B9 <> {} for x being
    Element of X st x in B \ B9 holds G.(B9 \/ {x}) = F.(G.B9,f.x) by A1,A3,A4
,Def3;
    for B9 being Element of Fin X st B9 c= B & B9 <> {} for x being
    Element of X st x in B holds G.(B9 \/ {x}) = F.(G.B9,f.x)
    proof
      let B9 be Element of Fin X such that
A8:   B9 c= B and
A9:   B9 <> {};
      let x be Element of X such that
A10:  x in B;
      per cases;
      suppose
        x in B9;
        then
A11:    B9 \/ {x} = B9 by ZFMISC_1:40;
        then
A12:    {x} c= B9 by XBOOLE_1:7;
        not x in B9 \ {x} by ZFMISC_1:56;
        then
A13:    x in B \ (B9 \ {x}) by A10,XBOOLE_0:def 5;
        per cases;
        suppose
A14:      B9 = {x};
          hence G.(B9 \/ {x}) = F.(f.x, f.x) by A2,A6
            .= F.(G.B9,f.x) by A6,A14;
        end;
        suppose
          B9 <> {x};
          then not B9 c= {x} by A12,XBOOLE_0:def 10;
          then B9 \ {x} <> {} by XBOOLE_1:37;
          then
A15:      G.(B9 \ {.x.} \/ {.x.}) = F.(G.(B9 \ {.x.}), f.x) by A7,A8,A13,
XBOOLE_1:1;
          thus G.(B9 \/ {x})
             = F.(G.(B9 \ {x}), F.(f.x,f.x)) by A2,A15,XBOOLE_1:39
            .= F.(F.(G.(B9 \ {.x.}),f.x), f.x) by A4,BINOP_1:def 3
            .= F.(G.B9,f.x) by A11,A15,XBOOLE_1:39;
        end;
      end;
      suppose
        not x in B9;
        then x in B \ B9 by A10,XBOOLE_0:def 5;
        hence thesis by A7,A8,A9;
      end;
    end;
    hence thesis by A5,A6;
  end;
  given G being Function of Fin X, Y such that
A16: ( IT = G.B & for e being Element of Y st e is_a_unity_wrt F holds G
  .{} = e ) & for x being Element of X holds G.{x} = f.x and
A17: for B9 being Element of Fin X st B9 c= B & B9 <> {} for x being
  Element of X st x in B holds G.(B9 \/ {x}) = F.(G.B9,f.x);
  for B9 being Element of Fin X st B9 c= B & B9 <> {} for x being Element
  of X st x in B \ B9 holds G.(B9 \/ {x}) = F.(G.B9,f.x)
  proof
    let B9 be Element of Fin X such that
A18: B9 c= B & B9 <> {};
    let x be Element of X;
    assume x in B \ B9;
    then x in B by XBOOLE_0:def 5;
    hence thesis by A17,A18;
  end;
  hence thesis by A1,A3,A4,A16,Def3;
end;

reserve X, Y for non empty set,
  F for (BinOp of Y),
  B for (Element of Fin X),
  f for Function of X,Y;

theorem Th17:
  F is commutative & F is associative implies for b being Element
  of X holds F $$ ({.b.},f) = f.b
proof
  assume
A1: F is commutative & F is associative;
  let b be Element of X;
  ex G being Function of Fin X, Y st F $$ ({.b.},f) = G.{b} & (for e being
Element of Y st e is_a_unity_wrt F holds G.{} = e) & ( for x being Element of X
holds G.{x} = f.x) & for B9 being Element of Fin X st B9 c= {b} & B9 <> {} for
x being Element of X st x in {b} \ B9 holds G.(B9 \/ {x}) = F.(G.B9,f.x) by A1
,Def3;
  hence thesis;
end;

theorem Th18:
  F is idempotent & F is commutative & F is associative implies
  for a,b being Element of X holds F $$ ({.a,b.},f) = F.(f.a, f.b)
proof
  assume
A1: F is idempotent & F is commutative & F is associative;
  let a,b be Element of X;
  consider G being Function of Fin X, Y such that
A2: F $$ ({.a,b.},f) = G.{a,b} and
  for e being Element of Y st e is_a_unity_wrt F holds G.{} = e and
A3: for x being Element of X holds G.{x} = f.x and
A4: for B9 being Element of Fin X st B9 c= {a,b} & B9 <> {} for x being
  Element of X st x in {a,b} holds G.(B9 \/ {x}) = F.(G.B9,f.x) by A1,Th16;
A5: b in {a,b} by TARSKI:def 2;
  thus F $$ ({.a,b.},f) = G.({.a.} \/ {.b.}) by A2,ENUMSET1:1
    .= F.(G.{.a.}, f.b) by A4,A5,ZFMISC_1:7
    .= F.(f.a, f.b) by A3;
end;

theorem Th19:
  F is idempotent & F is commutative & F is associative implies
  for a,b,c being Element of X holds F $$ ({.a,b,c.},f) = F.(F.(f.a, f.b), f.c)
proof
  assume
A1: F is idempotent & F is commutative & F is associative;
  let a,b,c be Element of X;
  consider G being Function of Fin X, Y such that
A2: F $$ ({.a,b,c.},f) = G.{a,b,c} and
  for e being Element of Y st e is_a_unity_wrt F holds G.{} = e and
A3: for x being Element of X holds G.{x} = f.x and
A4: for B9 being Element of Fin X st B9 c= {a,b,c} & B9 <> {} for x
  being Element of X st x in {a,b,c} holds G.(B9 \/ {x}) = F.(G.B9,f.x) by A1
,Th16;
A5: b in {a,b,c} by ENUMSET1:def 1;
A6: G.{a,b} = G.({a} \/ {b}) by ENUMSET1:1
    .= F.(G.{.a.}, f.b) by A4,A5,Th1
    .= F.(f.a, f.b) by A3;
A7: c in {a,b,c} by ENUMSET1:def 1;
  thus F $$ ({.a,b,c.},f) = G.({.a,b.} \/ {.c.}) by A2,ENUMSET1:3
    .= F.(F.(f.a, f.b), f.c) by A4,A6,A7,Th2;
end;



theorem Th20:
  F is idempotent & F is commutative & F is associative & B <> {}
implies for x being Element of X holds F $$(B \/ {.x.}, f) = F.(F $$(B,f),f.x)
proof
  assume
A1: F is idempotent & F is commutative & F is associative;
  assume
A2: B <> {};
  then consider G9 being Function of Fin X, Y such that
A3: F $$ (B,f) = G9.B and
  for e being Element of Y st e is_a_unity_wrt F holds G9.{} = e and
A4: for x being Element of X holds G9.{x} = f.x and
A5: for B9 being Element of Fin X st B9 c= B & B9 <> {} for x being
  Element of X st x in B holds G9.(B9 \/ {x}) = F.(G9.B9,f.x) by A1,Th16;
  let x be Element of X;
  consider G being Function of Fin X, Y such that
A6: F $$ (B \/ {.x.},f) = G.(B \/ {.x.}) and
  for e being Element of Y st e is_a_unity_wrt F holds G.{} = e and
A7: for x being Element of X holds G.{x} = f.x and
A8: for B9 being Element of Fin X st B9 c= B \/ {x} & B9 <> {} for x9
being Element of X st x9 in B \/ {x} holds G.(B9 \/ {x9}) = F.(G.B9,f.x9) by A1
,Th16;
  defpred X[set] means $1 <> {} & $1 c= B implies G.$1 = G9.$1;
A9: for B9 being Element of Fin X, b being Element of X holds X[B9] implies
  X[B9 \/ {b}]
  proof
A10: B c= B \/ {x} by XBOOLE_1:7;
    let C be (Element of Fin X), y be Element of X such that
A11: C <> {} & C c= B implies G.C = G9.C;
    assume that
    C \/ {y} <> {} and
A12: C \/ {y} c= B;
A13: C c= B & y in B by A12,ZFMISC_1:137;
    per cases;
    suppose
A14:  C = {};
      hence G.(C \/ {y}) = f.y by A7
        .= G9.(C \/ {y}) by A4,A14;
    end;
    suppose
A15:  C <> {};
      hence G.(C \/ {y}) = F.(G9.C, f.y) by A8,A11,A13,A10,XBOOLE_1:1
        .= G9.(C \/ {y}) by A5,A13,A15;
    end;
  end;
A16: X[{}.X];
A17: for C being Element of Fin X holds X[C] from FinSubInd3(A16,A9);
  x in B \/ {x} by ZFMISC_1:136;
  hence F $$(B \/ {.x.}, f) = F.(G.B, f.x) by A2,A6,A8,XBOOLE_1:7
    .= F.(F $$(B,f),f.x) by A2,A3,A17;
end;

theorem Th21:
  F is idempotent & F is commutative & F is associative implies
for B1,B2 being Element of Fin X st B1 <> {} & B2 <> {} holds F $$ (B1 \/ B2,f)
  = F.(F $$ (B1,f),F $$ (B2,f))
proof
  assume that
A1: F is idempotent and
A2: F is commutative and
A3: F is associative;
  let B1,B2 be Element of Fin X;
  defpred X[Element of Fin X] means $1 <> {} implies F $$ (B1 \/ $1,f) = F.(F
  $$ (B1,f),F $$ ($1,f));
  assume
A4: B1 <> {};
A5: for B9 being Element of Fin X, b being Element of X holds X[B9] implies
  X[B9 \/ {.b.}]
  proof
    let B be (Element of Fin X), x be Element of X such that
A6: B <> {} implies F $$ (B1 \/ B,f) = F.(F $$ (B1,f),F $$ (B,f)) and
    B \/ {x} <> {};
    per cases;
    suppose
A7:   B = {};
      hence
      F $$ (B1 \/ (B \/ {.x.}),f) = F.(F $$ (B1,f), f.x) by A1,A2,A3,A4,Th20
        .= F.(F $$ (B1,f),F $$ (B \/ {.x.},f)) by A2,A3,A7,Th17;
    end;
    suppose
A8:   B <> {};
      thus F $$ (B1 \/ (B \/ {.x.}),f) = F $$ (B1 \/ B \/ {.x.},f) by
XBOOLE_1:4
        .= F.(F.(F $$ (B1,f),F $$ (B,f)), f.x) by A1,A2,A3,A6,A8,Th20
        .= F.(F $$ (B1,f),F.(F $$ (B,f), f.x)) by A3,BINOP_1:def 3
        .= F.(F $$ (B1,f),F $$ (B \/ {.x.},f)) by A1,A2,A3,A8,Th20;
    end;
  end;
A9: X[{}.X];
  for B2 being Element of Fin X holds X[B2] from FinSubInd3(A9,A5);
  hence thesis;
end;

theorem
  F is commutative & F is associative & F is idempotent implies for x
  being Element of X st x in B holds F.(f.x,F$$(B,f)) = F$$(B,f)
proof
  assume that
A1: F is commutative & F is associative and
A2: F is idempotent;
  let x be Element of X;
  assume
A3: x in B;
  thus F.(f.x,F$$(B,f)) = F.(F$$({.x.},f), F$$(B,f)) by A1,Th17
    .= F$$({.x.} \/ B, f) by A1,A2,A3,Th21
    .= F$$(B,f) by A3,ZFMISC_1:40;
end;

theorem
  F is commutative & F is associative & F is idempotent implies for B,C
being Element of Fin X st B <> {} & B c= C holds F.(F$$(B,f),F$$(C,f)) = F$$(C,
  f)
proof
  assume
A1: F is commutative & F is associative & F is idempotent;
  let B,C be Element of Fin X such that
A2: B <> {} and
A3: B c= C;
  C <> {} by A2,A3;
  hence F.(F$$(B,f),F$$(C,f)) = F$$(B \/ C,f) by A1,A2,Th21
    .= F$$(C,f) by A3,XBOOLE_1:12;
end;

theorem Th24:
  B <> {} & F is commutative & F is associative & F is idempotent
implies for a being Element of Y st for b being Element of X st b in B holds f.
  b = a holds F$$(B,f) = a
proof
  assume that
A1: B <> {} and
A2: F is commutative & F is associative and
A3: F is idempotent;
  let a be Element of Y;
  defpred X[Element of Fin X] means (for b being Element of X st b in $1 holds
  f.b = a) implies F$$($1,f) = a;
A4: for B1,B2 being non empty Element of Fin X holds X[B1] & X[B2] implies X
  [B1 \/ B2]
  proof
    let B1,B2 be non empty Element of Fin X;
    assume that
A5: ( (for b being Element of X st b in B1 holds f.b = a) implies F$$
(B1,f) = a)&( (for b being Element of X st b in B2 holds f.b = a) implies F$$(
    B2,f) = a) and
A6: for b being Element of X st b in B1 \/ B2 holds f.b = a;
A7: now
      let b be Element of X;
      assume b in B2;
      then b in B1 \/ B2 by XBOOLE_0:def 3;
      hence f.b = a by A6;
    end;
    now
      let b be Element of X;
      assume b in B1;
      then b in B1 \/ B2 by XBOOLE_0:def 3;
      hence f.b = a by A6;
    end;
    hence F$$(B1 \/ B2,f) = F.(a,a) by A2,A3,A5,A7,Th21
      .= a by A3;
  end;
A8: for x being Element of X holds X[{.x.}]
  proof
    let x be Element of X such that
A9: for b being Element of X st b in {x} holds f.b = a;
A10: x in { x } by TARSKI:def 1;
    thus F$$({.x.},f) = f.x by A2,Th17
      .= a by A9,A10;
  end;
  for B being non empty Element of Fin X holds X[B] from FinSubInd2(A8,
  A4);
  hence thesis by A1;
end;

theorem Th25:
  F is commutative & F is associative & F is idempotent implies
  for a being Element of Y st f.:B = {a} holds F$$(B,f) = a
proof
  assume
A1: F is commutative & F is associative & F is idempotent;
  let a be Element of Y;
  assume
A2: f.:B = {a};
A3: for b being Element of X st b in B holds f.b = a
  proof
    let b be Element of X;
    assume b in B;
    then f.b in {a} by A2,FUNCT_2:35;
    hence thesis by TARSKI:def 1;
  end;
  B <> {} by A2;
  hence thesis by A1,A3,Th24;
end;

theorem Th26:
  F is commutative & F is associative & F is idempotent implies
for f,g being Function of X,Y for A,B being Element of Fin X st A <> {} & f.:A
  = g.:B holds F$$(A,f) = F$$(B,g)
proof
  assume that
A1: F is commutative and
A2: F is associative and
A3: F is idempotent;
  let f,g be Function of X,Y;
  defpred X[Element of Fin X] means $1 <> {} implies for B being Element of
  Fin X st f.:$1 = g.:B holds F$$($1,f) = F$$(B,g);
A4: for B9 being Element of Fin X, b being Element of X holds X[B9] implies
  X[B9 \/ {.b.}]
  proof
    let A be (Element of Fin X), x be Element of X such that
A5: A <> {} implies for B being Element of Fin X st f.:A = g.:B holds
    F$$(A,f) = F$$(B,g);
    assume A \/ {x} <> {};
    let B be Element of Fin X such that
A6: f.:(A \/ {x}) = g.:B;
    per cases;
    suppose
      f.x in f.:A;
      then consider x9 being Element of X such that
A7:   x9 in A and
A8:   f.x9 = f.x by FUNCT_2:65;
A9:   g.:B = f.:A \/ Im(f,x) by A6,RELAT_1:120
        .= f.:A \/ {f.x} by Th8
        .= f.:A by A7,A8,FUNCT_2:35,ZFMISC_1:40;
      thus F$$(A \/ {.x.},f) = F.(F$$(A,f), f.x) by A1,A2,A3,A7,Th20
        .= F.(F$$(A \/ {.x9.},f), f.x) by A7,ZFMISC_1:40
        .= F.(F.(F$$(A,f), f.x9), f.x) by A1,A2,A3,A7,Th20
        .= F.(F$$(A,f), F.(f.x9, f.x9)) by A2,A8,BINOP_1:def 3
        .= F$$(A \/ {.x9.},f) by A1,A2,A3,A7,Th20
        .= F$$(A,f) by A7,ZFMISC_1:40
        .= F$$(B,g) by A5,A7,A9;
    end;
    suppose
A10:  not f.x in f.:A;
      per cases;
      suppose
A11:    A = {};
        then
A12:    g.:B = Im(f,x) by A6
          .= {f.x} by Th8;
        thus F$$(A \/ {.x.},f) = f.x by A1,A2,A11,Th17
          .= F$$(B,g) by A1,A2,A3,A12,Th25;
      end;
      suppose
A13:    A <> {};
        reconsider B1 = B \ g"{f.x} as Element of Fin X by Th11;
A14:    now
          assume B1 = {};
          then
A15:      g.:B c= g.:g"{f.x} by RELAT_1:123,XBOOLE_1:37;
          g.:g"{f.x} c= {f.x} by FUNCT_1:75;
          then f.:(A \/ {x}) c= {f.x} by A6,A15,XBOOLE_1:1;
          then f.:A \/ f.:{x} c= {f.x} by RELAT_1:120;
          then f.:A = f.:A /\ {f.x} by XBOOLE_1:11,28
            .= {} by A10,ZFMISC_1:50,XBOOLE_0:def 7;
          hence contradiction by A13,Th13;
        end;
        reconsider B2 = B /\ g"{f.x} as Element of Fin X by Th11,XBOOLE_1:17;
A17:    B = B1 \/ B2 by XBOOLE_1:51;
A18:    for b being Element of X st b in B2 holds g.b = f.x
        proof
          let b be Element of X;
          assume b in B2;
          then b in g"{f.x} by XBOOLE_0:def 4;
          then g.b in {f.x} by FUNCT_1:def 7;
          hence thesis by TARSKI:def 1;
        end;
A19:    f.:A = f.:A \ {f.x} by A10,ZFMISC_1:57
          .= f.:A \ Im(f,x) by Th8
          .= (f.:A \/ f.:{x}) \ f.:{x} by XBOOLE_1:40
          .= f.:(A \/ {x}) \ Im(f,x) by RELAT_1:120
          .= g.:B \ {f.x} by A6,Th8
          .= g.:B1 by Th6;
        x in A \/ {x} by ZFMISC_1:136;
        then consider x9 being Element of X such that
A20:    x9 in B and
A21:    g.x9 = f.x by A6,FUNCT_2:35,65;
        x9 in g"{f.x} by A21,Th7;
        then
A22:    B2 <> {} by A20,XBOOLE_0:def 4;
        thus F$$(A \/ {.x.},f) = F.(F$$(A,f), f.x) by A1,A2,A3,A13,Th20
          .= F.(F$$(B1,g), f.x) by A5,A13,A19
          .= F.(F$$(B1,g), F$$(B2,g)) by A1,A2,A3,A22,A18,Th24
          .= F$$(B,g) by A1,A2,A3,A17,A14,A22,Th21;
      end;
    end;
  end;
A23: X[{}.X];
A24: for A being Element of Fin X holds X[A] from FinSubInd3(A23,A4);
  let A,B be Element of Fin X;
  assume A <> {} & f.:A = g.:B;
  hence thesis by A24;
end;

theorem
  for F,G being BinOp of Y st F is idempotent & F is commutative & F is
  associative & G is_distributive_wrt F for B being Element of Fin X st B <> {}
for f being Function of X,Y for a being Element of Y holds G.(a,F$$(B,f)) = F$$
  (B,G[;](a,f))
proof
  let F,G be BinOp of Y such that
A1: F is idempotent and
A2: F is commutative & F is associative and
A3: G is_distributive_wrt F;
  let B be Element of Fin X such that
A4: B<> {};
  let f be Function of X,Y;
  let a be Element of Y;
  defpred X[Element of Fin X] means G.(a,F$$($1,f)) = F$$($1,G[;](a,f));
A5: for B1,B2 being non empty Element of Fin X holds X[B1] & X[B2] implies X
  [B1 \/ B2]
  proof
    let B1,B2 be non empty Element of Fin X;
    assume
A6: G.(a,F$$(B1,f)) = F$$(B1,G[;](a,f)) & G.(a,F$$(B2,f)) = F$$(B2,G [;](a,f));
    thus G.(a,F$$(B1 \/ B2,f)) = G.(a,F.(F$$(B1,f),F$$(B2,f))) by A1,A2,Th21
      .= F.(F$$(B1,G[;](a,f)), F$$(B2,G[;] (a,f))) by A3,A6,BINOP_1:11
      .= F$$(B1 \/ B2,G[;](a,f)) by A1,A2,Th21;
  end;
A7: for x being Element of X holds X[{.x.}]
  proof
    let x be Element of X;
    thus G.(a,F$$({.x.},f)) = G.(a,f.x) by A2,Th17
      .= (G[;](a,f)).x by FUNCOP_1:53
      .= F$$({.x.},G[;](a,f)) by A2,Th17;
  end;
  for B being non empty Element of Fin X holds X[B] from FinSubInd2(A7,
  A5);
  hence thesis by A4;
end;

theorem
  for F,G being BinOp of Y st F is idempotent & F is commutative & F is
  associative & G is_distributive_wrt F for B being Element of Fin X st B <> {}
for f being Function of X,Y for a being Element of Y holds G.(F$$(B,f),a) = F$$
  (B,G[:](f,a))
proof
  let F,G be BinOp of Y such that
A1: F is idempotent and
A2: F is commutative & F is associative and
A3: G is_distributive_wrt F;
  let B be Element of Fin X such that
A4: B<> {};
  let f be Function of X,Y;
  let a be Element of Y;
  defpred X[Element of Fin X] means G.(F$$($1,f),a) = F$$($1,G[:](f,a));
A5: for B1,B2 being non empty Element of Fin X holds X[B1] & X[B2] implies X
  [B1 \/ B2]
  proof
    let B1,B2 be non empty Element of Fin X;
    assume
A6: G.(F$$(B1,f),a) = F$$(B1,G[:](f,a)) & G.(F$$(B2,f),a) = F$$(B2,G [:](f,a));
    thus G.(F$$(B1 \/ B2,f),a) = G.(F.(F$$(B1,f),F$$(B2,f)),a) by A1,A2,Th21
      .= F.(F$$(B1,G[:](f,a)), F$$(B2,G[:] (f,a))) by A3,A6,BINOP_1:11
      .= F$$(B1 \/ B2,G[:](f,a)) by A1,A2,Th21;
  end;
A7: for x being Element of X holds X[{.x.}]
  proof
    let x be Element of X;
    thus G.(F$$({.x.},f),a) = G.(f.x,a) by A2,Th17
      .= (G[:](f,a)).x by FUNCOP_1:48
      .= F$$({.x.},G[:](f,a)) by A2,Th17;
  end;
  for B being non empty Element of Fin X holds X[B] from FinSubInd2(A7,
  A5);
  hence thesis by A4;
end;

definition
  let X, Y be non empty set;
  let f be Function of X,Y;
  let A be Element of Fin X;
  redefine func f.:A -> Element of Fin Y;
  coherence by Lm1;
end;

theorem Th29:
  for A, X, Y being non empty set for F being BinOp of A st F is
  idempotent & F is commutative & F is associative for B being Element of Fin X
st B <> {} for f being Function of X,Y holds for g being Function of Y,A holds
  F$$(f.:B,g) = F$$(B,g*f)
proof
  let A, X, Y be non empty set, F be BinOp of A such that
A1: F is idempotent and
A2: F is commutative & F is associative;
  let B be Element of Fin X such that
A3: B <> {};
  let f be Function of X,Y;
  let g be Function of Y,A;
  defpred X[Element of Fin X] means F$$(f.:$1,g) = F$$($1,g*f);
A4: dom f = X by FUNCT_2:def 1;
A5: for B1,B2 being non empty Element of Fin X holds X[B1] & X[B2] implies X
  [B1 \/ B2]
  proof
    let B1,B2 be non empty Element of Fin X;
    assume
A6: F$$(f.:B1,g) = F$$(B1,g*f) & F$$(f.:B2,g) = F$$(B2,g*f);
B7: B1 c= X by FINSUB_1:def 5;
B8: B2 c= X by FINSUB_1:def 5;
    thus F$$(f.:(B1 \/ B2),g) = F$$(f.:B1 \/ f.:B2, g) by RELAT_1:120
      .= F.(F$$(f.:B1,g), F$$(f.:B2,g)) by A1,A2,B7,B8,Th21,A4
      .= F$$(B1 \/ B2,g*f) by A1,A2,A6,Th21;
  end;
A9: for x being Element of X holds X[{.x.}]
  proof
    let x be Element of X;
    f.:{.x.} = Im(f,x);
    hence F$$(f.:{.x.},g) = F$$({.f.x.},g) by A4,FUNCT_1:59
      .= g.(f.x) by A2,Th17
      .= (g*f).x by FUNCT_2:15
      .= F$$({.x.},g*f) by A2,Th17;
  end;
  for B being non empty Element of Fin X holds X[B] from FinSubInd2(A9,
  A5);
  hence thesis by A3;
end;

theorem Th30:
  F is commutative & F is associative & F is idempotent implies
  for Z being non empty set for G being BinOp of Z st G is commutative & G is
associative & G is idempotent for f being Function of X, Y for g being Function
  of Y,Z st for x,y being Element of Y holds g.(F.(x,y)) = G.(g.x,g.y) for B
  being Element of Fin X st B <> {} holds g.(F$$(B,f)) = G$$(B,g*f)
proof
  assume that
A1: F is commutative & F is associative and
A2: F is idempotent;
  let Z be non empty set, G be BinOp of Z such that
A3: G is commutative & G is associative and
A4: G is idempotent;
  let f be Function of X,Y;
  let g be Function of Y,Z such that
A5: for x,y being Element of Y holds g.(F.(x, y)) = G.(g.x, g.y);
  defpred X[Element of Fin X] means $1 <> {} implies g.(F$$($1,f)) = G$$($1,g*
  f);
A6: for B9 being Element of Fin X, b being Element of X holds X[B9] implies
  X[B9 \/ {.b.}]
  proof
    let B be (Element of Fin X), x be Element of X;
    assume that
A7: B <> {} implies g.(F$$(B,f)) = G$$(B,g*f) and
    B \/ {x} <> {};
    per cases;
    suppose
A8:   B = {};
      hence g.(F$$(B \/ {.x.},f)) = g.(f.x) by A1,Th17
        .= (g*f).x by FUNCT_2:15
        .= G$$(B \/ {.x.},g*f) by A3,A8,Th17;
    end;
    suppose
A9:   B <> {};
      hence g.(F$$(B \/ {.x.},f)) = g.(F.(F$$(B,f), f.x)) by A1,A2,Th20
        .= G.(g.(F$$(B,f)), g.(f.x)) by A5
        .= G.(G$$(B,g*f), (g*f).x) by A7,A9,FUNCT_2:15
        .= G$$(B \/ {.x.},g*f) by A3,A4,A9,Th20;
    end;
  end;
A10: X[{}.X];
  thus for B being Element of Fin X holds X[B] from FinSubInd3(A10,A6 );
end;



theorem Th31:
  F is commutative & F is associative & F is having_a_unity
  implies for f holds F$$({}.X,f) = the_unity_wrt F
proof
  assume
A1: F is commutative & F is associative & F is having_a_unity;
  let f;
  the_unity_wrt F is_a_unity_wrt F & ex G being Function of Fin X, Y st F
$$ ( {}.X,f) = G.{}.X & (for e being Element of Y st e is_a_unity_wrt F holds G
.{} = e) & (for x being Element of X holds G.{x} = f.x) & for B9 being Element
  of Fin X st B9 c= {}.X & B9 <> {} for x being Element of X st x in {}.X \ B9
  holds G.( B9 \/ {x}) = F.(G.B9,f.x) by A1,Def3,Th14;
  hence thesis;
end;

theorem Th32:
  F is idempotent & F is commutative & F is associative & F is
having_a_unity implies for x being Element of X holds F $$(B \/ {.x.}, f) = F.(
  F $$(B,f),f.x)
proof
  assume that
A1: F is idempotent and
A2: F is commutative & F is associative;
  assume
A3: F is having_a_unity;
  let x be Element of X;
A4: {} = {}.X;
  now
    assume
A5: B = {};
    hence F $$(B \/ {.x.}, f) = f.x by A2,Th17
      .= F.(the_unity_wrt F, f.x) by A3,Th15
      .= F.(F $$(B,f),f.x) by A2,A3,A4,A5,Th31;
  end;
  hence thesis by A1,A2,Th20;
end;

theorem
  F is idempotent & F is commutative & F is associative & F is
having_a_unity implies for B1,B2 being Element of Fin X holds F $$ (B1 \/ B2,f)
  = F.(F $$ (B1,f),F $$ (B2,f))
proof
  assume that
A1: F is idempotent and
A2: F is commutative & F is associative and
A3: F is having_a_unity;
  let B1,B2 be Element of Fin X;
  now
A4: {} = {}.X;
    assume
A5: B1 = {} or B2 = {};
    per cases by A5;
    suppose
A6:   B2 = {};
      hence F $$ (B1 \/ B2,f) = F.(F$$(B1,f),the_unity_wrt F) by A3,Th15
        .= F.(F $$ (B1,f),F $$ (B2,f)) by A2,A3,A4,A6,Th31;
    end;
    suppose
A7:   B1 = {};
      hence F $$ (B1 \/ B2,f) = F.(the_unity_wrt F, F$$(B2,f)) by A3,Th15
        .= F.(F $$ (B1,f),F $$ (B2,f)) by A2,A3,A4,A7,Th31;
    end;
  end;
  hence thesis by A1,A2,Th21;
end;

theorem
  F is commutative & F is associative & F is idempotent & F is
  having_a_unity implies for f,g being Function of X,Y for A,B being Element of
  Fin X st f.:A = g.:B holds F$$(A,f) = F$$(B,g)
proof
  assume that
A1: F is commutative & F is associative and
A2: F is idempotent and
A3: F is having_a_unity;
  let f,g be Function of X,Y;
  let A,B be Element of Fin X such that
A4: f.:A = g.:B;
  now
    assume
A5: A = {};
    then A = {}.X;
    then
A6: F$$(A,f) = the_unity_wrt F by A1,A3,Th31;
    f.:A = {} by A5;
    then B = {}.X by A4,Th13;
    hence thesis by A1,A3,A6,Th31;
  end;
  hence thesis by A1,A2,A4,Th26;
end;

theorem Th35:
  for A, X, Y being non empty set for F being BinOp of A st F is
  idempotent & F is commutative & F is associative & F is having_a_unity for B
  being Element of Fin X for f being Function of X,Y holds for g being Function
  of Y,A holds F$$(f.:B,g) = F$$(B,g*f)
proof
  let A, X, Y be non empty set, F be BinOp of A such that
A1: F is idempotent and
A2: F is commutative & F is associative and
A3: F is having_a_unity;
  let B be Element of Fin X;
  let f be Function of X,Y;
  let g be Function of Y,A;
  now
    assume
A4: B = {};
    then f.:B = {}.Y;
    then
A5: F$$(f.:B,g) = the_unity_wrt F by A2,A3,Th31;
    B = {}.X by A4;
    hence thesis by A2,A3,A5,Th31;
  end;
  hence thesis by A1,A2,Th29;
end;

theorem
  F is commutative & F is associative & F is idempotent & F is
having_a_unity implies for Z being non empty set for G being BinOp of Z st G is
  commutative & G is associative & G is idempotent & G is having_a_unity for f
  being Function of X, Y for g being Function of Y,Z st g.the_unity_wrt F =
  the_unity_wrt G & for x,y being Element of Y holds g.(F.(x,y)) = G.(g.x,g.y)
  for B being Element of Fin X holds g.(F$$(B,f)) = G$$(B,g*f)
proof
  assume that
A1: F is commutative & F is associative and
A2: F is idempotent and
A3: F is having_a_unity;
  let Z be non empty set;
  let G be BinOp of Z such that
A4: G is commutative & G is associative and
A5: G is idempotent and
A6: G is having_a_unity;
  let f be Function of X, Y;
  let g be Function of Y,Z such that
A7: g.the_unity_wrt F = the_unity_wrt G and
A8: for x,y being Element of Y holds g.(F.(x,y)) = G.(g.x,g.y);
  let B be Element of Fin X;
  per cases;
  suppose
    B = {};
    then
A9: B = {}.X;
    hence g.(F$$(B,f)) = g.the_unity_wrt F by A1,A3,Th31
      .= G$$(B,g*f) by A4,A6,A7,A9,Th31;
  end;
  suppose
    B <> {};
    hence thesis by A1,A2,A4,A5,A8,Th30;
  end;
end;

definition
  let A be set;
  func FinUnion A -> BinOp of Fin A means
  :Def4:
  for x,y being Element of Fin A holds it.(x,y) = (x \/ y);
  existence
  proof
    deffunc U(Element of Fin A,Element of Fin A) = $1 \/ $2;
    ex IT being BinOp of Fin A st for x,y being Element of Fin A holds IT.
    (x,y) = U(x,y) from BINOP_1:sch 4;
    hence thesis;
  end;
  uniqueness
  proof
    let IT, IT9 be BinOp of Fin A such that
A1: for x,y being Element of Fin A holds IT.(x,y) = (x \/ y) and
A2: for x,y being Element of Fin A holds IT9.(x,y) = (x \/ y);
    now
      let x,y be Element of Fin A;
      thus IT.(x,y) = (x \/ y) by A1
        .= IT9.(x,y) by A2;
    end;
    hence IT = IT9 by BINOP_1:2;
  end;
end;

reserve A for set,
  x,y,z for Element of Fin A;

theorem Th37:
  FinUnion A is idempotent
proof
  let x;
  thus FinUnion A.(x,x) = x \/ x by Def4
    .= x;
end;

theorem Th38:
  FinUnion A is commutative
proof
  let x,y;
  thus FinUnion A.(x,y) = y \/ x by Def4
    .= FinUnion A.(y,x) by Def4;
end;

theorem Th39:
  FinUnion A is associative
proof
  let x,y,z;
  thus FinUnion A.(FinUnion A.(x,y), z) = FinUnion A.(x \/ y, z) by Def4
    .= x \/ y \/ z by Def4
    .= x \/ (y \/ z) by XBOOLE_1:4
    .= FinUnion A.(x, y \/ z) by Def4
    .= FinUnion A.(x, FinUnion A.(y,z)) by Def4;
end;

theorem Th40:
  {}.A is_a_unity_wrt FinUnion A
proof
  thus for x holds FinUnion A.({}.A, x) =x
  proof
    let x;
    thus FinUnion A.({}.A, x) = {} \/ x by Def4
      .= x;
  end;
  let x;
  thus FinUnion A.(x, {}.A) = x \/ {} by Def4
    .= x;
end;

theorem Th41:
  FinUnion A is having_a_unity
proof
  {}.A is_a_unity_wrt FinUnion A by Th40;
  hence thesis by Def2;
end;

theorem
  the_unity_wrt FinUnion A is_a_unity_wrt FinUnion A
proof
  thus thesis by Th14,Th41;
end;

theorem Th43:
  the_unity_wrt FinUnion A = {}
proof
  {}.A is_a_unity_wrt FinUnion A by Th40;
  hence thesis by BINOP_1:def 8;
end;

reserve X,Y for non empty set,
  A for set,
  f for (Function of X, Fin A),
  i,j,k for (Element of X);

definition
  let X be non empty set, A be set;
  let B be Element of Fin X;
  let f be Function of X, Fin A;
  func FinUnion(B,f) -> Element of Fin A equals
  FinUnion A $$(B,f);
  coherence;
end;

theorem
  FinUnion({.i.},f) = f.i
proof
  FinUnion A is commutative by Th38;
  hence thesis by Th17,Th39;
end;

theorem
  FinUnion({.i,j.},f) = f.i \/ f.j
proof
  FinUnion A is idempotent & FinUnion A is commutative by Th37,Th38;
  hence FinUnion({.i,j.},f) = FinUnion A.(f.i, f.j) by Th18,Th39
    .= f.i \/ f.j by Def4;
end;

theorem
  FinUnion({.i,j,k.},f) = f.i \/ f.j \/ f.k
proof
  FinUnion A is idempotent & FinUnion A is commutative by Th37,Th38;
  hence FinUnion({.i,j,k.},f) = FinUnion A.(FinUnion A.(f.i, f.j), f.k) by Th19
,Th39
    .= FinUnion A.(f.i \/ f.j, f.k) by Def4
    .= f.i \/ f.j \/ f.k by Def4;
end;

theorem Th47:
  FinUnion({}.X,f) = {}
proof
  FinUnion A is commutative & FinUnion A is associative by Th38,Th39;
  hence FinUnion({}.X,f) = the_unity_wrt FinUnion A by Th31,Th41
    .= {} by Th43;
end;

theorem Th48:
  for B being Element of Fin X holds FinUnion(B \/ {.i.}, f) =
  FinUnion(B,f) \/ f.i
proof
  let B be Element of Fin X;
A1: FinUnion A is associative by Th39;
  FinUnion A is idempotent & FinUnion A is commutative by Th37,Th38;
  hence FinUnion(B \/ {.i.}, f) = FinUnion A.(FinUnion(B,f), f.i) by A1,Th32
,Th41
    .= FinUnion(B,f) \/ f.i by Def4;
end;

theorem Th49:
  for B being Element of Fin X holds FinUnion(B,f) = union (f.:B)
proof
  defpred X[Element of Fin X] means FinUnion($1,f) = union (f.:$1);
A1: for B being (Element of Fin X), i st X[B] holds X[B \/ {.i.}]
  proof
    let B be (Element of Fin X), i;
    assume FinUnion(B,f) = union (f.:B);
    hence FinUnion(B \/ {.i.}, f) = union(f.:B) \/ f.i by Th48
      .= union (f.:B) \/ union {f.i} by ZFMISC_1:25
      .= union (f.:B \/ {f.i}) by ZFMISC_1:78
      .= union (f.:B \/ Im(f,i)) by Th8
      .= union (f.:(B \/ {i})) by RELAT_1:120;
  end;
  FinUnion({}.X,f) = union (f.:{}.X) by Th47,ZFMISC_1:2;
  then
A2: X[{}.X];
  thus for B being Element of Fin X holds X[B] from FinSubInd3(A2,A1 );
end;

theorem
  for B1,B2 being Element of Fin X holds FinUnion(B1 \/ B2, f) =
  FinUnion(B1,f) \/ FinUnion(B2,f)
proof
  let B1,B2 be Element of Fin X;
  thus FinUnion(B1 \/ B2, f) = union(f.:(B1 \/ B2)) by Th49
    .= union(f.:B1 \/ f .:B2) by RELAT_1:120
    .= union(f.:B1) \/ union(f.:B2) by ZFMISC_1:78
    .= FinUnion(B1,f) \/ union(f.:B2) by Th49
    .= FinUnion(B1,f) \/ FinUnion(B2,f) by Th49;
end;

theorem
  for B being Element of Fin X for f being Function of X,Y holds for g
  being Function of Y,Fin A holds FinUnion(f.:B,g) = FinUnion(B,g*f)
proof
  let B be Element of Fin X;
  let f be Function of X,Y;
  let g be Function of Y,Fin A;
  thus FinUnion(f.:B,g) = union (g.:(f.:B)) by Th49
    .= union ((g*f).:B) by RELAT_1:126
    .= FinUnion(B,g*f) by Th49;
end;

theorem Th52:
  for A,X being non empty set, Y being set for G being BinOp of A
st G is commutative & G is associative & G is idempotent for B being Element of
Fin X st B <> {} for f being (Function of X,Fin Y), g being Function of Fin Y,A
  st for x,y being Element of Fin Y holds g.(x \/ y) = G.(g.x,g.y) holds g.(
  FinUnion(B,f)) = G$$(B,g*f)
proof
  let A,X be non empty set, Y be set, G be BinOp of A such that
A1: G is commutative & G is associative & G is idempotent;
  let B be Element of Fin X such that
A2: B <> {};
  let f be (Function of X,Fin Y), g be Function of Fin Y,A such that
A3: for x,y being Element of Fin Y holds g.(x \/ y) = G.(g.x,g.y);
A4: now
    let x,y be Element of Fin Y;
    thus g.(FinUnion Y.(x,y)) = g.(x \/ y) by Def4
      .= G.(g.x,g.y) by A3;
  end;
A5: FinUnion Y is idempotent by Th37;
  FinUnion Y is commutative & FinUnion Y is associative by Th38,Th39;
  hence thesis by A1,A5,A2,A4,Th30;
end;

theorem Th53:
  for Z being non empty set, Y being set for G being BinOp of Z st
G is commutative & G is associative & G is idempotent & G is having_a_unity for
  f being Function of X, Fin Y for g being Function of Fin Y,Z st g.{}.Y =
the_unity_wrt G & for x,y being Element of Fin Y holds g.(x \/ y) = G.(g.x,g.y)
  for B being Element of Fin X holds g.(FinUnion(B,f)) = G$$(B,g*f)
proof
  let Z be non empty set, Y be set;
  let G be BinOp of Z such that
A1: G is commutative & G is associative and
A2: G is idempotent and
A3: G is having_a_unity;
  let f be Function of X, Fin Y;
  let g be Function of Fin Y,Z such that
A4: g.{}.Y = the_unity_wrt G and
A5: for x,y being Element of Fin Y holds g.(x \/ y) = G.(g.x,g.y);
  let B be Element of Fin X;
A6: {} = {}.X;
A7: {} = {}.Fin Y;
  per cases;
  suppose
A8: B = {};
    then
A9: f.:B = {};
    thus g.(FinUnion(B,f)) = g.{}.Y by A6,A8,Th47
      .= G$$(f.:B,g) by A1,A3,A4,A7,A9,Th31
      .= G$$(B,g*f) by A1,A2,A3,Th35;
  end;
  suppose
    B <> {};
    hence thesis by A1,A2,A5,Th52;
  end;
end;

definition
  let A be set;
  func singleton A -> Function of A, Fin A means
  :Def6:
  for x being set st x in A holds it.x = {x};
  existence
  proof
    deffunc U(set) = {$1};
A1: now
      let x be set;
      assume
A2:   x in A;
      then reconsider A9 = A as non empty set;
      reconsider x9 = x as Element of A9 by A2;
      {.x9.} in Fin A9;
      hence U(x) in Fin A;
    end;
    thus ex f being Function of A, Fin A st for x being set st x in A holds f.
    x = U(x) from FUNCT_2:sch 2(A1);
  end;
  uniqueness
  proof
    let IT,IT9 be Function of A, Fin A such that
A3: for x being set st x in A holds IT.x = {x} and
A4: for x being set st x in A holds IT9.x = {x};
    now
      let x be set;
      assume
A5:   x in A;
      then IT.x = {x} by A3;
      hence IT.x = IT9.x by A4,A5;
    end;
    hence thesis by FUNCT_2:12;
  end;
end;

theorem Th54:
  for A being non empty set for f being Function of A, Fin A holds
  f = singleton A iff for x being Element of A holds f.x = {x}
proof
  let A be non empty set;
  let f be Function of A, Fin A;
  thus f = singleton A implies for x being Element of A holds f.x = {x} by Def6
;
  assume for x being Element of A holds f.x = {x};
  then for x be set holds x in A implies f.x = {x};
  hence thesis by Def6;
end;

theorem Th55:
  for x being set, y being Element of X holds x in singleton X.y iff x = y
proof
  let x be set, y be Element of X;
  singleton X.y = {y} by Th54;
  hence thesis by TARSKI:def 1;
end;

theorem
  for x,y,z being Element of X st x in singleton X.z & y in singleton X.
  z holds x = y
proof
  let x,y,z be Element of X;
  assume that
A1: x in singleton X.z and
A2: y in singleton X.z;
  x = z by A1,Th55;
  hence thesis by A2,Th55;
end;

Lm2: for D being non empty set, X, P being set for f being Function of X,D
holds f.:P c= D;

theorem Th57:
  for B being Element of Fin X, x being set holds x in FinUnion(B,
  f) iff ex i being Element of X st i in B & x in f.i
proof
  let B be Element of Fin X, x be set;
A1: now
    assume x in union (f.:B);
    then consider Z being set such that
A2: x in Z and
A3: Z in f .:B by TARSKI:def 4;
    f.:B is Subset of Fin A by Lm2;
    then reconsider Z as Element of Fin A by A3;
    consider i being Element of X such that
A4: i in B & Z = f.i by A3,FUNCT_2:65;
    take i9 = i;
    thus i9 in B & x in f.i9 by A2,A4;
  end;
  now
    given i being Element of X such that
A5: i in B and
A6: x in f.i;
    f.i in f.:B by A5,FUNCT_2:35;
    hence x in union (f.:B) by A6,TARSKI:def 4;
  end;
  hence thesis by A1,Th49;
end;

theorem
  for B being Element of Fin X holds FinUnion(B, singleton X) = B
proof
  let B be Element of Fin X;
  now
    let x be set;
    thus x in FinUnion(B, singleton X) implies x in B
    proof
      assume x in FinUnion(B, singleton X);
      then ex i being Element of X st i in B & x in singleton X.i by Th57;
      hence thesis by Th55;
    end;
    assume
A1: x in B;
    then reconsider x9 = x as Element of X by Th9;
    x in singleton X.x9 by Th55;
    hence x in FinUnion(B, singleton X) by A1,Th57;
  end;
  hence thesis by TARSKI:1;
end;

theorem
  for Y,Z being set for f being Function of X, Fin Y for g being
  Function of Fin Y, Fin Z st g.{}.Y = {}.Z & for x,y being Element of Fin Y
holds g.(x \/ y) = g.x \/ g.y for B being Element of Fin X holds g.(FinUnion(B,
  f)) = FinUnion(B,g*f)
proof
  let Y,Z be set;
  let f be Function of X, Fin Y;
  let g be Function of Fin Y, Fin Z;
  assume that
A1: g.{}.Y = {}.Z and
A2: for x,y being Element of Fin Y holds g.(x \/ y) = g.x \/ g.y;
A3: g.{}.Y = the_unity_wrt FinUnion Z by A1,Th43;
A4: now
    let x,y be Element of Fin Y;
    thus g.(x \/ y) = g.x \/ g.y by A2
      .= FinUnion Z.(g.x,g.y) by Def4;
  end;
  let B be Element of Fin X;
A5: FinUnion Z is idempotent by Th37;
  FinUnion Z is associative & FinUnion Z is commutative by Th38,Th39;
  hence thesis by A5,A3,A4,Th41,Th53;
end;














begin

reserve B for non empty set,
  A,X,x for set;

scheme
  Fraenkel59{ A() -> set, F(set) -> set, P,Q[set] } :
  { F(v) where v is Element of A() : P[v] } c=
   { F(u) where u is Element of A() : Q[u] }
provided
A1: for v being Element of A() holds P[v] implies Q[v]
proof
  let x be set;
  assume x in { F(v9) where v9 is Element of A() : P[v9] };
  then consider v being Element of A() such that
A2: x = F(v) and
A3: P[v];
  Q[v] by A1,A3;
  hence thesis by A2;
end;

scheme
  Fraenkel599 { A,B() -> set, F(set,set) -> set, P,Q[set,set] } :
  { F(u1,v1) where u1 is Element of A(), v1 is Element of B() : P[u1,v1] } c=
  { F(u2,v2) where u2 is Element of A(), v2 is Element of B() : Q[u2,v2] }
provided
A1: for u being (Element of A()), v being Element of B() holds P[u,v]
implies Q[u,v]
proof
  let x be set;
  assume x in {F(u9,v9) where u9 is (Element of A()), v9 is Element of B() :
  P[u9,v9]};
  then consider u being (Element of A()), v being Element of B() such that
A2: x = F(u,v) and
A3: P[u,v];
  Q[u,v] by A1,A3;
  hence thesis by A2;
end;

scheme
  Fraenkel69{ B() -> set, F(set) -> set, P, Q[set] } :
  { F(v1) where v1 is Element of B() : P[v1] } =
  { F(v2) where v2 is Element of B() : Q[v2] }
provided
A1: for v being Element of B() holds P[v] iff Q[v]
proof
  set A = { F(v1) where v1 is Element of B() : P[v1] }, B = { F(v2) where v2
  is Element of B() : Q[v2] };
A2: for v being Element of B() holds P[v] implies Q[v] by A1;
  thus A c= B from Fraenkel59(A2);
A3: for v being Element of B() holds Q[v] implies P[v] by A1;
  thus B c= A from Fraenkel59(A3);
end;

scheme
  Fraenkel699 { A,B() -> set, F(set,set) -> set, P,Q[set,set] } :
  { F(u1,v1) where u1 is Element of A(), v1 is Element of B() : P[u1,v1] } =
  { F(u2,v2) where u2 is Element of A(), v2 is Element of B() : Q[u2,v2] }
provided
A1: for u being Element of A(), v being Element of B() holds P[u,v]
iff Q[u,v]
proof
  set B = { F(u2,v2) where u2 is (Element of A()), v2 is Element of B() : Q[u2
  ,v2] };
  set A = { F(u1,v1) where u1 is (Element of A()), v1 is Element of B() : P[u1
  ,v1] };
A2: for u being (Element of A()), v being Element of B() holds P[u,v]
  implies Q[u,v] by A1;
  thus A c= B from Fraenkel599(A2);
A3: for u being (Element of A()), v being Element of B() holds Q[u,v]
  implies P[u,v] by A1;
  thus B c= A from Fraenkel599(A3);
end;

scheme
  FraenkelF9{ B() -> set, F,G(set) -> set, P[set] } :
  { F(v1) where v1 is Element of B() : P[v1] } =
  { G(v2) where v2 is Element of B() : P[v2] }
provided
A1: for v being Element of B() holds F(v) = G(v)
proof
  set X = { F(v1) where v1 is Element of B() : P[v1] }, Y = { G(v2) where v2
  is Element of B() : P[v2] };
A2: Y c= X
  proof
    let x be set;
    assume x in Y;
    then consider v1 being Element of B() such that
A3: x = G(v1) and
A4: P[v1];
    x = F(v1) by A1,A3;
    hence thesis by A4;
  end;
  X c= Y
  proof
    let x be set;
    assume x in X;
    then consider v1 being Element of B() such that
A5: x = F(v1) and
A6: P[v1];
    x = G(v1) by A1,A5;
    hence thesis by A6;
  end;
  hence thesis by A2,XBOOLE_0:def 10;
end;

scheme
  FraenkelF9R{ B() -> set, F,G(set) -> set, P[set] } :
  { F(v1) where v1 is Element of B() : P[v1] } =
  { G(v2) where v2 is Element of B() : P[v2] }
provided
A1: for v being Element of B() st P[v] holds F(v) = G(v)
proof
  set X = { F(v1) where v1 is Element of B() : P[v1] }, Y = { G(v2) where v2
  is Element of B() : P[v2] };
A2: Y c= X
  proof
    let x be set;
    assume x in Y;
    then consider v1 being Element of B() such that
A3: x = G(v1) and
A4: P[v1];
    x = F(v1) by A1,A3,A4;
    hence thesis by A4;
  end;
  X c= Y
  proof
    let x be set;
    assume x in X;
    then consider v1 being Element of B() such that
A5: x = F(v1) and
A6: P[v1];
    x = G(v1) by A1,A5,A6;
    hence thesis by A6;
  end;
  hence thesis by A2,XBOOLE_0:def 10;
end;

scheme
  FraenkelF99 { A,B() -> set, F(set,set) -> set, G(set,set) -> set,
                P[set,set] } :
  { F(u1,v1) where u1 is Element of A(), v1 is Element of B() : P[u1,v1] } =
  { G(u2,v2) where u2 is Element of A(), v2 is Element of B() : P[u2,v2] }
provided
A1: for u being Element of A(), v being Element of B() holds F(u,v) = G(u,v)
proof
  set Y = { G(u2,v2) where u2 is (Element of A()), v2 is Element of B() : P[u2
  ,v2] };
  set X = { F(u1,v1) where u1 is (Element of A()), v1 is Element of B() : P[u1
  ,v1] };
A2: Y c= X
  proof
    let x be set;
    assume x in Y;
    then consider u1 being (Element of A()), v1 being Element of B() such that
A3: x = G(u1,v1) and
A4: P[u1,v1];
    x = F(u1,v1) by A1,A3;
    hence thesis by A4;
  end;
  X c= Y
  proof
    let x be set;
    assume x in X;
    then consider u1 being (Element of A()), v1 being Element of B() such that
A5: x = F(u1,v1) and
A6: P[u1,v1];
    x = G(u1,v1) by A1,A5;
    hence thesis by A6;
  end;
  hence thesis by A2,XBOOLE_0:def 10;
end;

scheme
  FraenkelF699 { A,B() -> set, F(set,set) -> set, P,Q[set,set] } :
  { F(u1,v1) where u1 is Element of A(), v1 is Element of B() : P[u1,v1] } =
  { F(v2,u2) where u2 is Element of A(), v2 is Element of B() : Q[u2,v2] }
provided
A1: for u being Element of A(), v being Element of B() holds P[u,v]
iff Q[u,v] and
A2: for u being Element of A(), v being Element of B() holds F(u,v) = F(v,u)
proof
A3: for u being Element of A(), v being Element of B() holds Q[u,v]
  implies P[u,v] by A1;
A4: { F(u1,v1) where u1 is (Element of A()), v1 is Element of B() : Q[u1,v1]
} c= { F(u2,v2) where u2 is (Element of A()), v2 is Element of B() : P[u2,v2] }
  from Fraenkel599(A3);
  deffunc H(set,set) = F($2,$1);
A5: for u being (Element of A()), v being Element of B() holds P[u,v]
  implies Q[u,v] by A1;
A6: { F(u1,v1) where u1 is (Element of A()), v1 is Element of B() : P[u1,v1]
} c= { F(u2,v2) where u2 is (Element of A()), v2 is Element of B() : Q[u2,v2] }
  from Fraenkel599(A5);
A7: for u being Element of A(), v being Element of B() holds F(u,v) = H(u,v)
  by A2;
  { F(u1,v1) where u1 is Element of A(), v1 is Element of B() : Q[u1,v1] }
= { H(u2,v2) where u2 is Element of A(), v2 is Element of B() : Q[u2,v2] } from
  FraenkelF99(A7);
  hence thesis by A6,A4,XBOOLE_0:def 10;
end;

theorem Th1:
  for A,B being  set, F,G being Function of A,B for X
  being set st F|X = G|X for x being Element of A st x in X holds F.x = G.x
proof
  let A,B be set, F,G be Function of A,B;
  let X be set such that
A1: F|X = G|X;
  let x be Element of A;
  assume
A2: x in X;
  hence F.x = (G|X).x by A1,FUNCT_1:49
    .= G.x by A2,FUNCT_1:49;
end;

theorem Th2:
  for A,B being set holds Funcs(A,B) c= bool [:A,B:]
proof
  let A,B be set, x be set;
  assume x in Funcs(A,B);
  then consider f being Function such that
A1: x = f and
A2: dom f = A and
A3: rng f c= B by FUNCT_2:def 2;
A4: f c= [: dom f, rng f:] by RELAT_1:7;
  [:dom f, rng f:] c= [:A,B:] by A2,A3,ZFMISC_1:95;
  then f c= [:A,B:] by A4,XBOOLE_1:1;
  hence thesis by A1;
end;

theorem Th3:
  for X,Y being set st Funcs(X,Y) <> {} & X c= A & Y c= B for f
  being Element of Funcs(X,Y) holds f is PartFunc of A,B
proof
  let X,Y be set such that
A1: Funcs(X,Y) <> {} and
A2: X c= A and
A3: Y c= B;
  let f be Element of Funcs(X,Y);
  consider g being Function such that
A4: f = g and
A5: dom g = X and
A6: rng g c= Y by A1,FUNCT_2:def 2;
  rng g c= B by A3,A6,XBOOLE_1:1;
  hence thesis by A2,A4,A5,RELSET_1:4;
end;

scheme
  RelevantArgs
  { A,B,X() -> set, f,g() -> Function of A(),B(), P[set], Q[set] } :
  { f().u9 where u9 is Element of A() : P[u9] & u9 in X() } =
  { g().v9 where v9 is Element of A() : Q[v9] & v9 in X() }
provided
A1: f()|X() = g()|X() and
A2: for u being Element of A() st u in X() holds P[u] iff Q[u]
proof
  deffunc F(set) = f().$1;
  deffunc G(set) = g().$1;
  defpred PP[set] means P[$1] & $1 in X();
  defpred QQ[set] means Q[$1] & $1 in X();
  set C = { G(v9) where v9 is Element of A() : PP[v9] };
A3: for v being Element of A() holds PP[v] iff QQ[v] by A2;
A4: C = { G(v9) where v9 is Element of A() : QQ[v9] } from Fraenkel69(A3);
A5: for v being Element of A() st PP[v] holds F(v) = G(v) by A1,Th1;
  { F(u9) where u9 is Element of A() : PP[u9] } = C from FraenkelF9R(A5);
  hence thesis by A4;
end;

scheme
  FrSet0{ A() -> non empty set, P[set] }:
  { x where x is Element of A(): P[x] } c= A()
proof
  { x where x is Element of A(): P[x] } is Subset of A() from DOMAIN_1:sch 7;
  hence thesis;
end;

scheme
  Gen199{A,B() -> set, F(set,set) -> set, P[set,set], Q[set] } :
  for s being Element of A(), t being Element of B() st P[s,t] holds Q[F(s,t)]
provided
A1: for st1 being set st st1 in { F(s1,t1) where s1 is (Element of A()),
t1 is Element of B(): P[s1,t1] } holds Q[st1]
proof
  let s be (Element of A()), t be Element of B();
  assume P[s,t];
  then
  F(s,t) in { F(s1,t1) where s1 is (Element of A()), t1 is Element of B():
  P[s1,t1] };
  hence thesis by A1;
end;

scheme
  Gen199A{A,B() -> set, F(set,set) -> set, P[set,set], Q[set] } :
  for st1 being set st st1 in { F(s1,t1) where s1 is Element of A(),
  t1 is Element of B(): P[s1,t1] } holds Q[st1]
provided
A1: for s being Element of A(), t being Element of B() st P[s,t] holds
    Q[F(s,t)]
proof
  let st1 be set;
  assume st1 in { F(s1,t1) where s1 is (Element of A()), t1 is Element of B()
  : P[s1,t1] };
  then
  ex s1 being (Element of A()), t1 being Element of B() st st1 = F(s1,t1)
  & P[s1,t1];
  hence thesis by A1;
end;

scheme
  Gen299{A,B,C() -> set, F(set,set) -> Element of C(), P[set, set], Q[set] } :
  { st1 where st1 is Element of C() :
    st1 in { F(s1,t1) where s1 is Element of A(), t1 is Element of B() :
       P[s1,t1] } & Q[st1] } =
  { F(s2,t2) where s2 is Element of A(), t2 is Element of B() :
    P[s2,t2] & Q[F(s2,t2)]}
proof
  thus { st1 where st1 is Element of C() : st1 in { F(s1,t1) where s1 is (
  Element of A()), t1 is Element of B() : P[s1,t1] } & Q[st1] } c= { F(s2,t2)
  where s2 is (Element of A()), t2 is Element of B() : P[s2,t2] & Q[F(s2,t2)]}
  proof
    let x be set;
    assume x in { st1 where st1 is Element of C() : st1 in { F(s1,t1) where
    s1 is (Element of A()), t1 is Element of B() : P[s1,t1] } & Q[st1] };
    then consider st1 being Element of C() such that
A1: x = st1 and
A2: st1 in { F(s1,t1) where s1 is (Element of A()), t1 is Element of B
    () : P[s1,t1] } and
A3: Q[st1];
    ex s1 being (Element of A()), t1 being Element of B() st st1 = F(s1,t1
    ) & P[s1,t1] by A2;
    hence thesis by A1,A3;
  end;
  let x be set;
  assume x in { F(s2,t2) where s2 is (Element of A()), t2 is Element of B() :
  P[s2,t2] & Q[F(s2,t2)]};
  then consider s2 being (Element of A()), t2 being Element of B() such that
A4: x = F(s2,t2) and
A5: P[s2,t2] and
A6: Q[F(s2,t2)];
  F(s2,t2) in { F(s1,t1) where s1 is (Element of A()), t1 is Element of B
  ( ) : P[s1,t1] } by A5;
  hence thesis by A4,A6;
end;

scheme
  Gen39{A() -> set, F(set) -> set, P[set], Q[set] } : { F(s) where s
is Element of A() : s in { s1 where s1 is Element of A(): Q[s1] } & P[s] } = {
  F(s2) where s2 is Element of A() : Q[s2] & P[s2] } proof
  defpred QQ[set] means Q[$1] & P[$1];
  defpred PP[set] means $1 in { s1 where s1 is Element of A(): Q[s1] } & P[$1];
A1: for s being Element of A() holds PP[s] iff QQ[s]
  proof
    let s be Element of A();
    now
      assume s in { s1 where s1 is Element of A(): Q[s1] };
      then ex s1 being Element of A() st s = s1 & Q[s1];
      hence Q[s];
    end;
    hence thesis;
  end;
  thus { F(s) where s is Element of A() : PP[s] } = { F(s2) where s2 is
  Element of A() : QQ[s2] } from Fraenkel69(A1);
end;

scheme
  Gen399{A,B() -> set, F(set,set) -> set, P[set,
set], Q[set] } : { F(s,t) where s is (Element of A()), t is Element of B() : s
in { s1 where s1 is Element of A(): Q[s1] } & P[s,t] } = { F(s2,t2) where s2 is
  (Element of A()), t2 is Element of B() : Q[s2] & P[s2,t2] } proof
  defpred QQ[set,set] means Q[$1] & P[$1,$2];
  defpred PP[set,set] means $1 in { s1 where s1 is Element of A(): Q[s1] } & P
  [$1,$2];
A1: for s being Element of A(), t being Element of B() holds PP[s,t] iff QQ[
  s,t]
  proof
    let s be Element of A(), t be Element of B();
    now
      assume s in { s1 where s1 is Element of A(): Q[s1] };
      then ex s1 being Element of A() st s = s1 & Q[s1];
      hence Q[s];
    end;
    hence thesis;
  end;
  thus { F(s,t) where s is (Element of A()), t is Element of B() : PP[s,t] } =
  { F(s2,t2) where s2 is (Element of A()), t2 is Element of B() : QQ[s2,t2] }
  from Fraenkel699(A1);
end;

scheme
  Gen499{A,B() -> set, F(set,set) -> set, P[set,
set], Q[set,set] } : { F(s,t) where s is (Element of A()), t is Element of B()
: P[s,t] } c= { F(s1,t1) where s1 is (Element of A()), t1 is Element of B() : Q
  [s1,t1] }
provided
A1: for s being (Element of A()), t being Element of B() st P[s,t] ex s9
being Element of A() st Q[s9,t] & F(s,t) = F(s9,t)
proof
  let x be set;
  assume
  x in { F(s,t) where s is (Element of A()), t is Element of B() : P[s ,t] };
  then consider s being (Element of A()), t being Element of B() such that
A2: x = F(s,t) and
A3: P[s,t];
  ex s9 being Element of A() st Q[s9,t] & F(s,t) = F(s9,t) by A1,A3;
  hence thesis by A2;
end;

scheme
  FrSet1{ D,A() -> set, P[set], F(set) -> set }:
  { F(y) where y is Element of D() : F(y) in A() & P[y] } c= A()
proof
  let x be set;
  assume x in { F(y) where y is Element of D() : F(y) in A() & P[y] };
  then ex y being Element of D() st x = F(y) & F(y) in A() & P[y];
  hence thesis;
end;

scheme
  FrSet2{ D,A() -> set, Q[set], F(set) -> set }:
  { F(y) where y is Element of D() : Q[y] & not F(y) in A() } misses A()
proof
  assume
  { F(y) where y is Element of D() : Q[y] & not F(y) in A() } meets A( );
  then consider x be set such that
A1: x in { F(y) where y is Element of D() : Q[y] & not F(y) in A() } and
A2: x in A() by XBOOLE_0:3;
  ex y being Element of D() st x = F(y) & Q[y] & not F(y) in A() by A1;
  hence thesis by A2;
end;

scheme
  FrEqua1{ A,B() -> set, F(set,set) -> set, x()
-> (Element of B()), P[set,set], Q[set,set] }: { F(s,t) where s is (Element of
A()), t is Element of B(): Q[s,t] } = { F(s9,x()) where s9 is Element of A(): P
  [s9,x()] }
provided
A1: for s being Element of A() for t being Element of B() holds Q[s,t]
iff t = x() & P[s,t]
proof
  thus { F(s,t) where s is (Element of A()), t is Element of B(): Q[s,t] } c=
  { F(s9,x()) where s9 is Element of A(): P[s9,x()] }
  proof
    let x be set;
    assume
    x in { F(s,t) where s is (Element of A()), t is Element of B(): Q[ s,t ] };
    then consider s being (Element of A()), t being Element of B() such that
A2: x = F(s,t) and
A3: Q[s,t];
A4: P[s,t] by A1,A3;
    t = x() by A1,A3;
    hence thesis by A2,A4;
  end;
  let x be set;
  assume x in { F(s9,x()) where s9 is Element of A(): P[s9,x()] };
  then consider s9 being Element of A() such that
A5: x = F(s9,x()) and
A6: P[s9,x()];
  Q[s9,x()] by A1,A6;
  hence thesis by A5;
end;

scheme
  FrEqua2{ A,B() -> set, F(set,set) -> set, x()
  -> (Element of B()), P[set,set] }: { F(s,t) where s is (Element of A()), t is
Element of B(): t = x() & P[s,t] } = { F(s9,x()) where s9 is Element of A(): P[
  s9,x()] } proof
  defpred Q[set,set] means $2 = x() & P[$1,$2];
A1: for s being Element of A() for t being Element of B() holds Q[s,t] iff t
  = x() & P[s,t];
  thus { F(s,t) where s is Element of A(), t is Element of B(): Q[s,t] } = { F
  (s9,x()) where s9 is Element of A(): P[s9,x()] } from FrEqua1(A1);
end;



reserve phi for Element of Funcs(A,B);

theorem Th4:
  for X,Y being set st Funcs(X,Y) <> {} & X c= A & Y c= B for f
  being Element of Funcs(X,Y) ex phi being Element of Funcs(A,B) st phi|X = f
proof
  let X,Y be set such that
A1: Funcs(X,Y) <> {} and
A2: X c= A and
A3: Y c= B;
  let f be Element of Funcs(X,Y);
  reconsider f9=f as PartFunc of A,B by A1,A2,A3,Th3;
  consider phi being Function of A,B such that
A4: for x st x in dom f9 holds phi.x = f9.x by FUNCT_2:71;
  reconsider phi as Element of Funcs(A,B) by FUNCT_2:8;
  take phi;
  ex g being Function st f = g & dom g = X & rng g c= Y by A1,FUNCT_2:def 2;
  then dom f9 = A /\ X by XBOOLE_1:28
    .= dom phi /\ X by FUNCT_2:def 1;
  hence thesis by A4,FUNCT_1:46;
end;

theorem Th5:
  phi|X = phi|(A /\ X)
proof
  dom phi = A by FUNCT_2:def 1;
  then
A1: dom (phi|X) = (dom phi /\ A) /\ X by RELAT_1:61
    .= dom phi /\ (A /\ X) by XBOOLE_1:16;
  for x st x in dom (phi|X) holds (phi|X).x = phi.x by FUNCT_1:47;
  hence thesis by A1,FUNCT_1:46;
end;



scheme
  FraenkelFin { A() -> set, X() -> set, F(set) -> set }:
  { F(w) where w is Element of A(): w in X() } is finite
provided
A1: X() is finite
proof
  set M = { F(w) where w is Element of A(): w in X() };
  per cases;
  suppose
A2: A() is empty;
    M c= {F({})}
    proof
      let x;
      assume x in M;
      then consider w being Element of A() such that
A3:   x = F(w) & w in X();
      w = {} by A2,SUBSET_1:def 1;
      hence thesis by A3,TARSKI:def 1;
    end;
    hence thesis;
  end;
  suppose
A4: A() is non empty;
  consider f being Function such that
A5: dom f = X() /\ A() and
A6: for y being set st y in X() /\ A() holds f.y = F(y) from FUNCT_1:sch 3;
  M = f.:X()
  proof
    thus M c= f.:X()
    proof
      let x be set;
      assume x in M;
      then consider u being Element of A() such that
A7:   x = F(u) and
A8:   u in X();
A9:   u in dom f by A4,A5,A8,XBOOLE_0:def 4;
      then f.u = F(u) by A5,A6;
      hence thesis by A7,A8,A9,FUNCT_1:def 6;
    end;
    let x be set;
    assume x in f.:X();
    then consider y being set such that
A10: y in dom f and
A11: y in X() and
A12: x = f.y by FUNCT_1:def 6;
    reconsider y as Element of A() by A5,A10,XBOOLE_0:def 4;
    x = F(y) by A5,A6,A10,A12;
    hence thesis by A11;
  end;
  hence thesis by A1;
  end;
end;

scheme
  CartFin { A, B() -> non empty set, X, Y() -> set, F(set, set) -> set }:
  { F(u,v) where u is Element of A(), v is Element of B() :
    u in X() & v in Y () } is finite
provided
A1: X() is finite and
A2: Y() is finite
proof
  deffunc G(set) = F($1`1,$1`2);
  set M = { F(u9,v9) where u9 is (Element of A()), v9 is Element of B() : u9
  in X() & v9 in Y() };
  consider f being Function such that
A3: dom f = [:X() /\ A(), Y() /\ B():] and
A4: for y being set st y in [:X() /\ A(), Y() /\ B():] holds f.y = G(y)
  from FUNCT_1:sch 3;
A5: dom f = [:X(), Y():] /\ [:A(), B():] by A3,ZFMISC_1:100;
  M = f.:[:X(), Y():]
  proof
    thus M c= f.:[:X(), Y():]
    proof
      let x be set;
      assume x in M;
      then consider u being Element of A(), v being Element of B() such that
A6:   x = F(u,v) and
A7:   u in X() and
A8:   v in Y();
A9:   [u,v] in [:X(), Y():] by A7,A8,ZFMISC_1:87;
      then
A10:  [u,v] in dom f by A5,XBOOLE_0:def 4;
A11:  [u,v]`2 = v;
      [u,v]`1 = u;
      then f.(u,v) = F(u,v) by A3,A4,A10,A11;
      hence thesis by A6,A9,A10,FUNCT_1:def 6;
    end;
    let x be set;
    assume x in f.:[:X(), Y():];
    then consider y being set such that
A12: y in dom f and
A13: y in [:X(), Y():] and
A14: x = f.y by FUNCT_1:def 6;
    reconsider y as Element of [:A(), B():] by A5,A12,XBOOLE_0:def 4;
A15: y = [y`1,y`2] by MCART_1:21;
    then
A16: y`1 in X() by A13,ZFMISC_1:87;
A17: y`2 in Y() by A13,A15,ZFMISC_1:87;
    x = F(y`1,y`2) by A3,A4,A12,A14;
    hence thesis by A16,A17;
  end;
  hence thesis by A1,A2;
end;

scheme
  Finiteness { A()->non empty set, B()->(Element of Fin A()), P[(Element of A(
  )), Element of A()] } : for x being Element of A() st x in B() ex y being
Element of A() st y in B() & P[y,x] & for z being Element of A() st z in B() &
  P[z,y] holds P[y,z]
provided
A1: for x being Element of A() holds P[x,x] and
A2: for x,y,z being Element of A() st P[x,y] & P[y,z] holds P[x,z]
proof
  defpred R[(Element of A()), Element of Fin A()] means ex y being Element of
A() st y in $2 & P[y,$1] & for z being Element of A() st z in $2 & P[z,y] holds
  P[y,z];
  defpred Q[Element of Fin A()] means for x being Element of A() st x in $1
  holds R[x, $1];
A3: now
    let B be (Element of Fin A()), x be Element of A();
    assume
A4: Q[B];
    thus Q[B \/ {.x.}]
    proof
      let z be Element of A() such that
A5:   z in B \/ {x};
      now
        per cases by A5,ZFMISC_1:136;
        suppose
          z in B;
          then consider y being Element of A() such that
A6:       y in B and
A7:       P[y,z] and
A8:       for x being Element of A() st x in B & P[x,y] holds P[y,x] by A4;
          now
            per cases;
            case
A9:           P[x,y];
              thus x in B \/ {x} by ZFMISC_1:136;
              thus P[x,z] by A2,A7,A9;
              let v be Element of A() such that
A10:          v in B \/ {x} and
A11:          P[v,x];
A12:          now
                assume
A13:            v in B;
                P[v,y] by A2,A9,A11;
                then P[y,v] by A8,A13;
                hence P[x,v] by A2,A9;
              end;
              v in B or v = x by A10,ZFMISC_1:136;
              hence P[x,v] by A1,A12;
            end;
            case
A14:          not P[x,y];
              thus y in B \/ {x} by A6,XBOOLE_0:def 3;
              thus P[y,z] by A7;
              let v be Element of A() such that
A15:          v in B \/ {x} and
A16:          P[v,y];
              v in B or v = x by A15,ZFMISC_1:136;
              hence P[y,v] by A8,A14,A16;
            end;
          end;
          hence thesis;
        end;
        suppose
A17:      z = x;
          now
            per cases;
            case
              ex w being Element of A() st w in B & P[w,x];
              then consider w being Element of A() such that
A18:          w in B and
A19:          P[w,z] by A17;
              consider y being Element of A() such that
A20:          y in B and
A21:          P[y,w] and
A22:          for x being Element of A() st x in B & P[x,y] holds P[y
              ,x] by A4,A18;
              take u = y;
              thus u in B \/ {x} by A20,XBOOLE_0:def 3;
              thus P[u,z] by A2,A19,A21;
              let v be Element of A() such that
A23:          v in B \/ {x} and
A24:          P[v,u];
              v in B or v = x by A23,ZFMISC_1:136;
              hence P[u,v] by A2,A17,A19,A21,A22,A24;
            end;
            case
A25:          for w being Element of A() st w in B holds not P[w,x];
              thus x in B \/ {x} by ZFMISC_1:136;
              thus
A26:          P[x,x] by A1;
              let v be Element of A() such that
A27:          v in B \/ {x} and
A28:          P[v,x];
              not v in B by A25,A28;
              hence P[x,v] by A26,A27,ZFMISC_1:136;
            end;
          end;
          hence thesis by A17;
        end;
      end;
      hence thesis;
    end;
  end;
A29: Q[{}.A()];
  for B being Element of Fin A() holds Q[B] from SETWISEO:sch 4(A29,A3 );
  hence thesis;
end;

scheme
  FinIm{A,B()->non empty set, x()-> (Element of Fin B()), F
  (set)->(Element of A()), P[set,set]}: ex c1 being Element of Fin A() st for t
being Element of A() holds t in c1 iff ex t9 being Element of B() st t9 in x()
  & t = F(t9) & P[t,t9] proof
  defpred R[set] means ex t9 being Element of B() st t9 in x() & $1 = F(t9) &
  P[$1,t9];
  set c = { F(t9) where t9 is Element of B() : t9 in x() }, c1 = { tt where tt
  is Element of A() : R[tt]};
A1: c1 c= c
  proof
    let x;
    assume x in c1;
    then ex tt being Element of A() st x = tt & ex t9 being Element of B() st
    t9 in x() & tt = F(t9) & P[tt,t9];
    hence thesis;
  end;
A2: c1 c= A() from FrSet0;
A3: x() is finite;
  c is finite from FraenkelFin(A3);
  then reconsider c1 as Element of Fin A() by A1,A2,FINSUB_1:def 5;
  take c1;
  let t be Element of A();
  t in c1 implies ex tt being Element of A() st t = tt & ex t9 being
  Element of B() st t9 in x() & tt = F(t9) & P[tt,t9];
  hence thesis;
end;

registration
  let A, B be finite set;
  cluster Funcs(A,B) -> finite;
  coherence
  proof
    bool [:A,B:] is finite;
    hence thesis by Th2,FINSET_1:1;
  end;
end;

theorem
  for A,B being set st A is finite & B is finite holds Funcs(A,B) is finite;

scheme
  ImFin { A() -> set, B() -> non empty set, X() -> set, Y() -> set,
F(set) -> set } : { F(phi9) where phi9 is Element of Funcs(A(),B()): phi9.:X()
  c= Y() } is finite
provided
A1: X() is finite and
A2: Y() is finite and
A3: for phi,psi being Element of Funcs(A(),B()) holds phi|X() = psi|X()
implies F(phi) = F(psi)
proof
  defpred P[set,set] means for phi being Element of Funcs(A(),B()) st phi|X()
  = $1 holds $2 = F(phi);
  set Z = { F(phi9) where phi9 is Element of Funcs(A(),B()): phi9.: X() c= Y()
  };
  set x = the Element of Z;
A4: B() /\ Y() c= B() by XBOOLE_1:17;
  assume
A5: not thesis;
  then Z <> {};
  then x in Z;
  then consider phi being Element of Funcs(A(),B()) such that
  x = F(phi) and
A6: phi.:X() c= Y();
  now
    assume B() /\ Y() = {};
    then
A7: phi.:X() = {} by A6,XBOOLE_1:3,19;
    A() = dom phi by FUNCT_2:def 1;
    then A() misses X() by A7,RELAT_1:118;
    hence A() /\ X() = {} by XBOOLE_0:def 7;
  end;
  then reconsider FF = Funcs(A() /\ X(),B() /\ Y()), Z as non empty set by A5,
FUNCT_2:8;
A8: B() /\ Y() c= Y() by XBOOLE_1:17;
A9: A() /\ X() c= A() by XBOOLE_1:17;
A10: now
    let f be Element of FF;
    consider phi being Element of Funcs(A(),B()) such that
A11: phi|(A() /\ X()) = f by A9,A4,Th4;
A12: phi|X() = f by A11,Th5;
    ex g being Function st f = g & dom g = A() /\ X() & rng g c= B() /\ Y
    () by FUNCT_2:def 2;
    then rng (phi|X()) c= Y() by A8,A12,XBOOLE_1:1;
    then phi.:X() c= Y() by RELAT_1:115;
    then F(phi) in Z;
    then reconsider g9 = F(phi) as Element of Z;
    take g = g9;
    thus P[f,g] by A3,A12;
  end;
  consider F being Function of FF, Z such that
A13: for f being Element of FF holds P[f,F.f] from FUNCT_2:sch 3(A10);
  Z c= F.:Funcs(A() /\ X(), B() /\ Y())
  proof
    let y be set;
    assume y in Z;
    then consider phi being Element of Funcs(A(),B()) such that
A14: y = F(phi) and
A15: phi.:X() c= Y();
    rng (phi|X()) c= Y() by A15,RELAT_1:115;
    then
A16: rng (phi|X()) c= B() /\ Y() by XBOOLE_1:19;
A17: dom (phi|X()) = dom phi /\ X() by RELAT_1:61
      .= A() /\ X() by FUNCT_2:def 1;
    then reconsider
x = phi|X() as Element of Funcs(A() /\ X(), B() /\ Y()) by A16,FUNCT_2:def 2;
    phi|X() in Funcs(A() /\ X(), B() /\ Y()) by A17,A16,FUNCT_2:def 2;
    then
A18: x in dom F by FUNCT_2:def 1;
    y = F.x by A13,A14;
    hence thesis by A18,FUNCT_1:def 6;
  end;
  hence contradiction by A1,A2,A5;
end;



scheme
  FunctChoice { A()->non empty set, B()->non empty set, P[(Element of A()),
Element of B()], x()->Element of Fin A() }: ex ff being Function of A(), B() st
  for t being Element of A() st t in x() holds P[t,ff.t]
provided
A1: for t being Element of A() st t in x() ex ff being Element of B() st
P[t,ff]
proof
  set b = the Element of B();
  set M = { { ff where ff is Element of B() : P[tt,ff] } where tt is Element
  of A() : tt in x() };
  set f = the Function of A(), B();
  assume
A2: not thesis;
  then consider t being Element of A() such that
A3: t in x() and
  not P[t,f.t];
  { ff where ff is Element of B() : P[t,ff] } in M by A3;
  then reconsider M as non empty set;
  now
    let X;
    assume X in M;
    then consider t being Element of A() such that
A4: X = { ff where ff is Element of B() : P[t,ff] } and
A5: t in x();
    consider ff being Element of B() such that
A6: P[t,ff] by A1,A5;
    ff in X by A4,A6;
    hence X <> {};
  end;
  then consider Choice being Function such that
  dom Choice = M and
A7: X in M implies Choice.X in X by FUNCT_1:111;
  defpred Q[(Element of A()),set] means ($1 in x() implies $2 = Choice.{ ff
  where ff is Element of B() : P[$1,ff] }) & ($1 in x() or $2 = b);
A8: now
    let t be Element of A();
A9: now
      set s = { ff where ff is Element of B() : P[t,ff] };
      assume t in x();
      then s in M;
      then Choice.s in s by A7;
      then ex ff being Element of B() st Choice.s = ff & P[t,ff];
      hence Choice.s is Element of B();
    end;
    t in x() implies t in x();
    hence ex y being Element of B() st Q[t,y] by A9;
  end;
  consider f being Function of A(), B() such that
A10: for x being Element of A() holds Q[x,f.x] from FUNCT_2:sch 3(A8);
  now
    let t be Element of A();
    assume
A11: t in x();
    then
A12: { ff where ff is Element of B() : P[t,ff] } in M;
    f.t = Choice.{ ff where ff is Element of B() : P[t,ff] } by A10,A11;
    then f.t in { ff where ff is Element of B() : P[t,ff] } by A7,A12;
    then ex ff being Element of B() st f.t = ff & P[t,ff];
    hence P[t,f.t];
  end;
  hence contradiction by A2;
end;

scheme
  FuncsChoice { A()->non empty set, B()->non empty set, P[Element of A(),
Element of B()], x()->Element of Fin A() }: ex ff being Element of Funcs(A(),B(
  )) st for t being Element of A() st t in x() holds P[t,ff.t]
provided
A1: for t being Element of A() st t in x() ex ff being Element of B() st
P[t,ff]
proof
A2: for t being Element of A() st t in x() ex ff being Element of B() st P[t
  ,ff] by A1;
  consider ff being Function of A(), B() such that
A3: for t being Element of A() st t in x() holds P[t,ff.t] from
  FunctChoice(A2);
  reconsider ff as Element of Funcs(A(),B()) by FUNCT_2:8;
  take ff;
  thus thesis by A3;
end;

scheme
  FraenkelFin9 { A,B() -> non empty set, X() -> set, P[set,set] }: { x where x
  is Element of B(): ex w being Element of A() st P[w,x] & w in X() } is finite
provided
A1: X() is finite and
A2: for w being Element of A(), x,y being Element of B() st P[w,x] & P[w
,y] holds x = y
proof
  set M = { x where x is Element of B(): ex w being Element of A() st P[w,x] &
  w in X() };
  defpred Q[set,set] means P[$1,$2] & $1 in X() & $2 in B();
A3: for x,y being set st x in A() & Q[x,y] holds y in B();
A4: for x,y1,y2 being set st x in A() & Q[x,y1] & Q[x,y2] holds y1 = y2 by A2;
  consider f being PartFunc of A(),B() such that
A5: for x holds x in dom f iff x in A() & ex y being set st Q[x,y] and
A6: for x st x in dom f holds Q[x,f.x] from PARTFUN1:sch 2(A3,A4);
  M = f.:X()
  proof
    thus M c= f.:X()
    proof
      let x be set;
      assume x in M;
      then consider u being Element of B() such that
A7:   x = u and
A8:   ex w being Element of A() st P[w,u] & w in X();
      consider w being Element of A() such that
A9:   P[w,u] and
A10:  w in X() by A8;
A11:  w in dom f by A5,A9,A10;
      then Q[w,f.w] by A6;
      then f.w = x by A2,A7,A9;
      hence thesis by A10,A11,FUNCT_1:def 6;
    end;
    let x be set;
    assume x in f.:X();
    then consider y being set such that
A12: y in dom f and
A13: y in X() and
A14: x = f.y by FUNCT_1:def 6;
    reconsider x as Element of B() by A6,A12,A14;
    P[y,x] by A6,A12,A14;
    hence thesis by A12,A13;
  end;
  hence thesis by A1;
end;














begin

reserve A,B,C for Ordinal,
  X,X1,Y,Y1,Z,a,b,b1,b2,x,y,z for set,
  R for Relation,
  f,g,h for Function,
  k,m,n for Nat;

notation
  synonym 0 for {};
end;

definition
  let IT be set;
  attr IT is cardinal means
  :Def1:
  ex B st IT = B & for A st A,B
  are_equipotent holds B c= A;
end;

registration
  cluster cardinal for set;
  existence
  proof
    set A = the Ordinal;
    defpred P[Ordinal] means $1,A are_equipotent;
A1: ex A st P[A];
    consider B such that
A2: P[B] & for C st P[C] holds B c= C from ORDINAL1:sch 1(A1);
    reconsider IT = B as set;
    take IT,B;
    thus IT = B;
    let C;
    assume C,B are_equipotent;
    hence thesis by A2,WELLORD2:15;
  end;
end;

definition
  mode Cardinal is cardinal set;
end;

registration
  cluster cardinal -> ordinal for set;
  coherence
  proof
    let M be set;
    assume M is cardinal;
    then ex B st M = B & for A st A,B are_equipotent holds B c= A by Def1;
    hence thesis;
  end;
end;

reserve M,N for Cardinal;

theorem Th1:
  for X ex A st X,A are_equipotent
proof
  let X;
  consider R such that
A1: R well_orders X by WELLORD2:17;
  set Q = R|_2 X;
  take A = order_type_of Q;
  Q is well-ordering by A1,WELLORD2:16;
  then Q,RelIncl A are_isomorphic by WELLORD2:def 2;
  then consider f such that
A2: f is_isomorphism_of Q,RelIncl A by WELLORD1:def 8;
  take f;
  dom f = field Q & rng f = field RelIncl A by A2,WELLORD1:def 7;
  hence thesis by A1,A2,WELLORD1:def 7,WELLORD2:16,def 1;
end;

theorem Th2:
  M = N iff M,N are_equipotent
proof
  thus M = N implies M,N are_equipotent;
  consider A such that
A1: M = A and
A2: for C st C,A are_equipotent holds A c= C by Def1;
  consider B such that
A3: N = B and
A4: for C st C,B are_equipotent holds B c= C by Def1;
  assume M,N are_equipotent;
  then B c= A & A c= B by A1,A2,A3,A4;
  hence thesis by A1,A3,XBOOLE_0:def 10;
end;

theorem
  M in N iff M c= N & M <> N
proof
  M c< N iff M c= N & M <> N by XBOOLE_0:def 8;
  hence thesis by ORDINAL1:11,def 2;
end;

theorem
  M in N iff not N c= M by ORDINAL1:5,16;

definition
  let X;
  func card X -> Cardinal means
  :Def2:
  X,it are_equipotent;
  existence
  proof
    defpred P[Ordinal] means $1,X are_equipotent;
A1: ex A st P[A] by Th1;
    consider A such that
A2: P[A] & for B st P[B] holds A c= B from ORDINAL1:sch 1(A1);
    A is cardinal
    proof
      take A;
      thus A = A;
      let B;
      assume B,A are_equipotent;
      hence thesis by A2,WELLORD2:15;
    end;
    then reconsider M = A as Cardinal;
    take M;
    thus thesis by A2;
  end;
  uniqueness
  proof
    let M,N;
    assume X,M are_equipotent & X,N are_equipotent;
    then M,N are_equipotent by WELLORD2:15;
    hence thesis by Th2;
  end;
  projectivity;
end;

registration
  cluster empty -> cardinal for set;
  coherence
  proof
    let S be set;
    assume
A1: S is empty;
    take {};
    thus S = {} by A1;
    let A such that
    A,{} are_equipotent;
    thus thesis;
  end;
end;

registration
  let X be empty set;
  cluster card X -> empty;
  coherence by Def2;
end;

registration
  let X be non empty set;
  cluster card X -> non empty;
  coherence
  proof
    assume card X is empty;
    then X,{} are_equipotent by Def2;
    then ex f st f is one-to-one & dom f = X & rng f = {} by WELLORD2:def 4;
    hence contradiction by RELAT_1:42;
  end;
end;

theorem Th5:
  X,Y are_equipotent iff card X = card Y
proof
A1: Y,card Y are_equipotent by Def2;
A2: X,card X are_equipotent by Def2;
  thus X,Y are_equipotent implies card X = card Y
  proof
    assume X,Y are_equipotent;
    then card X,Y are_equipotent by A2,WELLORD2:15;
    hence thesis by Def2;
  end;
  assume card X = card Y;
  hence thesis by A2,A1,WELLORD2:15;
end;

theorem Th6:
  R is well-ordering implies field R,order_type_of R
  are_equipotent
proof
  assume R is well-ordering;
  then R,RelIncl order_type_of R are_isomorphic by WELLORD2:def 2;
  then consider f such that
A1: f is_isomorphism_of R,RelIncl order_type_of R by WELLORD1:def 8;
  take f;
  field RelIncl order_type_of R = order_type_of R by WELLORD2:def 1;
  hence thesis by A1,WELLORD1:def 7;
end;

theorem Th7:
  X c= M implies card X c= M
proof
  defpred P[Ordinal] means $1 c= M & X,$1 are_equipotent;
  reconsider m = M as Ordinal;
  assume X c= M;
  then
A1: order_type_of RelIncl X c= m & RelIncl X is well-ordering by WELLORD2:8,14;
  field RelIncl X = X by WELLORD2:def 1;
  then
A2: ex A st P[A] by A1,Th6;
  consider A such that
A3: P[A] & for B st P[B] holds A c= B from ORDINAL1:sch 1(A2);
  A is cardinal
  proof
    take A;
    thus A = A;
    let B;
    assume
A4: B,A are_equipotent;
    assume
A5: not A c= B;
    then m c= B by A3,A4,WELLORD2:15;
    hence contradiction by A3,A5,XBOOLE_1:1;
  end;
  then reconsider A as Cardinal;
  card X = A by A3,Def2;
  hence thesis by A3;
end;

theorem Th8:
  card A c= A
proof
  ( ex B st card A = B & for C st C,B are_equipotent holds B c= C)& A,card
  A are_equipotent by Def1,Def2;
  hence thesis;
end;

theorem
  X in M implies card X in M by Th8,ORDINAL1:12;


theorem Th10:
  card X c= card Y iff ex f st f is one-to-one & dom f = X & rng f c= Y
proof
  thus card X c= card Y implies ex f st f is one-to-one & dom f = X & rng f c=
  Y
  proof
    X,card X are_equipotent by Def2;
    then consider f such that
A1: f is one-to-one and
A2: dom f = X & rng f = card X by WELLORD2:def 4;
    assume
A3: card X c= card Y;
    Y,card Y are_equipotent by Def2;
    then consider g such that
A4: g is one-to-one and
A5: dom g = Y & rng g = card Y by WELLORD2:def 4;
    take h = g"*f;
    thus h is one-to-one by A1,A4;
    rng g = dom(g") & dom g = rng(g") by A4,FUNCT_1:33;
    hence thesis by A3,A2,A5,RELAT_1:26,27;
  end;
  given f such that
A6: f is one-to-one and
A7: dom f = X & rng f c= Y;
  Y,card Y are_equipotent by Def2;
  then consider g such that
A8: g is one-to-one and
A9: dom g = Y and
A10: rng g = card Y by WELLORD2:def 4;
A11: X,rng(g*f) are_equipotent
  proof
    take g*f;
    thus g*f is one-to-one by A6,A8;
    thus dom(g*f) = X by A7,A9,RELAT_1:27;
    thus thesis;
  end;
  rng(g*f),card rng(g*f) are_equipotent by Def2;
  then
A12: X,card rng(g*f) are_equipotent by A11,WELLORD2:15;
  card rng(g*f) c= card Y by A10,Th7,RELAT_1:26;
  hence thesis by A12,Def2;
end;

theorem Th11:
  X c= Y implies card X c= card Y
proof
  assume
A1: X c= Y;
  ex f st f is one-to-one & dom f = X & rng f c= Y
  proof
    take id X;
    thus thesis by A1,RELAT_1:45;
  end;
  hence thesis by Th10;
end;

theorem Th12:
  card X c= card Y iff ex f st dom f = Y & X c= rng f
proof
  thus card X c= card Y implies ex f st dom f = Y & X c= rng f
  proof
    assume card X c= card Y;
    then consider f such that
A1: f is one-to-one and
A2: dom f = X and
A3: rng f c= Y by Th10;
    defpred P[set,set] means $1 in rng f & $2 = f".$1 or not $1 in rng f & $2
    = 0;
A4: for x st x in Y ex y st P[x,y]
    proof
      let x such that
      x in Y;
      not x in rng f implies thesis;
      hence thesis;
    end;
A5: for x,y,z st x in Y & P[x,y] & P[x,z] holds y = z;
    consider g such that
A6: dom g = Y & for y st y in Y holds P[y,g.y] from FUNCT_1:sch 2(A5,
    A4);
    take g;
    thus dom g = Y by A6;
    let x;
    assume
A7: x in X;
    then
A8: f.x in rng f by A2,FUNCT_1:def 3;
    f".(f.x) = x by A1,A2,A7,FUNCT_1:34;
    then x = g.(f.x) by A3,A6,A8;
    hence thesis by A3,A6,A8,FUNCT_1:def 3;
  end;
  given f such that
A9: dom f = Y and
A10: X c= rng f;
  deffunc f(set) = f"{$1};
  consider g such that
A11: dom g = X & for x st x in X holds g.x = f(x) from FUNCT_1:sch 3;
  X <> {} implies thesis
  proof
    assume X <> {};
    then reconsider M = rng g as non empty set by A11,RELAT_1:42;
    for Z st Z in M holds Z <> {}
    proof
      let Z;
      assume Z in M;
      then consider x such that
A12:  x in dom g & Z = g.x by FUNCT_1:def 3;
A13:  x in {x} by TARSKI:def 1;
      Z = f"{x} & ex y st y in dom f & x = f.y by A10,A11,A12,FUNCT_1:def 3;
      hence thesis by A13,FUNCT_1:def 7;
    end;
    then consider F being Function such that
A14: dom F = M and
A15: for Z st Z in M holds F.Z in Z by FUNCT_1:111;
A16: dom(F*g) = X by A11,A14,RELAT_1:27;
A17: F*g is one-to-one
    proof
      let x,y;
      assume that
A18:  x in dom(F*g) and
A19:  y in dom(F*g) and
A20:  (F*g).x = (F*g).y;
A21:  g.y = f"{y} by A11,A16,A19;
      then f"{y} in M by A11,A16,A19,FUNCT_1:def 3;
      then F.(f"{y}) in f"{y} by A15;
      then
A22:  f.(F.(f"{y})) in {y} by FUNCT_1:def 7;
A23:  g.x = f"{x} by A11,A16,A18;
      then f"{x} in M by A11,A16,A18,FUNCT_1:def 3;
      then F.(f"{x}) in f"{x} by A15;
      then f.(F.(f"{x})) in {x} by FUNCT_1:def 7;
      then
A24:  f.(F.(f"{x})) = x by TARSKI:def 1;
      (F*g).x = F.(g.x) & (F*g).y = F. ( g.y) by A11,A16,A18,A19,FUNCT_1:13;
      hence thesis by A20,A23,A21,A22,A24,TARSKI:def 1;
    end;
    rng(F*g) c= Y
    proof
      let x;
      assume x in rng(F*g);
      then consider y such that
A25:  y in dom(F*g) and
A26:  x = (F*g).y by FUNCT_1:def 3;
A27:  x = F.(g.y) by A11,A16,A25,A26,FUNCT_1:13;
A28:  g.y = f"{y} by A11,A16,A25;
      then f"{y} in M by A11,A16,A25,FUNCT_1:def 3;
      then x in f"{y} by A15,A28,A27;
      hence thesis by A9,FUNCT_1:def 7;
    end;
    hence thesis by A16,A17,Th10;
  end;
  hence thesis;
end;

theorem Th13:
  not X,bool X are_equipotent
proof
  given f such that
  f is one-to-one and
A1: dom f = X & rng f = bool X;
  defpred P[set] means for Y st Y = f.$1 holds not $1 in Y;
  consider Z such that
A2: a in Z iff a in X & P[a] from XBOOLE_0:sch 1;
  Z c= X
  proof
    let a;
    thus thesis by A2;
  end;
  then consider a such that
A3: a in X and
A4: Z = f.a by A1,FUNCT_1:def 3;
  not a in Z by A2,A4;
  then ex Y st Y = f.a & a in Y by A2,A3;
  hence contradiction by A2,A4;
end;


theorem Th14:
  card X in card bool X
proof
  deffunc f(set) = {$1};
  consider f such that
A1: dom f = X & for x st x in X holds f.x = f(x) from FUNCT_1:sch 3;
A2: rng f c= bool X
  proof
    let x;
    assume x in rng f;
    then consider y such that
A3: y in dom f and
A4: x = f.y by FUNCT_1:def 3;
A5: { y } c= X
    proof
      let z;
      assume z in { y };
      hence thesis by A1,A3,TARSKI:def 1;
    end;
    f.y = { y } by A1,A3;
    hence thesis by A4,A5;
  end;
  not X,bool X are_equipotent by Th13;
  then
A6: card X <> card bool X by Th5;
  f is one-to-one
  proof
    let x,y;
    assume that
A7: x in dom f & y in dom f and
A8: f.x = f.y;
    f.x = { x } & f.y = { y } by A1,A7;
    hence thesis by A8,ZFMISC_1:3;
  end;
  then card X c= card bool X by A1,A2,Th10;
  then card X c< card bool X by A6,XBOOLE_0:def 8;
  hence thesis by ORDINAL1:11;
end;

definition
  let X;
  func nextcard X -> Cardinal means
  :Def3:
  card X in it & for M st card X in M
  holds it c= M;
  existence
  proof
    defpred P[Ordinal] means ex M st $1 = M & card X in M;
    card X in card bool X by Th14;
    then
A1: ex A st P[A];
    consider A such that
A2: P[A] and
A3: for B st P[B] holds A c= B from ORDINAL1:sch 1(A1);
    consider M such that
A4: A = M and
A5: card X in M by A2;
    take M;
    thus card X in M by A5;
    let N;
    assume card X in N;
    hence thesis by A3,A4;
  end;
  uniqueness
  proof
    let M1,M2 be Cardinal;
    assume card X in M1 & ( for M st card X in M holds M1 c= M)& card X in M2
    & for M st card X in M holds M2 c= M;
    then M1 c= M2 & M2 c= M1;
    hence thesis by XBOOLE_0:def 10;
  end;
end;

theorem
  {} in nextcard X
proof
  card {} c= card X & card X in nextcard X by Def3;
  hence thesis by ORDINAL1:12;
end;

theorem Th16:
  card X = card Y implies nextcard X = nextcard Y
proof
  assume
A1: card X = card Y;
  card X in nextcard X & for N st card X in N holds nextcard X c= N by Def3;
  hence thesis by A1,Def3;
end;

theorem Th17:
  X,Y are_equipotent implies nextcard X = nextcard Y
proof
  assume X,Y are_equipotent;
  then card X = card Y by Th5;
  hence thesis by Th16;
end;

theorem Th18:
  A in nextcard A
proof
  assume not A in nextcard A;
  then
A1: card nextcard A c= card A by Th11,ORDINAL1:16;
  A,card A are_equipotent by Def2;
  then
A2: nextcard card A = nextcard A by Th17;
A3: card card A in nextcard card A by Def3;
  card nextcard A = nextcard A by Def2;
  hence contradiction by A1,A2,A3,ORDINAL1:5;
end;

reserve S for T-Sequence;

definition
  let M;
  attr M is limit_cardinal means
  :Def4:
  not ex N st M = nextcard N;
end;

definition
  let A;
  func alef A -> set means
  :Def5:
  ex S st it = last S & dom S = succ A & S.{}
= card omega & (for B st succ B in succ A holds S.succ B = nextcard union { S.B
}) & for B st B in succ A & B <> {} & B is limit_ordinal holds S.B = card sup(S
  |B);
  correctness
  proof
    set B = card omega;
    deffunc D(Ordinal,T-Sequence) = card sup $2;
    deffunc C(Ordinal,set) = nextcard union {$2};
    thus (ex x,S st x = last S & dom S = succ A & S.{} = B & (for C st succ C
in succ A holds S.succ C = C(C,S.C)) & for C st C in succ A & C <> {} & C is
    limit_ordinal holds S.C = D(C,S|C) ) & for x1,x2 being set st (ex S st x1 =
last S & dom S = succ A & S.{} = B & (for C st succ C in succ A holds S.succ C
= C(C,S.C)) & for C st C in succ A & C <> {} & C is limit_ordinal holds S.C = D
(C,S|C) ) & (ex S st x2 = last S & dom S = succ A & S.{} = B & (for C st succ C
in succ A holds S.succ C = C(C,S.C)) & for C st C in succ A & C <> {} & C is
    limit_ordinal holds S.C = D(C,S|C) ) holds x1 = x2 from ORDINAL2:sch 7;
  end;
end;

Lm1: now
  deffunc D(Ordinal,T-Sequence) = card sup $2;
  deffunc C(Ordinal,set) = nextcard union { $2 };
  deffunc F(Ordinal) = alef $1;

A1: for A,x holds x = F(A) iff ex S st x = last S & dom S = succ A & S.{} =

card omega & (for C st succ C in succ A holds S.succ C = C(C,S.C)) & for C st C
  in succ A & C <> {} & C is limit_ordinal holds S.C = D(C,S|C) by Def5;

  F({}) = card omega from ORDINAL2:sch 8(A1);
  hence alef 0 = card omega;
  thus for A holds F(succ A) = C(A,F(A)) from ORDINAL2:sch 9(A1);

  thus A <> {} & A is limit_ordinal implies for S st dom S = A & for B st B in
  A holds S.B = alef B holds alef A = card sup S

  proof
    assume
A2: A <> {} & A is limit_ordinal;
    let S such that
A3: dom S = A and
A4: for B st B in A holds S.B = F(B);
    thus F(A) = D(A,S) from ORDINAL2:sch 10(A1,A2,A3,A4);
  end;
end;
deffunc f(Ordinal) = alef $1;

registration
  let A;
  cluster alef A -> cardinal;
  coherence
  proof
A1: now
      consider S such that
A2:   dom S = A & for B st B in A holds S.B = f(B) from ORDINAL2:sch 2;
      assume that
A3:   A <> {} and
A4:   for B holds A <> succ B;
      A is limit_ordinal by A4,ORDINAL1:29;
      then alef A = card sup S by A3,A2,Lm1;
      hence alef A is Cardinal;
    end;
    now
      given B such that
A5:   A = succ B;
      alef A = nextcard union {alef B} by A5,Lm1;
      hence thesis;
    end;
    hence thesis by A1,Lm1;
  end;
end;

theorem Th19:
  alef succ A = nextcard alef A
proof
  thus alef succ A = nextcard union { alef A } by Lm1
    .= nextcard alef A by ZFMISC_1:25;
end;

theorem
  A <> {} & A is limit_ordinal implies for S st dom S = A & for B st B
  in A holds S.B = alef B holds alef A = card sup S by Lm1;

theorem Th21:
  A in B iff alef A in alef B
proof
  defpred P[Ordinal] means for A st A in $1 holds alef A in alef $1;
A1: for B st P[B] holds P[succ B]
  proof
    let B such that
A2: P[B];
    let A;
A3: alef succ B = nextcard alef B by Th19;
A4: now
      assume A in B;
      then
A5:   alef A in alef B by A2;
      alef B in nextcard alef B by Th18;
      hence thesis by A3,A5,ORDINAL1:10;
    end;
A6: A c< B iff A c= B & A <> B by XBOOLE_0:def 8;
    assume A in succ B;
    hence thesis by A3,A6,A4,Th18,ORDINAL1:11,22;
  end;
A7: for B st B <> {} & B is limit_ordinal & for C st C in B holds P[C]
  holds P[B]
  proof
    let B such that
A8: B <> {} and
A9: B is limit_ordinal and
    for C st C in B for A st A in C holds alef A in alef C;
    let A;
    consider S such that
A10: dom S = B & for C st C in B holds S.C = f(C) from ORDINAL2:sch 2;
    assume A in B;
    then succ A in B by A9,ORDINAL1:28;
    then S.succ A in rng S & S.succ A = alef succ A by A10,FUNCT_1:def 3;
    then
A11: alef succ A in sup rng S by ORDINAL2:19;
    sup rng S = sup S by ORDINAL2:26;
    then
A12: alef succ A c= sup S by A11,ORDINAL1:def 2;
A13: card alef succ A = alef succ A by Def2;
A14: alef succ A = nextcard alef A & alef A in nextcard alef A by Th18,Th19;
    alef B = card sup S by A8,A9,A10,Lm1;
    then alef succ A c= alef B by A12,A13,Th11;
    hence thesis by A14;
  end;
A15: P[{}];
A16: for B holds P[B] from ORDINAL2:sch 1(A15,A1,A7);
  hence A in B implies alef A in alef B;
  assume
A17: alef A in alef B;
  then
A18: alef A <> alef B;
  not B in A by A16,A17;
  hence thesis by A18,ORDINAL1:14;
end;

theorem Th22:
  alef A = alef B implies A = B
proof
  assume
A1: alef A = alef B;
A2: now
    assume B in A;
    then alef B in alef A by Th21;
    hence contradiction by A1;
  end;
  now
    assume A in B;
    then alef A in alef B by Th21;
    hence contradiction by A1;
  end;
  hence thesis by A2,ORDINAL1:14;
end;

theorem
  A c= B iff alef A c= alef B
proof
A1: A c< B iff A <> B & A c= B by XBOOLE_0:def 8;
A2: alef A c< alef B iff alef A <> alef B & alef A c= alef B by XBOOLE_0:def 8;
  A in B iff alef A in alef B by Th21;
  hence thesis by A1,A2,Th22,ORDINAL1:11,def 2;
end;

theorem
  X c= Y & Y c= Z & X,Z are_equipotent implies X,Y are_equipotent & Y,Z
  are_equipotent
proof
  assume that
A1: X c= Y & Y c= Z and
A2: X,Z are_equipotent;
A3: card X = card Z by A2,Th5;
  card X c= card Y & card Y c= card Z by A1,Th11;
  then card X = card Y by A3,XBOOLE_0:def 10;
  hence thesis by A3,Th5;
end;

theorem
  bool Y c= X implies card Y in card X & not Y,X are_equipotent
proof
  assume bool Y c= X;
  then
A1: card bool Y c= card X by Th11;
  card Y in card bool Y by Th14;
  hence card Y in card X by A1;
  then card Y <> card X;
  hence thesis by Th5;
end;

theorem Th26:
  X,{} are_equipotent implies X = {}
proof
  given f such that
  f is one-to-one and
A1: dom f = X & rng f = {};
  thus thesis by A1,RELAT_1:42;
end;

theorem
  card {} = {};

theorem Th28:
  X,{x} are_equipotent iff ex x st X = { x }
proof
  thus X,{x} are_equipotent implies ex x st X = { x }
  proof
    assume X,{x} are_equipotent;
    then consider f such that
    f is one-to-one and
A1: dom f = { x } and
A2: rng f = X by WELLORD2:def 4;
    rng f = { f.x } by A1,FUNCT_1:4;
    hence thesis by A2;
  end;
  given y such that
A3: X = { y };
  take f = X --> x;
A4: dom f = X by FUNCOP_1:13;
  thus f is one-to-one
  proof
    let a,b;
    assume that
A5: a in dom f and
A6: b in dom f and
    f.a = f.b;
    a = y by A3,A4,A5,TARSKI:def 1;
    hence thesis by A3,A4,A6,TARSKI:def 1;
  end;
  thus dom f = X by FUNCOP_1:13;
  thus rng f c= { x } by FUNCOP_1:13;
  let a;
  assume a in { x };
  then
A7: a = x by TARSKI:def 1;
A8: y in { y } by TARSKI:def 1;
  then f.y = x by A3,FUNCOP_1:7;
  hence thesis by A3,A4,A7,A8,FUNCT_1:def 3;
end;

theorem Th29:
  card X = card { x } iff ex x st X = { x }
proof
  card X = card { x } iff X,{x} are_equipotent by Th5;
  hence thesis by Th28;
end;

theorem Th30:
  card { x } = 1
proof
A1: 1 = succ 0;
  1 is cardinal
  proof
    take IT = 1;
    thus 1 = IT;
    let A;
    assume A,IT are_equipotent;
    then ex y st A = { y } by A1,Th28;
    hence thesis by A1,ZFMISC_1:33;
  end;
  then reconsider M = 1 as Cardinal;
  { x },M are_equipotent by A1,Th28;
  hence thesis by Def2;
end;

theorem Th31:
  X misses X1 & Y misses Y1 & X,Y are_equipotent & X1,Y1
  are_equipotent implies X \/ X1,Y \/ Y1 are_equipotent
proof
  assume that
A1: X /\ X1 = {} and
A2: Y /\ Y1 = {};
  given f such that
A3: f is one-to-one and
A4: dom f = X and
A5: rng f = Y;
  given g such that
A6: g is one-to-one and
A7: dom g = X1 and
A8: rng g = Y1;
  defpred P[set,set] means $1 in X & $2 = f.$1 or $1 in X1 & $2 = g.$1;
A9: for x st x in X \/ X1 ex y st P[x,y]
  proof
    let x;
    assume x in X \/ X1;
    then x in X or x in X1 by XBOOLE_0:def 3;
    hence thesis;
  end;
A10: for x,y,z st x in X \/ X1 & P[x,y] & P[x,z] holds y = z by A1,
XBOOLE_0:def 4;
  consider h such that
A11: dom h = X \/ X1 and
A12: for x st x in X \/ X1 holds P[x,h.x] from FUNCT_1:sch 2(A10,A9);
  take h;
  thus h is one-to-one
  proof
    let x,y;
    assume that
A13: x in dom h and
A14: y in dom h and
A15: h.x = h.y;
A16: y in X & h.y = f.y or y in X1 & h.y = g.y by A11,A12,A14;
A17: x in X & h.x = f.x or x in X1 & h.x = g.x by A11,A12,A13;
A18: now
      assume
A19:  y in X & x in X1;
      then f.y in Y & g.x in Y1 by A4,A5,A7,A8,FUNCT_1:def 3;
      hence contradiction by A1,A2,A15,A17,A16,A19,XBOOLE_0:def 4;
    end;
    now
      assume
A20:  x in X & y in X1;
      then f.x in Y & g.y in Y1 by A4,A5,A7,A8,FUNCT_1:def 3;
      hence contradiction by A1,A2,A15,A17,A16,A20,XBOOLE_0:def 4;
    end;
    hence thesis by A3,A4,A6,A7,A15,A17,A16,A18,FUNCT_1:def 4;
  end;
  thus dom h = X \/ X1 by A11;
  thus rng h c= Y \/ Y1
  proof
    let x;
    assume x in rng h;
    then consider y such that
A21: y in dom h and
A22: x = h.y by FUNCT_1:def 3;
A23: y in X & x = f.y or y in X1 & x = g.y by A11,A12,A21,A22;
A24: now
      assume y in X1;
      then x in Y1 by A1,A7,A8,A23,FUNCT_1:def 3,XBOOLE_0:def 4;
      hence thesis by XBOOLE_0:def 3;
    end;
    now
      assume y in X;
      then x in Y by A1,A4,A5,A23,FUNCT_1:def 3,XBOOLE_0:def 4;
      hence thesis by XBOOLE_0:def 3;
    end;
    hence thesis by A11,A21,A24,XBOOLE_0:def 3;
  end;
  let x such that
A25: x in Y \/ Y1;
A26: now
    assume x in Y1;
    then consider y such that
A27: y in dom g and
A28: x = g.y by A8,FUNCT_1:def 3;
A29: y in X \/ X1 by A7,A27,XBOOLE_0:def 3;
    then P[y,h.y] by A12;
    hence thesis by A1,A7,A11,A27,A28,A29,FUNCT_1:def 3,XBOOLE_0:def 4;
  end;
  now
    assume x in Y;
    then consider y such that
A30: y in dom f and
A31: x = f.y by A5,FUNCT_1:def 3;
A32: y in X \/ X1 by A4,A30,XBOOLE_0:def 3;
    then P[y,h.y] by A12;
    hence thesis by A1,A4,A11,A30,A31,A32,FUNCT_1:def 3,XBOOLE_0:def 4;
  end;
  hence thesis by A25,A26,XBOOLE_0:def 3;
end;

theorem Th32:
  x in X & y in X implies X \ { x },X \ { y } are_equipotent
proof
  assume that
A1: x in X and
A2: y in X;
  defpred P[set,set] means $1 = y & $2 = x or $1 <> y & $1 = $2;
A3: for a st a in X \ { x } ex b st P[a,b]
  proof
    let a such that
    a in X \ { x };
    a = y implies thesis;
    hence thesis;
  end;
A4: for a,b1,b2 st a in X \ { x } & P[a,b1] & P[a,b2] holds b1 = b2;
  consider f such that
A5: dom f = X \ { x } & for a st a in X \ { x } holds P[a,f.a] from
  FUNCT_1:sch 2(A4, A3);
  take f;
  thus f is one-to-one
  proof
    let b1,b2;
    assume that
A6: b1 in dom f & b2 in dom f and
A7: f.b1 = f.b2 & b1 <> b2;
A8: ( not b1 in { x })& not b2 in { x } by A5,A6,XBOOLE_0:def 5;
    ( P[b1,f.b1])& P[b2,f.b2] by A5,A6;
    hence thesis by A7,A8,TARSKI:def 1;
  end;
  thus dom f = X \ { x } by A5;
  thus rng f c= X \ { y }
  proof
    let z;
    assume z in rng f;
    then consider a such that
A9: a in dom f and
A10: z = f.a by FUNCT_1:def 3;
A11: now
      assume
A12:  a = y;
      then ( not y in { x })& z = x by A5,A9,A10,XBOOLE_0:def 5;
      then y <> z by TARSKI:def 1;
      then
A13:  not z in { y } by TARSKI:def 1;
      z in X by A1,A5,A9,A10,A12;
      hence thesis by A13,XBOOLE_0:def 5;
    end;
    now
      assume a <> y;
      then ( not a in { y })& a = z by A5,A9,A10,TARSKI:def 1;
      hence thesis by A5,A9,XBOOLE_0:def 5;
    end;
    hence thesis by A11;
  end;
  let z;
  assume
A14: z in X \ { y };
  then not z in { y } by XBOOLE_0:def 5;
  then
A15: z <> y by TARSKI:def 1;
A16: now
    assume z <> x;
    then not z in { x } by TARSKI:def 1;
    then
A17: z in X \ { x } by A14,XBOOLE_0:def 5;
    then z = f.z by A5,A15;
    hence thesis by A5,A17,FUNCT_1:def 3;
  end;
  now
    assume
A18: z = x;
    then not y in { x } by A15,TARSKI:def 1;
    then
A19: y in dom f by A2,A5,XBOOLE_0:def 5;
    then z = f.y by A5,A18;
    hence thesis by A19,FUNCT_1:def 3;
  end;
  hence thesis by A16;
end;

theorem Th33:
  X c= dom f & f is one-to-one implies X,f.:X are_equipotent
proof
  assume that
A1: X c= dom f and
A2: f is one-to-one;
  take g = f|X;
  thus g is one-to-one by A2,FUNCT_1:52;
  thus dom g = X by A1,RELAT_1:62;
  thus thesis by RELAT_1:115;
end;

theorem Th34:
  X,Y are_equipotent & x in X & y in Y implies X \ { x },Y \ { y }
  are_equipotent
proof
  given f such that
A1: f is one-to-one and
A2: dom f = X and
A3: rng f = Y;
A4: X \ { x },f.:(X \ { x }) are_equipotent by A1,A2,Th33;
  assume that
A5: x in X and
A6: y in Y;
  f.x in Y by A2,A3,A5,FUNCT_1:def 3;
  then
A7: Y \ { f.x },Y \ { y } are_equipotent by A6,Th32;
  f.:(X \ { x }) = f.:X \ Im(f,x) by A1,FUNCT_1:64
    .= Y \ Im(f,x) by A2,A3,RELAT_1:113
    .= Y \ { f.x } by A2,A5,FUNCT_1:59;
  hence thesis by A4,A7,WELLORD2:15;
end;

theorem Th35:
  succ X, succ Y are_equipotent implies X, Y are_equipotent
proof
A1: X in succ X & Y in succ Y by ORDINAL1:6;
  X = succ X \ {X} & Y = succ Y \ {Y} by ORDINAL1:37;
  hence thesis by A1,Th34;
end;

theorem Th36:
  n = {} or ex m st n = succ m
proof
  defpred P[Nat] means $1 = {} or ex m st $1 = succ m;
A1: for a being Nat st P[a] holds P[succ a];
A2: P[{}];
  thus P[n] from ORDINAL2:sch 17(A2,A1);
end;

Lm2: n,m are_equipotent implies n = m
proof
  defpred P[Nat] means for n st n,$1 are_equipotent holds n = $1;
A1: for a being Nat st P[a] holds P[succ a]
  proof
    let a be Nat such that
A2: P[a];
    let n;
    assume
A3: n,succ a are_equipotent;
    per cases;
    suppose
      n = {};
      hence thesis by A3,Th26;
    end;
    suppose
      n <> {};
      then ex m st n = succ m by Th36;
      hence thesis by A2,A3,Th35;
    end;
  end;
A4: P[{}] by Th26;
  P[m] from ORDINAL2:sch 17(A4,A1);
  hence thesis;
end;

theorem Th37:
  x in omega implies x is cardinal
proof
  assume that
A1: x in omega and
A2: for B st x = B ex C st C,B are_equipotent & not B c= C;
  reconsider A = x as Ordinal by A1;
  consider B such that
A3: B,A are_equipotent and
A4: not A c= B by A2;
  B in A by A4,ORDINAL1:16;
  then B in omega by A1,ORDINAL1:10;
  then reconsider n = A, m = B as Nat by A1;
  n,m are_equipotent by A3;
  hence contradiction by A4,Lm2;
end;

registration
  cluster natural -> cardinal for number;
  correctness
  proof
    let n be number;
    assume n is natural;
    then n in omega by ORDINAL1:def 12;
    hence thesis by Th37;
  end;
end;

theorem Th38:
  X,Y are_equipotent & X is finite implies Y is finite
proof
  assume X,Y are_equipotent;
  then consider f such that
  f is one-to-one and
A1: dom f = X and
A2: rng f = Y by WELLORD2:def 4;
  given p being Function such that
A3: rng p = X and
A4: dom p in omega;
  take f*p;
  thus rng(f*p) = Y by A1,A2,A3,RELAT_1:28;
  thus thesis by A1,A3,A4,RELAT_1:27;
end;

theorem Th39:
  n is finite & card n is finite
proof
  reconsider n as Element of omega by ORDINAL1:def 12;
  rng id n = n & dom id n = n by RELAT_1:45;
  then n is finite by FINSET_1:def 1;
  hence thesis by Def2;
end;

theorem
  card n = card m implies n = m
proof
  assume
A1: card n = card m;
  reconsider n as Element of omega by ORDINAL1:def 12;
  card n = n by Def2;
  hence thesis by A1,Def2;
end;

theorem
  card n c= card m iff n c= m
proof
  card n = n by Def2;
  hence thesis by Def2;
end;

theorem Th42:
  card n in card m iff n in m
proof
  card n = n by Def2;
  hence thesis by Def2;
end;

theorem Th43:
  X is finite implies ex n st X,n are_equipotent
proof
  defpred P[set] means ex n st $1,n are_equipotent;
A1: P[{}];
A2: for Z,Y st Z in X & Y c= X & P[Y] holds P[Y \/ {Z}]
  proof
    let Z,Y such that
    Z in X and
    Y c= X;
    given n such that
A3: Y,n are_equipotent;
A4: not Z in Y implies thesis
    proof
      assume
A5:   not Z in Y;
      now
        set x = the Element of Y /\ { Z };
        assume Y /\ { Z } <> {};
        then x in Y & x in { Z } by XBOOLE_0:def 4;
        hence contradiction by A5,TARSKI:def 1;
      end;
      then
A6:   Y misses { Z } by XBOOLE_0:def 7;
A7:   now
        assume n meets { n };
        then consider x being set such that
A8:     x in n and
A9:     x in { n } by XBOOLE_0:3;
        x = n by A9,TARSKI:def 1;
        hence contradiction by A8;
      end;
      take succ n;
      { Z },{ n } are_equipotent by Th28;
      hence thesis by A3,A7,A6,Th31;
    end;
    Z in Y implies thesis
    proof
      assume
A10:  Z in Y;
      take n;
      { Z } c= Y by A10,ZFMISC_1:31;
      hence thesis by A3,XBOOLE_1:12;
    end;
    hence thesis by A4;
  end;
  assume
A11: X is finite;
  thus P[X] from FINSET_1:sch 2(A11,A1,A2);
end;

theorem Th44:
  nextcard card n = card succ n
proof
  reconsider sn = succ n as Nat;
A1: for M st card card n in M holds card succ n c= M
  proof
A2: card n = n by Def2;
    let M;
    assume card card n in M;
    then succ n c= M by A2,ORDINAL1:21;
    hence thesis by Def2;
  end;
  n in succ n by ORDINAL1:6;
  then card n in card sn by Th42;
  hence thesis by A1,Def3;
end;

definition
  let n be Nat;
  redefine func succ n -> Element of omega;
  coherence by ORDINAL1:def 12;
end;

definition
  let X be finite set;
  redefine func card X -> Element of omega;
  coherence
  proof
    consider n such that
A1: X,n are_equipotent by Th43;
    reconsider n as Element of omega by ORDINAL1:def 12;
    X,n are_equipotent by A1;
    hence thesis by Def2;
  end;
end;

theorem
  X is finite implies nextcard X is finite
proof
  assume X is finite;
  then reconsider X as finite set;
  card X = card card X;
  then
A1: card(succ card X) = nextcard card X by Th44;
  X,card X are_equipotent by Def2;
  then nextcard card X = nextcard X by Th17;
  hence thesis by A1,Th39;
end;

scheme
  CardinalInd { Sigma[set] }: for M holds Sigma[M]
provided
A1: Sigma[{}] and
A2: for M st Sigma[M] holds Sigma[nextcard M] and
A3: for M st M <> {} & M is limit_cardinal & for N st N in M holds Sigma
[N] holds Sigma[M]
proof
  let M;
  defpred P[Ordinal] means $1 is Cardinal implies Sigma[$1];
A4: for A st for B st B in A holds P[B] holds P[A]
  proof
    let A such that
A5: for B st B in A holds P[B] and
A6: A is Cardinal;
    reconsider M = A as Cardinal by A6;
A7: now
      assume not M is limit_cardinal;
      then consider N such that
A8:   M = nextcard N by Def4;
      N in M by A8,Th18;
      hence Sigma[M] by A2,A5,A8;
    end;
    now
      assume
A9:   M <> {} & M is limit_cardinal;
      for N st N in M holds Sigma[N] by A5;
      hence Sigma[M] by A3,A9;
    end;
    hence thesis by A1,A7;
  end;
  for A holds P[A] from ORDINAL1:sch 2(A4);
  hence thesis;
end;

scheme
  CardinalCompInd { Sigma[set] }: for M holds Sigma[M]
provided
A1: for M st for N st N in M holds Sigma[N] holds Sigma[M]
proof
  let M;
  defpred P[Ordinal] means $1 is Cardinal implies Sigma[$1];
A2: for A st for B st B in A holds P[B] holds P[A]
  proof
    let A such that
A3: for B st B in A holds B is Cardinal implies Sigma[B];
    assume A is Cardinal;
    then reconsider M = A as Cardinal;
    for N st N in M holds Sigma[N] by A3;
    hence thesis by A1;
  end;
  for A holds P[A] from ORDINAL1:sch 2(A2);
  hence thesis;
end;

theorem Th46:
  alef 0 = omega
proof
  thus alef 0 c= omega by Lm1,Th8;
  thus omega c= alef 0
  proof
    let x;
    assume
A1: x in omega;
    then reconsider A = x as Ordinal;
    consider n being Element of omega such that
A2: A = n by A1;
    succ n c= omega by ORDINAL1:21;
    then
A3: card succ n c= card omega by Th11;
    n in succ n & card succ n = succ n by Def2,ORDINAL1:6;
    hence thesis by A2,A3,Lm1;
  end;
end;

registration
  cluster omega -> cardinal for number;
  coherence by Th46;
end;

theorem
  card omega = omega by Lm1,Th46;

registration
  cluster omega -> limit_cardinal;
  coherence
  proof
    given N such that
A1: omega = nextcard N;
    N in omega by A1,Th18;
    then
A2: succ N in omega by ORDINAL1:def 12;
    reconsider n=N as Element of omega by A1,Th18;
A3: card succ n = succ n by Def2;
    nextcard card n = card succ n & card n = n by Def2,Th44;
    hence contradiction by A1,A2,A3;
  end;
end;

registration
  cluster -> finite for Element of omega;
  coherence by Th39;
end;

registration
  cluster finite for Cardinal;
  existence
  proof
    set n = the Element of omega;
    take n;
    thus thesis;
  end;
end;

theorem
  for M being finite Cardinal ex n st M = card n
proof
  let M be finite Cardinal;
  card M = M by Def2;
  hence thesis;
end;

registration
  let X be finite set;
  cluster card X -> finite;
  coherence;
end;

Lm3: A,n are_equipotent implies A = n
proof
  defpred P[Nat] means for A st A,$1 are_equipotent holds A = $1;
A1: for n st P[n] holds P[succ n]
  proof
    let n such that
A2: P[n];
    let A;
A3: n in (succ n) & (succ n) \ { n} = n by ORDINAL1:6,37;
    assume
A4: A,succ n are_equipotent;
    then A <> {} by Th26;
    then
A5: {} in A by ORDINAL3:8;
    now
      assume A is limit_ordinal;
      then
A6:   omega c= A by A5,ORDINAL1:def 11;
      card A = card (succ n) by A4,Th5;
      hence contradiction by A6,Lm1,Th11,Th46,ORDINAL1:5;
    end;
    then consider B such that
A7: A = succ B by ORDINAL1:29;
    B in A & A\{B} = B by A7,ORDINAL1:6,37;
    hence thesis by A2,A4,A7,A3,Th34;
  end;
A8: P[{}] by Th26;
  P[n] from ORDINAL2:sch 17(A8,A1);
  hence thesis;
end;

Lm4: A is finite iff A in omega
proof
  thus A is finite implies A in omega
  proof
    assume A is finite;
    then consider n such that
A1: A,n are_equipotent by Th43;
    A = n by A1,Lm3;
    hence thesis by ORDINAL1:def 12;
  end;
  assume A in omega;
  hence thesis;
end;

registration
  cluster omega -> infinite;
  coherence
  proof
    not omega in omega;
    hence thesis by Lm4;
  end;
end;

registration
  cluster infinite for set;
  existence
  proof
    take omega;
    thus thesis;
  end;
end;

registration
  let X be infinite set;
  cluster card X -> infinite;
  coherence
  proof
    X,card X are_equipotent by Def2;
    hence thesis by Th38;
  end;
end;

begin 

theorem Th49:
  1 = { 0 }
proof
  thus 1 = succ 0
    .= { 0 };
end;

theorem Th50:
  2 = { 0,1 }
proof
  thus 2 = succ 1
    .= { 0,1 } by Th49,ENUMSET1:1;
end;

theorem Th51:
  3 = { 0,1,2 }
proof
  thus 3 = succ 2
    .= { 0,1,2 } by Th50,ENUMSET1:3;
end;

theorem Th52:
  4 = { 0,1,2,3 }
proof
  thus 4 = succ 3
    .= { 0,1,2,3 } by Th51,ENUMSET1:6;
end;

theorem Th53:
  5 = { 0,1,2,3,4 }
proof
  thus 5 = succ 4
    .= { 0,1,2,3,4 } by Th52,ENUMSET1:10;
end;

theorem Th54:
  6 = { 0,1,2,3,4,5 }
proof
  thus 6 = succ 5
    .= { 0,1,2,3,4,5 } by Th53,ENUMSET1:15;
end;

theorem Th55:
  7 = { 0,1,2,3,4,5,6 }
proof
  thus 7 = succ 6
    .= { 0,1,2,3,4,5,6 } by Th54,ENUMSET1:21;
end;

theorem Th56:
  8 = { 0,1,2,3,4,5,6,7 }
proof
  thus 8 = succ 7
    .= { 0,1,2,3,4,5,6,7 } by Th55,ENUMSET1:28;
end;

theorem Th57:
  9 = { 0,1,2,3,4,5,6,7,8 }
proof
  thus 9 = succ 8
    .= { 0,1,2,3,4,5,6,7,8 } by Th56,ENUMSET1:84;
end;

theorem
  10 = { 0,1,2,3,4,5,6,7,8,9 }
proof
  thus 10 = succ 9
    .= { 0,1,2,3,4,5,6,7,8,9 } by Th57,ENUMSET1:85;
end;



theorem
  for f being Function st dom f is infinite & f is one-to-one holds rng
  f is infinite
proof
  let f be Function;
  assume that
A1: dom f is infinite and
A2: f is one-to-one;
  dom f,rng f are_equipotent by A2,WELLORD2:def 4;
  hence thesis by A1,Th38;
end;



reserve k,n,m for Nat;

definition

  let n;
  func Segm n -> set equals
  n;
  coherence;
end;

reserve l for Element of omega;

definition
  let n;
  redefine func Segm n -> Subset of omega;
  coherence
  proof
    n in omega by ORDINAL1:def 12;
    hence thesis by ORDINAL1:16;
  end;
end;



theorem
  A,n are_equipotent implies A = n by Lm3;

theorem
  A is finite iff A in omega by Lm4;

registration
  cluster natural -> finite for set;
  coherence
  proof
    let a be set;
    assume a in omega;
    hence thesis;
  end;
end;



registration
  let A be infinite set;
  cluster bool A -> infinite;
  coherence
  proof
    defpred P[set] means ex y st $1={y};
    consider X being set such that
A1: for x holds x in X iff x in bool A & P[x] from XBOOLE_0:sch 1;
    for x holds x in union X iff x in A
    proof
      let x;
      thus x in union X implies x in A
      proof
        assume x in union X;
        then consider B being set such that
A2:     x in B and
A3:     B in X by TARSKI:def 4;
        B in bool A by A1,A3;
        hence thesis by A2;
      end;
      assume x in A;
      then {x} c= A by ZFMISC_1:31;
      then
A4:   {x} in X by A1;
      x in {x} by TARSKI:def 1;
      hence thesis by A4,TARSKI:def 4;
    end;
    then
A5: union X = A by TARSKI:1;
A6: for B being set st B in X holds B is finite
    proof
      let B be set;
      assume B in X;
      then ex y st B = {y} by A1;
      hence thesis;
    end;
A7: X c= bool A
    proof
      let x;
      assume x in X;
      hence thesis by A1;
    end;
    assume bool A is finite;
    hence thesis by A5,A6,A7,FINSET_1:7;
  end;
  let B be non empty set;
  cluster [:A,B:] ->infinite;
  coherence
  proof
    deffunc F(set) = $1`1;
    consider f such that
A8: dom f = [:A,B:] and
A9: for x st x in [:A,B:] holds f.x = F(x) from FUNCT_1:sch 3;
    x in rng f iff x in A
    proof
      thus x in rng f implies x in A
      proof
        assume x in rng f;
        then consider y such that
A10:    y in dom f and
A11:    f.y = x by FUNCT_1:def 3;
        x = y`1 by A8,A9,A10,A11;
        hence thesis by A8,A10,MCART_1:10;
      end;
      set y = the Element of B;
      assume x in A;
      then
A12:  [x,y] in dom f by A8,ZFMISC_1:87;
      then f.([x,y]) = [x,y]`1 by A8,A9
        .= x;
      hence thesis by A12,FUNCT_1:def 3;
    end;
    then rng f = A by TARSKI:1;
    then
A13: f.:[:A,B:] = A by A8,RELAT_1:113;
    assume [:A,B:] is finite;
    hence contradiction by A13;
  end;
  cluster [:B,A:] ->infinite;
  coherence
  proof
    deffunc F(set) = $1`2;
    consider f such that
A14: dom f = [:B,A:] and
A15: for x st x in [:B,A:] holds f.x = F(x) from FUNCT_1:sch 3;
    y in rng f iff y in A
    proof
      thus y in rng f implies y in A
      proof
        assume y in rng f;
        then consider x such that
A16:    x in dom f and
A17:    f.x = y by FUNCT_1:def 3;
        y = x`2 by A14,A15,A16,A17;
        hence thesis by A14,A16,MCART_1:10;
      end;
      set x = the Element of B;
      assume y in A;
      then
A18:  [x,y] in dom f by A14,ZFMISC_1:87;
      [x,y]`2 = y;
      then f.([x,y]) = y by A14,A15,A18;
      hence thesis by A18,FUNCT_1:def 3;
    end;
    then rng f = A by TARSKI:1;
    then
A19: f.:[:B,A:] = A by A14,RELAT_1:113;
    assume [:B,A:] is finite;
    hence contradiction by A19;
  end;
end;

registration
  let X be infinite set;
  cluster infinite for Subset of X;
  existence
  proof
    X c= X;
    hence thesis;
  end;
end;



registration
  cluster finite ordinal -> natural for number;
  coherence
  proof
    let n be number;
    assume n is finite ordinal;
    then n in omega by Lm4;
    hence thesis;
  end;
end;

theorem Th62:
  for f being Function holds card f = card dom f
proof
  let f be Function;
  dom f,f are_equipotent
  proof
    deffunc F(set) = [$1,f.$1];
    consider g being Function such that
A1: dom g = dom f and
A2: for x st x in dom f holds g.x = F(x) from FUNCT_1:sch 3;
    take g;
    thus g is one-to-one
    proof
      let x,y;
      assume that
A3:   x in dom g and
A4:   y in dom g;
      assume g.x = g.y;
      then [x,f.x] = g.y by A1,A2,A3
        .= [y,f.y] by A1,A2,A4;
      hence thesis by XTUPLE_0:1;
    end;
    thus dom g = dom f by A1;
    thus rng g c= f
    proof
      let i be set;
      assume i in rng g;
      then consider x such that
A5:   x in dom g and
A6:   g.x = i by FUNCT_1:def 3;
      g.x = [x,f.x] by A1,A2,A5;
      hence thesis by A1,A5,A6,FUNCT_1:1;
    end;
    let x,y;
    assume
A7: [x,y] in f;
    then
A8: x in dom f by FUNCT_1:1;
    y = f.x by A7,FUNCT_1:1;
    then g.x = [x,y] by A2,A8;
    hence thesis by A1,A8,FUNCT_1:def 3;
  end;
  hence thesis by Th5;
end;

registration
  let X be finite set;
  cluster RelIncl X -> finite;
  coherence
  proof
    RelIncl X c= [:X,X:] by WELLORD2:23;
    hence thesis;
  end;
end;



theorem
  RelIncl X is finite implies X is finite
proof
  set R = RelIncl X;
  assume R is finite;
  then reconsider A = R as finite Relation;
  field A is finite;
  hence thesis by WELLORD2:def 1;
end;

theorem
 card(k -->x) = k
proof
  dom(k -->x) = k by FUNCOP_1:13;
 hence card(k -->x) = card k by Th62
     .= k by Def2;
end;

begin 

definition let N,X be set;
 attr X is N-element means
:Def7: card X = N;
end;

registration let N be Cardinal;
 cluster N-element for set;
 existence
  proof
   take N;
   thus card N = N by Def2;
  end;
end;

registration
 cluster 0-element -> empty for set;
 coherence
  proof let X be set;
   assume X is 0-element;
    then card X = 0 by Def7;
   hence X is empty;
  end;
 cluster empty -> 0-element for set;
 coherence
  proof let X be set;
   assume X is empty;
    then card X = {};
   hence X is 0-element by Def7;
  end;
end;

registration let x be set;
 cluster {x} -> 1-element;
 coherence
proof
   1 = succ 0;
   then { x },1 are_equipotent by Th28;
  hence card { x } = 1 by Def2;
end;
end;

registration let N be Cardinal;
 cluster N-element for Function;
 existence
  proof
   take f = N --> {{}};
    dom f = N by FUNCOP_1:13;
    then card dom f = N by Def2;
   hence card f = N by Th62;
  end;
end;

registration let N be Cardinal; let f be N-element Function;
 cluster dom f -> N-element;
 coherence
  proof
    card f = N by Def7;
   hence card dom f = N by Th62;
  end;
end;

registration
 cluster 1-element -> trivial non empty for set;
 coherence
  proof let X be set;
   assume X is 1-element;
    then card X = 1 by Def7
       .= card {0} by Th30;
    then ex x st X = {x} by Th29;
   hence thesis;
  end;
 cluster trivial non empty -> 1-element for set;
 coherence
  proof let X be set;
   assume X is trivial non empty;
    then ex x st X = {x} by ZFMISC_1:131;
   hence thesis;
  end;
end;

registration let X be non empty set;
 cluster 1-element for Subset of X;
 existence
  proof
   take the non empty trivial Subset of X;
   thus thesis;
  end;
end;

definition let X be non empty set;
 mode Singleton of X is 1-element Subset of X;
end;

theorem
 for X being non empty set, A being Singleton of X
  ex x being Element of X st A = {x}
  proof let X be non empty set, A be Singleton of X;
    consider x such that
A1:   A = {x} by ZFMISC_1:131;
    x in A by A1,TARSKI:def 1;
    then reconsider x as Element of X;
   take x;
   thus thesis by A1;
  end;

theorem Th66:
  card X c= card Y iff ex f st X c= f.:Y
proof
  deffunc G(set) = $1;
  thus card X c= card Y implies ex f st X c= f.:Y
  proof
    assume card X c= card Y;
    then consider f such that
A1: dom f = Y & X c= rng f by Th12;
    take f;
    thus thesis by A1,RELAT_1:113;
  end;
  given f such that
A2: X c= f.:Y;
  defpred C[set] means $1 in dom f;
  deffunc F(set) = f.$1;
  consider g such that
A3: dom g = Y & for x st x in Y holds (C[x] implies g.x = F(x)) & (not C
  [x] implies g.x = G(x)) from PARTFUN1:sch 1;
  X c= rng g
  proof
    let x;
    assume x in X;
    then consider y such that
A4: y in dom f and
A5: y in Y and
A6: x = f.y by A2,FUNCT_1:def 6;
    x = g.y by A3,A4,A5,A6;
    hence thesis by A3,A5,FUNCT_1:def 3;
  end;
  hence thesis by A3,Th12;
end;

theorem
  card (f.:X) c= card X by Th66;

theorem
  card X in card Y implies Y \ X <> {}
proof
  assume that
A1: card X in card Y and
A2: Y \ X = {};
  Y c= X by A2,XBOOLE_1:37;
  then card Y c= card X by Th11;
  hence contradiction by A1,ORDINAL1:5;
end;

theorem
  X,[:X,{x}:] are_equipotent & card X = card [:X,{x}:]
proof
  deffunc f(set) = [$1,x];
  consider f such that
A1: dom f = X & for y st y in X holds f.y = f(y) from FUNCT_1:sch 3;
  thus X,[:X,{x}:] are_equipotent
  proof
    take f;
    thus f is one-to-one
    proof
      let y,z;
      assume that
A2:   y in dom f & z in dom f and
A3:   f.y = f.z;
A4:   [y,x]`1 = y;
      f.y = [y,x] & f.z = [z,x] by A1,A2;
      hence thesis by A3,A4,MCART_1:7;
    end;
    thus dom f = X by A1;
    thus rng f c= [:X,{x}:]
    proof
      let y;
A5:   x in {x} by TARSKI:def 1;
      assume y in rng f;
      then consider z such that
A6:   z in dom f and
A7:   y = f.z by FUNCT_1:def 3;
      y = [z,x] by A1,A6,A7;
      hence thesis by A1,A6,A5,ZFMISC_1:87;
    end;
    let y;
    assume y in [:X,{x}:];
    then consider y1,y2 being set such that
A8: y1 in X and
A9: y2 in {x} and
A10: y = [y1,y2] by ZFMISC_1:84;
    y2 = x by A9,TARSKI:def 1;
    then y = f.y1 by A1,A8,A10;
    hence thesis by A1,A8,FUNCT_1:def 3;
  end;
  hence thesis by Th5;
end;



theorem
  for f being Function st f is one-to-one holds card dom f = card rng f
proof
  let f be Function;
  assume f is one-to-one;
  then
A1: dom f, f .: dom f are_equipotent by Th33;
  f .: dom f = rng f by RELAT_1:113;
  hence thesis by A1,Th5;
end;















begin

reserve X,Y,Z,X1,X2,Y1,Y2,x,y,z,t,x1,x2 for set,
  f,g,h,f1,f2,g1,g2 for Function;

scheme
  LambdaFS { FS()->set, f(set)->set }: ex f st dom f = FS() & for g st g in FS
  () holds f.g = f(g) proof
   consider f such that
A1: dom f = FS() & for x st x in FS() holds f.x = f(x) from FUNCT_1:sch
  3;
  take f;
  thus thesis by A1;
end;

theorem Th1:
  ~{} = {}
proof
  [:{} qua set,{}:] = {} & dom {} = {} by ZFMISC_1:90;
  then dom ~{} = {} by FUNCT_4:46;
  hence thesis;
end;

canceled 6;

theorem Th8:
  proj1 {} = {} & proj2 {} = {};

theorem Th9:
  Y <> {} or [:X,Y:] <> {} or [:Y,X:] <> {} implies proj1 [:X,Y:]
  = X & proj2 [:Y,X:] = X
proof
  set y = the Element of Y;
  assume Y <> {} or [:X,Y:] <> {} or [:Y,X:] <> {};
  then
A1: Y <> {} by ZFMISC_1:90;
  now
    let x;
    x in X implies [x,y] in [:X,Y:] by A1,ZFMISC_1:87;
    hence x in X iff ex y st [x,y] in [:X,Y:] by ZFMISC_1:87;
  end;
  hence proj1 [:X,Y:] = X by XTUPLE_0:def 12;
  now
    let x;
    x in X implies [y,x] in [:Y,X:] by A1,ZFMISC_1:87;
    hence x in X iff ex y st [y,x] in [:Y,X:] by ZFMISC_1:87;
  end;
  hence thesis by XTUPLE_0:def 13;
end;

theorem Th10:
  proj1 [:X,Y:] c= X & proj2 [:X,Y:] c= Y
proof
  thus proj1 [:X,Y:] c= X
  proof
    let x;
    assume x in proj1 [:X,Y:];
    then ex y st [x,y] in [:X,Y:] by XTUPLE_0:def 12;
    hence thesis by ZFMISC_1:87;
  end;
  let y;
  assume y in proj2 [:X,Y:];
  then ex x st [x,y] in [:X,Y:] by XTUPLE_0:def 13;
  hence thesis by ZFMISC_1:87;
end;

theorem Th11:
  Z c= [:X,Y:] implies proj1 Z c= X & proj2 Z c= Y
proof
  assume Z c= [:X,Y:];
  then
A1: proj1 Z c= proj1 [:X,Y:] & proj2 Z c= proj2 [:X,Y:] by XTUPLE_0:8,9;
  proj1 [:X,Y:] c= X & proj2 [:X,Y:] c= Y by Th10;
  hence thesis by A1,XBOOLE_1:1;
end;

theorem Th12:
  proj1 {[x,y]} = {x} & proj2 {[x,y]} = {y}
proof
  thus proj1 {[x,y]} c= {x}
  proof
    let z;
    assume z in proj1 {[x,y]};
    then consider t such that
A1: [z,t] in {[x,y]} by XTUPLE_0:def 12;
    [z,t] = [x,y] by A1,TARSKI:def 1;
    then z = x by XTUPLE_0:1;
    hence thesis by TARSKI:def 1;
  end;
  thus {x} c= proj1 {[x,y]}
  proof
    let z;
    assume z in {x};
    then z = x by TARSKI:def 1;
    then [z,y] in {[x,y]} by TARSKI:def 1;
    hence thesis by XTUPLE_0:def 12;
  end;
  thus proj2 {[x,y]} c= {y}
  proof
    let z;
    assume z in proj2 {[x,y]};
    then consider t such that
A2: [t,z] in {[x,y]} by XTUPLE_0:def 13;
    [t,z] = [x,y] by A2,TARSKI:def 1;
    then z = y by XTUPLE_0:1;
    hence thesis by TARSKI:def 1;
  end;
  thus {y} c= proj2 {[x,y]}
  proof
    let z;
    assume z in {y};
    then z = y by TARSKI:def 1;
    then [x,z] in {[x,y]} by TARSKI:def 1;
    hence thesis by XTUPLE_0:def 13;
  end;
end;

theorem
  proj1 {[x,y],[z,t]} = {x,z} & proj2 {[x,y],[z,t]} = {y,t}
proof
A1: proj2 {[x,y]} = {y} & proj2 {[z,t]} = {t} by Th12;
  {[x,y],[z,t]} = {[x,y]} \/ {[z,t]} by ENUMSET1:1;
  then
A2: proj1 {[x,y],[z,t]} = proj1 {[x,y]} \/ proj1 {[z,t]} & proj2 {[x,y],[z,t
  ]} = proj2 {[x,y]} \/ proj2 {[z,t]} by XTUPLE_0:23,27;
  proj1 {[x,y]} = {x} & proj1 {[z,t]} = {z} by Th12;
  hence thesis by A2,A1,ENUMSET1:1;
end;

theorem Th14:
  not (ex x,y st [x,y] in X) implies proj1 X = {} & proj2 X = {}
proof
  set x = the Element of proj2 X;
  assume
A1: not (ex x,y st [x,y] in X);
  hereby
    set x = the Element of proj1 X;
    assume proj1 X <> {};
    then ex y st [x,y] in X by XTUPLE_0:def 12;
    hence contradiction by A1;
  end;
  assume proj2 X <> {};
  then ex y st [y,x] in X by XTUPLE_0:def 13;
  hence thesis by A1;
end;

theorem
  proj1 X = {} or proj2 X = {} implies not ex x,y st [x,y] in X by
XTUPLE_0:def 12,def 13;

theorem
  proj1 X = {} iff proj2 X = {}
proof
  proj1 X = {} or proj2 X = {} implies not ex x,y st [x,y] in X by
XTUPLE_0:def 12,def 13;
  hence thesis by Th14;
end;

theorem Th17:
  proj1 dom f = proj2 dom ~f & proj2 dom f = proj1 dom ~f
proof
  thus proj1 dom f c= proj2 dom ~f
  proof
    let x;
    assume x in proj1 dom f;
    then consider y such that
A1: [x,y] in dom f by XTUPLE_0:def 12;
    [y,x] in dom ~f by A1,FUNCT_4:42;
    hence thesis by XTUPLE_0:def 13;
  end;
  thus proj2 dom ~f c= proj1 dom f
  proof
    let y;
    assume y in proj2 dom ~f;
    then consider x such that
A2: [x,y] in dom ~f by XTUPLE_0:def 13;
    [y,x] in dom f by A2,FUNCT_4:42;
    hence thesis by XTUPLE_0:def 12;
  end;
  thus proj2 dom f c= proj1 dom ~f
  proof
    let y;
    assume y in proj2 dom f;
    then consider x such that
A3: [x,y] in dom f by XTUPLE_0:def 13;
    [y,x] in dom ~f by A3,FUNCT_4:42;
    hence thesis by XTUPLE_0:def 12;
  end;
  thus proj1 dom ~f c= proj2 dom f
  proof
    let x;
    assume x in proj1 dom ~f;
    then consider y such that
A4: [x,y] in dom ~f by XTUPLE_0:def 12;
    [y,x] in dom f by A4,FUNCT_4:42;
    hence thesis by XTUPLE_0:def 13;
  end;
end;

theorem
  for f being Relation holds proj1 f = dom f & proj2 f = rng f;

definition

  let f;
  func curry f -> Function means
  :Def1:
  dom it = proj1 dom f & for x st x in
  proj1 dom f ex g st it.x = g & dom g = proj2 (dom f /\ [:{x},proj2 dom f:]) &
  for y st y in dom g holds g.y = f.(x,y);
  existence
  proof
    defpred F[set,Function] means dom $2 = proj2 (dom f /\ [:{$1},proj2 dom f
    :]) & for y st y in dom $2 holds $2.y = f.($1,y);
    defpred P[set,set] means ex g st $2 = g & F[ $1,g];
A1: for x,y,z st x in proj1 dom f & P[x,y] & P[x,z] holds y = z
    proof
      let x,y,z such that
      x in proj1 dom f;
      given g1 being Function such that
A2:   y = g1 and
A3:   F[x,g1];
      given g2 being Function such that
A4:   z = g2 and
A5:   F[x,g2];
      now
        let t be set;
        assume
A6:     t in proj2 (dom f /\ [:{x},proj2 dom f:]);
        then g1.t = f.(x,t) by A3;
        hence g1.t = g2.t by A5,A6;
      end;
      hence thesis by A2,A3,A4,A5,FUNCT_1:2;
    end;
A7: for x st x in proj1 dom f ex y st P[x,y]
    proof
      let x such that
      x in proj1 dom f;
      deffunc F(set) = f.[x,$1];
      consider g such that
A8:   dom g = proj2 (dom f /\ [:{x},proj2 dom f:]) & for y st y in
      proj2 (dom f /\ [:{x},proj2 dom f:]) holds g.y = F(y) from FUNCT_1:sch 3;
      reconsider y = g as set;
      take y,g;
      thus thesis by A8;
    end;
    ex g being Function st dom g = proj1 dom f & for x st x in proj1 dom
    f holds P[x,g.x] from FUNCT_1:sch 2(A1,A7);
    hence thesis;
  end;
  uniqueness
  proof
    let f1,f2 be Function such that
A9: dom f1 = proj1 dom f and
A10: for x st x in proj1 dom f ex g st f1.x = g & dom g = proj2 (dom f
    /\ [:{x},proj2 dom f:]) & for y st y in dom g holds g.y = f.(x,y) and
A11: dom f2 = proj1 dom f and
A12: for x st x in proj1 dom f ex g st f2.x = g & dom g = proj2 (dom f
    /\ [:{x},proj2 dom f:]) & for y st y in dom g holds g.y = f.(x,y);
    now
      let x;
      assume
A13:  x in proj1 dom f;
      then consider g1 being Function such that
A14:  f1.x = g1 and
A15:  dom g1 = proj2 (dom f /\ [:{x},proj2 dom f:]) and
A16:  for y st y in dom g1 holds g1.y = f.(x,y) by A10;
      consider g2 being Function such that
A17:  f2.x = g2 and
A18:  dom g2 = proj2 (dom f /\ [:{x},proj2 dom f:]) and
A19:  for y st y in dom g2 holds g2.y = f.(x,y) by A12,A13;
      now
        let y;
        assume
A20:    y in proj2 (dom f /\ [:{x},proj2 dom f:]);
        then g1.y = f.(x,y) by A15,A16;
        hence g1.y = g2.y by A18,A19,A20;
      end;
      hence f1.x = f2.x by A14,A15,A17,A18,FUNCT_1:2;
    end;
    hence thesis by A9,A11,FUNCT_1:2;
  end;
  func uncurry f -> Function means
  : Def2:
  (for t holds t in dom it iff ex x,g
,y st t = [x,y] & x in dom f & g = f.x & y in dom g) & for x,g st x in dom it &
  g = f.x`1 holds it.x = g.x`2;
  existence
  proof
    defpred P[set,set] means ex g st g = f.$1`1 & $2 = g.$1`2;
    deffunc F(Function) = dom $1;
    defpred P[set] means f.$1 is Function;
    consider X such that
A21: x in X iff x in dom f & P[ x ] from XBOOLE_0:sch 1;
    defpred P[set] means for g1 st g1 = f.$1`1 holds $1`2 in dom g1;
    consider g such that
A22: dom g = f.:X & for g1 st g1 in f.:X holds g.g1 = F(g1) from
    LambdaFS;
    consider Y such that
A23: x in Y iff x in [:X,union rng g:] & P[ x ] from XBOOLE_0:sch 1;
A24: for x st x in Y ex y st P[x,y]
    proof
      let x;
      assume x in Y;
      then x in [:X,union rng g:] by A23;
      then x`1 in X by MCART_1:10;
      then reconsider h = f.x`1 as Function by A21;
      take h.x`2,h;
      thus thesis;
    end;
A25: for x,x1,x2 st x in Y & P[x,x1] & P[x,x2] holds x1 = x2;
    consider F being Function such that
A26: dom F = Y & for x st x in Y holds P[x,F.x] from FUNCT_1:sch 2(A25
    ,A24);
    take F;
    thus for t holds t in dom F iff ex x,g,y st t = [x,y] & x in dom f & g = f
    .x & y in dom g
    proof
      let t;
      thus t in dom F implies ex x,g,y st t = [x,y] & x in dom f & g = f.x & y
      in dom g
      proof
        assume
A27:    t in dom F;
        take x = t`1;
    t in [:X,union rng g:] by A23,A26,A27;
        then
A29:    x in X by MCART_1:10;
        then reconsider h = f.x as Function by A21;
        take h, t`2;
        thus thesis by A21,A23,A26,A27,A29,MCART_1:21;
      end;
      given x be set, h be Function, y be set such that
A30:  t = [x,y] and
A31:  x in dom f and
A32:  h = f.x and
A33:  y in dom h;
A34:  x in X by A21,A31,A32;
      then h in f.:X by A31,A32,FUNCT_1:def 6;
      then g.h = dom h & g.h in rng g by A22,FUNCT_1:def 3;
      then dom h c= union rng g by ZFMISC_1:74;
      then
A35:  t in [:X,union rng g:] by A30,A33,A34,ZFMISC_1:87;
      t`1 = x by A30,MCART_1:7;
      then for g1 st g1 = f.t`1 holds t`2 in dom g1 by A30,A32,A33,MCART_1:7;
      hence thesis by A23,A26,A35;
    end;
    let x;
    let h be Function;
    assume that
A36: x in dom F and
A37: h = f.x`1;
    ex g st g = f.x`1 & F.x = g.x`2 by A26,A36;
    hence thesis by A37;
  end;
  uniqueness
  proof
    defpred P[set] means ex x,g,y st $1 = [x,y] & x in dom f & g = f.x & y in
    dom g;
    let f1,f2;
    assume that
A38: t in dom f1 iff P[t] and
A39: for x,g st x in dom f1 & g = f.x`1 holds f1.x = g.x`2 and
A40: t in dom f2 iff P[t] and
A41: for x,g st x in dom f2 & g = f.x`1 holds f2.x = g.x`2;
A42: dom f1 = dom f2 from XBOOLE_0:sch 2(A38,A40);
    now
      let x;
      assume
A43:  x in dom f1;
      then consider z,g,y such that
A44:  x = [z,y] and
      z in dom f and
A45:  g = f.z and
      y in dom g by A38;
A46:  z = x`1 & y = x`2 by A44,MCART_1:7;
      then f1.x = g.y by A39,A43,A45;
      hence f1.x = f2.x by A41,A42,A43,A45,A46;
    end;
    hence thesis by A42,FUNCT_1:2;
  end;
end;

definition
  let f;
  func curry' f -> Function equals
  curry ~f;
  correctness;
  func uncurry' f -> Function equals
  ~(uncurry f);
  correctness;
end;

theorem Th19:
  [x,y] in dom f implies x in dom curry f & (curry f).x is Function
proof
  assume [x,y] in dom f;
  then
A1: x in proj1 dom f by XTUPLE_0:def 12;
  hence x in dom curry f by Def1;
  ex g st (curry f).x = g & dom g = proj2 (dom f /\ [:{x},proj2 dom f:]) &
  for y st y in dom g holds g.y = f.(x,y) by A1,Def1;
  hence thesis;
end;

theorem Th20:
  [x,y] in dom f & g = (curry f).x implies y in dom g & g.y = f.(x ,y)
proof
  assume that
A1: [x,y] in dom f and
A2: g = (curry f).x;
  x in proj1 dom f by A1,XTUPLE_0:def 12;
  then
A3: ex h st (curry f).x = h & dom h = proj2 (dom f /\ [:{x},proj2 dom f :])
  & for y st y in dom h holds h.y = f.(x,y) by Def1;
  y in proj2 dom f by A1,XTUPLE_0:def 13;
  then [x,y] in [:{x},proj2 dom f:] by ZFMISC_1:105;
  then [x,y] in dom f /\ [:{x},proj2 dom f:] by A1,XBOOLE_0:def 4;
  hence y in dom g by A2,A3,XTUPLE_0:def 13;
  hence thesis by A2,A3;
end;

theorem
  [x,y] in dom f implies y in dom curry' f & (curry' f).y is Function
proof
  assume [x,y] in dom f;
  then [y,x] in dom ~f by FUNCT_4:42;
  hence thesis by Th19;
end;

theorem Th22:
  [x,y] in dom f & g = (curry' f).y implies x in dom g & g.x = f.(x,y)
proof
  assume [x,y] in dom f;
  then
A1: [y,x] in dom ~f by FUNCT_4:42;
  assume
A2: g = (curry' f).y;
  then g.x = (~f).(y,x) by A1,Th20;
  hence thesis by A1,A2,Th20,FUNCT_4:43;
end;

theorem
  dom curry' f = proj2 dom f
proof
  dom curry ~f = proj1 dom ~f by Def1;
  hence thesis by Th17;
end;

theorem Th24:
  [:X,Y:] <> {} & dom f = [:X,Y:] implies dom curry f = X & dom curry' f = Y
proof
  assume that
A1: [:X,Y:] <> {} and
A2: dom f = [:X,Y:];
  dom curry f = proj1 dom f by Def1;
  hence dom curry f = X by A1,A2,Th9;
  thus dom curry' f = proj1 dom ~f by Def1
    .= proj1 [:Y,X:] by A2,FUNCT_4:46
    .= Y by A1,Th9;
end;

theorem Th25:
  dom f c= [:X,Y:] implies dom curry f c= X & dom curry' f c= Y
proof
  assume
A1: dom f c= [:X,Y:];
  dom curry f = proj1 dom f by Def1;
  hence dom curry f c= X by A1,Th11;
A2: dom curry' f = proj1 dom ~f by Def1;
  dom ~f c= [:Y,X:] by A1,FUNCT_4:45;
  hence thesis by A2,Th11;
end;

theorem Th26:
  rng f c= Funcs(X,Y) implies dom uncurry f = [:dom f,X:] & dom
  uncurry' f = [:X,dom f:]
proof
  defpred P[set] means ex x,g,y st $1 = [x,y] & x in dom f & g = f.x & y in
  dom g;
  assume
A1: rng f c= Funcs(X,Y);
A2: t in [:dom f,X:] iff P[t]
  proof
    thus t in [:dom f,X:] implies ex x,g,y st t = [x,y] & x in dom f & g = f.x
    & y in dom g
    proof
      assume
A3:   t in [:dom f,X:];
      then t`1 in dom f by MCART_1:10;
      then f.t`1 in rng f by FUNCT_1:def 3;
      then consider g such that
A4:   f.t`1 = g & dom g = X and
      rng g c= Y by A1,FUNCT_2:def 2;
      take t`1, g, t`2;
      thus thesis by A3,A4,MCART_1:10,21;
    end;
    given x,g,y such that
A5: t = [x,y] and
A6: x in dom f and
A7: g = f.x and
A8: y in dom g;
    g in rng f by A6,A7,FUNCT_1:def 3;
    then ex g1 st g = g1 & dom g1 = X & rng g1 c= Y by A1,FUNCT_2:def 2;
    hence thesis by A5,A6,A8,ZFMISC_1:87;
  end;
A9: t in dom uncurry f iff P[t] by Def2;
  thus dom uncurry f = [:dom f,X:] from XBOOLE_0:sch 2(A9,A2);
  hence thesis by FUNCT_4:46;
end;

theorem
  not (ex x,y st [x,y] in dom f) implies curry f = {} & curry' f = {}
proof
  assume
A1: not ex x,y st [x,y] in dom f;
  then proj1 dom f = {} by Th14;
  then dom curry f = {} by Def1;
  hence curry f = {};
  now
    let x,y;
    assume [x,y] in dom ~f;
    then [y,x] in dom f by FUNCT_4:42;
    hence contradiction by A1;
  end;
  then proj1 dom ~f = {} by Th14;
  then dom curry ~f = {} by Def1;
  hence thesis;
end;

theorem
  not (ex x st x in dom f & f.x is Function) implies uncurry f = {} &
  uncurry' f = {}
proof
  assume
A1: not (ex x st x in dom f & f.x is Function);
A2: now
    set t = the Element of dom uncurry f;
    assume dom uncurry f <> {};
    then ex x,g,y st t = [x,y] & x in dom f & g = f.x & y in dom g by Def2;
    hence contradiction by A1;
  end;
  hence uncurry f = {};
  thus thesis by A2,Th1,RELAT_1:41;
end;

theorem Th29:
  [:X,Y:] <> {} & dom f = [:X,Y:] & x in X implies ex g st (curry
  f).x = g & dom g = Y & rng g c= rng f & for y st y in Y holds g.y = f.(x,y)
proof
  assume that
A1: [:X,Y:] <> {} and
A2: dom f = [:X,Y:] and
A3: x in X;
  {x} c= X by A3,ZFMISC_1:31;
  then
A4: [:{x},Y:] /\ dom f = [:{x},Y:] by A2,ZFMISC_1:101;
  x in proj1 dom f by A1,A2,A3,Th9;
  then consider g such that
A5: (curry f).x = g and
A6: dom g = proj2 (dom f /\ [:{x},proj2 dom f:]) & for y st y in dom g
  holds g.y = f.(x,y) by Def1;
  take g;
A7: proj2 [:{x},Y:] = Y by Th9;
A8: proj2 dom f = Y by A2,A3,Th9;
  rng g c= rng f
  proof
    let z;
    assume z in rng g;
    then consider y such that
A9: y in dom g & z = g.y by FUNCT_1:def 3;
    [x,y] in dom f & z = f.(x,y) by A2,A3,A6,A8,A4,A7,A9,ZFMISC_1:87;
    hence thesis by FUNCT_1:def 3;
  end;
  hence thesis by A5,A6,A8,A4,A7;
end;

theorem Th30:
  x in dom curry f implies (curry f).x is Function
proof
  assume
A1: x in dom curry f;
  dom curry f = proj1 dom f by Def1;
  then
  ex g st (curry f).x = g & dom g = proj2 (dom f /\ [:{x},proj2 dom f:] )
  & for y st y in dom g holds g.y = f.(x,y) by A1,Def1;
  hence thesis;
end;

theorem Th31:
  x in dom curry f & g = (curry f).x implies dom g = proj2 (dom f
/\ [:{x},proj2 dom f:]) & dom g c= proj2 dom f & rng g c= rng f & for y st y in
  dom g holds g.y = f.(x,y) & [x,y] in dom f
proof
  assume that
A1: x in dom curry f and
A2: g = (curry f).x;
  dom curry f = proj1 dom f by Def1;
  then consider h such that
A3: (curry f).x = h and
A4: dom h = proj2 (dom f /\ [:{x},proj2 dom f:]) and
A5: for y st y in dom h holds h.y = f.(x,y) by A1,Def1;
  thus dom g = proj2 (dom f /\ [:{x},proj2 dom f:]) by A2,A3,A4;
  dom f /\ [:{x},proj2 dom f:] c= dom f by XBOOLE_1:17;
  hence dom g c= proj2 dom f by A2,A3,A4,XTUPLE_0:9;
  thus rng g c= rng f
  proof
    let y;
    assume y in rng g;
    then consider z such that
A6: z in dom g and
A7: y = g.z by FUNCT_1:def 3;
    consider t such that
A8: [t,z] in dom f /\ [:{x},proj2 dom f:] by A2,A3,A4,A6,XTUPLE_0:def 13;
    [t,z] in dom f & [t,z] in [:{x},proj2 dom f:] by A8,XBOOLE_0:def 4;
    then
A9: [x,z] in dom f by ZFMISC_1:105;
    h.z = f.(x,z) by A2,A3,A5,A6;
    hence thesis by A2,A3,A7,A9,FUNCT_1:def 3;
  end;
  let y;
  assume
A10: y in dom g;
  then consider t such that
A11: [t,y] in dom f /\ [:{x},proj2 dom f:] by A2,A3,A4,XTUPLE_0:def 13;
  [t,y] in dom f & [t,y] in [:{x},proj2 dom f:] by A11,XBOOLE_0:def 4;
  hence thesis by A2,A3,A5,A10,ZFMISC_1:105;
end;

theorem Th32:
  [:X,Y:] <> {} & dom f = [:X,Y:] & y in Y implies ex g st (curry'
  f).y = g & dom g = X & rng g c= rng f & for x st x in X holds g.x = f.(x,y)
proof
  assume that
A1: [:X,Y:] <> {} and
A2: dom f = [:X,Y:] and
A3: y in Y;
A4: dom ~f = [:Y,X:] by A2,FUNCT_4:46;
  X <> {} by A1,ZFMISC_1:90;
  then consider g such that
A5: (curry ~f).y = g & dom g = X & rng g c= rng ~f and
A6: for x st x in X holds g.x = (~f).(y,x) by A3,A4,Th29;
  take g;
  rng ~f c= rng f by FUNCT_4:41;
  hence (curry' f).y = g & dom g = X & rng g c= rng f by A5,XBOOLE_1:1;
  let x;
  assume
A7: x in X;
  then
A8: g.x = (~f).(y,x) by A6;
  [y,x] in dom ~f by A3,A4,A7,ZFMISC_1:87;
  hence thesis by A8,FUNCT_4:43;
end;

theorem
  x in dom curry' f implies (curry' f).x is Function by Th30;

theorem
  x in dom curry' f & g = (curry' f).x implies dom g = proj1 (dom f /\
  [:proj1 dom f,{x}:]) & dom g c= proj1 dom f & rng g c= rng f & for y st y in
  dom g holds g.y = f.(y,x) & [y,x] in dom f
proof
A1: rng ~f c= rng f by FUNCT_4:41;
  assume
A2: x in dom curry' f & g = (curry' f).x;
  then dom g = proj2 (dom ~f /\ [:{x},proj2 dom ~f:]) by Th31;
  then
A3: dom g = proj2 (dom ~f /\ [:{x},proj1 dom f:]) by Th17;
  thus
A4: dom g c= proj1 (dom f /\ [:proj1 dom f,{x}:])
  proof
    let z;
    assume z in dom g;
    then consider y such that
A5: [y,z] in dom ~f /\ [:{x},proj1 dom f:] by A3,XTUPLE_0:def 13;
    [y,z] in [:{x},proj1 dom f:] by A5,XBOOLE_0:def 4;
    then
A6: [z,y] in [:proj1 dom f,{x}:] by ZFMISC_1:88;
    [y,z] in dom ~f by A5,XBOOLE_0:def 4;
    then [z,y] in dom f by FUNCT_4:42;
    then [z,y] in dom f /\ [:proj1 dom f,{x}:] by A6,XBOOLE_0:def 4;
    hence thesis by XTUPLE_0:def 12;
  end;
  thus proj1 (dom f /\ [:proj1 dom f,{x}:]) c= dom g
  proof
    let z;
    assume z in proj1 (dom f /\ [:proj1 dom f,{x}:]);
    then consider y such that
A7: [z,y] in dom f /\ [:proj1 dom f,{x}:] by XTUPLE_0:def 12;
    [z,y] in [:proj1 dom f,{x}:] by A7,XBOOLE_0:def 4;
    then
A8: [y,z] in [:{x},proj1 dom f:] by ZFMISC_1:88;
    [z,y] in dom f by A7,XBOOLE_0:def 4;
    then [y,z] in dom ~f by FUNCT_4:42;
    then [y,z] in dom ~f /\ [:{x},proj1 dom f:] by A8,XBOOLE_0:def 4;
    hence thesis by A3,XTUPLE_0:def 13;
  end;
  dom g c= proj2 dom ~f & rng g c= rng ~f by A2,Th31;
  hence dom g c= proj1 dom f & rng g c= rng f by A1,Th17,XBOOLE_1:1;
  let y;
  assume
A9: y in dom g;
  then consider z such that
A10: [y,z] in dom f /\ [:proj1 dom f,{x}:] by A4,XTUPLE_0:def 12;
  [y,z] in [:proj1 dom f,{x}:] by A10,XBOOLE_0:def 4;
  then
A11: z = x by ZFMISC_1:106;
  [y,z] in dom f by A10,XBOOLE_0:def 4;
  then [z,y] in dom ~f by FUNCT_4:42;
  then (~f).(x,y) = f.(y,x) by A11,FUNCT_4:43;
  hence thesis by A2,A9,A10,A11,Th31,XBOOLE_0:def 4;
end;

theorem Th35:
  dom f = [:X,Y:] implies rng curry f c= Funcs(Y,rng f) & rng
  curry' f c= Funcs(X,rng f)
proof
  assume
A1: dom f = [:X,Y:];
A2: now
    assume
A3: [:X,Y:] <> {};
    then
A4: dom curry f = X by A1,Th24;
    thus rng curry f c= Funcs(Y,rng f)
    proof
      let z;
      assume z in rng curry f;
      then consider x such that
A5:   x in dom curry f and
A6:   z = (curry f).x by FUNCT_1:def 3;
      ex g st (curry f).x = g & dom g = Y & rng g c= rng f & for y st y
      in Y holds g.y = f.(x,y) by A1,A3,A4,A5,Th29;
      hence thesis by A6,FUNCT_2:def 2;
    end;
A7: dom curry' f = Y by A1,A3,Th24;
    thus rng curry' f c= Funcs(X,rng f)
    proof
      let z;
      assume z in rng curry' f;
      then consider y such that
A8:   y in dom curry' f and
A9:   z = (curry' f).y by FUNCT_1:def 3;
      ex g st (curry' f).y = g & dom g = X & rng g c= rng f & for x st x
      in X holds g.x = f.(x,y) by A1,A3,A7,A8,Th32;
      hence thesis by A9,FUNCT_2:def 2;
    end;
  end;
  now
    assume
A10: dom f = {};
    then X = {} or Y = {} by A1;
    then
A11: [:Y,X:] = {} by ZFMISC_1:90;
    {} = dom curry f by A10,Def1,Th8;
    then
A12: rng curry f = {} by RELAT_1:42;
    dom ~f = [:Y,X:] & dom curry ~f = proj1 dom ~f by A1,Def1,FUNCT_4:46;
    then rng curry' f = {} by A11,Th8,RELAT_1:42;
    hence thesis by A12,XBOOLE_1:2;
  end;
  hence thesis by A1,A2;
end;

theorem
  rng curry f c= PFuncs(proj2 dom f,rng f) & rng curry' f c= PFuncs(
  proj1 dom f,rng f)
proof
A1: rng ~f c= rng f by FUNCT_4:41;
  thus rng curry f c= PFuncs(proj2 dom f,rng f)
  proof
    let t;
    assume t in rng curry f;
    then consider z such that
A2: z in dom curry f and
A3: t = (curry f).z by FUNCT_1:def 3;
    dom curry f = proj1 dom f by Def1;
    then consider g such that
A4: (curry f).z = g and
    dom g = proj2 (dom f /\ [:{z},proj2 dom f:]) and
    for y st y in dom g holds g.y = f.(z,y) by A2,Def1;
    dom g c= proj2 dom f & rng g c= rng f by A2,A4,Th31;
    hence thesis by A3,A4,PARTFUN1:def 3;
  end;
  let t;
  assume t in rng curry' f;
  then consider z such that
A5: z in dom curry' f and
A6: t = (curry' f).z by FUNCT_1:def 3;
  dom curry ~f = proj1 dom ~f by Def1;
  then consider g such that
A7: (curry ~f).z = g and
  dom g = proj2 (dom ~f /\ [:{z},proj2 dom ~f:]) and
  for y st y in dom g holds g.y = (~f).(z,y) by A5,Def1;
  rng g c= rng ~f by A5,A7,Th31;
  then
A8: rng g c= rng f by A1,XBOOLE_1:1;
  dom g c= proj2 dom ~f by A5,A7,Th31;
  then dom g c= proj1 dom f by Th17;
  hence thesis by A6,A7,A8,PARTFUN1:def 3;
end;

theorem Th37:
  rng f c= PFuncs(X,Y) implies dom uncurry f c= [:dom f,X:] & dom
  uncurry' f c= [:X,dom f:]
proof
  assume
A1: rng f c= PFuncs(X,Y);
  thus
A2: dom uncurry f c= [:dom f,X:]
  proof
    let x;
    assume x in dom uncurry f;
    then consider y,g,z such that
A3: x = [y,z] and
A4: y in dom f and
A5: g = f.y and
A6: z in dom g by Def2;
    g in rng f by A4,A5,FUNCT_1:def 3;
    then g is PartFunc of X,Y by A1,PARTFUN1:46;
    then dom g c= X by RELAT_1:def 18;
    hence thesis by A3,A4,A6,ZFMISC_1:87;
  end;
  let x;
  assume x in dom uncurry' f;
  then ex y,z st x = [z,y] & [y,z] in dom uncurry f by FUNCT_4:def 2;
  hence thesis by A2,ZFMISC_1:88;
end;

theorem Th38:
  x in dom f & g = f.x & y in dom g implies [x,y] in dom uncurry f
  & (uncurry f).(x,y) = g.y & g.y in rng uncurry f
proof
  assume that
A1: x in dom f and
A2: g = f.x and
A3: y in dom g;
  thus
A4: [x,y] in dom uncurry f by A1,A2,A3,Def2;
  [x,y]`1 = x & [x,y]`2 = y;
  hence (uncurry f).(x,y) = g.y by A2,A4,Def2;
  hence thesis by A4,FUNCT_1:def 3;
end;

theorem
  x in dom f & g = f.x & y in dom g implies [y,x] in dom uncurry' f & (
  uncurry' f).(y,x) = g.y & g.y in rng uncurry' f
proof
  assume
A1: x in dom f & g = f.x & y in dom g;
  then [x,y] in dom uncurry f by Th38;
  hence
A2: [y,x] in dom uncurry' f by FUNCT_4:42;
  (uncurry f).(x,y) = g.y by A1,Th38;
  hence (uncurry' f).(y,x) = g.y by A2,FUNCT_4:43;
  hence thesis by A2,FUNCT_1:def 3;
end;

theorem Th40:
  rng f c= PFuncs(X,Y) implies rng uncurry f c= Y & rng uncurry' f c= Y
proof
  assume
A1: rng f c= PFuncs(X,Y);
  thus
A2: rng uncurry f c= Y
  proof
    let x;
    assume x in rng uncurry f;
    then consider y such that
A3: y in dom uncurry f and
A4: x = (uncurry f).y by FUNCT_1:def 3;
    consider z,g,t such that
A5: y = [z,t] and
A6: z in dom f & g = f.z and
A7: t in dom g by A3,Def2;
    g in rng f by A6,FUNCT_1:def 3;
    then
A8: ex g1 st g = g1 & dom g1 c= X & rng g1 c= Y by A1,PARTFUN1:def 3;
    (uncurry f).(z,t) = g.t & g.t in rng g by A6,A7,Th38,FUNCT_1:def 3;
    hence thesis by A4,A5,A8;
  end;
  rng uncurry' f c= rng uncurry f by FUNCT_4:41;
  hence thesis by A2,XBOOLE_1:1;
end;

theorem Th41:
  rng f c= Funcs(X,Y) implies rng uncurry f c= Y & rng uncurry' f c= Y
proof
A1: Funcs(X,Y) c= PFuncs(X,Y) by FUNCT_2:72;
  assume rng f c= Funcs(X,Y);
  then rng f c= PFuncs(X,Y) by A1,XBOOLE_1:1;
  hence thesis by Th40;
end;

theorem Th42:
  curry {} = {} & curry' {} = {}
proof
A1: dom curry {} = proj1 dom {} by Def1;
  hence curry {} = {};
  thus thesis by A1,Th1;
end;

theorem Th43:
  uncurry {} = {} & uncurry' {} = {}
proof
A1: now
    set t = the Element of dom uncurry {};
    assume dom uncurry {} <> {};
    then ex x,g,y st t = [x,y] & x in dom {} & g = {} .x & y in dom g by Def2;
    hence contradiction;
  end;
  hence uncurry {} = {};
  thus thesis by A1,Th1,RELAT_1:41;
end;

theorem Th44:
  dom f1 = [:X,Y:] & dom f2 = [:X,Y:] & curry f1 = curry f2 implies f1 = f2
proof
  assume that
A1: dom f1 = [:X,Y:] and
A2: dom f2 = [:X,Y:] and
A3: curry f1 = curry f2;
A4: now
    assume [:X,Y:] <> {};
    now
      let z;
      assume
A5:   z in [:X,Y:];
      then consider g1 being Function such that
A6:   (curry f1).z`1 = g1 and
      dom g1 = Y and
      rng g1 c= rng f1 and
A7:   for y st y in Y holds g1.y = f1.(z`1,y) by A1,Th29,MCART_1:10;
A8:   z = [z`1,z`2] by A5,MCART_1:21;
A9:   z`2 in Y by A5,MCART_1:10;
      then
A10:  f1.(z`1,z`2) = g1.z`2 by A7;
      ex g2 being Function st (curry f2).z`1 = g2 & dom g2 = Y & rng g2 c=
rng f2 & for y st y in Y holds g2.y = f2.(z`1,y) by A2,A5,Th29,MCART_1:10;
      then f2.(z`1,z`2) = g1.z`2 by A3,A9,A6;
      hence f1.z = f2.z by A8,A10;
    end;
    hence thesis by A1,A2,FUNCT_1:2;
  end;
  [:X,Y:] = {} implies f1 = {} & f2 = {} by A1,A2;
  hence thesis by A4;
end;

theorem Th45:
  dom f1 = [:X,Y:] & dom f2 = [:X,Y:] & curry' f1 = curry' f2 implies f1 = f2
proof
  assume that
A1: dom f1 = [:X,Y:] and
A2: dom f2 = [:X,Y:] and
A3: curry' f1 = curry' f2;
  dom ~f1 = [:Y,X:] & dom ~f2 = [:Y,X:] by A1,A2,FUNCT_4:46;
  then
A4: ~f1 = ~f2 by A3,Th44;
  ~~f1 = f1 by A1,FUNCT_4:52;
  hence thesis by A2,A4,FUNCT_4:52;
end;

theorem Th46:
  rng f1 c= Funcs(X,Y) & rng f2 c= Funcs(X,Y) & X <> {} & uncurry
  f1 = uncurry f2 implies f1 = f2
proof
  assume that
A1: rng f1 c= Funcs(X,Y) and
A2: rng f2 c= Funcs(X,Y) and
A3: X <> {} and
A4: uncurry f1 = uncurry f2;
A5: dom uncurry f1 = [:dom f1,X:] & dom uncurry f2 = [:dom f2,X:] by A1,A2,Th26
;
  then dom f1 = {} implies dom f1 = dom f2 by A3,A4,ZFMISC_1:90;
  then
A6: dom f1 = dom f2 by A3,A4,A5,ZFMISC_1:110;
  now
    let x;
    assume
A7: x in dom f1;
    then f1.x in rng f1 by FUNCT_1:def 3;
    then consider g1 such that
A8: f1.x = g1 and
A9: dom g1 = X and
    rng g1 c= Y by A1,FUNCT_2:def 2;
    f2.x in rng f2 by A6,A7,FUNCT_1:def 3;
    then consider g2 such that
A10: f2.x = g2 and
A11: dom g2 = X and
    rng g2 c= Y by A2,FUNCT_2:def 2;
    now
      let y;
A12:  [x,y]`1 = x & [x,y]`2 = y;
      assume
A13:  y in X;
      then
A14:  [x,y] in dom uncurry f2 by A4,A7,A8,A9,Def2;
      [x,y] in dom uncurry f1 by A7,A8,A9,A13,Def2;
      then (uncurry f1).[x,y] = g1.y by A8,A12,Def2;
      hence g1.y = g2.y by A4,A10,A14,A12,Def2;
    end;
    hence f1.x = f2.x by A8,A9,A10,A11,FUNCT_1:2;
  end;
  hence thesis by A6,FUNCT_1:2;
end;

theorem
  rng f1 c= Funcs(X,Y) & rng f2 c= Funcs(X,Y) & X <> {} & uncurry' f1 =
  uncurry' f2 implies f1 = f2
proof
  assume that
A1: rng f1 c= Funcs(X,Y) and
A2: rng f2 c= Funcs(X,Y) and
A3: X <> {} & uncurry' f1 = uncurry' f2;
  dom uncurry f1 = [:dom f1,X:] by A1,Th26;
  then
A4: uncurry f1 = ~~(uncurry f1) by FUNCT_4:52;
  dom uncurry f2 = [:dom f2,X:] by A2,Th26;
  hence thesis by A1,A2,A3,A4,Th46,FUNCT_4:52;
end;

theorem Th48:
  rng f c= Funcs(X,Y) & X <> {} implies curry uncurry f = f &
  curry' uncurry' f = f
proof
  assume that
A1: rng f c= Funcs(X,Y) and
A2: X <> {};
A3: dom uncurry f = [:dom f,X:] by A1,Th26;
A4: now
A5: now
      let x;
      assume
A6:   x in dom f;
      then consider h such that
A7:   (curry uncurry f).x = h & dom h = X & rng h c= rng uncurry f and
A8:   y in X implies h.y = (uncurry f).(x,y) by A2,A3,Th29;
      f.x in rng f by A6,FUNCT_1:def 3;
      then consider g such that
A9:   f.x = g & dom g = X and
      rng g c= Y by A1,FUNCT_2:def 2;
      now
        let y;
        assume
A10:    y in X;
        then (uncurry f).(x,y) = g.y by A6,A9,Th38;
        hence g.y = h.y by A8,A10;
      end;
      hence f.x = (curry uncurry f).x by A9,A7,FUNCT_1:2;
    end;
    assume dom f <> {};
    then dom curry uncurry f = dom f by A2,A3,Th24;
    hence curry uncurry f = f by A5,FUNCT_1:2;
  end;
A11: now
    assume dom f = {};
    then dom uncurry f = {} & f = {} by A3,ZFMISC_1:90;
    hence curry uncurry f = f by Th42,RELAT_1:41;
  end;
  hence curry uncurry f = f by A4;
  thus thesis by A3,A11,A4,FUNCT_4:52;
end;

theorem
  dom f = [:X,Y:] implies uncurry curry f = f & uncurry' curry' f = f
proof
  assume
A1: dom f = [:X,Y:];
A2: now
A3: rng curry' f c= Funcs(X,rng f) by A1,Th35;
    assume
A4: dom f <> {};
    then X <> {} by A1,ZFMISC_1:90;
    then
A5: curry' uncurry' curry' f = curry' f by A3,Th48;
    dom curry' f = Y by A1,A4,Th24;
    then
A6: dom uncurry' curry' f = [:X,Y:] by A3,Th26;
A7: rng curry f c= Funcs(Y,rng f) by A1,Th35;
    Y <> {} by A1,A4,ZFMISC_1:90;
    then
A8: curry uncurry curry f = curry f by A7,Th48;
    dom curry f = X by A1,A4,Th24;
    then dom uncurry curry f = [:X,Y:] by A7,Th26;
    hence thesis by A1,A8,A5,A6,Th44,Th45;
  end;
  now
    assume dom f = {};
    then f = {};
    hence thesis by Th42,Th43;
  end;
  hence thesis by A2;
end;

theorem Th50:
  dom f c= [:X,Y:] implies uncurry curry f = f & uncurry' curry' f = f
proof
  assume
A1: dom f c= [:X,Y:];
A2: now
    let X,Y,f such that
A3: dom f c= [:X,Y:];
A4: dom uncurry curry f = dom f
    proof
      thus dom uncurry curry f c= dom f
      proof
        let x;
        assume x in dom uncurry curry f;
        then
        ex y,g,z st x = [y,z] & y in dom curry f & g = (curry f).y & z in
        dom g by Def2;
        hence thesis by Th31;
      end;
      let x;
      assume
A5:   x in dom f;
      then
A6:   x = [x`1,x`2] by A3,MCART_1:21;
      then reconsider g = (curry f).x`1 as Function by A5,Th19;
      x`2 in dom g & x`1 in dom curry f by A5,A6,Th19,Th20;
      hence thesis by A6,Th38;
    end;
    now
      let x;
      assume
A7:   x in dom f;
      then
A8:   x = [x`1,x`2] by A3,MCART_1:21;
      then reconsider g = (curry f).x`1 as Function by A7,Th19;
      (uncurry curry f).x = g.x`2 by A4,A7,Def2;
      then f.(x`1,x`2) = (uncurry curry f).(x`1,x`2) by A7,A8,Th20;
      hence f.x = (uncurry curry f).x by A8;
    end;
    hence uncurry curry f = f by A4,FUNCT_1:2;
  end;
  hence uncurry curry f = f by A1;
  dom ~f c= [:Y,X:] by A1,FUNCT_4:45;
  then uncurry curry ~f = ~f by A2;
  hence thesis by A1,FUNCT_4:52;
end;

theorem Th51:
  rng f c= PFuncs(X,Y) & not {} in rng f implies curry uncurry f =
  f & curry' uncurry' f = f
proof
  assume that
A1: rng f c= PFuncs(X,Y) and
A2: not {} in rng f;
A3: dom curry uncurry f = dom f
  proof
    dom uncurry f c= [:dom f,X:] by A1,Th37;
    hence dom curry uncurry f c= dom f by Th25;
    let x;
    assume
A4: x in dom f;
    then f.x in rng f by FUNCT_1:def 3;
    then consider g such that
A5: f.x = g and
    dom g c= X and
    rng g c= Y by A1,PARTFUN1:def 3;
    set y = the Element of dom g;
    g <> {} by A2,A4,A5,FUNCT_1:def 3;
    then
A6: [x,y] in dom uncurry f by A4,A5,Th38;
    dom curry uncurry f = proj1 dom uncurry f by Def1;
    hence thesis by A6,XTUPLE_0:def 12;
  end;
  now
    let x;
    assume
A7: x in dom f;
    then reconsider h = (curry uncurry f).x as Function by A3,Th30;
    f.x in rng f by A7,FUNCT_1:def 3;
    then consider g such that
A8: f.x = g and
    dom g c= X and
    rng g c= Y by A1,PARTFUN1:def 3;
A9: dom h = proj2 (dom uncurry f /\ [:{x},proj2 dom uncurry f:]) by A3,A7,Th31;
A10: dom h = dom g
    proof
      thus dom h c= dom g
      proof
        let z;
        assume z in dom h;
        then consider t such that
A11:    [t,z] in dom uncurry f /\ [:{x},proj2 dom uncurry f:] by A9,
XTUPLE_0:def 13;
        [t,z] in [:{x},proj2 dom uncurry f:] by A11,XBOOLE_0:def 4;
        then
A12:    t = x by ZFMISC_1:105;
        [t,z] in dom uncurry f by A11,XBOOLE_0:def 4;
        then consider x1,g1,x2 such that
A13:    [t,z] = [x1,x2] and
        x1 in dom f and
A14:    g1 = f.x1 & x2 in dom g1 by Def2;
        t = x1 by A13,XTUPLE_0:1;
        hence thesis by A8,A13,A14,A12,XTUPLE_0:1;
      end;
      let y;
      assume y in dom g;
      then [x,y] in dom uncurry f by A7,A8,Th38;
      hence thesis by Th20;
    end;
    now
      let y;
      assume
A15:  y in dom h;
      hence h.y = (uncurry f).(x,y) by A3,A7,Th31
        .= g.y by A7,A8,A10,A15,Th38;
    end;
    hence f.x = (curry uncurry f).x by A8,A10,FUNCT_1:2;
  end;
  hence
A16: curry uncurry f = f by A3,FUNCT_1:2;
  dom uncurry f c= [:dom f,X:] by A1,Th37;
  hence thesis by A16,FUNCT_4:52;
end;

theorem
  dom f1 c= [:X,Y:] & dom f2 c= [:X,Y:] & curry f1 = curry f2 implies f1 = f2
proof
  assume that
A1: dom f1 c= [:X,Y:] and
A2: dom f2 c= [:X,Y:];
  uncurry curry f1 = f1 by A1,Th50;
  hence thesis by A2,Th50;
end;

theorem
  dom f1 c= [:X,Y:] & dom f2 c= [:X,Y:] & curry' f1 = curry' f2 implies f1 = f2
proof
  assume that
A1: dom f1 c= [:X,Y:] and
A2: dom f2 c= [:X,Y:];
  uncurry' curry' f1 = f1 by A1,Th50;
  hence thesis by A2,Th50;
end;

theorem
  rng f1 c= PFuncs(X,Y) & rng f2 c= PFuncs(X,Y) & not {} in rng f1 & not
  {} in rng f2 & uncurry f1 = uncurry f2 implies f1 = f2
proof
  assume that
A1: rng f1 c= PFuncs(X,Y) and
A2: rng f2 c= PFuncs(X,Y) and
A3: not {} in rng f1 and
A4: not {} in rng f2;
  curry uncurry f1 = f1 by A1,A3,Th51;
  hence thesis by A2,A4,Th51;
end;

theorem
  rng f1 c= PFuncs(X,Y) & rng f2 c= PFuncs(X,Y) & not {} in rng f1 & not
  {} in rng f2 & uncurry' f1 = uncurry' f2 implies f1 = f2
proof
  assume that
A1: rng f1 c= PFuncs(X,Y) and
A2: rng f2 c= PFuncs(X,Y) and
A3: not {} in rng f1 and
A4: not {} in rng f2;
  curry' uncurry' f1 = f1 by A1,A3,Th51;
  hence thesis by A2,A4,Th51;
end;

theorem Th56:
  X c= Y implies Funcs(Z,X) c= Funcs(Z,Y)
proof
  assume
A1: X c= Y;
  let x;
  assume x in Funcs(Z,X);
  then consider f such that
A2: x = f & dom f = Z and
A3: rng f c= X by FUNCT_2:def 2;
  rng f c= Y by A1,A3,XBOOLE_1:1;
  hence thesis by A2,FUNCT_2:def 2;
end;

theorem Th57:
  Funcs({},X) = {{}}
proof
  thus Funcs({},X) c= {{}}
  proof
    let x;
    assume x in Funcs({},X);
    then ex f st x = f & dom f = {} & rng f c= X by FUNCT_2:def 2;
    then x = {};
    hence thesis by TARSKI:def 1;
  end;
  let x;
A1: {} c= X by XBOOLE_1:2;
  assume x in {{}};
  then
A2: x = {} by TARSKI:def 1;
  dom {} = {} & rng {} = {};
  hence thesis by A2,A1,FUNCT_2:def 2;
end;

theorem
  X,Funcs({x},X) are_equipotent & card X = card Funcs({x},X)
proof
  deffunc F(set) = {x} --> $1;
  consider f such that
A1: dom f = X & for y st y in X holds f.y = F(y) from FUNCT_1:sch 3;
A2: x in {x} by TARSKI:def 1;
  thus X,Funcs({x},X) are_equipotent
  proof
    take f;
    thus f is one-to-one
    proof
      let y,z;
      assume y in dom f & z in dom f;
      then
A3:   f.y = {x} --> y & f.z = {x} --> z by A1;
      ({x} --> y).x = y by A2,FUNCOP_1:7;
      hence thesis by A2,A3,FUNCOP_1:7;
    end;
    thus dom f = X by A1;
    thus rng f c= Funcs({x},X)
    proof
      let y;
      assume y in rng f;
      then consider z such that
A4:   z in dom f & y = f.z by FUNCT_1:def 3;
A5:   dom({x} --> z) = {x} & rng({x} --> z) = {z} by FUNCOP_1:8,13;
      y = {x} --> z & {z} c= X by A1,A4,ZFMISC_1:31;
      hence thesis by A5,FUNCT_2:def 2;
    end;
    let y;
    assume y in Funcs({x},X);
    then consider g such that
A6: y = g and
A7: dom g = {x} and
A8: rng g c= X by FUNCT_2:def 2;
A9: g.x in {g.x} by TARSKI:def 1;
A10: rng g = {g.x} by A7,FUNCT_1:4;
    then g = {x} --> (g.x) by A7,FUNCOP_1:9;
    then f.(g.x) = g by A1,A8,A10,A9;
    hence thesis by A1,A6,A8,A10,A9,FUNCT_1:def 3;
  end;
  hence thesis by CARD_1:5;
end;

theorem Th59:
  Funcs(X,{x}) = {X --> x}
proof
  thus Funcs(X,{x}) c= {X --> x}
  proof
    let y;
    assume y in Funcs(X,{x});
    then consider f such that
A1: y = f and
A2: dom f = X and
A3: rng f c= {x} by FUNCT_2:def 2;
A4: now
      set z = the Element of X;
A5:   X <> {} implies z in X;
A6:   dom({} --> x) = {};
      assume for z holds not z in X;
      hence f = X --> x by A2,A5,A6;
    end;
    now
      given z such that
A7:   z in X;
      f.z in rng f by A2,A7,FUNCT_1:def 3;
      then f.z = x & {f.z} c= rng f by A3,TARSKI:def 1,ZFMISC_1:31;
      then rng f = {x} by A3,XBOOLE_0:def 10;
      hence f = X --> x by A2,FUNCOP_1:9;
    end;
    hence thesis by A1,A4,TARSKI:def 1;
  end;
  let y;
  assume y in {X --> x};
  then
A8: y = X --> x by TARSKI:def 1;
  dom(X --> x) = X & rng(X --> x) c= {x} by FUNCOP_1:13;
  hence thesis by A8,FUNCT_2:def 2;
end;

theorem Th60:
  X1,Y1 are_equipotent & X2,Y2 are_equipotent implies Funcs(X1,X2)
  ,Funcs(Y1,Y2) are_equipotent & card Funcs(X1,X2) = card Funcs(Y1,Y2)
proof
  assume that
A1: X1,Y1 are_equipotent and
A2: X2,Y2 are_equipotent;
  consider f1 such that
A3: f1 is one-to-one and
A4: dom f1 = Y1 and
A5: rng f1 = X1 by A1,WELLORD2:def 4;
  consider f2 such that
A6: f2 is one-to-one and
A7: dom f2 = X2 and
A8: rng f2 = Y2 by A2,WELLORD2:def 4;
  Funcs(X1,X2),Funcs(Y1,Y2) are_equipotent
  proof
A9: rng(f1") = Y1 by A3,A4,FUNCT_1:33;
    deffunc F(Function) = f2*($1*f1);
    consider F being Function such that
A10: dom F = Funcs(X1,X2) & for g st g in Funcs(X1,X2) holds F.g = F(g
    ) from LambdaFS;
    take F;
    thus F is one-to-one
    proof
      let x,y;
      assume that
A11:  x in dom F and
A12:  y in dom F and
A13:  F.x = F.y;
      consider g1 being Function such that
A14:  x = g1 and
A15:  dom g1 = X1 and
A16:  rng g1 c= X2 by A10,A11,FUNCT_2:def 2;
A17:  F.x = f2*(g1*f1) by A10,A11,A14;
A18:  rng(g1*f1) c= X2 & dom(g1*f1) = Y1 by A4,A5,A15,A16,RELAT_1:27,28;
      consider g2 being Function such that
A19:  y = g2 and
A20:  dom g2 = X1 and
A21:  rng g2 c= X2 by A10,A12,FUNCT_2:def 2;
A22:  rng(g2*f1) c= X2 & dom(g2*f1) = Y1 by A4,A5,A20,A21,RELAT_1:27,28;
      F.x = f2*(g2*f1) by A10,A12,A13,A19;
      then
A23:  g1*f1 = g2*f1 by A6,A7,A17,A18,A22,FUNCT_1:27;
      now
        let z;
        assume z in X1;
        then consider z9 being set such that
A24:    z9 in Y1 & f1.z9 = z by A4,A5,FUNCT_1:def 3;
        thus g1.z = (g1*f1).z9 by A4,A24,FUNCT_1:13
          .= g2.z by A4,A23,A24,FUNCT_1:13;
      end;
      hence thesis by A14,A15,A19,A20,FUNCT_1:2;
    end;
    thus dom F = Funcs(X1,X2) by A10;
    thus rng F c= Funcs(Y1,Y2)
    proof
      let x;
      assume x in rng F;
      then consider y such that
A25:  y in dom F and
A26:  x = F.y by FUNCT_1:def 3;
      consider g such that
A27:  y = g and
A28:  dom g = X1 & rng g c= X2 by A10,A25,FUNCT_2:def 2;
      dom(g*f1) = Y1 & rng(g*f1) c= X2 by A4,A5,A28,RELAT_1:27,28;
      then
A29:  dom(f2*(g*f1)) = Y1 by A7,RELAT_1:27;
A30:  rng(f2*(g*f1)) c= Y2 by A8,RELAT_1:26;
      x = f2*(g*f1) by A10,A25,A26,A27;
      hence thesis by A29,A30,FUNCT_2:def 2;
    end;
    let x;
    assume x in Funcs(Y1,Y2);
    then consider g such that
A31: x = g and
A32: dom g = Y1 and
A33: rng g c= Y2 by FUNCT_2:def 2;
A34: f2*((f2"*g*(f1"))*f1) = f2*(f2"*g*(f1"*f1)) by RELAT_1:36
      .= f2*(f2"*g)*(f1"*f1) by RELAT_1:36
      .= f2*(f2")*g*(f1"*f1) by RELAT_1:36
      .= (id Y2)*g*(f1"*f1) by A6,A8,FUNCT_1:39
      .= (id Y2)*g*(id Y1) by A3,A4,FUNCT_1:39
      .= g*(id Y1) by A33,RELAT_1:53
      .= x by A31,A32,RELAT_1:52;
    dom(f2") = Y2 by A6,A8,FUNCT_1:33;
    then
A35: dom(f2"*g) = Y1 by A32,A33,RELAT_1:27;
    rng(f2") = X2 by A6,A7,FUNCT_1:33;
    then rng(f2"*g) c= X2 by RELAT_1:26;
    then
A36: rng(f2"*g*(f1")) c= X2 by A9,A35,RELAT_1:28;
    dom(f1") = X1 by A3,A5,FUNCT_1:33;
    then dom(f2"*g*(f1")) = X1 by A9,A35,RELAT_1:27;
    then
A37: f2"*g*(f1") in dom F by A10,A36,FUNCT_2:def 2;
    then F.(f2"*g*(f1")) = f2*((f2"*g*(f1"))*f1) by A10;
    hence thesis by A37,A34,FUNCT_1:def 3;
  end;
  hence thesis by CARD_1:5;
end;

theorem
  card X1 = card Y1 & card X2 = card Y2 implies card Funcs(X1,X2) = card
  Funcs(Y1,Y2)
proof
  assume card X1 = card Y1 & card X2 = card Y2;
  then X1,Y1 are_equipotent & X2,Y2 are_equipotent by CARD_1:5;
  hence thesis by Th60;
end;

theorem
  X1 misses X2 implies Funcs(X1 \/ X2,X),[:Funcs(X1,X),Funcs(X2,X):]
  are_equipotent & card Funcs(X1 \/ X2,X) = card [:Funcs(X1,X),Funcs(X2,X):]
proof
  deffunc F(Function) = [ $1|X1,$1|X2];
  consider f such that
A1: dom f = Funcs(X1 \/ X2,X) & for g st g in Funcs(X1 \/ X2,X) holds f.
  g = F(g) from LambdaFS;
  assume
A2: X1 misses X2;
  thus Funcs(X1 \/ X2,X),[:Funcs(X1,X),Funcs(X2,X):] are_equipotent
  proof
    take f;
    thus f is one-to-one
    proof
      let x,y;
      assume that
A3:   x in dom f and
A4:   y in dom f;
      consider g2 being Function such that
A5:   y = g2 and
A6:   dom g2 = X1 \/ X2 and
      rng g2 c= X by A1,A4,FUNCT_2:def 2;
A7:   f.y = [g2|X1,g2|X2] by A1,A4,A5;
      assume
A8:   f.x = f.y;
      consider g1 being Function such that
A9:   x = g1 and
A10:  dom g1 = X1 \/ X2 and
      rng g1 c= X by A1,A3,FUNCT_2:def 2;
A11:  f.x = [g1|X1,g1|X2] by A1,A3,A9;
      now
        let x;
        assume x in X1 \/ X2;
        then x in X1 or x in X2 by XBOOLE_0:def 3;
        then
        g1.x = g1|X1.x & g2.x = g2|X1.x or g1.x = g1|X2.x & g2.x = g2|X2.
        x by FUNCT_1:49;
        hence g1.x = g2.x by A11,A7,A8,XTUPLE_0:1;
      end;
      hence thesis by A9,A10,A5,A6,FUNCT_1:2;
    end;
    thus dom f = Funcs(X1 \/ X2,X) by A1;
    thus rng f c= [:Funcs(X1,X),Funcs(X2,X):]
    proof
      let x;
      assume x in rng f;
      then consider y such that
A12:  y in dom f and
A13:  x = f.y by FUNCT_1:def 3;
      consider g such that
A14:  y = g and
A15:  dom g = X1 \/ X2 and
A16:  rng g c= X by A1,A12,FUNCT_2:def 2;
      rng(g|X2) c= rng g by RELAT_1:70;
      then
A17:  rng(g|X2) c= X by A16,XBOOLE_1:1;
      rng(g|X1) c= rng g by RELAT_1:70;
      then
A18:  rng(g|X1) c= X by A16,XBOOLE_1:1;
      dom(g|X2) = X2 by A15,RELAT_1:62,XBOOLE_1:7;
      then
A19:  g|X2 in Funcs(X2,X) by A17,FUNCT_2:def 2;
      dom(g|X1) = X1 by A15,RELAT_1:62,XBOOLE_1:7;
      then g|X1 in Funcs(X1,X) by A18,FUNCT_2:def 2;
      then [g|X1,g|X2] in [:Funcs(X1,X),Funcs(X2,X):] by A19,ZFMISC_1:87;
      hence thesis by A1,A12,A13,A14;
    end;
    let x;
    assume
A20: x in [:Funcs(X1,X),Funcs(X2,X):];
    then
A21: x = [x`1,x`2] by MCART_1:21;
    x`1 in Funcs(X1,X) by A20,MCART_1:10;
    then consider g1 being Function such that
A22: x`1 = g1 and
A23: dom g1 = X1 and
A24: rng g1 c= X by FUNCT_2:def 2;
    x`2 in Funcs(X2,X) by A20,MCART_1:10;
    then consider g2 being Function such that
A25: x`2 = g2 and
A26: dom g2 = X2 and
A27: rng g2 c= X by FUNCT_2:def 2;
    rng(g1+*g2) c= rng g1 \/ rng g2 & rng g1 \/ rng g2 c= X \/ X by A24,A27,
FUNCT_4:17,XBOOLE_1:13;
    then
A28: rng(g1+*g2) c= X by XBOOLE_1:1;
    dom(g1+*g2) = X1 \/ X2 by A23,A26,FUNCT_4:def 1;
    then
A29: g1+*g2 in dom f by A1,A28,FUNCT_2:def 2;
    then f.(g1+*g2) = [(g1+*g2)|X1,(g1+*g2)|X2] by A1
      .= [(g1+*g2)|X1,g2] by A26,FUNCT_4:23
      .= x by A2,A21,A22,A23,A25,A26,FUNCT_4:33;
    hence thesis by A29,FUNCT_1:def 3;
  end;
  hence thesis by CARD_1:5;
end;

theorem
  Funcs([:X,Y:],Z),Funcs(X,Funcs(Y,Z)) are_equipotent & card Funcs([:X,Y
  :],Z) = card Funcs(X,Funcs(Y,Z))
proof
  deffunc F(Function) = curry $1;
  consider f such that
A1: dom f = Funcs([:X,Y:],Z) & for g st g in Funcs([:X,Y:],Z) holds f.g
  = F(g) from LambdaFS;
A2: now
    assume
A3: [:X,Y:] <> {};
    thus Funcs([:X,Y:],Z),Funcs(X,Funcs(Y,Z)) are_equipotent
    proof
      take f;
      thus f is one-to-one
      proof
        let x1,x2;
        assume that
A4:     x1 in dom f and
A5:     x2 in dom f and
A6:     f.x1 = f.x2;
        consider g2 such that
A7:     x2 = g2 and
A8:     dom g2 = [:X,Y:] and
        rng g2 c= Z by A1,A5,FUNCT_2:def 2;
A9:     f.x2 = curry g2 by A1,A5,A7;
        consider g1 such that
A10:    x1 = g1 and
A11:    dom g1 = [:X,Y:] and
        rng g1 c= Z by A1,A4,FUNCT_2:def 2;
        f.x1 = curry g1 by A1,A4,A10;
        hence thesis by A6,A10,A11,A7,A8,A9,Th44;
      end;
      thus dom f = Funcs([:X,Y:],Z) by A1;
      thus rng f c= Funcs(X,Funcs(Y,Z))
      proof
        let y;
        assume y in rng f;
        then consider x such that
A12:    x in dom f and
A13:    y = f.x by FUNCT_1:def 3;
        consider g such that
A14:    x = g and
A15:    dom g = [:X,Y:] and
A16:    rng g c= Z by A1,A12,FUNCT_2:def 2;
A17:    dom curry g = X by A3,A15,Th24;
        rng curry g c= Funcs(Y,rng g) & Funcs(Y,rng g) c= Funcs(Y,Z) by A15,A16
,Th35,Th56;
        then
A18:    rng curry g c= Funcs(Y,Z) by XBOOLE_1:1;
        y = curry g by A1,A12,A13,A14;
        hence thesis by A17,A18,FUNCT_2:def 2;
      end;
      let y;
      assume y in Funcs(X,Funcs(Y,Z));
      then consider g such that
A19:  y = g and
A20:  dom g = X and
A21:  rng g c= Funcs(Y,Z) by FUNCT_2:def 2;
      dom uncurry g = [:X,Y:] & rng uncurry g c= Z by A20,A21,Th26,Th41;
      then
A22:  uncurry g in Funcs([:X,Y:],Z) by FUNCT_2:def 2;
      Y <> {} by A3,ZFMISC_1:90;
      then curry uncurry g = g by A21,Th48;
      then f.(uncurry g) = y by A1,A19,A22;
      hence thesis by A1,A22,FUNCT_1:def 3;
    end;
    hence card Funcs([:X,Y:],Z) = card Funcs(X,Funcs(Y,Z)) by CARD_1:5;
  end;
  now
    assume
A23: [:X,Y:] = {};
    then
A24: Funcs([:X,Y:],Z) = {{}} by Th57;
A25: now
      assume Y = {};
      then Funcs(Y,Z) = {{}} by Th57;
      then Funcs(X,Funcs(Y,Z)) = {X --> {}} by Th59;
      hence Funcs([:X,Y:],Z),Funcs(X,Funcs(Y,Z)) are_equipotent by A24,
CARD_1:28;
    end;
    X = {} or Y = {} by A23;
    hence Funcs([:X,Y:],Z),Funcs(X,Funcs(Y,Z)) are_equipotent by A24,A25,Th57;
    X = {} implies Funcs(X,Funcs(Y,Z)) = {{}} by Th57;
    hence card Funcs([:X,Y:],Z) = card Funcs(X,Funcs(Y,Z)) by A23,A25,Th57,
CARD_1:5;
  end;
  hence thesis by A2;
end;

theorem
  Funcs(Z,[:X,Y:]),[:Funcs(Z,X),Funcs(Z,Y):] are_equipotent & card Funcs
  (Z,[:X,Y:]) = card [:Funcs(Z,X),Funcs(Z,Y):]
proof
  deffunc F(Function) = [pr1(X,Y)*$1,pr2(X,Y)*$1];
  consider f such that
A1: dom f = Funcs(Z,[:X,Y:]) & for g st g in Funcs(Z,[:X,Y:]) holds f.g
  = F(g) from LambdaFS;
  thus Funcs(Z,[:X,Y:]),[:Funcs(Z,X),Funcs(Z,Y):] are_equipotent
  proof
    take f;
    thus f is one-to-one
    proof
      let x1,x2;
      assume that
A2:   x1 in dom f and
A3:   x2 in dom f and
A4:   f.x1 = f.x2;
      consider g1 such that
A5:   x1 = g1 and
A6:   dom g1 = Z and
A7:   rng g1 c= [:X,Y:] by A1,A2,FUNCT_2:def 2;
      consider g2 such that
A8:   x2 = g2 and
A9:   dom g2 = Z and
A10:  rng g2 c= [:X,Y:] by A1,A3,FUNCT_2:def 2;
      [pr1(X,Y)*g1,pr2(X,Y)*g1] = f.x1 by A1,A2,A5
        .= [pr1(X,Y)*g2,pr2(X,Y)*g2] by A1,A3,A4,A8;
      then
A11:  pr1(X,Y)*g1 = pr1(X,Y)*g2 & pr2(X,Y)*g1 = pr2(X,Y)*g2 by XTUPLE_0:1;
      now
        let x;
        assume
A12:    x in Z;
        then
A13:    (pr2(X,Y)*g1).x = pr2(X,Y).(g1.x) by A6,FUNCT_1:13;
A14:    g2.x in rng g2 by A9,A12,FUNCT_1:def 3;
        then
A15:    g2.x = [(g2.x)`1,(g2.x)`2] by A10,MCART_1:21;
A16:    g1.x in rng g1 by A6,A12,FUNCT_1:def 3;
        then
A17:    g1.x = [(g1.x)`1,(g1.x)`2] by A7,MCART_1:21;
        (g2.x)`1 in X & (g2.x)`2 in Y by A10,A14,MCART_1:10;
        then
A18:    pr1(X,Y).((g2.x)`1,(g2.x)`2) = (g2.x)`1 & pr2(X,Y).((g2.x)`1,(g2.
        x)`2) = (g2 .x)`2 by FUNCT_3:def 4,def 5;
        (g1.x)`1 in X & (g1.x)`2 in Y by A7,A16,MCART_1:10;
        then
A19:    pr1(X,Y).((g1.x)`1,(g1.x)`2) = (g1.x)`1 & pr2(X,Y).((g1.x)`1,(g1.
        x)`2) = (g1 .x)`2 by FUNCT_3:def 4,def 5;
        (pr1(X,Y)*g1).x = pr1(X,Y).(g1.x) & (pr1(X,Y)*g2).x = pr1(X,Y).(
        g2.x) by A6,A9,A12,FUNCT_1:13;
        hence g1.x = g2.x by A9,A11,A12,A13,A17,A15,A19,A18,FUNCT_1:13;
      end;
      hence thesis by A5,A6,A8,A9,FUNCT_1:2;
    end;
    thus dom f = Funcs(Z,[:X,Y:]) by A1;
    thus rng f c= [:Funcs(Z,X),Funcs(Z,Y):]
    proof
      let x;
      assume x in rng f;
      then consider y such that
A20:  y in dom f and
A21:  x = f.y by FUNCT_1:def 3;
      consider g such that
A22:  y = g and
A23:  dom g = Z & rng g c= [:X,Y:] by A1,A20,FUNCT_2:def 2;
A24:  rng(pr1(X,Y)*g) c= X;
A25:  rng(pr2(X,Y)*g) c= Y;
      dom pr2(X,Y) = [:X,Y:] by FUNCT_3:def 5;
      then dom(pr2(X,Y)*g) = Z by A23,RELAT_1:27;
      then
A26:  pr2(X,Y)*g in Funcs(Z,Y) by A25,FUNCT_2:def 2;
      dom pr1(X,Y) = [:X,Y:] by FUNCT_3:def 4;
      then dom(pr1(X,Y)*g) = Z by A23,RELAT_1:27;
      then
A27:  pr1(X,Y)*g in Funcs(Z,X) by A24,FUNCT_2:def 2;
      x = [pr1(X,Y)*g,pr2(X,Y)*g] by A1,A20,A21,A22;
      hence thesis by A27,A26,ZFMISC_1:87;
    end;
    let x;
    assume
A28: x in [:Funcs(Z,X),Funcs(Z,Y):];
    then
A29: x = [x`1,x`2] by MCART_1:21;
    x`2 in Funcs(Z,Y) by A28,MCART_1:10;
    then consider g2 such that
A30: x`2 = g2 and
A31: dom g2 = Z and
A32: rng g2 c= Y by FUNCT_2:def 2;
    x`1 in Funcs(Z,X) by A28,MCART_1:10;
    then consider g1 such that
A33: x`1 = g1 and
A34: dom g1 = Z and
A35: rng g1 c= X by FUNCT_2:def 2;
    rng <:g1,g2:> c= [:rng g1,rng g2:] & [:rng g1,rng g2:] c= [:X,Y:] by A35
,A32,FUNCT_3:51,ZFMISC_1:96;
    then
A36: rng <:g1,g2:> c= [:X,Y:] by XBOOLE_1:1;
    dom <:g1,g2:> = Z /\ Z by A34,A31,FUNCT_3:def 7;
    then
A37: <:g1,g2:> in Funcs(Z,[:X,Y:]) by A36,FUNCT_2:def 2;
    pr1(X,Y)*<:g1,g2:> = g1 & pr2(X,Y)*<:g1,g2:> = g2 by A34,A35,A31,A32,
FUNCT_3:52;
    then f.<:g1,g2:> = [g1,g2] by A1,A37;
    hence thesis by A1,A29,A33,A30,A37,FUNCT_1:def 3;
  end;
  hence thesis by CARD_1:5;
end;

theorem
  x <> y implies Funcs(X,{x,y}),bool X are_equipotent & card Funcs(X,{x,
  y}) = card bool X
proof
  deffunc F(Function) = $1"{x};
  consider f such that
A1: dom f = Funcs(X,{x,y}) & for g st g in Funcs(X,{x,y}) holds f.g = F(
  g) from LambdaFS;
  assume
A2: x <> y;
  thus Funcs(X,{x,y}),bool X are_equipotent
  proof
    deffunc F(set) = x;
    take f;
    thus f is one-to-one
    proof
      let x1,x2;
      assume that
A3:   x1 in dom f and
A4:   x2 in dom f and
A5:   f.x1 = f.x2;
      consider g2 being Function such that
A6:   x2 = g2 and
A7:   dom g2 = X and
A8:   rng g2 c= {x,y} by A1,A4,FUNCT_2:def 2;
A9:   f.x2 = g2"{x} by A1,A4,A6;
      consider g1 being Function such that
A10:  x1 = g1 and
A11:  dom g1 = X and
A12:  rng g1 c= {x,y} by A1,A3,FUNCT_2:def 2;
A13:  f.x1 = g1"{x} by A1,A3,A10;
      now
        let z such that
A14:    z in X;
A15:    now
          assume
A16:      not g1.z in {x};
          then
A17:      g1.z <> x by TARSKI:def 1;
          not z in g2"{x} by A5,A13,A9,A16,FUNCT_1:def 7;
          then not g2.z in {x} by A7,A14,FUNCT_1:def 7;
          then
A18:      g2.z <> x by TARSKI:def 1;
          g1.z in rng g1 by A11,A14,FUNCT_1:def 3;
          then
A19:      g1.z = y by A12,A17,TARSKI:def 2;
          g2.z in rng g2 by A7,A14,FUNCT_1:def 3;
          hence g1.z = g2.z by A8,A18,A19,TARSKI:def 2;
        end;
        now
          assume
A20:      g1.z in {x};
          then z in g2"{x} by A5,A11,A13,A9,A14,FUNCT_1:def 7;
          then g2.z in {x} by FUNCT_1:def 7;
          then g2.z = x by TARSKI:def 1;
          hence g1.z = g2.z by A20,TARSKI:def 1;
        end;
        hence g1.z = g2.z by A15;
      end;
      hence thesis by A10,A11,A6,A7,FUNCT_1:2;
    end;
    thus dom f = Funcs(X,{x,y}) by A1;
    thus rng f c= bool X
    proof
      let z;
      assume z in rng f;
      then consider t being set such that
A21:  t in dom f and
A22:  z = f.t by FUNCT_1:def 3;
      consider g such that
A23:  t = g and
A24:  dom g = X and
      rng g c= {x,y} by A1,A21,FUNCT_2:def 2;
A25:  g"{x} c= X by A24,RELAT_1:132;
      z = g"{x} by A1,A21,A22,A23;
      hence thesis by A25;
    end;
    deffunc G(set) = y;
    let z;
    defpred P[set] means $1 in z;
    consider g such that
A26: dom g = X & for t st t in X holds (P[t] implies g.t = F(t)) & (
    not P[t] implies g.t = G(t)) from PARTFUN1:sch 1;
    assume
A27: z in bool X;
A28: g"{x} = z
    proof
      thus g"{x} c= z
      proof
        let t;
        assume
A29:    t in g"{x};
        then g.t in {x} by FUNCT_1:def 7;
        then
A30:    g.t = x by TARSKI:def 1;
        t in dom g by A29,FUNCT_1:def 7;
        hence thesis by A2,A26,A30;
      end;
      let t;
      assume
A31:  t in z;
      then g.t = x by A27,A26;
      then g.t in {x} by TARSKI:def 1;
      hence thesis by A27,A26,A31,FUNCT_1:def 7;
    end;
    rng g c= {x,y}
    proof
      let t;
      assume t in rng g;
      then ex v being set st v in dom g & t = g.v by FUNCT_1:def 3;
      then t = x or t = y by A26;
      hence thesis by TARSKI:def 2;
    end;
    then
A32: g in Funcs(X,{x,y}) by A26,FUNCT_2:def 2;
    then f.g = g"{x} by A1;
    hence thesis by A1,A32,A28,FUNCT_1:def 3;
  end;
  hence thesis by CARD_1:5;
end;

theorem
  x <> y implies Funcs({x,y},X),[:X,X:] are_equipotent & card Funcs({x,y
  },X) = card [:X,X:]
proof
  deffunc F(Function) = [ $1.x,$1.y];
  consider f such that
A1: dom f = Funcs({x,y},X) & for g st g in Funcs({x,y},X) holds f.g = F(
  g) from LambdaFS;
  assume
A2: x <> y;
  thus Funcs({x,y},X),[:X,X:] are_equipotent
  proof
    defpred P[set] means $1 = x;
    take f;
    thus f is one-to-one
    proof
      let x1,x2;
      assume that
A3:   x1 in dom f and
A4:   x2 in dom f and
A5:   f.x1 = f.x2;
      consider g2 such that
A6:   x2 = g2 and
A7:   dom g2 = {x,y} and
      rng g2 c= X by A1,A4,FUNCT_2:def 2;
      consider g1 such that
A8:   x1 = g1 and
A9:   dom g1 = {x,y} and
      rng g1 c= X by A1,A3,FUNCT_2:def 2;
A10:  [g1.x,g1.y] = f.x1 by A1,A3,A8
        .= [g2.x,g2.y] by A1,A4,A5,A6;
      now
        let z;
        assume z in {x,y};
        then z = x or z = y by TARSKI:def 2;
        hence g1.z = g2.z by A10,XTUPLE_0:1;
      end;
      hence thesis by A8,A9,A6,A7,FUNCT_1:2;
    end;
    thus dom f = Funcs({x,y},X) by A1;
    thus rng f c= [:X,X:]
    proof
      let z;
      assume z in rng f;
      then consider t such that
A11:  t in dom f and
A12:  z = f.t by FUNCT_1:def 3;
      consider g such that
A13:  t = g and
A14:  dom g = {x,y} and
A15:  rng g c= X by A1,A11,FUNCT_2:def 2;
      x in {x,y} by TARSKI:def 2;
      then
A16:  g.x in rng g by A14,FUNCT_1:def 3;
      y in {x,y} by TARSKI:def 2;
      then
A17:  g.y in rng g by A14,FUNCT_1:def 3;
      z = [g.x,g.y] by A1,A11,A12,A13;
      hence thesis by A15,A16,A17,ZFMISC_1:87;
    end;
    let z;
    deffunc F(set) = z`1;
    deffunc G(set) = z`2;
    consider g such that
A18: dom g = {x,y} & for t st t in {x,y} holds (P[t] implies g.t = F(t
    )) & (not P[t] implies g.t = G(t)) from PARTFUN1:sch 1;
    x in {x,y} by TARSKI:def 2;
    then
A19: g.x = z`1 by A18;
    y in {x,y} by TARSKI:def 2;
    then
A20: g.y = z`2 by A2,A18;
    assume
A21: z in [:X,X:];
    then
A22: z = [z`1,z`2] by MCART_1:21;
A23: z`1 in X & z`2 in X by A21,MCART_1:10;
    rng g c= X
    proof
      let t;
      assume t in rng g;
      then ex a being set st a in dom g & t = g.a by FUNCT_1:def 3;
      hence thesis by A23,A18;
    end;
    then
A24: g in Funcs({x,y},X) by A18,FUNCT_2:def 2;
    then f.g = [g.x,g.y] by A1;
    hence thesis by A1,A22,A24,A19,A20,FUNCT_1:def 3;
  end;
  hence thesis by CARD_1:5;
end;

begin 


notation
  synonym op0 for 0;
end;

definition
  redefine func op0 -> Element of 1;
  coherence by CARD_1:49,TARSKI:def 1;
end;

definition
  func op1 equals
  0 .--> 0;
  coherence;
  func op2 equals
  (0,0) :-> 0;
  coherence;
end;

definition
  redefine func op1 -> UnOp of 1;
  coherence by CARD_1:49;
  redefine func op2 -> BinOp of 1;
  coherence by CARD_1:49;
end;



reserve C,D,E for non empty set;
reserve c for Element of C,
  d for Element of D;

definition
  let D,X,E;
  let F be FUNCTION_DOMAIN of X,E;
  let f be Function of D,F;
  let d be Element of D;
  redefine func f.d -> Element of F;
  coherence
  proof
    thus f.d is Element of F;
  end;
end;

reserve f for Function of [:C,D:],E;

theorem Th1:
  curry f is Function of C,Funcs(D,E)
proof
A1: dom f = [:C,D:] by FUNCT_2:def 1;
A2: rng curry f c= Funcs(D,E)
  proof
A3: rng curry f c= Funcs(D,rng f) by A1,Th35;
    let x be set;
    assume x in rng curry f;
    then consider g being Function such that
A4: x = g and
A5: dom g = D and
A6: rng g c= rng f by A3,FUNCT_2:def 2;
    rng g c= E by A6,XBOOLE_1:1;
    then g is Function of D,E by A5,FUNCT_2:def 1,RELSET_1:4;
    hence thesis by A4,FUNCT_2:8;
  end;
  dom curry f = C by A1,Th24;
  hence thesis by A2,FUNCT_2:def 1,RELSET_1:4;
end;

theorem Th2:
  curry' f is Function of D,Funcs(C,E)
proof
A1: dom f = [:C,D:] by FUNCT_2:def 1;
A2: rng curry' f c= Funcs(C,E)
  proof
A3: rng curry' f c= Funcs(C,rng f) by A1,Th35;
    let x be set;
    assume x in rng curry' f;
    then consider g being Function such that
A4: x = g and
A5: dom g = C and
A6: rng g c= rng f by A3,FUNCT_2:def 2;
    rng g c= E by A6,XBOOLE_1:1;
    then g is Function of C,E by A5,FUNCT_2:def 1,RELSET_1:4;
    hence thesis by A4,FUNCT_2:8;
  end;
  dom curry' f = D by A1,Th24;
  hence thesis by A2,FUNCT_2:def 1,RELSET_1:4;
end;

definition
  let C,D,E,f;
  redefine func curry f -> Function of C,Funcs(D,E);
  coherence by Th1;
  redefine func curry' f -> Function of D,Funcs(C,E);
  coherence by Th2;
end;

theorem
  f.(c,d) = ((curry f).c).d
proof
  [c,d] in [:C,D:];
  then [c,d] in dom f by FUNCT_2:def 1;
  hence thesis by Th20;
end;

theorem
  f.(c,d) = ((curry' f).d).c
proof
  [c,d] in [:C,D:];
  then [c,d] in dom f by FUNCT_2:def 1;
  hence thesis by Th22;
end;



definition
  let A,B,C be non empty set;
  let f be Function of A, Funcs(B,C);
  redefine func uncurry f -> Function of [:A,B:],C;
  coherence
  proof
A1: rng f c= Funcs(B,C);
    then
A2: rng uncurry f c= C by Th41;
    dom uncurry f = [:dom f,B:] by A1,Th26
      .= [:A,B:] by FUNCT_2:def 1;
    hence thesis by A2,FUNCT_2:def 1,RELSET_1:4;
  end;
end;

theorem
  for A,B,C being non empty set, f being Function of A,Funcs(B,C)
  holds curry uncurry f = f
proof
  let A,B,C be non empty set, f be Function of A,Funcs(B,C);
  rng f c= Funcs(B,C);
  hence thesis by Th48;
end;

theorem
  for A,B,C being non empty set, f being Function of A, Funcs(B,C)
for a being Element of A, b being Element of B
      holds (uncurry f).(a,b) = f.a.b
proof
  let A,B,C be non empty set, f be Function of A, Funcs(B,C);
  let a be Element of A, b be Element of B;
  dom f = A & dom(f.a) = B by FUNCT_2:def 1;
  hence thesis by Th38;
end;















begin

reserve x,y,X,Y for set;
reserve C,D,E for non empty set;
reserve SC for Subset of C;
reserve SD for Subset of D;
reserve SE for Subset of E;
reserve c,c1,c2 for Element of C;
reserve d,d1,d2 for Element of D;
reserve e for Element of E;
reserve f,f1,g for PartFunc of C,D;
reserve t for PartFunc of D,C;
reserve s for PartFunc of D,E;
reserve h for PartFunc of C,E;
reserve F for PartFunc of D,D;

theorem Th1:
  dom f = dom g & (for c st c in dom f holds f/.c = g/.c) implies f = g
proof
  assume that
A1: dom f = dom g and
A2: for c st c in dom f holds f/.c = g/.c;
  now
    let x;
    assume
A3: x in dom f;
    then reconsider y=x as Element of C;
    f/.y=g/.y by A2,A3;
    then (f qua Function).y = g/.y by A3,PARTFUN1:def 6;
    hence (f qua Function).x = (g qua Function).x by A1,A3,PARTFUN1:def 6;
  end;
  hence thesis by A1,FUNCT_1:2;
end;

theorem Th2:
  y in rng f iff ex c st c in dom f & y = f/.c
proof
  thus y in rng f implies ex c st c in dom f & y = f/.c
  proof
    assume y in rng f;
    then consider x such that
A1: x in dom f and
A2: y = (f qua Function).x by FUNCT_1:def 3;
    reconsider x as Element of C by A1;
    take x;
    thus thesis by A1,A2,PARTFUN1:def 6;
  end;
  given c such that
A3: c in dom f and
A4: y = f/.c;
  (f qua Function).c in rng f by A3,FUNCT_1:def 3;
  hence thesis by A3,A4,PARTFUN1:def 6;
end;

theorem Th3:
  h = s*f iff (for c holds c in dom h iff c in dom f & f/.c in dom
  s) & for c st c in dom h holds h/.c = s/.(f/.c)
proof
  thus h = s*f implies (for c holds c in dom h iff c in dom f & f/.c in dom s)
  & for c st c in dom h holds h/.c = s/.(f/.c)
  proof
    assume
A1: h = s*f;
A2: now
      let c;
      thus c in dom h implies c in dom f & f/.c in dom s
      proof
        assume c in dom h;
        then c in dom f & (f qua Function).c in dom s by A1,FUNCT_1:11;
        hence thesis by PARTFUN1:def 6;
      end;
      assume that
A3:   c in dom f and
A4:   f/.c in dom s;
      (f qua Function).c in dom s by A3,A4,PARTFUN1:def 6;
      hence c in dom h by A1,A3,FUNCT_1:11;
    end;
    hence for c holds c in dom h iff c in dom f & f/.c in dom s;
    let c;
    assume
A5: c in dom h;
    then (h qua Function).c = (s qua Function).((f qua Function).c) by A1,
FUNCT_1:12;
    then
A6: h/.c = (s qua Function).((f qua Function).c) by A5,PARTFUN1:def 6;
    c in dom f by A2,A5;
    then
A7: h/.c = (s qua Function).(f/.c) by A6,PARTFUN1:def 6;
    f/.c in dom s by A2,A5;
    hence thesis by A7,PARTFUN1:def 6;
  end;
  assume that
A8: for c holds c in dom h iff c in dom f & f/.c in dom s and
A9: for c st c in dom h holds h/.c = s/.(f/.c);
A10: now
    let x;
    thus x in dom h implies x in dom f & (f qua Function).x in dom s
    proof
      assume
A11:  x in dom h;
      then reconsider y=x as Element of C;
      y in dom f & f/.y in dom s by A8,A11;
      hence thesis by PARTFUN1:def 6;
    end;
    thus x in dom f & (f qua Function).x in dom s implies x in dom h
    proof
      assume that
A12:  x in dom f and
A13:  (f qua Function).x in dom s;
      reconsider y=x as Element of C by A12;
      f/.y in dom s by A12,A13,PARTFUN1:def 6;
      hence thesis by A8,A12;
    end;
  end;
  now
    let x;
    assume
A14: x in dom h;
    then reconsider y=x as Element of C;
    h/.y = s/.(f/.y) by A9,A14;
    then
A15: (h qua Function).y = s/.(f/.y) by A14,PARTFUN1:def 6;
    f/.y in dom s by A8,A14;
    then
A16: (h qua Function).x = (s qua Function).(f/.y) by A15,PARTFUN1:def 6;
    y in dom f by A8,A14;
    hence (h qua Function).x = (s qua Function).((f qua Function).x) by A16,
PARTFUN1:def 6;
  end;
  hence thesis by A10,FUNCT_1:10;
end;

theorem Th4:
  c in dom f & f/.c in dom s implies (s*f)/.c = s/.(f/.c)
proof
  assume c in dom f & f/.c in dom s;
  then c in dom (s*f) by Th3;
  hence thesis by Th3;
end;

theorem
  rng f c= dom s & c in dom f implies (s*f)/.c = s/.(f/.c)
proof
  assume that
A1: rng f c= dom s and
A2: c in dom f;
  f/.c in rng f by A2,Th2;
  hence thesis by A1,A2,Th4;
end;

definition
  let D;
  let SD;
  redefine func id SD -> PartFunc of D,D;
  coherence
  proof
    dom id SD = SD & rng id SD = SD by RELAT_1:45;
    hence thesis by RELSET_1:4;
  end;
end;

theorem Th6:
  F = id SD iff dom F = SD & for d st d in SD holds F/.d = d
proof
  thus F = id SD implies dom F = SD & for d st d in SD holds F/.d = d
  proof
    assume
A1: F = id SD;
    hence
A2: dom F = SD by RELAT_1:45;
    let d;
    assume
A3: d in SD;
    then (F qua Function).d = d by A1,FUNCT_1:18;
    hence thesis by A2,A3,PARTFUN1:def 6;
  end;
  assume that
A4: dom F = SD and
A5: for d st d in SD holds F/.d = d;
  now
    let x;
    assume
A6: x in SD;
    then reconsider x1=x as Element of D;
    F/.x1=x1 by A5,A6;
    hence (F qua Function).x = x by A4,A6,PARTFUN1:def 6;
  end;
  hence thesis by A4,FUNCT_1:17;
end;

theorem
  d in dom F /\ SD implies F/.d = (F*id SD)/.d
proof
  assume
A1: d in dom F /\ SD;
  then
A2: d in dom F by XBOOLE_0:def 4;
  (F qua Function).d = ((F*(id SD)) qua Function).d by A1,FUNCT_1:20;
  then
A3: F/.d = ((F*(id SD)) qua Function).d by A2,PARTFUN1:def 6;
A4: d in SD by A1,XBOOLE_0:def 4;
  then
A5: d in dom id SD by RELAT_1:45;
  (id SD)/.d in dom F by A2,A4,Th6;
  then d in dom (F*(id SD)) by A5,Th3;
  hence thesis by A3,PARTFUN1:def 6;
end;

theorem
  d in dom((id SD)*F) iff d in dom F & F/.d in SD
proof
  thus d in dom((id SD)*F) implies d in dom F & F/.d in SD
  proof
    assume
A1: d in dom((id SD)*F);
    then F/.d in dom (id SD) by Th3;
    hence thesis by A1,Th3,RELAT_1:45;
  end;
  assume that
A2: d in dom F and
A3: F/.d in SD;
  F/.d in dom (id SD) by A3,RELAT_1:45;
  hence thesis by A2,Th3;
end;

theorem
  (for c1,c2 st c1 in dom f & c2 in dom f & f/.c1 = f/.c2 holds c1 = c2)
  implies f is one-to-one
proof
  assume
A1: for c1,c2 st c1 in dom f & c2 in dom f & f/.c1 = f/.c2 holds c1 = c2;
  now
    let x,y;
    assume that
A2: x in dom f and
A3: y in dom f and
A4: (f qua Function).x = (f qua Function).y;
    reconsider y1 = y as Element of C by A3;
    reconsider x1 = x as Element of C by A2;
    f/.x1 = (f qua Function).y1 by A2,A4,PARTFUN1:def 6;
    then f/.x1 = f/.y1 by A3,PARTFUN1:def 6;
    hence x=y by A1,A2,A3;
  end;
  hence thesis by FUNCT_1:def 4;
end;

theorem
  f is one-to-one & x in dom f & y in dom f & f/.x = f/.y implies x = y
proof
  assume that
A1: f is one-to-one and
A2: x in dom f and
A3: y in dom f;
  assume f/.x = f/.y;
  then f.x = f/.y by A2,PARTFUN1:def 6
    .= f.y by A3,PARTFUN1:def 6;
  hence thesis by A1,A2,A3,FUNCT_1:def 4;
end;

definition
  let X,Y;
  let f be one-to-one PartFunc of X,Y;
  redefine func f" -> PartFunc of Y,X;
  coherence by PARTFUN1:9;
end;

theorem Th11:
  for f being one-to-one PartFunc of C,D holds for g be PartFunc
of D,C holds g = f" iff dom g = rng f & for d,c holds d in rng f & c = g/.d iff
  c in dom f & d = f/.c
proof
  let f be one-to-one PartFunc of C,D;
  let g be PartFunc of D,C;
  thus g = f" implies dom g = rng f & for d,c holds d in rng f & c = g/.d iff
  c in dom f & d = f/.c
  proof
    assume
A1: g = f";
    hence dom g = rng f by FUNCT_1:32;
    let d,c;
A2: dom g = rng f by A1,FUNCT_1:32;
    thus d in rng f & c = g/.d implies c in dom f & d = f/.c
    proof
      assume that
A3:   d in rng f and
A4:   c = g/.d;
      c = (g qua Function).d by A2,A3,A4,PARTFUN1:def 6;
      then c in dom f & d = (f qua Function).c by A1,A3,FUNCT_1:32;
      hence thesis by PARTFUN1:def 6;
    end;
    assume that
A5: c in dom f and
A6: d = f/.c;
    d = (f qua Function).c by A5,A6,PARTFUN1:def 6;
    then d in rng f & c = (g qua Function).d by A1,A5,FUNCT_1:32;
    hence thesis by A2,PARTFUN1:def 6;
  end;
  assume that
A7: dom g = rng f and
A8: for d,c holds d in rng f & c = g/.d iff c in dom f & d = f/.c and
A9: g <> f";
  now
    per cases by A9,Th1;
    suppose
      dom (f") <> dom g;
      hence contradiction by A7,FUNCT_1:33;
    end;
    suppose
      ex d st d in dom (f") & f"/.d <> g/.d;
      then consider d such that
A10:  d in dom (f") and
A11:  f"/.d <> g/.d;
      f"/.d in rng (f") by A10,Th2;
      then
A12:  f"/.d in dom f by FUNCT_1:33;
      d in rng f by A10,FUNCT_1:33;
      then d = (f qua Function).(((f") qua Function).d) by FUNCT_1:35;
      then d = (f qua Function).(f"/.d) by A10,PARTFUN1:def 6;
      then d = f/.(f"/.d) by A12,PARTFUN1:def 6;
      hence contradiction by A8,A11,A12;
    end;
  end;
  hence contradiction;
end;

theorem
  for f being one-to-one PartFunc of C,D st c in dom f holds c = f"/.(f
  /.c) & c = (f"*f)/.c
proof
  let f be one-to-one PartFunc of C,D;
  assume
A1: c in dom f;
  hence
A2: c = f"/.(f/.c) by Th11;
  f" = f";
  then f/.c in rng f by A1,Th11;
  then f/.c in dom (f") by FUNCT_1:33;
  hence thesis by A1,A2,Th4;
end;

theorem
  for f being one-to-one PartFunc of C,D st d in rng f holds d = f/.(f"
  /.d) & d = (f*(f"))/.d
proof
  let f be one-to-one PartFunc of C,D;
  assume
A1: d in rng f;
  then d = ((f*f") qua Function).d & d in dom (f*f") by FUNCT_1:35,37;
  hence thesis by A1,Th11,PARTFUN1:def 6;
end;

theorem
  f is one-to-one & dom f = rng t & rng f = dom t & (for c,d st c in dom
  f & d in dom t holds f/.c = d iff t/.d = c) implies t = f"
proof
  assume that
A1: f is one-to-one & dom f = rng t & rng f = dom t and
A2: for c,d st c in dom f & d in dom t holds f/.c = d iff t/.d = c;
  now
    let x,y;
    assume that
A3: x in dom f and
A4: y in dom t;
    reconsider y1=y as Element of D by A4;
    reconsider x1=x as Element of C by A3;
    thus (f qua Function).x = y implies (t qua Function).y = x
    proof
      assume (f qua Function).x = y;
      then f/.x1 = y1 by A3,PARTFUN1:def 6;
      then t/.y1 = x1 by A2,A3,A4;
      hence thesis by A4,PARTFUN1:def 6;
    end;
    assume (t qua Function).y = x;
    then t/.y1 = x1 by A4,PARTFUN1:def 6;
    then f/.x1 = y1 by A2,A3,A4;
    hence (f qua Function).x = y by A3,PARTFUN1:def 6;
  end;
  hence thesis by A1,FUNCT_1:38;
end;

theorem Th15:
  g = f|X iff dom g = dom f /\ X & for c st c in dom g holds g/.c = f/.c
proof
  thus g = f|X implies dom g = dom f /\ X & for c st c in dom g holds g/.c = f
  /.c
  proof
    assume
A1: g = f|X;
    hence dom g = dom f /\ X by RELAT_1:61;
    let c;
    assume
A2: c in dom g;
    then (g qua Function).c = (f qua Function).c by A1,FUNCT_1:47;
    then
A3: g/.c = (f qua Function).c by A2,PARTFUN1:def 6;
    dom g = dom f /\ X by A1,RELAT_1:61;
    then c in dom f by A2,XBOOLE_0:def 4;
    hence thesis by A3,PARTFUN1:def 6;
  end;
  assume that
A4: dom g = dom f /\ X and
A5: for c st c in dom g holds g/.c = f/.c;
  now
    let x;
    assume
A6: x in dom g;
    then reconsider y=x as Element of C;
    g/.y = f/.y by A5,A6;
    then
A7: (g qua Function).y = f/.y by A6,PARTFUN1:def 6;
    x in dom f by A4,A6,XBOOLE_0:def 4;
    hence (g qua Function).x = (f qua Function).x by A7,PARTFUN1:def 6;
  end;
  hence thesis by A4,FUNCT_1:46;
end;

theorem Th16:
  c in dom f /\ X implies f|X/.c = f/.c
proof
  assume c in dom f /\ X;
  then c in dom (f|X) by RELAT_1:61;
  hence thesis by Th15;
end;

theorem
  c in dom f & c in X implies f|X/.c = f/.c
proof
  assume c in dom f & c in X;
  then c in dom f /\ X by XBOOLE_0:def 4;
  hence thesis by Th16;
end;

theorem
  c in dom f & c in X implies f/.c in rng (f|X)
proof
  assume that
A1: c in dom f and
A2: c in X;
  (f qua Function).c in rng (f|X) by A1,A2,FUNCT_1:50;
  hence thesis by A1,PARTFUN1:def 6;
end;

definition
  let C,D;
  let X,f;
  redefine func X|`f -> PartFunc of C,D;
  coherence by PARTFUN1:13;
end;

theorem Th19:
  g = X|`f iff (for c holds c in dom g iff c in dom f & f/.c in X)
  & for c st c in dom g holds g/.c = f/.c
proof
  thus g = X|`f implies (for c holds c in dom g iff c in dom f & f/.c in X) &
  for c st c in dom g holds g/.c = f/.c
  proof
    assume
A1: g = X|`f;
    now
      let c;
      thus c in dom g implies c in dom f & f/.c in X
      proof
        assume c in dom g;
        then c in dom f & (f qua Function).c in X by A1,FUNCT_1:54;
        hence thesis by PARTFUN1:def 6;
      end;
      assume that
A2:   c in dom f and
A3:   f/.c in X;
      (f qua Function).c in X by A2,A3,PARTFUN1:def 6;
      hence c in dom g by A1,A2,FUNCT_1:54;
    end;
    hence for c holds c in dom g iff c in dom f & f/.c in X;
    let c;
    assume
A4: c in dom g;
    then (g qua Function).c = (f qua Function).c by A1,FUNCT_1:55;
    then
A5: g/.c = (f qua Function).c by A4,PARTFUN1:def 6;
    c in dom f by A1,A4,FUNCT_1:54;
    hence thesis by A5,PARTFUN1:def 6;
  end;
  assume that
A6: for c holds c in dom g iff c in dom f & f/.c in X and
A7: for c st c in dom g holds g/.c = f/.c;
A8: now
    let x;
    thus x in dom g implies x in dom f & (f qua Function).x in X
    proof
      assume
A9:   x in dom g;
      then reconsider x1=x as Element of C;
      x1 in dom f & f/.x1 in X by A6,A9;
      hence thesis by PARTFUN1:def 6;
    end;
    assume that
A10: x in dom f and
A11: (f qua Function).x in X;
    reconsider x1=x as Element of C by A10;
    f/.x1 in X by A10,A11,PARTFUN1:def 6;
    hence x in dom g by A6,A10;
  end;
  now
    let x;
    assume
A12: x in dom g;
    then reconsider x1=x as Element of C;
    g/.x1 = f/.x1 by A7,A12;
    then
A13: (g qua Function).x1 = f/.x1 by A12,PARTFUN1:def 6;
    x1 in dom f by A6,A12;
    hence (g qua Function).x = (f qua Function).x by A13,PARTFUN1:def 6;
  end;
  hence thesis by A8,FUNCT_1:53;
end;

theorem
  c in dom (X|`f) iff c in dom f & f/.c in X by Th19;

theorem
  c in dom (X|`f) implies X|`f/.c = f/.c by Th19;

theorem Th22:
  SD = f.: X iff for d holds d in SD iff ex c st c in dom f & c in X & d = f/.c
proof
  thus SD = f.:X implies for d holds d in SD iff ex c st c in dom f & c in X &
  d = f/.c
  proof
    assume
A1: SD = f.:X;
    let d;
    thus d in SD implies ex c st c in dom f & c in X & d = f/.c
    proof
      assume d in SD;
      then consider x such that
A2:   x in dom f and
A3:   x in X and
A4:   d = (f qua Function).x by A1,FUNCT_1:def 6;
      reconsider x as Element of C by A2;
      take x;
      thus x in dom f & x in X by A2,A3;
      thus thesis by A2,A4,PARTFUN1:def 6;
    end;
    given c such that
A5: c in dom f and
A6: c in X & d = f/.c;
    f/.c = (f qua Function).c by A5,PARTFUN1:def 6;
    hence thesis by A1,A5,A6,FUNCT_1:def 6;
  end;
  assume that
A7: for d holds d in SD iff ex c st c in dom f & c in X & d = f/.c and
A8: SD <> f.:X;
  consider x such that
A9: not (x in SD iff x in f.:X) by A8,TARSKI:1;
  now
    per cases by A9;
    suppose
A10:  x in SD & not x in f.:X;
      then reconsider x as Element of D;
      consider c such that
A11:  c in dom f and
A12:  c in X and
A13:  x = f/.c by A7,A10;
      x = (f qua Function).c by A11,A13,PARTFUN1:def 6;
      hence contradiction by A10,A11,A12,FUNCT_1:def 6;
    end;
    suppose
A14:  x in f.:X & not x in SD;
      then consider y such that
A15:  y in dom f and
A16:  y in X and
A17:  x = (f qua Function).y by FUNCT_1:def 6;
      reconsider y as Element of C by A15;
      x = f/.y by A15,A17,PARTFUN1:def 6;
      hence contradiction by A7,A14,A15,A16;
    end;
  end;
  hence contradiction;
end;

theorem
  d in (f qua Relation of C,D).:X iff ex c st c in dom f & c in X & d =
  f/.c by Th22;

theorem
  c in dom f implies Im(f,c) = {f/.c}
proof
  assume
A1: c in dom f;
  hence Im(f,c) = {(f qua Function).c} by FUNCT_1:59
    .= {f/.c} by A1,PARTFUN1:def 6;
end;

theorem
  c1 in dom f & c2 in dom f implies f.:{c1,c2} = {f/.c1,f/.c2}
proof
  assume that
A1: c1 in dom f and
A2: c2 in dom f;
  thus f.:{c1,c2} = {(f qua Function).c1,(f qua Function).c2} by A1,A2,
FUNCT_1:60
    .= {f/.c1,(f qua Function).c2} by A1,PARTFUN1:def 6
    .= {f/.c1,f/.c2} by A2,PARTFUN1:def 6;
end;

theorem Th26:
  SC = f"X iff for c holds c in SC iff c in dom f & f/.c in X
proof
  thus SC = f"X implies for c holds c in SC iff c in dom f & f/.c in X
  proof
    assume
A1: SC = f"X;
    let c;
    thus c in SC implies c in dom f & f/.c in X
    proof
      assume c in SC;
      then c in dom f & (f qua Function).c in X by A1,FUNCT_1:def 7;
      hence thesis by PARTFUN1:def 6;
    end;
    assume that
A2: c in dom f and
A3: f/.c in X;
    (f qua Function).c in X by A2,A3,PARTFUN1:def 6;
    hence thesis by A1,A2,FUNCT_1:def 7;
  end;
  assume
A4: for c holds c in SC iff c in dom f & f/.c in X;
  now
    let x;
    thus x in SC implies x in dom f & (f qua Function).x in X
    proof
      assume
A5:   x in SC;
      then reconsider x1=x as Element of C;
      x1 in dom f & f/.x1 in X by A4,A5;
      hence thesis by PARTFUN1:def 6;
    end;
    assume that
A6: x in dom f and
A7: (f qua Function).x in X;
    reconsider x1=x as Element of C by A6;
    f/.x1 in X by A6,A7,PARTFUN1:def 6;
    hence x in SC by A4,A6;
  end;
  hence thesis by FUNCT_1:def 7;
end;

theorem
  for f ex g being Function of C,D st for c st c in dom f holds g.c = f /.c
proof
  let f;
  consider g being Function of C,D such that
A1: for x st x in dom f holds g.x = (f qua Function).x by FUNCT_2:71;
  take g;
  let c;
  assume
A2: c in dom f;
  then g.c = (f qua Function).c by A1;
  hence thesis by A2,PARTFUN1:def 6;
end;

theorem
  f tolerates g iff for c st c in dom f /\ dom g holds f/.c = g/.c
proof
  thus f tolerates g implies for c st c in dom f /\ dom g holds f/.c = g/.c
  proof
    assume
A1: f tolerates g;
    let c;
    assume
A2: c in dom f /\ dom g;
    then
A3: c in dom f by XBOOLE_0:def 4;
    (f qua Function).c = (g qua Function).c by A1,A2,PARTFUN1:def 4;
    then
A4: f/.c = (g qua Function).c by A3,PARTFUN1:def 6;
    c in dom g by A2,XBOOLE_0:def 4;
    hence thesis by A4,PARTFUN1:def 6;
  end;
  assume
A5: for c st c in dom f /\ dom g holds f/.c = g/.c;
  now
    let x;
    assume
A6: x in dom f /\ dom g;
    then reconsider x1=x as Element of C;
    x in dom f & f/.x1 = g/.x1 by A5,A6,XBOOLE_0:def 4;
    then
A7: (f qua Function).x = g/.x1 by PARTFUN1:def 6;
    x in dom g by A6,XBOOLE_0:def 4;
    hence (f qua Function).x = (g qua Function).x by A7,PARTFUN1:def 6;
  end;
  hence thesis by PARTFUN1:def 4;
end;

scheme
  PartFuncExD{D,C()->non empty set, P[set,set]}: ex f being PartFunc of D(),C(
) st (for d be Element of D() holds d in dom f iff (ex c be Element of C() st P
  [d,c])) & for d be Element of D() st d in dom f holds P[d,f/.d] proof
  defpred R[set] means ex c be Element of C() st P[$1,c];
  set x = the Element of C();
  defpred Q[set,set] means ((ex c be Element of C() st P[$1,c]) implies P[$1,
  $2]) & ((for c be Element of C() holds not P[$1,c]) implies $2=x);
  consider X be set such that
A1: for x holds x in X iff x in D() & R[x] from XBOOLE_0:sch 1;
  for x holds x in X implies x in D() by A1;
  then reconsider X as Subset of D() by TARSKI:def 3;
A2: for d be Element of D() ex z be Element of C() st Q[d,z]
  proof
    let d be Element of D();
    (for c be Element of C() holds not P[d,c]) implies ex z be Element of
C() st ((ex c be Element of C() st P[d,c]) implies P[d,z]) & ((for c be Element
    of C() holds not P[d,c]) implies z=x);
    hence thesis;
  end;
  consider g being Function of D(),C() such that
A3: for d be Element of D() holds Q[d,g.d] from FUNCT_2:sch 3(A2);
  reconsider f=g|X as PartFunc of D(),C();
  take f;
A4: dom g = D() by FUNCT_2:def 1;
  thus for d be Element of D() holds d in dom f iff ex c be Element of C() st
  P[d,c]
  proof
    let d be Element of D();
    dom f c= X by RELAT_1:58;
    hence d in dom f implies ex c be Element of C() st P[d,c] by A1;
    assume ex c be Element of C() st P[d,c];
    then d in X by A1;
    then d in dom g /\ X by A4,XBOOLE_0:def 4;
    hence thesis by RELAT_1:61;
  end;
  let d be Element of D();
  assume
A5: d in dom f;
  dom f c= X by RELAT_1:58;
  then ex c be Element of C() st P[d,c] by A1,A5;
  then P[d,g.d] by A3;
  then P[d,(f qua Function).d] by A5,FUNCT_1:47;
  hence thesis by A5,PARTFUN1:def 6;
end;

scheme
  LambdaPFD{D,C()->non empty set, F(set)->Element of C(), P[set]}: ex f being
  PartFunc of D(),C() st (for d be Element of D() holds d in dom f iff P[d]) &
  for d be Element of D() st d in dom f holds f/.d = F(d) proof
  defpred Q[set,set] means P[$1] & $2 = F($1);
  consider f being PartFunc of D(),C() such that
A1: for d be Element of D() holds d in dom f iff ex c be Element of C()
  st Q[d,c] and
A2: for d be Element of D() st d in dom f holds Q[d,f/.d] from
  PartFuncExD;
  take f;
  thus for d be Element of D() holds d in dom f iff P[d]
  proof
    let d be Element of D();
    thus d in dom f implies P[d]
    proof
      assume d in dom f;
      then ex c be Element of C() st P[d] & c = F(d) by A1;
      hence thesis;
    end;
    assume P[d];
    then ex c be Element of C() st P[d] & c = F(d);
    hence thesis by A1;
  end;
  thus thesis by A2;
end;

scheme
  UnPartFuncD{C,D()->non empty set, X()->set, F(set)->Element of D()}: for f,g
being PartFunc of C(),D() st (dom f=X() & for c be Element of C() st c in dom f
holds f/.c = F(c)) & (dom g=X() & for c be Element of C() st c in dom g holds g
  /.c = F(c)) holds f = g proof
  let f,g be PartFunc of C(),D();
  assume that
A1: dom f=X() and
A2: for c be Element of C() st c in dom f holds f/.c = F(c) and
A3: dom g=X() and
A4: for c be Element of C() st c in dom g holds g/.c = F(c);
  now
    let c be Element of C();
    assume
A5: c in dom f;
    hence f/.c = F(c) by A2
      .= g/.c by A1,A3,A4,A5;
  end;
  hence thesis by A1,A3,Th1;
end;

definition
  let C,D;
  let SC,d;
  redefine func SC --> d -> PartFunc of C,D;
  coherence
  proof
    dom (SC --> d) = SC by FUNCOP_1:13;
    hence thesis by RELSET_1:7;
  end;
end;

theorem Th29:
  c in SC implies (SC --> d)/.c = d
proof
  assume
A1: c in SC;
  then dom (SC --> d) = SC & ((SC --> d) qua Function).c = d by FUNCOP_1:7,13;
  hence thesis by A1,PARTFUN1:def 6;
end;

theorem
  (for c st c in dom f holds f/.c = d) implies f = dom f --> d
proof
  assume
A1: for c st c in dom f holds f/.c = d;
  now
    let x;
    assume
A2: x in dom f;
    then reconsider x1=x as Element of C;
    f/.x1 = d by A1,A2;
    hence ( f qua Function).x = d by A2,PARTFUN1:def 6;
  end;
  hence thesis by FUNCOP_1:11;
end;

theorem
  c in dom f implies f*(SE --> c) = SE --> f/.c
proof
  assume
A1: c in dom f;
  then f*(SE --> c) = SE --> (f qua Function).c by FUNCOP_1:17;
  hence thesis by A1,PARTFUN1:def 6;
end;

theorem
  (id SC) is total iff SC = C
proof
  thus (id SC) is total implies SC = C
  proof
    assume (id SC) is total;
    then dom (id SC) = C by PARTFUN1:def 2;
    hence thesis by RELAT_1:45;
  end;
  assume SC = C;
  then dom (id SC) = C by RELAT_1:45;
  hence thesis by PARTFUN1:def 2;
end;

theorem
  (SC --> d) is total implies SC <> {}
proof
  assume that
A1: (SC --> d) is total and
A2: SC = {};
  dom (SC --> d) = C by A1,PARTFUN1:def 2;
  hence contradiction by A2,FUNCOP_1:10;
end;

theorem
  (SC --> d) is total iff SC = C
proof
  thus (SC --> d) is total implies SC = C
  proof
    assume (SC --> d) is total;
    then dom (SC --> d) = C by PARTFUN1:def 2;
    hence thesis by FUNCOP_1:13;
  end;
  assume SC = C;
  then dom (SC --> d) = C by FUNCOP_1:13;
  hence thesis by PARTFUN1:def 2;
end;





definition
  let C,D,f;
  redefine attr f is constant means
  ex d st for c st c in dom f holds f.c = d;
  compatibility
  proof
    thus f is constant implies ex d st for c st c in dom f holds f.c = d
    proof
      assume
A1:   f is constant;
      per cases;
      suppose
A2:     dom f = {};
        set d = the Element of D;
        take d;
        thus thesis by A2;
      end;
      suppose
        dom f <> {};
        then consider c0 being set such that
A3:     c0 in dom f by XBOOLE_0:def 1;
        reconsider c0 as Element of C by A3;
        take d = f/.c0;
        let c;
        assume c in dom f;
        hence f.c = f.c0 by A1,A3,FUNCT_1:def 10
          .= d by A3,PARTFUN1:def 6;
      end;
    end;
    given d such that
A4: for c st c in dom f holds f.c = d;
    let x,y be set such that
A5: x in dom f and
A6: y in dom f;
    thus f.x = d by A4,A5
      .=f.y by A4,A6;
  end;
end;

theorem Th35:
  f|X is constant iff ex d st for c st c in X /\ dom f holds f/.c = d
proof
  thus f|X is constant implies ex d st for c st c in X /\ dom f holds f/.c = d
  proof
    given d such that
A1: for c st c in dom(f|X) holds (f|X).c = d;
    take d;
    let c;
    assume
A2: c in X /\ dom f;
    then
A3: c in dom(f|X) by RELAT_1:61;
    c in dom f by A2,XBOOLE_0:def 4;
    hence f/.c = f.c by PARTFUN1:def 6
      .= (f|X).c by A3,FUNCT_1:47
      .= d by A1,A3;
  end;
  given d such that
A4: for c st c in X /\ dom f holds f/.c = d;
  take d;
  let c;
  assume
A5: c in dom(f|X);
  then
A6: c in X /\ dom f by RELAT_1:61;
  then
A7: c in dom f by XBOOLE_0:def 4;
  thus (f|X).c = f.c by A5,FUNCT_1:47
    .= f/.c by A7,PARTFUN1:def 6
    .= d by A4,A6;
end;

theorem
  f|X is constant iff for c1,c2 st c1 in X /\ dom f & c2 in X /\ dom f
  holds f/.c1=f/.c2
proof
  thus f|X is constant implies for c1,c2 st c1 in X /\ dom f & c2 in X /\ dom
  f holds f/.c1=f/.c2
  proof
    assume f|X is constant;
    then consider d such that
A1: for c st c in X /\ dom f holds f/.c = d by Th35;
    let c1,c2;
    assume that
A2: c1 in X /\ dom f and
A3: c2 in X /\ dom f;
    f/.c1 = d by A1,A2;
    hence thesis by A1,A3;
  end;
  assume
A4: for c1,c2 st c1 in X /\ dom f & c2 in X /\ dom f holds f/.c1=f/.c2;
  now
    per cases;
    suppose
A5:   X /\ dom f = {};
      now
        set d = the Element of D;
        take d;
        let c;
        thus c in X /\ dom f implies f/.c = d by A5;
      end;
      hence thesis by Th35;
    end;
    suppose
A6:   X /\ dom f <> {};
      set x = the Element of X /\ dom f;
      x in dom f by A6,XBOOLE_0:def 4;
      then reconsider x as Element of C;
      for c holds c in X /\ dom f implies f/.c = f/.x by A4;
      hence thesis by Th35;
    end;
  end;
  hence thesis;
end;

theorem
  X meets dom f implies (f|X is constant iff ex d st rng (f|X) = {d})
proof
  assume
A1: X /\ dom f <> {};
  thus f|X is constant implies ex d st rng (f|X) = {d}
  proof
    assume f|X is constant;
    then consider d such that
A2: for c st c in X /\ dom f holds f/.c = d by Th35;
    take d;
    thus rng (f|X) c= {d}
    proof
      let x;
      assume x in rng (f|X);
      then consider y such that
A3:   y in dom (f|X) and
A4:   ((f|X) qua Function).y = x by FUNCT_1:def 3;
      reconsider y as Element of C by A3;
      dom (f|X) = X /\ dom f by RELAT_1:61;
      then d = f/.y by A2,A3
        .= f|X/.y by A3,Th15
        .= x by A3,A4,PARTFUN1:def 6;
      hence thesis by TARSKI:def 1;
    end;
    thus {d} c= rng (f|X)
    proof
      set y = the Element of X /\ dom f;
      y in dom f by A1,XBOOLE_0:def 4;
      then reconsider y as Element of C;
      let x such that
A5:   x in {d};
A6:   dom (f|X) = X /\ dom f by RELAT_1:61;
      then f|X/.y = f/.y by A1,Th15
        .= d by A1,A2
        .= x by A5,TARSKI:def 1;
      hence thesis by A1,A6,Th2;
    end;
  end;
  given d such that
A7: rng (f|X) = {d};
  take d;
  let c;
  assume
A8: c in dom (f|X);
  then f|X/.c in {d} by A7,Th2;
  then (f|X).c in {d} by A8,PARTFUN1:def 6;
  hence thesis by TARSKI:def 1;
end;

theorem
  f|X is constant & Y c= X implies f|Y is constant
proof
  assume that
A1: f|X is constant and
A2: Y c= X;
  consider d such that
A3: for c st c in X /\ dom f holds f/.c = d by A1,Th35;
  now
    let c;
    assume c in Y /\ dom f;
    then c in Y & c in dom f by XBOOLE_0:def 4;
    then c in X /\ dom f by A2,XBOOLE_0:def 4;
    hence f/.c = d by A3;
  end;
  hence thesis by Th35;
end;

theorem Th39:
  X misses dom f implies f|X is constant
proof
  assume
A1: X /\ dom f = {};
  now
    set d = the Element of D;
    take d;
    let c;
    thus c in X /\ dom f implies f/.c = d by A1;
  end;
  hence thesis by Th35;
end;

theorem
  f|SC = dom (f|SC) --> d implies f|SC is constant
proof
  assume
A1: f|SC = dom (f|SC) --> d;
  now
    let c;
    assume c in SC /\ dom f;
    then
A2: c in dom (f|SC) by RELAT_1:61;
    then f|SC/.c = d by A1,Th29;
    hence f/.c = d by A2,Th15;
  end;
  hence thesis by Th35;
end;

theorem
  f|{x} is constant
proof
  now
    per cases;
    suppose
      {x} /\ dom f = {};
      then {x} misses dom f by XBOOLE_0:def 7;
      hence thesis by Th39;
    end;
    suppose
A1:   {x} /\ dom f <> {};
      set y = the Element of {x} /\ dom f;
      y in {x} & y in dom f by A1,XBOOLE_0:def 4;
      then reconsider x1=x as Element of C by TARSKI:def 1;
      now
        take d = f/.x1;
        let c;
        assume c in {x} /\ dom f;
        then c in {x} by XBOOLE_0:def 4;
        hence f/.c = f/.x1 by TARSKI:def 1;
      end;
      hence thesis by Th35;
    end;
  end;
  hence thesis;
end;

theorem
  f|X is constant & f|Y is constant & X /\ Y meets dom f implies f|(X \/
  Y) is constant
proof
  assume that
A1: f|X is constant and
A2: f|Y is constant and
A3: X /\ Y /\ dom f <> {};
  consider d1 such that
A4: for c st c in X /\ dom f holds f/.c = d1 by A1,Th35;
  set x = the Element of X /\ Y /\ dom f;
A5: x in X /\ Y by A3,XBOOLE_0:def 4;
A6: x in dom f by A3,XBOOLE_0:def 4;
  then reconsider x as Element of C;
  x in Y by A5,XBOOLE_0:def 4;
  then
A7: x in Y /\ dom f by A6,XBOOLE_0:def 4;
  consider d2 such that
A8: for c st c in Y /\ dom f holds f/.c = d2 by A2,Th35;
  x in X by A5,XBOOLE_0:def 4;
  then x in X /\ dom f by A6,XBOOLE_0:def 4;
  then f/.x = d1 by A4;
  then
A9: d1 = d2 by A8,A7;
  take d1;
  let c;
  assume
A10: c in dom(f|(X \/ Y));
  then
A11: c in (X \/ Y) /\ dom f by RELAT_1:61;
  then
A12: c in dom f by XBOOLE_0:def 4;
A13: c in X \/ Y by A11,XBOOLE_0:def 4;
  now
    per cases by A13,XBOOLE_0:def 3;
    suppose
      c in X;
      then c in X /\ dom f by A12,XBOOLE_0:def 4;
      hence f/.c = d1 by A4;
    end;
    suppose
      c in Y;
      then c in Y /\ dom f by A12,XBOOLE_0:def 4;
      hence f/.c = d1 by A8,A9;
    end;
  end;
  then (f|(X \/ Y))/.c = d1 by A11,Th16;
  hence thesis by A10,PARTFUN1:def 6;
end;

theorem
  f|Y is constant implies f|X|Y is constant
proof
  assume f|Y is constant;
  then consider d such that
A1: for c st c in Y /\ dom f holds f/.c = d by Th35;
  take d;
  let c;
  assume
A2: c in dom (f|X|Y);
  then
A3: c in Y /\ dom (f|X) by RELAT_1:61;
  then
A4: c in Y by XBOOLE_0:def 4;
A5: c in dom (f|X) by A3,XBOOLE_0:def 4;
  then c in dom f /\ X by RELAT_1:61;
  then c in dom f by XBOOLE_0:def 4;
  then c in Y /\ dom f by A4,XBOOLE_0:def 4;
  then f/.c = d by A1;
  then (f|X)/.c = d by A5,Th15;
  then (f|X|Y)/.c = d by A3,Th16;
  hence thesis by A2,PARTFUN1:def 6;
end;

theorem
  (SC --> d)|SC is constant
proof
  take d;
  let c;
  assume
A1: c in dom((SC --> d)|SC);
  then
A2: c in SC /\ dom (SC --> d) by RELAT_1:61;
  then c in SC by XBOOLE_0:def 4;
  then (SC --> d)/.c = d by Th29;
  then ((SC --> d)|SC)/.c = d by A2,Th16;
  hence thesis by A1,PARTFUN1:def 6;
end;





theorem
 dom f c= dom g & (for c st c in dom f holds f/.c = g/.c) implies f c= g
proof
  assume that
A1: dom f c= dom g and
A2: for c st c in dom f holds f/.c = g/.c;
  now
    let x;
    assume
A3: x in dom f;
    then reconsider x1=x as Element of C;
    f/.x1 = g/.x1 by A2,A3;
    then (f qua Function).x = g/.x1 by A3,PARTFUN1:def 6;
    hence (f qua Function).x = (g qua Function).x by A1,A3,PARTFUN1:def 6;
  end;
  hence thesis by A1,GRFUNC_1:2;
end;

theorem Th46:
  c in dom f & d = f/.c iff [c,d] in f
proof
  thus c in dom f & d = f/.c implies [c,d] in f
  proof
    assume that
A1: c in dom f and
A2: d = f/.c;
    d = (f qua Function).c by A1,A2,PARTFUN1:def 6;
    hence thesis by A1,FUNCT_1:1;
  end;
  assume [c,d] in f;
  then c in dom f & d = (f qua Function).c by FUNCT_1:1;
  hence thesis by PARTFUN1:def 6;
end;

theorem
  [c,e] in (s*f) implies [c,f/.c] in f & [f/.c,e] in s
proof
  assume
A1: [c,e] in (s*f);
  then
A2: [(f qua Function).c,e] in s by GRFUNC_1:4;
A3: [c,(f qua Function).c] in f by A1,GRFUNC_1:4;
  then c in dom f by FUNCT_1:1;
  hence thesis by A3,A2,PARTFUN1:def 6;
end;

theorem
  f = {[c,d]} implies f/.c = d
proof
  assume
A1: f = {[c,d]};
  then [c,d] in f by TARSKI:def 1;
  then
A2: c in dom f by FUNCT_1:1;
  (f qua Function).c = d by A1,GRFUNC_1:6;
  hence thesis by A2,PARTFUN1:def 6;
end;

theorem
  dom f = {c} implies f = {[c,f/.c]}
proof
  assume dom f = {c};
  then c in dom f & f = {[c,(f qua Function).c]} by GRFUNC_1:7,TARSKI:def 1;
  hence thesis by PARTFUN1:def 6;
end;

theorem
  f1 = f /\ g & c in dom f1 implies f1/.c = f/.c & f1/.c = g/.c
proof
  assume that
A1: f1 = f /\ g and
A2: c in dom f1;
  (f1 qua Function).c = (g qua Function).c by A1,A2,GRFUNC_1:11;
  then
A3: f1/.c = (g qua Function).c by A2,PARTFUN1:def 6;
A4: [c,(f1 qua Function).c] in f1 by A2,FUNCT_1:1;
  then [c,(f1 qua Function).c] in f by A1,XBOOLE_0:def 4;
  then
A5: c in dom f by FUNCT_1:1;
  [c,(f1 qua Function).c] in g by A1,A4,XBOOLE_0:def 4;
  then
A6: c in dom g by FUNCT_1:1;
  (f1 qua Function).c = (f qua Function).c by A1,A2,GRFUNC_1:11;
  then f1/.c = (f qua Function).c by A2,PARTFUN1:def 6;
  hence thesis by A5,A6,A3,PARTFUN1:def 6;
end;

theorem
  c in dom f & f1 = f \/ g implies f1/.c = f/.c
proof
  assume that
A1: c in dom f and
A2: f1 = f \/ g;
  [c,(f qua Function).c] in f by A1,FUNCT_1:1;
  then [c,(f qua Function).c] in f1 by A2,XBOOLE_0:def 3;
  then
A3: c in dom f1 by FUNCT_1:1;
  (f1 qua Function).c = (f qua Function).c by A1,A2,GRFUNC_1:15;
  then f1/.c = (f qua Function).c by A3,PARTFUN1:def 6;
  hence thesis by A1,PARTFUN1:def 6;
end;

theorem
  c in dom g & f1 = f \/ g implies f1/.c = g/.c
proof
  assume that
A1: c in dom g and
A2: f1 = f \/ g;
  [c,(g qua Function).c] in g by A1,FUNCT_1:1;
  then [c,(g qua Function).c] in f1 by A2,XBOOLE_0:def 3;
  then
A3: c in dom f1 by FUNCT_1:1;
  (f1 qua Function).c = (g qua Function).c by A1,A2,GRFUNC_1:15;
  then f1/.c = (g qua Function).c by A3,PARTFUN1:def 6;
  hence thesis by A1,PARTFUN1:def 6;
end;

theorem
  c in dom f1 & f1 = f \/ g implies f1/.c = f/.c or f1/.c = g/.c
proof
  assume that
A1: c in dom f1 and
A2: f1 = f \/ g;
  [c,f1/.c] in f1 by A1,Th46;
  then
A3: [c,f1/.c] in f or [c,f1/.c] in g by A2,XBOOLE_0:def 3;
  now
    per cases by A3,FUNCT_1:1;
    suppose
      c in dom f;
      then [c,f/.c] in f by Th46;
      then [c,f/.c] in f1 by A2,XBOOLE_0:def 3;
      hence thesis by Th46;
    end;
    suppose
      c in dom g;
      then [c,g/.c] in g by Th46;
      then [c,g/.c] in f1 by A2,XBOOLE_0:def 3;
      hence thesis by Th46;
    end;
  end;
  hence thesis;
end;

theorem
  c in dom f & c in SC iff [c,f/.c] in (f|SC)
proof
  thus c in dom f & c in SC implies [c,f/.c] in (f|SC)
  proof
    assume that
A1: c in dom f and
A2: c in SC;
    [c,(f qua Function).c] in (f|SC) by A1,A2,GRFUNC_1:22;
    hence thesis by A1,PARTFUN1:def 6;
  end;
  assume [c,f/.c] in (f|SC);
  then c in dom (f|SC) by FUNCT_1:1;
  then c in dom f /\ SC by RELAT_1:61;
  hence thesis by XBOOLE_0:def 4;
end;

theorem
  c in dom f & f/.c in SD iff [c,f/.c] in (SD|`f)
proof
  thus c in dom f & f/.c in SD implies [c,f/.c] in (SD|`f)
  proof
    assume that
A1: c in dom f and
A2: f/.c in SD;
    (f qua Function).c in SD by A1,A2,PARTFUN1:def 6;
    then [c,(f qua Function).c] in (SD|`f) by A1,GRFUNC_1:24;
    hence thesis by A1,PARTFUN1:def 6;
  end;
  assume [c,f/.c] in (SD|`f);
  then c in dom (SD|`f) by FUNCT_1:1;
  then c in dom f & (f qua Function).c in SD by FUNCT_1:54;
  hence thesis by PARTFUN1:def 6;
end;

theorem
  c in f"SD iff [c,f/.c] in f & f/.c in SD
proof
  thus c in f"SD implies [c,f/.c] in f & f/.c in SD
  proof
    assume
A1: c in f"SD;
    then
A2: (f qua Function).c in SD by GRFUNC_1:26;
A3: [c,(f qua Function).c] in f by A1,GRFUNC_1:26;
    then c in dom f by FUNCT_1:1;
    hence thesis by A3,A2,PARTFUN1:def 6;
  end;
  assume that
A4: [c,f/.c] in f and
A5: f/.c in SD;
  c in dom f by A4,Th46;
  hence thesis by A5,Th26;
end;

theorem Th57:
  f|X is constant iff ex d st for c st c in X /\ dom f holds f.c = d
proof
  hereby
    assume f|X is constant;
    then consider d such that
A1: for c st c in X /\ dom f holds f/.c = d by Th35;
    take d;
    let c;
    assume
A2: c in X /\ dom f;
    then c in dom f by XBOOLE_0:def 4;
    hence f.c = f/.c by PARTFUN1:def 6
      .= d by A1,A2;
  end;
  given d such that
A3: for c st c in X /\ dom f holds f.c = d;
  take d;
  let c;
  assume
A4: c in dom(f|X);
  then
A5: c in X /\ dom f by RELAT_1:61;
  then
A6: c in dom f by XBOOLE_0:def 4;
  thus (f|X).c = (f|X)/.c by A4,PARTFUN1:def 6
    .= f/.c by A5,Th16
    .= f.c by A6,PARTFUN1:def 6
    .= d by A3,A5;
end;

theorem
  f|X is constant iff for c1,c2 st c1 in X /\ dom f & c2 in X /\ dom f
  holds f.c1=f.c2
proof
  thus f|X is constant implies for c1,c2 st c1 in X /\ dom f & c2 in X /\ dom
  f holds f.c1=f.c2
  proof
    assume f|X is constant;
    then consider d such that
A1: for c st c in X /\ dom f holds f.c = d by Th57;
    let c1,c2;
    assume that
A2: c1 in X /\ dom f and
A3: c2 in X /\ dom f;
    f.c1 = d by A1,A2;
    hence thesis by A1,A3;
  end;
  assume
A4: for c1,c2 st c1 in X /\ dom f & c2 in X /\ dom f holds f.c1=f.c2;
  now
    per cases;
    suppose
A5:   X /\ dom f = {};
      now
        set d = the Element of D;
        take d;
        let c;
        thus c in X /\ dom f implies f.c = d by A5;
      end;
      hence thesis by Th57;
    end;
    suppose
A6:   X /\ dom f <> {};
      set x = the Element of X /\ dom f;
      now
        let c;
A7:     x in dom f by A6,XBOOLE_0:def 4;
        assume c in X /\ dom f;
        hence f.c = f.x by A4,A7
          .= f/.x by A7,PARTFUN1:def 6;
      end;
      hence thesis by Th57;
    end;
  end;
  hence thesis;
end;

theorem
  d in f.:X implies ex c st c in dom f & c in X & d = f.c
proof
  assume d in f.:X;
  then consider x such that
A1: x in dom f and
A2: x in X & d = (f qua Function).x by FUNCT_1:def 6;
  reconsider x as Element of C by A1;
  take x;
  thus thesis by A1,A2;
end;

theorem
  f is one-to-one implies (d in rng f & c = (f").d iff c in dom f & d = f.c)
proof
A1: f" = (f qua Function)";
  assume f is one-to-one;
  hence thesis by A1,FUNCT_1:32;
end;

theorem
  for Y for f,g be Y-valued Function st f c= g
  for x st x in dom f holds f/.x = g/.x
proof let Y; let f,g be Y-valued Function;
    assume
A1: f c= g;
    then
A2: dom f c= dom g by GRFUNC_1:2;
    let x;
    assume
A3: x in dom f;
    then f.x = g.x by A1,GRFUNC_1:2;
    then f/.x = (g qua Function).x by A3,PARTFUN1:def 6;
    hence thesis by A2,A3,PARTFUN1:def 6;
end;














begin

reserve W,X,Y,Z for set,
  f,g for Function,
  a,x,y,z for set;

definition
  let B be set;
  attr B is subset-closed means
  :Def1:
  for X,Y holds X in B & Y c= X implies Y in B;
end;

definition
  let B be set;
  attr B is Tarski means
  :Def2:
  B is subset-closed & (for X holds X in B implies bool X in B) &
  for X holds X c= B implies X,B are_equipotent or X in B;
end;

definition
  let A,B be set;
  pred B is_Tarski-Class_of A means
  :Def3:
  A in B & B is Tarski;
end;

definition
  let A be set;
  func Tarski-Class A -> set means
  :Def4:
  it is_Tarski-Class_of A &
  for D being set st D is_Tarski-Class_of A holds it c= D;
  existence
  proof
    consider Big being set such that
A1: A in Big and
A2: for X,Y holds X in Big & Y c= X implies Y in Big and
    A3: (
 for X holds X in Big implies bool X in Big)& for X holds X c= Big implies X,
    Big are_equipotent or X in Big
    by ZFMISC_1:112;
 Big is subset-closed by A2,Def1;
then A4: Big is Tarski by A3,Def2;
    defpred P[set] means $1 is_Tarski-Class_of A;
    consider Classes being set such that
A5: X in Classes iff X in bool Big & P[X] from XBOOLE_0:sch 1;
    set IT = meet Classes;
    A6: Big
 in bool Big & Big is_Tarski-Class_of A by A1,A4,Def3,ZFMISC_1:def 1;
then A7: Big in Classes by A5;
A8: Classes <> {} by A5,A6;
A9: now
      let X;
      assume X in Classes;
then   X is_Tarski-Class_of A by A5;
      hence A in X by Def3;
    end;
then A10: A in IT by A8,SETFAM_1:def 1;
    take IT;
    thus A in IT by A8,A9,SETFAM_1:def 1;
    thus
A11: X in IT & Y c= X implies Y in IT
    proof
      assume that
A12:  X in IT and
A13:  Y c= X;
  now
        let Z;
        assume
A14:    Z in Classes;
then     Z is_Tarski-Class_of A by A5;
then     Z is Tarski by Def3;
then A15:    Z is subset-closed by Def2;
    X in Z by A12,A14,SETFAM_1:def 1;
        hence Y in Z by A13,A15,Def1;
      end;
      hence thesis by A8,SETFAM_1:def 1;
    end;
    thus
A16: X in IT implies bool X in IT
    proof
      assume
A17:  X in IT;
  now
        let Z;
        assume
A18:    Z in Classes;
then     Z is_Tarski-Class_of A by A5;
then A19:    Z is Tarski by Def3;
    X in Z by A17,A18,SETFAM_1:def 1;
        hence bool X in Z by A19,Def2;
      end;
      hence thesis by A8,SETFAM_1:def 1;
    end;
    thus
A20: X c= IT implies X,IT are_equipotent or X in IT
    proof
      assume that
A21:  X c= IT and
A22:  not X,IT are_equipotent;
  now
        let Z;
        assume
A23:    Z in Classes;
then     Z is_Tarski-Class_of A by A5;
then A24:    Z is Tarski by Def3;
A25:    IT c= Z by A23,SETFAM_1:3;
then     X c= Z by A21,XBOOLE_1:1;
then     X,Z are_equipotent or X in Z by A24,Def2;
        hence X in Z by A21,A22,A25,CARD_1:24;
      end;
      hence thesis by A8,SETFAM_1:def 1;
    end;
    let D be set;
    assume
A26: D is_Tarski-Class_of A;
then A27: A in D by Def3;
A28: D is Tarski by A26,Def3;
then A29: D is subset-closed by Def2;
A30: IT /\ D is_Tarski-Class_of A
    proof
      thus A in IT /\ D by A10,A27,XBOOLE_0:def 4;
      thus X in IT /\ D & Y c= X implies Y in IT /\ D
      proof
        assume that
A31:    X in IT /\ D and
A32:    Y c= X;
A33:    X in IT by A31,XBOOLE_0:def 4;
A34:    X in D by A31,XBOOLE_0:def 4;
A35:    Y in IT by A11,A32,A33;
    Y in D by A29,A32,A34,Def1;
        hence thesis by A35,XBOOLE_0:def 4;
      end;
      thus X in IT /\ D implies bool X in IT /\ D
      proof
        assume
A36:    X in IT /\ D;
then A37:    X in IT by XBOOLE_0:def 4;
A38:    X in D by A36,XBOOLE_0:def 4;
A39:    bool X in IT by A16,A37;
    bool X in D by A28,A38,Def2;
        hence thesis by A39,XBOOLE_0:def 4;
      end;
      let X such that
A40:  X c= IT /\ D and
A41:  not X,IT /\ D are_equipotent;
A42:  IT /\ D c= IT by XBOOLE_1:17;
A43:  IT /\ D c= D by XBOOLE_1:17;
A44:  not X,IT are_equipotent by A40,A41,A42,CARD_1:24;
      A45:  X c= D & not X,D are_equipotent by A40,A41,A43,CARD_1:24,XBOOLE_1:1
      ;
A46:  X in IT by A20,A40,A42,A44,XBOOLE_1:1;
  X in D by A28,A45,Def2;
      hence thesis by A46,XBOOLE_0:def 4;
    end;
 IT /\ D c= Big
    proof
      let x;
      assume x in IT /\ D;
then   x in IT by XBOOLE_0:def 4;
      hence thesis by A7,SETFAM_1:def 1;
    end;
then  IT /\ D in Classes by A5,A30;
then A47: IT c= IT /\ D by SETFAM_1:3;
 IT /\ D c= D by XBOOLE_1:17;
    hence thesis by A47,XBOOLE_1:1;
  end;
  uniqueness
  proof
    let D1,D2 be set such that
    A48: (
 D1 is_Tarski-Class_of A & for D being set st D is_Tarski-Class_of A holds
    D1 c= D )&( D2 is_Tarski-Class_of A & for D being set st D
    is_Tarski-Class_of A holds D2 c= D );
    thus D1 c= D2 & D2 c= D1 by A48;
  end;
end;

registration
  let A be set;
  cluster Tarski-Class A -> non empty;
  coherence
  proof
 Tarski-Class A is_Tarski-Class_of A by Def4;
    hence thesis by Def3;
  end;
end;

theorem
  W is Tarski iff W is subset-closed & (for X st X in W holds bool X in W) &
  for X st X c= W & card X in card W holds X in W
proof
  hereby
    assume
A1: W is Tarski;
    hence
W is subset-closed & for X st X in W holds bool X in W by Def2;
    let X;
    assume that
A2: X c= W and
A3: card X in card W;
 card X <> card W by A3;
then  not X,W are_equipotent by CARD_1:5;
    hence X in W by A1,A2,Def2;
  end;
 now
    assume
A4: for X st X c= W & card X in card W holds X in W;
    let X;
    assume X c= W;
then  card X c= card W & not card X in card W or X in W by A4,CARD_1:11;
then  card X = card W or X in W by CARD_1:3;
    hence X,W are_equipotent or X in W by CARD_1:5;
  end;
  hence thesis by Def2;
end;

theorem Th2:
  X in Tarski-Class X
proof
 Tarski-Class X is_Tarski-Class_of X by Def4;
  hence thesis by Def3;
end;

theorem Th3:
  Y in Tarski-Class X & Z c= Y implies Z in Tarski-Class X
proof
 Tarski-Class X is_Tarski-Class_of X by Def4;
then  Tarski-Class X is Tarski by Def3;
then  Tarski-Class X is subset-closed by Def2;
  hence thesis by Def1;
end;

theorem Th4:
  Y in Tarski-Class X implies bool Y in Tarski-Class X
proof
 Tarski-Class X is_Tarski-Class_of X by Def4;
then  Tarski-Class X is Tarski by Def3;
  hence thesis by Def2;
end;

theorem Th5:
  Y c= Tarski-Class X implies
  Y,Tarski-Class X are_equipotent or Y in Tarski-Class X
proof
 Tarski-Class X is_Tarski-Class_of X by Def4;
then  Tarski-Class X is Tarski by Def3;
  hence thesis by Def2;
end;

theorem
  Y c= Tarski-Class X & card Y in card Tarski-Class X implies
  Y in Tarski-Class X
proof
  assume that
A1: Y c= Tarski-Class X and
A2: card Y in card Tarski-Class X;
 card Y <> card Tarski-Class X by A2;
then  not Y,Tarski-Class X are_equipotent by CARD_1:5;
  hence thesis by A1,Th5;
end;

reserve u,v for Element of Tarski-Class(X),
  A,B,C for Ordinal,
  L for T-Sequence;

definition
  let X,A;
  func Tarski-Class(X,A) means
  :Def5:
  ex L st it = last L & dom L = succ A & L.{} = { X } &
  (for C st succ C in succ A holds
  L.succ C = { u : ex v st v in L.C & u c= v } \/ { bool v : v in L.C } \/
  bool(L.C) /\ Tarski-Class X) &
  for C st C in succ A & C <> {} & C is limit_ordinal
  holds L.C = (union rng(L|C)) /\ Tarski-Class X;
  correctness
  proof
    deffunc C(Ordinal,set) = { u : ex v st v in $2 & u c= v } \/
    { bool v : v in $2 } \/ bool $2 /\ Tarski-Class X;
    deffunc D(Ordinal,T-Sequence) = (union rng $2) /\ Tarski-Class X;
    thus (ex x,L st x = last L & dom L = succ A & L.{} = {X} &
    (for C st succ C in succ A holds L.succ C = C(C,L.C)) &
    for C st C in succ A & C <> {} & C is limit_ordinal
    holds L.C = D(C,L|C) ) & for x1,x2 being set st
    (ex L st x1 = last L & dom L = succ A & L.{} = {X} &
    (for C st succ C in succ A holds L.succ C = C(C,L.C)) &
    for C st C in succ A & C <> {} & C is limit_ordinal
    holds L.C = D(C,L|C) ) &
    (ex L st x2 = last L & dom L = succ A & L.{} = {X} &
    (for C st succ C in succ A holds L.succ C = C(C,L.C)) &
    for C st C in succ A & C <> {} & C is limit_ordinal holds L.C = D(C,L|C) )
    holds x1 = x2 from ORDINAL2:sch 7;
  end;
end;

Lm1: now
  let X;
  deffunc F(Ordinal) = Tarski-Class(X,$1);
  deffunc C(Ordinal,set) = { u : ex v st v in $2 & u c= v } \/
  { bool v : v in $2 } \/ bool $2 /\ Tarski-Class X;
  deffunc D(Ordinal,T-Sequence) = (union rng $2) /\ Tarski-Class X;
A1: for A,x holds x = F(A) iff
  ex L st x = last L & dom L = succ A & L.{} = { X } &
  (for C st succ C in succ A holds L.succ C = C(C,L.C)) &
  for C st C in succ A & C <> {} & C is limit_ordinal
  holds L.C = D(C,L|C) by Def5;
  thus F({}) = { X } from ORDINAL2:sch 8(A1);
  thus for A holds F(succ A) = C(A,F(A)) from ORDINAL2:sch 9(A1);
  thus A <> {} & A is limit_ordinal & dom L = A &
  (for B st B in A holds L.B = Tarski-Class(X,B)) implies
  Tarski-Class(X,A) = (union rng L) /\ Tarski-Class X
  proof
    assume that
A2: A <> {} & A is limit_ordinal and
A3: dom L = A and
A4: for B st B in A holds L.B = F(B);
    thus F(A) = D(A,L) from ORDINAL2:sch 10(A1,A2,A3,A4);
  end;
end;

definition
  let X,A;
  redefine func Tarski-Class(X,A) -> Subset of Tarski-Class X;
  coherence
  proof
    defpred P[Ordinal] means Tarski-Class(X,$1) is Subset of Tarski-Class X;
 { X } c= Tarski-Class X
    proof
      let x;
      assume x in { X };
then    x = X by TARSKI:def 1;
      hence thesis by Th2;
    end;
then A1: P[{}] by Lm1;
A2: P[B] implies P[succ B]
    proof
      assume Tarski-Class(X,B) is Subset of Tarski-Class X;
      then reconsider S = Tarski-Class(X,B) as Subset of Tarski-Class X;
      set Y = Tarski-Class(X,succ B);
   Y c= Tarski-Class X
      proof
        let x;
        assume x in Y;
then      x in { u : ex v st v in S & u c= v } \/ { bool v : v in S } \/
        bool S /\ Tarski-Class X by Lm1;
        then
A3:    x in { u : ex v st v in S & u c= v } \/ { bool v : v in S } or
        x in bool S /\ Tarski-Class X by XBOOLE_0:def 3;
A4:    now
          assume x in { u : ex v st v in S & u c= v };
then       ex u st x = u & ex v st v in S & u c= v;
          hence thesis;
        end;
    now
          assume x in { bool v : v in S };
then       ex v st x = bool v & v in S;
          hence thesis by Th4;
        end;
        hence thesis by A3,A4,XBOOLE_0:def 3,def 4;
      end;
      hence thesis;
    end;
A5: for B st B <> {} & B is limit_ordinal &
    for C st C in B holds P[C] holds P[B]
    proof
      let B such that
A6:  B <> {} & B is limit_ordinal and
      for C st C in B holds Tarski-Class(X,C) is Subset of Tarski-Class X;
      deffunc f(Ordinal) = Tarski-Class(X,$1);
      consider L such that
A7:  dom L = B & for C st C in B holds L.C = f(C) from ORDINAL2:sch 2;
  Tarski-Class(X,B) = (union rng L) /\ Tarski-Class X by A6,A7,Lm1;
      hence thesis by XBOOLE_1:17;
    end;
 for A holds P[A] from ORDINAL2:sch 1(A1,A2,A5);
    hence thesis;
  end;
end;

theorem
  Tarski-Class(X,{}) = { X } by Lm1;

theorem
  Tarski-Class(X,succ A) = { u : ex v st v in Tarski-Class(X,A) & u c= v } \/
  { bool v : v in Tarski-Class(X,A) } \/
  bool Tarski-Class(X,A) /\ Tarski-Class X by Lm1;

theorem Th9:
  A <> {} & A is limit_ordinal implies
  Tarski-Class(X,A) = { u : ex B st B in A & u in Tarski-Class(X,B) }
proof
  assume
A1: A <> {} & A is limit_ordinal;
  deffunc f(Ordinal) = Tarski-Class(X,$1);
  consider L such that
A2: dom L = A & for B st B in A holds L.B = f(B) from ORDINAL2:sch 2;
A3: Tarski-Class(X,A) = (union rng L) /\ Tarski-Class X by A1,A2,Lm1;
  thus
  Tarski-Class(X,A) c= { u : ex B st B in A & u in Tarski-Class(X,B) }
  proof
    let x;
    assume x in Tarski-Class(X,A);
then  x in union rng L by A3,XBOOLE_0:def 4;
    then consider Y such that
A4: x in Y and
A5: Y in rng L by TARSKI:def 4;
    consider y such that
A6: y in dom L and
A7: Y = L.y by A5,FUNCT_1:def 3;
    reconsider y as Ordinal by A6;
 Y = Tarski-Class(X,y) by A2,A6,A7;
    hence thesis by A2,A4,A6;
  end;
  let x;
  assume x in { u : ex B st B in A & u in Tarski-Class(X,B) };
  then consider u such that
A8: x = u and
A9: ex B st B in A & u in Tarski-Class(X,B);
  consider B such that
A10: B in A and
A11: u in Tarski-Class(X,B) by A9;
 L.B = Tarski-Class(X,B) by A2,A10;
then  Tarski-Class(X,B) in rng L by A2,A10,FUNCT_1:def 3;
then  u in union rng L by A11,TARSKI:def 4;
  hence thesis by A3,A8,XBOOLE_0:def 4;
end;

theorem Th10:
  Y in Tarski-Class(X,succ A) iff
  Y c= Tarski-Class(X,A) & Y in Tarski-Class X or
  ex Z st Z in Tarski-Class(X,A) & (Y c= Z or Y = bool Z)
proof
  set T1 = { u : ex v st v in Tarski-Class(X,A) & u c= v };
  set T2 = { bool v : v in Tarski-Class(X,A) };
  set T3 = bool Tarski-Class(X,A) /\ Tarski-Class X;
A1: Tarski-Class(X,succ A) = T1 \/ T2 \/ T3 by Lm1;
  thus Y in Tarski-Class(X,succ A) implies
  Y c= Tarski-Class(X,A) & Y in Tarski-Class X or
  ex Z st Z in Tarski-Class(X,A) & (Y c= Z or Y = bool Z)
  proof
    assume Y in Tarski-Class(X,succ A);
then A2: Y in T1 \/ T2 or Y in T3 by A1,XBOOLE_0:def 3;
A3: now
      assume Y in T1;
then    ex u st Y = u & ex v st v in Tarski-Class(X,A) & u c= v;
      hence ex Z st Z in Tarski-Class(X,A) & (Y c= Z or Y = bool Z);
    end;
A4: now
      assume Y in T2;
then    ex v st Y = bool v & v in Tarski-Class(X,A);
      hence ex Z st Z in Tarski-Class(X,A) & (Y c= Z or Y = bool Z);
    end;
     Y in T3 implies thesis by XBOOLE_0:def 4;
   hence thesis by A2,A3,A4,XBOOLE_0:def 3;
  end;
  assume
A5: Y c= Tarski-Class(X,A) & Y in Tarski-Class X or
  ex Z st Z in Tarski-Class(X,A) & (Y c= Z or Y = bool Z);
A6: now
    assume Y c= Tarski-Class(X,A) & Y in Tarski-Class X;
then  Y in T3 by XBOOLE_0:def 4;
    hence thesis by A1,XBOOLE_0:def 3;
  end;
 now
    given Z such that
A7: Z in Tarski-Class(X,A) and
A8: Y c= Z or Y = bool Z;
    reconsider Z as Element of Tarski-Class X by A7;
    reconsider y = Y as Element of Tarski-Class X by A7,A8,Th3,Th4;
A9: now
      assume Y c= Z;
then   y in T1 by A7;
then   Y in T1 \/ T2 by XBOOLE_0:def 3;
      hence thesis by A1,XBOOLE_0:def 3;
    end;
 now
      assume Y = bool Z;
then   y in T2 by A7;
then   Y in T1 \/ T2 by XBOOLE_0:def 3;
      hence thesis by A1,XBOOLE_0:def 3;
    end;
    hence thesis by A8,A9;
  end;
  hence thesis by A5,A6;
end;

theorem
  Y c= Z & Z in Tarski-Class(X,A) implies Y in Tarski-Class(X,succ A) by Th10;

theorem
  Y in Tarski-Class(X,A) implies bool Y in Tarski-Class(X,succ A) by Th10;

theorem Th13:
  A <> {} & A is limit_ordinal implies
  (x in Tarski-Class(X,A) iff ex B st B in A & x in Tarski-Class(X,B))
proof
  assume
A1: A <> {} & A is limit_ordinal;
then A2: Tarski-Class(X,A) = { u : ex B st B in A & u in Tarski-Class(X,B) }
  by Th9;
  thus
  x in Tarski-Class(X,A) implies ex B st B in A & x in Tarski-Class(X,B)
  proof
    assume x in Tarski-Class(X,A);
then  ex u st x = u & ex B st B in A & u in Tarski-Class(X,B) by A2;
    hence thesis;
  end;
  given B such that
A3: B in A and
A4: x in Tarski-Class(X,B);
  reconsider u = x as Element of Tarski-Class X by A4;
 u in { v : ex B st B in A & v in Tarski-Class(X,B) } by A3,A4;
  hence thesis by A1,Th9;
end;

theorem
  A <> {} & A is limit_ordinal & Y in Tarski-Class(X,A) &
  (Z c= Y or Z = bool Y) implies Z in Tarski-Class(X,A)
proof
  assume that
A1: A <> {} and
A2: A is limit_ordinal and
A3: Y in Tarski-Class(X,A);
  consider B such that
A4: B in A and
A5: Y in Tarski-Class(X,B) by A1,A2,A3,Th13;
A6: bool Y in Tarski-Class(X,succ B) by A5,Th10;
A7: Z c= Y implies Z in Tarski-Class(X,succ B) by A5,Th10;
A8: succ B in A by A2,A4,ORDINAL1:28;
  assume Z c= Y or Z = bool Y;
  hence thesis by A2,A6,A7,A8,Th13;
end;

theorem Th15:
  Tarski-Class(X,A) c= Tarski-Class(X,succ A)
proof
  let x;
  assume x in Tarski-Class(X,A);
then  x in { u : ex v st v in Tarski-Class(X,A) & u c= v };
then A1: x in { u : ex v st v in Tarski-Class(X,A) & u c= v } \/
  { bool v : v in Tarski-Class(X,A) } by XBOOLE_0:def 3;
   Tarski-Class
(X,succ A) = { u : ex v st v in Tarski-Class(X,A) & u c= v } \/
  { bool v : v in Tarski-Class(X,A) } \/
  bool Tarski-Class(X,A) /\ Tarski-Class X by Lm1;
  hence thesis by A1,XBOOLE_0:def 3;
end;

theorem Th16:
  A c= B implies Tarski-Class(X,A) c= Tarski-Class(X,B)
proof
  defpred OnP[Ordinal] means
  A c= $1 implies Tarski-Class(X,A) c= Tarski-Class(X,$1);
A1: for B st for C st C in B holds OnP[C] holds OnP[B]
  proof
    let B such that
A2: for C st C in B holds OnP[C] and
A3: A c= B;
    let x;
    assume
A4: x in Tarski-Class(X,A);
 now
      assume
A5:   A <> B;
then    A c< B by A3,XBOOLE_0:def 8;
then A6:   A in B by ORDINAL1:11;
A7:   B <> {} by A3,A5;
A8:  now
        given C such that
A9:    B = succ C;
    A c= C & C in B by A6,A9,ORDINAL1:22;
then A10:    Tarski-Class(X,A) c= Tarski-Class(X,C) by A2;
    Tarski-Class(X,C) c= Tarski-Class(X,B) by A9,Th15;
then     Tarski-Class(X,A) c= Tarski-Class(X,B) by A10,XBOOLE_1:1;
        hence thesis by A4;
      end;
  now
        assume for C holds B <> succ C;
then     B is limit_ordinal by ORDINAL1:29;
then     Tarski-Class(X,B) = { v : ex C st C in B & v in Tarski-Class(X,C)
        } by A7,Th9;
        hence thesis by A4,A6;
      end;
      hence thesis by A8;
    end;
    hence thesis by A4;
  end;
 for B holds OnP[B] from ORDINAL1:sch 2(A1);
  hence thesis;
end;

theorem Th17:
  ex A st Tarski-Class(X,A) = Tarski-Class(X,succ A)
proof
  assume
A1: for A holds Tarski-Class(X,A) <> Tarski-Class(X,succ A);
  defpred P[set] means ex A st $1 in Tarski-Class(X,A);
  consider Z such that
A2: x in Z iff x in Tarski-Class X & P[x] from XBOOLE_0:sch 1;
  defpred P[set,set] means
  ex A st $2 = A & $1 in Tarski-Class(X,succ A) & not $1 in Tarski-Class(X,A);
A3: for x,y,z st P[x,y] & P[x,z] holds y = z
  proof
    let x,y,z;
    given A such that
A4: y = A and
A5: x in Tarski-Class(X,succ A) and
A6: not x in Tarski-Class(X,A);
    given B such that
A7: z = B and
A8: x in Tarski-Class(X,succ B) and
A9: not x in Tarski-Class(X,B);
    assume
A10: y <> z;
 A c= B or B c= A;
then A11: A c< B or B c< A by A4,A7,A10,XBOOLE_0:def 8;
 now
      assume A c< B;
then   A in B by ORDINAL1:11;
then   succ A c= B by ORDINAL1:21;
then   Tarski-Class(X,succ A) c= Tarski-Class(X,B) by Th16;
      hence contradiction by A5,A9;
    end;
then  B in A by A11,ORDINAL1:11;
then  succ B c= A by ORDINAL1:21;
then  Tarski-Class(X,succ B) c= Tarski-Class(X,A) by Th16;
    hence contradiction by A6,A8;
  end;
  consider Y such that
A12: x in Y iff ex y st y in Z & P[y,x] from TARSKI:sch 1(A3);
 now
    let A;
A13: Tarski-Class(X,A) <> Tarski-Class(X,succ A) by A1;
A14: Tarski-Class(X,A) c= Tarski-Class(X,succ A) by Th15;
    consider x such that
A15: not (x in Tarski-Class(X,A) iff x in Tarski-Class(X,succ A)) by A13,
TARSKI:1;
 x in Z by A2,A15;
    hence A in Y by A12,A14,A15;
  end;
  hence contradiction by ORDINAL1:26;
end;

theorem Th18:
  Tarski-Class(X,A) = Tarski-Class(X,succ A) implies
  Tarski-Class(X,A) = Tarski-Class X
proof
  assume
A1: Tarski-Class(X,A) = Tarski-Class(X,succ A);
 {} c= A;
then A2: Tarski-Class(X,{}) c= Tarski-Class(X,A) by Th16;
A3: Tarski-Class(X,{}) = { X } & X in { X } by Lm1,TARSKI:def 1;
 Tarski-Class(X,A) is_Tarski-Class_of X
  proof
    thus X in Tarski-Class(X,A) by A2,A3;
A4: Tarski-Class(X,succ A) = { u : ex v st v in Tarski-Class(X,A) & u c= v } \/
    { bool v : v in Tarski-Class(X,A) } \/
    bool Tarski-Class(X,A) /\ Tarski-Class X by Lm1;
 Tarski-Class X is_Tarski-Class_of X by Def4;
then A5: Tarski-Class X is Tarski by Def3;
    thus for Z,Y being set st
    Z in Tarski-Class(X,A) & Y c= Z holds Y in Tarski-Class(X,A)
    proof
      let Z, Y be set;
      assume
A6:   Z in Tarski-Class(X,A) & Y c= Z;
  Tarski-Class X is_Tarski-Class_of X by Def4;
then   Tarski-Class X is Tarski by Def3;
then   Tarski-Class X is subset-closed by Def2;
      then reconsider y = Y as Element of Tarski-Class X by A6,Def1;
  ex v st v in Tarski-Class(X,A) & y c= v by A6;
then   Y in { u : ex v st v in Tarski-Class(X,A) & u c= v };
then   Y in { u : ex v st v in Tarski-Class(X,A) & u c= v } \/
      { bool v : v in Tarski-Class(X,A) } by XBOOLE_0:def 3;
      hence thesis by A1,A4,XBOOLE_0:def 3;
    end;
    thus Y in Tarski-Class(X,A) implies bool Y in Tarski-Class(X,A)
    proof
      assume Y in Tarski-Class(X,A);
then   bool Y in { bool u : u in Tarski-Class(X,A) };
then   bool Y in { u : ex v st v in Tarski-Class(X,A) & u c= v } \/
      { bool v : v in Tarski-Class(X,A) } by XBOOLE_0:def 3;
      hence thesis by A1,A4,XBOOLE_0:def 3;
    end;
    let Y;
    assume that
A7: Y c= Tarski-Class(X,A) and
A8: not Y,Tarski-Class(X,A) are_equipotent;
 Y c= Tarski-Class X by A7,XBOOLE_1:1;
then
 Y,Tarski-Class X are_equipotent or Y in Tarski-Class X by A5,Def2;
    hence thesis by A1,A7,A8,Th10,CARD_1:24;
  end;
  hence Tarski-Class(X,A) c= Tarski-Class X &
  Tarski-Class X c= Tarski-Class(X,A) by Def4;
end;

theorem Th19:
  ex A st Tarski-Class(X,A) = Tarski-Class X
proof
  consider A such that
A1: Tarski-Class(X,A) = Tarski-Class(X,succ A) by Th17;
  take A;
  thus thesis by A1,Th18;
end;

theorem
  ex A st Tarski-Class(X,A) = Tarski-Class X &
  for B st B in A holds Tarski-Class(X,B) <> Tarski-Class X
proof
  defpred P[Ordinal] means Tarski-Class(X,$1) = Tarski-Class X;
A1: ex A st P[A] by Th19;
  consider A such that
A2: P[A] & for B st P[B] holds A c= B from ORDINAL1:sch 1(A1);
  take A;
  thus Tarski-Class(X,A) = Tarski-Class X by A2;
  let B;
  assume B in A;
  hence thesis by A2,ORDINAL1:5;
end;

theorem
  Y <> X & Y in Tarski-Class X implies
  ex A st not Y in Tarski-Class(X,A) & Y in Tarski-Class(X,succ A)
proof
  assume that
A1: Y <> X and
A2: Y in Tarski-Class X;
  defpred P[Ordinal] means Y in Tarski-Class(X,$1);
 ex A st Tarski-Class(X,A) = Tarski-Class X by Th19;
then A3: ex A st P[A] by A2;
  consider A such that
A4: P[A] & for B st P[B] holds A c= B from ORDINAL1:sch 1(A3);
A5: not Y in { X } by A1,TARSKI:def 1;
A6: Tarski-Class(X,{}) = { X } by Lm1;
 now
    assume A is limit_ordinal;
then  ex B st B in A & Y in Tarski-Class(X,B) by A4,A5,A6,Th13;
    hence contradiction by A4,ORDINAL1:5;
  end;
  then consider B such that
A7: A = succ B by ORDINAL1:29;
  take B;
 not A c= B by A7,ORDINAL1:5,6;
  hence thesis by A4,A7;
end;

theorem Th22:
  X is epsilon-transitive implies
  for A st A <> {} holds Tarski-Class(X,A) is epsilon-transitive
proof
  assume
A1: Y in X implies Y c= X;
  defpred OnP[Ordinal] means
  $1 <> {} implies Tarski-Class(X,$1) is epsilon-transitive;
A2: for A st for B st B in A holds OnP[B] holds OnP[A]
  proof
    let A such that
A3: for B st B in A holds OnP[B] and
A4: A <> {};
    let Y such that
A5: Y in Tarski-Class(X,A);
A6: now
      given B such that
A7:   A = succ B;
   B in A by A7,ORDINAL1:6;
then A8:   B c= A by ORDINAL1:def 2;
A9:  OnP[B] by A3,A7,ORDINAL1:6;
A10:  Tarski-Class(X,B) c= Tarski-Class(X,A) by A8,Th16;
  now
        assume not Y c= Tarski-Class(X,B);
        then consider Z such that
A11:    Z in Tarski-Class(X,B) and
A12:    Y c= Z or Y = bool Z by A5,A7,Th10;
A13:    now
          assume
A14:      Y = bool Z;
          thus thesis
          proof
            let x;
            assume x in Y;
            hence thesis by A7,A11,A14,Th10;
          end;
        end;
    now
          assume
A15:      Y c= Z;
          thus thesis
          proof
            let x;
            assume
A16:        x in Y;
then A17:        x in Z by A15;
A18:        now
              assume B = {};
then           Tarski-Class(X,B) = { X } by Lm1;
then A19:          Z = X by A11,TARSKI:def 1;
then           x c= X by A1,A15,A16;
              hence thesis by A7,A11,A19,Th10;
            end;
        now
              assume B <> {};
then           Z c= Tarski-Class(X,B) by A9,A11,ORDINAL1:def 2;
then           x in Tarski-Class(X,B) by A17;
              hence thesis by A10;
            end;
            hence thesis by A18;
          end;
        end;
        hence thesis by A12,A13;
      end;
      hence thesis by A10,XBOOLE_1:1;
    end;
 now
      assume
A20:  for B holds A <> succ B;
then   A is limit_ordinal by ORDINAL1:29;
      then consider B such that
A21:  B in A and
A22:  Y in Tarski-Class(X,B) by A4,A5,Th13;
A23:  succ B c= A by A21,ORDINAL1:21;
A24:  succ B <> A by A20;
A25:  Tarski-Class(X,B) c= Tarski-Class(X,succ B) by Th15;
A26:  succ B c< A by A23,A24,XBOOLE_0:def 8;
A27:  Tarski-Class(X,succ B) c= Tarski-Class(X,A) by A23,Th16;
  Tarski-Class(X,succ B) is epsilon-transitive by A3,A26,ORDINAL1:11;
then   Y c= Tarski-Class(X,succ B) by A22,A25,ORDINAL1:def 2;
      hence thesis by A27,XBOOLE_1:1;
    end;
    hence thesis by A6;
  end;
  thus for A holds OnP[A] from ORDINAL1:sch 2(A2);
end;

theorem Th23:
  X is epsilon-transitive implies Tarski-Class X is epsilon-transitive
proof
  consider A such that
A1: Tarski-Class(X,A) = Tarski-Class X by Th19;
 Tarski-Class(X,A) c= Tarski-Class(X,succ A) by Th15;
then A2: Tarski-Class(X,A) = Tarski-Class(X,succ A) by A1,XBOOLE_0:def 10;
  assume X is epsilon-transitive;
  hence thesis by A1,A2,Th22;
end;

theorem Th24:
  Y in Tarski-Class X implies card Y in card Tarski-Class X
proof
  assume
A1: Y in Tarski-Class X;
 bool Y c= Tarski-Class X
  proof
    let x;
    assume x in bool Y;
    hence thesis by A1,Th3;
  end;
  then  card
 Y in card bool Y & card bool Y c= card Tarski-Class X by CARD_1:11,14;
  hence thesis;
end;

theorem Th25:
  Y in Tarski-Class X implies not Y,Tarski-Class X are_equipotent
proof
  assume Y in Tarski-Class X;
then  card Y in card Tarski-Class X by Th24;
then  card Y <> card Tarski-Class X;
  hence thesis by CARD_1:5;
end;

theorem Th26:
  x in Tarski-Class X & y in Tarski-Class X implies
  {x} in Tarski-Class X & {x,y} in Tarski-Class X
proof
  assume that
A1: x in Tarski-Class X and
A2: y in Tarski-Class X;
 bool x in Tarski-Class X by A1,Th4;
  thus
then A3: {x} in Tarski-Class X by Th3,ZFMISC_1:68;
 bool {x} = {{},{x}} by ZFMISC_1:24;
then A4: not {{},{x}},Tarski-Class X are_equipotent by A3,Th4,Th25;
 now
    assume
A5: x <> y;
 {{},{x}},{x,y} are_equipotent
    proof
      defpred C[set] means $1 = {};
      deffunc f(set) = x;
      deffunc g(set) = y;
      consider f such that
A6:  dom f = {{},{x}} & for z st z in {{},{x}} holds
      (C[z] implies f.z = f(z)) & (not C[z] implies f.z = g(z))
      from PARTFUN1:sch 1;
      take f;
      thus f is one-to-one
      proof
        let x1,x2 be set;
        assume that
A7:    x1 in dom f and
A8:    x2 in dom f;
A9:    x2 = {} or x2 = {x} by A6,A8,TARSKI:def 2;
A10:    x1 = {} implies f.x1 = x by A6,A7;
    x1 <> {} implies f.x1 = y by A6,A7;
        hence thesis by A5,A6,A7,A8,A9,A10,TARSKI:def 2;
      end;
      thus dom f = {{},{x}} by A6;
      thus rng f c= {x,y}
      proof
        let z;
        assume z in rng f;
then     ex u being set st u in dom f & z = f.u by FUNCT_1:def 3;
then     z = x or z = y by A6;
        hence thesis by TARSKI:def 2;
      end;
      let z;
      assume z in {x,y};
then A11:  z = x or z = y by TARSKI:def 2;
A12:  {} in dom f by A6,TARSKI:def 2;
A13:  {x} in dom f by A6,TARSKI:def 2;
A14:  f.{} = x by A6,A12;
  f.{x} = y by A6,A13;
      hence thesis by A11,A12,A13,A14,FUNCT_1:def 3;
    end;
then A15: not {x,y},Tarski-Class X are_equipotent by A4,WELLORD2:15;
 {x,y} c= Tarski-Class X by A1,A2,ZFMISC_1:32;
    hence thesis by A15,Th5;
  end;
  hence thesis by A3,ENUMSET1:29;
end;

theorem Th27:
  x in Tarski-Class X & y in Tarski-Class X implies [x,y] in Tarski-Class X
proof
  assume x in Tarski-Class X & y in Tarski-Class X;
  then  {x,y} in Tarski-Class X & {x} in Tarski-Class X by Th26;
  hence thesis by Th26;
end;

theorem
  Y c= Tarski-Class X & Z c= Tarski-Class X implies [:Y,Z:] c= Tarski-Class X
proof
  assume
A1: Y c= Tarski-Class X & Z c= Tarski-Class X;
  let x,y;
  assume [x,y] in [:Y,Z:];
  then x in Y & y in Z by ZFMISC_1:87;
  hence thesis by A1,Th27;
end;

definition
  let A;
  func Rank(A) means
  :Def6:
  ex L st it = last L & dom L = succ A & L.{} = {} &
  (for C st succ C in succ A holds L.succ C = bool(L.C)) &
  for C st C in succ A & C <> {} & C is limit_ordinal
  holds L.C = union rng(L|C);
  correctness
  proof
    deffunc C(Ordinal,set) = bool $2;
    deffunc D(Ordinal,T-Sequence) = union rng $2;
    thus (ex x,L st x = last L & dom L = succ A & L.{} = {} &
    (for C st succ C in succ A holds L.succ C = C(C,L.C)) &
    for C st C in succ A & C <> {} & C is limit_ordinal
    holds L.C = D(C,L|C) ) & for x1,x2 being set st
    (ex L st x1 = last L & dom L = succ A & L.{} = {} &
    (for C st succ C in succ A holds L.succ C = C(C,L.C)) &
    for C st C in succ A & C <> {} & C is limit_ordinal
    holds L.C = D(C,L|C) ) &
    (ex L st x2 = last L & dom L = succ A & L.{} = {} &
    (for C st succ C in succ A holds L.succ C = C(C,L.C)) &
    for C st C in succ A & C <> {} & C is limit_ordinal holds L.C = D(C,L|C) )
    holds x1 = x2 from ORDINAL2:sch 7;
  end;
end;

deffunc F(Ordinal) = Rank $1;

Lm2: now
  deffunc C(Ordinal,set) = bool $2;
  deffunc D(Ordinal,T-Sequence) = union rng $2;
A1: for A,x holds x = F(A) iff
  ex L st x = last L & dom L = succ A & L.{} = {} &
  (for C st succ C in succ A holds L.succ C = C(C,L.C)) &
  for C st C in succ A & C <> {} & C is limit_ordinal
  holds L.C = D(C,L|C) by Def6;
  thus F({}) = {} from ORDINAL2:sch 8(A1);
  thus for A holds F(succ A) = C(A,F(A)) from ORDINAL2:sch 9(A1);
  thus A <> {} & A is limit_ordinal & dom L = A &
  (for B st B in A holds L.B = Rank B) implies Rank A = union rng L
  proof
    assume that
A2: A <> {} & A is limit_ordinal and
A3: dom L = A and
A4: for B st B in A holds L.B = F(B);
    thus F(A) = D(A,L) from ORDINAL2:sch 10(A1,A2,A3,A4);
  end;
end;

theorem
  Rank {} = {} by Lm2;

theorem
  Rank succ A = bool Rank A by Lm2;

theorem Th31:
  A <> {} & A is limit_ordinal implies
  for x holds x in Rank A iff ex B st B in A & x in Rank B
proof
  assume
A1: A <> {} & A is limit_ordinal;
  consider L such that
A2: dom L = A & for B st B in A holds L.B = F(B) from ORDINAL2:sch 2;
A3: Rank A = union rng L by A1,A2,Lm2;
  let x;
  thus x in Rank A implies ex B st B in A & x in Rank B
  proof
    assume x in Rank A;
    then consider Y such that
A4: x in Y and
A5: Y in rng L by A3,TARSKI:def 4;
    consider y such that
A6: y in dom L and
A7: Y = L.y by A5,FUNCT_1:def 3;
    reconsider y as Ordinal by A6;
    take y;
    thus thesis by A2,A4,A6,A7;
  end;
  given B such that
A8: B in A and
A9: x in Rank B;
 L.B = Rank B by A2,A8;
then  Rank B in rng L by A2,A8,FUNCT_1:def 3;
  hence thesis by A3,A9,TARSKI:def 4;
end;

theorem Th32:
  X c= Rank A iff X in Rank succ A
proof
  thus X c= Rank A implies X in Rank succ A
  proof
    assume X c= Rank A;
then  X in bool Rank A;
    hence thesis by Lm2;
  end;
  assume X in Rank succ A;
then  X in bool Rank A by Lm2;
  hence thesis;
end;

registration
  let A;
  cluster Rank A -> epsilon-transitive;
  coherence
  proof
    defpred P[Ordinal] means X in Rank $1 implies X c= Rank $1;
A1: for A st for B st B in A holds P[B] holds P[A]
    proof
      let A such that
A2:   for B st B in A holds P[B];
      let X such that
A3:   X in Rank A;
      let x such that
A4:   x in X;
A5:   now
        assume
A6:     A is limit_ordinal;
        then consider B such that
A7:     B in A and
A8:     X in Rank B by A3,Lm2,Th31;
     X c= Rank B by A2,A7,A8;
        hence thesis by A4,A6,A7,Th31;
      end;
  now
        assume not A is limit_ordinal;
        then consider B such that
A9:    A = succ B by ORDINAL1:29;
    X c= Rank B by A3,A9,Th32;
then     x c= Rank B by A2,A4,A9,ORDINAL1:6;
        hence thesis by A9,Th32;
      end;
      hence thesis by A5;
    end;
 for A holds P[A] from ORDINAL1:sch 2(A1);
    hence P[A];
  end;
end;

theorem
  Rank A c= Rank succ A
proof
 Rank A in bool Rank A by ZFMISC_1:def 1;
then  Rank A in Rank succ A by Lm2;
  hence thesis by ORDINAL1:def 2;
end;

theorem Th34:
  union Rank A c= Rank A
proof
  let x;
  assume x in union Rank A;
  then consider X such that
A1: x in X and
A2: X in Rank A by TARSKI:def 4;
 X c= Rank A by A2,ORDINAL1:def 2;
  hence thesis by A1;
end;

theorem
  X in Rank A implies union X in Rank A
proof
  assume
A1: X in Rank A;
A2: now
    given B such that
A3: A = succ B;
 X c= Rank B by A1,A3,Th32;
then A4: union X c= union Rank B by ZFMISC_1:77;
 union Rank B c= Rank B by Th34;
then  union X c= Rank B by A4,XBOOLE_1:1;
    hence thesis by A3,Th32;
  end;
 now
    assume that
A5: A <> {} and
A6: for B holds A <> succ B;
A7: A is limit_ordinal by A6,ORDINAL1:29;
    then consider B such that
A8: B in A and
A9: X in Rank B by A1,A5,Th31;
 X c= Rank B by A9,ORDINAL1:def 2;
then A10: union X c= union Rank B by ZFMISC_1:77;
 union Rank B c= Rank B by Th34;
then A11: union X c= Rank B by A10,XBOOLE_1:1;
A12: succ B c= A by A8,ORDINAL1:21;
 succ B <> A by A6;
then A13: succ B c< A by A12,XBOOLE_0:def 8;
A14: union X in Rank succ B by A11,Th32;
 succ B in A by A13,ORDINAL1:11;
    hence thesis by A7,A14,Th31;
  end;
  hence thesis by A1,A2,Lm2;
end;

theorem Th36:
  A in B iff Rank A in Rank B
proof
  defpred OnP[Ordinal,Ordinal] means $1 in $2 implies Rank $1 in Rank $2;
A1: now
    let A;
    defpred P[Ordinal] means OnP[A,$1];
A2: for B st for C st C in B holds P[C] holds P[B]
    proof
      let B such that
A3:   for C st C in B holds OnP[A,C] and
A4:   A in B;
A5:   now
        given C such that
A6:     B = succ C;
A7:     A in C implies Rank A in Rank C by A3,A6,ORDINAL1:6;
     now
          assume
A8:       not A in C;
      A c= C & A <> C iff A c< C by XBOOLE_0:def 8;
          hence Rank A = Rank C by A4,A6,A8,ORDINAL1:11,22;
        end;
then     Rank A c= Rank C by A7,ORDINAL1:def 2;
        hence thesis by A6,Th32;
      end;
  now
        assume
A9:    for C holds B <> succ C;
then A10:    B is limit_ordinal by ORDINAL1:29;
A11:    B <> succ A by A9;
    succ A c= B by A4,ORDINAL1:21;
then     succ A c< B by A11,XBOOLE_0:def 8;
then A12:    succ A in B by ORDINAL1:11;
    Rank A in Rank succ A by Th32;
        hence thesis by A10,A12,Th31;
      end;
      hence thesis by A5;
    end;
    thus for B holds P[B] from ORDINAL1:sch 2(A2);
  end;
  hence OnP[A,B];
  assume that
A13: Rank A in Rank B and
A14: not A in B;
 B in A or B = A by A14,ORDINAL1:14;
  hence contradiction by A1,A13;
end;

theorem Th37:
  A c= B iff Rank A c= Rank B
proof
  thus A c= B implies Rank A c= Rank B
  proof
A1: A c< B iff A c= B & A <> B by XBOOLE_0:def 8;
    assume A c= B;
then  A = B or A in B by A1,ORDINAL1:11;
then  Rank A = Rank B or Rank A in Rank B by Th36;
    hence thesis by ORDINAL1:def 2;
  end;
  assume that
A2: Rank A c= Rank B and
A3: not A c= B;
 B in A by A3,ORDINAL1:16;
then  Rank B in Rank A by Th36;
  hence contradiction by A2,ORDINAL1:5;
end;

theorem Th38:
  A c= Rank A
proof
  defpred P[Ordinal] means $1 c= Rank $1;
A1: P[{}] by XBOOLE_1:2;
A2: P[B] implies P[succ B]
  proof
    assume B c= Rank B;
then  B in Rank succ B by Th32;
    then
 B c= Rank succ B & {B} c= Rank succ B by ORDINAL1:def 2,ZFMISC_1:31;
    hence thesis by XBOOLE_1:8;
  end;
A3: for A st A <> {} & A is limit_ordinal & for B st B in A holds P[B]
  holds P[A]
  proof
    let A such that
    A <> {} and
A4: A is limit_ordinal and
A5: for B st B in A holds B c= Rank B;
    let x;
    assume
A6: x in A;
    then reconsider B = x as Ordinal;
A7: succ B in A by A4,A6,ORDINAL1:28;
A8: B c= Rank B by A5,A6;
A9: succ B c= A by A7,ORDINAL1:def 2;
A10: B in Rank succ B by A8,Th32;
 Rank succ B c= Rank A by A9,Th37;
    hence thesis by A10;
  end;
 for B holds P[B] from ORDINAL2:sch 1(A1,A2,A3);
  hence thesis;
end;

theorem
  for A,X st X in Rank A holds
  not X,Rank A are_equipotent & card X in card Rank A
proof
  defpred OnP[Ordinal] means
  for X st X in Rank $1 holds not X,Rank $1 are_equipotent;
A1: for A st for B st B in A holds OnP[B] holds OnP[A]
  proof
    let A such that
A2: for B st B in A holds OnP[B];
    let X;
    assume
A3: X in Rank A;
A4: now
      given B such that
A5:   A = succ B;
   B in A by A5,ORDINAL1:6;
then A6:   B c= A by ORDINAL1:def 2;
A7:   Rank succ B = bool Rank B by Lm2;
then A8:   not Rank B,Rank A are_equipotent by A5,CARD_1:13;
  Rank B c= Rank A by A6,Th37;
      hence thesis by A3,A5,A7,A8,CARD_1:24;
    end;
 now
      assume that
A9:  A <> {} and
A10:  for B holds A <> succ B;
  A is limit_ordinal by A10,ORDINAL1:29;
      then consider B such that
A11:  B in A and
A12:  X in Rank B by A3,A9,Th31;
A13:  Rank B in Rank A by A11,Th36;
      A14:  (
 not X,Rank B are_equipotent)& X c= Rank B by A2,A11,A12,ORDINAL1:def 2;
  Rank B c= Rank A by A13,ORDINAL1:def 2;
      hence thesis by A14,CARD_1:24;
    end;
    hence thesis by A3,A4,Lm2;
  end;
A15: for A holds OnP[A] from ORDINAL1:sch 2(A1);
  let A,X;
  assume
A16: X in Rank A;
then A17: X c= Rank A by ORDINAL1:def 2;
A18: not X,Rank A are_equipotent by A15,A16;
A19: card X c= card Rank A by A17,CARD_1:11;
 card X <> card Rank A by A18,CARD_1:5;
  hence thesis by A15,A16,A19,CARD_1:3;
end;

theorem
  X c= Rank A iff bool X c= Rank succ A
proof
  thus X c= Rank A implies bool X c= Rank succ A
  proof
    assume
A1: X c= Rank A;
    let x;
    assume x in bool X;
then A2: x c= Rank A by A1,XBOOLE_1:1;
 Rank succ A = bool Rank A by Lm2;
    hence thesis by A2;
  end;
  assume
A3: bool X c= Rank succ A;
  let x;
  assume x in X;
then  { x } c= X by ZFMISC_1:31;
then A4: { x } in bool X;
 Rank succ A = bool Rank A & x in { x } by Lm2,TARSKI:def 1;
  hence thesis by A3,A4;
end;

theorem Th41:
  X c= Y & Y in Rank A implies X in Rank A
proof
  assume that
A1: X c= Y and
A2: Y in Rank A;
A3: now
    given B such that
A4: A = succ B;
A5: Rank succ B = bool Rank B by Lm2;
then  X c= Rank B by A1,A2,A4,XBOOLE_1:1;
    hence thesis by A4,A5;
  end;
 now
    assume for B holds A <> succ B;
then A6: A is limit_ordinal by ORDINAL1:29;
    then consider B such that
A7: B in A and
A8: Y in Rank B by A2,Lm2,Th31;
 Y c= Rank B by A8,ORDINAL1:def 2;
then A9: X c= Rank B by A1,XBOOLE_1:1;
A10: bool Rank B = Rank succ B by Lm2;
 succ B in A by A6,A7,ORDINAL1:28;
    hence thesis by A6,A9,A10,Th31;
  end;
  hence thesis by A3;
end;

theorem Th42:
  X in Rank A iff bool X in Rank succ A
proof
  thus X in Rank A implies bool X in Rank succ A
  proof
    assume
A1: X in Rank A;
 bool X c= Rank A
    proof
      let x;
      assume x in bool X;
      hence thesis by A1,Th41;
    end;
    hence thesis by Th32;
  end;
  assume bool X in Rank succ A;
then  X in bool X & bool X c= Rank A by Th32,ZFMISC_1:def 1;
  hence thesis;
end;

theorem Th43:
  x in Rank A iff {x} in Rank succ A
proof
 x in Rank A iff {x} c= Rank A by ZFMISC_1:31;
  hence thesis by Th32;
end;

theorem Th44:
  x in Rank A & y in Rank A iff {x,y} in Rank succ A
proof
 x in Rank A & y in Rank A iff {x,y} c= Rank A by ZFMISC_1:32;
  hence thesis by Th32;
end;

theorem
  x in Rank A & y in Rank A iff [x,y] in Rank succ succ A
proof
A1: x in Rank A iff {x} in Rank succ A by Th43;
 x in Rank A & y in Rank A iff {x,y} in Rank succ A by Th44;
  hence thesis by A1,Th44;
end;

theorem Th46:
  X is epsilon-transitive &
  Rank A /\ Tarski-Class X = Rank succ A /\ Tarski-Class X implies
  Tarski-Class X c= Rank A
proof
  assume that
A1: X is epsilon-transitive and
A2: Rank A /\ Tarski-Class X = Rank succ A /\ Tarski-Class X;
  given x such that
A3: x in Tarski-Class X & not x in Rank A;
 x in (Tarski-Class X) \ Rank A by A3,XBOOLE_0:def 5;
  then consider Y such that
A4: Y in (Tarski-Class X) \ Rank A and
A5: not ex x st x in (Tarski-Class X) \ Rank A & x in Y by TARSKI:2;
 Tarski-Class X is epsilon-transitive by A1,Th23;
then A6: Y c= Tarski-Class X by A4,ORDINAL1:def 2;
 Y c= Rank A
  proof
    let x;
    assume
A7: x in Y;
then  not x in (Tarski-Class X) \ Rank A by A5;
    hence thesis by A6,A7,XBOOLE_0:def 5;
  end;
then  Y in Rank succ A by Th32;
then A8: Y in Rank succ A /\ Tarski-Class X by A4,XBOOLE_0:def 4;
 not Y in Rank A by A4,XBOOLE_0:def 5;
  hence contradiction by A2,A8,XBOOLE_0:def 4;
end;

theorem Th47:
  X is epsilon-transitive implies ex A st Tarski-Class X c= Rank A
proof
  assume
A1: X is epsilon-transitive;
  assume
A2: not Tarski-Class X c= Rank A;
  defpred P[set] means ex A st $1 in Rank A;
  consider Power being set such that
A3: x in Power iff x in Tarski-Class X & P[x] from XBOOLE_0:sch 1;
  defpred P[set,set] means
  ex A st $2 = A & not $1 in Rank A & $1 in Rank succ A;
A4: for x,y,z st P[x,y] & P[x,z] holds y = z
  proof
    let x,y,z;
    given A1 being Ordinal such that
A5: y = A1 and
A6: not x in Rank A1 and
A7: x in Rank succ A1;
    given A2 being Ordinal such that
A8: z = A2 and
A9: not x in Rank A2 and
A10: x in Rank succ A2;
    assume y <> z;
then  A1 in A2 or A2 in A1 by A5,A8,ORDINAL1:14;
then A11: succ A1 c= A2 or succ A2 c= A1 by ORDINAL1:21;
 now
      assume succ A1 c= A2;
then   Rank succ A1 c= Rank A2 by Th37;
      hence contradiction by A7,A9;
    end;
then  Rank succ A2 c= Rank A1 by A11,Th37;
    hence contradiction by A6,A10;
  end;
  consider Y such that
A12: x in Y iff ex y st y in Power & P[y,x] from TARSKI:sch 1(A4);
 now
    let A;
 Rank A /\ Tarski-Class X <> Rank succ A /\ Tarski-Class X by A1,A2,Th46;
    then consider y such that
    A13: not
 (y in Rank A /\ Tarski-Class X iff y in Rank succ A /\ Tarski-Class X)
    by TARSKI:1;
 A in succ A by ORDINAL1:6;
then  A c= succ A by ORDINAL1:def 2;
then A14: Rank A c= Rank succ A by Th37;
then  Rank A /\ Tarski-Class X c= Rank succ A /\
    Tarski-Class X by XBOOLE_1:26;
then A15: y in Rank succ A by A13,XBOOLE_0:def 4;
A16: y in Tarski-Class X by A13,XBOOLE_0:def 4;
A17: not y in Rank A or not y in Tarski-Class X by A13,A14,XBOOLE_0:def 4;
 y in Power by A3,A15,A16;
    hence A in Y by A12,A13,A15,A17,XBOOLE_0:def 4;
  end;
  hence contradiction by ORDINAL1:26;
end;

theorem Th48:
  X is epsilon-transitive implies union X c= X
proof
  assume
A1: Y in X implies Y c= X;
  let x;
  assume x in union X;
  then consider Y such that
A2: x in Y and
A3: Y in X by TARSKI:def 4;
 Y c= X by A1,A3;
  hence thesis by A2;
end;

theorem Th49:
  X is epsilon-transitive & Y is epsilon-transitive implies
  X \/ Y is epsilon-transitive
proof
  assume that
A1: ( Z in X implies Z c= X) and
A2: ( Z in Y implies Z c= Y);
  let Z;
  assume Z in X \/ Y;
then  Z in X or Z in Y by XBOOLE_0:def 3;
then A3: Z c= X or Z c= Y by A1,A2;
 X c= X \/ Y & Y c= X \/ Y by XBOOLE_1:7;
  hence thesis by A3,XBOOLE_1:1;
end;

theorem
  X is epsilon-transitive & Y is epsilon-transitive implies
  X /\ Y is epsilon-transitive
proof
  assume that
A1: Z in X implies Z c= X and
A2: Z in Y implies Z c= Y;
  let Z;
  assume
A3: Z in X /\ Y;
then A4: Z in X by XBOOLE_0:def 4;
A5: Z in Y by A3,XBOOLE_0:def 4;
A6: Z c= X by A1,A4;
 Z c= Y by A2,A5;
  hence thesis by A6,XBOOLE_1:19;
end;

reserve n for Element of omega;

deffunc f(set,set) = union $2;

definition
  let X;
  func the_transitive-closure_of X -> set means
  :Def7:
  x in it iff ex f,n st x in f.n & dom f = omega & f.0 = X &
  for k being Nat holds f.(succ k) = union(f.k);
  existence
  proof
    consider f such that
A1: dom f = omega & f.0 = X &
    for n being Nat holds f.(succ n) = f(n,f.n)
     from ORDINAL2:sch 18;
    take UNI = union rng f;
    let x;
    thus x in UNI implies ex f,n st x in f.n & dom f = omega & f.0 = X &
    for k being Nat holds f.(succ k) = union(f.k)
    proof
      assume x in UNI;
      then consider Y such that
A2:   x in Y and
A3:   Y in rng f by TARSKI:def 4;
      consider y such that
A4:   y in dom f and
A5:   Y = f.y by A3,FUNCT_1:def 3;
      reconsider y as Element of omega by A1,A4;
      take f,y;
      thus thesis by A1,A2,A5;
    end;
    deffunc f(set,set) = union $2;
    given g,n such that
A6: x in g.n and
A7: dom g = omega and
A8: g.0 = X and
A9: for k being Nat holds g.(succ k) = f(k,g.k);
A10: dom f = omega by A1;
A11: f.0 = X by A1;
A12: for n being Nat holds f.(succ n) = f(n,f.n) by A1;
 g = f from ORDINAL2:sch 20(A7,A8,A9,A10,A11,A12);
then  g.n in rng f by A1,FUNCT_1:def 3;
    hence thesis by A6,TARSKI:def 4;
  end;
  uniqueness
  proof
    defpred P[set] means ex f,n st $1 in f.n & dom f = omega & f.0 = X &
    for k being Nat holds f.(succ k) = union(f.k);
    let U1,U2 be set such that
A13: x in U1 iff P[x] and
A14: x in U2 iff P[x];
    thus U1 = U2 from XBOOLE_0:sch 2(A13,A14);
  end;
end;

theorem Th51:
  the_transitive-closure_of X is epsilon-transitive
proof
  let Y;
  assume Y in the_transitive-closure_of X;
  then consider f,n such that
A1: Y in f.n and
A2: dom f = omega & f.0 = X and
A3: for k being Nat holds f.(succ k) = union(f.k) by Def7;
A4: f.(succ n) = union(f.n) by A3;
  let x;
  assume x in Y;
then A5: x in union(f.n) by A1,TARSKI:def 4;
   reconsider m = succ n as Element of omega by ORDINAL1:def 12;
   x in f.m by A4,A5;
  hence x in the_transitive-closure_of X by A2,A3,Def7;
end;

theorem Th52:
  X c= the_transitive-closure_of X
proof
  let x such that
A1: x in X;
  consider f such that
A2: dom f = omega and
A3: f.0 = X and
A4: for n being Nat holds f.(succ n) = f(n,f.n)
        from ORDINAL2:sch 18;
   reconsider z = 0 as Element of omega by ORDINAL1:def 12;
    x in f.z by A1,A3;
  hence x in the_transitive-closure_of X by A2,A3,A4,Def7;
end;

theorem Th53:
  X c= Y & Y is epsilon-transitive implies the_transitive-closure_of X c= Y
proof
  assume that
A1: X c= Y and
A2: Y is epsilon-transitive;
  let x;
  assume x in the_transitive-closure_of X;
  then consider f,n such that
A3: x in f.n and dom f = omega and
A4: f.0 = X and
A5: for k being Nat holds f.(succ k) = union(f.k) by Def7;
  defpred P[Nat] means f.$1 c= Y;
A6: P[0] by A1,A4;
A7: for k being Nat st P[k] holds P[succ k]
  proof
    let k be Nat;
    assume f.k c= Y;
then A8: union (f.k) c= union Y by ZFMISC_1:77;
 f.(succ k) = union (f.k) & union Y c= Y by A2,A5,Th48;
    hence thesis by A8,XBOOLE_1:1;
  end;
 P[n] from ORDINAL2:sch 17(A6,A7);
then  f.n c= Y;
  hence thesis by A3;
end;

theorem Th54:
  (for Z st X c= Z & Z is epsilon-transitive holds Y c= Z) & X c= Y &
  Y is epsilon-transitive implies the_transitive-closure_of X = Y
proof
  assume
A1: for Z st X c= Z & Z is epsilon-transitive holds Y c= Z;
  assume X c= Y & Y is epsilon-transitive;
  hence the_transitive-closure_of X c= Y by Th53;
 the_transitive-closure_of X is epsilon-transitive by Th51;
  hence thesis by A1,Th52;
end;

theorem Th55:
  X is epsilon-transitive implies the_transitive-closure_of X = X
proof
 for Z st X c= Z & Z is epsilon-transitive holds X c= Z;
  hence thesis by Th54;
end;

theorem
  the_transitive-closure_of {} = {} by Th55;

theorem
  the_transitive-closure_of A = A by Th55;

theorem Th58:
  X c= Y implies the_transitive-closure_of X c= the_transitive-closure_of Y
proof
  assume
A1: X c= Y;
 Y c= the_transitive-closure_of Y by Th52;
then  X c= the_transitive-closure_of Y by A1,XBOOLE_1:1;
  hence thesis by Th51,Th53;
end;

theorem
  the_transitive-closure_of the_transitive-closure_of X =
  the_transitive-closure_of X by Th51,Th55;

theorem
  the_transitive-closure_of (X \/ Y) =
  the_transitive-closure_of X \/ the_transitive-closure_of Y
proof
   X
 c= the_transitive-closure_of X & Y c= the_transitive-closure_of Y by Th52;
then
A1: X \/ Y c= the_transitive-closure_of X \/ the_transitive-closure_of Y by
XBOOLE_1:13;
  A2: the_transitive-closure_of X is epsilon-transitive &
  the_transitive-closure_of Y is epsilon-transitive by Th51;
 the_transitive-closure_of (X \/ Y) c= the_transitive-closure_of (
  the_transitive-closure_of X \/ the_transitive-closure_of Y) by A1,Th58;
  hence the_transitive-closure_of (X \/ Y) c=
  the_transitive-closure_of X \/ the_transitive-closure_of Y by A2,Th49,Th55
  ;
   the_transitive-closure_of X c= the_transitive-closure_of (X \/ Y) &
  the_transitive-closure_of Y c= the_transitive-closure_of (X \/ Y) by Th58,
XBOOLE_1:7;
  hence thesis by XBOOLE_1:8;
end;

theorem
  the_transitive-closure_of (X /\ Y) c=
  the_transitive-closure_of X /\ the_transitive-closure_of Y
proof
   the_transitive-closure_of (X /\ Y) c= the_transitive-closure_of X &
  the_transitive-closure_of (X /\ Y) c= the_transitive-closure_of Y by Th58,
XBOOLE_1:17;
  hence thesis by XBOOLE_1:19;
end;

theorem Th62:
  ex A st X c= Rank A
proof
  consider A such that
A1: Tarski-Class the_transitive-closure_of X c= Rank A by Th47,Th51;
  take A;
 the_transitive-closure_of X in Tarski-Class the_transitive-closure_of X
  by Th2;
then  X in Tarski-Class the_transitive-closure_of X by Th3,Th52;
  hence thesis by A1,ORDINAL1:def 2;
end;

definition
  let X;
  func the_rank_of X -> Ordinal means
  :Def8:
  X c= Rank it & for B st X c= Rank B holds it c= B;
  existence
  proof
    defpred P[Ordinal] means X c= Rank $1;
A1: ex A st P[A] by Th62;
    thus ex A st P[A] & for B st P[B] holds A c= B from ORDINAL1:sch 1(
    A1);
  end;
  uniqueness
  proof
    let A1,A2 be Ordinal such that
    A2: ( X c= Rank A1 & for B st X c= Rank B holds A1 c= B )&( X c= Rank A2 &
    for B st X c= Rank B holds A2 c= B );
    thus A1 c= A2 & A2 c= A1 by A2;
  end;
end;

theorem Th63:
  the_rank_of bool X = succ the_rank_of X
proof
A1: X c= Rank the_rank_of X by Def8;
A2: bool X c= Rank succ the_rank_of X
  proof
    let x;
    assume x in bool X;
then A3: x c= Rank the_rank_of X by A1,XBOOLE_1:1;
 bool Rank the_rank_of X = Rank succ the_rank_of X by Lm2;
    hence thesis by A3;
  end;
 for A st bool X c= Rank A holds succ the_rank_of X c= A
  proof
    let A such that
A4: bool X c= Rank A;
    defpred P[Ordinal] means X in Rank $1;
A5: X in bool X by ZFMISC_1:def 1;
then A6: ex A st P[A] by A4;
    consider B such that
A7: P[B] & for C st P[C] holds B c= C from ORDINAL1:sch 1(A6);
 now
      assume for C holds B <> succ C;
then   B is limit_ordinal by ORDINAL1:29;
then   ex C st C in B & X in Rank C by A7,Lm2,Th31;
      hence contradiction by A7,ORDINAL1:5;
    end;
    then consider C such that
A8: B = succ C;
 X c= Rank C by A7,A8,Th32;
then  the_rank_of X c= C by Def8;
then A9: the_rank_of X in B by A8,ORDINAL1:22;
 B c= A by A4,A5,A7;
    hence thesis by A9,ORDINAL1:21;
  end;
  hence thesis by A2,Def8;
end;

theorem
  the_rank_of Rank A = A
proof
 for B st Rank A c= Rank B holds A c= B by Th37;
  hence thesis by Def8;
end;

theorem Th65:
  X c= Rank A iff the_rank_of X c= A
proof
  thus X c= Rank A implies the_rank_of X c= A by Def8;
  assume the_rank_of X c= A;
then A1: Rank the_rank_of X c= Rank A by Th37;
 X c= Rank the_rank_of X by Def8;
  hence thesis by A1,XBOOLE_1:1;
end;

theorem Th66:
  X in Rank A iff the_rank_of X in A
proof
  thus X in Rank A implies the_rank_of X in A
  proof
    assume X in Rank A;
then  bool X in Rank succ A by Th42;
then A1: bool X c= Rank A by Th32;
 the_rank_of bool X = succ the_rank_of X by Th63;
then A2: the_rank_of X in the_rank_of bool X by ORDINAL1:6;
 the_rank_of bool X c= A by A1,Def8;
    hence thesis by A2;
  end;
  assume the_rank_of X in A;
then A3: succ the_rank_of X c= A by ORDINAL1:21;
 X c= Rank the_rank_of X by Def8;
then A4: X in Rank succ the_rank_of X by Th32;
 Rank succ the_rank_of X c= Rank A by A3,Th37;
  hence thesis by A4;
end;

theorem
  X c= Y implies the_rank_of X c= the_rank_of Y
proof
  assume
A1: X c= Y;
 Y c= Rank the_rank_of Y by Def8;
then  X c= Rank the_rank_of Y by A1,XBOOLE_1:1;
  hence thesis by Def8;
end;

theorem Th68:
  X in Y implies the_rank_of X in the_rank_of Y
proof
  assume
A1: X in Y;
 Y c= Rank the_rank_of Y by Def8;
  hence thesis by A1,Th66;
end;

theorem Th69:
  the_rank_of X c= A iff for Y st Y in X holds the_rank_of Y in A
proof
  set R = the_rank_of X;
A1: X c= Rank R by Def8;
  thus
  the_rank_of X c= A implies for Y st Y in X holds the_rank_of Y in A
  proof
    assume
A2: the_rank_of X c= A;
    let Y;
    assume Y in X;
then A3: Y in Rank R by A1;
 Rank R c= Rank A by A2,Th37;
    hence thesis by A3,Th66;
  end;
  assume
A4: for Y st Y in X holds the_rank_of Y in A;
 X c= Rank A
  proof
    let x;
    assume x in X;
then  the_rank_of x in A by A4;
    hence thesis by Th66;
  end;
  hence thesis by Def8;
end;

theorem Th70:
  A c= the_rank_of X iff for B st B in A ex Y st Y in X & B c= the_rank_of Y
proof
  thus A c= the_rank_of X implies
  for B st B in A ex Y st Y in X & B c= the_rank_of Y
  proof
    assume
A1: A c= the_rank_of X;
    let B;
    assume B in A;
then  not the_rank_of X c= B by A1,ORDINAL1:5;
then  not X c= Rank B by Def8;
then A2: X \ Rank B <> {} by XBOOLE_1:37;
    set x = the Element of X \ Rank B;
    take x;
A3: not x in Rank B by A2,XBOOLE_0:def 5;
    thus x in X by A2,XBOOLE_0:def 5;
 not the_rank_of x in B by A3,Th66;
    hence thesis by ORDINAL1:16;
  end;
  assume
A4: for B st B in A ex Y st Y in X & B c= the_rank_of Y;
  let x;
  assume
A5: x in A;
  then reconsider x as Ordinal;
  consider Y such that
A6: Y in X and
A7: x c= the_rank_of Y by A4,A5;
 the_rank_of Y in the_rank_of X by A6,Th68;
  hence thesis by A7,ORDINAL1:12;
end;

theorem
  the_rank_of X = {} iff X = {}
proof
  thus the_rank_of X = {} implies X = {}
  proof
    assume the_rank_of X = {};
then  X c= Rank {} by Def8;
    hence thesis by Lm2;
  end;
  assume X = {};
then  for Y st Y in X holds the_rank_of Y in {};
  hence the_rank_of X c= {} by Th69;
  thus {} c= the_rank_of X;
end;

theorem Th72:
  the_rank_of X = succ A implies ex Y st Y in X & the_rank_of Y = A
proof
  assume
A1: the_rank_of X = succ A;
 A in succ A by ORDINAL1:6;
  then consider Y such that
A2: Y in X and
A3: A c= the_rank_of Y by A1,Th70;
  take Y;
 the_rank_of Y in the_rank_of X by A2,Th68;
then  the_rank_of Y c= A by A1,ORDINAL1:22;
  hence thesis by A2,A3,XBOOLE_0:def 10;
end;

theorem
  the_rank_of A = A
proof
 A c= Rank A by Th38;
  hence the_rank_of A c= A by Th65;
  defpred P[Ordinal] means $1 c= the_rank_of $1;
A1: for A st for B st B in A holds P[B] holds P[A]
  proof
    let A such that
A2: for B st B in A holds B c= the_rank_of B;
 now
      let B such that
A3:   B in A;
      reconsider Y = B as set;
      take Y;
      thus Y in A & B c= the_rank_of Y by A2,A3;
    end;
    hence thesis by Th70;
  end;
 P[B] from ORDINAL1:sch 2(A1);
  hence thesis;
end;

theorem
  the_rank_of Tarski-Class X <> {} &
  the_rank_of Tarski-Class X is limit_ordinal
proof
A1: Tarski-Class X c= Rank the_rank_of Tarski-Class X by Def8;
  thus the_rank_of Tarski-Class X <> {}
  proof
    assume the_rank_of Tarski-Class X = {};
then  Tarski-Class X c= {} by Def8,Lm2;
    hence contradiction;
  end;
  assume not thesis;
  then consider A such that
A2: the_rank_of Tarski-Class X = succ A by ORDINAL1:29;
  consider Y such that
A3: Y in Tarski-Class X and
A4: the_rank_of Y = A by A2,Th72;
A5: bool Y in Tarski-Class X by A3,Th4;
A6: the_rank_of bool Y = succ A by A4,Th63;
 bool Y c= Rank A by A1,A2,A5,Th32;
then  succ A c= A by A6,Def8;
then  A in A by ORDINAL1:21;
  hence contradiction;
end;

begin 


reserve e,u for set;

scheme NonUniqFuncEx { X() -> set, P[set,set] }:
  ex f being Function st dom f = X() & for e st e in X() holds P[e,f.e]
provided
A1: for e st e in X() ex u st P[e,u]
proof
  per cases;
  suppose
 X() = {};
then A2: for e st e in X() ex u st u in {} & P[e,u];
 ex f being Function st dom f = X() & rng f c= {} &
    for e st e in X() holds P[e,f.e] from FUNCT_1:sch 5(A2);
    hence thesis;
  end;
  suppose
 X() <> {};
    then reconsider D = X() as non empty set;
    defpred Q[set,Ordinal] means ex u st u in Rank $2 & P[$1,u];
A3: for e being Element of D ex A st Q[e,A]
    proof
      let e be Element of D;
      consider u such that
A4:   P[e,u] by A1;
   u c= Rank the_rank_of u by Def8;
then    u in Rank succ the_rank_of u by Th32;
      hence thesis by A4;
    end;
    consider F be Function such that
A5: dom F = D & for d being Element of D ex A st A = F.d & Q[d,A] &
    for B st Q[d,B] holds A c= B from ORDINAL1:sch 6(A3);
A6: for e st e in X() ex u st u in Rank sup rng F & P[e,u]
    proof
      let e;
      assume
A7:  e in X();
      then consider A such that
A8:  A = F.e and
A9:  ex u st u in Rank A & P[e,u]
      and for B st ex u st u in Rank B & P[e,u] holds A c= B by A5;
      consider u such that
A10:  u in Rank A & P[e,u] by A9;
      take u;
  A in rng F by A5,A7,A8,FUNCT_1:def 3;
then   A in sup rng F by ORDINAL2:19;
then   A c= sup rng F by ORDINAL1:def 2;
then   Rank A c= Rank sup rng F by Th37;
      hence thesis by A10;
    end;
 ex f being Function st dom f = X() & rng f c= Rank sup rng F &
    for e st e in X() holds P[e, f.e] from FUNCT_1:sch 5(A6);
    hence thesis;
  end;
end;



definition
  let F,G be Relation;
  pred F,G are_fiberwise_equipotent means
  :Def9:
  for x be set holds card Coim(F,x) = card Coim(G,x);
  reflexivity;
  symmetry;
end;

Lm3: for F be Function, x be set st not x in rng F holds Coim(F,x) = {}
proof
  let F be Function,x;
  assume
A1: not x in rng F;
 now
    assume rng F meets {x};
then  ex y being set st y in rng F & y in {x} by XBOOLE_0:3;
    hence contradiction by A1,TARSKI:def 1;
  end;
  hence thesis by RELAT_1:138;
end;

theorem Th75:
  for F,G be Function st F,G are_fiberwise_equipotent holds rng F = rng G
proof
  let F,G be Function;
  assume
A1: F,G are_fiberwise_equipotent;
  thus rng F c= rng G
  proof
    let x;
    assume that
A2: x in rng F and
A3: not x in rng G;
 card Coim(F,x) = card Coim(G,x) by A1,Def9;
then A4: Coim(F,x), Coim(G,x) are_equipotent by CARD_1:5;
A5: ex y st y in dom F & F.y = x by A2,FUNCT_1:def 3;
 Coim(G,x) = {} by A3,Lm3;
then  x in {x} & F"{x} = {} by A4,CARD_1:26,TARSKI:def 1;
    hence contradiction by A5,FUNCT_1:def 7;
  end;
  let x;
  assume that
A6: x in rng G and
A7: not x in rng F;
 card Coim(G,x) = card Coim(F,x) by A1,Def9;
then A8: G"{x},F"{x} are_equipotent by CARD_1:5;
A9: ex y st y in dom G & G.y = x by A6,FUNCT_1:def 3;
 Coim(F,x) = {} by A7,Lm3;
then  x in {x} & Coim(G,x) = {} by A8,CARD_1:26,TARSKI:def 1;
  hence contradiction by A9,FUNCT_1:def 7;
end;

theorem
  for F,G,H be Function
  st F,G are_fiberwise_equipotent & F,H are_fiberwise_equipotent holds
  G,H are_fiberwise_equipotent
proof
  let F,G,H be Function;
  assume that
A1: F,G are_fiberwise_equipotent and
A2: F,H are_fiberwise_equipotent;
  let x;
  thus card Coim(G,x) = card Coim(F,x) by A1,Def9
    .= card Coim(H,x) by A2,Def9;
end;

theorem Th77:
  for F,G be Function holds F,G are_fiberwise_equipotent iff
  ex H be Function st dom H = dom F & rng H = dom G & H is one-to-one & F = G*H
proof
  let F,G be Function;
  thus F,G are_fiberwise_equipotent implies
  ex H be Function st dom H = dom F & rng H = dom G & H is one-to-one & F = G*H
  proof
    assume
A1: F,G are_fiberwise_equipotent;
    defpred P[set,set] means $2 is Function &
    for f be Function st $2 = f holds dom f = F"{$1} & rng f = G"{$1} &
    f is one-to-one;
A2: for x st x in rng F ex y st P[x,y]
    proof
      let x;
      assume x in rng F;
   card Coim(F,x) = card Coim(G,x) by A1,Def9;
then    F"{x},G"{x} are_equipotent by CARD_1:5;
      then consider H be Function such that
A3:   H is one-to-one & dom H = F"{x} & rng H = G"{x} by WELLORD2:def 4;
      take H;
      thus H is Function;
      let g be Function;
      assume g = H;
      hence thesis by A3;
    end;
    consider W be Function such that
A4: dom W = rng F & for x st x in rng F holds P[x,W.x] from NonUniqFuncEx(A2);
    defpred Q[set,set] means for H be Function st H = W.(F.$1) holds $2 = H.$1;
    set df = dom F, dg = dom G;
A5: for x st x in df ex y st y in dg & Q[x,y]
    proof
      let x;
      assume
A6:   x in df;
then A7:  F.x in rng F by FUNCT_1:def 3;
      then reconsider f = W.(F.x) as Function by A4;
A8:  dom f = F"{F.x} & rng f = G"{F.x} by A4,A7;
      take y = f.x;
  F.x in {F.x} by TARSKI:def 1;
then   x in F"{F.x} by A6,FUNCT_1:def 7;
then   f.x in G"{F.x} by A8,FUNCT_1:def 3;
      hence y in dg by FUNCT_1:def 7;
      let g be Function;
      assume g = W.(F.x);
      hence thesis;
    end;
    consider IT be Function such that
A9: dom IT = df & rng IT c= dg & for x st x in df holds Q[x,IT.x]
    from FUNCT_1:sch 5(A5);
    take IT;
    thus dom IT = df by A9;
 dom G c= rng IT
    proof
      let x;
      assume
A10:  x in dg;
then A11:  G.x in rng G by FUNCT_1:def 3;
A12:  rng F = rng G by A1,Th75;
      then reconsider f = W.(G.x) as Function by A4,A11;
A13:  dom f = F"{G.x} by A4,A11,A12;
A14:  rng f = G"{G.x} by A4,A11,A12;
  G.x in {G.x} by TARSKI:def 1;
then   x in rng f by A10,A14,FUNCT_1:def 7;
      then consider z be set such that
A15:  z in dom f and
A16:  f.z = x by FUNCT_1:def 3;
A17:  z in df by A13,A15,FUNCT_1:def 7;
  F.z in {G.x} by A13,A15,FUNCT_1:def 7;
then   F.z = G.x by TARSKI:def 1;
then   IT.z = x by A9,A16,A17;
      hence thesis by A9,A17,FUNCT_1:def 3;
    end;
    hence rng IT = dg by A9,XBOOLE_0:def 10;
 now
      let x,y;
      assume that
A18:  x in dom IT and
A19:  y in dom IT and
A20:  IT.x = IT.y;
A21:  F.x in rng F by A9,A18,FUNCT_1:def 3;
A22:  F.y in rng F by A9,A19,FUNCT_1:def 3;
      then reconsider f1 = W.(F.x), f2 = W.(F.y) as Function by A4,A21;
A23:  IT.x = f1.x & IT.y = f2.y by A9,A18,A19;
A24:  dom f1 = F"{F.x} by A4,A21;
A25:  rng f1 = G"{F.x} by A4,A21;
A26:  f1 is one-to-one by A4,A21;
A27:  dom f2 = F"{F.y} by A4,A22;
A28:  rng f2 = G"{F.y} by A4,A22;
A29:  F.x in {F.x} by TARSKI:def 1;
A30:  F.y in {F.y} by TARSKI:def 1;
A31:  x in F"{F.x} by A9,A18,A29,FUNCT_1:def 7;
A32:  y in F"{F.y} by A9,A19,A30,FUNCT_1:def 7;
A33:  f1.x in rng f1 by A24,A31,FUNCT_1:def 3;
  f2.y in rng f2 by A27,A32,FUNCT_1:def 3;
then
  f1.x in G"{F.x} /\ G"{F.y} by A20,A23,A25,A28,A33,XBOOLE_0:def 4;
then   f1.x in G"({F.x} /\ {F.y}) by FUNCT_1:68;
then A34:  G.(f1.x) in {F.x} /\ {F.y} by FUNCT_1:def 7;
then A35:  G.(f1.x) in {F.x} by XBOOLE_0:def 4;
A36:  G.(f1.x) in {F.y} by A34,XBOOLE_0:def 4;
A37:  G.(f1.x) = F.x by A35,TARSKI:def 1;
  G.(f1.x) = F.y by A36,TARSKI:def 1;
      hence x = y by A20,A23,A24,A26,A31,A32,A37,FUNCT_1:def 4;
    end;
    hence IT is one-to-one by FUNCT_1:def 4;
A38: dom(G*IT) = df by A9,RELAT_1:27;
 now
      let x;
      assume
A39:  x in df;
then A40:  F.x in rng F by FUNCT_1:def 3;
      then reconsider f = W.(F.x) as Function by A4;
A41:  dom f = F"{F.x} & rng f = G"{F.x} by A4,A40;
  F.x in {F.x} by TARSKI:def 1;
then   x in F"{F.x} by A39,FUNCT_1:def 7;
then   f.x in G"{F.x} by A41,FUNCT_1:def 3;
then   G.(f.x) in {F.x} by FUNCT_1:def 7;
then A42:  G.(f.x) = F.x by TARSKI:def 1;
  IT.x = f.x by A9,A39;
      hence F.x = (G*IT).x by A9,A39,A42,FUNCT_1:13;
    end;
    hence thesis by A38,FUNCT_1:2;
  end;
  given H be Function such that
A43: dom H = dom F and
A44: rng H = dom G and
A45: H is one-to-one and
A46: F = G*H;
  let x;
  set t = H|(F"{x});
A47: t is one-to-one by A45,FUNCT_1:52;
A48: dom t = F"{x} by A43,RELAT_1:62,132;
 rng t = G"{x}
  proof
    thus rng t c= G"{x}
    proof
      let z be set;
      assume z in rng t;
      then consider y such that
A49:  y in dom t and
A50:  t.y = z by FUNCT_1:def 3;
  F.y in {x} by A48,A49,FUNCT_1:def 7;
then A51:  F.y = x by TARSKI:def 1;
A52:  z = H.y by A49,A50,FUNCT_1:47;
  dom t = dom H /\ F"{x} by RELAT_1:61;
then A53:  y in dom H by A49,XBOOLE_0:def 4;
then A54:  z in dom G by A44,A52,FUNCT_1:def 3;
  x = G.z by A46,A51,A52,A53,FUNCT_1:13;
then   G.z in {x} by TARSKI:def 1;
      hence thesis by A54,FUNCT_1:def 7;
    end;
    let z be set;
    assume
A55: z in G"{x};
then A56: z in dom G by FUNCT_1:def 7;
 G.z in {x} by A55,FUNCT_1:def 7;
then A57: G.z = x by TARSKI:def 1;
    consider y such that
A58: y in dom H and
A59: H.y = z by A44,A56,FUNCT_1:def 3;
 F.y = x by A46,A57,A58,A59,FUNCT_1:13;
then  F.y in {x} by TARSKI:def 1;
then A60: y in dom t by A43,A48,A58,FUNCT_1:def 7;
then  t.y in rng t by FUNCT_1:def 3;
    hence thesis by A59,A60,FUNCT_1:47;
  end;
then  F"{x},G"{x} are_equipotent by A47,A48,WELLORD2:def 4;
  hence thesis by CARD_1:5;
end;

theorem Th78:
  for F,G be Function holds
  F,G are_fiberwise_equipotent iff for X be set holds card (F"X) = card (G"X)
proof
  let F,G be Function;
  thus F,G are_fiberwise_equipotent implies
  for X be set holds card(F"X) = card(G"X)
  proof
    assume F,G are_fiberwise_equipotent;
    then consider H be Function such that
A1: dom H = dom F and
A2: rng H = dom G and
A3: H is one-to-one and
A4: F = G*H by Th77;
    let X be set;
    set t = H|(F"X);
A5: t is one-to-one by A3,FUNCT_1:52;
A6: dom t = F"X by A1,RELAT_1:62,132;
 rng t = G"X
    proof
      thus rng t c= G"X
      proof
        let z be set;
        assume z in rng t;
        then consider y such that
A7:    y in dom t and
A8:    t.y = z by FUNCT_1:def 3;
A9:    F.y in X by A6,A7,FUNCT_1:def 7;
A10:    z = H.y by A7,A8,FUNCT_1:47;
    dom t = dom H /\ F"X by RELAT_1:61;
then A11:    y in dom H by A7,XBOOLE_0:def 4;
then A12:    z in dom G by A2,A10,FUNCT_1:def 3;
    G.z in X by A4,A9,A10,A11,FUNCT_1:13;
        hence thesis by A12,FUNCT_1:def 7;
      end;
      let z be set;
      assume
A13:  z in G"X;
then A14:  z in dom G by FUNCT_1:def 7;
A15:  G.z in X by A13,FUNCT_1:def 7;
      consider y such that
A16:  y in dom H and
A17:  H.y = z by A2,A14,FUNCT_1:def 3;
  F.y in X by A4,A15,A16,A17,FUNCT_1:13;
then A18:  y in dom t by A1,A6,A16,FUNCT_1:def 7;
then   t.y in rng t by FUNCT_1:def 3;
      hence thesis by A17,A18,FUNCT_1:47;
    end;
then  F"X,G"X are_equipotent by A5,A6,WELLORD2:def 4;
    hence thesis by CARD_1:5;
  end;
  assume for X be set holds card(F"X) = card(G"X);
  hence for x holds card Coim(F,x) = card Coim(G,x);
end;

theorem
  for D be non empty set, F,G be Function st rng F c= D & rng G c= D &
  for d be Element of D holds card Coim(F,d) = card Coim(G,d)
  holds F,G are_fiberwise_equipotent
proof
  let D be non empty set, F,G be Function;
  assume that
A1: rng F c= D and
A2: rng G c= D;
  assume
A3: for d be Element of D holds card Coim(F,d) = card Coim(G,d);
  let x;
  per cases;
  suppose
 not x in rng F;
then A4: Coim(F,x) = {} by Lm3;
 now
      assume
A5:   x in rng G;
      then reconsider d = x as Element of D by A2;
   card Coim(G,d) = card Coim(F,d) by A3;
then    G"{d},F"{d} are_equipotent by CARD_1:5;
then A6:  G"{x} = {} by A4,CARD_1:26;
      consider y such that
A7:  y in dom G and
A8:  G.y = x by A5,FUNCT_1:def 3;
  G.y in {x} by A8,TARSKI:def 1;
      hence contradiction by A6,A7,FUNCT_1:def 7;
    end;
    hence thesis by A4,Lm3;
  end;
  suppose
 x in rng F;
    then reconsider d = x as Element of D by A1;
 card Coim(F,d) = card Coim(G,d) by A3;
    hence thesis;
  end;
end;

theorem Th80:
  for F,G be Function st dom F = dom G holds
  F,G are_fiberwise_equipotent iff ex P be Permutation of dom F st F = G*P
proof
  let F,G be Function;
  assume
A1: dom F = dom G;
  thus F,G are_fiberwise_equipotent implies
  ex P be Permutation of dom F st F = G*P
  proof
    assume F,G are_fiberwise_equipotent;
    then consider I be Function such that
A2: dom I = dom F and
A3: rng I = dom G and
A4: I is one-to-one and
A5: F = G*I by Th77;
    reconsider I as Function of dom F,dom F by A1,A2,A3,FUNCT_2:2;
    reconsider I as Permutation of dom F by A1,A3,A4,FUNCT_2:57;
    take I;
    thus thesis by A5;
  end;
  given P be Permutation of dom F such that
A6: F = G*P;
 P is Function of dom F,dom F & dom F = {} implies dom F ={};
then  rng P = dom F & dom P = dom F by FUNCT_2:def 1,def 3;
  hence thesis by A1,A6,Th77;
end;

theorem
  for F,G be Function
  st F,G are_fiberwise_equipotent holds card dom F = card dom G
proof
  let F,G be Function;
  assume F,G are_fiberwise_equipotent;
then  card(F"(rng F)) = card(G"(rng F)) & rng F = rng G by Th75,Th78;
  hence card(dom F) = card(G"(rng G)) by RELAT_1:134
    .= card(dom G) by RELAT_1:134;
end;



theorem
  for f being set, p being Relation
  for x being set st x in rng p holds the_rank_of x in the_rank_of [p,f]
proof
  let f be set;
  let p be Relation;
  let y be set;
  assume y in rng p;
  then consider x being set such that
A1: [x,y] in p by XTUPLE_0:def 13;
A2: p in {p,f} by TARSKI:def 2;
A3: {p,f} in {{p,f},{p}} by TARSKI:def 2;
A4: y in {x,y} by TARSKI:def 2;
A5: {x,y} in {{x,y},{x}} by TARSKI:def 2;
A6: the_rank_of y in the_rank_of {x,y} by A4,Th68;
A7: the_rank_of {x,y} in the_rank_of [x,y] by A5,Th68;
A8: the_rank_of p in the_rank_of {p,f} by A2,Th68;
A9: the_rank_of {p,f} in the_rank_of [p,f] by A3,Th68;
A10: the_rank_of y in the_rank_of [x,y] by A6,A7,ORDINAL1:10;
A11: the_rank_of [x,y] in the_rank_of p by A1,Th68;
A12: the_rank_of p in the_rank_of [p,f] by A8,A9,ORDINAL1:10;
 the_rank_of y in the_rank_of p by A10,A11,ORDINAL1:10;
  hence thesis by A12,ORDINAL1:10;
end;



theorem
  for f, g, h being Function
  st dom f = dom g & rng f c= dom h & rng g c= dom h &
  f, g are_fiberwise_equipotent holds h*f, h*g are_fiberwise_equipotent
proof
  let f, g, h be Function such that
A1: dom f = dom g and
A2: rng f c= dom h and
A3: rng g c= dom h and
A4: f, g are_fiberwise_equipotent;
  consider p being Permutation of dom f such that
A5: f = g*p by A1,A4,Th80;
A6: dom (h*f) = dom f by A2,RELAT_1:27;
A7: dom (h*g) = dom g by A3,RELAT_1:27;
  h*f = h*g*p by A5,RELAT_1:36;
  hence thesis by A1,A6,A7,Th80;
end;















begin

reserve i,j,e,u for set;

theorem
  for f being Function st f is non-empty holds rng f is
  with_non-empty_elements;

theorem
  for f being Function holds f is empty-yielding iff f = {} or rng f = {
  {} }
proof
  let f be Function;
  hereby
    assume that
A1: f is empty-yielding and
A2: f <> {};
    thus rng f = { {} }
    proof
      hereby
        let i;
        assume i in rng f;
        then ex e st e in dom f & f.e = i by FUNCT_1:def 3;
        then i = {} by A1;
        hence i in { {} } by TARSKI:def 1;
      end;
      set e = the Element of dom f;
      let i;
      assume i in { {} };
      then
A3:   i = {} by TARSKI:def 1;
A4:   dom f <> {} by A2,RELAT_1:41;
      f.e is empty by A1;
      hence thesis by A4,A3,FUNCT_1:def 3;
    end;
  end;
  assume
A5: f = {} or rng f = { {} };
  per cases by A5;
  suppose
    f = {};
    hence for i st i in dom f holds f.i is empty by RELAT_1:38;
  end;
  suppose
A6: rng f = { {} };
    let i;
    assume i in dom f;
    then f.i in rng f by FUNCT_1:def 3;
    hence thesis by A6,TARSKI:def 1;
  end;
end;

reserve I for set; 

registration
  let I;
  cluster total for I-defined Function;
  existence
  proof
    take I --> {};
    thus thesis;
  end;
end;

definition
  let I;
  mode ManySortedSet of I is total I-defined Function;
end;

reserve x,X,Y,Z,V for ManySortedSet of I;

scheme
  KuratowskiFunction{A()-> set, F(set) -> set}:
 ex f being ManySortedSet of A() st for e st e in A() holds f.e in F(e)
provided
A1: for e st e in A() holds F(e) <> {}
proof
  defpred P[set,set] means $2 in F($1);
A2: now
    let e;
    set fe = the Element of F(e);
    assume
A3: e in A();
    reconsider fe as set;
    take fe;
    F(e) <> {} by A1,A3;
    hence P[e,fe];
  end;
  consider f being Function such that
A4: dom f = A() and
A5: for e st e in A() holds P[e,f.e] from CLASSES1:sch 1(A2);
  reconsider f as ManySortedSet of A() by A4,PARTFUN1:def 2,RELAT_1:def 18;
  take f;
  thus thesis by A5;
end;

definition
  let I,X,Y;
  pred X in Y means
  :Def1:
  for i st i in I holds X.i in Y.i;
  pred X c= Y means
  :Def2:
  for i st i in I holds X.i c= Y.i;
  reflexivity;
end;

definition
  let I be non empty set,X,Y be ManySortedSet of I;
  redefine pred X in Y;
  asymmetry
  proof
    let X,Y be ManySortedSet of I;
    assume
A1: for i st i in I holds X.i in Y.i;
    not for i st i in I holds Y.i in X.i
    proof
      assume
A2:   for i st i in I holds Y.i in X.i;
      consider i such that
A3:   i in I by XBOOLE_0:def 1;
      X.i in Y.i by A1,A3;
      hence thesis by A2,A3;
    end;
    hence thesis by Def1;
  end;
end;

scheme
  PSeparation { I()-> set, A() -> ManySortedSet of I(), P[set,set] } : ex X
being ManySortedSet of I() st for i being set st i in I() for e holds e in X.i
  iff e in A().i & P[i,e] proof
  defpred R[set,set] means for e holds e in $2 iff e in A().$1 & P[ $1,e];
A1: now
    let i such that
    i in I();
    defpred Q[set] means P[i,$1];
    ex u st for e holds e in u iff e in A().i & Q[e] from XBOOLE_0:sch 1;
    hence ex u st R[i,u];
  end;
  consider f being Function such that
A2: dom f = I() and
A3: for i st i in I() holds R[i,f.i] from CLASSES1:sch 1(A1);
  f is ManySortedSet of I() by A2,PARTFUN1:def 2,RELAT_1:def 18;
  hence thesis by A3;
end;

theorem Th3:
  (for i st i in I holds X.i = Y.i) implies X = Y
proof
  dom X = I & dom Y = I by PARTFUN1:def 2;
  hence thesis by FUNCT_1:2;
end;

definition
  let I;
  func [[0]]I -> ManySortedSet of I equals
  I --> {};
  coherence;
  correctness;
  let X,Y;
  func X \/ Y -> ManySortedSet of I means
  :Def4:
  for i st i in I holds it.i = X.i \/ Y.i;
  existence
  proof
    deffunc F(set) = X.$1 \/ Y.$1;
    consider f being Function such that
A1: dom f = I and
A2: for i st i in I holds f.i = F(i) from FUNCT_1:sch 3;
    f is ManySortedSet of I by A1,PARTFUN1:def 2,RELAT_1:def 18;
    hence thesis by A2;
  end;
  uniqueness
  proof
    let A1, A2 be ManySortedSet of I such that
A3: for i st i in I holds A1.i = X.i \/ Y.i and
A4: for i st i in I holds A2.i = X.i \/ Y.i;
    now
      let i;
      assume
A5:   i in I;
      hence A1.i = X.i \/ Y.i by A3
        .= A2.i by A4,A5;
    end;
    hence thesis by Th3;
  end;
  commutativity;
  idempotence;
  func X /\ Y -> ManySortedSet of I means
  :Def5:
  for i st i in I holds it.i = X.i /\ Y.i;
  existence
  proof
    deffunc F(set) = X.$1 /\ Y.$1;
    consider f being Function such that
A6: dom f = I and
A7: for i st i in I holds f.i = F(i) from FUNCT_1:sch 3;
    f is ManySortedSet of I by A6,PARTFUN1:def 2,RELAT_1:def 18;
    hence thesis by A7;
  end;
  uniqueness
  proof
    let A1, A2 be ManySortedSet of I such that
A8: for i st i in I holds A1.i = X.i /\ Y.i and
A9: for i st i in I holds A2.i = X.i /\ Y.i;
    now
      let i;
      assume
A10:  i in I;
      hence A1.i = X.i /\ Y.i by A8
        .= A2.i by A9,A10;
    end;
    hence thesis by Th3;
  end;
  commutativity;
  idempotence;
  func X \ Y -> ManySortedSet of I means
  :Def6:
  for i st i in I holds it.i = X.i \ Y.i;
  existence
  proof
    deffunc F(set) = X.$1 \ Y.$1;
    consider f being Function such that
A11: dom f = I and
A12: for i st i in I holds f.i = F(i) from FUNCT_1:sch 3;
    f is ManySortedSet of I by A11,PARTFUN1:def 2,RELAT_1:def 18;
    hence thesis by A12;
  end;
  uniqueness
  proof
    let A1, A2 be ManySortedSet of I such that
A13: for i st i in I holds A1.i = X.i \ Y.i and
A14: for i st i in I holds A2.i = X.i \ Y.i;
    now
      let i;
      assume
A15:  i in I;
      hence A1.i = X.i \ Y.i by A13
        .= A2.i by A14,A15;
    end;
    hence thesis by Th3;
  end;
  pred X overlaps Y means
  :Def7:
  for i st i in I holds X.i meets Y.i;
  symmetry;
  pred X misses Y means
  :Def8:
  for i st i in I holds X.i misses Y.i;
  symmetry;
end;

notation
  let I;
  let X,Y;
  antonym X meets Y for X misses Y;
end;

definition
  let I,X,Y;
  func X \+\ Y -> ManySortedSet of I equals
  (X \ Y) \/ (Y \ X);
  correctness;
  commutativity;
end;

theorem Th4:
  for i st i in I holds (X \+\ Y).i = X.i \+\ Y.i
proof
  let i such that
A1: i in I;
  thus (X \+\ Y).i = (X \ Y).i \/ (Y \ X).i by A1,Def4
    .= X.i \ Y.i \/ (Y \ X).i by A1,Def6
    .= X.i \+\ Y.i by A1,Def6;
end;

theorem Th5:
  [[0]]I.i = {}
proof
  per cases;
  suppose
    i in I;
    hence thesis by FUNCOP_1:7;
  end;
  suppose
    not i in I;
    then not i in dom [[0]]I by FUNCOP_1:13;
    hence thesis by FUNCT_1:def 2;
  end;
end;

theorem Th6:
  (for i st i in I holds X.i = {}) implies X = [[0]]I
proof
  assume
A1: for i st i in I holds X.i = {};
  now
    let i;
    assume i in I;
    hence X.i = {} by A1
      .= [[0]]I.i by Th5;
  end;
  hence thesis by Th3;
end;

theorem Th7:
  x in X or x in Y implies x in X \/ Y
proof
  assume
A1: x in X or x in Y;
  let i;
  assume
A2: i in I;
  then x.i in X.i or x.i in Y.i by A1,Def1;
  then x.i in X.i \/ Y.i by XBOOLE_0:def 3;
  hence thesis by A2,Def4;
end;

theorem Th8:
  x in X /\ Y iff x in X & x in Y
proof
  hereby
    assume
A1: x in X /\ Y;
    thus x in X
    proof
      let i;
      assume
A2:   i in I;
      then x.i in (X /\ Y).i by A1,Def1;
      then x.i in X.i /\ Y.i by A2,Def5;
      hence thesis by XBOOLE_0:def 4;
    end;
    thus x in Y
    proof
      let i;
      assume
A3:   i in I;
      then x.i in (X /\ Y).i by A1,Def1;
      then x.i in X.i /\ Y.i by A3,Def5;
      hence thesis by XBOOLE_0:def 4;
    end;
  end;
  assume
A4: x in X & x in Y;
  let i;
  assume
A5: i in I;
  then x.i in X.i & x.i in Y.i by A4,Def1;
  then x.i in X.i /\ Y.i by XBOOLE_0:def 4;
  hence thesis by A5,Def5;
end;

theorem Th9:
  x in X & X c= Y implies x in Y
proof
  assume
A1: x in X & X c= Y;
  let i;
  assume i in I;
  then x.i in X.i & X.i c= Y.i by A1,Def1,Def2;
  hence thesis;
end;

theorem Th10:
  x in X & x in Y implies X overlaps Y
proof
  assume
A1: x in X & x in Y;
  let i;
  assume i in I;
  then x.i in X.i & x.i in Y.i by A1,Def1;
  hence thesis by XBOOLE_0:3;
end;

theorem Th11:
  X overlaps Y implies ex x st x in X & x in Y
proof
  deffunc F(set) = X.$1 /\ Y.$1;
  assume
A1: X overlaps Y;
A2: now
    let i;
    assume i in I;
    then X.i meets Y.i by A1,Def7;
    hence F(i) <> {} by XBOOLE_0:def 7;
  end;
  consider x being ManySortedSet of I such that
A3: for i st i in I holds x.i in F(i) from KuratowskiFunction(A2);
  take x;
  now
    let i;
    assume i in I;
    then x.i in X.i /\ Y.i by A3;
    hence x.i in X.i by XBOOLE_0:def 4;
  end;
  hence x in X by Def1;
  now
    let i;
    assume i in I;
    then x.i in X.i /\ Y.i by A3;
    hence x.i in Y.i by XBOOLE_0:def 4;
  end;
  hence thesis by Def1;
end;

theorem
  x in X \ Y implies x in X
proof
  assume
A1: x in X \ Y;
  thus x in X
  proof
    let i;
    assume
A2: i in I;
    then x.i in (X \ Y).i by A1,Def1;
    then x.i in X.i \ Y.i by A2,Def6;
    hence thesis;
  end;
end;

begin 

Lm1:
 X c= Y & Y c= X implies X = Y
  proof
    assume
A1: X c= Y & Y c= X;
    now
      let i;
      assume i in I;
      then X.i c= Y.i & Y.i c= X.i by A1,Def2;
      hence X.i = Y.i by XBOOLE_0:def 10;
    end;
    hence thesis by Th3;
  end;

definition
  let I,X,Y;
  redefine pred X = Y means
  for i st i in I holds X.i = Y.i;
  compatibility by Th3;
end;

theorem Th13:
  X c= Y & Y c= Z implies X c= Z
proof
  assume that
A1: X c= Y and
A2: Y c= Z;
  let i such that
A3: i in I;
  let e;
  assume
A4: e in X.i;
  X.i c= Y.i by A1,A3,Def2;
  then
A5: e in Y.i by A4;
  Y.i c= Z.i by A2,A3,Def2;
  hence thesis by A5;
end;

theorem Th14:
  X c= X \/ Y
proof
    let i such that
A1: i in I;
    let e;
    assume e in X.i;
    then e in X.i \/ Y.i by XBOOLE_0:def 3;
    hence thesis by A1,Def4;
end;

theorem Th15:
  X /\ Y c= X
proof
    let i such that
A1: i in I;
    let e;
    assume e in (X /\ Y).i;
    then e in X.i /\ Y.i by A1,Def5;
    hence thesis by XBOOLE_0:def 4;
end;

theorem Th16:
  X c= Z & Y c= Z implies X \/ Y c= Z
proof
  assume
A1: X c= Z & Y c= Z;
  let i;
  assume
A2: i in I;
  then X.i c= Z.i & Y.i c= Z.i by A1,Def2;
  then X.i \/ Y.i c= Z.i by XBOOLE_1:8;
  hence thesis by A2,Def4;
end;

theorem Th17:
  Z c= X & Z c= Y implies Z c= X /\ Y
proof
  assume
A1: Z c= X & Z c= Y;
  let i;
  assume
A2: i in I;
  then Z.i c= X.i & Z.i c= Y.i by A1,Def2;
  then Z.i c= X.i /\ Y.i by XBOOLE_1:19;
  hence thesis by A2,Def5;
end;

theorem
  X c= Y implies X \/ Z c= Y \/ Z
proof
  assume
A1: X c= Y;
A2: Z c= Y \/ Z by Th14;
  Y c= Y \/ Z by Th14;
  then X c= Y \/ Z by A1,Th13;
  hence X \/ Z c= Y \/ Z by A2,Th16;
end;

theorem Th19:
  X c= Y implies X /\ Z c= Y /\ Z
proof
  assume
A1: X c= Y;
A2: X /\ Z c= Z by Th15;
  X /\ Z c= X by Th15;
  then X /\ Z c= Y by A1,Th13;
  hence X /\ Z c= Y /\ Z by A2,Th17;
end;

theorem Th20:
  X c= Y & Z c= V implies X \/ Z c= Y \/ V
proof
  assume that
A1: X c= Y and
A2: Z c= V;
  V c= Y \/ V by Th14;
  then
A3: Z c= Y \/ V by A2,Th13;
  Y c= Y \/ V by Th14;
  then X c= Y \/ V by A1,Th13;
  hence thesis by A3,Th16;
end;

theorem
  X c= Y & Z c= V implies X /\ Z c= Y /\ V
proof
  assume that
A1: X c= Y and
A2: Z c= V;
  X /\ Z c= Z by Th15;
  then
A3: X /\ Z c= V by A2,Th13;
  X /\ Z c= X by Th15;
  then X /\ Z c= Y by A1,Th13;
  hence thesis by A3,Th17;
end;

theorem Th22:
  X c= Y implies X \/ Y = Y
proof
  assume X c= Y;
  then
A1: X \/ Y c= Y by Th16;
  Y c= X \/ Y by Th14;
 hence thesis by A1,Lm1;
end;

theorem Th23:
  X c= Y implies X /\ Y = X
proof
  assume
A1: X c= Y;
A2: X /\ Y c= X by Th15;
  X c= X /\ Y by A1,Th17;
 hence thesis by A2,Lm1;
end;

theorem
  X /\ Y c= X \/ Z
proof
  X /\ Y c= X & X c= X \/ Z by Th14,Th15;
  hence thesis by Th13;
end;

theorem
  X c= Z implies X \/ Y /\ Z = (X \/ Y) /\ Z
proof
  assume
A1: X c= Z;
    let i;
    assume
A2: i in I;
    then
A3: X.i c= Z.i by A1,Def2;
    thus (X \/ Y /\ Z).i = X.i \/ (Y /\ Z).i by A2,Def4
      .= X.i \/ Y.i /\ Z.i by A2,Def5
      .= (X.i \/ Y.i) /\ Z.i by A3,XBOOLE_1:30
      .= (X \/ Y).i /\ Z.i by A2,Def4
      .= ((X \/ Y) /\ Z).i by A2,Def5;
end;

theorem
  X = Y \/ Z iff Y c= X & Z c= X & for V st Y c= V & Z c= V holds X c= V
proof
  thus X = Y \/ Z implies Y c= X & Z c= X & for V st Y c= V & Z c= V holds X
  c= V by Th14,Th16;
  assume that
A1: Y c= X & Z c= X and
A2: Y c= V & Z c= V implies X c= V;
  Y c= Y \/ Z & Z c= Y \/ Z by Th14;
  then
A3: X c= Y \/ Z by A2;
   Y \/ Z c= X by A1,Th16;
 hence thesis by A3,Lm1;
end;

theorem
  X = Y /\ Z iff X c= Y & X c= Z & for V st V c= Y & V c= Z holds V c= X
proof
  thus X = Y /\ Z implies X c= Y & X c= Z & for V st V c= Y & V c= Z holds V
  c= X by Th15,Th17;
  assume that
A1: X c= Y & X c= Z and
A2: V c= Y & V c= Z implies V c= X;
A3:  X c= Y /\ Z by A1,Th17;
  Y /\ Z c= Y & Y /\ Z c= Z implies Y /\ Z c= X by A2;
  hence thesis by A3,Lm1,Th15;
end;

theorem Th28:
  (X \/ Y) \/ Z = X \/ (Y \/ Z)
proof
    let i;
    assume
A1: i in I;
    hence (X \/ Y \/ Z).i = (X \/ Y).i \/ Z.i by Def4
      .= X.i \/ Y.i \/ Z.i by A1,Def4
      .= X.i \/ (Y.i \/ Z.i) by XBOOLE_1:4
      .= X.i \/ (Y \/ Z).i by A1,Def4
      .= (X \/ (Y \/ Z)).i by A1,Def4;
end;

theorem Th29:
  (X /\ Y) /\ Z = X /\ (Y /\ Z)
proof
    let i;
    assume
A1: i in I;
    hence (X /\ Y /\ Z).i = (X /\ Y).i /\ Z.i by Def5
      .= X.i /\ Y.i /\ Z.i by A1,Def5
      .= X.i /\ (Y.i /\ Z.i) by XBOOLE_1:16
      .= X.i /\ (Y /\ Z).i by A1,Def5
      .= (X /\ (Y /\ Z)).i by A1,Def5;
end;

theorem Th30:
  X /\ (X \/ Y) = X
proof
  X c= X \/ Y by Th14;
  then
A1: X c= X /\ (X \/ Y) by Th17;
 X /\ (X \/ Y) c= X by Th15;
  hence X /\ (X \/ Y) = X by A1,Lm1;
end;

theorem Th31:
  X \/ (X /\ Y) = X
proof
  X /\ Y c= X by Th15;
  then
A1: X \/ (X /\ Y) c= X by Th16;
 X c= X \/ (X /\ Y) by Th14;
  hence X \/ (X /\ Y) = X by A1,Lm1;
end;

theorem Th32:
  X /\ (Y \/ Z) = X /\ Y \/ X /\ Z
proof
    let i;
    assume
A1: i in I;
    hence (X /\ (Y \/ Z)).i = X.i /\ (Y \/ Z).i by Def5
      .= X.i /\ (Y.i \/ Z.i) by A1,Def4
      .= X.i /\ Y.i \/ X.i /\ Z.i by XBOOLE_1:23
      .= (X /\ Y).i \/ X.i /\ Z.i by A1,Def5
      .= (X /\ Y).i \/ (X /\ Z).i by A1,Def5
      .= (X /\ Y \/ X /\ Z).i by A1,Def4;
end;

theorem Th33:
  X \/ Y /\ Z = (X \/ Y) /\ (X \/ Z)
proof
  thus X \/ Y /\ Z = X \/ X /\ Z \/ Y /\ Z by Th31
    .= X \/ (X /\ Z \/ Y /\ Z) by Th28
    .= X \/ (X \/ Y) /\ Z by Th32
    .= (X \/ Y) /\ X \/ (X \/ Y) /\ Z by Th30
    .= (X \/ Y) /\ (X \/ Z) by Th32;
end;

theorem
  (X /\ Y) \/ (X /\ Z) = X implies X c= Y \/ Z
proof
  assume (X /\ Y) \/ (X /\ Z) = X;
  then X = X /\ (Y \/ Z) by Th32;
  hence thesis by Th15;
end;

theorem
  (X \/ Y) /\ (X \/ Z) = X implies Y /\ Z c= X
proof
  assume (X \/ Y) /\ (X \/ Z) = X;
  then X = X \/ (Y /\ Z) by Th33;
  hence thesis by Th14;
end;

theorem
  (X /\ Y) \/ (Y /\ Z) \/ (Z /\ X) = (X \/ Y) /\ (Y \/ Z) /\ (Z \/ X)
proof
  thus X /\ Y \/ Y /\ Z \/ Z /\ X = (X /\ Y \/ Y /\ Z \/ Z) /\ (X /\ Y \/ Y /\
  Z \/ X) by Th33
    .= (X /\ Y \/ (Y /\ Z \/ Z)) /\ (X /\ Y \/ Y /\ Z \/ X) by Th28
    .= (X /\ Y \/ Z) /\ (X /\ Y \/ Y /\ Z \/ X) by Th31
    .= (X /\ Y \/ Z) /\ (X /\ Y \/ X \/ Y /\ Z) by Th28
    .= (X /\ Y \/ Z) /\ (X \/ Y /\ Z) by Th31
    .= (X \/ Z) /\ (Y \/ Z) /\ (X \/ Y /\ Z) by Th33
    .= (X \/ Z) /\ (Y \/ Z) /\ ((X \/ Y) /\ (X \/ Z)) by Th33
    .= (X \/ Y) /\ ((Y \/ Z) /\ (X \/ Z) /\ (X \/ Z)) by Th29
    .= (X \/ Y) /\ ((Y \/ Z) /\ ((X \/ Z) /\ (X \/ Z))) by Th29
    .= (X \/ Y) /\ (Y \/ Z) /\ (Z \/ X) by Th29;
end;

theorem
  X \/ Y c= Z implies X c= Z
proof
  X c= X \/ Y by Th14;
  hence thesis by Th13;
end;

theorem
  X c= Y /\ Z implies X c= Y
proof
  Y /\ Z c= Y by Th15;
  hence thesis by Th13;
end;

theorem
  X \/ Y \/ Z = (X \/ Z) \/ (Y \/ Z)
proof
  thus X \/ Y \/ Z = X \/ (Y \/ (Z \/ Z)) by Th28
    .= X \/ (Z \/ Y \/ Z) by Th28
    .= (X \/ Z) \/ (Y \/ Z) by Th28;
end;

theorem
  X /\ Y /\ Z = (X /\ Z) /\ (Y /\ Z)
proof
  thus X /\ Y /\ Z = X /\ (Y /\ (Z /\ Z)) by Th29
    .= X /\ (Z /\ Y /\ Z) by Th29
    .= (X /\ Z) /\ (Y /\ Z) by Th29;
end;

theorem
  X \/ (X \/ Y) = X \/ Y
proof
  thus X \/ (X \/ Y) = X \/ X \/ Y by Th28
    .= X \/ Y;
end;

theorem
  X /\ (X /\ Y) = X /\ Y
proof
  thus X /\ (X /\ Y) = X /\ X /\ Y by Th29
    .= X /\ Y;
end;

begin 

theorem Th43:
  [[0]]I c= X
proof
  let i such that
A1: i in I;
  let e;
  assume e in [[0]]I.i;
  hence thesis by A1,FUNCOP_1:7;
end;

theorem Th44:
  X c= [[0]]I implies X = [[0]]I
proof
  assume X c= [[0]]I;
   then X c= [[0]]I & [[0]]I c= X by Th43;
  hence thesis by Lm1;
end;

theorem
  X c= Y & X c= Z & Y /\ Z = [[0]]I implies X = [[0]]I by Th17,Th44;

theorem
  X c= Y & Y /\ Z = [[0]]I implies X /\ Z = [[0]]I by Th44,Th19;

theorem
  X \/ [[0]]I = X & [[0]]I \/ X = X by Th22,Th43;

theorem
  X \/ Y = [[0]]I implies X = [[0]]I by Th44,Th14;

theorem
  X /\ [[0]]I = [[0]]I by Th23,Th43;

theorem
  X c= (Y \/ Z) & X /\ Z = [[0]]I implies X c= Y
proof
  assume that
A1: X c= (Y \/ Z) and
A2: X /\ Z = [[0]]I;
  X /\ (Y \/ Z)= X by A1,Th23;
  then Y /\ X \/ [[0]]I = X by A2,Th32;
  then Y /\ X = X by Th22,Th43;
  hence thesis by Th15;
end;

theorem
  Y c= X & X /\ Y = [[0]]I implies Y = [[0]]I by Th23;

begin 

theorem Th52:
  X \ Y = [[0]]I iff X c= Y
proof
  hereby
    assume
A1: X \ Y = [[0]]I;
    now
      let i;
      assume i in I;
      then X.i \ Y.i = (X \ Y).i by Def6
        .= {} by A1,Th5;
      hence X.i c= Y.i by XBOOLE_1:37;
    end;
    hence X c= Y by Def2;
  end;
  assume
A2: X c= Y;
  now
    let i;
    assume
A3: i in I;
    then
A4: X.i c= Y.i by A2,Def2;
    thus (X \ Y).i = X.i \ Y.i by A3,Def6
      .= {} by A4,XBOOLE_1:37;
  end;
  hence thesis by Th6;
end;

theorem Th53:
  X c= Y implies X \ Z c= Y \ Z
proof
  assume
A1: X c= Y;
  now
    let i;
    assume
A2: i in I;
    then
A3: (X \ Z).i = X.i \ Z.i & (Y \ Z).i = Y.i \ Z.i by Def6;
    X.i c= Y.i by A1,A2,Def2;
    hence (X \ Z).i c= (Y \ Z).i by A3,XBOOLE_1:33;
  end;
  hence thesis by Def2;
end;

theorem Th54:
  X c= Y implies Z \ Y c= Z \ X
proof
  assume
A1: X c= Y;
  now
    let i;
    assume
A2: i in I;
    then
A3: (Z \ X).i = Z.i \ X.i & (Z \ Y).i = Z.i \ Y.i by Def6;
    X.i c= Y.i by A1,A2,Def2;
    hence (Z \ Y).i c= (Z \ X).i by A3,XBOOLE_1:34;
  end;
  hence thesis by Def2;
end;

theorem
  X c= Y & Z c= V implies X \ V c= Y \ Z
proof
  assume X c= Y & Z c= V;
  then X \ V c= Y \ V & Y \ V c= Y \ Z by Th53,Th54;
  hence thesis by Th13;
end;

theorem Th56:
  X \ Y c= X
proof
  now
    let i such that
A1: i in I;
    X.i \ Y.i c= X.i;
    hence (X \ Y).i c= X.i by A1,Def6;
  end;
  hence thesis by Def2;
end;

theorem
  X c= Y \ X implies X = [[0]]I
proof
  assume
A1: X c= Y \ X;
    let i;
    assume
A2: i in I;
    then X.i c= (Y \ X).i by A1,Def2;
    then X.i c= Y.i \ X.i by A2,Def6;
    hence X.i = {} by XBOOLE_1:38
      .= [[0]]I.i by Th5;
end;

theorem
  X \ X = [[0]]I by Th52;

theorem Th59:
  X \ [[0]]I = X
proof
    let i;
    assume i in I;
    hence (X \ [[0]]I).i = X.i \ [[0]]I.i by Def6
      .= X.i \ {} by Th5
      .= X.i;
end;

theorem Th60:
  [[0]]I \ X = [[0]]I
proof
  [[0]]I c= X by Th43;
  hence thesis by Th52;
end;

theorem
  X \ (X \/ Y) = [[0]]I
proof
  X c= X \/ Y by Th14;
  hence thesis by Th52;
end;

theorem Th62:
  X /\ (Y \ Z) = (X /\ Y) \ Z
proof
    let i;
    assume
A1: i in I;
    hence (X /\ (Y \ Z)).i = X.i /\ (Y \ Z).i by Def5
      .= X.i /\ (Y.i \ Z.i) by A1,Def6
      .= X.i /\ Y.i \ Z.i by XBOOLE_1:49
      .= (X /\ Y).i \ Z.i by A1,Def5
      .= ((X /\ Y) \ Z).i by A1,Def6;
end;

theorem Th63:
  (X \ Y) /\ Y = [[0]]I
proof
A1: Y /\ X c= Y by Th15;
  thus (X \ Y) /\ Y = (Y /\ X) \ Y by Th62
    .= [[0]]I by A1,Th52;
  thus thesis;
end;

theorem Th64:
  X \ (Y \ Z) = (X \ Y) \/ X /\ Z
proof
    let i;
    assume
A1: i in I;
    hence (X \ (Y \ Z)).i = X.i \ (Y \ Z).i by Def6
      .= X.i \ (Y.i \ Z.i) by A1,Def6
      .= X.i \ Y.i \/ X.i /\ Z.i by XBOOLE_1:52
      .= X.i \ Y.i \/ (X /\ Z).i by A1,Def5
      .= (X \ Y).i \/ (X /\ Z).i by A1,Def6
      .= ((X \ Y) \/ X /\ Z).i by A1,Def4;
end;

theorem Th65:
  (X \ Y) \/ X /\ Y = X
proof
  thus X \ Y \/ X /\ Y = X \ (Y \ Y) by Th64
    .= X \ [[0]]I by Th52
    .= X by Th59;
end;

theorem
  X c= Y implies Y = X \/ (Y \ X)
proof
  assume
A1: X c= Y;
  thus Y = (Y \ X) \/ Y /\ X by Th65
    .= X \/ (Y \ X) by A1,Th23;
end;

theorem Th67:
  X \/ (Y \ X) = X \/ Y
proof
  thus X \/ (Y \ X) = X \/ Y /\ X \/ (Y \ X) by Th31
    .= X \/ (Y /\ X \/ (Y \ X)) by Th28
    .= X \/ Y by Th65;
end;

theorem Th68:
  X \ (X \ Y) = X /\ Y
proof
  thus X \ (X \ Y) = (X \ X) \/ X /\ Y by Th64
    .= [[0]]I \/ X /\ Y by Th52
    .= X /\ Y by Th22,Th43;
end;

theorem Th69:
  X \ (Y /\ Z) = (X \ Y) \/ (X \ Z)
proof
    let i;
    assume
A1: i in I;
    hence (X \ (Y /\ Z)).i = X.i \ (Y /\ Z).i by Def6
      .= X.i \ Y.i /\ Z.i by A1,Def5
      .= X.i \ Y.i \/ (X.i \ Z.i) by XBOOLE_1:54
      .= X.i \ Y.i \/ (X \ Z).i by A1,Def6
      .= (X \ Y).i \/ (X \ Z).i by A1,Def6
      .= ((X \ Y) \/ (X \ Z)).i by A1,Def4;
end;

theorem Th70:
  X \ X /\ Y = X \ Y
proof
  thus X \ X /\ Y = (X \ X) \/ (X \ Y) by Th69
    .= [[0]]I \/ (X \ Y) by Th52
    .= X \ Y by Th22,Th43;
end;

theorem
  X /\ Y = [[0]]I iff X \ Y = X
proof
  hereby
    assume
A1: X /\ Y = [[0]]I;
    thus X \ Y = X \ X /\ Y by Th70
      .= X by A1,Th59;
  end;
  thus thesis by Th63;
end;

theorem Th72:
  (X \/ Y) \ Z = (X \ Z) \/ (Y \ Z)
proof
    let i;
    assume
A1: i in I;
    hence ((X \/ Y) \ Z).i = (X \/ Y).i \ Z.i by Def6
      .= X.i \/ Y.i \ Z.i by A1,Def4
      .= X.i \ Z.i \/ (Y.i \ Z.i) by XBOOLE_1:42
      .= X.i \ Z.i \/ (Y \ Z).i by A1,Def6
      .= (X \ Z).i \/ (Y \ Z).i by A1,Def6
      .= ((X \ Z) \/ (Y \ Z)).i by A1,Def4;
end;

theorem Th73:
  (X \ Y) \ Z = X \ (Y \/ Z)
proof
    let i;
    assume
A1: i in I;
    hence ((X \ Y) \ Z).i = (X \ Y).i \ Z.i by Def6
      .= X.i \ Y.i \ Z.i by A1,Def6
      .= X.i \ (Y.i \/ Z.i) by XBOOLE_1:41
      .= X.i \ (Y \/ Z).i by A1,Def4
      .= (X \ (Y \/ Z)).i by A1,Def6;
end;

theorem
  (X /\ Y) \ Z = (X \ Z) /\ (Y \ Z)
proof
A1: X \ Z c= X by Th56;
  thus (X /\ Y) \ Z = (X \ Z) /\ Y by Th62
    .= (X \ Z) \ ((X \ Z) \ Y) by Th68
    .= (X \ Z) \ (X \ (Z \/ Y)) by Th73
    .= ((X \ Z) \ X) \/ ((X \ Z) /\ (Z \/ Y)) by Th64
    .= [[0]]I \/ ((X \ Z) /\ (Z \/ Y)) by A1,Th52
    .= (X \ Z) /\ (Y \/ Z) by Th22,Th43
    .= (X \ Z) /\ ((Y \ Z) \/ Z) by Th67
    .= (X \ Z) /\ (Y \ Z) \/ (X \ Z) /\ Z by Th32
    .= (X \ Z) /\ (Y \ Z) \/ [[0]]I by Th63
    .= (X \ Z) /\ (Y \ Z) by Th22,Th43;
end;

theorem Th75:
  (X \/ Y) \ Y = X \ Y
proof
  thus (X \/ Y) \ Y = X \ Y \/ (Y \ Y) by Th72
    .= X \ Y \/ [[0]]I by Th52
    .= X \ Y by Th22,Th43;
end;

theorem
  X c= Y \/ Z implies X \ Y c= Z & X \ Z c= Y
proof
  assume
A1: X c= Y \/ Z;
  then X \ Y c= Z \/ Y \ Y by Th53;
  then
A2: X \ Y c= Z \ Y by Th75;
  Z \ Y c= Z by Th56;
  hence X \ Y c= Z by A2,Th13;
  X \ Z c= Y \/ Z \ Z by A1,Th53;
  then
A3: X \ Z c= Y \ Z by Th75;
  Y \ Z c= Y by Th56;
  hence thesis by A3,Th13;
end;

theorem Th77:
  (X \/ Y) \ (X /\ Y) = (X \ Y) \/ (Y \ X)
proof
  thus (X \/ Y) \ (X /\ Y) = X \ (X /\ Y) \/ (Y \ (X /\ Y)) by Th72
    .= X \ Y \/ (Y \ (X /\ Y)) by Th70
    .= (X \ Y) \/ (Y \ X) by Th70;
end;

theorem Th78:
  (X \ Y) \ Y = X \ Y
proof
  thus (X \ Y) \ Y = X \ (Y \/ Y) by Th73
    .= X \ Y;
end;

theorem
  X \ (Y \/ Z) = (X \ Y) /\ (X \ Z)
proof
  thus X \ (Y \/ Z) = X \ Y \ Z by Th73
    .= (X \ Y) /\ X \ Z by Th56,Th23
    .= (X \ Y) /\ (X \ Z) by Th62;
end;

theorem
  X \ Y = Y \ X implies X = Y
proof
  assume X \ Y = Y \ X;
  hence X = Y \ X \/ Y /\ X by Th65
    .= Y by Th65;
end;

theorem
  X /\ (Y \ Z) = X /\ Y \ X /\ Z
proof
A1: X /\ Y c= X by Th15;
  X /\ Y \ X /\ Z = ((X /\ Y) \ X) \/ ((X /\ Y) \ Z) by Th69
    .= [[0]]I \/ ((X /\ Y) \ Z) by A1,Th52
    .= (X /\ Y) \ Z by Th22,Th43;
  hence thesis by Th62;
end;

theorem
  X \ Y c= Z implies X c= Y \/ Z
proof
  assume
A1: X \ Y c= Z;
  X /\ Y c= Y by Th15;
  then X /\ Y \/ (X \ Y) c= Y \/ Z by A1,Th20;
  hence thesis by Th65;
end;

theorem
  X \ Y c= X \+\ Y by Th14;

theorem
  X \+\ [[0]]I = X
proof
  thus X \+\ [[0]]I = X \/ ([[0]]I \ X) by Th59
    .= X \/ [[0]]I by Th60
    .= X by Th22,Th43;
end;

theorem
  X \+\ X = [[0]]I by Th52;

theorem
  X \/ Y = (X \+\ Y) \/ X /\ Y
proof
  thus X \/ Y = ((X \ Y) \/ X /\ Y) \/ Y by Th65
    .= (X \ Y) \/ (X /\ Y \/ Y) by Th28
    .= (X \ Y) \/ Y by Th31
    .= (X \ Y) \/ ((Y \ X) \/ (Y /\ X)) by Th65
    .= (X \+\ Y) \/ X /\ Y by Th28;
end;

theorem Th87:
  X \+\ Y = (X \/ Y) \ X /\ Y
proof
  thus X \+\ Y = (X \ X /\ Y) \/ (Y \ X) by Th70
    .= (X \ X /\ Y) \/ (Y \ X /\ Y) by Th70
    .= (X \/ Y) \ X /\ Y by Th72;
end;

theorem
  (X \+\ Y) \ Z = (X \ (Y \/ Z)) \/ (Y \ (X \/ Z))
proof
  thus (X \+\ Y) \ Z = (X \ Y \ Z) \/ (Y \ X \ Z) by Th72
    .= (X \ (Y \/ Z)) \/ (Y \ X \ Z) by Th73
    .= (X \ (Y \/ Z)) \/ (Y \ (X \/ Z)) by Th73;
end;

theorem
  X \ (Y \+\ Z) = X \ (Y \/ Z) \/ X /\ Y /\ Z
proof
  thus X \ (Y \+\ Z) = X \ ((Y \/ Z) \ Y /\ Z) by Th87
    .= X \ (Y \/ Z) \/ X /\ (Y /\ Z) by Th64
    .= X \ (Y \/ Z) \/ X /\ Y /\ Z by Th29;
end;

theorem Th90:
  (X \+\ Y) \+\ Z = X \+\ (Y \+\ Z)
proof
  set S1 = X \ (Y \/ Z), S2 = Y \ (X \/ Z), S3 = Z \ (X \/ Y), S4 = X /\ Y /\
  Z;
  thus (X \+\ Y) \+\ Z = (((X \ Y) \ Z) \/ ((Y \ X) \ Z)) \/ (Z \ ((X \ Y) \/
  (Y \ X))) by Th72
    .= ( S1 \/ ((Y \ X) \ Z)) \/ (Z \ ((X \ Y) \/ (Y \ X))) by Th73
    .= ( S1 \/ S2) \/ (Z \ ((X \ Y) \/ (Y \ X))) by Th73
    .= ( S1 \/ S2) \/ (Z \ ((X \/ Y) \ (X /\ Y))) by Th77
    .= (S1 \/ S2) \/ (S4 \/ S3) by Th64
    .= (S1 \/ S2 \/ S4) \/ S3 by Th28
    .= (S1 \/ S4 \/ S2) \/ S3 by Th28
    .= (S1 \/ S4) \/ (S2 \/ S3) by Th28
    .= (S1 \/ X /\ (Y /\ Z)) \/ (S2 \/ S3) by Th29
    .= X \ ((Y \/ Z) \ (Y /\ Z)) \/ (S2 \/ S3) by Th64
    .= X \ ((Y \ Z) \/ (Z \ Y)) \/ (S2 \/ (Z \ (Y \/ X))) by Th77
    .= X \ ((Y \ Z) \/ (Z \ Y)) \/ ((Y \ (Z \/ X)) \/ (Z \ Y \ X)) by Th73
    .= X \ ((Y \ Z) \/ (Z \ Y)) \/ ((Y \ Z \ X) \/ (Z \ Y \ X)) by Th73
    .= X \+\ (Y \+\ Z) by Th72;
end;

theorem
  X \ Y c= Z & Y \ X c= Z implies X \+\ Y c= Z by Th16;

theorem Th92:
  X \/ Y = X \+\ (Y \ X)
proof
  thus X \/ Y = Y \/ (X \/ X)
    .= X \/ Y \/ X by Th28
    .= (X \ Y) \/ Y \/ X by Th67
    .= (X \ Y) \/ (X \/ Y) by Th28
    .= (X \ Y) \/ (X \/ (Y \ X)) by Th67
    .= (X \ Y) \/ X /\ X \/ (Y \ X) by Th28
    .= (X \ (Y \ X)) \/ (Y \ X) by Th64
    .= X \+\ (Y \ X) by Th78;
end;

theorem Th93:
  X /\ Y = X \+\ (X \ Y)
proof
A1: X \ Y c= X by Th56;
  thus X /\ Y = X \ (X \ Y) by Th68
    .= (X \ (X \ Y)) \/ [[0]]I by Th22,Th43
    .= X \+\ (X \ Y) by A1,Th52;
end;

theorem Th94:
  X \ Y = X \+\ (X /\ Y)
proof
A1: X /\ Y c= X by Th15;
  thus X \ Y = X \ (X /\ Y) by Th70
    .= (X \ (X /\ Y)) \/ [[0]]I by Th22,Th43
    .= X \+\ (X /\ Y) by A1,Th52;
end;

theorem Th95:
  Y \ X = X \+\ (X \/ Y)
proof
A1: X c= Y \/ X by Th14;
  thus Y \ X = ((Y \/ X) \ X) by Th75
    .= [[0]]I \/ ((Y \/ X) \ X) by Th22,Th43
    .= X \+\ (X \/ Y) by A1,Th52;
end;

theorem
  X \/ Y = X \+\ Y \+\ X /\ Y
proof
  thus X \/ Y = X \+\ (Y \ X) by Th92
    .= X \+\ (Y \+\ X /\ Y) by Th94
    .= X \+\ Y \+\ X /\ Y by Th90;
end;

theorem
  X /\ Y = X \+\ Y \+\ (X \/ Y)
proof
  thus X /\ Y = X \+\ (X \ Y) by Th93
    .= X \+\ (Y \+\ (X \/ Y)) by Th95
    .= X \+\ Y \+\ (X \/ Y) by Th90;
end;

begin 

theorem
  X overlaps Y or X overlaps Z implies X overlaps Y \/ Z
proof
  assume
A1: X overlaps Y or X overlaps Z;
  per cases by A1;
  suppose
    X overlaps Y;
    then consider x such that
A2: x in X and
A3: x in Y by Th11;
    x in Y \/ Z by A3,Th7;
    hence thesis by A2,Th10;
  end;
  suppose
    X overlaps Z;
    then consider x such that
A4: x in X and
A5: x in Z by Th11;
    x in Y \/ Z by A5,Th7;
    hence thesis by A4,Th10;
  end;
end;

theorem Th99:
  X overlaps Y & Y c= Z implies X overlaps Z
proof
  assume that
A1: X overlaps Y and
A2: Y c= Z;
  consider x such that
A3: x in X and
A4: x in Y by A1,Th11;
  x in Z by A2,A4,Th9;
  hence thesis by A3,Th10;
end;

theorem Th100:
  X overlaps Y & X c= Z implies Z overlaps Y
proof
  assume that
A1: X overlaps Y and
A2: X c= Z;
  consider x such that
A3: x in X and
A4: x in Y by A1,Th11;
  x in Z by A2,A3,Th9;
  hence thesis by A4,Th10;
end;

theorem Th101:
  X c= Y & Z c= V & X overlaps Z implies Y overlaps V
proof
  assume that
A1: X c= Y and
A2: Z c= V;
  assume X overlaps Z;
  then Y overlaps Z by A1,Th100;
  hence thesis by A2,Th99;
end;

theorem
  X overlaps Y /\ Z implies X overlaps Y & X overlaps Z
proof
  assume X overlaps Y /\ Z;
  then consider x such that
A1: x in X and
A2: x in Y /\ Z by Th11;
  x in Y & x in Z by A2,Th8;
  hence thesis by A1,Th10;
end;

theorem
  X overlaps Z & X c= V implies X overlaps Z /\ V
proof
  assume that
A1: X overlaps Z and
A2: X c= V;
  consider x such that
A3: x in X and
A4: x in Z by A1,Th11;
  x in V by A2,A3,Th9;
  then x in Z /\ V by A4,Th8;
  hence thesis by A3,Th10;
end;

theorem
  X overlaps Y \ Z implies X overlaps Y by Th56,Th99;

theorem
  Y does not overlap Z implies X /\ Y does not overlap X /\ Z
proof
  assume
A1: Y does not overlap Z;
  X /\ Y c= Y & X /\ Z c= Z by Th15;
  hence thesis by A1,Th101;
end;

theorem
  X overlaps Y \ Z implies Y overlaps X \ Z
proof
  assume
A1: X overlaps Y \ Z;
  let i;
  assume
A2: i in I;
  then X.i meets (Y \ Z).i by A1,Def7;
  then X.i meets Y.i \ Z.i by A2,Def6;
  then Y.i meets X.i \ Z.i by XBOOLE_1:81;
  hence thesis by A2,Def6;
end;

theorem Th107:
  X meets Y & Y c= Z implies X meets Z
proof
  assume that
A1: X meets Y and
A2: Y c= Z;
  consider i such that
A3: i in I and
A4: X.i meets Y.i by A1,Def8;
  take i;
  Y.i c= Z.i by A2,A3,Def2;
  hence thesis by A3,A4,XBOOLE_1:63;
end;

theorem Th108:
  Y misses X \ Y
proof
  now
    let i;
    assume i in I;
    then (X \ Y).i = X.i \ Y.i by Def6;
    hence Y.i misses (X \ Y).i by XBOOLE_1:79;
  end;
  hence thesis by Def8;
end;

theorem
  X /\ Y misses X \ Y
proof
  X /\ Y c= Y & X \ Y misses Y by Th15,Th108;
  hence thesis by Th107;
end;

theorem
  X /\ Y misses X \+\ Y
proof
  now
    let i;
    assume i in I;
    then (X /\ Y).i = X.i /\ Y.i & (X \+\ Y).i = X.i \+\ Y.i by Def5,Th4;
    hence (X /\ Y).i misses (X \+\ Y).i by XBOOLE_1:103;
  end;
  hence thesis by Def8;
end;

theorem Th111:
  X misses Y implies X /\ Y = [[0]]I
proof
  assume
A1: X misses Y;
  now
    let i;
    assume
A2: i in I;
    then
A3: X.i misses Y.i by A1,Def8;
    thus (X /\ Y).i = X.i /\ Y.i by A2,Def5
      .= {} by A3,XBOOLE_0:def 7;
  end;
  hence thesis by Th6;
end;

theorem
  X <> [[0]]I implies X meets X
proof
  X = X /\ X;
  hence thesis by Th111;
end;

theorem
  X c= Y & X c= Z & Y misses Z implies X = [[0]]I
proof
  assume
A1: X c= Y & X c= Z;
  assume Y misses Z;
  then Y /\ Z = [[0]]I by Th111;
  hence thesis by A1,Th17,Th44;
end;

theorem
  Z \/ V = X \/ Y & X misses Z & Y misses V implies X = V & Y = Z
proof
  assume
A1: Z \/ V = X \/ Y;
  assume X misses Z & Y misses V;
  then
A2: X /\ Z = [[0]]I & Y /\ V = [[0]]I by Th111;
  thus X = X /\ (Z \/ V) by Th23,A1,Th14
    .= X /\ Z \/ X /\ V by Th32
    .= (X \/ Y) /\ V by A2,Th32
    .= V by A1,Th14,Th23;
  thus Y = Y /\ (Z \/ V) by Th23,A1,Th14
    .= Y /\ Z \/ Y /\ V by Th32
    .= (X \/ Y) /\ Z by A2,Th32
    .= Z by A1,Th14,Th23;
end;

theorem Th115:
  X misses Y implies X \ Y = X
proof
  assume
A1: X misses Y;
    let i;
    assume
A2: i in I;
    then
A3: (X \ Y).i = X.i \ Y.i by Def6;
    X.i misses Y.i by A1,A2,Def8;
    hence (X \ Y).i = X.i by A3,XBOOLE_1:83;
end;

theorem
  X misses Y implies (X \/ Y) \ Y = X
proof
  (X \/ Y) \ Y = X \ Y by Th75;
  hence thesis by Th115;
end;

theorem
  X \ Y = X implies X misses Y
proof
  assume
A1: X \ Y = X;
  let i;
  assume i in I;
  then X.i \ Y.i = X.i by A1,Def6;
  hence thesis by XBOOLE_1:83;
end;

theorem
  X \ Y misses Y \ X
proof
  let i;
  assume i in I;
  then (X \ Y).i = X.i \ Y.i & (Y \ X).i = Y.i \ X.i by Def6;
  hence thesis by XBOOLE_1:82;
end;

begin 

definition
  let I,X,Y;
  pred X [= Y means
  :Def11:
  for x st x in X holds x in Y;
  reflexivity;
end;

theorem
  X c= Y implies X [= Y
proof
  assume
A1: X c= Y;
  let x such that
A2: x in X;
  let i;
  assume
A3: i in I;
  then
A4: X.i c= Y.i by A1,Def2;
  x.i in X.i by A2,A3,Def1;
  hence thesis by A4;
end;

theorem
  X [= Y & Y [= Z implies X [= Z
proof
  assume that
A1: X [= Y and
A2: Y [= Z;
  let x;
  assume x in X;
  then x in Y by A1,Def11;
  hence thesis by A2,Def11;
end;

begin 

theorem
  [[0]]{} in [[0]]{}
proof
  let i;
  thus thesis;
end;

theorem
  for X being ManySortedSet of {} holds X = {}
proof
  let X be ManySortedSet of {};
  dom X = {} by PARTFUN1:def 2;
  hence thesis by RELAT_1:41;
end;

reserve I for non empty set,
  x,X,Y for ManySortedSet of I;

theorem
  X overlaps Y implies X meets Y
proof
  set i = the Element of I;
  assume X overlaps Y;
  then X.i meets Y.i by Def7;
  hence thesis by Def8;
end;

theorem Th124:
  not ex x st x in [[0]]I
proof
  set i = the Element of I;
  given x such that
A1: x in [[0]]I;
  x.i in [[0]]I.i by A1,Def1;
  hence contradiction by FUNCOP_1:7;
end;

theorem
  x in X & x in Y implies X /\ Y <> [[0]]I
proof
  assume x in X & x in Y;
  then x in X /\ Y by Th8;
  hence thesis by Th124;
end;

theorem
  X does not overlap [[0]]I
proof
  assume X overlaps [[0]]I;
  then ex x st x in X & x in [[0]]I by Th11;
  hence contradiction by Th124;
end;

theorem Th127:
  X /\ Y = [[0]]I implies X does not overlap Y
proof
  assume
A1: X /\ Y = [[0]]I;
  assume X overlaps Y;
  then consider x such that
A2: x in X & x in Y by Th11;
  x in X /\ Y by A2,Th8;
  hence contradiction by A1,Th124;
end;

theorem
  X overlaps X implies X <> [[0]]I
proof
  X = X /\ X;
  hence thesis by Th127;
end;

begin 

reserve I for set,
  x,X,Y,Z for ManySortedSet of I;

definition
  let I be set;
  let X be ManySortedSet of I;
  redefine attr X is empty-yielding means

  for i st i in I holds X.i is
  empty;
  compatibility
  proof
    dom X = I by PARTFUN1:def 2;
    hence thesis by FUNCT_1:def 8;
  end;
  redefine attr X is non-empty means
  :Def13:
  for i st i in I holds X.i is non empty;
  compatibility
  proof
    dom X = I by PARTFUN1:def 2;
    hence thesis by FUNCT_1:def 9;
  end;
end;

registration
  let I be set;
  cluster empty-yielding for ManySortedSet of I;
  existence
  proof
    take [[0]]I;
    let i;
    assume i in I;
    thus thesis by Th5;
  end;
  cluster non-empty for ManySortedSet of I;
  existence
  proof
    set e = the set;
    reconsider M = I --> {e} as ManySortedSet of I;
    take M;
    let i;
    assume i in I;
    hence thesis by FUNCOP_1:7;
  end;
end;

registration
  let I be non empty set;
  cluster non-empty -> non empty-yielding for ManySortedSet of I;
  coherence
  proof
    set i = the Element of I;
    let M be ManySortedSet of I;
    assume
A1: M is non-empty;
    take i;
    thus i in I;
    thus thesis by A1,Def13;
  end;
  cluster empty-yielding -> non non-empty for ManySortedSet of I;
  coherence;
end;

theorem
  X is empty-yielding iff X = [[0]]I
proof
  hereby
    assume X is empty-yielding;
    then for i st i in I holds X.i = {};
    hence X = [[0]]I by Th6;
  end;
  assume
A1: X = [[0]]I;
  let i;
  assume i in I;
  thus thesis by A1,Th5;
end;

theorem
  Y is empty-yielding & X c= Y implies X is empty-yielding
proof
  assume
A1: Y is empty-yielding & X c= Y;
  let i;
  assume i in I;
  then X.i c= Y.i & Y.i is empty by A1,Def2;
  hence thesis by XBOOLE_1:3;
end;

theorem Th131:
  X is non-empty & X c= Y implies Y is non-empty
proof
  assume
A1: X is non-empty & X c= Y;
  let i;
  assume i in I;
  then X.i c= Y.i & X.i is non empty by A1,Def2,Def13;
  hence thesis by XBOOLE_1:3;
end;

theorem Th132:
  X is non-empty & X [= Y implies X c= Y
proof
  assume that
A1: X is non-empty and
A2: X [= Y;
  deffunc F(set) = X.$1;
A3: for i st i in I holds F(i) <> {} by A1,Def13;
  consider f being ManySortedSet of I such that
A4: for i st i in I holds f.i in F(i) from KuratowskiFunction(A3);
  let i such that
A5: i in I;
  let e;
  deffunc G(set) = IFEQ(i,$1,e,f.$1);
  consider g being Function such that
A6: dom g = I and
A7: for u st u in I holds g.u = G(u) from FUNCT_1:sch 3;
  reconsider g as ManySortedSet of I by A6,PARTFUN1:def 2,RELAT_1:def 18;
A8: g.i = IFEQ(i,i,e,f.i) by A5,A7
    .= e by FUNCOP_1:def 8;
  assume
A9: e in X.i;
  g in X
  proof
    let u such that
A10: u in I;
    per cases;
    suppose
      u = i;
      hence thesis by A8,A9;
    end;
    suppose
A11:  u <> i;
      g.u = IFEQ(i,u,e,f.u) by A7,A10
        .= f.u by A11,FUNCOP_1:def 8;
      hence thesis by A4,A10;
    end;
  end;
  then g in Y by A2,Def11;
  hence thesis by A5,A8,Def1;
end;

theorem
  X is non-empty & X [= Y implies Y is non-empty
proof
  assume
A1: X is non-empty;
  assume X [= Y;
  then X c= Y by A1,Th132;
  hence thesis by A1,Th131;
end;

reserve X for non-empty ManySortedSet of I;

theorem
  ex x st x in X
proof
  deffunc F(set) = X.$1;
A1: for i st i in I holds F(i) <> {} by Def13;
  consider f being ManySortedSet of I such that
A2: for i st i in I holds f.i in F(i) from KuratowskiFunction(A1);
  take f;
  let i;
  thus thesis by A2;
end;

theorem Th135:
  (for x holds x in X iff x in Y) implies X = Y
proof
  deffunc F(set) = X.$1;
A1: for i st i in I holds F(i) <> {} by Def13;
  consider f being ManySortedSet of I such that
A2: for i st i in I holds f.i in F(i) from KuratowskiFunction(A1);
  assume
A3: for x holds x in X iff x in Y;
  now
    let i such that
A4: i in I;
    now
      let e;
      deffunc G(set) = IFEQ(i,$1,e,f.$1);
      consider g being Function such that
A5:   dom g = I and
A6:   for u st u in I holds g.u = G(u) from FUNCT_1:sch 3;
      reconsider g as ManySortedSet of I by A5,PARTFUN1:def 2,RELAT_1:def 18;
A7:   g.i = IFEQ(i,i,e,f.i) by A4,A6
        .= e by FUNCOP_1:def 8;
      hereby
        assume
A8:     e in X.i;
        g in X
        proof
          let u such that
A9:       u in I;
          per cases;
          suppose
            u = i;
            hence thesis by A7,A8;
          end;
          suppose
A10:        u <> i;
            g.u = IFEQ(i,u,e,f.u) by A6,A9
              .= f.u by A10,FUNCOP_1:def 8;
            hence thesis by A2,A9;
          end;
        end;
        then g in Y by A3;
        hence e in Y.i by A4,A7,Def1;
      end;
      assume
A11:  e in Y.i;
      g in Y
      proof
        let u such that
A12:    u in I;
        per cases;
        suppose
          u = i;
          hence thesis by A7,A11;
        end;
        suppose
A13:      u <> i;
          f in X by A2,Def1;
          then
A14:      f in Y by A3;
          g.u = IFEQ(i,u,e,f.u) by A6,A12
            .= f.u by A13,FUNCOP_1:def 8;
          hence thesis by A12,A14,Def1;
        end;
      end;
      then g in X by A3;
      hence e in X.i by A4,A7,Def1;
    end;
    hence X.i = Y.i by TARSKI:1;
  end;
  hence thesis by Th3;
end;

theorem
  (for x holds x in X iff x in Y & x in Z) implies X = Y /\ Z
proof
  assume
A1: for x holds x in X iff x in Y & x in Z;
  now
    let x;
    hereby
      assume x in X;
      then x in Y & x in Z by A1;
      hence x in Y /\ Z by Th8;
    end;
    assume x in Y /\ Z;
    then x in Y & x in Z by Th8;
    hence x in X by A1;
  end;
  hence thesis by Th135;
end;

begin 

scheme
  MSSEx { I() -> set, P[set,set] }: ex f being ManySortedSet of I() st for i
  st i in I() holds P[i,f.i]
provided
A1: for i st i in I() ex j st P[i,j]
proof
A2: for i st i in I() ex j st P[i,j] by A1;
  consider f being Function such that
A3: dom f = I() and
A4: for i st i in I() holds P[i,f.i] from CLASSES1:sch 1(A2);
  reconsider f as ManySortedSet of I() by A3,PARTFUN1:def 2,RELAT_1:def 18;
  take f;
  thus thesis by A4;
end;

scheme
  MSSLambda { I() -> set, F(set) -> set }: ex f being ManySortedSet of I() st
  for i st i in I() holds f.i = F(i) proof
  consider f being Function such that
A1: dom f = I() and
A2: for i st i in I() holds f.i = F(i) from FUNCT_1:sch 3;
  reconsider f as ManySortedSet of I() by A1,PARTFUN1:def 2,RELAT_1:def 18;
  take f;
  thus thesis by A2;
end;

registration
  let I be set;
  cluster Function-yielding for ManySortedSet of I;
  existence
  proof
    set f = the Function;
    reconsider F = I --> f as ManySortedSet of I;
    take F;
    let x be set;
    assume x in dom F;
    thus thesis;
  end;
end;

definition
  let I be set;
  mode ManySortedFunction of I is Function-yielding ManySortedSet of I;
end;

theorem
  not ex M being non-empty ManySortedSet of I st {} in rng M
proof
  let M be non-empty ManySortedSet of I;
A1: dom M = I by PARTFUN1:def 2;
  assume {} in rng M;
  then ex i st i in I & M.i = {} by A1,FUNCT_1:def 3;
  hence contradiction by Def13;
end;

definition
  let I be set;
  let M be ManySortedSet of I;
  mode Component of M is Element of rng M;
end;

theorem
  for I being non empty set for M being ManySortedSet of I, A being
  Component of M ex i st i in I & A = M.i
proof
  let I be non empty set;
  let M be ManySortedSet of I, A be Component of M;
A1: dom M = I by PARTFUN1:def 2;
  then rng M <> {} by RELAT_1:42;
  hence thesis by A1,FUNCT_1:def 3;
end;

theorem
  for M being ManySortedSet of I, i st i in I holds M.i is Component of M
proof
  let M be ManySortedSet of I, i;
  assume
A1: i in I;
  dom M = I by PARTFUN1:def 2;
  hence thesis by A1,FUNCT_1:def 3;
end;

definition
  let I;
  let B be ManySortedSet of I;
  mode Element of B -> ManySortedSet of I means
    for i st i in I holds it.i is
    Element of B.i;
  existence
  proof
    defpred Q[set,set] means $2 is Element of B.$1;
A1: for i st i in I ex j st Q[i,j]
    proof
      let i such that
      i in I;
      set j = the Element of B.i;
      reconsider j as set;
      take j;
      thus thesis;
    end;
    thus ex f being ManySortedSet of I st for i st i in I holds Q[i,f.i] from
    MSSEx(A1);
  end;
end;

begin 

definition
  let I;
  let A be ManySortedSet of I, B be ManySortedSet of I;
  mode ManySortedFunction of A,B -> ManySortedSet of I means
:Def15: for i st i in I holds it.i is Function of A.i, B.i;
  correctness
  proof
    defpred P[set,set] means $2 is Function of A.$1,B.$1;
A1: for i st i in I ex j st P[i,j]
    proof
      let i such that
      i in I;
      set j = the Function of A.i,B.i;
      reconsider j as set;
      take j;
      thus thesis;
    end;
    consider f being ManySortedSet of I such that
A2: for i st i in I holds P[i,f.i] from MSSEx(A1);
    f is Function-yielding
    proof
      let i;
      assume i in dom f;
      then i in I by PARTFUN1:def 2;
      hence thesis by A2;
    end;
    then reconsider f as ManySortedFunction of I;
    take f;
    let i;
    assume i in I;
    hence thesis by A2;
  end;
end;

registration
  let I;
  let A be ManySortedSet of I, B be ManySortedSet of I;
  cluster ->Function-yielding for ManySortedFunction of A,B;
  coherence
  proof
    let f be ManySortedFunction of A,B;
    let i be set;
    assume i in dom f;
    then i in I by PARTFUN1:def 2;
    hence thesis by Def15;
  end;
end;

registration
  let I be set;
  let J be non empty set;
  let O be Function of I,J;
  let F be ManySortedSet of J;
  cluster F*O -> total for I-defined Function;
  coherence
  proof
A1: dom F = J by PARTFUN1:def 2;
    rng O c= J & dom O = I by FUNCT_2:def 1,RELAT_1:def 19;
    then dom(F*O) = I by A1,RELAT_1:27;
    hence thesis by PARTFUN1:def 2;
  end;
end;

reserve D for non empty set,
  n for Nat;

scheme
  LambdaDMS {D()->non empty set, F(set)->set}: ex X be ManySortedSet of D() st
  for d be Element of D() holds X.d = F(d) proof
  consider f be Function such that
A1: dom f = D() and
A2: for d be Element of D() holds f.d = F(d) from FUNCT_1:sch 4;
  f is ManySortedSet of D() by A1,PARTFUN1:def 2,RELAT_1:def 18;
  hence thesis by A2;
end;

registration
  let J be non empty set, B be non-empty ManySortedSet of J, j be Element of J;
  cluster B.j -> non empty;
  coherence by Def13;
end;

reserve X,Y for ManySortedSet of I;

definition
  let I, X,Y;
  func [|X,Y|] -> ManySortedSet of I means
  for i st i in I holds it.i = [:X.i,Y.i:];
  existence
  proof
    deffunc F(set) = [:X.$1,Y.$1:];
    consider f be Function such that
A1: dom f = I & for i st i in I holds f.i = F(i) from FUNCT_1:sch 3;
    reconsider f as ManySortedSet of I by A1,PARTFUN1:def 2,RELAT_1:def 18;
    take f;
    thus thesis by A1;
  end;
  uniqueness
  proof
    let f,g be ManySortedSet of I;
    assume that
A2: for i st i in I holds f.i = [:X.i,Y.i:] and
A3: for i st i in I holds g.i = [:X.i,Y.i:];
    for x be set st x in I holds f.x = g.x
    proof
      let x be set;
      assume
A4:   x in I;
      then f.x = [:X.x,Y.x:] by A2;
      hence thesis by A3,A4;
    end;
    hence thesis by Th3;
  end;
end;

definition
  let I, X, Y;
  deffunc F(set) = Funcs(X.$1,Y.$1);
  func (Funcs) (X,Y) -> ManySortedSet of I means
  for i be set st i in I holds it.i = Funcs(X.i,Y.i);
  existence
  proof
    thus ex M being ManySortedSet of I st for i be set st i in I holds M.i = F
    (i) from MSSLambda;
  end;
  uniqueness
  proof
    let F1, F2 be ManySortedSet of I such that
A1: for i be set st i in I holds F1.i = Funcs(X.i, Y.i) and
A2: for i be set st i in I holds F2.i = Funcs(X.i, Y.i);
    now
      let i be set;
      assume
A3:   i in I;
      hence F1.i = Funcs(X.i, Y.i) by A1
        .= F2.i by A2,A3;
    end;
    hence thesis by Th3;
  end;
end;

definition
  let I be set, M be ManySortedSet of I;
  mode ManySortedSubset of M -> ManySortedSet of I means
  :Def18:
  it c= M;
  existence;
end;

registration
  let I be set, M be non-empty ManySortedSet of I;
  cluster non-empty for ManySortedSubset of M;
  existence
  proof
    reconsider N= M as ManySortedSubset of M by Def18;
    take N;
    thus thesis;
  end;
end;

definition
  let F,G be Function-yielding Function;
  func G**F -> Function means
  :Def19:
  dom it = (dom F) /\ (dom G) &
  for i be set st i in dom it holds it.i = (G.i)*(F.i);
  existence
  proof
    defpred P[set,set] means $2 = (G.$1)*(F.$1);
A1: for i be set st i in dom F /\ dom G ex u be set st P[i,u];
    ex H being Function st
    dom H = dom F /\ dom G & for i be set st i in dom F /\ dom G holds
    P[i,H.i] from CLASSES1:sch 1(A1);
    hence thesis;
  end;
  uniqueness
  proof
    let H1,H2 be Function;
    assume that
A2: dom H1 = dom F /\ dom G and
A3: for i be set st i in dom H1 holds H1.i = (G.i)*(F.i) and
A4: dom H2 = dom F /\ dom G and
A5: for i be set st i in dom H2 holds H2.i = (G.i)*(F.i);
    now
      let i be set;
      reconsider f = F.i as Function;
      reconsider g = G.i as Function;
      assume
A6:   i in dom F /\ dom G;
      then H1.i = g*f by A2,A3;
      hence H1.i = H2.i by A4,A5,A6;
    end;
    hence thesis by A2,A4,FUNCT_1:2;
  end;
end;

registration
  let F,G be Function-yielding Function;
  cluster G**F -> Function-yielding;
  coherence
  proof
    set H = G**F;
    let x be set;
    reconsider f = F.x as Function;
    reconsider g = G.x as Function;
    assume x in dom H;
    then H.x = g*f by Def19;
    hence thesis;
  end;
end;

definition
  let I be set, A be ManySortedSet of I, F be ManySortedFunction of I;
  func F.:.:A -> ManySortedSet of I means
  for i be set st i in I holds it.i = (F.i).:(A.i);
  existence
  proof
    defpred P[set,set] means $2 = (F.$1).:(A.$1);
A1: for i be set st i in I ex u be set st P[i,u];
    consider g being Function such that
A2: dom g = I & for i be set st i in I holds P[i,g.i] from CLASSES1:
    sch 1( A1);
    reconsider g as ManySortedSet of I by A2,PARTFUN1:def 2,RELAT_1:def 18;
    take g;
    thus thesis by A2;
  end;
  uniqueness
  proof
    let B,B1 be ManySortedSet of I;
    assume that
A3: for i be set st i in I holds B.i = (F.i).:(A.i) and
A4: for i be set st i in I holds B1.i = (F.i).:(A.i);
    now
      let i be set;
      reconsider f = F.i as Function;
      assume
A5:   i in I;
      then B.i = f.:(A.i) by A3;
      hence B.i = B1.i by A4,A5;
    end;
    hence thesis by Th3;
  end;
end;

registration
  let I;
  cluster [[0]]I -> empty-yielding;
  coherence
  proof
    rng [[0]]I c= {{}} by FUNCOP_1:13;
    hence thesis by RELAT_1:def 15;
  end;
end;

scheme
  MSSExD { I() -> non empty set, P[set,set] }: ex f being ManySortedSet of I()
  st for i being Element of I() holds P[i,f.i]
provided
A1: for i being Element of I() ex j being set st P[i,j]
proof
  defpred Z[set,set] means P[ $1,$2];
A2: for i being set st i in I() ex j being set st Z[i,j] by A1;
  consider f being ManySortedSet of I() such that
A3: for i being set st i in I() holds Z[i,f.i] from MSSEx(A2);
  take f;
  let i be Element of I();
  thus thesis by A3;
end;

registration
  let A be non empty set;
  cluster non-empty -> non empty-yielding for ManySortedSet of A;
  coherence;
end;

registration
  let X be non empty set;
  cluster -> non empty for ManySortedSet of X;
  coherence
  proof
    set x = the Element of X;
    let f be ManySortedSet of X;
    dom f = X by PARTFUN1:def 2;
    then [x,f.x] in f by FUNCT_1:def 2;
    hence thesis;
  end;
end;

theorem
  for F, G, H be Function-yielding Function holds
  (H ** G) ** F = H ** (G ** F)
proof
  let F, G, H be Function-yielding Function;
  set f = (H ** G) ** F, g = H ** (G ** F);
  now
A1: dom f = (dom (H ** G)) /\ dom F by Def19
      .= (dom H) /\ (dom G) /\ (dom F) by Def19;
    then
A2: dom f = (dom H) /\ ((dom G) /\ dom F) by XBOOLE_1:16;
    hence
A3: dom f = (dom H) /\ (dom (G ** F)) by Def19
      .= dom g by Def19;
    let x be set;
    assume
A4: x in dom f;
    then x in (dom H) /\ (dom G) by A1,XBOOLE_0:def 4;
    then
A5: x in dom (H ** G) by Def19;
    x in (dom G) /\ (dom F) by A2,A4,XBOOLE_0:def 4;
    then
A6: x in dom (G ** F) by Def19;
    thus f.x = ((H**G).x) * (F.x) by A4,Def19
      .= (H.x)*(G.x)*(F.x) by A5,Def19
      .= (H.x)*((G.x)*(F.x)) by RELAT_1:36
      .= (H.x)*((G**F).x) by A6,Def19
      .= g.x by A3,A4,Def19;
  end;
  hence thesis by FUNCT_1:2;
end;

registration
 let I be set, f be non-empty ManySortedSet of I;
 cluster total for I-defined f-compatible Function;
  existence
  proof
    deffunc F(set) = f.$1;
A1: for e st e in I holds F(e) <> {};
   consider g being ManySortedSet of I such that
A2:  for e st e in I holds g.e in F(e) from KuratowskiFunction(A1);
    g is f-compatible
    proof let x be set;
     assume x in dom g;
      then x in I by PARTFUN1:def 2;
     hence g.x in f.x by A2;
    end;
    then reconsider g as I-defined f-compatible Function;
   take g;
   thus g is total;
  end;
end;

theorem
 for I being set, f being non-empty ManySortedSet of I
 for p being f-compatible I-defined Function
  ex s being f-compatible ManySortedSet of I st p c= s
 proof
  let I be set, f being non-empty ManySortedSet of I;
  let p be f-compatible I-defined Function;
   set h = the (f)-compatible ManySortedSet of I;
  take h+*p;
  thus p c= h+*p by FUNCT_4:25;
 end;

theorem
  for I,A be set
  for s,ss being ManySortedSet of I
  holds (ss +* s | A) | A = s | A
proof
  let I,A be set;
  let s,ss be ManySortedSet of I;
  dom s = I by PARTFUN1:def 2
    .= dom ss by PARTFUN1:def 2;
  then A /\ dom ss c= A /\ dom s;
  hence thesis by FUNCT_4:88;
end;

registration
 let X be non empty set, Y be set;
 cluster X-valued for ManySortedSet of Y;
 existence
  proof
   take M = Y --> the Element of X;
A1:  {the Element of X} c= X by ZFMISC_1:31;
    rng M c= {the Element of X} by FUNCOP_1:13;
   hence rng M c= X by A1,XBOOLE_1:1;
  end;
end;

theorem
  for I,Y being non empty set, M be Y-valued ManySortedSet of I,
     x be Element of I
  holds M.x = M/.x
proof
  let I,Y be non empty set, M be Y-valued ManySortedSet of I,
     x be Element of I;
  dom M = I by PARTFUN1:def 2;
  hence M.x = M/.x by PARTFUN1:def 6;
end;

theorem
 for f being Function, M being ManySortedSet of I
   holds (f+*M)|I = M
proof
 let f be Function, M be ManySortedSet of I;
A1: dom(f|I) c= I by RELAT_1:58;
A2: dom M = I by PARTFUN1:def 2;
 thus (f+*M)|I = f|I +* M|I by FUNCT_4:71
   .= f|I +* M by A2,RELAT_1:69
   .= M by A1,A2,FUNCT_4:19;
end;

theorem
 for I being set, Y being non empty set
 for p being Y-valued I-defined Function
  ex s being Y-valued ManySortedSet of I st p c= s
 proof
  let I be set, Y being non empty set;
  let p be Y-valued I-defined Function;
   set h = the Y-valued ManySortedSet of I;
  take h+*p;
  thus p c= h+*p by FUNCT_4:25;
 end;

theorem
 X c= Y & Y c= X implies X = Y by Lm1;

definition let I be non empty set, A,B be ManySortedSet of I;
 redefine pred A = B means
  for i being Element of I holds A.i = B.i;
 compatibility
 proof
   thus A = B implies for i being Element of I holds A.i = B.i;
   assume for i being Element of I holds A.i = B.i;
    then for i being set st i in I holds A.i = B.i;
   hence thesis by Th3;
  end;
end;














begin

reserve A,B,C for Ordinal,
  K,L,M,N for Cardinal,
  x,y,y1,y2,z,u,X,Y,Z,Z1,Z2 for set,
  n for Nat,
  f,f1,g,h for Function,
  Q,R for Relation;

definition
  let IT be Function;
  attr IT is Cardinal-yielding means
  :Def1: for x st x in dom IT holds IT.x is Cardinal;
end;

registration
  cluster empty -> Cardinal-yielding for Function;
  coherence
  proof
    let f such that
A1: f is empty;
    let x;
    thus thesis by A1;
  end;
end;

registration
  cluster Cardinal-yielding for Function;
  existence
  proof
    take {};
    thus thesis;
  end;
end;

definition
  mode Cardinal-Function is Cardinal-yielding Function;
end;

reserve ff for Cardinal-Function;

registration
  let ff,X;
  cluster ff|X -> Cardinal-yielding;
  coherence
  proof
    ff|X is Cardinal-yielding
    proof
      let x;
      assume
A1:   x in dom(ff|X);
      then
A2:   (ff|X).x = ff.x by FUNCT_1:47;
      dom(ff|X) = dom ff /\ X by RELAT_1:61;
      then x in dom ff by A1,XBOOLE_0:def 4;
      hence thesis by A2,Def1;
    end;
    hence thesis;
  end;
end;

registration
  let X,K;
  cluster X --> K -> Cardinal-yielding;
  coherence
  proof
    let x;
    assume x in dom(X --> K);
    hence thesis by FUNCOP_1:7;
  end;
end;

registration
  let X,K;
  cluster X .--> K -> Cardinal-yielding;
  coherence;
end;

scheme CFLambda { A()->set, F(set)->Cardinal } :
  ex ff st dom ff = A() & for x st x in A() holds ff.x = F(x)
proof
  consider f such that
A1: dom f = A() & for x st x in A() holds f.x = F(x) from FUNCT_1:sch 3;
  f is Cardinal-yielding
  proof
    let x;
    assume x in dom f;
    then f.x = F(x) by A1;
    hence thesis;
  end;
  then reconsider f as Cardinal-Function;
  take f;
  thus thesis by A1;
end;

definition
  let f;
  func Card f -> Cardinal-Function means
  :
  Def2: dom it = dom f & for x st x in dom f holds it.x = card(f.x);
  existence
  proof
    deffunc f(set) = card (f.$1);
    thus ex g being Cardinal-Function st
    dom g = dom f & for x st x in dom f holds g.x = f(x) from CFLambda;
  end;
  uniqueness
  proof
    let a1,a2 be Cardinal-Function such that
A1: dom a1 = dom f and
A2: for x st x in dom f holds a1.x = card(f.x) and
A3: dom a2 = dom f and
A4: for x st x in dom f holds a2.x = card(f.x);
    now
      let x;
      assume
A5:   x in dom f;
      then a1.x = card(f.x) by A2;
      hence a1.x = a2.x by A4,A5;
    end;
    hence thesis by A1,A3,FUNCT_1:2;
  end;
  func disjoin f -> Function means
  :
  Def3: dom it = dom f & for x st x in dom f holds it.x = [:f.x,{x}:];
  existence
  proof
    deffunc f(set) = [:f.$1,{$1}:];
    thus ex g being Function st
    dom g = dom f & for x st x in dom f holds g.x = f(x) from FUNCT_1:sch 3;
  end;
  uniqueness
  proof
    let a1,a2 be Function such that
A6: dom a1 = dom f and
A7: for x st x in dom f holds a1.x = [:f.x,{x}:] and
A8: dom a2 = dom f and
A9: for x st x in dom f holds a2.x = [:f.x,{x}:];
    now
      let x;
      assume
A10:  x in dom f;
      then a1.x = [:f.x,{x}:] by A7;
      hence a1.x = a2.x by A9,A10;
    end;
    hence thesis by A6,A8,FUNCT_1:2;
  end;
  func Union f -> set equals
  union rng f;
  correctness;
  defpred P[set] means
  ex g st $1 = g & dom g = dom f & for x st x in dom f holds g.x in f.x;
  func product f -> set means
  :
  Def5: x in it iff ex g st x = g & dom g = dom f &
  for y st y in dom f holds g.y in f.y;
  existence
  proof
    consider X such that
A11: x in X iff x in Funcs(dom f, union rng f) & P[x] from XBOOLE_0:sch 1;
    take X;
    let x;
    thus x in X implies ex g st x = g & dom g = dom f &
    for x st x in dom f holds g.x in f.x by A11;
    given g such that
A12: x = g and
A13: dom g = dom f and
A14: for x st x in dom f holds g.x in f.x;
    rng g c= union rng f
    proof
      let y;
      assume y in rng g;
      then consider z such that
A15:  z in dom g and
A16:  y = g.z by FUNCT_1:def 3;
A17:  y in f.z by A13,A14,A15,A16;
      f.z in rng f by A13,A15,FUNCT_1:def 3;
      hence thesis by A17,TARSKI:def 4;
    end;
    then x in Funcs(dom f, union rng f) by A12,A13,FUNCT_2:def 2;
    hence thesis by A11,A12,A13,A14;
  end;
  uniqueness
  proof
    let X1,X2 be set such that
A18: x in X1 iff P[x] and
A19: x in X2 iff P[x];
    thus thesis from XBOOLE_0:sch 2(A18,A19);
  end;
end;

theorem Th1:
  Card ff = ff
proof
  now
    let x;
    assume x in dom ff;
    then reconsider M = ff.x as Cardinal by Def1;
    card M = M by CARD_1:def 2;
    hence ff.x = card (ff.x);
  end;
  hence thesis by Def2;
end;

theorem
  Card (X --> Y) = X --> card Y
proof
A1: dom Card (X --> Y) = dom(X --> Y) by Def2;
  then
A2: dom Card (X --> Y) = X by FUNCOP_1:13;
A3: dom (X --> card Y) = X by FUNCOP_1:13;
  now
    let x;
    assume
A4: x in X;
    then
A5: Card (X --> Y).x = card ((X --> Y).x) by A1,A2,Def2;
    (X --> card Y).x = card Y by A4,FUNCOP_1:7;
    hence Card (X --> Y).x = (X --> card Y).x by A4,A5,FUNCOP_1:7;
  end;
  hence thesis by A2,A3,FUNCT_1:2;
end;

theorem
  disjoin {} = {}
proof
  dom disjoin {} = {} by Def3,RELAT_1:38;
  hence thesis;
end;

theorem Th4:
  disjoin (x .--> X) = x .--> [:X,{x}:]
proof
A1: dom disjoin ({x} --> X) = dom ({x} --> X) by Def3;
A2: dom ({x} --> X) = {x} by FUNCOP_1:13;
A3: dom ({x} --> [:X,{x}:]) = {x} by FUNCOP_1:13;
  now
    let y;
    assume
A4: y in {x};
    then
A5: disjoin ({x} --> X).y = [:({x} --> X).y,{y}:] by A2,Def3;
A6: ({x} --> X).y = X by A4,FUNCOP_1:7;
    ({x} --> [:X,{x}:]).y = [:X,{x}:] by A4,FUNCOP_1:7;
    hence disjoin ({x} --> X).y = ({x} --> [:X,{x}:]).y by A4,A5,A6,
TARSKI:def 1;
  end;
  hence thesis by A1,A2,A3,FUNCT_1:2;
end;

theorem
  x in dom f & y in dom f & x <> y implies (disjoin f).x misses (disjoin f).y
proof
  assume that
A1: x in dom f and
A2: y in dom f and
A3: x <> y;
  set z = the Element of ((disjoin f).x) /\ ((disjoin f).y);
  assume
A4: ((disjoin f).x) /\ ((disjoin f).y) <> {};
A5: (disjoin f).x = [:f.x,{x}:] by A1,Def3;
A6: (disjoin f).y = [:f.y,{y}:] by A2,Def3;
A7: z in (disjoin f).x by A4,XBOOLE_0:def 4;
A8: z in (disjoin f).y by A4,XBOOLE_0:def 4;
A9: z`2 in {x} by A5,A7,MCART_1:10;
A10: z`2 in {y} by A6,A8,MCART_1:10;
  z`2 = x by A9,TARSKI:def 1;
  hence contradiction by A3,A10,TARSKI:def 1;
end;

theorem Th6:
  Union (X --> Y) c= Y
proof
  let x;
  assume x in Union (X --> Y);
  then consider Z such that
A1: x in Z and
A2: Z in rng (X --> Y) by TARSKI:def 4;
  ex z st z in dom (X --> Y) & Z = (X --> Y).z by A2,FUNCT_1:def 3;
  hence thesis by A1,FUNCOP_1:7;
end;

theorem Th7:
  X <> {} implies Union (X --> Y) = Y
proof
  assume
A1: X <> {};
  set x = the Element of X;
  thus Union (X --> Y) c= Y by Th6;
A2: dom (X --> Y) = X by FUNCOP_1:13;
  (X --> Y).x = Y by A1,FUNCOP_1:7;
  then Y in rng (X --> Y) by A1,A2,FUNCT_1:def 3;
  hence thesis by ZFMISC_1:74;
end;

theorem
  Union (x .--> Y) = Y by Th7;

theorem Th9:
  g in product f iff dom g = dom f & for x st x in dom f holds g.x in f.x
proof
  thus g in product f implies
  dom g = dom f & for x st x in dom f holds g.x in f.x
  proof
    assume g in product f;
    then ex h being Function st
    g = h & dom h = dom f & for x st x in dom f holds h.x in f.x by Def5;
    hence thesis;
  end;
  thus thesis by Def5;
end;

theorem Th10:
  product {} = {{}}
proof
  thus product {} c= {{}}
  proof
    let x;
    assume x in product {};
    then ex g st
    x = g & dom g = dom {} & for y st y in dom {} holds g.y in {} .y by Def5;
    then x = {};
    hence thesis by TARSKI:def 1;
  end;
  let x;
  assume x in {{}};
  then
A1: x = {} by TARSKI:def 1;
  for y st y in dom {} holds {} .y in {} .y;
  hence thesis by A1,Th9;
end;

theorem Th11:
  Funcs(X,Y) = product (X --> Y)
proof
  set f = (X --> Y);
A1: dom f = X by FUNCOP_1:13;
  thus Funcs(X,Y) c= product f
  proof
    let x;
    assume x in Funcs(X,Y);
    then consider g such that
A2: x = g and
A3: dom g = X and
A4: rng g c= Y by FUNCT_2:def 2;
    now
      let y;
      assume
A5:   y in dom f;
      then
A6:   f.y = Y by FUNCOP_1:7;
      g.y in rng g by A3,A5,FUNCT_1:def 3;
      hence g.y in f.y by A4,A6;
    end;
    hence thesis by A1,A2,A3,Def5;
  end;
  let x;
  assume x in product f;
  then consider g such that
A7: x = g and
A8: dom g = dom f and
A9: for x st x in dom f holds g.x in f.x by Def5;
  rng g c= Y
  proof
    let y;
    assume y in rng g;
    then consider z such that
A10: z in dom g and
A11: y = g.z by FUNCT_1:def 3;
    y in f.z by A8,A9,A10,A11;
    hence thesis by A8,A10,FUNCOP_1:7;
  end;
  hence thesis by A1,A7,A8,FUNCT_2:def 2;
end;
defpred P[set] means $1 is Function;

definition
  let x,X;
  defpred R[set,set] means ex g st $1 = g & $2 = g.x;
  func pi(X,x) -> set means
  : Def6:
  y in it iff ex f st f in X & y = f.x;
  existence
  proof
    consider Y such that
A1: y in Y iff y in X & P[y] from XBOOLE_0:sch 1;
A2: for y st y in Y ex z st R[y,z]
    proof
      let y;
      assume y in Y;
      then reconsider y as Function by A1;
      take y.x, y;
      thus thesis;
    end;
    consider f such that
A3: dom f = Y & for y st y in Y holds R[y,f.y] from CLASSES1:sch 1(A2);
    take rng f;
    let y;
    thus y in rng f implies ex f st f in X & y = f.x
    proof
      assume y in rng f;
      then consider z such that
A4:   z in dom f and
A5:   y = f.z by FUNCT_1:def 3;
      consider g such that
A6:   z = g and
A7:   f.z = g.x by A3,A4;
      take g;
      thus thesis by A1,A3,A4,A5,A6,A7;
    end;
    given g such that
A8: g in X and
A9: y = g.x;
A10: g in Y by A1,A8;
    then ex f1 st g = f1 & f.g = f1.x by A3;
    hence thesis by A3,A9,A10,FUNCT_1:def 3;
  end;
  uniqueness
  proof
    defpred Z[set] means ex f st f in X & $1 = f.x;
    let X1,X2 be set such that
A11: y in X1 iff Z[y] and
A12: y in X2 iff Z[y];
    thus thesis from XBOOLE_0:sch 2(A11,A12);
  end;
end;

theorem
  x in dom f & product f <> {} implies pi(product f,x) = f.x
proof
  assume that
A1: x in dom f and
A2: product f <> {};
A3: pi(product f,x) c= f.x
  proof
    let y;
    assume y in pi(product f,x);
    then ex g st g in product f & y = g.x by Def6;
    hence thesis by A1,Th9;
  end;
  f.x c= pi(product f,x)
  proof set z = the Element of product f;
    consider g such that
    z = g and
A4: dom g = dom f and
A5: for x st x in dom f holds g.x in f.x by A2,Def5;
    let y;
    deffunc f(set) = y;
    deffunc g(set) = g.$1;
    defpred C[set] means x = $1;
    consider h being Function such that
A6: dom h = dom g & for z st z in dom g holds
    (C[z] implies h.z = f(z)) & (not C[z] implies h.z = g(z)) from
    PARTFUN1:sch 1;
    assume
A7: y in f.x;
    now
      let z;
      assume
A8:   z in dom f;
      then x <> z implies g.z = h.z by A4,A6;
      hence h.z in f.z by A4,A5,A6,A7,A8;
    end;
    then
A9: h in product f by A4,A6,Th9;
    h.x = y by A1,A4,A6;
    hence thesis by A9,Def6;
  end;
  hence thesis by A3,XBOOLE_0:def 10;
end;

theorem
  pi({},x) = {}
proof
  set y = the Element of pi({},x);
  assume not thesis;
  then ex f st f in {} & y = f.x by Def6;
  hence contradiction;
end;

theorem
  pi({g},x) = {g.x}
proof
  thus pi({g},x) c= {g.x}
  proof
    let y;
    assume y in pi({g},x);
    then consider f such that
A1: f in {g} and
A2: y = f.x by Def6;
    f = g by A1,TARSKI:def 1;
    hence thesis by A2,TARSKI:def 1;
  end;
  let y;
  assume
A3: y in {g.x};
A4: g in {g} by TARSKI:def 1;
  y = g.x by A3,TARSKI:def 1;
  hence thesis by A4,Def6;
end;

theorem
  pi({f,g},x) = {f.x,g.x}
proof
  thus pi({f,g},x) c= {f.x,g.x}
  proof
    let y;
    assume y in pi({f,g},x);
    then consider f1 such that
A1: f1 in {f,g} and
A2: y = f1.x by Def6;
    f1 = f or f1 = g by A1,TARSKI:def 2;
    hence thesis by A2,TARSKI:def 2;
  end;
  let y;
  assume
A3: y in {f.x,g.x};
A4: f in {f,g} by TARSKI:def 2;
A5: g in {f,g} by TARSKI:def 2;
  y = g.x or y = f.x by A3,TARSKI:def 2;
  hence thesis by A4,A5,Def6;
end;

theorem Th16:
  pi(X \/ Y,x) = pi(X,x) \/ pi(Y,x)
proof
  thus pi(X \/ Y,x) c= pi(X,x) \/ pi(Y,x)
  proof
    let y;
    assume y in pi(X \/ Y,x);
    then consider f such that
A1: f in X \/ Y and
A2: y = f.x by Def6;
    f in X or f in Y by A1,XBOOLE_0:def 3;
    then y in pi(X,x) or y in pi(Y,x) by A2,Def6;
    hence thesis by XBOOLE_0:def 3;
  end;
  let y;
  assume y in pi(X,x) \/ pi(Y,x);
  then
A3: y in pi(X,x) or y in pi(Y,x) by XBOOLE_0:def 3;
A4: now
    assume y in pi(X,x);
    then consider f such that
A5: f in X and
A6: y = f.x by Def6;
    f in X \/ Y by A5,XBOOLE_0:def 3;
    hence thesis by A6,Def6;
  end;
  now
    assume not y in pi(X,x);
    then consider f such that
A7: f in Y and
A8: y = f.x by A3,Def6;
    f in X \/ Y by A7,XBOOLE_0:def 3;
    hence thesis by A8,Def6;
  end;
  hence thesis by A4;
end;

theorem
  pi(X /\ Y,x) c= pi(X,x) /\ pi(Y,x)
proof
  let y;
  assume y in pi(X /\ Y,x);
  then consider f such that
A1: f in X /\ Y and
A2: y = f.x by Def6;
A3: f in X by A1,XBOOLE_0:def 4;
A4: f in Y by A1,XBOOLE_0:def 4;
A5: y in pi(X,x) by A2,A3,Def6;
  y in pi(Y,x) by A2,A4,Def6;
  hence thesis by A5,XBOOLE_0:def 4;
end;

theorem Th18:
  pi(X,x) \ pi(Y,x) c= pi(X \ Y,x)
proof
  let y;
  assume
A1: y in pi(X,x) \ pi(Y,x);
  then consider f such that
A2: f in X and
A3: y = f.x by Def6;
  not y in pi(Y,x) by A1,XBOOLE_0:def 5;
  then not f in Y by A3,Def6;
  then f in X \ Y by A2,XBOOLE_0:def 5;
  hence thesis by A3,Def6;
end;

theorem
  pi(X,x) \+\ pi(Y,x) c= pi(X \+\ Y,x)
proof
A1: pi(X,x) \ pi(Y,x) c= pi(X\Y,x) by Th18;
A2: pi(Y,x) \ pi(X,x) c= pi(Y\X,x) by Th18;
  pi(X\Y,x) \/ pi(Y\X,x) = pi((X\Y) \/ (Y\X),x) by Th16;
  hence thesis by A1,A2,XBOOLE_1:13;
end;

theorem Th20:
  card pi(X,x) c= card X
proof
  consider Y such that
A1: y in Y iff y in X & P[y] from XBOOLE_0:sch 1;
  defpred R[set,set] means ex g st $1 = g & $2 = g.x;
A2: for y st y in Y ex z st R[y,z]
  proof
    let y;
    assume y in Y;
    then reconsider y as Function by A1;
    take y.x, y;
    thus thesis;
  end;
  consider f such that
A3: dom f = Y & for y st y in Y holds R[y,f.y] from CLASSES1:sch 1(A2);
  now
    let y;
    thus y in rng f implies ex f st f in X & y = f.x
    proof
      assume y in rng f;
      then consider z such that
A4:   z in dom f and
A5:   y = f.z by FUNCT_1:def 3;
      consider g such that
A6:   z = g and
A7:   f.z = g.x by A3,A4;
      take g;
      thus thesis by A1,A3,A4,A5,A6,A7;
    end;
    given g such that
A8: g in X and
A9: y = g.x;
A10: g in Y by A1,A8;
    then ex f1 st g = f1 & f.g = f1.x by A3;
    hence y in rng f by A3,A9,A10,FUNCT_1:def 3;
  end;
  then rng f = pi(X,x) by Def6;
  then
A11: card pi(X,x) c= card Y by A3,CARD_1:12;
  Y c= X
  proof
    let x;
    thus thesis by A1;
  end;
  then card Y c= card X by CARD_1:11;
  hence thesis by A11,XBOOLE_1:1;
end;

theorem Th21:
  x in Union disjoin f implies ex y,z st x = [y,z]
proof
  assume x in Union disjoin f;
  then consider X such that
A1: x in X and
A2: X in rng disjoin f by TARSKI:def 4;
  consider y such that
A3: y in dom disjoin f and
A4: X = (disjoin f).y by A2,FUNCT_1:def 3;
  y in dom f by A3,Def3;
  then X = [:f.y,{y}:] by A4,Def3;
  hence thesis by A1,RELAT_1:def 1;
end;

theorem Th22:
  x in Union disjoin f iff x`2 in dom f & x`1 in f.(x`2) & x = [x`1,x`2]
proof
  thus x in Union disjoin f implies
  x`2 in dom f & x`1 in f.(x`2) & x = [x`1,x`2]
  proof
    assume x in Union disjoin f;
    then consider X such that
A1: x in X and
A2: X in rng disjoin f by TARSKI:def 4;
    consider y such that
A3: y in dom disjoin f and
A4: X = (disjoin f).y by A2,FUNCT_1:def 3;
A5: y in dom f by A3,Def3;
    then
A6: X = [:f.y,{y}:] by A4,Def3;
    then
A7: x`1 in f.y by A1,MCART_1:10;
    x`2 in {y} by A1,A6,MCART_1:10;
    hence thesis by A1,A5,A6,A7,MCART_1:21,TARSKI:def 1;
  end;
  assume that
A8: x`2 in dom f and
A9: x`1 in f.(x`2) and
A10: x = [x`1,x`2];
A11: (disjoin f).(x`2) = [:f.(x`2),{x`2}:] by A8,Def3;
A12: dom f = dom disjoin f by Def3;
  x`2 in {x`2} by TARSKI:def 1;
  then
A13: x in [:f.(x`2),{x`2}:] by A9,A10,ZFMISC_1:87;
  [:f.(x`2),{x`2}:] in rng disjoin f by A8,A11,A12,FUNCT_1:def 3;
  hence thesis by A13,TARSKI:def 4;
end;

theorem Th23:
  f c= g implies disjoin f c= disjoin g
proof
  assume
A1: f c= g;
  then
A2: dom f c= dom g by GRFUNC_1:2;
A3: dom disjoin f = dom f by Def3;
A4: dom disjoin g = dom g by Def3;
  now
    let x;
    assume
A5: x in dom disjoin f;
    then
A6: (disjoin f).x = [:f.x,{x}:] by A3,Def3;
    f.x = g.x by A1,A3,A5,GRFUNC_1:2;
    hence (disjoin f).x = (disjoin g).x by A2,A3,A5,A6,Def3;
  end;
  hence thesis by A2,A3,A4,GRFUNC_1:2;
end;

theorem Th24:
  f c= g implies Union f c= Union g
proof
  assume
A1: f c= g;
  then
A2: dom f c= dom g by GRFUNC_1:2;
  let x;
  assume x in Union f;
  then consider X such that
A3: x in X and
A4: X in rng f by TARSKI:def 4;
  consider y such that
A5: y in dom f and
A6: X = f.y by A4,FUNCT_1:def 3;
  f.y = g.y by A1,A5,GRFUNC_1:2;
  then X in rng g by A2,A5,A6,FUNCT_1:def 3;
  hence thesis by A3,TARSKI:def 4;
end;

theorem
  Union disjoin (Y --> X) = [:X,Y:]
proof
  set f = Y --> X;
A1: dom f = Y by FUNCOP_1:13;
  thus Union disjoin f c= [:X,Y:]
  proof
    let x;
    assume x in Union disjoin f;
    then consider Z such that
A2: x in Z and
A3: Z in rng disjoin f by TARSKI:def 4;
    consider y such that
A4: y in dom disjoin f and
A5: Z = (disjoin f).y by A3,FUNCT_1:def 3;
A6: y in Y by A1,A4,Def3;
    then
A7: Z = [:f.y,{y}:] by A1,A5,Def3;
A8: f.y = X by A6,FUNCOP_1:7;
    {y} c= Y by A6,ZFMISC_1:31;
    then Z c= [:X,Y:] by A7,A8,ZFMISC_1:95;
    hence thesis by A2;
  end;
  let x1,x2 be set;
  assume
A9: [x1,x2] in [:X,Y:];
  then
A10: x1 in X by ZFMISC_1:87;
A11: x2 in Y by A9,ZFMISC_1:87;
  then
A12: f.x2 = X by FUNCOP_1:7;
A13: (disjoin f).(x2) = [:f.x2,{x2}:] by A1,A11,Def3;
A14: x2 in dom disjoin f by A1,A11,Def3;
  x2 in {x2} by TARSKI:def 1;
  then
A15: [x1,x2] in [:f.x2,{x2}:] by A10,A12,ZFMISC_1:87;
  [:f.x2,{x2}:] in rng disjoin f by A13,A14,FUNCT_1:def 3;
  hence thesis by A15,TARSKI:def 4;
end;

theorem Th26:
  product f = {} iff {} in rng f
proof
  thus product f = {} implies {} in rng f
  proof
    assume that
A1: product f = {} and
A2: not {} in rng f;
A3: now
      assume dom f = {};
      then for x st x in dom f holds f.x in f.x;
      hence thesis by A1,Def5;
    end;
    now
      assume dom f <> {};
      then reconsider M = rng f as non empty set by RELAT_1:42;
      X in M implies X <> {} by A2;
      then consider f1 such that
      dom f1 = M and
A4:   for X st X in M holds f1.X in X by FUNCT_1:111;
      deffunc g(set) = f1.(f.$1);
      consider g such that
      A5:   dom
 g = dom f & for x st x in dom f holds g.x = g(x) from FUNCT_1:sch 3;
      now
        let x;
        assume
A6:     x in dom f;
        then
A7:     f.x in M by FUNCT_1:def 3;
        g.x = f1.(f.x) by A5,A6;
        hence g.x in f.x by A4,A7;
      end;
      hence thesis by A1,A5,Def5;
    end;
    hence thesis by A3;
  end;
  assume {} in rng f;
  then
A8: ex x st ( x in dom f)&( {} = f.x) by FUNCT_1:def 3;
  assume
A9: product f <> {};
  set y = the Element of product f;
  ex g st ( y = g)&( dom g = dom f)&( for x st x in dom f
  holds g.x in f.x) by A9,Def5;
  hence contradiction by A8;
end;

theorem Th27:
  dom f = dom g & (for x st x in dom f holds f.x c= g.x) implies
  product f c= product g
proof
  assume that
A1: dom f = dom g and
A2: for x st x in dom f holds f.x c= g.x;
  let x;
  assume x in product f;
  then consider f1 such that
A3: x = f1 and
A4: dom f1 = dom f and
A5: for x st x in dom f holds f1.x in f.x by Def5;
  now
    let x;
    assume
A6: x in dom g;
    then
A7: f1.x in f.x by A1,A5;
    f.x c= g.x by A1,A2,A6;
    hence f1.x in g.x by A7;
  end;
  hence thesis by A1,A3,A4,Def5;
end;

reserve F,G for Cardinal-Function;

theorem
  for x st x in dom F holds card (F.x) = F.x
proof
  let x;
  assume x in dom F;
  then reconsider M = F.x as Cardinal by Def1;
  card M = M by CARD_1:def 2;
  hence thesis;
end;

theorem Th29:
  for x st x in dom F holds card ((disjoin F).x) = F.x
proof
  let x;
  assume
A1: x in dom F;
  then reconsider M = F.x as Cardinal by Def1;
  M,[:M,{x}:] are_equipotent by CARD_1:69;
  then M = card [:M,{x}:] by CARD_1:def 2;
  hence thesis by A1,Def3;
end;

definition
  let F;
  func Sum F -> Cardinal equals
  card Union disjoin F;
  correctness;
  func Product F -> Cardinal equals
  card product F;
  correctness;
end;

theorem
  dom F = dom G & (for x st x in dom F holds F.x c= G.x) implies Sum F c= Sum
  G
proof
  assume that
A1: dom F = dom G and
A2: for x st x in dom F holds F.x c= G.x;
  Union disjoin F c= Union disjoin G
  proof
    let x;
    assume x in Union disjoin F;
    then consider X such that
A3: x in X and
A4: X in rng disjoin F by TARSKI:def 4;
    consider y such that
A5: y in dom disjoin F and
A6: X = (disjoin F).y by A4,FUNCT_1:def 3;
A7: y in dom F by A5,Def3;
    then
A8: F.y c= G.y by A2;
A9: X = [:F.y,{y}:] by A6,A7,Def3;
A10: (disjoin G).y = [:G.y,{y}:] by A1,A7,Def3;
A11: y in dom disjoin G by A1,A7,Def3;
A12: X c= [:G.y,{y}:] by A8,A9,ZFMISC_1:95;
    [:G.y,{y}:] in rng disjoin G by A10,A11,FUNCT_1:def 3;
    hence thesis by A3,A12,TARSKI:def 4;
  end;
  hence thesis by CARD_1:11;
end;

theorem
  {} in rng F iff Product F = 0
proof
  thus {} in rng F implies Product F = 0 by Th26,CARD_1:27;
  assume Product F = 0;
  then product F = {};
  hence thesis by Th26;
end;

theorem
  dom F = dom G & (for x st x in dom F holds F.x c= G.x) implies
  Product F c= Product G by Th27,CARD_1:11;

theorem
  F c= G implies Sum F c= Sum G
proof
  assume F c= G;
  then disjoin F c= disjoin G by Th23;
  hence thesis by Th24,CARD_1:11;
end;

theorem
  F c= G & not 0 in rng G implies Product F c= Product G
proof
  assume
A1: F c= G;
  then
A2: dom F c= dom G by GRFUNC_1:2;
  assume
A3: not 0 in rng G;
  deffunc f(Function) = $1|dom F;
  consider f such that
A4: dom f = product G & for g st g in product G holds f.g = f(g)
  from FUNCT_5:sch 1;
  product F c= rng f
  proof
    let x;
    assume x in product F;
    then consider g such that
A5: x = g and
A6: dom g = dom F and
A7: for x st x in dom F holds g.x in F.x by Def5;
A8: product G <> {} by A3,Th26;
    set y = the Element of product G;
    consider h such that
    y = h
    and dom h = dom G and
A9: for x st x in dom G holds h.x in G.x by A8,Def5;
    deffunc f(set) = g.$1;
    deffunc g(set) = h.$1;
    defpred C[set] means $1 in dom F;
    consider f1 such that
A10: dom f1 = dom G & for x st x in dom G holds
    (C[x] implies f1.x = f(x)) & (not C[x] implies f1.x = g(x))
    from PARTFUN1:sch 1;
    now
      let z such that
A11:  z in dom G;
A12:  now
        assume
A13:    z in dom F;
        then
A14:    f1.z = g.z by A10,A11;
        g.z in F.z by A7,A13;
        hence f1.z in G.z by A1,A13,A14,GRFUNC_1:2;
      end;
      not z in dom F implies f1.z = h.z by A10,A11;
      hence f1.z in G.z by A9,A11,A12;
    end;
    then
A15: f1 in product G by A10,Def5;
    then
A16: f.f1 = f1|dom F by A4;
A17: dom(f1|dom F) = dom F by A2,A10,RELAT_1:62;
    now
      let z;
      assume
A18:  z in dom F;
      then (f1|dom F).z = f1.z by A17,FUNCT_1:47;
      hence (f1|dom F).z = g.z by A2,A10,A18;
    end;
    then f.f1 = g by A6,A16,A17,FUNCT_1:2;
    hence thesis by A4,A5,A15,FUNCT_1:def 3;
  end;
  hence thesis by A4,CARD_1:12;
end;

theorem
  Sum({} --> K) = 0
proof dom ({} --> K) = {};
  then dom disjoin ({} --> K) = {} by Def3;
  hence thesis by CARD_1:27,RELAT_1:42,ZFMISC_1:2;
end;

theorem
  Product ({} --> K) = 1 by Th10,CARD_1:30;

theorem
  Sum(x .--> K) = K
proof
  thus Sum(x .--> K) = card Union (x .--> [:K,{x}:]) by Th4
    .= card [:K,{x}:] by Th7
    .= card K by CARD_1:69
    .= K by CARD_1:def 2;
end;

theorem
  Product(x .--> K) = K
proof
  thus Product(x .--> K) = card Funcs({x},K) by Th11
    .= card K by FUNCT_5:58
    .= K by CARD_1:def 2;
end;

theorem Th39:
  card Union f c= Sum Card f
proof
A1: now
    assume dom f = {};
    then {} = Union f by RELAT_1:42,ZFMISC_1:2;
    hence thesis;
  end;
  now
    assume
A2: dom f <> {};
    defpred P[set,set] means
    x in $2 iff x in Funcs(card $1,$1) & ex g st x = g & rng g = $1;
    defpred W[set,set] means P[f.$1,$2];
A3: for x st x in dom f ex y st W[x,y]
    proof
      let x such that x in dom f;
      defpred A[set] means ex g st $1 = g & rng g = f.x;
      consider Y such that
A4:   z in Y iff z in Funcs(card (f.x),f.x) & A[z] from XBOOLE_0:sch 1;
      take Y;
      thus thesis by A4;
    end;
    consider k being Function such that
A5: dom k = dom f & for x st x in dom f holds W[x,k.x] from CLASSES1:sch 1
    (A3);
    reconsider M = rng k as non empty set by A2,A5,RELAT_1:42;
    now
      let X;
      assume X in M;
      then consider x such that
A6:   x in dom k and
A7:   X = k.x by FUNCT_1:def 3;
      card(f.x),f.x are_equipotent by CARD_1:def 2;
      then consider g such that
      g is one-to-one and
A8:   dom g = card(f.x) and
A9:   rng g = f.x by WELLORD2:def 4;
      g in Funcs(card(f.x),f.x) by A8,A9,FUNCT_2:def 2;
      hence X <> {} by A5,A6,A7,A9;
    end;
    then consider FF being Function such that
    dom FF = M and
A10: for X st X in M holds FF.X in X by FUNCT_1:111;
    set DD = union rng disjoin Card f;
    defpred S[set,set] means ex g st g = FF.(k.$1`2) & $2 = g.$1`1;
A11: for x st x in DD ex y st S[x,y]
    proof
      let x;
      assume x in DD;
      then consider X such that
A12:  x in X and
A13:  X in rng disjoin Card f by TARSKI:def 4;
      consider y such that
A14:  y in dom disjoin Card f and
A15:  X = (disjoin Card f).y by A13,FUNCT_1:def 3;
A16:  dom disjoin Card f = dom Card f by Def3;
A17:  dom Card f = dom f by Def2;
      X = [:(Card f).y,{y}:] by A14,A15,A16,Def3;
      then x`2 in {y} by A12,MCART_1:10;
      then
A18:  x`2 in dom f by A14,A16,A17,TARSKI:def 1;
      then k.x`2 in M by A5,FUNCT_1:def 3;
      then FF.(k.x`2) in k.x`2 by A10;
      then FF.(k.x`2) in Funcs(card (f.x`2),f.x`2) by A5,A18;
      then consider g such that
A19:  FF.(k.x`2) = g
      and dom g = card (f.x`2)
      and rng g c= f.x`2 by FUNCT_2:def 2;
      take g.x`1, g;
      thus thesis by A19;
    end;
    consider t being Function such that
A20: dom t = DD & for x st x in DD holds S[x,t.x] from CLASSES1:sch 1
    (A11);
    union rng f c= rng t
    proof
      let x;
      assume x in union rng f;
      then consider X such that
A21:  x in X and
A22:  X in rng f by TARSKI:def 4;
      consider y such that
A23:  y in dom f and
A24:  X = f.y by A22,FUNCT_1:def 3;
      k.y in M by A5,A23,FUNCT_1:def 3;
      then
A25:  FF.(k.y) in k.y by A10;
      then FF.(k.y) in Funcs(card(f.y),f.y) by A5,A23;
      then consider g such that
A26:  FF.(k.y) = g and
A27:  dom g = card(f.y)
      and rng g c= f.y by FUNCT_2:def 2;
      ex g st FF.(k.y) = g & rng g = f.y by A5,A23,A25;
      then consider z such that
A28:  z in dom g and
A29:  x = g.z by A21,A24,A26,FUNCT_1:def 3;
A30:  (Card f).y = card(f.y) by A23,Def2;
A31:  dom Card f = dom f by Def2;
      then
A32:  (disjoin Card f).y = [:dom g,{y}:] by A23,A27,A30,Def3;
A33:  y in {y} by TARSKI:def 1;
A34:  dom disjoin Card f = dom f by A31,Def3;
A35:  [z,y] in [:dom g,{y}:] by A28,A33,ZFMISC_1:87;
      [:dom g,{y}:] in rng disjoin Card f by A23,A32,A34,FUNCT_1:def 3;
      then
A36:  [z,y] in DD by A35,TARSKI:def 4;
A37:  [z,y]`1 = z;
      [z,y]`2 = y;
      then ex g st g = FF.(k.y) & t.[z,y] = g.z by A20,A36,A37;
      hence thesis by A20,A26,A29,A36,FUNCT_1:def 3;
    end;
    hence thesis by A20,CARD_1:12;
  end;
  hence thesis by A1;
end;

theorem
  card Union F c= Sum F
proof Card F = F by Th1;
  hence thesis by Th39;
end;


theorem
  dom F = dom G & (for x st x in dom F holds F.x in G.x) implies
  Sum F in Product G
proof
  assume that
A1: dom F = dom G and
A2: for x st x in dom F holds F.x in G.x;
  deffunc f(set) = (G.$1)\(F.$1);
  consider f such that
A3: dom f = dom F & for x st x in dom F holds f.x = f(x) from FUNCT_1:sch 3;
  now
    assume {} in rng f;
    then consider x such that
A4: x in dom f and
A5: {} = f.x by FUNCT_1:def 3;
    reconsider Fx = F.x, Gx = G.x as Cardinal by A1,A3,A4,Def1;
A6: Fx in Gx by A2,A3,A4;
    not Fx in Fx;
    then Fx in Gx \ Fx by A6,XBOOLE_0:def 5;
    hence contradiction by A3,A4,A5;
  end;
  then
A7: product f <> {} by Th26;
  set a = the Element of product f;
  consider h being Function such that
  a = h
  and dom h = dom f and
A8: for x st x in dom f holds h.x in f.x by A7,Def5;
  defpred P[set,Function] means dom $2 = dom F & for x st x in dom F holds
  ($1`2 = x implies $2.x = $1`1) & ($1`2 <> x implies $2.x = h.x);
  defpred R[set,set] means ex g st $2 = g & P[$1,g];
A9: for x st x in Union disjoin F ex y st R[x,y]
  proof
    let x such that x in Union disjoin F;
    deffunc f(set) = x`1;
    deffunc g(set) = h.$1;
    defpred C[set] means $1 = x`2;
    consider g such that
A10: dom g = dom F & for u st u in dom F holds
    (C[u] implies g.u = f(u)) & (not C[u] implies g.u = g(u))
    from PARTFUN1:sch 1;
    reconsider y = g as set;
    take y,g;
    thus thesis by A10;
  end;
  consider f1 such that
A11: dom f1 = Union disjoin F & for x st x in Union disjoin F holds R[x,f1.x]
  from CLASSES1:sch 1(A9);
A12: f1 is one-to-one
  proof
    let x,y;
    assume that
A13: x in dom f1 and
A14: y in dom f1 and
A15: f1.x = f1.y and
A16: x <> y;
    consider g1 being Function such that
A17: f1.x = g1 and
A18: P[x,g1] by A11,A13;
    consider g2 being Function such that
A19: f1.y = g2 and
A20: P[y,g2] by A11,A14;
A21: x`2 in dom F by A11,A13,Th22;
A22: y`2 in dom F by A11,A14,Th22;
A23: y`1 in F.(y`2) by A11,A14,Th22;
A24: ex x1,x2 being set st x = [x1,x2] by A11,A13,Th21;
A25: ex x1,x2 being set st y = [x1,x2] by A11,A14,Th21;
A26: x = [x`1,x`2] by A24,MCART_1:8;
A27: y = [y`1,y`2] by A25,MCART_1:8;
A28: now
      assume
A29:  x`1 = y`1;
A30:  g2.(y`2) = y`1 by A20,A22;
A31:  g1.(y`2) = h.(y`2) by A16,A18,A22,A26,A27,A29;
A32:  f.(y`2) = (G.(y`2))\(F.(y`2)) by A3,A22;
      h.(y`2) in f.(y`2) by A3,A8,A22;
      hence contradiction by A15,A17,A19,A23,A30,A31,A32,XBOOLE_0:def 5;
    end;
A33: x`2 = y`2 implies g1.(x`2) = x`1 & g2.(x`2) = y`1 by A18,A20,A21;
A34: g1.(y`2) = y`1 by A15,A17,A19,A20,A22;
A35: g1.(y`2) = h.(y`2) by A15,A17,A18,A19,A22,A28,A33;
A36: f.(y`2) = (G.(y`2))\(F.(y`2) ) by A3,A22;
    h.(y`2) in f.(y`2) by A3,A8,A22;
    hence contradiction by A23,A34,A35,A36,XBOOLE_0:def 5;
  end;
  rng f1 c= product G
  proof
    let x;
    assume x in rng f1;
    then consider y such that
A37: y in dom f1 and
A38: x = f1.y by FUNCT_1:def 3;
    consider g such that
A39: f1.y = g and
A40: P[y,g] by A11,A37;
    now
      let x;
      assume
A41:  x in dom G;
      then reconsider Gx = G.x, Fx = F.x as Cardinal by A1,Def1;
A42:  Fx in Gx by A1,A2,A41;
A43:  y`2 = x implies g.x = y`1 by A1,A40,A41;
A44:  y`2 <> x implies g.x = h.x by A1,A40,A41;
A45:  h.x in f.x by A1,A3,A8,A41;
A46:  f.x = Gx \ Fx by A1,A3,A41;
A47:  y`1 in F.(y`2) by A11,A37,Th22;
      Fx c= Gx by A42,CARD_1:3;
      hence g.x in G.x by A43,A44,A45,A46,A47;
    end;
    hence thesis by A1,A38,A39,A40,Def5;
  end;
  then
A48: Sum F c= Product G by A11,A12,CARD_1:10;
  now
    assume Sum F = Product G;
    then Union disjoin F,product G are_equipotent by CARD_1:5;
    then consider f such that
    f is one-to-one and
A49: dom f = Union disjoin F and
A50: rng f = product G
    by WELLORD2:def 4;
    deffunc f(set) = G.$1 \ pi(f.:((disjoin F).$1),$1);
    consider K being Function such that
A51: dom K = dom F & for x st x in dom F holds K.x = f(x) from FUNCT_1:sch 3;
    now
      assume {} in rng K;
      then consider x such that
A52:  x in dom F and
A53:  {} = K.x by A51,FUNCT_1:def 3;
A54:  K.x = G.x \ pi(f.:((disjoin F).x),x) by A51,A52;
      reconsider Fx = F.x, Gx = G.x as Cardinal by A1,A52,Def1;
A55:  card pi(f.:((disjoin F).x),x) c= card (f.:((disjoin F).x)) by Th20;
A56:  card (f.:((disjoin F).x)) c= card ((disjoin F).x) by CARD_1:67;
      card ((disjoin F).x) = Fx by A52,Th29;
      then
A57:  card pi(f.:((disjoin F).x),x) c= Fx by A55,A56,XBOOLE_1:1;
A58:  Fx in Gx by A2,A52;
      card Gx = Gx by CARD_1:def 2;
      hence contradiction by A53,A54,A57,A58,CARD_1:68,ORDINAL1:12;
    end;
    then
A59: product K <> {} by Th26;
    set t = the Element of product K;
    consider g such that
A60: t = g
    and dom g = dom K and
A61: for x st x in dom K holds g.x in K.x by A59,Def5;
    now
      let x;
      assume x in dom K;
      then K.x = G.x \ pi(f.:((disjoin F).x),x) by A51;
      hence K.x c= G.x;
    end;
    then product K c= product G by A1,A51,Th27;
    then t in product G by A59,TARSKI:def 3;
    then consider y such that
A62: y in dom f and
A63: t = f.y by A50,FUNCT_1:def 3;
    consider X such that
A64: y in X and
A65: X in rng disjoin F by A49,A62,TARSKI:def 4;
    consider x such that
A66: x in dom disjoin F and
A67: X = (disjoin F).x by A65,FUNCT_1:def 3;
    g in f.:X by A60,A62,A63,A64,FUNCT_1:def 6;
    then
A68: g.x in pi(f.:((disjoin F).x),x) by A67,Def6;
A69: x in dom F by A66,Def3;
A70: not g.x in G.x \ pi(f.:((disjoin F).x),x) by A68,XBOOLE_0:def 5;
    g.x in (K.x) by A51,A61,A69;
    hence contradiction by A51,A69,A70;
  end;
  hence thesis by A48,CARD_1:3;
end;

scheme FuncSeparation { X()->set, F(set)->set, P[set,set] }:
  ex f st dom f = X() &
  for x st x in X() for y holds y in f.x iff y in F(x) & P[x,y]
proof
  defpred Q[set,set] means y in $2 iff y in F($1) & P[$1,y];
A1: for x st x in X() ex y st Q[x,y]
  proof
    let x such that x in X();
    defpred R[set] means P[x,$1];
    consider Y such that
A2: y in Y iff y in F(x) & R[y] from XBOOLE_0:sch 1;
    take Y;
    thus thesis by A2;
  end;
  thus ex f st dom f = X() &
  for x st x in X() holds Q[x,f.x] from CLASSES1:sch 1(A1);
end;

Lm1: x in field R implies ex y st [x,y] in R or [y,x] in R
proof
  assume x in field R;
  then x in dom R or x in rng R by XBOOLE_0:def 3;
  hence thesis by XTUPLE_0:def 12,def 13;
end;

theorem Th42:
  X is finite implies card X in card omega
proof
  assume X is finite;
  then ex n being Nat st ( card X = card n) by CARD_1:48;
  hence thesis by CARD_1:47;
end;

theorem Th43:
  card A in card B implies A in B
proof
  assume that
A1: card A in card B and
A2: not A in B;
  not card B c= card A by A1,CARD_1:4;
  hence contradiction by A2,CARD_1:11,ORDINAL1:16;
end;

theorem Th44:
  card A in M implies A in M
proof
  card M = M by CARD_1:def 2;
  hence thesis by Th43;
end;

theorem Th45:
  X is c=-linear implies
  ex Y st Y c= X & union Y = union X & for Z st Z c= Y & Z <> {}
  ex Z1 st Z1 in Z & for Z2 st Z2 in Z holds Z1 c= Z2
proof
  assume
A1: X is c=-linear;
  consider R such that
A2: R well_orders X by WELLORD2:17;
A3: R|_2 X is well-ordering by A2,WELLORD2:16;
A4: field(R|_2 X) = X by A2,WELLORD2:16;
  R|_2 X, RelIncl order_type_of R|_2 X are_isomorphic by A3,WELLORD2:def 2;
  then RelIncl order_type_of R|_2 X, R|_2 X are_isomorphic by WELLORD1:40;
  then consider f such that
A5: f is_isomorphism_of RelIncl order_type_of R|_2 X, R|_2 X
  by WELLORD1:def 8;
  field RelIncl order_type_of R|_2 X = order_type_of R|_2 X by WELLORD2:def 1;
  then
A6: dom f = order_type_of R|_2 X by A5,WELLORD1:def 7;
A7: rng f = X by A4,A5,WELLORD1:def 7;
  defpred P[set] means for A,B st B in A & $1 = A holds f.B c= f.A;
  consider Y such that
A8: x in Y iff x in dom f & P[x] from XBOOLE_0:sch 1;
  take Z = f.:Y;
  thus Z c= X by A7,RELAT_1:111;
  thus union Z c= union X by A7,RELAT_1:111,ZFMISC_1:77;
  thus union X c= union Z
  proof
    let x;
    assume x in union X;
    then consider Z1 such that
A9: x in Z1 and
A10: Z1 in X by TARSKI:def 4;
    consider y such that
A11: y in dom f and
A12: Z1 = f.y by A7,A10,FUNCT_1:def 3;
    reconsider y as Ordinal by A6,A11;
    defpred P[Ordinal] means $1 c= y & x in f.$1;
A13: ex A st P[A] by A9,A12;
    consider A such that
A14: P[A] & for B st P[B] holds A c= B from ORDINAL1:sch 1(A13);
A15: A in dom f by A6,A11,A14,ORDINAL1:12;
    now
      let B,C;
      assume that
A16:  C in B and
A17:  A = B;
A18:  C in dom f by A6,A11,A14,A16,A17,ORDINAL1:10;
A19:  not C c= y or not x in f.C by A14,A16,A17,ORDINAL1:5;
A20:  f.A in X by A7,A15,FUNCT_1:def 3;
      f.C in X by A7,A18,FUNCT_1:def 3;
      then f.C,f.A are_c=-comparable by A1,A20,ORDINAL1:def 8;
      then f.C c= f.A or f.A c= f.C by XBOOLE_0:def 9;
      hence f.C c= f.B by A14,A16,A17,A19,ORDINAL1:def 2;
    end;
    then A in Y by A8,A15;
    then f.A in Z by A15,FUNCT_1:def 6;
    hence thesis by A14,TARSKI:def 4;
  end;
  let V be set;
  assume that
A21: V c= Z and
A22: V <> {};
  set x = the Element of V;
  x in Z by A21,A22,TARSKI:def 3;
  then consider y such that
A23: y in dom f and
A24: y in Y and
A25: x = f.y by FUNCT_1:def 6;
  reconsider y as Ordinal by A6,A23;
  defpred P[Ordinal] means $1 in Y & f.$1 in V;
  y = y;
  then
A26: ex A st P[A] by A22,A24,A25;
  consider A such that
A27: P[A] & for B st P[B] holds A c= B from ORDINAL1:sch 1(A26);
  take Z1 = f.A;
  thus Z1 in V by A27;
  let Z2;
  assume
A28: Z2 in V;
  then consider y such that
A29: y in dom f and
A30: y in Y and
A31: Z2 = f.y by A21,FUNCT_1:def 6;
  reconsider y as Ordinal by A6,A29;
  A c< y iff A c= y & A <> y by XBOOLE_0:def 8;
  then A = y or A in y by A27,A28,A30,A31,ORDINAL1:11;
  hence thesis by A8,A30,A31;
end;

theorem
  (for Z st Z in X holds card Z in M) & X is c=-linear implies
  card union X c= M
proof
  assume that
A1: Z in X implies card Z in M and
A2: X is c=-linear;
  consider XX being set such that
A3: XX c= X and
A4: union XX = union X and
A5: for Z st Z c= XX & Z <> {}
  ex Z1 st Z1 in Z & for Z2 st Z2 in Z holds Z1 c= Z2 by A2,Th45;
A6: now
    let Z1,Z2;
    assume that
A7: Z1 in XX and
A8: Z2 in XX;
    Z1,Z2 are_c=-comparable by A2,A3,A7,A8,ORDINAL1:def 8;
    hence Z1 c= Z2 or Z2 c= Z1 by XBOOLE_0:def 9;
  end;
  consider R such that
A9: R well_orders union X by WELLORD2:17;
A10: R is_reflexive_in union X by A9,WELLORD1:def 5;
A11: R is_transitive_in union X by A9,WELLORD1:def 5;
A12: R is_antisymmetric_in union X by A9,WELLORD1:def 5;
A13: R is_connected_in union X by A9,WELLORD1:def 5;
  defpred P[set,set] means ((ex Z1 st Z1 in XX & $1 in Z1 & not $2 in Z1) or
  (for Z1 st Z1 in XX holds $1 in Z1 iff $2 in Z1) & [$1,$2] in R);
  consider Q such that
A14: [x,y] in Q iff x in union X & y in union X & P[x,y] from RELAT_1:sch 1;
A15: field Q = union X
  proof
    thus field Q c= union X
    proof
      let x;
      assume x in field Q;
      then ex y st [x,y] in Q or [y,x] in Q by Lm1;
      hence thesis by A14;
    end;
    let x;
    assume
A16: x in union X;
    then
A17: [x,x] in R by A10,RELAT_2:def 1;
    for Z1 st Z1 in XX holds x in Z1 iff x in Z1;
    then [x,x] in Q by A14,A16,A17;
    hence thesis by RELAT_1:15;
  end;
A18: Q is reflexive
  proof
    let x;
    assume
A19: x in field Q;
    then
A20: [x,x] in R by A10,A15,RELAT_2:def 1;
    for Z1 st Z1 in XX holds x in Z1 iff x in Z1;
    hence thesis by A14,A15,A19,A20;
  end;
A21: Q is transitive
  proof
    let x,y,z such that
A22: x in field Q and
A23: y in field Q and
A24: z in field Q and
A25: [x,y] in Q and
A26: [y,z] in Q;
A27: now
      given Z1 such that
A28:  Z1 in XX and
A29:  x in Z1 and
A30:  not y in Z1;
      given Z2 such that
A31:  Z2 in XX and
A32:  y in Z2 and
A33:  not z in Z2;
      Z1 c= Z2 or Z2 c= Z1 by A6,A28,A31;
      hence thesis by A14,A15,A22,A24,A29,A30,A31,A32,A33;
    end;
A34: now
      given Z1 such that
A35:  Z1 in XX and
A36:  x in Z1 and
A37:  not y in Z1;
      assume that
A38:  for Z1 st Z1 in XX holds y in Z1 iff z in Z1
      and [y,z] in R;
      not z in Z1 by A35,A37,A38;
      hence thesis by A14,A15,A22,A24,A35,A36;
    end;
A39: now
      given Z1 such that
A40:  Z1 in XX and
A41:  y in Z1 and
A42:  not z in Z1;
      assume that
A43:  for Z1 st Z1 in XX holds x in Z1 iff y in Z1
      and [x,y] in R;
      x in Z1 by A40,A41,A43;
      hence thesis by A14,A15,A22,A24,A40,A42;
    end;
    now
      assume that
A44:  for Z1 st Z1 in XX holds x in Z1 iff y in Z1 and
A45:  [x,y] in R and
A46:  for Z1 st Z1 in XX holds y in Z1 iff z in Z1 and
A47:  [y,z] in R;
A48:  [x,z] in R by A11,A15,A22,A23,A24,A45,A47,RELAT_2:def 8;
      now
        let Z;
        assume
A49:    Z in XX;
        then x in Z iff y in Z by A44;
        hence x in Z iff z in Z by A46,A49;
      end;
      hence thesis by A14,A15,A22,A24,A48;
    end;
    hence thesis by A14,A25,A26,A27,A34,A39;
  end;
A50: Q is antisymmetric
  proof
    let x,y;
    assume that
A51: x in field Q and
A52: y in field Q and
A53: [x,y] in Q and
A54: [y,x] in Q;
A55: (ex Z1 st Z1 in XX & x in Z1 & not y in Z1) or (for Z1 st Z1 in XX
    holds x in Z1 iff y in Z1) & [x,y] in R by A14,A53;
A56: (ex Z1 st Z1 in XX & y in Z1 & not x in Z1) or (for Z1 st Z1 in XX
    holds y in Z1 iff x in Z1) & [y,x] in R by A14,A54;
    now
      given Z1 such that
A57:  Z1 in XX and
A58:  x in Z1 and
A59:  not y in Z1;
      given Z2 such that
A60:  Z2 in XX and
A61:  y in Z2 and
A62:  not x in Z2;
      Z1 c= Z2 or Z2 c= Z1 by A6,A57,A60;
      hence thesis by A58,A59,A61,A62;
    end;
    hence thesis by A12,A15,A51,A52,A55,A56,RELAT_2:def 4;
  end;
A63: Q is connected
  proof
    let x,y such that
A64: x in field Q and
A65: y in field Q and
A66: x <> y;
    now
      assume
A67:  for Z st Z in XX holds x in Z iff y in Z;
A68:  [x,y] in R or [y,x] in R by A13,A15,A64,A65,A66,RELAT_2:def 6;
      for Z st Z in XX holds y in Z iff x in Z by A67;
      hence thesis by A14,A15,A64,A65,A67,A68;
    end;
    hence thesis by A14,A15,A64,A65;
  end;
  Q is well_founded
  proof
    let Y such that
A69: Y c= field Q and
A70: Y <> {};
    defpred P[set] means $1 /\ Y <> {};
    consider Z such that
A71: x in Z iff x in XX & P[x] from XBOOLE_0:sch 1;
A72: Z c= XX
    proof
      let x;
      thus thesis by A71;
    end;
    set x = the Element of Y;
    x in union XX by A4,A15,A69,A70,TARSKI:def 3;
    then consider Z1 such that
A73: x in Z1 and
A74: Z1 in XX by TARSKI:def 4;
    Z1 /\ Y <> {} by A70,A73,XBOOLE_0:def 4;
    then Z <> {} by A71,A74;
    then consider Z1 such that
A75: Z1 in Z and
A76: for Z2 st Z2 in Z holds Z1 c= Z2 by A5,A72;
A77: Z1 in XX by A71,A75;
A78: Z1 /\ Y c= Z1 by XBOOLE_1:17;
A79: Z1 c= union X by A3,A77,ZFMISC_1:74;
    Z1 /\ Y <> {} by A71,A75;
    then consider x such that
A80: x in Z1 /\ Y and
A81: for y st y in Z1 /\ Y holds [x,y] in R
    by A9,A78,A79,WELLORD1:5,XBOOLE_1:1;
    take x;
    thus x in Y by A80,XBOOLE_0:def 4;
    assume
A82: Q-Seg x /\ Y <> {};
    set y = the Element of Q-Seg x /\ Y;
A83: x in Z1 by A80,XBOOLE_0:def 4;
A84: y in Q-Seg x by A82,XBOOLE_0:def 4;
A85: y in Y by A82,XBOOLE_0:def 4;
A86: y <> x by A84,WELLORD1:1;
A87: [y,x] in Q by A84,WELLORD1:1;
A88: now
      given Z2 such that
A89:  Z2 in XX and
A90:  y in Z2 and
A91:  not x in Z2;
      Z2 /\ Y <> {} by A85,A90,XBOOLE_0:def 4;
      then Z2 in Z by A71,A89;
      then Z1 c= Z2 by A76;
      hence contradiction by A83,A91;
    end;
    then
A92: y in Z1 by A14,A77,A83,A87;
A93: [y,x] in R by A14,A87,A88;
    y in Z1 /\ Y by A85,A92,XBOOLE_0:def 4;
    then
A94: [x,y] in R by A81;
A95: x in union X by A14,A87;
    y in union X by A14,A87;
    hence contradiction by A12,A86,A93,A94,A95,RELAT_2:def 4;
  end;
  then Q is well-ordering by A18,A21,A50,A63,WELLORD1:def 4;
  then Q,RelIncl order_type_of Q are_isomorphic by WELLORD2:def 2;
  then RelIncl order_type_of Q,Q are_isomorphic by WELLORD1:40;
  then consider g such that
A96: g is_isomorphism_of RelIncl order_type_of Q,Q by WELLORD1:def 8;
A97: field RelIncl order_type_of Q = order_type_of Q by WELLORD2:def 1;
  then
A98: dom g = order_type_of Q by A96,WELLORD1:def 7;
A99: rng g = union X by A15,A96,WELLORD1:def 7;
A100: g is one-to-one by A96,WELLORD1:def 7;
A101: for Z,x st Z in XX & x in Z holds Q-Seg x c= Z
  proof
    let Z,x such that
A102: Z in XX and
A103: x in Z;
    let y;
    assume y in Q-Seg x;
    then
A104: [y,x] in Q by WELLORD1:1;
    now
      given Z1 such that
A105: Z1 in XX and
A106: y in Z1 and
A107: not x in Z1;
      Z1 c= Z or Z c= Z1 by A6,A102,A105;
      hence thesis by A103,A106,A107;
    end;
    hence thesis by A14,A102,A103,A104;
  end;
A108: for A st A in order_type_of Q holds card A = card (Q-Seg(g.A))
  proof
    let A such that
A109: A in order_type_of Q;
    A,Q-Seg(g.A) are_equipotent
    proof
      take f = g|A;
      A c= dom g by A98,A109,ORDINAL1:def 2;
      hence
A110: f is one-to-one & dom f = A by A100,FUNCT_1:52,RELAT_1:62;
      thus rng f c= Q-Seg(g.A)
      proof
        let x;
        assume x in rng f;
        then consider y such that
A111:   y in dom f and
A112:   x = f.y by FUNCT_1:def 3;
        reconsider B = y as Ordinal by A110,A111;
A113:   B in order_type_of Q by A109,A110,A111,ORDINAL1:10;
        B c= A by A110,A111,ORDINAL1:def 2;
        then
A114:   [B,A] in RelIncl order_type_of Q by A109,A113,WELLORD2:def 1;
A115:   x = g.B by A111,A112,FUNCT_1:47;
A116:   A <> B by A110,A111;
A117:   [x,g.A] in Q by A96,A114,A115,WELLORD1:def 7;
        x <> g.A by A98,A100,A109,A113,A115,A116,FUNCT_1:def 4;
        hence thesis by A117,WELLORD1:1;
      end;
      let x;
      assume
A118: x in Q-Seg(g.A);
      then
A119: [x,g.A] in Q by WELLORD1:1;
      then x in union X by A14;
      then consider y such that
A120: y in dom g and
A121: x = g.y by A99,FUNCT_1:def 3;
      reconsider B = y as Ordinal by A98,A120;
      [B,A] in RelIncl order_type_of Q by A96,A97,A98,A109,A119,A120,A121,
WELLORD1:def 7;
      then
A122: B c= A by A98,A109,A120,WELLORD2:def 1;
      B <> A by A118,A121,WELLORD1:1;
      then B c< A by A122,XBOOLE_0:def 8;
      hence thesis by A120,A121,FUNCT_1:50,ORDINAL1:11;
    end;
    hence thesis by CARD_1:5;
  end;
A123: order_type_of Q c= M
  proof
    let x be Ordinal;
    assume
A124: x in order_type_of Q;
    reconsider A = x as Ordinal;
    g.x in union X by A98,A99,A124,FUNCT_1:def 3;
    then consider Z such that
A125: g.x in Z and
A126: Z in XX by A4,TARSKI:def 4;
A127: card (Q-Seg(g.x)) c= card Z by A101,A125,A126,CARD_1:11;
A128: card (Q-Seg(g.x)) = card A by A108,A124;
    card (Q-Seg(g.x)) in M by A1,A3,A126,A127,ORDINAL1:12;
    hence thesis by A128,Th44;
  end;
  order_type_of Q,union X are_equipotent by A98,A99,A100,WELLORD2:def 4;
  then
A129: card order_type_of Q = card union X by CARD_1:5;
  card M = M by CARD_1:def 2;
  hence thesis by A123,A129,CARD_1:11;
end;

begin 



registration
  let f be Function;
  cluster product f -> functional;
  coherence
  proof
    set d = product f;
    let x be set;
    assume x in d;
    then ex g being Function
    st x = g & dom g= dom f & for x being set st x in dom f holds g.x in f.x
    by Def5;
    hence thesis;
  end;
end;

registration
  let A be set;
  let B be with_non-empty_elements set;
  cluster -> non-empty for Function of A,B;
  coherence
  proof
    let f be Function of A,B;
    thus not {} in rng f;
  end;
end;



registration
  let f be non-empty Function;
  cluster product f -> non empty;
  coherence
  proof
    not {} in rng f;
    hence thesis by Th26;
  end;
end;



theorem
  for a,b,c,d being set st a <> b holds
  product (a,b) --> ({c},{d}) = { (a,b) --> (c,d) }
proof
  let a,b,c,d be set such that
A1: a <> b;
  set X = { (a,b) --> (c,d) }, f = (a,b) --> ({c},{d});
A2: dom f = {a,b} by FUNCT_4:62;
  now
    let x;
    thus x in X implies ex g being Function st x = g & dom g = dom f &
    for x st x in dom f holds g.x in f.x
    proof
      assume
A3:   x in X;
      take g = (a,b) --> (c,d);
      thus x = g by A3,TARSKI:def 1;
      thus dom g = dom f by A2,FUNCT_4:62;
      let x;
      assume x in dom f;
      then
A4:   x = a or x = b by A2,TARSKI:def 2;
A5:   g.a = c by A1,FUNCT_4:63;
A6:   f.a = {c} by A1,FUNCT_4:63;
A7:   g.b = d by FUNCT_4:63;
      f.b = {d} by FUNCT_4:63;
      hence thesis by A4,A5,A6,A7,TARSKI:def 1;
    end;
    given g being Function such that
A8: x = g and
A9: dom g = dom f and
A10: for x st x in dom f holds g.x in f.x;
A11: a in dom f by A2,TARSKI:def 2;
A12: b in dom f by A2,TARSKI:def 2;
A13: g.a in f.a by A10,A11;
A14: g.b in f.b by A10,A12;
A15: f.a = {c} by A1,FUNCT_4:63;
A16: f.b = {d} by FUNCT_4:63;
A17: g.a = c by A13,A15,TARSKI:def 1;
    g.b = d by A14,A16,TARSKI:def 1;
    then g = (a,b) --> (c,d) by A2,A9,A17,FUNCT_4:66;
    hence x in X by A8,TARSKI:def 1;
  end;
  hence thesis by Def5;
end;



theorem
  x in product f implies x is Function;

begin 

reserve A,B for set;

definition
  let f be Function;
  func sproduct f -> set means
  :Def9:
  x in it iff ex g st x = g & dom g c= dom f &
  for x st x in dom g holds g.x in f.x;
  existence
  proof
    defpred P[set] means ex g st $1 = g & dom g c= dom f &
    for x st x in dom g holds g.x in f.x;
    consider A being set such that
A1: x in A iff x in PFuncs(dom f, union rng f) & P[x] from XBOOLE_0:sch 1;
    now
      let x;
      thus x in A implies ex g st x = g & dom g c= dom f &
      for x st x in dom g holds g.x in f.x by A1;
      given g such that
A2:   x = g and
A3:   dom g c= dom f and
A4:   for x st x in dom g holds g.x in f.x;
      rng g c= union rng f
      proof
        let y be set;
        assume y in rng g;
        then consider z being set such that
A5:     z in dom g and
A6:     y = g.z by FUNCT_1:def 3;
A7:     g.z in f.z by A4,A5;
        f.z in rng f by A3,A5,FUNCT_1:def 3;
        hence thesis by A6,A7,TARSKI:def 4;
      end;
      then x in PFuncs(dom f, union rng f) by A2,A3,PARTFUN1:def 3;
      hence x in A by A1,A2,A3,A4;
    end;
    hence thesis;
  end;
  uniqueness
  proof
    defpred P[set] means ex g st $1 = g & dom g c= dom f &
    for x st x in dom g holds g.x in f.x;
    let A,B be set such that
A8: x in A iff P[x] and
A9: x in B iff P[x];
    thus thesis from XBOOLE_0:sch 2(A8,A9);
  end;
end;

registration
  let f be Function;
  cluster sproduct f -> functional non empty;
  coherence
  proof
    defpred P[set] means ex g st $1 = g & dom g c= dom f &
    for x st x in dom g holds g.x in f.x;
    consider A being set such that
A1: x in A iff x in PFuncs(dom f, union rng f) & P[x] from XBOOLE_0:sch 1;
    {} is PartFunc of dom f, union rng f by RELSET_1:12;
    then
A2: {} in PFuncs(dom f, union rng f) by PARTFUN1:45;
A3: dom {} c= dom f by XBOOLE_1:2;
A4: for x st x in dom {} holds {} .x in f.x;
    now
      let x be set;
      assume x in A;
      then ex g st x = g & dom g c= dom f &
      for x st x in dom g holds g.x in f.x by A1;
      hence x is Function;
    end;
    then reconsider A as functional non empty set by A1,A2,A3,A4,FUNCT_1:def 13
;
    now
      let x;
      thus x in A implies ex g st x = g & dom g c= dom f &
      for x st x in dom g holds g.x in f.x by A1;
      given g such that
A5:   x = g and
A6:   dom g c= dom f and
A7:   for x st x in dom g holds g.x in f.x;
      rng g c= union rng f
      proof
        let y be set;
        assume y in rng g;
        then consider z being set such that
A8:     z in dom g and
A9:     y = g.z by FUNCT_1:def 3;
A10:    g.z in f.z by A7,A8;
        f.z in rng f by A6,A8,FUNCT_1:def 3;
        hence thesis by A9,A10,TARSKI:def 4;
      end;
      then x in PFuncs(dom f, union rng f) by A5,A6,PARTFUN1:def 3;
      hence x in A by A1,A5,A6,A7;
    end;
    hence thesis by Def9;
  end;
end;

theorem Th49:
  g in sproduct f implies dom g c= dom f &
  for x st x in dom g holds g.x in f.x
proof
  assume g in sproduct f;
  then ex h st g = h & dom h c= dom f &
  for x st x in dom h holds h.x in f.x by Def9;
  hence thesis;
end;

theorem Th50:
  {} in sproduct f
proof
A1: dom {} c= dom f by XBOOLE_1:2;
  for x st x in dom {} holds {} .x in f.x;
  hence thesis by A1,Def9;
end;

registration let f;
 cluster empty for Element of sproduct f;
 existence
  proof
    {} in sproduct f by Th50;
   hence thesis;
  end;
end;

theorem Th51:
  product f c= sproduct f
proof
  let x;
  assume x in product f;
  then ex g st x = g & dom g = dom f &
  for x st x in dom f holds g.x in f.x by Def5;
  hence thesis by Def9;
end;

theorem Th52:
  x in sproduct f implies x is PartFunc of dom f, union rng f
proof
  assume x in sproduct f;
  then consider g such that
A1: x = g and
A2: dom g c= dom f and
A3: for x st x in dom g holds g.x in f.x by Def9;
  rng g c= union rng f
  proof
    let y be set;
    assume y in rng g;
    then consider z being set such that
A4: z in dom g and
A5: y = g.z by FUNCT_1:def 3;
A6: g.z in f.z by A3,A4;
    f.z in rng f by A2,A4,FUNCT_1:def 3;
    hence thesis by A5,A6,TARSKI:def 4;
  end;
  hence thesis by A1,A2,RELSET_1:4;
end;

theorem Th53:
  g in product f & h in sproduct f implies g +* h in product f
proof
  assume
A1: g in product f;
  then
A2: dom g = dom f by Th9;
  assume
A3: h in sproduct f;
  then
A4: dom g \/ dom h = dom f by A2,Th49,XBOOLE_1:12;
  then
A5: dom(g +* h) = dom f by FUNCT_4:def 1;
  now
    let x;
    assume
A6: x in dom f;
A7: (dom g \ dom h) \/ dom h = dom f by A4,XBOOLE_1:39;
    now per cases by A6,A7,XBOOLE_0:def 3;
      case
A8:     x in dom g \ dom h;
        then not x in dom h by XBOOLE_0:def 5;
        hence x in dom f & (g +* h).x = g.x by A2,A8,FUNCT_4:11;
      end;
      case x in dom h;
        hence (g +* h).x = h.x by FUNCT_4:13;
      end;
    end;
    hence (g +* h).x in f.x by A1,A3,Th9,Th49;
  end;
  hence thesis by A5,Th9;
end;

theorem
  product f <> {} implies
  (g in sproduct f iff ex h st h in product f & g c= h)
proof
  assume
A1: product f <> {};
  thus g in sproduct f implies ex h st h in product f & g c= h
  proof
    assume
A2: g in sproduct f;
    set k = the Element of product f;
    reconsider k as Function;
    take k +* g;
    thus k +* g in product f by A1,A2,Th53;
    thus thesis by FUNCT_4:25;
  end;
  given h such that
A3: h in product f and
A4: g c= h;
A5: dom h = dom f by A3,Th9;
A6: dom g c= dom h by A4,RELAT_1:11;
  now
    let x;
    assume
A7: x in dom g;
    then g.x = h.x by A4,GRFUNC_1:2;
    hence g.x in f.x by A3,A5,A6,A7,Th9;
  end;
  hence thesis by A5,A6,Def9;
end;

theorem Th55:
  sproduct f c= PFuncs(dom f,union rng f)
proof
  let x;
  assume x in sproduct f;
  then x is PartFunc of dom f, union rng f by Th52;
  hence thesis by PARTFUN1:45;
end;

theorem Th56:
  f c= g implies sproduct f c= sproduct g
proof
  assume
A1: f c= g;
  then
A2: dom f c= dom g by GRFUNC_1:2;
  let y;
  assume y in sproduct f;
  then consider h such that
A3: y = h and
A4: dom h c= dom f and
A5: for x st x in dom h holds h.x in f.x by Def9;
A6: dom h c= dom g by A2,A4,XBOOLE_1:1;
  now
    let x;
    assume
A7: x in dom h;
    then f.x = g.x by A1,A4,GRFUNC_1:2;
    hence h.x in g.x by A5,A7;
  end;
  hence thesis by A3,A6,Def9;
end;

theorem Th57:
  sproduct {} = {{}}
proof
  sproduct {} c= PFuncs({},{}) by Th55,RELAT_1:38,ZFMISC_1:2;
  hence sproduct {} c= {{}} by PARTFUN1:48;
  let x be set;
  assume x in {{}};
  then x = {} by TARSKI:def 1;
  hence thesis by Th50;
end;

theorem
  PFuncs(A,B) = sproduct (A --> B)
proof
  now per cases;
    case
A1:   A = {};
      then A --> B = {} --> B;
      hence thesis by A1,Th57,PARTFUN1:48;
    end;
    case A <> {};
      then
A2:   rng (A --> B) = { B } by FUNCOP_1:8;
A3:   dom(A --> B) = A by FUNCOP_1:13;
A4:   B = union rng (A --> B) by A2,ZFMISC_1:25;
      thus PFuncs(A,B) c= sproduct (A --> B)
      proof
        let x;
        assume x in PFuncs(A,B);
        then consider f being Function such that
A5:     x = f and
A6:     dom f c= A and
A7:     rng f c= B by PARTFUN1:def 3;
A8:     dom f c= dom (A --> B) by A6,FUNCOP_1:13;
        now
          let x;
          assume
A9:       x in dom f;
          then f.x in rng f by FUNCT_1:def 3;
          then f.x in B by A7;
          hence f.x in (A --> B).x by A6,A9,FUNCOP_1:7;
        end;
        hence thesis by A5,A8,Def9;
      end;
      thus sproduct (A --> B) c= PFuncs(A,B) by A3,A4,Th55;
    end;
  end;
  hence thesis by XBOOLE_0:def 10;
end;

theorem
  for A, B being non empty set for f being Function of A,B
  holds sproduct f = sproduct(f|{x where x is Element of A: f.x <> {} })
proof
  let A, B be non empty set;
  let f be Function of A,B;
  set X = {x where x is Element of A: f.x <> {} };
  thus sproduct f c= sproduct(f|X)
  proof
    let x;
    assume x in sproduct f;
    then consider g such that
A1: x = g and
A2: dom g c= dom f and
A3: for x st x in dom g holds g.x in f.x by Def9;
A4: now
      let x;
      assume
A5:   x in dom g;
      then reconsider a = x as Element of A by A2,FUNCT_2:def 1;
      f.a <> {} by A3,A5;
      hence x in X;
    end;
A6: now
      let x;
      assume
A7:   x in dom g;
      then x in X by A4;
      hence x in (dom f)/\ X by A2,A7,XBOOLE_0:def 4;
    end;
A8: dom g c= dom(f|X)
    proof
      let x;
      assume x in dom g;
      then x in (dom f)/\ X by A6;
      hence thesis by RELAT_1:61;
    end;
    now
      let x;
      assume
A9:   x in dom g;
      then g.x in f.x by A3;
      hence g.x in (f|X).x by A6,A9,FUNCT_1:48;
    end;
    hence thesis by A1,A8,Def9;
  end;
  thus thesis by Th56,RELAT_1:59;
end;

theorem Th60:
  x in dom f & y in f.x implies x .--> y in sproduct f
proof
  assume that
A1: x in dom f and
A2: y in f.x;
 dom(x .--> y) = {x} by FUNCOP_1:13;
  then
A3: dom(x .--> y) c= dom f by A1,ZFMISC_1:31;
  now
    let z;
    assume z in dom(x .--> y);
    then z = x by TARSKI:def 1;
    hence (x .--> y).z in f.z by A2,FUNCOP_1:72;
  end;
  hence thesis by A3,Def9;
end;

theorem
  sproduct f = {{}} iff for x st x in dom f holds f.x = {}
proof
  thus sproduct f = {{}} implies for x st x in dom f holds f.x = {}
  proof
    assume
A1: sproduct f = {{}};
    let x;
    assume
A2: x in dom f;
    assume
A3: f.x <> {};
    set y = the Element of f.x;
    x .--> y in sproduct f by A2,A3,Th60;
    hence contradiction by A1,TARSKI:def 1;
  end;
  assume
A4: for x st x in dom f holds f.x = {};
  now
    let x;
    thus x in sproduct f implies x = {}
    proof
      assume x in sproduct f;
      then consider g such that
A5:   x = g and
A6:   dom g c= dom f and
A7:   for y st y in dom g holds g.y in f.y by Def9;
      assume
A8:   x <> {};
      set y = the Element of dom g;
A9:   f.y <> {} by A5,A7,A8;
      y in dom f by A5,A6,A8,TARSKI:def 3;
      hence contradiction by A4,A9;
    end;
    thus x = {} implies x in sproduct f by Th50;
  end;
  hence thesis by TARSKI:def 1;
end;

theorem Th62:
  A c= sproduct f &
  (for h1,h2 being Function st h1 in A & h2 in A holds h1 tolerates h2)
  implies union A in sproduct f
proof
  assume that
A1: A c= sproduct f and
A2: for h1,h2 being Function st h1 in A & h2 in A holds h1 tolerates h2;
  reconsider g = union A as Function by A1,A2,PARTFUN1:78;
A3: dom g c= dom f
  proof
    let x;
    assume x in dom g;
    then consider y such that
A4: [x,y] in g by XTUPLE_0:def 12;
    consider h being set such that
A5: [x,y] in h and
A6: h in A by A4,TARSKI:def 4;
    reconsider h as Function by A1,A6;
A7: x in dom h by A5,XTUPLE_0:def 12;
    dom h c= dom f by A1,A6,Th49;
    hence thesis by A7;
  end;
  now
    let x;
    assume x in dom g;
    then consider y such that
A8: [x,y] in g by XTUPLE_0:def 12;
    consider h being set such that
A9: [x,y] in h and
A10: h in A by A8,TARSKI:def 4;
    reconsider h as Function by A1,A10;
A11: x in dom h by A9,XTUPLE_0:def 12;
    h.x = y by A9,FUNCT_1:1
      .= g.x by A8,FUNCT_1:1;
    hence g.x in f.x by A1,A10,A11,Th49;
  end;
  hence thesis by A3,Def9;
end;

theorem
  g tolerates h & g in sproduct f & h in sproduct f
  implies g \/ h in sproduct f
proof
  assume that
A1: g tolerates h and
A2: g in sproduct f and
A3: h in sproduct f;
A4: {g,h} c= sproduct f by A2,A3,ZFMISC_1:32;
A5: now
    let h1,h2 be Function;
    assume that
A6: h1 in {g,h} and
A7: h2 in {g,h};
A8: h1 = g or h1 = h by A6,TARSKI:def 2;
    h2 = g or h2 = h by A7,TARSKI:def 2;
    hence h1 tolerates h2 by A1,A8;
  end;
  union {g,h} = g \/ h by ZFMISC_1:75;
  hence thesis by A4,A5,Th62;
end;

theorem Th64:
  x c= h & h in sproduct f implies x in sproduct f
proof
  assume that
A1: x c= h and
A2: h in sproduct f;
  reconsider g = x as Function by A1;
A3: dom g c= dom h by A1,GRFUNC_1:2;
  dom h c= dom f by A2,Th49;
  then
A4: dom g c= dom f by A3,XBOOLE_1:1;
  now
    let x;
    assume
A5: x in dom g;
    then h.x in f.x by A2,A3,Th49;
    hence g.x in f.x by A1,A5,GRFUNC_1:2;
  end;
  hence thesis by A4,Def9;
end;

theorem Th65:
  g in sproduct f implies g|A in sproduct f by Th64,RELAT_1:59;

theorem Th66:
  g in sproduct f implies g|A in sproduct f|A
proof
A1: dom(g|A) = dom g /\ A by RELAT_1:61;
A2: dom(f|A) = dom f /\ A by RELAT_1:61;
  assume
A3: g in sproduct f;
  then
A4: dom(g|A) c= dom(f|A) by A1,A2,Th49,XBOOLE_1:26;
  now
    let x;
    assume
A5: x in dom(g|A);
    then
A6: (g|A).x = g.x by FUNCT_1:47;
A7: (f|A).x = f.x by A4,A5,FUNCT_1:47;
    x in dom g by A1,A5,XBOOLE_0:def 4;
    hence (g|A).x in (f|A).x by A3,A6,A7,Th49;
  end;
  hence thesis by A4,Def9;
end;

theorem
  h in sproduct(f+*g) implies
  ex f9,g9 being Function st f9 in sproduct f & g9 in sproduct g & h = f9+*g9
proof
  assume
A1: h in sproduct(f+*g);
  take h|(dom f \ dom g), h|dom g;
A2: h|(dom f \ dom g) in sproduct (f +* g)|(dom f \ dom g) by A1,Th66;
  sproduct (f +* g)|(dom f \ dom g) c= sproduct f by Th56,FUNCT_4:24;
  hence h|(dom f \ dom g) in sproduct f by A2;
  (f +* g)|dom g = g by FUNCT_4:23;
  hence h|dom g in sproduct g by A1,Th66;
  dom h c= dom(f+*g) by A1,Th49;
  then dom h c= dom f \/ dom g by FUNCT_4:def 1;
  then dom h c= (dom f \ dom g) \/ dom g by XBOOLE_1:39;
  hence thesis by FUNCT_4:70;
end;

theorem Th68:
  for f9,g9 being Function st dom g misses dom f9 \ dom g9 &
  f9 in sproduct f & g9 in sproduct g holds f9+*g9 in sproduct(f+*g)
proof
  let f9,g9 be Function such that
A1: dom g misses dom f9 \ dom g9 and
A2: f9 in sproduct f and
A3: g9 in sproduct g;
  set h = f9+*g9;
A4: dom f9 c= dom f by A2,Th49;
A5: dom g9 c= dom g by A3,Th49;
  then
A6: dom f9 \/ dom g9 c= dom f \/ dom g by A4,XBOOLE_1:13;
A7: dom h = dom f9 \/ dom g9 by FUNCT_4:def 1;
  then
A8: dom h c= dom(f+*g) by A6,FUNCT_4:def 1;
  x in dom h implies h.x in (f+*g).x
  proof
    assume
A9: x in dom h;
    then x in dom(f+*g) by A8;
    then
A10: x in dom f \/ dom g by FUNCT_4:def 1;
    x in dom f9 \ dom g9 \/ dom g9 by A7,A9,XBOOLE_1:39;
    then
A11: x in dom f9 \ dom g9 or x in dom g9 by XBOOLE_0:def 3;
    now per cases;
      case
A12:    x in dom g;
        then h.x = g9.x by A1,A7,A9,A11,FUNCT_4:def 1,XBOOLE_0:3;
        hence h.x in g.x by A1,A3,A11,A12,Th49,XBOOLE_0:3;
      end;
      case not x in dom g;
        then
A13:    not x in dom g9 by A5;
        then
A14:    h.x = f9.x by A7,A9,FUNCT_4:def 1;
        x in dom f9 by A7,A9,A13,XBOOLE_0:def 3;
        hence h.x in f.x by A2,A14,Th49;
      end;
    end;
    hence thesis by A10,FUNCT_4:def 1;
  end;
  hence thesis by A8,Def9;
end;

theorem
  for f9,g9 being Function st dom f9 misses dom g \ dom g9 &
  f9 in sproduct f & g9 in sproduct g holds f9+*g9 in sproduct(f+*g)
proof
  let f9,g9 be Function;
  assume dom f9 misses dom g \ dom g9;
  then dom g misses dom f9 \ dom g9 by XBOOLE_1:81;
  hence thesis by Th68;
end;

theorem Th70:
  g in sproduct f & h in sproduct f implies g +* h in sproduct f
proof
  assume that
A1: g in sproduct f and
A2: h in sproduct f;
A3: dom g c= dom f by A1,Th49;
  dom h c= dom f by A2,Th49;
  then dom g \/ dom h c= dom f by A3,XBOOLE_1:8;
  then
A4: dom(g+*h) c= dom f by FUNCT_4:def 1;
  now
    let x;
    assume x in dom(g+*h);
    then x in dom g \/ dom h by FUNCT_4:def 1;
    then
A5: x in (dom g \ dom h \/ dom h) by XBOOLE_1:39;
    now per cases by A5,XBOOLE_0:def 3;
      suppose
A6:     x in dom h;
        then h.x in f.x by A2,Th49;
        hence (g+*h).x in f.x by A6,FUNCT_4:13;
      end;
      suppose
A7:     x in dom g \ dom h;
        then
A8:     g.x in f.x by A1,Th49;
        not x in dom h by A7,XBOOLE_0:def 5;
        hence (g+*h).x in f.x by A8,FUNCT_4:11;
      end;
    end;
    hence (g+*h).x in f.x;
  end;
  hence thesis by A4,Def9;
end;

theorem
  for x1,x2,y1,y2 being set holds
  x1 in dom f & y1 in f.x1 & x2 in dom f & y2 in f.x2
  implies (x1,x2)-->(y1,y2) in sproduct f
proof
  let x1,x2,y1,y2 be set;
  assume that
A1: x1 in dom f and
A2: y1 in f.x1;
A3: x1 .--> y1 in sproduct f by A1,A2,Th60;
  assume that
A4: x2 in dom f and
A5: y2 in f.x2;
A6: x2 .--> y2 in sproduct f by A4,A5,Th60;
  (x1,x2)-->(y1,y2) = (x1 .--> y1) +* (x2 .--> y2) by FUNCT_4:def 4;
  hence thesis by A3,A6,Th70;
end;

begin  

definition
  let IT be set;
  attr IT is with_common_domain means
  :Def10:
  for f,g be Function st f in IT & g in IT holds dom f = dom g;
end;

registration
  cluster with_common_domain functional non empty for set;
  existence
  proof
    set h = the Function;
    take A = {h};
    for f,g be Function st f in A & g in A holds dom f = dom g
    proof
      let f,g be Function;
      assume that
A1:   f in A and
A2:   g in A;
      f = h by A1,TARSKI:def 1;
      hence thesis by A2,TARSKI:def 1;
    end;
    hence A is with_common_domain by Def10;
    thus A is functional;
    thus thesis;
  end;
end;

registration
  let f;
  cluster {f} -> with_common_domain;
coherence
proof
  let g,h be Function;
  assume g in {f};
  then g = f by TARSKI:def 1;
  hence thesis by TARSKI:def 1;
end;
end;

definition
  let X be functional set;
  func DOM X -> set equals
 meet { dom f where f is Element of X: not contradiction };
  coherence;
end;

LmX: for X being functional with_common_domain set
   for f being Function st f in X holds dom f = DOM X
  proof let X be functional with_common_domain set;
    set A = { dom f where f is Element of X: not contradiction };
   let f be Function;
   assume
Z:  f in X;
    then dom f in A;
    then
X:  { dom f } c= A by ZFMISC_1:31;
     A c= { dom f }
     proof let e be set;
      assume e in A;
       then consider g being Element of X such that
W:      e = dom g;
       e = dom f by W,Def10,Z;
      hence e in { dom f } by TARSKI:def 1;
     end;
    then A = { dom f } by X,XBOOLE_0:def 10;
   hence dom f = DOM X by SETFAM_1:10;
  end;

theorem Th72:
  for X be with_common_domain functional set st X = {{}} holds DOM X = {}
proof
  let X be with_common_domain functional set;
  assume
A1: X = {{}};
  {} in {{}} by TARSKI:def 1;
  hence thesis by A1,LmX,RELAT_1:38;
end;

registration let X be empty set;
 cluster DOM X -> empty;
 coherence
  proof set A = { dom f where f is Element of X: not contradiction };
X: A c= {{}}
    proof let e be set;
     assume e in A;
      then ex f being Element of X st e = dom f;
      then e = dom {} by SUBSET_1:def 1;
    hence thesis by TARSKI:def 1;
    end;
    {{}} c= A
     proof let e be set;
      assume e in {{}};
       then e = {} by TARSKI:def 1;
       then e = dom {};
       then e = dom the Element of X by SUBSET_1:def 1;
      hence thesis;
     end;
    then A = {{}} by X,XBOOLE_0:def 10;
   hence thesis by SETFAM_1:10;
  end;
end;

begin  

definition
  let S be functional set;
  func product" S -> Function means  
  :Def12:  dom it = DOM S &
  for i being set st i in dom it holds it.i = pi(S,i);
  existence
  proof
   per cases;
   suppose
S:   S = {};
    take {};
    thus thesis by S;
   end;
   suppose
  S <> {};
      then reconsider S1 = S as non empty functional set;
      set D = {dom f where f is Element of S1: not contradiction};
      defpred P[set,set] means $2 = pi(S,$1);
A1:   for e being set st e in meet D ex u being set st P[e,u];
      consider g being Function such that
A2:   dom g = meet D and
A3:   for e being set st e in meet D holds P[e,g.e] from CLASSES1:sch 1(
      A1);
      take g;
      thus thesis by A2,A3;
    end;
  end;
  uniqueness
  proof
    let f, g be Function such that
Z1: dom f = DOM S and
Z2: for i being set st i in dom f holds f.i = pi(S,i) and
Z3: dom g = DOM S and
Z4: for i being set st i in dom g holds g.i = pi(S,i);
     for x st x in dom f holds f.x = g.x
      proof let x;
       assume
Z:      x in dom f;
       hence f.x = pi(S,x) by Z2 .= g.x by Z1,Z3,Z,Z4;
      end;
   hence f = g by Z1,Z3,FUNCT_1:def 11;
  end;
end;

canceled;

theorem Th74:
  for S being non empty functional set,
  i being set st i in dom product" S holds
  (product" S).i = {f.i where f is Element of S: not contradiction}
proof
  let S be non empty functional set, i be set;
  assume
A1: i in dom product" S;
  hereby
    let x be set;
    assume x in (product" S).i;
    then x in pi(S,i) by A1,Def12;
    then ex f being Function st f in S & x = f.i by Def6;
    hence x in {f.i where f is Element of S: not contradiction};
  end;
  let x be set;
  assume x in {f.i where f is Element of S: not contradiction};
  then ex f being Element of S st x = f.i & not contradiction;
  then x in pi(S,i) by Def6;
  hence thesis by A1,Def12;
end;

definition
  let S be set;
  attr S is product-like means
  :Def13:
  ex f being Function st S = product f;
end;

registration
  let f be Function;
  cluster product f -> product-like;
  coherence by Def13;
end;

registration
  cluster product-like -> functional with_common_domain for set;
  coherence
  proof
    let S be set;
    given f being Function such that
A1: S = product f;
    thus S is functional by A1;
    let h, g be Function such that
A2: h in S and
A3: g in S;
    thus dom h = dom f by A1,A2,Th9
      .= dom g by A1,A3,Th9;
  end;
end;

registration
  cluster product-like non empty for set;
  existence
  proof
    set B = the with_non-empty_elements set,f = the Function of 0,B;
    take product f;
    thus thesis;
  end;
end;

canceled 2;

theorem Th77:
  for S being functional with_common_domain set holds S c= product product" S
proof
  let S be functional with_common_domain set;
  let f be set;
  assume
A1: f in S;
  then reconsider f as Element of S;
A2: dom f = DOM S by A1,LmX
    .= dom product" S by Def12;
  for i being set st i in dom product" S holds f.i in (product" S).i
  proof
    let i be set;
    assume i in dom product" S;
    then (product" S).i = pi(S,i) by Def12;
    hence thesis by A1,Def6;
  end;
  hence thesis by A2,Th9;
end;

theorem
  for S being non empty product-like set holds S = product product" S
proof
  let S be non empty product-like set;
  thus S c= product product" S by Th77;
  let x be set;
  assume x in product product" S;
  then consider g being Function such that
A1: x = g and
A2: dom g = dom product" S and
A3: for z being set st z in dom product" S holds g.z in (product" S).z
  by Def5;
  consider p being Function such that
A4: S = product p by Def13;
  set s = the Element of S;
A5: dom g = DOM S by A2,Def12
    .= dom s by LmX
    .= dom p by A4,Th9;
  for z being set st z in dom p holds g.z in p.z
  proof
    let z be set;
    assume
A6: z in dom p;
    then g.z in (product" S).z by A2,A3,A5;
    then g.z in pi(S,z) by A2,A5,A6,Def12;
    then ex f being Function st f in S & g.z = f.z by Def6;
    hence thesis by A4,A6,Th9;
  end;
  hence thesis by A1,A4,A5,Th9;
end;



theorem
  for f being Function
  for s, t being Element of product f, A be set holds
  s +* t|A is Element of product f
proof
  let f be Function;
  let s, t be Element of product f, A be set;
  per cases;
  suppose f is non-empty;
    then product f <> {};
    then
A1: t in product f;
    product f c= sproduct f by Th51;
    hence thesis by A1,Th53,Th65;
  end;
  suppose f is not non-empty;
    then {} in rng f by RELAT_1:def 9;
    then
A2: product f = {} by Th26;
    then
A3: s = {} by SUBSET_1:def 1;
    t = {} by A2,SUBSET_1:def 1;
    then t|A = {};
    hence thesis by A3;
  end;
end;

theorem
  for f being non-empty Function
  for p being Element of sproduct f
  ex s being Element of product f st p c= s
proof
  let f be non-empty Function,
  p be Element of sproduct f;
  set h = the Element of product f;
  reconsider s = h +* p as Element of product f by Th53;
  take s;
  thus thesis by FUNCT_4:25;
end;

theorem Th81:
  g in product f implies g|A in sproduct f
proof
A1: product f c= sproduct f by Th51;
  assume g in product f;
  hence thesis by A1,Th64,RELAT_1:59;
end;



definition
  let f be non-empty Function;
  let g be Element of product f;
  let X;
  redefine func g|X -> Element of sproduct f;
  coherence by Th81;
end;

theorem
  for f being non-empty Function
  for s,ss being Element of product f, A being set
  holds (ss +* s | A) | A = s | A
proof
  let f be non-empty Function;
  let s,ss be Element of product f;
  let A be set;
  dom s = dom f by Th9
    .= dom ss by Th9;
  then A /\ dom ss c= A /\ dom s;
  hence thesis by FUNCT_4:88;
end;



theorem
  for M,x, g being Function st x in product M holds x * g in product (M * g)
proof
  let M, x, g be Function;
  assume
A1: x in product M;
  set xg = x * g;
  set Mg = M * g;
A2: ex gg being Function st ( x = gg)&( dom gg = dom M)&( for x
  being set st x in dom M holds gg.x in M.x) by A1,Def5;
  then
A3: dom xg = dom Mg by RELAT_1:163;
  now
    let y be set;
    assume
A4: y in dom Mg;
    then
A5: y in dom g by FUNCT_1:11;
A6: g.y in dom M by A4,FUNCT_1:11;
A7: xg.y = x.(g.y) by A5,FUNCT_1:13;
    Mg.y = M.(g.y) by A5,FUNCT_1:13;
    hence xg.y in Mg.y by A2,A6,A7;
  end;
  hence thesis by A3,Def5;
end;



theorem
  X is finite iff card X in omega by Th42,CARD_1:47;

reserve A,B for Ordinal;

theorem Th85:
  A is infinite iff omega c= A
proof omega c= A iff not A in omega by ORDINAL1:16;
  hence thesis;
end;

theorem
  N is finite & not M is finite implies N in M & N c= M
proof
  assume that
A1: N is finite and
A2: not M is finite;
A3: N in omega by A1,CARD_1:61;
  omega c= M by A2,Th85;
  hence N in M by A3;
  thus thesis by A1,A2;
end;

theorem
  not X is finite iff ex Y st Y c= X & card Y = omega
proof
  thus not X is finite implies ex Y st Y c= X & card Y = omega
  proof
    assume not X is finite;
    then not card X in omega;
    then
A1: omega c= card X by CARD_1:4;
    card X,X are_equipotent by CARD_1:def 2;
    then consider f such that
A2: f is one-to-one and
A3: dom f = card X and
A4: rng f = X by WELLORD2:def 4;
    take Y = f.:(omega);
    thus Y c= X by A4,RELAT_1:111;
    omega,Y are_equipotent
    proof
      take f|(omega);
      thus thesis by A1,A2,A3,FUNCT_1:52,RELAT_1:62,115;
    end;
    hence thesis by CARD_1:def 2;
  end;
  given Y such that
A5: Y c= X and
A6: card Y = omega;
  thus thesis by A5,A6;
end;

theorem Th88:
  card X = card Y iff nextcard X = nextcard Y
proof
  thus card X = card Y implies nextcard X = nextcard Y by CARD_1:16;
  assume that
A1: nextcard X = nextcard Y and
A2: card X <> card Y;
  card X in card Y or card Y in card X by A2,ORDINAL1:14;
  then nextcard X c= card Y & card Y in nextcard Y or
  nextcard Y c= card X & card X in nextcard X by CARD_1:def 3;
  hence thesis by A1,ORDINAL1:12;
end;

theorem
  nextcard N = nextcard M implies M = N
proof
A1: card N = N by CARD_1:def 2;
  card M = M by CARD_1:def 2;
  hence thesis by A1,Th88;
end;

theorem Th90:
  N in M iff nextcard N c= M
proof
A1: N in nextcard N by CARD_1:18;
  card N = N by CARD_1:def 2;
  hence thesis by A1,CARD_1:def 3;
end;

theorem
  N in nextcard M iff N c= M
proof
A1: not N c= M iff M in N by CARD_1:4;
  N in nextcard M iff not nextcard M c= N by CARD_1:4;
  hence thesis by A1,Th90;
end;

theorem
  M is finite & (N c= M or N in M) implies N is finite
proof
  assume that
A1: M is finite and
A2: N c= M or N in M;
  N c= M by A2,CARD_1:3;
  hence thesis by A1;
end;

reserve n,k for Nat;

definition
  let X;
  attr X is countable means
  : Def14:
  card X c= omega;
  attr X is denumerable means
  card X = omega;
end;

registration
  cluster denumerable -> countable infinite for set;
  coherence
  proof
    let X be set;
    assume
A1: card X = omega;
    hence card X c= omega;
    thus thesis by A1;
  end;
  cluster countable infinite -> denumerable for set;
  coherence
  proof
    let X be set;
    assume
A2: card X c= omega;
    assume X is infinite;
    then omega c= card X by Th85;
    hence card X = omega by A2,XBOOLE_0:def 10;
  end;
end;

registration
  cluster finite -> countable for set;
  coherence
  proof
    let X be set;
    assume X is finite;
    then consider n being Nat such that
A1: X,n are_equipotent by CARD_1:43;
    card n = card X by A1,CARD_1:5;
    hence card X c= omega;
  end;
end;

registration
  cluster omega -> denumerable;
  coherence
  proof
    thus card omega = omega by CARD_1:47;
  end;
end;

registration
  cluster denumerable for set;
  existence
  proof
    take omega;
    thus thesis;
  end;
end;

theorem Th93:
  X is countable iff ex f st dom f = omega & X c= rng f
proof
  thus X is countable implies ex f st dom f = omega & X c= rng f
  proof
    assume card X c= omega;
    hence thesis by CARD_1:12,47;
  end;
  assume ex f st dom f = omega & X c= rng f;
  hence card X c= omega by CARD_1:12,47;
end;

registration
  let X be countable set;
  cluster -> countable for Subset of X;
  coherence
  proof
    let Y be Subset of X;
A1: card Y c= card X by CARD_1:11;
    card X c= omega by Def14;
    hence card Y c= omega by A1,XBOOLE_1:1;
  end;
end;

Lm2: Y c= X & X is countable implies Y is countable;

theorem
  X is countable implies X /\ Y is countable
  by Lm2,XBOOLE_1:17;

theorem
  X is countable implies X \ Y is countable;

theorem
  for A being non empty countable set
  ex f being Function of omega, A st rng f = A
proof
  let A be non empty countable set;
  consider f being Function such that
A1: dom f = omega and
A2: A c= rng f by Th93;
  consider x being set such that
A3: x in A by XBOOLE_0:def 1;
  set F = f|(f"A) +* (omega \ f"A --> x);
A4: rng F = A & dom F = omega
  proof
A5: f"A c= omega by A1,RELAT_1:132;
A6: dom(f|(f"A)) = omega /\ (f"A) by A1,RELAT_1:61;
    per cases;
    suppose
A7:   omega = f"A;
      then
A8:   omega \ f"A = {} by XBOOLE_1:37;
      then dom(f|(f"A)) /\ dom(omega \ f"A --> x) = {};
      then dom(f|(f"A)) misses dom(omega \ f"A --> x) by XBOOLE_0:def 7;
      then F = (f|(f"A)) \/ (omega \ f"A --> x) by FUNCT_4:31;
      hence rng F = rng(f|(f"A)) \/ rng(omega \ f"A --> x) by RELAT_1:12
        .= rng(f|(f"A)) \/ {} by A8
        .= f.:(f"A) by RELAT_1:115
        .= A by A2,FUNCT_1:77;
      thus
      dom F = dom(f|(f"A)) \/ dom(omega \ f"A --> x) by FUNCT_4:def 1
        .= dom(f|(f"A)) \/ {} by A8
        .= omega by A6,A7;
    end;
    suppose
A9:   omega <> f"A;
A10:  now
        assume omega \ f"A is empty;
        then omega c= f"A by XBOOLE_1:37;
        hence contradiction by A5,A9,XBOOLE_0:def 10;
      end;
      dom(omega \ f"A --> x) = omega \ f"A by FUNCOP_1:13;
      then F = (f|(f"A)) \/ (omega \ f"A --> x) by A6,FUNCT_4:31,XBOOLE_1:89;
      hence rng F = rng(f|(f"A)) \/ rng(omega \ f"A --> x) by RELAT_1:12
        .= rng(f|(f"A)) \/ {x} by A10,FUNCOP_1:8
        .= f.:(f"A) \/ {x} by RELAT_1:115
        .= A \/ {x} by A2,FUNCT_1:77
        .= A by A3,ZFMISC_1:40;
      thus
      dom F = dom(f|(f"A)) \/ dom(omega \ f"A --> x) by FUNCT_4:def 1
        .= omega /\ (f"A) \/ (omega \ f"A) by A6,FUNCOP_1:13
        .= omega by XBOOLE_1:51;
    end;
  end;
  then reconsider F as Function of omega, A by FUNCT_2:def 1,RELSET_1:4;
  take F;
  thus thesis by A4;
end;



theorem
  for f,g being non-empty Function, x being Element of product f,
  y being Element of product g holds x+*y in product (f+*g)
proof
  let f,g be non-empty Function, x be Element of product f;
  let y be Element of product g;
A1: dom x = dom f by Th9;
A2: dom y = dom g by Th9;
  then
A3: dom (x+*y) = dom f \/ dom g by A1,FUNCT_4:def 1;
A4: dom (f+*g) = dom f \/ dom g by FUNCT_4:def 1;
  now
    let z be set;
    assume
A5: z in dom (f+*g);
    then z in dom g or not z in dom g & z in dom f by A4,XBOOLE_0:def 3;
    then (x+*y).z = x.z & (f+*g).z = f.z & x.z in f.z or
    (x+*y).z = y.z & (f+*g).z = g.z & y.z in g.z
    by A1,A2,A4,A5,Th9,FUNCT_4:def 1;
    hence (x+*y).z in (f+*g).z;
  end;
  hence thesis by A3,A4,Th9;
end;

theorem
  for f,g being non-empty Function
  for x being Element of product (f+*g) holds x|dom g in product g
proof
  let f,g be non-empty Function;
  let x be Element of product (f+*g);
A1: dom x = dom (f+*g) by Th9;
A2: dom (f+*g) = dom f \/ dom g by FUNCT_4:def 1;
  then
A3: dom g c= dom x by A1,XBOOLE_1:7;
A4: dom (x|dom g) = dom g by A1,A2,RELAT_1:62,XBOOLE_1:7;
  now
    let z be set;
    assume
A5: z in dom (x|dom g);
    then
A6: (x|dom g).z = x.z by FUNCT_1:47;
    (f+*g).z = g.z by A4,A5,FUNCT_4:13;
    hence (x|dom g).z in g.z by A1,A3,A4,A5,A6,Th9;
  end;
  hence thesis by A4,Th9;
end;

theorem
  for f,g being non-empty Function st f tolerates g
  for x being Element of product (f+*g) holds x|dom f in product f
proof
  let f,g be non-empty Function such that
A1: f tolerates g;
  let x be Element of product (f+*g);
A2: dom x = dom (f+*g) by Th9;
A3: dom (f+*g) = dom f \/ dom g by FUNCT_4:def 1;
  then
A4: dom f c= dom x by A2,XBOOLE_1:7;
A5: dom (x|dom f) = dom f by A2,A3,RELAT_1:62,XBOOLE_1:7;

now
    let z be set;
    assume
A6: z in dom (x|dom f);
    then
A7: (x|dom f).z = x.z by FUNCT_1:47;
    (f+*g).z = f.z by A1,A5,A6,FUNCT_4:15;
   hence (x|dom f).z in f.z by A2,A4,A5,A6,A7,Th9;
  end;
  hence thesis by A5,Th9;
end;



theorem Th100:
  for S being with_common_domain functional set, f be Function
     st f in S  holds dom f = dom product" S
  proof let S be with_common_domain functional set, f be Function such that
A1: f in S;
   thus dom f = DOM S by A1,LmX
     .= dom product" S by Def12;
  end;

theorem Th101:
  for S being functional set, f be Function, i be set
     st f in S & i in dom product" S holds f.i in (product" S).i
proof
  let S being functional set, F be Function, i be set such that
A1: F in S;
  assume i in dom product" S;
   then (product" S).i = {f.i where f is Element of S: not contradiction}
      by A1,Th74;
 hence F.i in (product" S).i by A1;
end;

theorem
  for S being with_common_domain functional set, f be Function, i be set
     st f in S & i in dom f holds f.i in (product" S).i
 proof let S be with_common_domain functional set, f be Function, i be set;
  assume that
A1: f in S and
A2: i in dom f;
   dom f = dom(product" S) by A1,Th100;
  hence f.i in (product" S).i by A1,A2,Th101;
 end;



registration let X be with_common_domain set;
 cluster -> with_common_domain for Subset of X;
 coherence
  proof let Y be Subset of X;
   let f,g be Function;
   thus thesis by Def10;
  end;
end;



definition
  let f be Function, x be set;
  func proj(f,x) -> Function means
:Def16:
  dom it = product f & for y being
  Function st y in dom it holds it.y = y.x;
  existence
  proof
    defpred P[set,set] means for g be Function st $1 = g holds $2 = g.x;
A1: now
      let q be set;
      assume q in product f;
      then reconsider q1 = q as Function;
      take y = q1.x;
      thus P[q,y];
    end;
    consider F be Function such that
A2: dom F = product f & for a being set st a in product f holds P[a,F.
    a] from CLASSES1:sch 1(A1);
    take F;
    thus thesis by A2;
  end;
  uniqueness
  proof
    let F,G be Function such that
A3: dom F = product f and
A4: for y being Function st y in dom F holds F.y = y.x and
A5: dom G = product f and
A6: for y being Function st y in dom G holds G.y = y.x;
    now
      let y be set;
      assume
A7:   y in product f;
      then reconsider x1 = y as Function;
      thus F.y = x1.x by A3,A4,A7
        .= G.y by A5,A6,A7;
    end;
    hence thesis by A3,A5,FUNCT_1:2;
  end;
end;

registration
  let f be Function, x be set;
  cluster proj(f,x) -> (product f)-defined;
  coherence
   proof
     dom proj(f,x) = product f by Def16;
    hence thesis by RELAT_1:def 18;
   end;
end;

registration
  let f be Function, x be set;
  cluster proj(f,x) -> total;
  coherence
   proof
     dom proj(f,x) = product f by Def16;
    hence thesis by PARTFUN1:def 2;
   end;
end;

registration
  let f be non-empty Function;
  cluster -> f-compatible for Element of product f;
  coherence
   proof let e be Element of product f;
    let x;
    assume x in dom e;
     then x in dom f by Th9;
    hence e.x in f.x by Th9;
   end;
end;

registration
  let I be set; let f be I-defined non-empty Function;
  cluster -> I-defined for Element of product f;
  coherence;
end;

registration
  let f be Function;
  cluster -> f-compatible for Element of sproduct f;
  coherence
   proof let e be Element of sproduct f;
    let x;
    thus thesis by Th49;
   end;
end;

registration
  let I be set; let f be I-defined Function;
  cluster -> I-defined for Element of sproduct f;
  coherence;
end;

registration
  let I be set; let f be total I-defined non-empty Function;
  cluster -> total for Element of product f;
  coherence
   proof let e be Element of product f;
    thus dom e = dom f by Th9
           .= I by PARTFUN1:def 2;
   end;
end;

theorem Th103:
 for I being set, f being non-empty I-defined Function
 for p being f-compatible I-defined Function
  holds p in sproduct f
 proof let I be set, f be non-empty I-defined Function;
  let p be f-compatible I-defined Function;
A1: dom p c= dom f by FUNCT_1:105;
   for x being set st x in dom p holds p.x in f.x by FUNCT_1:def 14;
  hence p in sproduct f by A1,Def9;
 end;

theorem
 for I being set, f being non-empty I-defined Function
 for p being f-compatible I-defined Function
  ex s being Element of product f st p c= s
proof let I be set;
  let f be non-empty I-defined Function,
  p be f-compatible I-defined Function;
  reconsider p as Element of sproduct f by Th103;
  set h = the Element of product f;
  reconsider s = h +* p as Element of product f by Th53;
  take s;
  thus thesis by FUNCT_4:25;
end;



registration
  let X be infinite set, a be set;
  cluster X --> a -> infinite;
  coherence;
end;

registration
  cluster infinite for Function;
  existence
  proof
    take omega --> 0;
    thus thesis;
  end;
end;

registration
  let R be infinite Relation;
  cluster field R -> infinite;
  coherence by ORDERS_1:86;
end;

registration
  let X be infinite set;
  cluster RelIncl X -> infinite;
  coherence by CARD_1:63;
end;

theorem
for R,S being Relation st R,S are_isomorphic & R is finite
 holds S is finite
proof
 let R,S be Relation;
  given F being Function such that
A1: F is_isomorphism_of R,S;
  assume R is finite;
  then field R is finite;
  then dom F is finite by A1,WELLORD1:def 7;
  then rng F is finite by FINSET_1:8;
  then field S is finite by A1,WELLORD1:def 7;
  hence thesis;
end;

theorem
 product" {{}} = {}
 proof
   dom product" {{}} = DOM {{}} by Def12
       .= {} by Th72;
  hence thesis;
 end;

theorem Th107:
 for I being set, f being non-empty ManySortedSet of I
 for s being f-compatible ManySortedSet of I
  holds s in product f
 proof let I be set, f be non-empty ManySortedSet of I;
  let s be f-compatible ManySortedSet of I;
A1: dom s = I by PARTFUN1:def 2 .= dom f by PARTFUN1:def 2;
   then for x being set st x in dom f holds s.x in f.x by FUNCT_1:def 14;
  hence s in product f by A1,Th9;
 end;

registration
 let I be set, f be non-empty ManySortedSet of I;
 cluster -> total for Element of product f;
 coherence;
end;

definition
 let I be set, f be non-empty ManySortedSet of I;
 let M be f-compatible ManySortedSet of I;
 func down M -> Element of product f equals M;
 coherence by Th107;
end;

theorem
 for X being functional with_common_domain set
   for f being Function st f in X holds dom f = DOM X by LmX;

theorem
 for X being non empty functional set
  st for f being Function st f in X holds x in dom f
  holds x in DOM X
 proof let X be non empty functional set such that
Z: for f being Function st f in X holds x in dom f;
  set A = { dom f where f is Element of X: not contradiction };
   consider Y being set such that
W:  Y in X by XBOOLE_0:def 1;
    reconsider Y as Function by W;
S:  dom Y in A by W;
   for Y holds Y in A implies x in Y
    proof let Y;
     assume Y in A;
      then ex f being Element of X st Y = dom f;
     hence x in Y by Z;
   end;
  hence x in DOM X by S,SETFAM_1:def 1;
 end;
















begin

reserve x,y,y1,y2,z,a,b,X,Y,Z,V1,V2 for set,
  f,g,h,h9,f1,f2 for Function,
  i for Nat,
  P for Permutation of X,
  D,D1,D2,D3 for non empty set,
  d1 for Element of D1,
  d2 for Element of D2,
  d3 for Element of D3;

theorem Th1:
  product f c= Funcs(dom f, Union f)
proof
  let x;
  assume x in product f;
  then consider g such that
A1: x = g and
A2: dom g = dom f and
A3: for x st x in dom f holds g.x in f.x by CARD_3:def 5;
  rng g c= Union f
  proof
    let y;
A4: Union f = union rng f by CARD_3:def 4;
    assume y in rng g;
    then consider z such that
A5: z in dom g & y = g.z by FUNCT_1:def 3;
    y in f.z & f.z in rng f by A2,A3,A5,FUNCT_1:def 3;
    hence thesis by A4,TARSKI:def 4;
  end;
  hence thesis by A1,A2,FUNCT_2:def 2;
end;

begin

theorem Th2:
  x in dom ~f implies ex y,z st x = [y,z]
proof
  assume
A1: x in dom ~f;
  ex X,Y st dom ~f c= [:X,Y:] by FUNCT_4:44;
  hence thesis by A1,RELAT_1:def 1;
end;

theorem Th3:
  ~([:X,Y:] --> z) = [:Y,X:] --> z
proof
A1: dom ([:X,Y:] --> z) = [:X,Y:] by FUNCOP_1:13;
  then
A2: dom ~([:X,Y:] --> z) = [:Y,X:] by FUNCT_4:46;
A3: now
    let x;
    assume
A4: x in [:Y,X:];
    then consider y1,y2 such that
A5: x = [y2,y1] and
A6: [y1,y2] in [:X,Y:] by A1,A2,FUNCT_4:def 2;
A7: ([:X,Y:] --> z).(y1,y2) = z by A6,FUNCOP_1:7;
    ([:Y,X:] --> z).(y2,y1) = z by A4,A5,FUNCOP_1:7;
    then (~([:X,Y:] --> z)).(y2,y1) = ([:Y,X:] --> z).(y2,y1) by A1,A6,A7,
FUNCT_4:def 2;
    hence (~([:X,Y:] --> z)).x = ([:Y,X:] --> z).x by A5;
  end;
  dom ([:Y,X:] --> z) = [:Y,X:] by FUNCOP_1:13;
  hence thesis by A2,A3,FUNCT_1:2;
end;

theorem Th4:
  curry f = curry' ~f & uncurry f = ~uncurry' f
proof
A1: dom curry ~~f = proj1 dom ~~f by FUNCT_5:def 1;
A2: dom curry f = proj1 dom f by FUNCT_5:def 1;
A3: dom curry ~~f = dom curry f
  proof
    thus dom curry ~~f c= dom curry f
    proof
      let x;
      assume x in dom curry ~~f;
      then consider y such that
A4:   [x,y] in dom ~~f by A1,XTUPLE_0:def 12;
      [y,x] in dom ~f by A4,FUNCT_4:42;
      then [x,y] in dom f by FUNCT_4:42;
      hence thesis by A2,XTUPLE_0:def 12;
    end;
    let x;
    assume x in dom curry f;
    then consider y such that
A5: [x,y] in dom f by A2,XTUPLE_0:def 12;
    [y,x] in dom ~f by A5,FUNCT_4:42;
    then [x,y] in dom ~~f by FUNCT_4:42;
    hence thesis by A1,XTUPLE_0:def 12;
  end;
A6: curry' ~f = curry ~~f by FUNCT_5:def 3;
  now
    let x;
    assume
A7: x in dom curry f;
    then reconsider g = (curry f).x, h = (curry' ~f).x as Function by A6,A3,
FUNCT_5:30;
A8: dom g = proj2 (dom f /\ [:{x},proj2 dom f:]) by A7,FUNCT_5:31;
A9: dom h = proj1 (dom ~f /\ [:proj1 dom ~f,{x}:]) by A6,A3,A7,FUNCT_5:34;
A10: dom g = dom h
    proof
      thus dom g c= dom h
      proof
        let a;
        assume a in dom g;
        then consider b such that
A11:    [b,a] in dom f /\ [:{x},proj2 dom f:] by A8,XTUPLE_0:def 13;
        [b,a] in [:{x},proj2 dom f:] by A11,XBOOLE_0:def 4;
        then
A12:    [a,b] in [:proj2 dom f,{x}:] by ZFMISC_1:88;
        [b,a] in dom f by A11,XBOOLE_0:def 4;
        then
A13:    [a,b] in dom ~f by FUNCT_4:42;
        proj2 dom f = proj1 dom ~f by FUNCT_5:17;
        then [a,b] in dom ~f /\ [:proj1 dom ~f,{x}:] by A13,A12,XBOOLE_0:def 4;
        hence thesis by A9,XTUPLE_0:def 12;
      end;
      let a;
      assume a in dom h;
      then consider b such that
A14:  [a,b] in dom ~f /\ [:proj1 dom ~f,{x}:] by A9,XTUPLE_0:def 12;
      [a,b] in [:proj1 dom ~f,{x}:] by A14,XBOOLE_0:def 4;
      then
A15:  [b,a] in [:{x},proj1 dom ~f:] by ZFMISC_1:88;
      [a,b] in dom ~f by A14,XBOOLE_0:def 4;
      then
A16:  [b,a] in dom f by FUNCT_4:42;
      proj2 dom f = proj1 dom ~f by FUNCT_5:17;
      then [b,a] in dom f /\ [:{x},proj2 dom f:] by A16,A15,XBOOLE_0:def 4;
      hence thesis by A8,XTUPLE_0:def 13;
    end;
    now
      let a;
      assume
A17:  a in dom g;
      then
A18:  [x,a] in dom f & g.a = f.(x,a) by A7,FUNCT_5:31;
      h.a = (~f).(a,x) by A6,A3,A7,A10,A17,FUNCT_5:34;
      hence g.a = h.a by A18,FUNCT_4:def 2;
    end;
    hence (curry f).x = (curry' ~f).x by A10,FUNCT_1:2;
  end;
  hence curry f = curry' ~f by A6,A3,FUNCT_1:2;
A19: dom uncurry f = dom ~~uncurry f
  proof
    thus dom uncurry f c= dom ~~uncurry f
    proof
      let a;
      assume
A20:  a in dom uncurry f;
      then consider x,g,y such that
A21:  a = [x,y] and
      x in dom f and
      g = f.x and
      y in dom g by FUNCT_5:def 2;
      [y,x] in dom ~uncurry f by A20,A21,FUNCT_4:42;
      hence thesis by A21,FUNCT_4:42;
    end;
    let a;
    assume a in dom ~~uncurry f;
    then ex x,y st a = [y,x] & [x,y] in dom ~uncurry f by FUNCT_4:def 2;
    hence thesis by FUNCT_4:42;
  end;
A22: now
    let a;
    assume a in dom ~~uncurry f;
    then consider x,y such that
A23: a = [y,x] and
A24: [x,y] in dom ~uncurry f by FUNCT_4:def 2;
    (~uncurry f).(x,y) = (uncurry f).(y,x) & (~uncurry f).(x,y) = (~~
    uncurry f). (y,x) by A24,FUNCT_4:43,def 2;
    hence (uncurry f).a = (~~uncurry f).a by A23;
  end;
  uncurry' f = ~uncurry f by FUNCT_5:def 4;
  hence thesis by A19,A22,FUNCT_1:2;
end;

theorem
  [:X,Y:] <> {} implies curry ([:X,Y:] --> z) = X --> (Y --> z) & curry'
  ([:X,Y:] --> z) = Y --> (X --> z)
proof
  assume
A1: [:X,Y:] <> {};
A2: dom ([:X,Y:] --> z) = [:X,Y:] by FUNCOP_1:13;
A3: dom (X --> z) = X by FUNCOP_1:13;
A4: now
    let x;
    assume
A5: x in Y;
    then consider f such that
A6: (curry' ([:X,Y:] --> z)).x = f & dom f = X and
    rng f c= rng ([:X,Y:] --> z) and
A7: for y st y in X holds f.y = ([:X,Y:] --> z).(y,x) by A1,A2,FUNCT_5:32;
A8: now
      let y;
      assume
A9:   y in X;
      then
A10:  f.y = ([:X,Y:] --> z).(y,x) by A7;
      (X --> z).y = z & [y,x] in [:X,Y:] by A5,A9,FUNCOP_1:7,ZFMISC_1:87;
      hence f.y = (X --> z).y by A10,FUNCOP_1:7;
    end;
    (Y --> (X --> z)).x = X --> z by A5,FUNCOP_1:7;
    hence (curry' ([:X,Y:] --> z)).x = (Y --> (X --> z)).x by A3,A6,A8,
FUNCT_1:2;
  end;
A11: dom (Y --> z) = Y by FUNCOP_1:13;
A12: now
    let x;
    assume
A13: x in X;
    then consider f such that
A14: (curry ([:X,Y:] --> z)).x = f & dom f = Y and
    rng f c= rng ([:X,Y:] --> z) and
A15: for y st y in Y holds f.y = ([:X,Y:] --> z).(x,y) by A1,A2,FUNCT_5:29;
A16: now
      let y;
      assume
A17:  y in Y;
      then
A18:  f.y = ([:X,Y:] --> z).(x,y) by A15;
      (Y --> z).y = z & [x,y] in [:X,Y:] by A13,A17,FUNCOP_1:7,ZFMISC_1:87;
      hence f.y = (Y --> z).y by A18,FUNCOP_1:7;
    end;
    (X --> (Y --> z)).x = Y --> z by A13,FUNCOP_1:7;
    hence (curry ([:X,Y:] --> z)).x = (X --> (Y --> z)).x by A11,A14,A16,
FUNCT_1:2;
  end;
  dom (X --> (Y --> z)) = X & dom curry ([:X,Y:] --> z) = X by A1,A2,
FUNCOP_1:13,FUNCT_5:24;
  hence curry ([:X,Y:] --> z) = X --> (Y --> z) by A12,FUNCT_1:2;
  dom (Y --> (X --> z)) = Y & dom curry' ([:X,Y:] --> z) = Y by A1,A2,
FUNCOP_1:13,FUNCT_5:24;
  hence thesis by A4,FUNCT_1:2;
end;

theorem
  uncurry (X --> (Y --> z)) = [:X,Y:] --> z & uncurry' (X --> (Y --> z))
  = [:Y,X:] --> z
proof
A1: dom (X --> (Y --> z)) = X by FUNCOP_1:13;
A2: dom (Y --> z) = Y by FUNCOP_1:13;
  rng (Y --> z) c= {z} by FUNCOP_1:13;
  then (Y --> z) in Funcs (Y,{z}) by A2,FUNCT_2:def 2;
  then rng (X --> (Y --> z)) c= {Y --> z} & {(Y --> z)} c= Funcs (Y,{z}) by
FUNCOP_1:13,ZFMISC_1:31;
  then rng (X --> (Y --> z)) c= Funcs (Y,{z}) by XBOOLE_1:1;
  then
A3: dom uncurry (X --> (Y --> z)) = [:X,Y:] by A1,FUNCT_5:26;
A4: now
    let x;
    assume
A5: x in [:X,Y:];
    then consider y1,g,y2 such that
A6: x = [y1,y2] and
A7: y1 in X & g = (X --> (Y --> z)).y1 and
A8: y2 in dom g by A1,A3,FUNCT_5:def 2;
A9: g = Y --> z by A7,FUNCOP_1:7;
    then (Y --> z).y2 = z by A2,A8,FUNCOP_1:7;
    then z = (uncurry (X --> (Y --> z))).(y1,y2) by A1,A7,A8,A9,FUNCT_5:38;
    hence (uncurry (X --> (Y --> z))).x = ([:X,Y:] --> z).x by A5,A6,FUNCOP_1:7
;
  end;
  dom ([:X,Y:] --> z) = [:X,Y:] by FUNCOP_1:13;
  hence uncurry (X --> (Y --> z)) = [:X,Y:] --> z by A3,A4,FUNCT_1:2;
  then ~uncurry (X --> (Y --> z)) = [:Y,X:] --> z by Th3;
  hence thesis by FUNCT_5:def 4;
end;

theorem Th7:
  x in dom f & g = f.x implies rng g c= rng uncurry f & rng g c= rng uncurry' f
proof
  assume
A1: x in dom f & g = f.x;
  thus rng g c= rng uncurry f
  proof
    let y;
    assume y in rng g;
    then ex z st z in dom g & y = g.z by FUNCT_1:def 3;
    hence thesis by A1,FUNCT_5:38;
  end;
  let y;
  assume y in rng g;
  then ex z st z in dom g & y = g.z by FUNCT_1:def 3;
  hence thesis by A1,FUNCT_5:39;
end;

theorem Th8:
  dom uncurry (X --> f) = [:X,dom f:] & rng uncurry (X --> f) c=
rng f & dom uncurry' (X --> f) = [:dom f,X:] & rng uncurry' (X --> f) c= rng f
proof
  f in Funcs(dom f, rng f) by FUNCT_2:def 2;
  then rng (X --> f) c= {f} & {f} c= Funcs(dom f, rng f) by FUNCOP_1:13
,ZFMISC_1:31;
  then dom (X --> f) = X & rng (X --> f) c= Funcs(dom f, rng f) by FUNCOP_1:13
,XBOOLE_1:1;
  hence thesis by FUNCT_5:26,41;
end;

theorem
  X <> {} implies rng uncurry (X --> f) = rng f & rng uncurry' (X --> f)
  = rng f
proof
  set x = the Element of X;
  assume
A1: X <> {};
  then dom (X --> f) = X & (X --> f).x = f by FUNCOP_1:7,13;
  hence rng uncurry (X --> f) c= rng f & rng f c= rng uncurry (X --> f) & rng
  uncurry' (X --> f) c= rng f & rng f c= rng uncurry' (X --> f) by A1,Th7,Th8
;
end;

theorem Th10:
  [:X,Y:] <> {} & f in Funcs([:X,Y:],Z) implies curry f in Funcs(X
  ,Funcs(Y,Z)) & curry' f in Funcs(Y,Funcs(X,Z))
proof
  assume
A1: [:X,Y:] <> {};
  assume f in Funcs([:X,Y:],Z);
  then
A2: ex g st f = g & dom g = [:X,Y:] & rng g c= Z by FUNCT_2:def 2;
  then rng curry f c= Funcs(Y,rng f) & Funcs(Y,rng f) c= Funcs(Y,Z) by
FUNCT_5:35,56;
  then
A3: rng curry f c= Funcs(Y,Z) by XBOOLE_1:1;
  rng curry' f c= Funcs(X,rng f) & Funcs(X,rng f) c= Funcs(X,Z) by A2,
FUNCT_5:35,56;
  then
A4: rng curry' f c= Funcs(X,Z) by XBOOLE_1:1;
  dom curry f = X & dom curry' f = Y by A1,A2,FUNCT_5:24;
  hence thesis by A3,A4,FUNCT_2:def 2;
end;

theorem Th11:
  f in Funcs(X,Funcs(Y,Z)) implies uncurry f in Funcs([:X,Y:],Z) &
  uncurry' f in Funcs([:Y,X:],Z)
proof
  assume f in Funcs(X,Funcs(Y,Z));
  then
A1: ex g st f = g & dom g = X & rng g c= Funcs(Y,Z) by FUNCT_2:def 2;
  then
A2: dom uncurry f = [:X,Y:] & dom uncurry' f = [:Y,X:] by FUNCT_5:26;
  rng uncurry f c= Z & rng uncurry' f c= Z by A1,FUNCT_5:41;
  hence thesis by A2,FUNCT_2:def 2;
end;

theorem
  (curry f in Funcs(X,Funcs(Y,Z)) or curry' f in Funcs(Y,Funcs(X,Z))) &
  dom f c= [:V1,V2:] implies f in Funcs([:X,Y:],Z)
proof
  assume curry f in Funcs(X,Funcs(Y,Z)) or curry' f in Funcs(Y,Funcs(X,Z));
  then
A1: uncurry curry f in Funcs([:X,Y:],Z) or uncurry' curry' f in Funcs([:X,Y
  :],Z) by Th11;
  assume dom f c= [:V1,V2:];
  hence thesis by A1,FUNCT_5:50;
end;

theorem
  (uncurry f in Funcs([:X,Y:],Z) or uncurry' f in Funcs([:Y,X:],Z)) &
  rng f c= PFuncs(V1,V2) & dom f = X implies f in Funcs(X,Funcs(Y,Z))
proof
  assume that
A1: uncurry f in Funcs([:X,Y:],Z) or uncurry' f in Funcs([:Y,X:],Z) and
A2: rng f c= PFuncs(V1,V2) and
A3: dom f = X;
A4: uncurry' f = ~uncurry f by FUNCT_5:def 4;
A5: (ex g st uncurry f = g & dom g = [:X,Y:] & rng g c= Z) or ex g st
  uncurry' f = g & dom g = [:Y,X:] & rng g c= Z by A1,FUNCT_2:def 2;
  then
A6: dom uncurry' f = [:Y,X:] by A4,FUNCT_4:46;
  rng f c= Funcs(Y,Z)
  proof
    let y;
    assume
A7: y in rng f;
    then consider x such that
A8: x in dom f and
A9: y = f.x by FUNCT_1:def 3;
    ex g st y = g & dom g c= V1 & rng g c= V2 by A2,A7,PARTFUN1:def 3;
    then reconsider h = y as Function;
A10: dom h = Y
    proof
      thus dom h c= Y
      proof
        let z;
        assume z in dom h;
        then [z,x] in dom uncurry' f by A8,A9,FUNCT_5:39;
        hence thesis by A6,ZFMISC_1:87;
      end;
      let z;
      assume z in Y;
      then [z,x] in [:Y,X:] by A3,A8,ZFMISC_1:87;
      then [x,z] in dom uncurry f by A4,A6,FUNCT_4:42;
      then consider y1,f1,y2 such that
A11:  [x,z] = [y1,y2] and
      y1 in dom f and
A12:  f1 = f.y1 & y2 in dom f1 by FUNCT_5:def 2;
      x = y1 by A11,XTUPLE_0:1;
      hence thesis by A9,A11,A12,XTUPLE_0:1;
    end;
    rng h c= Z
    proof
      let z;
      assume z in rng h;
      then ex y1 st y1 in dom h & z = h.y1 by FUNCT_1:def 3;
      then z in rng uncurry f & z in rng uncurry' f by A8,A9,FUNCT_5:38,39;
      hence thesis by A5;
    end;
    hence thesis by A10,FUNCT_2:def 2;
  end;
  hence thesis by A3,FUNCT_2:def 2;
end;

theorem
  f in PFuncs([:X,Y:],Z) implies curry f in PFuncs(X,PFuncs(Y,Z)) &
  curry' f in PFuncs(Y,PFuncs(X,Z))
proof
  assume f in PFuncs([:X,Y:],Z);
  then
A1: ex g st f = g & dom g c= [:X,Y:] & rng g c= Z by PARTFUN1:def 3;
  then proj1 [:X,Y:] c= X & proj1 dom f c= proj1 [:X,Y:] by FUNCT_5:10
,XTUPLE_0:8;
  then proj1 dom f c= X by XBOOLE_1:1;
  then
A2: PFuncs(proj1 dom f,rng f) c= PFuncs(X,Z) by A1,PARTFUN1:50;
  proj2 [:X,Y:] c= Y & proj2 dom f c= proj2 [:X,Y:] by A1,FUNCT_5:10,XTUPLE_0:9
;
  then proj2 dom f c= Y by XBOOLE_1:1;
  then
  rng curry f c= PFuncs(proj2 dom f,rng f) & PFuncs(proj2 dom f,rng f) c=
  PFuncs(Y,Z) by A1,FUNCT_5:36,PARTFUN1:50;
  then
A3: rng curry f c= PFuncs(Y,Z) by XBOOLE_1:1;
  rng curry' f c= PFuncs(proj1 dom f,rng f) by FUNCT_5:36;
  then
A4: rng curry' f c= PFuncs(X,Z) by A2,XBOOLE_1:1;
  dom curry f c= X & dom curry' f c= Y by A1,FUNCT_5:25;
  hence thesis by A3,A4,PARTFUN1:def 3;
end;

theorem Th15:
  f in PFuncs(X,PFuncs(Y,Z)) implies uncurry f in PFuncs([:X,Y:],Z
  ) & uncurry' f in PFuncs([:Y,X:],Z)
proof
  assume f in PFuncs(X,PFuncs(Y,Z));
  then
A1: ex g st f = g & dom g c= X & rng g c= PFuncs(Y,Z) by PARTFUN1:def 3;
  then dom uncurry f c= [:dom f,Y:] & [:dom f,Y:] c= [:X,Y:] by FUNCT_5:37
,ZFMISC_1:96;
  then
A2: dom uncurry f c= [:X,Y:] by XBOOLE_1:1;
  dom uncurry' f c= [:Y,dom f:] & [:Y,dom f:] c= [:Y,X:] by A1,FUNCT_5:37
,ZFMISC_1:96;
  then
A3: dom uncurry' f c= [:Y,X:] by XBOOLE_1:1;
  rng uncurry f c= Z & rng uncurry' f c= Z by A1,FUNCT_5:40;
  hence thesis by A2,A3,PARTFUN1:def 3;
end;

theorem
  (curry f in PFuncs(X,PFuncs(Y,Z)) or curry' f in PFuncs(Y,PFuncs(X,Z))
  ) & dom f c= [:V1,V2:] implies f in PFuncs([:X,Y:],Z)
proof
  assume
  curry f in PFuncs(X,PFuncs(Y,Z)) or curry' f in PFuncs(Y,PFuncs(X,Z) );
  then
A1: uncurry curry f in PFuncs([:X,Y:],Z) or uncurry' curry' f in PFuncs([:X,
  Y:],Z) by Th15;
  assume dom f c= [:V1,V2:];
  hence thesis by A1,FUNCT_5:50;
end;

theorem
  (uncurry f in PFuncs([:X,Y:],Z) or uncurry' f in PFuncs([:Y,X:],Z)) &
  rng f c= PFuncs(V1,V2) & dom f c= X implies f in PFuncs(X,PFuncs(Y,Z))
proof
  assume that
A1: uncurry f in PFuncs([:X,Y:],Z) or uncurry' f in PFuncs([:Y,X:],Z) and
A2: rng f c= PFuncs(V1,V2) and
A3: dom f c= X;
A4: (ex g st uncurry f = g & dom g c= [:X,Y:] & rng g c= Z) or ex g st
  uncurry' f = g & dom g c= [:Y,X:] & rng g c= Z by A1,PARTFUN1:def 3;
  uncurry' f = ~uncurry f by FUNCT_5:def 4;
  then
A5: dom uncurry' f c= [:Y,X:] by A4,FUNCT_4:45;
  rng f c= PFuncs(Y,Z)
  proof
    let y;
    assume
A6: y in rng f;
    then consider x such that
A7: x in dom f & y = f.x by FUNCT_1:def 3;
    ex g st y = g & dom g c= V1 & rng g c= V2 by A2,A6,PARTFUN1:def 3;
    then reconsider h = y as Function;
A8: rng h c= Z
    proof
      let z;
      assume z in rng h;
      then ex y1 st y1 in dom h & z = h.y1 by FUNCT_1:def 3;
      then z in rng uncurry f & z in rng uncurry' f by A7,FUNCT_5:38,39;
      hence thesis by A4;
    end;
    dom h c= Y
    proof
      let z;
      assume z in dom h;
      then [z,x] in dom uncurry' f by A7,FUNCT_5:39;
      hence thesis by A5,ZFMISC_1:87;
    end;
    hence thesis by A8,PARTFUN1:def 3;
  end;
  hence thesis by A3,PARTFUN1:def 3;
end;

begin

definition
  let X be set;
  func SubFuncs X means
  :Def1:
  x in it iff x in X & x is Function;
  existence
  proof
    defpred P[set] means $1 is Function;
    ex Y being set st for x being set holds x in Y iff x in X & P[x] from
    XBOOLE_0:sch 1;
    hence thesis;
  end;
  uniqueness
  proof
    defpred P[set] means $1 in X & $1 is Function;
    let X1,X2 be set such that
A1: for x being set holds x in X1 iff P[x] and
A2: for x being set holds x in X2 iff P[x];
    thus X1 = X2 from XBOOLE_0:sch 2(A1,A2);
  end;
end;

theorem Th18:
  SubFuncs X c= X
proof
  let x;
  thus thesis by Def1;
end;

theorem Th19:
  x in f"SubFuncs rng f iff x in dom f & f.x is Function
proof
  f.x in rng f & f.x is Function iff f.x in SubFuncs rng f by Def1;
  hence thesis by FUNCT_1:def 3,def 7;
end;

Lm1: (for x st x in X holds x is Function) implies SubFuncs X = X
proof
  assume x in X implies x is Function;
  then x in X iff x in X & x is Function;
  hence thesis by Def1;
end;

theorem
  SubFuncs {} = {} & SubFuncs {f} = {f} & SubFuncs {f,g} = {f,g} &
  SubFuncs {f,g,h} = {f,g,h}
proof
  thus SubFuncs {} = {} by Th18,XBOOLE_1:3;
  for x st x in {f} holds x is Function;
  hence SubFuncs {f} = {f} by Lm1;
  for x st x in {f,g} holds x is Function;
  hence SubFuncs {f,g} = {f,g} by Lm1;
  for x st x in {f,g,h} holds x is Function by ENUMSET1:def 1;
  hence thesis by Lm1;
end;

theorem
  Y c= SubFuncs X implies SubFuncs Y = Y
proof
  assume Y c= SubFuncs X;
  then for x st x in Y holds x is Function by Def1;
  hence thesis by Lm1;
end;

definition
  let f be Function;
  func doms f -> Function means
  : Def2:
  dom it = f"SubFuncs rng f & for x st x
  in f"SubFuncs rng f holds it.x = proj1 (f.x);
  existence
  proof
    deffunc F(set) = proj1 (f.$1);
    ex F be Function st dom F = f"SubFuncs rng f & for x st x in f"
    SubFuncs rng f holds F.x = F(x) from FUNCT_1:sch 3;
    hence thesis;
  end;
  uniqueness
  proof
    let f1,f2 be Function such that
A1: dom f1 = f"SubFuncs rng f and
A2: for x st x in f"SubFuncs rng f holds f1.x = proj1 (f.x) and
A3: dom f2 = f"SubFuncs rng f and
A4: for x st x in f"SubFuncs rng f holds f2.x = proj1 (f.x);
    now
      let x;
      assume
A5:   x in f"SubFuncs rng f;
      then f1.x = proj1 (f.x) by A2;
      hence f1.x = f2.x by A4,A5;
    end;
    hence thesis by A1,A3,FUNCT_1:2;
  end;
  func rngs f -> Function means
  : Def3:
  dom it = f"SubFuncs rng f & for x st x
  in f"SubFuncs rng f holds it.x = proj2 (f.x);
  existence
  proof
    deffunc F(set) = proj2 (f.$1);
    ex F be Function st dom F = f"SubFuncs rng f & for x st x in f"
    SubFuncs rng f holds F.x = F(x) from FUNCT_1:sch 3;
    hence thesis;
  end;
  uniqueness
  proof
    let f1,f2 be Function such that
A6: dom f1 = f"SubFuncs rng f and
A7: for x st x in f"SubFuncs rng f holds f1.x = proj2 (f.x) and
A8: dom f2 = f"SubFuncs rng f and
A9: for x st x in f"SubFuncs rng f holds f2.x = proj2 (f.x);
    now
      let x;
      assume
A10:  x in f"SubFuncs rng f;
      then f1.x = proj2 (f.x) by A7;
      hence f1.x = f2.x by A9,A10;
    end;
    hence thesis by A6,A8,FUNCT_1:2;
  end;
  func meet f equals
  meet rng f;
  correctness;
end;

theorem Th22:
  x in dom f & g = f.x implies x in dom doms f & (doms f).x = dom
  g & x in dom rngs f & (rngs f).x = rng g
proof
  assume that
A1: x in dom f and
A2: g = f.x;
  g in rng f by A1,A2,FUNCT_1:def 3;
  then g in SubFuncs rng f by Def1;
  then x in f"SubFuncs rng f by A1,A2,FUNCT_1:def 7;
  hence thesis by A2,Def2,Def3;
end;

theorem
  doms {} = {} & rngs {} = {}
proof
  dom doms {} = {}"SubFuncs {} & dom rngs {} = {}"SubFuncs {} by Def2,Def3,
RELAT_1:38;
  hence thesis;
end;

theorem Th24:
  doms (X --> f) = X --> dom f & rngs (X --> f) = X --> rng f
proof
A1: dom (X --> dom f) = X & dom doms (X --> f) = (X --> f)"SubFuncs rng (X
  --> f ) by Def2,FUNCOP_1:13;
A2: rng (X --> f) c= {f} by FUNCOP_1:13;
A3: SubFuncs rng (X --> f) = rng (X --> f)
  proof
    thus SubFuncs rng (X --> f) c= rng (X --> f) by Th18;
    let x;
    assume x in rng (X --> f);
    hence thesis by A2,Def1;
  end;
A4: dom (X --> f) = X & (X --> f)"rng (X --> f) = dom (X --> f) by FUNCOP_1:13
,RELAT_1:134;
  now
    let x;
    assume
A5: x in X;
    then (X --> f).x = f & (X --> dom f).x = dom f by FUNCOP_1:7;
    hence (doms (X --> f)).x = (X --> dom f).x by A4,A3,A5,Def2;
  end;
  hence doms (X --> f) = X --> dom f by A1,A4,A3,FUNCT_1:2;
A6: now
    let x;
    assume
A7: x in X;
    then (X --> f).x = f & (X --> rng f).x = rng f by FUNCOP_1:7;
    hence (rngs (X --> f)).x = (X --> rng f).x by A4,A3,A7,Def3;
  end;
  dom (X --> rng f) = X & dom rngs (X --> f) = (X --> f)"SubFuncs rng (X
  --> f ) by Def3,FUNCOP_1:13;
  hence thesis by A4,A3,A6,FUNCT_1:2;
end;

theorem Th25:
  f <> {} implies (x in meet f iff for y st y in dom f holds x in f.y)
proof
  assume
A1: f <> {};
  thus x in meet f implies for y st y in dom f holds x in f.y
  proof
    assume
A2: x in meet f;
    let y;
    assume y in dom f;
    then f.y in rng f by FUNCT_1:def 3;
    then meet f c= f.y by SETFAM_1:3;
    hence thesis by A2;
  end;
  assume
A3: for y st y in dom f holds x in f.y;
  now
    let z;
    assume z in rng f;
    then ex y st y in dom f & z = f.y by FUNCT_1:def 3;
    hence x in z by A3;
  end;
  hence thesis by A1,SETFAM_1:def 1;
end;

theorem
  Union ({} --> Y) = {} & meet ({} --> Y) = {}
proof
  union rng ({} --> Y) = {} by ZFMISC_1:2;
  hence thesis by CARD_3:def 4,SETFAM_1:def 1;
end;

theorem Th27:
  X <> {} implies Union (X --> Y) = Y & meet (X --> Y) = Y
proof
  assume X <> {};
  then
A1: rng (X --> Y) = {Y} by FUNCOP_1:8;
  then union rng (X --> Y) = Y by ZFMISC_1:25;
  hence thesis by A1,CARD_3:def 4,SETFAM_1:10;
end;

definition
  let f be Function, x, y be set;
  func f..(x,y) -> set equals
  (uncurry f).(x,y);
  correctness;
end;

theorem
  x in X & y in dom f implies (X --> f)..(x,y) = f.y
proof
A1: x in X implies (X --> f).x = f by FUNCOP_1:7;
  dom (X --> f) = X by FUNCOP_1:13;
  hence thesis by A1,FUNCT_5:38;
end;

begin

definition
  let f be Function;
  func <:f:> -> Function equals
  curry ((uncurry' f)|([:meet doms f, dom f:]
  qua set));
  correctness;
end;

theorem Th29:
  dom <:f:> = meet doms f & rng <:f:> c= product rngs f
proof
  set f9 = (uncurry' f)|([:meet doms f, dom f:] qua set);
  dom ((uncurry' f)|([:meet doms f, dom f:] qua set)) c= [:meet doms f,
  dom f:] by RELAT_1:58;
  hence
A1: dom <:f:> c= meet doms f by FUNCT_5:25;
A2: dom doms f = f"SubFuncs rng f by Def2;
  thus meet doms f c= dom <:f:>
  proof
    set y = the Element of rng doms f;
    let x;
    assume
A3: x in meet doms f;
    then
A4: rng doms f <> {} by SETFAM_1:def 1;
    then
A5: x in y by A3,SETFAM_1:def 1;
    consider z such that
A6: z in dom doms f and
A7: y = (doms f).z by A4,FUNCT_1:def 3;
    f.z in SubFuncs rng f by A2,A6,FUNCT_1:def 7;
    then reconsider g = f.z as Function by Def1;
A8: z in dom f by A2,A6,FUNCT_1:def 7;
    then y = dom g by A7,Th22;
    then
A9: [x,z] in dom uncurry' f by A8,A5,FUNCT_5:39;
    [x,z] in [:meet doms f, dom f:] by A3,A8,ZFMISC_1:87;
    then [x,z] in dom (uncurry' f) /\ [:meet doms f, dom f:] by A9,
XBOOLE_0:def 4;
    then [x,z] in dom ((uncurry' f)|([:meet doms f, dom f:] qua set)) by
RELAT_1:61;
    then
x in proj1 dom ((uncurry' f)|([:meet doms f, dom f:] qua set))
                  by XTUPLE_0:def 12;
    hence thesis by FUNCT_5:def 1;
  end;
A10: dom rngs f = f"SubFuncs rng f by Def3;
  thus rng <:f:> c= product rngs f
  proof
    let y;
A11: dom f9 = dom (uncurry' f) /\ [:meet doms f, dom f:] by RELAT_1:61;
A12: uncurry' f = ~(uncurry f) by FUNCT_5:def 4;
    assume y in rng <:f:>;
    then consider x such that
A13: x in dom <:f:> and
A14: y = <:f:>.x by FUNCT_1:def 3;
    reconsider g = y as Function by A13,A14,FUNCT_5:30;
A15: dom g = proj2 (dom f9 /\ [:{x},proj2 dom f9:]) by A13,A14,FUNCT_5:31;
A16: dom g = dom rngs f
    proof
      thus dom g c= dom rngs f
      proof
        let z;
        assume z in dom g;
        then [x,z] in dom f9 by A13,A14,FUNCT_5:31;
        then [x,z] in dom uncurry' f by A11,XBOOLE_0:def 4;
        then [z,x] in dom uncurry f by A12,FUNCT_4:42;
        then consider y1, h, y2 such that
A17:    [z,x] = [y1,y2] and
A18:    y1 in dom f and
A19:    h = f.y1 and
        y2 in dom h by FUNCT_5:def 2;
A20:    z = y1 by A17,XTUPLE_0:1;
        h in rng f by A18,A19,FUNCT_1:def 3;
        then f.z in SubFuncs rng f by A19,A20,Def1;
        hence thesis by A10,A18,A20,FUNCT_1:def 7;
      end;
      let z;
A21:  x in {x} by TARSKI:def 1;
      assume
A22:  z in dom rngs f;
      then f.z in SubFuncs rng f by A10,FUNCT_1:def 7;
      then reconsider h = f.z as Function by Def1;
A23:  z in dom f by A10,A22,FUNCT_1:def 7;
      then dom h = (doms f).z by Th22;
      then dom h in rng doms f by A2,A10,A22,FUNCT_1:def 3;
      then x in dom h by A1,A13,SETFAM_1:def 1;
      then [z,x] in dom uncurry f by A23,FUNCT_5:def 2;
      then
A24:  [x,z] in dom uncurry' f by A12,FUNCT_4:42;
      [x,z] in [:meet doms f, dom f:] by A1,A13,A23,ZFMISC_1:87;
      then
A25:  [x,z] in dom f9 by A11,A24,XBOOLE_0:def 4;
      then z in proj2 dom f9 by XTUPLE_0:def 13;
      then [x,z] in [:{x},proj2 dom f9:] by A21,ZFMISC_1:87;
      then [x,z] in dom f9 /\ [:{x},proj2 dom f9:] by A25,XBOOLE_0:def 4;
      hence thesis by A15,XTUPLE_0:def 13;
    end;
    now
      let z;
      assume
A26:  z in dom rngs f;
      then f.z in SubFuncs rng f by A10,FUNCT_1:def 7;
      then reconsider h = f.z as Function by Def1;
A27:  z in dom f by A10,A26,FUNCT_1:def 7;
      then dom h = (doms f).z by Th22;
      then dom h in rng doms f by A2,A10,A26,FUNCT_1:def 3;
      then
A28:  x in dom h by A1,A13,SETFAM_1:def 1;
      then
A29:  h.x in rng h by FUNCT_1:def 3;
      g.z = f9.(x,z) & [x,z] in dom f9 by A13,A14,A16,A26,FUNCT_5:31;
      then (uncurry' f).(x,z) = g.z by FUNCT_1:47;
      then g.z = h.x by A27,A28,FUNCT_5:39;
      hence g.z in (rngs f).z by A27,A29,Th22;
    end;
    hence thesis by A16,CARD_3:def 5;
  end;
end;

theorem Th30:
  x in dom <:f:> implies <:f:>.x is Function
proof
  assume x in dom <:f:>;
  then
A1: <:f:>.x in rng <:f:> by FUNCT_1:def 3;
  rng <:f:> c= product rngs f by Th29;
  then ex g st <:f:>.x = g & dom g = dom rngs f & for x st x in dom rngs f
  holds g.x in (rngs f).x by A1,CARD_3:def 5;
  hence thesis;
end;

theorem Th31:
  x in dom <:f:> & g = <:f:>.x implies dom g = f"SubFuncs rng f &
  for y st y in dom g holds [y,x] in dom uncurry f & g.y = (uncurry f).(y,x)
proof
A1: rng <:f:> c= product rngs f & dom rngs f = f"SubFuncs rng f by Def3,Th29;
  assume
A2: x in dom <:f:> & g = <:f:>.x;
  then g in rng <:f:> by FUNCT_1:def 3;
  hence dom g = f"SubFuncs rng f by A1,CARD_3:9;
  let y such that
A3: y in dom g;
A4: [x,y] in dom ((uncurry' f)|([:meet doms f, dom f:] qua set)) by A2,A3,
FUNCT_5:31;
  then [x,y] in dom (uncurry' f) /\ [:meet doms f, dom f:] by RELAT_1:61;
  then
A5: [x,y] in dom uncurry' f by XBOOLE_0:def 4;
  g.y = ((uncurry' f)|([:meet doms f, dom f:] qua set)).(x,y) by A2,A3,
FUNCT_5:31;
  then
A6: g.y = (uncurry' f).(x,y) by A4,FUNCT_1:47;
  ~(uncurry f) = uncurry' f by FUNCT_5:def 4;
  hence thesis by A6,A5,FUNCT_4:42,43;
end;

theorem Th32:
  x in dom <:f:> implies for g st g in rng f holds x in dom g
proof
  assume
A1: x in dom <:f:>;
  let g;
  assume g in rng f;
  then consider y such that
A2: y in dom f & g = f.y by FUNCT_1:def 3;
  y in dom doms f & (doms f).y = dom g by A2,Th22;
  then dom g in rng doms f by FUNCT_1:def 3;
  then
A3: meet rng doms f c= dom g by SETFAM_1:3;
  meet doms f = dom <:f:> by Th29;
  hence thesis by A1,A3;
end;

theorem
  g in rng f & (for g st g in rng f holds x in dom g) implies x in dom <:f :>
proof
  assume that
A1: g in rng f and
A2: for g st g in rng f holds x in dom g;
  ex y st y in dom f & g = f.y by A1,FUNCT_1:def 3;
  then
A3: doms f <> {} by Th22,RELAT_1:38;
A4: now
    let y;
    assume y in dom doms f;
    then
A5: y in f"SubFuncs rng f by Def2;
    then f.y in SubFuncs rng f by FUNCT_1:def 7;
    then reconsider g = f.y as Function by Def1;
A6: y in dom f by A5,FUNCT_1:def 7;
    then g in rng f by FUNCT_1:def 3;
    then x in dom g by A2;
    hence x in (doms f).y by A6,Th22;
  end;
  dom <:f:> = meet doms f by Th29;
  hence thesis by A3,A4,Th25;
end;

theorem Th34:
  x in dom f & g = f.x & y in dom <:f:> & h = <:f:>.y implies g.y = h.x
proof
  assume that
A1: x in dom f & g = f.x and
A2: y in dom <:f:> and
A3: h = <:f:>.y;
  dom h = f"SubFuncs rng f by A2,A3,Th31;
  then x in dom h by A1,Th19;
  then
A4: h.x = (uncurry f).(x,y) by A2,A3,Th31;
  g in rng f by A1,FUNCT_1:def 3;
  then y in dom g by A2,Th32;
  hence thesis by A1,A4,FUNCT_5:38;
end;

theorem
  x in dom f & f.x is Function & y in dom <:f:> implies f..(x,y) = <:f:>
  ..( y, x )
proof
  assume that
A1: x in dom f and
A2: f.x is Function and
A3: y in dom <:f:>;
  reconsider g = f.x, h = <:f:>.y as Function by A2,A3,Th30;
A4: dom h = f"SubFuncs rng f by A3,Th31;
A5: g in rng f by A1,FUNCT_1:def 3;
  then g in SubFuncs rng f by Def1;
  then
A6: x in dom h by A1,A4,FUNCT_1:def 7;
  y in dom g by A3,A5,Th32;
  hence f..(x,y) = g.y by A1,FUNCT_5:38
    .= h.x by A1,A3,Th34
    .= <:f:>..(y,x) by A3,A6,FUNCT_5:38;
end;

definition
  let f be Function;
  func Frege f -> Function means
  : Def7:
  dom it = product doms f & for g st g
  in product doms f ex h st it.g = h & dom h = f"SubFuncs rng f & for x st x in
  dom h holds h.x = (uncurry f).(x,g.x);
  existence
  proof
    defpred P[set,set] means ex g,h st $1 = g & $2 = h & dom h = f"SubFuncs
    rng f & for z st z in dom h holds h.z = (uncurry f).(z,g.z);
A1: for x st x in product doms f ex y st P[x,y]
    proof
      let x;
      assume x in product doms f;
      then consider g such that
A2:   x = g and
      dom g = dom doms f and
      for x st x in dom doms f holds g.x in (doms f).x by CARD_3:def 5;
      deffunc F(set) = (uncurry f).[$1,g.$1];
      consider h such that
A3:   dom h = f"SubFuncs rng f & for z st z in f"SubFuncs rng f holds
      h.z = F(z) from FUNCT_1:sch 3;
      reconsider y = h as set;
      take y, g, h;
      thus thesis by A2,A3;
    end;
    consider F being Function such that
A4: dom F = product doms f & for x st x in product doms f holds P[x,F.
    x] from CLASSES1:sch 1(A1);
    take F;
    thus dom F = product doms f by A4;
    let g;
    assume g in product doms f;
    then
    ex g1,h being Function st g = g1 & F.g = h & dom h = f"SubFuncs rng f
    & for z st z in dom h holds h.z = (uncurry f).(z,g1.z) by A4;
    hence thesis;
  end;
  uniqueness
  proof
    let f1,f2 be Function such that
A5: dom f1 = product doms f and
A6: for g st g in product doms f ex h st f1.g = h & dom h = f"SubFuncs
    rng f & for x st x in dom h holds h.x = (uncurry f).(x,g.x) and
A7: dom f2 = product doms f and
A8: for g st g in product doms f ex h st f2.g = h & dom h = f"
    SubFuncs rng f & for x st x in dom h holds h.x = (uncurry f).(x,g.x);
    now
      let x;
      assume
A9:   x in product doms f;
      then consider g such that
A10:  x = g and
      dom g = dom doms f and
      for x st x in dom doms f holds g.x in (doms f).x by CARD_3:def 5;
      consider h2 being Function such that
A11:  f2.g = h2 and
A12:  dom h2 = f"SubFuncs rng f and
A13:  for y st y in dom h2 holds h2.y = (uncurry f).(y,g.y) by A8,A9,A10;
      consider h1 being Function such that
A14:  f1.g = h1 and
A15:  dom h1 = f"SubFuncs rng f and
A16:  for y st y in dom h1 holds h1.y = (uncurry f).(y,g.y) by A6,A9,A10;
      now
        let z;
        assume
A17:    z in f"SubFuncs rng f;
        then h1.z = (uncurry f).(z,g.z) by A15,A16;
        hence h1.z = h2.z by A12,A13,A17;
      end;
      hence f1.x = f2.x by A10,A14,A15,A11,A12,FUNCT_1:2;
    end;
    hence thesis by A5,A7,FUNCT_1:2;
  end;
end;

theorem
  g in product doms f & x in dom g implies (Frege f)..(g,x) = f..(x,g.x)
proof
  assume that
A1: g in product doms f and
A2: x in dom g;
A3: dom g = dom doms f by A1,CARD_3:9;
A4: dom doms f = f"SubFuncs rng f by Def2;
  consider h such that
A5: (Frege f).g = h and
A6: dom h = f"SubFuncs rng f and
A7: for x st x in dom h holds h.x = (uncurry f).(x,g.x) by A1,Def7;
  dom Frege f = product doms f by Def7;
  hence (Frege f)..(g,x) = h.x by A1,A2,A5,A6,A3,A4,FUNCT_5:38
    .= f..(x,g.x) by A2,A6,A7,A3,A4;
end;

Lm2: rng Frege f c= product rngs f
proof
  let x;
  assume x in rng Frege f;
  then consider y such that
A1: y in dom Frege f and
A2: x = (Frege f).y by FUNCT_1:def 3;
A3: dom Frege f = product doms f by Def7;
  then consider g such that
A4: y = g and
  dom g = dom doms f and
A5: for z st z in dom doms f holds g.z in (doms f).z by A1,CARD_3:def 5;
  consider h such that
A6: (Frege f).g = h and
A7: dom h = f"SubFuncs rng f and
A8: for z st z in dom h holds h.z = (uncurry f).(z,g.z) by A1,A3,A4,Def7;
A9: dom rngs f = f"SubFuncs rng f by Def3;
A10: dom doms f = f"SubFuncs rng f by Def2;
  now
    let z;
    assume
A11: z in dom rngs f;
    then f.z in SubFuncs rng f by A9,FUNCT_1:def 7;
    then reconsider t = f.z as Function by Def1;
A12: g.z in (doms f). z by A5,A9,A10,A11;
A13: z in dom f by A9,A11,FUNCT_1:def 7;
    then
A14: (rngs f).z = rng t by Th22;
A15: (doms f).z = dom t by A13,Th22;
    then
A16: t.(g.z) in rng t by A12,FUNCT_1:def 3;
    t.(g.z) = (uncurry f).(z,g.z) by A13,A12,A15,FUNCT_5:38;
    hence h.z in (rngs f).z by A7,A8,A9,A11,A14,A16;
  end;
  hence thesis by A2,A4,A6,A7,A9,CARD_3:def 5;
end;

theorem
  x in dom f & g = f.x & h in product doms f & h9 = (Frege f).h
  implies h.x in dom g & h9.x = g.(h.x) & h9 in product rngs f
proof
  assume that
A1: x in dom f & g = f.x and
A2: h in product doms f and
A3: h9 = (Frege f).h;
A4: x in f"SubFuncs rng f by A1,Th19;
A5: dom doms f = f"SubFuncs rng f by Def2;
  ( ex f2 st h = f2 & dom f2 = dom doms f & for x st x in dom doms f holds
  f2.x in (doms f).x)& (doms f).x = dom g by A1,A2,Th22,CARD_3:def 5;
  hence
A6: h.x in dom g by A5,A4;
  ex f1 st (Frege f).h = f1 & dom f1 = f"SubFuncs rng f & for x st x in
  dom f1 holds f1.x = (uncurry f).(x,h.x) by A2,Def7;
  hence h9.x = (uncurry f).(x,h.x) by A3,A4
    .= g.(h.x) by A1,A6,FUNCT_5:38;
A7: rng Frege f c= product rngs f by Lm2;
  dom Frege f = product doms f by Def7;
  then h9 in rng Frege f by A2,A3,FUNCT_1:def 3;
  hence thesis by A7;
end;

Lm3: product rngs f c= rng Frege f
proof
  let x;
  deffunc F(set) = (doms f).$1;
  assume x in product rngs f;
  then consider g such that
A1: x = g and
A2: dom g = dom rngs f and
A3: for y st y in dom rngs f holds g.y in (rngs f).y by CARD_3:def 5;
  defpred P[set,set] means ex f1 st f1 = f.$1 & f1.$2 = g.$1;
  consider h such that
A4: dom h = f"SubFuncs rng f & for x st x in f"SubFuncs rng f for y
  holds y in h.x iff y in F(x) & P[x,y] from CARD_3:sch 2;
A5: product doms f = dom Frege f by Def7;
A6: dom doms f = f"SubFuncs rng f by Def2;
A7: dom rngs f = f"SubFuncs rng f by Def3;
A8: now
    let X;
    assume X in rng h;
    then consider x such that
A9: x in dom h and
A10: X = h.x by FUNCT_1:def 3;
    f.x in SubFuncs rng f by A4,A9,FUNCT_1:def 7;
    then reconsider fx = f.x as Function by Def1;
A11: x in dom f by A4,A9,FUNCT_1:def 7;
    then
A12: (rngs f).x = rng fx by Th22;
    g.x in (rngs f).x by A3,A7,A4,A9;
    then
A13: ex y st y in dom fx & g.x = fx.y by A12,FUNCT_1:def 3;
    (doms f).x = dom fx by A11,Th22;
    hence X <> {} by A4,A9,A10,A13;
  end;
A14: now
    assume f"SubFuncs rng f <> {};
    then reconsider D = rng h as non empty set by A4,RELAT_1:42;
    consider Ch being Function such that
A15: dom Ch = D and
A16: for x st x in D holds Ch.x in x by A8,FUNCT_1:111;
A17: dom (Ch*h) = dom h by A15,RELAT_1:27;
    now
      let y;
      assume
A18:  y in dom doms f;
      then (Ch*h).y = Ch.(h.y) & h.y in rng h by A6,A4,A17,FUNCT_1:12,def 3;
      then (Ch*h).y in h.y by A16;
      hence (Ch*h).y in (doms f).y by A6,A4,A18;
    end;
    then
A19: Ch*h in product doms f by A6,A4,A17,CARD_3:def 5;
    then consider h1 being Function such that
A20: (Frege f).(Ch*h) = h1 and
A21: dom h1 = f"SubFuncs rng f and
A22: for x st x in dom h1 holds h1.x = (uncurry f).(x,(Ch*h).x) by Def7;
    now
      let z;
      assume
A23:  z in f"SubFuncs rng f;
      then
A24:  h1.z = (uncurry f).(z,(Ch*h).z) by A21,A22;
      (Ch*h).z = Ch.( h.z ) & h.z in rng h by A4,A17,A23,FUNCT_1:12,def 3;
      then
A25:  (Ch*h).z in h.z by A16;
      then consider f1 such that
A26:  f1 = f.z and
A27:  f1.((Ch*h).z) = g.z by A4,A23;
A28:  (Ch*h).z in (doms f).z by A4,A23,A25;
A29:  z in dom f by A23,FUNCT_1:def 7;
      then (doms f).z = dom f1 by A26,Th22;
      hence h1.z = g.z by A29,A24,A26,A27,A28,FUNCT_5:38;
    end;
    then x = h1 by A1,A2,A7,A21,FUNCT_1:2;
    hence thesis by A5,A19,A20,FUNCT_1:def 3;
  end;
  now
    assume
A30: f"SubFuncs rng f = {};
    then
A31: g = {} by A2,Def3;
    dom Frege f = {{}} by A6,A5,A30,CARD_3:10,RELAT_1:41;
    then
A32: {} in dom Frege f by TARSKI:def 1;
    then consider h such that
A33: (Frege f).{} = h and
A34: dom h = f"SubFuncs rng f and
    for x st x in dom h holds h.x = (uncurry f).(x,{} .x) by A5,Def7;
    h = {} by A30,A34;
    hence thesis by A1,A31,A32,A33,FUNCT_1:def 3;
  end;
  hence thesis by A14;
end;

theorem Th38:
  rng Frege f = product rngs f
proof
  thus rng Frege f c= product rngs f & product rngs f c= rng Frege f by Lm2,Lm3
;
end;

theorem Th39:
  not {} in rng f implies (Frege f is one-to-one iff for g st g in
  rng f holds g is one-to-one)
proof
  set h0 = the Element of product doms f;
A1: dom doms f = f"SubFuncs rng f by Def2;
  assume
A2: not {} in rng f;
  now
    assume {} in rng doms f;
    then consider x such that
A3: x in dom doms f and
A4: {} = (doms f).x by FUNCT_1:def 3;
A5: x in f"SubFuncs rng f by A3,Def2;
    then f.x in SubFuncs rng f by FUNCT_1:def 7;
    then reconsider g = f.x as Function by Def1;
A6: x in dom f by A5,FUNCT_1:def 7;
    then
A7: g in rng f by FUNCT_1:def 3;
    {} = dom g by A4,A6,Th22;
    hence contradiction by A2,A7,RELAT_1:41;
  end;
  then product doms f <> {} by CARD_3:26;
  then consider h such that
  h0 = h and
  dom h = dom doms f and
A8: for x st x in dom doms f holds h.x in (doms f).x by CARD_3:def 5;
A9: dom Frege f = product doms f by Def7;
  thus Frege f is one-to-one implies for g st g in rng f holds g is one-to-one
  proof
    deffunc G(set) = h.$1;
    assume
A10: for x,y st x in dom Frege f & y in dom Frege f & (Frege f).x = (
    Frege f).y holds x = y;
    let g;
    assume g in rng f;
    then consider z such that
A11: z in dom f & g = f.z by FUNCT_1:def 3;
    defpred P[set] means $1 = z;
    let x,y;
    deffunc F(set) = x;
    consider h1 being Function such that
A12: dom h1 = f"SubFuncs rng f & for a st a in f"SubFuncs rng f holds
(P[a] implies h1.a = F(a)) & (not P[a] implies h1.a = G(a)) from PARTFUN1:sch 1
    ;
    assume that
A13: x in dom g and
A14: y in dom g and
A15: g.x = g.y;
    now
      let a;
      assume
A16:  a in dom doms f;
      then
A17:  a <> z implies h1.a = h.a by A1,A12;
      a = z implies h1.a = x by A1,A12,A16;
      hence h1.a in (doms f).a by A8,A11,A13,A16,A17,Th22;
    end;
    then
A18: h1 in product doms f by A1,A12,CARD_3:def 5;
    then consider g1 being Function such that
A19: (Frege f).h1 = g1 and
A20: dom g1 = f"SubFuncs rng f and
A21: for x st x in dom g1 holds g1.x = (uncurry f).(x,h1.x) by Def7;
    defpred P[set] means $1 = z;
    deffunc G(set) = h.$1;
    deffunc F(set) = y;
    consider h2 being Function such that
A22: dom h2 = f"SubFuncs rng f & for a st a in f"SubFuncs rng f holds
(P[a] implies h2.a = F(a)) & (not P[a] implies h2.a = G(a)) from PARTFUN1:sch 1
    ;
    now
      let a;
      assume
A23:  a in dom doms f;
      then
A24:  a <> z implies h2.a = h.a by A1,A22;
      a = z implies h2.a = y by A1,A22,A23;
      hence h2.a in (doms f).a by A8,A11,A14,A23,A24,Th22;
    end;
    then
A25: h2 in product doms f by A1,A22,CARD_3:def 5;
    then consider g2 being Function such that
A26: (Frege f).h2 = g2 and
A27: dom g2 = f"SubFuncs rng f and
A28: for x st x in dom g2 holds g2.x = (uncurry f).(x,h2.x) by Def7;
    now
      let a;
      assume
A29:  a in f"SubFuncs rng f;
      then
A30:  g2.a = (uncurry f).(a,h2.a) by A27,A28;
A31:  g1.a = (uncurry f).(a,h1.a) by A20,A21,A29;
      per cases;
      suppose
A32:    a <> z;
        then h1.a = h.a by A12,A29;
        hence g1.a = g2.a by A22,A29,A30,A31,A32;
      end;
      suppose
A33:    a = z;
        then h1.a = x by A12,A29;
        then
A34:    g1.a = g.x by A11,A13,A31,A33,FUNCT_5:38;
        h2.a = y by A22,A29,A33;
        hence g1.a = g2.a by A11,A14,A15,A30,A33,A34,FUNCT_5:38;
      end;
    end;
    then g1 = g2 by A20,A27,FUNCT_1:2;
    then
A35: h1 = h2 by A9,A10,A18,A19,A25,A26;
    g in rng f by A11,FUNCT_1:def 3;
    then g in SubFuncs rng f by Def1;
    then
A36: z in f"SubFuncs rng f by A11,FUNCT_1:def 7;
    then h1.z = x by A12;
    hence thesis by A36,A22,A35;
  end;
  assume
A37: for g st g in rng f holds g is one-to-one;
  let x,y;
  assume that
A38: x in dom Frege f and
A39: y in dom Frege f and
A40: (Frege f).x = (Frege f).y;
  consider g2 being Function such that
A41: y = g2 and
A42: dom g2 = dom doms f and
A43: for x st x in dom doms f holds g2.x in (doms f).x by A9,A39,CARD_3:def 5;
  consider g1 being Function such that
A44: x = g1 and
A45: dom g1 = dom doms f and
A46: for x st x in dom doms f holds g1.x in (doms f).x by A9,A38,CARD_3:def 5;
  consider h2 being Function such that
A47: (Frege f).g2 = h2 and
A48: dom h2 = f"SubFuncs rng f & for x st x in dom h2 holds h2.x = (
  uncurry f).(x,g2.x) by A9,A39,A41,Def7;
  consider h1 being Function such that
A49: (Frege f).g1 = h1 and
A50: dom h1 = f"SubFuncs rng f & for x st x in dom h1 holds h1.x = (
  uncurry f).(x,g1.x) by A9,A38,A44,Def7;
  now
    let a;
    assume
A51: a in f"SubFuncs rng f;
    then f.a in SubFuncs rng f by FUNCT_1:def 7;
    then reconsider g = f.a as Function by Def1;
A52: a in dom f by A51,FUNCT_1:def 7;
    then
A53: dom g = (doms f).a by Th22;
    then
A54: g2.a in dom g by A1,A43,A51;
A55: g1.a in dom g by A1,A46,A51,A53;
    h2.a = (uncurry f).(a,g2.a) by A48,A51;
    then
A56: h2.a = g.(g2.a) by A52,A54,FUNCT_5:38;
    g in rng f by A52,FUNCT_1:def 3;
    then
A57: g is one-to-one by A37;
    h1.a = (uncurry f).(a,g1.a) by A50,A51;
    then h1.a = g.(g1.a) by A52,A55,FUNCT_5:38;
    hence g1.a = g2.a by A40,A44,A41,A49,A47,A55,A54,A56,A57,FUNCT_1:def 4;
  end;
  hence thesis by A1,A44,A45,A41,A42,FUNCT_1:2;
end;

begin

theorem
  <:{}:> = {} & Frege{} = {} .--> {}
proof
A1: dom doms {} = {}"SubFuncs {} by Def2,RELAT_1:38;
  then
A2: product doms {} = {{}} by CARD_3:10,RELAT_1:41;
A3: now
    let x;
    assume
A4: x in {{}};
    then
A5: x = {} by TARSKI:def 1;
    then
    ex h st (Frege{}).{} = h & dom h = {}"SubFuncs rng {} & for x st x in
    dom h holds h.x = (uncurry {}).(x,{} .x) by A2,A4,Def7;
    hence (Frege{}).x = {} by A5;
  end;
A6: meet({} qua set) = {} by SETFAM_1:def 1;
  rng doms {} = {} by A1,RELAT_1:38,41;
  then dom <:{}:> = {} by A6,Th29;
  hence <:{}:> = {};
  dom Frege{} = product doms {} by Def7;
  hence thesis by A2,A3,FUNCOP_1:11;
end;

theorem
  X <> {} implies dom <:X --> f:> = dom f & for x st x in dom f holds <:
  X --> f:>.x = X --> f.x
proof
  assume
A1: X <> {};
  thus
A2: dom <:X --> f:> = meet doms (X --> f) by Th29
    .= meet (X --> dom f) by Th24
    .= dom f by A1,Th27;
A3: rng <:X --> f:> c= product rngs (X --> f) & rngs (X --> f) = X --> rng f
  by Th24,Th29;
  let x;
  assume
A4: x in dom f;
  then <:X --> f:>.x in rng <:X --> f:> by A2,FUNCT_1:def 3;
  then consider g such that
A5: <:X --> f:>.x = g and
A6: dom g = dom (X --> rng f) and
  for y st y in dom (X --> rng f) holds g.y in (X --> rng f).y by A3,
CARD_3:def 5;
A7: dom g = X by A6,FUNCOP_1:13;
A8: dom (X --> f) = X by FUNCOP_1:13;
A9: now
    let y;
    assume
A10: y in X;
    then g.y = (uncurry (X --> f)).(y,x) & (X --> f).y = f by A2,A4,A5,A7,Th31,
FUNCOP_1:7;
    then g.y = f.x by A4,A8,A10,FUNCT_5:38;
    hence g.y = (X --> f.x).y by A10,FUNCOP_1:7;
  end;
  dom (X --> f.x) = X by FUNCOP_1:13;
  hence thesis by A5,A7,A9,FUNCT_1:2;
end;

theorem
  dom Frege(X-->f) = Funcs(X,dom f) & rng Frege(X-->f) = Funcs(X,rng f)
  & for g st g in Funcs(X,dom f) holds (Frege(X-->f)).g = f*g
proof
A1: product (X --> dom f) = Funcs(X, dom f) by CARD_3:11;
A2: doms (X --> f) = X --> dom f by Th24;
  rngs (X --> f) = X --> rng f & product (X --> rng f) = Funcs(X, rng f)
  by Th24,CARD_3:11;
  hence dom Frege(X-->f) = Funcs(X,dom f) & rng Frege(X-->f) = Funcs(X,rng f)
  by A2,A1,Def7,Th38;
  let g;
  assume
A3: g in Funcs(X,dom f);
  then consider h such that
A4: (Frege(X-->f)).g = h and
A5: dom h = (X --> f)"SubFuncs rng (X --> f) and
A6: for x st x in dom h holds h.x = (uncurry (X --> f)).(x,g.x) by A2,A1,Def7;
  dom (X --> dom f) = X by FUNCOP_1:13;
  then
A7: dom h = X by A2,A5,Def2;
A8: ex g9 being Function st g = g9 & dom g9 = X & rng g9 c= dom f by A3,
FUNCT_2:def 2;
  then
A9: dom (f*g) = X by RELAT_1:27;
A10: dom (X --> f) = X by FUNCOP_1:13;
  now
    let x;
    assume
A11: x in X;
    then
A12: h.x = (uncurry (X --> f)).(x,g.x) & (X --> f).x = f by A6,A7,FUNCOP_1:7;
    g.x in rng g & (f*g).x = f.(g.x) by A8,A9,A11,FUNCT_1:12,def 3;
    hence (f*g).x = h.x by A8,A10,A11,A12,FUNCT_5:38;
  end;
  hence thesis by A4,A7,A9,FUNCT_1:2;
end;

theorem Th43:
  dom f = X & dom g = X & (for x st x in X holds f.x,g.x
  are_equipotent) implies product f,product g are_equipotent
proof
  assume that
A1: dom f = X and
A2: dom g = X and
A3: for x st x in X holds f.x,g.x are_equipotent;
  defpred P[set,set] means ex f1 st $2 = f1 & f1 is one-to-one & dom f1 = f.$1
  & rng f1 = g.$1;
A4: for x st x in X ex y st P[x,y]
  proof
    let x;
    assume x in X;
    then f.x,g.x are_equipotent by A3;
    then ex h st h is one-to-one & dom h = f.x & rng h = g.x by WELLORD2:def 4;
    hence thesis;
  end;
  consider h such that
A5: dom h = X & for x st x in X holds P[x,h.x] from CLASSES1:sch 1(A4);
A6: now
    let x;
    assume
A7: x in X;
    then
    ex f1 st h.x = f1 & f1 is one-to-one & dom f1 = f.x & rng f1 = g.x by A5;
    hence (rngs h).x = g.x by A5,A7,Th22;
  end;
  now
    let x;
    assume x in rng h;
    then consider a such that
A8: a in X & x = h.a by A5,FUNCT_1:def 3;
    ex f1 st x = f1 & f1 is one-to-one & dom f1 = f.a & rng f1 = g.a by A5,A8;
    hence x is Function;
  end;
  then SubFuncs rng h = rng h by Lm1;
  then
A9: h"SubFuncs rng h = X by A5,RELAT_1:134;
  then dom rngs h = X by Def3;
  then
A10: rngs h = g by A2,A6,FUNCT_1:2;
A11: now
    let x;
    assume
A12: x in X;
    then
    ex f1 st h.x = f1 & f1 is one-to-one & dom f1 = f.x & rng f1 = g.x by A5;
    hence (doms h).x = f.x by A5,A12,Th22;
  end;
  dom doms h = X by A9,Def2;
  then
A13: doms h = f by A1,A11,FUNCT_1:2;
  now
    per cases;
    suppose
      {} in rng h;
      then consider x such that
A14:  x in X and
A15:  {} = h.x by A5,FUNCT_1:def 3;
A16:  ex f1 st {} = f1 & f1 is one-to-one & dom f1 = f.x & rng f1 = g.x
      by A5,A14,A15;
      then {} in rng f by A1,A14,FUNCT_1:def 3,RELAT_1:38;
      then
A17:  product f = {} by CARD_3:26;
      {} in rng g by A2,A14,A16,FUNCT_1:def 3,RELAT_1:38;
      hence thesis by A17,CARD_3:26;
    end;
    suppose
A18:  not {} in rng h;
A19:  now
        let f1;
        assume f1 in rng h;
        then consider x such that
A20:    x in X and
A21:    f1 = h.x by A5,FUNCT_1:def 3;
        ex f1 st h.x = f1 & f1 is one-to-one & dom f1 = f.x & rng f1 = g.
        x by A5,A20;
        hence f1 is one-to-one by A21;
      end;
      thus thesis
      proof
        take Frege h;
        thus thesis by A13,A10,A18,A19,Def7,Th38,Th39;
      end;
    end;
  end;
  hence thesis;
end;

theorem Th44:
  dom f = dom h & dom g = rng h & h is one-to-one & (for x st x in
  dom h holds f.x, g.(h.x) are_equipotent) implies product f,product g
  are_equipotent
proof
  set X = dom f;
  assume that
A1: dom f = dom h and
A2: dom g = rng h and
A3: h is one-to-one and
A4: for x st x in dom h holds f.x,g.(h.x) are_equipotent;
A5: h*(h") = id dom g by A2,A3,FUNCT_1:39;
A6: dom (g*h) = dom f by A1,A2,RELAT_1:27;
  now
    let x;
    assume
A7: x in dom f;
    then f.x,g.(h.x) are_equipotent by A1,A4;
    hence f.x,(g*h).x are_equipotent by A6,A7,FUNCT_1:12;
  end;
  then
A8: product f,product (g*h) are_equipotent by A6,Th43;
  defpred P[set,set] means ex f1 st $1 = f1 & $2 = f1*(h");
A9: for x st x in product (g*h) ex y st P[x,y]
  proof
    let x;
    assume x in product (g*h);
    then ex f1 st x = f1 & dom f1 = dom (g*h) & for y st y in dom (g*h) holds
    f1.y in (g*h).y by CARD_3:def 5;
    then consider f1 such that
A10: x = f1;
    f1*(h") = f1*(h");
    hence thesis by A10;
  end;
  consider F being Function such that
A11: dom F = product (g*h) & for x st x in product (g*h) holds P[x,F.x]
  from CLASSES1:sch 1(A9);
A12: rng (h") = X by A1,A3,FUNCT_1:33;
A13: F is one-to-one
  proof
    let x,y;
    assume that
A14: x in dom F and
A15: y in dom F and
A16: F.x = F.y;
    consider g2 being Function such that
A17: y = g2 and
A18: F.y = g2*(h") by A11,A15;
A19: dom g2 = X by A6,A11,A15,A17,CARD_3:9;
    consider g1 being Function such that
A20: x = g1 and
A21: F.x = g1*(h") by A11,A14;
    dom g1 = X by A6,A11,A14,A20,CARD_3:9;
    hence thesis by A12,A16,A20,A21,A17,A18,A19,FUNCT_1:86;
  end;
A22: (g*h)*(h") = g*(h*(h")) & g*id dom g = g by RELAT_1:36,52;
A23: product g c= rng F
  proof
    let x;
    assume x in product g;
    then consider f1 such that
A24: x = f1 and
A25: dom f1 = dom g and
A26: for z st z in dom g holds f1.z in g.z by CARD_3:def 5;
A27: dom (f1*h) = X by A1,A2,A25,RELAT_1:27;
    now
      let z;
      assume
A28:  z in X;
      then
A29:  h.z in dom g by A1,A2,FUNCT_1:def 3;
A30:  (h").(h.z) = z by A1,A3,A28,FUNCT_1:34;
      (f1*h).z = f1.(h.z) by A27,A28,FUNCT_1:12;
      then (f1*h).z in ((g*h)*(h")).(h.z) by A5,A22,A26,A29;
      hence (f1*h).z in (g*h).z by A5,A22,A29,A30,FUNCT_1:12;
    end;
    then
A31: f1*h in product (g*h) by A6,A27,CARD_3:def 5;
    then ex f2 st f1*h = f2 & F.(f1*h) = f2*(h") by A11;
    then F.(f1*h) = f1*id dom g by A5,RELAT_1:36
      .= x by A24,A25,RELAT_1:51;
    hence thesis by A11,A31,FUNCT_1:def 3;
  end;
A32: dom (h") = rng h by A3,FUNCT_1:33;
  rng F c= product g
  proof
    let x;
    assume x in rng F;
    then consider y such that
A33: y in dom F and
A34: x = F.y by FUNCT_1:def 3;
    consider f1 such that
A35: y = f1 and
A36: dom f1 = X and
A37: for z st z in X holds f1.z in (g*h).z by A6,A11,A33,CARD_3:def 5;
A38: dom (f1*(h")) = dom g by A2,A12,A32,A36,RELAT_1:27;
A39: now
      let z;
      assume
A40:  z in dom g;
      then
A41:  (h").z in X by A2,A12,A32,FUNCT_1:def 3;
      g.z = (g*h).((h").z) & (f1*(h")).z = f1.((h").z) by A5,A22,A38,A40,
FUNCT_1:12;
      hence (f1*(h")).z in g.z by A37,A41;
    end;
    ex f1 st y = f1 & F.y = f1*(h") by A11,A33;
    hence thesis by A34,A35,A38,A39,CARD_3:def 5;
  end;
  then rng F = product g by A23,XBOOLE_0:def 10;
  then product (g*h),product g are_equipotent by A11,A13,WELLORD2:def 4;
  hence thesis by A8,WELLORD2:15;
end;

theorem
  dom f = X implies product f,product (f*P) are_equipotent
proof
  assume
A1: dom f = X;
A2: rng P = X by FUNCT_2:def 3;
  dom P = X by FUNCT_2:52;
  then
A3: dom (f*P) = X by A1,A2,RELAT_1:27;
A4: rng (P") = X by FUNCT_2:def 3;
A5: dom (P") = X by FUNCT_2:52;
  now
    let x;
    assume
A6: x in dom (P");
    then P".x in X by A4,FUNCT_1:def 3;
    then (f*P).(P".x) = f.(P.(P".x)) by A3,FUNCT_1:12
      .= f.x by A5,A2,A6,FUNCT_1:35;
    hence f.x,(f*P).(P".x) are_equipotent;
  end;
  hence thesis by A1,A5,A4,A3,Th44;
end;

begin

definition
  let f,X;
  func Funcs(f,X) -> Function means
  :Def8:
  dom it = dom f & for x st x in dom f holds it.x = Funcs(f.x,X);
  existence
  proof
    deffunc F(set) = Funcs(f.$1,X);
    ex F be Function st dom F = dom f & for x st x in dom f holds F.x = F(
    x) from FUNCT_1:sch 3;
    hence thesis;
  end;
  uniqueness
  proof
    let f1,f2 such that
A1: dom f1 = dom f and
A2: for x st x in dom f holds f1.x = Funcs(f.x,X) and
A3: dom f2 = dom f and
A4: for x st x in dom f holds f2.x = Funcs(f.x,X);
    now
      let x;
      assume
A5:   x in dom f;
      then f1.x = Funcs(f.x,X) by A2;
      hence f1.x = f2.x by A4,A5;
    end;
    hence thesis by A1,A3,FUNCT_1:2;
  end;
end;

theorem
  not {} in rng f implies Funcs(f,{}) = dom f --> {}
proof
  assume
A1: not {} in rng f;
A2: now
    let x;
    assume
A3: x in dom f;
    then
A4: f.x <> {} by A1,FUNCT_1:def 3;
    thus (dom f --> {}).x = {} by A3,FUNCOP_1:7
      .= Funcs(f.x,{}) by A4;
  end;
  dom (dom f --> {}) = dom f by FUNCOP_1:13;
  hence thesis by A2,Def8;
end;

theorem
  Funcs({},X) = {}
proof
  dom Funcs({},X) = dom {} by Def8;
  hence thesis;
end;

theorem
  Funcs(X --> Y, Z) = X --> Funcs(Y,Z)
proof
A1: X = dom (X --> Y) by FUNCOP_1:13;
A2: now
    let x;
    assume
A3: x in X;
    then Funcs(X --> Y, Z).x = Funcs((X --> Y).x,Z) by A1,Def8;
    hence Funcs(X --> Y, Z).x = Funcs(Y,Z) by A3,FUNCOP_1:7;
  end;
  dom Funcs(X --> Y, Z) = dom (X --> Y) by Def8;
  hence thesis by A1,A2,FUNCOP_1:11;
end;

Lm4: [x,y] in dom f & g = (curry f).x & z in dom g implies [x,z] in dom f
proof
  assume [x,y] in dom f;
  then x in proj1 dom f by XTUPLE_0:def 12;
  then
A1: ex g9 being Function st (curry f).x = g9 & dom g9 = proj2 (dom f /\ [:{x
  },proj2 dom f:]) & for y st y in dom g9 holds g9.y = f.(x,y) by FUNCT_5:def 1
;
  assume g = (curry f).x & z in dom g;
  then consider y such that
A2: [y,z] in dom f /\ [:{x}, proj2 dom f:] by A1,XTUPLE_0:def 13;
  [y,z] in [:{x}, proj2 dom f:] by A2,XBOOLE_0:def 4;
  then
A3: y in {x} by ZFMISC_1:87;
  [y,z] in dom f by A2,XBOOLE_0:def 4;
  hence thesis by A3,TARSKI:def 1;
end;

theorem
  Funcs(Union disjoin f,X),product Funcs(f,X) are_equipotent
proof
  defpred P[set,set] means ex g,h st $1 = g & $2 = h & dom h = dom f & for y
  st y in dom f holds (f.y = {} implies h.y = {}) & (f.y <> {} implies h.y = (
  curry' g).y);
A1: for x st x in Funcs(Union disjoin f, X) ex z st P[x,z]
  proof
    defpred P[set] means f.$1 = {};
    deffunc F(set) = {};
    let x;
    assume x in Funcs(Union disjoin f, X);
    then consider g such that
A2: x = g and
    dom g = Union disjoin f and
    rng g c= X by FUNCT_2:def 2;
    deffunc G(set) = (curry' g).$1;
    ex h st dom h = dom f & for y st y in dom f holds (P[y] implies h.y =
    F(y)) & (not P[y] implies h.y = G(y)) from PARTFUN1:sch 1;
    hence thesis by A2;
  end;
  consider F being Function such that
A3: dom F = Funcs(Union disjoin f, X) & for x st x in Funcs(Union
  disjoin f, X) holds P[x,F.x] from CLASSES1:sch 1(A1);
  take F;
  thus F is one-to-one
  proof
    let x,y;
    assume that
A4: x in dom F and
A5: y in dom F and
A6: F.x = F.y;
A7: ex f2 being Function st y = f2 & dom f2 = Union disjoin f & rng f2 c=
    X by A3,A5,FUNCT_2:def 2;
    consider g1,h1 being Function such that
A8: x = g1 and
A9: F.x = h1 and
    dom h1 = dom f and
A10: for y st y in dom f holds (f.y = {} implies h1.y = {}) & (f.y <>
    {} implies h1.y = (curry' g1).y) by A3,A4;
    consider g2,h2 being Function such that
A11: y = g2 and
A12: F.y = h2 and
    dom h2 = dom f and
A13: for y st y in dom f holds (f.y = {} implies h2.y = {}) & (f.y <>
    {} implies h2.y = (curry' g2).y) by A3,A5;
A14: ex f1 being Function st x = f1 & dom f1 = Union disjoin f & rng f1 c=
    X by A3,A4,FUNCT_2:def 2;
    now
      let z;
      assume
A15:  z in Union disjoin f;
      then
A16:  z = [z`1,z`2] by CARD_3:22;
A17:  z`2 in dom f & z`1 in f.(z`2) by A15,CARD_3:22;
      then
A18:  h2.z`2 = (curry' g2).z`2 by A13;
      then reconsider
      g91 = h1.z`2, g92 = h2.z`2 as Function by A6,A7,A9,A11,A12,A15,A16,
FUNCT_5:21;
A19:  g2.(z`1,z`2) = g92.z`1 by A7,A11,A15,A16,A18,FUNCT_5:22;
      h1.z`2 = (curry' g1).z`2 by A10,A17;
      then g1.(z`1,z`2) = g91.z`1 by A14,A8,A15,A16,FUNCT_5:22;
      hence g1.z = g2.z by A6,A9,A12,A16,A19;
    end;
    hence thesis by A14,A7,A8,A11,FUNCT_1:2;
  end;
  thus dom F = Funcs(Union disjoin f, X) by A3;
  thus rng F c= product Funcs(f,X)
  proof
    let y;
    assume y in rng F;
    then consider x such that
A20: x in dom F and
A21: y = F.x by FUNCT_1:def 3;
    consider g,h such that
A22: x = g and
A23: F.x = h and
A24: dom h = dom f and
A25: for y st y in dom f holds (f.y = {} implies h.y = {}) & (f.y <>
    {} implies h.y = (curry' g).y) by A3,A20;
A26: ex f1 being Function st x = f1 & dom f1 = Union disjoin f & rng f1 c=
    X by A3,A20,FUNCT_2:def 2;
A27: now
      let z;
      assume z in dom Funcs(f,X);
      then
A28:  z in dom f by Def8;
      then
A29:  Funcs(f,X).z = Funcs(f.z,X) by Def8;
A30:  now
        set a = the Element of f.z;
        assume
A31:    f.z <> {};
        [a,z]`1 = a & [a,z]`2 = z;
        then
A32:    [a,z] in Union disjoin f by A28,A31,CARD_3:22;
        then reconsider k = (curry' g).z as Function by A22,A26,FUNCT_5:21;
A33:    z in dom curry' g by A22,A26,A32,FUNCT_5:21;
        then rng k c= rng g by FUNCT_5:34;
        then
A34:    rng k c= X by A22,A26,XBOOLE_1:1;
A35:    dom k = proj1 (dom g /\ [:proj1 dom g,{z}:]) by A33,FUNCT_5:34;
A36:    dom k = f.z
        proof
          thus dom k c= f.z
          proof
            let b be set;
            assume b in dom k;
            then consider c being set such that
A37:        [b,c] in dom g /\ [:proj1 dom g,{z}:] by A35,XTUPLE_0:def 12;
            [b,c] in [:proj1 dom g,{z}:] by A37,XBOOLE_0:def 4;
            then
A38:        c in {z} by ZFMISC_1:87;
A39:        [b,c]`1 = b & [b,c]`2 = c;
            [b,c] in dom g by A37,XBOOLE_0:def 4;
            then b in f.c by A22,A26,A39,CARD_3:22;
            hence thesis by A38,TARSKI:def 1;
          end;
          let b be set such that
A40:      b in f.z;
A41:      z in {z} by TARSKI:def 1;
          [b,z]`1 = b & [b,z]`2 = z;
          then
A42:      [b,z] in dom g by A22,A26,A28,A40,CARD_3:22;
          then b in proj1 dom g by XTUPLE_0:def 12;
          then [b,z] in [:proj1 dom g,{z}:] by A41,ZFMISC_1:87;
          then [b,z] in dom g /\ [:proj1 dom g,{z}:] by A42,XBOOLE_0:def 4;
          hence thesis by A35,XTUPLE_0:def 12;
        end;
        h.z = k by A25,A28,A31;
        hence h.z in Funcs(f,X).z by A29,A34,A36,FUNCT_2:def 2;
      end;
      now
        assume f.z = {};
        then h.z = {} & Funcs(f,X).z = {{}} by A25,A28,A29,FUNCT_5:57;
        hence h.z in Funcs(f,X).z by TARSKI:def 1;
      end;
      hence h.z in Funcs(f,X).z by A30;
    end;
    dom h = dom Funcs(f,X) by A24,Def8;
    hence thesis by A21,A23,A27,CARD_3:def 5;
  end;
  let x;
  assume x in product Funcs(f,X);
  then consider s being Function such that
A43: x = s and
A44: dom s = dom Funcs(f,X) and
A45: for z st z in dom Funcs(f,X) holds s.z in Funcs(f,X).z by CARD_3:def 5;
A46: dom s = dom f by A44,Def8;
A47: uncurry' s = ~uncurry s by FUNCT_5:def 4;
A48: dom uncurry' s = Union disjoin f
  proof
    thus dom uncurry' s c= Union disjoin f
    proof
      let z;
      assume
A49:  z in dom uncurry' s;
      then consider z1, z2 being set such that
A50:  z = [z1,z2] by A47,Th2;
A51:  z`1 = z1 & z`2 = z2 by A50,MCART_1:7;
      [z2,z1] in dom uncurry s by A47,A49,A50,FUNCT_4:42;
      then consider a being set, u being Function, b being set such that
A52:  [z2,z1] = [a,b] and
A53:  a in dom s and
A54:  u = s.a and
A55:  b in dom u by FUNCT_5:def 2;
A56:  [a,b]`1 = a & [z2,z1]`1 = z2;
      u in Funcs(f,X).a & Funcs(f,X).a = Funcs(f.a,X) by A44,A45,A46,A53,A54
,Def8;
      then
A57:  ex j being Function st u = j & dom j = f.z2 & rng j c= X by A52,A56,
FUNCT_2:def 2;
      [a,b]`2 = b & [z2,z1]`2 = z1;
      hence thesis by A46,A50,A52,A53,A55,A51,A56,A57,CARD_3:22;
    end;
    let z;
    assume
A58: z in Union disjoin f;
    then
A59: z`1 in f.z`2 & z = [z`1,z`2] by CARD_3:22;
A60: z`2 in dom f by A58,CARD_3:22;
    then
    s.z`2 in Funcs(f,X).z`2 & Funcs(f,X).z`2 = Funcs(f.z`2,X) by A44,A45,A46
,Def8;
    then ex j being Function st s.z`2 = j & dom j = f.z`2 & rng j c= X by
FUNCT_2:def 2;
    hence thesis by A46,A60,A59,FUNCT_5:39;
  end;
  rng uncurry' s c= X
  proof
    let d be set;
    assume d in rng uncurry' s;
    then consider z such that
A61: z in dom uncurry' s and
A62: d = (uncurry' s).z by FUNCT_1:def 3;
    consider z1, z2 being set such that
A63: z = [z1,z2] by A47,A61,Th2;
    [z2,z1] in dom uncurry s by A47,A61,A63,FUNCT_4:42;
    then consider a being set, u being Function, b being set such that
A64: [z2,z1] = [a,b] and
A65: a in dom s & u = s.a and
A66: b in dom u by FUNCT_5:def 2;
A67: [a,b]`1 = a & [z2,z1]`1 = z2;
    u in Funcs(f,X).a & Funcs(f,X).a = Funcs(f.a,X) by A44,A45,A46,A65,Def8;
    then
A68: ex j being Function st u = j & dom j = f.z2 & rng j c= X by A64,A67,
FUNCT_2:def 2;
A69: [a,b]`2 = b & [z2,z1]`2 = z1;
    (uncurry' s).(b,a) = u.b by A65,A66,FUNCT_5:39;
    then d in rng u by A62,A63,A64,A66,A67,A69,FUNCT_1:def 3;
    hence thesis by A68;
  end;
  then
A70: uncurry' s in dom F by A3,A48,FUNCT_2:def 2;
  then consider g,h such that
A71: uncurry' s = g and
A72: F.uncurry' s = h and
A73: dom h = dom f and
A74: for y st y in dom f holds (f.y = {} implies h.y = {}) & (f.y <> {}
  implies h.y = (curry' g).y) by A3;
  now
    let z;
    assume
A75: z in dom f;
    then s.z in Funcs(f,X).z & Funcs(f,X).z = Funcs(f.z,X) by A44,A45,A46,Def8;
    then consider v being Function such that
A76: s.z = v and
A77: dom v = f.z and
    rng v c= X by FUNCT_2:def 2;
A78: now
      set a = the Element of f.z;
      assume
A79:  f.z <> {};
      then
A80:  [a,z] in dom uncurry' s by A46,A75,A76,A77,FUNCT_5:39;
      then reconsider j = (curry' uncurry' s).z as Function by FUNCT_5:21;
A81:  j = (curry ~uncurry' s).z by FUNCT_5:def 3;
A82:  ~uncurry' s = uncurry s by Th4;
      then
A83:  [z,a] in dom uncurry s by A80,FUNCT_4:42;
A84:  dom j = dom v
      proof
        thus dom j c= dom v
        proof
          let b be set;
          assume b in dom j;
          then [z,b] in dom uncurry s by A81,A82,A83,Lm4;
          then consider
          a1 being set, g1 being Function, a2 being set such that
A85:      [z,b] = [a1,a2] and
          a1 in dom s and
A86:      g1 = s.a1 & a2 in dom g1 by FUNCT_5:def 2;
          z = a1 by A85,XTUPLE_0:1;
          hence thesis by A76,A85,A86,XTUPLE_0:1;
        end;
        let b be set;
        assume b in dom v;
        then [z,b] in dom uncurry s by A46,A75,A76,FUNCT_5:38;
        hence thesis by A81,A82,FUNCT_5:20;
      end;
      now
        let b be set;
        assume b in f.z;
        then
A87:    [z,b] in dom uncurry s by A77,A81,A82,A83,A84,Lm4;
        then consider a1 being set, g1 being Function, a2 being set such that
A88:    [z,b] = [a1,a2] and
A89:    a1 in dom s & g1 = s.a1 & a2 in dom g1 by FUNCT_5:def 2;
A90:    j.b = (uncurry s).(z,b) by A81,A82,A87,FUNCT_5:20;
        z = a1 & b = a2 by A88,XTUPLE_0:1;
        hence j.b = v.b by A76,A89,A90,FUNCT_5:38;
      end;
      then v = j by A77,A84,FUNCT_1:2;
      hence s.z = h.z by A71,A74,A75,A76,A79;
    end;
    f.z = {} implies s.z = {} & h.z = {} by A74,A75,A76,A77;
    hence s.z = h.z by A78;
  end;
  then h = s by A46,A73,FUNCT_1:2;
  hence thesis by A43,A70,A72,FUNCT_1:def 3;
end;

definition
  let X,f;
  func Funcs(X,f) -> Function means
  : Def9:
  dom it = dom f & for x st x in dom f holds it.x = Funcs(X,f.x);
  existence
  proof
    deffunc F(set) = Funcs(X,f.$1);
    ex F be Function st dom F = dom f & for x st x in dom f holds F.x = F(
    x) from FUNCT_1:sch 3;
    hence thesis;
  end;
  uniqueness
  proof
    let f1,f2 such that
A1: dom f1 = dom f and
A2: for x st x in dom f holds f1.x = Funcs(X,f.x) and
A3: dom f2 = dom f and
A4: for x st x in dom f holds f2.x = Funcs(X,f.x);
    now
      let x;
      assume
A5:   x in dom f;
      then f1.x = Funcs(X,f.x) by A2;
      hence f1.x = f2.x by A4,A5;
    end;
    hence thesis by A1,A3,FUNCT_1:2;
  end;
end;

Lm5: f <> {} & X <> {} implies product Funcs(X,f),Funcs(X,product f)
are_equipotent
proof
  defpred P[set,set] means ex g st $1 = g & $2 = <:g:>;
  assume that
A1: f <> {} and
A2: X <> {};
A3: for x st x in product Funcs(X,f) ex y st P[x,y]
  proof
    let x;
    assume x in product Funcs(X,f);
    then
    ex g st x = g & dom g = dom Funcs(X,f) & for x st x in dom Funcs(X,f)
    holds g.x in Funcs(X,f).x by CARD_3:def 5;
    then reconsider g = x as Function;
    reconsider y = <:g:> as set;
    take y, g;
    thus thesis;
  end;
  consider F being Function such that
A4: dom F = product Funcs(X,f) & for x st x in product Funcs(X,f) holds
  P[x,F.x] from CLASSES1:sch 1(A3);
  take F;
A5: g in product Funcs(X,f) implies dom <:g:> = X & SubFuncs rng g = rng g &
  rng <:g:> c= product f
  proof
A6: g"rng g = dom g by RELAT_1:134;
A7: dom Funcs(X,f) = dom f by Def9;
    assume
A8: g in product Funcs(X,f);
    then
A9: dom g = dom Funcs(X,f) by CARD_3:9;
A10: now
      let a be set;
      assume a in rng g;
      then consider z such that
A11:  z in dom g and
A12:  a = g.z by FUNCT_1:def 3;
      g.z in Funcs(X,f).z & Funcs(X,f).z = Funcs(X,f.z) by A8,A9,A7,A11,Def9,
CARD_3:9;
      hence a is Function by A12;
    end;
    then
A13: SubFuncs rng g = rng g by Lm1;
A14: now
      let z;
      assume
A15:  z in dom f;
      then g.z in Funcs(X,f).z & Funcs(X,f).z = Funcs(X,f.z) by A8,A7,Def9,
CARD_3:9;
      then ex h st g.z = h & dom h = X & rng h c= f.z by FUNCT_2:def 2;
      hence (dom f --> X).z = proj1 (g.z) by A15,FUNCOP_1:7;
    end;
    dom (dom f --> X) = dom f by FUNCOP_1:13;
    then doms g = dom f --> X by A9,A7,A13,A6,A14,Def2;
    then meet doms g = X by A1,Th27;
    hence
A16: dom <:g:> = X & SubFuncs rng g = rng g by A10,Lm1,Th29;
    let y;
    assume y in rng <:g:>;
    then consider x such that
A17: x in dom <:g:> and
A18: y = <:g:>.x by FUNCT_1:def 3;
    reconsider h = y as Function by A17,A18,Th30;
A19: dom h = dom f by A9,A7,A13,A6,A17,A18,Th31;
    now
      let z;
      assume
A20:  z in dom f;
      then
A21:  (uncurry g).(z,x) = h.z by A17,A18,A19,Th31;
      g.z in Funcs(X,f).z & Funcs(X,f).z = Funcs(X,f.z) by A8,A7,A20,Def9,
CARD_3:9;
      then consider j being Function such that
A22:  g.z = j and
A23:  dom j = X and
A24:  rng j c= f.z by FUNCT_2:def 2;
A25:  j.x in rng j by A16,A17,A23,FUNCT_1:def 3;
      (uncurry g).(z,x) = j.x by A9,A7,A16,A17,A20,A22,A23,FUNCT_5:38;
      hence h.z in f.z by A24,A21,A25;
    end;
    hence thesis by A19,CARD_3:def 5;
  end;
  thus F is one-to-one
  proof
    let x,y;
    assume that
A26: x in dom F and
A27: y in dom F and
A28: F.x = F.y and
A29: x <> y;
    consider gy being Function such that
A30: y = gy and
A31: F.y = <:gy:> by A4,A27;
    consider gx being Function such that
A32: x = gx and
A33: F.x = <:gx:> by A4,A26;
A34: dom gx = dom Funcs(X,f) by A4,A26,A32,CARD_3:9;
A35: dom Funcs(X,f) = dom f by Def9;
A36: dom gy = dom Funcs(X,f) by A4,A27,A30,CARD_3:9;
    now
      let z;
      assume
A37:  z in dom f;
      then
A38:  Funcs(X,f).z = Funcs(X,f.z) by Def9;
      gy.z in Funcs(X,f).z by A4,A27,A30,A35,A37,CARD_3:9;
      then consider hy being Function such that
A39:  gy.z = hy & dom hy = X and
      rng hy c= f.z by A38,FUNCT_2:def 2;
      gx.z in Funcs(X,f).z by A4,A26,A32,A35,A37,CARD_3:9;
      then consider hx being Function such that
A40:  gx.z = hx & dom hx = X and
      rng hx c= f.z by A38,FUNCT_2:def 2;
A41:  dom <:gx:> = X by A4,A5,A26,A32;
A42:  SubFuncs rng gx = rng gx & gx"rng gx = dom gx by A4,A5,A26,A32,
RELAT_1:134;
      now
        let b be set;
        assume
A43:    b in X;
        then reconsider fx = <:gx:>.b as Function by A41,Th30;
A44:    (uncurry gx).(z,b) = hx.b & (uncurry gy).(z,b) = hy.b by A34,A36,A35
,A37,A40,A39,A43,FUNCT_5:38;
A45:    dom fx = dom gx by A41,A42,A43,Th31;
        then fx.z = (uncurry gx).(z,b) by A34,A35,A37,A41,A43,Th31;
        hence hx.b = hy.b by A28,A33,A31,A34,A35,A37,A41,A43,A45,A44,Th31;
      end;
      hence gx.z = gy.z by A40,A39,FUNCT_1:2;
    end;
    hence thesis by A29,A32,A30,A34,A36,A35,FUNCT_1:2;
  end;
  thus dom F = product Funcs(X,f) by A4;
  thus rng F c= Funcs(X, product f)
  proof
    let y;
    assume y in rng F;
    then consider x such that
A46: x in dom F and
A47: y = F.x by FUNCT_1:def 3;
    consider g such that
A48: x = g and
A49: y = <:g:> by A4,A46,A47;
    dom <:g:> = X & rng <:g:> c= product f by A4,A5,A46,A48;
    hence thesis by A49,FUNCT_2:def 2;
  end;
  let y;
  assume y in Funcs(X, product f);
  then consider h such that
A50: y = h and
A51: dom h = X and
A52: rng h c= product f by FUNCT_2:def 2;
  set g = <:h:>;
A53: h"rng h = dom h by RELAT_1:134;
A54: now
    let z;
    assume
A55: z in X;
    then h.z in rng h by A51,FUNCT_1:def 3;
    then
A56: ex j being Function st h.z = j & dom j = dom f & for x st x in dom f
    holds j.x in f.x by A52,CARD_3:def 5;
    thus (X --> dom f).z = dom f by A55,FUNCOP_1:7
      .= (doms h).z by A51,A55,A56,Th22;
  end;
A57: meet doms h = dom g by Th29;
  now
    let z;
    assume z in rng h;
    then ex j being Function st z = j & dom j = dom f & for x st x in dom f
    holds j.x in f.x by A52,CARD_3:def 5;
    hence z is Function;
  end;
  then
A58: SubFuncs rng h = rng h by Lm1;
  dom doms h = h"SubFuncs rng h & dom (dom h --> dom f) = dom h by Def2,
FUNCOP_1:13;
  then X --> dom f = doms h by A51,A58,A53,A54,FUNCT_1:2;
  then
A59: dom g = dom f by A2,A57,Th27;
A60: now
    let z;
    assume
A61: z in dom f;
    then reconsider i = g.z as Function by A59,Th30;
A62: dom i = X by A51,A58,A53,A59,A61,Th31;
    rng i c= f.z
    proof
      let x;
      assume x in rng i;
      then consider a being set such that
A63:  a in dom i and
A64:  x = i.a by FUNCT_1:def 3;
      h.a in rng h by A51,A62,A63,FUNCT_1:def 3;
      then consider j being Function such that
A65:  h.a = j & dom j = dom f and
A66:  for x st x in dom f holds j.x in f.x by A52,CARD_3:def 5;
      i.a = (uncurry h).(a,z) by A59,A61,A63,Th31
        .= j.z by A51,A61,A62,A63,A65,FUNCT_5:38;
      hence thesis by A61,A64,A66;
    end;
    then g.z in Funcs(X,f.z) by A62,FUNCT_2:def 2;
    hence g.z in Funcs(X,f).z by A61,Def9;
  end;
A67: meet doms g = dom <:g:> by Th29;
  product f c= Funcs(dom f, Union f) by Th1;
  then
A68: rng h c= Funcs(dom f, Union f) by A52,XBOOLE_1:1;
  then
A69: dom uncurry h = [:dom h, dom f:] by FUNCT_5:26;
  dom f = dom Funcs(X,f) by Def9;
  then
A70: g in product Funcs(X,f) by A59,A60,CARD_3:def 5;
  then
A71: ex g9 being Function st g = g9 & F.g = <:g9:> by A4;
  dom uncurry' h = [:dom f, dom h:] by A68,FUNCT_5:26;
  then
A72: (uncurry' h)|([:meet doms h, dom h:] qua set) = uncurry' h by A57,A59,
RELAT_1:68;
A73: uncurry' h = ~uncurry h & curry ~uncurry h = curry' uncurry h by
FUNCT_5:def 3,def 4;
  dom <:g:> = X by A5,A70;
  then
A74: (uncurry h)|([:meet doms g, dom g:] qua set) = uncurry h by A51,A59,A69
,A67,RELAT_1:68;
  uncurry' curry' uncurry h = uncurry h by A69,FUNCT_5:49;
  then <:g:> = h by A1,A68,A72,A74,A73,FUNCT_5:48;
  hence thesis by A4,A50,A70,A71,FUNCT_1:def 3;
end;

theorem Th50:
  Funcs({},f) = dom f --> {{}}
proof
A1: now
    let x;
    assume x in dom f;
    hence Funcs({},f).x = Funcs({} qua set,f.x) by Def9
      .= {{}} by FUNCT_5:57;
  end;
  dom Funcs({},f) = dom f by Def9;
  hence thesis by A1,FUNCOP_1:11;
end;

theorem Th51:
  Funcs(X,{}) = {}
proof
  dom Funcs(X,{}) = dom {} by Def9;
  hence thesis;
end;

theorem
  Funcs(X, Y --> Z) = Y --> Funcs(X,Z)
proof
A1: Y = dom (Y --> Z) by FUNCOP_1:13;
A2: now
    let x;
    assume
A3: x in Y;
    then Funcs(X, Y --> Z).x = Funcs(X, (Y --> Z).x) by A1,Def9;
    hence Funcs(X, Y --> Z).x = Funcs(X,Z) by A3,FUNCOP_1:7;
  end;
  dom Funcs(X, Y --> Z) = dom (Y --> Z) by Def9;
  hence thesis by A1,A2,FUNCOP_1:11;
end;

theorem
  product Funcs(X,f),Funcs(X, product f) are_equipotent
proof
A1: Funcs(X, product {}) = {X --> {}} by CARD_3:10,FUNCT_5:59;
A2: product Funcs({},f) = product (dom f --> {{}}) by Th50
    .= Funcs(dom f, {{}}) by CARD_3:11
    .= {dom f --> {}} by FUNCT_5:59;
A3: Funcs({} qua set, product f) = {{}} & product Funcs(X,{}) = {{}} by Th51,
CARD_3:10,FUNCT_5:57;
  X <> {} & f <> {} implies thesis by Lm5;
  hence thesis by A2,A3,A1,CARD_1:28;
end;

begin 


definition
  let f be Function;

  func commute f -> Function-yielding Function equals
  curry' uncurry f;
  coherence
  proof
    set g = curry' uncurry f;
    for x st x in dom g holds g.x is Function by FUNCT_5:33;
    hence thesis by FUNCOP_1:def 6;
  end;
end;

theorem
  for f be Function, x be set st x in dom (commute f) holds (commute f).
  x is Function;

theorem Th55:
  for A,B,C be set, f be Function st A <> {} & B <> {} & f in
  Funcs(A,Funcs(B,C)) holds commute f in Funcs(B,Funcs(A,C))
proof
  let A,B,C be set, f be Function;
  assume A <> {} & B <> {} & f in Funcs(A,Funcs(B,C));
  then [:A,B:] <> {} & uncurry f in Funcs([:A,B:],C) by Th11,ZFMISC_1:90;
  hence thesis by Th10;
end;

theorem
  for A,B,C be set, f be Function st A <> {} & B <> {} & f in Funcs(A,
  Funcs(B,C)) for g,h be Function, x,y be set st x in A & y in B & f.x = g & (
  commute f).y = h holds h.x = g.y & dom h = A & dom g = B & rng h c= C & rng g
  c= C
proof
  let A,B,C be set, f be Function;
  assume that
A1: A <> {} & B <> {} and
A2: f in Funcs(A,Funcs(B,C));
  set uf = uncurry f;
  set cf = commute f;
  let g,h be Function, x,y be set;
  assume that
A3: x in A and
A4: y in B and
A5: f.x = g and
A6: (commute f).y = h;
  commute f in Funcs(B,Funcs(A,C)) by A1,A2,Th55;
  then
A7: ex g be Function st g = cf & dom g = B & rng g c= Funcs(A,C) by
FUNCT_2:def 2;
  then cf.y in rng cf by A4,FUNCT_1:def 3;
  then
A8: ex t be Function st t = h & dom t = A & rng t c= C by A6,A7,FUNCT_2:def 2;
A9: ex g be Function st g = f & dom g = A & rng g c= Funcs(B,C) by A2,
FUNCT_2:def 2;
  then f.x in rng f by A3,FUNCT_1:def 3;
  then
A10: ex t be Function st t = g & dom t = B & rng t c= C by A5,A9,FUNCT_2:def 2;
  then [x,y] in dom uf & uf.(x,y) = g.y by A3,A4,A5,A9,FUNCT_5:38;
  hence thesis by A6,A10,A8,FUNCT_5:22;
end;

theorem
  for A,B,C be set, f be Function st A <> {} & B <> {} & f in Funcs(A,
  Funcs(B,C)) holds commute commute f = f
proof
  let A,B,C be set, f be Function;
  assume that
A1: A <> {} & B <> {} and
A2: f in Funcs(A,Funcs(B,C));
A3: ex g be Function st g = f & dom g = A & rng g c= Funcs(B,C) by A2,
FUNCT_2:def 2;
  set cf = commute f;
A4: commute f in Funcs(B,Funcs(A,C)) by A1,A2,Th55;
  then commute cf in Funcs(A,Funcs(B,C)) by A1,Th55;
  then
A5: ex h be Function st h = commute cf & dom h = A & rng h c= Funcs(B,C) by
FUNCT_2:def 2;
A6: ex g be Function st g = cf & dom g = B & rng g c= Funcs(A,C) by A4,
FUNCT_2:def 2;
  for x st x in A holds f.x = (commute cf).x
  proof
    set ccf = commute cf, uf = uncurry f, ucf = uncurry cf;
    let x;
    assume
A7: x in A;
    then f.x in rng f by A3,FUNCT_1:def 3;
    then consider g be Function such that
A8: g = f.x & dom g = B and
    rng g c= C by A3,FUNCT_2:def 2;
    ccf.x in rng ccf by A5,A7,FUNCT_1:def 3;
    then consider h be Function such that
A9: h = ccf.x and
A10: dom h = B and
    rng h c= C by A5,FUNCT_2:def 2;
A11: for y st y in B for h be Function st h = cf.y holds x in dom h & h.x = g.y
    proof
      let y;
      assume y in B;
      then
A12:  [x,y] in dom uf & uf.(x,y) = g.y by A3,A7,A8,FUNCT_5:38;
      let h be Function;
      assume h = cf.y;
      hence thesis by A12,FUNCT_5:22;
    end;
A13: for y st y in B holds [y,x] in dom ucf & ucf.(y,x) = g.y
    proof
      let y;
      reconsider h = cf.y as Function;
      assume
A14:  y in B;
      then x in dom h & h.x = g.y by A11;
      hence thesis by A6,A14,FUNCT_5:38;
    end;
    for y st y in B holds h.y = g.y
    proof
      let y;
      assume y in B;
      then [y,x] in dom ucf & ucf.(y,x) = g.y by A13;
      hence thesis by A9,FUNCT_5:22;
    end;
    hence thesis by A8,A9,A10,FUNCT_1:2;
  end;
  hence thesis by A3,A5,FUNCT_1:2;
end;

theorem
  commute {} = {} by FUNCT_5:42,43;



theorem
  for f be Function-yielding Function holds dom doms f = dom f
proof
  let f be Function-yielding Function;
  thus dom doms f c= dom f
  proof
    let i be set;
    dom doms f = f"SubFuncs rng f by Def2;
    hence thesis by Th19;
  end;
  thus dom f c= dom doms f
  proof
    let i be set;
    assume
A1: i in dom f;
    f.i is Function & dom doms f = f"SubFuncs rng f by Def2;
    hence thesis by A1,Th19;
  end;
end;

theorem
  for f be Function-yielding Function holds dom rngs f = dom f
proof
  let f be Function-yielding Function;
  thus dom rngs f c= dom f
  proof
    let i be set;
    dom rngs f = f"SubFuncs rng f by Def3;
    hence thesis by Th19;
  end;
  thus dom f c= dom rngs f
  proof
    let i be set;
    assume
A1: i in dom f;
    f.i is Function & dom rngs f = f"SubFuncs rng f by Def3;
    hence thesis by A1,Th19;
  end;
end;















begin 

reserve A,B,C for Ordinal;

Lm1: {} in omega by ORDINAL1:def 11;

Lm2: 1 in omega;

definition
  func one equals
  1;
  correctness;
end;

begin 

definition
  let a,b be Ordinal;
  pred a,b are_relative_prime means
  :Def2:
  for c,d1,d2 being Ordinal st a = c *^ d1 & b = c *^ d2 holds c = 1;
  symmetry;
end;

theorem
  not {},{} are_relative_prime
proof
  take {}, {}, {};
  thus thesis by ORDINAL2:35;
end;

theorem Th2:
  1,A are_relative_prime
proof
  let c,d1,d2 be Ordinal;
  thus thesis by ORDINAL3:37;
end;

theorem Th3:
  {},A are_relative_prime implies A = 1
proof
A1: {} = A*^{} & A = A*^1 by ORDINAL2:38,39;
  assume {},A are_relative_prime & A <> 1;
  hence contradiction by A1,Def2;
end;

reserve a,b,c,d for natural Ordinal;

defpred PP[set] means ex B st B c= $1 & $1 in omega & $1 <> {} & not ex c,d1,
d2 being natural Ordinal st d1,d2 are_relative_prime & $1 = c *^ d1 & B = c *^
d2;

theorem
  a <> {} or b <> {} implies ex c,d1,d2 being natural Ordinal st d1,d2
  are_relative_prime & a = c *^ d1 & b = c *^ d2
proof
  assume that
A1: a <> {} or b <> {} and
A2: not ex c,d1,d2 being natural Ordinal st d1,d2 are_relative_prime & a
  = c *^ d1 & b = c *^ d2;
A3: ex A st PP[A]
  proof
    per cases;
    suppose
A4:   a c= b;
      take A = b, B = a;
      thus B c= A & A in omega & A <> {} by A1,A4,ORDINAL1:def 12;
      thus thesis by A2;
    end;
    suppose
A5:   b c= a;
      take A = a, B = b;
      thus B c= A & A in omega & A <> {} by A1,A5,ORDINAL1:def 12;
      thus thesis by A2;
    end;
  end;
  consider A such that
A6: PP[A] and
A7: for C st PP[C] holds A c= C from ORDINAL1:sch 1(A3);
  consider B such that
A8: B c= A and
A9: A in omega and
A10: A <> {} and
A11: not ex c,d1,d2 being natural Ordinal st d1,d2 are_relative_prime &
  A = c *^ d1 & B = c *^ d2 by A6;
  reconsider A,B as Element of omega by A8,A9,ORDINAL1:12;
  A = 1 *^ A & B = 1 *^ B by ORDINAL2:39;
  then not A,B are_relative_prime by A11;
  then consider c,d1,d2 being Ordinal such that
A12: A = c *^ d1 and
A13: B = c *^ d2 and
A14: c <> 1 by Def2;
A15: c <> {} by A10,A12,ORDINAL2:35;
  then
A16: d1 c= A & d2 c= B by A12,A13,ORDINAL3:36;
A17: d1 <> {} by A10,A12,ORDINAL2:38;
  then c c= A by A12,ORDINAL3:36;
  then reconsider c,d1,d2 as Element of omega by A16,ORDINAL1:12;
  1 in c or c in 1 by A14,ORDINAL1:14;
  then 1*^d1 in A by A12,A17,A15,ORDINAL3:14,19;
  then
A18: d1 in A by ORDINAL2:39;
A19: now
    let c9,d19,d29 be natural Ordinal;
    assume that
A20: d19,d29 are_relative_prime and
A21: d1 = c9 *^ d19 & d2 = c9 *^ d29;
    A = c*^c9*^d19 & B = c*^c9*^d29 by A12,A13,A21,ORDINAL3:50;
    hence contradiction by A11,A20;
  end;
  A = d1*^c & B = d2*^c by A12,A13;
  then d2 c= d1 by A8,A15,ORDINAL3:35;
  hence contradiction by A7,A17,A19,A18,ORDINAL1:5;
end;

reserve l,m,n for natural Ordinal;

registration
  let m,n;
  cluster m div^ n -> natural;
  coherence
  proof
A1: n = {} implies m div^ n = {} & {}*^1 = {} by ORDINAL2:35,ORDINAL3:def 6;
    n in 1 or 1 c= n by ORDINAL1:16;
    then (m div^ n)*^1 c= (m div^ n)*^n by A1,ORDINAL3:14,20;
    then
A2: m div^ n c= (m div^ n)*^n by ORDINAL2:39;
    (m div^ n)*^n c= m & m in omega by ORDINAL1:def 12,ORDINAL3:64;
    hence m div^ n in omega by A2,ORDINAL1:12,XBOOLE_1:1;
  end;
  cluster m mod^ n -> natural;
  coherence
  proof
    (m div^ n)*^n+^(m mod^ n) = m by ORDINAL3:65;
    then
A3: m mod^ n c= m by ORDINAL3:24;
    m in omega by ORDINAL1:def 12;
    hence m mod^ n in omega by A3,ORDINAL1:12;
  end;
end;

definition
  let k,n be Ordinal;
  pred k divides n means
  : Def3:
  ex a being Ordinal st n = k*^a;
  reflexivity
  proof
    let n be Ordinal;
    take 1;
    thus thesis by ORDINAL2:39;
  end;
end;

theorem Th5:
  a divides b iff ex c st b = a*^c
proof
  thus a divides b implies ex c st b = a*^c
  proof
    given c being Ordinal such that
A1: b = a*^c;
    per cases;
    suppose
      b = {};
      then b = a*^{} by ORDINAL2:38;
      hence thesis;
    end;
    suppose
      b <> {};
      then c is Element of omega by A1,ORDINAL3:75;
      hence thesis by A1;
    end;
  end;
  thus thesis by Def3;
end;

theorem Th6:
  for m,n st {} in m holds n mod^ m in m
proof
  let m,n;
  assume {} in m;
  then
A1: ex C st n = (n div^ m)*^m+^C & C in m by ORDINAL3:def 6;
  n mod^ m = n-^(n div^ m)*^m by ORDINAL3:def 7;
  hence thesis by A1,ORDINAL3:52;
end;

theorem Th7:
  for n,m holds m divides n iff n = m *^ (n div^ m)
proof
  let n,m;
  assume
A1: not thesis;
  then consider a such that
A2: n = m*^a by Th5;
  {}*^a = {} by ORDINAL2:35;
  then {} <> m by A1,A2,ORDINAL2:35;
  then
A3: {} in m by ORDINAL3:8;
  n = a*^m+^{} by A2,ORDINAL2:27;
  hence thesis by A1,A2,A3,Def3,ORDINAL3:def 6;
end;

theorem Th8:
  for n,m st n divides m & m divides n holds n = m
proof
  let n,m;
  assume that
A1: n divides m and
A2: m divides n;
A3: m = n *^ (m div^ n) by A1,Th7;
A4: (m div^ n) *^ (n div^ m) = 1 implies n = m
  proof
    assume (m div^ n) *^ (n div^ m) = 1;
    then m div^ n = 1 by ORDINAL3:37;
    hence thesis by A3,ORDINAL2:39;
  end;
  n = m *^ (n div^ m) by A2,Th7;
  then
A5: n *^ 1 = n & n = n *^ ((m div^ n) *^ (n div^ m)) by A3,ORDINAL2:39
,ORDINAL3:50;
  n = {} implies n = m by A3,ORDINAL2:35;
  hence thesis by A5,A4,ORDINAL3:33;
end;

theorem Th9:
  n divides {} & 1 divides n
proof
  {} = n *^ {} by ORDINAL2:35;
  hence n divides {} by Def3;
  n = 1 *^ n by ORDINAL2:39;
  hence thesis by Def3;
end;

theorem Th10:
  for n,m st {} in m & n divides m holds n c= m
proof
  let n,m such that
A1: {} in m;
  given a being Ordinal such that
A2: m = n*^a;
  a <> {} by A1,A2,ORDINAL2:38;
  hence thesis by A2,ORDINAL3:36;
end;

theorem Th11:
  for n,m,l st n divides m & n divides m +^ l holds n divides l
proof
  let n,m,l;
  assume n divides m;
  then consider a such that
A1: m = n*^a by Th5;
  assume n divides m +^ l;
  then consider b such that
A2: m +^ l = n*^b by Th5;
  assume
A3: not n divides l;
  l = n*^b -^ n*^a by A1,A2,ORDINAL3:52
    .= (b-^a)*^n by ORDINAL3:63;
  hence thesis by A3,Def3;
end;

Lm3: 1 = succ {};

definition
  let k,n be natural Ordinal;
  func k lcm n -> Element of omega means
  : Def4:
  k divides it & n divides it &
  for m st k divides m & n divides m holds it divides m;
  existence
  proof
    per cases;
    suppose
A1:   k = {} or n = {};
      reconsider w = {} as Element of omega by ORDINAL1:def 12;
      take w;
      thus k divides w & n divides w by Th9;
      let m;
      assume k divides m & n divides m;
      hence thesis by A1;
    end;
    suppose
A2:   k <> {} & n <> {};
      {} c= k;
      then {} c< k by A2,XBOOLE_0:def 8;
      then
A3:   {} in k by ORDINAL1:11;
      {} c= n;
      then {} c< n by A2,XBOOLE_0:def 8;
      then {} in n by ORDINAL1:11;
      then
A4:   1 c= n by Lm3,ORDINAL1:21;
      defpred P[Ordinal] means ex m st $1 = m & k divides m & n divides m & k
      c= m;
A5:   k*^1 = k by ORDINAL2:39;
      k divides k *^ n & n divides k *^ n by Def3;
      then
A6:   ex A st P[A] by A4,A5,ORDINAL2:41;
      consider A such that
A7:   P[A] and
A8:   for B st P[B] holds A c= B from ORDINAL1:sch 1(A6);
      consider l such that
A9:   A = l and
A10:  k divides l and
A11:  n divides l and
A12:  k c= l by A7;
      reconsider l as Element of omega by ORDINAL1:def 12;
      take l;
      thus k divides l & n divides l by A10,A11;
      let m such that
A13:  k divides m and
A14:  n divides m;
A15:  m = l*^(m div^ l)+^(m mod^ l) by ORDINAL3:65;
      l = k*^(l div^ k) by A10,Th7;
      then l*^(m div^ l) = k*^((l div^ k)*^(m div^ l)) by ORDINAL3:50;
      then k divides l*^(m div^ l) by Def3;
      then
A16:  k divides m mod^ l by A13,A15,Th11;
      l = n*^(l div^ n) by A11,Th7;
      then l*^(m div^ l) = n*^((l div^ n)*^(m div^ l)) by ORDINAL3:50;
      then n divides l*^(m div^ l) by Def3;
      then
A17:  n divides m mod^ l by A14,A15,Th11;
      now
A18:    {} c= m mod^ l;
        assume m mod^ l <> {};
        then {} c< m mod^ l by A18,XBOOLE_0:def 8;
        then k c= m mod^ l by A16,Th10,ORDINAL1:11;
        then l c= m mod^ l by A8,A9,A16,A17;
        hence contradiction by A12,A3,Th6,ORDINAL1:5;
      end;
      then m = l*^(m div^ l) by A15,ORDINAL2:27;
      hence thesis by Th7;
    end;
  end;
  uniqueness
  proof
    let m1,m2 be Element of omega;
    assume k divides m1 & n divides m1 & ( for m st k divides m & n divides
m holds m1 divides m)& k divides m2 &( n divides m2 & for m st k divides m & n
    divides m holds m2 divides m );
    then m1 divides m2 & m2 divides m1;
    hence thesis by Th8;
  end;
  commutativity;
end;

theorem Th12:
  m lcm n divides m*^n
proof
  m divides m*^n & n divides m*^n by Def3;
  hence thesis by Def4;
end;

theorem Th13:
  n <> {} implies (m*^n) div^ (m lcm n) divides m
proof
  assume
A1: n <> {};
  take ((m lcm n) div^ n);
  n divides m lcm n by Def4;
  then
A2: m lcm n = n*^((m lcm n) div^ n) by Th7;
  m lcm n divides m*^n by Th12;
  then m*^n = (m lcm n)*^ ((m*^n) div^ (m lcm n)) by Th7;
  then n*^m = n*^(((m lcm n) div^ n)*^ ((m*^n) div^ (m lcm n))) by A2,
ORDINAL3:50;
  hence m = ((m*^n) div^ (m lcm n))*^((m lcm n) div^ n) by A1,ORDINAL3:33;
end;

definition
  let k,n be natural Ordinal;
  func k hcf n -> Element of omega means
  : Def5:
  it divides k & it divides n &
  for m st m divides k & m divides n holds m divides it;
  existence
  proof
    per cases;
    suppose
A1:   k = {} or n = {};
      then reconsider m = k \/ n as Element of omega by ORDINAL1:def 12;
      take m;
      thus m divides k & m divides n by A1,Th9;
      thus thesis by A1;
    end;
    suppose
A2:   k <> {} & n <> {};
      reconsider l = (k*^n) div^ (k lcm n) as Element of omega by
ORDINAL1:def 12;
      take l;
      thus l divides k & l divides n by A2,Th13;
      let m;
      assume that
A3:   m divides k and
A4:   m divides n;
A5:   n = m*^(n div^ m) by A4,Th7;
      then
A6:   n div^ m <> {} by A2,ORDINAL2:35;
      m*^(k div^ m)*^(n div^ m) = n*^(k div^ m) by A5,ORDINAL3:50;
      then
A7:   n divides m*^(k div^ m)*^(n div^ m) by Def3;
      set mm = m*^((k div^ m)*^(n div^ m));
A8:   mm = m*^(k div^ m)*^(n div^ m) by ORDINAL3:50;
A9:   k = m*^(k div^ m) by A3,Th7;
      then k div^ m <> {} by A2,ORDINAL2:35;
      then
A10:  (k div^ m)*^(n div^ m) <> {} by A6,ORDINAL3:31;
      k divides m*^(k div^ m)*^(n div^ m) by A9,Def3;
      then k lcm n divides m*^(k div^ m)*^(n div^ m) by A7,Def4;
      then
A11:  mm = (k lcm n)*^(mm div^ (k lcm n)) by A8,Th7;
      m <> {} by A2,A9,ORDINAL2:35;
      then k lcm n <> {} by A11,A10,ORDINAL2:35,ORDINAL3:31;
      then
A12:  k*^n = k*^n+^{} & {} in k lcm n by ORDINAL2:27,ORDINAL3:8;
      mm*^m = (k lcm n)*^(m*^(mm div^ (k lcm n))) by A11,ORDINAL3:50;
      then k*^n = (k lcm n)*^(m*^(mm div^ (k lcm n))) by A9,A5,A8,ORDINAL3:50;
      then l = m*^(mm div^ (k lcm n)) by A12,ORDINAL3:66;
      hence thesis by Def3;
    end;
  end;
  uniqueness
  proof
    let m1,m2 be Element of omega;
    assume m1 divides k & m1 divides n & ( for m st m divides k & m divides
n holds m divides m1)& m2 divides k &( m2 divides n & for m st m divides k & m
    divides n holds m divides m2 );
    then m1 divides m2 & m2 divides m1;
    hence thesis by Th8;
  end;
  commutativity;
end;

theorem Th14:
  a hcf {} = a & a lcm {} = {}
proof
  reconsider e = a, c = {} as Element of omega by ORDINAL1:def 12;
A1: for b st a divides b & {} divides b holds c divides b;
  ( for n st n divides a & n divides {} holds n divides e)& a divides c by Th9
;
  hence thesis by A1,Def4,Def5;
end;

theorem Th15:
  a hcf b = {} implies a = {}
proof
  a hcf b divides a by Def5;
  then a = (a hcf b)*^(a div^ (a hcf b)) by Th7;
  hence thesis by ORDINAL2:35;
end;

theorem Th16:
  a hcf a = a & a lcm a = a
proof
  reconsider c = a as Element of omega by ORDINAL1:def 12;
  for b st b divides a & b divides a holds b divides c;
  hence a hcf a = a by Def5;
  for b st a divides b & a divides b holds c divides b;
  hence thesis by Def4;
end;

theorem Th17:
  (a*^c) hcf (b*^c) = (a hcf b)*^c
proof
  per cases;
  suppose
A1: c = {};
    then
A2: (a hcf b)*^c = {} by ORDINAL2:35;
    a*^c = {} & b*^c = {} by A1,ORDINAL2:35;
    hence thesis by A2,Th16;
  end;
  suppose
A3: c <> {} & a <> {};
    reconsider d = (a hcf b)*^c as Element of omega by ORDINAL1:def 12;
    set e = ((a*^c) hcf (b*^c)) div^ d;
    a hcf b divides b by Def5;
    then b = (a hcf b)*^(b div^ (a hcf b)) by Th7;
    then b*^c = d*^(b div^ (a hcf b)) by ORDINAL3:50;
    then
A4: d divides b*^c by Def3;
    a hcf b divides a by Def5;
    then a = (a hcf b)*^(a div^ (a hcf b)) by Th7;
    then a*^c = d*^(a div^ (a hcf b)) by ORDINAL3:50;
    then d divides a*^c by Def3;
    then d divides (a*^c) hcf (b*^c) by A4,Def5;
    then
A5: (a*^c) hcf (b*^c) = d*^e by Th7
      .= (a hcf b)*^e*^c by ORDINAL3:50;
    then (a hcf b)*^e*^c divides b*^c by Def5;
    then (a hcf b)*^e*^c*^((b*^c)div^((a hcf b)*^e*^c)) = b*^c by Th7;
    then (a hcf b)*^e*^((b*^c)div^((a hcf b)*^e*^c))*^c = b*^c by ORDINAL3:50;
    then (a hcf b)*^e*^((b*^c)div^((a hcf b)*^e*^c)) = b by A3,ORDINAL3:33;
    then
A6: (a hcf b)*^e divides b by Def3;
    (a hcf b)*^e*^c divides a*^c by A5,Def5;
    then (a hcf b)*^e*^c*^((a*^c)div^((a hcf b)*^e*^c)) = a*^c by Th7;
    then (a hcf b)*^e*^((a*^c)div^((a hcf b)*^e*^c))*^c = a*^c by ORDINAL3:50;
    then (a hcf b)*^e*^((a*^c)div^((a hcf b)*^e*^c)) = a by A3,ORDINAL3:33;
    then (a hcf b)*^e divides a by Def3;
    then (a hcf b)*^e divides a hcf b by A6,Def5;
    then (a hcf b)*^e*^((a hcf b) div^ ((a hcf b)*^e)) = a hcf b by Th7;
    then (a hcf b)*^(e*^((a hcf b) div^ ((a hcf b)*^e))) = a hcf b by
ORDINAL3:50
      .= (a hcf b)*^1 by ORDINAL2:39;
    then a hcf b = {} or e*^((a hcf b) div^ ((a hcf b)*^e)) = 1 by ORDINAL3:33;
    then e = 1 by A3,Th15,ORDINAL3:37;
    hence thesis by A5,ORDINAL2:39;
  end;
  suppose
    a = {};
    then a hcf b = b & a*^c = {} by Th14,ORDINAL2:35;
    hence thesis by Th14;
  end;
end;

theorem Th18:
  b <> {} implies a hcf b <> {} & b div^ (a hcf b) <> {}
proof
  a hcf b divides b by Def5;
  then b = (a hcf b)*^(b div^ (a hcf b)) by Th7;
  hence thesis by ORDINAL2:35;
end;

theorem Th19:
  a <> {} or b <> {} implies a div^ (a hcf b), b div^ (a hcf b)
  are_relative_prime
proof
  assume
A1: a <> {} or b <> {};
  set ab = a hcf b;
A2: 1*^a = a & 1*^b = b by ORDINAL2:39;
  per cases;
  suppose
    a = {} or b = {};
    then ab = b & b div^ b = 1 or ab = a & a div^ a = 1 by A1,A2,Th14,
ORDINAL3:68;
    hence thesis by Th2;
  end;
  suppose
A3: a <> {} & b <> {};
    ab divides b by Def5;
    then
A4: b = ab*^(b div^ ab) by Th7;
    then
A5: b div^ ab <> {} by A3,ORDINAL2:35;
    let c,d1,d2 be Ordinal such that
A6: a div^ (a hcf b) = c *^ d1 and
A7: b div^ (a hcf b) = c *^ d2;
    ab divides a by Def5;
    then
A8: a = ab*^(a div^ ab) by Th7;
    then a div^ ab <> {} by A3,ORDINAL2:35;
    then reconsider c,d1,d2 as Element of omega by A6,A7,A5,ORDINAL3:75;
    b = ab*^c*^d2 by A4,A7,ORDINAL3:50;
    then
A9: ab*^c divides b by Def3;
    a = ab*^c*^d1 by A8,A6,ORDINAL3:50;
    then ab*^c divides a by Def3;
    then ab*^c divides ab by A9,Def5;
    then ab = (ab*^c)*^(ab div^ (ab*^c)) by Th7;
    then ab = ab*^(c*^(ab div^ (ab*^c))) by ORDINAL3:50;
    then
A10: ab*^1 = ab*^(c*^(ab div^ (ab*^c))) by ORDINAL2:39;
    ab <> {} by A3,A8,ORDINAL2:35;
    then 1 = c*^(ab div^ (ab*^c)) by A10,ORDINAL3:33;
    hence thesis by ORDINAL3:37;
  end;
end;

theorem Th20:
  a,b are_relative_prime iff a hcf b = 1
proof
  a hcf b divides b by Def5;
  then
A1: b = (a hcf b)*^(b div^ (a hcf b)) by Th7;
  a hcf b divides a by Def5;
  then a = (a hcf b)*^(a div^ (a hcf b)) by Th7;
  hence a,b are_relative_prime implies a hcf b = 1 by A1,Def2;
  assume
A2: a hcf b = 1;
  let c,d1,d2 be Ordinal such that
A3: a = c*^d1 & b = c*^d2;
  a <> {} or b <> {} by A2,Th14;
  then reconsider c as Element of omega by A3,ORDINAL3:75;
  c divides a & c divides b by A3,Def3;
  then c divides 1 by A2,Def5;
  then ex d st 1 = c*^d by Th5;
  hence thesis by ORDINAL3:37;
end;

definition
  let a,b be natural Ordinal;
  func RED(a,b) -> Element of omega equals
  a div^ (a hcf b);
  coherence by ORDINAL1:def 12;
end;

theorem Th21:
  RED(a,b)*^(a hcf b) = a
proof
  a hcf b divides a by Def5;
  hence thesis by Th7;
end;

theorem
  a <> {} or b <> {} implies RED(a,b), RED(b,a) are_relative_prime by Th19;

theorem Th23:
  a,b are_relative_prime implies RED(a,b) = a
proof
  assume a,b are_relative_prime;
  then a hcf b = 1 by Th20;
  hence thesis by ORDINAL3:71;
end;

theorem Th24:
  RED(a,1) = a & RED(1,a) = 1
proof
  a,1 are_relative_prime by Th2;
  then a hcf 1 = 1 by Th20;
  hence thesis by ORDINAL3:71;
end;

theorem
  b <> {} implies RED(b,a) <> {} by Th18;

theorem Th26:
  RED({},a) = {} & (a <> {} implies RED(a,{}) = 1)
proof
  thus RED({},a) = {} by ORDINAL3:70;
  assume
A1: a <> {};
  thus RED(a,{}) = a div^ a by Th14
    .= a*^1 div^ a by ORDINAL2:39
    .= 1 by A1,ORDINAL3:68;
end;

theorem Th27:
  a <> {} implies RED(a,a) = 1
proof
  assume
A1: a <> {};
  thus RED(a,a) = a div^ a by Th16
    .= a*^1 div^ a by ORDINAL2:39
    .= 1 by A1,ORDINAL3:68;
end;

theorem Th28:
  c <> {} implies RED(a*^c, b*^c) = RED(a, b)
proof
  assume
A1: c <> {};
  a <> {} implies a hcf b <> {} by Th18;
  then
A2: a <> {} implies (a hcf b)*^c <> {} by A1,ORDINAL3:31;
A3: RED({},b) = {} & {}*^((a hcf b)*^c) = {} by ORDINAL2:35,ORDINAL3:70;
A4: a hcf b divides a by Def5;
  thus RED(a*^c, b*^c) = (a*^c)div^((a hcf b)*^c) by Th17
    .= (((a div^(a hcf b))*^(a hcf b))*^c)div^((a hcf b)*^c) by A4,Th7
    .= (RED(a,b)*^((a hcf b)*^c))div^((a hcf b)*^c) by ORDINAL3:50
    .= RED(a, b) by A2,A3,ORDINAL3:68,70;
end;

set RATplus = {[a,b] where a,b is Element of omega: a,b are_relative_prime & b
<> {}};

1,1 are_relative_prime by Th2;

then [1,1] in RATplus;

then reconsider RATplus as non empty set;

Lm4: [a,b] in RATplus implies a,b are_relative_prime & b <> {}
proof
  assume [a,b] in RATplus;
  then consider c,d being Element of omega such that
A1: [a,b] = [c,d] and
A2: c,d are_relative_prime & d <> {};
  a = c by A1,XTUPLE_0:1;
  hence thesis by A1,A2,XTUPLE_0:1;
end;

begin 

reserve i,j,k for Element of omega;

definition
  func RAT+ equals
  ({[i,j]: i,j are_relative_prime & j <> {}} \ {[k,1]: not
  contradiction}) \/ omega;
  correctness;
end;

Lm5: omega c= RAT+ by XBOOLE_1:7;

reserve x,y,z for Element of RAT+;

registration
  cluster RAT+ -> non empty;
  coherence;
end;

registration
  cluster non empty ordinal for Element of RAT+;
  existence by Lm2,Lm5;
end;

theorem Th29:
  x in omega or ex i,j st x = [i,j] & i,j are_relative_prime & j <> {} & j <> 1
proof
  assume not x in omega;
  then
A1: x in RATplus \ {[k,1]: not contradiction} by XBOOLE_0:def 3;
  then
A2: not x in {[k,1]: not contradiction} by XBOOLE_0:def 5;
  x in RATplus by A1;
  then consider a,b being Element of omega such that
A3: x = [a,b] & a,b are_relative_prime & b <> {};
  [a,1] in {[k,1]: not contradiction};
  hence thesis by A2,A3;
end;

theorem Th30:
  not ex i,j being set st [i,j] is Ordinal
proof
  given i,j being set such that
A1: [i,j] is Ordinal;
  {} in [i,j] by A1,ORDINAL3:8;
  hence thesis by TARSKI:def 2;
end;

theorem Th31:
  A in RAT+ implies A in omega
proof
  assume A in RAT+ & not A in omega;
  then ex i,j st A = [i,j] & i,j are_relative_prime & j <> {} & j <> 1 by Th29;
  hence thesis by Th30;
end;

registration
  cluster -> natural for ordinal Element of RAT+;
  coherence
  proof
    let x be ordinal Element of RAT+;
    assume not x in omega;
    then
A1: ex i,j st x = [i,j] & i,j are_relative_prime & j <> {} & j <> 1 by Th29;
    then {} in x by ORDINAL3:8;
    hence thesis by A1,TARSKI:def 2;
  end;
end;

theorem Th32:
  not ex i,j being set st [i,j] in omega
proof
  given i,j being set such that
A1: [i,j] in omega;
  reconsider a = [i,j] as Element of omega by A1;
  {} in a by ORDINAL3:8;
  hence thesis by TARSKI:def 2;
end;

theorem Th33:
  [i,j] in RAT+ iff i,j are_relative_prime & j <> {} & j <> 1
proof
A1: not [i,j] in omega by Th32;
  hereby
    assume [i,j] in RAT+;
    then
A2: [i,j] in RATplus \ {[k,1]: not contradiction} by A1,XBOOLE_0:def 3;
    hence i,j are_relative_prime & j <> {} by Lm4;
    not [i,j] in {[k,1]: not contradiction} by A2,XBOOLE_0:def 5;
    hence j <> 1;
  end;
  assume i,j are_relative_prime & j <> {};
  then
A3: [i,j] in RATplus;
  assume j <> 1;
  then not ex k st [i,j] = [k,1] by XTUPLE_0:1;
  then not [i,j] in {[k,1]: not contradiction};
  then [i,j] in RATplus \ {[k,1]: not contradiction} by A3,XBOOLE_0:def 5;
  hence thesis by XBOOLE_0:def 3;
end;

definition
  let x be Element of RAT+;
  func numerator x -> Element of omega means
  :Def8:
  it = x if x in omega otherwise ex a st x = [it,a];
  existence
  proof
    thus x in omega implies ex a being Element of omega st a = x;
    assume not x in omega;
    then x in RATplus \ {[k,1]: not contradiction} by XBOOLE_0:def 3;
    then x in RATplus;
    then consider a,b being Element of omega such that
A1: x = [a,b] and
    a,b are_relative_prime and
    b <> {};
    take a,b;
    thus thesis by A1;
  end;
  correctness by XTUPLE_0:1;
  func denominator x -> Element of omega means
  : Def9:
  it = 1 if x in omega otherwise ex a st x = [a,it];
  existence
  proof
    thus x in omega implies ex a being Element of omega st a = 1;
    assume not x in omega;
    then x in RATplus \ {[k,1]: not contradiction} by XBOOLE_0:def 3;
    then x in RATplus;
    then consider a,b being Element of omega such that
A2: x = [a,b] and
    a,b are_relative_prime and
    b <> {};
    take b,a;
    thus thesis by A2;
  end;
  correctness by XTUPLE_0:1;
end;

theorem Th34:
  numerator x, denominator x are_relative_prime
proof
  per cases;
  suppose
    x in omega;
    then denominator x = 1 by Def9;
    hence thesis by Th2;
  end;
  suppose
A1: not x in omega;
    then consider i,j such that
A2: x = [i,j] and
A3: i,j are_relative_prime and
    j <> {} and
    j <> 1 by Th29;
    i = numerator x by A1,A2,Def8;
    hence thesis by A1,A2,A3,Def9;
  end;
end;

theorem Th35:
  denominator x <> {}
proof
  per cases;
  suppose
    x in omega;
    hence thesis by Def9;
  end;
  suppose
A1: not x in omega;
    then ex i,j st x = [i,j] & i,j are_relative_prime & j <> {} & j <> 1 by
Th29;
    hence thesis by A1,Def9;
  end;
end;

theorem Th36:
  not x in omega implies x = [numerator x, denominator x] & denominator x <> 1
proof
  assume
A1: not x in omega;
  then consider i,j such that
A2: x = [i,j] and
  i,j are_relative_prime and
  j <> {} and
A3: j <> 1 by Th29;
  i = numerator x by A1,A2,Def8;
  hence thesis by A1,A2,A3,Def9;
end;

theorem Th37:
  x <> {} iff numerator x <> {}
proof
  hereby
    assume that
A1: x <> {} and
A2: numerator x = {};
A3: not x in omega by A1,A2,Def8;
    then consider i,j such that
A4: x = [i,j] and
A5: i,j are_relative_prime and
    j <> {} and
A6: j <> 1 by Th29;
    i = {} by A2,A3,A4,Def8;
    hence contradiction by A5,A6,Th3;
  end;
  {} in omega by ORDINAL1:def 11;
  hence thesis by Def8;
end;

theorem
  x in omega iff denominator x = 1 by Def9,Th36;

definition
  let i,j be natural Ordinal;
  func i/j -> Element of RAT+ equals
  :Def10:
  {} if j = {}, RED(i,j) if RED(j,i
  ) = 1 otherwise [RED(i,j), RED(j,i)];
  coherence
  proof
A1: now
      assume j <> {};
      then RED(i,j), RED(j,i) are_relative_prime & RED(j,i) <> {} by Th18,Th19;
      hence RED(j,i) <> 1 implies [RED(i,j), RED(j,i)] in RAT+ by Th33;
    end;
    RED(i,j) in omega;
    hence thesis by A1,Lm1,Lm5;
  end;
  consistency by Th26;
end;

notation
  let i,j be natural Ordinal;
  synonym quotient(i,j) for i/j;
end;

theorem Th39:
  (numerator x)/(denominator x) = x
proof
A1: denominator x <> {} by Th35;
A2: RED(numerator x, denominator x) = numerator x by Th23,Th34;
  numerator x, denominator x are_relative_prime by Th34;
  then
A3: RED(denominator x, numerator x) = denominator x by Th23;
  per cases;
  suppose
A4: denominator x = 1;
    then x in omega by Th36;
    then numerator x = x by Def8;
    hence thesis by A2,A3,A4,Def10;
  end;
  suppose
A5: denominator x <> 1;
    then not x in omega by Def9;
    then x = [numerator x, denominator x] by Th36;
    hence thesis by A1,A2,A3,A5,Def10;
  end;
end;

theorem Th40:
  {}/b = {} & a/1 = a
proof
A1: b <> {} implies RED(b,{}) = 1 by Th26;
  RED({},b) = {} by Th26;
  hence {}/b = {} by A1,Def10;
  RED(1,a) = 1 & RED(a,1) = a by Th24;
  hence thesis by Def10;
end;

theorem Th41:
  a <> {} implies a/a = 1
proof
  assume a <> {};
  then RED(a,a) = 1 by Th27;
  hence thesis by Def10;
end;

theorem Th42:
  b <> {} implies numerator (a/b) = RED(a,b) & denominator (a/b) = RED(b,a)
proof
  assume
A1: b <> {};
  per cases;
  suppose
A2: RED(b,a) = 1;
    then a/b = RED(a,b) by Def10;
    hence thesis by A2,Def8,Def9;
  end;
  suppose
A3: RED(b,a) <> 1;
A4: not [RED(a,b), RED(b,a)] in omega by Th32;
    a/b = [RED(a,b), RED(b,a)] by A1,A3,Def10;
    hence thesis by A4,Def8,Def9;
  end;
end;

theorem Th43:
  i,j are_relative_prime & j <> {} implies numerator (i/j) = i &
  denominator (i/j) = j
proof
  assume i,j are_relative_prime;
  then RED(i,j) = i & RED(j,i) = j by Th23;
  hence thesis by Th42;
end;

theorem Th44:
  c <> {} implies (a*^c)/(b*^c) = a/b
proof
  assume
A1: c <> {};
  per cases;
  suppose
    b = {};
    then a/b = {} & b*^c = {} by Def10,ORDINAL2:35;
    hence thesis by Def10;
  end;
  suppose
A2: b <> {};
    then
A3: b*^c <> {} by A1,ORDINAL3:31;
    then
A4: denominator ((a*^c)/(b*^c)) = RED(b*^c,a*^c) by Th42
      .= RED(b,a) by A1,Th28
      .= denominator (a/b) by A2,Th42;
    numerator ((a*^c)/(b*^c)) = RED(a*^c,b*^c) by A3,Th42
      .= RED(a,b) by A1,Th28
      .= numerator (a/b) by A2,Th42;
    hence (a*^c)/(b*^c) = (numerator (a/b))/denominator (a/b) by A4,Th39
      .= a/b by Th39;
  end;
end;

reserve i,j,k for natural Ordinal;

theorem Th45:
  j <> {} & l <> {} implies (i/j = k/l iff i*^l = j*^k)
proof
  assume that
A1: j <> {} and
A2: l <> {};
  set x = i/j, y = k/l;
  set ny = numerator y, dy = denominator y;
A3: ny = RED(k,l) by A2,Th42;
  set nx = numerator x, dx = denominator x;
A4: dx = RED(j,i) by A1,Th42;
A5: dy = RED(l,k) by A2,Th42;
A6: nx = RED(i,j) by A1,Th42;
  hereby
    assume i/j = k/l;
    then i = ny*^(i hcf j) & l = dx*^(l hcf k) by A6,A5,Th21;
    hence i*^l = ny*^(i hcf j)*^dx*^(l hcf k) by ORDINAL3:50
      .= ny*^((i hcf j)*^dx)*^(l hcf k) by ORDINAL3:50
      .= ny*^j*^(l hcf k) by A4,Th21
      .= j*^(ny*^(l hcf k)) by ORDINAL3:50
      .= j*^k by A3,Th21;
  end;
  assume
A7: i*^l = j*^k;
  then dx = RED(j*^l,j*^k) by A2,A4,Th28;
  then
A8: dx = dy by A1,A5,Th28;
  nx = RED(j*^k,j*^l) by A2,A6,A7,Th28;
  then nx = ny by A1,A3,Th28;
  then x = ny/dy by A8,Th39;
  hence thesis by Th39;
end;

definition
  let x,y be Element of RAT+;
  func x+y -> Element of RAT+ equals
  ((numerator x)*^(denominator y)+^(
  numerator y)*^(denominator x)) / ((denominator x)*^(denominator y));
  coherence;
  commutativity;
  func x*'y -> Element of RAT+ equals
  ((numerator x)*^(numerator y)) / ((
  denominator x)*^(denominator y));
  coherence;
  commutativity;
end;

theorem Th46:
  j <> {} & l <> {} implies (i/j)+(k/l) = (i*^l+^j*^k)/(j*^l)
proof
  assume that
A1: j <> {} and
A2: l <> {};
A3: denominator (k/l) = RED(l,k) & denominator (i/j) = RED(j,i) by A1,A2,Th42;
  numerator (k/l) = RED(k,l) & numerator (i/j) = RED(i,j) by A1,A2,Th42;
  hence
  (i/j)+(k/l) = (RED(i,j)*^RED(l,k)+^RED(j,i)*^RED(k,l))*^(i hcf j) /(RED
  (j,i)*^RED(l,k)*^(i hcf j)) by A1,A3,Th15,Th44
    .= (RED(i,j)*^RED(l,k)+^RED(j,i)*^RED(k,l))*^(i hcf j) /(RED(l,k)*^(RED(
  j,i)*^(i hcf j))) by ORDINAL3:50
    .= (RED(i,j)*^RED(l,k)+^RED(j,i)*^RED(k,l))*^(i hcf j)/(RED(l,k)*^j) by
Th21
    .= (RED(i,j)*^RED(l,k)*^(i hcf j)+^RED(j,i)*^RED(k,l)*^(i hcf j)) /(RED(
  l,k)*^j) by ORDINAL3:46
    .= (RED(l,k)*^(RED(i,j)*^(i hcf j))+^RED(j,i)*^RED(k,l)*^(i hcf j)) /(
  RED(l,k)*^j) by ORDINAL3:50
    .= (RED(l,k)*^i+^RED(j,i)*^RED(k,l)*^(i hcf j))/(RED(l,k)*^j) by Th21
    .= (RED(l,k)*^i+^RED(k,l)*^(RED(j,i)*^(i hcf j)))/(RED(l,k)*^j) by
ORDINAL3:50
    .= (RED(l,k)*^i+^RED(k,l)*^j)/(RED(l,k)*^j) by Th21
    .= ((RED(l,k)*^i+^RED(k,l)*^j)*^(k hcf l))/(RED(l,k)*^j*^(k hcf l)) by A2
,Th15,Th44
    .= ((RED(l,k)*^i+^RED(k,l)*^j)*^(k hcf l))/(j*^(RED(l,k)*^(k hcf l))) by
ORDINAL3:50
    .= ((RED(l,k)*^i+^RED(k,l)*^j)*^(k hcf l))/(j*^l) by Th21
    .= (RED(l,k)*^i*^(k hcf l)+^RED(k,l)*^j*^(k hcf l))/(j*^l) by ORDINAL3:46
    .= (i*^(RED(l,k)*^(k hcf l))+^RED(k,l)*^j*^(k hcf l))/(j*^l) by ORDINAL3:50
    .= (i*^l+^RED(k,l)*^j*^(k hcf l))/(j*^l) by Th21
    .= (i*^l+^j*^(RED(k,l)*^(k hcf l)))/(j*^l) by ORDINAL3:50
    .= (i*^l+^j*^k)/(j*^l) by Th21;
end;

theorem Th47:
  k <> {} implies (i/k)+(j/k) = (i+^j)/k
proof
  assume
A1: k <> {};
  hence (i/k)+(j/k) = (i*^k+^j*^k)/(k*^k) by Th46
    .= ((i+^j)*^k)/(k*^k) by ORDINAL3:46
    .= (i+^j)/k by A1,Th44;
end;

registration
  cluster empty for Element of RAT+;
  existence by Lm1,Lm5;
end;

definition
  redefine func {} -> Element of RAT+;
  coherence by Lm1,Lm5;
  redefine func one -> non empty ordinal Element of RAT+;
  coherence by Lm2,Lm5;
end;

theorem Th48:
  x*'{} = {}
proof
  numerator {} = {} & (numerator x)*^{} = {} by Def8,Lm1,ORDINAL2:35;
  hence thesis by Th40;
end;

theorem Th49:
  (i/j)*'(k/l) = (i*^k)/(j*^l)
proof
  per cases;
  suppose
A1: j <> {} & l <> {};
    then
A2: denominator (k/l) = RED(l,k) & denominator (i/j) = RED(j,i) by Th42;
    numerator (k/l) = RED(k,l) & numerator (i/j) = RED(i,j) by A1,Th42;
    hence
    (i/j)*'(k/l) = (RED(i,j)*^RED(k,l)*^(i hcf j))/(RED(j,i)*^RED(l,k)*^(
    i hcf j)) by A1,A2,Th15,Th44
      .= (RED(k,l)*^(RED(i,j)*^(i hcf j)))/(RED(j,i)*^RED(l,k)*^(i hcf j))
    by ORDINAL3:50
      .= (RED(k,l)*^i)/(RED(j,i)*^RED(l,k)*^(i hcf j)) by Th21
      .= (RED(k,l)*^i)/(RED(l,k)*^(RED(j,i)*^(i hcf j))) by ORDINAL3:50
      .= (RED(k,l)*^i)/(RED(l,k)*^j) by Th21
      .= (RED(k,l)*^i*^(l hcf k))/(RED(l,k)*^j*^(l hcf k)) by A1,Th15,Th44
      .= (i*^(RED(k,l)*^(l hcf k)))/(RED(l,k)*^j*^(l hcf k)) by ORDINAL3:50
      .= (i*^k)/(RED(l,k)*^j*^(l hcf k)) by Th21
      .= (i*^k)/(j*^(RED(l,k)*^(l hcf k))) by ORDINAL3:50
      .= (i*^k)/(j*^l) by Th21;
  end;
  suppose
A3: j = {} or l = {};
    then i/j = {} or k/l = {} by Def10;
    then
A4: (i/j)*'(k/l) = {} by Th48;
    j*^l = {} by A3,ORDINAL2:35;
    hence thesis by A4,Def10;
  end;
end;

theorem Th50:
  x+{} = x
proof
A1: (numerator x)*^1 = numerator x & {}*^(denominator x) = {} by ORDINAL2:35,39
;
A2: (denominator x)*^1 = denominator x & (numerator x)+^{} = numerator x by
ORDINAL2:27,39;
  denominator {} = 1 & numerator {} = {} by Def8,Def9,Lm1;
  hence thesis by A1,A2,Th39;
end;

theorem Th51:
  (x+y)+z = x+(y+z)
proof
  set nx = numerator x, ny = numerator y, nz = numerator z;
  set dx = denominator x, dy = denominator y, dz = denominator z;
A1: dy <> {} by Th35;
A2: dz <> {} by Th35;
  then
A3: dy*^dz <> {} by A1,ORDINAL3:31;
A4: dx <> {} by Th35;
  then
A5: dx*^dy <> {} by A1,ORDINAL3:31;
  thus x+y+z = (nx*^dy+^dx*^ny)/(dx*^dy)+nz/dz by Th39
    .= ((nx*^dy+^dx*^ny)*^dz+^(dx*^dy)*^nz)/(dx*^dy*^dz) by A2,A5,Th46
    .= (nx*^dy*^dz+^dx*^ny*^dz+^dx*^dy*^nz)/(dx*^dy*^dz) by ORDINAL3:46
    .= (nx*^(dy*^dz)+^dx*^ny*^dz+^dx*^dy*^nz)/(dx*^dy*^dz) by ORDINAL3:50
    .= (nx*^(dy*^dz)+^dx*^(ny*^dz)+^dx*^dy*^nz)/(dx*^dy*^dz) by ORDINAL3:50
    .= (nx*^(dy*^dz)+^dx*^(ny*^dz)+^dx*^(dy*^nz))/(dx*^dy*^dz) by ORDINAL3:50
    .= (nx*^(dy*^dz)+^(dx*^(ny*^dz)+^dx*^(dy*^nz)))/(dx*^dy*^dz) by ORDINAL3:30
    .= (nx*^(dy*^dz)+^dx*^(ny*^dz+^dy*^nz))/(dx*^dy*^dz) by ORDINAL3:46
    .= (nx*^(dy*^dz)+^dx*^(ny*^dz+^dy*^nz))/(dx*^(dy*^dz)) by ORDINAL3:50
    .= (ny*^dz+^dy*^nz)/(dy*^dz)+nx/dx by A4,A3,Th46
    .= x+(y+z) by Th39;
end;

theorem Th52:
  (x*'y)*'z = x*'(y*'z)
proof
  set nx = numerator x, ny = numerator y, nz = numerator z;
  set dx = denominator x, dy = denominator y, dz = denominator z;
A1: x = nx/dx by Th39;
  z = nz/dz by Th39;
  hence (x*'y)*'z = (nx*^ny*^nz)/(dx*^dy*^dz) by Th49
    .= (nx*^(ny*^nz))/(dx*^dy*^dz) by ORDINAL3:50
    .= (nx*^(ny*^nz))/(dx*^(dy*^dz)) by ORDINAL3:50
    .= x*'(y*'z) by A1,Th49;
end;

theorem Th53:
  x*'one = x
proof
  set y = one;
A1: (numerator x)*^1 = numerator x & (denominator x)*^1 = denominator x by
ORDINAL2:39;
  numerator y = 1 & denominator y = 1 by Def8,Def9;
  hence thesis by A1,Th39;
end;

theorem Th54:
  x <> {} implies ex y st x*'y = 1
proof
  set nx = numerator x, dx = denominator x;
A1: dx <> {} by Th35;
  assume x <> {};
  then
A2: nx <> {} by Th37;
  take y = dx/nx;
  nx, dx are_relative_prime by Th34;
  then denominator y = nx & numerator y = dx by A2,Th43;
  hence thesis by A2,A1,Th41,ORDINAL3:31;
end;

theorem Th55:
  x <> {} implies ex z st y = x*'z
proof
  reconsider o = one as Element of RAT+;
  assume x <> {};
  then consider z such that
A1: x*'z = 1 by Th54;
  take z*'y;
  thus y = y*'o by Th53
    .= x*'(z*'y) by A1,Th52;
end;

theorem Th56:
  x <> {} & x*'y = x*'z implies y = z
proof
  assume x <> {};
  then consider r being Element of RAT+ such that
A1: x*'r = 1 by Th54;
  r*'(x*'y) = one*'y by A1,Th52;
  then
A2: r*'(x*'y) = y by Th53;
  r*'(x*'z) = one*'z by A1,Th52;
  hence thesis by A2,Th53;
end;

theorem Th57:
  x*'(y+z) = x*'y+x*'z
proof
  set nx = numerator x, ny = numerator y, nz = numerator z;
  set dx = denominator x, dy = denominator y, dz = denominator z;
A1: dx <> {} by Th35;
  dz <> {} by Th35;
  then
A2: dx*^dz <> {} by A1,ORDINAL3:31;
  dy <> {} by Th35;
  then
A3: dx*^dy <> {} by A1,ORDINAL3:31;
  x = nx/dx by Th39;
  hence x*'(y+z) = (nx*^(ny*^dz+^dy*^nz))/(dx*^(dy*^dz)) by Th49
    .= (nx*^(ny*^dz)+^nx*^(dy*^nz))/(dx*^(dy*^dz)) by ORDINAL3:46
    .= (nx*^ny*^dz+^nx*^(dy*^nz))/(dx*^(dy*^dz)) by ORDINAL3:50
    .= (nx*^ny*^dz+^dy*^(nx*^nz))/(dx*^(dy*^dz)) by ORDINAL3:50
    .= (nx*^ny*^dz+^dy*^(nx*^nz))/(dy*^(dx*^dz)) by ORDINAL3:50
    .= (dx*^((nx*^ny)*^dz+^(dy*^(nx*^nz))))/(dx*^(dy*^(dx*^dz))) by Th35,Th44
    .= (dx*^((nx*^ny)*^dz)+^dx*^(dy*^(nx*^nz)))/(dx*^(dy*^(dx*^dz))) by
ORDINAL3:46
    .= (dx*^((nx*^ny)*^dz)+^(dx*^dy)*^(nx*^nz))/(dx*^(dy*^(dx*^dz))) by
ORDINAL3:50
    .= ((nx*^ny)*^(dx*^dz)+^(dx*^dy)*^(nx*^nz))/(dx*^(dy*^(dx*^dz))) by
ORDINAL3:50
    .= ((nx*^ny)*^(dx*^dz)+^(dx*^dy)*^(nx*^nz))/((dx*^dy)*^(dx*^dz)) by
ORDINAL3:50
    .= x*'y+x*'z by A2,A3,Th46;
end;

theorem Th58:
  for i,j being ordinal Element of RAT+ holds i+j = i+^j
proof
  let i,j be ordinal Element of RAT+;
  set ni = numerator i, nj = numerator j;
A1: j in omega by ORDINAL1:def 12;
  then
A2: denominator j = 1 by Def9;
A3: i in omega by ORDINAL1:def 12;
  then denominator i = 1 by Def9;
  hence i+j = (ni*^1+^1*^nj)/1 by A2,ORDINAL2:39
    .= (ni+^1*^nj)/1 by ORDINAL2:39
    .= (ni+^nj)/1 by ORDINAL2:39
    .= ni+^nj by Th40
    .= i+^nj by A3,Def8
    .= i+^j by A1,Def8;
end;

theorem
  for i,j being ordinal Element of RAT+ holds i*'j = i*^j
proof
  let i,j be ordinal Element of RAT+;
  set ni = numerator i, nj = numerator j;
A1: j in omega by ORDINAL1:def 12;
  then
A2: denominator j = 1 by Def9;
A3: i in omega by ORDINAL1:def 12;
  then denominator i = 1 by Def9;
  hence i*'j = (ni*^nj)/1 by A2,ORDINAL2:39
    .= ni*^nj by Th40
    .= i*^nj by A3,Def8
    .= i*^j by A1,Def8;
end;

theorem Th60:
  ex y st x = y+y
proof
  set 02 = one+one;
  02 = 1+^1 by Th58;
  then 02 <> {} by ORDINAL3:26;
  then consider z such that
A1: 02*'z = 1 by Th54;
  take y = z*'x;
  thus x = one*'x by Th53
    .= 02*'y by A1,Th52
    .= one*'y+one*'y by Th57
    .= y+one*'y by Th53
    .= y+y by Th53;
end;

definition
  let x,y be Element of RAT+;
  pred x <=' y means
  : Def13:
  ex z being Element of RAT+ st y = x+z;
  connectedness
  proof
    let x,y be Element of RAT+;
    set nx = numerator x, ny = numerator y;
    set dx = denominator x, dy = denominator y;
A1: dx <> {} & dy <> {} by Th35;
A2: nx/dx = x & ny/dy = y by Th39;
    nx*^dy c= ny*^dx or ny*^dx c= nx*^dy;
    then
A3: ny*^dx = nx*^dy+^(ny*^dx -^ nx*^dy) or nx*^dy = ny*^dx+^(nx*^dy -^ ny
    *^dx) by ORDINAL3:def 5;
    nx*^dy/(dx*^dy) = nx/dx & ny*^dx/(dx*^dy) = ny/dy by Th35,Th44;
    then x = y+((nx*^dy -^ ny*^dx)/(dx*^dy)) or y = x+((ny*^dx -^ nx*^dy)/(dx
    *^dy)) by A1,A2,A3,Th47,ORDINAL3:31;
    hence thesis;
  end;
end;

notation
  let x,y be Element of RAT+;
  antonym y < x for x <=' y;
end;

reserve r,s,t for Element of RAT+;

theorem
  not ex y being set st [{},y] in RAT+
proof
  given y being set such that
A1: [{},y] in RAT+;
  consider i,j being Element of omega such that
A2: [{},y] = [i,j] and
A3: i,j are_relative_prime and
  j <> {} and
A4: j <> 1 by A1,Th29,Th32;
  i = {} by A2,XTUPLE_0:1;
  hence thesis by A3,A4,Th3;
end;

theorem Th62:
  s + t = r + t implies s = r
proof
  assume
A1: s + t = r + t;
  set r1 = numerator r, r2 = denominator r;
  set t1 = numerator t, t2 = denominator t;
  set s1 = numerator s, s2 = denominator s;
A2: t2 <> {} by Th35;
A3: s2 <> {} by Th35;
  then
A4: s2*^t2 <> {} by A2,ORDINAL3:31;
A5: r2 <> {} by Th35;
  then r2*^t2 <> {} by A2,ORDINAL3:31;
  then (s1*^t2+^s2*^t1)*^(r2*^t2) = (r1*^t2+^r2*^t1)*^(s2*^t2) by A1,A4,Th45
    .= (r1*^t2+^r2*^t1)*^s2*^t2 by ORDINAL3:50;
  then (s1*^t2+^s2*^t1)*^r2*^t2 = (r1*^t2+^r2*^t1)*^s2*^t2 by ORDINAL3:50;
  then (s1*^t2+^s2*^t1)*^r2 = (r1*^t2+^r2*^t1)*^s2 by A2,ORDINAL3:33
    .= r1*^t2*^s2+^r2*^t1*^s2 by ORDINAL3:46;
  then s1*^t2*^r2+^s2*^t1*^r2 = r1*^t2*^s2+^r2*^t1*^s2 by ORDINAL3:46
    .= r1*^t2*^s2+^s2*^t1*^r2 by ORDINAL3:50;
  then s1*^t2*^r2 = r1*^t2*^s2 by ORDINAL3:21
    .= r1*^s2*^t2 by ORDINAL3:50;
  then s1*^r2*^t2 = r1*^s2*^t2 by ORDINAL3:50;
  then s1*^r2 = r1*^s2 by A2,ORDINAL3:33;
  then s1/s2 = r1/r2 by A3,A5,Th45
    .= r by Th39;
  hence thesis by Th39;
end;

theorem Th63:
  r+s = {} implies r = {}
proof
  set nr = numerator r, dr = denominator r;
  set ns = numerator s, ds = denominator s;
  assume
A1: r+s = {};
  denominator {} = 1 & numerator {} = {} by Def8,Def9,Lm1;
  then
A2: (nr*^ds+^ns*^dr)/(dr*^ds) = {}/1 by A1,Th39;
A3: ds <> {} by Th35;
  dr <> {} by Th35;
  then dr*^ds <> {} by A3,ORDINAL3:31;
  then (nr*^ds+^ns*^dr)*^1 = (dr*^ds)*^{} by A2,Th45
    .= {} by ORDINAL2:35;
  then nr*^ds+^ns*^dr = {} by ORDINAL3:31;
  then nr*^ds = {} by ORDINAL3:26;
  then nr = {} by A3,ORDINAL3:31;
  hence thesis by Th37;
end;

theorem Th64:
  {} <=' s
proof
  take s;
  thus thesis by Th50;
end;

theorem
  s <=' {} implies s = {}
proof
  assume ex r st {} = s+r;
  hence thesis by Th63;
end;

theorem Th66:
  r <=' s & s <=' r implies r = s
proof
  given x such that
A1: s = r+x;
  given y such that
A2: r = s+y;
  r+{} = r by Th50
    .= r+(x+y) by A1,A2,Th51;
  then x = {} by Th62,Th63;
  hence thesis by A1,Th50;
end;

theorem Th67:
  r <=' s & s <=' t implies r <=' t
proof
  given x such that
A1: s = r+x;
  given y such that
A2: t = s+y;
  take x+y;
  thus thesis by A1,A2,Th51;
end;

theorem
  r < s iff r <=' s & r <> s by Th66;

theorem
  r < s & s <=' t or r <=' s & s < t implies r < t by Th67;

theorem
  r < s & s < t implies r < t by Th67;

theorem Th71:
  x in omega & x+y in omega implies y in omega
proof
  assume that
A1: x in omega and
A2: x+y in omega;
A3: denominator (x+y) = 1 by A2,Def9;
  set nx = numerator x, dx = denominator x;
A4: dx = 1 by A1,Def9;
  set ny = numerator y, dy = denominator y;
A5: x+y = numerator (x+y)/denominator (x+y) & (nx*^dy+^ny*^dx)*^1 = nx*^dy+^
  ny*^ dx by Th39,ORDINAL2:39;
  dy <> {} by Th35;
  then dx*^dy <> {} by A4,ORDINAL3:31;
  then nx*^dy+^ny*^dx = dx*^dy*^numerator (x+y) by A3,A5,Th45
    .= dy*^numerator (x+y) by A4,ORDINAL2:39;
  then nx*^dy+^ny = dy*^numerator (x+y) by A4,ORDINAL2:39;
  then ny = (dy*^numerator (x+y))-^nx*^dy by ORDINAL3:52;
  then ny = dy*^((numerator (x+y))-^nx) by ORDINAL3:63;
  then
A6: dy divides ny by Def3;
A7: ny,dy are_relative_prime by Th34;
  for m being natural Ordinal st m divides dy & m divides ny holds m
  divides dy;
  then dy hcf ny = dy by A6,Def5;
  then
A8: dy = 1 by A7,Th20;
  y = ny/dy by Th39;
  then y = ny by A8,Th40;
  hence thesis;
end;

theorem Th72:
  for i being ordinal Element of RAT+ st i < x & x < i+one holds not x in omega
proof
  let i be ordinal Element of RAT+;
  assume that
A1: i < x and
A2: x < i+one and
A3: x in omega;
  consider z such that
A4: i+one = x+z by A2,Def13;
  i+one = i+^1 by Th58;
  then i+one in omega by Th31;
  then reconsider z9 = z as Element of omega by A3,A4,Th71;
  consider y such that
A5: x = i+y by A1,Def13;
  i in omega by Th31;
  then reconsider y9 = y as Element of omega by A3,A5,Th71;
  i+one = i+(y+z) by A5,A4,Th51;
  then 1 = y+z by Th62
    .= y9+^z9 by Th58;
  then y9 c= 1 by ORDINAL3:24;
  then y = {} or y = 1 by ORDINAL3:16;
  then i = x or i+one = x by A5,Th50;
  hence contradiction by A1,A2;
end;

theorem Th73:
  t <> {} implies ex r st r < t & not r in omega
proof
  assume
A1: t <> {};
A2: 1+^1 = succ 1 by ORDINAL2:31;
  per cases;
  suppose
A3: one <=' t;
    consider r such that
A4: one = r+r by Th60;
    take r;
    r <=' one & r <> 1 by A2,A4,Def13,Th58;
    then r < one by Th66;
    hence r < t by A3,Th67;
A5: 1*^1 = 1 by ORDINAL2:39;
    assume r in omega;
    then
A6: denominator r = 1 by Def9;
    then (numerator r)*^denominator r = numerator r by ORDINAL2:39;
    then
A7: 1 = (numerator r)+^numerator r by A4,A6,A5,Th40;
    then numerator r c= 1 by ORDINAL3:24;
    then numerator r = {} or numerator r = 1 by ORDINAL3:16;
    hence contradiction by A2,A7,ORDINAL2:27;
  end;
  suppose
A8: t < one;
    consider r such that
A9: t = r+r by Th60;
A10: {} <=' r by Th64;
    r <> {} by A1,A9,Th50;
    then
A11: {} < r by A10,Th66;
    take r;
A12: 1 = {}+one by Th50;
A13: r <=' t by A9,Def13;
    now
      assume r = t;
      then t+{} = t+t by A9,Th50;
      hence contradiction by A1,Th62;
    end;
    hence r < t by A13,Th66;
    r < one by A8,A13,Th67;
    hence thesis by A11,A12,Th72;
  end;
end;

theorem
  {s: s < t} in RAT+ iff t = {}
proof
  hereby
    assume
A1: {s: s < t} in RAT+ & t <> {};
    then consider r such that
A2: r < t and
A3: not r in omega by Th73;
    {} <=' t by Th64;
    then {} < t by A1,Th66;
    then
A4: {} in {s: s < t};
    r in {s: s < t} by A2;
    then {s: s < t} in omega implies r is Ordinal;
    then ex i,j being Element of omega st {s: s < t} = [i,j] & i,j
    are_relative_prime & j <> {} & j <> 1 by A1,A3,Th29,Th31;
    hence contradiction by A4,TARSKI:def 2;
  end;
  assume
A5: t = {};
  {s: s < t} c= {}
  proof
    let a be set;
    assume a in {s: s < t};
    then ex s st a = s & s < t;
    hence thesis by A5,Th64;
  end;
  then {s: s < t} = {};
  hence thesis;
end;

theorem
  for A being Subset of RAT+ st (ex t st t in A & t <> {}) & for r,s st
r in A & s <=' r holds s in A ex r1,r2,r3 being Element of RAT+ st r1 in A & r2
  in A & r3 in A & r1 <> r2 & r2 <> r3 & r3 <> r1
proof
  let A be Subset of RAT+;
  given t such that
A1: t in A and
A2: t <> {};
  assume
A3: for r,s st r in A & s <=' r holds s in A;
  consider x such that
A4: t = x+x by Th60;
  take {}, x, t;
  x <=' t by A4,Def13;
  hence {} in A & x in A & t in A by A1,A3,Th64;
  thus {} <> x by A2,A4,Th50;
  hereby
    assume x = t;
    then t+{} = t+t by A4,Th50;
    hence contradiction by A2,Th62;
  end;
  thus thesis by A2;
end;

theorem Th76:
  s + t <=' r + t iff s <=' r
proof
  thus s + t <=' r + t implies s <=' r
  proof
    given z such that
A1: r+t = s+t+z;
    take z;
    r+t = s+z+t by A1,Th51;
    hence thesis by Th62;
  end;
  given z such that
A2: r = s+z;
  take z;
  thus thesis by A2,Th51;
end;

theorem
  s <=' s + t
proof
  take t;
  thus thesis;
end;

theorem
  r*'s = {} implies r = {} or s = {}
proof
  set nr = numerator r, dr = denominator r;
  set ns = numerator s, ds = denominator s;
  assume
A1: r*'s = {};
  dr <> {} & ds <> {} by Th35;
  then
A2: dr*^ds <> {} by ORDINAL3:31;
  denominator {} = 1 & numerator {} = {} by Def8,Def9,Lm1;
  then (nr*^ns)/(dr*^ds) = {}/1 by A1,Th39;
  then (nr*^ns)*^1 = (dr*^ds)*^{} by A2,Th45
    .= {} by ORDINAL2:35;
  then nr*^ns = {} by ORDINAL3:31;
  then nr = {} or ns = {} by ORDINAL3:31;
  hence thesis by Th37;
end;

theorem Th79:
  r <=' s *' t implies ex t0 being Element of RAT+ st r = s *' t0 & t0 <=' t
proof
  given x such that
A1: s*'t = r+x;
  per cases;
  suppose
A2: s = {};
    take t;
    s*'t = {} by A2,Th48;
    hence thesis by A1,Th63;
  end;
  suppose
A3: s <> {};
    then consider t1 being Element of RAT+ such that
A4: x = s*'t1 by Th55;
    consider t0 being Element of RAT+ such that
A5: r = s*'t0 by A3,Th55;
    take t0;
    thus r = s*'t0 by A5;
    take t1;
    s*'t = s*'(t0+t1) by A1,A5,A4,Th57;
    hence thesis by A3,Th56;
  end;
end;

theorem
  t <> {} & s *' t <=' r *' t implies s <=' r
proof
  assume that
A1: t <> {} and
A2: s *' t <=' r *' t;
  ex x st s*'t = t*'x & x <=' r by A2,Th79;
  hence thesis by A1,Th56;
end;

theorem
  for r1,r2,s1,s2 being Element of RAT+ st r1+r2 = s1+s2 holds r1 <=' s1
  or r2 <=' s2
proof
  let r1,r2,s1,s2 be Element of RAT+ such that
A1: r1+r2 = s1+s2;
  assume that
A2: s1 < r1 and
A3: s2 < r2;
  s1+s2 < r1+s2 by A2,Th76;
  hence thesis by A1,A3,Th76;
end;

theorem Th82:
  s <=' r implies s *' t <=' r *' t
proof
  given x such that
A1: r = s+x;
  take x*'t;
  thus thesis by A1,Th57;
end;

theorem
  for r1,r2,s1,s2 being Element of RAT+ st r1*'r2 = s1*'s2 holds r1 <='
  s1 or r2 <=' s2
proof
  let r1,r2,s1,s2 be Element of RAT+ such that
A1: r1*'r2 = s1*'s2;
A2: {} <=' s1 by Th64;
  assume that
A3: s1 < r1 and
A4: s2 < r2;
  s2 <> r2 & s1*'s2 <=' r1*'s2 by A3,A4,Th82;
  then s1*'s2 < r1*'s2 by A1,A3,A2,Th56,Th66;
  hence thesis by A1,A4,Th82;
end;

theorem
  r = {} iff r + s = s
proof
  s+{} = s by Th50;
  hence thesis by Th62;
end;

theorem
  for s1,t1,s2,t2 being Element of RAT+ st s1 + t1 = s2 + t2 & s1 <=' s2
  holds t2 <=' t1
proof
  let s1,t1,s2,t2 be Element of RAT+ such that
A1: s1 + t1 = s2 + t2;
  given x such that
A2: s2 = s1+x;
  take x;
  s1+t1 = s1+(x+t2) by A1,A2,Th51;
  hence thesis by Th62;
end;

theorem Th86:
  r <=' s & s <=' r + t implies ex t0 being Element of RAT+ st s =
  r + t0 & t0 <=' t
proof
  given t0 being Element of RAT+ such that
A1: s = r+t0;
  assume s <=' r+t;
  then t0 <=' t by A1,Th76;
  hence thesis by A1;
end;

theorem
  r <=' s + t implies ex s0,t0 being Element of RAT+ st r = s0 + t0 & s0
  <=' s & t0 <=' t
proof
  assume
A1: r <=' s + t;
  per cases;
  suppose
    s <=' r;
    then
    s <=' s & ex t0 being Element of RAT+ st r = s + t0 & t0 <=' t by A1,Th86;
    hence thesis;
  end;
  suppose
A2: r <=' s;
    r = r+{} & {} <=' t by Th50,Th64;
    hence thesis by A2;
  end;
end;

theorem
  r < s & r < t implies ex t0 being Element of RAT+ st t0 <=' s & t0 <='
  t & r < t0
proof
  s <=' t & s <=' s or t <=' s & t <=' t;
  hence thesis;
end;

theorem
  r <=' s & s <=' t & s <> t implies r <> t by Th66;

theorem
  s < r + t & t <> {} implies ex r0,t0 being Element of RAT+ st s = r0 +
  t0 & r0 <=' r & t0 <=' t & t0 <> t
proof
  assume that
A1: s < r + t and
A2: t <> {};
  per cases;
  suppose
    r <=' s;
    then consider t0 being Element of RAT+ such that
A3: s = r + t0 & t0 <=' t by A1,Th86;
    take r,t0;
    thus thesis by A1,A3;
  end;
  suppose
A4: s <=' r;
    s = s+{} & {} <=' t by Th50,Th64;
    hence thesis by A2,A4;
  end;
end;

theorem
  for A being non empty Subset of RAT+ st A in RAT+ ex s st s in A & for
  r st r in A holds r <=' s
proof
  let A be non empty Subset of RAT+;
A1: now
    given i,j being Element of omega such that
A2: A = [i,j] and
A3: i,j are_relative_prime and
    j <> {} and
A4: j <> 1;
A5: now
      assume {i} in omega;
      then {} in {i} by ORDINAL3:8;
      then {} = i by TARSKI:def 1;
      hence contradiction by A3,A4,Th3;
    end;
    {i} in A by A2,TARSKI:def 2;
    then consider i1,j1 being Element of omega such that
A6: {i} = [i1,j1] and
A7: i1,j1 are_relative_prime and
    j1 <> {} and
A8: j1 <> 1 by A5,Th29;
    {i1,j1} in {i} by A6,TARSKI:def 2;
    then
A9: i = {i1,j1} by TARSKI:def 1;
    {i1} in {i} by A6,TARSKI:def 2;
    then i = {i1} by TARSKI:def 1;
    then j1 in {i1} by A9,TARSKI:def 2;
    then
A10: j1 = i1 by TARSKI:def 1;
    j1 = j1*^1 by ORDINAL2:39;
    hence contradiction by A7,A8,A10,Def2;
  end;
  assume A in RAT+;
  then reconsider B = A as Element of omega by A1,Th29;
A11: {} in B by ORDINAL3:8;
  now
    assume B is limit_ordinal;
    then omega c= B by A11,ORDINAL1:def 11;
    hence contradiction by ORDINAL1:5;
  end;
  then consider C such that
A12: B = succ C by ORDINAL1:29;
  C in B by A12,ORDINAL1:6;
  then reconsider C as ordinal Element of RAT+;
  take C;
  thus C in A by A12,ORDINAL1:6;
  let r;
  assume
A13: r in A;
  then r in B;
  then reconsider r as ordinal Element of RAT+;
  C-^r in omega by ORDINAL1:def 12;
  then reconsider x = C-^r as ordinal Element of RAT+ by Lm5;
  r c= C by A12,A13,ORDINAL1:22;
  then C = r+^x by ORDINAL3:def 5
    .= r+x by Th58;
  hence thesis by Def13;
end;

theorem
  ex t st r + t = s or s + t = r
proof
  r <=' s or s <=' r;
  hence thesis by Def13;
end;

theorem
  r < s implies ex t st r < t & t < s
proof
  assume
A1: r < s;
  then consider x such that
A2: s = r+x by Def13;
  consider y such that
A3: x = y+y by Th60;
  take t = r+y;
A4: s = t+y by A2,A3,Th51;
  then
A5: t <=' s by Def13;
  r <=' t by Def13;
  hence r < t by A1,A4,Th66;
  r <> s by A1;
  then s <> t by A4,Th62;
  hence thesis by A5,Th66;
end;

theorem
  ex s st r < s
proof
  take s = r+one;
  s+{} = s by Th50;
  then
A1: r <> s by Th62;
  r <=' s by Def13;
  hence thesis by A1,Th66;
end;

theorem
  t <> {} implies ex s st s in omega & r <=' s *' t
proof
  set y=((numerator r)*^(((numerator t)*^denominator r)-^1))/denominator r;
A1: denominator r <> {} by Th35;
  (denominator t)*^(numerator r) in omega by ORDINAL1:def 12;
  then reconsider
  s = (denominator t)*^(numerator r) as ordinal Element of RAT+ by Lm5;
  assume t <> {};
  then numerator t <> {} by Th37;
  then {} in (numerator t)*^denominator r by A1,ORDINAL3:8,31;
  then one c= (numerator t)*^denominator r by Lm3,ORDINAL1:21;
  then (numerator t)*^denominator r = 1+^(((numerator t)*^denominator r)-^ 1)
  by ORDINAL3:def 5;
  then
  s*^((numerator t)*^denominator r) = (denominator t)*^ ((numerator r)*^(1
  +^(((numerator t)*^denominator r)-^1))) by ORDINAL3:50
    .= (denominator t)*^((numerator r)*^1+^ (numerator r)*^(((numerator t)*^
  denominator r)-^1)) by ORDINAL3:46;
  then
A2: s*^(numerator t)*^denominator r = (denominator t)*^((numerator r)*^1+^
  (numerator r)*^(((numerator t)*^denominator r)-^1)) by ORDINAL3:50;
  take s;
  thus s in omega by ORDINAL1:def 12;
  take y;
  denominator t <> {} by Th35;
  then
  (s*^(numerator t))/denominator t = ((numerator r)*^1+^ (numerator r)*^(
  ((numerator t)*^denominator r)-^1))/denominator r by A1,A2,Th45
    .= (((numerator r)*^1)/denominator r)+y by Th35,Th47
    .= ((numerator r)/denominator r)+y by ORDINAL2:39
    .= r+y by Th39;
  then r+y = (s*^(numerator t))/(1*^denominator t) by ORDINAL2:39
    .= (s/1)*'((numerator t)/denominator t) by Th49
    .= s*'((numerator t)/denominator t) by Th40;
  hence thesis by Th39;
end;

scheme
  DisNat { n0,n1,n2() -> Element of RAT+, P[set] }: ex s st s in omega & P[s]
  & not P[s + n1()]
provided
A1: n1() = 1 and
A2: n0() = {} and
A3: n2() in omega and
A4: P[n0()] and
A5: not P[n2()]
proof
  defpred P1[Ordinal] means not P[$1] & $1 in omega;
A6: ex A st P1[A] by A3,A5;
  consider A such that
A7: P1[A] and
A8: for B st P1[B] holds A c= B from ORDINAL1:sch 1(A6);
A9: {} in A by A2,A4,A7,ORDINAL3:8;
  now
    assume A is limit_ordinal;
    then omega c= A by A9,ORDINAL1:def 11;
    then A in A by A7;
    hence contradiction;
  end;
  then consider B being Ordinal such that
A10: A = succ B by ORDINAL1:29;
A11: B in A by A10,ORDINAL1:8;
  then
A12: B in omega by A7,ORDINAL1:10;
  then reconsider s = B as ordinal Element of RAT+ by Lm5;
  take s;
  thus s in omega & P[s] by A8,A11,A12,ORDINAL1:5;
  A = B+^1 by A10,ORDINAL2:31;
  hence thesis by A1,A7,Th58;
end;














begin

reserve r,s,t,x9,y9,z9,p,q for Element of RAT+;

definition
  func DEDEKIND_CUTS -> Subset-Family of RAT+ equals
  { A where A is Subset of
RAT+: r in A implies (for s st s <=' r holds s in A) & ex s st s in A & r < s }
  \ { RAT+};
  coherence
  proof
    set C = { A where A is Subset of RAT+: r in A implies (for s st s <=' r
    holds s in A) & ex s st s in A & r < s };
    C c= bool RAT+
    proof
      let e be set;
      assume e in C;
      then
      ex A being Subset of RAT+ st e = A & for r holds r in A implies (for
      s st s <=' r holds s in A) & ex s st s in A & r < s;
      hence thesis;
    end;
    hence thesis by XBOOLE_1:1;
  end;
end;

registration
  cluster DEDEKIND_CUTS -> non empty;
  coherence
  proof
    set X = { s: s < one };
A1: X c= RAT+
    proof
      let e be set;
      assume e in X;
      then ex s st s = e & s < one;
      hence thesis;
    end;
    now
      assume one in X;
      then ex s st s = one & s < one;
      hence contradiction;
    end;
    then
A2: X <> RAT+;
    {} <=' one by ARYTM_3:64;
    then {} < one by ARYTM_3:68;
    then {} in X;
    then reconsider X as non empty Subset of RAT+ by A1;
    r in X implies (for s st s <=' r holds s in X) & ex s st s in X & r < s
    proof
      assume r in X;
      then
A3:   ex s st s = r & s < one;
      thus for s st s <=' r holds s in X
      proof
        let s;
        assume s <=' r;
        then s < one by A3,ARYTM_3:67;
        hence thesis;
      end;
      consider s such that
A4:   r < s and
A5:   s < one by A3,ARYTM_3:93;
      take s;
      thus s in X by A5;
      thus thesis by A4;
    end;
    then X in { A where A is Subset of RAT+: r in A implies (for s st s <=' r
    holds s in A) & ex s st s in A & r < s };
    hence thesis by A2,ZFMISC_1:56;
  end;
end;

definition
  func REAL+ equals
  RAT+ \/ DEDEKIND_CUTS \ {{ s: s < t}: t <> {}};
  coherence;
end;

reserve x,y,z for Element of REAL+;

set IR = { A where A is Subset of RAT+: r in A implies (for s st s <=' r holds
s in A) & ex s st s in A & r < s}, RA = {{ s: s < t}: t <> {}};

Lm1: not ex x,y being set st [x,y] in IR
proof
  given x,y being set such that
A1: [x,y] in IR;
  consider A being Subset of RAT+ such that
A2: [x,y] = A and
A3: r in A implies (for s st s <=' r holds s in A) & ex s st s in A & r
  < s by A1;
  {x} in A & for r,s st r in A & s <=' r holds s in A by A2,A3,TARSKI:def 2;
  then consider r1,r2,r3 being Element of RAT+ such that
A4: r1 in A and
A5: r2 in A and
A6: r3 in A & r1 <> r2 & r2 <> r3 & r3 <> r1 by ARYTM_3:75;
A7: r2 = { x,y } or r2 = { x } by A2,A5,TARSKI:def 2;
  r1 = { x,y } or r1 = { x } by A2,A4,TARSKI:def 2;
  hence contradiction by A2,A6,A7,TARSKI:def 2;
end;

Lm2: RAT+ misses RA
proof
  assume RAT+ meets RA;
  then consider e be set such that
A1: e in RAT+ and
A2: e in RA by XBOOLE_0:3;
  consider t such that
A3: e = { s: s < t } and
A4: t <> {} by A2;
  {} <=' t by ARYTM_3:64;
  then {} < t by A4,ARYTM_3:68;
  then
A5: {} in e by A3;
  e c= RAT+
  proof
    let u be set;
    assume u in e;
    then ex s st s = u & s < t by A3;
    hence thesis;
  end;
  then reconsider e as non empty Subset of RAT+ by A5;
  consider s such that
A6: s in e and
A7: for r st r in e holds r <=' s by A1,ARYTM_3:91;
  ex r st r = s & r < t by A3,A6;
  then consider r such that
A8: s < r and
A9: r < t by ARYTM_3:93;
  r in e by A3,A9;
  hence contradiction by A7,A8;
end;

theorem Th1:
  RAT+ c= REAL+ by Lm2,XBOOLE_1:7,86;

RAT+ \/ DEDEKIND_CUTS c= RAT+ \/ IR by XBOOLE_1:9;
then

Lm3: REAL+ c= RAT+ \/ IR by XBOOLE_1:1;

REAL+ = RAT+ \/ (IR \ { RAT+} \ RA) by Lm2,XBOOLE_1:87;
then

Lm4: DEDEKIND_CUTS \ RA c= REAL+ by XBOOLE_1:7;

Lm5: omega c= ({[c,d] where c,d is Element of omega: c,d are_relative_prime &
d <> {}} \ {[k,1] where k is Element of omega: not contradiction}) \/ omega by
XBOOLE_1:7;

theorem
  omega c= REAL+ by Lm5,Th1,XBOOLE_1:1;

registration
  cluster REAL+ -> non empty;
  coherence by Th1;
end;

definition
  let x;
  func DEDEKIND_CUT x -> Element of DEDEKIND_CUTS means
  :Def3:
  ex r st x = r & it = { s : s < r } if x in RAT+ otherwise it = x;
  existence
  proof
    thus x in RAT+ implies ex IT being Element of DEDEKIND_CUTS, r st x = r &
    IT = { s : s < r }
    proof
      assume x in RAT+;
      then reconsider r = x as Element of RAT+;
      set IT = { s : s < r };
A1:   IT c= RAT+
      proof
        let e be set;
        assume e in IT;
        then ex s st s = e & s < r;
        hence thesis;
      end;
      not ex s st s= r & s < r;
      then not r in IT;
      then
A2:   IT <> RAT+;
      reconsider IT as Subset of RAT+ by A1;
      t in IT implies (for s st s <=' t holds s in IT) & ex s st s in IT &
      t < s
      proof
        assume t in IT;
        then
A3:     ex s st t = s & s < r;
        then consider s0 being Element of RAT+ such that
A4:     t < s0 and
A5:     s0 < r by ARYTM_3:93;
        thus for s st s <=' t holds s in IT
        proof
          let s;
          assume s <=' t;
          then s < r by A3,ARYTM_3:69;
          hence thesis;
        end;
        take s0;
        thus s0 in IT by A5;
        thus thesis by A4;
      end;
      then IT in IR;
      then reconsider IT as Element of DEDEKIND_CUTS by A2,ZFMISC_1:56;
      take IT,r;
      thus thesis;
    end;
A6: x in REAL+;
    assume not x in RAT+;
    then x in DEDEKIND_CUTS by A6,XBOOLE_0:def 3;
    hence thesis;
  end;
  uniqueness;
  consistency;
end;

theorem
  not ex y being set st [{},y] in REAL+
proof
  given y being set such that
A1: [{},y] in REAL+;
  per cases by A1,Lm3,XBOOLE_0:def 3;
  suppose
    [{},y] in RAT+;
    hence contradiction by ARYTM_3:61;
  end;
  suppose
    [{},y] in IR;
    hence contradiction by Lm1;
  end;
end;

Lm6: (for r holds r < s iff r < t) implies s = t
proof
  assume
A1: for r holds r < s iff r < t;
  s <=' s;
  then
A2: t <=' s by A1;
  t <=' t;
  then s <=' t by A1;
  hence thesis by A2,ARYTM_3:66;
end;

definition
  let x be Element of DEDEKIND_CUTS;
  func GLUED x -> Element of REAL+ means
  :Def4:
  ex r st it = r & for s holds s
  in x iff s < r if ex r st for s holds s in x iff s < r otherwise it = x;
  uniqueness
  proof
    let IT1,IT2 be Element of REAL+;
    hereby
      assume ex r st for s holds s in x iff s < r;
      given r1 being Element of RAT+ such that
A1:   IT1 = r1 and
A2:   for s holds s in x iff s < r1;
      given r2 being Element of RAT+ such that
A3:   IT2 = r2 and
A4:   for s holds s in x iff s < r2;
      for s holds s < r1 iff s < r2
      proof
        let s;
        s in x iff s < r1 by A2;
        hence thesis by A4;
      end;
      hence IT1 = IT2 by A1,A3,Lm6;
    end;
    thus thesis;
  end;
  existence
  proof
    hereby
      given r such that
A5:   for s holds s in x iff s < r;
      r in RAT+;
      then reconsider y = r as Element of REAL+ by Th1;
      take y,r;
      thus y = r;
      thus for s holds s in x iff s < r by A5;
    end;
    assume
A6: not ex r st for s holds s in x iff s < r;
    now
      assume x in RA;
      then consider t such that
A7:   x = { s: s < t} and
      t <> {};
      for s holds s in x iff s < t
      proof
        let s;
        hereby
          assume s in x;
          then ex r st s = r & r < t by A7;
          hence s < t;
        end;
        thus thesis by A7;
      end;
      hence contradiction by A6;
    end;
    then x in DEDEKIND_CUTS \ RA by XBOOLE_0:def 5;
    then reconsider y = x as Element of REAL+ by Lm4;
    take y;
    thus thesis;
  end;
  consistency;
end;

Lm7: for B being set st B in DEDEKIND_CUTS & r in B ex s st s in B & r < s
proof
  let B be set;
  assume B in DEDEKIND_CUTS;
  then B in IR;
  then ex A being Subset of RAT+ st B = A & for t st t in A holds (for s st s
  <=' t holds s in A) & ex s st s in A & t < s;
  hence thesis;
end;

Lm8: {} in DEDEKIND_CUTS
proof
  reconsider B = {} as Subset of RAT+ by XBOOLE_1:2;
  r in B implies (for s st s <=' r holds s in B) & ex s st s in B & r < s;
  then {} in { A where A is Subset of RAT+: r in A implies (for s st s <=' r
  holds s in A) & ex s st s in A & r < s };
  hence thesis by ZFMISC_1:56;
end;

Lm9: DEDEKIND_CUTS /\ RAT+ = {{}}
proof
  now
    let e be set;
    thus e in DEDEKIND_CUTS /\ RAT+ implies e = {}
    proof
      assume that
A1:   e in DEDEKIND_CUTS /\ RAT+ and
A2:   e <> {};
      reconsider A = e as non empty Subset of RAT+ by A1,A2;
      A in RAT+ by A1,XBOOLE_0:def 4;
      then
A3:   ex s st s in A & for r st r in A holds r <=' s by ARYTM_3:91;
      e in DEDEKIND_CUTS by A1,XBOOLE_0:def 4;
      hence contradiction by A3,Lm7;
    end;
    thus e = {} implies e in DEDEKIND_CUTS /\ RAT+ by Lm8,XBOOLE_0:def 4;
  end;
  hence thesis by TARSKI:def 1;
end;

Lm10: DEDEKIND_CUT x = {} iff x = {}
proof
A1: not ex e being set st e in { s : s < {} }
  proof
    given e being set such that
A2: e in { s : s < {} };
    ex s st s = e & s < {} by A2;
    hence contradiction by ARYTM_3:64;
  end;
  thus DEDEKIND_CUT x = {} implies x = {}
  proof
    assume
A3: DEDEKIND_CUT x = {};
    per cases;
    suppose
A4:   x in RAT+;
      assume
A5:   x <> {};
      consider r such that
A6:   x = r and
A7:   DEDEKIND_CUT x = { s : s < r } by A4,Def3;
      {} <=' r by ARYTM_3:64;
      then {} < r by A6,A5,ARYTM_3:68;
      then {} in DEDEKIND_CUT x by A7;
      hence contradiction by A3;
    end;
    suppose
      not x in RAT+;
      hence thesis by A3,Def3;
    end;
  end;
  assume x = {};
  then ex r st {} = r & DEDEKIND_CUT x = { s : s < r } by Def3;
  hence thesis by A1,XBOOLE_0:def 1;
end;

Lm11: for A being Element of DEDEKIND_CUTS holds GLUED A = {} iff A = {}
proof
  let A be Element of DEDEKIND_CUTS;
  thus GLUED A = {} implies A = {}
  proof
    assume
A1: GLUED A = {};
    per cases;
    suppose
A2:   ex r st for s holds s in A iff s < r;
      assume A <> {};
      then consider r such that
A3:   r in A by SUBSET_1:4;
      ex r st GLUED A = r & for s holds s in A iff s < r by A2,Def4;
      then r < {} by A1,A3;
      hence contradiction by ARYTM_3:64;
    end;
    suppose
      not ex r st for s holds s in A iff s < r;
      hence thesis by A1,Def4;
    end;
  end;
  assume
A4: A = {};
  then for s holds s in A iff s < {} by ARYTM_3:64;
  then consider r such that
A5: GLUED A = r and
A6: for s holds s in A iff s < r by Def4;
  assume
A7: GLUED A <> {};
  {} <=' r by ARYTM_3:64;
  then {} < r by A5,A7,ARYTM_3:68;
  hence contradiction by A4,A6;
end;

Lm12: for A being Element of DEDEKIND_CUTS holds DEDEKIND_CUT GLUED A = A
proof
  let A be Element of DEDEKIND_CUTS;
  per cases;
  suppose
    ex r st for s holds s in A iff s < r;
    then consider r such that
A1: r = GLUED A and
A2: for s holds s in A iff s < r by Def4;
    consider s such that
A3: r = s and
A4: DEDEKIND_CUT GLUED A = { t : t < s } by A1,Def3;
    thus DEDEKIND_CUT GLUED A c= A
    proof
      let e be set;
      assume e in DEDEKIND_CUT GLUED A;
      then ex t st t = e & t < s by A4;
      hence thesis by A2,A3;
    end;
    let e be set;
    assume
A5: e in A;
    then reconsider s = e as Element of RAT+;
    s < r by A2,A5;
    hence thesis by A3,A4;
  end;
  suppose
A6: A = {};
    then GLUED A = {} by Lm11;
    hence thesis by A6,Lm10;
  end;
  suppose that
A7: A <> {} and
A8: not ex r st for s holds s in A iff s < r;
A9: GLUED A = A by A8,Def4;
    now
      assume GLUED A in RAT+;
      then GLUED A in RAT+ /\ DEDEKIND_CUTS by A9,XBOOLE_0:def 4;
      then GLUED A = {} by Lm9,TARSKI:def 1;
      hence contradiction by A7,Lm11;
    end;
    hence thesis by A9,Def3;
  end;
end;

Lm13: for x,y be set st x in IR & y in IR holds x c= y or y c= x
proof
  let x,y be set;
  assume x in IR;
  then
A1: ex A9 being Subset of RAT+ st x = A9 & for r holds r in A9 implies (for s
  st s <=' r holds s in A9) & ex s st s in A9 & r < s;
  assume y in IR;
  then
A2: ex B9 being Subset of RAT+ st y = B9 & for r holds r in B9 implies (for s
  st s <=' r holds s in B9) & ex s st s in B9 & r < s;
  assume not x c= y;
  then consider s being set such that
A3: s in x and
A4: not s in y by TARSKI:def 3;
  reconsider s as Element of RAT+ by A1,A3;
  let e be set;
  assume
A5: e in y;
  then reconsider r = e as Element of RAT+ by A2;
  r <=' s by A2,A4,A5;
  hence thesis by A1,A3;
end;

definition
  let x,y be Element of REAL+;
  pred x <=' y means
  :Def5:
  ex x9,y9 st x = x9 & y = y9 & x9 <=' y9 if x in
  RAT+ & y in RAT+, x in y if x in RAT+ & not y in RAT+, not y in x if not x in
  RAT+ & y in RAT+ otherwise x c= y;
  consistency;
  connectedness
  proof
    let x,y;
    assume
A1: not(x in RAT+ & y in RAT+ implies ex x9,y9 st x = x9 & y = y9 & x9
<=' y9) or not(x in RAT+ & not y in RAT+ implies x in y) or not(not x in RAT+ &
y in RAT+ implies not y in x) or not(not(x in RAT+ & y in RAT+ or x in RAT+ &
    not y in RAT+ or not x in RAT+ & y in RAT+) implies x c= y);
    thus y in RAT+ & x in RAT+ implies ex y9,x9 st y = y9 & x = x9 & y9 <=' x9
    proof
      assume y in RAT+ & x in RAT+;
      then reconsider y9 = y, x9 = x as Element of RAT+;
      y9 <=' x9 by A1;
      hence thesis;
    end;
    thus y in RAT+ & not x in RAT+ implies y in x by A1;
    thus not y in RAT+ & x in RAT+ implies not x in y by A1;
    assume
A2: not(y in RAT+ & x in RAT+ or y in RAT+ & not x in RAT+ or not y in
    RAT+ & x in RAT+);
    y in REAL+;
    then
A3: y in IR by A2,Lm3,XBOOLE_0:def 3;
    x in REAL+;
    then x in IR by A2,Lm3,XBOOLE_0:def 3;
    hence thesis by A1,A2,A3,Lm13;
  end;
end;

notation
  let x,y be Element of REAL+;
  antonym y < x for x <=' y;
end;

Lm14: x = x9 & y = y9 implies (x <=' y iff x9 <=' y9)
proof
  assume
A1: x = x9 & y = y9;
  hereby
    assume x <=' y;
    then ex x9,y9 st x = x9 & y = y9 & x9 <=' y9 by A1,Def5;
    hence x9 <=' y9 by A1;
  end;
  thus thesis by A1,Def5;
end;

Lm15: for B being set st B in DEDEKIND_CUTS & B <> {} ex r st r in B & r <> {}
proof
  let B be set such that
A1: B in DEDEKIND_CUTS and
A2: B <> {};
  B in IR by A1;
  then consider A being Subset of RAT+ such that
A3: B = A and
A4: r in A implies (for s st s <=' r holds s in A) & ex s st s in A & r < s;
  reconsider A as non empty Subset of RAT+ by A2,A3;
  consider r0 being Element of RAT+ such that
A5: r0 in A by SUBSET_1:4;
  consider r1 being Element of RAT+ such that
A6: r1 in A and
A7: r0 < r1 by A4,A5;
A8: r1 <> {} or r0 <> {} by A7;
  for r,s st r in A & s <=' r holds s in A by A4;
  then consider r1,r2,r3 being Element of RAT+ such that
A9: r1 in A & r2 in A and
  r3 in A and
A10: r1 <> r2 and
  r2 <> r3 and
  r3 <> r1 by A5,A6,A8,ARYTM_3:75;
  r1 <> {} or r2 <> {} by A10;
  hence thesis by A3,A9;
end;

Lm16: for B being set holds B in DEDEKIND_CUTS & r in B & s <=' r implies s in
B
proof
  let B be set such that
A1: B in DEDEKIND_CUTS and
A2: r in B & s <=' r;
  B in IR by A1;
  then ex A being Subset of RAT+ st B = A & for t st t in A holds (for s st s
  <=' t holds s in A) & ex s st s in A & t < s;
  hence thesis by A2;
end;

Lm17: for B being set st B in DEDEKIND_CUTS & B <> {} holds {} in B
proof
  let B be set such that
A1: B in DEDEKIND_CUTS and
A2: B <> {};
  reconsider B as non empty Subset of RAT+ by A1,A2;
  ex r st r in B by SUBSET_1:4;
  hence thesis by A1,Lm16,ARYTM_3:64;
end;

Lm18: for B being set st B in DEDEKIND_CUTS \ RA & not r in B & B <> {} ex s
st not s in B & s < r
proof
  let B be set such that
A1: B in DEDEKIND_CUTS \ RA and
A2: not r in B and
A3: B <> {};
A4: B in DEDEKIND_CUTS by A1,XBOOLE_0:def 5;
  assume
A5: for s st not s in B holds not s < r;
A6: B = { s: s < r}
  proof
    thus B c= { s: s < r}
    proof
      let e be set;
      assume
A7:   e in B;
      reconsider B as Element of DEDEKIND_CUTS by A1,XBOOLE_0:def 5;
      B c= RAT+;
      then reconsider t = e as Element of RAT+ by A7;
      not r <=' t by A2,A4,A7,Lm16;
      hence thesis;
    end;
    let e be set;
    assume e in { s: s < r};
    then ex s st s = e & s < r;
    hence thesis by A5;
  end;
  r <> {} by A2,A3,A4,Lm17;
  then B in RA by A6;
  hence contradiction by A1,XBOOLE_0:def 5;
end;

Lm19: DEDEKIND_CUT x in RA implies x in RAT+
proof
  assume
A1: DEDEKIND_CUT x in RA;
  assume not x in RAT+;
  then DEDEKIND_CUT x = x by Def3;
  hence contradiction by A1,XBOOLE_0:def 5;
end;

Lm20: DEDEKIND_CUT x c= DEDEKIND_CUT y implies x <=' y
proof
  assume
A1: DEDEKIND_CUT x c= DEDEKIND_CUT y;
  per cases;
  suppose that
A2: x = {} and
A3: not y in RAT+;
    DEDEKIND_CUT y = y & y <> {} by A3,Def3;
    then x in y by A2,Lm17;
    hence thesis by A2,A3,Def5;
  end;
  suppose
A4: y = {};
    then DEDEKIND_CUT y = {} by Lm10;
    then DEDEKIND_CUT x = {} by A1;
    then x = {} by Lm10;
    hence thesis by A4;
  end;
  suppose that
A5: x in RAT+ and
A6: y in RAT+;
    consider ry being Element of RAT+ such that
A7: y = ry and
A8: DEDEKIND_CUT y = { s : s < ry } by A6,Def3;
    consider rx being Element of RAT+ such that
A9: x = rx and
A10: DEDEKIND_CUT x = { s : s < rx } by A5,Def3;
    assume y < x;
    then ry < rx by A9,A7,Lm14;
    then ry in DEDEKIND_CUT x by A10;
    then ry in DEDEKIND_CUT y by A1;
    then ex s st s = ry & s < ry by A8;
    hence contradiction;
  end;
  suppose that
A11: x in RAT+ and
A12: not y in RAT+ and
A13: x <> {};
A14: DEDEKIND_CUT y = y by A12,Def3;
    consider rx being Element of RAT+ such that
A15: x = rx and
A16: DEDEKIND_CUT x = { s : s < rx } by A11,Def3;
    DEDEKIND_CUT x in RA by A13,A15,A16;
    then DEDEKIND_CUT x <> y by A12,A14,Lm19;
    then not DEDEKIND_CUT y c= DEDEKIND_CUT x by A1,A14,XBOOLE_0:def 10;
    then consider r0 being Element of RAT+ such that
A17: r0 in y and
A18: not r0 in DEDEKIND_CUT x by A14,SUBSET_1:2;
    rx <=' r0 by A16,A18;
    then x in y by A15,A14,A17,Lm16;
    hence thesis by A11,A12,Def5;
  end;
  suppose that
A19: not x in RAT+ and
A20: y in RAT+ and
A21: y <> {};
    consider ry being Element of RAT+ such that
A22: y = ry and
A23: DEDEKIND_CUT y = { s : s < ry } by A20,Def3;
A24: DEDEKIND_CUT y in RA by A21,A22,A23;
A25: DEDEKIND_CUT x = x by A19,Def3;
    then not x in RA by A19,Lm19;
    then not DEDEKIND_CUT y c= DEDEKIND_CUT x by A1,A24,A25,XBOOLE_0:def 10;
    then consider r0 being Element of RAT+ such that
A26: r0 in DEDEKIND_CUT y and
A27: not r0 in x by A25,SUBSET_1:2;
    ex s st s = r0 & s < ry by A23,A26;
    then not y in x by A22,A25,A27,Lm16;
    hence thesis by A19,A20,Def5;
  end;
  suppose that
A28: ( not x in RAT+)& not y in RAT+;
    x = DEDEKIND_CUT x & y = DEDEKIND_CUT y by A28,Def3;
    hence thesis by A1,A28,Def5;
  end;
end;

Lm21: x <=' y & y <=' x implies x = y
proof
  assume that
A1: x <=' y and
A2: y <=' x;
  per cases;
  suppose
    x in RAT+ & y in RAT+;
    then reconsider x9 = x, y9 = y as Element of RAT+;
    x9 <=' y9 & y9 <=' x9 by A1,A2,Lm14;
    hence thesis by ARYTM_3:66;
  end;
  suppose that
A3: x in RAT+ & not y in RAT+;
    x in y by A1,A3,Def5;
    hence thesis by A2,A3,Def5;
  end;
  suppose that
A4: ( not x in RAT+)& y in RAT+;
    y in x by A2,A4,Def5;
    hence thesis by A1,A4,Def5;
  end;
  suppose
    ( not x in RAT+)& not y in RAT+;
    then x c= y & y c= x by A1,A2,Def5;
    hence thesis by XBOOLE_0:def 10;
  end;
end;

Lm22: DEDEKIND_CUT x = DEDEKIND_CUT y implies x = y
proof
  assume DEDEKIND_CUT x = DEDEKIND_CUT y;
  then x <=' y & y <=' x by Lm20;
  hence thesis by Lm21;
end;

Lm23: GLUED DEDEKIND_CUT x = x
proof
  DEDEKIND_CUT GLUED DEDEKIND_CUT x = DEDEKIND_CUT x by Lm12;
  hence thesis by Lm22;
end;

definition
  let A,B be Element of DEDEKIND_CUTS;
  func A + B -> Element of DEDEKIND_CUTS equals
  :Def6:
  { r + s : r in A & s in
  B};
  coherence
  proof
    { r + s : r in A & s in B} in DEDEKIND_CUTS
    proof
      set C = { s + t: s in A & t in B};
      C c= RAT+
      proof
        let e be set;
        assume e in C;
        then ex s,t st e = s + t & s in A & t in B;
        hence thesis;
      end;
      then reconsider C as Subset of RAT+;
      A <> RAT+ by ZFMISC_1:56;
      then consider a0 being Element of RAT+ such that
A1:   not a0 in A by SUBSET_1:28;
      r in C implies (for s st s <=' r holds s in C) & ex s st s in C & r < s
      proof
        assume r in C;
        then consider s0,t0 being Element of RAT+ such that
A2:     r = s0 + t0 and
A3:     s0 in A and
A4:     t0 in B;
        thus for s st s <=' r holds s in C
        proof
          let s;
          assume s <=' r;
          then consider s1,t1 being Element of RAT+ such that
A5:       s = s1 + t1 and
A6:       s1 <=' s0 & t1 <=' t0 by A2,ARYTM_3:87;
          s1 in A & t1 in B by A3,A4,A6,Lm16;
          hence thesis by A5;
        end;
        consider s1 being Element of RAT+ such that
A7:     s1 in A and
A8:     s0 < s1 by A3,Lm7;
        take t2 = s1 + t0;
        thus t2 in C by A4,A7;
        thus thesis by A2,A8,ARYTM_3:76;
      end;
      then
A9:   C in IR;
      B <> RAT+ by ZFMISC_1:56;
      then consider b0 being Element of RAT+ such that
A10:  not b0 in B by SUBSET_1:28;
      now
        assume a0 + b0 in C;
        then consider s,t such that
A11:    a0 + b0 = s + t and
A12:    s in A & t in B;
        a0 <=' s or b0 <=' t by A11,ARYTM_3:81;
        hence contradiction by A1,A10,A12,Lm16;
      end;
      then C <> RAT+;
      hence thesis by A9,ZFMISC_1:56;
    end;
    hence thesis;
  end;
  commutativity
  proof
    let IT,A,B be Element of DEDEKIND_CUTS;
    set C = { r + s : r in A & s in B}, D = { r + s : r in B & s in A};
A13: D c= C
    proof
      let e be set;
      assume e in D;
      then ex r,s st e = r + s & r in B & s in A;
      hence thesis;
    end;
    C c= D
    proof
      let e be set;
      assume e in C;
      then ex r,s st e = r + s & r in A & s in B;
      hence thesis;
    end;
    hence thesis by A13,XBOOLE_0:def 10;
  end;
end;

Lm24: for B being set st B in DEDEKIND_CUTS ex r st not r in B
proof
  let B be set;
  assume
A1: B in DEDEKIND_CUTS;
  then reconsider B as Subset of RAT+;
  B <> RAT+ by A1,ZFMISC_1:56;
  hence thesis by SUBSET_1:28;
end;

definition
  let A,B be Element of DEDEKIND_CUTS;
  func A *' B -> Element of DEDEKIND_CUTS equals
  { r *' s : r in A & s in B};
  coherence
  proof
    per cases;
    suppose
A1:   A = {};
      not ex e being set st e in { r *' s : r in A & s in B}
      proof
        given e being set such that
A2:     e in { r *' s : r in A & s in B};
        ex r,s st e = r *' s & r in A & s in B by A2;
        hence contradiction by A1;
      end;
      hence thesis by Lm8,XBOOLE_0:def 1;
    end;
    suppose
A3:   A <> {};
      set C = { r *' s : r in A & s in B};
      C c= RAT+
      proof
        let e be set;
        assume e in C;
        then ex r,s st r*'s = e & r in A & s in B;
        hence thesis;
      end;
      then reconsider C as Subset of RAT+;
      r in C implies (for s st s <=' r holds s in C) & ex s st s in C & r < s
      proof
        assume r in C;
        then consider r0,s0 being Element of RAT+ such that
A4:     r = r0 *' s0 and
A5:     r0 in A and
A6:     s0 in B;
        thus for s st s <=' r holds s in C
        proof
          let s;
          assume s <=' r;
          then consider t0 being Element of RAT+ such that
A7:       s = r0 *' t0 and
A8:       t0 <=' s0 by A4,ARYTM_3:79;
          t0 in B by A6,A8,Lm16;
          hence thesis by A5,A7;
        end;
        consider t0 being Element of RAT+ such that
A9:     t0 in B and
A10:    s0 < t0 by A6,Lm7;
        per cases;
        suppose
A11:      r0 = {};
          consider r1 being Element of RAT+ such that
A12:      r1 in A and
A13:      r1 <> {} by A3,Lm15;
          take r1 *' t0;
          thus r1 *' t0 in C by A9,A12;
          t0 <> {} by A10,ARYTM_3:64;
          then
A14:      r1 *' t0 <> {} by A13,ARYTM_3:78;
A15:      r = {} by A4,A11,ARYTM_3:48;
          then r <=' r1 *' t0 by ARYTM_3:64;
          hence thesis by A15,A14,ARYTM_3:68;
        end;
        suppose
A16:      r0 <> {};
          s0 <> t0 by A10;
          then
A17:      r <> r0 *' t0 by A4,A16,ARYTM_3:56;
          take r0 *' t0;
          thus r0 *' t0 in C by A5,A9;
          r <=' r0 *' t0 by A4,A10,ARYTM_3:82;
          hence thesis by A17,ARYTM_3:68;
        end;
      end;
      then
A18:  C in IR;
      consider r0 being Element of RAT+ such that
A19:  not r0 in A by Lm24;
      consider s0 being Element of RAT+ such that
A20:  not s0 in B by Lm24;
      now
        assume r0 *' s0 in C;
        then consider r1,s1 being Element of RAT+ such that
A21:    r0 *' s0 = r1 *' s1 and
A22:    r1 in A & s1 in B;
        r0 <=' r1 or s0 <=' s1 by A21,ARYTM_3:83;
        hence contradiction by A19,A20,A22,Lm16;
      end;
      then C <> RAT+;
      hence thesis by A18,ZFMISC_1:56;
    end;
  end;
  commutativity
  proof
    let D,A,B be Element of DEDEKIND_CUTS;
    assume
A23: D = { r *' s : r in A & s in B};
    now
      let e be set;
      e in D iff ex r,s st e = r*'s & r in A & s in B by A23;
      then e in D iff ex r,s st e = r*'s & r in B & s in A;
      hence e in D iff e in { r *' s : r in B & s in A};
    end;
    hence thesis by TARSKI:1;
  end;
end;

definition
  let x,y be Element of REAL+;
  func x + y -> Element of REAL+ equals
  :Def8:
  x if y = {}, y if x = {}
  otherwise GLUED(DEDEKIND_CUT x + DEDEKIND_CUT y);
  coherence;
  consistency;
  commutativity;
  func x *' y -> Element of REAL+ equals
  GLUED(DEDEKIND_CUT x *' DEDEKIND_CUT
  y);
  coherence;
  commutativity;
end;

theorem Th4:
  x = {} implies x *' y = {}
proof
  set A = DEDEKIND_CUT x, B = DEDEKIND_CUT y;
  assume
A1: x = {};
  not ex e being set st e in { r *' s : r in A & s in B}
  proof
    given e being set such that
A2: e in { r *' s : r in A & s in B};
    ex r,s st e = r *' s & r in A & s in B by A2;
    hence contradiction by A1,Lm10;
  end;
  then { r *' s : r in A & s in B} = {} by XBOOLE_0:def 1;
  hence thesis by Lm11;
end;

theorem Th5:
  x + y = {} implies x = {}
proof
  assume
A1: x + y = {};
  per cases;
  suppose
    y = {};
    hence thesis by A1,Def8;
  end;
  suppose
A2: y <> {};
    then DEDEKIND_CUT y <> {} by Lm10;
    then consider s0 being Element of RAT+ such that
A3: s0 in DEDEKIND_CUT y by SUBSET_1:4;
    assume
A4: x <> {};
    then DEDEKIND_CUT x <> {} by Lm10;
    then consider r0 being Element of RAT+ such that
A5: r0 in DEDEKIND_CUT x by SUBSET_1:4;
A6: r0 + s0 in { r + s : r in DEDEKIND_CUT x & s in DEDEKIND_CUT y } by A5,A3;
    GLUED(DEDEKIND_CUT x + DEDEKIND_CUT y) = {} by A1,A2,A4,Def8;
    hence contradiction by A6,Lm11;
  end;
end;

Lm25: for A,B,C being Element of DEDEKIND_CUTS holds A + (B + C) c= A + B + C
proof
  let A,B,C be Element of DEDEKIND_CUTS;
  let e be set;
  assume e in A + (B + C);
  then consider r0,s0 being Element of RAT+ such that
A1: e = r0 + s0 & r0 in A and
A2: s0 in B + C;
  consider r1,s1 being Element of RAT+ such that
A3: s0 = r1 + s1 & r1 in B and
A4: s1 in C by A2;
  e = r0 + r1 + s1 & r0 + r1 in A + B by A1,A3,ARYTM_3:51;
  hence thesis by A4;
end;

Lm26: for A,B,C being Element of DEDEKIND_CUTS holds A + (B + C) = A + B + C
proof
  let A,B,C be Element of DEDEKIND_CUTS;
  A + (B + C) c= A + B + C & A + B + C c= A + (B + C) by Lm25;
  hence thesis by XBOOLE_0:def 10;
end;

Lm27: for A,B being Element of DEDEKIND_CUTS st A + B = {} holds A = {} or B =
{}
proof
  let A,B be Element of DEDEKIND_CUTS such that
A1: A + B = {};
  assume A <> {};
  then consider r0 being Element of RAT+ such that
A2: r0 in A by SUBSET_1:4;
  assume B <> {};
  then consider s0 being Element of RAT+ such that
A3: s0 in B by SUBSET_1:4;
  r0 + s0 in { r + s : r in A & s in B} by A2,A3;
  hence contradiction by A1;
end;

theorem Th6:
  x + (y + z) = (x + y) + z
proof
  per cases;
  suppose
A1: x = {};
    hence x + (y + z) = y + z by Def8
      .= (x + y) + z by A1,Def8;
  end;
  suppose
A2: y = {};
    hence x + (y + z) = x + z by Def8
      .= (x + y) + z by A2,Def8;
  end;
  suppose
A3: z = {};
    hence x + (y + z) = x + y by Def8
      .= (x + y) + z by A3,Def8;
  end;
  suppose that
A4: x <> {} and
A5: y <> {} and
A6: z <> {};
A7: now
      assume GLUED(DEDEKIND_CUT y + DEDEKIND_CUT z) = {};
      then DEDEKIND_CUT y + DEDEKIND_CUT z = {} by Lm11;
      then DEDEKIND_CUT y = {} or DEDEKIND_CUT z = {} by Lm27;
      hence contradiction by A5,A6,Lm10;
    end;
A8: now
      assume GLUED(DEDEKIND_CUT x + DEDEKIND_CUT y) = {};
      then DEDEKIND_CUT x + DEDEKIND_CUT y = {} by Lm11;
      then DEDEKIND_CUT x = {} or DEDEKIND_CUT y = {} by Lm27;
      hence contradiction by A4,A5,Lm10;
    end;
    thus x + (y + z) = x + GLUED(DEDEKIND_CUT y + DEDEKIND_CUT z) by A5,A6,Def8
      .= GLUED(DEDEKIND_CUT x + DEDEKIND_CUT GLUED(DEDEKIND_CUT y +
    DEDEKIND_CUT z)) by A4,A7,Def8
      .= GLUED(DEDEKIND_CUT x + (DEDEKIND_CUT y + DEDEKIND_CUT z)) by Lm12
      .= GLUED((DEDEKIND_CUT x + DEDEKIND_CUT y) + DEDEKIND_CUT z) by Lm26
      .= GLUED(DEDEKIND_CUT GLUED(DEDEKIND_CUT x + DEDEKIND_CUT y) +
    DEDEKIND_CUT z) by Lm12
      .= GLUED(DEDEKIND_CUT x + DEDEKIND_CUT y) + z by A6,A8,Def8
      .= (x + y) + z by A4,A5,Def8;
  end;
end;

theorem
  { A where A is Subset of RAT+: r in A implies (for s st s <=' r holds
  s in A) & ex s st s in A & r < s} is c=-linear
proof
  set IR = { A where A is Subset of RAT+: r in A implies (for s st s <=' r
  holds s in A) & ex s st s in A & r < s};
  let x,y be set;
  assume x in IR;
  then
A1: ex A9 being Subset of RAT+ st x = A9 & for r holds r in A9 implies (for s
  st s <=' r holds s in A9) & ex s st s in A9 & r < s;
  assume y in IR;
  then
A2: ex B9 being Subset of RAT+ st y = B9 & for r holds r in B9 implies (for s
  st s <=' r holds s in B9) & ex s st s in B9 & r < s;
  assume not x c= y;
  then consider s being set such that
A3: s in x and
A4: not s in y by TARSKI:def 3;
  reconsider s as Element of RAT+ by A1,A3;
  let e be set;
  assume
A5: e in y;
  then reconsider r = e as Element of RAT+ by A2;
  r <=' s by A2,A4,A5;
  hence thesis by A1,A3;
end;

Lm28: for e being set st e in REAL+ holds e <> RAT+
proof
  let e be set;
  assume e in REAL+;
  then e in RAT+ or e in DEDEKIND_CUTS by XBOOLE_0:def 3;
  hence thesis by ZFMISC_1:56;
end;

Lm29: for B being set holds B in IR & r in B & s <=' r implies s in B
proof
  let B be set such that
A1: B in IR and
A2: r in B & s <=' r;
  ex A being Subset of RAT+ st B = A & for t st t in A holds (for s st s
  <=' t holds s in A) & ex s st s in A & t < s by A1;
  hence thesis by A2;
end;

Lm30: y < x implies ex z st z in RAT+ & z < x & y < z
proof
  assume
A1: y < x;
  per cases;
  suppose
    x in RAT+ & y in RAT+;
    then reconsider x9 = x, y9 = y as Element of RAT+;
    y9 < x9 by A1,Lm14;
    then consider z9 being Element of RAT+ such that
A2: y9 < z9 & z9 < x9 by ARYTM_3:93;
    z9 in RAT+;
    then reconsider z = z9 as Element of REAL+ by Th1;
    take z;
    thus thesis by A2,Lm14;
  end;
  suppose that
A3: not x in RAT+ and
A4: y in RAT+;
    reconsider y9 = y as Element of RAT+ by A4;
    x in REAL+;
    then x in IR by A3,Lm3,XBOOLE_0:def 3;
    then consider A being Subset of RAT+ such that
A5: x = A and
A6: r in A implies (for s st s <=' r holds s in A) & ex s st s in A & r < s;
    y9 in x by A1,A3,Def5;
    then consider s such that
A7: s in A and
A8: y9 < s by A5,A6;
    s in RAT+;
    then reconsider z = s as Element of REAL+ by Th1;
    take z;
    thus z in RAT+;
    thus z < x by A3,A5,A7,Def5;
    thus thesis by A8,Lm14;
  end;
  suppose that
A9: x in RAT+ and
A10: not y in RAT+;
    reconsider x9 = x as Element of RAT+ by A9;
A11: not x9 in y by A1,A10,Def5;
    y in REAL+;
    then y in IR by A10,Lm3,XBOOLE_0:def 3;
    then consider B being Subset of RAT+ such that
A12: y = B and
A13: r in B implies (for s st s <=' r holds s in B) & ex s st s in B & r < s;
    B <> {} by A10,A12;
    then consider y1 being Element of RAT+ such that
A14: y1 in B by SUBSET_1:4;
    {} <=' y1 by ARYTM_3:64;
    then
A15: x9 <> {} by A12,A13,A11,A14;
    ex z9 st z9 < x9 & not z9 in y
    proof
      set C = { s: s < x9 };
      assume
A16:  not ex z9 st z9 < x9 & not z9 in y;
      y = C
      proof
        thus y c= C
        proof
          let e be set;
          assume
A17:      e in y;
          then reconsider z9 = e as Element of RAT+ by A12;
          not x9 <=' z9 by A12,A13,A11,A17;
          hence thesis;
        end;
        let e be set;
        assume e in C;
        then ex s st e = s & s < x9;
        hence thesis by A16;
      end;
      then y in RA by A15;
      hence contradiction by XBOOLE_0:def 5;
    end;
    then consider z9 such that
A18: z9 < x9 and
A19: not z9 in y;
    z9 in RAT+;
    then reconsider z = z9 as Element of REAL+ by Th1;
    take z;
    thus z in RAT+;
    thus z < x by A18,Lm14;
    thus thesis by A10,A19,Def5;
  end;
  suppose that
A20: not x in RAT+ and
A21: not y in RAT+;
    y in REAL+;
    then y in IR by A21,Lm3,XBOOLE_0:def 3;
    then consider B being Subset of RAT+ such that
A22: y = B and
    r in B implies (for s st s <=' r holds s in B) & ex s st s in B & r < s;
    x in REAL+;
    then x in IR by A20,Lm3,XBOOLE_0:def 3;
    then consider A being Subset of RAT+ such that
A23: x = A and
    r in A implies (for s st s <=' r holds s in A) & ex s st s in A & r < s;
    not x c= y by A1,A20,A21,Def5;
    then consider z9 being Element of RAT+ such that
A24: z9 in A and
A25: not z9 in B by A23,A22,SUBSET_1:2;
    z9 in RAT+;
    then reconsider z = z9 as Element of REAL+ by Th1;
    take z;
    thus z in RAT+;
    thus z < x by A20,A23,A24,Def5;
    thus thesis by A21,A22,A25,Def5;
  end;
end;

Lm31: x <=' y & y <=' z implies x <=' z
proof
  assume that
A1: x <=' y and
A2: y <=' z;
  per cases;
  suppose that
A3: x in RAT+ and
A4: y in RAT+ and
A5: z in RAT+;
    reconsider z9 = z as Element of RAT+ by A5;
    reconsider y9 = y as Element of RAT+ by A4;
    reconsider x9 = x as Element of RAT+ by A3;
    x9 <=' y9 & y9 <=' z9 by A1,A2,Lm14;
    then x9 <=' z9 by ARYTM_3:67;
    hence thesis by Lm14;
  end;
  suppose that
A6: x in RAT+ and
A7: y in RAT+ and
A8: not z in RAT+;
    reconsider y9 = y as Element of RAT+ by A7;
    reconsider x9 = x as Element of RAT+ by A6;
A9: x9 <=' y9 by A1,Lm14;
    z in REAL+;
    then z in IR by A8,Lm3,XBOOLE_0:def 3;
    then consider C being Subset of RAT+ such that
A10: z = C and
A11: r in C implies (for s st s <=' r holds s in C) & ex s st s in C & r < s;
    y in C by A2,A7,A8,A10,Def5;
    then x9 in C by A11,A9;
    hence thesis by A8,A10,Def5;
  end;
  suppose that
A12: x in RAT+ and
A13: not y in RAT+ and
A14: z in RAT+;
    reconsider z9 = z as Element of RAT+ by A14;
    reconsider x9 = x as Element of RAT+ by A12;
    y in REAL+;
    then y in IR by A13,Lm3,XBOOLE_0:def 3;
    then consider B being Subset of RAT+ such that
A15: y = B and
A16: r in B implies (for s st s <=' r holds s in B) & ex s st s in B & r < s;
    x9 in B & not z9 in B by A1,A2,A13,A15,Def5;
    then x9 <=' z9 by A16;
    hence thesis by Lm14;
  end;
  suppose that
A17: x in RAT+ and
A18: not y in RAT+ and
A19: not z in RAT+;
    reconsider x9 = x as Element of RAT+ by A17;
    y in REAL+;
    then y in IR by A18,Lm3,XBOOLE_0:def 3;
    then consider B being Subset of RAT+ such that
A20: y = B and
    r in B implies (for s st s <=' r holds s in B) & ex s st s in B & r < s;
A21: x9 in B by A1,A18,A20,Def5;
    z in REAL+;
    then z in IR by A19,Lm3,XBOOLE_0:def 3;
    then consider C being Subset of RAT+ such that
A22: z = C and
    r in C implies (for s st s <=' r holds s in C) & ex s st s in C & r < s;
    B c= C by A2,A18,A19,A20,A22,Def5;
    hence thesis by A19,A22,A21,Def5;
  end;
  suppose that
A23: not x in RAT+ and
A24: y in RAT+ and
A25: z in RAT+;
    reconsider z9 = z as Element of RAT+ by A25;
    reconsider y9 = y as Element of RAT+ by A24;
A26: y9 <=' z9 by A2,Lm14;
    x in REAL+;
    then x in IR by A23,Lm3,XBOOLE_0:def 3;
    then consider A being Subset of RAT+ such that
A27: x = A and
A28: r in A implies (for s st s <=' r holds s in A) & ex s st s in A & r < s;
    not y9 in A by A1,A23,A27,Def5;
    then not z9 in A by A28,A26;
    hence thesis by A23,A27,Def5;
  end;
  suppose that
A29: not x in RAT+ and
A30: y in RAT+ and
A31: not z in RAT+;
    x in REAL+;
    then x in IR by A29,Lm3,XBOOLE_0:def 3;
    then consider A being Subset of RAT+ such that
A32: x = A and
A33: r in A implies (for s st s <=' r holds s in A) & ex s st s in A & r < s;
A34: not y in A by A1,A29,A32,Def5;
    reconsider y9 = y as Element of RAT+ by A30;
    z in REAL+;
    then z in IR by A31,Lm3,XBOOLE_0:def 3;
    then consider C being Subset of RAT+ such that
A35: z = C and
A36: r in C implies (for s st s <=' r holds s in C) & ex s st s in C & r < s;
A37: y in C by A2,A30,A31,A35,Def5;
    A c= C
    proof
      let e be set;
      assume
A38:  e in A;
      then reconsider x9 = e as Element of RAT+;
      x9 <=' y9 by A33,A34,A38;
      hence thesis by A36,A37;
    end;
    hence thesis by A29,A31,A32,A35,Def5;
  end;
  suppose that
A39: not x in RAT+ and
A40: not y in RAT+ and
A41: z in RAT+;
    reconsider z9 = z as Element of RAT+ by A41;
    x in REAL+;
    then x in IR by A39,Lm3,XBOOLE_0:def 3;
    then consider A being Subset of RAT+ such that
A42: x = A and
    r in A implies (for s st s <=' r holds s in A) & ex s st s in A & r < s;
    y in REAL+;
    then y in IR by A40,Lm3,XBOOLE_0:def 3;
    then consider B being Subset of RAT+ such that
A43: y = B and
    r in B implies (for s st s <=' r holds s in B) & ex s st s in B & r < s;
    A c= B by A1,A39,A40,A42,A43,Def5;
    then not z9 in A by A2,A40,A43,Def5;
    hence thesis by A39,A42,Def5;
  end;
  suppose that
A44: not x in RAT+ and
A45: not y in RAT+ and
A46: not z in RAT+;
    z in REAL+;
    then z in IR by A46,Lm3,XBOOLE_0:def 3;
    then consider C being Subset of RAT+ such that
A47: z = C and
    r in C implies (for s st s <=' r holds s in C) & ex s st s in C & r < s;
    y in REAL+;
    then y in IR by A45,Lm3,XBOOLE_0:def 3;
    then consider B being Subset of RAT+ such that
A48: y = B and
    r in B implies (for s st s <=' r holds s in B) & ex s st s in B & r < s;
A49: B c= C by A2,A45,A46,A48,A47,Def5;
    x in REAL+;
    then x in IR by A44,Lm3,XBOOLE_0:def 3;
    then consider A being Subset of RAT+ such that
A50: x = A and
    r in A implies (for s st s <=' r holds s in A) & ex s st s in A & r < s;
    A c= B by A1,A44,A45,A50,A48,Def5;
    then A c= C by A49,XBOOLE_1:1;
    hence thesis by A44,A46,A50,A47,Def5;
  end;
end;

theorem
  for X,Y being Subset of REAL+ st (ex x st x in Y) & for x,y st x in X
& y in Y holds x <=' y ex z st for x,y st x in X & y in Y holds x <=' z & z <='
  y
proof
  let X,Y be Subset of REAL+;
  given x1 being Element of REAL+ such that
A1: x1 in Y;
  set Z = {z9 : ex x,z st z = z9 & x in X & z < x};
  assume
A2: for x,y st x in X & y in Y holds x <=' y;
  per cases;
  suppose
    ex z9 st for x9 holds x9 in Z iff x9 < z9;
    then consider z9 such that
A3: for x9 holds x9 in Z iff x9 < z9;
    z9 in RAT+;
    then reconsider z = z9 as Element of REAL+ by Th1;
    take z;
    let x,y such that
A4: x in X and
A5: y in Y;
    thus x <=' z
    proof
      assume z < x;
      then consider x0 being Element of REAL+ such that
A6:   x0 in RAT+ and
A7:   x0 < x & z < x0 by Lm30;
      reconsider x9 = x0 as Element of RAT+ by A6;
      z9 < x9 & x9 in Z by A4,A7,Lm14;
      hence contradiction by A3;
    end;
    assume y < z;
    then consider y0 being Element of REAL+ such that
A8: y0 in RAT+ and
A9: y0 < z and
A10: y < y0 by Lm30;
    reconsider y9 = y0 as Element of RAT+ by A8;
    y9 < z9 by A9,Lm14;
    then y9 in Z by A3;
    then
    ex y99 being Element of RAT+ st y9 = y99 & ex x,z st z = y99 & x in X
    & z < x;
    then consider x1,y1 being Element of REAL+ such that
A11: y1 = y9 and
A12: x1 in X and
A13: y1 < x1;
    y < x1 by A10,A11,A13,Lm31;
    hence contradiction by A2,A5,A12;
  end;
  suppose
A14: not ex z9 st for x9 holds x9 in Z iff x9 < z9;
A15: now
      assume Z in RA;
      then consider t such that
A16:  Z = { s: s < t } and
      t <> {};
      for x9 holds x9 in Z iff x9 < t
      proof
        let x9;
        hereby
          assume x9 in Z;
          then ex s st s = x9 & s < t by A16;
          hence x9 < t;
        end;
        thus thesis by A16;
      end;
      hence contradiction by A14;
    end;
A17: Z c= RAT+
    proof
      let e be set;
      assume e in Z;
      then ex z9 st e = z9 & ex x,z st z = z9 & x in X & z < x;
      hence thesis;
    end;
    now
      assume Z = {};
      then
A18:  for x9 st x9 in Z holds x9 < {};
      for x9 st x9 < {} holds x9 in Z by ARYTM_3:64;
      hence contradiction by A14,A18;
    end;
    then reconsider Z as non empty Subset of RAT+ by A17;
A19: now
      assume
A20:  Z = RAT+;
      per cases;
      suppose
        x1 in RAT+;
        then reconsider x9 = x1 as Element of RAT+;
        x9 in Z by A20;
        then ex z9 st x9 = z9 & ex x,z st z = z9 & x in X & z < x;
        hence contradiction by A1,A2;
      end;
      suppose
A21:    not x1 in RAT+;
        x1 in REAL+;
        then x1 in IR by A21,Lm3,XBOOLE_0:def 3;
        then consider A being Subset of RAT+ such that
A22:    x1 = A and
        r in A implies (for s st s <=' r holds s in A) & ex s st s in A &
        r < s;
        x1 <> RAT+ by Lm28;
        then consider x9 being Element of RAT+ such that
A23:    not x9 in A by A22,SUBSET_1:28;
        x9 in RAT+;
        then reconsider x2 = x9 as Element of REAL+ by Th1;
        x2 in Z by A20;
        then ex z9 st x9 = z9 & ex x,z st z = z9 & x in X & z < x;
        then consider x such that
A24:    x in X and
A25:    x2 < x;
        x1 < x2 by A21,A22,A23,Def5;
        then x1 < x by A25,Lm31;
        hence contradiction by A1,A2,A24;
      end;
    end;
    t in Z implies (for s st s <=' t holds s in Z) & ex s st s in Z & t < s
    proof
      t in RAT+;
      then reconsider y0 = t as Element of REAL+ by Th1;
      assume t in Z;
      then ex z9 st z9 = t & ex x,z st z = z9 & x in X & z < x;
      then consider x0 being Element of REAL+ such that
A26:  x0 in X and
A27:  y0 < x0;
      thus for s st s <=' t holds s in Z
      proof
        let s;
        s in RAT+;
        then reconsider z = s as Element of REAL+ by Th1;
        assume s <=' t;
        then z <=' y0 by Lm14;
        then z < x0 by A27,Lm31;
        hence thesis by A26;
      end;
      consider z such that
A28:  z in RAT+ and
A29:  z < x0 and
A30:  y0 < z by A27,Lm30;
      reconsider z9 = z as Element of RAT+ by A28;
      take z9;
      thus z9 in Z by A26,A29;
      thus thesis by A30,Lm14;
    end;
    then Z in IR;
    then
A31: Z in IR \ {RAT+} by A19,ZFMISC_1:56;
    then Z in IR \ {RAT+} \ RA by A15,XBOOLE_0:def 5;
    then reconsider z = Z as Element of REAL+ by Lm4;
    take z;
    let x,y such that
A32: x in X and
A33: y in Y;
A34: now
      assume z in RAT+;
      then z in {{}} by A31,Lm9,XBOOLE_0:def 4;
      hence contradiction by TARSKI:def 1;
    end;
    hereby
      assume z < x;
      then consider x0 being Element of REAL+ such that
A35:  x0 in RAT+ and
A36:  x0 < x and
A37:  z < x0 by Lm30;
      reconsider x9 = x0 as Element of RAT+ by A35;
      x9 in z by A32,A36;
      hence contradiction by A34,A37,Def5;
    end;
    assume y < z;
    then consider y0 being Element of REAL+ such that
A38: y0 in RAT+ and
A39: y0 < z and
A40: y < y0 by Lm30;
    reconsider y9 = y0 as Element of RAT+ by A38;
    y9 in z by A34,A39,Def5;
    then ex z9 st y9 = z9 & ex x,z st z = z9 & x in X & z < x;
    then consider x0 being Element of REAL+ such that
A41: x0 in X and
A42: y0 < x0;
    y < x0 by A40,A42,Lm31;
    hence contradiction by A2,A33,A41;
  end;
end;

Lm32: one = 1;

Lm33: {} = {};

Lm34: for A,B being Element of DEDEKIND_CUTS st A + B = A & A <> {} holds B =
{}
proof
  let A,B be Element of DEDEKIND_CUTS such that
A1: A + B = A and
A2: A <> {} and
A3: B <> {};
A4: ex A0 being Element of RAT+ st A0 in A by A2,SUBSET_1:4;
  consider y9 such that
A5: y9 in B and
A6: y9 <> {} by A3,Lm15;
  defpred P[Element of RAT+] means $1 *' y9 in A;
  {} *' y9 = {} by ARYTM_3:48;
  then
A7: P[{}] by A4,Lm16,ARYTM_3:64;
  A <> RAT+ by ZFMISC_1:56;
  then consider r such that
A8: not r in A by SUBSET_1:28;
  consider n being Element of RAT+ such that
A9: n in omega and
A10: r <=' n *' y9 by A6,ARYTM_3:95;
A11: not P[n] by A8,A10,Lm16;
  consider n0 being Element of RAT+ such that
  n0 in omega and
A12: ( P[n0])& not P[n0 + one] from ARYTM_3:sch 1(Lm32,Lm33,A9,A7,A11);
  (n0 + one) *' y9 = n0 *' y9 + one *' y9 by ARYTM_3:57
    .= n0 *' y9 + y9 by ARYTM_3:53;
  hence contradiction by A1,A5,A12;
end;

Lm35: x + y = x implies y = {}
proof
  assume that
A1: x + y = x and
A2: y <> {};
A3: x <> {} by A1,A2,Th5;
  then
A4: DEDEKIND_CUT x <> {} by Lm10;
  DEDEKIND_CUT x = DEDEKIND_CUT GLUED(DEDEKIND_CUT x + DEDEKIND_CUT y) by A1,A2
,A3,Def8
    .= DEDEKIND_CUT x + DEDEKIND_CUT y by Lm12;
  then DEDEKIND_CUT y = {} by A4,Lm34;
  hence contradiction by A2,Lm10;
end;

Lm36: for A,B being Element of DEDEKIND_CUTS st A <> {} & A c= B & A <> B ex C
being Element of DEDEKIND_CUTS st A + C = B
proof
  let A,B be Element of DEDEKIND_CUTS such that
A1: A <> {} and
A2: A c= B & A <> B;
  not B c= A by A2,XBOOLE_0:def 10;
  then consider s1 being Element of RAT+ such that
A3: s1 in B & not s1 in A by SUBSET_1:2;
  set DIF = { t : ex r,s st not r in A & s in B & r + t = s};
A4: DIF c= RAT+
  proof
    let e be set;
    assume e in DIF;
    then ex t st t = e & ex r,s st not r in A & s in B & r + t = s;
    hence thesis;
  end;
  s1 + {} = s1 by ARYTM_3:84;
  then
A5: {} in DIF by A3;
  then reconsider DIF as non empty Subset of RAT+ by A4;
  t in DIF implies (for s st s <=' t holds s in DIF) & ex s st s in DIF & t < s
  proof
    assume t in DIF;
    then ex x9 st x9 = t & ex r,s st not r in A & s in B & r + x9 = s;
    then consider r0,s0 being Element of RAT+ such that
A6: not r0 in A and
A7: s0 in B and
A8: r0 + t = s0;
    thus for s st s <=' t holds s in DIF
    proof
      let s;
      assume s <=' t;
      then consider p such that
A9:   s + p = t by ARYTM_3:def 13;
A10:  t <=' s0 by A8,ARYTM_3:77;
      p <=' t by A9,ARYTM_3:77;
      then p <=' s0 by A10,ARYTM_3:67;
      then consider q such that
A11:  p + q = s0 by ARYTM_3:def 13;
      r0 + s + p = q + p by A8,A9,A11,ARYTM_3:51;
      then
A12:  r0 + s = q by ARYTM_3:62;
      q in B by A7,A11,Lm16,ARYTM_3:77;
      hence thesis by A6,A12;
    end;
    consider s1 being Element of RAT+ such that
A13: s1 in B and
A14: s0 < s1 by A7,Lm7;
    consider q such that
A15: s0 + q = s1 by A14,ARYTM_3:def 13;
    take t + q;
A16: r0 + (t + q) = s1 by A8,A15,ARYTM_3:51;
    hence t + q in DIF by A6,A13;
    t <=' t + q & t <> t + q by A8,A14,A16,ARYTM_3:77;
    hence thesis by ARYTM_3:68;
  end;
  then
A17: DIF in IR;
  B <> RAT+ by ZFMISC_1:56;
  then consider s2 being Element of RAT+ such that
A18: not s2 in B by SUBSET_1:28;
  now
    assume s2 in DIF;
    then ex t st t = s2 & ex r,s st not r in A & s in B & r + t = s;
    hence contradiction by A18,Lm16,ARYTM_3:77;
  end;
  then DIF <> RAT+;
  then reconsider DIF as Element of DEDEKIND_CUTS by A17,ZFMISC_1:56;
  take DIF;
  set C = { r + t : r in A & t in DIF};
  B = C
  proof
    thus B c= C
    proof
      let e be set;
      assume
A19:  e in B;
      then reconsider y9 = e as Element of RAT+;
      per cases;
      suppose
A20:    y9 in A;
        y9 = y9 + {} by ARYTM_3:84;
        hence thesis by A5,A20;
      end;
      suppose
A21:    not y9 in A;
        consider s0 being Element of RAT+ such that
A22:    s0 in B and
A23:    y9 < s0 by A19,Lm7;
        set Z = { r : ex x9 st not x9 in A & x9 + r = s0 };
A24:    not s0 in A by A21,A23,Lm16;
A25:    s0 + {} = s0 by ARYTM_3:84;
        for r2 being Element of RAT+ st r2 < s0 ex s,t st s in A & t in Z
        & r2 = s + t
        proof
          let r2 be Element of RAT+;
          assume
A26:      r2 < s0;
          then
A27:      r2 <> s0;
          per cases;
          suppose
A28:        r2 in A;
            take r2,{};
            thus r2 in A by A28;
            thus {} in Z by A24,A25;
            thus thesis by ARYTM_3:84;
          end;
          suppose
A29:        not r2 in A;
            consider q being Element of RAT+ such that
A30:        r2 + q = s0 by A26,ARYTM_3:def 13;
            defpred P[Element of RAT+] means $1 *' q in A;
            {} *' q = {} by ARYTM_3:48;
            then
A31:        P[{}] by A1,Lm17;
            q <> {} by A27,A30,ARYTM_3:84;
            then consider n being Element of RAT+ such that
A32:        n in omega and
A33:        s0 <=' n *' q by ARYTM_3:95;
A34:        not P[n] by A24,A33,Lm16;
            consider n0 being Element of RAT+ such that
            n0 in omega and
A35:        P[n0] and
A36:        not P[n0 + one] from ARYTM_3:sch 1(Lm32,Lm33,A32,A31,A34);
            n0 *' q <=' r2 by A29,A35,Lm16;
            then n0 *' q + q <=' s0 by A30,ARYTM_3:76;
            then consider t such that
A37:        n0 *' q + q + t = s0 by ARYTM_3:def 13;
            take n0 *' q, t;
            thus n0 *' q in A by A35;
            (n0 + one) *' q = n0 *' q + one *' q by ARYTM_3:57
              .= n0 *' q + q by ARYTM_3:53;
            hence t in Z by A36,A37;
            n0 *' q + t + q = r2 + q by A30,A37,ARYTM_3:51;
            hence thesis by ARYTM_3:62;
          end;
        end;
        then consider s,t such that
A38:    s in A and
A39:    t in Z and
A40:    y9 = s + t by A23;
        ex r st t= r & ex x9 st not x9 in A & x9 + r = s0 by A39;
        then t in DIF by A22;
        hence thesis by A38,A40;
      end;
    end;
    let e be set;
    assume e in C;
    then consider s3,t3 being Element of RAT+ such that
A41: e = s3 + t3 and
A42: s3 in A and
A43: t3 in DIF;
    ex t st t3 = t & ex r,s st not r in A & s in B & r + t = s by A43;
    then consider r4,s4 being Element of RAT+ such that
A44: not r4 in A and
A45: s4 in B and
A46: r4 + t3 = s4;
    s3 <=' r4 by A42,A44,Lm16;
    then s3 + t3 <=' s4 by A46,ARYTM_3:76;
    hence thesis by A41,A45,Lm16;
  end;
  hence thesis;
end;

Lm37: x <=' y implies DEDEKIND_CUT x c= DEDEKIND_CUT y
proof
  assume
A1: x <=' y;
  assume
A2: not DEDEKIND_CUT x c= DEDEKIND_CUT y;
  DEDEKIND_CUT x in IR & DEDEKIND_CUT y in IR by XBOOLE_0:def 5;
  then DEDEKIND_CUT y c= DEDEKIND_CUT x by A2,Lm13;
  then y <=' x by Lm20;
  hence thesis by A1,A2,Lm21;
end;

theorem Th9:
  x <=' y implies ex z st x + z = y
proof
  assume
A1: x <=' y;
  per cases;
  suppose
A2: x = {};
    take y;
    thus thesis by A2,Def8;
  end;
  suppose
A3: x = y;
    {} in RAT+;
    then reconsider z = {} as Element of REAL+ by Th1;
    take z;
    thus thesis by A3,Def8;
  end;
  suppose that
A4: x <> {} and
A5: x <> y;
A6: DEDEKIND_CUT x <> {} by A4,Lm10;
    DEDEKIND_CUT x <> DEDEKIND_CUT y by A5,Lm22;
    then consider C being Element of DEDEKIND_CUTS such that
A7: DEDEKIND_CUT x + C = DEDEKIND_CUT y by A1,A6,Lm36,Lm37;
    take GLUED C;
    now
      assume
A8:   C = {};
      not ex e being set st e in { r + s : r in C & s in DEDEKIND_CUT x}
      proof
        given e being set such that
A9:     e in { r + s : r in C & s in DEDEKIND_CUT x};
        ex r,s st e = r + s & r in C & s in DEDEKIND_CUT x by A9;
        hence contradiction by A8;
      end;
      then { r + s : r in C & s in DEDEKIND_CUT x} = {} by XBOOLE_0:def 1;
      then DEDEKIND_CUT y = {} by A7,Def6;
      hence contradiction by A1,A6,Lm37,XBOOLE_1:3;
    end;
    then GLUED C <> {} by Lm11;
    hence x + GLUED C = GLUED(DEDEKIND_CUT x + DEDEKIND_CUT GLUED C) by A4,Def8
      .= GLUED DEDEKIND_CUT y by A7,Lm12
      .= y by Lm23;
  end;
end;

theorem Th10:
  ex z st x + z = y or y + z = x
proof
  x <=' y or y <=' x;
  hence thesis by Th9;
end;

theorem Th11:
  x + y = x + z implies y = z
proof
  assume
A1: x + y = x + z;
  consider q being Element of REAL+ such that
A2: z + q = y or y + q = z by Th10;
  per cases by A2;
  suppose
A3: z + q = y;
    then x + y = x + y + q by A1,Th6;
    then q = {} by Lm35;
    hence thesis by A3,Def8;
  end;
  suppose
A4: y + q = z;
    then x + z = x + z + q by A1,Th6;
    then q = {} by Lm35;
    hence thesis by A4,Def8;
  end;
end;

Lm38: for A,B,C being Element of DEDEKIND_CUTS holds A *' (B *' C) c= A *' B
*' C
proof
  let A,B,C be Element of DEDEKIND_CUTS;
  let e be set;
  assume e in A *' (B *' C);
  then consider r0,s0 being Element of RAT+ such that
A1: e = r0 *' s0 & r0 in A and
A2: s0 in B *' C;
  consider r1,s1 being Element of RAT+ such that
A3: s0 = r1 *' s1 & r1 in B and
A4: s1 in C by A2;
  e = r0 *' r1 *' s1 & r0 *' r1 in A *' B by A1,A3,ARYTM_3:52;
  hence thesis by A4;
end;

Lm39: for A,B,C being Element of DEDEKIND_CUTS holds A *' (B *' C) = A *' B *'
C
proof
  let A,B,C be Element of DEDEKIND_CUTS;
  A *' (B *' C) c= A *' B *' C & A *' B *' C c= A *' (B *' C) by Lm38;
  hence thesis by XBOOLE_0:def 10;
end;

theorem
  x *' (y *' z) = x *' y *' z
proof
  thus x *' (y *' z) = GLUED(DEDEKIND_CUT x *' (DEDEKIND_CUT y *' DEDEKIND_CUT
  z)) by Lm12
    .= GLUED((DEDEKIND_CUT x *' DEDEKIND_CUT y) *' DEDEKIND_CUT z) by Lm39
    .= x *' y *' z by Lm12;
end;

Lm40: x *' y = {} implies x = {} or y = {}
proof
  assume
A1: x *' y = {};
  DEDEKIND_CUT x = {} or DEDEKIND_CUT y = {}
  proof
    assume DEDEKIND_CUT x <> {};
    then consider r0 being Element of RAT+ such that
A2: r0 in DEDEKIND_CUT x by SUBSET_1:4;
    assume DEDEKIND_CUT y <> {};
    then consider s0 being Element of RAT+ such that
A3: s0 in DEDEKIND_CUT y by SUBSET_1:4;
    r0 *' s0 in { r *' s: r in DEDEKIND_CUT x & s in DEDEKIND_CUT y} by A2,A3;
    hence contradiction by A1,Lm11;
  end;
  hence thesis by Lm10;
end;

Lm41: for A,B,C being Element of DEDEKIND_CUTS holds A *' (B + C) = A *' B + A
*' C
proof
  let A,B,C be Element of DEDEKIND_CUTS;
  thus A *' (B + C) c= A *' B + A *' C
  proof
    let e be set;
    assume e in A *' (B + C);
    then consider r0, v0 being Element of RAT+ such that
A1: e = r0 *' v0 and
A2: r0 in A and
A3: v0 in B + C;
    consider s0,t0 being Element of RAT+ such that
A4: v0 = s0 + t0 and
A5: s0 in B & t0 in C by A3;
A6: e = r0 *' s0 + r0 *' t0 by A1,A4,ARYTM_3:57;
    r0 *' s0 in A *' B & r0 *' t0 in A *' C by A2,A5;
    hence thesis by A6;
  end;
  let e be set;
  assume e in A *' B + A *' C;
  then consider s1,t1 being Element of RAT+ such that
A7: e = s1 + t1 and
A8: s1 in A *' B and
A9: t1 in A *' C;
  consider r0,s0 being Element of RAT+ such that
A10: s1 = r0 *' s0 and
A11: r0 in A and
A12: s0 in B by A8;
  consider r09,t0 being Element of RAT+ such that
A13: t1 = r09 *' t0 and
A14: r09 in A and
A15: t0 in C by A9;
  per cases;
  suppose
    r0 <=' r09;
    then r0 *' s0 <=' r09 *' s0 by ARYTM_3:82;
    then consider s09 being Element of RAT+ such that
A16: r0 *' s0 = r09 *' s09 and
A17: s09 <=' s0 by ARYTM_3:79;
    s09 in B by A12,A17,Lm16;
    then
A18: s09 + t0 in B + C by A15;
    e = r09 *' (s09 + t0) by A7,A10,A13,A16,ARYTM_3:57;
    hence thesis by A14,A18;
  end;
  suppose
    r09 <=' r0;
    then r09 *' t0 <=' r0 *' t0 by ARYTM_3:82;
    then consider t09 being Element of RAT+ such that
A19: r09 *' t0 = r0 *' t09 and
A20: t09 <=' t0 by ARYTM_3:79;
    t09 in C by A15,A20,Lm16;
    then
A21: s0 + t09 in B + C by A12;
    e = r0 *' (s0 + t09) by A7,A10,A13,A19,ARYTM_3:57;
    hence thesis by A11,A21;
  end;
end;

theorem
  x *' (y + z) = (x *' y) + (x *' z)
proof
  per cases;
  suppose
A1: x = {};
    hence x *' (y + z) = x by Th4
      .= x + x by A1,Def8
      .= x + (x *' z) by A1,Th4
      .= (x *' y) + (x *' z) by A1,Th4;
  end;
  suppose
A2: y = {};
    hence x *' (y + z) = x *' z by Def8
      .= y + x *' z by A2,Def8
      .= (x *' y) + (x *' z) by A2,Th4;
  end;
  suppose
A3: z = {};
    hence x *' (y + z) = x *' y by Def8
      .= x *' y + z by A3,Def8
      .= (x *' y) + (x *' z) by A3,Th4;
  end;
  suppose that
A4: x <> {} and
A5: y <> {} & z <> {};
A6: x *' y <> {} & x *' z <> {} by A4,A5,Lm40;
    thus x *' (y + z) = GLUED(DEDEKIND_CUT x *' DEDEKIND_CUT GLUED(
    DEDEKIND_CUT y + DEDEKIND_CUT z)) by A5,Def8
      .= GLUED(DEDEKIND_CUT x *' (DEDEKIND_CUT y + DEDEKIND_CUT z)) by Lm12
      .= GLUED((DEDEKIND_CUT x *' DEDEKIND_CUT y) + (DEDEKIND_CUT x *'
    DEDEKIND_CUT z)) by Lm41
      .= GLUED((DEDEKIND_CUT x *' DEDEKIND_CUT y) + DEDEKIND_CUT (x*'z)) by
Lm12
      .= GLUED(DEDEKIND_CUT(x*'y) + DEDEKIND_CUT (x*'z)) by Lm12
      .= (x *' y) + (x *' z) by A6,Def8;
  end;
end;

one in RAT+;

then reconsider rone = one as Element of REAL+ by Th1;

Lm42: for B being set st B in IR & B <> {} ex r st r in B & r <> {}
proof
  let B be set such that
A1: B in IR and
A2: B <> {};
  consider A being Subset of RAT+ such that
A3: B = A and
A4: r in A implies (for s st s <=' r holds s in A) & ex s st s in A & r
  < s by A1;
  consider r0 being Element of RAT+ such that
A5: r0 in A by A2,A3,SUBSET_1:4;
  consider r1 being Element of RAT+ such that
A6: r1 in A and
A7: r0 < r1 by A4,A5;
A8: r1 <> {} or r0 <> {} by A7;
  for r,s st r in A & s <=' r holds s in A by A4;
  then consider r1,r2,r3 being Element of RAT+ such that
A9: r1 in A & r2 in A and
  r3 in A and
A10: r1 <> r2 and
  r2 <> r3 and
  r3 <> r1 by A5,A6,A8,ARYTM_3:75;
  r1 <> {} or r2 <> {} by A10;
  hence thesis by A3,A9;
end;

Lm43: for A being Element of DEDEKIND_CUTS st A <> {} ex B being Element of
DEDEKIND_CUTS st A *' B = DEDEKIND_CUT rone
proof
  let A be Element of DEDEKIND_CUTS such that
A1: A <> {};
  per cases;
  suppose
    A in RA;
    then consider r0 being Element of RAT+ such that
A2: A = { s: s < r0} and
A3: r0 <> {};
    consider s0 being Element of RAT+ such that
A4: r0 *' s0 = one by A3,ARYTM_3:54;
    set B = { s : s < s0 };
    B c= RAT+
    proof
      let e be set;
      assume e in B;
      then ex s st s = e & s < s0;
      hence thesis;
    end;
    then reconsider B as Subset of RAT+;
    r in B implies (for s st s <=' r holds s in B) & ex s st s in B & r < s
    proof
      assume r in B;
      then
A5:   ex s st s = r & s < s0;
      then consider t such that
A6:   r < t and
A7:   t < s0 by ARYTM_3:93;
      thus for s st s <=' r holds s in B
      proof
        let s;
        assume s <=' r;
        then s < s0 by A5,ARYTM_3:69;
        hence thesis;
      end;
      take t;
      thus t in B by A7;
      thus thesis by A6;
    end;
    then
A8: B in IR;
    not ex s st s = s0 & s < s0;
    then not s0 in B;
    then B <> RAT+;
    then reconsider B as Element of DEDEKIND_CUTS by A8,ZFMISC_1:56;
A9: A *' B = { s : s < r0 *' s0 }
    proof
      thus A *' B c= { s : s < r0 *' s0 }
      proof
        let e be set;
        assume e in A *' B;
        then consider r1,s1 being Element of RAT+ such that
A10:    e = r1 *' s1 and
A11:    r1 in A and
A12:    s1 in B;
        ex s st s = r1 & s < r0 by A2,A11;
        then
A13:    r1 *' s1 <=' r0 *' s1 by ARYTM_3:82;
A14:    ex s st s = s1 & s < s0 by A12;
        then s1 <> s0;
        then
A15:    r0 *' s1 <> r0 *' s0 by A3,ARYTM_3:56;
        r0 *' s1 <=' r0 *' s0 by A14,ARYTM_3:82;
        then r0 *' s1 < r0 *' s0 by A15,ARYTM_3:68;
        then r1 *' s1 < r0 *' s0 by A13,ARYTM_3:69;
        hence thesis by A10;
      end;
      let e be set;
      assume e in { s : s < r0 *' s0 };
      then consider s1 being Element of RAT+ such that
A16:  e = s1 and
A17:  s1 < r0 *' s0;
      consider t0 being Element of RAT+ such that
A18:  s1 = r0 *' t0 and
A19:  t0 <=' s0 by A17,ARYTM_3:79;
      t0 <> s0 by A17,A18;
      then t0 < s0 by A19,ARYTM_3:68;
      then t0 in B;
      then consider t1 being Element of RAT+ such that
A20:  t1 in B and
A21:  t0 < t1 by Lm7;
      r0 *' t0 <=' t1 *' r0 by A21,ARYTM_3:82;
      then consider r1 being Element of RAT+ such that
A22:  r0 *' t0 = t1 *' r1 and
A23:  r1 <=' r0 by ARYTM_3:79;
      t0 <> t1 by A21;
      then r1 <> r0 by A3,A22,ARYTM_3:56;
      then r1 < r0 by A23,ARYTM_3:68;
      then r1 in A by A2;
      hence thesis by A16,A18,A20,A22;
    end;
    ex t0 being Element of RAT+ st t0 = rone & DEDEKIND_CUT rone = { s :
    s < t0 } by Def3;
    hence thesis by A4,A9;
  end;
  suppose
A24: not A in RA;
    set B = { y9 : ex x9 st not x9 in A & x9 *' y9 <=' one };
A25: B c= RAT+
    proof
      let e be set;
      assume e in B;
      then ex y9 st y9 = e & ex x9 st not x9 in A & x9 *' y9 <=' one;
      hence thesis;
    end;
A26: A <> RAT+ by ZFMISC_1:56;
    then consider x0 being Element of RAT+ such that
A27: not x0 in A by SUBSET_1:28;
    x0 *' {} = {} by ARYTM_3:48;
    then x0 *' {} <=' one by ARYTM_3:64;
    then
A28: {} in B by A27;
    then reconsider B as non empty Subset of RAT+ by A25;
A29: A in IR by ZFMISC_1:56;
    ex x9 st x9 in A by A1,SUBSET_1:4;
    then
A30: {} in A by A29,Lm29,ARYTM_3:64;
    r in B implies (for s st s <=' r holds s in B) & ex s st s in B & r < s
    proof
      assume r in B;
      then ex y9 st r = y9 & ex x9 st not x9 in A & x9 *' y9 <=' one;
      then consider x9 such that
A31:  not x9 in A and
A32:  x9 *' r <=' one;
      thus for s st s <=' r holds s in B
      proof
        let s;
        assume s <=' r;
        then x9 *' s <=' x9 *' r by ARYTM_3:82;
        then x9 *' s <=' one by A32,ARYTM_3:67;
        hence thesis by A31;
      end;
      A in DEDEKIND_CUTS \ RA by A24,XBOOLE_0:def 5;
      then consider x99 being Element of RAT+ such that
A33:  not x99 in A and
A34:  x99 < x9 by A1,A31,Lm18;
      consider s such that
A35:  one = x99 *' s by A30,A33,ARYTM_3:55;
      take s;
      x99 *' s <=' one by A35;
      hence s in B by A33;
A36:  s <> {} by A35,ARYTM_3:48;
      x99 *' r <=' x9 *' r by A34,ARYTM_3:82;
      then x99 *' r <=' x99 *' s by A32,A35,ARYTM_3:67;
      then
A37:  r <=' s by A30,A33,ARYTM_3:80;
      x99 <> x9 & x99 *' s <=' x9 *' s by A34,ARYTM_3:82;
      then r <> s by A32,A35,A36,ARYTM_3:56,66;
      hence thesis by A37,ARYTM_3:68;
    end;
    then
A38: B in IR;
    consider x9 such that
A39: x9 in A and
A40: x9 <> {} by A1,A29,Lm42;
    consider y9 such that
A41: x9 *' y9 = one by A40,ARYTM_3:55;
    now
      assume y9 in B;
      then
A42:  ex s st s = y9 & ex x9 st not x9 in A & x9 *' s <=' one;
      y9 <> {} by A41,ARYTM_3:48;
      hence contradiction by A29,A39,A41,A42,Lm29,ARYTM_3:80;
    end;
    then B <> RAT+;
    then not B in {RAT+} by TARSKI:def 1;
    then reconsider B as Element of DEDEKIND_CUTS by A38,XBOOLE_0:def 5;
    take B;
    for r holds r in A *' B iff r < one
    proof
      let r;
      hereby
        assume r in A *' B;
        then consider s,t such that
A43:    r = s *' t and
A44:    s in A and
A45:    t in B;
        ex z9 st z9 = t & ex x9 st not x9 in A & x9 *' z9 <=' one by A45;
        then consider x9 such that
A46:    not x9 in A and
A47:    x9 *' t <=' one;
        s <=' x9 by A29,A44,A46,Lm29;
        then
A48:    s *' t <=' x9 *' t by ARYTM_3:82;
A49:    now
          assume
A50:      r = one;
          then t <> {} by A43,ARYTM_3:48;
          hence contradiction by A43,A44,A46,A47,A48,A50,ARYTM_3:56,66;
        end;
        r <=' one by A43,A47,A48,ARYTM_3:67;
        hence r < one by A49,ARYTM_3:68;
      end;
      assume
A51:  r < one;
      then
A52:  r <> one;
      per cases;
      suppose
        r = {};
        then r = {}*'{} by ARYTM_3:48;
        hence thesis by A28,A30;
      end;
      suppose
        r <> {};
        then consider r9 being Element of RAT+ such that
A53:    r *' r9 = one by ARYTM_3:55;
        { r *' s : s in A } c= RAT+
        proof
          let e be set;
          assume e in { r *' s : s in A };
          then ex s st e = r *' s & s in A;
          hence thesis;
        end;
        then reconsider rA = { r *' s : s in A } as Subset of RAT+;
        consider dr being Element of RAT+ such that
A54:    r + dr = one by A51,ARYTM_3:def 13;
        consider xx being Element of RAT+ such that
A55:    not xx in A by A26,SUBSET_1:28;
        set rr = x9 *' dr;
        dr <> {} by A52,A54,ARYTM_3:50;
        then consider n0 being Element of RAT+ such that
A56:    n0 in omega and
A57:    xx <=' n0 *' rr by A40,ARYTM_3:78,95;
        defpred P[Element of RAT+] means $1 *' rr in A;
A58:    P[{}] by A30,ARYTM_3:48;
A59:    not P[n0] by A29,A55,A57,Lm29;
        consider n1 being Element of RAT+ such that
        n1 in omega and
A60:    P[n1] and
A61:    not P[n1 + one] from ARYTM_3:sch 1(Lm32,Lm33,A56,A58,A59);
        set s0 = n1 *' rr;
A62:    now
          assume n1 *' rr in rA;
          then consider s0 being Element of RAT+ such that
A63:      r *' s0 = n1 *' rr and
A64:      s0 in A;
A65:      (n1 + one) *' rr = n1 *' rr + one *' rr by ARYTM_3:57
            .= r *' s0 + dr *' x9 by A63,ARYTM_3:53;
          s0 <=' x9 or x9 <=' s0;
          then consider s1 being Element of RAT+ such that
A66:      s1 = s0 & x9 <=' s1 or s1 = x9 & s0 <=' s1;
          dr *' x9 <=' dr *' s1 by A66,ARYTM_3:82;
          then
A67:      r *' s1 + dr *' x9 <=' r *' s1 + dr *' s1 by ARYTM_3:76;
          r *' s0 <=' r *' s1 by A66,ARYTM_3:82;
          then
A68:      r *' s0 + dr *' x9 <=' r *' s1 + dr *' x9 by ARYTM_3:76;
          r *' s1 + dr *' s1 = (r + dr) *' s1 by ARYTM_3:57
            .= s1 by A54,ARYTM_3:53;
          hence contradiction by A29,A39,A61,A64,A65,A66,A68,A67,Lm29;
        end;
A69:    now
          assume
A70:      s0 *' r9 in A;
          r *' (s0 *' r9) = one *' s0 by A53,ARYTM_3:52
            .= s0 by ARYTM_3:53;
          hence contradiction by A62,A70;
        end;
        r *' {} = {} by ARYTM_3:48;
        then {} in rA by A30;
        then consider s9 being Element of RAT+ such that
A71:    s0 *' s9 = one by A62,ARYTM_3:55;
A72:    s0 *' (r *' s9) = s0 *' s9 *' r by ARYTM_3:52
          .= r by A71,ARYTM_3:53;
        s0 *' r9 *' (r *' s9) = s0 *' r9 *' r *' s9 by ARYTM_3:52
          .= s0 *' one *' s9 by A53,ARYTM_3:52
          .= one by A71,ARYTM_3:53;
        then s0 *' r9 *' (r *' s9) <=' one;
        then r *' s9 in B by A69;
        hence thesis by A60,A72;
      end;
    end;
    then DEDEKIND_CUT GLUED(A *' B) = DEDEKIND_CUT rone by Def4;
    hence thesis by Lm12;
  end;
end;

theorem
  x <> {} implies ex y st x *' y = one
proof
  assume x <> {};
  then DEDEKIND_CUT x <> {} by Lm10;
  then consider B being Element of DEDEKIND_CUTS such that
A1: DEDEKIND_CUT x *' B = DEDEKIND_CUT rone by Lm43;
  take y = GLUED B;
  thus x *' y = GLUED DEDEKIND_CUT rone by A1,Lm12
    .= one by Lm23;
end;

Lm44: for A,B being Element of DEDEKIND_CUTS st A = { r: r < one } holds A *'
B = B
proof
  let A,B be Element of DEDEKIND_CUTS such that
A1: A = { r: r < one };
  thus A *' B c= B
  proof
    let e be set;
    assume e in A *' B;
    then consider r,s such that
A2: e = r *' s and
A3: r in A and
A4: s in B;
    ex t st t = r & t < one by A1,A3;
    then r *' s <=' one *' s by ARYTM_3:82;
    then r *' s <=' s by ARYTM_3:53;
    hence thesis by A2,A4,Lm16;
  end;
  let e be set;
  assume
A5: e in B;
  then reconsider s = e as Element of RAT+;
  consider s1 being Element of RAT+ such that
A6: s1 in B and
A7: s < s1 by A5,Lm7;
  s1 <> {} by A7,ARYTM_3:64;
  then consider s2 being Element of RAT+ such that
A8: s1 *' s2 = s by ARYTM_3:55;
A9: now
    assume s2 = one;
    then s = s1 by A8,ARYTM_3:53;
    hence contradiction by A7;
  end;
  one *' s = s2 *' s1 by A8,ARYTM_3:53;
  then s2 <=' one by A7,ARYTM_3:83;
  then s2 < one by A9,ARYTM_3:68;
  then s2 in A by A1;
  hence thesis by A6,A8;
end;

theorem
  x = one implies x *' y = y
proof
  assume
A1: x = one;
  then ex r st x = r & DEDEKIND_CUT x = { s : s < r } by Def3;
  hence x *' y = GLUED DEDEKIND_CUT y by A1,Lm44
    .= y by Lm23;
end;

Lm45: for i,j being Element of omega, x9,y9 st i = x9 & j = y9 holds i +^ j =
x9 + y9
proof
  let i,j be Element of omega, x9,y9;
  set a = (denominator x9)*^(denominator y9), b = (numerator x9)*^(denominator
  y9)+^(numerator y9)*^(denominator x9);
  assume
A1: i = x9;
  then
A2: denominator x9 = one by ARYTM_3:def 9;
  assume
A3: j = y9;
  then
A4: denominator y9 = one by ARYTM_3:def 9;
  then
A5: a = one by A2,ORDINAL2:39;
  then
A6: RED(a,b) = one by ARYTM_3:24;
  b = (numerator x9)*^one+^(numerator y9) by A2,A4,ORDINAL2:39
    .= (numerator x9)+^(numerator y9) by ORDINAL2:39
    .= i+^(numerator y9) by A1,ARYTM_3:def 8
    .= i +^ j by A3,ARYTM_3:def 8;
  hence i +^ j = RED(b,a) by A5,ARYTM_3:24
    .= x9 + y9 by A6,ARYTM_3:def 10;
end;

Lm46: z9 < x9 + y9 & x9 <> {} & y9 <> {} implies ex r,s st z9 = r + s & r < x9
& s < y9
proof
  assume that
A1: z9 < x9 + y9 and
A2: x9 <> {} and
A3: y9 <> {};
  consider r0,t0 being Element of RAT+ such that
A4: z9 = r0 + t0 and
A5: r0 <=' x9 and
A6: t0 <=' y9 & t0 <> y9 by A1,A3,ARYTM_3:90;
  per cases;
  suppose
A7: r0 = {};
    take {},t0;
    thus z9 = {} + t0 by A4,A7;
    {} <=' x9 by ARYTM_3:64;
    hence {} < x9 by A2,ARYTM_3:68;
    thus thesis by A6,ARYTM_3:68;
  end;
  suppose
A8: r0 <> {};
    t0 < y9 by A6,ARYTM_3:68;
    then consider t1 being Element of RAT+ such that
A9: t0 < t1 and
A10: t1 < y9 by ARYTM_3:93;
    z9 < t1 + r0 by A4,A9,ARYTM_3:76;
    then consider t2,r1 being Element of RAT+ such that
A11: z9 = t2 + r1 and
A12: t2 <=' t1 and
A13: r1 <=' r0 & r1 <> r0 by A8,ARYTM_3:90;
    take r1,t2;
    thus z9 = r1 + t2 by A11;
    r1 < r0 by A13,ARYTM_3:68;
    hence r1 < x9 by A5,ARYTM_3:69;
    thus thesis by A10,A12,ARYTM_3:69;
  end;
end;

Lm47: x in RAT+ & y in RAT+ implies ex x9,y9 st x = x9 & y = y9 & x + y = x9 +
y9
proof
  assume that
A1: x in RAT+ and
A2: y in RAT+;
  per cases;
  suppose
A3: x = {};
    reconsider y9 = y as Element of RAT+ by A2;
    take {},y9;
    thus x = {} by A3;
    thus y = y9;
    thus x + y = y by A3,Def8
      .= {} + y9 by ARYTM_3:50;
  end;
  suppose
A4: y = {};
    reconsider x9 = x as Element of RAT+ by A1;
    take x9,{};
    thus x = x9;
    thus y = {} by A4;
    thus x + y = x by A4,Def8
      .= x9 + {} by ARYTM_3:50;
  end;
  suppose that
A5: y <> {} & x <> {};
    set A = DEDEKIND_CUT x, B = DEDEKIND_CUT y;
    consider x9 such that
A6: x = x9 and
A7: DEDEKIND_CUT x = { s : s < x9 } by A1,Def3;
    consider y9 such that
A8: y = y9 and
A9: DEDEKIND_CUT y = { s : s < y9 } by A2,Def3;
A10: for s holds s in DEDEKIND_CUT x + DEDEKIND_CUT y iff s < x9 + y9
    proof
      let s2 be Element of RAT+;
      thus s2 in A + B implies s2 < x9 + y9
      proof
        assume s2 in A + B;
        then consider r1,s1 being Element of RAT+ such that
A11:    s2 = r1 + s1 and
A12:    r1 in A and
A13:    s1 in B;
        ex s st s = r1 & s < x9 by A7,A12;
        then
A14:    r1 + s1 <=' x9 + s1 by ARYTM_3:76;
        ex s st s = s1 & s < y9 by A9,A13;
        then x9 + s1 < x9 + y9 by ARYTM_3:76;
        hence thesis by A11,A14,ARYTM_3:69;
      end;
      assume s2 < x9 + y9;
      then consider t2,t0 being Element of RAT+ such that
A15:  s2 = t2 + t0 and
A16:  t2 < x9 & t0 < y9 by A5,A6,A8,Lm46;
      t0 in B & t2 in A by A7,A9,A16;
      hence thesis by A15;
    end;
    then consider r such that
A17: GLUED(DEDEKIND_CUT x + DEDEKIND_CUT y) = r and
A18: for s holds s in DEDEKIND_CUT x + DEDEKIND_CUT y iff s < r by Def4;
A19: for s holds s < x9 + y9 iff s < r
    proof
      let s;
      s in DEDEKIND_CUT x + DEDEKIND_CUT y iff s < x9 + y9 by A10;
      hence thesis by A18;
    end;
    take x9,y9;
    thus x = x9 & y = y9 by A6,A8;
    thus x + y = GLUED(DEDEKIND_CUT x + DEDEKIND_CUT y) by A5,Def8
      .= x9 + y9 by A17,A19,Lm6;
  end;
end;

theorem
  x in omega & y in omega implies y + x in omega
proof
  assume
A1: x in omega & y in omega;
  then reconsider x0 = x, y0 = y as Element of omega;
  consider x9,y9 being Element of RAT+ such that
A2: x = x9 & y = y9 and
A3: x + y = x9 + y9 by A1,Lm5,Lm47;
  x9 + y9 = x0 +^ y0 by A2,Lm45;
  hence thesis by A3,ORDINAL1:def 12;
end;

theorem
  for A being Subset of REAL+ st {} in A & for x,y st x in A & y = one
  holds x + y in A holds omega c= A
proof
  let A be Subset of REAL+;
  defpred P[set] means $1 in A;
  assume that
A1: P[{}] and
A2: for x,y st x in A & y = one holds x + y in A;
  let e be set;
  assume e in omega;
  then reconsider a = e as natural Ordinal;
A3: for a being natural Ordinal st P[a] holds P[succ a]
  proof
    one in RAT+;
    then reconsider rone = one as Element of REAL+ by Th1;
    let a be natural Ordinal;
    assume
A4: a in A;
    reconsider i = a as Element of omega by ORDINAL1:def 12;
A5: a in omega by ORDINAL1:def 12;
    then a in RAT+ by Lm5;
    then reconsider x = a as Element of REAL+ by Th1;
    consider x9, y9 being Element of RAT+ such that
A6: x = x9 & rone = y9 and
A7: x + rone = x9 + y9 by A5,Lm5,Lm47;
    x9 + y9 = i +^ 1 by A6,Lm45
      .= succ i by ORDINAL2:31;
    hence thesis by A2,A4,A7;
  end;
  P[a] from ORDINAL2:sch 17(A1,A3);
  hence thesis;
end;

theorem
  for x st x in omega holds for y holds y in x iff y in omega & y <> x &
  y <=' x
proof
  let x;
  assume
A1: x in omega;
  then reconsider m = x as Element of omega;
  reconsider x9 = x as Element of RAT+ by A1,Lm5;
  let y;
A2: x c= omega by A1,ORDINAL1:def 2;
  hereby
    assume
A3: y in x;
    then reconsider n = y as Element of omega by A2;
    thus y in omega by A2,A3;
    then reconsider y9 = y as Element of RAT+ by Lm5;
    thus y <> x by A3;
    n c= m by A3,ORDINAL1:def 2;
    then consider C being Ordinal such that
A4: m = n +^ C by ORDINAL3:27;
    C c= m by A4,ORDINAL3:24;
    then reconsider C as Element of omega by ORDINAL1:12;
    C in omega;
    then reconsider z9 = C as Element of RAT+ by Lm5;
    x9 = y9 + z9 by A4,Lm45;
    then y9 <=' x9 by ARYTM_3:def 13;
    hence y <=' x by Lm14;
  end;
  assume
A5: y in omega;
  then reconsider y9 = y as Element of RAT+ by Lm5;
  reconsider n = y as Element of omega by A5;
  assume
A6: y <> x;
  assume y <=' x;
  then y9 <=' x9 by Lm14;
  then consider z9 such that
A7: y9 + z9 = x9 by ARYTM_3:def 13;
  reconsider k = z9 as Element of omega by A1,A5,A7,ARYTM_3:71;
  n +^ k = m by A7,Lm45;
  then n c= m by ORDINAL3:24;
  then n c< m by A6,XBOOLE_0:def 8;
  hence thesis by ORDINAL1:11;
end;

theorem
  x = y + z implies z <=' x
proof
  {} in RAT+;
  then reconsider zz = {} as Element of REAL+ by Th1;
  assume
A1: x = y + z;
  assume
A2: not z <=' x;
  then consider y0 being Element of REAL+ such that
A3: x + y0 = z by Th9;
  x = x + (y + y0) by A1,A3,Th6;
  then x + zz = x + (y + y0) by Def8;
  then y0 = {} by Th5,Th11;
  then z = x by A3,Def8;
  hence thesis by A2;
end;

theorem
  {} in REAL+ & one in REAL+
proof
  {} in RAT+ & one in RAT+;
  hence thesis by Th1;
end;

theorem
  x in RAT+ & y in RAT+ implies ex x9,y9 st x = x9 & y = y9 & x *' y = x9 *' y9
proof
  assume that
A1: x in RAT+ and
A2: y in RAT+;
  per cases;
  suppose
A3: x = {};
    reconsider y9 = y as Element of RAT+ by A2;
    take {},y9;
    thus x = {} by A3;
    thus y = y9;
    thus x *' y = {} by A3,Th4
      .= {} *' y9 by ARYTM_3:48;
  end;
  suppose
A4: y = {};
    reconsider x9 = x as Element of RAT+ by A1;
    take x9,{};
    thus x = x9;
    thus y = {} by A4;
    thus x *' y = {} by A4,Th4
      .= x9 *' {} by ARYTM_3:48;
  end;
  suppose that
    y <> {} and
A5: x <> {};
    consider y9 such that
A6: y = y9 and
A7: DEDEKIND_CUT y = { s : s < y9 } by A2,Def3;
    set A = DEDEKIND_CUT x, B = DEDEKIND_CUT y;
    consider x9 such that
A8: x = x9 and
A9: DEDEKIND_CUT x = { s : s < x9 } by A1,Def3;
A10: for s holds s in DEDEKIND_CUT x *' DEDEKIND_CUT y iff s < x9 *' y9
    proof
      let s2 be Element of RAT+;
      thus s2 in A *' B implies s2 < x9 *' y9
      proof
        assume s2 in A *' B;
        then consider r1,s1 being Element of RAT+ such that
A11:    s2 = r1 *' s1 and
A12:    r1 in A and
A13:    s1 in B;
        ex s st s = r1 & s < x9 by A9,A12;
        then
A14:    r1 *' s1 <=' x9 *' s1 by ARYTM_3:82;
A15:    ex s st s = s1 & s < y9 by A7,A13;
        then s1 <> y9;
        then
A16:    x9 *' s1 <> x9 *' y9 by A5,A8,ARYTM_3:56;
        x9 *' s1 <=' x9 *' y9 by A15,ARYTM_3:82;
        then x9 *' s1 < x9 *' y9 by A16,ARYTM_3:68;
        hence thesis by A11,A14,ARYTM_3:69;
      end;
      assume
A17:  s2 < x9 *' y9;
      then consider t0 being Element of RAT+ such that
A18:  s2 = x9 *' t0 and
A19:  t0 <=' y9 by ARYTM_3:79;
      t0 <> y9 by A17,A18;
      then t0 < y9 by A19,ARYTM_3:68;
      then consider t1 being Element of RAT+ such that
A20:  t0 < t1 and
A21:  t1 < y9 by ARYTM_3:93;
      s2 <=' t1 *' x9 by A18,A20,ARYTM_3:82;
      then consider t2 being Element of RAT+ such that
A22:  s2 = t1 *' t2 and
A23:  t2 <=' x9 by ARYTM_3:79;
      now
        assume t2 = x9;
        then t0 = t1 by A5,A8,A18,A22,ARYTM_3:56;
        hence contradiction by A20;
      end;
      then t2 < x9 by A23,ARYTM_3:68;
      then
A24:  t2 in A by A9;
      t1 in B by A7,A21;
      hence thesis by A22,A24;
    end;
    then consider r such that
A25: GLUED(DEDEKIND_CUT x *' DEDEKIND_CUT y) = r and
A26: for s holds s in DEDEKIND_CUT x *' DEDEKIND_CUT y iff s < r by Def4;
    take x9,y9;
    thus x = x9 & y = y9 by A8,A6;
    for s holds s < x9 *' y9 iff s < r
    proof
      let s;
      s in DEDEKIND_CUT x *' DEDEKIND_CUT y iff s < x9 *' y9 by A10;
      hence thesis by A26;
    end;
    hence thesis by A25,Lm6;
  end;
end;














begin

reserve x,y,z for Element of REAL+;

theorem Th1:
  x + y = y implies x = {}
proof
  reconsider o = {} as Element of REAL+ by ARYTM_2:20;
  assume x + y = y;
  then x + y = y + o by ARYTM_2:def 8;
  hence thesis by ARYTM_2:11;
end;

reconsider u = one as Element of REAL+ by ARYTM_2:20;

Lm1: x *' y = x *' z & x <> {} implies y = z
proof
  assume
A1: x *' y = x *' z;
  assume x <> {};
  then consider x1 being Element of REAL+ such that
A2: x *' x1 = one by ARYTM_2:14;
  thus y = x *' x1 *' y by A2,ARYTM_2:15
    .= x1 *' (x *' z) by A1,ARYTM_2:12
    .= x *' x1 *' z by ARYTM_2:12
    .= z by A2,ARYTM_2:15;
end;

theorem
  x *' y = {} implies x = {} or y = {}
proof
  assume
A1: x *' y = {};
  assume x <> {};
  then consider x1 being Element of REAL+ such that
A2: x *' x1 = one by ARYTM_2:14;
  thus y = x *' x1 *' y by A2,ARYTM_2:15
    .= x *' y *' x1 by ARYTM_2:12
    .= {} by A1,ARYTM_2:4;
end;

theorem Th3:
  x <=' y & y <=' z implies x <=' z
proof
  assume x <=' y;
  then consider z1 being Element of REAL+ such that
A1: x + z1 = y by ARYTM_2:9;
  assume y <=' z;
  then consider z2 being Element of REAL+ such that
A2: y + z2 = z by ARYTM_2:9;
  z = x + (z1 + z2) by A1,A2,ARYTM_2:6;
  hence thesis by ARYTM_2:19;
end;

theorem Th4:
  x <=' y & y <=' x implies x = y
proof
  assume x <=' y;
  then consider z1 being Element of REAL+ such that
A1: x + z1 = y by ARYTM_2:9;
  assume y <=' x;
  then consider z2 being Element of REAL+ such that
A2: y + z2 = x by ARYTM_2:9;
  x = x + (z1 + z2) by A1,A2,ARYTM_2:6;
  then z1 = {} by Th1,ARYTM_2:5;
  hence thesis by A1,ARYTM_2:def 8;
end;

theorem Th5:
  x <=' y & y = {} implies x = {}
proof
  assume x <=' y;
  then ex z being Element of REAL+ st x + z = y by ARYTM_2:9;
  hence thesis by ARYTM_2:5;
end;

theorem Th6:
  x = {} implies x <=' y
proof
  assume x = {};
  then x + y = y by ARYTM_2:def 8;
  hence thesis by ARYTM_2:19;
end;

theorem Th7:
  x <=' y iff x + z <=' y + z
proof
  thus x <=' y implies x + z <=' y + z
  proof
    assume x <=' y;
    then consider z0 being Element of REAL+ such that
A1: x + z0 = y by ARYTM_2:9;
    x + z + z0 = y + z by A1,ARYTM_2:6;
    hence thesis by ARYTM_2:19;
  end;
  assume x + z <=' y + z;
  then consider z0 being Element of REAL+ such that
A2: x + z + z0 = y + z by ARYTM_2:9;
  y + z = x + z0 + z by A2,ARYTM_2:6;
  hence thesis by ARYTM_2:11,19;
end;

theorem Th8:
  x <=' y implies x *' z <=' y *' z
proof
  assume x <=' y;
  then consider z0 being Element of REAL+ such that
A1: x + z0 = y by ARYTM_2:9;
  y *' z = x *' z + z0 *' z by A1,ARYTM_2:13;
  hence thesis by ARYTM_2:19;
end;

Lm2: x *' y <=' x *' z & x <> {} implies y <=' z
proof
  assume x *' y <=' x *' z;
  then consider z0 being Element of REAL+ such that
A1: x *' y + z0 = x *' z by ARYTM_2:9;
  assume
A2: x <> {};
  then consider x1 being Element of REAL+ such that
A3: x *' x1 = one by ARYTM_2:14;
  x *' z = x *' y + u *' z0 by A1,ARYTM_2:15
    .= x *' y + x *' (x1 *' z0) by A3,ARYTM_2:12
    .= x *' (y + x1 *' z0) by ARYTM_2:13;
  hence thesis by A2,Lm1,ARYTM_2:19;
end;

definition
  let x,y be Element of REAL+;
  func x -' y -> Element of REAL+ means
  :Def1:
  it + y = x if y <=' x otherwise it = {};
  existence
  proof
    hereby
      assume y <=' x;
      then ex IT being Element of REAL+ st y + IT = x by ARYTM_2:9;
      hence ex IT being Element of REAL+ st IT + y = x;
    end;
    thus thesis by ARYTM_2:20;
  end;
  correctness by ARYTM_2:11;
end;

Lm3: x -' x = {}
proof
  x <=' x;
  then x -' x + x = x by Def1;
  hence thesis by Th1;
end;

theorem Th9:
  x <=' y or x -' y <> {}
proof
  assume
A1: not x <=' y;
  assume
A2: x -' y = {};
  x -' y + y = x by A1,Def1;
  then x = y by A2,ARYTM_2:def 8;
  hence contradiction by A1;
end;

theorem
  x <=' y & y -' x = {} implies x = y
proof
  assume
A1: x <=' y;
  assume y -' x = {};
  then y <=' x by Th9;
  hence thesis by A1,Th4;
end;

theorem Th11:
  x -' y <=' x
proof
  per cases;
  suppose
    y <=' x;
    then x -' y + y = x by Def1;
    hence thesis by ARYTM_2:19;
  end;
  suppose
    not y <=' x;
    then x -' y = {} by Def1;
    hence thesis by Th6;
  end;
end;

Lm4: x = {} implies y -' x = y
proof
  assume
A1: x = {};
  then
A2: x <=' y by Th6;
  thus y -' x = y -' x + x by A1,ARYTM_2:def 8
    .= y by A2,Def1;
end;

Lm5: x + y -' y = x
proof
  y <=' x + y by ARYTM_2:19;
  hence thesis by Def1;
end;

Lm6: x <=' y implies y -' (y -' x) = x
proof
  assume
A1: x <=' y;
  y -' x <=' y by Th11;
  then y -' (y -' x) + (y -' x) = y by Def1
    .= y -' x + x by A1,Def1;
  hence thesis by ARYTM_2:11;
end;

Lm7: z -' y <=' x iff z <=' x + y
proof
  per cases;
  suppose
    y <=' z;
    then z -' y + y = z by Def1;
    hence thesis by Th7;
  end;
  suppose
A1: not y <=' z;
A2: y <=' x + y by ARYTM_2:19;
    z -' y = {} by A1,Def1;
    hence thesis by A1,A2,Th3,Th6;
  end;
end;

Lm8: y <=' x implies (z + y <=' x iff z <=' x -' y)
proof
  assume y <=' x;
  then x -' y + y = x by Def1;
  hence thesis by Th7;
end;

Lm9: z -' y -' x = z -' (x + y)
proof
  per cases;
  suppose
A1: x + y <=' z;
    y <=' x + y by ARYTM_2:19;
    then
A2: y <=' z by A1,Th3;
    then
A3: x <=' z -' y by A1,Lm8;
    z -' y -' x + (x + y) = z -' y -' x + x + y by ARYTM_2:6
      .= z -' y + y by A3,Def1
      .= z by A2,Def1;
    hence thesis by A1,Def1;
  end;
  suppose
A4: x = {};
    hence z -' y -' x = z -' y by Lm4
      .= z -' (x + y) by A4,ARYTM_2:def 8;
  end;
  suppose that
A5: not y <=' z and
A6: x <> {};
    y <=' y + x by ARYTM_2:19;
    then
A7: not x + y <=' z by A5,Th3;
    now
      assume
A8:   x <=' z -' y;
      z -' y = {} by A5,Def1;
      hence contradiction by A6,A8,Th5;
    end;
    hence z -' y -' x = {} by Def1
      .= z -' (x + y) by A7,Def1;
  end;
  suppose that
A9: not x + y <=' z and
A10: y <=' z;
    not x <=' z -' y by A9,A10,Lm8;
    hence z -' y -' x = {} by Def1
      .= z -' (x + y) by A9,Def1;
  end;
end;

Lm10: y -' z -' x = y -' x -' z
proof
  thus y -' z -' x = y -' (x + z) by Lm9
    .= y -' x -' z by Lm9;
end;

theorem
  y <=' x & y <=' z implies x + (z -' y) = x -' y + z
proof
  assume that
A1: y <=' x and
A2: y <=' z;
  x + (z -' y) + y = x + ((z -' y) + y) by ARYTM_2:6
    .= x + z by A2,Def1
    .= x -' y + y + z by A1,Def1
    .= x -' y + z + y by ARYTM_2:6;
  hence thesis by ARYTM_2:11;
end;

theorem Th13:
  z <=' y implies x + (y -' z) = x + y -' z
proof
  assume
A1: z <=' y;
  y <=' x + y by ARYTM_2:19;
  then
A2: z <=' x + y by A1,Th3;
  x + (y -' z) + z = x + ((y -' z) + z) by ARYTM_2:6
    .= x + y by A1,Def1
    .= x + y -' z + z by A2,Def1;
  hence thesis by ARYTM_2:11;
end;

Lm11: y <=' z implies x -' (z -' y) = x + y -' z
proof
  assume
A1: y <=' z;
  per cases;
  suppose
A2: z -' y <=' x;
    then
A3: z <=' x + y by Lm7;
    x -' (z -' y) + (z -' y) = x by A2,Def1
      .= (x + z) -' z by Lm5
      .= (x + (y + (z -' y))) -' z by A1,Def1
      .= (x + y + (z -' y)) -' z by ARYTM_2:6
      .= x + y -' z + (z -' y) by A3,Th13;
    hence thesis by ARYTM_2:11;
  end;
  suppose
A4: not z -' y <=' x;
    then
A5: not z <=' x + y by Lm7;
    thus x -' (z -' y) = {} by A4,Def1
      .= x + y -' z by A5,Def1;
  end;
end;

Lm12: z <=' x & y <=' z implies x -' (z -' y) = x -' z + y
proof
  assume that
A1: z <=' x and
A2: y <=' z;
  thus x -' (z -' y) = x + y -' z by A2,Lm11
    .= x -' z + y by A1,Th13;
end;

Lm13: x <=' z & y <=' z implies x -' (z -' y) = y -' (z -' x)
proof
  assume that
A1: x <=' z and
A2: y <=' z;
  thus x -' (z -' y) = x + y -' z by A2,Lm11
    .= y -' (z -' x) by A1,Lm11;
end;

theorem
  z <=' x & y <=' z implies x -' z + y = x -' (z -' y)
proof
  assume that
A1: z <=' x and
A2: y <=' z;
  thus x -' (z -' y) = x + y -' z by A2,Lm11
    .= x -' z + y by A1,Th13;
end;

theorem
  y <=' x & y <=' z implies z -' y + x = x -' y + z
proof
  assume that
A1: y <=' x and
A2: y <=' z;
  z -' y + x + y = z -' y + y + x by ARYTM_2:6
    .= z + x by A2,Def1
    .= x -' y + y + z by A1,Def1
    .= x -' y + z + y by ARYTM_2:6;
  hence thesis by ARYTM_2:11;
end;

theorem
  x <=' y implies z -' y <=' z -' x
proof
  assume
A1: x <=' y;
  per cases;
  suppose
A2: y <=' z;
    z -' y + x <=' z -' y + y by A1,Th7;
    then
A3: z -' y + x <=' z by A2,Def1;
    x <=' z by A1,A2,Th3;
    then z -' y + x <=' z -' x + x by A3,Def1;
    hence thesis by Th7;
  end;
  suppose
    not y <=' z;
    then z -' y = {} by Def1;
    hence thesis by Th6;
  end;
end;

theorem
  x <=' y implies x -' z <=' y -' z
proof
  assume
A1: x <=' y;
  per cases;
  suppose
A2: z <=' x;
    then z <=' y by A1,Th3;
    then
A3: y -' z + z = y by Def1;
    x -' z + z = x by A2,Def1;
    hence thesis by A1,A3,Th7;
  end;
  suppose
    not z <=' x;
    then x -' z = {} by Def1;
    hence thesis by Th6;
  end;
end;

Lm14: x *' (y -' z) = (x *' y) -' (x *' z)
proof
  per cases;
  suppose
A1: z <=' y;
    then
A2: x *' z <=' x *' y by Th8;
    x *' (y -' z) + (x *' z) = x *' (y -' z + z) by ARYTM_2:13
      .= x *' y by A1,Def1
      .= (x *' y) -' (x *' z) + (x *' z) by A2,Def1;
    hence thesis by ARYTM_2:11;
  end;
  suppose
A3: x = {};
    then x *' y = {} by ARYTM_2:4;
    hence x *' (y -' z) = x *' y by A3,ARYTM_2:4
      .= (x *' y) -' (x *' z) by A3,Lm4,ARYTM_2:4;
  end;
  suppose
A4: not z <=' y & x <> {};
    then
A5: not x *' z <=' x *' y by Lm2;
    y -' z = {} by A4,Def1;
    hence x *' (y -' z) = {} by ARYTM_2:4
      .= (x *' y) -' (x *' z) by A5,Def1;
  end;
end;

definition
  let x,y be Element of REAL+;
  func x - y equals
  :Def2:
  x -' y if y <=' x otherwise [{},y -' x];
  correctness;
end;

theorem
  x - x = {}
proof
  x <=' x;
  then x - x = x -' x by Def2;
  hence thesis by Lm3;
end;

theorem
  x = {} & y <> {} implies x - y = [{},y]
proof
  assume that
A1: x = {} and
A2: y <> {};
  x <=' y by A1,Th6;
  then not y <=' x by A1,A2,Th4;
  hence x - y =[{},y -' x] by Def2
    .= [{},y] by A1,Lm4;
end;

theorem
  z <=' y implies x + (y -' z) = x + y - z
proof
  assume
A1: z <=' y;
  y <=' x + y by ARYTM_2:19;
  then z <=' x + y by A1,Th3;
  then x + y - z = x + y -' z by Def2;
  hence thesis by A1,Th13;
end;

theorem
  not z <=' y implies x - (z -' y) = x + y - z
proof
  assume
A1: not z <=' y;
  per cases;
  suppose
A2: z -' y <=' x;
    then z <=' x + y by Lm7;
    then
A3: x + y - z = x + y -' z by Def2;
    x - (z -' y) = x -' (z -' y) by A2,Def2;
    hence thesis by A1,A3,Lm11;
  end;
  suppose
A4: not z -' y <=' x;
    then
A5: not z <=' x + y by Lm7;
    (z -' y) -' x = z -' (x + y) by Lm9;
    hence x - (z -' y) = [{},z -' (x + y)] by A4,Def2
      .= x + y - z by A5,Def2;
  end;
end;

theorem
  y <=' x & not y <=' z implies x - (y -' z) = x -' y + z
proof
  assume that
A1: y <=' x and
A2: not y <=' z;
  y -' z <=' y by Th11;
  then y -' z <=' x by A1,Th3;
  then x - (y -' z) = x -' (y -' z) by Def2;
  hence thesis by A1,A2,Lm12;
end;

theorem
  not y <=' x & not y <=' z implies x - (y -' z) = z - (y -' x)
proof
  assume
A1: ( not y <=' x)& not y <=' z;
  per cases;
  suppose
A2: y <=' x + z;
    then y -' x <=' z by Lm7;
    then
A3: z - (y -' x) = z -' (y -' x) by Def2;
    y -' z <=' x by A2,Lm7;
    then x - (y -' z) = x -' (y -' z) by Def2;
    hence thesis by A1,A3,Lm13;
  end;
  suppose
A4: not y <=' x + z;
    then
A5: not y -' x <=' z by Lm7;
A6: y -' z -' x = y -' x -' z by Lm10;
    not y -' z <=' x by A4,Lm7;
    hence x - (y -' z) = [{},y -' x -' z] by A6,Def2
      .= z - (y -' x) by A5,Def2;
  end;
end;

theorem
  y <=' x implies x - (y + z) = x -' y - z
proof
  assume
A1: y <=' x;
  per cases;
  suppose
A2: y + z <=' x;
    then z <=' x -' y by A1,Lm8;
    then
A3: x -' y - z = x -' y -' z by Def2;
    x - (y + z) = x -' (y + z) by A2,Def2;
    hence thesis by A3,Lm9;
  end;
  suppose that
A4: not y + z <=' x and
A5: x <=' y;
A6: not z <=' x -' y by A1,A4,Lm8;
A7: (x + z) -' x = z by Lm5
      .= z -' (x -' x) by Lm3,Lm4;
    x = y by A1,A5,Th4;
    hence x - (y + z) = [{},z -' (x -' y)] by A4,A7,Def2
      .= x -' y - z by A6,Def2;
  end;
  suppose that
A8: not y + z <=' x and
A9: not x <=' y;
A10: not z <=' x -' y by A1,A8,Lm8;
    y + z -' x = z -' (x -' y) by A9,Lm11;
    hence x - (y + z) = [{},z -' (x -' y)] by A8,Def2
      .= x -' y - z by A10,Def2;
  end;
end;

theorem
  x <=' y & z <=' y implies y -' z - x = y -' x - z
proof
  assume that
A1: x <=' y and
A2: z <=' y;
  per cases;
  suppose
A3: x + z <=' y;
    then z <=' y -' x by A1,Lm8;
    then
A4: y -' x -' z = y -' x - z by Def2;
    x <=' y -' z by A2,A3,Lm8;
    then y -' z -' x = y -' z - x by Def2;
    hence thesis by A4,Lm10;
  end;
  suppose that
A5: not x + z <=' y and
A6: y <=' x;
A7: not x <=' y -' z by A2,A5,Lm8;
A8: not z <=' y -' x by A1,A5,Lm8;
A9: x = y by A1,A6,Th4;
    then x -' (x -' z) = z by A2,Lm6
      .= z -' (x -' x) by Lm3,Lm4;
    hence y -' z - x = [{},z -' (y -' x)] by A7,A9,Def2
      .= y -' x - z by A8,Def2;
  end;
  suppose that
A10: not x + z <=' y and
A11: y <=' z;
A12: not x <=' y -' z by A2,A10,Lm8;
A13: not z <=' y -' x by A1,A10,Lm8;
A14: z = y by A2,A11,Th4;
    x -' (z -' z) = x by Lm3,Lm4
      .= z -' (z -' x) by A1,A14,Lm6;
    hence y -' z - x = [{},z -' (y -' x)] by A12,A14,Def2
      .= y -' x - z by A13,Def2;
  end;
  suppose that
A15: not x + z <=' y and
A16: not y <=' x & not y <=' z;
A17: not z <=' y -' x by A1,A15,Lm8;
    ( not x <=' y -' z)& x -' (y -' z) = z -' (y -' x) by A15,A16,Lm8,Lm13;
    hence y -' z - x = [{},z -' (y -' x)] by Def2
      .= y -' x - z by A17,Def2;
  end;
end;

theorem
  z <=' y implies x *' (y -' z) = (x *' y) - (x *' z)
proof
  assume z <=' y;
  then x *' z <=' x *' y by Th8;
  then (x *' y) - (x *' z) = (x *' y) -' (x *' z) by Def2;
  hence thesis by Lm14;
end;

theorem Th27:
  not z <=' y & x <> {} implies [{},x *' (z -' y)] = (x *' y) - (x *' z)
proof
  assume ( not z <=' y)& x <> {};
  then
A1: not x *' z <=' x *' y by Lm2;
  thus [{},x *' (z -' y)] = [{},(x *' z) -' (x *' y)] by Lm14
    .= (x *' y) - (x *' z) by A1,Def2;
end;

theorem
  y -' z <> {} & z <=' y & x <> {} implies (x *' z) - (x *' y) = [{},x
  *' (y -' z)]
proof
  assume y -' z <> {};
  then
A1: y <> z by Lm3;
  assume z <=' y;
  then not y <=' z by A1,Th4;
  hence thesis by Th27;
end;














begin

Lm1: omega c= ({[c,d] where c,d is Element of omega: c,d are_relative_prime &
d <> {}} \ {[k,1] where k is Element of omega: not contradiction}) \/ omega by
XBOOLE_1:7;

notation
  synonym NAT for omega;
  synonym 0 for {};
end;

Lm2: 1 = succ 0;

definition
  func REAL equals
  REAL+ \/ [:{0},REAL+:] \ {[0,0]};
  coherence;
end;

Lm3: REAL+ c= REAL
proof
  REAL+ c= REAL+ \/ [:{0},REAL+:] by XBOOLE_1:7;
  hence thesis by ARYTM_2:3,ZFMISC_1:34;
end;

registration
  cluster REAL -> non empty;
  coherence by Lm3,XBOOLE_1:3;
end;

definition
  func COMPLEX equals
  Funcs({0,1},REAL) \ { x where x is Element of Funcs({0,1
  },REAL): x.1 = 0} \/ REAL;
  coherence;
  func RAT equals
  RAT+ \/ [:{0},RAT+:] \ {[0,0]};
  coherence;
  func INT equals
  NAT \/ [:{0},NAT:] \ {[0,0]};
  coherence;
  redefine func NAT -> Subset of REAL;
  coherence by Lm3,ARYTM_2:2,XBOOLE_1:1;
end;

Lm4: RAT+ c= RAT
proof
  RAT+ c= RAT+ \/ [:{0},RAT+:] by XBOOLE_1:7;
  hence thesis by ARYTM_3:61,ZFMISC_1:34;
end;

Lm5: NAT c= INT
proof
  NAT c= NAT \/ [:{0},NAT:] by XBOOLE_1:7;
  hence thesis by ARYTM_3:32,ZFMISC_1:34;
end;

registration
  cluster COMPLEX -> non empty;
  coherence;
  cluster RAT -> non empty;
  coherence by Lm4,XBOOLE_1:3;
  cluster INT -> non empty;
  coherence by Lm5,XBOOLE_1:3;
end;

reserve i,j,k for Element of NAT;
reserve a,b for Element of REAL;

Lm6: for x,y,z being set st [x,y] = {z} holds z = {x} & x = y
proof
  let x,y,z be set;
  assume
A1: [x,y] = {z};
  then {x} in {z} by TARSKI:def 2;
  hence
A2: z = {x} by TARSKI:def 1;
  {x,y} in {z} by A1,TARSKI:def 2;
  then {x,y} = z by TARSKI:def 1;
  hence thesis by A2,ZFMISC_1:5;
end;

Lm7: not (0,one)-->(a,b) in REAL
proof
  set IR = { A where A is Subset of RAT+: for r being Element of RAT+ st r in
  A holds (for s being Element of RAT+ st s <=' r holds s in A) & ex s being
  Element of RAT+ st s in A & r < s};
  set f = (0,one)-->(a,b);
A1: now
    f = {[0,a],[one,b]} by FUNCT_4:67;
    then
A2: [one,b] in f by TARSKI:def 2;
    assume f in [:{{}},REAL+:];
    then consider x,y being set such that
A3: x in {{}} and
    y in REAL+ and
A4: f = [x,y] by ZFMISC_1:84;
A5: x = 0 by A3,TARSKI:def 1;
    per cases by A4,A5,A2,TARSKI:def 2;
    suppose
      {{one,b},{one}} = {0};
      then 0 in {{one,b},{one}} by TARSKI:def 1;
      hence contradiction by TARSKI:def 2;
    end;
    suppose
      {{one,b},{one}} = {0,y};
      then 0 in {{one,b},{one}} by TARSKI:def 2;
      hence contradiction by TARSKI:def 2;
    end;
  end;
A6: f = {[0,a],[one,b]} by FUNCT_4:67;
  now
    assume f in {[i,j]: i,j are_relative_prime & j <> {}};
    then consider i,j such that
A7: f = [i,j] and
    i,j are_relative_prime and
    j <> {};
A8: {i} in f & {i,j} in f by A7,TARSKI:def 2;
A9: now
      assume i = j;
      then {i} = {i,j} by ENUMSET1:29;
      then
A10:  [i,j] = {{i}} by ENUMSET1:29;
      then [one,b] in {{i}} by A6,A7,TARSKI:def 2;
      then
A11:  [one,b] = {i} by TARSKI:def 1;
      [0,a] in {{i}} by A6,A7,A10,TARSKI:def 2;
      then [0,a] = {i} by TARSKI:def 1;
      hence contradiction by A11,XTUPLE_0:1;
    end;
    per cases by A6,A8,TARSKI:def 2;
    suppose
      {i,j} = [0,a] & {i} = [0,a];
      hence contradiction by A9,ZFMISC_1:5;
    end;
    suppose that
A12:  {i,j} = [0,a] and
A13:  {i} = [one,b];
      i in [0,a] by A12,TARSKI:def 2;
      then i = {0,a} or i = {0} by TARSKI:def 2;
      then
A14:  0 in i by TARSKI:def 1,def 2;
      i = {one} by A13,Lm6;
      hence contradiction by A14,TARSKI:def 1;
    end;
    suppose that
A15:  {i,j} = [one,b] and
A16:  {i} = [0,a];
      i in [one,b] by A15,TARSKI:def 2;
      then i = {one,b} or i = {one} by TARSKI:def 2;
      then
A17:  one in i by TARSKI:def 1,def 2;
      i = {0} by A16,Lm6;
      hence contradiction by A17,TARSKI:def 1;
    end;
    suppose
      {i,j} = [one,b] & {i} = [one,b];
      hence contradiction by A9,ZFMISC_1:5;
    end;
  end;
  then
A18: not f in {[i,j]: i,j are_relative_prime & j <> {}} \ {[k,one]: not
  contradiction};
  not ex x,y being set st {[0,x],[one,y]} in IR
  proof
    given x,y being set such that
A19: {[0,x],[one,y]} in IR;
    consider A being Subset of RAT+ such that
A20: {[0,x],[one,y]} = A and
A21: for r being Element of RAT+ st r in A holds (for s being Element
of RAT+ st s <=' r holds s in A) & ex s being Element of RAT+ st s in A & r < s
    by A19;
    [0,x] in A & for r,s being Element of RAT+ st r in A & s <=' r holds
    s in A by A20,A21,TARSKI:def 2;
    then consider r1,r2,r3 being Element of RAT+ such that
A22: r1 in A and
A23: r2 in A and
A24: r3 in A & r1 <> r2 & r2 <> r3 & r3 <> r1 by ARYTM_3:75;
A25: r2 = [0,x] or r2 = [one,y] by A20,A23,TARSKI:def 2;
    r1 = [0,x] or r1 = [one,y] by A20,A22,TARSKI:def 2;
    hence contradiction by A20,A24,A25,TARSKI:def 2;
  end;
  then
A26: not f in DEDEKIND_CUTS by A6,ARYTM_2:def 1;
  now
    assume f in omega;
    then {} in f by ORDINAL3:8;
    hence contradiction by A6,TARSKI:def 2;
  end;
  then not f in RAT+ by A18,XBOOLE_0:def 3;
  then not f in REAL+ by A26,ARYTM_2:def 2,XBOOLE_0:def 3;
  hence thesis by A1,XBOOLE_0:def 3;
end;

definition
  redefine func 0 -> Element of NAT;
  coherence by ORDINAL1:def 11;
end;

theorem Th1:
  REAL c< COMPLEX
proof
  set X = { x where x is Element of Funcs({0,one},REAL): x.one = 0};
  thus REAL c= COMPLEX by XBOOLE_1:7;
A1: now
    assume (0,1) --> (0,1) in X;
    then
    ex x being Element of Funcs({0,one},REAL) st x = (0,1) --> (0,1) & x.
    one = 0;
    hence contradiction by FUNCT_4:63;
  end;
  REAL+ c= REAL+ \/ [:{{}},REAL+:] by XBOOLE_1:7;
  then
A2: REAL+ c= REAL by ARYTM_2:3,ZFMISC_1:34;
  then reconsider z = 0, j = 1 as Element of REAL by ARYTM_2:20;
A3: not (0,1) --> (z,j) in REAL by Lm7;
  rng (0,1) --> (0,1) c= {0,1} & {0,1} c= REAL by A2,ARYTM_2:20,FUNCT_4:62
,ZFMISC_1:32;
  then dom (0,1) --> (0,1) = {0,1} & rng (0,1) --> (0,1) c= REAL by FUNCT_4:62
,XBOOLE_1:1;
  then (0,1) --> (0,1) in Funcs({0,one},REAL) by FUNCT_2:def 2;
  then (0,1) --> (0,1) in Funcs({0,one},REAL) \ X by A1,XBOOLE_0:def 5;
  hence thesis by A3,XBOOLE_0:def 3;
end;

Lm8: RAT c= REAL
proof
  [:{0},RAT+:] c= [:{0},REAL+:] by ARYTM_2:1,ZFMISC_1:95;
  then RAT+ \/ [:{0},RAT+:] c= REAL+ \/ [:{0},REAL+:] by ARYTM_2:1,XBOOLE_1:13;
  hence thesis by XBOOLE_1:33;
end;

reserve r,s,t for Element of RAT+;
reserve i,j,k for Element of omega;

Lm9: for i,j being ordinal Element of RAT+ st i in j holds i < j
proof
  let i,j be ordinal Element of RAT+;
A1: j in omega by ARYTM_3:31;
  i in omega by ARYTM_3:31;
  then reconsider x = i, y = j as Element of REAL+ by A1,ARYTM_2:2;
  assume
A2: i in j;
  then x <=' y by A1,ARYTM_2:18;
  then
A3: ex x9,y9 being Element of RAT+ st x = x9 & y = y9 & x9 <=' y9 by
ARYTM_2:def 5;
  i <> j by A2;
  hence thesis by A3,ARYTM_3:66;
end;

Lm10: for i,j being ordinal Element of RAT+ st i c= j holds i <='j
proof
  let i,j be ordinal Element of RAT+;
  assume i c= j;
  then consider C being ordinal number such that
A1: j = i+^C by ORDINAL3:27;
  C in omega by A1,ORDINAL3:74;
  then reconsider C as Element of RAT+ by Lm1;
  j = i + C by A1,ARYTM_3:58;
  hence thesis by ARYTM_3:def 13;
end;

Lm11: 2 = succ 1 .= succ 0 \/ {1} by ORDINAL1:def 1
  .= 0 \/ {0} \/ {1} by ORDINAL1:def 1
  .= {0,1} by ENUMSET1:1;

Lm12: for i,k being natural Ordinal st i *^ i = 2 *^ k ex k being natural
Ordinal st i = 2 *^ k
proof
  let i,k be natural Ordinal;
  assume
A1: i *^ i = 2 *^ k;
  set id2 = i div^ 2;
  {} in 2 by ORDINAL1:14;
  then
A2: i mod^ 2 in 2 by ARYTM_3:6;
  per cases by A2,Lm11,TARSKI:def 2;
  suppose
A3: i mod^ 2 = 0;
    take k = id2;
    thus i = k*^2+^0 by A3,ORDINAL3:65
      .= 2 *^ k by ORDINAL2:27;
  end;
  suppose
    i mod^ 2 = 1;
    then i = id2*^2+^1 by ORDINAL3:65;
    then
A4: i *^ i = id2*^2*^ (id2*^2+^1) +^ one *^ (id2*^2+^1) by ORDINAL3:46
      .= id2*^2*^ (id2*^2+^1) +^ (one *^(id2*^2)+^one *^1) by ORDINAL3:46
      .= id2*^2*^ (id2*^2+^1) +^ (one *^(id2*^2)+^ 1)by ORDINAL2:39
      .= id2*^2*^ (id2*^2+^1) +^ one *^(id2*^2)+^ 1 by ORDINAL3:30
      .= id2*^2*^ (id2*^2+^1 +^ one) +^ 1 by ORDINAL3:46
      .= id2*^ (id2*^2+^1 +^ one)*^2 +^ 1 by ORDINAL3:50;
A5: 1 divides 2 by ARYTM_3:9;
    2 divides id2*^ (id2*^2+^1 +^ one)*^2 & 2 divides i *^ i by A1,
ARYTM_3:def 3;
    then 2 divides 1 by A4,ARYTM_3:11;
    hence thesis by A5,ARYTM_3:8;
  end;
end;

1 in omega;

then reconsider a9 = 1 as Element of RAT+ by Lm1;

2 in omega;

then reconsider two = 2 as ordinal Element of RAT+ by Lm1;

Lm13: one + one = two
proof
  1 +^ 1 = succ(1 +^ {}) by Lm2,ORDINAL2:28
    .= succ 1 by ORDINAL2:27
    .= two;
  hence thesis by ARYTM_3:58;
end;

Lm14: for i being Element of RAT+ holds i + i = two *' i
proof
  let i be Element of RAT+;
  thus i + i = one *' i + i by ARYTM_3:53
    .= one *' i + one *' i by ARYTM_3:53
    .= two *' i by Lm13,ARYTM_3:57;
end;

theorem Th2:
  RAT c< REAL
proof
  defpred P[Element of RAT+] means $1 *' $1 < two;
  set X = { s : P[s] };
  reconsider X as Subset of RAT+ from DOMAIN_1:sch 7;
A1: 2 *^ 2 = two *' two & 1*^2 = 2 by ARYTM_3:59,ORDINAL2:39;
  2 = succ 1 .= 1 \/ {1} by ORDINAL1:def 1;
  then
A2: a9 <=' two by Lm10,XBOOLE_1:7;
  then
A3: a9 < two by ARYTM_3:68;
A4: a9 *' a9 = a9 by ARYTM_3:53;
  then
A5: 1 in X by A3;
A6: for r,t st r in X & t <=' r holds t in X
  proof
    let r,t;
    assume r in X;
    then
A7: ex s st r = s & s *' s < two;
    assume t <=' r;
    then t *' t <=' t *' r & t *' r <=' r *' r by ARYTM_3:82;
    then t *' t <=' r *' r by ARYTM_3:67;
    then t *' t < two by A7,ARYTM_3:69;
    hence thesis;
  end;
  then
A8: 0 in X by A5,ARYTM_3:64;
  now
    assume X = [0,0];
    then X = {{0}, {0}} by ENUMSET1:29
      .= {{0}} by ENUMSET1:29;
    hence contradiction by A8,TARSKI:def 1;
  end;
  then
A9: not X in {[0,0]} by TARSKI:def 1;
  reconsider 09 = 0 as Element of RAT+ by Lm1;
  set DD = { A where A is Subset of RAT+: r in A implies (for s st s <=' r
  holds s in A) & ex s st s in A & r < s };
  consider half being Element of RAT+ such that
A10: a9 = two*'half by ARYTM_3:55;
A11: one <=' two by Lm13,ARYTM_3:def 13;
  then
A12: one < two by ARYTM_3:68;
A13: now
    assume X in {{ s: s < t}: t <> 0};
    then consider t0 being Element of RAT+ such that
A14: X = { s: s < t0} and
A15: t0 <> 0;
    set n = numerator t0, d = denominator t0;
    now
      assume
A16:  t0 *' t0 <> two;
      per cases by A16,ARYTM_3:66;
      suppose
        t0 *' t0 < two;
        then t0 in X;
        then ex s st s = t0 & s < t0 by A14;
        hence contradiction;
      end;
      suppose
A17:    two < t0 *' t0;
        consider es being Element of RAT+ such that
A18:    two + es = t0 *' t0 or t0 *' t0 + es = two by ARYTM_3:92;
A19:    now
          assume 09 = es;
          then two + es = two by ARYTM_3:50;
          hence contradiction by A17,A18,ARYTM_3:def 13;
        end;
        09 <=' es by ARYTM_3:64;
        then 09 < es by A19,ARYTM_3:68;
        then consider s such that
A20:    09 < s and
A21:    s < es by ARYTM_3:93;
        now
          per cases;
          suppose
A22:        s < one;
A23:        s <> 0 by A20;
            then s *' s < s *' one by A22,ARYTM_3:80;
            then
A24:        s *' s < s by ARYTM_3:53;
A25:        now
              assume
A26:          t0 <=' one;
              then t0 *' t0 <=' t0 *' one by ARYTM_3:82;
              then t0 *' t0 <=' t0 by ARYTM_3:53;
              then t0 *' t0 <=' one by A26,ARYTM_3:67;
              hence contradiction by A11,A17,ARYTM_3:69;
            end;
            then
A27:        one *' one < one *' t0 by ARYTM_3:80;
            one *' t0 < two *' t0 by A12,A15,ARYTM_3:80;
            then
A28:        one *' one < two *' t0 by A27,ARYTM_3:70;
            consider t02s2 being Element of RAT+ such that
A29:        s *' s + t02s2 = t0 *' t0 or t0 *' t0 + t02s2 = s *' s by
ARYTM_3:92;
            s < t0 by A22,A25,ARYTM_3:70;
            then
A30:        s *' s < t0 by A24,ARYTM_3:70;
            consider 2t9 being Element of RAT+ such that
A31:        (two *' t0) *' 2t9 = one by A15,ARYTM_3:55,78;
            set x = s *' s *' 2t9;
            consider t0x being Element of RAT+ such that
A32:        x + t0x = t0 or t0 + t0x = x by ARYTM_3:92;
            x *' (two *' t0) = s *' s *' one by A31,ARYTM_3:52;
            then x <=' s *' s or two *' t0 <=' one by ARYTM_3:83;
            then
A33:        x < t0 by A28,A30,ARYTM_3:53,69;
            then
A34:        t0x <=' t0 by A32,ARYTM_3:def 13;
A35:        x *' t0x + x *' t0 + x *' x = x *' t0x + x *' x + x *' t0 by
ARYTM_3:51
              .= x *' t0 + x *' t0 by A32,A33,ARYTM_3:57,def 13
              .= x *' t0 *' one + x *' t0 by ARYTM_3:53
              .= x *' t0 *' one + x *' t0 *' one by ARYTM_3:53
              .= t0 *' x *' two by Lm13,ARYTM_3:57
              .= x *' (t0 *' two) by ARYTM_3:52
              .= s *' s *' one by A31,ARYTM_3:52
              .= s *' s by ARYTM_3:53;
            es <=' t0 *' t0 by A17,A18,ARYTM_3:def 13;
            then s < t0 *' t0 by A21,ARYTM_3:69;
            then
A36:        s *' s < t0 *' t0 by A24,ARYTM_3:70;
            then t02s2 + x *' x + s *' s = (t0x + x) *' t0 + x *' x by A29,A32
,A33,ARYTM_3:51,def 13
              .= t0x *' (t0x + x) + x *' t0 + x *' x by A32,A33,ARYTM_3:57
,def 13
              .= t0x *' t0x + x *' t0x + x *' t0 + x *' x by ARYTM_3:57
              .= t0x *' t0x + x *' t0x + (x *' t0 + x *' x) by ARYTM_3:51
              .= t0x *' t0x + (x *' t0x + (x *' t0 + x *' x)) by ARYTM_3:51
              .= t0x *' t0x + s *' s by A35,ARYTM_3:51;
            then t0x *' t0x = t02s2 + x *' x by ARYTM_3:62;
            then
A37:        t02s2 <=' t0x *' t0x by ARYTM_3:def 13;
            now
              assume
A38:          x = 0;
              per cases by A38,ARYTM_3:78;
              suppose
                s *' s = 0;
                hence contradiction by A23,ARYTM_3:78;
              end;
              suppose
                2t9 = 0;
                hence contradiction by A31,ARYTM_3:48;
              end;
            end;
            then t0x <> t0 by A32,A33,ARYTM_3:84,def 13;
            then t0x < t0 by A34,ARYTM_3:68;
            then t0x in X by A14;
            then
A39:        ex s st s = t0x & s *' s < two;
            s *' s < es by A21,A24,ARYTM_3:70;
            then two + s *' s < two + es by ARYTM_3:76;
            then two < t02s2 by A17,A18,A29,A36,ARYTM_3:76,def 13;
            hence contradiction by A37,A39,ARYTM_3:69;
          end;
          suppose
A40:        one <=' s;
            half *' two = one *' one by A10,ARYTM_3:53;
            then
A41:        half <=' one by A12,ARYTM_3:83;
            half <> one by A10,ARYTM_3:53;
            then
A42:        half < one by A41,ARYTM_3:68;
            then half < s by A40,ARYTM_3:69;
            then
A43:        half < es by A21,ARYTM_3:70;
            one <=' two by Lm13,ARYTM_3:def 13;
            then one < two by ARYTM_3:68;
            then
A44:        one *' t0 < two *' t0 by A15,ARYTM_3:80;
A45:        now
              assume
A46:          t0 <=' one;
              then t0 *' t0 <=' t0 *' one by ARYTM_3:82;
              then t0 *' t0 <=' t0 by ARYTM_3:53;
              then t0 *' t0 <=' one by A46,ARYTM_3:67;
              hence contradiction by A11,A17,ARYTM_3:69;
            end;
            then one *' one < one *' t0 by ARYTM_3:80;
            then
A47:        one *' one < two *' t0 by A44,ARYTM_3:70;
            set s = half;
            consider t02s2 being Element of RAT+ such that
A48:        s *' s + t02s2 = t0 *' t0 or t0 *' t0 + t02s2 = s *' s by
ARYTM_3:92;
A49:        half <> 0 by A10,ARYTM_3:48;
            then half *' half < half *' one by A42,ARYTM_3:80;
            then
A50:        half *' half < half by ARYTM_3:53;
            s < t0 by A42,A45,ARYTM_3:70;
            then
A51:        s *' s < t0 by A50,ARYTM_3:70;
            consider 2t9 being Element of RAT+ such that
A52:        (two *' t0) *' 2t9 = one by A15,ARYTM_3:55,78;
            set x = s *' s *' 2t9;
            consider t0x being Element of RAT+ such that
A53:        x + t0x = t0 or t0 + t0x = x by ARYTM_3:92;
            x *' (two *' t0) = s *' s *' one by A52,ARYTM_3:52;
            then x <=' s *' s or two *' t0 <=' one by ARYTM_3:83;
            then
A54:        x < t0 by A47,A51,ARYTM_3:53,69;
            then
A55:        t0x <=' t0 by A53,ARYTM_3:def 13;
A56:        x *' t0x + x *' t0 + x *' x = x *' t0x + x *' x + x *' t0 by
ARYTM_3:51
              .= x *' t0 + x *' t0 by A53,A54,ARYTM_3:57,def 13
              .= x *' t0 *' one + x *' t0 by ARYTM_3:53
              .= x *' t0 *' one + x *' t0 *' one by ARYTM_3:53
              .= t0 *' x *' two by Lm13,ARYTM_3:57
              .= x *' (t0 *' two) by ARYTM_3:52
              .= s *' s *' one by A52,ARYTM_3:52
              .= s *' s by ARYTM_3:53;
            es <=' t0 *' t0 by A17,A18,ARYTM_3:def 13;
            then s < t0 *' t0 by A43,ARYTM_3:69;
            then
A57:        s *' s < t0 *' t0 by A50,ARYTM_3:70;
            then t02s2 + x *' x + s *' s = t0 *' t0 + x *' x by A48,ARYTM_3:51
,def 13
              .= t0x *' (t0x + x) + x *' t0 + x *' x by A53,A54,ARYTM_3:57
,def 13
              .= t0x *' t0x + x *' t0x + x *' t0 + x *' x by ARYTM_3:57
              .= t0x *' t0x + x *' t0x + (x *' t0 + x *' x) by ARYTM_3:51
              .= t0x *' t0x + (x *' t0x + (x *' t0 + x *' x)) by ARYTM_3:51
              .= t0x *' t0x + s *' s by A56,ARYTM_3:51;
            then t0x *' t0x = t02s2 + x *' x by ARYTM_3:62;
            then
A58:        t02s2 <=' t0x *' t0x by ARYTM_3:def 13;
            now
              assume
A59:          x = 0;
              per cases by A59,ARYTM_3:78;
              suppose
                s *' s = 0;
                hence contradiction by A49,ARYTM_3:78;
              end;
              suppose
                2t9 = 0;
                hence contradiction by A52,ARYTM_3:48;
              end;
            end;
            then t0x <> t0 by A53,A54,ARYTM_3:84,def 13;
            then t0x < t0 by A55,ARYTM_3:68;
            then t0x in X by A14;
            then
A60:        ex s st s = t0x & s *' s < two;
            s *' s < es by A50,A43,ARYTM_3:70;
            then two + s *' s < two + es by ARYTM_3:76;
            then two < t02s2 by A17,A18,A48,A57,ARYTM_3:76,def 13;
            hence contradiction by A58,A60,ARYTM_3:69;
          end;
        end;
        hence contradiction;
      end;
    end;
    then
A61: two/1 = (n *^ n)/(d *^ d) by ARYTM_3:40;
    d <> 0 by ARYTM_3:35;
    then d *^ d <> {} by ORDINAL3:31;
    then
A62: two*^(d *^ d) = 1*^(n *^ n) by A61,ARYTM_3:45
      .= n *^ n by ORDINAL2:39;
    then consider k being natural Ordinal such that
A63: n = 2 *^ k by Lm12;
    two*^(d *^ d) = 2 *^ (k *^ (2 *^ k)) by A62,A63,ORDINAL3:50;
    then d *^ d = k *^ (2 *^ k) by ORDINAL3:33
      .= 2 *^ (k *^ k)by ORDINAL3:50;
    then
A64: ex p being natural Ordinal st d = 2 *^ p by Lm12;
    n, d are_relative_prime by ARYTM_3:34;
    hence contradiction by A63,A64,ARYTM_3:def 2;
  end;
  2 = succ 1;
  then 1 in 2 by ORDINAL1:6;
  then
A65: 1 *^ 2 in 2 *^ 2 by ORDINAL3:19;
A66: 09 <=' a9 by ARYTM_3:64;
  now
    let r;
    assume
A67: r in X;
    then
A68: ex s st r = s & s *' s < two;
    thus for t st t <=' r holds t in X by A6,A67;
    per cases;
    suppose
A69:  r = 0;
      take a9;
      thus a9 in X by A4,A3;
      thus r < a9 by A66,A69,ARYTM_3:68;
    end;
    suppose
A70:  r <> 0;
      then consider 3r9 being Element of RAT+ such that
A71:  (r + r + r) *' 3r9 = one by ARYTM_3:54,63;
      consider rr being Element of RAT+ such that
A72:  r *' r + rr = two or two + rr = r *' r by ARYTM_3:92;
      set eps = rr *' 3r9;
A73:  now
        assume
A74:    eps = 0;
        per cases by A74,ARYTM_3:78;
        suppose
          rr = 09;
          then r *' r = two by A72,ARYTM_3:50;
          hence contradiction by A68;
        end;
        suppose
          3r9 = 09;
          hence contradiction by A71,ARYTM_3:48;
        end;
      end;
      now
        per cases;
        suppose
          eps < r;
          then eps *' eps < r *' eps by A73,ARYTM_3:80;
          then
A75:      r *' eps + eps *' r + eps *' eps < r *' eps + eps *' r + r *'
          eps by ARYTM_3:76;
          take t = r + eps;
A76:      t *' t = r *' t + eps *' t by ARYTM_3:57
            .= r *' r + r *' eps + eps *' t by ARYTM_3:57
            .= r *' r + r *' eps + (eps *' r + eps *' eps) by ARYTM_3:57
            .= r *' r + (r *' eps + (eps *' r + eps *' eps)) by ARYTM_3:51
            .= r *' r + (r *' eps + eps *' r + eps *' eps) by ARYTM_3:51;
          r *' eps + eps *' r + r *' eps = eps *' (r + r) + r *' eps by
ARYTM_3:57
            .= eps *' (r + r + r) by ARYTM_3:57
            .= rr *' one by A71,ARYTM_3:52
            .= rr by ARYTM_3:53;
          then t *' t < two by A68,A72,A75,A76,ARYTM_3:76,def 13;
          hence t in X;
          09 <=' eps by ARYTM_3:64;
          then 09 < eps by A73,ARYTM_3:68;
          then r + 09 < r + eps by ARYTM_3:76;
          hence r < t by ARYTM_3:50;
        end;
        suppose
A77:      r <=' eps;
          eps *' (r + r + r) = rr *' one by A71,ARYTM_3:52
            .= rr by ARYTM_3:53;
          then
A78:      r *' (r + r + r) <=' rr by A77,ARYTM_3:82;
          take t = (a9 + half) *' r;
          a9 < two *' one by A3,ARYTM_3:53;
          then half < one by A10,ARYTM_3:82;
          then one + half < two by Lm13,ARYTM_3:76;
          then
A79:      t < two *' r by A70,ARYTM_3:80;
          then
A80:      two *' r *' t < two *' r *' (two *' r) by A70,ARYTM_3:78,80;
          a9 + half <> 0 by ARYTM_3:63;
          then t *' t < two *' r *' t by A70,A79,ARYTM_3:78,80;
          then
A81:      t *' t < two *' r *' (two *' r) by A80,ARYTM_3:70;
          r *' (r + r + r) + r *' r = r *' (r + r) + r *' r + r *' r by
ARYTM_3:57
            .= r *' (r + r) + (r *' r + r *' r) by ARYTM_3:51
            .= r *' (two *' r) + (r *' r + r *' r) by Lm14
            .= r *' (two *' r) + two *' (r *' r) by Lm14
            .= two *' (r *' r) + two *' (r *' r) by ARYTM_3:52
            .= two *' (two *' (r *' r)) by Lm14
            .= two *' (two *' r *' r) by ARYTM_3:52
            .= (two *' r) *' (two *' r) by ARYTM_3:52;
          then two *' r *' (two *' r) <=' two by A68,A72,A78,ARYTM_3:76,def 13;
          then t *' t < two by A81,ARYTM_3:69;
          hence t in X;
          09 <> half & 09 <=' half by A10,ARYTM_3:48,64;
          then 09 < half by ARYTM_3:68;
          then one + 09 < one + half by ARYTM_3:76;
          then one < one + half by ARYTM_3:50;
          then one *' r < t by A70,ARYTM_3:80;
          hence r < t by ARYTM_3:53;
        end;
      end;
      hence ex t st t in X & r < t;
    end;
  end;
  then
A82: X in DD;
  a9 *' half = half by ARYTM_3:53;
  then
A83: half in X by A10,A6,A2,A5,ARYTM_3:82;
A84: now
    assume
A85: X in RAT;
    per cases by A85,XBOOLE_0:def 3;
    suppose
A86:  X in RAT+;
      now
        per cases by A86,XBOOLE_0:def 3;
        suppose
          X in {[i,j]: i,j are_relative_prime & j <> {}} \ {[k,one]:
          not contradiction};
          then X in {[i,j]: i,j are_relative_prime & j <> {}};
          then ex i,j st X = [i,j] & i,j are_relative_prime & j <> {};
          hence contradiction by A8,TARSKI:def 2;
        end;
        suppose
A87:      X in omega;
          2 c= X by A5,A8,Lm11,ZFMISC_1:32;
          then
A88:      not X in 2 by ORDINAL1:5;
          now
            per cases by A87,A88,ORDINAL1:14;
            suppose
              X = two;
              then half = 0 or half = 1 by A83,Lm11,TARSKI:def 2;
              hence contradiction by A10,ARYTM_3:48,53;
            end;
            suppose
              two in X;
              then ex s st s = two & s *' s < two;
              hence contradiction by A1,A65,Lm9;
            end;
          end;
          hence contradiction;
        end;
      end;
      hence contradiction;
    end;
    suppose
      X in [:{0},RAT+:];
      then ex x,y being set st X = [x,y] by RELAT_1:def 1;
      hence contradiction by A8,TARSKI:def 2;
    end;
  end;
  now
    assume two in X;
    then ex s st two = s & s *' s < two;
    hence contradiction by A1,A65,Lm9;
  end;
  then X <> RAT+;
  then not X in {RAT+} by TARSKI:def 1;
  then X in DEDEKIND_CUTS by A82,ARYTM_2:def 1,XBOOLE_0:def 5;
  then X in RAT+ \/ DEDEKIND_CUTS by XBOOLE_0:def 3;
  then X in REAL+ by A13,ARYTM_2:def 2,XBOOLE_0:def 5;
  then X in REAL+ \/ [:{0},REAL+:] by XBOOLE_0:def 3;
  then X in REAL by A9,XBOOLE_0:def 5;
  hence thesis by A84,Lm8,XBOOLE_0:def 8;
end;

theorem Th3:
  RAT c< COMPLEX by Th1,Th2,XBOOLE_1:56;

Lm15: INT c= RAT
proof
  [:{0},NAT:] c= [:{0},RAT+:] by Lm1,ZFMISC_1:95;
  then NAT \/ [:{0},NAT:] c= RAT+ \/ [:{0},RAT+:] by Lm1,XBOOLE_1:13;
  hence thesis by XBOOLE_1:33;
end;

theorem Th4:
  INT c< RAT
proof
  1,2 are_relative_prime
  proof
    let c,d1,d2 be Ordinal such that
A1: 1 = c *^ d1 and
    2 = c *^ d2;
    thus thesis by A1,ORDINAL3:37;
  end;
  then
A2: [1,2] in RAT+ by ARYTM_3:33;
  not 1 in {0} by TARSKI:def 1;
  then ( not [1,2] in NAT)& not [1,2] in [:{0},NAT:] by ARYTM_3:32,ZFMISC_1:87
;
  then not [1,2] in NAT \/ [:{0},NAT:] by XBOOLE_0:def 3;
  then INT <> RAT by A2,Lm4,XBOOLE_0:def 5;
  hence thesis by Lm15,XBOOLE_0:def 8;
end;

theorem Th5:
  INT c< REAL by Th2,Th4,XBOOLE_1:56;

theorem Th6:
  INT c< COMPLEX by Th1,Th5,XBOOLE_1:56;

theorem Th7:
  NAT c< INT
proof
  0 in {0} by TARSKI:def 1;
  then [0,1] in [:{0},NAT:] by ZFMISC_1:87;
  then
A1: [0,1] in NAT \/ [:{0},NAT:] by XBOOLE_0:def 3;
A2: not [0,1] in NAT by ARYTM_3:32;
  [0,1] <> [0,0] by XTUPLE_0:1;
  then not [0,1] in {[0,0]} by TARSKI:def 1;
  then [0,1] in INT by A1,XBOOLE_0:def 5;
  hence thesis by A2,Lm5,XBOOLE_0:def 8;
end;

theorem Th8:
  NAT c< RAT by Th4,Th7,XBOOLE_1:56;

theorem Th9:
  NAT c< REAL by Th2,Th8,XBOOLE_1:56;

theorem Th10:
  NAT c< COMPLEX by Th1,Th9,XBOOLE_1:56;

begin 

theorem
  REAL c= COMPLEX by Th1,XBOOLE_0:def 8;

theorem
  RAT c= REAL by Th2,XBOOLE_0:def 8;

theorem
  RAT c= COMPLEX by Th3,XBOOLE_0:def 8;

theorem
  INT c= RAT by Th4,XBOOLE_0:def 8;

theorem
  INT c= REAL by Th5,XBOOLE_0:def 8;

theorem
  INT c= COMPLEX by Th6,XBOOLE_0:def 8;

theorem
  NAT c= INT by Lm5;

theorem Th18:
  NAT c= RAT by Th8,XBOOLE_0:def 8;

theorem Th19:
  NAT c= REAL;

theorem Th20:
  NAT c= COMPLEX by Th10,XBOOLE_0:def 8;

theorem
  REAL <> COMPLEX by Th1;

theorem
  RAT <> REAL by Th2;

theorem
  RAT <> COMPLEX by Th1,Th2;

theorem
  INT <> RAT by Th4;

theorem
  INT <> REAL by Th2,Th4;

theorem
  INT <> COMPLEX by Th1,Th2,Th4,XBOOLE_1:56;

theorem
  NAT <> INT by Th7;

theorem
  NAT <> RAT by Th4,Th7;

theorem
  NAT <> REAL by Th2,Th4,Th7,XBOOLE_1:56;

theorem
  NAT <> COMPLEX by Th1,Th2,Th8,XBOOLE_1:56;

definition
  func ExtREAL equals
  REAL \/ { REAL, [0,REAL] };
  coherence;
end;

registration
  cluster ExtREAL -> non empty;
  coherence;
end;

theorem Th31:
  REAL c= ExtREAL by XBOOLE_1:7;

theorem Th32:
  REAL <> ExtREAL
proof
  REAL in { REAL, [0,REAL] } by TARSKI:def 2;
  then REAL in ExtREAL by XBOOLE_0:def 3;
  hence thesis;
end;

theorem
  REAL c< ExtREAL by Th31,Th32,XBOOLE_0:def 8;

registration
  cluster INT -> infinite;
  coherence by Lm5,FINSET_1:1;
  cluster RAT -> infinite;
  coherence by Th18,FINSET_1:1;
  cluster REAL -> infinite;
  coherence by Th19,FINSET_1:1;
  cluster COMPLEX -> infinite;
  coherence by Th20,FINSET_1:1;
end;














begin 

Lm1: {} in {{}} by TARSKI:def 1;

theorem Th1:
  REAL+ c= REAL
proof
  REAL+ c= REAL+ \/ [:{{}},REAL+:] by XBOOLE_1:7;
  hence thesis by ARYTM_2:3,NUMBERS:def 1,ZFMISC_1:34;
end;

theorem Th2:
  for x being Element of REAL+ st x <> {} holds [{},x] in REAL
proof
  let x be Element of REAL+ such that
A1: x <> {};
A2: now
    assume [{},x] in {[{},{}]};
    then [{},x] = [{},{}] by TARSKI:def 1;
    hence contradiction by A1,XTUPLE_0:1;
  end;
  {} in {{}} by TARSKI:def 1;
  then [{},x] in [:{{}},REAL+:] by ZFMISC_1:87;
  then [{},x] in REAL+ \/ [:{{}},REAL+:] by XBOOLE_0:def 3;
  hence thesis by A2,NUMBERS:def 1,XBOOLE_0:def 5;
end;

theorem Th3:
  for y being set st [{},y] in REAL holds y <> {}
proof
  let y be set such that
A1: [{},y] in REAL and
A2: y = {};
  [{},y] in {[{},{}]} by A2,TARSKI:def 1;
  hence contradiction by A1,NUMBERS:def 1,XBOOLE_0:def 5;
end;

theorem Th4:
  for x,y being Element of REAL+ holds x - y in REAL
proof
  let x,y be Element of REAL+;
  per cases;
  suppose
    y <=' x;
    then x - y = x -' y by ARYTM_1:def 2;
    then x - y in REAL+;
    hence thesis by Th1;
  end;
  suppose
A1: not y <=' x;
    then x - y = [{},y -' x] by ARYTM_1:def 2;
    hence thesis by A1,Th2,ARYTM_1:9;
  end;
end;

theorem Th5:
  REAL+ misses [:{{}},REAL+:]
proof
  assume REAL+ meets [:{{}},REAL+:];
  then consider x being set such that
A1: x in REAL+ and
A2: x in [:{{}},REAL+:] by XBOOLE_0:3;
  consider x1,x2 being set such that
A3: x1 in {{}} and
  x2 in REAL+ and
A4: x = [x1,x2] by A2,ZFMISC_1:84;
  x1 = {} by A3,TARSKI:def 1;
  hence contradiction by A1,A4,ARYTM_2:3;
end;

begin  

theorem Th6:
  for x,y being Element of REAL+ st x - y = {} holds x = y
proof
  let x,y be Element of REAL+;
  assume
A1: x - y = {};
  {} <> [{},y -' x];
  then y <=' x & x -' y = {} by A1,ARYTM_1:def 2;
  hence thesis by ARYTM_1:10;
end;

Lm2:
  not ex a,b being set st 1 = [a,b]
proof
  let a,b be set;
  assume 1 = [a,b];
  then
A1: {{}} = {{a,b},{a}} by ORDINAL3:15,TARSKI:def 5;
  {a} in {{a,b},{a}} by TARSKI:def 2;
  hence contradiction by A1,TARSKI:def 1;
end;

theorem Th7:
  for x,y,z being Element of REAL+ st x <> {} & x *' y = x *' z holds y = z
proof
  let x,y,z be Element of REAL+;
  assume that
A1: x <> {} and
A2: x *' y = x *' z;
  per cases;
  suppose
A3: z <=' y;
    then x *' (y -' z) = (x *' y) - (x *' z) by ARYTM_1:26
      .= {} by A2,ARYTM_1:18;
    then {} = y -' z by A1,ARYTM_1:2
      .= y - z by A3,ARYTM_1:def 2;
    hence thesis by Th6;
  end;
  suppose
A4: y <=' z;
    then x *' (z -' y) = x *' z - x *' y by ARYTM_1:26
      .= {} by A2,ARYTM_1:18;
    then {} = z -' y by A1,ARYTM_1:2
      .= z - y by A4,ARYTM_1:def 2;
    hence thesis by Th6;
  end;
end;

begin 

definition
  let x,y be Element of REAL;

  func +(x,y) -> Element of REAL means
  :Def1:
  ex x9,y9 being Element of REAL+
  st x = x9 & y = y9 & it = x9 + y9 if x in REAL+ & y in REAL+, ex x9,y9 being
Element of REAL+ st x = x9 & y = [0,y9] & it = x9 - y9 if x in REAL+ & y in [:{
0},REAL+:], ex x9,y9 being Element of REAL+ st x = [0,x9] & y = y9 & it = y9 -
x9 if y in REAL+ & x in [:{0},REAL+:] otherwise ex x9,y9 being Element of REAL+
  st x = [0,x9] & y = [0,y9] & it = [0,x9+y9];
  existence
  proof
    hereby
      assume x in REAL+ & y in REAL+;
      then reconsider x9=x, y9=y as Element of REAL+;
      x9 + y9 in REAL+;
      then reconsider IT = x9 + y9 as Element of REAL by Th1;
      take IT,x9,y9;
      thus x = x9 & y = y9 & IT = x9 + y9;
    end;
A1: y in REAL+ \/ [:{0},REAL+:] by NUMBERS:def 1,XBOOLE_0:def 5;
    hereby
      assume x in REAL+;
      then reconsider x9=x as Element of REAL+;
      assume y in [:{0},REAL+:];
      then consider z,y9 being set such that
A2:   z in{0} and
A3:   y9 in REAL+ and
A4:   y = [z,y9] by ZFMISC_1:84;
      reconsider y9 as Element of REAL+ by A3;
      reconsider IT = x9 - y9 as Element of REAL by Th4;
      take IT,x9,y9;
      thus x = x9 & y = [0,y9] & IT = x9 - y9 by A2,A4,TARSKI:def 1;
    end;
    hereby
      assume y in REAL+;
      then reconsider y9=y as Element of REAL+;
      assume x in [:{0},REAL+:];
      then consider z,x9 being set such that
A5:   z in{0} and
A6:   x9 in REAL+ and
A7:   x = [z,x9] by ZFMISC_1:84;
      reconsider x9 as Element of REAL+ by A6;
      reconsider IT = y9 - x9 as Element of REAL by Th4;
      take IT,x9,y9;
      thus x = [0,x9] & y = y9 & IT = y9 - x9 by A5,A7,TARSKI:def 1;
    end;
    assume that
A8: not(x in REAL+ & y in REAL+) and
A9: not(x in REAL+ & y in [:{0},REAL+:]) and
A10: not(y in REAL+ & x in [:{0},REAL+:]);
A11: x in REAL+ \/ [:{0},REAL+:] by NUMBERS:def 1,XBOOLE_0:def 5;
    then x in REAL+ or x in [:{0},REAL+:] by XBOOLE_0:def 3;
    then consider z1,x9 being set such that
A12: z1 in{0} and
A13: x9 in REAL+ and
A14: x = [z1,x9] by A1,A8,A9,XBOOLE_0:def 3,ZFMISC_1:84;
    y in REAL+ or y in [:{0},REAL+:] by A1,XBOOLE_0:def 3;
    then consider z2,y9 being set such that
A15: z2 in{0} and
A16: y9 in REAL+ and
A17: y = [z2,y9] by A11,A8,A10,XBOOLE_0:def 3,ZFMISC_1:84;
    reconsider x9 as Element of REAL+ by A13;
    reconsider y9 as Element of REAL+ by A16;
    x = [0,x9] by A12,A14,TARSKI:def 1;
    then x9 + y9 <> 0 by Th3,ARYTM_2:5;
    then reconsider IT = [0,y9 + x9] as Element of REAL by Th2;
    take IT,x9,y9;
    thus thesis by A12,A14,A15,A17,TARSKI:def 1;
  end;
  uniqueness
  proof
    let IT1,IT2 be Element of REAL;
    thus x in REAL+ & y in REAL+ & (ex x9,y9 being Element of REAL+ st x = x9
& y = y9 & IT1 = x9 + y9) & (ex x9,y9 being Element of REAL+ st x = x9 & y = y9
    & IT2 = x9 + y9) implies IT1 = IT2;
    thus x in REAL+ & y in [:{0},REAL+:] & (ex x9,y9 being Element of REAL+ st
x = x9 & y = [0,y9] &IT1 = x9 - y9) & (ex x99,y99 being Element of REAL+ st x =
    x99 & y = [0,y99] & IT2 = x99 - y99) implies IT1 = IT2 by XTUPLE_0:1;
    thus y in REAL+ & x in [:{0},REAL+:] & (ex x9,y9 being Element of REAL+ st
x = [0,x9] & y = y9 & IT1 = y9 - x9) & (ex x99,y99 being Element of REAL+ st x
    = [0,x99] & y = y99 & IT2 = y99 - x99) implies IT1 = IT2 by XTUPLE_0:1;
    assume that
    not(x in REAL+ & y in REAL+) and
    not(x in REAL+ & y in [:{0},REAL+:]) and
    not(y in REAL+ & x in [:{0},REAL+:]);
    given x9,y9 being Element of REAL+ such that
A18: x = [0,x9] and
A19: y = [0,y9] & IT1 = [0,x9+y9];
    given x99,y99 being Element of REAL+ such that
A20: x = [0,x99] and
A21: y = [0,y99] & IT2 = [0,x99+y99];
    x9 = x99 by A18,A20,XTUPLE_0:1;
    hence thesis by A19,A21,XTUPLE_0:1;
  end;
  consistency by Th5,XBOOLE_0:3;
  commutativity;
  func *(x,y) -> Element of REAL means
  :Def2:
  ex x9,y9 being Element of REAL+
  st x = x9 & y = y9 & it = x9 *' y9 if x in REAL+ & y in REAL+, ex x9,y9 being
  Element of REAL+ st x = x9 & y = [0,y9] & it = [0,x9 *' y9] if x in REAL+ & y
in [:{0},REAL+:] & x <> 0, ex x9,y9 being Element of REAL+ st x = [0,x9] & y =
  y9 & it = [0,y9 *' x9] if y in REAL+ & x in [:{0},REAL+:] & y <> 0, ex x9,y9
being Element of REAL+ st x = [0,x9] & y = [0,y9] & it = y9 *' x9 if x in [:{0}
  ,REAL+:] & y in [:{0},REAL+:] otherwise it = 0;
  existence
  proof
    hereby
      assume x in REAL+ & y in REAL+;
      then reconsider x9=x, y9=y as Element of REAL+;
      x9 *' y9 in REAL+;
      then reconsider IT = x9 *' y9 as Element of REAL by Th1;
      take IT,x9,y9;
      thus x = x9 & y = y9 & IT = x9 *' y9;
    end;
    hereby
      assume x in REAL+;
      then reconsider x9=x as Element of REAL+;
      assume y in [:{0},REAL+:];
      then consider z,y9 being set such that
A22:  z in{0} and
A23:  y9 in REAL+ and
A24:  y = [z,y9] by ZFMISC_1:84;
      reconsider y9 as Element of REAL+ by A23;
      y = [0,y9] by A22,A24,TARSKI:def 1;
      then
A25:  y9 <> 0 by Th3;
      assume x <> 0;
      then reconsider IT = [0,x9 *' y9] as Element of REAL by A25,Th2,ARYTM_1:2
;
      take IT,x9,y9;
      thus x = x9 & y = [0,y9] & IT = [0,x9 *' y9] by A22,A24,TARSKI:def 1;
    end;
    hereby
      assume y in REAL+;
      then reconsider y9=y as Element of REAL+;
      assume x in [:{0},REAL+:];
      then consider z,x9 being set such that
A26:  z in{0} and
A27:  x9 in REAL+ and
A28:  x = [z,x9] by ZFMISC_1:84;
      reconsider x9 as Element of REAL+ by A27;
      x = [0,x9] by A26,A28,TARSKI:def 1;
      then
A29:  x9 <> 0 by Th3;
      assume y <> 0;
      then reconsider IT = [0,y9 *' x9] as Element of REAL by A29,Th2,ARYTM_1:2
;
      take IT,x9,y9;
      thus x = [0,x9] & y = y9 & IT = [0,y9 *' x9] by A26,A28,TARSKI:def 1;
    end;
    hereby
      assume x in [:{0},REAL+:];
      then consider z1,x9 being set such that
A30:  z1 in{0} and
A31:  x9 in REAL+ and
A32:  x = [z1,x9] by ZFMISC_1:84;
      reconsider x9 as Element of REAL+ by A31;
      assume y in [:{0},REAL+:];
      then consider z2,y9 being set such that
A33:  z2 in{0} and
A34:  y9 in REAL+ and
A35:  y = [z2,y9] by ZFMISC_1:84;
      reconsider y9 as Element of REAL+ by A34;
      x9 *' y9 in REAL+;
      then reconsider IT = y9 *' x9 as Element of REAL by Th1;
      take IT,x9,y9;
      thus x = [0,x9] & y = [0,y9] & IT = y9 *' x9 by A30,A32,A33,A35,
TARSKI:def 1;
    end;
    thus thesis;
  end;
  uniqueness
  proof
    let IT1,IT2 be Element of REAL;
    thus x in REAL+ & y in REAL+ & (ex x9,y9 being Element of REAL+ st x = x9
& y = y9 & IT1 = x9 *' y9) & (ex x9,y9 being Element of REAL+ st x = x9 & y =
    y9 & IT2 = x9 *' y9) implies IT1 = IT2;
    thus x in REAL+ & y in [:{0},REAL+:] & x <> 0 & (ex x9,y9 being Element of
REAL+ st x = x9 & y = [0,y9] & IT1 = [0,x9 *' y9]) & (ex x99,y99 being Element
of REAL+ st x = x99 & y = [0,y99] & IT2 = [0,x99 *' y99]) implies IT1 = IT2 by
XTUPLE_0:1;
    thus y in REAL+ & x in [:{0},REAL+:] & y <> 0 & (ex x9,y9 being Element of
REAL+ st x = [0,x9] & y = y9 & IT1 = [0,y9 *' x9]) & (ex x99,y99 being Element
of REAL+ st x = [0,x99] & y = y99 & IT2 = [0,y99 *' x99]) implies IT1 = IT2 by
XTUPLE_0:1;
    hereby
      assume that
      x in [:{0},REAL+:] and
      y in [:{0},REAL+:];
      given x9,y9 being Element of REAL+ such that
A36:  x = [0,x9] and
A37:  y = [0,y9] & IT1 = y9 *' x9;
      given x99,y99 being Element of REAL+ such that
A38:  x = [0,x99] and
A39:  y = [0,y99] & IT2 = y99 *' x99;
      x9 = x99 by A36,A38,XTUPLE_0:1;
      hence IT1 = IT2 by A37,A39,XTUPLE_0:1;
    end;
    thus thesis;
  end;
  consistency by Th5,XBOOLE_0:3;
  commutativity;
end;

reserve x,y for Element of REAL;

definition
  let x be Element of REAL;
  func opp x -> Element of REAL means
  :Def3:
  +(x,it) = 0;
  existence
  proof
    reconsider z9 = 0 as Element of REAL+ by ARYTM_2:20;
A1: x in REAL+ \/ [:{0},REAL+:] by NUMBERS:def 1,XBOOLE_0:def 5;
    per cases by A1,XBOOLE_0:def 3;
    suppose
A2:   x = 0;
      then reconsider x9 = x as Element of REAL+ by ARYTM_2:20;
      take x;
      x9 + x9 = 0 by A2,ARYTM_2:def 8;
      hence thesis by Def1;
    end;
    suppose that
A3:   x in REAL+ and
A4:   x <> 0;
A5:   [0,x] <> [0,0] by A4,XTUPLE_0:1;
      0 in {0} by TARSKI:def 1;
      then
A6:   [0,x] in [:{0},REAL+:] by A3,ZFMISC_1:87;
      then [0,x] in REAL+ \/ [:{0},REAL+:] by XBOOLE_0:def 3;
      then reconsider y = [0,x] as Element of REAL by A5,NUMBERS:def 1
,ZFMISC_1:56;
      reconsider x9 = x as Element of REAL+ by A3;
A7:   x9 <=' x9;
      take y;
      z9 + x9 = x9 by ARYTM_2:def 8;
      then z9 = x9 -' x9 by A7,ARYTM_1:def 1;
      then 0 = x9 - x9 by A7,ARYTM_1:def 2;
      hence thesis by A6,Def1;
    end;
    suppose
A8:   x in [:{0},REAL+:];
      then consider a,b being set such that
A9:   a in {0} and
A10:  b in REAL+ and
A11:  x = [a,b] by ZFMISC_1:84;
      reconsider y = b as Element of REAL by A10,Th1;
      take y;
      now
        per cases;
        case
          x in REAL+ & y in REAL+;
          hence ex x9,y9 being Element of REAL+ st x = x9 & y = y9 & 0 = x9 +
          y9 by A8,Th5,XBOOLE_0:3;
        end;
        case
          x in REAL+ & y in [:{0},REAL+:];
          hence
          ex x9,y9 being Element of REAL+ st x = x9 & y = [0,y9] & 0 = x9
          - y9 by A10,Th5,XBOOLE_0:3;
        end;
        case
          y in REAL+ & x in [:{0},REAL+:];
          reconsider x9 = b, y9 = y as Element of REAL+ by A10;
          take x9, y9;
          thus x = [0,x9] by A9,A11,TARSKI:def 1;
          thus y = y9;
          thus 0 = y9 - x9 by ARYTM_1:18;
        end;
        case
          not(x in REAL+ & y in REAL+) & not(x in REAL+ & y in [:{0},
          REAL+:]) & not (y in REAL+ & x in [:{0},REAL+:]);
          hence ex x9,y9 being Element of REAL+ st x = [0,x9] & y = [0,y9] & 0
          = [0,y9+x9] by A8,A10;
        end;
      end;
      hence thesis by Def1;
    end;
  end;
  uniqueness
  proof
    let y,z be Element of REAL such that
A12: +(x,y) = 0 and
A13: +(x,z) = 0;
    per cases;
    suppose
      x in REAL+ & y in REAL+ & z in REAL+;
      then
      ( ex x9,y9 being Element of REAL+ st x = x9 & y = y9 & 0 = x9 + y9)
& ex x9,y9 being Element of REAL+ st x = x9 & z = y9 & 0 = x9 + y9 by A12,A13
,Def1;
      hence thesis by ARYTM_2:11;
    end;
    suppose that
A14:  x in REAL+ and
A15:  y in REAL+ and
A16:  z in [:{0},REAL+:];
      consider x99,y99 being Element of REAL+ such that
A17:  x = x99 and
A18:  z = [0,y99] & 0 = x99 - y99 by A13,A14,A16,Def1;
      ex x9,y9 being Element of REAL+ st x = x9 & y = y9 & 0 = x9 + y9 by A12
,A14,A15,Def1;
      then
A19:  x99 = 0 by A17,ARYTM_2:5;
      [0,0] in {[0,0]} by TARSKI:def 1;
      then
A20:  not [0,0] in REAL by NUMBERS:def 1,XBOOLE_0:def 5;
      z in REAL;
      hence thesis by A18,A19,A20,ARYTM_1:19;
    end;
    suppose that
A21:  x in REAL+ and
A22:  z in REAL+ and
A23:  y in [:{0},REAL+:];
      consider x99,y9 being Element of REAL+ such that
A24:  x = x99 and
A25:  y = [0,y9] & 0 = x99 - y9 by A12,A21,A23,Def1;
      ex x9,z9 being Element of REAL+ st x = x9 & z = z9 & 0 = x9 + z9 by A13
,A21,A22,Def1;
      then
A26:  x99 = 0 by A24,ARYTM_2:5;
      [0,0] in {[0,0]} by TARSKI:def 1;
      then
A27:  not [0,0] in REAL by NUMBERS:def 1,XBOOLE_0:def 5;
      y in REAL;
      hence thesis by A25,A26,A27,ARYTM_1:19;
    end;
    suppose that
A28:  x in REAL+ and
A29:  y in [:{0},REAL+:] and
A30:  z in [:{0},REAL+:];
      consider x99,z9 being Element of REAL+ such that
A31:  x = x99 and
A32:  z = [0,z9] and
A33:  0 = x99 - z9 by A13,A28,A30,Def1;
      consider x9,y9 being Element of REAL+ such that
A34:  x = x9 and
A35:  y = [0,y9] and
A36:  0 = x9 - y9 by A12,A28,A29,Def1;
      y9 = x9 by A36,Th6
        .= z9 by A34,A31,A33,Th6;
      hence thesis by A35,A32;
    end;
    suppose that
A37:  z in REAL+ and
A38:  y in REAL+ and
A39:  x in [:{0},REAL+:];
      consider x99,z9 being Element of REAL+ such that
A40:  x = [0,x99] and
A41:  z = z9 and
A42:  0 = z9 - x99 by A13,A37,A39,Def1;
      consider x9,y9 being Element of REAL+ such that
A43:  x = [0,x9] and
A44:  y = y9 and
A45:  0 = y9 - x9 by A12,A38,A39,Def1;
      x9 = x99 by A43,A40,XTUPLE_0:1;
      then z9 = x9 by A42,Th6
        .= y9 by A45,Th6;
      hence thesis by A44,A41;
    end;
    suppose
      not(x in REAL+ & y in REAL+) & not(x in REAL+ & y in [:{0},
      REAL+:]) & not(y in REAL+ & x in [:{0},REAL+:]);
      then
      ex x9,y9 being Element of REAL+ st x = [0,x9] & y = [0,y9] & 0 = [0
      ,x9+y9] by A12,Def1;
      hence thesis;
    end;
    suppose
      not(x in REAL+ & z in REAL+) & not(x in REAL+ & z in [:{0},
      REAL+:]) & not(z in REAL+ & x in [:{0},REAL+:]);
      then
      ex x9,z9 being Element of REAL+ st x = [0,x9] & z = [0,z9] & 0 = [0
      ,x9+z9] by A13,Def1;
      hence thesis;
    end;
  end;
  involutiveness;
  func inv x -> Element of REAL means
  :Def4:
  *(x,it) = 1 if x <> 0 otherwise it = 0;
  existence
  proof
    thus x <> 0 implies ex y st *(x,y) = 1
    proof
A46:  x in REAL+ \/ [:{0},REAL+:] by NUMBERS:def 1,XBOOLE_0:def 5;
      assume
A47:  x <> 0;
      per cases by A46,XBOOLE_0:def 3;
      suppose
        x in REAL+;
        then reconsider x9 = x as Element of REAL+;
        consider y9 being Element of REAL+ such that
A48:    x9 *' y9 = 1 by A47,ARYTM_2:14;
        y9 in REAL+;
        then reconsider y = y9 as Element of REAL by Th1;
        take y;
        thus thesis by A48,Def2;
      end;
      suppose
A49:    x in [:{0},REAL+:];
        not x in {[0,0]} by NUMBERS:def 1,XBOOLE_0:def 5;
        then
A50:    x <> [0,0] by TARSKI:def 1;
        consider x1,x2 being set such that
        x1 in {0} and
A51:    x2 in REAL+ and
A52:    x = [x1,x2] by A49,ZFMISC_1:84;
        reconsider x2 as Element of REAL+ by A51;
        x1 in {0} by A49,A52,ZFMISC_1:87;
        then x2 <> 0 by A52,A50,TARSKI:def 1;
        then consider y2 being Element of REAL+ such that
A53:    x2 *' y2 = 1 by ARYTM_2:14;
A54:    now
          assume [0,y2] in {[0,0]};
          then [0,y2] = [0,0] by TARSKI:def 1;
          then y2 = 0 by XTUPLE_0:1;
          hence contradiction by A53,ARYTM_2:4;
        end;
A55:    [0,y2] in [:{0},REAL+:] by Lm1,ZFMISC_1:87;
        then [0,y2] in REAL+ \/ [:{0},REAL+:] by XBOOLE_0:def 3;
        then reconsider y = [0,y2] as Element of REAL by A54,NUMBERS:def 1
,XBOOLE_0:def 5;
        take y;
        consider x9,y9 being Element of REAL+ such that
A56:    x = [0,x9] and
A57:    y = [0,y9] and
A58:    *(x,y) = y9 *' x9 by A49,A55,Def2;
        y9 = y2 by A57,XTUPLE_0:1;
        hence thesis by A52,A53,A56,A58,XTUPLE_0:1;
      end;
    end;
    thus thesis;
  end;
  uniqueness
  proof
    let y,z be Element of REAL;
    thus x <> 0 & *(x,y) = 1 & *(x,z) = 1 implies y = z
    proof
      assume that
A59:  x <> 0 and
A60:  *(x,y) = 1 and
A61:  *(x,z) = 1;
A62:  for x,y being Element of REAL st *(x,y) =1 holds x <> 0
      proof
        let x,y be Element of REAL such that
A63:    *(x,y) =1 and
A64:    x = 0;
A65:    not x in [:{0},REAL+:] by A64,Th5,ARYTM_2:20,XBOOLE_0:3;
A66:    y in REAL+ \/ [:{0},REAL+:] by NUMBERS:def 1,XBOOLE_0:def 5;
        per cases by A66,XBOOLE_0:def 3;
        suppose
          y in REAL+;
          then ex x9,y9 being Element of REAL+ st x = x9 & y = y9 & 1 = x9 *'
          y9 by A63,A64,Def2,ARYTM_2:20;
          hence contradiction by A64,ARYTM_2:4;
        end;
        suppose
          y in [:{0},REAL+:];
          then not y in REAL+ by Th5,XBOOLE_0:3;
          hence contradiction by A63,A64,A65,Def2;
        end;
      end;
      then
A67:  y <> 0 by A60;
A68:  z <> 0 by A61,A62;
      per cases;
      suppose
        x in REAL+ & y in REAL+ & z in REAL+;
        then ( ex x9,y9 being Element of REAL+ st x = x9 & y = y9 & 1 = x9 *'
y9)& ex x9,y9 being Element of REAL+ st x = x9 & z = y9 & 1 = x9 *' y9 by A60
,A61,Def2;
        hence thesis by A59,Th7;
      end;
      suppose that
A69:    x in [:{0},REAL+:] and
A70:    y in [:{0},REAL+:] and
A71:    z in [:{0},REAL+:];
        consider x9,y9 being Element of REAL+ such that
A72:    x = [0,x9] and
A73:    y = [0,y9] & 1 = y9 *' x9 by A60,A69,A70,Def2;
        consider x99,z9 being Element of REAL+ such that
A74:    x = [0,x99] and
A75:    z = [0,z9] & 1 = z9 *' x99 by A61,A69,A71,Def2;
        x9 = x99 by A72,A74,XTUPLE_0:1;
        hence thesis by A72,A73,A75,Th3,Th7;
      end;
      suppose
        x in REAL+ & y in [:{0},REAL+:];
        then ex x9,y9 being Element of REAL+ st x = x9 & y = [0,y9] & 1 = [0,
        x9 *' y9] by A59,A60,Def2;
        hence thesis by Lm2;
      end;
      suppose
        x in [:{0},REAL+:] & y in REAL+;
        then ex x9,y9 being Element of REAL+ st x = [0,x9] & y = y9 & 1 = [0,
        y9 *' x9] by A60,A67,Def2;
        hence thesis by Lm2;
      end;
      suppose
        x in [:{0},REAL+:] & z in REAL+;
        then ex x9,z9 being Element of REAL+ st x = [0,x9] & z = z9 & 1 = [0,
        z9 *' x9] by A61,A68,Def2;
        hence thesis by Lm2;
      end;
      suppose
        x in REAL+ & z in [:{0},REAL+:];
        then ex x9,z9 being Element of REAL+ st x = x9 & z = [0,z9] & 1 = [0,
        x9 *' z9] by A59,A61,Def2;
        hence thesis by Lm2;
      end;
      suppose
        not (x in REAL+ & y in REAL+) & not (x in REAL+ & y in [:{0}
,REAL+:] & x <> 0) & not (y in REAL+ & x in [:{0},REAL+:] & y <> 0) & not (x in
        [:{0},REAL+:] & y in [:{0},REAL+:]);
        hence thesis by A60,Def2;
      end;
      suppose
        not (x in REAL+ & z in REAL+) & not (x in REAL+ & z in [:{0}
,REAL+:] & x <> 0) & not (z in REAL+ & x in [:{0},REAL+:] & z <> 0) & not (x in
        [:{0},REAL+:] & z in [:{0},REAL+:]);
        hence thesis by A61,Def2;
      end;
    end;
    thus thesis;
  end;
  consistency;
  involutiveness
  proof
    let x,y be Element of REAL;
    assume that
A76: y <> 0 implies *(y,x) = 1 and
A77: y = 0 implies x = 0;
    thus x <> 0 implies *(x,y) = 1 by A76,A77;
    assume
A78: x = 0;
A79: y in REAL+ \/ [:{0},REAL+:] by NUMBERS:def 1,XBOOLE_0:def 5;
    assume
A80: y <> 0;
    per cases by A79,XBOOLE_0:def 3;
    suppose
      y in REAL+;
      then ex x9,y9 being Element of REAL+ st x = x9 & y = y9 & 1 = x9 *' y9
      by A76,A78,A80,Def2,ARYTM_2:20;
      hence contradiction by A78,ARYTM_2:4;
    end;
    suppose
A81:  y in [:{0},REAL+:];
A82:  not x in [:{0},REAL+:] by A78,Th5,ARYTM_2:20,XBOOLE_0:3;
      not y in REAL+ by A81,Th5,XBOOLE_0:3;
      hence contradiction by A76,A78,A80,A82,Def2;
    end;
  end;
end;

begin 

Lm3: for x,y,z being set st [x,y] = {z} holds z = {x} & x = y
proof
  let x,y,z be set;
  assume [x,y] = {z};
  then
A1: {{x,y},{x}} = {z} by TARSKI:def 5;
  then {x} in {z} by TARSKI:def 2;
  hence
A2: z = {x} by TARSKI:def 1;
  {x,y} in {z} by A1,TARSKI:def 2;
  then {x,y} = z by TARSKI:def 1;
  hence thesis by A2,ZFMISC_1:5;
end;

reserve i,j,k for Element of NAT;
reserve a,b for Element of REAL;

theorem Th8:
  not (0,1)-->(a,b) in REAL
proof
  set IR = { A where A is Subset of RAT+: for r being Element of RAT+ st r in
  A holds (for s being Element of RAT+ st s <=' r holds s in A) & ex s being
  Element of RAT+ st s in A & r < s};
  set f = (0,1)-->(a,b);
A1: now
    f = {[0,a],[1,b]} by FUNCT_4:67;
    then
A2: [1,b] in f by TARSKI:def 2;
    assume f in [:{{}},REAL+:];
    then consider x,y being set such that
A3: x in {{}} and
    y in REAL+ and
A4: f = [x,y] by ZFMISC_1:84;
    x = 0 by A3,TARSKI:def 1;
    then f = {{0,y},{0}} by A4,TARSKI:def 5;
    then
A5: [1,b] = {0} or [1,b] = {0,y} by A2,TARSKI:def 2;
    per cases by A5,TARSKI:def 5;
    suppose
      {{1,b},{1}} = {0};
      then 0 in {{1,b},{1}} by TARSKI:def 1;
      hence contradiction by TARSKI:def 2;
    end;
    suppose
      {{1,b},{1}} = {0,y};
      then 0 in {{1,b},{1}} by TARSKI:def 2;
      hence contradiction by TARSKI:def 2;
    end;
  end;
A6: f = {[0,a],[1,b]} by FUNCT_4:67;
  now
    assume f in {[i,j]: i,j are_relative_prime & j <> {}};
    then consider i,j such that
A7: f = [i,j] and
    i,j are_relative_prime and
    j <> {};
A8: f = {{i,j},{i}} by A7,TARSKI:def 5;
    then
A9: {i} in f & {i,j} in f by TARSKI:def 2;
A10: now
      assume i = j;
      then {i} = {i,j} by ENUMSET1:29;
      then
A11:  [i,j] = {{i}} by A7,A8,ENUMSET1:29;
      then [1,b] in {{i}} by A6,A7,TARSKI:def 2;
      then
A12:  [1,b] = {i} by TARSKI:def 1;
      [0,a] in {{i}} by A6,A7,A11,TARSKI:def 2;
      then [0,a] = {i} by TARSKI:def 1;
      hence contradiction by A12,XTUPLE_0:1;
    end;
A13: [1,b] = {{1,b},{1}} by TARSKI:def 5;
A14: [0,a] = {{0,a},{0}} by TARSKI:def 5;
    per cases by A6,A9,TARSKI:def 2;
    suppose
      {i,j} = [0,a] & {i} = [0,a];
      hence contradiction by A10,ZFMISC_1:5;
    end;
    suppose that
A15:  {i,j} = [0,a] and
A16:  {i} = [1,b];
      i in [0,a] by A15,TARSKI:def 2;
      then i = {0,a} or i = {0} by A14,TARSKI:def 2;
      then
A17:  0 in i by TARSKI:def 1,def 2;
      i = {1} by A16,Lm3;
      hence contradiction by A17,TARSKI:def 1;
    end;
    suppose that
A18:  {i,j} = [1,b] and
A19:  {i} = [0,a];
      i in [1,b] by A18,TARSKI:def 2;
      then i = {1,b} or i = {1} by A13,TARSKI:def 2;
      then
A20:  1 in i by TARSKI:def 1,def 2;
      i = {0} by A19,Lm3;
      hence contradiction by A20,TARSKI:def 1;
    end;
    suppose
      {i,j} = [1,b] & {i} = [1,b];
      hence contradiction by A10,ZFMISC_1:5;
    end;
  end;
  then
A21: not f in {[i,j]: i,j are_relative_prime & j <> {}} \ {[k,1]: not
  contradiction};
  not ex x,y being set st {[0,x],[1,y]} in IR
  proof
    given x,y being set such that
A22: {[0,x],[1,y]} in IR;
    consider A being Subset of RAT+ such that
A23: {[0,x],[1,y]} = A and
A24: for r being Element of RAT+ st r in A holds (for s being Element
of RAT+ st s <=' r holds s in A) & ex s being Element of RAT+ st s in A & r < s
    by A22;
    [0,x] in A & for r,s being Element of RAT+ st r in A & s <=' r holds
    s in A by A23,A24,TARSKI:def 2;
    then consider r1,r2,r3 being Element of RAT+ such that
A25: r1 in A and
A26: r2 in A and
A27: r3 in A & r1 <> r2 & r2 <> r3 & r3 <> r1 by ARYTM_3:75;
A28: r2 = [0,x] or r2 = [1,y] by A23,A26,TARSKI:def 2;
    r1 = [0,x] or r1 = [1,y] by A23,A25,TARSKI:def 2;
    hence contradiction by A23,A27,A28,TARSKI:def 2;
  end;
  then
A29: not f in DEDEKIND_CUTS by A6,ARYTM_2:def 1;
  now
    assume f in omega;
    then {} in f by ORDINAL3:8;
    hence contradiction by A6,TARSKI:def 2;
  end;
  then not f in RAT+ by A21,XBOOLE_0:def 3;
  then not f in REAL+ by A29,ARYTM_2:def 2,XBOOLE_0:def 3;
  hence thesis by A1,NUMBERS:def 1,XBOOLE_0:def 3;
end;

definition
  let x,y be Element of REAL;

  func [*x,y*] -> Element of COMPLEX equals
  :Def5:
  x if y = 0 otherwise (0,1)
  --> (x,y);
  consistency;
  coherence
  proof
    set z = (0,1)-->(x,y);
    thus y = 0 implies x is Element of COMPLEX by NUMBERS:def 2,XBOOLE_0:def 3;
    assume
A1: y <> 0;
A2: now
      assume z in { r where r is Element of Funcs({0,1},REAL): r.1 = 0 };
      then ex r being Element of Funcs({0,1},REAL) st z = r & r.1 = 0;
      hence contradiction by A1,FUNCT_4:63;
    end;
    z in Funcs({0,1},REAL) by FUNCT_2:8;
    then
    z in Funcs({0,1},REAL) \ { r where r is Element of Funcs({0,1},REAL):
    r.1 = 0} by A2,XBOOLE_0:def 5;
    hence thesis by NUMBERS:def 2,XBOOLE_0:def 3;
  end;
end;

theorem
  for c being Element of COMPLEX ex r,s being Element of REAL st c = [*r ,s*]
proof
  let c be Element of COMPLEX;
  per cases;
  suppose
    c in REAL;
    then reconsider r=c, z=0 as Element of REAL;
    take r,z;
    thus thesis by Def5;
  end;
  suppose
    not c in REAL;
    then
A1: c in Funcs({0,1},REAL) \ { x where x is Element of Funcs({0,1},REAL):
    x.1 = 0} by NUMBERS:def 2,XBOOLE_0:def 3;
    then consider f being Function such that
A2: c = f and
A3: dom f = {0,1} and
A4: rng f c= REAL by FUNCT_2:def 2;
    1 in {0,1} by TARSKI:def 2;
    then
A5: f.1 in rng f by A3,FUNCT_1:3;
    0 in {0,1} by TARSKI:def 2;
    then f.0 in rng f by A3,FUNCT_1:3;
    then reconsider r = f.0, s = f.1 as Element of REAL by A4,A5;
    take r,s;
A6: c = (0,1)-->(r,s) by A2,A3,FUNCT_4:66;
    now
      assume s = 0;
      then (0,1)-->(r,s).1 = 0 by FUNCT_4:63;
      then c in { x where x is Element of Funcs({0,1},REAL): x.1 = 0} by A1,A6;
      hence contradiction by A1,XBOOLE_0:def 5;
    end;
    hence thesis by A6,Def5;
  end;
end;

theorem
  for x1,x2,y1,y2 being Element of REAL st [*x1,x2*] = [*y1,y2*] holds
  x1 = y1 & x2 = y2
proof
  let x1,x2,y1,y2 be Element of REAL such that
A1: [*x1,x2*] = [*y1,y2*];
  per cases;
  suppose
A2: x2 = 0;
    then
A3: [*x1,x2*] = x1 by Def5;
A4: now
      assume y2 <> 0;
      then x1 = (0,1) --> (y1,y2) by A1,A3,Def5;
      hence contradiction by Th8;
    end;
    hence x1 = y1 by A1,A3,Def5;
    thus thesis by A2,A4;
  end;
  suppose
    x2 <> 0;
    then
A5: [*y1,y2*] = (0,1) --> (x1,x2) by A1,Def5;
    now
      assume y2 = 0;
      then [*y1,y2*] = y1 by Def5;
      hence contradiction by A5,Th8;
    end;
    then [*y1,y2*] = (0,1) --> (y1,y2) by Def5;
    hence thesis by A5,FUNCT_4:68;
  end;
end;

set RR = [:{0},REAL+:] \ {[0,0]};
reconsider o = 0 as Element of REAL;

theorem Th11:
  for x,o being Element of REAL st o = 0 holds +(x,o) = x
proof
  reconsider y9 = 0 as Element of REAL+ by ARYTM_2:20;
  let x,o being Element of REAL such that
A1: o = 0;
  per cases;
  suppose
    x in REAL+;
    then reconsider x9 = x as Element of REAL+;
    x9 = x9 + y9 by ARYTM_2:def 8;
    hence thesis by A1,Def1;
  end;
  suppose
A2: not x in REAL+;
    x in REAL+ \/ [:{{}},REAL+:] by NUMBERS:def 1,XBOOLE_0:def 5;
    then
A3: x in [:{{}},REAL+:] by A2,XBOOLE_0:def 3;
    then consider x1,x2 being set such that
A4: x1 in {{}} and
A5: x2 in REAL+ and
A6: x = [x1,x2] by ZFMISC_1:84;
    reconsider x9 = x2 as Element of REAL+ by A5;
A7: x1 = 0 by A4,TARSKI:def 1;
    then x = y9 - x9 by A6,Th3,ARYTM_1:19;
    hence thesis by A1,A3,A6,A7,Def1;
  end;
end;

theorem Th12:
  for x,o being Element of REAL st o = 0 holds *(x,o) = 0
proof
  let x,o being Element of REAL such that
A1: o = 0;
  per cases;
  suppose
    x in REAL+;
    then reconsider x9 = x, y9 = 0 as Element of REAL+ by ARYTM_2:20;
    0 = x9 *' y9 by ARYTM_2:4;
    hence thesis by A1,Def2;
  end;
  suppose
A2: not x in REAL+;
    not o in [:{{}},REAL+:] by A1,Th5,ARYTM_2:20,XBOOLE_0:3;
    hence thesis by A1,A2,Def2;
  end;
end;

theorem Th13:
  for x,y,z being Element of REAL holds *(x,*(y,z)) = *(*(x,y),z)
proof
  let x,y,z be Element of REAL;
  per cases;
  suppose that
A1: x in REAL+ and
A2: y in REAL+ and
A3: z in REAL+;
A4: ex x99,y99 being Element of REAL+ st x = x99 & y = y99 & *(x,y) = x99
    *' y99 by A1,A2,Def2;
    then
A5: ex xy99,z99 being Element of REAL+ st *(x,y) = xy99 & z = z99 & *(*(x,y
    ),z) = xy99 *' z99 by A3,Def2;
A6: ex y9,z9 being Element of REAL+ st y = y9 & z = z9 & *( y,z) = y9 *' z9
    by A2,A3,Def2;
    then
    ex x9,yz9 being Element of REAL+ st x = x9 & *(y,z) = yz9 & *(x,*(y,z))
    = x9 *' yz9 by A1,Def2;
    hence thesis by A6,A4,A5,ARYTM_2:12;
  end;
  suppose that
A7: x in REAL+ & x <> 0 and
A8: y in RR and
A9: z in REAL+ & z <> 0;
    consider y9,z9 being Element of REAL+ such that
A10: y = [0,y9] and
A11: z = z9 and
A12: *(y,z) = [0,z9 *' y9] by A8,A9,Def2;
    *(y,z) in [:{0},REAL+:] by A12,Lm1,ZFMISC_1:87;
    then consider x9,yz9 being Element of REAL+ such that
A13: x = x9 and
A14: *(y,z) = [0,yz9] & *(x,*(y,z) ) = [0,x9 *' yz9] by A7,Def2;
    consider x99,y99 being Element of REAL+ such that
A15: x = x99 and
A16: y = [0,y99] and
A17: *(x,y) = [0,x99 *' y99] by A7,A8,Def2;
A18: y9 = y99 by A10,A16,XTUPLE_0:1;
    *(x,y) in [:{0},REAL+:] by A17,Lm1,ZFMISC_1:87;
    then consider xy99,z99 being Element of REAL+ such that
A19: *(x,y) = [0,xy99] and
A20: z = z99 and
A21: *(*(x,y),z) = [0,z99 *' xy99] by A9,Def2;
    thus *(x,*(y,z)) = [0,x9 *' (y9 *' z9)] by A12,A14,XTUPLE_0:1
      .= [0,(x99 *' y99) *' z99] by A11,A13,A15,A20,A18,ARYTM_2:12
      .= *(*(x,y),z) by A17,A19,A21,XTUPLE_0:1;
  end;
  suppose that
A22: x in RR and
A23: y in REAL+ & y <> 0 and
A24: z in REAL+ & z <> 0;
    consider y9,z9 being Element of REAL+ such that
A25: y = y9 & z = z9 and
A26: *(y,z) = y9 *' z9 by A23,A24,Def2;
    y9 *' z9 <> 0 by A23,A24,A25,ARYTM_1:2;
    then consider x9,yz9 being Element of REAL+ such that
A27: x = [0,x9] and
A28: *(y,z) = yz9 & *(x,*(y,z)) = [0,yz9 *' x9] by A22,A26,Def2;
    consider x99,y99 being Element of REAL+ such that
A29: x = [0,x99] and
A30: y = y99 and
A31: *(x,y) = [0,y99 *' x99] by A22,A23,Def2;
    *(x,y) in [:{0},REAL+:] by A31,Lm1,ZFMISC_1:87;
    then consider xy99,z99 being Element of REAL+ such that
A32: *(x,y) = [0,xy99] and
A33: z = z99 and
A34: *(*(x,y),z) = [0,z99 *' xy99] by A24,Def2;
    x9 = x99 by A27,A29,XTUPLE_0:1;
    hence *(x,*(y,z)) = [0,(x99 *' y99) *' z99] by A25,A26,A28,A30,A33,
ARYTM_2:12
      .= *(*(x,y),z) by A31,A32,A34,XTUPLE_0:1;
  end;
  suppose that
A35: x in RR and
A36: y in RR and
A37: z in REAL+ & z <> 0;
    consider x99,y99 being Element of REAL+ such that
A38: x = [0,x99] and
A39: y = [0,y99] and
A40: *(x,y) = y99 *' x99 by A35,A36,Def2;
    consider y9,z9 being Element of REAL+ such that
A41: y = [0,y9] and
A42: z = z9 and
A43: *(y,z) = [0,z9 *' y9] by A36,A37,Def2;
A44: y9 = y99 by A41,A39,XTUPLE_0:1;
    *(y,z) in [:{0},REAL+:] by A43,Lm1,ZFMISC_1:87;
    then consider x9,yz9 being Element of REAL+ such that
A45: x = [0,x9] and
A46: *(y,z) = [0,yz9] & *(x,*(y,z)) = yz9 *' x9 by A35,Def2;
A47: x9 = x99 by A45,A38,XTUPLE_0:1;
A48: ex xy99,z99 being Element of REAL+ st *(x,y) = xy99 & z = z99 & *(*(x,
    y),z) = xy99 *' z99 by A37,A40,Def2;
    thus *(x,*(y,z)) = x9 *' (y9 *' z9) by A43,A46,XTUPLE_0:1
      .= *(*(x,y),z) by A42,A40,A48,A47,A44,ARYTM_2:12;
  end;
  suppose that
A49: x in REAL+ & x <> 0 and
A50: y in REAL+ & y <> 0 and
A51: z in RR;
A52: ex x99,y99 being Element of REAL+ st x = x99 & y = y99 & *(x,y) = x99
    *' y99 by A49,A50,Def2;
    then *(x,y) <> 0 by A49,A50,ARYTM_1:2;
    then consider xy99,z99 being Element of REAL+ such that
A53: *(x,y) = xy99 and
A54: z = [0,z99] and
A55: *(*(x,y),z) = [0,xy99 *' z99] by A51,A52,Def2;
    consider y9,z9 being Element of REAL+ such that
A56: y = y9 and
A57: z = [0,z9] and
A58: *(y,z) = [0,y9 *' z9] by A50,A51,Def2;
A59: z9 = z99 by A57,A54,XTUPLE_0:1;
    *(y,z) in [:{0},REAL+:] by A58,Lm1,ZFMISC_1:87;
    then consider x9,yz9 being Element of REAL+ such that
A60: x = x9 and
A61: *(y,z) = [0,yz9] & *(x,*(y,z)) = [0,x9 *' yz9] by A49,Def2;
    thus *(x,*(y,z)) = [0,x9 *' (y9 *' z9)] by A58,A61,XTUPLE_0:1
      .= *(*(x,y),z) by A56,A60,A52,A53,A55,A59,ARYTM_2:12;
  end;
  suppose that
A62: x in REAL+ & x <> 0 and
A63: y in RR and
A64: z in RR;
    consider y9,z9 being Element of REAL+ such that
A65: y = [0,y9] and
A66: z = [0,z9] and
A67: *(y,z) = z9 *' y9 by A63,A64,Def2;
A68: ex x9,yz9 being Element of REAL+ st x = x9 & *(y,z) = yz9 & *(x,*(y,z)
    ) = x9 *' yz9 by A62,A67,Def2;
    consider x99,y99 being Element of REAL+ such that
A69: x = x99 and
A70: y = [0,y99] and
A71: *(x,y) = [0,x99 *' y99] by A62,A63,Def2;
A72: y9 = y99 by A65,A70,XTUPLE_0:1;
    *(x,y) in [:{0},REAL+:] by A71,Lm1,ZFMISC_1:87;
    then consider xy99,z99 being Element of REAL+ such that
A73: *(x,y) = [0,xy99] and
A74: z = [0,z99] and
A75: *(*(x,y),z) = z99 *' xy99 by A64,Def2;
    z9 = z99 by A66,A74,XTUPLE_0:1;
    hence *(x,*(y,z)) = (x99 *' y99) *' z99 by A67,A68,A69,A72,ARYTM_2:12
      .= *(*(x,y),z) by A71,A73,A75,XTUPLE_0:1;
  end;
  suppose that
A76: y in REAL+ & y <> 0 and
A77: x in RR and
A78: z in RR;
    consider x99,y99 being Element of REAL+ such that
A79: x = [0,x99] and
A80: y = y99 and
A81: *(x,y) = [0,y99 *' x99] by A76,A77,Def2;
    consider y9,z9 being Element of REAL+ such that
A82: y = y9 and
A83: z = [0,z9] and
A84: *(y,z) = [0,y9 *' z9] by A76,A78,Def2;
    [0,y9 *' z9] in [:{0},REAL+:] by Lm1,ZFMISC_1:87;
    then consider x9,yz9 being Element of REAL+ such that
A85: x = [0,x9] and
A86: *(y,z) = [0,yz9] & *(x,*(y,z)) = yz9 *' x9 by A77,A84,Def2;
A87: x9 = x99 by A85,A79,XTUPLE_0:1;
    *(x,y) in [:{0},REAL+:] by A81,Lm1,ZFMISC_1:87;
    then consider xy99,z99 being Element of REAL+ such that
A88: *(x,y) = [0,xy99] and
A89: z = [0,z99] and
A90: *(*(x,y),z) = z99 *' xy99 by A78,Def2;
A91: z9 = z99 by A83,A89,XTUPLE_0:1;
    thus *(x,*(y,z)) = x9 *' (y9 *' z9) by A84,A86,XTUPLE_0:1
      .= (x99 *' y99) *' z99 by A82,A80,A87,A91,ARYTM_2:12
      .= *(*(x,y),z) by A81,A88,A90,XTUPLE_0:1;
  end;
  suppose that
A92: x in RR and
A93: y in RR and
A94: z in RR;
    consider y9,z9 being Element of REAL+ such that
A95: y = [0,y9] and
A96: z = [0,z9] and
A97: *(y,z) = z9 *' y9 by A93,A94,Def2;
    not y in {[0,0]} by A93,XBOOLE_0:def 5;
    then
A98: y9 <> 0 by A95,TARSKI:def 1;
    not z in {[0,0]} by A94,XBOOLE_0:def 5;
    then z9 <> 0 by A96,TARSKI:def 1;
    then *(z,y) <> 0 by A97,A98,ARYTM_1:2;
    then consider x9,yz9 being Element of REAL+ such that
A99: x = [0,x9] and
A100: *(y,z) = yz9 & *(x,*(y,z)) = [0,yz9 *' x9] by A92,A97,Def2;
    consider x99,y99 being Element of REAL+ such that
A101: x = [0,x99] and
A102: y = [0,y99] and
A103: *(x,y) = y99 *' x99 by A92,A93,Def2;
A104: x9 = x99 by A99,A101,XTUPLE_0:1;
A105: y9 = y99 by A95,A102,XTUPLE_0:1;
    not y in {[0,0]} by A93,XBOOLE_0:def 5;
    then
A106: y9 <> 0 by A95,TARSKI:def 1;
    not x in {[0,0]} by A92,XBOOLE_0:def 5;
    then x9 <> 0 by A99,TARSKI:def 1;
    then *(x,y) <> 0 by A103,A104,A105,A106,ARYTM_1:2;
    then consider xy99,z99 being Element of REAL+ such that
A107: *(x,y) = xy99 and
A108: z = [0,z99] and
A109: *(*(x,y),z) = [0,xy99 *' z99] by A94,A103,Def2;
    z9 = z99 by A96,A108,XTUPLE_0:1;
    hence thesis by A97,A100,A103,A104,A105,A107,A109,ARYTM_2:12;
  end;
  suppose
A110: x = 0;
    hence *(x,*(y,z)) = 0 by Th12
      .= *(o,z) by Th12
      .= *(*(x,y),z) by A110,Th12;
  end;
  suppose
A111: y = 0;
    hence *(x,*(y,z)) = *(x,o) by Th12
      .= 0 by Th12
      .= *(o,z) by Th12
      .= *(*(x,y),z) by A111,Th12;
  end;
  suppose
A112: z = 0;
    hence *(x,*(y,z)) = *(x,o) by Th12
      .= 0 by Th12
      .= *(*(x,y),z) by A112,Th12;
  end;
  suppose
A113: not( x in REAL+ & y in REAL+ & z in REAL+) & not(x in REAL+ & y
in RR & z in REAL+) & not(y in REAL+ & x in RR & z in REAL+) & not(x in RR & y
in RR & z in REAL+) & not( x in REAL+ & y in REAL+ & z in RR) & not(x in REAL+
& y in RR & z in RR) & not(y in REAL+ & x in RR & z in RR) & not(x in RR & y in
    RR & z in RR);
A114: not [0,0] in REAL+ by ARYTM_2:3;
    REAL = (REAL+ \ {[{},{}]}) \/ ([:{{}},REAL+:] \ {[{},{}]}) by NUMBERS:def 1
,XBOOLE_1:42
      .= REAL+ \/ RR by A114,ZFMISC_1:57;
    hence thesis by A113,XBOOLE_0:def 3;
  end;
end;

theorem Th14:
  for x,y,z being Element of REAL holds *(x,+(y,z)) = +(*(x,y),*(x ,z))
proof
  let x,y,z be Element of REAL;
  per cases;
  suppose
A1: x = 0;
    hence *(x,+(y,z)) = 0 by Th12
      .= +(o,o) by Th11
      .= +(*(x,y),o) by A1,Th12
      .= +(*(x,y),*(x,z)) by A1,Th12;
  end;
  suppose that
A2: x in REAL+ and
A3: y in REAL+ & z in REAL+;
A4: (ex x9,y9 being Element of REAL+ st x = x9 & y = y9 & *( x,y) = x9 *'
y9 )& ex x99,z9 being Element of REAL+ st x = x99 & z = z9 & *(x,z) = x99 *' z9
    by A2,A3,Def2;
    then
A5: ex xy9,xz9 being Element of REAL+ st *(x,y) = xy9 & *(x, z) = xz9 & +(*
    (x,y),*(x,z)) = xy9 + xz9 by Def1;
A6: ex y99,z99 being Element of REAL+ st y = y99 & z = z99 & +(y,z) = y99 +
    z99 by A3,Def1;
    then
    ex x999,yz9 being Element of REAL+ st x = x999 & +(y,z) = yz9 & *(x,+(y
    ,z)) = x999 *' yz9 by A2,Def2;
    hence thesis by A4,A5,A6,ARYTM_2:13;
  end;
  suppose that
A7: x in REAL+ & x <> 0 and
A8: y in REAL+ and
A9: z in RR;
    consider y99,z99 being Element of REAL+ such that
A10: y = y99 and
A11: z = [0,z99] and
A12: +(y,z) = y99 - z99 by A8,A9,Def1;
    consider x9,y9 being Element of REAL+ such that
A13: x = x9 & y = y9 and
A14: *(x,y) = x9 *' y9 by A7,A8,Def2;
    consider x99,z9 being Element of REAL+ such that
A15: x = x99 and
A16: z = [0,z9] and
A17: *(x,z) = [0,x99 *' z9] by A7,A9,Def2;
    *(x,z) in [:{0},REAL+:] by A17,Lm1,ZFMISC_1:87;
    then
A18: ex xy9,xz9 being Element of REAL+ st *(x,y) = xy9 & *(x, z) = [0,xz9]
    & +(*(x,y),*(x,z)) = xy9 - xz9 by A14,Def1;
A19: z9 = z99 by A16,A11,XTUPLE_0:1;
    now
      per cases;
      suppose
A20:    z99 <=' y99;
        then
A21:    +(y,z) = y99 -' z99 by A12,ARYTM_1:def 2;
        then
        ex x999,yz9 being Element of REAL+ st x = x999 & +(y,z) = yz9 & *(
        x,+(y,z)) = x999 *' yz9 by A7,Def2;
        hence
        *(x,+(y,z)) = (x9 *' y9) - (x99 *' z9) by A13,A15,A10,A19,A20,A21,
ARYTM_1:26
          .= +(*(x,y),*(x,z)) by A14,A17,A18,XTUPLE_0:1;
      end;
      suppose
A22:    not z99 <=' y99;
        then
A23:    +(y,z) = [0,z99 -' y99] by A12,ARYTM_1:def 2;
        then +(y,z) in [:{0},REAL+:] by Lm1,ZFMISC_1:87;
        then consider x999,yz9 being Element of REAL+ such that
A24:    x = x999 and
A25:    +(y,z) = [0,yz9] & *(x,+(y,z)) = [0,x999 *' yz9] by A7,Def2;
        thus *(x,+(y,z)) = [0,x999 *' (z99 -' y99)] by A23,A25,XTUPLE_0:1
          .= (x9 *' y9) - (x99 *' z9) by A7,A13,A15,A10,A19,A22,A24,ARYTM_1:27
          .= +(*(x,y),*(x,z)) by A14,A17,A18,XTUPLE_0:1;
      end;
    end;
    hence thesis;
  end;
  suppose that
A26: x in REAL+ & x <> 0 and
A27: z in REAL+ and
A28: y in RR;
    consider z99,y99 being Element of REAL+ such that
A29: z = z99 and
A30: y = [0,y99] and
A31: +(z,y) = z99 - y99 by A27,A28,Def1;
    consider x9,z9 being Element of REAL+ such that
A32: x = x9 & z = z9 and
A33: *(x,z) = x9 *' z9 by A26,A27,Def2;
    consider x99,y9 being Element of REAL+ such that
A34: x = x99 and
A35: y = [0,y9] and
A36: *(x,y) = [0,x99 *' y9] by A26,A28,Def2;
    *(x,y) in [:{0},REAL+:] by A36,Lm1,ZFMISC_1:87;
    then
A37: ex xz9,xy9 being Element of REAL+ st *(x,z) = xz9 & *(x, y) = [0,xy9]
    & +(*(x,z),*(x,y)) = xz9 - xy9 by A33,Def1;
A38: y9 = y99 by A35,A30,XTUPLE_0:1;
    now
      per cases;
      suppose
A39:    y99 <=' z99;
        then
A40:    +(z,y) = z99 -' y99 by A31,ARYTM_1:def 2;
        then
        ex x999,zy9 being Element of REAL+ st x = x999 & +(z,y) = zy9 & *(
        x,+(z,y)) = x999 *' zy9 by A26,Def2;
        hence
        *(x,+(z,y)) = (x9 *' z9) - (x99 *' y9) by A32,A34,A29,A38,A39,A40,
ARYTM_1:26
          .= +(*(x,z),*(x,y)) by A33,A36,A37,XTUPLE_0:1;
      end;
      suppose
A41:    not y99 <=' z99;
        then
A42:    +(z,y) = [0,y99 -' z99] by A31,ARYTM_1:def 2;
        then +(z,y) in [:{0},REAL+:] by Lm1,ZFMISC_1:87;
        then consider x999,zy9 being Element of REAL+ such that
A43:    x = x999 and
A44:    +(z,y) = [0,zy9] & *(x,+(z,y)) = [0,x999 *' zy9] by A26,Def2;
        thus *(x,+(z,y)) = [0,x999 *' (y99 -' z99)] by A42,A44,XTUPLE_0:1
          .= (x9 *' z9) - (x99 *' y9) by A26,A32,A34,A29,A38,A41,A43,ARYTM_1:27
          .= +(*(x,z),*(x,y)) by A33,A36,A37,XTUPLE_0:1;
      end;
    end;
    hence thesis;
  end;
  suppose that
A45: x in REAL+ & x <> 0 and
A46: y in RR and
A47: z in RR;
    ( not(y in REAL+ & z in [:{0},REAL+:]))& not(y in [:{0},REAL+:] & z
    in REAL+) by A46,A47,Th5,XBOOLE_0:3;
    then consider y99,z99 being Element of REAL+ such that
A48: y = [0,y99] and
A49: z = [0,z99] and
A50: +(y,z) = [0,y99 + z99] by A46,Def1;
    +(y,z) in [:{0},REAL+:] by A50,Lm1,ZFMISC_1:87;
    then consider x999,yz9 being Element of REAL+ such that
A51: x = x999 and
A52: +(y,z) = [0,yz9] & *(x,+(y,z)) = [0,x999 *' yz9] by A45,Def2;
    consider x9,y9 being Element of REAL+ such that
A53: x = x9 and
A54: y = [0,y9] and
A55: *(x,y) = [0,x9 *' y9] by A45,A46,Def2;
A56: y9 = y99 by A54,A48,XTUPLE_0:1;
    consider x99,z9 being Element of REAL+ such that
A57: x = x99 and
A58: z = [0,z9] and
A59: *(x,z) = [0,x99 *' z9] by A45,A47,Def2;
    *(x,z) in [:{0},REAL+:] by A59,Lm1,ZFMISC_1:87;
    then
A60: not(*(x,y) in [:{0},REAL+:] & *(x,z) in REAL+) by Th5,XBOOLE_0:3;
    *(x,y) in [:{0},REAL+:] by A55,Lm1,ZFMISC_1:87;
    then not(*(x,y) in REAL+ & *(x,z) in [:{0},REAL+:]) by Th5,XBOOLE_0:3;
    then consider xy9,xz9 being Element of REAL+ such that
A61: *(x,y) = [0,xy9] and
A62: *(x,z) = [0,xz9] & +(*(x,y),*(x,z)) = [0,xy9 + xz9] by A55,A60,Def1,Lm1,
ZFMISC_1:87;
A63: xy9 = x9 *' y9 by A55,A61,XTUPLE_0:1;
A64: z9 = z99 by A58,A49,XTUPLE_0:1;
    thus *(x,+(y,z)) = [0,x999 *' (y99 + z99)] by A50,A52,XTUPLE_0:1
      .= [0,(x9 *' y9) + (x9 *' z9)] by A53,A51,A56,A64,ARYTM_2:13
      .= +(*(x,y),*(x,z)) by A53,A57,A59,A62,A63,XTUPLE_0:1;
  end;
  suppose that
A65: x in RR and
A66: y in REAL+ and
A67: z in REAL+;
    consider y99,z99 being Element of REAL+ such that
A68: y = y99 and
A69: z = z99 and
A70: +(y,z) = y99 + z99 by A66,A67,Def1;
    now
      per cases;
      suppose that
A71:    y <> 0 and
A72:    z <> 0;
        consider x99,z9 being Element of REAL+ such that
A73:    x = [0,x99] and
A74:    z = z9 and
A75:    *(x,z) = [0,z9 *' x99] by A65,A67,A72,Def2;
        y99 + z99 <> 0 by A69,A72,ARYTM_2:5;
        then consider x999,yz9 being Element of REAL+ such that
A76:    x = [0,x999] and
A77:    +(y,z) = yz9 & *(x,+(y,z)) = [0,yz9 *' x999] by A65,A70,Def2;
        consider x9,y9 being Element of REAL+ such that
A78:    x = [0,x9] and
A79:    y = y9 and
A80:    *(x,y) = [0,y9 *' x9] by A65,A66,A71,Def2;
A81:    x99 = x999 by A73,A76,XTUPLE_0:1;
        *(x,z) in [:{0},REAL+:] by A75,Lm1,ZFMISC_1:87;
        then
A82:    not(*(x,y) in [:{0},REAL+:] & *(x,z) in REAL+) by Th5,XBOOLE_0:3;
        *(x,y) in [:{0},REAL+:] by A80,Lm1,ZFMISC_1:87;
        then not(*(x,y) in REAL+ & *(x,z) in [:{0},REAL+:]) by Th5,XBOOLE_0:3;
        then consider xy9,xz9 being Element of REAL+ such that
A83:    *(x,y) = [0,xy9] and
A84:    *(x,z) = [0,xz9] & +(*(x,y),*(x,z)) = [0,xy9 + xz9] by A80,A82,Def1,Lm1
,ZFMISC_1:87;
A85:    xy9 = x9 *' y9 by A80,A83,XTUPLE_0:1;
        x9 = x99 by A78,A73,XTUPLE_0:1;
        hence *(x,+(y,z)) = [0,(x9 *' y9) + (x99 *' z9)] by A68,A69,A70,A79,A74
,A77,A81,ARYTM_2:13
          .= +(*(x,y),*(x,z)) by A75,A84,A85,XTUPLE_0:1;
      end;
      suppose
A86:    x = 0;
        hence *(x,+(y,z)) = 0 by Th12
          .= +(o,o) by Th11
          .= +(*(x,y),o) by A86,Th12
          .= +(*(x,y),*(x,z)) by A86,Th12;
      end;
      suppose
A87:    z = 0;
        hence *(x,+(y,z)) = *(x,y) by Th11
          .= +(*(x,y),*(x,z)) by A87,Th11,Th12;
      end;
      suppose
A88:    y = 0;
        hence *(x,+(y,z)) = *(x,z) by Th11
          .= +(*(x,y),*(x,z)) by A88,Th11,Th12;
      end;
    end;
    hence thesis;
  end;
  suppose that
A89: x in RR and
A90: y in REAL+ and
A91: z in RR;
    consider x99,z9 being Element of REAL+ such that
A92: x = [0,x99] and
A93: z = [0,z9] and
A94: *(x,z) = z9 *' x99 by A89,A91,Def2;
    now
      per cases;
      suppose
        y <> 0;
        then consider x9,y9 being Element of REAL+ such that
A95:    x = [0,x9] and
A96:    y = y9 and
A97:    *(x,y) = [0,y9 *' x9] by A89,A90,Def2;
        *(x,y) in [:{0},REAL+:] by A97,Lm1,ZFMISC_1:87;
        then consider xy9,xz9 being Element of REAL+ such that
A98:    *(x,y) = [0,xy9] and
A99:    *(x,z) = xz9 & +(*(x,y),*(x,z)) = xz9 - xy9 by A94,Def1;
        consider y99,z99 being Element of REAL+ such that
A100:   y = y99 and
A101:   z = [0,z99] and
A102:   +(y,z) = y99 - z99 by A91,A96,Def1;
A103:   z9 = z99 by A93,A101,XTUPLE_0:1;
        now
          per cases;
          suppose
A104:       z99 <=' y99;
            then
A105:       +(y,z) = y99 -' z99 by A102,ARYTM_1:def 2;
            now
              per cases;
              suppose
A106:           +(y,z) <> 0;
                then consider x999,yz9 being Element of REAL+ such that
A107:           x = [0,x999] and
A108:           +(y,z) = yz9 & *(x,+(y,z)) = [0,yz9 *' x999] by A89,A105,Def2;
                not x in {[0,0]} by NUMBERS:def 1,XBOOLE_0:def 5;
                then
A109:           x999 <> 0 by A107,TARSKI:def 1;
A110:           z9 = z99 by A93,A101,XTUPLE_0:1;
A111:           x9 = x99 by A92,A95,XTUPLE_0:1;
                x99 = x999 by A92,A107,XTUPLE_0:1;
                hence *(x,+(y,z)) = (x9 *' z9) - (x9 *' y9) by A96,A100,A104
,A105,A106,A108,A111,A110,A109,ARYTM_1:28
                  .= +(*(x,y),*(x,z)) by A94,A97,A98,A99,A111,XTUPLE_0:1;
              end;
              suppose
A112:           +(y,z) = 0;
                then
A113:           y99 = z99 by A104,A105,ARYTM_1:10;
A114:           xy9 = x9 *' y9 & x9 = x99 by A92,A95,A97,A98,XTUPLE_0:1;
                thus *(x,+(y,z)) = 0 by A112,Th12
                  .= +(*(x,y),*(x,z)) by A94,A96,A100,A99,A103,A113,A114,
ARYTM_1:18;
              end;
            end;
            hence thesis;
          end;
          suppose
A115:       not z99 <=' y99;
            then
A116:       +(y,z) = [0,z99 -' y99] by A102,ARYTM_1:def 2;
            then +(y,z) in [:{0},REAL+:] by Lm1,ZFMISC_1:87;
            then consider x999,yz9 being Element of REAL+ such that
A117:       x = [0,x999] and
A118:       +(y,z) = [0,yz9] & *(x,+(y,z)) = yz9 *' x999 by A89,Def2;
A119:       x99 = x999 by A92,A117,XTUPLE_0:1;
A120:       x9 = x99 by A92,A95,XTUPLE_0:1;
            thus *(x,+(y,z)) = x999 *' (z99 -' y99) by A116,A118,XTUPLE_0:1
              .= (x99 *' z9) - (x9 *' y9) by A96,A100,A103,A115,A120,A119,
ARYTM_1:26
              .= +(*(x,y),*(x,z)) by A94,A97,A98,A99,XTUPLE_0:1;
          end;
        end;
        hence thesis;
      end;
      suppose
A121:   y = 0;
        hence *(x,+(y,z)) = *(x,z) by Th11
          .= +(*(x,y),*(x,z)) by A121,Th11,Th12;
      end;
    end;
    hence thesis;
  end;
  suppose that
A122: x in RR and
A123: z in REAL+ and
A124: y in RR;
    consider x99,y9 being Element of REAL+ such that
A125: x = [0,x99] and
A126: y = [0,y9] and
A127: *(x,y) = y9 *' x99 by A122,A124,Def2;
    now
      per cases;
      suppose
        z <> 0;
        then consider x9,z9 being Element of REAL+ such that
A128:   x = [0,x9] and
A129:   z = z9 and
A130:   *(x,z) = [0,z9 *' x9] by A122,A123,Def2;
        *(x,z) in [:{0},REAL+:] by A130,Lm1,ZFMISC_1:87;
        then consider xz9,xy9 being Element of REAL+ such that
A131:   *(x,z) = [0,xz9] and
A132:   *(x,y) = xy9 & +(*(x,z),*(x,y)) = xy9 - xz9 by A127,Def1;
        consider z99,y99 being Element of REAL+ such that
A133:   z = z99 and
A134:   y = [0,y99] and
A135:   +(z,y) = z99 - y99 by A124,A129,Def1;
A136:   y9 = y99 by A126,A134,XTUPLE_0:1;
        now
          per cases;
          suppose
A137:       y99 <=' z99;
            then
A138:       +(z,y) = z99 -' y99 by A135,ARYTM_1:def 2;
            now
              per cases;
              suppose
A139:           +(z,y) <> 0;
                then consider x999,zy9 being Element of REAL+ such that
A140:           x = [0,x999] and
A141:           +(z,y) = zy9 & *(x,+(z,y)) = [0,zy9 *' x999] by A122,A138,Def2;
                not x in {[0,0]} by NUMBERS:def 1,XBOOLE_0:def 5;
                then
A142:           x999 <> 0 by A140,TARSKI:def 1;
A143:           y9 = y99 by A126,A134,XTUPLE_0:1;
A144:           x9 = x99 by A125,A128,XTUPLE_0:1;
                x99 = x999 by A125,A140,XTUPLE_0:1;
                hence *(x,+(z,y)) = (x9 *' y9) - (x9 *' z9) by A129,A133,A137
,A138,A139,A141,A144,A143,A142,ARYTM_1:28
                  .= +(*(x,z),*(x,y)) by A127,A130,A131,A132,A144,XTUPLE_0:1;
              end;
              suppose
A145:           +(z,y) = 0;
                then
A146:           z99 = y99 by A137,A138,ARYTM_1:10;
A147:           xz9 = x9 *' z9 & x9 = x99 by A125,A128,A130,A131,XTUPLE_0:1;
                thus *(x,+(z,y)) = 0 by A145,Th12
                  .= +(*(x,z),*(x,y)) by A127,A129,A133,A132,A136,A146,A147,
ARYTM_1:18;
              end;
            end;
            hence thesis;
          end;
          suppose
A148:       not y99 <=' z99;
            then
A149:       +(z,y) = [0,y99 -' z99] by A135,ARYTM_1:def 2;
            then +(z,y) in [:{0},REAL+:] by Lm1,ZFMISC_1:87;
            then consider x999,zy9 being Element of REAL+ such that
A150:       x = [0,x999] and
A151:       +(z,y) = [0,zy9] & *(x,+(z,y)) = zy9 *' x999 by A122,Def2;
A152:       x99 = x999 by A125,A150,XTUPLE_0:1;
A153:       x9 = x99 by A125,A128,XTUPLE_0:1;
            thus *(x,+(y,z)) = x999 *' (y99 -' z99) by A149,A151,XTUPLE_0:1
              .= (x99 *' y9) - (x9 *' z9) by A129,A133,A136,A148,A153,A152,
ARYTM_1:26
              .= +(*(x,y),*(x,z)) by A127,A130,A131,A132,XTUPLE_0:1;
          end;
        end;
        hence thesis;
      end;
      suppose
A154:   z = 0;
        hence *(x,+(y,z)) = *(x,y) by Th11
          .= +(*(x,y),*(x,z)) by A154,Th11,Th12;
      end;
    end;
    hence thesis;
  end;
  suppose that
A155: x in RR and
A156: y in RR and
A157: z in RR;
    ( not(y in REAL+ & z in [:{0},REAL+:]))& not(y in [:{0},REAL+:] & z
    in REAL+) by A156,A157,Th5,XBOOLE_0:3;
    then consider y99,z99 being Element of REAL+ such that
A158: y = [0,y99] and
A159: z = [0,z99] and
A160: +(y,z) = [0,y99 + z99] by A156,Def1;
    consider x99,z9 being Element of REAL+ such that
A161: x = [0,x99] and
A162: z = [0,z9] and
A163: *(x,z) = z9 *' x99 by A155,A157,Def2;
A164: z9 = z99 by A162,A159,XTUPLE_0:1;
    consider x9,y9 being Element of REAL+ such that
A165: x = [0,x9] and
A166: y = [0,y9] and
A167: *(x,y) = y9 *' x9 by A155,A156,Def2;
A168: y9 = y99 by A166,A158,XTUPLE_0:1;
    +(y,z) in [:{0},REAL+:] by A160,Lm1,ZFMISC_1:87;
    then consider x999,yz9 being Element of REAL+ such that
A169: x = [0,x999] and
A170: +(y,z) = [0,yz9] & *(x,+(y,z)) = yz9 *' x999 by A155,Def2;
A171: x9 = x999 by A165,A169,XTUPLE_0:1;
A172: (ex xy9,xz9 being Element of REAL+ st *(x,y) = xy9 & *(x, z) = xz9 &
+(*(x, y),*(x,z)) = xy9 + xz9 )& x9 = x99 by A165,A167,A161,A163,Def1,
XTUPLE_0:1;
    thus *(x,+(y,z)) = x999 *' (y99 + z99) by A160,A170,XTUPLE_0:1
      .= +(*(x,y),*(x,z)) by A167,A163,A172,A171,A168,A164,ARYTM_2:13;
  end;
  suppose
A173: not(x in REAL+ & y in REAL+ & z in REAL+) & not(x in REAL+ & y
in REAL+ & z in RR) & not(x in REAL+ & y in RR & z in REAL+) & not(x in REAL+ &
y in RR & z in RR) & not(x in RR & y in REAL+ & z in REAL+) & not(x in RR & y
in REAL+ & z in RR) & not(x in RR & y in RR & z in REAL+) & not(x in RR & y in
    RR & z in RR);
A174: not [0,0] in REAL+ by ARYTM_2:3;
    REAL = (REAL+ \ {[0,0]}) \/ ([:{0},REAL+:] \ {[0,0]}) by NUMBERS:def 1
,XBOOLE_1:42
      .= REAL+ \/ RR by A174,ZFMISC_1:57;
    hence thesis by A173,XBOOLE_0:def 3;
  end;
end;

theorem
  for x,y being Element of REAL holds *(opp x,y) = opp *(x,y)
proof
  let x,y be Element of REAL;
  +(*(opp x,y),*(x,y)) = *(+(opp x,x), y) by Th14
    .= *(o,y) by Def3
    .= 0 by Th12;
  hence thesis by Def3;
end;

theorem Th16:
  for x being Element of REAL holds *(x,x) in REAL+
proof
  let x be Element of REAL;
A1: x in REAL+ \/ [:{{}},REAL+:] by NUMBERS:def 1,XBOOLE_0:def 5;
  per cases by A1,XBOOLE_0:def 3;
  suppose
    x in REAL+;
    then
    ex x9,y9 being Element of REAL+ st x = x9 & x = y9 & *(x,x) = x9 *' y9
    by Def2;
    hence thesis;
  end;
  suppose
    x in [:{0},REAL+:];
    then
    ex x9,y9 being Element of REAL+ st x = [0,x9] & x = [0,y9] & *(x,x) =
    y9 *' x9 by Def2;
    hence thesis;
  end;
end;

theorem
  for x,y st +(*(x,x),*(y,y)) = 0 holds x = 0
proof
  let x,y such that
A1: +(*(x,x),*(y,y)) = 0;
  *(x,x) in REAL+ & *(y,y) in REAL+ by Th16;
  then consider x9,y9 being Element of REAL+ such that
A2: *(x,x) = x9 and
  *(y,y) = y9 and
A3: 0 = x9 + y9 by A1,Def1;
A4: x9 = 0 by A3,ARYTM_2:5;
A5: x in REAL+ \/ [:{{}},REAL+:] by NUMBERS:def 1,XBOOLE_0:def 5;
  per cases by A5,XBOOLE_0:def 3;
  suppose
    x in REAL+;
    then ex x9,y9 being Element of REAL+ st x = x9 & x = y9 & 0 = x9 *' y9 by
A2,A4,Def2;
    hence thesis by ARYTM_1:2;
  end;
  suppose
    x in [:{0},REAL+:];
    then consider x9,y9 being Element of REAL+ such that
A6: x = [0,x9] and
A7: x = [0,y9] and
A8: 0 = y9 *' x9 by A2,A4,Def2;
    x9 = y9 by A6,A7,XTUPLE_0:1;
    then x9 = 0 by A8,ARYTM_1:2;
    then x in {[0,0]} by A6,TARSKI:def 1;
    hence thesis by NUMBERS:def 1,XBOOLE_0:def 5;
  end;
end;

theorem Th18:
  for x,y,z being Element of REAL st x <> 0 & *(x,y) = 1 & *(x,z)
  = 1 holds y = z
proof
  let x,y,z be Element of REAL;
  assume that
A1: x <> 0 and
A2: *(x,y) = 1 and
A3: *(x,z) = 1;
  thus y = inv x by A1,A2,Def4
    .= z by A1,A3,Def4;
end;

theorem Th19:
  for x,y st y = 1 holds *(x,y) = x
proof
  let x,y such that
A1: y = 1;
  per cases;
  suppose
    x = 0;
    hence thesis by Th12;
  end;
  suppose
A2: x <> 0;
A3: now
      assume
A4:   inv x = 0;
      thus 1 = *(x, inv x) by A2,Def4
        .= 0 by A4,Th12;
    end;
A5: ex x9,y9 being Element of REAL+ st y = x9 & y = y9 & *(y,y) = x9 *' y9
    by A1,Def2,ARYTM_2:20;
A6: *(x,inv x) = 1 by A2,Def4;
    *(*(x,y), inv x) = *(*(x,inv x), y) by Th13
      .= *(y,y) by A1,A2,Def4
      .= 1 by A1,A5,ARYTM_2:15;
    hence thesis by A3,A6,Th18;
  end;
end;

reconsider j = 1 as Element of REAL;

theorem
  for x,y st y <> 0 holds *(*(x,y),inv y) = x
proof
  let x,y such that
A1: y <> 0;
  thus *(*(x,y),inv y) = *(x,*(y,inv y)) by Th13
    .= *(x,j) by A1,Def4
    .= x by Th19;
end;

theorem Th21:
  for x,y st *(x,y) = 0 holds x = 0 or y = 0
proof
  let x,y such that
A1: *(x,y) = 0 and
A2: x <> 0;
A3: *(x, inv x) = 1 by A2,Def4;
  thus y = *(j,y) by Th19
    .= *(*(x,y),inv x) by A3,Th13
    .= 0 by A1,Th12;
end;

theorem
  for x,y holds inv *(x,y) = *(inv x, inv y)
proof
  let x,y;
  per cases;
  suppose
A1: x = 0 or y = 0;
    then
A2: inv x = 0 or inv y = 0 by Def4;
    *(x,y) = 0 by A1,Th12;
    hence inv *(x,y) = 0 by Def4
      .= *(inv x, inv y) by A2,Th12;
  end;
  suppose that
A3: x <> 0 and
A4: y <> 0;
A5: *(x,y) <> 0 by A3,A4,Th21;
A6: *(x,inv x) = 1 by A3,Def4;
A7: *(y,inv y) = 1 by A4,Def4;
    *(*(x,y),*(inv x, inv y)) = *(*(*(x,y),inv x), inv y) by Th13
      .= *(*(j,y), inv y) by A6,Th13
      .= 1 by A7,Th19;
    hence thesis by A5,Def4;
  end;
end;

theorem Th23:
  for x,y,z being Element of REAL holds +(x,+(y,z)) = +(+(x,y),z)
proof
  let x,y,z be Element of REAL;
A1: x in REAL+ \/ [:{0},REAL+:] & y in REAL+ \/ [:{0},REAL+:] by NUMBERS:def 1
,XBOOLE_0:def 5;
A2: z in REAL+ \/ [:{0},REAL+:] by NUMBERS:def 1,XBOOLE_0:def 5;
  per cases by A1,A2,XBOOLE_0:def 3;
  suppose that
A3: x in REAL+ and
A4: y in REAL+ and
A5: z in REAL+;
A6: ex x99,y99 being Element of REAL+ st x = x99 & y = y99 & +(x,y) = x99 +
    y99 by A3,A4,Def1;
    then
A7: ex xy99,z99 being Element of REAL+ st +(x,y) = xy99 & z = z99 & +(+(x,
    y),z) = xy99 + z99 by A5,Def1;
A8: ex y9,z9 being Element of REAL+ st y = y9 & z = z9 & +( y,z) = y9 + z9
    by A4,A5,Def1;
    then
    ex x9,yz9 being Element of REAL+ st x = x9 & +(y,z) = yz9 & +(x,+(y,z))
    = x9 + yz9 by A3,Def1;
    hence thesis by A8,A6,A7,ARYTM_2:6;
  end;
  suppose that
A9: x in REAL+ and
A10: y in REAL+ and
A11: z in [:{0},REAL+:];
A12: ex x99,y99 being Element of REAL+ st x = x99 & y = y99 & +(x,y) = x99
    + y99 by A9,A10,Def1;
    then consider xy99,z99 being Element of REAL+ such that
A13: +(x,y) = xy99 and
A14: z = [0,z99] and
A15: +(+(x,y),z) = xy99 - z99 by A11,Def1;
    consider y9,z9 being Element of REAL+ such that
A16: y = y9 and
A17: z = [0,z9] and
A18: +(y,z) = y9 - z9 by A10,A11,Def1;
A19: z9 = z99 by A17,A14,XTUPLE_0:1;
    now
      per cases;
      suppose
A20:    z9 <=' y9;
        then
A21:    +(y,z) = y9 -' z9 by A18,ARYTM_1:def 2;
        then
        ex x9,yz9 being Element of REAL+ st x = x9 & +(y,z) = yz9 & +(x,+(
        y,z)) = x9 + yz9 by A9,Def1;
        hence thesis by A16,A12,A13,A15,A19,A20,A21,ARYTM_1:20;
      end;
      suppose
A22:    not z9 <=' y9;
        then
A23:    +(y,z) = [0,z9 -' y9] by A18,ARYTM_1:def 2;
        then +(y,z) in [:{0},REAL+:] by Lm1,ZFMISC_1:87;
        then consider x9,yz9 being Element of REAL+ such that
A24:    x = x9 and
A25:    +(y,z) = [0,yz9] & +(x,+(y,z)) = x9 - yz9 by A9,Def1;
        thus +(x,+(y,z)) = x9 - (z9 -' y9) by A23,A25,XTUPLE_0:1
          .= +(+(x,y),z) by A16,A12,A13,A15,A19,A22,A24,ARYTM_1:21;
      end;
    end;
    hence thesis;
  end;
  suppose that
A26: x in REAL+ and
A27: y in [:{0},REAL+:] and
A28: z in REAL+;
    consider x99,y99 being Element of REAL+ such that
A29: x = x99 and
A30: y = [0,y99] and
A31: +(x,y) = x99 - y99 by A26,A27,Def1;
    consider z9,y9 being Element of REAL+ such that
A32: z = z9 and
A33: y = [0,y9] and
A34: +(y,z) = z9 - y9 by A27,A28,Def1;
A35: y9 = y99 by A33,A30,XTUPLE_0:1;
    now
      per cases;
      suppose that
A36:    y9 <=' x99 and
A37:    y9 <=' z9;
A38:    +(y,z) = z9 -' y9 by A34,A37,ARYTM_1:def 2;
        then consider x9,yz9 being Element of REAL+ such that
A39:    x = x9 and
A40:    +(y,z) = yz9 & +(x,+(y,z)) = x9 + yz9 by A26,Def1;
A41:    +(x,y) = x9 -' y9 by A29,A31,A35,A36,A39,ARYTM_1:def 2;
        then
        ex z99,xy99 being Element of REAL+ st z = z99 & +(x,y) = xy99 & +(
        z,+(x,y)) = z99 + xy99 by A28,Def1;
        hence thesis by A32,A29,A36,A37,A38,A39,A40,A41,ARYTM_1:12;
      end;
      suppose that
A42:    y9 <=' x99 and
A43:    not y9 <=' z9;
A44:    +(x,y) = x99 -' y9 by A31,A35,A42,ARYTM_1:def 2;
        then
A45:    ex z99,xy99 being Element of REAL+ st z = z99 & +(x,y) = xy99 & +(
        z,+(x,y)) = z99 + xy99 by A28,Def1;
A46:    +(y,z) = [0,y9 -' z9] by A34,A43,ARYTM_1:def 2;
        then +(y,z) in [:{0},REAL+:] by Lm1,ZFMISC_1:87;
        then consider x9,yz9 being Element of REAL+ such that
A47:    x = x9 and
A48:    +(y,z) = [0,yz9] & +(x,+(y,z)) = x9 - yz9 by A26,Def1;
        thus +(x,+(y,z)) = x9 - (y9 -' z9) by A46,A48,XTUPLE_0:1
          .= +(+(x,y),z) by A32,A29,A42,A43,A47,A44,A45,ARYTM_1:22;
      end;
      suppose that
A49:    not y9 <=' x99 and
A50:    y9 <=' z9;
A51:    +(y,x) = [0,y9 -' x99] by A31,A35,A49,ARYTM_1:def 2;
        then +(y,x) in [:{0},REAL+:] by Lm1,ZFMISC_1:87;
        then consider z99,yx99 being Element of REAL+ such that
A52:    z = z99 and
A53:    +(y,x) = [0,yx99] & +(z,+(y,x)) = z99 - yx99 by A28,Def1;
A54:    +(z,y) = z9 -' y9 by A34,A50,ARYTM_1:def 2;
        then
        ex x9,zy99 being Element of REAL+ st x = x9 & +(z,y) = zy99 & +(x,
        +(z,y)) = x9 + zy99 by A26,Def1;
        hence +(x,+(y,z)) = z99 - (y9 -' x99) by A32,A29,A49,A50,A52,A54,
ARYTM_1:22
          .= +(+(x,y),z) by A51,A53,XTUPLE_0:1;
      end;
      suppose that
A55:    not y9 <=' x99 and
A56:    not y9 <=' z9;
A57:    +(y,z) = [0,y9 -' z9] by A34,A56,ARYTM_1:def 2;
        then +(y,z) in [:{0},REAL+:] by Lm1,ZFMISC_1:87;
        then consider x9,yz9 being Element of REAL+ such that
A58:    x = x9 and
A59:    +(y,z) = [0,yz9] & +(x,+(y,z)) = x9 - yz9 by A26,Def1;
A60:    +(y,x) = [0,y9 -' x99] by A31,A35,A55,ARYTM_1:def 2;
        then +(y,x) in [:{0},REAL+:] by Lm1,ZFMISC_1:87;
        then consider z99,yx99 being Element of REAL+ such that
A61:    z = z99 and
A62:    +(y,x) = [0,yx99] & +(z,+(y,x)) = z99 - yx99 by A28,Def1;
        thus +(x,+(y,z)) = x9 - (y9 -' z9) by A57,A59,XTUPLE_0:1
          .= z99 - (y9 -' x99) by A32,A29,A55,A56,A61,A58,ARYTM_1:23
          .= +(+(x,y),z) by A60,A62,XTUPLE_0:1;
      end;
    end;
    hence thesis;
  end;
  suppose that
A63: x in REAL+ and
A64: y in [:{0},REAL+:] and
A65: z in [:{0},REAL+:];
    ( not(z in REAL+ & y in [:{0},REAL+:]))& not(y in REAL+ & z in [:{0},
    REAL+:]) by A64,A65,Th5,XBOOLE_0:3;
    then consider y9,z9 being Element of REAL+ such that
A66: y = [0,y9] and
A67: z = [0,z9] and
A68: +(y,z) = [0,y9 + z9] by A64,Def1;
    +(y,z) in [:{0},REAL+:] by A68,Lm1,ZFMISC_1:87;
    then consider x9,yz9 being Element of REAL+ such that
A69: x = x9 and
A70: +(y,z) = [0,yz9] and
A71: +(x,+(y,z)) = x9 - yz9 by A63,Def1;
    consider x99,y99 being Element of REAL+ such that
A72: x = x99 and
A73: y = [0,y99] and
A74: +(x,y) = x99 - y99 by A63,A64,Def1;
A75: y9 = y99 by A66,A73,XTUPLE_0:1;
    now
      per cases;
      suppose
A76:    y99 <=' x99;
        then
A77:    +(x,y) = x99 -' y99 by A74,ARYTM_1:def 2;
        then consider xy99,z99 being Element of REAL+ such that
A78:    +(x,y) = xy99 and
A79:    z = [0,z99] and
A80:    +(+(x,y),z) = xy99 - z99 by A65,Def1;
A81:    z9 = z99 by A67,A79,XTUPLE_0:1;
        thus +(x,+(y,z)) = x9 - (y9 + z9) by A68,A70,A71,XTUPLE_0:1
          .= +(+(x,y),z) by A72,A69,A75,A76,A77,A78,A80,A81,ARYTM_1:24;
      end;
      suppose
A82:    not y99 <=' x99;
A83:    not(z in REAL+ & +(x,y) in [:{0},REAL+:]) by A65,Th5,XBOOLE_0:3;
A84:    +(x,y) = [0,y99 -' x99] by A74,A82,ARYTM_1:def 2;
        then +(x,y) in [:{0},REAL+:] by Lm1,ZFMISC_1:87;
        then not(+(x,y) in REAL+ & z in [:{0},REAL+:]) by Th5,XBOOLE_0:3;
        then consider xy99,z99 being Element of REAL+ such that
A85:    +(x,y) = [0,xy99] and
A86:    z = [0,z99] and
A87:    +(+(x,y),z) = [0,xy99 + z99] by A84,A83,Def1,Lm1,ZFMISC_1:87;
A88:    z9 = z99 by A67,A86,XTUPLE_0:1;
A89:    yz9 = z9 + y9 by A68,A70,XTUPLE_0:1;
        then y99 <=' yz9 by A75,ARYTM_2:19;
        then not yz9 <=' x9 by A72,A69,A82,ARYTM_1:3;
        hence +(x,+(y,z)) = [0,z9 + y9 -' x9] by A71,A89,ARYTM_1:def 2
          .= [0,z99 + (y99 -' x99)] by A72,A69,A75,A82,A88,ARYTM_1:13
          .= +(+(x,y),z) by A84,A85,A87,XTUPLE_0:1;
      end;
    end;
    hence thesis;
  end;
  suppose that
A90: z in REAL+ and
A91: y in REAL+ and
A92: x in [:{0},REAL+:];
A93: ex z99,y99 being Element of REAL+ st z = z99 & y = y99 & +(z,y) = z99
    + y99 by A90,A91,Def1;
    then consider zy99,x99 being Element of REAL+ such that
A94: +(z,y) = zy99 and
A95: x = [0,x99] and
A96: +(+(z,y),x) = zy99 - x99 by A92,Def1;
    consider y9,x9 being Element of REAL+ such that
A97: y = y9 and
A98: x = [0,x9] and
A99: +(y,x) = y9 - x9 by A91,A92,Def1;
A100: x9 = x99 by A98,A95,XTUPLE_0:1;
    now
      per cases;
      suppose
A101:   x9 <=' y9;
        then
A102:   +(y,x) = y9 -' x9 by A99,ARYTM_1:def 2;
        then ex z9,yx9 being Element of REAL+ st z = z9 & +(y,x) = yx9 & +(z,+
        (y,x)) = z9 + yx9 by A90,Def1;
        hence +(z,+(y,x)) = +(+(z,y),x) by A97,A93,A94,A96,A100,A101,A102,
ARYTM_1:20;
      end;
      suppose
A103:   not x9 <=' y9;
        then
A104:   +(y,x) = [0,x9 -' y9] by A99,ARYTM_1:def 2;
        then +(y,x) in [:{0},REAL+:] by Lm1,ZFMISC_1:87;
        then consider z9,yx9 being Element of REAL+ such that
A105:   z = z9 and
A106:   +(y,x) = [0,yx9] & +(z,+(y,x)) = z9 - yx9 by A90,Def1;
        thus +(z,+(y,x)) = z9 - (x9 -' y9) by A104,A106,XTUPLE_0:1
          .= +(+(z,y),x) by A97,A93,A94,A96,A100,A103,A105,ARYTM_1:21;
      end;
    end;
    hence thesis;
  end;
  suppose that
A107: x in [:{0},REAL+:] and
A108: y in REAL+ and
A109: z in [:{0},REAL+:];
    consider y9,z9 being Element of REAL+ such that
A110: y = y9 and
A111: z = [0,z9] and
A112: +(y,z) = y9 - z9 by A108,A109,Def1;
    consider x99,y99 being Element of REAL+ such that
A113: x = [0,x99] and
A114: y = y99 and
A115: +(x,y) = y99 - x99 by A107,A108,Def1;
    now
      per cases;
      suppose that
A116:   x99 <=' y99 and
A117:   z9 <=' y9;
A118:   +(y,z) = y9 -' z9 by A112,A117,ARYTM_1:def 2;
        then consider x9,yz9 being Element of REAL+ such that
A119:   x = [0,x9] and
A120:   +(y,z) = yz9 & +(x,+(y,z)) = yz9 - x9 by A107,Def1;
A121:   x9 = x99 by A113,A119,XTUPLE_0:1;
        then
A122:   +(x,y) = y9 -' x9 by A110,A114,A115,A116,ARYTM_1:def 2;
        then consider z99,xy99 being Element of REAL+ such that
A123:   z = [0,z99] and
A124:   +(x,y) = xy99 & +(z,+(x,y)) = xy99 - z99 by A109,Def1;
        z9 = z99 by A111,A123,XTUPLE_0:1;
        hence thesis by A110,A114,A116,A117,A118,A120,A121,A122,A124,ARYTM_1:25
;
      end;
      suppose that
A125:   not x99 <=' y99 and
A126:   z9 <=' y9;
A127:   not(z in REAL+ & +(x,y) in [:{0},REAL+:]) by A109,Th5,XBOOLE_0:3;
A128:   +(y,x) = [0,x99 -' y99] by A115,A125,ARYTM_1:def 2;
        then +(y,x) in [:{0},REAL+:] by Lm1,ZFMISC_1:87;
        then not(+(x,y) in REAL+ & z in [:{0},REAL+:]) by Th5,XBOOLE_0:3;
        then consider z99,yx9 being Element of REAL+ such that
A129:   z = [0,z99] and
A130:   +(y,x) = [0,yx9] & +(z,+(y,x)) = [0,z99 + yx9] by A128,A127,Def1,Lm1,
ZFMISC_1:87;
A131:   z9 = z99 by A111,A129,XTUPLE_0:1;
A132:   +(y,z) = y9 -' z9 by A112,A126,ARYTM_1:def 2;
        then consider x9,yz9 being Element of REAL+ such that
A133:   x = [0,x9] and
A134:   +(y,z) = yz9 and
A135:   +(x,+(y,z)) = yz9 - x9 by A107,Def1;
A136:   x9 = x99 by A113,A133,XTUPLE_0:1;
        yz9 <=' y9 by A132,A134,ARYTM_1:11;
        then not x9 <=' yz9 by A110,A114,A125,A136,ARYTM_1:3;
        hence +(x,+(y,z)) = [0,x9 -' (y9 -' z9)] by A132,A134,A135,
ARYTM_1:def 2
          .= [0,x99 -' y99 + z99] by A110,A114,A125,A126,A136,A131,ARYTM_1:14
          .= +(+(x,y),z) by A128,A130,XTUPLE_0:1;
      end;
      suppose that
A137:   not z9 <=' y9 and
A138:   x99 <=' y99;
A139:   not(x in REAL+ & +(z,y) in [:{0},REAL+:]) by A107,Th5,XBOOLE_0:3;
A140:   +(y,z) = [0,z9 -' y9] by A112,A137,ARYTM_1:def 2;
        then +(y,z) in [:{0},REAL+:] by Lm1,ZFMISC_1:87;
        then not(+(z,y) in REAL+ & x in [:{0},REAL+:]) by Th5,XBOOLE_0:3;
        then consider x9,yz99 being Element of REAL+ such that
A141:   x = [0,x9] and
A142:   +(y,z) = [0,yz99] & +(x,+(y,z)) = [0,x9 + yz99] by A140,A139,Def1,Lm1,
ZFMISC_1:87;
A143:   x99 = x9 by A113,A141,XTUPLE_0:1;
A144:   +(y,x) = y99 -' x99 by A115,A138,ARYTM_1:def 2;
        then consider z99,yx99 being Element of REAL+ such that
A145:   z = [0,z99] and
A146:   +(y,x) = yx99 and
A147:   +(z,+(y,x)) = yx99 - z99 by A109,Def1;
A148:   z99 = z9 by A111,A145,XTUPLE_0:1;
        yx99 <=' y99 by A144,A146,ARYTM_1:11;
        then
A149:   not z99 <=' yx99 by A110,A114,A137,A148,ARYTM_1:3;
        thus +(x,+(y,z)) = [0,z9 -' y9 + x9] by A140,A142,XTUPLE_0:1
          .= [0,z99 -' (y99 -' x99)] by A110,A114,A137,A138,A148,A143,
ARYTM_1:14
          .= +(+(x,y),z) by A144,A146,A147,A149,ARYTM_1:def 2;
      end;
      suppose that
A150:   not x99 <=' y99 and
A151:   not z9 <=' y9;
A152:   not(x in REAL+ & +(z,y) in [:{0},REAL+:]) by A107,Th5,XBOOLE_0:3;
A153:   not(z in REAL+ & +(x,y) in [:{0},REAL+:]) by A109,Th5,XBOOLE_0:3;
A154:   +(y,x) = [0,x99 -' y99] by A115,A150,ARYTM_1:def 2;
        then +(y,x) in [:{0},REAL+:] by Lm1,ZFMISC_1:87;
        then not(+(x,y) in REAL+ & z in [:{0},REAL+:]) by Th5,XBOOLE_0:3;
        then consider z99,yx9 being Element of REAL+ such that
A155:   z = [0,z99] and
A156:   +(y,x) = [0,yx9] & +(z,+(y,x)) = [0,z99 + yx9] by A154,A153,Def1,Lm1,
ZFMISC_1:87;
A157:   z9 = z99 by A111,A155,XTUPLE_0:1;
A158:   +(y,z) = [0,z9 -' y9] by A112,A151,ARYTM_1:def 2;
        then +(y,z) in [:{0},REAL+:] by Lm1,ZFMISC_1:87;
        then not(+(z,y) in REAL+ & x in [:{0},REAL+:]) by Th5,XBOOLE_0:3;
        then consider x9,yz99 being Element of REAL+ such that
A159:   x = [0,x9] and
A160:   +(y,z) = [0,yz99] & +(x,+(y,z)) = [0,x9 + yz99] by A158,A152,Def1,Lm1,
ZFMISC_1:87;
A161:   x9 = x99 by A113,A159,XTUPLE_0:1;
        thus +(x,+(y,z)) = [0,z9 -' y9 + x9] by A158,A160,XTUPLE_0:1
          .= [0,x99 -' y99 + z99] by A110,A114,A150,A151,A157,A161,ARYTM_1:15
          .= +(+(x,y),z) by A154,A156,XTUPLE_0:1;
      end;
    end;
    hence thesis;
  end;
  suppose that
A162: z in REAL+ and
A163: y in [:{0},REAL+:] and
A164: x in [:{0},REAL+:];
    ( not(x in REAL+ & y in [:{0},REAL+:]))& not(y in REAL+ & x in [:{0}
    ,REAL+:]) by A163,A164,Th5,XBOOLE_0:3;
    then consider y9,x9 being Element of REAL+ such that
A165: y = [0,y9] and
A166: x = [0,x9] and
A167: +(y,x) = [0,y9 + x9] by A163,Def1;
    +(y,x) in [:{0},REAL+:] by A167,Lm1,ZFMISC_1:87;
    then consider z9,yx9 being Element of REAL+ such that
A168: z = z9 and
A169: +(y,x) = [0,yx9] and
A170: +(z,+(y,x)) = z9 - yx9 by A162,Def1;
    consider z99,y99 being Element of REAL+ such that
A171: z = z99 and
A172: y = [0,y99] and
A173: +(z,y) = z99 - y99 by A162,A163,Def1;
A174: y9 = y99 by A165,A172,XTUPLE_0:1;
    now
      per cases;
      suppose
A175:   y99 <=' z99;
        then
A176:   +(z,y) = z99 -' y99 by A173,ARYTM_1:def 2;
        then consider zy99,x99 being Element of REAL+ such that
A177:   +(z,y) = zy99 and
A178:   x = [0,x99] and
A179:   +(+(z,y),x) = zy99 - x99 by A164,Def1;
A180:   x9 = x99 by A166,A178,XTUPLE_0:1;
        thus +(z,+(y,x)) = z9 - (y9 + x9) by A167,A169,A170,XTUPLE_0:1
          .= +(+(z,y),x) by A171,A168,A174,A175,A176,A177,A179,A180,ARYTM_1:24;
      end;
      suppose
A181:   not y99 <=' z99;
A182:   not(x in REAL+ & +(z,y) in [:{0},REAL+:]) by A164,Th5,XBOOLE_0:3;
A183:   +(z,y) = [0,y99 -' z99] by A173,A181,ARYTM_1:def 2;
        then +(z,y) in [:{0},REAL+:] by Lm1,ZFMISC_1:87;
        then not(+(z,y) in REAL+ & x in [:{0},REAL+:]) by Th5,XBOOLE_0:3;
        then consider zy99,x99 being Element of REAL+ such that
A184:   +(z,y) = [0,zy99] and
A185:   x = [0,x99] and
A186:   +(+(z,y),x) = [0,zy99 + x99] by A183,A182,Def1,Lm1,ZFMISC_1:87;
A187:   x9 = x99 by A166,A185,XTUPLE_0:1;
A188:   yx9 = x9 + y9 by A167,A169,XTUPLE_0:1;
        then y99 <=' yx9 by A174,ARYTM_2:19;
        then not yx9 <=' z9 by A171,A168,A181,ARYTM_1:3;
        hence +(z,+(y,x)) = [0,x9 + y9 -' z9] by A170,A188,ARYTM_1:def 2
          .= [0,x99 + (y99 -' z99)] by A171,A168,A174,A181,A187,ARYTM_1:13
          .= +(+(z,y),x) by A183,A184,A186,XTUPLE_0:1;
      end;
    end;
    hence thesis;
  end;
  suppose that
A189: x in [:{0},REAL+:] and
A190: y in [:{0},REAL+:] and
A191: z in [:{0},REAL+:];
A192: not(x in REAL+ & +(z,y) in [:{0},REAL+:]) by A189,Th5,XBOOLE_0:3;
    ( not(z in REAL+ & y in [:{0},REAL+:]))& not(y in REAL+ & z in [:{0}
    ,REAL+:]) by A190,A191,Th5,XBOOLE_0:3;
    then consider y9,z9 being Element of REAL+ such that
A193: y = [0,y9] and
A194: z = [0,z9] and
A195: +(y,z) = [0,y9 + z9] by A190,Def1;
    +(z,y) in [:{0},REAL+:] by A195,Lm1,ZFMISC_1:87;
    then not(+(z,y) in REAL+ & x in [:{0},REAL+:]) by Th5,XBOOLE_0:3;
    then consider x9,yz9 being Element of REAL+ such that
A196: x = [0,x9] and
A197: +(y,z) = [0,yz9] & +(x,+(y,z)) = [0,x9 + yz9] by A195,A192,Def1,Lm1,
ZFMISC_1:87;
A198: not(z in REAL+ & +(x,y) in [:{0},REAL+:]) by A191,Th5,XBOOLE_0:3;
    ( not(x in REAL+ & y in [:{0},REAL+:]))& not(y in REAL+ & x in [:{0}
    ,REAL+:]) by A189,A190,Th5,XBOOLE_0:3;
    then consider x99,y99 being Element of REAL+ such that
A199: x = [0,x99] and
A200: y = [0,y99] and
A201: +(x,y) = [0,x99 + y99] by A189,Def1;
A202: x9 = x99 by A199,A196,XTUPLE_0:1;
    +(x,y) in [:{0},REAL+:] by A201,Lm1,ZFMISC_1:87;
    then not(+(x,y) in REAL+ & z in [:{0},REAL+:]) by Th5,XBOOLE_0:3;
    then consider xy99,z99 being Element of REAL+ such that
A203: +(x,y) = [0,xy99] and
A204: z = [0,z99] and
A205: +(+(x,y),z) = [0,xy99 + z99] by A201,A198,Def1,Lm1,ZFMISC_1:87;
A206: z9 = z99 by A194,A204,XTUPLE_0:1;
A207: y9 = y99 by A193,A200,XTUPLE_0:1;
    thus +(x,+(y,z)) = [0,z9 + y9 + x9] by A195,A197,XTUPLE_0:1
      .= [0,z99 + (y99 + x99)] by A206,A202,A207,ARYTM_2:6
      .= +(+(x,y),z) by A201,A203,A205,XTUPLE_0:1;
  end;
end;

theorem
  [*x,y*] in REAL implies y = 0
proof
  assume
A1: [*x,y*] in REAL;
  assume y <> 0;
  then [*x,y*] = (0,1) --> (x,y) by Def5;
  hence contradiction by A1,Th8;
end;

theorem
  for x,y being Element of REAL holds opp +(x,y) = +(opp x, opp y)
proof
  let x,y be Element of REAL;
  +(+(x,y),+(opp x, opp y)) = +(x,+(y,+(opp x, opp y))) by Th23
    .= +(x,+(opp x,+(y, opp y))) by Th23
    .= +(x,+(opp x,o)) by Def3
    .= +(x,opp x) by Th11
    .= 0 by Def3;
  hence thesis by Def3;
end;














begin

definition
  func <i> equals
  (0,1) --> (0,1);
  coherence;
  let c be number;
  attr c is complex means
  :Def2:
  c in COMPLEX;
end;

registration
  cluster <i> -> complex;
  coherence
  proof
    set X = { x where x is Element of Funcs({0,1},REAL): x.1 = 0};
A1: now
      assume <i> in X;
then    ex x being Element of Funcs({0,1},REAL) st <i> = x & x.1 = 0;
      hence contradiction by FUNCT_4:63;
    end;
 <i> in Funcs({0,1},REAL) by FUNCT_2:8;
then  <i> in Funcs({0,1},REAL) \ X by A1,XBOOLE_0:def 5;
    hence <i> in COMPLEX by NUMBERS:def 2,XBOOLE_0:def 3;
  end;
end;

registration
  cluster complex for number;
  existence
  proof
    take <i>;
    thus thesis;
  end;
end;

registration
  sethood of complex number
  proof
    take COMPLEX;
    thus thesis by Def2;
  end;
end;

notation
  let x be complex number;
  synonym x is zero for x is empty;
end;

definition
  let x be complex number;
  redefine attr x is zero means
  x = 0;
  compatibility;
end;

definition
  let x,y be complex number;
 x in COMPLEX by Def2;
  then consider x1,x2 being Element of REAL such that
A1: x = [*x1,x2*] by ARYTM_0:9;
 y in COMPLEX by Def2;
  then consider y1,y2 being Element of REAL such that
A2: y = [*y1,y2*] by ARYTM_0:9;
  func x+y means
  :Def4:
  ex x1,x2,y1,y2 being Element of REAL st
  x = [*x1,x2*] & y = [*y1,y2*] & it = [*+(x1,y1),+(x2,y2)*];
  existence
  proof
    take [*+(x1,y1),+(x2,y2)*];
    thus thesis by A1,A2;
  end;
  uniqueness
  proof
    let c1,c2 be number;
    given x1,x2,y1,y2 being Element of REAL such that
A3: x = [*x1,x2*] and
A4: y = [*y1,y2*] and
A5: c1 = [*+(x1,y1),+(x2,y2)*];
    given x19,x29,y19,y29 being Element of REAL such that
A6: x = [*x19,x29*] and
A7: y = [*y19,y29*] and
A8: c2 = [*+(x19,y19),+(x29,y29)*];
A9: x1 = x19 & x2 = x29 by A3,A6,ARYTM_0:10;
 y1 = y19 by A4,A7,ARYTM_0:10;
    hence thesis by A4,A5,A7,A8,A9,ARYTM_0:10;
  end;
  commutativity;
  func x*y means
  :Def5:
  ex x1,x2,y1,y2 being Element of REAL st x = [*x1,x2*] & y = [*y1,y2*] &
  it = [* +(*(x1,y1),opp*(x2,y2)), +(*(x1,y2),*(x2,y1)) *];
  existence
  proof
    take [* +(*(x1,y1),opp*(x2,y2)), +(*(x1,y2),*(x2,y1)) *];
    thus thesis by A1,A2;
  end;
  uniqueness
  proof
    let c1,c2 be number;
    given x1,x2,y1,y2 being Element of REAL such that
A10: x = [*x1,x2*] and
A11: y = [*y1,y2*] and
A12: c1 = [*+(*(x1,y1),opp*(x2,y2)), +(*(x1,y2),*(x2,y1))*];
    given x19,x29,y19,y29 being Element of REAL such that
A13: x = [*x19,x29*] and
A14: y = [*y19,y29*] and
A15: c2 = [*+(*(x19,y19),opp*(x29,y29)), +(*(x19,y29),*(x29,y19))*];
A16: x1 = x19 & x2 = x29 by A10,A13,ARYTM_0:10;
 y1 = y19 & y2 = y29 by A11,A14,ARYTM_0:10;
    hence thesis by A12,A15,A16;
  end;
  commutativity;
end;

Lm1: 0 = [*0,0*] by ARYTM_0:def 5;

reconsider j = 1 as Element of REAL;

Lm2: for x,y,z being Element of REAL st +(x,y) = 0 & +(x,z) = 0 holds y = z
proof
  let x,y,z be Element of REAL such that
A1: +(x,y) = 0 and
A2: +(x,z) = 0;
  per cases;
  suppose
 x in REAL+ & y in REAL+ & z in REAL+;
    then  (
 ex x9,y9 being Element of REAL+ st x = x9 & y = y9 & 0 = x9 + y9)& ex x9,y9
    being Element of REAL+ st x = x9 & z = y9 & 0 = x9 + y9 by A1,A2,
ARYTM_0:def 1;
    hence thesis by ARYTM_2:11;
  end;
  suppose that
A3: x in REAL+ and
A4: y in REAL+ and
A5: z in [:{0},REAL+:];
A6: ex x9,y9 being Element of REAL+ st x = x9 & y = y9 & 0 =
    x9 + y9 by A1,A3,A4,ARYTM_0:def 1;
    consider x99,y99 being Element of REAL+ such that
A7: x = x99 and
A8: z = [0,y99] & 0 = x99 - y99 by A2,A3,A5,ARYTM_0:def 1;
A9: x99 = 0 by A6,A7,ARYTM_2:5;
 [{},{}] in {[{},{}]} by TARSKI:def 1;
then A10: not [{},{}] in REAL by NUMBERS:def 1,XBOOLE_0:def 5;
 z in REAL;
    hence thesis by A8,A9,A10,ARYTM_1:19;
  end;
  suppose that
A11: x in REAL+ and
A12: z in REAL+ and
A13: y in [:{0},REAL+:];
A14: ex x9,z9 being Element of REAL+ st x = x9 & z = z9 & 0 =
    x9 + z9 by A2,A11,A12,ARYTM_0:def 1;
    consider x99,y9 being Element of REAL+ such that
A15: x = x99 and
A16: y = [0,y9] & 0 = x99 - y9 by A1,A11,A13,ARYTM_0:def 1;
A17: x99 = 0 by A14,A15,ARYTM_2:5;
 [0,0] in {[0,0]} by TARSKI:def 1;
then A18: not [0,0] in REAL by NUMBERS:def 1,XBOOLE_0:def 5;
 y in REAL;
    hence thesis by A16,A17,A18,ARYTM_1:19;
  end;
  suppose that
A19: x in REAL+ and
A20: y in [:{0},REAL+:] and
A21: z in [:{0},REAL+:];
    consider x9,y9 being Element of REAL+ such that
A22: x = x9 and
A23: y = [0,y9] and
A24: 0 = x9 - y9 by A1,A19,A20,ARYTM_0:def 1;
    consider x99,z9 being Element of REAL+ such that
A25: x = x99 and
A26: z = [0,z9] and
A27: 0 = x99 - z9 by A2,A19,A21,ARYTM_0:def 1;
 y9 = x9 by A24,ARYTM_0:6
      .= z9 by A22,A25,A27,ARYTM_0:6;
    hence thesis by A23,A26;
  end;
  suppose that
A28: z in REAL+ and
A29: y in REAL+ and
A30: x in [:{0},REAL+:];
    consider x9,y9 being Element of REAL+ such that
A31: x = [0,x9] and
A32: y = y9 and
A33: 0 = y9 - x9 by A1,A29,A30,ARYTM_0:def 1;
    consider x99,z9 being Element of REAL+ such that
A34: x = [0,x99] and
A35: z = z9 and
A36: 0 = z9 - x99 by A2,A28,A30,ARYTM_0:def 1;
 x9 = x99 by A31,A34,XTUPLE_0:1;
then  z9 = x9 by A36,ARYTM_0:6
      .= y9 by A33,ARYTM_0:6;
    hence thesis by A32,A35;
  end;
  suppose
 not(x in REAL+ & y in REAL+) & not(x in REAL+ & y in [:{0},REAL+:]) &
    not(y in REAL+ & x in [:{0},REAL+:]);
then  ex x9,y9 being Element of REAL+
    st x = [0,x9] & y = [0,y9] & 0 = [0,x9+y9] by A1,ARYTM_0:def 1;
    hence thesis;
  end;
  suppose
 not(x in REAL+ & z in REAL+) & not(x in REAL+ & z in [:{0},REAL+:]) &
    not(z in REAL+ & x in [:{0},REAL+:]);
then  ex x9,z9 being Element of REAL+
    st x = [0,x9] & z = [0,z9] & 0 = [0,x9+z9] by A2,ARYTM_0:def 1;
    hence thesis;
  end;
end;

registration
  let z,z9 be complex number;
  cluster z+z9 -> complex;
  coherence
  proof
 ex x1,x2,y1,y2 being Element of REAL st
    z = [*x1,x2*] & z9 = [*y1,y2*] & z+z9 = [*+(x1,y1),+(x2,y2)*] by Def4;
    hence z+z9 in COMPLEX;
  end;
  cluster z*z9 -> complex;
  coherence
  proof
     ex
 x1,x2,y1,y2 being Element of REAL st z = [*x1,x2*] & z9 = [*y1,y2*] &
    z*z9 = [* +(*(x1,y1),opp*(x2,y2)), +(*(x1,y2),*(x2,y1)) *] by Def5;
    hence z*z9 in COMPLEX;
  end;
end;

definition
  let z be complex number;
 z in COMPLEX by Def2;
  then consider x,y being Element of REAL such that
A1: z = [*x,y*] by ARYTM_0:9;
  func -z -> complex number means
  :Def6:
  z + it = 0;
  existence
  proof
    reconsider z9 = [*opp x, opp y*] as complex number by Def2;
    take z9;
 0 = +(x,opp x) & 0 = +(y,opp y) by ARYTM_0:def 3;
    hence thesis by A1,Def4,Lm1;
  end;
  uniqueness
  proof
    let c1,c2 be complex number such that
A2: z+c1 = 0 and
A3: z+c2 = 0;
    consider x1,x2,y1,y2 being Element of REAL such that
A4: z = [*x1,x2*] and
A5: c1 = [*y1,y2*] and
A6: 0 = [*+(x1,y1),+(x2,y2)*] by A2,Def4;
    consider x19,x29,y19,y29 being Element of REAL such that
A7: z = [*x19,x29*] and
A8: c2 = [*y19,y29*] and
A9: 0 = [*+(x19,y19),+(x29,y29)*] by A3,Def4;
A10: x1 = x19 by A4,A7,ARYTM_0:10;
A11: x2 = x29 by A4,A7,ARYTM_0:10;
A12: +(x1,y1) = 0 by A6,Lm1,ARYTM_0:10;
 +(x1,y19) = 0 by A9,A10,Lm1,ARYTM_0:10;
then A13: y1 = y19 by A12,Lm2;
A14: +(x2,y2) = 0 by A6,Lm1,ARYTM_0:10;
 +(x2,y29) = 0 by A9,A11,Lm1,ARYTM_0:10;
    hence thesis by A5,A8,A13,A14,Lm2;
  end;
  involutiveness;
  func z" -> complex number means
  :Def7:
  z*it = 1 if z <> 0 otherwise it = 0;
  existence
  proof
    thus z <> 0 implies ex z9 being complex number st z*z9 = 1
    proof
      set y1 = *(x,inv +(*(x,x),*(y,y))), y2 = *(opp y,inv +(*(x,x),*(y,y)));
      set z9 = [* y1, y2 *];
      reconsider z9 as complex number by Def2;
      assume
A15:  z <> 0;
      take z9;
A16:  opp *(y,y2) = opp *(y,opp *(y,inv +(*(x,x),*(y,y)))) by ARYTM_0:15
        .= opp opp *(y,*(y,inv +(*(x,x),*(y,y)))) by ARYTM_0:15
        .= *(*(y,y),inv +(*(x,x),*(y,y))) by ARYTM_0:13;
A17:  *(x,y1) = *(*(x,x),inv +(*(x,x),*(y,y))) by ARYTM_0:13;
A18:  now
        assume +(*(x,x),*(y,y)) = 0;
then     x = 0 & y = 0 by ARYTM_0:17;
        hence contradiction by A1,A15,ARYTM_0:def 5;
      end;
  *(x,y2) = *(opp y,y1) by ARYTM_0:13
        .= opp *(y,y1) by ARYTM_0:15;
then   +(*(x,y2),*(y,y1)) = 0 by ARYTM_0:def 3;
then   [* +(*(x,y1),opp*(y,y2)), +(*(x,y2),*(y,y1)) *]
      = +(*(x,y1),opp*(y,y2)) by ARYTM_0:def 5
        .=*(+(*(x,x),*(y,y)),inv +(*(x,x),*(y,y))) by A16,A17,ARYTM_0:14
        .= 1 by A18,ARYTM_0:def 4;
      hence thesis by A1,Def5;
    end;
    assume z = 0;
    hence thesis;
  end;
  uniqueness
  proof
    let c1,c2 be complex number;
    thus z <> 0 & z*c1 = 1 & z*c2 = 1 implies c1 = c2
    proof
      assume that
A19:  z <> 0 and
A20:  z*c1 = 1 and
A21:  z*c2 = 1;
A22:  for z9 being complex number st z*z9 = 1
holds z9 = [* *(x,inv +(*(x,x),*(y,y))) , *(opp y,inv +(*(x,x),*(y,y))) *]
      proof
        let z9 being complex number such that
A23:    z*z9 = 1;
        consider x1,x2,x9,y9 being Element of REAL such that
A24:    z = [*x1,x2*] and
A25:    z9 = [*x9,y9*] and
A26:    1 = [* +(*(x1,x9),opp*(x2,y9)), +(*(x1,y9),*(x2,x9)) *] by A23,Def5;
A27:    x = x1 & y = x2 by A1,A24,ARYTM_0:10;
        per cases by A1,A19,ARYTM_0:def 5;
        suppose that
A28:      x = 0 and
A29:      y <> 0;
      +(y, opp y) = 0 by ARYTM_0:def 3;
then A30:      opp y <> 0 by A29,ARYTM_0:11;
      *(x,y9) = 0 & *(x,x9) = 0 by A28,ARYTM_0:12;
then A31:      1 = [* opp*(y,y9), +(0,*(y,x9)) *] by A26,A27,ARYTM_0:11
            .= [* opp*(y,y9), *(y,x9) *] by ARYTM_0:11;
A32:      1 = [*j,0*] by ARYTM_0:def 5;
      *(opp y,y9) = opp*(y,y9) by ARYTM_0:15
            .= 1 by A31,A32,ARYTM_0:10;
then A33:      y9 = inv opp y by A30,ARYTM_0:def 4;
A34:      *(x,x) = 0 by A28,ARYTM_0:12;
      *(opp y,opp inv y) = opp *(y,opp inv y) by ARYTM_0:15
            .= opp opp *(y,inv y) by ARYTM_0:15
            .= 1 by A29,ARYTM_0:def 4;
then A35:      inv opp y = opp inv y by A30,ARYTM_0:def 4;
      *(y,x9) = 0 by A31,A32,ARYTM_0:10;
          hence z9 = [*0,inv opp y*] by A25,A29,A33,ARYTM_0:21
            .= [* 0 , opp *(j,inv y) *] by A35,ARYTM_0:19
            .= [* 0 , opp *(*(y,inv y),inv y) *] by A29,ARYTM_0:def 4
            .= [* 0 , opp *(y,*(inv y,inv y)) *] by ARYTM_0:13
            .= [* 0 , *(opp y,*(inv y,inv y)) *] by ARYTM_0:15
            .= [* 0 , *(opp y,inv *(y,y)) *] by ARYTM_0:22
            .= [* 0 , *(opp y,inv +(0,*(y,y))) *] by ARYTM_0:11
            .= [* *(x,inv +(*(x,x),*(y,y))) , *(opp y,inv +(*(x,x),*(y,y))) *]
          by A28,A34,ARYTM_0:12;
        end;
        suppose that
A36:      opp y = 0 and
A37:      x <> 0;
      +(y,opp y) = 0 by ARYTM_0:def 3;
then A38:      y = 0 by A36,ARYTM_0:11;
then A39:      *(y,x9) = 0 by ARYTM_0:12;
      opp*(y,y9) = *(opp y,y9) by ARYTM_0:15
            .= 0 by A36,ARYTM_0:12;
then A40:      1 = [* *(x,x9), +(*(x,y9),0) *] by A26,A27,A39,ARYTM_0:11
            .= [* *(x,x9), *(x,y9) *] by ARYTM_0:11;
A41:      1 = [*j,0*] by ARYTM_0:def 5;
then       *(x,x9) = 1 by A40,ARYTM_0:10;
then A42:      x9 = inv x by A37,ARYTM_0:def 4;
      *(x,y9) = 0 by A40,A41,ARYTM_0:10;
then A43:      y9 = 0 by A37,ARYTM_0:21;
A44:      *(y,y) = 0 by A38,ARYTM_0:12;
      x9 = *(j,inv x) by A42,ARYTM_0:19
            .= *(*(x,inv x),inv x) by A37,ARYTM_0:def 4
            .= *(x,*(inv x,inv x)) by ARYTM_0:13
            .= *(x,inv *(x,x)) by ARYTM_0:22
            .= *(x,inv +(*(x,x),0)) by ARYTM_0:11;
          hence thesis by A25,A36,A43,A44,ARYTM_0:12;
        end;
        suppose that
A45:      opp y <> 0 and
A46:      x <> 0;
A47:      now
            assume +(*(*(x,x), inv opp y),opp y) = 0;
then         +(*(*(x,x), inv opp y),*(opp y,j)) = 0 by ARYTM_0:19;
then         +(*(*(x,x), inv opp y),*(opp y,*(opp y, inv opp y))) = 0
            by A45,ARYTM_0:def 4;
then         +(*(*(x,x), inv opp y),*(*(opp y,opp y), inv opp y)) = 0
            by ARYTM_0:13;
then A48:        *(inv opp y, +(*(x,x),*(opp y,opp y))) = 0 by ARYTM_0:14;
        +(*(x,x),*(opp y,opp y)) <> 0 by A46,ARYTM_0:17;
then A49:        inv opp y = 0 by A48,ARYTM_0:21;
        *(opp y,inv opp y) = 1 by A45,ARYTM_0:def 4;
            hence contradiction by A49,ARYTM_0:12;
          end;
          reconsider j = 1 as Element of REAL;
A50:      1 = [*j,0*] by ARYTM_0:def 5;
then       +(*(x1,y9),*(x2,x9)) = 0 by A26,ARYTM_0:10;
then       *(x,y9) = opp*(y,x9) by A27,ARYTM_0:def 3;
then       *(x,y9) = *(opp y,x9) by ARYTM_0:15;
then A51:      x9 = *(*(x,y9), inv opp y) by A45,ARYTM_0:20
            .= *(x,*(y9, inv opp y)) by ARYTM_0:13;
then       +
(*(x,*(x,*(y9, inv opp y))),opp*(y,y9)) = 1 by A26,A27,A50,ARYTM_0:10;
then       +(*(*(x,x),*(y9, inv opp y)),opp*(y,y9)) = 1 by ARYTM_0:13;
then       +(*(*(x,x),*(y9, inv opp y)),*(opp y,y9)) = 1 by ARYTM_0:15;
then       +(*(y9,*(*(x,x), inv opp y)),*(opp y,y9)) = 1 by ARYTM_0:13;
then       *(y9,+(*(*(x,x), inv opp y),opp y)) = 1 by ARYTM_0:14;
then A52:      y9 = inv +(*(*(x,x), inv opp y),opp y) by A47,ARYTM_0:def 4;
then A53:      x9 = *(x,inv *(+(*(*(x,x), inv opp y),opp y), opp y)) by A51,
ARYTM_0:22
.= *(x,inv +(*(*(*(x,x), inv opp y),opp y),*(opp y, opp y))) by ARYTM_0:14
.= *(x,inv +(*(*(*(x,x), inv opp y),opp y),opp*(y, opp y))) by ARYTM_0:15
.= *(x,inv +(*(*(*(x,x), inv opp y),opp y),opp opp*(y,y))) by ARYTM_0:15
            .= *(x,inv +(*(*(x,x), *(inv opp y,opp y)),*(y,y))) by ARYTM_0:13
            .= *(x,inv +(*(*(x,x), j),*(y,y))) by A45,ARYTM_0:def 4
            .= *(x,inv +(*(x,x),*(y,y))) by ARYTM_0:19;
      y9 = *(j,inv +(*(*(x,x), inv opp y),opp y)) by A52,ARYTM_0:19
            .= *(*(opp y, inv opp y),inv +(*(*(x,x), inv opp y),opp y))
          by A45,ARYTM_0:def 4
.= *(opp y, *(inv opp y,inv +(*(*(x,x), inv opp y),opp y))) by ARYTM_0:13
.= *(opp y, inv *(opp y,+(*(*(x,x), inv opp y),opp y))) by ARYTM_0:22
            .= *(opp y, inv +(*(opp y,*(*(x,x), inv opp y)),*(opp y,opp y)))
          by ARYTM_0:14
            .= *(opp y, inv +(*(*(x,x),*(opp y, inv opp y)),*(opp y,opp y)))
          by ARYTM_0:13
            .= *(opp y, inv +(*(*(x,x),j),*(opp y,opp y))) by A45,ARYTM_0:def 4
            .= *(opp y, inv +(*(x,x),*(opp y,opp y))) by ARYTM_0:19
            .= *(opp y, inv +(*(x,x),opp *(y,opp y))) by ARYTM_0:15
            .= *(opp y, inv +(*(x,x),opp opp *(y,y))) by ARYTM_0:15
            .= *(opp y, inv +(*(x,x),*(y,y)));
          hence thesis by A25,A53;
        end;
      end;
      thus
      then c1 = [* *(x,inv +(*(x,x),*(y,y))) , *(opp y,inv +(*(x,x),*(y,y)))
      *] by A20
        .= c2 by A21,A22;
    end;
    thus thesis;
  end;
  consistency;
  involutiveness
  proof
    let z,z9 be complex number;
    assume that
A54: z9 <> 0 implies z9*z = 1 and
A55: z9 = 0 implies z = 0;
    thus z <> 0 implies z*z9 = 1 by A54,A55;
    assume
A56: z = 0;
    assume z9 <> 0;
    then consider x1,x2,y1,y2 being Element of REAL such that
A57: z = [*x1,x2*] and z9 = [*y1,y2*] and
A58: 1 = [*+(*(x1,y1),opp*(x2,y2)), +(*(x1,y2),*(x2,y1))*] by A54,Def5;
A59: z = [*0,0*] by A56,ARYTM_0:def 5;
then A60: x1 = 0 by A57,ARYTM_0:10;
A61: x2 = 0 by A57,A59,ARYTM_0:10;
A62: *(x1,y1) = 0 by A60,ARYTM_0:12;
 *(x2,y2) = 0 by A61,ARYTM_0:12;
then A63: +(*(x1,y1),opp*(x2,y2)) = 0 by A62,ARYTM_0:def 3;
A64: *(x1,y2) = 0 by A60,ARYTM_0:12;
 *(x2,y1) = 0 by A61,ARYTM_0:12;
then  +(*(x1,y2),*(x2,y1)) = 0 by A64,ARYTM_0:11;
    hence contradiction by A58,A63,ARYTM_0:def 5;
  end;
end;

definition
  let x,y be complex number;
  func x-y equals
  x+(-y);
  coherence;
  func x/y equals
  x * y";
  coherence;
end;

registration
  let x,y be complex number;
  cluster x-y -> complex;
  coherence;
  cluster x/y -> complex;
  coherence;
end;

Lm3: for x, y, z being complex number holds x * (y * z) = (x * y) * z
proof
  let x, y, z be complex number;
  consider x1,x2,y1,y2 being Element of REAL such that
A1: x = [*x1,x2*] and
A2: y = [*y1,y2*] and
A3: x*y = [* +(*(x1,y1),opp*(x2,y2)), +(*(x1,y2),*(x2,y1)) *] by Def5;
  consider y3,y4,z1,z2 being Element of REAL such that
A4: y = [*y3,y4*] and
A5: z = [*z1,z2*] and
A6: y*z = [* +(*(y3,z1),opp*(y4,z2)), +(*(y3,z2),*(y4,z1)) *] by Def5;
A7: y1 = y3 by A2,A4,ARYTM_0:10;
A8: y2 = y4 by A2,A4,ARYTM_0:10;
  consider x3,x4,yz1,yz2 being Element of REAL such that
A9: x = [*x3,x4*] and
A10: y*z = [*yz1,yz2*] and
A11: x*(y*z) = [* +(*(x3,yz1),opp*(x4,yz2)), +(*(x3,yz2),*(x4,yz1)) *] by Def5;
A12: x1 = x3 by A1,A9,ARYTM_0:10;
A13: x2 = x4 by A1,A9,ARYTM_0:10;
  consider xy1,xy2,z3,z4 being Element of REAL such that
A14: x*y = [*xy1,xy2*] and
A15: z = [*z3,z4*] and
A16: (x*y)*z = [* +(*(xy1,z3),opp*(xy2,z4)), +(*(xy1,z4),*(xy2,z3)) *] by Def5;
A17: z1 = z3 by A5,A15,ARYTM_0:10;
A18: z2 = z4 by A5,A15,ARYTM_0:10;
A19: xy1 = +(*(x1,y1),opp*(x2,y2)) by A3,A14,ARYTM_0:10;
A20: xy2 = +(*(x1,y2),*(x2,y1)) by A3,A14,ARYTM_0:10;
A21: yz1 = +(*(y3,z1),opp*(y4,z2)) by A6,A10,ARYTM_0:10;
A22: yz2 = +(*(y3,z2),*(y4,z1)) by A6,A10,ARYTM_0:10;
 +(*(opp x4,*(y3,z2)),*(opp x4,*(y4,z1)))
  = +(*(opp x4,*(y4,z1)),*(*(opp x2,y1),z4)) by A7,A13,A18,ARYTM_0:13
    .= +(*(*(opp x2,y2),z3),*(*(opp x2,y1),z4)) by A8,A13,A17,ARYTM_0:13;
then A23: +(*(x3,*(opp y4,z2)),+(*(opp x4,*(y3,z2)),*(opp x4,*(y4,z1)) ))
  = +(*(*(x1,opp y2),z4), +(*(*(opp x2,y2),z3),*(*(opp x2,y1),z4)))
  by A8,A12,A18,ARYTM_0:13
    .= +(*(opp *(x1,y2),z4), +(*(*(opp x2,y2),z3),*(*(opp x2,y1),z4)))
  by ARYTM_0:15
    .= +(*(*(opp x1,y2),z4), +(*(*(opp x2,y2),z3),*(*(opp x2,y1),z4)))
  by ARYTM_0:15
    .= +(*(*(opp x2,y2),z3), +(*(*(opp x1,y2),z4),*(*(opp x2,y1),z4)))
  by ARYTM_0:23;
A24: +(*(x3,yz1),opp*(x4,yz2)) = +(*(x3,yz1),*(opp x4,yz2)) by ARYTM_0:15
    .= +(*(x3,+(*(y3,z1),*(opp y4,z2))),*(opp x4,yz2)) by A21,ARYTM_0:15
    .= +(+(*(x3,*(y3,z1)),*(x3,*(opp y4,z2))),
  *(opp x4,+(*(y3,z2),*(y4,z1)))) by A22,ARYTM_0:14
    .= +(+(*(x3,*(y3,z1)),*(x3,*(opp y4,z2))),
  +(*(opp x4,*(y3,z2)),*(opp x4,*(y4,z1)))) by ARYTM_0:14
    .= +(*(x3,*(y3,z1)),+(*(*(opp x2,y2),z3),
  +(*(*(opp x1,y2),z4),*(*(opp x2,y1),z4)))) by A23,ARYTM_0:23
    .= +(+(*(x3,*(y3,z1)),*(*(opp x2,y2),z3)),
  +(*(*(opp x1,y2),z4),*(*(opp x2,y1),z4))) by ARYTM_0:23
    .= +(+(*(*(x1,y1),z3),*(*(opp x2,y2),z3)),
  +(*(*(opp x1,y2),z4),*(*(opp x2,y1),z4))) by A7,A12,A17,ARYTM_0:13
    .= +(*(+(*(x1,y1),*(opp x2,y2)),z3),
  +(*(*(opp x1,y2),z4),*(*(opp x2,y1),z4))) by ARYTM_0:14
    .= +(*(+(*(x1,y1),*(opp x2,y2)),z3),
  +(*(*(opp x1,y2),z4),*(opp *(x2,y1),z4))) by ARYTM_0:15
    .= +(*(+(*(x1,y1),*(opp x2,y2)),z3),
  +(*(*(opp x1,y2),z4),opp *(*(x2,y1),z4))) by ARYTM_0:15
    .= +(*(+(*(x1,y1),*(opp x2,y2)),z3),
  +(*(opp *(x1,y2),z4),opp *(*(x2,y1),z4))) by ARYTM_0:15
    .= +(*(+(*(x1,y1),*(opp x2,y2)),z3),
  +(opp *(*(x1,y2),z4),opp *(*(x2,y1),z4))) by ARYTM_0:15
    .= +(*(+(*(x1,y1),*(opp x2,y2)),z3),
  opp +(*(*(x1,y2),z4),*(*(x2,y1),z4))) by ARYTM_0:25
    .= +(*(+(*(x1,y1),*(opp x2,y2)),z3),
  opp*( +(*(x1,y2),*(x2,y1)),z4)) by ARYTM_0:14
    .= +(*(+(*(x1,y1),*(opp x2,y2)),z3),*(opp xy2,z4)) by A20,ARYTM_0:15
    .= +(*(xy1,z3),*(opp xy2,z4)) by A19,ARYTM_0:15
    .= +(*(xy1,z3),opp*(xy2,z4)) by ARYTM_0:15;
A25: +(*(opp*(x2,y2),z4),*(*(x2,y1),z3))
  = +(opp*(*(x2,y2),z4),*(*(x2,y1),z3)) by ARYTM_0:15
    .= +(*(x4,*(y3,z1)),opp*(*(x2,y2),z4)) by A7,A13,A17,ARYTM_0:13
    .= +(*(x4,*(y3,z1)),opp*(x4,*(y4,z2))) by A8,A13,A18,ARYTM_0:13
    .= +(*(x4,*(y3,z1)),*(x4,opp*(y4,z2))) by ARYTM_0:15;
A26: +(*(opp*(x2,y2),z4),*(xy2,z3))
  = +(*(opp*(x2,y2),z4),+(*(*(x1,y2),z3),*(*(x2,y1),z3))) by A20,ARYTM_0:14
    .= +(*(*(x1,y2),z3),+(*(opp*(x2,y2),z4),*(*(x2,y1),z3))) by ARYTM_0:23
    .= +(*(x3,*(y4,z1)),+(*(x4,*(y3,z1)),*(x4,opp*(y4,z2))))
  by A8,A12,A17,A25,ARYTM_0:13
    .= +(*(x3,*(y4,z1)),*(x4,yz1)) by A21,ARYTM_0:14;
 +(*(xy1,z4),*(xy2,z3))
  = +(+(*(*(x1,y1),z4),*(opp*(x2,y2),z4)),*(xy2,z3)) by A19,ARYTM_0:14
    .= +(*(*(x1,y1),z4),+(*(opp*(x2,y2),z4),*(xy2,z3))) by ARYTM_0:23
    .= +(*(x3,*(y3,z2)),+(*(x3,*(y4,z1)),*(x4,yz1)))
  by A7,A12,A18,A26,ARYTM_0:13
    .= +(+(*(x3,*(y3,z2)),*(x3,*(y4,z1))),*(x4,yz1)) by ARYTM_0:23
    .= +(*(x3,yz2),*(x4,yz1)) by A22,ARYTM_0:14;
  hence thesis by A11,A16,A24;
end;

registration
  cluster non zero for complex number;
  existence
  proof
 REAL c= COMPLEX & 1 in REAL by NUMBERS:def 2,XBOOLE_1:7;
then  1 is complex number by Def2;
    hence thesis;
  end;
end;

Lm4: REAL c= COMPLEX by NUMBERS:def 2,XBOOLE_1:7;

registration
  let x be non zero complex number;
  cluster -x -> non zero;
  coherence
  proof
    assume
A1: -x = 0;
 x + -x = 0 by Def6;
    then consider x1,x2,y1,y2 being Element of REAL such that
A2: x = [*x1,x2*] and
A3: -x = [*y1,y2*] and
A4: 0 = [*+(x1,y1),+(x2,y2)*] by Def4;
A5: +(x2,y2) = 0 by A4,ARYTM_0:24;
then A6: +(x1,y1) = 0 by A4,ARYTM_0:def 5;
 y2 = 0 by A1,A3,ARYTM_0:24;
then A7: y1 = 0 by A1,A3,ARYTM_0:def 5;
 x2 = 0 by A1,A3,A5,ARYTM_0:11,24;
then  x = x1 by A2,ARYTM_0:def 5
      .= 0 by A6,A7,ARYTM_0:11;
    hence contradiction;
  end;
  cluster x" -> non zero;
  coherence
  proof
    assume
A8: x" = 0;
 x*x" = 1 by Def7;
    then consider x1,x2,y1,y2 being Element of REAL such that
    x = [*x1,x2*] and
A9: x" = [*y1,y2*] and
A10: 1 = [* +(*(x1,y1),opp*(x2,y2)), +(*(x1,y2),*(x2,y1)) *] by Def5;
 y2 = 0 by A8,A9,ARYTM_0:24;
then A11: y1 = 0 by A8,A9,ARYTM_0:def 5;
 +(*(x1,y2),*(x2,y1)) = 0 by A10,ARYTM_0:24;
then  1 = +(*(x1,y1),opp*(x2,y2)) by A10,ARYTM_0:def 5
      .= +(*(x1,y1),*(opp x2,y2)) by ARYTM_0:15
      .= *(opp x2,y2) by A11,ARYTM_0:11,12;
    hence contradiction by A8,A9,ARYTM_0:12,24;
  end;
  let y be non zero complex number;
  cluster x*y -> non zero;
  coherence
  proof
 1 in REAL;
    then reconsider j =1 as complex number by Def2,Lm4;
    consider u1,u2,v1,v2 being Element of REAL such that
A12: j = [*u1,u2*] and
    A13: y
 = [*v1,v2*] & j*y = [* +(*(u1,v1),opp*(u2,v2)), +(*(u1,v2),*(u2,v1)) *]
    by Def5;
A14: u2 = 0 by A12,ARYTM_0:24;
then A15: +(*(u1,v2),*(u2,v1)) = *(u1,v2) by ARYTM_0:11,12;
A16: u1 = 1 by A12,A14,ARYTM_0:def 5;
 +(0,opp 0) = 0 by ARYTM_0:def 3;
then A17: opp 0 = 0 by ARYTM_0:11;
A18: +(*(u1,v1),opp*(u2,v2)) = +(v1,opp*(u2,v2)) by A16,ARYTM_0:19
      .= +(v1,*(opp u2,v2)) by ARYTM_0:15
      .= +(v1,*(0,v2)) by A12,A17,ARYTM_0:24
      .= v1 by ARYTM_0:11,12;
 0 in REAL;
    then reconsider z =0 as complex number by Def2,Lm4;
    consider u1,u2,v1,v2 being Element of REAL such that
    x" = [*u1,u2*] and
A19: z = [*v1,v2*] and
A20: x"*z = [* +(*(u1,v1),opp*(u2,v2)), +(*(u1,v2),*(u2,v1)) *] by Def5;
 v2 = 0 by A19,ARYTM_0:24;
then A21: v1 = 0 by A19,ARYTM_0:def 5;
then A22: +(*(u1,v1),opp*(u2,v2)) = opp*(u2,v2) by ARYTM_0:11,12
      .= 0 by A17,A19,ARYTM_0:12,24;
A23: +(*(u1,v2),*(u2,v1)) = +(0,*(u2,v1)) by A19,ARYTM_0:12,24
      .= *(u2,v1) by ARYTM_0:11
      .= 0 by A21,ARYTM_0:12;
    assume
A24: x*y = 0;
A25: x"*x*y = x"*(x*y) by Lm3
      .= 0 by A20,A22,A23,A24,ARYTM_0:def 5;
 x"*x*y = j * y by Def7
      .= y by A13,A15,A16,A18,ARYTM_0:19;
    hence contradiction by A25;
  end;
end;

registration
  let x,y be non zero complex number;
  cluster x/y -> non zero;
  coherence;
end;

registration
  cluster -> complex for Element of REAL;
  coherence
  proof
    let n be Element of REAL;
 n in REAL;
    hence thesis by Def2,Lm4;
  end;
end;

registration
  cluster natural -> complex for number;
  coherence
  proof
    let n be set;
    assume n is natural;
then  n in NAT by ORDINAL1:def 12;
    hence thesis;
  end;
end;

registration
  cluster -> complex for Element of COMPLEX;
  coherence by Def2;
end;














begin









reserve x for complex number;

theorem Th1:
  x + 0 = x
proof
  x in COMPLEX by XCMPLX_0:def 2;
  then consider x1,x2 being Element of REAL such that
A1: x = [*x1,x2*] by ARYTM_0:9;
  0 = [*0,0*] by ARYTM_0:def 5;
  then x + 0 = [*+(x1,0),+(x2,0)*] by A1,XCMPLX_0:def 4
    .= [* x1,+(x2,0)*] by ARYTM_0:11
    .= x by A1,ARYTM_0:11;
  hence thesis;
end;

Lm1: -0 = 0
proof
  0 + -0 = -0 by Th1;
  hence thesis by XCMPLX_0:def 6;
end;

Lm2: opp 0 = 0
proof
  +(0,0) = 0 by ARYTM_0:11;
  hence thesis by ARYTM_0:def 3;
end;

theorem Th2:
  x * 0 = 0
proof
  x in COMPLEX by XCMPLX_0:def 2;
  then consider x1,x2 being Element of REAL such that
A1: x = [*x1,x2*] by ARYTM_0:9;
  0 = [*0,0*] by ARYTM_0:def 5;
  then x * 0 = [* +(*(x1,0),opp*(x2,0)), +(*(x1,0),*(x2,0)) *] by A1,
XCMPLX_0:def 5
    .= [* +(*(x1,0),opp 0), +(*(x1,0),*(x2,0)) *] by ARYTM_0:12
    .= [* +(*(x1,0),opp 0), +(*(x1,0),0) *] by ARYTM_0:12
    .= [* +(0,opp 0), +(*(x1,0),0) *] by ARYTM_0:12
    .= [* +(0,opp 0), +(0,0) *] by ARYTM_0:12
    .= [* +(0,opp 0), 0 *] by ARYTM_0:11
    .= [* opp 0, 0 *] by ARYTM_0:11
    .= 0 by Lm2,ARYTM_0:def 5;
  hence thesis;
end;

theorem Th3:
  1 * x = x
proof
  x in COMPLEX by XCMPLX_0:def 2;
  then consider x1,x2 being Element of REAL such that
A1: x = [*x1,x2*] by ARYTM_0:9;
  1 = [*1,0*] by ARYTM_0:def 5;
  then x * 1 = [* +(*(x1,1),opp*(x2,0)), +(*(x1,0),*(x2,1)) *] by A1,
XCMPLX_0:def 5
    .= [* +(*(x1,1),opp 0), +(*(x1,0),*(x2,1)) *] by ARYTM_0:12
    .= [* +(x1,opp 0), +(*(x1,0),*(x2,1)) *] by ARYTM_0:19
    .= [* +(x1,opp 0), +(*(x1,0),x2) *] by ARYTM_0:19
    .= [* +(x1,0), +(0,x2) *] by Lm2,ARYTM_0:12
    .= [* x1, +(0,x2) *] by ARYTM_0:11
    .= x by A1,ARYTM_0:11;
  hence thesis;
end;

theorem
  x - 0 = x
proof
  x - 0 = x + 0 by Lm1,XCMPLX_0:def 8;
  hence thesis by Th1;
end;

theorem
  0 / x = 0
proof
  0 / x = 0 * x" by XCMPLX_0:def 9;
  hence thesis by Th2;
end;

Lm3: 1" = 1
proof
  1 * 1" = 1" by Th3;
  hence thesis by XCMPLX_0:def 7;
end;

theorem
  x / 1 = x
proof
  x / 1 = x * 1 by Lm3,XCMPLX_0:def 9;
  hence thesis by Th3;
end;














begin

reserve x for set;

definition
  let x;
  attr x is ext-real means
  :Def1:
  x in ExtREAL;
end;

registration
  cluster ext-real for number;
  existence
  proof
    take 0;
    thus 0 in ExtREAL by XBOOLE_0:def 3;
  end;
  cluster -> ext-real for Element of ExtREAL;
  coherence by Def1;
end;

registration
  sethood of ext-real number
  proof
    take ExtREAL;
    thus thesis by Def1;
  end;
end;

definition
  func +infty equals
  REAL;
  coherence;
  func -infty equals
  [0,REAL];
  coherence;
end;

definition
  redefine func ExtREAL equals
  REAL \/ {+infty,-infty};
  compatibility;
end;

registration
  cluster +infty -> ext-real;
  coherence
  proof
    +infty in {REAL,[0,REAL]} by TARSKI:def 2;
    then +infty in ExtREAL by XBOOLE_0:def 3;
    hence thesis;
  end;
  cluster -infty -> ext-real;
  coherence
  proof
    -infty in {REAL,[0,REAL]} by TARSKI:def 2;
    then -infty in ExtREAL by XBOOLE_0:def 3;
    hence thesis;
  end;
end;

definition
  let x,y be ext-real number;
  pred x <= y means
  :Def5:
  ex x9,y9 being Element of REAL+ st x = x9 & y = y9
& x9 <=' y9 if x in REAL+ & y in REAL+, ex x9,y9 being Element of REAL+ st x =
  [0,x9] & y = [0,y9] & y9 <=' x9 if x in [:{0},REAL+:] & y in [:{0},REAL+:]
  otherwise y in REAL+ & x in [:{0},REAL+:] or x = -infty or y = +infty;
  consistency by ARYTM_0:5,XBOOLE_0:3;
  reflexivity
  proof
    let x be ext-real number such that
A1: not((x in REAL+ & x in REAL+ implies ex x9,y9 being Element of
REAL+ st x = x9 & x = y9 & x9 <=' y9) & (x in [:{0},REAL+:] & x in [:{0},REAL+
:] implies ex x9,y9 being Element of REAL+ st x = [0,x9] & x = [0,y9] & y9 <='
x9) & (not(x in REAL+ & x in REAL+) & not(x in [:{0},REAL+:] & x in [:{0},REAL+
    :]) implies x in REAL+ & x in [:{0},REAL+:] or x = -infty or x = +infty));
    x in ExtREAL by Def1;
    then
A2: x in REAL+ \/ [:{0},REAL+:] \ {[0,0]} or x in {+infty,-infty} by
XBOOLE_0:def 3;
    per cases by A1;
    suppose that
A3:   x in REAL+ and
A4:   not ex x9,y9 being Element of REAL+ st x = x9 & x = y9 & x9 <=' y9;
      reconsider x9 = x as Element of REAL+ by A3;
      not x9 <=' x9 by A4;
      hence thesis;
    end;
    suppose that
A5:   x in [:{0},REAL+:] and
A6:   not ex x9,y9 being Element of REAL+ st x = [0,x9] & x = [0,y9] &
      y9 <=' x9;
      consider z,x9 being set such that
A7:   z in {0} and
A8:   x9 in REAL+ and
A9:   x = [z,x9] by A5,ZFMISC_1:84;
      reconsider x9 as Element of REAL+ by A8;
      x = [0,x9] by A7,A9,TARSKI:def 1;
      then not x9 <=' x9 by A6;
      hence thesis;
    end;
    suppose
      not(not x in REAL+ & not x in [:{0},REAL+:] implies x in REAL+
      & x in [:{0},REAL+:] or x = -infty or x = +infty);
      hence thesis by A2,TARSKI:def 2,XBOOLE_0:def 3;
    end;
  end;
  connectedness
  proof
    let x,y be ext-real number such that
A10: not((x in REAL+ & y in REAL+ implies ex x9,y9 being Element of
REAL+ st x = x9 & y = y9 & x9 <=' y9) & (x in [:{0},REAL+:] & y in [:{0},REAL+
:] implies ex x9,y9 being Element of REAL+ st x = [0,x9] & y = [0,y9] & y9 <='
x9) & (not(x in REAL+ & y in REAL+) & not(x in [:{0},REAL+:] & y in [:{0},REAL+
    :]) implies y in REAL+ & x in [:{0},REAL+:] or x = -infty or y = +infty));
    x in ExtREAL by Def1;
    then
A11: x in REAL+ \/ [:{0},REAL+:] \ {[0,0]} or x in {+infty,-infty} by
XBOOLE_0:def 3;
    y in ExtREAL by Def1;
    then
A12: y in REAL+ \/ [:{0},REAL+:] \ {[0,0]} or y in {+infty,-infty} by
XBOOLE_0:def 3;
    per cases by A10;
    suppose that
A13:  x in REAL+ & y in REAL+ and
A14:  not ex x9,y9 being Element of REAL+ st x = x9 & y = y9 & x9 <=' y9;
      hereby
        assume y in REAL+ & x in REAL+;
        then reconsider x9 = x, y9 = y as Element of REAL+;
        take y9,x9;
        thus y = y9 & x = x9;
        thus y9 <=' x9 by A14;
      end;
      thus thesis by A13,ARYTM_0:5,XBOOLE_0:3;
    end;
    suppose that
A15:  x in [:{0},REAL+:] & y in [:{0},REAL+:] and
A16:  not ex x9,y9 being Element of REAL+ st x = [0,x9] & y = [0,y9]
      & y9 <=' x9;
      now
        assume y in [:{0},REAL+:];
        then consider z,y9 being set such that
A17:    z in {0} and
A18:    y9 in REAL+ and
A19:    y = [z,y9] by ZFMISC_1:84;
A20:    z = 0 by A17,TARSKI:def 1;
        assume x in [:{0},REAL+:];
        then consider z,x9 being set such that
A21:    z in {0} and
A22:    x9 in REAL+ and
A23:    x = [z,x9] by ZFMISC_1:84;
        reconsider x9,y9 as Element of REAL+ by A18,A22;
        take y9,x9;
        thus y = [0,y9] & x = [0,x9] by A17,A19,A21,A23,TARSKI:def 1;
        z = 0 by A21,TARSKI:def 1;
        hence x9 <=' y9 by A16,A19,A20,A23;
      end;
      hence thesis by A15,ARYTM_0:5,XBOOLE_0:3;
    end;
    suppose
      not(not(x in REAL+ & y in REAL+) & not(x in [:{0},REAL+:] & y
in [:{0},REAL+:]) implies y in REAL+ & x in [:{0},REAL+:] or x = -infty or y =
      +infty);
      hence thesis by A11,A12,TARSKI:def 2,XBOOLE_0:def 3;
    end;
  end;
end;

reserve a,b,c,d for ext-real number;

notation
  let a,b;
  synonym b >= a for a <= b;
  antonym b < a for a <= b;
  antonym a > b for a <= b;
end;

Lm1: +infty <> [0,0]
proof
  assume +infty = [0,0];
  then +infty = {{0},{0}} by ENUMSET1:29
    .= {{0}} by ENUMSET1:29;
  hence contradiction by TARSKI:def 1;
end;

Lm2: not +infty in REAL+ by ARYTM_0:1,ORDINAL1:5;

Lm3: not -infty in REAL+
proof
A1: {0,REAL} in -infty & REAL in {0,REAL} by TARSKI:def 2;
  assume -infty in REAL+;
  hence contradiction by A1,ARYTM_0:1,XREGULAR:7;
end;

Lm4: not +infty in [:{0},REAL+:]
proof
  assume +infty in [:{0},REAL+:];
  then +infty in REAL+ \/ [:{0},REAL+:] by XBOOLE_0:def 3;
  then +infty in REAL by Lm1,ZFMISC_1:56;
  hence contradiction;
end;

Lm5: not -infty in [:{0},REAL+:]
proof
  assume -infty in [:{0},REAL+:];
  then REAL in REAL+ by ZFMISC_1:87;
  hence contradiction by ARYTM_0:1,ORDINAL1:5;
end;

Lm6: -infty < +infty
proof
  -infty <> +infty by TARSKI:def 2;
  hence thesis by Def5,Lm3,Lm5;
end;

theorem Th1:
  a <= b & b <= a implies a = b
proof
  assume that
A1: a <= b and
A2: b <= a;
  per cases;
  suppose
    a in REAL+ & b in REAL+;
    then
    (ex a9,b9 being Element of REAL+ st a = a9 & b = b9 & a9 <=' b9 )& ex
b99, a99 being Element of REAL+ st b = b99 & a = a99 & b99 <=' a99 by A1,A2
,Def5;
    hence thesis by ARYTM_1:4;
  end;
  suppose
A3: a in REAL+ & b in [:{0},REAL+:];
    then ( not b in REAL+)& not a in [:{0},REAL+:] by ARYTM_0:5,XBOOLE_0:3;
    hence thesis by A1,A3,Def5,Lm3,Lm4;
  end;
  suppose
A4: b in REAL+ & a in [:{0},REAL+:];
    then ( not a in REAL+)& not b in [:{0},REAL+:] by ARYTM_0:5,XBOOLE_0:3;
    hence thesis by A2,A4,Def5,Lm3,Lm4;
  end;
  suppose that
A5: a in [:{0},REAL+:] & b in [:{0},REAL+:];
    consider a9,b9 being Element of REAL+ such that
A6: a = [0,a9] & b = [0,b9] and
A7: b9 <=' a9 by A1,A5,Def5;
    consider b99,a99 being Element of REAL+ such that
A8: b = [0,b99] & a = [0,a99] and
A9: a99 <=' b99 by A2,A5,Def5;
    a9 = a99 & b9 = b99 by A6,A8,XTUPLE_0:1;
    hence thesis by A7,A8,A9,ARYTM_1:4;
  end;
  suppose
    (a = -infty or a = +infty) & (b = -infty or b = +infty);
    hence thesis by A1,A2,Lm6;
  end;
  suppose that
A10: ( not(a in REAL+ & b in REAL+))& not(a in [:{0},REAL+:] & b in [:
    {0},REAL+:]) and
A11: not(b in REAL+ & a in [:{0},REAL+:]) and
A12: not(a in REAL+ & b in [:{0},REAL+:]);
    a = -infty or b = +infty by A1,A10,A11,Def5;
    hence thesis by A2,A10,A12,Def5,Lm6;
  end;
end;

Lm7: -infty >= a implies a = -infty
proof
  a >= -infty by Def5,Lm3,Lm5;
  hence thesis by Th1;
end;

Lm8: +infty <= a implies a = +infty
proof
  a <= +infty by Def5,Lm2,Lm4;
  hence thesis by Th1;
end;

theorem Th2:
  a <= b & b <= c implies a <= c
proof
  assume that
A1: a <= b and
A2: b <= c;
  per cases;
  suppose that
A3: a in REAL+ and
A4: b in REAL+ and
A5: c in REAL+;
    consider b99,c9 being Element of REAL+ such that
A6: b = b99 and
A7: c = c9 and
A8: b99 <=' c9 by A2,A4,A5,Def5;
    consider a9,b9 being Element of REAL+ such that
A9: a = a9 and
A10: b = b9 & a9 <=' b9 by A1,A3,A4,Def5;
    a9 <=' c9 by A10,A6,A8,ARYTM_1:3;
    hence thesis by A5,A9,A7,Def5;
  end;
  suppose
A11: a in REAL+ & b in [:{0},REAL+:];
    then
    ( not(a in REAL+ & b in REAL+))& not(a in [:{0},REAL+:] & b in [:{0},
    REAL+:]) by ARYTM_0:5,XBOOLE_0:3;
    hence thesis by A1,A11,Def5,Lm3,Lm4;
  end;
  suppose
A12: b in REAL+ & c in [:{0},REAL+:];
    then
    ( not(c in REAL+ & b in REAL+))& not(c in [:{0},REAL+:] & b in [:{0},
    REAL+:]) by ARYTM_0:5,XBOOLE_0:3;
    hence thesis by A2,A12,Def5,Lm3,Lm4;
  end;
  suppose that
A13: a in [:{0},REAL+:] & c in REAL+;
    ( not(a in REAL+ & c in REAL+))& not(a in [:{0},REAL+:] & c in [:{0},
    REAL+:]) by A13,ARYTM_0:5,XBOOLE_0:3;
    hence thesis by A13,Def5;
  end;
  suppose that
A14: a in [:{0},REAL+:] and
A15: b in [:{0},REAL+:] and
A16: c in [:{0},REAL+:];
    consider b99,c9 being Element of REAL+ such that
A17: b = [0,b99] and
A18: c = [0,c9] and
A19: c9 <=' b99 by A2,A15,A16,Def5;
    consider a9,b9 being Element of REAL+ such that
A20: a = [0,a9] and
A21: b = [0,b9] and
A22: b9 <=' a9 by A1,A14,A15,Def5;
    b9 = b99 by A21,A17,XTUPLE_0:1;
    then c9 <=' a9 by A22,A19,ARYTM_1:3;
    hence thesis by A14,A16,A20,A18,Def5;
  end;
  suppose that
A23: not(a in REAL+ & b in REAL+ & c in REAL+) and
A24: not(a in REAL+ & b in [:{0},REAL+:]) and
A25: not(b in REAL+ & c in [:{0},REAL+:]) and
A26: not(a in [:{0},REAL+:] & c in REAL+) and
A27: not(a in [:{0},REAL+:] & b in [:{0},REAL+:] & c in [:{0},REAL+:]);
A28: b = +infty implies c = +infty by A2,Lm8;
A29: b = -infty implies a = -infty by A1,Lm7;
    a = -infty or b = +infty or b = -infty or c = +infty by A1,A2,A23,A25,A26
,A27,Def5;
    hence thesis by A1,A2,A23,A24,A25,A27,A28,A29,Def5;
  end;
end;

theorem
  a <= +infty by Def5,Lm2,Lm4;

theorem
  +infty <= a implies a = +infty by Lm8;

theorem
  a >= -infty by Def5,Lm3,Lm5;

theorem
  -infty >= a implies a = -infty by Lm7;

theorem
  -infty < +infty by Lm6;

theorem
  not +infty in REAL;

Lm9: a in REAL or a = +infty or a = -infty
proof
  a in ExtREAL by Def1;
  then a in REAL or a in { +infty, -infty } by XBOOLE_0:def 3;
  hence thesis by TARSKI:def 2;
end;

theorem Th9:
  a in REAL implies +infty > a
proof
  assume a in REAL;
  then
A1: a <> +infty;
  +infty >= a by Def5,Lm2,Lm4;
  hence thesis by A1,Th1;
end;

theorem Th10:
  a in REAL & b >= a implies b in REAL or b = +infty
proof
  assume that
A1: a in REAL and
A2: b >= a;
  assume
A3: not b in REAL;
  b = -infty implies a = -infty by A2,Lm7;
  hence thesis by A1,A3,Lm9;
end;

theorem Th11:
  not -infty in REAL
proof
A1: {0,REAL} in {{0,REAL},{0}} & REAL in {0,REAL} by TARSKI:def 2;
  assume -infty in REAL;
  hence contradiction by A1,XREGULAR:7;
end;

theorem Th12:
  a in REAL implies -infty < a
proof
  -infty <= a by Def5,Lm3,Lm5;
  hence thesis by Th1,Th11;
end;

theorem Th13:
  a in REAL & b <= a implies b in REAL or b = -infty
proof
  assume that
A1: a in REAL and
A2: b <= a;
  assume
A3: not b in REAL;
  b = +infty implies a = +infty by A2,Lm8;
  hence thesis by A1,A3,Lm9;
end;

theorem
  a in REAL or a = +infty or a = -infty by Lm9;

begin 

registration
  cluster natural -> ext-real for number;
  coherence
  proof
    let a be number;
    assume a is natural;
    then a in NAT by ORDINAL1:def 12;
    hence a in ExtREAL by XBOOLE_0:def 3;
  end;
end;

notation
  let a be number;
  synonym a is zero for a is empty;
end;

definition
  let a be ext-real number;
  attr a is positive means
  a > 0;
  attr a is negative means
  a < 0;
  redefine attr a is zero means
  a = 0;
  compatibility;
end;

registration
  cluster positive -> non negative non zero for ext-real number;
  coherence
  proof
    let r be ext-real number;
    assume r > 0;
    hence r >= 0 & r <> 0;
  end;
  cluster non negative non zero -> positive for ext-real number;
  coherence
  proof
    let r be ext-real number;
    assume r >= 0 & r <> 0;
    hence r > 0 by Th1;
  end;
  cluster negative -> non positive non zero for ext-real number;
  coherence
  proof
    let r be ext-real number;
    assume r < 0;
    hence r <= 0 & r <> {};
  end;
  cluster non positive non zero -> negative for ext-real number;
  coherence;
  cluster zero -> non negative non positive for ext-real number;
  coherence;
  cluster non negative non positive -> zero for ext-real number;
  coherence;
end;

registration
  cluster +infty -> positive;
  coherence
  proof
    thus 0 < +infty by Th9;
  end;
  cluster -infty -> negative;
  coherence
  proof
    thus 0 > -infty by Th12;
  end;
end;

registration
  cluster positive for ext-real number;
  existence
  proof
    take +infty;
    thus thesis;
  end;
  cluster negative for ext-real number;
  existence
  proof
    take -infty;
    thus thesis;
  end;
  cluster zero for ext-real number;
  existence
  proof
    reconsider z = 0 as ext-real number;
    take z;
    thus z = 0;
  end;
end;

begin 

definition
  let a,b;
  func min(a,b) equals
  :Def9:
  a if a <= b otherwise b;
  correctness;
  commutativity by Th1;
  idempotence;
  func max(a,b) equals
  :Def10:
  a if b <= a otherwise b;
  correctness;
  commutativity by Th1;
  idempotence;
end;

theorem
  min(a,b) = a or min(a,b) = b by Def9;

theorem
  max(a,b) = a or max(a,b) = b by Def10;

registration
  let a,b;
  cluster min(a,b) -> ext-real;
  coherence by Def9;
  cluster max(a,b) -> ext-real;
  coherence by Def10;
end;

theorem Th17:
  min(a,b) <= a
proof
  a <= b or not a <= b;
  hence thesis by Def9;
end;

theorem Th18:
  a <= b & c <= d implies min(a,c) <= min(b,d)
proof
  assume that
A1: a <= b and
A2: c <= d;
  min(a,c) <= c by Th17;
  then
A3: min(a,c) <= d by A2,Th2;
  min(a,c) <= a by Th17;
  then min(a,c) <= b by A1,Th2;
  hence thesis by A3,Def9;
end;

theorem
  a < b & c < d implies min(a,c) < min(b,d)
proof
  assume that
A1: a < b and
A2: c < d;
  min(a,c) <= c by Th17;
  then
A3: min(a,c) < d by A2,Th2;
  min(a,c) <= a by Th17;
  then min(a,c) < b by A1,Th2;
  hence thesis by A3,Def9;
end;

theorem
  a <= b & a <= c implies a <= min(b,c) by Def9;

theorem
  a < b & a < c implies a < min(b,c) by Def9;

theorem
  a <= min(b,c) implies a <= b
proof
  min(b,c) <= b by Th17;
  hence thesis by Th2;
end;

theorem
  a < min(b,c) implies a < b
proof
  min(b,c) <= b by Th17;
  hence thesis by Th2;
end;

theorem
  c <= a & c <= b & (for d st d <= a & d <= b holds d <= c) implies c =
  min(a,b)
proof
  assume that
A1: c <= a & c <= b and
A2: for d st d <= a & d <= b holds d <= c;
  min(a,b) <= a & min(a,b) <= b by Th17;
  then
A3: min(a,b) <= c by A2;
  c <= min(a,b) by A1,Def9;
  hence thesis by A3,Th1;
end;

theorem Th25:
  a <= max(a,b)
proof
  a <= b or not a <= b;
  hence thesis by Def10;
end;

theorem Th26:
  a <= b & c <= d implies max(a,c) <= max(b,d)
proof
  assume that
A1: a <= b and
A2: c <= d;
  d <= max(b,d) by Th25;
  then
A3: c <= max(b,d) by A2,Th2;
  b <= max(b,d) by Th25;
  then a <= max(b,d) by A1,Th2;
  hence thesis by A3,Def10;
end;

theorem
  a < b & c < d implies max(a,c) < max(b,d)
proof
  assume that
A1: a < b and
A2: c < d;
  d <= max(b,d) by Th25;
  then
A3: c < max(b,d) by A2,Th2;
  b <= max(b,d) by Th25;
  then a < max(b,d) by A1,Th2;
  hence thesis by A3,Def10;
end;

theorem
  b <= a & c <= a implies max(b,c) <= a by Def10;

theorem
  b < a & c < a implies max(b,c) < a by Def10;

theorem
  max(b,c) <= a implies b <= a
proof
  b <= max(b,c) by Th25;
  hence thesis by Th2;
end;

theorem
  max(b,c) < a implies b < a
proof
  b <= max(b,c) by Th25;
  hence thesis by Th2;
end;

theorem
  a <= c & b <= c & (for d st a <= d & b <= d holds c <= d) implies c =
  max(a,b)
proof
  assume that
A1: a <= c & b <= c and
A2: for d st a <= d & b <= d holds c <= d;
  a <= max(a,b) & b <= max(a,b) by Th25;
  then
A3: c <= max(a,b) by A2;
  max(a,b) <= c by A1,Def10;
  hence thesis by A3,Th1;
end;

theorem
  min(min(a,b),c) = min(a,min(b,c))
proof
  per cases by Th2;
  suppose
    a <= b & a <= c;
    then min(a,b) = a & min(a,c) = a by Def9;
    hence thesis by Def9;
  end;
  suppose
    b <= a & b <= c;
    then min(a,b) = b & min(b,c) = b by Def9;
    hence thesis;
  end;
  suppose
    c <= b & c <= a;
    then min(b,c) = c & min(a,c) = c by Def9;
    hence thesis by Def9;
  end;
end;

theorem
  max(max(a,b),c) = max(a,max(b,c))
proof
  per cases by Th2;
  suppose
A1: a <= b & a <= c;
A2: max(b,c) = b or max(b,c) = c by Def10;
    max(a,b) = b by A1,Def10;
    hence thesis by A1,A2,Def10;
  end;
  suppose
A3: b <= a & b <= c;
    then max(a,b) = a by Def10;
    hence thesis by A3,Def10;
  end;
  suppose
A4: c <= b & c <= a;
A5: max(a,b) = b or max(a,b) = a by Def10;
    max(b,c) = b by A4,Def10;
    hence thesis by A4,A5,Def10;
  end;
end;

theorem
  min(max(a,b),b) = b
proof
  b <= max(a,b) by Th25;
  hence thesis by Def9;
end;

theorem
  max(min(a,b),b) = b
proof
  min(a,b) <= b by Th17;
  hence thesis by Def10;
end;

theorem Th37:
  a <= c implies max(a,min(b,c)) = min(max(a,b),c)
proof
  assume
A1: a <= c;
  per cases;
  suppose
A2: a <= b;
    then a <= min(b,c) by A1,Def9;
    hence max(a,min(b,c)) = min(b,c) by Def10
      .= min(max(a,b),c) by A2,Def10;
  end;
  suppose
A3: b <= a;
    then b <= c by A1,Th2;
    hence max(a,min(b,c)) = max(a,b) by Def9
      .= a by A3,Def10
      .= min(a,c) by A1,Def9
      .= min(max(a,b),c) by A3,Def10;
  end;
end;

theorem
  min(a,max(b,c)) = max(min(a,b),min(a,c))
proof
  per cases;
  suppose
A1: b <= c;
    then
A2: min(a,b) <= min(a,c) by Th18;
    thus min(a,max(b,c)) = min(a,c) by A1,Def10
      .= max(min(a,b),min(a,c)) by A2,Def10;
  end;
  suppose
A3: c <= b;
    then
A4: min(a,c) <= min(a,b) by Th18;
    thus min(a,max(b,c)) = min(a,b) by A3,Def10
      .= max(min(a,b),min(a,c)) by A4,Def10;
  end;
end;

theorem
  max(a,min(b,c)) = min(max(a,b),max(a,c))
proof
  per cases;
  suppose
A1: b <= c;
    then
A2: max(a,b) <= max(a,c) by Th26;
    thus max(a,min(b,c)) = max(a,b) by A1,Def9
      .= min(max(a,b),max(a,c)) by A2,Def9;
  end;
  suppose
A3: c <= b;
    then
A4: max(a,c) <= max(a,b) by Th26;
    thus max(a,min(b,c)) = max(a,c) by A3,Def9
      .= min(max(a,b),max(a,c)) by A4,Def9;
  end;
end;

theorem
  max(max(min(a,b),min(b,c)),min(c,a)) = min(min(max(a,b),max(b,c)),max( c,a))
proof
  per cases;
  suppose
A1: a <= c;
    then
A2: max(a,b) <= max(b,c) by Th26;
    min(a,b) <= min(b,c) by A1,Th18;
    hence max(max(min(a,b),min(b,c)),min(c,a)) = max(min(b,c),min(c,a)) by
Def10
      .= max(min(b,c),a) by A1,Def9
      .= min(max(a,b),c) by A1,Th37
      .= min(max(a,b),max(c,a)) by A1,Def10
      .= min(min(max(a,b),max(b,c)),max(c,a)) by A2,Def9;
  end;
  suppose
A3: c <= a;
    then
A4: max(a,b) >= max(b,c) by Th26;
    min(a,b) >= min(b,c) by A3,Th18;
    hence max(max(min(a,b),min(b,c)),min(c,a)) = max(min(a,b),min(c,a)) by
Def10
      .= max(min(a,b),c) by A3,Def9
      .= min(max(c,b),a) by A3,Th37
      .= min(max(c,b),max(c,a)) by A3,Def10
      .= min(min(max(a,b),max(b,c)),max(c,a)) by A4,Def9;
  end;
end;

theorem
  max(a,+infty) = +infty
proof
  a <= +infty by Def5,Lm2,Lm4;
  hence thesis by Def10;
end;

theorem
  min(a,+infty) = a
proof
  a <= +infty by Def5,Lm2,Lm4;
  hence thesis by Def9;
end;

theorem
  max(a,-infty) = a
proof
  a >= -infty by Def5,Lm3,Lm5;
  hence thesis by Def10;
end;

theorem
  min(a,-infty) = -infty
proof
  a >= -infty by Def5,Lm3,Lm5;
  hence thesis by Def9;
end;

begin 

theorem
  a in REAL & c in REAL & a <= b & b <= c implies b in REAL
proof
  assume that
A1: a in REAL and
A2: c in REAL and
A3: a <= b and
A4: b <= c;
  b in REAL or b = +infty by A1,A3,Th10;
  hence thesis by A2,A4,Th13;
end;

theorem
  a in REAL & a <= b & b < c implies b in REAL
proof
  assume that
A1: a in REAL & a <= b and
A2: b < c;
  b in REAL or b = +infty by A1,Th10;
  hence thesis by A2,Lm8;
end;

theorem
  c in REAL & a < b & b <= c implies b in REAL
proof
  assume that
A1: c in REAL and
A2: a < b and
A3: b <= c;
  b in REAL or b = -infty by A1,A3,Th13;
  hence thesis by A2,Lm7;
end;

theorem
  a < b & b < c implies b in REAL
proof
  assume
A1: a < b & b < c;
  b in REAL or b = +infty or b = -infty by Lm9;
  hence thesis by A1,Lm7,Lm8;
end;



definition
  let x,y be ext-real number, a,b be set;
  func IFGT(x,y,a,b) equals
  :Def11:
  a if x > y otherwise b;
  correctness;
end;

registration
  let x,y be ext-real number, a,b be Nat;
  cluster IFGT(x,y,a,b) -> natural;
  coherence by Def11;
end;



theorem
  max(a,b) <= a implies max(a,b) = a
proof
  assume max(a,b) <= a;
  then max(a,b) < a or max(a,b) = a by Th1;
  hence thesis by Th25;
end;

theorem
  a <= min(a,b) implies min(a,b) = a
proof
  assume min(a,b) >= a;
  then min(a,b) > a or min(a,b) = a by Th1;
  hence thesis by Th17;
end;














begin

definition
  let r be number;
  attr r is real means
  :Def1:
  r in REAL;
end;

registration
  cluster -> real for Element of REAL;
  coherence
  proof
    let r be Element of REAL;
    thus r in REAL;
  end;
end;

registration
  cluster -infty -> non real;
  coherence
  proof
A1: {0,REAL} in {{0,REAL},{0}} & REAL in {0,REAL} by TARSKI:def 2;
    assume -infty in REAL;
    hence contradiction by A1,XREGULAR:7;
  end;
  cluster +infty -> non real;
  coherence
  proof
    thus not +infty in REAL;
  end;
end;

registration
  cluster natural -> real for number;
  coherence
  proof
    let x be number;
    assume x is natural;
    then x in NAT by ORDINAL1:def 12;
    hence x in REAL;
  end;
  cluster real -> complex for number;
  coherence
  proof
    let x be number;
    assume x is real;
    then x in REAL by Def1;
    hence x in COMPLEX by NUMBERS:def 2,XBOOLE_0:def 3;
  end;
end;

registration
  cluster real for number;
  existence
  proof
    take 0;
    thus 0 in REAL;
  end;
  cluster real -> ext-real for number;
  coherence
  proof
    let n be number;
    assume n in REAL;
    hence n in ExtREAL by XBOOLE_0:def 3;
  end;
end;

Lm1: for x being real number, x1,x2 being Element of REAL st x = [*x1,x2*]
holds x2 = 0 & x = x1
proof
  let x be real number, x1,x2 being Element of REAL;
  assume
A1: x = [*x1,x2*];
A2: x in REAL by Def1;
  thus now
    assume x2 <> 0;
    then x = (0,1) --> (x1,x2) by A1,ARYTM_0:def 5;
    hence contradiction by A2,ARYTM_0:8;
  end;
  hence thesis by A1,ARYTM_0:def 5;
end;

registration
  let x be real number;
  cluster -x -> real;
  coherence
  proof
    x + -x = 0;
    then consider x1,x2,y1,y2 being Element of REAL such that
A1: x = [*x1,x2*] and
A2: -x = [*y1,y2*] and
A3: 0 = [*+(x1,y1),+(x2,y2)*] by XCMPLX_0:def 4;
    +(x2,y2) = 0 & x2 = 0 by A1,A3,Lm1;
    then y2 = 0 by ARYTM_0:11;
    hence thesis by A2,ARYTM_0:def 5;
  end;
  cluster x" -> real;
  coherence
  proof
    per cases;
    suppose x = 0;
      hence thesis by XCMPLX_0:def 7;
    end;
    suppose
A4:   x <> 0;
      then x * x" = 1 by XCMPLX_0:def 7;
      then consider x1,x2,y1,y2 being Element of REAL such that
A5:   x = [*x1,x2*] and
A6:   x" = [*y1,y2*] and
A7:   1 = [* +(*(x1,y1),opp*(x2,y2)), +(*(x1,y2),*(x2,y1)) *] by XCMPLX_0:def 5
;
      +(*(x1,y2),*(x2,y1)) = 0 & x2 = 0 by A5,A7,Lm1;
      then 0 = *(x1,y2) by ARYTM_0:11,12;
      then x1 = 0 or y2 = 0 by ARYTM_0:21;
      hence thesis by A4,A5,A6,Lm1,ARYTM_0:def 5;
    end;
  end;
  let y be real number;
  cluster x + y -> real;
  coherence
  proof
    consider x1,x2,y1,y2 being Element of REAL such that
A8: x = [*x1,x2*] & y = [*y1,y2*] and
A9: x+y = [*+(x1,y1),+(x2,y2)*] by XCMPLX_0:def 4;
    x2 = 0 & y2 = 0 by A8,Lm1;
    then +(x2,y2) = 0 by ARYTM_0:11;
    hence thesis by A9,ARYTM_0:def 5;
  end;
  cluster x * y -> real;
  coherence
  proof
    consider x1,x2,y1,y2 being Element of REAL such that
A10: x = [*x1,x2*] and
A11: y = [*y1,y2*] and
A12: x*y = [* +(*(x1,y1),opp*(x2,y2)), +(*(x1,y2),*(x2,y1)) *] by
XCMPLX_0:def 5;
    x2 = 0 by A10,Lm1;
    then
A13: *(x2,y1) = 0 by ARYTM_0:12;
A14: y2 = 0 by A11,Lm1;
    then *(opp x2,y2) = 0 by ARYTM_0:12;
    then
A15: opp*(x2,y2) = 0 by ARYTM_0:15;
    *(x1,y2) = 0 by A14,ARYTM_0:12;
    then +(*(x1,y2),*(x2,y1)) = 0 by A13,ARYTM_0:11;
    then x * y = +(*(x1,y1),0) by A12,A15,ARYTM_0:def 5
      .= *(x1,y1) by ARYTM_0:11;
    hence thesis;
  end;
end;

registration
  let x,y be real number;
  cluster x-y -> real;
  coherence;
  cluster x/y -> real;
  coherence;
end;

begin

reserve r,s,t for real number;

Lm2: (r in REAL+ & s in REAL+ & ex x9,y9 being Element of REAL+ st r = x9 & s
= y9 & x9 <=' y9) or (r in [:{0},REAL+:] & s in [:{0},REAL+:] & ex x9,y9 being
Element of REAL+ st r = [0,x9] & s = [0,y9] & y9 <=' x9) or s in REAL+ & r in
[:{0},REAL+:] implies r <= s
proof
  assume
A1: (r in REAL+ & s in REAL+ & ex x9,y9 being Element of REAL+ st r = x9
  & s = y9 & x9 <=' y9) or (r in [:{0},REAL+:] & s in [:{0},REAL+:] & ex x9,y9
being Element of REAL+ st r = [0,x9] & s = [0,y9] & y9 <=' x9) or s in REAL+ &
  r in [:{0},REAL+:];
  per cases;
  case
    r in REAL+ & s in REAL+;
    hence thesis by A1,ARYTM_0:5,XBOOLE_0:3;
  end;
  case
    r in [:{0},REAL+:] & s in [:{0},REAL+:];
    hence thesis by A1,ARYTM_0:5,XBOOLE_0:3;
  end;
  case
    not(r in REAL+ & s in REAL+) & not (r in [:{0},REAL+:] & s in [:{0
    },REAL+:]);
    hence thesis by A1;
  end;
end;

Lm3: {} in {{}} by TARSKI:def 1;

Lm4: r <= s & s <= r implies r = s
proof
  assume that
A1: r <= s and
A2: s <= r;
A3: r in REAL & s in REAL by Def1;
  per cases by A3,NUMBERS:def 1,XBOOLE_0:def 3;
  suppose r in REAL+ & s in REAL+;
    then
    (ex r9,s9 being Element of REAL+ st r = r9 & s = s9 & r9 <=' s9 )& ex
s99, r99 being Element of REAL+ st s = s99 & r = r99 & s99 <=' r99 by A1,A2,
XXREAL_0:def 5;
    hence thesis by ARYTM_1:4;
  end;
  suppose
A4: r in REAL+ & s in [:{0},REAL+:];
    then
    ( not(r in REAL+ & s in REAL+))& not(r in [:{0},REAL+:] & s in [:{0},
    REAL+:]) by ARYTM_0:5,XBOOLE_0:3;
    hence thesis by A1,A4,XXREAL_0:def 5;
  end;
  suppose
A5: s in REAL+ & r in [:{0},REAL+:];
    then
    ( not(r in REAL+ & s in REAL+))& not(r in [:{0},REAL+:] & s in [:{0},
    REAL+:]) by ARYTM_0:5,XBOOLE_0:3;
    hence thesis by A2,A5,XXREAL_0:def 5;
  end;
  suppose that
A6: r in [:{0},REAL+:] & s in [:{0},REAL+:];
    consider r9,s9 being Element of REAL+ such that
A7: r = [0,r9] & s = [0,s9] and
A8: s9 <=' r9 by A1,A6,XXREAL_0:def 5;
    consider s99,r99 being Element of REAL+ such that
A9: s = [0,s99] & r = [0,r99] and
A10: r99 <=' s99 by A2,A6,XXREAL_0:def 5;
    r9 = r99 & s9 = s99 by A7,A9,XTUPLE_0:1;
    hence thesis by A8,A9,A10,ARYTM_1:4;
  end;
end;

Lm5: r <= s implies r + t <= s + t
proof
  reconsider x1=r, y1=s, z1=t as Element of REAL by Def1;
A1: for x9 being Element of REAL, r st x9 = r holds +(x9,z1) = r + t
  proof
    let x9 be Element of REAL, r such that
A2: x9 = r;
    consider x1,x2,y1,y2 being Element of REAL such that
A3: r = [* x1,x2 *] & t = [*y1,y2*] and
A4: r+t = [*+(x1,y1),+(x2,y2)*] by XCMPLX_0:def 4;
    x2 = 0 & y2 = 0 by A3,Lm1;
    then
A5: +(x2,y2) = 0 by ARYTM_0:11;
    r = x1 & t = y1 by A3,Lm1;
    hence thesis by A2,A4,A5,ARYTM_0:def 5;
  end;
  then
A6: +(y1,z1) = s + t;
A7: +(x1,z1) = r + t by A1;
  assume
A8: r <= s;
  per cases by A8,XXREAL_0:def 5;
  suppose that
A9: r in REAL+ and
A10: s in REAL+ and
A11: t in REAL+;
    consider s9,t99 being Element of REAL+ such that
A12: s = s9 & t = t99 and
A13: +(y1,z1) = s9 + t99 by A10,A11,ARYTM_0:def 1;
    consider x9,t9 being Element of REAL+ such that
A14: r = x9 & t = t9 and
A15: +(x1,z1) = x9 + t9 by A9,A11,ARYTM_0:def 1;
    ex x99,s99 being Element of REAL+ st r = x99 & s = s99 & x99 <=' s99
    by A8,A9,A10,XXREAL_0:def 5;
    then x9 + t9 <=' s9 + t99 by A14,A12,ARYTM_1:7;
    hence thesis by A6,A7,A15,A13,Lm2;
  end;
  suppose that
A16: r in [:{0},REAL+:] and
A17: s in REAL+ and
A18: t in REAL+;
    consider s9,t99 being Element of REAL+ such that
    s = s9 and
A19: t = t99 and
A20: +(y1,z1) = s9 + t99 by A17,A18,ARYTM_0:def 1;
    consider x9,t9 being Element of REAL+ such that
    r = [0,x9] and
A21: t = t9 and
A22: +(x1,z1) = t9 - x9 by A16,A18,ARYTM_0:def 1;
    now
      per cases;
      suppose
A23:    x9 <=' t9;
        t9 -' x9 <=' t9 & t9 <=' s9 + t99 by A21,A19,ARYTM_1:11,ARYTM_2:19;
        then
A24:    t9 -' x9 <=' s9 + t99 by ARYTM_1:3;
        t9 - x9 = t9 -' x9 by A23,ARYTM_1:def 2;
        hence thesis by A6,A7,A22,A20,A24,Lm2;
      end;
      suppose
        not x9 <=' t9;
        then t9 - x9 = [0,x9 -' t9] by ARYTM_1:def 2;
        then t9 - x9 in [:{0},REAL+:] by Lm3,ZFMISC_1:87;
        then
A25:    not r + t in REAL+ by A7,A22,ARYTM_0:5,XBOOLE_0:3;
        not s + t in [:{0},REAL+:] by A6,A20,ARYTM_0:5,XBOOLE_0:3;
        hence thesis by A25,XXREAL_0:def 5;
      end;
    end;
    hence thesis;
  end;
  suppose that
A26: r in [:{0},REAL+:] and
A27: s in [:{0},REAL+:] and
A28: t in REAL+;
    consider s9,t99 being Element of REAL+ such that
A29: s = [0,s9] and
A30: t = t99 and
A31: +(y1,z1) = t99 - s9 by A27,A28,ARYTM_0:def 1;
    consider x99,s99 being Element of REAL+ such that
A32: r = [0,x99] and
A33: s = [0,s99] and
A34: s99 <=' x99 by A8,A26,A27,XXREAL_0:def 5;
    consider x9,t9 being Element of REAL+ such that
A35: r = [0,x9] and
A36: t = t9 and
A37: +(x1,z1) = t9 - x9 by A26,A28,ARYTM_0:def 1;
A38: x9 = x99 by A32,A35,XTUPLE_0:1;
A39: s9 = s99 by A33,A29,XTUPLE_0:1;
    now
      per cases;
      suppose
A40:    x9 <=' t9;
        then s9 <=' t9 by A34,A38,A39,ARYTM_1:3;
        then
A41:    t9 - s9 = t9 -' s9 by ARYTM_1:def 2;
A42:    t9 - x9 = t9 -' x9 by A40,ARYTM_1:def 2;
        t9 -' x9 <=' t99 -' s9 by A34,A36,A30,A38,A39,ARYTM_1:16;
        hence thesis by A6,A7,A36,A37,A30,A31,A42,A41,Lm2;
      end;
      suppose
        not x9 <=' t9;
        then
A43:    +(x1,z1) = [0,x9 -' t9] by A37,ARYTM_1:def 2;
        then
A44:    +(x1,z1) in [:{0},REAL+:] by Lm3,ZFMISC_1:87;
        now
          per cases;
          suppose
            s9 <=' t9;
            then t9 - s9 = t9 -' s9 by ARYTM_1:def 2;
            then
A45:        not +(y1,z1) in [:{0},REAL+:] by A36,A30,A31,ARYTM_0:5,XBOOLE_0:3;
            not +(x1,z1) in REAL+ by A44,ARYTM_0:5,XBOOLE_0:3;
            hence thesis by A6,A7,A45,XXREAL_0:def 5;
          end;
          suppose
A46:        not s9 <=' t9;
A47:        s9 -' t9 <=' x9 -' t9 by A34,A38,A39,ARYTM_1:17;
A48:        +(y1,z1) = [0,s9 -' t9] by A36,A30,A31,A46,ARYTM_1:def 2;
            then +(y1,z1) in [:{0},REAL+:] by Lm3,ZFMISC_1:87;
            hence thesis by A6,A7,A43,A44,A48,A47,Lm2;
          end;
        end;
        hence thesis;
      end;
    end;
    hence thesis;
  end;
  suppose that
A49: r in REAL+ and
A50: s in REAL+ and
A51: t in [:{0},REAL+:];
    consider s9,t99 being Element of REAL+ such that
A52: s = s9 and
A53: t = [0,t99] and
A54: +(y1,z1) = s9 - t99 by A50,A51,ARYTM_0:def 1;
    consider x9,t9 being Element of REAL+ such that
A55: r = x9 and
A56: t = [0,t9] and
A57: +(x1,z1) = x9 - t9 by A49,A51,ARYTM_0:def 1;
A58: t9 = t99 by A56,A53,XTUPLE_0:1;
A59: ex x99,s99 being Element of REAL+ st r = x99 & s = s99 & x99 <=' s99
    by A8,A49,A50,XXREAL_0:def 5;
    now
      per cases;
      suppose
A60:    t9 <=' x9;
        then t9 <=' s9 by A59,A55,A52,ARYTM_1:3;
        then
A61:    s9 - t9 = s9 -' t9 by ARYTM_1:def 2;
A62:    x9 - t9 = x9 -' t9 by A60,ARYTM_1:def 2;
        x9 -' t9 <=' s9 -' t99 by A59,A55,A52,A58,ARYTM_1:17;
        hence thesis by A6,A7,A57,A54,A58,A62,A61,Lm2;
      end;
      suppose
        not t9 <=' x9;
        then
A63:    +(x1,z1) = [0,t9 -' x9] by A57,ARYTM_1:def 2;
        then
A64:    +(x1,z1) in [:{0},REAL+:] by Lm3,ZFMISC_1:87;
        now
          per cases;
          suppose
            t9 <=' s9;
            then s9 - t9 = s9 -' t9 by ARYTM_1:def 2;
            then
A65:        not +(y1,z1) in [:{0},REAL+:] by A54,A58,ARYTM_0:5,XBOOLE_0:3;
            not +(x1,z1) in REAL+ by A64,ARYTM_0:5,XBOOLE_0:3;
            hence thesis by A6,A7,A65,XXREAL_0:def 5;
          end;
          suppose
A66:        not t9 <=' s9;
A67:        t9 -' s9 <=' t9 -' x9 by A59,A55,A52,ARYTM_1:16;
A68:        +(y1,z1) = [0,t9 -' s9] by A54,A58,A66,ARYTM_1:def 2;
            then +(y1,z1) in [:{0},REAL+:] by Lm3,ZFMISC_1:87;
            hence thesis by A6,A7,A63,A64,A68,A67,Lm2;
          end;
        end;
        hence thesis;
      end;
    end;
    hence thesis;
  end;
  suppose that
A69: r in [:{0},REAL+:] and
A70: s in REAL+ and
A71: t in [:{0},REAL+:];
    ( not r in REAL+)& not t in REAL+ by A69,A71,ARYTM_0:5,XBOOLE_0:3;
    then consider x9,t9 being Element of REAL+ such that
    r = [0,x9] and
A72: t = [0,t9] and
A73: +(x1,z1) = [0,x9 + t9] by ARYTM_0:def 1;
A74: +(x1,z1) in [:{0},REAL+:] by A73,Lm3,ZFMISC_1:87;
    consider s9,t99 being Element of REAL+ such that
    s = s9 and
A75: t = [0,t99] and
A76: +(y1,z1) = s9 - t99 by A70,A71,ARYTM_0:def 1;
A77: t9 = t99 by A72,A75,XTUPLE_0:1;
    now
      per cases;
      suppose
        t9 <=' s9;
        then s9 - t99 = s9 -' t99 by A77,ARYTM_1:def 2;
        then
A78:    not +(y1,z1) in [:{0},REAL+:] by A76,ARYTM_0:5,XBOOLE_0:3;
        not +(x1,z1) in REAL+ by A74,ARYTM_0:5,XBOOLE_0:3;
        hence thesis by A6,A7,A78,XXREAL_0:def 5;
      end;
      suppose
A79:    not t9 <=' s9;
        t9 -' s9 <=' t9 & t9 <=' t9 + x9 by ARYTM_1:11,ARYTM_2:19;
        then
A80:    t9 -' s9 <=' t9 + x9 by ARYTM_1:3;
A81:    +(y1,z1) = [0,t9 -' s9] by A76,A77,A79,ARYTM_1:def 2;
        then +(y1,z1) in [:{0},REAL+:] by Lm3,ZFMISC_1:87;
        hence thesis by A6,A7,A73,A74,A81,A80,Lm2;
      end;
    end;
    hence thesis;
  end;
  suppose that
A82: r in [:{0},REAL+:] and
A83: s in [:{0},REAL+:] and
A84: t in [:{0},REAL+:];
    ( not s in REAL+)& not t in REAL+ by A83,A84,ARYTM_0:5,XBOOLE_0:3;
    then consider s9,t99 being Element of REAL+ such that
A85: s = [0,s9] and
A86: t = [0,t99] and
A87: +(y1,z1) = [0,s9 + t99] by ARYTM_0:def 1;
A88: +(y1,z1) in [:{0},REAL+:] by A87,Lm3,ZFMISC_1:87;
    ( not r in REAL+)& not t in REAL+ by A82,A84,ARYTM_0:5,XBOOLE_0:3;
    then consider x9,t9 being Element of REAL+ such that
A89: r = [0,x9] and
A90: t = [0,t9] and
A91: +(x1,z1) = [0,x9 + t9] by ARYTM_0:def 1;
A92: +(x1,z1) in [:{0},REAL+:] by A91,Lm3,ZFMISC_1:87;
A93: t9 = t99 by A90,A86,XTUPLE_0:1;
    consider x99,s99 being Element of REAL+ such that
A94: r = [0,x99] and
A95: s = [0,s99] and
A96: s99 <=' x99 by A8,A82,A83,XXREAL_0:def 5;
A97: s9 = s99 by A95,A85,XTUPLE_0:1;
    x9 = x99 by A94,A89,XTUPLE_0:1;
    then s9 + t9 <=' x9 + t99 by A96,A97,A93,ARYTM_1:7;
    hence thesis by A6,A7,A91,A87,A93,A92,A88,Lm2;
  end;
end;

Lm6: r <= s & s <= t implies r <= t
proof
  assume that
A1: r <= s and
A2: s <= t;
A3: r in REAL & s in REAL by Def1;
A4: t in REAL by Def1;
  per cases by A3,A4,NUMBERS:def 1,XBOOLE_0:def 3;
  suppose that
A5: r in REAL+ and
A6: s in REAL+ and
A7: t in REAL+;
    consider s99,t9 being Element of REAL+ such that
A8: s = s99 and
A9: t = t9 and
A10: s99 <=' t9 by A2,A6,A7,XXREAL_0:def 5;
    consider x9,s9 being Element of REAL+ such that
A11: r = x9 and
A12: s = s9 & x9 <=' s9 by A1,A5,A6,XXREAL_0:def 5;
    x9 <=' t9 by A12,A8,A10,ARYTM_1:3;
    hence thesis by A11,A9,Lm2;
  end;
  suppose
A13: r in REAL+ & s in [:{0},REAL+:];
    then
    ( not(r in REAL+ & s in REAL+))& not(r in [:{0},REAL+:] & s in [:{0},
    REAL+:]) by ARYTM_0:5,XBOOLE_0:3;
    hence thesis by A1,A13,XXREAL_0:def 5;
  end;
  suppose
A14: s in REAL+ & t in [:{0},REAL+:];
    then
    ( not(t in REAL+ & s in REAL+))& not(t in [:{0},REAL+:] & s in [:{0},
    REAL+:]) by ARYTM_0:5,XBOOLE_0:3;
    hence thesis by A2,A14,XXREAL_0:def 5;
  end;
  suppose that
A15: r in [:{0},REAL+:] and
A16: t in REAL+;
    ( not(r in REAL+ & t in REAL+))& not(r in [:{0},REAL+:] & t in [:{0},
    REAL+:]) by A15,A16,ARYTM_0:5,XBOOLE_0:3;
    hence thesis by A16,XXREAL_0:def 5;
  end;
  suppose that
A17: r in [:{0},REAL+:] and
A18: s in [:{0},REAL+:] and
A19: t in [:{0},REAL+:];
    consider s99,t9 being Element of REAL+ such that
A20: s = [0,s99] and
A21: t = [0,t9] and
A22: t9 <=' s99 by A2,A18,A19,XXREAL_0:def 5;
    consider x9,s9 being Element of REAL+ such that
A23: r = [0,x9] and
A24: s = [0,s9] and
A25: s9 <=' x9 by A1,A17,A18,XXREAL_0:def 5;
    s9 = s99 by A24,A20,XTUPLE_0:1;
    then t9 <=' x9 by A25,A22,ARYTM_1:3;
    hence thesis by A17,A19,A23,A21,Lm2;
  end;
end;

reconsider z = 0 as Element of REAL+ by ARYTM_2:20;

Lm7: not 0 in [:{0},REAL+:] by ARYTM_0:5,ARYTM_2:20,XBOOLE_0:3;

reconsider j = 1 as Element of REAL+ by ARYTM_2:20;

z <=' j by ARYTM_1:6;
then

Lm8: 0 <= 1 by Lm2;

1 + -1 = 0;

then consider x1,x2,y1,y2 being Element of REAL such that
Lm9: 1 = [*x1,x2*] and
Lm10: -1 = [*y1,y2*] & 0 = [*+(x1,y1),+(x2,y2)*] by XCMPLX_0:def 4;

Lm11: x1 = 1 by Lm1,Lm9;

Lm12: y1 = -1 & +(x1,y1) = 0 by Lm1,Lm10;

Lm13: now
  assume -1 in REAL+;
  then ex x9,y9 being Element of REAL+ st x1 = x9 & y1 = y9 & z = x9 + y9 by
Lm11,Lm12,ARYTM_0:def 1,ARYTM_2:20;
  hence contradiction by Lm11,ARYTM_2:5;
end;

Lm14: r >= 0 & s > 0 implies r + s > 0
proof
  assume r >= 0;
  then r + s >= 0 + s by Lm5;
  hence thesis by Lm6;
end;

Lm15: r <= 0 & s < 0 implies r + s < 0
proof
  assume r <= 0;
  then r + s <= 0 + s by Lm5;
  hence thesis by Lm6;
end;

reconsider o = 0 as Element of REAL+ by ARYTM_2:20;

Lm16: r <= s & 0 <= t implies r * t <= s * t
proof
  reconsider x1=r, y1=s, z1=t as Element of REAL by Def1;
  assume that
A1: r <= s and
A2: 0 <= t;
  not o in [:{0},REAL+:] by ARYTM_0:5,XBOOLE_0:3;
  then
A3: t in REAL+ by A2,XXREAL_0:def 5;
  for x9 being Element of REAL, r st x9 = r holds *(x9,z1) = r * t
  proof
    let x9 be Element of REAL, r such that
A4: x9 = r;
    consider x1,x2,y1,y2 being Element of REAL such that
A5: r = [* x1,x2 *] and
A6: t = [*y1,y2*] and
A7: r*t = [* +(*(x1,y1),opp*(x2,y2)), +(*(x1,y2),*(x2,y1)) *] by XCMPLX_0:def 5
;
    x2 = 0 by A5,Lm1;
    then
A8: *(x2,y1) = 0 by ARYTM_0:12;
A9: y2 = 0 by A6,Lm1;
    then *(x1,y2) = 0 by ARYTM_0:12;
    then
A10: +(*(x1,y2),*(x2,y1)) = 0 by A8,ARYTM_0:11;
    r = x1 & t = y1 by A5,A6,Lm1;
    hence *(x9,z1) = +(*(x1,y1),*(opp x2,y2)) by A4,A9,ARYTM_0:11,12
      .= +(*(x1,y1),opp*(x2,y2)) by ARYTM_0:15
      .= r * t by A7,A10,ARYTM_0:def 5;
  end;
  then
A11: *(y1,z1) = s * t & *(x1,z1) = r * t;
  assume
A12: not thesis;
  then
A13: t <> 0;
  per cases by A1,XXREAL_0:def 5;
  suppose that
A14: r in REAL+ and
A15: s in REAL+;
    consider s9,t99 being Element of REAL+ such that
A16: s = s9 and
A17: t = t99 & *(y1,z1) = s9 *' t99 by A3,A15,ARYTM_0:def 2;
    consider x9,t9 being Element of REAL+ such that
A18: r = x9 and
A19: t = t9 & *(x1,z1) = x9 *' t9 by A3,A14,ARYTM_0:def 2;
    ex x99,s99 being Element of REAL+ st r = x99 & s = s99 & x99 <=' s99
    by A1,A14,A15,XXREAL_0:def 5;
    then x9 *' t9 <=' s9 *' t9 by A18,A16,ARYTM_1:8;
    hence contradiction by A11,A12,A19,A17,Lm2;
  end;
  suppose that
A20: r in [:{0},REAL+:] and
A21: s in REAL+;
    ex x9,t9 being Element of REAL+ st r = [0,x9] & t = t9 & *(x1,z1) = [0
    ,t9 *' x9] by A3,A13,A20,ARYTM_0:def 2;
    then *(x1,z1) in [:{0},REAL+:] by Lm3,ZFMISC_1:87;
    then
A22: not *(x1,z1) in REAL+ by ARYTM_0:5,XBOOLE_0:3;
    ex s9,t99 being Element of REAL+ st s = s9 & t = t99 & * (y1,z1) = s9
    *' t99 by A3,A21,ARYTM_0:def 2;
    then not *(y1,z1) in [:{0},REAL+:] by ARYTM_0:5,XBOOLE_0:3;
    hence contradiction by A11,A12,A22,XXREAL_0:def 5;
  end;
  suppose that
A23: r in [:{0},REAL+:] and
A24: s in [:{0},REAL+:];
    consider s9,t99 being Element of REAL+ such that
A25: s = [0,s9] and
A26: t = t99 and
A27: *(y1,z1) = [0,t99 *' s9] by A3,A13,A24,ARYTM_0:def 2;
A28: *(y1,z1) in [:{0},REAL+:] by A27,Lm3,ZFMISC_1:87;
    consider x99,s99 being Element of REAL+ such that
A29: r = [0,x99] and
A30: s = [0,s99] and
A31: s99 <=' x99 by A1,A23,A24,XXREAL_0:def 5;
A32: s9 = s99 by A30,A25,XTUPLE_0:1;
    consider x9,t9 being Element of REAL+ such that
A33: r = [0,x9] and
A34: t = t9 and
A35: *(x1,z1) = [0,t9 *' x9] by A3,A13,A23,ARYTM_0:def 2;
A36: *(x1,z1) in [:{0},REAL+:] by A35,Lm3,ZFMISC_1:87;
    x9 = x99 by A29,A33,XTUPLE_0:1;
    then s9 *' t9 <=' x9 *' t9 by A31,A32,ARYTM_1:8;
    hence contradiction by A11,A12,A34,A35,A26,A27,A36,A28,Lm2;
  end;
end;

Lm17: r > 0 & s > 0 implies r*s > 0
proof
  assume
A1: r > 0 & s > 0;
  then r * s >= 0 * s by Lm16;
  hence thesis by A1,Lm4;
end;

Lm18: r > 0 & s < 0 implies r*s < 0
proof
  assume
A1: r > 0 & s < 0;
  then r * s <= r * 0 by Lm16;
  hence thesis by A1,Lm4;
end;

Lm19: s<=t implies -t<=-s
proof
  assume s<=t;
  then s-t<=t-t by Lm5;
  then s-t-s<=0-s by Lm5;
  hence thesis;
end;

Lm20: r <= 0 & s >= 0 implies r*s <= 0
proof
  assume
A1: r <= 0 & s >= 0;
  per cases by A1,Lm4;
  suppose
    r = 0 or s = 0;
    hence thesis;
  end;
  suppose
    r < 0 & s > 0;
    hence thesis by Lm18;
  end;
end;

registration
  cluster positive for real number;
  existence
  proof
    take r = 1;
    thus 0 < r by Lm4,Lm8;
  end;
  cluster negative for real number;
  existence
  proof
    take r = -1;
    thus 0 > r by Lm7,Lm13,XXREAL_0:def 5;
  end;
  cluster zero for real number;
  existence
  proof
    take 0;
    thus thesis;
  end;
end;

registration
  let r,s be non negative real number;
  cluster r + s -> non negative;
  coherence
  proof
A1: s >= 0 by XXREAL_0:def 7;
A2: r >= 0 by XXREAL_0:def 7;
    per cases by A2,Lm4;
    suppose
      r = 0;
      hence r + s >= 0 by XXREAL_0:def 7;
    end;
    suppose
      r > 0;
      hence r+s >= 0 by A1,Lm14;
    end;
  end;
end;

registration
  let r,s be non positive real number;
  cluster r + s -> non positive;
  coherence
  proof
A1: s <= 0 by XXREAL_0:def 6;
A2: r <= 0 by XXREAL_0:def 6;
    per cases by A2,Lm4;
    suppose
      r = 0;
      hence r + s <= 0 by XXREAL_0:def 6;
    end;
    suppose
      r < 0;
      hence r + s <= 0 by A1,Lm15;
    end;
  end;
end;

registration
  let r be positive real number;
  let s be non negative real number;
  cluster r + s -> positive;
  coherence
  proof
    r > 0 & s >= 0 by XXREAL_0:def 6;
    hence r+s > 0 by Lm14;
  end;
  cluster s + r -> positive;
  coherence;
end;

registration
  let r be negative real number;
  let s be non positive real number;
  cluster r + s -> negative;
  coherence
  proof
    r < 0 & s <= 0 by XXREAL_0:def 7;
    hence r+s < 0 by Lm15;
  end;
  cluster s + r -> negative;
  coherence;
end;

registration
  let r be non positive real number;
  cluster -r -> non negative;
  coherence
  proof
    assume
A1: -r < 0;
    --r <= 0 by XXREAL_0:def 6;
    then -r+--r < 0 by A1,Lm15;
    hence contradiction;
  end;
end;

registration
  let r be non negative real number;
  cluster -r -> non positive;
  coherence
  proof
    assume
A1: -r > 0;
    --r >= 0 by XXREAL_0:def 7;
    then -r+--r > 0 by A1,Lm14;
    hence contradiction;
  end;
end;

registration
  let r be non negative real number, s be non positive real number;
  cluster r - s -> non negative;
  coherence;
  cluster s - r -> non positive;
  coherence;
end;

registration
  let r be positive real number;
  let s be non positive real number;
  cluster r - s -> positive;
  coherence;
  cluster s - r -> negative;
  coherence;
end;

registration
  let r be negative real number;
  let s be non negative real number;
  cluster r - s -> negative;
  coherence;
  cluster s - r -> positive;
  coherence;
end;

registration
  let r be non positive real number, s be non negative real number;
  cluster r*s -> non positive;
  coherence
  proof
    r <= 0 & s >= 0 by XXREAL_0:def 6;
    hence r*s <= 0 by Lm20;
  end;
  cluster s*r -> non positive;
  coherence;
end;

registration
  let r,s be non positive real number;
  cluster r*s -> non negative;
  coherence
  proof
A1: r <= 0 & s <= 0 by XXREAL_0:def 6;
    per cases by A1,Lm4;
    suppose
      r = 0 or s = 0;
      hence r * s >= 0;
    end;
    suppose
A2:   --r < --0 & s < 0;
      then 0<-r by Lm19;
      then s*(-r)<=0*(-r) by A2,Lm16;
      then s*(-r)<0*(-r) by A2,Lm4;
      then --0*r<--s*r by Lm19;
      hence r * s >= 0;
    end;
  end;
end;

registration
  let r,s be non negative real number;
  cluster r*s -> non negative;
  coherence
  proof
A1: r >= 0 & s >= 0 by XXREAL_0:def 7;
    per cases by A1,Lm4;
    suppose
      r = 0 or s = 0;
      hence r * s >= 0;
    end;
    suppose
      r > 0 & s > 0;
      hence r * s >= 0 by Lm17;
    end;
  end;
end;

registration
  let r be positive real number;
  cluster r" -> positive;
  coherence
  proof
    assume
A1: r" <= 0;
    r"" > 0 by XXREAL_0:def 6;
    then r"*r"" = 1 & r"*r"" <= 0 by A1,Lm20,XCMPLX_0:def 7;
    hence contradiction by Lm4,Lm8;
  end;
end;

registration
  let r be non positive real number;
  cluster r" -> non positive;
  coherence
  proof
A1: r"" <= 0 by XXREAL_0:def 6;
    assume
A2: r" > 0;
    per cases by A1,Lm4;
    suppose
      r"" = 0;
      hence contradiction by A2;
    end;
    suppose
A3:   r"" < 0;
      r"*r"" = 1 by A2,XCMPLX_0:def 7;
      hence contradiction by A2,A3,Lm8,Lm18;
    end;
  end;
end;

registration
  let r be negative real number;
  cluster r" -> negative;
  coherence;
end;

registration
  let r be non negative real number;
  cluster r" -> non negative;
  coherence
  proof
A1: r"" >= 0 by XXREAL_0:def 7;
    assume
A2: r" < 0;
    per cases by A1,Lm4;
    suppose
      r"" = 0;
      hence contradiction by A2;
    end;
    suppose
A3:   r"" > 0;
      r"*r"" = 1 by A2,XCMPLX_0:def 7;
      hence contradiction by A2,A3,Lm8,Lm18;
    end;
  end;
end;

registration
  let r be non negative real number, s be non positive real number;
  cluster r/s -> non positive;
  coherence;
  cluster s/r -> non positive;
  coherence;
end;

registration
  let r,s be non negative real number;
  cluster r/s -> non negative;
  coherence;
end;

registration
  let r,s be non positive real number;
  cluster r/s -> non negative;
  coherence;
end;

begin 

registration
  let r,s;
  cluster min(r,s) -> real;
  coherence by XXREAL_0:15;
  cluster max(r,s) -> real;
  coherence by XXREAL_0:16;
end;

definition
  let r,s be real number;
  func r -' s equals
  :Def2:
  r - s if r -s >= 0 otherwise 0;
  correctness;
end;

registration
  let r,s be real number;
  cluster r -' s -> real;
  coherence
  proof
    r -' s = r - s or r -' s = 0 by Def2;
    hence thesis;
  end;
end;

registration
  let r,s be real number;
  cluster r -' s -> non negative for real number;
  coherence
  proof
    r -' s = r - s & r - s >= 0 or r -' s = 0 by Def2;
    hence thesis by XXREAL_0:def 7;
  end;
end;














begin









reserve x, y, z for real number;

Lm1: x <= y & y <= x implies x = y
proof
  assume that
A1: x <= y and
A2: y <= x;
A3: x in REAL & y in REAL by XREAL_0:def 1;
  per cases by A3,NUMBERS:def 1,XBOOLE_0:def 3;
  suppose
    x in REAL+ & y in REAL+;
    then
    (ex x9,y9 being Element of REAL+ st x = x9 & y = y9 & x9 <=' y9 )& ex
y99, x99 being Element of REAL+ st y = y99 & x = x99 & y99 <=' x99 by A1,A2,
XXREAL_0:def 5;
    hence thesis by ARYTM_1:4;
  end;
  suppose
A4: x in REAL+ & y in [:{0},REAL+:];
    then
    ( not(x in REAL+ & y in REAL+))& not(x in [:{0},REAL+:] & y in [:{0},
    REAL+:]) by ARYTM_0:5,XBOOLE_0:3;
    hence thesis by A1,A4,XXREAL_0:def 5;
  end;
  suppose
A5: y in REAL+ & x in [:{0},REAL+:];
    then
    ( not(x in REAL+ & y in REAL+))& not(x in [:{0},REAL+:] & y in [:{0},
    REAL+:]) by ARYTM_0:5,XBOOLE_0:3;
    hence thesis by A2,A5,XXREAL_0:def 5;
  end;
  suppose that
A6: x in [:{0},REAL+:] & y in [:{0},REAL+:];
    consider x9,y9 being Element of REAL+ such that
A7: x = [0,x9] & y = [0,y9] and
A8: y9 <=' x9 by A1,A6,XXREAL_0:def 5;
    consider y99,x99 being Element of REAL+ such that
A9: y = [0,y99] & x = [0,x99] and
A10: x99 <=' y99 by A2,A6,XXREAL_0:def 5;
    x9 = x99 & y9 = y99 by A7,A9,XTUPLE_0:1;
    hence thesis by A8,A9,A10,ARYTM_1:4;
  end;
end;

Lm2: x <= y & y <= z implies x <= z
proof
  assume that
A1: x <= y and
A2: y <= z;
A3: x in REAL & y in REAL by XREAL_0:def 1;
A4: z in REAL by XREAL_0:def 1;
  per cases by A3,A4,NUMBERS:def 1,XBOOLE_0:def 3;
  suppose that
A5: x in REAL+ and
A6: y in REAL+ and
A7: z in REAL+;
    consider y99,z9 being Element of REAL+ such that
A8: y = y99 and
A9: z = z9 and
A10: y99 <=' z9 by A2,A6,A7,XXREAL_0:def 5;
    consider x9,y9 being Element of REAL+ such that
A11: x = x9 and
A12: y = y9 & x9 <=' y9 by A1,A5,A6,XXREAL_0:def 5;
    x9 <=' z9 by A12,A8,A10,ARYTM_1:3;
    hence thesis by A11,A9,XXREAL_0:def 5;
  end;
  suppose
A13: x in REAL+ & y in [:{0},REAL+:];
    then
    ( not(x in REAL+ & y in REAL+))& not(x in [:{0},REAL+:] & y in [:{0},
    REAL+:]) by ARYTM_0:5,XBOOLE_0:3;
    hence thesis by A1,A13,XXREAL_0:def 5;
  end;
  suppose
A14: y in REAL+ & z in [:{0},REAL+:];
    then
    ( not(z in REAL+ & y in REAL+))& not(z in [:{0},REAL+:] & y in [:{0},
    REAL+:]) by ARYTM_0:5,XBOOLE_0:3;
    hence thesis by A2,A14,XXREAL_0:def 5;
  end;
  suppose that
A15: x in [:{0},REAL+:] and
A16: z in REAL+;
    ( not(x in REAL+ & z in REAL+))& not(x in [:{0},REAL+:] & z in [:{0},
    REAL+:]) by A15,A16,ARYTM_0:5,XBOOLE_0:3;
    hence thesis by A16,XXREAL_0:def 5;
  end;
  suppose that
A17: x in [:{0},REAL+:] and
A18: y in [:{0},REAL+:] and
A19: z in [:{0},REAL+:];
    consider y99,z9 being Element of REAL+ such that
A20: y = [0,y99] and
A21: z = [0,z9] and
A22: z9 <=' y99 by A2,A18,A19,XXREAL_0:def 5;
    consider x9,y9 being Element of REAL+ such that
A23: x = [0,x9] and
A24: y = [0,y9] and
A25: y9 <=' x9 by A1,A17,A18,XXREAL_0:def 5;
    y9 = y99 by A24,A20,XTUPLE_0:1;
    then z9 <=' x9 by A25,A22,ARYTM_1:3;
    hence thesis by A17,A19,A23,A21,XXREAL_0:def 5;
  end;
end;

theorem
  x <= y & x is positive implies y is positive
proof
  assume that
A1: x <= y and
A2: x is positive;
  x > 0 by A2,XXREAL_0:def 6;
  then y > 0 by A1,Lm2;
  hence thesis by XXREAL_0:def 6;
end;

theorem
  x <= y & y is negative implies x is negative
proof
  assume that
A1: x <= y and
A2: y is negative;
  y < 0 by A2,XXREAL_0:def 7;
  then x < 0 by A1,Lm2;
  hence thesis by XXREAL_0:def 7;
end;

theorem
  x <= y & x is non negative implies y is non negative
proof
  assume that
A1: x <= y and
A2: x is non negative and
A3: y is negative;
  y < 0 by A3,XXREAL_0:def 7;
  then x < 0 by A1,Lm2;
  hence thesis by A2,XXREAL_0:def 7;
end;

theorem
  x <= y & y is non positive implies x is non positive
proof
  assume that
A1: x <= y and
A2: y is non positive and
A3: x is positive;
  x > 0 by A3,XXREAL_0:def 6;
  then y > 0 by A1,Lm2;
  hence thesis by A2,XXREAL_0:def 6;
end;

theorem
  x <= y & y is non zero & x is non negative implies y is positive
proof
  assume that
A1: x <= y and
A2: y is non zero and
A3: x is non negative and
A4: y is non positive;
  y <= 0 by A4,XXREAL_0:def 6;
  then
A5: y < 0 by A2,Lm1;
  x >= 0 by A3,XXREAL_0:def 7;
  hence thesis by A1,A5,Lm2;
end;

theorem
  x <= y & x is non zero & y is non positive implies x is negative
proof
  assume that
A1: x <= y and
A2: x is non zero and
A3: y is non positive and
A4: x is non negative;
  x >= 0 by A4,XXREAL_0:def 7;
  then
A5: x > 0 by A2,Lm1;
  y <= 0 by A3,XXREAL_0:def 6;
  hence thesis by A1,A5,Lm2;
end;

theorem
  not x <= y & x is non positive implies y is negative
proof
  assume that
A1: x > y and
A2: x is non positive & y is non negative;
  x <= 0 & y >= 0 by A2,XXREAL_0:def 6,def 7;
  hence thesis by A1,Lm2;
end;

theorem
  not x <= y & y is non negative implies x is positive
proof
  assume that
A1: x > y and
A2: y is non negative & x is non positive;
  x <= 0 & y >= 0 by A2,XXREAL_0:def 6,def 7;
  hence thesis by A1,Lm2;
end;
 
















begin

reserve a, b, c, d, e for complex number;



theorem 
  a + (b + c) = (a + b) + c;

theorem 
  a + c = b + c implies a = b;

theorem 
  a = a + b implies b = 0;



theorem Th4:
  a * (b * c) = (a * b) * c;

theorem 
  c <> 0 & a * c = b * c implies a = b
proof
  assume
A1: c<>0;
  assume a * c = b * c;
  then a * (c * c") = b * c * c" by Th4;
  then a * 1 = b * c * c" by A1,XCMPLX_0:def 7;
  then a = b * (c * c");
  then a = b * 1 by A1,XCMPLX_0:def 7;
  hence thesis;
end;

theorem 
  a*b=0 implies a=0 or b=0;

theorem Th7: 
  b <> 0 & a * b = b implies a = 1
proof
  assume that
A1: b<>0 and
A2: a*b=b;
  a*b*b"=1 by A1,A2,XCMPLX_0:def 7;
  then a*1=1 by A2,Th4;
  hence thesis;
end;



theorem 
  a * (b + c) = a * b + a * c;

theorem 
  (a + b + c) * d = a * d + b * d + c * d;

theorem 
  (a + b) * (c + d) = a * c + a * d + b * c + b * d;

theorem 
  2 * a = a + a;

theorem 
  3 * a = a + a + a;

theorem 
  4 * a = a + a + a + a;



theorem 
  a - a = 0;

theorem 
  a - b = 0 implies a = b;

theorem 
  b - a = b implies a = 0;



theorem 
  a = a - (b - b);

theorem 
  a - (a - b) = b;

theorem 
  a - c = b - c implies a = b;

theorem 
  c - a = c - b implies a = b;

theorem 
  a - b - c = a - c - b;

theorem 
  a - c = (a - b) - (c - b);

theorem 
  (c - a) - (c - b) = b - a;

theorem 
  a - b = c - d implies a - c = b - d;



Lm1: a"*b"=(a*b)"
proof
  per cases;
  suppose
A1: a = 0 or b = 0;
    then a" = 0 or b" = 0;
    hence thesis by A1;
  end;
  suppose that
A2: a<>0 and
A3: b<>0;
    thus a"*b"=a"*b"*1 .=a"*b"*((a*b)*(a*b)") by A2,A3,XCMPLX_0:def 7
      .=a"*a*(b"*b)*(a*b)"
      .=1*(b"*b)*(a*b)" by A2,XCMPLX_0:def 7
      .=1*(a*b)" by A3,XCMPLX_0:def 7
      .=(a*b)";
  end;
end;

Lm2: a/(b/c)=(a*c)/b
proof
  thus a/(b/c)=a/(b*c") by XCMPLX_0:def 9
    .=a*(b*c")" by XCMPLX_0:def 9
    .=a*(b"*c"") by Lm1
    .=a*c*b"
    .=(a*c)/b by XCMPLX_0:def 9;
end;

Lm3: b<>0 implies a/b*b=a
proof
  assume
A1: b<>0;
  thus a/b*b=a*b"*b by XCMPLX_0:def 9
    .=a*(b"*b)
    .=a*1 by A1,XCMPLX_0:def 7
    .=a;
end;

Lm4: 1/a=a"
proof
  thus 1/a= 1 * a" by XCMPLX_0:def 9
    .=a";
end;

Lm5: a<>0 implies a/a = 1
proof
  assume
A1: a<>0;
  thus a/a=a*a" by XCMPLX_0:def 9
    .=1 by A1,XCMPLX_0:def 7;
end;



theorem 
  a = a + (b - b);

theorem 
  a = a + b - b;

theorem 
  a = a - b + b;

theorem 
  a + c = a + b + (c - b);



theorem 
  a + b - c = a - c + b;

theorem 
  a - b + c = c - b + a;

theorem 
  a + c = a + b - (b - c);

theorem 
  a - c = a + b - (c + b);

theorem 
  a + b = c + d implies a - c = d - b;

theorem 
  a - c = d - b implies a + b = c + d;

theorem 
  a + b = c - d implies a + d = c - b;



theorem 
  a - (b + c) = a - b - c;

theorem 
  a - (b - c) = a - b + c;

theorem 
  a - (b - c) = a + (c - b);

theorem 
  a - c = (a - b) + (b - c);

theorem 
  a * (b - c) = a * b - a * c;

theorem 
  (a - b) * (c - d) = (b - a) * (d - c);

theorem 
  (a - b - c) * d = a * d - b * d - c * d;



theorem 
  (a + b - c) * d = a * d + b * d - c * d;

theorem 
  (a - b + c) * d = a * d - b * d + c * d;

theorem 
  (a + b) * (c - d) = a * c - a * d + b * c - b * d;

theorem 
  (a - b) * (c + d) = a * c + a * d - b * c - b * d;

theorem 
  (a - b) * (e - d) = a * e - a * d - b * e + b * d;



theorem 
  a / b / c = a / c / b
proof
  thus a/b/c =a*b"/c by XCMPLX_0:def 9
    .=a*b"*c" by XCMPLX_0:def 9
    .=a*c"*b"
    .=a/c*b" by XCMPLX_0:def 9
    .=a/c/b by XCMPLX_0:def 9;
end;



theorem Th49: 
  a / 0 = 0
proof
  thus a/0 = a*0" by XCMPLX_0:def 9
    .= 0;
end;

theorem 
  a <> 0 & b <> 0 implies a / b <> 0;



theorem 
  b <> 0 implies a = a / (b / b)
proof
A1: a=a/1;
  assume b<>0;
  hence thesis by A1,Lm5;
end;

Lm6: (a/b) * (c/d) = (a*c)/(b*d)
proof
  thus (a/b) * (c/d) =(a*b")*(c/d) by XCMPLX_0:def 9
    .=(a*b")*(c*d") by XCMPLX_0:def 9
    .=(a*c)*(b"*d")
    .=(a*c)*(b*d)" by Lm1
    .=(a*c)/(b*d) by XCMPLX_0:def 9;
end;

Lm7: (a/b)"=b/a
proof
  per cases;
  suppose
A1: a = 0;
    hence (a/b)" =b*0" .=b/a by A1,XCMPLX_0:def 9;
  end;
  suppose
A2: b = 0;
    hence (a/b)" =(a*0")" by XCMPLX_0:def 9
      .=b/a by A2;
  end;
  suppose
A3: a<>0 & b<>0;
    (a/b) * (b/a) =(a*b)/(a*b) by Lm6;
    then (a/b)*(b/a) = 1 by A3,Lm5;
    hence thesis by A3,XCMPLX_0:def 7;
  end;
end;

Lm8: a*(b/c) = (a*b)/c
proof
  thus a*(b/c) = (a/1)*(b/c) .= (a*b)/(1*c) by Lm6
    .= (a*b)/c;
end;

theorem 
  a <> 0 implies a / (a / b) = b
proof
  assume
A1: a <> 0;
  thus a/(a/b) = a * (a/b)" by XCMPLX_0:def 9
    .= a * (b/a) by Lm7
    .= a*b/a by Lm8
    .= a/a*b by Lm8
    .= 1 * b by A1,Lm5
    .= b;
end;

theorem 
  c <> 0 & a / c = b / c implies a = b
proof
  assume that
A1: c<>0 and
A2: a/c =b/c;
  a=b/c*c by A1,A2,Lm3;
  hence thesis by A1,Lm3;
end;



Lm9: b<>0 implies a=a*b/b
proof
A1: a=a*1;
  assume b<>0;
  then a=a*(b/b) by A1,Lm5;
  then a=a*(b*b") by XCMPLX_0:def 9;
  then a=a*b*b";
  hence thesis by XCMPLX_0:def 9;
end;

theorem 
  a / b <> 0 implies b = a / (a / b)
proof
  assume
A1: a/b<>0;
  then b<>0 by Th49;
  then a/b*b = a by Lm3;
  hence thesis by A1,Lm9;
end;



Lm10: c<>0 implies a/b=(a*c)/(b*c)
proof
  assume
A1: c<>0;
  thus a/b =a*b"*1 by XCMPLX_0:def 9
    .=a*b"*(c*c") by A1,XCMPLX_0:def 7
    .=(a*c)*(b"*c")
    .=(a*c)*(b*c)" by Lm1
    .=(a*c)/(b*c) by XCMPLX_0:def 9;
end;

theorem 
  c <> 0 implies a / b = (a / c) / (b / c)
proof
  assume c<>0;
  hence a/b=(a*c")/(b*c") by Lm10
    .=(a/c)/(b*c") by XCMPLX_0:def 9
    .=(a/c)/(b/c) by XCMPLX_0:def 9;
end;



theorem 
  1 / (1 / a) = a
proof
  thus 1/(1/a) = (1*a) /1 by Lm2
    .= a;
end;

Lm11: (a*b")"=a"*b
proof
  thus (a*b")"=a"*b"" by Lm1
    .=a"*b;
end;

theorem 
  1 / (a / b) = b / a
proof
  thus 1/(a/b)=1/(a*b") by XCMPLX_0:def 9
    .=(a*b")" by Lm4
    .=b*a" by Lm11
    .=b/a by XCMPLX_0:def 9;
end;

theorem Th58:
  a / b = 1 implies a = b
proof
  assume
A1: a/b = 1;
  then b <> 0 by Th49;
  then a=1*b by A1,Lm3;
  hence thesis;
end;

Lm12: a"=b" implies a=b
proof
  assume a"=b";
  then a=b"";
  hence thesis;
end;

theorem Th59: 
  1 / a = 1 / b implies a = b
proof
  assume 1/a=1/b;
  then a"=1/b by Lm4;
  then a" = b" by Lm4;
  hence thesis by Lm12;
end;



theorem 
  a <> 0 implies a / a = 1 by Lm5;

theorem 
  b <> 0 & b / a = b implies a = 1
proof
  assume that
A1: b<>0 and
A2: b/a=b;
  a <> 0 by A1,A2,Th49;
  then b=b*a by A2,Lm3;
  hence thesis by A1,Th7;
end;



theorem Th62: 
  a / c + b / c = (a + b) / c
proof
  thus a/c + b/c =a*c" + b/c by XCMPLX_0:def 9
    .= a*c" + b*c" by XCMPLX_0:def 9
    .= (a+b)*c"
    .= (a+b)/c by XCMPLX_0:def 9;
end;

theorem 
  (a + b + e) / d = a / d + b / d + e / d
proof
  thus (a+b+e)/d=(a+b)/d+e/d by Th62
    .=a/d+b/d+e/d by Th62;
end;



theorem 
  (a + a) / 2 = a;

theorem 
  a/2 + a/2 = a;

theorem 
  a = (a + b) / 2 implies a = b;



theorem 
  (a + a + a)/3 = a;

theorem 
  a/3 + a/3 + a/3 = a;



theorem 
  (a + a + a + a) / 4 = a;

theorem 
  a/4 + a/4 + a/4 + a/4 = a;

theorem 
  a / 4 + a / 4 = a / 2;

theorem 
  (a + a) / 4 = a / 2;



theorem 
  a * b = 1 implies a = 1 / b
proof
  assume
A1: a*b=1;
  then b<>0;
  then a*1=1*b" by A1,XCMPLX_0:def 7;
  hence thesis by Lm4;
end;

theorem 
  a * (b / c) = (a * b) / c by Lm8;

theorem 
  a / b * e = e / b * a
proof
  thus a/b*e=a*e/b by Lm8
    .=e/b*a by Lm8;
end;



theorem 
  (a / b) * (c / d) = (a * c) / (b * d) by Lm6;

theorem 
  a / (b / c) = (a * c) / b by Lm2;

Lm13: (a/b)/(c/d)=(a*d)/(b*c)
proof
  thus (a/b)/(c/d) = (a/b) * (c/d)" by XCMPLX_0:def 9
    .=(a/b) * (d/c) by Lm7
    .=(a*d)/(b*c) by Lm6;
end;

theorem Th78: 
  a / (b * c) = a / b / c
proof
  thus a/(b*c) = a*1/(b*c) .= a/b/(c/1) by Lm13
    .= a/b/c;
end;

theorem 
  a / (b / c) = a * (c / b)
proof
  thus a/(b/c)=(a*c)/b by Lm2
    .=a*c*b" by XCMPLX_0:def 9
    .=a*(c*b")
    .=a*(c/b) by XCMPLX_0:def 9;
end;

theorem 
  a / (b / c) = c / b * a
proof
  a/(b/c)=(a*c)/b by Lm2
    .=a*c*b" by XCMPLX_0:def 9
    .=a*(c*b")
    .=a*(c/b) by XCMPLX_0:def 9;
  hence thesis;
end;

theorem Th81: 
  a / (b / e) = e * (a / b)
proof
  thus a/(b/e)=(a*e)/b by Lm2
    .=e*a*b" by XCMPLX_0:def 9
    .=e*(a*b")
    .=e*(a/b) by XCMPLX_0:def 9;
end;

theorem 
  a / (b / c) = a / b * c
proof
  a/(b/c)=(a*c)/b by Lm2
    .=c*a*b" by XCMPLX_0:def 9
    .=c*(a*b")
    .=c*(a/b) by XCMPLX_0:def 9;
  hence thesis;
end;

Lm14: a*(1/b)=a/b
proof
  thus a/b=a*b" by XCMPLX_0:def 9
    .=a*(1/b) by Lm4;
end;

Lm15: 1/c*(a/b)=a/(b*c)
proof
  a/b/c =c"*(a/b) by XCMPLX_0:def 9
    .=1/c*(a/b) by Lm4;
  hence thesis by Th78;
end;

theorem 
  (a * b) / (c * d) = (a / c * b) / d
proof
  thus a*b/(c*d)=1/c*(a*b/d) by Lm15
    .=1/c*(a*b)/d by Lm8
    .=1/c*a*b/d
    .=a/c*b/d by Lm14;
end;



theorem 
  (a / b) / (c / d) = (a * d) / (b * c) by Lm13;

theorem 
  (a / c) * (b / d) = (a / d) * (b / c)
proof
  thus a/c*(b/d)=a*b/(d*c) by Lm6
    .=a/d*(b/c) by Lm6;
end;

theorem 
  a / (b * c * (d / e)) = (e / c) * (a / (b * d))
proof
  thus a/(b*c*(d/e)) = a/(b*c*(d*e")) by XCMPLX_0:def 9
    .= a/(c*(b*d*e"))
    .= a/(c*((b*d)/e)) by XCMPLX_0:def 9
    .= a/((b*d)/(e/c)) by Th81
    .= (e/c)*(a/(b*d)) by Th81;
end;



theorem 
  b <> 0 implies a / b * b = a by Lm3;

theorem 
  b <> 0 implies a = a * (b / b)
proof
A1: a=a*1;
  assume b<>0;
  hence thesis by A1,Lm5;
end;

theorem 
  b <> 0 implies a = a * b / b by Lm9;

theorem 
  b <> 0 implies a * c = a * b * (c / b)
proof
  assume
A1: b<>0;
  thus a*c =a*1*c .=a*(b*b")*c by A1,XCMPLX_0:def 7
    .=a*b*(b"*c)
    .=a*b*(c/b) by XCMPLX_0:def 9;
end;



theorem 
  c <> 0 implies a / b = (a * c) / (b * c) by Lm10;

theorem 
  c <> 0 implies a / b = a / (b * c) * c
proof
  assume
A1: c<>0;
  c*(a/(b*c))=c*((a*1)/(b*c)) .=c*(1/c*(a/b)) by Lm6
    .=1/c*c*(a/b)
    .=1*(a/b) by A1,Lm3
    .=a/b;
  hence thesis;
end;

theorem 
  b <> 0 implies a * c = a * b / (b / c)
proof
  assume
A1: b<>0;
  thus a*c =a*1*c .=a*(b*b")*c by A1,XCMPLX_0:def 7
    .=a*b*(b"*c)
    .=a*b*(b*c")" by Lm11
    .=a*b/(b*c") by XCMPLX_0:def 9
    .=a*b/(b/c) by XCMPLX_0:def 9;
end;

theorem Th94: 
  c <> 0 & d <> 0 & a * c = b * d implies a / d = b / c
proof
  assume that
A1: c<>0 and
A2: d<>0;
  assume a*c = b*d;
  then a=b*d/c by A1,Lm9;
  then a=d*(b/c) by Lm8;
  hence thesis by A2,Lm9;
end;

theorem Th95: 
  c <> 0 & d<>0 & a/d=b/c implies a*c = b*d
proof
  assume that
A1: c<>0 and
A2: d<>0 and
A3: a/d=b/c;
  c*(a/d)=b by A1,A3,Lm3;
  then a*c/d=b by Lm8;
  hence thesis by A2,Lm3;
end;

theorem 
  c <> 0 & d <> 0 & a * c = b / d implies a * d = b / c
proof
  assume that
A1: c<>0 and
A2: d<>0;
  assume a*c =b/d;
  then a*c*d=b by A2,Lm3;
  then a*d*c =b;
  hence thesis by A1,Lm9;
end;



theorem 
  c <> 0 implies a / b = c * (a / c / b)
proof
  assume
A1: c<>0;
  thus a/b=a*b" by XCMPLX_0:def 9
    .=c*(a/c)*b" by A1,Lm3
    .=c*(a/c*b")
    .=c*(a/c/b) by XCMPLX_0:def 9;
end;

theorem 
  c <> 0 implies a / b = a / c * (c / b)
proof
  assume
A1: c<>0;
  thus a/b=a*b" by XCMPLX_0:def 9
    .=a/c*c*b" by A1,Lm3
    .=a/c*(c*b")
    .=a/c*(c/b) by XCMPLX_0:def 9;
end;



theorem 
  a * (1 / b) = a / b by Lm14;

Lm16: 1/a"=a
proof
  1/a"=a"" by Lm4
    .=a;
  hence thesis;
end;

theorem 
  a / (1 / b) = a * b
proof
  thus a/(1/b)=a/b" by Lm4
    .=a*(1/b") by Lm14
    .=a*b by Lm16;
end;

theorem 
  a / b * c = 1 / b * c * a
proof
  a/b*c = 1/b*a*c by Lm14;
  hence thesis;
end;



theorem 
  (1 / a) * (1 / b) = 1 / (a * b)
proof
  thus (1/a)*(1/b)=a"*(1/b) by Lm4
    .=a"*b" by Lm4
    .=(a*b)" by Lm1
    .=1/(a*b) by Lm4;
end;

theorem 
  1 / c * (a / b) = a / (b * c) by Lm15;



theorem 
  a / b / c = 1 / b * (a / c)
proof
  a/b/c =a*b"/c by XCMPLX_0:def 9
    .=a*b"*c" by XCMPLX_0:def 9
    .=a*c"*b"
    .=a/c*b" by XCMPLX_0:def 9
    .=a/c/b by XCMPLX_0:def 9;
  hence a/b/c =b"*(a/c) by XCMPLX_0:def 9
    .=1/b*(a/c) by Lm4;
end;

theorem 
  a / b / c = 1 / c * (a / b)
proof
  thus a/b/c =c"*(a/b) by XCMPLX_0:def 9
    .=1/c*(a/b) by Lm4;
end;



theorem Th106: 
  a <> 0 implies a * (1 / a) = 1
proof
  assume
A1: a<>0;
  thus a*(1/a)=a*a" by Lm4
    .=1 by A1,XCMPLX_0:def 7;
end;

theorem 
  b <> 0 implies a = a * b * (1 / b)
proof
A1: a=a*1;
  assume b<>0;
  then a=a*(b/b) by A1,Lm5;
  then a=a*(b*b") by XCMPLX_0:def 9
    .=a*(b*(1/b)) by Lm4;
  hence thesis;
end;

theorem 
  b <> 0 implies a = a * (1 / b * b)
proof
  assume
A1: b<>0;
  thus a=a*1 .=a*(1/b*b) by A1,Lm3;
end;

theorem 
  b <> 0 implies a = a * (1 / b) * b
proof
  assume
A1: b<>0;
  a=a*1 .=a*(1/b*b) by A1,Lm3;
  hence thesis;
end;

theorem 
  b <> 0 implies a = a / (b * (1 / b))
proof
  assume
A1: b<>0;
  thus a=a/1 .=a/(b*(1/b)) by A1,Th106;
end;

theorem 
  a <> 0 & b <> 0 implies 1 / (a * b) <> 0;

theorem 
  a <> 0 & b <> 0 implies (a / b) * (b / a) = 1
proof
  assume
A1: a<>0 & b<>0;
  (b/a)=(a/b)" by Lm7;
  hence thesis by A1,XCMPLX_0:def 7;
end;



theorem Th113: 
  b <> 0 implies a / b + c = (a + b * c) / b
proof
  assume
A1: b<>0;
  a/b+c =a/b+1*c .=a/b+b*b"*c by A1,XCMPLX_0:def 7
    .=a/b+b*c*b"
    .=a/b+c*b/b by XCMPLX_0:def 9
    .=(a+c*b)/b by Th62;
  hence thesis;
end;

theorem Th114: 
  c <> 0 implies a + b = c * (a / c + b / c)
proof
  assume
A1: c<>0;
  hence a+b=c*(a/c)+b by Lm3
    .=c*(a/c)+c*(b/c) by A1,Lm3
    .=c*(a/c+b/c);
end;

theorem Th115: 
  c <> 0 implies a + b = (a * c + b * c) / c
proof
  assume
A1: c<>0;
  hence a+b=a*c/c+b by Lm9
    .=a*c/c+b*c/c by A1,Lm9
    .=(a*c+b*c)/c by Th62;
end;

theorem Th116: 
  b <> 0 & d <> 0 implies a / b + c / d =(a * d + c * b) / (b * d )
proof
  assume
A1: b<>0;
  assume d<>0;
  hence a/b + c/d=(a*d)/(b*d) + c/d by Lm10
    .=(a*d)/(b*d) + (c*b)/(b*d) by A1,Lm10
    .=(a*d + c*b)/(b*d) by Th62;
end;

theorem Th117: 
  a <> 0 implies a + b = a * (1 + b / a)
proof
  assume
A1: a<>0;
  hence a+b=a*(a/a+b/a) by Th114
    .=a*(1+b/a) by A1,Lm5;
end;



theorem 
  a / (2 * b) + a / (2 * b) = a / b
proof
  thus a/(2*b)+a/(2*b)=(a+a)/(2*b) by Th62
    .=2*a/(2*b)
    .=a/b by Lm10;
end;



theorem 
  a / (3 * b) + a / (3 * b) + a / (3 * b) = a / b
proof
  thus a/(3*b)+a/(3*b)+a/(3*b)=(a+a)/(3*b)+a/(3*b) by Th62
    .=(a+a+a)/(3*b) by Th62
    .=3*a/(3*b)
    .=a/b by Lm10;
end;



Lm17: -a/b=(-a)/b
proof
  thus -a/b=-(a*b") by XCMPLX_0:def 9
    .=(-a)*b"
    .=(-a)/b by XCMPLX_0:def 9;
end;

theorem Th120: 
  a / c - b / c = (a - b) / c
proof
  thus a/c - b/c = a/c+ -b/c .=a/c+(-b)/c by Lm17
    .=(a+ -b)/c by Th62
    .=(a-b)/c;
end;

theorem 
  a - a / 2 = a / 2;

theorem 
  (a - b - c) / d = a / d - b / d - c / d
proof
  thus (a-b-c)/d=(a-b)/d-c/d by Th120
    .=a/d-b/d-c/d by Th120;
end;

theorem 
  b <> 0 & d <> 0 & b <> d & a / b = e / d implies a / b = (a - e) / (b - d )
proof
  assume that
A1: b<>0 and
A2: d<>0 and
A3: b<>d and
A4: a/b=e/d;
  a*d=e*b by A1,A2,A4,Th95;
  then
A5: a*(b-d)=(a-e)*b;
  b-d<>0 by A3;
  hence thesis by A1,A5,Th94;
end;



theorem 
  (a + b - e) / d = a / d + b / d - e / d
proof
  thus (a+b-e)/d=(a+b)/d-e/d by Th120
    .=a/d+b/d-e/d by Th62;
end;

theorem 
  (a - b + e) / d = a / d - b / d + e / d
proof
  thus (a-b+e)/d=(a-b)/d+e/d by Th62
    .=a/d-b/d+e/d by Th120;
end;



theorem Th126: 
  b <> 0 implies a / b - e = (a - e * b) / b
proof
  assume
A1: b<>0;
  thus a/b-e=a/b+-e .=(a+(-e)*b)/b by A1,Th113
    .=(a-e*b)/b;
end;

theorem 
  b <> 0 implies c - a / b = (c * b - a) / b
proof
  assume
A1: b<>0;
  thus c-a/b = -(a/b-c) .=-(a-c*b)/b by A1,Th126
    .=(-(a-c*b))/b by Lm17
    .=(c*b-a)/b;
end;

theorem 
  c <> 0 implies a - b = c * (a / c - b / c)
proof
  assume
A1: c<>0;
  hence a-b=c*(a/c)-b by Lm3
    .=c*(a/c)-c*(b/c) by A1,Lm3
    .=c*(a/c-b/c);
end;

theorem 
  c <> 0 implies a - b = (a * c - b * c) / c
proof
  assume
A1: c<>0;
  thus a-b=a+-b .=(a*c+(-b)*c)/c by A1,Th115
    .=(a*c-b*c)/c;
end;

theorem 
  b <> 0 & d <> 0 implies a / b - c / d = (a * d - c * b) / (b * d)
proof
  assume
A1: b<>0;
  assume
A2: d<>0;
  thus a/b - c/d =a/b + -c/d .=a/b + (-c)/d by Lm17
    .=(a*d + (-c)*b)/(b*d) by A1,A2,Th116
    .=(a*d - c*b)/(b*d);
end;

theorem 
  a <> 0 implies a - b = a * (1 - b / a)
proof
  assume
A1: a<>0;
  thus a-b=a+-b .=a*(1+(-b)/a) by A1,Th117
    .=a*(1+-b/a) by Lm17
    .=a*(1-b/a);
end;



theorem 
  a <> 0 implies c = (a * c + b - b) / a by Lm9;



theorem 
  -a = -b implies a = b;

theorem 
  -a = 0 implies a = 0;

theorem 
  a + -b = 0 implies a = b;

theorem 
  a = a + b + -b;

theorem 
  a = a + (b + -b);

theorem 
  a = (- b + a) + b;

theorem 
  - (a + b) = -a + -b;

theorem 
  - (-a + b) = a + -b;

theorem 
  a+b=-(-a+-b);



theorem 
  -(a - b) = b - a;

theorem 
  - a - b = - b - a;

theorem 
  a = - b - (- a - b);



theorem 
  - a - b - c = - a - c - b;

theorem 
  - a - b - c = - b - c - a;

theorem 
  - a - b - c = - c - b - a;

theorem 
  (c - a) - (c - b) = - (a - b);



theorem 
  0 - a = - a;



theorem 
  a + b = a - - b;

theorem 
  a = a - (b + -b);

theorem 
  a - c = b + - c implies a = b;

theorem 
  c - a = c + - b implies a = b;



theorem 
  a + b - c = - c + a + b;

theorem 
  a - b + c = - b + c + a;

theorem 
  a - (- b - c) = a + b + c;



theorem 
  a - b - c = - b - c + a;

theorem 
  a - b - c = - c + a - b;

theorem 
  a - b - c = - c - b + a;



theorem 
  - (a + b) = - b - a;

theorem 
  - (a - b) = - a + b;

theorem 
  -(-a+b)=a-b;

theorem 
  a + b = -(- a - b);

theorem 
  - a + b - c = - c + b - a;



theorem 
  - a + b - c = - c - a + b;

theorem 
  - (a + b + c) = - a - b - c;

theorem 
  - (a + b - c) = - a - b + c;

theorem 
  - (a - b + c) = - a + b - c;

theorem 
  - (a - b - c) = - a + b + c;

theorem 
  - (- a + b + c) = a - b - c;

theorem 
  - (- a + b - c) = a - b + c;

theorem 
  - (- a - b + c) = a + b - c;

theorem 
  - (- a - b - c) = a + b + c;



theorem 
  (- a) * b = -(a * b);

theorem 
  (- a) * b = a * (- b);

theorem 
  (- a) * (- b) = a * b;

theorem 
  - a * (- b) = a * b;

theorem 
  -(-a) * b = a * b;

theorem 
  (-1) * a = -a;

theorem 
  (- a) * (- 1) = a;

theorem Th181: 
  b<>0 & a*b=-b implies a=-1
proof
  assume that
A1: b<>0 and
A2: a*b=-b;
  a*(b*b")=(-b)*b" by A2,Th4;
  then a*1=(-b)*b" by A1,XCMPLX_0:def 7;
  hence a=-b*b" .=-1 by A1,XCMPLX_0:def 7;
end;



theorem Th182:
  a * a = 1 implies a = 1 or a = -1
proof
  assume a*a=1;
  then (a-1)*(a+1) =0;
  then a-1=0 or a+1=0;
  hence thesis;
end;

theorem 
  -a + 2 * a = a;

theorem 
  (a - b) * c = (b - a) * (- c);

theorem 
  (a - b) * c = - (b - a) * c;

theorem 
  a - 2 * a = -a;



theorem 
  -a / b = (-a) / b by Lm17;

theorem Th188: 
  a / (- b) = -a / b
proof
  a/(-b)=(a*(-1))/((-b)*(-1)) by Lm10;
  then a/(-b)= (-a)/((-(-b))*1) .= -a/b by Lm17;
  hence thesis;
end;

theorem 
  - a / (- b) = a / b
proof
  thus -a/(-b)=--a/b by Th188
    .=a/b;
end;

theorem Th190: 
  -(- a) / b = a / b
proof
  thus -(-a)/b=--a/b by Lm17
    .=a/b;
end;

theorem 
  (- a) / (- b) = a / b
proof
  -(-a)/b=a/b by Th190;
  hence thesis by Th188;
end;

theorem 
  (-a) / b = a / (-b)
proof
  thus (-a)/b=-a/b by Lm17
    .=a/(-b) by Th188;
end;

theorem 
  -a = a / (-1);

theorem 
  a = (- a) / (-1);

theorem 
  a / b = - 1 implies a = - b & b = - a
proof
  assume
A1: a/b=-1;
  then b <> 0 by Th49;
  then a=(-1)*b by A1,Lm3;
  hence thesis;
end;

theorem 
  b <> 0 & b / a = - b implies a = -1
proof
  assume that
A1: b<>0 and
A2: b/a=-b;
  a <> 0 by A1,A2,Th49;
  then b=(-b)*a by A2,Lm3;
  then b=-b*a;
  hence thesis by A1,A2,Th181;
end;

theorem 
  a <> 0 implies (-a) / a = -1
proof
  assume
A1: a<>0;
  thus (-a)/a=-a/a by Lm17
    .=-1 by A1,Lm5;
end;

theorem 
  a <> 0 implies a / (- a) = -1
proof
  assume
A1: a<>0;
  thus a/(-a)=-a/a by Th188
    .=-1 by A1,Lm5;
end;

Lm18: a<>0 & a=a" implies a=1 or a=-1
proof
  assume
A1: a<>0;
  assume a=a";
  then a*a=1 by A1,XCMPLX_0:def 7;
  hence thesis by Th182;
end;

theorem 
  a <> 0 & a = 1 / a implies a = 1 or a = -1
proof
  assume a<>0;
  then a=a" implies a=1 or a=-1 by Lm18;
  hence thesis by Lm4;
end;

theorem 
  b <> 0 & d <> 0 & b <> -d & a / b = e / d implies a / b = (a + e) / (b + d )
proof
  assume that
A1: b<>0 and
A2: d<>0 and
A3: b<>-d and
A4: a/b=e/d;
  a*d=e*b by A1,A2,A4,Th95;
  then
A5: a*(b+d)=(a+e)*b;
  b+d<>0 by A3;
  hence thesis by A1,A5,Th94;
end;



theorem 
  a" = b" implies a = b by Lm12;

theorem 
  0" = 0;



theorem
  b <> 0 implies a = a*b*b"
proof
A1: a*(b*b") = a*b*b";
  assume b <> 0;
  then a*1 = a*b*b" by A1,XCMPLX_0:def 7;
  hence thesis;
end;

theorem 
  a" * b" = (a * b)" by Lm1;

theorem 
  (a * b")" = a" * b by Lm11;

theorem 
  (a" * b")" = a * b
proof
  thus (a"*b")"=a""*b"" by Lm1
    .=a*b;
end;

theorem 
  a <> 0 & b <> 0 implies a * b" <> 0;

theorem 
  a <> 0 & b <> 0 implies a" * b" <> 0;

theorem 
  a * b" = 1 implies a = b
proof
  assume a*b"=1;
  then a/b = 1 by XCMPLX_0:def 9;
  hence thesis by Th58;
end;

theorem 
  a * b = 1 implies a = b"
proof
  assume
A1: a*b=1;
  then b<>0;
  hence thesis by A1,XCMPLX_0:def 7;
end;



theorem Th211:
  a <> 0 & b <> 0 implies a" + b" = (a + b)*(a*b)"
proof
  assume that
A1: a <> 0 and
A2: b <> 0;
  b" = b"*1;
  then b" = b"*(a"*a) by A1,XCMPLX_0:def 7;
  then b" = (a"*b")*a;
  then
A3: b" = (a*b)"*a by Lm1;
  a" = a"*1;
  then a" = a"*(b"*b) by A2,XCMPLX_0:def 7;
  then a" = (a"*b")*b;
  then a" = (a*b)"*b by Lm1;
  hence thesis by A3;
end;

Lm19: (- a)" = -a"
proof
  thus (-a)"=1/(-a) by Lm4
    .=-1/a by Th188
    .=-a" by Lm4;
end;



theorem
  a <> 0 & b <> 0 implies a" - b" = (b - a)*(a*b)"
proof
  assume
A1: a <> 0 & b <> 0;
  thus a" - b" = a" + -(b") .= a" + (-b)" by Lm19
    .= (a + -b)*(a*-b)" by A1,Th211
    .= (a + -b)*(-a*b)"
    .= (a + -b)*-((a*b)") by Lm19
    .= (b - a)*(a*b)";
end;



theorem 
  (a / b)" = b / a by Lm7;

theorem
  (a"/b") = b/a
proof
  thus (a"/b") = a"*b"" by XCMPLX_0:def 9
    .= b/a by XCMPLX_0:def 9;
end;

theorem 
  1 / a = a" by Lm4;

theorem 
  1 / a" = a by Lm16;

theorem 
  (1 / a)" = a
proof
  1/a=a" implies (1/a)"=a;
  hence thesis by Lm4;
end;

theorem 
  1 / a = b" implies a = b
proof
  1/a=1/b implies a=b by Th59;
  hence thesis by Lm4;
end;



theorem
  a/b" = a*b
proof
  thus a/b" = a*b"" by XCMPLX_0:def 9
    .= a*b;
end;

theorem
  a"*(c/b) = c/(a*b)
proof
  thus a"*(c/b) = a"*(c*b") by XCMPLX_0:def 9
    .= c*(a"*b")
    .= c*(a*b)" by Lm1
    .= c/(a*b) by XCMPLX_0:def 9;
end;

theorem
  a"/b = (a*b)"
proof
  thus a"/b = a"*b" by XCMPLX_0:def 9
    .= (a*b)" by Lm1;
end;



theorem 
  (- a)" = -a" by Lm19;

theorem 
  a <> 0 & a = a" implies a = 1 or a = -1 by Lm18;

begin 


theorem
  a+b+c-b=a+c;

theorem
  a-b+c+b=a+c;

theorem
  a+b-c-b=a-c;

theorem
  a-b-c+b=a-c;

theorem
  a-a-b=-b;

theorem
  -a+a-b=-b;

theorem
  a-b-a=-b;

theorem
  -a-b+a=-b;

begin 


theorem
  for a,b st b<>0 ex e st a=b/e
proof
  let a,b;
  assume
A1: b<>0;
  per cases;
  suppose
A2: a = 0;
    take 0;
    b/0 = b*0" by XCMPLX_0:def 9;
    hence thesis by A2;
  end;
  suppose
A3: a <> 0;
    take e=b/a;
    thus a=a*1 .=a*(e*e") by A1,A3,XCMPLX_0:def 7
      .=a*e*e"
      .=a*(a"*b)*e" by XCMPLX_0:def 9
      .=a*a"*b*e"
      .=1*b*e" by A3,XCMPLX_0:def 7
      .=b/e by XCMPLX_0:def 9;
  end;
end;



theorem
  a/(b*c*(d/e))=(e/c)*(a/(b*d))
proof
  thus a/(b*c*(d/e)) = a/(b*c*(d*e")) by XCMPLX_0:def 9
    .= a/(c*(b*d*e"))
    .= a/(c*((b*d)/e)) by XCMPLX_0:def 9
    .= a/((b*d)/(e/c)) by Th81
    .= (e/c)*(a/(b*d)) by Th81;
end;



theorem
  ((d - c)/b) * a + c = (1 - a/b)*c + (a/b) * d
proof
  per cases;
  suppose
A1: b = 0;
A2: a/b = a*b" by XCMPLX_0:def 9
      .= a*0 by A1;
    thus ((d - c)/b) * a + c = ((d - c)*b") * a + c by XCMPLX_0:def 9
      .= ((d - c)*0) * a + c by A1
      .= (1 - (a/b))*c + (a/b) * d by A2;
  end;
  suppose
A3: b <> 0;
    ((d - c)/b) * a + c = ((d - c)/b) * a + c*1
      .= ((d - c)/b) * a + c*(b/b) by A3,Lm5
      .= ((d - c)/b) * a + c*b/b by Lm8
      .= (d - c) * a / b + c*b/b by Lm8
      .= ((d-c)*a + c*b)/ b by Th62
      .= ((b-a)*c + a * d) / b
      .= (b-a)*c/b + a*d/b by Th62
      .= (b-a)*c/b + (a/b) * d by Lm8
      .= ((b - a)/b)*c + (a/b) * d by Lm8
      .= (b/b - a/b)*c + (a/b) * d by Th120
      .= (1 - a/b)*c + (a/b) * d by A3,Lm5;
    hence thesis;
  end;
end;



theorem
  a <> 0 implies a * b + c = a * (b + (c/a))
proof
  assume a <> 0;
  hence a * b + c = a * b + a*(c/a) by Lm3
    .= a * (b + (c/a));
end;














begin

reserve a,b,c,d for real number;
reserve r,s for real number;

Lm1: (r in REAL+ & s in REAL+ & ex x9,y9 being Element of REAL+ st r = x9 & s
= y9 & x9 <=' y9) or (r in [:{0},REAL+:] & s in [:{0},REAL+:] & ex x9,y9 being
Element of REAL+ st r = [0,x9] & s = [0,y9] & y9 <=' x9) or s in REAL+ & r in
[:{0},REAL+:] implies r <= s

proof
  assume

A1: (r in REAL+ & s in REAL+ & ex x9,y9 being Element of REAL+ st r = x9
  & s = y9 & x9 <=' y9) or (r in [:{0},REAL+:] & s in [:{0},REAL+:] & ex x9,y9

being Element of REAL+ st r = [0,x9] & s = [0,y9] & y9 <=' x9) or s in REAL+ &
  r in [:{0},REAL+:];

  per cases;
  case
    r in REAL+ & s in REAL+;
    hence thesis by A1,ARYTM_0:5,XBOOLE_0:3;
  end;
  case
    r in [:{0},REAL+:] & s in [:{0},REAL+:];
    hence thesis by A1,ARYTM_0:5,XBOOLE_0:3;
  end;
  case

    not(r in REAL+ & s in REAL+) & not (r in [:{0},REAL+:] & s in [:{0
    },REAL+:]);

    hence thesis by A1;
  end;
end;

Lm2: for x1,x2 being Element of REAL st a = [*x1,x2*] holds x2 = 0 & a = x1
proof
  let x1,x2 being Element of REAL;
  assume
A1: a = [*x1,x2*];
A2: a in REAL by XREAL_0:def 1;
  thus now
    assume x2 <> 0;
    then a = (0,1) --> (x1,x2) by A1,ARYTM_0:def 5;
    hence contradiction by A2,ARYTM_0:8;
  end;
  hence thesis by A1,ARYTM_0:def 5;
end;

Lm3: for a9,b9 being Element of REAL, a,b st a9 = a & b9 = b holds +(a9,b9) =
a + b

proof
  let a9,b9 be Element of REAL, a,b such that
A1: a9 = a and
A2: b9 = b;
  consider x1,x2,y1,y2 being Element of REAL such that
A3: a = [* x1,x2 *] and
A4: b = [*y1,y2*] and
A5: a+b = [*+(x1,y1),+(x2,y2)*] by XCMPLX_0:def 4;
A6: y2 = 0 by A4,Lm2;
  x2 = 0 by A3,Lm2;
  then
A7: +(x2,y2) = 0 by A6,ARYTM_0:11;
A8: b = y1 by A4,Lm2;
  a = x1 by A3,Lm2;
  hence thesis by A1,A2,A5,A8,A7,ARYTM_0:def 5;
end;

Lm4: {} in {{}} by TARSKI:def 1;

Lm5: a <= b implies a + c <= b + c
proof
  reconsider x1=a, y1=b, z1=c as Element of REAL by XREAL_0:def 1;
A1: +(y1,z1) = b + c by Lm3;
A2: +(x1,z1) = a + c by Lm3;
  assume
A3: a <= b;
  per cases by A3,XXREAL_0:def 5;
  suppose that
A4: a in REAL+ and
A5: b in REAL+ and
A6: c in REAL+;
    consider b9,c99 being Element of REAL+ such that
A7: b = b9 and
A8: c = c99 and
A9: +(y1,z1) = b9 + c99 by A5,A6,ARYTM_0:def 1;
    consider a9,c9 being Element of REAL+ such that
A10: a = a9 and
A11: c = c9 and
A12: +(x1,z1) = a9 + c9 by A4,A6,ARYTM_0:def 1;

    ex a99,b99 being Element of REAL+ st a = a99 & b = b99 & a99 <='
    b99 by A3,A4,A5,XXREAL_0:def 5;

    then a9 + c9 <=' b9 + c99 by A10,A11,A7,A8,ARYTM_1:7;
    hence thesis by A1,A2,A12,A9,Lm1;
  end;
  suppose that
A13: a in [:{0},REAL+:] and
A14: b in REAL+ and
A15: c in REAL+;
    consider b9,c99 being Element of REAL+ such that
    b = b9 and
A16: c = c99 and
A17: +(y1,z1) = b9 + c99 by A14,A15,ARYTM_0:def 1;
    consider a9,c9 being Element of REAL+ such that
    a = [0,a9] and
A18: c = c9 and
A19: +(x1,z1) = c9 - a9 by A13,A15,ARYTM_0:def 1;
    now
      per cases;
      suppose
A20:    a9 <=' c9;
A21:    c9 -' a9 <=' c9 by ARYTM_1:11;
        c9 <=' b9 + c99 by A18,A16,ARYTM_2:19;
        then
A22:    c9 -' a9 <=' b9 + c99 by A21,ARYTM_1:3;
        c9 - a9 = c9 -' a9 by A20,ARYTM_1:def 2;
        hence thesis by A1,A2,A19,A17,A22,Lm1;
      end;
      suppose
        not a9 <=' c9;
        then c9 - a9 = [0,a9 -' c9] by ARYTM_1:def 2;
        then c9 - a9 in [:{0},REAL+:] by Lm4,ZFMISC_1:87;
        then
A23:    not a + c in REAL+ by A2,A19,ARYTM_0:5,XBOOLE_0:3;
        not b + c in [:{0},REAL+:] by A1,A17,ARYTM_0:5,XBOOLE_0:3;
        hence thesis by A23,XXREAL_0:def 5;
      end;
    end;
    hence thesis;
  end;
  suppose that
A24: a in [:{0},REAL+:] and
A25: b in [:{0},REAL+:] and
A26: c in REAL+;
    consider b9,c99 being Element of REAL+ such that
A27: b = [0,b9] and
A28: c = c99 and
A29: +(y1,z1) = c99 - b9 by A25,A26,ARYTM_0:def 1;
    consider a99,b99 being Element of REAL+ such that
A30: a = [0,a99] and
A31: b = [0,b99] and
A32: b99 <=' a99 by A3,A24,A25,XXREAL_0:def 5;
    consider a9,c9 being Element of REAL+ such that
A33: a = [0,a9] and
A34: c = c9 and
A35: +(x1,z1) = c9 - a9 by A24,A26,ARYTM_0:def 1;
A36: a9 = a99 by A30,A33,XTUPLE_0:1;
A37: b9 = b99 by A31,A27,XTUPLE_0:1;
    now
      per cases;
      suppose
A38:    a9 <=' c9;
        then b9 <=' c9 by A32,A36,A37,ARYTM_1:3;
        then
A39:    c9 - b9 = c9 -' b9 by ARYTM_1:def 2;
A40:    c9 - a9 = c9 -' a9 by A38,ARYTM_1:def 2;
        c9 -' a9 <=' c99 -' b9 by A32,A34,A28,A36,A37,ARYTM_1:16;
        hence thesis by A1,A2,A34,A35,A28,A29,A40,A39,Lm1;
      end;
      suppose
        not a9 <=' c9;
        then
A41:    +(x1,z1) = [0,a9 -' c9] by A35,ARYTM_1:def 2;
        then
A42:    +(x1,z1) in [:{0},REAL+:] by Lm4,ZFMISC_1:87;
        now
          per cases;
          suppose
            b9 <=' c9;
            then c9 - b9 = c9 -' b9 by ARYTM_1:def 2;

            then
A43:        not +(y1,z1) in [:{0},REAL+:] by A34,A28,A29,ARYTM_0:5,XBOOLE_0:3;

            not +(x1,z1) in REAL+ by A42,ARYTM_0:5,XBOOLE_0:3;
            hence thesis by A1,A2,A43,XXREAL_0:def 5;
          end;
          suppose
A44:        not b9 <=' c9;
A45:        b9 -' c9 <=' a9 -' c9 by A32,A36,A37,ARYTM_1:17;
A46:        +(y1,z1) = [0,b9 -' c9] by A34,A28,A29,A44,ARYTM_1:def 2;
            then +(y1,z1) in [:{0},REAL+:] by Lm4,ZFMISC_1:87;
            hence thesis by A1,A2,A41,A42,A46,A45,Lm1;
          end;
        end;
        hence thesis;
      end;
    end;
    hence thesis;
  end;
  suppose that
A47: a in REAL+ and
A48: b in REAL+ and
A49: c in [:{0},REAL+:];
    consider b9,c99 being Element of REAL+ such that
A50: b = b9 and
A51: c = [0,c99] and
A52: +(y1,z1) = b9 - c99 by A48,A49,ARYTM_0:def 1;
    consider a9,c9 being Element of REAL+ such that
A53: a = a9 and
A54: c = [0,c9] and
A55: +(x1,z1) = a9 - c9 by A47,A49,ARYTM_0:def 1;
A56: c9 = c99 by A54,A51,XTUPLE_0:1;

A57: ex a99,b99 being Element of REAL+ st a = a99 & b = b99 & a99 <='
    b99 by A3,A47,A48,XXREAL_0:def 5;

    now
      per cases;
      suppose
A58:    c9 <=' a9;
        then c9 <=' b9 by A57,A53,A50,ARYTM_1:3;
        then
A59:    b9 - c9 = b9 -' c9 by ARYTM_1:def 2;
A60:    a9 - c9 = a9 -' c9 by A58,ARYTM_1:def 2;
        a9 -' c9 <=' b9 -' c99 by A57,A53,A50,A56,ARYTM_1:17;
        hence thesis by A1,A2,A55,A52,A56,A60,A59,Lm1;
      end;
      suppose
        not c9 <=' a9;
        then
A61:    +(x1,z1) = [0,c9 -' a9] by A55,ARYTM_1:def 2;
        then
A62:    +(x1,z1) in [:{0},REAL+:] by Lm4,ZFMISC_1:87;
        now
          per cases;
          suppose
            c9 <=' b9;
            then b9 - c9 = b9 -' c9 by ARYTM_1:def 2;

            then
A63:        not +(y1,z1) in [:{0},REAL+:] by A52,A56,ARYTM_0:5,XBOOLE_0:3;

            not +(x1,z1) in REAL+ by A62,ARYTM_0:5,XBOOLE_0:3;
            hence thesis by A1,A2,A63,XXREAL_0:def 5;
          end;
          suppose
A64:        not c9 <=' b9;
A65:        c9 -' b9 <=' c9 -' a9 by A57,A53,A50,ARYTM_1:16;
A66:        +(y1,z1) = [0,c9 -' b9] by A52,A56,A64,ARYTM_1:def 2;
            then +(y1,z1) in [:{0},REAL+:] by Lm4,ZFMISC_1:87;
            hence thesis by A1,A2,A61,A62,A66,A65,Lm1;
          end;
        end;
        hence thesis;
      end;
    end;
    hence thesis;
  end;
  suppose that
A67: a in [:{0},REAL+:] and
A68: b in REAL+ and
A69: c in [:{0},REAL+:];
A70: not c in REAL+ by A69,ARYTM_0:5,XBOOLE_0:3;
    not a in REAL+ by A67,ARYTM_0:5,XBOOLE_0:3;
    then consider a9,c9 being Element of REAL+ such that
    a = [0,a9] and
A71: c = [0,c9] and
A72: +(x1,z1) = [0,a9 + c9] by A70,ARYTM_0:def 1;
A73: +(x1,z1) in [:{0},REAL+:] by A72,Lm4,ZFMISC_1:87;
    consider b9,c99 being Element of REAL+ such that
    b = b9 and
A74: c = [0,c99] and
A75: +(y1,z1) = b9 - c99 by A68,A69,ARYTM_0:def 1;
A76: c9 = c99 by A71,A74,XTUPLE_0:1;
    now
      per cases;
      suppose
        c9 <=' b9;
        then b9 - c99 = b9 -' c99 by A76,ARYTM_1:def 2;
        then
A77:    not +(y1,z1) in [:{0},REAL+:] by A75,ARYTM_0:5,XBOOLE_0:3;
        not +(x1,z1) in REAL+ by A73,ARYTM_0:5,XBOOLE_0:3;
        hence thesis by A1,A2,A77,XXREAL_0:def 5;
      end;
      suppose
A78:    not c9 <=' b9;
A79:    c9 <=' c9 + a9 by ARYTM_2:19;
        c9 -' b9 <=' c9 by ARYTM_1:11;
        then
A80:    c9 -' b9 <=' c9 + a9 by A79,ARYTM_1:3;
A81:    +(y1,z1) = [0,c9 -' b9] by A75,A76,A78,ARYTM_1:def 2;
        then +(y1,z1) in [:{0},REAL+:] by Lm4,ZFMISC_1:87;
        hence thesis by A1,A2,A72,A73,A81,A80,Lm1;
      end;
    end;
    hence thesis;
  end;
  suppose that
A82: a in [:{0},REAL+:] and
A83: b in [:{0},REAL+:] and
A84: c in [:{0},REAL+:];
A85: not c in REAL+ by A84,ARYTM_0:5,XBOOLE_0:3;
A86: not c in REAL+ by A84,ARYTM_0:5,XBOOLE_0:3;
    not a in REAL+ by A82,ARYTM_0:5,XBOOLE_0:3;
    then consider a9,c9 being Element of REAL+ such that
A87: a = [0,a9] and
A88: c = [0,c9] and
A89: +(x1,z1) = [0,a9 + c9] by A86,ARYTM_0:def 1;
A90: +(x1,z1) in [:{0},REAL+:] by A89,Lm4,ZFMISC_1:87;
    not b in REAL+ by A83,ARYTM_0:5,XBOOLE_0:3;
    then consider b9,c99 being Element of REAL+ such that
A91: b = [0,b9] and
A92: c = [0,c99] and
A93: +(y1,z1) = [0,b9 + c99] by A85,ARYTM_0:def 1;
A94: +(y1,z1) in [:{0},REAL+:] by A93,Lm4,ZFMISC_1:87;
A95: c9 = c99 by A88,A92,XTUPLE_0:1;
    consider a99,b99 being Element of REAL+ such that
A96: a = [0,a99] and
A97: b = [0,b99] and
A98: b99 <=' a99 by A3,A82,A83,XXREAL_0:def 5;
A99: b9 = b99 by A97,A91,XTUPLE_0:1;
    a9 = a99 by A96,A87,XTUPLE_0:1;
    then b9 + c9 <=' a9 + c99 by A98,A99,A95,ARYTM_1:7;
    hence thesis by A1,A2,A89,A93,A95,A90,A94,Lm1;
  end;
end;

Lm6: a <= b & c <= d implies a+c <= b+d
proof
  assume a<=b;
  then
A1: a+c <= b+c by Lm5;
  assume c <= d;
  then b+c <= b+d by Lm5;
  hence thesis by A1,XXREAL_0:2;
end;

Lm7: a <= b implies a-c <= b-c
proof
  a <= b implies a + -c <= b + -c by Lm5;
  hence thesis;
end;

Lm8: a < b & c <= d implies a+c < b+d
proof
  assume that
A1: a<b and
A2: c <= d;
  for a,b,c,d holds a<b & c <= d implies a+c < b+d
  proof
    let a,b,c,d;
    assume
A3: a<b;
    assume
A4: c <= d;
A5: a+c <> b+d
    proof
      a+c <= d+a by A4,Lm5;
      then
A6:   a+c-d<=a+d-d by Lm7;
      assume a+c = b+d;
      hence contradiction by A3,A6;
    end;
    a+c <= b+d by A3,A4,Lm6;
    hence thesis by A5,XXREAL_0:1;
  end;
  hence thesis by A1,A2;
end;

Lm9: 0 < a implies b < b+a
proof
  assume 0<a;
  then b+0<b+a by Lm8;
  hence thesis;
end;

theorem Th1:
  ex b st a < b
proof
  take a+1;
  a+0<a+1 by Lm8;
  hence thesis;
end;

theorem Th2:
  ex b st b < a
proof
  take a-1;
  a+ -1<a+ -0 by Lm8;
  hence thesis;
end;

theorem
  ex c st a < c & b < c
proof
  per cases;
  suppose
A1: a <= b;
    take z = b + 1;
    b < z by Lm9;
    hence thesis by A1,XXREAL_0:2;
  end;
  suppose
A2: b <= a;
    take z = a + 1;
    a < z by Lm9;
    hence thesis by A2,XXREAL_0:2;
  end;
end;

Lm10: a+c <= b+c implies a <= b
proof
  assume a+c <= b+c;
  then a+c+(-c)<=b+c+(-c) by Lm5;
  hence thesis;
end;

theorem
  ex c st c < a & c < b
proof
  per cases;
  suppose
A1: a <= b;
    take z = a+-1;
    z < a+0 by Lm10;
    hence thesis by A1,XXREAL_0:2;
  end;
  suppose
A2: b <= a;
    take z = b+-1;
    z < b+0 by Lm10;
    hence thesis by A2,XXREAL_0:2;
  end;
end;

Lm11: for a9,b9 being Element of REAL, a,b st a9 = a & b9 = b holds *(a9,b9) =
a * b

proof
  let a9,b9 be Element of REAL, a,b such that
A1: a9 = a and
A2: b9 = b;
  consider x1,x2,y1,y2 being Element of REAL such that
A3: a = [* x1,x2 *] and
A4: b = [*y1,y2*] and

A5: a*b = [* +(*(x1,y1),opp*(x2,y2)), +(*(x1,y2),*(x2,y1)) *] by XCMPLX_0:def 5
;

A6: b = y1 by A4,Lm2;
  x2 = 0 by A3,Lm2;
  then
A7: *(x2,y1) = 0 by ARYTM_0:12;
A8: y2 = 0 by A4,Lm2;
  then *(x1,y2) = 0 by ARYTM_0:12;
  then
A9: +(*(x1,y2),*(x2,y1)) = 0 by A7,ARYTM_0:11;
  a = x1 by A3,Lm2;
  hence *(a9,b9) = +(*(x1,y1),*(opp x2,y2)) by A1,A2,A6,A8,ARYTM_0:11,12
    .= +(*(x1,y1),opp*(x2,y2)) by ARYTM_0:15
    .= a * b by A5,A9,ARYTM_0:def 5;
end;
reconsider o = 0 as Element of REAL+ by ARYTM_2:20;

Lm12: a <= b & 0 <= c implies a*c <= b*c
proof
  assume that
A1: a <= b and
A2: 0 <= c;
  not o in [:{0},REAL+:] by ARYTM_0:5,XBOOLE_0:3;
  then
A3: c in REAL+ by A2,XXREAL_0:def 5;
  reconsider x1=a, y1=b, z1=c as Element of REAL by XREAL_0:def 1;
A4: *(y1,z1) = b * c by Lm11;
A5: *(x1,z1) = a * c by Lm11;
  assume
A6: not thesis;
  then
A7: c <> 0;
  per cases by A1,XXREAL_0:def 5;
  suppose that
A8: a in REAL+ and
A9: b in REAL+;
    consider b9,c99 being Element of REAL+ such that
A10: b = b9 and
A11: c = c99 and
A12: *(y1,z1) = b9 *' c99 by A3,A9,ARYTM_0:def 2;
    consider a9,c9 being Element of REAL+ such that
A13: a = a9 and
A14: c = c9 and
A15: *(x1,z1) = a9 *' c9 by A3,A8,ARYTM_0:def 2;

    ex a99,b99 being Element of REAL+ st a = a99 & b = b99 & a99 <='
    b99 by A1,A8,A9,XXREAL_0:def 5;

    then a9 *' c9 <=' b9 *' c9 by A13,A10,ARYTM_1:8;
    hence contradiction by A4,A5,A6,A14,A15,A11,A12,Lm1;
  end;
  suppose that
A16: a in [:{0},REAL+:] and
A17: b in REAL+;

    ex a9,c9 being Element of REAL+ st a = [0,a9] & c = c9 & *(x1,z1)
    = [0,c9 *' a9] by A3,A7,A16,ARYTM_0:def 2;

    then *(x1,z1) in [:{0},REAL+:] by Lm4,ZFMISC_1:87;
    then
A18: not *(x1,z1) in REAL+ by ARYTM_0:5,XBOOLE_0:3;

    ex b9,c99 being Element of REAL+ st b = b9 & c = c99 & * (y1,z1) =
    b9 *' c99 by A3,A17,ARYTM_0:def 2;

    then not *(y1,z1) in [:{0},REAL+:] by ARYTM_0:5,XBOOLE_0:3;
    hence contradiction by A4,A5,A6,A18,XXREAL_0:def 5;
  end;
  suppose that
A19: a in [:{0},REAL+:] and
A20: b in [:{0},REAL+:];
    consider b9,c99 being Element of REAL+ such that
A21: b = [0,b9] and
A22: c = c99 and
A23: *(y1,z1) = [0,c99 *' b9] by A3,A7,A20,ARYTM_0:def 2;
A24: *(y1,z1) in [:{0},REAL+:] by A23,Lm4,ZFMISC_1:87;
    consider a99,b99 being Element of REAL+ such that
A25: a = [0,a99] and
A26: b = [0,b99] and
A27: b99 <=' a99 by A1,A19,A20,XXREAL_0:def 5;
A28: b9 = b99 by A26,A21,XTUPLE_0:1;
    consider a9,c9 being Element of REAL+ such that
A29: a = [0,a9] and
A30: c = c9 and
A31: *(x1,z1) = [0,c9 *' a9] by A3,A7,A19,ARYTM_0:def 2;
A32: *(x1,z1) in [:{0},REAL+:] by A31,Lm4,ZFMISC_1:87;
    a9 = a99 by A25,A29,XTUPLE_0:1;
    then b9 *' c9 <=' a9 *' c9 by A27,A28,ARYTM_1:8;
    hence contradiction by A4,A5,A6,A30,A31,A22,A23,A32,A24,Lm1;
  end;
end;

Lm13: 0 < c & a < b implies a*c < b*c
proof
  assume
A1: 0<c;
  assume
A2: a<b;
A3: a*c <> b*c
  proof
    assume a*c = b*c;
    then a*(c*c")=b*c*c" by XCMPLX_1:4;
    then a*1=b*(c*c") by A1,XCMPLX_0:def 7;
    then a=b*1 by A1,XCMPLX_0:def 7;
    hence contradiction by A2;
  end;
  a*c <= b*c by A1,A2,Lm12;
  hence thesis by A3,XXREAL_0:1;
end;

theorem Th5:
  a < b implies ex c st a < c & c < b
proof
  assume
A1: a<b;
  take z=(a+b)/2;
  1*a+a<a+b by A1,Lm10;
  then 2"*(2*a)<2"*(a+b) by Lm13;
  hence a<z;
  a+b<1*b+b by A1,Lm10;
  then 2"*(a+b)<2"*(2*b) by Lm13;
  hence thesis;
end;

begin

theorem
  a <= b iff a + c <= b + c by Lm5,Lm10;

theorem
  a <= b & c <= d implies a+c <= b+d by Lm6;

theorem
  a < b & c <= d implies a+c < b+d by Lm8;

Lm14: a<=b implies -b<=-a
proof
  assume a<=b;
  then a-b<=b-b by Lm7;
  then a-b-a<=0-a by Lm7;
  hence thesis;
end;

Lm15: -b<=-a implies a<=b
proof
  assume -b<=-a;
  then -(-a)<=-(-b) by Lm14;
  hence thesis;
end;

begin

theorem Th9:
  a <= b iff a-c <= b-c
proof
  thus a <= b implies a-c <= b-c by Lm7;
  assume a-c <= b-c;
  then a+ (-c) + c <= b+ (-c)+c by Lm5;
  hence thesis;
end;

theorem
  a <= b iff c-b <= c-a
proof
  hereby
    assume
A1: a <= b;
    assume c-a < c-b;
    then -(c-b)<-(c-a) by Lm15;
    then b-c < a-c;
    hence contradiction by A1,Lm7;
  end;
  assume c-a>=c-b;
  then -(c-a)<=-(c-b) by Lm14;
  then a-c <= b-c;
  hence thesis by Th9;
end;

Lm16: a+b <= c implies a <= c-b
proof
  assume a+b<=c;
  then a+b-b<=c-b by Lm7;
  hence thesis;
end;

Lm17: a <= b-c implies a+c <= b
proof
  assume a <= b-c;
  then a+c <= b - c + c by Lm5;
  hence thesis;
end;

Lm18: a <= b+c implies a-b <= c
proof
  assume a<=b+c;
  then a-b <= c+b-b by Lm7;
  hence thesis;
end;

Lm19: a-b <= c implies a <= b+c
proof
  assume a-b <= c;
  then a + -b <= c;
  then a <= c - -b by Lm16;
  hence thesis;
end;

theorem
  a <= b-c implies c <= b-a
proof
  assume a<=b-c;
  then a+c <= b by Lm17;
  hence thesis by Lm16;
end;

theorem
  a-b <= c implies a-c <= b
proof
  assume c>=a-b;
  then c+b>=a by Lm19;
  hence thesis by Lm18;
end;

theorem Th13:
  a <= b & c <= d implies a-d <= b-c
proof
  assume that
A1: a <= b and
A2: c <= d;
  -d <= -c by A2,Lm14;
  then a+ -d <= b+ -c by A1,Lm6;
  hence thesis;
end;

theorem Th14:
  a < b & c <= d implies a-d < b-c
proof
  assume that
A1: a < b and
A2: c <= d;
  - d <= - c by A2,Lm14;
  then a+ -d < b+ -c by A1,Lm8;
  hence thesis;
end;

theorem Th15:
  a <= b & c < d implies a-d < b-c
proof
  assume that
A1: a <= b and
A2: c < d;
  c-b < d-a by A1,A2,Th14;
  then -(d-a) < -(c-b) by Lm15;
  hence thesis;
end;

Lm20: a+b<=c+d implies a-c <= d-b
proof
  assume a+b<=c+d;
  then a<=c+d-b by Lm16;
  then a<=c+(d-b);
  hence thesis by Lm18;
end;

theorem
  a-b <= c-d implies a-c <= b-d
proof
  assume a-b<=c-d;
  then a-b+d<=c by Lm17;
  then a+d-b<=c;
  then a+d<=c+b by Lm19;
  hence thesis by Lm20;
end;

theorem
  a-b <= c-d implies d-b <= c-a
proof
  assume a-b<=c-d;
  then a-b+d<=c by Lm17;
  then a+d-b<=c;
  then a+d<=c+b by Lm19;
  hence thesis by Lm20;
end;

theorem
  a-b <= c-d implies d-c <= b-a
proof
  assume a-b<=c-d;
  then a-b+d<=c by Lm17;
  then a+d-b<=c;
  then a+d<=c+b by Lm19;
  hence thesis by Lm20;
end;

begin

theorem
  a+b <= c iff a <= c-b by Lm16,Lm17;

theorem
  a <= b+c iff a-b <= c by Lm18,Lm19;

theorem
  a+b<=c+d iff a-c <= d-b
proof
  thus a+b<=c+d implies a-c <= d-b by Lm20;
  assume
A1: a-c <= d-b;
  assume c+d<a+b;
  then d<b+a-c by Lm19;
  then d<b+(a-c);
  hence thesis by A1,Lm17;
end;

theorem
  a+b<=c-d implies a+d<=c-b
proof
  assume
A1: a+b<=c-d;
  per cases by A1;
  suppose
    a+b<=c-d;
    then a+b+d<=c by Lm17;
    then a+d+b<=c;
    hence thesis by Lm16;
  end;
  suppose
    b+a<=c-d;
    then a+b+d<=c by Lm17;
    then a+d+b<=c;
    hence thesis by Lm16;
  end;
end;

theorem
  a-b<=c+d implies a-d<=c+b
proof
  assume
A1: c+d>=a-b;
  per cases by A1;
  suppose
    c+d>=a-b;
    then c+d+b>=a by Lm19;
    then c+b+d>=a;
    hence thesis by Lm18;
  end;
  suppose
    d+c>=a-b;
    then c+d+b>=a by Lm19;
    then c+b+d>=a;
    hence thesis by Lm18;
  end;
end;

begin

theorem
  a<=b iff -b<=-a by Lm14,Lm15;

Lm21: a < b implies 0 < b-a
proof
  assume a < b;
  then a + 0 < b;
  hence thesis by Lm19;
end;

theorem Th25:
  a<=-b implies b<=-a
proof
  assume a<=-b;
  then a+b<=-b+b by Lm5;
  then b--a<=0;
  hence thesis by Lm21;
end;

Lm22: a <= b implies 0 <= b-a
proof
  assume a <= b;
  then a - a <= b - a by Lm7;
  hence thesis;
end;

theorem Th26:
  -b<=a implies -a<=b
proof
  assume
A1: -b<=a;
  assume b<-a;
  then b+a<-a+a by Lm10;
  then a--b<0;
  hence thesis by A1,Lm22;
end;

begin

theorem
  0 <= a & 0 <= b & a+b = 0 implies a = 0;

theorem
  a <= 0 & b <= 0 & a+b = 0 implies a = 0;

begin

theorem
  0 < a implies b < b+a by Lm9;

theorem
  a < 0 implies a+b < b
proof
  assume a<0;
  then b+a<0+b by Lm10;
  hence thesis;
end;

Lm23: a < b implies a-b < 0
proof
  assume
A1: a < b;
  assume a-b>=0;
  then a-b+b>=0+b by Lm5;
  hence thesis by A1;
end;

theorem
  0 <= a implies b <= a+b
proof
  assume
A1: 0 <= a;
  assume
A2: a+b<b;
  per cases by A2;
  suppose
    a+b<b;
    then a+b-b<0 by Lm23;
    hence thesis by A1;
  end;
  suppose
    b-a>b;
    then b+-a-b>0 by Lm21;
    hence thesis by A1;
  end;
end;

theorem
  a <= 0 implies a+b <= b
proof
  assume a<=0;
  then b+a<=0+b by Lm6;
  hence thesis;
end;

begin

theorem
  0 <= a & 0 <= b implies 0 <= a+b;

theorem
  0 <= a & 0 < b implies 0 < a+b;

theorem Th35:
  a <= 0 & c <= b implies c+a <= b
proof
  assume that
A1: a <= 0 and
A2: c <= b;
  a+c <= 0+b by A1,A2,Lm6;
  hence thesis;
end;

theorem Th36:
  a <= 0 & c < b implies c+a < b
proof
  assume that
A1: a <= 0 and
A2: c < b;
  a+c < 0+b by A1,A2,Lm8;
  hence thesis;
end;

theorem Th37:
  a < 0 & c <= b implies c+a < b
proof
  assume that
A1: a < 0 and
A2: c <= b;
  a+c < 0+b by A1,A2,Lm8;
  hence thesis;
end;

theorem Th38:
  0 <= a & b <= c implies b <= a+c
proof
  assume that
A1: 0 <= a and
A2: b <= c;
  b+0 <= a+c by A1,A2,Lm6;
  hence thesis;
end;

theorem Th39:
  0 < a & b <= c implies b < a+c
proof
  assume that
A1: 0 < a and
A2: b <= c;
  b+0 < a+c by A1,A2,Lm8;
  hence thesis;
end;

theorem Th40:
  0 <= a & b < c implies b < a+c
proof
  assume that
A1: 0 <= a and
A2: b < c;
  b+0 < a+c by A1,A2,Lm8;
  hence thesis;
end;

Lm24: c < 0 & a < b implies b*c < a*c
proof
  assume
A1: c < 0;
  assume a<b;
  then a*(-c)<b*(-c) by A1,Lm13;
  then -(a*c)<-(b*c);
  hence thesis by Lm14;
end;

Lm25: 0 <= c & b <= a implies b/c <= a/c
proof
  assume that
A1: 0<=c and
A2: b <= a;
  per cases by A1;
  suppose
    c = 0;
    then
A3: c" = 0;
    then b*c" = 0;
    then
A4: b/c = 0 by XCMPLX_0:def 9;
    a*c" = 0 by A3;
    hence thesis by A4,XCMPLX_0:def 9;
  end;
  suppose
A5: 0<c;
    assume a/c < b/c;
    then a/c*c < b/c*c by A5,Lm13;
    then a<b/c*c by A5,XCMPLX_1:87;
    hence contradiction by A2,A5,XCMPLX_1:87;
  end;
end;

Lm26: 0 < c & a < b implies a/c < b/c
proof
  assume
A1: 0<c;
  a<b implies a/c < b/c
  proof
    assume a<b;
    then a*c"<b*c" by A1,Lm13;
    then a/c <b*c" by XCMPLX_0:def 9;
    hence thesis by XCMPLX_0:def 9;
  end;
  hence thesis;
end;

Lm27: 0 < a implies a/2 < a
proof
  assume 0<a;
  then 0+a/2<a/2+a/2 by Lm10;
  hence thesis;
end;

begin

theorem
  (for a being real number st a>0 holds c <= b+a) implies c <= b
proof
  assume
A1: for a st a>0 holds b+a>=c;
  set d=c-b;
  assume b<c;
  then
A2: 0<d by Lm21;
  then b+d/2<b+d by Lm10,Lm27;
  hence contradiction by A1,A2;
end;

theorem
  (for a being real number st a<0 holds b+a <= c) implies b <= c
proof
  assume
A1: for a st a<0 holds b+a<=c;
  set d=c-b;
  assume c < b;
  then
A2: 0>c-b by Lm23;
  then (-d)/2<-d by Lm27;
  then c+-d/2<c+-d by Lm10;
  then c-d/2<c-d;
  then c < b+d/2 by Lm16;
  hence contradiction by A1,A2;
end;

begin

theorem
  0 <= a implies b-a <= b
proof
  0 <= a implies b - a <= b - 0 by Th13;
  hence thesis;
end;

theorem Th44:
  0 < a implies b-a < b
proof
  assume 0 < a;
  then b + -a < b + 0 by Lm10;
  hence thesis;
end;

theorem
  a <= 0 implies b <= b-a
proof
  assume a<=0;
  then b+a<=0+b by Lm6;
  hence thesis by Lm16;
end;

theorem
  a < 0 implies b < b-a
proof
  assume a<0;
  then b+a<0+b by Lm10;
  hence thesis by Lm19;
end;

theorem
  a <= b implies a-b <= 0
proof
  assume
A1: a <= b;
  assume a-b>0;
  then a-b+b>0+b by Lm10;
  hence thesis by A1;
end;

theorem Th48:
  a <= b implies 0 <= b-a
proof
  assume a <= b;
  then a - a <= b - a by Lm7;
  hence thesis;
end;

theorem
  a < b implies a-b < 0 by Lm23;

theorem
  a < b implies 0 < b-a by Lm21;

theorem
  0 <= a & b < c implies b-a < c
proof
  assume that
A1: 0 <= a and
A2: b < c;
  b+0 < a+c by A1,A2,Th40;
  hence thesis by Lm17;
end;

theorem
  a <= 0 & b <= c implies b <= c-a
proof
  assume that
A1: a <= 0 and
A2: b <= c;
  b+a <= c by A1,A2,Th35;
  hence thesis by Lm16;
end;

theorem
  a <= 0 & b < c implies b < c-a
proof
  assume that
A1: a <= 0 and
A2: b < c;
  b+a < c by A1,A2,Th36;
  hence thesis by Lm19;
end;

theorem
  a < 0 & b <= c implies b < c-a
proof
  assume that
A1: a < 0 and
A2: b <= c;
  a+b < c by A1,A2,Th37;
  hence thesis by Lm19;
end;

theorem
  a <> b implies 0 < a-b or 0 < b-a
proof
  assume
A1: a <> b;
  per cases by A1,XXREAL_0:1;
  suppose
    a < b;
    then 0 + a < b;
    hence thesis by Lm19;
  end;
  suppose
    b < a;
    then 0 + b < a;
    hence thesis by Lm19;
  end;
end;

begin

theorem
  (for a being real number st a<0 holds c <= b-a) implies b>=c
proof
  assume
A1: for a st a<0 holds b-a>=c;
  set d=b-c;
  assume b<c;
  then
A2: 0>b-c by Lm23;
  then (-d)/2<-d by Lm27;
  then b+-d/2<b+-d by Lm10;
  then b-d/2<b-d;
  hence contradiction by A1,A2;
end;

theorem
  (for a being real number st a>0 holds b-a <= c) implies b<=c
proof
  assume
A1: for a st a>0 holds b-a<=c;
  set d=b-c;
  assume b>c;
  then
A2: 0<d by Lm21;
  then c+d/2<c+d by Lm10,Lm27;
  then c < b-d/2 by Lm19;
  hence contradiction by A1,A2;
end;

begin

theorem
  a < 0 iff 0 < -a;

theorem
  a <= -b implies a+b <= 0
proof
  assume a<=-b;
  then a+b<=-b+b by Lm5;
  hence thesis;
end;

theorem
  -a <= b implies 0 <= a+b
proof
  assume b>=-a;
  then b+a>=-a+a by Lm5;
  hence thesis;
end;

theorem
  a < -b implies a+b < 0
proof
  assume a<-b;
  then a+b<-b+b by Lm10;
  hence thesis;
end;

theorem
  -b < a implies 0 < a+b
proof
  assume a>-b;
  then a+b>-b+b by Lm10;
  hence thesis;
end;

Lm28: a <= b & c <= 0 implies b*c <= a*c
proof
  assume
A1: a<=b;
  assume c <= 0;
  then a*(-c)<=b*(-c) by A1,Lm12;
  then -(a*c)<=-(b*c);
  hence thesis by Lm15;
end;

begin

theorem
  0 <= a*a
proof
  per cases;
  suppose
    0 <= a;
    hence thesis;
  end;
  suppose
    not 0 <= a;
    hence thesis;
  end;
end;

theorem
  a <= b & 0 <= c implies a*c <= b*c by Lm12;

theorem
  a <= b & c <= 0 implies b*c <= a*c by Lm28;

theorem Th66:
  0 <= a & a <= b & 0 <= c & c <= d implies a*c <= b*d
proof
  assume that
A1: 0<=a and
A2: a<=b and
A3: 0<=c and
A4: c <= d;
A5: a*c <= b*c by A2,A3,Lm12;
  b*c <= b*d by A1,A2,A4,Lm12;
  hence thesis by A5,XXREAL_0:2;
end;

theorem
  a <= 0 & b <= a & c <= 0 & d <= c implies a*c <= b*d
proof
  assume that
A1: 0>=a and
A2: a>=b and
A3: 0>=c and
A4: c>=d;
A5: -c <= -d by A4,Lm14;
  -a<=-b by A2,Lm14;
  then (-a)*(-c)<=(-b)*(-d) by A1,A3,A5,Th66;
  hence thesis;
end;

theorem
  0 < c & a < b implies a*c < b*c by Lm13;

theorem
  c < 0 & a < b implies b*c < a*c by Lm24;

theorem
  a < 0 & b <= a & c < 0 & d < c implies a*c < b*d
proof
  assume that
A1: 0>a and
A2: a>=b and
A3: 0>c and
A4: c>d;
A5: a*c < a*d by A1,A4,Lm24;
  a*d<=b*d by A2,A3,A4,Lm28;
  hence thesis by A5,XXREAL_0:2;
end;

begin

theorem
  0 <= a & 0 <= b & 0 <= c & 0 <= d & a*c+b*d = 0 implies a = 0 or c = 0;

Lm29: c < 0 & a < b implies b/c < a/c
proof
  assume that
A1: c < 0 and
A2: a < b;
  a/(-c)<b/(-c) by A1,A2,Lm26;
  then -a/c < b/(-c) by XCMPLX_1:188;
  then -a/c < -b/c by XCMPLX_1:188;
  hence thesis by Lm14;
end;

Lm30: c <= 0 & b/c < a/c implies a < b
proof
  assume that
A1: c <= 0 and
A2: b/c < a/c;
A3: now
    assume c = 0;
    then
A4: c" = 0;
    a/c = a*c" by XCMPLX_0:def 9
      .= b*c" by A4
      .= b/c by XCMPLX_0:def 9;
    hence contradiction by A2;
  end;
  then a/c*c < b/c*c by A1,A2,Lm24;
  then a<b/c*c by A3,XCMPLX_1:87;
  hence thesis by A3,XCMPLX_1:87;
end;

begin

theorem
  0 <= c & b <= a implies b/c <= a/c by Lm25;

theorem
  c <= 0 & a <= b implies b/c <= a/c by Lm30;

theorem
  0 < c & a < b implies a/c < b/c by Lm26;

theorem
  c < 0 & a < b implies b/c < a/c by Lm29;

theorem
  0 < c & 0 < a & a < b implies c/b < c/a
proof
  assume that
A1: 0<c and
A2: 0<a and
A3: a<b;
  a*b"<b*b" by A2,A3,Lm13;
  then a*b"<1 by A2,A3,XCMPLX_0:def 7;
  then a"*(a*b")<a"*1 by A2,Lm13;
  then a"*a*b"<a";
  then 1*b"<a" by A2,XCMPLX_0:def 7;
  then c*b"<c*a" by A1,Lm13;
  then c/b<c*a" by XCMPLX_0:def 9;
  hence thesis by XCMPLX_0:def 9;
end;

Lm31: b < 0 & a < b implies b" < a"
proof
A1: a" = 1/a by XCMPLX_1:215;
  assume that
A2: 0>b and
A3: a<b;
  b*(a")<a*(a") by A2,A3,Lm24;
  then b*(a")<1 by A1,A2,A3,XCMPLX_1:87;
  then b"*(b*(a"))>1*(b") by A2,Lm24;
  then b"*b*(a")>1*(b");
  then 1*(a")>1*(b") by A2,XCMPLX_0:def 7;
  hence thesis;
end;

begin

theorem Th77:
  0 < b & a*b <= c implies a <= c/b
proof
  assume that
A1: b>0 and
A2: a*b<=c;
  a*b/b<=c/b by A1,A2,Lm25;
  hence thesis by A1,XCMPLX_1:89;
end;

theorem Th78:
  b < 0 & a*b <= c implies c/b <= a
proof
  assume that
A1: b<0 and
A2: a*b<=c;
  a*b/b>=c/b by A1,A2,Lm30;
  hence thesis by A1,XCMPLX_1:89;
end;

theorem Th79:
  0 < b & c <= a*b implies c/b <= a
proof
  assume that
A1: b>0 and
A2: a*b>=c;
  a*b/b>=c/b by A1,A2,Lm25;
  hence thesis by A1,XCMPLX_1:89;
end;

theorem Th80:
  b < 0 & c <= a*b implies a <= c/b
proof
  assume that
A1: b<0 and
A2: a*b>=c;
  a*b/b<=c/b by A1,A2,Lm30;
  hence thesis by A1,XCMPLX_1:89;
end;

theorem Th81:
  0 < b & a*b < c implies a< c/b
proof
  assume that
A1: b>0 and
A2: a*b<c;
  a*b/b<c/b by A1,A2,Lm26;
  hence thesis by A1,XCMPLX_1:89;
end;

theorem Th82:
  b < 0 & a*b < c implies c/b < a
proof
  assume that
A1: b<0 and
A2: a*b<c;
  a*b/b>c/b by A1,A2,Lm29;
  hence thesis by A1,XCMPLX_1:89;
end;

theorem Th83:
  0 < b & c < a*b implies c/b < a
proof
  assume that
A1: b>0 and
A2: a*b>c;
  a*b/b>c/b by A1,A2,Lm26;
  hence thesis by A1,XCMPLX_1:89;
end;

theorem Th84:
  b < 0 & c < a*b implies a < c/b
proof
  assume that
A1: b<0 and
A2: a*b>c;
  a*b/b<c/b by A1,A2,Lm29;
  hence thesis by A1,XCMPLX_1:89;
end;

Lm32: 0 < a & a <= b implies b" <= a"
proof
  assume that
A1: 0<a and
A2: a<=b;
  a*(b")<=b*(b") by A1,A2,Lm12;
  then a*(b")<=1 by A1,A2,XCMPLX_0:def 7;
  then a"*(a*b")<=1*a" by A1,Lm12;
  then a"*a*b"<=1*a";
  then 1*b"<=1*a" by A1,XCMPLX_0:def 7;
  hence thesis;
end;

Lm33: b < 0 & a <= b implies b" <= a"
proof
  assume that
A1: 0>b and
A2: a<=b;
  b*(a")<=a*(a") by A1,A2,Lm28;
  then b*(a")<=1 by A1,A2,XCMPLX_0:def 7;
  then b"*(b*(a"))>=1*(b") by A1,Lm28;
  then b"*b*a">=1*b";
  then 1*a">=1*b" by A1,XCMPLX_0:def 7;
  hence thesis;
end;

begin

theorem
  0 < a & a <= b implies b" <= a" by Lm32;

theorem
  b < 0 & a <= b implies b" <= a" by Lm33;

theorem
  b < 0 & a < b implies b" < a" by Lm31;

Lm34: 0 < a & a < b implies b" < a"
proof
A1: b" = 1/b by XCMPLX_1:215;
  assume that
A2: 0<a and
A3: a<b;
  a*(b")<b*(b") by A2,A3,Lm13;
  then a*(b")<1 by A1,A2,A3,XCMPLX_1:87;
  then a"*(a*(b"))<1*(a") by A2,Lm13;
  then a"*a*b"<1*(a");
  then 1*b"<1*(a") by A2,XCMPLX_0:def 7;
  hence thesis;
end;

theorem
  0 < a & a < b implies b" < a" by Lm34;

theorem
  0 < b" & b" <= a" implies a <= b
proof
  assume that
A1: b">0 and
A2: a">=b";
  a""<=b"" by A1,A2,Lm32;
  hence thesis;
end;

theorem
  a" < 0 & b" <= a" implies a <= b
proof
  assume that
A1: a"<0 and
A2: a">=b";
  a""<=b"" by A1,A2,Lm33;
  hence thesis;
end;

theorem
  0 < b" & b" < a" implies a < b
proof
  assume that
A1: b">0 and
A2: a">b";
  a""<b"" by A1,A2,Lm34;
  hence thesis;
end;

theorem
  a" < 0 & b" < a" implies a < b
proof
  assume that
A1: a"<0 and
A2: a">b";
  a""<b"" by A1,A2,Lm31;
  hence thesis;
end;

Lm35: for a being non negative non positive real number holds 0 = a*b;

begin

theorem
  0 <= a & (b-a)*(b+a) <= 0 implies -a <= b & b <= a
proof
  assume that
A1: a>=0 and
A2: (b-a)*(b+a)<=0;
  b+0<=b+a by A1,Lm6;
  then b+a>=0 by A1,A2;
  then
A3: b>=0-a by Lm18;
  b-a>=0 & b+a<=0 or b-a<=0 & b+a>=0 by A2;
  then b<=a+0 by A1,Lm19;
  hence thesis by A3;
end;

theorem
  0 <= a & (b-a)*(b+a) < 0 implies -a < b & b < a
proof
  assume that
A1: a>=0 and
A2: (b-a)*(b+a)<0;
A3: b-a>0 & b+a<0 or b-a<0 & b+a>0 by A2,Lm35;
  then
A4: b<a+0 by A1,Lm16;
  b+0<=b+a by A1,Lm6;
  then b>0-a by A1,A3,Lm17;
  hence thesis by A4;
end;

theorem
  0 <= (b-a)*(b+a) implies b <= -a or a <= b
proof
  assume (b-a)*(b+a)>=0;
  then b-a>=0 & b+a>=0 or b-a<=0 & b+a<=0;
  then b-a+a>=0+a or b+a<=0 by Lm6;
  then b>=0+a or b+a-a<=0-a by Lm7;
  hence thesis;
end;

theorem
  0 <= a & 0 <= c & a < b & c < d implies a*c < b*d
proof
  assume that
A1: 0<=a and
A2: 0<=c and
A3: a<b and
A4: c < d;
A5: a*c <= a*d by A1,A4,Lm12;
  a*d<b*d by A2,A3,A4,Lm13;
  hence thesis by A5,XXREAL_0:2;
end;

theorem
  0 <= a & 0 < c & a < b & c <= d implies a*c < b*d
proof
  assume that
A1: 0<=a and
A2: 0<c and
A3: a<b and
A4: c <=d;
A5: a*c <=a*d by A1,A4,Lm12;
  a*d<b*d by A2,A3,A4,Lm13;
  hence thesis by A5,XXREAL_0:2;
end;

theorem Th98:
  0 < a & 0 < c & a <= b & c < d implies a*c < b*d
proof
  assume that
A1: 0<a and
A2: 0<c and
A3: a<=b and
A4: c < d;
A5: a*c < a*d by A1,A4,Lm13;
  a*d<=b*d by A2,A3,A4,Lm12;
  hence thesis by A5,XXREAL_0:2;
end;

theorem
  0 < c & b < 0 & a < b implies c/b < c/a
proof
  assume that
A1: c>0 and
A2: b<0 and
A3: a<b;
  a">b" by A2,A3,Lm31;
  then a"*c>b"*c by A1,Lm13;
  then c/a>b"*c by XCMPLX_0:def 9;
  hence thesis by XCMPLX_0:def 9;
end;

theorem
  c < 0 & 0 < a & a < b implies c/a < c/b
proof
  assume that
A1: c < 0 and
A2: a>0 and
A3: a<b;
  a">b" by A2,A3,Lm34;
  then a"*c < b"*c by A1,Lm24;
  then c/a<b"*c by XCMPLX_0:def 9;
  hence thesis by XCMPLX_0:def 9;
end;

theorem
  c < 0 & b < 0 & a < b implies c/a < c/b
proof
  assume that
A1: c < 0 and
A2: b<0 and
A3: a<b;
  a">b" by A2,A3,Lm31;
  then a"*c < b"*c by A1,Lm24;
  then c/a<b"*c by XCMPLX_0:def 9;
  hence thesis by XCMPLX_0:def 9;
end;

theorem
  0 < b & 0 < d & a*d <= c*b implies a/b <= c/d
proof
  assume that
A1: b>0 and
A2: d>0 and
A3: a*d<=c*b;
  a*d/b<=c by A1,A3,Th79;
  then a/b*d<=c by XCMPLX_1:74;
  hence thesis by A2,Th77;
end;

theorem
  b < 0 & d < 0 & a*d <= c*b implies a/b <= c/d
proof
  assume that
A1: b<0 and
A2: d<0 and
A3: a*d<=c*b;
  a*d/b>=c by A1,A3,Th80;
  then a/b*d>=c by XCMPLX_1:74;
  hence thesis by A2,Th80;
end;

theorem
  0 < b & d < 0 & a*d <= c*b implies c/d <= a/b
proof
  assume that
A1: b>0 and
A2: d<0 and
A3: a*d<=c*b;
  a*d/b<=c by A1,A3,Th79;
  then a/b*d<=c by XCMPLX_1:74;
  hence thesis by A2,Th78;
end;

theorem
  b < 0 & 0 < d & a*d <= c*b implies c/d <= a/b
proof
  assume that
A1: b<0 and
A2: d>0 and
A3: a*d<=c*b;
  a*d/b>=c by A1,A3,Th80;
  then a/b*d>=c by XCMPLX_1:74;
  hence thesis by A2,Th79;
end;

theorem
  0 < b & 0 < d & a*d < c*b implies a/b < c/d
proof
  assume that
A1: b>0 and
A2: d>0 and
A3: a*d<c*b;
  a*d/b<c by A1,A3,Th83;
  then a/b*d<c by XCMPLX_1:74;
  hence thesis by A2,Th81;
end;

theorem
  b < 0 & d < 0 & a*d < c*b implies a/b < c/d
proof
  assume that
A1: b<0 and
A2: d<0 and
A3: a*d<c*b;
  a*d/b>c by A1,A3,Th84;
  then a/b*d>c by XCMPLX_1:74;
  hence thesis by A2,Th84;
end;

theorem
  0 < b & d < 0 & a*d < c*b implies c/d < a/b
proof
  assume that
A1: b>0 and
A2: d<0 and
A3: a*d<c*b;
  a*d/b<c by A1,A3,Th83;
  then a/b*d<c by XCMPLX_1:74;
  hence thesis by A2,Th82;
end;

theorem
  b < 0 & 0 < d & a*d < c*b implies c/d < a/b
proof
  assume that
A1: b<0 and
A2: d>0 and
A3: a*d<c*b;
  a*d/b>c by A1,A3,Th84;
  then a/b*d>c by XCMPLX_1:74;
  hence thesis by A2,Th83;
end;

theorem
  b < 0 & d < 0 & a*b < c/d implies a*d < c/b
proof
  assume that
A1: b<0 and
A2: d<0;
  assume a*b<c/d;
  then a*b*d>c by A2,Th78;
  then a*d*b>c;
  hence thesis by A1,Th84;
end;

theorem
  0 < b & 0 < d & a*b < c/d implies a*d < c/b
proof
  assume that
A1: b>0 and
A2: d>0;
  assume a*b<c/d;
  then a*b*d<c by A2,Th79;
  then a*d*b<c;
  hence thesis by A1,Th81;
end;

theorem
  b < 0 & d < 0 & c/d < a*b implies c/b < a*d
proof
  assume that
A1: b<0 and
A2: d<0;
  assume a*b>c/d;
  then a*b*d<c by A2,Th80;
  then a*d*b<c;
  hence thesis by A1,Th82;
end;

theorem
  0 < b & 0 < d & c/d < a*b implies c/b < a*d
proof
  assume that
A1: b>0 and
A2: d>0;
  assume a*b>c/d;
  then a*b*d>c by A2,Th77;
  then a*d*b>c;
  hence thesis by A1,Th83;
end;

theorem
  b < 0 & 0 < d & a*b < c/d implies c/b < a*d
proof
  assume that
A1: b<0 and
A2: d>0;
  assume a*b<c/d;
  then a*b*d<c by A2,Th79;
  then a*d*b<c;
  hence thesis by A1,Th82;
end;

theorem
  0 < b & d < 0 & a*b < c/d implies c/b < a*d
proof
  assume that
A1: b>0 and
A2: d<0;
  assume a*b<c/d;
  then a*b*d>c by A2,Th78;
  then a*d*b>c;
  hence thesis by A1,Th83;
end;

theorem
  b < 0 & 0 < d & c/d < a*b implies a*d < c/b
proof
  assume that
A1: b<0 and
A2: d>0;
  assume a*b>c/d;
  then a*b*d>c by A2,Th77;
  then a*d*b>c;
  hence thesis by A1,Th84;
end;

theorem
  0 < b & d < 0 & c/d < a*b implies a*d < c/b
proof
  assume that
A1: b>0 and
A2: d<0;
  assume a*b>c/d;
  then a*b*d<c by A2,Th80;
  then a*d*b<c;
  hence thesis by A1,Th81;
end;

theorem
  0 < a & 0 <= c & a <= b implies c/b <= c/a
proof
  assume that
A1: 0<a and
A2: 0<=c and
A3: a<=b;
  a*b"<=b*b" by A1,A3,Lm12;
  then a*b"<=1 by A1,A3,XCMPLX_0:def 7;
  then a"*(a*b")<=a"*1 by A1,Lm12;
  then a"*a*b"<=a";
  then 1*b"<=a" by A1,XCMPLX_0:def 7;
  then c*b"<=c*a" by A2,Lm12;
  then c/b<=c*a" by XCMPLX_0:def 9;
  hence thesis by XCMPLX_0:def 9;
end;

theorem
  0 <= c & b < 0 & a <= b implies c/b <= c/a
proof
  assume that
A1: c>=0 and
A2: b<0 and
A3: a<=b;
  a">=b" by A2,A3,Lm33;
  then a"*c>=b"*c by A1,Lm12;
  then c/a>=b"*c by XCMPLX_0:def 9;
  hence thesis by XCMPLX_0:def 9;
end;

theorem
  c <= 0 & 0 < a & a <= b implies c/a <= c/b
proof
  assume
A1: c <= 0;
  assume that
A2: a>0 and
A3: a<=b;
  a">=b" by A2,A3,Lm32;
  then a"*c <= b"*c by A1,Lm28;
  then c/a<=b"*c by XCMPLX_0:def 9;
  hence thesis by XCMPLX_0:def 9;
end;

theorem
  c <= 0 & b < 0 & a <= b implies c/a <= c/b
proof
  assume that
A1: c <= 0 and
A2: b<0 and
A3: a<=b;
  a">=b" by A2,A3,Lm33;
  then a"*c <= b"*c by A1,Lm28;
  then c/a<=b"*c by XCMPLX_0:def 9;
  hence thesis by XCMPLX_0:def 9;
end;

theorem
  0 < a iff 0 < a";

theorem
  a < 0 iff a" < 0;

theorem Th124:
  a < 0 & 0 < b implies a" < b"
proof
  assume that
A1: a<0 and
A2: b>0;
  a"<0 by A1;
  hence thesis by A2;
end;

theorem
  a" < 0 & 0 < b" implies a < b
proof
  assume that
A1: a"<0 and
A2: b">0;
  a""<b"" by A1,A2,Th124;
  hence thesis;
end;

begin

theorem
  a <= 0 & 0 <= a implies 0 = a*b by Lm35;

theorem
  0 <= a & 0 <= b implies 0 <= a*b;

theorem
  a <= 0 & b <= 0 implies 0 <= a*b;

theorem
  0 < a & 0 < b implies 0 < a*b;

theorem
  a < 0 & b < 0 implies 0 < a*b;

theorem
  0 <= a & b <= 0 implies a*b <= 0;

theorem
  0 < a & b < 0 implies a*b < 0;

theorem
  a*b<0 implies a>0 & b<0 or a<0 & b>0
proof
  assume
A1: a*b<0;
  then
A2: b <> 0;
  a <> 0 by A1;
  hence thesis by A1,A2;
end;

theorem
  a*b>0 implies a>0 & b>0 or a<0 & b<0
proof
  assume
A1: a*b>0;
  then
A2: b <> 0;
  a <> 0 by A1;
  hence thesis by A1,A2;
end;

theorem
  a <= 0 & b <= 0 implies 0 <= a/b;

theorem
  0 <= a & 0 <= b implies 0 <= a/b;

theorem
  0 <= a & b <= 0 implies a/b <= 0;

theorem
  a <= 0 & 0 <= b implies a/b <= 0;

theorem
  0 < a & 0 < b implies 0 < a/b;

theorem
  a < 0 & b < 0 implies 0 < a/b;

theorem
  a < 0 & 0 < b implies a/b < 0;

theorem
  a < 0 & 0 < b implies b/a < 0;

theorem
  a/b<0 implies b<0 & a>0 or b>0 & a<0
proof
  assume
A1: a/b<0;
  then
A2: a <> 0;
  a/b = a*b" by XCMPLX_0:def 9;
  then b" <> 0 by A1;
  hence thesis by A1,A2;
end;

theorem
  a/b>0 implies b>0 & a>0 or b<0 & a<0
proof
  assume
A1: a/b>0;
  then
A2: a <> 0;
  a/b = a*b" by XCMPLX_0:def 9;
  then b" <> 0 by A1;
  hence thesis by A1,A2;
end;

begin

theorem
  a <= b implies a < b+1
proof
  assume a<=b;
  then
A1: a-1<=b-1 by Lm7;
  b-1<b-1+1 by Lm9;
  then a-1<b by A1,XXREAL_0:2;
  then a-1+1<b+1 by Lm10;
  hence thesis;
end;

theorem
  a - 1 < a
proof
  a+- 1 < a + 0 by Lm10;
  hence thesis;
end;

theorem
  a <= b implies a-1 < b
proof
  assume a<=b;
  then
A1: a-1<=b-1 by Lm7;
  b-1<b-1+1 by Lm9;
  hence thesis by A1,XXREAL_0:2;
end;

theorem
  -1 < a implies 0 < 1+a
proof
  assume -1<a;
  then -1+1<a+1 by Lm8;
  hence thesis;
end;

theorem
  a<1 implies 1-a>0 by Lm21;

begin

theorem
  a <= 1 & 0 <= b & b <= 1 & a*b = 1 implies a=1
proof
  assume that
A1: a<=1 and
A2: 0<=b and
A3: b<=1 and
A4: a*b=1;
  now
    per cases by A2;
    case
      b=0;
      hence contradiction by A4;
    end;
    case
A5:   b>0;
      then a=b" by A4,XCMPLX_0:def 7;
      then a>=1" by A3,A5,Lm32;
      hence thesis by A1,XXREAL_0:1;
    end;
  end;
  hence thesis;
end;

theorem
  0 <= a & 1 <= b implies a <= a*b
proof
  assume that
A1: a>=0 and
A2: b>=1;
  a*b>=a*1 by A1,A2,Lm12;
  hence thesis;
end;

theorem
  a <= 0 & b <= 1 implies a <= a*b
proof
  assume that
A1: a<=0 and
A2: b<=1;
  a*b>=a*1 by A1,A2,Lm28;
  hence thesis;
end;

theorem
  0 <= a & b <= 1 implies a*b <= a
proof
  assume that
A1: a>=0 and
A2: b<=1;
  a*b<=a*1 by A1,A2,Lm12;
  hence thesis;
end;

theorem
  a <= 0 & 1 <= b implies a*b <= a
proof
  assume that
A1: a<=0 and
A2: b>=1;
  a*b<=a*1 by A1,A2,Lm28;
  hence thesis;
end;

theorem Th155:
  0 < a & 1 < b implies a < a*b
proof
  assume that
A1: a>0 and
A2: b>1;
  a*b>a*1 by A1,A2,Lm13;
  hence thesis;
end;

theorem
  a < 0 & b < 1 implies a < a*b
proof
  assume that
A1: a<0 and
A2: b<1;
  a*b>a*1 by A1,A2,Lm24;
  hence thesis;
end;

theorem
  0 < a & b < 1 implies a*b < a
proof
  assume that
A1: a>0 and
A2: b<1;
  a*b<a*1 by A1,A2,Lm13;
  hence thesis;
end;

theorem
  a < 0 & 1 < b implies a*b < a
proof
  assume that
A1: a<0 and
A2: b>1;
  a*b<a*1 by A1,A2,Lm24;
  hence thesis;
end;

theorem
  1 <= a & 1 <= b implies 1 <= a*b
proof
  assume that
A1: a>=1 and
A2: b>=1;
  a*b>=b*1 by A1,A2,Lm12;
  hence thesis by A2,XXREAL_0:2;
end;

theorem
  0 <= a & a <= 1 & b <= 1 implies a*b <= 1
proof
  assume that
A1: 0<=a and
A2: a<=1 and
A3: b<=1;
  a*b<=1*a by A1,A3,Lm12;
  hence thesis by A2,XXREAL_0:2;
end;

theorem
  1 < a & 1 <= b implies 1 < a*b
proof
  assume that
A1: a>1 and
A2: b>=1;
  a*b>b*1 by A1,A2,Lm13;
  hence thesis by A2,XXREAL_0:2;
end;

theorem
  0 <= a & a < 1 & b <= 1 implies a*b < 1
proof
  assume that
A1: 0<=a and
A2: a<1 and
A3: b<=1;
  a*b<=1*a by A1,A3,Lm12;
  hence thesis by A2,XXREAL_0:2;
end;

theorem
  a <= -1 & b <= -1 implies 1 <= a*b
proof
  assume that
A1: a<=-1 and
A2: b<=-1;
A3: -b>=--1 by A2,Lm14;
  a*b>=b*(-1) by A1,A2,Lm28;
  hence thesis by A3,XXREAL_0:2;
end;

theorem
  a < -1 & b <= -1 implies 1 < a*b
proof
  assume that
A1: a<-1 and
A2: b<=-1;
A3: -b>=--1 by A2,Lm14;
  a*b>b*(-1) by A1,A2,Lm24;
  hence thesis by A3,XXREAL_0:2;
end;

theorem
  a <= 0 & -1 <= a & -1 <= b implies a*b <= 1
proof
  assume that
A1: 0>=a and
A2: a>=-1 and
A3: b>=-1;
A4: -a<=--1 by A2,Lm14;
  a*b<=(-1)*a by A1,A3,Lm28;
  hence thesis by A4,XXREAL_0:2;
end;

theorem
  a <= 0 & -1 < a & -1 <= b implies a*b < 1
proof
  assume that
A1: 0>=a and
A2: a>-1 and
A3: b>=-1;
A4: -a<--1 by A2,Lm15;
  a*b<=(-1)*a by A1,A3,Lm28;
  hence thesis by A4,XXREAL_0:2;
end;

begin

theorem Th167:
  (for a st 1 < a holds c <= b*a) implies c <= b
proof
  assume
A1: for a st a>1 holds b*a>=c;
  assume
A2: not thesis;
A3: b>=0
  proof
A4: c <= b*2 by A1;
    assume b<0;
    then
A5: b*2<0;
    then c < 0 by A1;
    then b/c>c/c by A2,Lm29;
    then b/c>1 by A5,A4,XCMPLX_1:60;
    then
A6: b*(b/c)>=c by A1;
    b*(b/c)<c*(b/c) by A2,A5,A4,Lm13;
    then b*(b/c)<b by A5,A4,XCMPLX_1:87;
    hence contradiction by A2,A6,XXREAL_0:2;
  end;
  per cases by A3;
  suppose
A7: b>0;
    then b/b<c/b by A2,Lm26;
    then 1<c/b by A7,XCMPLX_1:60;
    then consider d being real number such that
A8: 1<d and
A9: d<c/b by Th5;
    b*d<b*(c/b) by A7,A9,Lm13;
    then b*d<c by A7,XCMPLX_1:87;
    hence contradiction by A1,A8;
  end;
  suppose
A10: b=0;
    b*2>=c by A1;
    hence contradiction by A2,A10;
  end;
end;

Lm36: 1 < a implies a" < 1
proof
  assume
A1: 1<a;
  then 1*a"<a*a" by Lm13;
  hence thesis by A1,XCMPLX_0:def 7;
end;

theorem
  (for a st 0 < a & a < 1 holds b*a <= c) implies b <= c
proof
  assume
A1: for a st 0<a & a<1 holds b*a<=c;
  now
    let d;
    assume
A2: d>1;
    then b*(d")<=c by A1,Lm36;
    then b/d<=c by XCMPLX_0:def 9;
    hence b<=c*d by A2,Th81;
  end;
  hence thesis by Th167;
end;

Lm37: a <= b & 0 <= a implies a/b <= 1
proof
  assume
A1: a <= b;
  assume
A2: 0 <= a;
  per cases by A2;
  suppose
    a = 0;
    hence thesis;
  end;
  suppose
A3: a > 0;
    then a/b<=b/b by A1,Lm25;
    hence thesis by A1,A3,XCMPLX_1:60;
  end;
end;

Lm38: b <= a & 0 <= a implies b/a <= 1
proof
  assume
A1: b <= a;
  assume
A2: a>=0;
  per cases by A2;
  suppose
    a = 0;
    then a" = 0;
    then b*a" < 1;
    hence thesis by XCMPLX_0:def 9;
  end;
  suppose
A3: a > 0;
    assume b/a>1;
    then b/a*a>1*a by A3,Lm13;
    hence thesis by A1,A3,XCMPLX_1:87;
  end;
end;

theorem
  (for a st 0 < a & a < 1 holds b <= a*c) implies b <= 0
proof
  assume that
A1: for a st 0 < a & a < 1 holds b <= a*c and
A2: b > 0;
A3: b*2 > b by A2,Th155;
  then
A4: b > b/2 by Th83;
  per cases;
  suppose
    c <= 0;
    then 1/2*c <= 0;
    hence contradiction by A1,A2;
  end;
  suppose that
A5: c <= b and
A6: c > 0;
    set a = c/(2*b);
    c/b*2 > c/b by A2,A6,Th155;
    then c/b > c/b/2 by Th83;
    then
A7: c/b > c/(b*2) by XCMPLX_1:78;
    c/b <= 1 by A5,A6,Lm37;
    then a < 1 by A7,XXREAL_0:2;
    then
A8: a*c >= b by A1,A2,A6;
    per cases;
    suppose
      c >= b;
      then b = c by A5,XXREAL_0:1;
      then a*c = b/2 by A6,XCMPLX_1:92;
      hence contradiction by A2,A8,Th83,Th155;
    end;
    suppose
A9:   c < b;
      then a*c < c/(2*b)*b by A6,Th98;
      then
A10:  a*c < c/2 by A2,XCMPLX_1:92;
      c/2 < b/2 by A9,Lm26;
      then a*c < b/2 by A10,XXREAL_0:2;
      hence contradiction by A4,A8,XXREAL_0:2;
    end;
  end;
  suppose that
A11: b <= c and
A12: c > 0;
    set a = b/(2*c);
    b/c*2 > b/c by A2,A12,Th155;
    then b/c > b/c/2 by Th83;
    then
A13: b/c > b/(c*2) by XCMPLX_1:78;
    b/c <= 1 by A11,A12,Lm38;
    then a < 1 by A13,XXREAL_0:2;
    then
A14: a*c >= b by A1,A2,A12;
    a*c = b/2 by A12,XCMPLX_1:92;
    hence contradiction by A3,A14,Th83;
  end;
end;

theorem
  0 <= d & d <= 1 & 0 <= a & 0 <= b & d*a+(1-d)*b=0 implies d=0 & b=0 or
  d=1 & a=0 or a=0 & b=0
proof
  assume that
A1: 0<=d and
A2: d<=1 and
A3: a>=0 and
A4: b>=0 and
A5: d*a+(1-d)*b=0;
  d-d<=1-d by A2,Lm7;
  then 1-d=0 or b=0 by A1,A3,A4,A5;
  hence thesis by A5;
end;

theorem
  0 <= d & a <= b implies a <= (1-d)*a+d*b
proof
  assume that
A1: 0<=d and
A2: a<=b;
  d*a<=d*b by A1,A2,Lm12;
  then (1-d)*a+d*a<=(1-d)*a+d*b by Lm6;
  hence thesis;
end;

theorem
  d <= 1 & a <= b implies (1-d)*a+d*b <= b
proof
  assume that
A1: d<=1 and
A2: a<=b;
  1-d>=0 by A1,Th48;
  then (1-d)*a<=(1-d)*b by A2,Lm12;
  then (1-d)*a+d*b<=(1-d)*b+d*b by Lm6;
  hence thesis;
end;

theorem
  0 <= d & d <= 1 & a <= b & a <= c implies a <= (1-d)*b+d*c
proof
  assume that
A1: 0<=d and
A2: d<=1 and
A3: a<=b and
A4: a<=c;
  1-d>=0 by A2,Th48;
  then
A5: (1-d)*a<=(1-d)*b by A3,Lm12;
A6: (1-d)*a+d*a = a;
  d*a<=d*c by A1,A4,Lm12;
  hence thesis by A5,A6,Lm6;
end;

theorem
  0 <= d & d <= 1 & b <= a & c <= a implies (1-d)*b+d*c <= a
proof
  assume that
A1: 0<=d and
A2: d<=1 and
A3: a>=b and
A4: a>=c;
  1-d>=0 by A2,Th48;
  then
A5: (1-d)*a>=(1-d)*b by A3,Lm12;
A6: (1-d)*a+d*a = a;
  d*a>=d*c by A1,A4,Lm12;
  hence thesis by A5,A6,Lm6;
end;

theorem
  0 <= d & d <= 1 & a < b & a < c implies a < (1-d)*b+d*c
proof
  assume that
A1: 0 <= d and
A2: d <= 1 and
A3: a<b and
A4: a<c;
  per cases;
  suppose
    d=0;
    hence thesis by A3;
  end;
  suppose
    d=1;
    hence thesis by A4;
  end;
  suppose
A5: not(d=0 or d=1);
    then d<1 by A2,XXREAL_0:1;
    then 1-d>0 by Lm21;
    then
A6: (1-d)*a<(1-d)*b by A3,Lm13;
A7: (1-d)*a+d*a=a;
    d*a<d*c by A1,A4,A5,Lm13;
    hence thesis by A6,A7,Lm8;
  end;
end;

theorem
  0 <= d & d <= 1 & b < a & c < a implies (1-d)*b+d*c < a
proof
  assume that
A1: 0 <= d and
A2: d <= 1 and
A3: a>b and
A4: a>c;
  per cases;
  suppose
    d=0;
    hence thesis by A3;
  end;
  suppose
    d=1;
    hence thesis by A4;
  end;
  suppose
A5: not(d=0 or d=1);
    then d<1 by A2,XXREAL_0:1;
    then 1-d>0 by Lm21;
    then
A6: (1-d)*a>(1-d)*b by A3,Lm13;
A7: (1-d)*a+d*a=a;
    d*a>d*c by A1,A4,A5,Lm13;
    hence thesis by A6,A7,Lm8;
  end;
end;

theorem
  0 < d & d < 1 & a <= b & a < c implies a < (1-d)*b+d*c
proof
  assume that
A1: 0 < d and
A2: d < 1 and
A3: a <= b and
A4: a < c;
  1-d > 0 by A2,Lm21;
  then
A5: (1-d)*a <= (1-d)*b by A3,Lm12;
A6: (1-d)*a+d*a = a;
  d*a < d*c by A1,A4,Lm13;
  hence thesis by A5,A6,Lm8;
end;

theorem
  0 < d & d < 1 & b < a & c <= a implies (1-d)*b+d*c < a
proof
  assume that
A1: 0 < d and
A2: d < 1 and
A3: a > b and
A4: a >= c;
  1-d > 0 by A2,Lm21;
  then
A5: (1-d)*a > (1-d)*b by A3,Lm13;
A6: (1-d)*a+d*a = a;
  d*a >= d*c by A1,A4,Lm12;
  hence thesis by A5,A6,Lm8;
end;

theorem
  0 <= d & d <= 1 & a <= (1-d)*a+d*b & b < (1-d)*a+d*b implies d = 0
proof
  assume that
A1: d >= 0 and
A2: d <= 1 and
A3: a <= (1-d)*a + d*b and
A4: b < (1-d)*a + d*b;
  set s = (1-d)*a + d*b;
  assume d <> 0;
  then
A5: d*b < d*s by A1,A4,Lm13;
  1-d >= 0 by A2,Th48;
  then
A6: (1-d)*a <= (1-d)*s by A3,Lm12;
  1*s = (1-d)*s + d*s;
  hence contradiction by A5,A6,Lm8;
end;

theorem
  0 <= d & d <= 1 & (1-d)*a+d*b <= a & (1-d)*a+d*b < b implies d = 0
proof
  assume that
A1: d >= 0 and
A2: d <= 1 and
A3: a >= (1-d)*a + d*b and
A4: b > (1-d)*a + d*b;
  set s = (1-d)*a + d*b;
  assume d <> 0;
  then
A5: d*b > d*s by A1,A4,Lm13;
  1-d >= 0 by A2,Th48;
  then
A6: (1-d)*a >= (1-d)*s by A3,Lm12;
  1*s = (1-d)*s + d*s;
  hence contradiction by A5,A6,Lm8;
end;

begin

theorem
  0 < a & a <= b implies 1 <= b/a
proof
  assume that
A1: 0<a and
A2: a<=b;
  b/a>=a/a by A1,A2,Lm25;
  hence thesis by A1,XCMPLX_1:60;
end;

theorem
  a < 0 & b <= a implies 1 <= b/a
proof
  assume that
A1: a < 0 and
A2: b<=a;
  b/a>=a/a by A1,A2,Lm30;
  hence thesis by A1,XCMPLX_1:60;
end;

theorem
  0 <= a & a <= b implies a/b <= 1 by Lm37;

theorem
  b <= a & a <= 0 implies a/b <= 1
proof
  assume
A1: b <= a;
  assume
A2: a <= 0;
  per cases by A2;
  suppose
    a = 0;
    hence thesis;
  end;
  suppose
A3: a < 0;
    then a/b<=b/b by A1,Lm30;
    hence thesis by A1,A3,XCMPLX_1:60;
  end;
end;

theorem
  0 <= a & b <= a implies b/a <= 1 by Lm38;

theorem
  a <= 0 & a <= b implies b/a <= 1
proof
  assume
A1: a<=0;
  per cases by A1;
  suppose
    a = 0;
    then a" = 0;
    then b*a" = 0;
    hence thesis by XCMPLX_0:def 9;
  end;
  suppose
A2: a<0;
    assume
A3: a <= b;
    assume b/a>1;
    then b/a*a<1*a by A2,Lm24;
    hence thesis by A2,A3,XCMPLX_1:87;
  end;
end;

theorem
  0 < a & a < b implies 1 < b/a
proof
  assume
A1: a>0;
  assume
A2: a < b;
  assume b/a<=1;
  then b/a*a<=1*a by A1,Lm12;
  hence thesis by A1,A2,XCMPLX_1:87;
end;

theorem
  a < 0 & b < a implies 1 < b/a
proof
  assume that
A1: a < 0 and
A2: b<a;
  b/a>a/a by A1,A2,Lm29;
  hence thesis by A1,XCMPLX_1:60;
end;

theorem
  0 <= a & a < b implies a/b < 1
proof
  assume that
A1: 0 <= a and
A2: a < b;
  a/b<b/b by A1,A2,Lm26;
  hence thesis by A1,A2,XCMPLX_1:60;
end;

theorem
  a <= 0 & b < a implies a/b < 1
proof
  assume that
A1: a <= 0 and
A2: b<a;
  a/b<b/b by A1,A2,Lm29;
  hence thesis by A1,A2,XCMPLX_1:60;
end;

theorem
  0 < a & b < a implies b/a < 1
proof
  assume
A1: a > 0;
  assume
A2: b < a;
  assume b/a>=1;
  then b/a*a>=1*a by A1,Lm12;
  hence thesis by A1,A2,XCMPLX_1:87;
end;

theorem
  a < 0 & a < b implies b/a < 1
proof
  assume
A1: a<0;
  assume
A2: a < b;
  assume b/a>=1;
  then b/a*a<=1*a by A1,Lm28;
  hence thesis by A1,A2,XCMPLX_1:87;
end;

begin

theorem
  0 <= b & -b <= a implies -1 <= a/b
proof
  assume
A1: b>=0;
  per cases by A1;
  suppose
    b = 0;
    then b" = 0;
    then a*b" = 0;
    hence thesis by XCMPLX_0:def 9;
  end;
  suppose
A2: b>0;
    assume
A3: -b <= a;
    assume a/b<-1;
    then a/b*b<(-1)*b by A2,Lm13;
    hence thesis by A2,A3,XCMPLX_1:87;
  end;
end;

theorem
  0 <= b & -a <= b implies -1 <= a/b
proof
  assume
A1: b >= 0;
  per cases by A1;
  suppose
    b = 0;
    then b" = 0;
    then a*b" = 0;
    hence thesis by XCMPLX_0:def 9;
  end;
  suppose
A2: b > 0;
    assume
A3: -a <= b;
    assume a/b<-1;
    then a/b*b<(-1)*b by A2,Lm13;
    then a<-b by A2,XCMPLX_1:87;
    hence thesis by A3,Th26;
  end;
end;

theorem
  b <= 0 & a <= -b implies -1 <= a/b
proof
  assume
A1: b<=0;
  per cases by A1;
  suppose
    b=0;
    then b" =0;
    then a*b" = 0;
    hence thesis by XCMPLX_0:def 9;
  end;
  suppose
A2: b < 0;
    assume
A3: a <= -b;
    assume a/b<-1;
    then a/b*b>(-1)*b by A2,Lm24;
    hence thesis by A2,A3,XCMPLX_1:87;
  end;
end;

theorem
  b <= 0 & b <= -a implies -1 <= a/b
proof
  assume
A1: b<=0;
  per cases by A1;
  suppose
    b = 0;
    then b" =0;
    then a*b" = 0;
    hence thesis by XCMPLX_0:def 9;
  end;
  suppose
A2: b < 0;
    assume
A3: b <= -a;
    assume a/b<-1;
    then a/b*b>(-1)*b by A2,Lm24;
    then a>-b by A2,XCMPLX_1:87;
    hence thesis by A3,Th25;
  end;
end;

theorem
  0 < b & -b < a implies -1 < a/b
proof
  assume
A1: b>0;
  assume
A2: -b < a;
  assume a/b<=-1;
  then a/b*b<=(-1)*b by A1,Lm12;
  hence thesis by A1,A2,XCMPLX_1:87;
end;

theorem
  0 < b & -a < b implies -1 < a/b
proof
  assume
A1: b>0;
  assume
A2: -a < b;
  assume a/b<=-1;
  then a/b*b<=(-1)*b by A1,Lm12;
  then a<=-b by A1,XCMPLX_1:87;
  hence thesis by A2,Th25;
end;

theorem
  b < 0 & a < -b implies -1 < a/b
proof
  assume
A1: b<0;
  assume
A2: a < -b;
  assume a/b<=-1;
  then a/b*b>=(-1)*b by A1,Lm28;
  hence thesis by A1,A2,XCMPLX_1:87;
end;

theorem
  b < 0 & b < -a implies -1 < a/b
proof
  assume
A1: b<0;
  assume
A2: b < -a;
  assume a/b<=-1;
  then a/b*b>=(-1)*b by A1,Lm28;
  then a>=-b by A1,XCMPLX_1:87;
  hence thesis by A2,Th26;
end;

theorem
  0 < b & a <= -b implies a/b <= -1
proof
  assume
A1: b>0;
  assume
A2: a <= -b;
  assume a/b>-1;
  then a/b*b>(-1)*b by A1,Lm13;
  hence thesis by A1,A2,XCMPLX_1:87;
end;

theorem
  0 < b & b <= -a implies a/b <= -1
proof
  assume
A1: b>0;
  assume
A2: b <= -a;
  assume a/b>-1;
  then a/b*b>(-1)*b by A1,Lm13;
  then a>-b by A1,XCMPLX_1:87;
  hence thesis by A2,Th25;
end;

theorem
  b < 0 & -b <= a implies a/b <= -1
proof
  assume
A1: b<0;
  assume
A2: -b <= a;
  assume a/b>-1;
  then a/b*b<(-1)*b by A1,Lm24;
  hence thesis by A1,A2,XCMPLX_1:87;
end;

theorem
  b < 0 & -a <= b implies a/b <= -1
proof
  assume
A1: b<0;
  assume
A2: -a <= b;
  assume a/b>-1;
  then a/b*b<(-1)*b by A1,Lm24;
  then a<-b by A1,XCMPLX_1:87;
  hence thesis by A2,Th26;
end;

theorem
  0 < b & a < -b implies a/b < -1
proof
  assume
A1: b>0;
  assume
A2: a < -b;
  assume a/b>=-1;
  then a/b*b>=(-1)*b by A1,Lm12;
  hence thesis by A1,A2,XCMPLX_1:87;
end;

theorem
  0 < b & b < -a implies a/b < -1
proof
  assume
A1: b>0;
  assume
A2: b < -a;
  assume a/b>=-1;
  then a/b*b>=(-1)*b by A1,Lm12;
  then a>=-b by A1,XCMPLX_1:87;
  hence thesis by A2,Th26;
end;

theorem
  b < 0 & -b < a implies a/b < -1
proof
  assume
A1: b<0;
  assume
A2: -b < a;
  assume a/b>=-1;
  then a/b*b<=(-1)*b by A1,Lm28;
  hence thesis by A1,A2,XCMPLX_1:87;
end;

theorem
  b < 0 & -a < b implies a/b < -1
proof
  assume
A1: b<0;
  assume
A2: -a < b;
  assume a/b>=-1;
  then a/b*b<=(-1)*b by A1,Lm28;
  then a<=-b by A1,XCMPLX_1:87;
  hence thesis by A2,Th25;
end;

begin

theorem Th209:
  (for a being real number st 0 < a & a < 1 holds c <= b/a)
  implies c <= b
proof
  assume
A1: for a st a>0 & a<1 holds b/a>=c;
  now
    let d;
    assume d>1;
    then
A2: b/(d")>=c by A1,Lm36;
    d" = 1/d by XCMPLX_1:215;
    then b*d/1>=c by A2,XCMPLX_1:77;
    hence b*d>=c;
  end;
  hence thesis by Th167;
end;

theorem
  (for a being real number st 1 < a holds b/a <= c) implies b <= c
proof
  (for a st a>1 holds b/a<=c) implies b<=c
  proof
    assume
A1: for a st a>1 holds b/a<=c;
    now
      let d;
A2:   d" = 1/d by XCMPLX_1:215;
      assume that
A3:   0<d and
A4:   d<1;
      d">1" by A3,A4,Lm34;
      then b/(d")<=c by A1;
      then b*d/1<=c by A2,XCMPLX_1:77;
      hence b<=c/d by A3,Th77;
    end;
    hence thesis by Th209;
  end;
  hence thesis;
end;

theorem
  1 <= a implies a" <= 1
proof
  1 <= a implies a" <= 1" by Lm32;
  hence thesis;
end;

theorem
  1 < a implies a" < 1 by Lm36;

theorem
  a <= -1 implies -1 <= a"
proof
  a <= -1 implies (-1)" <= a" by Lm33;
  hence thesis;
end;

theorem
  a < -1 implies -1 < a"
proof
  a < -1 implies (-1)" < a" by Lm31;
  hence thesis;
end;

begin

theorem
  0 < a implies 0 < a/2;

theorem
  0 < a implies a/2 < a by Lm27;

theorem
  a <= 1/2 implies 2 * a - 1 <= 0
proof
  assume a <= 1/2;
  then 2 * a <= 2 * (1/2) by Lm12;
  then 2 * a <= 1 + 0;
  hence thesis by Lm18;
end;

theorem
  a <= 1/2 implies 0 <= 1 - 2 * a
proof
  assume a <= 1/2;
  then 2 * a <= 2 * (1/2) by Lm12;
  then 2 * a + 0 <= 1;
  hence thesis by Lm16;
end;

theorem
  a >= 1/2 implies 2 * a - 1 >= 0
proof
  assume a >= 1/2;
  then 2 * a >= 2 * (1/2) by Lm12;
  then 2 * a >= 1 + 0;
  hence thesis by Lm16;
end;

theorem
  a >= 1/2 implies 0 >= 1 - 2 * a
proof
  assume a >= 1/2;
  then 2 * a >= 2 * (1/2) by Lm12;
  then 2 * a + 0 >= 1;
  hence thesis by Lm18;
end;

begin

theorem
  0 < a implies a/3 < a
proof
  assume
A1: 0<a;
  then
A2: a/3+a/3+0<a/3+a/3+a/3 by Lm10;
  0+a/3<a/3+a/3 by A1,Lm10;
  hence thesis by A2,XXREAL_0:2;
end;

theorem
  0 < a implies 0 < a/3;

begin

theorem
  0 < a implies a/4 < a
proof
  assume
A1: 0<a;
  then
A2: 0+a/4<a/4+a/4 by Lm10;
  then a/4< a/4+a/4+a/4 by A1,Lm10;
  then a/4+a/4<a/4+a/4+a/4+a/4 by Lm10;
  hence thesis by A2,XXREAL_0:2;
end;

theorem
  0 < a implies 0 < a/4;

theorem
  for a,b st b<>0 ex c st a=b/c
proof
  let a,b;
  assume
A1: b<>0;
  then consider c being complex number such that
A2: a=b/c by XCMPLX_1:232;
  per cases;
  suppose
    c = 0;
    hence thesis by A2;
  end;
  suppose
    c <> 0;
    then c = b/a by A1,A2,XCMPLX_1:54;
    then reconsider c as real number;
    take c;
    thus thesis by A2;
  end;
end;

begin 



theorem
  r<s implies r < (r+s)/2 & (r+s)/2 < s
proof
  assume r<s;
  then
A1: r/2 < s/2 by Lm26;
  then r/2 + r/2 < r/2 + s/2 by Lm10;
  hence r < (r+s)/2;
  r/2 + s/2 < s/2 + s/2 by A1,Lm10;
  hence thesis;
end;



registration
  cluster -> ext-real for Element of REAL;
  coherence;
end;

reserve p,q,r,s,t for ext-real number;

theorem Th227:
  r < t implies ex s st r < s & s < t
proof
  assume
A1: r < t;
  then
A2: r <> +infty by XXREAL_0:3;
A3: t <> -infty by A1,XXREAL_0:5;
  per cases by A2,A3,XXREAL_0:14;
  suppose that
A4: r = -infty and
A5: t = +infty;
    take 0;
    thus thesis by A4,A5;
  end;
  suppose that
A6: r = -infty and
A7: t in REAL;
    reconsider t as real number by A7;
    consider s being real number such that
A8: s < t by Th2;
    take s;
    s in REAL by XREAL_0:def 1;
    hence r < s by A6,XXREAL_0:12;
    thus thesis by A8;
  end;
  suppose that
A9: r in REAL and
A10: t = +infty;
    reconsider r9 = r as real number by A9;
    consider s being real number such that
A11: r9 < s by Th1;
    take s;
    thus r < s by A11;
    s in REAL by XREAL_0:def 1;
    hence thesis by A10,XXREAL_0:9;
  end;
  suppose that
A12: r in REAL and
A13: t in REAL;
    reconsider r,t as real number by A12,A13;
    consider s being real number such that
A14: r < s and
A15: s < t by A1,Th5;
    take s;
    thus thesis by A14,A15;
  end;
end;

theorem
  r < s & (for q st r < q & q < s holds t <= q) implies t <= r
proof
  assume that
A1: r < s and
A2: for q st r < q & q < s holds t <= q;
  for q st r < q holds t <= q
  proof
    let q such that
A3: r < q;
    per cases;
    suppose
      q < s;
      hence thesis by A2,A3;
    end;
    suppose
A4:   s <= q;
      consider p such that
A5:   r < p and
A6:   p < s by A1,Th227;
      t <= p by A2,A5,A6;
      then t <= s by A6,XXREAL_0:2;
      hence thesis by A4,XXREAL_0:2;
    end;
  end;
  hence thesis by Th227;
end;

theorem
  r < s & (for q st r < q & q < s holds q <= t) implies s <= t
proof
  assume that
A1: r < s and
A2: for q st r < q & q < s holds q <= t;
  for q st t < q holds s <= q
  proof
    let q such that
A3: t < q and
A4: q < s;
    per cases;
    suppose
      r < q;
      hence thesis by A2,A3,A4;
    end;
    suppose
A5:   q <= r;
      consider p such that
A6:   r < p and
A7:   p < s by A1,Th227;
      p <= t by A2,A6,A7;
      then r <= t by A6,XXREAL_0:2;
      hence thesis by A3,A5,XXREAL_0:2;
    end;
  end;
  hence thesis by Th227;
end;



theorem
  0 <= a & b <= c implies b-a <= c
proof
  assume that
A1: 0 <= a and
A2: b <= c;
  b+0 <= a+c by A1,A2,Th38;
  hence thesis by Lm18;
end;

theorem
  0 < a & b <= c implies b-a < c
proof
  assume that
A1: 0 < a and
A2: b <= c;
  b+0 < a+c by A1,A2,Th39;
  hence thesis by Lm17;
end;

begin

theorem
  a -' a = 0
proof
  a-a = 0;
  hence thesis by XREAL_0:def 2;
end;

theorem Th233:
  b <= a implies a -' b = a - b
proof
  assume a >= b;
  then a - b >= 0 by Th48;
  hence thesis by XREAL_0:def 2;
end;

theorem
  c <= a & c <= b & a -' c = b -' c implies a = b
proof
  assume that
A1: a >= c and
A2: b >= c and
A3: a -' c = b -' c;
  a - c >= 0 by A1,Th48;
  then
A4: a -' c = a - c by XREAL_0:def 2;
  b - c >= 0 by A2,Th48;
  then a + (-c) = b + (-c) by A3,A4,XREAL_0:def 2;
  hence thesis;
end;

theorem
  a >= b implies a -' b + b = a
proof
  assume a >= b;
  then a - b = a -' b by Th233;
  hence thesis;
end;



theorem
  a <= b & c < a implies b -' a < b -' c
proof
  assume that
A1: a <= b and
A2: c < a;
A3: b - a < b - c by A2,Th15;
  b -' c = b - c by A1,A2,Th233,XXREAL_0:2;
  hence thesis by A1,A3,Th233;
end;



theorem
 1 <= a implies a -' 1 < a
 proof
  assume 1 <= a;
   then a -' 1 = a - 1 by Th233;
  hence a -' 1 < a by Th44;
 end;














begin

reserve r,s for real number;

Lm1: (r in REAL+ & s in REAL+ & ex x9,y9 being Element of REAL+ st r = x9 & s
= y9 & x9 <=' y9) or (r in [:{0},REAL+:] & s in [:{0},REAL+:] & ex x9,y9 being
Element of REAL+ st r = [0,x9] & s = [0,y9] & y9 <=' x9) or s in REAL+ & r in
[:{0},REAL+:] implies r <= s
proof
  assume
A1: (r in REAL+ & s in REAL+ & ex x9,y9 being Element of REAL+ st r = x9
  & s = y9 & x9 <=' y9) or (r in [:{0},REAL+:] & s in [:{0},REAL+:] & ex x9,y9
being Element of REAL+ st r = [0,x9] & s = [0,y9] & y9 <=' x9) or s in REAL+ &
  r in [:{0},REAL+:];
  per cases;
  case
    r in REAL+ & s in REAL+;
    hence thesis by A1,ARYTM_0:5,XBOOLE_0:3;
  end;
  case
    r in [:{0},REAL+:] & s in [:{0},REAL+:];
    hence thesis by A1,ARYTM_0:5,XBOOLE_0:3;
  end;
  case
    not(r in REAL+ & s in REAL+) & not (r in [:{0},REAL+:] & s in [:{0
    },REAL+:]);
    hence thesis by A1;
  end;
end;

reserve x,y,z for real number,
  k for Nat,
  i for Element of NAT;

Lm2: for x being real number, x1,x2 being Element of REAL st x = [*x1,x2*]
holds x2 = 0 & x = x1
proof
  let x be real number, x1,x2 being Element of REAL;
  assume
A1: x = [*x1,x2*];
A2: x in REAL by XREAL_0:def 1;
  thus now
    assume x2 <> 0;
    then x = (0,1) --> (x1,x2) by A1,ARYTM_0:def 5;
    hence contradiction by A2,ARYTM_0:8;
  end;
  hence thesis by A1,ARYTM_0:def 5;
end;

Lm3: for x9,y9 being Element of REAL, x,y st x9 = x & y9 = y holds +(x9,y9) =
x + y
proof
  let x9,y9 be Element of REAL, x,y such that
A1: x9 = x & y9 = y;
  consider x1,x2,y1,y2 being Element of REAL such that
A2: x = [* x1,x2 *] & y = [*y1,y2*] and
A3: x+y = [*+(x1,y1),+(x2,y2)*] by XCMPLX_0:def 4;
  x2 = 0 & y2 = 0 by A2,Lm2;
  then
A4: +(x2,y2) = 0 by ARYTM_0:11;
  x = x1 & y = y1 by A2,Lm2;
  hence thesis by A1,A3,A4,ARYTM_0:def 5;
end;

Lm4: {} in {{}} by TARSKI:def 1;

reconsider o = 0 as Element of REAL+ by ARYTM_2:20;

reserve r,r1,r2 for Element of REAL+;

theorem
  for X,Y being Subset of REAL st for x,y st x in X & y in Y holds x <=
  y ex z st for x,y st x in X & y in Y holds x <= z & z <= y
proof
  let X,Y be Subset of REAL;
  assume
A1: for x,y st x in X & y in Y holds x <= y;
  per cases;
  suppose
A2: X = 0 or Y = 0;
    take 1;
    thus thesis by A2;
  end;
  suppose that
A3: X <> 0 and
A4: Y <> 0;
    consider x1 being Element of REAL such that
A5: x1 in X by A3,SUBSET_1:4;
A6: X c= REAL+ \/ [:{0},REAL+:] by NUMBERS:def 1,XBOOLE_1:1;
A7: Y c= REAL+ \/ [:{0},REAL+:] by NUMBERS:def 1,XBOOLE_1:1;
A8: ex x2 being Element of REAL st x2 in Y by A4,SUBSET_1:4;
    thus thesis
    proof
      per cases;
      suppose that
A9:     X misses REAL+ and
A10:    Y misses [:{0},REAL+:];
        take z = 0;
        let x,y such that
A11:    x in X and
A12:    y in Y;
        ( not z in [:{0},REAL+:])& not x in REAL+ by A9,A11,ARYTM_0:5
,ARYTM_2:20,XBOOLE_0:3;
        hence x <= z by XXREAL_0:def 5;
        Y c= REAL+ by A7,A10,XBOOLE_1:73;
        then reconsider y9 = y as Element of REAL+ by A12;
        o <=' y9 by ARYTM_1:6;
        hence thesis by Lm1;
      end;
      suppose
A13:    Y meets [:{0},REAL+:];
        {r: [0,r] in X} c= REAL+
        proof
          let u be set;
          assume u in {r: [0,r] in X};
          then ex r st u = r & [0,r] in X;
          hence thesis;
        end;
        then reconsider X9 = {r: [0,r] in X} as Subset of REAL+;
        {r: [0,r] in Y} c= REAL+
        proof
          let u be set;
          assume u in {r: [0,r] in Y};
          then ex r st u = r & [0,r] in Y;
          hence thesis;
        end;
        then reconsider Y9 = {r : [0,r] in Y} as Subset of REAL+;
        consider e being set such that
A14:    e in Y and
A15:    e in [:{0},REAL+:] by A13,XBOOLE_0:3;
        consider u,y1 being set such that
A16:    u in {0} and
A17:    y1 in REAL+ and
A18:    e = [u,y1] by A15,ZFMISC_1:84;
        reconsider y1 as Element of REAL+ by A17;
        e in REAL by A14;
        then
A19:    [0,y1] in REAL by A16,A18,TARSKI:def 1;
        then reconsider y0 = [0,y1] as real number;
A20:    y0 in Y by A14,A16,A18,TARSKI:def 1;
        then
A21:    y1 in Y9;
A22:    y0 in [:{0},REAL+:] by Lm4,ZFMISC_1:87;
A23:    X c= [:{0},REAL+:]
        proof
          let u be set;
          assume
A24:      u in X;
          then reconsider x = u as real number;
          now
            assume x in REAL+;
            then
A25:        not x in [:{0},REAL+:] by ARYTM_0:5,XBOOLE_0:3;
            x <= y0 & not y0 in REAL+ by A1,A22,A20,A24,ARYTM_0:5,XBOOLE_0:3;
            hence contradiction by A25,XXREAL_0:def 5;
          end;
          hence thesis by A6,A24,XBOOLE_0:def 3;
        end;
        then consider e,x3 being set such that
A26:    e in {0} and
A27:    x3 in REAL+ and
A28:    x1 = [e,x3] by A5,ZFMISC_1:84;
        reconsider x3 as Element of REAL+ by A27;
        x1 = [0,x3] by A26,A28,TARSKI:def 1;
        then
A29:    x3 in X9 by A5;
        for y9,x9 being Element of REAL+ st y9 in Y9 & x9 in X9 holds y9 <=' x9
        proof
          let y9,x9 be Element of REAL+;
          assume y9 in Y9;
          then
A30:      ex r1 st y9 = r1 & [0,r1] in Y;
          assume x9 in X9;
          then
A31:      ex r2 st x9 = r2 & [0,r2] in X;
          then reconsider x = [0,x9], y = [0,y9] as real number by A30;
A32:      x in [:{0},REAL+:] & y in [:{0},REAL+:] by Lm4,ZFMISC_1:87;
          x <= y by A1,A30,A31;
          then consider x99,y99 being Element of REAL+ such that
A33:      x = [0,x99] and
A34:      y = [0,y99] & y99 <=' x99 by A32,XXREAL_0:def 5;
          x9 = x99 by A33,XTUPLE_0:1;
          hence thesis by A34,XTUPLE_0:1;
        end;
        then consider z9 being Element of REAL+ such that
A35:    for y9,x9 being Element of REAL+ st y9 in Y9 & x9 in X9 holds
        y9 <=' z9 & z9 <=' x9 by A29,ARYTM_2:8;
A36:    y1 <> 0 by A19,ARYTM_0:3;
        y1 <=' z9 by A21,A29,A35;
        then [0,z9] in REAL by A36,ARYTM_0:2,ARYTM_1:5;
        then reconsider z = [0,z9] as real number;
        take z;
        let x,y such that
A37:    x in X and
A38:    y in Y;
        consider e,x9 being set such that
A39:    e in {0} and
A40:    x9 in REAL+ and
A41:    x = [e,x9] by A23,A37,ZFMISC_1:84;
        reconsider x9 as Element of REAL+ by A40;
A42:    z in [:{0},REAL+:] by Lm4,ZFMISC_1:87;
A43:    x = [0,x9] by A39,A41,TARSKI:def 1;
        then
A44:    x9 in X9 by A37;
        now
          per cases by A7,A38,XBOOLE_0:def 3;
          suppose
A45:        y in REAL+;
            z9 <=' x9 by A21,A35,A44;
            hence x <= z by A23,A42,A37,A43,Lm1;
A46:        not y in [:{0},REAL+:] by A45,ARYTM_0:5,XBOOLE_0:3;
            not z in REAL+ by A42,ARYTM_0:5,XBOOLE_0:3;
            hence z <= y by A46,XXREAL_0:def 5;
          end;
          suppose
A47:        y in [:{0},REAL+:];
            then consider e,y9 being set such that
A48:        e in {0} and
A49:        y9 in REAL+ and
A50:        y = [e,y9] by ZFMISC_1:84;
            reconsider y9 as Element of REAL+ by A49;
A51:        y = [0,y9] by A48,A50,TARSKI:def 1;
            then y9 in Y9 by A38;
            then y9 <=' z9 & z9 <=' x9 by A35,A44;
            hence thesis by A23,A42,A37,A43,A47,A51,Lm1;
          end;
        end;
        hence thesis;
      end;
      suppose
A52:    X meets REAL+;
        reconsider X9 = X /\ REAL+ as Subset of REAL+ by XBOOLE_1:17;
        consider x1 being set such that
A53:    x1 in X and
A54:    x1 in REAL+ by A52,XBOOLE_0:3;
        reconsider x1 as Element of REAL+ by A54;
        x1 in REAL+;
        then reconsider x0 = x1 as real number by ARYTM_0:1;
A55:    Y c= REAL+
        proof
          let u be set;
          assume
A56:      u in Y;
          then reconsider y = u as real number;
          now
            assume y in [:{0},REAL+:];
            then
A57:        not y in REAL+ by ARYTM_0:5,XBOOLE_0:3;
            x0 <= y & not x0 in [:{0},REAL+:] by A1,A53,A56,ARYTM_0:5
,XBOOLE_0:3;
            hence contradiction by A57,XXREAL_0:def 5;
          end;
          hence thesis by A7,A56,XBOOLE_0:def 3;
        end;
        then reconsider Y9 = Y as Subset of REAL+;
        for x9,y9 being Element of REAL+ st x9 in X9 & y9 in Y9 holds x9 <=' y9
        proof
          let x9,y9 be Element of REAL+;
A58:      X9 c= X by XBOOLE_1:17;
          x9 in REAL+ & y9 in REAL+;
          then reconsider x = x9, y = y9 as real number by ARYTM_0:1
;
          assume x9 in X9 & y9 in Y9;
          then x <= y by A1,A58;
          then ex x9,y9 being Element of REAL+ st x = x9 & y = y9 & x9 <=' y9
          by XXREAL_0:def 5;
          hence thesis;
        end;
        then consider z9 being Element of REAL+ such that
A59:    for x9,y9 being Element of REAL+ st x9 in X9 & y9 in Y9
        holds x9 <=' z9 & z9 <=' y9 by A8,ARYTM_2:8;
        z9 in REAL+;
        then reconsider z = z9 as real number by ARYTM_0:1;
        take z;
        let x,y such that
A60:    x in X and
A61:    y in Y;
        reconsider y9 = y as Element of REAL+ by A55,A61;
A62:    x0 in X9 by A53,XBOOLE_0:def 4;
        now
          per cases by A6,A60,XBOOLE_0:def 3;
          suppose
            x in REAL+;
            then reconsider x9 = x as Element of REAL+;
            x9 in X9 by A60,XBOOLE_0:def 4;
            then x9 <=' z9 & z9 <=' y9 by A59,A61;
            hence thesis by Lm1;
          end;
          suppose
A63:        x in [:{0},REAL+:];
A64:        not z in [:{0},REAL+:] by ARYTM_0:5,XBOOLE_0:3;
            not x in REAL+ by A63,ARYTM_0:5,XBOOLE_0:3;
            hence x <= z by A64,XXREAL_0:def 5;
            z9 <=' y9 by A62,A59,A61;
            hence z <= y by Lm1;
          end;
        end;
        hence thesis;
      end;
    end;
  end;
end;

theorem
  x in NAT & y in NAT implies x + y in NAT
proof
  reconsider x1 = x, y1 = y as Element of REAL by XREAL_0:def 1;
A1: +(x1,y1) = x + y by Lm3;
  assume
A2: x in NAT & y in NAT;
  then ex x9,y9 being Element of REAL+ st x1 = x9 & y1 = y9 & +(x1,y1) = x9 +
  y9 by ARYTM_0:def 1,ARYTM_2:2;
  hence thesis by A1,A2,ARYTM_2:16;
end;

theorem
  for A being Subset of REAL st 0 in A & for x st x in A holds x + 1 in
  A holds NAT c= A
proof
  let A be Subset of REAL such that
A1: 0 in A and
A2: for x st x in A holds x + 1 in A;
  reconsider B = A /\ REAL+ as Subset of REAL+ by XBOOLE_1:17;
A3: B c= A by XBOOLE_1:17;
A4: for x9,y9 being Element of REAL+ st x9 in B & y9 = 1 holds x9 + y9 in B
  proof
    let x9,y9 be Element of REAL+ such that
A5: x9 in B and
A6: y9 = 1;
    x9 in REAL+;
    then reconsider x = x9 as Element of REAL by ARYTM_0:1;
    reconsider xx = x as real number;
    xx+1 in A by A2,A3,A5;
    then (ex x99,y99 being Element of REAL+ st x = x99 & 1 = y99 & +(x,1) =
    x99 + y99 )& +(x,1) in A by Lm3,ARYTM_0:def 1,ARYTM_2:20;
    hence thesis by A6,XBOOLE_0:def 4;
  end;
  0 in B by A1,ARYTM_2:20,XBOOLE_0:def 4;
  then NAT c= B by A4,ARYTM_2:17;
  hence thesis by A3,XBOOLE_1:1;
end;

theorem
  k = { i: i < k }
proof
A1: k in NAT by ORDINAL1:def 12;
  thus k c= { i: i < k }
  proof
    reconsider k9 = k as Element of NAT by ORDINAL1:def 12;
    let e be set;
    assume
A2: e in k;
A3: k9 c= NAT by ORDINAL1:def 2;
    then reconsider j = e as Element of NAT by A2;
    e in NAT by A3,A2;
    then reconsider y9 = e as Element of REAL+ by ARYTM_2:2;
    k9 in NAT;
    then reconsider x9 = k9 as Element of REAL+ by ARYTM_2:2;
    y9 <=' x9 by A2,ARYTM_2:18;
    then
A4: j <= k by Lm1;
    y9 <> x9 by A2;
    then j < k by A4,XXREAL_0:1;
    hence thesis;
  end;
  let e be set;
  assume e in { i: i < k };
  then consider i be Element of NAT such that
A5: e = i and
A6: not k <= i;
  i in NAT;
  then reconsider x9 = e, y9 = k as Element of REAL+ by A5,A1,ARYTM_2:2;
  not y9 <=' x9 by A5,A6,Lm1;
  hence thesis by A5,A6,A1,ARYTM_2:18;
end;














begin

registration
  cluster -> real for Element of REAL;
  coherence;
end;

definition
  mode Real is Element of REAL;
end;

registration
  cluster positive for Real;
  existence
  proof
    take 1;
    thus thesis;
  end;
end;

definition
  let x be Real;
  redefine func -x -> Real;
  coherence by XREAL_0:def 1;
  redefine func x" -> Real;
  coherence by XREAL_0:def 1;
end;

definition
  let x be real number, y be Real;
  redefine func x+y -> Real;
  coherence by XREAL_0:def 1;
  redefine func x*y -> Real;
  coherence by XREAL_0:def 1;
  redefine func x-y -> Real;
  coherence by XREAL_0:def 1;
  redefine func x/y -> Real;
  coherence by XREAL_0:def 1;
end;

definition
  let x be Real, y be real number;
  redefine func x+y -> Real;
  coherence by XREAL_0:def 1;
  redefine func x*y -> Real;
  coherence by XREAL_0:def 1;
  redefine func x-y -> Real;
  coherence by XREAL_0:def 1;
  redefine func x/y -> Real;
  coherence by XREAL_0:def 1;
end;



reserve s,t for Element of RAT+;

theorem
  REAL+ = { r where r is Real: 0 <= r}
 proof set RP = { r where r is Real: 0 <= r};
  thus REAL+ c= RP
   proof let e be set;
    assume
A1:     e in REAL+;
      then reconsider r = e as Real by ARYTM_0:1;
      reconsider o = 0, s = r as Element of REAL+ by ARYTM_2:20,A1;
      o <=' s by ARYTM_1:6;
      then 0 <= r by ARYTM_2:20,XXREAL_0:def 5;
      hence e in RP;
   end;
  let e be set;
  assume e in RP;
   then
A2:   ex r being Real st e = r & 0 <= r;
   not 0 in [:{0},REAL+:] by ARYTM_0:5,XBOOLE_0:3,ARYTM_2:20;
  hence e in REAL+ by A2,XXREAL_0:def 5;
 end;















begin

reserve a,b,c,x,y,z for real number;

scheme
  RealContinuity { P[set], Q[set] } :
  ex z st for x,y st P[x] & Q[y] holds x <= z & z <= y
provided
A1: for x,y st P[x] & Q[y] holds x <= y
proof
  set Y = { z where z is Element of REAL : Q[z] };
  set X = { z where z is Element of REAL : P[z] };
A2: X c= REAL
  proof
    let x be set;
    assume x in X;
    then ex z being Element of REAL st z = x & P[z];
    hence thesis;
  end;
  Y c= REAL
  proof
    let x be set;
    assume x in Y;
    then ex z being Element of REAL st z = x & Q[z];
    hence thesis;
  end;
  then reconsider X, Y as Subset of REAL by A2;
  for x,y being real number st x in X & y in Y holds x <= y
  proof
    let x,y be real number;
    assume that
A3: x in X and
A4: y in Y;
A5: ex z being Element of REAL st z = y & Q[z] by A4;
    ex z being Element of REAL st z = x & P[z] by A3;
    hence thesis by A1,A5;
  end;
  then consider z being real number such that
A6: for x,y being real number st x in X & y in Y holds x <= z & z <= y
  by AXIOMS:1;
  take z;
  let x,y;
  assume that
A7: P[x] and
A8: Q[y];
  y is Element of REAL by XREAL_0:def 1;
  then
A9: y in Y by A8;
  x is Element of REAL by XREAL_0:def 1;
  then x in X by A7;
  hence thesis by A6,A9;
end;

definition
  let x,y be Element of REAL;
  redefine func min(x,y) -> Element of REAL;
  coherence by XREAL_0:def 1;
  redefine func max(x,y) -> Element of REAL;
  coherence by XREAL_0:def 1;
end;

theorem
  min(x,y) + max(x,y) = x + y
proof
  per cases;
  suppose
A1: x <= y;
    then min(x,y) = x by XXREAL_0:def 9;
    hence thesis by A1,XXREAL_0:def 10;
  end;
  suppose
A2: x > y;
    then min(x,y) = y by XXREAL_0:def 9;
    hence thesis by A2,XXREAL_0:def 10;
  end;
end;

theorem
  for x,y being real number st 0 <= x & 0 <= y holds max(x,y) <= x + y
proof
  let x,y be real number;
  assume that
A1: 0 <= x and
A2: 0 <= y;
  now
    per cases by XXREAL_0:16;
    suppose
A3:   max(x,y) = x;
      x + 0 <= x + y by A2,XREAL_1:7;
      hence thesis by A3;
    end;
    suppose
A4:   max(x,y) = y;
      y + 0 <= y + x by A1,XREAL_1:7;
      hence thesis by A4;
    end;
  end;
  hence thesis;
end;

definition
  let x be complex number;

  func x^2 equals
  x * x;
  correctness;
end;

registration
  let x be complex number;
  cluster x^2 -> complex;
  coherence;
end;

registration
  let x;
  cluster x^2 -> real;
  coherence;
end;

definition
  let x be Element of COMPLEX;
  redefine func x^2 -> Element of COMPLEX;
  coherence by XCMPLX_0:def 2;
end;

definition
  let x be Element of REAL;
  redefine func x^2 -> Element of REAL;
  coherence by XREAL_0:def 1;
end;

theorem
  for a being complex number holds a^2 = (-a)^2;

theorem
  for a, b being complex number holds (a + b)^2 = a^2 + 2*a*b + b^2;

theorem
  for a, b being complex number holds (a - b)^2 = a^2 - 2*a*b + b^2;

theorem
  for a being complex number holds (a + 1)^2 = a^2 + 2*a + 1;

theorem
  for a being complex number holds (a - 1)^2 = a^2 - 2*a + 1;

theorem
  for a, b being complex number holds (a - b)*(a + b) = a^2 - b^2;

theorem
  for a, b being complex number holds (a*b)^2 = a^2*b^2;

theorem Th10:
  for a, b being complex number st a^2 - b^2 <> 0 holds
    1/(a+b) = (a-b)/(a^2-b^2)
proof
  let a, b be complex number;
  assume a^2-b^2 <> 0;
  then (a-b) <> 0;
  hence 1/(a+b) = (1*(a-b))/((a+b)*(a-b)) by XCMPLX_1:91
    .= (a-b)/(a^2-b^2);
end;

theorem Th11:
  for a, b being complex number st a^2-b^2 <> 0 holds 1/(a-b) = (a+b)/(a^2-b^2)
proof
  let a, b be complex number;
  assume a^2-b^2 <> 0;
  then (a+b)*(a-b) <> 0;
  then (a+b) <> 0;
  hence 1/(a-b) = (1*(a+b))/((a-b)*(a+b)) by XCMPLX_1:91
    .= (a+b)/(a^2-b^2);
end;

theorem
  0 <> a implies 0 < a^2 by XREAL_1:63;

theorem Th13:
  0 < a & a < 1 implies a^2 < a
proof
  assume that
A1: 0 < a and
A2: a < 1;
  a*a < a*1 by A1,A2,XREAL_1:68;
  hence thesis;
end;

theorem Th14:
  1 < a implies a < a^2
proof
  assume 1 < a;
  then a*1 < a*a by XREAL_1:68;
  hence thesis;
end;

Lm1: 0 < a implies ex x st 0 < x & x^2 < a
proof
  assume
A1: 0 < a;
  per cases;
  suppose
A2: 1 <= a;
    reconsider x = 2" as real number;
    take x;
    thus 0 < x;
    thus thesis by A2,XXREAL_0:2;
  end;
  suppose
A3: a < 1;
    take x = a;
    thus 0 < x by A1;
    thus thesis by A1,A3,Th13;
  end;
end;

theorem Th15:
  0 <= x & x <= y implies x^2 <= y^2
proof
  assume that
A1: 0 <= x and
A2: x <= y;
A3: x*y <= y*y by A1,A2,XREAL_1:64;
  x*x <= x*y by A1,A2,XREAL_1:64;
  hence thesis by A3,XXREAL_0:2;
end;

theorem Th16:
  0 <= x & x < y implies x^2 < y^2
proof
  assume that
A1: 0 <= x and
A2: x < y;
A3: x*y < y*y by A1,A2,XREAL_1:68;
  x*x <= x*y by A1,A2,XREAL_1:64;
  hence thesis by A3,XXREAL_0:2;
end;

Lm2: 0 <= x & 0 <= y & x^2 = y^2 implies x = y
proof
  assume that
A1: 0 <= x and
A2: 0 <= y;
  assume
A3: x^2 = y ^2;
  then
A4: y <= x by A1,Th16;
  x <= y by A2,A3,Th16;
  hence thesis by A4,XXREAL_0:1;
end;

definition
  let a;
  assume
A1: 0 <= a;
  func sqrt a -> real number means
  :Def2:
  0 <= it & it^2 = a;
  existence
  proof
    defpred Y[real number] means 0 <= $1 & a <= $1^2;
    defpred X[real number] means $1 <= 0 or $1^2 <= a;
    a <= a + 1 by XREAL_1:29;
    then
A2: 0 + a <= a^2 + a + (a + 1) by A1,XREAL_1:7;
A3: now
      let x,y such that
A4:   X[x] and
A5:   Y[y];
      per cases;
      suppose
        x <= 0;
        hence x <= y by A5;
      end;
      suppose
        not x <= 0;
        then x^2 <= y^2 by A4,A5,XXREAL_0:2;
        hence x <= y by A5,Th16;
      end;
    end;
    consider z such that
A6: for x,y st X[x] & Y[y] holds x <= z & z <= y from RealContinuity(A3);
    take z;
A7: (a + 1)^2 = a^2 + a + (a + 1);
    hence 0 <= z by A1,A2,A6;
    assume
A8: z^2 <> a;
    now
      per cases by A8,XXREAL_0:1;
      suppose
A9:     z <= 0;
        then z = 0 by A1,A7,A2,A6;
        then ex c st 0 < c & c^2 < a by A1,A8,Lm1;
        hence contradiction by A1,A7,A2,A6,A9;
      end;
      suppose
A10:    z^2 < a & not z <= 0;
        set b = a - z^2;
A11:    0 < b by A10,XREAL_1:50;
        then consider c such that
A12:    0 < c and
A13:    c^2 < b/2 by Lm1;
        set eps = min(c, b/(4*z));
A14:    0 < eps by A10,A11,A12,XXREAL_0:15;
        then
A15:    z < z + eps by XREAL_1:29;
        eps*(2*z) <= b/(2*(2*z))*(2*z) by A10,XREAL_1:64,XXREAL_0:17;
        then eps*(2*z) <= b/2/(2*z)*(2*z) by XCMPLX_1:78;
        then
A16:    2*z*eps <= b/2 by A10,XCMPLX_1:87;
        eps^2 <= c^2 by A14,Th15,XXREAL_0:17;
        then eps^2 <= b/2 by A13,XXREAL_0:2;
        then
A17:    2*z*eps + eps^2 <= b/2 + b/2 by A16,XREAL_1:7;
A18:    (z + eps)^2 = z^2 + (2*z*eps + eps^2);
        a = z^2 + b;
        then (z + eps)^2 <= a by A18,A17,XREAL_1:6;
        hence contradiction by A1,A7,A2,A6,A15;
      end;
      suppose
A19:    a < z^2 & not z <= 0;
        set b = z^2 - a;
        set eps = min(b/(2*z),z);
A20:    (z - eps)^2 = z^2 - (2*z*eps - eps^2);
        0 < b by A19,XREAL_1:50;
        then 0 < eps by A19,XXREAL_0:15;
        then
A21:    z - eps < z by XREAL_1:44;
        0 <= eps^2 by XREAL_1:63;
        then
A22:    2*z*eps - eps^2 <= 2*z*eps - 0 by XREAL_1:13;
        eps*(2*z) <= b/(2*z)*(2*z) by A19,XREAL_1:64,XXREAL_0:17;
        then 2*z*eps <= b by A19,XCMPLX_1:87;
        then
A23:    2*z*eps - eps^2 <= b by A22,XXREAL_0:2;
A24:    0 <= z - eps by XREAL_1:48,XXREAL_0:17;
        a = z^2 - b;
        then a <= (z - eps)^2 by A20,A23,XREAL_1:13;
        hence contradiction by A6,A24,A21;
      end;
    end;
    hence contradiction;
  end;
  uniqueness by Lm2;
end;

definition
  let a be Element of REAL;
  redefine func sqrt a -> Element of REAL;
  coherence by XREAL_0:def 1;
end;

theorem Th17:
  sqrt 0 = 0
proof
  sqrt 0^2 = 0 by Def2;
  hence thesis;
end;

theorem Th18:
  sqrt 1 = 1
proof
  sqrt 1^2 = 1 by Def2;
  hence thesis;
end;

Lm3: 0 <= x & x < y implies sqrt x < sqrt y
proof
  assume that
A1: 0 <= x and
A2: x < y;
A3: (sqrt y)^2 = y by A1,A2,Def2;
A4: (sqrt x)^2 = x by A1,Def2;
  0 <= sqrt y by A1,A2,Def2;
  hence thesis by A2,A4,A3,Th15;
end;

theorem
  1 < sqrt 2 by Lm3,Th18;

Lm4: 2^2 = 2*2;

theorem Th20:
  sqrt 4 = 2 by Def2,Lm4;

theorem
  sqrt 2 < 2 by Lm3,Th20;

theorem Th22:
  0 <= a implies sqrt a^2 = a by Def2;

theorem
  a <= 0 implies sqrt a^2 = -a
proof
A1: a^2 = (-a)^2;
  assume a <= 0;
  hence thesis by A1,Def2;
end;

theorem Th24:
  0 <= a & sqrt a = 0 implies a = 0
proof
  0 <= a & sqrt a = 0 implies a = 0^2 by Def2;
  hence thesis;
end;

theorem Th25:
  0 < a implies 0 < sqrt a
proof
  assume
A1: 0 < a;
  then sqrt a <> 0^2 by Def2;
  hence thesis by A1,Def2;
end;

theorem Th26:
  0 <= x & x <= y implies sqrt x <= sqrt y
proof
  per cases;
  suppose
    x = y;
    hence thesis;
  end;
  suppose
A1: x <> y;
    assume
A2: 0 <= x;
    assume x <= y;
    then x < y by A1,XXREAL_0:1;
    hence thesis by A2,Lm3;
  end;
end;

theorem
  0 <= x & x < y implies sqrt x < sqrt y by Lm3;

theorem Th28:
  0 <= x & 0 <= y & sqrt x = sqrt y implies x = y
proof
  assume that
A1: 0 <= x and
A2: 0 <= y and
A3: sqrt x = sqrt y;
  assume x <> y;
  then x < y or y < x by XXREAL_0:1;
  hence contradiction by A1,A2,A3,Lm3;
end;

theorem Th29:
  0 <= a & 0 <= b implies sqrt (a*b) = sqrt a * sqrt b
proof
  assume that
A1: 0 <= a and
A2: 0 <= b;
A3: 0 <= sqrt a by A1,Def2;
A4: 0 <= sqrt b by A2,Def2;
  (sqrt(a*b))^2 = a*b by A1,A2,Def2
    .= (sqrt a)^2*b by A1,Def2
    .= (sqrt a)^2*(sqrt b)^2 by A2,Def2
    .= (sqrt a*sqrt b)^2;
  hence sqrt(a*b) = sqrt(sqrt a*sqrt b)^2 by A1,A2,Def2
    .= sqrt a*sqrt b by A3,A4,Def2;
end;

theorem Th30:
  0 <= a & 0 <= b implies sqrt (a/b) = sqrt a/sqrt b
proof
  assume that
A1: 0 <= a and
A2: 0 <= b;
A3: (sqrt b)^2 = b by A2,Def2;
  (sqrt a)^2 = a by A1,Def2;
  then
A4: (sqrt a/sqrt b)^2 = a/b by A3,XCMPLX_1:76;
A5: 0 <= sqrt b by A2,Def2;
  0 <= sqrt a by A1,Def2;
  hence thesis by A5,A4,Def2;
end;

theorem
  for a,b being real number st 0 <= a & 0 <= b holds
    sqrt(a + b) = 0 iff a = 0 & b = 0 by Th17,Th24;

theorem
  0 < a implies sqrt (1/a) = 1/sqrt a by Th18,Th30;

theorem
  0 < a implies sqrt a/a = 1/sqrt a
proof
  assume
A1: 0 < a;
  then sqrt a <> 0^2 by Def2;
  hence sqrt a/a = (sqrt a)^2/(a*sqrt a) by XCMPLX_1:91
    .= (1*a)/(sqrt a*a) by A1,Def2
    .= 1/sqrt a by A1,XCMPLX_1:91;
end;

theorem
  0 < a implies a / sqrt a = sqrt a
proof
  assume
A1: 0 < a;
  then sqrt a <> 0^2 by Def2;
  hence a /sqrt a = (a*sqrt a) /(sqrt a)^2 by XCMPLX_1:91
    .= (sqrt a*a) /(1*a) by A1,Def2
    .= sqrt a/1 by A1,XCMPLX_1:91
    .= sqrt a;
end;

theorem
  0 <= a & 0 <= b implies (sqrt a - sqrt b)*(sqrt a + sqrt b) = a - b
proof
  assume that
A1: 0 <= a and
A2: 0 <= b;
  thus (sqrt a - sqrt b)*(sqrt a + sqrt b) = (sqrt a)^2 - (sqrt b)^2
    .= a - (sqrt b)^2 by A1,Def2
    .= a - b by A2,Def2;
end;

Lm5: 0 <= a & 0 <= b & a <> b implies (sqrt a)^2-(sqrt b)^2 <> 0
proof
  assume that
A1: 0 <= a and
A2: 0 <= b and
A3: a <> b;
A4: 0 <= sqrt a by A1,Def2;
A5: 0 <= sqrt b by A2,Def2;
  sqrt a <> sqrt b by A1,A2,A3,Th28;
  hence thesis by A4,A5,Lm2;
end;

theorem
  0 <= a & 0 <= b & a <>b implies 1/(sqrt a+sqrt b) = (sqrt a - sqrt b)/ (a-b)
proof
  assume that
A1: 0 <= a and
A2: 0 <= b and
A3: a <>b;
  thus 1/(sqrt a+sqrt b) = (sqrt a - sqrt b)/((sqrt a)^2-(sqrt b)^2) by A1,A2
,A3,Lm5,Th10
    .= (sqrt a - sqrt b)/(a-(sqrt b)^2) by A1,Def2
    .= (sqrt a - sqrt b)/(a-b) by A2,Def2;
end;

theorem
  0 <= b & b < a implies 1/(sqrt a+sqrt b) = (sqrt a - sqrt b)/(a-b)
proof
  assume that
A1: 0 <= b and
A2: b < a;
  thus 1/(sqrt a+sqrt b) = (sqrt a - sqrt b)/((sqrt a)^2-(sqrt b)^2) by A1,A2
,Lm5,Th10
    .= (sqrt a - sqrt b)/(a-(sqrt b)^2) by A1,A2,Def2
    .= (sqrt a - sqrt b)/(a-b) by A1,Def2;
end;

theorem
  0 <= a & 0 <= b implies 1/(sqrt a-sqrt b) = (sqrt a + sqrt b)/(a-b)
proof
  assume that
A1: 0 <= a and
A2: 0 <= b;
  per cases;
  suppose
    a <> b;
    hence 1/(sqrt a-sqrt b) = (sqrt a + sqrt b)/((sqrt a)^2-(sqrt b)^2) by A1
,A2,Lm5,Th11
      .= (sqrt a + sqrt b)/(a-(sqrt b)^2) by A1,Def2
      .= (sqrt a + sqrt b)/(a-b) by A2,Def2;
  end;
  suppose
A3: a = b;
    then 1/(sqrt a-sqrt b) = 0;
    hence thesis by A3;
  end;
end;

theorem
  0 <= b & b < a implies 1/(sqrt a-sqrt b) = (sqrt a + sqrt b)/(a-b)
proof
  assume that
A1: 0 <= b and
A2: b < a;
  thus 1/(sqrt a-sqrt b) = (sqrt a + sqrt b)/((sqrt a)^2-(sqrt b)^2) by A1,A2
,Lm5,Th11
    .= (sqrt a + sqrt b)/(a-(sqrt b)^2) by A1,A2,Def2
    .= (sqrt a + sqrt b)/(a-b) by A1,Def2;
end;

theorem
  for x,y being complex number st x^2=y^2 holds x=y or x=-y
proof
  let x,y be complex number;
  assume x^2=y^2;
  then (x-y)*(x+y)=0;
  then x-y=0 or x+y=0;
  hence thesis;
end;

theorem
  for x being complex number st x^2=1 holds x=1 or x=-1
proof
  let x be complex number;
  assume x^2=1;
  then (x-1)*(x+1)=0;
  then x-1=0 or x+1=0;
  hence thesis;
end;

theorem
  0<=x & x<=1 implies x^2<=x
proof
  assume that
A1: 0<=x and
A2: x<=1;
  per cases by A1;
  suppose
    0=x;
    hence thesis;
  end;
  suppose
A3: 0<x;
    per cases by A2,XXREAL_0:1;
    suppose
      x=1;
      hence thesis;
    end;
    suppose
      x<1;
      hence thesis by A3,Th13;
    end;
  end;
end;

theorem
  x^2-1<=0 implies -1<=x & x<=1
proof
  assume x^2-1<=0;
  then (x-1)*(x+1)<=0;
  hence thesis by XREAL_1:93;
end;

begin 


theorem Th44:
  a <= 0 & x < a implies x^2 > a^2
proof
  assume that
A1: a<=0 and
A2: x<a;
  -x>-a by A2,XREAL_1:24;
  then (-x)^2>(-a)^2 by A1,Th16;
  hence thesis;
end;



theorem
  -1 >= a implies -a <= a^2
proof
  assume -1 >= a;
  then --1<=-a by XREAL_1:24;
  then -a<= (-a)^2 by XREAL_1:151;
  hence thesis;
end;

theorem
  -1 > a implies -a < a^2
proof
  assume -1 > a;
  then --1< -a by XREAL_1:24;
  then -a< (-a)^2 by Th14;
  hence thesis;
end;

theorem
  b^2 <= a^2 & a >= 0 implies -a <= b & b <= a
proof
  assume that
A1: b^2<= a^2 and
A2: a>=0;
  now
    assume
A3: -a>b or b>a;
    now
      per cases by A3;
      case
        -a>b;
        then --a<-b by XREAL_1:24;
        then a^2<(-b)^2 by A2,Th16;
        hence contradiction by A1;
      end;
      case
        b>a;
        hence contradiction by A1,A2,Th16;
      end;
    end;
    hence contradiction;
  end;
  hence thesis;
end;

theorem
  b^2 < a^2 & a >= 0 implies -a < b & b < a
proof
  assume that
A1: b^2< a^2 and
A2: a>=0;
  now
    assume
A3: -a>=b or b>=a;
    now
      per cases by A3;
      case
        -a>=b;
        then --a<= -b by XREAL_1:24;
        then a^2<=(-b)^2 by A2,Th15;
        hence contradiction by A1;
      end;
      case
        b>=a;
        hence contradiction by A1,A2,Th15;
      end;
    end;
    hence contradiction;
  end;
  hence thesis;
end;

theorem Th49:
  -a <= b & b <= a implies b^2 <= a^2
proof
  assume that
A1: -a<=b and
A2: b<=a;
  per cases;
  suppose
    b>=0;
    hence thesis by A2,Th15;
  end;
  suppose
A3: b<0;
    --a>=-b by A1,XREAL_1:24;
    then (-b)^2<= a^2 by A3,Th15;
    hence thesis;
  end;
end;

theorem
  -a < b & b < a implies b^2 < a^2
proof
  assume that
A1: -a<b and
A2: b<a;
  per cases;
  suppose
    b>=0;
    hence thesis by A2,Th16;
  end;
  suppose
A3: b<0;
    --a>-b by A1,XREAL_1:24;
    then (-b)^2< a^2 by A3,Th16;
    hence thesis;
  end;
end;



theorem
  a^2 <= 1 implies -1 <= a & a <= 1
proof
  assume a^2<=1;
  then a^2-1^2<=1^2-1^2 by XREAL_1:9;
  then (a-1)*(a+1)<=0;
  hence thesis by XREAL_1:93;
end;

theorem
  a^2 < 1 implies -1 < a & a < 1
proof
  assume a^2<1;
  then a^2-1^2<1^2-1^2 by XREAL_1:9;
  then (a-1)*(a+1)<0;
  hence thesis by XREAL_1:94;
end;



theorem Th53:
  -1 <= a & a <= 1 & -1 <= b & b <= 1 implies a^2*b^2 <= 1
proof
  assume that
A1: -1<=a and
A2: a<=1 and
A3: -1<=b and
A4: b<=1;
A5: 0<=b^2 by XREAL_1:63;
  a^2<=1^2 by A1,A2,Th49;
  then
A6: a^2*b^2 <= 1*b^2 by A5,XREAL_1:64;
  b^2<=1^2 by A3,A4,Th49;
  hence thesis by A6,XXREAL_0:2;
end;

theorem Th54:
  a >= 0 & b >= 0 implies a*sqrt(b) = sqrt(a^2*b)
proof
  assume that
A1: a>=0 and
A2: b>=0;
  sqrt(a^2)=a by A1,Def2;
  hence thesis by A1,A2,Th29;
end;

Lm6: -1<=a & a<=1 & -1<=b & b<=1 implies (1+a^2)*b^2<=1+b^2
proof
  assume that
A1: -1<=a and
A2: a<=1 and
A3: -1<=b and
A4: b<=1;
  a^2*b^2<=1 by A1,A2,A3,A4,Th53;
  then 1*b^2+a^2*b^2<=1+b^2 by XREAL_1:7;
  hence thesis;
end;

theorem Th55:
  -1 <= a & a <= 1 & -1 <= b & b <= 1 implies (-b)*sqrt(1+a^2) <=
  sqrt(1+b^2) & -sqrt(1+b^2) <= b*sqrt(1+a^2)
proof
  assume that
A1: -1<=a and
A2: a<=1 and
A3: -1<=b and
A4: b<=1;
A5: a^2>=0 by XREAL_1:63;
  then
A6: 1+a^2>=1+0 by XREAL_1:7;
  b^2>=0 by XREAL_1:63;
  then
A7: sqrt(1+b^2)>=0 by Def2;
A8: sqrt(1+a^2)>=0 by A5,Def2;
A9: now
    per cases;
    suppose
      b>=0;
      hence (-b)*sqrt(1+a^2)<=sqrt(1+b^2) by A8,A7;
    end;
    suppose
A10:  b<0;
A11:  (-b)^2>=0 by XREAL_1:63;
      (-b)*sqrt(1+a^2)=sqrt((-b)^2*(1+a^2)) by A5,A10,Th54;
      hence (-b)*sqrt(1+a^2)<=sqrt(1+b^2) by A1,A2,A3,A4,A6,A11,Lm6,Th26;
    end;
  end;
  then -((-b)*sqrt(1+a^2)) >= - sqrt(1+b^2) by XREAL_1:24;
  hence thesis by A9;
end;

theorem
  -1 <= a & a <= 1 & -1 <= b & b <= 1 implies b*sqrt(1+a^2) <= sqrt(1+b^2)
proof
  assume that
A1: -1<=a and
A2: a<=1 and
A3: -1<=b and
A4: b<=1;
A5: -1<=-b by A4,XREAL_1:24;
  --1>=-b by A3,XREAL_1:24;
  then (--b)*sqrt(1+a^2)<=sqrt(1+(-b)^2) by A1,A2,A5,Th55;
  hence thesis;
end;

Lm7: b<=0 & a<=b implies a*sqrt(1+b^2)<= b*sqrt(1+a^2)
proof
  assume that
A1: b<=0 and
A2: a<=b;
A3: (-a)*sqrt(1+b^2)=sqrt((-a)^2*(1+b^2)) by A1,A2,Th54;
  a<b or a=b by A2,XXREAL_0:1;
  then b^2<a^2 or a=b by A1,Th44;
  then
A4: b^2*1+b^2*a^2<=a^2*1+a^2*b^2 by XREAL_1:7;
A5: (b)^2>=0 by XREAL_1:63;
A6: a^2>=0 by XREAL_1:63;
  then (-b)*sqrt(1+a^2)=sqrt((-b)^2*(1+a^2)) by A1,Th54;
  then -(a*sqrt(1+b^2))>= -(b*sqrt(1+a^2)) by A6,A3,A4,A5,Th26;
  hence thesis by XREAL_1:24;
end;

Lm8: for a,b being real number st a<=0 & a<=b holds a*sqrt(1+b^2)<= b*sqrt(1+a
^2)
proof
  let a,b be real number;
  assume that
A1: a<=0 and
A2: a<=b;
  now
    per cases;
    case
      b<=0;
      hence thesis by A2,Lm7;
    end;
    case
A3:   b>0;
      (b)^2 >=0 by XREAL_1:63;
      then sqrt(1+(b)^2)>0 by Th25;
      then
A4:   a*sqrt(1+b^2)<=0 by A1;
      (a)^2 >=0 by XREAL_1:63;
      then sqrt(1+(a)^2)>0 by Th25;
      hence thesis by A3,A4;
    end;
  end;
  hence thesis;
end;

Lm9: for a,b being real number st a>=0 & a>=b holds a*sqrt(1+b^2)>= b*sqrt(1+a
^2)
proof
  let a,b be real number;
  assume that
A1: a>=0 and
A2: a>=b;
  -a <= -b by A2,XREAL_1:24;
  then (-a)*sqrt(1+(-b)^2)<= (-b)*sqrt(1+(-a)^2) by A1,Lm8;
  then -(a*sqrt(1+(b)^2))<= -(b*sqrt(1+(a)^2));
  hence thesis by XREAL_1:24;
end;

theorem
  a >= b implies a*sqrt(1+b^2) >= b*sqrt(1+a^2)
proof
  assume
A1: a>=b;
  per cases;
  suppose
    a>=0;
    hence thesis by A1,Lm9;
  end;
  suppose
    a<0;
    hence thesis by A1,Lm7;
  end;
end;

theorem
 a >= 0 implies sqrt(a+b^2) >= b
proof
 assume
A1: a >= 0;
 per cases;
 suppose b < 0;
  hence thesis by A1,Def2;
 end;
 suppose
A2: b >= 0;
A3: b^2 >= 0 by XREAL_1:63;
  a+b^2 >= 0+b^2 by A1,XREAL_1:6;
  then sqrt(a+b^2) >= sqrt(b^2) by A3,Th26;
  hence sqrt(a+b^2) >= b by A2,Def2;
 end;
end;



theorem
  0 <= a & 0 <= b implies sqrt(a+b) <= sqrt a + sqrt b
proof
  assume that
A1: 0 <= a and
A2: 0 <= b;
A3: 0 <= sqrt a by A1,Def2;
  0 <= sqrt(a*b) by A1,A2,Def2;
  then 0 <= sqrt a*sqrt b by A1,A2,Th29;
  then 0 <= 2*(sqrt a*sqrt b);
  then a + 0 <= a + 2*sqrt a*sqrt b by XREAL_1:6;
  then
A4: a + b <= a + 2*sqrt a*sqrt b + b by XREAL_1:6;
A5: 0 <= sqrt b by A2,Def2;
  sqrt(a + 2*sqrt a*sqrt b + b) = sqrt((sqrt a)^2 + 2*sqrt a*sqrt b + b)
  by A1,Def2
    .= sqrt((sqrt a)^2 + 2*sqrt a*sqrt b + (sqrt b)^2) by A2,Def2
    .= sqrt((sqrt a + sqrt b)^2)
    .= sqrt a + sqrt b by A3,A5,Th22;
  hence thesis by A1,A2,A4,Th26;
end;















begin

reserve x for Real,
  p,k,l,m,n,s,h,i,j,k1,t,t1 for Nat,
  X for Subset of REAL;



theorem Th1:
  for X st 0 in X & for x st x in X holds x + 1 in X for n holds n in X
proof
  let A be Subset of REAL such that
A1: 0 in A;
  assume x in A implies x + 1 in A;
  then for x being real number st x in A holds x + 1 in A;
  then
A2: NAT c= A by A1,AXIOMS:3;
  let n;
  n in NAT by ORDINAL1:def 12;
  hence thesis by A2;
end;





registration
  let n,k be Nat;
  cluster n + k -> natural;
  coherence
  proof
    defpred P[Real] means ex k st $1 = k & n + k is natural;
    consider X such that
A1: x in X iff P[x] from SUBSET_1:sch 3;
A2: now
      let x;
      assume x in X;
      then consider k such that
A3:   x = k and
A4:   n + k is natural by A1;
A5:   (n + k) + 1 = n + (k + 1);
      reconsider k as Element of NAT by ORDINAL1:def 12;
      n + k is Element of NAT by A4,ORDINAL1:def 12;
      then k + 1 is Element of NAT & (n + k) + 1 is Element of NAT by AXIOMS:2
;
      hence x + 1 in X by A1,A3,A5;
    end;
    n + 0 = n;
    then 0 in X by A1;
    then k in X by A2,Th1;
    then ex m st k = m & n + m is natural by A1;
    hence thesis;
  end;
end;

definition
  let n be Nat, k be Element of NAT;
  redefine func n + k -> Element of NAT;
  coherence by ORDINAL1:def 12;
end;

definition
  let n be Element of NAT, k be Nat;
  redefine func n + k -> Element of NAT;
  coherence by ORDINAL1:def 12;
end;




scheme
  Ind { P[Nat] } : for k being Element of NAT holds P[k]
provided
A1: P[0] and
A2: for k being Element of NAT st P[k] holds P[k + 1]
proof
  let k;
  defpred P1[Real] means ex k st P[k] & k = $1;
  consider X such that
A3: x in X iff P1[x] from SUBSET_1:sch 3;
A4: for x st x in X holds x + 1 in X
  proof
    let x;
    assume x in X;
    then consider k such that
A5: P[k] and
A6: k = x by A3;
    reconsider k as Element of NAT by ORDINAL1:def 12;
    P[k + 1] by A2,A5;
    hence thesis by A3,A6;
  end;
  0 in X by A1,A3;
  then k in X by A4,Th1;
  then ex n st P[n] & n = k by A3;
  hence thesis;
end;

scheme
  NatInd { P[Nat] } : for k being Nat holds P[k]
provided
A1: P[0] and
A2: for k be Nat st P[k] holds P[k + 1]
proof
A3: for k be Element of NAT st P[k] holds P[k + 1] by A2;
  let k be Nat;
A4: k is Element of NAT by ORDINAL1:def 12;
A5: P[0] by A1;
  for k be Element of NAT holds P[k] from Ind(A5,A3);
  hence thesis by A4;
end;




registration
  let n,k be Nat;
  cluster n * k -> natural;
  coherence
  proof
    defpred P[Nat] means n*$1 is natural;
A1: for m be Nat holds P[m] implies P[m+1]
    proof
      let m be Nat;
      assume P[m];
      then reconsider k = n * m as Nat;
      k + n is Nat;
      hence thesis;
    end;
A2: P[0];
    for m be Nat holds P[m] from NatInd(A2,A1);
    hence thesis;
  end;
end;

definition
  let n be Nat, k be Element of NAT;
  redefine func n * k -> Element of NAT;
  coherence by ORDINAL1:def 12;
end;

definition
  let n be Element of NAT, k be Nat;
  redefine func n * k -> Element of NAT;
  coherence by ORDINAL1:def 12;
end;





theorem Th2:
  0 <= i
proof
  defpred P[Nat] means 0 <= $1;
A1: for n st P[n] holds P[n+1];
A2: P[0];
  for k holds P[k] from NatInd(A2,A1);
  hence thesis;
end;

theorem
  0 <> i implies 0 < i by Th2;

theorem
  i <= j implies i * h <= j * h
proof
  assume
A1: i <= j;
  0 <= h by Th2;
  hence thesis by A1,XREAL_1:64;
end;

theorem
  0 < i + 1
proof
  assume
A1: 0 >= i + 1;
  0 <= i by Th2;
  hence contradiction by A1;
end;

theorem Th6:
  i = 0 or ex k st i = k + 1
proof
  defpred P[Nat] means $1 = 0 or ex k st $1 = k + 1;
A1: P[h] implies P[h + 1];
A2: P[0];
  for i holds P[i] from NatInd(A2,A1);
  hence thesis;
end;

theorem Th7:
  i + j = 0 implies i = 0 & j = 0
proof
  assume
A1: i + j = 0;
  0 <= i & 0 <= j by Th2;
  hence thesis by A1;
end;

registration
  cluster non zero for Nat;
  existence
  proof
    take 1;
    thus thesis;
  end;
end;

registration
  let m be Nat, n be non zero Nat;
  cluster m + n -> non zero;
  coherence by Th7;
  cluster n + m -> non zero;
  coherence;
end;

scheme
  DefbyInd { N()->Nat, F(Nat,Nat)->Nat, P[Nat,Nat] } : (for k being Nat ex n
  being Nat st P[k,n]) & for k,n,m being Nat st P[k,n] & P[k,m] holds n = m
provided
A1: for k,n being Nat holds P[k,n] iff k = 0 & n = N() or ex m,l being
Nat st k = m + 1 & P[m,l] & n = F(k,l)
proof
  reconsider N = N() as Element of NAT by ORDINAL1:def 12;
  defpred P[Nat] means ex n being Nat st P[$1,n];
A2: for k being Nat holds P[k] implies P[k+1]
  proof
    let k be Nat;
    given n being Nat such that
A3: P[k,n];
    reconsider F = F(k+1,n) as Element of NAT by ORDINAL1:def 12;
    take F;
    thus thesis by A1,A3;
  end;
  P[0,N] by A1;
  then
A4: P[0];
  thus for k being Nat holds P[k] from NatInd(A4,A2);
  defpred P[Nat] means for n,m being Nat st P[$1,n] & P[$1,m] holds n = m;
A5: for k being Nat st P[k] holds P[k+1]
  proof
    let k be Nat such that
A6: for n,m being Nat st P[k,n] & P[k,m] holds n = m;
    let n,m be Nat;
    assume ( P[k+1,n])& P[k+1,m];
    then (ex l,u being Nat st k + 1 = l + 1 &( P[l,u])& n = F(k + 1 ,u) )& ex
    v,w being Nat st k + 1 = v + 1 & P[v,w] & m = F(k + 1,w) by A1;
    hence thesis by A6;
  end;
A7: P[0]
  proof
    let n,m be Nat such that
A8: P[0,n] and
A9: P[0,m];
    not ex m,l being Nat st 0 = m + 1 & P[m,l] & n = F(0,l);
    then ( not ex n,l being Nat st 0 = n + 1 & P[n,l] & m = F(0,l))& n = N()
    by A1,A8;
    hence thesis by A1,A9;
  end;
  thus for k being Nat holds P[k] from NatInd(A7,A5);
end;

theorem Th8:
  i <= j + 1 implies i <= j or i = j + 1
proof
  defpred P[Nat] means for j holds $1 <= j + 1 implies $1 <= j or $1 = j+1;
A1: for i st P[i] holds P[i+1]
  proof
    let i such that
A2: for j holds i <= j + 1 implies i <= j or i = j + 1;
    let j;
    assume
A3: i + 1 <= j + 1;
A4: now
      given k such that
A5:   j = k + 1;
      i <= k + 1 by A3,A5,XREAL_1:6;
      then i <= k or i = k + 1 by A2;
      hence thesis by A5,XREAL_1:6;
    end;
    now
A6:   0 <= i by Th2;
      assume
A7:   j = 0;
      then i <= 0 by A3,XREAL_1:6;
      hence thesis by A7,A6;
    end;
    hence thesis by A4,Th6;
  end;
A8: P[0] by Th2;
  for i holds P[i] from NatInd(A8,A1);
  hence thesis;
end;

theorem Th9:
  i <= j & j <= i + 1 implies i = j or j = i + 1
proof
  assume that
A1: i <= j and
A2: j <= i + 1;
  j <= i or j = i + 1 by A2,Th8;
  hence thesis by A1,XXREAL_0:1;
end;

theorem Th10:
  i <= j implies ex k st j = i + k
proof
  defpred P[Nat] means i <= $1 implies ex k st $1 = i + k;
A1: for j st P[j] holds P[j+1]
  proof
    let j such that
A2: i <= j implies ex k st j = i + k;
A3: now
      assume i <= j;
      then consider k such that
A4:   j = i + k by A2;
      i + k + 1 = i + (k + 1);
      hence thesis by A4;
    end;
A5: now
      assume i = j + 1;
      then j + 1 = i + 0;
      hence thesis;
    end;
    assume i <= j + 1;
    hence thesis by A3,A5,Th8;
  end;
A6: P[0]
  proof
    assume
A7: i <= 0;
    take 0;
    thus thesis by A7,Th2;
  end;
  for j holds P[j] from NatInd(A6,A1);
  hence thesis;
end;

theorem Th11:
  i <= i + j
proof
  0 + i = i & 0 <= j by Th2;
  hence thesis by XREAL_1:7;
end;

scheme
  CompInd { P[Nat] } : for k being Nat holds P[k]
provided
A1: for k being Nat st for n being Nat st n < k holds P[n] holds P[k]
proof
  let k be Nat;
  defpred P1[Nat] means for n being Nat st n < $1 holds P[n];
A2: for k be Nat holds P1[k] implies P1[k+1]
  proof
    let k be Nat;
    assume
A3: for n being Nat st n < k holds P[n];
    let n be Nat;
    assume n < k + 1;
    then n <= k by Th8;
    then n < k or n = k & n <= k by XXREAL_0:1;
    hence thesis by A1,A3;
  end;
A4: P1[0] by Th2;
  for k being Nat holds P1[k] from NatInd(A4,A2);
  then for n being Nat st n < k holds P[n];
  hence thesis by A1;
end;



scheme
 Min { P[Nat] } : ex k being Nat st P[k] & for n being Nat st P[n] holds k <= n
provided
A1: ex k being Nat st P[k]
proof
  defpred P1[Nat] means not P[$1];
  assume
A2: not thesis;
A3: for k being Nat st for n being Nat st n < k holds P1[n] holds P1[k]
  proof
    let k be Nat;
    assume
A4: for n being Nat st n < k holds not P[n];
    not (ex n being Nat st P[n] & not k <= n) implies not P[k] by A2;
    hence thesis by A4;
  end;
  for k be Nat holds P1[k] from CompInd(A3);
  hence contradiction by A1;
end;



scheme
  Max { P[Nat], N()->Nat } : ex k being Nat st P[k] & for n being Nat st P[n]
  holds n <= k
provided
A1: for k being Nat st P[k] holds k <= N() and
A2: ex k being Nat st P[k]
proof
  defpred P1[Nat] means for n being Nat st P[n] holds n <= $1;
A3: ex k being Nat st P1[k] by A1;
  consider k being Nat such that
A4: P1[k] and
A5: for m being Nat st P1[m] holds k <= m from Min(A3);
  take k;
  thus P[k]
  proof
    consider n being Nat such that
A6: P[n] by A2;
A7: n <= k by A4,A6;
    assume
A8: not P[k];
    then n <> k by A6;
    then k <> 0 by A7,Th2;
    then consider m such that
A9: k = m + 1 by Th6;
    for n being Nat st P[n] holds n <= m by A4,A8,A9,Th8;
    then
A10: k <= m by A5;
    -m + m + 1 = -m + (m + 1);
    hence contradiction by A9,A10,XREAL_1:6;
  end;
  thus thesis by A4;
end;

theorem Th12:
  i <= j implies i <= j + h
proof
  assume i <= j;
  then
A1: i + h <= j + h by XREAL_1:7;
  0 <= h by Th2;
  then i + 0 <= i + h by XREAL_1:7;
  hence thesis by A1,XXREAL_0:2;
end;

theorem Th13:
  i < j + 1 iff i <= j
proof
  thus i < j + 1 implies i <= j by Th8;
  assume
A1: i <= j;
A2: now
A3: j + (- j) = 0 & j + 1 + (- j) = 1;
    assume i = j + 1;
    hence contradiction by A1,A3,XREAL_1:6;
  end;
  i <= j + 1 by A1,Th12;
  hence thesis by A2,XXREAL_0:1;
end;

theorem Th14:
  j < 1 implies j = 0
proof
  assume j < 1;
  then j < 0 + 1;
  then
A1: j <= 0 by Th13;
  assume j <> 0;
  hence thesis by A1,Th2;
end;

theorem
  i * j = 1 implies i = 1
proof
  assume
A1: i * j = 1;
  then i <> 0;
  then consider m such that
A2: i = m + 1 by Th6;
  j <> 0 by A1;
  then consider l such that
A3: j = l + 1 by Th6;
A4: m * l + m + l + 1 = 0 + 1 by A1,A2,A3;
  then m * l + m = 0;
  hence thesis by A2,A4;
end;

theorem Th16:
  k <> 0 implies n < n + k
proof
  assume k <> 0;
  then
A1: n <> n + k;
  n <= n + k by Th12;
  hence thesis by A1,XXREAL_0:1;
end;

scheme
  Regr { P[Nat] } : P[0]
provided
A1: ex k being Nat st P[k] and
A2: for k being Nat st k <> 0 & P[k] ex n being Nat st n < k & P[n]
proof
  consider k being Nat such that
A3: P[k] & for n being Nat st P[n] holds k <= n from Min(A1);
  now
    assume k <> 0;
    then ex n being Nat st n < k & P[n] by A2,A3;
    hence contradiction by A3;
  end;
  hence thesis by A3;
end;



theorem
  0 < m implies for n ex k,t st n = (m*k)+t & t < m
proof
  defpred P[Nat] means ex k,t st $1 = (m*k)+t & t < m;
  assume
A1: 0 < m;
A2: for n st P[n] holds P[n+1]
  proof
    let n;
    given k1,t1 such that
A3: n = (m*k1)+t1 and
A4: t1 < m;
A5: t1+1 < m implies ex k,t st n+1 = (m*k)+t & t < m
    proof
      assume
A6:   t1+1 < m;
      take k = k1, t = t1+1;
      thus n+1 = m*k+t by A3;
      thus thesis by A6;
    end;
A7: t1+1 = m implies ex k,t st n+1 = (m*k)+t & t < m
    proof
      assume
A8:   t1+1 = m;
      take k = k1+1 , t = 0;
      thus n+1 = m*k+t by A3,A8;
      thus thesis by A1;
    end;
    t1+1 <= m by A4,Th13;
    hence thesis by A5,A7,XXREAL_0:1;
  end;
  0 = m*0+0;
  then
A9: P[0] by A1;
  thus for n holds P[n] from NatInd(A9,A2);
end;

theorem
  n = m*k+t & t < m & n = m*k1+t1 & t1 < m implies k = k1 & t = t1
proof
  assume that
A1: n = m*k+t and
A2: t < m and
A3: n = m*k1+t1 and
A4: t1 < m;
A5: now
    assume k1 <= k;
    then consider u being Nat such that
A6: k = k1 + u by Th10;
    m * (k1 + u) + t = m * k1 + (m * u + t);
    then
A7: m * u + t = t1 by A1,A3,A6,XCMPLX_1:2;
    now
      given w being Nat such that
A8:   u = w + 1;
      m * u + t = m + (m * w + t) by A8;
      hence contradiction by A4,A7,Th11;
    end;
    then
A9: u = 0 by Th6;
    hence k = k1 by A6;
    thus t = t1 by A1,A3,A6,A9,XCMPLX_1:2;
  end;
  now
    assume k <= k1;
    then consider u being Nat such that
A10: k1 = k + u by Th10;
    m * (k + u) + t1 = m * k + (m * u + t1);
    then
A11: m * u + t1 = t by A1,A3,A10,XCMPLX_1:2;
    now
      given w being Nat such that
A12:  u = w + 1;
      m * u + t1 = m + (m * w + t1) by A12;
      hence contradiction by A2,A11,Th11;
    end;
    then
A13: u = 0 by Th6;
    hence k = k1 by A10;
    thus t = t1 by A1,A3,A10,A13,XCMPLX_1:2;
  end;
  hence thesis by A5;
end;

registration
  cluster -> ordinal for Nat;
  coherence;
end;

registration
  cluster non empty ordinal for Subset of REAL;
  existence
  proof
    take NAT;
    thus thesis;
  end;
end;

theorem
  k < k + n iff 1 <= n
proof
  thus k < k + n implies 1 <= n
  proof
    assume
A1: k < k + n;
    assume not 1 <= n;
    then n = 0 by Th14;
    hence thesis by A1;
  end;
  assume 1 <= n;
  hence thesis by Th16;
end;

theorem
  k < n implies n - 1 is Element of NAT
proof
  assume k < n;
  then k + 1 <= n by Th13;
  then consider j being Nat such that
A1: n = k + 1 + j by Th10;
  n - 1 = k + j by A1;
  hence thesis by ORDINAL1:def 12;
end;

theorem
  k <= n implies n - k is Element of NAT
proof
  assume
A1: k <= n;
  per cases by A1,XXREAL_0:1;
  suppose
    k < n;
    then k + 1 <= n by Th13;
    then consider j being Nat such that
A2: n = k + 1 + j by Th10;
    reconsider j as Element of NAT by ORDINAL1:def 12;
    n - k = 1 + j by A2;
    hence thesis;
  end;
  suppose
    k = n;
    then n - k = 0;
    hence thesis;
  end;
end;

begin 



theorem Th22:
  m < n+1 implies m < n or m = n
proof
  assume m<n+1;
  then m<=n by Th13;
  hence thesis by XXREAL_0:1;
end;

theorem
  k < 2 implies k = 0 or k = 1
proof
  assume k<2;
  then k<1+1;
  hence thesis by Th14,Th22;
end;

registration
  cluster non zero for Element of NAT;
  existence
  proof
    take 1;
    thus thesis;
  end;
end;

registration
  cluster -> non negative for Element of NAT;
  coherence by Th2;
end;

registration
  cluster -> non negative for Nat;
  coherence by Th2;
end;



theorem
  i <> 0 & h = j*i implies j <= h
proof
  assume i<>0;
  then consider k such that
A1: i = k+1 by Th6;
  assume h=j*i;
  then h=j*k+j*1 by A1;
  hence thesis by Th11;
end;





scheme
  Ind1{M() -> Nat, P[Nat]} : for i being Nat st M() <= i holds P[i]
provided
A1: P[M()] and
A2: for j being Nat st M() <= j holds P[j] implies P[j+1]
proof
  reconsider M = M() as Element of NAT by ORDINAL1:def 12;
  defpred Q[Nat] means P[M() + $1];
A3: now
    let m be Element of NAT;
    assume Q[m];
    then P[M+m+1] by A2,Th11;
    hence Q[m+1];
  end;
  let i;
  assume M() <= i;
  then consider m being Nat such that
A4: i = M() + m by Th10;
A5: Q[0] by A1;
A6: for m being Element of NAT holds Q[m] from Ind(A5,A3);
  m in NAT by ORDINAL1:def 12;
  hence thesis by A6,A4;
end;



scheme
  CompInd1 { a() -> Nat, P[Nat] } : for k being Nat st k >= a() holds P[k]
provided
A1: for k being Nat st k>=a() & (for n being Nat st n>=a() & n<k holds P
[n]) holds P[k]
proof
  defpred P1[Nat] means for n being Nat st (n>=a() & n<$1) holds P[n];
A2: for k being Nat st k>=a() holds P1[k] implies P1[k+1]
  proof
    let k be Nat;
    assume k>=a();
    assume
A3: for n being Nat st n>=a() & n<k holds P[n];
    let n be Nat;
    assume that
A4: n>=a() and
A5: n<k+1;
    n<=k by A5,Th13;
    then n<k or n = k by XXREAL_0:1;
    hence thesis by A1,A3,A4;
  end;
  let k be Nat;
  assume
A6: k>=a();
A7: P1[a()];
  for k being Nat st k>=a() holds P1[k] from Ind1(A7,A2);
  then for n being Nat st n>=a() & n<k holds P[n] by A6;
  hence thesis by A1,A6;
end;



theorem Th25:
  n <= 1 implies n = 0 or n = 1
proof
  assume
A1: n <= 1;
  assume that
A2: not n = 0 and
A3: not n = 1;
  n < 0+1 by A1,A3,XXREAL_0:1;
  hence contradiction by A2,Th13;
end;

theorem Th26:
  n <= 2 implies n = 0 or n = 1 or n = 2
proof
  assume n <= 2;
  then n <= 1+1;
  hence thesis by Th8,Th25;
end;

theorem Th27:
  n <= 3 implies n = 0 or n = 1 or n = 2 or n = 3
proof
  assume n <= 3;
  then n <= 2+1;
  hence thesis by Th8,Th26;
end;

theorem Th28:
  n <= 4 implies n = 0 or n = 1 or n = 2 or n = 3 or n = 4
proof
  assume n <= 4;
  then n <= 3+1;
  hence thesis by Th8,Th27;
end;

theorem Th29:
  n <= 5 implies n = 0 or n = 1 or n = 2 or n = 3 or n = 4 or n =
  5
proof
  assume n <= 5;
  then n <= 4+1;
  hence thesis by Th8,Th28;
end;

theorem Th30:
  n <= 6 implies n = 0 or n = 1 or n = 2 or n = 3 or n = 4 or n =
  5 or n = 6
proof
  assume n <= 6;
  then n <= 5+1;
  hence thesis by Th8,Th29;
end;

theorem Th31:
  n <= 7 implies n = 0 or n = 1 or n = 2 or n = 3 or n = 4 or n =
  5 or n = 6 or n = 7
proof
  assume n <= 7;
  then n <= 6+1;
  hence thesis by Th8,Th30;
end;

theorem Th32:
  n <= 8 implies n = 0 or n = 1 or n = 2 or n = 3 or n = 4 or n =
  5 or n = 6 or n = 7 or n = 8
proof
  assume n <= 8;
  then n <= 7+1;
  hence thesis by Th8,Th31;
end;

theorem Th33:
  n <= 9 implies n = 0 or n = 1 or n = 2 or n = 3 or n = 4 or n =
  5 or n = 6 or n = 7 or n = 8 or n = 9
proof
  assume n <= 9;
  then n <= 8+1;
  hence thesis by Th8,Th32;
end;

theorem Th34:
  n <= 10 implies n = 0 or n = 1 or n = 2 or n = 3 or n = 4 or n =
  5 or n = 6 or n = 7 or n = 8 or n = 9 or n = 10
proof
  assume n <= 10;
  then n <= 9+1;
  hence thesis by Th8,Th33;
end;

theorem Th35:
  n <= 11 implies n = 0 or n = 1 or n = 2 or n = 3 or n = 4 or n =
  5 or n = 6 or n = 7 or n = 8 or n = 9 or n = 10 or n = 11
proof
  assume n <= 11;
  then n <= 10+1;
  hence thesis by Th8,Th34;
end;

theorem Th36:
  n <= 12 implies n = 0 or n = 1 or n = 2 or n = 3 or n = 4 or n =
  5 or n = 6 or n = 7 or n = 8 or n = 9 or n = 10 or n = 11 or n = 12
proof
  assume n <= 12;
  then n <= 11+1;
  hence thesis by Th8,Th35;
end;

theorem Th37:
  n <= 13 implies n = 0 or n = 1 or n = 2 or n = 3 or n = 4 or n = 5 or
  n = 6 or n = 7 or n = 8 or n = 9 or n = 10 or n = 11 or n = 12 or n = 13
proof
  assume n <= 13;
  then n <= 12+1;
  hence thesis by Th8,Th36;
end;



scheme
  Indfrom1 { P[Nat] } : for k being non empty Nat holds P[k]
provided
A1: ( P[1]) and
A2: for k being non empty Nat st P[k] holds P[k + 1]
proof
  defpred P[Nat] means $1 is non empty implies P[$1];
A3: now
    let k be Nat;
    assume
A4: P[k];
    now
      assume k+1 is non empty;
      k is empty or k is non empty;
      hence P[k+1] by A1,A2,A4;
    end;
    hence P[k+1];
  end;
A5: P[0];
  for k being Nat holds P[k] from NatInd(A5,A3);
  hence thesis;
end;



definition
  let A be set;
  func min* A -> Element of NAT means :Def1:
  it in A & for k st k in A holds it <= k
  if A is non empty Subset of NAT otherwise it = 0;
  existence
  proof
    A is non empty Subset of NAT implies ex i st i in A & for k be Nat st
    k in A holds i <= k
    proof
      defpred P[Nat] means $1 in A;
      set x = the Element of A;
      assume
A1:   A is non empty Subset of NAT;
      then x is Element of NAT by TARSKI:def 3;
      then
A2:   ex k be Nat st P[k] by A1;
      ex k be Nat st P[k] & for i be Nat st P[i] holds k <= i from Min( A2
      );
      hence thesis;
    end;
    hence thesis;
  end;
  uniqueness
  proof
    let i,n be Element of NAT;
    A is non empty Subset of NAT & (i in A & for k st k in A holds i <= k)
    & (n in A & for k st k in A holds n <= k) implies i = n
    proof
      assume that
      A is non empty Subset of NAT and
A3:   i in A & ( for k st k in A holds i <= k)& n in A & for k st k in
      A holds n <= k;
      i <= n & n <= i by A3;
      hence thesis by XXREAL_0:1;
    end;
    hence thesis;
  end;
  consistency;
end;



reserve x,X,Y,Z for set;

theorem Th38:
  for n being Nat holds succ n = n + 1
proof
  let n be Nat;
A1: n+1 = {L where L is Element of NAT: L < n+1} by AXIOMS:4;
A2: n = {K where K is Element of NAT: K < n} by AXIOMS:4;
  thus succ n c= n+1
  proof
    let x such that
A3: x in succ n;
    per cases by A3,ORDINAL1:8;
    suppose
      x in n;
      then consider K being Element of NAT such that
A4:   x = K and
A5:   K < n by A2;
      K < n+1 by A5,Th13;
      hence thesis by A1,A4;
    end;
    suppose
A6:   x = n;
      reconsider n as Element of NAT by ORDINAL1:def 12;
      n < n+1 by Th13;
      hence thesis by A1,A6;
    end;
  end;
  let x;
  assume x in n+1;
  then consider K being Element of NAT such that
A7: x = K and
A8: K < n+1 by A1;
  K <= n by A8,Th13;
  then K = n or K < n by XXREAL_0:1;
  then x = n or x in n by A2,A7;
  hence thesis by ORDINAL1:8;
end;

theorem Th39:
  n <= m iff n c= m
proof
  defpred P[Nat] means for m holds $1 <= m iff $1 c= m;
A1: for n st P[n] holds P[n+1]
  proof
    let n such that
A2: P[n];
    let m;
    thus n+1 <= m implies (n+1) c= m
    proof
      assume n+1 <= m;
      then consider k being Nat such that
A3:   m = n+1+k by Th10;
      reconsider k as Element of NAT by ORDINAL1:def 12;
      n <= n+k by Th11;
      then n c= (n+k) by A2;
      then
A4:   succ n c= succ (n+k) by ORDINAL2:1;
      (n+k+1) = succ (n+k) by Th38;
      hence thesis by A3,A4,Th38;
    end;
    assume
A5: (n+1) c= m;
    (n+1) = succ n by Th38;
    then
A6: n in m by A5,ORDINAL1:21;
    then n c= m by ORDINAL1:def 2;
    then
A7: n <= m by A2;
    n <> m by A6;
    then n < m by A7,XXREAL_0:1;
    hence thesis by Th13;
  end;
A8: P[0];
  for n holds P[n] from NatInd(A8,A1);
  hence thesis;
end;

theorem Th40:
  card n c= card m iff n <= m
proof
  reconsider n,m as Element of NAT by ORDINAL1:def 12;
  card n = n & card m = m by CARD_1:def 2;
  hence thesis by Th39;
end;

theorem Th41:
  card n in card m iff n < m
proof
A1: n <= m & n <> m iff n < m by XXREAL_0:1;
  card n c< card m iff card n c= card m & card n <> card m by XBOOLE_0:def 8;
  hence thesis by A1,Th40,ORDINAL1:11,def 2;
end;

reserve M,N for Cardinal;

theorem
  nextcard card n = card(n+1)
proof
  reconsider n as Element of NAT by ORDINAL1:def 12;
A1: for M st card card n in M holds card(n+1) c= M
  proof
A2: card n = n by CARD_1:def 2;
    let M;
    assume card card n in M;
    then
A3: succ n c= M by A2,ORDINAL1:21;
    (n+1) = succ n by Th38;
    hence thesis by A3,CARD_1:def 2;
  end;
  n < n+1 by Th13;
  then card card n in card(n+1) by Th41;
  hence thesis by A1,CARD_1:def 3;
end;

definition
  let n be Nat;
  redefine func succ n equals
  n+1;
  compatibility by Th38;
end;

theorem
  for X,Y being finite set holds X c= Y implies card X <= card Y
proof
  let X,Y be finite set;
  assume X c= Y;
  then card X c= card Y by CARD_1:11;
  hence thesis by Th39;
end;

reserve l for Element of NAT;

theorem
  k in n iff k < n
proof
  hereby
    assume k in n;
    then k in { l : l < n } by AXIOMS:4;
    then ex l st k = l & l < n;
    hence k < n;
  end;
  assume
A1: k < n;
  reconsider k as Element of NAT by ORDINAL1:def 12;
  k in { l : l < n } by A1;
  hence thesis by AXIOMS:4;
end;

theorem
  n in n+1
proof
  reconsider n as Element of NAT by ORDINAL1:def 12;
  n<n+1 by XREAL_1:29;
  then n in {l:l<n+1};
  hence thesis by AXIOMS:4;
end;

theorem
  k <= n implies k = k /\ n
proof
  assume k <= n;
  then Segm k c= Segm n by Th39;
  hence thesis by XBOOLE_1:28;
end;



scheme
  LambdaRecEx{A() -> set,G(set,set) -> set}: ex f being Function st dom f =
  NAT & f.0 = A() & for n being Nat holds f.(n+1) = G(n,f.n) proof
  deffunc D(set,set) = {};
  consider L being T-Sequence such that
A1: dom L = NAT and
A2: {} in NAT implies L.{} = A() and
A3: for A being Ordinal st succ A in NAT holds L.(succ A) = G(A,L.A) and
  for A being Ordinal st A in NAT & A <> {} & A is limit_ordinal holds L.A
  = D(A,L|A) from ORDINAL2:sch 5;
  take L;
  thus dom L = NAT by A1;
  thus L.0 = A() by A2;
  let n be Nat;
  succ n in NAT;
  hence thesis by A3;
end;

scheme
  LambdaRecExD{D() -> non empty set, A() -> Element of D(), G(set,set) ->
Element of D()}: ex f being Function of NAT,D() st f.0 = A() & for n being Nat
  holds f.(n+1) = G(n,f.n) proof
  consider f being Function such that
A1: dom f = NAT and
A2: f.0 = A() and
A3: for n being Nat holds f.(n+1) = G(n,f.n) from LambdaRecEx;
  for x being set st x in NAT holds f.x in D()
  proof
    let x be set;
    assume x in NAT;
    then reconsider n = x as Nat;
    per cases by Th6;
    suppose
      n = 0;
      hence thesis by A2;
    end;
    suppose
      ex k st n = k + 1;
      then consider k such that
A4:   n = k+1;
      f.n = G(k,f.k) by A3,A4;
      hence thesis;
    end;
  end;
  then reconsider f as Function of NAT,D() by A1,FUNCT_2:3;
  take f;
  thus thesis by A2,A3;
end;

scheme
  RecUn{A() -> set, F, G() -> Function, P[set,set,set]}: F() = G()
provided
A1: dom F() = NAT and
A2: F().0 = A() and
A3: for n holds P[n,F().n,F().(n+1)] and
A4: dom G() = NAT and
A5: G().0 = A() and
A6: for n holds P[n,G().n,G().(n+1)] and
A7: for n for x,y1,y2 being set st P[n,x,y1] & P[n,x,y2] holds y1 = y2
proof
  defpred P[Nat] means F().$1 = G().$1;
A8: for n st P[n] holds P[n+1]
  proof
    let n;
    assume F().n = G().n;
    then
A9: P[n,F().n,G().(n+1)] by A6;
    P[n,F().n,F().(n+1)] by A3;
    hence thesis by A7,A9;
  end;
A10: P[0] by A2,A5;
  for n holds P[n] from NatInd(A10,A8);
  then for x st x in NAT holds F().x = G().x;
  hence thesis by A1,A4,FUNCT_1:2;
end;

scheme
  RecUnD{D() -> non empty set,A() -> Element of D(), P[set,set,set], F, G() ->
  Function of NAT,D()} : F() = G()
provided
A1: F().0 = A() and
A2: for n holds P[n,F().n,F().(n+1)] and
A3: G().0 = A() and
A4: for n holds P[n,G().n,G().(n+1)] and
A5: for n being Nat,x,y1,y2 being Element of D() st P[n,x,y1] & P[n,x,y2
] holds y1=y2
proof
  defpred Q[Nat] means F().$1 <> G().$1;
A6: dom F() = NAT & dom G() = NAT by FUNCT_2:def 1;
  assume F() <> G();
  then ex x st x in NAT & F().x <> G().x by A6,FUNCT_1:2;
  then
A7: ex k be Nat st Q[k];
  consider m be Nat such that
A8: Q[m] & for n be Nat st Q[n] holds m <= n from Min(A7);
  now
    assume m<>0;
    then consider k being Nat such that
A9: m=k+1 by Th6;
    reconsider k as Element of NAT by ORDINAL1:def 12;
    k < m by A9,Th13;
    then
A10: F().k = G().k by A8;
    ( P[k,F().k,F().(k+1)])& P[k,G().k,G().(k+1)] by A2,A4;
    hence contradiction by A5,A8,A9,A10;
  end;
  hence contradiction by A1,A3,A8;
end;

scheme
  LambdaRecUn{A() -> set, RecFun(set,set) -> set, F, G() -> Function}: F() = G
  ()
provided
A1: dom F() = NAT and
A2: F().0 = A() and
A3: for n holds F().(n+1) = RecFun(n,F().n) and
A4: dom G() = NAT and
A5: G().0 = A() and
A6: for n holds G().(n+1) = RecFun(n,G().n)
proof
  defpred P[Nat,set,set] means $3=RecFun($1,$2);
A7: for n holds P[n,G().n,G().(n+1)] by A6;
A8: for n being Nat,x,y1,y2 being set st P[n,x,y1] & P[n,x,y2] holds y1=y2;
A9: for n holds P[n,F().n,F().(n+1)] by A3;
  thus F() = G() from RecUn(A1,A2,A9,A4,A5,A7,A8);
end;

scheme
  LambdaRecUnD{D() -> non empty set,A() -> Element of D(), RecFun(set,set) ->
  Element of D(), F, G() -> Function of NAT,D()}: F() = G()
provided
A1: F().0 = A() and
A2: for n holds F().(n+1) = RecFun(n,F().n) and
A3: G().0 = A() and
A4: for n holds G().(n+1) = RecFun(n,G().n)
proof
  defpred P[Nat,set,set] means $3=RecFun($1,$2);
A5: for n holds P[n,G().n,G().(n+1)] by A4;
A6: for n being Nat,x,y1,y2 being Element of D() st P[n,x,y1] & P[n,x,y2]
  holds y1=y2;
A7: for n holds P[n,F().n,F().(n+1)] by A2;
  thus F() = G() from RecUnD(A1,A7,A3,A5,A6);
end;



registration
  let x,y be Nat;
  cluster min(x,y) -> natural;
  coherence by XXREAL_0:15;
  cluster max(x,y) -> natural;
  coherence by XXREAL_0:16;
end;

definition
  let x,y be Element of NAT;
  redefine func min(x,y) -> Element of NAT;
  coherence by XXREAL_0:15;
  redefine func max(x,y) -> Element of NAT;
  coherence by XXREAL_0:16;
end;



scheme
  MinIndex { F(Nat)->Nat} : ex k st F(k)=0 & for n st F(n)=0 holds k <= n
provided
A1: for k holds (F(k+1) < F(k) or F(k) = 0)
proof
  defpred X[Nat] means F($1)=0;
  defpred P[Nat] means ex n being Nat st $1 = F(n);
A2: for k be Nat st k <> 0 & P[k] ex m be Nat st m < k & P[m]
  proof
    let k be Nat;
    assume that
A3: k <> 0 and
A4: P[k];
    consider n being Nat such that
A5: k = F(n) by A4;
    take F(n + 1);
    thus thesis by A1,A3,A5;
  end;
  F(0) is Nat;
  then
A6: ex k be Nat st P[k];
  P[0] from Regr(A6,A2);
  then
A7: ex k be Nat st X[k];
  consider k be Nat such that
A8: ( X[k])& for n be Nat st X[n] holds k <= n from Min(A7);
  take k;
  thus thesis by A8;
end;



definition
  let D be set, f be Function of NAT,D;
  let n be Nat;
  redefine func f.n -> Element of D;
  coherence
  proof
    reconsider n as Element of NAT by ORDINAL1:def 12;
    f.n is Element of D;
    hence thesis;
  end;
end;



definition
  let X be set;
  mode sequence of X is Function of NAT, X;
end;




definition
  let s be ManySortedSet of NAT, k be Nat;
  func s ^\ k -> ManySortedSet of NAT means
  :Def3:
  for n being Nat holds it.n=s.(n+k);
  existence
  proof
    defpred P[set,set] means
     ex n being Element of NAT st n=$1 & $2=s.(n+k);
A1: for i being set st i in NAT ex j being set st P[i,j]
     proof let i be set;
      assume
     i in NAT;
       then reconsider n=i as Element of NAT;
      take j = s.(n+k);
      thus P[i,j];
     end;
    consider f being ManySortedSet of NAT such that
A2:  for i being set st i in NAT holds P[i,f.i] from PBOOLE:sch 3(A1);
    take f;
    let n be Nat;
     n in NAT by ORDINAL1:def 12;
     then P[n,f.n] by A2;
    hence thesis;
  end;
  uniqueness
  proof
    let s1,s2 be ManySortedSet of NAT;
    assume that
A3: for n holds s1.n=s.(n+k) and
A4: for n holds s2.n=s.(n+k);
    now
      let n be set;
      assume
      n in NAT;
        then reconsider nn = n as Element of NAT;
      thus s1.n=s.(nn+k) by A3
           .=s2.n by A4;
    end;
    hence s1=s2 by PBOOLE:3;
  end;
end;

Lm1:
  for s be ManySortedSet of NAT, k be Nat
   holds rng(s ^\ k) c= rng s
 proof
  let s be ManySortedSet of NAT, k be Nat;
  let i be set;
  assume i in rng(s ^\ k);
   then consider u being set such that
A1: u in dom(s ^\ k) and
A2: i = (s ^\ k).u by FUNCT_1:def 3;
   reconsider n = u as Element of NAT by A1,PARTFUN1:def 2;
A3:  dom s = NAT by PARTFUN1:def 2;
   i = s.(n+k) by A2,Def3;
  hence i in rng s by A3,FUNCT_1:3;
 end;

registration
 let X be non empty set, s be X-valued ManySortedSet of NAT, k be Nat;
 cluster s ^\ k -> X-valued;
 coherence
  proof
A1:  rng(s ^\ k) c= rng s by Lm1;
   rng s c= X by RELAT_1:def 19;
   hence rng(s ^\ k) c= X by A1,XBOOLE_1:1;
  end;
end;

definition
 let X be non empty set, s be sequence of X, k be Nat;
 redefine func s ^\ k -> sequence of X;
  coherence
   proof
     rng(s ^\ k) c= X & dom(s ^\ k) = NAT
          by PARTFUN1:def 2,RELAT_1:def 19;
    hence s ^\ k is sequence of X by RELSET_1:4;
   end;
end;

reserve X for non empty set,
  s for sequence of X,
  k,m for Nat;

theorem
  s^\0 =s
proof
  now
    let n be Element of NAT;
    thus (s ^\0).n=s.(n+0) by Def3
      .=s.n;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem Th48:
  s ^\k^\m = s^\(k+m)
proof
  now
    let n be Element of NAT;
    thus (s ^\k^\m).n=(s ^\k).(n+m) by Def3
      .=s.(n+m+k) by Def3
      .=s.(n+(k+m))
      .=(s ^\(k+m)).n by Def3;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  s^\k^\m = s^\m^\k
proof
  thus s^\k^\m = s^\(k+m) by Th48
    .= s^\m^\k by Th48;
end;

registration let N be sequence of NAT; let X,s;
 cluster s * N -> Function-like NAT-defined X-valued;
 coherence;
end;

registration let N be sequence of NAT; let X,s;
 cluster s * N -> total;
 coherence;
end;






theorem
  for N being sequence of NAT holds (s * N) ^\k = s * (N ^\k)
proof
  let N be sequence of NAT;
  now
    let n be Element of NAT;
    thus ((s * N) ^\k).n = (s * N).(n + k) by Def3
      .= s.(N.(n + k)) by FUNCT_2:15
      .= s.((N ^\k).n) by Def3
      .= (s * (N ^\k)).n by FUNCT_2:15;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  s.n in rng s
proof
  n in NAT by ORDINAL1:def 12;
  then n in dom s by FUNCT_2:def 1;
  hence thesis by FUNCT_1:3;
end;

theorem
  (for n holds s.n in Y) implies rng s c= Y
proof
  assume
A1: for n holds s.n in Y;
  let y be set;
  assume y in rng s;
  then consider x such that
A2: x in dom s and
A3: y = s.x by FUNCT_1:def 3;
  x in NAT by A2,FUNCT_2:def 1;
  hence thesis by A1,A3;
end;



theorem
  for n being Nat holds n is non zero implies n = 1 or n > 1
proof
  let n be Nat;
  assume n is non zero;
  then 0+1 <= n by Th13;
  hence n = 1 or n > 1 by XXREAL_0:1;
end;

theorem
 succ n = {l : l <= n}
  proof
    defpred P[Nat] means $1 <= n;
    defpred Q[Nat] means $1 < n+1;
    deffunc F(Nat) = $1;
A1:   for l holds Q[l] iff P[l] by Th13;
   thus succ n = n+1
     .= {F(l) : Q[l]} by AXIOMS:4
     .= {F(l) : P[l]} from FRAENKEL:sch 3(A1);
  end;

scheme MinPred { F(Element of NAT) -> Element of NAT, P[set]} :
  ex k be Element of NAT st P[k] &
    for n be Element of NAT st P[n] holds k <= n
provided
A1: for k be Element of NAT holds F(k+1) < F(k) or P[k]
proof
  now
    consider f being Function of NAT, NAT such that
A2: for k be Element of NAT holds f.k = F(k) from FUNCT_2:sch 4;
A3:  f.0 in rng f by FUNCT_2:4;
    rng f c= NAT
    proof
      let x be set; assume x in rng f; then
      consider y being set such that
A4:    y in dom f & x = f.y by FUNCT_1:def 3;
      reconsider y1 = y as Element of NAT by A4,FUNCT_2:def 1;
      x = F(y1) by A4,A2;
      hence thesis;
    end; then
    reconsider rf = rng f as non empty Subset of NAT by A3;
    set m = min* rf;
    m in rf by Def1;
    then consider x being set such that
A5: x in dom f and
A6: m = f.x by FUNCT_1:def 3;
    reconsider x as Element of NAT by A5,FUNCT_2:def 1;
    f.x = F(x) & f.(x+1) = F(x+1) by A2; then
A7: f.(x+1) < f.x or P[x] by A1;
    assume
A8: for k holds not P[k];
    f.(x+1) in rf by FUNCT_2:4;
    hence contradiction by Def1,A8,A6,A7;
  end; then
A9: ex k be Nat st P[k];
  consider k be Nat such that
A10: P[k] and
A11: for n be Nat st P[n] holds k <= n from Min(A9);
  k in NAT & for n be Element of NAT st P[n] holds k <= n
    by A11,ORDINAL1:def 12;
  hence thesis by A10;
end;

registration let k be Ordinal, x be set;
  cluster k --> x -> T-Sequence-like;
 coherence
 proof
   dom(k --> x) = k by FUNCOP_1:13;
   hence thesis by ORDINAL1:def 7;
 end;
end;

theorem
  for s be ManySortedSet of NAT, k be Nat
   holds rng(s ^\ k) c= rng s by Lm1;



theorem Th56:
  for n,m st n<=m & m<=n+2 holds m=n or m=n+1 or m=n+2
proof
  let n,m;
  assume that
A1: n<=m and
A2: m<=n+2;
  per cases;
  suppose
    m<=n+1;
    hence thesis by A1,Th9;
  end;
  suppose
    m>n+1;
    then m=n+1 or m=n+1+1 by A2,Th9;
    hence thesis;
  end;
end;

theorem Th57:
  for n,m st n<=m & m<=n+3 holds m=n or m=n+1 or m=n+2 or m=n+3
proof
  let n,m;
  assume that
A1: n<=m and
A2: m<=n+3;
  per cases;
  suppose
    m<=n+2;
    hence thesis by A1,Th56;
  end;
  suppose
    m>n+2;
    then m=n+2 or m=n+2+1 by A2,Th9;
    hence thesis;
  end;
end;

theorem
  for n,m st n<=m & m<=n+4 holds m=n or m=n+1 or m=n+2 or m=n+3 or m=n+4
proof
  let n,m;
  assume that
A1: n<=m and
A2: m<=n+4;
  per cases;
  suppose
    m<=n+3;
    hence thesis by A1,Th57;
  end;
  suppose
    m>n+3;
    then m=n+3 or m=n+3+1 by A2,Th9;
    hence thesis;
  end;
end;



theorem
  for X being finite set st 1 < card X holds ex x1,x2 being set st
  x1 in X & x2 in X & x1 <> x2
proof
  let X be finite set;
  set x1 = choose X;
  assume
A1: 1 < card X;
  then X <> {};
  then
A2: x1 in X;
  now
    assume
A3: for x2 being set st x2 in X holds x2 = x1;
    now
      let x be set;
      hereby
        assume x in X;
        then x = x1 by A3;
        hence x in {x1} by TARSKI:def 1;
      end;
      assume x in {x1};
      hence x in X by A2,TARSKI:def 1;
    end;
    then X = {x1} by TARSKI:1;
    hence contradiction by A1,CARD_1:30;
  end;
  hence thesis;
end;

theorem
  n <= 14 implies n = 0 or n = 1 or n = 2 or n = 3 or n = 4 or n = 5 or
  n = 6 or n = 7 or n = 8 or n = 9 or n = 10 or n = 11 or n = 12 or n = 13
   or n = 14
proof
  assume n <= 14;
  then n <= 13+1;
  hence thesis by Th8,Th37;
end;















begin

reserve X,x,y,z for set,
  k,l,n for Element of NAT,
  r for real number;

Lm1: z in [:{0},NAT:] \ {[0,0]} implies ex k st z = -k
proof
A1: [:{0},NAT:] c= [:{0},REAL+:] by ARYTM_2:2,ZFMISC_1:95;
  assume
A2: z in [:{0},NAT:] \ {[0,0]};
  then
A3: not z in {[0,0]} by XBOOLE_0:def 5;
  z in NAT \/ [:{0},NAT:] by A2,XBOOLE_0:def 3;
  then z in INT by A3,NUMBERS:def 4,XBOOLE_0:def 5;
  then reconsider z9 = z as Element of REAL by NUMBERS:15;
  consider x,y such that
  x in {0} and
A4: y in NAT and
A5: z = [x,y] by A2,ZFMISC_1:84;
  reconsider y as Element of NAT by A4;
  z in [:{0},NAT:] by A2;
  then consider x9,y9 being Element of REAL+ such that
A6: z9 = [0,x9] and
A7: y = y9 and
A8: +(z9,y) = y9 - x9 by A4,A1,ARYTM_0:def 1,ARYTM_2:2;
  x9 = y9 by A5,A6,A7,XTUPLE_0:1;
  then
A9: y9 - x9 = 0 by ARYTM_1:18;
  take y;
  consider x1,x2,y1,y2 being Element of REAL such that
A10: z9 = [*x1,x2*] and
A11: y = [*y1,y2*] and
A12: z9 + y = [*+(x1,y1),+(x2,y2)*] by XCMPLX_0:def 4;
A13: x2 = 0 by A10,ARYTM_0:24;
  then
A14: +(x2,y2) = 0 by A11,ARYTM_0:11,24;
  y2 = 0 by A11,ARYTM_0:24;
  then
A15: y = y1 by A11,ARYTM_0:def 5;
  z9 = x1 by A10,A13,ARYTM_0:def 5;
  then z9 + y = 0 by A12,A15,A8,A9,A14,ARYTM_0:def 5;
  hence thesis;
end;

Lm2: for k st x = -k & k <> x holds x in [:{0},NAT:] \ {[0,0]}
proof
  let k such that
A1: x = -k and
A2: k <> x;
  reconsider r = x as Element of REAL by A1;
  r + k = 0 by A1;
  then consider x1,x2,y1,y2 being Element of REAL such that
A3: r = [*x1,x2*] and
A4: k = [*y1,y2*] and
A5: 0 = [*+(x1,y1),+(x2,y2)*] by XCMPLX_0:def 4;
A6: y2 = 0 by A4,ARYTM_0:24;
  then
A7: y1 = k by A4,ARYTM_0:def 5;
  +(x2,y2) = 0 by A5,ARYTM_0:24;
  then
A8: +(x1,y1) = 0 by A5,ARYTM_0:def 5;
A9: k in omega;
A10: now
    assume x1 in REAL+;
    then consider x9,y9 being Element of REAL+ such that
A11: x1 = x9 and
A12: y1 = y9 and
A13: 0 = x9 + y9 by A9,A7,A8,ARYTM_0:def 1,ARYTM_2:2;
A14: y9 = 0 by A13,ARYTM_2:5;
    x9 = 0 by A13,ARYTM_2:5;
    hence contradiction by A2,A3,A4,A6,A11,A12,A14,ARYTM_0:24;
  end;
  x2 = 0 by A3,ARYTM_0:24;
  then
A15: x1 = r by A3,ARYTM_0:def 5;
  r in REAL+ \/ [:{0},REAL+:] by NUMBERS:def 1,XBOOLE_0:def 5;
  then x in [:{0},REAL+:] by A15,A10,XBOOLE_0:def 3;
  then consider x9,y9 being Element of REAL+ such that
A16: x1 = [0,x9] and
A17: y1 = y9 and
A18: 0 = y9 - x9 by A9,A15,A7,A8,ARYTM_0:def 1,ARYTM_2:2;
A19: 0 in {0} by TARSKI:def 1;
A20: not r in {[0,0]} by NUMBERS:def 1,XBOOLE_0:def 5;
  x9 = y9 by A18,ARYTM_0:6;
  then x in [:{0},NAT:] by A15,A7,A16,A17,A19,ZFMISC_1:87;
  hence thesis by A20,XBOOLE_0:def 5;
end;

definition
  redefine func INT means
  :Def1:
  x in it iff ex k st x = k or x = - k;
  compatibility
  proof
    let I be set;
    thus I = INT implies for x holds x in I iff ex k st x = k or x = - k
    proof
      assume
A1:   I = INT;
      let x;
      thus x in I implies ex k st x = k or x = - k
      proof
        assume
A2:     x in I;
        then
A3:     not x in {[0,0]} by A1,NUMBERS:def 4,XBOOLE_0:def 5;
        per cases by A1,A2,NUMBERS:def 4,XBOOLE_0:def 3;
        suppose
          x in NAT;
          hence thesis;
        end;
        suppose
          x in [:{0},NAT:];
          then x in [:{0},NAT:] \ {[0,0]} by A3,XBOOLE_0:def 5;
          hence thesis by Lm1;
        end;
      end;
      given k such that
A4:   x = k or x = - k;
      per cases by A4;
      suppose
A5:     x = k;
        then
A6:     not x in {[0,0]} by NUMBERS:def 1,XBOOLE_0:def 5;
        x in NAT \/ [:{0},NAT:] by A5,XBOOLE_0:def 3;
        hence thesis by A1,A6,NUMBERS:def 4,XBOOLE_0:def 5;
      end;
      suppose
        x = -k & k <> x;
        then
A7:     x in [:{0},NAT:] \ {[0,0]} by Lm2;
        then
A8:     not x in {[0,0]} by XBOOLE_0:def 5;
        x in NAT \/ [:{0},NAT:] by A7,XBOOLE_0:def 3;
        hence thesis by A1,A8,NUMBERS:def 4,XBOOLE_0:def 5;
      end;
    end;
    assume
A9: for x holds x in I iff ex k st x = k or x = - k;
    thus I c= INT
    proof
      let x;
      assume x in I;
      then consider k such that
A10:  x = k or x = - k by A9;
      per cases by A10;
      suppose
A11:    x = k;
        then
A12:    not x in {[0,0]} by NUMBERS:def 1,XBOOLE_0:def 5;
        x in NAT \/ [:{0},NAT:] by A11,XBOOLE_0:def 3;
        hence thesis by A12,NUMBERS:def 4,XBOOLE_0:def 5;
      end;
      suppose
        x = -k & k <> x;
        then
A13:    x in [:{0},NAT:] \ {[0,0]} by Lm2;
        then
A14:    not x in {[0,0]} by XBOOLE_0:def 5;
        x in NAT \/ [:{0},NAT:] by A13,XBOOLE_0:def 3;
        hence thesis by A14,NUMBERS:def 4,XBOOLE_0:def 5;
      end;
    end;
    let x;
    assume
A15: x in INT;
    then
A16: not x in {[0,0]} by NUMBERS:def 4,XBOOLE_0:def 5;
    per cases by A15,NUMBERS:def 4,XBOOLE_0:def 3;
    suppose
      x in NAT;
      hence thesis by A9;
    end;
    suppose
      x in [:{0},NAT:];
      then x in [:{0},NAT:] \ {[0,0]} by A16,XBOOLE_0:def 5;
      then ex k st x = k or x = - k by Lm1;
      hence thesis by A9;
    end;
  end;
end;

definition
  let i be number;
  attr i is integer means
  :Def2:
  i in INT;
end;

registration
  cluster integer for Real;
  existence
  proof
    take 0;
    thus 0 in INT by Def1;
  end;
  cluster integer for number;
  existence
  proof
    take 0;
    thus 0 in INT by Def1;
  end;
  cluster -> integer for Element of INT;
  coherence by Def2;
end;

definition
  mode Integer is integer number;
end;

theorem Th1:
  for k being Nat st r = k or r = -k holds r is Integer
proof
  let k be Nat;
  assume
A1: r = k or r = -k;
  k in NAT by ORDINAL1:def 12;
  then r in INT by A1,Def1;
  hence thesis;
end;

theorem Th2:
  r is Integer implies ex k st r = k or r = - k
proof
  assume r is Integer;
  then r is Element of INT by Def2;
  hence thesis by Def1;
end;



registration
  cluster natural -> integer for number;
  coherence by Th1;
end;

Lm3: x in INT implies x in REAL by NUMBERS:15;

registration
  cluster integer -> real for number;
  coherence
  proof
    let n be number;
    assume n is integer;
    then n is Element of INT by Def2;
    then reconsider n as Element of REAL by Lm3;
    n is real;
    hence thesis;
  end;
end;

reserve i,i0,i1,i2,i3,i4,i5,i8,i9,j for Integer;

registration
  let i1,i2 be Integer;
  cluster i1 + i2 -> integer;
  coherence
  proof
    consider l such that
A1: i2 = l or i2 = - l by Th2;
    consider k such that
A2: i1 = k or i1 = - k by Th2;
A3: now
A4:   now
        assume l - k <= 0;
        then l <= 0 + k by XREAL_1:20;
        then consider z being Nat such that
A5:     k = l + z by NAT_1:10;
        - z = - k + l by A5;
        hence l - k is Integer by Th1;
      end;
      assume that
A6:   i1 = - k and
A7:   i2 = l;
      now
        assume 0 <= l - k;
        then 0 + k <= l by XREAL_1:19;
        then ex z being Nat st l = k + z by NAT_1:10;
        hence l - k is Integer;
      end;
      hence i1 + i2 is Integer by A6,A7,A4;
    end;
A8: now
A9:   now
        assume k - l <= 0;
        then k <= 0 + l by XREAL_1:20;
        then consider z being Nat such that
A10:    l = k + z by NAT_1:10;
        - z = - l + k by A10;
        hence k - l is Integer by Th1;
      end;
      assume that
A11:  i1 = k and
A12:  i2 = - l;
      now
        assume 0 <= k - l;
        then 0 + l <= k by XREAL_1:19;
        then ex z being Nat st k = l + z by NAT_1:10;
        hence k - l is Integer;
      end;
      hence i1 + i2 is Integer by A11,A12,A9;
    end;
    now
      assume that
A13:  i1 = - k and
A14:  i2 = - l;
      i1 + i2 = - (k + l) by A13,A14;
      hence i1 + i2 is Integer by Th1;
    end;
    hence thesis by A2,A1,A8,A3;
  end;
  cluster i1 * i2 -> integer;
  coherence
  proof
    consider l such that
A15: i2 = l or i2 = - l by Th2;
    consider k such that
A16: i1 = k or i1 = - k by Th2;
A17: now
      assume that
A18:  i1 = - k and
A19:  i2 = - l;
      i1 * i2 = k * l by A18,A19;
      hence i1 * i2 is Integer;
    end;
    now
      assume i1 = - k & i2 = l or i1 = k & i2 = - l;
      then i1 * i2 = - (k * l);
      hence i1 * i2 is Integer by Th1;
    end;
    hence thesis by A16,A15,A17;
  end;
end;

registration
  let i0 be Integer;
  cluster - i0 -> integer;
  coherence
  proof
    ex k st i0 = k or i0 = - k by Th2;
    hence thesis by Th1;
  end;
end;

registration
  let i1,i2 be Integer;
  cluster i1 - i2 -> integer;
  coherence
  proof
    i1 - i2 = i1 + (- i2);
    hence thesis;
  end;
end;



theorem Th3:
  0 <= i0 implies i0 in NAT
proof
  consider k such that
A1: i0 = k or i0 = - k by Th2;
  assume 0 <= i0;
  then i0 = - k implies i0 is Element of NAT;
  hence thesis by A1;
end;

theorem
  r is Integer implies r + 1 is Integer & r - 1 is Integer;

theorem Th5:
  i2 <= i1 implies i1 - i2 in NAT
proof
  assume i2 <= i1;
  then i2 - i2 <= i1 - i2 by XREAL_1:9;
  hence thesis by Th3;
end;

theorem Th6:
  i1 + k = i2 implies i1 <= i2
proof
  reconsider i0 = k as Integer;
  assume i1 + k = i2;
  then 0 + (i1 + k) <= k + i2 by XREAL_1:6;
  then i1 + i0 - i0 <= i2 + i0 - i0 by XREAL_1:9;
  hence thesis;
end;

Lm4: for j being Element of NAT holds j < 1 implies j = 0
proof
  let j be Element of NAT;
  assume j < 1;
  then j < 0 + 1;
  hence thesis by NAT_1:13;
end;

Lm5: 0 < i1 implies 1 <= i1
proof
  assume
A1: 0 < i1;
  then reconsider i2 = i1 as Element of NAT by Th3;
  0 <> i2 by A1;
  hence thesis by Lm4;
end;

theorem Th7:
  i0 < i1 implies i0 + 1 <= i1
proof
  assume i0 < i1;
  then i0 + (- i0) < i1 + (- i0) by XREAL_1:6;
  then 1 <= i1 + (- i0) by Lm5;
  then 1 + i0 <= i1 + (- i0) + i0 by XREAL_1:6;
  hence thesis;
end;

theorem Th8:
  i1 < 0 implies i1 <= - 1
proof
  assume i1 < 0;
  then 0 < - i1 by XREAL_1:58;
  then 1 <= - i1 by Lm5;
  then - - i1 <= - 1 by XREAL_1:24;
  hence thesis;
end;

theorem Th9:
  i1 * i2 = 1 iff i1 = 1 & i2 = 1 or i1 = - 1 & i2 = - 1
proof
  thus i1 * i2 = 1 implies i1 = 1 & i2 = 1 or i1 = - 1 & i2 = - 1
  proof
    assume
A1: i1 * i2 = 1;
    then
A2: not i2 = 0;
A3: now
A4:   (- i1) * (- i2) = i1 * i2;
      assume that
A5:   i1 < 0 and
A6:   i2 < 0;
A7:   -i2 is Element of NAT by A6,Th3;
      -i1 is Element of NAT by A5,Th3;
      then - (- i1) = - 1 by A1,A7,A4,NAT_1:15;
      hence i1 = - 1 & i2 = - 1 by A1;
    end;
A8: now
      assume that
A9:   0 < i1 and
A10:  0 < i2;
A11:  i2 is Element of NAT by A10,Th3;
      i1 is Element of NAT by A9,Th3;
      hence i1 = 1 & i2 = 1 by A1,A11,NAT_1:15;
    end;
    not i1 = 0 by A1;
    hence thesis by A1,A2,A8,A3;
  end;
  thus thesis;
end;

theorem
  i1 * i2 = - 1 iff i1 = - 1 & i2 = 1 or i1 = 1 & i2 = - 1
proof
  thus i1 * i2 = - 1 implies i1 = - 1 & i2 = 1 or i1 = 1 & i2 = - 1
  proof
    assume i1 * i2 = - 1;
    then (- i1) * i2 = 1;
    then - (- i1) = - 1 & i2 = 1 or - i1 = - 1 & i2 = - 1 by Th9;
    hence thesis;
  end;
  assume i1 = - 1 & i2 = 1 or i1 = 1 & i2 = - 1;
  hence thesis;
end;

Lm6: i0 <= i1 implies ex k st i0 + k = i1
proof
  assume i0 <= i1;
  then reconsider k = i1 - i0 as Element of NAT by Th5;
  i0 + k = i1;
  hence thesis;
end;

Lm7: i0 <= i1 implies ex k st i1 - k = i0
proof
  assume i0 <= i1;
  then reconsider k = i1 - i0 as Element of NAT by Th5;
  i1 - k = i0;
  hence thesis;
end;

scheme
  SepInt { P[Integer] } : ex X being Subset of INT st for x being Integer
  holds x in X iff P[x] proof
  defpred P1[set] means ex i0 st i0 = $1 & P[i0];
  consider X such that
A1: for y holds y in X iff y in INT & P1[y] from XBOOLE_0:sch 1;
  y in X implies y in INT by A1;
  then reconsider X as Subset of INT by TARSKI:def 3;
  take X;
  let i0;
A2: i0 in X implies P[i0]
  proof
    assume i0 in X;
    then ex i1 st i0 = i1 & P[i1] by A1;
    hence thesis;
  end;
  P[i0] implies i0 in X
  proof
    assume
A3: P[i0];
    i0 in INT by Def2;
    hence thesis by A1,A3;
  end;
  hence thesis by A2;
end;

scheme
  IntIndUp { F() -> Integer, P[Integer] } : for i0 st F() <= i0 holds P[i0]
provided
A1: P[F()] and
A2: for i2 st F() <= i2 holds P[i2] implies P[i2 + 1]
proof
  defpred Q[Nat] means for i2 st $1 = i2 - F() holds P[i2];
A3: for k st Q[k] holds Q[k + 1]
  proof
    let k;
    reconsider i8 = k as Integer;
    assume
A4: Q[k];
    let i2;
    assume
A5: k + 1 = i2 - F();
    then i2 - 1 = i8 + F();
    then
A6: F() <= i2 - 1 by Th6;
    k = i2 - 1 - F() by A5;
    then P[i2 - 1 + 1] by A2,A4,A6;
    hence thesis;
  end;
  let i0;
  assume F() <= i0;
  then reconsider l = i0 - F() as Element of NAT by Th5;
A7: l = i0 - F();
A8: Q[0] by A1;
  for k holds Q[k] from NAT_1:sch 1(A8,A3);
  hence thesis by A7;
end;

scheme
  IntIndDown { F() -> Integer, P[Integer] } : for i0 st i0 <= F() holds P[i0]
provided
A1: P[F()] and
A2: for i2 st i2 <= F() holds P[i2] implies P[i2 - 1]
proof
  defpred Q[Integer] means for i2 st $1 = - i2 holds P[i2];
A3: for i2 st - F() <= i2 holds Q[i2] implies Q[i2 + 1]
  proof
    let i2;
    assume that
A4: - F() <= i2 and
A5: Q[i2];
    let i3;
    assume
A6: i2 + 1 = - i3;
    then
A7: i2 = - (i3 + 1);
    - (- i3 - 1) <= - (- F()) by A4,A6,XREAL_1:24;
    then P[i3 + 1 - 1] by A2,A5,A7;
    hence thesis;
  end;
  let i0;
  assume i0 <= F();
  then
A8: - F() <= - i0 by XREAL_1:24;
A9: Q[- F()] by A1;
  for i2 st - F() <= i2 holds Q[i2] from IntIndUp(A9,A3);
  hence thesis by A8;
end;

scheme
  IntIndFull { F() -> Integer, P[Integer] } : for i0 holds P[i0]
provided
A1: P[F()] and
A2: for i2 holds P[i2] implies P[i2 - 1] & P[i2 + 1]
proof
  let i0;
A3: P[F()] by A1;
A4: now
A5: for i2 st i2 <= F() holds P[i2] implies P[i2 - 1] by A2;
A6: for i2 st i2 <= F() holds P[i2] from IntIndDown(A3,A5);
    assume i0 <= F();
    hence thesis by A6;
  end;
  now
A7: for i2 st F() <= i2 holds P[i2] implies P[i2 + 1] by A2;
A8: for i2 st F() <= i2 holds P[i2] from IntIndUp(A3,A7);
    assume F() <= i0;
    hence thesis by A8;
  end;
  hence thesis by A4;
end;

scheme
  IntMin { F() -> Integer, P[Integer] } : ex i0 st P[i0] & for i1 st P[i1]
  holds i0 <= i1
provided
A1: for i1 st P[i1] holds F() <= i1 and
A2: ex i1 st P[i1]
proof
  defpred Q[Nat] means P[F() + $1];
  consider i1 such that
A3: P[i1] by A2;
  ex k st F() + k = i1 by A1,A3,Lm6;
  then
A4: ex k be Nat st Q[k] by A3;
  consider l be Nat such that
A5: Q[l] & for n be Nat st Q[n] holds l <= n from NAT_1:sch 5(A4);
  take i0 = F() + l;
  for i1 st P[i1] holds i0 <= i1
  proof
    let i1;
    assume
A6: P[i1];
    then ex n st F() + n = i1 by A1,Lm6;
    then i0 - F() <= i1 - F() by A5,A6;
    hence thesis by XREAL_1:9;
  end;
  hence thesis by A5;
end;

scheme
  IntMax { F() -> Integer, P[Integer] } : ex i0 st P[i0] & for i1 st P[i1]
  holds i1 <= i0
provided
A1: for i1 st P[i1] holds i1 <= F() and
A2: ex i1 st P[i1]
proof
  defpred Q[Nat] means P[F() - $1];
  consider i1 such that
A3: P[i1] by A2;
  ex k st i1 = F() - k by A1,A3,Lm7;
  then
A4: ex k be Nat st Q[k] by A3;
  consider l be Nat such that
A5: Q[l] & for n be Nat st Q[n] holds l <= n from NAT_1:sch 5(A4);
  take i0 = F() - l;
  for i1 st P[i1] holds i1 <= i0
  proof
    let i1;
    assume
A6: P[i1];
    then consider n such that
A7: F() - n = i1 by A1,Lm7;
    l <= n by A5,A6,A7;
    then F() + (- i0) - F() <= F() - i1 - F() by A7,XREAL_1:9;
    then - i0 <= F() + (- i1) - F();
    hence thesis by XREAL_1:24;
  end;
  hence thesis by A5;
end;



definition
  let i1,i2 be Integer;
  pred i1 divides i2 means
  :Def3:
  ex i3 st i2 = i1 * i3;
  reflexivity
  proof
    let a be Integer;
    take 1;
    thus thesis;
  end;
end;

definition
  let i1,i2,i3 be Integer;
  pred i1,i2 are_congruent_mod i3 means
  i3 divides i1 - i2;
end;

definition
  let i1,i2,i3 be Integer;
  redefine pred i1,i2 are_congruent_mod i3 means
  :Def5:
  ex i4 st i3 * i4 = i1 - i2;
  compatibility
  proof
    thus i1,i2 are_congruent_mod i3 implies
    ex i4 st i3 * i4 = i1 - i2
    proof
      assume i3 divides i1-i2;
      hence thesis by Def3;
    end;
    given i4 such that
A1: i3 * i4 = i1 - i2;
    take i4;
    thus thesis by A1;
  end;
end;

theorem
  i1,i1 are_congruent_mod i2
proof
  i2 * 0 = i1 - i1;
  hence thesis by Def5;
end;

theorem
  i1,0 are_congruent_mod i1 & 0,i1 are_congruent_mod i1
proof
A1: i1 * (- 1) = 0 - i1;
  i1 * 1 = i1 - 0;
  hence thesis by A1,Def5;
end;

theorem
  i1,i2 are_congruent_mod 1
proof
  i1 - i2 = 1 * (i1 - i2);
  hence thesis by Def5;
end;

theorem
  i1,i2 are_congruent_mod i3 implies i2,i1 are_congruent_mod i3
proof
  assume i1,i2 are_congruent_mod i3;
  then consider i0 such that
A1: (i1 - i2) = i3 * i0 by Def5;
  i2 - i1 = i3 * (- i0) by A1;
  hence thesis by Def5;
end;

theorem
  i1,i2 are_congruent_mod i5 & i2,i3 are_congruent_mod i5 implies
  i1,i3 are_congruent_mod i5
proof
  assume that
A1: i1,i2 are_congruent_mod i5 and
A2: i2,i3 are_congruent_mod i5;
  consider i8 such that
A3: i5 * i8 = i1 - i2 by A1,Def5;
  consider i9 such that
A4: i5 * i9 = i2 - i3 by A2,Def5;
  i5 * (i8 + i9) = i1 - i3 by A3,A4;
  hence thesis by Def5;
end;

theorem
  i1,i2 are_congruent_mod i5 & i3,i4 are_congruent_mod i5 implies
  (i1 + i3),(i2 + i4) are_congruent_mod i5
proof
  assume that
A1: i1,i2 are_congruent_mod i5 and
A2: i3,i4 are_congruent_mod i5;
  consider i8 such that
A3: i5 * i8 = i1 - i2 by A1,Def5;
  consider i9 such that
A4: i5 * i9 = i3 - i4 by A2,Def5;
  i5 * (i8 + i9) = (i1 + i3) - (i2 + i4) by A3,A4;
  hence thesis by Def5;
end;

theorem
  i1,i2 are_congruent_mod i5 & i3,i4 are_congruent_mod i5 implies
  (i1 - i3),(i2 - i4) are_congruent_mod i5
proof
  assume that
A1: i1,i2 are_congruent_mod i5 and
A2: i3,i4 are_congruent_mod i5;
  consider i8 such that
A3: i5 * i8 = i1 - i2 by A1,Def5;
  consider i9 such that
A4: i5 * i9 = i3 - i4 by A2,Def5;
  (i1 - i3) - (i2 - i4) = i5 * (i8 - i9) by A3,A4;
  hence thesis by Def5;
end;

theorem
  i1,i2 are_congruent_mod i5 & i3,i4 are_congruent_mod i5 implies
  (i1 * i3),(i2 * i4) are_congruent_mod i5
proof
  assume that
A1: i1,i2 are_congruent_mod i5 and
A2: i3,i4 are_congruent_mod i5;
  consider i8 such that
A3: i5 * i8 = i1 - i2 by A1,Def5;
  consider i9 such that
A4: i5 * i9 = i3 - i4 by A2,Def5;
  (i1 * i3) - (i2 * i4) = (i1 - i2) * i3 + (i3 - i4) * i2
    .= (i5 * i8) * i3 + (i5 * i9) * i2 by A3,A4
    .= i5 * ((i8 * i3) + (i9 * i2));
  hence thesis by Def5;
end;

theorem
  (i1 + i2),i3 are_congruent_mod i5 iff i1, (i3 - i2) are_congruent_mod i5
proof
  thus (i1 + i2),i3 are_congruent_mod i5 implies i1,(i3 - i2)
  are_congruent_mod i5
  proof
    assume (i1 + i2),i3 are_congruent_mod i5;
    then
A1: ex i0 st i5 * i0 = (i1 + i2) - i3 by Def5;
    (i1 + i2) - i3 = i1 - (i3 - i2);
    hence thesis by A1,Def5;
  end;
  assume i1, (i3 - i2) are_congruent_mod i5;
  then
A2: ex i0 st i5 * i0 = i1 - (i3 - i2) by Def5;
  i1 - (i3 - i2) = (i1 + i2) - i3;
  hence thesis by A2,Def5;
end;

theorem
  i4 * i5 = i3 implies (i1,i2 are_congruent_mod i3 implies
  i1,i2 are_congruent_mod i4)
proof
  assume
A1: i4 * i5 = i3;
  assume i1,i2 are_congruent_mod i3;
  then consider i0 such that
A2: i3 * i0 = i1 - i2 by Def5;
  i1 - i2 = i4 * (i5 * i0) by A1,A2;
  hence thesis by Def5;
end;

theorem
  i1,i2 are_congruent_mod i5 iff (i1 + i5),i2 are_congruent_mod i5
proof
  thus i1,i2 are_congruent_mod i5 implies (i1 + i5),i2 are_congruent_mod i5
  proof
    assume i1,i2 are_congruent_mod i5;
    then consider i0 such that
A1: i5 * i0 = i1 - i2 by Def5;
    i5 * (i0 + 1) = (i1 + i5) - i2 by A1;
    hence thesis by Def5;
  end;
  assume (i1 + i5),i2 are_congruent_mod i5;
  then consider i0 such that
A2: i5 * i0 = (i1 + i5) - i2 by Def5;
  i5 * (i0 - 1) = i1 - i2 by A2;
  hence thesis by Def5;
end;

theorem
  i1,i2 are_congruent_mod i5 iff (i1 - i5),i2 are_congruent_mod i5
proof
  thus i1,i2 are_congruent_mod i5 implies (i1 - i5),i2 are_congruent_mod i5
  proof
    assume i1,i2 are_congruent_mod i5;
    then consider i0 such that
A1: i5 * i0 = i1 - i2 by Def5;
    i5 * (i0 - 1) = (i1 - i5) - i2 by A1;
    hence thesis by Def5;
  end;
  assume (i1 - i5),i2 are_congruent_mod i5;
  then consider i0 such that
A2: i5 * i0 = (i1 - i5) - i2 by Def5;
  i5 * (i0 + 1) = i1 - i2 by A2;
  hence thesis by Def5;
end;

theorem Th23:
  i1 <= r & r - 1 < i1 & i2 <= r & r - 1 < i2 implies i1 = i2
proof
  assume that
A1: i1 <= r and
A2: r - 1 < i1 and
A3: i2 <= r and
A4: r - 1 < i2;
  i2 = i1 + (i2 - i1);
  then consider i0 such that
A5: i2 = i1 + i0;
A6: now
    assume that
A7: 0 < i0 and
    i1 <> i2;
    1 <= i0 by A7,Lm5;
    then r - 1 + 1 < i1 + i0 by A2,XREAL_1:8;
    hence contradiction by A3,A5;
  end;
A8: now
    assume that
A9: i0 < 0 and
    i1 <> i2;
    i0 <= - 1 by A9,Th8;
    then i1 + i0 <= r + (- 1) by A1,XREAL_1:7;
    hence contradiction by A4,A5;
  end;
  i0 = 0 implies i2 = i1 by A5;
  hence thesis by A8,A6;
end;

theorem Th24:
  r <= i1 & i1 < r + 1 & r <= i2 & i2 < r + 1 implies i1 = i2
proof
  assume that
A1: r <= i1 and
A2: i1 < r + 1 and
A3: r <= i2 and
A4: i2 < r + 1;
  i2 = i1 + (i2 - i1);
  then consider i0 such that
A5: i2 = i1 + i0;
A6: now
    assume that
A7: i0 < 0 and
    i1 <> i2;
    i0 <= - 1 by A7,Th8;
    then i1 + i0 < r + 1 + (- 1) by A2,XREAL_1:8;
    hence contradiction by A3,A5;
  end;
A8: now
    assume that
A9: 0 < i0 and
    i1 <> i2;
    1 <= i0 by A9,Lm5;
    hence contradiction by A1,A4,A5,XREAL_1:7;
  end;
  i0 = 0 implies i2 = i1 by A5;
  hence thesis by A6,A8;
end;

reserve r1,p,p1,g,g1,g2 for real number,
  Y for Subset of REAL;

Lm8: for r ex n st r<n
proof
  defpred P[Real] means for r st r in NAT holds not $1<r;
  let r;
  consider Y such that
A1: for p1 being Real holds p1 in Y iff P[p1] from SUBSET_1:sch 3;
  for r,p1 st r in NAT & p1 in Y holds r <= p1 by A1;
  then consider g2 such that
A2: for r,p st r in NAT & p in Y holds r <= g2 & g2 <= p by AXIOMS:1;
A3: g2+-1<g2+0 by XREAL_1:6;
  for g ex r st r in NAT & g<r
  proof
    given g1 such that
A4: for r st r in NAT holds not g1<r;
    g1 is Real by XREAL_0:def 1;
    then
A5: g1 in Y by A1,A4;
    now
      assume not g2-1 in Y;
      then consider r1 such that
A6:   r1 in NAT and
A7:   g2-1<r1 by A1;
A8:   r1+1 in NAT by A6,AXIOMS:2;
      g2-1+1<r1+1 by A7,XREAL_1:6;
      hence contradiction by A2,A5,A8;
    end;
    hence contradiction by A2,A3;
  end;
  then consider p such that
A9: p in NAT and
A10: r<p;
  reconsider p as Element of NAT by A9;
  take p;
  thus thesis by A10;
end;

definition
  let r be real number;
  func [\ r /] -> Integer means
  :Def6:
  it <= r & r - 1 < it;
  existence
  proof
    defpred P[Integer] means r < $1;
    consider n such that
A1: - r < n by Lm8;
A2: - n < -- r by A1,XREAL_1:24;
A3: for i1 st P[i1] holds -n <= i1
    proof
      let i1;
      assume r < i1;
      then r + -n < i1 + r by A2,XREAL_1:8;
      hence thesis by XREAL_1:6;
    end;
    ex n st r < n by Lm8;
    then
A4: ex i0 st P[i0];
    consider i1 such that
A5: P[i1] & for i2 st P[i2] holds i1 <= i2 from IntMin(A3,A4);
A6: r - 1 < i1 - 1 by A5,XREAL_1:9;
    r < i1 - 1 implies i1 <= i1-1 by A5;
    hence thesis by A6,XREAL_1:146;
  end;
  uniqueness by Th23;
  projectivity by XREAL_1:44;
end;

theorem Th25:
  [\ r /] = r iff r is integer
proof
  r is Integer implies [\ r /] = r
  proof
    r + 0 < r + 1 by XREAL_1:6;
    then
A1: r - 1 < r + 1 - 1 by XREAL_1:9;
    assume r is Integer;
    hence thesis by A1,Def6;
  end;
  hence thesis;
end;

theorem Th26:
  [\ r /] < r iff r is not integer
proof
  now
    assume
A1: not r is Integer;
    [\ r /] <= r by Def6;
    hence [\ r /] < r by A1,XXREAL_0:1;
  end;
  hence thesis by Th25;
end;

theorem
  [\ r /] - 1 < r & [\ r /] < r + 1
proof
  [\ r /] <= r by Def6;
  then
A1: [\ r /] + 0 < r + 1 by XREAL_1:8;
  then [\ r /] + (- 1) < r + 1 + (- 1) by XREAL_1:6;
  hence thesis by A1;
end;

theorem Th28:
  [\ r /] + i0 = [\ r + i0 /]
proof
  r - 1 < [\ r /] by Def6;
  then r - 1 + i0 < [\ r /] + i0 by XREAL_1:6;
  then
A1: r + i0 - 1 < [\ r /] + i0;
  [\ r /] <= r by Def6;
  then [\ r /] + i0 <= r + i0 by XREAL_1:6;
  hence thesis by A1,Def6;
end;

theorem Th29:
  r < [\ r /] + 1
proof
  r - 1 < [\ r /] by Def6;
  then r - 1 + 1 < [\ r /] + 1 by XREAL_1:6;
  hence thesis;
end;

definition
  let r be real number;
  func [/ r \] -> Integer means
  :Def7:
  r <= it & it < r + 1;
  existence
  proof
A1: now
A2:   r + 0 < r + 1 by XREAL_1:6;
      assume r is Integer;
      hence r <= [\ r /] & [\ r /] < r + 1 by A2,Th25;
    end;
    now
      assume not r is Integer;
      then [\ r /] < r by Th26;
      hence r <= [\ r /] + 1 & [\ r /] + 1 < r + 1 by Th29,XREAL_1:6;
    end;
    hence thesis by A1;
  end;
  uniqueness by Th24;
  projectivity by XREAL_1:29;
end;

theorem Th30:
  [/ r \] = r iff r is integer
proof
  r + 0 < r + 1 by XREAL_1:6;
  hence thesis by Def7;
end;

theorem Th31:
  r < [/ r \] iff r is not integer
proof
  now
    assume
A1: not r is Integer;
    r <= [/ r \] by Def7;
    hence r < [/ r \] by A1,XXREAL_0:1;
  end;
  hence thesis by Th30;
end;

theorem
  r - 1 < [/ r \] & r < [/ r \] + 1
proof
  r <= [/ r \] by Def7;
  then
A1: r + 0 < [/ r \] + 1 by XREAL_1:8;
  then r + (- 1) < [/ r \] + 1 + (- 1) by XREAL_1:6;
  hence thesis by A1;
end;

theorem
  [/ r \] + i0 = [/ r + i0 \]
proof
  [/ r \] < r + 1 by Def7;
  then [/ r \] + i0 < r + 1 + i0 by XREAL_1:6;
  then
A1: [/ r \] + i0 < r + i0 + 1;
  r <= [/ r \] by Def7;
  then r + i0 <= [/ r \] + i0 by XREAL_1:6;
  hence thesis by A1,Def7;
end;

theorem Th34:
  [\ r /] = [/ r \] iff r is integer
proof
A1: now
    assume
A2: not r is Integer;
    then [\ r /] < r by Th26;
    hence [\ r /] <> [/ r \] by A2,Th31;
  end;
  now
    assume r is Integer;
    hence [\ r /] = r & r = [/ r \] by Th25,Th30;
    hence [\ r /] = [/ r \];
  end;
  hence thesis by A1;
end;

theorem Th35:
  [\ r /] < [/ r \] iff r is not integer
proof
  now
    assume
A1: not r is Integer;
    then
A2: r < [/ r \] by Th31;
    [\ r /] < r by A1,Th26;
    hence [\ r /] < [/ r \] by A2,XXREAL_0:2;
  end;
  hence thesis by Th34;
end;

theorem
  [\ r /] <= [/ r \]
proof
A1: r <= [/ r \] by Def7;
  [\ r /] <= r by Def6;
  hence thesis by A1,XXREAL_0:2;
end;

theorem
  [\ ([/ r \]) /] = [/ r \] by Th25;

canceled 2;

theorem
  [/ ([\ r /]) \] = [\ r /] by Th30;

theorem
  [\ r /] = [/ r \] iff not [\ r /] + 1 = [/ r \]
proof
  set Diff = [/ r \] + (- [\ r /]);
A1: now
    assume not r is Integer;
    then [\ r /] < [/ r \] by Th35;
    then [\ r /] + (- [\ r /]) < Diff by XREAL_1:6;
    then
A2: 1 <= Diff by Lm5;
    r - 1 < [\ r /] by Def6;
    then
A3: - [\ r /] < - (r - 1) by XREAL_1:24;
    [/ r \] < r + 1 by Def7;
    then Diff < r + 1 + (- (r - 1)) by A3,XREAL_1:8;
    then Diff + 1 + (- 1) <= 1 + 1 + (- 1) by Th7;
    then [\ r /] + 1 = [\ r /] + Diff by A2,XXREAL_0:1;
    hence [\ r /] + 1 = [/ r \] & [\ r /] <> [/ r \];
  end;
  now
    assume r is Integer;
    then [\ r /] = [/ r \] by Th34;
    hence [\ r /] = [/ r \] & [\ r /] + 1 <> [/ r \];
  end;
  hence thesis by A1;
end;

definition
  let r be real number;
  func frac r equals
  r - [\ r /];
  coherence;
end;

registration
  let r be real number;
  cluster frac r -> real;
  coherence;
end;

definition
  let r be real number;
  redefine func frac r -> Real;
  coherence by XREAL_0:def 1;
end;

theorem
  r = [\ r /] + frac r;

theorem Th43:
  frac r < 1 & 0 <= frac r
proof
  r - 1 < [\ r /] by Def6;
  then frac r + (r - 1) < r - [\ r /] + [\ r /] by XREAL_1:6;
  then frac r + (- 1) + r - r < r - r by XREAL_1:9;
  then
A1: frac r - 1 + 1 < 0 + 1 by XREAL_1:6;
  [\ r /] <= r by Def6;
  then [\ r /] + (r - [\ r /]) <= r + frac r by XREAL_1:6;
  then r - r <= r + frac r - r by XREAL_1:9;
  hence thesis by A1;
end;

theorem
  [\ frac r /] = 0
proof
  [\ frac r /] = [\ (r + (- [\ r /])) /] .= [\ r /] + (- [\ r /]) by Th28
    .= 0;
  hence thesis;
end;

theorem Th45:
  frac r = 0 iff r is integer
proof
  now
    assume r is Integer;
    then r = [\ r /] by Th25;
    hence frac r = 0;
  end;
  hence thesis;
end;

theorem
  0 < frac r iff r is not integer
proof
  now
    assume not r is Integer;
    then 0 <> frac r;
    hence 0 < frac r by Th43;
  end;
  hence thesis by Th45;
end;



definition
  let i1,i2 be Integer;
  func i1 div i2 -> Integer equals
  [\ i1 / i2 /];
  correctness;
end;

definition
  let i1,i2 be Integer;
  func i1 mod i2 -> Integer equals
  :Def10:
  i1 - (i1 div i2) * i2 if i2 <> 0
  otherwise 0;
  correctness;
end;

theorem Th47:
  for r being real number st r <> 0 holds [\ r / r /] = 1
proof
  let r be real number;
  assume r <> 0;
  hence [\ r / r /] = [\ 1 /] by XCMPLX_1:60
    .= 1 by Th25;
end;

theorem
  for i being Integer holds i div 0 = 0
proof
  let i be Integer;
A1: i / 0 = i * 0" by XCMPLX_0:def 9
    .= i * 0;
  0 - 1 < 0;
  hence thesis by A1,Def6;
end;

theorem
  for i being Integer st i <> 0 holds i div i = 1 by Th47;

theorem
  for i being Integer holds i mod i = 0
proof
  let i be Integer;
  per cases;
  suppose
    i = 0;
    hence thesis by Def10;
  end;
  suppose
A1: i <> 0;
    hence i mod i = i - (i div i) * i by Def10
      .= i - 1 * i by A1,Th47
      .= 0;
  end;
end;

begin 


theorem
  for k, i being Integer holds
  k < i implies ex j being Element of NAT st j = i-k & 1 <= j
proof
  let k,i be Integer;
  assume k < i;
  then
A1: k - k < i - k by XREAL_1:9;
  then reconsider j = i - k as Element of NAT by Th3;
  reconsider j9 = j, 09 = 0 as Integer;
  take j;
  thus j = i - k;
  09 + 1 <= j9 by A1,Th7;
  hence thesis;
end;



theorem Th52:
  for a,b being Integer st a < b holds a <= b - 1
proof
  let a,b be Integer;
  assume a < b;
  then a - 1 < b - 1 by XREAL_1:9;
  then a - 1 + 1 <= b - 1 by Th7;
  hence thesis;
end;



theorem
  for r being real number st r>=0 holds [/ r \]>=0 & [\ r /]>=0 &
  [/ r \] is Element of NAT & [\ r /] is Element of NAT
proof
  let r be real number;
  assume
A1: r>=0;
A2: r<=[/ r \] by Def7;
  r-1<[\ r /] by Def6;
  then r-1+1<[\ r /]+1 by XREAL_1:6;
  then 0-1<[\ r /]+1-1 by A1,XREAL_1:9;
  then [\ r /]>=0 by Th8;
  hence thesis by A1,A2,Th3;
end;



theorem Th54:
  for i being Integer, r being real number st i <= r holds i <= [\ r /]
proof
  let i be Integer;
  let r be real number;
  assume i <= r;
  then
A1: i-1 <= r-1 by XREAL_1:9;
  r-1 < [\ r /] by Def6;
  then i-1 < [\ r /] by A1,XXREAL_0:2;
  then i-1+1 <= [\ r /] by Th7;
  hence thesis;
end;

theorem
  for m,n being Nat holds 0 <= m qua Integer div n by Th54;



theorem
  0 < i & 1 < j implies i div j < i
proof
  assume that
A1: 0 < i and
A2: 1 < j;
  assume
A3: i <= i div j;
  i div j <= i/j by Def6;
  then j * (i div j) <= j * (i/j) by A2,XREAL_1:64;
  then j * (i div j) <= i by A2,XCMPLX_1:87;
  then j * (i div j) <= i div j by A3,XXREAL_0:2;
  then j * (i div j) * (i div j)" <= (i div j) * (i div j)" by A1,A3,XREAL_1:64
;
  then j * ((i div j) * (i div j)") <= (i div j) * (i div j)";
  then j * 1 <= (i div j) * (i div j)" by A1,A3,XCMPLX_0:def 7;
  hence contradiction by A1,A2,A3,XCMPLX_0:def 7;
end;



theorem
  i2 >= 0 implies i1 mod i2 >= 0
proof
  assume
A1: i2 >= 0;
  per cases by A1;
  suppose
A2: i2 > 0;
    [\ i1/i2 /] <= i1/i2 by Def6;
    then (i1 div i2)*i2 <= (i1/i2)*i2 by A2,XREAL_1:64;
    then (i1 div i2)*i2 <= i1 by A2,XCMPLX_1:87;
    then i1 - (i1 div i2)*i2 >= 0 by XREAL_1:48;
    hence thesis by Def10;
  end;
  suppose
    i2 = 0;
    hence thesis by Def10;
  end;
end;

theorem
  i2 > 0 implies i1 mod i2 < i2
proof
  assume
A1: i2 > 0;
  i1/i2 -1 < [\ i1/i2 /] by Def6;
  then (i1/i2 -1)*i2 < (i1 div i2)*i2 by A1,XREAL_1:68;
  then i1/i2*i2 -1*i2 < (i1 div i2)*i2;
  then i1 -i2 < (i1 div i2)*i2-0 by A1,XCMPLX_1:87;
  then i1 -(i1 div i2)*i2 < i2-0 by XREAL_1:17;
  hence thesis by A1,Def10;
end;

theorem
  i2 <> 0 implies i1 = (i1 div i2) * i2 + (i1 mod i2)
proof
  assume i2 <> 0;
  then
  (i1 div i2) * i2 +(i1 mod i2) = (i1 div i2 )*i2 + (i1 - ( i1 div i2 )*i2
  ) by Def10
    .= i1;
  hence thesis;
end;



theorem
  for m,j being Integer holds m*j, 0 are_congruent_mod m
proof
  let m,j be Integer;
  take j;
  thus m*j = m*j - 0;
end;



theorem
  i >= 0 & j >= 0 implies i div j >= 0
proof
  assume that
A1: i >= 0 and
A2: j >= 0;
A3: i / j - 1 < [\ i / j /] by Def6;
  i / j - 1 >= 0-1 by A1,A2,XREAL_1:9;
  then - 1 < [\ i / j /] by A3,XXREAL_0:2;
  hence thesis by Th8;
end;



theorem
  for n being Nat st n > 0 for a being Integer holds a mod n
  = 0 iff n divides a
proof
  let n be Nat;
  assume
A1: n > 0;
  let a be Integer;
A2: now
A3: a mod n = a - (a div n) * n by A1,Def10
      .= a - ([\ a/n /]) * n;
    assume n divides a;
    then consider k being Integer such that
A4: n * k = a by Def3;
    -n <> 0 by A1;
    then -n + a < 0 + a by XREAL_1:6;
    then (-n + a) * n" < (n * k) * n" by A1,A4,XREAL_1:68;
    then (-n) * n" + a * n" < (n * k) * n";
    then (-n) * n" + a/n < (n * n") * k by XCMPLX_0:def 9;
    then (-n) * n" + a/n < 1 * k by A1,XCMPLX_0:def 7;
    then -n * n" + a/n < k;
    then -1 + a/n < k by A1,XCMPLX_0:def 7;
    then
A5: a/n - 1 < k;
    k <= a/n
    proof
      assume k > a/n;
      then k * n > a/n * n by A1,XREAL_1:68;
      then k * n > (a * n") * n by XCMPLX_0:def 9;
      then k * n > a * (n" * n);
      then n * k > a * 1 by A1,XCMPLX_0:def 7;
      hence thesis by A4;
    end;
    then ([\ a/n /]) = k by A5,Def6;
    hence a mod n = 0 by A4,A3;
  end;
  now
    assume a mod n = 0;
    then 0 = a - (a div n) * n by A1,Def10;
    hence n divides a by Def3;
  end;
  hence thesis by A2;
end;



reserve r, s for real number;

theorem
  r/s is not Integer implies - [\ r/s /] = [\ (-r) / s /] + 1
proof
  r/s - 1 < [\ r/s /] by Def6;
  then - (r/s - 1) > - [\ r/s /] by XREAL_1:24;
  then -r/s + 1 > - [\ r/s /];
  then - [\ r/s /] <= (-r) / s + 1 by XCMPLX_1:187;
  then
A1: - [\ r/s /] - 1 <= (-r) / s + 1 - 1 by XREAL_1:9;
  assume r/s is not Integer;
  then [\ r/s /] < r/s by Th26;
  then -r/s < - [\ r/s /] by XREAL_1:24;
  then -r/s - 1 < - [\ r/s /] - 1 by XREAL_1:9;
  then (-r)/s - 1 < - [\ r/s /] - 1 by XCMPLX_1:187;
  then - [\ r/s /] - 1 + 1 = [\ (-r) / s /] + 1 by A1,Def6;
  hence thesis;
end;

theorem
  r/s is Integer implies - [\ r/s /] = [\ (-r) / s /]
proof
  assume r/s is Integer;
  then
A1: [\ r/s /] = r/s by Th25;
A2: -r/s = (-r)/s by XCMPLX_1:187;
  then (-r) / s - 1 < - [\ r/s /] - 0 by A1,XREAL_1:15;
  hence thesis by A1,A2,Def6;
end;



theorem
  r <= i implies [/ r \] <= i
proof
  assume r <= i;
  then
A1: r+1 <= i+1 by XREAL_1:6;
  [/ r \] < r+1 by Def7;
  then [/ r \] < i+1 by A1,XXREAL_0:2;
  then [/ r \] <= i+1-1 by Th52;
  hence thesis;
end;



scheme
  FinInd{M, N() -> Element of NAT, P[Nat]} : for i being Element of NAT st M()
  <= i & i <= N() holds P[i]
provided
A1: P[M()] and
A2: for j being Element of NAT st M() <= j & j < N() holds P[j] implies P[j+1]
proof
  defpred Q[Nat] means M() <= $1 & $1 <= N() & not(P[$1]);
  assume not(for i being Element of NAT st M() <= i & i <= N() holds P[i]);
  then
A3: ex i being Nat st Q[i];
  consider k being Nat such that
A4: Q[k] & for k9 being Nat st Q[k9] holds k <= k9 from NAT_1:sch 5(A3);
  per cases;
  suppose
    k = M();
    hence thesis by A1,A4;
  end;
  suppose
    k <> M();
    then M() < k by A4,XXREAL_0:1;
    then M() + 1 <= k by NAT_1:13;
    then
A5: (M() + 1) - 1 <= k - 1 by XREAL_1:9;
    then reconsider k9 = k - 1 as Element of NAT by Th3;
A6: k9 <> k9 + 1;
    k9 <= k9 + 1 by NAT_1:11;
    then
A7: k9 < k by A6,XXREAL_0:1;
    then
A8: k9 < N() by A4,XXREAL_0:2;
A9: (k - 1) + 1 = k + 0;
    not(Q[k9]) by A4,A7;
    hence thesis by A2,A4,A5,A9,A8;
  end;
end;

scheme
  FinInd2{M,N() -> Element of NAT, P[Nat]} : for i being Element of NAT st M()
  <= i & i <= N() holds P[i]
provided
A1: P[M()] and
A2: for j being Element of NAT st M() <= j & j < N() holds (for j9 being
Element of NAT st M() <= j9 & j9 <= j holds P[j9]) implies P[j+1]
proof
  defpred Q[Element of NAT] means for j being Element of NAT st M() <= j & j
  <= ($1) holds P[j];
A3: for j being Element of NAT st M() <= j & j < N() holds Q[j] implies Q[j+ 1]
  proof
    let j be Element of NAT;
    assume that
A4: M() <= j and
A5: j < N();
    assume
A6: Q[j];
    thus Q[j+1]
    proof
      let i be Element of NAT;
      assume that
A7:   M() <= i and
A8:   i <= j + 1;
      per cases;
      suppose
        i = j + 1;
        hence thesis by A2,A4,A5,A6;
      end;
      suppose
        i <> j + 1;
        then i < j + 1 by A8,XXREAL_0:1;
        then i <= j by NAT_1:13;
        hence thesis by A6,A7;
      end;
    end;
  end;
A9: Q[M()] by A1,XXREAL_0:1;
  for i being Element of NAT st M() <= i & i <= N() holds Q[i] from
  FinInd(A9,A3);
  hence thesis;
end;



reserve i for Integer,
  a, b, r, s for real number;

theorem
  frac(r+i) = frac r
proof
  thus frac(r+i) = r+i-[\r+i/]
    .= r+i-([\r/]+i) by Th28
    .= r-[\r/]+i-i
    .= frac r;
end;

theorem Th67:
  r <= a & a < [\r/]+1 implies [\a/] = [\r/]
proof
  assume
A1: r <= a;
  assume a < [\r/]+1;
  then
A2: a-1 < [\r/]+1-1 by XREAL_1:9;
  [\r/] <= r by Def6;
  then [\r/] <= a by A1,XXREAL_0:2;
  hence thesis by A2,Def6;
end;

theorem Th68:
  r <= a & a < [\r/]+1 implies frac r <= frac a
proof
  assume
A1: r <= a;
  assume a < [\r/]+1;
  then [\a/] = [\r/] by A1,Th67;
  hence thesis by A1,XREAL_1:9;
end;

theorem
  r < a & a < [\r/]+1 implies frac r < frac a
proof
  assume
A1: r < a;
  assume a < [\r/]+1;
  then [\a/] = [\r/] by A1,Th67;
  hence thesis by A1,XREAL_1:9;
end;

theorem Th70:
  a >= [\r/]+1 & a <= r+1 implies [\a/] = [\r/]+1
proof
  assume
A1: a >= [\r/]+1;
  assume a <= r+1;
  then a-1 <= r+1-1 by XREAL_1:9;
  then a-1 < [\r/]+1 by Th29,XXREAL_0:2;
  hence thesis by A1,Def6;
end;

theorem Th71:
  a >= [\r/]+1 & a < r+1 implies frac a < frac r
proof
  assume
A1: a >= [\r/]+1;
  assume
A2: a < r+1;
  then a-1 < r by XREAL_1:19;
  then
A3: frac a = a-[\a/] & a-1-[\r/] < r-[\r/] by XREAL_1:14;
  [\a/] = [\r/]+1 by A1,A2,Th70;
  hence thesis by A3;
end;

theorem
  r <= a & a < r+1 & r <= b & b < r+1 & frac a = frac b implies a = b
proof
  assume that
A1: r <= a and
A2: a < r+1 and
A3: r <= b and
A4: b < r+1 and
A5: frac a = frac b;
A6: [\r/] <= r by Def6;
  then
A7: [\r/] <= a by A1,XXREAL_0:2;
A8: [\r/] <= b by A3,A6,XXREAL_0:2;
  per cases;
  suppose
A9: a < [\r/]+1 & b >= [\r/]+1;
    then frac a >= frac r by A1,Th68;
    hence thesis by A4,A5,A9,Th71;
  end;
  suppose
A10: a >= [\r/]+1 & b < [\r/]+1;
    then frac a < frac r by A2,Th71;
    hence thesis by A3,A5,A10,Th68;
  end;
  suppose
A11: a < [\r/]+1 & b < [\r/]+1;
    then b-1 < [\r/]+1-1 by XREAL_1:9;
    then
A12: [\b/] = [\r/] by A8,Def6;
    a-1 < [\r/]+1-1 by A11,XREAL_1:9;
    then [\a/] = [\r/] by A7,Def6;
    hence thesis by A5,A12;
  end;
  suppose
    a >= [\r/]+1 & b >= [\r/]+1;
    then [\a/] = [\r/]+1 & [\b/] = [\r/]+1 by A2,A4,Th70;
    hence thesis by A5;
  end;
end;















begin

reserve x for set,
  a,b for real number,
  k,k1,l,l1 for Element of NAT,
  m,m1,n, n1 for Integer;




reserve i1,j1 for Element of NAT;

Lm1: omega c= ({[c,d] where c,d is Element of omega: c,d are_relative_prime &
d <> {}} \ {[k,1] where k is Element of omega: not contradiction}) \/ omega by
XBOOLE_1:7;

Lm2: quotient(i1,j1) = i1/j1
proof
  set x = quotient(i1,j1);
  per cases;
  suppose
A1: j1 = {};
    hence x = {} by ARYTM_3:def 10
      .= i1/j1 by A1;
  end;
  suppose
A2: j1 <> 0;
    +(0,opp 0) = 0 by ARYTM_0:def 3;
    then
A3: opp 0 = 0 by ARYTM_0:11;
    j1 * j1" = 1 by A2,XCMPLX_0:def 7;
    then consider x1,x2,y1,y2 being Real such that
A4: j1 = [*x1,x2*] and
A5: j1" = [*y1,y2*] and
A6: 1 = [* +(*(x1,y1),opp*(x2,y2)), +(*(x1,y2),*(x2,y1)) *] by XCMPLX_0:def 5;
    +(*(x1,y2),*(x2,y1)) = 0 by A6,ARYTM_0:24;
    then
A7: 1 = +(*(x1,y1),opp*(x2,y2)) by A6,ARYTM_0:def 5
      .= +(*(x1,y1),opp 0) by A4,ARYTM_0:12,24
      .= *(x1,y1) by A3,ARYTM_0:11;
    x2 = 0 by A4,ARYTM_0:24;
    then
A8: j1 = x1 by A4,ARYTM_0:def 5;
    then x1 in omega;
    then reconsider xx = x1 as Element of RAT+ by Lm1;
    consider yy being Element of RAT+ such that
A9: xx *' yy = one by A2,A8,ARYTM_3:54;
A10: xx in RAT+;
    yy in RAT+;
    then reconsider xx1 = xx, yy1 = yy as Element of REAL+ by A10,ARYTM_2:1;
A11: ex x9,y9 being Element of RAT+ st xx1 = x9 & yy1 = y9 & xx1 *' yy1 =
    x9 *' y9 by ARYTM_2:21;
    yy1 in REAL+;
    then reconsider xx1, yy1 as Element of REAL by ARYTM_0:1;
    ex x9,y9 being Element of REAL+ st xx1 = x9 & yy1 = y9 & *(xx1,yy1) =
    x9 *' y9 by ARYTM_0:def 2;
    then
A12: yy = y1 by A2,A8,A7,A9,A11,ARYTM_0:18;
    then
A13: y1 in RAT+;
    y2 = 0 by A5,ARYTM_0:24;
    then
A14: j1" = y1 by A5,ARYTM_0:def 5;
A15: j1 in omega;
    then reconsider a = x, b = j1 as Element of RAT+ by Lm1;
    consider x91,x92,y91,y92 being Real such that
A16: i1 = [*x91,x92*] and
A17: j1" = [*y91,y92*] and
A18: i1*j1" = [* +(*(x91,y91),opp*(x92,y92)), +(*(x91,y92),*(x92,y91))
    *] by XCMPLX_0:def 5;
A19: +(*(x91,y92),*(x92,y91)) = 0 by A18,ARYTM_0:24;
    x1 in omega by A8;
    then consider x19,y19 being Element of REAL+ such that
A20: x1 = x19 and
A21: y1 = y19 and
A22: *(x1,y1) = x19 *' y19 by A13,ARYTM_0:def 2,ARYTM_2:1,2;
A23: ex x199,y199 being Element of RAT+ st x19 = x199 & y19 = y199 & x19
    *' y19 = x199 *' y199 by A8,A12,A15,A20,A21,Lm1,ARYTM_2:21;
A24: y92 = 0 by A17,ARYTM_0:24;
    then j1" = y91 by A17,ARYTM_0:def 5;
    then
A25: i1*j1" = +(*(x91,y1),opp*(x92,0)) by A14,A18,A24,A19,ARYTM_0:def 5
      .= +(*(x91,y1),0) by A3,ARYTM_0:12
      .= *(x91,y1) by ARYTM_0:11;
    x92 = 0 by A16,ARYTM_0:24;
    then
A26: i1 = x91 by A16,ARYTM_0:def 5;
    then
A27: x91 in omega;
    then x91 in RAT+ by Lm1;
    then consider x9,y9 being Element of REAL+ such that
A28: x91 = x9 and
A29: y1 = y9 and
A30: i1 * j1" = x9 *' y9 by A25,A13,ARYTM_0:def 2,ARYTM_2:1;
    consider x99,y99 being Element of RAT+ such that
A31: x9 = x99 and
A32: y9 = y99 and
A33: i1 * j1" = x99 *' y99 by A27,A12,A28,A29,A30,Lm1,ARYTM_2:21;
    a *' b = quotient(i1,j1) *' quotient(j1,one) by ARYTM_3:40
      .= quotient(i1*^j1,j1*^one) by ARYTM_3:49
      .= quotient(i1,one) *' quotient(j1,j1) by ARYTM_3:49
      .= quotient(i1,one) *' one by A2,ARYTM_3:41
      .= quotient(i1,one) by ARYTM_3:53
      .= i1 by ARYTM_3:40
      .= x99 *' one by A26,A28,A31,ARYTM_3:53
      .= x99 *' y99 *' b by A8,A7,A29,A32,A20,A21,A22,A23,ARYTM_3:52;
    hence thesis by A2,A33,ARYTM_3:56;
  end;
end;

0 in omega;

then reconsider 09 = 0 as Element of REAL+ by ARYTM_2:2;

Lm3: for x9 being Element of REAL+ st x9 = a holds 09 - x9 = -a
proof
  let xx being Element of REAL+ such that
A1: xx = a;
  per cases;
  suppose
    xx = 0;
    hence thesis by A1,ARYTM_1:18;
  end;
  suppose
A2: xx <> 0;
    set b = 09 - xx;
A3: 09 <=' xx by ARYTM_1:6;
    then not xx <=' 09 by A2,ARYTM_1:4;
    then
A4: b = [{},xx -' 09] by ARYTM_1:def 2;
    now
      assume b = [0,0];
      then xx -' 09 = 0 by A4,XTUPLE_0:1;
      hence contradiction by A2,A3,ARYTM_1:10;
    end;
    then
A5: not b in {[0,0]} by TARSKI:def 1;
A6: xx -' 09 = xx -' 09 + 09 by ARYTM_2:def 8
      .= xx by A3,ARYTM_1:def 1;
    0 in {0} by TARSKI:def 1;
    then
A7: b in [:{0},REAL+:] by A4,ZFMISC_1:87;
    then b in REAL+ \/ [:{0},REAL+:] by XBOOLE_0:def 3;
    then reconsider b as Element of REAL by A5,NUMBERS:def 1,XBOOLE_0:def 5;
    consider x1,x2,y1,y2 being Element of REAL such that
A8: a = [*x1,x2*] and
A9: b = [*y1,y2*] and
A10: a+b = [*+(x1,y1),+(x2,y2)*] by XCMPLX_0:def 4;
    +(x2,y2) = 0 by A10,ARYTM_0:24;
    then
A11: a+b = +(x1,y1) by A10,ARYTM_0:def 5;
    a in REAL by XREAL_0:def 1;
    then x2 = 0 by A8,ARYTM_0:24;
    then
A12: a = x1 by A8,ARYTM_0:def 5;
    y2 = 0 by A9,ARYTM_0:24;
    then
A13: b = y1 by A9,ARYTM_0:def 5;
    then consider x9,y9 being Element of REAL+ such that
A14: x1 = x9 and
A15: y1 = [0,y9] and
A16: +(x1,y1) = x9 - y9 by A1,A7,A12,ARYTM_0:def 1;
    y9 = xx -' 09 by A4,A13,A15,XTUPLE_0:1;
    then a + b = 0 by A1,A12,A11,A14,A16,A6,ARYTM_1:18;
    hence thesis;
  end;
end;

Lm4: x in RAT implies ex m,n st x = m/n
proof
  assume
A1: x in RAT;
  per cases by A1,NUMBERS:def 3,XBOOLE_0:def 3;
  suppose
    x in RAT+;
    then reconsider x as Element of RAT+;
    take numerator x,denominator x;
    quotient(numerator x,denominator x) = x by ARYTM_3:39;
    hence thesis by Lm2;
  end;
  suppose
A2: x in [:{0},RAT+:];
A3: not x in {[0,0]} by A1,NUMBERS:def 3,XBOOLE_0:def 5;
    consider x1,x2 being set such that
A4: x1 in {0} and
A5: x2 in RAT+ and
A6: x = [x1,x2] by A2,ZFMISC_1:84;
    reconsider x2 as Element of RAT+ by A5;
    reconsider x9 = x2 as Element of REAL+ by A5,ARYTM_2:1;
    x = [0,x9] by A4,A6,TARSKI:def 1;
    then
A7: x2 <> 0 by A3,TARSKI:def 1;
    take m = -numerator x2, n = denominator x2;
A8: x2 = quotient(numerator x2,denominator x2) by ARYTM_3:39
      .= numerator x2/n by Lm2;
    x1 = 0 by A4,TARSKI:def 1;
    hence x = 09 - x9 by A6,A7,ARYTM_1:19
      .= -numerator x2/n by A8,Lm3
      .= m/n;
  end;
end;

Lm5: x = m/l implies x in RAT
proof
A1: m is Element of INT by INT_1:def 2;
  assume
A2: x = m/l;
  then
A3: not x in {[0,0]} by NUMBERS:def 1,XBOOLE_0:def 5;
  per cases;
  suppose
    l = 0;
    then x = m * 0" by A2
      .= 0;
    then x in RAT+ \/ [:{0},RAT+:] by Lm1,XBOOLE_0:def 3;
    hence thesis by A3,NUMBERS:def 3,XBOOLE_0:def 5;
  end;
  suppose
    ex k st m = k;
    then consider k such that
A4: m = k;
    x = quotient(k,l) by A2,A4,Lm2;
    then x in RAT+ \/ [:{0},RAT+:] by XBOOLE_0:def 3;
    hence thesis by A3,NUMBERS:def 3,XBOOLE_0:def 5;
  end;
  suppose that
A5: l <> 0 and
A6: for k holds m <> k;
    consider k such that
A7: m = -k by A1,A6,INT_1:def 1;
A8: k/l = quotient(k,l) by Lm2;
    then k/l in RAT+;
    then reconsider x9 = k/l as Element of REAL+ by ARYTM_2:1;
A9: 0 in {0} by TARSKI:def 1;
A10: x = -k/l by A2,A7;
    m <> 0 by A6;
    then
A11: x9 <> 0 by A2,A5,A10,XCMPLX_1:50;
    x = 09 - x9 by A10,Lm3
      .= [0,x9] by A11,ARYTM_1:19;
    then x in [:{0},RAT+:] by A8,A9,ZFMISC_1:87;
    then x in RAT+ \/ [:{0},RAT+:] by XBOOLE_0:def 3;
    hence thesis by A3,NUMBERS:def 3,XBOOLE_0:def 5;
  end;
end;

Lm6: x = m/n implies x in RAT
proof
  assume
A1: x = m/n;
  n is Element of INT by INT_1:def 2;
  then consider k such that
A2: n = k or n = -k by INT_1:def 1;
  per cases by A2;
  suppose
    n = k;
    hence thesis by A1,Lm5;
  end;
  suppose
    n = -k;
    then x = (-m)/k by A1,XCMPLX_1:192;
    hence thesis by Lm5;
  end;
end;

definition
  redefine func RAT means
  :Def1:
  x in it iff ex m,n st x = m/n;
  compatibility
  proof
    let R be set;
    thus R = RAT implies for x holds x in R iff ex m,n st x = m/n by Lm4,Lm6;
    assume
A1: for x holds x in R iff ex m,n st x = m/n;
    thus R c= RAT
    proof
      let x;
      assume x in R;
      then ex m,n st x = m/n by A1;
      hence thesis by Lm6;
    end;
    let x;
    assume x in RAT;
    then ex m,n st x = m/n by Lm4;
    hence thesis by A1;
  end;
end;

definition
  let r be number;
  attr r is rational means
  :Def2:
  r in RAT;
end;

registration
  cluster rational for Real;
  existence
  proof
    reconsider a = 0, b = 1 as Integer;
    take x=a/b;
    thus x is Real;
    thus x in RAT by Def1;
  end;
end;

registration
  cluster rational for number;
  existence
  proof
    reconsider a = 0, b = 1 as Integer;
    take x=a/b;
    thus x in RAT by Def1;
  end;
end;

definition
  mode Rational is rational number;
end;

theorem Th1:
  x in RAT implies ex m,n st n<>0 & x = m/n
proof
  assume
A1: x in RAT;
  per cases;
  suppose
    x = 0;
    then x = 0/1;
    hence thesis;
  end;
  suppose
A2: x <> 0;
    consider m,n such that
A3: x = m/n by A1,Def1;
    take m,n;
    hereby
      assume n = 0;
      then n" = 0;
      hence contradiction by A2,A3;
    end;
    thus thesis by A3;
  end;
end;

theorem Th2:
  x is Rational implies ex m,n st n<>0 & x = m/n
proof
  assume x is Rational;
  then x in RAT by Def2;
  hence thesis by Th1;
end;

registration
  cluster rational -> real for number;
  coherence
  proof
    let n be number;
    assume n in RAT;
    hence n in REAL by NUMBERS:12;
  end;
end;

theorem Th3:
  m/n is rational
proof
  set x = m/n;
  x is Rational iff x in RAT by Def2;
  hence thesis by Def1;
end;

registration
  cluster integer -> rational for number;
  coherence
  proof
    let x be number;
    assume x is integer;
    then reconsider x as Integer;
    x = x/1;
    hence thesis by Th3;
  end;
end;

reserve p,q for Rational;




registration
  let p,q;
  cluster p*q -> rational;
  coherence
  proof
    consider m2,n2 being Integer such that
    n2<>0 and
A1: q = m2/n2 by Th2;
    consider m1,n1 being Integer such that
    n1<>0 and
A2: p = m1/n1 by Th2;
    p*q=(m1*m2)/(n1*n2) by A2,A1,XCMPLX_1:76;
    hence thesis by Th3;
  end;
  cluster p+q -> rational;
  coherence
  proof
    consider m2,n2 being Integer such that
A3: n2<>0 and
A4: q = m2/n2 by Th2;
    consider m1,n1 being Integer such that
A5: n1<>0 and
A6: p = m1/n1 by Th2;
    p+q=(m1*n2+m2*n1)/(n1*n2) by A5,A6,A3,A4,XCMPLX_1:116;
    hence thesis by Th3;
  end;
  cluster p-q -> rational;
  coherence
  proof
    consider m2,n2 being Integer such that
A7: n2<>0 and
A8: q = m2/n2 by Th2;
    consider m1,n1 being Integer such that
A9: n1<>0 and
A10: p = m1/n1 by Th2;
    p-q=(m1*n2-m2*n1)/(n1*n2) by A9,A10,A7,A8,XCMPLX_1:130;
    hence thesis by Th3;
  end;
  cluster p/q -> rational;
  coherence
  proof
    consider m1,n1 being Integer such that
    n1<>0 and
A11: p = m1/n1 by Th2;
    consider m2,n2 being Integer such that
    n2<>0 and
A12: q = m2/n2 by Th2;
    p/q = p*(1/(m2/n2)) by A12
    .= p*((1*n2)/m2) by XCMPLX_1:77
    .= (m1*n2)/(n1*m2) by A11,XCMPLX_1:76;
    hence thesis by Th3;
  end;
end;

registration
  let p;
  cluster -p -> rational;
  coherence
  proof
    consider m,n such that
    n<>0 and
A1: p=m/n by Th2;
    -p = (-m)/n by A1;
    hence thesis;
  end;
  cluster p" -> rational;
  coherence
  proof
    p" = 1/p;
    hence thesis;
  end;
end;

canceled 3;




theorem
  a<b implies ex p st a<p & p<b
proof
  set d=b-a;
  set m=[\d"/]+1;
  set l=[\m*a/]+1;
  set p=l/m;
  assume a<b;
  then
A1: a-a<b-a by XREAL_1:9;
  then
A2: 0<m by INT_1:29;
  d"-1<[\d"/] by INT_1:def 6;
  then d"<[\d"/]+1 by XREAL_1:19;
  then d"*d<m*d by A1,XREAL_1:68;
  then
A3: 1<m*d by A1,XCMPLX_0:def 7;
  take p;
A4: 0<>m by A1,INT_1:29;
  m*a-1<[\m*a/] by INT_1:def 6;
  then m*a<[\m*a/]+1 by XREAL_1:19;
  then m"*(m*a)<m"*l by A2,XREAL_1:68;
  then (m"*m)*a<m"*l;
  then 1*a<m"*l by A4,XCMPLX_0:def 7;
  hence a<p;
  [\m*a/]<=m*a by INT_1:def 6;
  then [\m*a/]+1<m*a+m*d by A3,XREAL_1:8;
  then m"*l<m"*(m*b) by A2,XREAL_1:68;
  then m"*l<(m"*m)*b;
  then m"*l<1*b by A4,XCMPLX_0:def 7;
  hence p<b;
end;

theorem Th8:
  ex m,k st k<>0 & p=m/k
proof
  consider m,n such that
A1: n<>0 and
A2: p=m/n by Th2;
  per cases by A1;
  suppose
    0<n;
    then n is Element of NAT by INT_1:3;
    hence thesis by A1,A2;
  end;
  suppose
A3: n<0;
A4: p = -(-m)/n by A2
      .= (-m)/(-n) by XCMPLX_1:188;
A5: -n<>0 by A1;
    -n is Element of NAT by A3,INT_1:3;
    hence thesis by A4,A5;
  end;
end;







theorem Th9:
  ex m,k st k<>0 & p=m/k & for n,l st l<>0 & p=n/l holds k<=l
proof
  defpred P[Nat] means ($1<>0 & ex m1 st p=m1/$1);
  ex m,k st k<>0 & p=m/k by Th8;
  then
A1: ex l be Nat st P[l];
  ex k1 be Nat st P[k1] & for l1 be Nat st P[l1] holds k1<=l1 from NAT_1:
  sch 5(A1);
  then consider k1 be Nat such that
A2: k1<>0 and
A3: ex m1 st p=m1/k1 and
A4: for l1 be Nat st (l1<>0 & ex n1 st p=n1/l1) holds k1<=l1;
  reconsider k1 as Element of NAT by ORDINAL1:def 12;
  consider m1 such that
A5: p=m1/k1 and
A6: for l1 be Nat st (l1<>0 & ex n1 st p=n1/l1) holds k1<=l1 by A3,A4;
  take m1, k1;
  thus k1<>0 by A2;
  thus m1/k1=p by A5;
  thus thesis by A6;
end;

definition
  let p;
  func denominator(p) -> Element of NAT means
  :Def3:
  it<>0 & (ex m st p=m/it) & for n,k st k<>0 & p=n/k holds it<=k;
  existence
  proof
    consider m,k such that
A1: k<>0 and
A2: p=m/k and
A3: for n,l st l<>0 & p=n/l holds k<=l by Th9;
    take k;
    thus thesis by A1,A2,A3;
  end;
  uniqueness
  proof
    let k,l such that
A4: k<>0 and
A5: ex m st p=m/k and
A6: for m1,k1 st k1<>0 & p=m1/k1 holds k<=k1 and
A7: l<>0 and
A8: ex n st p=n/l and
A9: for n1,l1 st l1<>0 & p=n1/l1 holds l<=l1;
A10: l<=k by A4,A5,A9;
    k<=l by A6,A7,A8;
    hence thesis by A10,XXREAL_0:1;
  end;
end;

definition
  let p;
  func numerator(p) -> Integer equals
  denominator(p)*p;
  coherence
  proof
A1: denominator(p)<>0 by Def3;
    ex m st p=m/denominator(p) by Def3;
    hence thesis by A1,XCMPLX_1:87;
  end;
end;



theorem Th10:
  0<denominator(p)
proof
  0 <> denominator p by Def3;
  hence thesis;
end;

theorem Th11:
  1<=denominator(p)
proof
  0<denominator(p) by Th10;
  then 0+1<=denominator(p) by NAT_1:13;
  hence thesis;
end;

theorem Th12:
  0<denominator(p)"
proof
  0<denominator(p) by Th10;
  hence thesis;
end;

theorem Th13:
  1>=denominator(p)"
proof
A1: 1*denominator(p)"<=denominator(p)*denominator(p)" by Th11,XREAL_1:64;
  0<>denominator(p) by Def3;
  hence thesis by A1,XCMPLX_0:def 7;
end;

theorem Th14:
  numerator(p)=0 iff p=0
proof
  denominator(p)<>0 by Def3;
  hence thesis by XCMPLX_1:6;
end;

theorem Th15:
  p=numerator(p)/denominator(p) & p=numerator(p)*denominator(p)" &
  p=denominator(p)"*numerator(p)
proof
  set x=denominator(p);
A1: x<>0 by Def3;
  thus
A2: numerator(p)/denominator(p)=p*(x*x") .=p*1 by A1,XCMPLX_0:def 7
    .=p;
  hence p=numerator(p)*denominator(p)";
  thus thesis by A2;
end;

theorem
  p<>0 implies denominator(p)=numerator(p)/p
proof
A1: (p"*p)*denominator(p)=p"*numerator(p);
  assume p<>0;
  then 1*denominator(p)=p"*numerator(p) by A1,XCMPLX_0:def 7;
  hence thesis;
end;

theorem Th17:
  p is Integer implies denominator(p)=1 & numerator(p)=p
proof
  assume p is Integer;
  then reconsider m=p as Integer;
  p =m/1;
  then
A1: denominator(p)<=1 by Def3;
  1<=denominator(p) by Th11;
  hence denominator(p)=1 by A1,XXREAL_0:1;
  hence thesis;
end;

theorem Th18:
  (numerator(p)=p or denominator(p)=1) implies p is Integer
proof
  assume
A1: numerator(p)=p or denominator(p)=1;
  per cases by A1;
  suppose
    numerator(p)=p;
    hence thesis;
  end;
  suppose
    denominator(p)=1;
    then numerator(p)=p;
    hence thesis;
  end;
end;

theorem
  numerator(p)=p iff denominator(p)=1 by Th17;

theorem
  (numerator(p)=p or denominator(p)=1) & 0<=p implies p is Element of NAT
proof
  assume that
A1: numerator(p)=p or denominator(p)=1 and
A2: 0<=p;
  p is Integer by A1,Th18;
  hence thesis by A2,INT_1:3;
end;

theorem
  1<denominator(p) iff p is not integer
proof
  now
    assume
A1: not p is Integer;
    denominator(p)>=1 by Th11;
    hence denominator(p)>1 by A1,Th18,XXREAL_0:1;
  end;
  hence thesis by Th17;
end;

Lm7: 1"=1;

theorem
  1>denominator(p)" iff p is not integer
proof
A1: denominator(p)"<=1 by Th13;
  now
    assume not p is Integer;
    then denominator(p)" <> 1 by Lm7,Th18;
    hence denominator(p)"<1 by A1,XXREAL_0:1;
  end;
  hence thesis by Lm7,Th17;
end;

theorem Th23:
  numerator(p)=denominator(p) iff p=1
proof
A1: denominator(p)<>0 by Def3;
  now
    assume numerator(p)=denominator(p);
    then numerator(p)/denominator(p)=1 by A1,XCMPLX_1:60;
    hence p=1 by Th15;
  end;
  hence thesis;
end;

theorem Th24:
  numerator(p)=-denominator(p) iff p=-1
proof
A1: 0<>denominator(p) by Def3;
  now
    assume numerator(p)=-denominator(p);
    hence p=(-denominator(p))/denominator(p) by Th15
      .=-denominator(p)/denominator(p)
      .=-1 by A1,XCMPLX_1:60;
  end;
  hence thesis;
end;

theorem
  -numerator(p)=denominator(p) iff p=-1
proof
  -numerator(p)=denominator(p) iff numerator(p)=-denominator(p);
  hence thesis by Th24;
end;




theorem Th26:
  m<>0 implies p=(numerator(p)*m)/(denominator(p)*m)
proof
  assume m<>0;
  then numerator(p)/denominator(p)=(numerator(p)*m)/(denominator(p)*m) by
XCMPLX_1:91;
  hence thesis by Th15;
end;

theorem Th27:
  k<>0 & p=m/k implies ex l st m=numerator(p)*l & k=denominator(p) *l
proof
  assume that
A1: k<>0 and
A2: p=m/k;
A3: p*k=m by A1,A2,XCMPLX_1:87;
  defpred P[Nat] means $1*denominator(p)<=k;
A4: denominator(p)<>0 by Def3;
A5: for l be Nat st P[l] holds l<=k
  proof
    0+1<=denominator(p) by A4,NAT_1:13;
    then
A6: k*1<=k*denominator(p) by NAT_1:4;
    let l be Nat such that
A7: l*denominator(p)<=k;
    assume not thesis;
    then k*denominator(p)<l*denominator(p) by A4,XREAL_1:68;
    hence contradiction by A7,A6,XXREAL_0:2;
  end;
A8: 1*denominator(p)<=k by A1,A2,Def3;
  then
A9: ex l1 be Nat st P[l1];
  consider l be Nat such that
A10: P[l] and
A11: for l1 be Nat st P[l1] holds l1<=l from NAT_1:sch 6(A5,A9);
  reconsider l as Element of NAT by ORDINAL1:def 12;
  take l;
A12: 0<>l by A8,A11;
  then
A13: l*denominator(p)<>0 by A4,XCMPLX_1:6;
A14: now
    assume
A15: l*denominator(p)<k;
    then 0+l*denominator(p)<k;
    then 0<=k-l*denominator(p) by XREAL_1:20;
    then reconsider x=k-l*denominator(p) as Element of NAT by INT_1:3;
A16: 0<>x by A15;
    m/k=(numerator(p)*l)/(l*denominator(p)) by A2,A12,Th26;
    then p=(m-numerator(p)*l)/x by A2,A13,A15,XCMPLX_1:123;
    then denominator(p)<=k-l*denominator(p) by A16,Def3;
    then l*denominator(p)+1*denominator(p)<=k by XREAL_1:19;
    then (l+1)*denominator(p)<=k;
    then l+1<=l by A11;
    hence contradiction by NAT_1:13;
  end;
  then k=denominator(p)*l by A10,XXREAL_0:1;
  hence m=numerator(p)/denominator(p)*(denominator(p)*l) by A3,Th15
    .=(numerator(p)/denominator(p)*denominator(p))*l
    .=numerator(p)*l by A4,XCMPLX_1:87;
  thus thesis by A10,A14,XXREAL_0:1;
end;

theorem
  p=m/n & n<>0 implies ex m1 st m=numerator(p)*m1 & n=denominator(p)*m1
proof
  assume that
A1: p=m/n and
A2: n<>0;
  per cases by A2;
  suppose
    n<0;
    then reconsider x=-n as Element of NAT by INT_1:3;
A3: p=-(-m)/n by A1
      .=(-m)/x by XCMPLX_1:188;
    x<>0 by A2;
    then consider k such that
A4: -m=numerator(p)*k and
A5: x=denominator(p)*k by A3,Th27;
    take y=-k;
    thus m=-numerator(p)*k by A4
      .=numerator(p)*y;
    thus n=-denominator(p)*k by A5
      .=denominator(p)*y;
  end;
  suppose
    0<n;
    then reconsider x=n as Element of NAT by INT_1:3;
    consider k such that
A6: m=numerator(p)*k and
A7: x=denominator(p)*k by A1,A2,Th27;
    reconsider y=k as Integer;
    take y;
    thus thesis by A6,A7;
  end;
end;



theorem Th29:
  not ex l st 1<l & ex m,k st numerator(p)=m*l & denominator(p)=k* l
proof
  assume not thesis;
  then consider l such that
A1: 1<l and
A2: ex m,k st numerator(p)=m*l & denominator(p)=k*l;
  consider m,k such that
A3: numerator(p)=m*l and
A4: denominator(p)=k*l by A2;
A5: p=(m*l)/(k*l) by A3,A4,Th15
    .=(m/k)*(l/l) by XCMPLX_1:76
    .=m/k*1 by A1,XCMPLX_1:60
    .=m/k;
A6: k<>0 by A4,Def3;
  then k*1<k*l by A1,XREAL_1:68;
  hence contradiction by A4,A6,A5,Def3;
end;

theorem Th30:
  (p=m/k & k<>0 & not ex l st 1<l & ex m1,k1 st m=m1*l & k=k1*l)
  implies k=denominator(p) & m=numerator(p)
proof
  assume that
A1: p=m/k and
A2: k<>0 and
A3: not ex l st 1<l & ex m1,k1 st m=m1*l & k=k1*l;
  consider l such that
A4: m=numerator(p)*l and
A5: k=denominator(p)*l by A1,A2,Th27;
  0<l by A2,A5;
  then
A6: 0+1<=l by NAT_1:13;
  l<=1 by A3,A4,A5;
  then
A7: l=1 by A6,XXREAL_0:1;
  hence k=denominator(p) by A5;
  thus thesis by A4,A7;
end;

theorem Th31:
  p<-1 iff numerator(p)<-denominator(p)
proof
A1: 0<>denominator(p) by Def3;
A2: 0<denominator(p)" by Th12;
A3: now
    assume numerator(p)<-denominator(p);
    then
    numerator(p)*denominator(p)"<((-1)*denominator(p))*denominator(p)" by A2,
XREAL_1:68;
    then numerator(p)*denominator(p)"<(-1)*(denominator(p)*denominator(p)");
    then numerator(p)*denominator(p)"<(-1)*1 by A1,XCMPLX_0:def 7;
    hence p<-1 by Th15;
  end;
A4: 0<denominator(p) by Th10;
  now
    assume p<-1;
    then numerator(p)/denominator(p)<-1 by Th15;
    then
    (numerator(p)/denominator(p))*denominator(p)<(-1)*denominator(p) by A4,
XREAL_1:68;
    hence numerator(p)<-denominator(p) by A1,XCMPLX_1:87;
  end;
  hence thesis by A3;
end;

theorem Th32:
  p<=-1 iff numerator(p)<=-denominator(p)
proof
A1: now
    assume
A2: p<=-1;
    per cases by A2,XXREAL_0:1;
    suppose
      p=-1;
      hence numerator(p)<=-denominator(p);
    end;
    suppose
      p<-1;
      hence numerator(p)<=-denominator(p) by Th31;
    end;
  end;
  now
    assume
A3: numerator(p)<=-denominator(p);
    per cases by A3,XXREAL_0:1;
    suppose
      numerator(p)=-denominator(p);
      hence p<=-1 by Th24;
    end;
    suppose
      numerator(p)<-denominator(p);
      hence p<=-1 by Th31;
    end;
  end;
  hence thesis by A1;
end;

theorem
  p<-1 iff denominator(p)<-numerator(p)
proof
  denominator(p)<-numerator(p) iff -(-numerator(p))<-denominator(p) by
XREAL_1:24;
  hence thesis by Th31;
end;

theorem
  p<=-1 iff denominator(p)<=-numerator(p)
proof
  denominator(p)<=-numerator(p) iff -(-numerator(p))<=-denominator(p) by
XREAL_1:24;
  hence thesis by Th32;
end;

theorem Th35:
  p<1 iff numerator(p)<denominator(p)
proof
A1: 0<>denominator(p) by Def3;
A2: 0<denominator(p)" by Th12;
A3: now
    assume numerator(p)<denominator(p);
    then numerator(p)*denominator(p)"<denominator(p)*denominator(p)" by A2,
XREAL_1:68;
    then numerator(p)*denominator(p)"<1 by A1,XCMPLX_0:def 7;
    hence p<1 by Th15;
  end;
A4: 0<denominator(p) by Th10;
  now
    assume p<1;
    then numerator(p)/denominator(p)<1 by Th15;
    then numerator(p)/denominator(p)*denominator(p)<1*denominator(p) by A4,
XREAL_1:68;
    hence numerator(p)<denominator(p) by A1,XCMPLX_1:87;
  end;
  hence thesis by A3;
end;

theorem
  p<=1 iff numerator(p)<=denominator(p)
proof
A1: now
    assume
A2: p<=1;
    per cases by A2,XXREAL_0:1;
    suppose
      p=1;
      hence numerator(p)<=denominator(p);
    end;
    suppose
      p<1;
      hence numerator(p)<=denominator(p) by Th35;
    end;
  end;
  now
    assume
A3: numerator(p)<=denominator(p);
    per cases by A3,XXREAL_0:1;
    suppose
      numerator(p)=denominator(p);
      hence p<=1 by Th23;
    end;
    suppose
      numerator(p)<denominator(p);
      hence p<=1 by Th35;
    end;
  end;
  hence thesis by A1;
end;

theorem
  p<0 iff numerator(p)<0
proof
  now
    assume p<0;
    then numerator(p)/denominator(p)<0 by Th15;
    hence numerator(p)<0;
  end;
  hence thesis;
end;

theorem Th38:
  p<=0 iff numerator(p)<=0
proof
  now
    assume
A1: numerator(p)<=0;
    per cases by A1;
    suppose
      numerator(p)=0;
      hence p<=0 by Th14;
    end;
    suppose
      numerator(p)<0;
      hence p<=0;
    end;
  end;
  hence thesis;
end;

theorem Th39:
  a<p iff a*denominator(p)<numerator(p)
proof
A1: 0<denominator(p)" by Th12;
A2: 0<>denominator(p) by Def3;
A3: now
    assume a*denominator(p)<numerator(p);
    then
    (a*denominator(p))*denominator(p)"<numerator(p)*denominator(p)" by A1,
XREAL_1:68;
    then a*(denominator(p)*denominator(p)")<numerator(p)*denominator(p)";
    then a*1<numerator(p)*denominator(p)" by A2,XCMPLX_0:def 7;
    hence a<p by Th15;
  end;
  0<denominator(p) by Th10;
  hence thesis by A3,XREAL_1:68;
end;

theorem
  a<=p iff a*denominator(p)<=numerator(p)
proof
A1: denominator(p)<>0 by Def3;
A2: now
    assume
A3: a*denominator(p)<=numerator(p);
    per cases by A3,XXREAL_0:1;
    suppose
      a*denominator(p)=numerator(p);
      then p=(a*denominator(p))/(1*denominator(p)) by Th15
        .=(a/1)*(denominator(p)/denominator(p))
        .=a by A1,XCMPLX_1:60;
      hence a<=p;
    end;
    suppose
      a*denominator(p)<numerator(p);
      hence a<=p by Th39;
    end;
  end;
  now
    assume
A4: a<=p;
    per cases by A4,XXREAL_0:1;
    suppose
      a=p;
      hence numerator(p)>=a*denominator(p);
    end;
    suppose
      a<p;
      hence a*denominator(p)<=numerator(p) by Th39;
    end;
  end;
  hence thesis by A2;
end;

theorem
  denominator p=denominator q & numerator p=numerator q implies p = q
proof
  assume that
A1: denominator(p)=denominator(q) and
A2: numerator(p)=numerator(q);
  thus p=numerator(q)/denominator(q) by A1,A2,Th15
    .=q by Th15;
end;

theorem
  p<q iff numerator(p)*denominator(q)<numerator(q)*denominator(p)
proof
A1: 0<denominator(q) by Th10;
  p<q iff numerator(p)/denominator(p)<q by Th15;
  then
A2: p<q iff numerator(p)/denominator(p)<numerator(q)/denominator(q) by Th15;
  0<denominator(p) by Th10;
  hence thesis by A1,A2,XREAL_1:102,106;
end;

theorem Th43:
  denominator(-p)=denominator(p) & numerator(-p)=-numerator(p)
proof
A1: denominator(p)<>0 by Def3;
A2: p=-(-p) .=-numerator(-p)/denominator(-p) by Th15
    .=(-numerator(-p))/denominator(-p);
  denominator(-p)<>0 by Def3;
  then consider l such that
  -numerator(-p)=numerator(p)*l and
A3: denominator(-p)=denominator(p)*l by A2,Th27;
  -p=-numerator(p)/denominator(p) by Th15
    .=(-numerator(p))/denominator(p);
  then consider k such that
A4: -numerator(p)=numerator(-p)*k and
A5: denominator(p)=denominator(-p)*k by A1,Th27;
  denominator p = denominator(p)*l*k by A5,A3
    .= denominator(p)*(l*k);
  then
A6: k = 1 by A1,NAT_1:15,XCMPLX_1:7;
  hence denominator(p)=denominator(-p) by A5;
  thus thesis by A4,A6;
end;

theorem Th44:
  0<p & q=1/p iff numerator(q)=denominator(p) & denominator(q)= numerator(p)
proof
A1: now
    set x=denominator(p);
    assume that
A2: 0<p and
A3: q=1/p;
A4: q=1/(numerator(p)/denominator(p)) by A3,Th15
      .=(1*denominator(p))/numerator(p) by XCMPLX_1:77
      .=denominator(p)/numerator(p);
    reconsider y=numerator(p) as Element of NAT by A2,INT_1:3;
A5: not ex k st 1<k & ex m,l st x=m*k & y=l*k
    proof
      assume not thesis;
      then consider k such that
A6:   1<k and
A7:   ex m,l st x=m*k & y=l*k;
      consider m,l such that
A8:   x=m*k and
A9:   y=l*k by A7;
      now
        assume not 0<=m;
        then m*k<0*m by A6,XREAL_1:69;
        hence contradiction by A8;
      end;
      then reconsider z=m as Element of NAT by INT_1:3;
      denominator(p)=z*k by A8;
      hence contradiction by A6,A9,Th29;
    end;
    0<>numerator(p) by A2,Th38;
    hence numerator(p)=denominator(q) & denominator(p)=numerator(q) by A4,A5
,Th30;
  end;
  now
    assume that
A10: numerator(q)=denominator(p) and
A11: denominator(q)=numerator(p);
    thus 0<p by A11,Th10;
    thus q =(1*denominator(p))/numerator(p) by A10,A11,Th15
      .=1/(numerator(p)/denominator(p)) by XCMPLX_1:77
      .=1/p by Th15;
  end;
  hence thesis by A1;
end;

theorem
  p<0 & q=1/p iff numerator(q)=-denominator(p) & denominator(q)=- numerator(p)
proof
A1: now
    set s=-q;
    set r=-p;
    assume that
A2: p<0 and
A3: q=1/p;
A4: s=1/r by A3,XCMPLX_1:188;
A5: 0<r by A2,XREAL_1:58;
    then numerator(s)=denominator(r) by A4,Th44;
    then -numerator(q)=denominator(r) by Th43;
    then
A6: -(-numerator(q))=-denominator(p) by Th43;
    denominator(s)=numerator(r) by A5,A4,Th44;
    then denominator(q)=numerator(r) by Th43;
    hence numerator(q)=-denominator(p) & denominator(q)=-numerator(p) by A6
,Th43;
  end;
  now
    assume that
A7: numerator(q)=-denominator(p) and
A8: denominator(q)=-numerator( p);
    thus p<0 by A8,Th10;
    thus q=(-denominator(p))/(-numerator(p)) by A7,A8,Th15
      .=-(denominator(p)/(-numerator(p)))
      .=-(-denominator(p)/numerator(p)) by XCMPLX_1:188
      .=(1*denominator(p))/numerator(p)
      .=1/(numerator(p)/denominator(p)) by XCMPLX_1:77
      .=1/p by Th15;
  end;
  hence thesis by A1;
end;














begin

reserve x, X, F for set;

definition
  let X be set;
  attr X is complex-membered means
  :Def1:
  x in X implies x is complex;
  attr X is ext-real-membered means
  :Def2:
  x in X implies x is ext-real;
  attr X is real-membered means
  :Def3:
  x in X implies x is real;
  attr X is rational-membered means
  :Def4:
  x in X implies x is rational;
  attr X is integer-membered means
  :Def5:
  x in X implies x is integer;
  attr X is natural-membered means
  :Def6:
  x in X implies x is natural;
end;

registration
  cluster natural-membered -> integer-membered for set;
  coherence
  proof
    let X;
    assume
A1: X is natural-membered;
    let x;
    assume x in X;
    then x is natural by A1,Def6;
    then x in omega by ORDINAL1:def 12;
    hence x in INT by NUMBERS:17;
  end;
  cluster integer-membered -> rational-membered for set;
  coherence
  proof
    let X;
    assume
A2: X is integer-membered;
    let x;
    assume x in X;
    then x is integer by A2,Def5;
    then x in INT by INT_1:def 2;
    hence x in RAT by NUMBERS:14;
  end;
  cluster rational-membered -> real-membered for set;
  coherence
  proof
    let X;
    assume
A3: X is rational-membered;
    let x;
    assume x in X;
    then x is rational by A3,Def4;
    then x in RAT by RAT_1:def 2;
    hence x in REAL by NUMBERS:12;
  end;
  cluster real-membered -> ext-real-membered for set;
  coherence
  proof
    let X;
    assume
A4: X is real-membered;
    let x;
    assume x in X;
    then x is real by A4,Def3;
    then x in REAL by XREAL_0:def 1;
    hence x in ExtREAL by NUMBERS:31;
  end;
  cluster real-membered -> complex-membered for set;
  coherence
  proof
    let X;
    assume
A5: X is real-membered;
    let x;
    assume x in X;
    then x is real by A5,Def3;
    then x in REAL by XREAL_0:def 1;
    hence x in COMPLEX by NUMBERS:11;
  end;
end;

registration
  cluster non empty natural-membered for set;
  existence
  proof
    take 1;
    thus 1 is non empty;
    let x;
    assume x in 1;
    hence thesis by CARD_1:49,TARSKI:def 1;
  end;
end;

registration
  cluster -> complex-membered for Subset of COMPLEX;
  coherence
  proof
    let S be Subset of COMPLEX;
    let x;
    thus thesis;
  end;
  cluster -> ext-real-membered for Subset of ExtREAL;
  coherence
  proof
    let S be Subset of ExtREAL;
    let x;
    thus thesis by XXREAL_0:def 1;
  end;
  cluster -> real-membered for Subset of REAL;
  coherence
  proof
    let S be Subset of REAL;
    let x;
    thus thesis;
  end;
  cluster -> rational-membered for Subset of RAT;
  coherence
  proof
    let S be Subset of RAT;
    let x;
    thus thesis by RAT_1:def 2;
  end;
  cluster -> integer-membered for Subset of INT;
  coherence
  proof
    let S be Subset of INT;
    let x;
    thus thesis;
  end;
  cluster -> natural-membered for Subset of NAT;
  coherence
  proof
    let S be Subset of NAT;
    let x;
    thus thesis;
  end;
end;

registration
  cluster COMPLEX -> complex-membered;
  coherence
  proof
    let x;
    thus thesis;
  end;
  cluster ExtREAL -> ext-real-membered;
  coherence
  proof
    let x;
    thus thesis by XXREAL_0:def 1;
  end;
  cluster REAL -> real-membered;
  coherence
  proof
    let x;
    thus thesis;
  end;
  cluster RAT -> rational-membered;
  coherence
  proof
    let x;
    thus thesis by RAT_1:def 2;
  end;
  cluster INT -> integer-membered;
  coherence
  proof
    let x;
    thus thesis;
  end;
  cluster NAT -> natural-membered;
  coherence
  proof
    let x;
    thus thesis;
  end;
end;

theorem Th1:
  X is complex-membered implies X c= COMPLEX
proof
  assume
A1: X is complex-membered;
  let x;
  assume x in X;
  then x is complex by A1,Def1;
  hence thesis by XCMPLX_0:def 2;
end;

theorem Th2:
  X is ext-real-membered implies X c= ExtREAL
proof
  assume
A1: X is ext-real-membered;
  let x;
  assume x in X;
  then x is ext-real by A1,Def2;
  hence thesis by XXREAL_0:def 1;
end;

theorem Th3:
  X is real-membered implies X c= REAL
proof
  assume
A1: X is real-membered;
  let x;
  assume x in X;
  then x is real by A1,Def3;
  hence thesis by XREAL_0:def 1;
end;

theorem Th4:
  X is rational-membered implies X c= RAT
proof
  assume
A1: X is rational-membered;
  let x;
  assume x in X;
  then x is rational by A1,Def4;
  hence thesis by RAT_1:def 2;
end;

theorem Th5:
  X is integer-membered implies X c= INT
proof
  assume
A1: X is integer-membered;
  let x;
  assume x in X;
  then x is integer by A1,Def5;
  hence thesis by INT_1:def 2;
end;

theorem Th6:
  X is natural-membered implies X c= NAT
proof
  assume
A1: X is natural-membered;
  let x;
  assume x in X;
  then x is natural by A1,Def6;
  hence thesis by ORDINAL1:def 12;
end;

registration
  let X be complex-membered set;
  cluster -> complex for Element of X;
  coherence
  proof
    let e be Element of X;
    per cases;
    suppose
      X is empty;
      hence thesis by SUBSET_1:def 1;
    end;
    suppose
      X is non empty;
      hence thesis by Def1;
    end;
  end;
end;

registration
  let X be ext-real-membered set;
  cluster -> ext-real for Element of X;
  coherence
  proof
    let e be Element of X;
    per cases;
    suppose
      X is empty;
      hence thesis by SUBSET_1:def 1;
    end;
    suppose
      X is non empty;
      hence thesis by Def2;
    end;
  end;
end;

registration
  let X be real-membered set;
  cluster -> real for Element of X;
  coherence
  proof
    let e be Element of X;
    per cases;
    suppose
      X is empty;
      hence thesis by SUBSET_1:def 1;
    end;
    suppose
      X is non empty;
      hence thesis by Def3;
    end;
  end;
end;

registration
  let X be rational-membered set;
  cluster -> rational for Element of X;
  coherence
  proof
    let e be Element of X;
    per cases;
    suppose
      X is empty;
      hence thesis by SUBSET_1:def 1;
    end;
    suppose
      X is non empty;
      hence thesis by Def4;
    end;
  end;
end;

registration
  let X be integer-membered set;
  cluster -> integer for Element of X;
  coherence
  proof
    let e be Element of X;
    per cases;
    suppose
      X is empty;
      hence thesis by SUBSET_1:def 1;
    end;
    suppose
      X is non empty;
      hence thesis by Def5;
    end;
  end;
end;

registration
  let X be natural-membered set;
  cluster -> natural for Element of X;
  coherence
  proof
    let e be Element of X;
    per cases;
    suppose
      X is empty;
      hence thesis by SUBSET_1:def 1;
    end;
    suppose
      X is non empty;
      hence thesis by Def6;
    end;
  end;
end;

reserve c, c1, c2, c3 for complex number,
  e, e1, e2, e3 for ext-real number,
  r , r1, r2, r3 for real number,
  w, w1, w2, w3 for rational number,
  i, i1, i2, i3 for integer number,
  n, n1, n2, n3 for Nat;

theorem
  for X being non empty complex-membered set ex c st c in X
proof
  let X be non empty complex-membered set;
  ex x being set st x in X by XBOOLE_0:def 1;
  hence thesis;
end;

theorem
  for X being non empty ext-real-membered set ex e st e in X
proof
  let X be non empty ext-real-membered set;
  ex x being set st x in X by XBOOLE_0:def 1;
  hence thesis;
end;

theorem
  for X being non empty real-membered set ex r st r in X
proof
  let X be non empty real-membered set;
  ex x being set st x in X by XBOOLE_0:def 1;
  hence thesis;
end;

theorem
  for X being non empty rational-membered set ex w st w in X
proof
  let X be non empty rational-membered set;
  ex x being set st x in X by XBOOLE_0:def 1;
  hence thesis;
end;

theorem
  for X being non empty integer-membered set ex i st i in X
proof
  let X be non empty integer-membered set;
  ex x being set st x in X by XBOOLE_0:def 1;
  hence thesis;
end;

theorem
  for X being non empty natural-membered set ex n st n in X
proof
  let X be non empty natural-membered set;
  ex x being set st x in X by XBOOLE_0:def 1;
  hence thesis;
end;

theorem
  for X being complex-membered set st for c holds c in X holds X = COMPLEX
proof
  let X be complex-membered set such that
A1: for c holds c in X;
  thus X c= COMPLEX by Th1;
  let e be set;
  assume e in COMPLEX;
  hence thesis by A1;
end;

theorem
  for X being ext-real-membered set st for e holds e in X holds X = ExtREAL
proof
  let X be ext-real-membered set such that
A1: for e holds e in X;
  thus X c= ExtREAL by Th2;
  let e be set;
  assume e in ExtREAL;
  hence thesis by A1;
end;

theorem
  for X being real-membered set st for r holds r in X holds X = REAL
proof
  let X be real-membered set such that
A1: for r holds r in X;
  thus X c= REAL by Th3;
  let e be set;
  assume e in REAL;
  hence thesis by A1;
end;

theorem
  for X being rational-membered set st for w holds w in X holds X = RAT
proof
  let X be rational-membered set such that
A1: for w holds w in X;
  thus X c= RAT by Th4;
  let e be set;
  assume e in RAT;
  hence thesis by A1;
end;

theorem
  for X being integer-membered set st for i holds i in X holds X = INT
proof
  let X be integer-membered set such that
A1: for i holds i in X;
  thus X c= INT by Th5;
  let e be set;
  assume e in INT;
  hence thesis by A1;
end;

theorem
  for X being natural-membered set st for n holds n in X holds X = NAT
proof
  let X be natural-membered set such that
A1: for n holds n in X;
  thus X c= NAT by Th6;
  let e be set;
  assume e in NAT;
  hence thesis by A1;
end;

theorem Th19:
  for Y being complex-membered set st X c= Y holds X is complex-membered
proof
  let Y be complex-membered set such that
A1: X c= Y;
  let x;
  thus thesis by A1;
end;

theorem Th20:
  for Y being ext-real-membered set st X c= Y holds X is ext-real-membered
proof
  let Y be ext-real-membered set such that
A1: X c= Y;
  let x;
  thus thesis by A1;
end;

theorem Th21:
  for Y being real-membered set st X c= Y holds X is real-membered
proof
  let Y be real-membered set such that
A1: X c= Y;
  let x;
  thus thesis by A1;
end;

theorem Th22:
  for Y being rational-membered set st X c= Y holds X is rational-membered
proof
  let Y be rational-membered set such that
A1: X c= Y;
  let x;
  thus thesis by A1;
end;

theorem Th23:
  for Y being integer-membered set st X c= Y holds X is integer-membered
proof
  let Y be integer-membered set such that
A1: X c= Y;
  let x;
  thus thesis by A1;
end;

theorem Th24:
  for Y being natural-membered set st X c= Y holds X is natural-membered
proof
  let Y be natural-membered set such that
A1: X c= Y;
  let x;
  thus thesis by A1;
end;

registration
  cluster empty -> natural-membered for set;
  coherence
  proof
    let X be set such that
A1: X is empty;
    let x;
    thus thesis by A1;
  end;
end;

registration
  let c;
  cluster {c} -> complex-membered;
  coherence
  proof
    let e be set;
    thus thesis by TARSKI:def 1;
  end;
end;

registration
  let e;
  cluster {e} -> ext-real-membered;
  coherence
  proof
    let e be set;
    thus thesis by TARSKI:def 1;
  end;
end;

registration
  let r;
  cluster {r} -> real-membered;
  coherence
  proof
    let e be set;
    thus thesis by TARSKI:def 1;
  end;
end;

registration
  let w;
  cluster {w} -> rational-membered;
  coherence
  proof
    let e be set;
    thus thesis by TARSKI:def 1;
  end;
end;

registration
  let i;
  cluster {i} -> integer-membered;
  coherence
  proof
    let e be set;
    thus thesis by TARSKI:def 1;
  end;
end;

registration
  let n;
  cluster {n} -> natural-membered;
  coherence
  proof
    let e be set;
    thus thesis by TARSKI:def 1;
  end;
end;

registration
  let c1,c2;
  cluster {c1,c2} -> complex-membered;
  coherence
  proof
    let e be set;
    thus thesis by TARSKI:def 2;
  end;
end;

registration
  let e1,e2;
  cluster {e1,e2} -> ext-real-membered;
  coherence
  proof
    let e be set;
    thus thesis by TARSKI:def 2;
  end;
end;

registration
  let r1,r2;
  cluster {r1,r2} -> real-membered;
  coherence
  proof
    let e be set;
    thus thesis by TARSKI:def 2;
  end;
end;

registration
  let w1,w2;
  cluster {w1,w2} -> rational-membered;
  coherence
  proof
    let e be set;
    thus thesis by TARSKI:def 2;
  end;
end;

registration
  let i1,i2;
  cluster {i1,i2} -> integer-membered;
  coherence
  proof
    let e be set;
    thus thesis by TARSKI:def 2;
  end;
end;

registration
  let n1,n2;
  cluster {n1,n2} -> natural-membered;
  coherence
  proof
    let e be set;
    thus thesis by TARSKI:def 2;
  end;
end;

registration
  let c1,c2,c3;
  cluster {c1,c2,c3} -> complex-membered;
  coherence
  proof
    let e be set;
    thus thesis by ENUMSET1:def 1;
  end;
end;

registration
  let e1,e2,e3;
  cluster {e1,e2,e3} -> ext-real-membered;
  coherence
  proof
    let e be set;
    thus thesis by ENUMSET1:def 1;
  end;
end;

registration
  let r1,r2,r3;
  cluster {r1,r2,r3} -> real-membered;
  coherence
  proof
    let e be set;
    thus thesis by ENUMSET1:def 1;
  end;
end;

registration
  let w1,w2,w3;
  cluster {w1,w2,w3} -> rational-membered;
  coherence
  proof
    let e be set;
    thus thesis by ENUMSET1:def 1;
  end;
end;

registration
  let i1,i2,i3;
  cluster {i1,i2,i3} -> integer-membered;
  coherence
  proof
    let e be set;
    thus thesis by ENUMSET1:def 1;
  end;
end;

registration
  let n1,n2,n3;
  cluster {n1,n2,n3} -> natural-membered;
  coherence
  proof
    let e be set;
    thus thesis by ENUMSET1:def 1;
  end;
end;

registration
  let X be complex-membered set;
  cluster -> complex-membered for Subset of X;
  coherence
  proof
    let S be Subset of X, x;
    thus thesis;
  end;
end;

registration
  let X be ext-real-membered set;
  cluster -> ext-real-membered for Subset of X;
  coherence
  proof
    let S be Subset of X, x;
    thus thesis;
  end;
end;

registration
  let X be real-membered set;
  cluster -> real-membered for Subset of X;
  coherence
  proof
    let S be Subset of X, x;
    thus thesis;
  end;
end;

registration
  let X be rational-membered set;
  cluster -> rational-membered for Subset of X;
  coherence
  proof
    let S be Subset of X, x;
    thus thesis;
  end;
end;

registration
  let X be integer-membered set;
  cluster -> integer-membered for Subset of X;
  coherence
  proof
    let S be Subset of X, x;
    thus thesis;
  end;
end;

registration
  let X be natural-membered set;
  cluster -> natural-membered for Subset of X;
  coherence
  proof
    let S be Subset of X, x;
    thus thesis;
  end;
end;

registration
  let X,Y be complex-membered set;
  cluster X \/ Y -> complex-membered;
  coherence
  proof
    let x;
    assume x in X \/ Y;
    then x in X or x in Y by XBOOLE_0:def 3;
    hence thesis;
  end;
end;

registration
  let X,Y be ext-real-membered set;
  cluster X \/ Y -> ext-real-membered;
  coherence
  proof
    let x;
    assume x in X \/ Y;
    then x in X or x in Y by XBOOLE_0:def 3;
    hence thesis;
  end;
end;

registration
  let X,Y be real-membered set;
  cluster X \/ Y -> real-membered;
  coherence
  proof
    let x;
    assume x in X \/ Y;
    then x in X or x in Y by XBOOLE_0:def 3;
    hence thesis;
  end;
end;

registration
  let X,Y be rational-membered set;
  cluster X \/ Y -> rational-membered;
  coherence
  proof
    let x;
    assume x in X \/ Y;
    then x in X or x in Y by XBOOLE_0:def 3;
    hence thesis;
  end;
end;

registration
  let X,Y be integer-membered set;
  cluster X \/ Y -> integer-membered;
  coherence
  proof
    let x;
    assume x in X \/ Y;
    then x in X or x in Y by XBOOLE_0:def 3;
    hence thesis;
  end;
end;

registration
  let X,Y be natural-membered set;
  cluster X \/ Y -> natural-membered;
  coherence
  proof
    let x;
    assume x in X \/ Y;
    then x in X or x in Y by XBOOLE_0:def 3;
    hence thesis;
  end;
end;

registration
  let X be complex-membered set, Y be set;
  cluster X /\ Y -> complex-membered;
  coherence by Th19,XBOOLE_1:17;
  cluster Y /\ X -> complex-membered;
  coherence;
end;

registration
  let X be ext-real-membered set, Y be set;
  cluster X /\ Y -> ext-real-membered;
  coherence by Th20,XBOOLE_1:17;
  cluster Y /\ X -> ext-real-membered;
  coherence;
end;

registration
  let X be real-membered set, Y be set;
  cluster X /\ Y -> real-membered;
  coherence by Th21,XBOOLE_1:17;
  cluster Y /\ X -> real-membered;
  coherence;
end;

registration
  let X be rational-membered set, Y be set;
  cluster X /\ Y -> rational-membered;
  coherence by Th22,XBOOLE_1:17;
  cluster Y /\ X -> rational-membered;
  coherence;
end;

registration
  let X be integer-membered set, Y be set;
  cluster X /\ Y -> integer-membered;
  coherence by Th23,XBOOLE_1:17;
  cluster Y /\ X -> integer-membered;
  coherence;
end;

registration
  let X be natural-membered set, Y be set;
  cluster X /\ Y -> natural-membered;
  coherence by Th24,XBOOLE_1:17;
  cluster Y /\ X -> natural-membered;
  coherence;
end;

registration
  let X be complex-membered set, Y be set;
  cluster X \ Y -> complex-membered;
  coherence;
end;

registration
  let X be ext-real-membered set, Y be set;
  cluster X \ Y -> ext-real-membered;
  coherence;
end;

registration
  let X be real-membered set, Y be set;
  cluster X \ Y -> real-membered;
  coherence;
end;

registration
  let X be rational-membered set, Y be set;
  cluster X \ Y -> rational-membered;
  coherence;
end;

registration
  let X be integer-membered set, Y be set;
  cluster X \ Y -> integer-membered;
  coherence;
end;

registration
  let X be natural-membered set, Y be set;
  cluster X \ Y -> natural-membered;
  coherence;
end;

registration
  let X,Y be complex-membered set;
  cluster X \+\ Y -> complex-membered;
  coherence;
end;

registration
  let X,Y be ext-real-membered set;
  cluster X \+\ Y -> ext-real-membered;
  coherence;
end;

registration
  let X,Y be real-membered set;
  cluster X \+\ Y -> real-membered;
  coherence;
end;

registration
  let X,Y be rational-membered set;
  cluster X \+\ Y -> rational-membered;
  coherence;
end;

registration
  let X,Y be integer-membered set;
  cluster X \+\ Y -> integer-membered;
  coherence;
end;

registration
  let X,Y be natural-membered set;
  cluster X \+\ Y -> natural-membered;
  coherence;
end;

definition
  let X be complex-membered set, Y be set;
  redefine pred X c= Y means
  :Def7:
  c in X implies c in Y;
  compatibility
  proof
    thus X c= Y implies for c st c in X holds c in Y;
    assume
A1: c in X implies c in Y;
    let x;
    assume x in X;
    hence thesis by A1;
  end;
end;

definition
  let X be ext-real-membered set, Y be set;
  redefine pred X c= Y means
  :Def8:
  e in X implies e in Y;
  compatibility
  proof
    thus X c= Y implies for e st e in X holds e in Y;
    assume
A1: e in X implies e in Y;
    let x;
    assume x in X;
    hence thesis by A1;
  end;
end;

definition
  let X be real-membered set, Y be set;
  redefine pred X c= Y means
  :Def9:
  r in X implies r in Y;
  compatibility
  proof
    thus X c= Y implies for r st r in X holds r in Y;
    assume
A1: r in X implies r in Y;
    let e;
    assume e in X;
    hence thesis by A1;
  end;
end;

definition
  let X be rational-membered set, Y be set;
  redefine pred X c= Y means
  :Def10:
  w in X implies w in Y;
  compatibility
  proof
    thus X c= Y implies for w st w in X holds w in Y;
    assume
A1: w in X implies w in Y;
    let r;
    assume r in X;
    hence thesis by A1;
  end;
end;

definition
  let X be integer-membered set, Y be set;
  redefine pred X c= Y means
  :Def11:
  i in X implies i in Y;
  compatibility
  proof
    thus X c= Y implies for i st i in X holds i in Y;
    assume
A1: i in X implies i in Y;
    let w;
    assume w in X;
    hence thesis by A1;
  end;
end;

definition
  let X be natural-membered set, Y be set;
  redefine pred X c= Y means
  :Def12:
  n in X implies n in Y;
  compatibility
  proof
    thus X c= Y implies for n st n in X holds n in Y;
    assume
A1: n in X implies n in Y;
    let i;
    assume i in X;
    hence thesis by A1;
  end;
end;

definition
  let X,Y be complex-membered set;
  redefine pred X = Y means
  c in X iff c in Y;
  compatibility
  proof
    thus X = Y implies for c holds c in X iff c in Y;
    assume c in X iff c in Y;
    hence X c= Y & Y c= X by Def7;
  end;
end;

definition
  let X,Y be ext-real-membered set;
  redefine pred X = Y means
  e in X iff e in Y;
  compatibility
  proof
    thus X = Y implies for e holds e in X iff e in Y;
    assume e in X iff e in Y;
    then X c= Y & Y c= X by Def8;
    hence thesis by XBOOLE_0:def 10;
  end;
end;

definition
  let X,Y be real-membered set;
  redefine pred X = Y means
  r in X iff r in Y;
  compatibility
  proof
    thus X = Y implies for r holds r in X iff r in Y;
    assume r in X iff r in Y;
    then X c= Y & Y c= X by Def9;
    hence thesis by XBOOLE_0:def 10;
  end;
end;

definition
  let X,Y be rational-membered set;
  redefine pred X = Y means
  w in X iff w in Y;
  compatibility
  proof
    thus X = Y implies for w holds w in X iff w in Y;
    assume w in X iff w in Y;
    then X c= Y & Y c= X by Def10;
    hence thesis by XBOOLE_0:def 10;
  end;
end;

definition
  let X,Y be integer-membered set;
  redefine pred X = Y means
  i in X iff i in Y;
  compatibility
  proof
    thus X = Y implies for i holds i in X iff i in Y;
    assume i in X iff i in Y;
    then X c= Y & Y c= X by Def11;
    hence thesis by XBOOLE_0:def 10;
  end;
end;

definition
  let X,Y be natural-membered set;
  redefine pred X = Y means
  n in X iff n in Y;
  compatibility
  proof
    thus X = Y implies for n holds n in X iff n in Y;
    assume n in X iff n in Y;
    then X c= Y & Y c= X by Def12;
    hence thesis by XBOOLE_0:def 10;
  end;
end;

definition
  let X,Y be complex-membered set;
  redefine pred X meets Y means
  ex c st c in X & c in Y;
  compatibility
  proof
    thus X misses Y implies not ex c st c in X & c in Y by XBOOLE_0:3;
    assume
A1: not ex c st c in X & c in Y;
    assume X meets Y;
    then ex x st x in X & x in Y by XBOOLE_0:3;
    hence thesis by A1;
  end;
end;

definition
  let X,Y be ext-real-membered set;
  redefine pred X meets Y means
  ex e st e in X & e in Y;
  compatibility
  proof
    thus X misses Y implies not ex e st e in X & e in Y by XBOOLE_0:3;
    assume
A1: not ex e st e in X & e in Y;
    assume X meets Y;
    then ex x st x in X & x in Y by XBOOLE_0:3;
    hence thesis by A1;
  end;
end;

definition
  let X,Y be real-membered set;
  redefine pred X meets Y means
  ex r st r in X & r in Y;
  compatibility
  proof
    thus X misses Y implies not ex r st r in X & r in Y by XBOOLE_0:3;
    assume
A1: not ex r st r in X & r in Y;
    assume X meets Y;
    then ex x st x in X & x in Y by XBOOLE_0:3;
    hence thesis by A1;
  end;
end;

definition
  let X,Y be rational-membered set;
  redefine pred X meets Y means
  ex w st w in X & w in Y;
  compatibility
  proof
    thus X misses Y implies not ex w st w in X & w in Y by XBOOLE_0:3;
    assume
A1: not ex w st w in X & w in Y;
    assume X meets Y;
    then ex x st x in X & x in Y by XBOOLE_0:3;
    hence thesis by A1;
  end;
end;

definition
  let X,Y be integer-membered set;
  redefine pred X meets Y means
  ex i st i in X & i in Y;
  compatibility
  proof
    thus X misses Y implies not ex i st i in X & i in Y by XBOOLE_0:3;
    assume
A1: not ex i st i in X & i in Y;
    assume X meets Y;
    then ex x st x in X & x in Y by XBOOLE_0:3;
    hence thesis by A1;
  end;
end;

definition
  let X,Y be natural-membered set;
  redefine pred X meets Y means
  ex n st n in X & n in Y;
  compatibility
  proof
    thus X misses Y implies not ex n st n in X & n in Y by XBOOLE_0:3;
    assume
A1: not ex n st n in X & n in Y;
    assume X meets Y;
    then ex x st x in X & x in Y by XBOOLE_0:3;
    hence thesis by A1;
  end;
end;

theorem
  (for X st X in F holds X is complex-membered) implies union F is
  complex-membered
proof
  assume
A1: for X st X in F holds X is complex-membered;
  let x;
  assume x in union F;
  then consider X such that
A2: x in X and
A3: X in F by TARSKI:def 4;
  X is complex-membered by A1,A3;
  hence thesis by A2;
end;

theorem
  (for X st X in F holds X is ext-real-membered) implies union F is
  ext-real-membered
proof
  assume
A1: for X st X in F holds X is ext-real-membered;
  let x;
  assume x in union F;
  then consider X such that
A2: x in X and
A3: X in F by TARSKI:def 4;
  X is ext-real-membered by A1,A3;
  hence thesis by A2;
end;

theorem
  (for X st X in F holds X is real-membered) implies union F is real-membered
proof
  assume
A1: for X st X in F holds X is real-membered;
  let x;
  assume x in union F;
  then consider X such that
A2: x in X and
A3: X in F by TARSKI:def 4;
  X is real-membered by A1,A3;
  hence thesis by A2;
end;

theorem
  (for X st X in F holds X is rational-membered) implies union F is
  rational-membered
proof
  assume
A1: for X st X in F holds X is rational-membered;
  let x;
  assume x in union F;
  then consider X such that
A2: x in X and
A3: X in F by TARSKI:def 4;
  X is rational-membered by A1,A3;
  hence thesis by A2;
end;

theorem
  (for X st X in F holds X is integer-membered) implies union F is
  integer-membered
proof
  assume
A1: for X st X in F holds X is integer-membered;
  let x;
  assume x in union F;
  then consider X such that
A2: x in X and
A3: X in F by TARSKI:def 4;
  X is integer-membered by A1,A3;
  hence thesis by A2;
end;

theorem
  (for X st X in F holds X is natural-membered) implies union F is
  natural-membered
proof
  assume
A1: for X st X in F holds X is natural-membered;
  let x;
  assume x in union F;
  then consider X such that
A2: x in X and
A3: X in F by TARSKI:def 4;
  X is natural-membered by A1,A3;
  hence thesis by A2;
end;

theorem
  for X st X in F & X is complex-membered holds meet F is
  complex-membered by Th19,SETFAM_1:3;

theorem
  for X st X in F & X is ext-real-membered holds meet F is
  ext-real-membered by Th20,SETFAM_1:3;

theorem
  for X st X in F & X is real-membered holds meet F is real-membered by Th21,
SETFAM_1:3;

theorem
  for X st X in F & X is rational-membered holds meet F is
  rational-membered by Th22,SETFAM_1:3;

theorem
  for X st X in F & X is integer-membered holds meet F is
  integer-membered by Th23,SETFAM_1:3;

theorem
  for X st X in F & X is natural-membered holds meet F is
  natural-membered by Th24,SETFAM_1:3;

scheme
  CMSeparation {P[set]}: ex X being complex-membered set st for c holds c in X
  iff P[c] proof
  consider X being set such that
A1: x in X iff x in COMPLEX & P[x] from XBOOLE_0:sch 1;
  X is complex-membered
  proof
    let x;
    assume x in X;
    then x in COMPLEX by A1;
    hence thesis;
  end;
  then reconsider X as complex-membered set;
  take X;
  let c;
  c in COMPLEX by XCMPLX_0:def 2;
  hence thesis by A1;
end;

scheme
  EMSeparation {P[set]}: ex X being ext-real-membered set st for e holds e in
  X iff P[e] proof
  consider X being set such that
A1: x in X iff x in ExtREAL & P[x] from XBOOLE_0:sch 1;
  X is ext-real-membered
  proof
    let x;
    assume x in X;
    then x in ExtREAL by A1;
    hence thesis;
  end;
  then reconsider X as ext-real-membered set;
  take X;
  let e;
  e in ExtREAL by XXREAL_0:def 1;
  hence thesis by A1;
end;

scheme
  RMSeparation {P[set]}: ex X being real-membered set st for r holds r in X
  iff P[r] proof
  consider X being set such that
A1: x in X iff x in REAL & P[x] from XBOOLE_0:sch 1;
  X is real-membered
  proof
    let x;
    assume x in X;
    then x in REAL by A1;
    hence thesis;
  end;
  then reconsider X as real-membered set;
  take X;
  let r;
  r in REAL by XREAL_0:def 1;
  hence thesis by A1;
end;

scheme
  WMSeparation {P[set]}: ex X being rational-membered set st for w holds w in
  X iff P[w] proof
  consider X being set such that
A1: x in X iff x in RAT & P[x] from XBOOLE_0:sch 1;
  X is rational-membered
  proof
    let x;
    assume x in X;
    then x in RAT by A1;
    hence thesis;
  end;
  then reconsider X as rational-membered set;
  take X;
  let w;
  w in RAT by RAT_1:def 2;
  hence thesis by A1;
end;

scheme
  IMSeparation {P[set]}: ex X being integer-membered set st for i holds i in X
  iff P[i] proof
  consider X being set such that
A1: x in X iff x in INT & P[x] from XBOOLE_0:sch 1;
  X is integer-membered
  proof
    let x;
    assume x in X;
    then x in INT by A1;
    hence thesis;
  end;
  then reconsider X as integer-membered set;
  take X;
  let i;
  i in INT by INT_1:def 2;
  hence thesis by A1;
end;

scheme
  NMSeparation {P[set]}: ex X being natural-membered set st for n holds n in X
  iff P[n] proof
  consider X being set such that
A1: x in X iff x in NAT & P[x] from XBOOLE_0:sch 1;
  X is natural-membered
  proof
    let x;
    assume x in X;
    then x in NAT by A1;
    hence thesis;
  end;
  then reconsider X as natural-membered set;
  take X;
  let n;
  n in NAT by ORDINAL1:def 12;
  hence thesis by A1;
end;

registration
  cluster non empty natural-membered for set;
  existence
  proof
    set X = the non empty natural-membered set;
    take X;
    thus thesis;
  end;
end;



registration
  cluster -> natural-membered for Element of NAT;
  coherence
  proof
    let n be Element of NAT;
    let x be set;
    n c= omega by ORDINAL1:def 2;
    hence thesis;
  end;
end;



reserve a,b,d for real number;

theorem
  for X,Y being real-membered set st X<>{} & Y<>{} & for a,b st a in X &
b in Y holds a<=b holds ex d st (for a st a in X holds a<=d) & for b st b in Y
  holds d<=b
proof
  let X,Y be real-membered set;
  set x = the Element of X;
  reconsider a=x as real number;
  set y = the Element of Y;
  reconsider b=y as real number;
  assume X<>{} & Y<>{};
  then
A1: a in X & b in Y;
A2: X c= REAL & Y c= REAL by Th3;
  assume for a,b st a in X & b in Y holds a<=b;
  then consider d being real number such that
A3: for a,b being real number st a in X & b in Y holds a<=d & d<=b by A2,
AXIOMS:1;
  reconsider d as real number;
  take d;
  thus thesis by A1,A3;
end;



definition
  let X be set;
  attr X is add-closed means
  for x, y being complex number st x in X & y in X holds x+y in X;
end;

registration
  cluster empty -> add-closed for set;
  coherence
  proof
    let X be set such that
A1: X is empty;
    let x be complex number;
    thus thesis by A1;
  end;
  cluster COMPLEX -> add-closed;
  coherence
  proof
    let x be complex number;
    thus thesis by XCMPLX_0:def 2;
  end;
  cluster REAL -> add-closed;
  coherence
  proof
    let x be complex number;
    thus thesis by XREAL_0:def 1;
  end;
  cluster RAT -> add-closed;
  coherence
  proof
    let x be complex number;
    thus thesis by RAT_1:def 2;
  end;
  cluster INT -> add-closed;
  coherence
  proof
    let x be complex number;
    thus thesis by INT_1:def 2;
  end;
  cluster NAT -> add-closed;
  coherence
  proof
    let x be complex number;
    thus thesis by ORDINAL1:def 12;
  end;
  cluster non empty add-closed natural-membered for set;
  existence
  proof
    take NAT;
    thus thesis;
  end;
end;














begin

definition
  let f be Relation;
  attr f is complex-valued means
  :Def1:
  rng f c= COMPLEX;
  attr f is ext-real-valued means
  :Def2:
  rng f c= ExtREAL;
  attr f is real-valued means
  :Def3:
  rng f c= REAL;
  canceled 2;
  attr f is natural-valued means
  :Def6:
  rng f c= NAT;
end;

registration
  cluster natural-valued -> INT-valued for Relation;
  coherence
  proof
    let R be Relation;
    assume rng R c= NAT;
    hence rng R c= INT by NUMBERS:17,XBOOLE_1:1;
  end;
  cluster INT-valued -> RAT-valued for Relation;
  coherence
  proof
    let R be Relation;
    assume rng R c= INT;
    hence rng R c= RAT by NUMBERS:14,XBOOLE_1:1;
  end;
  cluster INT-valued -> real-valued for Relation;
  coherence
  proof
    let R be Relation;
    assume rng R c= INT;
    hence rng R c= REAL by XBOOLE_1:1,NUMBERS:15;
  end;
  cluster RAT-valued -> real-valued for Relation;
  coherence
  proof
    let R be Relation;
    assume rng R c= RAT;
    hence rng R c= REAL by NUMBERS:12,XBOOLE_1:1;
  end;
  cluster real-valued -> ext-real-valued for Relation;
  coherence
  proof
    let R be Relation;
    assume rng R c= REAL;
    hence rng R c= ExtREAL by NUMBERS:31,XBOOLE_1:1;
  end;
  cluster real-valued -> complex-valued for Relation;
  coherence
  proof
    let R be Relation;
    assume rng R c= REAL;
    hence rng R c= COMPLEX by NUMBERS:11,XBOOLE_1:1;
  end;
  cluster natural-valued -> RAT-valued for Relation;
  coherence
  proof
    let R be Relation;
    assume rng R c= NAT;
    hence rng R c= RAT by NUMBERS:18,XBOOLE_1:1;
  end;
  cluster natural-valued -> real-valued for Relation;
  coherence
  proof
    let R be Relation;
    assume rng R c= NAT;
    hence rng R c= REAL by XBOOLE_1:1;
  end;
end;

registration
  cluster empty -> natural-valued for Relation;
  coherence
  proof
    let R be Relation;
    assume R is empty;
    hence rng R c= NAT by RELAT_1:38,XBOOLE_1:2;
  end;
end;

registration
  cluster natural-valued for Function;
  existence
  proof
    take {};
    thus thesis;
  end;
end;

registration
  let R be complex-valued Relation;
  cluster rng R -> complex-membered;
  coherence
  proof
    rng R c= COMPLEX by Def1;
    hence thesis;
  end;
end;

registration
  let R be ext-real-valued Relation;
  cluster rng R -> ext-real-membered;
  coherence
  proof
    rng R c= ExtREAL by Def2;
    hence thesis;
  end;
end;

registration
  let R be real-valued Relation;
  cluster rng R -> real-membered;
  coherence
  proof
    rng R c= REAL by Def3;
    hence thesis;
  end;
end;

registration
  let R be RAT-valued Relation;
  cluster rng R -> rational-membered;
  coherence;
end;

registration
  let R be INT-valued Relation;
  cluster rng R -> integer-membered;
  coherence;
end;

registration
  let R be natural-valued Relation;
  cluster rng R -> natural-membered;
  coherence
  proof
    rng R c= NAT by Def6;
    hence thesis;
  end;
end;

reserve x,y,X for set,
  f for Function,
  R,S for Relation;

theorem Th1:
  for S being complex-valued Relation st R c= S holds R is complex-valued
proof
  let S be complex-valued Relation;
  assume R c= S;
  then
A1: rng R c= rng S by RELAT_1:11;
  rng S c= COMPLEX by Def1;
  hence rng R c= COMPLEX by A1,XBOOLE_1:1;
end;

theorem Th2:
  for S being ext-real-valued Relation st R c= S holds R is ext-real-valued
proof
  let S be ext-real-valued Relation;
  assume R c= S;
  then
A1: rng R c= rng S by RELAT_1:11;
  rng S c= ExtREAL by Def2;
  hence rng R c= ExtREAL by A1,XBOOLE_1:1;
end;

theorem Th3:
  for S being real-valued Relation st R c= S holds R is real-valued
proof
  let S be real-valued Relation;
  assume R c= S;
  then
A1: rng R c= rng S by RELAT_1:11;
  rng S c= REAL by Def3;
  hence rng R c= REAL by A1,XBOOLE_1:1;
end;

theorem
  for S being RAT-valued Relation st R c= S holds R is RAT-valued;

theorem
  for S being INT-valued Relation st R c= S holds R is INT-valued;

theorem Th6:
  for S being natural-valued Relation st R c= S holds R is natural-valued
proof
  let S be natural-valued Relation;
  assume R c= S;
  then
A1: rng R c= rng S by RELAT_1:11;
  rng S c= NAT by Def6;
  hence rng R c= NAT by A1,XBOOLE_1:1;
end;

registration
  let R be complex-valued Relation;
  cluster -> complex-valued for Subset of R;
  coherence by Th1;
end;

registration
  let R be ext-real-valued Relation;
  cluster -> ext-real-valued for Subset of R;
  coherence by Th2;
end;

registration
  let R be real-valued Relation;
  cluster -> real-valued for Subset of R;
  coherence by Th3;
end;

registration
  let R be RAT-valued Relation;
  cluster -> RAT-valued for Subset of R;
  coherence;
end;

registration
  let R be INT-valued Relation;
  cluster -> INT-valued for Subset of R;
  coherence;
end;

registration
  let R be natural-valued Relation;
  cluster -> natural-valued for Subset of R;
  coherence by Th6;
end;

registration
  let R,S be complex-valued Relation;
  cluster R \/ S -> complex-valued;
  coherence
  proof
A1: rng(R \/ S) = rng R \/ rng S by RELAT_1:12;
    rng R c= COMPLEX & rng S c= COMPLEX by Def1;
    hence rng(R \/ S) c= COMPLEX by A1,XBOOLE_1:8;
  end;
end;

registration
  let R,S be ext-real-valued Relation;
  cluster R \/ S -> ext-real-valued;
  coherence
  proof
A1: rng(R \/ S) = rng R \/ rng S by RELAT_1:12;
    rng R c= ExtREAL & rng S c= ExtREAL by Def2;
    hence rng(R \/ S) c= ExtREAL by A1,XBOOLE_1:8;
  end;
end;

registration
  let R,S be real-valued Relation;
  cluster R \/ S -> real-valued;
  coherence
  proof
A1: rng(R \/ S) = rng R \/ rng S by RELAT_1:12;
    rng R c= REAL & rng S c= REAL by Def3;
    hence rng(R \/ S) c= REAL by A1,XBOOLE_1:8;
  end;
end;

registration
  let R,S be RAT-valued Relation;
  cluster R \/ S -> RAT-valued;
  coherence;
end;

registration
  let R,S be INT-valued Relation;
  cluster R \/ S -> INT-valued;
  coherence;
end;

registration
  let R,S be natural-valued Relation;
  cluster R \/ S -> natural-valued;
  coherence
  proof
A1: rng(R \/ S) = rng R \/ rng S by RELAT_1:12;
    rng R c= NAT & rng S c= NAT by Def6;
    hence rng(R \/ S) c= NAT by A1,XBOOLE_1:8;
  end;
end;

registration
  let R be complex-valued Relation;
  let S;
  cluster R /\ S -> complex-valued;
  coherence
  proof
    R /\ S c= R by XBOOLE_1:17;
    then
A1: rng(R /\ S) c= rng R by RELAT_1:11;
    rng R c= COMPLEX by Def1;
    hence rng(R /\ S) c= COMPLEX by A1,XBOOLE_1:1;
  end;
  cluster R \ S -> complex-valued;
  coherence;
end;

registration
  let R be ext-real-valued Relation;
  let S;
  cluster R /\ S -> ext-real-valued;
  coherence
  proof
    R /\ S c= R by XBOOLE_1:17;
    then
A1: rng(R /\ S) c= rng R by RELAT_1:11;
    rng R c= ExtREAL by Def2;
    hence rng(R /\ S) c= ExtREAL by A1,XBOOLE_1:1;
  end;
  cluster R \ S -> ext-real-valued;
  coherence;
end;

registration
  let R be real-valued Relation;
  let S;
  cluster R /\ S -> real-valued;
  coherence
  proof
    R /\ S c= R by XBOOLE_1:17;
    then
A1: rng(R /\ S) c= rng R by RELAT_1:11;
    rng R c= REAL by Def3;
    hence rng(R /\ S) c= REAL by A1,XBOOLE_1:1;
  end;
  cluster R \ S -> real-valued;
  coherence;
end;

registration
  let R be RAT-valued Relation;
  let S;
  cluster R /\ S -> RAT-valued;
  coherence;
  cluster R \ S -> RAT-valued;
  coherence;
end;

registration
  let R be INT-valued Relation;
  let S;
  cluster R /\ S -> INT-valued;
  coherence;
  cluster R \ S -> INT-valued;
  coherence;
end;

registration
  let R be natural-valued Relation;
  let S;
  cluster R /\ S -> natural-valued;
  coherence
  proof
    R /\ S c= R by XBOOLE_1:17;
    then
A1: rng(R /\ S) c= rng R by RELAT_1:11;
    rng R c= NAT by Def6;
    hence rng(R /\ S) c= NAT by A1,XBOOLE_1:1;
  end;
  cluster R \ S -> natural-valued;
  coherence;
end;

registration
  let R,S be complex-valued Relation;
  cluster R \+\ S -> complex-valued;
  coherence;
end;

registration
  let R,S be ext-real-valued Relation;
  cluster R \+\ S -> ext-real-valued;
  coherence;
end;

registration
  let R,S be real-valued Relation;
  cluster R \+\ S -> real-valued;
  coherence;
end;

registration
  let R,S be RAT-valued Relation;
  cluster R \+\ S -> RAT-valued;
  coherence;
end;

registration
  let R,S be INT-valued Relation;
  cluster R \+\ S -> INT-valued;
  coherence;
end;

registration
  let R,S be natural-valued Relation;
  cluster R \+\ S -> natural-valued;
  coherence;
end;

registration
  let R be complex-valued Relation;
  let X;
  cluster R.:X -> complex-membered;
  coherence
  proof
    R.:X c= rng R by RELAT_1:111;
    hence thesis;
  end;
end;

registration
  let R be ext-real-valued Relation;
  let X;
  cluster R.:X -> ext-real-membered;
  coherence
  proof
    R.:X c= rng R by RELAT_1:111;
    hence thesis;
  end;
end;

registration
  let R be real-valued Relation;
  let X;
  cluster R.:X -> real-membered;
  coherence
  proof
    R.:X c= rng R by RELAT_1:111;
    hence thesis;
  end;
end;

registration
  let R be RAT-valued Relation;
  let X;
  cluster R.:X -> rational-membered;
  coherence
  proof
    R.:X c= rng R by RELAT_1:111;
    hence thesis;
  end;
end;

registration
  let R be INT-valued Relation;
  let X;
  cluster R.:X -> integer-membered;
  coherence
  proof
    R.:X c= rng R by RELAT_1:111;
    hence thesis;
  end;
end;

registration
  let R be natural-valued Relation;
  let X;
  cluster R.:X -> natural-membered;
  coherence
  proof
    R.:X c= rng R by RELAT_1:111;
    hence thesis;
  end;
end;

registration
  let R be complex-valued Relation;
  let x;
  cluster Im(R,x) -> complex-membered;
  coherence;
end;

registration
  let R be ext-real-valued Relation;
  let x;
  cluster Im(R,x) -> ext-real-membered;
  coherence;
end;

registration
  let R be real-valued Relation;
  let x;
  cluster Im(R,x) -> real-membered;
  coherence;
end;

registration
  let R be RAT-valued Relation;
  let x;
  cluster Im(R,x) -> rational-membered;
  coherence;
end;

registration
  let R be INT-valued Relation;
  let x;
  cluster Im(R,x) -> integer-membered;
  coherence;
end;

registration
  let R be natural-valued Relation;
  let x;
  cluster Im(R,x) -> natural-membered;
  coherence;
end;

registration
  let R be complex-valued Relation;
  let X;
  cluster R|X -> complex-valued;
  coherence
  proof
    rng R c= COMPLEX & rng(R|X) c= rng R by Def1,RELAT_1:70;
    hence rng(R|X) c= COMPLEX by XBOOLE_1:1;
  end;
end;

registration
  let R be ext-real-valued Relation;
  let X;
  cluster R|X -> ext-real-valued;
  coherence
  proof
    rng R c= ExtREAL & rng(R|X) c= rng R by Def2,RELAT_1:70;
    hence rng(R|X) c= ExtREAL by XBOOLE_1:1;
  end;
end;

registration
  let R be real-valued Relation;
  let X;
  cluster R|X -> real-valued;
  coherence
  proof
    rng R c= REAL & rng(R|X) c= rng R by Def3,RELAT_1:70;
    hence rng(R|X) c= REAL by XBOOLE_1:1;
  end;
end;

registration
  let R be RAT-valued Relation;
  let X;
  cluster R|X -> RAT-valued;
  coherence;
end;

registration
  let R be INT-valued Relation;
  let X;
  cluster R|X -> INT-valued;
  coherence;
end;

registration
  let R be natural-valued Relation;
  let X;
  cluster R|X -> natural-valued;
  coherence
  proof
    rng R c= NAT & rng(R|X) c= rng R by Def6,RELAT_1:70;
    hence rng(R|X) c= NAT by XBOOLE_1:1;
  end;
end;

registration
  let X be complex-membered set;
  cluster id X -> complex-valued;
  coherence
  proof
    thus rng id X c= COMPLEX by MEMBERED:1;
  end;
end;

registration
  let X be ext-real-membered set;
  cluster id X -> ext-real-valued;
  coherence
  proof
    thus rng id X c= ExtREAL by MEMBERED:2;
  end;
end;

registration
  let X be real-membered set;
  cluster id X -> real-valued;
  coherence
  proof
    thus rng id X c= REAL by MEMBERED:3;
  end;
end;

registration
  let X be rational-membered set;
  cluster id X -> RAT-valued;
  coherence
  proof
    thus rng id X c= RAT by MEMBERED:4;
  end;
end;

registration
  let X be integer-membered set;
  cluster id X -> INT-valued;
  coherence
  proof
    thus rng id X c= INT by MEMBERED:5;
  end;
end;

registration
  let X be natural-membered set;
  cluster id X -> natural-valued;
  coherence
  proof
    thus rng id X c= NAT by MEMBERED:6;
  end;
end;

registration
  let R;
  let S be complex-valued Relation;
  cluster R*S -> complex-valued;
  coherence
  proof
    rng S c= COMPLEX & rng(R*S) c= rng S by Def1,RELAT_1:26;
    hence rng(R*S) c= COMPLEX by XBOOLE_1:1;
  end;
end;

registration
  let R;
  let S be ext-real-valued Relation;
  cluster R*S -> ext-real-valued;
  coherence
  proof
    rng S c= ExtREAL & rng(R*S) c= rng S by Def2,RELAT_1:26;
    hence rng(R*S) c= ExtREAL by XBOOLE_1:1;
  end;
end;

registration
  let R;
  let S be real-valued Relation;
  cluster R*S -> real-valued;
  coherence
  proof
    rng S c= REAL & rng(R*S) c= rng S by Def3,RELAT_1:26;
    hence rng(R*S) c= REAL by XBOOLE_1:1;
  end;
end;

registration
  let R;
  let S be RAT-valued Relation;
  cluster R*S -> RAT-valued;
  coherence;
end;

registration
  let R;
  let S be INT-valued Relation;
  cluster R*S -> INT-valued;
  coherence;
end;

registration
  let R;
  let S be natural-valued Relation;
  cluster R*S -> natural-valued;
  coherence
  proof
    rng S c= NAT & rng(R*S) c= rng S by Def6,RELAT_1:26;
    hence rng(R*S) c= NAT by XBOOLE_1:1;
  end;
end;

definition
  let f be Function;
  redefine attr f is complex-valued means
  :Def7:
  for x st x in dom f holds f.x is complex;
  compatibility
  proof
    thus f is complex-valued implies for x st x in dom f holds f.x is complex
    proof
      assume
A1:   f is complex-valued;
      let x;
      assume
A2:   x in dom f;
      reconsider f as complex-valued Function by A1;
      f.x in rng f by A2,FUNCT_1:3;
      hence thesis;
    end;
    assume
A3: for x st x in dom f holds f.x is complex;
    let y;
    assume y in rng f;
    then ex x st x in dom f & y = f.x by FUNCT_1:def 3;
    then y is complex by A3;
    hence thesis by XCMPLX_0:def 2;
  end;
  redefine attr f is ext-real-valued means
  :Def8:
  for x st x in dom f holds f.x is ext-real;
  compatibility
  proof
    thus f is ext-real-valued implies for x st x in dom f holds f.x is ext-real
    proof
      assume
A4:   f is ext-real-valued;
      let x;
      assume
A5:   x in dom f;
      reconsider f as ext-real-valued Function by A4;
      f.x in rng f by A5,FUNCT_1:3;
      hence thesis;
    end;
    assume
A6: for x st x in dom f holds f.x is ext-real;
    let y;
    assume y in rng f;
    then ex x st x in dom f & y = f.x by FUNCT_1:def 3;
    then y is ext-real by A6;
    hence thesis by XXREAL_0:def 1;
  end;
  redefine attr f is real-valued means
  :Def9:
  for x st x in dom f holds f.x is real;
  compatibility
  proof
    thus f is real-valued implies for x st x in dom f holds f.x is real
    proof
      assume
A7:   f is real-valued;
      let x;
      assume
A8:   x in dom f;
      reconsider f as real-valued Function by A7;
      f.x in rng f by A8,FUNCT_1:3;
      hence thesis;
    end;
    assume
A9: for x st x in dom f holds f.x is real;
    let y;
    assume y in rng f;
    then ex x st x in dom f & y = f.x by FUNCT_1:def 3;
    then y is real by A9;
    hence thesis by XREAL_0:def 1;
  end;
  canceled 2;
  redefine attr f is natural-valued means
  :Def12:
  for x st x in dom f holds f.x is natural;
  compatibility
  proof
    thus f is natural-valued implies for x st x in dom f holds f.x is natural
    proof
      assume
A10:  f is natural-valued;
      let x;
      assume
A11:  x in dom f;
      reconsider f as natural-valued Function by A10;
      f.x in rng f by A11,FUNCT_1:3;
      hence thesis;
    end;
    assume
A12: for x st x in dom f holds f.x is natural;
    let y;
    assume y in rng f;
    then ex x st x in dom f & y = f.x by FUNCT_1:def 3;
    then y is natural by A12;
    hence thesis by ORDINAL1:def 12;
  end;
end;

theorem Th7:
  f is complex-valued iff for x holds f.x is complex
proof
  hereby
    assume
A1: f is complex-valued;
    let x;
    per cases;
    suppose
      x in dom f;
      hence f.x is complex by A1,Def7;
    end;
    suppose
      not x in dom f;
      hence f.x is complex by FUNCT_1:def 2;
    end;
  end;
  assume for x holds f.x is complex;
  then for x st x in dom f holds f.x is complex;
  hence thesis by Def7;
end;

theorem Th8:
  f is ext-real-valued iff for x holds f.x is ext-real
proof
  hereby
    assume
A1: f is ext-real-valued;
    let x;
    per cases;
    suppose
      x in dom f;
      hence f.x is ext-real by A1,Def8;
    end;
    suppose
      not x in dom f;
      hence f.x is ext-real by FUNCT_1:def 2;
    end;
  end;
  assume for x holds f.x is ext-real;
  then for x st x in dom f holds f.x is ext-real;
  hence thesis by Def8;
end;

theorem Th9:
  f is real-valued iff for x holds f.x is real
proof
  hereby
    assume
A1: f is real-valued;
    let x;
    per cases;
    suppose
      x in dom f;
      hence f.x is real by A1,Def9;
    end;
    suppose
      not x in dom f;
      hence f.x is real by FUNCT_1:def 2;
    end;
  end;
  assume for x holds f.x is real;
  then for x st x in dom f holds f.x is real;
  hence thesis by Def9;
end;

theorem Th10:
  f is RAT-valued iff for x holds f.x is rational
proof
  hereby
    assume
A1: f is RAT-valued;
    let x;
    per cases;
    suppose x in dom f;
      then f.x in rng f by FUNCT_1:def 3;
      hence f.x is rational by A1;
    end;
    suppose not x in dom f;
      hence f.x is rational by FUNCT_1:def 2;
    end;
  end;
  assume
A2: for x holds f.x is rational;
  let y be set;
  assume y in rng f;
  then consider x such that x in dom f and
A3: f.x = y by FUNCT_1:def 3;
  f.x is rational by A2;
  hence thesis by A3,RAT_1:def 2;
end;

theorem Th11:
  f is INT-valued iff for x holds f.x is integer
proof
  hereby
    assume
A1: f is INT-valued;
    let x;
    per cases;
    suppose x in dom f;
      then f.x in rng f by FUNCT_1:def 3;
      hence f.x is integer by A1;
    end;
    suppose not x in dom f;
      hence f.x is integer by FUNCT_1:def 2;
    end;
  end;
  assume
A2: for x holds f.x is integer;
  let y be set;
  assume y in rng f;
  then consider x such that x in dom f and
A3: f.x = y by FUNCT_1:def 3;
  f.x is integer by A2;
  hence thesis by A3,INT_1:def 2;
end;

theorem Th12:
  f is natural-valued iff for x holds f.x is natural
proof
  hereby
    assume
A1: f is natural-valued;
    let x;
    per cases;
    suppose
      x in dom f;
      hence f.x is natural by A1,Def12;
    end;
    suppose
      not x in dom f;
      hence f.x is natural by FUNCT_1:def 2;
    end;
  end;
  assume for x holds f.x is natural;
  then for x st x in dom f holds f.x is natural;
  hence thesis by Def12;
end;

registration
  let f be complex-valued Function;
  let x;
  cluster f.x -> complex;
  coherence by Th7;
end;

registration
  let f be ext-real-valued Function;
  let x;
  cluster f.x -> ext-real;
  coherence by Th8;
end;

registration
  let f be real-valued Function;
  let x;
  cluster f.x -> real;
  coherence by Th9;
end;

registration
  let f be RAT-valued Function;
  let x;
  cluster f.x -> rational;
  coherence by Th10;
end;

registration
  let f be INT-valued Function;
  let x;
  cluster f.x -> integer;
  coherence by Th11;
end;

registration
  let f be natural-valued Function;
  let x;
  cluster f.x -> natural;
  coherence by Th12;
end;

registration
  let X;
  let x be complex number;
  cluster X --> x -> complex-valued;
  coherence
  proof
    rng(X --> x) c= COMPLEX by MEMBERED:1;
    hence thesis by Def1;
  end;
end;

registration
  let X;
  let x be ext-real number;
  cluster X --> x -> ext-real-valued;
  coherence
  proof
    rng(X --> x) c= ExtREAL by MEMBERED:2;
    hence thesis by Def2;
  end;
end;

registration
  let X;
  let x be real number;
  cluster X --> x -> real-valued;
  coherence
  proof
    rng(X --> x) c= REAL by MEMBERED:3;
    hence thesis by Def3;
  end;
end;

registration
  let X;
  let x be rational number;
  cluster X --> x -> RAT-valued;
  coherence
  proof
    rng(X --> x) c= RAT by MEMBERED:4;
    hence thesis by RELAT_1:def 19;
  end;
end;

registration
  let X;
  let x be integer number;
  cluster X --> x -> INT-valued;
  coherence
  proof
    rng(X --> x) c= INT by MEMBERED:5;
    hence thesis by RELAT_1:def 19;
  end;
end;

registration
  let X;
  let x be Nat;
  cluster X --> x -> natural-valued;
  coherence
  proof
    rng(X --> x) c= NAT by MEMBERED:6;
    hence thesis by Def6;
  end;
end;

registration
  let f,g be complex-valued Function;
  cluster f +* g -> complex-valued;
  coherence
  proof
    rng(f +* g) c= rng f \/ rng g by FUNCT_4:17;
    then rng(f +* g) c= COMPLEX by MEMBERED:1;
    hence thesis by Def1;
  end;
end;

registration
  let f,g be ext-real-valued Function;
  cluster f +* g -> ext-real-valued;
  coherence
  proof
    rng(f +* g) c= rng f \/ rng g by FUNCT_4:17;
    then rng(f +* g) c= ExtREAL by MEMBERED:2;
    hence thesis by Def2;
  end;
end;

registration
  let f,g be real-valued Function;
  cluster f +* g -> real-valued;
  coherence
  proof
    rng(f +* g) c= rng f \/ rng g by FUNCT_4:17;
    then rng(f +* g) c= REAL by MEMBERED:3;
    hence thesis by Def3;
  end;
end;

registration
  let f,g be RAT-valued Function;
  cluster f +* g -> RAT-valued;
  coherence
  proof
    rng(f +* g) c= rng f \/ rng g by FUNCT_4:17;
    then rng(f +* g) c= RAT by MEMBERED:4;
    hence thesis by RELAT_1:def 19;
  end;
end;

registration
  let f,g be INT-valued Function;
  cluster f +* g -> INT-valued;
  coherence
  proof
    rng(f +* g) c= rng f \/ rng g by FUNCT_4:17;
    then rng(f +* g) c= INT by MEMBERED:5;
    hence thesis by RELAT_1:def 19;
  end;
end;

registration
  let f,g be natural-valued Function;
  cluster f +* g -> natural-valued;
  coherence
  proof
    rng(f +* g) c= rng f \/ rng g by FUNCT_4:17;
    then rng(f +* g) c= NAT by MEMBERED:6;
    hence thesis by Def6;
  end;
end;

registration
  let x;
  let y be complex number;
  cluster x .--> y -> complex-valued;
  coherence;
end;

registration
  let x;
  let y be ext-real number;
  cluster x .--> y -> ext-real-valued;
  coherence;
end;

registration
  let x;
  let y be real number;
  cluster x .--> y -> real-valued;
  coherence;
end;

registration
  let x;
  let y be rational number;
  cluster x .--> y -> RAT-valued;
  coherence;
end;

registration
  let x;
  let y be integer number;
  cluster x .--> y -> INT-valued;
  coherence;
end;

registration
  let x;
  let y be Nat;
  cluster x .--> y -> natural-valued;
  coherence;
end;

registration
  let X;
  let Y be complex-membered set;
  cluster -> complex-valued for Relation of X,Y;
  coherence
  proof
    let R be Relation of X,Y;
    thus rng R c= COMPLEX by MEMBERED:1;
  end;
end;

registration
  let X;
  let Y be ext-real-membered set;
  cluster -> ext-real-valued for Relation of X,Y;
  coherence
  proof
    let R be Relation of X,Y;
    thus rng R c= ExtREAL by MEMBERED:2;
  end;
end;

registration
  let X;
  let Y be real-membered set;
  cluster -> real-valued for Relation of X,Y;
  coherence
  proof
    let R be Relation of X,Y;
    thus rng R c= REAL by MEMBERED:3;
  end;
end;

registration
  let X;
  let Y be rational-membered set;
  cluster -> RAT-valued for Relation of X,Y;
  coherence
  proof
    let R be Relation of X,Y;
    thus rng R c= RAT by MEMBERED:4;
  end;
end;

registration
  let X;
  let Y be integer-membered set;
  cluster -> INT-valued for Relation of X,Y;
  coherence
  proof
    let R be Relation of X,Y;
    thus rng R c= INT by MEMBERED:5;
  end;
end;

registration
  let X;
  let Y be natural-membered set;
  cluster -> natural-valued for Relation of X,Y;
  coherence
  proof
    let R be Relation of X,Y;
    thus rng R c= NAT by MEMBERED:6;
  end;
end;

registration
  let X;
  let Y be complex-membered set;
  cluster [:X,Y:] -> complex-valued;
  coherence
  proof
    rng [:X,Y:] c= Y by RELAT_1:159;
    hence rng [:X,Y:] c= COMPLEX by MEMBERED:1;
  end;
end;

registration
  let X;
  let Y be ext-real-membered set;
  cluster [:X,Y:] -> ext-real-valued;
  coherence
  proof
    rng [:X,Y:] c= Y by RELAT_1:159;
    hence rng [:X,Y:] c= ExtREAL by MEMBERED:2;
  end;
end;

registration
  let X;
  let Y be real-membered set;
  cluster [:X,Y:] -> real-valued;
  coherence
  proof
    rng [:X,Y:] c= Y by RELAT_1:159;
    hence rng [:X,Y:] c= REAL by MEMBERED:3;
  end;
end;

registration
  let X;
  let Y be rational-membered set;
  cluster [:X,Y:] -> RAT-valued;
  coherence
  proof
    rng [:X,Y:] c= Y by RELAT_1:159;
    hence rng [:X,Y:] c= RAT by MEMBERED:4;
  end;
end;

registration
  let X;
  let Y be integer-membered set;
  cluster [:X,Y:] -> INT-valued;
  coherence
  proof
    rng [:X,Y:] c= Y by RELAT_1:159;
    hence rng [:X,Y:] c= INT by MEMBERED:5;
  end;
end;

registration
  let X;
  let Y be natural-membered set;
  cluster [:X,Y:] -> natural-valued;
  coherence
  proof
    rng [:X,Y:] c= Y by RELAT_1:159;
    hence rng [:X,Y:] c= NAT by MEMBERED:6;
  end;
end;

notation
  let f be ext-real-valued Relation;
  synonym f is non-zero for f is non-empty;
end;

registration
  cluster non empty constant natural-valued INT-valued RAT-valued for Function;
  existence
  proof
    take 1 .--> 1;
    thus thesis;
  end;
end;

theorem
  for f being non empty constant complex-valued Function ex r being
  complex number st for x st x in dom f holds f.x = r
proof
  let f be non empty constant complex-valued Function;
  consider r being set such that
A1: for x st x in dom f holds f.x = r by FUNCOP_1:78;
  consider x being set such that
A2: x in dom f by XBOOLE_0:def 1;
  r = f.x by A1,A2;
  hence thesis by A1;
end;

theorem
  for f being non empty constant ext-real-valued Function ex r being
  ext-real number st for x st x in dom f holds f.x = r
proof
  let f be non empty constant ext-real-valued Function;
  consider r being set such that
A1: for x st x in dom f holds f.x = r by FUNCOP_1:78;
  consider x being set such that
A2: x in dom f by XBOOLE_0:def 1;
  r = f.x by A1,A2;
  hence thesis by A1;
end;

theorem
  for f being non empty constant real-valued Function ex r being real
  number st for x st x in dom f holds f.x = r
proof
  let f be non empty constant real-valued Function;
  consider r being set such that
A1: for x st x in dom f holds f.x = r by FUNCOP_1:78;
  consider x being set such that
A2: x in dom f by XBOOLE_0:def 1;
  r = f.x by A1,A2;
  hence thesis by A1;
end;

theorem
  for f being non empty constant RAT-valued Function ex r being
  rational number st for x st x in dom f holds f.x = r
proof
  let f be non empty constant RAT-valued Function;
  consider r being set such that
A1: for x st x in dom f holds f.x = r by FUNCOP_1:78;
  consider x being set such that
A2: x in dom f by XBOOLE_0:def 1;
  r = f.x by A1,A2;
  hence thesis by A1;
end;

theorem
  for f being non empty constant INT-valued Function ex r being
  integer number st for x st x in dom f holds f.x = r
proof
  let f be non empty constant INT-valued Function;
  consider r being set such that
A1: for x st x in dom f holds f.x = r by FUNCOP_1:78;
  consider x being set such that
A2: x in dom f by XBOOLE_0:def 1;
  r = f.x by A1,A2;
  hence thesis by A1;
end;

theorem
  for f being non empty constant natural-valued Function ex r being
  Nat st for x st x in dom f holds f.x = r
proof
  let f be non empty constant natural-valued Function;
  consider r being set such that
A1: for x st x in dom f holds f.x = r by FUNCOP_1:78;
  consider x being set such that
A2: x in dom f by XBOOLE_0:def 1;
  r = f.x by A1,A2;
  hence thesis by A1;
end;

begin 

reserve e1,e2 for ext-real number;

definition
  let f be ext-real-valued Function;
  attr f is increasing means
  :Def13:
  for e1,e2 st e1 in dom f & e2 in dom f & e1 < e2 holds f.e1 < f.e2;
  attr f is decreasing means
  :Def14:
  for e1,e2 st e1 in dom f & e2 in dom f & e1 < e2 holds f.e1 > f.e2;
  attr f is non-decreasing means
  :Def15:
  for e1,e2 st e1 in dom f & e2 in dom f & e1 <= e2 holds f.e1 <= f.e2;
  attr f is non-increasing means
  :Def16:
  for e1,e2 st e1 in dom f & e2 in dom f & e1 <= e2 holds f.e1 >= f.e2;
end;



registration
  cluster trivial -> increasing decreasing for ext-real-valued Function;
  coherence
  proof
    let f be ext-real-valued Function;
    assume
A1: f is trivial;
    thus f is increasing
    proof
      let e1,e2;
      assume e1 in dom f & e2 in dom f;
      hence thesis by A1,ZFMISC_1:def 10;
    end;
    let e1,e2;
    assume e1 in dom f & e2 in dom f;
    hence thesis by A1,ZFMISC_1:def 10;
  end;
end;

registration
  cluster increasing -> non-decreasing for ext-real-valued Function;
  coherence
  proof
    let f be ext-real-valued Function;
    assume
A1: f is increasing;
    let e1,e2 such that
A2: e1 in dom f & e2 in dom f and
A3: e1 <= e2;
    per cases by A3,XXREAL_0:1;
    suppose
      e1 = e2;
      hence thesis;
    end;
    suppose
      e1 < e2;
      hence thesis by A1,A2,Def13;
    end;
  end;
  cluster decreasing -> non-increasing for ext-real-valued Function;
  coherence
  proof
    let f be ext-real-valued Function;
    assume
A4: f is decreasing;
    let e1,e2 such that
A5: e1 in dom f & e2 in dom f and
A6: e1 <= e2;
    per cases by A6,XXREAL_0:1;
    suppose
      e1 = e2;
      hence thesis;
    end;
    suppose
      e1 < e2;
      hence thesis by A4,A5,Def14;
    end;
  end;
end;

registration
  let f,g be increasing ext-real-valued Function;
  cluster g*f -> increasing;
  coherence
  proof
    let e1,e2;
    assume that
A1: e1 in dom(g*f) and
A2: e2 in dom (g*f);
A3: e1 in dom f by A1,FUNCT_1:11;
    then
A4: (g*f).e1 = g.(f.e1) by FUNCT_1:13;
A5: e2 in dom f by A2,FUNCT_1:11;
    then
A6: (g*f).e2 = g.(f.e2) by FUNCT_1:13;
    assume e1 < e2;
    then
A7: f.e1 < f.e2 by A3,A5,Def13;
    f.e1 in dom g & f.e2 in dom g by A1,A2,FUNCT_1:11;
    hence thesis by A4,A6,A7,Def13;
  end;
end;

registration
  let f,g be non-decreasing ext-real-valued Function;
  cluster g*f -> non-decreasing;
  coherence
  proof
    let e1,e2;
    assume that
A1: e1 in dom(g*f) and
A2: e2 in dom (g*f);
A3: e1 in dom f by A1,FUNCT_1:11;
    then
A4: (g*f).e1 = g.(f.e1) by FUNCT_1:13;
A5: e2 in dom f by A2,FUNCT_1:11;
    then
A6: (g*f).e2 = g.(f.e2) by FUNCT_1:13;
    assume e1 <= e2;
    then
A7: f.e1 <= f.e2 by A3,A5,Def15;
    f.e1 in dom g & f.e2 in dom g by A1,A2,FUNCT_1:11;
    hence thesis by A4,A6,A7,Def15;
  end;
end;

registration
  let f,g be decreasing ext-real-valued Function;
  cluster g*f -> increasing;
  coherence
  proof
    let e1,e2;
    assume that
A1: e1 in dom(g*f) and
A2: e2 in dom (g*f);
A3: e1 in dom f by A1,FUNCT_1:11;
    then
A4: (g*f).e1 = g.(f.e1) by FUNCT_1:13;
A5: e2 in dom f by A2,FUNCT_1:11;
    then
A6: (g*f).e2 = g.(f.e2) by FUNCT_1:13;
    assume e1 < e2;
    then
A7: f.e1 > f.e2 by A3,A5,Def14;
    f.e1 in dom g & f.e2 in dom g by A1,A2,FUNCT_1:11;
    hence thesis by A4,A6,A7,Def14;
  end;
end;

registration
  let f,g be non-increasing ext-real-valued Function;
  cluster g*f -> non-decreasing;
  coherence
  proof
    let e1,e2;
    assume that
A1: e1 in dom(g*f) and
A2: e2 in dom (g*f);
A3: e1 in dom f by A1,FUNCT_1:11;
    then
A4: (g*f).e1 = g.(f.e1) by FUNCT_1:13;
A5: e2 in dom f by A2,FUNCT_1:11;
    then
A6: (g*f).e2 = g.(f.e2) by FUNCT_1:13;
    assume e1 <= e2;
    then
A7: f.e1 >= f.e2 by A3,A5,Def16;
    f.e1 in dom g & f.e2 in dom g by A1,A2,FUNCT_1:11;
    hence thesis by A4,A6,A7,Def16;
  end;
end;

registration
  let f be decreasing ext-real-valued Function;
  let g be increasing ext-real-valued Function;
  cluster g*f -> decreasing;
  coherence
  proof
    let e1,e2;
    assume that
A1: e1 in dom(g*f) and
A2: e2 in dom (g*f);
A3: e1 in dom f by A1,FUNCT_1:11;
    then
A4: (g*f).e1 = g.(f.e1) by FUNCT_1:13;
A5: e2 in dom f by A2,FUNCT_1:11;
    then
A6: (g*f).e2 = g.(f.e2) by FUNCT_1:13;
    assume e1 < e2;
    then
A7: f.e1 > f.e2 by A3,A5,Def14;
    f.e1 in dom g & f.e2 in dom g by A1,A2,FUNCT_1:11;
    hence thesis by A4,A6,A7,Def13;
  end;
  cluster f*g -> decreasing;
  coherence
  proof
    let e1,e2;
    assume that
A8: e1 in dom(f*g) and
A9: e2 in dom (f*g);
A10: e1 in dom g by A8,FUNCT_1:11;
    then
A11: (f*g).e1 = f.(g.e1) by FUNCT_1:13;
A12: e2 in dom g by A9,FUNCT_1:11;
    then
A13: (f*g).e2 = f.(g.e2) by FUNCT_1:13;
    assume e1 < e2;
    then
A14: g.e1 < g.e2 by A10,A12,Def13;
    g.e1 in dom f & g.e2 in dom f by A8,A9,FUNCT_1:11;
    hence thesis by A11,A13,A14,Def14;
  end;
end;

registration
  let f be non-increasing ext-real-valued Function;
  let g be non-decreasing ext-real-valued Function;
  cluster g*f -> non-increasing;
  coherence
  proof
    let e1,e2;
    assume that
A1: e1 in dom(g*f) and
A2: e2 in dom (g*f);
A3: e1 in dom f by A1,FUNCT_1:11;
    then
A4: (g*f).e1 = g.(f.e1) by FUNCT_1:13;
A5: e2 in dom f by A2,FUNCT_1:11;
    then
A6: (g*f).e2 = g.(f.e2) by FUNCT_1:13;
    assume e1 <= e2;
    then
A7: f.e1 >= f.e2 by A3,A5,Def16;
    f.e1 in dom g & f.e2 in dom g by A1,A2,FUNCT_1:11;
    hence thesis by A4,A6,A7,Def15;
  end;
  cluster f*g -> non-increasing;
  coherence
  proof
    let e1,e2;
    assume that
A8: e1 in dom(f*g) and
A9: e2 in dom (f*g);
A10: e1 in dom g by A8,FUNCT_1:11;
    then
A11: (f*g).e1 = f.(g.e1) by FUNCT_1:13;
A12: e2 in dom g by A9,FUNCT_1:11;
    then
A13: (f*g).e2 = f.(g.e2) by FUNCT_1:13;
    assume e1 <= e2;
    then
A14: g.e1 <= g.e2 by A10,A12,Def15;
    g.e1 in dom f & g.e2 in dom f by A8,A9,FUNCT_1:11;
    hence thesis by A11,A13,A14,Def16;
  end;
end;

registration
  let X be ext-real-membered set;
  cluster id X -> increasing for Function of X,X;
  coherence
  proof
    id X is increasing
    proof
      let e1,e2;
      assume that
A1:   e1 in dom id X and
A2:   e2 in dom id X;
      (id X).e1 = e1 by A1,FUNCT_1:18;
      hence thesis by A2,FUNCT_1:18;
    end;
    hence thesis;
  end;
end;

registration
  cluster increasing for sequence of NAT;
  existence
  proof
    take id NAT;
    thus thesis;
  end;
end;

definition
  let s be ManySortedSet of NAT;
  mode subsequence of s -> ManySortedSet of NAT means
  :Def17:
  ex N being increasing sequence of NAT st it = s * N;
  existence
  proof
    take s,id NAT;
    dom s = NAT by PARTFUN1:def 2;
    hence thesis by RELAT_1:52;
  end;
end;

Lm1:
 for x being non empty set, M be ManySortedSet of NAT,
     s be subsequence of M
   holds rng s c= rng M
 proof let x be non empty set, M be ManySortedSet of NAT,
     s be subsequence of M;
   ex N being increasing sequence of NAT st s = M * N by Def17;
  hence rng s c= rng M by RELAT_1:26;
 end;

registration
 let X be non empty set, s be X-valued ManySortedSet of NAT;
 cluster -> X-valued for subsequence of s;
 coherence
  proof let ss be subsequence of s;
    rng ss c= rng s by Lm1;
   hence rng ss c= X by XBOOLE_1:1;
  end;
end;

definition
  let X be non empty set, s be sequence of X;
  redefine mode subsequence of s -> sequence of X;
  coherence
  proof let ss be subsequence of s;
    rng ss c= X & dom ss = NAT by PARTFUN1:def 2;
   hence ss is sequence of X by RELSET_1:4;
  end;
end;

definition
  let X be non empty set, s be sequence of X, k be Nat;
  redefine func s ^\ k -> subsequence of s;
  coherence
  proof
    set N = (id NAT)^\k;
    N is increasing
    proof
      let e1,e2;
      assume e1 in dom N & e2 in dom N;
      then reconsider i=e1, j=e2 as Element of NAT;
A1:   N.j = (id NAT).(j+k) by NAT_1:def 3
        .= j+k by FUNCT_1:18;
      assume
A2:   e1 < e2;
      N.i = (id NAT).(i+k) by NAT_1:def 3
        .= i+k by FUNCT_1:18;
      hence thesis by A1,A2,XREAL_1:6;
    end;
    then reconsider N as increasing sequence of NAT;
    thus s ^\ k is subsequence of s
    proof take N;
     thus s^\k = (s* id NAT)^\k by FUNCT_2:17
      .= s * N by NAT_1:50;
    end;
  end;
end;

reserve s,s1,s2,s3 for sequence of X;

reserve XX for non empty set,
        ss,ss1,ss2,ss3 for sequence of XX;

theorem
  ss is subsequence of ss
proof
  take id NAT;
  thus thesis by FUNCT_2:17;
end;

theorem
  ss1 is subsequence of ss2 & ss2 is subsequence of ss3 implies
  ss1 is subsequence of ss3
proof
  given N1 being increasing sequence of NAT such that
A1: ss1 = ss2*N1;
  given N2 being increasing sequence of NAT such that
A2: ss2 = ss3*N2;
  take N2*N1;
  thus thesis by A1,A2,RELAT_1:36;
end;



registration
  let X;
  cluster constant for sequence of X;
  existence
  proof
    per cases;
    suppose X = {};
      then reconsider s = {} as sequence of X by FUNCT_2:def 1,RELSET_1:12;
      take s;
      thus thesis;
    end;
    suppose X <> {};
      then consider x such that
A1:   x in X by XBOOLE_0:def 1;
      reconsider s = NAT --> x as sequence of X by A1,FUNCOP_1:45;
      take s;
      thus thesis;
    end;
  end;
end;

theorem Th21:
  for ss1 being subsequence of ss holds rng ss1 c= rng ss
proof
  let ss1 be subsequence of ss;
  let x;
  consider N being increasing sequence of NAT such that
A1: ss1 = ss * N by Def17;
    assume x in rng ss1;
    then consider n being set such that
A2: n in NAT and
A3: x = ss1.n by FUNCT_2:11;
A4: N.n in NAT by A2,FUNCT_2:5;
    x = ss.(N.n) by A1,A2,A3,FUNCT_2:15;
    hence thesis by A4,FUNCT_2:4;
end;

registration
  let X be non empty set;
  let s be constant sequence of X;
  cluster -> constant for subsequence of s;
  coherence
  proof
    let s1 be subsequence of s;
    rng s1 c= rng s by Th21;
    hence thesis;
  end;
end;



definition
  let X be non empty set, N be increasing sequence of NAT, s be sequence of X;
  redefine func s*N -> subsequence of s;
  correctness by Def17;
end;



reserve X,Y for non empty set,
  Z for set;
reserve s,s1 for sequence of X,
  h,h1 for PartFunc of X,Y,
  h2 for PartFunc of Y ,Z,
  x for Element of X,
  N for increasing sequence of NAT;

theorem
  rng s c= dom h & s1 is subsequence of s implies h/*s1 is subsequence of h/*s
proof
  assume that
A1: rng s c= dom h and
A2: s1 is subsequence of s;
  consider N such that
A3: s1=s*N by A2,Def17;
  take N;
  thus thesis by A1,A3,FUNCT_2:110;
end;



registration
  let X be with_non-empty_element set;
  cluster non-empty for sequence of X;
  existence
  proof
    consider x being non empty set such that
A1: x in X by SETFAM_1:def 10;
    reconsider s = NAT --> x as sequence of X by A1,FUNCOP_1:45;
    take s;
    thus thesis;
  end;
end;

registration
  let X be with_non-empty_element set, s be non-empty sequence of X;
  cluster -> non-empty for subsequence of s;
  coherence
  proof
    let s1 be subsequence of s;
    rng s1 c= rng s by Th21;
    hence not {} in rng s1;
  end;
end;

reserve i,j for Nat;

definition
  let X be non empty set, s be sequence of X;
  redefine attr s is constant means
  ex x being Element of X st for n being Nat holds s.n=x;
  compatibility
  proof
    hereby
      assume s is constant;
      then consider x being Element of X such that
A1:   for n being Element of NAT st n in dom s holds s.n = x by PARTFUN2:def 1;
      take x;
      let n be Nat;
      dom s = NAT & n in NAT by FUNCT_2:def 1,ORDINAL1:def 12;
      hence s.n = x by A1;
    end;
    given x be Element of X such that
A2: for n being Nat holds s.n=x;
    take x;
    thus thesis by A2;
  end;
end;

theorem Th23:
  for X being set for s being constant sequence of X holds s.i = s.j
proof
  let X be set;
  let s be constant sequence of X;
  per cases;
  suppose
A1: X is empty;
    then s.i = {};
    hence thesis by A1;
  end;
  suppose X is not empty;
    then dom s = NAT by FUNCT_2:def 1;
    then i in dom s & j in dom s by ORDINAL1:def 12;
    hence thesis by FUNCT_1:def 10;
  end;
end;

theorem Th24:
  (for i,j holds s.i = s.j) implies s is constant
proof
  assume for i,j holds s.i = s.j;
  then for x,y being set st x in dom s & y in dom s holds s.x = s.y;
  hence thesis by FUNCT_1:def 10;
end;

theorem
  (for i holds s.i = s.(i+1)) implies s is constant
proof
  assume
A1: for i holds s.i = s.(i+1);
  now
    let i,j;
A2: now
      let i,j such that
A3:   i <= j;
      defpred P[Nat] means i <= $1 implies s.i = s.$1;
A4:   for j being Nat st P[j] holds P[j+1]
      proof
        let j being Nat such that
A5:     P[j];
        assume i <= j+1;
        then i < j+1 or i = j+1 by XXREAL_0:1;
        hence thesis by A1,A5,NAT_1:13;
      end;
A6:   P[0] by NAT_1:3;
      for j being Nat holds P[j] from NAT_1:sch 2(A6,A4);
      hence s.i = s.j by A3;
    end;
    i <= j or j <= i;
    hence s.i = s.j by A2;
  end;
  hence thesis by Th24;
end;

theorem
  s is constant & s1 is subsequence of s implies s = s1
proof
  assume that
A1: s is constant and
A2: s1 is subsequence of s;
  let n be Element of NAT;
  consider N such that
A3: s1=s*N by A2,Def17;
  thus s1.n=s.(N.n) by A3,FUNCT_2:15
    .=s.n by A1,Th23;
end;

reserve n for Nat;

theorem Th27:
  rng s c= dom h implies (h/*s)^\n=h/*(s^\n)
proof
  assume
A1: rng s c= dom h;
  let m be Element of NAT;
A2: rng (s^\n) c= rng s by Th21;
  thus ((h/*s)^\n).m = (h/*s).(m+n) by NAT_1:def 3
    .= h.(s.(m+n)) by A1,FUNCT_2:108
    .= h.((s^\n).m) by NAT_1:def 3
    .= (h/*(s^\n)).m by A1,A2,FUNCT_2:108,XBOOLE_1:1;
end;

theorem Th28:
  s.n in rng s
proof
  n in NAT by ORDINAL1:def 12;
  hence thesis by FUNCT_2:112;
end;

theorem
  h is total implies h/*(s^\n) = (h/*s)^\n
proof
  assume h is total;
  then dom h = X by PARTFUN1:def 2;
  then rng s c= dom h;
  hence thesis by Th27;
end;

theorem Th30:
  rng s c= dom h implies h.:(rng s) = rng (h/*s)
proof
  assume
A1: rng s c= dom h;
  now
    let r be Element of Y;
    thus r in h.:(rng s) implies r in rng (h/*s)
    proof
      assume r in h.:(rng s);
      then consider p being Element of X such that
      p in dom h and
A2:   p in rng s and
A3:   r=h.p by PARTFUN2:59;
      consider n being Element of NAT such that
A4:   p=s.n by A2,FUNCT_2:113;
      r = (h/*s).n by A1,A3,A4,FUNCT_2:108;
      hence thesis by Th28;
    end;
    assume r in rng (h/*s);
    then consider n being Element of NAT such that
A5: (h/*s).n=r by FUNCT_2:113;
A6: s.n in rng s by Th28;
    r = h.(s.n) by A1,A5,FUNCT_2:108;
    hence r in h.:(rng s) by A1,A6,FUNCT_1:def 6;
  end;
  hence thesis by SUBSET_1:3;
end;

theorem
  rng s c= dom (h2*h1) implies h2/*(h1/*s) = (h2*h1)/*s
proof
  assume
A1: rng s c= dom (h2*h1);
  now
    let n be Element of NAT;
A2: rng s c= dom h1 by A1,FUNCT_1:101;
    h1.:(rng s) c= dom h2 by A1,FUNCT_1:101;
    then
A3: rng (h1/*s) c= dom h2 by A2,Th30;
    s.n in rng s by Th28;
    then
A4: s.n in dom h1 by A1,FUNCT_1:11;
    thus ((h2*h1)/*s).n = (h2*h1).(s.n) by A1,FUNCT_2:108
      .= h2.(h1.(s.n)) by A4,FUNCT_1:13
      .= h2.((h1/*s).n) by A2,FUNCT_2:108
      .= (h2/*(h1/*s)).n by A3,FUNCT_2:108;
  end;
  hence thesis by FUNCT_2:63;
end;

definition
  let f be ext-real-valued Function;
  attr f is zeroed means
  f.{} = 0;
end;



registration
  cluster COMPLEX-valued -> complex-valued for Relation;
  coherence
  proof
    let R be Relation;
    assume R is COMPLEX-valued;
    hence rng R c= COMPLEX by RELAT_1:def 19;
  end;
  cluster ExtREAL-valued -> ext-real-valued for Relation;
  coherence
  proof
    let R be Relation;
    assume R is ExtREAL-valued;
    hence rng R c= ExtREAL by RELAT_1:def 19;
  end;
  cluster REAL-valued -> real-valued for Relation;
  coherence
  proof
    let R be Relation;
    assume R is REAL-valued;
    hence rng R c= REAL by RELAT_1:def 19;
  end;
  cluster NAT-valued -> natural-valued for Relation;
  coherence
  proof
    let R be Relation;
    assume R is NAT-valued;
    hence rng R c= NAT by RELAT_1:def 19;
  end;
end;

definition
  let s be ManySortedSet of NAT;
  redefine attr s is constant means
  ex x being set st for n being Nat holds s.n=x;
  compatibility
  proof
A1: dom s = NAT by PARTFUN1:def 2;
    hereby
      assume
A2:     s is constant;
      take x = s.0;
      let n be Nat;
       0 in dom s & n in dom s by A1,ORDINAL1:def 12;
      hence s.n = x by A2,FUNCT_1:def 10;
    end;
    given x being set such that
A3: for n being Nat holds s.n=x;
  let n1,n2 be set;
   assume
A4:  n1 in dom s & n2 in dom s;
   hence s.n1 = x by A3 .= s.n2 by A3,A4;
 end;
end;

theorem
 for x being non empty set, M be ManySortedSet of NAT,
     s be subsequence of M
   holds rng s c= rng M by Lm1;

registration
  let X be set;
  cluster natural-valued for ManySortedSet of X;
  existence
  proof
    take X --> 0;
    thus thesis;
  end;
end;














begin

reserve a,b,c,d for real number;



theorem Th1:
  a^2 + b^2 = 0 implies a = 0
proof
  0 <= a^2 & 0 <= b^2 by XREAL_1:63;
  hence thesis;
end;

Lm1: 0 <= a^2 + b^2
proof
  0<=a^2 & 0<=b^2 by XREAL_1:63;
  hence thesis;
end;



definition
  let z be complex number;

  func Re z means
  :Def1:
  it = z if z in REAL
  otherwise ex f being Function of 2,REAL st z = f & it = f.0;
  existence
  proof
    thus z in REAL implies ex IT being set st IT = z;
A1: z in COMPLEX by XCMPLX_0:def 2;
    assume not z in REAL;
    then
    z in Funcs(2,REAL) \ { x where x is Element of Funcs(2,REAL): x.1 = 0
    } by A1,CARD_1:50,NUMBERS:def 2,XBOOLE_0:def 3;
    then reconsider f = z as Function of 2, REAL by FUNCT_2:66;
    take f.0, f;
    thus thesis;
  end;
  uniqueness;
  consistency;
  func Im z means
  :Def2:
  it = 0 if z in REAL otherwise ex f being Function of
  2,REAL st z = f & it = f.1;
  existence
  proof
    thus z in REAL implies ex IT being set st IT = 0;
A2: z in COMPLEX by XCMPLX_0:def 2;
    assume not z in REAL;
    then
    z in Funcs(2,REAL) \ { x where x is Element of Funcs(2,REAL): x.1 = 0
    } by A2,CARD_1:50,NUMBERS:def 2,XBOOLE_0:def 3;
    then reconsider f = z as Function of 2, REAL by FUNCT_2:66;
    take f.1, f;
    thus thesis;
  end;
  uniqueness;
  consistency;
end;

registration
  let z be complex number;
  cluster Re z -> real;
  coherence
  proof
    per cases;
    suppose
      z in REAL;
      hence thesis by Def1;
    end;
    suppose
      not z in REAL;
      then consider f being Function of 2,REAL such that
      z = f and
A1:   Re z = f.0 by Def1;
      0 in 2 by CARD_1:50,TARSKI:def 2;
      then f.0 in REAL by FUNCT_2:5;
      hence thesis by A1;
    end;
  end;
  cluster Im z -> real;
  coherence
  proof
    per cases;
    suppose
      z in REAL;
      hence thesis by Def2;
    end;
    suppose
      not z in REAL;
      then consider f being Function of 2,REAL such that
      z = f and
A2:   Im z = f.1 by Def2;
      1 in 2 by CARD_1:50,TARSKI:def 2;
      then f.1 in REAL by FUNCT_2:5;
      hence thesis by A2;
    end;
  end;
end;

definition
  let z be complex number;
  redefine func Re z -> Real;
  coherence by XREAL_0:def 1;
  redefine func Im z -> Real;
  coherence by XREAL_0:def 1;
end;

theorem Th2:
  for f being Function of 2,REAL
  ex a,b being Element of REAL st f = (0,1)-->(a,b)
proof
  let f be Function of 2,REAL;
  0 in 2 & 1 in 2 by CARD_1:50,TARSKI:def 2;
  then reconsider a = f.0, b = f.1 as Element of REAL by FUNCT_2:5;
  take a,b;
  dom f = {0,1} by CARD_1:50,FUNCT_2:def 1;
  hence thesis by FUNCT_4:66;
end;

Lm2:
for a, b being Element of REAL holds Re [*a,b*] = a & Im [*a,b*] = b
proof
  let a, b be Element of REAL;
  reconsider a9 =a, b9 = b as Element of REAL;
  thus Re [*a,b*] = a
  proof
    per cases;
    suppose
      b = 0;
      then [*a,b*] = a by ARYTM_0:def 5;
      hence thesis by Def1;
    end;
    suppose
      b <> 0;
      then
A1:   [*a,b*] = (0,1)-->(a9,b9) by ARYTM_0:def 5;
      then reconsider f = [*a,b*] as Function of 2, REAL by CARD_1:50;
      ( not [*a,b*] in REAL)& f.0 = a by A1,ARYTM_0:8,FUNCT_4:63;
      hence thesis by Def1;
    end;
  end;
  per cases;
  suppose
A2: b = 0;
    then [*a,b*] = a by ARYTM_0:def 5;
    hence thesis by A2,Def2;
  end;
  suppose
    b <> 0;
    then
A3: [*a,b*] = (0,1)-->(a9,b9) by ARYTM_0:def 5;
    then reconsider f = [*a,b*] as Function of 2, REAL by CARD_1:50;
    ( not [*a,b*] in REAL)& f.1 = b by A3,ARYTM_0:8,FUNCT_4:63;
    hence thesis by Def2;
  end;
end;

reserve z,z1,z2 for complex number;

Lm3: [*Re z, Im z*] = z
proof
A1: z in COMPLEX by XCMPLX_0:def 2;
  per cases;
  suppose
A2: z in REAL;
    then Im z = 0 by Def2;
    hence [*Re z, Im z*] = Re z by ARYTM_0:def 5
      .= z by A2,Def1;
  end;
  suppose
A3: not z in REAL;
    then
A4: ex f being Function of 2,REAL st z = f & Im z = f.1 by Def2;
    then consider a,b being Element of REAL such that
A5: z = (0,1)-->(a,b) by Th2;
    reconsider f = z as Element of Funcs(2,REAL) by A5,CARD_1:50,FUNCT_2:8;
    z in Funcs(2,REAL) \ { x where x is Element of Funcs(2,REAL): x.1 = 0
    } by A1,A3,CARD_1:50,NUMBERS:def 2,XBOOLE_0:def 3;
    then not z in { x where x is Element of Funcs(2,REAL): x.1 = 0} by
XBOOLE_0:def 5;
    then f.1 <> 0;
    then
A6: b <> 0 by A5,FUNCT_4:63;
    ex f being Function of 2,REAL st z = f & Re z = f.0 by A3,Def1;
    then
A7: Re z = a by A5,FUNCT_4:63;
    Im z = b by A4,A5,FUNCT_4:63;
    hence thesis by A5,A7,A6,ARYTM_0:def 5;
  end;
end;

theorem Th3:
  Re z1 = Re z2 & Im z1 = Im z2 implies z1 = z2
proof
  assume Re z1 = Re z2 & Im z1 = Im z2;
  hence z1 = [*Re z2,Im z2*] by Lm3
    .= z2 by Lm3;
end;

definition
  let z1,z2 be complex number;

  redefine pred z1 = z2 means
  Re z1 = Re z2 & Im z1 = Im z2;
  compatibility by Th3;
end;

notation
  synonym 0c for 0;
end;

definition
  redefine func 0c -> Element of COMPLEX;
  correctness by XCMPLX_0:def 2;
end;

definition

  func 1r -> Element of COMPLEX equals
  1;
  correctness by XCMPLX_0:def 2;
  redefine func <i> -> Element of COMPLEX;
  coherence by XCMPLX_0:def 2;
end;

Lm4: 0 = [*0,0*] by ARYTM_0:def 5;

Lm5: 1r = [*1,0*] by ARYTM_0:def 5;

theorem Th4:
  Re 0 = 0 & Im 0 = 0 by Lm2,Lm4;

theorem Th5:
  z = 0 iff (Re z)^2 + (Im z)^2 = 0
proof
  set r = Re z, i = Im z;
  thus z=0 implies r^2 + i^2 = 0 by Th4;
  assume 0 = r^2+i^2;
  then r = 0 & i = 0 by Th1;
  hence thesis by Th3,Th4;
end;

theorem Th6:
  Re(1r) = 1 & Im(1r) = 0 by Lm2,Lm5;

Lm6: <i> = [*0,1*] by ARYTM_0:def 5,XCMPLX_0:def 1;

theorem Th7:
  Re(<i>) = 0 & Im(<i>) = 1 by Lm2,Lm6;

Lm7: for x being real number, x1,x2 being Element of REAL st x = [*x1,x2*]
holds x2 = 0 & x = x1
proof
  let x be real number, x1,x2 being Element of REAL;
  assume
A1: x = [*x1,x2*];
A2: x in REAL by XREAL_0:def 1;
  hereby
    assume x2 <> 0;
    then x = (0,1) --> (x1,x2) by A1,ARYTM_0:def 5;
    hence contradiction by A2,ARYTM_0:8;
  end;
  hence thesis by A1,ARYTM_0:def 5;
end;

Lm8: for x9,y9 being Element of REAL, x,y being real number st x9 = x & y9 = y
holds +(x9,y9) = x + y
proof
  let x9,y9 be Element of REAL, x,y be real number such that
A1: x9 = x & y9 = y;
  consider x1,x2,y1,y2 being Element of REAL such that
A2: x = [* x1,x2 *] & y = [*y1,y2*] and
A3: x+y = [*+(x1,y1),+(x2,y2)*] by XCMPLX_0:def 4;
  x2 = 0 & y2 = 0 by A2,Lm7;
  then
A4: +(x2,y2) = 0 by ARYTM_0:11;
  x = x1 & y = y1 by A2,Lm7;
  hence thesis by A1,A3,A4,ARYTM_0:def 5;
end;

Lm9: for x being Element of REAL holds opp x = -x
proof
  let x be Element of REAL;
  +(x,opp x) = 0 by ARYTM_0:def 3;
  then x + opp x = 0 by Lm8;
  hence thesis;
end;

Lm10: for x9,y9 being Element of REAL, x,y being real number st x9 = x & y9 =
y holds *(x9,y9) = x * y
proof
  let x9,y9 be Element of REAL, x,y be real number such that
A1: x9 = x & y9 = y;
  consider x1,x2,y1,y2 being Element of REAL such that
A2: x = [* x1,x2 *] and
A3: y = [*y1,y2*] and
A4: x*y = [* +(*(x1,y1),opp*(x2,y2)), +(*(x1,y2),*(x2,y1)) *] by XCMPLX_0:def 5
;
  x2 = 0 by A2,Lm7;
  then
A5: *(x2,y1) = 0 by ARYTM_0:12;
A6: y2 = 0 by A3,Lm7;
  then *(x1,y2) = 0 by ARYTM_0:12;
  then
A7: +(*(x1,y2),*(x2,y1)) = 0 by A5,ARYTM_0:11;
  x = x1 & y = y1 by A2,A3,Lm7;
  hence *(x9,y9) = +(*(x1,y1),*(opp x2,y2)) by A1,A6,ARYTM_0:11,12
    .= +(*(x1,y1),opp*(x2,y2)) by ARYTM_0:15
    .= x * y by A4,A7,ARYTM_0:def 5;
end;

Lm11: for x,y,z being complex number st z = x + y holds Re z = Re x + Re y
proof
  let x,y,z be complex number such that
A1: z = x + y;
  consider x1,x2,y1,y2 being Element of REAL such that
A2: x = [*x1,x2*] & y = [*y1,y2*] and
A3: x + y = [*+(x1,y1),+(x2,y2)*] by XCMPLX_0:def 4;
A4: Re x = x1 & Re y = y1 by A2,Lm2;
  thus Re z = +(x1,y1) by A1,A3,Lm2
    .= Re x + Re y by A4,Lm8;
end;

Lm12: for x,y,z being complex number st z = x+y holds Im z = Im x + Im y
proof
  let x,y,z be complex number such that
A1: z = x+y;
  consider x1,x2,y1,y2 being Element of REAL such that
A2: x = [*x1,x2*] & y = [*y1,y2*] and
A3: x + y = [*+(x1,y1),+(x2,y2)*] by XCMPLX_0:def 4;
A4: Im x = x2 & Im y = y2 by A2,Lm2;
  thus Im z = +(x2,y2) by A1,A3,Lm2
    .= Im x + Im y by A4,Lm8;
end;

Lm13: for x,y,z being complex number st z = x * y holds Re z = Re x * Re y -
Im x * Im y
proof
  let x,y,z be complex number such that
A1: z = x * y;
  consider x1,x2,y1,y2 being Element of REAL such that
A2: x = [*x1,x2*] & y = [*y1,y2*] and
A3: x * y = [* +(*(x1,y1),opp*(x2,y2)), +(*(x1,y2),*(x2,y1)) *] by
XCMPLX_0:def 5;
A4: Re x = x1 & Re y = y1 by A2,Lm2;
A5: Im x = x2 & Im y = y2 by A2,Lm2;
  thus Re z = +(*(x1,y1),opp*(x2,y2)) by A1,A3,Lm2
    .= *(x1,y1) + opp*(x2,y2) by Lm8
    .= x1*y1 + opp*(x2,y2) by Lm10
    .= x1*y1 + -*(x2,y2) by Lm9
    .= x1*y1 - *(x2,y2)
    .= Re x * Re y - Im x * Im y by A4,A5,Lm10;
end;

Lm14: for x,y,z being complex number st z = x*y holds Im z = Re x * Im y + Im
x * Re y
proof
  let x,y,z be complex number such that
A1: z = x*y;
  consider x1,x2,y1,y2 being Element of REAL such that
A2: x = [*x1,x2*] & y = [*y1,y2*] and
A3: x * y = [* +(*(x1,y1),opp*(x2,y2)), +(*(x1,y2),*(x2,y1)) *] by
XCMPLX_0:def 5;
A4: Im x = x2 & Im y = y2 by A2,Lm2;
A5: Re x = x1 & Re y = y1 by A2,Lm2;
  thus Im z = +(*(x1,y2),*(x2,y1)) by A1,A3,Lm2
    .= *(x1,y2) + *(x2,y1) by Lm8
    .= x1*y2 + *(x2,y1) by Lm10
    .= Re x * Im y + Im x * Re y by A4,A5,Lm10;
end;

Lm15: z1+z2 = [* Re z1 + Re z2, Im z1 + Im z2 *]
proof
  set z = [* Re z1 + Re z2, Im z1 + Im z2 *];
  reconsider z9 = z1 + z2 as Element of COMPLEX by XCMPLX_0:def 2;
A1: Im z = Im z1 + Im z2 by Lm2
    .= Im z9 by Lm12;
  Re z = Re z1 + Re z2 by Lm2
    .= Re z9 by Lm11;
  hence thesis by A1,Th3;
end;

Lm16: z1 * z2 = [* Re z1 * Re z2 - Im z1 * Im z2, Re z1 * Im z2 + Re z2 * Im
z1 *]
proof
  set z = [* Re z1 * Re z2 - Im z1 * Im z2, Re z1 * Im z2 + Re z2 * Im z1 *];
  reconsider z9 = z1 * z2 as Element of COMPLEX by XCMPLX_0:def 2;
A1: Im z = Re z1 * Im z2 + Re z2 * Im z1 by Lm2
    .= Im z9 by Lm14;
  Re z = Re z1 * Re z2 - Im z1 * Im z2 by Lm2
    .= Re z9 by Lm13;
  hence thesis by A1,Th3;
end;

Lm17: Re(z1 * z2) = Re z1 * Re z2 - Im
z1 * Im z2 & Im(z1 * z2) = Re z1 * Im z2 + Re z2 * Im z1
proof
  z1 * z2 = [*Re z1 * Re z2 - Im z1 * Im z2, Re z1 * Im z2 + Re z2 * Im z1
  *] by Lm16;
  hence thesis by Lm2;
end;

Lm18: Re(z1 + z2) = Re z1 + Re z2 & Im(z1 + z2) = Im z1 + Im z2
proof
  (z1 + z2) = [* Re z1 + Re z2, Im z1 + Im z2 *] by Lm15;
  hence thesis by Lm2;
end;

Lm19: for x being Real holds Re (x*<i>) = 0
proof
  let x be Real;
  thus Re (x*<i>) = Re x * Re (<i>) - Im x * Im (<i>) by Lm17
    .= Re x * 0 - 0 * Im (<i>) by Def2,Th7
    .= 0;
end;

Lm20: for x being Real holds Im (x*<i>) = x
proof
  let x be Real;
  thus Im (x*<i>) = Re x * Im (<i>) + Im x * Re (<i>) by Lm17
    .= x by Def1,Th7;
end;

Lm21: for x, y being Real holds [*x,y*] = x + y * <i>
proof
  let x, y be Real;
A1: Im (x + y*<i>) = Im x + Im (y*<i>) by Lm18
    .= 0 + Im (y*<i>) by Def2
    .= y by Lm20;
  Re (x + y*<i>) = Re x + Re (y*<i>) by Lm18
    .= Re x + 0 by Lm19
    .= x by Def1;
  hence thesis by A1,Lm3;
end;

definition

  let z1,z2 be Element of COMPLEX;
  redefine func z1 + z2 -> Element of COMPLEX equals
  Re z1 + Re z2 + (Im z1 + Im z2)*<i>;
  coherence by XCMPLX_0:def 2;
  compatibility
  proof
    z1 + z2 = [* Re z1 + Re z2, Im z1 + Im z2 *] by Lm15;
    hence thesis by Lm21;
  end;
end;

theorem Th8:
  Re(z1 + z2) = Re z1 + Re z2 & Im(z1 + z2) = Im z1 + Im z2
proof
  (z1 + z2) = [* Re z1 + Re z2, Im z1 + Im z2 *] by Lm15;
  hence thesis by Lm2;
end;

definition
  let z1,z2 be Element of COMPLEX;
  redefine func z1 * z2 -> Element of COMPLEX equals
  Re z1 * Re z2 - Im z1 *
  Im z2+(Re z1 * Im z2 + Re z2 * Im z1)*<i>;
  coherence by XCMPLX_0:def 2;
  compatibility
  proof
    z1 * z2 = [* Re z1 * Re z2 - Im z1 * Im z2, Re z1 * Im z2 + Re z2 * Im
    z1 *] by Lm16;
    hence thesis by Lm21;
  end;
end;

theorem Th9:
  Re(z1 * z2) = Re z1 * Re z2 - Im z1 * Im z2 &
  Im(z1 * z2) = Re z1 * Im z2 + Re z2 * Im z1
proof
  z1 * z2 = [*Re z1 * Re z2 - Im z1 * Im z2, Re z1 * Im z2 + Re z2 * Im z1
  *] by Lm16;
  hence thesis by Lm2;
end;

theorem
  Re (a*<i>) = 0
proof
A1: a in REAL by XREAL_0:def 1;
  thus Re (a*<i>) = Re a * Re (<i>) - Im a * Im (<i>) by Th9
  .= Re a * 0 - 0 * Im (<i>) by Def2,Th7,A1
  .= 0;
end;

theorem
  Im (a*<i>) = a
proof
A1: a in REAL by XREAL_0:def 1;
  thus Im (a*<i>) = Re a * Im (<i>) + Im a * Re (<i>) by Th9
  .= a by Def1,Th7,A1;
end;

theorem Th12:
  Re(a+b*<i>) = a & Im(a+b*<i>) = b
proof
  reconsider a,b as Real by XREAL_0:def 1;
  (a+b*<i>) = [*a,b*] by Lm21;
  hence thesis by Lm2;
end;

theorem Th13:
  Re z+(Im z)*<i> = z
proof
  [*Re z, Im z*] = z by Lm3;
  hence thesis by Lm21;
end;

theorem Th14:
  Im z1 = 0 & Im z2 = 0 implies Re(z1*z2) = Re z1 * Re z2 & Im(z1*z2) = 0
proof
  assume that
A1: Im z1 = 0 and
A2: Im z2 = 0;
  thus Re(z1*z2) = Re z1 * Re z2 - Im z1 * Im z2 by Th9
    .= Re z1 * Re z2 by A1;
  thus Im(z1*z2) = Re z1 * Im z2 + Re z2 * Im z1 by Th9
    .= 0 by A1,A2;
end;

theorem Th15:
  Re z1 = 0 & Re z2 = 0 implies Re(z1*z2) = - Im z1 * Im z2 & Im(z1*z2) = 0
proof
  assume that
A1: Re z1 = 0 and
A2: Re z2 = 0;
  thus Re(z1*z2) = Re z1 * Re z2 - Im z1 * Im z2 by Th9
    .= - Im z1 * Im z2 by A1;
  thus Im(z1*z2) = Re z1 * Im z2 + Re z2 * Im z1 by Th9
    .= 0 by A1,A2;
end;

theorem
  Re(z*z) = (Re z)^2 - (Im z)^2 & Im(z*z) = 2*(Re z *Im z)
proof
  thus Re(z*z) = (Re z)^2 - (Im z)^2 by Th9;
  thus Im(z*z) = Re z *Im z + Re z *Im z by Th9
    .= 2*(Re z *Im z);
end;

definition
  let z be Element of COMPLEX;
  redefine func -z -> Element of COMPLEX equals
  :Def7:
  -Re z+(-Im z)*<i>;
  coherence by XCMPLX_0:def 2;
  compatibility
  proof
    set z9 = [* -Re z, -Im z *];
    z9 + z = [* Re z9 + Re z, Im z9 + Im z *] by Lm15
      .= [* -Re z + Re z, Im z9 + Im z *] by Lm2
      .= [* 0, -Im z + Im z *] by Lm2
      .= 0 by ARYTM_0:def 5;
    hence thesis by Lm21;
  end;
end;

theorem Th17:
  Re(-z) = -(Re z) & Im(-z) = -(Im z)
proof
  z in COMPLEX by XCMPLX_0:def 2;
  then -z = -Re z+(-Im z)*<i> by Def7;
  hence thesis by Th12;
end;

theorem
  <i>*<i> = -1r;

definition
  let z1,z2 be Element of COMPLEX;
  redefine func z1 - z2 -> Element of COMPLEX equals
  :Def8:
  Re z1 - Re z2 + (Im z1 - Im z2)*<i>;
  coherence by XCMPLX_0:def 2;
  compatibility
  proof
A1: z1 = Re z1 + (Im z1)*<i> by Th13;
    z1 - z2 = z1 + -z2 .= z1 + (-Re z2+(-Im z2)*<i>) by Def7
      .= Re z1 - Re z2 + (Im z1 - Im z2)*<i> by A1;
    hence thesis;
  end;
end;

theorem Th19:
  Re(z1 - z2) = Re z1 - Re z2 & Im(z1 - z2) = Im z1 - Im z2
proof
A1: z1 in COMPLEX & z2 in COMPLEX by XCMPLX_0:def 2;
    hence Re(z1 - z2) = Re(Re z1 - Re z2 + (Im z1 - Im z2)*<i>) by Def8
    .= Re z1 - Re z2 by Th12;
  thus Im(z1 - z2) = Im(Re z1 - Re z2 + (Im z1 - Im z2)*<i>) by A1,Def8
    .= Im z1 - Im z2 by Th12;
end;

definition
  let z be Element of COMPLEX;
  redefine func z" -> Element of COMPLEX equals
  :Def9:
  Re z / ((Re z)^2+(Im z
  )^2)+((- Im z) / ((Re z)^2+(Im z)^2))*<i>;
  coherence by XCMPLX_0:def 2;
  compatibility
  proof
    reconsider z9 =Re z / ((Re z)^2+(Im z)^2)+((- Im z) / ((Re z)^2+(Im z)^2))
    *<i> as Element of COMPLEX by XCMPLX_0:def 2;
    per cases;
    suppose
      z = 0;
      hence thesis by Th4;
    end;
    suppose
A1:   z <> 0;
      then
A2:   (Re z)^2+(Im z)^2 <> 0 by Th5;
A3:   Im z9 = (-Im z) / ((Re z)^2+(Im z)^2) by Th12;
      then
A4:   Re z * Im z9 + Re z9 * Im z = (Re z)/1 * ((-Im z) / ((Re z)^2+(Im z)
      ^2)) + Re z / ((Re z)^2+(Im z)^2) * Im z by Th12
        .= Re z * (-Im z) / (1*((Re z)^2+(Im z)^2)) + Re z / ((Re z)^2+(Im z
      )^2) * (Im z)/1 by XCMPLX_1:76
        .= Re z * (-Im z) / (1*((Re z)^2+(Im z)^2)) + (Im z)/1 * Re z / ((Re
      z)^2+(Im z)^2) by XCMPLX_1:76
        .= (- Re z * Im z + Re z * Im z) / ((Re z)^2+(Im z)^2) by XCMPLX_1:62
        .= 0;
A5:   Re z * Re z9 - Im z * Im z9 = (Re z)/1 * (Re z / ((Re z)^2+(Im z)^2)
      ) - Im z *((-Im z)/((Re z)^2+(Im z)^2)) by A3,Th12
        .= Re z * Re z / (1*((Re z)^2+(Im z)^2)) - (Im z)/1 *((-Im z)/((Re z
      )^2+(Im z)^2)) by XCMPLX_1:76
        .= (Re z)^2 / ((Re z)^2+(Im z)^2) - Im z *(-Im z)/(1*((Re z)^2+(Im z
      )^2)) by XCMPLX_1:76
        .= (Re z)^2 / ((Re z)^2+(Im z)^2) - (-((Im z)^2))/(1*((Re z)^2+(Im z
      )^2))
        .= (Re z)^2 / ((Re z)^2+(Im z)^2) - -((Im z)^2)/((Re z)^2+(Im z)^2)
      by XCMPLX_1:187
        .= (Re z)^2 / ((Re z)^2+(Im z)^2) + (Im z)^2/(1*((Re z)^2+(Im z)^2))
        .= ((Re z)^2 + (Im z)^2)/((Re z)^2+(Im z)^2) by XCMPLX_1:62
        .= 1 by A2,XCMPLX_1:60;
      z * z9= [*Re z * Re z9 - Im z * Im z9,Re z * Im z9 + Re z9 * Im z*]
      by Lm16
        .= 1 by A5,A4,ARYTM_0:def 5;
      hence thesis by A1,XCMPLX_0:def 7;
    end;
  end;
end;

theorem Th20:
  Re(z") = Re z / ((Re z)^2+(Im z)^2) & Im(z") = (- Im z) / ((Re z)^2+(Im z)^2)
proof
  z in COMPLEX by XCMPLX_0:def 2;
  then
  z" = Re z / ((Re z)^2+(Im z)^2)+((- Im z) / ((Re z)^2+(Im z)^2))*<i>
  by Def9;
  hence thesis by Th12;
end;

theorem
  <i>" = -<i>;

theorem Th22:
  Re z <> 0 & Im z = 0 implies Re(z") = (Re z)" & Im(z") = 0
proof
  assume that
A1: Re z <> 0 and
A2: Im z = 0;
  Re(z") = Re z / ((Re z)^2+(Im z)^2) by Th20;
  hence Re(z") = (1*Re z) / (Re z * Re z) by A2
    .= 1/Re z by A1,XCMPLX_1:91
    .= (Re z)" by XCMPLX_1:215;
  Im(z") = (- Im z) / ((Re z)^2+(Im z)^2) by Th20;
  hence thesis by A2;
end;

theorem Th23:
  Re z = 0 & Im z <> 0 implies Re(z") = 0 & Im(z") = -(Im z)"
proof
  assume that
A1: Re z = 0 and
A2: Im z <> 0;
  Re(z") = Re z / ((Re z)^2+(Im z)^2) by Th20;
  hence Re(z") = 0 by A1;
  Im(z") = (- Im z) / ((Re z)^2+(Im z)^2) by Th20;
  hence Im(z") = -(1*Im z / (Im z * Im z)) by A1,XCMPLX_1:187
    .= -(1 / Im z) by A2,XCMPLX_1:91
    .= - (Im z)" by XCMPLX_1:215;
end;

definition
  let z1,z2 be complex number;
  redefine func z1 / z2 -> Element of COMPLEX equals
  :Def10:
  (Re z1 * Re z2 + Im z1 * Im z2) / ((Re z2)^2 + (Im z2)^2) +
  ((Re z2 * Im z1 - Re z1 * Im z2) / ((Re z2)^2 + (Im z2)^2))*<i>;
  coherence by XCMPLX_0:def 2;
  compatibility
  proof
    reconsider z1,z2 as Element of COMPLEX by XCMPLX_0:def 2;
    set k = (Re z2)^2 + (Im z2)^2;
A1: Re[*Re z2/k,(-Im z2)/k*] = Re(Re z2/k+(-Im z2)/k*<i>) by Lm21
      .= Re z2/k by Th12;
A2: Im[*Re z2/k,(-Im z2)/k*] = Im(Re z2/k+(-Im z2)/k*<i>) by Lm21
      .= (-Im z2)/k by Th12;
    z1 / z2 = z1 * z2" by XCMPLX_0:def 9
      .= z1*[* Re z2/k, (-Im z2)/k *] by Lm21
      .= [* (Re z1)/1 * (Re z2/k) - Im z1 * ((-Im z2)/k), Re z1 * ((-Im z2)/
    k) + (Re z2/k) * Im z1 *] by A1,A2,Lm16
      .= (Re z1)/1 * (Re z2/k) - Im z1 * ((-Im z2)/k)+ (Re z1 * ((-Im z2)/k)
    + (Re z2/k) * Im z1)*<i> by Lm21
      .= Re z1 * Re z2/(1*k) - (Im z1)/1 * ((-Im z2)/k)+ (Re z1 * ((-Im z2)/
    k) + (Re z2/k) * Im z1)*<i> by XCMPLX_1:76
      .= Re z1 * Re z2/k - Im z1 * (-Im z2)/(1*k)+ ((Re z1)/1 * ((-Im z2)/k)
    + (Re z2/k) * Im z1)*<i> by XCMPLX_1:76
      .= Re z1 * Re z2/k - Im z1 * (-Im z2)/k+ (Re z1 * (-Im z2)/(1*k) + (Re
    z2/k) * ((Im z1)/1))*<i> by XCMPLX_1:76
      .= Re z1 * Re z2/k - Im z1 * (-Im z2)/k+ (Re z1 * (-Im z2)/k + Im z1 *
    Re z2/(1*k))*<i> by XCMPLX_1:76
      .= (Re z1 * Re z2 - Im z1 * (-Im z2))/k+ (Re z1 * (-Im z2)/k + Im z1 *
    Re z2/(1*k))*<i> by XCMPLX_1:120
      .= (Re z1 * Re z2 + Im z1 * Im z2)/k+ ((-Re z1 * Im z2 + Im z1 * Re z2
    )/k)*<i> by XCMPLX_1:62
      .= (Re z1 * Re z2 + Im z1 * Im z2) / ((Re z2)^2 + (Im z2)^2) + ((Re z2
    * Im z1 - Re z1 * Im z2) / ((Re z2)^2 + (Im z2)^2))*<i>;
    hence thesis;
  end;
end;

theorem
  Re(z1 / z2) = (Re z1 * Re z2 + Im z1 * Im z2) / ((Re z2)^2 + (Im z2)^2) &
  Im(z1 / z2) = (Re z2 * Im z1 - Re z1 * Im z2) / ((Re z2)^2 + (Im z2)^2)
proof
  thus Re(z1 / z2) = Re((Re z1 * Re z2 + Im z1 * Im z2) / ((Re z2)^2 + (Im z2)
^2) + ((Re z2 * Im z1 - Re z1 * Im z2) / ((Re z2)^2 + (Im z2)^2))*<i>) by Def10
    .= (Re z1 * Re z2 + Im z1 * Im z2) / ((Re z2)^2 + (Im z2)^2) by Th12;
  thus Im(z1 / z2) = Im((Re z1 * Re z2 + Im z1 * Im z2) / ((Re z2)^2 + (Im z2)
^2) + ((Re z2 * Im z1 - Re z1 * Im z2) / ((Re z2)^2 + (Im z2)^2))*<i>) by Def10
    .= (Re z2 * Im z1 - Re z1 * Im z2) / ((Re z2)^2 + (Im z2)^2) by Th12;
end;

theorem
  Im z1 = 0 & Im z2 = 0 & Re z2 <> 0 implies Re(z1/z2) = (Re z1)/(Re z2)
  & Im(z1/z2) = 0
proof
  assume that
A1: Im z1 = 0 and
A2: Im z2 = 0 & Re z2 <> 0;
A3: z1/z2 = z1*z2" & Im(z2") = 0 by A2,Th22,XCMPLX_0:def 9;
  hence Re(z1/z2) = (Re z1)*Re(z2") by A1,Th14
    .= (Re z1)*(Re z2)" by A2,Th22
    .= (Re z1)/(Re z2) by XCMPLX_0:def 9;
  thus thesis by A1,A3,Th14;
end;

theorem
  Re z1 = 0 & Re z2 = 0 & Im z2 <> 0 implies Re(z1/z2) = (Im z1)/(Im z2)
  & Im(z1/z2) = 0
proof
  assume that
A1: Re z1 = 0 and
A2: Re z2 = 0 & Im z2 <> 0;
A3: z1/z2 = z1*z2" & Re(z2") = 0 by A2,Th23,XCMPLX_0:def 9;
  hence Re(z1/z2) = -(Im z1)*Im(z2") by A1,Th15
    .= -(Im z1)*-(Im z2)" by A2,Th23
    .= --(Im z1)*(Im z2)"
    .= (Im z1)/(Im z2) by XCMPLX_0:def 9;
  thus thesis by A1,A3,Th15;
end;

definition
  let z be complex number;
  func z*' -> complex number equals
  Re z-(Im z)*<i>;
  correctness;
  involutiveness
  proof
    let z,z9 be complex number;
    assume z = Re z9-(Im z9)*<i>;
    then z = Re z9+(-Im z9)*<i>;
    then Re z = Re z9 & -Im z = --Im z9 by Th12;
    hence z9= Re z+(-Im z)*<i> by Th13
      .= Re z-(Im z)*<i>;
  end;
end;

definition
  let z be complex number;
  redefine func z*' -> Element of COMPLEX;
  coherence by XCMPLX_0:def 2;
end;

theorem Th27:
  Re (z*') = Re z & Im (z*') = -Im z
proof
  z*' = Re z+(-Im z)*<i>;
  hence thesis by Th12;
end;

theorem
  0*' = 0 by Th4;

theorem
  z*' = 0 implies z = 0
proof
  assume z*' = 0;
  then 0 = Re z+(-Im z)*<i>;
  hence Re z = Re 0 & Im z = Im 0 by Th4,Th12;
end;

theorem
  1r*' = 1r by Th6;

theorem
  <i>*' = -<i> by Th7;

theorem Th32:
  (z1 + z2)*' = z1*' + z2*'
proof
  thus Re ((z1 + z2)*') = Re(z1 + z2) by Th27
    .= Re z1 + Re z2 by Th8
    .= Re (z1*') + Re z2 by Th27
    .= Re (z1*') + Re (z2*') by Th27
    .= Re (z1*'+ z2*') by Th8;
  thus Im ((z1 + z2)*') = -Im(z1 + z2) by Th27
    .= -(Im z1 + --Im z2) by Th8
    .= -Im z1 + -Im z2
    .= Im (z1*') + -Im z2 by Th27
    .= Im (z1*') + Im (z2*') by Th27
    .= Im (z1*' + z2*') by Th8;
end;

theorem Th33:
  (-z)*' = -(z*')
proof
  thus Re ((-z)*') = Re -z by Th27
    .= - Re z by Th17
    .= - Re (z*') by Th27
    .= Re -(z*') by Th17;
  thus Im ((-z)*') = -Im -z by Th27
    .= - -Im z by Th17
    .= - Im (z*') by Th27
    .= Im -(z*') by Th17;
end;

theorem
  (z1 - z2)*' = z1*' - z2*'
proof
  thus (z1 - z2)*' = (z1 + -z2)*' .= z1*' + (-z2)*' by Th32
    .= z1*' + -(z2*') by Th33
    .= z1*' - z2*';
end;

theorem Th35:
  (z1*z2)*' = z1*'*z2*'
proof
A1: Re(z1*') = Re z1 & Re(z2*') = Re z2 by Th27;
A2: Im(z1*') = -Im z1 & Im(z2*') = -Im z2 by Th27;
  thus Re((z1*z2)*') = Re(z1*z2) by Th27
    .= (Re (z1*') * Re (z2*')) - (-Im (z1*')) * -Im (z2*') by A1,A2,Th9
    .= (Re (z1*') * Re (z2*')) - --(Im (z1*') * Im (z2*'))
    .= Re(z1*'*z2*') by Th9;
  thus Im((z1*z2)*') = -Im(z1*z2) by Th27
    .= -((Re (z1*') * -Im (z2*')) + (Re (z2*') * -Im (z1*'))) by A1,A2,Th9
    .= (Re (z2*') * Im (z1*'))+--(Re (z1*') * Im (z2*'))
    .= Im(z1*'*z2*') by Th9;
end;

theorem Th36:
  z"*' = z*'"
proof
A1: Re z = Re (z*') & -Im z = Im (z*') by Th27;
  thus Re(z"*') = Re(z") by Th27
    .= Re z / ((Re z)^2+(Im z)^2) by Th20
    .= Re(z*'") by A1,Th20;
  thus Im(z"*') = -Im(z") by Th27
    .= -((- Im z) / ((Re z)^2+(Im z)^2)) by Th20
    .= (-Im (z*'))/((Re (z*'))^2+(Im (z*'))^2) by A1,XCMPLX_1:187
    .= Im(z*'") by Th20;
end;

theorem
  (z1/z2)*' = (z1*')/(z2*')
proof
  thus (z1/z2)*' = (z1*z2")*' by XCMPLX_0:def 9
    .= (z1*'*z2"*') by Th35
    .= (z1*'*z2*'") by Th36
    .= (z1*')/(z2*') by XCMPLX_0:def 9;
end;

theorem
  Im z = 0 implies z*' = z
proof
  Re (z*') = Re z & Im (z*') = -Im z by Th27;
  hence thesis by Th3;
end;

theorem
  Re z = 0 implies z*' = -z
proof
A1: z in COMPLEX by XCMPLX_0:def 2;
  assume
A2: Re z = 0;
  hence z*' = -0+(-Im z)*<i> .= -z by A1,A2,Def7;
end;

theorem
  Re(z*z*') = (Re z)^2 + (Im z)^2 & Im(z*z*') = 0
proof
  thus Re(z*(z*')) = Re z * Re (z*') - Im z * Im (z*') by Th9
    .= Re z * Re z - Im z * Im (z*') by Th27
    .= Re z * Re z - Im z * -Im z by Th27
    .= (Re z)^2 + (Im z)^2;
  thus Im(z*(z*')) = Re z * Im (z*') + Re (z*') * Im z by Th9
    .= Re z * -Im z + Re (z*') * Im z by Th27
    .= -Re z * Im z + Re z * Im z by Th27
    .= 0;
end;

theorem
  Re(z + z*') = 2*Re z & Im(z + z*') = 0
proof
  thus Re(z + z*') = Re z + Re (z*') by Th8
    .= Re z + Re z by Th27
    .= 2*Re z;
  thus Im(z + (z*')) = Im z + Im (z*') by Th8
    .= Im z + -Im z by Th27
    .= 0;
end;

theorem
  Re(z - z*') = 0 & Im(z - z*') = 2*Im z
proof
  thus Re(z - z*') = Re z - Re (z*') by Th19
    .= Re z - Re z by Th27
    .= 0;
  thus Im(z - z*') = Im z - Im (z*') by Th19
    .= Im z - -Im z by Th27
    .= 2*Im z;
end;

definition
  let z be complex number;
  func |.z.| -> real number equals
  sqrt ((Re z)^2 + (Im z)^2);
  coherence;
  projectivity
  proof
    let r be real number;
    let z be complex number;
    assume
A1: r = sqrt ((Re z)^2 + (Im z)^2);
    then
A2: Im r = 0 by Def2;
    (Re z)^2 >= 0 & (Im z)^2 >= 0 by XREAL_1:63;
    then r >= 0 by A1,SQUARE_1:def 2;
    then
A3: Re r >= 0 by A1,Def1;
    thus r = Re r by A1,Def1
      .= sqrt ((Re r)^2 + (Im r)^2) by A3,A2,SQUARE_1:22;
  end;
end;

definition
  let z;
  redefine func |.z.| -> Real;
  coherence;
end;

theorem Th43:
  a >= 0 implies |.a.| = a
proof
A1: a in REAL by XREAL_0:def 1;
  then
A2: Im a = 0 by Def2;
  assume a >= 0;
  then Re a >= 0 by A1,Def1;
  hence |.a.| = Re a by A2,SQUARE_1:22
  .= a by A1,Def1;
end;

registration
  cluster |.0.| -> zero;
  coherence by Th4,SQUARE_1:17;
end;

theorem
  |.0.| = 0;

registration
  let z be non zero complex number;
  cluster |.z.| -> non zero;
  coherence
  proof
    assume |.z.| is zero;
    then (Re z)^2 + (Im z)^2 = 0 by Lm1,SQUARE_1:24;
    hence thesis by Th5;
  end;
end;

theorem
  |.z.| = 0 implies z = 0;

registration
  let z;
  cluster |.z.| -> non negative;
  coherence
  proof
    0 <= (Re z)^2 + (Im z)^2 by Lm1;
    hence thesis by SQUARE_1:def 2;
  end;
end;

theorem
  0 <= |.z.|;

theorem
  z <> 0 iff 0 < |.z.|;

theorem Th48:
  |.1r.| = 1 by Th6,SQUARE_1:18;

theorem
  |.<i>.| = 1 by Th7,SQUARE_1:18;

Lm22: |.-z.| = |.z.|
proof
  thus |.-z.| = sqrt ((-Re z)^2 + (Im -z)^2) by Th17
    .= sqrt ((-Re z)^2 + (-Im z)^2) by Th17
    .= |.z.|;
end;

Lm23: a <= 0 implies |.a.| = -a
proof
  assume a <= 0;
  then |.-a.| = -a by Th43;
  hence thesis by Lm22;
end;

Lm24: sqrt a^2 = |.a.|
proof
  per cases;
  suppose
A1: 0 <= a;
    then sqrt a^2 = a by SQUARE_1:22;
    hence thesis by A1,Th43;
  end;
  suppose
A2: not 0 <= a;
    then |.a.| = -a by Lm23;
    hence thesis by A2,SQUARE_1:23;
  end;
end;

theorem
  Im z = 0 implies |.z.| = |.Re z.| by Lm24;

theorem
  Re z = 0 implies |.z.| = |.Im z.| by Lm24;

theorem
  |.-z.| = |.z.| by Lm22;

theorem Th53:
  |.z*'.| = |.z.|
proof
  thus |.z*'.| = sqrt ((Re z)^2 + (Im (z*'))^2) by Th27
    .= sqrt ((Re z)^2 + (-Im z)^2) by Th27
    .= |.z.|;
end;

Lm25: -|.a.| <= a & a <= |.a.|
proof
  a < 0 implies -|.a.| <= a & a <= |.a.|
  proof
    assume a < 0;
    then |.a.| = -a by Lm23;
    hence thesis;
  end;
  hence thesis by Th43;
end;

theorem
  Re z <= |.z.|
proof
  0<=(Im z)^2 by XREAL_1:63;
  then
A1: (Re z)^2+0 <= ((Re z)^2 + (Im z)^2) by XREAL_1:7;
  0<=(Re z)^2 by XREAL_1:63;
  then sqrt (Re z)^2 <= |.z.| by A1,SQUARE_1:26;
  then
A2: |.Re z.| <= |.z.| by Lm24;
  Re z <= |.Re z.| by Lm25;
  hence thesis by A2,XXREAL_0:2;
end;

theorem
  Im z <= |.z.|
proof
  0<=(Re z)^2 by XREAL_1:63;
  then
A1: (Im z)^2+0 <= ((Re z)^2 + (Im z)^2) by XREAL_1:7;
  0<=(Im z)^2 by XREAL_1:63;
  then sqrt (Im z)^2 <= |.z.| by A1,SQUARE_1:26;
  then
A2: |.Im z.| <= |.z.| by Lm24;
  Im z <= |.Im z.| by Lm25;
  hence thesis by A2,XXREAL_0:2;
end;

theorem Th56:
  |.z1 + z2.| <= |.z1.| + |.z2.|
proof
  set r1 = Re z1, r2 = Re z2, i1 = Im z1, i2 = Im z2;
A1: (Im(z1 + z2))^2 = (i1 + i2)^2 by Th8
    .= i1^2 + 2*i1*i2 + i2^2;
A2: 0 <= r1^2+i1^2 by Lm1;
  (r1^2+i1^2)*(r2^2+i2^2)-(r1*r2+i1*i2)^2 = (r1*i2-i1*r2)^2;
  then 0 <= (r1^2+i1^2)*(r2^2+i2^2)-(r1*r2+i1*i2)^2 by XREAL_1:63;
  then
A3: (r1*r2+i1*i2)^2+0 <= (r1^2+i1^2)*(r2^2+i2^2) by XREAL_1:19;
  r1*r2+i1*i2 <= |.r1*r2+i1*i2.| by Lm25;
  then
A4: r1*r2+i1*i2 <= sqrt (r1*r2+i1*i2)^2 by Lm24;
A5: 0 <= r2^2+i2^2 by Lm1;
  then
A6: (sqrt (r2^2+i2^2))^2 = r2^2+i2^2 by SQUARE_1:def 2;
  0<=(r1*r2+i1*i2)^2 by XREAL_1:63;
  then sqrt (r1*r2+i1*i2)^2 <= sqrt ((r1^2+i1^2)*(r2^2+i2^2)) by A3,SQUARE_1:26
;
  then sqrt (r1*r2+i1*i2)^2 <= sqrt (r1^2+i1^2)*sqrt (r2^2+i2^2) by A2,A5,
SQUARE_1:29;
  then
A7: r1*r2 + i1*i2 <= sqrt (r1^2+i1^2)*sqrt (r2^2+i2^2) by A4,XXREAL_0:2;
  (2*r1*r2 + 2*i1*i2) = 2*(r1*r2 + i1*i2);
  then (2*r1*r2 + 2*i1*i2) <= 2*(sqrt (r1^2+i1^2)*sqrt (r2^2+i2^2)) by A7,
XREAL_1:64;
  then
A8: (r1^2 + i1^2)+(2*r1*r2+2*i1*i2) <= (r1^2+i1^2)+2*sqrt (r1^2+i1^2)*sqrt
  (r2^2+i2^2) by XREAL_1:7;
  (Re(z1 + z2))^2 = (r1+ r2)^2 by Th8
    .= r1^2 + 2*r1*r2 + r2^2;
  then
  (Re(z1+z2))^2+(Im(z1+z2))^2 = r1^2 + i1^2 + (2*r1*r2 + 2*i1*i2) + (r2^2
  + i2^2) by A1;
  then
A9: (Re(z1+z2))^2+(Im(z1+z2))^2 <= (r1^2+i1^2)+2*sqrt (r1^2+i1^2)*sqrt (r2
  ^2+i2^2)+(r2^2+i2^2) by A8,XREAL_1:7;
A10: 0 <= (Re(z1 + z2))^2 + (Im(z1 + z2))^2 by Lm1;
  (sqrt (r1^2+i1^2))^2 = r1^2+i1^2 by A2,SQUARE_1:def 2;
  then sqrt ((Re(z1 + z2))^2 + (Im(z1 + z2))^2) <= sqrt ((sqrt (r1^2+i1^2) +
  sqrt (r2^2+i2^2))^2) by A6,A9,A10,SQUARE_1:26;
  hence thesis by SQUARE_1:22;
end;

theorem Th57:
  |.z1 - z2.| <= |.z1.| + |.z2.|
proof
  |.z1 - z2.| = |.z1 + - z2.|;
  then |.z1 - z2.| <= |.z1.| + |.-z2.| by Th56;
  hence thesis by Lm22;
end;

theorem
  |.z1.| - |.z2.| <= |.z1 + z2.|
proof
  z1 = z1 + z2 - z2;
  then |.z1.| <= |.z1 + z2.| + |.z2.| by Th57;
  hence thesis by XREAL_1:20;
end;

theorem Th59:
  |.z1.| - |.z2.| <= |.z1 - z2.|
proof
  z1 = z1 - z2 + z2;
  then |.z1.| <= |.z1 - z2.| + |.z2.| by Th56;
  hence thesis by XREAL_1:20;
end;

theorem Th60:
  |.z1 - z2.| = |.z2 - z1.|
proof
  thus |.z1 - z2.| = |.-(z2 - z1).| .= |.z2 - z1.| by Lm22;
end;

theorem Th61:
  |.z1 - z2.| = 0 iff z1 = z2
proof
  thus |.z1 - z2.| = 0 implies z1 = z2
  proof
    assume |.z1 - z2.| = 0;
    then z1 - z2 = 0;
    hence thesis;
  end;
  thus thesis;
end;

theorem
  z1 <> z2 iff 0 < |.z1 - z2.|
proof
  thus z1 <> z2 implies 0 < |.z1 - z2.|
  proof
    assume z1 <> z2;
    then |.z1 - z2.| <> 0 by Th61;
    hence thesis;
  end;
  thus thesis;
end;

theorem
  |.z1 - z2.| <= |.z1 - z.| + |.z - z2.|
proof
  |.z1 - z2.| = |.(z1 - z) + (z - z2).|;
  hence thesis by Th56;
end;

Lm26: -b <= a & a <= b iff |.a.| <= b
proof
A1: |.a.| <= b implies -b <= a & a <= b
  proof
    assume
A2: |.a.| <= b;
 a < -0 implies -b <= a & a <= b
    proof
      assume
A3:   a < -0;
      then -a <= b by A2,Lm23;
      then -b <= -(-a) by XREAL_1:24;
      hence thesis by A3;
    end;
    hence thesis by A2,Th43;
  end;
  -b <= a & a <= b implies |.a.| <= b
  proof
    assume that
A4: -b <= a and
A5: a <= b;
    -a <= -(-b) by A4,XREAL_1:24;
    then a < 0 implies |.a.| <= b by Lm23;
    hence thesis by A5,Th43;
  end;
  hence thesis by A1;
end;

theorem
  |.|.z1.| - |.z2.|.| <= |.z1 - z2.|
proof
  |.z2.| - |.z1.| <= |.z2 - z1.| by Th59;
  then -(|.z1.| - |.z2.|) <= |.z1 - z2.| by Th60;
  then
A1: -|.z1 - z2.| <= --(|.z1.| - |.z2.|) by XREAL_1:24;
  |.z1.| - |.z2.| <= |.z1 - z2.| by Th59;
  hence thesis by A1,Lm26;
end;

theorem Th65:
  |.z1*z2.| = |.z1.|*|.z2.|
proof
  set r1 = Re z1, r2 = Re z2, i1 = Im z1, i2 = Im z2;
A1: 0<=r1^2 + i1^2 & 0<=r2^2 + i2^2 by Lm1;
A2: (Im(z1*z2))^2 = (r1*i2 + r2*i1)^2 by Th9
    .= 2*(r1*r2)*(i1*i2) + ((r1*i2)^2 + (r2*i1)^2);
  (Re(z1*z2))^2 = (r1*r2 - i1*i2)^2 by Th9
    .= (r1*r2)^2 + (i1*i2)^2 + - 2*(r1*r2)*(i1*i2);
  then (Re(z1*z2))^2+(Im(z1*z2))^2 = (r1^2 + i1^2)*(r2^2 + i2^2) by A2;
  hence thesis by A1,SQUARE_1:29;
end;

theorem Th66:
  |.z".| = |.z.|"
proof
  per cases;
  suppose
A1: z <> 0;
    set r2i2 = (Re z)^2+(Im z)^2;
A2: r2i2 <> 0 by A1,Th5;
A3: 0 <= r2i2 by Lm1;
    thus |.z".| = sqrt ((Re z / r2i2)^2 + (Im(z"))^2) by Th20
      .= sqrt ((Re z / r2i2)^2 + ((-Im z) / r2i2)^2) by Th20
      .= sqrt ((Re z)^2 / r2i2^2 + ((-Im z) / r2i2)^2) by XCMPLX_1:76
      .= sqrt ((Re z)^2 / r2i2^2 + (-Im z)^2 / r2i2^2) by XCMPLX_1:76
      .= sqrt ((1*r2i2) / (r2i2*r2i2)) by XCMPLX_1:62
      .= sqrt (1 / r2i2) by A2,XCMPLX_1:91
      .= 1 / |.z.| by A3,SQUARE_1:18,30
      .= |.z.|" by XCMPLX_1:215;
  end;
  suppose
A4: z = 0;
    hence |.z".| = 0" .= |.z.|" by A4;
  end;
end;

theorem Th67:
  |.z1.| / |.z2.| = |.z1/z2.|
proof
  thus |.z1.|/|.z2.| = |.z1.|*|.z2.|" by XCMPLX_0:def 9
    .= |.z1.|*|.z2".| by Th66
    .= |.z1*z2".| by Th65
    .= |.z1/z2.| by XCMPLX_0:def 9;
end;

theorem
  |.z*z.| = (Re z)^2 + (Im z)^2
proof
  0<=(Re z)^2 + (Im z)^2 & |.z*z.| = |.z.|*|.z.| by Lm1,Th65;
  then |.z*z.| = sqrt (((Re z)^2 + (Im z)^2)^2) by SQUARE_1:29;
  hence thesis by Lm1,SQUARE_1:22;
end;

theorem
  |.z*z.| = |.z*z*'.|
proof
  thus |.z*z.| = |.z.|*|.z.| by Th65
    .= |.z.|*|.z*'.| by Th53
    .= |.z*z*'.| by Th65;
end;



theorem
  a <= 0 implies |.a.| = -a by Lm23;

theorem Th71:
  |.a.| = a or |.a.| = -a
proof
  a >= 0 or a < 0;
  hence thesis by Lm23,Th43;
end;

theorem 
  sqrt a^2 = |.a.| by Lm24;

theorem 
  min(a,b) = (a + b - |.a - b.|) / 2
proof
  per cases;
  suppose
A1: a <= b;
    |.a - b.| = |.-(b - a).| .= |.b - a.| by Lm22
      .= b - a by A1,Th43,XREAL_1:48;
    hence thesis by A1,XXREAL_0:def 9;
  end;
  suppose
A2: b <= a;
    hence min(a,b) = ((a+b)- (a - b))/2 by XXREAL_0:def 9
      .= ((a+b)-|.a-b.|)/2 by A2,Th43,XREAL_1:48;
  end;
end;

theorem 
  max(a,b) = (a + b + |.a - b.|) / 2
proof
  per cases;
  suppose
A1: b <= a;
    hence max(a,b) = ((a+b)+ (a - b))/2 by XXREAL_0:def 10
      .= ((a+b)+|.a-b.|)/2 by A1,Th43,XREAL_1:48;
  end;
  suppose
A2: a <= b;
    then
A3: 0 <= b - a by XREAL_1:48;
    thus max(a,b) = ((a+b)+ -(a - b))/2 by A2,XXREAL_0:def 10
      .= ((a+b)+|.-(a-b).|)/2 by A3,Th43
      .= ((a+b)+|.a-b.|)/2 by Lm22;
  end;
end;

theorem Th75: 
  |.a.|^2 = a^2
proof
  |.a.| = a or |.a.| = -a by Th71;
  hence thesis;
end;

theorem
  -|.a.| <= a & a <= |.a.| by Lm25;

notation
  let z be complex number;
  synonym abs z for |.z.|;
end;

definition
  let z be complex number;
  redefine func abs z -> Real;
  coherence;
end;

theorem
  a+b*<i> = c+d*<i> implies a = c & b = d
proof
  assume
A1: a+b*<i> = c+d*<i>;
  then a-c+(b-d)*<i> = 0;
  then a-c = 0 by Th4,Th12;
  hence thesis by A1;
end;



theorem
  sqrt(a^2+b^2) <= abs(a)+abs(b)
proof
A1: (sqrt(a^2+b^2))^2>=0 by XREAL_1:63;
  a^2>=0 & b^2>=0 by XREAL_1:63;
  then
A2: (sqrt(a^2+b^2))^2=a^2+b^2 by SQUARE_1:def 2;
  (abs(a)+abs(b))^2=(abs(a))^2+2*(abs(a))*(abs(b))+(abs(b))^2 & (abs
  (a) )^2=a^2 by Th75;
  then
  (abs(a)+abs(b))^2 - (sqrt(a^2+b^2))^2= a^2+2*(abs(a))*(abs(b))+b
  ^2-(a^2+b^2) by A2,Th75
    .=2*(abs(a))*(abs(b));
  then (abs(a)+abs(b))^2>= (sqrt(a^2+b^2))^2 by XREAL_1:49;
  then sqrt((abs(a)+abs(b))^2)>=sqrt((sqrt(a^2+b^2))^2) by A1,SQUARE_1:26;
  hence thesis by A2,SQUARE_1:22;
end;

theorem
  abs(a) <= sqrt(a^2+b^2)
proof
  a^2>=0 & a^2+0<= a^2+b^2 by XREAL_1:6,63;
  then sqrt(a^2)<= sqrt(a^2+b^2) by SQUARE_1:26;
  hence thesis by Lm24;
end;

theorem
  |. 1/z1 .| = 1 / |.z1.| by Th48,Th67;














begin 

reserve x, y, z, s, t for real number;

definition
  let x be real number;
  redefine func |.x.| equals
  :Def1:
  x if 0 <= x otherwise -x;
  correctness by COMPLEX1:43,70;
end;

theorem
  abs x = x or abs x = -x by Def1;

theorem Th2:
  x = 0 iff abs x = 0 by Def1,COMPLEX1:47;

theorem
  abs x = -x & x <> 0 implies x < 0 by Def1;

theorem
  -abs x <= x & x <= abs x
proof
  per cases;
  suppose
A1: x < 0;
    then abs x = -x by Def1;
    hence thesis by A1;
  end;
  suppose
A2: 0 <= x;
    then -x <= -0;
    hence thesis by A2,Def1;
  end;
end;

theorem
  -y <= x & x <= y iff abs x <= y
proof
  hereby
    assume that
A1: -y <= x and
A2: x <= y;
    -x <= --y by A1,XREAL_1:24;
    hence abs x <= y by A2,Def1;
  end;
  assume
A3: abs x <= y;
  then
A4: 0 <= y by COMPLEX1:46;
  per cases;
  suppose 0 < x;
    hence thesis by A3,A4,Def1;
  end;
  suppose
A5: x < 0;
    then -x <= y by A3,Def1;
    then -y <= --x by XREAL_1:24;
    hence thesis by A5;
  end;
  suppose x = -0;
    hence thesis by A4;
  end;
end;

theorem
  x <> 0 implies abs(x) * abs(1/x) = 1
proof
  assume x <> 0;
  then abs x * abs(1/x) = abs(x * (1/x)) & abs(x * (1/x)) = abs 1 by
COMPLEX1:65,XCMPLX_1:106;
  hence thesis by Def1;
end;

theorem
  abs(1/x) = 1/abs(x) by COMPLEX1:80;

theorem 
  0 <= x*y implies sqrt (x*y) = sqrt abs(x)*sqrt abs(y)
proof
  assume 0 <= x*y;
  then abs(x*y) = x*y by Def1;
  then
A1: abs(x)*abs(y) = x*y by COMPLEX1:65;
  0 <= abs(x) & 0 <= abs(y) by COMPLEX1:46;
  hence thesis by A1,SQUARE_1:29;
end;

theorem
  abs x <= z & abs y <= t implies abs(x+y) <= z + t
proof
  assume abs(x) <= z & abs(y) <= t;
  then abs(x+y) <= abs(x) + abs(y) & abs(x) + abs(y) <= z + t by COMPLEX1:56
,XREAL_1:7;
  hence thesis by XXREAL_0:2;
end;

theorem 
  0 < x/y implies sqrt (x/y) = sqrt abs(x) / sqrt abs(y)
proof
  assume 0 < x/y;
  then x/y = abs(x/y) by Def1;
  then
A1: x/y = abs(x)/abs(y) by COMPLEX1:67;
  0 <= abs(x) & 0 <= abs(y) by COMPLEX1:46;
  hence thesis by A1,SQUARE_1:30;
end;

theorem
  0 <= x * y implies abs (x+y) = abs(x) + abs(y)
proof
  assume
A1: 0 <= x * y;
  per cases by A1;
  suppose
A2: x * y = 0;
    per cases by A2,XCMPLX_1:6;
    suppose
A3:   x = 0;
      then abs(x) + abs(y) = 0 + abs(y) by Def1
        .= abs(y);
      hence thesis by A3;
    end;
    suppose
A4:   y = 0;
      then abs(x) + abs(y) = abs(x) + 0 by Def1
        .= abs(x);
      hence thesis by A4;
    end;
  end;
  suppose
A5: 0 < x * y;
    then
A6: x <> 0 & y <> 0;
    per cases by A5,A6;
    suppose that
A7:   0 < x & 0 < y;
      abs x = x & abs y = y by A7,Def1;
      hence thesis by A7,Def1;
    end;
    suppose that
A8:   x < 0 and
A9:   y < 0;
      abs(x) = -x by A8,Def1;
      then abs(x) + abs(y) = (-1) * x + -(1 * y) by A9,Def1
        .= - ( x + y );
      hence thesis by A8,A9,Def1;
    end;
  end;
end;

theorem
  abs(x+y) = abs(x) + abs(y) implies 0 <= x * y
proof
A1: x * y < 0 implies x < 0 & 0 < y or 0 < x & y < 0
  proof
    assume
A2: x * y < 0;
    then x <> 0 & y <> 0;
    hence thesis by A2;
  end;
A3: x < 0 & 0 < y & x + y < 0 implies abs(x+y) <> abs(x) + abs(y)
  proof
    assume that
A4: x < 0 and
A5: 0 < y and
A6: x + y < 0;
    -(1 * x) + -y < -x + y by A5,XREAL_1:6;
    then
A7: -( 1 * ( x + y ) ) < -x + y;
    abs x = -x & abs y = y by A4,A5,Def1;
    hence thesis by A6,A7,Def1;
  end;
A8: 0 < x & y < 0 & x + y < 0 implies abs(x+y) <> abs(x) + abs(y)
  proof
    assume that
A9: 0 < x and
A10: y < 0 and
A11: x + y < 0;
    -(1 * x) + -y < x + -y by A9,XREAL_1:6;
    then
A12: - (1 * ( x + y )) < x + -y;
    abs x = x & abs y = -y by A9,A10,Def1;
    hence thesis by A11,A12,Def1;
  end;
A13: 0 < x & y < 0 & 0 <= x + y implies abs(x+y) <> abs(x) + abs(y)
  proof
    assume that
A14: 0 < x and
A15: y < 0 and
A16: 0 <= x + y;
A17: abs y = -y by A15,Def1;
    x + y < x + -y & abs x = x by A14,A15,Def1,XREAL_1:6;
    hence thesis by A16,A17,Def1;
  end;
A18: x < 0 & 0 < y & 0 <= x + y implies abs(x+y) <> abs(x) + abs(y)
  proof
    assume that
A19: x < 0 and
A20: 0 < y and
A21: 0 <= x + y;
A22: abs y = y by A20,Def1;
    x + y < -x + y & abs x = -x by A19,Def1,XREAL_1:6;
    hence thesis by A21,A22,Def1;
  end;
  assume abs(x+y) = abs(x) + abs(y) & 0 > x * y;
  hence contradiction by A1,A3,A18,A8,A13;
end;

theorem
  abs(x + y)/(1 + abs(x+y)) <= abs(x)/(1 + abs(x)) + abs(y)/(1 + abs(y))
proof
A1: s <= t & 0 < 1 + s & 0 < 1 + t implies s/(1 + s) <= t/(1 + t)
  proof
    assume that
A2: s <= t and
A3: 0 < 1 + s and
A4: 0 < 1 + t;
    s * 1 + s * t <= t + s * t by A2,XREAL_1:6;
    then s * (1 + t) * (1 + s)" <= t * (1 + s) * (1 + s)" by A3,XREAL_1:64;
    then s * (1 + t) * (1 + s)" <= t * ((1 + s) * (1 + s)");
    then s * (1 + t) * (1 + s)" <= t * 1 by A3,XCMPLX_0:def 7;
    then (s * (1 + s)") * (1 + t) * (1 + t)" <= t * (1 + t)" by A4,XREAL_1:64;
    then (s * (1 + s)") * ((1 + t) * (1 + t)") <= t * (1 + t)";
    then (s * (1 + s)") * 1 <= t * (1 + t)" by A4,XCMPLX_0:def 7;
    then s/(1 + s) <= t * (1 + t)" by XCMPLX_0:def 9;
    hence thesis by XCMPLX_0:def 9;
  end;
  set a = abs x, b = abs y, c = abs (x+y);
A5: 0 <= a by COMPLEX1:46;
A6: 0 <= b by COMPLEX1:46;
A7: 0 + 0 < 1 + a by COMPLEX1:46,XREAL_1:8;
A8: 0 < 1 + a & 0 < 1 + a + b implies a/(1 + a + b) <= a/(1 + a)
  proof
    assume that
A9: 0 < 1 + a and
A10: 0 < 1 + a + b;
    0 + a <= a * b + a by A5,A6,XREAL_1:6;
    then a * 1 + a * a <= a * (1 + b) + a * a by XREAL_1:6;
    then a * (1 + a) * (1 + a)" <= a * (1 + a + b) * (1 + a)" by A9,XREAL_1:64;
    then a * ((1 + a) * (1 + a)") <= a * (1 + a + b) * (1 + a)";
    then a * 1 <= a * (1 + a + b) * (1 + a)" by A7,XCMPLX_0:def 7;
    then a * (1 + a + b)" <= (a * (1 + a)") * (1 + a + b) * (1 + a + b)" by A10
,XREAL_1:64;
    then a * (1 + a + b)" <= (a * (1 + a)") * ((1 + a + b) * (1 + a + b)");
    then a * (1 + a + b)" <= (a * (1 + a)") * 1 by A5,A6,XCMPLX_0:def 7;
    then a/(1 + a + b) <= a * (1 + a)" by XCMPLX_0:def 9;
    hence thesis by XCMPLX_0:def 9;
  end;
A11: 0 + 0 < 1 + b by COMPLEX1:46,XREAL_1:8;
A12: 0 < 1 + b & 0 < 1 + a + b implies b/(1 + a + b) <= b/(1 + b)
  proof
    assume that
A13: 0 < 1 + b and
A14: 0 < 1 + a + b;
    0 + b <= a * b + b by A5,A6,XREAL_1:6;
    then b * 1 + b * b <= (1 + a) * b + b * b by XREAL_1:6;
    then (b * (1 + b)) * (1 + b)" <= (b * (1 + a + b)) * (1 + b)" by A13,
XREAL_1:64;
    then b * ((1 + b) * (1 + b)") <= (b * (1 + a + b)) * (1 + b)";
    then b * 1 <= (b * (1 + a + b)) * (1 + b)" by A11,XCMPLX_0:def 7;
    then
    b * (1 + a + b)" <= ((b * (1 + b)") * (1 + a + b )) * (1 + a + b)" by A14,
XREAL_1:64;
    then b * (1 + a + b)" <= (b * (1 + b)") * ((1 + a + b) *(1 + a + b)");
    then b * (1 + a + b)" <= (b * (1 + b)") * 1 by A5,A6,XCMPLX_0:def 7;
    then b/(1 + a + b) <= b * (1 + b)" by XCMPLX_0:def 9;
    hence thesis by XCMPLX_0:def 9;
  end;
  0 + 0 < 1 + c by COMPLEX1:46,XREAL_1:8;
  then
A15: c/(1 + c) <= (a + b)/(1 + (a + b)) by A5,A6,A1,COMPLEX1:56;
  (a + b)/(1 + a + b) = a/(1 + a + b) + b/(1 + a + b) by XCMPLX_1:62;
  then (a + b)/(1 + a + b) <= a/(1 + a) + b/(1 + b) by A6,A7,A8,A12,XREAL_1:7;
  hence thesis by A15,XXREAL_0:2;
end;



definition
  let x;
  func sgn x -> real number equals
  :Def2:
  1 if 0 < x, -1 if x < 0 otherwise 0;
  coherence;
  consistency;
  projectivity;
end;

registration
  let x;
  cluster sgn x -> integer;
  coherence
  proof
    x = 0 or x > 0 or x < 0;
    hence thesis by Def2;
  end;
end;

definition
  let x be Real;
  redefine func sgn x -> Real;
  coherence by XREAL_0:def 1;
end;

theorem
  sgn x = 1 implies 0 < x
proof
  assume that
A1: sgn x = 1 and
A2: 0 >= x;
  x < 0 or x = 0 by A2;
  hence contradiction by A1,Def2;
end;

theorem
  sgn x = -1 implies x < 0
proof
  assume that
A1: sgn x = -1 and
A2: x >= 0;
  0 < x or x = 0 by A2;
  hence contradiction by A1,Def2;
end;

theorem Th16:
  sgn x = 0 implies x = 0
proof
  assume that
A1: sgn x = 0 and
A2: x <> 0;
  0 < x or x < 0 by A2;
  hence contradiction by A1,Def2;
end;

theorem
  x = abs(x) * sgn x
proof
A1: 0 < x implies x = abs(x) * sgn x
  proof
    assume
A2: 0 < x;
    then abs(x) = x by Def1;
    then abs(x) * sgn x = x * 1 by A2,Def2;
    hence thesis;
  end;
A3: x < 0 implies x = abs(x) * sgn x
  proof
    assume
A4: x < 0;
    then abs(x) = -x by Def1;
    then abs(x) * sgn x = (-x) * (-1) by A4,Def2
      .= x;
    hence thesis;
  end;
  x = 0 implies x = abs(x) * sgn x
  proof
    assume
A5: x = 0;
    then sgn x = 0 by Def2;
    hence thesis by A5;
  end;
  hence thesis by A1,A3;
end;

theorem Th18:
  sgn (x * y) = sgn x * sgn y
proof
A1: 0 < x & 0 < y implies sgn (x * y) = sgn x * sgn y
  proof
    assume that
A2: 0 < x and
A3: 0 < y;
A4: sgn y = 1 by A3,Def2;
    0 * y < x * y & sgn x = 1 by A2,A3,Def2,XREAL_1:68;
    hence thesis by A4,Def2;
  end;
A5: 0 < x & y < 0 implies sgn (x * y) = sgn x * sgn y
  proof
    assume that
A6: 0 < x and
A7: y < 0;
    sgn y = -1 by A7,Def2;
    then
A8: sgn x * sgn y = 1 * (-1) by A6,Def2
      .= -1;
    x * y < 0 * y by A6,A7,XREAL_1:69;
    hence thesis by A8,Def2;
  end;
A9: x < 0 & y < 0 implies sgn (x * y) = sgn x * sgn y
  proof
    assume that
A10: x < 0 and
A11: y < 0;
    sgn y = -1 by A11,Def2;
    then
A12: sgn x * sgn y = (-1) * (-1) by A10,Def2
      .= 1;
    x * 0 < x * y by A10,A11,XREAL_1:69;
    hence thesis by A12,Def2;
  end;
A13: x < 0 & 0 < y implies sgn (x * y) = sgn x * sgn y
  proof
    assume that
A14: x < 0 and
A15: 0 < y;
    sgn y = 1 by A15,Def2;
    then
A16: sgn x * sgn y = -1 by A14,Def2;
    x * y < 0 * y by A14,A15,XREAL_1:68;
    hence thesis by A16,Def2;
  end;
  x = 0 or y = 0 implies sgn (x * y) = sgn x * sgn y
  proof
    assume
A17: x = 0 or y = 0;
    then sgn x = 0 or sgn y = 0 by Def2;
    hence thesis by A17;
  end;
  hence thesis by A1,A5,A13,A9;
end;

canceled;

theorem
  sgn (x + y) <= sgn x + sgn y + 1
proof
A1: y = 0 implies sgn (x + y) <= sgn x + sgn y + 1
  proof
    assume
A2: y = 0;
    then sgn x + sgn y + 1 = sgn x + 0 + 1 by Def2
      .= sgn x + 1;
    hence thesis by A2,XREAL_1:29;
  end;
A3: 0 < x & 0 < y implies sgn (x + y) <= sgn x + sgn y + 1
  proof
    sgn x < sgn x + 1 by XREAL_1:29;
    then
A4: sgn x + 0 < sgn x + 1 + 1 by XREAL_1:8;
    assume
A5: 0 < x & 0 < y;
    then sgn x = 1 & sgn y = 1 by Def2;
    hence thesis by A5,A4,Def2;
  end;
A6: x < 0 & 0 < y implies sgn (x + y) <= sgn x + sgn y + 1
  proof
    assume that
A7: x < 0 and
A8: 0 < y;
    sgn x = -1 by A7,Def2;
    then
A9: sgn x + sgn y + 1 = 1 by A8,Def2;
    x + y < 0 or x + y = 0 or 0 < x + y;
    hence thesis by A9,Def2;
  end;
A10: 0 < x & y < 0 implies sgn (x + y) <= sgn x + sgn y +1
  proof
    assume that
A11: 0 < x and
A12: y < 0;
    sgn x = 1 by A11,Def2;
    then
A13: sgn x + sgn y + 1 = 1 + -1 + 1 by A12,Def2
      .= 1;
    x + y < 0 or x + y = 0 or 0 < x + y;
    hence thesis by A13,Def2;
  end;
A14: x < 0 & y < 0 implies sgn (x + y) <= sgn x + sgn y + 1
  proof
    assume that
A15: x < 0 and
A16: y < 0;
    sgn y = -1 by A16,Def2;
    then sgn x + sgn y + 1 = -1 by A15,Def2;
    hence thesis by A15,A16,Def2;
  end;
  x = 0 implies sgn (x + y) <= sgn x + sgn y + 1
  proof
    assume
A17: x = 0;
    then sgn x + sgn y + 1 = 0 + sgn y + 1 by Def2
      .= sgn y + 1;
    hence thesis by A17,XREAL_1:29;
  end;
  hence thesis by A3,A10,A6,A14,A1;
end;

theorem Th21:
  x <> 0 implies sgn x * sgn (1/x) = 1
proof
  assume x <> 0;
  then sgn ( x * (1/x) ) = sgn 1 by XCMPLX_1:106;
  then sgn ( x * (1/x) ) = 1 by Def2;
  hence thesis by Th18;
end;

theorem Th22:
  1/(sgn x) = sgn (1/x)
proof
  per cases;
  suppose
A1: x = 0;
    hence 1/(sgn x) = 1/0 by Def2
      .= sgn (1/x) by A1,Def2;
  end;
  suppose
A2: x <> 0;
    then (sgn x * sgn (1/x)) * (1/(sgn x)) = 1 * (1/(sgn x)) by Th21;
    then sgn (1/x) * (sgn x * (1/(sgn x))) = 1/(sgn x);
    then sgn (1/x) * 1 = 1 /(sgn x) by A2,Th16,XCMPLX_1:106;
    hence thesis;
  end;
end;

theorem
  sgn x + sgn y - 1 <= sgn ( x + y )
proof
A1: x = 0 or y = 0 implies sgn x + sgn y - 1 <= sgn (x + y)
  proof
A2: y = 0 implies sgn x + sgn y - 1 <= sgn (x + y)
    proof
A3:   sgn x - 1 < sgn x + -1 + 1 by XREAL_1:29;
      assume
A4:   y = 0;
      then sgn x + sgn y - 1 = sgn x + 0 - 1 by Def2
        .= sgn x - 1;
      hence thesis by A4,A3;
    end;
A5: x = 0 implies sgn x + sgn y - 1 <= sgn (x + y)
    proof
A6:   sgn y - 1 < sgn y + -1 + 1 by XREAL_1:29;
      assume
A7:   x = 0;
      then sgn x + sgn y - 1 = 0 + sgn y - 1 by Def2
        .= sgn y - 1;
      hence thesis by A7,A6;
    end;
    assume x = 0 or y = 0;
    hence thesis by A5,A2;
  end;
A8: x < 0 & y < 0 implies sgn x + sgn y - 1 <= sgn (x + y)
  proof
    assume that
A9: x < 0 and
A10: y < 0;
    sgn x = -1 by A9,Def2;
    then
A11: sgn x = sgn (x + y) by A9,A10,Def2;
A12: sgn (x + y) + -1 - 1 < sgn (x + y) + -1 - 1 + 1 & sgn (x + y) + -1 <
    sgn (x + y) + -1 + 1 by XREAL_1:29;
    sgn y = -1 by A10,Def2;
    hence thesis by A11,A12,XXREAL_0:2;
  end;
A13: 0 < x & y < 0 implies sgn x + sgn y - 1 <= sgn (x + y)
  proof
    assume that
A14: 0 < x and
A15: y < 0;
    sgn x = 1 by A14,Def2;
    then
A16: sgn x + sgn y = 1 + -1 by A15,Def2
      .= 0;
    x + y < 0 or x + y = 0 or 0 < x + y;
    hence thesis by A16,Def2;
  end;
A17: x < 0 & 0 < y implies sgn x + sgn y - 1 <= sgn (x + y)
  proof
    assume that
A18: x < 0 and
A19: 0 < y;
    sgn x = -1 by A18,Def2;
    then
A20: sgn x + sgn y = -1 + 1 by A19,Def2
      .= 0;
    x + y < 0 or x + y = 0 or 0 < x + y;
    hence thesis by A20,Def2;
  end;
  0 < x & 0 < y implies sgn x + sgn y - 1 <= sgn (x + y)
  proof
    assume that
A21: 0 < x and
A22: 0 < y;
    sgn y = 1 by A22,Def2;
    then sgn x + sgn y - 1 = 1 by A21,Def2;
    hence thesis by A21,A22,Def2;
  end;
  hence thesis by A8,A17,A13,A1;
end;

theorem
  sgn x = sgn (1/x)
proof
A1: 0 < x implies sgn x = sgn (1/x)
    proof
      assume
A2:   0 < x;
      sgn (1/x) = 1/(sgn x) by Th22;
      then sgn (1/x) = 1/1 by A2,Def2
        .= 1;
      hence thesis by A2,Def2;
    end;
    x < 0 implies sgn x = sgn (1/x)
    proof
      assume
A3:   x < 0;
      then sgn x = -1 by Def2;
      then sgn (1/x) = 1/(-1) by Th22;
      hence thesis by A3,Def2;
    end;
    hence thesis by A1;
end;

theorem
  sgn (x/y) = (sgn x)/(sgn y)
proof
  per cases;
  suppose
A1: y = 0;
    hence sgn (x/y) = sgn (x*0") by XCMPLX_0:def 9
      .= (sgn x)*0" by Def2
      .= (sgn x)/0 by XCMPLX_0:def 9
      .= (sgn x)/(sgn y) by A1,Def2;
  end;
  suppose
A2: y <> 0;
    x/y = (x/y) * 1 .= (x/y) * (y * (1/y)) by A2,XCMPLX_1:106
      .= ((x/y) * y) * (1/y)
      .= x * (1/y) by A2,XCMPLX_1:87;
    then sgn (x/y) = sgn x * sgn (1/y) by Th18
      .= ((sgn x)/1) * (1/(sgn y)) by Th22
      .= (sgn x * 1)/(1 * sgn y) by XCMPLX_1:76
      .= (sgn x)/(1 * sgn y);
    hence thesis;
  end;
end;



reserve r, s for real number;

theorem
  0 <= r + abs(r)
proof
A1: 0 <= abs(r) by COMPLEX1:46;
  abs(r) + abs(r) = r + abs(r) or abs(r) + r = -r + r by Def1;
  hence thesis by A1;
end;

theorem
  0 <= -r + abs(r)
proof
  r <= abs(r) by COMPLEX1:76;
  then -r >= -abs(r) by XREAL_1:24;
  then -r+abs(r) >= -abs(r)+abs(r) by XREAL_1:7;
  hence thesis;
end;

theorem
  abs(r) = abs(s) implies r = s or r = -s
proof
  assume
A1: abs(r) = abs(s);
  assume
A2: r <> s;
  per cases by Def1;
  suppose
    abs(r) = r & abs(s) = s;
    hence thesis by A1,A2;
  end;
  suppose
    abs(r) = r & abs(s) = -s;
    hence thesis by A1;
  end;
  suppose
    abs(r) = -r & abs(s) = s;
    hence thesis by A1;
  end;
  suppose
    abs(r) = -r & abs(s) = -s;
    hence thesis by A1,A2;
  end;
end;

theorem
  for m being Nat holds
  m = abs m
proof
  let m be Nat;
   m >= 0 by NAT_1:2;
  hence abs m= m by Def1;
end;



theorem
  r<=0 implies abs(r)=-r
proof
  assume
A1: r<=0;
  per cases by A1;
  suppose r<0;
    hence thesis by Def1;
  end;
  suppose r=0;
    hence thesis by Th2;
  end;
end;















begin

definition
  let a be integer number;
  redefine func abs a -> Element of NAT;
  coherence
  proof
    per cases;
    suppose
A1:   a >= 0;
      then abs a = a by ABSVALUE:def 1;
      hence thesis by A1,INT_1:3;
    end;
    suppose
      a < 0;
      then abs a = -a & -a > 0 by ABSVALUE:def 1,XREAL_1:58;
      hence thesis by INT_1:3;
    end;
  end;
end;

reserve a,b,c for Integer;

theorem Th1:
  a divides b & a divides b + c implies a divides c
proof
  given u being Integer such that
A1: b = a * u;
  given t being Integer such that
A2: b + c = a * t;
  c = a*(t - u) by A1,A2;
  hence thesis by INT_1:def 3;
end;

theorem Th2: 
  a divides b implies a divides b * c
proof
  assume a divides b;
  then consider l being Integer such that
A1: b = a * l by INT_1:def 3;
  a * l * c = a * (l * c);
  hence thesis by A1,INT_1:def 3;
end;

theorem Th3:
  0 divides a iff a = 0
proof
  0 divides a implies a = 0
  proof
    assume 0 divides a;
    then ex b being Integer st a = 0*b by INT_1:def 3;
    hence thesis;
  end;
  hence thesis;
end;

Lm1: a divides -a & -a divides a
proof
  -a = a*(-1);
  hence a divides -a by INT_1:def 3;
  a = (-a)*(-1);
  hence thesis by INT_1:def 3;
end;

Lm2: a divides b & b divides c implies a divides c
proof
  assume that
A1: a divides b and
A2: b divides c;
  consider k being Integer such that
A3: b = a*k by A1,INT_1:def 3;
  consider l being Integer such that
A4: c = b*l by A2,INT_1:def 3;
  c = a*(k*l) by A3,A4;
  hence thesis by INT_1:def 3;
end;

Lm3: a divides b iff a divides -b
proof
  thus a divides b implies a divides -b
  proof
    assume
A1: a divides b;
    b divides -b by Lm1;
    hence thesis by A1,Lm2;
  end;
    assume
A2: a divides -b;
    -b divides b by Lm1;
    hence thesis by A2,Lm2;
end;

Lm4: a divides b iff -a divides b
proof
  thus a divides b implies -a divides b
  proof
    assume
A1: a divides b;
    -a divides a by Lm1;
    hence thesis by A1,Lm2;
  end;
    assume
A2: -a divides b;
    a divides -a by Lm1;
    hence thesis by A2,Lm2;
end;

Lm5: a divides 0 & 1 divides a & -1 divides a
proof
  0 = a*0;
  hence a divides 0 by INT_1:def 3;
  a = 1*a;
  hence 1 divides a by INT_1:def 3;
  a = (-1)*(-a);
  hence thesis by INT_1:def 3;
end;

Lm6: a divides b & a divides c implies a divides b mod c
proof
  assume that
A1: a divides b and
A2: a divides c;
A3: now
    assume c <> 0;
    then
A4: b = c * (b div c) + (b mod c) by INT_1:59;
    a divides c * (b div c) by A2,Th2;
    hence thesis by A1,A4,Th1;
  end;
  now
    assume c = 0;
    then b mod c = 0 by INT_1:def 10;
    hence thesis by Lm5;
  end;
  hence thesis by A3;
end;

reserve i,j,k,l for Nat;

Lm7: k divides l iff ex t being Nat st l = k * t
proof
  hereby
    assume
A1: k divides l;
    thus ex t being Nat st l = k * t
    proof
A2:   k >= 0 by NAT_1:2;
      consider t being Integer such that
A3:   l = k * t by A1,INT_1:def 3;
      per cases by NAT_1:3;
      suppose
        0 < l;
        then 0 <= t by A3,A2,XREAL_1:131;
        then reconsider t as Element of NAT by INT_1:3;
        take t;
        thus l = k * t by A3;
      end;
      suppose
A4:     l = 0;
        take 0;
        thus l = k*0 by A4;
      end;
    end;
  end;
  assume ex t being Nat st l = k * t;
  hence thesis by INT_1:def 3;
end;

Lm8: i divides j & j divides i implies i = j
proof
  assume i divides j;
  then consider a such that
A1: j = i*a by INT_1:def 3;
  assume j divides i;
  then consider b such that
A2: i = j*b by INT_1:def 3;
  i<>0 implies i = j
  proof
A3: i >= 0 by NAT_1:2;
    assume
A4: i<>0;
    1*i = i*(a*b) by A1,A2;
    then a*b = 1 by A4,XCMPLX_1:5;
    then i = j*1 or i = j*(-1) by A2,INT_1:9;
    hence thesis by A4,A3;
  end;
  hence thesis by A1;
end;

definition
  let a,b be Integer;

  func a lcm b -> Nat means
  :Def1:
  a divides it & b divides it & for m being
  Integer st a divides m & b divides m holds it divides m;
  existence
  proof
    per cases;
    suppose
A1:   a = 0 or b = 0;
      take 0;
      thus a divides 0 & b divides 0 by Lm5;
      thus thesis by A1;
    end;
    suppose
A2:   a <> 0 & b <> 0;
      defpred P[Nat] means a divides $1 & b divides $1 & $1 <> 0;
      a*b in INT by INT_1:def 2;
      then consider k being Element of NAT such that
A3:   a*b = k or a*b = -k by INT_1:def 1;
      b divides a*b by INT_1:def 3;
      then
A4:   b divides k by A3,Lm3;
      a divides a*b by INT_1:def 3;
      then
A5:   a divides k by A3,Lm3;
      k <> 0 by A2,A3,XCMPLX_1:6;
      then
A6:   ex k being Nat st P[k] by A5,A4;
      consider k being Nat such that
A7:   P[k] and
A8:   for n being Nat st P[n] holds k <= n from NAT_1:sch 5(A6);
      take k;
      thus a divides k & b divides k by A7;
      let m be Integer;
      m in INT by INT_1:def 2;
      then consider n being Element of NAT such that
A9:   m = n or m = - n by INT_1:def 1;
      assume that
A10:  a divides m and
A11:  b divides m;
      b divides n by A9,A11,Lm3;
      then
A12:  b divides n mod k by A7,Lm6;
A13:  k > 0 by A7,NAT_1:3;
      then n mod k in NAT by INT_1:3,57;
      then reconsider i = n mod k as Nat;
      assume
A14:  not k divides m;
A15:  now
        assume i = 0;
        then n - (n div k) * k = 0 by A7,INT_1:def 10;
        then k divides n by INT_1:def 3;
        hence contradiction by A9,A14,Lm3;
      end;
      a divides n by A9,A10,Lm3;
      then a divides n mod k by A7,Lm6;
      then k divides n by A8,A13,A12,A15,INT_1:58;
      hence contradiction by A9,A14,Lm3;
    end;
  end;
  uniqueness
  proof
    let IT1,IT2 be Nat;
    assume a divides IT1 & b divides IT1 &
    ( for m being Integer st a divides m & b divides m holds IT1 divides m) &
    a divides IT2 & ( b divides IT2 &
    for m being Integer st a divides m & b divides m holds IT2 divides m );
    then IT1 divides IT2 & IT2 divides IT1;
    hence thesis by Lm8;
  end;
  commutativity;
end;

theorem Th4:
  a = 0 or b = 0 iff a lcm b = 0
proof
A1: b = 0 implies (a lcm b) = 0
  proof
    assume b = 0;
    then 0 divides (a lcm b) by Def1;
    hence thesis by Th3;
  end;
A2: (a lcm b) = 0 implies a = 0 or b = 0
  proof
A3: b divides b implies b divides b*a by Th2;
    assume
A4: a lcm b = 0;
    a divides a implies a divides a*b by Th2;
    then 0 divides a*b by A4,A3,Def1;
    then a*b = 0 by Th3;
    hence thesis by XCMPLX_1:6;
  end;
  a = 0 implies a lcm b = 0
  proof
    assume a = 0;
    then 0 divides (a lcm b) by Def1;
    hence thesis by Th3;
  end;
  hence thesis by A1,A2;
end;

Lm9: 0 < j & i divides j implies i <= j
proof
  assume that
A1: 0 < j and
A2: i divides j;
  consider l such that
A3: j = i * l by A2,Lm7;
  l <> 0 by A1,A3;
  then consider k such that
A4: l = k + 1 by NAT_1:6;
  i * (k + 1) = i + i * k;
  hence thesis by A3,A4,NAT_1:11;
end;

definition
  let a,b be Integer;
  func a gcd b -> Nat means
  :Def2:
  it divides a & it divides b &
  for m being Integer st m divides a & m divides b holds m divides it;
  existence
  proof
    per cases;
    suppose
A1:   a = 0;
      b in INT by INT_1:def 2;
      then consider k being Element of NAT such that
A2:   b = k or b = -k by INT_1:def 1;
      take k;
      thus k divides a & k divides b by A1,A2,Lm4,Lm5;
      let m be Integer;
      assume that
      m divides a and
A3:   m divides b;
      thus thesis by A2,A3,Lm3;
    end;
    suppose
A4:   b = 0;
      a in INT by INT_1:def 2;
      then consider k being Element of NAT such that
A5:   a = k or a = -k by INT_1:def 1;
      take k;
      thus k divides a & k divides b by A4,A5,Lm4,Lm5;
      let m be Integer;
      assume that
A6:   m divides a and
      m divides b;
      thus thesis by A5,A6,Lm3;
    end;
    suppose
A7:   a <> 0 & b <> 0;
      defpred P[Nat] means $1 divides a & $1 divides b & $1 <> 0;
A8:   a divides a*b by INT_1:def 3;
      a*b in INT by INT_1:def 2;
      then consider k being Element of NAT such that
A9:   a*b = k or a*b = -k by INT_1:def 1;
      k <> 0 by A7,A9,XCMPLX_1:6;
      then
A10:  k > 0 by NAT_1:3;
A11:  for i being Nat st P[i] holds i <= k
      proof
        let i be Nat;
        assume P[i];
        then i divides a*b by A8,Lm2;
        then i divides k by A9,Lm3;
        hence thesis by A10,Lm9;
      end;
      1 divides a & 1 divides b by Lm5;
      then
A12:  ex k being Nat st P[k];
      consider k being Nat such that
A13:  P[k] and
A14:  for n being Nat st P[n] holds n <= k from NAT_1:sch 6(A11,A12);
      take k;
      thus k divides a & k divides b by A13;
      let m be Integer;
      assume that
A15:  m divides a and
A16:  m divides b;
      m in INT by INT_1:def 2;
      then consider n being Element of NAT such that
A17:  m = n or m = - n by INT_1:def 1;
      set i = n lcm k;
A18:  k divides i by Def1;
A19:  now
        assume i = 0;
        then n = 0 or k = 0 by Th4;
        hence contradiction by A7,A13,A17,A15,Th3;
      end;
      then 0 < i by NAT_1:3;
      then
A20:  k <= i by A18,Lm9;
      n divides b by A17,A16,Lm4;
      then
A21:  n lcm k divides b by A13,Def1;
      n divides a by A17,A15,Lm4;
      then n lcm k divides a by A13,Def1;
      then k >= n lcm k by A14,A21,A19;
      then k = i by A20,XXREAL_0:1;
      then
A22:  n divides k by Def1;
      assume not m divides k;
      hence contradiction by A17,A22,Lm4;
    end;
  end;
  uniqueness
  proof
    let IT1,IT2 be Nat;
    assume IT1 divides a & IT1 divides b &
    ( for m being Integer st m divides a & m divides b holds m divides IT1)
    & IT2 divides a & (
    IT2 divides b & for m being Integer st m divides a & m divides b holds
    m divides IT2 );
    then IT1 divides IT2 & IT2 divides IT1;
    hence thesis by Lm8;
  end;
  commutativity;
end;

theorem Th5:
  a = 0 & b = 0 iff a gcd b = 0
proof
  0 divides 0 gcd 0 by Def2;
  hence a = 0 & b = 0 implies a gcd b = 0 by Th3;
  assume a gcd b = 0;
  then 0 divides a & 0 divides b by Def2;
  hence thesis by Th3;
end;

reserve n for Nat;
reserve a,b,c,d,a1,b1,a2,b2,k,l for Integer;

theorem Th6:
  -n is Element of NAT iff n = 0
proof
  thus -n is Element of NAT implies n = 0
  proof
    assume -n is Element of NAT;
    then -n>=0 & n+(-n)>=0+n by NAT_1:2;
    hence thesis;
  end;
  thus thesis;
end;

registration
  let n be non zero Nat;
  cluster -n -> non natural;
  coherence
  proof
    not -n is Element of NAT by Th6;
    hence thesis by ORDINAL1:def 12;
  end;
end;

theorem
  not -1 is Element of NAT;

theorem
  a divides -a & -a divides a by Lm1;

theorem
  a divides b & b divides c implies a divides c by Lm2;

theorem Th10:
  (a divides b iff a divides -b) & (a divides b iff -a divides b) &
  (a divides b iff -a divides -b) & (a divides -b iff -a divides b)
proof
A1: a divides b implies a divides -b
  proof
    assume
A2: a divides b;
    b divides -b by Lm1;
    hence thesis by A2,Lm2;
  end;
A3: a divides -b implies a divides b
  proof
    assume
A4: a divides -b;
    -b divides b by Lm1;
    hence thesis by A4,Lm2;
  end;
  hence a divides b iff a divides -b by A1;
A5: -a divides b implies a divides b
  proof
    assume
A6: -a divides b;
    a divides -a by Lm1;
    hence thesis by A6,Lm2;
  end;
A7: -a divides -b implies a divides b
  proof
    assume
A8: -a divides -b;
    -b divides b by Lm1;
    hence thesis by A5,A8,Lm2;
  end;
A9: a divides b implies -a divides b
  proof
    assume
A10: a divides b;
    -a divides a by Lm1;
    hence thesis by A10,Lm2;
  end;
  hence a divides b iff -a divides b by A5;
  a divides b implies -a divides -b
  proof
    assume
A11: a divides b;
    -a divides a by Lm1;
    hence thesis by A1,A11,Lm2;
  end;
  hence a divides b iff -a divides -b by A7;
  thus thesis by A1,A3,A9,A5;
end;

theorem
  a divides b & b divides a implies a = b or a = -b
proof
  assume that
A1: a divides b and
A2: b divides a;
  consider a1 such that
A3: b = a*a1 by A1,INT_1:def 3;
  consider b1 such that
A4: a = b*b1 by A2,INT_1:def 3;
  a<>0 implies a = b or a = -b
  proof
    assume
A5: a<>0;
    1*a = a*(a1*b1) by A3,A4;
    then a1*b1 = 1 by A5,XCMPLX_1:5;
    then a = b*1 or a = b*(-1) by A4,INT_1:9;
    hence thesis;
  end;
  hence thesis by A1,Th3;
end;

theorem
  a divides 0 & 1 divides a & -1 divides a by Lm5;

theorem Th13:
  a divides 1 or a divides -1 implies a = 1 or a = -1
proof
A1: a divides 1 implies a = 1 or a = -1
  proof
    assume a divides 1;
    then ex b st 1 = a*b by INT_1:def 3;
    hence thesis by INT_1:9;
  end;
A2: a divides -1 implies a = 1 or a = -1
  proof
    assume a divides -1;
    then ex b st -1 = a*b by INT_1:def 3;
    hence thesis by INT_1:10;
  end;
  assume a divides 1 or a divides -1;
  hence thesis by A1,A2;
end;

theorem
  a = 1 or a = -1 implies a divides 1 & a divides -1 by Lm5;

theorem
  a,b are_congruent_mod c iff c divides (a-b)
proof
  thus a,b are_congruent_mod c implies c divides (a-b)
  proof
    assume a,b are_congruent_mod c;
    then ex d st (a-b) = c*d by INT_1:def 5;
    hence thesis by INT_1:def 3;
  end;
  assume c divides (a-b);
  then ex d st (a-b) = c*d by INT_1:def 3;
  hence thesis by INT_1:def 5;
end;

theorem
  a divides b iff (abs a) divides (abs b)
proof
  thus a divides b implies (abs a) divides (abs b)
  proof
    assume a divides b;
    then consider c such that
A1: b = a*c by INT_1:def 3;
    abs(b) = abs(a)*abs(c) by A1,COMPLEX1:65;
    hence thesis by INT_1:def 3;
  end;
    assume (abs a) divides (abs b);
    then consider m being Integer such that
A2: abs(b) = abs(a)*m by INT_1:def 3;
A3: a>=0 implies a divides b
    proof
      assume a>=0;
      then
A4:   abs(b) = a*m by A2,ABSVALUE:def 1;
      per cases;
      suppose b<0;
        then -b = a*m by A4,ABSVALUE:def 1;
        then b = a*(-m);
        hence thesis by INT_1:def 3;
      end;
      suppose b>=0;
        then b = a*m by A4,ABSVALUE:def 1;
        hence thesis by INT_1:def 3;
      end;
    end;
    a<0 implies a divides b
    proof
      assume a<0;
      then
A5:   abs(b) = (-a)*m by A2,ABSVALUE:def 1;
      per cases;
      suppose b<0;
        then -b = -(a*m) by A5,ABSVALUE:def 1;
        hence thesis by INT_1:def 3;
      end;
      suppose b>=0;
        then b = a*(-m) by A5,ABSVALUE:def 1;
        hence thesis by INT_1:def 3;
      end;
    end;
    hence thesis by A3;
end;

theorem
  a lcm b is Element of NAT by ORDINAL1:def 12;

theorem
  a divides a lcm b by Def1;

theorem
  for c st a divides c & b divides c holds a lcm b divides c by Def1;

theorem
  a gcd b is Element of NAT by ORDINAL1:def 12;

theorem Th21:
  a gcd b divides a by Def2;

theorem
  for c st c divides a & c divides b holds c divides (a gcd b) by Def2;



definition
  let a,b;
  pred a,b are_relative_prime means
  :Def3:
  a gcd b = 1;
  symmetry;
end;

theorem
  a<>0 or b<>0 implies ex a1,b1 st a = (a gcd b)*a1 & b = (a gcd b)*b1 &
  a1,b1 are_relative_prime
proof
  assume a<>0 or b<>0;
  then
A1: a gcd b <>0 by Th5;
  (a gcd b) divides a by Def2;
  then consider a1 such that
A2: a = (a gcd b)*a1 by INT_1:def 3;
  (a gcd b) divides b by Def2;
  then consider b1 such that
A3: b = (a gcd b)*b1 by INT_1:def 3;
  (a1 gcd b1) divides b1 by Def2;
  then consider b2 such that
A4: b1 = (a1 gcd b1)*b2 by INT_1:def 3;
  b = ((a gcd b)*(a1 gcd b1))*b2 by A3,A4;
  then
A5: (a gcd b)*(a1 gcd b1) divides b by INT_1:def 3;
  (a1 gcd b1) divides a1 by Def2;
  then consider a2 such that
A6: a1 = (a1 gcd b1)*a2 by INT_1:def 3;
  a = ((a gcd b)*(a1 gcd b1))*a2 by A2,A6;
  then (a gcd b)*(a1 gcd b1) divides a by INT_1:def 3;
  then (a gcd b)*(a1 gcd b1) divides (a gcd b) by A5,Def2;
  then consider c such that
A7: a gcd b = ((a gcd b)*(a1 gcd b1))*c by INT_1:def 3;
  (a gcd b)*1 = (a gcd b)*((a1 gcd b1)*c) by A7;
  then 1 = (a1 gcd b1)*c by A1,XCMPLX_1:5;
  then a1 gcd b1 = 1 or a1 gcd b1 = -1 by INT_1:9;
  then a1,b1 are_relative_prime by Def3;
  hence thesis by A2,A3;
end;

theorem Th24:
  a,b are_relative_prime implies (c*a gcd c*b) = abs(c) & c*a gcd
  b*c = abs(c) & a*c gcd c*b = abs(c) & a*c gcd b*c = abs(c)
proof
  assume a,b are_relative_prime;
  then
A1: a gcd b = 1 by Def3;
  thus
A2: c*a gcd c*b = abs(c)
  proof
    (c*a gcd c*b) divides c*b by Def2;
    then consider l such that
A3: c*b = (c*a gcd c*b)*l by INT_1:def 3;
    (c*a gcd c*b) divides c*a by Def2;
    then consider k such that
A4: c*a = (c*a gcd c*b)*k by INT_1:def 3;
    c divides c*a & c divides c*b by Th2;
    then c divides (c*a gcd c*b) by Def2;
    then consider d such that
A5: c*a gcd c*b = c*d by INT_1:def 3;
A6: c*b = c*(d*l) by A5,A3;
A7: c*a = c*(d*k) by A5,A4;
A8: c <>0 implies c*a gcd c*b = abs(c)
    proof
      assume
A9:   c <>0;
      then b = d*l by A6,XCMPLX_1:5;
      then
A10:  d divides b by INT_1:def 3;
      a = d*k by A7,A9,XCMPLX_1:5;
      then d divides a by INT_1:def 3;
      then d divides 1 by A1,A10,Def2;
      then c*a gcd c*b = c*1 or c*a gcd c*b = c*(-1) by A5,Th13;
      then c*a gcd c*b = c*1 or c*a gcd c*b = (-c)*1;
      then
A11:  abs((c*a gcd c*b)) = abs(c) by COMPLEX1:52;
      c*a gcd c*b >=0 by NAT_1:2;
      hence thesis by A11,ABSVALUE:def 1;
    end;
    0*a gcd 0*b = 0 by Th5
      .= abs 0 by ABSVALUE:2;
    hence thesis by A8;
  end;
  hence c*a gcd b*c = abs(c);
  thus thesis by A2;
end;

theorem Th25:
  c divides a*b & a,c are_relative_prime implies c divides b
proof
  assume that
A1: c divides a*b and
A2: a,c are_relative_prime;
  c divides c*b by Th2;
  then
A3: c divides (a*b gcd c*b) by A1,Def2;
A4: a*b gcd c*b = abs(b) by A2,Th24;
  b<0 implies c divides b
  proof
    assume b<0;
    then c divides (-b) by A4,A3,ABSVALUE:def 1;
    hence thesis by Th10;
  end;
  hence thesis by A4,A3,ABSVALUE:def 1;
end;

theorem
  a,c are_relative_prime & b,c are_relative_prime implies
  a*b,c are_relative_prime
proof
  assume that
A1: a,c are_relative_prime and
A2: b,c are_relative_prime;
A3: (a gcd c) = 1 by A1,Def3;
A4: ((a*b gcd c) gcd a) divides a by Def2;
A5: (a*b gcd c) divides c by Def2;
  ((a*b gcd c) gcd a) divides (a*b gcd c) by Def2;
  then ((a*b gcd c) gcd a) divides c by A5,Lm2;
  then ((a*b gcd c) gcd a) divides 1 by A3,A4,Def2;
  then (a*b gcd c) gcd a = 1 or (a*b gcd c) gcd a = -1 by Th13;
  then a,(a*b gcd c) are_relative_prime by Def3;
  then
A6: (a*b gcd c) divides b by Th21,Th25;
  (b gcd c) = 1 by A2,Def3;
  then (a*b gcd c) divides 1 by A5,A6,Def2;
  then a*b gcd c = 1 or a*b gcd c = -1 by Th13;
  hence thesis by Def3;
end;





reserve p,p1,q,l for Nat;

definition
  let p be Nat;
  attr p is prime means
  :Def4:
  p > 1 & for n being Nat st n divides p holds n = 1 or n = p;
end;

theorem Th27:
  0 < b & a divides b implies a <= b
proof
  assume
A1: 0 < b;
  assume a divides b;
  then consider c such that
A2: b = a * c by INT_1:def 3;
  per cases;
  suppose
    a <= 0;
    hence thesis by A1,XXREAL_0:2;
  end;
  suppose
A3: a > 0;
    then c > 0 by A1,A2,XREAL_1:131;
    then c >= 0+1 by INT_1:7;
    hence thesis by A2,A3,XREAL_1:151;
  end;
end;

theorem Th28:
  2 is prime
proof
  thus 2>1;
  let n be Nat;
  assume
A1: n divides 2;
  then n <= 2 by Th27;
  then n = 0 or n = 1 or n = 2 by NAT_1:26;
  hence thesis by A1,Th3;
end;

theorem Th29:
  not 4 is prime
proof
  ex n st n divides 4 & n<>1 & n<>4
  proof
    take 2;
    4 = 2*2;
    hence thesis by INT_1:def 3;
  end;
  hence thesis by Def4;
end;

registration
  cluster prime for Nat;
  existence by Th28;
  cluster non zero non prime for Nat;
  existence by Th29;
end;

theorem
  p is prime & q is prime implies p,q are_relative_prime or p = q
proof
  assume that
A1: p is prime and
A2: q is prime;
A3: (p gcd q) divides q by Def2;
  assume not p,q are_relative_prime;
  then
A4: p gcd q<>1 by Def3;
  (p gcd q) divides p by Def2;
  then p gcd q = p by A1,A4,Def4;
  hence thesis by A2,A4,A3,Def4;
end;

theorem
  l>=2 implies ex p being Element of NAT st p is prime & p divides l
proof
  defpred P[Nat] means ex p st p is prime & p divides $1;
A1: for k being Nat st k>=2 holds (for n being Nat st n>=2 holds n<k implies
  P[n]) implies P[k]
  proof
    let k be Nat;
    assume
A2: k>=2;
    assume
A3: for n being Nat st n>=2 holds n<k implies ex p st p is prime & p divides n;
A4: k+1-1>1+1-1 by A2,NAT_1:13;
    not k is prime implies ex p being Element of NAT st p is prime & p
    divides k
    proof
      assume not k is prime;
      then consider m being Nat such that
A5:   m divides k and
A6:   m<>1 and
A7:   m<>k by A4,Def4;
      m<>0 by A5,A7,Th3;
      then m>0 by NAT_1:3;
      then m>=0+1 by NAT_1:13;
      then m>1 by A6,XXREAL_0:1;
      then
A8:   m>=1+1 by NAT_1:13;
      k>0 by A2,XXREAL_0:2;
      then m<=k by A5,Th27;
      then m<k by A7,XXREAL_0:1;
      then consider p1 such that
A9:   p1 is prime & p1 divides m by A3,A8;
      reconsider p1 as Element of NAT by ORDINAL1:def 12;
      take p1;
      thus thesis by A5,A9,Lm2;
    end;
    hence thesis;
  end;
A10: for k being Nat st k>=2 holds P[k] from NAT_1:sch 9(A1);
  assume l>=2;
  then consider p such that
A11: p is prime & p divides l by A10;
  reconsider p as Element of NAT by ORDINAL1:def 12;
  take p;
  thus thesis by A11;
end;

begin 


theorem
  for i,j being Integer st i >= 0 & j >= 0 holds
  abs(i) mod abs(j) = i mod j & abs(i) div abs(j) = i div j
proof
  let i,j be Integer;
  assume that
A1: i >= 0 and
A2: j >= 0;
  per cases by A2,XXREAL_0:1;
  suppose
A3: j > 0;
    i = abs(i) by A1,ABSVALUE:def 1;
    hence thesis by A3,ABSVALUE:def 1;
  end;
  suppose
A4: j = 0;
    abs(0) = 0 by ABSVALUE:def 1;
    then abs(i) mod abs(0) = 0 & abs(i) div abs(0) = 0 by INT_1:48,def 10;
    hence thesis by A4,INT_1:48,def 10;
  end;
end;



theorem
  a lcm b = abs(a) lcm abs(b)
proof
A1: abs b = b or abs b = -b by ABSVALUE:1;
A2: abs a = a or abs a = -a by ABSVALUE:1;
A3: now
    let m be Integer;
    assume abs a divides m & abs b divides m;
    then a divides m & b divides m by A2,A1,Th10;
    hence a lcm b divides m by Def1;
  end;
  b divides a lcm b by Def1;
  then
A4: abs b divides a lcm b by A1,Th10;
  a divides a lcm b by Def1;
  then abs a divides a lcm b by A2,Th10;
  hence thesis by A4,A3,Def1;
end;

theorem
  a gcd b = abs(a) gcd abs(b)
proof
A1: abs b = b or abs b = -b by ABSVALUE:1;
A2: abs a = a or abs a = -a by ABSVALUE:1;
A3: now
    let m be Integer;
    assume m divides abs a & m divides abs b;
    then m divides a & m divides b by A2,A1,Th10;
    hence m divides a gcd b by Def2;
  end;
  a gcd b divides b by Def2;
  then
A4: a gcd b divides abs b by A1,Th10;
  a gcd b divides a by Def2;
  then a gcd b divides abs a by A2,Th10;
  hence thesis by A4,A3,Def2;
end;















begin

reserve p,k,l,m,n,s,h,i,j,t for Nat;



Lm1: now
  let k,l;
  k div l in NAT by INT_1:3,55;
  hence k div l is Nat;
end;

Lm2: now
  let k,l;
  k mod l in NAT by INT_1:3,57;
  hence k mod l is Nat;
end;

definition
  let k,l be Nat;
  redefine func k div l -> Nat means       
  :Def1:
  ( ex t being Nat st k = l * it + t & t < l ) or it = 0 & l = 0;
  compatibility
  proof
    let r be Nat;
    per cases;
    suppose l = 0;
      hence thesis by INT_1:48;
    end;
    suppose
A1:   l > 0;
      then
A2:   k = l * (k div l) + (k mod l) by INT_1:59;
A3:   k mod l is Nat by Lm2;
      hence r = k div l implies
      (ex t being Nat st k = l * r + t & t < l) or r = 0 & l = 0
      by A1,A2,INT_1:58;
      assume
A4:   ( ex t being Nat st k = l * r + t & t < l ) or r = 0 & l = 0;
A5:   k mod l < l by A1,INT_1:58;
      k div l is Nat by Lm1;
      hence thesis by A2,A3,A4,A5,NAT_1:18;
    end;
  end;
  coherence by Lm1;
end;

definition
  let k,l be Nat;
  redefine func k mod l -> Nat means         
  :Def2:
  ( ex t being Nat st k = l * t + it & it < l ) or it = 0 & l = 0;
  compatibility
  proof
    let r be Nat;
    per cases;
    suppose l = 0;
      hence thesis by INT_1:def 10;
    end;
    suppose
A1:   l > 0;
      then
A2:   k = l * (k div l) + (k mod l) by INT_1:59;
      hence r = k mod l implies
      ( ex t being Nat st k = l * t + r & r < l ) or r = 0 & l = 0
      by A1,INT_1:58;
      assume
A3:   ( ex t being Nat st k = l * t + r & r < l ) or r = 0 & l = 0;
A4:   k mod l < l by A1,INT_1:58;
      k mod l is Nat by Lm2;
      hence thesis by A2,A3,A4,NAT_1:18;
    end;
  end;
  coherence by Lm2;
end;

definition
  let k,l be Nat;
  redefine func k div l -> Element of NAT;
  coherence
  proof
    k div l is Nat;
    hence thesis by ORDINAL1:def 12;
  end;
  redefine func k mod l -> Element of NAT;
  coherence
  proof
    k mod l is Nat;
    hence thesis by ORDINAL1:def 12;
  end;
end;

theorem Th1:
  0 < i implies j mod i < i
proof
  assume 0 < i;
  then ex t being Nat st j = i * t + (j mod i) & j mod i < i by Def2;
  hence thesis;
end;

theorem
  0 < i implies j = i * (j div i) + (j mod i) by INT_1:59;



definition
  let k,l be Nat;
  redefine pred k divides l means
  :Def3:
  ex t being Nat st l = k * t;
  compatibility
  proof
    hereby
      assume
A1:   k divides l;
      thus ex t being Nat st l = k * t
      proof
        consider t being Integer such that
A2:     l = k * t by A1,INT_1:def 3;
        per cases;
        suppose 0 < l;
          then 0 <= t by A2;
          then reconsider t as Element of NAT by INT_1:3;
          take t;
          thus l = k * t by A2;
        end;
        suppose
A3:       l = 0;
          take 0;
          thus l = k*0 by A3;
        end;
      end;
    end;
    assume ex t being Nat st l = k * t;
    hence thesis by INT_1:def 3;
  end;
  reflexivity
  proof
    let i;
    i = i * 1;
    hence thesis;
  end;
end;

theorem Th3:
  j divides i iff i = j * (i div j)
proof
  thus j divides i implies i = j * (i div j)
  proof
    assume j divides i;
    then consider k being Nat such that
A1: i = j * k by Def3;
A2: i = j * k + 0 by A1;
    now
      assume
A3:   j <> 0;
      then
A4:   i = j * (i div j) + (i mod j) by INT_1:59;
      i mod j < j by A3,Th1;
      hence thesis by A2,A4,NAT_1:18;
    end;
    hence thesis by A1;
  end;
  assume i = j * (i div j);
  hence thesis by Def3;
end;

theorem
  i divides j & j divides h implies i divides h
proof
  assume that
A1: i divides j and
A2: j divides h;
A3: j = i * (j div i) by A1,Th3;
  h = j * (h div j) by A2,Th3;
  then h = i * ((j div i)*(h div j)) by A3;
  hence thesis by Def3;
end;

theorem Th5:
  i divides j & j divides i implies i = j
proof
  assume that
A1: i divides j and
A2: j divides i;
A3: j = i * (j div i) by A1,Th3;
  i = j * (i div j) by A2,Th3;
  then
A4: i*1 = i * (j div i) * (i div j) by A3
    .= i * ((j div i) * (i div j));
A5: i = 0 implies i = j by A3;
  (j div i) * (i div j) = 1 implies i = j
  proof
    assume (j div i) * (i div j) = 1;
    then j div i = 1 by NAT_1:15;
    hence thesis by A3;
  end;
  hence thesis by A4,A5,XCMPLX_1:5;
end;

theorem Th6:
  i divides 0 & 1 divides i
proof
  0 = i * 0;
  hence i divides 0 by Def3;
  i = i * 1;
  hence thesis by Def3;
end;

theorem
  0 < j & i divides j implies i <= j
proof
  assume that
A1: 0 < j and
A2: i divides j;
A3: j = i * (j div i) by A2,Th3;
  then j div i <> 0 by A1;
  then consider m such that
A4: j div i = m + 1 by NAT_1:6;
  i * (m + 1) = i + i * m;
  hence thesis by A3,A4,NAT_1:11;
end;

theorem Th8:
  i divides j & i divides h implies i divides j+h
proof
  assume that
A1: i divides j and
A2: i divides h;
A3: j = i * (j div i) by A1,Th3;
  h = i * (h div i) by A2,Th3;
  then j + h = i * ((j div i) + (h div i)) by A3;
  hence thesis by Def3;
end;

theorem Th9:
  i divides j implies i divides j * h
proof
  assume i divides j;
  then consider l being Nat such that
A1: j = i * l by Def3;
  i * l * h = i * (l * h);
  hence thesis by A1,Def3;
end;

theorem Th10:
  i divides j & i divides j + h implies i divides h
proof
  assume that
A1: i divides j and
A2: i divides j + h;
  consider t being Nat such that
A3: j + h = i * t by A2,Def3;
  consider u being Nat such that
A4: j = i * u by A1,Def3;
  now
    assume
A5: i <> 0;
    then
A6: h = i * (h div i) + (h mod i) by INT_1:59;
A7: j + (i * (h div i) + (h mod i)) = i * (u + (h div i)) + (h mod i) by A4;
A8: h mod i < i by A5,Th1;
    j + h = i * t + 0 by A3;
    then h mod i = 0 by A6,A7,A8,NAT_1:18;
    hence thesis by A6,Th3;
  end;
  hence thesis by A3;
end;

theorem Th11:
  i divides j & i divides h implies i divides j mod h
proof
  assume that
A1: i divides j and
A2: i divides h;
A3: now
    assume h = 0;
    then j mod h = 0 by Def2;
    hence thesis by Th6;
  end;
  now
    assume h <> 0;
    then
A4: j = h * (j div h) + (j mod h) by INT_1:59;
    i divides h * (j div h) by A2,Th9;
    hence thesis by A1,A4,Th10;
  end;
  hence thesis by A3;
end;



definition
  let k,n be Nat;
  redefine func k lcm n means
  :Def4:
  k divides it & n divides it &
  for m being Nat st k divides m & n divides m holds it divides m;
  compatibility
  proof
    let IT be Nat;
    thus IT = k lcm n implies k divides IT & n divides IT &
    for m being Nat st k divides m & n divides m holds IT divides m
    by INT_2:def 1;
    assume that
A1: k divides IT and
A2: n divides IT and
A3: for m being Nat st k divides m & n divides m holds IT divides m;
    for m being Integer st k divides m & n divides m holds IT divides m
    proof
      let m be Integer;
      m in INT by INT_1:def 2;
      then consider i being Element of NAT such that
A4:   m = i or m = -i by INT_1:def 1;
      assume that
A5:   k divides m and
A6:   n divides m;
A7:   k divides i by A4,A5,INT_2:10;
      n divides i by A4,A6,INT_2:10;
      then IT divides i by A3,A7;
      hence thesis by A4,INT_2:10;
    end;
    hence thesis by A1,A2,INT_2:def 1;
  end;
end;

definition
  let k,n be Nat;
  redefine func k lcm n -> Element of NAT;
  coherence by ORDINAL1:def 12;
end;

definition
  let k,n be Nat;
  redefine func k gcd n means
  :Def5:
  it divides k & it divides n &
  for m being Nat st m divides k & m divides n holds m divides it;
  compatibility
  proof
    let IT be Nat;
    thus IT = k gcd n implies IT divides k & IT divides n &
    for m being Nat st m divides k & m divides n holds m divides IT
    by INT_2:def 2;
    assume that
A1: IT divides k and
A2: IT divides n and
A3: for m being Nat st m divides k & m divides n holds m divides IT;
    for m being Integer st m divides k & m divides n holds m divides IT
    proof
      let m be Integer;
      m in INT by INT_1:def 2;
      then consider i being Element of NAT such that
A4:   m = i or m = -i by INT_1:def 1;
      assume that
A5:   m divides k and
A6:   m divides n;
A7:   i divides k by A4,A5,INT_2:10;
      i divides n by A4,A6,INT_2:10;
      then i divides IT by A3,A7;
      hence thesis by A4,INT_2:10;
    end;
    hence thesis by A1,A2,INT_2:def 2;
  end;
end;

definition
  let k,n be Nat;
  redefine func k gcd n -> Element of NAT;
  coherence by ORDINAL1:def 12;
end;



scheme Euklides { Q(Nat)->Nat, a,b()->Nat } :
  ex n being Element of NAT st Q(n) = a() gcd b() & Q(n + 1) = 0
provided
A1: 0 < b() & b() < a() and
A2: Q(0) = a() & Q(1) = b() and
A3: for n being Element of NAT holds Q(n + 2) = Q(n) mod Q(n + 1)
proof
  defpred P[Nat] means ex n being Element of NAT st $1 = Q(n);
A4: ex k being Nat st P[k] by A2;
A5: for k being Nat st k <> 0 & P[k] ex n being Nat st n < k & P[n]
  proof
    let k be Nat;
    assume that
A6: k <> 0 and
A7: ex n being Element of NAT st k = Q(n);
    consider n being Element of NAT such that
A8: k = Q(n) by A7;
    reconsider Q = Q(n+1) as Element of NAT by ORDINAL1:def 12;
A9: n = 0 implies Q < k by A1,A2,A8;
    now
      given m such that
A10:  n = m + 1;
      reconsider m1=m as Element of NAT by ORDINAL1:def 12;
      Q(m1 + 2) = Q(m1) mod Q(m1 + 1) by A3;
      hence Q(n + 1) < k by A6,A8,A10,Th1;
      take m = n + 1;
      thus Q(n + 1) =Q(m);
    end;
    hence thesis by A9,NAT_1:6;
  end;
A11: P[0] from NAT_1:sch 7(A4,A5);
  defpred Q[Nat] means 0 = Q($1);
A12: ex n being Nat st Q[n] by A11;
  consider k being Nat such that
A13: Q[k] & for n being Nat st Q[n] holds k <= n from NAT_1:sch 5(A12);
  consider n such that
A14: k = n + 1 by A1,A2,A13,NAT_1:6;
  reconsider n as Element of NAT by ORDINAL1:def 12;
  take n;
  defpred PP[Nat] means Q(n) divides Q($1) & Q(n) divides Q($1 + 1);
  PP[n] by A13,A14,Th6;
  then
A15: ex k being Nat st PP[k];
A16: for k being Nat st k <> 0 & PP[k] ex m being Nat st m < k & PP[m]
  proof
    let l be Nat;
    assume that
A17: l <> 0 and
A18: Q(n) divides Q(l) and
A19: Q(n) divides Q(l + 1);
    consider m such that
A20: l = m + 1 by A17,NAT_1:6;
    reconsider m as Element of NAT by ORDINAL1:def 12;
    take m;
A21: m <= m + 1 by NAT_1:11;
    m <> m + 1;
    hence m < l by A20,A21,XXREAL_0:1;
A22: now
      assume
A23:  Q(m + 1) = 0;
      now
        assume
A24:    m + 1 <> k;
        k <= m + 1 by A13,A23;
        then k < m + 1 by A24,XXREAL_0:1;
        then
A25:    k <= m by NAT_1:13;
        defpred Q[Nat] means k <= $1 implies Q($1) = 0;
A26:    Q[0] by A13;
A27:    for m st Q[m] holds Q[m+1]
        proof
          let m such that
A28:      k <= m implies Q(m) = 0 and
A29:      k <= m + 1;
          now
            assume k <> m + 1;
            then
A30:        k < m + 1 by A29,XXREAL_0:1;
            then consider l such that
A31:        m = l + 1 by A1,A2,A28,NAT_1:6,13;
            reconsider l as Element of NAT by ORDINAL1:def 12;
            Q(l + 2) = Q(l) mod Q(l + 1) by A3;
            hence thesis by A28,A30,A31,Def2,NAT_1:13;
          end;
          hence thesis by A13;
        end;
        for m holds Q[m] from NAT_1:sch 2(A26,A27);
        then Q(m) = 0 by A25;
        hence Q(n) divides Q(m) by Th6;
      end;
      hence Q(n) divides Q(m) by A14;
    end;
    now
      assume
A32:  Q(m + 1) <> 0;
      Q(m + 2) = Q(m) mod Q(m + 1) by A3;
      then
A33:  Q(m) = Q(m + 1) * (Q(m) div Q(m + 1)) + Q(m + 2) by A32,INT_1:59;
      Q(n) divides Q(m + 1) * (Q(m) div Q(m + 1)) by A18,A20,Th9;
      hence Q(n) divides Q(m) by A19,A20,A33,Th8;
    end;
    hence Q(n) divides Q(m) by A22;
    thus Q(n) divides Q(m + 1) by A18,A20;
  end;
  now
    PP[0] from NAT_1:sch 7(A15,A16);
    hence Q(n) divides a() & Q(n) divides b() by A2;
    let m;
    defpred P1[Nat] means m divides Q($1) & m divides Q($1 + 1);
    assume that
A34: m divides a() and
A35: m divides b();
A36: P1[0] by A2,A34,A35;
A37: for k be Element of NAT st P1[k] holds P1[k+1]
    proof
      let k be Element of NAT;
      assume that
A38:  m divides Q(k) and
A39:  m divides Q(k + 1);
      thus m divides Q(k + 1) by A39;
      Q(k + 2) = Q(k) mod Q(k + 1) by A3;
      hence m divides Q(k + 1 + 1) by A38,A39,Th11;
    end;
    for k be Element of NAT holds P1[k] from NAT_1:sch 1(A36,A37);
    hence m divides Q(n);
  end;
  hence Q(n) = a() gcd b() by Def5;
  thus thesis by A13,A14;
end;

theorem
  n mod 2 = 0 or n mod 2 = 1
proof
  assume
A1: n mod 2 <> 0;
A2: 2 = 1 + 1;
  n mod 2 < 2 by Th1;
  then
A3: n mod 2 <= 1 by A2,NAT_1:13;
  n mod 2 >= 1 by A1,NAT_1:14;
  hence thesis by A3,XXREAL_0:1;
end;

theorem
  (k * n) mod k = 0
proof
  per cases;
  suppose k = 0;
    hence thesis by Def2;
  end;
  suppose
A1: k <> 0;
    k * n = k * n + 0;
    hence thesis by A1,Def2;
  end;
end;

theorem
  k > 1 implies 1 mod k = 1
proof
  assume
A1: k > 1;
  1 = k * 0 + 1;
  hence thesis by A1,Def2;
end;

theorem
  k mod n = 0 & l = k - m * n implies l mod n = 0
proof
  assume that
A1: k mod n = 0 and
A2: l = k - m * n;
  per cases;
  suppose n = 0;
    hence thesis by A1,A2;
  end;
  suppose n <> 0;
    then consider t being Nat such that
A3: k = n * t + 0 and
A4: 0 < n by A1,Def2;
A5: l = n * (t - m) + 0 by A2,A3;
    now
      assume t < m + 0;
      then t - m < 0 by XREAL_1:19;
      then l < n * 0 by A4,A5,XREAL_1:68;
      hence contradiction;
    end;
    then t - m is Element of NAT by NAT_1:21;
    hence thesis by A4,A5,Def2;
  end;
end;

theorem
  n <> 0 & k mod n = 0 & l < n implies k + l mod n = l
proof
  assume that
A1: n <> 0 and
A2: k mod n = 0 and
A3: l < n;
  ex t being Nat st k = n * t + 0 & 0 < n by A1,A2,Def2;
  hence thesis by A3,Def2;
end;

theorem
  k mod n = 0 implies k + l mod n = l mod n
proof
  assume that
A1: k mod n = 0;
  per cases;
  suppose
A2: n = 0;
    hence k + l mod n = 0 by Def2
      .= l mod n by A2,Def2;
  end;
  suppose
A3: n <> 0;
    then consider m being Nat such that
A4: k = n * m + 0 and 0 < n by A1,Def2;
    consider t being Nat such that
A5: l = n * t + (l mod n) and
A6: l mod n < n by A3,Def2;
    k + l = n * (m + t) + (l mod n) by A4,A5;
    hence thesis by A6,Def2;
  end;
end;

theorem
  k <> 0 implies (k * n) div k = n
proof
  assume
A1: k <> 0;
  k * n = k * n + 0;
  hence thesis by A1,Def1;
end;

theorem
  k mod n = 0 implies (k + l) div n = (k div n) + (l div n)
proof
  assume
A1: k mod n = 0;
  per cases;
  suppose
A2: n <> 0;
    then
A3: k = n * (k div n) + 0 by A1,INT_1:59;
A4: ex t being Nat st l = n * t + (l mod n) & l mod n < n by A2,Def2;
    l = n * (l div n) + (l mod n) by A2,INT_1:59;
    then k + l = n * ((k div n) + (l div n)) + (l mod n) by A3;
    hence thesis by A4,Def1;
  end;
  suppose
A5: n = 0;
    then
A6: (k + l) div n = 0 by Def1;
    k div n = 0 by A5,Def1;
    hence thesis by A5,A6,Def1;
  end;
end;

begin 


theorem
  k <> 0 implies m * k div k = m
proof
  assume k <> 0;
  then m*k = k * m + 0 & 0 < k or m = 0 & k = 0;
  hence thesis by Def1;
end;



theorem Th21:
  m mod n = (n*k + m) mod n
proof
  per cases;
  suppose
A1: n > 0;
    m mod n = t implies (n*k + m) mod n = t
    proof
      assume m mod n = t;
      then consider q being Nat such that
A2:   m = n * q + t and
A3:   t < n by A1,Def2;
      ex p st n*k + m = n*p + t & t < n
      proof
        reconsider p = k+q as Element of NAT by ORDINAL1:def 12;
        take p;
        thus thesis by A2,A3;
      end;
      hence thesis by Def2;
    end;
    hence thesis;
  end;
  suppose n = 0;
    hence thesis;
  end;
end;

theorem Th22:
  (p+s) mod n = ((p mod n)+s) mod n
proof
  per cases;
  suppose n > 0;
    then (p + s) mod n = (n*(p div n) + (p mod n) + s) mod n by INT_1:59
      .=(n*(p div n) + ((p mod n) + s)) mod n
      .=((p mod n) + s) mod n by Th21;
    hence thesis;
  end;
  suppose
A1: n = 0;
    hence (p+s) mod n = 0 by Def2
      .= ((p mod n)+s) mod n by A1,Def2;
  end;
end;

theorem
  (p+s) mod n = (p + ( s mod n)) mod n by Th22;

theorem Th24:
  k < n implies k mod n = k
proof
  assume
A1: k<n;
  k=n*0 + k;
  hence thesis by A1,Def2;
end;

theorem Th25:
  n mod n = 0
proof
  per cases;
  suppose
A1: n>0;
    n=n*1+0;
    hence thesis by A1,Def2;
  end;
  suppose n = 0;
    hence thesis by Def2;
  end;
end;

theorem
  0 = 0 mod n
proof
  n = 0 or n > 0;
  hence thesis by Def2,Th24;
end;



theorem Th27:
  i < j implies i div j = 0
proof
  assume
A1: i<j;
  per cases by Def1;
  suppose ex j1 being Nat st i=j*(i div j)+j1 & j1<j;
    then consider j1 being Nat such that
A2: i=j*(i div j)+j1 and j1<j;
    assume i div j<>0;
    then consider k such that
A3: i div j = k+1 by NAT_1:6;
    i = j + (j*k+j1) by A2,A3;
    hence contradiction by A1,NAT_1:11;
  end;
  suppose i div j=0 & j=0;
    hence thesis;
  end;
end;



theorem Th28:
  m > 0 implies n gcd m = m gcd (n mod m)
proof
  set r=n mod m, x=n gcd m, y=m gcd r;
  assume m>0;
  then consider t be Nat such that
A1: n = m * t + r and r < m by Def2;
  reconsider t as Element of NAT by ORDINAL1:def 12;
A2: x divides n by Def5;
A3: x divides m by Def5;
  then x divides r by A2,Th11;
  then
A4: x divides y by A3,Def5;
A5: y divides m by Def5;
A6: y divides r by Def5;
  y divides m*t by A5,Th9;
  then y divides n by A1,A6,Th8;
  then y divides x by A5,Def5;
  hence thesis by A4,Th5;
end;



scheme INDI{ k,n() -> Element of NAT, P[set] }: P[n()]
provided
A1: P[0] and
A2: k() > 0 and
A3: for i,j st P[k()*i] & j <> 0 & j <= k() holds P[k()*i+j]
proof
  defpred R[Nat] means P[k()*$1];
A4: R[0] by A1;
A5: now
    let i;
    assume
A6: R[i];
    k()*(i+1) = k()*i + k();
    hence R[i+1] by A2,A3,A6;
  end;
A7: for i holds R[i] from NAT_1:sch 2(A4,A5);
  per cases;
  suppose n() mod k() = 0;
    then n() = k() * (n() div k()) + 0 by A2,INT_1:59;
    hence thesis by A7;
  end;
  suppose
A8: n() mod k() <> 0;
A9: n() = k() * (n() div k()) + (n() mod k()) by A2,INT_1:59;
    n() mod k() <= k() by A2,Th1;
    hence thesis by A3,A7,A8,A9;
  end;
end;



theorem
  i*j = (i lcm j)*(i gcd j)
proof
A1: i<>0 & j<>0 implies i*j = (i lcm j)*(i gcd j)
  proof
    assume that
A2: i<>0 and
A3: j<>0;
A4: i divides i implies i divides i*j by Th9;
    j divides j implies j divides j*i by Th9;
    then (i lcm j) divides i*j by A4,Def4;
    then consider c being Nat such that
A5: i*j = (i lcm j)*c by Def3;
A6: i divides (i lcm j) by Def4;
A7: j divides (i lcm j) by Def4;
    consider d being Nat such that
A8: (i lcm j) = i*d by A6,Def3;
    consider e being Nat such that
A9: (i lcm j) = j*e by A7,Def3;
    i*j = i*(c*d) by A5,A8;
    then
A10: j = c*d by A2,XCMPLX_1:5;
    i*j = j*(c*e) by A5,A9;
    then i = c*e by A3,XCMPLX_1:5;
    then
A11: c divides i by Def3;
A12: c divides j by A10,Def3;
    for f being Nat holds f divides i & f divides j implies f divides c
    proof
      let f be Nat;
      assume that
A13:  f divides i and
A14:  f divides j;
      consider g being Nat such that
A15:  i = f*g by A13,Def3;
      consider h being Nat such that
A16:  j = f*h by A14,Def3;
A17:  j*g = g*h*f by A16;
      i*h = g*h*f by A15;
      then
A18:  i divides g*h*f by Def3;
      j divides g*h*f by A17,Def3;
      then (i lcm j) divides g*h*f by A18,Def4;
      then consider z being Nat such that
A19:  g*h*f = (i lcm j)*z by Def3;
A20:  c*(i lcm j) = (g*(h*f))*f by A5,A15,A16
        .= (i lcm j)*z*f by A19
        .= (z*f)*(i lcm j);
      (i lcm j) <> 0 by A2,A3,INT_2:4;
      then c = f*z by A20,XCMPLX_1:5;
      hence thesis by Def3;
    end;
    hence thesis by A5,A11,A12,Def5;
  end;
  i = 0 or j = 0 implies i*j = (i lcm j)*(i gcd j)
  proof
    assume
A21: i = 0 or j = 0;
    then i*j = 0*(i gcd j) .= (i lcm j)*(i gcd j) by A21,INT_2:4;
    hence thesis;
  end;
  hence thesis by A1;
end;



theorem
  for i,j being Integer st i>=0 & j>0 holds i gcd j= j gcd (i mod j)
proof
  let i,j be Integer;
  assume that
A1: i>=0 and
A2: j>0;
A3: abs(j)> 0 by A2,ABSVALUE:def 1;
  thus i gcd j = abs(i) gcd abs(j) by INT_2:34
    .=abs(j) gcd (abs(i) mod abs(j)) by A3,Th28
    .=abs(j) gcd abs(abs(i) mod abs(j)) by ABSVALUE:def 1
    .=j gcd (abs(i) mod abs(j)) by INT_2:34
    .=j gcd (i mod j) by A1,A2,INT_2:32;
end;



theorem
  i lcm i = i
proof
  for m being Nat st i divides m & i divides m holds i divides m;
  hence thesis by Def4;
end;

theorem
  i gcd i = i
proof
  for m being Nat st m divides i & m divides i holds m divides i;
  hence thesis by Def5;
end;



theorem
  for i, j being Nat st i < j & i <> 0 holds i/j is not integer
proof
  let i, j be Nat such that
A1: i < j and
A2: i <> 0;
  assume i/j is integer;
  then reconsider r = i/j as Integer;
  r = [\ r /] by INT_1:25
    .= i qua Integer div j by INT_1:def 9;
  hence thesis by A1,A2,Th27,XCMPLX_1:50;
end;



definition
  let i,j be Nat;
  redefine func i -' j -> Element of NAT;
  coherence
  proof
    i -' j = i - j & 0 <= i -' j or i -' j = 0 by XREAL_0:def 2;
    hence thesis by INT_1:3;
  end;
end;

theorem Th34:
  for i,j being Nat holds i + j -' j = i
proof
  let i, j be Nat;
  i + j - j >= 0;
  hence thesis by XREAL_0:def 2;
end;

reserve a,b for Nat;

theorem Th35:
  a -' b <= a
proof
  a-b <= a-0 by XREAL_1:230;
  hence thesis by XREAL_0:def 2;
end;

reserve n,i,j,k for Nat;

theorem Th36:
  n-'i=0 implies n<=i
proof
  assume
A1: n-'i=0;
  assume i<n;
  then i+1<=n by NAT_1:13;
  then i+1-i<=n-i by XREAL_1:9;
  hence contradiction by A1,XREAL_0:def 2;
end;

theorem
  i<=j implies j+k-'i=j+k-i by NAT_1:12,XREAL_1:233;

theorem
  i<=j implies j+k-'i=j-'i+k
proof
  assume
A1: i<=j;
  hence j+k-'i=j+k-i by NAT_1:12,XREAL_1:233
    .=j-i+k
    .=j-'i+k by A1,XREAL_1:233;
end;

reserve i,j,k,i1,i2,n for Nat;

theorem Th39:
  i-'i1>=1 or i-i1>=1 implies i-'i1=i-i1
proof
  assume
A1: i-'i1>=1 or i-i1>=1;
  per cases by A1;
  suppose i-'i1>=1;
    hence thesis by XREAL_0:def 2;
  end;
  suppose i-i1>=1;
    hence thesis by XREAL_0:def 2;
  end;
end;

theorem
  n-'0=n
proof
  n-'0=n-0 by XREAL_0:def 2
    .=n;
  hence thesis;
end;

theorem
  i1<=i2 implies n-'i2<=n-'i1
proof
  assume
A1: i1<=i2;
  per cases;
  suppose
A2: i2<=n;
    then
A3: n-'i1=n-i1 by A1,XREAL_1:233,XXREAL_0:2;
    n-'i2=n-i2 by A2,XREAL_1:233;
    hence thesis by A1,A3,XREAL_1:10;
  end;
  suppose i2>n;
    then n-i2<0 by XREAL_1:49;
    hence thesis by XREAL_0:def 2;
  end;
end;

theorem Th42:
  i1<=i2 implies i1-'n<=i2-'n
proof
  assume
A1: i1<=i2;
  per cases;
  suppose i1-n>=0;
    then
A2: i1-'n=i1-n by XREAL_0:def 2;
    i1-n<=i2-n by A1,XREAL_1:9;
    hence thesis by A2,XREAL_0:def 2;
  end;
  suppose i1-n<0;
    hence thesis by XREAL_0:def 2;
  end;
end;

theorem Th43:
  i-'i1>=1 or i-i1>=1 implies i-'i1+i1=i
proof
  assume i-'i1>=1 or i-i1>=1;
  then i-'i1+i1=i-i1+i1 by Th39
    .=i;
  hence thesis;
end;

theorem
  i1<=i2 implies i1-'1<=i2
proof
  assume
A1: i1<=i2;
  per cases;
  suppose i1-1>=0;
    then i1-'1=i1-1 by XREAL_0:def 2;
    then i1-'1<=i1-1+1 by NAT_1:12;
    hence thesis by A1,XXREAL_0:2;
  end;
  suppose i1-1<0;
    hence thesis by XREAL_0:def 2;
  end;
end;

theorem Th45:
  i-'2=i-'1-'1
proof
  per cases;
  suppose
A1: i>=2;
    then 1<=i by XXREAL_0:2;
    then i-1>=0 by XREAL_1:48;
    then
A2: i-'1=i-1 by XREAL_0:def 2;
    i-1>=1+1-1 by A1,XREAL_1:9;
    then i-1-1>=1-1 by XREAL_1:9;
    then i-'1-'1 =i-2 by A2,XREAL_0:def 2;
    hence thesis by XREAL_0:def 2;
  end;
  suppose
A3: i<2;
    then i-2<2-2 by XREAL_1:9;
    then
A4: i-'2=0 by XREAL_0:def 2;
A5: i+1-1<=1+1-1 by A3,NAT_1:13;
    now per cases;
      case 1<=i;
        then i=1 by A5,XXREAL_0:1;
        then i-'1-'1=0-'1 by XREAL_1:232;
        hence thesis by A4,Th35;
      end;
      case i<1;
        then i-1<1-1 by XREAL_1:9;
        then i-'1-'1=0-'1 by XREAL_0:def 2;
        hence thesis by A4,Th35;
      end;
    end;
    hence thesis;
  end;
end;

theorem Th46:
  i1+1<=i2 implies i1-'1<i2 & i1-'2<i2 & i1<=i2
proof
  assume
A1: i1+1<=i2;
  then
A2: i1<i2 by NAT_1:13;
  i1-'1<=i1 by Th35;
  hence
A3: i1-'1<i2 by A2,XXREAL_0:2;
A4: i1-'1-'1=i1-'2 by Th45;
  i1-'1-'1<=i1-'1 by Th35;
  hence i1-'2<i2 by A3,A4,XXREAL_0:2;
  thus thesis by A1,NAT_1:13;
end;

theorem
  i1+2<=i2 or i1+1+1<=i2 implies i1+1<i2 & i1+1-'1<i2 & i1+1-'2<i2 & i1+1<=i2 &
  i1-'1+1<i2 & i1-'1+1-'1<i2 & i1<i2 & i1-'1<i2 & i1-'2<i2 & i1<=i2
proof
  assume i1+2<=i2 or i1+1+1<=i2;
  then i1+1+1<=i2;
  hence
A1: i1+1<i2 & i1+1-'1<i2 & i1+1-'2<i2 & i1+1<=i2 by Th46,NAT_1:13;
  i1-'1<=i1 by Th35;
  then i1-'1+1<=i1+1 by XREAL_1:6;
  then
A2: i1-'1+1<i2 by A1,XXREAL_0:2;
  i1-'1+1-'1<=i1-'1+1 by Th35;
  hence i1-'1+1<i2 & i1-'1+1-'1<i2 by A2,XXREAL_0:2;
  thus thesis by A1,Th46,NAT_1:13;
end;

theorem
  i1<=i2 or i1<=i2-'1 implies
  i1<i2+1 & i1<=i2+1 & i1<i2+1+1 & i1<=i2+1+1 & i1<i2+2 & i1<=i2+2
proof
  assume
A1: i1<=i2 or i1<=i2-'1;
A2: now
    assume i1<=i2;
    then
A3: i1<i2+1 by NAT_1:13;
    i2+1+1=i2+(1+1);
    hence thesis by A3,NAT_1:13;
  end;
  now
    assume
A4: i1<=i2-'1;
    i2-'1<=i2 by Th35;
    hence thesis by A2,A4,XXREAL_0:2;
  end;
  hence thesis by A1,A2;
end;

theorem
  i1<i2 or i1+1<=i2 implies i1<=i2-'1
proof
  assume
A1: i1<i2 or i1+1<=i2;
  per cases by A1;
  suppose
A2: i1<i2;
    then i1+1<=i2 by NAT_1:13;
    then
A3: i1+1-1<=i2-1 by XREAL_1:9;
    0+1<=i2 by A2,NAT_1:13;
    hence thesis by A3,XREAL_1:233;
  end;
  suppose
A4: i1+1<=i2;
    then
A5: i1+1-1<=i2-1 by XREAL_1:9;
    0+1<=i2 by A4,NAT_1:13;
    hence thesis by A5,XREAL_1:233;
  end;
end;

theorem Th50:
  i>=i1 implies i>=i1-'i2
proof
  assume
A1: i>=i1;
  i1>=i1-'i2 by Th35;
  hence thesis by A1,XXREAL_0:2;
end;

theorem
  1<=i & 1<=i1-'i implies i1-'i<i1
proof
  assume that
A1: 1<=i and
A2: 1<=i1-'i;
  i1-i=(i1-'i)+i-i by A2,Th43
    .=i1-'i;
  hence thesis by A1,XREAL_1:231;
end;

theorem Th52:
  i-'k <= j implies i <= j + k
proof
  assume
A1: i-'k <= j;
  per cases;
  suppose
A2: i >= k;
    i-'k +k <= j + k by A1,XREAL_1:6;
    hence thesis by A2,XREAL_1:235;
  end;
  suppose
A3: i <= k;
    k <= j + k by NAT_1:11;
    hence thesis by A3,XXREAL_0:2;
  end;
end;

theorem
  i <= j + k implies i-'k <= j
proof
  assume i <= j + k;
  then i -' k <= j + k -' k by Th42;
  hence thesis by Th34;
end;

theorem
  i <= j -' k & k <= j implies i + k <= j
proof
  assume that
A1: i <= j -' k and
A2: j >= k;
  i + k <= j -' k + k by A1,XREAL_1:6;
  hence thesis by A2,XREAL_1:235;
end;

theorem
  j + k <= i implies k <= i -' j
proof
  assume
A1: j + k <= i;
  per cases by A1,XXREAL_0:1;
  suppose j + k = i;
    hence thesis by Th34;
  end;
  suppose j + k < i;
    hence thesis by Th52;
  end;
end;

theorem
  k <= i & i < j implies i -' k < j -' k
proof
  assume that
A1: k <= i and
A2: i < j;
  i -' k + k = i by A1,XREAL_1:235;
  then i -' k + k < j -' k + k by A1,A2,XREAL_1:235,XXREAL_0:2;
  hence thesis by XREAL_1:6;
end;

theorem
  i < j & k < j implies i -' k < j -' k
proof
  assume that
A1: i < j and
A2: k < j;
  per cases;
  suppose k <= i;
    then
A3: i -' k = i - k by XREAL_1:233;
    j -' k = j - k by A2,XREAL_1:233;
    hence thesis by A1,A3,XREAL_1:9;
  end;
  suppose k > i;
    then i - k < 0 by XREAL_1:49;
    then
A4: i -' k = 0 by XREAL_0:def 2;
    j -' k <> 0 by A2,Th36;
    hence thesis by A4;
  end;
end;



theorem
  i <= j implies j-'(j-'i) = i
proof
  assume
A1: i <= j;
  j-'(j-'i) + (j-'i) = j by Th50,XREAL_1:235
    .= i + (j-'i) by A1,XREAL_1:235;
  hence thesis;
end;



theorem
  for n,k being Nat st n < k holds k -' (n+1) + 1 = k -' n
proof
  let n,k be Nat such that
A1: n < k;
A2: k -' n = k - n by A1,XREAL_1:233;
  n+1 <= k by A1,NAT_1:13;
  then k -' (n+1) = k - (n+1) by XREAL_1:233;
  hence thesis by A2;
end;

theorem
  for A being finite set holds A is trivial iff card A < 2
proof
  let A be finite set;
  hereby
    assume
A1:  A is trivial;
    per cases;
    suppose
      A is empty;
      hence card A < 2;
    end;
    suppose
      A is non empty;
      then ex x being set st A = {x} by A1,ZFMISC_1:131;
      then card A = 1 by CARD_1:30;
      hence card A < 2;
    end;
  end;
  assume
A2: card A < 2;
  per cases by A2,NAT_1:23;
  suppose card A = 0;
    hence thesis;
  end;
  suppose card A = 1;
    then A is 1-element by CARD_1:def 7;
   hence thesis;
  end;
end;



theorem Th61:
  for n,a,k being Integer holds (n <> 0 implies (a + n * k) div n =
  (a div n) + k) & (a + n * k) mod n = a mod n
proof
  let n,a,k be Integer;
  thus
A1: now
    assume
A2: n <> 0;
    thus (a + n * k) div n = [\ (a + n * k)/n /] by INT_1:def 9
      .= [\ (a + n * k) * n" /] by XCMPLX_0:def 9
      .= [\ a * n" + (n * n") * k /]
      .= [\ a * n" + 1 * k /] by A2,XCMPLX_0:def 7
      .= [\ a * n" /] + k by INT_1:28
      .= [\ a/n /] + k by XCMPLX_0:def 9
      .= (a div n) + k by INT_1:def 9;
  end;
  per cases;
  suppose
A3: n <> 0;
    hence (a + n * k) mod n = (a + n * k) - ((a div n) + k) * n by A1,
INT_1:def 10
      .= a - (a div n) * n
      .= a mod n by A3,INT_1:def 10;
  end;
  suppose
    n = 0;
    hence thesis;
  end;
end;

theorem Th62:
  for n being Nat st n > 0 for a being Integer holds a
  mod n >= 0 & a mod n < n
proof
  let n be Nat;
  assume
A1: n > 0;
  let a be Integer;
  now
    a div n = [\ a/n /] by INT_1:def 9;
    then a div n <= a/n by INT_1:def 6;
    then (a div n) * n <= a/n * n by XREAL_1:64;
    then (a div n) * n <= (a * n") * n by XCMPLX_0:def 9;
    then (a div n) * n <= a * (n" * n);
    then (a div n) * n <= a * 1 by A1,XCMPLX_0:def 7;
    then (a div n) * n - (a div n) * n <= a - (a div n) * n by XREAL_1:9;
    hence 0 <= a mod n by INT_1:def 10;
    assume a mod n >= n;
    then a - (a div n) * n >= n by A1,INT_1:def 10;
    then (a + -(a div n) * n) + (a div n) * n >= n + (a div n) * n by XREAL_1:6
;
    then a - n >= (n + (a div n) * n) - n by XREAL_1:9;
    then (a - n) * n" >= ((a div n) * n) * n" by XREAL_1:64;
    then (a - n) * n" >= (a div n) * (n * n");
    then a * n" - n * n" >= (a div n) * 1 by A1,XCMPLX_0:def 7;
    then a * n" - 1 >= (a div n) by A1,XCMPLX_0:def 7;
    then
A2: a/n - 1 >= (a div n) by XCMPLX_0:def 9;
    a div n = [\ a/n /] by INT_1:def 9;
    hence contradiction by A2,INT_1:def 6;
  end;
  hence thesis;
end;

theorem Th63:
  for n,a being Integer holds (0 <= a & a < n implies a mod n = a)
  & (0 > a & a >= -n implies a mod n = n + a)
proof
  let n,a be Integer;
  per cases;
  suppose
    n = 0;
    hence thesis;
  end;
  suppose
A1: n <> 0;
    hereby
      assume that
A2:   0 <= a and
A3:   a < n;
      reconsider aa = a as Element of NAT by A2,INT_1:3;
      reconsider nn = n as Element of NAT by A2,A3,INT_1:3;
      consider t being Nat such that
A4:   aa = nn * t + (aa mod nn) and
      (aa mod nn) < nn by A1,Def2;
      t = 0
      proof
        assume t <> 0;
        then t >= 1 + 0 by INT_1:7;
        then
A5:     t * n >= 1 * n by A2,A3,XREAL_1:64;
        nn * t + (aa mod nn) >= nn * t by NAT_1:11;
        hence thesis by A3,A4,A5,XXREAL_0:2;
      end;
      hence a mod n = a by A4;
    end;
    assume that
A6: 0 > a and
A7: a >= -n;
A8: n >= 0 by A6,A7;
A9: a/n - 1 < -1
    proof
      assume a/n - 1 >= -1;
      then (a/n - 1) + 1 >= -1 + 1 by XREAL_1:6;
      then a * n" >= 0 by XCMPLX_0:def 9;
      then (a * n") * n >= 0 * n by A8;
      then a * (n" * n) >= 0;
      then a * 1 >= 0 by A1,XCMPLX_0:def 7;
      hence thesis by A6;
    end;
    a * n" >= (-n) * n" by A7,A8,XREAL_1:64;
    then a /n >= -(n * n") by XCMPLX_0:def 9;
    then -1 <= a/n by A1,XCMPLX_0:def 7;
    then [\ a/n /] = -1 by A9,INT_1:def 6;
    then
A10: a div n = -1 by INT_1:def 9;
    a mod n = a - (a div n) * n by A1,INT_1:def 10;
    hence thesis by A10;
  end;
end;

theorem Th64:
  for n,a,b being Integer holds (n <> 0 & a mod n = b mod n
implies a,b are_congruent_mod n) & (a,b are_congruent_mod n implies a mod n = b
  mod n)
proof
  let n,a,b be Integer;
  hereby
    assume
A1: n <> 0;
    assume a mod n = b mod n;
    then a - (a div n) * n = b mod n by A1,INT_1:def 10;
    then a - (a div n) * n = b - (b div n) * n by A1,INT_1:def 10;
    then a - b = (-(b div n) + (a div n)) * n;
    then n divides (a-b) by INT_1:def 3;
    hence a,b are_congruent_mod n by INT_2:15;
  end;
  assume a,b are_congruent_mod n;
  then n divides (a-b) by INT_2:15;
  then consider k being Integer such that
A2: n * k = a - b by INT_1:def 3;
  a = n * k + b by A2;
  hence thesis by Th61;
end;

theorem
  for n being Nat for a being Integer holds (a mod n)
  mod n = a mod n
proof
  let n be Nat;
  let a be Integer;
  per cases;
  suppose
A1: n = 0;
    hence (a mod n) mod n = 0 by INT_1:def 10
      .= a mod n by A1,INT_1:def 10;
  end;
  suppose
    n <> 0;
    then a mod n >= 0 & a mod n < n by Th62;
    hence thesis by Th63;
  end;
end;

theorem
  for n,a,b being Integer holds (a + b) mod n = ((a mod n) + (b mod n)) mod n
proof
  let n,a,b be Integer;
  per cases;
  suppose
A1: n = 0;
    hence (a + b) mod n = 0 by INT_1:def 10
      .= ((a mod n) + (b mod n)) mod n by A1,INT_1:def 10;
  end;
  suppose
    n <> 0;
    then
    a mod n + (a div n) * n = (a - (a div n) * n) + (a div n) * n & b mod
    n + (b div n) * n = (b - (b div n) * n) + (b div n) * n by INT_1:def 10;
    then (a + b) - ((a mod n) + (b mod n)) = ((a div n) + (b div n)) * n;
    then n divides (a + b) - ((a mod n) + (b mod n)) by INT_1:def 3;
    then a+b,(a mod n)+(b mod n) are_congruent_mod n by INT_2:15;
    hence thesis by Th64;
  end;
end;

theorem
  for n,a,b being Integer holds (a * b) mod n = ((a mod n) * (b mod n)) mod n
proof
  let n,a,b be Integer;
  per cases;
  suppose
A1: n = 0;
    hence (a * b) mod n = 0 by INT_1:def 10
      .= ((a mod n) * (b mod n)) mod n by A1,INT_1:def 10;
  end;
  suppose
    n <> 0;
    then
    a mod n + (a div n) * n = (a - (a div n) * n) + (a div n) * n & b mod
    n + (b div n) * n = (b - (b div n) * n) + (b div n) * n by INT_1:def 10;
    then (a * b) - ((a mod n) * (b mod n)) = (((a mod n) * (b div n)) + (((a
    div n) * (b mod n)) + (((a div n) * n) * (b div n)))) * n;
    then n divides ((a * b) - ((a mod n) * (b mod n))) by INT_1:def 3;
    then (a*b),((a mod n)*(b mod n)) are_congruent_mod n by INT_2:15;
    hence thesis by Th64;
  end;
end;

theorem
  for a,b being Integer ex s,t being Integer st a gcd b = s * a + t * b
proof
  let a,b be Integer;
A1: for a,b being Integer st a > 0 & b > 0 holds ex s,t being Integer st (a
  gcd b) = (s * a + t * b)
  proof
    let a,b be Integer;
    assume that
A2: a > 0 and
A3: b > 0;
    reconsider a,b as Element of NAT by A2,A3,INT_1:3;
    set M = {z where z is Element of NAT : ex s,t being Integer st z = s * a +
    t * b};
    defpred P[Nat] means ($1 in M & $1 <> 0);
    a = 1 * a + 0 * b;
    then
A4: a in M;
    then
A5: ex k being Nat st P[k] by A2;
    consider g being Nat such that
A6: P[g] & for n being Nat st P[n] holds g <= n from NAT_1:sch 5(A5);
    set G = {zz where zz is Element of NAT : ex s being Element of NAT st zz =
    s * g};
    ex z being Element of NAT st z = g & ex s,t being Integer st z = s * a
    + t * b by A6;
    then consider s,t being Integer such that
A7: g = s * a + t * b;
A8: for x being set holds x in M implies x in G
    proof
      let x be set;
      assume x in M;
      then consider x9 being Element of NAT such that
A9:   x9 = x and
A10:  ex u,v being Integer st x9 = u * a + v * b;
      consider u,v being Integer such that
A11:  x = u * a + v * b by A9,A10;
      consider r being Nat such that
A12:  x9 = g * (x9 div g) + r and
A13:  r < g by A6,Def1;
A14:  r in NAT by ORDINAL1:def 12;
      r = x9 - g * (x9 div g) by A12
        .= a * (u + -(s * (x9 div g))) + b * (v + -(t * (x9 div g)))
         by A7,A9,A11;
      then r in M by A14;
      then r = 0 by A6,A13;
      hence thesis by A9,A12;
    end;
    for x being set holds x in G implies x in M
    proof
      let x be set;
      assume x in G;
      then
A15:  ex x9 being Element of NAT st x9 = x & ex u being Element of NAT st
      x9 = u * g;
      then consider u being Integer such that
A16:  x = u * g;
      x = (u * s) * a + (u * t) * b by A7,A16;
      hence thesis by A15;
    end;
    then
A17: G = M by A8,TARSKI:1;
A18: abs(b) = b by ABSVALUE:def 1;
A19: abs(a) = a by ABSVALUE:def 1;
A20: for m being Nat st m divides abs(a) & m divides abs(b) holds m divides g
    proof
      ex g9 being Element of NAT st g9 = g & ex s,t being Integer st g9 =
      s * a + t * b by A6;
      then consider s,t being Integer such that
A21:  g = s * a + t * b;
      let m be Nat;
      assume that
A22:  m divides abs(a) and
A23:  m divides abs(b);
      consider u being Nat such that
A24:  a = m * u by A19,A22,Def3;
      consider v being Nat such that
A25:  b = m * v by A18,A23,Def3;
A26:  g = m * (s * u + t * v) by A24,A25,A21;
      then s * u + t * v >= 0 by A6;
      then s * u + t * v is Element of NAT by INT_1:3;
      hence thesis by A26,Def3;
    end;
    b = 0 * a + 1 * b;
    then b in M;
    then
    ex b9 being Element of NAT st b9 = b & ex t being Element of NAT st b9
    = t * g by A17;
    then
A27: g divides abs(b) by A18,Def3;
    ex a9 being Element of NAT st a9 = a & ex s being Element of NAT st a9
    = s * g by A4,A17;
    then g divides abs(a) by A19,Def3;
    then g = abs(a) gcd abs(b) by A27,A20,Def5
      .= a gcd b by INT_2:34;
    hence thesis by A7;
  end;
  now
    per cases;
    case
A28:  a = 0 or b = 0;
A29:  for a,b being Integer holds a = 0 implies a gcd b = abs(b)
      proof
        let a,b be Integer;
        assume a = 0;
        then abs(a) = 0 by ABSVALUE:def 1;
        then
A30:    abs(b) divides abs(a) by Th6;
        a gcd b = abs(a) gcd abs(b) & for m being Nat st m divides abs(a)
        & m divides abs(b) holds m divides abs(b) by INT_2:34;
        hence thesis by A30,Def5;
      end;
      now
        per cases by A28;
        case
          a = 0;
          then
A31:      a gcd b = abs(b) by A29;
          now
            per cases;
            case
              b >= 0;
              hence a gcd b = 0 * a + 1 * b by A31,ABSVALUE:def 1;
            end;
            case
              b < 0;
              hence a gcd b = -(b * 1) by A31,ABSVALUE:def 1
                .= 0 * a + (-1) * b;
            end;
          end;
          hence thesis;
        end;
        case
          b = 0;
          then
A32:      a gcd b = abs(a) by A29;
          now
            per cases;
            case
              a >= 0;
              hence a gcd b = 1 * a + 0 * b by A32,ABSVALUE:def 1;
            end;
            case
              a < 0;
              hence a gcd b = -(a * 1) by A32,ABSVALUE:def 1
                .= 0 * b + (-1) * a;
            end;
          end;
          hence thesis;
        end;
      end;
      hence thesis;
    end;
    case
A33:  a <> 0 & b <> 0;
      now
        per cases;
        case
          a >= 0 & b >= 0;
          hence thesis by A1,A33;
        end;
        case
A34:      a < 0 & b >= 0;
          then -a > 0 by XREAL_1:58;
          then consider s,t being Integer such that
A35:      -a gcd b = s * -a + t * b by A1,A33,A34;
A36:      a gcd b = abs(a) gcd abs(b) by INT_2:34
            .= abs((-a)) gcd abs(b) by COMPLEX1:52
            .= -a gcd b by INT_2:34;
          s * -a + t * b = (-s) * a + t * b;
          hence thesis by A35,A36;
        end;
        case
A37:      a >= 0 & b < 0;
          then -b > 0 by XREAL_1:58;
          then consider s,t being Integer such that
A38:      a gcd -b = s * a + t * -b by A1,A33,A37;
A39:      a gcd b = abs(a) gcd abs(b) by INT_2:34
            .= abs(a) gcd abs((-b)) by COMPLEX1:52
            .= a gcd -b by INT_2:34;
          s * a + t * -b = s * a + (-t) * b;
          hence thesis by A38,A39;
        end;
        case
          a < 0 & b < 0;
          then -a > 0 & -b > 0 by XREAL_1:58;
          then consider s,t being Integer such that
A40:      -a gcd -b = s * -a + t * -b by A1;
A41:      a gcd b = abs(a) gcd abs(b) by INT_2:34
            .= abs(a) gcd abs((-b)) by COMPLEX1:52
            .= abs((-a)) gcd abs((-b)) by COMPLEX1:52
            .= -a gcd -b by INT_2:34;
          s * -a + t * -b = (-s) * a + (-t) * b;
          hence thesis by A40,A41;
        end;
      end;
      hence thesis;
    end;
  end;
  hence thesis;
end;



theorem
  n mod k = k - 1 implies (n+1) mod k = 0
proof
  per cases;
  suppose
    k <> 0;
    then k >= 1 by NAT_1:14;
    then reconsider K = k - 1 as Element of NAT by INT_1:3,XREAL_1:48;
A1: K + 1 = k - 0;
    assume n mod k = k-1;
    then (n+1) mod k = k mod k by A1,Th22;
    hence thesis by Th25;
  end;
  suppose
    k=0;
    hence thesis;
  end;
end;

theorem
  n mod k < k - 1 implies (n+1) mod k = (n mod k) + 1
proof
  assume n mod k < k - 1;
  then
A1: (n mod k) + 1 < k by XREAL_1:20;
  (n+1) mod k = ((n mod k)+1) mod k by Th22;
  hence thesis by A1,Th24;
end;















begin

scheme
  FuncDefUniq{C, D()->non empty set, F(Element of C())->set}: for f1,f2 being
Function of C(),D() st (for x being Element of C() holds f1.x = F(x)) & (for x
  being Element of C() holds f2.x = F(x)) holds f1 = f2 proof
  let f1,f2 be Function of C(),D() such that
A1: for x being Element of C() holds f1.x = F(x) and
A2: for x being Element of C() holds f2.x = F(x);
  now
    let x be Element of C();
    thus f1.x = F(x) by A1
      .= f2.x by A2;
  end;
  hence thesis by FUNCT_2:63;
end;

scheme
  BinOpDefuniq{A()->non empty set, O(Element of A(),Element of A())->set}: for
o1,o2 being BinOp of A() st (for a,b being Element of A() holds o1.(a,b) = O(a,
  b)) & (for a,b being Element of A() holds o2.(a,b) = O(a,b)) holds o1 = o2
proof
  let o1,o2 be BinOp of A() such that
A1: for a,b being Element of A() holds o1.(a,b) = O(a,b) and
A2: for a,b being Element of A() holds o2.(a,b) = O(a,b);
  now
    let a,b be Element of A();
    thus o1.(a,b) = O(a,b) by A1
      .= o2.(a,b) by A2;
  end;
  hence thesis by BINOP_1:2;
end;

scheme
  CFuncDefUniq{F(complex number)->set}: for f1,f2 being
Function of COMPLEX,COMPLEX st
   (for x being complex number holds f1.x = F(x)) & (for x
  being complex number holds f2.x = F(x)) holds f1 = f2 proof
  let f1,f2 be Function of COMPLEX,COMPLEX such that
A1: for x being complex number holds f1.x = F(x) and
A2: for x being complex number holds f2.x = F(x);
  now
    let x be Element of COMPLEX;
    thus f1.x = F(x) by A1
      .= f2.x by A2;
  end;
  hence thesis by FUNCT_2:63;
end;

scheme
  RFuncDefUniq{F(real number)->set}: for f1,f2 being
Function of REAL,REAL st
   (for x being real number holds f1.x = F(x)) & (for x
  being real number holds f2.x = F(x)) holds f1 = f2 proof
  let f1,f2 be Function of REAL,REAL such that
A1: for x being real number holds f1.x = F(x) and
A2: for x being real number holds f2.x = F(x);
  now
    let x be Element of REAL;
    thus f1.x = F(x) by A1
      .= f2.x by A2;
  end;
  hence thesis by FUNCT_2:63;
end;

registration
  cluster -> rational for Element of RAT;
  coherence by RAT_1:def 2;
end;

scheme
  WFuncDefUniq{F(rational number)->set}: for f1,f2 being
Function of RAT,RAT st
   (for x being rational number holds f1.x = F(x)) & (for x
  being rational number holds f2.x = F(x)) holds f1 = f2 proof
  let f1,f2 be Function of RAT,RAT such that
A1: for x being rational number holds f1.x = F(x) and
A2: for x being rational number holds f2.x = F(x);
  now
    let x be Element of RAT;
    thus f1.x = F(x) by A1
      .= f2.x by A2;
  end;
  hence thesis by FUNCT_2:63;
end;

scheme
  IFuncDefUniq{F(integer number)->set}: for f1,f2 being
Function of INT,INT st
   (for x being integer number holds f1.x = F(x)) & (for x
  being integer number holds f2.x = F(x)) holds f1 = f2 proof
  let f1,f2 be Function of INT,INT such that
A1: for x being integer number holds f1.x = F(x) and
A2: for x being integer number holds f2.x = F(x);
  now
    let x be Element of INT;
    thus f1.x = F(x) by A1
      .= f2.x by A2;
  end;
  hence thesis by FUNCT_2:63;
end;

scheme
  NFuncDefUniq{F(Nat)->set}: for f1,f2 being
Function of NAT,NAT st
   (for x being Nat holds f1.x = F(x)) & (for x
  being Nat holds f2.x = F(x)) holds f1 = f2 proof
  let f1,f2 be Function of NAT,NAT such that
A1: for x being Nat holds f1.x = F(x) and
A2: for x being Nat holds f2.x = F(x);
  now
    let x be Element of NAT;
    thus f1.x = F(x) by A1
      .= f2.x by A2;
  end;
  hence thesis by FUNCT_2:63;
end;

scheme
  CBinOpDefuniq{O(complex number,complex number)->set}:
  for o1,o2 being BinOp of COMPLEX st
  (for a,b being complex number holds o1.(a,b) = O(a,b)) &
  (for a,b being complex number holds o2.(a,b) = O(a,b)) holds o1 = o2
proof
  let o1,o2 be BinOp of COMPLEX such that
A1: for a,b being complex number holds o1.(a,b) = O(a,b) and
A2: for a,b being complex number holds o2.(a,b) = O(a,b);
  now
    let a,b be Element of COMPLEX;
    thus o1.(a,b) = O(a,b) by A1
      .= o2.(a,b) by A2;
  end;
  hence thesis by BINOP_1:2;
end;

scheme
  RBinOpDefuniq{O(real number,real number)->set}:
  for o1,o2 being BinOp of REAL st
  (for a,b being real number holds o1.(a,b) = O(a,b)) &
  (for a,b being real number holds o2.(a,b) = O(a,b)) holds o1 = o2
proof
  let o1,o2 be BinOp of REAL such that
A1: for a,b being real number holds o1.(a,b) = O(a,b) and
A2: for a,b being real number holds o2.(a,b) = O(a,b);
  now
    let a,b be Element of REAL;
    thus o1.(a,b) = O(a,b) by A1
      .= o2.(a,b) by A2;
  end;
  hence thesis by BINOP_1:2;
end;

scheme
  WBinOpDefuniq{O(rational number,rational number)->set}:
  for o1,o2 being BinOp of RAT st
  (for a,b being rational number holds o1.(a,b) = O(a,b)) &
  (for a,b being rational number holds o2.(a,b) = O(a,b)) holds o1 = o2
proof
  let o1,o2 be BinOp of RAT such that
A1: for a,b being rational number holds o1.(a,b) = O(a,b) and
A2: for a,b being rational number holds o2.(a,b) = O(a,b);
  now
    let a,b be Element of RAT;
    thus o1.(a,b) = O(a,b) by A1
      .= o2.(a,b) by A2;
  end;
  hence thesis by BINOP_1:2;
end;

scheme
  IBinOpDefuniq{O(integer number,integer number)->set}:
  for o1,o2 being BinOp of INT st
  (for a,b being integer number holds o1.(a,b) = O(a,b)) &
  (for a,b being integer number holds o2.(a,b) = O(a,b)) holds o1 = o2
proof
  let o1,o2 be BinOp of INT such that
A1: for a,b being integer number holds o1.(a,b) = O(a,b) and
A2: for a,b being integer number holds o2.(a,b) = O(a,b);
  now
    let a,b be Element of INT;
    thus o1.(a,b) = O(a,b) by A1
      .= o2.(a,b) by A2;
  end;
  hence thesis by BINOP_1:2;
end;

scheme
  NBinOpDefuniq{O(Nat,Nat)->set}:
  for o1,o2 being BinOp of NAT st
  (for a,b being Nat holds o1.(a,b) = O(a,b)) &
  (for a,b being Nat holds o2.(a,b) = O(a,b)) holds o1 = o2
proof
  let o1,o2 be BinOp of NAT such that
A1: for a,b being Nat holds o1.(a,b) = O(a,b) and
A2: for a,b being Nat holds o2.(a,b) = O(a,b);
  now
    let a,b be Element of NAT;
    thus o1.(a,b) = O(a,b) by A1
      .= o2.(a,b) by A2;
  end;
  hence thesis by BINOP_1:2;
end;

scheme
  CLambda2D{F(complex number,complex number) -> complex number}:
 ex f being Function of [:COMPLEX,COMPLEX:],COMPLEX st
  for x,y being complex number holds f.(x,y)=F(x,y)
proof
  defpred P[complex number,complex number,set] means $3 = F($1,$2);
A1: for x,y being Element of COMPLEX ex z being Element of COMPLEX st P[x,y,z]
   proof let x,y being Element of COMPLEX;
    reconsider z = F(x,y) as Element of COMPLEX by XCMPLX_0:def 2;
    take z;
    thus P[x,y,z];
   end;
  consider f being Function of [:COMPLEX,COMPLEX:],COMPLEX such that
A2:   for x,y being Element of COMPLEX
   holds P[x,y,f.(x,y)] from BINOP_1:sch 3(A1);
 take f; let x,y be complex number;
  reconsider x,y as Element of COMPLEX by XCMPLX_0:def 2;
  P[x,y,f.(x,y)] by A2;
  then f.(x,y)=F(x,y);
 hence thesis;
end;

scheme
  RLambda2D{F(real number,real number) -> real number}:
 ex f being Function of [:REAL,REAL:],REAL st
  for x,y being real number holds f.(x,y)=F(x,y)
proof
  defpred P[real number,real number,set] means $3 = F($1,$2);
A1: for x,y being Element of REAL ex z being Element of REAL st P[x,y,z]
   proof let x,y being Element of REAL;
    reconsider z = F(x,y) as Element of REAL by XREAL_0:def 1;
    take z;
    thus P[x,y,z];
   end;
  consider f being Function of [:REAL,REAL:],REAL such that
A2:   for x,y being Element of REAL holds P[x,y,f.(x,y)] from BINOP_1:sch 3(A1
);
 take f; let x,y be real number;
  reconsider x,y as Element of REAL by XREAL_0:def 1;
  P[x,y,f.(x,y)] by A2;
  then f.(x,y)=F(x,y);
 hence thesis;
end;

scheme
  WLambda2D{F(rational number,rational number) -> rational number}:
 ex f being Function of [:RAT,RAT:],RAT st
  for x,y being rational number holds f.(x,y)=F(x,y)
proof
  defpred P[rational number,rational number,set] means $3 = F($1,$2);
A1: for x,y being Element of RAT ex z being Element of RAT st P[x,y,z]
   proof let x,y being Element of RAT;
    reconsider z = F(x,y) as Element of RAT by RAT_1:def 2;
    take z;
    thus P[x,y,z];
   end;
  consider f being Function of [:RAT,RAT:],RAT such that
A2:   for x,y being Element of RAT holds P[x,y,f.(x,y)] from BINOP_1:sch 3(A1);
 take f; let x,y be rational number;
  reconsider x,y as Element of RAT by RAT_1:def 2;
  P[x,y,f.(x,y)] by A2;
  then f.(x,y)=F(x,y);
 hence thesis;
end;

scheme
  ILambda2D{F(integer number,integer number) -> integer number}:
 ex f being Function of [:INT,INT:],INT st
  for x,y being integer number holds f.(x,y)=F(x,y)
proof
  defpred P[integer number,integer number,set] means $3 = F($1,$2);
A1: for x,y being Element of INT ex z being Element of INT st P[x,y,z]
   proof let x,y being Element of INT;
    reconsider z = F(x,y) as Element of INT by INT_1:def 2;
    take z;
    thus P[x,y,z];
   end;
  consider f being Function of [:INT,INT:],INT such that
A2:   for x,y being Element of INT holds P[x,y,f.(x,y)] from BINOP_1:sch 3(A1);
 take f; let x,y be integer number;
  reconsider x,y as Element of INT by INT_1:def 2;
  P[x,y,f.(x,y)] by A2;
  then f.(x,y)=F(x,y);
 hence thesis;
end;

scheme
  NLambda2D{F(Nat,Nat) -> Nat}:
 ex f being Function of [:NAT,NAT:],NAT st
  for x,y being Nat holds f.(x,y)=F(x,y)
proof
  defpred P[Nat,Nat,set] means $3 = F($1,$2);
A1: for x,y being Element of NAT ex z being Element of NAT st P[x,y,z]
   proof let x,y being Element of NAT;
    reconsider z = F(x,y) as Element of NAT by ORDINAL1:def 12;
    take z;
    thus P[x,y,z];
   end;
  consider f being Function of [:NAT,NAT:],NAT such that
A2:   for x,y being Element of NAT holds P[x,y,f.(x,y)] from BINOP_1:sch 3(A1);
 take f; let x,y be Nat;
  reconsider x,y as Element of NAT by ORDINAL1:def 12;
  P[x,y,f.(x,y)] by A2;
  then f.(x,y)=F(x,y);
 hence thesis;
end;

scheme
  CLambdaD{F(complex number) -> complex number }: ex f
being Function of COMPLEX,COMPLEX st
 for x being complex number holds f.x = F(x)
proof
  defpred P[Element of COMPLEX,set] means $2 = F($1);
A1: for x being Element of COMPLEX ex y being Element of COMPLEX st P[x,y]
  proof let x being Element of COMPLEX;
    reconsider y = F(x) as Element of COMPLEX by XCMPLX_0:def 2;
   take y;
   thus P[x,y];
  end;
  consider f being Function of COMPLEX,COMPLEX such that
A2: for x being Element of COMPLEX holds P[x,f.x] from FUNCT_2:sch 3(A1);
 take f; let c be complex number;
  reconsider c as Element of COMPLEX by XCMPLX_0:def 2;
  P[c,f.c] by A2;
 hence thesis;
end;

scheme
  RLambdaD{F(real number) -> real number }: ex f
being Function of REAL,REAL st
 for x being real number holds f.x = F(x)
proof
  defpred P[Element of REAL,set] means $2 = F($1);
A1: for x being Element of REAL ex y being Element of REAL st P[x,y]
  proof let x being Element of REAL;
    reconsider y = F(x) as Element of REAL by XREAL_0:def 1;
   take y;
   thus P[x,y];
  end;
  consider f being Function of REAL,REAL such that
A2: for x being Element of REAL holds P[x,f.x] from FUNCT_2:sch 3(A1);
 take f; let c be real number;
  reconsider c as Element of REAL by XREAL_0:def 1;
  P[c,f.c] by A2;
 hence thesis;
end;

scheme
  WLambdaD{F(rational number) -> rational number }: ex f
being Function of RAT,RAT st
 for x being rational number holds f.x = F(x)
proof
  defpred P[Element of RAT,set] means $2 = F($1);
A1: for x being Element of RAT ex y being Element of RAT st P[x,y]
  proof let x being Element of RAT;
    reconsider y = F(x) as Element of RAT by RAT_1:def 2;
   take y;
   thus P[x,y];
  end;
  consider f being Function of RAT,RAT such that
A2: for x being Element of RAT holds P[x,f.x] from FUNCT_2:sch 3(A1);
 take f; let c be rational number;
  reconsider c as Element of RAT by RAT_1:def 2;
  P[c,f.c] by A2;
 hence thesis;
end;

scheme
  ILambdaD{F(integer number) -> integer number }: ex f
being Function of INT,INT st
 for x being integer number holds f.x = F(x)
proof
  defpred P[Element of INT,set] means $2 = F($1);
A1: for x being Element of INT ex y being Element of INT st P[x,y]
  proof let x being Element of INT;
    reconsider y = F(x) as Element of INT by INT_1:def 2;
   take y;
   thus P[x,y];
  end;
  consider f being Function of INT,INT such that
A2: for x being Element of INT holds P[x,f.x] from FUNCT_2:sch 3(A1);
 take f; let c be integer number;
  reconsider c as Element of INT by INT_1:def 2;
  P[c,f.c] by A2;
 hence thesis;
end;

scheme
  NLambdaD{F(Nat) -> Nat }: ex f
being Function of NAT,NAT st
 for x being Nat holds f.x = F(x)
proof
  defpred P[Element of NAT,set] means $2 = F($1);
A1: for x being Element of NAT ex y being Element of NAT st P[x,y]
  proof let x being Element of NAT;
    reconsider y = F(x) as Element of NAT by ORDINAL1:def 12;
   take y;
   thus P[x,y];
  end;
  consider f being Function of NAT,NAT such that
A2: for x being Element of NAT holds P[x,f.x] from FUNCT_2:sch 3(A1);
 take f; let c be Nat;
  reconsider c as Element of NAT by ORDINAL1:def 12;
  P[c,f.c] by A2;
 hence thesis;
end;

reserve c,c1,c2 for complex number;

definition
  let c1;
  redefine func -c1 -> Element of COMPLEX;
  coherence by XCMPLX_0:def 2;
  redefine func c1" -> Element of COMPLEX;
  coherence by XCMPLX_0:def 2;
  let c2;
  redefine func c1+c2 -> Element of COMPLEX;
  coherence by XCMPLX_0:def 2;
  redefine func c1-c2 -> Element of COMPLEX;
  coherence by XCMPLX_0:def 2;
  redefine func c1*c2 -> Element of COMPLEX;
  coherence by XCMPLX_0:def 2;
  redefine func c1/c2 -> Element of COMPLEX;
  coherence by XCMPLX_0:def 2;
end;

reserve r,r1,r2 for real number;

definition
  let r1;
  redefine func -r1 -> Element of REAL;
  coherence by XREAL_0:def 1;
  redefine func r1" -> Element of REAL;
  coherence by XREAL_0:def 1;
  let r2;
  redefine func r1+r2 -> Element of REAL;
  coherence by XREAL_0:def 1;
  redefine func r1-r2 -> Element of REAL;
  coherence by XREAL_0:def 1;
  redefine func r1*r2 -> Element of REAL;
  coherence by XREAL_0:def 1;
  redefine func r1/r2 -> Element of REAL;
  coherence by XREAL_0:def 1;
end;

reserve w,w1,w2 for rational number;

definition
  let w1;
  redefine func -w1 -> Element of RAT;
  coherence by RAT_1:def 2;
  redefine func w1" -> Element of RAT;
  coherence by RAT_1:def 2;
  let w2;
  redefine func w1+w2 -> Element of RAT;
  coherence by RAT_1:def 2;
  redefine func w1-w2 -> Element of RAT;
  coherence by RAT_1:def 2;
  redefine func w1*w2 -> Element of RAT;
  coherence by RAT_1:def 2;
  redefine func w1/w2 -> Element of RAT;
  coherence by RAT_1:def 2;
end;

reserve i,i1,i2 for integer number;

definition
  let i1;
  redefine func -i1 -> Element of INT;
  coherence by INT_1:def 2;
  let i2;
  redefine func i1+i2 -> Element of INT;
  coherence by INT_1:def 2;
  redefine func i1-i2 -> Element of INT;
  coherence by INT_1:def 2;
  redefine func i1*i2 -> Element of INT;
  coherence by INT_1:def 2;
end;

reserve n1,n2 for Nat;

definition
  let n1,n2;
  redefine func n1+n2 -> Element of NAT;
  coherence by ORDINAL1:def 12;
  redefine func n1*n2 -> Element of NAT;
  coherence by ORDINAL1:def 12;
end;

definition
  func compcomplex -> UnOp of COMPLEX means
  for c holds it.c = -c;
  existence from CLambdaD;
  uniqueness from CFuncDefUniq;
  func invcomplex -> UnOp of COMPLEX means
  for c holds it.c = c";
  existence from CLambdaD;
  uniqueness from CFuncDefUniq;
  func addcomplex -> BinOp of COMPLEX means
  :Def3:
  for c1,c2 holds it.(c1,c2) = c1 + c2;
  existence from CLambda2D;
  uniqueness from CBinOpDefuniq;
  func diffcomplex -> BinOp of COMPLEX means
  for c1,c2 holds it.(c1,c2) = c1 - c2;
  existence from CLambda2D;
  uniqueness from CBinOpDefuniq;
  func multcomplex -> BinOp of COMPLEX means
  :Def5:
  for c1,c2 holds it.(c1,c2) = c1 * c2;
  existence from CLambda2D;
  uniqueness from CBinOpDefuniq;
  func divcomplex -> BinOp of COMPLEX means
  for c1,c2 holds it.(c1,c2) = c1 / c2;
  existence from CLambda2D;
  uniqueness from CBinOpDefuniq;
end;

definition
  func compreal -> UnOp of REAL means
  for r holds it.r = -r;
  existence from RLambdaD;
  uniqueness from RFuncDefUniq;
  func invreal -> UnOp of REAL means
  for r holds it.r = r";
  existence from RLambdaD;
  uniqueness from RFuncDefUniq;
  func addreal -> BinOp of REAL means
  :Def9:
  for r1,r2 holds it.(r1,r2) = r1 + r2;
  existence from RLambda2D;
  uniqueness from RBinOpDefuniq;
  func diffreal -> BinOp of REAL means
  for r1,r2 holds it.(r1,r2) = r1 - r2;
  existence from RLambda2D;
  uniqueness from RBinOpDefuniq;
  func multreal -> BinOp of REAL means
  :Def11:
  for r1,r2 holds it.(r1,r2) = r1 * r2;
  existence from RLambda2D;
  uniqueness from RBinOpDefuniq;
  func divreal -> BinOp of REAL means
  for r1,r2 holds it.(r1,r2) = r1 / r2;
  existence from RLambda2D;
  uniqueness from RBinOpDefuniq;
end;

definition
  func comprat -> UnOp of RAT means
  for w holds it.w = -w;
  existence from WLambdaD;
  uniqueness from WFuncDefUniq;
  func invrat -> UnOp of RAT means
  for w holds it.w = w";
  existence from WLambdaD;
  uniqueness from WFuncDefUniq;
  func addrat -> BinOp of RAT means
  :Def15:
  for w1,w2 holds it.(w1,w2) = w1 + w2;
  existence from WLambda2D;
  uniqueness from WBinOpDefuniq;
  func diffrat -> BinOp of RAT means
  for w1,w2 holds it.(w1,w2) = w1 - w2;
  existence from WLambda2D;
  uniqueness from WBinOpDefuniq;
  func multrat -> BinOp of RAT means
  :Def17:
  for w1,w2 holds it.(w1,w2) = w1 * w2;
  existence from WLambda2D;
  uniqueness from WBinOpDefuniq;
  func divrat -> BinOp of RAT means
  for w1,w2 holds it.(w1,w2) = w1 / w2;
  existence from WLambda2D;
  uniqueness from WBinOpDefuniq;
end;

definition
  func compint -> UnOp of INT means
  for i holds it.i = -i;
  existence from ILambdaD;
  uniqueness from IFuncDefUniq;
  func addint -> BinOp of INT means
  :Def20:
  for i1,i2 holds it.(i1,i2) = i1 + i2;
  existence from ILambda2D;
  uniqueness from IBinOpDefuniq;
  func diffint -> BinOp of INT means
  for i1,i2 holds it.(i1,i2) = i1 - i2;
  existence from ILambda2D;
  uniqueness from IBinOpDefuniq;
  func multint -> BinOp of INT means
  :Def22:
  for i1,i2 holds it.(i1,i2) = i1 * i2;
  existence from ILambda2D;
  uniqueness from IBinOpDefuniq;
end;

definition
  func addnat -> BinOp of NAT means
  :Def23:
  for n1,n2 holds it.(n1,n2) = n1 + n2;
  existence from NLambda2D;
  uniqueness from NBinOpDefuniq;
  func multnat -> BinOp of NAT means
  :Def24:
  for n1,n2 holds it.(n1,n2) = n1 * n2;
  existence from NLambda2D;
  uniqueness from NBinOpDefuniq;
end;

registration
  cluster addcomplex -> commutative associative;
  coherence
  proof
    thus addcomplex is commutative
    proof
      let c1,c2 be Element of COMPLEX;
      thus addcomplex.(c1,c2) = c1 + c2 by Def3
        .= addcomplex.(c2,c1) by Def3;
    end;
    let c1,c2,c3 be Element of COMPLEX;
    thus addcomplex.(c1,addcomplex.(c2,c3)) = c1 + addcomplex.(c2,c3) by Def3
      .= c1 + (c2 + c3) by Def3
      .= c1 + c2 + c3
      .= addcomplex.(c1,c2) + c3 by Def3
      .= addcomplex.(addcomplex.(c1,c2),c3) by Def3;
  end;
  cluster multcomplex -> commutative associative;
  coherence
  proof
    thus multcomplex is commutative
    proof
      let c1,c2 be Element of COMPLEX;
      thus multcomplex.(c1,c2) = c1 * c2 by Def5
        .= multcomplex.(c2,c1) by Def5;
    end;
    let c1,c2,c3 be Element of COMPLEX;
    thus multcomplex.(c1,multcomplex.(c2,c3)) = c1 * multcomplex.(c2,c3) by
Def5
      .= c1 * (c2 * c3) by Def5
      .= c1 * c2 * c3
      .= multcomplex.(c1,c2) * c3 by Def5
      .= multcomplex.(multcomplex.(c1,c2),c3) by Def5;
  end;
  cluster addreal -> commutative associative;
  coherence
  proof
    thus addreal is commutative
    proof
      let r1,r2 be Element of REAL;
      thus addreal.(r1,r2) = r1 + r2 by Def9
        .= addreal.(r2,r1) by Def9;
    end;
    let r1,r2,r3 be Element of REAL;
    thus addreal.(r1,addreal.(r2,r3)) = r1 + addreal.(r2,r3) by Def9
      .= r1 + (r2 + r3) by Def9
      .= r1 + r2 + r3
      .= addreal.(r1,r2) + r3 by Def9
      .= addreal.(addreal.(r1,r2),r3) by Def9;
  end;
  cluster multreal -> commutative associative;
  coherence
  proof
    thus multreal is commutative
    proof
      let r1,r2 be Element of REAL;
      thus multreal.(r1,r2) = r1 * r2 by Def11
        .= multreal.(r2,r1) by Def11;
    end;
    let r1,r2,r3 be Element of REAL;
    thus multreal.(r1,multreal.(r2,r3)) = r1 * multreal.(r2,r3) by Def11
      .= r1 * (r2 * r3) by Def11
      .= r1 * r2 * r3
      .= multreal.(r1,r2) * r3 by Def11
      .= multreal.(multreal.(r1,r2),r3) by Def11;
  end;
  cluster addrat -> commutative associative;
  coherence
  proof
    thus addrat is commutative
    proof
      let w1,w2 be Element of RAT;
      thus addrat.(w1,w2) = w1 + w2 by Def15
        .= addrat.(w2,w1) by Def15;
    end;
    let w1,w2,w3 be Element of RAT;
    thus addrat.(w1,addrat.(w2,w3)) = w1 + addrat.(w2,w3) by Def15
      .= w1 + (w2 + w3) by Def15
      .= w1 + w2 + w3
      .= addrat.(w1,w2) + w3 by Def15
      .= addrat.(addrat.(w1,w2),w3) by Def15;
  end;
  cluster multrat -> commutative associative;
  coherence
  proof
    thus multrat is commutative
    proof
      let w1,w2 be Element of RAT;
      thus multrat.(w1,w2) = w1 * w2 by Def17
        .= multrat.(w2,w1) by Def17;
    end;
    let w1,w2,w3 be Element of RAT;
    thus multrat.(w1,multrat.(w2,w3)) = w1 * multrat.(w2,w3) by Def17
      .= w1 * (w2 * w3) by Def17
      .= w1 * w2 * w3
      .= multrat.(w1,w2) * w3 by Def17
      .= multrat.(multrat.(w1,w2),w3) by Def17;
  end;
  cluster addint -> commutative associative;
  coherence
  proof
    thus addint is commutative
    proof
      let i1,i2 be Element of INT;
      thus addint.(i1,i2) = i1 + i2 by Def20
        .= addint.(i2,i1) by Def20;
    end;
    let i1,i2,i3 be Element of INT;
    thus addint.(i1,addint.(i2,i3)) = i1 + addint.(i2,i3) by Def20
      .= i1 + (i2 + i3) by Def20
      .= i1 + i2 + i3
      .= addint.(i1,i2) + i3 by Def20
      .= addint.(addint.(i1,i2),i3) by Def20;
  end;
  cluster multint -> commutative associative;
  coherence
  proof
    thus multint is commutative
    proof
      let i1,i2 be Element of INT;
      thus multint.(i1,i2) = i1 * i2 by Def22
        .= multint.(i2,i1) by Def22;
    end;
    let i1,i2,i3 be Element of INT;
    thus multint.(i1,multint.(i2,i3)) = i1 * multint.(i2,i3) by Def22
      .= i1 * (i2 * i3) by Def22
      .= i1 * i2 * i3
      .= multint.(i1,i2) * i3 by Def22
      .= multint.(multint.(i1,i2),i3) by Def22;
  end;
  cluster addnat -> commutative associative;
  coherence
  proof
    thus addnat is commutative
    proof
      let n1,n2 be Element of NAT;
      thus addnat.(n1,n2) = n1 + n2 by Def23
        .= addnat.(n2,n1) by Def23;
    end;
    let n1,n2,n3 be Element of NAT;
    thus addnat.(n1,addnat.(n2,n3)) = n1 + addnat.(n2,n3) by Def23
      .= n1 + (n2 + n3) by Def23
      .= n1 + n2 + n3
      .= addnat.(n1,n2) + n3 by Def23
      .= addnat.(addnat.(n1,n2),n3) by Def23;
  end;
  cluster multnat -> commutative associative;
  coherence
  proof
    thus multnat is commutative
    proof
      let n1,n2 be Element of NAT;
      thus multnat.(n1,n2) = n1 * n2 by Def24
        .= multnat.(n2,n1) by Def24;
    end;
    let n1,n2,n3 be Element of NAT;
    thus multnat.(n1,multnat.(n2,n3)) = n1 * multnat.(n2,n3) by Def24
      .= n1 * (n2 * n3) by Def24
      .= n1 * n2 * n3
      .= multnat.(n1,n2) * n3 by Def24
      .= multnat.(multnat.(n1,n2),n3) by Def24;
  end;
end;

Lm1: 0 in NAT;

then reconsider z = 0 as Element of COMPLEX by NUMBERS:20;

Lm2: z is_a_unity_wrt addcomplex
proof
  thus for c being Element of COMPLEX holds addcomplex.(z,c) = c
  proof
    let c be Element of COMPLEX;
    thus addcomplex.(z,c) = 0 + c by Def3
      .= c;
  end;
  let c be Element of COMPLEX;
  thus addcomplex.(c,z) = c + 0 by Def3
    .= c;
end;

Lm3: 0 is_a_unity_wrt addreal
proof
  thus for r being Element of REAL holds addreal.(0,r) = r
  proof
    let r be Element of REAL;
    thus addreal.(0,r) = 0 + r by Def9
      .= r;
  end;
  let r be Element of REAL;
  thus addreal.(r,0) = r + 0 by Def9
    .= r;
end;

reconsider z = 0 as Element of RAT by Lm1,NUMBERS:18;

Lm4: z is_a_unity_wrt addrat
proof
  thus for w being Element of RAT holds addrat.(z,w) = w
  proof
    let w be Element of RAT;
    thus addrat.(z,w) = 0 + w by Def15
      .= w;
  end;
  let w be Element of RAT;
  thus addrat.(w,z) = w + 0 by Def15
    .= w;
end;

reconsider z = 0 as Element of INT by Lm1,NUMBERS:17;

Lm5: z is_a_unity_wrt addint
proof
  thus for i being Element of INT holds addint.(z,i) = i
  proof
    let i be Element of INT;
    thus addint.(z,i) = 0 + i by Def20
      .= i;
  end;
  let i be Element of INT;
  thus addint.(i,z) = i + 0 by Def20
    .= i;
end;

Lm6: 0 is_a_unity_wrt addnat
proof
  thus for n being Element of NAT holds addnat.(0,n) = n
  proof
    let n be Element of NAT;
    thus addnat.(0,n) = 0 + n by Def23
      .= n;
  end;
  let n be Element of NAT;
  thus addnat.(n,0) = n + 0 by Def23
    .= n;
end;

Lm7: 1 in NAT;

then reconsider z = 1 as Element of COMPLEX by NUMBERS:20;

Lm8: z is_a_unity_wrt multcomplex
proof
  thus for c being Element of COMPLEX holds multcomplex.(z,c) = c
  proof
    let c be Element of COMPLEX;
    thus multcomplex.(z,c) = 1 * c by Def5
      .= c;
  end;
  let c be Element of COMPLEX;
  thus multcomplex.(c,z) = c * 1 by Def5
    .= c;
end;

Lm9: 1 is_a_unity_wrt multreal
proof
  thus for r being Element of REAL holds multreal.(1,r) = r
  proof
    let r be Element of REAL;
    thus multreal.(1,r) = 1 * r by Def11
      .= r;
  end;
  let r be Element of REAL;
  thus multreal.(r,1) = r * 1 by Def11
    .= r;
end;

reconsider z = 1 as Element of RAT by Lm7,NUMBERS:18;

Lm10: z is_a_unity_wrt multrat
proof
  thus for w being Element of RAT holds multrat.(z,w) = w
  proof
    let w be Element of RAT;
    thus multrat.(z,w) = 1 * w by Def17
      .= w;
  end;
  let w be Element of RAT;
  thus multrat.(w,z) = w * 1 by Def17
    .= w;
end;

reconsider z = 1 as Element of INT by Lm7,NUMBERS:17;

Lm11: z is_a_unity_wrt multint
proof
  thus for i being Element of INT holds multint.(z,i) = i
  proof
    let i be Element of INT;
    thus multint.(z,i) = 1 * i by Def22
      .= i;
  end;
  let i be Element of INT;
  thus multint.(i,z) = i * 1 by Def22
    .= i;
end;

Lm12: 1 is_a_unity_wrt multnat
proof
  thus for n being Element of NAT holds multnat.(1,n) = n
  proof
    let n be Element of NAT;
    thus multnat.(1,n) = 1 * n by Def24
      .= n;
  end;
  let n be Element of NAT;
  thus multnat.(n,1) = n * 1 by Def24
    .= n;
end;

registration
  cluster addcomplex -> having_a_unity;
  coherence by Lm2,SETWISEO:def 2;
  cluster addreal -> having_a_unity;
  coherence by Lm3,SETWISEO:def 2;
  cluster addrat -> having_a_unity;
  coherence by Lm4,SETWISEO:def 2;
  cluster addint -> having_a_unity;
  coherence by Lm5,SETWISEO:def 2;
  cluster addnat -> having_a_unity;
  coherence by Lm6,SETWISEO:def 2;
  cluster multcomplex -> having_a_unity;
  coherence by Lm8,SETWISEO:def 2;
  cluster multreal -> having_a_unity;
  coherence by Lm9,SETWISEO:def 2;
  cluster multrat -> having_a_unity;
  coherence by Lm10,SETWISEO:def 2;
  cluster multint -> having_a_unity;
  coherence by Lm11,SETWISEO:def 2;
  cluster multnat -> having_a_unity;
  coherence by Lm12,SETWISEO:def 2;
end;

theorem
  the_unity_wrt addcomplex = 0 by Lm2,BINOP_1:def 8;

theorem
  the_unity_wrt addreal = 0 by Lm3,BINOP_1:def 8;

theorem
  the_unity_wrt addrat = 0 by Lm4,BINOP_1:def 8;

theorem
  the_unity_wrt addint = 0 by Lm5,BINOP_1:def 8;

theorem
  the_unity_wrt addnat = 0 by Lm6,BINOP_1:def 8;

theorem
  the_unity_wrt multcomplex = 1 by Lm8,BINOP_1:def 8;

theorem
  the_unity_wrt multreal = 1 by Lm9,BINOP_1:def 8;

theorem
  the_unity_wrt multrat = 1 by Lm10,BINOP_1:def 8;

theorem
  the_unity_wrt multint = 1 by Lm11,BINOP_1:def 8;

theorem
  the_unity_wrt multnat = 1 by Lm12,BINOP_1:def 8;















begin

reserve x for set,
  p,q,r,s,t,u for ext-real number,
  a for Element of ExtREAL;

scheme Conti { P,Q[set] }:
  ex s st (for r st P[r] holds r <= s) & for r st Q[r] holds s <= r
provided
A1: for r,s st P[r] & Q[s] holds r <= s
proof
  set A = { a : P[a]}, B = { a : Q[a]};
  reconsider X = A /\ REAL, Y = B /\ REAL as Subset of REAL by XBOOLE_1:17;
  per cases;
  suppose
A2: X = {};
    per cases;
    suppose
A3:   +infty in A;
      take +infty;
      thus for r st P[r] holds r <= +infty by XXREAL_0:3;
      ex a st a = +infty & P[a] by A3;
      hence thesis by A1;
    end;
    suppose
A4:   not +infty in A;
      take -infty;
      thus for r st P[r] holds r <= -infty
      proof
        let r such that
A5:     P[r];
        r in ExtREAL by XXREAL_0:def 1;
        then
A6:     r in A by A5;
        per cases by A4,A6,XXREAL_0:14;
        suppose r = -infty;
          hence thesis;
        end;
        suppose r in REAL;
          hence thesis by A2,A6,XBOOLE_0:def 4;
        end;
      end;
      thus thesis by XXREAL_0:5;
    end;
  end;
  suppose
A7: Y = {};
    per cases;
    suppose
A8:   -infty in B;
      take -infty;
      ex a st a = -infty & Q[a] by A8;
      hence for r st P[r] holds r <= -infty by A1;
      thus thesis by XXREAL_0:5;
    end;
    suppose
A9:   not -infty in B;
      take +infty;
      thus for r st P[r] holds r <= +infty by XXREAL_0:3;
      let r such that
A10:  Q[r];
      r in ExtREAL by XXREAL_0:def 1;
      then
A11:  r in B by A10;
      per cases by A9,A11,XXREAL_0:14;
      suppose r = +infty;
        hence thesis;
      end;
      suppose r in REAL;
        hence thesis by A7,A11,XBOOLE_0:def 4;
      end;
    end;
  end;
  suppose that
A12: X <> {} and
A13: Y <> {};
    for x,y being real number st x in X & y in Y holds x <= y
    proof
      let x,y be real number;
      assume x in X;
      then x in A by XBOOLE_0:def 4;
      then
A14:  ex a st ( a = x)&( P[a]);
      assume y in Y;
      then y in B by XBOOLE_0:def 4;
      then ex a st a = y & Q[a];
      hence thesis by A1,A14;
    end;
    then consider s being real number such that
A15: for x,y being real number st x in X & y in Y holds x <= s & s <= y
    by AXIOMS:1;
    reconsider s as ext-real number;
    take s;
    thus for r st P[r] holds r <= s
    proof
      let r;
      consider x being Element of REAL such that
A16:  x in Y by A13,SUBSET_1:4;
      x in B by A16,XBOOLE_0:def 4;
      then
A17:  ex a st x = a & Q[a];
      assume
A18:  P[r];
      per cases by A1,A17,A18,XXREAL_0:13;
      suppose
A19:    r in REAL;
        then reconsider r as real number;
        r is Element of ExtREAL by XXREAL_0:def 1;
        then r in A by A18;
        then r in X by A19,XBOOLE_0:def 4;
        hence thesis by A15,A16;
      end;
      suppose r = -infty;
        hence thesis by XXREAL_0:5;
      end;
    end;
    let r;
    consider x being Element of REAL such that
A20: x in X by A12,SUBSET_1:4;
    x in A by A20,XBOOLE_0:def 4;
    then
A21: ex a st x = a & P[a];
    assume
A22: Q[r];
    per cases by A1,A21,A22,XXREAL_0:10;
    suppose
A23:  r in REAL;
      then reconsider r as real number;
      r is Element of ExtREAL by XXREAL_0:def 1;
      then r in B by A22;
      then r in Y by A23,XBOOLE_0:def 4;
      hence thesis by A15,A20;
    end;
    suppose r = +infty;
      hence thesis by XXREAL_0:3;
    end;
  end;
end;

begin 

definition
  let r,s;
  func [.r,s.] equals
  { a : r <= a & a <= s };
  correctness;
  func [.r,s.[ equals
  { a : r <= a & a < s };
  correctness;
  func ].r,s.] equals
  { a : r < a & a <= s };
  correctness;
  func ].r,s.[ equals
  { a : r < a & a < s };
  correctness;
end;

theorem Th1:
  t in [.r,s.] iff r <= t & t <= s
proof
  hereby
    assume t in [.r,s.];
    then ex a st a = t & r <= a & a <= s;
    hence r <= t & t <= s;
  end;
  t is Element of ExtREAL by XXREAL_0:def 1;
  hence thesis;
end;

theorem Th2:
  t in ].r,s.] iff r < t & t <= s
proof
  hereby
    assume t in ].r,s.];
    then ex a st a = t & r < a & a <= s;
    hence r < t & t <= s;
  end;
  t is Element of ExtREAL by XXREAL_0:def 1;
  hence thesis;
end;

theorem Th3:
  t in [.r,s.[ iff r <= t & t < s
proof
  hereby
    assume t in [.r,s.[;
    then ex a st a = t & r <= a & a < s;
    hence r <= t & t < s;
  end;
  t is Element of ExtREAL by XXREAL_0:def 1;
  hence thesis;
end;

theorem Th4:
  t in ].r,s.[ iff r < t & t < s
proof
  hereby
    assume t in ].r,s.[;
    then ex a st a = t & r < a & a < s;
    hence r < t & t < s;
  end;
  t is Element of ExtREAL by XXREAL_0:def 1;
  hence thesis;
end;

registration
  let r,s; 
  cluster [.r,s.] -> ext-real-membered;
  coherence
  proof
    let x;
    assume x in [.r,s.];
    then ex a st x = a & r <= a & a <= s;
    hence thesis;
  end;
  cluster [.r,s.[ -> ext-real-membered;
  coherence
  proof
    let x;
    assume x in [.r,s.[;
    then ex a st x = a & r <= a & a < s;
    hence thesis;
  end;
  cluster ].r,s.] -> ext-real-membered;
  coherence
  proof
    let x;
    assume x in ].r,s.];
    then ex a st x = a & r < a & a <= s;
    hence thesis;
  end;
  cluster ].r,s.[ -> ext-real-membered;
  coherence
  proof
    let x;
    assume x in ].r,s.[;
    then ex a st x = a & r < a & a < s;
    hence thesis;
  end;
end;

theorem Th5:
  x in [.p,q.] implies x in ].p,q.[ or x = p or x = q
proof
  assume
A1: x in [. p, q .];
  then reconsider s = x as ext-real number;
A2: p <= s by A1,Th1;
A3: s <= q by A1,Th1;
A4: p = s or p < s by A2,XXREAL_0:1;
  s = q or s < q by A3,XXREAL_0:1;
  hence thesis by A4,Th4;
end;

theorem Th6:
  x in [.p,q.] implies x in ].p,q.] or x = p
proof
  assume
A1: x in [.p,q.];
  then reconsider s = x as ext-real number;
A2: p <= s by A1,Th1;
A3: s <= q by A1,Th1;
  p = s or p < s by A2,XXREAL_0:1;
  hence thesis by A3,Th2;
end;

theorem Th7:
  x in [.p,q.] implies x in [.p,q.[ or x = q
proof
  assume
A1: x in [.p,q.];
  then reconsider s = x as ext-real number;
A2: p <= s by A1,Th1;
  s <= q by A1,Th1;
  then q = s or s < q by XXREAL_0:1;
  hence thesis by A2,Th3;
end;

theorem Th8:
  x in [.p,q.[ implies x in ].p,q.[ or x = p
proof
  assume
A1: x in [.p,q.[;
  then reconsider s = x as ext-real number;
A2: p <= s by A1,Th3;
A3: s < q by A1,Th3;
  p = s or p < s by A2,XXREAL_0:1;
  hence thesis by A3,Th4;
end;

theorem Th9:
  x in ].p,q.] implies x in ].p,q.[ or x = q
proof
  assume
A1: x in ].p,q.];
  then reconsider s = x as ext-real number;
A2: p < s by A1,Th2;
  s <= q by A1,Th2;
  then q = s or s < q by XXREAL_0:1;
  hence thesis by A2,Th4;
end;

theorem
  x in [.p,q.[ implies x in ].p,q.] & x <> q or x = p
proof
  assume
A1: x in [.p,q.[;
  then reconsider s = x as ext-real number;
A2: p <= s by A1,Th3;
A3: s < q by A1,Th3;
  p = s or p < s by A2,XXREAL_0:1;
  hence thesis by A3,Th2;
end;

theorem
  x in ].p,q.] implies x in [.p,q.[ & x <> p or x = q
proof
  assume
A1: x in ].p,q.];
  then reconsider s = x as ext-real number;
A2: p < s by A1,Th2;
  s <= q by A1,Th2;
  then q = s or s < q by XXREAL_0:1;
  hence thesis by A2,Th3;
end;

theorem Th12:
  x in ].p,q.] implies x in [.p,q.] & x <> p
proof
  assume
A1: x in ].p,q.];
  then reconsider s = x as ext-real number;
A2: p < s by A1,Th2;
  s <= q by A1,Th2;
  hence thesis by A2,Th1;
end;

theorem Th13:
  x in [.p,q.[ implies x in [.p,q.] & x <> q
proof
  assume
A1: x in [.p,q.[;
  then reconsider s = x as ext-real number;
A2: p <= s by A1,Th3;
  s < q by A1,Th3;
  hence thesis by A2,Th1;
end;

theorem Th14:
  x in ].p,q.[ implies x in [.p,q.[ & x <> p
proof
  assume
A1: x in ].p,q.[;
  then reconsider s = x as ext-real number;
A2: p < s by A1,Th4;
  s < q by A1,Th4;
  hence thesis by A2,Th3;
end;

theorem Th15:
  x in ].p,q.[ implies x in ].p,q.] & x <> q
proof
  assume
A1: x in ].p,q.[;
  then reconsider s = x as ext-real number;
A2: p < s by A1,Th4;
  s < q by A1,Th4;
  hence thesis by A2,Th2;
end;

theorem Th16:
  x in ].p,q.[ implies x in [.p,q.] & x <> p & x <> q
proof
  assume
A1: x in ].p,q.[;
  then x in ].p,q.] by Th15;
  hence thesis by A1,Th12,Th15;
end;

theorem Th17: 
  [.r,r.] = {r}
proof
  let s;
  thus s in [.r,r.] implies s in {r}
  proof
    assume s in [.r,r.];
    then ex a st s = a & r <= a & a <= r;
    then s = r by XXREAL_0:1;
    hence thesis by TARSKI:def 1;
  end;
  assume s in {r};
  then
A1: s = r by TARSKI:def 1;
  reconsider s as Element of ExtREAL by XXREAL_0:def 1;
  s <= s;
  hence thesis by A1;
end;

theorem Th18: 
  [.r,r.[ = {}
proof
  let p;
  not ex p st p in [.r,r.[
  proof
    given p such that
A1: p in [.r,r.[;
    ex a st p = a & r <= a & a < r by A1;
    hence thesis;
  end;
  hence thesis;
end;

theorem Th19: 
  ].r,r.] = {}
proof
  let p;
  thus p in ].r,r.] implies p in {}
  proof
    assume p in ].r,r.];
    then ex a st p = a & r < a & a <= r;
    hence thesis;
  end;
  thus thesis;
end;

theorem Th20: 
  ].r,r.[ = {}
proof
  let p;
  thus p in ].r,r.[ implies p in {}
  proof
    assume p in ].r,r.[;
    then ex a st p = a & r < a & a < r;
    hence thesis;
  end;
  thus thesis;
end;

registration
  let r;
  cluster [.r,r.] -> non empty;
  coherence
  proof
    [.r,r.] = {r} by Th17;
    hence thesis;
  end;
  cluster [.r,r.[ -> empty;
  coherence by Th18;
  cluster ].r,r.] -> empty;
  coherence by Th19;
  cluster ].r,r.[ -> empty;
  coherence by Th20;
end;

theorem Th21:
  ].p,q.[ c= ].p,q.]
proof
  let s;
  assume
A1: s in ].p,q.[;
  then
A2: p < s by Th4;
  s < q by A1,Th4;
  hence thesis by A2,Th2;
end;

theorem Th22:
  ].p,q.[ c= [.p,q.[
proof
  let s;
  assume
A1: s in ].p,q.[;
  then
A2: p < s by Th4;
  s < q by A1,Th4;
  hence thesis by A2,Th3;
end;

theorem Th23:
  ].p,q.] c= [.p,q.]
proof
  let s;
  assume
A1: s in ].p,q.];
  then
A2: p < s by Th2;
  s <= q by A1,Th2;
  hence thesis by A2,Th1;
end;

theorem Th24:
  [.p,q.[ c= [.p,q.]
proof
  let s;
  assume
A1: s in [.p,q.[;
  then
A2: p <= s by Th3;
  s < q by A1,Th3;
  hence thesis by A2,Th1;
end;

theorem Th25:
  ].p,q.[ c= [.p,q.]
proof
A1: ].p,q.[ c= [.p,q.[ by Th22;
  [.p,q.[ c= [.p,q.] by Th24;
  hence thesis by A1,XBOOLE_1:1;
end;

theorem 
  p <= q implies ].q,p.] = {}
proof
  assume
A1: p <= q;
  assume ].q,p.] <> {};
  then consider r such that
A2: r in ].q,p.] by MEMBERED:8;
A3: q < r by A2,Th2;
  r <= p by A2,Th2;
  hence contradiction by A1,A3,XXREAL_0:2;
end;

theorem Th27: 
  p <= q implies [.q,p.[ = {}
proof
  assume
A1: p <= q;
  assume [.q,p.[ <> {};
  then consider r such that
A2: r in [.q,p.[ by MEMBERED:8;
A3: q <= r by A2,Th3;
  r < p by A2,Th3;
  hence contradiction by A1,A3,XXREAL_0:2;
end;

theorem Th28: 
  p <= q implies ].q,p.[ = {}
proof
  assume p <= q;
  then [.q,p.[ = {} by Th27;
  hence thesis by Th22,XBOOLE_1:3;
end;

theorem Th29:  
  p < q implies [.q,p.] = {}
proof
  assume
A1: p < q;
  assume [.q,p.] <> {};
  then consider r such that
A2: r in [.q,p.] by MEMBERED:8;
A3: q <= r by A2,Th1;
  r <= p by A2,Th1;
  hence contradiction by A1,A3,XXREAL_0:2;
end;

theorem
  r <= s implies [.r,s.] is non empty by Th1;

theorem
  p < q implies [.p,q.[ is non empty by Th3;

theorem
  p < q implies ].p,q.] is non empty by Th2;

theorem
  p < q implies ].p,q.[ is non empty
proof
  assume p < q;
  then ex s st p < s & s < q by XREAL_1:227;
  hence thesis by Th4;
end;

theorem Th34:
  p <= r & s <= q implies [.r,s.] c= [.p,q.]
proof
  assume that
A1: p <= r and
A2: s <= q;
  let t;
  assume
A3: t in [.r,s.];
  then
A4: r <= t by Th1;
A5: t <= s by A3,Th1;
A6: p <= t by A1,A4,XXREAL_0:2;
  t <= q by A2,A5,XXREAL_0:2;
  hence thesis by A6,Th1;
end;

theorem Th35:
  p <= r & s <= q implies [.r,s.[ c= [.p,q.]
proof
A1: [.r,s.[ c= [.r,s.] by Th24;
  assume that
A2: p <= r and
A3: s <= q;
  [.r,s.] c= [.p,q.] by A2,A3,Th34;
  hence thesis by A1,XBOOLE_1:1;
end;

theorem Th36:
  p <= r & s <= q implies ].r,s.] c= [.p,q.]
proof
A1: ].r,s.] c= [.r,s.] by Th23;
  assume that
A2: p <= r and
A3: s <= q;
  [.r,s.] c= [.p,q.] by A2,A3,Th34;
  hence thesis by A1,XBOOLE_1:1;
end;

theorem Th37:
  p <= r & s <= q implies ].r,s.[ c= [.p,q.]
proof
A1: ].r,s.[ c= [.r,s.] by Th25;
  assume that
A2: p <= r and
A3: s <= q;
  [.r,s.] c= [.p,q.] by A2,A3,Th34;
  hence thesis by A1,XBOOLE_1:1;
end;

theorem Th38:
  p <= r & s <= q implies [.r,s.[ c= [.p,q.[
proof
  assume that
A1: p <= r and
A2: s <= q;
  let t;
  assume
A3: t in [.r,s.[;
  then
A4: r <= t by Th3;
A5: t < s by A3,Th3;
A6: p <= t by A1,A4,XXREAL_0:2;
  t < q by A2,A5,XXREAL_0:2;
  hence thesis by A6,Th3;
end;

theorem Th39:
  p < r & s <= q implies [.r,s.] c= ].p,q.]
proof
  assume that
A1: p < r and
A2: s <= q;
  let t;
  assume
A3: t in [.r,s.];
  then
A4: r <= t by Th1;
A5: t <= s by A3,Th1;
A6: p <t by A1,A4,XXREAL_0:2;
  t <= q by A2,A5,XXREAL_0:2;
  hence thesis by A6,Th2;
end;

theorem Th40:
  p < r & s <= q implies [.r,s.[ c= ].p,q.]
proof
A1: [.r,s.[ c= [.r,s.] by Th24;
  assume that
A2: p < r and
A3: s <= q;
  [.r,s.] c= ].p,q.] by A2,A3,Th39;
  hence thesis by A1,XBOOLE_1:1;
end;

theorem Th41:
  p <= r & s <= q implies ].r,s.[ c= ].p,q.]
proof
  assume that
A1: p <= r and
A2: s <= q;
  let t;
  assume
A3: t in ].r,s.[;
  then
A4: r < t by Th4;
A5: t < s by A3,Th4;
A6: p < t by A1,A4,XXREAL_0:2;
  t < q by A2,A5,XXREAL_0:2;
  hence thesis by A6,Th2;
end;

theorem Th42:
  p <= r & s <= q implies ].r,s.] c= ].p,q.]
proof
  assume that
A1: p <= r and
A2: s <= q;
  let t;
  assume
A3: t in ].r,s.];
  then
A4: r < t by Th2;
A5: t <= s by A3,Th2;
A6: p < t by A1,A4,XXREAL_0:2;
  t <= q by A2,A5,XXREAL_0:2;
  hence thesis by A6,Th2;
end;

theorem Th43:
  p <= r & s < q implies [.r,s.] c= [.p,q.[
proof
  assume that
A1: p <= r and
A2: s < q;
  let t;
  assume
A3: t in [.r,s.];
  then
A4: r <= t by Th1;
A5: t <= s by A3,Th1;
A6: p <= t by A1,A4,XXREAL_0:2;
  t < q by A2,A5,XXREAL_0:2;
  hence thesis by A6,Th3;
end;

theorem Th44:
  p <= r & s < q implies ].r,s.] c= [.p,q.[
proof
A1: ].r,s.] c= [.r,s.] by Th23;
  assume that
A2: p <= r and
A3: s < q;
  [.r,s.] c= [.p,q.[ by A2,A3,Th43;
  hence thesis by A1,XBOOLE_1:1;
end;

theorem Th45:
  p <= r & s <= q implies ].r,s.[ c= [.p,q.[
proof
  assume that
A1: p <= r and
A2: s <= q;
  let t;
  assume
A3: t in ].r,s.[;
  then
A4: r < t by Th4;
A5: t < s by A3,Th4;
A6: p <= t by A1,A4,XXREAL_0:2;
  t < q by A2,A5,XXREAL_0:2;
  hence thesis by A6,Th3;
end;

theorem Th46:
  p <= r & s <= q implies ].r,s.[ c= ].p,q.[
proof
  assume that
A1: p <= r and
A2: s <= q;
  let t;
  assume
A3: t in ].r,s.[;
  then
A4: r < t by Th4;
A5: t < s by A3,Th4;
A6: p < t by A1,A4,XXREAL_0:2;
  t < q by A2,A5,XXREAL_0:2;
  hence thesis by A6,Th4;
end;

theorem Th47:
  p < r & s < q implies [.r,s.] c= ].p,q.[
proof
  assume that
A1: p < r and
A2: s < q;
  let t;
  assume
A3: t in [.r,s.];
  then
A4: r <= t by Th1;
A5: t <= s by A3,Th1;
A6: p < t by A1,A4,XXREAL_0:2;
  t < q by A2,A5,XXREAL_0:2;
  hence thesis by A6,Th4;
end;

theorem Th48:
  p < r & s <= q implies [.r,s.[ c= ].p,q.[
proof
  assume that
A1: p < r and
A2: s <= q;
  let t;
  assume
A3: t in [.r,s.[;
  then
A4: r <= t by Th3;
A5: t < s by A3,Th3;
A6: p < t by A1,A4,XXREAL_0:2;
  t < q by A2,A5,XXREAL_0:2;
  hence thesis by A6,Th4;
end;

theorem Th49:
  p <= r & s < q implies ].r,s.] c= ].p,q.[
proof
  assume that
A1: p <= r and
A2: s < q;
  let t;
  assume
A3: t in ].r,s.];
  then
A4: r < t by Th2;
A5: t <= s by A3,Th2;
A6: p < t by A1,A4,XXREAL_0:2;
  t < q by A2,A5,XXREAL_0:2;
  hence thesis by A6,Th4;
end;

theorem Th50:
  r <= s & [.r,s.] c= [.p,q.] implies p <= r & s <= q
proof
  assume
A1: r <= s;
  then
A2: r in [.r,s.] by Th1;
  s in [.r,s.] by A1,Th1;
  hence thesis by A2,Th1;
end;

theorem Th51:
  r < s & ].r,s.[ c= [.p,q.] implies p <= r & s <= q
proof
  assume that
A1: r < s and
A2: ].r,s.[ c= [.p,q.];
  now
    let t;
    assume that
A3: r < t and
A4: t < s;
    t in ].r,s.[ by A3,A4,Th4;
    hence p <= t by A2,Th1;
  end;
  hence p <= r by A1,XREAL_1:228;
  now
    let t;
    assume that
A5: r < t and
A6: t < s;
    t in ].r,s.[ by A5,A6,Th4;
    hence t <= q by A2,Th1;
  end;
  hence thesis by A1,XREAL_1:229;
end;

theorem Th52:
  r < s & [.r,s.[ c= [.p,q.] implies p <= r & s <= q
proof
  assume that
A1: r < s and
A2: [.r,s.[ c= [.p,q.];
  ].r,s.[ c= [.r,s.[ by Th22;
  then ].r,s.[ c= [.p,q.] by A2,XBOOLE_1:1;
  hence thesis by A1,Th51;
end;

theorem Th53:
  r < s & ].r,s.] c= [.p,q.] implies p <= r & s <= q
proof
  assume that
A1: r < s and
A2: ].r,s.] c= [.p,q.];
  ].r,s.[ c= ].r,s.] by Th21;
  then ].r,s.[ c= [.p,q.] by A2,XBOOLE_1:1;
  hence thesis by A1,Th51;
end;

theorem Th54:
  r <= s & [.r,s.] c= [.p,q.[ implies p <= r & s < q
proof
  assume that
A1: r <= s and
A2: [.r,s.] c= [.p,q.[;
  [.p,q.[ c= [.p,q.] by Th24;
  then [.r,s.] c= [.p,q.] by A2,XBOOLE_1:1;
  hence p <= r by A1,Th50;
  s in [.r,s.] by A1,Th1;
  hence thesis by A2,Th3;
end;

theorem Th55:
  r < s & [.r,s.[ c= [.p,q.[ implies p <= r & s <= q
proof
  assume that
A1: r < s and
A2: [.r,s.[ c= [.p,q.[;
  ].r,s.[ c= [.r,s.[ by Th22;
  then
A3: ].r,s.[ c= [.p,q.[ by A2,XBOOLE_1:1;
  [.p,q.[ c= [.p,q.] by Th24;
  then ].r,s.[ c= [.p,q.] by A3,XBOOLE_1:1;
  hence thesis by A1,Th51;
end;

theorem Th56:
  r < s & ].r,s.[ c= [.p,q.[ implies p <= r & s <= q
proof
  assume that
A1: r < s and
A2: ].r,s.[ c= [.p,q.[;
  [.p,q.[ c= [.p,q.] by Th24;
  then ].r,s.[ c= [.p,q.] by A2,XBOOLE_1:1;
  hence thesis by A1,Th51;
end;

theorem Th57:
  r < s & ].r,s.] c= [.p,q.[ implies p <= r & s < q
proof
  assume that
A1: r < s and
A2: ].r,s.] c= [.p,q.[;
  [.p,q.[ c= [.p,q.] by Th24;
  then ].r,s.] c= [.p,q.] by A2,XBOOLE_1:1;
  hence p <= r by A1,Th53;
  s in ].r,s.] by A1,Th2;
  hence thesis by A2,Th3;
end;

theorem Th58:
  r <= s & [.r,s.] c= ].p,q.] implies p < r & s <= q
proof
  assume that
A1: r <= s and
A2: [.r,s.] c= ].p,q.];
  ].p,q.] c= [.p,q.] by Th23;
  then
A3: [.r,s.] c= [.p,q.] by A2,XBOOLE_1:1;
  r in [.r,s.] by A1,Th1;
  hence p < r by A2,Th2;
  thus thesis by A1,A3,Th50;
end;

theorem Th59:
  r < s & ].r,s.[ c= ].p,q.] implies p <= r & s <= q
proof
  assume that
A1: r < s and
A2: ].r,s.[ c= ].p,q.];
  ].p,q.] c= [.p,q.] by Th23;
  then ].r,s.[ c= [.p,q.] by A2,XBOOLE_1:1;
  hence thesis by A1,Th51;
end;

theorem Th60:
  r < s & [.r,s.[ c= ].p,q.] implies p < r & s <= q
proof
  assume that
A1: r < s and
A2: [.r,s.[ c= ].p,q.];
  ].p,q.] c= [.p,q.] by Th23;
  then
A3: [.r,s.[ c= [.p,q.] by A2,XBOOLE_1:1;
  r in [.r,s.[ by A1,Th3;
  hence p < r by A2,Th2;
  thus thesis by A1,A3,Th52;
end;

theorem Th61:
  r < s & ].r,s.] c= ].p,q.] implies p <= r & s <= q
proof
  assume that
A1: r < s and
A2: ].r,s.] c= ].p,q.];
  ].r,s.[ c= ].r,s.] by Th21;
  then
A3: ].r,s.[ c= ].p,q.] by A2,XBOOLE_1:1;
  ].p,q.] c= [.p,q.] by Th23;
  then ].r,s.[ c= [.p,q.] by A3,XBOOLE_1:1;
  hence thesis by A1,Th51;
end;

theorem Th62:
  r <= s & [.r,s.] c= ].p,q.[ implies p < r & s < q
proof
  assume that
A1: r <= s and
A2: [.r,s.] c= ].p,q.[;
  r in [.r,s.] by A1,Th1;
  hence p < r by A2,Th4;
  s in [.r,s.] by A1,Th1;
  hence thesis by A2,Th4;
end;

theorem Th63:
  r < s & ].r,s.[ c= ].p,q.[ implies p <= r & s <= q
proof
  assume that
A1: r < s and
A2: ].r,s.[ c= ].p,q.[;
  ].p,q.[ c= [.p,q.] by Th25;
  then ].r,s.[ c= [.p,q.] by A2,XBOOLE_1:1;
  hence thesis by A1,Th51;
end;

theorem Th64:
  r < s & [.r,s.[ c= ].p,q.[ implies p < r & s <= q
proof
  assume that
A1: r < s and
A2: [.r,s.[ c= ].p,q.[;
  ].p,q.[ c= [.p,q.] by Th25;
  then
A3: [.r,s.[ c= [.p,q.] by A2,XBOOLE_1:1;
  r in [.r,s.[ by A1,Th3;
  hence p < r by A2,Th4;
  thus thesis by A1,A3,Th52;
end;

theorem Th65:
  r < s & ].r,s.] c= ].p,q.[ implies p <= r & s < q
proof
  assume that
A1: r < s and
A2: ].r,s.] c= ].p,q.[;
  ].p,q.[ c= [.p,q.] by Th25;
  then ].r,s.] c= [.p,q.] by A2,XBOOLE_1:1;
  hence p <= r by A1,Th53;
  s in ].r,s.] by A1,Th2;
  hence thesis by A2,Th4;
end;

theorem 
  p <= q & [.p,q.] = [.r,s.] implies p = r & q = s
proof
  assume that
A1: p <= q and
A2: [.p,q.] = [.r,s.];
A3: r <= p by A1,A2,Th50;
A4: q <= s by A1,A2,Th50;
  r <= q by A1,A3,XXREAL_0:2;
  then
A5: r <= s by A4,XXREAL_0:2;
  then
A6: p <= r by A2,Th50;
  s <= q by A2,A5,Th50;
  hence thesis by A3,A4,A6,XXREAL_0:1;
end;

theorem
  p < q & ].p,q.[ = ].r,s.[ implies p = r & q = s
proof
  assume that
A1: p < q and
A2: ].p,q.[ = ].r,s.[;
A3: r <= p by A1,A2,Th63;
A4: q <= s by A1,A2,Th63;
  r < q by A1,A3,XXREAL_0:2;
  then
A5: r < s by A4,XXREAL_0:2;
  then
A6: p <= r by A2,Th63;
  s <= q by A2,A5,Th63;
  hence thesis by A3,A4,A6,XXREAL_0:1;
end;

theorem
  p < q & ].p,q.] = ].r,s.] implies p = r & q = s
proof
  assume that
A1: p < q and
A2: ].p,q.] = ].r,s.];
A3: r <= p by A1,A2,Th61;
A4: q <= s by A1,A2,Th61;
  r < q by A1,A3,XXREAL_0:2;
  then
A5: r < s by A4,XXREAL_0:2;
  then
A6: p <= r by A2,Th61;
  s <= q by A2,A5,Th61;
  hence thesis by A3,A4,A6,XXREAL_0:1;
end;

theorem
  p < q & [.p,q.[ = [.r,s.[ implies p = r & q = s
proof
  assume that
A1: p < q and
A2: [.p,q.[ = [.r,s.[;
A3: r <= p by A1,A2,Th55;
A4: q <= s by A1,A2,Th55;
  r < q by A1,A3,XXREAL_0:2;
  then
A5: r < s by A4,XXREAL_0:2;
  then
A6: p <= r by A2,Th55;
  s <= q by A2,A5,Th55;
  hence thesis by A3,A4,A6,XXREAL_0:1;
end;

theorem
  r <= s implies [.r,s.] <> ].p,q.]
proof
  assume that
A1: r <= s and
A2: [.r,s.] = ].p,q.];
  now
    assume r in ].p,q.];
    then
A3: p < r by Th2;
    s <= q by A1,A2,Th58;
    then r <= q by A1,XXREAL_0:2;
    then p < q by A3,XXREAL_0:2;
    hence contradiction by A2,A3,Th53;
  end;
  hence contradiction by A1,A2,Th1;
end;

theorem
  r <= s implies [.r,s.] <> [.p,q.[
proof
  assume that
A1: r <= s and
A2: [.r,s.] = [.p,q.[;
  now
    assume s in [.p,q.[;
    then
A3: s < q by Th3;
    p <= r by A1,A2,Th54;
    then p <= s by A1,XXREAL_0:2;
    then p < q by A3,XXREAL_0:2;
    hence contradiction by A2,A3,Th52;
  end;
  hence contradiction by A1,A2,Th1;
end;

theorem
  r <= s implies [.r,s.] <> ].p,q.[
proof
  assume that
A1: r <= s and
A2: [.r,s.] = ].p,q.[;
  now
    assume s in ].p,q.[;
    then
A3: s < q by Th4;
    p <= r by A1,A2,Th62;
    then p <= s by A1,XXREAL_0:2;
    then p < q by A3,XXREAL_0:2;
    hence contradiction by A2,A3,Th51;
  end;
  hence contradiction by A1,A2,Th1;
end;

theorem
  r < s implies [.r,s.[ <> [.p,q.]
proof
  assume that
A1: r < s and
A2: [.r,s.[ = [.p,q.];
A3: not s in [.r,s.[ by Th3;
  p <= r by A1,A2,Th52;
  then
A4: p <= s by A1,XXREAL_0:2;
  s <= q by A1,A2,Th52;
  hence contradiction by A2,A3,A4,Th1;
end;

theorem
  r < s implies [.r,s.[ <> ].p,q.]
proof
  assume that
A1: r < s and
A2: [.r,s.[ = ].p,q.];
A3: not s in [.r,s.[ by Th3;
  p <= r by A1,A2,Th60;
  then
A4: p < s by A1,XXREAL_0:2;
  s <= q by A1,A2,Th60;
  hence contradiction by A2,A3,A4,Th2;
end;

theorem
  r < s implies [.r,s.[ <> ].p,q.[
proof
  assume that
A1: r < s and
A2: [.r,s.[ = ].p,q.[;
  now
    assume r in ].p,q.[;
    then
A3: p < r by Th4;
    s <= q by A1,A2,Th64;
    then r < q by A1,XXREAL_0:2;
    then p < q by A3,XXREAL_0:2;
    hence contradiction by A2,A3,Th56;
  end;
  hence contradiction by A1,A2,Th3;
end;

theorem
  r < s implies ].r,s.] <> [.p,q.]
proof
  assume that
A1: r < s and
A2: ].r,s.] = [.p,q.];
A3: not r in ].r,s.] by Th2;
A4: p <= r by A1,A2,Th53;
  s <= q by A1,A2,Th53;
  then r <= q by A1,XXREAL_0:2;
  hence contradiction by A2,A3,A4,Th1;
end;

theorem
  r < s implies ].r,s.] <> [.p,q.[
proof
  assume that
A1: r < s and
A2: ].r,s.] = [.p,q.[;
A3: not r in ].r,s.] by Th2;
A4: p <= r by A1,A2,Th57;
  s <= q by A1,A2,Th57;
  then r < q by A1,XXREAL_0:2;
  hence contradiction by A2,A3,A4,Th3;
end;

theorem
  r < s implies ].r,s.] <> ].p,q.[
proof
  assume that
A1: r < s and
A2: ].r,s.] = ].p,q.[;
  now
    assume s in ].p,q.[;
    then
A3: s < q by Th4;
    p <= r by A1,A2,Th65;
    then p <= s by A1,XXREAL_0:2;
    then p < q by A3,XXREAL_0:2;
    hence contradiction by A2,A3,Th59;
  end;
  hence contradiction by A1,A2,Th2;
end;

theorem
  r < s implies ].r,s.[ <> [.p,q.]
proof
  assume that
A1: r < s and
A2: ].r,s.[ = [.p,q.];
A3: not r in ].r,s.[ by Th4;
A4: p <= r by A1,A2,Th51;
  s <= q by A1,A2,Th51;
  then r <= q by A1,XXREAL_0:2;
  hence contradiction by A2,A3,A4,Th1;
end;

theorem
  r < s implies ].r,s.[ <> ].p,q.]
proof
  assume that
A1: r < s and
A2: ].r,s.[ = ].p,q.];
A3: not s in ].r,s.[ by Th4;
  p <= r by A1,A2,Th59;
  then
A4: p < s by A1,XXREAL_0:2;
  s <= q by A1,A2,Th59;
  hence contradiction by A2,A3,A4,Th2;
end;

theorem
  r < s implies ].r,s.[ <> [.p,q.[
proof
  assume that
A1: r < s and
A2: ].r,s.[ = [.p,q.[;
A3: not r in ].r,s.[ by Th4;
A4: p <= r by A1,A2,Th56;
  s <= q by A1,A2,Th56;
  then r < q by A1,XXREAL_0:2;
  hence contradiction by A2,A3,A4,Th3;
end;

theorem
  r <= s & [.r,s.] c< [.p,q.] implies p < r or s < q
proof
  assume
A1: r <= s;
  assume
A2: [.r,s.] c< [.p,q.];
  then
A3: [.r,s.] c= [.p,q.] by XBOOLE_0:def 8;
  then
A4: p <= r by A1,Th50;
A5: s <= q by A1,A3,Th50;
  p <> r or s <> q by A2;
  hence thesis by A4,A5,XXREAL_0:1;
end;

theorem
  r < s & ].r,s.[ c= [.p,q.] implies [.r,s.] c= [.p,q.]
proof
  assume that
A1: r < s and
A2: ].r,s.[ c= [.p,q.];
  let t;
  assume
A3: t in [.r,s.];
  per cases by A3,Th5;
  suppose t in ].r,s.[;
    hence thesis by A2;
  end;
  suppose
A4: t = r;
    then
A5: p <= t by A1,A2,Th51;
    s <= q by A1,A2,Th51;
    then t <= q by A1,A4,XXREAL_0:2;
    hence thesis by A5,Th1;
  end;
  suppose
A6: t = s;
A7: s <= q by A1,A2,Th51;
    p <= r by A1,A2,Th51;
    then p <= t by A1,A6,XXREAL_0:2;
    hence thesis by A6,A7,Th1;
  end;
end;

theorem
  r < s implies [.s,p.[ c= ].r,p.[
proof
  assume
A1: r < s;
  let t;
  assume
A2: t in [.s,p.[;
  then s <= t by Th3;
  then
A3: r < t by A1,XXREAL_0:2;
  t < p by A2,Th3;
  hence thesis by A3,Th4;
end;

theorem Th85: 
  s <= r implies [.r,s.] c= {r} & [.r,s.] c= {s}
proof
  assume
A1: s <= r;
  thus [.r,s.] c= {r}
  proof
    let t;
    assume
A2: t in [.r,s.];
    then
A3: t <= s by Th1;
A4: r <= t by A2,Th1;
    t <= r by A1,A3,XXREAL_0:2;
    then r = t by A4,XXREAL_0:1;
    hence thesis by TARSKI:def 1;
  end;
  let t;
  assume
A5: t in [.r,s.];
  then r <= t by Th1;
  then
A6: s <= t by A1,XXREAL_0:2;
  t <= s by A5,Th1;
  then s = t by A6,XXREAL_0:1;
  hence thesis by TARSKI:def 1;
end;

theorem
  ].r,s.[ misses {r,s}
proof
  let t;
  assume
A1: t in ].r,s.[;
  then
A2: r < t by Th4;
  t < s by A1,Th4;
  hence thesis by A2,TARSKI:def 2;
end;

theorem
  [.r,s.[ misses {s}
proof
  let t;
  assume t in [.r,s.[;
  then t < s by Th3;
  hence thesis by TARSKI:def 1;
end;

theorem
  ].r,s.] misses {r}
proof
  let t;
  assume t in ].r,s.];
  then r < t by Th2;
  hence thesis by TARSKI:def 1;
end;

theorem
  s <= p implies [.r,s.] misses ].p,q.[
proof
  assume
A1: s <= p;
  let t;
  assume t in [.r,s.];
  then t <= s by Th1;
  then t <= p by A1,XXREAL_0:2;
  hence thesis by Th4;
end;

theorem
  s <= p implies [.r,s.] misses ].p,q.]
proof
  assume
A1: s <= p;
  let t;
  assume t in [.r,s.];
  then t <= s by Th1;
  then t <= p by A1,XXREAL_0:2;
  hence thesis by Th2;
end;

theorem
  s <= p implies ].r,s.] misses ].p,q.[
proof
  assume
A1: s <= p;
  let t;
  assume t in ].r,s.];
  then t <= s by Th2;
  then t <= p by A1,XXREAL_0:2;
  hence thesis by Th4;
end;

theorem
  s <= p implies ].r,s.] misses ].p,q.]
proof
  assume
A1: s <= p;
  let t;
  assume t in ].r,s.];
  then t <= s by Th2;
  then t <= p by A1,XXREAL_0:2;
  hence thesis by Th2;
end;

theorem
  s <= p implies ].r,s.[ misses [.p,q.]
proof
  assume
A1: s <= p;
  let t;
  assume t in ].r,s.[;
  then t < s by Th4;
  then t < p by A1,XXREAL_0:2;
  hence thesis by Th1;
end;

theorem
  s <= p implies ].r,s.[ misses [.p,q.[
proof
  assume
A1: s <= p;
  let t;
  assume t in ].r,s.[;
  then t < s by Th4;
  then t < p by A1,XXREAL_0:2;
  hence thesis by Th3;
end;

theorem
  s <= p implies [.r,s.[ misses [.p,q.]
proof
  assume
A1: s <= p;
  let t;
  assume t in [.r,s.[;
  then t < s by Th3;
  then t < p by A1,XXREAL_0:2;
  hence thesis by Th1;
end;

theorem
  s <= p implies [.r,s.[ misses [.p,q.[
proof
  assume
A1: s <= p;
  let t;
  assume t in [.r,s.[;
  then t < s by Th3;
  then t < p by A1,XXREAL_0:2;
  hence thesis by Th3;
end;

theorem 
  r < p & r < s implies not ].r,s.[ c= [.p,q.]
proof
  assume that
A1: r < p and
A2: r < s;
  per cases;
  suppose
A3: s <= p;
    consider t such that
A4: r < t and
A5: t < s by A2,XREAL_1:227;
    take t;
    thus t in ].r,s.[ by A4,A5,Th4;
    t < p by A3,A5,XXREAL_0:2;
    hence thesis by Th1;
  end;
  suppose
A6: p <= s;
    consider t such that
A7: r < t and
A8: t < p by A1,XREAL_1:227;
    take t;
    t < s by A6,A8,XXREAL_0:2;
    hence t in ].r,s.[ by A7,Th4;
    thus thesis by A8,Th1;
  end;
end;

theorem 
  r < p & r < s implies not [.r,s.[ c= [.p,q.]
proof
  assume that
A1: r < p and
A2: r < s;
  per cases;
  suppose
A3: s <= p;
    consider t such that
A4: r < t and
A5: t < s by A2,XREAL_1:227;
    take t;
    thus t in [.r,s.[ by A4,A5,Th3;
    t < p by A3,A5,XXREAL_0:2;
    hence thesis by Th1;
  end;
  suppose
A6: p <= s;
    consider t such that
A7: r < t and
A8: t < p by A1,XREAL_1:227;
    take t;
    t < s by A6,A8,XXREAL_0:2;
    hence t in [.r,s.[ by A7,Th3;
    thus thesis by A8,Th1;
  end;
end;

theorem 
  r < p & r < s implies not ].r,s.] c= [.p,q.]
proof
  assume that
A1: r < p and
A2: r < s;
  per cases;
  suppose
A3: s <= p;
    consider t such that
A4: r < t and
A5: t < s by A2,XREAL_1:227;
    take t;
    thus t in ].r,s.] by A4,A5,Th2;
    t < p by A3,A5,XXREAL_0:2;
    hence thesis by Th1;
  end;
  suppose
A6: p <= s;
    consider t such that
A7: r < t and
A8: t < p by A1,XREAL_1:227;
    take t;
    t < s by A6,A8,XXREAL_0:2;
    hence t in ].r,s.] by A7,Th2;
    thus thesis by A8,Th1;
  end;
end;

theorem 
  r < p & r <= s implies not [.r,s.] c= [.p,q.]
proof
  assume that
A1: r < p and
A2: r <= s;
  take t=r;
  thus t in [.r,s.] by A2,Th1;
  thus thesis by A1,Th1;
end;

theorem 
  r < p & r < s implies not ].r,s.[ c= [.p,q.[
proof
  assume that
A1: r < p and
A2: r < s;
  per cases;
  suppose
A3: s <= p;
    consider t such that
A4: r < t and
A5: t < s by A2,XREAL_1:227;
    take t;
    thus t in ].r,s.[ by A4,A5,Th4;
    t < p by A3,A5,XXREAL_0:2;
    hence thesis by Th3;
  end;
  suppose
A6: p <= s;
    consider t such that
A7: r < t and
A8: t < p by A1,XREAL_1:227;
    take t;
    t < s by A6,A8,XXREAL_0:2;
    hence t in ].r,s.[ by A7,Th4;
    thus thesis by A8,Th3;
  end;
end;

theorem 
  r < p & r < s implies not ].r,s.] c= [.p,q.[
proof
  assume that
A1: r < p and
A2: r < s;
  per cases;
  suppose
A3: s <= p;
    consider t such that
A4: r < t and
A5: t < s by A2,XREAL_1:227;
    take t;
    thus t in ].r,s.] by A4,A5,Th2;
    t < p by A3,A5,XXREAL_0:2;
    hence thesis by Th3;
  end;
  suppose
A6: p <= s;
    consider t such that
A7: r < t and
A8: t < p by A1,XREAL_1:227;
    take t;
    t < s by A6,A8,XXREAL_0:2;
    hence t in ].r,s.] by A7,Th2;
    thus thesis by A8,Th3;
  end;
end;

theorem
  r < p & r < s implies not [.r,s.[ c= [.p,q.[
proof
  assume that
A1: r < p and
A2: r < s;
  per cases;
  suppose
A3: s <= p;
    consider t such that
A4: r < t and
A5: t < s by A2,XREAL_1:227;
    take t;
    thus t in [.r,s.[ by A4,A5,Th3;
    t < p by A3,A5,XXREAL_0:2;
    hence thesis by Th3;
  end;
  suppose
A6: p <= s;
    consider t such that
A7: r < t and
A8: t < p by A1,XREAL_1:227;
    take t;
    t < s by A6,A8,XXREAL_0:2;
    hence t in [.r,s.[ by A7,Th3;
    thus thesis by A8,Th3;
  end;
end;

theorem 
  r < p & r <= s implies not [.r,s.] c= [.p,q.[
proof
  assume that
A1: r < p and
A2: r <= s;
  take t=r;
  thus t in [.r,s.] by A2,Th1;
  thus thesis by A1,Th3;
end;

theorem 
  r < p & r < s implies not ].r,s.[ c= ].p,q.]
proof
  assume that
A1: r < p and
A2: r < s;
  per cases;
  suppose
A3: s <= p;
    consider t such that
A4: r < t and
A5: t < s by A2,XREAL_1:227;
    take t;
    thus t in ].r,s.[ by A4,A5,Th4;
    t < p by A3,A5,XXREAL_0:2;
    hence thesis by Th2;
  end;
  suppose
A6: p <= s;
    consider t such that
A7: r < t and
A8: t < p by A1,XREAL_1:227;
    take t;
    t < s by A6,A8,XXREAL_0:2;
    hence t in ].r,s.[ by A7,Th4;
    thus thesis by A8,Th2;
  end;
end;

theorem 
  r <= p & r < s implies not [.r,s.[ c= ].p,q.]
proof
  assume that
A1: r <= p and
A2: r < s;
  take t=r;
  thus t in [.r,s.[ by A2,Th3;
  thus thesis by A1,Th2;
end;

theorem 
  r < p & r < s implies not ].r,s.] c= ].p,q.]
proof
  assume that
A1: r < p and
A2: r < s;
  per cases;
  suppose
A3: s <= p;
    consider t such that
A4: r < t and
A5: t < s by A2,XREAL_1:227;
    take t;
    thus t in ].r,s.] by A4,A5,Th2;
    t < p by A3,A5,XXREAL_0:2;
    hence thesis by Th2;
  end;
  suppose
A6: p <= s;
    consider t such that
A7: r < t and
A8: t < p by A1,XREAL_1:227;
    take t;
    t <= s by A6,A8,XXREAL_0:2;
    hence t in ].r,s.] by A7,Th2;
    thus thesis by A8,Th2;
  end;
end;

theorem 
  r <= p & r <= s implies not [.r,s.] c= ].p,q.]
proof
  assume that
A1: r <= p and
A2: r <= s;
  take r;
  thus r in [.r,s.] by A2,Th1;
  thus thesis by A1,Th2;
end;

theorem 
  r <= p & r <= s implies not [.r,s.] c= ].p,q.[
proof
  assume that
A1: r <= p and
A2: r <= s;
  take r;
  thus r in [.r,s.] by A2,Th1;
  thus thesis by A1,Th4;
end;

theorem 
  r <= p & r < s implies not [.r,s.[ c= ].p,q.[
proof
  assume that
A1: r <= p and
A2: r < s;
  take r;
  thus r in [.r,s.[ by A2,Th3;
  thus thesis by A1,Th4;
end;

theorem 
  r < p & r < s implies not ].r,s.] c= ].p,q.[
proof
  assume that
A1: r < p and
A2: r < s;
  per cases;
  suppose
A3: s <= p;
    consider t such that
A4: r < t and
A5: t < s by A2,XREAL_1:227;
    take t;
    thus t in ].r,s.] by A4,A5,Th2;
    t < p by A3,A5,XXREAL_0:2;
    hence thesis by Th4;
  end;
  suppose
A6: p <= s;
    consider t such that
A7: r < t and
A8: t < p by A1,XREAL_1:227;
    take t;
    t <= s by A6,A8,XXREAL_0:2;
    hence t in ].r,s.] by A7,Th2;
    thus thesis by A8,Th4;
  end;
end;

theorem 
  r < p & r < s implies not ].r,s.[ c= ].p,q.[
proof
  assume that
A1: r < p and
A2: r < s;
  per cases;
  suppose
A3: s <= p;
    consider t such that
A4: r < t and
A5: t < s by A2,XREAL_1:227;
    take t;
    thus t in ].r,s.[ by A4,A5,Th4;
    t < p by A3,A5,XXREAL_0:2;
    hence thesis by Th4;
  end;
  suppose
A6: p <= s;
    consider t such that
A7: r < t and
A8: t < p by A1,XREAL_1:227;
    take t;
    t < s by A6,A8,XXREAL_0:2;
    hence t in ].r,s.[ by A7,Th4;
    thus thesis by A8,Th4;
  end;
end;

theorem 
  q < s & r < s implies not ].r,s.[ c= [.p,q.]
proof
  assume that
A1: q < s and
A2: r < s;
  per cases;
  suppose
A3: r <= q;
    consider t such that
A4: q < t and
A5: t < s by A1,XREAL_1:227;
    take t;
    r < t by A3,A4,XXREAL_0:2;
    hence t in ].r,s.[ by A5,Th4;
    thus thesis by A4,Th1;
  end;
  suppose
A6: q <= r;
    consider t such that
A7: r < t and
A8: t < s by A2,XREAL_1:227;
    take t;
    thus t in ].r,s.[ by A7,A8,Th4;
    q < t by A6,A7,XXREAL_0:2;
    hence thesis by Th1;
  end;
end;

theorem
  q < s & r < s implies not [.r,s.[ c= [.p,q.]
proof
  assume that
A1: q < s and
A2: r < s;
  per cases;
  suppose
A3: r <= q;
    consider t such that
A4: q < t and
A5: t < s by A1,XREAL_1:227;
    take t;
    r < t by A3,A4,XXREAL_0:2;
    hence t in [.r,s.[ by A5,Th3;
    thus thesis by A4,Th1;
  end;
  suppose
A6: q <= r;
    consider t such that
A7: r < t and
A8: t < s by A2,XREAL_1:227;
    take t;
    thus t in [.r,s.[ by A7,A8,Th3;
    q < t by A6,A7,XXREAL_0:2;
    hence thesis by Th1;
  end;
end;

theorem
  q < s & r < s implies not ].r,s.] c= [.p,q.]
proof
  assume that
A1: q < s and
A2: r < s;
  per cases;
  suppose
A3: r <= q;
    consider t such that
A4: q < t and
A5: t < s by A1,XREAL_1:227;
    take t;
    r < t by A3,A4,XXREAL_0:2;
    hence t in ].r,s.] by A5,Th2;
    thus thesis by A4,Th1;
  end;
  suppose
A6: q <= r;
    consider t such that
A7: r < t and
A8: t < s by A2,XREAL_1:227;
    take t;
    thus t in ].r,s.] by A7,A8,Th2;
    q < t by A6,A7,XXREAL_0:2;
    hence thesis by Th1;
  end;
end;

theorem
  q < s & r <= s implies not [.r,s.] c= [.p,q.]
proof
  assume that
A1: q < s and
A2: r <= s;
  per cases;
  suppose
A3: r <= q;
    consider t such that
A4: q < t and
A5: t < s by A1,XREAL_1:227;
    take t;
    r < t by A3,A4,XXREAL_0:2;
    hence t in [.r,s.] by A5,Th1;
    thus thesis by A4,Th1;
  end;
  suppose
A6: q < r;
    take t=r;
    thus t in [.r,s.] by A2,Th1;
    thus thesis by A6,Th1;
  end;
end;

theorem
  q < s & r < s implies not ].r,s.[ c= [.p,q.[
proof
  assume that
A1: q < s and
A2: r < s;
  per cases;
  suppose
A3: r <= q;
    consider t such that
A4: q < t and
A5: t < s by A1,XREAL_1:227;
    take t;
    r < t by A3,A4,XXREAL_0:2;
    hence t in ].r,s.[ by A5,Th4;
    thus thesis by A4,Th3;
  end;
  suppose
A6: q < r;
    consider t such that
A7: r < t and
A8: t < s by A2,XREAL_1:227;
    take t;
    thus t in ].r,s.[ by A7,A8,Th4;
    q < t by A6,A7,XXREAL_0:2;
    hence thesis by Th3;
  end;
end;

theorem
  q <= s & r < s implies not ].r,s.] c= [.p,q.[
proof
  assume that
A1: q <= s and
A2: r < s;
  take t=s;
  thus t in ].r,s.] by A2,Th2;
  thus thesis by A1,Th3;
end;

theorem
  q < s & r < s implies not [.r,s.[ c= [.p,q.[
proof
  assume that
A1: q < s and
A2: r < s;
  per cases;
  suppose
A3: r <= q;
    consider t such that
A4: q < t and
A5: t < s by A1,XREAL_1:227;
    take t;
    r < t by A3,A4,XXREAL_0:2;
    hence t in [.r,s.[ by A5,Th3;
    thus thesis by A4,Th3;
  end;
  suppose
A6: q < r;
    consider t such that
A7: r < t and
A8: t < s by A2,XREAL_1:227;
    take t;
    thus t in [.r,s.[ by A7,A8,Th3;
    q < t by A6,A7,XXREAL_0:2;
    hence thesis by Th3;
  end;
end;

theorem
  q < s & r < s implies not ].r,s.[ c= ].p,q.]
proof
  assume that
A1: q < s and
A2: r < s;
  per cases;
  suppose
A3: r <= q;
    consider t such that
A4: q < t and
A5: t < s by A1,XREAL_1:227;
    take t;
    r < t by A3,A4,XXREAL_0:2;
    hence t in ].r,s.[ by A5,Th4;
    thus thesis by A4,Th2;
  end;
  suppose
A6: q < r;
    consider t such that
A7: r < t and
A8: t < s by A2,XREAL_1:227;
    take t;
    thus t in ].r,s.[ by A7,A8,Th4;
    q < t by A6,A7,XXREAL_0:2;
    hence thesis by Th2;
  end;
end;

theorem
  q < s & r <= s implies not [.r,s.] c= ].p,q.]
proof
  assume that
A1: q < s and
A2: r <= s;
  take t=s;
  thus t in [.r,s.] by A2,Th1;
  thus thesis by A1,Th2;
end;

theorem
  q < s & r < s implies not [.r,s.[ c= ].p,q.]
proof
  assume that
A1: q < s and
A2: r < s;
  per cases;
  suppose
A3: r <= q;
    consider t such that
A4: q < t and
A5: t < s by A1,XREAL_1:227;
    take t;
    r < t by A3,A4,XXREAL_0:2;
    hence t in [.r,s.[ by A5,Th3;
    thus thesis by A4,Th2;
  end;
  suppose
A6: q < r;
    consider t such that
A7: r < t and
A8: t < s by A2,XREAL_1:227;
    take t;
    thus t in [.r,s.[ by A7,A8,Th3;
    q < t by A6,A7,XXREAL_0:2;
    hence thesis by Th2;
  end;
end;

theorem
  q < s & r < s implies not ].r,s.] c= ].p,q.]
proof
  assume that
A1: q < s and
A2: r < s;
  per cases;
  suppose
A3: r <= q;
    consider t such that
A4: q < t and
A5: t < s by A1,XREAL_1:227;
    take t;
    r < t by A3,A4,XXREAL_0:2;
    hence t in ].r,s.] by A5,Th2;
    thus thesis by A4,Th2;
  end;
  suppose
A6: q < r;
    consider t such that
A7: r < t and
A8: t < s by A2,XREAL_1:227;
    take t;
    thus t in ].r,s.] by A7,A8,Th2;
    q < t by A6,A7,XXREAL_0:2;
    hence thesis by Th2;
  end;
end;

theorem
  q <= s & r <= s implies not [.r,s.] c= ].p,q.[
proof
  assume that
A1: q <= s and
A2: r <= s;
  take t=s;
  thus t in [.r,s.] by A2,Th1;
  thus thesis by A1,Th4;
end;

theorem
  q < s & r < s implies not [.r,s.[ c= ].p,q.[
proof
  assume that
A1: q < s and
A2: r < s;
  per cases;
  suppose
A3: r <= q;
    consider t such that
A4: q < t and
A5: t < s by A1,XREAL_1:227;
    take t;
    r < t by A3,A4,XXREAL_0:2;
    hence t in [.r,s.[ by A5,Th3;
    thus thesis by A4,Th4;
  end;
  suppose
A6: q < r;
    consider t such that
A7: r < t and
A8: t < s by A2,XREAL_1:227;
    take t;
    thus t in [.r,s.[ by A7,A8,Th3;
    q < t by A6,A7,XXREAL_0:2;
    hence thesis by Th4;
  end;
end;

theorem
  q <= s & r < s implies not ].r,s.] c= ].p,q.[
proof
  assume that
A1: q <= s and
A2: r < s;
  take t=s;
  thus t in ].r,s.] by A2,Th2;
  thus thesis by A1,Th4;
end;

theorem
  q < s & r < s implies not ].r,s.[ c= ].p,q.[
proof
  assume that
A1: q < s and
A2: r < s;
  per cases;
  suppose
A3: r <= q;
    consider t such that
A4: q < t and
A5: t < s by A1,XREAL_1:227;
    take t;
    r < t by A3,A4,XXREAL_0:2;
    hence t in ].r,s.[ by A5,Th4;
    thus thesis by A4,Th4;
  end;
  suppose
A6: q < r;
    consider t such that
A7: r < t and
A8: t < s by A2,XREAL_1:227;
    take t;
    thus t in ].r,s.[ by A7,A8,Th4;
    q < t by A6,A7,XXREAL_0:2;
    hence thesis by Th4;
  end;
end;

begin 

theorem Th128:
  r <= s implies [.r,s.] = ].r,s.[ \/ {r,s}
proof
  assume
A1: r <= s;
  let t;
  thus t in [.r,s.] implies t in ].r,s.[ \/ {r,s}
  proof
    assume t in [.r,s.];
    then t in ].r,s.[ or t = r or t = s by Th5;
    then t in ].r,s.[ or t in {r,s} by TARSKI:def 2;
    hence thesis by XBOOLE_0:def 3;
  end;
  assume t in ].r,s.[ \/ {r,s};
  then t in ].r,s.[ or t in {r,s} by XBOOLE_0:def 3;
  then t in ].r,s.[ or t = r or t = s by TARSKI:def 2;
  hence thesis by A1,Th1,Th16;
end;

theorem Th129:
  r <= s implies [.r,s.] = [.r,s.[ \/ {s}
proof
  assume
A1: r <= s;
  let t;
  thus t in [.r,s.] implies t in [.r,s.[ \/ {s}
  proof
    assume t in [.r,s.];
    then t in [.r,s.[ or t = s by Th7;
    then t in [.r,s.[ or t in {s} by TARSKI:def 1;
    hence thesis by XBOOLE_0:def 3;
  end;
  assume t in [.r,s.[ \/ {s};
  then t in [.r,s.[ or t in {s} by XBOOLE_0:def 3;
  then t in [.r,s.[ or t = s by TARSKI:def 1;
  hence thesis by A1,Th1,Th13;
end;

theorem Th130:
  r <= s implies [.r,s.] = {r} \/ ].r,s.]
proof
  assume
A1: r <= s;
  let t;
  thus t in [.r,s.] implies t in {r} \/ ].r,s.]
  proof
    assume t in [.r,s.];
    then t in ].r,s.] or t = r by Th6;
    then t in ].r,s.] or t in {r} by TARSKI:def 1;
    hence thesis by XBOOLE_0:def 3;
  end;
  assume t in {r} \/].r,s.];
  then t in ].r,s.] or t in {r} by XBOOLE_0:def 3;
  then t in ].r,s.] or t = r by TARSKI:def 1;
  hence thesis by A1,Th1,Th12;
end;

theorem Th131:
  r < s implies [.r,s.[ = {r} \/ ].r,s.[
proof
  assume
A1: r < s;
  let t;
  thus t in [.r,s.[ implies t in {r} \/ ].r,s.[
  proof
    assume t in [.r,s.[;
    then t in ].r,s.[ or t = r by Th8;
    then t in ].r,s.[ or t in {r} by TARSKI:def 1;
    hence thesis by XBOOLE_0:def 3;
  end;
  assume t in {r} \/].r,s.[;
  then t in ].r,s.[ or t in {r} by XBOOLE_0:def 3;
  then t in ].r,s.[ or t = r by TARSKI:def 1;
  hence thesis by A1,Th3,Th14;
end;

theorem Th132:
  r < s implies ].r,s.] = ].r,s.[ \/ {s}
proof
  assume
A1: r < s;
  let t;
  thus t in ].r,s.] implies t in ].r,s.[ \/ {s}
  proof
    assume t in ].r,s.];
    then t in ].r,s.[ or t = s by Th9;
    then t in ].r,s.[ or t in {s} by TARSKI:def 1;
    hence thesis by XBOOLE_0:def 3;
  end;
  assume t in ].r,s.[ \/ {s};
  then t in ].r,s.[ or t in {s} by XBOOLE_0:def 3;
  then t in ].r,s.[ or t = s by TARSKI:def 1;
  hence thesis by A1,Th2,Th15;
end;

theorem
  r <= s implies [.r,s.] \ {r,s} = ].r,s.[
proof
  assume r <= s;
  then
A1: [.r,s.] = ].r,s.[ \/ {r,s} by Th128;
A2: not r in ].r,s.[ by Th4;
  not s in ].r,s.[ by Th4;
  hence thesis by A1,A2,ZFMISC_1:121;
end;

theorem
  r <= s implies [.r,s.] \ {r} = ].r,s.]
proof
  assume r <= s;
  then
A1: [.r,s.] = {r} \/ ].r,s.] by Th130;
  not r in ].r,s.] by Th2;
  hence thesis by A1,ZFMISC_1:117;
end;

theorem
  r <= s implies [.r,s.] \ {s} = [.r,s.[
proof
  assume r <= s;
  then
A1: [.r,s.] = [.r,s.[ \/ {s} by Th129;
  not s in [.r,s.[ by Th3;
  hence thesis by A1,ZFMISC_1:117;
end;

theorem
  r < s implies [.r,s.[ \ {r} = ].r,s.[
proof
  assume r < s;
  then
A1: [.r,s.[= {r} \/ ].r,s.[ by Th131;
  not r in ].r,s.[ by Th4;
  hence thesis by A1,ZFMISC_1:117;
end;

theorem
  r < s implies ].r,s.] \ {s} = ].r,s.[
proof
  assume r < s;
  then
A1: ].r,s.] = ].r,s.[ \/ {s} by Th132;
  not s in ].r,s.[ by Th4;
  hence thesis by A1,ZFMISC_1:117;
end;

theorem
  r < s & s < t implies ].r,s.] /\ [.s,t.[ = {s}
proof
  assume that
A1: r < s and
A2: s < t;
  now
    let x;
    hereby
      assume
A3:   x in ].r,s.] /\ [.s,t.[;
      then reconsider p = x as ext-real number;
A4:   p in ].r,s.] by A3,XBOOLE_0:def 4;
      p in [.s,t.[ by A3,XBOOLE_0:def 4;
      then
A5:   s <= p by Th3;
      p <= s by A4,Th2;
      hence x = s by A5,XXREAL_0:1;
    end;
    assume
A6: x = s;
A7: s in ].r,s.] by A1,Th2;
    s in [.s,t.[ by A2,Th3;
    hence x in ].r,s.] /\ [.s,t.[ by A6,A7,XBOOLE_0:def 4;
  end;
  hence thesis by TARSKI:def 1;
end;

theorem
  [.r,s.[ /\ [.p,q.[ = [.max(r,p),min(s,q).[
proof
  let t;
  thus t in [.r,s.[ /\ [.p,q.[ implies t in [.max(r,p),min(s,q).[
  proof
    assume
A1: t in [.r,s.[ /\ [.p,q.[;
    then
A2: t in [.r,s.[ by XBOOLE_0:def 4;
A3: t in [.p,q.[ by A1,XBOOLE_0:def 4;
A4: r <= t by A2,Th3;
A5: t < s by A2,Th3;
A6: p <= t by A3,Th3;
A7: t < q by A3,Th3;
A8: max(r,p) <= t by A4,A6,XXREAL_0:28;
    t < min(s,q) by A5,A7,XXREAL_0:21;
    hence thesis by A8,Th3;
  end;
  assume
A9: t in [.max(r,p),min(s,q).[;
  then
A10: max(r,p) <= t by Th3;
A11: t < min(s,q) by A9,Th3;
A12: r <= t by A10,XXREAL_0:30;
A13: p <= t by A10,XXREAL_0:30;
A14: t < s by A11,XXREAL_0:23;
A15: t < q by A11,XXREAL_0:23;
A16: t in [.r,s.[ by A12,A14,Th3;
  t in [.p,q.[ by A13,A15,Th3;
  hence thesis by A16,XBOOLE_0:def 4;
end;

theorem
  [.r,s.] /\ [.p,q.] = [.max(r,p),min(s,q).]
proof
  let t;
  thus t in [.r,s.] /\ [.p,q.] implies t in [.max(r,p),min(s,q).]
  proof
    assume
A1: t in [.r,s.] /\ [.p,q.];
    then
A2: t in [.r,s.] by XBOOLE_0:def 4;
A3: t in [.p,q.] by A1,XBOOLE_0:def 4;
A4: r <= t by A2,Th1;
A5: t <= s by A2,Th1;
A6: p <= t by A3,Th1;
A7: t <= q by A3,Th1;
A8: max(r,p) <= t by A4,A6,XXREAL_0:28;
    t <= min(s,q) by A5,A7,XXREAL_0:20;
    hence thesis by A8,Th1;
  end;
  assume
A9: t in [.max(r,p),min(s,q).];
  then
A10: max(r,p) <= t by Th1;
A11: t <= min(s,q) by A9,Th1;
A12: r <= t by A10,XXREAL_0:30;
A13: p <= t by A10,XXREAL_0:30;
A14: t <= s by A11,XXREAL_0:22;
A15: t <= q by A11,XXREAL_0:22;
A16: t in [.r,s.] by A12,A14,Th1;
  t in [.p,q.] by A13,A15,Th1;
  hence thesis by A16,XBOOLE_0:def 4;
end;

theorem
  ].r,s.] /\ ].p,q.] = ].max(r,p),min(s,q).]
proof
  let t;
  thus t in ].r,s.] /\ ].p,q.] implies t in ].max(r,p),min(s,q).]
  proof
    assume
A1: t in ].r,s.] /\ ].p,q.];
    then
A2: t in ].r,s.] by XBOOLE_0:def 4;
A3: t in ].p,q.] by A1,XBOOLE_0:def 4;
A4: r < t by A2,Th2;
A5: t <= s by A2,Th2;
A6: p < t by A3,Th2;
A7: t <= q by A3,Th2;
A8: max(r,p) < t by A4,A6,XXREAL_0:29;
    t <= min(s,q) by A5,A7,XXREAL_0:20;
    hence thesis by A8,Th2;
  end;
  assume
A9: t in ].max(r,p),min(s,q).];
  then
A10: max(r,p) < t by Th2;
A11: t <= min(s,q) by A9,Th2;
A12: r < t by A10,XXREAL_0:31;
A13: p < t by A10,XXREAL_0:31;
A14: t <= s by A11,XXREAL_0:22;
A15: t <= q by A11,XXREAL_0:22;
A16: t in ].r,s.] by A12,A14,Th2;
  t in ].p,q.] by A13,A15,Th2;
  hence thesis by A16,XBOOLE_0:def 4;
end;

theorem
  ].r,s.[ /\ ].p,q.[ = ].max(r,p),min(s,q).[
proof
  let t;
  thus t in ].r,s.[ /\ ].p,q.[ implies t in ].max(r,p),min(s,q).[
  proof
    assume
A1: t in ].r,s.[ /\ ].p,q.[;
    then
A2: t in ].r,s.[ by XBOOLE_0:def 4;
A3: t in ].p,q.[ by A1,XBOOLE_0:def 4;
A4: r < t by A2,Th4;
A5: t < s by A2,Th4;
A6: p < t by A3,Th4;
A7: t < q by A3,Th4;
A8: max(r,p) < t by A4,A6,XXREAL_0:29;
    t < min(s,q) by A5,A7,XXREAL_0:21;
    hence thesis by A8,Th4;
  end;
  assume
A9: t in ].max(r,p),min(s,q).[;
  then
A10: max(r,p) < t by Th4;
A11: t < min(s,q) by A9,Th4;
A12: r < t by A10,XXREAL_0:31;
A13: p < t by A10,XXREAL_0:31;
A14: t < s by A11,XXREAL_0:23;
A15: t < q by A11,XXREAL_0:23;
A16: t in ].r,s.[ by A12,A14,Th4;
  t in ].p,q.[ by A13,A15,Th4;
  hence thesis by A16,XBOOLE_0:def 4;
end;

theorem
  r <= p & s <= q implies [.r,s.] /\ [.p,q.] = [.p,s.]
proof
  assume that
A1: r <= p and
A2: s <= q;
  let t;
  thus t in [.r,s.] /\ [.p,q.] implies t in [.p,s.]
  proof
    assume
A3: t in [.r,s.] /\ [.p,q.];
    then
A4: t in [.r,s.] by XBOOLE_0:def 4;
A5: t in [.p,q.] by A3,XBOOLE_0:def 4;
A6: t <= s by A4,Th1;
    p <= t by A5,Th1;
    hence thesis by A6,Th1;
  end;
  assume
A7: t in [.p,s.];
  then
A8: p <= t by Th1;
A9: t <= s by A7,Th1;
A10: r <= t by A1,A8,XXREAL_0:2;
A11: t <= q by A2,A9,XXREAL_0:2;
A12: t in [.r,s.] by A9,A10,Th1;
  t in [.p,q.] by A8,A11,Th1;
  hence thesis by A12,XBOOLE_0:def 4;
end;

theorem
  r <= p & s <= q implies [.r,s.[ /\ [.p,q.] = [.p,s.[
proof
  assume that
A1: r <= p and
A2: s <= q;
  let t;
  thus t in [.r,s.[ /\ [.p,q.] implies t in [.p,s.[
  proof
    assume
A3: t in [.r,s.[ /\ [.p,q.];
    then
A4: t in [.r,s.[ by XBOOLE_0:def 4;
A5: t in [.p,q.] by A3,XBOOLE_0:def 4;
A6: t < s by A4,Th3;
    p <= t by A5,Th1;
    hence thesis by A6,Th3;
  end;
  assume
A7: t in [.p,s.[;
  then
A8: p <= t by Th3;
A9: t < s by A7,Th3;
A10: r <= t by A1,A8,XXREAL_0:2;
A11: t <= q by A2,A9,XXREAL_0:2;
A12: t in [.r,s.[ by A9,A10,Th3;
  t in [.p,q.] by A8,A11,Th1;
  hence thesis by A12,XBOOLE_0:def 4;
end;

theorem
  r >= p & s > q implies [.r,s.[ /\ [.p,q.] = [.r,q.]
proof
  assume that
A1: r >= p and
A2: s > q;
  let t;
  thus t in [.r,s.[ /\ [.p,q.] implies t in [.r,q.]
  proof
    assume
A3: t in [.r,s.[ /\ [.p,q.];
    then
A4: t in [.r,s.[ by XBOOLE_0:def 4;
A5: t in [.p,q.] by A3,XBOOLE_0:def 4;
A6: r <= t by A4,Th3;
    t <= q by A5,Th1;
    hence thesis by A6,Th1;
  end;
  assume
A7: t in [.r,q.];
  then
A8: r <= t by Th1;
A9: t <= q by A7,Th1;
  then
A10: t < s by A2,XXREAL_0:2;
A11: p <= t by A1,A8,XXREAL_0:2;
A12: t in [.r,s.[ by A8,A10,Th3;
  t in [.p,q.] by A9,A11,Th1;
  hence thesis by A12,XBOOLE_0:def 4;
end;

theorem
  r < p & s <= q implies ].r,s.] /\ [.p,q.] = [.p,s.]
proof
  assume that
A1: r < p and
A2: s <= q;
  let t;
  thus t in ].r,s.] /\ [.p,q.] implies t in [.p,s.]
  proof
    assume
A3: t in ].r,s.] /\ [.p,q.];
    then
A4: t in ].r,s.] by XBOOLE_0:def 4;
A5: t in [.p,q.] by A3,XBOOLE_0:def 4;
A6: t <= s by A4,Th2;
    p <= t by A5,Th1;
    hence thesis by A6,Th1;
  end;
  assume
A7: t in [.p,s.];
  then
A8: p <= t by Th1;
A9: t <= s by A7,Th1;
A10: r < t by A1,A8,XXREAL_0:2;
A11: t <= q by A2,A9,XXREAL_0:2;
A12: t in ].r,s.] by A9,A10,Th2;
  t in [.p,q.] by A8,A11,Th1;
  hence thesis by A12,XBOOLE_0:def 4;
end;

theorem
  r >= p & s >= q implies ].r,s.] /\ [.p,q.] = ].r,q.]
proof
  assume that
A1: r >= p and
A2: s >= q;
  let t;
  thus t in ].r,s.] /\ [.p,q.] implies t in ].r,q.]
  proof
    assume
A3: t in ].r,s.] /\ [.p,q.];
    then
A4: t in ].r,s.] by XBOOLE_0:def 4;
A5: t in [.p,q.] by A3,XBOOLE_0:def 4;
A6: r < t by A4,Th2;
    t <= q by A5,Th1;
    hence thesis by A6,Th2;
  end;
  assume
A7: t in ].r,q.];
  then
A8: r < t by Th2;
A9: t <= q by A7,Th2;
  then
A10: t <= s by A2,XXREAL_0:2;
A11: p <= t by A1,A8,XXREAL_0:2;
A12: t in ].r,s.] by A8,A10,Th2;
  t in [.p,q.] by A9,A11,Th1;
  hence thesis by A12,XBOOLE_0:def 4;
end;

theorem
  r < p & s <= q implies ].r,s.[ /\ [.p,q.] = [.p,s.[
proof
  assume that
A1: r < p and
A2: s <= q;
  let t;
  thus t in ].r,s.[ /\ [.p,q.] implies t in [.p,s.[
  proof
    assume
A3: t in ].r,s.[ /\ [.p,q.];
    then
A4: t in ].r,s.[ by XBOOLE_0:def 4;
A5: t in [.p,q.] by A3,XBOOLE_0:def 4;
A6: t < s by A4,Th4;
    p <= t by A5,Th1;
    hence thesis by A6,Th3;
  end;
  assume
A7: t in [.p,s.[;
  then
A8: p <= t by Th3;
A9: t < s by A7,Th3;
A10: r < t by A1,A8,XXREAL_0:2;
A11: t <= q by A2,A9,XXREAL_0:2;
A12: t in ].r,s.[ by A9,A10,Th4;
  t in [.p,q.] by A8,A11,Th1;
  hence thesis by A12,XBOOLE_0:def 4;
end;

theorem
  r >= p & s > q implies ].r,s.[ /\ [.p,q.] = ].r,q.]
proof
  assume that
A1: r >= p and
A2: s > q;
  let t;
  thus t in ].r,s.[ /\ [.p,q.] implies t in ].r,q.]
  proof
    assume
A3: t in ].r,s.[ /\ [.p,q.];
    then
A4: t in ].r,s.[ by XBOOLE_0:def 4;
A5: t in [.p,q.] by A3,XBOOLE_0:def 4;
A6: r < t by A4,Th4;
    t <= q by A5,Th1;
    hence thesis by A6,Th2;
  end;
  assume
A7: t in ].r,q.];
  then
A8: r < t by Th2;
A9: t <= q by A7,Th2;
  then
A10: t < s by A2,XXREAL_0:2;
A11: p <= t by A1,A8,XXREAL_0:2;
A12: t in ].r,s.[ by A8,A10,Th4;
  t in [.p,q.] by A9,A11,Th1;
  hence thesis by A12,XBOOLE_0:def 4;
end;

theorem
  r <= p & s <= q implies [.r,s.[ /\ [.p,q.[ = [.p,s.[
proof
  assume that
A1: r <= p and
A2: s <= q;
  let t;
  thus t in [.r,s.[ /\ [.p,q.[ implies t in [.p,s.[
  proof
    assume
A3: t in [.r,s.[ /\ [.p,q.[;
    then
A4: t in [.r,s.[ by XBOOLE_0:def 4;
A5: t in [.p,q.[ by A3,XBOOLE_0:def 4;
A6: t < s by A4,Th3;
    p <= t by A5,Th3;
    hence thesis by A6,Th3;
  end;
  assume
A7: t in [.p,s.[;
  then
A8: p <= t by Th3;
A9: t < s by A7,Th3;
A10: r <= t by A1,A8,XXREAL_0:2;
A11: t < q by A2,A9,XXREAL_0:2;
A12: t in [.r,s.[ by A9,A10,Th3;
  t in [.p,q.[ by A8,A11,Th3;
  hence thesis by A12,XBOOLE_0:def 4;
end;

theorem
  r >= p & s >= q implies [.r,s.[ /\ [.p,q.[ = [.r,q.[
proof
  assume that
A1: r >= p and
A2: s >= q;
  let t;
  thus t in [.r,s.[ /\ [.p,q.[ implies t in [.r,q.[
  proof
    assume
A3: t in [.r,s.[ /\ [.p,q.[;
    then
A4: t in [.r,s.[ by XBOOLE_0:def 4;
A5: t in [.p,q.[ by A3,XBOOLE_0:def 4;
A6: r <= t by A4,Th3;
    t < q by A5,Th3;
    hence thesis by A6,Th3;
  end;
  assume
A7: t in [.r,q.[;
  then
A8: r <= t by Th3;
A9: t < q by A7,Th3;
  then
A10: t < s by A2,XXREAL_0:2;
A11: p <= t by A1,A8,XXREAL_0:2;
A12: t in [.r,s.[ by A8,A10,Th3;
  t in [.p,q.[ by A9,A11,Th3;
  hence thesis by A12,XBOOLE_0:def 4;
end;

theorem Th152:
  r < p & s < q implies ].r,s.] /\ [.p,q.[ = [.p,s.]
proof
  assume that
A1: r < p and
A2: s < q;
  let t;
  thus t in ].r,s.] /\ [.p,q.[ implies t in [.p,s.]
  proof
    assume
A3: t in ].r,s.] /\ [.p,q.[;
    then
A4: t in ].r,s.] by XBOOLE_0:def 4;
A5: t in [.p,q.[ by A3,XBOOLE_0:def 4;
A6: t <= s by A4,Th2;
    p <= t by A5,Th3;
    hence thesis by A6,Th1;
  end;
  assume
A7: t in [.p,s.];
  then
A8: p <= t by Th1;
A9: t <= s by A7,Th1;
A10: r < t by A1,A8,XXREAL_0:2;
A11: t < q by A2,A9,XXREAL_0:2;
A12: t in ].r,s.] by A9,A10,Th2;
  t in [.p,q.[ by A8,A11,Th3;
  hence thesis by A12,XBOOLE_0:def 4;
end;

theorem
  r >= p & s >= q implies ].r,s.] /\ [.p,q.[ = ].r,q.[
proof
  assume that
A1: r >= p and
A2: s >= q;
  let t;
  thus t in ].r,s.] /\ [.p,q.[ implies t in ].r,q.[
  proof
    assume
A3: t in ].r,s.] /\ [.p,q.[;
    then
A4: t in ].r,s.] by XBOOLE_0:def 4;
A5: t in [.p,q.[ by A3,XBOOLE_0:def 4;
A6: r < t by A4,Th2;
    t < q by A5,Th3;
    hence thesis by A6,Th4;
  end;
  assume
A7: t in ].r,q.[;
  then
A8: r < t by Th4;
A9: t < q by A7,Th4;
  then
A10: t <= s by A2,XXREAL_0:2;
A11: p <= t by A1,A8,XXREAL_0:2;
A12: t in ].r,s.] by A8,A10,Th2;
  t in [.p,q.[ by A9,A11,Th3;
  hence thesis by A12,XBOOLE_0:def 4;
end;

theorem Th154:
  r < p & s <= q implies ].r,s.[ /\ [.p,q.[ = [.p,s.[
proof
  assume that
A1: r < p and
A2: s <= q;
  let t;
  thus t in ].r,s.[ /\ [.p,q.[ implies t in [.p,s.[
  proof
    assume
A3: t in ].r,s.[ /\ [.p,q.[;
    then
A4: t in ].r,s.[ by XBOOLE_0:def 4;
A5: t in [.p,q.[ by A3,XBOOLE_0:def 4;
A6: t < s by A4,Th4;
    p <= t by A5,Th3;
    hence thesis by A6,Th3;
  end;
  assume
A7: t in [.p,s.[;
  then
A8: p <= t by Th3;
A9: t < s by A7,Th3;
A10: r < t by A1,A8,XXREAL_0:2;
A11: t < q by A2,A9,XXREAL_0:2;
A12: t in ].r,s.[ by A9,A10,Th4;
  t in [.p,q.[ by A8,A11,Th3;
  hence thesis by A12,XBOOLE_0:def 4;
end;

theorem
  r >= p & s >= q implies ].r,s.[ /\ [.p,q.[ = ].r,q.[
proof
  assume that
A1: r >= p and
A2: s >= q;
  let t;
  thus t in ].r,s.[ /\ [.p,q.[ implies t in ].r,q.[
  proof
    assume
A3: t in ].r,s.[ /\ [.p,q.[;
    then
A4: t in ].r,s.[ by XBOOLE_0:def 4;
A5: t in [.p,q.[ by A3,XBOOLE_0:def 4;
A6: r < t by A4,Th4;
    t < q by A5,Th3;
    hence thesis by A6,Th4;
  end;
  assume
A7: t in ].r,q.[;
  then
A8: r < t by Th4;
A9: t < q by A7,Th4;
  then
A10: t < s by A2,XXREAL_0:2;
A11: p <= t by A1,A8,XXREAL_0:2;
A12: t in ].r,s.[ by A8,A10,Th4;
  t in [.p,q.[ by A9,A11,Th3;
  hence thesis by A12,XBOOLE_0:def 4;
end;

theorem
  r <= p & s <= q implies ].r,s.] /\ ].p,q.] = ].p,s.]
proof
  assume that
A1: r <= p and
A2: s <= q;
  let t;
  thus t in ].r,s.] /\ ].p,q.] implies t in ].p,s.]
  proof
    assume
A3: t in ].r,s.] /\ ].p,q.];
    then
A4: t in ].r,s.] by XBOOLE_0:def 4;
A5: t in ].p,q.] by A3,XBOOLE_0:def 4;
A6: t <= s by A4,Th2;
    p < t by A5,Th2;
    hence thesis by A6,Th2;
  end;
  assume
A7: t in ].p,s.];
  then
A8: p < t by Th2;
A9: t <= s by A7,Th2;
A10: r < t by A1,A8,XXREAL_0:2;
A11: t <= q by A2,A9,XXREAL_0:2;
A12: t in ].r,s.] by A9,A10,Th2;
  t in ].p,q.] by A8,A11,Th2;
  hence thesis by A12,XBOOLE_0:def 4;
end;

theorem
  r >= p & s >= q implies ].r,s.] /\ ].p,q.] = ].r,q.]
proof
  assume that
A1: r >= p and
A2: s >= q;
  let t;
  thus t in ].r,s.] /\ ].p,q.] implies t in ].r,q.]
  proof
    assume
A3: t in ].r,s.] /\ ].p,q.];
    then
A4: t in ].r,s.] by XBOOLE_0:def 4;
A5: t in ].p,q.] by A3,XBOOLE_0:def 4;
A6: r < t by A4,Th2;
    t <= q by A5,Th2;
    hence thesis by A6,Th2;
  end;
  assume
A7: t in ].r,q.];
  then
A8: r < t by Th2;
A9: t <= q by A7,Th2;
  then
A10: t <= s by A2,XXREAL_0:2;
A11: p < t by A1,A8,XXREAL_0:2;
A12: t in ].r,s.] by A8,A10,Th2;
  t in ].p,q.] by A9,A11,Th2;
  hence thesis by A12,XBOOLE_0:def 4;
end;

theorem
  r <= p & s <= q implies ].r,s.[ /\ ].p,q.] = ].p,s.[
proof
  assume that
A1: r <= p and
A2: s <= q;
  let t;
  thus t in ].r,s.[ /\ ].p,q.] implies t in ].p,s.[
  proof
    assume
A3: t in ].r,s.[ /\ ].p,q.];
    then
A4: t in ].r,s.[ by XBOOLE_0:def 4;
A5: t in ].p,q.] by A3,XBOOLE_0:def 4;
A6: t < s by A4,Th4;
    p < t by A5,Th2;
    hence thesis by A6,Th4;
  end;
  assume
A7: t in ].p,s.[;
  then
A8: p < t by Th4;
A9: t < s by A7,Th4;
A10: r < t by A1,A8,XXREAL_0:2;
A11: t <= q by A2,A9,XXREAL_0:2;
A12: t in ].r,s.[ by A9,A10,Th4;
  t in ].p,q.] by A8,A11,Th2;
  hence thesis by A12,XBOOLE_0:def 4;
end;

theorem Th159:
  r >= p & s > q implies ].r,s.[ /\ ].p,q.] = ].r,q.]
proof
  assume that
A1: r >= p and
A2: s > q;
  let t;
  thus t in ].r,s.[ /\ ].p,q.] implies t in ].r,q.]
  proof
    assume
A3: t in ].r,s.[ /\ ].p,q.];
    then
A4: t in ].r,s.[ by XBOOLE_0:def 4;
A5: t in ].p,q.] by A3,XBOOLE_0:def 4;
A6: r < t by A4,Th4;
    t <= q by A5,Th2;
    hence thesis by A6,Th2;
  end;
  assume
A7: t in ].r,q.];
  then
A8: r < t by Th2;
A9: t <= q by A7,Th2;
  then
A10: t < s by A2,XXREAL_0:2;
A11: p < t by A1,A8,XXREAL_0:2;
A12: t in ].r,s.[ by A8,A10,Th4;
  t in ].p,q.] by A9,A11,Th2;
  hence thesis by A12,XBOOLE_0:def 4;
end;

theorem Th160:
  r <= p & s <= q implies ].r,s.[ /\ ].p,q.[ = ].p,s.[
proof
  assume that
A1: r <= p and
A2: s <= q;
  let t;
  thus t in ].r,s.[ /\ ].p,q.[ implies t in ].p,s.[
  proof
    assume
A3: t in ].r,s.[ /\ ].p,q.[;
    then
A4: t in ].r,s.[ by XBOOLE_0:def 4;
A5: t in ].p,q.[ by A3,XBOOLE_0:def 4;
A6: t < s by A4,Th4;
    p < t by A5,Th4;
    hence thesis by A6,Th4;
  end;
  assume
A7: t in ].p,s.[;
  then
A8: p < t by Th4;
A9: t < s by A7,Th4;
A10: r < t by A1,A8,XXREAL_0:2;
A11: t < q by A2,A9,XXREAL_0:2;
A12: t in ].r,s.[ by A9,A10,Th4;
  t in ].p,q.[ by A8,A11,Th4;
  hence thesis by A12,XBOOLE_0:def 4;
end;

theorem
  [.r,s.[ \/ [.p,q.[ c= [.min(r,p),max(s,q).[
proof
  let t;
  assume t in [.r,s.[ \/ [.p,q.[;
  then t in [.r,s.[ or t in [.p,q.[ by XBOOLE_0:def 3;
  then
A1: r <= t & t < s or p <= t & t < q by Th3;
  then
A2: min(r,p) <= t by XXREAL_0:23;
  t < max(s,q) by A1,XXREAL_0:30;
  hence thesis by A2,Th3;
end;

theorem
  [.r,s.[ meets [.p,q.[ implies [.r,s.[ \/ [.p,q.[ = [.min(r,p),max(s,q).[
proof
  assume [.r,s.[ meets [.p,q.[;
  then consider u such that
A1: u in [.r,s.[ and
A2: u in [.p,q.[ by MEMBERED:def 20;
  let t;
  thus t in [.r,s.[ \/ [.p,q.[ implies t in [.min(r,p),max(s,q).[
  proof
    assume t in [.r,s.[ \/ [.p,q.[;
    then t in [.r,s.[ or t in [.p,q.[ by XBOOLE_0:def 3;
    then
A3: r <= t & t < s or p <= t & t < q by Th3;
    then
A4: min(r,p) <= t by XXREAL_0:23;
    t < max(s,q) by A3,XXREAL_0:30;
    hence thesis by A4,Th3;
  end;
A5: r <= u by A1,Th3;
A6: u < s by A1,Th3;
A7: p <= u by A2,Th3;
A8: u < q by A2,Th3;
  assume
A9: t in [.min(r,p),max(s,q).[;
  then
A10: min(r,p) <= t by Th3;
A11: t < max(s,q) by A9,Th3;
  per cases by A10,A11,XXREAL_0:21,28;
  suppose r <= t & t < s or p <= t & t < q;
    then t in [.r,s.[ or t in [.p,q.[ by Th3;
    hence thesis by XBOOLE_0:def 3;
  end;
  suppose that
A12: p <= t and
A13: t < s;
    u <= t or t <= u;
    then r <= t or t < q by A5,A8,XXREAL_0:2;
    then t in [.r,s.[ or t in [.p,q.[ by A12,A13,Th3;
    hence thesis by XBOOLE_0:def 3;
  end;
  suppose that
A14: r <= t and
A15: t < q;
    u <= t or t <= u;
    then t < s or p <= t by A6,A7,XXREAL_0:2;
    then t in [.r,s.[ or t in [.p,q.[ by A14,A15,Th3;
    hence thesis by XBOOLE_0:def 3;
  end;
end;

theorem
  ].r,s.] \/ ].p,q.] c= ].min(r,p),max(s,q).]
proof
  let t;
  assume t in ].r,s.] \/ ].p,q.];
  then t in ].r,s.] or t in ].p,q.] by XBOOLE_0:def 3;
  then
A1: r < t & t <= s or p < t & t <= q by Th2;
  then
A2: min(r,p) < t by XXREAL_0:22;
  t <= max(s,q) by A1,XXREAL_0:31;
  hence thesis by A2,Th2;
end;

theorem
  ].r,s.] meets ].p,q.] implies ].r,s.] \/ ].p,q.] = ].min(r,p),max(s,q).]
proof
  assume ].r,s.] meets ].p,q.];
  then consider u such that
A1: u in ].r,s.] and
A2: u in ].p,q.] by MEMBERED:def 20;
  let t;
  thus t in ].r,s.] \/ ].p,q.] implies t in ].min(r,p),max(s,q).]
  proof
    assume t in ].r,s.] \/ ].p,q.];
    then t in ].r,s.] or t in ].p,q.] by XBOOLE_0:def 3;
    then
A3: r < t & t <= s or p < t & t <= q by Th2;
    then
A4: min(r,p) < t by XXREAL_0:22;
    t <= max(s,q) by A3,XXREAL_0:31;
    hence thesis by A4,Th2;
  end;
A5: r < u by A1,Th2;
A6: u <= s by A1,Th2;
A7: p < u by A2,Th2;
A8: u <= q by A2,Th2;
  assume
A9: t in ].min(r,p),max(s,q).];
  then
A10: min(r,p) < t by Th2;
A11: t <= max(s,q) by A9,Th2;
  per cases by A10,A11,XXREAL_0:20,29;
  suppose r < t & t <= s or p < t & t <= q;
    then t in ].r,s.] or t in ].p,q.] by Th2;
    hence thesis by XBOOLE_0:def 3;
  end;
  suppose that
A12: p < t and
A13: t <= s;
    u <= t or t <= u;
    then r < t or t <= q by A5,A8,XXREAL_0:2;
    then t in ].r,s.] or t in ].p,q.] by A12,A13,Th2;
    hence thesis by XBOOLE_0:def 3;
  end;
  suppose that
A14: r < t and
A15: t <= q;
    u <= t or t <= u;
    then t <= s or p < t by A6,A7,XXREAL_0:2;
    then t in ].r,s.] or t in ].p,q.] by A14,A15,Th2;
    hence thesis by XBOOLE_0:def 3;
  end;
end;

theorem
  r <= s & s <= t implies [.r,s.] \/ [.s,t.] = [.r,t.]
proof
  assume that
A1: r <= s and
A2: s <= t;
  let p;
  thus p in [.r,s.] \/ [.s,t.] implies p in [.r,t.]
  proof
    assume p in [.r,s.] \/ [.s,t.];
    then p in [.r,s.] or p in [.s,t.] by XBOOLE_0:def 3;
    then
A3: r <= p & p <= s or s <= p & p <= t by Th1;
    then
A4: r <= p by A1,XXREAL_0:2;
    p <= t by A2,A3,XXREAL_0:2;
    hence thesis by A4,Th1;
  end;
  assume p in [.r,t.];
  then r <= p & p <= s or s <= p & p <= t by Th1;
  then p in [.r,s.] or p in [.s,t.] by Th1;
  hence thesis by XBOOLE_0:def 3;
end;

theorem Th166:
  r <= s & s <= t implies [.r,s.[ \/ [.s,t.] = [.r,t.]
proof
  assume that
A1: r <= s and
A2: s <= t;
  let p;
  thus p in [.r,s.[ \/ [.s,t.] implies p in [.r,t.]
  proof
    assume p in [.r,s.[ \/ [.s,t.];
    then p in [.r,s.[ or p in [.s,t.] by XBOOLE_0:def 3;
    then
A3: r <= p & p < s or s <= p & p <= t by Th1,Th3;
    then
A4: r <= p by A1,XXREAL_0:2;
    p <= t by A2,A3,XXREAL_0:2;
    hence thesis by A4,Th1;
  end;
  assume p in [.r,t.];
  then r <= p & p < s or s <= p & p <= t by Th1;
  then p in [.r,s.[ or p in [.s,t.] by Th1,Th3;
  hence thesis by XBOOLE_0:def 3;
end;

theorem Th167:
  r <= s & s <= t implies [.r,s.] \/ ].s,t.] = [.r,t.]
proof
  assume that
A1: r <= s and
A2: s <= t;
  let p;
  thus p in [.r,s.] \/ ].s,t.] implies p in [.r,t.]
  proof
    assume p in [.r,s.] \/ ].s,t.];
    then p in [.r,s.] or p in ].s,t.] by XBOOLE_0:def 3;
    then
A3: r <= p & p <= s or s < p & p <= t by Th1,Th2;
    then
A4: r <= p by A1,XXREAL_0:2;
    p <= t by A2,A3,XXREAL_0:2;
    hence thesis by A4,Th1;
  end;
  assume p in [.r,t.];
  then r <= p & p <= s or s < p & p <= t by Th1;
  then p in [.r,s.] or p in ].s,t.] by Th1,Th2;
  hence thesis by XBOOLE_0:def 3;
end;

theorem
  r <= s & s <= t implies [.r,s.[ \/ [.s,t.[ = [.r,t.[
proof
  assume that
A1: r <= s and
A2: s <= t;
  let p;
  thus p in [.r,s.[ \/ [.s,t.[ implies p in [.r,t.[
  proof
    assume p in [.r,s.[ \/ [.s,t.[;
    then p in [.r,s.[ or p in [.s,t.[ by XBOOLE_0:def 3;
    then
A3: r <= p & p < s or s <= p & p < t by Th3;
    then
A4: r <= p by A1,XXREAL_0:2;
    p < t by A2,A3,XXREAL_0:2;
    hence thesis by A4,Th3;
  end;
  assume p in [.r,t.[;
  then r <= p & p < s or s <= p & p < t by Th3;
  then p in [.r,s.[ or p in [.s,t.[ by Th3;
  hence thesis by XBOOLE_0:def 3;
end;

theorem Th169:
  r <= s & s < t implies [.r,s.] \/ ].s,t.[ = [.r,t.[
proof
  assume that
A1: r <= s and
A2: s < t;
  let p;
  thus p in [.r,s.] \/ ].s,t.[ implies p in [.r,t.[
  proof
    assume p in [.r,s.] \/ ].s,t.[;
    then p in [.r,s.] or p in ].s,t.[ by XBOOLE_0:def 3;
    then
A3: r <= p & p <= s or s < p & p < t by Th1,Th4;
    then
A4: r <= p by A1,XXREAL_0:2;
    p < t by A2,A3,XXREAL_0:2;
    hence thesis by A4,Th3;
  end;
  assume p in [.r,t.[;
  then r <= p & p <= s or s < p & p < t by Th3;
  then p in [.r,s.] or p in ].s,t.[ by Th1,Th4;
  hence thesis by XBOOLE_0:def 3;
end;

theorem
  r <= s & s <= t implies ].r,s.] \/ ].s,t.] = ].r,t.]
proof
  assume that
A1: r <= s and
A2: s <= t;
  let p;
  thus p in ].r,s.] \/ ].s,t.] implies p in ].r,t.]
  proof
    assume p in ].r,s.] \/ ].s,t.];
    then p in ].r,s.] or p in ].s,t.] by XBOOLE_0:def 3;
    then
A3: r < p & p <= s or s < p & p <= t by Th2;
    then
A4: r < p by A1,XXREAL_0:2;
    p <= t by A2,A3,XXREAL_0:2;
    hence thesis by A4,Th2;
  end;
  assume p in ].r,t.];
  then r < p & p <= s or s < p & p <= t by Th2;
  then p in ].r,s.] or p in ].s,t.] by Th2;
  hence thesis by XBOOLE_0:def 3;
end;

theorem Th171:
  r <= s & s < t implies ].r,s.] \/ ].s,t.[ = ].r,t.[
proof
  assume that
A1: r <= s and
A2: s < t;
  let p;
  thus p in ].r,s.] \/ ].s,t.[ implies p in ].r,t.[
  proof
    assume p in ].r,s.] \/ ].s,t.[;
    then p in ].r,s.] or p in ].s,t.[ by XBOOLE_0:def 3;
    then
A3: r < p & p <= s or s < p & p < t by Th2,Th4;
    then
A4: r < p by A1,XXREAL_0:2;
    p < t by A2,A3,XXREAL_0:2;
    hence thesis by A4,Th4;
  end;
  assume p in ].r,t.[;
  then r < p & p <= s or s < p & p < t by Th4;
  then p in ].r,s.] or p in ].s,t.[ by Th2,Th4;
  hence thesis by XBOOLE_0:def 3;
end;

theorem
  r < s & s < t implies ].r,s.] \/ [.s,t.[ = ].r,t.[
proof
  assume that
A1: r < s and
A2: s < t;
  let p;
  thus p in ].r,s.] \/ [.s,t.[ implies p in ].r,t.[
  proof
    assume p in ].r,s.] \/ [.s,t.[;
    then p in ].r,s.] or p in [.s,t.[ by XBOOLE_0:def 3;
    then
A3: r < p & p <= s or s <= p & p < t by Th2,Th3;
    then
A4: r < p by A1,XXREAL_0:2;
    p < t by A2,A3,XXREAL_0:2;
    hence thesis by A4,Th4;
  end;
  assume p in ].r,t.[;
  then r < p & p <= s or s < p & p < t by Th4;
  then p in ].r,s.] or p in [.s,t.[ by Th2,Th3;
  hence thesis by XBOOLE_0:def 3;
end;

theorem Th173:
  r < s & s < t implies ].r,s.[ \/ [.s,t.[ = ].r,t.[
proof
  assume that
A1: r < s and
A2: s < t;
  let p;
  thus p in ].r,s.[ \/ [.s,t.[ implies p in ].r,t.[
  proof
    assume p in ].r,s.[ \/ [.s,t.[;
    then p in ].r,s.[ or p in [.s,t.[ by XBOOLE_0:def 3;
    then
A3: r < p & p < s or s <= p & p < t by Th3,Th4;
    then
A4: r < p by A1,XXREAL_0:2;
    p < t by A2,A3,XXREAL_0:2;
    hence thesis by A4,Th4;
  end;
  assume p in ].r,t.[;
  then r < p & p < s or s <= p & p < t by Th4;
  then p in ].r,s.[ or p in [.s,t.[ by Th3,Th4;
  hence thesis by XBOOLE_0:def 3;
end;

theorem Th174:
  p <= s & r <= q & s <= r implies [.p,r.] \/ [.s,q.] = [.p,q.]
proof
  assume that
A1: p <= s and
A2: r <= q and
A3: s <= r;
  let t;
  thus t in [.p,r.] \/ [.s,q.] implies t in [.p,q.]
  proof
    assume t in [.p,r.] \/ [.s,q.];
    then t in [.p,r.] or t in [.s,q.] by XBOOLE_0:def 3;
    then
A4: p <= t & t <= r or s <= t & t <= q by Th1;
    then
A5: p <= t by A1,XXREAL_0:2;
    t <= q by A2,A4,XXREAL_0:2;
    hence thesis by A5,Th1;
  end;
  assume t in [.p,q.];
  then p <= t & t <= r or s <= t & t <= q by A3,Th1,XXREAL_0:2;
  then t in [.p,r.] or t in [.s,q.] by Th1;
  hence thesis by XBOOLE_0:def 3;
end;

theorem Th175:
  p <= s & r <= q & s < r implies [.p,r.[ \/ ].s,q.] = [.p,q.]
proof
  assume that
A1: p <= s and
A2: r <= q and
A3: s < r;
  let t;
  thus t in [.p,r.[ \/ ].s,q.] implies t in [.p,q.]
  proof
    assume t in [.p,r.[ \/ ].s,q.];
    then t in [.p,r.[ or t in ].s,q.] by XBOOLE_0:def 3;
    then
A4: p <= t & t <= r or s <= t & t <= q by Th2,Th3;
    then
A5: p <= t by A1,XXREAL_0:2;
    t <= q by A2,A4,XXREAL_0:2;
    hence thesis by A5,Th1;
  end;
  assume t in [.p,q.];
  then p <= t & t < r or s < t & t <= q by A3,Th1,XXREAL_0:2;
  then t in [.p,r.[ or t in ].s,q.] by Th2,Th3;
  hence thesis by XBOOLE_0:def 3;
end;

theorem
  p <= s & s <= r & r < q implies [.p,r.] \/ [.s,q.[ = [.p,q.[
proof
  assume that
A1: p <= s and
A2: s <= r and
A3: r < q;
  let t;
  thus t in [.p,r.] \/ [.s,q.[ implies t in [.p,q.[
  proof
    assume t in [.p,r.] \/ [.s,q.[;
    then t in [.p,r.] or t in [.s,q.[ by XBOOLE_0:def 3;
    then
A4: p <= t & t <= r or s <= t & t < q by Th1,Th3;
    then
A5: p <= t by A1,XXREAL_0:2;
    t < q by A3,A4,XXREAL_0:2;
    hence thesis by A5,Th3;
  end;
  assume t in [.p,q.[;
  then p <= t & t <= r or s <= t & t < q by A2,Th3,XXREAL_0:2;
  then t in [.p,r.] or t in [.s,q.[ by Th1,Th3;
  hence thesis by XBOOLE_0:def 3;
end;

theorem
  p < s & r <= q & s <= r implies ].p,r.] \/ [.s,q.] = ].p,q.]
proof
  assume that
A1: p < s and
A2: r <= q and
A3: s <= r;
  let t;
  thus t in ].p,r.] \/ [.s,q.] implies t in ].p,q.]
  proof
    assume t in ].p,r.] \/ [.s,q.];
    then t in ].p,r.] or t in [.s,q.] by XBOOLE_0:def 3;
    then
A4: p < t & t <= r or s <= t & t <= q by Th1,Th2;
    then
A5: p < t by A1,XXREAL_0:2;
    t <= q by A2,A4,XXREAL_0:2;
    hence thesis by A5,Th2;
  end;
  assume t in ].p,q.];
  then p < t & t <= r or s <= t & t <= q by A3,Th2,XXREAL_0:2;
  then t in ].p,r.] or t in [.s,q.] by Th1,Th2;
  hence thesis by XBOOLE_0:def 3;
end;

theorem Th178:
  p < s & r < q & s <= r implies ].p,r.] \/ [.s,q.[ = ].p,q.[
proof
  assume that
A1: p < s and
A2: r < q and
A3: s <= r;
  let t;
  thus t in ].p,r.] \/ [.s,q.[ implies t in ].p,q.[
  proof
    assume t in ].p,r.] \/ [.s,q.[;
    then t in ].p,r.] or t in [.s,q.[ by XBOOLE_0:def 3;
    then
A4: p < t & t <= r or s <= t & t < q by Th2,Th3;
    then
A5: p < t by A1,XXREAL_0:2;
    t < q by A2,A4,XXREAL_0:2;
    hence thesis by A5,Th4;
  end;
  assume t in ].p,q.[;
  then p < t & t <= r or s <= t & t < q by A3,Th4,XXREAL_0:2;
  then t in ].p,r.] or t in [.s,q.[ by Th2,Th3;
  hence thesis by XBOOLE_0:def 3;
end;

theorem
  p<=r & p<=s & r<=q & s<=q implies [.p,r.[ \/ [.r,s.] \/ ].s,q.] = [.p,q.]
proof
  assume that
A1: p <= r and
A2: p <= s and
A3: r <= q and
A4: s <= q;
  per cases;
  suppose r <= s;
    hence [.p,r.[ \/ [.r,s.] \/ ].s,q.] = [.p,s.] \/ ].s,q.] by A1,Th166
      .= [.p,q.] by A2,A4,Th167;
  end;
  suppose
A5: s < r;
    hence [.p,r.[ \/ [.r,s.] \/ ].s,q.] = [.p,r.[ \/ {} \/ ].s,q.] by Th29
      .= [.p,q.] by A2,A3,A5,Th175;
  end;
end;

theorem
  p<r & p<s & r<q & s<q implies ].p,r.] \/ ].r,s.[ \/ [.s,q.[ = ].p,q.[
proof
  assume that
A1: p < r and
A2: p < s and
A3: r < q and
A4: s < q;
  per cases;
  suppose r < s;
    hence ].p,r.] \/ ].r,s.[ \/ [.s,q.[ = ].p,s.[ \/ [.s,q.[ by A1,Th171
      .= ].p,q.[ by A2,A4,Th173;
  end;
  suppose
A5: s <= r;
    hence ].p,r.] \/ ].r,s.[ \/ [.s,q.[ = ].p,r.] \/ {} \/ [.s,q.[ by Th28
      .= ].p,q.[ by A2,A3,A5,Th178;
  end;
end;

theorem
  p <= r & r <= s & s <= q implies [.p,r.] \/ ].r,s.[ \/ [.s,q.] = [.p,q.]
proof
  assume that
A1: p <= r and
A2: r <= s and
A3: s <= q;
A4: p <= s by A1,A2,XXREAL_0:2;
A5: r <= q by A2,A3,XXREAL_0:2;
  per cases;
  suppose r < s;
    hence [.p,r.] \/ ].r,s.[ \/ [.s,q.] = [.p,s.[ \/ [.s,q.] by A1,Th169
      .= [.p,q.] by A3,A4,Th166;
  end;
  suppose
A6: s <= r;
    hence [.p,r.] \/ ].r,s.[ \/ [.s,q.] = [.p,r.] \/ {} \/ [.s,q.] by Th28
      .= [.p,q.] by A4,A5,A6,Th174;
  end;
end;

theorem Th182:
  r <= s implies [.r,t.] \ [.r,s.] = ].s,t.]
proof
  assume that
A1: r <= s;
  let p;
  thus p in [.r,t.] \ [.r,s.] implies p in ].s,t.]
  proof
    assume
A2: p in [.r,t.] \ [.r,s.];
    then
A3: not p in [.r,s.] by XBOOLE_0:def 5;
A4: p <= t by A2,Th1;
    p < r or s < p by A3,Th1;
    hence thesis by A2,A4,Th1,Th2;
  end;
  assume
A5: p in ].s,t.];
  then
A6: s < p by Th2;
  then
A7: r <= p by A1,XXREAL_0:2;
  p <= t by A5,Th2;
  then
A8: p in [.r,t.] by A7,Th1;
  not p in [.r,s.] by A6,Th1;
  hence thesis by A8,XBOOLE_0:def 5;
end;

theorem Th183:
  r <= s implies [.r,t.[ \ [.r,s.] = ].s,t.[
proof
  assume that
A1: r <= s;
  let p;
  thus p in [.r,t.[ \ [.r,s.] implies p in ].s,t.[
  proof
    assume
A2: p in [.r,t.[ \ [.r,s.];
    then
A3: not p in [.r,s.] by XBOOLE_0:def 5;
A4: p < t by A2,Th3;
    p < r or s < p by A3,Th1;
    hence thesis by A2,A4,Th3,Th4;
  end;
  assume
A5: p in ].s,t.[;
  then
A6: s < p by Th4;
  then
A7: r <= p by A1,XXREAL_0:2;
  p < t by A5,Th4;
  then
A8: p in [.r,t.[ by A7,Th3;
  not p in [.r,s.] by A6,Th1;
  hence thesis by A8,XBOOLE_0:def 5;
end;

theorem Th184:
  r < s implies [.r,t.] \ [.r,s.[ = [.s,t.]
proof
  assume that
A1: r < s;
  let p;
  thus p in [.r,t.] \ [.r,s.[ implies p in [.s,t.]
  proof
    assume
A2: p in [.r,t.] \ [.r,s.[;
    then
A3: not p in [.r,s.[ by XBOOLE_0:def 5;
A4: p <= t by A2,Th1;
    p < r or s <= p by A3,Th3;
    hence thesis by A2,A4,Th1;
  end;
  assume
A5: p in [.s,t.];
  then
A6: s <= p by Th1;
  then
A7: r <= p by A1,XXREAL_0:2;
  p <= t by A5,Th1;
  then
A8: p in [.r,t.] by A7,Th1;
  not p in [.r,s.[ by A6,Th3;
  hence thesis by A8,XBOOLE_0:def 5;
end;

theorem Th185:
  r < s implies [.r,t.[ \ [.r,s.[ = [.s,t.[
proof
  assume that
A1: r < s;
  let p;
  thus p in [.r,t.[ \ [.r,s.[ implies p in [.s,t.[
  proof
    assume
A2: p in [.r,t.[ \ [.r,s.[;
    then
A3: not p in [.r,s.[ by XBOOLE_0:def 5;
A4: p < t by A2,Th3;
    p < r or s <= p by A3,Th3;
    hence thesis by A2,A4,Th3;
  end;
  assume
A5: p in [.s,t.[;
  then
A6: s <= p by Th3;
  then
A7: r <= p by A1,XXREAL_0:2;
  p < t by A5,Th3;
  then
A8: p in [.r,t.[ by A7,Th3;
  not p in [.r,s.[ by A6,Th3;
  hence thesis by A8,XBOOLE_0:def 5;
end;

theorem Th186:
  r <= s implies [.r,t.] \ [.r,s.] = ].s,t.]
proof
  assume that
A1: r <= s;
  let p;
  thus p in [.r,t.] \ [.r,s.] implies p in ].s,t.]
  proof
    assume
A2: p in [.r,t.] \ [.r,s.];
    then
A3: not p in [.r,s.] by XBOOLE_0:def 5;
A4: p <= t by A2,Th1;
    p < r or s < p by A3,Th1;
    hence thesis by A2,A4,Th1,Th2;
  end;
  assume
A5: p in ].s,t.];
  then
A6: s < p by Th2;
  then
A7: r <= p by A1,XXREAL_0:2;
  p <= t by A5,Th2;
  then
A8: p in [.r,t.] by A7,Th1;
  not p in [.r,s.] by A6,Th1;
  hence thesis by A8,XBOOLE_0:def 5;
end;

theorem Th187:
  r < s implies ].r,t.[ \ ].r,s.] = ].s,t.[
proof
  assume that
A1: r < s;
  let p;
  thus p in ].r,t.[ \ ].r,s.] implies p in ].s,t.[
  proof
    assume
A2: p in ].r,t.[ \ ].r,s.];
    then
A3: not p in ].r,s.] by XBOOLE_0:def 5;
A4: p < t by A2,Th4;
    p <= r or s < p by A3,Th2;
    hence thesis by A2,A4,Th4;
  end;
  assume
A5: p in ].s,t.[;
  then
A6: s < p by Th4;
  then
A7: r < p by A1,XXREAL_0:2;
  p < t by A5,Th4;
  then
A8: p in ].r,t.[ by A7,Th4;
  not p in ].r,s.] by A6,Th2;
  hence thesis by A8,XBOOLE_0:def 5;
end;

theorem Th188:
  r < s implies ].r,t.] \ ].r,s.[ = [.s,t.]
proof
  assume that
A1: r < s;
  let p;
  thus p in ].r,t.] \ ].r,s.[ implies p in [.s,t.]
  proof
    assume
A2: p in ].r,t.] \ ].r,s.[;
    then
A3: not p in ].r,s.[ by XBOOLE_0:def 5;
A4: p <= t by A2,Th2;
    p <= r or s <= p by A3,Th4;
    hence thesis by A2,A4,Th1,Th2;
  end;
  assume
A5: p in [.s,t.];
  then
A6: s <= p by Th1;
  then
A7: r < p by A1,XXREAL_0:2;
  p <= t by A5,Th1;
  then
A8: p in ].r,t.] by A7,Th2;
  not p in ].r,s.[ by A6,Th4;
  hence thesis by A8,XBOOLE_0:def 5;
end;

theorem Th189:
  r < s implies ].r,t.[ \ ].r,s.[ = [.s,t.[
proof
  assume that
A1: r < s;
  let p;
  thus p in ].r,t.[ \ ].r,s.[ implies p in [.s,t.[
  proof
    assume
A2: p in ].r,t.[ \ ].r,s.[;
    then
A3: not p in ].r,s.[ by XBOOLE_0:def 5;
A4: p < t by A2,Th4;
    p <= r or s <= p by A3,Th4;
    hence thesis by A2,A4,Th3,Th4;
  end;
  assume
A5: p in [.s,t.[;
  then
A6: s <= p by Th3;
  then
A7: r < p by A1,XXREAL_0:2;
  p < t by A5,Th3;
  then
A8: p in ].r,t.[ by A7,Th4;
  not p in ].r,s.[ by A6,Th4;
  hence thesis by A8,XBOOLE_0:def 5;
end;

theorem Th190:
  s <= t implies [.r,t.] \ [.s,t.] = [.r,s.[
proof
  assume that
A1: s <= t;
  let p;
  thus p in [.r,t.] \ [.s,t.] implies p in [.r,s.[
  proof
    assume
A2: p in [.r,t.] \ [.s,t.];
    then
A3: not p in [.s,t.] by XBOOLE_0:def 5;
A4: r <= p by A2,Th1;
    p < s or t < p by A3,Th1;
    hence thesis by A2,A4,Th1,Th3;
  end;
  assume
A5: p in [.r,s.[;
  then
A6: p < s by Th3;
A7: r <= p by A5,Th3;
  p <= t by A1,A6,XXREAL_0:2;
  then
A8: p in [.r,t.] by A7,Th1;
  not p in [.s,t.] by A6,Th1;
  hence thesis by A8,XBOOLE_0:def 5;
end;

theorem Th191:
  s <= t implies ].r,t.] \ [.s,t.] = ].r,s.[
proof
  assume that
A1: s <= t;
  let p;
  thus p in ].r,t.] \ [.s,t.] implies p in ].r,s.[
  proof
    assume
A2: p in ].r,t.] \ [.s,t.];
    then
A3: not p in [.s,t.] by XBOOLE_0:def 5;
A4: r < p by A2,Th2;
    p < s or t < p by A3,Th1;
    hence thesis by A2,A4,Th2,Th4;
  end;
  assume
A5: p in ].r,s.[;
  then
A6: p < s by Th4;
A7: r < p by A5,Th4;
  p <= t by A1,A6,XXREAL_0:2;
  then
A8: p in ].r,t.] by A7,Th2;
  not p in [.s,t.] by A6,Th1;
  hence thesis by A8,XBOOLE_0:def 5;
end;

theorem Th192:
  s < t implies [.r,t.] \ ].s,t.] = [.r,s.]
proof
  assume that
A1: s < t;
  let p;
  thus p in [.r,t.] \ ].s,t.] implies p in [.r,s.]
  proof
    assume
A2: p in [.r,t.] \ ].s,t.];
    then
A3: not p in ].s,t.] by XBOOLE_0:def 5;
A4: r <= p by A2,Th1;
    p <= s or t < p by A3,Th2;
    hence thesis by A2,A4,Th1;
  end;
  assume
A5: p in [.r,s.];
  then
A6: p <= s by Th1;
A7: r <= p by A5,Th1;
  p <= t by A1,A6,XXREAL_0:2;
  then
A8: p in [.r,t.] by A7,Th1;
  not p in ].s,t.] by A6,Th2;
  hence thesis by A8,XBOOLE_0:def 5;
end;

theorem Th193:
  s < t implies ].r,t.] \ ].s,t.] = ].r,s.]
proof
  assume that
A1: s < t;
  let p;
  thus p in ].r,t.] \ ].s,t.] implies p in ].r,s.]
  proof
    assume
A2: p in ].r,t.] \ ].s,t.];
    then
A3: not p in ].s,t.] by XBOOLE_0:def 5;
A4: r < p by A2,Th2;
    p <= s or t < p by A3,Th2;
    hence thesis by A2,A4,Th2;
  end;
  assume
A5: p in ].r,s.];
  then
A6: p <= s by Th2;
A7: r < p by A5,Th2;
  p <= t by A1,A6,XXREAL_0:2;
  then
A8: p in ].r,t.] by A7,Th2;
  not p in ].s,t.] by A6,Th2;
  hence thesis by A8,XBOOLE_0:def 5;
end;

theorem Th194:
  s < t implies [.r,t.[ \ [.s,t.[ = [.r,s.[
proof
  assume that
A1: s < t;
  let p;
  thus p in [.r,t.[ \ [.s,t.[ implies p in [.r,s.[
  proof
    assume
A2: p in [.r,t.[ \ [.s,t.[;
    then
A3: not p in [.s,t.[ by XBOOLE_0:def 5;
A4: r <= p by A2,Th3;
    p < s or t <= p by A3,Th3;
    hence thesis by A2,A4,Th3;
  end;
  assume
A5: p in [.r,s.[;
  then
A6: p < s by Th3;
A7: r <= p by A5,Th3;
  p < t by A1,A6,XXREAL_0:2;
  then
A8: p in [.r,t.[ by A7,Th3;
  not p in [.s,t.[ by A6,Th3;
  hence thesis by A8,XBOOLE_0:def 5;
end;

theorem Th195:
  s < t implies ].r,t.[ \ [.s,t.[ = ].r,s.[
proof
  assume that
A1: s < t;
  let p;
  thus p in ].r,t.[ \ [.s,t.[ implies p in ].r,s.[
  proof
    assume
A2: p in ].r,t.[ \ [.s,t.[;
    then
A3: not p in [.s,t.[ by XBOOLE_0:def 5;
A4: r < p by A2,Th4;
    p < s or t <= p by A3,Th3;
    hence thesis by A2,A4,Th4;
  end;
  assume
A5: p in ].r,s.[;
  then
A6: p < s by Th4;
A7: r < p by A5,Th4;
  p < t by A1,A6,XXREAL_0:2;
  then
A8: p in ].r,t.[ by A7,Th4;
  not p in [.s,t.[ by A6,Th3;
  hence thesis by A8,XBOOLE_0:def 5;
end;

theorem Th196:
  s < t implies [.r,t.[ \ ].s,t.[ = [.r,s.]
proof
  assume that
A1: s < t;
  let p;
  thus p in [.r,t.[ \ ].s,t.[ implies p in [.r,s.]
  proof
    assume
A2: p in [.r,t.[ \ ].s,t.[;
    then
A3: not p in ].s,t.[ by XBOOLE_0:def 5;
A4: r <= p by A2,Th3;
    p <= s or t <= p by A3,Th4;
    hence thesis by A2,A4,Th1,Th3;
  end;
  assume
A5: p in [.r,s.];
  then
A6: p <= s by Th1;
A7: r <= p by A5,Th1;
  p < t by A1,A6,XXREAL_0:2;
  then
A8: p in [.r,t.[ by A7,Th3;
  not p in ].s,t.[ by A6,Th4;
  hence thesis by A8,XBOOLE_0:def 5;
end;

theorem Th197:
  s < t implies ].r,t.[ \ ].s,t.[ = ].r,s.]
proof
  assume that
A1: s < t;
  let p;
  thus p in ].r,t.[ \ ].s,t.[ implies p in ].r,s.]
  proof
    assume
A2: p in ].r,t.[ \ ].s,t.[;
    then
A3: not p in ].s,t.[ by XBOOLE_0:def 5;
A4: r < p by A2,Th4;
    p <= s or t <= p by A3,Th4;
    hence thesis by A2,A4,Th2,Th4;
  end;
  assume
A5: p in ].r,s.];
  then
A6: p <= s by Th2;
A7: r < p by A5,Th2;
  p < t by A1,A6,XXREAL_0:2;
  then
A8: p in ].r,t.[ by A7,Th4;
  not p in ].s,t.[ by A6,Th4;
  hence thesis by A8,XBOOLE_0:def 5;
end;

theorem
  [.p,q.[ meets [.r,s.[ implies [.p,q.[ \ [.r,s.[ = [.p,r.[ \/ [.s,q.[
proof
  assume [.p,q.[ meets [.r,s.[;
  then consider u such that
A1: u in [.r,s.[ and
A2: u in [.p,q.[ by MEMBERED:def 20;
A3: r <= u by A1,Th3;
A4: u <= s by A1,Th3;
A5: p <= u by A2,Th3;
  u <= q by A2,Th3;
  then
A6: r <= q by A3,XXREAL_0:2;
A7: p <= s by A4,A5,XXREAL_0:2;
  let t;
  thus t in [.p,q.[ \ [.r,s.[ implies t in [.p,r.[ \/ [.s,q.[
  proof
    assume
A8: t in [.p,q.[ \ [.r,s.[;
    then
A9: not t in [.r,s.[ by XBOOLE_0:def 5;
A10: p <= t by A8,Th3;
A11: t < q by A8,Th3;
    t < r or s <= t by A9,Th3;
    then t in [.p,r.[ or t in [.s,q.[ by A10,A11,Th3;
    hence thesis by XBOOLE_0:def 3;
  end;
  assume t in [.p,r.[ \/ [.s,q.[;
  then t in [.p,r.[ or t in [.s,q.[ by XBOOLE_0:def 3;
  then
A12: p <= t & t < r or s <= t & t < q by Th3;
  then
A13: p <= t by A7,XXREAL_0:2;
  t < q by A6,A12,XXREAL_0:2;
  then
A14: t in [.p,q.[ by A13,Th3;
  not t in [.r,s.[ by A12,Th3;
  hence thesis by A14,XBOOLE_0:def 5;
end;

theorem
  ].p,q.] meets ].r,s.] implies ].p,q.] \ ].r,s.] = ].p,r.] \/ ].s,q.]
proof
  assume ].p,q.] meets ].r,s.];
  then consider u such that
A1: u in ].r,s.] and
A2: u in ].p,q.] by MEMBERED:def 20;
A3: r < u by A1,Th2;
A4: u <= s by A1,Th2;
A5: p < u by A2,Th2;
  u <= q by A2,Th2;
  then
A6: r <= q by A3,XXREAL_0:2;
A7: p <= s by A4,A5,XXREAL_0:2;
  let t;
  thus t in ].p,q.] \ ].r,s.] implies t in ].p,r.] \/ ].s,q.]
  proof
    assume
A8: t in ].p,q.] \ ].r,s.];
    then
A9: not t in ].r,s.] by XBOOLE_0:def 5;
A10: p < t by A8,Th2;
A11: t <= q by A8,Th2;
    t <= r or s < t by A9,Th2;
    then t in ].p,r.] or t in ].s,q.] by A10,A11,Th2;
    hence thesis by XBOOLE_0:def 3;
  end;
  assume t in ].p,r.] \/ ].s,q.];
  then t in ].p,r.] or t in ].s,q.] by XBOOLE_0:def 3;
  then
A12: p < t & t <= r or s < t & t <= q by Th2;
  then
A13: p < t by A7,XXREAL_0:2;
  t <= q by A6,A12,XXREAL_0:2;
  then
A14: t in ].p,q.] by A13,Th2;
  not t in ].r,s.] by A12,Th2;
  hence thesis by A14,XBOOLE_0:def 5;
end;

theorem
  p <= r & s <= q implies [.p,q.] \ ([.p,r.] \/ [.s,q.]) = ].r,s.[
proof
  assume that
A1: p <= r and
A2: s <= q;
  thus [.p,q.] \ ([.p,r.] \/ [.s,q.])
  = [.p,q.] \ [.p,r.] \ [.s,q.] by XBOOLE_1:41
    .= ].r,q.] \ [.s,q.] by A1,Th182
    .= ].r,s.[ by A2,Th191;
end;

theorem
  r <= s & s <= t implies [.r,t.] \ {s} = [.r,s.[ \/ ].s,t.]
proof
  assume that
A1: r <= s and
A2: s <= t;
  let p;
  thus p in [.r,t.] \ {s} implies p in [.r,s.[ \/ ].s,t.]
  proof
    assume
A3: p in [.r,t.] \ {s};
    then not p in {s} by XBOOLE_0:def 5;
    then p <> s by TARSKI:def 1;
    then r <= p & p < s or s < p & p <= t by A3,Th1,XXREAL_0:1;
    then p in [.r,s.[ or p in ].s,t.] by Th2,Th3;
    hence thesis by XBOOLE_0:def 3;
  end;
  assume p in [.r,s.[ \/ ].s,t.];
  then p in [.r,s.[ or p in ].s,t.] by XBOOLE_0:def 3;
  then
A4: r <= p & p < s or s < p & p <= t by Th2,Th3;
  then
A5: r <= p by A1,XXREAL_0:2;
  p <= t by A2,A4,XXREAL_0:2;
  then
A6: p in [.r,t.] by A5,Th1;
  not p in {s} by A4,TARSKI:def 1;
  hence thesis by A6,XBOOLE_0:def 5;
end;

theorem
  r <= s & s < t implies [.r,t.[ \ {s} = [.r,s.[ \/ ].s,t.[
proof
  assume that
A1: r <= s and
A2: s < t;
  let p;
  thus p in [.r,t.[ \ {s} implies p in [.r,s.[ \/ ].s,t.[
  proof
    assume
A3: p in [.r,t.[ \ {s};
    then not p in {s} by XBOOLE_0:def 5;
    then p <> s by TARSKI:def 1;
    then r <= p & p < s or s < p & p < t by A3,Th3,XXREAL_0:1;
    then p in [.r,s.[ or p in ].s,t.[ by Th3,Th4;
    hence thesis by XBOOLE_0:def 3;
  end;
  assume p in [.r,s.[ \/ ].s,t.[;
  then p in [.r,s.[ or p in ].s,t.[ by XBOOLE_0:def 3;
  then
A4: r <= p & p < s or s < p & p < t by Th3,Th4;
  then
A5: r <= p by A1,XXREAL_0:2;
  p < t by A2,A4,XXREAL_0:2;
  then
A6: p in [.r,t.[ by A5,Th3;
  not p in {s} by A4,TARSKI:def 1;
  hence thesis by A6,XBOOLE_0:def 5;
end;

theorem
  r < s & s <= t implies ].r,t.] \ {s} = ].r,s.[ \/ ].s,t.]
proof
  assume that
A1: r < s and
A2: s <= t;
  let p;
  thus p in ].r,t.] \ {s} implies p in ].r,s.[ \/ ].s,t.]
  proof
    assume
A3: p in ].r,t.] \ {s};
    then not p in {s} by XBOOLE_0:def 5;
    then p <> s by TARSKI:def 1;
    then r < p & p < s or s < p & p <= t by A3,Th2,XXREAL_0:1;
    then p in ].r,s.[ or p in ].s,t.] by Th2,Th4;
    hence thesis by XBOOLE_0:def 3;
  end;
  assume p in ].r,s.[ \/ ].s,t.];
  then p in ].r,s.[ or p in ].s,t.] by XBOOLE_0:def 3;
  then
A4: r < p & p < s or s < p & p <= t by Th2,Th4;
  then
A5: r < p by A1,XXREAL_0:2;
  p <= t by A2,A4,XXREAL_0:2;
  then
A6: p in ].r,t.] by A5,Th2;
  not p in {s} by A4,TARSKI:def 1;
  hence thesis by A6,XBOOLE_0:def 5;
end;

theorem
  r < s & s < t implies ].r,t.[ \ {s} = ].r,s.[ \/ ].s,t.[
proof
  assume that
A1: r < s and
A2: s < t;
  let p;
  thus p in ].r,t.[ \ {s} implies p in ].r,s.[ \/ ].s,t.[
  proof
    assume
A3: p in ].r,t.[ \ {s};
    then not p in {s} by XBOOLE_0:def 5;
    then p <> s by TARSKI:def 1;
    then r < p & p < s or s < p & p < t by A3,Th4,XXREAL_0:1;
    then p in ].r,s.[ or p in ].s,t.[ by Th4;
    hence thesis by XBOOLE_0:def 3;
  end;
  assume p in ].r,s.[ \/ ].s,t.[;
  then p in ].r,s.[ or p in ].s,t.[ by XBOOLE_0:def 3;
  then
A4: r < p & p < s or s < p & p < t by Th4;
  then
A5: r < p by A1,XXREAL_0:2;
  p < t by A2,A4,XXREAL_0:2;
  then
A6: p in ].r,t.[ by A5,Th4;
  not p in {s} by A4,TARSKI:def 1;
  hence thesis by A6,XBOOLE_0:def 5;
end;

theorem
  not s in ].r,s .[ \/ ].s,t.[
proof
  assume s in ].r,s .[ \/ ].s,t.[;
  then s in ].r,s .[ or s in ].s,t.[ by XBOOLE_0:def 3;
  hence contradiction by Th4;
end;

theorem
  not s in [.r,s .[ \/ ].s,t.[
proof
  assume s in [.r,s .[ \/ ].s,t.[;
  then s in [.r,s .[ or s in ].s,t.[ by XBOOLE_0:def 3;
  hence contradiction by Th3,Th4;
end;

theorem
  not s in ].r,s .[ \/ ].s,t.]
proof
  assume s in ].r,s .[ \/ ].s,t.];
  then s in ].r,s .[ or s in ].s,t.] by XBOOLE_0:def 3;
  hence contradiction by Th2,Th4;
end;

theorem
  not s in [.r,s .[ \/ ].s,t.]
proof
  assume s in [.r,s .[ \/ ].s,t.];
  then s in [.r,s .[ or s in ].s,t.] by XBOOLE_0:def 3;
  hence contradiction by Th2,Th3;
end;

begin 

theorem
  [.-infty,+infty.] = ExtREAL
proof
  let r;
  thus r in [.-infty,+infty.] implies r in ExtREAL by XXREAL_0:def 1;
  assume r in ExtREAL;
A1: -infty <= r by XXREAL_0:5;
  r <= +infty by XXREAL_0:3;
  hence thesis by A1,Th1;
end;

theorem 
  ].p,-infty.[ = {}
proof
  not ex x st x in ].p,-infty.[
  proof
    given x such that
A1: x in ].p,-infty.[;
    reconsider s = x as ext-real number by A1;
    s < -infty by A1,Th4;
    hence contradiction by XXREAL_0:5;
  end;
  hence thesis by XBOOLE_0:def 1;
end;

theorem 
  [.p,-infty.[ = {}
proof
  not ex x st x in [.p,-infty.[
  proof
    given x such that
A1: x in [.p,-infty.[;
    reconsider s = x as ext-real number by A1;
    s < -infty by A1,Th3;
    hence contradiction by XXREAL_0:5;
  end;
  hence thesis by XBOOLE_0:def 1;
end;

theorem 
  ].p,-infty.] = {}
proof
  not ex x st x in ].p,-infty.]
  proof
    given x such that
A1: x in ].p,-infty.];
    reconsider s = x as ext-real number by A1;
A2: p < s by A1,Th2;
    s <= -infty by A1,Th2;
    then p < -infty by A2,XXREAL_0:2;
    hence contradiction by XXREAL_0:5;
  end;
  hence thesis by XBOOLE_0:def 1;
end;

theorem 
  p <> -infty implies [.p,-infty.] = {}
proof
  assume
A1: p <> -infty;
  not ex x st x in [.p,-infty.]
  proof
    given x such that
A2: x in [.p,-infty.];
    reconsider s = x as ext-real number by A2;
A3: p <= s by A2,Th1;
    s <= -infty by A2,Th1;
    hence contradiction by A1,A3,XXREAL_0:2,6;
  end;
  hence thesis by XBOOLE_0:def 1;
end;

theorem 
  ].+infty,p.[ = {}
proof
  not ex x st x in ].+infty,p.[
  proof
    given x such that
A1: x in ].+infty,p.[;
    reconsider s = x as ext-real number by A1;
    +infty < s by A1,Th4;
    hence contradiction by XXREAL_0:3;
  end;
  hence thesis by XBOOLE_0:def 1;
end;

theorem 
  [.+infty,p.[ = {}
proof
  not ex x st x in [.+infty,p.[
  proof
    given x such that
A1: x in [.+infty,p.[;
    reconsider s = x as ext-real number by A1;
A2: +infty <= s by A1,Th3;
    s < p by A1,Th3;
    then p > +infty by A2,XXREAL_0:2;
    hence contradiction by XXREAL_0:3;
  end;
  hence thesis by XBOOLE_0:def 1;
end;

theorem 
  ].+infty,p.] = {}
proof
  not ex x st x in ].+infty,p.]
  proof
    given x such that
A1: x in ].+infty,p.];
    reconsider s = x as ext-real number by A1;
    +infty < s by A1,Th2;
    hence contradiction by XXREAL_0:3;
  end;
  hence thesis by XBOOLE_0:def 1;
end;

theorem 
  p <> +infty implies [.+infty,p.] = {}
proof
  assume
A1: p <> +infty;
  not ex x st x in [.+infty,p.]
  proof
    given x such that
A2: x in [.+infty,p.];
    reconsider s = x as ext-real number by A2;
A3: +infty <= s by A2,Th1;
    s <= p by A2,Th1;
    hence contradiction by A1,A3,XXREAL_0:2,4;
  end;
  hence thesis by XBOOLE_0:def 1;
end;

theorem
  p > q implies p in ].q, +infty.]
proof
  p <= +infty by XXREAL_0:3;
  hence thesis by Th2;
end;

theorem
  q <= p implies p in [.q, +infty.]
proof
  p <= +infty by XXREAL_0:3;
  hence thesis by Th1;
end;

theorem
  p <= q implies p in [.-infty,q.]
proof
  p >= -infty by XXREAL_0:5;
  hence thesis by Th1;
end;

theorem
  p < q implies p in [.-infty, q.[
proof
  p >= -infty by XXREAL_0:5;
  hence thesis by Th3;
end;

begin 

theorem
  p <= q implies [.p,q.] = [.p,q.] \/ [.q,p.]
proof
  assume
A1: p <= q;
  then
A2: [.q,p.] c= {p} by Th85;
  p in [.p,q.] by A1,Th1;
  then {p} c= [.p,q.] by ZFMISC_1:31;
  hence thesis by A2,XBOOLE_1:1,12;
end;

theorem
  r <= s & s <= t implies not r in ].s,t.[ \/ ].t,p.[
proof
  assume that
A1: r <= s and
A2: s <= t;
  assume r in ].s,t.[ \/ ].t,p.[;
  then r in ].s,t.[ or r in ].t,p.[ by XBOOLE_0:def 3;
  then s < r & r < t or t < r & r < p by Th4;
  hence contradiction by A1,A2,XXREAL_0:2;
end;



theorem Th224:
  REAL = ].-infty,+infty.[
proof
  let x be ext-real number;
  thus x in REAL implies x in ].-infty,+infty.[
  proof
    assume
A1: x in REAL;
    then
A2: -infty < x by XXREAL_0:12;
    x < +infty by A1,XXREAL_0:9;
    hence thesis by A2,Th4;
  end;
  assume
A3: x in ].-infty,+infty.[;
  then
A4: -infty < x by Th4;
  x < +infty by A3,Th4;
  hence thesis by A4,XXREAL_0:14;
end;

theorem Th225:
  ].p,q.[ c= REAL
proof
  let x be ext-real number;
  assume
A1: x in ].p,q.[;
  then
A2: p < x by Th4;
  x < q by A1,Th4;
  hence thesis by A2,XXREAL_0:48;
end;

theorem Th226:
  p in REAL implies [.p,q.[ c= REAL
proof
  assume
A1: p in REAL;
  let x be ext-real number;
  assume
A2: x in [.p,q.[;
  then
A3: p <= x by Th3;
  x < q by A2,Th3;
  hence thesis by A1,A3,XXREAL_0:46;
end;

theorem Th227:
  q in REAL implies ].p,q.] c= REAL
proof
  assume
A1: q in REAL;
  let x be ext-real number;
  assume
A2: x in ].p,q.];
  then
A3: p < x by Th2;
  x <= q by A2,Th2;
  hence thesis by A1,A3,XXREAL_0:47;
end;

theorem Th228:
  p in REAL & q in REAL implies [.p,q.] c= REAL
proof
  assume that
A1: p in REAL and
A2: q in REAL;
  let x be ext-real number;
  assume
A3: x in [.p,q.];
  then
A4: p <= x by Th1;
  x <= q by A3,Th1;
  hence thesis by A1,A2,A4,XXREAL_0:45;
end;

registration
  let p,q;
  cluster ].p,q.[ -> real-membered;
  coherence by Th225,MEMBERED:21;
end;

registration
  let p be real number, q;
  cluster [.p,q.[ -> real-membered;
  coherence
  proof
    p in REAL by XREAL_0:def 1;
    then [.p,q.[ c= REAL by Th226;
    hence thesis;
  end;
end;

registration
  let q be real number, p;
  cluster ].p,q.] -> real-membered;
  coherence
  proof
    q in REAL by XREAL_0:def 1;
    then ].p,q.] c= REAL by Th227;
    hence thesis;
  end;
end;

registration
  let p,q be real number;
  cluster [.p,q.] -> real-membered;
  coherence
  proof
A1: p in REAL by XREAL_0:def 1;
    q in REAL by XREAL_0:def 1;
    then [.p,q.] c= REAL by A1,Th228;
    hence thesis;
  end;
end;



theorem
  ].-infty,s.[ = {g where g is Real : g<s}
proof
  thus ].-infty,s.[ c= {g where g is Real : g<s}
  proof
    let x be real number;
    assume
A1: x in ].-infty,s.[;
    then
A2: -infty < x by Th4;
A3: x < s by A1,Th4;
    then x in REAL by A2,XXREAL_0:48;
    hence thesis by A3;
  end;
  let x be set;
  assume x in {g where g is Real : g<s};
  then consider g being Real such that
A4: x = g and
A5: g < s;
  -infty < g by XXREAL_0:12;
  hence thesis by A4,A5,Th4;
end;

theorem
  ].s,+infty.[ = {g where g is Real : s<g}
proof
  thus ].s,+infty.[ c= {g where g is Real : s<g}
  proof
    let x be real number;
    assume
A1: x in ].s,+infty.[;
    then
A2: s < x by Th4;
    x < +infty by A1,Th4;
    then x in REAL by A2,XXREAL_0:48;
    hence thesis by A2;
  end;
  let x;
  assume x in {g where g is Real : s<g};
  then consider g being Real such that
A3: x = g and
A4: s < g;
  g < +infty by XXREAL_0:9;
  hence thesis by A3,A4,Th4;
end;

theorem
  for s being real number holds ].-infty,s.] = {g where g is Real : g<=s}
proof
  let s be real number;
  thus ].-infty,s.] c= {g where g is Real : g<=s}
  proof
    let x be real number;
    assume x in ].-infty,s.];
    then
A1: x <= s by Th2;
    x in REAL by XREAL_0:def 1;
    hence thesis by A1;
  end;
  let x;
  assume x in {g where g is Real : g<=s};
  then consider g being Real such that
A2: x = g and
A3: g <= s;
  -infty < g by XXREAL_0:12;
  hence thesis by A2,A3,Th2;
end;

theorem
  for s being real number holds [.s,+infty.[ = {g where g is Real : s<=g}
proof
  let s be real number;
  thus [.s,+infty.[ c= {g where g is Real : s<=g}
  proof
    let x be real number;
    assume x in [.s,+infty.[;
    then
A1: s <= x by Th3;
    x in REAL by XREAL_0:def 1;
    hence thesis by A1;
  end;
  let x;
  assume x in {g where g is Real : s<=g};
  then consider g being Real such that
A2: x = g and
A3: s <= g;
  g < +infty by XXREAL_0:9;
  hence thesis by A2,A3,Th3;
end;

theorem
  for x being real number holds x in ].-infty,u.[ iff x < u
proof
  let x be real number;
  x in REAL by XREAL_0:def 1;
  then -infty < x by XXREAL_0:12;
  hence thesis by Th4;
end;

theorem
  for x being real number holds x in ].-infty,u.] iff x <= u
proof
  let x be real number;
  x in REAL by XREAL_0:def 1;
  then -infty < x by XXREAL_0:12;
  hence thesis by Th2;
end;

theorem
  for x being real number holds x in ].u,+infty.[ iff u < x
proof
  let x be real number;
  x in REAL by XREAL_0:def 1;
  then x < +infty by XXREAL_0:9;
  hence thesis by Th4;
end;

theorem
  for x being real number holds x in [.u,+infty.[ iff u <= x
proof
  let x be real number;
  x in REAL by XREAL_0:def 1;
  then x < +infty by XXREAL_0:9;
  hence thesis by Th3;
end;

theorem
  p <= r implies [.r,s.] c= [.p,+infty.]
proof
  s <= +infty by XXREAL_0:3;
  hence thesis by Th34;
end;

theorem
  p <= r implies [.r,s.[ c= [.p,+infty.]
proof
  s <= +infty by XXREAL_0:3;
  hence thesis by Th35;
end;

theorem
  p <= r implies ].r,s.] c= [.p,+infty.]
proof
  s <= +infty by XXREAL_0:3;
  hence thesis by Th36;
end;

theorem
  p <= r implies ].r,s.[ c= [.p,+infty.]
proof
  s <= +infty by XXREAL_0:3;
  hence thesis by Th37;
end;

theorem
  p <= r implies [.r,s.[ c= [.p,+infty.[
proof
  s <= +infty by XXREAL_0:3;
  hence thesis by Th38;
end;

theorem
  p < r implies [.r,s.] c= ].p,+infty.] by Th39,XXREAL_0:3;

theorem
  p < r implies [.r,s.[ c= ].p,+infty.] by Th40,XXREAL_0:3;

theorem
  p <= r implies ].r,s.[ c= ].p,+infty.]
proof
  s <= +infty by XXREAL_0:3;
  hence thesis by Th41;
end;

theorem
  p <= r implies ].r,s.] c= ].p,+infty.]
proof
  s <= +infty by XXREAL_0:3;
  hence thesis by Th42;
end;

theorem
  p <= r implies ].r,s.[ c= [.p,+infty.[
proof
  s <= +infty by XXREAL_0:3;
  hence thesis by Th45;
end;

theorem
  p <= r implies ].r,s.[ c= ].p,+infty.[
proof
  s <= +infty by XXREAL_0:3;
  hence thesis by Th46;
end;

theorem
  p < r implies [.r,s.[ c= ].p,+infty.[ by Th48,XXREAL_0:3;

theorem
  for s being real number st p < r holds [.r,s.] c= ].p,+infty.[
proof
  let s be real number;
  s in REAL by XREAL_0:def 1;
  then s < +infty by XXREAL_0:9;
  hence thesis by Th47;
end;

theorem
  for s being real number st p <= r holds ].r,s.] c= ].p,+infty.[
proof
  let s be real number;
  s in REAL by XREAL_0:def 1;
  hence thesis by Th49,XXREAL_0:9;
end;

theorem
  for s being real number st p <= r holds [.r,s.] c= [.p,+infty.[
proof
  let s be real number;
  s in REAL by XREAL_0:def 1;
  hence thesis by Th43,XXREAL_0:9;
end;

theorem
  for s being real number st p <= r holds ].r,s.] c= [.p,+infty.[
proof
  let s be real number;
  s in REAL by XREAL_0:def 1;
  hence thesis by Th44,XXREAL_0:9;
end;

theorem
  s <= q implies [.r,s.] c= [.-infty,q.]
proof
  -infty <= r by XXREAL_0:5;
  hence thesis by Th34;
end;

theorem
  s <= q implies [.r,s.[ c= [.-infty,q.]
proof
  -infty <= r by XXREAL_0:5;
  hence thesis by Th35;
end;

theorem
  s <= q implies ].r,s.] c= [.-infty,q.]
proof
  -infty <= r by XXREAL_0:5;
  hence thesis by Th36;
end;

theorem
  s <= q implies ].r,s.[ c= [.-infty,q.]
proof
  -infty <= r by XXREAL_0:5;
  hence thesis by Th37;
end;

theorem
  s <= q implies [.r,s.[ c= [.-infty,q.[
proof
  -infty <= r by XXREAL_0:5;
  hence thesis by Th38;
end;

theorem
  s <= q implies ].r,s.[ c= ].-infty,q.]
proof
  -infty <= r by XXREAL_0:5;
  hence thesis by Th41;
end;

theorem
  s <= q implies ].r,s.] c= ].-infty,q.]
proof
  -infty <= r by XXREAL_0:5;
  hence thesis by Th42;
end;

theorem
  s < q implies [.r,s.] c= [.-infty,q.[ by Th43,XXREAL_0:5;

theorem
  s < q implies ].r,s.] c= [.-infty,q.[ by Th44,XXREAL_0:5;

theorem
  s <= q implies ].r,s.[ c= [.-infty,q.[
proof
  -infty <= r by XXREAL_0:5;
  hence thesis by Th45;
end;

theorem
  s <= q implies ].r,s.[ c= ].-infty,q.[
proof
  -infty <= r by XXREAL_0:5;
  hence thesis by Th46;
end;

theorem
  s < q implies ].r,s.] c= ].-infty,q.[ by Th49,XXREAL_0:5;

theorem
  for r being real number st s <= q holds [.r,s.] c= ].-infty,q.]
proof
  let r be real number;
  r in REAL by XREAL_0:def 1;
  hence thesis by Th39,XXREAL_0:12;
end;

theorem
  for r being real number st s <= q holds [.r,s.[ c= ].-infty,q.]
proof
  let r be real number;
  r in REAL by XREAL_0:def 1;
  hence thesis by Th40,XXREAL_0:12;
end;

theorem
  for r being real number st s < q holds [.r,s.] c= ].-infty,q.[
proof
  let r be real number;
  r in REAL by XREAL_0:def 1;
  then -infty < r by XXREAL_0:12;
  hence thesis by Th47;
end;

theorem
  for r being real number st s <= q holds [.r,s.[ c= ].-infty,q.[
proof
  let r be real number;
  r in REAL by XREAL_0:def 1;
  hence thesis by Th48,XXREAL_0:12;
end;

theorem  
  for a,b being ext-real number holds ].-infty,b.[ /\ ].a,+infty.[ = ].a,b.[
proof
  let a,b be ext-real number;
A1: -infty <= a by XXREAL_0:5;
  b <= +infty by XXREAL_0:3;
  hence thesis by A1,Th160;
end;

theorem 
  for b being real number holds ].-infty,b.] /\ ].p,+infty.[ = ].p,b.]
proof
  let b be real number;
A1: b in REAL by XREAL_0:def 1;
  -infty <= p by XXREAL_0:5;
  hence thesis by A1,Th159,XXREAL_0:9;
end;

theorem
  for a being real number holds ].-infty,p.[ /\ [.a,+infty.[ = [.a,p.[
proof
  let a be real number;
  a in REAL by XREAL_0:def 1;
  then -infty < a by XXREAL_0:12;
  hence thesis by Th154,XXREAL_0:3;
end;

theorem 
  for a,b being real number holds ].-infty,a.] /\ [.b,+infty.[ = [.b,a.]
proof
  let a,b be real number;
A1: a in REAL by XREAL_0:def 1;
  b in REAL by XREAL_0:def 1;
  then
A2: -infty < b by XXREAL_0:12;
  a < +infty by A1,XXREAL_0:9;
  hence thesis by A2,Th152;
end;

theorem
  s <= p implies [.r,s.[ misses ].p,q.[
proof
  assume
A1: s <= p;
  let t;
  assume t in [.r,s.[;
  then t <= s by Th3;
  then t <= p by A1,XXREAL_0:2;
  hence thesis by Th4;
end;

theorem
  s <= p implies [.r,s.[ misses ].p,q.]
proof
  assume
A1: s <= p;
  let t;
  assume t in [.r,s.[;
  then t <= s by Th3;
  then t <= p by A1,XXREAL_0:2;
  hence thesis by Th2;
end;

theorem
  s <= p implies ].r,s.[ misses ].p,q.[
proof
  assume
A1: s <= p;
  let t be real number;
  assume t in ].r,s.[;
  then t <= s by Th4;
  then t <= p by A1,XXREAL_0:2;
  hence thesis by Th4;
end;

theorem
  s <= p implies ].r,s.[ misses ].p,q.]
proof
  assume
A1: s <= p;
  let t;
  assume t in ].r,s.[;
  then t <= s by Th4;
  then t <= p by A1,XXREAL_0:2;
  hence thesis by Th2;
end;

theorem
  s < p implies [.r,s.] misses [.p,q.[
proof
  assume
A1: s < p;
  let t;
  assume t in [.r,s.];
  then t <= s by Th1;
  then t < p by A1,XXREAL_0:2;
  hence thesis by Th3;
end;

theorem
  s < p implies [.r,s.] misses [.p,q.]
proof
  assume
A1: s < p;
  let t;
  assume t in [.r,s.];
  then t <= s by Th1;
  then t < p by A1,XXREAL_0:2;
  hence thesis by Th1;
end;

theorem
  s < p implies ].r,s.] misses [.p,q.[
proof
  assume
A1: s < p;
  let t;
  assume t in ].r,s.];
  then t <= s by Th2;
  then t < p by A1,XXREAL_0:2;
  hence thesis by Th3;
end;

theorem
  s < p implies ].r,s.] misses [.p,q.]
proof
  assume
A1: s < p;
  let t;
  assume t in ].r,s.];
  then t <= s by Th2;
  then t < p by A1,XXREAL_0:2;
  hence thesis by Th1;
end;

theorem
  [.-infty,t.] \ [.-infty,s.] = ].s,t.] by Th182,XXREAL_0:5;

theorem
  [.-infty,t.[ \ [.-infty,s.] = ].s,t.[ by Th183,XXREAL_0:5;

theorem
  [.-infty,t.] \ [.-infty,s.] = ].s,t.] by Th186,XXREAL_0:5;

theorem
  [.r,+infty.] \ [.s,+infty.] = [.r,s.[ by Th190,XXREAL_0:3;

theorem
  ].r,+infty.] \ [.s,+infty.] = ].r,s.[ by Th191,XXREAL_0:3;

theorem
  for s being real number holds [.-infty,t.] \ [.-infty,s.[ = [.s,t.]
proof
  let s be real number;
  s in REAL by XREAL_0:def 1;
  hence thesis by Th184,XXREAL_0:12;
end;

theorem
  for s being real number holds [.-infty,t.[ \ [.-infty,s.[ = [.s,t.[
proof
  let s be real number;
  s in REAL by XREAL_0:def 1;
  hence thesis by Th185,XXREAL_0:12;
end;

theorem
  for s being real number holds ].-infty,t.[ \ ].-infty,s.] = ].s,t.[
proof
  let s be real number;
  s in REAL by XREAL_0:def 1;
  hence thesis by Th187,XXREAL_0:12;
end;

theorem
  for s being real number holds ].-infty,t.] \ ].-infty,s.[ = [.s,t.]
proof
  let s be real number;
  s in REAL by XREAL_0:def 1;
  hence thesis by Th188,XXREAL_0:12;
end;

theorem
  for s being real number holds ].-infty,t.[ \ ].-infty,s.[ = [.s,t.[
proof
  let s be real number;
  s in REAL by XREAL_0:def 1;
  hence thesis by Th189,XXREAL_0:12;
end;

theorem
  for s being real number holds [.r,+infty.] \ ].s,+infty.] = [.r,s.]
proof
  let s be real number;
  s in REAL by XREAL_0:def 1;
  hence thesis by Th192,XXREAL_0:9;
end;

theorem
  for s being real number holds ].r,+infty.] \ ].s,+infty.] = ].r,s.]
proof
  let s be real number;
  s in REAL by XREAL_0:def 1;
  hence thesis by Th193,XXREAL_0:9;
end;

theorem
  for s being real number holds [.r,+infty.[ \ [.s,+infty.[ = [.r,s.[
proof
  let s be real number;
  s in REAL by XREAL_0:def 1;
  hence thesis by Th194,XXREAL_0:9;
end;

theorem
  for s being real number holds ].r,+infty.[ \ [.s,+infty.[ = ].r,s.[
proof
  let s be real number;
  s in REAL by XREAL_0:def 1;
  hence thesis by Th195,XXREAL_0:9;
end;

theorem
  for s being real number holds [.r,+infty.[ \ ].s,+infty.[ = [.r,s.]
proof
  let s be real number;
  s in REAL by XREAL_0:def 1;
  hence thesis by Th196,XXREAL_0:9;
end;

theorem
  for s being real number holds ].r,+infty.[ \ ].s,+infty.[ = ].r,s.]
proof
  let s be real number;
  s in REAL by XREAL_0:def 1;
  hence thesis by Th197,XXREAL_0:9;
end;

theorem Th297:
  r < s & p < q implies ].r,q.[ \ ].p,s.[ = ].r,p.] \/ [.s,q.[
proof
  assume that
A1: r < s and
A2: p < q;
  let x be ext-real number;
  thus x in ].r,q.[ \ ].p,s.[ implies x in ].r,p.] \/ [.s,q.[
  proof
    assume
A3: x in ].r,q.[ \ ].p,s.[;
    then
A4: not x in ].p,s.[ by XBOOLE_0:def 5;
A5: r < x by A3,Th4;
A6: x < q by A3,Th4;
    not(p < x & x < s) by A4,Th4;
    then x in ].r,p.] or x in [.s,q.[ by A5,A6,Th2,Th3;
    hence thesis by XBOOLE_0:def 3;
  end;
  assume x in ].r,p.] \/ [.s,q.[;
  then x in ].r,p.] or x in [.s,q.[ by XBOOLE_0:def 3;
  then
A7: r < x & x <= p or s <= x & x < q by Th2,Th3;
  then
A8: r < x by A1,XXREAL_0:2;
  x < q by A2,A7,XXREAL_0:2;
  then
A9: x in ].r,q.[ by A8,Th4;
  not x in ].p,s.[ by A7,Th4;
  hence thesis by A9,XBOOLE_0:def 5;
end;

theorem Th298:
  r <= s & p < q implies [.r,q.[ \ ].p,s.[ = [.r,p.] \/ [.s,q.[
proof
  assume that
A1: r <= s and
A2: p < q;
  let x be ext-real number;
  thus x in [.r,q.[ \ ].p,s.[ implies x in [.r,p.] \/ [.s,q.[
  proof
    assume
A3: x in [.r,q.[ \ ].p,s.[;
    then
A4: not x in ].p,s.[ by XBOOLE_0:def 5;
A5: r <= x by A3,Th3;
A6: x < q by A3,Th3;
    not(p < x & x < s) by A4,Th4;
    then x in [.r,p.] or x in [.s,q.[ by A5,A6,Th1,Th3;
    hence thesis by XBOOLE_0:def 3;
  end;
  assume x in [.r,p.] \/ [.s,q.[;
  then x in [.r,p.] or x in [.s,q.[ by XBOOLE_0:def 3;
  then
A7: r <= x & x <= p or s <= x & x < q by Th1,Th3;
  then
A8: r <= x by A1,XXREAL_0:2;
  x < q by A2,A7,XXREAL_0:2;
  then
A9: x in [.r,q.[ by A8,Th3;
  not x in ].p,s.[ by A7,Th4;
  hence thesis by A9,XBOOLE_0:def 5;
end;

theorem Th299:
  r < s & p <= q implies ].r,q.] \ ].p,s.[ = ].r,p.] \/ [.s,q.]
proof
  assume that
A1: r < s and
A2: p <= q;
  let x be ext-real number;
  thus x in ].r,q.] \ ].p,s.[ implies x in ].r,p.] \/ [.s,q.]
  proof
    assume
A3: x in ].r,q.] \ ].p,s.[;
    then
A4: not x in ].p,s.[ by XBOOLE_0:def 5;
A5: r < x by A3,Th2;
A6: x <= q by A3,Th2;
    not(p < x & x < s) by A4,Th4;
    then x in ].r,p.] or x in [.s,q.] by A5,A6,Th1,Th2;
    hence thesis by XBOOLE_0:def 3;
  end;
  assume x in ].r,p.] \/ [.s,q.];
  then x in ].r,p.] or x in [.s,q.] by XBOOLE_0:def 3;
  then
A7: r < x & x <= p or s <= x & x <= q by Th1,Th2;
  then
A8: r < x by A1,XXREAL_0:2;
  x <= q by A2,A7,XXREAL_0:2;
  then
A9: x in ].r,q.] by A8,Th2;
  not x in ].p,s.[ by A7,Th4;
  hence thesis by A9,XBOOLE_0:def 5;
end;

theorem Th300:
  r <= s & p <= q implies [.r,q.] \ ].p,s.[ = [.r,p.] \/ [.s,q.]
proof
  assume that
A1: r <= s and
A2: p <= q;
  let x be ext-real number;
  thus x in [.r,q.] \ ].p,s.[ implies x in [.r,p.] \/ [.s,q.]
  proof
    assume
A3: x in [.r,q.] \ ].p,s.[;
    then
A4: not x in ].p,s.[ by XBOOLE_0:def 5;
A5: r <= x by A3,Th1;
A6: x <= q by A3,Th1;
    not(p < x & x < s) by A4,Th4;
    then x in [.r,p.] or x in [.s,q.] by A5,A6,Th1;
    hence thesis by XBOOLE_0:def 3;
  end;
  assume x in [.r,p.] \/ [.s,q.];
  then x in [.r,p.] or x in [.s,q.] by XBOOLE_0:def 3;
  then
A7: r <= x & x <= p or s <= x & x <= q by Th1;
  then
A8: r <= x by A1,XXREAL_0:2;
  x <= q by A2,A7,XXREAL_0:2;
  then
A9: x in [.r,q.] by A8,Th1;
  not x in ].p,s.[ by A7,Th4;
  hence thesis by A9,XBOOLE_0:def 5;
end;

theorem Th301:
  r < s & p <= q implies ].r,q.[ \ [.p,s.[ = ].r,p.[ \/ [.s,q.[
proof
  assume that
A1: r < s and
A2: p <= q;
  let x be ext-real number;
  thus x in ].r,q.[ \ [.p,s.[ implies x in ].r,p.[ \/ [.s,q.[
  proof
    assume
A3: x in ].r,q.[ \ [.p,s.[;
    then
A4: not x in [.p,s.[ by XBOOLE_0:def 5;
A5: r < x by A3,Th4;
A6: x < q by A3,Th4;
    not(p <= x & x < s) by A4,Th3;
    then x in ].r,p.[ or x in [.s,q.[ by A5,A6,Th3,Th4;
    hence thesis by XBOOLE_0:def 3;
  end;
  assume x in ].r,p.[ \/ [.s,q.[;
  then x in ].r,p.[ or x in [.s,q.[ by XBOOLE_0:def 3;
  then
A7: r < x & x < p or s <= x & x < q by Th3,Th4;
  then
A8: r < x by A1,XXREAL_0:2;
  x < q by A2,A7,XXREAL_0:2;
  then
A9: x in ].r,q.[ by A8,Th4;
  not x in [.p,s.[ by A7,Th3;
  hence thesis by A9,XBOOLE_0:def 5;
end;

theorem Th302:
  r <= s & p <= q implies [.r,q.[ \ [.p,s.[ = [.r,p.[ \/ [.s,q.[
proof
  assume that
A1: r <= s and
A2: p <= q;
  let x be ext-real number;
  thus x in [.r,q.[ \ [.p,s.[ implies x in [.r,p.[ \/ [.s,q.[
  proof
    assume
A3: x in [.r,q.[ \ [.p,s.[;
    then
A4: not x in [.p,s.[ by XBOOLE_0:def 5;
A5: r <= x by A3,Th3;
A6: x < q by A3,Th3;
    not(p <= x & x < s) by A4,Th3;
    then x in [.r,p.[ or x in [.s,q.[ by A5,A6,Th3;
    hence thesis by XBOOLE_0:def 3;
  end;
  assume x in [.r,p.[ \/ [.s,q.[;
  then x in [.r,p.[ or x in [.s,q.[ by XBOOLE_0:def 3;
  then
A7: r <= x & x < p or s <= x & x < q by Th3;
  then
A8: r <= x by A1,XXREAL_0:2;
  x < q by A2,A7,XXREAL_0:2;
  then
A9: x in [.r,q.[ by A8,Th3;
  not x in [.p,s.[ by A7,Th3;
  hence thesis by A9,XBOOLE_0:def 5;
end;

theorem Th303:
  r < s & p <= q implies ].r,q.] \ [.p,s.[ = ].r,p.[ \/ [.s,q.]
proof
  assume that
A1: r < s and
A2: p <= q;
  let x be ext-real number;
  thus x in ].r,q.] \ [.p,s.[ implies x in ].r,p.[ \/ [.s,q.]
  proof
    assume
A3: x in ].r,q.] \ [.p,s.[;
    then
A4: not x in [.p,s.[ by XBOOLE_0:def 5;
A5: r < x by A3,Th2;
A6: x <= q by A3,Th2;
    not(p <= x & x < s) by A4,Th3;
    then x in ].r,p.[ or x in [.s,q.] by A5,A6,Th1,Th4;
    hence thesis by XBOOLE_0:def 3;
  end;
  assume x in ].r,p.[ \/ [.s,q.];
  then x in ].r,p.[ or x in [.s,q.] by XBOOLE_0:def 3;
  then
A7: r < x & x < p or s <= x & x <= q by Th1,Th4;
  then
A8: r < x by A1,XXREAL_0:2;
  x <= q by A2,A7,XXREAL_0:2;
  then
A9: x in ].r,q.] by A8,Th2;
  not x in [.p,s.[ by A7,Th3;
  hence thesis by A9,XBOOLE_0:def 5;
end;

theorem Th304:
  r <= s & p <= q implies [.r,q.] \ [.p,s.[ = [.r,p.[ \/ [.s,q.]
proof
  assume that
A1: r <= s and
A2: p <= q;
  let x be ext-real number;
  thus x in [.r,q.] \ [.p,s.[ implies x in [.r,p.[ \/ [.s,q.]
  proof
    assume
A3: x in [.r,q.] \ [.p,s.[;
    then
A4: not x in [.p,s.[ by XBOOLE_0:def 5;
A5: r <= x by A3,Th1;
A6: x <= q by A3,Th1;
    not(p <= x & x < s) by A4,Th3;
    then x in [.r,p.[ or x in [.s,q.] by A5,A6,Th1,Th3;
    hence thesis by XBOOLE_0:def 3;
  end;
  assume x in [.r,p.[ \/ [.s,q.];
  then x in [.r,p.[ or x in [.s,q.] by XBOOLE_0:def 3;
  then
A7: r <= x & x < p or s <= x & x <= q by Th1,Th3;
  then
A8: r <= x by A1,XXREAL_0:2;
  x <= q by A2,A7,XXREAL_0:2;
  then
A9: x in [.r,q.] by A8,Th1;
  not x in [.p,s.[ by A7,Th3;
  hence thesis by A9,XBOOLE_0:def 5;
end;

theorem Th305:
  r < s & p < q implies ].r,q.[ \ ].p,s.] = ].r,p.] \/ ].s,q.[
proof
  assume that
A1: r < s and
A2: p < q;
  let x be ext-real number;
  thus x in ].r,q.[ \ ].p,s.] implies x in ].r,p.] \/ ].s,q.[
  proof
    assume
A3: x in ].r,q.[ \ ].p,s.];
    then
A4: not x in ].p,s.] by XBOOLE_0:def 5;
A5: r < x by A3,Th4;
A6: x < q by A3,Th4;
    not(p < x & x <= s) by A4,Th2;
    then x in ].r,p.] or x in ].s,q.[ by A5,A6,Th2,Th4;
    hence thesis by XBOOLE_0:def 3;
  end;
  assume x in ].r,p.] \/ ].s,q.[;
  then x in ].r,p.] or x in ].s,q.[ by XBOOLE_0:def 3;
  then
A7: r < x & x <= p or s < x & x < q by Th2,Th4;
  then
A8: r < x by A1,XXREAL_0:2;
  x < q by A2,A7,XXREAL_0:2;
  then
A9: x in ].r,q.[ by A8,Th4;
  not x in ].p,s.] by A7,Th2;
  hence thesis by A9,XBOOLE_0:def 5;
end;

theorem Th306:
  r <= s & p < q implies [.r,q.[ \ ].p,s.] = [.r,p.] \/ ].s,q.[
proof
  assume that
A1: r <= s and
A2: p < q;
  let x be ext-real number;
  thus x in [.r,q.[ \ ].p,s.] implies x in [.r,p.] \/ ].s,q.[
  proof
    assume
A3: x in [.r,q.[ \ ].p,s.];
    then
A4: not x in ].p,s.] by XBOOLE_0:def 5;
A5: r <= x by A3,Th3;
A6: x < q by A3,Th3;
    not(p < x & x <= s) by A4,Th2;
    then x in [.r,p.] or x in ].s,q.[ by A5,A6,Th1,Th4;
    hence thesis by XBOOLE_0:def 3;
  end;
  assume x in [.r,p.] \/ ].s,q.[;
  then x in [.r,p.] or x in ].s,q.[ by XBOOLE_0:def 3;
  then
A7: r <= x & x <= p or s < x & x < q by Th1,Th4;
  then
A8: r <= x by A1,XXREAL_0:2;
  x < q by A2,A7,XXREAL_0:2;
  then
A9: x in [.r,q.[ by A8,Th3;
  not x in ].p,s.] by A7,Th2;
  hence thesis by A9,XBOOLE_0:def 5;
end;

theorem Th307:
  r < s & p <= q implies ].r,q.] \ ].p,s.] = ].r,p.] \/ ].s,q.]
proof
  assume that
A1: r < s and
A2: p <= q;
  let x be ext-real number;
  thus x in ].r,q.] \ ].p,s.] implies x in ].r,p.] \/ ].s,q.]
  proof
    assume
A3: x in ].r,q.] \ ].p,s.];
    then
A4: not x in ].p,s.] by XBOOLE_0:def 5;
A5: r < x by A3,Th2;
A6: x <= q by A3,Th2;
    not(p < x & x <= s) by A4,Th2;
    then x in ].r,p.] or x in ].s,q.] by A5,A6,Th2;
    hence thesis by XBOOLE_0:def 3;
  end;
  assume x in ].r,p.] \/ ].s,q.];
  then x in ].r,p.] or x in ].s,q.] by XBOOLE_0:def 3;
  then
A7: r < x & x <= p or s < x & x <= q by Th2;
  then
A8: r < x by A1,XXREAL_0:2;
  x <= q by A2,A7,XXREAL_0:2;
  then
A9: x in ].r,q.] by A8,Th2;
  not x in ].p,s.] by A7,Th2;
  hence thesis by A9,XBOOLE_0:def 5;
end;

theorem Th308:
  r <= s & p <= q implies [.r,q.] \ ].p,s.] = [.r,p.] \/ ].s,q.]
proof
  assume that
A1: r <= s and
A2: p <= q;
  let x be ext-real number;
  thus x in [.r,q.] \ ].p,s.] implies x in [.r,p.] \/ ].s,q.]
  proof
    assume
A3: x in [.r,q.] \ ].p,s.];
    then
A4: not x in ].p,s.] by XBOOLE_0:def 5;
A5: r <= x by A3,Th1;
A6: x <= q by A3,Th1;
    not(p < x & x <= s) by A4,Th2;
    then x in [.r,p.] or x in ].s,q.] by A5,A6,Th1,Th2;
    hence thesis by XBOOLE_0:def 3;
  end;
  assume x in [.r,p.] \/ ].s,q.];
  then x in [.r,p.] or x in ].s,q.] by XBOOLE_0:def 3;
  then
A7: r <= x & x <= p or s < x & x <= q by Th1,Th2;
  then
A8: r <= x by A1,XXREAL_0:2;
  x <= q by A2,A7,XXREAL_0:2;
  then
A9: x in [.r,q.] by A8,Th1;
  not x in ].p,s.] by A7,Th2;
  hence thesis by A9,XBOOLE_0:def 5;
end;

theorem Th309:
  r <= s & p <= q implies ].r,q.[ \ [.p,s.] = ].r,p.[ \/ ].s,q.[
proof
  assume that
A1: r <= s and
A2: p <= q;
  let x be real number;
  thus x in ].r,q.[ \ [.p,s.] implies x in ].r,p.[ \/ ].s,q.[
  proof
    assume
A3: x in ].r,q.[ \ [.p,s.];
    then
A4: not x in [.p,s.] by XBOOLE_0:def 5;
A5: r < x by A3,Th4;
A6: x < q by A3,Th4;
    not(p <= x & x <= s) by A4,Th1;
    then x in ].r,p.[ or x in ].s,q.[ by A5,A6,Th4;
    hence thesis by XBOOLE_0:def 3;
  end;
  assume x in ].r,p.[ \/ ].s,q.[;
  then x in ].r,p.[ or x in ].s,q.[ by XBOOLE_0:def 3;
  then
A7: r < x & x < p or s < x & x < q by Th4;
  then
A8: r < x by A1,XXREAL_0:2;
  x < q by A2,A7,XXREAL_0:2;
  then
A9: x in ].r,q.[ by A8,Th4;
  not x in [.p,s.] by A7,Th1;
  hence thesis by A9,XBOOLE_0:def 5;
end;

theorem Th310:
  r <= s & p <= q implies [.r,q.[ \ [.p,s.] = [.r,p.[ \/ ].s,q.[
proof
  assume that
A1: r <= s and
A2: p <= q;
  let x be ext-real number;
  thus x in [.r,q.[ \ [.p,s.] implies x in [.r,p.[ \/ ].s,q.[
  proof
    assume
A3: x in [.r,q.[ \ [.p,s.];
    then
A4: not x in [.p,s.] by XBOOLE_0:def 5;
A5: r <= x by A3,Th3;
A6: x < q by A3,Th3;
    not(p <= x & x <= s) by A4,Th1;
    then x in [.r,p.[ or x in ].s,q.[ by A5,A6,Th3,Th4;
    hence thesis by XBOOLE_0:def 3;
  end;
  assume x in [.r,p.[ \/ ].s,q.[;
  then x in [.r,p.[ or x in ].s,q.[ by XBOOLE_0:def 3;
  then
A7: r <= x & x < p or s < x & x < q by Th3,Th4;
  then
A8: r <= x by A1,XXREAL_0:2;
  x < q by A2,A7,XXREAL_0:2;
  then
A9: x in [.r,q.[ by A8,Th3;
  not x in [.p,s.] by A7,Th1;
  hence thesis by A9,XBOOLE_0:def 5;
end;

theorem Th311:
  r < s & p <= q implies ].r,q.] \ [.p,s.] = ].r,p.[ \/ ].s,q.]
proof
  assume that
A1: r < s and
A2: p <= q;
  let x be ext-real number;
  thus x in ].r,q.] \ [.p,s.] implies x in ].r,p.[ \/ ].s,q.]
  proof
    assume
A3: x in ].r,q.] \ [.p,s.];
    then
A4: not x in [.p,s.] by XBOOLE_0:def 5;
A5: r < x by A3,Th2;
A6: x <= q by A3,Th2;
    not(p <= x & x <= s) by A4,Th1;
    then x in ].r,p.[ or x in ].s,q.] by A5,A6,Th2,Th4;
    hence thesis by XBOOLE_0:def 3;
  end;
  assume x in ].r,p.[ \/ ].s,q.];
  then x in ].r,p.[ or x in ].s,q.] by XBOOLE_0:def 3;
  then
A7: r < x & x < p or s < x & x <= q by Th2,Th4;
  then
A8: r < x by A1,XXREAL_0:2;
  x <= q by A2,A7,XXREAL_0:2;
  then
A9: x in ].r,q.] by A8,Th2;
  not x in [.p,s.] by A7,Th1;
  hence thesis by A9,XBOOLE_0:def 5;
end;

theorem Th312:
  r <= s & p <= q implies [.r,q.] \ [.p,s.] = [.r,p.[ \/ ].s,q.]
proof
  assume that
A1: r <= s and
A2: p <= q;
  let x be ext-real number;
  thus x in [.r,q.] \ [.p,s.] implies x in [.r,p.[ \/ ].s,q.]
  proof
    assume
A3: x in [.r,q.] \ [.p,s.];
    then
A4: not x in [.p,s.] by XBOOLE_0:def 5;
A5: r <= x by A3,Th1;
A6: x <= q by A3,Th1;
    not(p <= x & x <= s) by A4,Th1;
    then x in [.r,p.[ or x in ].s,q.] by A5,A6,Th2,Th3;
    hence thesis by XBOOLE_0:def 3;
  end;
  assume x in [.r,p.[ \/ ].s,q.];
  then x in [.r,p.[ or x in ].s,q.] by XBOOLE_0:def 3;
  then
A7: r <= x & x < p or s < x & x <= q by Th2,Th3;
  then
A8: r <= x by A1,XXREAL_0:2;
  x <= q by A2,A7,XXREAL_0:2;
  then
A9: x in [.r,q.] by A8,Th1;
  not x in [.p,s.] by A7,Th1;
  hence thesis by A9,XBOOLE_0:def 5;
end;

theorem Th313:
  r <= p & p <= q implies ].r,q.[ \ {p} = ].r,p.[ \/ ].p,q.[
proof
  [.p,p.] = {p} by Th17;
  hence thesis by Th309;
end;

theorem Th314:
  r <= p & p <= q implies [.r,q.[ \ {p} = [.r,p.[ \/ ].p,q.[
proof
  [.p,p.] = {p} by Th17;
  hence thesis by Th310;
end;

theorem Th315:
  r < p & p <= q implies ].r,q.] \ {p} = ].r,p.[ \/ ].p,q.]
proof
  [.p,p.] = {p} by Th17;
  hence thesis by Th311;
end;

theorem Th316:
  r <= p & p <= q implies [.r,q.] \ {p} = [.r,p.[ \/ ].p,q.]
proof
  [.p,p.] = {p} by Th17;
  hence thesis by Th312;
end;

theorem Th317:
  r < q & p <= q implies ].r,q.] \ ].p,q.[ = ].r,p.] \/ {q}
proof
  [.q,q.] = {q} by Th17;
  hence thesis by Th299;
end;

theorem Th318:
  r <= q & p <= q implies [.r,q.] \ ].p,q.[ = [.r,p.] \/ {q}
proof
  [.q,q.] = {q} by Th17;
  hence thesis by Th300;
end;

theorem Th319:
  r < q & p <= q implies ].r,q.] \ [.p,q.[ = ].r,p.[ \/ {q}
proof
  [.q,q.] = {q} by Th17;
  hence thesis by Th303;
end;

theorem Th320:
  r <= q & p <= q implies [.r,q.] \ [.p,q.[ = [.r,p.[ \/ {q}
proof
  [.q,q.] = {q} by Th17;
  hence thesis by Th304;
end;

theorem Th321:
  p <= s & p < q implies [.p,q.[ \ ].p,s.[ = {p} \/ [.s,q.[
proof
  [.p,p.] = {p} by Th17;
  hence thesis by Th298;
end;

theorem Th322:
  p <= s & p <= q implies [.p,q.] \ ].p,s.[ = {p} \/ [.s,q.]
proof
  [.p,p.] = {p} by Th17;
  hence thesis by Th300;
end;

theorem Th323:
  p <= s & p < q implies [.p,q.[ \ ].p,s.] = {p} \/ ].s,q.[
proof
  [.p,p.] = {p} by Th17;
  hence thesis by Th306;
end;

theorem Th324:
  p <= s & p <= q implies [.p,q.] \ ].p,s.] = {p} \/ ].s,q.]
proof
  [.p,p.] = {p} by Th17;
  hence thesis by Th308;
end;

theorem Th325:
  p < q implies [.-infty,q.[ \ ].p,s.[ = [.-infty,p.] \/ [.s,q.[
  by Th298,XXREAL_0:5;

theorem Th326:
  p <= q implies [.-infty,q.] \ ].p,s.[ = [.-infty,p.] \/ [.s,q.]
proof
  -infty <= s by XXREAL_0:5;
  hence thesis by Th300;
end;

theorem Th327:
  p <= q implies [.-infty,q.[ \ [.p,s.[ = [.-infty,p.[ \/ [.s,q.[
proof
  -infty <= s by XXREAL_0:5;
  hence thesis by Th302;
end;

theorem Th328:
  p <= q implies [.-infty,q.] \ [.p,s.[ = [.-infty,p.[ \/ [.s,q.]
proof
  -infty <= s by XXREAL_0:5;
  hence thesis by Th304;
end;

theorem Th329:
  p < q implies [.-infty,q.[ \ ].p,s.] = [.-infty,p.] \/ ].s,q.[
  by Th306,XXREAL_0:5;

theorem Th330:
  p <= q implies [.-infty,q.] \ ].p,s.] = [.-infty,p.] \/ ].s,q.]
proof
  -infty <= s by XXREAL_0:5;
  hence thesis by Th308;
end;

theorem Th331:
  p <= q implies [.-infty,q.[ \ [.p,s.] = [.-infty,p.[ \/ ].s,q.[
proof
  -infty <= s by XXREAL_0:5;
  hence thesis by Th310;
end;

theorem Th332:
  p <= q implies [.-infty,q.] \ [.p,s.] = [.-infty,p.[ \/ ].s,q.]
proof
  -infty <= s by XXREAL_0:5;
  hence thesis by Th312;
end;

theorem Th333:
  for s being real number holds p < q
  implies ].-infty,q.[ \ ].p,s.[ = ].-infty,p.] \/ [.s,q.[
proof
  let s be real number;
  s in REAL by XREAL_0:def 1;
  then -infty < s by XXREAL_0:12;
  hence thesis by Th297;
end;

theorem Th334:
  for s being real number holds p <= q
  implies ].-infty,q.] \ ].p,s.[ = ].-infty,p.] \/ [.s,q.]
proof
  let s be real number;
  s in REAL by XREAL_0:def 1;
  hence thesis by Th299,XXREAL_0:12;
end;

theorem Th335:
  for s being real number holds p <= q
  implies ].-infty,q.[ \ [.p,s.[ = ].-infty,p.[ \/ [.s,q.[
proof
  let s be real number;
  s in REAL by XREAL_0:def 1;
  hence thesis by Th301,XXREAL_0:12;
end;

theorem Th336:
  for s being real number holds p <= q
  implies ].-infty,q.] \ [.p,s.[ = ].-infty,p.[ \/ [.s,q.]
proof
  let s be real number;
  s in REAL by XREAL_0:def 1;
  hence thesis by Th303,XXREAL_0:12;
end;

theorem Th337:
  for s being real number holds p < q
  implies ].-infty,q.[ \ ].p,s.] = ].-infty,p.] \/ ].s,q.[
proof
  let s be real number;
  s in REAL by XREAL_0:def 1;
  then -infty < s by XXREAL_0:12;
  hence thesis by Th305;
end;

theorem Th338:
  for s being real number holds p <= q
  implies ].-infty,q.] \ ].p,s.] = ].-infty,p.] \/ ].s,q.]
proof
  let s be real number;
  s in REAL by XREAL_0:def 1;
  hence thesis by Th307,XXREAL_0:12;
end;

theorem Th339:
  p <= q implies ].-infty,q.[ \ [.p,s.] = ].-infty,p.[ \/ ].s,q.[
proof
  -infty <= s by XXREAL_0:5;
  hence thesis by Th309;
end;

theorem Th340:
  for s being real number holds p <= q
  implies ].-infty,q.] \ [.p,s.] = ].-infty,p.[ \/ ].s,q.]
proof
  let s be real number;
  s in REAL by XREAL_0:def 1;
  hence thesis by Th311,XXREAL_0:12;
end;

theorem Th341:
  p <= q implies [.-infty,q.[ \ {p} = [.-infty,p.[ \/ ].p,q.[
proof
  -infty <= p by XXREAL_0:5;
  hence thesis by Th314;
end;

theorem Th342:
  p <= q implies [.-infty,q.] \ {p} = [.-infty,p.[ \/ ].p,q.]
proof
  -infty <= p by XXREAL_0:5;
  hence thesis by Th316;
end;

theorem
  p <= q implies [.-infty,q.] \ ].p,q.[ = [.-infty,p.] \/ {q}
proof
  -infty <= q by XXREAL_0:5;
  hence thesis by Th318;
end;

theorem
  p <= q implies [.-infty,q.] \ [.p,q.[ = [.-infty,p.[ \/ {q}
proof
  -infty <= q by XXREAL_0:5;
  hence thesis by Th320;
end;

theorem
  [.-infty,q.] \ ].-infty,s.[ = {-infty} \/ [.s,q.]
proof
A1: -infty <= s by XXREAL_0:5;
  -infty <= q by XXREAL_0:5;
  hence thesis by A1,Th322;
end;

theorem
  [.-infty,q.] \ ].-infty,s.] = {-infty} \/ ].s,q.]
proof
A1: -infty <= s by XXREAL_0:5;
  -infty <= q by XXREAL_0:5;
  hence thesis by A1,Th324;
end;

theorem
  for q being real number holds
  [.-infty,q.[ \ ].-infty,s.[ = {-infty} \/ [.s,q.[
proof
  let q be real number;
A1: q in REAL by XREAL_0:def 1;
  -infty <= s by XXREAL_0:5;
  hence thesis by A1,Th321,XXREAL_0:12;
end;

theorem
  for q being real number holds
  [.-infty,q.[ \ ].-infty,s.] = {-infty} \/ ].s,q.[
proof
  let q be real number;
A1: q in REAL by XREAL_0:def 1;
  -infty <= s by XXREAL_0:5;
  hence thesis by A1,Th323,XXREAL_0:12;
end;

theorem Th349:
  p <= q implies ].-infty,q.[ \ {p} = ].-infty,p.[ \/ ].p,q.[
proof
  -infty <= p by XXREAL_0:5;
  hence thesis by Th313;
end;

theorem Th350:
  for p being real number holds p <= q
  implies ].-infty,q.] \ {p} = ].-infty,p.[ \/ ].p,q.]
proof
  let p be real number;
  p in REAL by XREAL_0:def 1;
  hence thesis by Th315,XXREAL_0:12;
end;

theorem
  for q being real number holds p <= q
  implies ].-infty,q.] \ ].p,q.[ = ].-infty,p.] \/ {q}
proof
  let q be real number;
  q in REAL by XREAL_0:def 1;
  hence thesis by Th317,XXREAL_0:12;
end;

theorem
  for q being real number holds p <= q
  implies ].-infty,q.] \ [.p,q.[ = ].-infty,p.[ \/ {q}
proof
  let q be real number;
  q in REAL by XREAL_0:def 1;
  hence thesis by Th319,XXREAL_0:12;
end;

theorem
  r < s implies ].r,+infty.] \ ].p,s.[ = ].r,p.] \/ [.s,+infty.]
  by Th299,XXREAL_0:3;

theorem
  r <= s implies [.r,+infty.] \ ].p,s.[ = [.r,p.] \/ [.s,+infty.]
proof
  p <= +infty by XXREAL_0:3;
  hence thesis by Th300;
end;

theorem
  r < s implies ].r,+infty.[ \ [.p,s.[ = ].r,p.[ \/ [.s,+infty.[
  by Th301,XXREAL_0:3;

theorem
  r <= s implies [.r,+infty.[ \ [.p,s.[ = [.r,p.[ \/ [.s,+infty.[
proof
  p <= +infty by XXREAL_0:3;
  hence thesis by Th302;
end;

theorem
  r < s implies ].r,+infty.] \ [.p,s.[ = ].r,p.[ \/ [.s,+infty.]
  by Th303,XXREAL_0:3;

theorem
  r <= s implies [.r,+infty.] \ [.p,s.[ = [.r,p.[ \/ [.s,+infty.]
proof
  p <= +infty by XXREAL_0:3;
  hence thesis by Th304;
end;

theorem
  r < s implies ].r,+infty.] \ ].p,s.] = ].r,p.] \/ ].s,+infty.]
  by Th307,XXREAL_0:3;

theorem
  r <= s implies [.r,+infty.] \ ].p,s.] = [.r,p.] \/ ].s,+infty.]
proof
  p <= +infty by XXREAL_0:3;
  hence thesis by Th308;
end;

theorem
  r <= s implies ].r,+infty.[ \ [.p,s.] = ].r,p.[ \/ ].s,+infty.[
proof
  p <= +infty by XXREAL_0:3;
  hence thesis by Th309;
end;

theorem
  r <= s implies [.r,+infty.[ \ [.p,s.] = [.r,p.[ \/ ].s,+infty.[
proof
  p <= +infty by XXREAL_0:3;
  hence thesis by Th310;
end;

theorem
  r < s implies ].r,+infty.] \ [.p,s.] = ].r,p.[ \/ ].s,+infty.]
  by Th311,XXREAL_0:3;

theorem
  r <= s implies [.r,+infty.] \ [.p,s.] = [.r,p.[ \/ ].s,+infty.]
proof
  p <= +infty by XXREAL_0:3;
  hence thesis by Th312;
end;

theorem
  r <= p implies ].r,+infty.[ \ {p} = ].r,p.[ \/ ].p,+infty.[
proof
  p <= +infty by XXREAL_0:3;
  hence thesis by Th313;
end;

theorem
  r <= p implies [.r,+infty.[ \ {p} = [.r,p.[ \/ ].p,+infty.[
proof
  p <= +infty by XXREAL_0:3;
  hence thesis by Th314;
end;

theorem
  r < p implies ].r,+infty.] \ {p} = ].r,p.[ \/ ].p,+infty.]
  by Th315,XXREAL_0:3;

theorem
  r <= p implies [.r,+infty.] \ {p} = [.r,p.[ \/ ].p,+infty.]
proof
  p <= +infty by XXREAL_0:3;
  hence thesis by Th316;
end;

theorem
  [.r,+infty.] \ ].p,+infty.[ = [.r,p.] \/ {+infty}
proof
A1: r <= +infty by XXREAL_0:3;
  p <= +infty by XXREAL_0:3;
  hence thesis by A1,Th318;
end;

theorem
  [.r,+infty.] \ [.p,+infty.[ = [.r,p.[ \/ {+infty}
proof
A1: r <= +infty by XXREAL_0:3;
  p <= +infty by XXREAL_0:3;
  hence thesis by A1,Th320;
end;

theorem
  for r being real number
  holds ].r,+infty.] \ ].p,+infty.[ = ].r,p.] \/ {+infty}
proof
  let r be real number;
  r in REAL by XREAL_0:def 1;
  then r < +infty by XXREAL_0:9;
  hence thesis by Th317,XXREAL_0:3;
end;

theorem
  for r being real number
  holds ].r,+infty.] \ [.p,+infty.[ = ].r,p.[ \/ {+infty}
proof
  let r be real number;
  r in REAL by XREAL_0:def 1;
  then r < +infty by XXREAL_0:9;
  hence thesis by Th319,XXREAL_0:3;
end;

theorem
  p <= s implies [.p,+infty.] \ ].p,s.[ = {p} \/ [.s,+infty.]
proof
  p <= +infty by XXREAL_0:3;
  hence thesis by Th322;
end;

theorem
  p <= s implies [.p,+infty.] \ ].p,s.] = {p} \/ ].s,+infty.]
proof
  p <= +infty by XXREAL_0:3;
  hence thesis by Th324;
end;

theorem
  [.-infty,+infty.] \ ].p,s.[ = [.-infty,p.] \/ [.s,+infty.]
  by Th326,XXREAL_0:3;

theorem
  [.-infty,+infty.[ \ [.p,s.[ = [.-infty,p.[ \/ [.s,+infty.[
  by Th327,XXREAL_0:3;

theorem
  [.-infty,+infty.] \ [.p,s.[ = [.-infty,p.[ \/ [.s,+infty.]
  by Th328,XXREAL_0:3;

theorem
  [.-infty,+infty.] \ ].p,s.] = [.-infty,p.] \/ ].s,+infty.]
  by Th330,XXREAL_0:3;

theorem
  [.-infty,+infty.[ \ [.p,s.] = [.-infty,p.[ \/ ].s,+infty.[
  by Th331,XXREAL_0:3;

theorem
  [.-infty,+infty.] \ [.p,s.] = [.-infty,p.[ \/ ].s,+infty.]
  by Th332,XXREAL_0:3;

theorem
  for s being real number holds
  ].-infty,+infty.] \ ].p,s.[ = ].-infty,p.] \/ [.s,+infty.]
  by Th334,XXREAL_0:3;

theorem
  for s being real number holds REAL \ [.p,s.[ = ].-infty,p.[ \/ [.s,+infty.[
  by Th224,Th335,XXREAL_0:3;

theorem
  for s being real number holds
  ].-infty,+infty.] \ [.p,s.[ = ].-infty,p.[ \/ [.s,+infty.]
  by Th336,XXREAL_0:3;

theorem
  for s being real number holds
  ].-infty,+infty.] \ ].p,s.] = ].-infty,p.] \/ ].s,+infty.]
  by Th338,XXREAL_0:3;

theorem
  REAL \ [.p,s.] = ].-infty,p.[ \/ ].s,+infty.[ by Th224,Th339,XXREAL_0:3;

theorem
  for s being real number holds
  ].-infty,+infty.] \ [.p,s.] = ].-infty,p.[ \/ ].s,+infty.]
  by Th340,XXREAL_0:3;

theorem
  [.-infty,+infty.[ \ {p} = [.-infty,p.[ \/ ].p,+infty.[ by Th341,XXREAL_0:3;

theorem
  [.-infty,+infty.] \ {p} = [.-infty,p.[ \/ ].p,+infty.] by Th342,XXREAL_0:3;

theorem
  REAL \ {p} = ].-infty,p.[ \/ ].p,+infty.[ by Th224,Th349,XXREAL_0:3;

theorem
  for p being real number holds
  ].-infty,+infty.] \ {p} = ].-infty,p.[ \/ ].p,+infty.]
  by Th350,XXREAL_0:3;

theorem
  for p being real number holds
  r < s implies ].r,+infty.[ \ ].p,s.[ = ].r,p.] \/ [.s,+infty.[
proof
  let p be real number;
  p in REAL by XREAL_0:def 1;
  then p < +infty by XXREAL_0:9;
  hence thesis by Th297;
end;

theorem
  for p being real number holds
  r <= s implies [.r,+infty.[ \ ].p,s.[ = [.r,p.] \/ [.s,+infty.[
proof
  let p be real number;
  p in REAL by XREAL_0:def 1;
  hence thesis by Th298,XXREAL_0:9;
end;

theorem
  for p being real number holds
  r < s implies ].r,+infty.[ \ ].p,s.] = ].r,p.] \/ ].s,+infty.[
proof
  let p be real number;
  p in REAL by XREAL_0:def 1;
  then p < +infty by XXREAL_0:9;
  hence thesis by Th305;
end;

theorem
  for p being real number holds
  r <= s implies [.r,+infty.[ \ ].p,s.] = [.r,p.] \/ ].s,+infty.[
proof
  let p be real number;
  p in REAL by XREAL_0:def 1;
  hence thesis by Th306,XXREAL_0:9;
end;

theorem
  for p being real number holds
  p <= s implies [.p,+infty.[ \ ].p,s.] = {p} \/ ].s,+infty.[
proof
  let p be real number;
  p in REAL by XREAL_0:def 1;
  hence thesis by Th323,XXREAL_0:9;
end;

theorem
  for p being real number holds
  [.-infty,+infty.[ \ ].p,s.[ = [.-infty,p.] \/ [.s,+infty.[
proof
  let p be real number;
  p in REAL by XREAL_0:def 1;
  hence thesis by Th325,XXREAL_0:9;
end;

theorem
  for p being real number holds
  [.-infty,+infty.[ \ ].p,s.] = [.-infty,p.] \/ ].s,+infty.[
proof
  let p be real number;
  p in REAL by XREAL_0:def 1;
  hence thesis by Th329,XXREAL_0:9;
end;

theorem
  for s,p being real number holds REAL \ ].p,s.[ = ].-infty,p.] \/ [.s,+infty.[
proof
  let s,p be real number;
  p in REAL by XREAL_0:def 1;
  hence thesis by Th224,Th333,XXREAL_0:9;
end;

theorem
  for s,p being real number holds REAL \ ].p,s.] = ].-infty,p.] \/ ].s,+infty.[
proof
  let s,p be real number;
  p in REAL by XREAL_0:def 1;
  hence thesis by Th224,Th337,XXREAL_0:9;
end;

theorem
  for p being real number holds
  p <= s implies [.p,+infty.[ \ ].p,s.[ = {p} \/ [.s,+infty.[
proof
  let p be real number;
  p in REAL by XREAL_0:def 1;
  hence thesis by Th321,XXREAL_0:9;
end;

theorem Th401:
  r < s implies [.r,s.] \ [.r,s.[ = {s}
proof
  [.s,s.] = {s} by Th17;
  hence thesis by Th184;
end;

theorem Th402:
  r < s implies ].r,s.] \ ].r,s.[ = {s}
proof
  [.s,s.] = {s} by Th17;
  hence thesis by Th188;
end;

theorem Th403:
  r < t implies [.r,t.] \ ].r,t.] = {r}
proof
  [.r,r.] = {r} by Th17;
  hence thesis by Th192;
end;

theorem Th404:
  r < t implies [.r,t.[ \ ].r,t.[ = {r}
proof
  [.r,r.] = {r} by Th17;
  hence thesis by Th196;
end;

theorem
  for s being real number holds [.-infty,s.] \ [.-infty,s.[ = {s}
proof
  let s be real number;
  s in REAL by XREAL_0:def 1;
  hence thesis by Th401,XXREAL_0:12;
end;

theorem
  for s being real number holds ].-infty,s.] \ ].-infty,s.[ = {s}
proof
  let s be real number;
  s in REAL by XREAL_0:def 1;
  hence thesis by Th402,XXREAL_0:12;
end;

theorem
  for s being real number holds [.-infty,s.] \ ].-infty,s.] = {-infty}
proof
  let s be real number;
  s in REAL by XREAL_0:def 1;
  hence thesis by Th403,XXREAL_0:12;
end;

theorem
  for s being real number holds [.-infty,s.[ \ ].-infty,s.[ = {-infty}
proof
  let s be real number;
  s in REAL by XREAL_0:def 1;
  hence thesis by Th404,XXREAL_0:12;
end;

theorem
  for s being real number holds [.s,+infty.] \ [.s,+infty.[ = {+infty}
proof
  let s be real number;
  s in REAL by XREAL_0:def 1;
  hence thesis by Th401,XXREAL_0:9;
end;

theorem
  for s being real number holds ].s,+infty.] \ ].s,+infty.[ = {+infty}
proof
  let s be real number;
  s in REAL by XREAL_0:def 1;
  hence thesis by Th402,XXREAL_0:9;
end;

theorem
  for s being real number holds [.s,+infty.] \ ].s,+infty.] = {s}
proof
  let s be real number;
  s in REAL by XREAL_0:def 1;
  hence thesis by Th403,XXREAL_0:9;
end;

theorem
  for s being real number holds [.s,+infty.[ \ ].s,+infty.[ = {s}
proof
  let s be real number;
  s in REAL by XREAL_0:def 1;
  hence thesis by Th404,XXREAL_0:9;
end;



theorem
  r <= s & s < t implies [.r,s.] \/ [.s,t.[ = [.r,t.[
proof
  assume that
A1: r <= s and
A2: s < t;
  let p;
  thus p in [.r,s.] \/ [.s,t.[ implies p in [.r,t.[
  proof
    assume p in [.r,s.] \/ [.s,t.[;
    then p in [.r,s.] or p in [.s,t.[ by XBOOLE_0:def 3;
    then
A3: r <= p & p <= s or s <= p & p < t by Th1,Th3;
    then
A4: r <= p by A1,XXREAL_0:2;
    p < t by A2,A3,XXREAL_0:2;
    hence thesis by A4,Th3;
  end;
  assume p in [.r,t.[;
  then r <= p & p <= s or s <= p & p < t by Th3;
  then p in [.r,s.] or p in [.s,t.[ by Th1,Th3;
  hence thesis by XBOOLE_0:def 3;
end;

theorem
  r < s & s <= t implies ].r,s.] \/ [.s,t.] = ].r,t.]
proof
  assume that
A1: r < s and
A2: s <= t;
  let p;
  thus p in ].r,s.] \/ [.s,t.] implies p in ].r,t.]
  proof
    assume p in ].r,s.] \/ [.s,t.];
    then p in ].r,s.] or p in [.s,t.] by XBOOLE_0:def 3;
    then
A3: r < p & p <= s or s <= p & p <= t by Th1,Th2;
    then
A4: r < p by A1,XXREAL_0:2;
    p <= t by A2,A3,XXREAL_0:2;
    hence thesis by A4,Th2;
  end;
  assume p in ].r,t.];
  then r < p & p <= s or s <= p & p <= t by Th2;
  then p in ].r,s.] or p in [.s,t.] by Th1,Th2;
  hence thesis by XBOOLE_0:def 3;
end;

theorem
  r < s & s < t implies ].r,s.] \/ [.s,t.[ = ].r,t.[
proof
  assume that
A1: r < s and
A2: s < t;
  let p;
  thus p in ].r,s.] \/ [.s,t.[ implies p in ].r,t.[
  proof
    assume p in ].r,s.] \/ [.s,t.[;
    then p in ].r,s.] or p in [.s,t.[ by XBOOLE_0:def 3;
    then
A3: r < p & p <= s or s <= p & p < t by Th2,Th3;
    then
A4: r < p by A1,XXREAL_0:2;
    p < t by A2,A3,XXREAL_0:2;
    hence thesis by A4,Th4;
  end;
  assume p in ].r,t.[;
  then r < p & p <= s or s <= p & p < t by Th4;
  then p in ].r,s.] or p in [.s,t.[ by Th2,Th3;
  hence thesis by XBOOLE_0:def 3;
end;



theorem
  r <= s & s < t implies [.r,s.] /\ [.s,t.[ = {s}
proof
  assume that
A1: r <= s and
A2: s < t;
  now
    let x;
    hereby
      assume
A3:   x in [.r,s.] /\ [.s,t.[;
      then reconsider p = x as ext-real number;
A4:   p in [.r,s.] by A3,XBOOLE_0:def 4;
      p in [.s,t.[ by A3,XBOOLE_0:def 4;
      then
A5:   s <= p by Th3;
      p <= s by A4,Th1;
      hence x = s by A5,XXREAL_0:1;
    end;
    assume
A6: x = s;
A7: s in [.r,s.] by A1,Th1;
    s in [.s,t.[ by A2,Th3;
    hence x in [.r,s.] /\ [.s,t.[ by A6,A7,XBOOLE_0:def 4;
  end;
  hence thesis by TARSKI:def 1;
end;

theorem
  r < s & s <= t implies ].r,s.] /\ [.s,t.] = {s}
proof
  assume that
A1: r < s and
A2: s <= t;
  now
    let x;
    hereby
      assume
A3:   x in ].r,s.] /\ [.s,t.];
      then reconsider p = x as ext-real number;
A4:   p in ].r,s.] by A3,XBOOLE_0:def 4;
      p in [.s,t.] by A3,XBOOLE_0:def 4;
      then
A5:   s <= p by Th1;
      p <= s by A4,Th2;
      hence x = s by A5,XXREAL_0:1;
    end;
    assume
A6: x = s;
A7: s in ].r,s.] by A1,Th2;
    s in [.s,t.] by A2,Th1;
    hence x in ].r,s.] /\ [.s,t.] by A6,A7,XBOOLE_0:def 4;
  end;
  hence thesis by TARSKI:def 1;
end;

theorem
  r <= s & s <= t implies [.r,s.] /\ [.s,t.] = {s}
proof
  assume that
A1: r <= s and
A2: s <= t;
  now
    let x;
    hereby
      assume
A3:   x in [.r,s.] /\ [.s,t.];
      then reconsider p = x as ext-real number;
A4:   p in [.r,s.] by A3,XBOOLE_0:def 4;
      p in [.s,t.] by A3,XBOOLE_0:def 4;
      then
A5:   s <= p by Th1;
      p <= s by A4,Th1;
      hence x = s by A5,XXREAL_0:1;
    end;
    assume
A6: x = s;
A7: s in [.r,s.] by A1,Th1;
    s in [.s,t.] by A2,Th1;
    hence x in [.r,s.] /\ [.s,t.] by A6,A7,XBOOLE_0:def 4;
  end;
  hence thesis by TARSKI:def 1;
end;



theorem
  [.-infty,s.] = ].-infty,s.[ \/ {-infty,s} by Th128,XXREAL_0:5;

theorem
  [.-infty,s.] = [.-infty,s.[ \/ {s} by Th129,XXREAL_0:5;

theorem
  [.-infty,s.] = {-infty} \/ ].-infty,s.] by Th130,XXREAL_0:5;

theorem
  for s being real number holds [.-infty,s.[ = {-infty} \/ ].-infty,s.[
proof
  let s be real number;
  s in REAL by XREAL_0:def 1;
  hence thesis by Th131,XXREAL_0:12;
end;

theorem
  for s being real number holds ].-infty,s.] = ].-infty,s.[ \/ {s}
proof
  let s be real number;
  s in REAL by XREAL_0:def 1;
  hence thesis by Th132,XXREAL_0:12;
end;

theorem
  [.r,+infty.] = ].r,+infty.[ \/ {r,+infty} by Th128,XXREAL_0:3;

theorem
  [.r,+infty.] = [.r,+infty.[ \/ {+infty} by Th129,XXREAL_0:3;

theorem
  [.r,+infty.] = {r} \/ ].r,+infty.] by Th130,XXREAL_0:3;

theorem
  for r being real number holds [.r,+infty.[ = {r} \/ ].r,+infty.[
proof
  let r be real number;
  r in REAL by XREAL_0:def 1;
  hence thesis by Th131,XXREAL_0:9;
end;

theorem
  for r being real number holds ].r,+infty.] = ].r,+infty.[ \/ {+infty}
proof
  let r be real number;
  r in REAL by XREAL_0:def 1;
  hence thesis by Th132,XXREAL_0:9;
end;















begin

reserve A,B for Ordinal,
  K,M,N for Cardinal,
  x,x1,x2,y,y1,y2,z,u,X,Y,Z,X1,X2, Y1,Y2 for set,
  f,g for Function;

theorem
  x in X & X,Y are_equipotent implies Y <> {} & ex x st x in Y
proof
  assume
A1: x in X;
  given f such that
  f is one-to-one and
A2: dom f = X & rng f = Y;
  f.x in Y by A1,A2,FUNCT_1:def 3;
  hence Y <> {};
  take f.x;
  thus thesis by A1,A2,FUNCT_1:def 3;
end;

theorem
  bool X,bool card X are_equipotent & card bool X = card bool card X
proof
  consider f such that
A1: f is one-to-one and
A2: dom f = X and
A3: rng f = card X by CARD_1:def 2,WELLORD2:def 4;
  deffunc g(set) = f.:$1;
  consider g such that
A4: dom g = bool X & for x st x in bool X holds g.x = g(x) from FUNCT_1:
  sch 3;
  thus bool X,bool card X are_equipotent
  proof
    take g;
    thus g is one-to-one
    proof
      let x,y;
      assume that
A5:   x in dom g and
A6:   y in dom g and
A7:   g.x = g.y;
A8:   g.x = f.:x & g.y = f.:y by A4,A5,A6;
A9:   y c= x
      proof
        let z;
        assume
A10:    z in y;
        then f.z in f.:y by A2,A4,A6,FUNCT_1:def 6;
        then ex u st u in dom f & u in x & f.z = f.u by A7,A8,FUNCT_1:def 6;
        hence thesis by A1,A2,A4,A6,A10,FUNCT_1:def 4;
      end;
      x c= y
      proof
        let z;
        assume
A11:    z in x;
        then f.z in f.:x by A2,A4,A5,FUNCT_1:def 6;
        then ex u st u in dom f & u in y & f.z = f.u by A7,A8,FUNCT_1:def 6;
        hence thesis by A1,A2,A4,A5,A11,FUNCT_1:def 4;
      end;
      hence thesis by A9,XBOOLE_0:def 10;
    end;
    thus dom g = bool X by A4;
    thus rng g c= bool card X
    proof
      let x;
      assume x in rng g;
      then consider y such that
A12:  y in dom g and
A13:  x = g.y by FUNCT_1:def 3;
A14:  f.:y c= rng f by RELAT_1:111;
      g.y = f.:y by A4,A12;
      hence thesis by A3,A13,A14;
    end;
    let x;
A15: f"x c= dom f by RELAT_1:132;
    assume x in bool card X;
    then f.:(f"x) = x by A3,FUNCT_1:77;
    then g.(f"x) = x by A2,A4,A15;
    hence thesis by A2,A4,A15,FUNCT_1:def 3;
  end;
  hence thesis by CARD_1:5;
end;

deffunc g(set) = $1`1;

theorem
  Z in Funcs(X,Y) implies Z,X are_equipotent & card Z = card X
proof
  assume Z in Funcs(X,Y);
  then consider f such that
A1: Z = f and
A2: dom f = X and
  rng f c= Y by FUNCT_2:def 2;
  thus Z,X are_equipotent
  proof
    consider g such that
A3: dom g = Z & for x st x in Z holds g.x = g(x) from FUNCT_1:sch 3;
    take g;
    thus g is one-to-one
    proof
      let x,y;
      assume that
A4:   x in dom g and
A5:   y in dom g;
A6:   g.x = x`1 & g.y = y`1 by A3,A4,A5;
      ex x1,x2 being set st [x1,x2] = y by A1,A3,A5,RELAT_1:def 1;
      then
A7:   y = [y`1,y`2] by MCART_1:8;
      ex x1,x2 being set st [x1,x2] = x by A1,A3,A4,RELAT_1:def 1;
      then
A8:   x = [x`1,x`2] by MCART_1:8;
      then x`2 = f.(x`1) by A1,A3,A4,FUNCT_1:1;
      hence thesis by A1,A3,A5,A8,A7,A6,FUNCT_1:1;
    end;
    thus dom g = Z by A3;
    thus rng g c= X
    proof
      let x;
      assume x in rng g;
      then consider y such that
A9:   y in dom g and
A10:  x = g.y by FUNCT_1:def 3;
      ex x1,x2 being set st [x1,x2] = y by A1,A3,A9,RELAT_1:def 1;
      then
A11:  y = [y`1,y`2] by MCART_1:8;
      x = y`1 by A3,A9,A10;
      hence thesis by A1,A2,A3,A9,A11,FUNCT_1:1;
    end;
    let x;
    assume x in X;
    then
A12: [x,f.x] in Z by A1,A2,FUNCT_1:def 2;
    then g.[x,f.x] = [x,f.x]`1 by A3
      .= x;
    hence thesis by A3,A12,FUNCT_1:def 3;
  end;
  hence thesis by CARD_1:5;
end;

Lm1: x1 <> x2 implies A+^B,[:A,{x1}:] \/ [:B,{x2}:] are_equipotent & card(A+^B
) = card([:A,{x1}:] \/ [:B,{x2}:])
proof
  defpred C[set] means $1 in A;
  deffunc F(set) = [$1,x1];
  deffunc G(Ordinal) = [$1-^A,x2];
  consider f such that
A1: dom f = A+^B and
A2: for x being Ordinal holds x in A+^B implies (C[x] implies f.x = F(x)
  ) & (not C[x] implies f.x = G(x)) from FINSET_1:sch 1;
  assume
A3: x1 <> x2;
  thus A+^B,[:A,{x1}:] \/ [:B,{x2}:] are_equipotent
  proof
    take f;
    thus f is one-to-one
    proof
      let x,y;
      assume that
A4:   x in dom f & y in dom f and
A5:   f.x = f.y;
      reconsider C1 = x, C2 = y as Ordinal by A1,A4;
      per cases;
      suppose
A6:     x in A & y in A;
A7:     [x,x1]`1 = C1;
        f.C1 = [x,x1] & f.C2 = [y,x1] by A1,A2,A4,A6;
        hence thesis by A5,A7,MCART_1:7;
      end;
      suppose
A8:     not x in A & not y in A;
A9:     [C1-^A,x2]`1 = C1-^A;
        f.x = [C1-^A,x2] & f.y = [C2-^A,x2] by A1,A2,A4,A8;
        then
A10:    C1-^A = C2-^A by A5,A9,MCART_1:7;
        A c= C1 by A8,ORDINAL1:16;
        then
A11:    C1 = A+^(C1-^A) by ORDINAL3:def 5;
        A c= C2 by A8,ORDINAL1:16;
        hence thesis by A10,A11,ORDINAL3:def 5;
      end;
      suppose
A12:    x in A & not y in A;
A13:    [x,x1]`2 = x1;
        f.x = [x,x1] & f.y = [C2-^A,x2] by A1,A2,A4,A12;
        hence thesis by A3,A5,A13,MCART_1:7;
      end;
      suppose
A14:    not x in A & y in A;
A15:    [y,x1]`2 = x1;
        f.y = [y,x1] & f.x = [C1-^A,x2] by A1,A2,A4,A14;
        hence thesis by A3,A5,A15,MCART_1:7;
      end;
    end;
    thus dom f = A+^B by A1;
    thus rng f c= [:A,{x1}:] \/ [:B,{x2}:]
    proof
      let x;
A16:  x1 in {x1} by TARSKI:def 1;
A17:  x2 in {x2} by TARSKI:def 1;
      assume x in rng f;
      then consider y such that
A18:  y in dom f and
A19:  x = f.y by FUNCT_1:def 3;
      reconsider C = y as Ordinal by A1,A18;
      per cases;
      suppose
        y in A;
        then x = [C,x1] & [C,x1] in [:A,{x1}:] by A1,A2,A18,A19,A16,ZFMISC_1:87
;
        hence thesis by XBOOLE_0:def 3;
      end;
      suppose
A20:    not y in A;
        then A c= C by ORDINAL1:16;
        then A+^(C-^A) = C by ORDINAL3:def 5;
        then C-^A in B by A1,A18,ORDINAL3:22;
        then
A21:    [C-^A,x2] in [:B,{x2}:] by A17,ZFMISC_1:87;
        x = [C-^A,x2] by A1,A2,A18,A19,A20;
        hence thesis by A21,XBOOLE_0:def 3;
      end;
    end;
    let x such that
A22: x in [:A,{x1}:] \/ [:B,{x2}:];
A23: now
      assume x in [:B,{x2}:];
      then consider y,z such that
A24:  y in B and
A25:  z in {x2} and
A26:  x = [y,z] by ZFMISC_1:84;
      reconsider y as Ordinal by A24;
A27:  A+^y in A+^B by A24,ORDINAL2:32;
A28:  not A+^y in A by ORDINAL1:5,ORDINAL3:24;
      A+^y-^A = y & z = x2 by A25,ORDINAL3:52,TARSKI:def 1;
      then x = f.(A+^y) by A2,A26,A27,A28;
      hence thesis by A1,A27,FUNCT_1:def 3;
    end;
    now
      assume x in [:A,{x1}:];
      then consider y,z such that
A29:  y in A and
A30:  z in {x1} and
A31:  x = [y,z] by ZFMISC_1:84;
A32:  A c= A+^B by ORDINAL3:24;
      z = x1 by A30,TARSKI:def 1;
      then x = f.y by A2,A29,A31,A32;
      hence thesis by A1,A29,A32,FUNCT_1:def 3;
    end;
    hence thesis by A22,A23,XBOOLE_0:def 3;
  end;
  hence thesis by CARD_1:5;
end;

deffunc plus(set,set) = [:$1,{0}:] \/ [:$2,{1}:];

Lm2: [:X,Y:],[:Y,X:] are_equipotent & card [:X,Y:] = card [:Y,X:]
proof
  deffunc f(set) = [$1`2,$1`1];
  consider f such that
A1: dom f = [:X,Y:] & for x st x in [:X,Y:] holds f.x = f(x) from
  FUNCT_1:sch 3;
  thus [:X,Y:],[:Y,X:] are_equipotent
  proof
    take f;
    thus f is one-to-one
    proof
      let x,y;
      assume
A2:   x in dom f & y in dom f;
      then
A3:   x = [x`1,x`2] & y = [y`1,y`2] by A1,MCART_1:21;
      assume
A4:   f.x = f.y;
A5:   f.x = [x`2,x`1] & f.y = [y`2,y `1] by A1,A2;
      then x`1 = y`1 by A4,XTUPLE_0:1;
      hence thesis by A3,A5,A4,XTUPLE_0:1;
    end;
    thus dom f = [:X,Y:] by A1;
    thus rng f c= [:Y,X:]
    proof
      let x;
      assume x in rng f;
      then consider y such that
A6:   y in dom f and
A7:   x = f.y by FUNCT_1:def 3;
A8:   y`2 in Y by A1,A6,MCART_1:10;
      x = [y`2,y`1] & y`1 in X by A1,A6,A7,MCART_1:10;
      hence thesis by A8,ZFMISC_1:87;
    end;
    let x;
A9: [x`2,x`1]`1 = x`2 & [x`2,x`1]`2 = x`1;
    assume
A10: x in [:Y,X:];
    then
A11: x = [x`1,x`2] by MCART_1:21;
A12:  x`1 in Y & x`2 in X by A10,MCART_1:10;
    then
 [x`2,x`1] in [:X,Y:] by ZFMISC_1:87;
    then f.[x`2,x`1] in rng f by A1,FUNCT_1:def 3;
    hence thesis by A1,A11,A12,A9,ZFMISC_1:87;
  end;
  hence thesis by CARD_1:5;
end;

definition
  let M,N;
  func M +` N -> Cardinal equals
  card( M +^ N);
  coherence;
  commutativity
  proof
    let C be Cardinal;
    let M,N;
    assume C = card( M +^ N);
    hence C = card plus( N, M) by Lm1
      .= card( N +^ M) by Lm1;
  end;
  func M *` N -> Cardinal equals
  card [:M,N:];
  coherence;
  commutativity by Lm2;
  func exp(M,N) -> Cardinal equals
  card Funcs(N,M);
  coherence;
end;

theorem Th4:
  [:X,Y:],[:Y,X:] are_equipotent & card [:X,Y:] = card [:Y,X:] by Lm2;

theorem Th5:
  [:[:X,Y:],Z:],[:X,[:Y,Z:]:] are_equipotent & card [:[:X,Y:],Z:]
  = card [:X,[:Y,Z:]:]
proof
  deffunc f(set) = [$1`1`1,[$1`1`2,$1`2]];
  consider f such that
A1: dom f = [:[:X,Y:],Z:] & for x st x in [:[:X,Y:],Z:] holds f.x = f(x)
  from FUNCT_1:sch 3;
  thus [:[:X,Y:],Z:],[:X,[:Y,Z:]:] are_equipotent
  proof
    take f;
    thus f is one-to-one
    proof
      let x,y;
      assume that
A2:   x in dom f and
A3:   y in dom f;
      assume
A4:   f.x = f.y;
A5:   x = [x`1,x`2] & y = [y`1,y`2] by A1,A2,A3,MCART_1:21;
      x`1 in [:X,Y:] by A1,A2,MCART_1:10;
      then
A6:   x`1 = [x`1`1,x`1`2 ] by MCART_1:21;
A7:   f.x = [x`1`1,[x`1`2,x`2]] & f.y = [y`1`1,[y`1`2,y`2]] by A1,A2,A3;
      then
A8:   x`1`1 = y`1`1 by A4,XTUPLE_0:1;
      y`1 in [:X,Y :] by A1,A3,MCART_1:10;
      then
A9:   y`1 = [y`1`1,y`1`2] by MCART_1:21;
A10:  [x`1`2,x`2] = [y`1`2,y`2] by A7,A4,XTUPLE_0:1;
      then x`1`2 = y`1`2 by XTUPLE_0:1;
      hence thesis by A5,A8,A10,A6,A9,XTUPLE_0:1;
    end;
    thus dom f = [:[:X,Y:],Z:] by A1;
    thus rng f c= [:X,[:Y,Z:]:]
    proof
      let x;
      assume x in rng f;
      then consider y such that
A11:  y in dom f and
A12:  x = f.y by FUNCT_1:def 3;
A13:  y`1 in [:X,Y:] by A1,A11,MCART_1:10;
      then
A14:  y`1`2 in Y by MCART_1:10;
      y`2 in Z by A1,A11,MCART_1:10;
      then
A15:  [y`1`2,y`2] in [:Y,Z:] by A14,ZFMISC_1:87;
A16:  y`1`1 in X by A13,MCART_1:10;
      x = [y`1`1,[y`1`2,y`2]] by A1,A11,A12;
      hence thesis by A16,A15,ZFMISC_1:87;
    end;
    let x;
A17: [x`1,x`2`1]`1 = x`1 & [x`1,x`2`1]`2 = x`2 `1;
A18: [[x`1,x`2`1],x`2`2]`1 = [x`1,x`2`1] & [[x`1,x`2`1],x`2`2]`2 = x`2`2;
    assume
A19: x in [:X,[:Y,Z:]:];
    then
A20: x`2 in [:Y,Z:] by MCART_1:10;
    then
A21: x`2`1 in Y by MCART_1:10;
A22: x`2`2 in Z by A20,MCART_1:10;
    x`1 in X by A19,MCART_1:10;
    then
A23: [x`1,x`2`1] in [:X,Y:] by A21,ZFMISC_1:87;
    then
A24: [[x`1,x`2`1],x`2`2] in [:[:X,Y:],Z:] by A22,ZFMISC_1:87;
A25: x`2 = [x`2`1,x`2`2] by A20,MCART_1:21;
    x = [x`1,x`2] by A19,MCART_1:21;
    then x = f.[[x`1,x`2`1],x`2`2] by A1,A25,A17,A23,A18,A22,ZFMISC_1:87;
    hence thesis by A1,A24,FUNCT_1:def 3;
  end;
  hence thesis by CARD_1:5;
end;

theorem Th6:
  X,[:X,{x}:] are_equipotent & card X = card [:X,{x}:]
proof
  deffunc f(set) = [$1,x];
  consider f such that
A1: dom f = X & for y st y in X holds f.y = f(y) from FUNCT_1:sch 3;
  thus X,[:X,{x}:] are_equipotent
  proof
    take f;
    thus f is one-to-one
    proof
      let y,z;
      assume that
A2:   y in dom f & z in dom f and
A3:   f.y = f.z;
A4:   [y,x]`1 = y;
      f.y = [y,x] & f.z = [z,x] by A1,A2;
      hence thesis by A3,A4,MCART_1:7;
    end;
    thus dom f = X by A1;
    thus rng f c= [:X,{x}:]
    proof
      let y;
A5:   x in {x} by TARSKI:def 1;
      assume y in rng f;
      then consider z such that
A6:   z in dom f and
A7:   y = f.z by FUNCT_1:def 3;
      y = [z,x] by A1,A6,A7;
      hence thesis by A1,A6,A5,ZFMISC_1:87;
    end;
    let y;
    assume y in [:X,{x}:];
    then consider y1,y2 being set such that
A8: y1 in X and
A9: y2 in {x} and
A10: y = [y1,y2] by ZFMISC_1:84;
    y2 = x by A9,TARSKI:def 1;
    then y = f.y1 by A1,A8,A10;
    hence thesis by A1,A8,FUNCT_1:def 3;
  end;
  hence thesis by CARD_1:5;
end;

Lm3: [:X,Y:],[:card X,Y:] are_equipotent
proof
  consider f such that
A1: f is one-to-one and
A2: dom f = X and
A3: rng f = card X by CARD_1:def 2,WELLORD2:def 4;
  deffunc g(set) = [f.$1`1,$1`2];
  consider g such that
A4: dom g = [:X,Y:] & for x st x in [:X,Y:] holds g.x = g(x) from
  FUNCT_1:sch 3;
  take g;
  thus g is one-to-one
  proof
    let x,y;
    assume
A5: x in dom g & y in dom g;
    then
A6: x`1 in X & y`1 in X by A4,MCART_1:10;
    assume
A7: g.x = g.y;
    g.x = [f.x`1,x`2] & g.y = [f.y`1,y `2] by A4,A5;
    then
A8: f.x`1 = f.y`1 & x`2 = y`2 by A7,XTUPLE_0:1;
    x = [x`1,x`2] & y = [y`1,y`2] by A4,A5,MCART_1:21;
    hence thesis by A1,A2,A6,A8,FUNCT_1:def 4;
  end;
  thus dom g = [:X,Y:] by A4;
  thus rng g c= [:card X,Y:]
  proof
    let y;
    assume y in rng g;
    then consider x such that
A9: x in dom g and
A10: y = g.x by FUNCT_1:def 3;
    x`1 in X by A4,A9,MCART_1:10;
    then
A11: f.x`1 in card X by A2,A3,FUNCT_1:def 3;
    y = [f.x`1,x`2] & x`2 in Y by A4,A9,A10,MCART_1:10;
    hence thesis by A11,ZFMISC_1:87;
  end;
  let y;
  assume
A12: y in [:card X,Y:];
  then y`1 in card X by MCART_1:10;
  then consider x such that
A13: x in X and
A14: y`1 = f.x by A2,A3,FUNCT_1:def 3;
A15: y = [y`1,y`2] by A12,MCART_1:21;
A16: y`2 in Y by A12,MCART_1:10;
  then
A17: [x,y`2] in [:X,Y:] by A13,ZFMISC_1:87;
  [x,y`2]`1 = x & [x,y`2]`2 = y`2;
  then g.[x,y`2] = y by A4,A15,A14,A16,A13,ZFMISC_1:87;
  hence thesis by A4,A17,FUNCT_1:def 3;
end;

theorem Th7:
  [:X,Y:],[:card X,Y:] are_equipotent & [:X,Y:],[:X,card Y:]
are_equipotent & [:X,Y:],[:card X,card Y:] are_equipotent & card [:X,Y:] = card
[:card X,Y:] & card [:X,Y:] = card [:X,card Y:] & card [:X,Y:] = card [:card X,
  card Y:]
proof
  [:Y,X:],[:card Y,X:] are_equipotent & [:X,Y:],[:Y,X:] are_equipotent by Lm2
,Lm3;
  then
A1: [:X,Y:],[:card Y,X:] are_equipotent by WELLORD2:15;
A2: [:card Y,X:],[:X,card Y:] are_equipotent by Lm2;
  hence
A3: [:X,Y:],[:card X,Y:] are_equipotent & [:X,Y:],[:X,card Y:]
  are_equipotent by A1,Lm3,WELLORD2:15;
  [:X,card Y:],[:card X,card Y:] are_equipotent by Lm3;
  hence [:X,Y:],[:card X,card Y:] are_equipotent by A3,WELLORD2:15;
  hence thesis by A2,A1,Lm3,CARD_1:5,WELLORD2:15;
end;

theorem Th8:
  X1,Y1 are_equipotent & X2,Y2 are_equipotent implies [:X1,X2:],[:
  Y1,Y2:] are_equipotent & card [:X1,X2:] = card [:Y1,Y2:]
proof
  assume X1,Y1 are_equipotent & X2,Y2 are_equipotent;
  then
A1: card X1 = card Y1 & card X2 = card Y2 by CARD_1:5;
  [:X1,X2:],[:card X1,card X2:] are_equipotent & [:Y1,Y2:],[:card Y1,card
  Y2:] are_equipotent by Th7;
  hence [:X1,X2:],[:Y1,Y2:] are_equipotent by A1,WELLORD2:15;
  hence thesis by CARD_1:5;
end;

theorem
  x1 <> x2 implies A+^B,[:A,{x1}:] \/ [:B,{x2}:] are_equipotent & card(A
  +^B) = card([:A,{x1}:] \/ [:B,{x2}:]) by Lm1;

theorem Th10:
  x1 <> x2 implies K+`M,[:K,{x1}:] \/ [:M,{x2}:] are_equipotent &
  K+`M = card([:K,{x1}:] \/ [:M,{x2}:])
proof
  assume x1 <> x2;
  then card([:K,{x1}:] \/ [:M,{x2}:]) = K+`M by Lm1;
  hence thesis by CARD_1:def 2;
end;

theorem Th11:
  A*^B,[:A,B:] are_equipotent & card(A*^B) = card [:A,B:]
proof
  defpred P[set,set] means ex O1, O2 being Ordinal st O1 = $1`1 & O2 = $1`2 &
  $2 = O1*^B+^O2;
A1: for x st x in [:A,B:] ex y st P[x,y]
  proof
    let x;
    assume x in [:A,B:];
    then x`1 in A & x`2 in B by MCART_1:10;
    then reconsider x1 = x`1, x2 = x`2 as Ordinal;
    take x1*^B+^x2;
    take x1, x2;
    thus thesis;
  end;
  consider f such that
A2: dom f = [:A,B:] & for x st x in [:A,B:] holds P[x,f.x] from CLASSES1
  :sch 1(A1);
A3: [:A,B:],A*^B are_equipotent
  proof
    take f;
    thus f is one-to-one
    proof
      let x,y;
      assume
A4:   x in dom f & y in dom f;
      then
A5:   x`2 in B & y`2 in B by A2,MCART_1:10;
      x`1 in A & y`1 in A by A2,A4,MCART_1:10;
      then reconsider x1 = x`1, y1 = y`1 as Ordinal;
      assume
A6:   f.x = f.y;
A7:   x = [x`1,x`2] & y = [y`1,y`2] by A2,A4,MCART_1:21;
A8:   (ex O1, O2 being Ordinal st O1 = x`1 & O2 = x`2 & f.x = O1*^B+^O2 )
& ex O3, O4 being Ordinal st O3 = y`1 & O4 = y`2 & f.y = O3*^B+^O4 by A2,A4;
      then x1 = y1 by A5,A6,ORDINAL3:48;
      hence thesis by A7,A5,A8,A6,ORDINAL3:48;
    end;
    thus dom f = [:A,B:] by A2;
    thus rng f c= A*^B
    proof
      let y;
A9:   1*^B = B by ORDINAL2:39;
      assume y in rng f;
      then consider x such that
A10:  x in dom f and
A11:  y = f.x by FUNCT_1:def 3;
      consider x1, x2 being Ordinal such that
A12:  x1 = x`1 and
A13:  x2 = x`2 & f.x = x1*^B+^x2 by A2,A10;
      x1+^1 = succ x1 by ORDINAL2:31;
      then
A14:  x1*^B+^1*^B = (succ x1)*^B by ORDINAL3:46;
      succ x1 c= A by A12,A2,A10,MCART_1:10,ORDINAL1:21;
      then
A15:  x1*^B+^1*^B c= A*^B by A14,ORDINAL2:41;
      y in x1*^B+^1*^B by A11,A13,A9,A2,A10,MCART_1:10,ORDINAL2:32;
      hence thesis by A15;
    end;
    let y;
    assume
A16: y in A*^B;
    then reconsider C = y as Ordinal;
A17: C = (C div^ B)*^B+^(C mod^ B) & [C div^ B,C mod^ B]`1 = C div^ B by
ORDINAL3:65;
    C div^ B in A & C mod^ B in B by A16,ORDINAL3:67;
    then
A18: [C div^ B,C mod^ B] in [:A,B:] by ZFMISC_1:87;
    then
    [C div^ B,C mod^ B]`2 = C mod^ B & ex O1, O2 being Ordinal st O1 = [C
div^ B,C mod^ B]`1 & O2 = [C div^ B,C mod^ B]`2 & f.[C div^ B,C mod^ B] = O1*^B
    +^ O2 by A2;
    hence thesis by A2,A18,A17,FUNCT_1:def 3;
  end;
  hence A*^B,[:A,B:] are_equipotent;
  thus thesis by A3,CARD_1:5;
end;

deffunc plus(set,set) = [:$1,{0}:] \/ [:$2,{1}:];
deffunc plus(set,set,set,set) = [:$1,{$3}:] \/ [:$2,{$4}:];

theorem Th12:
  X1,Y1 are_equipotent & X2,Y2 are_equipotent & x1 <> x2 & y1 <>
y2 implies [:X1,{x1}:] \/ [:X2,{x2}:],[:Y1,{y1}:] \/ [:Y2,{y2}:] are_equipotent
  & card ([:X1,{x1}:] \/ [:X2,{x2}:]) = card ([:Y1,{y1}:] \/ [:Y2,{y2}:])
proof
  assume that
A1: X1,Y1 are_equipotent and
A2: X2,Y2 are_equipotent and
A3: x1 <> x2 and
A4: y1 <> y2;
  {x2},{y2} are_equipotent by CARD_1:28;
  then
A5: [:X2,{x2}:],[:Y2,{y2}:] are_equipotent by A2,Th8;
A6: now
    assume [:Y1,{y1}:] meets [:Y2,{y2}:];
    then consider y being set such that
A7: y in [:Y1,{y1}:] and
A8: y in [:Y2,{y2}:] by XBOOLE_0:3;
    y`2 in {y1} by A7,MCART_1:10;
    then
A9: y`2 = y1 by TARSKI:def 1;
    y`2 in {y2} by A8,MCART_1:10;
    hence contradiction by A4,A9,TARSKI:def 1;
  end;
A10: now
    assume [:X1,{x1}:] meets [:X2,{x2}:];
    then consider x being set such that
A11: x in [:X1,{x1}:] and
A12: x in [:X2,{x2}:] by XBOOLE_0:3;
    x`2 in {x1} by A11,MCART_1:10;
    then
A13: x`2 = x1 by TARSKI:def 1;
    x`2 in {x2} by A12,MCART_1:10;
    hence contradiction by A3,A13,TARSKI:def 1;
  end;
  {x1},{y1} are_equipotent by CARD_1:28;
  then [:X1,{x1}:],[:Y1,{y1}:] are_equipotent by A1,Th8;
  hence [:X1,{x1}:] \/ [:X2,{x2}:],[:Y1,{y1}:] \/ [:Y2,{y2}:]
  are_equipotent by A5,A10,A6,CARD_1:31;
  hence thesis by CARD_1:5;
end;

theorem Th13:
  card(A+^B) = card A +` card B
proof
A1: A+^B,plus(A,B) are_equipotent by Lm1;
  A,card A are_equipotent & B,card B are_equipotent by CARD_1:def 2;
  then
A2: plus(A,B),plus( card A, card B) are_equipotent by Th12;
  plus( card A, card B),card A +^ card B are_equipotent by Lm1;
  then plus(A,B),card A +^ card B are_equipotent by A2,WELLORD2:15;
  then A+^B,card A +^ card B are_equipotent by A1,WELLORD2:15;
  hence thesis by CARD_1:5;
end;

theorem Th14:
  card(A*^B) = card A *` card B
proof
  thus card (A*^B) = card [:A,B:] by Th11
    .= card A *` card B by Th7;
end;

theorem
  [:X,{0}:] \/ [:Y,{1}:],[:Y,{0}:] \/ [:X,{1}:] are_equipotent & card([:
  X,{0}:] \/ [:Y,{1}:]) = card([:Y,{0}:] \/ [:X,{1}:]) by Th12;

theorem Th16:
  [:X1,X2:] \/ [:Y1,Y2:],[:X2,X1:] \/ [:Y2,Y1:] are_equipotent &
  card ([:X1,X2:] \/ [:Y1,Y2:]) = card ([:X2,X1:] \/ [:Y2,Y1:])
proof
  deffunc f(set) = [$1`2,$1`1];
  consider f such that
A1: dom f = [:X1,X2:] \/ [:Y1,Y2:] & for x st x in [:X1,X2:] \/ [:Y1,Y2
  :] holds f.x = f(x) from FUNCT_1:sch 3;
  thus [:X1,X2:] \/ [:Y1,Y2:],[:X2,X1:] \/ [:Y2,Y1:] are_equipotent
  proof
    take f;
    thus f is one-to-one
    proof
      let x1,x2;
      assume that
A2:   x1 in dom f and
A3:   x2 in dom f and
A4:   f.x1 = f.x2;
      x1 in [:X1,X2:] or x1 in [:Y1,Y2:] by A1,A2,XBOOLE_0:def 3;
      then
A5:   x1 = [x1`1,x1`2] by MCART_1:21;
      x2 in [:X1,X2:] or x2 in [:Y1,Y2:] by A1,A3,XBOOLE_0:def 3;
      then
A6:   x2 = [x2`1,x2`2] by MCART_1:21;
A7:   f.x1 = [x1`2,x1`1] & f.x2 = [x2`2,x2`1] by A1,A2,A3;
      then x1`1 = x2`1 by A4,XTUPLE_0:1;
      hence thesis by A4,A7,A5,A6,XTUPLE_0:1;
    end;
    thus dom f = [:X1,X2:] \/ [:Y1,Y2:] by A1;
    thus rng f c= [:X2,X1:] \/ [:Y2,Y1:]
    proof
      let x;
      assume x in rng f;
      then consider y such that
A8:   y in dom f and
A9:   x = f.y by FUNCT_1:def 3;
      y in [:X1,X2:] or y in [:Y1,Y2:] by A1,A8,XBOOLE_0:def 3;
      then
A10:  y`1 in X1 & y`2 in X2 or y`1 in Y1 & y`2 in Y2 by MCART_1:10;
      x = [y`2,y`1] by A1,A8,A9;
      then x in [:X2,X1:] or x in [:Y2,Y1:] by A10,ZFMISC_1:87;
      hence thesis by XBOOLE_0:def 3;
    end;
    let x;
A11: [x`2,x`1]`1 = x`2 & [x`2,x`1]`2 = x`1;
    assume x in [:X2,X1:] \/ [:Y2,Y1:];
    then
A12: x in [:X2,X1:] or x in [:Y2,Y1:] by XBOOLE_0:def 3;
    then x`1 in X2 & x`2 in X1 or x`1 in Y2 & x`2 in Y1 by MCART_1:10;
    then [x`2,x`1] in [:X1,X2:] or [x`2,x`1] in [:Y1,Y2:] by ZFMISC_1:87;
    then
A13: [x`2,x`1] in [:X1,X2:] \/ [:Y1,Y2:] by XBOOLE_0:def 3;
    x = [x`1,x`2] by A12,MCART_1:21;
    then f.[x`2,x`1] = x by A1,A13,A11;
    hence thesis by A1,A13,FUNCT_1:def 3;
  end;
  hence thesis by CARD_1:5;
end;

theorem Th17:
  x <> y implies card X +` card Y = card([:X,{x}:] \/ [:Y,{y}:])
proof
  assume
A1: x <> y;
  X,card X are_equipotent & Y,card Y are_equipotent by CARD_1:def 2;
  then card plus(X,Y,x,y) = card plus(card X,card Y,x,y) by A1,Th12;
  hence thesis by A1,Lm1;
end;

theorem
  M+`0 = M
proof
  thus M+`0 = card plus(M,{}) by Lm1
    .= card M by Th6
    .= M by CARD_1:def 2;
end;

Lm4: x <> y implies [:X,{x}:] misses [:Y,{y}:]
proof
  assume
A1: x <> y;
  assume not thesis;
  then consider z being set such that
A2: z in [:X,{x}:] and
A3: z in [:Y,{y}:] by XBOOLE_0:3;
  z`2 = x by A2,MCART_1:13;
  hence contradiction by A1,A3,MCART_1:13;
end;

theorem Th19:
  (K+`M)+`N = K+`(M+`N)
proof
A1: card(K+`M+`N) = K+`M+`N;
A2: K+`M+`N,[:K+`M,{0}:] \/ [:N,{2}:] are_equipotent by Th10;
  [:M,{1}:] misses [:N,{2}:] by Lm4;
  then
A3: [:M,{1}:] /\ [:N,{2}:] = {} by XBOOLE_0:def 7;
  [:K,{0}:] misses [:N,{2}:] by Lm4;
  then [:K,{0}:] /\ [:N,{2}:] = {} by XBOOLE_0:def 7;
  then ([:K,{0}:] \/ [:M,{1}:]) /\ [:N,{2}:] = {} \/ {} by A3,XBOOLE_1:23
    .= {};
  then
A4: ([:K,{0}:] \/ [:M,{1}:]) misses [:N,{2}:] by XBOOLE_0:def 7;
  K+`M,[:K,{0}:] \/ [:M,{1}:] are_equipotent & K+`M,[:K+`M,{0}:]
  are_equipotent by Th6,Th10;
  then
A5: [:K+`M,{0}:],[:K,{0}:] \/ [:M,{1}:] are_equipotent by WELLORD2:15;
  [:K,{0}:] misses [:N,{2}:] by Lm4;
  then
A6: [:K,{0}:] /\ [:N,{2}:] = {} by XBOOLE_0:def 7;
  [:K,{0}:] misses [:M,{1}:] by Lm4;
  then [:K,{0}:] /\ [:M,{1}:] = {} by XBOOLE_0:def 7;
  then [:K,{0}:] /\ ([:M,{1}:] \/ [:N,{2}:]) = {} \/ {} by A6,XBOOLE_1:23
    .= {};
  then
A7: [:K,{0}:] misses ([:M,{1}:] \/ [:N,{2}:]) by XBOOLE_0:def 7;
  M+`N,[:M,{1}:] \/ [:N,{2}:] are_equipotent & M+`N,[:M+`N,{2}:]
  are_equipotent by Th6,Th10;
  then
A8: [:M+`N,{2}:],[:M,{1}:] \/ [:N,{2}:] are_equipotent by WELLORD2:15;
  [:K,{0}:] misses [:M+`N,{2}:] by Lm4;
  then
A9: [:K,{0}:] \/ ([:M,{1}:] \/ [:N,{2}:]),[:K,{0}:] \/ [:M+`N,{2}:]
  are_equipotent by A7,A8,CARD_1:31;
  [:K+`M,{0}:] misses [:N,{2}:] by Lm4;
  then
A10: [:K+`M,{0}:] \/ [:N,{2}:], [:K,{0}:] \/ [:M,{1}:] \/ [:N,{2}:]
  are_equipotent by A4,A5,CARD_1:31;
  [:K,{0}:] \/ ([:M,{1}:] \/ [:N,{2}:]) = [:K,{0}:] \/ [:M,{1}:] \/ [:N,{
  2}:] by XBOOLE_1:4;
  then [:K+`M,{0}:] \/ [:N,{2}:],[:K,{0}:] \/ [:M+`N,{2}:] are_equipotent by
A10,A9,WELLORD2:15;
  then
A11: K+`M+`N,[:K,{0}:] \/ [:M+`N,{2}:] are_equipotent by A2,WELLORD2:15;
  [:K,{0}:] \/ [:M+`N,{2}:],K+`(M+`N) are_equipotent by Th10;
  then K+`M+`N,K+`(M+`N) are_equipotent by A11,WELLORD2:15;
  hence thesis by A1,CARD_1:def 2;
end;

theorem
  K*`0 = 0;

theorem Th21:
  K*`1 = K
proof
  K = card K by CARD_1:def 2;
  hence thesis by Th6,ORDINAL3:15;
end;

theorem Th22:
  (K*`M)*`N = K*`(M*`N)
proof
  thus (K*`M)*`N = card [:[:K,M:],N:] by Th7
    .= card [:K,[:M,N:]:] by Th5
    .= K*`(M*`N) by Th7;
end;

theorem Th23:
  2*`K = K+`K
proof
  thus 2*`K = card([:{{}},K:] \/ [:{1},K:]) by CARD_1:50,ZFMISC_1:109
    .= card([:K,{{}}:] \/ [:K,{1}:]) by Th16
    .= card K +` card K by Th17
    .= K +` card K by CARD_1:def 2
    .= K +` K by CARD_1:def 2;
end;

theorem Th24:
  K*`(M+`N) = K*`M +` K*`N
proof
A1: [:card [:K,M:],{0}:],[:[:K,M:],{0}:] are_equipotent by Th7;
  M,[:M,{0}:] are_equipotent by Th6;
  then
A2: [:K,M:],[:K,[:M,{0}:]:] are_equipotent by Th8;
  [:[:K,M:],{0}:],[:K,M:] are_equipotent by Th6;
  then [:[:K,M:],{0}:],[:K,[:M,{0}:]:] are_equipotent by A2,WELLORD2:15;
  then
A3: [:card [:K,M:],{0}:],[:K,[:M,{0}:]:] are_equipotent by A1,WELLORD2:15;
A4: [:card [:K,N:],{1}:],[:[:K,N:],{1}:] are_equipotent by Th7;
  [:M,{0}:] misses [:N,{1}:] by Lm4;
  then
A5: [:K,[:M,{0}:]:] misses [:K,[:N,{1}:]:] by ZFMISC_1:104;
  N,[:N,{1}:] are_equipotent by Th6;
  then
A6: [:K,N:],[:K,[:N,{1}:]:] are_equipotent by Th8;
A7: K*`(M+`N) = card [:K,card plus(M,N):] by Th10
    .= card [:K,plus(M,N):] by Th7
    .= card ([:K,[:M,{0}:]:] \/ [:K,[:N,{1}:]:]) by ZFMISC_1:97;
  [:[:K,N:],{1}:],[:K,N:] are_equipotent by Th6;
  then [:[:K,N:],{1}:],[:K,[:N,{1}:]:] are_equipotent by A6,WELLORD2:15;
  then
A8: [:card [:K,N:],{1}:],[:K,[:N,{1}:]:] are_equipotent by A4,WELLORD2:15;
  [:card [:K,M:],{0}:] misses [:card [:K,N:],{1}:] by Lm4;
  then
  [:card [:K,M:],{0}:] \/ [:card [:K,N:],{1}:], [:K,[:M,{0}:]:] \/ [:K,[:
  N,{1}:]:] are_equipotent by A3,A8,A5,CARD_1:31;
  hence K*`(M+`N) = card([:card [:K,M:],{0}:] \/ [:card [:K,N:],{1}:]) by A7,
CARD_1:5
    .= K*`M +` K*`N by Th10;
end;

theorem
  exp(K,0) = 1
proof
  thus exp(K,0) = card {{}} by FUNCT_5:57
    .= 1 by CARD_1:30;
end;

theorem
  K <> 0 implies exp(0,K) = 0;

theorem
  exp(K,1) = K & exp(1,K) = 1
proof
  thus exp(K,1) = card K by FUNCT_5:58,ORDINAL3:15
    .= K by CARD_1:def 2;
  thus exp(1,K)
     = card {K --> {}} by FUNCT_5:59,ORDINAL3:15
    .= 1 by CARD_1:30;
end;

theorem
  exp(K,M+`N) = exp(K,M)*`exp(K,N)
proof
A1: [:M,{0}:] misses [:N,{1}:] by ZFMISC_1:108;
  [:M,{0}:],M are_equipotent by Th6;
  then
A2: Funcs([:M,{0}:],K),Funcs(M,K) are_equipotent by FUNCT_5:60;
  [:N,{1}:],N are_equipotent by Th6;
  then
A3: Funcs([:N,{1}:],K),Funcs(N,K) are_equipotent by FUNCT_5:60;
  M+`N,[:M,{0}:] \/ [:N,{1}:] are_equipotent by Th10;
  hence exp(K,M+`N) = card Funcs([:M,{0}:] \/ [:N,{1}:],K) by FUNCT_5:60
    .= card [:Funcs([:M,{0}:],K),Funcs([:N,{1}:],K):] by A1,FUNCT_5:62
    .= card [:Funcs(M,K),Funcs(N,K):] by A2,A3,Th8
    .= exp(K,M)*`exp(K,N) by Th7;
end;

theorem
  exp(K*`M,N) = exp(K,N)*`exp(M,N)
proof
  card(K*`M) = K*`M & card N = card N;
  hence exp(K*`M,N) = card Funcs(N,[:K,M:]) by FUNCT_5:61
    .= card [:Funcs(N,K),Funcs(N,M):] by FUNCT_5:64
    .= exp(K,N)*`exp(M,N) by Th7;
end;

theorem
  exp(K,M*`N) = exp(exp(K,M),N)
proof
A1: Funcs(M,K),card Funcs(M,K) are_equipotent by CARD_1:def 2;
  [:M,N:],M*`N are_equipotent & [:N,M:],[:M,N:] are_equipotent by Lm2,
CARD_1:def 2;
  then [:N,M:],M*`N are_equipotent by WELLORD2:15;
  hence exp(K,M*`N) = card Funcs([:N,M:],K) by FUNCT_5:60
    .= card Funcs(N,Funcs(M,K)) by FUNCT_5:63
    .= exp(exp(K,M),N) by A1,FUNCT_5:60;
end;


theorem
  exp(2,card X) = card bool X
proof
  card card X = card X & card 2 = card {{},1} by CARD_1:50;
  hence exp(2,card X) = card Funcs(X,{{},1}) by FUNCT_5:61
    .= card bool X by FUNCT_5:65;
end;

theorem
  exp(K,2) = K*`K by CARD_1:50,FUNCT_5:66;

theorem
  exp(K+`M,2) = K*`K +` 2*`K*`M +` M*`M
proof
  thus exp(K+`M,2) = (K+`M)*`(K+`M) by CARD_1:50,FUNCT_5:66
    .= K*`(K+`M) +` M*`(K+`M) by Th24
    .= K*`K +` K*`M +` M*`(K+`M) by Th24
    .= K*`K +` K*`M +` (M*`K +` M*`M) by Th24
    .= K*`K +` K*`M +` K*`M +` M*`M by Th19
    .= K*`K +` (K*`M +` K*`M) +` M*`M by Th19
    .= K*`K +` 2*`(K*`M) +` M*`M by Th23
    .= K*`K +` 2*`K*`M +` M*`M by Th22;
end;

theorem Th34:
  card(X \/ Y) c= card X +` card Y
proof
  consider f such that
A1: dom f = plus(X,Y) & for x st x in plus(X,Y) holds f.x = g(x) from
  FUNCT_1:sch 3;
  X \/ Y c= rng f
  proof
    let x;
    assume x in X \/ Y;
    then
A2: x in X or x in Y by XBOOLE_0:def 3;
    per cases;
    suppose
      x in X;
      then [x,0] in [:X,{0}:] by ZFMISC_1:106;
      then
A3:   [x,0] in plus(X,Y) by XBOOLE_0:def 3;
      [x,0]`1 = x;
      then x = f.[x,0] by A1,A3;
      hence thesis by A1,A3,FUNCT_1:def 3;
    end;
    suppose
      not x in X;
      then [x,1] in [:Y,{1}:] by A2,ZFMISC_1:106;
      then
A4:   [x,1] in plus(X,Y) by XBOOLE_0:def 3;
      [x,1]`1 = x;
      then x = f.[x,1] by A1,A4;
      hence thesis by A1,A4,FUNCT_1:def 3;
    end;
  end;
  then card(X \/ Y) c= card plus(X,Y) by A1,CARD_1:12;
  hence thesis by Th17;
end;

theorem Th35:
  X misses Y implies card (X \/ Y) = card X +` card Y
proof
  assume
A1: X misses Y;
  X,[:X,{0}:] are_equipotent & [:X,{0}:],[:card X,{0}:] are_equipotent by Th6
,Th7;
  then
A2: X,[:card X,{0}:] are_equipotent by WELLORD2:15;
  Y,[:Y,{1}:] are_equipotent & [:Y,{1}:],[:card Y,{1}:] are_equipotent by Th6
,Th7;
  then
A3: Y,[:card Y,{1}:] are_equipotent by WELLORD2:15;
  [:card X,{0}:] misses [:card Y,{1}:] by Lm4;
  then X \/ Y,[:card X,{0}:] \/ [:card Y,{1}:] are_equipotent by A1,A2,A3,
CARD_1:31;
  hence card (X \/ Y) = card ([:card X,{0}:] \/ [:card Y,{1}:]) by CARD_1:5
    .= card X +` card Y by Th10;
end;

reserve m,n for Element of NAT;

theorem Th36:
  n+m = n +^ m
proof
  defpred P[Element of NAT] means n+$1 = n +^ $1;
A1: for m st P[m] holds P[m+1]
  proof
    let m such that
A2: P[m];
    thus n+(m+1) = n+m+1 .= succ( n +^ m) by A2,NAT_1:38
      .= n +^ succ m by ORDINAL2:28
      .= n +^ (m+1) by NAT_1:38;
  end;
A3: P[0] by ORDINAL2:27;
  for m holds P[m] from NAT_1:sch 1(A3,A1);
  hence thesis;
end;

theorem Th37:
  n*m = n *^ m
proof
  defpred P[Element of NAT] means $1*m = $1 *^ m;
A1: for n st P[n] holds P[n+1]
  proof
    let n such that
A2: P[n];
    thus (n+1)*m = n*m+1*m .= n *^ m +^ m by A2,Th36
      .= n *^ m +^ 1 *^ m by ORDINAL2:39
      .= ( n +^ 1) *^ m by ORDINAL3:46
      .= (succ n) *^ m by ORDINAL2:31
      .= (n+1) *^ m by NAT_1:38;
  end;
A3: P[0] by ORDINAL2:35;
  for n holds P[n] from NAT_1:sch 1(A3,A1);
  hence thesis;
end;

theorem Th38:
  card(n+m) = card n +` card m
proof
  thus card(n+m) = card( n +^ m) by Th36
    .= card n +` card m by Th13;
end;

theorem Th39:
  card(n*m) = card n *` card m
proof
  thus card(n*m) = card( n *^ m) by Th37
    .= card n *` card m by Th14;
end;

theorem Th40:
  for X,Y being finite set st X misses Y holds card (X \/ Y) = card X + card Y
proof
  let X,Y be finite set;
  assume X misses Y;
  then card card (X \/ Y) = card card X +` card card Y by Th35
    .= card (card X + card Y) by Th38;
  hence thesis by CARD_1:40;
end;

theorem Th41:
  for X being finite set st not x in X holds card (X \/ {x}) = card X + 1
proof
  let X be finite set;
A1: card {x} = 1 by CARD_1:30;
  assume not x in X;
  hence thesis by A1,Th40,ZFMISC_1:50;
end;

theorem Th42:
  for X being set holds card X = 1 iff ex x st X = {x}
proof
  let X be set;
  card {0} = 1 by CARD_1:30;
  hence card X = 1 implies ex x st X = {x} by CARD_1:29;
  given x such that
A1: X = {x};
  thus thesis by A1,CARD_1:30;
end;

theorem Th43:
  for X,Y being finite set holds card(X \/ Y) <= card X + card Y
proof
  let X,Y be finite set;
  card X = card card X & card Y = card card Y;
  then card X +` card Y = card(card X + card Y) by Th38;
  then
  card card(X \/ Y) = card(X \/ Y) & card (X \/ Y) c= card(card X + card Y
  ) by Th34;
  hence thesis by NAT_1:40;
end;

theorem Th44:
  for X,Y being finite set st Y c= X holds card (X \ Y) = card X - card Y
proof
  let X,Y be finite set;
  defpred P[set] means ex S being finite set st S = $1 & card (X \ S) = card X
  - card S;
  card X - 0 = card X & X \ {} = X;
  then
A1: P[{}] by CARD_1:27;
  assume
A2: Y c= X;
A3: for X1,Z being set st X1 in Y & Z c= Y & P[Z] holds P[Z \/ {X1}]
  proof
    let X1,Z be set such that
A4: X1 in Y and
A5: Z c= Y and
A6: P[Z] and
A7: not P[Z \/ {X1}];
A8: card {X1} = 1 by CARD_1:30;
A9: now
      assume X1 in Z;
      then {X1} c= Z by ZFMISC_1:31;
      then Z = Z \/ {X1} by XBOOLE_1:12;
      hence P[Z \/ {X1}] by A6;
    end;
    then
A10: X1 in X \ Z by A2,A4,A7,XBOOLE_0:def 5;
    then consider m being Nat such that
A11: card (X \ Z) = m+1 by NAT_1:6;
    reconsider Z1 = Z as finite set by A5;
A12: X \ Z, card (X \ Z) are_equipotent & X \ (Z \/ {X1}) = X \ Z \ {X1}
    by CARD_1:def 2,XBOOLE_1:41;
    card { X1} = 1 by CARD_1:30;
    then
A13: card Z1 + card {X1} = card (Z1 \/ {X1}) by A7,A9,Th41;
    m+1 = succ m by NAT_1:38;
    then m in m+1 & m = m+1 \ {m} by ORDINAL1:6,37;
    then X \ (Z \/ {X1}), m are_equipotent by A10,A11,A12,CARD_1:34;
    then card (X \ (Z \/ {X1})) = card X - card (Z1 \/ {X1}) by A6,A13,A11,A8,
CARD_1:def 2;
    hence contradiction by A7;
  end;
A14: Y is finite;
  P[Y] from FINSET_1:sch 2(A14,A1,A3);
  hence thesis;
end;

theorem
  for X,Y being finite set holds card (X \/ Y) = card X + card Y - card
  (X /\ Y)
proof
  let X,Y be finite set;
  Y \ X = Y \ X /\ Y by XBOOLE_1:47;
  then
A1: card (Y \ X) = card Y - card (X /\ Y) by Th44,XBOOLE_1:17;
  card (X \/ (Y \ X)) = card X + card (Y \ X) by Th40,XBOOLE_1:79;
  hence thesis by A1,XBOOLE_1:39;
end;

theorem
  for X,Y being finite set holds card [:X,Y:] = card X * card Y
proof
  let X,Y be finite set;
  card card [:X,Y:] = card card X *` card card Y by Th7
    .= card (card X * card Y) by Th39;
  hence thesis by CARD_1:40;
end;

theorem 
  for f being finite Function holds card rng f <= card dom f
     by CARD_1:12,NAT_1:39;

theorem
  for X,Y being finite set st X c< Y holds card X < card Y & card X in card Y
proof
  let X,Y be finite set;
  assume
A1: X c< Y;
  then X c= Y by XBOOLE_0:def 8;
  then
A2: Y = X \/ (Y\X) by XBOOLE_1:45;
  then
A3: card Y = card X + card (Y\X) by Th40,XBOOLE_1:79;
  then
A4: card X <= card Y by NAT_1:11;
  now
    assume card (Y\X) = 0;
    then Y \ X = {};
    hence contradiction by A1,A2;
  end;
  then card X <> card Y by A3;
  hence card X < card Y by A4,XXREAL_0:1;
  hence thesis by NAT_1:44;
end;

theorem
  (card X c= card Y or card X in card Y) & Y is finite implies X is finite
proof
  assume that
A1: card X c= card Y or card X in card Y and
A2: Y is finite;
  card X c= card Y by A1,ORDINAL1:def 2;
  hence thesis by A2;
end;

reserve x1,x2,x3,x4,x5,x6,x7,x8 for set;

theorem Th50:
  card {x1,x2} <= 2
proof
A1: {x1,x2} = {x1} \/ {x2} & 1+1 = 2 by ENUMSET1:1;
  card {x1} = 1 & card {x2} = 1 by CARD_1:30;
  hence thesis by A1,Th43;
end;

theorem Th51:
  card {x1,x2,x3} <= 3
proof
  card {x2,x3} <= 2 by Th50;
  then
A1: 1 + card {x2,x3} <= 1+2 by XREAL_1:7;
  card {x1} = 1 & {x1,x2,x3} = {x1} \/ {x2,x3} by CARD_1:30,ENUMSET1:2;
  then card {x1,x2,x3} <= 1 + card {x2,x3} by Th43;
  hence thesis by A1,XXREAL_0:2;
end;

theorem Th52:
  card {x1,x2,x3,x4} <= 4
proof
  card {x2,x3,x4} <= 3 by Th51;
  then
A1: 1 + card {x2,x3,x4} <= 1+3 by XREAL_1:7;
  card {x1} = 1 & {x1,x2,x3,x4} = {x1} \/ {x2,x3,x4} by CARD_1:30,ENUMSET1:4;
  then card {x1,x2,x3,x4} <= 1 + card {x2,x3,x4 } by Th43;
  hence thesis by A1,XXREAL_0:2;
end;

theorem Th53:
  card {x1,x2,x3,x4,x5} <= 5
proof
  card {x2,x3,x4,x5} <= 4 by Th52;
  then
A1: 1 + card {x2,x3,x4,x5} <= 1+4 by XREAL_1:7;
  card {x1} = 1 & {x1,x2,x3,x4,x5} = {x1} \/ {x2,x3,x4,x5} by CARD_1:30
,ENUMSET1:7;
  then card {x1,x2,x3,x4,x5} <= 1 + card {x2,x3,x4,x5} by Th43;
  hence thesis by A1,XXREAL_0:2;
end;

theorem Th54:
  card {x1,x2,x3,x4,x5,x6} <= 6
proof
  card {x2,x3,x4,x5,x6} <= 5 by Th53;
  then
A1: 1 + card {x2,x3,x4,x5,x6} <= 1+5 by XREAL_1:7;
  card {x1} = 1 & {x1,x2,x3,x4,x5,x6} = {x1} \/ {x2,x3,x4,x5,x6} by CARD_1:30
,ENUMSET1:11;
  then card {x1,x2,x3,x4,x5,x6} <= 1 + card {x2,x3,x4,x5,x6} by Th43;
  hence thesis by A1,XXREAL_0:2;
end;

theorem Th55:
  card {x1,x2,x3,x4,x5,x6,x7} <= 7
proof
  card {x2,x3,x4,x5,x6,x7} <= 6 by Th54;
  then
A1: 1 + card {x2,x3,x4,x5,x6,x7} <= 1+6 by XREAL_1:7;
  card {x1} = 1 & {x1,x2,x3,x4,x5,x6,x7} = {x1} \/ {x2,x3,x4,x5,x6,x7} by
CARD_1:30,ENUMSET1:16;
  then card {x1,x2,x3,x4,x5,x6,x7} <= 1 + card {x2,x3,x4,x5,x6,x7} by Th43;
  hence thesis by A1,XXREAL_0:2;
end;

theorem
  card {x1,x2,x3,x4,x5,x6,x7,x8} <= 8
proof
  card {x2,x3,x4,x5,x6,x7,x8} <= 7 by Th55;
  then
A1: 1 + card {x2,x3,x4,x5,x6,x7,x8} <= 1+7 by XREAL_1:7;
  {x1,x2,x3,x4,x5,x6,x7,x8} = {x1} \/ {x2,x3,x4,x5,x6,x7,x8} & card {x1} =
  1 by CARD_1:30,ENUMSET1:22;
  then card {x1,x2,x3,x4,x5,x6,x7,x8} <= 1 + card {x2,x3,x4,x5,x6,x7,x8} by
Th43;
  hence thesis by A1,XXREAL_0:2;
end;

theorem Th57:
  x1 <> x2 implies card {x1,x2} = 2
proof
A1: card {x1} = 1 & card {x2} = 1 by CARD_1:30;
A2: {x1,x2} = {x1} \/ {x2} & 1+1 = 2 by ENUMSET1:1;
  assume x1 <> x2;
  hence thesis by A1,A2,Th40,ZFMISC_1:11;
end;

theorem Th58:
  x1 <> x2 & x1 <> x3 & x2 <> x3 implies card {x1,x2,x3} = 3
proof
  assume x1 <> x2 & x1 <> x3 & x2 <> x3;
  then
A1: card {x1,x2} = 2 & not x3 in {x1,x2} by Th57,TARSKI:def 2;
  {x1,x2,x3} = {x1,x2} \/ {x3} by ENUMSET1:3;
  hence card {x1,x2,x3} = 2+1 by A1,Th41
    .= 3;
end;

theorem Th59:
  x1 <> x2 & x1 <> x3 & x1 <> x4 & x2 <> x3 & x2 <> x4 & x3 <> x4
  implies card {x1,x2,x3,x4} = 4
proof
  assume x1 <> x2 & x1 <> x3 & x1 <> x4 & x2 <> x3 & x2 <> x4 & x3 <> x4;
  then
A1: card {x1,x2,x3} = 3 & not x4 in {x1,x2,x3} by Th58,ENUMSET1:def 1;
  {x1,x2,x3,x4} = {x1,x2,x3} \/ {x4} by ENUMSET1:6;
  hence card {x1,x2,x3,x4} = 3+1 by A1,Th41
    .= 4;
end;

begin 


theorem
  for X being set st card X = 2 ex x,y st x <> y & X = {x,y}
proof
  let X be set;
  assume
A1: card X = 2;
  then consider x being set such that
A2: x in X by CARD_1:27,XBOOLE_0:def 1;
  X is finite by A1;
  then reconsider Y = X as finite set;
  {x} c= X by A2,ZFMISC_1:31;
  then card(X \ {x}) = card Y - card{x} by Th44
    .= 2 - 1 by A1,CARD_1:30;
  then consider y such that
A3: X \ {x} = {y} by Th42;
  take x,y;
  x in {x} by TARSKI:def 1;
  hence x <> y by A3,XBOOLE_0:def 5;
  thus X c= {x,y}
  proof
    let z be set;
    assume
A4: z in X;
    per cases;
    suppose
      z = x;
      hence thesis by TARSKI:def 2;
    end;
    suppose
      z <> x;
      then not z in {x} by TARSKI:def 1;
      then z in {y} by A3,A4,XBOOLE_0:def 5;
      then z = y by TARSKI:def 1;
      hence thesis by TARSKI:def 2;
    end;
  end;
  let z be set;
  assume z in {x,y};
  then
A5: z = x or z = y by TARSKI:def 2;
  y in X \ {x} by A3,TARSKI:def 1;
  hence thesis by A2,A5;
end;



theorem
  for f being Function holds card rng f c= card dom f
proof
  let f be Function;
  rng f = f.:dom f by RELAT_1:113;
  hence thesis by CARD_1:66;
end;



Lm5: now
  let n;
  assume
A1: for Z being finite set holds card Z = n & Z <> {} & (for X,Y st X in
  Z & Y in Z holds X c= Y or Y c= X) implies union Z in Z;
  let Z be finite set;
  assume that
A2: card Z = n + 1 and
A3: Z <> {} and
A4: for X,Y st X in Z & Y in Z holds X c= Y or Y c= X;
  set y = the Element of Z;
  per cases;
  suppose
    n = 0;
    then consider x such that
A5: Z = {x} by A2,Th42;
    union Z = x by A5,ZFMISC_1:25;
    hence union Z in Z by A5,TARSKI:def 1;
  end;
  suppose
A6: n <> 0;
    set Y = Z \ {y};
    reconsider Y as finite set;
    {y} c= Z by A3,ZFMISC_1:31;
    then
A7: card Y = (n + 1) - card{y} by A2,Th44
      .= n + 1 - 1 by CARD_1:30
      .= n;
    for a,b being set st a in Y & b in Y holds a c= b or b c= a by A4;
    then
A8: union Y in Y by A1,A6,A7,CARD_1:27;
    then
A9: union Y in Z;
    Z = (Z \ {y}) \/ {y}
    proof
      thus Z c= (Z \ {y}) \/ {y}
      proof
        let x;
        assume x in Z;
        then x in Z \ {y} or x in {y} by XBOOLE_0:def 5;
        hence thesis by XBOOLE_0:def 3;
      end;
      let x;
      assume x in (Z \ {y}) \/ {y};
      then
A10:  x in (Z \ {y}) or x in {y} by XBOOLE_0:def 3;
      {y} c= Z by A3,ZFMISC_1:31;
      hence thesis by A10;
    end;
    then
A11: union Z = union Y \/ union {y} by ZFMISC_1:78
      .= union Y \/ y by ZFMISC_1:25;
A12: y in Z by A3;
    y c= union Y or union Y c= y by A4,A8;
    hence union Z in Z by A9,A12,A11,XBOOLE_1:12;
  end;
end;

Lm6: for Z being finite set holds Z <> {} & (for X,Y st X in Z & Y in Z holds
X c= Y or Y c= X) implies union Z in Z
proof
  defpred P[Element of NAT] means for Z being finite set st card Z = $1 & Z <>
  {} & (for X,Y st X in Z & Y in Z holds X c= Y or Y c= X) holds union Z in Z;
  let Z be finite set;
A1: card Z = card Z;
A2: P[0];
A3: for k be Element of NAT st P[k] holds P[k+1] by Lm5;
  for n holds P[n] from NAT_1:sch 1(A2,A3);
  hence thesis by A1;
end;

theorem
  Z <> {} & Z is finite & (for X,Y st X in Z & Y in Z holds X c= Y or Y
  c= X) implies union Z in Z by Lm6;

theorem
  x1,x2,x3,x4,x5 are_mutually_different implies card {x1,x2,x3,x4,x5} = 5
proof
A1: {x1,x2,x3,x4,x5} = {x1,x2,x3,x4} \/ {x5} by ENUMSET1:10;
  assume
A2: x1,x2,x3,x4,x5 are_mutually_different;
  then
A3: x3 <> x5 & x4 <> x5 by ZFMISC_1:def 7;
A4: x2 <> x4 & x3 <> x4 by A2,ZFMISC_1:def 7;
A5: x1 <> x4 & x2 <> x3 by A2,ZFMISC_1:def 7;
  x1 <> x5 & x2 <> x5 by A2,ZFMISC_1:def 7;
  then
A6: not x5 in {x1,x2,x3,x4} by A3,ENUMSET1:def 2;
  x1 <> x2 & x1 <> x3 by A2,ZFMISC_1:def 7;
  then card {x1,x2,x3,x4} = 4 by A5,A4,Th59;
  hence card {x1,x2,x3,x4,x5} = 4+1 by A6,A1,Th41
    .= 5;
end;



theorem
  for M1, M2 being set st card M1 = 0 & card M2 = 0 holds M1 = M2
proof
  let M1, M2 be set;
  assume that
A1: card M1 = {} and
A2: card M2 = {};
  M1 = {} by A1;
  hence thesis by A2;
end;



registration
  let x,y;
  cluster [x,y] -> non natural;
  coherence
  proof
    assume [x,y] is natural;
    then reconsider n = [x,y] as Nat;
    card n <= 2 by Th50;
    then
A1: n <= 2 by CARD_1:def 2;
    per cases by A1,NAT_1:26;
    suppose
      n = 0;
      hence contradiction;
    end;
    suppose
      n = 1;
      hence contradiction by CARD_1:49,ZFMISC_1:4;
    end;
    suppose
      n = 2;
      hence contradiction by CARD_1:50,ZFMISC_1:6;
    end;
  end;
end;

begin 

reserve A,B,C for Ordinal,
  K,L,M,N for Cardinal,
  x,y,y1,y2,z,u,X,Y,Z,Z1,Z2 for set,
  n for Nat,
  f,f1,g,h for Function,
  Q,R for Relation;

theorem
  Sum(M --> N) = M*`N
proof
  thus Sum(M --> N) = card [:N,M:] by CARD_3:25
    .= M*`N by Lm2;
end;

theorem
  Product(N --> M) = exp(M,N) by CARD_3:11;

scheme FinRegularity { X()->finite set, P[set,set] }:
  ex x st x in X() & for y st y in X() & y <> x holds not P[y,x]
provided
A1: X() <> {} and
A2: for x,y st P[x,y] & P[y,x] holds x = y and
A3: for x,y,z st P[x,y] & P[y,z] holds P[x,z]
proof
  defpred Q[Nat] means
  for X being finite set st card X = $1 & X <> {}
  ex x st x in X & for y st y in X & y <> x holds not P[y,x];
A4: Q[0];
A5: Q[n] implies Q[n+1]
  proof
    assume
A6: for X being finite set st card X = n & X <> {}
    ex x st x in X & for y st y in X & y <> x holds not P[y,x];
    let X be finite set;
    assume that
A7: card X = n+1 and
A8: X <> {};
    set x = the Element of X;
A9: now
      assume X\{x} = {};
      then
A10:  X c= {x} by XBOOLE_1:37;
      thus thesis
      proof
        take x;
        thus x in X by A8;
        thus thesis by A10,TARSKI:def 1;
      end;
    end;
    now
      assume
A11:  X\{x} <> {};
      {x} c= X by A8,ZFMISC_1:31;
      then
A12:  card (X\{x}) = (n+1) - card {x} by A7,Th44;
      card {x} = 1 by CARD_1:30;
      then consider y such that
A13:  y in X\{x} and
A14:  for z st z in X\{x} & z <> y holds not P[z,y] by A6,A11,A12;
A15:  now
        assume
A16:    P[x,y];
        thus thesis
        proof
          take x;
          thus x in X by A8;
          let z;
          assume that
A17:      z in X and
A18:      z <> x and
A19:      P[z,x];
          not z in {x} by A18,TARSKI:def 1;
          then
A20:      z in X \ {x} by A17,XBOOLE_0:def 5;
A21:      not y in {x} by A13,XBOOLE_0:def 5;
A22:      z = y by A3,A14,A16,A19,A20;
          y <> x by A21,TARSKI:def 1;
          hence contradiction by A2,A16,A19,A22;
        end;
      end;
      now
        assume
A23:    not P[x,y];
        thus thesis
        proof
          take y;
          thus y in X by A13;
          let z such that
A24:      z in X and
A25:      z <> y;
          z in {x} or not z in {x};
          then z = x or z in X \ {x} by A24,TARSKI:def 1,XBOOLE_0:def 5;
          hence thesis by A14,A23,A25;
        end;
      end;
      hence thesis by A15;
    end;
    hence thesis by A9;
  end;
A26: Q[n] from NAT_1:sch 2(A4,A5);
  card X() = card X();
  hence thesis by A1,A26;
end;

scheme MaxFinSetElem { X()->finite set, P[set,set] }:
  ex x st x in X() & for y st y in X() holds P[x,y]
provided
A1: X() <> {} and
A2: for x,y holds P[x,y] or P[y,x] and
A3: for x,y,z st P[x,y] & P[y,z] holds P[x,z]
proof
  defpred Q[Nat] means
  for X being finite set st card X = $1 & X <> {}
  ex x st x in X & for y st y in X holds P[x,y];
A4: Q[0];
A5: Q[n] implies Q[n+1]
  proof
    assume
A6: for X being finite set st card X = n & X <> {}
    ex x st x in X & for y st y in X holds P[x,y];
    let X be finite set;
    assume that
A7: card X = n+1 and
A8: X <> {};
    set x = the Element of X;
A9: now
      assume X\{x} = {};
      then
A10:  X c= {x} by XBOOLE_1:37;
      thus thesis
      proof
        take x;
        thus x in X by A8;
        let y;
        assume y in X;
        then y = x by A10,TARSKI:def 1;
        hence thesis by A2;
      end;
    end;
    now
      assume
A11:  X\{x} <> {};
      {x} c= X by A8,ZFMISC_1:31;
      then
A12:  card (X\{x}) = (n+1) - card {x} by A7,Th44;
      card {x} = 1 by CARD_1:30;
      then consider y such that
A13:  y in X\{x} and
A14:  for z st z in X\{x} holds P[y,z] by A6,A11,A12;
A15:  P[x,y] or P[y,x] by A2;
A16:  P[x,x] by A2;
      P[y,y] by A2;
      then consider z such that
A17:  z = x or z = y and
A18:  P[z,x] and
A19:  P[z,y] by A15,A16;
      thus thesis
      proof
        take z;
        thus z in X by A13,A17;
        let u;
A20:    u in {x} or not u in {x};
        assume u in X;
        then u = x or u in X\{x} by A20,TARSKI:def 1,XBOOLE_0:def 5;
        then P[z,u] or P[y,u] by A14,A18;
        hence thesis by A3,A19;
      end;
    end;
    hence thesis by A9;
  end;
A21: Q[n] from NAT_1:sch 2(A4,A5);
  card X() = card X();
  hence thesis by A1,A21;
end;

Lm7: Rank n is finite implies Rank (n+1) is finite
proof
  n+1 = succ n by NAT_1:38;
  then Rank (n+1) = bool Rank n by CLASSES1:30;
  hence thesis;
end;

Lm8: 1 = card 1 by CARD_1:def 2;
Lm9: 2 = card 2 by CARD_1:def 2;

theorem
  Rank n is finite
proof
  defpred P[Nat] means Rank $1 is finite;
A1: P[0] by CLASSES1:29;
A2: for n st P[n] holds P[n+1] by Lm7;
  for n holds P[n] from NAT_1:sch 2(A1,A2);
  hence thesis;
end;

theorem
  0 in M iff 1 c= M
proof 0+1 = 1;
  then nextcard card 0 = card 1 by NAT_1:42;
  hence thesis by Lm8,CARD_3:90;
end;

theorem
  1 in M iff 2 c= M
proof 1+1 = 2;
  then nextcard card 1 = card 2 by NAT_1:42;
  hence thesis by Lm8,Lm9,CARD_3:90;
end;

reserve n,k for Nat;

theorem Th70:
  A is limit_ordinal iff for B,n st B in A holds B+^ n in A
proof
  thus A is limit_ordinal implies for B,n st B in A holds B+^ n in A
  proof
    assume
A1: A is limit_ordinal;
    let B,n;
    defpred P[Nat] means B+^ $1 in A;
    assume B in A;
    then
A2: P[0] by ORDINAL2:27;
A3: P[k] implies P[k+1]
    proof (k+1) = succ k by NAT_1:38;
      then B+^(k+1) = succ (B+^ k) by ORDINAL2:28;
      hence thesis by A1,ORDINAL1:28;
    end;
    P[k] from NAT_1:sch 2(A2,A3);
    hence thesis;
  end;
  assume
A4: for B,n st B in A holds B+^ n in A;
  now
    let B;
    assume B in A;
    then B+^ 1 in A by A4;
    hence succ B in A by ORDINAL2:31;
  end;
  hence thesis by ORDINAL1:28;
end;

theorem Th71:
  A+^succ n = succ A +^ n & A +^ (n+1) = succ A +^ n
proof
  defpred P[Nat] means A+^succ $1 = succ A +^ $1;
  A+^succ 0 = succ A by ORDINAL2:31
    .= succ A +^ 0 by ORDINAL2:27;
  then
A1: P[0];
A2: P[k] implies P[k+1]
  proof
    assume
A3: P[k];
A4: k+1 = succ k by NAT_1:38;
    hence A+^succ (k+1) = succ (succ A +^ k) by A3,ORDINAL2:28
      .= succ A +^ k +^ 1 by ORDINAL2:31
      .= succ A +^ ( k +^ 1) by ORDINAL3:30
      .= succ A +^ (k+1) by A4,ORDINAL2:31;
  end;
  P[k] from NAT_1:sch 2(A1,A2);
  hence A+^succ n = succ A +^ n;
  hence thesis by NAT_1:38;
end;

theorem Th72:
  ex n st A*^succ 1 = A +^ n
proof
  defpred P[Ordinal] means ex n st $1*^2 = $1+^ n;
  {}+^{} = {} by ORDINAL2:27;
  then
A1: P[0] by ORDINAL2:35;
A2: for A st P[A] holds P[succ A]
  proof
    let A;
    given n such that
A3: A*^2 = A+^ n;
    take n+1;
    (succ A)*^2 = A*^2+^2 by ORDINAL2:36
      .= succ(A*^succ 1+^1) by ORDINAL2:28
      .= succ succ(A+^ n) by A3,ORDINAL2:31
      .= succ (A+^succ n) by ORDINAL2:28
      .= succ (A+^ (n+1)) by NAT_1:38
      .= A+^succ (n+1) by ORDINAL2:28;
    hence thesis by Th71;
  end;
A4: for A st A <> {} & A is limit_ordinal & for B st B in A holds P[B]
  holds P[A]
  proof
    let A;
    assume that
A5: A <> {} and
A6: A is limit_ordinal and
A7: for B st B in A holds P[B];
    take 0;
    deffunc f(Ordinal) = $1*^2;
    consider phi being Ordinal-Sequence such that
A8: dom phi = A and
A9: for B st B in A holds phi.B = f(B) from ORDINAL2:sch 3;
A10: A*^2 = union sup phi by A5,A6,A8,A9,ORDINAL2:37
      .= union sup rng phi by ORDINAL2:26;
    thus A*^2 c= A+^ 0
    proof
      let B;
      assume B in A*^2;
      then consider X such that
A11:  B in X and
A12:  X in sup rng phi by A10,TARSKI:def 4;
      reconsider X as Ordinal by A12;
      consider C being Ordinal such that
A13:  C in rng phi and
A14:  X c= C by A12,ORDINAL2:21;
      consider x such that
A15:  x in dom phi and
A16:  C = phi.x by A13,FUNCT_1:def 3;
      reconsider x as Ordinal by A15;
A17:  ex n st x*^2 = x+^ n by A7,A8,A15;
      C = x*^2 by A8,A9,A15,A16;
      then
A18:  C in A by A6,A8,A15,A17,Th70;
      A+^{} = A by ORDINAL2:27;
      hence thesis by A11,A14,A18,ORDINAL1:10;
    end;
A19: 1 in succ 1 by ORDINAL1:6;
A20: A+^ 0 = A by ORDINAL2:27;
A21: A = A*^1 by ORDINAL2:39;
    1 c= 2 by A19,ORDINAL1:def 2;
    hence thesis by A20,A21,ORDINAL2:42;
  end;
  for A holds P[A] from ORDINAL2:sch 1(A1,A2,A4);
  hence thesis;
end;

theorem Th73:
  A is limit_ordinal implies A *^ succ 1 = A
proof consider n such that
A1: A*^2 = A+^ n by Th72;
  assume A is limit_ordinal;
  then
A2: A+^ n is limit_ordinal by A1,ORDINAL3:40;
  now
    assume n <> 0;
    then consider k being Nat such that
A3: n = k+1 by NAT_1:6;
    reconsider k as Element of NAT by ORDINAL1:def 12;
    n = succ k by A3,NAT_1:38;
    then A+^ n = succ(A+^ k) by ORDINAL2:28;
    hence contradiction by A2,ORDINAL1:29;
  end;
  hence thesis by A1,ORDINAL2:27;
end;

Lm10: omega is limit_ordinal by ORDINAL1:def 11;

theorem Th74:
  omega c= A implies 1+^A = A
proof
  deffunc f(Ordinal) = 1+^$1;
  consider phi being Ordinal-Sequence such that
A1: dom phi = omega & for B st B in omega holds phi.B = f(B)
  from ORDINAL2:sch 3;
A2: 1+^omega = sup phi by A1,Lm10,ORDINAL2:29
    .= sup rng phi by ORDINAL2:26;
A3: 1+^omega c= omega
  proof
    let B;
    assume B in 1+^omega;
    then consider C being Ordinal such that
A4: C in rng phi and
A5: B c= C by A2,ORDINAL2:21;
    consider x such that
A6: x in dom phi and
A7: C = phi.x by A4,FUNCT_1:def 3;
    reconsider x as Ordinal by A6;
    reconsider x9 = x as Cardinal by A1,A6;
    reconsider x as finite Ordinal by A1,A6;
A8: C = 1+^x by A1,A6,A7;
    succ x in omega by A1,A6,Lm10,ORDINAL1:28;
    then C in omega by A8,ORDINAL2:31;
    hence thesis by A5,ORDINAL1:12;
  end;
  assume omega c= A;
  then
A9: ex B st A = omega+^B by ORDINAL3:27;
  omega c= 1+^omega by ORDINAL3:24;
  then omega = 1+^omega by A3,XBOOLE_0:def 10;
  hence thesis by A9,ORDINAL3:30;
end;

registration
  cluster infinite -> limit_ordinal for Cardinal;
  coherence
  proof
    let M be Cardinal;
    assume M is infinite;
    then
A1: not M in omega;
    assume not thesis;
    then consider A such that
A2: M = succ A by ORDINAL1:29;
    omega <> M & omega c= M iff omega c< M by XBOOLE_0:def 8;
    then omega = succ A or omega in succ A by A1,A2,CARD_1:4,ORDINAL1:11;
    then
A3: omega c= A by Lm10,ORDINAL1:22,29;
    card (A+^1) = card 1 +` card A by Th13
      .= card (1+^A) by Th13
      .= card A by A3,Th74;
    then card succ A = card A by ORDINAL2:31;
    then
A4: A,succ A are_equipotent by CARD_1:5;
A5: not succ A c= A by ORDINAL1:5,6;
    ex B st succ A = B & for A st A,B are_equipotent holds B c= A by A2,
CARD_1:def 1;
    hence contradiction by A4,A5;
  end;
end;

theorem Th75:
  not M is finite implies M+`M = M
proof
  assume not M is finite;
  then M*^succ 1 = M by Th73;
  then card M = (card 2)*`card M by Th14
    .= card (succ 1*^ M) by Th14
    .= card (1*^ M+^ M) by ORDINAL2:36
    .= M+`M by ORDINAL2:39;
  hence thesis by CARD_1:def 2;
end;

theorem Th76:
  not M is finite & (N c= M or N in M) implies M+`N = M & N+`M = M
proof
  assume that
A1: not M is finite and
A2: N c= M or N in M;
A3: M+`M = M by A1,Th75;
  N c= M by A2,CARD_1:3;
  then
A4: M +^ N c= M +^ M by ORDINAL2:33;
A5: M c= M +^ N by ORDINAL3:24;
A6: card M = M by A3;
A7: M+`N c= M by A3,A4,CARD_1:11;
  M c= M+`N by A5,A6,CARD_1:11;
  hence thesis by A7,XBOOLE_0:def 10;
end;

theorem
  not X is finite & (X,Y are_equipotent or Y,X are_equipotent) implies
  X \/ Y,X are_equipotent & card (X \/ Y) = card X
proof
  assume that
A1: not X is finite and
A2: X,Y are_equipotent or Y,X are_equipotent;
A3: card X = card Y by A2,CARD_1:5;
A4: card X c= card (X \/ Y) by CARD_1:11,XBOOLE_1:7;
A5: card (X \/ Y) c= card X +` card Y by Th34;
  card X +` card Y = card X by A1,A3,Th75;
  then card X = card (X \/ Y) by A4,A5,XBOOLE_0:def 10;
  hence thesis by CARD_1:5;
end;

theorem
  not X is finite & Y is finite implies
  X \/ Y,X are_equipotent & card (X \/ Y) = card X
proof
  assume that
A1: not X is finite and
A2: Y is finite;
  card Y in card X by A1,A2,CARD_3:86;
  then
A3: card X +` card Y = card X by A1,Th76;
A4: card (X \/ Y) c= card X +` card Y by Th34;
  card X c= card (X \/ Y) by CARD_1:11,XBOOLE_1:7;
  then card X = card (X \/ Y) by A3,A4,XBOOLE_0:def 10;
  hence thesis by CARD_1:5;
end;

theorem
  not X is finite & (card Y in card X or card Y c= card X) implies
  X \/ Y,X are_equipotent & card (X \/ Y) = card X
proof
  assume that
A1: not X is finite and
A2: card Y in card X or card Y c= card X;
A3: card X +` card Y = card X by A1,A2,Th76;
A4: card (X \/ Y) c= card X +` card Y by Th34;
  card X c= card (X \/ Y) by CARD_1:11,XBOOLE_1:7;
  then card X = card (X \/ Y) by A3,A4,XBOOLE_0:def 10;
  hence thesis by CARD_1:5;
end;

theorem
  for M,N being finite Cardinal holds M+`N is finite
proof
  let M,N be finite Cardinal;
A1: card M = M by CARD_1:def 2;
  card N = N by CARD_1:def 2;
  then M+`N = card (card M + card N) by A1,Th38;
  hence thesis;
end;

theorem
  not M is finite implies not M+`N is finite & not N+`M is finite
proof
  assume
A1: not M is finite;
  M c= N or N c= M;
  then M c= N & not N is finite or M+`N = M & N+`M = M by A1,Th76;
  hence thesis by A1,Th76;
end;

theorem
  for M,N being finite Cardinal holds M*`N is finite;

theorem Th83:
  K in L & M in N or K c= L & M in N or K in L & M c= N or K c= L & M c= N
  implies K+`M c= L+`N & M+`K c= L+`N
proof
  assume that
A1: K in L & M in N or K c= L & M in N or K in L & M c= N or K c= L & M c=
  N;
A2: K c= L by A1,CARD_1:3;
A3: M c= N by A1,CARD_1:3;
A4: K+`M = card ( K +^ M);
A5: K +^ M c= L +^ N by A2,A3,ORDINAL3:18;
  hence K+`M c= L+`N by CARD_1:11;
  thus thesis by A4,A5,CARD_1:11;
end;

theorem
  M in N or M c= N implies K+`M c= K+`N & K+`M c= N+`K &
  M+`K c= K+`N & M+`K c= N+`K by Th83;

theorem Th85:
  X is countable & Y is countable implies X \/ Y is countable
proof
  assume that
A1: card X c= omega and
A2: card Y c= omega;
A3: card (X \/ Y) c= card X +` card Y by Th34;
A4: omega +` omega = omega by Th75;
  card X +` card Y c= omega +` omega by A1,A2,Th83;
  hence card (X \/ Y) c= omega by A3,A4,XBOOLE_1:1;
end;

theorem Th86:
  (card dom f c= M & for x st x in dom f holds card (f.x) c= N) implies
  card Union f c= M*`N
proof
  assume that
A1: card dom f c= M and
A2: for x st x in dom f holds card (f.x) c= N;
A3: card Union f c= Sum Card f by CARD_3:39;
A4: dom Card f = dom f by CARD_3:def 2;
A5: dom(dom f --> N) = dom f by FUNCOP_1:13;
  now
    let x;
    assume
A6: x in dom Card f;
    then
A7: (Card f).x = card (f.x) by A4,CARD_3:def 2;
    (dom f --> N).x = N by A4,A6,FUNCOP_1:7;
    hence (Card f).x c= (dom f --> N).x by A2,A4,A6,A7;
  end;
  then Sum Card f c= Sum(dom f --> N) by A4,A5,CARD_3:30;
  then
A8: card Union f c= Sum(dom f --> N) by A3,XBOOLE_1:1;
A9: [:card dom f,N:] c= [:M,N:] by A1,ZFMISC_1:95;
  Sum(dom f --> N) = card [:N,dom f:] by CARD_3:25
    .= card [:N,card dom f:] by Th7
    .= card [:card dom f,N:] by Th4;
  then
A10: Sum(dom f --> N) c= card [:M,N:] by A9,CARD_1:11;
  thus thesis by A8,A10,XBOOLE_1:1;
end;

theorem
  (card X c= M & for Y st Y in X holds card Y c= N) implies card union X c=
  M*`N
proof
  assume that
A1: card X c= M and
A2: for Y st Y in X holds card Y c= N;
  now
    let x;
    assume x in dom id X;
    then (id X).x in X by FUNCT_1:18;
    hence card ((id X).x) c= N by A2;
  end;
  then card Union id X c= M*`N by A1,Th86;
  hence thesis;
end;

theorem Th88:
  for f st dom f is finite &
  for x st x in dom f holds f.x is finite holds Union f is finite
proof
  let f;
  assume that
A1: dom f is finite and
A2: for x st x in dom f holds f.x is finite;
  reconsider df = dom f as finite set by A1;
  now
    assume dom f <> {};
    then
A3: df <> {};
    defpred P[set,set] means card (f.$2) c= card (f.$1);
A4: for x,y holds P[x,y] or P[y,x];
A5: for x,y,z being set st P[x,y] & P[y,z] holds P[x,z] by XBOOLE_1:1;
    consider x such that
A6: x in df & for y st y in df holds P[x,y] from MaxFinSetElem(A3,A4,
    A5);
    reconsider fx = f.x as finite set by A2,A6;
A7: card Union f c= (card card df) *` (card (f.x)) by A6,Th86;
    card (f.x) = card card fx;
    then card Union f c= card ((card df) * (card fx)) by A7,Th39;
    hence thesis;
  end;
  hence thesis by RELAT_1:42,ZFMISC_1:2;
end;

theorem
  (omega)*`(card n) c= omega & (card n)*`(omega) c= omega
proof
  defpred P[Nat] means (omega)*`(card $1) c= omega;
A1: P[0];
A2: for k being Nat holds P[k] implies P[k+1]
  proof
    let k be Nat;
    assume
A3: P[k];
    card (k+1) = (k+1) by CARD_1:def 2
      .= succ k by NAT_1:38;
    then card (k+1) = card succ k;
    then (omega)*`(card (k+1)) =
    card (succ k *^ omega) by Th14,CARD_1:47
      .= card ( k *^ omega +^ omega) by ORDINAL2:36
      .= card ( k *^ omega) +` omega by Th13,CARD_1:47
      .= (omega)*`(card k) +` omega by Th14,CARD_1:47
      .= omega by A3,Th76;
    hence thesis;
  end;
A4: for k being Nat holds P[k] from NAT_1:sch 2(A1,A2);
  hence (omega)*`(card n) c= omega;
  thus thesis by A4;
end;

theorem Th90:
  K in L & M in N or K c= L & M in N or K in L & M c= N or K c= L & M c= N
  implies K*`M c= L*`N & M*`K c= L*`N
proof
  assume that
A1: K in L & M in N or K c= L & M in N or K in L & M c= N or K c= L & M c=
  N;
A2: K c= L by A1,CARD_1:3;
A3: M c= N by A1,CARD_1:3;
A4: K*`M = card [:K,M:];
A5: [:K,M:] c= [:L,N:] by A2,A3,ZFMISC_1:96;
  hence K*`M c= L*`N by CARD_1:11;
  thus thesis by A4,A5,CARD_1:11;
end;

theorem
  M in N or M c= N implies
  K*`M c= K*`N & K*`M c= N*`K & M*`K c= K*`N & M*`K c= N*`K by Th90;

theorem Th92:
  K in L & M in N or K c= L & M in N or K in L & M c= N or K c= L & M c= N
  implies K = 0 or exp(K,M) c= exp(L,N)
proof
  assume that
A1: K in L & M in N or K c= L & M in N or K in L & M c= N or K c= L & M c=
  N;
A2: K c= L by A1,CARD_1:3;
A3: M c= N by A1,CARD_1:3;
  now
    assume L <> {};
    then
A4: Funcs(N\M,L) <> {};
    0 c= card Funcs(N\M,L);
    then 0 in card Funcs(N\M,L) by A4,CARD_1:3;
    then
A5: nextcard card 0 c= card Funcs(N\M,L) by CARD_1:def 3;
    0+1 = 1;
    then
A6: 1 c= card Funcs(N\M,L) by A5,Lm8,NAT_1:42;
A7: M misses (N\M) by XBOOLE_1:79;
A8: N = M \/ (N\M) by A3,XBOOLE_1:45;
    Funcs(M,K) c= Funcs(M,L) by A2,FUNCT_5:56;
    then
A9: exp(K,M) c= card Funcs(M,L) by CARD_1:11;
A10: exp(L,N) = card [:Funcs(M,L),Funcs(N\M,L):] by A7,A8,FUNCT_5:62;
A11: card [:Funcs(M,L),Funcs(N\M,L):] = card [:card Funcs(M,L),card Funcs(N\
    M,L):] by Th7;
    card Funcs(M,L) *` card Funcs(N\M,L) = card [:card Funcs(M,L),card
    Funcs(N\M,L):];
    then 1 *` card Funcs(M,L) c= exp(L,N) by A6,A10,A11,Th90;
    then card Funcs(M,L) c= exp(L,N) by Th21;
    hence thesis by A9,XBOOLE_1:1;
  end;
  hence thesis by A1;
end;

theorem
  M in N or M c= N implies K = 0 or
  exp(K,M) c= exp(K,N) & exp(M,K) c= exp(N,K)
proof
  assume that
A1: M in N or M c= N and
A2: K <> 0;
  thus exp(K,M) c= exp(K,N) by A1,A2,Th92;
  M = 0 implies exp(M,K) = 0 by A2;
  hence thesis by A1,Th92;
end;

theorem Th94:
  M c= M+`N & N c= M+`N
proof
A1: M c= M +^ N by ORDINAL3:24;
A2: N c= M +^ N by ORDINAL3:24;
A3: card N = N by CARD_1:def 2;
  card M = M by CARD_1:def 2;
  then
A4: M c= card ( M +^ N) by A1,CARD_1:11;
  N c= card ( M +^ N) by A2,A3,CARD_1:11;
  hence thesis by A4;
end;

theorem
  N <> 0 implies M c= M*`N & M c= N*`M
proof
  assume
A1: N <> 0;
A2: card M = M by CARD_1:def 2;
  card N = N by CARD_1:def 2;
  then
A3: M*`N = card ( M *^ N) by A2,Th14;
  M c= M *^ N by A1,ORDINAL3:36;
  hence thesis by A2,A3,CARD_1:11;
end;

theorem Th96:
  K in L & M in N implies K+`M in L+`N & M+`K in L+`N
proof
A1: for K,L,M,N st K in L & M in N & L c= N holds K+`M in L+`N
  proof
    let K,L,M,N such that
A2: K in L and
A3: M in N and
A4: L c= N;
A5: now
      assume
A6:   N is finite;
      then reconsider N as finite Cardinal;
      reconsider L,M,K as finite Cardinal by A2,A3,A4,A6,CARD_3:92;
A7:   card K = K by CARD_1:def 2;
A8:   card L = L by CARD_1:def 2;
A9:   card M = M by CARD_1:def 2;
A10:  card N = N by CARD_1:def 2;
A11:  K+`M = card (card K + card M) by A7,A9,Th38;
A12:  L+`N = card (card L + card N) by A8,A10,Th38;
A13:  card K < card L by A2,A7,A8,NAT_1:41;
      card M < card N by A3,A9,A10,NAT_1:41;
      then card K + card M < card L + card N by A13,XREAL_1:8;
      hence thesis by A11,A12,NAT_1:41;
    end;
    now
      assume
A14:  not N is finite;
      then
A15:  L+`N = N by A4,Th76;
A16:  omega c= N by A14,CARD_3:85;
      K c= M & (M is finite or not M is finite) or
      M c= K & (K is finite or not K is finite);
      then
A17:  K is finite & M is finite or K+`M = M or K+`M = K & K in N by A2,A4,Th76
;
      now
        assume that
A18:    K is finite and
A19:    M is finite;
        reconsider K,M as finite Cardinal by A18,A19;
A20:    card K = K by CARD_1:def 2;
        card M = M by CARD_1:def 2;
        then K+`M = card (card K + card M) by A20,Th38
          .= (card K + card M) by CARD_1:def 2;
        hence thesis by A15,A16,TARSKI:def 3;
      end;
      hence thesis by A3,A4,A14,A17,Th76;
    end;
    hence thesis by A5;
  end;
  L c= N or N c= L;
  hence thesis by A1;
end;

theorem
  K+`M in K+`N implies M in N
proof
  assume that
A1: K+`M in K+`N and
A2: not M in N;
  N c= M by A2,CARD_1:4;
  then K+`N c= K+`M by Th83;
  hence thesis by A1,CARD_1:4;
end;

theorem
  card X +` card Y = card X & card Y in card X implies card (X \ Y) = card X
proof
  assume that
A1: card X +` card Y = card X and
A2: card Y in card X;
A3: card X c= card (X \/ Y) by CARD_1:11,XBOOLE_1:7;
  card (X \/ Y) c= card X by A1,Th34;
  then
A4: card (X \/ Y) = card X by A3,XBOOLE_0:def 10;
  (X \ Y) \/ Y = X \/ Y by XBOOLE_1:39;
  then
A5: card X = card (X \ Y) +` card Y by A4,Th35,XBOOLE_1:79;
  then
A6: card (X \ Y) c= card X by Th94;
A7: now
    assume not card X is finite;
    then
A8: card X +` card X = card X by Th75;
    assume not thesis;
    then card (X \ Y) in card X by A6,CARD_1:3;
    then card X in card X by A2,A5,A8,Th96;
    hence contradiction;
  end;
  now
    assume card X is finite;
    then reconsider X,Y as finite set by A2,CARD_3:92;
    card Y = card card Y;
    then card (card X + card Y) = card card X by A1,Th38;
    then card X + card Y = card X + 0 by CARD_1:40;
    then Y = {};
    hence thesis;
  end;
  hence thesis by A7;
end;

theorem
  K*`M in K*`N implies M in N
proof
  assume that
A1: K*`M in K*`N and
A2: not M in N;
  N c= M by A2,CARD_1:4;
  then K*`N c= K*`M by Th90;
  hence thesis by A1,CARD_1:4;
end;

theorem
  X is countable & Y is countable implies X \+\ Y is countable
proof
  assume that
A1: X is countable and
A2: Y is countable;
A3: X \ Y is countable by A1;
  Y \ X is countable by A2;
  hence thesis by A3,Th85;
end;

registration
  let A be finite set, B be set, f be Function of A, Fin B;
  cluster Union f -> finite;
  coherence
  proof
    now
      let x be set;
      assume
A1:   x in dom f;
      then reconsider A as non empty set;
      reconsider x9 = x as Element of A by A1;
      reconsider f9 = f as Function of A, Fin B;
      f9.x9 is finite;
      hence f.x is finite;
    end;
    hence thesis by Th88;
  end;
end;

registration let f be finite finite-yielding Function;
 cluster product f -> finite;
 coherence
proof
  Funcs(dom f, Union f) is finite by FRAENKEL:6;
  hence thesis by FINSET_1:1,FUNCT_6:1;
end;
end;



theorem
  for F be Function st dom F is infinite & rng F is finite ex x st
  x in rng F & F"{x} is infinite
proof
  let F be Function such that
A1: dom F is infinite and
A2: rng F is finite;
  deffunc f(set)=F"{$1};
  consider g be Function such that
A3: dom g = rng F & for x st x in rng F holds g.x = f(x) from FUNCT_1:
  sch 3;
A4: dom F c= Union g
  proof
    let x such that
A5: x in dom F;
A6: F.x in rng F by A5,FUNCT_1:def 3;
    then
A7: g.(F.x) in rng g by A3,FUNCT_1:def 3;
    F.x in {F.x} by TARSKI:def 1;
    then
A8: x in F"{F.x} by A5,FUNCT_1:def 7;
    g.(F.x)=F"{F.x} by A3,A6;
    then x in union rng g by A8,A7,TARSKI:def 4;
    hence thesis;
  end;
  assume
A9: for x st x in rng F holds F"{x} is finite;
  now
    let x such that
A10: x in dom g;
    g.x=F"{x} by A3,A10;
    hence g.x is finite by A9,A3,A10;
  end;
  then Union g is finite by A2,A3,Th88;
  hence thesis by A1,A4;
end;















begin

scheme
  FinInter{m, n() -> Integer, F(set)->set, P[set]}: {F(i) where i is Element
  of INT: m()<=i & i<=n() & P[i]} is finite proof
  set F = {F(i) where i is Element of INT: m()<=i & i<=n() & P[i]};
  per cases;
  suppose
A1: n() < m();
    now
      assume F <> {};
      then consider x being set such that
A2:   x in F by XBOOLE_0:def 1;
      ex i being Element of INT st x = F(i) & m()<=i & i<=n() & P[i] by A2;
      hence contradiction by A1,XXREAL_0:2;
    end;
    then reconsider F as empty set;
    F is finite;
    hence thesis;
  end;
  suppose
    m() <= n();
    then reconsider k = n()-m() as Element of NAT by INT_1:5;
    set F = {F(i) where i is Element of INT: m()<=i & i<=n() & P[i]};
    defpred Q[set,set] means ex i being Integer st $1 = i & $2 = F(i+m());
A3: for e being set st e in k+1 ex u being set st Q[e,u]
    proof
      let e be set;
      assume e in k+1;
      then reconsider i = e as Nat;
      take F(i+m()),i;
      thus thesis;
    end;
    consider f being Function such that
A4: dom f = k+1 and
A5: for i being set st i in k+1 holds Q[i,f.i] from CLASSES1:sch 1(A3);
A6: F c= rng f
    proof
      let x be set;
      assume x in F;
      then consider j being Element of INT such that
A7:   x=F(j) and
A8:   m()<=j and
A9:   j<=n() and
      P[j];
      reconsider l = j-m() as Element of NAT by A8,INT_1:5;
      l <= k by A9,XREAL_1:9;
      then l < k+1 by NAT_1:13;
      then
A10:  l in k+1 by NAT_1:44;
      then Q[j-m(),f.(j-m())] by A5;
      then f.(j-m()) = F(j+m()-m()) .= F(j);
      hence thesis by A4,A7,A10,FUNCT_1:def 3;
    end;
    rng f is finite by A4,FINSET_1:8;
    hence thesis by A6;
  end;
end;

reserve x,y,z,r,s for ext-real number;

definition
  let X be ext-real-membered set;
  mode UpperBound of X -> ext-real number means
    :Def1:
    x in X implies x <= it;
  existence
  proof
    take +infty;
    thus thesis by XXREAL_0:4;
  end;
  mode LowerBound of X -> ext-real number means
    :Def2:
    x in X implies it <= x;
  existence
  proof
    take -infty;
    thus thesis by XXREAL_0:5;
  end;
end;

definition
  let X be ext-real-membered set;
  func sup X -> ext-real number means
  :Def3:
  it is UpperBound of X & for x being UpperBound of X holds it <= x;
  existence
  proof
    defpred Q[set] means $1 is UpperBound of X;
    defpred P[set] means $1 in X;
A1: for r,s st P[r] & Q[s] holds r <= s by Def1;
    consider s such that
A2: for r st P[r] holds r <= s and
A3: for r st Q[r] holds s <= r from XXREAL_1:sch 1(A1);
    take s;
    thus s is UpperBound of X by A2,Def1;
    thus thesis by A3;
  end;
  uniqueness
  proof
    let y,z such that
A4: y is UpperBound of X and
A5: for x being UpperBound of X holds y <= x and
A6: z is UpperBound of X and
A7: for x being UpperBound of X holds z <= x;
A8: y <= z by A5,A6;
    z <= y by A4,A7;
    hence thesis by A8,XXREAL_0:1;
  end;
  func inf X -> ext-real number means
  :Def4:
  it is LowerBound of X & for x being LowerBound of X holds x <= it;
  existence
  proof
    defpred Q[set] means $1 in X;
    defpred P[set] means $1 is LowerBound of X;
A9: for r,s st P[r] & Q[s] holds r <= s by Def2;
    consider s such that
A10: for r st P[r] holds r <= s and
A11: for r st Q[r] holds s <= r from XXREAL_1:sch 1(A9);
    take s;
    thus s is LowerBound of X by A11,Def2;
    thus thesis by A10;
  end;
  uniqueness
  proof
    let y,z such that
A12: y is LowerBound of X and
A13: for x being LowerBound of X holds x <= y and
A14: z is LowerBound of X and
A15: for x being LowerBound of X holds x <= z;
A16: z <= y by A13,A14;
    y <= z by A12,A15;
    hence thesis by A16,XXREAL_0:1;
  end;
end;

definition
  let X be ext-real-membered set;
  attr X is left_end means
  :Def5:
  inf X in X;
  attr X is right_end means
  :Def6:
  sup X in X;
end;

theorem Th1:
  y is UpperBound of {x} iff x <= y
proof
  x in {x} by TARSKI:def 1;
  hence y is UpperBound of {x} implies x <= y by Def1;
  assume
A1: x <= y;
  let z;
  assume z in {x};
  hence thesis by A1,TARSKI:def 1;
end;

theorem Th2:
  y is LowerBound of {x} iff y <= x
proof
  x in {x} by TARSKI:def 1;
  hence y is LowerBound of {x} implies y <= x by Def2;
  assume
A1: y <= x;
  let z;
  assume z in {x};
  hence thesis by A1,TARSKI:def 1;
end;

Lm1: sup {x} = x
proof
A1: for z being UpperBound of {x} holds x <= z
  proof
    let z be UpperBound of {x};
    x in {x} by TARSKI:def 1;
    hence thesis by Def1;
  end;
  x is UpperBound of {x} by Th1;
  hence thesis by A1,Def3;
end;

Lm2: inf {x} = x
proof
A1: for z being LowerBound of {x} holds z <= x
  proof
    let z be LowerBound of {x};
    x in {x} by TARSKI:def 1;
    hence thesis by Def2;
  end;
  x is LowerBound of {x} by Th2;
  hence thesis by A1,Def4;
end;

registration
  cluster -> ext-real-membered for Element of Fin ExtREAL;
  coherence
  proof
    let X be Element of Fin ExtREAL;
    X c= ExtREAL by FINSUB_1:def 5;
    hence thesis;
  end;
end;

reserve A,B for ext-real-membered set;

theorem Th3:
  x in A implies inf A <= x
proof
  inf A is LowerBound of A by Def4;
  hence thesis by Def2;
end;

theorem Th4:
  x in A implies x <= sup A
proof
  sup A is UpperBound of A by Def3;
  hence thesis by Def1;
end;

theorem Th5:
  B c= A implies for x being LowerBound of A holds x is LowerBound of B
proof
  assume
A1: B c= A;
  let x be LowerBound of A;
  let y;
  assume y in B;
  hence thesis by A1,Def2;
end;

theorem Th6:
  B c= A implies for x being UpperBound of A holds x is UpperBound of B
proof
  assume
A1: B c= A;
  let x be UpperBound of A;
  let y;
  assume y in B;
  hence thesis by A1,Def1;
end;

theorem Th7:
  for x being LowerBound of A, y being LowerBound of B holds min(x,
  y) is LowerBound of A\/ B
proof
  let x be LowerBound of A, y be LowerBound of B;
  set m = min(x,y);
  let z;
  assume
A1: z in A \/ B;
  per cases by A1,XBOOLE_0:def 3;
  suppose
A2: z in A;
A3: m <= x by XXREAL_0:17;
    x <= z by A2,Def2;
    hence thesis by A3,XXREAL_0:2;
  end;
  suppose
A4: z in B;
A5: m <= y by XXREAL_0:17;
    y <= z by A4,Def2;
    hence thesis by A5,XXREAL_0:2;
  end;
end;

theorem Th8:
  for x being UpperBound of A, y being UpperBound of B holds max(x,
  y) is UpperBound of A\/ B
proof
  let x be UpperBound of A, y be UpperBound of B;
  set m = max(x,y);
  let z;
  assume
A1: z in A \/ B;
  per cases by A1,XBOOLE_0:def 3;
  suppose
A2: z in A;
A3: x <= m by XXREAL_0:25;
    z <= x by A2,Def1;
    hence thesis by A3,XXREAL_0:2;
  end;
  suppose
A4: z in B;
A5: y <= m by XXREAL_0:25;
    z <= y by A4,Def1;
    hence thesis by A5,XXREAL_0:2;
  end;
end;

theorem Th9:
  inf(A \/ B) = min(inf A,inf B)
proof
  set m = min(inf A,inf B);
A1: inf B is LowerBound of B by Def4;
A2: for x being LowerBound of A \/ B holds x <= m
  proof
    let x be LowerBound of A \/ B;
    x is LowerBound of B by Th5,XBOOLE_1:7;
    then
A3: x <= inf B by Def4;
    x is LowerBound of A by Th5,XBOOLE_1:7;
    then x <= inf A by Def4;
    hence thesis by A3,XXREAL_0:20;
  end;
  inf A is LowerBound of A by Def4;
  then m is LowerBound of A \/ B by A1,Th7;
  hence thesis by A2,Def4;
end;

theorem Th10:
  sup(A \/ B) = max(sup A,sup B)
proof
  set m = max(sup A,sup B);
A1: sup B is UpperBound of B by Def3;
A2: for x being UpperBound of A \/ B holds m <= x
  proof
    let x be UpperBound of A \/ B;
    x is UpperBound of B by Th6,XBOOLE_1:7;
    then
A3: sup B <= x by Def3;
    x is UpperBound of A by Th6,XBOOLE_1:7;
    then sup A <= x by Def3;
    hence thesis by A3,XXREAL_0:28;
  end;
  sup A is UpperBound of A by Def3;
  then m is UpperBound of A \/ B by A1,Th8;
  hence thesis by A2,Def3;
end;

registration
  cluster finite -> left_end right_end for non empty ext-real-membered set;
  coherence
  proof
    let X be non empty ext-real-membered set;
    defpred P[non empty ext-real-membered set] means $1 is left_end right_end;
    assume
A1: X is finite;
    X c= ExtREAL by MEMBERED:2;
    then
A2: X is non empty Element of Fin ExtREAL by A1,FINSUB_1:def 5;
A3: for B1,B2 being non empty Element of Fin ExtREAL holds P[B1] & P[B2]
    implies P[B1 \/ B2]
    proof
      let B1,B2 be non empty Element of Fin ExtREAL;
      assume
A4:   P[B1];
      inf(B1 \/ B2) = min(inf B1,inf B2) by Th9;
      then
A5:   inf(B1 \/ B2) = inf B1 or inf(B1 \/ B2) = inf B2 by XXREAL_0:15;
      assume
A6:   P[B2];
      then
A7:   inf B2 in B2 by Def5;
      inf B1 in B1 by A4,Def5;
      hence inf(B1 \/ B2) in B1 \/ B2 by A7,A5,XBOOLE_0:def 3;
      sup(B1 \/ B2) = max(sup B1,sup B2) by Th10;
      then
A8:   sup(B1 \/ B2) = sup B1 or sup(B1 \/ B2) = sup B2 by XXREAL_0:16;
A9:   sup B2 in B2 by A6,Def6;
      sup B1 in B1 by A4,Def6;
      hence sup(B1 \/ B2) in B1 \/ B2 by A9,A8,XBOOLE_0:def 3;
    end;
A10: for x being Element of ExtREAL holds P[{.x.}]
    proof
      let x being Element of ExtREAL;
      sup {x} = x by Lm1;
      then
A11:  sup {x} in {x} by TARSKI:def 1;
      inf {x} = x by Lm2;
      then inf {x} in {x} by TARSKI:def 1;
      hence thesis by A11,Def5,Def6;
    end;
    for B being non empty Element of Fin ExtREAL holds P[B] from SETWISEO
    :sch 3(A10,A3);
    hence thesis by A2;
  end;
end;

registration
  cluster -> left_end for non empty natural-membered set;
  coherence
  proof
    let X be non empty natural-membered set;
    defpred P[set] means $1 in X;
A1: ex k being Nat st P[k] by MEMBERED:12;
    consider k being Nat such that
A2: P[k] and
A3: for n being Nat st P[n] holds k <= n from NAT_1:sch 5(A1);
A4: k is LowerBound of X
    proof
      let y;
      assume y in X;
      hence thesis by A3;
    end;
    for x being LowerBound of X holds x <= k by A2,Def2;
    hence inf X in X by A2,A4,Def4;
  end;
end;

registration
  cluster right_end for non empty natural-membered set;
  existence
  proof
    take {0};
    thus thesis;
  end;
end;

notation
  let X be left_end ext-real-membered set;
  synonym min X for inf X;
end;

notation
  let X be right_end ext-real-membered set;
  synonym max X for sup X;
end;

definition
  let X be left_end ext-real-membered set;
  redefine func min X means
  :Def7:
  it in X & for x st x in X holds it <= x;
  compatibility
  proof
    let y;
    thus y = min X implies y in X & for x st x in X holds y <= x by Def5,Th3;
    assume that
A1: y in X and
A2: for x st x in X holds y <= x;
A3: for x being LowerBound of X holds x <= y by A1,Def2;
    y is LowerBound of X by A2,Def2;
    hence thesis by A3,Def4;
  end;
end;

definition
  let X be right_end ext-real-membered set;
  redefine func max X means
  :Def8:
  it in X & for x st x in X holds x <= it;
  compatibility
  proof
    let y;
    thus y = max X implies y in X & for x st x in X holds x <= y by Def6,Th4;
    assume that
A1: y in X and
A2: for x st x in X holds x <= y;
A3: for x being UpperBound of X holds y <= x by A1,Def1;
    y is UpperBound of X by A2,Def1;
    hence thesis by A3,Def3;
  end;
end;

theorem
  max{x} = x by Lm1;

Lm3: x <= y implies y is UpperBound of {x,y}
proof
  assume
A1: x <= y;
  let z;
  assume z in {x,y};
  hence thesis by A1,TARSKI:def 2;
end;

Lm4: for z being UpperBound of {x,y} holds y <= z
proof
  let z be UpperBound of {x,y};
  y in {x,y} by TARSKI:def 2;
  hence thesis by Def1;
end;

theorem
  max(x,y) = max{x,y}
proof
  now
    per cases;
    case
A1:   x <= y;
A2:   for z being UpperBound of {x,y} holds y <= z by Lm4;
      y is UpperBound of {x,y} by A1,Lm3;
      hence max{x,y} = y by A2,Def3;
    end;
    case
A3:   y < x;
A4:   for z being UpperBound of {x,y} holds x <= z by Lm4;
      x is UpperBound of {x,y} by A3,Lm3;
      hence max{x,y} = x by A4,Def3;
    end;
  end;
  hence thesis by XXREAL_0:def 10;
end;

theorem
  min{x} = x by Lm2;

Lm5: y <= x implies y is LowerBound of {x,y}
proof
  assume
A1: y <= x;
  let z;
  assume z in {x,y};
  hence thesis by A1,TARSKI:def 2;
end;

Lm6: for z being LowerBound of {x,y} holds z <= y
proof
  let z be LowerBound of {x,y};
  y in {x,y} by TARSKI:def 2;
  hence thesis by Def2;
end;

theorem
  min{x,y} = min(x,y)
proof
  now
    per cases;
    case
A1:   y <= x;
A2:   for z being LowerBound of {x,y} holds z <= y by Lm6;
      y is LowerBound of {x,y} by A1,Lm5;
      hence min{x,y} = y by A2,Def4;
    end;
    case
A3:   x < y;
A4:   for z being LowerBound of {x,y} holds z <= x by Lm6;
      x is LowerBound of {x,y} by A3,Lm5;
      hence min{x,y} = x by A4,Def4;
    end;
  end;
  hence thesis by XXREAL_0:def 9;
end;

definition
  let X be ext-real-membered set;
  attr X is bounded_below means
  :Def9:
  ex r being real number st r is LowerBound of X;
  attr X is bounded_above means
  :Def10:
  ex r being real number st r is UpperBound of X;
end;

registration
  cluster non empty finite natural-membered for set;
  existence
  proof
    take {0};
    thus thesis;
  end;
end;

definition
  let X be ext-real-membered set;
  attr X is real-bounded means
:Def11:
  X is bounded_below bounded_above;
end;

registration
  cluster real-bounded -> bounded_above bounded_below
          for ext-real-membered set;
  coherence by Def11;
  cluster bounded_above bounded_below -> real-bounded
          for ext-real-membered set;
  coherence by Def11;
end;

registration
  cluster finite -> real-bounded for real-membered set;
  coherence
  proof
    let X be real-membered set;
    assume
A1: X is finite;
    per cases;
    suppose
A2:   X is empty;
      thus X is bounded_below
      proof
        take 0;
        let x;
        thus thesis by A2;
      end;
      take 0;
      let x;
      thus thesis by A2;
    end;
    suppose
      X is non empty;
      then reconsider Y=X as non empty finite real-membered set by A1;
      inf Y in X by Def5;
      then reconsider m = inf X as real number;
      thus X is bounded_below
      proof
        take m;
        let x;
        assume x in X;
        hence thesis by Th3;
      end;
      sup Y in X by Def6;
      then reconsider m = sup X as real number;
      take m;
      let x;
      assume x in X;
      hence thesis by Th4;
    end;
  end;
end;

registration
  cluster real-bounded for non empty natural-membered set;
  existence
  proof
    take {0};
    thus thesis;
  end;
end;

theorem Th15:
  for X being non empty real-membered set st X is bounded_below
  holds inf X in REAL
proof
  let X be non empty real-membered set;
  given r being real number such that
A1: r is LowerBound of X;
  consider s being real number such that
A2: s in X by MEMBERED:9;
A3: inf X <= s by A2,Th3;
A4: r in REAL by XREAL_0:def 1;
A5: s in REAL by XREAL_0:def 1;
  r <= inf X by A1,Def4;
  hence thesis by A4,A5,A3,XXREAL_0:45;
end;

theorem Th16:
  for X being non empty real-membered set st X is bounded_above
  holds sup X in REAL
proof
  let X be non empty real-membered set;
  given r being real number such that
A1: r is UpperBound of X;
  consider s being real number such that
A2: s in X by MEMBERED:9;
A3: s <= sup X by A2,Th4;
A4: r in REAL by XREAL_0:def 1;
A5: s in REAL by XREAL_0:def 1;
  sup X <= r by A1,Def3;
  hence thesis by A4,A5,A3,XXREAL_0:45;
end;

registration
  let X be bounded_above non empty real-membered set;
  cluster sup X -> real;
  coherence
  proof
    sup X in REAL by Th16;
    hence thesis;
  end;
end;

registration
  let X be bounded_below non empty real-membered set;
  cluster inf X -> real;
  coherence
  proof
    inf X in REAL by Th15;
    hence thesis;
  end;
end;

registration
  cluster bounded_above -> right_end for non empty integer-membered set;
  coherence
  proof
    let X be non empty integer-membered set;
    assume X is bounded_above;
    then reconsider Y=X as bounded_above non empty integer-membered set;
    set s = sup Y;
A1: [\ s /] <= s by INT_1:def 6;
    [\ s /] is UpperBound of X
    proof
      let x;
      assume x in X;
      hence thesis by Th4,INT_1:54;
    end;
    then s <= [\ s /] by Def3;
    then reconsider s as Integer by A1,XXREAL_0:1;
    assume
A2: not sup X in X;
    s - 1 is UpperBound of X
    proof
      let x;
      assume
A3:   x in X;
      then reconsider i=x as Integer;
      i <= s by A3,Th4;
      then i < s by A2,A3,XXREAL_0:1;
      hence thesis by INT_1:52;
    end;
    then s - 1 >= s by Def3;
    hence contradiction by XREAL_1:146;
  end;
end;

registration
  cluster bounded_below -> left_end for non empty integer-membered set;
  coherence
  proof
    let X be non empty integer-membered set;
    assume X is bounded_below;
    then reconsider Y=X as bounded_below non empty integer-membered set;
    set s = inf Y;
A1: [/ s \] >= s by INT_1:def 7;
    [/ s \] is LowerBound of X
    proof
      let x;
      assume x in X;
      hence thesis by Th3,INT_1:65;
    end;
    then [/ s \] <= s by Def4;
    then reconsider s as Integer by A1,XXREAL_0:1;
    assume
A2: not inf X in X;
    s + 1 is LowerBound of X
    proof
      let x;
      assume
A3:   x in X;
      then reconsider i=x as Integer;
      i >= s by A3,Th3;
      then s < i by A2,A3,XXREAL_0:1;
      hence thesis by INT_1:7;
    end;
    then s + 1 <= s by Def4;
    hence contradiction by XREAL_1:145;
  end;
end;

registration
  cluster -> bounded_below for natural-membered set;
  coherence
  proof
    let X be natural-membered set;
    take 0;
    let x;
    assume x in X;
    hence thesis;
  end;
end;

registration
  let X be left_end real-membered set;
  cluster min X -> real;
  coherence
  proof
    min X in X by Def7;
    hence thesis;
  end;
end;

registration
  let X be left_end rational-membered set;
  cluster min X -> rational;
  coherence
  proof
    min X in X by Def7;
    hence thesis;
  end;
end;

registration
  let X be left_end integer-membered set;
  cluster min X -> integer;
  coherence
  proof
    min X in X by Def7;
    hence thesis;
  end;
end;

registration
  let X be left_end natural-membered set;
  cluster min X -> natural;
  coherence
  proof
    min X in X by Def7;
    hence thesis;
  end;
end;

registration
  let X be right_end real-membered set;
  cluster max X -> real;
  coherence
  proof
    max X in X by Def8;
    hence thesis;
  end;
end;

registration
  let X be right_end rational-membered set;
  cluster max X -> rational;
  coherence
  proof
    max X in X by Def8;
    hence thesis;
  end;
end;

registration
  let X be right_end integer-membered set;
  cluster max X -> integer;
  coherence
  proof
    max X in X by Def8;
    hence thesis;
  end;
end;

registration
  let X be right_end natural-membered set;
  cluster max X -> natural;
  coherence
  proof
    max X in X by Def8;
    hence thesis;
  end;
end;

registration
  cluster left_end -> bounded_below for real-membered set;
  coherence
  proof
    let X be real-membered set;
    assume inf X in X;
    then reconsider r = inf X as real number;
    take r;
    let x;
    thus thesis by Th3;
  end;
  cluster right_end -> bounded_above for real-membered set;
  coherence
  proof
    let X be real-membered set;
    assume sup X in X;
    then reconsider r = sup X as real number;
    take r;
    let x;
    thus thesis by Th4;
  end;
end;

theorem Th17:
  x is LowerBound of [.x,y.]
proof
  let z;
  assume z in [.x,y.];
  hence thesis by XXREAL_1:1;
end;

theorem Th18:
  x is LowerBound of ].x,y.]
proof
  let z;
  assume z in ].x,y.];
  hence thesis by XXREAL_1:2;
end;

theorem Th19:
  x is LowerBound of [.x,y.[
proof
  let z;
  assume z in [.x,y.[;
  hence thesis by XXREAL_1:3;
end;

theorem Th20:
  x is LowerBound of ].x,y.[
proof
  let z;
  assume z in ].x,y.[;
  hence thesis by XXREAL_1:4;
end;

theorem Th21:
  y is UpperBound of [.x,y.]
proof
  let z;
  assume z in [.x,y.];
  hence thesis by XXREAL_1:1;
end;

theorem Th22:
  y is UpperBound of ].x,y.]
proof
  let z;
  assume z in ].x,y.];
  hence thesis by XXREAL_1:2;
end;

theorem Th23:
  y is UpperBound of [.x,y.[
proof
  let z;
  assume z in [.x,y.[;
  hence thesis by XXREAL_1:3;
end;

theorem Th24:
  y is UpperBound of ].x,y.[
proof
  let z;
  assume z in ].x,y.[;
  hence thesis by XXREAL_1:4;
end;

theorem Th25:
  x <= y implies inf [.x,y.] = x
proof
  assume
A1: x <= y;
A2: for z being LowerBound of [.x,y.] holds z <= x
  proof
    let z be LowerBound of [.x,y.];
    x in [.x,y.] by A1,XXREAL_1:1;
    hence thesis by Def2;
  end;
  x is LowerBound of [.x,y.] by Th17;
  hence thesis by A2,Def4;
end;

theorem Th26:
  x < y implies inf [.x,y.[ = x
proof
  assume
A1: x < y;
A2: for z being LowerBound of [.x,y.[ holds z <= x
  proof
    let z be LowerBound of [.x,y.[;
    x in [.x,y.[ by A1,XXREAL_1:3;
    hence thesis by Def2;
  end;
  x is LowerBound of [.x,y.[ by Th19;
  hence thesis by A2,Def4;
end;

theorem Th27:
  x < y implies inf ].x,y.] = x
proof
  assume
A1: x < y;
A2: for z being LowerBound of ].x,y.] holds z <= x
  proof
    let z be LowerBound of ].x,y.];
    for r st x < r & r < y holds z <= r
    proof
      let r;
      assume that
A3:   x < r and
A4:   r < y;
      r in ].x,y.] by A3,A4,XXREAL_1:2;
      hence thesis by Def2;
    end;
    hence thesis by A1,XREAL_1:228;
  end;
  x is LowerBound of ].x,y.] by Th18;
  hence thesis by A2,Def4;
end;

theorem Th28:
  x < y implies inf ].x,y.[ = x
proof
  assume
A1: x < y;
A2: for z being LowerBound of ].x,y.[ holds z <= x
  proof
    let z be LowerBound of ].x,y.[;
    for r st x < r & r < y holds z <= r
    proof
      let r;
      assume that
A3:   x < r and
A4:   r < y;
      r in ].x,y.[ by A3,A4,XXREAL_1:4;
      hence thesis by Def2;
    end;
    hence thesis by A1,XREAL_1:228;
  end;
  x is LowerBound of ].x,y.[ by Th20;
  hence thesis by A2,Def4;
end;

theorem Th29:
  x <= y implies sup [.x,y.] = y
proof
  assume
A1: x <= y;
A2: for z being UpperBound of [.x,y.] holds y <= z
  proof
    let z be UpperBound of [.x,y.];
    y in [.x,y.] by A1,XXREAL_1:1;
    hence thesis by Def1;
  end;
  y is UpperBound of [.x,y.] by Th21;
  hence thesis by A2,Def3;
end;

theorem Th30:
  x < y implies sup ].x,y.] = y
proof
  assume
A1: x < y;
A2: for z being UpperBound of ].x,y.] holds y <= z
  proof
    let z be UpperBound of ].x,y.];
    y in ].x,y.] by A1,XXREAL_1:2;
    hence thesis by Def1;
  end;
  y is UpperBound of ].x,y.] by Th22;
  hence thesis by A2,Def3;
end;

theorem Th31:
  x < y implies sup [.x,y.[ = y
proof
  assume
A1: x < y;
A2: for z being UpperBound of [.x,y.[ holds y <= z
  proof
    let z be UpperBound of [.x,y.[;
    for r st x < r & r < y holds r <= z
    proof
      let r;
      assume that
A3:   x < r and
A4:   r < y;
      r in [.x,y.[ by A3,A4,XXREAL_1:3;
      hence thesis by Def1;
    end;
    hence thesis by A1,XREAL_1:229;
  end;
  y is UpperBound of [.x,y.[ by Th23;
  hence thesis by A2,Def3;
end;

theorem Th32:
  x < y implies sup ].x,y.[ = y
proof
  assume
A1: x < y;
A2: for z being UpperBound of ].x,y.[ holds y <= z
  proof
    let z be UpperBound of ].x,y.[;
    for r st x < r & r < y holds r <= z
    proof
      let r;
      assume that
A3:   x < r and
A4:   r < y;
      r in ].x,y.[ by A3,A4,XXREAL_1:4;
      hence thesis by Def1;
    end;
    hence thesis by A1,XREAL_1:229;
  end;
  y is UpperBound of ].x,y.[ by Th24;
  hence thesis by A2,Def3;
end;

theorem Th33:
  x <= y implies [.x,y.] is left_end right_end
proof
  assume
A1: x <= y;
  then x in [.x,y.] by XXREAL_1:1;
  then inf [.x,y.] in [.x,y.] by A1,Th25;
  hence [.x,y.] is left_end by Def5;
  y in [.x,y.] by A1,XXREAL_1:1;
  hence sup [.x,y.] in [.x,y.] by A1,Th29;
end;

theorem Th34:
  x < y implies [.x,y.[ is left_end
proof
  assume
A1: x < y;
  then x in [.x,y.[ by XXREAL_1:3;
  hence inf [.x,y.[ in [.x,y.[ by A1,Th26;
end;

theorem Th35:
  x < y implies ].x,y.] is right_end
proof
  assume
A1: x < y;
  then y in ].x,y.] by XXREAL_1:2;
  hence sup ].x,y.] in ].x,y.] by A1,Th30;
end;

theorem Th36:
  x is LowerBound of {}
proof
  let y;
  thus thesis;
end;

theorem Th37:
  x is UpperBound of {}
proof
  let y;
  thus thesis;
end;

theorem Th38:
  inf {} = +infty
proof
A1: for x being LowerBound of {} holds x <= +infty by XXREAL_0:3;
  +infty is LowerBound of {} by Th36;
  hence thesis by A1,Def4;
end;

theorem Th39:
  sup {} = -infty
proof
A1: for x being UpperBound of {} holds -infty <= x by XXREAL_0:5;
  -infty is UpperBound of {} by Th37;
  hence thesis by A1,Def3;
end;

theorem Th40:
  for X being ext-real-membered set holds X is non empty iff inf X <= sup X
proof
  let X be ext-real-membered set;
  thus X is non empty implies inf X <= sup X
  proof
    assume X is non empty;
    then consider x such that
A1: x in X by MEMBERED:8;
A2: x <= sup X by A1,Th4;
    inf X <= x by A1,Th3;
    hence thesis by A2,XXREAL_0:2;
  end;
  assume inf X <= sup X;
  hence thesis by Th38,Th39;
end;

registration
  cluster real-bounded -> finite for integer-membered set;
  coherence
  proof
    let X be integer-membered set;
    per cases;
    suppose
      X is empty;
      hence thesis;
    end;
    suppose
A1:   X is non empty;
      assume X is bounded_below bounded_above;
      then reconsider Z = X as bounded_below bounded_above non empty
      integer-membered set by A1;
      set m = inf Z, n = sup Z;
      defpred P[set] means $1 in X;
      deffunc F(set) = $1;
      set Y = {F(i) where i is Element of INT: m<=i & i<=n & P[i]};
A2:   X c= Y
      proof
        let i be Integer;
A3:     i in INT by INT_1:def 2;
        assume
A4:     P[i];
        then
A5:     i<=n by Th4;
        m<=i by A4,Th3;
        hence thesis by A3,A4,A5;
      end;
      Y is finite from FinInter;
      hence thesis by A2;
    end;
  end;
end;

registration
  cluster -> finite for bounded_above natural-membered set;
  coherence;
end;

theorem
  for X being ext-real-membered set holds +infty is UpperBound of X
proof
  let X be ext-real-membered set;
  let x such that
  x in X;
  thus thesis by XXREAL_0:3;
end;

theorem
  for X being ext-real-membered set holds -infty is LowerBound of X
proof
  let X be ext-real-membered set;
  let x such that
  x in X;
  thus thesis by XXREAL_0:5;
end;

theorem Th43:
  for X,Y being ext-real-membered set st X c= Y & Y is
  bounded_above holds X is bounded_above
proof
  let X,Y be ext-real-membered set;
  assume
A1: X c= Y;
  given r being real number such that
A2: r is UpperBound of Y;
  take r;
  thus thesis by A1,A2,Th6;
end;

theorem Th44:
  for X,Y being ext-real-membered set st X c= Y & Y is
  bounded_below holds X is bounded_below
proof
  let X,Y be ext-real-membered set;
  assume
A1: X c= Y;
  given r being real number such that
A2: r is LowerBound of Y;
  take r;
  thus thesis by A1,A2,Th5;
end;

theorem
  for X,Y being ext-real-membered set st X c= Y & Y is real-bounded holds
  X is real-bounded
proof
  let X,Y be ext-real-membered set;
  assume
A1: X c= Y;
  assume Y is bounded_below bounded_above;
  hence X is bounded_below bounded_above by A1,Th43,Th44;
end;

theorem Th46:
  REAL is non bounded_below non bounded_above
proof
  thus REAL is non bounded_below
  proof
    given r being real number such that
A1: r is LowerBound of REAL;
    consider s being real number such that
A2: s < r by XREAL_1:2;
    s in REAL by XREAL_0:def 1;
    hence contradiction by A1,A2,Def2;
  end;
  given r being real number such that
A3: r is UpperBound of REAL;
  consider s being real number such that
A4: r < s by XREAL_1:1;
  s in REAL by XREAL_0:def 1;
  hence contradiction by A3,A4,Def1;
end;

registration
  cluster REAL -> non bounded_below non bounded_above;
  coherence by Th46;
end;

theorem
  {+infty} is bounded_below
proof
  take 0;
  let x;
  assume x in {+infty};
  hence thesis by TARSKI:def 1;
end;

theorem
  {-infty} is bounded_above
proof
  take 0;
  let x;
  assume x in {-infty};
  hence thesis by TARSKI:def 1;
end;

theorem Th49:
  for X being bounded_above non empty ext-real-membered set st X
  <> {-infty} holds ex x being Element of REAL st x in X
proof
  let X be bounded_above non empty ext-real-membered set;
  assume X <> {-infty};
  then consider x being set such that
A1: x in X and
A2: x <> -infty by ZFMISC_1:35;
  reconsider x as ext-real number by A1;
  consider r being real number such that
A3: r is UpperBound of X by Def10;
A4: r in REAL by XREAL_0:def 1;
  x <= r by A3,A1,Def1;
  then x in REAL by A4,A2,XXREAL_0:13;
  hence thesis by A1;
end;

theorem Th50:
  for X being bounded_below non empty ext-real-membered set st X
  <> {+infty} holds ex x being Element of REAL st x in X
proof
  let X be bounded_below non empty ext-real-membered set;
  assume X <> {+infty};
  then consider x being set such that
A1: x in X and
A2: x <> +infty by ZFMISC_1:35;
  reconsider x as ext-real number by A1;
  consider r being real number such that
A3: r is LowerBound of X by Def9;
A4: r in REAL by XREAL_0:def 1;
  r <= x by A3,A1,Def2;
  then x in REAL by A4,A2,XXREAL_0:10;
  hence thesis by A1;
end;

theorem Th51:
  for X being ext-real-membered set st -infty is UpperBound of X
  holds X c= {-infty}
proof
  let X be ext-real-membered set such that
A1: -infty is UpperBound of X;
  let x;
  assume x in X;
  then x = -infty by A1,Def1,XXREAL_0:6;
  hence thesis by TARSKI:def 1;
end;

theorem Th52:
  for X being ext-real-membered set st +infty is LowerBound of X
  holds X c= {+infty}
proof
  let X be ext-real-membered set such that
A1: +infty is LowerBound of X;
  let x;
  assume x in X;
  then x = +infty by A1,Def2,XXREAL_0:4;
  hence thesis by TARSKI:def 1;
end;

theorem
  for X being non empty ext-real-membered set st ex y being UpperBound
  of X st y <> +infty holds X is bounded_above
proof
  let X be non empty ext-real-membered set;
  given y being UpperBound of X such that
A1: y <> +infty;
  per cases;
  suppose
A2: y = -infty;
    take 0;
    let x;
    assume
A3: x in X;
    X c= {-infty} by A2,Th51;
    hence thesis by A3,TARSKI:def 1;
  end;
  suppose
    y <> -infty;
    then y in REAL by A1,XXREAL_0:14;
    then reconsider y as real number;
    take y;
    thus thesis;
  end;
end;

theorem
  for X being non empty ext-real-membered set st ex y being LowerBound
  of X st y <> -infty holds X is bounded_below
proof
  let X be non empty ext-real-membered set;
  given y being LowerBound of X such that
A1: y <> -infty;
  per cases;
  suppose
A2: y = +infty;
    take 0;
    let x;
    assume
A3: x in X;
    X c= {+infty} by A2,Th52;
    hence thesis by A3,TARSKI:def 1;
  end;
  suppose
    y <> +infty;
    then y in REAL by A1,XXREAL_0:14;
    then reconsider y as real number;
    take y;
    thus thesis;
  end;
end;

theorem
  for X being non empty ext-real-membered set, x being UpperBound of X
  st x in X holds x = sup X
proof
  let X be non empty ext-real-membered set, x be UpperBound of X;
  assume x in X;
  then for y being UpperBound of X holds x <= y by Def1;
  hence thesis by Def3;
end;

theorem
  for X being non empty ext-real-membered set, x being LowerBound of X
  st x in X holds x = inf X
proof
  let X be non empty ext-real-membered set, x be LowerBound of X;
  assume x in X;
  then for y being LowerBound of X holds y <= x by Def2;
  hence thesis by Def4;
end;

theorem Th57:
  for X being non empty ext-real-membered set st X is
  bounded_above & X <> {-infty} holds sup X in REAL
proof
  let X be non empty ext-real-membered set;
  assume
A1: X is bounded_above;
  then consider r being real number such that
A2: r is UpperBound of X by Def10;
  assume X <> {-infty};
  then
A3: ex x being Element of REAL st x in X by A1,Th49;
  sup X is UpperBound of X by Def3;
  then
A4: not sup X = -infty by A3,Def1,XXREAL_0:12;
A5: r in REAL by XREAL_0:def 1;
  sup X <= r by A2,Def3;
  hence thesis by A5,A4,XXREAL_0:13;
end;

theorem Th58:
  for X being non empty ext-real-membered set st X is
  bounded_below & X <> {+infty} holds inf X in REAL
proof
  let X be non empty ext-real-membered set;
  assume
A1: X is bounded_below;
  then consider r being real number such that
A2: r is LowerBound of X by Def9;
  assume X <> {+infty};
  then
A3: ex x being Element of REAL st x in X by A1,Th50;
  inf X is LowerBound of X by Def4;
  then
A4: inf X <> +infty by A3,Def2,XXREAL_0:9;
A5: r in REAL by XREAL_0:def 1;
  r <= inf X by A2,Def4;
  hence thesis by A5,A4,XXREAL_0:10;
end;

theorem
  for X,Y being ext-real-membered set st X c= Y holds sup X <= sup Y
proof
  let X,Y be ext-real-membered set;
  assume
A1: X c= Y;
  sup Y is UpperBound of Y by Def3;
  then sup Y is UpperBound of X by A1,Th6;
  hence thesis by Def3;
end;

theorem
  for X,Y being ext-real-membered set st X c= Y holds inf Y <= inf X
proof
  let X,Y be ext-real-membered set;
  assume
A1: X c= Y;
  inf Y is LowerBound of Y by Def4;
  then inf Y is LowerBound of X by A1,Th5;
  hence thesis by Def4;
end;

theorem
  for X being ext-real-membered set, x being ext-real number st (ex y
  being ext-real number st y in X & x <= y) holds x <= sup X
proof
  let X be ext-real-membered set;
  let x be ext-real number;
  given y being ext-real number such that
A1: y in X and
A2: x <= y;
  y <= sup X by A1,Th4;
  hence thesis by A2,XXREAL_0:2;
end;

theorem
  for X being ext-real-membered set, x being ext-real number st (ex y
  being ext-real number st y in X & y <= x) holds inf X <= x
proof
  let X be ext-real-membered set;
  let x be ext-real number;
  given y being ext-real number such that
A1: y in X and
A2: y <= x;
  inf X <= y by A1,Th3;
  hence thesis by A2,XXREAL_0:2;
end;

theorem
  for X,Y being ext-real-membered set st for x being ext-real number st
  x in X ex y being ext-real number st y in Y & x <= y holds sup X <= sup Y
proof
  let X,Y be ext-real-membered set;
  assume
A1: for x being ext-real number st x in X ex y being ext-real number st
  y in Y & x <= y;
  sup Y is UpperBound of X
  proof
    let x be ext-real number;
    assume x in X;
    then consider y being ext-real number such that
A2: y in Y and
A3: x <= y by A1;
    y <= sup Y by A2,Th4;
    hence thesis by A3,XXREAL_0:2;
  end;
  hence thesis by Def3;
end;

theorem
  for X,Y being ext-real-membered set st for y being ext-real number st
  y in Y ex x being ext-real number st x in X & x <= y holds inf X <= inf Y
proof
  let X,Y be ext-real-membered set;
  assume
A1: for y being ext-real number st y in Y ex x being ext-real number st
  x in X & x <= y;
  inf X is LowerBound of Y
  proof
    let y be ext-real number;
    assume y in Y;
    then consider x being ext-real number such that
A2: x in X and
A3: x <= y by A1;
    inf X <= x by A2,Th3;
    hence thesis by A3,XXREAL_0:2;
  end;
  hence thesis by Def4;
end;

theorem Th65:
  for X,Y being ext-real-membered set, x being UpperBound of X, y
  being UpperBound of Y holds min(x,y) is UpperBound of X /\ Y
proof
  let X,Y be ext-real-membered set, x be UpperBound of X, y be UpperBound of Y;
  let a be ext-real number;
  assume
A1: a in X /\ Y;
  then a in Y by XBOOLE_0:def 4;
  then
A2: a <= y by Def1;
  a in X by A1,XBOOLE_0:def 4;
  then a <= x by Def1;
  hence thesis by A2,XXREAL_0:20;
end;

theorem Th66:
  for X,Y being ext-real-membered set, x being LowerBound of X, y
  being LowerBound of Y holds max(x,y) is LowerBound of X /\ Y
proof
  let X,Y be ext-real-membered set, x be LowerBound of X, y be LowerBound of Y;
  let a be ext-real number;
  assume
A1: a in X /\ Y;
  then a in Y by XBOOLE_0:def 4;
  then
A2: y <= a by Def2;
  a in X by A1,XBOOLE_0:def 4;
  then x <= a by Def2;
  hence thesis by A2,XXREAL_0:28;
end;

theorem
  for X,Y being ext-real-membered set holds sup(X /\ Y) <= min(sup X,sup Y)
proof
  let X,Y be ext-real-membered set;
A1: sup Y is UpperBound of Y by Def3;
  sup X is UpperBound of X by Def3;
  then min(sup X,sup Y) is UpperBound of X /\ Y by A1,Th65;
  hence thesis by Def3;
end;

theorem
  for X,Y being ext-real-membered set holds max(inf X,inf Y) <= inf(X /\ Y)
proof
  let X,Y be ext-real-membered set;
A1: inf Y is LowerBound of Y by Def4;
  inf X is LowerBound of X by Def4;
  then max(inf X,inf Y) is LowerBound of X /\ Y by A1,Th66;
  hence thesis by Def4;
end;

registration
  cluster real-bounded -> real-membered for ext-real-membered set;
  coherence
  proof
    let X be ext-real-membered set such that
A1: X is real-bounded;
    let x be set;
    assume
A2: x in X;
    then reconsider X as non empty ext-real-membered set;
    consider s being real number such that
A3: s is UpperBound of X by A1,Def10;
    reconsider x as ext-real number by A2;
A4: s in REAL by XREAL_0:def 1;
A5: x <= s by A2,A3,Def1;
    consider r being real number such that
A6: r is LowerBound of X by A1,Def9;
A7: r in REAL by XREAL_0:def 1;
    r <= x by A2,A6,Def2;
    then x in REAL by A5,A7,A4,XXREAL_0:45;
    hence thesis;
  end;
end;

theorem Th69:
  A c= [.inf A, sup A.]
proof
  let x be ext-real number;
  assume
A1: x in A;
  then
A2: x <= sup A by Th4;
  inf A <= x by A1,Th3;
  hence thesis by A2,XXREAL_1:1;
end;

theorem
  sup A = inf A implies A = {inf A}
proof
  assume
A1: sup A = inf A;
  then A c= [.inf A,inf A.] by Th69;
  then
A2: A c= {inf A} by XXREAL_1:17;
  A <> {} by A1,Th38,Th39;
  hence thesis by A2,ZFMISC_1:33;
end;

theorem Th71:
  x <= y & x is UpperBound of A implies y is UpperBound of A
proof
  assume that
A1: x <= y and
A2: x is UpperBound of A;
  let z;
  assume z in A;
  then z <= x by A2,Def1;
  hence thesis by A1,XXREAL_0:2;
end;

theorem Th72:
  y <= x & x is LowerBound of A implies y is LowerBound of A
proof
  assume that
A1: y <= x and
A2: x is LowerBound of A;
  let z;
  assume z in A;
  then x <= z by A2,Def2;
  hence thesis by A1,XXREAL_0:2;
end;

theorem
  A is bounded_above iff sup A <> +infty
proof
  hereby
    assume
A1: A is bounded_above;
    per cases by A1,Th57;
    suppose
      A = {};
      hence sup A <> +infty by Th39;
    end;
    suppose
      A = {-infty};
      hence sup A <> +infty by Lm1;
    end;
    suppose
      sup A in REAL;
      hence sup A <> +infty;
    end;
  end;
  assume
A2: sup A <> +infty;
  per cases by A2,XXREAL_0:14;
  suppose
A3: sup A = -infty;
    take 0;
    -infty is UpperBound of A by A3,Def3;
    hence thesis by Th71;
  end;
  suppose
    sup A in REAL;
    then reconsider r = sup A as real number;
    take r;
    thus thesis by Def3;
  end;
end;

theorem
  A is bounded_below iff inf A <> -infty
proof
  hereby
    assume
A1: A is bounded_below;
    per cases by A1,Th58;
    suppose
      A = {};
      hence inf A <> -infty by Th38;
    end;
    suppose
      A = {+infty};
      hence inf A <> -infty by Lm2;
    end;
    suppose
      inf A in REAL;
      hence inf A <> -infty;
    end;
  end;
  assume
A2: inf A <> -infty;
  per cases by A2,XXREAL_0:14;
  suppose
A3: inf A = +infty;
    take 0;
    +infty is LowerBound of A by A3,Def4;
    hence thesis by Th72;
  end;
  suppose
    inf A in REAL;
    then reconsider r = inf A as real number;
    take r;
    thus thesis by Def4;
  end;
end;

begin 

definition
  let A be ext-real-membered set;
  attr A is interval means
  :Def12:
  for r,s being ext-real number st r in A & s in A holds [.r,s.] c= A;
end;

registration
  cluster ExtREAL -> interval;
  coherence
  proof
    let r,s;
    thus thesis by MEMBERED:2;
  end;
  cluster empty -> interval for ext-real-membered set;
  coherence
  proof
    let A be ext-real-membered set;
    assume
A1: A is empty;
    let r,s;
    thus thesis by A1;
  end;
  let r,s;
  cluster [.r,s.] -> interval;
  coherence
  proof
    let x,y;
    assume x in [.r,s.];
    then
A2: r <= x by XXREAL_1:1;
    assume y in [.r,s.];
    then
A3: y <= s by XXREAL_1:1;
    let z;
    assume
A4: z in [.x,y.];
    then x <= z by XXREAL_1:1;
    then
A5: r <= z by A2,XXREAL_0:2;
    z <= y by A4,XXREAL_1:1;
    then z <= s by A3,XXREAL_0:2;
    hence thesis by A5,XXREAL_1:1;
  end;
  cluster ].r,s.] -> interval;
  coherence
  proof
    let x,y;
    assume x in ].r,s.];
    then
A6: r < x by XXREAL_1:2;
    assume y in ].r,s.];
    then
A7: y <= s by XXREAL_1:2;
    let z;
    assume
A8: z in [.x,y.];
    then x <= z by XXREAL_1:1;
    then
A9: r < z by A6,XXREAL_0:2;
    z <= y by A8,XXREAL_1:1;
    then z <= s by A7,XXREAL_0:2;
    hence thesis by A9,XXREAL_1:2;
  end;
  cluster [.r,s.[ -> interval;
  coherence
  proof
    let x,y;
    assume x in [.r,s.[;
    then
A10: r <= x by XXREAL_1:3;
    assume y in [.r,s.[;
    then
A11: y < s by XXREAL_1:3;
    let z;
    assume
A12: z in [.x,y.];
    then x <= z by XXREAL_1:1;
    then
A13: r <= z by A10,XXREAL_0:2;
    z <= y by A12,XXREAL_1:1;
    then z < s by A11,XXREAL_0:2;
    hence thesis by A13,XXREAL_1:3;
  end;
  cluster ].r,s.[ -> interval;
  coherence
  proof
    let x,y;
    assume x in ].r,s.[;
    then
A14: r < x by XXREAL_1:4;
    assume y in ].r,s.[;
    then
A15: y < s by XXREAL_1:4;
    let z;
    assume
A16: z in [.x,y.];
    then x <= z by XXREAL_1:1;
    then
A17: r < z by A14,XXREAL_0:2;
    z <= y by A16,XXREAL_1:1;
    then z < s by A15,XXREAL_0:2;
    hence thesis by A17,XXREAL_1:4;
  end;
end;

registration
  cluster REAL -> interval;
  coherence by XXREAL_1:224;
end;

registration
  cluster interval for non empty ext-real-membered set;
  existence
  proof
    take REAL;
    thus thesis;
  end;
end;

registration
  let A,B be interval ext-real-membered set;
  cluster A /\ B -> interval;
  coherence
  proof
    let r,s be ext-real number;
    assume that
A1: r in A /\ B and
A2: s in A /\ B;
A3: s in B by A2,XBOOLE_0:def 4;
    r in B by A1,XBOOLE_0:def 4;
    then
A4: [.r,s.] c= B by A3,Def12;
A5: s in A by A2,XBOOLE_0:def 4;
    r in A by A1,XBOOLE_0:def 4;
    then [.r,s.] c= A by A5,Def12;
    hence thesis by A4,XBOOLE_1:19;
  end;
end;

reserve A,B for ext-real-membered set;

registration
  let r,s;
  cluster ].r,s.] -> non left_end;
  coherence
  proof
    assume ].r,s.] is left_end;
    then
A1: inf ].r,s.] in ].r,s.] by Def5;
    then
A2: inf ].r,s.] <= s by XXREAL_1:2;
    r < inf ].r,s.] by A1,XXREAL_1:2;
    then r < s by A2,XXREAL_0:2;
    then r = inf ].r,s.] by Th27;
    hence contradiction by A1,XXREAL_1:2;
  end;
  cluster [.r,s.[ -> non right_end;
  coherence
  proof
    assume [.r,s.[ is right_end;
    then
A3: sup [.r,s.[ in [.r,s.[ by Def6;
    then
A4: sup [.r,s.[ < s by XXREAL_1:3;
    r <= sup [.r,s.[ by A3,XXREAL_1:3;
    then r < s by A4,XXREAL_0:2;
    then s = sup [.r,s.[ by Th31;
    hence contradiction by A3,XXREAL_1:3;
  end;
  cluster ].r,s.[ -> non left_end non right_end;
  coherence
  proof
    thus ].r,s.[ is non left_end
    proof
      assume ].r,s.[ is left_end;
      then
A5:   inf ].r,s.[ in ].r,s.[ by Def5;
      then
A6:   inf ].r,s.[ < s by XXREAL_1:4;
      r < inf ].r,s.[ by A5,XXREAL_1:4;
      then r < s by A6,XXREAL_0:2;
      then r = inf ].r,s.[ by Th28;
      hence contradiction by A5,XXREAL_1:4;
    end;
    assume ].r,s.[ is right_end;
    then
A7: sup ].r,s.[ in ].r,s.[ by Def6;
    then
A8: sup ].r,s.[ < s by XXREAL_1:4;
    r < sup ].r,s.[ by A7,XXREAL_1:4;
    then r < s by A8,XXREAL_0:2;
    then s = sup ].r,s.[ by Th32;
    hence contradiction by A7,XXREAL_1:4;
  end;
end;

registration
  cluster left_end right_end interval for ext-real-membered set;
  existence
  proof
    take [.0,1.];
    thus thesis by Th33;
  end;
  cluster non left_end right_end interval for ext-real-membered set;
  existence
  proof
    take ].0,1.];
    thus thesis by Th35;
  end;
  cluster left_end non right_end interval for ext-real-membered set;
  existence
  proof
    take [.0,1.[;
    thus thesis by Th34;
  end;
  cluster non left_end non right_end interval non empty for ext-real-membered
    set;
  existence
  proof
    take ].0,2.[;
    1 in ].0,2.[ by XXREAL_1:4;
    hence thesis;
  end;
end;

theorem
  for A being left_end right_end interval ext-real-membered set holds
  A = [.min A, max A.]
proof
  let A be left_end right_end interval ext-real-membered set;
  let x;
A1: max A in A by Def6;
  thus x in A implies x in [.min A, max A.]
  proof
    assume
A2: x in A;
    then
A3: x <= max A by Th4;
    min A <= x by A2,Th3;
    hence thesis by A3,XXREAL_1:1;
  end;
  min A in A by Def5;
  then [.min A, max A.] c= A by A1,Def12;
  hence thesis;
end;

theorem
  for A being non left_end right_end interval ext-real-membered set
  holds A = ].inf A, max A.]
proof
  let A be non left_end right_end interval ext-real-membered set;
  let x;
  defpred P[ext-real number] means $1 in A & $1 < x;
  thus x in A implies x in ].inf A, max A.]
  proof
A1: not inf A in A by Def5;
    assume
A2: x in A;
    then
A3: x <= max A by Th4;
    inf A <= x by A2,Th3;
    then inf A < x by A2,A1,XXREAL_0:1;
    hence thesis by A3,XXREAL_1:2;
  end;
  assume
A4: x in ].inf A, max A.];
  per cases;
  suppose
    not ex r st P[r];
    then x is LowerBound of A by Def2;
    then x <= inf A by Def4;
    hence thesis by A4,XXREAL_1:2;
  end;
  suppose
    ex r st P[r];
    then consider r such that
A5: r in A and
A6: r < x;
    x <= max A by A4,XXREAL_1:2;
    then
A7: x in [.r,max A.] by A6,XXREAL_1:1;
    max A in A by Def6;
    then [.r,max A.] c= A by A5,Def12;
    hence thesis by A7;
  end;
end;

theorem
  for A being left_end non right_end interval ext-real-membered set
  holds A = [.min A, sup A.[
proof
  let A be left_end non right_end interval ext-real-membered set;
  let x;
  defpred P[ext-real number] means $1 in A & $1 > x;
  thus x in A implies x in [.min A, sup A.[
  proof
A1: not sup A in A by Def6;
    assume
A2: x in A;
    then
A3: min A <= x by Th3;
    x <= sup A by A2,Th4;
    then x < sup A by A2,A1,XXREAL_0:1;
    hence thesis by A3,XXREAL_1:3;
  end;
  assume
A4: x in [.min A, sup A.[;
  per cases;
  suppose
    not ex r st P[r];
    then x is UpperBound of A by Def1;
    then sup A <= x by Def3;
    hence thesis by A4,XXREAL_1:3;
  end;
  suppose
    ex r st P[r];
    then consider r such that
A5: r in A and
A6: r > x;
    inf A <= x by A4,XXREAL_1:3;
    then
A7: x in [.inf A,r.] by A6,XXREAL_1:1;
    min A in A by Def5;
    then [.inf A,r.] c= A by A5,Def12;
    hence thesis by A7;
  end;
end;

theorem Th78:
  for A being non left_end non right_end non empty interval
  ext-real-membered set holds A =].inf A,sup A.[
proof
  let A be non left_end non right_end non empty interval ext-real-membered
  set;
  let x;
  defpred P[ext-real number] means $1 in A & $1 < x;
  defpred Q[ext-real number] means $1 in A & $1 > x;
  thus x in A implies x in ].inf A,sup A.[
  proof
    assume
A1: x in A;
    then
A2: x <> sup A by Def6;
    x <= sup A by A1,Th4;
    then
A3: x < sup A by A2,XXREAL_0:1;
A4: x <> inf A by A1,Def5;
    inf A <= x by A1,Th3;
    then inf A < x by A4,XXREAL_0:1;
    hence thesis by A3,XXREAL_1:4;
  end;
  assume
A5: x in ].inf A,sup A.[;
  per cases;
  suppose
    not ex r st P[r];
    then x is LowerBound of A by Def2;
    then x <= inf A by Def4;
    hence thesis by A5,XXREAL_1:4;
  end;
  suppose
    not ex r st Q[r];
    then x is UpperBound of A by Def1;
    then sup A <= x by Def3;
    hence thesis by A5,XXREAL_1:4;
  end;
  suppose that
A6: ex r st P[r] and
A7: ex r st Q[r];
    consider r such that
A8: r in A and
A9: r < x by A6;
    consider s such that
A10: s in A and
A11: s > x by A7;
A12: x in [.r,s.] by A9,A11,XXREAL_1:1;
    [.r,s.] c= A by A8,A10,Def12;
    hence thesis by A12;
  end;
end;

theorem
  for A being non left_end non right_end interval ext-real-membered
  set ex r,s st r <= s & A =].r,s.[
proof
  let A be non left_end non right_end interval ext-real-membered set;
  per cases;
  suppose
A1: A is empty;
    take -infty,-infty;
    thus thesis by A1;
  end;
  suppose
A2: A is not empty;
    take inf A, sup A;
    thus inf A <= sup A by A2,Th40;
    thus thesis by A2,Th78;
  end;
end;

theorem Th80:
  A is interval implies for x,y,r st x in A & y in A & x <= r & r
  <= y holds r in A
proof
  assume
A1: A is interval;
  let x,y,r such that
A2: x in A and
A3: y in A and
A4: x <= r and
A5: r <= y;
A6: r in [.x,y.] by A4,A5,XXREAL_1:1;
  [.x,y.] c= A by A1,A2,A3,Def12;
  hence thesis by A6;
end;

theorem Th81:
  A is interval implies for x,r st x in A & x <= r & r < sup A holds r in A
proof
  assume
A1: A is interval;
  let x,r such that
A2: x in A and
A3: x <= r and
A4: r < sup A;
  per cases;
  suppose
    ex y st y in A & r < y;
    hence thesis by A1,A2,A3,Th80;
  end;
  suppose
    not ex y st y in A & r < y;
    then r is UpperBound of A by Def1;
    hence thesis by A4,Def3;
  end;
end;

theorem Th82:
  A is interval implies for x,r st x in A & inf A < r & r <= x holds r in A
proof
  assume
A1: A is interval;
  let x,r such that
A2: x in A and
A3: inf A < r and
A4: r <= x;
  per cases;
  suppose
    ex y st y in A & r > y;
    hence thesis by A1,A2,A4,Th80;
  end;
  suppose
    not ex y st y in A & r > y;
    then r is LowerBound of A by Def2;
    hence thesis by A3,Def4;
  end;
end;

theorem
  A is interval implies for r st inf A < r & r < sup A holds r in A
proof
  assume
A1: A is interval;
  let r such that
A2: inf A < r and
A3: r < sup A;
  per cases;
  suppose
    ex y st y in A & r > y;
    hence thesis by A1,A3,Th81;
  end;
  suppose
    not ex y st y in A & r > y;
    then r is LowerBound of A by Def2;
    hence thesis by A2,Def4;
  end;
end;

theorem Th84:
  (for x,y,r st x in A & y in A & x < r & r < y holds r in A)
  implies A is interval
proof
  assume
A1: for x,y,r st x in A & y in A & x < r & r < y holds r in A;
  let x,y such that
A2: x in A and
A3: y in A;
  let r;
  assume r in [.x,y.];
  then r in ].x,y.[ \/ {x,y} by XXREAL_1:29,128;
  then
A4: r in ].x,y.[ or r in {x,y} by XBOOLE_0:def 3;
  per cases by A4,TARSKI:def 2;
  suppose
    r = x;
    hence thesis by A2;
  end;
  suppose
    r = y;
    hence thesis by A3;
  end;
  suppose
A5: r in ].x,y.[;
    then
A6: r < y by XXREAL_1:4;
    x < r by A5,XXREAL_1:4;
    hence thesis by A1,A2,A3,A6;
  end;
end;

theorem
  (for x,r st x in A & x < r & r < sup A holds r in A) implies A is interval
proof
  assume
A1: for x,r st x in A & x < r & r < sup A holds r in A;
  let x,y such that
A2: x in A and
A3: y in A;
  let r;
  assume r in [.x,y.];
  then r in ].x,y.[ \/ {x,y} by XXREAL_1:29,128;
  then
A4: r in ].x,y.[ or r in {x,y} by XBOOLE_0:def 3;
  per cases by A4,TARSKI:def 2;
  suppose
    r = x;
    hence thesis by A2;
  end;
  suppose
    r = y;
    hence thesis by A3;
  end;
  suppose
A5: r in ].x,y.[;
    then
A6: r < y by XXREAL_1:4;
    y <= sup A by A3,Th4;
    then
A7: r < sup A by A6,XXREAL_0:2;
    x < r by A5,XXREAL_1:4;
    hence thesis by A1,A2,A7;
  end;
end;

theorem
  (for y,r st y in A & inf A < r & r < y holds r in A) implies A is interval
proof
  assume
A1: for y,r st y in A & inf A < r & r < y holds r in A;
  let x,y such that
A2: x in A and
A3: y in A;
  let r;
  assume r in [.x,y.];
  then r in ].x,y.[ \/ {x,y} by XXREAL_1:29,128;
  then
A4: r in ].x,y.[ or r in {x,y} by XBOOLE_0:def 3;
  per cases by A4,TARSKI:def 2;
  suppose
    r = x;
    hence thesis by A2;
  end;
  suppose
    r = y;
    hence thesis by A3;
  end;
  suppose
A5: r in ].x,y.[;
    then
A6: x < r by XXREAL_1:4;
    inf A <= x by A2,Th3;
    then
A7: inf A < r by A6,XXREAL_0:2;
    r < y by A5,XXREAL_1:4;
    hence thesis by A1,A3,A7;
  end;
end;

theorem
  (for r st inf A < r & r < sup A holds r in A) implies A is interval
proof
  assume
A1: for r st inf A < r & r < sup A holds r in A;
  let x,y such that
A2: x in A and
A3: y in A;
  let r;
  assume r in [.x,y.];
  then r in ].x,y.[ \/ {x,y} by XXREAL_1:29,128;
  then
A4: r in ].x,y.[ or r in {x,y} by XBOOLE_0:def 3;
  per cases by A4,TARSKI:def 2;
  suppose
    r = x;
    hence thesis by A2;
  end;
  suppose
    r = y;
    hence thesis by A3;
  end;
  suppose
A5: r in ].x,y.[;
    then
A6: r < y by XXREAL_1:4;
    y <= sup A by A3,Th4;
    then
A7: r < sup A by A6,XXREAL_0:2;
A8: x < r by A5,XXREAL_1:4;
    inf A <= x by A2,Th3;
    then inf A < r by A8,XXREAL_0:2;
    hence thesis by A1,A7;
  end;
end;

theorem Th88:
  (for x,y,r st x in A & y in A & x <= r & r <= y holds r in A)
  implies A is interval
proof
  assume for x,y,r st x in A & y in A & x <= r & r <= y holds r in A;
  then for x,y,r st x in A & y in A & x < r & r < y holds r in A;
  hence thesis by Th84;
end;

theorem
  A is interval & B is interval & A meets B implies A \/ B is interval
proof
  assume that
A1: A is interval and
A2: B is interval;
  given z such that
A3: z in A and
A4: z in B;
  for x,y,r st x in A \/ B & y in A \/ B & x <= r & r <= y holds r in A \/ B
  proof
    let x,y,r such that
A5: x in A \/ B and
A6: y in A \/ B and
A7: x <= r and
A8: r <= y;
    per cases by A5,A6,XBOOLE_0:def 3;
    suppose
      x in A & y in A;
      then r in A by A1,A7,A8,Th80;
      hence thesis by XBOOLE_0:def 3;
    end;
    suppose that
A9:   x in A and
A10:  y in B;
      per cases;
      suppose
        r <= z;
        then r in A by A1,A3,A7,A9,Th80;
        hence thesis by XBOOLE_0:def 3;
      end;
      suppose
        z <= r;
        then r in B by A2,A4,A8,A10,Th80;
        hence thesis by XBOOLE_0:def 3;
      end;
    end;
    suppose that
A11:  x in B and
A12:  y in A;
      per cases;
      suppose
        z <= r;
        then r in A by A1,A3,A8,A12,Th80;
        hence thesis by XBOOLE_0:def 3;
      end;
      suppose
        r <= z;
        then r in B by A2,A4,A7,A11,Th80;
        hence thesis by XBOOLE_0:def 3;
      end;
    end;
    suppose
      x in B & y in B;
      then r in B by A2,A7,A8,Th80;
      hence thesis by XBOOLE_0:def 3;
    end;
  end;
  hence thesis by Th88;
end;

theorem
  A is interval & B is left_end interval & sup A = inf B implies A \/
  B is interval
proof
  assume that
A1: A is interval and
A2: B is left_end interval and
A3: sup A = inf B;
  set z = inf B;
A4: z in B by A2,Def5;
  for x,y,r st x in A \/ B & y in A \/ B & x < r & r < y holds r in A \/ B
  proof
    let x,y,r such that
A5: x in A \/ B and
A6: y in A \/ B and
A7: x < r and
A8: r < y;
    per cases by A5,A6,XBOOLE_0:def 3;
    suppose
      x in A & y in A;
      then r in A by A1,A7,A8,Th80;
      hence thesis by XBOOLE_0:def 3;
    end;
    suppose that
A9:   x in A and
A10:  y in B;
      per cases;
      suppose
        r < z;
        then r in A by A1,A3,A7,A9,Th81;
        hence thesis by XBOOLE_0:def 3;
      end;
      suppose
        z <= r;
        then r in B by A2,A4,A8,A10,Th80;
        hence thesis by XBOOLE_0:def 3;
      end;
    end;
    suppose that
A11:  x in B and
A12:  y in A;
      per cases;
      suppose
A13:    z < r;
        y <= z by A3,A12,Th4;
        hence thesis by A8,A13,XXREAL_0:2;
      end;
      suppose
        r <= z;
        then r in B by A2,A4,A7,A11,Th80;
        hence thesis by XBOOLE_0:def 3;
      end;
    end;
    suppose
      x in B & y in B;
      then r in B by A2,A7,A8,Th80;
      hence thesis by XBOOLE_0:def 3;
    end;
  end;
  hence thesis by Th84;
end;

theorem
  A is right_end interval & B is interval & sup A = inf B implies A \/
  B is interval
proof
  assume that
A1: A is right_end interval and
A2: B is interval and
A3: sup A = inf B;
  set z = inf B;
A4: z in A by A1,A3,Def6;
  for x,y,r st x in A \/ B & y in A \/ B & x < r & r < y holds r in A \/ B
  proof
    let x,y,r such that
A5: x in A \/ B and
A6: y in A \/ B and
A7: x < r and
A8: r < y;
    per cases by A5,A6,XBOOLE_0:def 3;
    suppose
      x in A & y in A;
      then r in A by A1,A7,A8,Th80;
      hence thesis by XBOOLE_0:def 3;
    end;
    suppose that
A9:   x in A and
A10:  y in B;
      per cases;
      suppose
        r <= z;
        then r in A by A1,A4,A7,A9,Th80;
        hence thesis by XBOOLE_0:def 3;
      end;
      suppose
        z < r;
        then r in B by A2,A8,A10,Th82;
        hence thesis by XBOOLE_0:def 3;
      end;
    end;
    suppose that
A11:  x in B and
A12:  y in A;
      per cases;
      suppose
        z <= r;
        then r in A by A1,A4,A8,A12,Th80;
        hence thesis by XBOOLE_0:def 3;
      end;
      suppose
A13:    r < z;
        z <= x by A11,Th3;
        hence thesis by A7,A13,XXREAL_0:2;
      end;
    end;
    suppose
      x in B & y in B;
      then r in B by A2,A7,A8,Th80;
      hence thesis by XBOOLE_0:def 3;
    end;
  end;
  hence thesis by Th84;
end;

registration
 cluster left_end -> non empty for ext-real-membered set;
 coherence
  proof let A be ext-real-membered set;
   assume inf A in A;
   hence thesis;
  end;
 cluster right_end -> non empty for ext-real-membered set;
 coherence
  proof let A be ext-real-membered set;
   assume sup A in A;
   hence thesis;
  end;
end;



theorem
  for A being non empty Subset of ExtREAL st
   for r being Element of ExtREAL st r in A
  holds r <= -infty holds A = {-infty}
proof
  let A be non empty Subset of ExtREAL such that
A1: for r being Element of ExtREAL st r in A holds r <= -infty;
  assume A <> {-infty};
  then ex a being Element of A st a <> -infty by SETFAM_1:49;
  hence contradiction by A1,XXREAL_0:6;
end;

theorem
  for A being non empty Subset of ExtREAL st
   for r being Element of ExtREAL st r in A
  holds +infty <= r holds A = {+infty}
proof
  let A be non empty Subset of ExtREAL such that
A1: for r being Element of ExtREAL st r in A holds +infty <= r;
  assume A <> {+infty};
  then ex a being Element of A st a <> +infty by SETFAM_1:49;
  hence contradiction by A1,XXREAL_0:4;
end;

theorem Th94:
  for A being non empty Subset of ExtREAL, r being ext-real number
  st r < sup A ex s being Element of ExtREAL st s in A & r < s
proof
  let A be non empty Subset of ExtREAL, r be ext-real number;
  assume
A1: r < sup A;
  assume
A2: for s being Element of ExtREAL st s in A holds not r < s;
  r is UpperBound of A
  proof
    let x be ext-real number;
    assume x in A;
    hence thesis by A2;
  end;
  hence contradiction by A1,Def3;
end;

theorem Th95:
  for A being non empty Subset of ExtREAL, r being Element of ExtREAL
    st inf A < r
   ex s being Element of ExtREAL st s in A & s < r
proof
  let A be non empty Subset of ExtREAL, r be Element of ExtREAL;
  assume
A1: inf A < r;
  assume
A2: for s being Element of ExtREAL st s in A holds not s < r;
  r is LowerBound of A
  proof
    let x be ext-real number;
    assume x in A;
    hence thesis by A2;
  end;
  hence contradiction by A1,Def4;
end;

theorem
  for A,B being non empty Subset of ExtREAL st for r,s being Element of ExtREAL
  st r in A & s in B holds r <= s holds sup A <= inf B
proof
  let A,B be non empty Subset of ExtREAL;
  assume
A1: for r,s being Element of ExtREAL st r in A & s in B holds r <= s;
  assume not sup A <= inf B;
  then consider s1 being Element of ExtREAL such that
A2: s1 in A and
A3: inf B < s1 by Th94;
  ex s2 being Element of ExtREAL st s2 in B & s2 < s1 by A3,Th95;
  hence contradiction by A1,A2;
end;















begin

reserve x,y,z,w for ext-real number,
  r for real number;

definition
  let x,y;
  redefine pred x <= y means
  :Def1:
  ex p,q being Element of REAL st p = x & q = y & p <= q
  if x in REAL & y in REAL otherwise x = -infty or y = +infty;
  consistency;
  compatibility
  proof
    thus x in REAL & y in REAL implies (x <= y iff ex p,q being Element of
    REAL st p = x & q = y & p <= q);
    thus not(x in REAL & y in REAL) implies (x <= y iff x = -infty or y =
    +infty)
    proof
      assume
A1:   not(x in REAL & y in REAL);
      per cases by A1;
      suppose
A2:     not x in REAL;
        thus x <= y implies x = -infty or y = +infty
        proof
          assume
A3:       x <= y;
          assume x <> -infty;
          then x = +infty by A2,XXREAL_0:14;
          hence thesis by A3,XXREAL_0:4;
        end;
        thus thesis by XXREAL_0:3,5;
      end;
      suppose
A4:     not y in REAL;
        thus x <= y implies x = -infty or y = +infty
        proof
          assume
A5:       x <= y;
          assume
A6:       x <> -infty;
          assume y <> +infty;
          then y = -infty by A4,XXREAL_0:14;
          hence thesis by A5,A6,XXREAL_0:6;
        end;
        thus thesis by XXREAL_0:3,5;
      end;
    end;
  end;
end;

registration
  cluster non real positive for ext-real number;
  existence
  proof
    take +infty;
    thus thesis;
  end;
  cluster non real negative for ext-real number;
  existence
  proof
    take -infty;
    thus thesis;
  end;
end;

theorem Th1:
  for x being non real positive ext-real number holds x = +infty
proof
  let x be non real positive ext-real number;
  not x in REAL;
  hence thesis by XXREAL_0:14;
end;

theorem Th2:
  for x being non real negative ext-real number holds x = -infty
proof
  let x be non real negative ext-real number;
  not x in REAL;
  hence thesis by XXREAL_0:14;
end;

registration
  cluster non real non negative -> positive for ext-real number;
  coherence;
  cluster non real non positive -> negative for ext-real number;
  coherence;
end;

theorem Th3:
  x < z implies ex y being real number st x < y & y < z
proof
  assume
A1: x < z;
  per cases by XXREAL_0:14;
  suppose x in REAL & z in REAL;
    hence thesis by A1,XREAL_1:5;
  end;
  suppose x = +infty or z = -infty;
    hence thesis by A1,XXREAL_0:4,6;
  end;
  suppose
A2: z = +infty;
    per cases by A1,A2,XXREAL_0:14;
    suppose x = -infty;
      hence thesis by A2;
    end;
    suppose x in REAL;
      then reconsider x1 = x as real number;
      take x1+1;
A3:   x1+1 in REAL by XREAL_0:def 1;
      x1+0 < x1+1 by XREAL_1:8;
      hence thesis by A2,A3,XXREAL_0:9;
    end;
  end;
  suppose
A4: x = -infty;
    per cases by A1,A4,XXREAL_0:14;
    suppose z = +infty;
      hence thesis by A4;
    end;
    suppose z in REAL;
      then reconsider z1 = z as real number;
      take z1-1;
A5:   z1-1 in REAL by XREAL_0:def 1;
      z1-1 < z1-0 by XREAL_1:10;
      hence thesis by A4,A5,XXREAL_0:12;
    end;
  end;
end;

begin 

definition
  let x,y;
  func x + y -> ext-real number means
  :Def2:
  ex a,b being complex number st x = a & y = b & it = a + b
  if x is real & y is real,
  it = +infty if x = +infty & y <> -infty or y = +infty & x <> -infty,
  it = -infty if x = -infty & y <> +infty or y = -infty & x <> +infty
  otherwise it = 0;
  existence
  proof
    thus x is real & y is real implies ex c being ext-real number, a, b being
    complex number st x = a & y = b & c = a + b
    proof
      assume x is real & y is real;
      then reconsider a = x, b = y as real number;
      take a+b, a, b;
      thus thesis;
    end;
    thus thesis;
  end;
  uniqueness;
  consistency;
  commutativity;
end;

definition
  let x;
  func -x -> ext-real number means
  :Def3:
  ex a being complex number st x = a & it = -a if x is real,
  it = -infty if x = +infty otherwise it = +infty;
  existence
  proof
    thus x is real implies ex c being ext-real number, a being complex number
    st x = a & c = -a
    proof
      assume x is real;
      then reconsider a = x as real number;
      take -a, a;
      thus thesis;
    end;
    thus thesis;
  end;
  uniqueness;
  consistency;
  involutiveness
  proof
    let y,x be ext-real number such that
A1: x is real implies ex a being complex number st x = a & y = - a and
A2: x = +infty implies y = -infty and
A3: x is not real & x <> +infty implies y = +infty;
    thus y is real implies ex a being complex number st y = a & x = - a
    proof
      assume y is real;
      then consider a being complex number such that
A4:   x = a & y = - a by A1,A2,A3;
      take -a;
      thus thesis by A4;
    end;
    x in REAL implies x is real;
    hence y = +infty implies x = -infty by A1,A2,XXREAL_0:14;
    thus thesis by A1,A3;
  end;
end;

definition
  let x,y;
  func x - y -> ext-real number equals
  x + -y;
  coherence;
end;

registration
  let x,y be real number, a,b be complex number;
  identify x+y with a+b when x = a, y = b;
  compatibility by Def2;
end;

registration
  let x be real number, a be complex number;
  identify -x with -a when x = a;
  compatibility by Def3;
end;



registration
  let r be real number;
  cluster -r -> real;
  coherence;
end;



registration
  let r,s be real number;
  cluster r+s -> real;
  coherence;
  cluster r-s -> real;
  coherence;
end;

registration
  let x be real number, y be non real ext-real number;
  cluster x+y -> non real for number;
  coherence
  proof
    not y in REAL;
    then
A1: y = -infty or y = +infty by XXREAL_0:14;
    let z be number;
    assume z = x+y;
    hence thesis by A1,Def2;
  end;
end;

registration
  let x,y be non real positive ext-real number;
  cluster x+y -> non real;
  coherence
  proof
    x = +infty by Th1;
    hence thesis by Def2;
  end;
end;

registration
  let x,y be non real negative ext-real number;
  cluster x+y -> non real;
  coherence
  proof
    x = -infty by Th2;
    hence thesis by Def2;
  end;
end;

registration
  let x be non real negative ext-real number, y be non real positive
  ext-real number;
  cluster x+y -> zero;
  coherence
  proof
    x = -infty & y = +infty by Th1,Th2;
    hence thesis by Def2;
  end;
end;

registration
  let x,y be real number, a,b be complex number;
  identify x-y with a-b when x = a, y = b;
  compatibility;
end;

theorem Th4:
  x + 0 = x
proof
  per cases by XXREAL_0:14;
  suppose
    x in REAL;
    then reconsider r=x as real number;
    r + 0 = r;
    hence thesis;
  end;
  suppose
    x = -infty or x = +infty;
    hence thesis by Def2;
  end;
end;

theorem Th5:
  --infty = +infty by Def3;

theorem Th6:
  -+infty = -infty by Def3;

Lm1: +infty + +infty = +infty by Def2;

Lm2: -infty + -infty = -infty by Def2;

theorem Th7:
  x + -x = 0
proof
  per cases by XXREAL_0:14;
  suppose
    x = -infty;
    hence thesis by Th5;
  end;
  suppose
    x in REAL;
    then reconsider x as real number;
    reconsider y=-x as real number;
    x + y = 0;
    hence thesis;
  end;
  suppose
    x = +infty;
    hence thesis by Th6;
  end;
end;

theorem
  x+y = 0 implies x = -y
proof
  assume
A1: x+y = 0;
  per cases by XXREAL_0:14;
  suppose
    x in REAL;
    then reconsider x as real number;
    x+y is real by A1;
    then reconsider y as real number;
    x+y = 0 by A1;
    then x = -y;
    hence thesis;
  end;
  suppose
A2: x = -infty;
    then y = +infty by A1,Def2;
    hence thesis by A2,Def3;
  end;
  suppose
A3: x = +infty;
    then y = -infty by A1,Def2;
    hence thesis by A3,Def3;
  end;
end;

Lm3: for x holds -x in REAL iff x in REAL
proof
  let x;
  hereby
    assume -x in REAL;
    then reconsider a = -x as real number;
    -a in REAL by XREAL_0:def 1;
    hence x in REAL;
  end;
  assume x in REAL;
  then reconsider a = x as real number;
  -a is real;
  hence thesis by XREAL_0:def 1;
end;

Lm4: -(+infty + -infty) = +infty - +infty by Def3
  .= --infty - +infty by Def3;

Lm5: -+infty = -infty by Def3;

Lm6: x in REAL implies -(x + +infty) = -+infty + -x
proof
  assume
A1: x in REAL;
  then x + +infty = +infty by Def2;
  hence thesis by A1,Def2,Th6;
end;

Lm7: x in REAL implies -(x + -infty) = --infty + -x
proof
  assume
A1: x in REAL;
  then x + -infty = -infty by Def2;
  hence thesis by A1,Def2,Th5;
end;

theorem Th9:
  -(x+y) = -x + -y
proof
  per cases by XXREAL_0:14;
  suppose
A1: x = +infty & y = +infty;
    hence -(x + y) = -+infty by Def2
      .= -x +- y by A1,Def2,Lm5;
  end;
  suppose
    x = +infty & y = -infty;
    hence thesis by Lm4;
  end;
  suppose
    x = +infty & y in REAL;
    hence thesis by Lm6;
  end;
  suppose
    x = -infty & y = +infty;
    hence thesis by Lm4;
  end;
  suppose
A2: x = -infty & y = -infty;
    hence -(x + y) = --infty by Def2
      .= -x +- y by A2,Def2,Th5;
  end;
  suppose
    x = -infty & y in REAL;
    hence thesis by Lm7;
  end;
  suppose
    x in REAL & y = +infty;
    hence thesis by Lm6;
  end;
  suppose
    x in REAL & y = -infty;
    hence thesis by Lm7;
  end;
  suppose
    x in REAL & y in REAL;
    then reconsider a=x, b=y as real number;
    -(x + y) = -(a + b) .= -a + - b;
    hence thesis;
  end;
end;

reserve f,g for ext-real number;

theorem Th10: 
  -f = -g implies f = g
proof
  assume
A1: -f = -g;
  per cases by XXREAL_0:14;
  suppose
A2: f in REAL;
    now
      assume not g in REAL;
      then g = +infty or g = -infty by XXREAL_0:14;
      hence contradiction by A1,A2,Def3;
    end;
    then
A3: ex a being complex number st g = a & -g = -a by Def3;
    ex a being complex number st f = a & -f = -a by A2,Def3;
    hence thesis by A1,A3;
  end;
  suppose
    f = +infty;
    hence thesis by A1,Th5;
  end;
  suppose
    f = -infty;
    then - -g = -infty by A1;
    hence thesis by A1;
  end;
end;

Lm8: x + y = +infty implies x = +infty or y = +infty
proof
  assume
A1: x + y = +infty;
  assume ( not x = +infty)& not y = +infty;
  then x in REAL & y in REAL or x in REAL & y = -infty or x = -infty & y in
  REAL or x = -infty & y = -infty by XXREAL_0:14;
  hence thesis by A1,Def2;
end;

Lm9: x + y = -infty implies (x = -infty or y = -infty)
proof
  assume
A1: x + y = -infty;
  assume ( not x = -infty)& not y = -infty;
  then x in REAL & y in REAL or x in REAL & y = +infty or x = +infty & y in
  REAL or x = +infty & y = +infty by XXREAL_0:14;
  hence thesis by A1,Def2;
end;

theorem Th11: 
  r+f = r+g implies f = g
proof
  assume
A1: r+f = r+g;
  per cases by XXREAL_0:14;
  suppose
A2: f in REAL;
    now
      assume not g in REAL;
      then g = +infty or g = -infty by XXREAL_0:14;
      hence contradiction by A1,A2;
    end;
    then
A3: ex c, d being complex number st r = c & g = d & r+g = c+d by Def2;
    ex a, b being complex number st r = a & f = b & r+f = a+ b by A2,Def2;
    hence thesis by A1,A3;
  end;
  suppose
A4: f = +infty;
    then r+f = +infty by Def2;
    hence thesis by A1,A4,Lm8;
  end;
  suppose
A5: f = -infty;
    then r+f = -infty by Def2;
    hence thesis by A1,A5,Lm9;
  end;
end;

theorem 
  r-f = r-g implies f = g
proof
  assume r-f = r-g;
  then -f = -g by Th11;
  hence thesis by Th10;
end;

theorem Th13:
  x <> +infty implies +infty - x = +infty & x - +infty = -infty
proof
  assume
A1: x <> +infty;
  then -x <> -infty by Th5;
  hence +infty - x = +infty by Def2;
  - +infty = -infty by Def3;
  hence thesis by A1,Def2;
end;

theorem Th14:
  x <> -infty implies -infty - x = -infty & x - -infty = +infty
proof
  assume
A1: x <> -infty;
  now
    assume -x = +infty;
    then --x = -infty by Def3;
    hence contradiction by A1;
  end;
  hence -infty - x = -infty by Def2;
  thus thesis by A1,Def2,Th5;
end;

theorem Th15:
  x - 0 = x
proof
  per cases by XXREAL_0:14;
  suppose
    x in REAL;
    then reconsider a = x as real number;
    x - 0 = a - 0 .= x;
    hence thesis;
  end;
  suppose
    x = -infty;
    hence thesis by Th14;
  end;
  suppose
    x = +infty;
    hence thesis by Th13;
  end;
end;

theorem
  x + y = +infty implies x = +infty or y = +infty by Lm8;

theorem
  x + y = -infty implies (x = -infty or y = -infty) by Lm9;

theorem Th18:
  x - y = +infty implies (x = +infty or y = -infty)
proof
  assume
A1: x - y = +infty;
  assume ( not x = +infty)& not y = -infty;
  then x in REAL & y in REAL or x in REAL & y = +infty or x = -infty & y in
  REAL or x = -infty & y = +infty by XXREAL_0:14;
  hence thesis by A1,Th13,Th14;
end;

theorem Th19:
  x - y = -infty implies (x = -infty or y = +infty)
proof
  assume
A1: x - y = -infty;
  assume ( not x = -infty)& not y = +infty;
  then x in REAL & y in REAL or x in REAL & y = -infty or x = +infty & y in
  REAL or x = +infty & y = -infty by XXREAL_0:14;
  hence thesis by A1,Th13,Th14;
end;

theorem Th20:
  not ( x = +infty & y = -infty or x = -infty & y = +infty ) & x +
  y in REAL implies x in REAL & y in REAL
proof
  assume
A1: ( not ( x = +infty & y = -infty or x = -infty & y = +infty ))& x + y
  in REAL;
A2: x in REAL or x = -infty or x = +infty by XXREAL_0:14;
A3: y in REAL or y = -infty or y = +infty by XXREAL_0:14;
  assume not (x in REAL & y in REAL);
  hence thesis by A1,A2,A3;
end;

theorem Th21:
  not ( x = +infty & y = +infty or x = -infty & y = -infty ) & x -
  y in REAL implies x in REAL & y in REAL
proof
  assume
A1: ( not ( x = +infty & y = +infty or x = -infty & y = -infty ))& x - y
  in REAL;
A2: x in REAL or x = -infty or x = +infty by XXREAL_0:14;
A3: y in REAL or y = -infty or y = +infty by XXREAL_0:14;
  assume not (x in REAL & y in REAL);
  hence thesis by A1,A2,A3,Th13,Th14;
end;

theorem Th22:
  x is real implies y - x + x = y & y + x - x = y
proof
  assume
A1: x is real;
  per cases by A1,XXREAL_0:14;
  suppose
    x in REAL & y in REAL;
    then reconsider a = x, b = y as real number;
A2: (y + x) - x = b + a - a .= y;
    y - x + x = (b - a) + a .= y;
    hence thesis by A2;
  end;
  suppose
A3: x in REAL & y = -infty;
    then y - x = -infty by Def2;
    hence thesis by A3,Def2;
  end;
  suppose
A4: x in REAL & y = +infty;
    then y - x = +infty by Def2;
    hence thesis by A4,Def2;
  end;
end;

theorem Th23:
  (x = +infty iff -x = -infty) & (x = -infty iff -x = +infty)
proof
  -x = +infty implies x = -infty
  proof
    assume -x = +infty;
    then --x = -infty by Def3;
    hence thesis;
  end;
  hence thesis by Th5;
end;

theorem
  z is real implies x = x + z - z
proof
  assume
A1: z is real;
  per cases by XXREAL_0:14;
  suppose
A2: x = +infty;
    then x + z = +infty by A1,Def2;
    hence thesis by A1,A2,Th13;
  end;
  suppose
A3: x = -infty;
    then x + z = -infty by A1,Def2;
    hence thesis by A1,A3,Th14;
  end;
  suppose
    x is Element of REAL;
    then reconsider a = x, c = z as real number by A1;
    x + z - z = a + c - c;
    hence thesis;
  end;
end;

Lm10: -+infty = -infty by Def3;

Lm11: x in REAL implies -(x + -infty) = --infty + -x
proof
A1: --infty = +infty by Th23;
  assume
A2: x in REAL;
  then x + -infty = -infty by Def2;
  hence thesis by A2,A1,Def2;
end;

theorem Th25:
  -(x + y) = -y - x
proof
  per cases by XXREAL_0:14;
  suppose
A1: x = +infty & y = +infty;
    hence -(x + y) = -+infty by Def2
      .= -y - x by A1,Def2,Lm10;
  end;
  suppose
    x = +infty & y = -infty;
    hence thesis by Lm4;
  end;
  suppose
    x = +infty & y in REAL;
    hence thesis by Lm6;
  end;
  suppose
    x = -infty & y = +infty;
    hence thesis by Lm4;
  end;
  suppose
A2: x = -infty & y = -infty;
    hence -(x + y) = --infty by Def2
      .= -y - x by A2,Def2,Th5;
  end;
  suppose
    x = -infty & y in REAL;
    hence thesis by Lm11;
  end;
  suppose
    x in REAL & y = +infty;
    hence thesis by Lm6;
  end;
  suppose
    x in REAL & y = -infty;
    hence thesis by Lm11;
  end;
  suppose
    x in REAL & y in REAL;
    then reconsider a=x, b=y as real number;
    -(x + y) = -(a + b) .= -a + - b;
    hence thesis;
  end;
end;

theorem
  -(x - y) = -x + y & -(x - y) = y - x
proof
  -(x - y) = -(-y) - x by Th25;
  hence thesis;
end;

theorem Th27:
  -(-x + y) = x - y & -(-x + y) = x + -y
proof
  -(-x + y) = -(-x) - y by Th25
    .= x - y;
  hence thesis;
end;

theorem
  0 <= y & y < +infty implies z = z + y - y
proof
  assume 0 <= y & y < +infty;
  then y in REAL by XXREAL_0:14;
  hence thesis by Th22;
end;

theorem Th29:
  not (x = +infty & y = -infty) & not (x = -infty & y = +infty) &
  not ( y = +infty & z = -infty or y = -infty & z = +infty ) &
  not ( x = +infty & z = -infty or x = -infty & z = +infty ) implies
  x + y + z = x + (y + z)
proof
  assume
A1: ( not (x = +infty & y = -infty or x = -infty & y = +infty))& not ( y
  = +infty & z = -infty or y = -infty & z = +infty ) & not ( x = +infty & z =
  -infty or x = -infty & z = +infty );
  per cases by A1,XXREAL_0:14;
  suppose
    x in REAL & y in REAL & z in REAL;
    then reconsider A = x, B = y, C = z as real number;
    A + B + C = x + y + z & A + (B + C) = x + (y + z);
    hence thesis;
  end;
  suppose
A2: x in REAL & y = +infty & z in REAL;
    then x + y = +infty & y + z = +infty by Def2;
    hence thesis by A2;
  end;
  suppose
A3: x in REAL & y = -infty & z in REAL;
    then x + y = -infty & y + z = -infty by Def2;
    hence thesis by A3;
  end;
  suppose
A4: x = -infty & y in REAL & z in REAL;
    then x + y = -infty by Def2;
    then x + y + z = -infty by A4,Def2;
    hence thesis by A4,Def2;
  end;
  suppose
A5: x = +infty & y in REAL & z in REAL;
    then x + y = +infty by Def2;
    then x + y + z = +infty by A5,Def2;
    hence thesis by A5,Def2;
  end;
  suppose
A6: x in REAL & y in REAL & z = +infty;
    then y + z = +infty by Def2;
    then x + (y + z) = +infty by A6,Def2;
    hence thesis by A6,Def2;
  end;
  suppose
A7: x in REAL & y in REAL & z = -infty;
    then y + z = -infty by Def2;
    then x + (y + z) = -infty by A7,Def2;
    hence thesis by A7,Def2;
  end;
  suppose
A8: x = +infty & y = +infty & z in REAL;
    then x + y = +infty by Def2;
    then
A9: x + y + z = +infty by A8,Def2;
    y + z <> -infty by A8,Def2;
    hence thesis by A8,A9,Def2;
  end;
  suppose
A10: x in REAL & y = -infty & z = -infty;
    then
A11: x + y = -infty by Def2;
    then x + y + z = -infty by A10,Def2;
    hence thesis by A10,A11;
  end;
  suppose
A12: x = -infty & y = -infty & z in REAL;
    then
A13: x + y = -infty by Def2;
    then x + y + z = -infty by A12,Def2;
    hence thesis by A12,A13;
  end;
  suppose
    x = -infty & y in REAL & z = -infty or x = -infty & y = -infty &
z = -infty or x = +infty & y in REAL & z = +infty or x = +infty & y = +infty &
    z = +infty;
    hence thesis;
  end;
  suppose
A14: x in REAL & y = +infty & z = +infty;
    then x + y = +infty & y + z = +infty by Def2;
    hence thesis by A14;
  end;
end;

theorem Th30:
  not (x = +infty & y = -infty) & not (x = -infty & y = +infty) &
not (y = +infty & z = +infty) & not (y = -infty & z = -infty) & not (x = +infty
  & z = +infty) & not (x = -infty & z = -infty) implies x + y - z = x + (y - z)
proof
  assume that
A1: ( not (x = +infty & y = -infty))& not (x = -infty & y = +infty) and
A2: ( not (y = +infty & z = +infty))& not (y = -infty & z = -infty) and
A3: ( not (x = +infty & z = +infty))& not (x = -infty & z = -infty);
A4: ( not (x = +infty & -z = -infty))& not (x = -infty & -z = +infty) by A3
,Th23;
  ( not (y = +infty & -z = -infty))& not (y = -infty & -z = +infty) by A2,Th23;
  hence thesis by A1,A4,Th29;
end;

theorem 
  not (x = +infty & y = +infty) & not (x = -infty & y = -infty) & not (y
= +infty & z = -infty) & not (y = -infty & z = +infty) & not (x = +infty & z =
  +infty) & not (x = -infty & z = -infty) implies x - y - z = x - (y + z)
proof
  assume that
A1: not(x=+infty & y=+infty) and
A2: not(x=-infty & y=-infty) and
A3: not (y = +infty & z = -infty) and
A4: not (y = -infty & z = +infty) and
A5: not (x = +infty & z = +infty) and
A6: not (x = -infty & z = -infty);
  per cases;
  suppose
A7: x = +infty;
    then x - y = +infty by A1,Th13;
    then
A8: x - y - z = +infty by A5,A7,Th13;
    y + z <> +infty by A1,A5,A7,Lm8;
    hence thesis by A7,A8,Th13;
  end;
  suppose
A9: x = -infty;
    then x - y = -infty by A2,Th14;
    then
A10: x - y - z = -infty by A6,A9,Th14;
    y + z <> -infty by A2,A6,A9,Lm9;
    hence thesis by A9,A10,Th14;
  end;
  suppose
A11: x <> +infty & x <> -infty;
    then x in REAL by XXREAL_0:14;
    then reconsider a = x as real number;
    per cases;
    suppose
A12:  y = +infty;
      then x - y = -infty & y + z = +infty by A3,A11,Def2,Th13;
      hence thesis by A3,A12,Th14;
    end;
    suppose
A13:  y = -infty;
      then x - y = +infty & y + z = -infty by A4,A11,Def2,Th14;
      hence thesis by A4,A13,Th13;
    end;
    suppose
      y <> +infty & y <> -infty;
      then y in REAL by XXREAL_0:14;
      then reconsider b = y as real number;
A14:  x - y = a - b;
      per cases;
      suppose
        z = +infty;
        then x - y - z = -infty & y + z = +infty by A14,Def2,Th13;
        hence thesis by A11,Th13;
      end;
      suppose
        z = -infty;
        then x - y - z = +infty & y + z = -infty by A14,Def2,Th14;
        hence thesis by A11,Th14;
      end;
      suppose
        z <> +infty & z <> -infty;
        then z in REAL by XXREAL_0:14;
        then reconsider c = z as real number;
        x - y - z = a - b - c & x - (y + z) = a - (b + c);
        hence thesis;
      end;
    end;
  end;
end;

theorem Th32:
  not (x = +infty & y = +infty) & not (x = -infty & y = -infty) &
not (y = +infty & z = +infty) & not (y = -infty & z = -infty) & not (x = +infty
  & z = -infty) & not (x = -infty & z = +infty) implies x - y + z = x - (y - z)
proof
  assume that
A1: not(x=+infty & y=+infty) and
A2: not(x=-infty & y=-infty) and
A3: not (y = +infty & z = +infty) and
A4: not (y = -infty & z = -infty) and
A5: not (x = +infty & z = -infty) and
A6: not (x = -infty & z = +infty);
  per cases;
  suppose
A7: x = +infty;
    then x - y = +infty by A1,Th13;
    then
A8: x - y + z = +infty by A5,A7,Def2;
    y - z <> +infty by A1,A5,A7,Th18;
    hence thesis by A7,A8,Th13;
  end;
  suppose
A9: x = -infty;
    then x - y = -infty by A2,Th14;
    then
A10: x - y + z = -infty by A6,A9,Def2;
    y - z <> -infty by A2,A6,A9,Th19;
    hence thesis by A9,A10,Th14;
  end;
  suppose
A11: x <> +infty & x <> -infty;
    then x in REAL by XXREAL_0:14;
    then reconsider a = x as real number;
    per cases;
    suppose
A12:  y = +infty;
      then x - y = -infty & y - z = +infty by A3,A11,Th13;
      hence thesis by A3,A12,Def2;
    end;
    suppose
A13:  y = -infty;
      then x - y = +infty & y - z = -infty by A4,A11,Th14;
      hence thesis by A4,A13,Def2;
    end;
    suppose
      y <> +infty & y <> -infty;
      then y in REAL by XXREAL_0:14;
      then reconsider b = y as real number;
A14:  x - y = a - b;
      per cases;
      suppose
        z = +infty;
        then x - y + z = +infty & y - z = -infty by A14,Def2,Th13;
        hence thesis by A11,Th14;
      end;
      suppose
        z = -infty;
        then x - y + z = -infty & y - z = +infty by A14,Def2,Th14;
        hence thesis by A11,Th13;
      end;
      suppose
        z <> +infty & z <> -infty;
        then z in REAL by XXREAL_0:14;
        then reconsider c = z as real number;
        x - y + z = a - b + c & x - (y - z) = a - (b - c);
        hence thesis;
      end;
    end;
  end;
end;

theorem Th33:
  z is real implies (z + x) - (z + y) = x - y
proof
  assume
A1: z is real;
  per cases by A1,XXREAL_0:14;
  suppose
A2: x = -infty;
    per cases by XXREAL_0:14;
    suppose
A3:   y = -infty;
      (z + -infty) - (z + -infty) = -infty - (z + -infty) by A1,Def2
        .= -infty - -infty by A1,Def2;
      hence thesis by A2,A3;
    end;
    suppose
A4:   y = +infty;
      (z + -infty) - (z + +infty) = -infty - (z + +infty) by A1,Def2
        .= -infty - +infty by A1,Def2;
      hence thesis by A2,A4;
    end;
    suppose
A5:   y in REAL;
      then consider a,b being complex number such that
A6:   z = a & y = b and
      z+y = a + b by A1,Def2;
      reconsider a,b as real number by A1,A5,A6;
A7:   a+b in REAL by XREAL_0:def 1;
      (z + -infty) - (z + y) = -infty - (z + y) by A1,Def2
        .= -infty by A6,A7,Def2
        .= -infty - y by A5,Def2;
      hence thesis by A2;
    end;
  end;
  suppose
A8: y = +infty;
    per cases by XXREAL_0:14;
    suppose
A9:   x = -infty;
      (z + -infty) - (z + +infty) = z + -infty - +infty by A1,Def2
        .= -infty - +infty by A1,Def2;
      hence thesis by A8,A9;
    end;
    suppose
A10:  x = +infty;
      (z + +infty) - (z + +infty) = z + +infty - +infty by A1,Def2
        .= +infty - +infty by A1,Def2;
      hence thesis by A8,A10;
    end;
    suppose
A11:  x in REAL;
      then consider a,b being complex number such that
A12:  z = a & x = b and
      z+x = a + b by A1,Def2;
      reconsider a,b as real number by A1,A11,A12;
A13:  a+b in REAL by XREAL_0:def 1;
A14:  - +infty = -infty by Def3;
      (z + x) - (z + +infty) = z + x - +infty by A1,Def2
        .= z + x + -infty by Def3
        .= -infty by A12,A13,Def2
        .= x + - +infty by A11,A14,Def2;
      hence thesis by A8;
    end;
  end;
  suppose
    x in REAL & y in REAL & z in REAL;
    then reconsider a = x, b = y, c = z as real number;
    (z + x) - (z + y) = (a + c) - (c + b) .= a - b
      .= x - y;
    hence thesis;
  end;
  suppose
A15: x = +infty & y in REAL & z in REAL;
    then reconsider c = z, b = y as real number;
A16: z + y = c + b;
    z + x = +infty & x - y = +infty by A15,Def2;
    hence thesis by A16,Th13;
  end;
  suppose
A17: x in REAL & y = -infty & z in REAL;
    then reconsider c = z, a = x as real number;
    z + x = c + a & z + y = -infty by A17,Def2;
    then (z + x) - (z + y) = +infty by Th14
      .= x - y by A17,Th14;
    hence thesis;
  end;
  suppose
A18: x = +infty & y = -infty & z in REAL;
    then z + y = -infty & not z + x = -infty by Def2;
    then (z + x) - (z + y) = +infty by Th14
      .= x - y by A18,Th14;
    hence thesis;
  end;
end;

theorem
  y is real implies (z - y) + (y - x) = z - x
proof
  assume
A1: y is real;
  thus (z - y) + (y - x) = (z - y) - (x - y) by Th27
    .= z - x by A1,Th33;
end;

begin 

Lm12: x <= y implies x + z <= y + z
proof
  assume
A1: x <= y;
  per cases by XXREAL_0:14;
  suppose
A2: x in REAL & y in REAL;
    per cases by XXREAL_0:14;
    suppose
A3:   z = -infty;
      -infty <= y + z by XXREAL_0:5;
      hence thesis by A2,A3,Def2;
    end;
    suppose
      z in REAL;
      then reconsider x, y, z as real number by A2;
      x + z <= y + z by A1,XREAL_1:6;
      hence thesis;
    end;
    suppose
A4:   z = +infty;
      x + z <= +infty by XXREAL_0:3;
      hence thesis by A2,A4,Def2;
    end;
  end;
  suppose
A5: x in REAL & y = +infty;
    per cases by XXREAL_0:14;
    suppose
A6:   z = -infty;
      -infty <= y+z by XXREAL_0:5;
      hence thesis by A5,A6,Def2;
    end;
    suppose
A7:   z in REAL;
      x + z <= +infty by XXREAL_0:3;
      hence thesis by A5,A7,Def2;
    end;
    suppose
      z = +infty;
      hence thesis by A5,Lm1,XXREAL_0:3;
    end;
  end;
  suppose
A8: x = -infty & y in REAL;
    per cases by XXREAL_0:14;
    suppose
      z = -infty;
      hence thesis by A8,Lm2,XXREAL_0:5;
    end;
    suppose
A9:   z in REAL;
      -infty <= y + z by XXREAL_0:5;
      hence thesis by A8,A9,Def2;
    end;
    suppose
A10:  z = +infty;
      x + z <= +infty by XXREAL_0:3;
      hence thesis by A8,A10,Def2;
    end;
  end;
  suppose
A11: x = -infty & y = +infty;
    per cases by XXREAL_0:14;
    suppose
      z = -infty;
      hence thesis by A11,Lm2;
    end;
    suppose
A12:  z in REAL;
      -infty <= y + z by XXREAL_0:5;
      hence thesis by A11,A12,Def2;
    end;
    suppose
      z = +infty;
      hence thesis by A11,Lm1;
    end;
  end;
  suppose
    x = +infty;
    hence thesis by A1,XXREAL_0:4;
  end;
  suppose
    y = -infty;
    hence thesis by A1,XXREAL_0:6;
  end;
end;

Lm13: x >= 0 & y > 0 implies x + y > 0
proof
  assume x >= 0;
  then x + y >= 0 + y by Lm12;
  hence thesis by Th4;
end;

Lm14: x <= 0 & y < 0 implies x + y < 0
proof
  assume x <= 0;
  then x + y <= 0 + y by Lm12;
  hence thesis by Th4;
end;

registration
  let x,y be non negative ext-real number;
  cluster x + y -> non negative;
  coherence
  proof
    per cases;
    suppose
      x = 0;
      hence x + y >= 0 by Th4;
    end;
    suppose
      x > 0;
      hence x+y >= 0 by Lm13;
    end;
  end;
end;

registration
  let x,y be non positive ext-real number;
  cluster x + y -> non positive;
  coherence
  proof
    per cases;
    suppose
      x = 0;
      hence x + y <= 0 by Th4;
    end;
    suppose
      x < 0;
      hence x + y <= 0 by Lm14;
    end;
  end;
end;

registration
  let x be positive ext-real number;
  let y be non negative ext-real number;
  cluster x + y -> positive;
  coherence by Lm13;
  cluster y + x -> positive;
  coherence;
end;

registration
  let x be negative ext-real number;
  let y be non positive ext-real number;
  cluster x + y -> negative;
  coherence by Lm14;
  cluster y + x -> negative;
  coherence;
end;

registration
  let x be non positive ext-real number;
  cluster -x -> non negative;
  coherence
  proof
    assume -x < 0;
    then -x - -x < 0;
    hence contradiction by Th7;
  end;
end;

registration
  let x be non negative ext-real number;
  cluster -x -> non positive;
  coherence
  proof
    assume -x > 0;
    then -x - -x > 0;
    hence contradiction by Th7;
  end;
end;

registration
  let x be positive ext-real number;
  cluster -x -> negative;
  coherence
  proof
    assume -x >= 0;
    then -x- -x > 0;
    hence contradiction by Th7;
  end;
end;

registration
  let x be negative ext-real number;
  cluster -x -> positive;
  coherence
  proof
    assume -x <= 0;
    then -x- -x < 0;
    hence contradiction by Th7;
  end;
end;

registration
  let x be non negative ext-real number, y be non positive ext-real number;
  cluster x - y -> non negative;
  coherence;
  cluster y - x -> non positive;
  coherence;
end;

registration
  let x be positive ext-real number;
  let y be non positive ext-real number;
  cluster x - y -> positive;
  coherence;
  cluster y - x -> negative;
  coherence;
end;

registration
  let x be negative ext-real number;
  let y be non negative ext-real number;
  cluster x - y -> negative;
  coherence;
  cluster y - x -> positive;
  coherence;
end;

Lm15: for x,y st x <= y holds - y <= - x
proof
  let x,y;
  assume
A1: x <= y;
  per cases;
  case that
A2: -y in REAL and
A3: -x in REAL;
    x in REAL by A3,Lm3;
    then consider a being complex number such that
A4: x = a and
A5: -x = - a by Def3;
    y in REAL by A2,Lm3;
    then consider b being complex number such that
A6: y = b and
A7: -y = - b by Def3;
    x in REAL & y in REAL by A2,A3,Lm3;
    then reconsider a,b as real number by A6,A4;
    reconsider mb=-b, ma=-a as Element of REAL by XREAL_0:def 1;
    take mb,ma;
    thus mb = -y & ma = -x by A7,A5;
    thus thesis by A1,A6,A4,XREAL_1:24;
  end;
  case
    not(-y in REAL & -x in REAL);
    then not(y in REAL & x in REAL) by Lm3;
    then x = -infty or y = +infty by A1,Def1;
    hence thesis by Def3;
  end;
end;

theorem
  x <= y implies x + z <= y + z by Lm12;

theorem Th36:
  x <= y & z <= w implies x + z <= y + w
proof
  assume that
A1: x <= y and
A2: z <= w;
  per cases;
  suppose
A3: x = +infty & z = -infty;
A4: w <> -infty implies +infty + w = +infty by Def2;
    y = +infty by A1,A3,XXREAL_0:4;
    hence thesis by A3,A4;
  end;
  suppose
A5: x = -infty & z = +infty;
A6: y <> -infty implies +infty + y = +infty by Def2;
    w = +infty by A2,A5,XXREAL_0:4;
    hence thesis by A5,A6;
  end;
  suppose
A7: y = +infty & w = -infty;
A8: x <> +infty implies -infty + x = -infty by Def2;
    z = -infty by A2,A7,XXREAL_0:6;
    hence thesis by A7,A8;
  end;
  suppose
A9: y = -infty & w = +infty;
A10: z <> +infty implies -infty + z = -infty by Def2;
    x = -infty by A1,A9,XXREAL_0:6;
    hence thesis by A9,A10;
  end;
  suppose
A11: not ( x = +infty & z = -infty or x = -infty & z = +infty ) & not
    ( y = +infty & w = -infty or y = -infty & w = +infty );
    reconsider a = x + z, b = y + w as Element of ExtREAL by XXREAL_0:def 1;
A12: not(a = +infty & b = -infty)
    proof
      assume that
A13:  a = +infty and
A14:  b = -infty;
      x = +infty or z = +infty by A13,Lm8;
      hence thesis by A1,A2,A11,A14,Lm9,XXREAL_0:4;
    end;
A15: a in REAL & b in REAL implies a <= b
    proof
      assume
A16:  a in REAL & b in REAL;
      then
A17:  z in REAL & w in REAL by A11,Th20;
      x in REAL & y in REAL by A11,A16,Th20;
      then consider Ox,Oy,Os,Ot being real number such that
A18:  Ox = x & Oy = y & Os = z & Ot = w and
A19:  Ox <= Oy & Os <= Ot by A1,A2,A17;
      Ox + Os <= Os + Oy & Os + Oy <= Ot + Oy by A19,XREAL_1:6;
      hence thesis by A18,XXREAL_0:2;
    end;
A20: a = +infty & b in REAL implies a <= b
    proof
      assume that
A21:  a = +infty and
      b in REAL;
      x = +infty or z = +infty by A21,Lm8;
      then y = +infty or w = +infty by A1,A2,XXREAL_0:4;
      then b = +infty by A11,Def2;
      hence thesis by XXREAL_0:4;
    end;
A22: a in REAL & b = -infty implies a <= b
    proof
      assume that
      a in REAL and
A23:  b = -infty;
      y = -infty or w = -infty by A23,Lm9;
      then x = -infty or z = -infty by A1,A2,XXREAL_0:6;
      then a = -infty by A11,Def2;
      hence thesis by XXREAL_0:5;
    end;
    a in REAL & b in REAL or a in REAL & b = +infty or a in REAL & b =
-infty or a = +infty & b in REAL or a = +infty & b = +infty or a = +infty & b =
-infty or a = -infty & b in REAL or a = -infty & b = +infty or a = -infty & b =
    -infty by XXREAL_0:14;
    hence thesis by A15,A22,A20,A12,XXREAL_0:4,5;
  end;
end;

theorem Th37:
  x <= y & z <= w implies x - w <= y - z
proof
  assume that
A1: x <= y and
A2: z <= w;
  -w <= -z by A2,Lm15;
  hence thesis by A1,Th36;
end;

theorem Th38:
  x <= y iff - y <= - x
proof
  thus x <= y implies - y <= - x by Lm15;
  - y <= - x implies --x <= --y by Lm15;
  hence thesis;
end;

theorem Th39:
  0 <= z implies x <= x + z
proof
  assume 0 <= z;
  then x + 0 <= x + z by Th36;
  hence thesis by Th4;
end;

theorem
  x <= y implies y-x >= 0
proof
  assume x <= y;
  then -y <= -x by Lm15;
  then -y +y <= y + -x by Th36;
  hence thesis by Th7;
end;

theorem
  (z = -infty & y = +infty implies x <= 0) & (z = +infty & y = -infty
  implies x <= 0) implies (x - y <= z implies x <= z + y)
proof
  assume that
A1: z = -infty & y = +infty implies x <= 0 and
A2: z = +infty & y = -infty implies x <= 0 and
A3: x - y <= z;
  per cases by XXREAL_0:14;
  suppose
A4: z = -infty;
    per cases by A3,A4,Th19,XXREAL_0:6;
    suppose
      x = -infty;
      hence thesis by XXREAL_0:5;
    end;
    suppose
      y = +infty;
      hence thesis by A1,A4;
    end;
  end;
  suppose
A5: z = +infty;
    per cases;
    suppose
      y = -infty;
      hence thesis by A2,A5;
    end;
    suppose
      y <> -infty;
      then z + y = +infty by A5,Def2;
      hence thesis by XXREAL_0:3;
    end;
  end;
  suppose
A6: z in REAL;
    per cases by A3,A6,XXREAL_0:13;
    suppose
A7:   x-y in REAL;
      per cases by A7,Th21;
      suppose
        y = +infty;
        then z + y = +infty by A6,Def2;
        hence thesis by XXREAL_0:3;
      end;
      suppose
        x = -infty;
        hence thesis by XXREAL_0:5;
      end;
      suppose
        x in REAL & y in REAL;
        then reconsider a = x, b = y, c = z as Element of REAL by A6;
        a - b <= c by A3;
        then a <= b + c by XREAL_1:20;
        hence thesis;
      end;
    end;
    suppose
A8:   x-y = -infty;
      per cases by A8,Th19;
      suppose
        x = -infty;
        hence thesis by XXREAL_0:5;
      end;
      suppose
        y = +infty;
        then z + y = +infty by A6,Def2;
        hence thesis by XXREAL_0:3;
      end;
    end;
  end;
end;

theorem
  (x = +infty & y = +infty implies 0 <= z) & (x = -infty & y = -infty
  implies 0 <= z) implies (x <= z + y implies x - y <= z)
proof
  assume that
A1: x = +infty & y = +infty implies 0 <= z and
A2: x = -infty & y = -infty implies 0 <= z and
A3: x <= z + y;
  per cases by XXREAL_0:14;
  suppose
A4: x = +infty;
    per cases by A3,A4,Lm8,XXREAL_0:4;
    suppose
      z = +infty;
      hence thesis by XXREAL_0:3;
    end;
    suppose
      y = +infty;
      hence thesis by A1,A4,Th7;
    end;
  end;
  suppose
A5: x = -infty;
    per cases;
    suppose
      y = -infty;
      hence thesis by A2,A5,Th7;
    end;
    suppose
      --y <> -infty;
      then -y <> +infty by Def3;
      then x - y = -infty by A5,Def2;
      hence thesis by XXREAL_0:5;
    end;
  end;
  suppose
A6: x in REAL;
    per cases by A3,A6,XXREAL_0:10;
    suppose
A7:   z + y in REAL;
      per cases by A7,Th20;
      suppose
        y = +infty;
        then x - y = -infty by A6,Def2,Th5;
        hence thesis by XXREAL_0:5;
      end;
      suppose
        z = +infty;
        hence thesis by XXREAL_0:3;
      end;
      suppose
        z in REAL & y in REAL;
        then reconsider a = x, b = y, c = z as Element of REAL by A6;
        a <= b + c by A3;
        then a - b <= c by XREAL_1:20;
        hence thesis;
      end;
    end;
    suppose
A8:   z + y = +infty;
      per cases by A8,Lm8;
      suppose
        z = +infty;
        hence thesis by XXREAL_0:3;
      end;
      suppose
        y = +infty;
        then x - y = -infty by A6,Def2,Th5;
        hence thesis by XXREAL_0:5;
      end;
    end;
  end;
end;

theorem Th43:
  z in REAL & x < y implies x + z < y + z & x - z < y - z
proof
  assume that
A1: z in REAL and
A2: x < y;
A3: x + z <> y + z
  proof
    assume x + z = y + z;
    then x = (y + z) - z by A1,Th22
      .= y by A1,Th22;
    hence thesis by A2;
  end;
A4: x - z <> y - z
  proof
    assume x - z = y - z;
    then x = (y - z) + z by A1,Th22
      .= y by A1,Th22;
    hence thesis by A2;
  end;
  x + z <= y + z & x - z <= y - z by A2,Th36;
  hence thesis by A3,A4,XXREAL_0:1;
end;

theorem
  0<= x & 0<= y & 0<= z implies (x + y) + z = x + (y + z)
proof
  assume that
A1: 0<= x and
A2: 0<= y and
A3: 0<= z;
  per cases by A1,A2,A3,XXREAL_0:14;
  suppose
    x in REAL & y in REAL & z in REAL;
    then consider a,b,c,d,e being real number such that
A4: x = a & y = b & z = c and
    x + y = d and
    y + z = e;
    (x + y) + z = (a + b) + c by A4
      .= a + (b + c)
      .= x + (y + z) by A4;
    hence thesis;
  end;
  suppose
A5: x = +infty;
    then (x + y) + z = +infty + z by A2,Def2
      .= +infty by A3,Def2
      .= +infty + (y + z) by A2,A3,Def2;
    hence thesis by A5;
  end;
  suppose
A6: y = +infty;
    then (x + y) + z = +infty + z by A1,Def2
      .= +infty by A3,Def2
      .= x + (+infty) by A1,Def2
      .= x + (+infty + z) by A3,Def2;
    hence thesis by A6;
  end;
  suppose
A7: z = +infty;
    then (x + y) + z = +infty by A1,A2,Def2
      .= x + (+infty) by A1,Def2
      .= x + (y + (+infty)) by A2,Def2;
    hence thesis by A7;
  end;
end;

theorem Th45:
  x is real implies (y + x <= z iff y <= z - x)
proof
  assume
A1: x is real;
A2: (z - x) + x = z
  proof
    per cases by XXREAL_0:14;
    suppose
      z in REAL;
      then reconsider a = x, b = z as real number by A1;
      thus (z - x) + x = (b - a) + a .= z;
    end;
    suppose
A3:   z = -infty;
      hence (z - x) + x = -infty + x by A1,Th14
        .= z by A1,A3,Def2;
    end;
    suppose
A4:   z = +infty;
      hence (z - x) + x = +infty + x by A1,Th13
        .= z by A1,A4,Def2;
    end;
  end;
  hereby
A5: (y + x) - x = y
    proof
      per cases by XXREAL_0:14;
      suppose
        y in REAL;
        then reconsider a = x, b = y as Element of REAL by A1,XXREAL_0:14;
        (y + x) - x = (b + a) - a .= y;
        hence thesis;
      end;
      suppose
A6:     y = -infty or y = +infty;
        per cases by A6;
        suppose
A7:       y = -infty;
          hence (y + x) - x = -infty - x by A1,Def2
            .= y by A1,A7,Th14;
        end;
        suppose
A8:       y = +infty;
          hence (y + x) - x = +infty - x by A1,Def2
            .= y by A1,A8,Th13;
        end;
      end;
    end;
    assume y + x <= z;
    hence y <= z - x by A5,Th37;
  end;
  assume y <= z - x;
  hence thesis by A2,Th36;
end;

theorem Th46:
  0 < x & x < y implies 0 < y - x
proof
  assume that
A1: 0 < x and
A2: x < y;
A3: x in REAL by A1,A2,XXREAL_0:48;
A4: 0 <> y - x
  proof
    assume 0= y - x;
    then x = (y - x) + x by Th4
      .= y by A3,Th22;
    hence thesis by A2;
  end;
  0 + x < y by A2,Th4;
  hence thesis by A3,A4,Th45;
end;

theorem
  0 <= x & 0 <= z & z + x < y implies z < y - x
proof
  assume that
A1: 0 <= x and
A2: 0 <= z and
A3: z + x < y;
  x in REAL
  proof
A4: x <> +infty
    proof
      assume x = +infty;
      then +infty < y by A2,A3,Def2;
      hence thesis by XXREAL_0:4;
    end;
    assume not x in REAL;
    hence thesis by A1,A4,XXREAL_0:10;
  end;
  then z <= y - x & z <> y - x by A3,Th22,Th45;
  hence thesis by XXREAL_0:1;
end;

theorem
  0 <= x & 0 <= z & z + x < y implies z <= y
proof
  assume that
A1: 0 <= x and
A2: 0 <= z and
A3: z + x < y;
  x in REAL
  proof
A4: x <> +infty
    proof
      assume x = +infty;
      then +infty < y by A2,A3,Def2;
      hence thesis by XXREAL_0:4;
    end;
    assume not x in REAL;
    hence thesis by A1,A4,XXREAL_0:10;
  end;
  then
A5: (z + x) - x = z by Th22;
  y - 0 = y by Th15;
  hence thesis by A1,A3,A5,Th37;
end;

theorem Th49:
  0 <= x & x < z implies ex y being real number st 0 < y & x + y < z
proof
  assume that
A1: 0 <= x and
A2: x < z;
  per cases by A1;
  suppose
A3: 0 < x;
    then 0 < z - x by A2,Th46;
    then consider y being real number such that
A4: 0 < y and
A5: y < z - x by Th3;
    take y;
A6: x + y <= x + (z - x) by A5,Th36;
A7: x in REAL by A2,A3,XXREAL_0:48;
    then
A8: x + (z - x) = z by Th22;
    x + y <> z by A7,A5,Th22;
    hence thesis by A4,A6,A8,XXREAL_0:1;
  end;
  suppose
A9: 0 = x;
    consider y being real number such that
A10: 0 < y & y < z by A1,A2,Th3;
    take y;
    thus thesis by A9,A10,Th4;
  end;
end;

theorem
  0 < x implies ex y being real number st 0 < y & y + y < x
proof
  assume 0 < x;
  then consider x1 being real number such that
A1: 0 < x1 and
A2: x1 < x by Th3;
  consider x2 being real number such that
A3: 0 < x2 and
A4: x1 + x2 < x by A1,A2,Th49;
  take y = min(x1,x2);
  per cases;
  suppose
A5: x1 <= x2;
    hence 0 < y by A1,XXREAL_0:def 9;
    y = x1 by A5,XXREAL_0:def 9;
    then y + y <= x1 + x2 by A5,Th36;
    hence thesis by A4,XXREAL_0:2;
  end;
  suppose
A6: x2 <= x1;
    hence 0 < y by A3,XXREAL_0:def 9;
    y = x2 by A6,XXREAL_0:def 9;
    then y + y <= x1 + x2 by A6,Th36;
    hence thesis by A4,XXREAL_0:2;
  end;
end;

theorem Th51:
  x < y & x < +infty & -infty < y implies 0 < y- x
proof
  assume that
A1: x < y and
A2: x < +infty and
A3: -infty < y;
  per cases;
  suppose
    y = +infty;
    hence thesis by A2,Th13;
  end;
  suppose
A4: y <> +infty;
    per cases;
    suppose
      x = -infty;
      hence thesis by A3,Th14;
    end;
    suppose
A5:   x <> -infty;
A6:   y in REAL by A3,A4,XXREAL_0:14;
      x in REAL by A2,A5,XXREAL_0:14;
      then reconsider a = x, b = y as real number by A6;
      b - a > 0 by A1,XREAL_1:50;
      hence thesis;
    end;
  end;
end;

theorem
  not ( x = +infty & y = -infty or x = -infty & y = +infty ) & x + y < z
  implies x <> +infty & y <> +infty & z <> -infty & x < z - y
proof
  assume that
A1: not ( x=+infty & y=-infty or x=-infty & y=+infty ) and
A2: x + y < z;
  per cases;
  suppose
A3: z = +infty;
    then x<>+infty by A1,A2,Def2;
    then
A4: x < +infty by XXREAL_0:4;
    y<>+infty by A1,A2,A3,Def2;
    hence thesis by A3,A4,Th13;
  end;
  suppose
A5: z <> +infty;
A6: -infty <= x + y by XXREAL_0:5;
    then z in REAL by A2,A5,XXREAL_0:14;
    then reconsider c = z as real number;
A7: x + y < +infty by A2,XXREAL_0:2,4;
    then
A8: x <> +infty by A1,Def2;
A9: y <> +infty by A1,A7,Def2;
    per cases;
    suppose
A10:  y = -infty;
      then x < +infty by A1,XXREAL_0:4;
      hence thesis by A2,A6,A10,Th14;
    end;
    suppose
      y <> -infty;
      then y in REAL by A9,XXREAL_0:14;
      then reconsider b = y as real number;
      per cases;
      suppose
A11:    x = -infty;
        hence x <> +infty;
A12:    z - y = c - b;
        hence y <> +infty;
        thus z <> -infty by A12;
        c-b in REAL by XREAL_0:def 1;
        hence thesis by A11,XXREAL_0:12;
      end;
      suppose
        x <> -infty;
        then x in REAL by A8,XXREAL_0:14;
        then reconsider a = x as real number;
        x + y = a + b;
        then a < c - b by A2,XREAL_1:20;
        hence thesis;
      end;
    end;
  end;
end;

theorem
  not ( z = +infty & y = +infty or z = -infty & y = -infty ) & x < z - y
  implies x <> +infty & y <> +infty & z <> -infty & x + y < z
proof
  assume that
A1: not ( z=+infty & y=+infty or z=-infty & y=-infty ) and
A2: x < z - y;
  per cases;
  suppose
A3: x = -infty;
A4: -infty <= z by XXREAL_0:5;
    z<>-infty by A1,A2,A3,Th14;
    then
A5: -infty < z by A4,XXREAL_0:1;
    y<>+infty by A1,A2,A3,Th13;
    hence thesis by A3,A5,Def2;
  end;
  suppose
A6: x <> -infty;
A7: z - y <= +infty by XXREAL_0:4;
    then x in REAL by A2,A6,XXREAL_0:14;
    then reconsider a = x as real number;
A8: -infty <= x by XXREAL_0:5;
    then
A9: z <> -infty by A1,A2,Th14;
A10: y <> +infty by A1,A2,A8,Th13;
    per cases;
    suppose
A11:  y = -infty;
      -infty <= z by XXREAL_0:5;
      then -infty < z by A1,A11,XXREAL_0:1;
      hence thesis by A2,A7,A11,Def2;
    end;
    suppose
      y <> -infty;
      then y in REAL by A10,XXREAL_0:14;
      then reconsider b = y as real number;
      per cases;
      suppose
A12:    z = +infty;
        a+b in REAL by XREAL_0:def 1;
        hence thesis by A12,XXREAL_0:9;
      end;
      suppose
        z <> +infty;
        then z in REAL by A9,XXREAL_0:14;
        then reconsider c = z as real number;
        z - y = c - b;
        then a + b < c by A2,XREAL_1:20;
        hence thesis;
      end;
    end;
  end;
end;

theorem
  not ( x = +infty & y = +infty or x = -infty & y = -infty ) & x - y < z
  implies x <> +infty & y <> -infty & z <> -infty & x < z + y
proof
  assume that
A1: not ( x=+infty & y=+infty or x=-infty & y=-infty ) and
A2: x - y < z;
  per cases;
  suppose
A3: z = +infty;
    then x<>+infty by A1,A2,Th13;
    then
A4: x < +infty by XXREAL_0:4;
    y<>-infty by A1,A2,A3,Th14;
    hence thesis by A3,A4,Def2;
  end;
  suppose
A5: z <> +infty;
A6: -infty <= x - y by XXREAL_0:5;
    then z in REAL by A2,A5,XXREAL_0:14;
    then reconsider c = z as real number;
A7: x - y < +infty by A2,XXREAL_0:2,4;
    then
A8: x <> +infty by A1,Th13;
A9: y <> -infty by A1,A7,Th14;
    per cases;
    suppose
A10:  y = +infty;
      then x < +infty by A1,XXREAL_0:4;
      hence thesis by A2,A6,A10,Def2;
    end;
    suppose
      y <> +infty;
      then y in REAL by A9,XXREAL_0:14;
      then reconsider b = y as real number;
      per cases;
      suppose
A11:    x = -infty;
        c+b in REAL by XREAL_0:def 1;
        hence thesis by A11,XXREAL_0:12;
      end;
      suppose
        x <> -infty;
        then x in REAL by A8,XXREAL_0:14;
        then reconsider a = x as real number;
        x - y = a - b;
        then a < c + b by A2,XREAL_1:19;
        hence thesis;
      end;
    end;
  end;
end;

theorem
  not ( z = +infty & y = -infty or z = -infty & y = +infty ) & x < z + y
  implies x <> +infty & y <> -infty & z <> -infty & x - y < z
proof
  assume that
A1: not ( z=+infty & y=-infty or z=-infty & y=+infty ) and
A2: x < z + y;
  per cases;
  suppose
A3: x = -infty;
A4: -infty <= z by XXREAL_0:5;
    z<>-infty by A1,A2,A3,Def2;
    then
A5: -infty < z by A4,XXREAL_0:1;
    y<>-infty by A1,A2,A3,Def2;
    hence thesis by A3,A5,Th14;
  end;
  suppose
A6: x <> -infty;
A7: z + y <= +infty by XXREAL_0:4;
    then x in REAL by A2,A6,XXREAL_0:14;
    then reconsider a = x as real number;
A8: -infty <= x by XXREAL_0:5;
    then
A9: z <> -infty by A1,A2,Def2;
A10: y <> -infty by A1,A2,A8,Def2;
    per cases;
    suppose
A11:  y = +infty;
      -infty <= z by XXREAL_0:5;
      then -infty < z by A1,A11,XXREAL_0:1;
      hence thesis by A2,A7,A11,Th13;
    end;
    suppose
      y <> +infty;
      then y in REAL by A10,XXREAL_0:14;
      then reconsider b = y as real number;
      per cases;
      suppose
A12:    z = +infty;
        a-b in REAL by XREAL_0:def 1;
        hence thesis by A12,XXREAL_0:9;
      end;
      suppose
        z <> +infty;
        then z in REAL by A9,XXREAL_0:14;
        then reconsider c = z as real number;
        z + y = c + b;
        then a - b < c by A2,XREAL_1:19;
        hence thesis;
      end;
    end;
  end;
end;

theorem
  not ( x = +infty & y = -infty or x = -infty & y = +infty or y = +infty
& z = +infty or y = -infty & z = -infty ) & x + y <= z implies y <> +infty & x
  <= z - y
proof
  assume that
A1: not ( x=+infty & y=-infty or x=-infty & y=+infty or y = +infty & z =
  +infty or y = -infty & z = -infty ) and
A2: x + y <= z;
  thus y <> +infty
  proof
    assume
A3: y = +infty;
    then x + y = +infty by A1,Def2;
    hence contradiction by A1,A2,A3,XXREAL_0:4;
  end;
  per cases;
  suppose
    z = +infty;
    then z - y = +infty by A1,Th13;
    hence thesis by XXREAL_0:4;
  end;
  suppose
A4: z <> +infty;
    then
A5: x + y < +infty by A2,XXREAL_0:2,4;
    then
A6: x <> +infty by A1,Def2;
A7: y <> +infty by A1,A5,Def2;
    per cases;
    suppose
      x = -infty;
      hence thesis by XXREAL_0:5;
    end;
    suppose
      x <> -infty;
      then x in REAL by A6,XXREAL_0:14;
      then reconsider a = x as real number;
      per cases;
      suppose
        y = -infty;
        then z - y = +infty by A1,Th14;
        hence thesis by XXREAL_0:4;
      end;
      suppose
        y <> -infty;
        then y in REAL by A7,XXREAL_0:14;
        then reconsider b = y as real number;
        a+b in REAL by XREAL_0:def 1;
        then z <> -infty by A2,XXREAL_0:12;
        then z in REAL by A4,XXREAL_0:14;
        then reconsider c = z as real number;
        x + y = a + b;
        then a <= c - b by A2,XREAL_1:19;
        hence thesis;
      end;
    end;
  end;
end;

theorem
  not (x = +infty & y = -infty) & not (x = -infty & y = +infty) & not (y
  = +infty & z = +infty) & x <= z - y implies y <> +infty & x + y <= z
proof
  assume that
A1: not (x = +infty & y = -infty) and
A2: not (x = -infty & y = +infty) and
A3: not (y = +infty & z = +infty) and
A4: x <= z - y;
  thus
A5: y <> +infty
  proof
    assume
A6: y = +infty;
    then z - y = -infty by A3,Th13;
    hence contradiction by A2,A4,A6,XXREAL_0:6;
  end;
  per cases by A5;
  suppose
    y = -infty;
    then x + y = -infty by A1,Def2;
    hence thesis by XXREAL_0:5;
  end;
  suppose
A7: y <> +infty & y <> -infty;
    - -infty = +infty by Def3;
    then
A8: -y <> -infty by A7;
    - +infty = -infty by Def3;
    then -y <> +infty by A7;
    then z - y + y = z + (-y + y) by A7,A8,Th29
      .= z + 0 by Th7
      .= z by Th4;
    hence thesis by A4,Th36;
  end;
end;

theorem
  not ( x = +infty & y = +infty or x = -infty & y = -infty or y = +infty
  & z = -infty or y = -infty & z = +infty ) & x - y <= z implies y <> -infty
proof
  assume that
A1: not ( x=+infty & y=+infty or x=-infty & y=-infty or y = +infty & z =
  -infty or y = -infty & z = +infty ) and
A2: x - y <= z;
    assume
A3: y = -infty;
    then x - y = +infty by A1,Th14;
    hence contradiction by A1,A2,A3,XXREAL_0:4;
end;

theorem
  not (x = -infty & y = -infty) & not (y = -infty & z = +infty) & x <= z
  + y implies y <> -infty
proof
  assume that
A1: not(x=-infty & y=-infty) and
A2: not(y = -infty & z = +infty) and
A3: x <= z + y;
    assume
A4: y = -infty;
    then z + y = -infty by A2,Def2;
    hence contradiction by A1,A3,A4,XXREAL_0:6;
end;

theorem
  (x <= -y implies y <= -x) & (-x <= y implies -y <= x)
proof
  x <= -y implies --y <= -x by Th38;
  hence x <= -y implies y <= -x;
  -x <= y implies -y <= --x by Th38;
  hence thesis;
end;

theorem
  (for e be real number st 0 < e holds x < y + e) implies x <= y
proof
  assume
A1: for e be real number st 0 < e holds x < y + e;
  per cases;
  suppose
A2: y = +infty or y=-infty;
    per cases by A2;
    suppose
      y=+infty;
      hence thesis by XXREAL_0:4;
    end;
    suppose
A3:   y=-infty;
      x < y + 1 by A1;
      hence thesis by A3,Def2;
    end;
  end;
  suppose
A4: y <> +infty & y <>-infty;
    per cases;
    suppose
A5:   x = +infty;
      x < y+ 1 by A1;
      hence thesis by A5,XXREAL_0:4;
    end;
    suppose
A6:   x <> +infty;
      now
        assume
A7:     x <> -infty;
        then x in REAL by A6,XXREAL_0:14;
        then reconsider x1=x as real number;
        -infty <= x by XXREAL_0:5;
        then
A8:     -infty < x by A7,XXREAL_0:1;
        y in REAL by A4,XXREAL_0:14;
        then reconsider y1=y as real number;
        assume
A9:     not x <= y;
        y < +infty by A4,XXREAL_0:4;
        then 0 < x- y by A9,A8,Th51;
        then x < y + (x1-y1) by A1;
        then x < y + x-y by Th30;
        then x < x +(y-y) by A4,Th30;
        then x < x +0 by Th7;
        hence contradiction by Th4;
      end;
      hence thesis by XXREAL_0:5;
    end;
  end;
end;

reserve t for ext-real number;

theorem
  t <> -infty & t <> +infty & x < y implies x + t < y + t
proof
  assume that
A1: t <> -infty & t <> +infty and
A2: x < y;
A3: t-t = 0 by Th7;
A4: now
    assume x + t = y + t;
    then x + (t -t) = y + t-t by A1,Th30;
    then x+ 0 = y + (t-t) by A1,A3,Th30;
    then x=y+ 0 by A3,Th4;
    hence contradiction by A2,Th4;
  end;
  x + t <= y + t by A2,Th36;
  hence thesis by A4,XXREAL_0:1;
end;

theorem
  t <> -infty & t <> +infty & x < y implies x - t < y - t
proof
  assume that
A1: t <> -infty & t <> +infty and
A2: x < y;
A3: t-t = 0 by Th7;
A4: now
    assume x - t = y - t;
    then x - (t -t) = y - t+t by A1,Th32;
    then x - 0 = y - (t-t) by A1,A3,Th32;
    then x= y+ 0 by A3,Th4;
    hence contradiction by A2,Th4;
  end;
  x - t <= y - t by A2,Th37;
  hence thesis by A4,XXREAL_0:1;
end;

theorem
  x < y & w < z implies x + w < y + z
proof
  assume that
A2: x < y and
A3: w < z;
  -infty <= w by XXREAL_0:5;
  then per cases by XXREAL_0:1;
  suppose
B2: w = -infty;
B3: y <> -infty & z <> -infty by A2,A3,XXREAL_0:5;
    x <> +infty by A2,XXREAL_0:3;
    then x + w = -infty by B2,Def2;
    hence thesis by B3,Lm9,XXREAL_0:6;
  end;
  suppose
A1: -infty < w;
  per cases;
  suppose
A4: y = +infty;
A5: z <= +infty by XXREAL_0:4;
    y + z = +infty by A1,A3,A4,Def2;
    hence thesis by A2,A3,A4,A5,Lm8,XXREAL_0:4;
  end;
  suppose
A6: y <> +infty;
    -infty <= x by XXREAL_0:5;
    then y in REAL by A2,A6,XXREAL_0:14;
    then
A7: y + w < y + z by A3,Th43;
    z <= +infty by XXREAL_0:4;
    then w in REAL by A1,A3,XXREAL_0:14;
    then x + w < y + w by A2,Th43;
    hence thesis by A7,XXREAL_0:2;
  end;
  end;
end;

theorem
  0 <= x & z + x <= y implies z <= y
proof
  assume 0 <= x;
  then z <= z + x by Th39;
  hence thesis by XXREAL_0:2;
end;

begin 

definition
  let x,y be ext-real number;
  func x * y -> ext-real number means
  :Def5:
  ex a,b being complex number st x = a & y = b & it = a * b
    if x is real & y is real,
  it = +infty if (x is not real or y is not real) &
    (x is positive & y is positive or x is negative & y is negative),
  it = -infty if (x is not real or y is not real) &
    (x is positive & y is negative or x is negative & y is positive)
  otherwise it = 0;
  existence
  proof
    thus x is real & y is real implies ex c being ext-real number, a, b being
    complex number st x = a & y = b & c = a * b
    proof
      assume x is real & y is real;
      then reconsider a = x, b = y as real number;
      take a*b, a, b;
      thus thesis;
    end;
    thus thesis;
  end;
  uniqueness;
  consistency;
  commutativity;
end;

registration
  let x,y be real number, a,b be complex number;
  identify x*y with a*b when x = a, y = b;
  compatibility by Def5;
end;

definition
  let x be ext-real number;
  func x" -> ext-real number means
  :Def6:
  ex a being complex number st x = a & it = a" if x is real otherwise it = 0;
  existence
  proof
    thus x is real implies ex c being ext-real number, a being complex number
    st x = a & c = a"
    proof
      assume x is real;
      then reconsider a = x as real number;
      take a", a;
      thus thesis;
    end;
    thus thesis;
  end;
  uniqueness;
  consistency;
end;

registration
  let x be real number, a be complex number;
  identify x" with a" when x = a;
  compatibility by Def6;
end;

definition
  let x,y be ext-real number;
  func x / y -> ext-real number equals
  x * y";
  coherence;
end;

registration
  let x,y be real number, a,b be complex number;
  identify x/y with a/b when x = a, y = b;
  compatibility
  proof
    reconsider y1=y" as real number;
    assume x = a & y = b;
    hence a / b = x * y1 .= x / y;
  end;
end;

Lm16: x * 0 = 0
proof
  per cases by XXREAL_0:14;
  suppose
    x in REAL;
    then ex a,b being complex number st x = a & 0 = b & x * 0 = a * b by Def5;
    hence thesis;
  end;
  suppose
    x = +infty;
    hence thesis by Def5;
  end;
  suppose
    x = -infty;
    hence thesis by Def5;
  end;
end;

registration
  let x be positive ext-real number, y be negative ext-real number;
  cluster x*y -> negative;
  coherence
  proof
    per cases;
    suppose
      x in REAL & y in REAL;
      then reconsider x,y as real number;
      x*y < 0;
      hence thesis;
    end;
    suppose
      not x in REAL or not y in REAL;
      then x is not real or y is not real by XREAL_0:def 1;
      hence thesis by Def5;
    end;
  end;
end;

registration
  let x,y be negative ext-real number;
  cluster x*y -> positive;
  coherence
  proof
    per cases;
    suppose
      x in REAL & y in REAL;
      then reconsider x,y as real number;
      x*y > 0;
      hence thesis;
    end;
    suppose
      not x in REAL or not y in REAL;
      then x is not real or y is not real by XREAL_0:def 1;
      hence thesis by Def5;
    end;
  end;
end;

registration
  let x,y be positive ext-real number;
  cluster x*y -> positive;
  coherence
  proof
    per cases;
    suppose
      x in REAL & y in REAL;
      then reconsider x,y as real number;
      x*y > 0;
      hence thesis;
    end;
    suppose
      not x in REAL or not y in REAL;
      then x is not real or y is not real by XREAL_0:def 1;
      hence thesis by Def5;
    end;
  end;
end;

registration
  let x be non positive ext-real number, y be non negative ext-real number;
  cluster x*y -> non positive;
  coherence
  proof
    per cases;
    suppose
      x = 0 or y = 0;
      hence x * y <= 0 by Lm16;
    end;
    suppose
      x < 0 & y > 0;
      hence x * y <= 0;
    end;
  end;
end;

registration
  let x,y be non positive ext-real number;
  cluster x*y -> non negative;
  coherence
  proof
    per cases;
    suppose
      x = 0 or y = 0;
      hence x * y >= 0 by Lm16;
    end;
    suppose
      x < 0 & y < 0;
      hence x * y >= 0;
    end;
  end;
end;

registration
  let x,y be non negative ext-real number;
  cluster x*y -> non negative;
  coherence
  proof
    per cases;
    suppose
      x = 0 or y = 0;
      hence x * y >= 0 by Lm16;
    end;
    suppose
      x > 0 & y > 0;
      hence x * y >= 0;
    end;
  end;
end;

registration
  let x be non positive ext-real number;
  cluster x" -> non positive;
  coherence
  proof
    per cases;
    suppose
      x is real;
      then reconsider x as real number;
      x" is non positive;
      hence thesis;
    end;
    suppose
      x is not real;
      hence thesis by Def6;
    end;
  end;
end;

registration
  let x be non negative ext-real number;
  cluster x" -> non negative;
  coherence
  proof
    per cases;
    suppose
      x is real;
      then reconsider x as real number;
      x" is non negative;
      hence thesis;
    end;
    suppose
      x is not real;
      hence thesis by Def6;
    end;
  end;
end;

registration
  let x be non negative ext-real number, y be non positive ext-real number;
  cluster x/y -> non positive;
  coherence;
  cluster y/x -> non positive;
  coherence;
end;

registration
  let x,y be non negative ext-real number;
  cluster x/y -> non negative;
  coherence;
end;

registration
  let x,y be non positive ext-real number;
  cluster x/y -> non negative;
  coherence;
end;

Lm17: x is not real & x*y = 0 implies y = 0
proof
  assume that
A1: x is not real and
A2: x*y = 0;
  not(x is positive & y is positive or x is negative & y is negative) by A2;
  hence thesis by A1,A2;
end;

registration
  let x,y be non zero ext-real number;
  cluster x*y -> non zero;
  coherence
  proof
    per cases;
    suppose
      x is real & y is real;
      then reconsider r=x, s=y as real number;
      assume x*y is zero;
      then r*s = 0;
      hence contradiction;
    end;
    suppose
      x is not real or y is not real;
      hence thesis by Lm17;
    end;
  end;
end;

registration
  let x be zero ext-real number, y be ext-real number;
  cluster x*y -> zero for ext-real number;
  coherence by Lm16;
end;

Lm18: x = 0 implies x*(y*z) = (x*y)*z
proof
  assume
A1: x = 0;
  hence x*(y*z) = 0 .= (x*y)*z by A1;
end;

Lm19: y = 0 implies x*(y*z) = (x*y)*z
proof
  assume
A1: y = 0;
  hence x*(y*z) = 0 .= (x*y)*z by A1;
end;

Lm20: y is not real implies x*(y*z) = (x*y)*z
proof
  assume y is not real;
  then
A1: not y in REAL;
  assume
A2: not thesis;
  then
A3: x <> 0 & z <> 0 by Lm18;
  per cases by A1,XXREAL_0:14;
  suppose
A4: y = -infty;
    per cases by A3;
    suppose
A5:   x is positive & z is positive;
      then x*(-infty*z) = x*-infty by Def5
        .= -infty by A5,Def5
        .= -infty*z by A5,Def5
        .= (x*-infty)*z by A5,Def5;
      hence thesis by A2,A4;
    end;
    suppose
A6:   x is positive & z is negative;
      then x*(-infty*z) = x*+infty by Def5
        .= +infty by A6,Def5
        .= -infty*z by A6,Def5
        .= (x*-infty)*z by A6,Def5;
      hence thesis by A2,A4;
    end;
    suppose
A7:   x is negative & z is positive;
      then x*(-infty*z) = x*-infty by Def5
        .= +infty by A7,Def5
        .= +infty*z by A7,Def5
        .= (x*-infty)*z by A7,Def5;
      hence thesis by A2,A4;
    end;
    suppose
A8:   x is negative & z is negative;
      then x*(-infty*z) = x*+infty by Def5
        .= -infty by A8,Def5
        .= +infty*z by A8,Def5
        .= (x*-infty)*z by A8,Def5;
      hence thesis by A2,A4;
    end;
  end;
  suppose
A9: y = +infty;
    per cases by A3;
    suppose
A10:  x is positive & z is positive;
      then x*(+infty*z) = x*+infty by Def5
        .= +infty by A10,Def5
        .= +infty*z by A10,Def5
        .= (x*+infty)*z by A10,Def5;
      hence thesis by A2,A9;
    end;
    suppose
A11:  x is positive & z is negative;
      then x*(+infty*z) = x*-infty by Def5
        .= -infty by A11,Def5
        .= +infty*z by A11,Def5
        .= (x*+infty)*z by A11,Def5;
      hence thesis by A2,A9;
    end;
    suppose
A12:  x is negative & z is positive;
      then x*(+infty*z) = x*+infty by Def5
        .= -infty by A12,Def5
        .= -infty*z by A12,Def5
        .= (x*+infty)*z by A12,Def5;
      hence thesis by A2,A9;
    end;
    suppose
A13:  x is negative & z is negative;
      then x*(+infty*z) = x*-infty by Def5
        .= +infty by A13,Def5
        .= -infty*z by A13,Def5
        .= (x*+infty)*z by A13,Def5;
      hence thesis by A2,A9;
    end;
  end;
end;

Lm21: x is not real implies x*(y*z) = (x*y)*z
proof
  assume x is not real;
  then
A1: not x in REAL;
  assume
A2: not thesis;
  then
A3: y <> 0 & z <> 0 by Lm18,Lm19;
  per cases by A1,XXREAL_0:14;
  suppose
A4: x = -infty;
    per cases by A3;
    suppose
A5:   y is positive & z is positive;
      then -infty*(y*z) = -infty by Def5
        .= -infty*z by A5,Def5
        .= (-infty*y)*z by A5,Def5;
      hence thesis by A2,A4;
    end;
    suppose
A6:   y is positive & z is negative;
      then -infty*(y*z) = +infty by Def5
        .= -infty*z by A6,Def5
        .= (-infty*y)*z by A6,Def5;
      hence thesis by A2,A4;
    end;
    suppose
A7:   y is negative & z is positive;
      then -infty*(y*z) = +infty by Def5
        .= +infty*z by A7,Def5
        .= (-infty*y)*z by A7,Def5;
      hence thesis by A2,A4;
    end;
    suppose
A8:   y is negative & z is negative;
      then -infty*(y*z) = -infty by Def5
        .= +infty*z by A8,Def5
        .= (-infty*y)*z by A8,Def5;
      hence thesis by A2,A4;
    end;
  end;
  suppose
A9: x = +infty;
    per cases by A3;
    suppose
A10:  y is positive & z is positive;
      then +infty*(y*z) = +infty by Def5
        .= +infty*z by A10,Def5
        .= (+infty*y)*z by A10,Def5;
      hence thesis by A2,A9;
    end;
    suppose
A11:  y is positive & z is negative;
      then +infty*(y*z) = -infty by Def5
        .= +infty*z by A11,Def5
        .= (+infty*y)*z by A11,Def5;
      hence thesis by A2,A9;
    end;
    suppose
A12:  y is negative & z is positive;
      then +infty*(y*z) = -infty by Def5
        .= -infty*z by A12,Def5
        .= (+infty*y)*z by A12,Def5;
      hence thesis by A2,A9;
    end;
    suppose
A13:  y is negative & z is negative;
      then +infty*(y*z) = +infty by Def5
        .= -infty*z by A13,Def5
        .= (+infty*y)*z by A13,Def5;
      hence thesis by A2,A9;
    end;
  end;
end;

theorem Th66:
  x*(y*z) = (x*y)*z
proof
  per cases;
  suppose
    x is real & y is real & z is real;
    then reconsider r=x, s=y, t=z as real number;
    reconsider rs = r*s, sx = s*t as real number;
    thus x*(y*z) = r*sx .= rs*t
      .= (x*y)*z;
  end;
  suppose
    x is not real or z is not real;
    hence thesis by Lm21;
  end;
  suppose
    y is not real;
    hence thesis by Lm20;
  end;
end;



registration
  let r be real number;
  cluster r" -> real;
  coherence;
end;



registration
  let r,s be real number;
  cluster r*s -> real;
  coherence;
  cluster r/s -> real;
  coherence;
end;

registration
  cluster -infty" -> zero;
  coherence by Def6;
  cluster +infty" -> zero;
  coherence by Def6;
end;

theorem 
  (f*g)" = f"*g"
proof
  per cases by XXREAL_0:14;
  suppose
    f in REAL & g in REAL;
    then reconsider f1 = f, g1 = g as real number;
A1: (ex a being complex number st f1 = a & f" = a" )& ex b being complex
    number st g1 = b & g" = b" by Def6;
    then ex a,b being complex number st f" = a & g" = b & f"*g" = a * b by Def5
;
    then f"*g" = (f1*g1)" by A1,XCMPLX_1:204;
    hence thesis;
  end;
  suppose
A2: f = +infty;
    g is positive or g is negative or g = 0;
    then (f*g)" = +infty" or (f*g)" = -infty" or (f*g)" = 0" by A2,Def5;
    hence thesis by A2;
  end;
  suppose
A3: f = -infty;
    g is positive or g is negative or g = 0;
    then (f*g)" = +infty" or (f*g)" = -infty" or (f*g)" = 0" by A3,Def5;
    hence thesis by A3;
  end;
  suppose
A4: g = +infty;
    f is positive or f is negative or f = 0;
    then (f*g)" = +infty" or (f*g)" = -infty" or (f*g)" = 0" by A4,Def5;
    hence thesis by A4;
  end;
  suppose
A5: g = -infty;
    f is positive or f is negative or f = 0;
    then (f*g)" = +infty" or (f*g)" = -infty" or (f*g)" = 0" by A5,Def5;
    hence thesis by A5;
  end;
end;

theorem 
  r <> 0 & r*f = r*g implies f = g
proof
  assume that
A1: r <> 0 and
A2: r*f = r*g;
A3: r is positive or r is negative by A1;
  per cases by XXREAL_0:14;
  suppose
A4: f in REAL;
    now
      assume not g in REAL;
      then g = +infty or g = -infty by XXREAL_0:14;
      hence contradiction by A2,A3,A4,Def5;
    end;
    then
A5: ex c, d being complex number st r = c & g = d & r*g = c*d by Def5;
    ex a, b being complex number st r = a & f = b & r*f = a *b by A4,Def5;
    hence thesis by A1,A2,A5,XCMPLX_1:5;
  end;
  suppose
A6: f = +infty;
    assume f <> g;
    then g in REAL or g = -infty by A6,XXREAL_0:14;
    hence thesis by A2,A3,A6,Def5;
  end;
  suppose
A7: f = -infty;
    assume f <> g;
    then g in REAL or g = +infty by A7,XXREAL_0:14;
    hence thesis by A2,A3,A7,Def5;
  end;
end;

theorem Th69:
  x <> +infty & x <> -infty & x * y = +infty implies y = +infty or y = -infty
proof
  assume that
A1: x <> +infty & x <> -infty and
A2: x * y = +infty;
  assume y <> +infty & y <> -infty;
  then
A3: y in REAL by XXREAL_0:14;
  x in REAL by A1,XXREAL_0:14;
  then reconsider a=x, b=y as real number by A3;
  x * y = a * b;
  hence contradiction by A2;
end;

theorem Th70:
  x <> +infty & x <> -infty & x * y = -infty implies y = +infty or y = -infty
proof
  assume that
A1: x <> +infty & x <> -infty and
A2: x * y = -infty;
  assume y <> +infty & y <> -infty;
  then
A3: y in REAL by XXREAL_0:14;
  x in REAL by A1,XXREAL_0:14;
  then reconsider a=x, b=y as real number by A3;
  x * y = a * b;
  hence contradiction by A2;
end;

Lm22: x*y in REAL implies x in REAL & y in REAL or x*y = 0
proof
  assume
A1: x*y in REAL;
  assume not x in REAL or not y in REAL;
  then
A2: x is not real or y is not real by XREAL_0:def 1;
  assume
A3: x*y <> 0;
  per cases by A3;
  suppose
    x is positive & y is positive or x is negative & y is negative;
    hence contradiction by A1,A2,Def5;
  end;
  suppose
    x is positive & y is negative or x is negative & y is positive;
    hence contradiction by A1,A2,Def5;
  end;
end;

theorem Th71:
  x <= y & 0 <= z implies x*z <= y*z
proof
  assume that
A1: x <= y and
A2: 0 <= z;
  per cases by A2;
  suppose
    z = 0;
    hence thesis;
  end;
  suppose
A3: z > 0;
    per cases;
    suppose
A4:   x = 0;
      then z*y >= 0 by A1,A2;
      hence thesis by A4;
    end;
    suppose
A5:   x <> 0;
      per cases;
      case that
A6:     x*z in REAL & y*z in REAL;
        y*z = 0 implies y = 0 by A3;
        then reconsider x,y,z as Element of REAL by A3,A5,A6,Lm22;
        reconsider p = x*z, q = y*z as Element of REAL by XREAL_0:def 1;
        take p, q;
        thus thesis by A1,A2,XREAL_1:64;
      end;
      case
A7:     not x*z in REAL or not y*z in REAL;
        per cases by A7;
        suppose
A8:       not x*z in REAL;
          per cases by A8,XXREAL_0:14;
          suppose
            x*z = -infty;
            hence thesis;
          end;
          suppose
            x*z = +infty;
            then
A9:         x <> -infty by A3;
A10:        not x in REAL or not y in REAL or not z in REAL by A8,XREAL_0:def 1
;
            per cases by A9,A10,XXREAL_0:14;
            suppose
              y = +infty;
              hence thesis by A3,Def5;
            end;
            suppose
              x = +infty;
              then y = +infty by A1,XXREAL_0:4;
              hence thesis by A3,Def5;
            end;
            suppose
              y = -infty;
              then x = -infty by A1,XXREAL_0:6;
              hence thesis by A3,Def5;
            end;
            suppose that
A11:          not z in REAL and
              x in REAL & y in REAL;
A12:          z = +infty by A3,A11,XXREAL_0:14;
              per cases by A5;
              suppose
                x < 0;
                hence thesis by A12,Def5;
              end;
              suppose
                0 < x;
                hence thesis by A1,A12,Def5;
              end;
            end;
          end;
        end;
        suppose
          not y*z in REAL;
          then
A13:      not x in REAL or not y in REAL or not z in REAL by XREAL_0:def 1;
          per cases by A13,XXREAL_0:14;
          suppose
            x = -infty;
            hence thesis by A3,Def5;
          end;
          suppose
            y = +infty;
            hence thesis by A3,Def5;
          end;
          suppose
            x = +infty;
            then y = +infty by A1,XXREAL_0:4;
            hence thesis by A3,Def5;
          end;
          suppose
            y = -infty;
            then x = -infty by A1,XXREAL_0:6;
            hence thesis by A3,Def5;
          end;
          suppose that
A14:        not z in REAL and
            x in REAL & y in REAL;
A15:        z = +infty by A3,A14,XXREAL_0:14;
            per cases by A5;
            suppose
              x < 0;
              hence thesis by A15,Def5;
            end;
            suppose
              0 < x;
              hence thesis by A1,A15,Def5;
            end;
          end;
        end;
      end;
    end;
  end;
end;

theorem Th72:
  x < y & 0 < z & z <> +infty implies x*z < y*z
proof
  assume that
A1: x < y and
A2: 0 < z and
A3: z <> +infty;
A4: now
A5: x <> +infty & y <> -infty by A1,XXREAL_0:3,5;
    assume
A6: x*z = y*z;
    per cases by A5,XXREAL_0:14;
    suppose
A7:   x in REAL & y in REAL;
      z in REAL by A2,A3,XXREAL_0:14;
      then reconsider x,y,z as real number by A7;
      x*z < y*z by A1,A2,XREAL_1:68;
      hence contradiction by A6;
    end;
    suppose
A8:   y = +infty;
      then y*z = +infty by A2,Def5;
      then x = +infty or x = -infty by A2,A3,A6,Th69;
      hence contradiction by A1,A2,A6,A8;
    end;
    suppose
A9:   x = -infty;
      then x*z = -infty by A2,Def5;
      then y = +infty or y = -infty by A2,A3,A6,Th70;
      hence contradiction by A1,A2,A6,A9;
    end;
  end;
  x*z <= y*z by A1,A2,Th71;
  hence thesis by A4,XXREAL_0:1;
end;

theorem
  x*y in REAL implies x in REAL & y in REAL or x*y = 0 by Lm22;

theorem
  +infty" = 0;

theorem
  -infty" = 0;

theorem
  x/+infty = 0;

theorem
  x/-infty = 0;

theorem Th78:
  x <> -infty & x <> +infty & x <> 0 implies x / x = 1
proof
  assume x <> -infty & x <> +infty;
  then x in REAL by XXREAL_0:14;
  then reconsider a=x as real number;
  assume x <> 0;
  then a/a =1 by XCMPLX_1:60;
  hence thesis;
end;

theorem
  x <= y & 0 < z implies x/z <= y/z by Th71;

theorem
  x < y & 0 < z & z <> +infty implies x/z < y/z
proof
  assume that
A1: x < y and
A2: 0 < z and
A3: z <> +infty;
  per cases by A3,XXREAL_0:14;
  suppose
    z = -infty;
    hence thesis by A2;
  end;
  suppose
    z in REAL;
    then reconsider z as real number;
    z" > 0 by A2;
    hence thesis by A1,Th72;
  end;
end;

theorem Th81:
  1*x = x
proof
  per cases by XXREAL_0:14;
  suppose
    x in REAL;
    then reconsider x, y=1 as real number;
    y*x = x;
    hence thesis;
  end;
  suppose
    x = -infty or x = +infty;
    hence thesis by Def5;
  end;
end;

theorem Th82:
  y" = 0 implies y = +infty or y = -infty or y = 0
proof
  assume
A1: y" = 0;
  assume y <> +infty & y <> -infty;
  then y in REAL by XXREAL_0:14;
  then reconsider y as real number;
  y"" = 0 by A1;
  hence thesis;
end;

theorem Th83:
  0 < y & y <> +infty implies +infty / y = +infty
proof
  assume that
A1: 0<y and
A2: y<>+infty;
  y" <> 0 by A1,A2,Th82;
  hence thesis by A1,Def5;
end;

theorem Th84:
  y < 0 & -infty <> y implies -infty / y = +infty
proof
  assume that
A1: y<0 and
A2: -infty<>y;
  y" <> 0 by A1,A2,Th82;
  hence thesis by A1,Def5;
end;

theorem Th85:
  y < 0 & -infty <> y implies +infty / y = -infty
proof
  assume that
A1: y<0 and
A2: -infty<>y;
  y" <> 0 by A1,A2,Th82;
  hence thesis by A1,Def5;
end;

theorem Th86:
  0 < y & y <> +infty implies -infty / y = -infty
proof
  assume that
A1: 0<y and
A2: y<>+infty;
  y" <> 0 by A1,A2,Th82;
  hence thesis by A1,Def5;
end;

theorem 
  x <> +infty & x <> -infty & x <> 0 implies x*(1/x) = 1 & (1/x)*x = 1
proof
  assume that
A1: x <> +infty & x <> -infty and
A2: x <> 0;
  x in REAL by A1,XXREAL_0:14;
  then reconsider a = x as real number;
  x * (1 / x) = a * (1/a) .= 1 by A2,XCMPLX_1:106;
  hence thesis;
end;

theorem
  -infty <> y & y <> +infty & y <> 0 implies x * y / y = x & x * (y / y) = x
proof
  assume that
A1: -infty <> y and
A2: y <> +infty and
A3: y <> 0;
  reconsider b = y as Element of REAL by A1,A2,XXREAL_0:14;
A4: x * y / y = x
  proof
    per cases;
    suppose
A5:   x = +infty;
      per cases by A3;
      suppose
A6:     0 < y;
        then x * y = +infty by A5,Def5;
        hence thesis by A2,A5,A6,Th83;
      end;
      suppose
A7:     y < 0;
        then x * y = -infty by A5,Def5;
        hence thesis by A1,A5,A7,Th84;
      end;
    end;
    suppose
A8:   x = -infty;
      per cases by A3;
      suppose
A9:     0 <y;
        then x * y = -infty by A8,Def5;
        hence thesis by A2,A8,A9,Th86;
      end;
      suppose
A10:    y < 0;
        then x * y = +infty by A8,Def5;
        hence thesis by A1,A8,A10,Th85;
      end;
    end;
    suppose
      x <> +infty & x <> -infty;
      then x in REAL by XXREAL_0:14;
      then reconsider a = x as real number;
      (x * y)/y = (a * b)/b .= a by A3,XCMPLX_1:89;
      hence thesis;
    end;
  end;
  y / y = 1 by A1,A2,A3,Th78;
  hence thesis by A4,Th81;
end;

theorem
  +infty * y <> 1 & -infty * y <> 1
proof
  y = 0 or 0 < y or y < 0;
  hence thesis by Def5;
end;

theorem
  x * y <> +infty & x * y <> -infty implies x in REAL or y in REAL
proof
  assume that
A1: x * y <> +infty and
A2: x * y <> -infty;
  assume
A3: ( not x in REAL)& not y in REAL;
  per cases by A3,XXREAL_0:14;
  suppose
    x = +infty & y = +infty;
    hence contradiction by A1,Def5;
  end;
  suppose
    x = +infty & y = -infty;
    hence contradiction by A2,Def5;
  end;
  suppose
    x = -infty & y = +infty;
    hence contradiction by A2,Def5;
  end;
  suppose
    x = -infty & y = -infty;
    hence contradiction by A1,Def5;
  end;
end;

begin 

theorem Th91:
  (-1)*x = -x
proof
  per cases by XXREAL_0:14;
  suppose
    x in REAL;
    then reconsider x, y=-1 as real number;
    y*x = -x;
    hence thesis;
  end;
  suppose
A1: x = -infty;
    hence (-1)*x = +infty by Def5
      .= -x by A1,Def3;
  end;
  suppose
A2: x = +infty;
    hence (-1)*x = -infty by Def5
      .= -x by A2,Def3;
  end;
end;

theorem Th92:
  - x*y = (-x)*y
proof
  thus - x*y = (-1)*(x*y) by Th91
    .= (-1)*x*y by Th66
    .= (-x)*y by Th91;
end;

theorem Th93:
  y = -z implies x*(y+z) = x*y +x*z
proof
  assume
A1: y = -z;
  hence x*(y+z) = x*0 by Th7
    .= x*y - x*y by Th7
    .= x*y + x*(-y) by Th92
    .= x*y + x*z by A1;
end;

theorem Th94:
  2*x = x+x
proof
  per cases by XXREAL_0:14;
  suppose
    x in REAL;
    then reconsider x as real number;
    2*x = x+x;
    hence thesis;
  end;
  suppose
A1: x = -infty;
    hence 2*x = -infty by Def5
      .= x+x by A1,Def2;
  end;
  suppose
A2: x = +infty;
    hence 2*x = +infty by Def5
      .= x+x by A2,Def2;
  end;
end;

Lm23: x*(y+y) = x*y+x*y
proof
  thus x*(y+y) = x*(2*y) by Th94
    .= 2*(x*y) by Th66
    .= x*y+x*y by Th94;
end;

Lm24: x*(0+z) = x*0 +x*z
proof
  thus x*(0+z) = x*z by Th4
    .= x*0 +x*z by Th4;
end;

Lm25: 0*(y+z) = 0*y +0*z;

Lm26: x is real & y is real implies x*(y+z) = x*y +x*z
proof
  assume that
A1: x is real and
A2: y is real;
  reconsider r=x, s=y as real number by A1,A2;
A3: x*y = r*s;
  per cases by XXREAL_0:14;
  suppose
    z in REAL;
    then reconsider t=z as real number;
    reconsider u=s+t, v=r*s, w=r*t as real number;
    r*u = v + w;
    hence thesis;
  end;
  suppose
A4: z = -infty;
    then
A5: y+z = -infty by A2,Def2;
    per cases;
    suppose
      x is zero;
      hence thesis by Lm25;
    end;
    suppose
A6:   x is positive;
      hence x*(y+z) = -infty by A5,Def5
        .= x*y + -infty by A3,Def2
        .= x*y +x*z by A4,A6,Def5;
    end;
    suppose
A7:   x is negative;
      hence x*(y+z) = +infty by A5,Def5
        .= x*y + +infty by A3,Def2
        .= x*y +x*z by A4,A7,Def5;
    end;
  end;
  suppose
A8: z = +infty;
    then
A9: y+z = +infty by A2,Def2;
    per cases;
    suppose
      x is zero;
      hence thesis by Lm25;
    end;
    suppose
A10:  x is positive;
      hence x*(y+z) = +infty by A9,Def5
        .= x*y + +infty by A3,Def2
        .= x*y +x*z by A8,A10,Def5;
    end;
    suppose
A11:  x is negative;
      hence x*(y+z) = -infty by A9,Def5
        .= x*y + -infty by A3,Def2
        .= x*y +x*z by A8,A11,Def5;
    end;
  end;
end;

Lm27: x is real & y is not real implies x*(y+z) = x*y +x*z
proof
  assume that
A1: x is real and
A2: y is not real;
  per cases;
  suppose
    z is real;
    hence thesis by A1,Lm26;
  end;
  suppose
    z is not real;
    then
A3: not z in REAL;
A4: not y in REAL by A2;
    per cases by A4,A3,XXREAL_0:14;
    suppose
      y = -infty & z = -infty;
      hence thesis by Lm23;
    end;
    suppose
      y = -infty & z = +infty;
      hence thesis by Th5,Th93;
    end;
    suppose
      y = +infty & z = -infty;
      hence thesis by Th6,Th93;
    end;
    suppose
      y = +infty & z = +infty;
      hence thesis by Lm23;
    end;
  end;
end;

theorem Th95:
  x is real implies x*(y+z) = x*y +x*z
proof
  assume
A1: x is real;
  per cases;
  suppose
    y is real & z is real;
    hence thesis by A1,Lm26;
  end;
  suppose
    y is not real or z is not real;
    hence thesis by A1,Lm27;
  end;
end;

theorem Th96:
  y>=0 & z>=0 implies x*(y+z) = x*y +x*z
proof
  assume
A1: y>=0 & z>=0;
  per cases by A1;
  suppose
    y = 0 or z = 0;
    hence thesis by Lm24;
  end;
  suppose
A2: y>0 & z>0;
    per cases;
    suppose
      x is real;
      hence thesis by Th95;
    end;
    suppose
      x is not real;
      then
A3:   not x in REAL;
      per cases by A3,XXREAL_0:14;
      suppose
A4:     x = -infty;
        hence x*(y+z) = -infty by A2,Def5
          .= -infty + x*z by A2,A4,Def2
          .= x*y + x*z by A2,A4,Def5;
      end;
      suppose
A5:     x = +infty;
        hence x*(y+z) = +infty by A2,Def5
          .= +infty + x*z by A2,A5,Def2
          .= x*y + x*z by A2,A5,Def5;
      end;
    end;
  end;
end;

theorem
  y<=0 & z<=0 implies x*(y+z) = x*y +x*z
proof
  assume
A1: y<=0 & z<=0;
  thus x*(y+z) = --(x*(y+z)) .= -(x*-(y+z)) by Th92
    .= -(x*(-y+-z)) by Th9
    .= -(x*(-y)+x*(-z)) by A1,Th96
    .= -(-x*y+x*(-z)) by Th92
    .= -(-x*y+-x*z) by Th92
    .= --(x*y +x*z) by Th9
    .= x*y +x*z;
end;

theorem
  x*(0+z) = x*0 +x*z by Lm24;

theorem 
  (-f)" = -f"
proof
  per cases by XXREAL_0:14;
  suppose
A1: f in REAL;
    then reconsider g = f as real number;
A2: -f = -g;
    consider a being complex number such that
A3: f = a and
A4: f" = a" by A1,Def6;
A5: (-a)" = -a" by XCMPLX_1:222;
    ex m being complex number st -f = m & -f" = m"
    proof
      take -a;
      thus -f = -a by A3,A2;
      thus thesis by A4,A5,A2,Def3;
    end;
    hence thesis by A2,Def6;
  end;
  suppose
A6: f = +infty;
    hence (-f)" = (-infty)" by Def3
      .= -f" by A6;
  end;
  suppose
A7: f = -infty;
    hence (-f)" = (+infty)" by Def3
      .= -f" by A7;
  end;
end;

theorem
  x is real implies x * (y - z) = x * y - x * z
proof
  assume x is real;
  then x * (y - z) = x * y + x * (-z) by Th95
    .= x * y + -(x * z) by Th92;
  hence thesis;
end;

theorem Th101:
  x <= y & z <= 0 implies y*z <= x*z
proof
  assume x <= y & z <= 0;
  then
A1: x*(-z) <= y*(-z) by Th71;
  -x*z = x*(-z) & -y*z = y*(-z) by Th92;
  hence thesis by A1,Th38;
end;

theorem Th102:
  x < y & z < 0 & z <> -infty implies y*z < x*z
proof
  assume x < y & z < 0 & z <> -infty;
  then
A1: x*(-z) < y*(-z) by Th5,Th10,Th72;
  -x*z = x*(-z) & -y*z = y*(-z) by Th92;
  hence thesis by A1,Th38;
end;

theorem
  x <= y & z < 0 implies y/z <= x/z by Th101;

theorem
  x < y & z < 0 & z <> -infty implies y/z < x/z
proof
  assume that
A1: x < y and
A2: 0 > z and
A3: z <> -infty;
  per cases by A3,XXREAL_0:14;
  suppose
    z = +infty;
    hence thesis by A2;
  end;
  suppose
    z in REAL;
    then reconsider z as real number;
    z" < 0 by A2;
    hence thesis by A1,Th102;
  end;
end;

theorem
  x/2 + x/2 = x
proof
  x/2 + x/2 = (x+x)/2 by Th95
    .= 2*x/2 by Th94
    .= x*(2/2) by Th66
    .= x by Th81;
  hence thesis;
end;














begin

reserve w, w1, w2 for Element of ExtREAL;
reserve c, c1, c2 for Element of COMPLEX;
reserve A, B, C, D for complex-membered set;
reserve F, G, H, I for ext-real-membered set;
reserve a, b, s, t, z for complex number;
reserve f, g, h, i, j for ext-real number;
reserve r for real number;
reserve e for set;

definition
  let w;
  redefine func -w -> Element of ExtREAL;
  coherence by XXREAL_0:def 1;
  redefine func w" -> Element of ExtREAL;
  coherence by XXREAL_0:def 1;
  let w1;
  redefine func w*w1 -> Element of ExtREAL;
  coherence by XXREAL_0:def 1;
end;

registration
  let a, b, c, d be complex number;
  cluster {a,b,c,d} -> complex-membered;
  coherence
  proof
    let e;
    thus thesis by ENUMSET1:def 2;
  end;
end;

registration
  let a, b, c, d be ext-real number;
  cluster {a,b,c,d} -> ext-real-membered;
  coherence
  proof
    let e;
    thus thesis by ENUMSET1:def 2;
  end;
end;

definition
  let F be ext-real-membered set;
  func --F -> ext-real-membered set equals
  {-w: w in F};
  coherence
  proof
    {-w: w in F} is ext-real-membered
    proof
      let e;
      assume e in {-w: w in F};
      then ex w st e = -w & w in F;
      hence thesis;
    end;
    hence thesis;
  end;
  involutiveness
  proof
    let A, B be ext-real-membered set;
    assume
A1: A = {-w: w in B};
    thus B c= {-w: w in A}
    proof
      let z be ext-real number;
A2:   z in ExtREAL by XXREAL_0:def 1;
A3:   -z in ExtREAL & z = - -z by XXREAL_0:def 1;
      assume z in B;
      then -z in A by A1,A2;
      hence thesis by A3;
    end;
    let e;
    assume e in {-w: w in A};
    then consider w1 such that
A4: -w1 = e and
A5: w1 in A;
    ex w st -w = w1 & w in B by A1,A5;
    hence thesis by A4;
  end;
end;

theorem Th1:
  f in F iff -f in --F
proof
  f in ExtREAL by XXREAL_0:def 1;
  hence f in F implies -f in --F;
  assume -f in --F;
  then
A1: ex w st -w = -f & w in F;
  - -f = f;
  hence thesis by A1;
end;

theorem Th2:
  -f in F iff f in --F
proof
  - -f = f;
  hence thesis by Th1;
end;

registration
  let F be empty set;
  cluster --F -> empty;
  coherence
  proof
    assume --F is non empty;
    then the Element of --F in --F;
    then ex w st the Element of --F = -w & w in F;
    hence thesis;
  end;
end;

registration
  let F be ext-real-membered non empty set;
  cluster --F -> non empty;
  coherence
  proof
    -the Element of F in --F by Th1;
    hence thesis;
  end;
end;

Lm1: F c= G implies --F c= --G
proof
  assume
A1: F c= G;
  let j;
  assume j in --F;
  then -j in F by Th2;
  hence thesis by A1,Th2;
end;

theorem Th3:
  F c= G iff --F c= --G
proof
  ----F = F & ----G = G;
  hence thesis by Lm1;
end;

theorem
  --F = --G implies F = G
proof
  assume --F = --G;
  then F c= G & G c= F by Th3;
  hence thesis by XBOOLE_0:def 10;
end;

theorem Th5:
  -- (F \/ G) = (--F) \/ (--G)
proof
  let i;
  hereby
    assume i in --(F\/G);
    then -i in F \/ G by Th2;
    then -i in F or -i in G by XBOOLE_0:def 3;
    then i in --F or i in --G by Th2;
    hence i in --F \/ --G by XBOOLE_0:def 3;
  end;
  assume i in --F \/ --G;
  then i in --F or i in --G by XBOOLE_0:def 3;
  then -i in F or -i in G by Th2;
  then -i in F \/ G by XBOOLE_0:def 3;
  hence thesis by Th2;
end;

theorem Th6:
  -- (F /\ G) = (--F) /\ (--G)
proof
  let i;
  hereby
    assume i in --(F/\G);
    then
A1: -i in F /\ G by Th2;
    then -i in G by XBOOLE_0:def 4;
    then
A2: i in --G by Th2;
    -i in F by A1,XBOOLE_0:def 4;
    then i in --F by Th2;
    hence i in (--F) /\ --G by A2,XBOOLE_0:def 4;
  end;
  assume
A3: i in (--F) /\ --G;
  then i in --G by XBOOLE_0:def 4;
  then
A4: -i in G by Th2;
  i in --F by A3,XBOOLE_0:def 4;
  then -i in F by Th2;
  then -i in F /\ G by A4,XBOOLE_0:def 4;
  hence thesis by Th2;
end;

theorem Th7:
  -- (F \ G) = (--F) \ (--G)
proof
  let i;
  hereby
    assume i in --(F\G);
    then
A1: -i in F \ G by Th2;
    then not -i in G by XBOOLE_0:def 5;
    then
A2: not i in --G by Th2;
    i in --F by A1,Th2;
    hence i in (--F) \ --G by A2,XBOOLE_0:def 5;
  end;
  assume
A3: i in (--F) \ --G;
  then not i in --G by XBOOLE_0:def 5;
  then
A4: not -i in G by Th2;
  -i in F by A3,Th2;
  then -i in F \ G by A4,XBOOLE_0:def 5;
  hence thesis by Th2;
end;

theorem Th8:
  -- (F \+\ G) = (--F) \+\ (--G)
proof
  thus -- (F \+\ G) = --(F\G) \/ --(G\F) by Th5
    .= --F \ --G \/ --(G\F) by Th7
    .= (--F) \+\ --G by Th7;
end;

theorem Th9:
  --{f} = {-f}
proof
  let i;
  hereby
    assume i in --{f};
    then consider w such that
A1: i = -w and
A2: w in {f};
    w = f by A2,TARSKI:def 1;
    hence i in {-f} by A1,TARSKI:def 1;
  end;
  assume i in {-f};
  then
A3: i = -f by TARSKI:def 1;
  f in ExtREAL & f in {f} by TARSKI:def 1,XXREAL_0:def 1;
  hence thesis by A3;
end;

theorem Th10:
  --{f,g} = {-f,-g}
proof
  thus --{f,g} = --({f}\/{g}) by ENUMSET1:1
    .= --{f} \/ --{g} by Th5
    .= {-f} \/ --{g} by Th9
    .= {-f} \/ {-g} by Th9
    .= {-f,-g} by ENUMSET1:1;
end;

definition
  let A be complex-membered set;
  func --A -> complex-membered set equals
  {-c: c in A};
  coherence
  proof
    {-c: c in A} is complex-membered
    proof
      let e;
      assume e in {-c: c in A};
      then ex c st e = -c & c in A;
      hence thesis;
    end;
    hence thesis;
  end;
  involutiveness
  proof
    let A, B;
    assume
A1: A = {-c: c in B};
    thus B c= {-c: c in A}
    proof
      let z;
A2:   z in COMPLEX by XCMPLX_0:def 2;
A3:   -z in COMPLEX & z = - -z by XCMPLX_0:def 2;
      assume z in B;
      then -z in A by A1,A2;
      hence thesis by A3;
    end;
    let e;
    assume e in {-c: c in A};
    then consider r0 being Element of COMPLEX such that
A4: -r0 = e and
A5: r0 in A;
    ex c st -c = r0 & c in B by A1,A5;
    hence thesis by A4;
  end;
end;

theorem Th11:
  a in A iff -a in --A
proof
  a in COMPLEX by XCMPLX_0:def 2;
  hence a in A implies -a in --A;
  assume -a in --A;
  then ex c st -c = -a & c in A;
  hence thesis;
end;

theorem Th12:
  -a in A iff a in --A
proof
  - -a = a;
  hence thesis by Th11;
end;

registration
  let A be empty set;
  cluster --A -> empty;
  coherence
  proof
    assume --A is non empty;
    then the Element of --A in --A;
    then ex c st the Element of --A = -c & c in A;
    hence thesis;
  end;
end;

registration
  let A be complex-membered non empty set;
  cluster --A -> non empty;
  coherence
  proof
    -the Element of A in --A by Th11;
    hence thesis;
  end;
end;

registration
  let A be real-membered set;
  cluster --A -> real-membered;
  coherence
  proof
    let e;
    assume e in --A;
    then ex c st e = -c & c in A;
    hence thesis;
  end;
end;

registration
  let A be rational-membered set;
  cluster --A -> rational-membered;
  coherence
  proof
    let e;
    assume e in --A;
    then ex c st e = -c & c in A;
    hence thesis;
  end;
end;

registration
  let A be integer-membered set;
  cluster --A -> integer-membered;
  coherence
  proof
    let e;
    assume e in --A;
    then ex c st e = -c & c in A;
    hence thesis;
  end;
end;

registration
  let A be real-membered set, F be ext-real-membered set;
  identify --A with --F when A = F;
  compatibility
  proof
    assume
A1: A = F;
    let a be ext-real number;
    hereby
      assume
A2:   a in --A;
      then reconsider b = a as real number;
      -b in A by A2,Th12;
      hence a in --F by A1,Th2;
    end;
    assume a in --F;
    then consider w such that
A3: a = -w and
A4: w in F;
    reconsider b = w as real number by A1,A4;
    -b in COMPLEX by XCMPLX_0:def 2;
    then -a in COMPLEX by A3,XCMPLX_0:def 2;
    then -b in --A by A1,A3,A4;
    hence thesis by A3;
  end;
end;

Lm2: A c= B implies --A c= --B
proof
  assume
A1: A c= B;
  let a;
  assume a in --A;
  then -a in A by Th12;
  hence thesis by A1,Th12;
end;

theorem Th13:
  A c= B iff --A c= --B
proof
  ----A = A & ----B = B;
  hence thesis by Lm2;
end;

theorem
  --A = --B implies A = B
proof
  assume --A = --B;
  then A c= B & B c= A by Th13;
  hence thesis by XBOOLE_0:def 10;
end;

theorem Th15:
  -- (A \/ B) = (--A) \/ (--B)
proof
  let z;
  hereby
    assume z in --(A\/B);
    then -z in A \/ B by Th12;
    then -z in A or -z in B by XBOOLE_0:def 3;
    then z in --A or z in --B by Th12;
    hence z in --A \/ --B by XBOOLE_0:def 3;
  end;
  assume z in --A \/ --B;
  then z in --A or z in --B by XBOOLE_0:def 3;
  then -z in A or -z in B by Th12;
  then -z in A \/ B by XBOOLE_0:def 3;
  hence thesis by Th12;
end;

theorem Th16:
  -- (A /\ B) = (--A) /\ (--B)
proof
  let z;
  hereby
    assume z in --(A/\B);
    then
A1: -z in A /\ B by Th12;
    then -z in B by XBOOLE_0:def 4;
    then
A2: z in --B by Th12;
    -z in A by A1,XBOOLE_0:def 4;
    then z in --A by Th12;
    hence z in (--A) /\ --B by A2,XBOOLE_0:def 4;
  end;
  assume
A3: z in (--A) /\ --B;
  then z in --B by XBOOLE_0:def 4;
  then
A4: -z in B by Th12;
  z in --A by A3,XBOOLE_0:def 4;
  then -z in A by Th12;
  then -z in A /\ B by A4,XBOOLE_0:def 4;
  hence thesis by Th12;
end;

theorem Th17:
  -- (A \ B) = (--A) \ (--B)
proof
  let z;
  hereby
    assume z in --(A\B);
    then
A1: -z in A \ B by Th12;
    then not -z in B by XBOOLE_0:def 5;
    then
A2: not z in --B by Th12;
    z in --A by A1,Th12;
    hence z in (--A) \ --B by A2,XBOOLE_0:def 5;
  end;
  assume
A3: z in (--A) \ --B;
  then not z in --B by XBOOLE_0:def 5;
  then
A4: not -z in B by Th12;
  -z in A by A3,Th12;
  then -z in A \ B by A4,XBOOLE_0:def 5;
  hence thesis by Th12;
end;

theorem Th18:
  -- (A \+\ B) = (--A) \+\ (--B)
proof
  thus -- (A \+\ B) = --(A\B) \/ --(B\A) by Th15
    .= --A \ --B \/ --(B\A) by Th17
    .= (--A) \+\ --B by Th17;
end;

theorem Th19:
  --{a} = {-a}
proof
  let z;
  hereby
    assume z in --{a};
    then consider c such that
A1: z = -c and
A2: c in {a};
    c = a by A2,TARSKI:def 1;
    hence z in {-a} by A1,TARSKI:def 1;
  end;
  assume z in {-a};
  then
A3: z = -a by TARSKI:def 1;
  a in COMPLEX & a in {a} by TARSKI:def 1,XCMPLX_0:def 2;
  hence thesis by A3;
end;

theorem Th20:
  --{a,b} = {-a,-b}
proof
  thus --{a,b} = --({a}\/{b}) by ENUMSET1:1
    .= --{a} \/ --{b} by Th15
    .= {-a} \/ --{b} by Th19
    .= {-a} \/ {-b} by Th19
    .= {-a,-b} by ENUMSET1:1;
end;

definition
  let F be ext-real-membered set;
  func F"" -> ext-real-membered set equals
  {w": w in F};
  coherence
  proof
    {w": w in F} is ext-real-membered
    proof
      let e;
      assume e in {w": w in F};
      then ex w st e = w" & w in F;
      hence thesis;
    end;
    hence thesis;
  end;
end;

theorem Th21:
  f in F implies f" in F""
proof
  f in ExtREAL by XXREAL_0:def 1;
  hence thesis;
end;

registration
  let F be empty set;
  cluster F"" -> empty;
  coherence
  proof
    assume F"" is non empty;
    then the Element of F"" in F"";
    then ex w st the Element of F"" = w" & w in F;
    hence thesis;
  end;
end;

registration
  let F be ext-real-membered non empty set;
  cluster F"" -> non empty;
  coherence
  proof
    (the Element of F)" in F"" by Th21;
    hence thesis;
  end;
end;

theorem
  F c= G implies F"" c= G""
proof
  assume
A1: F c= G;
  let i;
  assume i in F"";
  then ex w st i = w" & w in F;
  hence thesis by A1;
end;

theorem Th23:
  (F \/ G)"" = (F"") \/ (G"")
proof
  let i;
  hereby
    assume i in (F\/G)"";
    then consider w such that
A1: i = w" and
A2: w in F \/ G;
    w in F or w in G by A2,XBOOLE_0:def 3;
    then w" in F"" or w" in G"";
    hence i in F"" \/ G"" by A1,XBOOLE_0:def 3;
  end;
  assume
A3: i in F"" \/ G"";
  per cases by A3,XBOOLE_0:def 3;
  suppose
    i in F"";
    then consider w such that
A4: i = w" and
A5: w in F;
    w in F\/G by A5,XBOOLE_0:def 3;
    hence thesis by A4;
  end;
  suppose
    i in G"";
    then consider w such that
A6: i = w" and
A7: w in G;
    w in F\/G by A7,XBOOLE_0:def 3;
    hence thesis by A6;
  end;
end;

theorem Th24:
  (F /\ G)"" c= (F"") /\ (G"")
proof
  let i;
  assume i in (F/\G)"";
  then consider w such that
A1: i = w" and
A2: w in F /\ G;
  w in G by A2,XBOOLE_0:def 4;
  then
A3: w" in G"";
  w in F by A2,XBOOLE_0:def 4;
  then w" in F"";
  hence thesis by A1,A3,XBOOLE_0:def 4;
end;

theorem
  --(F"") = (--F)""
proof
  let i;
  hereby
    assume i in --F"";
    then -i in F"" by Th2;
    then consider w such that
A1: -i = w" and
A2: w in F;
    (-w)" = -w" & -w in --F by A2,XXREAL_3:99;
    hence i in (--F)"" by A1;
  end;
  assume i in (--F)"";
  then consider w such that
A3: i = w" and
A4: w in --F;
  (-w)" = -w" & -w in F by A4,Th2,XXREAL_3:99;
  then -i in F"" by A3;
  hence thesis by Th2;
end;

theorem Th26:
  {f}"" = {f"}
proof
  let i;
  hereby
    assume i in {f}"";
    then consider w such that
A1: i = w" and
A2: w in {f};
    w = f by A2,TARSKI:def 1;
    hence i in {f"} by A1,TARSKI:def 1;
  end;
  assume i in {f"};
  then
A3: i = f" by TARSKI:def 1;
  f in ExtREAL & f in {f} by TARSKI:def 1,XXREAL_0:def 1;
  hence thesis by A3;
end;

theorem Th27:
  {f,g}"" = {f",g"}
proof
  thus {f,g}"" = ({f}\/{g})"" by ENUMSET1:1
    .= ({f}"") \/ ({g}"") by Th23
    .= {f"} \/ ({g}"") by Th26
    .= ({f"}) \/ {g"} by Th26
    .= {f",g"} by ENUMSET1:1;
end;

definition
  let A be complex-membered set;
  func A"" -> complex-membered set equals
  {c": c in A};
  coherence
  proof
    {c": c in A} is complex-membered
    proof
      let e;
      assume e in {c": c in A};
      then ex c st e = c" & c in A;
      hence thesis;
    end;
    hence thesis;
  end;
  involutiveness
  proof
    let A, B;
    assume
A1: A = {c": c in B};
    thus B c= {c": c in A}
    proof
      let z;
A2:   z in COMPLEX by XCMPLX_0:def 2;
A3:   z" in COMPLEX & z = z" " by XCMPLX_0:def 2;
      assume z in B;
      then z" in A by A1,A2;
      hence thesis by A3;
    end;
    let e;
    assume e in {c": c in A};
    then consider r0 being Element of COMPLEX such that
A4: r0" = e and
A5: r0 in A;
    ex c st c" = r0 & c in B by A1,A5;
    hence thesis by A4;
  end;
end;

theorem Th28:
  a in A iff a" in A""
proof
  a in COMPLEX by XCMPLX_0:def 2;
  hence a in A implies a" in A"";
  assume a" in A"";
  then ex c st c" = a" & c in A;
  hence thesis by XCMPLX_1:201;
end;

theorem Th29:
  a" in A iff a in A""
proof
  a" " = a;
  hence thesis by Th28;
end;

registration
  let A be empty set;
  cluster A"" -> empty;
  coherence
  proof
    assume A"" is non empty;
    then the Element of A"" in A"";
    then ex c st the Element of A"" = c" & c in A;
    hence thesis;
  end;
end;

registration
  let A be complex-membered non empty set;
  cluster A"" -> non empty;
  coherence
  proof
    (the Element of A)" in A"" by Th28;
    hence thesis;
  end;
end;

registration
  let A be real-membered set;
  cluster A"" -> real-membered;
  coherence
  proof
    let e;
    assume e in A"";
    then ex c st e = c" & c in A;
    hence thesis;
  end;
end;

registration
  let A be rational-membered set;
  cluster A"" -> rational-membered;
  coherence
  proof
    let e;
    assume e in A"";
    then ex c st e = c" & c in A;
    hence thesis;
  end;
end;

registration
  let A be real-membered set, F be ext-real-membered set;
  identify A"" with F"" when A = F;
  compatibility
  proof
    assume
A1: A = F;
    let a be ext-real number;
    hereby
      assume a in A"";
      then consider c such that
A2:   a = c" and
A3:   c in A;
      reconsider w = c as Element of ExtREAL by A3,XXREAL_0:def 1;
      ex m being complex number st w = m & w" = m" by A3,XXREAL_3:def 6;
      hence a in F"" by A1,A2,A3;
    end;
    assume a in F"";
    then consider w such that
A4: a = w" and
A5: w in F;
    reconsider c = w as Element of COMPLEX by A1,A5,XCMPLX_0:def 2;
    w" = c" by A1,A5,XXREAL_3:def 6;
    hence thesis by A1,A4,A5;
  end;
end;

Lm3: A c= B implies A"" c= B""
proof
  assume
A1: A c= B;
  let a;
  assume a in A"";
  then a" in A by Th29;
  hence thesis by A1,Th29;
end;

theorem Th30:
  A c= B iff A"" c= B""
proof
  A"""" = A & B"""" = B;
  hence thesis by Lm3;
end;

theorem
  A"" = B"" implies A = B
proof
  assume A"" = B"";
  then A c= B & B c= A by Th30;
  hence thesis by XBOOLE_0:def 10;
end;

theorem Th32:
  (A \/ B)"" = (A"") \/ (B"")
proof
  let a;
  hereby
    assume a in (A\/B)"";
    then a" in A\/B by Th29;
    then a" in A or a" in B by XBOOLE_0:def 3;
    then a in A"" or a in B"" by Th29;
    hence a in A"" \/ B"" by XBOOLE_0:def 3;
  end;
  assume a in A"" \/ B"";
  then a in A"" or a in B"" by XBOOLE_0:def 3;
  then a" in A or a" in B by Th29;
  then a" in A\/B by XBOOLE_0:def 3;
  hence thesis by Th29;
end;

theorem Th33:
  (A /\ B)"" = (A"") /\ (B"")
proof
  let a;
  hereby
    assume a in (A/\B)"";
    then
A1: a" in A/\B by Th29;
    then a" in B by XBOOLE_0:def 4;
    then
A2: a in B"" by Th29;
    a" in A by A1,XBOOLE_0:def 4;
    then a in A"" by Th29;
    hence a in (A"") /\ (B"") by A2,XBOOLE_0:def 4;
  end;
  assume
A3: a in (A"") /\ (B"");
  then a in B"" by XBOOLE_0:def 4;
  then
A4: a" in B by Th29;
  a in A"" by A3,XBOOLE_0:def 4;
  then a" in A by Th29;
  then a" in A/\B by A4,XBOOLE_0:def 4;
  hence thesis by Th29;
end;

theorem Th34:
  (A \ B)"" = (A"") \ (B"")
proof
  let a;
  hereby
    assume a in (A\B)"";
    then
A1: a" in A\B by Th29;
    then not a" in B by XBOOLE_0:def 5;
    then
A2: not a in B"" by Th29;
    a in A"" by A1,Th29;
    hence a in (A"") \ (B"") by A2,XBOOLE_0:def 5;
  end;
  assume
A3: a in (A"") \ (B"");
  then not a in B"" by XBOOLE_0:def 5;
  then
A4: not a" in B by Th29;
  a" in A by A3,Th29;
  then a" in A\B by A4,XBOOLE_0:def 5;
  hence thesis by Th29;
end;

theorem Th35:
  (A \+\ B)"" = (A"") \+\ (B"")
proof
  thus (A \+\ B)"" = ((A\B)"") \/ ((B\A)"") by Th32
    .= ((A"")\(B"")) \/ ((B\A)"") by Th34
    .= (A"") \+\ (B"") by Th34;
end;

theorem Th36:
  --(A"") = (--A)""
proof
  let a;
A1: (-a)" = -a" by XCMPLX_1:222;
  hereby
    assume a in --A"";
    then -a in A"" by Th12;
    then (-a)" in A by Th29;
    then a" in --A by A1,Th12;
    hence a in (--A)"" by Th29;
  end;
  assume a in (--A)"";
  then a" in --A by Th29;
  then -a" in A by Th12;
  then -a in A"" by A1,Th29;
  hence thesis by Th12;
end;

theorem Th37:
  {a}"" = {a"}
proof
  let z;
  hereby
    assume z in {a}"";
    then consider c such that
A1: z = c" and
A2: c in {a};
    c = a by A2,TARSKI:def 1;
    hence z in {a"} by A1,TARSKI:def 1;
  end;
  assume z in {a"};
  then
A3: z = a" by TARSKI:def 1;
  a in COMPLEX & a in {a} by TARSKI:def 1,XCMPLX_0:def 2;
  hence thesis by A3;
end;

theorem Th38:
  {a,b}"" = {a",b"}
proof
  let z;
  hereby
    assume z in {a,b}"";
    then consider c such that
A1: z = c" and
A2: c in {a,b};
    c = a or c = b by A2,TARSKI:def 2;
    hence z in {a",b"} by A1,TARSKI:def 2;
  end;
A3: a in {a,b} & b in {a,b} by TARSKI:def 2;
  assume z in {a",b"};
  then
A4: z = a" or z = b" by TARSKI:def 2;
  a in COMPLEX & b in COMPLEX by XCMPLX_0:def 2;
  hence thesis by A4,A3;
end;

definition
  let F, G be ext-real-membered set;
  func F++G equals
  {w1+w2: w1 in F & w2 in G};
  coherence;
  commutativity
  proof
    let X be set;
    let F, G;
    assume
A1: X = {w1+w2: w1 in F & w2 in G};
    thus X c= {w1+w2: w1 in G & w2 in F}
    proof
      let e;
      assume e in X;
      then ex w1,w2 st e = w1+w2 & w1 in F & w2 in G by A1;
      hence thesis;
    end;
    let e;
    assume e in {w1+w2: w1 in G & w2 in F};
    then ex w1,w2 st e = w1+w2 & w1 in G & w2 in F;
    hence thesis by A1;
  end;
end;

theorem Th39:
  f in F & g in G implies f+g in F++G
proof
  f in ExtREAL & g in ExtREAL by XXREAL_0:def 1;
  hence thesis;
end;

registration
  let F be empty set;
  let G be ext-real-membered set;
  cluster F++G -> empty;
  coherence
  proof
    assume F++G is non empty;
    then the Element of (F++G) in F++G;
    then ex w1,w2 st the Element of (F++G) = w1+w2 & w1 in F & w2 in G;
    hence thesis;
  end;
  cluster G++F -> empty;
  coherence;
end;

registration
  let F, G be ext-real-membered non empty set;
  cluster F++G -> non empty;
  coherence
  proof
    (the Element of F)+the Element of G in F++G by Th39;
    hence thesis;
  end;
end;

registration
  let F, G be ext-real-membered set;
  cluster F++G -> ext-real-membered;
  coherence
  proof
    let e;
    assume e in F++G;
    then ex w1,w2 st e = w1+w2 & w1 in F & w2 in G;
    hence thesis;
  end;
end;

theorem
  F c= G & H c= I implies F++H c= G++I
proof
  assume
A1: F c= G & H c= I;
  let i;
  assume i in F++H;
  then ex w,w1 st i = w+w1 & w in F & w1 in H;
  hence thesis by A1;
end;

theorem Th41:
  F ++ (G \/ H) = (F++G) \/ (F++H)
proof
  let j;
  hereby
    assume j in F++(G\/H);
    then consider w,w1 such that
A1: j = w+w1 and
A2: w in F and
A3: w1 in G\/H;
    w1 in G or w1 in H by A3,XBOOLE_0:def 3;
    then w+w1 in F++G or w+w1 in F++H by A2;
    hence j in (F++G)\/(F++H) by A1,XBOOLE_0:def 3;
  end;
  assume
A4: j in (F++G)\/(F++H);
  per cases by A4,XBOOLE_0:def 3;
  suppose
    j in F++G;
    then consider w,w1 such that
A5: j = w+w1 & w in F and
A6: w1 in G;
    w1 in G\/H by A6,XBOOLE_0:def 3;
    hence thesis by A5;
  end;
  suppose
    j in F++H;
    then consider w,w1 such that
A7: j = w+w1 & w in F and
A8: w1 in H;
    w1 in G\/H by A8,XBOOLE_0:def 3;
    hence thesis by A7;
  end;
end;

theorem Th42:
  F ++ (G /\ H) c= (F++G) /\ (F++H)
proof
  let j;
  assume j in F++(G/\H);
  then consider w,w1 such that
A1: j = w+w1 and
A2: w in F and
A3: w1 in G/\H;
  w1 in H by A3,XBOOLE_0:def 4;
  then
A4: w+w1 in F++H by A2;
  w1 in G by A3,XBOOLE_0:def 4;
  then w+w1 in F++G by A2;
  hence thesis by A1,A4,XBOOLE_0:def 4;
end;

theorem Th43:
  {f}++{g} = {f+g}
proof
  let j;
  hereby
    assume j in {f}++{g};
    then consider w1,w2 such that
A1: j = w1+w2 and
A2: w1 in {f} & w2 in {g};
    w1 = f & w2 = g by A2,TARSKI:def 1;
    hence j in {f+g} by A1,TARSKI:def 1;
  end;
A3: f in {f} & g in {g} by TARSKI:def 1;
  assume j in {f+g};
  then
A4: j = f+g by TARSKI:def 1;
  f in ExtREAL & g in ExtREAL by XXREAL_0:def 1;
  hence thesis by A4,A3;
end;

theorem Th44:
  {f}++{g,h} = {f+g,f+h}
proof
  thus {f}++{g,h} = {f}++({g}\/{h}) by ENUMSET1:1
    .= ({f}++{g}) \/ ({f}++{h}) by Th41
    .= {f+g} \/ ({f}++{h}) by Th43
    .= {f+g} \/ {f+h} by Th43
    .= {f+g,f+h} by ENUMSET1:1;
end;

theorem Th45:
  {f,g}++{h,i} = {f+h,f+i,g+h,g+i}
proof
  thus {f,g}++{h,i} = ({f}\/{g})++{h,i} by ENUMSET1:1
    .= ({f}++{h,i}) \/ ({g}++{h,i}) by Th41
    .= {f+h,f+i} \/ ({g}++{h,i}) by Th44
    .= {f+h,f+i} \/ {g+h,g+i} by Th44
    .= {f+h,f+i,g+h,g+i} by ENUMSET1:5;
end;

definition
  let A, B be complex-membered set;
  func A++B equals
  {c1+c2: c1 in A & c2 in B};
  coherence;
  commutativity
  proof
    let X be set;
    let A, B;
    assume
A1: X = {c1+c2: c1 in A & c2 in B};
    thus X c= {c1+c2: c1 in B & c2 in A}
    proof
      let e;
      assume e in X;
      then ex c1,c2 st e = c1+c2 & c1 in A & c2 in B by A1;
      hence thesis;
    end;
    let e;
    assume e in {c1+c2: c1 in B & c2 in A};
    then ex c1,c2 st e = c1+c2 & c1 in B & c2 in A;
    hence thesis by A1;
  end;
end;

theorem Th46:
  a in A & b in B implies a+b in A++B
proof
  a in COMPLEX & b in COMPLEX by XCMPLX_0:def 2;
  hence thesis;
end;

registration
  let A be empty set;
  let B be complex-membered set;
  cluster A++B -> empty;
  coherence
  proof
    assume A++B is non empty;
    then the Element of (A++B) in A++B;
    then ex c1,c2 st the Element of (A++B) = c1+c2 & c1 in A & c2 in B;
    hence thesis;
  end;
  cluster B++A -> empty;
  coherence;
end;

registration
  let A, B be complex-membered non empty set;
  cluster A++B -> non empty;
  coherence
  proof
    (the Element of A)+the Element of B in A++B by Th46;
    hence thesis;
  end;
end;

registration
  let A, B be complex-membered set;
  cluster A++B -> complex-membered;
  coherence
  proof
    let e;
    assume e in A++B;
    then ex c1,c2 st e = c1+c2 & c1 in A & c2 in B;
    hence thesis;
  end;
end;

registration
  let A, B be real-membered set;
  cluster A++B -> real-membered;
  coherence
  proof
    let e;
    assume e in A++B;
    then ex c1,c2 st e = c1+c2 & c1 in A & c2 in B;
    hence thesis;
  end;
end;

registration
  let A, B be rational-membered set;
  cluster A++B -> rational-membered;
  coherence
  proof
    let e;
    assume e in A++B;
    then ex c1,c2 st e = c1+c2 & c1 in A & c2 in B;
    hence thesis;
  end;
end;

registration
  let A, B be integer-membered set;
  cluster A++B -> integer-membered;
  coherence
  proof
    let e;
    assume e in A++B;
    then ex c1,c2 st e = c1+c2 & c1 in A & c2 in B;
    hence thesis;
  end;
end;

registration
  let A, B be natural-membered set;
  cluster A++B -> natural-membered;
  coherence
  proof
    let e;
    assume e in A++B;
    then ex c1,c2 st e = c1+c2 & c1 in A & c2 in B;
    hence thesis;
  end;
end;

registration
  let A, B be real-membered set, F, G be ext-real-membered set;
  identify A++B with F++G when A = F, B = G;
  compatibility
  proof
    assume
A1: A = F & B = G;
    let a be ext-real number;
    hereby
      assume a in A++B;
      then consider c,c1 such that
A2:   a = c+c1 and
A3:   c in A & c1 in B;
      reconsider d1 = c, d2 = c1 as Element of ExtREAL by A3,XXREAL_0:def 1;
      a = d1+d2 by A2,A3,XXREAL_3:def 2;
      hence a in F++G by A1,A3;
    end;
    assume a in F++G;
    then consider w,w1 such that
A4: a = w+w1 and
A5: w in F & w1 in G;
    reconsider d1 = w, d2 = w1 as Element of COMPLEX by A1,A5,XCMPLX_0:def 2;
    a = d1+d2 by A1,A4,A5,XXREAL_3:def 2;
    hence thesis by A1,A5;
  end;
end;

theorem Th47:
  A c= B & C c= D implies A++C c= B++D
proof
  assume
A1: A c= B & C c= D;
  let a;
  assume a in A++C;
  then ex c,c1 st a = c+c1 & c in A & c1 in C;
  hence thesis by A1;
end;

theorem Th48:
  A ++ (B \/ C) = (A++B) \/ (A++C)
proof
  let a;
  hereby
    assume a in A++(B\/C);
    then consider c,c1 such that
A1: a = c+c1 and
A2: c in A and
A3: c1 in B\/C;
    c1 in B or c1 in C by A3,XBOOLE_0:def 3;
    then c+c1 in A++B or c+c1 in A++C by A2;
    hence a in (A++B)\/(A++C) by A1,XBOOLE_0:def 3;
  end;
  assume
A4: a in (A++B)\/(A++C);
  per cases by A4,XBOOLE_0:def 3;
  suppose
    a in A++B;
    then consider c,c1 such that
A5: a = c+c1 & c in A and
A6: c1 in B;
    c1 in B\/C by A6,XBOOLE_0:def 3;
    hence thesis by A5;
  end;
  suppose
    a in A++C;
    then consider c,c1 such that
A7: a = c+c1 & c in A and
A8: c1 in C;
    c1 in B\/C by A8,XBOOLE_0:def 3;
    hence thesis by A7;
  end;
end;

theorem Th49:
  A ++ (B /\ C) c= (A++B) /\ (A++C)
proof
  let a;
  assume a in A++(B/\C);
  then consider c,c1 such that
A1: a = c+c1 and
A2: c in A and
A3: c1 in B/\C;
  c1 in C by A3,XBOOLE_0:def 4;
  then
A4: c+c1 in A++C by A2;
  c1 in B by A3,XBOOLE_0:def 4;
  then c+c1 in A++B by A2;
  hence thesis by A1,A4,XBOOLE_0:def 4;
end;

theorem Th50:
  (A++B)++C = A++(B++C)
proof
  let a;
  hereby
    assume a in A++B++C;
    then consider c,c1 such that
A1: a = c+c1 and
A2: c in A++B and
A3: c1 in C;
    consider c2,c3 being Element of COMPLEX such that
A4: c = c2+c3 and
A5: c2 in A and
A6: c3 in B by A2;
    a = c2+(c3+c1) & c3+c1 in B++C by A1,A3,A4,A6;
    hence a in A++(B++C) by A5;
  end;
  assume a in A++(B++C);
  then consider c,c1 such that
A7: a = c+c1 & c in A and
A8: c1 in B++C;
  consider c2,c3 being Element of COMPLEX such that
A9: c1 = c2+c3 & c2 in B and
A10: c3 in C by A8;
  a = c+c2+c3 & c+c2 in A++B by A7,A9;
  hence thesis by A10;
end;

theorem Th51:
  {a}++{b} = {a+b}
proof
  let z;
  hereby
    assume z in {a}++{b};
    then consider c1,c2 such that
A1: z = c1+c2 and
A2: c1 in {a} & c2 in {b};
    c1 = a & c2 = b by A2,TARSKI:def 1;
    hence z in {a+b} by A1,TARSKI:def 1;
  end;
A3: a in {a} & b in {b} by TARSKI:def 1;
  assume z in {a+b};
  then
A4: z = a+b by TARSKI:def 1;
  a in COMPLEX & b in COMPLEX by XCMPLX_0:def 2;
  hence thesis by A4,A3;
end;

theorem Th52:
  {a}++{s,t} = {a+s,a+t}
proof
  thus {a}++{s,t} = {a}++({s}\/{t}) by ENUMSET1:1
    .= ({a}++{s}) \/ ({a}++{t}) by Th48
    .= {a+s} \/ ({a}++{t}) by Th51
    .= {a+s} \/ {a+t} by Th51
    .= {a+s,a+t} by ENUMSET1:1;
end;

theorem Th53:
  {a,b}++{s,t} = {a+s,a+t,b+s,b+t}
proof
  thus {a,b}++{s,t} = ({a}\/{b})++{s,t} by ENUMSET1:1
    .= ({a}++{s,t}) \/ ({b}++{s,t}) by Th48
    .= {a+s,a+t} \/ ({b}++{s,t}) by Th52
    .= {a+s,a+t} \/ {b+s,b+t} by Th52
    .= {a+s,a+t,b+s,b+t} by ENUMSET1:5;
end;

definition
  let F, G be ext-real-membered set;
  func F--G equals
  F ++ --G;
  coherence;
end;

theorem Th54:
  F--G = {w1-w2: w1 in F & w2 in G}
proof
  thus F--G c= {w1-w2: w1 in F & w2 in G}
  proof
    let e;
    assume e in F--G;
    then consider w1,w2 such that
A1: e = w1+w2 and
A2: w1 in F and
A3: w2 in --G;
    e = w1- -w2 & -w2 in G by A1,A3,Th2;
    hence thesis by A2;
  end;
  let e;
  assume e in {w1-w2: w1 in F & w2 in G};
  then consider w1,w2 such that
A4: e = w1-w2 & w1 in F and
A5: w2 in G;
  -w2 in --G by A5;
  hence thesis by A4;
end;

theorem Th55:
  f in F & g in G implies f-g in F--G
proof
A1: f in ExtREAL & g in ExtREAL by XXREAL_0:def 1;
  F--G = {w1-w2: w1 in F & w2 in G} by Th54;
  hence thesis by A1;
end;

registration
  let F be empty set;
  let G be ext-real-membered set;
  cluster F--G -> empty;
  coherence;
  cluster G--F -> empty;
  coherence;
end;

registration
  let F, G be ext-real-membered non empty set;
  cluster F--G -> non empty;
  coherence;
end;

registration
  let F, G be ext-real-membered set;
  cluster F--G -> ext-real-membered;
  coherence;
end;

theorem
  F c= G & H c= I implies F--H c= G--I
proof
  assume
A1: F c= G & H c= I;
  let i;
A2: F--H = {w1-w2: w1 in F & w2 in H} by Th54;
  assume i in F--H;
  then ex w,w1 st i = w-w1 & w in F & w1 in H by A2;
  hence thesis by A1,Th55;
end;

theorem
  F -- (G \/ H) = (F--G) \/ (F--H)
proof
  thus F -- (G \/ H) = F++(--G\/--H) by Th5
    .= (F--G) \/ (F--H) by Th41;
end;

theorem
  F -- (G /\ H) c= (F--G) /\ (F--H)
proof
  F -- (G /\ H) = F++((--G)/\--H) by Th6;
  hence thesis by Th42;
end;

Lm4: --(F++G) = (--F) ++ --G
proof
  let j;
  hereby
    assume j in --(F++G);
    then consider w such that
A1: j = -w and
A2: w in F++G;
    consider w1,w2 such that
A3: w = w1+w2 and
A4: w1 in F & w2 in G by A2;
A5: -(w1+w2) = -w1-w2 by XXREAL_3:9;
    -w1 in --F & -w2 in --G by A4;
    hence j in (--F) ++ --G by A1,A3,A5;
  end;
  assume j in (--F) ++ --G;
  then consider w1,w2 such that
A6: j = w1+w2 and
A7: w1 in --F & w2 in --G;
  -w1 in F & -w2 in G by A7,Th2;
  then -(w1+w2) = -w1-w2 & -w1+-w2 in F++G by XXREAL_3:9;
  hence thesis by A6,Th2;
end;

theorem
  --(F++G) = (--F) -- G by Lm4;

theorem Th60:
  --(F--G) = (--F) ++ G
proof
  thus --(F--G) = (--F)----G by Lm4
    .= (--F) ++ G;
end;

theorem
  {f}--{g} = {f-g}
proof
  thus {f}--{g} = {f}++{-g} by Th9
    .= {f-g} by Th43;
end;

theorem
  {f}--{h,i} = {f-h,f-i}
proof
  thus {f}--{h,i} = {f}++{-h,-i} by Th10
    .= {f-h,f-i} by Th44;
end;

theorem
  {f,g}--{h} = {f-h,g-h}
proof
  thus {f,g}--{h} = {f,g}++{-h} by Th9
    .= {f-h,g-h} by Th44;
end;

theorem
  {f,g}--{h,i} = {f-h,f-i,g-h,g-i}
proof
  thus {f,g}--{h,i} = {f,g}++{-h,-i} by Th10
    .= {f-h,f-i,g-h,g-i} by Th45;
end;

definition
  let A, B be complex-membered set;
  func A--B equals
  A ++ --B;
  coherence;
end;

theorem Th65:
  A--B = {c1-c2: c1 in A & c2 in B}
proof
  thus A--B c= {c1-c2: c1 in A & c2 in B}
  proof
    let e;
    assume e in A--B;
    then consider c1,c2 such that
A1: e = c1+c2 and
A2: c1 in A and
A3: c2 in --B;
    e = c1- -c2 & -c2 in B by A1,A3,Th12;
    hence thesis by A2;
  end;
  let e;
  assume e in {c1-c2: c1 in A & c2 in B};
  then consider c1,c2 such that
A4: e = c1-c2 & c1 in A and
A5: c2 in B;
  -c2 in --B by A5;
  hence thesis by A4;
end;

theorem Th66:
  a in A & b in B implies a-b in A--B
proof
A1: a in COMPLEX & b in COMPLEX by XCMPLX_0:def 2;
  A--B = {c1-c2: c1 in A & c2 in B} by Th65;
  hence thesis by A1;
end;

registration
  let A be empty set;
  let B be complex-membered set;
  cluster A--B -> empty;
  coherence;
  cluster B--A -> empty;
  coherence;
end;

registration
  let A, B be complex-membered non empty set;
  cluster A--B -> non empty;
  coherence;
end;

registration
  let A, B be complex-membered set;
  cluster A--B -> complex-membered;
  coherence;
end;

registration
  let A, B be real-membered set;
  cluster A--B -> real-membered;
  coherence;
end;

registration
  let A, B be rational-membered set;
  cluster A--B -> rational-membered;
  coherence;
end;

registration
  let A, B be integer-membered set;
  cluster A--B -> integer-membered;
  coherence;
end;

registration
  let A, B be real-membered set, F, G be ext-real-membered set;
  identify A--B with F--G when A = F, B = G;
  compatibility;
end;

theorem Th67:
  A c= B & C c= D implies A--C c= B--D
proof
  assume
A1: A c= B & C c= D;
  let z;
A2: A--C = {c1-c2: c1 in A & c2 in C} by Th65;
  assume z in A--C;
  then ex c,c1 st z = c-c1 & c in A & c1 in C by A2;
  hence thesis by A1,Th66;
end;

Lm5: --(A++B) = (--A) ++ --B
proof
  let a;
  hereby
    assume a in --(A++B);
    then consider c such that
A1: a = -c and
A2: c in A++B;
    consider c1,c2 such that
A3: c = c1+c2 and
A4: c1 in A & c2 in B by A2;
A5: -c1 in --A & -c2 in --B by A4;
    a = -c1+-c2 by A1,A3;
    hence a in (--A) ++ --B by A5;
  end;
  assume a in (--A) ++ --B;
  then consider c1,c2 such that
A6: a = c1+c2 and
A7: c1 in --A & c2 in --B;
  -c1 in A & -c2 in B by A7,Th12;
  then -c1+-c2 in A++B;
  then -a in A++B by A6;
  hence thesis by Th12;
end;

theorem
  A -- (B \/ C) = (A--B) \/ (A--C)
proof
  thus A -- (B \/ C) = A++(--B\/--C) by Th15
    .= (A--B) \/ (A--C) by Th48;
end;

theorem
  A -- (B /\ C) c= (A--B) /\ (A--C)
proof
  A -- (B /\ C) = A++((--B)/\--C) by Th16;
  hence thesis by Th49;
end;

theorem
  --(A++B) = (--A) -- B by Lm5;

theorem Th71:
  --(A--B) = (--A) ++ B
proof
  thus --(A--B) = (--A)----B by Lm5
    .= (--A) ++ B;
end;

theorem
  A++(B--C) = A++B--C by Th50;

theorem
  A--(B++C) = A--B--C
proof
  thus A--(B++C) = A++(--B--C) by Lm5
    .= A--B--C by Th50;
end;

theorem
  A--(B--C) = A--B++C
proof
  thus A--(B--C) = A++((--B)++C) by Th71
    .= A--B++C by Th50;
end;

theorem Th75:
  {a}--{b} = {a-b}
proof
  thus {a}--{b} = {a}++{-b} by Th19
    .= {a-b} by Th51;
end;

theorem
  {a}--{s,t} = {a-s,a-t}
proof
  thus {a}--{s,t} = {a}++{-s,-t} by Th20
    .= {a-s,a-t} by Th52;
end;

theorem
  {a,b}--{s} = {a-s,b-s}
proof
  thus {a,b}--{s} = {a,b}++{-s} by Th19
    .= {a-s,b-s} by Th52;
end;

theorem
  {a,b}--{s,t} = {a-s,a-t,b-s,b-t}
proof
  thus {a,b}--{s,t} = {a,b}++{-s,-t} by Th20
    .= {a-s,a-t,b-s,b-t} by Th53;
end;

definition
  let F, G be ext-real-membered set;
  func F**G equals
  {w1*w2: w1 in F & w2 in G};
  coherence;
  commutativity
  proof
    let X be set;
    let F, G;
    assume
A1: X = {w1*w2: w1 in F & w2 in G};
    thus X c= {w1*w2: w1 in G & w2 in F}
    proof
      let e;
      assume e in X;
      then ex w1,w2 st e = w1*w2 & w1 in F & w2 in G by A1;
      hence thesis;
    end;
    let e;
    assume e in {w1*w2: w1 in G & w2 in F};
    then ex w1,w2 st e = w1*w2 & w1 in G & w2 in F;
    hence thesis by A1;
  end;
end;

registration
  let F be empty set;
  let G be ext-real-membered set;
  cluster F**G -> empty;
  coherence
  proof
    assume F**G is non empty;
    then the Element of (F**G) in F**G;
    then ex w1,w2 st the Element of (F**G) = w1*w2 & w1 in F & w2 in G;
    hence thesis;
  end;
  cluster G**F -> empty;
  coherence;
end;

registration
  let F, G be ext-real-membered set;
  cluster F**G -> ext-real-membered;
  coherence
  proof
    let e;
    assume e in F**G;
    then ex w1,w2 st e = w1*w2 & w1 in F & w2 in G;
    hence thesis;
  end;
end;

theorem Th79:
  f in F & g in G implies f*g in F**G
proof
  f in ExtREAL & g in ExtREAL by XXREAL_0:def 1;
  hence thesis;
end;

registration
  let F, G be ext-real-membered non empty set;
  cluster F**G -> non empty;
  coherence
  proof
    (the Element of F)*the Element of G in F**G by Th79;
    hence thesis;
  end;
end;

theorem Th80:
  (F**G)**H = F**(G**H)
proof
  let i;
  hereby
    assume i in F**G**H;
    then consider w,w1 such that
A1: i = w*w1 and
A2: w in F**G and
A3: w1 in H;
    consider w2,w3 being Element of ExtREAL such that
A4: w = w2*w3 and
A5: w2 in F and
A6: w3 in G by A2;
    i = w2*(w3*w1) & w3*w1 in G**H by A1,A3,A4,A6,XXREAL_3:66;
    hence i in F**(G**H) by A5;
  end;
  assume i in F**(G**H);
  then consider w,w1 such that
A7: i = w*w1 & w in F and
A8: w1 in G**H;
  consider w2,w3 being Element of ExtREAL such that
A9: w1 = w2*w3 & w2 in G and
A10: w3 in H by A8;
  i = w*w2*w3 & w*w2 in F**G by A7,A9,XXREAL_3:66;
  hence thesis by A10;
end;

theorem Th81:
  F c= G & H c= I implies F**H c= G**I
proof
  assume
A1: F c= G & H c= I;
  let i;
  assume i in F**H;
  then ex w,w1 st i = w*w1 & w in F & w1 in H;
  hence thesis by A1;
end;

theorem Th82:
  F ** (G \/ H) = (F**G) \/ (F**H)
proof
  let j;
  hereby
    assume j in F**(G\/H);
    then consider w,w1 such that
A1: j = w*w1 and
A2: w in F and
A3: w1 in G\/H;
    w1 in G or w1 in H by A3,XBOOLE_0:def 3;
    then w*w1 in F**G or w*w1 in F**H by A2;
    hence j in (F**G)\/(F**H) by A1,XBOOLE_0:def 3;
  end;
  assume
A4: j in (F**G)\/(F**H);
  per cases by A4,XBOOLE_0:def 3;
  suppose
    j in F**G;
    then consider w,w1 such that
A5: j = w*w1 & w in F and
A6: w1 in G;
    w1 in G\/H by A6,XBOOLE_0:def 3;
    hence thesis by A5;
  end;
  suppose
    j in F**H;
    then consider w,w1 such that
A7: j = w*w1 & w in F and
A8: w1 in H;
    w1 in G\/H by A8,XBOOLE_0:def 3;
    hence thesis by A7;
  end;
end;

theorem Th83:
  F ** (G /\ H) c= (F**G) /\ (F**H)
proof
  let j;
  assume j in F**(G/\H);
  then consider w,w1 such that
A1: j = w*w1 and
A2: w in F and
A3: w1 in G/\H;
  w1 in H by A3,XBOOLE_0:def 4;
  then
A4: w*w1 in F**H by A2;
  w1 in G by A3,XBOOLE_0:def 4;
  then w*w1 in F**G by A2;
  hence thesis by A1,A4,XBOOLE_0:def 4;
end;

theorem
  F**--G = --(F**G)
proof
  let i;
  hereby
    assume i in F**--G;
    then consider w,w1 such that
A1: i = w*w1 and
A2: w in F and
A3: w1 in --G;
    w*-w1 = -(w*w1) by XXREAL_3:92;
    then
A4: i = -(w*-w1) by A1;
    -w1 in G by A3,Th2;
    then w*-w1 in F**G by A2;
    hence i in --(F**G) by A4;
  end;
  assume i in --(F**G);
  then -i in F**G by Th2;
  then consider w,w1 such that
A5: -i = w*w1 & w in F and
A6: w1 in G;
  w*-w1 = -(w*w1) & -w1 in --G by A6,XXREAL_3:92;
  hence thesis by A5;
end;

theorem Th85:
  (F**G)"" = (F"") ** (G"")
proof
  let i;
  hereby
    assume i in (F**G)"";
    then consider w such that
A1: i = w" and
A2: w in F**G;
    consider w1,w2 such that
A3: w = w1*w2 and
A4: w1 in F & w2 in G by A2;
A5: w1" in F"" & w2" in G"" by A4;
    i = w1"*w2" by A1,A3,XXREAL_3:67;
    hence i in (F"")**(G"") by A5;
  end;
  assume i in (F"")**(G"");
  then consider w,w1 such that
A6: i = w*w1 and
A7: w in F"" and
A8: w1 in G"";
  consider w3 being Element of ExtREAL such that
A9: w1 = w3" and
A10: w3 in G by A8;
  consider w2 such that
A11: w = w2" and
A12: w2 in F by A7;
A13: w2*w3 in F**G by A12,A10;
  i = (w2*w3)" by A6,A11,A9,XXREAL_3:67;
  hence thesis by A13;
end;

theorem Th86:
  {f}**{g} = {f*g}
proof
  let i;
  hereby
    assume i in {f}**{g};
    then consider w1,w2 such that
A1: i = w1*w2 and
A2: w1 in {f} & w2 in {g};
    w1 = f & w2 = g by A2,TARSKI:def 1;
    hence i in {f*g} by A1,TARSKI:def 1;
  end;
A3: f in {f} & g in {g} by TARSKI:def 1;
  assume i in {f*g};
  then
A4: i = f*g by TARSKI:def 1;
  f in ExtREAL & g in ExtREAL by XXREAL_0:def 1;
  hence thesis by A4,A3;
end;

theorem Th87:
  {f}**{h,i} = {f*h,f*i}
proof
  thus {f}**{h,i} = {f}**({h}\/{i}) by ENUMSET1:1
    .= ({f}**{h}) \/ ({f}**{i}) by Th82
    .= {f*h} \/ ({f}**{i}) by Th86
    .= {f*h} \/ {f*i} by Th86
    .= {f*h,f*i} by ENUMSET1:1;
end;

theorem Th88:
  {f,g}**{h,i} = {f*h,f*i,g*h,g*i}
proof
  thus {f,g}**{h,i} = ({f}\/{g})**{h,i} by ENUMSET1:1
    .= ({f}**{h,i}) \/ ({g}**{h,i}) by Th82
    .= {f*h,f*i} \/ ({g}**{h,i}) by Th87
    .= {f*h,f*i} \/ {g*h,g*i} by Th87
    .= {f*h,f*i,g*h,g*i} by ENUMSET1:5;
end;

definition
  let A, B be complex-membered set;
  func A**B equals
  {c1*c2: c1 in A & c2 in B};
  coherence;
  commutativity
  proof
    let X be set;
    let A, B;
    assume
A1: X = {c1*c2: c1 in A & c2 in B};
    thus X c= {c1*c2: c1 in B & c2 in A}
    proof
      let e;
      assume e in X;
      then ex c1,c2 st e = c1*c2 & c1 in A & c2 in B by A1;
      hence thesis;
    end;
    let e;
    assume e in {c1*c2: c1 in B & c2 in A};
    then ex c1,c2 st e = c1*c2 & c1 in B & c2 in A;
    hence thesis by A1;
  end;
end;

theorem Th89:
  a in A & b in B implies a*b in A**B
proof
  a in COMPLEX & b in COMPLEX by XCMPLX_0:def 2;
  hence thesis;
end;

registration
  let A be empty set;
  let B be complex-membered set;
  cluster A**B -> empty;
  coherence
  proof
    assume A**B is non empty;
    then the Element of (A**B) in A**B;
    then ex c1,c2 st the Element of (A**B) = c1*c2 & c1 in A & c2 in B;
    hence thesis;
  end;
  cluster B**A -> empty;
  coherence;
end;

registration
  let A, B be complex-membered non empty set;
  cluster A**B -> non empty;
  coherence
  proof
    (the Element of A)*the Element of B in A**B by Th89;
    hence thesis;
  end;
end;

registration
  let A, B be complex-membered set;
  cluster A**B -> complex-membered;
  coherence
  proof
    let e;
    assume e in A**B;
    then ex c1,c2 st e = c1*c2 & c1 in A & c2 in B;
    hence thesis;
  end;
end;

registration
  let A, B be real-membered set;
  cluster A**B -> real-membered;
  coherence
  proof
    let e;
    assume e in A**B;
    then ex c1,c2 st e = c1*c2 & c1 in A & c2 in B;
    hence thesis;
  end;
end;

registration
  let A, B be rational-membered set;
  cluster A**B -> rational-membered;
  coherence
  proof
    let e;
    assume e in A**B;
    then ex c1,c2 st e = c1*c2 & c1 in A & c2 in B;
    hence thesis;
  end;
end;

registration
  let A, B be integer-membered set;
  cluster A**B -> integer-membered;
  coherence
  proof
    let e;
    assume e in A**B;
    then ex c1,c2 st e = c1*c2 & c1 in A & c2 in B;
    hence thesis;
  end;
end;

registration
  let A, B be natural-membered set;
  cluster A**B -> natural-membered;
  coherence
  proof
    let e;
    assume e in A**B;
    then ex c1,c2 st e = c1*c2 & c1 in A & c2 in B;
    hence thesis;
  end;
end;

registration
  let A, B be real-membered set, F, G be ext-real-membered set;
  identify A**B with F**G when A = F, B = G;
  compatibility
  proof
    assume
A1: A = F & B = G;
    let a be ext-real number;
    hereby
      assume a in A**B;
      then consider c,c1 such that
A2:   a = c*c1 and
A3:   c in A & c1 in B;
      reconsider d1 = c, d2 = c1 as Element of ExtREAL by A3,XXREAL_0:def 1;
      a = d1*d2 by A2,A3,XXREAL_3:def 5;
      hence a in F**G by A1,A3;
    end;
    assume a in F**G;
    then consider w,w1 such that
A4: a = w*w1 and
A5: w in F & w1 in G;
    reconsider d1 = w, d2 = w1 as Element of COMPLEX by A1,A5,XCMPLX_0:def 2;
    a = d1*d2 by A1,A4,A5,XXREAL_3:def 5;
    hence thesis by A1,A5;
  end;
end;

theorem Th90:
  (A**B)**C = A**(B**C)
proof
  let a;
  hereby
    assume a in A**B**C;
    then consider c,c1 such that
A1: a = c*c1 and
A2: c in A**B and
A3: c1 in C;
    consider c2,c3 being Element of COMPLEX such that
A4: c = c2*c3 and
A5: c2 in A and
A6: c3 in B by A2;
    a = c2*(c3*c1) & c3*c1 in B**C by A1,A3,A4,A6;
    hence a in A**(B**C) by A5;
  end;
  assume a in A**(B**C);
  then consider c,c1 such that
A7: a = c*c1 & c in A and
A8: c1 in B**C;
  consider c2,c3 being Element of COMPLEX such that
A9: c1 = c2*c3 & c2 in B and
A10: c3 in C by A8;
  a = c*c2*c3 & c*c2 in A**B by A7,A9;
  hence thesis by A10;
end;

theorem
  A c= B & C c= D implies A**C c= B**D
proof
  assume
A1: A c= B & C c= D;
  let a;
  assume a in A**C;
  then ex c,c1 st a = c*c1 & c in A & c1 in C;
  hence thesis by A1;
end;

theorem Th92:
  A ** (B \/ C) = (A**B) \/ (A**C)
proof
  let a;
  hereby
    assume a in A**(B\/C);
    then consider c,c1 such that
A1: a = c*c1 and
A2: c in A and
A3: c1 in B\/C;
    c1 in B or c1 in C by A3,XBOOLE_0:def 3;
    then c*c1 in A**B or c*c1 in A**C by A2;
    hence a in (A**B)\/(A**C) by A1,XBOOLE_0:def 3;
  end;
  assume
A4: a in (A**B)\/(A**C);
  per cases by A4,XBOOLE_0:def 3;
  suppose
    a in A**B;
    then consider c,c1 such that
A5: a = c*c1 & c in A and
A6: c1 in B;
    c1 in B\/C by A6,XBOOLE_0:def 3;
    hence thesis by A5;
  end;
  suppose
    a in A**C;
    then consider c,c1 such that
A7: a = c*c1 & c in A and
A8: c1 in C;
    c1 in B\/C by A8,XBOOLE_0:def 3;
    hence thesis by A7;
  end;
end;

theorem Th93:
  A ** (B /\ C) c= (A**B) /\ (A**C)
proof
  let a;
  assume a in A**(B/\C);
  then consider c,c1 such that
A1: a = c*c1 and
A2: c in A and
A3: c1 in B/\C;
  c1 in C by A3,XBOOLE_0:def 4;
  then
A4: c*c1 in A**C by A2;
  c1 in B by A3,XBOOLE_0:def 4;
  then c*c1 in A**B by A2;
  hence thesis by A1,A4,XBOOLE_0:def 4;
end;

theorem Th94:
  A**--B = --(A**B)
proof
  let a;
  hereby
    assume a in A**--B;
    then consider c,c1 such that
A1: a = c*c1 and
A2: c in A and
A3: c1 in --B;
    -c1 in B by A3,Th12;
    then
A4: c*-c1 in A**B by A2;
    a = -(c*-c1) by A1;
    hence a in --(A**B) by A4;
  end;
  assume a in --(A**B);
  then -a in A**B by Th12;
  then consider c,c1 such that
A5: -a = c*c1 and
A6: c in A and
A7: c1 in B;
  a = c*-c1 & -c1 in --B by A5,A7;
  hence thesis by A6;
end;

theorem Th95:
  A**(B++C) c= A**B ++ A**C
proof
  let a;
  assume a in A**(B++C);
  then consider c,c1 such that
A1: a = c*c1 and
A2: c in A and
A3: c1 in B++C;
  consider c2,c3 being Element of COMPLEX such that
A4: c1 = c2+c3 & c2 in B and
A5: c3 in C by A3;
A6: c*c3 in A**C by A2,A5;
  a = c*c2+c*c3 & c*c2 in A**B by A1,A2,A4;
  hence thesis by A6;
end;

theorem Th96:
  A**(B--C) c= A**B -- A**C
proof
  A**(B++--C) c= A**B++A**--C by Th95;
  hence thesis by Th94;
end;

theorem Th97:
  (A**B)"" = (A"") ** (B"")
proof
  let a;
  hereby
    assume a in (A**B)"";
    then consider c such that
A1: a = c" and
A2: c in A**B;
    consider c1,c2 such that
A3: c = c1*c2 and
A4: c1 in A & c2 in B by A2;
A5: c1" in A"" & c2" in B"" by A4;
    a = c1"*c2" by A1,A3,XCMPLX_1:204;
    hence a in (A"")**(B"") by A5;
  end;
  assume a in (A"")**(B"");
  then consider c,c1 such that
A6: a = c*c1 and
A7: c in A"" and
A8: c1 in B"";
  consider c3 being Element of COMPLEX such that
A9: c1 = c3" and
A10: c3 in B by A8;
  consider c2 such that
A11: c = c2" and
A12: c2 in A by A7;
A13: c2*c3 in A**B by A12,A10;
  a = (c2*c3)" by A6,A11,A9,XCMPLX_1:204;
  hence thesis by A13;
end;

theorem Th98:
  {a}**{b} = {a*b}
proof
  let z;
  hereby
    assume z in {a}**{b};
    then consider c1,c2 such that
A1: z = c1*c2 and
A2: c1 in {a} & c2 in {b};
    c1 = a & c2 = b by A2,TARSKI:def 1;
    hence z in {a*b} by A1,TARSKI:def 1;
  end;
A3: a in {a} & b in {b} by TARSKI:def 1;
  assume z in {a*b};
  then
A4: z = a*b by TARSKI:def 1;
  a in COMPLEX & b in COMPLEX by XCMPLX_0:def 2;
  hence thesis by A4,A3;
end;

theorem Th99:
  {a}**{s,t} = {a*s,a*t}
proof
  thus {a}**{s,t} = {a}**({s}\/{t}) by ENUMSET1:1
    .= ({a}**{s}) \/ ({a}**{t}) by Th92
    .= {a*s} \/ ({a}**{t}) by Th98
    .= {a*s} \/ {a*t} by Th98
    .= {a*s,a*t} by ENUMSET1:1;
end;

theorem Th100:
  {a,b}**{s,t} = {a*s,a*t,b*s,b*t}
proof
  thus {a,b}**{s,t} = ({a}\/{b})**{s,t} by ENUMSET1:1
    .= ({a}**{s,t}) \/ ({b}**{s,t}) by Th92
    .= {a*s,a*t} \/ ({b}**{s,t}) by Th99
    .= {a*s,a*t} \/ {b*s,b*t} by Th99
    .= {a*s,a*t,b*s,b*t} by ENUMSET1:5;
end;

definition
  let F, G be ext-real-membered set;
  func F///G equals
  F**(G"");
  coherence;
end;

theorem Th101:
  F///G = {w1/w2: w1 in F & w2 in G}
proof
  thus F///G c= {w1/w2: w1 in F & w2 in G}
  proof
    let e;
    assume e in F///G;
    then consider w1,w2 such that
A1: e = w1*w2 and
A2: w1 in F and
A3: w2 in G"";
    consider w such that
A4: w2 = w" and
A5: w in G by A3;
    e = w1/w by A1,A4;
    hence thesis by A2,A5;
  end;
  let e;
  assume e in {w1/w2: w1 in F & w2 in G};
  then consider w1,w2 such that
A6: e = w1/w2 & w1 in F and
A7: w2 in G;
  w2" in G"" by A7;
  hence thesis by A6;
end;

theorem Th102:
  f in F & g in G implies f/g in F///G
proof
A1: f in ExtREAL & g in ExtREAL by XXREAL_0:def 1;
  F///G = {w1/w2: w1 in F & w2 in G} by Th101;
  hence thesis by A1;
end;

registration
  let F be empty set;
  let G be ext-real-membered set;
  cluster F///G -> empty;
  coherence;
  cluster G///F -> empty;
  coherence;
end;

registration
  let F, G be ext-real-membered non empty set;
  cluster F///G -> non empty;
  coherence;
end;

registration
  let F, G be ext-real-membered set;
  cluster F///G -> ext-real-membered;
  coherence;
end;

theorem
  F c= G & H c= I implies F///H c= G///I
proof
  assume
A1: F c= G & H c= I;
  let i;
A2: F///H = {w1/w2: w1 in F & w2 in H} by Th101;
  assume i in F///H;
  then ex w,w1 st i = w/w1 & w in F & w1 in H by A2;
  hence thesis by A1,Th102;
end;

theorem
  (F \/ G) /// H = (F///H) \/ (G///H) by Th82;

theorem
  (F /\ G) /// H c= (F///H) /\ (G///H) by Th83;

theorem
  F /// (G \/ H) = (F///G) \/ (F///H)
proof
  thus F /// (G \/ H) = F**((G"")\/(H"")) by Th23
    .= (F///G) \/ (F///H) by Th82;
end;

theorem
  F /// (G /\ H) c= (F///G) /\ (F///H)
proof
  (G /\ H)"" c= (G"")/\(H"") by Th24;
  then
A1: F**((G /\ H)"") c= F**((G"")/\(H"")) by Th81;
  F**((G"")/\(H"")) c= (F**(G""))/\(F**(H"")) by Th83;
  hence thesis by A1,XBOOLE_1:1;
end;

theorem
  (F**G)///H = F**(G///H) by Th80;

theorem
  (F///G)**H = (F**H)///G by Th80;

theorem
  (F///G)///H = F///(G**H)
proof
  thus (F///G)///H = F**((G"")**(H"")) by Th80
    .= F///(G**H) by Th85;
end;

theorem
  {f}///{g} = {f/g}
proof
  thus {f}///{g} = {f}**{g"} by Th26
    .= {f/g} by Th86;
end;

theorem
  {f}///{h,i} = {f/h,f/i}
proof
  thus {f}///{h,i} = {f}**{h",i"} by Th27
    .= {f/h,f/i} by Th87;
end;

theorem
  {f,g}///{h} = {f/h,g/h}
proof
  thus {f,g}///{h} = {f,g}**{h"} by Th26
    .= {f/h,g/h} by Th87;
end;

theorem
  {f,g}///{h,i} = {f/h,f/i,g/h,g/i}
proof
  thus {f,g}///{h,i} = {f,g}**{h",i"} by Th27
    .= {f/h,f/i,g/h,g/i} by Th88;
end;

definition
  let A, B be complex-membered set;
  func A///B equals
  A**(B"");
  coherence;
end;

theorem Th115:
  A///B = {c1/c2: c1 in A & c2 in B}
proof
  thus A///B c= {c1/c2: c1 in A & c2 in B}
  proof
    let e;
    assume e in A///B;
    then consider c1,c2 such that
A1: e = c1*c2 and
A2: c1 in A and
A3: c2 in B"";
    e = c1/(c2") & c2" in B by A1,A3,Th29;
    hence thesis by A2;
  end;
  let e;
  assume e in {c1/c2: c1 in A & c2 in B};
  then consider c1,c2 such that
A4: e = c1/c2 & c1 in A and
A5: c2 in B;
  c2" in B"" by A5;
  hence thesis by A4;
end;

theorem Th116:
  a in A & b in B implies a/b in A///B
proof
A1: a in COMPLEX & b in COMPLEX by XCMPLX_0:def 2;
  A///B = {c1/c2: c1 in A & c2 in B} by Th115;
  hence thesis by A1;
end;

registration
  let A be empty set;
  let B be complex-membered set;
  cluster A///B -> empty;
  coherence;
  cluster B///A -> empty;
  coherence;
end;

registration
  let A, B be complex-membered non empty set;
  cluster A///B -> non empty;
  coherence;
end;

registration
  let A, B be complex-membered set;
  cluster A///B -> complex-membered;
  coherence;
end;

registration
  let A, B be real-membered set;
  cluster A///B -> real-membered;
  coherence;
end;

registration
  let A, B be rational-membered set;
  cluster A///B -> rational-membered;
  coherence;
end;

registration
  let A, B be real-membered set, F, G be ext-real-membered set;
  identify A///B with F///G when A = F, B = G;
  compatibility;
end;

theorem
  A c= B & C c= D implies A///C c= B///D
proof
  assume
A1: A c= B & C c= D;
  let z;
A2: A///C = {c1/c2: c1 in A & c2 in C} by Th115;
  assume z in A///C;
  then ex c,c1 st z = c/c1 & c in A & c1 in C by A2;
  hence thesis by A1,Th116;
end;

theorem
  A /// (B \/ C) = (A///B) \/ (A///C)
proof
  thus A /// (B \/ C) = A**((B"")\/(C"")) by Th32
    .= (A///B) \/ (A///C) by Th92;
end;

theorem
  A /// (B /\ C) c= (A///B) /\ (A///C)
proof
  A /// (B /\ C) = A**((B"")/\(C"")) by Th33;
  hence thesis by Th93;
end;

theorem
  A///--B = --(A///B)
proof
  thus A///--B = A**--(B"") by Th36
    .= --(A///B) by Th94;
end;

theorem
  (--A)///B = --(A///B) by Th94;

theorem
  (A++B)///C c= A///C ++ B///C by Th95;

theorem
  (A--B)///C c= A///C -- B///C
proof
  (A++--B)///C c= A///C++(--B)///C by Th95;
  hence thesis by Th94;
end;

theorem
  (A**B)///C = A**(B///C) by Th90;

theorem
  (A///B)**C = (A**C)///B by Th90;

theorem
  (A///B)///C = A///(B**C)
proof
  thus (A///B)///C = A**((B"")**(C"")) by Th90
    .= A///(B**C) by Th97;
end;

theorem
  A///(B///C) = (A**C)///B
proof
  thus A///(B///C) = A**((B"")**(C"""")) by Th97
    .= (A**C)///B by Th90;
end;

theorem
  {a}///{b} = {a/b}
proof
  thus {a}///{b} = {a}**{b"} by Th37
    .= {a/b} by Th98;
end;

theorem
  {a}///{s,t} = {a/s,a/t}
proof
  thus {a}///{s,t} = {a}**{s",t"} by Th38
    .= {a/s,a/t} by Th99;
end;

theorem
  {a,b}///{s} = {a/s,b/s}
proof
  thus {a,b}///{s} = {a,b}**{s"} by Th37
    .= {a/s,b/s} by Th99;
end;

theorem
  {a,b}///{s,t} = {a/s,a/t,b/s,b/t}
proof
  thus {a,b}///{s,t} = {a,b}**{s",t"} by Th38
    .= {a/s,a/t,b/s,b/t} by Th100;
end;

definition
  let F be ext-real-membered set;
  let f be ext-real number;
  func f++F equals
  {f}++F;
  coherence;
end;

theorem Th132:
  g in G implies f+g in f++G
proof
  f in {f} by TARSKI:def 1;
  hence thesis by Th39;
end;

theorem Th133:
  f++F = {f+w: w in F}
proof
  thus f++F c= {f+w: w in F}
  proof
    let e;
    assume e in f++F;
    then consider w1,w2 such that
A1: e = w1+w2 and
A2: w1 in {f} and
A3: w2 in F;
    w1 = f by A2,TARSKI:def 1;
    hence thesis by A1,A3;
  end;
  let e;
  assume e in {f+w: w in F};
  then ex w st e = f+w & w in F;
  hence thesis by Th132;
end;

theorem Th134:
  e in f++F implies ex w st e = f+w & w in F
proof
  f++F = {f+w: w in F} by Th133;
  hence thesis;
end;

registration
  let F be empty set;
  let f be ext-real number;
  cluster f++F -> empty;
  coherence;
end;

registration
  let F be ext-real-membered non empty set;
  let f be ext-real number;
  cluster f++F -> non empty;
  coherence;
end;

registration
  let F be ext-real-membered set;
  let f be ext-real number;
  cluster f++F -> ext-real-membered;
  coherence;
end;

theorem Th135:
  r++F c= r++G implies F c= G
proof
  assume
A1: r++F c= r++G;
  let i;
  assume i in F;
  then r+i in r++F by Th132;
  then ex w st r+i = r+w & w in G by A1,Th134;
  hence thesis by XXREAL_3:11;
end;

theorem
  r++F = r++G implies F = G
proof
  assume r++F = r++G;
  then F c= G & G c= F by Th135;
  hence thesis by XBOOLE_0:def 10;
end;

theorem Th137:
  r ++ (F /\ G) = (r++F) /\ (r++G)
proof
A1: (r++F) /\ (r++G) c= r ++ (F /\ G)
  proof
    let j;
    assume
A2: j in (r++F) /\ (r++G);
    then j in r++F by XBOOLE_0:def 4;
    then consider w such that
A3: j = r+w and
A4: w in F by Th134;
    j in r++G by A2,XBOOLE_0:def 4;
    then consider w1 such that
A5: j = r+w1 and
A6: w1 in G by Th134;
    w = w1 by A3,A5,XXREAL_3:11;
    then w in F /\ G by A4,A6,XBOOLE_0:def 4;
    hence thesis by A3,Th132;
  end;
  r ++ (F /\ G) c= (r++F) /\ (r++G) by Th42;
  hence thesis by A1,XBOOLE_0:def 10;
end;

theorem Th138:
  (f++F) \ (f++G) c= f ++ (F \ G)
proof
  let i;
  assume
A1: i in (f++F) \ (f++G);
  then consider w such that
A2: i = f+w and
A3: w in F by Th134;
  now
    assume not w in F\G;
    then w in G by A3,XBOOLE_0:def 5;
    then f+w in f++G by Th132;
    hence contradiction by A1,A2,XBOOLE_0:def 5;
  end;
  hence thesis by A2,Th132;
end;

theorem Th139:
  r ++ (F \ G) = (r++F) \ (r++G)
proof
A1: r ++ (F \ G) c= (r++F) \ (r++G)
  proof
    let i;
    assume i in r ++ (F \ G);
    then consider w such that
A2: i = r+w and
A3: w in F\G by Th134;
A4: now
      assume r+w in r++G;
      then consider w1 such that
A5:   r+w = r+w1 and
A6:   w1 in G by Th134;
      w = w1 by A5,XXREAL_3:11;
      hence contradiction by A3,A6,XBOOLE_0:def 5;
    end;
    r+w in r++F by A3,Th132;
    hence thesis by A2,A4,XBOOLE_0:def 5;
  end;
  (r++F) \ (r++G) c= r ++ (F \ G) by Th138;
  hence thesis by A1,XBOOLE_0:def 10;
end;

theorem Th140:
  r ++ (F \+\ G) = (r++F) \+\ (r++G)
proof
  thus r ++ (F \+\ G) = (r++(F\G)) \/ (r++(G\F)) by Th41
    .= ((r++F)\(r++G)) \/ (r++(G\F)) by Th139
    .= (r++F) \+\ (r++G) by Th139;
end;

definition
  let A be complex-membered set;
  let a be complex number;
  func a++A equals
  {a}++A;
  coherence;
end;

theorem Th141:
  b in A implies a+b in a++A
proof
  a in {a} by TARSKI:def 1;
  hence thesis by Th46;
end;

theorem Th142:
  a++A = {a+c: c in A}
proof
  thus a++A c= {a+c: c in A}
  proof
    let e;
    assume e in a++A;
    then consider c1,c2 such that
A1: e = c1+c2 and
A2: c1 in {a} and
A3: c2 in A;
    c1 = a by A2,TARSKI:def 1;
    hence thesis by A1,A3;
  end;
  let e;
  assume e in {a+c: c in A};
  then ex c st e = a+c & c in A;
  hence thesis by Th141;
end;

theorem Th143:
  e in a++A implies ex c st e = a+c & c in A
proof
  a++A = {a+c: c in A} by Th142;
  hence thesis;
end;

registration
  let A be empty set;
  let a be complex number;
  cluster a++A -> empty;
  coherence;
end;

registration
  let A be complex-membered non empty set;
  let a be complex number;
  cluster a++A -> non empty;
  coherence;
end;

registration
  let A be complex-membered set;
  let a be complex number;
  cluster a++A -> complex-membered;
  coherence;
end;

registration
  let A be real-membered set;
  let a be real number;
  cluster a++A -> real-membered;
  coherence;
end;

registration
  let A be rational-membered set;
  let a be rational number;
  cluster a++A -> rational-membered;
  coherence;
end;

registration
  let A be integer-membered set;
  let a be integer number;
  cluster a++A -> integer-membered;
  coherence;
end;

registration
  let A be natural-membered set;
  let a be Nat;
  cluster a++A -> natural-membered;
  coherence;
end;

registration
  let A be real-membered set, F be ext-real-membered set;
  let a be real number, f be ext-real number;
  identify a++A with f++F when a = f, A = F;
  compatibility;
end;

theorem Th144:
  A c= B iff a++A c= a++B
proof
  thus A c= B implies a++A c= a++B by Th47;
  assume
A1: a++A c= a++B;
  let z;
  assume z in A;
  then a+z in a++A by Th141;
  then ex c st a+z = a+c & c in B by A1,Th143;
  hence thesis;
end;

theorem
  a++A = a++B implies A = B
proof
  assume a++A = a++B;
  then A c= B & B c= A by Th144;
  hence thesis by XBOOLE_0:def 10;
end;

theorem
  0++A = A
proof
  let a;
  hereby
    assume a in 0++A;
    then consider c1,c2 such that
A1: a = c1+c2 and
A2: c1 in {0} and
A3: c2 in A;
    c1 = 0 by A2,TARSKI:def 1;
    hence a in A by A1,A3;
  end;
  assume a in A;
  then 0+a in {0+c: c in A};
  hence thesis by Th142;
end;

theorem
  (a+b)++A = a++(b++A)
proof
  thus (a+b)++A = {a}++{b}++A by Th51
    .= a++(b++A) by Th50;
end;

theorem
  a++(A++B) = (a++A)++B by Th50;

theorem Th149:
  a ++ (A /\ B) = (a++A) /\ (a++B)
proof
A1: (a++A) /\ (a++B) c= a ++ (A /\ B)
  proof
    let z;
    assume
A2: z in (a++A) /\ (a++B);
    then z in a++A by XBOOLE_0:def 4;
    then consider c such that
A3: z = a+c and
A4: c in A by Th143;
    z in a++B by A2,XBOOLE_0:def 4;
    then ex c1 st z = a+c1 & c1 in B by Th143;
    then c in A/\B by A3,A4,XBOOLE_0:def 4;
    hence thesis by A3,Th141;
  end;
  a ++ (A /\ B) c= (a++A) /\ (a++B) by Th49;
  hence thesis by A1,XBOOLE_0:def 10;
end;

theorem Th150:
  a ++ (A \ B) = (a++A) \ (a++B)
proof
  let z;
  hereby
    assume z in a ++ (A \ B);
    then consider c such that
A1: z = a+c and
A2: c in A\B by Th143;
A3: now
      assume a+c in a++B;
      then ex c1 st a+c = a+c1 & c1 in B by Th143;
      hence contradiction by A2,XBOOLE_0:def 5;
    end;
    a+c in a++A by A2,Th141;
    hence z in (a++A) \ (a++B) by A1,A3,XBOOLE_0:def 5;
  end;
  assume
A4: z in (a++A) \ (a++B);
  then consider c such that
A5: z = a+c and
A6: c in A by Th143;
  now
    assume not c in A\B;
    then c in B by A6,XBOOLE_0:def 5;
    then a+c in a++B by Th141;
    hence contradiction by A4,A5,XBOOLE_0:def 5;
  end;
  hence thesis by A5,Th141;
end;

theorem Th151:
  a ++ (A \+\ B) = (a++A) \+\ (a++B)
proof
  thus a ++ (A \+\ B) = (a++(A\B)) \/ (a++(B\A)) by Th48
    .= ((a++A)\(a++B)) \/ (a++(B\A)) by Th150
    .= (a++A) \+\ (a++B) by Th150;
end;

definition
  let F be ext-real-membered set;
  let f be ext-real number;
  func f--F equals
  {f}--F;
  coherence;
end;

theorem Th152:
  g in G implies f-g in f--G
proof
  f in {f} by TARSKI:def 1;
  hence thesis by Th55;
end;

theorem Th153:
  f--F = {f-w: w in F}
proof
  thus f--F c= {f-w: w in F}
  proof
    let e;
    assume e in f--F;
    then consider w1,w2 such that
A1: e = w1+w2 and
A2: w1 in {f} & w2 in --F;
A3: w1- -w2 = w1+w2;
    -w2 in F & w1 = f by A2,Th2,TARSKI:def 1;
    hence thesis by A1,A3;
  end;
  let e;
  assume e in {f-w: w in F};
  then ex w st e = f-w & w in F;
  hence thesis by Th152;
end;

theorem Th154:
  e in f--F implies ex w st e = f-w & w in F
proof
  f--F = {f-w: w in F} by Th153;
  hence thesis;
end;

registration
  let F be empty set;
  let f be ext-real number;
  cluster f--F -> empty;
  coherence;
end;

registration
  let F be ext-real-membered non empty set;
  let f be ext-real number;
  cluster f--F -> non empty;
  coherence;
end;

registration
  let F be ext-real-membered set;
  let f be ext-real number;
  cluster f--F -> ext-real-membered;
  coherence;
end;

theorem Th155:
  r--F c= r--G implies F c= G
proof
  assume
A1: r--F c= r--G;
  let i;
  assume i in F;
  then r-i in r--F by Th152;
  then ex w st r-i = r-w & w in G by A1,Th154;
  hence thesis by XXREAL_3:12;
end;

theorem
  r--F = r--G implies F = G
proof
  assume r--F = r--G;
  then F c= G & G c= F by Th155;
  hence thesis by XBOOLE_0:def 10;
end;

theorem Th157:
  r -- (F/\G) = (r--F) /\ (r--G)
proof
  thus r -- (F/\G) = r ++ ((--F)/\(--G)) by Th6
    .= (r++--F) /\ (r++--G) by Th137
    .= (r--F) /\ (r--G);
end;

theorem Th158:
  r -- (F\G) = (r--F) \ (r--G)
proof
  thus r -- (F\G) = r ++ ((--F)\(--G)) by Th7
    .= (r++--F) \ (r++--G) by Th139
    .= (r--F) \ (r--G);
end;

theorem Th159:
  r -- (F\+\G) = (r--F) \+\ (r--G)
proof
  thus r -- (F\+\G) = r ++ ((--F)\+\(--G)) by Th8
    .= (r++--F) \+\ (r++--G) by Th140
    .= (r--F) \+\ (r--G);
end;

definition
  let A be complex-membered set;
  let a be complex number;
  func a--A equals
  {a}--A;
  coherence;
end;

theorem Th160:
  b in A implies a-b in a--A
proof
  a in {a} by TARSKI:def 1;
  hence thesis by Th66;
end;

theorem Th161:
  a--A = {a-c: c in A}
proof
A1: a--A = {c1-c2: c1 in {a} & c2 in A} by Th65;
  thus a--A c= {a-c: c in A}
  proof
    let e;
    assume e in a--A;
    then consider c1,c2 such that
A2: e = c1-c2 and
A3: c1 in {a} and
A4: c2 in A by A1;
    c1 = a by A3,TARSKI:def 1;
    hence thesis by A2,A4;
  end;
  let e;
  assume e in {a-c: c in A};
  then ex c st e = a-c & c in A;
  hence thesis by Th160;
end;

theorem Th162:
  e in a--A implies ex c st e = a-c & c in A
proof
  a--A = {a-c: c in A} by Th161;
  hence thesis;
end;

registration
  let A be empty set;
  let a be complex number;
  cluster a--A -> empty;
  coherence;
end;

registration
  let A be complex-membered non empty set;
  let a be complex number;
  cluster a--A -> non empty;
  coherence;
end;

registration
  let A be complex-membered set;
  let a be complex number;
  cluster a--A -> complex-membered;
  coherence;
end;

registration
  let A be real-membered set;
  let a be real number;
  cluster a--A -> real-membered;
  coherence;
end;

registration
  let A be rational-membered set;
  let a be rational number;
  cluster a--A -> rational-membered;
  coherence;
end;

registration
  let A be integer-membered set;
  let a be integer number;
  cluster a--A -> integer-membered;
  coherence;
end;

registration
  let A be real-membered set, F be ext-real-membered set;
  let a be real number, f be ext-real number;
  identify a--A with f--F when a = f, A = F;
  compatibility;
end;

theorem Th163:
  A c= B iff a--A c= a--B
proof
  thus A c= B implies a--A c= a--B by Th67;
  assume
A1: a--A c= a--B;
  let z;
  assume z in A;
  then a-z in a--A by Th160;
  then ex c st a-z = a-c & c in B by A1,Th162;
  hence thesis;
end;

theorem
  a--A = a--B implies A = B
proof
  assume a--A = a--B;
  then A c= B & B c= A by Th163;
  hence thesis by XBOOLE_0:def 10;
end;

theorem Th165:
  a -- (A/\B) = (a--A) /\ (a--B)
proof
  thus a -- (A/\B) = a ++ ((--A)/\(--B)) by Th16
    .= (a++--A) /\ (a++--B) by Th149
    .= (a--A) /\ (a--B);
end;

theorem Th166:
  a -- (A\B) = (a--A) \ (a--B)
proof
  thus a -- (A\B) = a ++ ((--A)\(--B)) by Th17
    .= (a++--A) \ (a++--B) by Th150
    .= (a--A) \ (a--B);
end;

theorem Th167:
  a -- (A\+\B) = (a--A) \+\ (a--B)
proof
  thus a -- (A\+\B) = a ++ ((--A)\+\(--B)) by Th18
    .= (a++--A) \+\ (a++--B) by Th151
    .= (a--A) \+\ (a--B);
end;

definition
  let F be ext-real-membered set;
  let f be ext-real number;
  func F--f equals
  F--{f};
  coherence;
end;

theorem Th168:
  g in G implies g-f in G--f
proof
  f in {f} by TARSKI:def 1;
  hence thesis by Th55;
end;

theorem Th169:
  F--f = {w-f: w in F}
proof
  thus F--f c= {w-f: w in F}
  proof
    let e;
    assume e in F--f;
    then consider w1,w2 such that
A1: e = w1+w2 & w1 in F and
A2: w2 in --{f};
    -w2 in {f} by A2,Th2;
    then w1- -w2 = w1+w2 & -w2 = f by TARSKI:def 1;
    hence thesis by A1;
  end;
  let e;
  assume e in {w-f: w in F};
  then ex w st e = w-f & w in F;
  hence thesis by Th168;
end;

theorem
  e in F--f implies ex w st e = w-f & w in F
proof
  F--f = {w-f: w in F} by Th169;
  hence thesis;
end;

registration
  let F be empty set;
  let f be ext-real number;
  cluster F--f -> empty;
  coherence;
end;

registration
  let F be ext-real-membered non empty set;
  let f be ext-real number;
  cluster F--f -> non empty;
  coherence;
end;

registration
  let F be ext-real-membered set;
  let f be ext-real number;
  cluster F--f -> ext-real-membered;
  coherence;
end;

theorem
  F -- f = -- (f -- F) by Th60;

theorem
  f -- F = -- (F -- f) by Th60;

theorem
  (F/\G) -- r = (F--r) /\ (G--r)
proof
  thus (F/\G) -- r = --(r--(F/\G)) by Th60
    .= --((r--F) /\ (r--G)) by Th157
    .= (--(r--F)) /\ (--(r--G)) by Th6
    .= (--(r--F)) /\ (G--r) by Th60
    .= (F--r) /\ (G--r) by Th60;
end;

theorem
  (F\G) -- r = (F--r) \ (G--r)
proof
  thus (F\G) -- r = --(r--(F\G)) by Th60
    .= --((r--F) \ (r--G)) by Th158
    .= (--(r--F)) \ (--(r--G)) by Th7
    .= (--(r--F)) \ (G--r) by Th60
    .= (F--r) \ (G--r) by Th60;
end;

theorem
  (F\+\G) -- r = (F--r) \+\ (G--r)
proof
  thus (F\+\G) -- r = --(r--(F\+\G)) by Th60
    .= --((r--F) \+\ (r--G)) by Th159
    .= (--(r--F)) \+\ (--(r--G)) by Th8
    .= (--(r--F)) \+\ (G--r) by Th60
    .= (F--r) \+\ (G--r) by Th60;
end;

definition
  let A be complex-membered set;
  let a be complex number;
  func A--a equals
  A--{a};
  coherence;
end;

theorem Th176:
  b in A implies b-a in A--a
proof
  a in {a} by TARSKI:def 1;
  hence thesis by Th66;
end;

theorem Th177:
  A--a = {c-a: c in A}
proof
A1: A--a = {c1-c2: c1 in A & c2 in {a}} by Th65;
  thus A--a c= {c-a: c in A}
  proof
    let e;
    assume e in A--a;
    then consider c1,c2 such that
A2: e = c1-c2 & c1 in A and
A3: c2 in {a} by A1;
    c2 = a by A3,TARSKI:def 1;
    hence thesis by A2;
  end;
  let e;
  assume e in {c-a: c in A};
  then ex c st e = c-a & c in A;
  hence thesis by Th176;
end;

theorem Th178:
  e in A--a implies ex c st e = c-a & c in A
proof
  A--a = {c-a: c in A} by Th177;
  hence thesis;
end;

registration
  let A be empty set;
  let a be complex number;
  cluster A--a -> empty;
  coherence;
end;

registration
  let A be complex-membered non empty set;
  let a be complex number;
  cluster A--a -> non empty;
  coherence;
end;

registration
  let A be complex-membered set;
  let a be complex number;
  cluster A--a -> complex-membered;
  coherence;
end;

registration
  let A be real-membered set;
  let a be real number;
  cluster A--a -> real-membered;
  coherence;
end;

registration
  let A be rational-membered set;
  let a be rational number;
  cluster A--a -> rational-membered;
  coherence;
end;

registration
  let A be integer-membered set;
  let a be integer number;
  cluster A--a -> integer-membered;
  coherence;
end;

registration
  let A be real-membered set, F be ext-real-membered set;
  let a be real number, f be ext-real number;
  identify A--a with F--f when a = f, A = F;
  compatibility;
end;

theorem Th179:
  A c= B iff A--a c= B--a
proof
  thus A c= B implies A--a c= B--a by Th67;
  assume
A1: A--a c= B--a;
  let z;
  assume z in A;
  then z-a in A--a by Th176;
  then ex c st z-a = c-a & c in B by A1,Th178;
  hence thesis;
end;

theorem
  A--a = B--a implies A = B
proof
  assume A--a = B--a;
  then A c= B & B c= A by Th179;
  hence thesis by XBOOLE_0:def 10;
end;

theorem
  A -- a = -- (a -- A) by Th71;

theorem
  a -- A = -- (A -- a) by Th71;

theorem
  (A/\B) -- a = (A--a) /\ (B--a)
proof
  thus (A/\B) -- a = --(a--(A/\B)) by Th71
    .= --((a--A) /\ (a--B)) by Th165
    .= (--(a--A)) /\ (--(a--B)) by Th16
    .= (--(a--A)) /\ (B--a) by Th71
    .= (A--a) /\ (B--a) by Th71;
end;

theorem
  (A\B) -- a = (A--a) \ (B--a)
proof
  thus (A\B) -- a = --(a--(A\B)) by Th71
    .= --((a--A) \ (a--B)) by Th166
    .= (--(a--A)) \ (--(a--B)) by Th17
    .= (--(a--A)) \ (B--a) by Th71
    .= (A--a) \ (B--a) by Th71;
end;

theorem
  (A\+\B) -- a = (A--a) \+\ (B--a)
proof
  thus (A\+\B) -- a = --(a--(A\+\B)) by Th71
    .= --((a--A) \+\ (a--B)) by Th167
    .= (--(a--A)) \+\ (--(a--B)) by Th18
    .= (--(a--A)) \+\ (B--a) by Th71
    .= (A--a) \+\ (B--a) by Th71;
end;

definition
  let F be ext-real-membered set;
  let f be ext-real number;
  func f**F equals
  {f}**F;
  coherence;
end;

theorem Th186:
  g in G implies f*g in f**G
proof
  f in {f} by TARSKI:def 1;
  hence thesis by Th79;
end;

theorem Th187:
  f**F = {f*w: w in F}
proof
  thus f**F c= {f*w: w in F}
  proof
    let e;
    assume e in f**F;
    then consider w1,w2 such that
A1: e = w1*w2 and
A2: w1 in {f} and
A3: w2 in F;
    w1 = f by A2,TARSKI:def 1;
    hence thesis by A1,A3;
  end;
  let e;
  assume e in {f*w: w in F};
  then ex w st e = f*w & w in F;
  hence thesis by Th186;
end;

theorem Th188:
  e in f**F implies ex w st e = f*w & w in F
proof
  f**F = {f*w: w in F} by Th187;
  hence thesis;
end;

registration
  let F be empty set;
  let f be ext-real number;
  cluster f**F -> empty;
  coherence;
end;

registration
  let F be ext-real-membered non empty set;
  let f be ext-real number;
  cluster f**F -> non empty;
  coherence;
end;

registration
  let F be ext-real-membered set;
  let f be ext-real number;
  cluster f**F -> ext-real-membered;
  coherence;
end;

theorem
  r <> 0 implies r ** (F/\G) = (r**F) /\ (r**G)
proof
  assume
A1: r <> 0;
A2: (r**F) /\ (r**G) c= r ** (F /\ G)
  proof
    let j;
    assume
A3: j in (r**F) /\ (r**G);
    then j in r**F by XBOOLE_0:def 4;
    then consider w such that
A4: j = r*w and
A5: w in F by Th188;
    j in r**G by A3,XBOOLE_0:def 4;
    then consider w1 such that
A6: j = r*w1 and
A7: w1 in G by Th188;
    w = w1 by A1,A4,A6,XXREAL_3:68;
    then w in F /\ G by A5,A7,XBOOLE_0:def 4;
    hence thesis by A4,Th186;
  end;
  r ** (F /\ G) c= (r**F) /\ (r**G) by Th83;
  hence thesis by A2,XBOOLE_0:def 10;
end;

theorem Th190:
  (f**F) \ (f**G) c= f ** (F \ G)
proof
  let i;
  assume
A1: i in (f**F) \ (f**G);
  then consider w such that
A2: i = f*w and
A3: w in F by Th188;
  now
    assume not w in F\G;
    then w in G by A3,XBOOLE_0:def 5;
    then f*w in f**G by Th186;
    hence contradiction by A1,A2,XBOOLE_0:def 5;
  end;
  hence thesis by A2,Th186;
end;

theorem Th191:
  r <> 0 implies r ** (F\G) = (r**F) \ (r**G)
proof
  assume
A1: r <> 0;
A2: r ** (F \ G) c= (r**F) \ (r**G)
  proof
    let i;
    assume i in r ** (F \ G);
    then consider w such that
A3: i = r*w and
A4: w in F\G by Th188;
A5: now
      assume r*w in r**G;
      then consider w1 such that
A6:   r*w = r*w1 and
A7:   w1 in G by Th188;
      w = w1 by A1,A6,XXREAL_3:68;
      hence contradiction by A4,A7,XBOOLE_0:def 5;
    end;
    r*w in r**F by A4,Th186;
    hence thesis by A3,A5,XBOOLE_0:def 5;
  end;
  (r**F) \ (r**G) c= r ** (F \ G) by Th190;
  hence thesis by A2,XBOOLE_0:def 10;
end;

theorem
  r <> 0 implies r ** (F\+\G) = (r**F) \+\ (r**G)
proof
  assume
A1: r <> 0;
  thus r ** (F \+\ G) = (r**(F\G)) \/ (r**(G\F)) by Th82
    .= ((r**F)\(r**G)) \/ (r**(G\F)) by A1,Th191
    .= (r**F) \+\ (r**G) by A1,Th191;
end;

definition
  let A be complex-membered set;
  let a be complex number;
  func a**A equals
  {a}**A;
  coherence;
end;

theorem Th193:
  b in A implies a*b in a**A
proof
  a in {a} by TARSKI:def 1;
  hence thesis by Th89;
end;

theorem Th194:
  a**A = {a*c: c in A}
proof
  thus a**A c= {a*c: c in A}
  proof
    let e;
    assume e in a**A;
    then consider c1,c2 such that
A1: e = c1*c2 and
A2: c1 in {a} and
A3: c2 in A;
    c1 = a by A2,TARSKI:def 1;
    hence thesis by A1,A3;
  end;
  let e;
  assume e in {a*c: c in A};
  then ex c st e = a*c & c in A;
  hence thesis by Th193;
end;

theorem Th195:
  e in a**A implies ex c st e = a*c & c in A
proof
  a**A = {a*c: c in A} by Th194;
  hence thesis;
end;

registration
  let A be empty set;
  let a be complex number;
  cluster a**A -> empty;
  coherence;
end;

registration
  let A be complex-membered non empty set;
  let a be complex number;
  cluster a**A -> non empty;
  coherence;
end;

registration
  let A be complex-membered set;
  let a be complex number;
  cluster a**A -> complex-membered;
  coherence;
end;

registration
  let A be real-membered set;
  let a be real number;
  cluster a**A -> real-membered;
  coherence;
end;

registration
  let A be rational-membered set;
  let a be rational number;
  cluster a**A -> rational-membered;
  coherence;
end;

registration
  let A be integer-membered set;
  let a be integer number;
  cluster a**A -> integer-membered;
  coherence;
end;

registration
  let A be natural-membered set;
  let a be Nat;
  cluster a**A -> natural-membered;
  coherence;
end;

registration
  let A be real-membered set, F be ext-real-membered set;
  let a be real number, f be ext-real number;
  identify a**A with f**F when a = f, A = F;
  compatibility;
end;

theorem Th196:
  a <> 0 & a**A c= a**B implies A c= B
proof
  assume that
A1: a <> 0 and
A2: a**A c= a**B;
  let z;
  assume z in A;
  then a*z in a**A by Th193;
  then ex c st a*z = a*c & c in B by A2,Th195;
  hence thesis by A1,XCMPLX_1:5;
end;

theorem
  a <> 0 & a**A = a**B implies A = B
proof
  assume a <> 0 & a**A = a**B;
  then A c= B & B c= A by Th196;
  hence thesis by XBOOLE_0:def 10;
end;

theorem Th198:
  a <> 0 implies a ** (A/\B) = (a**A) /\ (a**B)
proof
  assume
A1: a <> 0;
A2: (a**A) /\ (a**B) c= a ** (A /\ B)
  proof
    let z;
    assume
A3: z in (a**A) /\ (a**B);
    then z in a**A by XBOOLE_0:def 4;
    then consider c such that
A4: z = a*c and
A5: c in A by Th195;
    z in a**B by A3,XBOOLE_0:def 4;
    then consider c1 such that
A6: z = a*c1 and
A7: c1 in B by Th195;
    c = c1 by A1,A4,A6,XCMPLX_1:5;
    then c in A /\ B by A5,A7,XBOOLE_0:def 4;
    hence thesis by A4,Th193;
  end;
  a ** (A /\ B) c= (a**A) /\ (a**B) by Th93;
  hence thesis by A2,XBOOLE_0:def 10;
end;

theorem Th199:
  a <> 0 implies a ** (A\B) = (a**A) \ (a**B)
proof
  assume
A1: a <> 0;
  let z;
  hereby
    assume z in a ** (A \ B);
    then consider c such that
A2: z = a*c and
A3: c in A\B by Th195;
A4: now
      assume a*c in a**B;
      then consider c1 such that
A5:   a*c = a*c1 and
A6:   c1 in B by Th195;
      c = c1 by A1,A5,XCMPLX_1:5;
      hence contradiction by A3,A6,XBOOLE_0:def 5;
    end;
    a*c in a**A by A3,Th193;
    hence z in (a**A) \ (a**B) by A2,A4,XBOOLE_0:def 5;
  end;
  assume
A7: z in (a**A) \ (a**B);
  then consider c such that
A8: z = a*c and
A9: c in A by Th195;
  now
    assume not c in A\B;
    then c in B by A9,XBOOLE_0:def 5;
    then a*c in a**B by Th193;
    hence contradiction by A7,A8,XBOOLE_0:def 5;
  end;
  hence thesis by A8,Th193;
end;

theorem Th200:
  a <> 0 implies a ** (A\+\B) = (a**A) \+\ (a**B)
proof
  assume
A1: a <> 0;
  thus a ** (A \+\ B) = (a**(A\B)) \/ (a**(B\A)) by Th92
    .= ((a**A)\(a**B)) \/ (a**(B\A)) by A1,Th199
    .= (a**A) \+\ (a**B) by A1,Th199;
end;

theorem Th201:
  0**A c= {0}
proof
  let a;
  assume a in 0**A;
  then consider c1,c2 such that
A1: a = c1*c2 and
A2: c1 in {0} and
  c2 in A;
  c1 = 0 by A2,TARSKI:def 1;
  hence thesis by A1,TARSKI:def 1;
end;

theorem
  A <> {} implies 0**A = {0}
proof
  assume
A1: A <> {};
A2: {0} c= 0**A
  proof
    set x = the Element of A;
    let a be Nat;
    assume
A3: a in {0};
    x in COMPLEX by XCMPLX_0:def 2;
    then
A4: 0*x in {0*c: c in A} by A1;
    0**A = {0*c: c in A} by Th194;
    hence thesis by A3,A4,TARSKI:def 1;
  end;
  0**A c= {0} by Th201;
  hence thesis by A2,XBOOLE_0:def 10;
end;

theorem
  1**A = A
proof
  let a;
  hereby
    assume a in 1**A;
    then consider c1,c2 such that
A1: a = c1*c2 and
A2: c1 in {1} and
A3: c2 in A;
    c1 = 1 by A2,TARSKI:def 1;
    hence a in A by A1,A3;
  end;
  assume a in A;
  then 1*a in {1*c: c in A};
  hence thesis by Th194;
end;

theorem
  (a*b)**A = a**(b**A)
proof
  thus (a*b)**A = {a}**{b}**A by Th98
    .= a**(b**A) by Th90;
end;

theorem
  a**(A**B) = (a**A)**B by Th90;

theorem
  (a+b)**A c= a**A ++ b**A
proof
  {a}++{b} = {a+b} by Th51;
  hence thesis by Th95;
end;

theorem
  (a-b)**A c= a**A -- b**A
proof
  {a}--{b} = {a-b} by Th75;
  hence thesis by Th96;
end;

theorem Th208:
  a**(B++C) = a**B++a**C
proof
A1: a**B++a**C c= a**(B++C)
  proof
    let z;
    assume z in a**B++a**C;
    then consider c,c1 such that
A2: z = c+c1 and
A3: c in a**B and
A4: c1 in a**C;
    consider c3 being Element of COMPLEX such that
A5: c1 = a*c3 and
A6: c3 in C by A4,Th195;
    consider c2 such that
A7: c = a*c2 and
A8: c2 in B by A3,Th195;
A9: c2+c3 in B++C by A8,A6;
    z = a*(c2+c3) by A2,A7,A5;
    hence thesis by A9,Th193;
  end;
  a**(B++C) c= a**B++a**C by Th95;
  hence thesis by A1,XBOOLE_0:def 10;
end;

theorem
  a**(B--C) = a**B--a**C
proof
  thus a**(B--C) = a**B++a**--C by Th208
    .= a**B--a**C by Th94;
end;

definition
  let F be ext-real-membered set;
  let f be ext-real number;
  func f///F equals
  {f}///F;
  coherence;
end;

theorem Th210:
  g in G implies f/g in f///G
proof
  f in {f} by TARSKI:def 1;
  hence thesis by Th102;
end;

theorem Th211:
  f///F = {f/w: w in F}
proof
  thus f///F c= {f/w: w in F}
  proof
    let e;
    assume e in f///F;
    then consider w1,w2 such that
A1: e = w1*w2 and
A2: w1 in {f} and
A3: w2 in F"";
    consider w3 being Element of ExtREAL such that
A4: w2 = w3" & w3 in F by A3;
    w1*w3" = w1/w3 & w1 = f by A2,TARSKI:def 1;
    hence thesis by A1,A4;
  end;
  let e;
  assume e in {f/w: w in F};
  then ex w st e = f/w & w in F;
  hence thesis by Th210;
end;

theorem
  e in f///F implies ex w st e = f/w & w in F
proof
  f///F = {f/w: w in F} by Th211;
  hence thesis;
end;

registration
  let F be empty set;
  let f be ext-real number;
  cluster f///F -> empty;
  coherence;
end;

registration
  let F be ext-real-membered non empty set;
  let f be ext-real number;
  cluster f///F -> non empty;
  coherence;
end;

registration
  let F be ext-real-membered set;
  let f be ext-real number;
  cluster f///F -> ext-real-membered;
  coherence;
end;

definition
  let A be complex-membered set;
  let a be complex number;
  func a///A equals
  {a}///A;
  coherence;
end;

theorem Th213:
  b in A implies a/b in a///A
proof
  a in {a} by TARSKI:def 1;
  hence thesis by Th116;
end;

theorem Th214:
  a///A = {a/c: c in A}
proof
  thus a///A c= {a/c: c in A}
  proof
    let e;
    assume e in a///A;
    then consider c1,c2 such that
A1: e = c1*c2 and
A2: c1 in {a} & c2 in A"";
A3: c1*c2 = c1/(c2");
    c2" in A & c1 = a by A2,Th29,TARSKI:def 1;
    hence thesis by A1,A3;
  end;
  let e;
  assume e in {a/c: c in A};
  then ex c st e = a/c & c in A;
  hence thesis by Th213;
end;

theorem Th215:
  e in a///A implies ex c st e = a/c & c in A
proof
  a///A = {a/c: c in A} by Th214;
  hence thesis;
end;

registration
  let A be empty set;
  let a be complex number;
  cluster a///A -> empty;
  coherence;
end;

registration
  let A be complex-membered non empty set;
  let a be complex number;
  cluster a///A -> non empty;
  coherence;
end;

registration
  let A be complex-membered set;
  let a be complex number;
  cluster a///A -> complex-membered;
  coherence;
end;

registration
  let A be real-membered set;
  let a be real number;
  cluster a///A -> real-membered;
  coherence;
end;

registration
  let A be rational-membered set;
  let a be rational number;
  cluster a///A -> rational-membered;
  coherence;
end;

registration
  let A be real-membered set, F be ext-real-membered set;
  let a be real number, f be ext-real number;
  identify a///A with f///F when a = f, A = F;
  compatibility;
end;

theorem Th216:
  a <> 0 & a///A c= a///B implies A c= B
proof
  assume that
A1: a <> 0 and
A2: a///A c= a///B;
  let z;
  assume z in A;
  then a/z in a///A by Th213;
  then consider c such that
A3: a/z = a/c and
A4: c in B by A2,Th215;
  z" = c" by A1,A3,XCMPLX_1:5;
  hence thesis by A4,XCMPLX_1:201;
end;

theorem
  a <> 0 & a///A = a///B implies A = B
proof
  assume a <> 0 & a///A = a///B;
  then A c= B & B c= A by Th216;
  hence thesis by XBOOLE_0:def 10;
end;

theorem
  a <> 0 implies a /// (A/\B) = (a///A) /\ (a///B)
proof
  assume
A1: a <> 0;
  thus a /// (A/\B) = a ** ((A"")/\(B"")) by Th33
    .= (a**(A"")) /\ (a**(B"")) by A1,Th198
    .= (a///A) /\ (a///B);
end;

theorem
  a <> 0 implies a /// (A\B) = (a///A) \ (a///B)
proof
  assume
A1: a <> 0;
  thus a /// (A\B) = a ** ((A"")\(B"")) by Th34
    .= (a**(A"")) \ (a**(B"")) by A1,Th199
    .= (a///A) \ (a///B);
end;

theorem
  a <> 0 implies a /// (A\+\B) = (a///A) \+\ (a///B)
proof
  assume
A1: a <> 0;
  thus a /// (A\+\B) = a ** ((A"")\+\(B"")) by Th35
    .= (a**(A"")) \+\ (a**(B"")) by A1,Th200
    .= (a///A) \+\ (a///B);
end;

theorem
  (a+b)///A c= a///A ++ b///A
proof
  {a}++{b} = {a+b} by Th51;
  hence thesis by Th95;
end;

theorem
  (a-b)///A c= a///A -- b///A
proof
  {a}--{b} = {a-b} by Th75;
  hence thesis by Th96;
end;

definition
  let F be ext-real-membered set;
  let f be ext-real number;
  func F///f equals
  F///{f};
  coherence;
end;

theorem Th223:
  g in G implies g/f in G///f
proof
  f in {f} by TARSKI:def 1;
  hence thesis by Th102;
end;

theorem Th224:
  F///f = {w/f: w in F}
proof
  thus F///f c= {w/f: w in F}
  proof
    let e;
    assume e in F///f;
    then consider w1,w2 such that
A1: e = w1*w2 & w1 in F and
A2: w2 in {f}"";
    consider w3 being Element of ExtREAL such that
A3: w2 = w3" and
A4: w3 in {f} by A2;
    w1*w3" = w1/w3 & w3 = f by A4,TARSKI:def 1;
    hence thesis by A1,A3;
  end;
  let e;
  assume e in {w/f: w in F};
  then ex w st e = w/f & w in F;
  hence thesis by Th223;
end;

theorem
  e in F///f implies ex w st e = w/f & w in F
proof
  F///f = {w/f: w in F} by Th224;
  hence thesis;
end;

registration
  let F be empty set;
  let f be ext-real number;
  cluster F///f -> empty;
  coherence;
end;

registration
  let F be ext-real-membered non empty set;
  let f be ext-real number;
  cluster F///f -> non empty;
  coherence;
end;

registration
  let F be ext-real-membered set;
  let f be ext-real number;
  cluster F///f -> ext-real-membered;
  coherence;
end;

definition
  let A be complex-membered set;
  let a be complex number;
  func A///a equals
  A///{a};
  coherence;
end;

theorem Th226:
  b in A implies b/a in A///a
proof
  a in {a} by TARSKI:def 1;
  hence thesis by Th116;
end;

theorem Th227:
  A///a = {c/a: c in A}
proof
  thus A///a c= {c/a: c in A}
  proof
    let e;
    assume e in A///a;
    then consider c1,c2 such that
A1: e = c1*c2 & c1 in A and
A2: c2 in {a}"";
    {a}"" = {a"} by Th37;
    then c1*c2 = c1/(c2") & c2 = a" by A2,TARSKI:def 1;
    hence thesis by A1;
  end;
  let e;
  assume e in {c/a: c in A};
  then ex c st e = c/a & c in A;
  hence thesis by Th226;
end;

theorem Th228:
  e in A///a implies ex c st e = c/a & c in A
proof
  A///a = {c/a: c in A} by Th227;
  hence thesis;
end;

registration
  let A be empty set;
  let a be complex number;
  cluster A///a -> empty;
  coherence;
end;

registration
  let A be complex-membered non empty set;
  let a be complex number;
  cluster A///a -> non empty;
  coherence;
end;

registration
  let A be complex-membered set;
  let a be complex number;
  cluster A///a -> complex-membered;
  coherence;
end;

registration
  let A be real-membered set;
  let a be real number;
  cluster A///a -> real-membered;
  coherence;
end;

registration
  let A be rational-membered set;
  let a be rational number;
  cluster A///a -> rational-membered;
  coherence;
end;

registration
  let A be real-membered set, F be ext-real-membered set;
  let a be real number, f be ext-real number;
  identify A///a with F///f when a = f, A = F;
  compatibility;
end;

theorem Th229:
  a <> 0 & A///a c= B///a implies A c= B
proof
  assume that
A1: a <> 0 and
A2: A///a c= B///a;
  let z;
  assume z in A;
  then z/a in A///a by Th226;
  then ex c st z/a = c/a & c in B by A2,Th228;
  hence thesis by A1,XCMPLX_1:5;
end;

theorem
  a <> 0 & A///a = B///a implies A = B
proof
  assume a <> 0 & A///a = B///a;
  then A c= B & B c= A by Th229;
  hence thesis by XBOOLE_0:def 10;
end;

theorem
  a <> 0 implies (A/\B) /// a = (A///a) /\ (B///a)
proof
  assume
A1: a <> 0;
A2: {a}"" = {a"} by Th37;
  thus (A/\B)///a = a"**(A/\B) by Th37
    .= (a"**A)/\(a"**B) by A1,Th198
    .= (A///a) /\ (B///a) by A2;
end;

theorem
  a <> 0 implies (A\B) /// a = (A///a) \ (B///a)
proof
  assume
A1: a <> 0;
A2: {a}"" = {a"} by Th37;
  thus (A\B)///a = a"**(A\B) by Th37
    .= (a"**A)\(a"**B) by A1,Th199
    .= (A///a) \ (B///a) by A2;
end;

theorem
  a <> 0 implies (A\+\B) /// a = (A///a) \+\ (B///a)
proof
  assume
A1: a <> 0;
A2: {a}"" = {a"} by Th37;
  thus (A\+\B)///a = a"**(A\+\B) by Th37
    .= (a"**A)\+\(a"**B) by A1,Th200
    .= (A///a) \+\ (B///a) by A2;
end;

theorem Th234:
  (A++B)///a = A///a ++ B///a
proof
  thus (A++B)///a = a"**(A++B) by Th37
    .= a"**A++a"**B by Th208
    .= A///a++a"**B by Th37
    .= A///a ++ B///a by Th37;
end;

theorem
  (A--B)///a = A///a -- B///a
proof
  thus (A--B)///a = A///a ++ (--B)///a by Th234
    .= A///a -- B///a by Th94;
end;














begin

definition
  mode R_eal is Element of ExtREAL;
end;

definition
  redefine func +infty -> R_eal;
  coherence by XXREAL_0:def 1;
  redefine func -infty -> R_eal;
  coherence by XXREAL_0:def 1;
end;





definition

  let X be ext-real-membered set;
  func SetMajorant(X) -> ext-real-membered set means
  :Def1:
  for x being ext-real number holds x in it iff x is UpperBound of X;
  existence
  proof
    defpred P[ext-real number] means $1 is UpperBound of X;
    consider Y being ext-real-membered set such that
A1: for x being ext-real number holds x in Y iff P[x] from MEMBERED:
    sch 2;
    take Y;
    thus thesis by A1;
  end;
  uniqueness
  proof
    let Y1,Y2 be ext-real-membered set such that
A2: for x being ext-real number holds x in Y1 iff x is UpperBound of X and
A3: for x being ext-real number holds x in Y2 iff x is UpperBound of X;
    let x be ext-real number;
    x in Y1 iff x is UpperBound of X by A2;
    hence thesis by A3;
  end;
end;

registration
  let X be ext-real-membered set;
  cluster SetMajorant(X) -> non empty;
  coherence
  proof
    set x = the UpperBound of X;
    x in SetMajorant(X) by Def1;
    hence thesis;
  end;
end;

theorem
  for X,Y being ext-real-membered set st X c= Y holds for x being
  ext-real number holds x in SetMajorant Y implies x in SetMajorant(X)
proof
  let X,Y be ext-real-membered set;
  assume
A1: X c= Y;
  let x be ext-real number;
  assume x in SetMajorant(Y);
  then x is UpperBound of Y by Def1;
  then x is UpperBound of X by A1,XXREAL_2:6;
  hence thesis by Def1;
end;

definition
  let X be ext-real-membered set;
  func SetMinorant(X) -> ext-real-membered set means
  :Def2:
  for x being ext-real number holds x in it iff x is LowerBound of X;
  existence
  proof
    defpred P[ext-real number] means $1 is LowerBound of X;
    consider Y being ext-real-membered set such that
A1: for x being ext-real number holds x in Y iff P[x] from MEMBERED:
    sch 2;
    take Y;
    thus thesis by A1;
  end;
  uniqueness
  proof
    let Y1,Y2 be ext-real-membered set such that
A2: for x being ext-real number holds x in Y1 iff x is LowerBound of X and
A3: for x being ext-real number holds x in Y2 iff x is LowerBound of X;
    let x be ext-real number;
    x in Y1 iff x is LowerBound of X by A2;
    hence thesis by A3;
  end;
end;

registration
  let X be ext-real-membered set;
  cluster SetMinorant(X) -> non empty;
  coherence
  proof
    set x = the LowerBound of X;
    x in SetMinorant(X) by Def2;
    hence thesis;
  end;
end;

theorem
  for X,Y being ext-real-membered set st X c= Y holds for x being
  ext-real number holds x in SetMinorant(Y) implies x in SetMinorant(X)
proof
  let X,Y be ext-real-membered set;
  assume
A1: X c= Y;
  let x be ext-real number;
  assume x in SetMinorant(Y);
  then x is LowerBound of Y by Def2;
  then x is LowerBound of X by A1,XXREAL_2:5;
  hence thesis by Def2;
end;





theorem
  for X being non empty ext-real-membered set holds sup X = inf
  SetMajorant(X) & inf X = sup SetMinorant(X)
proof
  let X be non empty ext-real-membered set;
  for y being ext-real number st y in SetMajorant X holds sup X <= y
  proof
    let y be ext-real number;
    assume y in SetMajorant X;
    then y is UpperBound of X by Def1;
    hence thesis by XXREAL_2:def 3;
  end;
  then
A1: sup X is LowerBound of SetMajorant X by XXREAL_2:def 2;
  inf X is LowerBound of X by XXREAL_2:def 4;
  then
A2: inf X in SetMinorant X by Def2;
  for y being ext-real number st y in SetMinorant X holds y <= inf X
  proof
    let y be ext-real number;
    assume y in SetMinorant X;
    then y is LowerBound of X by Def2;
    hence thesis by XXREAL_2:def 4;
  end;
  then
A3: inf X is UpperBound of SetMinorant X by XXREAL_2:def 1;
  sup X is UpperBound of X by XXREAL_2:def 3;
  then sup X in SetMajorant X by Def1;
  hence thesis by A1,A2,A3,XXREAL_2:55,56;
end;

registration
  let X be non empty set;
  cluster non empty with_non-empty_elements for Subset-Family of X;
  existence
  proof
    take {[#]X};
    thus {[#]X} is non empty;
    assume {} in {[#]X};
    hence contradiction by TARSKI:def 1;
  end;
end;

definition
  let X be non empty set;
  mode bool_DOMAIN of X is non empty with_non-empty_elements Subset-Family of
    X;
end;

definition
  let F be bool_DOMAIN of ExtREAL;

  func SUP(F) -> ext-real-membered set means
  :Def3:
  for a being ext-real
number holds a in it iff ex A being non empty ext-real-membered set st A in F &
  a = sup A;
  existence
  proof
    defpred P[ext-real number] means ex A being non empty ext-real-membered
    set st A in F & $1 = sup A;
    consider S being ext-real-membered set such that
A1: for a being ext-real number holds a in S iff P[a] from MEMBERED:
    sch 2;
    reconsider S as ext-real-membered set;
    take S;
    thus thesis by A1;
  end;
  uniqueness
  proof
    let S1,S2 be ext-real-membered set such that
A2: for a being ext-real number holds a in S1 iff ex A being non empty
    ext-real-membered set st A in F & a = sup A and
A3: for a being ext-real number holds a in S2 iff ex A being non empty
    ext-real-membered set st A in F & a = sup A;
    let a be ext-real number;
    hereby
      assume a in S1;
      then
      ex A being non empty ext-real-membered set st A in F & a = sup A by A2;
      hence a in S2 by A3;
    end;
    assume a in S2;
    then
    ex A being non empty ext-real-membered set st A in F & a = sup A by A3;
    hence thesis by A2;
  end;
end;

registration
  let F be bool_DOMAIN of ExtREAL;
  cluster SUP(F) -> non empty;
  coherence
  proof
    set A = the Element of F;
    reconsider A as non empty ext-real-membered set by SETFAM_1:def 8;
    sup A = sup A;
    hence thesis by Def3;
  end;
end;

theorem Th4:
  for F being bool_DOMAIN of ExtREAL, S being non empty
  ext-real-membered number st S = union F holds sup S is UpperBound of SUP(F)
proof
  let F be bool_DOMAIN of ExtREAL, S be non empty ext-real-membered set;
  assume
A1: S = union F;
  for x being ext-real number st x in SUP(F) holds x <= sup S
  proof
    let x be ext-real number;
    assume x in SUP(F);
    then consider A being non empty ext-real-membered set such that
A2: A in F and
A3: x = sup A by Def3;
    A c= S
    proof
      let a be ext-real number;
      assume a in A;
      hence thesis by A1,A2,TARSKI:def 4;
    end;
    hence thesis by A3,XXREAL_2:59;
  end;
  hence thesis by XXREAL_2:def 1;
end;

theorem Th5:
  for F being bool_DOMAIN of ExtREAL, S being ext-real-membered
  set st S = union F holds sup SUP(F) is UpperBound of S
proof
  let F be bool_DOMAIN of ExtREAL, S be ext-real-membered set;
  assume
A1: S = union F;
  for x being ext-real number st x in S holds x <= sup SUP(F)
  proof
    let x be ext-real number;
    assume x in S;
    then consider Z being set such that
A2: x in Z and
A3: Z in F by A1,TARSKI:def 4;
    reconsider Z as non empty ext-real-membered set by A2,A3;
    consider a being set such that
A4: a = sup Z;
    reconsider a as ext-real number by A4;
    sup Z is UpperBound of Z & a in SUP(F) by A3,A4,Def3,XXREAL_2:def 3;
    hence thesis by A2,A4,XXREAL_2:61,def 1;
  end;
  hence thesis by XXREAL_2:def 1;
end;

theorem
  for F being bool_DOMAIN of ExtREAL, S being non empty
  ext-real-membered set st S = union F holds sup S = sup SUP(F)
proof
  let F be bool_DOMAIN of ExtREAL, S be non empty ext-real-membered set;
  set a = sup S;
  set b = sup SUP(F);
  assume
A1: S = union F;
  then sup S is UpperBound of SUP(F) by Th4;
  then
A2: b <= a by XXREAL_2:def 3;
  sup SUP(F) is UpperBound of S by A1,Th5;
  then a <= b by XXREAL_2:def 3;
  hence thesis by A2,XXREAL_0:1;
end;

definition
  let F be bool_DOMAIN of ExtREAL;
  func INF F -> ext-real-membered set means
  :Def4:
  for a being ext-real
number holds a in it iff ex A being non empty ext-real-membered set st A in F &
  a = inf A;
  existence
  proof
    set A = the Element of F;
    defpred P[ext-real number] means ex A being non empty ext-real-membered
    set st A in F & $1 = inf A;
    reconsider A as non empty ext-real-membered set by SETFAM_1:def 8;
    consider S being ext-real-membered set such that
A1: for a being ext-real number holds a in S iff P[a] from MEMBERED:
    sch 2;
    inf A = inf A;
    then reconsider S as non empty ext-real-membered set by A1;
    take S;
    thus thesis by A1;
  end;
  uniqueness
  proof
    let S1,S2 be ext-real-membered set such that
A2: for a being ext-real number holds a in S1 iff ex A being non empty
    ext-real-membered set st A in F & a = inf A and
A3: for a being ext-real number holds a in S2 iff ex A being non empty
    ext-real-membered set st A in F & a = inf A;
    for a being set holds a in S1 iff a in S2
    proof
      let a be set;
      hereby
        assume
A4:     a in S1;
        then reconsider a9 = a as ext-real number;
        ex A being non empty ext-real-membered set st A in F & a9 = inf A
        by A2,A4;
        hence a in S2 by A3;
      end;
      assume
A5:   a in S2;
      then reconsider a as ext-real number;
      ex A being non empty ext-real-membered set st A in F & a = inf A by A3,A5
;
      hence thesis by A2;
    end;
    hence thesis by TARSKI:1;
  end;
end;

registration
  let F be bool_DOMAIN of ExtREAL;
  cluster INF(F) -> non empty;
  coherence
  proof
    set A = the Element of F;
    reconsider A as non empty ext-real-membered set by SETFAM_1:def 8;
    inf A = inf A;
    hence thesis by Def4;
  end;
end;

theorem Th7:
  for F being bool_DOMAIN of ExtREAL, S being non empty
  ext-real-membered set st S = union F holds inf S is LowerBound of INF(F)
proof
  let F be bool_DOMAIN of ExtREAL, S be non empty ext-real-membered set;
  assume
A1: S = union F;
  for x being ext-real number st x in INF(F) holds inf S <= x
  proof
    let x be ext-real number;
    assume x in INF(F);
    then consider A being non empty ext-real-membered set such that
A2: A in F and
A3: x = inf A by Def4;
    A c= S
    proof
      let a be ext-real number;
      assume a in A;
      hence thesis by A1,A2,TARSKI:def 4;
    end;
    hence thesis by A3,XXREAL_2:60;
  end;
  hence thesis by XXREAL_2:def 2;
end;

theorem Th8:
  for F being bool_DOMAIN of ExtREAL, S being ext-real-membered
  set st S = union F holds inf INF(F) is LowerBound of S
proof
  let F be bool_DOMAIN of ExtREAL, S be ext-real-membered set;
  assume
A1: S = union F;
  for x being ext-real number st x in S holds inf INF(F) <= x
  proof
    let x be ext-real number;
    assume x in S;
    then consider Z being set such that
A2: x in Z and
A3: Z in F by A1,TARSKI:def 4;
    reconsider Z as non empty ext-real-membered set by A2,A3;
    consider a being set such that
A4: a = inf Z;
    reconsider a as ext-real number by A4;
    inf Z is LowerBound of Z & a in INF(F) by A3,A4,Def4,XXREAL_2:def 4;
    hence thesis by A2,A4,XXREAL_2:62,def 2;
  end;
  hence thesis by XXREAL_2:def 2;
end;

theorem
  for F being bool_DOMAIN of ExtREAL, S being non empty
  ext-real-membered set st S = union F holds inf S = inf INF(F)
proof
  let F be bool_DOMAIN of ExtREAL, S be non empty ext-real-membered set;
  set a = inf S;
  set b = inf INF(F);
  assume
A1: S = union F;
  then inf S is LowerBound of INF(F) by Th7;
  then
A2: a <= b by XXREAL_2:def 4;
  inf INF(F) is LowerBound of S by A1,Th8;
  then b <= a by XXREAL_2:def 4;
  hence thesis by A2,XXREAL_0:1;
end;














begin

reserve x, a, b, c for real number;

definition
  let a,b,c be complex number;
  func delta(a,b,c) equals
  b^2 - 4 * a * c;
  coherence;
end;

registration
  let a,b,c be complex number;
  cluster delta(a,b,c) -> complex;
  coherence;
end;

registration
  let a,b,c;
  cluster delta(a,b,c) -> real;
  coherence;
end;

definition
  let a,b,c be Real;
  redefine func delta(a,b,c) -> Real;
  coherence by XREAL_0:def 1;
end;

theorem Th1:
  for a, b, c, x being complex number holds a <> 0 implies a * x^2
  + b * x + c = a * (x + b/(2 * a))^2 - delta(a,b,c)/(4 * a)
proof
  let a, b, c, x be complex number;
  assume
A1: a <> 0;
  then
A2: 4 * a <> 0;
  a * x^2 + b * x + c = a * x^2 + (b * x) * 1 + c
    .= a * x^2 + (b * x) * (a * (1/a)) + c by A1,XCMPLX_1:106
    .= a * (x^2 + ((b * x) * (1/a))) + c
    .= a * (x^2 + ((b * x)/a)) + c by XCMPLX_1:99
    .= a * (x^2 + x * (b/a)) + c by XCMPLX_1:74
    .= a * (x^2 + x * ((2 * b)/(2 * a))) + c by XCMPLX_1:91
    .= a * (x^2 + x * (2 * (b/(2 * a)))) + c by XCMPLX_1:74
    .= (a * (x^2 + 2 * x * (b/(2 * a))) + b^2/(4 * a)) + (c - b^2/(4 * a))
    .= (a * (x^2 + 2 * x * (b/(2 * a))) + a * (b^2/(4 * a) * (1/a))) + (c -
  b^2/(4 * a)) by A1,XCMPLX_1:109
    .= (a * (x^2 + 2 * x * (b/(2 * a))) + a * ((b^2 * 1)/((4 * a) * a))) + (
  c - b^2/(4 * a)) by XCMPLX_1:76
    .= (a * (x^2 + 2 * x * (b/(2 * a))) + a * (b^2/(2 * a)^2)) + (c - b^2/(4
  * a))
    .= (a * (x^2 + 2 * x * (b/(2 * a))) + a * (b/(2 * a))^2) + (c - b^2/(4 *
  a)) by XCMPLX_1:76
    .= a * (x + b/(2 * a))^2 - (b^2/(4 * a) - c)
    .= a * (x + b/(2 * a))^2 - (b^2/(4 * a) - ((4 * a * c)/(4 * a))) by A2,
XCMPLX_1:89
    .= a * (x + b/(2 * a))^2 - (b^2 - (4 * a * c))/(4 * a) by XCMPLX_1:120;
  hence thesis;
end;

theorem
  a > 0 & delta(a,b,c) <= 0 implies a * x^2 + b * x + c >= 0
proof
  assume that
A1: a > 0 and
A2: delta(a,b,c) <= 0;
  - delta(a,b,c) >= -0 & 4 * a > 0 by A1,A2,XREAL_1:25,129;
  then (- delta(a,b,c))/(4 * a) >= 0 by XREAL_1:136;
  then - delta(a,b,c)/(4 * a) >= 0 by XCMPLX_1:187;
  then
A3: a * (x + b/(2 * a))^2 + - delta(a,b,c)/(4 * a) >= a * (x + b/(2 * a) )^2
  + 0 by XREAL_1:7;
  (x +b/(2 * a))^2 >= 0 by XREAL_1:63;
  then a * (x +b/(2 * a))^2 >= 0 by A1,XREAL_1:127;
  then a * (x + b/(2 * a))^2 - delta(a,b,c)/(4 * a) >= 0 by A3,XXREAL_0:2;
  hence thesis by A1,Th1;
end;

theorem
  a > 0 & delta(a,b,c) < 0 implies a * x^2 + b * x + c > 0
proof
  assume that
A1: a > 0 and
A2: delta(a,b,c) < 0;
  - delta(a,b,c) > -0 & 4 * a > 0 by A1,A2,XREAL_1:26,129;
  then (- delta(a,b,c))/(4 * a) > 0 by XREAL_1:139;
  then - delta(a,b,c)/(4 * a) > 0 by XCMPLX_1:187;
  then
A3: a * (x + b/(2 * a))^2 + - delta(a,b,c)/(4 * a) > a * (x + b/(2 * a)) ^2
  by XREAL_1:29;
  (x +b/(2 * a))^2 >= 0 by XREAL_1:63;
  then a * (x +b/(2 * a))^2 >= 0 by A1,XREAL_1:127;
  then a * (x + b/(2 * a))^2 - delta(a,b,c)/(4 * a) > 0 by A3,XXREAL_0:2;
  hence thesis by A1,Th1;
end;

theorem
  a < 0 & delta(a,b,c) <= 0 implies a * x^2 + b * x + c <= 0
proof
  assume that
A1: a < 0 and
A2: delta(a,b,c) <= 0;
  - delta(a,b,c) >= -0 & 4 * a < 0 by A1,A2,XREAL_1:25,132;
  then (- delta(a,b,c))/(4 * a) <= 0 by XREAL_1:137;
  then - delta(a,b,c)/(4 * a) <= 0 by XCMPLX_1:187;
  then
A3: a * (x + b/(2 * a))^2 + - delta(a,b,c)/(4 * a) <= a * (x + b/(2 * a) )^2
  + 0 by XREAL_1:7;
  a * (x +b/(2 * a))^2 <= 0 by A1,XREAL_1:63,131;
  then a * (x + b/(2 * a))^2 - delta(a,b,c)/(4 * a) <= 0 by A3,XXREAL_0:2;
  hence thesis by A1,Th1;
end;

theorem
  a < 0 & delta(a,b,c) < 0 implies a * x^2 + b * x + c < 0
proof
  assume that
A1: a < 0 and
A2: delta(a,b,c) < 0;
  - delta(a,b,c) > 0 & 4 * a < 0 by A1,A2,XREAL_1:58,132;
  then (- delta(a,b,c))/(4 * a) < 0 by XREAL_1:142;
  then - delta(a,b,c)/(4 * a) < 0 by XCMPLX_1:187;
  then
A3: a * (x + b/(2 * a))^2 + - delta(a,b,c)/(4 * a) < a * (x + b/(2 * a)) ^2
  + 0 by XREAL_1:6;
  a * (x +b/(2 * a))^2 <= 0 by A1,XREAL_1:63,131;
  then a * (x + b/(2 * a))^2 - delta(a,b,c)/(4 * a) < 0 by A3,XXREAL_0:2;
  hence thesis by A1,Th1;
end;

theorem Th6:
  a > 0 & a * x^2 + b * x + c >= 0 implies (2 * a * x + b)^2 -
  delta(a,b,c) >= 0
proof
  assume that
A1: a > 0 and
A2: a * x^2 + b * x + c >= 0;
A3: 4 * a <> 0 by A1;
  4 * a > 0 & a * (x + b/(2 * a))^2 - delta(a,b,c)/(4 * a) >= 0 by A1,A2,Th1,
XREAL_1:129;
  then (4 * a) * (a * (x + b/(2 * a))^2 - delta(a,b,c)/(4 * a)) >= 0 by
XREAL_1:127;
  then
A4: ((2 * a) * x + (2 * a) * (b/(2 * a)))^2 - (4 * a) * (delta(a,b,c)/(4 * a
  )) >= 0;
  2 * a <> 0 by A1;
  then (2 * a * x + b)^2 - (4 * a) * (delta(a,b,c)/(4 * a)) >= 0 by A4,
XCMPLX_1:87;
  hence thesis by A3,XCMPLX_1:87;
end;

theorem Th7:
  a > 0 & a * x^2 + b * x + c > 0 implies (2 * a * x + b)^2 - delta (a,b,c) > 0
proof
  assume that
A1: a > 0 and
A2: a * x^2 + b * x + c > 0;
A3: 4 * a <> 0 by A1;
  4 * a > 0 & a * (x + b/(2 * a))^2 - delta(a,b,c)/(4 * a) > 0 by A1,A2,Th1,
XREAL_1:129;
  then (4 * a) * (a * (x + b/(2 * a))^2 - delta(a,b,c)/(4 * a)) > 0 by
XREAL_1:129;
  then
A4: ((2 * a) * x + (2 * a) * (b/(2 * a)))^2 - (4 * a) * (delta(a,b,c)/(4 * a
  )) > 0;
  2 * a <> 0 by A1;
  then (2 * a * x + b)^2 - (4 * a) * (delta(a,b,c)/(4 * a)) > 0 by A4,
XCMPLX_1:87;
  hence thesis by A3,XCMPLX_1:87;
end;

theorem Th8:
  a < 0 & a * x^2 + b * x + c <= 0 implies (2 * a * x + b)^2 -
  delta(a,b,c) >= 0
proof
  assume that
A1: a < 0 and
A2: a * x^2 + b * x + c <= 0;
A3: 4 * a <> 0 by A1;
  4 * a < 0 & a * (x + b/(2 * a))^2 - delta(a,b,c)/(4 * a) <= 0 by A1,A2,Th1,
XREAL_1:132;
  then (4 * a) * (a * (x + b/(2 * a))^2 - delta(a,b,c)/(4 * a)) >= 0 by
XREAL_1:128;
  then
A4: ((2 * a) * x + (2 * a) * (b/(2 * a)))^2 - (4 * a) * (delta(a,b,c)/(4 * a
  )) >= 0;
  2 * a <> 0 by A1;
  then (2 * a * x + b)^2 - (4 * a) * (delta(a,b,c)/(4 * a)) >= 0 by A4,
XCMPLX_1:87;
  hence thesis by A3,XCMPLX_1:87;
end;

theorem Th9:
  a < 0 & a * x^2 + b * x + c < 0 implies (2 * a * x + b)^2 - delta (a,b,c) > 0
proof
  assume that
A1: a < 0 and
A2: a * x^2 + b * x + c < 0;
A3: 4 * a <> 0 by A1;
  4 * a < 0 & a * (x + b/(2 * a))^2 - delta(a,b,c)/(4 * a) < 0 by A1,A2,Th1,
XREAL_1:132;
  then (4 * a) * (a * (x + b/(2 * a))^2 - delta(a,b,c)/(4 * a)) > 0 by
XREAL_1:130;
  then
A4: ((2 * a) * x + (2 * a) * (b/(2 * a)))^2 - (4 * a) * (delta(a,b,c)/(4 * a
  )) > 0;
  2 * a <> 0 by A1;
  then (2 * a * x + b)^2 - (4 * a) * (delta(a,b,c)/(4 * a)) > 0 by A4,
XCMPLX_1:87;
  hence thesis by A3,XCMPLX_1:87;
end;

theorem
  ( for x holds a * x^2 + b * x + c >= 0 ) & a > 0 implies delta(a,b,c) <= 0
proof
  assume that
A1: for x holds a * x^2 + b * x + c >= 0 and
A2: a > 0;
  a * (- b/(2 * a))^2 + b * (- b/(2 * a)) + c >= 0 by A1;
  then (2 * a * (- b/(2 * a)) + b)^2 - delta(a,b,c) >= 0 by A2,Th6;
  then
A3: (- (2 * a) * (b/(2 * a)) + b)^2 - delta(a,b,c) >= 0;
  2 * a <> 0 by A2;
  then (- b + b)^2 - delta(a,b,c) >= 0 by A3,XCMPLX_1:87;
  then - delta(a,b,c) >= -0;
  hence thesis by XREAL_1:24;
end;

theorem
  ( for x holds a * x^2 + b * x + c <= 0 ) & a < 0 implies delta(a,b,c) <= 0
proof
  assume that
A1: for x holds a * x^2 + b * x + c <= 0 and
A2: a < 0;
  a * (- b/(2 * a))^2 + b * (- b/(2 * a)) + c <= 0 by A1;
  then (2 * a * (- b/(2 * a)) + b)^2 - delta(a,b,c) >= 0 by A2,Th8;
  then
A3: (- (2 * a) * (b/(2 * a)) + b)^2 - delta(a,b,c) >= 0;
  2 * a <> 0 by A2;
  then (- b + b)^2 - delta(a,b,c) >= 0 by A3,XCMPLX_1:87;
  then - delta(a,b,c) >= -0;
  hence thesis by XREAL_1:24;
end;

theorem
  ( for x holds a * x^2 + b * x + c > 0 ) & a > 0 implies delta(a,b,c) < 0
proof
  assume that
A1: for x holds a * x^2 + b * x + c > 0 and
A2: a > 0;
  a * (- b/(2 * a))^2 + b * (- b/(2 * a)) + c > 0 by A1;
  then (2 * a * (- b/(2 * a)) + b)^2 - delta(a,b,c) > 0 by A2,Th7;
  then
A3: (- (2 * a) * (b/(2 * a)) + b)^2 - delta(a,b,c) > 0;
  2 * a <> 0 by A2;
  then (- b + b)^2 - delta(a,b,c) > 0 by A3,XCMPLX_1:87;
  then - delta(a,b,c) > 0;
  hence thesis by XREAL_1:58;
end;

theorem
  ( for x holds a * x^2 + b * x + c < 0 ) & a < 0 implies delta(a,b,c) < 0
proof
  assume that
A1: for x holds a * x^2 + b * x + c < 0 and
A2: a < 0;
  a * (- b/(2 * a))^2 + b * (- b/(2 * a)) + c < 0 by A1;
  then (2 * a * (- b/(2 * a)) + b)^2 - delta(a,b,c) > 0 by A2,Th9;
  then
A3: (- (2 * a) * (b/(2 * a)) + b)^2 - delta(a,b,c) > 0;
  2 * a <> 0 by A2;
  then (- b + b)^2 - delta(a,b,c) > 0 by A3,XCMPLX_1:87;
  then - delta(a,b,c) > -0;
  hence thesis by XREAL_1:24;
end;

theorem Th14:
  for a, b, c, x being complex number holds a <> 0 & a * x^2 + b *
  x + c = 0 implies (2 * a * x + b)^2 - delta(a,b,c) = 0
proof
  let a, b, c, x be complex number;
  assume that
A1: a <> 0 and
A2: a * x^2 + b * x + c = 0;
A3: 4 * a <> 0 by A1;
  a * (x + b/(2 * a))^2 - delta(a,b,c)/(4 * a) = 0 by A1,A2,Th1;
  then
A4: ((2 * a) * x + (2 * a) * (b/(2 * a)))^2 - (4 * a) * (delta(a,b,c)/(4 * a
  )) = 0;
  2 * a <> 0 by A1;
  then (2 * a * x + b)^2 - (4 * a) * (delta(a,b,c)/(4 * a)) = 0 by A4,
XCMPLX_1:87;
  hence thesis by A3,XCMPLX_1:87;
end;

Lm1: for a, b being complex number holds a^2 = b^2 implies a = b or a = - b
proof
  let a, b be complex number;
  assume a^2 = b^2;
  then (a + b) * (a - b) = 0;
  then a + b = 0 or a - b = 0 by XCMPLX_1:6;
  hence thesis;
end;

theorem
  a <> 0 & delta(a,b,c) >= 0 & a * x^2 + b * x + c = 0 implies x = (- b
  - sqrt delta(a,b,c))/(2 * a) or x = (- b + sqrt delta(a,b,c))/(2 * a)
proof
  assume that
A1: a <> 0 and
A2: delta(a,b,c) >= 0 and
A3: a * x^2 + b * x + c = 0;
  (2 * a * x + b)^2 - delta(a,b,c) = 0 by A1,A3,Th14;
  then (2 * a * x + b)^2 = (sqrt delta(a,b,c))^2 by A2,SQUARE_1:def 2;
  then
A4: 2 * a * x + b = sqrt delta(a,b,c) or 2 * a * x + b = - sqrt delta(a, b,c
  ) by Lm1;
  2 * a <> 0 by A1;
  hence thesis by A4,XCMPLX_1:89;
end;

theorem Th16:
  a <> 0 & delta(a,b,c) >= 0 implies a * x^2 + b * x + c = a * (x
- (- b - sqrt delta(a,b,c))/(2 * a)) * (x - (- b + sqrt delta(a,b,c))/(2 * a))
proof
  assume that
A1: a <> 0 and
A2: delta(a,b,c) >= 0;
  a * x^2 + b * x + c = a * (x + b/(2 * a))^2 - 1 * (delta(a,b,c)/(4 * a))
  by A1,Th1
    .= a * (x + b/(2 * a))^2 - (a * (1/a)) * (delta(a,b,c)/(4 * a)) by A1,
XCMPLX_1:106
    .= a * ((x + b/(2 * a))^2 - (1/a) * (delta(a,b,c)/(4 * a)))
    .= a * ((x + b/(2 * a))^2 - (delta(a,b,c) * 1)/((4 * a) * a)) by
XCMPLX_1:76
    .= a * ((x + b/(2 * a))^2 - (sqrt delta(a,b,c))^2/(2 * a)^2) by A2,
SQUARE_1:def 2
    .= a * ((x + b/(2 * a))^2 - (sqrt delta(a,b,c)/(2 * a))^2) by XCMPLX_1:76
    .= a * (x - (- b/(2 * a) + sqrt delta(a,b,c)/(2 * a))) * (x - (- b/(2 *
  a) - sqrt delta(a,b,c)/(2 * a)))
    .= a * (x - ((- b)/(2 * a) + sqrt delta(a,b,c)/(2 * a))) * (x - (- b/(2
  * a) - sqrt delta(a,b,c)/(2 * a))) by XCMPLX_1:187
    .= a * (x - ((- b)/(2 * a) + sqrt delta(a,b,c)/(2 * a))) * (x - ((- b)/(
  2 * a) - sqrt delta(a,b,c)/(2 * a))) by XCMPLX_1:187
    .= a * (x - (- b + sqrt delta(a,b,c))/(2 * a)) * (x - ((- b)/(2 * a) -
  sqrt delta(a,b,c)/(2 * a))) by XCMPLX_1:62
    .= a * (x - (- b + sqrt delta(a,b,c))/(2 * a)) * (x - (- b - sqrt delta(
  a,b,c))/(2 * a)) by XCMPLX_1:120;
  hence thesis;
end;

theorem Th17:
  a < 0 & delta(a,b,c) > 0 implies (- b + sqrt delta(a,b,c))/(2 *
  a) < (- b - sqrt delta(a,b,c))/(2 * a)
proof
  assume that
A1: a < 0 and
A2: delta(a,b,c) > 0;
  sqrt delta(a,b,c) > 0 by A2,SQUARE_1:25;
  then 2 * sqrt delta(a,b,c) > 0 by XREAL_1:129;
  then sqrt delta(a,b,c) + sqrt delta(a,b,c) > 0;
  then sqrt delta(a,b,c) > - sqrt delta(a,b,c) by XREAL_1:59;
  then
A3: - b + sqrt delta(a,b,c) > - b + - sqrt delta(a,b,c) by XREAL_1:6;
  2 * a < 0 by A1,XREAL_1:132;
  hence thesis by A3,XREAL_1:75;
end;

theorem
  a < 0 & delta(a,b,c) > 0 implies ( a * x^2 + b * x + c > 0 iff (- b +
  sqrt delta(a,b,c))/(2 * a) < x & x < (- b - sqrt delta(a,b,c))/(2 * a) )
proof
  assume that
A1: a < 0 and
A2: delta(a,b,c) > 0;
  thus a * x^2 + b * x + c > 0 implies (- b + sqrt delta(a,b,c))/(2 * a) < x &
  x < (- b - sqrt delta(a,b,c))/(2 * a)
  proof
    assume a * x^2 + b * x + c > 0;
    then
    a * (x - (- b - sqrt delta(a,b,c))/(2 * a)) * (x - (- b + sqrt delta(a
    ,b,c))/(2 * a)) > 0 by A1,A2,Th16;
    then
    a * ((x - (- b - sqrt delta(a,b,c))/(2 * a)) * (x - (- b + sqrt delta(
    a,b,c))/(2 * a))) > 0;
    then
    (x - (- b - sqrt delta(a,b,c))/(2 * a)) * (x - (- b + sqrt delta(a,b,c
    ))/(2 * a)) < 0/a by A1,XREAL_1:84;
    then
    x - (- b - sqrt delta(a,b,c))/(2 * a) > 0 & x - (- b + sqrt delta(a,b,
c))/(2 * a) < 0 or x - (- b - sqrt delta(a,b,c))/(2 * a) < 0 & x - (- b + sqrt
    delta(a,b,c))/(2 * a) > 0 by XREAL_1:133;
    then
    x > (- b - sqrt delta(a,b,c))/(2 * a) & x < (- b + sqrt delta(a,b,c))/
(2 * a) & (- b + sqrt delta(a,b,c))/(2 * a) < (- b - sqrt delta(a,b,c))/(2 * a)
or x < (- b - sqrt delta(a,b,c))/(2 * a) & x > (- b + sqrt delta(a,b,c))/(2 * a
    ) by A1,A2,Th17,XREAL_1:47,48;
    hence thesis by XXREAL_0:2;
  end;
  assume (- b + sqrt delta(a,b,c))/(2 * a) < x & x < (- b - sqrt delta(a,b,c
  ))/(2 * a );
  then
  x - (- b + sqrt delta(a,b,c))/(2 * a) > 0 & x - (- b - sqrt delta(a,b,c
  ))/(2 * a) < 0 by XREAL_1:49,50;
  then
  (x - (- b + sqrt delta(a,b,c))/(2 * a)) * (x - (- b - sqrt delta(a,b,c)
  )/(2 * a)) < 0 by XREAL_1:132;
  then
  a * ((x - (- b + sqrt delta(a,b,c))/(2 * a)) * (x - (- b - sqrt delta(a
  ,b,c))/(2 * a))) > 0 by A1,XREAL_1:130;
  then
  a * (x - (- b - sqrt delta(a,b,c))/(2 * a)) * (x - (- b + sqrt delta(a,
  b,c))/(2 * a)) > 0;
  hence thesis by A1,A2,Th16;
end;

theorem
  a < 0 & delta(a,b,c) > 0 implies ( a * x^2 + b * x + c < 0 iff x < (-
  b + sqrt delta(a,b,c))/(2 * a) or x > (- b - sqrt delta(a,b,c))/(2 * a) )
proof
  assume that
A1: a < 0 and
A2: delta(a,b,c) > 0;
  thus a * x^2 + b * x + c < 0 implies x < (- b + sqrt delta(a,b,c))/(2 * a)
  or x > (- b - sqrt delta(a,b,c))/(2 * a)
  proof
    assume a * x^2 + b * x + c < 0;
    then
    a * (x - (- b - sqrt delta(a,b,c))/(2 * a)) * (x - (- b + sqrt delta(a
    ,b,c))/(2 * a)) < 0 by A1,A2,Th16;
    then
    a * ((x - (- b - sqrt delta(a,b,c))/(2 * a)) * (x - (- b + sqrt delta(
    a,b,c))/(2 * a))) < 0;
    then
    (x - (- b - sqrt delta(a,b,c))/(2 * a)) * (x - (- b + sqrt delta(a,b,c
    ))/(2 * a)) > 0/a by A1,XREAL_1:82;
    then
    x - (- b - sqrt delta(a,b,c))/(2 * a) > 0 & x - (- b + sqrt delta(a,b,
c))/(2 * a) > 0 or x - (- b - sqrt delta(a,b,c))/(2 * a) < 0 & x - (- b + sqrt
    delta(a,b,c))/(2 * a) < 0 by XREAL_1:134;
    hence thesis by XREAL_1:47,48;
  end;
  assume x < (- b + sqrt delta(a,b,c))/(2 * a) or x > (- b - sqrt delta(a,b,
  c))/(2 * a);
  then
A3: x - (- b + sqrt delta(a,b,c))/(2 * a) < 0 or x - (- b - sqrt delta(a,b,
  c))/(2 * a) > 0 by XREAL_1:49,50;
  (- b + sqrt delta(a,b,c))/(2 * a) < (- b - sqrt delta(a,b,c))/(2 * a)
  by A1,A2,Th17;
  then
  x - (- b + sqrt delta(a,b,c))/(2 * a) > x - (- b - sqrt delta(a,b,c))/(
  2 * a) by XREAL_1:10;
  then
  x - (- b + sqrt delta(a,b,c))/(2 * a) < 0 & x - (- b - sqrt delta(a,b,c
  ))/(2 * a) < 0 or x - (- b - sqrt delta(a,b,c))/(2 * a) > 0 & x - (- b + sqrt
  delta(a,b,c))/(2 * a) > 0 by A3,XXREAL_0:2;
  then
  (x - (- b + sqrt delta(a,b,c))/(2 * a)) * (x - (- b - sqrt delta(a,b,c)
  )/(2 * a)) > 0 by XREAL_1:129,130;
  then
  a * ((x - (- b + sqrt delta(a,b,c))/(2 * a)) * (x - (- b - sqrt delta(a
  ,b,c))/(2 * a))) < 0 by A1,XREAL_1:132;
  then
  a * (x - (- b - sqrt delta(a,b,c))/(2 * a)) * (x - (- b + sqrt delta(a,
  b,c))/(2 * a)) < 0;
  hence thesis by A1,A2,Th16;
end;

theorem
  for a, b, c, x being complex number holds a <> 0 & delta(a,b,c) = 0 &
  a * x^2 + b * x + c = 0 implies x = - b/(2 * a)
proof
  let a, b, c, x be complex number;
  assume that
A1: a <> 0 and
A2: delta(a,b,c) = 0 & a * x^2 + b * x + c = 0;
  (2 * a * x + b)^2 - 0 = 0 by A1,A2,Th14;
  then
A3: 2 * a * x + b = 0 by XCMPLX_1:6;
  2 * a <> 0 by A1;
  then x = (- b)/(2 * a) by A3,XCMPLX_1:89;
  hence thesis by XCMPLX_1:187;
end;

theorem Th21:
  a > 0 & (2 * a * x + b)^2 - delta(a,b,c) > 0 implies a * x^2 + b * x + c > 0
proof
  assume that
A1: a > 0 and
A2: (2 * a * x + b)^2 - delta(a,b,c) > 0;
  4 * a <> 0 by A1;
  then
A3: (2 * a * x + b)^2 - (4 * a) * (delta(a,b,c)/(4 * a)) > 0 by A2,XCMPLX_1:87;
  2 * a <> 0 by A1;
  then
  (2 * a * x + (2 * a) * (b/(2 * a)))^2 - (4 * a) * (delta(a,b,c)/(4 * a))
  > 0 by A3,XCMPLX_1:87;
  then
A4: (4 * a) * (a * (x + b/(2 * a))^2 - delta(a,b,c)/(4 * a)) > 0;
  4 * a > 0 by A1,XREAL_1:129;
  then a * (x + b/(2 * a))^2 - delta(a,b,c)/(4 * a) > 0/(4 * a) by A4,
XREAL_1:83;
  hence thesis by A1,Th1;
end;

theorem
  a > 0 & delta(a,b,c) = 0 implies ( a * x^2 + b * x + c > 0 iff x <> -
  b/(2 * a) )
proof
  assume that
A1: a > 0 and
A2: delta(a,b,c) = 0;
A3: 2 * a <> 0 by A1;
  thus a * x^2 + b * x + c > 0 implies x <> - b/(2 * a)
  proof
    assume a * x^2 + b * x + c > 0;
    then (2 * a * x + b)^2 - 0 > 0 by A1,A2,Th7;
    then 2 * a * x <> - b;
    then x <> (- b)/(2 * a) by A3,XCMPLX_1:87;
    hence thesis by XCMPLX_1:187;
  end;
  assume x <> - b/(2 * a);
  then x <> (- b)/(2 * a) by XCMPLX_1:187;
  then 2 * a * x + b <> 0 by A3,XCMPLX_1:89;
  then (2 * a * x + b)^2 - delta(a,b,c) > 0 by A2,SQUARE_1:12;
  hence thesis by A1,Th21;
end;

theorem Th23:
  a < 0 & (2 * a * x + b)^2 - delta(a,b,c) > 0 implies a * x^2 + b * x + c < 0
proof
  assume that
A1: a < 0 and
A2: (2 * a * x + b)^2 - delta(a,b,c) > 0;
  4 * a <> 0 by A1;
  then
A3: (2 * a * x + b)^2 - (4 * a) * (delta(a,b,c)/(4 * a)) > 0 by A2,XCMPLX_1:87;
  2 * a <> 0 by A1;
  then
  (2 * a * x + (2 * a) * (b/(2 * a)))^2 - (4 * a) * (delta(a,b,c)/(4 * a))
  > 0 by A3,XCMPLX_1:87;
  then
A4: (4 * a) * (a * (x + b/(2 * a))^2 - delta(a,b,c)/(4 * a)) > 0;
  4 * a < 0 by A1,XREAL_1:132;
  then a * (x + b/(2 * a))^2 - delta(a,b,c)/(4 * a) < 0/(4 * a) by A4,
XREAL_1:84;
  hence thesis by A1,Th1;
end;

theorem
  a < 0 & delta(a,b,c) = 0 implies ( a * x^2 + b * x + c < 0 iff x <> -
  b/(2 * a) )
proof
  assume that
A1: a < 0 and
A2: delta(a,b,c) = 0;
A3: 2 * a <> 0 by A1;
  thus a * x^2 + b * x + c < 0 implies x <> - b/(2 * a)
  proof
    assume a * x^2 + b * x + c < 0;
    then (2 * a * x + b)^2 - 0 > 0 by A1,A2,Th9;
    then 2 * a * x <> - b;
    then x <> (- b)/(2 * a) by A3,XCMPLX_1:87;
    hence thesis by XCMPLX_1:187;
  end;
  assume x <> - b/(2 * a);
  then x <> (- b)/(2 * a) by XCMPLX_1:187;
  then 2 * a * x + b <> 0 by A3,XCMPLX_1:89;
  then (2 * a * x + b)^2 - delta(a,b,c) > 0 by A2,SQUARE_1:12;
  hence thesis by A1,Th23;
end;

theorem Th25:
  a > 0 & delta(a,b,c) > 0 implies (- b + sqrt delta(a,b,c))/(2 *
  a) > (- b - sqrt delta(a,b,c))/(2 * a)
proof
  assume that
A1: a > 0 and
A2: delta(a,b,c) > 0;
  sqrt delta(a,b,c) > 0 by A2,SQUARE_1:25;
  then 2 * sqrt delta(a,b,c) > 0 by XREAL_1:129;
  then sqrt delta(a,b,c) + sqrt delta(a,b,c) > 0;
  then sqrt delta(a,b,c) > - sqrt delta(a,b,c) by XREAL_1:59;
  then
A3: - b + sqrt delta(a,b,c) > - b + - sqrt delta(a,b,c) by XREAL_1:6;
  2 * a > 0 by A1,XREAL_1:129;
  hence thesis by A3,XREAL_1:74;
end;

theorem
  a > 0 & delta(a,b,c) > 0 implies ( a * x^2 + b * x + c < 0 iff (- b -
  sqrt delta(a,b,c))/(2 * a) < x & x < (- b + sqrt delta(a,b,c))/(2 * a) )
proof
  assume that
A1: a > 0 and
A2: delta(a,b,c) > 0;
  thus a * x^2 + b * x + c < 0 implies (- b - sqrt delta(a,b,c))/(2 * a) < x &
  x < (- b + sqrt delta(a,b,c))/(2 * a)
  proof
    assume a * x^2 + b * x + c < 0;
    then
    a * (x - (- b - sqrt delta(a,b,c))/(2 * a)) * (x - (- b + sqrt delta(a
    ,b,c))/(2 * a)) < 0 by A1,A2,Th16;
    then
    a * ((x - (- b - sqrt delta(a,b,c))/(2 * a)) * (x - (- b + sqrt delta(
    a,b,c))/(2 * a))) < 0;
    then
    (x - (- b - sqrt delta(a,b,c))/(2 * a)) * (x - (- b + sqrt delta(a,b,c
    ))/(2 * a)) < 0/a by A1,XREAL_1:81;
    then
    x - (- b - sqrt delta(a,b,c))/(2 * a) > 0 & x - (- b + sqrt delta(a,b,
c))/(2 * a) < 0 or x - (- b - sqrt delta(a,b,c))/(2 * a) < 0 & x - (- b + sqrt
    delta(a,b,c))/(2 * a) > 0 by XREAL_1:133;
    then
    x > (- b - sqrt delta(a,b,c))/(2 * a) & x < (- b + sqrt delta(a,b,c))/
(2 * a) or (- b + sqrt delta(a,b,c))/(2 * a) > (- b - sqrt delta(a,b,c))/(2 * a
) & x < (- b - sqrt delta(a,b,c))/(2 * a) & x > (- b + sqrt delta(a,b,c))/(2 *
    a) by A1,A2,Th25,XREAL_1:47,48;
    hence thesis by XXREAL_0:2;
  end;
  assume (- b - sqrt delta(a,b,c))/(2 * a) < x & x < (- b + sqrt delta(a,b,c
  ))/(2 * a );
  then
  x - (- b - sqrt delta(a,b,c))/(2 * a) > 0 & x - (- b + sqrt delta(a,b,c
  ))/(2 * a) < 0 by XREAL_1:49,50;
  then
  (x - (- b - sqrt delta(a,b,c))/(2 * a)) * (x - (- b + sqrt delta(a,b,c)
  )/(2 * a)) < 0 by XREAL_1:132;
  then
  a * ((x - (- b - sqrt delta(a,b,c))/(2 * a)) * (x - (- b + sqrt delta(a
  ,b,c))/(2 * a))) < 0 by A1,XREAL_1:132;
  then
  a * (x - (- b - sqrt delta(a,b,c))/(2 * a)) * (x - (- b + sqrt delta(a,
  b,c))/(2 * a)) < 0;
  hence thesis by A1,A2,Th16;
end;

theorem
  a > 0 & delta(a,b,c) > 0 implies ( a * x^2 + b * x + c > 0 iff x < (-
  b - sqrt delta(a,b,c))/(2 * a) or x > (- b + sqrt delta(a,b,c))/(2 * a) )
proof
  assume that
A1: a > 0 and
A2: delta(a,b,c) > 0;
  thus a * x^2 + b * x + c > 0 implies x < (- b - sqrt delta(a,b,c))/(2 * a)
  or x > (- b + sqrt delta(a,b,c))/(2 * a)
  proof
    assume a * x^2 + b * x + c > 0;
    then
    a * (x - (- b - sqrt delta(a,b,c))/(2 * a)) * (x - (- b + sqrt delta(a
    ,b,c))/(2 * a)) > 0 by A1,A2,Th16;
    then
    a * ((x - (- b - sqrt delta(a,b,c))/(2 * a)) * (x - (- b + sqrt delta(
    a,b,c))/(2 * a))) > 0;
    then
    (x - (- b - sqrt delta(a,b,c))/(2 * a)) * (x - (- b + sqrt delta(a,b,c
    ))/(2 * a)) > 0/a by A1,XREAL_1:83;
    then
    x - (- b - sqrt delta(a,b,c))/(2 * a) > 0 & x - (- b + sqrt delta(a,b,
c))/(2 * a) > 0 or x - (- b - sqrt delta(a,b,c))/(2 * a) < 0 & x - (- b + sqrt
    delta(a,b,c))/(2 * a) < 0 by XREAL_1:134;
    hence thesis by XREAL_1:47,48;
  end;
  assume x < (- b - sqrt delta(a,b,c))/(2 * a) or x > (- b + sqrt delta(a,b,
  c))/(2 * a);
  then
A3: x - (- b - sqrt delta(a,b,c))/(2 * a) < 0 or x - (- b + sqrt delta(a,b,
  c))/(2 * a) > 0 by XREAL_1:49,50;
  (- b + sqrt delta(a,b,c))/(2 * a) > (- b - sqrt delta(a,b,c))/(2 * a)
  by A1,A2,Th25;
  then
  x - (- b + sqrt delta(a,b,c))/(2 * a) < x - (- b - sqrt delta(a,b,c))/(
  2 * a) by XREAL_1:10;
  then
  x - (- b - sqrt delta(a,b,c))/(2 * a) < 0 & x - (- b + sqrt delta(a,b,c
  ))/(2 * a) < 0 or x - (- b - sqrt delta(a,b,c))/(2 * a) > 0 & x - (- b + sqrt
  delta(a,b,c))/(2 * a) > 0 by A3,XXREAL_0:2;
  then
  (x - (- b + sqrt delta(a,b,c))/(2 * a)) * (x - (- b - sqrt delta(a,b,c)
  )/(2 * a)) > 0 by XREAL_1:129,130;
  then
  a * ((x - (- b + sqrt delta(a,b,c))/(2 * a)) * (x - (- b - sqrt delta(a
  ,b,c))/(2 * a))) > 0 by A1,XREAL_1:129;
  then
  a * (x - (- b - sqrt delta(a,b,c))/(2 * a)) * (x - (- b + sqrt delta(a,
  b,c))/(2 * a)) > 0;
  hence thesis by A1,A2,Th16;
end;














begin

theorem Th1:
  for X,x being set holds for F being Function of [:X,X:],X holds x
  in [:X,X:] implies F.x in X
proof
  let X,x be set;
  let F be Function of [:X,X:],X;
  X={} implies [:X,X:]={};
  then
A1: dom F = [:X,X:] by FUNCT_2:def 1;
  assume x in [:X,X:];
  then rng F c= X & F.x in rng F by A1,FUNCT_1:def 3,RELAT_1:def 19;
  hence thesis;
end;

definition

  let X be set;
  let F be BinOp of X;
  mode Preserv of F -> Subset of X means
    :Def1:
    for x being set holds x in [: it,it:] implies F.x in it;
  existence
  proof
    X c= X;
    then reconsider Z = X as Subset of X;
    take Z;
    thus thesis by Th1;
  end;
end;

definition
  let R be Relation;
  let A be set;
  func R||A equals
  R | [:A,A:];
  coherence;
end;

registration
  let R be Relation;
  let A be set;
  cluster R||A -> Relation-like;
  coherence;
end;

registration
  let R be Function;
  let A be set;
  cluster R||A -> Function-like;
  coherence;
end;

theorem Th2:
  for X being set, F being BinOp of X, A being Preserv of F holds F
  ||A is BinOp of A
proof
  let X be set, F be BinOp of X, A be Preserv of F;
  X={} implies [:X,X:]={};
  then dom F = [:X,X:] by FUNCT_2:def 1;
  then
A1: dom (F||A) = [:A,A:] by RELAT_1:62,ZFMISC_1:96;
  for x being set holds x in [:A,A:] implies F||A.x in A
  proof
    let x be set;
    assume
A2: x in [:A,A:];
    then F||A.x=F.x by A1,FUNCT_1:47;
    hence thesis by A2,Def1;
  end;
  hence thesis by A1,FUNCT_2:3;
end;

definition
  let X be set;
  let F be BinOp of X;
  let A be Preserv of F;
  redefine func F||A -> BinOp of A;
  coherence by Th2;
end;

theorem Th3:
  for X being set holds X is non trivial iff for x being
  set holds X\{x} is non empty set
proof
  let X be set;
  hereby
    assume
A1: X is non trivial;
    let x be set;
    X <> {x} by A1;
    then consider y being set such that
A2: y in X and
A3: x <> y by A1,ZFMISC_1:35;
    not y in {x} by A3,TARSKI:def 1;
    hence X\{x} is non empty set by A2,XBOOLE_0:def 5;
  end;
  assume
A4: for x being set holds X\{x} is non empty set;
   X\{{}} c= X;
   then X is non empty by A4;
   then consider x being set such that
A5:  x in X by XBOOLE_0:def 1;
   X\{x} is non empty by A4;
   then consider y being set such that
A6:  y in X\{x} by XBOOLE_0:def 1;
  take x,y;
  thus x in X by A5;
  thus y in X by A6;
   x in {x} by TARSKI:def 1;
   then not x in X\{x} by XBOOLE_0:def 5;
  hence x <> y by A6;
end;

theorem
  ex A being non empty set st for z being Element of A holds A \ {z} is
  non empty set
proof
  set B = {0,1};
  take B;
  for z being Element of B holds B\{z} is non empty set
  proof
    let z be Element of B;
    0 in B & not 0 in {1} by TARSKI:def 1,def 2;
    then
A1: B\{1} is non empty set by XBOOLE_0:def 5;
    1 in B & not 1 in {0} by TARSKI:def 1,def 2;
    then B\{0} is non empty set by XBOOLE_0:def 5;
    hence thesis by A1,TARSKI:def 2;
  end;
  hence thesis;
end;

definition

  let X be non trivial set;
  let F be BinOp of X;
  let x be Element of X;
  pred F is_Bin_Op_Preserv x means
  X\{x} is Preserv of F & F||X\{x} is BinOp of X\{x};
  correctness;
end;

theorem Th5:
  for X being set holds for A being Subset of X holds ex F being
  BinOp of X st for x being set holds x in [:A,A:] implies F.x in A
proof
  let X be set;
  let A be Subset of X;
  set S = pr1(X,X);
  take S;
  for x being set holds x in [:A,A:] implies S.x in A
  proof
    let x be set;
    assume x in [:A,A:];
    then consider p,q being set such that
A1: p in A & q in A and
A2: x = [p,q] by ZFMISC_1:def 2;
    S.x = S.(p,q) by A2;
    hence thesis by A1,FUNCT_3:def 4;
  end;
  hence thesis;
end;

definition
  let X be set;
  let A be Subset of X;
  mode Presv of X,A -> BinOp of X means
    :Def4:
    for x being set holds x in [:A, A:] implies it.x in A;
  existence by Th5;
end;

theorem Th6:
  for X being set, A being Subset of X, F being Presv of X,A holds
  F||A is BinOp of A
proof
  let X be set;
  let A be Subset of X;
  let F be Presv of X,A;
  X={} implies [:X,X:]={};
  then dom F = [:X,X:] by FUNCT_2:def 1;
  then
A1: dom (F||A) = [:A,A:] by RELAT_1:62,ZFMISC_1:96;
  for x being set holds x in [:A,A:] implies F||A.x in A
  proof
    let x be set;
    assume
A2: x in [:A,A:];
    then F||A.x=F.x by A1,FUNCT_1:47;
    hence thesis by A2,Def4;
  end;
  hence thesis by A1,FUNCT_2:3;
end;

definition
  let X be set;
  let A be Subset of X;
  let F be Presv of X,A;
  func F|||A -> BinOp of A equals
  F||A;
  coherence by Th6;
end;

definition
  let A be set;
  let x be Element of A;
  mode DnT of x,A -> BinOp of A means
    :Def6:
    for y being set holds y in [:A\{x },A\{x}:] implies it.y in A\{x};
  existence by Th5;
end;

theorem Th7:
  for A being non trivial set holds for x being Element of A holds
  for F being DnT of x,A holds F||(A\{x}) is BinOp of A\{x}
proof
  let A be non trivial set;
  let x be Element of A;
  let F be DnT of x,A;
  dom F = [:A,A:] by FUNCT_2:def 1;
  then
A1: dom(F||(A\{x})) = [:A\{x},A\{x}:] by RELAT_1:62,ZFMISC_1:96;
  for y being set holds y in [:A\{x},A\{x}:] implies F||(A\{x}).y in A\{x}
  proof
    let y be set;
    assume
A2: y in [:A\{x},A\{x}:];
    then F||(A\{x}).y=F.y by A1,FUNCT_1:47;
    hence thesis by A2,Def6;
  end;
  hence thesis by A1,FUNCT_2:3;
end;

definition
  let A be non trivial set;
  let x be Element of A;
  let F be DnT of x,A;
  func F!(A,x) -> BinOp of A\{x} equals
  F||(A\{x});
  coherence by Th7;
end;

theorem Th8:
  for F being non trivial set holds for A being Singleton of F
  holds F\A is non empty set
proof
  let F be non trivial set;
  let A be Singleton of F;
  ex x being Element of F st A = {x} by CARD_1:65;
  hence thesis by Th3;
end;

registration
  let F be non trivial set;
  let A be Singleton of F;
  cluster F\A -> non empty;
  coherence by Th8;
end;















begin

reserve m for Cardinal,
  A,B,C for Ordinal,
  x,y,z,X,Y,Z,W for set,
  f for Function;

Lm1: Tarski-Class X is Tarski
proof
  Tarski-Class X is_Tarski-Class_of X by CLASSES1:def 4;
  hence thesis by CLASSES1:def 3;
end;

registration
  cluster Tarski -> subset-closed for set;
  coherence by CLASSES1:def 2;
end;

registration
  let X be set;
  cluster Tarski-Class X -> Tarski;
  coherence by Lm1;
end;

theorem Th1:
  W is subset-closed & X in W implies not X,W are_equipotent & card X in card W
proof
  assume
A1: W is subset-closed;
  assume
A2: X in W;
  bool X c= W
  proof
    let x;
    assume x in bool X;
    hence thesis by A1,A2,CLASSES1:def 1;
  end;
  then
A3: card bool X c= card W by CARD_1:11;
A4: card X in card bool X by CARD_1:14;
  then card X in card W by A3;
  then card X <> card W;
  hence thesis by A4,A3,CARD_1:5;
end;

theorem Th2:
  W is Tarski & x in W & y in W implies {x} in W & {x,y} in W
proof
  defpred C[set] means $1 = {};
  assume that
A1: W is Tarski and
A2: x in W and
A3: y in W;
A4: {x} c= bool x by ZFMISC_1:68;
  bool x in W by A1,A2,CLASSES1:def 2;
  hence {x} in W by A1,A4,CLASSES1:def 1;
  then
A5: bool {x} in W by A1,CLASSES1:def 2;
  deffunc g(set) = x;
  deffunc f(set) = y;
  consider f such that
A6: dom f = {{},{x}} & for z st z in {{},{x}} holds (C[z] implies f.z =
  f(z)) & (not C[z] implies f.z = g(z)) from PARTFUN1:sch 1;
  {x,y} c= rng f
  proof
    let z;
A7: {} in {{},{x}} by TARSKI:def 2;
A8: {x} in {{},{x}} by TARSKI:def 2;
    assume z in {x,y};
    then z = x or z = y by TARSKI:def 2;
    then f.{} = z or f.{x} = z by A6,A7,A8;
    hence thesis by A6,A7,A8,FUNCT_1:def 3;
  end;
  then
A9: card {x,y} c= card {{},{x}} by A6,CARD_1:12;
A10: {x,y} c= W
  proof
    let z;
    assume z in {x,y};
    hence thesis by A2,A3,TARSKI:def 2;
  end;
  bool {x} = {{},{x}} by ZFMISC_1:24;
  then card {{},{x}} in card W by A1,A5,Th1;
  then card {x,y} in card W by A9,ORDINAL1:12;
  hence thesis by A1,A10,CLASSES1:1;
end;

theorem Th3:
  W is Tarski & x in W & y in W implies [x,y] in W
proof
  assume
A1: W is Tarski;
  assume that
A2: x in W and
A3: y in W;
A4: {x} in W by A1,A2,Th2;
  {x,y} in W by A1,A2,A3,Th2;
  hence thesis by A1,A4,Th2;
end;

theorem Th4:
  W is Tarski & X in W implies Tarski-Class X c= W
proof
  assume that
A1: W is Tarski and
A2: X in W;
  reconsider D = W as non empty set by A2;
  D is_Tarski-Class_of X by A1,A2,CLASSES1:def 3;
  hence thesis by CLASSES1:def 4;
end;

theorem Th5:
  W is Tarski & A in W implies succ A in W & A c= W
proof
  assume that
A1: for X,Y st X in W & Y c= X holds Y in W and
A2: for X st X in W holds bool X in W and
  for X st X c= W holds X,W are_equipotent or X in W and
A3: A in W;
  bool A in W by A2,A3;
  hence succ A in W by A1,ORDINAL2:3;
  let x;
  assume
A4: x in A;
  then reconsider B = x as Ordinal;
  B c= A by A4,ORDINAL1:def 2;
  hence thesis by A1,A3;
end;

theorem
  A in Tarski-Class W implies succ A in Tarski-Class W & A c=
  Tarski-Class W by Th5;

theorem Th7:
  W is subset-closed & X is epsilon-transitive & X in W implies X c= W
proof
  assume that
A1: for X,Y st X in W & Y c= X holds Y in W and
A2: Y in X implies Y c= X and
A3: X in W;
  let x;
  thus thesis by A1,A2,A3;
end;

theorem
  X is epsilon-transitive & X in Tarski-Class W implies X c=
  Tarski-Class W by Th7;

theorem Th9:
  W is Tarski implies On W = card W
proof
  assume
A1: W is Tarski;
  now
    let X;
    assume
A2: X in On W;
    hence X is Ordinal by ORDINAL1:def 9;
    reconsider A = X as Ordinal by A2,ORDINAL1:def 9;
A3: X in W by A2,ORDINAL1:def 9;
    thus X c= On W
    proof
      let x;
      assume
A4:   x in X;
      then x in A;
      then reconsider B = x as Ordinal;
      B c= A by A4,ORDINAL1:def 2;
      then B in W by A1,A3,CLASSES1:def 1;
      hence thesis by ORDINAL1:def 9;
    end;
  end;
  then reconsider ON = On W as Ordinal by ORDINAL1:19;
A5: now
    assume ON in W;
    then ON in ON by ORDINAL1:def 9;
    hence contradiction;
  end;
  ON c= W by ORDINAL2:7;
  then
A6: ON,W are_equipotent or ON in W by A1,CLASSES1:def 2;
  now
    let A;
    assume that
A7: A,ON are_equipotent and
A8: not ON c= A;
    A in ON by A8,ORDINAL1:16;
    then A in W by ORDINAL1:def 9;
    hence contradiction by A1,A6,A5,A7,Th1,WELLORD2:15;
  end;
  then reconsider ON as Cardinal by CARD_1:def 1;
  ON = card ON by CARD_1:def 2;
  hence thesis by A6,A5,CARD_1:5;
end;

theorem
  On Tarski-Class W = card Tarski-Class W by Th9;

theorem Th11:
  W is Tarski & X in W implies card X in W
proof
  assume that
A1: W is Tarski and
A2: X in W;
A3: card W = On W by A1,Th9;
  card X in card W by A1,A2,Th1;
  hence thesis by A3,ORDINAL1:def 9;
end;

theorem
  X in Tarski-Class W implies card X in Tarski-Class W by Th11;

theorem Th13:
  W is Tarski & x in card W implies x in W
proof
  assume
A1: W is Tarski;
  assume x in card W;
  then x in On W by A1,Th9;
  hence thesis by ORDINAL1:def 9;
end;

theorem
  x in card Tarski-Class W implies x in Tarski-Class W by Th13;

theorem
  W is Tarski & m in card W implies m in W
proof
  assume
A1: W is Tarski;
  assume m in card W;
  then m in On W by A1,Th9;
  hence thesis by ORDINAL1:def 9;
end;

theorem
  m in card Tarski-Class W implies m in Tarski-Class W
proof
  assume m in card Tarski-Class W;
  then m in On Tarski-Class W by Th9;
  hence thesis by ORDINAL1:def 9;
end;

theorem
  W is Tarski & m in W implies m c= W by Th5;

theorem
  m in Tarski-Class W implies m c= Tarski-Class W by Th5;

theorem Th19:
  W is Tarski implies card W is limit_ordinal
proof
  assume
A1: W is Tarski;
  now
    let A;
    assume A in card W;
    then A in W by A1,Th13;
    then succ A in W by A1,Th5;
    then succ A in On W by ORDINAL1:def 9;
    hence succ A in card W by A1,Th9;
  end;
  hence thesis by ORDINAL1:28;
end;

theorem
  W is Tarski & W <> {} implies card W <> 0 & card W <> {} & card W is
  limit_ordinal by Th19;

theorem Th21:
  card Tarski-Class W <> 0 & card Tarski-Class W <> {} & card
  Tarski-Class W is limit_ordinal
proof
  thus card Tarski-Class W <> 0;
  thus card Tarski-Class W <> {};
  now
    let A;
    assume A in card Tarski-Class W;
    then A in Tarski-Class W by Th13;
    then succ A in Tarski-Class W by Th5;
    then succ A in On Tarski-Class W by ORDINAL1:def 9;
    hence succ A in card Tarski-Class W by Th9;
  end;
  hence thesis by ORDINAL1:28;
end;

reserve f,g for Function,
  L for T-Sequence,
  F for Cardinal-Function;

theorem Th22:
  W is Tarski & (X in W & W is epsilon-transitive or X in W & X c=
  W or card X in card W & X c= W) implies Funcs(X,W) c= W
proof
  assume
A1: W is Tarski;
  assume that
A2: X in W & W is epsilon-transitive or X in W & X c= W or card X in
  card W & X c= W;
A3: card X in card W by A1,A2,Th1;
  let x;
  assume
A4: x in Funcs(X,W);
  then consider f such that
A5: x = f and
A6: dom f = X and
A7: rng f c= W by FUNCT_2:def 2;
A8: X c= W by A2,ORDINAL1:def 2;
A9: f c= W
  proof
    let y;
    assume
A10: y in f;
    then consider y1,y2 being set such that
A11: [y1,y2] = y by RELAT_1:def 1;
A12: y1 in dom f by A10,A11,FUNCT_1:1;
    y2 = f.y1 by A10,A11,FUNCT_1:1;
    then y2 in rng f by A12,FUNCT_1:def 3;
    hence thesis by A1,A8,A6,A7,A11,A12,Th3;
  end;
  card f = card X by A4,A5,CARD_2:3;
  hence thesis by A1,A3,A5,A9,CLASSES1:1;
end;

theorem
  X in Tarski-Class W & W is epsilon-transitive or X in Tarski-Class W &
  X c= Tarski-Class W or card X in card Tarski-Class W & X c= Tarski-Class W
  implies Funcs(X,Tarski-Class W) c= Tarski-Class W
proof
  assume that
A1: X in Tarski-Class W & W is epsilon-transitive or X in Tarski-Class W
  & X c= Tarski-Class W or card X in card Tarski-Class W & X c= Tarski-Class W;
A2: card X in card Tarski-Class W by A1,CLASSES1:24;
  let x;
  assume
A3: x in Funcs(X,Tarski-Class W);
  then consider f such that
A4: x = f and
A5: dom f = X and
A6: rng f c= Tarski-Class W by FUNCT_2:def 2;
  W is epsilon-transitive implies Tarski-Class W is epsilon-transitive by
CLASSES1:23;
  then
A7: X c= Tarski-Class W by A1,ORDINAL1:def 2;
A8: f c= Tarski-Class W
  proof
    let y;
    assume
A9: y in f;
    then consider y1,y2 being set such that
A10: [y1,y2] = y by RELAT_1:def 1;
A11: y1 in dom f by A9,A10,FUNCT_1:1;
    y2 = f.y1 by A9,A10,FUNCT_1:1;
    then y2 in rng f by A11,FUNCT_1:def 3;
    hence thesis by A7,A5,A6,A10,A11,CLASSES1:27;
  end;
  card f = card X by A3,A4,CARD_2:3;
  hence thesis by A2,A4,A8,CLASSES1:6;
end;

theorem Th24:
  dom L is limit_ordinal & (for A st A in dom L holds L.A = Rank A
  ) implies Rank dom L = Union L
proof
  assume that
A1: dom L is limit_ordinal and
A2: for A st A in dom L holds L.A = Rank A;
A3: union rng L = Union L by CARD_3:def 4;
  now
    assume
A4: dom L <> {};
    thus Rank dom L c= Union L
    proof
      let x;
      assume x in Rank dom L;
      then consider A such that
A5:   A in dom L and
A6:   x in Rank A by A1,A4,CLASSES1:31;
      L.A = Rank A by A2,A5;
      then Rank A in rng L by A5,FUNCT_1:def 3;
      hence thesis by A3,A6,TARSKI:def 4;
    end;
    thus Union L c= Rank dom L
    proof
      let x;
      assume x in Union L;
      then consider X such that
A7:   x in X and
A8:   X in rng L by A3,TARSKI:def 4;
      consider y such that
A9:   y in dom L and
A10:  X = L.y by A8,FUNCT_1:def 3;
      reconsider y as Ordinal by A9;
      X = Rank y by A2,A9,A10;
      hence thesis by A1,A7,A9,CLASSES1:31;
    end;
  end;
  hence thesis by A3,CLASSES1:29,RELAT_1:42,XBOOLE_0:def 10,ZFMISC_1:2;
end;

defpred PQ[Ordinal] means W is Tarski & $1 in On W implies card Rank $1 in
card W & Rank $1 in W;

Lm2: PQ[{}] by Th9,CARD_1:27,CLASSES1:29,ORDINAL1:def 9;

Lm3: PQ[A] implies PQ[succ A]
proof
  assume
A1: PQ[A];
A2: A in succ A by ORDINAL1:6;
  let W;
  assume that
A3: W is Tarski and
A4: succ A in On W;
  card W = On W by A3,Th9;
  then A in On W by A4,A2,ORDINAL1:10;
  then Rank A in W by A1,A3;
  then
A5: bool Rank A in W by A3,CLASSES1:def 2;
  Rank succ A = bool Rank A by CLASSES1:30;
  hence thesis by A3,A5,Th1;
end;

Lm4: A <> {} & A is limit_ordinal & (for B st B in A holds PQ[B]) implies PQ[A
]
proof
  deffunc f(Ordinal) = Rank $1;
  assume that
A1: A <> {} and
A2: A is limit_ordinal and
A3: for B st B in A holds PQ[B];
  let W;
  assume that
A4: W is Tarski and
A5: A in On W;
  consider L such that
A6: dom L = A & for B st B in A holds L.B = f(B) from ORDINAL2:sch 2;
  deffunc g(set) = card bool (L.$1);
  consider F such that
A7: dom F = A & for x st x in A holds F.x = g(x) from CARD_3:sch 1;
A8: product F c= Funcs(A,W)
  proof
    let x;
    assume x in product F;
    then consider f such that
A9: x = f and
A10: dom f = dom F and
A11: for x st x in dom F holds f.x in F.x by CARD_3:def 5;
    rng f c= W
    proof
A12:  A in W by A5,ORDINAL1:def 9;
      let z;
      assume z in rng f;
      then consider y such that
A13:  y in dom f and
A14:  z = f.y by FUNCT_1:def 3;
      reconsider y as Ordinal by A7,A10,A13;
A15:  f.y in F.y by A10,A11,A13;
      y c= A by A7,A10,A13,ORDINAL1:def 2;
      then y in W by A4,A12,CLASSES1:def 1;
      then
A16:  y in On W by ORDINAL1:def 9;
      L.y = Rank y by A6,A7,A10,A13;
      then L.y in W by A3,A4,A7,A10,A13,A16;
      then bool (L.y) in W by A4,CLASSES1:def 2;
      then
A17:  card bool (L.y) in W by A4,Th11;
      F.y = card bool (L.y) by A7,A10,A13;
      then F.y c= W by A4,A17,Th5;
      hence thesis by A14,A15;
    end;
    hence thesis by A7,A9,A10,FUNCT_2:def 2;
  end;
A18: Product F = card product F by CARD_3:def 8;
A19: A in W by A5,ORDINAL1:def 9;
  then A c= W by A4,Th5;
  then Funcs(A,W) c= W by A4,A19,Th22;
  then product F c= W by A8,XBOOLE_1:1;
  then
A20: Product F c= card W by A18,CARD_1:11;
A21: for x st x in dom Card L holds (Card L).x in F.x
  proof
    let x;
A22: card (L.x) in card bool (L.x) by CARD_1:14;
    assume x in dom Card L;
    then
A23: x in dom L by CARD_3:def 2;
    then F.x = card bool (L.x) by A6,A7;
    hence thesis by A23,A22,CARD_3:def 2;
  end;
  dom Card L = dom L by CARD_3:def 2;
  then
A24: Sum Card L in Product F by A6,A7,A21,CARD_3:41;
A25: Rank A c= W
  proof
    let x;
    assume x in Rank A;
    then consider B such that
A26: B in A and
A27: x in Rank B by A1,A2,CLASSES1:31;
    B c= A by A26,ORDINAL1:def 2;
    then B in W by A4,A19,CLASSES1:def 1;
    then B in On W by ORDINAL1:def 9;
    then Rank B c= W by A3,A4,A26,Th7;
    hence thesis by A27;
  end;
A28: Rank A = Union L by A2,A6,Th24;
  hence card Rank A in card W by A24,A20,CARD_3:39,ORDINAL1:12;
  card Rank A in Product F by A28,A24,CARD_3:39,ORDINAL1:12;
  hence thesis by A4,A20,A25,CLASSES1:1;
end;

theorem Th25:
  W is Tarski & A in On W implies card Rank A in card W & Rank A in W
proof
  PQ[B] from ORDINAL2:sch 1(Lm2,Lm3,Lm4);
  hence thesis;
end;

theorem
  A in On Tarski-Class W implies card Rank A in card Tarski-Class W &
  Rank A in Tarski-Class W by Th25;

theorem Th27:
  W is Tarski implies Rank card W c= W
proof
  assume
A1: W is Tarski;
  now
    assume
A2: W <> {};
    thus thesis
    proof
A3:   card W is limit_ordinal by A1,Th19;
      let x;
      assume x in Rank card W;
      then consider A such that
A4:   A in card W and
A5:   x in Rank A by A2,A3,CLASSES1:31;
      A in On W by A1,A4,Th9;
      then Rank A c= W by A1,Th7,Th25;
      hence thesis by A5;
    end;
  end;
  hence thesis by CLASSES1:29;
end;

theorem Th28:
  Rank card Tarski-Class W c= Tarski-Class W
proof
A1: card Tarski-Class W is limit_ordinal by Th21;
  let x;
  assume x in Rank card Tarski-Class W;
  then consider A such that
A2: A in card Tarski-Class W and
A3: x in Rank A by A1,CLASSES1:31;
  A in On Tarski-Class W by A2,Th9;
  then Rank A c= Tarski-Class W by Th7,Th25;
  hence thesis by A3;
end;

deffunc f(set) = the_rank_of $1;
deffunc g(set) = card bool $1;

theorem Th29:
  W is Tarski & W is epsilon-transitive & X in W implies the_rank_of X in W
proof
  assume that
A1: W is Tarski and
A2: W is epsilon-transitive;
A3: On W = card W by A1,Th9;
  defpred P[Ordinal] means ex X st $1 = the_rank_of X & X in W & not $1 in W;
  assume that
A4: X in W and
A5: not the_rank_of X in W;
A6: ex A st P[A] by A4,A5;
  consider A such that
A7: P[A] and
A8: for B st P[B] holds A c= B from ORDINAL1:sch 1(A6);
  consider X such that
A9: A = the_rank_of X and
A10: X in W and
A11: not A in W by A7;
  defpred P[set] means ex Y st Y in X & $1 = the_rank_of Y;
  consider LL being set such that
A12: x in LL iff x in On W & P[x] from XBOOLE_0:sch 1;
  consider ff being Cardinal-Function such that
A13: dom ff = LL & for x st x in LL holds ff.x = g(x) from CARD_3:sch 1;
A14: LL c= On W
  proof
    let x;
    thus thesis by A12;
  end;
A15: product ff c= Funcs(LL,W)
  proof
    let x;
    assume x in product ff;
    then consider g such that
A16: x = g and
A17: dom g = dom ff and
A18: for x st x in dom ff holds g.x in ff.x by CARD_3:def 5;
    rng g c= W
    proof
      let y;
      assume y in rng g;
      then consider x such that
A19:  x in dom g and
A20:  y = g.x by FUNCT_1:def 3;
A21:  ff.x = card bool x by A13,A17,A19;
      x in W by A14,A13,A17,A19,ORDINAL1:def 9;
      then bool x in W by A1,CLASSES1:def 2;
      then card bool x in W by A1,Th11;
      then
A22:  card bool x c= W by A1,Th5;
      y in ff.x by A17,A18,A19,A20;
      hence thesis by A21,A22;
    end;
    hence thesis by A13,A16,A17,FUNCT_2:def 2;
  end;
  now
    let Z;
    assume Z in union LL;
    then consider Y such that
A23: Z in Y and
A24: Y in LL by TARSKI:def 4;
    Y in On W by A12,A24;
    then reconsider Y as Ordinal by ORDINAL1:def 9;
A25: Y c= union LL by A24,ZFMISC_1:74;
A26: Z in Y by A23;
    hence Z is Ordinal;
    reconsider A = Z as Ordinal by A26;
    A c= Y by A23,ORDINAL1:def 2;
    hence Z c= union LL by A25,XBOOLE_1:1;
  end;
  then reconsider ULL = union LL as Ordinal by ORDINAL1:19;
A27: dom Card id LL = dom id LL by CARD_3:def 2;
A28: dom id LL = LL by RELAT_1:45;
  now
    let x;
    assume
A29: x in dom Card id LL;
    then
A30: (Card id LL).x = card ((id LL).x) by A27,CARD_3:def 2;
A31: (id LL).x = x by A28,A27,A29,FUNCT_1:18;
    ff.x = card bool x by A13,A28,A27,A29;
    hence (Card id LL).x in ff.x by A31,A30,CARD_1:14;
  end;
  then
A32: Sum Card id LL in Product ff by A13,A28,A27,CARD_3:41;
  Union id LL = union rng id LL by CARD_3:def 4
    .= ULL by RELAT_1:45;
  then
A33: card ULL in Product ff by A32,CARD_3:39,ORDINAL1:12;
  consider f such that
A34: dom f = X & for x st x in X holds f.x = f(x) from FUNCT_1:sch 3;
  LL c= rng f
  proof
    let x;
    assume x in LL;
    then consider Y such that
A35: Y in X and
A36: x = the_rank_of Y by A12;
    f.Y = x by A34,A35,A36;
    hence thesis by A34,A35,FUNCT_1:def 3;
  end;
  then
A37: card LL c= card X by A34,CARD_1:12;
  card X in card W by A1,A10,Th1;
  then card LL <> card W by A37,ORDINAL1:12;
  then
A38: not LL,W are_equipotent by CARD_1:5;
A39: card product ff = Product ff by CARD_3:def 8;
A40: X c= W by A2,A10,ORDINAL1:def 2;
  X c= Rank card W
  proof
    let x;
    assume
A41: x in X;
    then not A c= the_rank_of x by A9,CLASSES1:68,ORDINAL1:5;
    then the_rank_of x in W by A8,A40,A41;
    then the_rank_of x in card W by A3,ORDINAL1:def 9;
    then
A42: Rank the_rank_of x in Rank card W by CLASSES1:36;
    x c= Rank the_rank_of x by CLASSES1:def 8;
    hence thesis by A42,CLASSES1:41;
  end;
  then
A43: A c= On W by A9,A3,CLASSES1:65;
  On W c= ULL
  proof
    let x;
    assume
A44: x in On W;
    then reconsider B = x as Ordinal by ORDINAL1:def 9;
    now
      assume
A45:  for Y st Y in X holds the_rank_of Y c= B;
      X c= Rank succ B
      proof
        let y;
        assume y in X;
        then the_rank_of y c= B by A45;
        then the_rank_of y in succ B by ORDINAL1:22;
        hence thesis by CLASSES1:66;
      end;
      then
A46:  A c= succ B by A9,CLASSES1:65;
      B in W by A44,ORDINAL1:def 9;
      then succ B in W by A1,Th5;
      hence contradiction by A1,A11,A46,CLASSES1:def 1;
    end;
    then consider Y such that
A47: Y in X and
A48: not the_rank_of Y c= B;
    the_rank_of Y in A by A9,A47,CLASSES1:68;
    then the_rank_of Y in LL by A43,A12,A47;
    then
A49: the_rank_of Y c= ULL by ZFMISC_1:74;
    B in the_rank_of Y by A48,ORDINAL1:16;
    hence thesis by A49;
  end;
  then
A50: card On W c= card ULL by CARD_1:11;
  On W c= W by ORDINAL2:7;
  then LL c= W by A14,XBOOLE_1:1;
  then LL in W by A1,A38,CLASSES1:def 2;
  then Funcs(LL,W) c= W by A1,A2,Th22;
  then product ff c= W by A15,XBOOLE_1:1;
  then
A51: Product ff c= card W by A39,CARD_1:11;
  On W = card W by A1,Th9;
  hence contradiction by A33,A51,A50,CARD_1:4;
end;

theorem Th30:
  W is Tarski & W is epsilon-transitive implies W c= Rank card W
proof
  assume that
A1: W is Tarski and
A2: W is epsilon-transitive;
  let x;
  assume x in W;
  then the_rank_of x in W by A1,A2,Th29;
  then
A3: the_rank_of x in On W by ORDINAL1:def 9;
  On W = card W by A1,Th9;
  then
A4: Rank the_rank_of x in Rank card W by A3,CLASSES1:36;
  x c= Rank the_rank_of x by CLASSES1:def 8;
  hence thesis by A4,CLASSES1:41;
end;

theorem Th31:
  W is Tarski & W is epsilon-transitive implies Rank card W = W
proof
  assume that
A1: W is Tarski and
A2: W is epsilon-transitive;
  thus Rank card W c= W & W c= Rank card W by A1,A2,Th27,Th30;
end;

theorem
  W is Tarski & A in On W implies card Rank A c= card W
proof
  assume that
A1: W is Tarski and
A2: A in On W;
  card Rank A in card W by A1,A2,Th25;
  hence thesis by CARD_1:3;
end;

theorem
  A in On Tarski-Class W implies card Rank A c= card Tarski-Class W
proof
  assume A in On Tarski-Class W;
  then card Rank A in card Tarski-Class W by Th25;
  hence thesis by CARD_1:3;
end;

theorem Th34:
  W is Tarski implies card W = card Rank card W
proof
  assume W is Tarski;
  then
A1: card Rank card W c= card W by Th27,CARD_1:11;
  card card W c= card Rank card W by CARD_1:11,CLASSES1:38;
  hence thesis by A1,XBOOLE_0:def 10;
end;

theorem
  card Tarski-Class W = card Rank card Tarski-Class W by Th34;

theorem Th36:
  W is Tarski & X c= Rank card W implies X,Rank card W
  are_equipotent or X in Rank card W
proof
  assume that
A1: W is Tarski and
A2: X c= Rank card W and
A3: not X,Rank card W are_equipotent;
  defpred P[set] means ex Y st Y in X & $1 = the_rank_of Y;
  consider LL being set such that
A4: x in LL iff x in On W & P[x] from XBOOLE_0:sch 1;
  consider ff being Cardinal-Function such that
A5: dom ff = LL & for x st x in LL holds ff.x = g(x) from CARD_3:sch 1;
A6: LL c= On W
  proof
    let x;
    thus thesis by A4;
  end;
A7: product ff c= Funcs(LL,W)
  proof
    let x;
    assume x in product ff;
    then consider g such that
A8: x = g and
A9: dom g = dom ff and
A10: for x st x in dom ff holds g.x in ff.x by CARD_3:def 5;
    rng g c= W
    proof
      let y;
      assume y in rng g;
      then consider x such that
A11:  x in dom g and
A12:  y = g.x by FUNCT_1:def 3;
A13:  ff.x = card bool x by A5,A9,A11;
      x in W by A6,A5,A9,A11,ORDINAL1:def 9;
      then bool x in W by A1,CLASSES1:def 2;
      then card bool x in W by A1,Th11;
      then
A14:  card bool x c= W by A1,Th5;
      y in ff.x by A9,A10,A11,A12;
      hence thesis by A13,A14;
    end;
    hence thesis by A5,A8,A9,FUNCT_2:def 2;
  end;
A15: card W = card Rank card W by A1,Th34;
  then
A16: card X <> card W by A3,CARD_1:5;
  On W c= W by ORDINAL2:7;
  then
A17: LL c= W by A6,XBOOLE_1:1;
  now
    let Z;
    assume Z in union LL;
    then consider Y such that
A18: Z in Y and
A19: Y in LL by TARSKI:def 4;
    Y in On W by A4,A19;
    then reconsider Y as Ordinal by ORDINAL1:def 9;
A20: Y c= union LL by A19,ZFMISC_1:74;
A21: Z in Y by A18;
    hence Z is Ordinal;
    reconsider A = Z as Ordinal by A21;
    A c= Y by A18,ORDINAL1:def 2;
    hence Z c= union LL by A20,XBOOLE_1:1;
  end;
  then reconsider ULL = union LL as Ordinal by ORDINAL1:19;
A22: dom Card id LL = dom id LL by CARD_3:def 2;
A23: dom id LL = LL by RELAT_1:45;
  now
    let x;
    assume
A24: x in dom Card id LL;
    then
A25: (Card id LL).x = card ((id LL).x) by A22,CARD_3:def 2;
A26: (id LL).x = x by A23,A22,A24,FUNCT_1:18;
    ff.x = card bool x by A5,A23,A22,A24;
    hence (Card id LL).x in ff.x by A26,A25,CARD_1:14;
  end;
  then
A27: Sum Card id LL in Product ff by A5,A23,A22,CARD_3:41;
  consider f such that
A28: dom f = X & for x st x in X holds f.x = f(x) from FUNCT_1:sch 3;
  LL c= rng f
  proof
    let x;
    assume x in LL;
    then consider Y such that
A29: Y in X and
A30: x = the_rank_of Y by A4;
    f.Y = x by A28,A29,A30;
    hence thesis by A28,A29,FUNCT_1:def 3;
  end;
  then
A31: card LL c= card X by A28,CARD_1:12;
A32: card product ff = Product ff by CARD_3:def 8;
  card X c= card W by A2,A15,CARD_1:11;
  then card X in card W by A16,CARD_1:3;
  then card LL <> card W by A31,ORDINAL1:12;
  then not LL,W are_equipotent by CARD_1:5;
  then LL in W by A1,A17,CLASSES1:def 2;
  then Funcs(LL,W) c= W by A1,A17,Th22;
  then product ff c= W by A7,XBOOLE_1:1;
  then
A33: Product ff c= card W by A32,CARD_1:11;
A34: card W is limit_ordinal by A1,Th19;
A35: card W = On W by A1,Th9;
  X c= Rank succ ULL
  proof
    let x;
    defpred P[Ordinal] means $1 in card W & x c= Rank $1;
    assume
A36: x in X;
    then
A37: f.x = the_rank_of x by A28;
    consider A such that
A38: A in card W and
A39: x in Rank A by A2,A34,A36,CLASSES1:29,31;
    P[A] by A38,A39,ORDINAL1:def 2;
    then
A40: ex A st P[A];
    consider A such that
A41: P[A] and
A42: for B st P[B] holds A c= B from ORDINAL1:sch 1(A40);
    now
      let B;
      assume x c= Rank B;
      then A c= card W & card W c= B or A c= B by A41,A42,ORDINAL1:16;
      hence A c= B by XBOOLE_1:1;
    end;
    then A = the_rank_of x by A41,CLASSES1:def 8;
    then f.x in LL by A4,A35,A36,A37,A41;
    then the_rank_of x c= ULL by A37,ZFMISC_1:74;
    then
A43: Rank the_rank_of x c= Rank ULL by CLASSES1:37;
    x c= Rank the_rank_of x by CLASSES1:def 8;
    then x c= Rank ULL by A43,XBOOLE_1:1;
    hence thesis by CLASSES1:32;
  end;
  then
A44: X in Rank succ succ ULL by CLASSES1:32;
  Union id LL = union rng id LL by CARD_3:def 4
    .= ULL by RELAT_1:45;
  then card ULL in card W by A27,A33,CARD_3:39,ORDINAL1:12;
  then
A45: ULL <> On W by A35;
  union card W = card W by A34,ORDINAL1:def 6;
  then ULL c= On W by A6,A35,ZFMISC_1:77;
  then ULL c< On W by A45,XBOOLE_0:def 8;
  then ULL in card W by A35,ORDINAL1:11;
  then succ ULL in card W by A34,ORDINAL1:28;
  then succ succ ULL in card W by A34,ORDINAL1:28;
  hence thesis by A34,A44,CLASSES1:31;
end;

theorem
  X c= Rank card Tarski-Class W implies X,Rank card Tarski-Class W
  are_equipotent or X in Rank card Tarski-Class W by Th36;

theorem Th38:
  W is Tarski implies Rank card W is Tarski
proof
  assume
A1: W is Tarski;
  set B = Rank card W;
  thus for X,Y holds X in B & Y c= X implies Y in B by CLASSES1:41;
  now
A2: card W is limit_ordinal by A1,Th19;
    assume
A3: W <> {};
    thus for X holds X in B implies bool X in B
    proof
      let X;
      assume X in B;
      then consider A such that
A4:   A in card W and
A5:   X in Rank A by A3,A2,CLASSES1:31;
A6:   bool X in Rank succ A by A5,CLASSES1:42;
      succ A in card W by A2,A4,ORDINAL1:28;
      hence thesis by A2,A6,CLASSES1:31;
    end;
  end;
  hence thesis by A1,Th36,CLASSES1:29;
end;

theorem
  Rank card Tarski-Class W is Tarski by Th38;

theorem Th40:
  X is epsilon-transitive & A in the_rank_of X implies ex Y st Y
  in X & the_rank_of Y = A
proof
  assume that
A1: X is epsilon-transitive and
A2: A in the_rank_of X;
  defpred P[Ordinal] means ex Y st A in $1 & Y in X & the_rank_of Y = $1;
  assume
A3: not thesis;
A4: ex B st P[B]
  proof
    assume
A5: for B,Y st A in B & Y in X holds the_rank_of Y <> B;
    X c= Rank A
    proof
      let x;
      assume
A6:   x in X;
      then
A7:   the_rank_of x <> A by A3;
      the_rank_of x c= A by A5,A6,ORDINAL1:16;
      then the_rank_of x c< A by A7,XBOOLE_0:def 8;
      then the_rank_of x in A by ORDINAL1:11;
      hence thesis by CLASSES1:66;
    end;
    then the_rank_of X c= A by CLASSES1:65;
    hence contradiction by A2,ORDINAL1:5;
  end;
  consider B such that
A8: P[B] and
A9: for C st P[C] holds B c= C from ORDINAL1:sch 1(A4);
  consider Y such that
A10: A in B and
A11: Y in X and
A12: the_rank_of Y = B by A8;
  Y c= Rank A
  proof
    let x;
A13: Y c= X by A1,A11,ORDINAL1:def 2;
    assume
A14: x in Y;
    then the_rank_of x in B by A12,CLASSES1:68;
    then not A in the_rank_of x by A9,A14,A13,ORDINAL1:5;
    then
A15: the_rank_of x c= A by ORDINAL1:16;
    the_rank_of x <> A by A3,A14,A13;
    then the_rank_of x c< A by A15,XBOOLE_0:def 8;
    then the_rank_of x in A by ORDINAL1:11;
    hence thesis by CLASSES1:66;
  end;
  then the_rank_of Y c= A by CLASSES1:65;
  hence contradiction by A10,A12,ORDINAL1:5;
end;

theorem Th41:
  X is epsilon-transitive implies card the_rank_of X c= card X
proof
  consider f such that
A1: dom f = X & for x st x in X holds f.x = f(x) from FUNCT_1:sch 3;
  assume
A2: X is epsilon-transitive;
  the_rank_of X c= rng f
  proof
    let x;
    assume
A3: x in the_rank_of X;
    then reconsider x9 = x as Ordinal;
    consider Y such that
A4: Y in X and
A5: the_rank_of Y = x9 by A2,A3,Th40;
    f.Y = x by A1,A4,A5;
    hence thesis by A1,A4,FUNCT_1:def 3;
  end;
  hence thesis by A1,CARD_1:12;
end;

theorem Th42:
  W is Tarski & X is epsilon-transitive & X in W implies X in Rank card W
proof
  assume
A1: W is Tarski;
  assume
A2: X is epsilon-transitive;
  assume X in W;
  then card X in card W by A1,Th1;
  then
A3: card the_rank_of X in card W by A2,Th41,ORDINAL1:12;
  card card W = card W;
  then the_rank_of X in card W by A3,CARD_3:43;
  then
A4: Rank the_rank_of X in Rank card W by CLASSES1:36;
  X c= Rank the_rank_of X by CLASSES1:def 8;
  hence thesis by A4,CLASSES1:41;
end;

theorem
  X is epsilon-transitive & X in Tarski-Class W implies X in Rank card
  Tarski-Class W by Th42;

theorem Th44:
  W is epsilon-transitive implies Rank card Tarski-Class W is_Tarski-Class_of W
proof
A1: W in Tarski-Class W by CLASSES1:2;
  assume W is epsilon-transitive;
  hence W in Rank card Tarski-Class W & Rank card Tarski-Class W is Tarski by
A1,Th38,Th42;
end;

theorem
  W is epsilon-transitive implies Rank card Tarski-Class W = Tarski-Class W
proof
  assume W is epsilon-transitive;
  then Rank card Tarski-Class W is_Tarski-Class_of W by Th44;
  hence Rank card Tarski-Class W c= Tarski-Class W & Tarski-Class W c= Rank
  card Tarski-Class W by Th28,CLASSES1:def 4;
end;

definition
  let IT be set;
  attr IT is universal means
  :Def1:
  IT is epsilon-transitive & IT is Tarski;
end;

registration
  cluster universal -> epsilon-transitive Tarski for set;
  coherence by Def1;
  cluster epsilon-transitive Tarski -> universal for set;
  coherence by Def1;
end;

registration
  cluster universal non empty for set;
  existence
  proof
    set X = the set;
    take V = Tarski-Class the_transitive-closure_of X;
    thus V is epsilon-transitive by CLASSES1:23,51;
    thus thesis;
  end;
end;

definition
  mode Universe is universal non empty set;
end;

reserve U1,U2,U for Universe;

theorem Th46:
  On U is Ordinal
proof
  On U = card U by Th9;
  hence thesis;
end;

theorem Th47:
  X is epsilon-transitive implies Tarski-Class X is universal
proof
  assume X is epsilon-transitive;
  hence Tarski-Class X is epsilon-transitive & Tarski-Class X is Tarski by
CLASSES1:23;
end;

theorem
  Tarski-Class U is Universe by Th47;

registration
  let U;
  cluster On U -> ordinal;
  coherence by Th46;
  cluster Tarski-Class U -> universal;
  coherence by Th47;
end;

theorem Th49:
  Tarski-Class A is universal
proof
  set M = Tarski-Class A;
  thus M is epsilon-transitive & M is Tarski by CLASSES1:23;
end;

registration
  let A;
  cluster Tarski-Class A -> universal;
  coherence by Th49;
end;

theorem Th50:
  U = Rank On U
proof
  card U = On U by Th9;
  hence thesis by Th31;
end;

theorem Th51:
  On U <> {} & On U is limit_ordinal
proof
  card U = On U by Th9;
  hence thesis by Th19;
end;

theorem
  U1 in U2 or U1 = U2 or U2 in U1
proof
A1: On U1 in On U2 or On U1 = On U2 or On U2 in On U1 by ORDINAL1:14;
A2: U2 = Rank On U2 by Th50;
  U1 = Rank On U1 by Th50;
  hence thesis by A1,A2,CLASSES1:36;
end;

theorem Th53:
  U1 c= U2 or U2 in U1
proof
A1: On U1 c= On U2 or On U2 in On U1 by ORDINAL1:16;
A2: U2 = Rank On U2 by Th50;
  U1 = Rank On U1 by Th50;
  hence thesis by A1,A2,CLASSES1:36,37;
end;

theorem Th54:
  U1,U2 are_c=-comparable
proof
A1: On U1 c= On U2 or On U2 c= On U1;
A2: U2 = Rank On U2 by Th50;
  U1 = Rank On U1 by Th50;
  hence U1 c= U2 or U2 c= U1 by A1,A2,CLASSES1:37;
end;

theorem
  U1 \/ U2 is Universe & U1 /\ U2 is Universe
proof
  U1,U2 are_c=-comparable by Th54;
  then U1 c= U2 or U2 c= U1 by XBOOLE_0:def 9;
  hence thesis by XBOOLE_1:12,28;
end;

theorem Th56:
  {} in U
proof
  {} c= the Element of U by XBOOLE_1:2;
  hence thesis by CLASSES1:def 1;
end;

theorem
  x in U implies {x} in U by Th2;

theorem
  x in U & y in U implies {x,y} in U & [x,y] in U by Th2,Th3;

theorem Th59:
  X in U implies bool X in U & union X in U & meet X in U
proof
  assume
A1: X in U;
  hence bool X in U by CLASSES1:def 2;
  U = Rank On U by Th50;
  hence
A2: union X in U by A1,CLASSES1:35;
  meet X c= union X by SETFAM_1:2;
  hence thesis by A2,CLASSES1:def 1;
end;

theorem Th60:
  X in U & Y in U implies X \/ Y in U & X /\ Y in U & X \ Y in U & X \+\ Y in U
proof
  assume that
A1: X in U and
A2: Y in U;
A3: union {X,Y} = X \/ Y by ZFMISC_1:75;
A4: meet {X,Y} = X /\ Y by SETFAM_1:11;
  {X,Y} in U by A1,A2,Th2;
  hence
A5: X \/ Y in U & X /\ Y in U by A3,A4,Th59;
  X \+\ Y = (X \/ Y)\(X /\ Y) by XBOOLE_1:101;
  hence thesis by A1,A5,CLASSES1:def 1;
end;

theorem Th61:
  X in U & Y in U implies [:X,Y:] in U & Funcs(X,Y) in U
proof
  assume that
A1: X in U and
A2: Y in U;
  X \/ Y in U by A1,A2,Th60;
  then bool(X \/ Y) in U by Th59;
  then
A3: bool bool(X \/ Y) in U by Th59;
  [:X,Y:] c= bool bool(X \/ Y) by ZFMISC_1:86;
  hence [:X,Y:] in U by A3,CLASSES1:def 1;
  then
A4: bool [:X,Y:] in U by Th59;
  Funcs(X,Y) c= bool [:X,Y:] by FRAENKEL:2;
  hence thesis by A4,CLASSES1:def 1;
end;

reserve u,v for Element of U;

registration
  let U1;
  cluster non empty for Element of U1;
  existence
  proof
    {} in U1 by Th56;
    then reconsider x = bool {} as Element of U1 by Th59;
    take x;
    thus thesis;
  end;
end;

definition
  let U,u;
  redefine func {u} -> Element of U;
  coherence by Th2;
  redefine func bool u -> Element of U;
  coherence by Th59;
  redefine func union u -> Element of U;
  coherence by Th59;
  redefine func meet u -> Element of U;
  coherence by Th59;
  let v;
  redefine func {u,v} -> Element of U;
  coherence by Th2;
  redefine func [u,v] -> Element of U;
  coherence by Th3;
  redefine func u \/ v -> Element of U;
  coherence by Th60;
  redefine func u /\ v -> Element of U;
  coherence by Th60;
  redefine func u \ v -> Element of U;
  coherence by Th60;
  redefine func u \+\ v -> Element of U;
  coherence by Th60;
  redefine func [:u,v:] -> Element of U;
  coherence by Th61;
  redefine func Funcs(u,v) -> Element of U;
  coherence by Th61;
end;

definition
  func FinSETS -> Universe equals
  Tarski-Class {};
  correctness;
end;

Lm5: omega is limit_ordinal by ORDINAL1:def 11;

theorem Th62:
  card Rank omega = card omega
proof
  deffunc h(Ordinal) = Rank $1;
  consider L such that
A1: dom L = omega & for A st A in omega holds L.A = h(A) from ORDINAL2:
  sch 2;
  now
    let X,Y;
    assume X in rng L;
    then consider x such that
A2: x in dom L and
A3: X = L.x by FUNCT_1:def 3;
    assume Y in rng L;
    then consider y such that
A4: y in dom L and
A5: Y = L.y by FUNCT_1:def 3;
    reconsider x,y as Ordinal by A2,A4;
A6: Y = Rank y by A1,A4,A5;
A7: x c= y or y c= x;
    X = Rank x by A1,A2,A3;
    then X c= Y or Y c= X by A6,A7,CLASSES1:37;
    hence X,Y are_c=-comparable by XBOOLE_0:def 9;
  end;
  then
A8: rng L is c=-linear by ORDINAL1:def 8;
A9: card omega c= card Rank omega by CARD_1:11,CLASSES1:38;
A10: now
    let X;
    assume X in rng L;
    then consider x such that
A11: x in dom L and
A12: X = L.x by FUNCT_1:def 3;
    reconsider x as Ordinal by A11;
    X = Rank x by A1,A11,A12;
    hence card X in card omega by A1,A11,CARD_2:67,CARD_3:42;
  end;
  Rank omega = Union L by A1,Lm5,Th24
    .= union rng L by CARD_3:def 4;
  then card Rank omega c= card omega by A8,A10,CARD_3:46;
  hence thesis by A9,XBOOLE_0:def 10;
end;

theorem Th63:
  Rank omega is Tarski
proof
  thus X in Rank omega & Y c= X implies Y in Rank omega by CLASSES1:41;
  thus X in Rank omega implies bool X in Rank omega
  proof
    assume X in Rank omega;
    then consider A such that
A1: A in omega and
A2: X in Rank A by Lm5,CLASSES1:31;
A3: bool X in Rank succ A by A2,CLASSES1:42;
    succ A in omega by A1,Lm5,ORDINAL1:28;
    hence thesis by A3,Lm5,CLASSES1:31;
  end;
  thus X c= Rank omega implies X,Rank omega are_equipotent or X in Rank omega
  proof
A4: 0 in omega;
    defpred P[set,set] means the_rank_of $2 c= the_rank_of $1;
    assume that
A5: X c= Rank omega and
A6: not X,Rank omega are_equipotent and
A7: not X in Rank omega;
A8: card X <> card omega by A6,Th62,CARD_1:5;
    card X c= card omega by A5,Th62,CARD_1:11;
    then card X in omega by A8,CARD_1:3,47;
    then reconsider X as finite set;
A9: for x,y holds P[x,y] or P[y,x];
A10: for x,y,z st P[x,y] & P[y,z] holds P[x,z] by XBOOLE_1:1;
    omega c= Rank omega by CLASSES1:38;
    then
A11: X <> {} by A7,A4;
    consider x such that
A12: x in X & for y st y in X holds P[x,y] from CARD_2:sch 2(A11,A9,A10);
    set A = the_rank_of x;
    now
      let Y;
      assume Y in X;
      then the_rank_of Y c= A by A12;
      hence the_rank_of Y in succ A by ORDINAL1:22;
    end;
    then
A13: the_rank_of X c= succ A by CLASSES1:69;
    A in omega by A5,A12,CLASSES1:66;
    then succ A in omega by Lm5,ORDINAL1:28;
    then the_rank_of X in omega by A13,ORDINAL1:12;
    hence thesis by A7,CLASSES1:66;
  end;
end;

theorem
  FinSETS = Rank omega
proof
A1: omega c= Rank omega by CLASSES1:38;
  then reconsider M = Rank omega as non empty set;
  0 in omega;
  then M is_Tarski-Class_of {} by A1,Th63,CLASSES1:def 3;
  hence FinSETS c= Rank omega by CLASSES1:def 4;
A2: {} in On FinSETS by Th51,ORDINAL3:8;
A3: FinSETS = Rank On FinSETS by Th50;
  On FinSETS is limit_ordinal by Th51;
  then omega c= On FinSETS by A2,ORDINAL1:def 11;
  hence thesis by A3,CLASSES1:37;
end;

definition
  func SETS -> Universe equals
  Tarski-Class FinSETS;
  correctness;
end;

registration
  let X be set;
  cluster the_transitive-closure_of X -> epsilon-transitive;
  coherence by CLASSES1:51;
end;

registration
  let X be epsilon-transitive set;
  cluster Tarski-Class X -> epsilon-transitive;
  coherence by CLASSES1:23;
end;

definition
  let X be set;
  func Universe_closure X -> Universe means
  :Def4:
  X c= it & for Y being Universe st X c= Y holds it c= Y;
  uniqueness
  proof
    let T1, T2 be Universe;
    assume
A1: not thesis;
    then
A2: T2 c= T1;
    T1 c= T2 by A1;
    hence thesis by A1,A2,XBOOLE_0:def 10;
  end;
  existence
  proof
    per cases;
    suppose
      Rank the_rank_of X is Universe;
      then reconsider R = Rank the_rank_of X as Universe;
      take R;
      thus X c= R by CLASSES1:def 8;
      let Y be Universe;
      assume X c= Y;
      then the_rank_of X c= the_rank_of Y by CLASSES1:67;
      then
A3:   R c= Rank the_rank_of Y by CLASSES1:37;
A4:   Rank card Y = Y by Th31;
      then the_rank_of Y c= card Y by CLASSES1:def 8;
      then Rank the_rank_of Y c= Y by A4,CLASSES1:37;
      hence thesis by A3,XBOOLE_1:1;
    end;
    suppose
A5:   Rank the_rank_of X is not Universe;
      take R = Tarski-Class Rank the_rank_of X;
A6:   Rank the_rank_of X in R by CLASSES1:2;
      X c= Rank the_rank_of X by CLASSES1:def 8;
      then X in R by A6,CLASSES1:def 1;
      hence X c= R by ORDINAL1:def 2;
      let Y be Universe;
      assume X c= Y;
      then
A7:   the_rank_of X c= the_rank_of Y by CLASSES1:67;
A8:   Rank card Y = Y by Th31;
      then
A9:   the_rank_of Y c= card Y by CLASSES1:def 8;
      Y c= Rank the_rank_of Y by CLASSES1:def 8;
      then card Y c= the_rank_of Y by A8,CLASSES1:37;
      then card Y = the_rank_of Y by A9,XBOOLE_0:def 10;
      then the_rank_of X c< card Y by A5,A7,A8,XBOOLE_0:def 8;
      then the_rank_of X in card Y by ORDINAL1:11;
      then Rank the_rank_of X in Y by A8,CLASSES1:36;
      then Y is_Tarski-Class_of Rank the_rank_of X by CLASSES1:def 3;
      hence thesis by CLASSES1:def 4;
    end;
  end;
end;

deffunc C(Ordinal,set) = Tarski-Class $2;
deffunc D(Ordinal,T-Sequence) = Universe_closure Union $2;

definition
  mode FinSet is Element of FinSETS;
  mode Set is Element of SETS;
  let A;
  func UNIVERSE A means
  : Def5:
  ex L st it = last L & dom L = succ A & L.{} =
FinSETS & (for C st succ C in succ A holds L.succ C = Tarski-Class(L.C)) & for
  C st C in succ A & C <> {} & C is limit_ordinal holds L.C = Universe_closure
  Union(L|C);
  correctness
  proof
    thus (ex x,L st x = last L & dom L = succ A & L.{} = FinSETS & (for C st
succ C in succ A holds L.succ C = C(C,L.C)) & for C st C in succ A & C <> {} &
C is limit_ordinal holds L.C = D(C,L|C) ) & for x1,x2 being set st (ex L st x1
= last L & dom L = succ A & L.{} = FinSETS & (for C st succ C in succ A holds L
.succ C = C(C,L.C)) & for C st C in succ A & C <> {} & C is limit_ordinal holds
L.C = D(C,L|C) ) & (ex L st x2 = last L & dom L = succ A & L.{} = FinSETS & (
for C st succ C in succ A holds L.succ C = C(C,L.C)) & for C st C in succ A & C
<> {} & C is limit_ordinal holds L.C = D(C,L|C) ) holds x1 = x2 from ORDINAL2:
    sch 7;
  end;
end;

deffunc u(Ordinal) = UNIVERSE $1;

Lm6: now
A1: for A,x holds x = u(A) iff ex L st x = last L & dom L = succ A & L.{} =
FinSETS & (for C st succ C in succ A holds L.succ C = C(C,L.C)) & for C st C in
  succ A & C <> {} & C is limit_ordinal holds L.C = D(C,L|C) by Def5;
  thus u({}) = FinSETS from ORDINAL2:sch 8(A1);
  thus for A holds u(succ A) = C(A,u(A)) from ORDINAL2:sch 9(A1);
  let A,L;
  assume that
A2: A <> {} & A is limit_ordinal and
A3: dom L = A and
A4: for B st B in A holds L.B = u(B);
  thus u(A) = D(A,L) from ORDINAL2:sch 10(A1,A2,A3,A4);
end;

registration
  let A;
  cluster UNIVERSE A -> universal non empty;
  coherence
  proof
    defpred P[Ordinal] means UNIVERSE $1 is Universe;
A1: P[B] implies P[succ B]
    proof
      assume P[B];
      then reconsider UU = UNIVERSE B as Universe;
      UNIVERSE succ B = Tarski-Class UU by Lm6;
      hence thesis;
    end;
A2: for A st A <> {} & A is limit_ordinal & for B st B in A holds P[B]
    holds P[A]
    proof
      let A such that
A3:   A <> {} and
A4:   A is limit_ordinal and
      for B st B in A holds P[B];
      consider L such that
A5:   dom L = A & for B st B in A holds L.B = u(B) from ORDINAL2:sch 2;
      UNIVERSE A = Universe_closure Union L by A3,A4,A5,Lm6
        .= Universe_closure union rng L by CARD_3:def 4;
      hence thesis;
    end;
A6: P[{}] by Lm6;
    for A holds P[A] from ORDINAL2:sch 1(A6,A1,A2);
    hence thesis;
  end;
end;

theorem
  UNIVERSE {} = FinSETS by Lm6;

theorem
  UNIVERSE succ A = Tarski-Class UNIVERSE A by Lm6;

Lm7: 1 = succ 0;

theorem
  UNIVERSE 1 = SETS by Lm6,Lm7;

theorem
  A <> {} & A is limit_ordinal & dom L = A & (for B st B in A holds L.B
  = UNIVERSE B) implies UNIVERSE A = Universe_closure Union L by Lm6;

theorem
  FinSETS c= U & Tarski-Class {} c= U & UNIVERSE {} c= U
proof
A1: On U c= Rank On U by CLASSES1:38;
A2: Rank On U = U by Th50;
  {} in On U by Th51,ORDINAL3:8;
  hence thesis by A2,A1,Lm6,Th4;
end;

theorem Th70:
  A in B iff UNIVERSE A in UNIVERSE B
proof
  defpred P[Ordinal] means for A st A in $1 holds UNIVERSE A in UNIVERSE $1;
A1: for B st P[B] holds P[succ B]
  proof
    let B such that
A2: P[B];
    let A;
    assume
A3: A in succ B;
    A c< B iff A c= B & A <> B by XBOOLE_0:def 8;
    then A in B or A = B by A3,ORDINAL1:11,22;
    then
A4: UNIVERSE A in UNIVERSE B or UNIVERSE A = UNIVERSE B by A2;
    UNIVERSE succ B = Tarski-Class UNIVERSE B by Lm6;
    then UNIVERSE B in UNIVERSE succ B by CLASSES1:2;
    hence thesis by A4,ORDINAL1:10;
  end;
A5: for A st A <> {} & A is limit_ordinal & for B st B in A holds P[B]
  holds P[A]
  proof
    let B;
    assume that
A6: B <> {} and
A7: B is limit_ordinal and
    for C st C in B for A st A in C holds UNIVERSE A in UNIVERSE C;
    let A;
    consider L such that
A8: dom L = B & for C st C in B holds L.C = u(C) from ORDINAL2:sch 2;
    assume A in B;
    then
A9: succ A in B by A7,ORDINAL1:28;
    then L.succ A = UNIVERSE succ A by A8;
    then UNIVERSE succ A in rng L by A9,A8,FUNCT_1:def 3;
    then
A10: UNIVERSE succ A c= union rng L by ZFMISC_1:74;
    UNIVERSE B = Universe_closure Union L by A6,A7,A8,Lm6
      .= Universe_closure union rng L by CARD_3:def 4;
    then union rng L c= UNIVERSE B by Def4;
    then
A11: UNIVERSE succ A c= UNIVERSE B by A10,XBOOLE_1:1;
A12: UNIVERSE A in Tarski-Class UNIVERSE A by CLASSES1:2;
    UNIVERSE succ A = Tarski-Class UNIVERSE A by Lm6;
    hence thesis by A12,A11;
  end;
A13: P[{}];
A14: for B holds P[B] from ORDINAL2:sch 1(A13,A1,A5);
  hence A in B implies UNIVERSE A in UNIVERSE B;
  assume that
A15: UNIVERSE A in UNIVERSE B and
A16: not A in B;
  B c< A iff B c= A & B <> A by XBOOLE_0:def 8;
  then B in A or B = A by A16,ORDINAL1:11,16;
  hence contradiction by A14,A15;
end;

theorem
  UNIVERSE A = UNIVERSE B implies A = B
proof
  assume that
A1: UNIVERSE A = UNIVERSE B and
A2: A <> B;
  A in B or B in A by A2,ORDINAL1:14;
  then UNIVERSE A in UNIVERSE B or UNIVERSE B in UNIVERSE A by Th70;
  hence contradiction by A1;
end;

theorem
  A c= B iff UNIVERSE A c= UNIVERSE B
proof
  thus A c= B implies UNIVERSE A c= UNIVERSE B
  proof
    assume
A1: A c= B;
    assume not UNIVERSE A c= UNIVERSE B;
    then UNIVERSE B in UNIVERSE A by Th53;
    then B in A by Th70;
    hence contradiction by A1,ORDINAL1:5;
  end;
  assume
A2: UNIVERSE A c= UNIVERSE B;
  assume not A c= B;
  then B in A by ORDINAL1:16;
  then UNIVERSE B in UNIVERSE A by Th70;
  hence contradiction by A2,ORDINAL1:5;
end;

reserve u,v for Element of Tarski-Class(X);

theorem
  Tarski-Class(X,{}) in Tarski-Class(X,1) &
  Tarski-Class(X,{}) <> Tarski-Class(X,1)
proof
  deffunc F(Ordinal) = Tarski-Class(X,$1);
  deffunc C(Ordinal,set) = { u : ex v st v in $2 & u c= v } \/
  { bool v : v in $2 } \/ bool $2 /\ Tarski-Class X;
  deffunc D(Ordinal,T-Sequence) = (union rng $2) /\ Tarski-Class X;
A1: for A,x holds x = F(A) iff
  ex L st x = last L & dom L = succ A & L.{} = { X } &
  (for C st succ C in succ A holds L.succ C = C(C,L.C)) &
  for C st C in succ A & C <> {} & C is limit_ordinal
  holds L.C = D(C,L|C) by CLASSES1:def 5;
A2: F({}) = { X } from ORDINAL2:sch 8(A1);
  then X in Tarski-Class(X,{}) by TARSKI:def 1;
  then
A3: bool X in Tarski-Class X by CLASSES1:4;
 { X } c= bool X
  proof
    let x;
    assume x in { X };
then  x = X by TARSKI:def 1;
    hence thesis by ZFMISC_1:def 1;
  end;
then  1 = succ 0 & { X } in Tarski-Class X by A3,CLASSES1:3;
  thus
then A4: Tarski-Class(X,{}) in Tarski-Class(X,1) by A2,CLASSES1:10;
  assume Tarski-Class(X,{}) = Tarski-Class(X,1);
  hence contradiction by A4;
end;















begin

reserve phi,fi,psi for Ordinal-Sequence,
  A,A1,B,C,D for Ordinal,
  f,g for Function,
  X for set,
  x,y,z for set;

Lm1: {} in omega by ORDINAL1:def 11;

Lm2: omega is limit_ordinal by ORDINAL1:def 11;

Lm3: 1 = succ {};

registration
  let L be Ordinal-Sequence;
  cluster last L -> ordinal;
  coherence;
end;

theorem
  dom fi = succ A implies last fi is_limes_of fi & lim fi = last fi
proof
  assume
A1: dom fi = succ A;
  then
A2: last fi = fi.A by ORDINAL2:6;
  thus last fi is_limes_of fi
  proof
    per cases;
    case
A3:   last fi = {};
      take A;
      thus A in dom fi by A1,ORDINAL1:6;
      let B;
      assume that
A4:   A c= B and
A5:   B in dom fi;
      B c= A by A1,A5,ORDINAL1:22;
      hence thesis by A2,A3,A4,XBOOLE_0:def 10;
    end;
    case
      last fi <> {};
      let B,C such that
A6:   B in last fi and
A7:   last fi in C;
      take A;
      thus A in dom fi by A1,ORDINAL1:6;
      let D;
      assume that
A8:   A c= D and
A9:   D in dom fi;
      D c= A by A1,A9,ORDINAL1:22;
      hence thesis by A2,A6,A7,A8,XBOOLE_0:def 10;
    end;
  end;
  hence thesis by ORDINAL2:def 10;
end;

definition
  let fi,psi be T-Sequence;
  func fi^psi -> T-Sequence means
  : Def1:
  dom it = (dom fi)+^(dom psi) & (for
A st A in dom fi holds it.A = fi.A) & for A st A in dom psi holds it.((dom fi)
  +^A) = psi.A;
  existence
  proof
    defpred P[set] means $1 in dom fi;
    deffunc G(Ordinal) = psi.(($1)-^dom fi);
    deffunc F(set) = fi.$1;
    consider f such that
A1: dom f = (dom fi)+^(dom psi) and
A2: for x being Ordinal st x in (dom fi)+^(dom psi) holds (P[x]
    implies f.x = F(x)) & (not P[x] implies f.x = G(x)) from FINSET_1:sch 1;
    reconsider f as T-Sequence by A1,ORDINAL1:def 7;
    take f;
    thus dom f = (dom fi)+^(dom psi) by A1;
    thus for A st A in dom fi holds f.A = fi.A
    proof
A3:   dom fi c= dom f by A1,ORDINAL3:24;
      let A;
      assume A in dom fi;
      hence thesis by A1,A2,A3;
    end;
    let A;
    dom fi c= (dom fi)+^A by ORDINAL3:24;
    then
A4: not (dom fi)+^A in dom fi by ORDINAL1:5;
    assume A in dom psi;
    then (dom fi)+^A in dom f by A1,ORDINAL2:32;
    then f.((dom fi)+^A) = psi.((((dom fi)+^A))-^dom fi) by A1,A2,A4;
    hence thesis by ORDINAL3:52;
  end;
  uniqueness
  proof
    let f1,f2 be T-Sequence such that
A5: dom f1 = (dom fi)+^(dom psi) and
A6: for A st A in dom fi holds f1.A = fi.A and
A7: for A st A in dom psi holds f1.((dom fi)+^A) = psi.A and
A8: dom f2 = (dom fi)+^(dom psi) and
A9: for A st A in dom fi holds f2.A = fi.A and
A10: for A st A in dom psi holds f2.((dom fi)+^A) = psi.A;
    now
      let x;
      assume
A11:  x in (dom fi)+^(dom psi);
      then reconsider A = x as Ordinal;
      now
        per cases;
        suppose
A12:      x in dom fi;
          then f1.A = fi.A by A6;
          hence f1.x = f2.x by A9,A12;
        end;
        suppose
          not x in dom fi;
          then dom fi c= A by ORDINAL1:16;
          then
A13:      (dom fi)+^(A-^dom fi) = A by ORDINAL3:def 5;
          then f1.A = psi.(A-^dom fi) by A7,A11,ORDINAL3:22;
          hence f1.x = f2.x by A10,A11,A13,ORDINAL3:22;
        end;
      end;
      hence f1.x = f2.x;
    end;
    hence thesis by A5,A8,FUNCT_1:2;
  end;
end;

theorem Th2:
 rng(fi^psi) c= rng fi \/ rng psi
   proof set f = fi^psi, A1 = rng fi, A2 = rng psi;
A1:  dom(fi^psi) = (dom fi)+^(dom psi) by Def1;
    let y;
      assume y in rng f;
      then consider x such that
A2:   x in dom f and
A3:   y = f.x by FUNCT_1:def 3;
      reconsider x as Ordinal by A2;
        per cases;
        suppose
A4:       x in dom fi;
          then
A5:       fi.x in rng fi by FUNCT_1:def 3;
          y = fi.x by A3,A4,Def1;
          then y in A1 by A5;
          hence thesis by XBOOLE_0:def 3;
        end;
        suppose
          not x in dom fi;
          then dom fi c= x by ORDINAL1:16;
          then
A6:      (dom fi)+^(x-^dom fi) = x by ORDINAL3:def 5;
          then
A7:      x-^dom fi in dom psi by A1,A2,ORDINAL3:22;
          then y = psi.(x-^dom fi) by A3,A6,Def1;
          then y in rng psi by A7,FUNCT_1:def 3;
          then y in A2;
          hence thesis by XBOOLE_0:def 3;
        end;
    end;

registration
  let fi,psi;
  cluster fi^psi -> Ordinal-yielding;
  coherence
  proof
    set f = fi^psi;
    consider A1 being Ordinal such that
A1: rng fi c= A1 by ORDINAL2:def 4;
    consider A2 being Ordinal such that
A2: rng psi c= A2 by ORDINAL2:def 4;
A3:   A2 c= A1+^A2 by ORDINAL3:24;
   A1 c= A1+^A2 by ORDINAL3:24;
    then
A4:   A1 \/ A2 c= A1+^A2 by A3,XBOOLE_1:8;
A5:  rng f c= rng fi \/ rng psi by Th2;
    rng fi \/ rng psi c= A1 \/ A2 by A1,A2,XBOOLE_1:13;
    then rng f c= A1 \/ A2 by A5,XBOOLE_1:1;
    then rng f c= A1+^A2 by A4,XBOOLE_1:1;
    hence thesis by ORDINAL2:def 4;
  end;
end;

theorem Th3:
  A is_limes_of psi implies A is_limes_of fi^psi
proof
  assume that
A1: A = {} & (ex B st B in dom psi & for C st B c= C & C in dom psi
holds psi.C = {}) or A <> {} & for B,C st B in A & A in C ex D st D in dom psi
  & for E being Ordinal st D c= E & E in dom psi holds B in psi.E & psi.E in C;
A2: dom(fi^psi) = (dom fi)+^(dom psi) by Def1;
  per cases;
  case
    A = {};
    then consider B such that
A3: B in dom psi and
A4: for C st B c= C & C in dom psi holds psi.C = {} by A1;
    take B1 = (dom fi)+^B;
    thus B1 in dom(fi^psi) by A2,A3,ORDINAL2:32;
    let C;
    assume that
A5: B1 c= C and
A6: C in dom(fi^psi);
A7: C = B1+^(C-^B1) by A5,ORDINAL3:def 5
      .= (dom fi)+^(B+^(C-^B1)) by ORDINAL3:30;
    then
A8: B+^(C-^B1) in dom psi by A2,A6,ORDINAL3:22;
    B c= B+^(C-^B1) by ORDINAL3:24;
    then psi.(B+^(C-^B1)) = {} by A2,A4,A6,A7,ORDINAL3:22;
    hence thesis by A7,A8,Def1;
  end;
  case
    A <> {};
    let B,C;
    assume that
A9: B in A and
A10: A in C;
    consider D such that
A11: D in dom psi and
A12: for E being Ordinal st D c= E & E in dom psi holds B in psi.E &
    psi.E in C by A1,A9,A10;
    take D1 = (dom fi)+^D;
    thus D1 in dom(fi^psi) by A2,A11,ORDINAL2:32;
    let E be Ordinal;
    assume that
A13: D1 c= E and
A14: E in dom(fi^psi);
A15: D c= D+^(E-^D1) by ORDINAL3:24;
A16: E = D1+^(E-^D1) by A13,ORDINAL3:def 5
      .= (dom fi)+^(D+^(E-^D1)) by ORDINAL3:30;
    then
A17: D+^(E-^D1) in dom psi by A2,A14,ORDINAL3:22;
    then (fi^psi).E = psi.(D+^(E-^D1)) by A16,Def1;
    hence thesis by A12,A15,A17;
  end;
end;

theorem
  A is_limes_of fi implies B+^A is_limes_of B+^fi
proof
  assume that
A1: A = {} & (ex B st B in dom fi & for C st B c= C & C in dom fi holds
fi.C = {}) or A <> {} & for B,C st B in A & A in C ex D st D in dom fi & for E
  being Ordinal st D c= E & E in dom fi holds B in fi.E & fi.E in C;
A2: dom fi = dom(B+^fi) by ORDINAL3:def 1;
  per cases;
  case
A3: B+^A = {};
    then consider A1 such that
A4: A1 in dom fi and
A5: for C st A1 c= C & C in dom fi holds fi.C = {} by A1,ORDINAL3:26;
    take A1;
    thus A1 in dom(B+^fi) by A4,ORDINAL3:def 1;
    let C;
    assume that
A6: A1 c= C and
A7: C in dom(B+^fi);
A8: (B+^fi).C = B+^(fi.C) by A2,A7,ORDINAL3:def 1;
    fi.C = {} by A2,A5,A6,A7;
    hence thesis by A3,A8,ORDINAL3:26;
  end;
  case
    B+^A <> {};
    now
      per cases;
      suppose
A9:     A = {};
        then consider A1 such that
A10:    A1 in dom fi and
A11:    for C st A1 c= C & C in dom fi holds fi.C = {} by A1;
        let B1,B2 be Ordinal such that
A12:    B1 in B+^A and
A13:    B+^A in B2;
        take A1;
        thus A1 in dom(B+^fi) by A10,ORDINAL3:def 1;
        let C;
        assume that
A14:    A1 c= C and
A15:    C in dom(B+^fi);
        (B+^fi).C = B+^(fi.C) by A2,A15,ORDINAL3:def 1;
        hence B1 in (B+^fi).C & (B+^fi).C in B2 by A2,A9,A11,A12,A13,A14,A15;
      end;
      suppose
A16:    A <> {};
        let B1,B2 be Ordinal;
        assume that
A17:    B1 in B+^A and
A18:    B+^A in B2;
        B1-^B in A by A16,A17,ORDINAL3:60;
        then consider A1 such that
A19:    A1 in dom fi and
A20:    for C st A1 c= C & C in dom fi holds B1-^B in fi.C & fi.C in
        B2-^B by A1,A18,ORDINAL3:61;
A21:    B1 c= B+^(B1-^B) by ORDINAL3:62;
A22:    B c= B+^A by ORDINAL3:24;
        B+^A c= B2 by A18,ORDINAL1:def 2;
        then B c= B2 by A22,XBOOLE_1:1;
        then
A23:    B+^(B2-^B) = B2 by ORDINAL3:def 5;
        take A1;
        thus A1 in dom(B+^fi) by A19,ORDINAL3:def 1;
        let C;
        assume that
A24:    A1 c= C and
A25:    C in dom(B+^fi);
A26:    (B+^fi).C = B+^(fi.C) by A2,A25,ORDINAL3:def 1;
        reconsider E = fi.C as Ordinal;
        B1-^B in E by A2,A20,A24,A25;
        then
A27:    B+^(B1-^B) in B+^E by ORDINAL2:32;
        E in B2-^B by A2,A20,A24,A25;
        hence B1 in (B+^fi).C & (B+^fi).C in B2 by A21,A26,A23,A27,ORDINAL1:12
,ORDINAL2:32;
      end;
    end;
    hence thesis;
  end;
end;

Lm4: A is_limes_of fi implies dom fi <> {}
proof
  assume that
A1: A = {} & (ex B st B in dom fi & for C st B c= C & C in dom fi holds
fi.C = {}) or A <> {} & for B,C st B in A & A in C ex D st D in dom fi & for E
  being Ordinal st D c= E & E in dom fi holds B in fi.E & fi.E in C;
  now
    per cases;
    suppose
      A = {};
      hence thesis by A1;
    end;
    suppose
      A <> {};
      then {} in A by ORDINAL3:8;
      then
      ex B st B in dom fi & for C st B c= C & C in dom fi holds {} in fi.C
      & fi.C in succ A by A1,ORDINAL1:6;
      hence thesis;
    end;
  end;
  hence thesis;
end;

theorem Th5:
  A is_limes_of fi implies A*^B is_limes_of fi*^B
proof
A1: dom fi = dom(fi*^B) by ORDINAL3:def 4;
  assume
A2: A is_limes_of fi;
  then
A3: dom fi <> {} by Lm4;
  per cases;
  case
    A*^B = {};
    then
A4: B = {} or A = {} by ORDINAL3:31;
    now
      per cases;
      suppose
A5:     B = {};
        set x = the Element of dom fi;
        reconsider x as Ordinal;
        take A1 = x;
        thus A1 in dom(fi*^B) by A1,A3;
        let C;
        assume that
        A1 c= C and
A6:     C in dom(fi*^B);
        thus (fi*^B).C = (fi.C)*^B by A1,A6,ORDINAL3:def 4
          .= {} by A5,ORDINAL2:38;
      end;
      suppose
        B <> {};
        then consider A1 such that
A7:     A1 in dom fi and
A8:     for C st A1 c= C & C in dom fi holds fi.C = {} by A2,A4,ORDINAL2:def 9
;
        take A1;
        thus A1 in dom(fi*^B) by A7,ORDINAL3:def 4;
        let C;
        assume that
A9:     A1 c= C and
A10:    C in dom(fi*^B);
A11:    (fi*^B).C = (fi.C)*^B by A1,A10,ORDINAL3:def 4;
        fi.C = {} by A1,A8,A9,A10;
        hence (fi*^B).C = {} by A11,ORDINAL2:35;
      end;
    end;
    hence thesis;
  end;
  case
A12: A*^B <> {};
    let B1,B2 be Ordinal such that
A13: B1 in A*^B and
A14: A*^B in B2;
A15: B <> {} by A12,ORDINAL2:38;
A16: now
      assume not ex A1 st A = succ A1;
      then A is limit_ordinal by ORDINAL1:29;
      then consider C such that
A17:  C in A and
A18:  B1 in C*^B by A13,ORDINAL3:41;
      A in succ A by ORDINAL1:6;
      then consider D such that
A19:  D in dom fi and
A20:  for A1 st D c= A1 & A1 in dom fi holds C in fi.A1 & fi.A1 in
      succ A by A2,A17,ORDINAL2:def 9;
      take D;
      thus D in dom(fi*^B) by A19,ORDINAL3:def 4;
      let A1;
      assume that
A21:  D c= A1 and
A22:  A1 in dom(fi*^B);
      reconsider E = fi.A1 as Ordinal;
      fi.A1 in succ A by A1,A20,A21,A22;
      then
A23:  E c= A by ORDINAL1:22;
      C in fi.A1 by A1,A20,A21,A22;
      then C*^B in E*^B by A15,ORDINAL2:40;
      then
A24:  B1 in E*^B by A18,ORDINAL1:10;
      (fi*^B).A1 = (fi.A1)*^B by A1,A22,ORDINAL3:def 4;
      hence B1 in (fi*^B).A1 & (fi*^B).A1 in B2 by A14,A23,A24,ORDINAL1:12
,ORDINAL2:41;
    end;
    now
A25:  A in succ A by ORDINAL1:6;
      given A1 such that
A26:  A = succ A1;
      A1 in A by A26,ORDINAL1:6;
      then consider D such that
A27:  D in dom fi and
A28:  for C st D c= C & C in dom fi holds A1 in fi.C & fi.C in succ A
      by A2,A25,ORDINAL2:def 9;
      take D;
      thus D in dom(fi*^B) by A27,ORDINAL3:def 4;
      let C;
      assume that
A29:  D c= C and
A30:  C in dom(fi*^B);
      reconsider E = fi.C as Ordinal;
      A1 in E by A1,A28,A29,A30;
      then
A31:  A c= E by A26,ORDINAL1:21;
      E in succ A by A1,A28,A29,A30;
      then
A32:  E c= A by ORDINAL1:22;
      (fi*^B).C = E*^B by A1,A30,ORDINAL3:def 4;
      hence B1 in (fi*^B).C & (fi*^B).C in B2 by A13,A14,A31,A32,
XBOOLE_0:def 10;
    end;
    hence thesis by A16;
  end;
end;

theorem Th6:
  dom fi = dom psi & B is_limes_of fi & C is_limes_of psi & ((for A
st A in dom fi holds fi.A c= psi.A) or for A st A in dom fi holds fi.A in psi.A
  ) implies B c= C
proof
  assume that
A1: dom fi = dom psi and
A2: B = {} & (ex A1 st A1 in dom fi & for C st A1 c= C & C in dom fi
holds fi.C = {}) or B <> {} & for A1,C st A1 in B & B in C ex D st D in dom fi
  & for E being Ordinal st D c= E & E in dom fi holds A1 in fi.E & fi.E in C
  and
A3: C = {} & (ex B st B in dom psi & for A1 st B c= A1 & A1 in dom psi
  holds psi.A1 = {}) or C <> {} & for B,A1 st B in C & C in A1 ex D st D in dom
psi & for E being Ordinal st D c= E & E in dom psi holds B in psi.E & psi.E in
  A1 and
A4: (for A st A in dom fi holds fi.A c= psi.A) or for A st A in dom fi
  holds fi.A in psi.A;
A5: now
    let A;
    reconsider A1 = fi.A, A2 = psi.A as Ordinal;
    assume A in dom fi;
    then A1 c= A2 or A1 in A2 by A4;
    hence fi.A c= psi.A by ORDINAL1:def 2;
  end;
  now
    per cases;
    suppose
      B = {} & C = {};
      hence thesis;
    end;
    suppose
      B = {} & C <> {};
      then B in C by ORDINAL3:8;
      hence thesis by ORDINAL1:def 2;
    end;
    suppose
A6:   B <> {} & C = {};
      then {} in B by ORDINAL3:8;
      then consider A2 being Ordinal such that
A7:   A2 in dom fi and
A8:   for A st A2 c= A & A in dom fi holds {} in fi.A & fi.A in succ
      B by A2,ORDINAL1:6;
      consider A1 such that
A9:   A1 in dom psi and
A10:  for A st A1 c= A & A in dom psi holds psi.A = {} by A3,A6;
A11:  A1 \/ A2 = A1 or A1 \/ A2 = A2 by ORDINAL3:12;
      then
A12:  fi.(A1 \/ A2) c= psi.(A1 \/ A2) by A1,A5,A9,A7;
      A2 c= A1 \/ A2 by XBOOLE_1:7;
      then {} in fi.(A1 \/ A2) by A1,A9,A7,A8,A11;
      hence thesis by A1,A9,A10,A7,A11,A12,XBOOLE_1:7;
    end;
    suppose
A13:  B <> {} & C <> {};
      assume not B c= C;
      then C in B by ORDINAL1:16;
      then consider A1 such that
A14:  A1 in dom fi and
A15:  for A st A1 c= A & A in dom fi holds C in fi.A & fi.A in succ B
      by A2,ORDINAL1:6;
      {} in C by A13,ORDINAL3:8;
      then consider A2 being Ordinal such that
A16:  A2 in dom psi and
A17:  for A st A2 c= A & A in dom psi holds {} in psi.A & psi.A in
      succ C by A3,ORDINAL1:6;
A18:  A1 \/ A2 = A1 or A1 \/ A2 = A2 by ORDINAL3:12;
      reconsider A3 = psi.(A1 \/ A2) as Ordinal;
      A2 c= A1 \/ A2 by XBOOLE_1:7;
      then psi.(A1 \/ A2) in succ C by A1,A14,A16,A17,A18;
      then
A19:  A3 c= C by ORDINAL1:22;
      A1 c= A1 \/ A2 by XBOOLE_1:7;
      then
A20:  C in fi.(A1 \/ A2) by A1,A14,A15,A16,A18;
      fi.(A1 \/ A2) c= psi.(A1 \/ A2) by A1,A5,A14,A16,A18;
      hence contradiction by A20,A19,ORDINAL1:5;
    end;
  end;
  hence thesis;
end;

reserve f1,f2 for Ordinal-Sequence;

theorem
  dom f1 = dom fi & dom fi = dom f2 & A is_limes_of f1 & A is_limes_of
  f2 & (for A st A in dom fi holds f1.A c= fi.A & fi.A c= f2.A) implies A
  is_limes_of fi
proof
  assume that
A1: dom f1 = dom fi and
A2: dom fi = dom f2 and
A3: A = {} & (ex B st B in dom f1 & for C st B c= C & C in dom f1 holds
f1.C = {}) or A <> {} & for B,C st B in A & A in C ex D st D in dom f1 & for E
  being Ordinal st D c= E & E in dom f1 holds B in f1.E & f1.E in C and
A4: A = {} & (ex B st B in dom f2 & for C st B c= C & C in dom f2 holds
f2.C = {}) or A <> {} & for B,C st B in A & A in C ex D st D in dom f2 & for E
  being Ordinal st D c= E & E in dom f2 holds B in f2.E & f2.E in C and
A5: for A st A in dom fi holds f1.A c= fi.A & fi.A c= f2.A;
  per cases;
  case
    A = {};
    then consider B being Ordinal such that
A6: B in dom f2 and
A7: for C st B c= C & C in dom f2 holds f2.C = {} by A4;
    take B;
    thus B in dom fi by A2,A6;
    let C;
    assume that
A8: B c= C and
A9: C in dom fi;
    f2.C = {} by A2,A7,A8,A9;
    hence thesis by A5,A9,XBOOLE_1:3;
  end;
  case
    A <> {};
    let B,C;
    assume that
A10: B in A and
A11: A in C;
    consider D2 being Ordinal such that
A12: D2 in dom f2 and
A13: for A1 st D2 c= A1 & A1 in dom f2 holds B in f2.A1 & f2.A1 in C
    by A4,A10,A11;
    consider D1 being Ordinal such that
A14: D1 in dom f1 and
A15: for A1 st D1 c= A1 & A1 in dom f1 holds B in f1.A1 & f1.A1 in C
    by A3,A10,A11;
    take D = D1 \/ D2;
    thus D in dom fi by A1,A2,A14,A12,ORDINAL3:12;
    let A1;
    assume that
A16: D c= A1 and
A17: A1 in dom fi;
    reconsider B1 = fi.A1, B2 = f2.A1 as Ordinal;
A18: B1 c= B2 by A5,A17;
    D2 c= D by XBOOLE_1:7;
    then D2 c= A1 by A16,XBOOLE_1:1;
    then
A19: B2 in C by A2,A13,A17;
    D1 c= D by XBOOLE_1:7;
    then D1 c= A1 by A16,XBOOLE_1:1;
    then
A20: B in f1.A1 by A1,A15,A17;
    f1.A1 c= fi.A1 by A5,A17;
    hence thesis by A20,A18,A19,ORDINAL1:12;
  end;
end;

theorem Th8:
  dom fi <> {} & dom fi is limit_ordinal & fi is increasing implies
  sup fi is_limes_of fi & lim fi = sup fi
proof
  assume that
A1: dom fi <> {} & dom fi is limit_ordinal and
A2: A in B & B in dom fi implies fi.A in fi.B;
  reconsider x = fi.{} as Ordinal;
  {} in dom fi by A1,ORDINAL3:8;
  then
A3: x in rng fi by FUNCT_1:def 3;
  thus sup fi is_limes_of fi
  proof
    per cases;
    case
      sup fi = {};
      hence thesis by A3,ORDINAL2:19;
    end;
    case
      sup fi <> {};
      let A,B;
      assume that
A4:   A in sup fi and
A5:   sup fi in B;
      consider C such that
A6:   C in rng fi and
A7:   A c= C by A4,ORDINAL2:21;
      consider x such that
A8:   x in dom fi and
A9:   C = fi.x by A6,FUNCT_1:def 3;
      reconsider x as Ordinal by A8;
      take M = succ x;
      thus M in dom fi by A1,A8,ORDINAL1:28;
      let D;
      assume that
A10:  M c= D and
A11:  D in dom fi;
      reconsider E = fi.D as Ordinal;
      x in M by ORDINAL1:6;
      then C in E by A2,A9,A10,A11;
      hence A in fi.D by A7,ORDINAL1:12;
      fi.D in rng fi by A11,FUNCT_1:def 3;
      then E in sup fi by ORDINAL2:19;
      hence thesis by A5,ORDINAL1:10;
    end;
  end;
  hence thesis by ORDINAL2:def 10;
end;

theorem Th9:
  fi is increasing & A c= B & B in dom fi implies fi.A c= fi.B
proof
  assume that
A1: for A,B st A in B & B in dom fi holds fi.A in fi.B and
A2: A c= B and
A3: B in dom fi;
  reconsider C = fi.B as Ordinal;
  now
    per cases;
    suppose
      A = B;
      hence thesis;
    end;
    suppose
      A <> B;
      then A c< B by A2,XBOOLE_0:def 8;
      then A in B by ORDINAL1:11;
      then fi.A in C by A1,A3;
      hence thesis by ORDINAL1:def 2;
    end;
  end;
  hence thesis;
end;

theorem Th10:
  fi is increasing & A in dom fi implies A c= fi.A
proof
  assume that
A1: for A,B st A in B & B in dom fi holds fi.A in fi.B and
A2: A in dom fi and
A3: not A c= fi.A;
  defpred P[set] means $1 in dom fi & not $1 c= fi.$1;
A4: ex A st P[A] by A2,A3;
  consider A such that
A5: P[A] and
A6: for B st P[B] holds A c= B from ORDINAL1:sch 1(A4);
  reconsider B = fi.A as Ordinal;
A7: B in A by A5,ORDINAL1:16;
  then not B c= fi.B by A1,A5,ORDINAL1:5;
  hence contradiction by A5,A6,A7,ORDINAL1:10;
end;

theorem Th11:
  phi is increasing implies phi"A is Ordinal
proof
  assume
A1: for A,B st A in B & B in dom phi holds phi.A in phi.B;
  now
    let X;
    assume
A2: X in phi"A;
    then
A3: X in dom phi by FUNCT_1:def 7;
    hence X is Ordinal;
    reconsider B = X as Ordinal by A3;
A4: phi.X in A by A2,FUNCT_1:def 7;
    thus X c= phi"A
    proof
      let x;
      assume
A5:   x in X;
      then x in B;
      then reconsider C = x, D = phi.B as Ordinal;
      reconsider E = phi.C as Ordinal;
      E in D by A1,A3,A5;
      then
A6:   phi.x in A by A4,ORDINAL1:10;
      C in dom phi by A3,A5,ORDINAL1:10;
      hence thesis by A6,FUNCT_1:def 7;
    end;
  end;
  hence thesis by ORDINAL1:19;
end;

theorem Th12:
  f1 is increasing implies f2*f1 is Ordinal-Sequence
proof
A1: dom(f2*f1) = f1"dom f2 by RELAT_1:147;
  assume f1 is increasing;
  then dom(f2*f1) is Ordinal by A1,Th11;
  then reconsider f = f2*f1 as T-Sequence by ORDINAL1:def 7;
  consider A such that
A2: rng f2 c= A by ORDINAL2:def 4;
  rng f c= rng f2 by RELAT_1:26;
  then rng f c= A by A2,XBOOLE_1:1;
  hence thesis by ORDINAL2:def 4;
end;

theorem Th13:
  f1 is increasing & f2 is increasing implies ex phi st phi = f1*
  f2 & phi is increasing
proof
  assume that
A1: f1 is increasing and
A2: f2 is increasing;
  reconsider f = f1*f2 as Ordinal-Sequence by A2,Th12;
  take f;
  thus f = f1*f2;
  let A,B;
  assume that
A3: A in B and
A4: B in dom f;
  reconsider A1 = f2.A, B1 = f2.B as Ordinal;
A5: B1 in dom f1 by A4,FUNCT_1:11;
  dom f c= dom f2 by RELAT_1:25;
  then
A6: A1 in B1 by A2,A3,A4,ORDINAL2:def 12;
A7: f.B = f1.B1 by A4,FUNCT_1:12;
  f.A = f1.A1 by A3,A4,FUNCT_1:12,ORDINAL1:10;
  hence thesis by A1,A6,A5,A7,ORDINAL2:def 12;
end;

theorem Th14:
  f1 is increasing & A is_limes_of f2 & sup rng f1 = dom f2 & fi =
  f2*f1 implies A is_limes_of fi
proof
  assume that
A1: f1 is increasing and
A2: A = {} & (ex B st B in dom f2 & for C st B c= C & C in dom f2 holds
f2.C = {}) or A <> {} & for B,C st B in A & A in C ex D st D in dom f2 & for E
  being Ordinal st D c= E & E in dom f2 holds B in f2.E & f2.E in C and
A3: sup rng f1 = dom f2 and
A4: fi = f2*f1;
  per cases;
  case
    A = {};
    then consider B such that
A5: B in dom f2 and
A6: for C st B c= C & C in dom f2 holds f2.C = {} by A2;
    consider B1 being Ordinal such that
A7: B1 in rng f1 and
A8: B c= B1 by A3,A5,ORDINAL2:21;
    consider x such that
A9: x in dom f1 and
A10: B1 = f1.x by A7,FUNCT_1:def 3;
    reconsider x as Ordinal by A9;
    take x;
    B1 in dom f2 by A3,A7,ORDINAL2:19;
    hence x in dom fi by A4,A9,A10,FUNCT_1:11;
    let C such that
A11: x c= C and
A12: C in dom fi;
    reconsider C1 = f1.C as Ordinal;
A13: dom fi c= dom f1 by A4,RELAT_1:25;
    then B1 c= C1 by A1,A10,A11,A12,Th9;
    then
A14: B c= C1 by A8,XBOOLE_1:1;
    C1 in rng f1 by A12,A13,FUNCT_1:def 3;
    then f2.C1 = {} by A3,A6,A14,ORDINAL2:19;
    hence thesis by A4,A12,FUNCT_1:12;
  end;
  case
    A <> {};
    let B,C;
    assume that
A15: B in A and
A16: A in C;
    consider D such that
A17: D in dom f2 and
A18: for A1 st D c= A1 & A1 in dom f2 holds B in f2.A1 & f2.A1 in C by A2,A15
,A16;
    consider B1 being Ordinal such that
A19: B1 in rng f1 and
A20: D c= B1 by A3,A17,ORDINAL2:21;
    consider x such that
A21: x in dom f1 and
A22: B1 = f1.x by A19,FUNCT_1:def 3;
    reconsider x as Ordinal by A21;
    take x;
    B1 in dom f2 by A3,A19,ORDINAL2:19;
    hence x in dom fi by A4,A21,A22,FUNCT_1:11;
    let E be Ordinal such that
A23: x c= E and
A24: E in dom fi;
    reconsider E1 = f1.E as Ordinal;
A25: dom fi c= dom f1 by A4,RELAT_1:25;
    then E1 in rng f1 by A24,FUNCT_1:def 3;
    then
A26: E1 in dom f2 by A3,ORDINAL2:19;
    B1 c= E1 by A1,A22,A23,A24,A25,Th9;
    then
A27: D c= E1 by A20,XBOOLE_1:1;
    then
A28: f2.E1 in C by A18,A26;
    B in f2.E1 by A18,A27,A26;
    hence thesis by A4,A24,A28,FUNCT_1:12;
  end;
end;

theorem Th15:
  phi is increasing implies phi|A is increasing
proof
  assume
A1: for A,B st A in B & B in dom phi holds phi.A in phi.B;
  let B,C such that
A2: B in C and
A3: C in dom (phi|A);
A4: phi.B = (phi|A).B by A2,A3,FUNCT_1:47,ORDINAL1:10;
  dom (phi|A) c= dom phi by RELAT_1:60;
  then phi.B in phi.C by A1,A2,A3;
  hence thesis by A3,A4,FUNCT_1:47;
end;

theorem Th16:
  phi is increasing & dom phi is limit_ordinal implies sup phi is limit_ordinal
proof
  assume that
A1: phi is increasing and
A2: dom phi is limit_ordinal;
  now
    let A;
    assume A in sup phi;
    then consider B such that
A3: B in rng phi and
A4: A c= B by ORDINAL2:21;
    consider x such that
A5: x in dom phi and
A6: B = phi.x by A3,FUNCT_1:def 3;
    reconsider x as Ordinal by A5;
A7: succ x in dom phi by A2,A5,ORDINAL1:28;
    reconsider C = phi.succ x as Ordinal;
    x in succ x by ORDINAL1:6;
    then B in C by A1,A6,A7,ORDINAL2:def 12;
    then
A8: succ B c= C by ORDINAL1:21;
A9: succ A c= succ B by A4,ORDINAL2:1;
    C in rng phi by A7,FUNCT_1:def 3;
    then C in sup phi by ORDINAL2:19;
    then succ B in sup phi by A8,ORDINAL1:12;
    hence succ A in sup phi by A9,ORDINAL1:12;
  end;
  hence thesis by ORDINAL1:28;
end;

Lm5: rng f c= X implies (g|X)*f = g*f
proof
A1: f"rng f = dom f by RELAT_1:134;
  assume rng f c= X;
  then
A2: f"rng f c= f"X by RELAT_1:143;
  f"X c= dom f by RELAT_1:132;
  then
A3: f"X = dom f by A2,A1,XBOOLE_0:def 10;
  dom ((g|X)*f) = f"(dom (g|X)) by RELAT_1:147
    .= f"(dom g /\ X) by RELAT_1:61
    .= f"(dom g) /\ f"X by FUNCT_1:68
    .= f"(dom g) by A3,RELAT_1:132,XBOOLE_1:28
    .= dom (g*f) by RELAT_1:147;
  hence thesis by GRFUNC_1:3,RELAT_1:64;
end;

theorem
  fi is increasing & fi is continuous & psi is continuous & phi = psi*fi
  implies phi is continuous
proof
  assume that
A1: fi is increasing and
A2: for A,B st A in dom fi & A <> {} & A is limit_ordinal & B = fi.A
  holds B is_limes_of fi|A and
A3: for A,B st A in dom psi & A <> {} & A is limit_ordinal & B = psi.A
  holds B is_limes_of psi|A and
A4: phi = psi*fi;
  let A,B such that
A5: A in dom phi and
A6: A <> {} and
A7: A is limit_ordinal and
A8: B = phi.A;
  reconsider A1 = fi.A as Ordinal;
A9: fi|A is increasing by A1,Th15;
A10: dom phi c= dom fi by A4,RELAT_1:25;
  then A c= dom fi by A5,ORDINAL1:def 2;
  then
A11: dom (fi|A) = A by RELAT_1:62;
  A1 is_limes_of fi|A by A2,A5,A6,A7,A10;
  then lim (fi|A) = A1 by ORDINAL2:def 10;
  then
A12: sup (fi|A) = A1 by A6,A7,A11,A9,Th8;
A13: B = psi.A1 by A4,A5,A8,FUNCT_1:12;
A14: {} in A by A6,ORDINAL3:8;
A15: A1 in dom psi by A4,A5,FUNCT_1:11;
  then A1 c= dom psi by ORDINAL1:def 2;
  then
A16: dom (psi|A1) = A1 by RELAT_1:62;
A17: rng (fi|A) c= sup rng (fi|A)
  proof
    let x;
    assume
A18: x in rng (fi|A);
    then ex y st y in dom (fi|A) & x = (fi|A).y by FUNCT_1:def 3;
    hence thesis by A18,ORDINAL2:19;
  end;
  phi|A = psi*(fi| A ) by A4,RELAT_1:83;
  then
A19: phi|A = (psi|A1)*(fi|A) by A17,A12,Lm5;
  A c= A1 by A1,A5,A10,Th10;
  then B is_limes_of psi|A1 by A3,A7,A13,A15,A11,A14,A9,A12,Th16;
  hence thesis by A9,A16,A12,A19,Th14;
end;

theorem
  (for A st A in dom fi holds fi.A = C+^A) implies fi is increasing
proof
  assume
A1: for A st A in dom fi holds fi.A = C+^A;
  let A,B;
  assume that
A2: A in B and
A3: B in dom fi;
A4: fi.B = C+^B by A1,A3;
  fi.A = C+^A by A1,A2,A3,ORDINAL1:10;
  hence thesis by A2,A4,ORDINAL2:32;
end;

theorem Th19:
  C <> {} & (for A st A in dom fi holds fi.A = A*^C) implies fi is increasing
proof
  assume that
A1: C <> {} and
A2: for A st A in dom fi holds fi.A = A*^C;
  let A,B;
  assume that
A3: A in B and
A4: B in dom fi;
A5: fi.B = B*^C by A2,A4;
  fi.A = A*^C by A2,A3,A4,ORDINAL1:10;
  hence thesis by A1,A3,A5,ORDINAL2:40;
end;

theorem Th20:
  A <> {} implies exp({},A) = {}
proof
  defpred FF[Ordinal] means $1 <> {} implies exp({},$1) = {};
A1: FF[B] implies FF[succ B]
  proof
    assume that
    FF[B] and
    succ B <> {};
    thus exp({},succ B) = {}*^exp({},B) by ORDINAL2:44
      .= {} by ORDINAL2:35;
  end;
A2: for B st B <> {} & B is limit_ordinal & for C st C in B holds FF[C]
  holds FF[B]
  proof
    deffunc F(Ordinal) = exp({},$1);
    let A such that
A3: A <> {} and
A4: A is limit_ordinal and
A5: for C st C in A holds FF[C] and
    A <> {};
    consider fi such that
A6: dom fi = A & for B st B in A holds fi.B = F(B) from ORDINAL2:sch 3;
    {} is_limes_of fi
    proof
      per cases;
      case
        {} = {};
        take B = 1;
        {} in A by A3,ORDINAL3:8;
        hence B in dom fi by A4,A6,Lm3,ORDINAL1:28;
        let D;
        assume
A7:     B c= D;
        assume
A8:     D in dom fi;
        then FF[D] by A5,A6;
        hence thesis by A6,A7,A8,Lm3,ORDINAL1:21;
      end;
      case
        {} <> {};
        thus thesis;
      end;
    end;
    then lim fi = {} by ORDINAL2:def 10;
    hence thesis by A3,A4,A6,ORDINAL2:45;
  end;
A9: FF[{}];
  FF[B] from ORDINAL2:sch 1(A9,A1,A2);
  hence thesis;
end;

Lm6: A <> {} & A is limit_ordinal implies for fi st dom fi = A & for B st B in
A holds fi.B = exp({},B) holds {} is_limes_of fi
proof
  assume that
A1: A <> {} and
A2: A is limit_ordinal;
  let fi;
  assume that
A3: dom fi = A and
A4: for B st B in A holds fi.B = exp({},B);
  per cases;
  case
    {} = {};
    take B = 1;
    {} in A by A1,ORDINAL3:8;
    hence B in dom fi by A2,A3,Lm3,ORDINAL1:28;
    let D;
    assume that
A5: B c= D and
A6: D in dom fi;
A7: D <> {} by A5,Lm3,ORDINAL1:21;
    exp({},D) = fi.D by A3,A4,A6;
    hence thesis by A7,Th20;
  end;
  case
    {} <> {};
    thus thesis;
  end;
end;

Lm7: A <> {} implies for fi st dom fi = A & for B st B in A holds fi.B = exp(1
,B) holds 1 is_limes_of fi
proof
  assume that
A1: A <> {};
  let fi;
  assume that
A2: dom fi = A and
A3: for B st B in A holds fi.B = exp(1,B);
  per cases;
  case
    1 = {};
    hence thesis;
  end;
  case
    1 <> {};
    let A1,A2 be Ordinal such that
A4: A1 in 1 and
A5: 1 in A2;
    take B = {};
    thus B in dom fi by A1,A2,ORDINAL3:8;
    let D;
    assume that
    B c= D and
A6: D in dom fi;
    exp(1,D) = fi.D by A2,A3,A6;
    hence thesis by A4,A5,ORDINAL2:46;
  end;
end;

Lm8: for A st A <> {} & A is limit_ordinal holds ex fi st dom fi = A & (for B
st B in A holds fi.B = exp(C,B)) & ex D st D is_limes_of fi
proof
  defpred P[Ordinal] means $1 <> {} & $1 is limit_ordinal & for fi st dom fi =
  $1 & for B st B in $1 holds fi.B = exp(C,B) for D holds not D is_limes_of fi;
  let A such that
A1: A <> {} and
A2: A is limit_ordinal and
A3: for fi st dom fi = A & for B st B in A holds fi.B = exp(C,B) for D
  holds not D is_limes_of fi;
  deffunc F(Ordinal) = exp(C,$1);
A4: ex A st P[A] by A1,A2,A3;
  consider A such that
A5: P[A] and
A6: for A1 st P[A1] holds A c= A1 from ORDINAL1:sch 1(A4);
  consider fi such that
A7: dom fi = A & for B st B in A holds fi.B = F(B) from ORDINAL2:sch 3;
A8: now
    assume C = {} or C = 1;
    then {} is_limes_of fi or 1 is_limes_of fi by A5,A7,Lm6,Lm7;
    hence contradiction by A5,A7;
  end;
  then {} in C by ORDINAL3:8;
  then 1 c= C by Lm3,ORDINAL1:21;
  then
A9: 1 c< C by A8,XBOOLE_0:def 8;
A10: for B2,B1 being Ordinal st B1 in B2 & B2 in A holds exp(C,B1) in exp(C,
  B2)
  proof
    defpred V[Ordinal] means for B1 being Ordinal st B1 in $1 & $1 in A holds
    exp(C,B1) in exp(C,$1);
A11: V[B] implies V[succ B]
    proof
      assume
A12:  for B1 being Ordinal st B1 in B & B in A holds exp(C,B1) in exp (C,B);
      let B1 be Ordinal such that
A13:  B1 in succ B and
A14:  succ B in A;
A15:  B1 c= B by A13,ORDINAL1:22;
      B in succ B by ORDINAL1:6;
      then
A16:  B in A by A14,ORDINAL1:10;
A17:  1*^exp(C,B) = exp(C,B) by ORDINAL2:39;
A18:  exp(C,B) <> {}
      proof
        now
          per cases;
          suppose
            B = {};
            hence thesis by ORDINAL2:43;
          end;
          suppose
A19:        B <> {};
            B in succ B by ORDINAL1:6;
            then B in A by A14,ORDINAL1:10;
            hence thesis by A12,A19,ORDINAL3:8;
          end;
        end;
        hence thesis;
      end;
A20:  exp(C,succ B) = C*^exp(C,B) by ORDINAL2:44;
      then
A21:  exp(C,B) in exp(C,succ B) by A9,A18,A17,ORDINAL1:11,ORDINAL2:40;
      now
        assume B1 <> B;
        then B1 c< B by A15,XBOOLE_0:def 8;
        then B1 in B by ORDINAL1:11;
        then exp(C,B1) in exp(C,B) by A12,A16;
        hence thesis by A21,ORDINAL1:10;
      end;
      hence thesis by A9,A18,A20,A17,ORDINAL1:11,ORDINAL2:40;
    end;
A22: for B st B <> {} & B is limit_ordinal & for D st D in B holds V[D]
    holds V[B]
    proof
      let B such that
A23:  B <> {} and
A24:  B is limit_ordinal and
A25:  for D st D in B holds V[D];
      let B1 be Ordinal;
      assume that
A26:  B1 in B and
A27:  B in A;
      consider psi such that
A28:  dom psi = B and
A29:  for D st D in B holds psi.D = exp(C,D) and
      ex D st D is_limes_of psi by A6,A24,A26,A27,ORDINAL1:5;
      psi.B1 = exp(C,B1) by A26,A29;
      then
A30:  exp(C,B1) in rng psi by A26,A28,FUNCT_1:def 3;
      psi is increasing
      proof
        let B1,B2 be Ordinal;
        assume that
A31:    B1 in B2 and
A32:    B2 in dom psi;
        B2 in A by A27,A28,A32,ORDINAL1:10;
        then
A33:    exp(C,B1) in exp(C,B2) by A25,A28,A31,A32;
        psi.B1 = exp(C,B1) by A28,A29,A31,A32,ORDINAL1:10;
        hence thesis by A28,A29,A32,A33;
      end;
      then
A34:  lim psi = sup psi by A23,A24,A28,Th8;
      exp(C,B) = lim psi by A23,A24,A28,A29,ORDINAL2:45;
      hence thesis by A34,A30,ORDINAL2:19;
    end;
A35: V[{}];
    thus for B holds V[B] from ORDINAL2:sch 1(A35,A11,A22);
  end;
  fi is increasing
  proof
    let B1,B2 be Ordinal;
    assume that
A36: B1 in B2 and
A37: B2 in dom fi;
A38: fi.B1 = exp(C,B1) by A7,A36,A37,ORDINAL1:10;
    exp(C,B1) in exp(C,B2) by A7,A10,A36,A37;
    hence thesis by A7,A37,A38;
  end;
  then sup fi is_limes_of fi by A5,A7,Th8;
  hence contradiction by A5,A7;
end;

theorem Th21:
  A <> {} & A is limit_ordinal implies for fi st dom fi = A & for
  B st B in A holds fi.B = exp(C,B) holds exp(C,A) is_limes_of fi
proof
  assume that
A1: A <> {} and
A2: A is limit_ordinal;
  consider psi such that
A3: dom psi = A and
A4: for B st B in A holds psi.B = exp(C,B) and
A5: ex D st D is_limes_of psi by A1,A2,Lm8;
  let fi such that
A6: dom fi = A and
A7: for B st B in A holds fi.B = exp(C,B);
  now
    let x;
    assume
A8: x in A;
    then reconsider B = x as Ordinal;
    thus fi.x = exp(C,B) by A7,A8
      .= psi.x by A4,A8;
  end;
  then fi = psi by A6,A3,FUNCT_1:2;
  then consider D such that
A9: D is_limes_of fi by A5;
  D = lim fi by A9,ORDINAL2:def 10
    .= exp(C,A) by A1,A2,A6,A7,ORDINAL2:45;
  hence thesis by A9;
end;

theorem Th22:
  C <> {} implies exp(C,A) <> {}
proof
  defpred P[Ordinal] means exp(C,$1) <> {};
  assume
A1: C <> {};
A2: for A st P[A] holds P[succ A]
  proof
    let A such that
A3: exp(C,A) <> {};
    exp(C,succ A) = C*^exp(C,A) by ORDINAL2:44;
    hence thesis by A1,A3,ORDINAL3:31;
  end;
A4: for A st A <> {} & A is limit_ordinal & for B st B in A holds P[B] holds
  P[A]
  proof
    let A such that
A5: A <> {} and
A6: A is limit_ordinal and
A7: for B st B in A holds exp(C,B) <> {};
    consider fi such that
A8: dom fi = A and
A9: for B st B in A holds fi.B = exp(C,B) and
A10: ex D st D is_limes_of fi by A5,A6,Lm8;
A11: exp(C,A) = lim fi by A5,A6,A8,A9,ORDINAL2:45;
    assume
A12: exp(C,A) = {};
    consider D such that
A13: D is_limes_of fi by A10;
    lim fi = D by A13,ORDINAL2:def 10;
    then consider B such that
A14: B in dom fi and
A15: for D st B c= D & D in dom fi holds fi.D = {} by A11,A13,A12,
ORDINAL2:def 9;
    fi.B = exp(C,B) by A8,A9,A14;
    hence contradiction by A7,A8,A14,A15;
  end;
A16: P[{}] by ORDINAL2:43;
  for A holds P[A] from ORDINAL2:sch 1(A16,A2,A4);
  hence thesis;
end;

theorem Th23:
  1 in C implies exp(C,A) in exp(C,succ A)
proof
A1: 1*^exp(C,A) = exp(C,A) by ORDINAL2:39;
  assume 1 in C;
  then exp(C,A) in C*^exp(C,A) by A1,Th22,ORDINAL2:40;
  hence thesis by ORDINAL2:44;
end;

theorem Th24:
  1 in C & A in B implies exp(C,A) in exp(C,B)
proof
  defpred OO[Ordinal] means for A st A in $1 holds exp(C,A) in exp(C,$1);
A1: for B st B <> {} & B is limit_ordinal & for D st D in B holds OO[D]
  holds OO[B]
  proof
    deffunc F(Ordinal) = exp(C,$1);
    let B such that
A2: B <> {} and
A3: B is limit_ordinal and
A4: for D st D in B holds OO[D];
    consider fi such that
A5: dom fi = B & for D st D in B holds fi.D = F(D) from ORDINAL2:sch
    3;
    fi is increasing
    proof
      let B1,B2 be Ordinal;
      assume that
A6:   B1 in B2 and
A7:   B2 in dom fi;
A8:   fi.B1 = exp(C,B1) by A5,A6,A7,ORDINAL1:10;
      exp(C,B1) in exp(C,B2) by A4,A5,A6,A7;
      hence thesis by A5,A7,A8;
    end;
    then
A9: sup fi = lim fi by A2,A3,A5,Th8;
    let A such that
A10: A in B;
    fi.A = exp(C,A) by A10,A5;
    then
A11: exp(C,A) in rng fi by A10,A5,FUNCT_1:def 3;
    exp(C,B) = lim fi by A2,A3,A5,ORDINAL2:45;
    hence thesis by A9,A11,ORDINAL2:19;
  end;
  assume
A12: 1 in C;
A13: for B st OO[B] holds OO[succ B]
  proof
    let B such that
A14: for A st A in B holds exp(C,A) in exp(C,B);
    let A;
    assume A in succ B;
    then
A15: A c= B by ORDINAL1:22;
A16: now
      assume A <> B;
      then A c< B by A15,XBOOLE_0:def 8;
      hence exp(C,A) in exp(C,B) by A14,ORDINAL1:11;
    end;
    exp(C,B) in exp(C,succ B) by A12,Th23;
    hence thesis by A16,ORDINAL1:10;
  end;
A17: OO[{}];
  for B holds OO[B] from ORDINAL2:sch 1(A17,A13,A1);
  hence thesis;
end;

theorem Th25:
  1 in C & (for A st A in dom fi holds fi.A = exp(C,A)) implies fi
  is increasing
proof
  assume that
A1: 1 in C and
A2: for A st A in dom fi holds fi.A = exp(C,A);
  let A,B;
  assume that
A3: A in B and
A4: B in dom fi;
A5: fi.B = exp(C,B) by A2,A4;
  fi.A = exp(C,A) by A2,A3,A4,ORDINAL1:10;
  hence thesis by A1,A3,A5,Th24;
end;

theorem
  1 in C & A <> {} & A is limit_ordinal implies for fi st dom fi = A &
  for B st B in A holds fi.B = exp(C,B) holds exp(C,A) = sup fi
proof
  assume that
A1: 1 in C and
A2: A <> {} and
A3: A is limit_ordinal;
  let fi;
  assume that
A4: dom fi = A and
A5: for B st B in A holds fi.B = exp(C,B);
  fi is increasing by A1,A4,A5,Th25;
  then lim fi = sup fi by A2,A3,A4,Th8;
  hence thesis by A2,A3,A4,A5,ORDINAL2:45;
end;

theorem
  C <> {} & A c= B implies exp(C,A) c= exp(C,B)
proof
A1: A c< B iff A c= B & A <> B by XBOOLE_0:def 8;
  assume C <> {};
  then {} in C by ORDINAL3:8;
  then
A2: 1 c= C by Lm3,ORDINAL1:21;
  assume A c= B;
  then
A3: A in B or A = B by A1,ORDINAL1:11;
  now
    per cases;
    suppose
A4:   C = 1;
      then exp(C,A) = 1 by ORDINAL2:46;
      hence thesis by A4,ORDINAL2:46;
    end;
    suppose
      C <> 1;
      then 1 c< C by A2,XBOOLE_0:def 8;
      then 1 in C by ORDINAL1:11;
      then exp(C,A) in exp(C,B) or exp(C,A) = exp(C,B) by A3,Th24;
      hence thesis by ORDINAL1:def 2;
    end;
  end;
  hence thesis;
end;

theorem
  A c= B implies exp(A,C) c= exp(B,C)
proof
  defpred P[Ordinal] means exp(A,$1) c= exp(B,$1);
  assume
A1: A c= B;
A2: for C st P[C] holds P[succ C]
  proof
    let C;
A3: exp(B,succ C) = B*^exp(B,C) by ORDINAL2:44;
    exp(A,succ C) = A*^exp(A,C) by ORDINAL2:44;
    hence thesis by A1,A3,ORDINAL3:20;
  end;
A4: for C st C <> {} & C is limit_ordinal & for D st D in C holds P[D] holds
  P[C]
  proof
    deffunc F(Ordinal) = exp(A,$1);
    let C;
    assume that
A5: C <> {} and
A6: C is limit_ordinal and
A7: for D st D in C holds exp(A,D) c= exp(B,D);
    consider f1 such that
A8: dom f1 = C & for D st D in C holds f1.D = F(D) from ORDINAL2:sch
    3;
    deffunc F(Ordinal) = exp(B,$1);
    consider f2 such that
A9: dom f2 = C & for D st D in C holds f2.D = F(D) from ORDINAL2:sch
    3;
A10: now
      let D;
      assume
A11:  D in dom f1;
      then
A12:  f1.D = exp(A,D) by A8;
      f2.D = exp(B,D) by A8,A9,A11;
      hence f1.D c= f2.D by A7,A8,A11,A12;
    end;
A13: exp(A,C) is_limes_of f1 by A5,A6,A8,Th21;
    exp(B,C) is_limes_of f2 by A5,A6,A9,Th21;
    hence thesis by A8,A9,A13,A10,Th6;
  end;
  exp(A,{}) = 1 by ORDINAL2:43;
  then
A14: P[{}] by ORDINAL2:43;
  for C holds P[C] from ORDINAL2:sch 1(A14,A2,A4);
  hence thesis;
end;

theorem
  1 in C & A <> {} implies 1 in exp(C,A)
proof
  assume that
A1: 1 in C and
A2: A <> {};
  exp(C,{}) = 1 by ORDINAL2:43;
  hence thesis by A1,A2,Th24,ORDINAL3:8;
end;

theorem Th30:
  exp(C,A+^B) = exp(C,B)*^exp(C,A)
proof
  defpred P[Ordinal] means exp(C,A+^$1) = exp(C,$1)*^exp(C,A);
A1: 1 = exp(C,{}) by ORDINAL2:43;
A2: for B st P[B] holds P[succ B]
  proof
    let B such that
A3: exp(C,A+^B) = exp(C,B)*^exp(C,A);
    thus exp(C,A+^succ B) = exp(C,succ(A+^B)) by ORDINAL2:28
      .= C*^exp(C,A+^B) by ORDINAL2:44
      .= C*^exp(C,B)*^exp(C,A) by A3,ORDINAL3:50
      .= exp(C,succ B)*^exp(C,A) by ORDINAL2:44;
  end;
A4: for B st B <> {} & B is limit_ordinal & for D st D in B holds P[D] holds
  P[B]
  proof
    deffunc F(Ordinal) = exp(C,$1);
    let B such that
A5: B <> {} and
A6: B is limit_ordinal and
A7: for D st D in B holds exp(C,A+^D) = exp(C,D)*^exp(C,A);
    consider fi such that
A8: dom fi = B & for D st D in B holds fi.D = F(D) from ORDINAL2:sch
    3;
    consider psi such that
A9: dom psi = A+^B & for D st D in A+^B holds psi.D = F(D) from
    ORDINAL2:sch 3;
    deffunc F(Ordinal) = exp(C,$1);
    consider f1 such that
A10: dom f1 = A & for D st D in A holds f1.D = F(D) from ORDINAL2:sch
    3;
A11: now
      let D;
      assume D in dom(fi*^exp(C,A));
      then
A12:  D in dom fi by ORDINAL3:def 4;
      hence psi.((dom f1)+^D) = exp(C,A+^D) by A8,A9,A10,ORDINAL2:32
        .= (exp(C,D))*^exp(C,A) by A7,A8,A12
        .= (fi.D)*^exp(C,A) by A8,A12
        .= (fi*^exp(C,A)).D by A12,ORDINAL3:def 4;
    end;
A13: now
      let D such that
A14:  D in dom f1;
      A c= A+^B by ORDINAL3:24;
      hence psi.D = exp(C,D) by A9,A10,A14
        .= f1.D by A10,A14;
    end;
    dom psi = (dom f1)+^(dom(fi*^exp(C,A))) by A8,A9,A10,ORDINAL3:def 4;
    then
A15: psi = f1^(fi*^exp(C,A)) by A13,A11,Def1;
    exp(C,B)*^exp(C,A) is_limes_of fi*^exp(C,A) by A5,A6,A8,Th5,Th21;
    then
A16: exp(C,B)*^exp(C,A) is_limes_of psi by A15,Th3;
A17: A+^B <> {} by A5,ORDINAL3:26;
    A+^B is limit_ordinal by A5,A6,ORDINAL3:29;
    then lim psi = exp(C,A+^B) by A9,A17,ORDINAL2:45;
    hence thesis by A16,ORDINAL2:def 10;
  end;
  exp(C,A) = 1*^exp(C,A) by ORDINAL2:39;
  then
A18: P[{}] by A1,ORDINAL2:27;
  for B holds P[B] from ORDINAL2:sch 1(A18,A2,A4);
  hence thesis;
end;

theorem
  exp(exp(C,A),B) = exp(C,B*^A)
proof
  defpred P[Ordinal] means exp(exp(C,A),$1) = exp(C,$1*^A);
A1: exp(C,{}) = 1 by ORDINAL2:43;
A2: for B st P[B] holds P[succ B]
  proof
    let B;
    assume exp(exp(C,A),B) = exp(C,B*^A);
    hence exp(exp(C,A),succ B) = exp(C,A)*^exp(C,B*^A) by ORDINAL2:44
      .= exp(C,B*^A+^A) by Th30
      .= exp(C,(succ B)*^A) by ORDINAL2:36;
  end;
A3: for B st B <> {} & B is limit_ordinal & for D st D in B holds P[D] holds
  P[B]
  proof
    deffunc F(Ordinal) = exp(exp(C,A),$1);
    let B;
    assume that
A4: B <> {} and
A5: B is limit_ordinal and
A6: for D st D in B holds exp(exp(C,A),D) = exp(C,D*^A);
    consider fi such that
A7: dom fi = B & for D st D in B holds fi.D = F(D) from ORDINAL2:sch
    3;
    deffunc F(Ordinal) = $1*^A;
    consider f1 such that
A8: dom f1 = B & for D st D in B holds f1.D = F(D) from ORDINAL2:sch
    3;
    deffunc F(Ordinal) = exp(C,$1);
    consider f2 such that
A9: dom f2 = B*^A & for D st D in B*^A holds f2.D = F(D) from
    ORDINAL2:sch 3;
A10: now
      assume
A11:  A <> {};
      then B*^A <> {} by A4,ORDINAL3:31;
      then
A12:  exp(C,B*^A) is_limes_of f2 by A5,A9,Th21,ORDINAL3:40;
A13:  rng f1 c= dom f2
      proof
        let x;
        assume x in rng f1;
        then consider y such that
A14:    y in dom f1 and
A15:    x = f1.y by FUNCT_1:def 3;
        reconsider y as Ordinal by A14;
        x = y*^A by A8,A14,A15;
        hence thesis by A8,A9,A11,A14,ORDINAL2:40;
      end;
A16:  sup rng f1 = dom f2
      proof
        sup rng f1 c= sup dom f2 by A13,ORDINAL2:22;
        hence sup rng f1 c= dom f2 by ORDINAL2:18;
        let x;
        assume
A17:    x in dom f2;
        then reconsider D = x as Ordinal;
        consider A1 such that
A18:    A1 in B and
A19:    D in A1*^A by A5,A9,A17,ORDINAL3:41;
        f1.A1 = A1*^A by A8,A18;
        then A1*^A in rng f1 by A8,A18,FUNCT_1:def 3;
        then A1*^A in sup rng f1 by ORDINAL2:19;
        hence thesis by A19,ORDINAL1:10;
      end;
A20:  dom(f2*f1) = B
      proof
        thus dom(f2*f1) c= B by A8,RELAT_1:25;
        let x;
        assume
A21:    x in B;
        then reconsider E = x as Ordinal;
A22:    f1.E = E*^A by A8,A21;
        E*^A in B*^A by A11,A21,ORDINAL2:40;
        hence thesis by A8,A9,A21,A22,FUNCT_1:11;
      end;
      now
        let x;
        assume
A23:    x in B;
        then reconsider D = x as Ordinal;
A24:    f1.D = D*^A by A8,A23;
        thus fi.x = exp(exp(C,A),D) by A7,A23
          .= exp(C,D*^A) by A6,A23
          .= f2.(f1.D) by A9,A11,A23,A24,ORDINAL2:40
          .= (f2*f1).x by A8,A23,FUNCT_1:13;
      end;
      then fi = f2*f1 by A7,A20,FUNCT_1:2;
      then exp(C,B*^A) is_limes_of fi by A8,A11,A12,A16,Th14,Th19;
      then exp(C,B*^A) = lim fi by ORDINAL2:def 10;
      hence thesis by A4,A5,A7,ORDINAL2:45;
    end;
A25: B*^{} = {} by ORDINAL2:38;
    exp(C,{}) = 1 by ORDINAL2:43;
    hence thesis by A25,A10,ORDINAL2:46;
  end;
  exp(exp(C,A),{}) = 1 by ORDINAL2:43;
  then
A26: P[{}] by A1,ORDINAL2:35;
  for B holds P[B] from ORDINAL2:sch 1(A26,A2,A3);
  hence thesis;
end;

theorem
  1 in C implies A c= exp(C,A)
proof
  defpred P[Ordinal] means $1 c= exp(C,$1);
  assume
A1: 1 in C;
A2: P[B] implies P[succ B]
  proof
    assume
A3: B c= exp(C,B);
A4: exp(C,B) = 1*^exp(C,B) by ORDINAL2:39;
    exp(C,succ B) = C*^exp(C,B) by ORDINAL2:44;
    then exp(C,B) in exp(C,succ B) by A1,A4,Th22,ORDINAL2:40;
    then B in exp(C,succ B) by A3,ORDINAL1:12;
    hence thesis by ORDINAL1:21;
  end;
A5: for A st A <> {} & A is limit_ordinal & for B st B in A holds P[B] holds
  P[A]
  proof
    deffunc F(Ordinal) = exp(C,$1);
    let A such that
A6: A <> {} and
A7: A is limit_ordinal and
A8: for B st B in A holds B c= exp(C,B);
    consider fi such that
A9: dom fi = A & for B st B in A holds fi.B = F(B) from ORDINAL2:sch
    3;
    let x;
    assume
A10: x in A;
    then reconsider B = x as Ordinal;
    fi.B = exp(C,B) by A9,A10;
    then exp(C,B) in rng fi by A9,A10,FUNCT_1:def 3;
    then
A11: exp(C,B) in sup fi by ORDINAL2:19;
    fi is increasing by A1,A9,Th25;
    then
A12: sup fi = lim fi by A6,A7,A9,Th8
      .= exp(C,A) by A6,A7,A9,ORDINAL2:45;
    B c= exp(C,B) by A8,A10;
    hence thesis by A12,A11,ORDINAL1:12;
  end;
A13: P[{}] by XBOOLE_1:2;
  P[B] from ORDINAL2:sch 1(A13,A2,A5);
  hence thesis;
end;


scheme
  CriticalNumber { phi(Ordinal) -> Ordinal } : ex A st phi(A) = A
provided
A1: for A,B st A in B holds phi(A) in phi(B) and
A2: for A st A <> {} & A is limit_ordinal for phi st dom phi = A & for B
st B in A holds phi.B = phi(B) holds phi(A) is_limes_of phi
proof
A3: now
    defpred P[Ordinal] means not $1 c= phi($1);
    assume
A4: ex A st P[A];
    consider A such that
A5: P[A] and
A6: for B st P[B] holds A c= B from ORDINAL1:sch 1(A4);
    phi(phi(A)) in phi(A) by A1,A5,ORDINAL1:16;
    then not phi(A) c= phi(phi(A)) by ORDINAL1:5;
    hence contradiction by A5,A6;
  end;
  deffunc G(Ordinal,T-Sequence) = {};
  deffunc F(Ordinal,Ordinal) = phi($2);
  consider phi such that
A7: dom phi = omega and
A8: {} in omega implies phi.{} = phi({}) and
A9: for A st succ A in omega holds phi.(succ A) = F(A,phi.A) and
  for A st A in omega & A <> {} & A is limit_ordinal holds phi.A = G(A,phi
  |A) from ORDINAL2:sch 11;
  assume
A10: not thesis;
A11: now
    let A;
A12: A <> phi(A) by A10;
    A c= phi(A) by A3;
    then A c< phi(A) by A12,XBOOLE_0:def 8;
    hence A in phi(A) by ORDINAL1:11;
  end;
A13: phi is increasing
  proof
    let A,B;
    assume that
A14: A in B and
A15: B in dom phi;
    defpred R[Ordinal] means A+^$1 in omega & $1 <> {} implies phi.A in phi.(A
    +^$1);
A16: for B st B <> {} & B is limit_ordinal & for C st C in B holds R[C]
    holds R[B]
    proof
      let B such that
A17:  B <> {} and
A18:  B is limit_ordinal and
      for C st C in B holds A+^C in omega & C <> {} implies phi.A in phi.(
      A+^C) and
A19:  A+^B in omega and
A20:  B <> {};
      A+^B <> {} by A20,ORDINAL3:26;
      then
A21:  {} in A+^B by ORDINAL3:8;
      A+^B is limit_ordinal by A17,A18,ORDINAL3:29;
      then omega c= A+^B by A21,ORDINAL1:def 11;
      hence thesis by A19,ORDINAL1:5;
    end;
A22: for C st R[C] holds R[succ C]
    proof
      let C such that
A23:  A+^C in omega & C <> {} implies phi.A in phi.(A+^C) and
A24:  A+^succ C in omega and
      succ C <> {};
      reconsider D = phi.(A+^C) as Ordinal;
A25:  A+^C in succ(A+^C) by ORDINAL1:6;
A26:  D in phi(D) by A11;
A27:  A+^succ C = succ(A+^C) by ORDINAL2:28;
      then phi.(A+^succ C) = phi(D) by A9,A24;
      hence thesis by A23,A24,A25,A27,A26,ORDINAL1:10,ORDINAL2:27;
    end;
A28: R[{}];
A29: for C holds R[C] from ORDINAL2:sch 1(A28,A22,A16);
    ex C st B = A+^C & C <> {} by A14,ORDINAL3:28;
    hence thesis by A7,A15,A29;
  end;
  deffunc phi(Ordinal) = phi ($1);
  consider fi such that
A30: dom fi = sup phi & for A st A in sup phi holds fi.A = phi(A) from
  ORDINAL2:sch 3;
  phi({}) in rng phi by A7,A8,Lm1,FUNCT_1:def 3;
  then
A31: sup phi <> {} by ORDINAL2:19;
  then
A32: phi(sup phi) is_limes_of fi by A2,A7,A13,A30,Lm2,Th16;
A33: sup fi c= sup phi
  proof
    let x;
    assume
A34: x in sup fi;
    then reconsider A = x as Ordinal;
    consider B such that
A35: B in rng fi and
A36: A c= B by A34,ORDINAL2:21;
    consider y such that
A37: y in dom fi and
A38: B = fi.y by A35,FUNCT_1:def 3;
    reconsider y as Ordinal by A37;
    consider C such that
A39: C in rng phi and
A40: y c= C by A30,A37,ORDINAL2:21;
    y c< C iff y <> C & y c= C by XBOOLE_0:def 8;
    then
A41: phi(y) in phi(C) or y = C by A1,A40,ORDINAL1:11;
    B = phi(y) by A30,A37,A38;
    then
A42: B c= phi(C) by A41,ORDINAL1:def 2;
    consider z such that
A43: z in dom phi and
A44: C = phi.z by A39,FUNCT_1:def 3;
    reconsider z as Ordinal by A43;
A45: succ z in omega by A7,A43,Lm2,ORDINAL1:28;
    then
A46: phi.succ z in rng phi by A7,FUNCT_1:def 3;
    phi.succ z = phi(C) by A9,A44,A45;
    then phi(C) in sup phi by A46,ORDINAL2:19;
    then B in sup phi by A42,ORDINAL1:12;
    hence thesis by A36,ORDINAL1:12;
  end;
A47: fi is increasing
  proof
    let A,B;
    assume that
A48: A in B and
A49: B in dom fi;
A50: fi.B = phi(B) by A30,A49;
    fi.A = phi(A) by A30,A48,A49,ORDINAL1:10;
    hence thesis by A1,A48,A50;
  end;
  sup phi is limit_ordinal by A7,A13,Lm2,Th16;
  then sup fi = lim fi by A30,A31,A47,Th8
    .= phi(sup phi) by A32,ORDINAL2:def 10;
  hence contradiction by A11,A33,ORDINAL1:5;
end;

reserve W for Universe;

registration
  let W;
  cluster ordinal for Element of W;
  existence
  proof
A1: On W c= W by ORDINAL2:7;
    {} in On W by CLASSES2:51,ORDINAL3:8;
    hence thesis by A1;
  end;
end;

definition
  let W;
  mode Ordinal of W is ordinal Element of W;
  mode Ordinal-Sequence of W is Function of On W, On W;
end;

registration
  let W;
  cluster non empty for Ordinal of W;
  existence
  proof
A1: On W c= W by ORDINAL2:7;
    {} in On W by CLASSES2:51,ORDINAL3:8;
    then 1 in W by A1,Lm3,CLASSES2:5;
    hence thesis;
  end;
end;

registration
  let W;
  cluster On W -> non empty;
  coherence by CLASSES2:51;
end;

registration
  let W;
  cluster -> T-Sequence-like Ordinal-yielding for Ordinal-Sequence of W;
  coherence
  proof
    let s be Ordinal-Sequence of W;
    thus dom s is ordinal by FUNCT_2:def 1;
    take On W;
    thus rng s c= On W by RELAT_1:def 19;
  end;
end;

reserve A1,B1 for Ordinal of W,
  phi for Ordinal-Sequence of W;

scheme
  UOSLambda { W() -> Universe, F(set) -> Ordinal of W() } : ex phi being
Ordinal-Sequence of W() st for a being Ordinal of W() holds phi.a = F(a) proof
  consider psi such that
A1: dom psi = On W() & for A st A in On W() holds psi.A = F(A) from
  ORDINAL2:sch 3;
  rng psi c= On W()
  proof
    let x;
    assume x in rng psi;
    then consider y such that
A2: y in dom psi and
A3: x = psi.y by FUNCT_1:def 3;
    reconsider y as Ordinal by A2;
    x = F(y) by A1,A2,A3;
    hence thesis by ORDINAL1:def 9;
  end;
  then reconsider psi as Ordinal-Sequence of W() by A1,FUNCT_2:def 1,RELSET_1:4
;
  take psi;
  let a be Ordinal of W();
  a in On W() by ORDINAL1:def 9;
  hence thesis by A1;
end;

definition
  let W;

  func 0-element_of W -> Ordinal of W equals
  {};
  correctness
  proof
A1: On W c= W by ORDINAL2:7;
    {} in On W by ORDINAL3:8;
    then reconsider A = {} as Ordinal of W by A1;
    A = {};
    hence thesis;
  end;
  func 1-element_of W -> non empty Ordinal of W equals
  1;
  correctness
  proof
A2: On W c= W by ORDINAL2:7;
    {} in On W by ORDINAL3:8;
    then reconsider A = 1 as Ordinal of W by A2,Lm3,CLASSES2:5;
    A = 1;
    hence thesis;
  end;
  let phi,A1;
  redefine func phi.A1 -> Ordinal of W;
  coherence
  proof
    reconsider B = phi.A1 as Ordinal;
A3: dom phi = On W by FUNCT_2:def 1;
A4: rng phi c= On W by RELAT_1:def 19;
    A1 in On W by ORDINAL1:def 9;
    then B in rng phi by A3,FUNCT_1:def 3;
    then
A5: B in On W by A4;
    On W c= W by ORDINAL2:7;
    hence thesis by A5;
  end;
end;

definition
  let W;
  let p2,p1 be Ordinal-Sequence of W;
  redefine func p1*p2 -> Ordinal-Sequence of W;
  coherence
  proof
A1: rng p2 c= On W by RELAT_1:def 19;
A2: dom p2 = On W by FUNCT_2:def 1;
    dom p1 = On W by FUNCT_2:def 1;
    then
A3: dom(p1*p2) = On W by A2,A1,RELAT_1:27;
    then reconsider f = p1*p2 as T-Sequence by ORDINAL1:def 7;
A4: rng p1 c= On W by RELAT_1:def 19;
    rng(p1*p2) c= rng p1 by RELAT_1:26;
    then rng f c= On W by A4,XBOOLE_1:1;
    hence thesis by A3,FUNCT_2:def 1,RELSET_1:4;
  end;
end;

theorem
  0-element_of W = {} & 1-element_of W = 1;

definition
  let W,A1;
  redefine func succ A1 -> non empty Ordinal of W;
  coherence by CLASSES2:5;
  let B1;
  redefine func A1 +^ B1 -> Ordinal of W;
  coherence
  proof
    defpred P[Ordinal] means $1 in W implies A1+^$1 in W;
A1: for B st for C st C in B holds P[C] holds P[B]
    proof
      let B such that
A2:   for C st C in B holds C in W implies A1+^C in W and
A3:   B in W;
      [:B,{1-element_of W}:] in W by A3,CLASSES2:61;
      then [:A1,{0-element_of W}:] \/ [:B,{1-element_of W}:] in W by
CLASSES2:60;
      then
A4:   card([:A1,{0-element_of W}:] \/ [:B,{1-element_of W}:]) in card W by
CLASSES2:1;
A5:   A1+^B c= W
      proof
        let x;
        assume
A6:     x in A1+^B;
        then reconsider A = x as Ordinal;
A7:     now
A8:       B c= W by A3,ORDINAL1:def 2;
          assume A1 c= A;
          then consider C such that
A9:       A = A1+^C by ORDINAL3:27;
          C in B by A6,A9,ORDINAL3:22;
          hence thesis by A2,A9,A8;
        end;
A10:    A in A1 or A1 c= A by ORDINAL1:16;
        A1 c= W by ORDINAL1:def 2;
        hence thesis by A10,A7;
      end;
      card(A1+^B) = card([:A1,{0-element_of W}:] \/ [:B,{ 1-element_of W}
      :] ) by CARD_2:9;
      hence thesis by A4,A5,CLASSES1:1;
    end;
    for B holds P[B] from ORDINAL1:sch 2(A1);
    hence thesis;
  end;
end;

definition
  let W,A1,B1;
  redefine func A1 *^ B1 -> Ordinal of W;
  coherence
  proof
    defpred P[Ordinal] means $1 in W implies $1*^B1 in W;
A1: for A st for C st C in A holds P[C] holds P[A]
    proof
      let A such that
A2:   for C st C in A holds C in W implies C*^B1 in W and
A3:   A in W;
      [:A,B1:] in W by A3,CLASSES2:61;
      then
A4:   card [:A,B1:] in card W by CLASSES2:1;
A5:   A*^B1 c= W
      proof
        let x;
A6:     B1 c= W by ORDINAL1:def 2;
        assume
A7:     x in A*^B1;
        then reconsider B = x as Ordinal;
        B1 <> {} by A7,ORDINAL2:38;
        then consider C,D such that
A8:     B = C*^B1+^D and
A9:     D in B1 by ORDINAL3:47;
        C*^B1 c= B by A8,ORDINAL3:24;
        then C*^B1 in A*^B1 by A7,ORDINAL1:12;
        then
A10:    C in A by ORDINAL3:34;
        A c= W by A3,ORDINAL1:def 2;
        then reconsider CB = C*^B1, D as Ordinal of W by A2,A9,A6,A10;
        x = CB+^D by A8;
        hence thesis;
      end;
      card(A*^B1) = card [:A,B1:] by CARD_2:11;
      hence thesis by A4,A5,CLASSES1:1;
    end;
    for A holds P[A] from ORDINAL1:sch 2(A1);
    hence thesis;
  end;
end;

theorem Th34:
  A1 in dom phi
proof
  dom phi = On W by FUNCT_2:def 1;
  hence thesis by ORDINAL1:def 9;
end;

theorem Th35:
  dom fi in W & rng fi c= W implies sup fi in W
proof
  assume that
A1: dom fi in W and
A2: rng fi c= W;
  ex A st rng fi c= A by ORDINAL2:def 4;
  then for x st x in rng fi holds x is Ordinal;
  then reconsider B = union rng fi as Ordinal by ORDINAL1:23;
A3: rng fi = fi.:(dom fi) by RELAT_1:113;
A4: sup fi c= succ B
  proof
    let x;
    assume
A5: x in sup fi;
    then reconsider A = x as Ordinal;
    consider C such that
A6: C in rng fi and
A7: A c= C by A5,ORDINAL2:21;
    C c= union rng fi by A6,ZFMISC_1:74;
    then A c= B by A7,XBOOLE_1:1;
    hence thesis by ORDINAL1:22;
  end;
  card dom fi in card W by A1,CLASSES2:1;
  then card rng fi in card W by A3,CARD_1:67,ORDINAL1:12;
  then rng fi in W by A2,CLASSES1:1;
  then union rng fi in W by CLASSES2:59;
  then succ B in W by CLASSES2:5;
  hence thesis by A4,CLASSES1:def 1;
end;

reserve L for T-Sequence;

theorem
  phi is increasing & phi is continuous & omega in W implies ex A st A
  in dom phi & phi.A = A
proof
  deffunc D(set,set) = {};
  assume that
A1: phi is increasing and
A2: phi is continuous and
A3: omega in W;
  deffunc C(set,set) = phi.$2;
  reconsider N = phi.(0-element_of W) as Ordinal;
  consider L such that
A4: dom L = omega and
A5: {} in omega implies L.{} = N and
A6: for A st succ A in omega holds L.(succ A) = C(A,L.A) and
  for A st A in omega & A <> {} & A is limit_ordinal holds L.A = D(A,L|A)
  from ORDINAL2:sch 5;
  defpred P[Ordinal] means $1 in dom L implies L.$1 is Ordinal of W;
A7: for A st P[A] holds P[succ A]
  proof
    let A such that
A8: A in dom L implies L.A is Ordinal of W and
A9: succ A in dom L;
    A in succ A by ORDINAL1:6;
    then reconsider x = L.A as Ordinal of W by A8,A9,ORDINAL1:10;
    L.succ A = phi.x by A4,A6,A9;
    hence thesis;
  end;
A10: for A st A <> {} & A is limit_ordinal & for B st B in A holds P[B]
  holds P[A]
  proof
    let A such that
A11: A <> {} and
A12: A is limit_ordinal and
    for B st B in A holds B in dom L implies L.B is Ordinal of W and
A13: A in dom L;
    {} in A by A11,ORDINAL3:8;
    then omega c= A by A12,ORDINAL1:def 11;
    hence thesis by A4,A13,ORDINAL1:5;
  end;
A14: P[{}] by A4,A5;
A15: for A holds P[A] from ORDINAL2:sch 1(A14,A7,A10);
  rng L c= sup rng L
  proof
    let x;
    assume
A16: x in rng L;
    then consider y such that
A17: y in dom L and
A18: x = L.y by FUNCT_1:def 3;
    reconsider y as Ordinal by A17;
    reconsider A = L.y as Ordinal of W by A15,A17;
    A in sup rng L by A16,A18,ORDINAL2:19;
    hence thesis by A18;
  end;
  then reconsider L as Ordinal-Sequence by ORDINAL2:def 4;
A19: dom phi = On W by FUNCT_2:def 1;
  assume
A20: not thesis;
A21: now
    let A1;
    A1 in dom phi by Th34;
    then
A22: A1 c= phi.A1 by A1,Th10;
    A1 <> phi.A1 by A20,Th34;
    then A1 c< phi.A1 by A22,XBOOLE_0:def 8;
    hence A1 in phi.A1 by ORDINAL1:11;
  end;
  L is increasing
  proof
    let A,B;
    assume that
A23: A in B and
A24: B in dom L;
    defpred P[Ordinal] means A+^$1 in omega & $1 <> {} implies L.A in L.(A+^$1
    );
A25: for B st B <> {} & B is limit_ordinal & for C st C in B holds P[C]
    holds P[B]
    proof
      let B such that
A26:  B <> {} and
A27:  B is limit_ordinal and
      for C st C in B holds A+^C in omega & C <> {} implies L.A in L.(A+^C
      ) and
A28:  A+^B in omega and
A29:  B <> {};
      A+^B <> {} by A29,ORDINAL3:26;
      then
A30:  {} in A+^B by ORDINAL3:8;
      A+^B is limit_ordinal by A26,A27,ORDINAL3:29;
      then omega c= A+^B by A30,ORDINAL1:def 11;
      hence thesis by A28,ORDINAL1:5;
    end;
A31: for C st P[C] holds P[succ C]
    proof
      let C such that
A32:  A+^C in omega & C <> {} implies L.A in L.(A+^C) and
A33:  A+^succ C in omega and
      succ C <> {};
A34:  A+^succ C = succ(A+^C) by ORDINAL2:28;
A35:  A+^C in succ(A+^C) by ORDINAL1:6;
      then reconsider D = L.(A+^C) as Ordinal of W by A4,A15,A33,A34,
ORDINAL1:10;
A36:  D in phi.D by A21;
      L.(A+^succ C) = phi.D by A6,A33,A34;
      hence thesis by A32,A33,A35,A34,A36,ORDINAL1:10,ORDINAL2:27;
    end;
A37: P[{}];
A38: for C holds P[C] from ORDINAL2:sch 1(A37,A31,A25);
    ex C st B = A+^C & C <> {} by A23,ORDINAL3:28;
    hence thesis by A4,A24,A38;
  end;
  then
A39: sup L is limit_ordinal by A4,Lm2,Th16;
A40: rng L c= W
  proof
    let x;
    assume x in rng L;
    then consider y such that
A41: y in dom L and
A42: x = L.y by FUNCT_1:def 3;
    reconsider y as Ordinal by A41;
    L.y is Ordinal of W by A15,A41;
    hence thesis by A42;
  end;
  then reconsider S = sup L as Ordinal of W by A3,A4,Th35;
  set fi = phi|sup L;
  N in rng L by A4,A5,Lm1,FUNCT_1:def 3;
  then
A43: sup L <> {} by ORDINAL2:19;
A44: S in On W by ORDINAL1:def 9;
  then
A45: phi.S is_limes_of fi by A2,A43,A39,A19,ORDINAL2:def 13;
  S c= dom phi by A44,A19,ORDINAL1:def 2;
  then
A46: dom fi = S by RELAT_1:62;
A47: sup fi c= sup L
  proof
    let x;
    assume
A48: x in sup fi;
    then reconsider A = x as Ordinal;
    consider B such that
A49: B in rng fi and
A50: A c= B by A48,ORDINAL2:21;
    consider y such that
A51: y in dom fi and
A52: B = fi.y by A49,FUNCT_1:def 3;
    reconsider y as Ordinal by A51;
    consider C such that
A53: C in rng L and
A54: y c= C by A46,A51,ORDINAL2:21;
    reconsider C1 = C as Ordinal of W by A40,A53;
    y c< C1 iff y c= C1 & y <> C1 by XBOOLE_0:def 8;
    then y in C1 & C1 in dom phi or y = C by A19,A54,ORDINAL1:11,def 9;
    then
A55: phi.y in phi.C1 or y = C1 by A1,ORDINAL2:def 12;
    B = phi.y by A51,A52,FUNCT_1:47;
    then
A56: B c= phi.C1 by A55,ORDINAL1:def 2;
    consider z such that
A57: z in dom L and
A58: C = L.z by A53,FUNCT_1:def 3;
    reconsider z as Ordinal by A57;
A59: succ z in omega by A4,A57,Lm2,ORDINAL1:28;
    then
A60: L.succ z in rng L by A4,FUNCT_1:def 3;
    L.succ z = phi.C by A6,A58,A59;
    then phi.C1 in sup L by A60,ORDINAL2:19;
    then B in sup L by A56,ORDINAL1:12;
    hence thesis by A50,ORDINAL1:12;
  end;
  fi is increasing
  proof
A61: dom fi c= dom phi by RELAT_1:60;
    let A,B;
    assume that
A62: A in B and
A63: B in dom fi;
A64: fi.B = phi.B by A63,FUNCT_1:47;
    fi.A = phi.A by A62,A63,FUNCT_1:47,ORDINAL1:10;
    hence thesis by A1,A62,A63,A61,A64,ORDINAL2:def 12;
  end;
  then sup fi = lim fi by A43,A39,A46,Th8
    .= phi.(sup L) by A45,ORDINAL2:def 10;
  then not S in phi.S by A47,ORDINAL1:5;
  hence contradiction by A21;
end;

begin 


reserve e,u for set;

theorem
  A is_cofinal_with B & B is_cofinal_with C implies A is_cofinal_with C
proof
  given xi1 being Ordinal-Sequence such that
A1: dom xi1 = B and
A2: rng xi1 c= A and
A3: xi1 is increasing and
A4: A = sup xi1;
  given xi2 being Ordinal-Sequence such that
A5: dom xi2 = C and
A6: rng xi2 c= B and
A7: xi2 is increasing and
A8: B = sup xi2;
  consider xi being Ordinal-Sequence such that
A9: xi = xi1*xi2 and
A10: xi is increasing by A3,A7,Th13;
  take xi;
  thus
A11: dom xi = C by A1,A5,A6,A9,RELAT_1:27;
  rng xi c= rng xi1 by A9,RELAT_1:26;
  hence
A12: rng xi c= A & xi is increasing by A2,A10,XBOOLE_1:1;
  thus A c= sup xi
  proof
    let a be set;
    assume
A13: a in A;
    then reconsider a as Ordinal;
    consider b being Ordinal such that
A14: b in rng xi1 and
A15: a c= b by A4,A13,ORDINAL2:21;
    consider e such that
A16: e in B and
A17: b = xi1.e by A1,A14,FUNCT_1:def 3;
    reconsider e as Ordinal by A16;
    consider c being Ordinal such that
A18: c in rng xi2 and
A19: e c= c by A8,A16,ORDINAL2:21;
    consider u such that
A20: u in C and
A21: c = xi2.u by A5,A18,FUNCT_1:def 3;
    reconsider u as Ordinal by A20;
A22: xi1.c = xi.u by A9,A11,A20,A21,FUNCT_1:12;
    xi.u in rng xi by A11,A20,FUNCT_1:def 3;
    then
A23: xi.u in sup xi by ORDINAL2:19;
    xi1.e c= xi1.c by A1,A3,A6,A18,A19,Th9;
    hence thesis by A15,A17,A22,A23,ORDINAL1:12,XBOOLE_1:1;
  end;
  sup rng xi c= sup A by A12,ORDINAL2:22;
  hence thesis by ORDINAL2:18;
end;

theorem
  A is_cofinal_with B implies (A is limit_ordinal iff B is limit_ordinal )
proof
  given psi such that
A1: dom psi = B and
A2: rng psi c= A and
A3: psi is increasing and
A4: A = sup psi;
  thus A is limit_ordinal implies B is limit_ordinal
  proof
    assume
A5: A is limit_ordinal;
    now
      let C;
      reconsider c = psi.C as Ordinal;
      assume
A6:   C in B;
      then psi.C in rng psi by A1,FUNCT_1:def 3;
      then succ c in A by A2,A5,ORDINAL1:28;
      then consider b being Ordinal such that
A7:   b in rng psi and
A8:   succ c c= b by A4,ORDINAL2:21;
      consider e such that
A9:   e in B and
A10:  b = psi.e by A1,A7,FUNCT_1:def 3;
      reconsider e as Ordinal by A9;
      c in succ c by ORDINAL1:6;
      then not e c= C by A1,A3,A6,A8,A10,Th9,ORDINAL1:5;
      then C in e by ORDINAL1:16;
      then succ C c= e by ORDINAL1:21;
      hence succ C in B by A9,ORDINAL1:12;
    end;
    hence thesis by ORDINAL1:28;
  end;
  thus thesis by A1,A3,A4,Th16;
end;



registration let D;
  let f,g be T-Sequence of D;
  cluster f^g -> D-valued;
  coherence
  proof
    rng f c= D & rng g c= D by RELAT_1:def 19;
    then
A1:  rng f \/ rng g c= D by XBOOLE_1:8;
    rng(f^g) c= rng f \/ rng g by Th2;
   hence rng(f^g) c= D by A1,XBOOLE_1:1;
  end;
end;














begin

reserve k,l,m,n,k1,a,b,c,i for Nat,
  x,y,z,y1,y2,X,Y for set,
  f,g for Function;



definition
  let n be Nat;
  func Seg n -> set equals
  { k where k is Element of NAT: 1 <= k & k <= n };
  correctness;
end;

definition
  let n be Nat;
  redefine func Seg n -> Subset of NAT;
  coherence
  proof
    set X = Seg n;
    X c= NAT
    proof
      let x;
      assume x in X;
      then ex k being Element of NAT st x = k & 1 <= k & k <= n;
      hence thesis;
    end;
    hence thesis;
  end;
end;

theorem Th1:
  a in Seg b iff 1 <= a & a <= b
proof
A1: a in NAT by ORDINAL1:def 12;
  a in { m where m is Element of NAT: 1 <= m & m <= b } iff
  ex m being Element of NAT st a = m & 1 <= m & m <= b;
  hence thesis by A1;
end;

registration let n be zero Nat;
  cluster Seg n -> empty;
  coherence
   proof
    assume Seg n is not empty;
     then consider x being set such that
A1:    x in Seg n by XBOOLE_0:def 1;
     ex k being Element of NAT st k = x & 1 <= k & k <= 0 by A1;
    hence contradiction;
   end;
end;

theorem Th2:
  Seg 1 = { 1 } & Seg 2 = { 1,2 }
proof
  now
    let x;
    thus x in Seg 1 implies x in { 1 }
    proof
      assume x in Seg 1;
      then consider k being Element of NAT such that
A1:   x = k and
A2:   1 <= k & k <= 1;
      k = 1 by A2,XXREAL_0:1;
      hence thesis by A1,TARSKI:def 1;
    end;
    assume x in { 1 };
    then x = 1 by TARSKI:def 1;
    hence x in Seg 1;
  end;
  hence Seg 1 = { 1 } by TARSKI:1;
  now
    let x;
    thus x in Seg 2 implies x in { 1,2 }
    proof
      assume x in Seg 2;
      then consider k being Element of NAT such that
A3:   x = k and
A4:   1 <= k and
A5:   k <= 2;
      k <= 1 + 1 by A5;
      then k = 1 or k = 2 by A4,NAT_1:9;
      hence thesis by A3,TARSKI:def 2;
    end;
    assume x in { 1,2 };
    then x = 1 or x = 2 by TARSKI:def 2;
    hence x in Seg 2;
  end;
  hence thesis by TARSKI:1;
end;

theorem Th3:
  a = 0 or a in Seg a
proof
  assume a <> 0;
  then ex b be Nat st a = b + 1 by NAT_1:6;
  then a in NAT & 1 <= a by NAT_1:11;
  hence thesis;
end;

registration let n be non zero Nat;
  cluster Seg n -> non empty;
  coherence by Th3;
end;

theorem
  a+1 in Seg(a+1) by Th3;

theorem Th5:
  a <= b iff Seg a c= Seg b
proof
  thus a <= b implies Seg a c= Seg b
  proof
    assume
A1: a <= b;
    let x;
    assume
A2: x in Seg a;
    then reconsider c = x as Element of NAT;
A3: 1 <= c by A2,Th1;
    c <= a by A2,Th1;
    then c <= b by A1,XXREAL_0:2;
    hence thesis by A3;
  end;
  assume
A4: Seg a c= Seg b;
  now
    assume a <> 0;
    then a in Seg a by Th3;
    hence thesis by A4,Th1;
  end;
  hence thesis;
end;

theorem Th6:
  Seg a = Seg b implies a = b
proof
  assume Seg a = Seg b;
  then a <= b & b <= a by Th5;
  hence thesis by XXREAL_0:1;
end;

theorem Th7:
  c <= a implies Seg c = Seg a /\ Seg c
proof
  assume c <= a;
  then Seg c c= Seg a by Th5;
  hence thesis by XBOOLE_1:28;
end;

theorem
  Seg c = Seg c /\ Seg a implies c <= a by Th6,Th7;

theorem Th9:
  Seg a \/ { a+1 } = Seg (a+1)
proof
  thus Seg a \/ { a+1 } c= Seg (a+1)
  proof
    a+0<=a+1 by XREAL_1:7;
    then
A1: Seg a c= Seg(a+1) by Th5;
    let x;
    assume x in Seg a \/ { a+1 };
    then x in Seg a or x in { a+1 } by XBOOLE_0:def 3;
    then x in Seg (a+1) or x = a+1 by A1,TARSKI:def 1;
    hence thesis by Th3;
  end;
  let x;
  assume
A2: x in Seg (a+1);
  then reconsider x as Element of NAT;
A3: x<=a+1 by A2,Th1;
A4: 1<=x by A2,Th1;
  x<=a or x=a+1 by A3,NAT_1:8;
  then x in Seg a or x in {a+1} by A4,TARSKI:def 1;
  hence thesis by XBOOLE_0:def 3;
end;

theorem
  for k being Nat holds Seg k = Seg(k + 1) \ {k + 1}
proof
  let k be Nat;
A1: Seg(k + 1) = Seg k \/ {k + 1} by Th9;
  Seg k misses {k + 1}
  proof
    assume not thesis;
    then
A2: Seg k /\ {k + 1} <> {} by XBOOLE_0:def 7;
    set x = the Element of Seg k /\ {k + 1};
A3: x in Seg k by A2,XBOOLE_0:def 4;
    then reconsider x as Element of NAT;
    x in {k + 1} by A2,XBOOLE_0:def 4;
    then
A4: x = k + 1 by TARSKI:def 1;
    x <= k by A3,Th1;
    hence thesis by A4,XREAL_1:29;
  end;
  hence thesis by A1,XBOOLE_1:88;
end;



definition
  let IT be Relation;
  attr IT is FinSequence-like means
  :Def2:
  ex n st dom IT = Seg n;
end;

registration
  cluster empty -> FinSequence-like for Relation;
  coherence
  proof
    let R be Relation;
    assume
A1: R is empty;
    take 0;
    thus thesis by A1;
  end;
end;

definition
  mode FinSequence is FinSequence-like Function;
end;

reserve p,q,r,s,t for FinSequence;

defpred P[set,set] means ex k st $1 = k & $2 = k+1;

registration
  let n be Nat;
  cluster Seg n -> finite;
  coherence
  proof
    Seg n is finite
    proof
A1:   n = { k where k is Element of NAT: k < n } by AXIOMS:4;
A2:   for x st x in n ex y st P[x,y]
      proof
        let x;
        assume x in n;
        then ex k being Element of NAT st x = k & k < n by A1;
        then reconsider k = x as Nat;
        take k+1;
        thus thesis;
      end;
      consider f such that
A3:   dom f = n and
A4:   for x st x in n holds P[x,f.x] from CLASSES1:sch 1(A2);
      take f;
      thus rng f = Seg n
      proof
        thus rng f c= Seg n
        proof
          let x;
          assume x in rng f;
          then consider y being set such that
A5:       y in dom f and
A6:       x = f.y by FUNCT_1:def 3;
          consider k such that
A7:       y = k and
A8:       x = k+1 by A3,A4,A5,A6;
          ex m being Element of NAT st m = y & m < n by A1,A3,A5;
          then 1 <= k+1 & k+1 <= n by A7,NAT_1:11,13;
          hence thesis by A8;
        end;
        let x;
        assume x in Seg n;
        then consider k being Element of NAT such that
A9:     x = k and
A10:    1 <= k and
A11:    k <= n;
        consider i being Nat such that
A12:    1+i = k by A10,NAT_1:10;
        i in NAT & i < n by A11,A12,NAT_1:13,ORDINAL1:def 12;
        then
A13:    i in n by A1;
        then P[i,f.i] by A4;
        hence thesis by A3,A9,A12,A13,FUNCT_1:def 3;
      end;
      thus thesis by A3,ORDINAL1:def 12;
    end;
    hence thesis;
  end;
end;

registration
  cluster FinSequence-like -> finite for Function;
  coherence
  proof
    let f be Function;
    given n such that
A1: dom f = Seg n;
    rng f is finite by A1,FINSET_1:8;
    then [:dom f, rng f:] is finite by A1;
    hence thesis by FINSET_1:1,RELAT_1:7;
  end;
end;

Lm1: Seg n,n are_equipotent
proof
  defpred P[Nat] means Seg $1,$1 are_equipotent;
A1: P[0];
A2: for n st P[n] holds P[n+1]
  proof
    let n such that
A3: Seg n,n are_equipotent;
A4: (n+1) = succ n by NAT_1:38
      .= n \/ { n } by ORDINAL1:def 1;
    A5: Seg(n+1) = Seg n \/ { n+1 } & { n+1 },{ n } are_equipotent by Th9,
CARD_1:28;
A6: now
      assume n meets { n };
      then consider x being set such that
A7:   x in n and
A8:   x in { n } by XBOOLE_0:3;
      x = n by A8,TARSKI:def 1;
      hence contradiction by A7;
    end;
    now
      assume Seg n meets { n+1 };
      then consider x being set such that
A9:   x in Seg n and
A10:  x in { n+1 } by XBOOLE_0:3;
A11:  x = n+1 by A10,TARSKI:def 1;
      not n+1 <= n by NAT_1:13;
      hence contradiction by A9,A11,Th1;
    end;
    hence thesis by A3,A4,A5,A6,CARD_1:31;
  end;
  for n holds P[n] from NAT_1:sch 2(A1,A2);
  hence thesis;
end;

Lm2: card Seg n = card n proof Seg n,n are_equipotent by Lm1;
  hence thesis by CARD_1:5;
end;

registration let n;
 cluster Seg n -> n-element;
 coherence
  proof
    card Seg n = card n by Lm2;
   hence card Seg n = n by CARD_1:def 2;
  end;
end;

notation
  let p;
  synonym len p for card p;
end;

definition
  let p;
  redefine func len p -> Element of NAT means
  :Def3:
  Seg it = dom p;
  coherence
  proof card p = card p;
    hence thesis;
  end;
  compatibility
  proof
    let k be Element of NAT;
    consider n such that
A1: dom p = Seg n by Def2;
    dom p,p are_equipotent
    proof
      deffunc F(set) = [$1,p.$1];
      consider g being Function such that
A2:   dom g = dom p and
A3:   for x st x in dom p holds g.x = F(x) from FUNCT_1:sch 3;
      take g;
      thus g is one-to-one
      proof
        let x,y;
        assume that
A4:     x in dom g and
A5:     y in dom g;
        assume g.x = g.y;
        then [x,p.x] = g.y by A2,A3,A4
          .= [y,p.y] by A2,A3,A5;
        hence thesis by XTUPLE_0:1;
      end;
      thus dom g = dom p by A2;
      thus rng g c= p
      proof
        let i be set;
        assume i in rng g;
        then consider x such that
A6:     x in dom g and
A7:     g.x = i by FUNCT_1:def 3;
        g.x = [x,p.x] by A2,A3,A6;
        hence thesis by A2,A6,A7,FUNCT_1:1;
      end;
      let x,y be set;
      assume
A8:   [x,y] in p;
      then
A9:   x in dom p by FUNCT_1:1;
      y = p.x by A8,FUNCT_1:1;
      then g.x = [x,y] by A3,A9;
      hence thesis by A2,A9,FUNCT_1:def 3;
    end;
    then
A10: card p = card dom p by CARD_1:5;
    thus k = card p implies Seg k = dom p
    proof
      assume k = card p;
      then k = card n by A1,A10,Lm2;
      hence thesis by A1,CARD_1:def 2;
    end;
    assume
A11: Seg k = dom p;
    thus k = card k by CARD_1:def 2
      .= card p by A10,A11,Lm2;
  end;
end;

definition
  let p;
  redefine func dom p -> Subset of NAT;
  coherence
  proof dom p = Seg len p by Def3;
    hence thesis;
  end;
end;

theorem
  (ex k st dom f c= Seg k) implies ex p st f c= p
proof
  given k such that
A1: dom f c= Seg k;
  deffunc F(set) = f.$1;
  consider g such that
A2: dom g = Seg k & for x st x in Seg k holds g.x = F(x) from FUNCT_1:sch 3;
  reconsider g as FinSequence by A2,Def2;
  take g;
  let y,z be set;
  assume
A3: [y,z] in f;
  then
A4: y in dom f by XTUPLE_0:def 12;
  then
A5: f.y = z by A3,FUNCT_1:def 2;
  [y,g.y] in g by A1,A2,A4,FUNCT_1:1;
  hence thesis by A1,A2,A4,A5;
end;

scheme SeqEx{A()->Nat,P[set,set]}:
  ex p st dom p = Seg A() & for k st k in Seg A() holds P[k,p.k]
provided
A1: for k st k in Seg A() ex x st P[k,x]
proof
A2: for x st x in Seg A() ex y st P[x,y] by A1;
  consider f being Function such that
A3: dom f = Seg A() &
  for x st x in Seg A() holds P[x,f.x] from CLASSES1:sch 1(A2);
  reconsider p=f as FinSequence by A3,Def2;
  take p;
  thus thesis by A3;
end;

scheme SeqLambda{A()->Nat,F(set) -> set}:
  ex p being FinSequence st len p = A() & for k st k in dom p holds p.k=F(k)
proof
  consider f being Function such that
  A1: dom
 f = Seg A() & for x st x in Seg A() holds f.x=F(x) from FUNCT_1:sch 3;
A2: A() in NAT by ORDINAL1:def 12;
  reconsider p=f as FinSequence by A1,Def2;
  take p;
  thus thesis by A1,A2,Def3;
end;

theorem
  z in p implies ex k st k in dom p & z=[k,p.k]
proof
  assume
A1: z in p;
  then consider x,y such that
A2: z=[x,y] by RELAT_1:def 1;
  x in dom p by A1,A2,FUNCT_1:1;
  then reconsider k = x as Nat;
  take k;
  thus thesis by A1,A2,FUNCT_1:1;
end;

theorem Th13:
  dom p = dom q & (for k st k in dom p holds p.k = q.k) implies p=q
proof
  assume that
A1: dom p = dom q and
A2: for k st k in dom p holds p.k = q.k;
  for x st x in dom p holds p.x=q.x by A2;
  hence thesis by A1,FUNCT_1:2;
end;

theorem Th14:
  len p = len q & (for k st 1 <= k & k <= len p holds p.k = q.k) implies p=q
proof
  assume that
A1: len p = len q and
A2: for k st 1<=k & k<=len p holds p.k = q.k;
A3: dom p = Seg len p by Def3;
A4: dom q = Seg len q by Def3;
  now
    let x;
    assume
A5: x in dom p;
    then reconsider k=x as Nat;
    1 <= k & k <= len p by A3,A5,Th1;
    hence p.x = q.x by A2;
  end;
  hence thesis by A1,A3,A4,FUNCT_1:2;
end;

theorem Th15:
  p|(Seg a) is FinSequence
proof
A1: dom(p|Seg a) = dom p /\ Seg a by RELAT_1:61
    .=Seg len p /\ Seg a by Def3;
  len p <= a or a <= len p;
  then dom(p|Seg a) = Seg len p or dom(p|Seg a) = Seg a by A1,Th7;
  hence thesis by Def2;
end;

theorem
  rng p c= dom f implies f*p is FinSequence
proof
  assume rng p c= dom f;
  then dom(f*p) = dom p by RELAT_1:27
    .= Seg len p by Def3;
  hence thesis by Def2;
end;

theorem Th17:
  a <= len p & q = p|(Seg a) implies len q = a & dom q = Seg a
proof
  assume that
A1: a <= len p and
A2: q = p|(Seg a);
  Seg a c= Seg len p by A1,Th5;
  then Seg a c= dom p by Def3;
  then a in NAT & dom q = Seg a by A2,ORDINAL1:def 12,RELAT_1:62;
  hence thesis by Def3;
end;





definition
  let D be set;
  mode FinSequence of D -> FinSequence means
    :Def4:
    rng it c= D;
  existence
  proof
    rng {} c= D by XBOOLE_1:2;
    hence thesis;
  end;
end;

registration
  let D be set;
  cluster -> D-valued for FinSequence of D;
  coherence
  proof
    let f be FinSequence of D;
    thus rng f c= D by Def4;
  end;
end;

Lm3: for D being set, f being FinSequence of D holds f is PartFunc of NAT,D
proof
  let D be set, f be FinSequence of D;
  dom f c= NAT & rng f c= D by Def4;
  hence thesis by RELSET_1:4;
end;

registration
  cluster empty -> FinSequence-like for Relation;
  coherence;
end;

registration
  let D be set;
  cluster FinSequence-like for PartFunc of NAT,D;
  existence
  proof {} is PartFunc of NAT,D by RELSET_1:12;
    hence thesis;
  end;
end;

definition
  let D be set;
  redefine mode FinSequence of D -> FinSequence-like PartFunc of NAT,D;
  coherence by Lm3;
end;

reserve D for set;

theorem Th18:
  for p being FinSequence of D holds p|(Seg a) is FinSequence of D
proof
  let p be FinSequence of D;
A1: p|(Seg a) is FinSequence by Th15;
  rng(p|(Seg a)) c= rng p & rng p c= D by Def4,RELAT_1:70;
  then rng(p|(Seg a)) c= D by XBOOLE_1:1;
  hence thesis by A1,Def4;
end;

theorem
  for D being non empty set ex p being FinSequence of D st len p = a
proof
  let D be non empty set;
  reconsider a as Element of NAT by ORDINAL1:def 12;
  set y = the Element of D;
  set p = Seg a --> y;
A1: dom p = Seg a by FUNCOP_1:13;
  then reconsider p as FinSequence by Def2;
  rng p c= {y} & {y} c= D by FUNCOP_1:13,ZFMISC_1:31;
  then rng p c= D by XBOOLE_1:1;
  then reconsider q=p as FinSequence of D by Def4;
  take q;
  thus thesis by A1,Def3;
end;





Lm4: q = {} iff len q = 0;

theorem
  p <> {} iff len p >= 1
proof
  hereby
    assume p <> {};
    then len p+1 > 0+1 by XREAL_1:8;
    hence len p >=1 by NAT_1:13;
  end;
  assume len p >= 1;
  hence thesis;
end;

definition
  let x;
  func <*x*> -> set equals
  { [1,x] };
  coherence;
end;

definition
  let D be set;
  func <*>D -> FinSequence of D equals
  {};
  coherence
  proof
    rng {} c= D by XBOOLE_1:2;
    hence thesis by Def4;
  end;
end;

registration
  let D be set;
  cluster <*>D -> empty;
  coherence;
end;

registration
  let D be set;
  cluster empty for FinSequence of D;
  existence
  proof
    take <*>D;
    thus thesis;
  end;
end;

definition
  let p,q;
  func p^q -> FinSequence means
  :Def7:
  dom it = Seg (len p + len q) & (for k st k in dom p holds it.k=p.k) &
  for k st k in dom q holds it.(len p + k) = q.k;
  existence
  proof
    defpred P[set,set] means
    (for k st k=$1 & 1 <= k & k <= len p holds $2=p.k) &
    (for k st k=$1 & len p + 1 <= k & k <= len p + len q
    holds $2=q.(k-len p));
A1: for x st x in Seg(len p + len q) ex y st P[x,y]
    proof
      let x;
      assume x in Seg(len p + len q);
      then reconsider m=x as Element of NAT;
A2:   now
        assume
A3:     len p + 1 <= m;
        set y = q.(m-len p);
A4:     for k st k=x & 1 <= k & k <= len p holds y=p.k
        proof
          let k;
          assume that
A5:       k=x and 1 <= k and
A6:       k <= len p;
          m + (len p + 1) <= m + len p by A3,A5,A6,XREAL_1:7;
          then m + len p + 1 <= m + len p + 0;
          hence thesis by XREAL_1:6;
        end;
        for k st k=x & len p + 1 <= k & k <= len p + len q holds y=q.(k-len p);
        hence thesis by A4;
      end;
      now
        assume
A7:     not len p + 1 <= m;
        set y=p.m;
( for k st k=x & 1 <= k & k <= len p holds y=p.k)& for k st k=x & len p + 1 <=
        k & k <= len p + len q holds y=q.(k-len p) by A7;
        hence thesis;
      end;
      hence thesis by A2;
    end;
    consider f such that
A8: dom f=Seg(len p + len q) &
    for x st x in Seg(len p + len q) holds P[x,f.x] from CLASSES1:sch 1
    (A1);
A9: for k st k in dom p holds f.k=p.k
    proof
      let k;
      assume k in dom p;
      then
A10:  k in Seg len p by Def3;
      then
A11:  1 <= k & k <= len p by Th1;
      len p <= len p + len q by NAT_1:11;
      then Seg len p c= Seg(len p + len q) by Th5;
      hence thesis by A8,A10,A11;
    end;
A12: for n st n in dom q holds f.(len p+n)=q.n
    proof
      let n;
      assume n in dom q;
      then
A13:  n in Seg len q by Def3;
      then
A14:  1 <= n by Th1;
A15:  n <= len q by A13,Th1;
A16:  len p + 1 <= len p + n by A14,XREAL_1:7;
A17:  len p + n <= len p + len q by A15,XREAL_1:7;
      reconsider n as Element of NAT by ORDINAL1:def 12;
      len p + n <= len p + n + len p by NAT_1:11;
      then len p + 1 <= len p + n + len p by A16,XXREAL_0:2;
      then 1 <= len p + n by XREAL_1:6;
      then (len p + n) in Seg(len p + len q) by A17;
      then f.(len p + n)=q.(n + len p - len p) by A8,A16,A17;
      hence thesis;
    end;
    reconsider r=f as FinSequence by A8,Def2;
    take r;
    thus thesis by A8,A9,A12;
  end;
  uniqueness
  proof
    let f,g be FinSequence such that
A18: dom f = Seg(len p + len q) and
A19: for k st k in dom p holds f.k=p.k and
A20: for k st k in dom q holds f.(len p + k)=q.k and
A21: dom g = Seg(len p + len q) and
A22: for k st k in dom p holds g.k=p.k and
A23: for k st k in dom q holds g.(len p + k)=q.k;
    for x st x in Seg(len p + len q) holds f.x=g.x
    proof
      let x;
      assume
A24:  x in Seg(len p + len q);
      then reconsider k=x as Element of NAT;
A25:  1 <= k by A24,Th1;
A26:  now
        assume len p + 1 <= k;
        then consider m be Nat such that
A27:    len p + 1 + m = k by NAT_1:10;
        reconsider m as Element of NAT by ORDINAL1:def 12;
        len p + (1 + m) <= len p + len q by A24,A27,Th1;
        then 1+0<=1+m & 1 + m <= len q by XREAL_1:6;
        then (1+m) in Seg len q;
        then
A28:    (1+m) in dom q by Def3;
        then g.(len p +(1+m)) = q.(1+m) by A23;
        hence thesis by A20,A27,A28;
      end;
      now
        assume not len p + 1 <= k;
        then k <= len p by NAT_1:8;
        then k in Seg len p by A25;
        then
A29:    k in dom p by Def3;
        then f.k=p.k by A19;
        hence thesis by A22,A29;
      end;
      hence thesis by A26;
    end;
    hence thesis by A18,A21,FUNCT_1:2;
  end;
end;

theorem Th21:
  p = (p ^ q) | (dom p)
proof
A1: dom(p ^ q) = Seg(len p + len q) by Def7;
A2: dom p = Seg len p by Def3;
  then
A3: dom p = dom(p ^ q) /\ Seg len p by A1,Th7,NAT_1:12;
  for x st x in dom p holds p.x = (p ^ q).x by Def7;
  hence thesis by A2,A3,FUNCT_1:46;
end;

theorem Th22:
  len(p^q) = len p + len q
proof
  dom(p^q) = Seg(len p + len q) by Def7;
  hence thesis by Def3;
end;

theorem Th23:
  for k being Nat st
  len p + 1 <= k & k <= len p + len q holds (p^q).k=q.(k-len p)
proof
  let k be Nat;
  assume that
A1: len p + 1 <= k and
A2: k <= len p + len q;
  consider m be Nat such that
A3: (len p + 1)+m=k by A1,NAT_1:10;
  reconsider m as Element of NAT by ORDINAL1:def 12;
A4: len p+(1+m) = k by A3;
  1+m = k - len p by A3;
  then
A5: 1 <= 1+m by A1,XREAL_1:19;
  k-len p <= len q by A2,XREAL_1:20;
  then 1+m in Seg len q by A3,A5;
  then 1+m in dom q by Def3;
  hence thesis by A4,Def7;
end;

theorem Th24:
  for k being Nat st len p < k & k <= len(p^q) holds (p^q).k = q.(k - len p)
proof
  let k be Nat;
  assume len p < k & k <= len(p^q);
  then len p + 1 <= k & k <= len p + len q by Th22,NAT_1:13;
  hence thesis by Th23;
end;

theorem Th25:
  for k being Nat st k in dom (p^q) holds
  (k in dom p or ex n st n in dom q & k=len p + n )
proof
  let k be Nat;
  assume k in dom(p^q);
  then
A1: k in Seg len (p^q) by Def3;
  then
A2: k in Seg(len p + len q) by Th22;
A3: k in NAT & 1 <= k by A1,Th1;
A4: now
    assume not len p+1 <= k;
    then k <= len p by NAT_1:8;
    then k in Seg len p by A3;
    hence thesis by Def3;
  end;
  now
    assume len p + 1 <= k;
    then consider n be Nat such that
A5: k=len p + 1 + n by NAT_1:10;
    reconsider n as Element of NAT by ORDINAL1:def 12;
    len p + (1 + n) <= len p + len q by A2,A5,Th1;
    then
A6: 1+n <= len q by XREAL_1:6;
    1 <= 1+n by NAT_1:11;
    then 1+n in Seg len q by A6;
    then
A7: 1+n in dom q by Def3;
    k=len p + (1+n) by A5;
    hence thesis by A7;
  end;
  hence thesis by A4;
end;

theorem Th26:
  dom p c= dom(p^q)
proof
  len p <= len p + len q by NAT_1:11;
  then Seg len p c= Seg(len p + len q) by Th5;
  then Seg len p c= dom(p^q) by Def7;
  hence thesis by Def3;
end;

theorem Th27:
  x in dom q implies ex k st k=x & len p + k in dom(p^q)
proof
  assume
A1: x in dom q;
  then
A2: x in Seg len q by Def3;
  reconsider k=x as Element of NAT by A1;
  take k;
A3: 1 <= k by A2,Th1;
A4: k <= len q by A2,Th1;
  k <= len p + k by NAT_1:11;
  then
A5: 1 <= len p + k by A3,XXREAL_0:2;
  len p + k <= len p + len q by A4,XREAL_1:7;
  then len p + k in Seg(len p + len q) by A5;
  hence thesis by Def7;
end;

theorem Th28:
  k in dom q implies len p + k in dom(p^q)
proof
  assume k in dom q;
  then ex n st n=k & len p + n in dom(p^q) by Th27;
  hence thesis;
end;

theorem Th29:
  rng p c= rng(p^q)
proof
  now
    let x;
    assume x in rng p;
    then consider y such that
A1: y in dom p and
A2: x=p.y by FUNCT_1:def 3;
    reconsider k=y as Element of NAT by A1;
A3: dom p c= dom(p^q) by Th26;
    (p^q).k=p.k by A1,Def7;
    hence x in rng(p^q) by A1,A2,A3,FUNCT_1:def 3;
  end;
  hence thesis by TARSKI:def 3;
end;

theorem Th30:
  rng q c= rng(p^q)
proof
  now
    let x;
    assume x in rng q;
    then consider y such that
A1: y in dom q and
A2: x=q.y by FUNCT_1:def 3;
    reconsider k=y as Element of NAT by A1;
    len p + k in dom(p^q) & (p^q).(len p + k) = q.k by A1,Def7,Th28;
    hence x in rng(p^q) by A2,FUNCT_1:def 3;
  end;
  hence thesis by TARSKI:def 3;
end;

theorem Th31:
  rng(p^q) = rng p \/ rng q
proof
  now
    let x;
    assume x in rng(p^q);
    then consider y such that
A1: y in dom(p^q) and
A2: x=(p^q).y by FUNCT_1:def 3;
A3: y in Seg(len p + len q) by A1,Def7;
    reconsider k=y as Element of NAT by A1;
A4: 1 <= k by A3,Th1;
A5: k <= len p + len q by A3,Th1;
A6: now
      assume
A7:   len p + 1 <= k;
      then
A8:   q.(k-len p) = x by A2,A5,Th23;
      consider m be Nat such that
A9:   (len p+1)+m = k by A7,NAT_1:10;
      reconsider m as Element of NAT by ORDINAL1:def 12;
      len p +(1+m) = k by A9;
      then 1+0<=1+m & 1+m <= len q by A5,XREAL_1:6;
      then 1+m in Seg len q;
      then k-len p in dom q by A9,Def3;
      hence x in rng q by A8,FUNCT_1:def 3;
    end;
    now
      assume not len p + 1 <= k;
      then k <= len p by NAT_1:8;
      then k in Seg len p by A4;
      then
A10:  k in dom p by Def3;
      then p.k = x by A2,Def7;
      hence x in rng p by A10,FUNCT_1:def 3;
    end;
    hence x in rng p \/ rng q by A6,XBOOLE_0:def 3;
  end;
  then
A11: rng(p^q) c= rng p \/ rng q by TARSKI:def 3;
  rng p c= rng(p^q) & rng q c= rng(p^q) by Th29,Th30;
  then (rng p \/ rng q) c= rng(p^q) by XBOOLE_1:8;
  hence thesis by A11,XBOOLE_0:def 10;
end;

theorem Th32:
  p^q^r = p^(q^r)
proof
A1: dom ((p^q)^r) = Seg(len (p^q) + len r) by Def7
    .= Seg(len p + len q + len r) by Th22
    .= Seg(len p + (len q + len r))
    .= Seg(len p + len(q^r)) by Th22;
A2: for k st k in dom p holds ((p^q)^r).k=p.k
  proof
    let k;
    assume
A3: k in dom p;
    dom p c= dom(p^q) by Th26;
    hence (p^q^r).k=(p^q).k by A3,Def7
      .=p.k by A3,Def7;
  end;
  for k st k in dom(q^r) holds ((p^q)^r).(len p + k)=(q^r).k
  proof
    let k;
    assume
A4: k in dom(q^r);
A5: now
      assume
A6:   k in dom q;
      then (len p + k) in dom(p^q) by Th28;
      hence (p^q^r).(len p + k) = (p^q).(len p + k) by Def7
        .=q.k by A6,Def7
        .=(q^r).k by A6,Def7;
    end;
    now
      assume not k in dom q;
      then consider n such that
A7:   n in dom r and
A8:   k=len q + n by A4,Th25;
      thus (p^q^r).(len p + k) =(p^q^r).(len p + len q + n) by A8
        .=(p^q^r).(len(p^q) + n) by Th22
        .=r.n by A7,Def7
        .=(q^r).k by A7,A8,Def7;
    end;
    hence thesis by A5;
  end;
  hence thesis by A1,A2,Def7;
end;

theorem
  p^r = q^r or r^p = r^q implies p = q
proof
  assume
A1: p^r = q^r or r^p = r^q;
A2: now
    assume
A3: p^r = q^r;
    then len p + len r = len(q^r) by Th22;
    then len p + len r = len q + len r by Th22;
    then
A4: dom p = Seg len q by Def3
      .= dom q by Def3;
    for k st k in dom p holds p.k=q.k
    proof
      let k;
      assume
A5:   k in dom p;
      hence p.k=(q^r).k by A3,Def7
        .=q.k by A4,A5,Def7;
    end;
    hence thesis by A4,Th13;
  end;
  now
    assume
A6: r^p=r^q;
    then len r + len p = len(r^q) by Th22
      .=len r + len q by Th22;
    then
A7: dom p = Seg len q by Def3
      .= dom q by Def3;
    for k st k in dom p holds p.k=q.k
    proof
      let k;
      assume
A8:   k in dom p;
      hence p.k = (r^q).(len r + k) by A6,Def7
        .= q.k by A7,A8,Def7;
    end;
    hence thesis by A7,Th13;
  end;
  hence thesis by A1,A2;
end;

theorem Th34:
  p^{} = p & {}^p = p
proof
A1: dom(p^{}) = Seg len (p^{}) by Def3
    .= Seg (len p + len {}) by Th22
    .= dom p by Def3;
  for k st k in dom p holds p.k=(p^{}).k by Def7;
  hence p^{} = p by A1,Th13;
A2: dom({}^p) = Seg len ({}^p) by Def3
    .= Seg (len {} + len p) by Th22
    .= dom p by Def3;
  for k st k in dom p holds p.k = ({}^p).k
  proof
    let k;
    assume
A3: k in dom p;
    thus ({}^p).k =({}^p).(len {} + k) .=p.k by A3,Def7;
  end;
  hence {}^p=p by A2,Th13;
end;

theorem Th35:
  p^q = {} implies p={} & q={}
proof
  assume p^q={};
  then 0 = len (p^q) .= len p + len q by Th22;
  hence thesis;
end;

definition
  let D be set;
  let p,q be FinSequence of D;
  redefine func p^q -> FinSequence of D;
  coherence
  proof
A1: rng(p^q) = rng p \/ rng q & rng p c= D by Def4,Th31;
    rng q c= D by Def4;
    hence rng(p^q) c= D by A1,XBOOLE_1:8;
  end;
end;

Lm5: for x1, y1 being set holds [x,y] in {[x1,y1]} implies x = x1 & y = y1
proof
  let x1, y1 be set;
  assume [x,y] in {[x1,y1]};
  then [x,y] = [x1,y1] by TARSKI:def 1;
  hence thesis by XTUPLE_0:1;
end;

definition
  let x;
  redefine func <*x*> -> Function means
  :Def8:
  dom it = Seg 1 & it.1 = x;
  coherence;
  compatibility
  proof
    let f be Function;
    hereby
      assume
A1:   f = <*x*>;
      hence dom f = Seg 1 by Th2,RELAT_1:9;
      [1,x] in f by A1,TARSKI:def 1;
      hence f.1 = x by FUNCT_1:1;
    end;
    assume that
A2: dom f = Seg 1 and
A3: f.1 = x;
    reconsider g = { [1,f.1] } as Function;
    [y,z] in f iff [y,z] in g
    proof
      hereby
        assume
A4:     [y,z] in f;
        then
A5:     y in {1} by A2,Th2,XTUPLE_0:def 12;
A6:     z in rng f by A4,XTUPLE_0:def 13;
A7:     rng f = {f.1} by A2,Th2,FUNCT_1:4;
A8:     y = 1 by A5,TARSKI:def 1;
        z = f.1 by A6,A7,TARSKI:def 1;
        hence [y,z] in g by A8,TARSKI:def 1;
      end;
      assume [y,z] in g;
      then
A9:   y = 1 & z = f.1 by Lm5;
      1 in dom f by A2;
      hence thesis by A9,FUNCT_1:def 2;
    end;
    hence thesis by A3,RELAT_1:def 2;
  end;
end;

registration
  let x;
  cluster <*x*> -> Function-like Relation-like;
  coherence;
end;

registration
  let x;
  cluster <*x*> -> FinSequence-like;
  coherence
  proof
    take 1;
    thus thesis by Def8;
  end;
end;

theorem Th36:
  p^q is FinSequence of D implies p is FinSequence of D & q is FinSequence of D
proof
  assume p^q is FinSequence of D;
  then rng(p^q) c= D by Def4;
  then
A1: rng p \/ rng q c= D by Th31;
  rng p c= rng p \/ rng q by XBOOLE_1:7;
  then rng p c= D by A1,XBOOLE_1:1;
  hence p is FinSequence of D by Def4;
  rng q c= rng p \/ rng q by XBOOLE_1:7;
  then rng q c= D by A1,XBOOLE_1:1;
  hence thesis by Def4;
end;

definition
  let x,y;
  func <*x,y*> -> set equals
  <*x*>^<*y*>;
  correctness;
  let z;
  func <*x,y,z*> -> set equals
  <*x*>^<*y*>^<*z*>;
  correctness;
end;

registration
  let x,y;
  cluster <*x,y*> -> Function-like Relation-like;
  coherence;
  let z;
  cluster <*x,y,z*> -> Function-like Relation-like;
  coherence;
end;

registration
  let x,y;
  cluster <*x,y*> -> FinSequence-like;
  coherence;
  let z;
  cluster <*x,y,z*> -> FinSequence-like;
  coherence;
end;

theorem
  <*x*> = { [1,x] };

theorem Th38:
  p=<*x*> iff dom p = Seg 1 & rng p = {x}
proof
  thus p = <*x*> implies dom p = Seg 1 & rng p = {x}
  proof
    assume
A1: p = <*x*>;
    hence dom p = Seg 1 by Def8;
    dom p = {1} by A1,Def8,Th2;
    then rng p = {p.1} by FUNCT_1:4;
    hence thesis by A1,Def8;
  end;
  assume that
A2: dom p = Seg 1 and
A3: rng p = {x};
  1 in dom p by A2;
  then p.1 in {x} by A3,FUNCT_1:def 3;
  then p.1 = x by TARSKI:def 1;
  hence thesis by A2,Def8;
end;

theorem Th39:
  p=<*x*> iff len p = 1 & rng p = {x}
proof
  len p = 1 iff dom p = Seg 1 by Def3;
  hence thesis by Th38;
end;

theorem Th40:
  p = <*x*> iff len p = 1 & p.1 = x
proof
  len p = 1 iff dom p = Seg 1 by Def3;
  hence thesis by Def8;
end;

theorem
  (<*x*>^p).1 = x
proof
  1 in Seg 1;
  then 1 in dom <*x*> by Def8;
  then (<*x*>^p).1 = <*x*>.1 by Def7;
  hence thesis by Def8;
end;

theorem Th42:
  (p^<*x*>).(len p + 1)=x
proof
  dom <*x*> = Seg 1 by Def8;
  then 1 in dom <*x*>;
  hence (p^<*x*>).(len p + 1) = <*x*>.1 by Def7
    .=x by Def8;
end;

theorem
  <*x,y,z*>=<*x*>^<*y,z*> & <*x,y,z*>=<*x,y*>^<*z*> by Th32;

theorem Th44:
  p = <*x,y*> iff len p = 2 & p.1=x & p.2=y
proof
  thus
  p = <*x,y*> implies len p = 2 & p.1=x & p.2=y
  proof
    assume
A1: p=<*x,y*>;
    hence len p = len <*x*> + len <*y*> by Th22
      .= 1 + len <*y*> by Th39
      .= 1 + 1 by Th39
      .=2;
A2: 1 in {1} by TARSKI:def 1;
    then 1 in dom <*x*> by Def8,Th2;
    hence p.1 = <*x*>.1 by A1,Def7
      .= x by Def8;
A3: 1 in dom <*y*> by A2,Def8,Th2;
    thus p.2 = (<*x*>^<*y*>).(1+1) by A1
      .= (<*x*>^<*y*>).(len <*x*> + 1) by Th39
      .= <*y*>.1 by A3,Def7
      .= y by Def8;
  end;
  assume that
A4: len p = 2 and
A5: p.1=x and
A6: p.2=y;
A7: dom p = Seg(1+1) by A4,Def3
    .= Seg(len <*x*> + 1) by Th39
    .= Seg(len <*x*> + len <*y*>) by Th39;
A8: for k st k in dom <*x*> holds p.k=<*x*>.k
  proof
    let k;
    assume k in dom <*x*>;
    then k in {1} by Def8,Th2;
    then k=1 by TARSKI:def 1;
    hence thesis by A5,Def8;
  end;
  for k st k in dom <*y*> holds p.((len <*x*>)+k)=<*y*>.k
  proof
    let k;
    assume k in dom <*y*>;
    then
A9: k in {1} by Def8,Th2;
    thus p.((len <*x*>) + k) = p.(1+k) by Th39
      .=p.(1+1) by A9,TARSKI:def 1
      .=<*y*>.1 by A6,Def8
      .= <*y*>.k by A9,TARSKI:def 1;
  end;
  hence thesis by A7,A8,Def7;
end;

theorem Th45:
  p = <*x,y,z*> iff len p = 3 & p.1 = x & p.2 = y & p.3 = z
proof
  thus p = <*x,y,z*> implies len p = 3 & p.1 = x & p.2 = y & p.3 = z
  proof
    assume
A1: p =<*x,y,z*>;
    hence len p =len <*x,y*> + len <*z*> by Th22
      .=2 + len <*z*> by Th44
      .=2+1 by Th39
      .=3;
A2: 1 in {1} by TARSKI:def 1;
    then
A3: 1 in dom <*x*> by Def8,Th2;
    thus p.1 = (<*x*>^<*y,z*>).1 by A1,Th32
      .=<*x*>.1 by A3,Def7
      .=x by Def8;
    2 in Seg 2 & len <*x,y*> = 2 by Th44;
    then 2 in dom <*x,y*> by Def3;
    hence p.2 =<*x,y*>.2 by A1,Def7
      .=y by Th44;
A4: 1 in dom <*z*> by A2,Def8,Th2;
    thus p.3 = (<*x,y*>^<*z*>).(2+1) by A1
      .=(<*x,y*>^<*z*>).(len (<*x,y*>) + 1) by Th44
      .= <*z*>.1 by A4,Def7
      .= z by Def8;
  end;
  assume that
A5: len p = 3 and
A6: p.1 = x and
A7: p.2 = y and
A8: p.3 = z;
A9: dom p = Seg(2+1) by A5,Def3
    .= Seg((len <*x,y*>) + 1) by Th44
    .= Seg((len <*x,y*>) + len <*z*>) by Th39;
A10: for k st k in dom <*x,y*> holds p.k=<*x,y*>.k
  proof
    let k such that
A11: k in dom <*x,y*>;
    len <*x,y*> = 2 by Th44;
    then
A12: k in Seg 2 by A11,Def3;
A13: k=1 implies p.k=<*x,y*>.k by A6,Th44;
    k=2 implies p.k=<*x,y*>.k by A7,Th44;
    hence thesis by A12,A13,Th2,TARSKI:def 2;
  end;
  for k st k in dom <*z*> holds p.( (len <*x,y*>) + k) = <*z*>.k
  proof
    let k;
    assume k in dom <*z*>;
    then k in {1} by Def8,Th2;
    then
A14: k = 1 by TARSKI:def 1;
    hence p.( (len <*x,y*>) + k) = p.(2+1) by Th44
      .=<*z*>.k by A8,A14,Def8;
  end;
  hence thesis by A9,A10,Def7;
end;

theorem Th46:
  p <> {} implies ex q,x st p=q^<*x*>
proof
  assume p <> {};
  then consider n be Nat such that
A1: len p = n + 1 by NAT_1:6;
  reconsider n as Element of NAT by ORDINAL1:def 12;
  reconsider q=p|Seg n as FinSequence by Th15;
  take q;
  take p.(len p);
A2: dom q = dom p /\ Seg n by RELAT_1:61
    .= Seg len p /\ Seg n by Def3;
  n <=len p by A1,NAT_1:11;
  then Seg n c= Seg len p by Th5;
  then
A3: dom q = Seg n by A2,XBOOLE_1:28;
A4: dom(q^<*p.(len p)*>) = Seg len (q^<*p.(len p)*>) by Def3
    .= Seg(len q + len <*p.(len p)*>) by Th22
    .= Seg(len q + 1) by Th39
    .= Seg len p by A1,A3,Def3
    .= dom p by Def3;
  for x st x in dom p holds p.x = (q^<*p.(len p)*>).x
  proof
    let x;
    assume
A5: x in dom p;
    then reconsider k = x as Element of NAT;
    k in Seg(n+1) by A1,A5,Def3;
    then
A6: k in Seg n \/ {n+1} by Th9;
A7: now
      assume
A8:   k in Seg n;
      hence p.k=q.k by A3,FUNCT_1:47
        .=(q^<*p.(len p)*>).k by A3,A8,Def7;
    end;
    now
      assume
A9:   k in {n+1};
      1 in Seg 1;
      then
A10:  1 in dom <*p.(len p)*> by Def8;
      thus (q^<*p.(len p)*>).k =(q^<*p.(len p)*>).(n+1) by A9,TARSKI:def 1
        .=(q^<*p.(len p)*>).(len q + 1) by A3,Def3
        .=<*p.(len p)*>.1 by A10,Def7
        .=p.(n+1) by A1,Def8
        .=p.k by A9,TARSKI:def 1;
    end;
    hence thesis by A6,A7,XBOOLE_0:def 3;
  end;
  hence q^<*p.(len p)*>=p by A4,FUNCT_1:2;
end;

definition
  let D be non empty set;
  let x be Element of D;
  redefine func <*x*> -> FinSequence of D;
  coherence
  proof
    let y;
    assume y in rng <*x*>;
    then y in {x} by Th39;
    then y = x by TARSKI:def 1;
    hence thesis;
  end;
end;



scheme IndSeq{P[FinSequence]}: for p holds P[p]
provided
A1: P[{}] and
A2: for p,x st P[p] holds P[p^<*x*>]
proof
  let p;
  defpred R[set] means for p st len p = $1 holds P[p];
  consider X being Subset of REAL such that
A3: for x being Real holds x in X iff R[x] from SUBSET_1:sch 3;
  for k holds k in X
  proof
    defpred S[Nat] means $1 in X;
    for q st len q = 0 holds P[q] by A1,Lm4;
    then
A4: S[0] by A3;
    now
      let n such that
A5:   n in X;
      now
        let q;
        assume
A6:     len q = n+1;
        then q <> {};
        then consider r,x such that
A7:     q=r^<*x*> by Th46;
        len q = len r + len <*x*> by A7,Th22
          .= len r + 1 by Th39;
        then P[r] by A3,A5,A6;
        hence P[q] by A2,A7;
      end;
      hence n+1 in X by A3;
    end;
    then
A8: for n st S[n] holds S[n+1];
    thus for n holds S[n] from NAT_1:sch 2(A4,A8);
  end;
  then len p in X;
  hence thesis by A3;
end;

theorem
  for p,q,r,s being FinSequence st p^q = r^s & len p <= len r
  ex t being FinSequence st p^t = r
proof
  defpred S[FinSequence] means
  for p,q,s st p^q=$1^s & len p <= len $1 holds ex t st p^t=$1;
A1: S[{}]
  proof
    let p,q,s;
    assume that
    p^q={}^s and
A2: len p <= len {};
    take {};
    thus p^{} = p by Th34
      .= {} by A2;
  end;
A3: for r,x st S[r] holds S[r^<*x*>]
  proof
    let r,x;
    assume
A4: for p,q,s st p^q=r^s & len p <= len r ex t st p^t=r;
    let p,q,s;
    assume that
A5: p^q=(r^<*x*>)^s and
A6: len p <= len (r^<*x*>);
A7: now
      assume len p = len(r^<*x*>);
      then
A8:   dom p = Seg len(r^<*x*>) by Def3
        .= dom(r^<*x*>) by Def3;
A9:   for k st k in dom p holds p.k=(r^<*x*>).k
      proof
        let k;
        assume
A10:    k in dom p;
        hence p.k = (r^<*x*>^s).k by A5,Def7
          .=(r^<*x*>).k by A8,A10,Def7;
      end;
      p^{} = p by Th34
        .=r^<*x*> by A8,A9,Th13;
      hence thesis;
    end;
    now
      assume len p <> len(r^<*x*>);
      then len p <> len r + len <*x*> by Th22;
      then
A11:  len p <> len r + 1 by Th39;
      len p <= len r + len <*x*> by A6,Th22;
      then
A12:  len p <= len r + 1 by Th39;
      p^q=r^(<*x*>^s) by A5,Th32;
      then consider t being FinSequence such that
A13:  p^t = r by A4,A11,A12,NAT_1:8;
      p^(t^<*x*>) = r^<*x*> by A13,Th32;
      hence thesis;
    end;
    hence thesis by A7;
  end;
  for r holds S[r] from IndSeq(A1,A3);
  hence thesis;
end;

registration
 cluster  -> NAT-defined for FinSequence;
 coherence
  proof let f be FinSequence;
   thus dom f c= NAT;
  end;
end;

definition
  let D be set;
  func D* -> set means
  :Def11:
  x in it iff x is FinSequence of D;
  existence
  proof
    defpred P[set] means $1 is FinSequence of D;
    consider X such that
A1: x in X iff x in bool [:NAT,D:] & P[x] from XBOOLE_0:sch 1;
    take X;
    let x;
    thus x in X implies x is FinSequence of D by A1;
    assume x is FinSequence of D;
    then reconsider p = x as FinSequence of D;
    p c= [:NAT,D:];
    hence thesis by A1;
  end;
  uniqueness
  proof
    let D1,D2 be set such that
A2: x in D1 iff x is FinSequence of D and
A3: x in D2 iff x is FinSequence of D;
    now
      let x;
      thus x in D1 implies x in D2
      proof
        assume x in D1;
        then x is FinSequence of D by A2;
        hence thesis by A3;
      end;
      assume x in D2;
      then x is FinSequence of D by A3;
      hence x in D1 by A2;
    end;
    hence thesis by TARSKI:1;
  end;
end;

registration
  let D be set;
  cluster D* -> non empty;
  coherence
  proof
    set f = the FinSequence of D;
    f in D* by Def11;
    hence thesis;
  end;
end;

theorem
  rng p = rng q & p is one-to-one & q is one-to-one implies len p = len q
proof
  defpred P[FinSequence] means $1 is one-to-one implies
  for q st rng $1 = rng q & q is one-to-one holds len $1 = len q;
A1: P[{}] by RELAT_1:41;
  now
    let p,x;
    assume
A2: p is one-to-one implies
    for q st rng p = rng q & q is one-to-one holds len p = len q;
    assume
A3: p ^ <* x *> is one-to-one;
    let q;
    assume that
A4: rng(p ^ <* x *>) = rng q and
A5: q is one-to-one;
A6: rng(p ^ <* x *>) = rng p \/ rng<* x *> by Th31
      .= rng p \/ {x} by Th38;
    x in {x} by TARSKI:def 1;
    then x in rng q by A4,A6,XBOOLE_0:def 3;
    then consider y such that
A7: y in dom q and
A8: x = q.y by FUNCT_1:def 3;
A9: y in Seg(len q) by A7,Def3;
    reconsider y as Element of NAT by A7;
A10: 1 <= y by A9,Th1;
    then consider k be Nat such that
A11: 1 + k = y by NAT_1:10;
A12: y <= len q by A9,Th1;
    then consider n be Nat such that
A13: y + n = len q by NAT_1:10;
    reconsider k,n as Element of NAT by ORDINAL1:def 12;
    reconsider q1 = q | (Seg k) as FinSequence by Th15;
    defpred P[Nat,set] means $2 = q.(y + $1);
A14: for j being Nat st j in Seg n ex x st P[j,x];
    consider q2 being FinSequence such that
A15: dom q2 = Seg n and
A16: for j being Nat st j in Seg n holds P[j,q2.j] from SeqEx(A14);
    k <= y by A11,NAT_1:12;
    then
A17: k <= len q by A12,XXREAL_0:2;
    then
A18: len q1 = k by Th17;
    len(q1 ^ <* x *>) + len q2 = len q1 + len<* x *> + len q2 by Th22
      .= k + 1 + len q2 by A18,Th39
      .= len q by A11,A13,A15,Def3;
    then
A19: dom q = Seg(len(q1 ^ <* x *>) + len q2) by Def3;
A20: rng(q1 ^ q2) = rng q \ {x}
    proof
      thus rng(q1 ^ q2) c= rng q \ {x}
      proof
        let z be set;
        assume z in rng(q1 ^ q2);
        then
A21:    z in rng q1 \/ rng q2 by Th31;
A22:    now
          assume
A23:      z in rng q1;
          A24:      rng
 q1 = q .: (Seg k) & q .: (Seg k) c= rng q by RELAT_1:111,115;
          consider y1 such that
A25:      y1 in dom q1 and
A26:      q1.y1 = z by A23,FUNCT_1:def 3;
A27:      q1.y1 = q.y1 by A25,FUNCT_1:47;
A28:      y1 in Seg(len q1) by A25,Def3;
          reconsider y1 as Element of NAT by A25;
A29:      y1 <= k by A18,A28,Th1;
A30:      k < y by A11,XREAL_1:29;
          Seg k c= Seg(len q) by A17,Th5;
          then dom q1 c= Seg(len q) by A17,Th17;
          then dom q1 c= dom q by Def3;
          then x <> z by A5,A7,A8,A25,A26,A27,A29,A30,FUNCT_1:def 4;
          then not z in {x} by TARSKI:def 1;
          hence thesis by A23,A24,XBOOLE_0:def 5;
        end;
        now
          assume z in rng q2;
          then consider y1 such that
A31:      y1 in dom q2 and
A32:      q2.y1 = z by FUNCT_1:def 3;
          reconsider y1 as Element of NAT by A31;
A33:      q2.y1 = q.(y + y1) by A15,A16,A31;
A34:      1 <= y + y1 by A10,NAT_1:12;
          y1 <= n by A15,A31,Th1;
          then y + y1 <= len q by A13,XREAL_1:7;
          then y + y1 in Seg(len q) by A34;
          then
A35:      y + y1 in dom q by Def3;
          then
A36:      z in rng q by A32,A33,FUNCT_1:def 3;
          y1 <> 0 by A15,A31,Th1;
          then y <> y + y1;
          then x <> z by A5,A7,A8,A32,A33,A35,FUNCT_1:def 4;
          then not z in {x} by TARSKI:def 1;
          hence thesis by A36,XBOOLE_0:def 5;
        end;
        hence thesis by A21,A22,XBOOLE_0:def 3;
      end;
      let z be set;
      assume
A37:  z in rng q \ {x};
      then consider y1 such that
A38:  y1 in dom q and
A39:  z = q.y1 by FUNCT_1:def 3;
A40:  y1 in Seg(len q) by A38,Def3;
      reconsider y1 as Element of NAT by A38;
      not z in {x} by A37,XBOOLE_0:def 5;
      then
A41:  y <> y1 by A8,A39,TARSKI:def 1;
A42:  now
        assume
A43:    y < y1;
        then consider j being Nat such that
A44:    y1 = y + j by NAT_1:10;
        reconsider j as Element of NAT by ORDINAL1:def 12;
A45:    1 <= j by A43,A44,NAT_1:19;
        y + j <= len q by A40,A44,Th1;
        then j <= n by A13,XREAL_1:6;
        then
A46:    j in Seg n by A45;
        then z = q2.j by A16,A39,A44;
        hence z in rng q2 by A15,A46,FUNCT_1:def 3;
      end;
      now
        assume
A47:    y1 < y;
A48:    1 <= y1 by A40,Th1;
        y1 <= k by A11,A47,NAT_1:13;
        then y1 in Seg k by A48;
        then
A49:    y1 in dom q1 by A17,Th17;
        then q1.y1 = z by A39,FUNCT_1:47;
        hence z in rng q1 by A49,FUNCT_1:def 3;
      end;
      then z in rng q1 \/ rng q2 by A41,A42,XBOOLE_0:def 3,XXREAL_0:1;
      hence thesis by Th31;
    end;
A50: p = (p ^ <* x *>) | (dom p) by Th21;
A51: rng p = rng(p ^ <* x *>) \ {x}
    proof
      thus rng p c= rng(p ^ <* x *>) \ {x}
      proof
        let z be set;
        assume
A52:    z in rng p;
A53:    rng p c= rng(p ^ <* x *>) by A50,RELAT_1:70;
        consider y1 such that
A54:    y1 in dom p and
A55:    p.y1 = z by A52,FUNCT_1:def 3;
A56:    y1 in Seg(len p) by A54,Def3;
        reconsider y1 as Element of NAT by A54;
A57:    (p ^ <* x *>).(len p + 1) = x by Th42;
A58:    (p ^ <* x *>).y1 = p.y1 by A50,A54,FUNCT_1:47;
A59:    y1 <= len p by A56,Th1;
A60:    len p < len p +1 by XREAL_1:29;
A61:    1 <= y1 by A56,Th1;
        y1 <= len p + 1 by A59,A60,XXREAL_0:2;
        then
A62:    y1 in Seg(len p + 1) by A61;
A63:    len p + 1 in Seg(len p + 1) by Th3;
A64:    y1 in Seg(len p + len<* x *>) by A62,Th40;
A65:    len p + 1 in Seg(len p + len<* x *>) by A63,Th40;
A66:    y1 in dom(p ^ <* x *>) by A64,Def7;
        len p + 1 in dom(p ^ <* x *>) by A65,Def7;
        then x <> z by A3,A55,A57,A58,A59,A60,A66,FUNCT_1:def 4;
        then not z in {x} by TARSKI:def 1;
        hence thesis by A52,A53,XBOOLE_0:def 5;
      end;
      let z be set;
      assume
A67:  z in rng(p ^ <* x *>) \ {x};
      then
A68:  z in rng(p ^ <* x *>);
A69:  not z in {x} by A67,XBOOLE_0:def 5;
      z in rng p \/ rng<* x *> by A68,Th31;
      then z in rng p or z in rng<* x *> by XBOOLE_0:def 3;
      hence thesis by A69,Th38;
    end;
A70: q1 ^ q2 is one-to-one
    proof
      let y1,y2;
      assume that
A71:  y1 in dom(q1 ^ q2) & y2 in dom(q1 ^ q2) and
A72:  (q1 ^ q2).y1 = (q1 ^ q2).y2;
      reconsider m1 = y1, m2 = y2 as Element of NAT by A71;
A73:  q1 is one-to-one by A5,FUNCT_1:52;
A74:  now
        assume
A75:    m1 in dom q1 & m2 in dom q1;
        then (q1 ^ q2).m1 = q1.m1 & (q1 ^ q2).m2 = q1.m2 by Def7;
        hence thesis by A72,A73,A75,FUNCT_1:def 4;
      end;
A76:  now
        assume
A77:    m1 in dom q1;
        given j being Nat such that
A78:    j in dom q2 and
A79:    m2 = len q1 + j;
        reconsider j as Element of NAT by ORDINAL1:def 12;
A80:    (q1 ^ q2).m2 = q2.j by A78,A79,Def7;
        (q1 ^ q2).m1 = q1.m1 & q1.m1 = q.m1 by A77,Def7,FUNCT_1:47;
        then
A81:    q.m1 = q.(y + j) by A15,A16,A72,A78,A80;
        1 <= j by A15,A78,Th1;
        then
A82:    1 <= y + j by NAT_1:12;
        j <= n by A15,A78,Th1;
        then y + j <= len q by A13,XREAL_1:7;
        then y + j in Seg(len q) by A82;
        then
A83:    y + j in dom q by Def3;
        m1 in Seg k by A17,A77,Th17;
        then
A84:    m1 <= k by Th1;
        k < y by A11,XREAL_1:29;
        then
A85:    m1 < y by A84,XXREAL_0:2;
        dom q1 c= dom q & y <= y + j by NAT_1:12,RELAT_1:60;
        hence thesis by A5,A77,A81,A83,A85,FUNCT_1:def 4;
      end;
A86:  now
        assume
A87:    m2 in dom q1;
        given j be Nat such that
A88:    j in dom q2 and
A89:    m1 = len q1 + j;
        reconsider j as Element of NAT by ORDINAL1:def 12;
A90:    (q1 ^ q2).m1 = q2.j by A88,A89,Def7;
        (q1 ^ q2).m2 = q1.m2 & q1.m2 = q.m2 by A87,Def7,FUNCT_1:47;
        then
A91:    q.m2 = q.(y + j) by A15,A16,A72,A88,A90;
        1 <= j by A15,A88,Th1;
        then
A92:    1 <= y + j by NAT_1:12;
        j <= n by A15,A88,Th1;
        then y + j <= len q by A13,XREAL_1:7;
        then y + j in Seg(len q) by A92;
        then
A93:    y + j in dom q by Def3;
        m2 in Seg k by A17,A87,Th17;
        then
A94:    m2 <= k by Th1;
        k < y by A11,XREAL_1:29;
        then
A95:    m2 < y by A94,XXREAL_0:2;
        dom q1 c= dom q & y <= y + j by NAT_1:12,RELAT_1:60;
        hence thesis by A5,A87,A91,A93,A95,FUNCT_1:def 4;
      end;
      now
        given j1 being Nat such that
A96:    j1 in dom q2 and
A97:    m1 = len q1 + j1;
        given j2 being Nat such that
A98:    j2 in dom q2 and
A99:    m2 = len q1 + j2;
        reconsider j1,j2 as Element of NAT by ORDINAL1:def 12;
A100:   (q1 ^ q2).m1 = q2.j1 by A96,A97,Def7;
A101:   (q1 ^ q2).m2 = q2.j2 by A98,A99,Def7;
A102:   (q1 ^ q2).m1 = q.(y + j1) by A15,A16,A96,A100;
A103:   (q1 ^ q2).m2 = q.(y + j2) by A15,A16,A98,A101;
A104:   1 <= j1 by A15,A96,Th1;
A105:   1 <= j2 by A15,A98,Th1;
A106:   1 <= y + j1 by A104,NAT_1:12;
A107:   1 <= y + j2 by A105,NAT_1:12;
A108:   j1 <= n by A15,A96,Th1;
A109:   j2 <= n by A15,A98,Th1;
A110:   y + j1 <= len q by A13,A108,XREAL_1:7;
A111:   y + j2 <= len q by A13,A109,XREAL_1:7;
A112:   y + j1 in Seg(len q) by A106,A110;
A113:   y + j2 in Seg(len q) by A107,A111;
A114:   y + j1 in dom q by A112,Def3;
        y + j2 in dom q by A113,Def3;
        then y + j1 = y + j2 by A5,A72,A102,A103,A114,FUNCT_1:def 4;
        hence thesis by A97,A99;
      end;
      hence thesis by A71,A74,A76,A86,Th25;
    end;
A115: now
      let j be Nat;
      assume
A116: j in dom(q1 ^ <* x *>);
A117: now
        assume
A118:   j in dom q1;
        then (q1 ^ <* x *>).j = q1.j by Def7;
        hence q.j = (q1 ^ <* x *>).j by A118,FUNCT_1:47;
      end;
      now
        given i be Nat such that
A119:   i in dom<* x *> and
A120:   j = len q1 + i;
        i in {1} by A119,Th2,Th38;
        then i = 1 by TARSKI:def 1;
        hence q.j = (q1 ^ <* x *>).j by A8,A11,A18,A120,Th42;
      end;
      hence q.j = (q1 ^ <* x *>).j by A116,A117,Th25;
    end;
    now
      let j be Nat;
      assume j in dom q2;
      hence q2.j = q.(len q1 + 1 + j) by A11,A15,A16,A18
        .= q.(len q1 + len<* x *> + j) by Th39
        .= q.(len(q1 ^ <* x *>) + j) by Th22;
    end;
    then q1 ^ <* x *> ^ q2 = q by A19,A115,Def7;
    then
A121: len q = len(q1 ^ <* x *>) + len q2 by Th22
      .= len <* x *> + len q1 + len q2 by Th22
      .= 1 + len q1 + len q2 by Th40
      .= 1 + (len q1 + len q2)
      .= len(q1 ^ q2) + 1 by Th22;
    len(p ^ <* x *>) = len p + len<* x *> by Th22
      .= len p + 1 by Th40;
    hence len(p ^ <* x *>) = len q by A2,A3,A4,A20,A50,A51,A70,A121,FUNCT_1:52;
  end;
  then
A122: for p,x st P[p] holds P[p^<*x*>];
  for p holds P[p] from IndSeq(A1,A122);
  hence thesis;
end;

theorem
  {} in D*
proof
  {} = <*>D;
  hence thesis by Def11;
end;

scheme SepSeq{D()->non empty set, P[FinSequence]}:
  ex X st for x holds x in X iff ex p st p in D()* & P[p] & x=p
proof
  defpred R[set] means ex p st P[p] & $1=p;
  consider Y such that
A1: x in Y iff x in D()* & R[x] from XBOOLE_0:sch 1;
  take Y;
  x in Y iff ex p st p in D()* & P[p] & x=p
  proof
    now
      assume x in Y;
      then x in D()* & ex p st P[p] & x=p by A1;
      hence ex p st p in D()* & P[p] & x=p;
    end;
    hence thesis by A1;
  end;
  hence thesis;
end;





definition
  let IT be Function;
  attr IT is FinSubsequence-like means
  :Def12:
  ex k st dom IT c= Seg k;
end;

registration
  cluster FinSubsequence-like for Function;
  existence
  proof
    take {},len {};
    thus thesis;
  end;
end;

definition
  mode FinSubsequence is FinSubsequence-like Function;
end;

registration
 cluster FinSequence-like -> FinSubsequence-like for Function;
 coherence
proof
  let p be Function;
  assume p is FinSequence-like;
   then reconsider p as FinSequence;
  dom p = Seg len p by Def3;
  hence thesis by Def12;
end;
 let p be FinSubsequence, X be set;
 cluster p|X -> FinSubsequence-like;
 coherence
proof
  consider k such that
A1: dom p c= Seg k by Def12;
   dom(p|X) c= dom p by RELAT_1:60;
   then  dom(p|X) c= Seg k by A1,XBOOLE_1:1;
  hence thesis by Def12;
end;
 cluster X|`p -> FinSubsequence-like;
 coherence
proof
  consider k such that
A2: dom p c= Seg k by Def12;
   dom(X|`p) c= dom p by FUNCT_1:56;
   then  dom(X|`p) c= Seg k by A2,XBOOLE_1:1;
  hence thesis by Def12;
end;
end;

registration
 cluster  -> NAT-defined for FinSubsequence;
 coherence
  proof let f be FinSubsequence;
    ex n st dom f c= Seg n by Def12;
   hence dom f c= NAT by XBOOLE_1:1;
  end;
end;

reserve p9 for FinSubsequence;

definition
  let X;
  given k being Nat such that
A1: X c= Seg k;
  func Sgm X -> FinSequence of NAT means
  :Def13:
  rng it = X & for l,m,k1,k2 being Nat st 1 <= l & l < m & m <= len it &
  k1=it.l & k2=it.m holds k1 < k2;
  existence
  proof
    defpred P[Nat] means for X st X c= Seg $1 ex p being FinSequence of NAT st
    rng p = X & for l,m,k1,k2 being Nat st ( 1 <= l & l < m & m <= len p &
    k1=p.l & k2=p.m) holds k1 < k2;
A2: P[0]
    proof
      let X;
      assume
A3:   X c= Seg 0;
      take <*>(NAT);
      thus rng <*>(NAT) = X by A3;
      thus thesis;
    end;
A4: for k being Nat st P[k] holds P[k+1]
    proof
      let k be Nat such that
A5:   for X st X c= Seg k holds ex p being FinSequence of NAT st
      rng p = X & for l,m,k1,k2 be Nat st 1 <= l & l < m & m <= len p &
      k1=p.l & k2=p.m holds k1 < k2;
      let X;
      assume
A6:   X c= Seg(k+1);
      now
        assume not X c= Seg k;
        then consider x such that
A7:     x in X and
A8:     not x in Seg k by TARSKI:def 3;
        x in Seg(k+1) by A6,A7;
        then reconsider n=x as Element of NAT;
A9:     n=k+1
        proof
          assume
A10:      n <> k+1;
A11:      n <= k+1 by A6,A7,Th1;
A12:      1 <= n by A6,A7,Th1;
          n <= k by A10,A11,NAT_1:8;
          hence contradiction by A8,A12;
        end;
        set Y=X\{k+1};
A13:    Y c= Seg k
        proof
          let x;
          assume
A14:      x in Y;
          then
A15:      x in X;
A16:      not x in {k+1} by A14,XBOOLE_0:def 5;
A17:      x in Seg(k+1) by A6,A15;
A18:      x<>k+1 by A16,TARSKI:def 1;
          reconsider m=x as Element of NAT by A17;
A19:      m <= k+1 by A6,A15,Th1;
A20:      1 <= m by A6,A15,Th1;
          m <= k by A18,A19,NAT_1:8;
          hence thesis by A20;
        end;
        then consider p being FinSequence of NAT such that
A21:    rng p = Y and
A22:    for l,m,k1,k2 be Nat st 1 <= l & l < m & m <= len p & k1=p.l & k2
        =p.m holds k1 < k2
        by A5;
        reconsider k as Element of NAT by ORDINAL1:def 12;
        consider q such that
A23:    q=p^<*k+1*>;
        reconsider q as FinSequence of NAT by A23;
A24:    {k+1} c= X by A7,A9,ZFMISC_1:31;
A25:    rng q = rng p \/ rng <*k+1*> by A23,Th31
          .= Y \/ {k+1} by A21,Th38
          .= X \/ {k+1} by XBOOLE_1:39
          .= X by A24,XBOOLE_1:12;
        for l,m,k1,k2 be Nat st 1 <= l & l < m & m <= len q &
        k1=q.l & k2=q.m holds k1 < k2
        proof
          let l,m,k1,k2 be Nat;
          assume that
A26:      1 <= l and
A27:      l < m and
A28:      m <= len q and
A29:      k1=q.l and
A30:      k2=q.m;
          l < len (p^<*k+1*>) by A23,A27,A28,XXREAL_0:2;
          then l < len p + len <*k+1*> by Th22;
          then l < len p + 1 by Th39;
          then l <= len p by NAT_1:13;
          then l in Seg len p by A26,Th1;
          then
A31:      l in dom p by Def3;
A32:      1 <= m by A26,A27,XXREAL_0:2;
A33:      now
            assume
A34:        m=len q;
            k1 = p.l by A23,A29,A31,Def7;
            then k1 in Y by A21,A31,FUNCT_1:def 3;
            then
A35:        k1 <= k by A13,Th1;
            q.m = (p^<*k+1*>).(len p + len <*k+1*>) by A23,A34,Th22
              .=(p^<*k+1*>).(len p + 1) by Th39
              .= k+1 by Th42;
            hence thesis by A30,A35,NAT_1:13;
          end;
          now
            assume m <> len q;
            then m < len (p^<*k+1*>) by A23,A28,XXREAL_0:1;
            then m < len p + len <*k+1*> by Th22;
            then m < len p + 1 by Th39;
            then
A36:        m <= len p by NAT_1:13;
            then m in Seg len p by A32,Th1;
            then m in dom p by Def3;
            then
A37:        k2 = p.m by A23,A30,Def7;
            k1 = p.l by A23,A29,A31,Def7;
            hence thesis by A22,A26,A27,A36,A37;
          end;
          hence thesis by A33;
        end;
        hence thesis by A25;
      end;
      hence thesis by A5;
    end;
    for k being Nat holds P[k] from NAT_1:sch 2(A2,A4);
    hence thesis by A1;
  end;
  uniqueness
  proof
    let p,q be FinSequence of NAT such that
    A38: (
 rng p = X & for l,m,k1,k2 being Nat st 1 <= l & l < m & m <= len p & k1=p.
l & k2=p.m holds k1 < k2 )&( rng q = X & for l,m,k1,k2 being Nat st 1 <= l &
    l < m & m <= len q & k1=q.l & k2=q.m holds k1 < k2 );
    defpred S[FinSequence] means for X st ex k being Nat st X c= Seg k holds
    ($1 is FinSequence of NAT & rng $1 = X &
    for l,m,k1,k2 being Nat st ( 1 <= l & l < m & m <= len $1 &
    k1=$1.l & k2=$1.m) holds k1 < k2) implies
    for q being FinSequence of NAT st rng q = X &
    for l,m,k1,k2 being Nat st ( 1 <= l & l < m & m <= len q &
    k1=q.l & k2=q.m) holds k1 < k2 holds q=$1;
A39: S[{}];
A40: for p,x st S[p] holds S[p^<*x*>]
    proof
      let p,x;
      assume
A41:  S[p];
      let X;
      given k being Nat such that
A42:  X c= Seg k;
      assume that
A43:  p^<*x*> is FinSequence of NAT and
A44:  rng (p^<*x*>) = X and
A45:  for l,m,k1,k2 being Nat st 1 <= l & l < m & m <= len(p^<*x*>) &
      k1=(p^<*x*>).l & k2=(p^<*x*>).m holds k1 < k2;
      let q be FinSequence of NAT;
      assume that
A46:  rng q = X and
A47:  for l,m,k1,k2 being Nat st 1 <= l & l < m & m <= len q & k1=q.l &
      k2=q.m holds k1 < k2;
      1 in Seg 1;
      then
A48:  1 in dom <*x*> by Def8;
A49:  ex m st m=x & for l st l in X & l <> x holds l < m
      proof
        <*x*> is FinSequence of NAT by A43,Th36;
        then rng <*x*> c= NAT by Def4;
        then {x} c= NAT by Th38;
        then reconsider m=x as Element of NAT by ZFMISC_1:31;
        take m;
        thus m=x;
        thus for l st l in X & l <> x holds l < m
        proof
          let l;
          assume that
A50:      l in X and
A51:      l <> x;
          consider y such that
A52:      y in dom (p^<*x*>) and
A53:      l=(p^<*x*>).y by A44,A50,FUNCT_1:def 3;
A54:      y in Seg len (p^<*x*>) by A52,Def3;
          reconsider k=y as Element of NAT by A52;
          k <= len (p^<*x*>) by A54,Th1;
          then k <= len p + len <*x*> by Th22;
          then
A55:      k <= len p + 1 by Th39;
A56:      k <> len p + 1
          proof
            assume k = len p + 1;
            then (p^<*x*>).k = <*x*>.1 by A48,Def7
              .= x by Def8;
            hence contradiction by A51,A53;
          end;
A57:      1 <= k by A54,Th1;
          k < len p + 1 by A55,A56,XXREAL_0:1;
          then k < len p + len <*x*> by Th39;
          then
A58:      k < len(p^<*x*>) by Th22;
          m=(p^<*x*>).(len p + 1) by Th42
            .= (p^<*x*>).(len p + len <*x*>) by Th39
            .= (p^<*x*>).(len(p^<*x*>)) by Th22;
          hence thesis by A45,A53,A57,A58;
        end;
      end;
      then reconsider m = x as Nat;
      len q <> 0
      proof
        assume len q = 0;
        then p^<*x*> = {} by A44,A46,Lm4,RELAT_1:38;
        then 0 = len (p^<*x*>) .= len p + len <*x*> by Th22
          .= 1 + len p by Th39;
        hence contradiction;
      end;
      then consider n be Nat such that
A59:  len q = n+1 by NAT_1:6;
      deffunc F(Nat) = q.$1;
      ex q9 being FinSequence st len q9 = n &
      for m st m in dom q9 holds q9.m = F(m) from SeqLambda;
      then consider q9 being FinSequence such that
A60:  len q9 = n and
A61:  for m st m in dom q9 holds q9.m = q.m;
      now
        let x;
        assume x in rng q9;
        then consider y such that
A62:    y in dom q9 and
A63:    x=q9.y by FUNCT_1:def 3;
A64:    y in Seg len q9 by A62,Def3;
        reconsider y as Element of NAT by A62;
A65:    y <= n by A60,A64,Th1;
A66:    n <= n + 1 by NAT_1:11;
A67:    1 <= y by A64,Th1;
        y <= n+1 by A65,A66,XXREAL_0:2;
        then y in Seg (n+1) by A67;
        then y in dom q by A59,Def3;
        then
A68:    q.y in rng q by FUNCT_1:def 3;
        rng q c= NAT by Def4;
        then q.y in NAT by A68;
        hence x in NAT by A61,A62,A63;
      end;
      then rng q9 c= NAT by TARSKI:def 3;
      then reconsider f=q9 as FinSequence of NAT by Def4;
A69:  dom q = Seg (n+1) by A59,Def3
        .= Seg (len q9 + len <*x*>) by A60,Th39;
A70:  for m st m in dom <*x*> holds q.(len q9 + m) = <*x*>.m
      proof
        let m;
        assume m in dom <*x*>;
        then
A71:    m in {1} by Def8,Th2;
        then
A72:    m=1 by TARSKI:def 1;
        q.(len q9 + m) = x
        proof
          assume q.(len q9 + m) <> x;
          then
A73:      q.len q <> x by A59,A60,A71,TARSKI:def 1;
          consider d being Nat such that
A74:      d=x and
A75:      for l st l in X & l<>x holds l<d by A49;
          x in {x} by TARSKI:def 1;
          then x in rng <*x*> by Th38;
          then x in rng p \/ rng <*x*> by XBOOLE_0:def 3;
          then x in rng q by A44,A46,Th31;
          then consider y such that
A76:      y in dom q and
A77:      x=q.y by FUNCT_1:def 3;
A78:      y in Seg len q by A76,Def3;
          reconsider y as Element of NAT by A76;
A79:      1 <= y by A78,Th1;
A80:      y <= len q by A78,Th1;
          then
A81:      y < len q by A73,A77,XXREAL_0:1;
          1 <= len q by A79,A80,XXREAL_0:2;
          then len q in Seg len q;
          then
A82:      len q in dom q by Def3;
A83:      q.len q in X by A46,A82,FUNCT_1:def 3;
          reconsider k = q.len q as Nat;
          k < d by A73,A75,A83;
          hence contradiction by A47,A74,A77,A79,A81;
        end;
        hence thesis by A72,Th40;
      end;
      then
A84:  q9^<*x*> = q by A61,A69,Def7;
A85:  ex m st X\{x} c= Seg m by A42,XBOOLE_1:1;
A86:  not x in rng p
      proof
        assume x in rng p;
        then consider y such that
A87:    y in dom p and
A88:    x=p.y by FUNCT_1:def 3;
A89:    y in Seg len p by A87,Def3;
        reconsider y as Element of NAT by A87;
A90:    y <= len p by A89,Th1;
A91:    len p + 1 = len p + len <*x*> by Th39
          .= len (p^<*x*>) by Th22;
A92:    1 <= y by A89,Th1;
A93:    y < len p + 1 by A90,NAT_1:13;
A94:    m = (p^<*x*>).y by A87,A88,Def7;
        m = (p^<*x*>).(len p + 1 ) by Th42;
        hence contradiction by A45,A91,A92,A93,A94;
      end;
A95:  X = rng p \/ rng <*x*> by A44,Th31
        .= rng p \/ {x} by Th39;
A96:  for z holds z in rng p \/ {x} \ {x} iff z in rng p
      proof
        let z;
        thus z in rng p \/ {x} \ {x} implies z in rng p
        proof
          assume
A97:      z in rng p \/ {x} \ {x};
          then not z in {x} by XBOOLE_0:def 5;
          hence thesis by A97,XBOOLE_0:def 3;
        end;
        assume z in rng p;
        then
        ( not z in {x})& z in rng p \/ {x} by A86,TARSKI:def 1,XBOOLE_0:def 3;
        hence thesis by XBOOLE_0:def 5;
      end;
A98: for l,m,k1,k2 being Nat st 1 <= l & l < m & m <= len p &
      k1=p.l & k2=p.m holds k1 < k2
      proof
        let l,m,k1,k2 be Nat;
        assume that
A99:   1 <= l and
A100:   l < m and
A101:   m <= len p and
A102:   k1=p.l and
A103:   k2=p.m;
        l <= len p by A100,A101,XXREAL_0:2;
        then l in Seg len p by A99,Th1;
        then l in dom p by Def3;
        then
A104:   k1 = (p^<*x*>).l by A102,Def7;
        1 <= m by A99,A100,XXREAL_0:2;
        then m in Seg len p by A101,Th1;
        then m in dom p by Def3;
        then
A105:   k2 = (p^<*x*>).m by A103,Def7;
        len p <= len p + 1 by NAT_1:11;
        then m <= len p + 1 by A101,XXREAL_0:2;
        then m <= len p + len <*x*> by Th39;
        then m <= len (p^<*x*>) by Th22;
        hence thesis by A45,A99,A100,A104,A105;
      end;
A106: p is FinSequence of NAT by A43,Th36;
A107: rng p = X\{x} by A95,A96,TARSKI:1;
A108: not x in rng f
      proof
        assume x in rng f;
        then consider y such that
A109:   y in dom f and
A110:   x=f.y by FUNCT_1:def 3;
A111:   y in Seg len f by A109,Def3;
        reconsider y as Element of NAT by A109;
A112:   y <= len f by A111,Th1;
A113:   len f + 1 = len f + len <*x*> by Th39
          .= len (f^<*x*>) by Th22;
A114:   1 <= y by A111,Th1;
A115:   y < len f + 1 by A112,NAT_1:13;
A116:   m = q.y by A61,A109,A110;
        m = q.(len f + 1) by A84,Th42;
        hence contradiction by A47,A84,A113,A114,A115,A116;
      end;
A117: X = rng f \/ rng <*x*> by A46,A84,Th31
        .= rng f \/ {x} by Th39;
A118: for z holds z in rng f \/ {x} \ {x} iff z in rng f
      proof
        let z;
        thus z in rng f \/ {x} \ {x} implies z in rng f
        proof
          assume
A119:     z in rng f \/ {x} \ {x};
          then not z in {x} by XBOOLE_0:def 5;
          hence thesis by A119,XBOOLE_0:def 3;
        end;
        assume z in rng f;
        then
        ( not z in {x})& z in rng f \/ {x} by A108,TARSKI:def 1,XBOOLE_0:def 3;
        hence thesis by XBOOLE_0:def 5;
      end;
      A120: for l,m,k1,k2 being Nat st 1 <= l & l < m & m <= len f &
      k1=f.l & k2=f.m holds k1 < k2
      proof
        let l,m,k1,k2 be Nat;
        assume that
A121:   1 <= l and
A122:   l < m and
A123:   m <= len f and
A124:   k1=f.l and
A125:   k2=f.m;
A126:   m <= len q by A59,A60,A123,NAT_1:13;
        l <= n by A60,A122,A123,XXREAL_0:2;
        then
A127:   l in Seg n by A121,Th1;
        1 <= m by A121,A122,XXREAL_0:2;
        then
A128:   m in Seg n by A60,A123,Th1;
A129:   l in dom q9 by A60,A127,Def3;
A130:   m in dom q9 by A60,A128,Def3;
A131:   k1 = q.l by A61,A124,A129;
        k2 = q.m by A61,A125,A130;
        hence thesis by A47,A121,A122,A126,A131;
      end;
      rng f = X\{x} by A117,A118,TARSKI:1;
      then q9 = p by A41,A85,A98,A106,A107,A120;
      hence thesis by A61,A69,A70,Def7;
    end;
    for p holds S[p] from IndSeq(A39,A40);
    hence thesis by A1,A38;
  end;
end;

theorem Th50:
  rng Sgm dom p9 = dom p9
proof
  ex k st dom p9 c= Seg k by Def12;
  hence thesis by Def13;
end;

definition
  let p9;
  func Seq p9 -> Function equals
  p9* Sgm(dom p9);
  coherence;
end;

registration
  let p9;
  cluster Seq p9 -> FinSequence-like;
  coherence
  proof
    rng Sgm dom p9 = dom p9 by Th50;
    then dom(p9*Sgm(dom p9)) = dom Sgm dom p9 by RELAT_1:27
      .=Seg len Sgm dom p9 by Def3;
    hence thesis by Def2;
  end;
end;

theorem
  for X st ex k st X c= Seg k holds Sgm X = {} iff X = {}
proof
  let X;
  given k such that
A1: X c= Seg k;
  thus Sgm X = {} implies X = {}
  proof
    assume Sgm X = {};
    hence X = rng {} by A1,Def13
      .= {};
  end;
  assume X={};
  then rng Sgm X = {} by A1,Def13;
  hence thesis;
end;

begin 

theorem 
  D is finite iff ex p st D = rng p
proof
  thus D is finite implies ex p st D = rng p
  proof
    given g being Function such that
A1: rng g = D and
A2: dom g in omega;
    reconsider n = dom g as Element of NAT by A2;
    defpred R[set,set] means P[$2,$1];
A3: for x st x in Seg n ex y st R[x,y]
    proof
      let x;
      assume
A4:   x in Seg n;
      then reconsider x as Element of NAT;
      x >= 1 by A4,Th1;
      then ex k be Nat st x = 1+k by NAT_1:10;
      hence thesis;
    end;
    consider f such that
A5: dom f = Seg n and
A6: for x st x in Seg n holds R[x,f.x] from CLASSES1:sch 1(A3);
A7: rng f = dom g
    proof
A8:   n = { k where k is Element of NAT: k < n } by AXIOMS:4;
      thus rng f c= dom g
      proof
        let x;
        assume x in rng f;
        then consider y such that
A9:     y in dom f and
A10:    x = f.y by FUNCT_1:def 3;
        consider k such that
A11:    x = k and
A12:    y = k+1 by A5,A6,A9,A10;
        k + 1 <= n by A5,A9,A12,Th1;
        then k in NAT & k < n by NAT_1:13,ORDINAL1:def 12;
        hence thesis by A8,A11;
      end;
      let x;
      assume x in dom g;
      then consider k being Element of NAT such that
A13:  x = k and
A14:  k < n by A8;
      1 <= k+1 & k+1 <= n by A14,NAT_1:11,13;
      then
A15:  k+1 in Seg n;
      then ex k1 st f.(k+1) = k1 & k+1 = k1+1 by A6;
      hence thesis by A5,A13,A15,FUNCT_1:def 3;
    end;
    then dom(g*f) = Seg n by A5,RELAT_1:27;
    then reconsider p = g*f as FinSequence by Def2;
    take p;
    thus thesis by A1,A7,RELAT_1:28;
  end;
  given p such that
A16: D = rng p;
  consider n such that
A17: dom p = Seg n by Def2;
A18: n = { k where k is Element of NAT: k < n } by AXIOMS:4;
A19: for x st x in n ex y st P[x,y]
  proof
    let x;
    assume x in n;
    then ex k being Element of NAT st x = k & k < n by A18;
    then reconsider k = x as Nat;
    take k+1;
    thus thesis;
  end;
  consider f such that
A20: dom f = n and
A21: for x st x in n holds P[x,f.x] from CLASSES1:sch 1(A19);
  take p*f;
A22: rng f = dom p
  proof
    thus rng f c= dom p
    proof
      let x;
      assume x in rng f;
      then consider y such that
A23:  y in dom f and
A24:  x = f.y by FUNCT_1:def 3;
      consider k such that
A25:  y = k and
A26:  x = k+1 by A20,A21,A23,A24;
      ex m being Element of NAT st k = m & m < n by A18,A20,A23,A25;
      then 1 <= k+1 & k+1 <= n by NAT_1:11,13;
      hence thesis by A17,A26;
    end;
    let x;
    assume
A27: x in dom p;
    then reconsider x as Element of NAT;
    1 <= x by A17,A27,Th1;
    then consider m be Nat such that
A28: x = 1+m by NAT_1:10;
    reconsider m as Element of NAT by ORDINAL1:def 12;
    x <= n by A17,A27,Th1;
    then m < n by A28,NAT_1:13;
    then
A29: m in n by A18;
    then ex k st m = k & f.m = k+1 by A21;
    hence thesis by A20,A28,A29,FUNCT_1:def 3;
  end;
  hence rng(p*f) = D by A16,RELAT_1:28;
  dom(p*f) = dom f by A22,RELAT_1:27;
  hence thesis by A20,ORDINAL1:def 12;
end;

begin 

theorem
  Seg n,Seg m are_equipotent implies n = m
proof
  defpred P[Nat] means ex n st Seg n,Seg $1 are_equipotent & n <> $1;
  assume Seg n,Seg m are_equipotent & n <> m;
  then
A1: ex m being Nat st P[m];
  consider m being Nat such that
A2: P[m] and
A3: for k being Nat st P[k] holds m <= k from NAT_1:sch 5(A1);
  consider n such that
A4: Seg n,Seg m are_equipotent and
A5: n <> m by A2;
A6: ex f st
  f is one-to-one & dom f = Seg n & rng f = Seg m by A4,WELLORD2:def 4;
A7: now
    assume m = 0;
    then Seg m = {};
    then Seg m = Seg n by A6,RELAT_1:42;
    hence contradiction by A5,Th6;
  end;
  then consider m1 being Nat such that
A8: m = m1+1 by NAT_1:6;
A9: now
    assume n = 0;
    then Seg n = {};
    then Seg m = Seg n by A6,RELAT_1:42;
    hence contradiction by A5,Th6;
  end;
  then consider n1 being Nat such that
A10: n = n1+1 by NAT_1:6;
  reconsider m1,n1 as Element of NAT by ORDINAL1:def 12;
A11: n in Seg n by A9,Th3;
A12: m in Seg m by A7,Th3;
A13: not n1+1 <= n1 by NAT_1:13;
A14: not m1+1 <= m1 by NAT_1:13;
A15: not n in Seg n1 by A10,A13,Th1;
A16: not m in Seg m1 by A8,A14,Th1;
A17: (Seg n1) /\ { n } c= {}
  proof
    let x;
    assume x in (Seg n1) /\ { n };
    then x in Seg n1 & x in { n } by XBOOLE_0:def 4;
    hence thesis by A15,TARSKI:def 1;
  end;
A18: (Seg m1) /\ { m } c= {}
  proof
    let x;
    assume x in (Seg m1) /\ { m };
    then x in Seg m1 & x in { m } by XBOOLE_0:def 4;
    hence thesis by A16,TARSKI:def 1;
  end;
A19: Seg n = (Seg n1) \/ { n } by A10,Th9;
A20: Seg m = (Seg m1) \/ { m } by A8,Th9;
A21: (Seg n1) /\ { n } = {} by A17;
A22: (Seg m1) /\ { m } = {} by A18;
  A23: (
Seg n1) \ { n } = ((Seg n1) \/ { n }) \ { n } & (Seg n1) misses { n } by A21,
XBOOLE_0:def 7,XBOOLE_1:40;
  A24: (
Seg m1) \ { m } = ((Seg m1) \/ { m }) \ { m } & (Seg m1) misses { m } by A22,
XBOOLE_0:def 7,XBOOLE_1:40;
A25: (Seg n) \ { n } = Seg n1 by A19,A23,XBOOLE_1:83;
  (Seg m) \ { m } = Seg m1 by A20,A24,XBOOLE_1:83;
  hence contradiction by A3,A4,A5,A8,A10,A11,A12,A14,A25,CARD_1:34;
end;

theorem
  Seg n,n are_equipotent by Lm1;

theorem
  card Seg n = card n by Lm2;

theorem
  X is finite implies ex n st X,Seg n are_equipotent
proof
  assume X is finite;
  then consider n being Nat such that
A1: X,n are_equipotent by CARD_1:43;
  take n;
  n,Seg n are_equipotent by Lm1;
  hence thesis by A1,WELLORD2:15;
end;

theorem
  for n being Nat holds card Seg n = n
proof
  let n be Nat;
  Seg n,n are_equipotent by Lm1;
  hence thesis by CARD_1:def 2;
end;

begin 


registration
  let x be set;
  cluster <*x*> -> non empty;
  coherence;
end;



registration
  cluster non empty for FinSequence;
  existence
  proof
    take <* 0 *>;
    thus thesis;
  end;
end;



registration
  let f1 be FinSequence, f2 be non empty FinSequence;
  cluster f1^f2 -> non empty;
  coherence by Th35;
  cluster f2^f1 -> non empty;
  coherence by Th35;
end;

registration
  let x,y be set;
  cluster <*x,y*> -> non empty;
  coherence;
  let z be set;
  cluster <*x,y,z*> -> non empty;
  coherence;
end;



scheme SeqDEx{D()->non empty set,A()->Nat,P[set,set]}:
  ex p being FinSequence of D() st dom p = Seg A() &
  for k st k in Seg A() holds P[k,p.k]
provided
A1: for k st k in Seg A() ex x being Element of D() st P[k,x]
proof
A2: for y be set st y in Seg A() ex x st x in D() & P[y,x]
  proof
    let y;
    assume
A3: y in Seg A();
    then reconsider k=y as Element of NAT;
    consider x being Element of D() such that
A4: P[k,x] by A1,A3;
    take x;
    thus thesis by A4;
  end;
  consider f being Function such that
A5: dom f = Seg A() and
A6: rng f c= D() and
A7: for y be set st y in Seg A() holds P[y,f.y] from FUNCT_1:sch 5(A2);
  reconsider f as FinSequence by A5,Def2;
  reconsider p=f as FinSequence of D() by A6,Def4;
  take p;
  thus thesis by A5,A7;
end;



reserve D for set,
  p for FinSequence of D,
  m for Nat;

definition
  let m;
  let p be FinSequence;
  func p|m -> FinSequence equals
  p|Seg m;
  coherence by Th15;
end;

definition
  let D,m,p;
  redefine func p|m -> FinSequence of D;
  coherence by Th18;
end;

registration
  let f be FinSequence;
  cluster f|0 -> empty;
  coherence;
end;

theorem
  len q <= i implies q|i = q
proof
  assume len q <= i;
  then Seg len q c= Seg i by Th5;
  then dom q c= Seg i by Def3;
  hence thesis by RELAT_1:68;
end;

theorem
  i <= len q implies len(q|i) = i
proof
  assume i <= len q;
  then Seg i c= Seg(len q) by Th5;
  then Seg i c= dom q by Def3;
  then i in NAT & Seg i = dom(q|i) by ORDINAL1:def 12,RELAT_1:62;
  hence thesis by Def3;
end;



theorem
  i in Seg n implies i+m in Seg (n+m)
proof
  assume
A1: i in Seg n;
  then
A2: 1 <= i by Th1;
A3: i <= n by A1,Th1;
  i <= i+m by NAT_1:11;
  then
A4: 1 <= i+m by A2,XXREAL_0:2;
  i+m <= n+m by A3,XREAL_1:7;
  hence thesis by A4,Th1;
end;

theorem
  i>0 & i+m in Seg (n+m) implies i in Seg n & i in Seg (n+m)
proof
  assume that
A1: i > 0 and
A2: i+m in Seg (n+m);
  1 = 0+1;
  then
A3: 1 <= i by A1,NAT_1:13;
A4: i+m <= n+m by A2,Th1;
  then i <= n by XREAL_1:6;
  hence i in Seg n by A3,Th1;
  i <= i+m by NAT_1:11;
  then i <= n+m by A4,XXREAL_0:2;
  hence thesis by A3,Th1;
end;



definition
  let R be Relation;
  func R[*] -> Relation means
  for x,y being set holds [x,y] in it iff x in field R & y in field R &
  ex p being FinSequence st len p >= 1 & p.1 = x & p.(len p) = y &
  for i being Nat st i >= 1 & i < len p holds [p.i,p.(i+1)] in R;
  existence
  proof
    defpred X[set,set] means
    ex p being FinSequence st len p >= 1 & p.1 = $1 & p.(len p) = $2 &
    for i being Nat st i >= 1 & i < len p holds [p.i,p.(i+1)] in R;
    thus ex S being Relation st for x,y being set holds
    [x,y] in S iff x in field R & y in field R & X[x,y] from RELAT_1:sch 1;
  end;
  uniqueness
  proof
    let R1,R2 be Relation such that
A1: for x,y being set holds [x,y] in R1 iff x in field R & y in field R &
    ex p being FinSequence st len p >= 1 & p.1 = x & p.(len p) = y &
    for i being Nat st i >= 1 & i < len p holds [p.i,p.(i+1)] in R and
A2: for x,y being set holds [x,y] in R2 iff x in field R & y in field R &
    ex p being FinSequence st len p >= 1 & p.1 = x & p.(len p) = y &
    for i being Nat st i >= 1 & i < len p holds [p.i,p.(i+1)] in R;
    let x,y be set;
    thus [x,y] in R1 implies [x,y] in R2
    proof
      assume
A3:   [x,y] in R1;
      then
A4:   x in field R & y in field R by A1;
      ex p being FinSequence st len p >= 1 & p.1 = x & p.(len p) = y & for i
      being Nat st i >= 1 & i < len p holds [p.i,p.(i+1)] in R by A1,A3;
      hence thesis by A2,A4;
    end;
    assume
A5: [x,y] in R2;
    then
A6: x in field R & y in field R by A2;
    ex p being FinSequence st len p >= 1 & p.1 = x & p.(len p) = y & for i
    being Nat st i >= 1 & i < len p holds [p.i,p.(i+1)] in R by A2,A5;
    hence thesis by A1,A6;
  end;
end;

theorem
  for D1,D2 being set st D1 c= D2 holds D1* c= D2*
proof
  let D1,D2 be set such that
A1: D1 c= D2;
  let x be set;
  assume x in D1*;
  then reconsider p = x as FinSequence of D1 by Def11;
  rng p c= D1 by Def4;
  then rng p c= D2 by A1,XBOOLE_1:1;
  then x is FinSequence of D2 by Def4;
  hence thesis by Def11;
end;



registration
  let D be set;
  cluster D* -> functional;
  coherence
  proof
    let x be set;
    thus thesis by Def11;
  end;
end;



theorem
  for p,q being FinSequence st p c= q holds len p <= len q
proof
  let p,q be FinSequence;
  assume p c= q;
  then card p c= card q by CARD_1:11;
  hence thesis by NAT_1:39;
end;

theorem
  for p,q being FinSequence, i being Nat st
  1 <= i & i <= len p holds (p ^ q).i = p.i
proof
  let p,q be FinSequence, i be Nat;
  assume
A1: 1 <= i & i <= len p;
  i in NAT & Seg len p = dom p by Def3,ORDINAL1:def 12;
  then i in dom p by A1;
  hence thesis by Def7;
end;

theorem
  for p,q being FinSequence, i being Nat st
  1 <= i & i <= len q holds (p ^ q).(len p + i) = q.i
proof
  let p,q be FinSequence, i be Nat;
  assume 1 <= i & i <= len q;
  then
  len p + 1 <= len p + i & len p + i <= len p + len q by XREAL_1:6;
  hence (p ^ q).(len p + i) = q.(len p + i - len p) by Th23
    .= q.i;
end;



scheme FinSegRng{m, n() -> Nat, F(set)->set, P[set]}:
  {F(i) where i is Element of NAT: m()<=i & i<=n() & P[i]} is finite
proof
  set F = {F(i) where i is Element of NAT: m()<=i & i<=n() & P[i]};
  deffunc G(Nat) = F($1-1);
  consider f being FinSequence such that
A1: len f = n()+1 and
A2: for k st k in dom f holds f.k = G(k) from SeqLambda;
  F c= rng f
  proof
    let x be set;
    assume x in F;
    then consider j being Element of NAT such that
A3: x=F(j) and m()<=j and
A4: j<=n() and P[j];
    1<=j+1 & j+1<=n()+1 by A4,NAT_1:11,XREAL_1:6;
    then j+1 in Seg (n()+1);
    then
A5: j+1 in dom f by A1,Def3;
    then f.(j+1) = F(j+1-1) by A2
      .= F(j);
    hence thesis by A3,A5,FUNCT_1:def 3;
  end;
  hence thesis;
end;



Lm6: 1 in Seg 3 & 2 in Seg 3;

Lm7: 3 in Seg 3;

Lm8: 1 in Seg 4 & 2 in Seg 4;

Lm9: 3 in Seg 4 & 4 in Seg 4;

Lm10: 1 in Seg 5 & 2 in Seg 5;

Lm11: 3 in Seg 5 & 4 in Seg 5;

Lm12: 5 in Seg 5;

Lm13: 1 in Seg 6 & 2 in Seg 6;

Lm14: 3 in Seg 6 & 4 in Seg 6;

Lm15: 5 in Seg 6 & 6 in Seg 6;

Lm16: 1 in Seg 7 & 2 in Seg 7;

Lm17: 3 in Seg 7 & 4 in Seg 7;

Lm18: 5 in Seg 7 & 6 in Seg 7;

Lm19: 7 in Seg 7;

Lm20: 1 in Seg 8 & 2 in Seg 8;

Lm21: 3 in Seg 8 & 4 in Seg 8;

Lm22: 5 in Seg 8 & 6 in Seg 8;

Lm23: 7 in Seg 8 & 8 in Seg 8;

theorem Th66:
  for x1, x2, x3, x4 being set, p being FinSequence
  st p = <*x1*>^<*x2*>^<*x3*>^<*x4*>
  holds len p = 4 & p.1 = x1 & p.2 = x2 & p.3 = x3 & p.4 = x4
proof
  let x1, x2, x3, x4 be set, p be FinSequence;
  assume
A1: p = <*x1*>^<*x2*>^<*x3*>^<*x4*>;
  set p13 = <*x1*>^<*x2*>^<*x3*>;
A2: p13 = <*x1, x2, x3*>;
  then
A3: len p13 = 3 by Th45;
A4: p13.1 = x1 & p13.2 = x2 by A2,Th45;
A5: p13.3 = x3 by A2,Th45;
  thus len p = len p13 + len <*x4*> by A1,Th22
    .= 3 + 1 by A3,Th40
    .= 4;
  dom p13 = Seg 3 by A3,Def3;
  hence p.1 = x1 & p.2 = x2 & p.3 = x3 by A1,A4,A5,Def7,Lm6,Lm7;
  thus p.4 = p.(len p13 + 1) by A3
    .= x4 by A1,Th42;
end;

theorem Th67:
  for x1, x2, x3, x4, x5 being set, p being FinSequence
  st p = <*x1*>^<*x2*>^<*x3*>^<*x4*>^<*x5*>
  holds len p = 5 & p.1 = x1 & p.2 = x2 & p.3 = x3 & p.4 = x4 & p.5 = x5
proof
  let x1, x2, x3, x4, x5 be set, p be FinSequence;
  assume
A1: p = <*x1*>^<*x2*>^<*x3*>^<*x4*>^<*x5*>;
  set p14 = <*x1*>^<*x2*>^<*x3*>^<*x4*>;
A2: len p14 = 4 by Th66;
A3: p14.1 = x1 & p14.2 = x2 by Th66;
A4: p14.3 = x3 & p14.4 = x4 by Th66;
  thus len p = len p14 + len <*x5*> by A1,Th22
    .= 4 + 1 by A2,Th40
    .= 5;
  dom p14 = Seg 4 by A2,Def3;
  hence p.1 = x1 & p.2 = x2 & p.3 = x3 & p.4 = x4 by A1,A3,A4,Def7,Lm8,Lm9;
  thus p.5 = p.(len p14 + 1) by A2
    .= x5 by A1,Th42;
end;

theorem Th68:
  for x1, x2, x3, x4, x5, x6 being set, p being FinSequence
  st p = <*x1*>^<*x2*>^<*x3*>^<*x4*>^<*x5*>^<*x6*>
  holds len p = 6 & p.1 = x1 & p.2 = x2 & p.3 = x3 & p.4 = x4 & p.5 = x5 &
  p.6 = x6
proof
  let x1, x2, x3, x4, x5, x6 be set, p be FinSequence;
  assume
A1: p = <*x1*>^<*x2*>^<*x3*>^<*x4*>^<*x5*>^<*x6*>;
  set p15 = <*x1*>^<*x2*>^<*x3*>^<*x4*>^<*x5*>;
A2: len p15 = 5 by Th67;
A3: p15.1 = x1 & p15.2 = x2 by Th67;
A4: p15.3 = x3 & p15.4 = x4 by Th67;
A5: p15.5 = x5 by Th67;
  thus len p = len p15 + len <*x6*> by A1,Th22
    .= 5 + 1 by A2,Th40
    .= 6;
  dom p15 = Seg 5 by A2,Def3;
  hence p.1 = x1 & p.2 = x2 & p.3 = x3 & p.4 = x4 & p.5 = x5
  by A1,A3,A4,A5,Def7,Lm10,Lm11,Lm12;
  thus p.6 = p.(len p15 + 1) by A2
    .= x6 by A1,Th42;
end;

theorem Th69:
  for x1, x2, x3, x4, x5, x6, x7 being set, p being FinSequence
  st p = <*x1*>^<*x2*>^<*x3*>^<*x4*>^<*x5*>^<*x6*>^<*x7*>
  holds len p = 7 & p.1 = x1 & p.2 = x2 & p.3 = x3 & p.4 = x4 & p.5 = x5 &
  p.6 = x6 & p.7 = x7
proof
  let x1, x2, x3, x4, x5, x6, x7 be set, p be FinSequence;
  assume
A1: p = <*x1*>^<*x2*>^<*x3*>^<*x4*>^<*x5*>^<*x6*>^<*x7*>;
  set p16 = <*x1*>^<*x2*>^<*x3*>^<*x4*>^<*x5*>^<*x6*>;
A2: len p16 = 6 by Th68;
A3: p16.1 = x1 & p16.2 = x2 by Th68;
A4: p16.3 = x3 & p16.4 = x4 by Th68;
A5: p16.5 = x5 & p16.6 = x6 by Th68;
  thus len p = len p16 + len <*x7*> by A1,Th22
    .= 6 + 1 by A2,Th40
    .= 7;
  dom p16 = Seg 6 by A2,Def3;
  hence p.1 = x1 & p.2 = x2 & p.3 = x3 & p.4 = x4 & p.5 = x5 & p.6 = x6
  by A1,A3,A4,A5,Def7,Lm13,Lm14,Lm15;
  thus p.7 = p.(len p16 + 1) by A2
    .= x7 by A1,Th42;
end;

theorem Th70:
  for x1,x2,x3,x4, x5, x6, x7, x8 being set, p being FinSequence
  st p = <*x1*>^<*x2*>^<*x3*>^<*x4*>^<*x5*>^<*x6*>^<*x7*>^<*x8*>
  holds len p = 8 & p.1 = x1 & p.2 = x2 & p.3 = x3 & p.4 = x4 & p.5 = x5 &
  p.6 = x6 & p.7 = x7 & p.8 = x8
proof
  let x1, x2, x3, x4, x5, x6, x7, x8 be set, p be FinSequence;
  assume
A1: p = <*x1*>^<*x2*>^<*x3*>^<*x4*>^<*x5*>^<*x6*>^<*x7*>^<*x8*>;
  set p17 = <*x1*>^<*x2*>^<*x3*>^<*x4*>^<*x5*>^<*x6*>^<*x7*>;
A2: len p17 = 7 by Th69;
A3: p17.1 = x1 & p17.2 = x2 by Th69;
A4: p17.3 = x3 & p17.4 = x4 by Th69;
A5: p17.5 = x5 & p17.6 = x6 by Th69;
A6: p17.7 = x7 by Th69;
  thus len p = len p17 + len <*x8*> by A1,Th22
    .= 7 + 1 by A2,Th40
    .= 8;
  dom p17 = Seg 7 by A2,Def3;
  hence p.1 = x1 & p.2 = x2 & p.3 = x3 & p.4 = x4 & p.5 = x5 & p.6 = x6 &
  p.7 = x7 by A1,A3,A4,A5,A6,Def7,Lm16,Lm17,Lm18,Lm19;
  thus p.8 = p.(len p17 + 1) by A2
    .= x8 by A1,Th42;
end;

theorem
  for x1,x2,x3,x4,x5,x6,x7, x8, x9 being set, p being FinSequence
  st p = <*x1*>^<*x2*>^<*x3*>^<*x4*>^<*x5*>^<*x6*>^<*x7*>^<*x8*>^<*x9*>
  holds len p = 9 & p.1 = x1 & p.2 = x2 & p.3 = x3 & p.4 = x4 & p.5 = x5 &
  p.6 = x6 & p.7 = x7 & p.8 = x8 & p.9 = x9
proof
  let x1, x2, x3, x4, x5, x6, x7, x8, x9 be set, p be FinSequence;
  assume
A1: p = <*x1*>^<*x2*>^<*x3*>^<*x4*>^<*x5*>^<*x6*>^<*x7*>^<*x8*>^<*x9*>;
  set p17 = <*x1*>^<*x2*>^<*x3*>^<*x4*>^<*x5*>^<*x6*>^<*x7*>^<*x8*>;
A2: len p17 = 8 by Th70;
A3: p17.1 = x1 & p17.2 = x2 by Th70;
A4: p17.3 = x3 & p17.4 = x4 by Th70;
A5: p17.5 = x5 & p17.6 = x6 by Th70;
A6: p17.7 = x7 & p17.8 = x8 by Th70;
  thus len p = len p17 + len <*x9*> by A1,Th22
    .= 8 + 1 by A2,Th40
    .= 9;
  dom p17 = Seg 8 by A2,Def3;
  hence p.1 = x1 & p.2 = x2 & p.3 = x3 & p.4 = x4 & p.5 = x5 & p.6 = x6 &
  p.7 = x7 & p.8 = x8 by A1,A3,A4,A5,A6,Def7,Lm20,Lm21,Lm22,Lm23;
  thus p.9 = p.(len p17 + 1) by A2
    .= x9 by A1,Th42;
end;



theorem
  for p being FinSequence holds p | Seg 0 = {};

theorem
  for f,g being FinSequence holds f | Seg 0 = g | Seg 0;

theorem
  for D being non empty set, x being Element of D holds
  <* x *> is FinSequence of D;

theorem
  for D being set, p,q being FinSequence of D holds p ^ q is FinSequence of D;



reserve a, b, c, d, e, f for set;

theorem
  <*a*> = <*b*> implies a = b
proof
  assume
A1: <*a*> = <*b*>;
  thus a = <*a*>.1 by Def8
    .= b by A1,Def8;
end;

theorem
  <*a,b*> = <*c,d*> implies a = c & b = d
proof
  assume
A1: <*a,b*> = <*c,d*>;
  thus a = <*a,b*>.1 by Th44
    .= c by A1,Th44;
  thus b = <*a,b*>.2 by Th44
    .= d by A1,Th44;
end;

theorem
  <*a,b,c*> = <*d,e,f*> implies a = d & b = e & c = f
proof
  assume
A1: <*a,b,c*> = <*d,e,f*>;
  thus a = <*a,b,c*>.1 by Th45
    .= d by A1,Th45;
  thus b = <*a,b,c*>.2 by Th45
    .= e by A1,Th45;
  thus c = <*a,b,c*>.3 by Th45
    .= f by A1,Th45;
end;



registration
  cluster non empty non-empty for FinSequence;
  existence
  proof
    take <*{{}}*>;
    thus <*{{}}*> is non empty;
    assume {} in rng <*{{}}*>;
    then {} in {{{}}} by Th38;
    hence contradiction by TARSKI:def 1;
  end;
end;

theorem Th79:
  for p,q being FinSequence st q = p|Seg n holds len q <= len p
proof
  let p,q be FinSequence;
  assume q = p|Seg n;
  then
A1: dom q = dom p /\ Seg n by RELAT_1:61;
  Seg len q = dom q & Seg len p = dom p by Def3;
  then Seg len q c= Seg len p by A1,XBOOLE_1:17;
  hence thesis by Th5;
end;

theorem
  for p,r being FinSequence st r = p|Seg n ex q being FinSequence st p = r^q
proof
  let p,r be FinSequence;
  assume
A1: r = p|Seg n;
  then consider m be Nat such that
A2: len p = len r + m by Th79,NAT_1:10;
  deffunc U(Nat) = p.(len r + $1);
  consider q being FinSequence such that
A3: len q = m & for k st k in dom q holds q.k = U(k) from SeqLambda;
  take q;
A4: len p = len(r^q) by A2,A3,Th22;
  now
    let k such that
A5: 1 <= k and
A6: k <= len p;
A7: k in NAT by ORDINAL1:def 12;
A8: now
      assume k <= len r;
      then
A9:   k in Seg len r by A5,A7;
A10:  dom r = Seg len r by Def3;
      then (r^q).k = r.k by A9,Def7;
      hence p.k = (r^q).k by A1,A9,A10,FUNCT_1:47;
    end;
    now
      assume
A11:  not k <= len r;
      then consider j being Nat such that
A12:  k = len r + j by NAT_1:10;
      k-len r = j by A12;
      then
A13:  (r^q).k = q.j by A4,A6,A11,Th24;
      j <> 0 by A11,A12;
      then
A14:  0+1 <= j by NAT_1:13;
      j <= len q by A2,A3,A6,A12,XREAL_1:6;
      then j in Seg m by A3,A14,Th1;
      then j in dom q by A3,Def3;
      hence p.k = (r^q).k by A3,A12,A13;
    end;
    hence p.k = (r^q).k by A8;
  end;
  hence thesis by A4,Th14;
end;



registration
  let D be non empty set;
  cluster non empty for FinSequence of D;
  existence
  proof set x = the Element of D;
    take <*x*>;
    thus thesis;
  end;
end;



definition
  let p,q be FinSequence;
  redefine pred p = q means
  len p = len q & for k st 1 <= k & k <= len p holds p.k = q.k;
  compatibility by Th14;
end;



theorem
  for M1, M2 being set st card M1 = 0 & card M2 = 0 holds M1 = M2
proof
  let M1, M2 be set;
  assume that
A1: card M1 = {} and
A2: card M2 = {};
  M1 = {} by A1;
  hence thesis by A2;
end;



registration
  let n be non zero Nat;
  cluster Seg n -> non empty;
  coherence;
end;



theorem
  for p being FinSequence, n,m being Nat st m <= n holds p|n|m = p|m
proof
  let p be FinSequence, n,m be Nat;
  assume m <= n;
  then Seg m c= Seg n by Th5;
  hence thesis by RELAT_1:74;
end;



reserve m for Element of NAT;

theorem
  Seg n = (n+1) \ {0}
proof
A1: n+1 = {m: m < n+1} by AXIOMS:4;
  thus Seg n c= (n+1) \ {0}
  proof
    let x;
    assume x in Seg n;
    then consider k being Element of NAT such that
A2: x = k and
A3: 1 <= k and
A4: k <= n;
    k < n+1 by A4,NAT_1:13;
    then
A5: x in n+1 by A1,A2;
    not x in {0} by A2,A3,TARSKI:def 1;
    hence thesis by A5,XBOOLE_0:def 5;
  end;
  let x;
  assume
A6: x in (n+1) \ {0};
  then
A7: x in n+1;
A8: not x in {0} by A6,XBOOLE_0:def 5;
  consider m such that
A9: x = m and
A10: m < n+1 by A1,A7;
A11: x <> 0 by A8,TARSKI:def 1;
  0+1 = 1;
  then
A12: 1 <= m by A9,A11,NAT_1:13;
  m <= n by A10,NAT_1:13;
  hence thesis by A9,A12;
end;



registration
  let n be Nat;
  cluster n-element for FinSequence;
  existence
  proof
    set p = Seg n --> {};
    dom p = Seg n by FUNCOP_1:13;
    then reconsider p as FinSequence by Def2;
    take p;
    Seg len p = dom p by Def3;
    hence len p = n by Th6,FUNCOP_1:13;
  end;
end;



registration
  let x be set;
  cluster <*x*> -> 1-element;
  coherence;
  let y be set;
  cluster <*x,y*> -> 2-element;
  coherence
  proof
    thus len<*x,y*> = 2 by Th44;
  end;
  let z be set;
  cluster <*x,y,z*> -> 3-element;
  coherence
  proof
   thus len<*x,y,z*> = 3 by Th45;
  end;
end;



definition let X be set;
 attr X is FinSequence-membered means
:Def18: x in X implies x is FinSequence;
end;

registration
 cluster empty -> FinSequence-membered for set;
 coherence
  proof let X be set such that
A1: X is empty;
   let x;
   assume x in X;
   hence thesis by A1;
  end;
end;

registration
 cluster non empty FinSequence-membered for set;
 existence
  proof
   take X = {the FinSequence};
   thus X is non empty;
   let x;
   thus thesis by TARSKI:def 1;
  end;
end;

registration let X be set;
 cluster X* -> FinSequence-membered;
 coherence
  proof
   let x;
   assume x in X*;
   hence x is FinSequence by Def11;
  end;
end;

theorem
 for f being Function st f in D* & x in dom f holds f.x in D
proof let f be Function such that
A1: f in D* and
A2: x in dom f;
  f is FinSequence of D by A1,Def11;
  then
A3:   rng f c= D by Def4;
  f.x in rng f by A2,FUNCT_1:3;
 hence f.x in D by A3;
end;

registration
 cluster FinSequence-membered -> functional for set;
 coherence
  proof let X such that
A1:  X is FinSequence-membered;
   let x be set;
   assume x in X;
   hence x is Function by A1,Def18;
  end;
end;

theorem
 for X being FinSequence-membered set ex Y being non empty set st X c= Y*
proof
 let X be FinSequence-membered set;
  set Z = {rng f where f is Element of X: f in X};
 take Y = succ union Z;
 let x;
 assume
A1: x in X;
  then reconsider x as FinSequence by Def18;
  rng x in {rng f where f is Element of X: f in X} by A1;
  then rng x c= Y by ORDINAL3:1,SETFAM_1:41;
  then x is FinSequence of Y by Def4;
 hence thesis by Def11;
end;

registration
  let X be FinSequence-membered set;
  cluster -> FinSequence-like for Element of X;
  coherence
  proof
    let e be Element of X;
    X is empty or X is non empty;
    hence thesis by Def18,SUBSET_1:def 1;
  end;
end;

registration let X be FinSequence-membered set;
 cluster -> FinSequence-membered for Subset of X;
 coherence
  proof let Y be Subset of X;
   let x;
   assume x in Y;
   hence x is FinSequence;
  end;
end;



theorem
  for p,q being FinSequence st q = p|Seg n holds len q <= n
proof
  let p,q be FinSequence;
  assume q = p|Seg n;
then A1: dom q = dom p /\ Seg n by RELAT_1:61;
 Seg len q = dom q by Def3;
then  Seg len q c= Seg n by A1,XBOOLE_1:17;
  hence thesis by Th5;
end;

theorem
  for p,q being FinSequence st p = p^q or p = q^p holds q = {}
proof
  let p,q be FinSequence such that
A1: p = p^q or p = q^p;
   len(p^q) = len p + len q & len(q^p) = len q + len p by Th22;
  hence thesis by A1;
end;

theorem
  for p,q being FinSequence st p^q = <*x*> holds
  p = <*x*> & q = {} or p = {} & q = <*x*>
proof
  let p,q be FinSequence;
  assume
A1: p^q = <*x*>;
then A2: 1 = len(p^q) by Th40
    .= len p + len q by Th22;
A3: now
    assume len p = 0;
    thus
then  p = {};
    hence q = <*x*> by A1,Th34;
  end;
 now
    assume len p <> 0;
then A4: 0+1 <= len p by NAT_1:13;
 len p <= 1 by A2,NAT_1:11;
then  len p = 1 by A4,XXREAL_0:1;
    thus
then  q = {} by A2;
    hence p = <*x*> by A1,Th34;
  end;
  hence thesis by A3;
end;



theorem
 for f being n-element FinSequence holds dom f = Seg n
  proof let f be n-element FinSequence;
    len f = n by CARD_1:def 7;
   hence dom f = Seg n by Def3;
  end;

registration let n,m;
 let f be n-element FinSequence, g be m-element FinSequence;
 cluster f^g -> (n+m)-element;
 coherence
  proof
   len f = n & len g = m by CARD_1:def 7;
   hence len(f^g) = n+m by Th22;
  end;
end;



registration
  cluster increasing -> one-to-one for real-valued FinSequence;
  coherence
  proof
    let f be real-valued FinSequence;
    assume
A1: f is increasing;
    let x,y be set;
    assume that
A2: x in dom f & y in dom f and
A3: f.x=f.y;
    reconsider nx=x,ny=y as Element of NAT by A2;
    nx<=ny & nx>=ny by A1,A2,A3,VALUED_0:def 13;
    hence thesis by XXREAL_0:1;
  end;
end;



theorem
 for x, y being set st x in dom <*y*> holds x = 1
proof
  let x, y be set;
  assume x in dom <*y*>;
  then x in Seg 1 by Th38;
  hence thesis by Th2,TARSKI:def 1;
end;



registration let X;
 cluster X-valued for FinSequence;
 existence
  proof
   take <*>X;
   thus thesis;
  end;
end;



registration
  let D be FinSequence-membered set;
  let f be D-valued Function;
  let x be set;
  cluster f.x -> FinSequence-like;
  coherence
  proof
    per cases;
    suppose x in dom f;
      then
A1:   f.x in rng f by FUNCT_1:def 3;
      rng f c= D by RELAT_1:def 19;
      hence thesis by A1;
    end;
    suppose not x in dom f;
      hence thesis by FUNCT_1:def 2;
    end;
  end;
end;














begin

reserve n,m,k for Element of NAT,
  D for non empty set,
  Z,x,y,z,y1,y2 for set,
  p,q for FinSequence;

Lm1: for p being FinSequence of D st 1 <= n & n <= len p holds p.n is Element
of D
proof
  let p be FinSequence of D;
  assume 1 <= n & n <= len p;
  then n in Seg len p by FINSEQ_1:1;
  then n in dom p by FINSEQ_1:def 3;
  then
A1: p.n in rng p by FUNCT_1:def 3;
  rng p c= D by FINSEQ_1:def 4;
  hence thesis by A1;
end;

definition
  let p be natural-valued Function;
  let n be set;
  redefine func p.n -> Element of NAT;
  coherence by ORDINAL1:def 12;
end;



scheme
  RecEx{A() -> set,P[set,set,set]}: ex f being Function st dom f = NAT & f.0 =
  A() & for n being Element of NAT holds P[n,f.n,f.(n+1)]
provided
A1: for n being Element of NAT for x being set ex y being set st P[n,x,y ]
proof
  defpred Q[set,set,set] means ex O2 being Ordinal st O2 = $3 & (for X being
set, n being Element of NAT st X c= Rank the_rank_of $2 ex Y being set st Y c=
Rank O2 & P[n,X,Y]) & (for O being Ordinal st for X being set, n being Element
  of NAT st X c= Rank the_rank_of $2 ex Y being set st Y c= Rank O & P[n,X,Y]
  holds O2 c= O);
A2: for n being Element of NAT for x being set ex y being set st Q[n,x,y]
  proof
    defpred Y[set,set] means for m being Element of NAT ex y being set st $2
    is Ordinal & y c= Rank the_rank_of $2 & P[m,$1,y];
    let n be Element of NAT;
    let x be set;
    defpred X[Ordinal] means for X being set, n being Element of NAT st X c=
    Rank the_rank_of x ex Y being set st Y c= Rank $1 & P[n,X,Y];
A3: for x9 being set st x9 in bool Rank the_rank_of x ex o being set st Y[
    x9,o]
    proof
      let x9 be set;
      assume x9 in bool Rank the_rank_of x;
      defpred X[set,set] means ex y being set st $2 is Ordinal & y c= Rank
      the_rank_of $2 & P[$1,x9,y];
A4:   for e being set st e in NAT ex u being set st X[e,u]
      proof
        let i be set;
        assume i in NAT;
        then reconsider i9 = i as Element of NAT;
        thus ex o being set, y being set st o is Ordinal & y c= Rank
        the_rank_of o & P[i,x9,y]
        proof
          consider y being set such that
A5:       P[i9,x9,y] by A1;
          take o = the_rank_of y, y;
          thus o is Ordinal;
          the_rank_of the_rank_of y = the_rank_of y by CLASSES1:73;
          hence y c= Rank the_rank_of o by CLASSES1:65;
          thus thesis by A5;
        end;
      end;
      consider h being Function such that
A6:   dom h = NAT and
A7:   for i being set st i in NAT holds X[i,h.i] from CLASSES1:sch 1(
      A4);
      take o = sup rng h;
      thus for m being Element of NAT ex y being set st o is Ordinal & y c=
      Rank the_rank_of o & P[m,x9,y]
      proof
        let m be Element of NAT;
        consider y being set such that
A8:     h.m is Ordinal and
A9:     y c= Rank the_rank_of (h.m) and
A10:    P[m,x9,y] by A7;
        reconsider O = h.m as Ordinal by A8;
        h.m in rng h by A6,FUNCT_1:def 3;
        then h.m in sup rng h by A8,ORDINAL2:19;
        then h.m c= o by ORDINAL1:def 2;
        then
A11:    Rank O c= Rank o by CLASSES1:37;
        take y;
        thus o is Ordinal;
        y c= Rank O by A9,CLASSES1:73;
        then y c= Rank o by A11,XBOOLE_1:1;
        hence y c= Rank the_rank_of o by CLASSES1:73;
        thus thesis by A10;
      end;
    end;
    consider f being Function such that
A12: dom f = bool Rank the_rank_of x and
A13: for x9 being set st x9 in bool Rank the_rank_of x holds Y[x9,f.x9
    ] from CLASSES1:sch 1(A3);
A14: ex O being Ordinal st X[O]
    proof
      take O2 = sup rng f;
      thus for X being set, m being Element of NAT st X c= Rank the_rank_of x
      ex Y being set st Y c= Rank O2 & P[m,X,Y]
      proof
        let X be set;
        let m be Element of NAT;
        assume
A15:    X c= Rank the_rank_of x;
        then consider Y being set such that
A16:    f.X is Ordinal and
A17:    Y c= Rank the_rank_of (f.X) and
A18:    P[m,X,Y] by A13;
        reconsider O = f.X as Ordinal by A16;
        f.X in rng f by A12,A15,FUNCT_1:def 3;
        then f.X in sup rng f by A16,ORDINAL2:19;
        then f.X c= O2 by ORDINAL1:def 2;
        then
A19:    Rank O c= Rank O2 by CLASSES1:37;
        take Y;
        the_rank_of O = O by CLASSES1:73;
        hence Y c= Rank O2 by A17,A19,XBOOLE_1:1;
        thus thesis by A18;
      end;
    end;
    consider O2 being Ordinal such that
A20: X[O2] & for O being Ordinal st X[O] holds O2 c= O from ORDINAL1:
    sch 1(A14);
    take O2;
    thus thesis by A20;
  end;
A21: for n being Element of NAT for x,y1,y2 being set st Q[n,x,y1] & Q[n,x,
  y2] holds y1 = y2
  proof
    let n be Element of NAT;
    let x,y1,y2 be set;
    assume that
A22: Q[n,x,y1] and
A23: Q[n,x,y2];
    consider O2 being Ordinal such that
A24: O2 = y2 and
A25: ( for X being set, n being Element of NAT st X c= Rank
the_rank_of x ex Y being set st Y c= Rank O2 & P[n,X,Y])& for O being Ordinal
    st for X being set, n being Element of NAT st X c= Rank the_rank_of x ex Y
    being set st Y c= Rank O & P[n,X,Y] holds O2 c= O by A23;
    consider O1 being Ordinal such that
A26: O1 = y1 and
A27: ( for X being set, n being Element of NAT st X c= Rank
the_rank_of x ex Y being set st Y c= Rank O1 & P[n,X,Y])& for O being Ordinal
    st for X being set, n being Element of NAT st X c= Rank the_rank_of x ex Y
    being set st Y c= Rank O & P[n,X,Y] holds O1 c= O by A22;
    O1 c= O2 & O2 c= O1 by A27,A25;
    hence thesis by A26,A24,XBOOLE_0:def 10;
  end;
  ex f being Function st dom f = NAT & f.0 = the_rank_of A() & for n
  being Element of NAT holds Q[n,f.n,f.(n+1)]
  proof
    deffunc S(Element of NAT) = {k : k <= $1};
A28: for p,q being Function,k st dom p = S(k) & dom q = S(k+1) & p.0 = q.0
    & for n st n < k holds Q[n,p.n,p.(n+1)] & Q[n,q.n,q.(n+1)] holds p.k = q.k
    proof
      let p,q be Function;
      let k;
      defpred Z[Element of NAT] means $1 <= k implies p.$1 = q.$1;
      assume that
      dom p = S(k) and
      dom q = S(k+1) and
A29:  p.0 = q.0 and
A30:  for n st n < k holds Q[n,p.n,p.(n+1)] & Q[n,q.n,q.(n+1)];
A31:  for n st Z[n] holds Z[n+1]
      proof
        let n;
        assume
A32:    n <= k implies p.n = q.n;
        assume n+1 <= k;
        then
A33:    n < k by NAT_1:13;
        then
A34:    Q[n,p.n,p.(n+1)] by A30;
        Q[n,p.n,q.(n+1)] by A30,A32,A33;
        hence thesis by A21,A34;
      end;
A35:  Z[0] by A29;
      for n holds Z[n] from NAT_1:sch 1(A35,A31);
      hence thesis;
    end;
A36: for n ex p being Function st dom p = S(n) & p.0 = the_rank_of A() &
    for k st k < n holds Q[k,p.k,p.(k+1)]
    proof
      defpred Z[Element of NAT] means ex p being Function st dom p = S($1) & p
      .0 = the_rank_of A() & for k st k < $1 holds Q[k,p.k,p.(k+1)];
A37:  for n st Z[n] holds Z[n+1]
      proof
        let n;
        given p being Function such that
        dom p = S(n) and
A38:    p.0 = the_rank_of A() and
A39:    for k st k < n holds Q[k,p.k,p.(k+1)];
        consider z such that
A40:    Q[n,p.n,z] by A2;
        defpred ST[set,set] means (for k st k=$1 holds (k <= n implies $2 = p.
        k) & (k=n+1 implies $2 = z ));
A41:    for x st x in S(n+1) ex y st ST[x,y]
        proof
          let x;
          assume x in S(n+1);
          then
A42:      ex m st m=x & m <= n+1;
          then reconsider t=x as Element of NAT;
A43:      t <= n implies thesis
          proof
            assume
A44:        t <= n;
            take p.x;
            let k;
            assume
A45:        k=x;
            hence k <= n implies p.x = p.k;
            assume k=n+1;
            then n+1 <= n+0 by A44,A45;
            hence thesis by XREAL_1:6;
          end;
          t = n+1 implies thesis
          proof
            assume
A46:        t=n+1;
            take z;
            let k such that
A47:        k=x;
            thus k <= n implies z=p.k
            proof
              assume k <= n;
              then n+1 <= n+0 by A46,A47;
              hence thesis by XREAL_1:6;
            end;
            thus thesis;
          end;
          hence thesis by A42,A43,NAT_1:8;
        end;
        consider q being Function such that
A48:    dom q = S(n+1) & for x st x in S(n+1) holds ST[x,q.x] from
        CLASSES1:sch 1 (A41);
        take q;
        thus dom q = S(n+1) by A48;
        0 in S(n+1);
        hence q.0 = the_rank_of A() by A38,A48;
        let k such that
A49:    k < n+1;
A50:    now
A51:      k+1 <= n+1 by A49,NAT_1:13;
          assume k <> n;
          then k+1 <> n+1;
          then
A52:      k+1 <= n by A51,NAT_1:8;
          then
A53:      k < n by NAT_1:13;
          k+1 in S(n+1) by A51;
          then
A54:      q.(k+1) = p.(k+1) by A48,A52;
          k in S(n+1) by A49;
          then p.k = q.k by A48,A53;
          hence thesis by A39,A54,A53;
        end;
        now
          assume
A55:      k = n;
          then k <= n+1 by NAT_1:11;
          then k in S(n+1);
          then
A56:      q.k = p.k by A48,A55;
          k+1 in S(n+1) by A55;
          then q.(k+1) = z by A48,A55;
          hence thesis by A40,A55,A56;
        end;
        hence thesis by A50;
      end;
A57:  Z[0]
      proof
        set s = S(0) --> the_rank_of A();
        take s;
        thus dom s = S(0) by FUNCOP_1:13;
        0 in S(0);
        hence s.0 = the_rank_of A() by FUNCOP_1:7;
        thus thesis;
      end;
      thus for n holds Z[n] from NAT_1:sch 1(A57,A37);
    end;
    ex f being Function st dom f = NAT & for n ex p being Function st dom
p = S(n) & p.0 = the_rank_of A() & (for k st k < n holds Q[k,p.k,p.(k+1)]) & f.
    n = p.n
    proof
      defpred Z[set,set] means for n st n=$1 ex p being Function st dom p = S(
n) & p.0 = the_rank_of A() & (for k st k < n holds Q[k,p.k,p.(k+1)]) & $2 = p.n
      ;
A58:  for x st x in NAT ex y st Z[x,y]
      proof
        let x;
        assume x in NAT;
        then reconsider n=x as Element of NAT;
        consider p being Function such that
A59:    dom p = S(n) & p.0 = the_rank_of A() & for k st k < n holds Q
        [k,p.k,p.(k +1)] by A36;
        take p.n;
        let k such that
A60:    k = x;
        take p;
        thus thesis by A59,A60;
      end;
      consider f being Function such that
A61:  dom f = NAT & for x st x in NAT holds Z[x,f.x] from CLASSES1:
      sch 1 (A58);
      take f;
      thus dom f = NAT by A61;
      let n;
      consider p being Function such that
A62:  dom p = S(n) & p.0 = the_rank_of A() and
A63:  for k st k < n holds Q[k,p.k,p.(k+1)] and
A64:  f.n = p.n by A61;
      take p;
      thus dom p = S(n) & p.0 = the_rank_of A() by A62;
      thus for k st k < n holds Q[k,p.k,p.(k+1)] by A63;
      thus thesis by A64;
    end;
    then consider f being Function such that
A65: dom f = NAT and
A66: for n ex p being Function st dom p = S(n) & p.0 = the_rank_of A(
    ) & (for k st k < n holds Q[k,p.k,p.(k+1)]) & f.n = p.n;
    take f;
    thus dom f = NAT by A65;
    ex p being Function st dom p = S(0) & p.0 = the_rank_of A() & (for k
    st k < 0 holds Q[k,p.k,p.(k+1)]) & f.0 = p.0 by A66;
    hence f.0 = the_rank_of A();
    let d be Element of NAT;
    consider p being Function such that
A67: dom p = S(d+1) and
A68: p.0 = the_rank_of A() and
A69: for k st k < d+1 holds Q[k,p.k,p.(k+1)] and
A70: f.(d+1) = p.(d+1) by A66;
    consider q being Function such that
A71: dom q = S(d) and
A72: q.0 = the_rank_of A() and
A73: for k st k < d holds Q[k,q.k,q.(k+1)] and
A74: f.d = q.d by A66;
    dom p = S(d+1) & dom q = S(d) & p.0 = q.0 & for k st k < d holds Q[k
    ,q.k,q.(k+1)] & Q[k,p.k,p.(k+1)]
    proof
      thus dom p = S(d+1) by A67;
      thus dom q = S(d) by A71;
      thus p.0 = q.0 by A68,A72;
      let k;
      assume
A75:  k < d;
      hence Q[k,q.k,q.(k+1)] by A73;
      d <= d+1 by NAT_1:11;
      then k < d+1 by A75,XXREAL_0:2;
      hence thesis by A69;
    end;
    then p.d = q.d by A28;
    hence thesis by A69,A70,A74,XREAL_1:29;
  end;
  then consider g being Function such that
A76: dom g = NAT and
A77: g.0 = the_rank_of A() and
A78: for n being Element of NAT holds Q[n,g.n,g.(n+1)];
  defpred X[set,set] means ex i being Element of NAT st i = $1`2 & P[$1`2,$1`1
  ,$2`1] & $2`2 = i+1 & not ex y being set st P[$1`2,$1`1,y] & the_rank_of y in
  the_rank_of $2`1;
A79: [A(),0]`1 = A() & [A(),0]`2 = 0;
  set beta = sup rng g;
A80: for x being set st x in [:Rank beta, NAT:] ex u being set st X[x,u]
  proof
    let x be set;
    defpred X[Ordinal] means ex y being set st y c= Rank $1 & P[x`2,x`1,y];
    assume x in [:Rank beta, NAT:];
    then consider a,b being set such that
A81: a in Rank beta and
A82: b in NAT and
A83: x = [a,b] by ZFMISC_1:def 2;
    reconsider b as Element of NAT by A82;
A84: x`2 = b & x`1 = a by A83,MCART_1:7;
    the_rank_of a in beta by A81,CLASSES1:66;
    then consider alfa being Ordinal such that
A85: alfa in rng g and
A86: the_rank_of a c= alfa by ORDINAL2:21;
    consider k being set such that
A87: k in dom g and
A88: alfa = g.k by A85,FUNCT_1:def 3;
    reconsider k as Element of NAT by A76,A87;
A89: Q[k,alfa,g.(k+1)] by A78,A88;
    then reconsider O2 = g.(k+1) as Ordinal;
    a c= Rank alfa by A86,CLASSES1:65;
    then a c= Rank the_rank_of alfa by CLASSES1:73;
    then ex y being set st y c= Rank O2 & P[x`2,x`1,y] by A89,A84;
    then
A90: ex O being Ordinal st X[O];
    consider O being Ordinal such that
A91: X[O] and
A92: for O9 being Ordinal st X[O9] holds O c= O9 from ORDINAL1:sch 1(
    A90);
    consider Y being set such that
A93: Y c= Rank O and
A94: P[b,a,Y] by A84,A91;
A95: the_rank_of Y c= O by A93,CLASSES1:65;
    take u = [Y,b+1], i = b;
    thus i = x`2 by A83,MCART_1:7;
    thus P[x`2,x`1,u`1] by A84,A94,MCART_1:7;
    thus u`2 = i+1 by MCART_1:7;
    given y being set such that
A96: P[x`2,x`1,y] and
A97: the_rank_of y in the_rank_of u`1;
A98: y c= Rank the_rank_of y by CLASSES1:def 8;
    the_rank_of y in the_rank_of Y by A97,MCART_1:7;
    hence contradiction by A92,A96,A95,A98,ORDINAL1:5;
  end;
  consider F being Function such that
  dom F = [:Rank beta, NAT:] and
A99: for x being set st x in [:Rank beta, NAT:] holds X[x,F.x] from
  CLASSES1:sch 1(A80);
  deffunc U(Nat,set) = (F.[$2,$1])`1;
  consider f being Function such that
A100: dom f = NAT and
A101: f.0 = A() and
A102: for n being Nat holds f.(n+1) = U(n,f.n) from NAT_1:sch 11;
  defpred X[Element of NAT] means the_rank_of (f.$1) in sup rng g;
  g.0 in rng g by A76,FUNCT_1:def 3;
  then
A103: X[0] by A77,A101,ORDINAL2:19;
A104: for n being Element of NAT st X[n] holds X[n+1]
  proof
    let n be Element of NAT;
    assume
A105: the_rank_of (f.n) in sup rng g;
    then consider o1 being Ordinal such that
A106: o1 in rng g and
A107: the_rank_of (f.n) c= o1 by ORDINAL2:21;
    f.n in Rank sup rng g by A105,CLASSES1:66;
    then
A108: [f.n,n] in [:Rank beta, NAT:] by ZFMISC_1:87;
    consider m being set such that
A109: m in dom g and
A110: g.m = o1 by A106,FUNCT_1:def 3;
    reconsider m as Element of NAT by A76,A109;
    consider o2 being Ordinal such that
A111: o2 = g.(m+1) and
A112: for X being set, n being Element of NAT st X c= Rank the_rank_of
    (g.m) ex Y being set st Y c= Rank o2 & P[n,X,Y] and
    for o being Ordinal st for X being set, n being Element of NAT st X c=
Rank the_rank_of (g.m) ex Y being set st Y c= Rank o & P[n,X,Y] holds o2 c= o
    by A78;
    the_rank_of (f.n) c= the_rank_of (g.m) by A107,A110,CLASSES1:73;
    then f.n c= Rank the_rank_of (g.m) by CLASSES1:65;
    then consider YY being set such that
A113: YY c= Rank o2 and
A114: P[n,f.n,YY] by A112;
A115: the_rank_of YY c= o2 by A113,CLASSES1:65;
    [f.n,n]`1 = f.n & [f.n,n]`2 = n;
    then ex i being Element of NAT st i = n &( P[n,f.n,(F.[f.n,n]) `1])& (F.[f
    .n,n])`2 = i+1 & not ex y being set st P[n,f.n,y] & the_rank_of y in
    the_rank_of (F.[f.n,n])`1 by A99,A108;
    then
A116: the_rank_of (F.[f.n,n])`1 c= the_rank_of YY by A114,ORDINAL1:16;
    g.(m+1) in rng g by A76,FUNCT_1:def 3;
    then
A117: o2 in sup rng g by A111,ORDINAL2:19;
    f.(n+1) = (F.[f.n,n])`1 by A102;
    hence thesis by A116,A115,A117,ORDINAL1:12,XBOOLE_1:1;
  end;
A118: for n being Element of NAT holds X[n] from NAT_1:sch 1(A103,A104);
  defpred X[Element of NAT] means P[$1,f.$1,f.($1+1)];
A119: for n being Element of NAT st X[n] holds X[n+1]
  proof
    let n be Element of NAT;
    assume P[n,f.n,f.(n+1)];
    the_rank_of (f.(n+1)) in sup rng g by A118;
    then f.(n+1) in Rank sup rng g by CLASSES1:66;
    then
A120: [f.(n+1),n+1] in [:Rank beta, NAT:] by ZFMISC_1:87;
    [f.(n+1),n+1]`1 = f.(n+1) & [f.(n+1),n+1]`2 = n+1;
    then ex i being Element of NAT st i = n+1 &( P[n+1,f.(n+1),(F.[ f.(n+1),n+
1])`1])& (F.[f.(n+1),n+1])`2 = i+1 & not ex y being set st P[(n+1), f.(n+1),y]
    & the_rank_of y in the_rank_of (F.[f.(n+1),n+1])`1 by A99,A120;
    hence thesis by A102;
  end;
  take f;
  thus dom f = NAT by A100;
  thus f.0 = A() by A101;
  A() in Rank sup rng g by A101,A103,CLASSES1:66;
  then [A(),0] in [:Rank beta, NAT:] by ZFMISC_1:87;
  then ex i being Element of NAT st i = [A(),0]`2 &( P[0,A(),(F.[ A(),0])`1])&
  (F.[A(),0])`2 = i+1 & not ex y being set st P[0,A(),y] & the_rank_of y in
  the_rank_of (F.[A(),0])`1 by A99,A79;
  then
A121: X[0] by A101,A102;
  thus for n being Element of NAT holds X[n] from NAT_1:sch 1(A121,A119);
end;

scheme
  RecExD{D()->non empty set,A() -> Element of D(), P[set,set,set]}: ex f being
Function of NAT,D() st f.0 = A() & for n being Element of NAT holds P[n,f.n,f.(
  n+1)]
provided
A1: for n being Element of NAT for x being Element of D() ex y being
Element of D() st P[n,x,y]
proof
  defpred Q[Element of NAT qua non empty set,set,set] means P[$1,$2,$3];
A2: for x being Element of NAT qua non empty set for y being Element of D()
  ex z being Element of D() st Q[x,y,z] by A1;
  consider f being Function of [:(NAT qua non empty set),D():],D() such that
A3: for x being Element of NAT qua non empty set for y being Element of
  D() holds Q[x,y,f.(x,y)] from BINOP_1:sch 3(A2);
  defpred P[FinSequence] means ($1.1=A() & for n st n+2 <= len $1 holds $1.(n+
  2) = f.[n,$1.(n+1)]);
  consider X being set such that
A4: for x holds x in X iff ex p st p in D()* & P[p] & x=p from FINSEQ_1:
  sch 4;
  set Y = union X;
A5: x in X implies x in D()*
  proof
    assume x in X;
    then
    ex p st p in D()* & (p.1=A() & for n st n + 2 <= len p holds p.(n+2) =
    f.[n,p.(n+1)]) & x=p by A4;
    hence thesis;
  end;
A6: for p,q st p in X & q in X & len p <= len q holds p c= q
  proof
    let p,q;
    assume that
A7: p in X and
A8: q in X and
A9: len p <= len q;
A10: ex q9 being FinSequence st q9 in D()* & (q9.1 = A() & for n st n+2<=
    len q9 holds q9.(n+2) = f.[n,q9.(n+1)]) & q=q9 by A4,A8;
A11: ex p9 being FinSequence st p9 in D()* & (p9.1 = A() & for n st n+2<=
    len p9 holds p9.(n+2) = f.[n,p9.(n+1)]) & p=p9 by A4,A7;
A12: for n st 1 <= n & n <= len p holds p.n = q.n
    proof
      defpred P[Nat] means 1 <= $1 & $1 <= len p & p.$1 <> q.$1;
      assume ex n st P[n];
      then
A13:  ex n be Nat st P[n];
      consider k be Nat such that
A14:  P[k] & for n be Nat st P[n] holds k <= n from NAT_1:sch 5(A13);
      k = 1
      proof
        0 <> k by A14;
        then consider n being Nat such that
A15:    k = n+1 by NAT_1:6;
        n+0 <= n+1 by XREAL_1:7;
        then
A16:    n <= len p by A14,A15,XXREAL_0:2;
A17:    n+0 < n+1 by XREAL_1:6;
        assume
A18:    k <> 1;
        then 1 < n+1 by A14,A15,XXREAL_0:1;
        then
A19:    1 <= n by NAT_1:13;
        n <> 0 by A18,A15;
        then consider m being Nat such that
A20:    n=m+1 by NAT_1:6;
        reconsider m as Element of NAT by ORDINAL1:def 12;
A21:    m+2 <= len q by A9,A14,A15,A20,XXREAL_0:2;
        p.k = p.(m+(1+1)) by A15,A20
          .=f.[m,p.n] by A11,A14,A15,A20
          .=f.[m,q.(m+1)] by A14,A15,A19,A16,A17,A20
          .=q.k by A10,A15,A20,A21;
        hence thesis by A14;
      end;
      hence contradiction by A11,A10,A14;
    end;
    now
      let x;
      assume x in p;
      then consider n be Nat such that
A22:  n in dom p and
A23:  x = [n,p.n] by FINSEQ_1:12;
A24:  n in Seg len p by A22,FINSEQ_1:def 3;
      then
A25:  1 <= n by FINSEQ_1:1;
A26:  n <= len p by A24,FINSEQ_1:1;
      then n <= len q by A9,XXREAL_0:2;
      then n in Seg len q by A25,FINSEQ_1:1;
      then
A27:  n in dom q by FINSEQ_1:def 3;
      x = [n,q.n] by A12,A22,A23,A25,A26;
      hence x in q by A27,FUNCT_1:1;
    end;
    hence thesis by TARSKI:def 3;
  end;
  ex f being Function st f = Y
  proof
    defpred Z[set,set] means [$1,$2] in Y;
A28: for x,y,z st Z[x,y] & Z[x,z] holds y=z
    proof
      let x,y,z;
      assume that
A29:  [x,y] in Y and
A30:  [x,z] in Y;
      consider Z2 being set such that
A31:  [x,z] in Z2 and
A32:  Z2 in X by A30,TARSKI:def 4;
      Z2 in D()* by A5,A32;
      then reconsider q=Z2 as FinSequence of D() by FINSEQ_1:def 11;
      consider Z1 being set such that
A33:  [x,y] in Z1 and
A34:  Z1 in X by A29,TARSKI:def 4;
      Z1 in D()* by A5,A34;
      then reconsider p=Z1 as FinSequence of D() by FINSEQ_1:def 11;
A35:  now
        assume len q <= len p;
        then
A36:    q c= Z1 by A6,A34,A32;
        [x,y] in p by A33;
        hence thesis by A31,A36,FUNCT_1:def 1;
      end;
      now
        assume len p <= len q;
        then p c= Z2 by A6,A34,A32;
        then [x,y] in q by A33;
        hence thesis by A31,FUNCT_1:def 1;
      end;
      hence thesis by A35;
    end;
    consider h being Function such that
A37: for x,y holds [x,y] in h iff x in NAT & Z[x,y] from FUNCT_1:sch 1
    (A28);
    take h;
A38: for x,y holds [x,y] in h iff [x,y] in Y
    proof
      let x,y;
      thus [x,y] in h implies [x,y] in Y by A37;
      thus [x,y] in Y implies [x,y] in h
      proof
        assume
A39:    [x,y] in Y;
        then consider Z such that
A40:    [x,y] in Z and
A41:    Z in X by TARSKI:def 4;
        Z in D()* by A5,A41;
        then reconsider p=Z as FinSequence of D() by FINSEQ_1:def 11;
        x in dom p by A40,XTUPLE_0:def 12;
        hence thesis by A37,A39;
      end;
    end;
    for x holds x in h iff x in Y
    proof
      let x;
      thus x in h implies x in Y
      proof
        assume
A42:    x in h;
        then ex y,z st [y,z] = x by RELAT_1:def 1;
        hence thesis by A38,A42;
      end;
      assume
A43:  x in Y;
      then consider Z such that
A44:  x in Z and
A45:  Z in X by TARSKI:def 4;
      Z in D()* by A5,A45;
      then reconsider p=Z as FinSequence of D() by FINSEQ_1:def 11;
      x in p by A44;
      then ex y,z st [y,z] = x by RELAT_1:def 1;
      hence thesis by A38,A43;
    end;
    hence thesis by TARSKI:1;
  end;
  then consider g being Function such that
A46: g = Y;
A47: for x st x in rng g holds x in D()
  proof
    let x;
    assume x in rng g;
    then consider y such that
A48: y in dom g & x = g.y by FUNCT_1:def 3;
    [y,x] in Y by A46,A48,FUNCT_1:1;
    then consider Z such that
A49: [y,x] in Z and
A50: Z in X by TARSKI:def 4;
    Z in D()* by A5,A50;
    then reconsider p=Z as FinSequence of D() by FINSEQ_1:def 11;
    y in dom p & x = p.y by A49,FUNCT_1:1;
    then
A51: x in rng p by FUNCT_1:def 3;
    rng p c= D() by FINSEQ_1:def 4;
    hence thesis by A51;
  end;
  then rng g c= D() by TARSKI:def 3;
  then reconsider h = g as Function of dom g,D() by FUNCT_2:2;
A52: for n holds (n+1) in dom h
  proof
    defpred P[Element of NAT] means $1+1 in dom h;
A53: for n st n+2 <= len <*A()*> holds <*A()*>.(n+2) = f.[n,<*A()*>.(n+1)]
    proof
      let n;
      assume n+2 <= len <*A()*>;
      then n+2 <= 1 by FINSEQ_1:39;
      then n+(1+1) <= n+1 by NAT_1:12;
      hence thesis by XREAL_1:6;
    end;
    <*A()*>.1 = A() & <*A()*> in D()* by FINSEQ_1:def 8,def 11;
    then <*A()*> in X by A4,A53;
    then
A54: {[1,A()]} in X by FINSEQ_1:37;
A55: for k st P[k] holds P[k+1]
    proof
      let k;
      assume k+1 in dom h;
      then [k+1,h.(k+1)] in Y by A46,FUNCT_1:1;
      then consider Z such that
A56:  [k+1,h.(k+1)] in Z and
A57:  Z in X by TARSKI:def 4;
      Z in D()* by A5,A57;
      then reconsider Z as FinSequence of D() by FINSEQ_1:def 11;
A58:  k+1 = len Z implies thesis
      proof
        set p=Z^<*f.[k,Z.(k+1)]*>;
A59:    1 <= k+1+1 by NAT_1:12;
        assume
A60:    k+1 = len Z;
        then 1 <= len Z by NAT_1:12;
        then 1 in Seg len Z by FINSEQ_1:1;
        then
A61:    1 in dom Z by FINSEQ_1:def 3;
A62:    for n st n+2 <= len p holds p.(n+2) = f.[n,p.(n+1)]
        proof
          let n;
          assume n+2 <= len p;
          then
A63:      n+2 <= len Z + len <*f.[k,Z.(k+1)]*> by FINSEQ_1:22;
          then
A64:      n+2 <= len Z + 1 by FINSEQ_1:40;
A65:      n+2 <> len Z + 1 implies thesis
          proof
            n+1+1 <= len Z + 1 by A63,FINSEQ_1:40;
            then 1 <= n + 1 & n+1 <= len Z by NAT_1:12,XREAL_1:6;
            then n+1 in Seg len Z by FINSEQ_1:1;
            then
A66:        n+1 in dom Z by FINSEQ_1:def 3;
A67:        1 <= n+(1+1) by NAT_1:12;
            assume
A68:        n+2 <> len Z + 1;
            then n+2 <= len Z by A64,NAT_1:8;
            then n+2 in Seg len Z by A67,FINSEQ_1:1;
            then
A69:        n+2 in dom Z by FINSEQ_1:def 3;
            ex q st q in D()* & q.1 = A() & (for n st n+2 <= len q holds
            q.(n+2) = f.[n,q.(n+1)]) & Z = q by A4,A57;
            then Z.(n+2) = f.[n,Z.(n+1)] by A64,A68,NAT_1:8;
            then p.(n+2) = f.[n,Z.(n+1)] by A69,FINSEQ_1:def 7;
            hence thesis by A66,FINSEQ_1:def 7;
          end;
          n+2 = len Z + 1 implies thesis
          proof
            n+1+1 <= len Z + 1 by A63,FINSEQ_1:40;
            then 1 <= n + 1 & n+1 <= len Z by NAT_1:12,XREAL_1:6;
            then n+1 in Seg len Z by FINSEQ_1:1;
            then
A70:        n+1 in dom Z by FINSEQ_1:def 3;
            assume
A71:        n+2 = len Z + 1;
            then p.(n+2) =<*f.[k,Z.(k+1)]*>.(n+2-(n+2-1 )) by A63,FINSEQ_1:23
              .=f.[n,Z.(n+1)] by A60,A71,FINSEQ_1:40;
            hence thesis by A70,FINSEQ_1:def 7;
          end;
          hence thesis by A65;
        end;
A72:    p in D()*
        proof
          1 <= k + 1 by NAT_1:12;
          then k+1 in Seg len Z by A60,FINSEQ_1:1;
          then k+1 in dom Z by FINSEQ_1:def 3;
          then
A73:      Z.(k+1) in rng Z by FUNCT_1:def 3;
          rng Z c= D() by FINSEQ_1:def 4;
          then reconsider z=Z.(k+1) as Element of D() by A73;
          reconsider n=k as Element of (NAT qua non empty set);
          p= Z^<*f.[n,z]*>;
          hence thesis by FINSEQ_1:def 11;
        end;
        len p = len Z + len <*f.[k,Z.(k+1)]*> by FINSEQ_1:22
          .=k+1+1 by A60,FINSEQ_1:39;
        then k+1+1 in Seg len p by A59,FINSEQ_1:1;
        then k+1+1 in dom p by FINSEQ_1:def 3;
        then
A74:    [k+1+1,p.(k+1+1)] in p by FUNCT_1:1;
        ex p being FinSequence st p in D()* & p.1 = A() & (for n st n+2
        <= len p holds p.(n+2) = f.[n,p.(n+1)]) & Z=p by A4,A57;
        then p.1 = A() by A61,FINSEQ_1:def 7;
        then p in X by A4,A72,A62;
        then [k+1+1,p.(k+1+1)] in h by A46,A74,TARSKI:def 4;
        hence thesis by FUNCT_1:1;
      end;
      k+1 <> len Z implies thesis
      proof
        k+1 in dom Z by A56,FUNCT_1:1;
        then k+1 in Seg len Z by FINSEQ_1:def 3;
        then
A75:    k+1 <= len Z by FINSEQ_1:1;
        assume k+1 <> len Z;
        then k+1 < len Z by A75,XXREAL_0:1;
        then
A76:    k+1+1 <= len Z by NAT_1:13;
        1 <= k+1+1 by NAT_1:12;
        then k+1+1 in Seg len Z by A76,FINSEQ_1:1;
        then k+1+1 in dom Z by FINSEQ_1:def 3;
        then [k+1+1,Z.(k+1+1)] in Z by FUNCT_1:1;
        then [k+1+1,Z.(k+1+1)] in h by A46,A57,TARSKI:def 4;
        hence thesis by FUNCT_1:1;
      end;
      hence thesis by A58;
    end;
    [1,A()] in {[1,A()]} by TARSKI:def 1;
    then [1,A()] in h by A46,A54,TARSKI:def 4;
    then
A77: P[0] by FUNCT_1:1;
    thus for k holds P[k] from NAT_1:sch 1(A77,A55);
  end;
A78: for n holds h.(n+2) = f.[n,h.(n+1)]
  proof
    let n;
    (n+1)+1 in dom h by A52;
    then [n+2,h.(n+2)] in h by FUNCT_1:def 2;
    then consider Z being set such that
A79: [n+2,h.(n+2)] in Z and
A80: Z in X by A46,TARSKI:def 4;
A81: ex p st p in D()* & (p.1=A() & for n st n+2 <= len p holds p.(n+2) =
    f.[n,p.(n+1)]) & Z=p by A4,A80;
    Z in D()* by A5,A80;
    then reconsider Z as FinSequence of D() by FINSEQ_1:def 11;
    n+2 in dom Z by A79,FUNCT_1:1;
    then n+2 in Seg len Z by FINSEQ_1:def 3;
    then
A82: n+2 <= len Z by FINSEQ_1:1;
    n+1 <= n+2 by XREAL_1:7;
    then 1 <= n+1 & n+1 <= len Z by A82,NAT_1:12,XXREAL_0:2;
    then n+1 in Seg len Z by FINSEQ_1:1;
    then n+1 in dom Z by FINSEQ_1:def 3;
    then [n+1,Z.(n+1)] in Z by FUNCT_1:1;
    then
A83: [n+1,Z.(n+1)] in h by A46,A80,TARSKI:def 4;
    thus h.(n+2) = Z.(n+2) by A79,FUNCT_1:1
      .= f.[n,Z.(n+1)] by A81,A82
      .= f.[n,h.(n+1)] by A83,FUNCT_1:1;
  end;
  ex g being Function of NAT,D() st for n holds g.n = h.(n+1)
  proof
    ex g being Function st dom g = NAT & for n holds g.n = h.(n+1)
    proof
      defpred P[set,set] means ex n st n = $1 & $2=h.(n+1);
A84:  for x st x in NAT ex y st P[x,y]
      proof
        let x;
        assume x in NAT;
        then reconsider n=x as Element of NAT;
        take h.(n+1);
        take n;
        thus thesis;
      end;
      consider g being Function such that
A85:  dom g = NAT & for x st x in NAT holds P[x,g.x] from CLASSES1:
      sch 1 (A84);
      take g;
      thus dom g = NAT by A85;
      thus for n holds g.n = h.(n+1)
      proof
        let n;
        ex m st m=n & g.n = h.(m+1) by A85;
        hence thesis;
      end;
    end;
    then consider g being Function such that
A86: dom g = NAT and
A87: for n holds g.n = h.(n+1);
    rng g c= D()
    proof
      let x;
      assume x in rng g;
      then consider y such that
A88:  y in dom g and
A89:  x = g.y by FUNCT_1:def 3;
      reconsider k=y as Element of NAT by A86,A88;
      k+1 in dom h by A52;
      then
A90:  h.(k+1) in rng h by FUNCT_1:def 3;
      x=h.(k+1) by A87,A89;
      hence thesis by A47,A90;
    end;
    then reconsider g as Function of NAT,D() by A86,FUNCT_2:2;
    take g;
    thus thesis by A87;
  end;
  then consider g being Function of NAT,D() such that
A91: for n holds g.n = h.(n+1);
A92: for n st n + 2 <= len <*A()*> holds <*A()*>.(n+2) = f.[n,<*A()*>.(n+1) ]
  proof
    let n;
    assume n + 2 <= len <*A()*>;
    then n +1+1 <= 0 + 1 by FINSEQ_1:39;
    then n + 1 <= 0 by XREAL_1:6;
    then n + 1 <= 0 + n;
    hence thesis by XREAL_1:6;
  end;
  <*A()*> in D()* & <*A()*>.1 = A() by FINSEQ_1:def 8,def 11;
  then <*A()*> in X by A4,A92;
  then
A93: {[1,A()]} in X by FINSEQ_1:37;
  take g;
  [1,A()] in {[1,A()]} by TARSKI:def 1;
  then [1,A()] in h by A46,A93,TARSKI:def 4;
  then A() = h.(0+1) by FUNCT_1:1
    .= g.0 by A91;
  hence g.0 = A();
  let n be Element of NAT;
A94: h.(n+(1+1)) =f.(n,h.(n+1)) by A78;
  P[n,g.n,f.(n,g.n)] by A3;
  then P[n,g.n,h.(n+1+1)] by A91,A94;
  hence thesis by A91;
end;

scheme
  FinRecEx{A() -> set,N() -> Nat,P[set,set,set]}: ex p being FinSequence st
len p = N() & (p.1 = A() or N() = 0) & for n st 1 <= n & n < N() holds P[n,p.n,
  p.(n+1)]
provided
A1: for n being Element of NAT st 1 <= n & n < N() for x being set ex y
being set st P[n,x,y]
proof
  defpred Q[Element of NAT,set,set] means ($1 < N()-1 implies P[$1+1,$2,$3]) &
  (not $1 < N()-1 implies $3=0);
A2: for n being Element of NAT for x being set ex y being set st Q[n,x,y]
  proof
    let n be Element of NAT,x be set;
    n < N()-1 implies thesis
    proof
      assume
A3:   n < N()-1;
      then n+1 < N() by XREAL_1:20;
      then consider y such that
A4:   P[n+1,x,y] by A1,NAT_1:11;
      take y;
      thus n < N()-1 implies P[n+1,x,y] by A4;
      thus thesis by A3;
    end;
    hence thesis;
  end;
  consider f being Function such that
A5: dom f = NAT & f.0 = A() & for n being Element of NAT holds Q[n,f.n,f
  .(n+1)] from RecEx(A2);
  defpred Q[set,set] means for r being Real st r=$1 holds $2=f.(r-1);
A6: for x st x in REAL ex y st Q[x,y]
  proof
    let x;
    assume x in REAL;
    then reconsider r=x as Real;
    take f.(r-1);
    thus thesis;
  end;
  consider g being Function such that
A7: dom g = REAL & for x st x in REAL holds Q[x,g.x] from CLASSES1:sch
  1(A6);
A8: dom (g|Seg N()) = Seg N() by A7,RELAT_1:62,XBOOLE_1:1;
  then reconsider p = g|Seg N() as FinSequence by FINSEQ_1:def 2;
  take p;
  N() in NAT by ORDINAL1:def 12;
  hence len p = N() by A8,FINSEQ_1:def 3;
  N() <> 0 implies (p.1 = A() or N() = 0)
  proof
    assume N() <> 0;
    then consider k being Nat such that
A9: N() = k+1 by NAT_1:6;
    0 + 1 <= k +1 by XREAL_1:7;
    then 1 in Seg N() by A9,FINSEQ_1:1;
    then p.1 = g.1 by FUNCT_1:49
      .= f.(1-1) by A7
      .= A() by A5;
    hence thesis;
  end;
  hence p.1 = A() or N() = 0;
  let n;
  assume that
A10: 1 <= n and
A11: n < N();
  0 <> n by A10;
  then consider k being Nat such that
A12: n = k+1 by NAT_1:6;
A13: for n being Element of NAT st n < N() holds p.(n+1) = f.n
  proof
    let n be Element of NAT;
    assume n < N();
    then 1 <= n+1 & n+1 <= N() by NAT_1:11,13;
    then
A14: n+1 in Seg N() by FINSEQ_1:1;
    g.(n+1) = f.(n+1-1) by A7
      .= f.n;
    hence thesis by A14,FUNCT_1:49;
  end;
  reconsider k as Element of NAT by ORDINAL1:def 12;
  k <= k+1 by NAT_1:11;
  then
A15: k < N() by A11,A12,XXREAL_0:2;
  k < N()-1 by A11,A12,XREAL_1:20;
  then P[k+1,f.k,f.(k+1)] by A5;
  then P[k+1,f.k,p.(k+1+1)] by A13,A11,A12;
  hence thesis by A13,A12,A15;
end;

scheme
  FinRecExD{D() -> non empty set,A() -> Element of D(), N() -> Nat, P[set,set,
set]}: ex p being FinSequence of D() st len p = N() & (p.1 = A() or N() = 0) &
  for n st 1 <= n & n < N() holds P[n,p.n,p.(n+1)]
provided
A1: for n being Element of NAT st 1 <= n & n < N() for x being Element
of D() ex y being Element of D() st P[n,x,y]
proof
  set 00 = the Element of D();
  defpred Z[Element of NAT,set,set] means ($1 < N()-1 implies P[$1+1,$2,$3]) &
  (not $1 < N()-1 implies $3=00);
A2: for n being Element of NAT for x being Element of D() ex y being Element
  of D() st Z[n,x,y]
  proof
    let n be Element of NAT,x be Element of D();
    n < N()-1 implies thesis
    proof
      assume
A3:   n < N()-1;
      then n+1 < N() by XREAL_1:20;
      then consider y being Element of D() such that
A4:   P[n+1,x,y] by A1,NAT_1:11;
      take y;
      thus n < N()-1 implies P[n+1,x,y] by A4;
      thus thesis by A3;
    end;
    hence thesis;
  end;
  consider f being Function of NAT,D() such that
A5: f.0 = A() & for n being Element of NAT holds Z[n,f.n,f.(n+1)] from
  RecExD(A2);
  defpred P[set,set] means for r being Real st r = $1 holds $2 = f.(r-1);
A6: for x st x in REAL ex y st P[x,y]
  proof
    let x;
    assume x in REAL;
    then reconsider r=x as Real;
    take f.(r-1);
    thus thesis;
  end;
  consider g being Function such that
A7: dom g = REAL & for x st x in REAL holds P[x,g.x] from CLASSES1:sch
  1(A6);
A8: dom (g|Seg N()) = Seg N() by A7,RELAT_1:62,XBOOLE_1:1;
  then reconsider p = g|Seg N() as FinSequence by FINSEQ_1:def 2;
  rng p c= D()
  proof
    let x;
    assume x in rng p;
    then consider y such that
A9: y in dom p and
A10: x = p.y by FUNCT_1:def 3;
    reconsider y as Element of NAT by A9;
A11: f.(y-1) in D()
    proof
      y <> 0 by A8,A9,FINSEQ_1:1;
      then consider k being Nat such that
A12:  y = k+1 by NAT_1:6;
      reconsider k as Element of NAT by ORDINAL1:def 12;
      f.k in D();
      hence thesis by A12;
    end;
    p.y = g.y by A9,FUNCT_1:47;
    hence thesis by A7,A10,A11;
  end;
  then reconsider p as FinSequence of D() by FINSEQ_1:def 4;
  take p;
  N() in NAT by ORDINAL1:def 12;
  hence len p = N() by A8,FINSEQ_1:def 3;
  N() <> 0 implies (p.1 = A() or N() = 0)
  proof
    assume N() <> 0;
    then consider k being Nat such that
A13: N() = k+1 by NAT_1:6;
    0 + 1 <= k +1 by XREAL_1:7;
    then 1 in Seg N() by A13,FINSEQ_1:1;
    then p.1 = g.1 by FUNCT_1:49
      .= f.(1-1) by A7
      .= A() by A5;
    hence thesis;
  end;
  hence p.1 = A() or N() = 0;
  let n;
  assume that
A14: 1 <= n and
A15: n < N();
  0 <> n by A14;
  then consider k being Nat such that
A16: n = k+1 by NAT_1:6;
A17: for n being Element of NAT st n < N() holds p.(n+1) = f.n
  proof
    let n be Element of NAT;
    assume n < N();
    then 1 <= n+1 & n+1 <= N() by NAT_1:11,13;
    then
A18: n+1 in Seg N() by FINSEQ_1:1;
    g.(n+1) = f.(n+1-1) by A7
      .= f.n;
    hence thesis by A18,FUNCT_1:49;
  end;
  reconsider k as Element of NAT by ORDINAL1:def 12;
  k <= k+1 by NAT_1:11;
  then
A19: k < N() by A15,A16,XXREAL_0:2;
  k < N() - 1 by A15,A16,XREAL_1:20;
  then P[k+1,f.k,f.(k+1)] by A5;
  then P[k+1,f.k,p.(k+1+1)] by A17,A15,A16;
  hence thesis by A17,A16,A19;
end;

scheme
  SeqBinOpEx{S() -> FinSequence,P[set,set,set]}: ex x st ex p being
FinSequence st x = p.(len p) & len p = len S() & p.1 = S().1 & for k st 1 <= k
  & k < len S() holds P[S().(k+1),p.k,p.(k+1)]
provided
A1: for k,x st 1 <= k & k < len S() ex y st P[S().(k+1),x,y]
proof
  defpred Q[Element of NAT,set,set] means P[S().($1+1),$2,$3];
A2: for k st 1 <= k & k < len S() for x ex y st Q[k,x,y] by A1;
  consider p such that
A3: len p = len S() & (p.1 = S().1 or len S() = 0) & for k st 1 <= k & k
  < len S() holds Q[k,p.k,p.(k+1)] from FinRecEx(A2);
A4: len S() <> 0 implies thesis
  proof
    assume
A5: len S() <> 0;
    take p.(len p),p;
    thus p.(len p) = p.(len p) & len p = len S() & p.1 = S().1 by A3,A5;
    let k;
    assume 1 <= k & k < len S();
    hence thesis by A3;
  end;
  len S() = 0 implies thesis
  proof
    assume
A6: len S() = 0;
    take S().0,S();
    thus S().0 = S().(len S()) & len S() = len S() & S().1 = S(). 1 by A6;
    thus thesis by A6;
  end;
  hence thesis by A4;
end;

scheme
  LambdaSeqBinOpEx{S() -> FinSequence,F(set,set) -> set}: ex x st ex p being
FinSequence st x = p.(len p) & len p = len S() & p.1 = S().1 & for k st 1 <= k
  & k < len S() holds p.(k+1) = F(S().(k+1),p.k) proof
  defpred P[set,set,set] means $3 = F($1,$2);
A1: for k,x st 1 <= k & k < len S() ex y st P[S().(k+1),x,y];
  consider x such that
A2: ex p st x = p.(len p) & len p = len S() & p.1 = S().1 & for k st 1
  <= k & k < len S() holds P[S().(k+1),p.k,p.(k+1)] from SeqBinOpEx(A1);
  take x;
  consider p such that
A3: x = p.(len p) & len p = len S() & p.1 = S().1 and
A4: for k st 1 <= k & k < len S() holds p.(k+1) = F(S().(k+1),p.k) by A2;
  take p;
  thus x = p.(len p) & len p = len S() & p.1 = S().1 by A3;
  let k;
  assume 1 <= k & k < len S();
  hence thesis by A4;
end;



scheme
  FinRecUn{A() -> set,N() -> Nat, F, G() -> FinSequence, P[set,set,set]}: F()
  = G()
provided
A1: for n st 1 <= n & n < N() for x,y1,y2 being set st P[n,x,y1] & P[n,x
,y2] holds y1 = y2 and
A2: len F() = N() & (F().1 = A() or N() = 0) & for n st 1 <= n & n < N()
holds P[n,F().n,F().(n+1)] and
A3: len G() = N() & (G().1 = A() or N() = 0) & for n st 1 <= n & n < N()
holds P[n,G().n,G().(n+1)]
proof
  defpred P[Nat] means 1 <= $1 & $1 <= N() & F().$1 <> G().$1;
  assume
A4: F() <> G();
  dom F() = Seg len G() by A2,A3,FINSEQ_1:def 3
    .= dom G() by FINSEQ_1:def 3;
  then consider x such that
A5: x in dom F() and
A6: F().x <> G().x by A4,FUNCT_1:2;
A7: x in Seg len F() by A5,FINSEQ_1:def 3;
  reconsider x as Element of NAT by A5;
A8: 1 <= x by A7,FINSEQ_1:1;
  x <= N() by A2,A7,FINSEQ_1:1;
  then
A9: ex n be Nat st P[n] by A6,A8;
  consider n be Nat such that
A10: P[n]& for k be Nat st P[k] holds n <= k from NAT_1:sch 5(A9);
  0 <> n by A10;
  then consider k being Nat such that
A11: n = k+1 by NAT_1:6;
  reconsider k as Element of NAT by ORDINAL1:def 12;
  n <> 1 by A2,A3,A10;
  then 1 < n by A10,XXREAL_0:1;
  then
A12: 1 <= k by A11,NAT_1:13;
  k < n by A11,XREAL_1:29;
  then
A13: k < N() by A10,XXREAL_0:2;
  n > k by A11,NAT_1:13;
  then F().k = G().k by A10,A12,A13;
  then
A14: P[k,F().k,G().(k+1)] by A3,A12,A13;
  P[k,F().k,F().(k+1)] by A2,A12,A13;
  hence contradiction by A1,A10,A11,A12,A13,A14;
end;

scheme
  FinRecUnD{D() -> non empty set, A() -> Element of D(), N() -> Nat, F, G() ->
  FinSequence of D(), P[set,set,set]}: F() = G()
provided
A1: for n st 1 <= n & n < N() for x,y1,y2 being Element of D() st P[n,x,
y1] & P[n,x,y2] holds y1 = y2 and
A2: len F() = N() & (F().1 = A() or N() = 0) & for n st 1 <= n & n < N()
holds P[n,F().n,F().(n+1)] and
A3: len G() = N() & (G().1 = A() or N() = 0) & for n st 1 <= n & n < N()
holds P[n,G().n,G().(n+1)]
proof
  defpred P[Nat] means 1 <= $1 & $1 <= N() & F().$1 <> G().$1;
  assume
A4: F() <> G();
  dom F() = Seg len G() by A2,A3,FINSEQ_1:def 3
    .= dom G() by FINSEQ_1:def 3;
  then consider x such that
A5: x in dom F() and
A6: F().x <> G().x by A4,FUNCT_1:2;
A7: x in Seg len F() by A5,FINSEQ_1:def 3;
  reconsider x as Element of NAT by A5;
A8: 1 <= x by A7,FINSEQ_1:1;
  x <= N() by A2,A7,FINSEQ_1:1;
  then
A9: ex n be Nat st P[n] by A6,A8;
  consider n be Nat such that
A10: P[n] & for k be Nat st P[k] holds n <= k from NAT_1:sch 5(A9);
  0 <> n by A10;
  then consider k being Nat such that
A11: n = k+1 by NAT_1:6;
  reconsider k as Element of NAT by ORDINAL1:def 12;
  reconsider Gk1 = G().(k+1) as Element of D() by A3,A10,A11,Lm1;
  n <> 1 by A2,A3,A10;
  then 1 < n by A10,XXREAL_0:1;
  then
A12: 1 <= k by A11,NAT_1:13;
  k < n by A11,XREAL_1:29;
  then
A13: k < N() by A10,XXREAL_0:2;
  then reconsider Fk = F().k as Element of D() by A2,A12,Lm1;
  n > k by A11,NAT_1:13;
  then F().k = G().k by A10,A12,A13;
  then
A14: P[k,Fk,Gk1] by A3,A12,A13;
  reconsider Fk1 = F().(k+1) as Element of D() by A2,A10,A11,Lm1;
  P[k,Fk,Fk1] by A2,A12,A13;
  hence contradiction by A1,A10,A11,A12,A13,A14;
end;

scheme
  SeqBinOpUn{S() -> FinSequence,P[set,set,set],x, y() -> set}: x() = y()
provided
A1: for k,x,y1,y2,z st 1 <= k & k < len S() & z = S().(k+1) & P[z,x,y1]
& P[z,x,y2] holds y1 = y2 and
A2: ex p being FinSequence st x() = p.(len p) & len p = len S() & p.1 =
S().1 & for k st 1 <= k & k < len S() holds P[S().(k+1),p.k,p.(k+1)] and
A3: ex p being FinSequence st y() = p.(len p) & len p = len S() & p.1 =
S().1 & for k st 1 <= k & k < len S() holds P[S().(k+1),p.k,p.(k+1)]
proof
  defpred Q[Element of NAT,set,set] means P[S().($1+1),$2,$3];
A4: for k st 1 <= k & k < len S() for x,y1,y2 st Q[k,x,y1] & Q[k,x,y2] holds
  y1 = y2 by A1;
  consider q such that
A5: y() = q.(len q) and
A6: len q = len S() & q.1 = S().1 & for k st 1 <= k & k < len S() holds
  Q[k, q.k,q.(k+1)] by A3;
A7: len q = len S() & (q.1 = S().1 or len S() = 0) & for k st 1 <= k & k <
  len S() holds Q[k,q.k,q.(k+1)] by A6;
  consider p such that
A8: x() = p.(len p) and
A9: len p = len S() & p.1 = S().1 & for k st 1 <= k & k < len S() holds
  Q[k, p.k,p.(k+1)] by A2;
A10: len p = len S() & (p.1 = S().1 or len S() = 0) & for k st 1 <= k & k <
  len S() holds Q[k,p.k,p.(k+1)] by A9;
  p = q from FinRecUn(A4,A10,A7);
  hence thesis by A8,A5;
end;

scheme
  LambdaSeqBinOpUn{S() -> FinSequence, F(set,set) -> set, x, y() -> set}: x()
  = y()
provided
A1: ex p being FinSequence st x() = p.(len p) & len p = len S() & p.1 =
S().1 & for k st 1 <= k & k < len S() holds p.(k+1) = F(S().(k+1),p.k) and
A2: ex p being FinSequence st y() = p.(len p) & len p = len S() & p.1 =
S().1 & for k st 1 <= k & k < len S() holds p.(k+1) = F(S().(k+1),p.k)
proof
  defpred P[set,set,set] means $3 = F($1,$2);
A3: ex p st y() = p.(len p) & len p = len S() & p.1 = S().1 & for k st 1 <=
  k & k < len S() holds P[S().(k+1),p.k,p.(k+1)] by A2;
A4: for k,x,y1,y2,z st 1 <= k & k < len S() & z = S().(k+1) & P[z,x,y1] & P[
  z,x,y2] holds y1 = y2;
A5: ex p st x() = p.(len p) & len p = len S() & p.1 = S().1 & for k st 1 <=
  k & k < len S() holds P[S().(k+1),p.k,p.(k+1)] by A1;
  thus x() = y() from SeqBinOpUn(A4,A5,A3);
end;



scheme
  DefRec{A() -> set,n() -> Nat,P[set,set,set]}: (ex y being set st ex f being
Function st y = f.n() & dom f = NAT & f.0 = A() & for n holds P[n,f.n,f.(n+1)])
& for y1,y2 being set st (ex f being Function st y1 = f.n() & dom f = NAT & f.0
  = A() & for n holds P[n,f.n,f.(n+1)]) & (ex f being Function st y2 = f.n() &
  dom f = NAT & f.0 = A() & for n holds P[n,f.n,f.(n+1)]) holds y1 = y2
provided
A1: for n,x ex y st P[n,x,y] and
A2: for n,x,y1,y2 st P[n,x,y1] & P[n,x,y2] holds y1 = y2
proof
A3: for n,x ex y st P[n,x,y] by A1;
  consider f being Function such that
A4: dom f = NAT & f.0 = A() & for n being Element of NAT holds P[n,f.n,f
  .(n+1)] from RecEx(A3);
A5: for n being Nat,x,y1,y2 st P[n,x,y1] & P[n,x,y2] holds y1 = y2
  proof
    let n be Nat;
    n in NAT by ORDINAL1:def 12;
    hence thesis by A2;
  end;
  thus ex y being set st ex f being Function st y = f.n() & dom f = NAT & f.0
  = A() & for n holds P[n,f.n,f.(n+1)]
  proof
    take f.n(),f;
    thus thesis by A4;
  end;
  let y1,y2 be set;
  given f1 being Function such that
A6: y1 = f1.n() and
A7: dom f1 = NAT and
A8: f1.0 = A() and
A9: for n holds P[n,f1.n,f1.(n+1)];
A10: for n being Nat holds P[n,f1.n,f1.(n+1)]
  proof
    let n be Nat;
    n in NAT by ORDINAL1:def 12;
    hence thesis by A9;
  end;
  given f2 being Function such that
A11: y2 = f2.n() and
A12: dom f2 = NAT and
A13: f2.0 = A() and
A14: for n holds P[n,f2.n,f2.(n+1)];
A15: for n being Nat holds P[n,f2.n,f2.(n+1)]
  proof
    let n be Nat;
    n in NAT by ORDINAL1:def 12;
    hence thesis by A14;
  end;
  f1 = f2 from NAT_1:sch 13(A7,A8,A10,A12,A13,A15,A5);
  hence thesis by A6,A11;
end;

scheme
  LambdaDefRec{A() -> set,n() -> Nat,RecFun(set,set) -> set}: (ex y being set
st ex f being Function st y = f.n() & dom f = NAT & f.0 = A() & for n holds f.(
n+1) = RecFun(n,f.n)) & for y1,y2 being set st (ex f being Function st y1 = f.n
  () & dom f = NAT & f.0 = A() & for n holds f.(n+1) = RecFun(n,f.n)) & (ex f
being Function st y2 = f.n() & dom f = NAT & f.0 = A() & for n holds f.(n+1) =
  RecFun(n,f.n)) holds y1 = y2 proof
  defpred P[set,set,set] means for z st z = $2 holds $3 = RecFun($1,z);
A1: for n,x ex y st P[n,x,y]
  proof
    let n,x;
    take RecFun(n,x);
    let z;
    assume z = x;
    hence thesis;
  end;
A2: for n,x,y1,y2 st P[n,x,y1] & P[n,x,y2] holds y1 = y2
  proof
    let n,x,y1,y2;
    assume that
A3: for z st z = x holds y1 = RecFun(n,z) and
A4: for z st z = x holds y2 = RecFun(n,z);
    thus y1 = RecFun(n,x) by A3
      .= y2 by A4;
  end;
A5: (ex y being set st ex f being Function st y = f.n() & dom f = NAT & f.0
  = A() & for n holds P[n,f.n,f.(n+1)] ) & for y1,y2 being set st (ex f being
Function st y1 = f.n() & dom f = NAT & f.0 = A() & for n holds P[n,f.n,f.(n+1)]
) & (ex f being Function st y2 = f.n() & dom f = NAT & f.0 = A() & for n holds
  P[n,f.n,f.(n+1)] ) holds y1 = y2 from DefRec(A1,A2);
  then consider y being set, f being Function such that
A6: y = f.n() & dom f = NAT &( f.0 = A() & for n holds P[n,f.n,f.(n+1)] );
  thus ex y being set st ex f being Function st y = f.n() & dom f = NAT & f.0
  = A() & for n holds f.(n+1) = RecFun(n,f.n)
  proof
    take y,f;
    thus thesis by A6;
  end;
  let y1,y2 being set;
  given f1 being Function such that
A7: y1 = f1.n() & dom f1 = NAT & f1.0 = A() and
A8: for n holds f1.(n+1) = RecFun(n,f1.n);
A9: for n holds P[n,f1.n,f1.(n+1)] by A8;
  given f2 being Function such that
A10: y2 = f2.n() & dom f2 = NAT & f2.0 = A() and
A11: for n holds f2.(n+1) = RecFun(n,f2.n);
  for n holds P[n,f2.n,f2.(n+1)] by A11;
  hence thesis by A5,A7,A10,A9;
end;

scheme
  DefRecD{D() -> non empty set,A() -> (Element of D()), n() -> Nat,P[set,set,
set]}: (ex y being Element of D() st ex f being Function of NAT,D() st y = f.n(
) & f.0 = A() & for n holds P[n,f.n,f.(n+1)]) & for y1,y2 being Element of D()
st (ex f being Function of NAT,D() st y1 = f.n() & f.0 = A() & for n holds P[n,
f.n,f.(n+1)]) & (ex f being Function of NAT,D() st y2 = f.n() & f.0 = A() & for
  n holds P[n,f.n,f.(n+1)]) holds y1 = y2
provided
A1: for n being Element of NAT,x being Element of D() ex y being Element
of D() st P[n,x,y] and
A2: for n being Element of NAT, x,y1,y2 being Element of D() st P[n,x,y1
] & P[n,x,y2] holds y1 = y2
proof
A3: for n being Element of NAT,x being Element of D() ex y being Element of
  D() st P[n,x,y] by A1;
  consider f being Function of NAT,D() such that
A4: f.0 = A() & for n being Element of NAT holds P[n,f.n,f.(n+1)] from
  RecExD(A3);
A5: for n being Nat, x,y1,y2 being Element of D() st P[n,x,y1] & P[n,x,y2]
  holds y1 = y2
  proof
    let n be Nat;
    n in NAT by ORDINAL1:def 12;
    hence thesis by A2;
  end;
  thus ex y being Element of D() st ex f being Function of NAT,D() st y = f.n(
  ) & f.0 = A() & for n holds P[n,f.n,f.(n+1)]
  proof
    reconsider n = n() as Element of NAT by ORDINAL1:def 12;
    take f.n,f;
    thus thesis by A4;
  end;
  let y1,y2 be Element of D();
  given f1 being Function of NAT,D() such that
A6: y1 = f1.n() and
A7: f1.0 = A() and
A8: for n holds P[n,f1.n,f1.(n+1)];
A9: f1.0 = A() by A7;
A10: for n being Nat holds P[n,f1.n,f1.(n+1)]
  proof
    let n be Nat;
    n in NAT by ORDINAL1:def 12;
    hence thesis by A8;
  end;
  given f2 being Function of NAT,D() such that
A11: y2 = f2.n() and
A12: f2.0 = A() and
A13: for n holds P[n,f2.n,f2.(n+1)];
A14: for n being Nat holds P[n,f2.n,f2.(n+1)]
  proof
    let n be Nat;
    n in NAT by ORDINAL1:def 12;
    hence thesis by A13;
  end;
A15: f2.0 = A() by A12;
  f1 = f2 from NAT_1:sch 14(A9,A10,A15,A14,A5);
  hence thesis by A6,A11;
end;

scheme
  LambdaDefRecD{D() -> non empty set, A() -> Element of D(), n() -> Nat,
  RecFun(set,set) -> Element of D()}: (ex y being Element of D() st ex f being
Function of NAT,D() st y = f.n() & f.0 = A() & for n being Nat holds f.(n+1) =
RecFun(n,f.n)) & for y1,y2 being Element of D() st (ex f being Function of NAT,
D() st y1 = f.n() & f.0 = A() & for n being Nat holds f.(n+1) = RecFun(n,f.n))
  & (ex f being Function of NAT,D() st y2 = f.n() & f.0 = A() & for n being Nat
  holds f.(n+1) = RecFun(n,f.n)) holds y1 = y2 proof
  defpred Q[set,set,set] means for z being Element of D() st z=$2 holds $3 =
  RecFun($1,z);
A1: for n being Element of NAT,x being Element of D() ex y being Element of
  D() st Q[n,x,y]
  proof
    let n be Element of NAT,x be Element of D();
    take RecFun(n,x);
    let z be Element of D();
    assume z = x;
    hence thesis;
  end;
A2: for n being Element of NAT,x,y1,y2 being Element of D() st Q[n,x,y1] & Q
  [n,x,y2] holds y1 = y2
  proof
    let n be Element of NAT,x,y1,y2 be Element of D();
    assume that
A3: for z being Element of D() st z = x holds y1 = RecFun(n,z) and
A4: for z being Element of D() st z = x holds y2 = RecFun(n,z);
    thus y1 = RecFun(n,x) by A3
      .= y2 by A4;
  end;
A5: (ex y being Element of D() st ex f being Function of NAT,D() st y = f.n(
) & f.0 = A() & for n being Element of NAT holds Q[n,f.n,f.(n+1)]) & for y1,y2
being Element of D() st (ex f being Function of NAT,D() st y1 = f.n() & f.0 = A
() & for n being Element of NAT holds Q[n,f.n,f.(n+1)]) & (ex f being Function
of NAT,D() st y2 = f.n() & f.0 = A() & for n being Element of NAT holds Q[n,f.n
  ,f.(n+1)]) holds y1 = y2 from DefRecD(A1,A2);
  then consider y being Element of D(), f being Function of NAT,D() such that
A6: y = f.n() & f.0 = A() and
A7: for n being Element of NAT holds Q[n,f.n,f.(n+1)];
  thus ex y being Element of D() st ex f being Function of NAT,D() st y = f.n(
  ) & f.0 = A() & for n being Nat holds f.(n+1) = RecFun(n,f.n)
  proof
    take y,f;
    thus y = f.n() & f.0 = A() by A6;
    let n be Nat;
    reconsider n as Element of NAT by ORDINAL1:def 12;
    Q[n,f.n,f.(n+1)] by A7;
    hence thesis;
  end;
  let y1,y2 being Element of D();
  given f being Function of NAT,D() such that
A8: y1 = f.n() & f.0 = A() and
A9: for n being Nat holds f.(n+1) = RecFun(n,f.n);
A10: for n being Element of NAT holds Q[n,f.n,f.(n+1)] by A9;
  given f2 being Function of NAT,D() such that
A11: y2 = f2.n() & f2.0 = A() and
A12: for n being Nat holds f2.(n+1) = RecFun(n,f2.n);
  for n being Element of NAT holds Q[n,f2.n,f2.(n+1)] by A12;
  hence thesis by A5,A8,A11,A10;
end;

scheme
  SeqBinOpDef{S() -> FinSequence,P[set,set,set]}: (ex x st ex p being
FinSequence st x = p.(len p) & len p = len S() & p.1 = S().1 & for k st 1 <= k
  & k < len S() holds P[S().(k+1),p.k,p.(k+1)]) & for x,y st (ex p being
FinSequence st x = p.(len p) & len p = len S() & p.1 = S().1 & for k st 1 <= k
& k < len S() holds P[S().(k+1),p.k,p.(k+1)]) & (ex p being FinSequence st y =
p.(len p) & len p = len S() & p.1 = S().1 & for k st 1 <= k & k < len S() holds
  P[S().(k+1),p.k,p.(k+1)]) holds x = y
provided
A1: for k,y st 1 <= k & k < len S() ex z st P[S().(k+1),y,z] and
A2: for k,x,y1,y2,z st 1 <= k & k < len S() & z = S().(k+1) & P[z,x,y1]
& P[z,x,y2] holds y1 = y2
proof
A3: for k,x,y1,y2,z st 1 <= k & k < len S() & z = S().(k+1) & P[z,x,y1] & P[
  z,x,y2] holds y1 = y2 by A2;
A4: for k,y st 1 <= k & k < len S() ex z st P[S().(k+1),y,z] by A1;
  thus ex x st ex p being FinSequence st x = p.(len p) & len p = len S() & p.1
  = S().1 & for k st 1 <= k & k < len S() holds P[S().(k+1),p.k,p.(k+1)] from
  SeqBinOpEx(A4);
  let x,y;
  assume
A5: ex p being FinSequence st x = p.(len p) & len p = len S() & p.1 = S(
  ).1 & for k st 1 <= k & k < len S() holds P[S().(k+1),p.k,p.(k+1)];
  assume
A6: ex p being FinSequence st y = p.(len p) & len p = len S() & p.1 = S(
  ).1 & for k st 1 <= k & k < len S() holds P[S().(k+1),p.k,p.(k+1)];
  thus x = y from SeqBinOpUn(A3,A5,A6);
end;

scheme
  LambdaSeqBinOpDef{S() -> FinSequence,F(set,set) -> set}: (ex x st ex p being
FinSequence st x = p.(len p) & len p = len S() & p.1 = S().1 & for k st 1 <= k
  & k < len S() holds p.(k+1) = F(S().(k+1),p.k)) & for x,y st (ex p being
FinSequence st x = p.(len p) & len p = len S() & p.1 = S().1 & for k st 1 <= k
& k < len S() holds p.(k+1) = F(S().(k+1),p.k)) & (ex p being FinSequence st y
  = p.(len p) & len p = len S() & p.1 = S().1 & for k st 1 <= k & k < len S()
  holds p.(k+1) = F(S().(k+1),p.k)) holds x = y proof
  thus ex x st ex p being FinSequence st x = p.(len p) & len p = len S() & p.1
  = S().1 & for k st 1 <= k & k < len S() holds p.(k+1) = F(S().(k+1),p.k) from
  LambdaSeqBinOpEx;
  let x,y;
  assume
A1: ex p being FinSequence st x = p.(len p) & len p = len S() & p.1 = S(
  ).1 & for k st 1 <= k & k < len S() holds p.(k+1) = F(S().(k+1),p.k);
  assume
A2: ex p being FinSequence st y = p.(len p) & len p = len S() & p.1 = S(
  ).1 & for k st 1 <= k & k < len S() holds p.(k+1) = F(S().(k+1),p.k);
  thus x = y from LambdaSeqBinOpUn(A1,A2);
end;



scheme
  SeqExD{D() -> non empty set, N() -> Element of NAT, P[set,set]}: ex p being
FinSequence of D() st dom p = Seg N() & for k being Element of NAT st k in Seg
  N() holds P[k,p/.k]
provided
A1: for k being Element of NAT st k in Seg N() ex x being Element of D()
st P[k,x]
proof
  per cases;
  suppose
A2: N() = 0;
    take <*>(D());
    thus thesis by A2;
  end;
  suppose
A3: N() <> 0;
    now
      assume
A4:   Seg N() = {};
      now
        per cases;
        case
          N() = 0;
          hence contradiction by A3;
        end;
        case
          N() <> 0;
          hence contradiction by A4;
        end;
      end;
      hence contradiction;
    end;
    then reconsider M = Seg N() as non empty set;
A5: for x being Element of M ex y being Element of D() st P[x,y]
    proof
      let x be Element of M;
      x in Seg N();
      hence thesis by A1;
    end;
    consider f being Function of M,D() such that
A6: for x being Element of M holds P[x,f.x] from FUNCT_2:sch 3(A5);
    dom f = Seg N() by FUNCT_2:def 1;
    then reconsider q = f as FinSequence by FINSEQ_1:def 2;
    now
      let u be set;
A7:   rng q c= D() by RELAT_1:def 19;
      assume u in rng q;
      hence u in D() by A7;
    end;
    then rng q c= D() by TARSKI:def 3;
    then reconsider q as FinSequence of D() by FINSEQ_1:def 4;
    take q;
    now
      let k be Element of NAT;
      assume
A8:   k in Seg N();
      then k in dom q by FUNCT_2:def 1;
      then q.k = q/.k by PARTFUN1:def 6;
      hence P[k,q/.k] by A6,A8;
    end;
    hence thesis by FUNCT_2:def 1;
  end;
end;

scheme
  FinRecExD2{D() -> non empty set,A() -> (Element of D()), N() -> Element of
NAT, P[set,set,set]}: ex p being FinSequence of D() st len p = N() & (p/.1 = A(
) or N() = 0) & for n being Element of NAT st 1 <= n & n <= N()-1 holds P[n,p/.
  n,p/.(n+1)]
provided
A1: for n being Element of NAT st 1 <= n & n <= N()-1 holds for x being
Element of D() ex y being Element of D() st P[n,x,y]
proof
  set 00 = the Element of D();
  defpred Q[Element of NAT,set,set] means (0 <= $1 & $1 <= N()-2 implies P[$1+
  1,$2,$3]) & (not (0 <= $1 & $1 <= N()-2) implies $3=00);
A2: for n be Element of NAT for x be Element of D() ex y be Element of D()
  st Q[n,x,y]
  proof
    let n be Element of NAT,x be Element of D();
    0 <= n & n <= N()-2 implies thesis
    proof
A3:   0+1 <= n+1 by XREAL_1:6;
      assume that
      0 <= n and
A4:   n <= N()-2;
      n+1 <= N()-2+1 by A4,XREAL_1:6;
      then consider y being Element of D() such that
A5:   P[n+1,x,y] by A1,A3;
      take y;
      thus 0 <= n & n <= N()-2 implies P[n+1,x,y] by A5;
      thus thesis by A4;
    end;
    hence thesis;
  end;
  consider f being Function of NAT,D() such that
A6: f.0 = A() & for n being Element of NAT holds Q[n,f.n,f.(n+1)]
   from RecExD(A2);
  defpred Q[set,set] means for r being Real st r = $1 holds $2 = f.(r-1);
A7: for x being set st x in REAL ex y being set st Q[x,y]
  proof
    let x be set;
    assume x in REAL;
    then reconsider r=x as Real;
    take f.(r-1);
    thus thesis;
  end;
  consider g being Function such that
A8: dom g = REAL & for x being set st x in REAL holds Q[x,g.x] from
  CLASSES1:sch 1(A7);
A9: dom (g|Seg N()) = Seg N() by A8,RELAT_1:62,XBOOLE_1:1;
  then reconsider p = g|Seg N() as FinSequence by FINSEQ_1:def 2;
  now
    let x be set;
    assume x in rng p;
    then consider y being set such that
A10: y in dom p and
A11: x = p.y by FUNCT_1:def 3;
    reconsider y as Element of NAT by A10;
A12: f.(y-1) in D()
    proof
      y <> 0 by A9,A10,FINSEQ_1:1;
      then consider k being Nat such that
A13:  y = k+1 by NAT_1:6;
      reconsider k as Element of NAT by ORDINAL1:def 12;
      f.k in D();
      hence thesis by A13;
    end;
    p.y = g.y by A10,FUNCT_1:47;
    hence x in D() by A8,A11,A12;
  end;
  then rng p c= D() by TARSKI:def 3;
  then reconsider p as FinSequence of D() by FINSEQ_1:def 4;
  take p;
  thus len p = N() by A9,FINSEQ_1:def 3;
  N() <> 0 implies p/.1 = A()
  proof
    assume N() <> 0;
    then consider k being Nat such that
A14: N() = k+1 by NAT_1:6;
    0 + 1 <= k +1 by XREAL_1:7;
    then
A15: 1 in Seg N() by A14,FINSEQ_1:1;
    then p/.1 = p.1 by A9,PARTFUN1:def 6
      .= g.1 by A15,FUNCT_1:49
      .= f.(1-1) by A8
      .= A() by A6;
    hence thesis;
  end;
  hence p/.1 = A() or N() = 0;
  let n be Element of NAT;
  assume that
A16: 1 <= n and
A17: n <= N()-1;
  consider k being Nat such that
A18: n = k+1 by A16,NAT_1:6;
A19: for n being Element of NAT st n <= N()-1 holds p/.(n+1) = f.n
  proof
    let n be Element of NAT;
    assume n <= N() - 1;
    then
A20: n+1 <= N()-1+1 by XREAL_1:6;
A21: g.(n+1) = f.(n+1-1) by A8
      .= f.n;
    1 <= n+1 by NAT_1:11;
    then
A22: n+1 in Seg N() by A20,FINSEQ_1:1;
    then p/.(n+1) = p.(n+1) by A9,PARTFUN1:def 6;
    hence thesis by A22,A21,FUNCT_1:49;
  end;
A23: k <= k+1 by NAT_1:11;
A24: k in NAT by ORDINAL1:def 12;
  k <= N()-1-1 by A17,A18,XREAL_1:19;
  then P[k+1,f.k,f.(k+1)] by A6,A24;
  then P[k+1,f.k,p/.(k+1+1)] by A19,A17,A18;
  hence thesis by A19,A17,A18,A24,A23,XXREAL_0:2;
end;














begin 

Lm1: for f being FinSequence, h being Function st dom h = dom f holds h is
FinSequence
proof
  let f be FinSequence, h be Function such that
A1: dom h = dom f;
  h is FinSequence-like
  proof
    take len f;
    thus thesis by A1,FINSEQ_1:def 3;
  end;
  hence thesis;
end;

Lm2: for f, g being FinSequence, h being Function st dom h = dom f /\ dom g
holds h is FinSequence
proof
  let f, g be FinSequence, h be Function such that
A1: dom h = dom f /\ dom g;
  consider n being Nat such that
A2: dom f = Seg n by FINSEQ_1:def 2;
  consider m being Nat such that
A3: dom g = Seg m by FINSEQ_1:def 2;
  h is FinSequence-like
  proof
    per cases;
    suppose
A4:   n <= m;
      take n;
      thus thesis by A1,A2,A3,A4,FINSEQ_1:7;
    end;
    suppose
A5:   m <= n;
      take m;
      thus thesis by A1,A2,A3,A5,FINSEQ_1:7;
    end;
  end;
  hence thesis;
end;

registration
  cluster complex-valued for FinSequence;
  existence
  proof
    take <*>COMPLEX;
    thus thesis;
  end;
end;


registration
  let r be rational number;
  cluster |. r .| -> rational;
  coherence
  proof
    |. r .| = -r or |. r .| = r by COMPLEX1:71;
    hence thesis;
  end;
end;

definition
  let f1,f2 be complex-valued Function;
  deffunc F(set) = f1.$1 + f2.$1;
  set X = dom f1 /\ dom f2;
  func f1 + f2 -> Function means
  :Def1:
  dom it = dom f1 /\ dom f2 &
  for c being set st c in dom it holds it.c = f1.c + f2.c;
  existence
  proof
    ex f being Function st dom f = X & for x being set st x in X holds f.x
    = F(x) from FUNCT_1:sch 3;
    hence thesis;
  end;
  uniqueness
  proof
    let f, g be Function such that
A1: dom f = X and
A2: for c being set st c in dom f holds f.c = F(c) and
A3: dom g = X and
A4: for c being set st c in dom g holds g.c = F(c);
    now
      let x be set;
      assume
A5:   x in dom f;
      hence f.x = F(x) by A2
        .= g.x by A1,A3,A4,A5;
    end;
    hence thesis by A1,A3,FUNCT_1:2;
  end;
  commutativity;
end;

registration
  let f1,f2 be complex-valued Function;
  cluster f1+f2 -> complex-valued;
  coherence
  proof
    let x be set;
    assume x in dom (f1+f2);
    then (f1+f2).x = f1.x + f2.x by Def1;
    hence thesis;
  end;
end;

registration
  let f1,f2 be real-valued Function;
  cluster f1+f2 -> real-valued;
  coherence
  proof
    let x be set;
    assume x in dom (f1+f2);
    then (f1+f2).x = f1.x + f2.x by Def1;
    hence thesis;
  end;
end;

registration
  let f1,f2 be RAT-valued Function;
  cluster f1+f2 -> RAT-valued;
  coherence
  proof
    let y be set;
    assume y in rng(f1+f2);
    then consider x being set such that
A1: x in dom (f1+f2) and
A2: (f1+f2).x = y by FUNCT_1:def 3;
    (f1+f2).x = f1.x + f2.x by A1,Def1;
    hence thesis by A2,RAT_1:def 2;
  end;
end;

registration
  let f1,f2 be INT-valued Function;
  cluster f1+f2 -> INT-valued;
  coherence
  proof
    let y be set;
    assume y in rng(f1+f2);
    then consider x being set such that
A1: x in dom (f1+f2) and
A2: (f1+f2).x = y by FUNCT_1:def 3;
    (f1+f2).x = f1.x + f2.x by A1,Def1;
    hence thesis by A2,INT_1:def 2;
  end;
end;

registration
  let f1,f2 be natural-valued Function;
  cluster f1+f2 -> natural-valued;
  coherence
  proof
    let x be set;
    assume x in dom (f1+f2);
    then (f1+f2).x = f1.x + f2.x by Def1;
    hence thesis;
  end;
end;

definition
  let C be set;
  let D1,D2 be complex-membered set;
  let f1 be PartFunc of C,D1;
  let f2 be PartFunc of C,D2;
  redefine func f1+f2 -> PartFunc of C,COMPLEX;
  coherence
  proof
    dom (f1+f2) = dom f1 /\ dom f2 & rng (f1+f2) c= COMPLEX by Def1,
VALUED_0:def 1;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D1,D2 be real-membered set;
  let f1 be PartFunc of C,D1;
  let f2 be PartFunc of C,D2;
  redefine func f1+f2 -> PartFunc of C,REAL;
  coherence
  proof
    dom (f1+f2) = dom f1 /\ dom f2 & rng (f1+f2) c= REAL by Def1,VALUED_0:def 3
;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D1,D2 be rational-membered set;
  let f1 be PartFunc of C,D1;
  let f2 be PartFunc of C,D2;
  redefine func f1+f2 -> PartFunc of C,RAT;
  coherence
  proof
    dom (f1+f2) = dom f1 /\ dom f2 & rng (f1+f2) c= RAT by Def1,RELAT_1:def 19;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D1,D2 be integer-membered set;
  let f1 be PartFunc of C,D1;
  let f2 be PartFunc of C,D2;
  redefine func f1+f2 -> PartFunc of C,INT;
  coherence
  proof
    dom (f1+f2) = dom f1 /\ dom f2 & rng (f1+f2) c= INT by Def1,RELAT_1:def 19;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D1,D2 be natural-membered set;
  let f1 be PartFunc of C,D1;
  let f2 be PartFunc of C,D2;
  redefine func f1+f2 -> PartFunc of C,NAT;
  coherence
  proof
    dom (f1+f2) = dom f1 /\ dom f2 & rng (f1+f2) c= NAT by Def1,VALUED_0:def 6;
    hence thesis by RELSET_1:4;
  end;
end;

registration
  let C be set;
  let D1,D2 be complex-membered non empty set;
  let f1 be Function of C,D1;
  let f2 be Function of C,D2;
  cluster f1+f2 -> total for PartFunc of C,COMPLEX;
  coherence
  proof
    dom f1 = C & dom f2 = C by FUNCT_2:def 1;
    then dom(f1+f2) = C /\ C by Def1
      .= C;
    hence thesis by PARTFUN1:def 2;
  end;
end;

registration
  let C be set;
  let D1,D2 be real-membered non empty set;
  let f1 be Function of C,D1;
  let f2 be Function of C,D2;
  cluster f1+f2 -> total for PartFunc of C,REAL;
  coherence
  proof
    dom f1 = C & dom f2 = C by FUNCT_2:def 1;
    then dom(f1+f2) = C /\ C by Def1
      .= C;
    hence thesis by PARTFUN1:def 2;
  end;
end;

registration
  let C be set;
  let D1,D2 be rational-membered non empty set;
  let f1 be Function of C,D1;
  let f2 be Function of C,D2;
  cluster f1+f2 -> total for PartFunc of C,RAT;
  coherence
  proof
    dom f1 = C & dom f2 = C by FUNCT_2:def 1;
    then dom(f1+f2) = C /\ C by Def1
      .= C;
    hence thesis by PARTFUN1:def 2;
  end;
end;

registration
  let C be set;
  let D1,D2 be integer-membered non empty set;
  let f1 be Function of C,D1;
  let f2 be Function of C,D2;
  cluster f1+f2 -> total for PartFunc of C,INT;
  coherence
  proof
    dom f1 = C & dom f2 = C by FUNCT_2:def 1;
    then dom(f1+f2) = C /\ C by Def1
      .= C;
    hence thesis by PARTFUN1:def 2;
  end;
end;

registration
  let C be set;
  let D1,D2 be natural-membered non empty set;
  let f1 be Function of C,D1;
  let f2 be Function of C,D2;
  cluster f1+f2 -> total for PartFunc of C,NAT;
  coherence
  proof
    dom f1 = C & dom f2 = C by FUNCT_2:def 1;
    then dom(f1+f2) = C /\ C by Def1
      .= C;
    hence thesis by PARTFUN1:def 2;
  end;
end;

theorem
  for C being set, D1,D2 being complex-membered non empty set for f1
  being Function of C,D1, f2 being Function of C,D2 for c being Element of C
  holds (f1+f2).c = f1.c + f2.c
proof
  let C be set;
  let D1,D2 be complex-membered non empty set;
  let f1 be Function of C,D1, f2 be Function of C,D2;
A1: dom(f1+f2) = C by FUNCT_2:def 1;
  let c be Element of C;
  per cases;
  suppose
    C is non empty;
    hence thesis by A1,Def1;
  end;
  suppose
A2: C is empty;
    then dom f1 = {};
    then f1.c = 0 by FUNCT_1:def 2;
    hence thesis by A2;
  end;
end;

registration
  let f1, f2 be complex-valued FinSequence;
  cluster f1+f2 -> FinSequence-like;
  coherence
  proof
    dom(f1+f2) = dom f1 /\ dom f2 by Def1;
    hence thesis by Lm2;
  end;
end;

begin 

definition
  let f be complex-valued Function, r be complex number;
  deffunc F(set) = r + f.$1;
  func r + f -> Function means
  :Def2:
  dom it = dom f &
  for c being set st c in dom it holds it.c = r + f.c;
  existence
  proof
    ex g being Function st dom g = dom f & for x being set st x in dom f
    holds g.x = F(x) from FUNCT_1:sch 3;
    hence thesis;
  end;
  uniqueness
  proof
    let h, g be Function such that
A1: dom h = dom f and
A2: for c being set st c in dom h holds h.c = F(c) and
A3: dom g = dom f and
A4: for c being set st c in dom g holds g.c = F(c);
    now
      let x be set;
      assume
A5:   x in dom h;
      hence h.x = F(x) by A2
        .= g.x by A1,A3,A4,A5;
    end;
    hence thesis by A1,A3,FUNCT_1:2;
  end;
end;

notation
  let f be complex-valued Function, r be complex number;
  synonym f + r for r + f;
end;

registration
  let f be complex-valued Function, r be complex number;
  cluster r+f -> complex-valued;
  coherence
  proof
    let x be set;
    assume x in dom (r+f);
    then (r+f).x = r + f.x by Def2;
    hence thesis;
  end;
end;

registration
  let f be real-valued Function, r be real number;
  cluster r+f -> real-valued;
  coherence
  proof
    let x be set;
    assume x in dom (r+f);
    then (r+f).x = r + f.x by Def2;
    hence thesis;
  end;
end;

registration
  let f be RAT-valued Function, r be rational number;
  cluster r+f -> RAT-valued;
  coherence
  proof
    let y be set;
    assume y in rng (r+f);
    then consider x being set such that
A1: x in dom (r+f) and
A2: (r+f).x = y by FUNCT_1:def 3;
    (r+f).x = r + f.x by A1,Def2;
    hence thesis by A2,RAT_1:def 2;
  end;
end;

registration
  let f be INT-valued Function, r be integer number;
  cluster r+f -> INT-valued;
  coherence
  proof
    let y be set;
    assume y in rng (r+f);
    then consider x being set such that
A1: x in dom (r+f) and
A2: (r+f).x = y by FUNCT_1:def 3;
    (r+f).x = r + f.x by A1,Def2;
    hence thesis by A2,INT_1:def 2;
  end;
end;

registration
  let f be natural-valued Function, r be Nat;
  cluster r+f -> natural-valued;
  coherence
  proof
    let x be set;
    assume x in dom (r+f);
    then (r+f).x = r + f.x by Def2;
    hence thesis;
  end;
end;

definition
  let C be set;
  let D be complex-membered set;
  let f be PartFunc of C,D;
  let r be complex number;
  redefine func r+f -> PartFunc of C,COMPLEX;
  coherence
  proof
    dom (r+f) = dom f & rng (r+f) c= COMPLEX by Def2,VALUED_0:def 1;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D be real-membered set;
  let f be PartFunc of C,D;
  let r be real number;
  redefine func r+f -> PartFunc of C,REAL;
  coherence
  proof
    dom (r+f) = dom f & rng (r+f) c= REAL by Def2,VALUED_0:def 3;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D be rational-membered set;
  let f be PartFunc of C,D;
  let r be rational number;
  redefine func r+f -> PartFunc of C,RAT;
  coherence
  proof
    dom (r+f) = dom f & rng (r+f) c= RAT by Def2,RELAT_1:def 19;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D be integer-membered set;
  let f be PartFunc of C,D;
  let r be integer number;
  redefine func r+f -> PartFunc of C,INT;
  coherence
  proof
    dom (r+f) = dom f & rng (r+f) c= INT by Def2,RELAT_1:def 19;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D be natural-membered set;
  let f be PartFunc of C,D;
  let r be Nat;
  redefine func r+f -> PartFunc of C,NAT;
  coherence
  proof
    dom (r+f) = dom f & rng (r+f) c= NAT by Def2,VALUED_0:def 6;
    hence thesis by RELSET_1:4;
  end;
end;

registration
  let C be set;
  let D be complex-membered non empty set;
  let f be Function of C,D;
  let r be complex number;
  cluster r+f -> total for PartFunc of C,COMPLEX;
  coherence
  proof
    dom(r+f) = dom f by Def2
      .= C by FUNCT_2:def 1;
    hence thesis by PARTFUN1:def 2;
  end;
end;

registration
  let C be set;
  let D be real-membered non empty set;
  let f be Function of C,D;
  let r be real number;
  cluster r+f -> total for PartFunc of C,REAL;
  coherence
  proof
    dom(r+f) = dom f by Def2
      .= C by FUNCT_2:def 1;
    hence thesis by PARTFUN1:def 2;
  end;
end;

registration
  let C be set;
  let D be rational-membered non empty set;
  let f be Function of C,D;
  let r be rational number;
  cluster r+f -> total for PartFunc of C,RAT;
  coherence
  proof
    dom(r+f) = dom f by Def2
      .= C by FUNCT_2:def 1;
    hence thesis by PARTFUN1:def 2;
  end;
end;

registration
  let C be set;
  let D be integer-membered non empty set;
  let f be Function of C,D;
  let r be integer number;
  cluster r+f -> total for PartFunc of C,INT;
  coherence
  proof
    dom(r+f) = dom f by Def2
      .= C by FUNCT_2:def 1;
    hence thesis by PARTFUN1:def 2;
  end;
end;

registration
  let C be set;
  let D be natural-membered non empty set;
  let f be Function of C,D;
  let r be Nat;
  cluster r+f -> total for PartFunc of C,NAT;
  coherence
  proof
    dom(r+f) = dom f by Def2
      .= C by FUNCT_2:def 1;
    hence thesis by PARTFUN1:def 2;
  end;
end;

theorem
  for C being non empty set, D being complex-membered non empty set
  for f being Function of C,D, r being complex number
  for c being Element of C holds
  (r+f).c = r + f.c
proof
  let C be non empty set;
  let D be complex-membered non empty set;
  let f be Function of C,D, r be complex number;
  dom(r+f) = C by FUNCT_2:def 1;
  hence thesis by Def2;
end;

registration
  let f be complex-valued FinSequence, r be complex number;
  cluster r+f -> FinSequence-like;
  coherence
  proof
    dom (r+f) = dom f by Def2;
    hence thesis by Lm1;
  end;
end;

begin 

definition
  let f be complex-valued Function, r be complex number;
  func f - r -> Function equals
  -r + f;
  coherence;
end;

theorem
  for f being complex-valued Function, r being complex number holds dom
  (f-r) = dom f & for c being set st c in dom f holds (f-r).c = f.c - r
proof
  let f be complex-valued Function, r be complex number;
  dom (f-r) = dom f by Def2;
  hence thesis by Def2;
end;

registration
  let f be complex-valued Function, r be complex number;
  cluster f-r -> complex-valued;
  coherence;
end;

registration
  let f be real-valued Function, r be real number;
  cluster f-r -> real-valued;
  coherence;
end;

registration
  let f be RAT-valued Function, r be rational number;
  cluster f-r -> RAT-valued;
  coherence;
end;

registration
  let f be INT-valued Function, r be integer number;
  cluster f-r -> INT-valued;
  coherence;
end;

definition
  let C be set;
  let D be complex-membered set;
  let f be PartFunc of C,D;
  let r be complex number;
  redefine func f-r -> PartFunc of C,COMPLEX;
  coherence
  proof
    dom (f-r) = dom f & rng (f-r) c= COMPLEX by Def2,VALUED_0:def 1;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D be real-membered set;
  let f be PartFunc of C,D;
  let r be real number;
  redefine func f-r -> PartFunc of C,REAL;
  coherence
  proof
    dom (f-r) = dom f & rng (f-r) c= REAL by Def2,VALUED_0:def 3;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D be rational-membered set;
  let f be PartFunc of C,D;
  let r be rational number;
  redefine func f-r -> PartFunc of C,RAT;
  coherence
  proof
    dom (f-r) = dom f & rng (f-r) c= RAT by Def2,RELAT_1:def 19;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D be integer-membered set;
  let f be PartFunc of C,D;
  let r be integer number;
  redefine func f-r -> PartFunc of C,INT;
  coherence
  proof
    dom (f-r) = dom f & rng (f-r) c= INT by Def2,RELAT_1:def 19;
    hence thesis by RELSET_1:4;
  end;
end;

registration
  let C be set;
  let D be complex-membered non empty set;
  let f be Function of C,D;
  let r be complex number;
  cluster f-r -> total for PartFunc of C,COMPLEX;
  coherence;
end;

registration
  let C be set;
  let D be real-membered non empty set;
  let f be Function of C,D;
  let r be real number;
  cluster f-r -> total for PartFunc of C,REAL;
  coherence;
end;

registration
  let C be set;
  let D be rational-membered non empty set;
  let f be Function of C,D;
  let r be rational number;
  cluster f-r -> total for PartFunc of C,RAT;
  coherence;
end;

registration
  let C be set;
  let D be integer-membered non empty set;
  let f be Function of C,D;
  let r be integer number;
  cluster f-r -> total for PartFunc of C,INT;
  coherence;
end;

theorem
  for C being non empty set, D being complex-membered non empty set for
f being Function of C,D, r being complex number for c being Element of C holds
  (f-r).c = f.c - r
proof
  let C be non empty set;
  let D be complex-membered non empty set;
  let f be Function of C,D, r be complex number;
  dom (f-r) = C by FUNCT_2:def 1;
  hence thesis by Def2;
end;

registration
  let f be complex-valued FinSequence, r be complex number;
  cluster f-r -> FinSequence-like;
  coherence;
end;

begin 

definition
  let f1,f2 be complex-valued Function;
  deffunc F(set) = f1.$1 * f2.$1;
  set X = dom f1 /\ dom f2;
  func f1 (#) f2 -> Function means
  :Def4:
  dom it = dom f1 /\ dom f2 & for c
  being set st c in dom it holds it.c = f1.c * f2.c;
  existence
  proof
    ex f being Function st dom f = X & for x being set st x in X holds f.x
    = F(x) from FUNCT_1:sch 3;
    hence thesis;
  end;
  uniqueness
  proof
    let f, g be Function such that
A1: dom f = X and
A2: for c being set st c in dom f holds f.c = F(c) and
A3: dom g = X and
A4: for c being set st c in dom g holds g.c = F(c);
    now
      let x be set;
      assume
A5:   x in dom f;
      hence f.x = F(x) by A2
        .= g.x by A1,A3,A4,A5;
    end;
    hence thesis by A1,A3,FUNCT_1:2;
  end;
  commutativity;
end;

theorem
  for f1,f2 being complex-valued Function for c being set holds (f1(#)f2
  ).c = f1.c * f2.c
proof
  let f1,f2 be complex-valued Function;
  let c be set;
A1: dom (f1(#)f2) = dom f1 /\ dom f2 by Def4;
  per cases;
  suppose
    c in dom (f1(#)f2);
    hence thesis by Def4;
  end;
  suppose
A2: not c in dom (f1(#)f2);
    then not c in dom f1 or not c in dom f2 by A1,XBOOLE_0:def 4;
    then f1.c = 0 or f2.c = 0 by FUNCT_1:def 2;
    hence thesis by A2,FUNCT_1:def 2;
  end;
end;

registration
  let f1,f2 be complex-valued Function;
  cluster f1(#)f2 -> complex-valued;
  coherence
  proof
    let x be set;
    assume x in dom (f1(#)f2);
    then (f1(#)f2).x = f1.x * f2.x by Def4;
    hence thesis;
  end;
end;

registration
  let f1,f2 be real-valued Function;
  cluster f1(#)f2 -> real-valued;
  coherence
  proof
    let x be set;
    assume x in dom (f1(#)f2);
    then (f1(#)f2).x = f1.x * f2.x by Def4;
    hence thesis;
  end;
end;

registration
  let f1,f2 be RAT-valued Function;
  cluster f1(#)f2 -> RAT-valued;
  coherence
  proof
    let y be set;
    assume y in rng (f1(#)f2);
    then consider x being set such that
A1: x in dom (f1(#)f2) and
A2: (f1(#)f2).x = y by FUNCT_1:def 3;
    (f1(#)f2).x = f1.x * f2.x by A1,Def4;
    hence thesis by A2,RAT_1:def 2;
  end;
end;

registration
  let f1,f2 be INT-valued Function;
  cluster f1(#)f2 -> INT-valued;
  coherence
  proof
    let y be set;
    assume y in rng (f1(#)f2);
    then consider x being set such that
A1: x in dom (f1(#)f2) and
A2: (f1(#)f2).x = y by FUNCT_1:def 3;
    (f1(#)f2).x = f1.x * f2.x by A1,Def4;
    hence thesis by A2,INT_1:def 2;
  end;
end;

registration
  let f1,f2 be natural-valued Function;
  cluster f1(#)f2 -> natural-valued;
  coherence
  proof
    let x be set;
    assume x in dom (f1(#)f2);
    then (f1(#)f2).x = f1.x * f2.x by Def4;
    hence thesis;
  end;
end;

definition
  let C be set;
  let D1,D2 be complex-membered set;
  let f1 be PartFunc of C,D1;
  let f2 be PartFunc of C,D2;
  redefine func f1(#)f2 -> PartFunc of C,COMPLEX;
  coherence
  proof
    dom (f1(#)f2) = dom f1 /\ dom f2 & rng (f1(#)f2) c= COMPLEX by Def4,
VALUED_0:def 1;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D1,D2 be real-membered set;
  let f1 be PartFunc of C,D1;
  let f2 be PartFunc of C,D2;
  redefine func f1(#)f2 -> PartFunc of C,REAL;
  coherence
  proof
    dom (f1(#)f2) = dom f1 /\ dom f2 & rng (f1(#)f2) c= REAL by Def4,
VALUED_0:def 3;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D1,D2 be rational-membered set;
  let f1 be PartFunc of C,D1;
  let f2 be PartFunc of C,D2;
  redefine func f1(#)f2 -> PartFunc of C,RAT;
  coherence
  proof
    dom (f1(#)f2) = dom f1 /\ dom f2 & rng (f1(#)f2) c= RAT by Def4,
RELAT_1:def 19;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D1,D2 be integer-membered set;
  let f1 be PartFunc of C,D1;
  let f2 be PartFunc of C,D2;
  redefine func f1(#)f2 -> PartFunc of C,INT;
  coherence
  proof
    dom (f1(#)f2) = dom f1 /\ dom f2 & rng (f1(#)f2) c= INT by Def4,
RELAT_1:def 19;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D1,D2 be natural-membered set;
  let f1 be PartFunc of C,D1;
  let f2 be PartFunc of C,D2;
  redefine func f1(#)f2 -> PartFunc of C,NAT;
  coherence
  proof
    dom (f1(#)f2) = dom f1 /\ dom f2 & rng (f1(#)f2) c= NAT by Def4,
VALUED_0:def 6;
    hence thesis by RELSET_1:4;
  end;
end;

registration
  let C be set;
  let D1,D2 be complex-membered non empty set;
  let f1 be Function of C,D1;
  let f2 be Function of C,D2;
  cluster f1(#)f2 -> total for PartFunc of C,COMPLEX;
  coherence
  proof
    dom f1 = C & dom f2 = C by FUNCT_2:def 1;
    then dom(f1(#)f2) = C /\ C by Def4
      .= C;
    hence thesis by PARTFUN1:def 2;
  end;
end;

registration
  let C be set;
  let D1,D2 be real-membered non empty set;
  let f1 be Function of C,D1;
  let f2 be Function of C,D2;
  cluster f1(#)f2 -> total for PartFunc of C,REAL;
  coherence
  proof
    dom f1 = C & dom f2 = C by FUNCT_2:def 1;
    then dom(f1(#)f2) = C /\ C by Def4
      .= C;
    hence thesis by PARTFUN1:def 2;
  end;
end;

registration
  let C be set;
  let D1,D2 be rational-membered non empty set;
  let f1 be Function of C,D1;
  let f2 be Function of C,D2;
  cluster f1(#)f2 -> total for PartFunc of C,RAT;
  coherence
  proof
    dom f1 = C & dom f2 = C by FUNCT_2:def 1;
    then dom(f1(#)f2) = C /\ C by Def4
      .= C;
    hence thesis by PARTFUN1:def 2;
  end;
end;

registration
  let C be set;
  let D1,D2 be integer-membered non empty set;
  let f1 be Function of C,D1;
  let f2 be Function of C,D2;
  cluster f1(#)f2 -> total for PartFunc of C,INT;
  coherence
  proof
    dom f1 = C & dom f2 = C by FUNCT_2:def 1;
    then dom(f1(#)f2) = C /\ C by Def4
      .= C;
    hence thesis by PARTFUN1:def 2;
  end;
end;

registration
  let C be set;
  let D1,D2 be natural-membered non empty set;
  let f1 be Function of C,D1;
  let f2 be Function of C,D2;
  cluster f1(#)f2 -> total for PartFunc of C,NAT;
  coherence
  proof
    dom f1 = C & dom f2 = C by FUNCT_2:def 1;
    then dom(f1(#)f2) = C /\ C by Def4
      .= C;
    hence thesis by PARTFUN1:def 2;
  end;
end;

registration
  let f1, f2 be complex-valued FinSequence;
  cluster f1(#)f2 -> FinSequence-like;
  coherence
  proof
    dom(f1(#)f2) = dom f1 /\ dom f2 by Def4;
    hence thesis by Lm2;
  end;
end;

begin 

definition
  let f be complex-valued Function, r be complex number;
  deffunc F(set) = r * f.$1;
  func r (#) f -> Function means
  :Def5:
  dom it = dom f & for c being set st c
  in dom it holds it.c = r * f.c;
  existence
  proof
    ex g being Function st dom g = dom f & for x being set st x in dom f
    holds g.x = F(x) from FUNCT_1:sch 3;
    hence thesis;
  end;
  uniqueness
  proof
    let h, g be Function such that
A1: dom h = dom f and
A2: for c being set st c in dom h holds h.c = F(c) and
A3: dom g = dom f and
A4: for c being set st c in dom g holds g.c = F(c);
    now
      let x be set;
      assume
A5:   x in dom h;
      hence h.x = F(x) by A2
        .= g.x by A1,A3,A4,A5;
    end;
    hence thesis by A1,A3,FUNCT_1:2;
  end;
end;

notation
  let f be complex-valued Function, r be complex number;
  synonym f (#) r for r (#) f;
end;

theorem Th6:
  for f being complex-valued Function, r being complex number for c
  being set holds (r(#)f).c = r * f.c
proof
  let f be complex-valued Function, r be complex number;
  let c be set;
A1: dom f = dom (r(#)f) by Def5;
  per cases;
  suppose
    c in dom f;
    hence thesis by A1,Def5;
  end;
  suppose
A2: not c in dom f;
    hence (r(#)f).c = r*0 by A1,FUNCT_1:def 2
      .= r * f.c by A2,FUNCT_1:def 2;
  end;
end;

registration
  let f be complex-valued Function, r be complex number;
  cluster r(#)f -> complex-valued;
  coherence
  proof
    let x be set;
    assume x in dom (r(#)f);
    then (r(#)f).x = r * f.x by Def5;
    hence thesis;
  end;
end;

registration
  let f be real-valued Function, r be real number;
  cluster r(#)f -> real-valued;
  coherence
  proof
    let x be set;
    assume x in dom (r(#)f);
    then (r(#)f).x = r * f.x by Def5;
    hence thesis;
  end;
end;

registration
  let f be RAT-valued Function, r be rational number;
  cluster r(#)f -> RAT-valued;
  coherence
  proof
    let y be set;
    assume y in rng (r(#)f);
    then consider x being set such that
A1: x in dom (r(#)f) and
A2: (r(#)f).x = y by FUNCT_1:def 3;
    (r(#)f).x = r * f.x by A1,Def5;
    hence thesis by A2,RAT_1:def 2;
  end;
end;

registration
  let f be INT-valued Function, r be integer number;
  cluster r(#)f -> INT-valued;
  coherence
  proof
    let y be set;
    assume y in rng (r(#)f);
    then consider x being set such that
A1: x in dom (r(#)f) and
A2: (r(#)f).x = y by FUNCT_1:def 3;
    (r(#)f).x = r * f.x by A1,Def5;
    hence thesis by A2,INT_1:def 2;
  end;
end;

registration
  let f be natural-valued Function, r be Nat;
  cluster r(#)f -> natural-valued;
  coherence
  proof
    let x be set;
    assume x in dom (r(#)f);
    then (r(#)f).x = r * f.x by Def5;
    hence thesis;
  end;
end;

definition
  let C be set;
  let D be complex-membered set;
  let f be PartFunc of C,D;
  let r be complex number;
  redefine func r(#)f -> PartFunc of C,COMPLEX;
  coherence
  proof
    dom (r(#)f) = dom f & rng (r(#)f) c= COMPLEX by Def5,VALUED_0:def 1;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D be real-membered set;
  let f be PartFunc of C,D;
  let r be real number;
  redefine func r(#)f -> PartFunc of C,REAL;
  coherence
  proof
    dom (r(#)f) = dom f & rng (r(#)f) c= REAL by Def5,VALUED_0:def 3;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D be rational-membered set;
  let f be PartFunc of C,D;
  let r be rational number;
  redefine func r(#)f -> PartFunc of C,RAT;
  coherence
  proof
    dom (r(#)f) = dom f & rng (r(#)f) c= RAT by Def5,RELAT_1:def 19;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D be integer-membered set;
  let f be PartFunc of C,D;
  let r be integer number;
  redefine func r(#)f -> PartFunc of C,INT;
  coherence
  proof
    dom (r(#)f) = dom f & rng (r(#)f) c= INT by Def5,RELAT_1:def 19;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D be natural-membered set;
  let f be PartFunc of C,D;
  let r be Nat;
  redefine func r(#)f -> PartFunc of C,NAT;
  coherence
  proof
    dom (r(#)f) = dom f & rng (r(#)f) c= NAT by Def5,VALUED_0:def 6;
    hence thesis by RELSET_1:4;
  end;
end;

registration
  let C be set;
  let D be complex-membered non empty set;
  let f be Function of C,D;
  let r be complex number;
  cluster r(#)f -> total for PartFunc of C,COMPLEX;
  coherence
  proof
    dom(r(#)f) = dom f by Def5
      .= C by FUNCT_2:def 1;
    hence thesis by PARTFUN1:def 2;
  end;
end;

registration
  let C be set;
  let D be real-membered non empty set;
  let f be Function of C,D;
  let r be real number;
  cluster r(#)f -> total for PartFunc of C,REAL;
  coherence
  proof
    dom(r(#)f) = dom f by Def5
      .= C by FUNCT_2:def 1;
    hence thesis by PARTFUN1:def 2;
  end;
end;

registration
  let C be set;
  let D be rational-membered non empty set;
  let f be Function of C,D;
  let r be rational number;
  cluster r(#)f -> total for PartFunc of C,RAT;
  coherence
  proof
    dom(r(#)f) = dom f by Def5
      .= C by FUNCT_2:def 1;
    hence thesis by PARTFUN1:def 2;
  end;
end;

registration
  let C be set;
  let D be integer-membered non empty set;
  let f be Function of C,D;
  let r be integer number;
  cluster r(#)f -> total for PartFunc of C,INT;
  coherence
  proof
    dom(r(#)f) = dom f by Def5
      .= C by FUNCT_2:def 1;
    hence thesis by PARTFUN1:def 2;
  end;
end;

registration
  let C be set;
  let D be natural-membered non empty set;
  let f be Function of C,D;
  let r be Nat;
  cluster r(#)f -> total for PartFunc of C,NAT;
  coherence
  proof
    dom(r(#)f) = dom f by Def5
      .= C by FUNCT_2:def 1;
    hence thesis by PARTFUN1:def 2;
  end;
end;

theorem
  for C being non empty set, D being complex-membered non empty set for
  f being Function of C,D, r being complex number for g being Function of C,
  COMPLEX st for c being Element of C holds g.c = r * f.c holds g = r(#)f
proof
  let C be non empty set, D be complex-membered non empty set;
  let f be Function of C,D, r be complex number;
  let g be Function of C,COMPLEX such that
A1: for c being Element of C holds g.c = r * f.c;
  let x be Element of C;
  thus g.x = r*f.x by A1
    .= (r(#)f).x by Th6;
end;

registration
  let f be complex-valued FinSequence, r be complex number;
  cluster r(#)f -> FinSequence-like;
  coherence
  proof
    dom (r(#)f) = dom f by Def5;
    hence thesis by Lm1;
  end;
end;

begin 

definition
  let f be complex-valued Function;
  func -f -> complex-valued Function equals
  (-1) (#) f;
  coherence;
  involutiveness
  proof
    let r, h be complex-valued Function;
    assume
A1: r = (-1)(#)h;
    thus dom ((-1)(#)r) = dom r by Def5
      .= dom h by A1,Def5;
    let c be set;
    assume c in dom h;
    reconsider a = (-1)*(h.c) as complex number;
    thus h.c = (-1) * a
      .= (-1)*r.c by A1,Th6
      .= ((-1)(#)r).c by Th6;
  end;
end;

theorem Th8:
  for f being complex-valued Function holds dom -f = dom f & for c
  being set holds (-f).c = -(f.c)
proof
  let f be complex-valued Function;
  thus
A1: dom -f = dom f by Def5;
  let c be set;
  per cases;
  suppose
    c in dom f;
    hence (-f).c = (-1)*f.c by A1,Def5
      .= -(f.c);
  end;
  suppose
A2: not c in dom f;
    hence (-f).c = -(0 qua complex number) by A1,FUNCT_1:def 2
      .= -(f.c) by A2,FUNCT_1:def 2;
  end;
end;

theorem
  for f being complex-valued Function, g being Function st dom f = dom g
  & for c being set st c in dom f holds g.c = -(f.c) holds g = -f
proof
  let f be complex-valued Function, g be Function;
  assume that
A1: dom f = dom g and
A2: for c being set st c in dom f holds g.c = -(f.c);
  thus dom -f = dom g by A1,Def5;
  let c be set;
  assume
A3: c in dom g;
  thus (-f).c = -f.c by Th8
    .= g.c by A1,A2,A3;
end;

registration
  let f be complex-valued Function;
  cluster -f -> complex-valued;
  coherence;
end;

registration
  let f be real-valued Function;
  cluster -f -> real-valued;
  coherence;
end;

registration
  let f be RAT-valued Function;
  cluster -f -> RAT-valued;
  coherence;
end;

registration
  let f be INT-valued Function;
  cluster -f -> INT-valued;
  coherence;
end;

definition
  let C be set;
  let D be complex-membered set;
  let f be PartFunc of C,D;
  redefine func -f -> PartFunc of C,COMPLEX;
  coherence
  proof
    dom -f = dom f & rng -f c= COMPLEX by Def5,VALUED_0:def 1;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D be real-membered set;
  let f be PartFunc of C,D;
  redefine func -f -> PartFunc of C,REAL;
  coherence
  proof
    dom -f = dom f & rng -f c= REAL by Def5,VALUED_0:def 3;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D be rational-membered set;
  let f be PartFunc of C,D;
  redefine func -f -> PartFunc of C,RAT;
  coherence
  proof
    dom -f = dom f & rng -f c= RAT by Def5,RELAT_1:def 19;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D be integer-membered set;
  let f be PartFunc of C,D;
  redefine func -f -> PartFunc of C,INT;
  coherence
  proof
    dom -f = dom f & rng -f c= INT by Def5,RELAT_1:def 19;
    hence thesis by RELSET_1:4;
  end;
end;

registration
  let C be set;
  let D be complex-membered non empty set;
  let f be Function of C,D;
  cluster -f -> total for PartFunc of C,COMPLEX;
  coherence;
end;

registration
  let C be set;
  let D be real-membered non empty set;
  let f be Function of C,D;
  cluster -f -> total for PartFunc of C,REAL;
  coherence;
end;

registration
  let C be set;
  let D be rational-membered non empty set;
  let f be Function of C,D;
  cluster -f -> total for PartFunc of C,RAT;
  coherence;
end;

registration
  let C be set;
  let D be integer-membered non empty set;
  let f be Function of C,D;
  cluster -f -> total for PartFunc of C,INT;
  coherence;
end;

registration
  let f be complex-valued FinSequence;
  cluster -f -> FinSequence-like;
  coherence;
end;

begin 

definition
  let f be complex-valued Function;
  deffunc F(set) = (f.$1)";
  func f" -> complex-valued Function means
  :Def7:
  dom it = dom f & for c being
  set st c in dom it holds it.c = (f.c)";
  existence
  proof
    consider g being Function such that
A1: dom g = dom f & for x being set st x in dom f holds g.x = F(x)
    from FUNCT_1:sch 3;
    g is complex-valued
    proof
      let x be set;
      assume x in dom g;
      then g.x = (f.x)" by A1;
      hence thesis;
    end;
    hence thesis by A1;
  end;
  uniqueness
  proof
    let h, g be complex-valued Function such that
A2: dom h = dom f and
A3: for c being set st c in dom h holds h.c = F(c) and
A4: dom g = dom f and
A5: for c being set st c in dom g holds g.c = F(c);
    now
      let x be set;
      assume
A6:   x in dom h;
      hence h.x = F(x) by A3
        .= g.x by A2,A4,A5,A6;
    end;
    hence thesis by A2,A4,FUNCT_1:2;
  end;
  involutiveness
  proof
    let r, h be complex-valued Function;
    assume that
A7: dom r = dom h and
A8: for c being set st c in dom r holds r.c = (h.c)";
    thus dom r = dom h by A7;
    let c be set;
    assume
A9: c in dom h;
    thus h.c = (h.c)""
      .= (r.c)" by A7,A8,A9;
  end;
end;


theorem Th10:
  for f being complex-valued Function holds for c being set holds
  f".c = (f.c)"
proof
  let f be complex-valued Function;
  let c be set;
A1: dom (f") = dom f by Def7;
  per cases;
  suppose
    c in dom f;
    hence thesis by A1,Def7;
  end;
  suppose
A2: not c in dom f;
    hence f".c = 0 qua complex number" by A1,FUNCT_1:def 2
      .= (f.c)" by A2,FUNCT_1:def 2;
  end;
end;

registration
  let f be real-valued Function;
  cluster f" -> real-valued;
  coherence
  proof
    let x be set;
    assume x in dom (f");
    then f".x = (f.x)" by Def7;
    hence thesis;
  end;
end;

registration
  let f be RAT-valued Function;
  cluster f" -> RAT-valued;
  coherence
  proof
    let y be set;
    assume y in rng (f");
    then consider x being set such that
A1: x in dom (f") and
A2: (f").x = y by FUNCT_1:def 3;
    f".x = (f.x)" by A1,Def7;
    hence thesis by A2,RAT_1:def 2;
  end;
end;

definition
  let C be set;
  let D be complex-membered set;
  let f be PartFunc of C,D;
  redefine func f" -> PartFunc of C,COMPLEX;
  coherence
  proof
    dom (f") = dom f & rng (f") c= COMPLEX by Def7,VALUED_0:def 1;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D be real-membered set;
  let f be PartFunc of C,D;
  redefine func f" -> PartFunc of C,REAL;
  coherence
  proof
    dom (f") = dom f & rng (f") c= REAL by Def7,VALUED_0:def 3;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D be rational-membered set;
  let f be PartFunc of C,D;
  redefine func f" -> PartFunc of C,RAT;
  coherence
  proof
    dom (f") = dom f & rng (f") c= RAT by Def7,RELAT_1:def 19;
    hence thesis by RELSET_1:4;
  end;
end;

registration
  let C be set;
  let D be complex-membered non empty set;
  let f be Function of C,D;
  cluster f" -> total for PartFunc of C,COMPLEX;
  coherence
  proof
    dom (f") = dom f by Def7
      .= C by FUNCT_2:def 1;
    hence thesis by PARTFUN1:def 2;
  end;
end;

registration
  let C be set;
  let D be real-membered non empty set;
  let f be Function of C,D;
  cluster f" -> total for PartFunc of C,REAL;
  coherence
  proof
    dom (f") = dom f by Def7
      .= C by FUNCT_2:def 1;
    hence thesis by PARTFUN1:def 2;
  end;
end;

registration
  let C be set;
  let D be rational-membered non empty set;
  let f be Function of C,D;
  cluster f" -> total for PartFunc of C,RAT;
  coherence
  proof
    dom (f") = dom f by Def7
      .= C by FUNCT_2:def 1;
    hence thesis by PARTFUN1:def 2;
  end;
end;

registration
  let f be complex-valued FinSequence;
  cluster f" -> FinSequence-like;
  coherence
  proof
    dom (f") = dom f by Def7;
    hence thesis by Lm1;
  end;
end;

begin 

definition
  let f be complex-valued Function;
  func f^2 -> Function equals
  f (#) f;
  coherence;
end;

theorem Th11:
  for f being complex-valued Function holds dom (f^2) = dom f &
  for c being set holds f^2.c = (f.c)^2
proof
  let f be complex-valued Function;
  thus
A1: dom (f^2) = dom f /\ dom f by Def4
    .= dom f;
  let c be set;
  per cases;
  suppose
    c in dom f;
    hence thesis by A1,Def4;
  end;
  suppose
A2: not c in dom f;
    hence f^2.c = 0 qua complex number^2 by A1,FUNCT_1:def 2
      .= (f.c)^2 by A2,FUNCT_1:def 2;
  end;
end;

registration
  let f be complex-valued Function;
  cluster f^2 -> complex-valued;
  coherence;
end;

registration
  let f be real-valued Function;
  cluster f^2 -> real-valued;
  coherence;
end;

registration
  let f be RAT-valued Function;
  cluster f^2 -> RAT-valued;
  coherence;
end;

registration
  let f be INT-valued Function;
  cluster f^2 -> INT-valued;
  coherence;
end;

registration
  let f be natural-valued Function;
  cluster f^2 -> natural-valued;
  coherence;
end;

definition
  let C be set;
  let D be complex-membered set;
  let f be PartFunc of C,D;
  redefine func f^2 -> PartFunc of C,COMPLEX;
  coherence
  proof
    dom (f^2) = dom f & rng (f^2) c= COMPLEX by Th11,VALUED_0:def 1;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D be real-membered set;
  let f be PartFunc of C,D;
  redefine func f^2 -> PartFunc of C,REAL;
  coherence
  proof
    dom (f^2) = dom f & rng (f^2) c= REAL by Th11,VALUED_0:def 3;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D be rational-membered set;
  let f be PartFunc of C,D;
  redefine func f^2 -> PartFunc of C,RAT;
  coherence
  proof
    dom (f^2) = dom f & rng (f^2) c= RAT by Th11,RELAT_1:def 19;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D be integer-membered set;
  let f be PartFunc of C,D;
  redefine func f^2 -> PartFunc of C,INT;
  coherence
  proof
    dom (f^2) = dom f & rng (f^2) c= INT by Th11,RELAT_1:def 19;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D be natural-membered set;
  let f be PartFunc of C,D;
  redefine func f^2 -> PartFunc of C,NAT;
  coherence
  proof
    dom (f^2) = dom f & rng (f^2) c= NAT by Th11,VALUED_0:def 6;
    hence thesis by RELSET_1:4;
  end;
end;

registration
  let C be set;
  let D be complex-membered non empty set;
  let f be Function of C,D;
  cluster f^2 -> total for PartFunc of C,COMPLEX;
  coherence;
end;

registration
  let C be set;
  let D be real-membered non empty set;
  let f be Function of C,D;
  cluster f^2 -> total for PartFunc of C,REAL;
  coherence;
end;

registration
  let C be set;
  let D be rational-membered non empty set;
  let f be Function of C,D;
  cluster f^2 -> total for PartFunc of C,RAT;
  coherence;
end;

registration
  let C be set;
  let D be integer-membered non empty set;
  let f be Function of C,D;
  cluster f^2 -> total for PartFunc of C,INT;
  coherence;
end;

registration
  let C be set;
  let D be natural-membered non empty set;
  let f be Function of C,D;
  cluster f^2 -> total for PartFunc of C,NAT;
  coherence;
end;

registration
  let f be complex-valued FinSequence;
  cluster f^2 -> FinSequence-like;
  coherence;
end;

begin 

definition
  let f1,f2 be complex-valued Function;
  func f1 - f2 -> Function equals
  f1 + - f2;
  coherence;
end;

registration
  let f1,f2 be complex-valued Function;
  cluster f1-f2 -> complex-valued;
  coherence;
end;

registration
  let f1,f2 be real-valued Function;
  cluster f1-f2 -> real-valued;
  coherence;
end;

registration
  let f1,f2 be RAT-valued Function;
  cluster f1-f2 -> RAT-valued;
  coherence;
end;

registration
  let f1,f2 be INT-valued Function;
  cluster f1-f2 -> INT-valued;
  coherence;
end;

theorem Th12:
  for f1,f2 being complex-valued Function holds dom (f1-f2) = dom
  f1 /\ dom f2
proof
  let f1,f2 be complex-valued Function;
  thus dom (f1-f2) = dom f1 /\ dom -f2 by Def1
    .= dom f1 /\ dom f2 by Def5;
end;

theorem
  for f1,f2 being complex-valued Function for c being set st c in dom (
  f1-f2) holds (f1-f2).c = f1.c - f2.c
proof
  let f1,f2 be complex-valued Function;
  let c be set;
  assume c in dom (f1-f2);
  hence (f1-f2).c = f1.c+(-f2).c by Def1
    .= f1.c-f2.c by Th8;
end;

theorem
  for f1,f2 being complex-valued Function, f being Function st dom f =
  dom (f1-f2) & for c being set st c in dom f holds f.c = f1.c - f2.c holds f =
  f1-f2
proof
  let f1,f2 be complex-valued Function, f be Function such that
A1: dom f = dom (f1-f2) and
A2: for c being set st c in dom f holds f.c = f1.c - f2.c;
  thus dom f = dom (f1-f2) by A1;
  let c be set;
  assume
A3: c in dom f;
  hence f.c = f1.c - f2.c by A2
    .= f1.c+(-f2).c by Th8
    .= (f1-f2).c by A1,A3,Def1;
end;

definition
  let C be set;
  let D1,D2 be complex-membered set;
  let f1 be PartFunc of C,D1;
  let f2 be PartFunc of C,D2;
  redefine func f1-f2 -> PartFunc of C,COMPLEX;
  coherence
  proof
    dom (f1-f2) = dom f1 /\ dom f2 & rng (f1-f2) c= COMPLEX by Th12,
VALUED_0:def 1;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D1,D2 be real-membered set;
  let f1 be PartFunc of C,D1;
  let f2 be PartFunc of C,D2;
  redefine func f1-f2 -> PartFunc of C,REAL;
  coherence
  proof
    dom (f1-f2) = dom f1 /\ dom f2 & rng (f1-f2) c= REAL by Th12,VALUED_0:def 3
;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D1,D2 be rational-membered set;
  let f1 be PartFunc of C,D1;
  let f2 be PartFunc of C,D2;
  redefine func f1-f2 -> PartFunc of C,RAT;
  coherence
  proof
    dom (f1-f2) = dom f1 /\ dom f2 & rng (f1-f2) c= RAT by Th12,RELAT_1:def 19;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D1,D2 be integer-membered set;
  let f1 be PartFunc of C,D1;
  let f2 be PartFunc of C,D2;
  redefine func f1-f2 -> PartFunc of C,INT;
  coherence
  proof
    dom (f1-f2) = dom f1 /\ dom f2 & rng (f1-f2) c= INT by Th12,RELAT_1:def 19;
    hence thesis by RELSET_1:4;
  end;
end;

Lm3: for C being set, D1,D2 being complex-membered non empty set, f1 being
Function of C,D1, f2 being Function of C,D2 holds dom(f1-f2) = C

proof
  let C be set;
  let D1,D2 be complex-membered non empty set;
  let f1 be Function of C,D1;
  let f2 be Function of C,D2;
  thus dom(f1-f2) = dom f1 /\ dom -f2 by Def1
    .= C /\ dom -f2 by FUNCT_2:def 1
    .= C /\ C by FUNCT_2:def 1
    .= C;
end;

registration
  let C be set;
  let D1,D2 be complex-membered non empty set;
  let f1 be Function of C,D1;
  let f2 be Function of C,D2;
  cluster f1-f2 -> total for PartFunc of C,COMPLEX;
  coherence
  proof
    dom(f1-f2) = C by Lm3;
    hence thesis by PARTFUN1:def 2;
  end;
end;

registration
  let C be set;
  let D1,D2 be real-membered non empty set;
  let f1 be Function of C,D1;
  let f2 be Function of C,D2;
  cluster f1-f2 -> total for PartFunc of C,REAL;
  coherence
  proof
    dom(f1-f2) = C by Lm3;
    hence thesis by PARTFUN1:def 2;
  end;
end;

registration
  let C be set;
  let D1,D2 be rational-membered non empty set;
  let f1 be Function of C,D1;
  let f2 be Function of C,D2;
  cluster f1-f2 -> total for PartFunc of C,RAT;
  coherence
  proof
    dom(f1-f2) = C by Lm3;
    hence thesis by PARTFUN1:def 2;
  end;
end;

registration
  let C be set;
  let D1,D2 be integer-membered non empty set;
  let f1 be Function of C,D1;
  let f2 be Function of C,D2;
  cluster f1-f2 -> total for PartFunc of C,INT;
  coherence
  proof
    dom(f1-f2) = C by Lm3;
    hence thesis by PARTFUN1:def 2;
  end;
end;

theorem
  for C being set, D1,D2 being complex-membered non empty set for f1
  being Function of C,D1, f2 being Function of C,D2 for c being Element of C
  holds (f1-f2).c = f1.c - f2.c
proof
  let C be set;
  let D1,D2 be complex-membered non empty set;
  let f1 be Function of C,D1, f2 be Function of C,D2;
  let c be Element of C;
A1: dom(f1-f2) = C by FUNCT_2:def 1;
  per cases;
  suppose
    C is non empty;
    hence (f1-f2).c = f1.c + (-f2).c by A1,Def1
      .= f1.c - f2.c by Th8;
  end;
  suppose
A2: C is empty;
    then dom f2 = {};
    then f2.c = 0 by FUNCT_1:def 2;
    hence thesis by A2;
  end;
end;

registration
  let f1, f2 be complex-valued FinSequence;
  cluster f1-f2 -> FinSequence-like;
  coherence;
end;

begin 

definition
  let f1,f2 be complex-valued Function;
  func f1 /" f2 -> Function equals
  f1 (#) (f2");
  coherence;
end;

theorem Th16:
  for f1,f2 being complex-valued Function holds dom (f1/"f2) = dom
  f1 /\ dom f2
proof
  let f1,f2 be complex-valued Function;
  thus dom (f1/"f2) = dom f1 /\ dom (f2") by Def4
    .= dom f1 /\ dom f2 by Def7;
end;

theorem
  for f1,f2 being complex-valued Function for c being set holds (f1/"f2)
  .c = f1.c / f2.c
proof
  let f1,f2 be complex-valued Function;
  let c be set;
A1: dom (f1/"f2) = dom f1 /\ dom f2 by Th16;
  per cases;
  suppose
    c in dom (f1/"f2);
    hence (f1/"f2).c = f1.c * (f2").c by Def4
      .= f1.c / f2.c by Th10;
  end;
  suppose
A2: not c in dom (f1/"f2);
    then not c in dom f1 or not c in dom f2 by A1,XBOOLE_0:def 4;
    then
A3: f1.c = 0 or f2.c = 0 by FUNCT_1:def 2;
    thus (f1/"f2).c = 0 / 0 by A2,FUNCT_1:def 2
      .= f1.c / f2.c by A3;
  end;
end;

registration
  let f1,f2 be complex-valued Function;
  cluster f1/"f2 -> complex-valued;
  coherence;
end;

registration
  let f1,f2 be real-valued Function;
  cluster f1/"f2 -> real-valued;
  coherence;
end;

registration
  let f1,f2 be RAT-valued Function;
  cluster f1/"f2 -> RAT-valued;
  coherence;
end;

definition
  let C be set;
  let D1,D2 be complex-membered set;
  let f1 be PartFunc of C,D1;
  let f2 be PartFunc of C,D2;
  redefine func f1/"f2 -> PartFunc of C,COMPLEX;
  coherence
  proof
    dom (f1/"f2) = dom f1 /\ dom f2 & rng (f1/"f2) c= COMPLEX by Th16,
VALUED_0:def 1;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D1,D2 be real-membered set;
  let f1 be PartFunc of C,D1;
  let f2 be PartFunc of C,D2;
  redefine func f1/"f2 -> PartFunc of C,REAL;
  coherence
  proof
    dom (f1/"f2) = dom f1 /\ dom f2 & rng (f1/"f2) c= REAL by Th16,
VALUED_0:def 3;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D1,D2 be rational-membered set;
  let f1 be PartFunc of C,D1;
  let f2 be PartFunc of C,D2;
  redefine func f1/"f2 -> PartFunc of C,RAT;
  coherence
  proof
    dom (f1/"f2) = dom f1 /\ dom f2 & rng (f1/"f2) c= RAT by Th16,
RELAT_1:def 19;
    hence thesis by RELSET_1:4;
  end;
end;

Lm4: for C being set, D1,D2 being complex-membered non empty set for f1 being
Function of C,D1, f2 being Function of C,D2 holds dom(f1/"f2) = C

proof
  let C be set;
  let D1,D2 be complex-membered non empty set;
  let f1 be Function of C,D1, f2 be Function of C,D2;
  thus dom(f1/"f2) = dom f1 /\ dom f2 by Th16
    .= C /\ dom f2 by FUNCT_2:def 1
    .= C /\ C by FUNCT_2:def 1
    .= C;
end;

registration
  let C be set;
  let D1,D2 be complex-membered non empty set;
  let f1 be Function of C,D1;
  let f2 be Function of C,D2;
  cluster f1/"f2 -> total for PartFunc of C,COMPLEX;
  coherence
  proof
    dom(f1/"f2) = C by Lm4;
    hence thesis by PARTFUN1:def 2;
  end;
end;

registration
  let C be set;
  let D1,D2 be real-membered non empty set;
  let f1 be Function of C,D1;
  let f2 be Function of C,D2;
  cluster f1/"f2 -> total for PartFunc of C,REAL;
  coherence
  proof
    dom(f1/"f2) = C by Lm4;
    hence thesis by PARTFUN1:def 2;
  end;
end;

registration
  let C be set;
  let D1,D2 be rational-membered non empty set;
  let f1 be Function of C,D1;
  let f2 be Function of C,D2;
  cluster f1/"f2 -> total for PartFunc of C,RAT;
  coherence
  proof
    dom(f1/"f2) = C by Lm4;
    hence thesis by PARTFUN1:def 2;
  end;
end;

registration
  let f1, f2 be complex-valued FinSequence;
  cluster f1/"f2 -> FinSequence-like;
  coherence;
end;

begin 

definition
  let f be complex-valued Function;
  deffunc F(set) = |.f.$1.|;
  func |. f .| -> real-valued Function means
  :Def11:
  dom it = dom f & for c
  being set st c in dom it holds it.c = |. f.c .|;
  existence
  proof
    consider g being Function such that
A1: dom g = dom f & for x being set st x in dom f holds g.x = F(x)
    from FUNCT_1:sch 3;
    g is real-valued
    proof
      let x be set;
      assume x in dom g;
      then g.x = |. f.x .| by A1;
      hence thesis;
    end;
    hence thesis by A1;
  end;
  uniqueness
  proof
    let h, g be real-valued Function such that
A2: dom h = dom f and
A3: for c being set st c in dom h holds h.c = F(c) and
A4: dom g = dom f and
A5: for c being set st c in dom g holds g.c = F(c);
    now
      let x be set;
      assume
A6:   x in dom h;
      hence h.x = F(x) by A3
        .= g.x by A2,A4,A5,A6;
    end;
    hence thesis by A2,A4,FUNCT_1:2;
  end;
  projectivity
  proof
    let r be real-valued Function;
    let h be complex-valued Function;
    assume that
    dom r = dom h and
A7: for c being set st c in dom r holds r.c = |.h.c.|;
    thus dom r = dom r;
    let c be set;
    assume
A8: c in dom r;
    hence r.c = |.|.h.c.|.| by A7
      .= |.r.c.| by A7,A8;
  end;
end;

notation
  let f be complex-valued Function;
  synonym abs f for |. f .|;
end;

theorem
  for f being complex-valued Function holds for c being set holds |.f.|.
  c = |.f.c.|
proof
  let f be complex-valued Function;
  let c be set;
A1: dom |.f.| = dom f by Def11;
  per cases;
  suppose
    c in dom f;
    hence thesis by A1,Def11;
  end;
  suppose
A2: not c in dom f;
    hence |.f.|.c = |.0 qua complex number.| by A1,COMPLEX1:44,FUNCT_1:def 2
      .= |.f.c.| by A2,FUNCT_1:def 2;
  end;
end;

registration
  let f be RAT-valued Function;
  cluster |.f.| -> RAT-valued;
  coherence
  proof
    let y be set;
    assume y in rng |.f.|;
    then consider x being set such that
A1: x in dom |.f.| and
A2: (|.f.|).x = y by FUNCT_1:def 3;
    |.f.|.x = |.f.x.| by A1,Def11;
    hence thesis by A2,RAT_1:def 2;
  end;
end;

registration
  let f be INT-valued Function;
  cluster |.f.| -> natural-valued;
  coherence
  proof
    let x be set;
    abs(f.x) is natural;
    hence thesis by Def11;
  end;
end;

definition
  let C be set;
  let D be complex-membered set;
  let f be PartFunc of C,D;
  redefine func |.f.| -> PartFunc of C,REAL;
  coherence
  proof
    dom |.f.| = dom f & rng |.f.| c= REAL by Def11,VALUED_0:def 3;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D be complex-membered set;
  let f be PartFunc of C,D;
  redefine func abs(f) -> PartFunc of C,REAL;
  coherence
  proof
    abs(f) = |.f.|;
    hence thesis;
  end;
end;

definition
  let C be set;
  let D be rational-membered set;
  let f be PartFunc of C,D;
  redefine func |.f.| -> PartFunc of C,RAT;
  coherence
  proof
    dom |.f.| = dom f & rng |.f.| c= RAT by Def11,RELAT_1:def 19;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D be rational-membered set;
  let f be PartFunc of C,D;
  redefine func abs(f) -> PartFunc of C,RAT;
  coherence
  proof
    abs(f) = |.f.|;
    hence thesis;
  end;
end;

definition
  let C be set;
  let D be integer-membered set;
  let f be PartFunc of C,D;
  redefine func |.f.| -> PartFunc of C,NAT;
  coherence
  proof
    dom |.f.| = dom f & rng |.f.| c= NAT by Def11,VALUED_0:def 6;
    hence thesis by RELSET_1:4;
  end;
end;

definition
  let C be set;
  let D be integer-membered set;
  let f be PartFunc of C,D;
  redefine func abs(f) -> PartFunc of C,NAT;
  coherence
  proof
    abs(f) = |.f.|;
    hence thesis;
  end;
end;

registration
  let C be set;
  let D be complex-membered non empty set;
  let f be Function of C,D;
  cluster |.f.| -> total for PartFunc of C,REAL;
  coherence
  proof
    dom |.f.| = dom f by Def11
      .= C by FUNCT_2:def 1;
    hence thesis by PARTFUN1:def 2;
  end;
end;

registration
  let C be set;
  let D be rational-membered non empty set;
  let f be Function of C,D;
  cluster |.f.| -> total for PartFunc of C,RAT;
  coherence
  proof
    dom |.f.| = dom f by Def11
      .= C by FUNCT_2:def 1;
    hence thesis by PARTFUN1:def 2;
  end;
end;

registration
  let C be set;
  let D be integer-membered non empty set;
  let f be Function of C,D;
  cluster |.f.| -> total for PartFunc of C,NAT;
  coherence
  proof
    dom |.f.| = dom f by Def11
      .= C by FUNCT_2:def 1;
    hence thesis by PARTFUN1:def 2;
  end;
end;

registration
  let f be complex-valued FinSequence;
  cluster |.f.| -> FinSequence-like;
  coherence
  proof
    dom abs f = dom f by Def11;
    hence thesis by Lm1;
  end;
end;

theorem
  for f, g being FinSequence, h being Function st dom h = dom f /\ dom g
  holds h is FinSequence by Lm2;

begin 



reserve m,j,p,q,n,l for Element of NAT;

definition
  let p be Function, k be Nat;
  func Shift(p,k) -> Function means
  :Def12:
  dom it = { m+k:m in dom p } & for
  m st m in dom p holds it.(m+k) = p.m;
  existence
  proof
    defpred P [set,set] means ex m st $1 = m+k & $2 = p.m;
    set A = { m+k:m in dom p };
A1: for e being set st e in A ex u being set st P[e,u]
    proof
      let e be set;
      assume e in A;
      then consider m such that
A2:   e = m+k and
      m in dom p;
      take p.m;
      thus thesis by A2;
    end;
    consider f being Function such that
A3: dom f = A and
A4: for e being set st e in A holds P[e,f.e] from CLASSES1:sch 1(A1);
    take f;
    thus dom f = { m+k:m in dom p } by A3;
    let m;
    assume m in dom p;
    then m+k in A;
    then ex j st m+k = j+k & f.(m+k) = p.j by A4;
    hence thesis;
  end;
  uniqueness
  proof
    let IT1,IT2 be Function such that
A5: dom IT1 = { m+k:m in dom p } and
A6: for m st m in dom p holds IT1.(m+k) = p.m and
A7: dom IT2 = { m+k:m in dom p } and
A8: for m st m in dom p holds IT2.(m+k) = p.m;
    for x being set st x in dom IT1 holds IT1.x = IT2.x
    proof
      let x be set;
      assume x in dom IT1;
      then consider m such that
A9:   x = m+k & m in dom p by A5;
      thus IT1.x = p.m by A6,A9
        .= IT2.x by A8,A9;
    end;
    hence thesis by A5,A7,FUNCT_1:2;
  end;
end;

registration
  let p be Function, k be Nat;
  cluster Shift(p,k) -> NAT-defined;
  coherence
  proof
A1: dom Shift(p,k) = { m+k:m in dom p } by Def12;
    Shift(p,k) is NAT-defined
    proof
      let x be set;
      assume x in dom Shift(p,k);
      then ex m st x = m+k & m in dom p by A1;
      hence thesis;
    end;
    hence thesis;
  end;
end;

theorem 
  for P,Q being Function, k being Nat st P c= Q holds Shift(P
  ,k) c= Shift(Q,k)
proof
  let P,Q be Function;
  let k be Nat;
  assume
A1: P c= Q;
  then
A2: dom P c= dom Q by GRFUNC_1:2;
A3: dom Shift(P,k) = {m + k: m in dom P} by Def12;
A4: dom Shift(Q,k) = {m + k: m in dom Q} by Def12;
  now
    let x be set;
    assume x in dom Shift(P,k);
    then ex m1 being Element of NAT st x = m1 + k & m1 in dom P by A3;
    hence x in dom Shift(Q,k) by A2,A4;
  end;
  then
A5: dom Shift(P,k) c= dom Shift(Q,k) by TARSKI:def 3;
  now
    let x be set;
    assume x in dom Shift(P,k);
    then consider m1 being Element of NAT such that
A6: x = m1 + k and
A7: m1 in dom P by A3;
    thus Shift(P,k).x = Shift(P,k).(m1 + k) by A6
      .= P.m1 by A7,Def12
      .= Q.m1 by A1,A7,GRFUNC_1:2
      .= Shift(Q,k).(m1 + k) by A2,A7,Def12
      .= Shift(Q,k).x by A6;
  end;
  hence thesis by A5,GRFUNC_1:2;
end;

theorem 
  for n,m being Nat
  for I being Function holds Shift(Shift(I,m),n) = Shift(I,m+n)
proof let n,m be Nat;
  let I be Function;
  set A = {l+m:l in dom I };
A1: dom Shift(I,m) = A by Def12;
A2: now
    let l;
    assume
A3: l in dom I;
    then
A4: l+m in dom Shift(I,m) by A1;
    thus Shift(Shift(I,m),n).(l+(m+n)) = Shift(Shift(I,m),n).(l+m+n)
      .= Shift(I,m).(l+m) by A4,Def12
      .= I.l by A3,Def12;
  end;
  {p+n: p in A } = { q+(m+n): q in dom I}
  proof
    thus {p+n: p in A } c= { q+(m+n): q in dom I}
    proof
      let x be set;
      assume x in {p+n: p in A };
      then consider p such that
A5:   x = p+n and
A6:   p in A;
      consider l such that
A7:   p = l+m and
A8:   l in dom I by A6;
      x = l+(m+n) by A5,A7;
      hence thesis by A8;
    end;
    let x be set;
    assume x in { q+(m+n): q in dom I};
    then consider q such that
A9: x = q+(m+n) & q in dom I;
    x = (q+m)+n & q+m in A by A9;
    hence thesis;
  end;
  then dom Shift(Shift(I,m),n) = { l+(m+n): l in dom I} by A1,Def12;
  hence thesis by A2,Def12;
end;

theorem 
  for s,f be Function for n being Nat holds Shift(f*s,n) = f*Shift(s,n)
proof
  let s,f be Function;
  let n be Nat;
A1: dom(f*s) c= dom s by RELAT_1:25;
A2: dom Shift(s,n)= { m+n: m in dom s } by Def12;
  now
    let e be set;
    thus e in { m+n: m in dom(f*s) } implies e in dom Shift(s,n) & Shift(s,n).
    e in dom f
    proof
      assume e in { m+n: m in dom(f*s) };
      then consider m such that
A3:   e = m+n and
A4:   m in dom(f*s);
      thus e in dom Shift(s,n) by A2,A1,A3,A4;
      Shift(s,n).e = s.m by A1,A3,A4,Def12;
      hence thesis by A4,FUNCT_1:11;
    end;
    assume e in dom Shift(s,n);
    then consider m0 being Element of NAT such that
A5: e = m0+n and
A6: m0 in dom s by A2;
    assume Shift(s,n).e in dom f;
    then s.m0 in dom f by A5,A6,Def12;
    then m0 in dom(f*s) by A6,FUNCT_1:11;
    hence e in { m+n: m in dom(f*s) } by A5;
  end;
  then Shift(s,n)"dom f = { m+n: m in dom(f*s) } by FUNCT_1:def 7;
  then
A7: dom(f*Shift(s,n)) = { m+n: m in dom(f*s) } by RELAT_1:147;
  now
    let m;
    assume
A8: m in dom(f*s);
    then m+n in dom Shift(s,n) by A2,A1;
    hence (f*Shift(s,n)).(m+n) = f.(Shift(s,n).(m+n)) by FUNCT_1:13
      .= f.(s.m) by A1,A8,Def12
      .= (f*s).m by A8,FUNCT_1:12;
  end;
  hence thesis by A7,Def12;
end;

theorem 
  for I,J being Function, n being Nat
   holds Shift(I +* J, n) = Shift(I,n) +* Shift(J,n)
proof
  let I,J be Function, n be Nat;
A1: dom Shift(J,n) = { m+n: m in dom J } by Def12;
A2: now
    let m such that
A3: m in dom(I +* J);
    per cases;
    suppose
A4:   m in dom J;
      then m+n in dom Shift(J,n) by A1;
      hence (Shift(I,n) +* Shift(J,n)).(m+n) = Shift(J,n).(m+n) by FUNCT_4:13
        .= J.m by A4,Def12
        .= (I +* J).m by A4,FUNCT_4:13;
    end;
    suppose
A5:   not m in dom J;
      m in dom I \/ dom J by A3,FUNCT_4:def 1;
      then
A6:   m in dom I by A5,XBOOLE_0:def 3;
      not ex l st m+n = l+n & l in dom J by A5;
      then not m+n in dom Shift(J,n) by A1;
      hence (Shift(I,n) +* Shift(J,n)).(m+n) = Shift(I,n).(m+n) by FUNCT_4:11
        .= I.m by A6,Def12
        .= (I +* J).m by A5,FUNCT_4:11;
    end;
  end;
A7: dom Shift(I,n) = { m+n: m in dom I } by Def12;
A8: dom Shift(I,n) \/ dom Shift(J,n) = { m+n: m in dom I \/ dom J }
  proof
    hereby
      let x be set;
      assume x in dom Shift(I,n) \/ dom Shift(J,n);
      then x in dom Shift(I,n) or x in dom Shift(J,n) by XBOOLE_0:def 3;
      then consider m such that
A9:   x = m+n & m in dom J or x = m+n & m in dom I by A1,A7;
      m in dom I \/ dom J by A9,XBOOLE_0:def 3;
      hence x in { l+n: l in dom I \/ dom J } by A9;
    end;
    let x be set;
    assume x in { m+n: m in dom I \/ dom J };
    then consider m such that
A10: x = m+n and
A11: m in dom I \/ dom J;
    m in dom I or m in dom J by A11,XBOOLE_0:def 3;
    then x in dom Shift(I,n) or x in dom Shift(J,n) by A1,A7,A10;
    hence thesis by XBOOLE_0:def 3;
  end;
  dom(I +* J) = dom I \/ dom J by FUNCT_4:def 1;
  then dom(Shift(I,n) +* Shift(J,n)) = { m+n: m in dom(I +* J ) } by A8,
FUNCT_4:def 1;
  hence thesis by A2,Def12;
end;



theorem Th24:
  for p being Function,k,il being Nat st il in dom p
  holds il+k in dom Shift(p,k)
proof
  let p be Function, k,il be Nat such that
A1: il in dom p;
A2:  dom Shift(p,k) = { loc+k where loc is Element of NAT : loc in dom p}
      by Def12;
   reconsider loc = il as Element of NAT by ORDINAL1:def 12;
   loc+k in dom Shift(p,k) by A2,A1;
  hence thesis;
end;



theorem Th25:
  for p being Function, k being Nat holds rng Shift(p,k) c= rng p
proof
  let p be Function, k being Nat;
  let y be set;
  assume y in rng Shift(p,k);
  then consider x being set such that
A1: x in dom Shift(p,k) and
A2: y = Shift(p,k).x by FUNCT_1:def 3;
  x in { m+k:m in dom p } by A1,Def12;
  then consider m such that
A3: x = m+k and
A4: m in dom p;
  p.m = Shift(p,k).x by A3,A4,Def12;
  hence thesis by A2,A4,FUNCT_1:def 3;
end;

theorem
  for p being Function st dom p c= NAT for k being Nat holds
  rng Shift(p,k) = rng p
proof
  let p be Function such that
A1: dom p c= NAT;
  let k be Nat;
  thus rng Shift(p,k) c= rng p by Th25;
  let y be set;
  assume y in rng p;
  then consider x being set such that
A2: x in dom p and
A3: y = p.x by FUNCT_1:def 3;
  reconsider x as Element of NAT by A1,A2;
  x+k in dom Shift(p,k) & Shift(p,k).(x+k) = y by A2,A3,Def12,Th24;
  hence thesis by FUNCT_1:def 3;
end;

registration
  let p be finite Function, k be Nat;
  cluster Shift(p,k) -> finite;
  coherence
  proof
    deffunc F(Element of NAT) = $1+k;
A1: dom p is finite;
    { F(w) where w is Element of NAT: w in dom p } is finite from FRAENKEL
    :sch 21(A1);
    then dom Shift(p,k) is finite by Def12;
    hence thesis by FINSET_1:10;
  end;
end;

reserve e1,e2 for ext-real number;

definition
  let X be non empty ext-real-membered set, s be sequence of X;
  redefine attr s is increasing means
  for n being Nat holds s.n < s.(n+1);
  compatibility
  proof
    thus s is increasing implies for n being Nat holds s.n < s.(n+1 )
    proof
      assume
A1:   s is increasing;
      let n be Nat;
A2:   n < n+1 by NAT_1:13;
      dom s = NAT & n in NAT by FUNCT_2:def 1,ORDINAL1:def 12;
      hence thesis by A1,A2,VALUED_0:def 13;
    end;
    assume
A3: for n being Nat holds s.n < s.(n+1);
    let e1,e2;
    assume e1 in dom s & e2 in dom s;
    then reconsider m=e1, n=e2 as Nat;
    defpred P[Nat] means m < $1 implies s.m < s.$1;
A4: for j being Nat st P[j] holds P[j+1]
    proof
      let j being Nat such that
A5:   P[j];
      assume m < j+1;
      then m <= j by NAT_1:13;
      then s.m < s.j or m = j by A5,XXREAL_0:1;
      hence thesis by A3,XXREAL_0:2;
    end;
    assume
A6: e1 < e2;
A7: P[0];
    for j being Nat holds P[j] from NAT_1:sch 2(A7,A4);
    then s.m < s.n by A6;
    hence thesis;
  end;
  redefine attr s is decreasing means
  for n being Nat holds s.n > s.(n+1);
  compatibility
  proof
    thus s is decreasing implies for n being Nat holds s.n > s.(n+1 )
    proof
      assume
A8:   s is decreasing;
      let n be Nat;
A9:   n < n+1 by NAT_1:13;
      dom s = NAT & n in NAT by FUNCT_2:def 1,ORDINAL1:def 12;
      hence thesis by A8,A9,VALUED_0:def 14;
    end;
    assume
A10: for n being Nat holds s.n > s.(n+1);
    let e1,e2;
    assume e1 in dom s & e2 in dom s;
    then reconsider m=e1, n=e2 as Nat;
    defpred P[Nat] means m < $1 implies s.m > s.$1;
A11: for j being Nat st P[j] holds P[j+1]
    proof
      let j being Nat such that
A12:  P[j];
      assume m < j+1;
      then m <= j by NAT_1:13;
      then s.m > s.j or m = j by A12,XXREAL_0:1;
      hence thesis by A10,XXREAL_0:2;
    end;
    assume
A13: e1 < e2;
A14: P[0];
    for j being Nat holds P[j] from NAT_1:sch 2(A14,A11);
    then s.m > s.n by A13;
    hence thesis;
  end;
  redefine attr s is non-decreasing means
  for n being Nat holds s.n <= s.(n+1);
  compatibility
  proof
    thus s is non-decreasing implies for n being Nat holds s.n <= s.(n+1 )
    proof
      assume
A15:  s is non-decreasing;
      let n be Nat;
A16:  n < n+1 by NAT_1:13;
      dom s = NAT & n in NAT by FUNCT_2:def 1,ORDINAL1:def 12;
      hence thesis by A15,A16,VALUED_0:def 15;
    end;
    assume
A17: for n being Nat holds s.n <= s.(n+1);
    let e1,e2;
    assume e1 in dom s & e2 in dom s;
    then reconsider m=e1, n=e2 as Nat;
    defpred P[Nat] means m <= $1 implies s.m <= s.$1;
A18: for j being Nat st P[j] holds P[j+1]
    proof
      let j being Nat such that
A19:  P[j];
      assume m <= j+1;
      then
A20:  m < j+1 or m = j+1 by XXREAL_0:1;
      s.j <= s.(j+1) by A17;
      hence thesis by A19,A20,NAT_1:13,XXREAL_0:2;
    end;
    assume
A21: e1 <= e2;
A22: P[0];
    for j being Nat holds P[j] from NAT_1:sch 2(A22,A18);
    then s.m <= s.n by A21;
    hence thesis;
  end;
  redefine attr s is non-increasing means
  for n being Nat holds s.n >= s.(n+1);
  compatibility
  proof
    thus s is non-increasing implies for n being Nat holds s.n >= s.(n+1 )
    proof
      assume
A23:  s is non-increasing;
      let n be Nat;
A24:  n < n+1 by NAT_1:13;
      dom s = NAT & n in NAT by FUNCT_2:def 1,ORDINAL1:def 12;
      hence thesis by A23,A24,VALUED_0:def 16;
    end;
    assume
A25: for n being Nat holds s.n >= s.(n+1);
    let e1,e2;
    assume e1 in dom s & e2 in dom s;
    then reconsider m=e1, n=e2 as Nat;
    defpred P[Nat] means m <= $1 implies s.m >= s.$1;
A26: for j being Nat st P[j] holds P[j+1]
    proof
      let j being Nat such that
A27:  P[j];
      assume m <= j+1;
      then
A28:  m < j+1 or m = j+1 by XXREAL_0:1;
      s.j >= s.(j+1) by A25;
      hence thesis by A27,A28,NAT_1:13,XXREAL_0:2;
    end;
    assume
A29: e1 <= e2;
A30: P[0];
    for j being Nat holds P[j] from NAT_1:sch 2(A30,A26);
    then s.m >= s.n by A29;
    hence thesis;
  end;
end;



scheme
  SubSeqChoice { X() -> non empty set, S() -> sequence of X(), P[set]} :
   ex S1 being subsequence of S() st for n being Element of NAT holds P[S1.n]
provided
A1: for n being Element of NAT ex m being Element of NAT st n <= m & P[S
().m]
proof
  defpred P1[set,set,set] means $3 in NAT & (for m,k being Element of NAT st m
  = $2 & k = $3 holds m < k & P[S().k]);
  consider n0 being Element of NAT such that
  0 <= n0 and
A2: P[S().n0] by A1;
A3: for n being Element of NAT for x being set ex y being set st P1[n,x,y]
  proof
    let n be Element of NAT, x be set;
    per cases;
    suppose
      x in NAT;
      then reconsider mx = x as Element of NAT;
      consider my being Element of NAT such that
A4:   mx + 1 <= my & P[S().my] by A1;
      take my;
      thus my in NAT;
      thus thesis by A4,NAT_1:13;
    end;
    suppose
A5:   not x in NAT;
      take 0;
      set y = 0;
      thus y in NAT;
      let m, k be Element of NAT;
      assume that
A6:   m = x and
      k = y;
      thus thesis by A5,A6;
    end;
  end;
  consider g being Function such that
A7: dom g = NAT and
A8: g.0 = n0 and
A9: for n being Element of NAT holds P1[n,g.n,g.(n+1)] from RECDEF_1:
  sch 1(A3);
  rng g c= NAT
  proof
    defpred P[Element of NAT] means g.$1 in NAT;
    let y be set;
    assume y in rng g;
    then consider x being set such that
A10: x in dom g and
A11: g.x = y by FUNCT_1:def 3;
    reconsider n = x as Element of NAT by A7,A10;
A12: for k being Element of NAT holds P[k] implies P[k+1] by A9;
A13: P[0] by A8;
    for k being Element of NAT holds P[k] from NAT_1:sch 1(A13,A12);
    then g.n in NAT;
    hence thesis by A11;
  end;
  then reconsider g as sequence of NAT by A7,FUNCT_2:2;
  g is increasing
  proof
    let n be Nat;
    reconsider n as Element of NAT by ORDINAL1:def 12;
    g.n < g.(n+1) by A9;
    hence thesis;
  end;
  then reconsider g as increasing sequence of NAT;
  reconsider S1 = S() * g as sequence of X();
A14: dom S1 = NAT by FUNCT_2:def 1;
  reconsider S1 as subsequence of S();
  take S1;
  thus for n being Element of NAT holds P[S1.n]
  proof
    let n be Element of NAT;
    per cases;
    suppose
      n = 0;
      hence thesis by A2,A8,A14,FUNCT_1:12;
    end;
    suppose
      n <> 0;
      then n >= 0 qua Nat + 1 by NAT_1:13;
      then reconsider n9 = n-1 as Element of NAT by INT_1:5;
      reconsider k = g.(n9+1) as Element of NAT;
      for m,k being Element of NAT st m = g.n9 & k = g.(n9+1) holds P[S()
      .k] by A9;
      then P[S().k];
      hence thesis by A14,FUNCT_1:12;
    end;
  end;
end;



theorem
  for k being Nat
  for F being NAT-defined Function
  holds dom F,dom Shift(F,k) are_equipotent
proof
  let k be Nat;
  let F be NAT-defined Function;
  defpred P[set,set] means
  ex il being Element of NAT
   st $1 = il & $2 = k+il;
A1: for e being set st e in dom F ex u being set st P[e,u]
  proof
    let e be set;
    assume e in dom F;
    then reconsider e as Element of NAT;
    take k+e, e;
    thus thesis;
  end;
  consider f being Function such that
A2: dom f = dom F and
A3: for x being set st x in dom F holds P[x,f.x] from CLASSES1:sch 1(A1);
  take f;
  hereby
    let x1, x2 be set such that
A4: x1 in dom f and
A5: x2 in dom f and
A6: f.x1 = f.x2;
    consider i1 being Element of NAT such that
A7: x1 = i1 and
A8: f.x1 = k+i1 by A2,A3,A4;
    consider i2 being Element of NAT such that
A9: x2 = i2 and
A10: f.x2 = k+i2 by A2,A3,A5;
    thus x1 = x2 by A7,A6,A8,A10,A9;
  end;
  thus dom f = dom F by A2;
A11: dom Shift(F,k) =
  { (m+k) where m is Element of NAT: m in dom F } by Def12;
  hereby
    let y be set;
    assume y in rng f;
    then consider x being set such that
A12: x in dom f and
A13: f.x = y by FUNCT_1:def 3;
    consider il being Element of NAT such that
A14: x = il and
A15: f.x = k+il by A2,A3,A12;
   thus y in dom Shift(F,k) by A2,A11,A12,A13,A14,A15;
  end;
  let y be set;
  assume y in dom Shift(F,k);
  then consider m being Element of NAT such that
A16: y = (m+k) and
A17: m in dom F by A11;
  consider il being Element of NAT such that
A18: m = il and
A19: f.m = k+il by A3,A17;
  thus thesis by A2,A16,A17,A18,A19,FUNCT_1:def 3;
end;

registration
 let F be NAT-defined Function;
 reduce Shift(F,0) to F;
 reducibility
proof
A1: dom F = { m+(0 qua complex number) where m is Element of NAT: m in dom F }
  proof
    hereby
     let a be set;
     assume
A2:   a in dom F;
      then reconsider l=a as Element of NAT;
      a = (l qua complex number)+(0 qua complex number);
     hence a in { m+(0 qua complex number)
      where m is Element of NAT: m in dom F } by A2;
    end;
    let a be set;
    assume
    a in { m+(0 qua complex number) where m is Element of NAT: m in dom F };
    then ex m being Element of NAT
     st a = m+(0 qua complex number) & m in dom F;
    hence thesis;
  end;
  for m being Element of NAT st m in dom F
   holds F.(m+(0 qua complex number)) = F.m;
  hence thesis by A1,Def12;
end;
end;

theorem
  for F being NAT-defined Function holds Shift(F,0) = F;

registration
  let X be non empty set;
  let F be X-valued Function, k be Nat;
  cluster Shift(F,k) -> X-valued;
  coherence
  proof
A1: dom Shift(F,k) =
    { (m+k) where m is Element of NAT: m in dom F } by Def12;
    thus rng Shift(F,k) c= X
     proof let x be set;
      assume x in rng Shift(F,k);
       then consider y being set such that
A2:     y in dom Shift(F,k) and
A3:     x = Shift(F,k).y by FUNCT_1:def 3;
       consider m being Element of NAT such that
A4:     y = (m+k) and
A5:     m in dom F by A2,A1;
       Shift(F,k).(m+k) = F.m by A5,Def12
            .= F/.m by A5,PARTFUN1:def 6;
      hence x in X by A3,A4;
     end;
  end;
end;

registration  
 cluster non empty NAT-defined for Function;
 existence
  proof
   take id NAT;
   thus thesis;
  end;
end;

registration
  let F be empty Function, k be Nat;
  cluster Shift(F,k) -> empty;
  coherence
  proof
A1: dom Shift(F,k) =
    { (m+k) where m is Element of NAT: m in dom F } by Def12;
    assume Shift(F,k) is non empty;
    then reconsider f = Shift(F,k) as non empty Function;
    dom f is non empty;
    then consider a being set such that
A2: a in dom Shift(F,k) by XBOOLE_0:def 1;
    ex m being Element of NAT st a = (m+k) & m in dom F by A1,A2;
    hence thesis;
  end;
end;

registration
  let F be non empty NAT-defined Function, k be Nat;
  cluster Shift(F,k) -> non empty;
  coherence
  proof
A1: dom Shift(F,k) =
    { (m+k) where m is Element of NAT: m in dom F } by Def12;
    consider a being set such that
A2: a in dom F by XBOOLE_0:def 1;
    reconsider a as Element of NAT by A2;
    consider m being Nat such that
A3: a = m;
    reconsider m as Element of NAT by ORDINAL1:def 12;
    (m+k) in dom Shift(F,k) by A1,A2,A3;
    hence thesis;
  end;
end;

theorem
  for F being Function, k being Nat st k > 0
  holds not 0 in dom Shift(F,k)
proof
  let F be Function, k be Nat such that
A1: k > 0 and
A2: 0 in dom Shift(F,k);
  dom Shift(F,k) = { m+k where m is Element of NAT: m in dom F } by Def12;
  then ex m being Element of NAT st 0 = m+k &  m in dom F by A2;
  hence contradiction by A1;
end;

registration
 cluster NAT-defined finite non empty for Function;
 existence
  proof
   take f = 0 .--> 0;
    dom f = {0} by FUNCOP_1:13;
   hence dom f c= NAT by ZFMISC_1:31;
   thus thesis;
  end;
end;

registration
  let F be NAT-defined Relation;
  cluster dom F -> natural-membered;
 coherence;
end;

definition
  let F be non empty NAT-defined finite Function;
  func LastLoc F -> Element of NAT equals
   max dom F;
 coherence by ORDINAL1:def 12;
end;

definition
  let F be non empty NAT-defined finite Function;
  func CutLastLoc F -> Function equals
  F \ ( LastLoc F .--> F.LastLoc F );
  coherence;
end;

registration
 let F be non empty NAT-defined finite Function;
 cluster CutLastLoc F -> NAT-defined finite;
 coherence;
end;

theorem Th30:
  for F being non empty NAT-defined finite Function holds
  LastLoc F in dom F by XXREAL_2:def 8;

theorem
  for F, G being non empty NAT-defined finite Function st F c= G holds
  LastLoc F <= LastLoc G by RELAT_1:11,XXREAL_2:59;

theorem
  for F being non empty NAT-defined finite Function, l being
  Element of NAT st l in dom F holds l <= LastLoc F by XXREAL_2:def 8;

definition
  let F be non empty NAT-defined Function;
  func FirstLoc F -> Element of NAT equals
   min dom F;
  coherence by ORDINAL1:def 12;
end;

theorem
  for F being non empty NAT-defined finite Function holds
  FirstLoc F in dom F by XXREAL_2:def 7;

theorem
  for F, G being non empty NAT-defined finite Function st F c= G holds
  FirstLoc G <= FirstLoc F by RELAT_1:11,XXREAL_2:60;

theorem
  for l1 being Element of NAT
  for F being non empty NAT-defined finite Function st l1 in dom
  F holds FirstLoc F <= l1 by XXREAL_2:def 7;

theorem Th36:
  for F being non empty NAT-defined finite Function
  holds dom CutLastLoc F = (dom F) \ {LastLoc F}
proof
  let F be non empty NAT-defined finite Function;
A1: dom (LastLoc F .--> (F.LastLoc F)) = {LastLoc F} by FUNCOP_1:13;
  reconsider R = {[LastLoc F, F.LastLoc F]} as Relation;
A2: R = LastLoc F .--> (F.LastLoc F) by FUNCT_4:82;
  then
A3: dom R = {LastLoc F} by FUNCOP_1:13;
  thus dom CutLastLoc F c= (dom F) \ {LastLoc F}
  proof
    let x be set;
    assume x in dom CutLastLoc F;
    then consider y being set such that
A4: [x,y] in F \ R by A2,XTUPLE_0:def 12;
A5: not [x,y] in R by A4,XBOOLE_0:def 5;
A6: x in dom F by A4,XTUPLE_0:def 12;
    per cases by A5,TARSKI:def 1;
    suppose x <> LastLoc F;
      then not x in dom R by A3,TARSKI:def 1;
      hence thesis by A1,A2,A6,XBOOLE_0:def 5;
    end;
    suppose
A7:   y <> F.LastLoc F;
      now
        assume x in dom R;
        then x = LastLoc F by A3,TARSKI:def 1;
        hence contradiction by A4,A7,FUNCT_1:1;
      end;
      hence thesis by A1,A2,A6,XBOOLE_0:def 5;
    end;
  end;
  thus thesis by A1,RELAT_1:3;
end;

theorem
  for F being non empty NAT-defined finite Function
  holds dom F = dom CutLastLoc F \/ {LastLoc F}
proof
  let F be non empty NAT-defined finite Function;
  LastLoc F in dom F by Th30;
  then
A1: {LastLoc F} c= dom F by ZFMISC_1:31;
  dom CutLastLoc F = (dom F) \ {LastLoc F} by Th36;
  hence thesis by A1,XBOOLE_1:45;
end;

registration
 cluster 1-element NAT-defined finite for Function;
 existence
  proof
   take 0 .--> 0;
   thus thesis;
  end;
end;

registration
  let F be 1-element NAT-defined finite Function;
  cluster CutLastLoc F -> empty;
  coherence
  proof
    LastLoc F in dom F by Th30;
    then
A1: [LastLoc F,F.LastLoc F] in F by FUNCT_1:def 2;
    assume not thesis;
    then consider a being set such that
A2: a in CutLastLoc F by XBOOLE_0:def 1;
A3: a = [LastLoc F,F.LastLoc F] by A1,A2,ZFMISC_1:def 10;
    not a in LastLoc F .--> F.LastLoc F by A2,XBOOLE_0:def 5;
    then not a in {[LastLoc F,F.LastLoc F]} by FUNCT_4:82;
    hence thesis by A3,TARSKI:def 1;
  end;
end;

theorem
  for F being non empty NAT-defined finite Function
  holds card CutLastLoc F = card F - 1
proof
  let F be non empty NAT-defined finite Function;
  LastLoc F .--> F.LastLoc F c= F
  proof
    let a, b be set;
    assume [a,b] in LastLoc F .--> F.LastLoc F;
    then [a,b] in {[LastLoc F,F.LastLoc F]} by FUNCT_4:82;
    then
A1: [a,b] = [LastLoc F,F.LastLoc F] by TARSKI:def 1;
    LastLoc F in dom F by Th30;
    hence thesis by A1,FUNCT_1:def 2;
  end;
  hence card CutLastLoc F = card F - card (LastLoc F .--> F.LastLoc F)
  by CARD_2:44
    .= card F - card {[LastLoc F,F.LastLoc F]} by FUNCT_4:82
    .= card F - 1 by CARD_1:30;
end;

begin 

registration let X be set, f be X-defined complex-valued Function;
 cluster -f -> X-defined;
 coherence
  proof
A1:  dom -f = dom f by Def5;
   thus dom -f c= X by A1;
  end;
 cluster f" -> X-defined;
 coherence
  proof
A2:  dom(f") = dom f by Def7;
   thus dom(f") c= X by A2;
  end;
 cluster f^2 -> X-defined;
 coherence
  proof
A3:  dom(f^2) = dom f by Th11;
   thus dom(f^2) c= X by A3;
  end;
 cluster |.f.| -> X-defined;
 coherence
  proof
A4:  dom |.f.| = dom f by Def11;
   thus dom |.f.| c= X by A4;
  end;
end;

registration let X be set;
 cluster total for X-defined natural-valued Function;
 existence
  proof
   take the total PartFunc of X, NAT;
   thus thesis;
  end;
end;

registration let X be set,
   f be total X-defined complex-valued Function;
 cluster -f -> total;
 coherence
  proof
A1:  dom -f = dom f by Def5;
    dom f = X by PARTFUN1:def 2;
   hence thesis by A1,PARTFUN1:def 2;
  end;
 cluster f" -> total;
 coherence
  proof
A2:  dom(f") = dom f by Def7;
    dom f = X by PARTFUN1:def 2;
   hence thesis by A2,PARTFUN1:def 2;
  end;
 cluster f^2 -> total;
 coherence
  proof
A3:  dom(f^2) = dom f by Th11;
    dom f = X by PARTFUN1:def 2;
   hence thesis by A3,PARTFUN1:def 2;
  end;
 cluster |.f.| -> total;
 coherence
  proof
A4:  dom |.f.| = dom f by Def11;
    dom f = X by PARTFUN1:def 2;
   hence thesis by A4,PARTFUN1:def 2;
  end;
end;

registration
 let X be set, f be X-defined complex-valued Function,
     r be complex number;
 cluster r+f -> X-defined;
 coherence
  proof
A1:  dom(r+f) = dom f by Def2;
   thus dom(r+f) c= X by A1;
  end;
 cluster f-r -> X-defined;
 coherence;
 cluster r(#)f -> X-defined;
 coherence
  proof
A2:  dom(r(#)f) = dom f by Def5;
   thus dom(r(#)f) c= X by A2;
  end;
end;

registration let X be set,
   f be total X-defined complex-valued Function,
     r be complex number;
 cluster r+f -> total;
 coherence
  proof
A1:  dom(r+f) = dom f by Def2;
    dom f = X by PARTFUN1:def 2;
   hence thesis by A1,PARTFUN1:def 2;
  end;
 cluster f-r -> total;
 coherence;
 cluster r(#)f -> total;
 coherence
  proof
A2:  dom(r(#)f) = dom f by Def5;
    dom f = X by PARTFUN1:def 2;
   hence thesis by A2,PARTFUN1:def 2;
  end;
end;

registration
 let X be set, f1 be complex-valued Function;
 let f2 be X-defined complex-valued Function;
 cluster f1 + f2 -> X-defined;
 coherence
  proof
    dom(f1 + f2) = dom f1 /\ dom f2 by Def1;
    then
A1:  dom(f1 + f2) c= dom f2 by XBOOLE_1:17;
   thus dom(f1 + f2) c= X by A1,XBOOLE_1:1;
  end;
 cluster f1 - f2 -> X-defined;
 coherence;
 cluster f1(#)f2 -> X-defined;
 coherence
  proof
    dom(f1(#)f2) = dom f1 /\ dom f2 by Def4;
    then
A2:  dom(f1(#)f2) c= dom f2 by XBOOLE_1:17;
   thus dom(f1(#)f2) c= X by A2,XBOOLE_1:1;
  end;
 cluster f1/"f2 -> X-defined;
 coherence;
end;

registration let X be set;
 let f1,f2 be total X-defined complex-valued Function;
 cluster f1 + f2 -> total;
 coherence
  proof
A1:  dom(f1 + f2) = dom f1 /\ dom f2 by Def1;
    dom f1 = X & dom f2 = X by PARTFUN1:def 2;
   hence thesis by A1,PARTFUN1:def 2;
  end;
 cluster f1 - f2 -> total;
 coherence;
 cluster f1(#)f2 -> total;
 coherence
  proof
A2:  dom(f1(#)f2) = dom f1 /\ dom f2 by Def4;
    dom f1 = X & dom f2 = X by PARTFUN1:def 2;
   hence thesis by A2,PARTFUN1:def 2;
  end;
 cluster f1/"f2 -> total;
 coherence;
end;

registration let X be non empty set;
 let F be X-valued non empty NAT-defined finite Function;
 cluster CutLastLoc F -> X-valued;
 coherence;
end;

theorem
 for f being Function
 for i,n being Nat st i in dom Shift(f,n)
  ex j being Nat st j in dom f & i = j + n
 proof let f be Function;
  let i,n be Nat;
A: dom Shift(f,n) = { m+n:m in dom f } by Def12;
  assume i in dom Shift(f,n);
   then ex m being Element of NAT st i = m + n & m in dom f by A;
  hence ex j being Nat st j in dom f & i = j + n;
 end;














begin

reserve i,j,k,l for Nat;
reserve A,a,b,x,x1,x2,x3 for set;
reserve D,D9,E for non empty set;
reserve d,d1,d2,d3 for Element of D;
reserve d9,d19,d29,d39 for Element of D9;
reserve p,q,r for FinSequence;



theorem
  min(i,j) is Element of NAT & max(i,j) is Element of NAT by ORDINAL1:def 12;

theorem Th2:
  l = min(i,j) implies Seg i /\ Seg j = Seg l
proof
  i <= j or j <= i;
  then i <= j & Seg i /\ Seg j = Seg i or j <= i & Seg i /\ Seg j = Seg j by
FINSEQ_1:7;
  hence thesis by XXREAL_0:def 9;
end;

theorem Th3:
  i <= j implies max(0,i-j) = 0
proof
  assume i <= j;
  then i-i <= j-i by XREAL_1:9;
  then -(j-i) <= -0;
  hence thesis by XXREAL_0:def 10;
end;

theorem Th4:
  j <= i implies max(0,i-j) = i-j
proof
  assume j <= i;
  then j-j <= i-j by XREAL_1:9;
  hence thesis by XXREAL_0:def 10;
end;

theorem
  max(0,i-j) is Element of NAT
proof
  per cases;
  suppose
    i <= j;
    hence thesis by Th3;
  end;
  suppose
A1: j <= i;
    then i - j is Element of NAT by NAT_1:21;
    hence thesis by A1,Th4;
  end;
end;

theorem
  min(0,i) = 0 & min(i,0) = 0 & max(0,i) = i & max(i,0) = i by XXREAL_0:def 9
,def 10;



theorem
  i in Seg (l+1) implies i in Seg l or i = l+1
proof
  assume
A1: i in Seg(l+1);
  then i <= l+1 by FINSEQ_1:1;
  then
A2: 1 <= i & i <= l or i = l+1 by A1,FINSEQ_1:1,NAT_1:8;
  i is Element of NAT by ORDINAL1:def 12;
  hence thesis by A2;
end;

theorem
  i in Seg l implies i in Seg(l+j)
proof
  l <= l+j by NAT_1:11;
  then Seg l c= Seg(l+j) by FINSEQ_1:5;
  hence thesis;
end;



theorem
  len p = len q & (for j st j in dom p holds p.j = q.j) implies p = q
proof
  assume that
A1: len p = len q and
A2: for j st j in dom p holds p.j = q.j;
  dom p = Seg len p & dom q = Seg len p by A1,FINSEQ_1:def 3;
  hence thesis by A2,FINSEQ_1:13;
end;

theorem Th10:
  b in rng p implies ex i st i in dom p & p.i = b
proof
  assume b in rng p;
  then ex a st a in dom p & b = p.a by FUNCT_1:def 3;
  hence thesis;
end;

theorem Th11:
  for D being set for p being FinSequence of D st i in dom p holds p.i in D
proof
  let D be set;
  let p be FinSequence of D;
  assume i in dom p;
  then
A1: p.i in rng p by FUNCT_1:def 3;
  rng p c= D by FINSEQ_1:def 4;
  hence thesis by A1;
end;

theorem Th12:
  for D being set holds (for i st i in dom p holds p.i in D)
  implies p is FinSequence of D
proof
  let D be set;
  assume
A1: for i st i in dom p holds p.i in D;
  let b;
  assume b in rng p;
  then ex i st i in dom p & p.i = b by Th10;
  hence thesis by A1;
end;

Lm1: rng <*x1,x2*> = {x1,x2}
proof
  thus rng<* x1,x2 *> = rng<* x1 *> \/ rng <* x2 *> by FINSEQ_1:31
    .= rng<* x1 *> \/ {x2} by FINSEQ_1:38
    .= {x1} \/ {x2} by FINSEQ_1:39
    .= {x1,x2} by ENUMSET1:1;
end;

theorem Th13:
  <*d1,d2*> is FinSequence of D
proof
  rng <*d1,d2*> = {d1,d2} by Lm1;
  hence thesis by FINSEQ_1:def 4;
end;

Lm2: rng <*x1,x2,x3*> = {x1,x2,x3}
proof
  thus rng <* x1,x2,x3 *> = rng(<* x1 *> ^ <* x2,x3 *>) by FINSEQ_1:43
    .= rng<* x1 *> \/ rng <* x2,x3 *> by FINSEQ_1:31
    .= {x1} \/ rng<* x2,x3 *> by FINSEQ_1:38
    .= {x1} \/ {x2,x3} by Lm1
    .= {x1,x2,x3} by ENUMSET1:2;
end;

theorem Th14:
  <*d1,d2,d3*> is FinSequence of D
proof
  rng <*d1,d2,d3*> = {d1,d2,d3} by Lm2;
  hence thesis by FINSEQ_1:def 4;
end;

theorem Th15:
  i in dom p implies i in dom(p^q)
proof
A1: dom p c= dom(p^q) by FINSEQ_1:26;
  assume i in dom p;
  hence thesis by A1;
end;

theorem Th16:
  len(p^<*a*>) = len p + 1
proof
  len(p^<*a*>) = len p + len <*a*> by FINSEQ_1:22;
  hence thesis by FINSEQ_1:39;
end;

theorem
  p^<*a*> = q^<*b*> implies p = q & a = b
proof
  assume
A1: p^<*a*> = q^<*b*>;
A2: (p^<*a*>).(len p + 1) = a & (q^<*b*>).(len q + 1) = b by FINSEQ_1:42;
  len(p^<*a*>) = len p + 1 & len(q^<*b*>) = len q + 1 by Th16;
  hence thesis by A1,A2,FINSEQ_1:33;
end;

theorem
  len p = i + 1 implies ex q,a st p = q^<*a*> by CARD_1:27,FINSEQ_1:46;

theorem Th19:
  for p being FinSequence of A st len p <> 0 ex q being
  FinSequence of A, d being Element of A st p = q^<*d*>
proof
  let p be FinSequence of A;
  assume
A1: len p <> 0;
  then p <> {};
  then consider q being FinSequence,d being set such that
A2: p = q^<*d*> by FINSEQ_1:46;
  for i st i in dom q holds q.i in A
  proof
    let i;
    assume i in dom q;
    then p.i = q.i & i in dom p by A2,Th15,FINSEQ_1:def 7;
    hence thesis by Th11;
  end;
  then
A3: q is FinSequence of A by Th12;
  len p in Seg len p by A1,FINSEQ_1:3;
  then
A4: len p in dom p by FINSEQ_1:def 3;
  len p = len q + 1 by A2,Th16;
  then p.(len p) = d by A2,FINSEQ_1:42;
  then d is Element of A by A4,Th11;
  hence thesis by A2,A3;
end;

theorem Th20:
  q = p|(Seg i) & len p <= i implies p = q
proof
  assume
A1: q = p|(Seg i);
  assume len p <= i;
  then Seg len p c= Seg i by FINSEQ_1:5;
  then dom p c= Seg i by FINSEQ_1:def 3;
  hence thesis by A1,RELAT_1:68;
end;

theorem
  q = p|(Seg i) implies len q = min(i,len p)
proof
  assume
A1: q = p|(Seg i);
  now
    per cases;
    case
      i <= len p;
      hence len q = i by A1,FINSEQ_1:17;
    end;
    case
      not i <= len p;
      hence len q = len p by A1,Th20;
    end;
  end;
  hence thesis by XXREAL_0:def 9;
end;

theorem Th22:
  len r = i+j implies ex p,q st len p = i & len q = j & r = p^q
proof
  assume
A1: len r = i+j;
  reconsider z=i as Element of NAT by ORDINAL1:def 12;
  reconsider p = r|(Seg z) as FinSequence by FINSEQ_1:15;
  consider q being FinSequence such that
A2: r = p^q by FINSEQ_1:80;
  take p,q;
  i <= len r by A1,NAT_1:11;
  hence len p = i by FINSEQ_1:17;
  then len(p^q) = i + len q by FINSEQ_1:22;
  hence len q = j by A1,A2;
  thus thesis by A2;
end;

theorem Th23:
  for r being FinSequence of D st len r = i+j ex p,q being
  FinSequence of D st len p = i & len q = j & r = p^q
proof
  let r be FinSequence of D;
  assume len r = i+j;
  then consider p,q being FinSequence such that
A1: len p = i & len q = j and
A2: r = p^q by Th22;
  p is FinSequence of D & q is FinSequence of D by A2,FINSEQ_1:36;
  hence thesis by A1,A2;
end;

scheme
  SeqLambdaD{i()->Nat,D()->non empty set,F(set)->Element of D()}: ex z being
FinSequence of D() st len z = i() & for j being Nat st j in dom z holds z.j = F
  (j) proof
  consider z being FinSequence such that
A1: len z = i() and
A2: for i being Nat st i in dom z holds z.i = F(i) from FINSEQ_1:sch 2;
A3: Seg i() = dom z by A1,FINSEQ_1:def 3;
  for j be Nat st j in Seg i() holds z.j in D()
  proof
    let j be Nat;
    assume j in Seg i();
    then z.j = F(j) by A2,A3;
    hence thesis;
  end;
  then reconsider z as FinSequence of D() by A3,Th12;
  take z;
  thus len z = i() by A1;
  let j be Nat;
  thus thesis by A2;
end;

scheme
  IndSeqD{D()->set, P[set]}: for p being FinSequence of D() holds P[p]
provided
A1: P[<*> D()] and
A2: for p being FinSequence of D() for x being Element of D() st P[p]
holds P[p^<*x*>]
proof
  defpred R[set] means for p being FinSequence of D() st len p = $1 holds P[p];
A3: for i st R[i] holds R[i+1]
  proof
    let i such that
A4: for p being FinSequence of D() st len p = i holds P[p];
    let p be FinSequence of D();
    assume
A5: len p = i + 1;
    then consider q being FinSequence of D(), x being Element of D() such that
A6: p = q^<*x*> by Th19;
    len p = len q + 1 by A6,Th16;
    hence thesis by A2,A4,A5,A6;
  end;
  let p be FinSequence of D();
A7: len p = len p;
A8: R[0] proof let p be FinSequence of D();
     assume len p = 0;
      then p = <*>D();
     hence thesis by A1;
    end;
  for i holds R[i] from NAT_1:sch 2(A8,A3);
  hence thesis by A7;
end;

theorem Th24:
  for D being set, D1 being Subset of D for p being FinSequence of D1
  holds p is FinSequence of D
proof
  let D be set, D1 be Subset of D;
  let p be FinSequence of D1;
  rng p c= D1 by FINSEQ_1:def 4;
  hence rng p c= D by XBOOLE_1:1;
end;

theorem Th25:
  for f being Function of Seg i, D holds f is FinSequence of D
proof
  let f be Function of Seg i, D;
  reconsider i as Element of NAT by ORDINAL1:def 12;
  dom f = Seg i by FUNCT_2:def 1;
  then
A1: f is FinSequence by FINSEQ_1:def 2;
  rng f c= D by RELAT_1:def 19;
  hence thesis by A1,FINSEQ_1:def 4;
end;

theorem
  for p being FinSequence of D holds p is Function of dom p, D
proof
  let p be FinSequence of D;
  rng p c= D by FINSEQ_1:def 4;
  hence thesis by FUNCT_2:2;
end;

theorem
  for f being Function of NAT,D holds f|(Seg i) is FinSequence of D
proof
  let f be Function of NAT,D;
  f|(Seg i) is Function of Seg i, D by FUNCT_2:32;
  hence thesis by Th25;
end;

theorem
  for f being Function of NAT,D st q = f|(Seg i) holds len q = i
proof
  let f be Function of NAT,D;
  reconsider i as Element of NAT by ORDINAL1:def 12;
  f|(Seg i) is Function of Seg i, D by FUNCT_2:32;
  then dom(f|(Seg i)) = Seg i by FUNCT_2:def 1;
  hence thesis by FINSEQ_1:def 3;
end;

theorem Th29:
  for f being Function st rng p c= dom f & q = f*p holds len q = len p
proof
  let f be Function;
  assume rng p c= dom f;
  then dom(f*p) = dom p by RELAT_1:27;
  then dom(f*p) = Seg len p by FINSEQ_1:def 3;
  hence thesis by FINSEQ_1:def 3;
end;

theorem Th30:
  D = Seg i implies for p being FinSequence for q being
  FinSequence of D st i <= len p holds p*q is FinSequence
proof
  assume
A1: D = Seg i;
  let p be FinSequence;
  let q be FinSequence of D;
  assume i <= len p;
  then Seg i c= Seg len p by FINSEQ_1:5;
  then
A2: Seg i c= dom p by FINSEQ_1:def 3;
  rng q c= Seg i by A1,FINSEQ_1:def 4;
  then dom (p*q) = dom q by A2,RELAT_1:27,XBOOLE_1:1;
  then dom (p*q) = Seg len q by FINSEQ_1:def 3;
  hence thesis by FINSEQ_1:def 2;
end;

theorem
  D = Seg i implies for p being FinSequence of D9 for q being
  FinSequence of D st i <= len p holds p*q is FinSequence of D9
proof
  assume
A1: D = Seg i;
  let p be FinSequence of D9;
  let q be FinSequence of D;
  assume i <= len p;
  then reconsider pq = p*q as FinSequence by A1,Th30;
  rng pq c= rng p & rng p c= D9 by FINSEQ_1:def 4,RELAT_1:26;
  then rng pq c= D9 by XBOOLE_1:1;
  hence thesis by FINSEQ_1:def 4;
end;

theorem Th32:
  for A,D being set for p being FinSequence of A for f being
  Function of A,D holds f*p is FinSequence of D
proof
  let A,D be set;
  let p be FinSequence of A;
  let f be Function of A,D;
  per cases;
  suppose D = {};
    then f*p = <*>D;
    hence thesis;
  end;
  suppose
A1: D <> {};
A2: rng p c= A by RELAT_1:def 19;
A3: rng(f*p) c= D by RELAT_1:def 19;
    dom f = A by A1,FUNCT_2:def 1;
    then f*p is FinSequence by A2,FINSEQ_1:16;
    hence thesis by A3,FINSEQ_1:def 4;
  end;
end;

theorem Th33:
  for p being FinSequence of A
  for f being Function of A,D9 st q = f*p holds len q = len p
proof
  let p be FinSequence of A;
  let f be Function of A,D9;
  dom f = A & rng p c= A by FINSEQ_1:def 4,FUNCT_2:def 1;
  hence thesis by Th29;
end;

theorem
  for x being set, f being Function st x in dom f holds f*<*x*> = <*f.x*>
proof
  let x be set, f be Function;
  assume A1: x in dom f;
  then reconsider D = dom f, E = rng f as non empty set by FUNCT_1:3;
  rng <*x*> = {x} by FINSEQ_1:38;
  then rng <*x*> c= D by A1,ZFMISC_1:31;
  then reconsider p = <*x*> as FinSequence of D by FINSEQ_1:def 4;
  reconsider f as Function of D, E by FUNCT_2:def 1,RELSET_1:4;
  reconsider q = f*p as FinSequence of E by Th32;
A2: p.1 = x by FINSEQ_1:40;
A3: len q = len p by Th33 .= 1 by FINSEQ_1:39;
  then 1 in Seg len q;
  then 1 in dom q by FINSEQ_1:def 3;
  then q.1 = f.x by A2,FUNCT_1:12;
  hence thesis by A3,FINSEQ_1:40;
end;

theorem
  for p being FinSequence of D
  for f being Function of D,D9 st p = <*x1*> holds f*p = <*f.x1*>
proof
  let p be FinSequence of D;
  let f be Function of D,D9 such that
A1: p = <*x1*>;
A2: p.1 = x1 by A1,FINSEQ_1:40;
  reconsider q = f*p as FinSequence of D9 by Th32;
  len p = 1 by A1,FINSEQ_1:39;
  then
A3: len q = 1 by Th33;
  then 1 in Seg len q;
  then 1 in dom q by FINSEQ_1:def 3;
  then q.1 = f.x1 by A2,FUNCT_1:12;
  hence thesis by A3,FINSEQ_1:40;
end;

theorem Th36:
  for p being FinSequence of D for f being Function of D,D9 st p =
  <*x1,x2*> holds f*p = <*f.x1,f.x2*>
proof
  let p be FinSequence of D;
  let f be Function of D,D9 such that
A1: p = <*x1,x2*>;
A2: p.2 = x2 by A1,FINSEQ_1:44;
  reconsider q = f*p as FinSequence of D9 by Th32;
  len p = 2 by A1,FINSEQ_1:44;
  then
A3: len q = 2 by Th33;
  then 2 in Seg len q;
  then 2 in dom q by FINSEQ_1:def 3;
  then
A4: q.2 = f.x2 by A2,FUNCT_1:12;
  1 in Seg len q by A3;
  then
A5: 1 in dom q by FINSEQ_1:def 3;
  p.1 = x1 by A1,FINSEQ_1:44;
  then q.1 = f.x1 by A5,FUNCT_1:12;
  hence thesis by A3,A4,FINSEQ_1:44;
end;

theorem Th37:
  for p being FinSequence of D for f being Function of D,D9 st p =
  <*x1,x2,x3*> holds f*p = <*f.x1,f.x2,f.x3*>
proof
  let p be FinSequence of D;
  let f be Function of D,D9 such that
A1: p = <*x1,x2,x3*>;
A2: p.2 = x2 by A1,FINSEQ_1:45;
  reconsider q = f*p as FinSequence of D9 by Th32;
  len p = 3 by A1,FINSEQ_1:45;
  then
A3: len q = 3 by Th33;
  then 2 in Seg len q;
  then 2 in dom q by FINSEQ_1:def 3;
  then
A4: q.2 = f.x2 by A2,FUNCT_1:12;
A5: p.3 = x3 by A1,FINSEQ_1:45;
A6: p.1 = x1 by A1,FINSEQ_1:45;
  3 in Seg len q by A3;
  then 3 in dom q by FINSEQ_1:def 3;
  then
A7: q.3 = f.x3 by A5,FUNCT_1:12;
  1 in Seg len q by A3;
  then 1 in dom q by FINSEQ_1:def 3;
  then q.1 = f.x1 by A6,FUNCT_1:12;
  hence thesis by A3,A4,A7,FINSEQ_1:45;
end;

theorem Th38:
  for f being Function of Seg i,Seg j st (j = 0 implies i = 0) & j
  <= len p holds p*f is FinSequence
proof
  let f be Function of Seg i,Seg j;
  assume j = 0 implies i = 0;
  then Seg j = {} implies Seg i = {};
  then
A1: dom f = Seg i by FUNCT_2:def 1;
  assume j <= len p;
  then rng f c= Seg j & Seg j c= Seg len p by FINSEQ_1:5,RELAT_1:def 19;
  then rng f c= Seg len p by XBOOLE_1:1;
  then rng f c= dom p by FINSEQ_1:def 3;
  then dom(p*f) = dom f by RELAT_1:27;
  hence thesis by A1,FINSEQ_1:def 2;
end;

theorem
  for f being Function of Seg i,Seg i st i <= len p holds p*f is
  FinSequence by Th38;

theorem
  for f being Function of dom p,dom p holds p*f is FinSequence
proof
  dom p = Seg len p by FINSEQ_1:def 3;
  hence thesis by Th38;
end;

theorem Th41:
  for f being Function of Seg i,Seg i st rng f = Seg i & i <= len
  p & q = p*f holds len q = i
proof
  let f be Function of Seg i,Seg i;
  assume rng f = Seg i & i <= len p;
  then rng f c= Seg len p by FINSEQ_1:5;
  then rng f c= dom p by FINSEQ_1:def 3;
  then
A1: dom(p*f) = dom f by RELAT_1:27;
  i is Element of NAT & dom f = Seg i by FUNCT_2:52,ORDINAL1:def 12;
  hence thesis by A1,FINSEQ_1:def 3;
end;

theorem Th42:
  for f being Function of dom p,dom p st rng f = dom p & q = p*f
  holds len q = len p
proof
  Seg len p = dom p by FINSEQ_1:def 3;
  hence thesis by Th41;
end;

theorem Th43:
  for f being Permutation of Seg i st i <= len p & q = p*f holds len q = i
proof
  let f be Permutation of Seg i;
  rng f = Seg i by FUNCT_2:def 3;
  hence thesis by Th41;
end;

theorem
  for f being Permutation of dom p st q = p*f holds len q = len p
proof
  Seg len p = dom p by FINSEQ_1:def 3;
  hence thesis by Th43;
end;

theorem Th45:
  for p being FinSequence of D for f being Function of Seg i,Seg j
  st (j = 0 implies i = 0) & j <= len p holds p*f is FinSequence of D
proof
  let p be FinSequence of D;
  let f be Function of Seg i,Seg j such that
A1: ( j = 0 implies i = 0)& j <= len p;
  set q = p*f;
  rng p c= D & rng q c= rng p by FINSEQ_1:def 4,RELAT_1:26;
  then
A2: rng q c= D by XBOOLE_1:1;
  q is FinSequence by A1,Th38;
  hence thesis by A2,FINSEQ_1:def 4;
end;

theorem
  for p being FinSequence of D for f being Function of Seg i,Seg i st i
  <= len p holds p*f is FinSequence of D by Th45;

theorem Th47:
  for p being FinSequence of D for f being Function of dom p,dom p
  holds p*f is FinSequence of D
proof
  let p be FinSequence of D;
  Seg len p = dom p by FINSEQ_1:def 3;
  hence thesis by Th45;
end;

theorem Th48:
  id Seg k is FinSequence of NAT
proof
  set I = id Seg k;
  reconsider k as Element of NAT by ORDINAL1:def 12;
  dom I = Seg k;
  then rng I = Seg k & I is FinSequence by FINSEQ_1:def 2;
  hence thesis by FINSEQ_1:def 4;
end;

definition
  let i be Nat;
  func idseq i -> FinSequence equals
  id Seg i;
  coherence by Th48;
end;

registration
  let k;
  cluster idseq k -> k-element;
  coherence
  proof
    k in NAT & dom idseq k = Seg k by ORDINAL1:def 12,RELAT_1:45;
    hence len idseq k = k by FINSEQ_1:def 3;
  end;
end;

registration
  cluster idseq 0 -> empty;
  coherence;
end;

theorem
  for k being Element of Seg i holds (idseq i).k = k
proof
  let k be Element of Seg i;
  per cases;
  suppose
    i <> 0;
    hence thesis by FUNCT_1:18;
  end;
  suppose
    i = 0;
    then k = 0 & dom(idseq i) = 0 by SUBSET_1:def 1;
    hence thesis by FUNCT_1:def 2;
  end;
end;

theorem Th50:
  idseq 1 = <*1*>
proof
  1 in Seg 1;
  then len idseq 1 = 1 & (idseq 1).1 = 1 by CARD_1:def 7,FUNCT_1:18;
  hence thesis by FINSEQ_1:40;
end;

theorem Th51:
  idseq (i+1) = (idseq i) ^ <*i+1*>
proof
  set p = idseq (i+1);
  consider q being FinSequence , a being set such that
A1: p = q^<*a*> by FINSEQ_1:46;
A2: len p = i + 1 & len p = len q + 1 by A1,Th16,CARD_1:def 7;
A3: dom q = Seg len q by FINSEQ_1:def 3;
A4: for a st a in Seg i holds q.a = a
  proof
    i <= i+1 by NAT_1:11;
    then
A5: Seg i c= Seg (i+1) by FINSEQ_1:5;
    let a;
    assume
A6: a in Seg i;
    then reconsider j = a as Element of NAT;
    p.j = q.j by A1,A2,A3,A6,FINSEQ_1:def 7;
    hence thesis by A6,A5,FUNCT_1:18;
  end;
  p.(i+1) = i+1 by FINSEQ_1:4,FUNCT_1:18;
  then a = i+1 by A1,A2,FINSEQ_1:42;
  hence thesis by A1,A2,A3,A4,FUNCT_1:17;
end;

theorem Th52:
  idseq 2 = <*1,2*> by Th50,Th51;

theorem
  idseq 3 = <*1,2,3*> by Th51,Th52;

theorem Th54:
  len p <= k implies p*(idseq k) = p
proof
  assume
A1: len p <= k;
  reconsider k as Element of NAT by ORDINAL1:def 12;
  dom p = Seg len p by FINSEQ_1:def 3;
  then dom p c= Seg k by A1,FINSEQ_1:5;
  hence thesis by RELAT_1:51;
end;

theorem
  idseq k is Permutation of Seg k;

theorem Th56:
  (Seg k) --> a is FinSequence
proof
  set p = Seg k --> a;
  reconsider k as Element of NAT by ORDINAL1:def 12;
  dom p = Seg k by FUNCOP_1:13;
  hence thesis by FINSEQ_1:def 2;
end;

registration
  let k, a;
  cluster (Seg k) --> a -> FinSequence-like;
  coherence by Th56;
end;

definition
  let i be Nat, a be set;
  func i |-> a -> FinSequence equals
  Seg i --> a;
  coherence;
end;

registration
  let k be Nat, a be set;
  cluster k |-> a -> k-element;
  coherence
  proof
    k in NAT & dom(k |-> a) = Seg k by FUNCOP_1:13,ORDINAL1:def 12;
    hence len(k |-> a) = k by FINSEQ_1:def 3;
  end;
end;

theorem Th57:
  for d being set, w being set st w in Seg k holds (k |-> d).w = d by
FUNCOP_1:7;

theorem
  0 |-> a = {};

theorem Th59:
  1 |-> a = <*a*>
proof
  1 in Seg 1;
  then len(1 |-> a) = 1 & (1 |-> a).1 = a by CARD_1:def 7,FUNCOP_1:7;
  hence thesis by FINSEQ_1:40;
end;

theorem Th60:
  (i+1) |-> a = (i |-> a) ^ <*a*>
proof
  set p = (i+1) |-> a;
  consider q being FinSequence , x being set such that
A1: p = q^<*x*> by FINSEQ_1:46;
A2: len p = i + 1 & len p = len q + 1 by A1,Th16,CARD_1:def 7;
A3: now
    per cases;
    suppose
A4:   i = 0;
      then q = {} by A2;
      hence q = i |-> a by A4;
    end;
    suppose
A5:   i <> 0;
A6:   rng q c= rng p & rng p = {a} by A1,FINSEQ_1:29,FUNCOP_1:8;
A7:   dom q = Seg len q by FINSEQ_1:def 3;
      then i in dom q by A2,A5,FINSEQ_1:3;
      then q.i in rng q by FUNCT_1:def 3;
      then rng q = {a} by A6,ZFMISC_1:33;
      hence q = i |-> a by A2,A7,FUNCOP_1:9;
    end;
  end;
  p.(i+1) = a by FINSEQ_1:4,FUNCOP_1:7;
  hence thesis by A1,A2,A3,FINSEQ_1:42;
end;

theorem Th61:
  2 |-> a = <*a,a*>
proof
  thus 2 |-> a = (1+1) |-> a .= (1|->a)^<*a*> by Th60
    .= <*a,a*> by Th59;
end;

theorem
  3 |-> a = <*a,a,a*>
proof
  thus 3 |-> a = (2+1) |-> a .= (2|->a)^<*a*> by Th60
    .= <*a,a*>^<*a*> by Th61
    .= <*a,a,a*>;
end;

theorem Th63:
  k |-> d is FinSequence of D
proof
  set s = k |-> d;
  rng s c= {d} by FUNCOP_1:13;
  then rng s c= D by XBOOLE_1:1;
  hence thesis by FINSEQ_1:def 4;
end;

Lm3: for F being Function st [:rng p,rng q:] c= dom F & i = min(len p,len q)
holds dom(F.:(p,q)) = Seg i
proof
  let F be Function such that
A1: [:rng p,rng q:] c= dom F and
A2: i = min(len p,len q);
A3: rng <:p,q:> c= [:rng p,rng q:] & dom <:p,q:> = dom p /\ dom q by FUNCT_3:51
,def 7;
  dom p = Seg len p & dom q = Seg len q by FINSEQ_1:def 3;
  then dom p /\ dom q = Seg i by A2,Th2;
  hence thesis by A1,A3,RELAT_1:27,XBOOLE_1:1;
end;

theorem Th64:
  for F being Function st [:rng p,rng q:] c= dom F holds F.:(p,q)
  is FinSequence
proof
  let F be Function;
  reconsider k = min(len p,len q) as Element of NAT by XXREAL_0:15;
  assume [:rng p,rng q:] c= dom F;
  then dom(F.:(p,q)) = Seg k by Lm3;
  hence thesis by FINSEQ_1:def 2;
end;

theorem Th65:
  for F being Function st [:rng p,rng q:] c= dom F & r = F.:(p,q)
  holds len r = min(len p,len q)
proof
  let F be Function;
  reconsider k = min(len p,len q) as Element of NAT by XXREAL_0:15;
  assume [:rng p,rng q:] c= dom F;
  then dom(F.:(p,q)) = Seg k by Lm3;
  hence thesis by FINSEQ_1:def 3;
end;

Lm4: for F being Function st [:{a},rng p:] c= dom F holds dom(F[;](a,p)) = dom
p
proof
  let F be Function such that
A1: [:{a},rng p:] c= dom F;
  set q = dom p --> a;
  dom q = dom p by FUNCOP_1:13;
  then
A2: dom <:q,p:> = dom p by FUNCT_3:50;
  rng q c= {a} by FUNCOP_1:13;
  then rng <:q,p:> c= [:rng q,rng p:] & [:rng q,rng p:] c= [:{a},rng p:] by
FUNCT_3:51,ZFMISC_1:95;
  then
A3: rng <:q,p:> c= [:{a},rng p:] by XBOOLE_1:1;
  thus thesis by A1,A3,A2,RELAT_1:27,XBOOLE_1:1;
end;

theorem Th66:
  for F being Function st [:{a},rng p:] c= dom F holds F[;](a,p) is FinSequence
proof
  let F be Function;
  assume [:{a},rng p:] c= dom F;
  then dom(F[;](a,p)) = dom p by Lm4;
  then dom(F[;](a,p)) = Seg len p by FINSEQ_1:def 3;
  hence thesis by FINSEQ_1:def 2;
end;

theorem Th67:
  for F being Function st [:{a},rng p:] c= dom F & r = F[;](a,p)
  holds len r = len p
proof
  let F be Function;
  assume [:{a},rng p:] c= dom F;
  then dom(F[;](a,p)) = dom p by Lm4;
  then dom(F[;](a,p)) = Seg len p by FINSEQ_1:def 3;
  hence thesis by FINSEQ_1:def 3;
end;

Lm5: for F being Function st [:rng p,{a}:] c= dom F holds dom(F[:](p,a)) = dom
p
proof
  let F be Function such that
A1: [:rng p,{a}:] c= dom F;
  set q = dom p --> a;
  dom q = dom p by FUNCOP_1:13;
  then
A2: dom <:p,q:> = dom p by FUNCT_3:50;
  rng q c= {a} by FUNCOP_1:13;
  then rng <:p,q:> c= [:rng p,rng q:] & [:rng p,rng q:] c= [:rng p,{a}:] by
FUNCT_3:51,ZFMISC_1:95;
  then
A3: rng <:p,q:> c= [:rng p,{a}:] by XBOOLE_1:1;
  thus thesis by A1,A3,A2,RELAT_1:27,XBOOLE_1:1;
end;

theorem Th68:
  for F being Function st [:rng p,{a}:] c= dom F holds F[:](p,a) is FinSequence
proof
  let F be Function;
  assume [:rng p,{a}:] c= dom F;
  then dom(F[:](p,a)) = dom p by Lm5;
  then dom(F[:](p,a)) = Seg len p by FINSEQ_1:def 3;
  hence thesis by FINSEQ_1:def 2;
end;

theorem Th69:
  for F being Function st [:rng p,{a}:] c= dom F & r = F[:](p,a)
  holds len r = len p
proof
  let F be Function;
  assume [:rng p,{a}:] c= dom F;
  then dom(F[:](p,a)) = dom p by Lm5;
  then dom(F[:](p,a)) = Seg len p by FINSEQ_1:def 3;
  hence thesis by FINSEQ_1:def 3;
end;

theorem Th70:
  for F being Function of [:D,D9:],E for p being FinSequence of D
  for q being FinSequence of D9 holds F.:(p,q) is FinSequence of E
proof
  let F be Function of [:D,D9:],E;
  let p be FinSequence of D;
  let q be FinSequence of D9;
A1: rng(F.:(p,q)) c= rng F by RELAT_1:26;
  rng p c= D & rng q c= D9 by FINSEQ_1:def 4;
  then [:rng p,rng q:] c= [:D,D9:] by ZFMISC_1:96;
  then [:rng p,rng q:] c= dom F by FUNCT_2:def 1;
  then
A2: F.:(p,q) is FinSequence by Th64;
  rng F c= E by RELAT_1:def 19;
  then rng(F.:(p,q)) c= E by A1,XBOOLE_1:1;
  hence thesis by A2,FINSEQ_1:def 4;
end;

theorem Th71:
  for F being Function of [:D,D9:],E for p being FinSequence of D
  for q being FinSequence of D9 st r = F.:(p,q) holds len r = min(len p,len q)
proof
  let F be Function of [:D,D9:],E;
  let p be FinSequence of D;
  let q be FinSequence of D9;
  rng p c= D & rng q c= D9 by FINSEQ_1:def 4;
  then [:rng p,rng q:] c= [:D,D9:] by ZFMISC_1:96;
  then [:rng p,rng q:] c= dom F by FUNCT_2:def 1;
  hence thesis by Th65;
end;

theorem Th72:
  for F being Function of [:D,D9:],E for p being FinSequence of D
for q being FinSequence of D9 st len p = len q & r = F.:(p,q) holds len r = len
  p & len r = len q
proof
  let F be Function of [:D,D9:],E;
  let p be FinSequence of D;
  let q be FinSequence of D9;
  assume that
A1: len p = len q and
A2: r = F.:(p,q);
  len r = min(len p,len q) by A2,Th71;
  hence thesis by A1;
end;

theorem
  for F being Function of [:D,D9:],E for p being FinSequence of D for p9
  being FinSequence of D9 holds F.:(<*>D,p9) = <*>E & F.:(p,<*>D9) = <*>E
proof
  let F be Function of [:D,D9:],E;
  let p be FinSequence of D;
  let p9 be FinSequence of D9;
  reconsider r = F.:(<*>D,p9),r9 = F.:(p,<*>D9) as FinSequence of E by Th70;
  len(<*>D) = 0;
  then len r = min(0,len p9) & len r9 = min(len p,0) by Th71;
  hence thesis by XXREAL_0:def 9;
end;

theorem
  for F being Function of [:D,D9:],E for p being FinSequence of D for q
  being FinSequence of D9 st p = <*d1*> & q = <*d19*> holds F.:(p,q) = <*F.(d1,
  d19)*>
proof
  let F be Function of [:D,D9:],E;
  let p be FinSequence of D;
  let q be FinSequence of D9 such that
A1: p = <*d1*> & q = <*d19*>;
A2: p.1 = d1 & q.1 = d19 by A1,FINSEQ_1:40;
  reconsider r = F.:(p,q) as FinSequence of E by Th70;
  len p = 1 & len q = 1 by A1,FINSEQ_1:39;
  then
A3: len r = 1 by Th72;
  then 1 in Seg len r;
  then 1 in dom r by FINSEQ_1:def 3;
  then r.1 = F.(d1,d19) by A2,FUNCOP_1:22;
  hence thesis by A3,FINSEQ_1:40;
end;

theorem
  for F being Function of [:D,D9:],E for p being FinSequence of D for q
being FinSequence of D9 st p = <*d1,d2*> & q = <*d19,d29*> holds F.:(p,q) = <*F
  .(d1,d19),F.(d2,d29)*>
proof
  let F be Function of [:D,D9:],E;
  let p be FinSequence of D;
  let q be FinSequence of D9 such that
A1: p = <*d1,d2*> & q = <*d19,d29*>;
A2: p.2 = d2 & q.2 = d29 by A1,FINSEQ_1:44;
  reconsider r = F.:(p,q) as FinSequence of E by Th70;
  len p = 2 & len q = 2 by A1,FINSEQ_1:44;
  then
A3: len r = 2 by Th72;
  then 2 in Seg len r;
  then 2 in dom r by FINSEQ_1:def 3;
  then
A4: r.2 = F.(d2,d29) by A2,FUNCOP_1:22;
  1 in Seg len r by A3;
  then
A5: 1 in dom r by FINSEQ_1:def 3;
  p.1 = d1 & q.1 = d19 by A1,FINSEQ_1:44;
  then r.1 = F.(d1,d19) by A5,FUNCOP_1:22;
  hence thesis by A3,A4,FINSEQ_1:44;
end;

theorem
  for F being Function of [:D,D9:],E for p being FinSequence of D for q
being FinSequence of D9 st p = <*d1,d2,d3*> & q = <*d19,d29,d39*> holds F.:(p,q
  ) = <*F.(d1,d19),F.(d2,d29),F.(d3,d39)*>
proof
  let F be Function of [:D,D9:],E;
  let p be FinSequence of D;
  let q be FinSequence of D9 such that
A1: p = <*d1,d2,d3*> & q = <*d19,d29,d39*>;
A2: p.2 = d2 & q.2 = d29 by A1,FINSEQ_1:45;
  reconsider r = F.:(p,q) as FinSequence of E by Th70;
  len p = 3 & len q = 3 by A1,FINSEQ_1:45;
  then
A3: len r = 3 by Th72;
  then 2 in Seg len r;
  then 2 in dom r by FINSEQ_1:def 3;
  then
A4: r.2 = F.(d2,d29) by A2,FUNCOP_1:22;
A5: p.3 = d3 & q.3 = d39 by A1,FINSEQ_1:45;
A6: p.1 = d1 & q.1 = d19 by A1,FINSEQ_1:45;
  3 in Seg len r by A3;
  then 3 in dom r by FINSEQ_1:def 3;
  then
A7: r.3 = F.(d3,d39) by A5,FUNCOP_1:22;
  1 in Seg len r by A3;
  then 1 in dom r by FINSEQ_1:def 3;
  then r.1 = F.(d1,d19) by A6,FUNCOP_1:22;
  hence thesis by A3,A4,A7,FINSEQ_1:45;
end;

theorem Th77:
  for F being Function of [:D,D9:],E for p being FinSequence of D9
  holds F[;](d,p) is FinSequence of E
proof
  let F be Function of [:D,D9:],E;
  let p be FinSequence of D9;
A1: rng(F[;](d,p)) c= rng F by RELAT_1:26;
  rng p c= D9 by FINSEQ_1:def 4;
  then [:{d},rng p:] c= [:D,D9:] by ZFMISC_1:96;
  then [:{d},rng p:] c= dom F by FUNCT_2:def 1;
  then
A2: F[;](d,p) is FinSequence by Th66;
  rng F c= E by RELAT_1:def 19;
  then rng(F[;](d,p)) c= E by A1,XBOOLE_1:1;
  hence thesis by A2,FINSEQ_1:def 4;
end;

theorem Th78:
  for F being Function of [:D,D9:],E for p being FinSequence of D9
  st r = F[;](d,p) holds len r = len p
proof
  let F be Function of [:D,D9:],E;
  let p be FinSequence of D9;
  rng p c= D9 by FINSEQ_1:def 4;
  then [:{d},rng p:] c= [:D,D9:] by ZFMISC_1:96;
  then [:{d},rng p:] c= dom F by FUNCT_2:def 1;
  hence thesis by Th67;
end;

theorem
  for F being Function of [:D,D9:],E holds F[;](d,<*>D9) = <*>E
proof
  let F be Function of [:D,D9:],E;
  reconsider r = F[;](d,<*>D9) as FinSequence of E by Th77;
  len(<*>D9) = 0;
  then len r = 0 by Th78;
  hence thesis;
end;

theorem
  for F being Function of [:D,D9:],E for p being FinSequence of D9 st p
  = <*d19*> holds F[;](d,p) = <*F.(d,d19)*>
proof
  let F be Function of [:D,D9:],E;
  let p be FinSequence of D9 such that
A1: p = <*d19*>;
A2: p.1 = d19 by A1,FINSEQ_1:40;
  reconsider r = F[;](d,p) as FinSequence of E by Th77;
  len p = 1 by A1,FINSEQ_1:39;
  then
A3: len r = 1 by Th78;
  then 1 in Seg len r;
  then 1 in dom r by FINSEQ_1:def 3;
  then r.1 = F.(d,d19) by A2,FUNCOP_1:32;
  hence thesis by A3,FINSEQ_1:40;
end;

theorem
  for F being Function of [:D,D9:],E for p being FinSequence of D9 st p
  = <*d19,d29*> holds F[;](d,p) = <*F.(d,d19),F.(d,d29)*>
proof
  let F be Function of [:D,D9:],E;
  let p be FinSequence of D9 such that
A1: p = <*d19,d29*>;
A2: p.2 = d29 by A1,FINSEQ_1:44;
  reconsider r = F[;](d,p) as FinSequence of E by Th77;
  len p = 2 by A1,FINSEQ_1:44;
  then
A3: len r = 2 by Th78;
  then 2 in Seg len r;
  then 2 in dom r by FINSEQ_1:def 3;
  then
A4: r.2 = F.(d,d29) by A2,FUNCOP_1:32;
  1 in Seg len r by A3;
  then
A5: 1 in dom r by FINSEQ_1:def 3;
  p.1 = d19 by A1,FINSEQ_1:44;
  then r.1 = F.(d,d19) by A5,FUNCOP_1:32;
  hence thesis by A3,A4,FINSEQ_1:44;
end;

theorem
  for F being Function of [:D,D9:],E for p being FinSequence of D9 st p
  = <*d19,d29,d39*> holds F[;](d,p) = <*F.(d,d19),F.(d,d29),F.(d,d39)*>
proof
  let F be Function of [:D,D9:],E;
  let p be FinSequence of D9 such that
A1: p = <*d19,d29,d39*>;
A2: p.2 = d29 by A1,FINSEQ_1:45;
  reconsider r = F[;](d,p) as FinSequence of E by Th77;
  len p = 3 by A1,FINSEQ_1:45;
  then
A3: len r = 3 by Th78;
  then 2 in Seg len r;
  then 2 in dom r by FINSEQ_1:def 3;
  then
A4: r.2 = F.(d,d29) by A2,FUNCOP_1:32;
A5: p.3 = d39 by A1,FINSEQ_1:45;
A6: p.1 = d19 by A1,FINSEQ_1:45;
  3 in Seg len r by A3;
  then 3 in dom r by FINSEQ_1:def 3;
  then
A7: r.3 = F.(d,d39) by A5,FUNCOP_1:32;
  1 in Seg len r by A3;
  then 1 in dom r by FINSEQ_1:def 3;
  then r.1 = F.(d,d19) by A6,FUNCOP_1:32;
  hence thesis by A3,A4,A7,FINSEQ_1:45;
end;

theorem Th83:
  for F being Function of [:D,D9:],E for p being FinSequence of D
  holds F[:](p,d9) is FinSequence of E
proof
  let F be Function of [:D,D9:],E;
  let p be FinSequence of D;
A1: rng(F[:](p,d9)) c= rng F by RELAT_1:26;
  rng p c= D by FINSEQ_1:def 4;
  then [:rng p,{d9}:] c= [:D,D9:] by ZFMISC_1:96;
  then [:rng p,{d9}:] c= dom F by FUNCT_2:def 1;
  then
A2: F[:](p,d9) is FinSequence by Th68;
  rng F c= E by RELAT_1:def 19;
  then rng(F[:](p,d9)) c= E by A1,XBOOLE_1:1;
  hence thesis by A2,FINSEQ_1:def 4;
end;

theorem Th84:
  for F being Function of [:D,D9:],E for p being FinSequence of D
  st r = F[:](p,d9) holds len r = len p
proof
  let F be Function of [:D,D9:],E;
  let p be FinSequence of D;
  rng p c= D by FINSEQ_1:def 4;
  then [:rng p,{d9}:] c= [:D,D9:] by ZFMISC_1:96;
  then [:rng p,{d9}:] c= dom F by FUNCT_2:def 1;
  hence thesis by Th69;
end;

theorem
  for F being Function of [:D,D9:],E holds F[:](<*>D,d9) = <*>E
proof
  let F be Function of [:D,D9:],E;
  reconsider r = F[:](<*>D,d9) as FinSequence of E by Th83;
  len(<*>D) = 0;
  then len r = 0 by Th84;
  hence thesis;
end;

theorem
  for F being Function of [:D,D9:],E for p being FinSequence of D st p =
  <*d1*> holds F[:](p,d9) = <*F.(d1,d9)*>
proof
  let F be Function of [:D,D9:],E;
  let p be FinSequence of D such that
A1: p = <*d1*>;
A2: p.1 = d1 by A1,FINSEQ_1:40;
  reconsider r = F[:](p,d9) as FinSequence of E by Th83;
  len p = 1 by A1,FINSEQ_1:39;
  then
A3: len r = 1 by Th84;
  then 1 in Seg len r;
  then 1 in dom r by FINSEQ_1:def 3;
  then r.1 = F.(d1,d9) by A2,FUNCOP_1:27;
  hence thesis by A3,FINSEQ_1:40;
end;

theorem
  for F being Function of [:D,D9:],E for p being FinSequence of D st p =
  <*d1,d2*> holds F[:](p,d9) = <*F.(d1,d9),F.(d2,d9)*>
proof
  let F be Function of [:D,D9:],E;
  let p be FinSequence of D such that
A1: p = <*d1,d2*>;
A2: p.2 = d2 by A1,FINSEQ_1:44;
  reconsider r = F[:](p,d9) as FinSequence of E by Th83;
  len p = 2 by A1,FINSEQ_1:44;
  then
A3: len r = 2 by Th84;
  then 2 in Seg len r;
  then 2 in dom r by FINSEQ_1:def 3;
  then
A4: r.2 = F.(d2,d9) by A2,FUNCOP_1:27;
  1 in Seg len r by A3;
  then
A5: 1 in dom r by FINSEQ_1:def 3;
  p.1 = d1 by A1,FINSEQ_1:44;
  then r.1 = F.(d1,d9) by A5,FUNCOP_1:27;
  hence thesis by A3,A4,FINSEQ_1:44;
end;

theorem
  for F being Function of [:D,D9:],E for p being FinSequence of D st p =
  <*d1,d2,d3*> holds F[:](p,d9) = <*F.(d1,d9),F.(d2,d9),F.(d3,d9)*>
proof
  let F be Function of [:D,D9:],E;
  let p be FinSequence of D such that
A1: p = <*d1,d2,d3*>;
A2: p.2 = d2 by A1,FINSEQ_1:45;
  reconsider r = F[:](p,d9) as FinSequence of E by Th83;
  len p = 3 by A1,FINSEQ_1:45;
  then
A3: len r = 3 by Th84;
  then 2 in Seg len r;
  then 2 in dom r by FINSEQ_1:def 3;
  then
A4: r.2 = F.(d2,d9) by A2,FUNCOP_1:27;
A5: p.3 = d3 by A1,FINSEQ_1:45;
A6: p.1 = d1 by A1,FINSEQ_1:45;
  3 in Seg len r by A3;
  then 3 in dom r by FINSEQ_1:def 3;
  then
A7: r.3 = F.(d3,d9) by A5,FUNCOP_1:27;
  1 in Seg len r by A3;
  then 1 in dom r by FINSEQ_1:def 3;
  then r.1 = F.(d1,d9) by A6,FUNCOP_1:27;
  hence thesis by A3,A4,A7,FINSEQ_1:45;
end;



definition
  let D be set;
  mode FinSequenceSet of D -> set means
  :Def3:
  a in it implies a is FinSequence of D;
  existence
  proof
    take D*;
    thus thesis by FINSEQ_1:def 11;
  end;
end;

definition
  let D be set, S be FinSequenceSet of D;
  redefine mode Element of S -> FinSequence of D;
  coherence
  proof
    per cases;
    suppose S is non empty;
      hence thesis by Def3;
    end;
    suppose S is empty;
      then S = <*>D;
      hence thesis by SUBSET_1:def 1;
    end;
  end;
end;

registration
  let D be set;
  cluster non empty for FinSequenceSet of D;
  existence
  proof
    {<*>D} is FinSequenceSet of D
    proof
      let a;
      thus thesis by TARSKI:def 1;
    end;
    hence thesis;
  end;
end;

theorem Th89:
  for D being set holds D* is FinSequenceSet of D
proof
  let D be set;
  D* is FinSequenceSet of D
  proof
    let a;
    thus thesis by FINSEQ_1:def 11;
  end;
  hence thesis;
end;

definition
  let D be set;
  redefine func D* -> FinSequenceSet of D;
  coherence by Th89;
end;

theorem
  for D being set, D9 being FinSequenceSet of D holds D9 c= D*
proof
  let D be set, D9 be FinSequenceSet of D;
  let a;
  assume a in D9;
  then a is FinSequence of D by Def3;
  hence thesis by FINSEQ_1:def 11;
end;

theorem
  for D9 being Subset of D, S being FinSequenceSet of D9
  holds S is FinSequenceSet of D
proof
  let D9 be Subset of D, S be FinSequenceSet of D9;
  S is FinSequenceSet of D
  proof
    let a;
    assume a in S;
    then reconsider p = a as FinSequence of D9 by Def3;
    rng p c= D9 by FINSEQ_1:def 4;
    then rng p c= D by XBOOLE_1:1;
    hence thesis by FINSEQ_1:def 4;
  end;
  hence thesis;
end;

reserve s for Element of D*;

registration
  let i be Nat, D;
  cluster i-element for FinSequence of D;
  existence
   proof
    take i |-> the Element of D;
    thus thesis by Th63;
   end;
end;

definition
  let i be Nat, D be non empty set;
  mode Tuple of i,D is i-element FinSequence of D;
end;

definition
  let i be Nat;
  let D be set;
  func i-tuples_on D -> FinSequenceSet of D equals
  { s where s is Element of D*: len s = i };
  coherence
  proof
    now
      let a;
      assume a in { s where s is Element of D*: len s = i };
      then ex s being Element of D* st s = a & len s = i;
      hence a is FinSequence of D;
    end;
    hence thesis by Def3;
  end;
end;

registration
  let i be Nat, D;
  cluster i-tuples_on D -> non empty;
  coherence
  proof
    set d = the Element of D;
    i |-> d is FinSequence of D by Th63;
    then reconsider S = i |-> d as Element of D* by FINSEQ_1:def 11;
    len S = i by CARD_1:def 7;
    then S in { s : len s = i };
    hence thesis;
  end;
end;

registration
  let D;
  let i be Nat;
  cluster -> i-element for Element of i-tuples_on D;
  coherence
  proof
    let z be Element of i-tuples_on D;
    z in { p9 where p9 is Element of D*: len p9 = i };
    then ex p9 being Element of D* st p9 = z & len p9 = i;
    hence len z = i;
  end;
end;

theorem Th92:
  for D be set, z being FinSequence of D holds z is Element of (
  len z)-tuples_on D
proof
  let D be set, z be FinSequence of D;
  z is Element of D* by FINSEQ_1:def 11;
  then z in { z9 where z9 is Element of D*: len z9 = len z};
  hence thesis;
end;

theorem
 for D being set holds i-tuples_on D = Funcs(Seg i,D)
proof let D be set;
  now
    reconsider j=i as Element of NAT by ORDINAL1:def 12;
    let z be set;
    thus z in i-tuples_on D implies z in Funcs(Seg i,D)
    proof
      assume z in i-tuples_on D;
      then consider s being Element of D* such that
A1:   z = s and
A2:   len s = i;
A3:   rng s c= D by FINSEQ_1:def 4;
      dom s = Seg i by A2,FINSEQ_1:def 3;
      hence thesis by A1,A3,FUNCT_2:def 2;
    end;
    assume z in Funcs(Seg i,D);
    then consider p being Function such that
A4: z = p and
A5: dom p = Seg j and
A6: rng p c= D by FUNCT_2:def 2;
    p is FinSequence by A5,FINSEQ_1:def 2;
    then p is FinSequence of D by A6,FINSEQ_1:def 4;
    then reconsider p as Element of D* by FINSEQ_1:def 11;
    len p = i by A5,FINSEQ_1:def 3;
    hence z in i-tuples_on D by A4;
  end;
  hence thesis by TARSKI:1;
end;

theorem
  for D being set holds 0-tuples_on D = { <*>D }
proof
  let D be set;
  now
    let z be set;
    thus z in 0-tuples_on D implies z = <*>D
    proof
      assume z in 0-tuples_on D;
      then ex s being Element of D* st z = s & len s = 0;
      hence thesis;
    end;
    <*>D is Element of D* & len <*>D = 0 by FINSEQ_1:def 11;
    hence z = <*>D implies z in 0-tuples_on D;
  end;
  hence thesis by TARSKI:def 1;
end;

theorem
  for z being Tuple of 0,D for t being Tuple of i, D
   holds z^t = t & t^z = t by FINSEQ_1:34;

theorem Th96:
  1-tuples_on D = { <*d*>: not contradiction }
proof
  now
    let x be set;
    thus x in 1-tuples_on D implies x in { <*d*>: not contradiction }
    proof
      assume x in 1-tuples_on D;
      then consider s such that
A1:   x = s and
A2:   len s = 1;
      1 in Seg 1;
      then 1 in dom s by A2,FINSEQ_1:def 3;
      then reconsider d1 = s.1 as Element of D by Th11;
      s = <*d1*> by A2,FINSEQ_1:40;
      hence thesis by A1;
    end;
    assume x in { <*d*>: not contradiction };
    then consider d such that
A3: x = <*d*>;
    len <*d*> = 1 & <*d*> is Element of D* by FINSEQ_1:40,def 11;
    hence x in 1-tuples_on D by A3;
  end;
  hence thesis by TARSKI:1;
end;

Lm6:
 for z being Tuple of i,D holds z in i-tuples_on D proof let z be Tuple of i,D;
A1: z is Element of D* by FINSEQ_1:def 11;
  len z = i by CARD_1:def 7;
  hence z in i-tuples_on D by A1;
 end;

theorem Th97:
  for z being Tuple of 1,D ex d st z = <*d*>
proof
  let z be Tuple of 1,D;
  z in 1-tuples_on D by Lm6;
  then z in { <*d*>: not contradiction } by Th96;
  hence thesis;
end;

theorem Th98:
  <*d*> in 1-tuples_on D
proof
  <*d*> in { <*d1*>: not contradiction };
  hence thesis by Th96;
end;

theorem Th99:
  2-tuples_on D = { <*d1,d2*>: not contradiction }
proof
  now
    let x be set;
    thus x in 2-tuples_on D implies x in { <*d1,d2*>: not contradiction }
    proof
      assume x in 2-tuples_on D;
      then consider s such that
A1:   x = s and
A2:   len s = 2;
      2 in Seg 2;
      then
A3:   2 in dom s by A2,FINSEQ_1:def 3;
      1 in Seg 2;
      then 1 in dom s by A2,FINSEQ_1:def 3;
      then reconsider d19 = s.1, d29 = s.2 as Element of D by A3,Th11;
      s = <*d19,d29*> by A2,FINSEQ_1:44;
      hence thesis by A1;
    end;
    assume x in { <*d1,d2*>: not contradiction };
    then consider d1,d2 such that
A4: x = <*d1,d2*>;
    <*d1,d2*> is FinSequence of D by Th13;
    then len <*d1,d2*> = 2 & <*d1,d2*> is Element of D* by FINSEQ_1:44,def 11;
    hence x in 2-tuples_on D by A4;
  end;
  hence thesis by TARSKI:1;
end;

theorem
  for z being Tuple of 2,D ex d1,d2 st z = <*d1,d2*>
proof
  let z be Tuple of 2,D;
A1: z is Element of D* by FINSEQ_1:def 11;
  len z = 2 by CARD_1:def 7;
  then z in 2-tuples_on D by A1;
  then z in { <*d1,d2*>: not contradiction } by Th99;
  hence thesis;
end;

theorem Th101:
  <*d1,d2*> in 2-tuples_on D
proof
  <*d1,d2*> in { <*c1,c2*> where c1 is (Element of D), c2 is Element of D:
  not contradiction };
  hence thesis by Th99;
end;

theorem Th102:
  3-tuples_on D = { <*d1,d2,d3*>: not contradiction }
proof
  now
    let x be set;
    thus x in 3-tuples_on D implies x in { <*d1,d2,d3*>: not contradiction }
    proof
      assume x in 3-tuples_on D;
      then consider s such that
A1:   x = s and
A2:   len s = 3;
      2 in Seg(3);
      then
A3:   2 in dom s by A2,FINSEQ_1:def 3;
      3 in Seg(3);
      then
A4:   3 in dom s by A2,FINSEQ_1:def 3;
      1 in Seg(3);
      then 1 in dom s by A2,FINSEQ_1:def 3;
      then reconsider d19 = s.1, d29 = s.2, d39 = s.3 as Element of D by A3,A4
,Th11;
      s = <*d19,d29,d39*> by A2,FINSEQ_1:45;
      hence thesis by A1;
    end;
    assume x in { <*d1,d2,d3*>: not contradiction };
    then consider d1,d2,d3 such that
A5: x = <*d1,d2,d3*>;
    <*d1,d2,d3*> is FinSequence of D by Th14;
    then len <*d1,d2,d3*> = 3 & <*d1,d2,d3*> is Element of D* by FINSEQ_1:45
,def 11;
    hence x in 3-tuples_on D by A5;
  end;
  hence thesis by TARSKI:1;
end;

theorem
  for z being Tuple of 3,D ex d1,d2,d3 st z = <*d1,d2,d3*>
proof
  let z be Tuple of 3,D;
A1: z is Element of D* by FINSEQ_1:def 11;
  len z = 3 by CARD_1:def 7;
  then z in 3-tuples_on D by A1;
  then z in { <*d1,d2,d3*>: not contradiction } by Th102;
  hence thesis;
end;

theorem Th104:
  <*d1,d2,d3*> in 3-tuples_on D
proof
  <*d1,d2,d3*> in { <*c1,c2,c3*> where c1 is (Element of D),c2 is (Element
  of D),c3 is Element of D : not contradiction };
  hence thesis by Th102;
end;

theorem Th105:
  (i+j)-tuples_on D = {z^t where z is Tuple of i,D,
  t is Tuple of j,D: not contradiction}
proof
  set T = {z^t where z is Tuple of i,D, t is Tuple of j,D: not contradiction};
  now
    let x be set;
    thus x in (i+j)-tuples_on D implies x in T
    proof
      assume x in (i+j)-tuples_on D;
      then consider s such that
A1:   x = s and
A2:   len s = i + j;
      consider z,t being FinSequence of D such that
A3:   len z = i & len t = j and
A4:   s = z^t by A2,Th23;
      z is Tuple of i,D & t is Tuple of j,D by A3,CARD_1:def 7;
      hence thesis by A1,A4;
    end;
    assume x in T;
    then consider z being Tuple of i,D, t being Tuple of j,D such that
A5: x = z^t;
    len z = i & len t = j by CARD_1:def 7;
    then
A6: len(z^t) = i+j by FINSEQ_1:22;
    z^t is Element of D* by FINSEQ_1:def 11;
    hence x in (i+j)-tuples_on D by A5,A6;
  end;
  hence thesis by TARSKI:1;
end;

theorem Th106:
  for s being Tuple of i+j,D
   ex z being Element of i-tuples_on D, t being Element of j-tuples_on D
    st s = z^t
proof
  let s be Tuple of i+j,D;
A1: s is Element of D* by FINSEQ_1:def 11;
  len s = i+j by CARD_1:def 7;
  then s in (i+j)-tuples_on D by A1;
  then s in {z^t where z is Tuple of i,D, t is Tuple of j,D:not contradiction}
       by Th105;
  then consider z being Tuple of i,D, t being Tuple of j,D such that
A2:  s = z^t;
   reconsider z as Element of i-tuples_on D by Lm6;
   reconsider t as Element of j-tuples_on D by Lm6;
   s = z^t by A2;
  hence thesis;
end;

theorem
  for z being Tuple of i,D for t being Tuple of j,D holds z^t is Tuple of i+j,D
proof
  let z be Tuple of i,D;
  let t be Tuple of j,D;
  z^t in
   {z9^t9 where z9 is Tuple of i,D, t9 is Tuple of j,D: not contradiction};
  then z^t in (i+j)-tuples_on D by Th105;
  hence thesis;
end;

theorem
  D* = union {i-tuples_on D where i is Element of NAT: not contradiction}
proof
  a in D* iff a in union {i-tuples_on D where i is Element of NAT : not
  contradiction}
  proof
    thus a in D* implies a in union {i-tuples_on D where i is Element of NAT :
    not contradiction}
    proof
      assume a in D*;
      then reconsider a as FinSequence of D by FINSEQ_1:def 11;
      a is Element of (len a)-tuples_on D & (len a)-tuples_on D in {i
      -tuples_on D where i is Element of NAT : not contradiction} by Th92;
      hence thesis by TARSKI:def 4;
    end;
    assume a in union {i-tuples_on D where i is Element of NAT: not
    contradiction};
    then consider Z being set such that
A1: a in Z and
A2: Z in {i-tuples_on D where i is Element of NAT: not contradiction}
    by TARSKI:def 4;
    consider i being Element of NAT such that
A3: Z = i-tuples_on D by A2;
     ex s being Element of D* st s = a & len s = i by A1,A3;
    hence thesis;
  end;
  hence thesis by TARSKI:1;
end;

theorem
  for D9 being non empty Subset of D
  for z being Tuple of i,D9 holds z is Element of i-tuples_on D
proof let D9 be non empty Subset of D;
 let z being Tuple of i,D9;
  z is Tuple of i, D by Th24;
 hence thesis by Lm6;
end;

Lm7:
 x in i-tuples_on A implies x is i-element FinSequence
 proof
  assume x in i-tuples_on A;
   then x in { s where s is Element of A*: len s = i };
   then ex s being Element of A* st x = s & len s = i;
  hence thesis by CARD_1:def 7;
 end;

theorem
  i-tuples_on D = j-tuples_on A implies i = j
proof
  set f = i |-> the Element of D;
  f is Tuple of i,D by Th63;
  then
A1:  f in i-tuples_on D by Lm6;
  assume i-tuples_on D = j-tuples_on A;
   then f in j-tuples_on A by A1;
   then f is j-element by Lm7;
   then len f = j by CARD_1:def 7;
  hence thesis by CARD_1:def 7;
end;

theorem
  idseq i is Element of i-tuples_on NAT
proof
  idseq i is FinSequence of NAT & len idseq i = i by Th48,CARD_1:def 7;
  hence thesis by Th92;
end;

theorem Th112:
  i |-> d is Element of i-tuples_on D
proof
  i |-> d is FinSequence of D & len(i |-> d) = i by Th63,CARD_1:def 7;
  hence thesis by Lm6;
end;

theorem
  for z being Tuple of i,D for f being Function of D,D9
  holds f*z is Element of i-tuples_on D9
proof
  let z be Tuple of i,D;
  let f be Function of D,D9;
  reconsider s = f*z as FinSequence of D9 by Th32;
  len z = i by CARD_1:def 7;
  then len s = i by Th33;
  hence thesis by Th92;
end;

theorem Th114:
  for z being Tuple of i,D for f being Function of
  Seg i,Seg i st rng f = Seg i holds z*f is Element of i-tuples_on D
proof
  let z be Tuple of i,D;
  let f be Function of Seg i,Seg i;
A1: len z = i by CARD_1:def 7;
  then
A2: Seg i = dom z by FINSEQ_1:def 3;
  then reconsider t = z*f as FinSequence of D by Th47;
  assume rng f = Seg i;
  then len t = len z by A2,Th42;
  hence thesis by A1,Th92;
end;

theorem
  for z being Tuple of i,D for f being Permutation of Seg i
  holds z*f is Tuple of i,D
proof
  let z be Tuple of i,D;
  let f be Permutation of Seg i;
  rng f = Seg i by FUNCT_2:def 3;
  hence thesis by Th114;
end;

theorem
  for z being Tuple of i,D for d holds (z^<*d*>).(i+1) = d
proof
  let z be Tuple of i,D;
  let d;
  len z = i by CARD_1:def 7;
  hence thesis by FINSEQ_1:42;
end;

theorem
  for z being Tuple of i+1,D ex t being Element of i-tuples_on D, d
   st z = t^<*d*>
proof
  let z be Tuple of i+1,D;
  consider t being Element of i-tuples_on D, s being Element of 1-tuples_on D
   such that
A1: z = t^s by Th106;
  ex d st s = <*d*> by Th97;
  hence thesis by A1;
end;

theorem
  for z being Tuple of i,D holds z*(idseq i) = z
proof
  let z be Tuple of i,D;
  len z = i by CARD_1:def 7;
  hence thesis by Th54;
end;

theorem
  for z1,z2 being Tuple of i,D st for j st j in Seg i holds
  z1.j = z2.j holds z1 = z2
proof
  let z1,z2 be Tuple of i,D;
  len z2 = i by CARD_1:def 7;
  then
A1: dom z2 = Seg i by FINSEQ_1:def 3;
  len z1 = i by CARD_1:def 7;
  then dom z1 = Seg i by FINSEQ_1:def 3;
  hence thesis by A1,FINSEQ_1:13;
end;

theorem
  for F being Function of [:D,D9:],E for z1 being Tuple of i,D
   for z2 being Tuple of i,D9 holds F.:(z1,z2) is Element of i
  -tuples_on E
proof
  let F be Function of [:D,D9:],E;
  let z1 be Tuple of i,D;
  let z2 be Tuple of i,D9;
  reconsider r = F.:(z1,z2) as FinSequence of E by Th70;
  len z1 = i & len z2 = i by CARD_1:def 7;
  then len r = i by Th72;
  hence thesis by Th92;
end;

theorem
  for F being Function of [:D,D9:],E for z being Tuple of i,D9
   holds F[;](d,z) is Element of i-tuples_on E
proof
  let F be Function of [:D,D9:],E;
  let z be Tuple of i,D9;
  reconsider r = F[;](d,z) as FinSequence of E by Th77;
  len z = i by CARD_1:def 7;
  then len r = i by Th78;
  hence thesis by Th92;
end;

theorem
  for F being Function of [:D,D9:],E for z being Tuple of i,D
    holds F[:](z,d9) is Element of i-tuples_on E
proof
  let F be Function of [:D,D9:],E;
  let z be Tuple of i,D;
  reconsider r = F[:](z,d9) as FinSequence of E by Th83;
  len z = i by CARD_1:def 7;
  then len r = i by Th84;
  hence thesis by Th92;
end;

theorem
  (i+j)|->x = (i|->x)^(j|->x)
proof
  defpred P[Nat] means (i+$1)|->x = (i|->x)^($1|->x);
A1: for j st P[j] holds P[j+1]
  proof
    let j such that
A2: (i+j)|->x = (i|->x)^(j|->x);
    thus (i+(j+1))|->x = (i+j+1)|->x .= ((i+j)|->x) ^ <*x*> by Th60
      .= (i|->x)^((j|->x) ^ <*x*>) by A2,FINSEQ_1:32
      .= (i|->x)^((j+1)|->x) by Th60;
  end;
  (i+0)|->x = (i|->x)^{} by FINSEQ_1:34
    .= (i|->x)^(0|->x);
  then
A3: P[0];
  for j holds P[j] from NAT_1:sch 2(A3,A1);
  hence thesis;
end;



theorem
  for i, D for x being Tuple of i,D holds dom x = Seg i
proof
  let i, D;
  let x be Tuple of i,D;
  len x = i by CARD_1:def 7;
  hence thesis by FINSEQ_1:def 3;
end;

theorem
  for f being Function, x, y being set st x in dom f & y in dom f holds
  f*<*x,y*> = <*f.x,f.y*>
proof
  let f be Function;
  let x,y be set;
  assume that
A1: x in dom f and
A2: y in dom f;
  reconsider D = dom f, E = rng f as non empty set by A1,FUNCT_1:3;
  rng <*x,y*> = {x,y} by Lm1;
  then rng <*x,y*> c= D by A1,A2,ZFMISC_1:32;
  then reconsider p = <*x,y*> as FinSequence of D by FINSEQ_1:def 4;
  reconsider g = f as Function of D,E by FUNCT_2:def 1,RELSET_1:4;
  thus f*<*x,y*> = g*p .= <*f.x,f.y*> by Th36;
end;

theorem
  for f being Function, x, y, z being set st x in dom f & y in dom f & z
  in dom f holds f*<*x,y,z*> = <*f.x,f.y,f.z*>
proof
  let f be Function;
  let x,y,z be set;
  assume that
A1: x in dom f and
A2: y in dom f & z in dom f;
  reconsider D = dom f, E = rng f as non empty set by A1,FUNCT_1:3;
  rng <*x,y,z*> = {x,y,z} by Lm2;
  then
A3: rng <*x,y,z*> = {x,y} \/ {z} by ENUMSET1:3;
  {x,y} c= D & {z} c= D by A1,A2,ZFMISC_1:31,32;
  then rng <*x,y,z*> c= D by A3,XBOOLE_1:8;
  then reconsider p = <*x,y,z*> as FinSequence of D by FINSEQ_1:def 4;
  reconsider g = f as Function of D,E by FUNCT_2:def 1,RELSET_1:4;
  thus f*<*x,y,z*> = g*p .= <*f.x,f.y,f.z*> by Th37;
end;

theorem
  rng <*x1,x2*> = {x1,x2} by Lm1;

theorem
  rng <*x1,x2,x3*> = {x1,x2,x3} by Lm2;

begin 


theorem
  for p1,p2,q being FinSequence st p1 c= q & p2 c= q & len p1 = len p2
  holds p1 = p2
proof
  let p1,p2,q be FinSequence;
  assume that
A1: p1 c= q and
A2: p2 c= q and
A3: len p1 = len p2;
  reconsider i = len p1 as Element of NAT;
A4: dom p1 = Seg i & dom p2 = Seg i by A3,FINSEQ_1:def 3;
  now
    let j;
    assume
A5: j in dom p1;
    hence p1.j = q.j by A1,GRFUNC_1:2
      .= p2.j by A2,A4,A5,GRFUNC_1:2;
  end;
  hence thesis by A4,FINSEQ_1:13;
end;



reserve m,n for Nat,
  s,w for FinSequence of NAT;

theorem
  for D being non empty set,
      s being FinSequence of D st s <> {}
   ex w being FinSequence of D, n being Element of D st s = <*n*>^w
proof
  let D be non empty set,
      s be FinSequence of D;
  defpred P[FinSequence of D] means
    $1 <> {} implies ex w being FinSequence of D,
    n being Element of D st $1 = <*n*>^w;
A1: for s being FinSequence of D
     for m being Element of D st P[s] holds P[s^<*m*>]
  proof
    let s be FinSequence of D;
    let m be Element of D such that
A2: s <> {} implies ex w being FinSequence of D,
     n be Element of D st s = <*n*>^w;
    assume s^<*m*> <> {};
    per cases;
    suppose
A3:   s = {};
      reconsider w = <*> D as FinSequence of D;
      take w, n = m;
      thus s^<*m*> = <*m*> by A3,FINSEQ_1:34
        .= <*n*>^w by FINSEQ_1:34;
    end;
    suppose
      s <> {};
      then consider w be FinSequence of D,
         n be Element of D such that
A4:   s = <*n*>^w by A2;
      take w^<*m*>,n;
      thus thesis by A4,FINSEQ_1:32;
    end;
  end;
A5: P[<*> D];
  for p being FinSequence of D holds P[p] from IndSeqD(A5,A1);
  hence thesis;
end;



registration
  let D be set;
  cluster -> functional for FinSequenceSet of D;
  coherence
  proof
    let f be FinSequenceSet of D;
    let x be set;
    thus thesis by Def3;
  end;
end;



definition
  let D,n,d;
  redefine func n |-> d -> Element of n-tuples_on D;
  coherence by Th112;
end;



theorem
 for z being set holds z is Tuple of i,D iff z in i-tuples_on D
 proof let z be set;
  thus z is Tuple of i,D implies z in i-tuples_on D by Lm6;
  assume z in i-tuples_on D;
   then ex s being Element of D* st z = s & len s = i;
  hence z is Tuple of i,D by CARD_1:def 7;
 end;



theorem Th132:
  for A being set, i being (Element of NAT), p being FinSequence
  holds p in i-tuples_on A iff len p = i & rng p c= A
proof
  let A be set, i be (Element of NAT), p be FinSequence;
  hereby
    assume p in i-tuples_on A;
    then ex q being Element of A* st p = q & len q = i;
    hence len p = i & rng p c= A by FINSEQ_1:def 4;
  end;
  assume
A1: len p = i;
  assume rng p c= A;
  then p is FinSequence of A by FINSEQ_1:def 4;
  then p in A* by FINSEQ_1:def 11;
  hence thesis by A1;
end;

theorem
  for A being set, i being (Element of NAT), p being FinSequence of
  A holds p in i-tuples_on A iff len p = i
proof
  let A be set, i be (Element of NAT), p be FinSequence of A;
  rng p c= A by RELAT_1:def 19;
  hence thesis by Th132;
end;

theorem
  for A being set, i being Element of NAT holds i-tuples_on A c= A*
proof
  let A be set, i be (Element of NAT), x be set;
  assume x in i-tuples_on A;
  then x is FinSequence of A by Def3;
  hence thesis by FINSEQ_1:def 11;
end;

theorem Th135:
  for A,x being set holds x in 1-tuples_on A iff ex a being set st
  a in A & x = <*a*>
proof
  let A,x be set;
  hereby
    assume x in 1-tuples_on A;
    then x in {s where s is Element of A*: len s = 1};
    then consider s being Element of A* such that
A1: x = s and
A2: len s = 1;
    take a = s.1;
A3: rng <*a*> = {a} & a in {a} by FINSEQ_1:39,TARSKI:def 1;
A4: rng s c= A by RELAT_1:def 19;
    x = <*a*> by A1,A2,FINSEQ_1:40;
    hence a in A & x = <*a*> by A1,A3,A4;
  end;
  given a being set such that
A5: a in A and
A6: x = <*a*>;
  reconsider A as non empty set by A5;
  reconsider a as Element of A by A5;
  <*a*> is Element of 1-tuples_on A by Th98;
  hence thesis by A6;
end;

theorem
  for A,a being set st <*a*> in 1-tuples_on A holds a in A
proof
  let A,a be set;
  assume <*a*> in 1-tuples_on A;
  then
A1: ex a9 being set st a9 in A & <*a*> = <*a9*> by Th135;
  <*a*>.1 = a by FINSEQ_1:40;
  hence thesis by A1,FINSEQ_1:40;
end;

theorem Th137:
  for A,x being set holds x in 2-tuples_on A iff
  ex a,b being set st a in A & b in A & x = <*a,b*>
proof
  let A,x be set;
  hereby
    assume x in 2-tuples_on A;
    then x in {s where s is Element of A*: len s = 2};
    then consider s being Element of A* such that
A1: x = s and
A2: len s = 2;
    take a = s.1, b = s.2;
A3: rng <*a,b*> = {a,b} & a in {a,b} by Lm1,TARSKI:def 2;
A4: b in {a,b} & rng s c= A by RELAT_1:def 19,TARSKI:def 2;
    x = <*a,b*> by A1,A2,FINSEQ_1:44;
    hence a in A & b in A & x = <*a,b*> by A1,A3,A4;
  end;
  given a,b being set such that
A5: a in A and
A6: b in A and
A7: x = <*a,b*>;
  reconsider A as non empty set by A5;
  reconsider a,b as Element of A by A5,A6;
  <*a,b*> is Element of 2-tuples_on A by Th101;
  hence thesis by A7;
end;

theorem
  for A,a,b being set st <*a,b*> in 2-tuples_on A holds a in A & b in A
proof
  let A,a,b be set;
  assume <*a,b*> in 2-tuples_on A;
  then
A1: ex a9,b9 being set st a9 in A & b9 in A & <*a,b*> = <*a9,b9*> by Th137;
  <*a,b*>.1 = a & <*a,b*>.2 = b by FINSEQ_1:44;
  hence thesis by A1,FINSEQ_1:44;
end;

theorem Th139:
  for A,x being set holds x in 3-tuples_on A iff
  ex a,b,c being set st a in A & b in A & c in A & x = <*a,b,c*>
proof
  let A,x be set;
  hereby
    assume x in 3-tuples_on A;
    then x in {s where s is Element of A*: len s = 3};
    then consider s being Element of A* such that
A1: x = s and
A2: len s = 3;
    take a = s.1, b = s.2, c = s.3;
A3: rng <*a,b,c*> = {a,b,c} & a in {a,b,c} by Lm2,ENUMSET1:def 1;
A4: rng s c= A by RELAT_1:def 19;
A5: b in {a,b,c} & c in {a,b,c} by ENUMSET1:def 1;
    x = <*a,b,c*> by A1,A2,FINSEQ_1:45;
    hence a in A & b in A & c in A & x = <*a,b,c*> by A1,A3,A5,A4;
  end;
  given a,b,c being set such that
A6: a in A and
A7: b in A & c in A and
A8: x = <*a,b,c*>;
  reconsider A as non empty set by A6;
  reconsider a,b,c as Element of A by A6,A7;
  <*a,b,c*> is Element of 3-tuples_on A by Th104;
  hence thesis by A8;
end;

theorem
  for A,a,b,c being set st <*a,b,c*> in 3-tuples_on A holds
  a in A & b in A & c in A
proof
  let A,a,b,c be set;
A1: <*a,b,c*>.3 = c by FINSEQ_1:45;
  assume <*a,b,c*> in 3-tuples_on A;
  then
A2: ex a9,b9,c9 being set st a9 in A & b9 in A & c9 in A & <*a,b,c*> = <*a9,
  b9,c9*> by Th139;
  <*a,b,c*>.1 = a & <*a,b,c*>.2 = b by FINSEQ_1:45;
  hence thesis by A2,A1,FINSEQ_1:45;
end;

theorem
  x in i-tuples_on A implies x is i-element FinSequence by Lm7;



theorem
  for A being non empty set, n holds n-tuples_on A c= A*
proof
  let A be non empty set, n;
  defpred P[Element of A*] means len $1 = n;
  { s where s is Element of A*: P[s] } c= A* from FRAENKEL:sch 10;
  hence thesis;
end;



theorem
  n |-> x = m |-> x implies n = m
proof
  len(n |-> x) = n by CARD_1:def 7;
  hence thesis by CARD_1:def 7;
end;



reserve i,j,e,u for set,
        n for Element of NAT;

definition
  let I be set;
  let M be ManySortedSet of I;
  func M# -> ManySortedSet of I* means
  :Def5:
  for i being Element of I* holds it.i = product(M*i);
  existence
  proof
    defpred P[set,set] means ex j being Element of I* st j = $1 & $2 = product
    (M*j);
A1: for i st i in I* ex j st P[i,j]
    proof
      let i;
      assume i in I*;
      then reconsider j = i as Element of I*;
      reconsider e = product(M*j) as set;
      take e,j;
      thus thesis;
    end;
    consider f being ManySortedSet of I* such that
A2: for i st i in I* holds P[i,f.i] from PBOOLE:sch 3(A1);
    take f;
    let i be Element of I*;
    ex j being Element of I* st j = i & f.i = product(M*j) by A2;
    hence thesis;
  end;
  uniqueness
  proof
    let N1,N2 be ManySortedSet of I* such that
A3: for i being Element of I* holds N1.i = product(M*i) and
A4: for i being Element of I* holds N2.i = product(M*i);
    now
      let i;
      assume i in I*;
      then reconsider e = i as Element of I*;
      thus N1.i = product(M*e) by A3
        .= N2.i by A4;
    end;
    hence thesis by PBOOLE:3;
  end;
end;

registration
  let I be set;
  let M be non-empty ManySortedSet of I;
  cluster M# -> non-empty;
  coherence
  proof
    M# is non-empty
    proof
      let i;
      assume i in I*;
      then reconsider f = i as Element of I*;
      product(M*f) <> {};
      hence thesis by Def5;
    end;
    hence thesis;
  end;
end;

definition
  let a be set;
  func *-->a -> Function of NAT,{a}* means
  :Def6:
  for n being Element of NAT holds it.n = n |-> a;
  existence
  proof
    defpred P[Element of NAT,set] means $2 = $1 |-> a;
A1: for x being Element of NAT ex y being Element of {a}* st P[x,y]
    proof
      let n be Element of NAT;
      a is Element of {a} by TARSKI:def 1;
      then n |-> a is FinSequence of {a} by Th63;
      then reconsider u = n |-> a as Element of {a}* by FINSEQ_1:def 11;
      take u;
      thus thesis;
    end;
    ex f being Function of NAT, {a}* st for x being Element of NAT holds P
    [x,f.x] from FUNCT_2:sch 3(A1);
    hence thesis;
  end;
  uniqueness
  proof
    let f1,f2 be Function of NAT,{a}* such that
A2: for n being Element of NAT holds f1.n = n |-> a and
A3: for n being Element of NAT holds f2.n = n |-> a;
    now
      let k be Element of NAT;
      thus f1.k = k |-> a by A2
        .= f2.k by A3;
    end;
    hence f1 = f2 by FUNCT_2:63;
  end;
end;

theorem Th144:
  for a,b being set holds (a .--> b)*(n|->a) = n |-> b
proof
  let a,b be set;
A1: now
    let x be set;
    hereby
      assume x in dom (n |-> b);
      then
A2:   x in Seg n by FUNCOP_1:13;
      hence x in dom(n|->a) by FUNCOP_1:13;
      dom(a .--> b) = {a} & (n|->a).x = a by A2,FUNCOP_1:7,13;
      hence (n|->a).x in dom(a .--> b) by TARSKI:def 1;
    end;
    assume that
A3: x in dom(n|->a) and
    (n|->a).x in dom(a .--> b);
    x in Seg n by A3,FUNCOP_1:13;
    hence x in dom (n |-> b) by FUNCOP_1:13;
  end;
  now
    let x be set;
A4: a in {a} by TARSKI:def 1;
    assume x in dom (n |-> b);
    then
A5: x in Seg n by FUNCOP_1:13;
    hence (n |-> b).x = b by FUNCOP_1:7
      .= (a .--> b).a by A4,FUNCOP_1:7
      .= (a .--> b).((n|->a).x) by A5,FUNCOP_1:7;
  end;
  hence thesis by A1,FUNCT_1:10;
end;

theorem
  for a being set for M being ManySortedSet of {a} st M = a .--> D holds
  (M#* *-->a).n = Funcs(Seg n, D)
proof
  let a be set;
  let M be ManySortedSet of {a} such that
A1: M = a .--> D;
  a is Element of {a} by TARSKI:def 1;
  then n |-> a is FinSequence of {a} by Th63;
  then
A2: n |-> a in {a}* by FINSEQ_1:def 11;
  dom(*-->a) = NAT by FUNCT_2:def 1;
  hence (M#* *-->a).n = M#.((*-->a).n) by FUNCT_1:13
    .= M#.(n|->a) by Def6
    .= product((a .--> D)*(n|->a)) by A1,A2,Def5
    .= product(n |-> D) by Th144
    .= Funcs(Seg n, D) by CARD_3:11;
end;

theorem
 for F being total NAT-defined Function, p be NAT-defined Function,
     n be Element of NAT st Shift(p,n) c= F
 for i being Element of NAT st i in dom p holds F.(n+i) = p.i
 proof
  let F be total NAT-defined Function, p be NAT-defined Function,
     n be Element of NAT such that
A1: Shift(p,n) c= F;
  let i be Element of NAT;
  assume
A2: i in dom p;
   then n+i in dom Shift(p,n) by VALUED_1:24;
  hence F.(n+i) = Shift(p,n).(n+i) by A1,GRFUNC_1:2
      .= p.i by A2,VALUED_1:def 12;
 end;

registration
  let i be Nat;
  cluster i |-> 0 -> empty-yielding;
  coherence
  proof
    let x;
    dom (i|->0) = Seg i by FUNCOP_1:13;
    hence thesis by Th57;
  end;
end;



registration
  let D be set;
  cluster -> FinSequence-membered for FinSequenceSet of D;
  coherence
  proof
    let A be FinSequenceSet of D;
    let x be set;
    thus thesis by Def3;
  end;
end;














begin

reserve f for Function;
reserve n,k,n1 for Element of NAT;
reserve r,p for real number;
reserve x,y,z for set;

definition
  mode Real_Sequence is 
     sequence of REAL;
end;

reserve seq,seq1,seq2,seq3,seq9,seq19 for Real_Sequence;

theorem Th1:
  f is Real_Sequence iff (dom f=NAT & for x st x in NAT holds f.x is real)
proof
  thus f is Real_Sequence implies (dom f=NAT & for x st x in NAT holds f.x is
  real) by FUNCT_2:def 1;
  assume that
A1: dom f= NAT and
A2: for x st x in NAT holds f.x is real;
  now
    let y;
    assume y in rng f;
    then consider x such that
A3: x in dom f and
A4: y=f.x by FUNCT_1:def 3;
    f.x is real by A1,A2,A3;
    hence y in REAL by A4,XREAL_0:def 1;
  end;
  then rng f c=REAL by TARSKI:def 3;
  hence thesis by A1,FUNCT_2:def 1,RELSET_1:4;
end;

theorem Th2:
  f is Real_Sequence iff (dom f=NAT & for n holds f.n is real)
proof
  thus f is Real_Sequence implies (dom f=NAT & for n holds f.n is real) by Th1;
  assume that
A1: dom f=NAT and
A2: for n holds f.n is real;
  for x holds x in NAT implies f.x is real by A2;
  hence thesis by A1,Th1;
end;

definition
  let f be real-valued Function, x be set;
  redefine func f.x -> Real;
  coherence by XREAL_0:def 1;
end;

registration
  cluster non-zero for PartFunc of NAT,REAL;
  existence
  proof
    reconsider p = NAT --> 1 as PartFunc of NAT,REAL by FUNCOP_1:46;
    take p;
    rng p = {1} by FUNCOP_1:8;
    hence not 0 in rng p by TARSKI:def 1;
  end;
end;

theorem
  for f being non-zero PartFunc of NAT,REAL holds rng f c= REAL \ {0}
proof
  let f be non-zero PartFunc of NAT,REAL;
  not {} in rng f by RELAT_1:def 9;
  hence thesis by ZFMISC_1:34;
end;

theorem Th4:
  seq is non-zero iff for x st x in NAT holds seq.x<>0
proof
  thus seq is non-zero implies for x st x in NAT holds seq.x<>0
  proof
    assume
A1: seq is non-zero;
    let x;
    assume x in NAT;
    then x in dom seq by Th2;
    then seq.x in rng seq by FUNCT_1:def 3;
    hence thesis by A1,RELAT_1:def 9;
  end;
  assume
A2: for x st x in NAT holds seq.x<>0;
  assume 0 in rng seq;
  then ex x st x in dom seq & seq.x=0 by FUNCT_1:def 3;
  hence contradiction by A2;
end;

theorem Th5:
  seq is non-zero iff for n holds seq.n<>0
proof
  thus seq is non-zero implies for n holds seq.n<>0 by Th4;
  assume for n holds seq.n<>0;
  then for x holds x in NAT implies seq.x<>0;
  hence thesis by Th4;
end;

theorem
  for r ex seq st rng seq={r}
proof
  let r;
  consider f such that
A1: dom f=NAT and
A2: rng f={r} by FUNCT_1:5;
  for x st x in {r} holds x in REAL by XREAL_0:def 1;
  then rng f c= REAL by A2,TARSKI:def 3;
  then reconsider f as Real_Sequence by A1,FUNCT_2:def 1,RELSET_1:4;
  take f;
  thus thesis by A2;
end;

scheme
  ExRealSeq{F(set)->real number}: ex seq st for n holds seq.n=F(n) proof
  defpred P[set,set] means ex n st n=$1 & $2=F(n);
A1: now
    let x;
    assume x in NAT;
    then consider n such that
A2: n=x;
    reconsider r2=F(n) as set;
    take y=r2;
    thus P[x,y] by A2;
  end;
  consider f such that
A3: dom f=NAT and
A4: for x st x in NAT holds P[x,f.x] from CLASSES1:sch 1(A1);
  now
    let x;
    assume x in NAT;
    then ex n st n=x & f.x=F(n) by A4;
    hence f.x is real;
  end;
  then reconsider f as Real_Sequence by A3,Th1;
  take seq=f;
  let n;
  ex k st k=n & seq.n=F(k) by A4;
  hence thesis;
end;





scheme
  PartFuncExD9{D,C()->non empty set, P[set,set]}: ex f being PartFunc of D(),C
() st (for d be Element of D() holds d in dom f iff (ex c be Element of C() st
  P[d,c])) & for d be Element of D() st d in dom f holds P[d,f.d] proof
  defpred X[set] means ex c be Element of C() st P[$1,c];
  set x = the Element of C();
  defpred X[Element of D(),Element of C()] means ((ex c be Element of C() st P
  [$1,c]) implies P[$1,$2]) & ((for c be Element of C() holds not P[$1,c])
  implies $2=x);
  consider X be set such that
A1: for x holds x in X iff x in D() & X[x] from XBOOLE_0:sch 1;
  for x holds x in X implies x in D() by A1;
  then reconsider X as Subset of D() by TARSKI:def 3;
A2: for d be Element of D() ex z be Element of C() st X[d,z]
  proof
    let d be Element of D();
    (for c be Element of C() holds not P[d,c]) implies ex z st ((ex c be
Element of C() st P[d,c]) implies P[d,z]) & ((for c be Element of C() holds not
    P[d,c]) implies z=x);
    hence thesis;
  end;
  consider g being Function of D(),C() such that
A3: for d be Element of D() holds X[d,g.d] from FUNCT_2:sch 3(A2);
  reconsider f=g|X as PartFunc of D(),C();
  take f;
A4: dom g = D() by FUNCT_2:def 1;
  thus for d be Element of D() holds d in dom f iff ex c be Element of C() st
  P[d,c]
  proof
    let d be Element of D();
    dom f c= X by RELAT_1:58;
    hence d in dom f implies ex c be Element of C() st P[d,c] by A1;
    assume ex c be Element of C() st P[d,c];
    then d in X by A1;
    then d in dom g /\ X by A4,XBOOLE_0:def 4;
    hence thesis by RELAT_1:61;
  end;
  let d be Element of D();
  assume
A5: d in dom f;
  dom f c= X by RELAT_1:58;
  then ex c be Element of C() st P[d,c] by A1,A5;
  then P[d,g.d] by A3;
  hence thesis by A5,FUNCT_1:47;
end;

scheme
  LambdaPFD9{D,C()->non empty set, F(set)->Element of C(), P[set]}: ex f being
  PartFunc of D(),C() st (for d be Element of D() holds d in dom f iff P[d]) &
  for d be Element of D() st d in dom f holds f.d = F(d) proof
  defpred X[Element of D(),set] means P[$1] & $2 = F($1);
  consider f being PartFunc of D(),C() such that
A1: for d be Element of D() holds d in dom f iff ex c be Element of C()
  st X[d,c] and
A2: for d be Element of D() st d in dom f holds X[d,f.d] from
  PartFuncExD9;
  take f;
  thus for d be Element of D() holds d in dom f iff P[d]
  proof
    let d be Element of D();
    thus d in dom f implies P[d]
    proof
      assume d in dom f;
      then ex c be Element of C() st P[d] & c = F(d) by A1;
      hence thesis;
    end;
    assume P[d];
    then ex c be Element of C() st P[d] & c = F(d);
    hence thesis by A1;
  end;
  thus thesis by A2;
end;

scheme
  UnPartFuncD9{C,D,X() -> set, F(set)->set}: for f,g being PartFunc of C(),D()
st (dom f=X() & for c be Element of C() st c in dom f holds f.c = F(c)) & (dom
  g=X() & for c be Element of C() st c in dom g holds g.c = F(c)) holds f = g
proof
  let f,g be PartFunc of C(),D();
  assume that
A1: dom f=X() and
A2: for c be Element of C() st c in dom f holds f.c = F(c) and
A3: dom g=X() and
A4: for c be Element of C() st c in dom g holds g.c = F(c);
  now
    let c be Element of C();
    assume
A5: c in dom f;
    hence f.c = F(c) by A2
      .= g.c by A1,A3,A4,A5;
  end;
  hence thesis by A1,A3,PARTFUN1:5;
end;

theorem Th7:
  seq = seq1 + seq2 iff for n holds seq.n =seq1.n + seq2.n
proof
  thus seq = seq1 + seq2 implies for n holds seq.n =seq1.n + seq2.n
  proof
    assume
A1: seq = seq1 + seq2;
    let n;
    dom seq = NAT by FUNCT_2:def 1;
    hence thesis by A1,VALUED_1:def 1;
  end;
  assume for n holds seq.n =seq1.n + seq2.n;
  then
A2: for n being set st n in dom seq holds seq.n = seq1.n + seq2.n;
  dom seq = NAT /\ NAT by FUNCT_2:def 1
    .= NAT /\ dom seq2 by FUNCT_2:def 1
    .= dom seq1 /\ dom seq2 by FUNCT_2:def 1;
  hence thesis by A2,VALUED_1:def 1;
end;

theorem Th8:
  seq = seq1 (#) seq2 iff for n holds seq.n =seq1.n * seq2.n
proof
  thus seq = seq1 (#) seq2 implies for n holds seq.n =seq1.n * seq2.n
  proof
    assume
A1: seq = seq1 (#) seq2;
    let n;
    dom seq = NAT by FUNCT_2:def 1;
    hence thesis by A1,VALUED_1:def 4;
  end;
  assume for n holds seq.n =seq1.n * seq2.n;
  then
A2: for n being set st n in dom seq holds seq.n = seq1.n * seq2.n;
  dom seq = NAT /\ NAT by FUNCT_2:def 1
    .= NAT /\ dom seq2 by FUNCT_2:def 1
    .= dom seq1 /\ dom seq2 by FUNCT_2:def 1;
  hence thesis by A2,VALUED_1:def 4;
end;

theorem Th9:
  seq1 = r(#)seq2 iff for n holds seq1.n=r*seq2.n
proof
  thus seq1 = r(#)seq2 implies for n holds seq1.n=r*seq2.n by VALUED_1:6;
  assume for n holds seq1.n=r*seq2.n;
  then
A1: for n being set st n in dom seq1 holds seq1.n = r * seq2.n;
  dom seq1 = NAT by FUNCT_2:def 1
    .= dom seq2 by FUNCT_2:def 1;
  hence thesis by A1,VALUED_1:def 5;
end;

theorem
  seq1 = -seq2 iff for n holds seq1.n= -seq2.n
proof
  thus seq1 = -seq2 implies for n holds seq1.n=-seq2.n by VALUED_1:8;
  assume for n holds seq1.n= -seq2.n;
  then
A1: for n being set st n in dom seq1 holds seq1.n = - seq2.n;
  dom seq1 = NAT by FUNCT_2:def 1
    .= dom seq2 by FUNCT_2:def 1;
  hence thesis by A1,VALUED_1:9;
end;

theorem
  seq1 - seq2 = seq1 +- seq2;

theorem Th12:
  seq1 = abs seq iff for n holds seq1.n= abs(seq.n)
proof
  thus seq1 = abs seq implies for n holds seq1.n=abs(seq.n) by VALUED_1:18;
  assume for n holds seq1.n= abs(seq.n);
  then
A1: for n being set st n in dom seq1 holds seq1.n = abs(seq.n);
  dom seq1 = NAT by FUNCT_2:def 1
    .= dom seq by FUNCT_2:def 1;
  hence thesis by A1,VALUED_1:def 11;
end;

theorem Th13:
  (seq1 + seq2) + seq3 = seq1 + (seq2 + seq3)
proof
  now
    let n;
    thus ((seq1+seq2)+seq3).n=(seq1+seq2).n+ seq3.n by Th7
      .=seq1.n+seq2.n+seq3.n by Th7
      .=seq1.n+(seq2.n+seq3.n)
      .=seq1.n+(seq2+seq3).n by Th7
      .=(seq1+(seq2+seq3)).n by Th7;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem Th14:
  (seq1 (#) seq2) (#) seq3 = seq1 (#) (seq2 (#) seq3)
proof
  now
    let n;
    thus ((seq1(#)seq2)(#)seq3).n=(seq1(#)seq2).n*seq3.n by Th8
      .=seq1.n*seq2.n*seq3.n by Th8
      .=seq1.n*(seq2.n*seq3.n)
      .=seq1.n*(seq2(#)seq3).n by Th8
      .=(seq1(#)(seq2(#)seq3)).n by Th8;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem Th15:
  (seq1 + seq2) (#) seq3 = seq1 (#) seq3 + seq2 (#) seq3
proof
  now
    let n;
    thus ((seq1+seq2)(#)seq3).n=(seq1+seq2).n*seq3.n by Th8
      .=(seq1.n+seq2.n)*seq3.n by Th7
      .=seq1.n*seq3.n+seq2.n*seq3.n
      .=(seq1(#)seq3).n+seq2.n*seq3.n by Th8
      .=(seq1(#)seq3).n+(seq2(#)seq3).n by Th8
      .=((seq1(#)seq3)+(seq2(#)seq3)).n by Th7;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  seq3 (#) (seq1 + seq2) = seq3 (#) seq1 + seq3 (#) seq2 by Th15;

theorem
  -seq = (-1) (#) seq;

theorem Th18:
  r(#)(seq1(#)seq2)=r(#)seq1(#)seq2
proof
  now
    let n;
    thus (r(#)(seq1(#)seq2)).n=r*(seq1(#)seq2).n by Th9
      .=r*(seq1.n*seq2.n) by Th8
      .=(r*seq1.n)*seq2.n
      .=(r(#)seq1).n*seq2.n by Th9
      .=(r(#)seq1 (#) seq2).n by Th8;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem Th19:
  r (#) (seq1 (#) seq2) = seq1 (#) (r (#) seq2)
proof
  now
    let n;
    thus (r(#)(seq1(#)seq2)).n=r*(seq1(#)seq2).n by Th9
      .=r*(seq1.n*seq2.n) by Th8
      .=seq1.n*(r*seq2.n)
      .=seq1.n*(r(#)seq2).n by Th9
      .=(seq1(#)(r(#)seq2)).n by Th8;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem Th20:
  (seq1 - seq2) (#) seq3 = seq1 (#) seq3 - seq2 (#) seq3
proof
  thus (seq1-seq2)(#)seq3=seq1(#)seq3+(-seq2)(#)seq3 by Th15
    .=seq1(#)seq3-seq2(#)seq3 by Th18;
end;

theorem
  seq3(#)seq1-seq3(#)seq2=seq3(#)(seq1-seq2) by Th20;

theorem Th22:
  r(#)(seq1+seq2)=r(#)seq1+r(#)seq2
proof
  now
    let n;
    thus (r(#)(seq1 + seq2)).n=r*(seq1+seq2).n by Th9
      .=r*(seq1.n+seq2.n) by Th7
      .=r*seq1.n+r*seq2.n
      .=(r(#)seq1).n+r*seq2.n by Th9
      .=(r(#)seq1).n+(r(#)seq2).n by Th9
      .=((r(#)seq1)+(r(#)seq2)).n by Th7;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem Th23:
  (r*p)(#)seq=r(#)(p(#)seq)
proof
  now
    let n;
    thus ((r*p)(#)seq).n=(r*p)*seq.n by Th9
      .=r*(p*seq.n)
      .=r*(p(#)seq).n by Th9
      .=(r(#)(p(#)seq)).n by Th9;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem Th24:
  r(#)(seq1-seq2)=r(#)seq1-r(#)seq2
proof
  thus r(#)(seq1-seq2)=r(#)seq1+r(#)(-seq2) by Th22
    .=r(#)seq1+((-1)*r)(#)seq2 by Th23
    .=r(#)seq1-(r(#)seq2) by Th23;
end;

theorem
  r(#)(seq1/"seq)=(r(#)seq1)/"seq
proof
  thus r(#)(seq1/"seq)= r(#)(seq1(#)(seq")) .= (r(#)seq1)/"seq by Th18;
end;

theorem
  seq1-(seq2+seq3)=seq1-seq2-seq3
proof
  thus seq1-(seq2+seq3)=seq1+(-seq2+(-1)(#)seq3) by Th22
    .=seq1-seq2-seq3 by Th13;
end;

theorem
  1(#)seq=seq
proof
  now
    let n;
    thus (1(#)seq).n=1*seq.n by Th9
      .=seq.n;
  end;
  hence thesis by FUNCT_2:63;
end;

canceled;

theorem
  seq1 - (-seq2) = seq1 + seq2;

theorem
  seq1-(seq2-seq3)=seq1-seq2+seq3
proof
  thus seq1-(seq2-seq3)=seq1+(-seq2-(-seq3)) by Th24
    .=seq1-seq2+seq3 by Th13;
end;

theorem
  seq1+(seq2-seq3)=seq1+seq2-seq3
proof
  thus seq1+(seq2-seq3)=seq1+seq2+-seq3 by Th13
    .=seq1+seq2-seq3;
end;

theorem
  (-seq1)(#)seq2=-(seq1(#)seq2) & seq1(#)(-seq2)=-(seq1(#)seq2) by Th18;

theorem Th33:
  seq is non-zero implies seq" is non-zero
proof
  assume that
A1: seq is non-zero and
A2: not seq" is non-zero;
  consider n1 such that
A3: (seq").n1=0 by A2,Th5;
  (seq.n1)"=(seq").n1 by VALUED_1:10;
  hence contradiction by A1,A3,Th5,XCMPLX_1:202;
end;

canceled;

theorem Th35:
  seq is non-zero & seq1 is non-zero iff seq(#)seq1 is non-zero
proof
  thus seq is non-zero & seq1 is non-zero implies seq(#)seq1 is non-zero
  proof
    assume
A1: seq is non-zero & seq1 is non-zero;
    now
      let n;
A2:   (seq(#)seq1).n=(seq.n)*(seq1.n) by Th8;
      seq.n<>0 & seq1.n<>0 by A1,Th5;
      hence (seq(#)seq1).n<>0 by A2,XCMPLX_1:6;
    end;
    hence thesis by Th5;
  end;
  assume
A3: seq(#)seq1 is non-zero;
  now
    let n;
    (seq(#)seq1).n=(seq.n)*(seq1.n) by Th8;
    hence seq.n<>0 by A3,Th5;
  end;
  hence seq is non-zero by Th5;
  now
    let n;
    (seq(#)seq1).n=(seq.n)*(seq1.n) by Th8;
    hence seq1.n<>0 by A3,Th5;
  end;
  hence thesis by Th5;
end;

theorem Th36:
  seq"(#)seq1"=(seq(#)seq1)"
proof
  now
    let n;
    thus ((seq")(#)(seq1")).n=((seq").n)*((seq1").n) by Th8
      .=((seq").n)*(seq1.n)" by VALUED_1:10
      .=(seq.n)"*(seq1.n)" by VALUED_1:10
      .=((seq.n)*(seq1.n))" by XCMPLX_1:204
      .=((seq(#)seq1).n)" by Th8
      .=((seq(#)seq1)").n by VALUED_1:10;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  seq is non-zero implies (seq1/"seq)(#)seq=seq1
proof
  assume
A1: seq is non-zero;
  now
    let n;
A2: seq.n<>0 by A1,Th5;
    thus ((seq1/"seq)(#)seq).n=((seq1(#)seq").n)*seq.n by Th8
      .=(seq1.n)*(seq".n)*seq.n by Th8
      .=(seq1.n)*(seq.n)"*seq.n by VALUED_1:10
      .=(seq1.n)*((seq.n)"*seq.n)
      .=(seq1.n)*1 by A2,XCMPLX_0:def 7
      .=seq1.n;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  (seq9/"seq)(#)(seq19/"seq1)=(seq9(#)seq19)/"(seq(#)seq1)
proof
  now
    let n;
    thus ((seq9/"seq)(#)(seq19/"seq1)).n=((seq9(#)seq").n)*(seq19/"seq1).n by
Th8
      .=(seq9.n)*(seq".n)*(seq19(#)seq1").n by Th8
      .=(seq9.n)*(seq".n)*((seq19.n)*seq1".n) by Th8
      .=(seq9.n)*((seq19.n)*((seq".n)*seq1".n))
      .=(seq9.n)*((seq19.n)*((seq"(#)seq1").n)) by Th8
      .=(seq9.n)*(seq19.n)*((seq"(#)seq1").n)
      .=(seq9.n)*(seq19.n)*((seq(#)seq1)".n) by Th36
      .=((seq9(#)seq19).n)*(seq(#)seq1)".n by Th8
      .=((seq9(#)seq19)/"(seq(#)seq1)).n by Th8;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  seq is non-zero & seq1 is non-zero implies seq/"seq1 is non-zero
proof
  assume that
A1: seq is non-zero and
A2: seq1 is non-zero;
  seq1" is non-zero by A2,Th33;
  hence thesis by A1,Th35;
end;

theorem Th40:
  (seq/"seq1)"=seq1/"seq
proof
  now
    let n;
    thus (seq/"seq1)".n=(seq"(#)seq1"").n by Th36
      .=(seq1/"seq).n;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  seq2(#)(seq1/"seq)=(seq2(#)seq1)/"seq
proof
  thus seq2(#)(seq1/"seq) = seq2(#)(seq1(#)(seq"))
    .=(seq2(#)seq1)/"seq by Th14;
end;

theorem
  seq2/"(seq/"seq1)=(seq2(#)seq1)/"seq
proof
  thus seq2/"(seq/"seq1) = seq2(#)(seq1/"seq) by Th40
    .= seq2(#)(seq1(#)(seq"))
    .= (seq2(#)seq1)/"seq by Th14;
end;

theorem Th43:
  seq1 is non-zero implies seq2/"seq=(seq2(#)seq1)/"(seq(#)seq1)
proof
  assume
A1: seq1 is non-zero;
  now
    let n;
A2: seq1.n<>0 by A1,Th5;
    thus (seq2/"seq).n=(seq2.n)*1*seq".n by Th8
      .=(seq2.n)*((seq1.n)*(seq1.n)")*seq".n by A2,XCMPLX_0:def 7
      .=(seq2.n)*(seq1.n)*((seq1.n)"*seq".n)
      .=((seq2(#)seq1).n)*((seq1.n)"*seq".n) by Th8
      .=((seq2(#)seq1).n)*((seq1".n)*seq".n) by VALUED_1:10
      .=((seq2(#)seq1).n)*(seq"(#)seq1").n by Th8
      .=((seq2(#)seq1).n)*(seq(#)seq1)".n by Th36
      .=((seq2(#)seq1)/"(seq(#)seq1)).n by Th8;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem Th44:
  r<>0 & seq is non-zero implies r(#)seq is non-zero
proof
  assume that
A1: r<>0 and
A2: seq is non-zero and
A3: not r(#)seq is non-zero;
  consider n1 such that
A4: (r(#)seq).n1=0 by A3,Th5;
A5: seq.n1 <> 0 by A2,Th5;
  r*seq.n1=0 by A4,Th9;
  hence contradiction by A1,A5,XCMPLX_1:6;
end;

theorem
  seq is non-zero implies -seq is non-zero by Th44;

theorem Th46:
  (r(#)seq)"=r"(#)seq"
proof
  now
    let n;
    thus (r(#)seq)".n=((r(#)seq).n)" by VALUED_1:10
      .=(r*(seq.n))" by Th9
      .=r"*(seq.n)" by XCMPLX_1:204
      .=r"*seq".n by VALUED_1:10
      .=(r"(#)seq").n by Th9;
  end;
  hence thesis by FUNCT_2:63;
end;

Lm1: (-1)"=-1;

theorem
  (-seq)" = (-1)(#)seq" by Lm1,Th46;

theorem
  -seq1/"seq=(-seq1)/"seq & seq1/"(-seq)=-seq1/"seq
proof
  thus -seq1/"seq=(-1)(#)(seq1(#)(seq")) .=(-seq1)/"seq by Th18;
  thus seq1/"(-seq)=seq1(#)((-1)(#)seq") by Lm1,Th46
    .=-(seq1/"seq) by Th19;
end;

theorem
  seq1/"seq + seq19/"seq = (seq1 + seq19) /" seq & seq1/"seq - seq19/"
  seq = (seq1 - seq19) /" seq
proof
  thus seq1/"seq + seq19/"seq = (seq1 + seq19) (#) (seq") by Th15
    .= (seq1 + seq19) /" seq;
  thus seq1/"seq - seq19/"seq = (seq1 - seq19) (#) (seq") by Th20
    .= (seq1 - seq19) /" seq;
end;

theorem
  seq is non-zero & seq9 is non-zero implies seq1/"seq + seq19/"seq9=(
seq1(#)seq9+seq19(#)seq)/"(seq(#)seq9) & seq1/"seq - seq19/"seq9=(seq1(#)seq9-
  seq19(#)seq)/"(seq(#)seq9)
proof
  assume that
A1: seq is non-zero and
A2: seq9 is non-zero;
  thus seq1/"seq + seq19/"seq9=(seq1(#)seq9)/"(seq(#)seq9)+seq19/"seq9 by A2
,Th43
    .=(seq1(#)seq9)/"(seq(#)seq9)+(seq19(#)seq)/"(seq(#)seq9) by A1,Th43
    .=(seq1(#)seq9+seq19(#)seq)(#)((seq(#)seq9)") by Th15
    .=(seq1(#)seq9+seq19(#)seq)/"(seq(#)seq9);
  thus seq1/"seq - seq19/"seq9=(seq1(#)seq9)/"(seq(#)seq9)-seq19/"seq9 by A2
,Th43
    .=(seq1(#)seq9)/"(seq(#)seq9)-(seq19(#)seq)/"(seq(#)seq9) by A1,Th43
    .=(seq1(#)seq9-seq19(#)seq)(#)((seq(#)seq9)") by Th20
    .=(seq1(#)seq9-seq19(#)seq)/"(seq(#)seq9);
end;

theorem
  (seq19/"seq)/"(seq9/"seq1)=(seq19(#)seq1)/"(seq(#)seq9)
proof
  thus (seq19/"seq)/"(seq9/"seq1)=(seq19/"seq)(#)(seq9"(#)seq1"") by Th36
    .=seq19(#)seq"(#)seq1(#)seq9" by Th14
    .=seq19(#)(seq1(#)seq")(#)seq9" by Th14
    .=seq19(#)((seq1(#)seq")(#)seq9") by Th14
    .=seq19(#)(seq1(#)(seq"(#)seq9")) by Th14
    .=seq19(#)seq1(#)(seq"(#)seq9") by Th14
    .=(seq19(#)seq1)/"(seq(#)seq9) by Th36;
end;

theorem Th52:
  abs(seq(#)seq9)=abs(seq)(#)abs(seq9)
proof
  now
    let n;
    thus (abs(seq(#)seq9)).n=abs(((seq(#)seq9).n)) by Th12
      .=abs((seq.n)*(seq9.n)) by Th8
      .=abs((seq.n))*abs((seq9.n)) by COMPLEX1:65
      .=((abs(seq)).n)*abs((seq9.n)) by Th12
      .=((abs(seq)).n)*(abs(seq9)).n by Th12
      .=(abs(seq)(#)abs(seq9)).n by Th8;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  seq is non-zero implies abs(seq) is non-zero
proof
  assume
A1: seq is non-zero;
  now
    let n;
    seq.n<>0 by A1,Th5;
    then abs(seq.n)<>0 by COMPLEX1:47;
    hence (abs(seq)).n<>0 by Th12;
  end;
  hence thesis by Th5;
end;

theorem Th54:
  abs(seq)"=abs(seq")
proof
  now
    let n;
    thus (abs(seq")).n=abs(seq".n) by Th12
      .=abs((seq.n)") by VALUED_1:10
      .=abs(1/(seq.n)) by XCMPLX_1:215
      .=1/abs(seq.n) by ABSVALUE:7
      .=(abs(seq.n))" by XCMPLX_1:215
      .=(abs(seq).n)" by Th12
      .=(abs(seq))".n by VALUED_1:10;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  abs(seq9/"seq)=abs(seq9)/"abs(seq)
proof
  thus abs(seq9/"seq)=abs(seq9)(#)abs(seq") by Th52
    .=abs(seq9)/"abs(seq) by Th54;
end;

theorem
  abs(r(#)seq)=abs(r)(#)abs(seq)
proof
  now
    let n;
    thus abs(r(#)seq).n=abs((r(#)seq).n) by Th12
      .=abs(r*(seq.n)) by Th9
      .=abs(r)*abs(seq.n) by COMPLEX1:65
      .=abs(r)*(abs(seq)).n by Th12
      .=(abs(r)(#)abs(seq)).n by Th9;
  end;
  hence thesis by FUNCT_2:63;
end;














begin

reserve f for Function;
reserve n,k,n1 for Element of NAT;
reserve r,p for Element of COMPLEX;
reserve x,y for set;

definition
  mode Complex_Sequence is sequence of COMPLEX;
end;

reserve seq,seq1,seq2,seq3,seq9,seq19 for Complex_Sequence;

theorem Th1:
  f is Complex_Sequence iff (dom f=NAT & for x st x in NAT holds f.
  x is Element of COMPLEX)
proof
  thus f is Complex_Sequence implies (dom f=NAT & for x st x in NAT holds f.x
  is Element of COMPLEX)
  proof
    assume
A1: f is Complex_Sequence;
    hence
A2: dom f=NAT by FUNCT_2:def 1;
    let x;
    assume x in NAT;
    then
A3: f.x in rng f by A2,FUNCT_1:def 3;
    rng f c=COMPLEX by A1,RELAT_1:def 19;
    hence thesis by A3;
  end;
  assume that
A4: dom f= NAT and
A5: for x st x in NAT holds f.x is Element of COMPLEX;
  now
    let y;
    assume y in rng f;
    then consider x such that
A6: x in dom f and
A7: y=f.x by FUNCT_1:def 3;
    f.x is Element of COMPLEX by A4,A5,A6;
    hence y in COMPLEX by A7;
  end;
  then rng f c=COMPLEX by TARSKI:def 3;
  hence thesis by A4,FUNCT_2:def 1,RELSET_1:4;
end;

theorem Th2:
  f is Complex_Sequence iff (dom f=NAT & for n holds f.n is Element of COMPLEX)
proof
  thus f is Complex_Sequence implies (dom f=NAT & for n holds f.n is Element
  of COMPLEX) by Th1;
  assume that
A1: dom f=NAT and
A2: for n holds f.n is Element of COMPLEX;
  for x holds x in NAT implies f.x is Element of COMPLEX by A2;
  hence thesis by A1,Th1;
end;

scheme
  ExComplexSeq{F(set) -> complex number}: ex seq st for n holds seq.n=F(n)
proof
  defpred P[set,set] means ex n st n=$1 & $2=F(n);
A1: now
    let x;
    assume x in NAT;
    then consider n such that
A2: n=x;
    reconsider r2=F(n) as set;
    take y=r2;
    thus P[x,y] by A2;
  end;
  consider f such that
A3: dom f=NAT and
A4: for x st x in NAT holds P[x,f.x] from CLASSES1:sch 1(A1);
  now
    let x;
    assume x in NAT;
    then ex n st n=x & f.x=F(n) by A4;
    hence f.x is Element of COMPLEX by XCMPLX_0:def 2;
  end;
  then reconsider f as Complex_Sequence by A3,Th1;
  take seq=f;
  let n;
  ex k st k=n & seq.n=F(k) by A4;
  hence thesis;
end;

notation
  let f be complex-valued Relation;
  synonym f is non-zero for f is non-empty;
end;

registration
  cluster non-zero for Complex_Sequence;
  existence
  proof
    take s = NAT --> 1r;
    rng s = {1} by FUNCOP_1:8;
    hence not 0 in rng s by TARSKI:def 1;
  end;
end;

theorem Th3:
  seq is non-zero iff for x st x in NAT holds seq.x<>0c
proof
  thus seq is non-zero implies for x st x in NAT holds seq.x<>0c
  proof
    assume
A1: seq is non-zero;
    let x;
    assume x in NAT;
    then x in dom seq by Th2;
    then seq.x in rng seq by FUNCT_1:def 3;
    hence thesis by A1,RELAT_1:def 9;
  end;
  assume
A2: for x st x in NAT holds seq.x<>0c;
  assume 0 in rng seq;
  then ex x st x in dom seq & seq.x=0 by FUNCT_1:def 3;
  hence contradiction by A2;
end;

theorem Th4:
  seq is non-zero iff for n holds seq.n<>0c
proof
  thus seq is non-zero implies for n holds seq.n<>0c by Th3;
  assume for n holds seq.n<>0c;
  then for x holds x in NAT implies seq.x<>0c;
  hence thesis by Th3;
end;

theorem
  for IT being non-zero Complex_Sequence holds rng IT c= COMPLEX \ {0c}
proof
  let IT be non-zero Complex_Sequence;
  not 0c in rng IT by RELAT_1:def 9;
  hence thesis by ZFMISC_1:34;
end;

theorem
  for r ex seq st rng seq={r}
proof
  let r;
  consider f such that
A1: dom f=NAT and
A2: rng f={r} by FUNCT_1:5;
  now
    let x;
    assume x in {r};
    then x=r by TARSKI:def 1;
    hence x in COMPLEX;
  end;
  then rng f c= COMPLEX by A2,TARSKI:def 3;
  then reconsider f as Complex_Sequence by A1,FUNCT_2:def 1,RELSET_1:4;
  take f;
  thus thesis by A2;
end;

theorem Th7:
  (seq1+seq2)+seq3=seq1+(seq2+seq3)
proof
  now
    let n;
    thus ((seq1+seq2)+seq3).n=(seq1+seq2 qua Complex_Sequence).n+ seq3.n by
VALUED_1:1
      .=seq1.n+seq2.n+seq3.n by VALUED_1:1
      .=seq1.n+(seq2.n+seq3.n)
      .=seq1.n+(seq2+seq3).n by VALUED_1:1
      .=(seq1+(seq2+seq3)).n by VALUED_1:1;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem Th8:
  (seq1(#)seq2)(#)seq3=seq1(#)(seq2(#)seq3)
proof
  now
    let n;
    thus ((seq1(#)seq2)(#)seq3).n=(seq1(#)seq2).n*seq3.n by VALUED_1:5
      .=seq1.n*seq2.n*seq3.n by VALUED_1:5
      .=seq1.n*(seq2.n*seq3.n)
      .=seq1.n*(seq2(#)seq3).n by VALUED_1:5
      .=(seq1(#)(seq2(#)seq3)).n by VALUED_1:5;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem Th9:
  (seq1+seq2)(#)seq3=seq1(#)seq3+seq2(#)seq3
proof
  now
    let n;
    thus ((seq1+seq2)(#)seq3).n=(seq1+seq2).n*seq3.n by VALUED_1:5
      .=(seq1.n+seq2.n)*seq3.n by VALUED_1:1
      .=seq1.n*seq3.n+seq2.n*seq3.n
      .=(seq1(#)seq3).n+seq2.n*seq3.n by VALUED_1:5
      .=(seq1(#)seq3).n+(seq2(#)seq3).n by VALUED_1:5
      .=((seq1(#)seq3)+(seq2(#)seq3)).n by VALUED_1:1;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  seq3(#)(seq1+seq2)=seq3(#)seq1+seq3(#)seq2 by Th9;

theorem
  -seq=(-1r)(#)seq;

theorem Th12:
  r(#)(seq1(#)seq2)=r(#)seq1(#)seq2
proof
  now
    let n;
    thus (r(#)(seq1(#)seq2)).n=r*(seq1(#)seq2).n by VALUED_1:6
      .=r*(seq1.n*seq2.n) by VALUED_1:5
      .=(r*seq1.n)*seq2.n
      .=(r(#)seq1).n*seq2.n by VALUED_1:6
      .=(r(#)seq1 (#) seq2).n by VALUED_1:5;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem Th13:
  r(#)(seq1(#)seq2)=seq1(#)(r(#)seq2)
proof
  now
    let n;
    thus (r(#)(seq1(#)seq2)).n=r*(seq1(#)seq2).n by VALUED_1:6
      .=r*(seq1.n*seq2.n) by VALUED_1:5
      .=seq1.n*(r*seq2.n)
      .=seq1.n*(r(#)seq2).n by VALUED_1:6
      .=(seq1(#)(r(#)seq2)).n by VALUED_1:5;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem Th14:
  (seq1-seq2)(#)seq3=seq1(#)seq3-seq2(#)seq3
proof
  thus (seq1-seq2)(#)seq3=seq1(#)seq3+(-seq2)(#)seq3 by Th9
    .=seq1(#)seq3+((-1r)(#)seq2)(#)seq3
    .=seq1(#)seq3+(-1r)(#)(seq2(#)seq3) by Th12
    .=seq1(#)seq3-seq2(#)seq3;
end;

theorem
  seq3(#)seq1-seq3(#)seq2=seq3(#)(seq1-seq2) by Th14;

theorem Th16:
  r(#)(seq1+seq2)=r(#)seq1+r(#)seq2
proof
  now
    let n;
    thus (r(#)(seq1 + seq2)).n=r*(seq1+seq2).n by VALUED_1:6
      .=r*(seq1.n+seq2.n) by VALUED_1:1
      .=r*seq1.n+r*seq2.n
      .=(r(#)seq1).n+r*seq2.n by VALUED_1:6
      .=(r(#)seq1).n+(r(#)seq2).n by VALUED_1:6
      .=((r(#)seq1)+(r(#)seq2)).n by VALUED_1:1;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem Th17:
  (r*p)(#)seq=r(#)(p(#)seq)
proof
  now
    let n;
    thus ((r*p)(#)seq).n=(r*p)*seq.n by VALUED_1:6
      .=r*(p*seq.n)
      .=r*(p(#)seq).n by VALUED_1:6
      .=(r(#)(p(#)seq)).n by VALUED_1:6;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem Th18:
  r(#)(seq1-seq2)=r(#)seq1-r(#)seq2
proof
  thus r(#)(seq1-seq2)=r(#)seq1+r(#)(-seq2) by Th16
    .=r(#)seq1+r(#)((-1r)(#)seq2)
    .=r(#)seq1+((-1r)*r)(#)seq2 by Th17
    .=r(#)seq1+(-1r)(#)(r(#)seq2) by Th17
    .=r(#)seq1-(r(#)seq2);
end;

theorem
  r(#)(seq1/"seq)=(r(#)seq1)/"seq
proof
  thus r(#)(seq1/"seq) = r(#)(seq1(#)(seq)") .= (r(#)seq1)/"seq by Th12;
end;

theorem
  seq1-(seq2+seq3)=seq1-seq2-seq3
proof
  thus seq1-(seq2+seq3)=seq1+(-1r)(#)(seq2+seq3)
    .=seq1+((-1r)(#)seq2+(-1r)(#)seq3) by Th16
    .=seq1+(-seq2+(-1r)(#)seq3)
    .=seq1+(-seq2+-seq3)
    .=seq1-seq2-seq3 by Th7;
end;

theorem
  1r(#)seq=seq
proof
  now
    let n;
    thus (1r(#)seq).n=1r*seq.n by VALUED_1:6
      .=seq.n;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  --seq = seq;

theorem
  seq1 - (-seq2) = seq1 + seq2;

theorem
  seq1 - (seq2 - seq3) = seq1 - seq2 + seq3
proof
  thus seq1-(seq2-seq3)=seq1+(-1r)(#)(seq2-seq3)
    .=seq1+((-1r)(#)seq2-((-1r)(#)seq3)) by Th18
    .=seq1+(-seq2-((-1r)(#)seq3))
    .=seq1+(-seq2-(-seq3))
    .=seq1-seq2+seq3 by Th7;
end;

theorem
  seq1 + (seq2 - seq3) = seq1 + seq2 - seq3
proof
  thus seq1 + (seq2-seq3) = seq1 + (seq2+-seq3) .= seq1+seq2-seq3 by Th7;
end;

theorem
  (-seq1)(#)seq2=-(seq1(#)seq2) & seq1(#)(-seq2)=-(seq1(#)seq2)
proof
  thus (-seq1)(#)seq2=(-1r)(#)seq1(#)seq2 .=(-1r)(#)(seq1(#)seq2) by Th12
    .=-(seq1(#)seq2);
  thus seq1(#)(-seq2)=seq1(#)((-1r)(#)seq2) .=(-1r)(#)(seq1(#)seq2) by Th13
    .=-(seq1(#)seq2);
end;

theorem Th27:
  seq is non-zero implies seq" is non-zero
proof
  assume that
A1: seq is non-zero and
A2: not seq" is non-zero;
  consider n1 such that
A3: (seq").n1=0c by A2,Th4;
  (seq.n1)"=(seq").n1 by VALUED_1:10;
  hence contradiction by A1,A3,Th4,XCMPLX_1:202;
end;

canceled;

theorem Th29:
  seq is non-zero & seq1 is non-zero iff seq(#)seq1 is non-zero
proof
  thus seq is non-zero & seq1 is non-zero implies seq(#)seq1 is non-zero
  proof
    assume
A1: seq is non-zero & seq1 is non-zero;
    now
      let n;
A2:   (seq(#)seq1).n=(seq.n)*(seq1.n) by VALUED_1:5;
      seq.n<>0c & seq1.n<>0c by A1,Th4;
      hence (seq(#)seq1).n<>0c by A2;
    end;
    hence thesis by Th4;
  end;
  assume
A3: seq(#)seq1 is non-zero;
  now
    let n;
    (seq(#)seq1).n=(seq.n)*(seq1.n) by VALUED_1:5;
    hence seq.n<>0c by A3,Th4;
  end;
  hence seq is non-zero by Th4;
  now
    let n;
    (seq(#)seq1).n=(seq.n)*(seq1.n) by VALUED_1:5;
    hence seq1.n<>0c by A3,Th4;
  end;
  hence thesis by Th4;
end;

theorem Th30:
  seq"(#)seq1"=(seq(#)seq1)"
proof
  now
    let n;
    thus ((seq")(#)(seq1")).n=((seq").n)*((seq1").n) by VALUED_1:5
      .=((seq").n)*(seq1.n)" by VALUED_1:10
      .=(seq.n)"*(seq1.n)" by VALUED_1:10
      .=((seq.n)*(seq1.n))" by XCMPLX_1:204
      .=((seq(#)seq1).n)" by VALUED_1:5
      .=((seq(#)seq1)").n by VALUED_1:10;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  seq is non-zero implies (seq1/"seq)(#)seq=seq1
proof
  assume
A1: seq is non-zero;
  now
    let n;
A2: seq.n<>0c by A1,Th4;
    thus ((seq1/"seq)(#)seq).n=((seq1(#)seq").n)*seq.n by VALUED_1:5
      .=(seq1.n)*(seq".n)*seq.n by VALUED_1:5
      .=(seq1.n)*(seq.n)"*seq.n by VALUED_1:10
      .=(seq1.n)*((seq.n)"*seq.n)
      .=(seq1.n)*1r by A2,XCMPLX_0:def 7
      .=seq1.n;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  (seq9/"seq)(#)(seq19/"seq1)=(seq9(#)seq19)/"(seq(#)seq1)
proof
  now
    let n;
    thus ((seq9/"seq)(#)(seq19/"seq1)).n=((seq9(#)seq").n)*(seq19/"seq1).n by
VALUED_1:5
      .=(seq9.n)*(seq".n)*(seq19(#)seq1").n by VALUED_1:5
      .=(seq9.n)*(seq".n)*((seq19.n)*seq1".n) by VALUED_1:5
      .=(seq9.n)*((seq19.n)*((seq".n)*seq1".n))
      .=(seq9.n)*((seq19.n)*((seq"(#)seq1").n)) by VALUED_1:5
      .=(seq9.n)*(seq19.n)*((seq"(#)seq1").n)
      .=(seq9.n)*(seq19.n)*((seq(#)seq1)".n) by Th30
      .=((seq9(#)seq19).n)*(seq(#)seq1)".n by VALUED_1:5
      .=((seq9(#)seq19)/"(seq(#)seq1)).n by VALUED_1:5;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  seq is non-zero & seq1 is non-zero implies seq/"seq1 is non-zero
proof
  assume that
A1: seq is non-zero and
A2: seq1 is non-zero;
  seq1" is non-zero by A2,Th27;
  hence thesis by A1,Th29;
end;

theorem Th34:
  (seq/"seq1)"=seq1/"seq
proof
  now
    let n;
    thus (seq/"seq1)".n=(seq"(#)seq1"").n by Th30
      .=(seq1/"seq).n;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  seq2 (#) (seq1 /" seq) = (seq2 (#) seq1) /" seq
proof
  thus seq2 (#) (seq1 /" seq) = seq2 (#) seq1 (#) (seq") by Th8
    .= (seq2 (#) seq1) /" seq;
end;

theorem
  seq2/"(seq/"seq1)=(seq2(#)seq1)/"seq
proof
  now
    let n;
    thus (seq2/"(seq/"seq1)).n=((seq2(#)(seq1/"seq))).n by Th34
      .=((seq2(#)seq1)(#)(seq")).n by Th8
      .=((seq2(#)seq1)/"seq).n;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem Th37:
  seq1 is non-zero implies seq2/"seq=(seq2(#)seq1)/"(seq(#)seq1)
proof
  assume
A1: seq1 is non-zero;
  now
    let n;
A2: seq1.n<>0c by A1,Th4;
    thus (seq2/"seq).n=(seq2.n)*1r*seq".n by VALUED_1:5
      .=(seq2.n)*((seq1.n)*(seq1.n)")*seq".n by A2,XCMPLX_0:def 7
      .=(seq2.n)*(seq1.n)*((seq1.n)"*seq".n)
      .=((seq2(#)seq1).n)*((seq1.n)"*seq".n) by VALUED_1:5
      .=((seq2(#)seq1).n)*((seq1".n)*seq".n) by VALUED_1:10
      .=((seq2(#)seq1).n)*(seq"(#)seq1").n by VALUED_1:5
      .=((seq2(#)seq1).n)*(seq(#)seq1)".n by Th30
      .=((seq2(#)seq1)/"(seq(#)seq1)).n by VALUED_1:5;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem Th38:
  r<>0c & seq is non-zero implies r(#)seq is non-zero
proof
  assume that
A1: r<>0c and
A2: seq is non-zero and
A3: not r(#)seq is non-zero;
  consider n1 such that
A4: (r(#)seq).n1=0c by A3,Th4;
A5: r*seq.n1=0c by A4,VALUED_1:6;
  seq.n1 <> 0c by A2,Th4;
  hence contradiction by A1,A5;
end;

theorem
  seq is non-zero implies -seq is non-zero
proof
A1: --1r=1r;
  assume seq is non-zero;
  hence thesis by A1,Th38;
end;

theorem Th40:
  (r(#)seq)"=r"(#)seq"
proof
  now
    let n;
    thus (r(#)seq)".n=((r(#)seq).n)" by VALUED_1:10
      .=(r*(seq.n))" by VALUED_1:6
      .=r"*(seq.n)" by XCMPLX_1:204
      .=r"*seq".n by VALUED_1:10
      .=(r"(#)seq").n by VALUED_1:6;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem Th41:
  seq is non-zero implies (-seq)"=(-1r)(#)seq"
proof
  (-1)" = -1;
  hence thesis by Th40;
end;

theorem
  seq is non-zero implies -seq1/"seq=(-seq1)/"seq & seq1/"(-seq)=-seq1/" seq
proof
  assume
A1: seq is non-zero;
  thus -seq1/"seq=(-1r)(#)(seq1/"seq) .=((-1r)(#)seq1)(#)(seq") by Th12
    .=(-seq1)/"seq;
  thus seq1/"(-seq)=seq1(#)((-1r)(#)seq") by A1,Th41
    .=(-1r)(#)(seq1(#)(seq")) by Th13
    .=-(seq1/"seq);
end;

theorem
  seq1/"seq + seq19/"seq = (seq1+seq19) /" seq & seq1/"seq - seq19/"seq
  = (seq1-seq19) /" seq
proof
  thus seq1/"seq + seq19/"seq = (seq1+seq19)(#)(seq") by Th9
    .= (seq1+seq19) /" seq;
  thus seq1/"seq - seq19/"seq = (seq1-seq19)(#)(seq") by Th14
    .= (seq1-seq19) /" seq;
end;

theorem
  seq is non-zero & seq9 is non-zero implies seq1/"seq + seq19/"seq9=(
seq1(#)seq9+seq19(#)seq)/"(seq(#)seq9) & seq1/"seq - seq19/"seq9=(seq1(#)seq9-
  seq19(#)seq)/"(seq(#)seq9)
proof
  assume that
A1: seq is non-zero and
A2: seq9 is non-zero;
  thus seq1/"seq + seq19/"seq9=(seq1(#)seq9)/"(seq(#)seq9)+seq19/"seq9 by A2
,Th37
    .=(seq1(#)seq9)/"(seq(#)seq9)+(seq19(#)seq)/"(seq(#)seq9) by A1,Th37
    .=(seq1(#)seq9+seq19(#)seq)(#)((seq(#)seq9)") by Th9
    .=(seq1(#)seq9+seq19(#)seq)/"(seq(#)seq9);
  thus seq1/"seq - seq19/"seq9=(seq1(#)seq9)/"(seq(#)seq9)-seq19/"seq9 by A2
,Th37
    .=(seq1(#)seq9)/"(seq(#)seq9)-(seq19(#)seq)/"(seq(#)seq9) by A1,Th37
    .=(seq1(#)seq9-seq19(#)seq)(#)((seq(#)seq9)") by Th14
    .=(seq1(#)seq9-seq19(#)seq)/"(seq(#)seq9);
end;

theorem
  (seq19/"seq)/"(seq9/"seq1)=(seq19(#)seq1)/"(seq(#)seq9)
proof
  thus (seq19/"seq)/"(seq9/"seq1)= (seq19/"seq)(#)(seq9"(#)seq1"") by Th30
    .=seq19(#)seq"(#)seq1(#)seq9" by Th8
    .=seq19(#)(seq1(#)seq")(#)seq9" by Th8
    .=seq19(#)((seq1(#)seq")(#)seq9") by Th8
    .=seq19(#)(seq1(#)(seq"(#)seq9")) by Th8
    .=seq19(#)seq1(#)(seq"(#)seq9") by Th8
    .=(seq19(#)seq1)/"(seq(#)seq9) by Th30;
end;

theorem Th46:
  |.seq(#)seq9.|=|.seq.|(#)|.seq9.|
proof
  now
    let n;
    thus (|.seq(#)seq9.|).n=|.((seq(#)seq9).n).| by VALUED_1:18
      .=|.(seq.n)*(seq9.n).| by VALUED_1:5
      .=|.(seq.n).|*|.(seq9.n).| by COMPLEX1:65
      .=((|.seq.|).n)*|.(seq9.n).| by VALUED_1:18
      .=((|.seq.|).n)*(|.seq9.|).n by VALUED_1:18
      .=(|.seq.|(#)|.seq9.|).n by VALUED_1:5;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  seq is non-zero implies |.seq.| is non-zero
proof
  assume
A1: seq is non-zero;
  now
    let n;
    seq.n<>0c by A1,Th4;
    then |.seq.n.|<>0 by COMPLEX1:45;
    hence (|.seq.|).n<>0 by VALUED_1:18;
  end;
  hence thesis by SEQ_1:5;
end;

theorem Th48:
  |. seq .| " = |. seq" .|
proof
  now
    let n;
    thus (|.seq".|).n=|.seq".n.| by VALUED_1:18
      .=|.(seq.n)".| by VALUED_1:10
      .=|.1r/(seq.n).| by XCMPLX_1:215
      .=1/|.seq.n.| by COMPLEX1:48,67
      .=|.seq.n.|" by XCMPLX_1:215
      .=(|.seq.|.n)" by VALUED_1:18
      .=(|.seq.|)".n by VALUED_1:10;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  |.seq9/"seq.|=|.seq9.|/"|.seq.|
proof
  thus |.seq9/"seq.|=|.seq9.|(#)|.seq".| by Th46
    .=|.seq9.|/"|.seq.| by Th48;
end;

theorem
  |.r(#)seq.|=|.r.|(#)|.seq.|
proof
  now
    let n;
    thus |.r(#)seq.|.n=|.(r(#)seq).n.| by VALUED_1:18
      .=|.r*(seq.n).| by VALUED_1:6
      .=|.r.|*|.seq.n.| by COMPLEX1:65
      .=|.r.|*(|.seq.|).n by VALUED_1:18
      .=(|.r.|(#)|.seq.|).n by VALUED_1:6;
  end;
  hence thesis by FUNCT_2:63;
end;














begin

definition
  func FALSE equals
  0;
  coherence;
  func TRUE equals
  1;
  coherence;
end;

definition
  let p be set;
  attr p is boolean means
  :Def3:
  p = FALSE or p = TRUE;
end;

registration
  cluster FALSE -> boolean;
  coherence by Def3;
  cluster TRUE -> boolean;
  coherence by Def3;
  cluster boolean for number;
  existence by Def3;
  cluster boolean -> natural for number;
  coherence
  proof
    let x be number;
    assume x = FALSE or x = TRUE;
    hence thesis;
  end;
end;

reserve p,q,r,s for boolean number;

definition
  let p;
  func 'not' p -> boolean number equals
  1 - p;
  coherence
  proof
    p = FALSE or p = TRUE by Def3;
    hence thesis by Def3;
  end;
  involutiveness;
  let q;
  func p '&' q equals
  p*q;
  correctness;
  commutativity;
  idempotence
  proof
    let p;
    p = FALSE or p = TRUE by Def3;
    hence thesis;
  end;
end;

registration
  let p,q;
  cluster p '&' q -> boolean;
  coherence
  proof
    p = FALSE or p = TRUE by Def3;
    hence thesis;
  end;
end;

definition
  let p,q;
  func p 'or' q equals
  'not' ('not' p '&' 'not' q);
  coherence;
  commutativity;
  idempotence;
end;

definition
  let p,q;
  func p => q equals
  'not' p 'or' q;
  coherence;
end;

registration
  let p,q;
  cluster p 'or' q -> boolean;
  coherence;
  cluster p => q -> boolean;
  coherence;
end;

definition
  let p,q;
  func p <=> q equals
  (p => q) '&' (q => p);
  coherence;
  commutativity;
end;

registration
  let p,q;
  cluster p <=> q -> boolean;
  coherence;
end;

definition
  let p,q;
  func p 'nand' q equals
  'not'(p '&' q);
  coherence;
  commutativity;
  func p 'nor' q equals
  'not'(p 'or' q);
  coherence;
  commutativity;
  func p 'xor' q equals
  'not'(p <=> q);
  coherence;
  commutativity;
  func p '\' q equals
  p '&' 'not' q;
  coherence;
end;

registration
  let p,q;
  cluster p 'nand' q -> boolean;
  coherence;
  cluster p 'nor' q -> boolean;
  coherence;
  cluster p 'xor' q -> boolean;
  coherence;
  cluster p '\' q -> boolean;
  coherence;
end;

begin

theorem
  p '&' p = p;

theorem
  p '&' (p '&' q) = p '&' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'or' p = p;

theorem
  p 'or' (p 'or' q) = p 'or' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'or' p '&' q = p
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p '&' (p 'or' q) = p
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p '&' (p 'or' q) = p 'or' (p '&' q)
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p '&' (q 'or' r) = p '&' q 'or' p '&' r
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'or' q '&' r = (p 'or' q) '&' (p 'or' r)
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p '&' q 'or' q '&' r 'or' r '&' p = (p 'or' q) '&' (q 'or' r) '&' (r 'or' p)
proof
A1: q = FALSE or q = TRUE by Def3;
  p = FALSE or p = TRUE by Def3;
  hence thesis by A1;
end;

theorem
  p '&' ('not' p 'or' q) = p '&' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'or' 'not' p '&' q = p 'or' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p => (p => q) = p => q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p '&' (p => q) = p '&' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p => (p '&' q) = p => q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p '&' 'not' (p => q) = p '&' 'not' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  'not' p 'or' (p => q) = p => q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  'not' p '&' (p => q) = 'not' p 'or' 'not' p '&' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p <=> q <=> r = p <=> (q <=> r)
proof
  q = FALSE or q = TRUE by Def3;
  hence thesis;
end;

theorem
  p '&' (p <=> q) = p '&' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  'not' p '&' (p <=> q) ='not' p '&' 'not' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p '&' (q <=> r) = p '&' ('not' q 'or' r) '&' ('not' r 'or' q);

theorem
  p 'or' (q <=> r) = (p 'or' 'not' q 'or' r) '&' (p 'or' 'not' r 'or' q)
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  'not' p '&' (p <=> q) =('not' p '&' 'not' q) '&' ('not' p 'or' q)
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  'not' p '&' (q <=> r) = 'not' p '&' ('not' q 'or' r) '&' ('not' r 'or' q);

theorem
  p => (p <=> q) = p => q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p => (p <=> q) = p => (p => q)
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'or' (p <=> q) = q => p
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  'not' p 'or' (p <=> q) = p => q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p => (q <=> r) = ('not' p 'or' 'not' q 'or' r) '&' ('not' p 'or' q
  'or' 'not' r)
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'nor' p = 'not' p;

theorem
  p 'nor' (p '&' q) = 'not' p
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'nor' (p 'or' q) = 'not' p '&' 'not' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'nor' (p 'nor' q) = 'not' p '&' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'nor' (p '&' q) = 'not' p
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'nor' (p 'or' q) = p 'nor' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  'not' p '&' (p 'nor' q) = p 'nor' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'or' (q 'nor' r) = (p 'or' 'not' q) '&' (p 'or' 'not' r)
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'nor' (q 'nor' r) = 'not' p '&' q 'or' 'not' p '&' r
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'nor' (q '&' r) = 'not' (p 'or' q) 'or' 'not' (p 'or' r)
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p '&' (q 'nor' r) = p '&' 'not' q '&' 'not' r
proof
  thus p '&' (q 'nor' r) = p '&' ('not' q '&' 'not' r)
    .= p '&' 'not' q '&' 'not' r;
end;

theorem
  p => (p 'nor' q) = 'not' p
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p => (q 'nor' r) = (p => 'not' q) '&' (p => 'not' r)
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'or' (p 'nor' q) = q => p
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'or' (q 'nor' r) = (q => p) '&' (r => p)
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p => (q 'nor' r) = ('not' p 'or' 'not' q) '&' ('not' p 'or' 'not' r)
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'nor' (p <=> q) = 'not' p '&' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  'not' p '&' (p <=> q) = p 'nor' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'nor' (q <=> r) = 'not' ((p 'or' 'not' q 'or' r) '&' (p 'or' 'not' r
  'or' q))
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p <=> q = p '&' q 'or' (p 'nor' q)
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'nand' p = 'not' p;

theorem
  p '&' (p 'nand' q) = p '&' 'not' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'nand' (p '&' q) = 'not' (p '&' q)
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'nand' (q 'nand' r) = ('not' p 'or' q) '&' ('not' p 'or' r)
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'nand' (q 'or' r) = 'not' (p '&' q) '&' 'not' (p '&' r)
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p => (p 'nand' q) = p 'nand' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'nand' (p 'nand' q) = p => q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'nand' (q 'nand' r) = (p => q) '&' (p => r)
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'nand' (p => q) = 'not' (p '&' q)
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'nand' (q => r) = (p => q) '&' (p => 'not' r)
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  'not' p 'or' (p 'nand' q) = p 'nand' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'nand' (q => r) = ('not' p 'or' q) '&' ('not' p 'or' 'not' r)
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  'not' p '&' (p 'nand' q) = 'not' p 'or' 'not' p '&' 'not' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p '&' (q 'nand' r) = p '&' 'not' q 'or' p '&' 'not' r
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'nand' (p <=> q) = 'not' (p '&' q)
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;



theorem
  p 'nand' (q <=> r) = 'not' (p '&' ('not' q 'or' r) '&' ('not' r 'or' q ));

theorem
  p 'nand' (q 'nor' r) = 'not' p 'or' q 'or' r
proof
  thus p 'nand' (q 'nor' r) = 'not' p 'or' (q 'or' r) .= 'not' p 'or' q 'or' r;
end;

theorem
  p '\' q '\' q = p '\' q
proof
  q = FALSE or q = TRUE by Def3;
  hence thesis;
end;

theorem
  p '&' (p '\' q) = p '\' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'nor' (p <=> q) = q '\' p
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'nor' (p 'nor' q) = q '\' p
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'xor' (p 'xor' q) = q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  (p 'xor' q) 'xor' r = p 'xor' (q 'xor' r)
proof
  q = FALSE or q = TRUE by Def3;
  hence thesis;
end;

theorem
  'not' (p 'xor' q) = 'not' p 'xor' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p '&' (q 'xor' r) = (p '&' q) 'xor' (p '&' r)
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p '&' (p 'xor' q) = p '&' 'not' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'xor' (p '&' q) = p '&' 'not' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  'not' p '&' (p 'xor' q) ='not' p '&' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'or' (p 'xor' q) = p 'or' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'or' ('not' p 'xor' q) =p 'or' 'not' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;



theorem
  p 'xor' ('not' p '&' q) = p 'or' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'xor' (p 'or' q) = 'not' p '&' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'xor' (q '&' r) = (p 'or' (q '&' r)) '&' ('not' p 'or' 'not' (q '&' r))
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  'not' p 'xor' (p => q) = p '&' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p => (p 'xor' q) = 'not' p 'or' 'not' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'xor' (p => q) = 'not' p 'or' 'not' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  'not' p 'xor' (q => p) = (p '&' (p 'or' 'not' q)) 'or' ('not' p '&' q)
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'xor' (p <=> q) = 'not' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  'not' p 'xor' (p <=> q) = q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'nor' (p 'xor' q) = 'not' p '&' 'not' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'nor' (p 'xor' q) = p 'nor' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'xor' (p 'nor' q) = q => p
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'nand' (p 'xor' q) = p => q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'xor' (p 'nand' q) = p => q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'xor' (p => q) = p 'nand' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'nand' (q 'xor' r) = (p '&' q) <=> (p '&' r)
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'xor' (p '&' q) = p '\' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p '&' (p 'xor' q) = p '\' q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  'not' p '&' (p 'xor' q) = q '\' p
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'xor' (p 'or' q) = q '\' p
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

begin

theorem
  p '&' q = TRUE implies p = TRUE & q = TRUE
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  'not'(p '&' 'not' p) = TRUE
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p => p = TRUE
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p => (q => p) = TRUE
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p => ((p => q) => q) = TRUE
proof
A1: q = FALSE or q = TRUE by Def3;
  p = FALSE or p = TRUE by Def3;
  hence thesis by A1;
end;

theorem
  (p => q) => ((q => r) => (p => r)) = TRUE
proof
A1: q = FALSE or q = TRUE by Def3;
A2: p = FALSE or p = TRUE by Def3;
  r = FALSE or r = TRUE by Def3;
  hence thesis by A1,A2;
end;

theorem
  (p => q) => ((r => p) => (r => q)) = TRUE
proof
A1: q = FALSE or q = TRUE by Def3;
A2: p = FALSE or p = TRUE by Def3;
  r = FALSE or r = TRUE by Def3;
  hence thesis by A1,A2;
end;

theorem
  (p => (p => q)) => (p => q) = TRUE
proof
A1: q = FALSE or q = TRUE by Def3;
  p = FALSE or p = TRUE by Def3;
  hence thesis by A1;
end;

theorem
  (p => (q => r)) => ((p => q) => (p => r)) = TRUE
proof
A1: q = FALSE or q = TRUE by Def3;
A2: p = FALSE or p = TRUE by Def3;
  r = FALSE or r = TRUE by Def3;
  hence thesis by A1,A2;
end;

theorem
  (p => (q => r)) => (q => (p => r)) = TRUE
proof
A1: q = FALSE or q = TRUE by Def3;
A2: p = FALSE or p = TRUE by Def3;
  r = FALSE or r = TRUE by Def3;
  hence thesis by A1,A2;
end;

theorem
  ((p => q) => r) => (q => r) = TRUE
proof
A1: q = FALSE or q = TRUE by Def3;
  r = FALSE or r = TRUE by Def3;
  hence thesis by A1;
end;

theorem
  (TRUE => p) => p = TRUE
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p => q = TRUE implies (q => r) => (p => r) = TRUE
proof
  r = FALSE or r = TRUE by Def3;
  hence thesis;
end;

theorem
  p => (p => q) = TRUE implies p => q = TRUE
proof
  q = FALSE or q = TRUE by Def3;
  hence thesis;
end;

theorem
  p => (q => r) = TRUE implies (p => q) => (p =>r) = TRUE
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p => q = TRUE & q => p = TRUE implies p = q
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p => q = TRUE & q => r = TRUE implies p => r = TRUE
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  ('not' p => p) => p = TRUE
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  'not' p = TRUE implies p => q = TRUE;

theorem
  p => q = TRUE & p => 'not' q = TRUE implies 'not' p = TRUE
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  (p => q) => ('not' (q '&' r) => 'not' (p '&' r)) = TRUE
proof
A1: q = FALSE or q = TRUE by Def3;
A2: p = FALSE or p = TRUE by Def3;
  r = FALSE or r = TRUE by Def3;
  hence thesis by A1,A2;
end;

theorem
  p 'or' (p => q) = TRUE
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p => (p 'or' q) = TRUE
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  'not' q 'or' (q => p => p) = TRUE
proof
A1: q = FALSE or q = TRUE by Def3;
  p = FALSE or p = TRUE by Def3;
  hence thesis by A1;
end;

theorem
  p <=> p = TRUE
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p <=> q <=> r <=> p <=> (q <=> r) = TRUE
proof
A1: q = FALSE or q = TRUE by Def3;
A2: p = FALSE or p = TRUE by Def3;
  r = FALSE or r = TRUE by Def3;
  hence thesis by A1,A2;
end;

theorem
  p <=> q = TRUE & q <=> r = TRUE implies p <=> r = TRUE
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p <=> q = TRUE & r <=> s = TRUE implies (p <=> r) <=> (q <=> s) = TRUE
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  'not' (p <=> 'not' p) = TRUE
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p <=> q = TRUE & r <=> s = TRUE implies (p '&' r) <=> (q '&' s) = TRUE
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p <=> q = TRUE & r <=> s = TRUE implies (p 'or' r) <=> (q 'or' s) = TRUE
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p <=> q = TRUE iff p => q = TRUE & q => p = TRUE
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p <=> q = TRUE & r <=> s = TRUE implies (p => r) <=> (q => s) = TRUE
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  'not' (p 'nor' 'not' p) = TRUE
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'nand' 'not' p = TRUE
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'or' (p 'nand' q) = TRUE
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'nand' (p 'nor' q) = TRUE
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p '&' 'not' p = FALSE
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p '&' p = FALSE implies p = FALSE;

theorem
  p '&' q = FALSE implies p = FALSE or q = FALSE
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  'not' (p => p) = FALSE
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p <=> 'not' p = FALSE
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  'not'(p <=> p) = FALSE
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p '&' (p 'nor' q) = FALSE
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'nor' (p => q) = FALSE
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'nor' (p 'nand' q) = FALSE
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;

theorem
  p 'xor' p = FALSE
proof
  p = FALSE or p = TRUE by Def3;
  hence thesis;
end;














begin

reserve X,Y,Z,x,y,z for set;
reserve i,j for Element of NAT;
reserve A,B,C for Subset of X;
reserve R,R1,R2 for Relation of X;
reserve AX for Subset of [:X,X:];
reserve SFXX for Subset-Family of [:X,X:];



definition
  let X;
  func nabla X -> Relation of X equals
  [:X,X:];
  coherence
  proof
    [:X,X:] c=[:X,X:];
    hence thesis;
  end;
end;

registration
  let X;
  cluster nabla X -> total reflexive;
  coherence
  proof
    thus dom nabla X c= X;
    thus X c= dom nabla X
    proof
      let x be set;
      assume x in X;
      then [x,x] in [:X,X:] by ZFMISC_1:87;
      hence thesis by XTUPLE_0:def 12;
    end;
    let x;
    assume x in field nabla X;
    then x in dom nabla X \/ rng nabla X;
    hence thesis by ZFMISC_1:87;
  end;
end;

definition
  let X;
  let R1,R2;
  redefine func R1 /\ R2 -> Relation of X;
  coherence
  proof
    R1 /\ R2 c= [:X,X:];
    hence thesis;
  end;
  redefine func R1 \/ R2 -> Relation of X;
  coherence
  proof
    R1 \/ R2 c= [:X,X:];
    hence thesis;
  end;
end;

theorem
  nabla X \/ R1 = nabla X by XBOOLE_1:12;

theorem
  id X is_reflexive_in X & id X is_symmetric_in X & id X is_transitive_in X
proof
  thus for x st x in X holds [x,x] in id X by RELAT_1:def 10;
  thus for x,y st x in X & y in X & [x,y] in id X holds [y,x] in id X by
RELAT_1:def 10;
  thus for x,y,z st x in X & y in X & z in X & [x,y] in id X & [y,z] in id X
  holds [x,z] in id X by RELAT_1:def 10;
end;

definition
  let X;
  mode Tolerance of X is total reflexive symmetric Relation of X;
  mode Equivalence_Relation of X is total symmetric transitive Relation of X;
end;

theorem
  id X is Equivalence_Relation of X;

theorem Th4:
  nabla X is Equivalence_Relation of X
proof
  for x,y holds x in X & y in X & [x,y] in nabla X implies [y,x] in nabla
  X by ZFMISC_1:88;
  then
A1: nabla X is_symmetric_in X by RELAT_2:def 3;
  for x,y,z st x in X & y in X & z in X & [x,y] in nabla X & [y,z] in
  nabla X holds [x,z] in nabla X by ZFMISC_1:87;
  then
A2: nabla X is_transitive_in X by RELAT_2:def 8;
  field nabla X = X by ORDERS_1:12;
  hence thesis by A1,A2,RELAT_2:def 11,def 16;
end;

registration
  let X;
  cluster nabla X -> total symmetric transitive;
  coherence by Th4;
end;

reserve EqR,EqR1,EqR2,EqR3 for Equivalence_Relation of X;

Lm1: [x,y] in R implies x in X & y in X
proof
  assume [x,y] in R;
  then x in dom R & y in rng R by XTUPLE_0:def 12,def 13;
  hence thesis;
end;

theorem Th5:
  for R being total reflexive Relation of X holds x in X implies [ x,x] in R
proof
  let R be total reflexive Relation of X;
  field R = X by ORDERS_1:12;
  then R is_reflexive_in X by RELAT_2:def 9;
  hence thesis by RELAT_2:def 1;
end;

theorem Th6:
  for R being total symmetric Relation of X holds [x,y] in R implies [y,x] in R
proof
  let R be total symmetric Relation of X;
  field R = X by ORDERS_1:12;
  then
A1: R is_symmetric_in X by RELAT_2:def 11;
  assume
A2: [x,y] in R;
  then x in X & y in X by Lm1;
  hence thesis by A2,A1,RELAT_2:def 3;
end;

theorem Th7:
  for R being total transitive Relation of X holds [x,y] in R & [y
  ,z] in R implies [x,z] in R
proof
  let R be total transitive Relation of X;
  assume that
A1: [x,y] in R and
A2: [y,z] in R;
A3: z in X by A2,Lm1;
  field R = X by ORDERS_1:12;
  then
A4: R is_transitive_in X by RELAT_2:def 16;
  x in X & y in X by A1,Lm1;
  hence thesis by A1,A2,A3,A4,RELAT_2:def 8;
end;

theorem
  for R being total reflexive Relation of X holds (ex x being set st x
  in X) implies R <> {};

theorem Th9:
  R is Equivalence_Relation of X iff R is reflexive symmetric
  transitive & field R = X
proof
  thus R is Equivalence_Relation of X implies R is reflexive symmetric
  transitive & field R = X by ORDERS_1:12;
  assume that
A1: R is reflexive and
A2: R is symmetric & R is transitive and
A3: field R = X;
  R is_reflexive_in X by A1,A3,RELAT_2:def 9;
  then dom R = X by ORDERS_1:13;
  hence thesis by A2,PARTFUN1:def 2;
end;

definition
  let X;
  let EqR1,EqR2;
  redefine func EqR1 /\ EqR2 -> Equivalence_Relation of X;
  coherence
  proof
    for x st x in X holds [x,x] in EqR2 by Th5;
    then
A1: id X c= EqR2 by RELAT_1:47;
    for x st x in X holds [x,x] in EqR1 by Th5;
    then id X c= EqR1 by RELAT_1:47;
    then id X c= EqR1 /\ EqR2 by A1,XBOOLE_1:19;
    then dom(EqR1 /\ EqR2) = X & rng(EqR1 /\ EqR2) = X by RELSET_1:16,17;
    then field (EqR1 /\ EqR2) = X;
    hence thesis by Th9;
  end;
end;

theorem
  id X /\ EqR = id X
proof
  now
    let x,y;
    assume [x,y] in id X;
    then x in X & x = y by RELAT_1:def 10;
    hence [x,y] in EqR by Th5;
  end;
  then id X c= EqR by RELAT_1:def 3;
  hence thesis by XBOOLE_1:28;
end;

theorem Th11:
  for SFXX st (SFXX <> {} & for Y st Y in SFXX holds Y is
  Equivalence_Relation of X) holds meet SFXX is Equivalence_Relation of X
proof
  let SFXX such that
A1: SFXX <> {} and
A2: for Y st Y in SFXX holds Y is Equivalence_Relation of X;
  reconsider XX = meet SFXX as Relation of X;
A3: XX is_symmetric_in X
  proof
    let x,y;
    assume that
    x in X and
    y in X and
A4: [x,y] in XX;
    now
      let Y;
      assume Y in SFXX;
      then Y is Equivalence_Relation of X & [x,y] in Y by A2,A4,SETFAM_1:def 1;
      hence [y,x] in Y by Th6;
    end;
    hence thesis by A1,SETFAM_1:def 1;
  end;
A5: XX is_transitive_in X
  proof
    let x,y,z;
    assume that
    x in X and
    y in X and
    z in X and
A6: [x,y] in XX and
A7: [y,z] in XX;
    now
      let Y;
      assume
A8:   Y in SFXX;
      then
A9:   [y,z] in Y by A7,SETFAM_1:def 1;
      Y is Equivalence_Relation of X & [x,y] in Y by A2,A6,A8,SETFAM_1:def 1;
      hence [x,z] in Y by A9,Th7;
    end;
    hence thesis by A1,SETFAM_1:def 1;
  end;
  XX is_reflexive_in X
  proof
    let x such that
A10: x in X;
    for Y st Y in SFXX holds [x,x] in Y
    proof
      let Y;
      assume Y in SFXX;
      then Y is Equivalence_Relation of X by A2;
      hence thesis by A10,Th5;
    end;
    hence thesis by A1,SETFAM_1:def 1;
  end;
  then field XX = X & dom XX = X by ORDERS_1:13;
  hence thesis by A3,A5,PARTFUN1:def 2,RELAT_2:def 11,def 16;
end;

theorem Th12:
  for R holds ex EqR st R c= EqR & for EqR2 st R c= EqR2 holds EqR c= EqR2
proof
  let R;
  defpred P[set] means $1 is Equivalence_Relation of X & R c= $1;
  consider F being Subset-Family of [:X,X:] such that
A1: for AX holds AX in F iff P[AX] from SUBSET_1:sch 3;
  R c= nabla X;
  then
A2: F <> {} by A1;
  for Y st Y in F holds Y is Equivalence_Relation of X by A1;
  then reconsider EqR = meet F as Equivalence_Relation of X by A2,Th11;
A3: now
    let EqR2;
    assume R c= EqR2;
    then EqR2 in F by A1;
    hence EqR c= EqR2 by SETFAM_1:3;
  end;
  take EqR;
  for Y st Y in F holds R c= Y by A1;
  hence thesis by A2,A3,SETFAM_1:5;
end;

definition
  let X;
  let EqR1,EqR2;

  func EqR1 "\/" EqR2 -> Equivalence_Relation of X means
  :Def2:
  EqR1 \/ EqR2 c= it & for EqR st EqR1 \/ EqR2 c= EqR holds it c= EqR;
  existence by Th12;
  uniqueness
  proof
    let R1,R2 be Equivalence_Relation of X;
    assume EqR1 \/ EqR2 c= R1 & ( for EqR st EqR1 \/ EqR2 c= EqR holds R1 c=
    EqR)& EqR1 \/ EqR2 c= R2 & for EqR st EqR1 \/ EqR2 c= EqR holds R2 c= EqR;
    then R1 c= R2 & R2 c= R1;
    hence thesis by XBOOLE_0:def 10;
  end;
  commutativity;
  idempotence;
end;

theorem
  (EqR1 "\/" EqR2) "\/" EqR3 = EqR1 "\/" (EqR2 "\/" EqR3)
proof
  for EqR4 be Equivalence_Relation of X holds EqR4 = EqR1 "\/" (EqR2 "\/"
  EqR3) implies (EqR1 "\/" EqR2) "\/" EqR3 c= EqR4
  proof
    let EqR4 be Equivalence_Relation of X;
A1: EqR2 \/ EqR3 c= EqR2 "\/" EqR3 by Def2;
    assume EqR4 = EqR1 "\/" (EqR2 "\/" EqR3);
    then
A2: EqR1 \/ (EqR2 "\/" EqR3) c= EqR4 by Def2;
    EqR2 "\/" EqR3 c= EqR1 \/ (EqR2 "\/" EqR3) by XBOOLE_1:7;
    then EqR2 "\/" EqR3 c= EqR4 by A2,XBOOLE_1:1;
    then
A3: EqR2 \/ EqR3 c= EqR4 by A1,XBOOLE_1:1;
    EqR2 c= EqR2 \/ EqR3 by XBOOLE_1:7;
    then
A4: EqR2 c= EqR4 by A3,XBOOLE_1:1;
    EqR1 c= EqR1 \/ (EqR2 "\/" EqR3) by XBOOLE_1:7;
    then EqR1 c= EqR4 by A2,XBOOLE_1:1;
    then EqR1 \/ EqR2 c= EqR4 by A4,XBOOLE_1:8;
    then
A5: EqR1 "\/" EqR2 c= EqR4 by Def2;
    EqR3 c= EqR2 \/ EqR3 by XBOOLE_1:7;
    then EqR3 c= EqR4 by A3,XBOOLE_1:1;
    then (EqR1 "\/" EqR2) \/ EqR3 c= EqR4 by A5,XBOOLE_1:8;
    hence thesis by Def2;
  end;
  then
A6: (EqR1 "\/" EqR2) "\/" EqR3 c= EqR1 "\/" (EqR2 "\/" EqR3);
  for EqR4 be Equivalence_Relation of X holds EqR4 = (EqR1 "\/" EqR2) "\/"
  EqR3 implies EqR1 "\/" (EqR2 "\/" EqR3) c= EqR4
  proof
    let EqR4 be Equivalence_Relation of X;
A7: EqR1 \/ EqR2 c= EqR1 "\/" EqR2 by Def2;
    assume EqR4 = (EqR1 "\/" EqR2) "\/" EqR3;
    then
A8: (EqR1 "\/" EqR2) \/ EqR3 c= EqR4 by Def2;
    EqR1 "\/" EqR2 c= (EqR1 "\/" EqR2) \/ EqR3 by XBOOLE_1:7;
    then EqR1 "\/" EqR2 c= EqR4 by A8,XBOOLE_1:1;
    then
A9: EqR1 \/ EqR2 c= EqR4 by A7,XBOOLE_1:1;
    EqR3 c= (EqR1 "\/" EqR2) \/ EqR3 by XBOOLE_1:7;
    then
A10: EqR3 c= EqR4 by A8,XBOOLE_1:1;
    EqR2 c= EqR1 \/ EqR2 by XBOOLE_1:7;
    then EqR2 c= EqR4 by A9,XBOOLE_1:1;
    then EqR2 \/ EqR3 c= EqR4 by A10,XBOOLE_1:8;
    then
A11: EqR2 "\/" EqR3 c= EqR4 by Def2;
    EqR1 c= EqR1 \/ EqR2 by XBOOLE_1:7;
    then EqR1 c= EqR4 by A9,XBOOLE_1:1;
    then EqR1 \/ (EqR2 "\/" EqR3) c= EqR4 by A11,XBOOLE_1:8;
    hence thesis by Def2;
  end;
  then EqR1 "\/" (EqR2 "\/" EqR3) c= (EqR1 "\/" EqR2) "\/" EqR3;
  hence thesis by A6,XBOOLE_0:def 10;
end;

theorem
  EqR "\/" EqR = EqR;

theorem
  EqR1 "\/" EqR2 = EqR2 "\/" EqR1;

theorem
  EqR1 /\ (EqR1 "\/" EqR2) = EqR1
proof
  thus EqR1 /\ (EqR1 "\/" EqR2) c= EqR1 by XBOOLE_1:17;
  EqR1 c= EqR1 \/ EqR2 & EqR1 \/ EqR2 c= EqR1 "\/" EqR2 by Def2,XBOOLE_1:7;
  then EqR1 c= EqR1 "\/" EqR2 by XBOOLE_1:1;
  hence thesis by XBOOLE_1:19;
end;

theorem
  EqR1 "\/" (EqR1 /\ EqR2) = EqR1
proof
  EqR1 = EqR1 \/ (EqR1 /\ EqR2) & for EqR st EqR1 \/ (EqR1 /\ EqR2) c= EqR
  holds EqR1 c= EqR by XBOOLE_1:22;
  hence thesis by Def2;
end;

scheme
  ExEqRel {X() -> set,P[set,set]}: ex EqR being Equivalence_Relation of X() st
  for x,y holds [x,y] in EqR iff x in X() & y in X() & P[x,y]
provided
A1: for x st x in X() holds P[x,x] and
A2: for x,y st P[x,y] holds P[y,x] and
A3: for x,y,z st P[x,y] & P[y,z] holds P[x,z]
proof
  consider Y being Relation of X(),X() such that
A4: for x,y holds [x,y] in Y iff x in X() & y in X() & P[x,y] from
  RELSET_1:sch 1;
A5: Y is_transitive_in X()
  proof
    let x,y,z;
    assume that
A6: x in X() and
    y in X() and
A7: z in X() and
A8: [x,y] in Y & [y,z] in Y;
    ( P[x,y])& P[y,z] by A4,A8;
    then P[x,z] by A3;
    hence thesis by A4,A6,A7;
  end;
A9: Y is_symmetric_in X()
  proof
    let x,y;
    assume that
A10: x in X() & y in X() and
A11: [x,y] in Y;
    P[x,y] by A4,A11;
    then P[y,x] by A2;
    hence thesis by A4,A10;
  end;
  Y is_reflexive_in X()
  proof
    let x;
    assume
A12: x in X();
    then P[x,x] by A1;
    hence thesis by A4,A12;
  end;
  then field Y = X() & dom Y = X() by ORDERS_1:13;
  then reconsider R1 = Y as Equivalence_Relation of X() by A9,A5,PARTFUN1:def 2
,RELAT_2:def 11,def 16;
  take R1;
  thus thesis by A4;
end;



notation
  let R be Relation, x be set;
  synonym Class(R,x) for Im(R,x);
end;

definition
  let X, Y be set, R be Relation of X, Y, x be set;

  redefine func Class(R,x) -> Subset of Y;
  coherence
  proof
    R.:{x} c= Y;
    hence thesis;
  end;
end;

theorem
  for R being Relation holds y in Class (R,x) iff [x,y] in R
proof
  let R be Relation;
  thus y in Class(R,x) implies [x,y] in R
  proof
    assume y in Class(R,x);
    then ex z being set st [z,y] in R & z in {x} by RELAT_1:def 13;
    hence thesis by TARSKI:def 1;
  end;
A1: x in {x} by TARSKI:def 1;
  assume [x,y] in R;
  hence thesis by A1,RELAT_1:def 13;
end;

theorem Th19:
  for R being total symmetric Relation of X holds y in Class (R,x)
  iff [y,x] in R
proof
  let R be total symmetric Relation of X;
  thus y in Class(R,x) implies [y,x] in R
  proof
    assume y in Class(R,x);
    then ex z being set st [z,y] in R & z in {x} by RELAT_1:def 13;
    then [x,y] in R by TARSKI:def 1;
    hence thesis by Th6;
  end;
  assume [y,x] in R;
  then
A1: [x,y] in R by Th6;
  x in {x} by TARSKI:def 1;
  hence thesis by A1,RELAT_1:def 13;
end;

theorem Th20:
  for R being Tolerance of X holds for x st x in X holds x in Class (R,x)
proof
  let R be Tolerance of X;
  let x;
  assume x in X;
  then [x,x] in R by Th5;
  hence thesis by Th19;
end;

theorem
  for R being Tolerance of X holds for x st x in X holds ex y st x in
  Class(R,y)
proof
  let R be Tolerance of X;
  let x;
  assume x in X;
  then x in Class(R,x) by Th20;
  hence thesis;
end;

theorem
  for R being transitive Tolerance of X holds y in Class(R,x) & z in
  Class(R,x) implies [y,z] in R
proof
  let R be transitive Tolerance of X;
  assume that
A1: y in Class(R,x) and
A2: z in Class(R,x);
  [z,x] in R by A2,Th19;
  then
A3: [x,z] in R by Th6;
  [y,x] in R by A1,Th19;
  hence thesis by A3,Th7;
end;

Lm2: for x st x in X holds [x,y] in EqR iff Class(EqR,x) = Class(EqR,y)
proof
  let x such that
A1: x in X;
  thus [x,y] in EqR implies Class(EqR,x) = Class(EqR,y)
  proof
    assume
A2: [x,y] in EqR;
    now
      let z;
      assume z in Class(EqR,y);
      then
A3:   [z,y] in EqR by Th19;
      [y,x] in EqR by A2,Th6;
      then [z,x] in EqR by A3,Th7;
      hence z in Class(EqR,x) by Th19;
    end;
    then
A4: Class(EqR,y) c= Class(EqR,x) by TARSKI:def 3;
    now
      let z;
      assume z in Class(EqR,x);
      then [z,x] in EqR by Th19;
      then [z,y] in EqR by A2,Th7;
      hence z in Class(EqR,y) by Th19;
    end;
    then Class(EqR,x) c= Class(EqR,y) by TARSKI:def 3;
    hence thesis by A4,XBOOLE_0:def 10;
  end;
  assume Class(EqR,x) = Class(EqR,y);
  then x in Class(EqR,y) by A1,Th20;
  hence thesis by Th19;
end;

theorem Th23:
  for x st x in X holds y in Class(EqR,x) iff Class(EqR,x) = Class (EqR,y)
proof
  let x such that
A1: x in X;
  thus y in Class(EqR,x) implies Class(EqR,x) = Class(EqR,y)
  proof
    assume y in Class(EqR,x);
    then [y,x] in EqR by Th19;
    then [x,y] in EqR by Th6;
    hence thesis by A1,Lm2;
  end;
  assume Class(EqR,x) = Class(EqR,y);
  then [x,y] in EqR by A1,Lm2;
  then [y,x] in EqR by Th6;
  hence thesis by Th19;
end;

theorem Th24:
  for x,y st y in X holds Class(EqR,x) = Class(EqR,y) or Class(EqR
  ,x) misses Class(EqR,y)
proof
  let x,y;
A1: not [x,y] in EqR implies Class(EqR,x) misses Class(EqR,y)
  proof
    assume
A2: not [x,y] in EqR;
    assume Class(EqR,x) meets Class(EqR,y);
    then consider z such that
A3: z in Class(EqR,x) and
A4: z in Class(EqR,y) by XBOOLE_0:3;
    [z,x] in EqR by A3,Th19;
    then
A5: [x,z] in EqR by Th6;
    [z,y] in EqR by A4,Th19;
    hence contradiction by A2,A5,Th7;
  end;
  assume
A6: y in X;
  [x,y] in EqR implies Class(EqR,x) = Class(EqR,y)
  proof
    assume [x,y] in EqR;
    then x in Class(EqR,y) by Th19;
    hence thesis by A6,Th23;
  end;
  hence thesis by A1;
end;

theorem Th25:
  for x st x in X holds Class(id X,x) = {x}
proof
  let x;
A1: now
    let y;
    assume y in Class(id X,x);
    then [y,x] in id X by Th19;
    hence y = x by RELAT_1:def 10;
  end;
  assume x in X;
  then for y holds y in Class(id X,x) iff y = x by A1,Th20;
  hence thesis by TARSKI:def 1;
end;

theorem Th26:
  for x st x in X holds Class(nabla X,x) = X
proof
  let x such that
A1: x in X;
  now
    let y;
    assume y in X;
    then [y,x] in nabla X by A1,ZFMISC_1:87;
    hence y in Class(nabla X,x) by Th19;
  end;
  then for y holds y in X iff y in Class(nabla X,x);
  hence thesis by TARSKI:1;
end;

theorem Th27:
  (ex x st Class(EqR,x) = X) implies EqR = nabla X
proof
  given x such that
A1: Class(EqR,x) = X;
  [:X,X:] c= EqR
  proof
    let y,z;
    assume
A2: [y,z] in [:X,X:];
    then z in Class(EqR,x) by A1,ZFMISC_1:87;
    then [z,x] in EqR by Th19;
    then
A3: [x,z] in EqR by Th6;
    y in Class(EqR,x) by A1,A2,ZFMISC_1:87;
    then [y,x] in EqR by Th19;
    hence thesis by A3,Th7;
  end;
  hence thesis by XBOOLE_0:def 10;
end;

theorem
  x in X implies ([x,y] in EqR1 "\/" EqR2 iff ex f being FinSequence st
1 <= len f & x = f.1 & y = f.(len f) & for i st 1 <= i & i < len f holds [f.i,f
  .(i+1)] in EqR1 \/ EqR2)
proof
  assume
A1: x in X;
  thus [x,y] in EqR1 "\/" EqR2 implies ex f being FinSequence st 1 <= len f &
  x = f.1 & y = f.(len f) & for i st 1 <= i & i < len f holds [f.i,f.(i+1)] in
  EqR1 \/ EqR2
  proof
    defpred P[set,set] means ex f being FinSequence st (1 <= len f & $1 = f.1
& $2 = f.(len f) & for i st 1 <= i & i < len f holds [f.i,f.(i+1)] in EqR1 \/
    EqR2);
    consider Y being Relation of X,X such that
A2: for x,y holds [x,y] in Y iff x in X & y in X & P[x,y] from
    RELSET_1:sch 1;
A3: Y is_transitive_in X
    proof
      let x,y,z;
      assume that
A4:   x in X and
A5:   y in X and
A6:   z in X and
A7:   [x,y] in Y and
A8:   [y,z] in Y;
      consider g being FinSequence such that
A9:   1 <= len g and
A10:  y = g.1 and
A11:  z = g.(len g) and
A12:  for i st 1 <= i & i < len g holds [g.i,g.(i+1)] in EqR1 \/ EqR2 by A2,A8;
      consider f being FinSequence such that
A13:  1 <= len f and
A14:  x = f.1 and
A15:  y = f.(len f) and
A16:  for i st 1 <= i & i < len f holds [f.i,f.(i+1)] in EqR1 \/ EqR2 by A2,A7;
      set h = f^g;
A17:  len h = len f + len g by FINSEQ_1:22;
A18:  len f + 1 <= len f + len g by A9,XREAL_1:7;
      then
A19:  h.(len h) = g.(len g + len f - len f) by A17,FINSEQ_1:23
        .= g.(len g);
A20:  for i st 1 <= i & i < len h holds [h.i,h.(i+1)] in EqR1 \/ EqR2
      proof
        let i;
        assume that
A21:    1 <= i and
A22:    i < len h;
A23:    1 <= i & i < len f or i = len f or len f < i by A21,XXREAL_0:1;
        now
          per cases by A22,A23,NAT_1:13;
          suppose
A24:        1 <= i & i < len f;
            then 1 <= i + 1 & i + 1 <= len f by NAT_1:13;
            then i + 1 in Seg len f by FINSEQ_1:1;
            then i + 1 in dom f by FINSEQ_1:def 3;
            then
A25:        h.(i + 1) = f.(i + 1) by FINSEQ_1:def 7;
            i in Seg(len f) by A24,FINSEQ_1:1;
            then i in dom f by FINSEQ_1:def 3;
            then h.i = f.i by FINSEQ_1:def 7;
            hence thesis by A16,A24,A25;
          end;
          suppose
A26:        i = len f;
            len f in Seg len f by A13,FINSEQ_1:1;
            then len f in dom f by FINSEQ_1:def 3;
            then
A27:        h.i = y by A15,A26,FINSEQ_1:def 7;
A28:        [y,y] in EqR1 by A5,Th5;
            h.(i + 1) = g.(1 + len f - len f) by A18,A26,FINSEQ_1:23
              .= y by A10;
            hence thesis by A27,A28,XBOOLE_0:def 3;
          end;
          suppose
A29:        len f + 1 <= i & i < len h;
            then len f < i by NAT_1:13;
            then reconsider k = i - len f as Element of NAT by NAT_1:21;
            i < len f + len g by A29,FINSEQ_1:22;
            then
A30:        i - len f < len g by XREAL_1:19;
            len f + 1 <= i + 1 & i + 1 <= len f + len g by A17,A29,NAT_1:13;
            then
A31:        h.(i + 1) = g.(1 + i - len f) by FINSEQ_1:23
              .= g.((i - len f) + 1);
            1 + len f - len f <= i - len f by A29,XREAL_1:9;
            then [g.k,g.(k + 1)] in EqR1 \/ EqR2 by A12,A30;
            hence thesis by A17,A29,A31,FINSEQ_1:23;
          end;
        end;
        hence thesis;
      end;
      1 in Seg(len f) by A13,FINSEQ_1:1;
      then 1 in dom f by FINSEQ_1:def 3;
      then
A32:  x = h.1 by A14,FINSEQ_1:def 7;
      1 <= len h by A13,A17,NAT_1:12;
      hence thesis by A2,A4,A6,A11,A32,A19,A20;
    end;
A33: Y is_symmetric_in X
    proof
      let x,y;
      assume that
A34:  x in X & y in X and
A35:  [x,y] in Y;
      consider f being FinSequence such that
A36:  1 <= len f and
A37:  x = f.1 and
A38:  y = f.(len f) and
A39:  for i st 1 <= i & i < len f holds [f.i,f.(i+1)] in EqR1 \/ EqR2
      by A2,A35;
      defpred P[Nat,set] means $2 = f.(1 + (len f) - $1);
A40:  for k be Nat st k in Seg(len f) ex z st P[k,z];
      consider g being FinSequence such that
A41:  dom g = Seg(len f) & for k being Nat st k in Seg(len f) holds P
      [k,g.k] from FINSEQ_1:sch 1(A40);
A42:  len f = len g by A41,FINSEQ_1:def 3;
A43:  for j st 1 <= j & j < len g holds [g.j,g.(j+1)] in EqR1 \/ EqR2
      proof
        let j;
        assume that
A44:    1 <= j and
A45:    j < len g;
        reconsider k = (len f) - j as Element of NAT by A42,A45,NAT_1:21;
        j - (len f) < (len f) - (len f) by A42,A45,XREAL_1:9;
        then - ((len f) - j) < - 0;
        then 0 < k;
        then
A46:    0 + 1 <= k by NAT_1:13;
        - j < -0 by A44,XREAL_1:24;
        then (len f) + -j < 0 + (len f) by XREAL_1:6;
        then
A47:    [f.k,f.(k + 1)] in EqR1 \/ EqR2 by A39,A46;
A48:    now
          per cases by A47,XBOOLE_0:def 3;
          suppose
            [f.k,f.(k + 1)] in EqR1;
            then [f.(k + 1),f.k] in EqR1 by Th6;
            hence [f.(k + 1),f.k] in EqR1 \/ EqR2 by XBOOLE_0:def 3;
          end;
          suppose
            [f.k,f.(k + 1)] in EqR2;
            then [f.(k + 1),f.k] in EqR2 by Th6;
            hence [f.(k + 1),f.k] in EqR1 \/ EqR2 by XBOOLE_0:def 3;
          end;
        end;
        1 <= (j + 1) & (j + 1) <= len f by A42,A45,NAT_1:12,13;
        then (j + 1) in Seg(len f) by FINSEQ_1:1;
        then
A49:    g.(j + 1) = f.((len f) + 1 - (1 + j)) by A41
          .= f.((len f) - j);
        j in Seg(len f) by A42,A44,A45,FINSEQ_1:1;
        then g.j = f.(1 + (len f) - j) by A41
          .= f.(((len f) - j) + 1);
        hence thesis by A49,A48;
      end;
      len f in Seg(len f) by A36,FINSEQ_1:1;
      then g.(len f) = f.(1 + (len f) - len f) by A41
        .= f.(1 + 0);
      then
A50:  x = g.(len g) by A37,A41,FINSEQ_1:def 3;
      1 in Seg(len f) by A36,FINSEQ_1:1;
      then
A51:  g.1 = f.((len f) + 1 - 1) by A41
        .= f.(len f);
      1 <= len g by A36,A41,FINSEQ_1:def 3;
      hence thesis by A2,A34,A38,A51,A50,A43;
    end;
    Y is_reflexive_in X
    proof
      let x such that
A52:  x in X;
      set g = <*x*>;
A53:  for i st 1 <= i & i < len g holds [g.i,g.(i+1)] in EqR1 \/ EqR2 by
FINSEQ_1:40;
      len g = 1 & g.1 = x by FINSEQ_1:40;
      hence thesis by A2,A52,A53;
    end;
    then field Y = X & dom Y = X by ORDERS_1:13;
    then reconsider R1 = Y as Equivalence_Relation of X by A33,A3,
PARTFUN1:def 2,RELAT_2:def 11,def 16;
    for x,y holds [x,y] in EqR1 \/ EqR2 implies [x,y] in R1
    proof
      let x,y;
      assume
A54:  [x,y] in EqR1 \/ EqR2;
      set g = <*x,y*>;
A55:  len g = 1 + 1 by FINSEQ_1:44;
A56:  g.1 = x by FINSEQ_1:44;
A57:  for i st 1 <= i & i < len g holds [g.i,g.(i+1)] in EqR1 \/ EqR2
      proof
        let i;
        assume that
A58:    1 <= i and
A59:    i < len g;
        i <= 1 by A55,A59,NAT_1:13;
        then 1 = i by A58,XXREAL_0:1;
        hence thesis by A54,A56,FINSEQ_1:44;
      end;
      len g = 2 by FINSEQ_1:44;
      then
A60:  g.1 = x & g.(len g) = y by FINSEQ_1:44;
      x in X & y in X by A54,Lm1;
      hence thesis by A2,A55,A60,A57;
    end;
    then EqR1 \/ EqR2 c= R1 by RELAT_1:def 3;
    then
A61: EqR1 "\/" EqR2 c= R1 by Def2;
    assume [x,y] in EqR1 "\/" EqR2;
    then consider f being FinSequence such that
A62: 1 <= len f & x = f.1 &( y = f.(len f) & for i st 1 <= i & i <
    len f holds [f.i,f.(i+1)] in EqR1 \/ EqR2 ) by A2,A61;
    take f;
    thus thesis by A62;
  end;
  given f being FinSequence such that
A63: 1 <= len f and
A64: x = f.1 and
A65: y = f.(len f) and
A66: for i st 1 <= i & i < len f holds [f.i,f.(i+1)] in EqR1 \/ EqR2;
  defpred P[Element of NAT] means 1 <= $1 & $1 <= len f implies [f.1,f.$1] in
  EqR1 "\/" EqR2;
A67: now
    let i;
    assume
A68: P[i];
    thus P[i+1]
    proof
      assume that
A69:  1 <= i + 1 and
A70:  i + 1 <= len f;
A71:  1 <= i or 1 = i + 1 by A69,NAT_1:8;
A72:  EqR1 \/ EqR2 c= EqR1 "\/" EqR2 by Def2;
A73:  i < len f by A70,NAT_1:13;
      now
        per cases by A71;
        suppose
A74:      1 <= i;
          then [f.i,f.(i+1)] in EqR1 \/ EqR2 by A66,A73;
          hence thesis by A68,A70,A72,A74,Th7,NAT_1:13;
        end;
        suppose
A75:      0 = i;
          [f.1,f.1] in EqR1 by A1,A64,Th5;
          then [f.1,f.1] in EqR1 \/ EqR2 by XBOOLE_0:def 3;
          hence thesis by A72,A75;
        end;
      end;
      hence thesis;
    end;
  end;
A76: P[0];
  for i holds P[i] from NAT_1:sch 1(A76,A67);
  hence thesis by A63,A64,A65;
end;

theorem Th29:
  for E being Equivalence_Relation of X st E = EqR1 \/ EqR2 holds
for x st x in X holds Class(E,x) = Class(EqR1,x) or Class(E,x) = Class(EqR2,x)
proof
  let E be Equivalence_Relation of X such that
A1: E = EqR1 \/ EqR2;
  for x st x in X holds Class(E,x) = Class(EqR1,x) or Class(E,x) = Class(
  EqR2,x)
  proof
    let x such that
    x in X;
    assume that
A2: not Class(E,x) = Class(EqR1,x) and
A3: not Class(E,x) = Class(EqR2,x);
    consider y such that
A4: y in Class(E,x) & not y in Class(EqR1,x) or y in Class(EqR1,x) &
    not y in Class(E,x) by A2,TARSKI:1;
A5: now
      assume that
A6:   y in Class(EqR1,x) and
A7:   not y in Class(E,x);
      [y,x] in EqR1 by A6,Th19;
      then [y,x] in E by A1,XBOOLE_0:def 3;
      hence contradiction by A7,Th19;
    end;
    then
A8: [y,x] in E by A4,Th19;
    consider z such that
A9: z in Class(E,x) & not z in Class(EqR2,x) or z in Class(EqR2,x) &
    not z in Class(E,x) by A3,TARSKI:1;
A10: now
      assume that
A11:  z in Class(EqR2,x) and
A12:  not z in Class(E,x);
      [z,x] in EqR2 by A11,Th19;
      then [z,x] in E by A1,XBOOLE_0:def 3;
      hence contradiction by A12,Th19;
    end;
    then
A13: [z,x] in E by A9,Th19;
    not [z,x] in EqR2 by A9,A10,Th19;
    then
A14: [z,x] in EqR1 by A1,A13,XBOOLE_0:def 3;
A15: now
      assume [y,z] in EqR1;
      then
A16:  [z,y] in EqR1 by Th6;
      [x,z] in EqR1 by A14,Th6;
      then [x,y] in EqR1 by A16,Th7;
      then [y,x] in EqR1 by Th6;
      hence contradiction by A4,A5,Th19;
    end;
    not [y,x] in EqR1 by A4,A5,Th19;
    then
A17: [y,x] in EqR2 by A1,A8,XBOOLE_0:def 3;
A18: now
      assume
A19:  [y,z] in EqR2;
      [x,y] in EqR2 by A17,Th6;
      then [x,z] in EqR2 by A19,Th7;
      then [z,x] in EqR2 by Th6;
      hence contradiction by A9,A10,Th19;
    end;
    [x,z] in E by A13,Th6;
    then [y,z] in E by A8,Th7;
    hence contradiction by A1,A18,A15,XBOOLE_0:def 3;
  end;
  hence thesis;
end;

theorem
  EqR1 \/ EqR2 = nabla X implies EqR1 = nabla X or EqR2 = nabla X
proof
  assume
A1: EqR1 \/ EqR2 = nabla X;
  (not X = {}) implies EqR1 = nabla X or EqR2 = nabla X
  proof
    set y = the Element of X;
A2: now
      let x;
      assume
A3:   x in X;
      then
      Class(nabla X,x) = Class(EqR1,x) or Class(nabla X,x) = Class(EqR2,x)
      by A1,Th29;
      hence Class(EqR1,x) = X or Class(EqR2,x) = X by A3,Th26;
    end;
    assume X <> {};
    then Class(EqR1,y) = X or Class(EqR2,y) = X by A2;
    hence thesis by Th27;
  end;
  hence thesis;
end;

definition
  let X;
  let EqR;
  func Class EqR -> Subset-Family of X means
  :Def3:
  A in it iff ex x st x in X & A = Class(EqR,x);
  existence
  proof
    defpred P[set] means ex x st x in X & $1 = Class(EqR,x);
    consider F being Subset-Family of X such that
A1: for A holds A in F iff P[A] from SUBSET_1:sch 3;
    take F;
    thus thesis by A1;
  end;
  uniqueness
  proof
    let F1,F2 be Subset-Family of X;
    assume that
A2: for A holds A in F1 iff ex x st x in X & A = Class(EqR,x) and
A3: for A holds A in F2 iff ex x st x in X & A = Class(EqR,x);
    now
      let A;
      A in F1 iff ex x st x in X & A = Class(EqR,x) by A2;
      hence A in F1 iff A in F2 by A3;
    end;
    hence thesis by SETFAM_1:31;
  end;
end;

theorem Th31:
  X = {} implies Class EqR = {}
proof
  assume that
A1: X = {} and
A2: Class EqR <> {};
  set z = the Element of Class EqR;
  z is Subset of X by A2,TARSKI:def 3;
  then ex x st x in X & z = Class(EqR,x) by A2,Def3;
  hence contradiction by A1;
end;

definition
  let X;
  mode a_partition of X -> Subset-Family of X means
    :Def4:
    union it = X & for
    A st A in it holds A<>{} & for B st B in it holds A = B or A misses B;
  existence
  proof
    defpred P[set] means ex x st x in X & $1 = {x};
    consider F being Subset-Family of X such that
A1: for A being Subset of X holds A in F iff P[A] from SUBSET_1:sch 3;
A2: for A st A in F holds A<>{} & for B st B in F holds A = B or A misses B
    proof
      let A;
      assume A in F;
      then consider x such that
      x in X and
A3:   A = {x} by A1;
      thus A <> {} by A3;
      let B;
      assume B in F;
      then consider y such that
      y in X and
A4:   B = {y} by A1;
      now
        assume
A5:     not x = y;
        for z st z in A holds not z in B
        proof
          let z;
          assume z in A;
          then not z = y by A3,A5,TARSKI:def 1;
          hence thesis by A4,TARSKI:def 1;
        end;
        hence A misses B by XBOOLE_0:3;
      end;
      hence thesis by A3,A4;
    end;
    take F;
    now
      let y;
      now
        set Z = {y};
        assume
A6:     y in X;
        then Z is Subset of X by ZFMISC_1:31;
        then y in Z & Z in F by A1,A6,TARSKI:def 1;
        hence ex Z st y in Z & Z in F;
      end;
      hence y in X iff ex Z st y in Z & Z in F;
    end;
    hence thesis by A2,TARSKI:def 4;
  end;
end;

theorem Th32:
  for P being a_partition of {} holds P = {}
proof
  let P be a_partition of {};
  assume not thesis;
  then consider Z being set such that
A1: Z in P by XBOOLE_0:def 1;
  Z <> {} by A1,Def4;
  hence thesis by A1;
end;

theorem Th33:
  Class EqR is a_partition of X
proof
  now
    let x;
    now
      consider Z such that
A1:   Z = Class(EqR,x);
      assume
A2:   x in X;
      then Z in Class EqR by A1,Def3;
      hence ex Z st x in Z & Z in Class EqR by A2,A1,Th20;
    end;
    hence x in X iff ex Z st x in Z & Z in Class EqR;
  end;
  hence union(Class EqR) = X by TARSKI:def 4;
  let A;
  assume A in Class EqR;
  then
A3: ex x st x in X & A = Class(EqR,x) by Def3;
  hence A <> {} by Th20;
  let B;
  assume B in Class EqR;
  then ex y st y in X & B = Class(EqR,y) by Def3;
  hence thesis by A3,Th24;
end;

theorem Th34:
  for P being a_partition of X holds ex EqR st P = Class EqR
proof
  let P be a_partition of X;
A1: X <> {} implies ex EqR st P = Class EqR
  proof
    defpred P[set,set] means ex A st A in P & $1 in A & $2 in A;
    assume X <> {};
A2: for x,y,z st P[x,y] & P[y,z] holds P[x,z]
    proof
      let x,y,z;
      given A such that
A3:   A in P and
A4:   x in A & y in A;
      given B such that
A5:   B in P and
A6:   y in B & z in B;
      A = B or A misses B by A3,A5,Def4;
      hence thesis by A3,A4,A6,XBOOLE_0:3;
    end;
A7: union P = X by Def4;
A8: for x st x in X holds P[x,x]
    proof
      let x;
      assume x in X;
      then consider Z such that
A9:   x in Z and
A10:  Z in P by A7,TARSKI:def 4;
      reconsider A = Z as Subset of X by A10;
      take A;
      thus thesis by A9,A10;
    end;
A11: for x,y st P[x,y] holds P[y,x];
    consider R being Equivalence_Relation of X such that
A12: for x,y holds [x,y] in R iff x in X & y in X & P[x,y] from
    ExEqRel(A8,A11,A2);
    take R;
    now
      let A;
A13:  now
        set x = the Element of A;
        assume
A14:    A in P;
        then
A15:    A <> {} by Def4;
        then
A16:    x in X by TARSKI:def 3;
        now
          let y;
A17:      now
            assume y in Class(R,x);
            then [y,x] in R by Th19;
            then consider B such that
A18:        B in P & y in B and
A19:        x in B by A12;
            A meets B by A15,A19,XBOOLE_0:3;
            hence y in A by A14,A18,Def4;
          end;
          now
            assume y in A;
            then [y,x] in R by A12,A14,A16;
            hence y in Class(R,x) by Th19;
          end;
          hence y in A iff y in Class(R,x) by A17;
        end;
        then A = Class(R,x) by TARSKI:1;
        hence A in Class R by A16,Def3;
      end;
      now
        assume A in Class R;
        then consider x such that
A20:    x in X and
A21:    A = Class(R,x) by Def3;
        x in Class(R,x) by A20,Th20;
        then [x,x] in R by Th19;
        then consider B such that
A22:    B in P and
A23:    x in B and
        x in B by A12;
        now
          let y;
A24:      now
            assume y in A;
            then [y,x] in R by A21,Th19;
            then consider C such that
A25:        C in P & y in C and
A26:        x in C by A12;
            B meets C by A23,A26,XBOOLE_0:3;
            hence y in B by A22,A25,Def4;
          end;
          now
            assume y in B;
            then [y,x] in R by A12,A22,A23;
            hence y in A by A21,Th19;
          end;
          hence y in A iff y in B by A24;
        end;
        hence A in P by A22,TARSKI:1;
      end;
      hence A in P iff A in Class R by A13;
    end;
    hence thesis by SETFAM_1:31;
  end;
  X = {} implies ex EqR st P = Class EqR
  proof
    set EqR1 = the Equivalence_Relation of X;
    assume
A27: X = {};
    take EqR1;
    P = {} by A27,Th32;
    hence thesis by A27,Th31;
  end;
  hence thesis by A1;
end;

theorem
  for x st x in X holds [x,y] in EqR iff Class(EqR,x) = Class(EqR,y) by Lm2;

theorem
  x in Class EqR implies ex y being Element of X st x = Class(EqR,y)
proof
  assume
A1: x in Class EqR;
  then reconsider x as Subset of X;
  consider y such that
A2: y in X and
A3: x = Class(EqR,y) by A1,Def3;
  reconsider y as Element of X by A2;
  take y;
  thus thesis by A3;
end;

begin 


registration
  let X be non empty set;
  cluster -> non empty for a_partition of X;
  coherence
  proof
    set x = the Element of X;
    let P be a_partition of X;
    union P = X by Def4;
    then ex A being set st x in A & A in P by TARSKI:def 4;
    hence thesis;
  end;
end;



registration
  let X be set;
  cluster -> with_non-empty_elements for a_partition of X;
  coherence
  proof
    let P be a_partition of X;
    assume {} in P;
    hence contradiction by Def4;
  end;
end;

definition
  let X be set, R be Equivalence_Relation of X;
  redefine func Class R -> a_partition of X;
  coherence by Th33;
end;



registration
  let I be non empty set, R be Equivalence_Relation of I;
  cluster Class R -> non empty;
  coherence;
end;

registration
  let I be non empty set, R be Equivalence_Relation of I;
  cluster Class R -> with_non-empty_elements;
  coherence;
end;

notation
  let I be non empty set, R be Equivalence_Relation of I;
  let x be Element of I;
  synonym EqClass(R,x) for Class(R,x);
end;

definition
  let I be non empty set, R be Equivalence_Relation of I;
  let x be Element of I;
  redefine func EqClass(R,x) -> Element of Class R;
  coherence
  proof
    thus Class(R,x) is Element of Class R by Def3;
  end;
end;

definition
  let X be set;
  func SmallestPartition X -> a_partition of X equals
  Class id X;
  coherence;
end;

theorem
  for X being non empty set holds SmallestPartition X = {{x} where x is
  Element of X: not contradiction}
proof
  let X be non empty set;
  set Y = {{x} where x is Element of X: not contradiction};
  hereby
    let x be set;
    assume x in SmallestPartition X;
    then consider y being set such that
A1: y in X and
A2: x = Class(id X, y) by Def3;
    x = {y} by A1,A2,Th25;
    hence x in Y by A1;
  end;
  let x be set;
  assume x in Y;
  then consider y being Element of X such that
A3: x = {y};
  Class(id X, y) = x by A3,Th25;
  hence thesis by Def3;
end;




reserve X for non empty set,
  x for Element of X;

definition
  let X be non empty set,x be Element of X,S1 be a_partition of X;
  func EqClass(x,S1) -> Subset of X means
  :Def6:
  x in it & it in S1;
  existence
  proof
    consider EQR being Equivalence_Relation of X such that
A1: S1 = Class EQR by Th34;
    take Class(EQR,x);
    thus x in Class(EQR,x) by Th20;
    thus thesis by A1,Def3;
  end;
  uniqueness
  proof
    let A,B be Subset of X;
    assume that
A2: x in A and
A3: A in S1 and
A4: x in B and
A5: B in S1;
    A meets B by A2,A4,XBOOLE_0:3;
    hence thesis by A3,A5,Def4;
  end;
end;

theorem Th38:
  for S1,S2 being a_partition of X st (for x being Element of X
  holds EqClass(x,S1) = EqClass(x,S2)) holds S1=S2
proof
  let S1,S2 be a_partition of X;
  assume
A1: for x being Element of X holds EqClass(x,S1) = EqClass(x,S2);
  now
    let P be Subset of X;
    thus P in S1 implies P in S2
    proof
      set x = the Element of P;
      assume
A2:   P in S1;
      then
A3:   P is non empty by Def4;
      then x in P;
      then reconsider x as Element of X;
      P = EqClass(x,S1) by A2,A3,Def6;
      then P = EqClass(x,S2) by A1;
      hence thesis by Def6;
    end;
    thus P in S2 implies P in S1
    proof
      set x = the Element of P;
      assume
A4:   P in S2;
      then
A5:   P <> {} by Def4;
      then x in P;
      then reconsider x as Element of X;
      P = EqClass(x,S2) by A4,A5,Def6;
      then P = EqClass(x,S1) by A1;
      hence thesis by Def6;
    end;
  end;
  hence thesis by SETFAM_1:31;
end;

theorem Th39:
  for X being non empty set holds {X} is a_partition of X
proof
  let X be non empty set;
  reconsider A1 = {X} as Subset-Family of X by ZFMISC_1:68;
A1: for A being Subset of X st A in A1 holds A <> {} & for B being Subset of
  X st B in A1 holds A = B or A misses B
  proof
    let A be Subset of X;
    assume
A2: A in A1;
    hence A <> {} by TARSKI:def 1;
    let B be Subset of X;
    assume B in A1;
    then B = X by TARSKI:def 1;
    hence thesis by A2,TARSKI:def 1;
  end;
  union A1 = X by ZFMISC_1:25;
  hence thesis by A1,Def4;
end;

definition
  let X be set;
  mode Family-Class of X is Subset-Family of bool X;
end;

definition

  let X be set;
  let F be Family-Class of X;
  attr F is partition-membered means
  :Def7:
  for S being set st S in F holds S is a_partition of X;
end;

registration
  let X be set;
  cluster partition-membered for Family-Class of X;
  existence
  proof
    reconsider E = {} as Family-Class of X by XBOOLE_1:2;
    take E;
    let S be set;
    assume S in E;
    hence thesis;
  end;
end;

definition
  let X be set;
  mode Part-Family of X is partition-membered Family-Class of X;
end;

reserve F for Part-Family of X;

registration
  let X be non empty set;
  cluster non empty for a_partition of X;
  existence
  proof
    take {X};
    thus thesis by Th39;
  end;
end;

theorem Th40:
  for X being set, p being a_partition of X holds {p} is Part-Family of X
proof
  let X be set;
  let p be a_partition of X;
  for x be set st x in {p} holds x is a_partition of X by TARSKI:def 1;
  hence thesis by Def7;
end;

registration
  let X be set;
  cluster non empty for Part-Family of X;
  existence
  proof
    set p = the a_partition of X;
    reconsider P = {p} as Part-Family of X by Th40;
    take P;
    thus thesis;
  end;
end;

theorem Th41:
  for S1 being a_partition of X, x,y being Element of X holds
  EqClass(x,S1) meets EqClass(y,S1) implies EqClass(x,S1) = EqClass(y,S1)
proof
  let S1 be a_partition of X;
  let x,y be Element of X;
  consider EQR being Equivalence_Relation of X such that
A1: S1 = Class EQR by Th34;
A2: y in Class(EQR,y) by Th20;
  Class(EQR,y) in S1 by A1,Def3;
  then
A3: Class(EQR,y) = EqClass(y,S1) by A2,Def6;
A4: x in Class(EQR,x) by Th20;
  Class(EQR,x) in S1 by A1,Def3;
  then
A5: Class(EQR,x) = EqClass(x,S1) by A4,Def6;
  assume EqClass(x,S1) meets EqClass(y,S1);
  hence thesis by A5,A3,Th24;
end;

Lm3: for A being set holds A in {EqClass(x,S) where S is a_partition of X : S
in F} implies ex T being a_partition of X st T in F & A = EqClass(x,T)
proof
  let A be set;
  assume A in {EqClass(x,S) where S is a_partition of X : S in F};
  then consider S being a_partition of X such that
A1: A = EqClass(x,S) & S in F;
  take S;
  thus thesis by A1;
end;

theorem
  for A being set,X being non empty set,S being a_partition of X holds A
  in S implies ex x being Element of X st A = EqClass(x,S)
proof
  let A be set,X be non empty set,S be a_partition of X;
  assume
A1: A in S;
  then A is non empty by Def4;
  then consider x being set such that
A2: x in A by XBOOLE_0:def 1;
  take x;
  thus thesis by A1,A2,Def6;
end;

definition
  let X be non empty set,F be non empty Part-Family of X;
  func Intersection F -> non empty a_partition of X means
  for x being Element
of X holds EqClass(x,it) = meet{EqClass(x,S) where S is a_partition of X : S in
  F};
  uniqueness
  proof
    given S1,S2 being a_partition of X such that
A1: for x being Element of X holds EqClass(x,S1) = meet{EqClass(x,S)
    where S is a_partition of X : S in F} and
A2: for x being Element of X holds EqClass(x,S2) = meet{EqClass(x,S)
    where S is a_partition of X : S in F} and
A3: not S1 = S2;
    now
      let x be Element of X;
      EqClass(x,S1) = meet{EqClass(x,S) where S is a_partition of X : S in
      F} by A1;
      hence EqClass(x,S1) = EqClass(x,S2) by A2;
    end;
    hence contradiction by A3,Th38;
  end;
  existence
  proof
    thus ex G being non empty a_partition of X st for x being Element of X
holds EqClass(x,G) = meet{EqClass(x,S) where S is a_partition of X : S in F}
    proof
      set G = {meet{EqClass(x,S) where S is a_partition of X : S in F} where x
      is Element of X : not contradiction };
      G c= bool X
      proof
        let y be set;
        assume y in G;
        then consider x being Element of X such that
A4:     y = meet{EqClass(x,S) where S is a_partition of X : S in F};
        y c= X
        proof
          let e be set;
          consider T being set such that
A5:       T in F by XBOOLE_0:def 1;
          reconsider T as a_partition of X by A5,Def7;
          EqClass(x,T) in {EqClass(x,S) where S is a_partition of X : S
          in F } by A5;
          then consider f being set such that
A6:       f in {EqClass(x,S) where S is a_partition of X : S in F};
          consider S being a_partition of X such that
A7:       f = EqClass(x,S) and
          S in F by A6;
          assume e in y;
          then e in EqClass(x,S) by A4,A6,A7,SETFAM_1:def 1;
          hence thesis;
        end;
        hence thesis;
      end;
      then reconsider G as Subset-Family of X;
      G is a_partition of X
      proof
        X c= union G
        proof
          let a be set;
          consider T being set such that
A8:       T in F by XBOOLE_0:def 1;
          assume a in X;
          then reconsider a1=a as Element of X;
          reconsider T as a_partition of X by A8,Def7;
A9:       meet{EqClass(a1,S) where S is a_partition of X : S in F} in G;
A10:      for T being set st T in {EqClass(a1,S) where S is a_partition
          of X : S in F} holds a1 in T
          proof
            let T be set;
            assume T in {EqClass(a1,S) where S is a_partition of X : S in F};
            then ex A being a_partition of X st T = EqClass(a1,A) & A in F;
            hence thesis by Def6;
          end;
          EqClass(a1,T) in {EqClass(a1,S) where S is a_partition of X : S
          in F} by A8;
          then
          a in meet{EqClass(a1,S) where S is a_partition of X: S in F} by A10,
SETFAM_1:def 1;
          hence thesis by A9,TARSKI:def 4;
        end;
        hence union G = X by XBOOLE_0:def 10;
        let A be Subset of X;
        consider T being set such that
A11:    T in F by XBOOLE_0:def 1;
        reconsider T as a_partition of X by A11,Def7;
        assume
A12:    A in G;
        then consider x being Element of X such that
A13:    A = meet{EqClass(x,S) where S is a_partition of X : S in F};
A14:    for Y being set st Y in {EqClass(x,S) where S is a_partition of X
        : S in F} holds x in Y
        proof
          let Y be set;
          assume Y in {EqClass(x,S) where S is a_partition of X : S in F};
          then ex T being a_partition of X st Y = EqClass(x,T) & T in F;
          hence thesis by Def6;
        end;
        EqClass(x,T) in {EqClass(x,S) where S is a_partition of X : S in
        F} by A11;
        hence A<>{} by A13,A14,SETFAM_1:def 1;
        consider x being Element of X such that
A15:    A = meet{EqClass(x,S) where S is a_partition of X : S in F} by A12;
        let B be Subset of X;
        assume B in G;
        then consider y being Element of X such that
A16:    B = meet{EqClass(y,S) where S is a_partition of X : S in F};
        thus A = B or A misses B
        proof
A17:      {EqClass(y,S) where S is a_partition of X : S in F} is non empty
          proof
            consider T being set such that
A18:        T in F by XBOOLE_0:def 1;
            reconsider T as a_partition of X by A18,Def7;
            EqClass(y,T) in {EqClass(y,S) where S is a_partition of X : S
            in F} by A18;
            hence thesis;
          end;
A19:      {EqClass(x,S) where S is a_partition of X : S in F} is non empty
          proof
            consider T being set such that
A20:        T in F by XBOOLE_0:def 1;
            reconsider T as a_partition of X by A20,Def7;
            EqClass(x,T) in {EqClass(x,S) where S is a_partition of X : S
            in F} by A20;
            hence thesis;
          end;
          now
            assume A meets B;
            then consider c being set such that
A21:        c in A & c in B by XBOOLE_0:3;
            c in meet{EqClass(x,S) where S is a_partition of X : S in F}
/\ meet{EqClass(y,S) where S is a_partition of X : S in F} by A15,A16,A21,
XBOOLE_0:def 4;
            then
A22:        c in meet({EqClass(x,S) where S is a_partition of X: S in F }
\/ {EqClass(y,S) where S is a_partition of X : S in F}) by A19,A17,SETFAM_1:9;
A23:        now
              let T be a_partition of X;
              assume T in F;
              then
              EqClass(y,T) in {EqClass(y,S) where S is a_partition of X :
              S in F};
              then
              EqClass(y,T) in {EqClass(x,S) where S is a_partition of X :
S in F} \/ {EqClass(y,S) where S is a_partition of X : S in F} by
XBOOLE_0:def 3;
              hence c in EqClass(y,T) by A22,SETFAM_1:def 1;
            end;
A24:        now
              let T be a_partition of X;
              assume T in F;
              then
              EqClass(x,T) in {EqClass(x,S) where S is a_partition of X :
              S in F};
              then
              EqClass(x,T) in {EqClass(x,S) where S is a_partition of X :
S in F} \/ {EqClass(y,S) where S is a_partition of X : S in F} by
XBOOLE_0:def 3;
              hence c in EqClass(x,T) by A22,SETFAM_1:def 1;
            end;
A25:        for T being a_partition of X st T in F holds ex A being set
            st A in EqClass(x,T) & A in EqClass(y,T)
            proof
              let T be a_partition of X;
              assume
A26:          T in F;
              take c;
              thus thesis by A24,A23,A26;
            end;
A27:        for T being a_partition of X st T in F holds (EqClass(x,T))
            meets (EqClass(y,T))
            proof
              let T be a_partition of X;
              assume T in F;
              then ex A being set st A in EqClass(x,T) & A in EqClass(y,T) by
A25;
              hence thesis by XBOOLE_0:3;
            end;
A28:        {EqClass(y,S) where S is a_partition of X : S in F} c= {
            EqClass(x,S) where S is a_partition of X : S in F}
            proof
              let A be set;
              assume A in {EqClass(y,S) where S is a_partition of X : S in F };
              then consider T being a_partition of X such that
A29:          T in F and
A30:          A = EqClass(y,T) by Lm3;
              A = EqClass(x,T) by A27,A29,A30,Th41;
              hence thesis by A29;
            end;
            {EqClass(x,S) where S is a_partition of X : S in F} c= {
            EqClass(y,S) where S is a_partition of X : S in F}
            proof
              let A be set;
              assume A in {EqClass(x,S) where S is a_partition of X : S in F };
              then consider T being a_partition of X such that
A31:          T in F and
A32:          A = EqClass(x,T) by Lm3;
              A = EqClass(y,T) by A27,A31,A32,Th41;
              hence thesis by A31;
            end;
            hence thesis by A15,A16,A28,XBOOLE_0:def 10;
          end;
          hence thesis;
        end;
      end;
      then reconsider G as non empty a_partition of X;
      take G;
      for x being Element of X holds EqClass(x,G) = meet{EqClass(x,S)
      where S is a_partition of X : S in F}
      proof
        let x be Element of X;
A33:    now
          let Y be set;
          assume Y in {EqClass(x,S) where S is a_partition of X : S in F};
          then ex T being a_partition of X st Y = EqClass(x,T) & T in F;
          hence x in Y by Def6;
        end;
        {EqClass(x,S) where S is a_partition of X : S in F} is non empty
        proof
          consider T being set such that
A34:      T in F by XBOOLE_0:def 1;
          reconsider T as a_partition of X by A34,Def7;
          EqClass(x,T) in {EqClass(x,S) where S is a_partition of X : S
          in F} by A34;
          hence thesis;
        end;
        then meet{EqClass(x,S) where S is a_partition of X : S in F} in G & x
in meet{ EqClass(x,S) where S is a_partition of X : S in F} by A33,
SETFAM_1:def 1;
        hence thesis by Def6;
      end;
      hence thesis;
    end;
  end;
end;

theorem Th43:
  for X being non empty set, S being a_partition of X, A being
  Subset of S holds (union S) \ (union A) = union (S \ A)
proof
  let X be non empty set;
  let S be a_partition of X;
  let A be Subset of S;
  thus (union S) \ (union A) c= union (S \ A)
  proof
    let y be set;
    assume
A1: y in (union S) \ (union A);
    then y in (union S) by XBOOLE_0:def 5;
    then consider z being set such that
A2: y in z and
A3: z in S by TARSKI:def 4;
    not y in (union A) by A1,XBOOLE_0:def 5;
    then not z in A by A2,TARSKI:def 4;
    then z in S \ A by A3,XBOOLE_0:def 5;
    hence thesis by A2,TARSKI:def 4;
  end;
  thus union (S \ A) c= (union S) \ (union A)
  proof
    let y be set;
    assume y in union(S \ A);
    then consider z being set such that
A4: y in z and
A5: z in S \ A by TARSKI:def 4;
A6: z in S by A5,XBOOLE_0:def 5;
A7: not z in A by A5,XBOOLE_0:def 5;
A8: now
      let w be set;
      assume
A9:   w in A;
      then w in S;
      then z misses w by A6,A7,A9,Def4;
      hence not y in w by A4,XBOOLE_0:3;
    end;
A10: now
      assume y in union A;
      then ex v being set st y in v & v in A by TARSKI:def 4;
      hence contradiction by A8;
    end;
    y in union S by A4,A6,TARSKI:def 4;
    hence thesis by A10,XBOOLE_0:def 5;
  end;
end;

theorem
  for X being non empty set,A being Subset of X, S being a_partition of
  X holds A in S implies union(S \ {A}) = X \ A
proof
  let X be non empty set;
  let A be Subset of X;
  let S be a_partition of X;
  assume
A1: A in S;
  {A} c= S
  proof
    let y be set;
    assume y in {A};
    hence thesis by A1,TARSKI:def 1;
  end;
  then reconsider AA = {A} as Subset of S;
  thus union (S \ {A}) = (union S) \ (union AA) by Th43
    .= X \ (union {A}) by Def4
    .= X \ A by ZFMISC_1:25;
end;



theorem
  {} is a_partition of {}
proof
  reconsider A = {} as Subset-Family of {} by XBOOLE_1:2;
  union A = {} & for a be Subset of {} st a in A holds a<>{} & for b be
  Subset of {} st b in A holds a = b or a misses b;
  hence thesis by Def4;
end;

begin 

reserve e,u,v,X,Y,X1 for set;

theorem
  for F being Function st X c= F"X1 holds F.:X c= X1
proof
  let F be Function;
  assume X c= F"X1;
  then
A1: F.:X c= F.:F"X1 by RELAT_1:123;
  F.:F"X1 c= X1 by FUNCT_1:75;
  hence thesis by A1,XBOOLE_1:1;
end;

theorem Th47:
  e c= [:X,Y:] implies (.:pr1(X,Y)).e = pr1(X,Y).:e
proof
  assume e c= [:X,Y:];
  then e c= dom pr1(X,Y) by FUNCT_3:def 4;
  hence thesis by FUNCT_3:def 1;
end;

theorem Th48:
  e c= [:X,Y:] implies (.:pr2(X,Y)).e = pr2(X,Y).:e
proof
  assume e c= [:X,Y:];
  then e c= dom pr2(X,Y) by FUNCT_3:def 5;
  hence thesis by FUNCT_3:def 1;
end;

theorem Th49:
  for X1 being Subset of X, Y1 being Subset of Y st [:X1,Y1:] <>
  {} holds pr1(X,Y).:[:X1,Y1:] = X1 & pr2(X,Y).:[:X1,Y1:] = Y1
proof
  let X1 be Subset of X, Y1 be Subset of Y;
  assume
A1: [:X1,Y1:] <> {};
  then
A2: X1 <> {};
A3: Y1 <> {} by A1;
A4: X <> {} by A2;
  now
    set y = the Element of Y1;
    let x be set;
    thus x in pr1(X,Y).:[:X1,Y1:] implies x in X1
    proof
      assume x in pr1(X,Y).:[:X1,Y1:];
      then consider u such that
A5:   u in [:X,Y:] and
A6:   u in [:X1,Y1:] & x = pr1(X,Y).u by FUNCT_2:64;
A7:   u`2 in Y by A5,MCART_1:10;
      u`1 in X1 & x = pr1(X,Y).(u`1,u`2) by A6,MCART_1:10,21;
      hence thesis by A7,FUNCT_3:def 4;
    end;
    assume
A8: x in X1;
    y in Y by A3,TARSKI:def 3;
    then
A9: x = pr1(X,Y).(x,y) by A8,FUNCT_3:def 4;
    [x,y] in [:X1,Y1:] by A3,A8,ZFMISC_1:87;
    hence x in pr1(X,Y).:[:X1,Y1:] by A4,A9,FUNCT_2:35;
  end;
  hence pr1(X,Y).:[:X1,Y1:] = X1 by TARSKI:1;
A10: Y <> {} by A3;
  now
    set x = the Element of X1;
    let y be set;
    thus y in pr2(X,Y).:[:X1,Y1:] implies y in Y1
    proof
      assume y in pr2(X,Y).:[:X1,Y1:];
      then consider u such that
A11:  u in [:X,Y:] and
A12:  u in [:X1,Y1:] & y = pr2(X,Y).u by FUNCT_2:64;
A13:  u`1 in X by A11,MCART_1:10;
      u`2 in Y1 & y = pr2(X,Y).(u`1,u`2) by A12,MCART_1:10,21;
      hence thesis by A13,FUNCT_3:def 5;
    end;
    assume
A14: y in Y1;
    x in X by A2,TARSKI:def 3;
    then
A15: y = pr2(X,Y).(x,y) by A14,FUNCT_3:def 5;
    [x,y] in [:X1,Y1:] by A2,A14,ZFMISC_1:87;
    hence y in pr2(X,Y).:[:X1,Y1:] by A10,A15,FUNCT_2:35;
  end;
  hence thesis by TARSKI:1;
end;

theorem Th50:
  for X1 being Subset of X, Y1 being Subset of Y st [:X1,Y1:] <>
  {} holds .:pr1(X,Y). [:X1,Y1:] = X1 & .:pr2(X,Y). [:X1,Y1:] = Y1
proof
  let X1 be Subset of X, Y1 be Subset of Y;
  assume
A1: [:X1,Y1:] <> {};
  thus .:pr1(X,Y). [:X1,Y1:] = pr1(X,Y).:[:X1,Y1:] by Th47
    .= X1 by A1,Th49;
  thus .:pr2(X,Y). [:X1,Y1:] = pr2(X,Y).:[:X1,Y1:] by Th48
    .= Y1 by A1,Th49;
end;

theorem
  for A being Subset of [:X,Y:], H being Subset-Family of [:X,Y:]
st for e st e in H holds e c= A & ex X1 being Subset of X, Y1 being Subset of Y
  st e =[:X1,Y1:] holds [:union(.:pr1(X,Y).:H), meet(.:pr2(X,Y).:H):] c= A
proof
  let A be Subset of [:X,Y:], H be Subset-Family of [:X,Y:] such that
A1: for e st e in H holds e c= A & ex X1 being Subset of X, Y1 being
  Subset of Y st e =[:X1,Y1:];
  let u,v;
  assume
A2: [u,v] in [:union(.:pr1(X,Y).:H), meet(.:pr2(X,Y).:H):];
  then
A3: v in meet(.:pr2(X,Y).:H) by ZFMISC_1:87;
  u in union(.:pr1(X,Y).:H) by A2,ZFMISC_1:87;
  then consider x being set such that
A4: u in x and
A5: x in .:pr1(X,Y).:H by TARSKI:def 4;
  consider y being set such that
  y in dom .:pr1(X,Y) and
A6: y in H and
A7: x = .:pr1(X,Y).y by A5,FUNCT_1:def 6;
  consider X1 being Subset of X, Y1 being Subset of Y such that
A8: y =[:X1,Y1:] by A1,A6;
A9: y <> {} by A4,A7,FUNCT_3:8;
  y in bool[:X,Y:] by A6;
  then y in bool dom pr2(X,Y) by FUNCT_3:def 5;
  then y in dom .:pr2(X,Y) by FUNCT_3:def 1;
  then .:pr2(X,Y).y in .:pr2(X,Y).:H by A6,FUNCT_1:def 6;
  then Y1 in .:pr2(X,Y).:H by A8,A9,Th50;
  then
A10: v in Y1 by A3,SETFAM_1:def 1;
  u in X1 by A4,A7,A8,A9,Th50;
  then
A11: [u,v] in y by A8,A10,ZFMISC_1:87;
  y c= A by A1,A6;
  hence thesis by A11;
end;

theorem
  for A being Subset of [:X,Y:], H being Subset-Family of [:X,Y:] st for
e st e in H holds e c= A & ex X1 being Subset of X, Y1 being Subset of Y st e =
  [:X1,Y1:] holds [:meet(.:pr1(X,Y).:H), union(.:pr2(X,Y).:H):] c= A
proof
  let A be Subset of [:X,Y:], H be Subset-Family of [:X,Y:] such that
A1: for e st e in H holds e c= A & ex X1 being Subset of X, Y1 being
  Subset of Y st e =[:X1,Y1:];
  let u,v;
  assume
A2: [u,v] in [:meet(.:pr1(X,Y).:H), union(.:pr2(X,Y).:H):];
  then
A3: u in meet(.:pr1(X,Y).:H) by ZFMISC_1:87;
  v in union(.:pr2(X,Y).:H) by A2,ZFMISC_1:87;
  then consider x being set such that
A4: v in x and
A5: x in .:pr2(X,Y).:H by TARSKI:def 4;
  consider y being set such that
  y in dom .:pr2(X,Y) and
A6: y in H and
A7: x = .:pr2(X,Y).y by A5,FUNCT_1:def 6;
  consider X1 being Subset of X, Y1 being Subset of Y such that
A8: y =[:X1,Y1:] by A1,A6;
A9: y <> {} by A4,A7,FUNCT_3:8;
  y in bool[:X,Y:] by A6;
  then y in bool dom pr1(X,Y) by FUNCT_3:def 4;
  then y in dom .:pr1(X,Y) by FUNCT_3:def 1;
  then .:pr1(X,Y).y in .:pr1(X,Y).:H by A6,FUNCT_1:def 6;
  then X1 in .:pr1(X,Y).:H by A8,A9,Th50;
  then
A10: u in X1 by A3,SETFAM_1:def 1;
  v in Y1 by A4,A7,A8,A9,Th50;
  then
A11: [u,v] in y by A8,A10,ZFMISC_1:87;
  y c= A by A1,A6;
  hence thesis by A11;
end;

theorem
  for X being set, Y being non empty set, f being Function of X,Y
  for H being Subset-Family of X holds union(.:f.:H) = f.: union H
proof
  let X be set, Y be non empty set, f be Function of X,Y;
  let H be Subset-Family of X;
  dom f = X by FUNCT_2:def 1;
  hence thesis by FUNCT_3:14;
end;

reserve X,Y,Z for non empty set;

theorem
  for X being set, a being Subset-Family of X holds union union a
  = union { union A where A is Subset of X: A in a }
proof
  let X be set, a be Subset-Family of X;
  thus union union a c= union { union A where A is Subset of X: A in a }
  proof
    let e;
    assume e in union union a;
    then consider B being set such that
A1: e in B and
A2: B in union a by TARSKI:def 4;
    consider C being set such that
A3: B in C and
A4: C in a by A2,TARSKI:def 4;
A5: union C in { union A where A is Subset of X: A in a } by A4;
    e in union C by A1,A3,TARSKI:def 4;
    hence thesis by A5,TARSKI:def 4;
  end;
  let e;
  assume e in union { union A where A is Subset of X: A in a };
  then consider c being set such that
A6: e in c and
A7: c in { union A where A is Subset of X: A in a } by TARSKI:def 4;
  consider A being Subset of X such that
A8: c = union A and
A9: A in a by A7;
  consider b being set such that
A10: e in b and
A11: b in A by A6,A8,TARSKI:def 4;
  b in union a by A9,A11,TARSKI:def 4;
  hence thesis by A10,TARSKI:def 4;
end;

theorem Th55:
  for X being set for D being Subset-Family of X st union D = X
for A being Subset of D, B being Subset of X st B = union A holds B` c= union A
  `
proof
  let X be set;
  let D be Subset-Family of X such that
A1: union D = X;
  let A be Subset of D, B be Subset of X such that
A2: B = union A;
  let e;
  assume
A3: e in B`;
  then consider u such that
A4: e in u and
A5: u in D by A1,TARSKI:def 4;
  not e in B by A3,XBOOLE_0:def 5;
  then not u in A by A2,A4,TARSKI:def 4;
  then u in A` by A5,SUBSET_1:29;
  hence thesis by A4,TARSKI:def 4;
end;

theorem
  for F being Function of X,Y, G being Function of X,Z st for x,x9
being Element of X st F.x=F.x9 holds G.x=G.x9 ex H being Function of Y,Z st H*F
  =G
proof
  let F be Function of X,Y, G be Function of X,Z;
  defpred P[set,set] means not (ex x being Element of X st $1 = F.x) or for x
  being Element of X st $1 = F.x holds $2 = G.x;
  assume
A1: for x,x9 being Element of X st F.x=F.x9 holds G.x=G.x9;
A2: now
    let e such that
    e in Y;
    now
      per cases;
      case
        ex x being Element of X st e = F.x;
        then consider x being Element of X such that
A3:     e = F.x;
        take u = G.x;
        thus u in Z & ((ex x being Element of X st e = F.x) implies ex x being
        Element of X st e = F.x & u = G.x) by A3;
      end;
      case
A4:     not ex x being Element of X st e = F.x;
        set u = the Element of Z;
        u in Z;
        hence
        ex u st u in Z & ((ex x being Element of X st e = F.x) implies ex
        x being Element of X st e = F.x & u = G.x) by A4;
      end;
    end;
    then consider u such that
A5: u in Z and
A6: (ex x being Element of X st e = F.x) implies ex x being Element of
    X st e = F.x & u = G.x;
    take u;
    thus u in Z by A5;
    thus P[e,u] by A1,A6;
  end;
  consider H being Function of Y,Z such that
A7: for e st e in Y holds P[e,H.e] from FUNCT_2:sch 1(A2);
  take H;
  now
    let x be Element of X;
    thus (H*F).x = H.(F.x) by FUNCT_2:15
      .= G.x by A7;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  for X,Y,Z for y being Element of Y, F being (Function of X,Y), G
  being Function of Y,Z holds F"{y} c= (G*F)"{G.y}
proof
  let X,Y,Z;
  let y be Element of Y, F be (Function of X,Y), G be Function of Y,Z;
  F"{y} c= (G*F)"Im(G,y) by FUNCT_2:44;
  hence thesis by SETWISEO:8;
end;

theorem
  for F being Function of X,Y, x being Element of X, z being
  Element of Z holds [:F,id Z:].(x,z) = [F.x,z]
proof
  let F be Function of X,Y, x be Element of X, z be Element of Z;
  thus [:F,id Z:].(x,z) = [F.x, (id Z).z] by FUNCT_3:75
    .= [F.x,z] by FUNCT_1:18;
end;

theorem
  for F being Function of X,Y, A being Subset of X, B being Subset
  of Z holds [:F,id Z:].:[:A,B:] = [:F.:A,B:]
proof
  let F be Function of X,Y, A be Subset of X, B be Subset of Z;
  thus [:F,id Z:].:[:A,B:] = [:F.:A, (id Z).:B:] by FUNCT_3:72
    .= [:F.:A,B:] by FUNCT_1:92;
end;

theorem
  for F being Function of X,Y, y being Element of Y, z being
  Element of Z holds [:F,id Z:]"{[y,z]} = [:F"{y},{z}:]
proof
  let F be Function of X,Y, y be Element of Y, z be Element of Z;
  thus [:F,id Z:]"{[y,z]} = [:F,id Z:]"[:{y},{z}:] by ZFMISC_1:29
    .= [:F"{y},(id Z)"{z}:] by FUNCT_3:73
    .= [:F"{y},{z}:] by FUNCT_2:94;
end;

theorem
  for D being Subset-Family of X, A being Subset of D holds union
  A is Subset of X
proof
  let D be Subset-Family of X, A be Subset of D;
  union A c= X
  proof
    let e;
    assume e in union A;
    then ex u st e in u & u in A by TARSKI:def 4;
    then e in union D by TARSKI:def 4;
    hence thesis;
  end;
  hence thesis;
end;

theorem
  for X being set, D being a_partition of X, A,B being Subset of D
  holds union(A /\ B) = union A /\ union B
proof
  let X be set, D be a_partition of X, A,B be Subset of D;
  thus union(A/\B) c= union A /\ union B by ZFMISC_1:79;
  let e;
  assume
A1: e in union A /\ union B;
  then e in union A by XBOOLE_0:def 4;
  then consider a being set such that
A2: e in a and
A3: a in A by TARSKI:def 4;
A4: a in D by A3;
  e in union B by A1,XBOOLE_0:def 4;
  then consider b being set such that
A5: e in b and
A6: b in B by TARSKI:def 4;
A7: b in D by A6;
  not a misses b by A2,A5,XBOOLE_0:3;
  then a = b by A4,A7,Def4;
  then a in A/\B by A3,A6,XBOOLE_0:def 4;
  hence thesis by A2,TARSKI:def 4;
end;

theorem
  for D being a_partition of X, A being Subset of D, B being
  Subset of X st B = union A holds B` = union A`
proof
  let D be a_partition of X, A be Subset of D, B be Subset of X;
  assume
A1: B = union A;
  union D = X by Def4;
  hence B` c= union A` by A1,Th55;
  let e;
  assume e in union A`;
  then consider u such that
A2: e in u and
A3: u in A` by TARSKI:def 4;
A4: u in D by A3;
  assume not e in B`;
  then e in B by A2,A4,SUBSET_1:29;
  then consider v being set such that
A5: e in v and
A6: v in A by A1,TARSKI:def 4;
A7: v in D by A6;
  not u misses v by A2,A5,XBOOLE_0:3;
  then u = v by A4,A7,Def4;
  hence contradiction by A3,A6,XBOOLE_0:def 5;
end;

theorem 
  for E being Equivalence_Relation of X holds Class(E) is non empty;

registration
  let X be non empty set;
  cluster non empty for a_partition of X;
  existence
  proof

    reconsider P = Class nabla X as a_partition of X;
    take P;
    thus thesis;
  end;
end;

definition
  let X;
  let D be non empty a_partition of X;
  func proj D -> Function of X, D means
  :Def9:
  for p being Element of X holds p in it.p;
  existence
  proof
    defpred P[set,set] means $1 in $2;
A1: now
A2:   union D = X by Def4;
      let e;
      assume e in X;
      then ex u st e in u & u in D by A2,TARSKI:def 4;
      hence ex u st u in D & P[e,u];
    end;
    consider F being Function of X, D such that
A3: for e st e in X holds P[e,F.e] from FUNCT_2:sch 1(A1);
    take F;
    let p be Element of X;
    thus thesis by A3;
  end;
  correctness
  proof
    let F,G be Function of X,D such that
A4: ( for p being Element of X holds p in F.p)& for p being Element of
    X holds p in G.p;
    now
      let x be Element of X;
      x in F.x & x in G.x by A4;
      then
A5:   not F.x misses G.x by XBOOLE_0:3;
      F.x is Subset of X & G.x is Subset of X by TARSKI:def 3;
      hence F.x = G.x by A5,Def4;
    end;
    hence F = G by FUNCT_2:63;
  end;
end;

theorem Th65:
  for D being non empty a_partition of X, p being Element of X, A
  being Element of D st p in A holds A = (proj D).p
proof
  let D be non empty a_partition of X, p be Element of X, A be Element of D
  such that
A1: p in A;
  p in (proj D).p by Def9;
  then (proj D).p is Subset of X & not (proj D).p misses A by A1,TARSKI:def 3
,XBOOLE_0:3;
  hence thesis by Def4;
end;

theorem
  for D being non empty a_partition of X, p being Element of D
  holds p = proj D " {p}
proof
  let D be non empty a_partition of X, p be Element of D;
  thus p c= proj D " {p}
  proof
    let e;
    assume
A1: e in p;
    then (proj D).e = p by Th65;
    then (proj D).e in {p} by TARSKI:def 1;
    hence thesis by A1,FUNCT_2:38;
  end;
  let e;
  assume
A2: e in proj D " {p};
  then (proj D).e in {p} by FUNCT_1:def 7;
  then (proj D).e = p by TARSKI:def 1;
  hence e in p by A2,Def9;
end;

theorem
  for D being non empty a_partition of X, A being Subset of D
  holds (proj D)"A = union A
proof
  let D be non empty a_partition of X, A be Subset of D;
  thus (proj D)"A c= union A
  proof
    let e;
    assume e in (proj D)"A;
    then (proj D).e in A & e in (proj D).e by Def9,FUNCT_2:38;
    hence thesis by TARSKI:def 4;
  end;
  let e;
  assume e in union A;
  then consider u such that
A1: e in u and
A2: u in A by TARSKI:def 4;
A3: u in D by A2;
  then (proj D).e = u by A1,Th65;
  hence thesis by A1,A2,A3,FUNCT_2:38;
end;

theorem
  for D being non empty a_partition of X, W being Element of D ex
  W9 being Element of X st proj(D).W9=W
proof
  let D be non empty a_partition of X, W be Element of D;
  reconsider p = W as Subset of X;
  p <> {} by Def4;
  then consider W9 being Element of X such that
A1: W9 in p by SUBSET_1:4;
  take W9;
  thus thesis by A1,Th65;
end;

theorem
  for D being non empty a_partition of X, W being Subset of X st
for B being Subset of X st B in D & B meets W holds B c= W holds W = proj D " (
  proj D .: W)
proof
  let D be non empty a_partition of X, W be Subset of X such that
A1: for B being Subset of X st B in D & B meets W holds B c= W;
  W c= X;
  then W c= dom proj D by FUNCT_2:def 1;
  hence W c= proj D " (proj D .: W) by FUNCT_1:76;
  let e;
  assume
A2: e in proj D " (proj D .: W);
  then reconsider d = e as Element of X;
  (proj D).e in proj D .: W by A2,FUNCT_1:def 7;
  then consider c being Element of X such that
A3: c in W and
A4: (proj D).d = (proj D).c by FUNCT_2:65;
  reconsider B = (proj D).c as Subset of X by TARSKI:def 3;
  c in (proj D).c by Def9;
  then B meets W by A3,XBOOLE_0:3;
  then
A5: B c= W by A1;
  d in B by A4,Def9;
  hence thesis by A5;
end;

theorem
  for X being set, P being a_partition of X, x, a, b being set st
    x in a & a in P & x in b & b in P holds a = b
  proof
    let X be set, P be a_partition of X, x,a,b be set;
    assume that
A1: x in a and
A2: a in P and
A3: x in b and
A4: b in P;
    a meets b by A1,A3,XBOOLE_0:3;
    hence thesis by A2,A4,Def4;
  end;















begin 

reserve n,n1,n2,m for Element of NAT;
reserve r,g1,g2,g,g9 for Element of COMPLEX;
reserve R,R2 for Real;
reserve s,s9,s1 for Complex_Sequence;

theorem Th1:
  g<>0c & r<>0c implies |.g"-r".|=|.g-r.|/(|.g.|*|.r.|)
proof
  assume
A1: g<>0c & r<>0c;
  thus |.g"-r".|=|.1r/g-r".| by COMPLEX1:def 4,XCMPLX_1:215
    .=|.1r/g-1r/r.| by COMPLEX1:def 4,XCMPLX_1:215
    .=|.1r/g+-1r/r.|
    .=|.1r/g+-1r*r".| by XCMPLX_0:def 9
    .=|.1r/g+(-1r)*r".|
    .=|.1r/g+(-1r)/r.| by XCMPLX_0:def 9
    .=|.(1r*r+(-1r)*g)/(r*g).| by A1,XCMPLX_1:116
    .=|.r-g.|/|.g*r.| by COMPLEX1:67,def 4
    .=|.-(g-r).|/(|.g.|*|.r.|) by COMPLEX1:65
    .=|.g-r.|/(|.g.|*|.r.|) by COMPLEX1:52;
end;

theorem Th2:
  for n ex r being Real st 0<r & for m st m<=n holds |.s.m.|<r
proof
  let n;
  defpred P[Element of NAT] means ex r being real number st 0<r & for m st m<=
  $1 holds |.s.m.|<r;
A1: for n st P[n] holds P[n+1]
  proof
    let n;
    given R1 be real number such that
A2: 0<R1 and
A3: for m st m <= n holds |.s.m.| < R1;
A4: now
      assume
A5:   R1 <= |.s.(n+1).|;
      take R= |.s.(n+1).| +1;
      0+0<R by COMPLEX1:46,XREAL_1:8;
      hence 0<R;
      let m such that
A6:   m<=n+1;
A7:   now
        assume m<=n;
        then |.s.m.| <R1 by A3;
        then |.s.m.| < |.s.(n+1).| by A5,XXREAL_0:2;
        then |.s.m.| + 0 < R by XREAL_1:8;
        hence |.s.m.| < R;
      end;
      now
        assume m=n+1;
        then |.s.m.| +0 <R by XREAL_1:8;
        hence |.s.m.| < R;
      end;
      hence |.s.m.| <R by A6,A7,NAT_1:8;
    end;
    now
      assume
A8:   |.s.(n+1).| <= R1;
      take R=R1+1;
      thus R>0 by A2;
      let m such that
A9:   m <= n+1;
A10:  now
        assume m <= n;
        then
A11:    |.s.m.| < R1 by A3;
        R1+0<R by XREAL_1:8;
        hence |.s.m.| < R by A11,XXREAL_0:2;
      end;
      now
A12:    R1+0 <R by XREAL_1:8;
        assume m=n+1;
        hence |.s.m.| < R by A8,A12,XXREAL_0:2;
      end;
      hence |.s.m.| < R by A9,A10,NAT_1:8;
    end;
    hence thesis by A4;
  end;
A13: P[0]
  proof
    take R=|.s.0.|+1;
    0+0 < |.s.0.| +1 by COMPLEX1:46,XREAL_1:8;
    hence 0<R;
    let m;
    assume m <= 0;
    then m=0 by NAT_1:2;
    then |.s.m.| +0 < R by XREAL_1:8;
    hence thesis;
  end;
  for n holds P[n] from NAT_1:sch 1(A13,A1);
  then consider R being real number such that
A14: R>0 & for m st m<=n holds |.s.m.|<R;
  R is Real by XREAL_0:def 1;
  hence thesis by A14;
end;

begin 

definition
  let f be complex-valued Function;
  deffunc F(set) = (f.$1)*';
  func f*' -> complex-valued Function means
  :Def1:
  dom it = dom f &
  for c being set st c in dom it holds it.c = (f.c)*';
  existence
  proof
    consider F being Function such that
A1:  dom F = dom f and
A2:  for x being set st x in dom f holds F.x = F(x) from FUNCT_1:sch 3;
     for x being set st x in dom F holds F.x is complex
      proof let x be set;
       assume x in dom F;
        then F.x = F(x) by A1,A2;
       hence thesis;
      end;
     then reconsider F as complex-valued Function by VALUED_0:def 7;
    take F;
    thus dom f = dom F by A1;
    thus thesis by A1,A2;
  end;
  uniqueness
  proof
   let h,g be complex-valued Function such that
A3: dom h=dom f and
A4: for c be set st c in dom h holds h.c = F(c) and
A5: dom g=dom f and
A6: for c be set st c in dom g holds g.c = F(c);
   thus dom h=dom g by A3,A5;
   let x be set;
   assume
A7:  x in dom h;
   hence h.x = F(x) by A4
      .= g.x by A3,A5,A6,A7;
  end;
 involutiveness
  proof let g,f be complex-valued Function;
   assume that
A8:  dom g = dom f and
A9: for c being set st c in dom g holds g.c = (f.c)*';
  thus dom f = dom g by A8;
  let c be set;
  assume
A10: c in dom f;
   thus f.c = (f.c)*'*'
     .= (g.c)*' by A8,A10,A9;
  end;
end;

definition
  let C be non empty set;
  let f be Function of C,COMPLEX;
  redefine func f*' -> Function of C,COMPLEX means
  :Def2:
  dom it = C & for n being Element of C holds it.n=(f.n)*';
  coherence
   proof
A1:  dom(f*') = dom f by Def1
      .= C by PARTFUN1:def 2;
     for x being set st x in C holds (f*').x in COMPLEX by XCMPLX_0:def 2;
    hence f*' is Function of C,COMPLEX by A1,FUNCT_2:3;
   end;
  compatibility
  proof
    let IT be Function of C,COMPLEX;
    thus IT = f*' implies dom IT = C & for c being Element of C holds IT.c = (
    f.c)*'
    proof
      assume
A2:   IT = f*';
      thus
A3:   dom IT = C by FUNCT_2:def 1;
      let c be Element of C;
      thus thesis by A2,A3,Def1;
    end;
    assume that
A4:  dom IT = C and
A5: for c being Element of C holds IT.c = (f.c)*';
A6:  dom IT = dom f by A4,FUNCT_2:def 1;
     for c being set st c in dom IT holds IT.c = (f.c)*' by A5;
    hence IT = f*' by A6,Def1;
  end;
end;

registration
 let C be non empty set;
 let s be complex-valued ManySortedSet of C;
 cluster s*' -> C-defined;
 coherence
  proof
    dom s c= C;
   hence dom(s*') c= C by Def1;
  end;
end;

registration
  let C be non empty set;
  let seq be complex-valued ManySortedSet of C;
  cluster seq*' -> total for C-defined Function;
  coherence
  proof let f be C-defined Function such that
A1: f = seq*';
    thus dom f = dom seq by Def1,A1
      .= C by PARTFUN1:def 2;
  end;
end;

theorem
  s is non-zero implies s*' is non-zero
proof
  assume
A1: s is non-zero;
  now
    let n;
    (s.n)*' <>0c by A1,COMPLEX1:29;
    hence s*'.n <>0c by Def2;
  end;
  hence thesis by COMSEQ_1:4;
end;

theorem
  (r(#)s)*' = (r*')(#)(s*')
proof
  now
    let n;
    thus (r(#)s)*'.n = ((r(#)s).n)*' by Def2
      .= (r*s.n)*' by VALUED_1:6
      .= (r*')*(s.n)*' by COMPLEX1:35
      .= (r*')*(s*'.n) by Def2
      .= ((r*')(#)(s*')).n by VALUED_1:6;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  (s (#) s9)*' = (s*') (#) (s9*')
proof
  now
    let n;
    thus (s (#) s9)*'.n = ((s (#) s9).n)*' by Def2
      .= (s.n * s9.n)*' by VALUED_1:5
      .= (s.n)*' * (s9.n)*' by COMPLEX1:35
      .= (s*'.n) * (s9.n)*' by Def2
      .= (s*'.n) * (s9*'.n) by Def2
      .= (s*' (#) s9*').n by VALUED_1:5;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  (s*')" = (s")*'
proof
  now
    let n;
    thus (s*')".n = (s*'.n)" by VALUED_1:10
      .= ((s.n)*')" by Def2
      .= ((s.n)")*' by COMPLEX1:36
      .= (s".n)*' by VALUED_1:10
      .= (s")*'.n by Def2;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  (s9/"s)*' = (s9*') /" (s*')
proof
  now
    let n;
    thus (s9/"s)*'.n = ((s9 (#) s").n)*' by Def2
      .= ((s9.n) * (s".n))*' by VALUED_1:5
      .= ((s9.n) * (s.n)")*' by VALUED_1:10
      .= (s9.n)*' * ((s.n)")*' by COMPLEX1:35
      .= (s9.n)*' * ((s.n)*')" by COMPLEX1:36
      .= (s9*').n * ((s.n)*')" by Def2
      .= (s9*').n * ((s*').n)" by Def2
      .= (s9*').n * ((s*')").n by VALUED_1:10
      .= ((s9*') /" (s*')).n by VALUED_1:5;
  end;
  hence thesis by FUNCT_2:63;
end;

begin 

definition
  let f be complex-valued Function;
  attr f is bounded means

  ex r being real number st
   for y being set st y in dom f holds abs(f.y) <r;
end;

definition
  let s;
  redefine attr s is bounded means
  :Def4:
  ex r being Real st for n holds |.s.n .|<r;
  compatibility
  proof
    thus s is bounded implies ex r being Real st for n holds |.s.n.|<r
    proof
      given r being real number such that
A1:   for y being set st y in dom s holds abs(s.y) < r;
      reconsider r as Element of REAL by XREAL_0:def 1;
      take r;
      let n;
      n in NAT;
      then n in dom s by FUNCT_2:def 1;
      hence thesis by A1;
    end;
    given r being Real such that
A2: for n holds |.s.n.|<r;
    take r;
    let y being set;
    assume y in dom s;
    hence thesis by A2;
  end;
end;

reconsider sc = NAT --> 0c as Complex_Sequence;

registration
  cluster bounded for Complex_Sequence;
  existence
  proof
    take sc, 1;
    let n;
    thus thesis by COMPLEX1:44,FUNCOP_1:7;
  end;
end;

theorem Th8:
  s is bounded iff ex r being Real st (0<r & for n holds |.s.n.|<r)
proof
  thus s is bounded implies ex r being Real st (0<r & for n holds |.s.n.|<r)
  proof
    assume s is bounded;
    then consider r being Real such that
A1: for n holds |.s.n.|<r by Def4;
    take r;
    now
      let n;
      0 <= |.s.n.| by COMPLEX1:46;
      hence 0< r by A1;
    end;
    hence thesis by A1;
  end;
  thus thesis by Def4;
end;

begin




definition
  let s be complex-valued ManySortedSet of NAT;
  attr s is convergent means
  :Def5:
  ex g st for p be Real st 0<p ex n st for m st n<=m holds |.s.m-g.|<p;
end;

definition
  let s;
  assume
A1: s is convergent;
  func lim s -> Element of COMPLEX means
  :Def6:
  for p be Real st 0<p ex n st for m st n<=m holds |.s.m-it.|<p;
  existence by A1,Def5;
  uniqueness
  proof
    let g1,g2;
    assume that
A2: for p be Real st 0<p ex n1 st for m st n1<=m holds |.s.m-g1.|<p and
A3: for p be Real st 0<p ex n2 st for m st n2<=m holds |.s.m-g2.|<p and
A4: g1<>g2;
    reconsider p = |.g1 - g2.|/2 as Real;
A5: |.g1-g2.| > 0 by A4,COMPLEX1:62;
    then consider n1 such that
A6: for m st n1<=m holds |.s.m-g1.|<p by A2;
    consider n2 such that
A7: for m st n2<=m holds |.s.m-g2.|<p by A3,A5;
    reconsider n = max(n1,n2) as Element of NAT by FINSEQ_2:1;
    for m st n <= m holds 2*p<2*p
    proof
      let m;
      assume
A8:   n <= m;
      n2<=n by XXREAL_0:25;
      then n + n2 <= n + m by A8,XREAL_1:7;
      then n2<=m by XREAL_1:6;
      then
A9:   |.s.m - g2.|< p by A7;
      |.g1 - g2.| = |.g1 - s.m - (g2 - s.m).|;
      then |.g1 - g2.| <= |.g1-s.m.| + |.g2-s.m.| by COMPLEX1:57;
      then
A10:  |.g1 - g2.| <= |.s.m - g1.| + |.g2-s.m.| by COMPLEX1:60;
      n1<=n by XXREAL_0:25;
      then n + n1 <= n + m by A8,XREAL_1:7;
      then n1<=m by XREAL_1:6;
      then |.s.m - g1.|< p by A6;
      then |.s.m - g1.| + |.s.m - g2.| <p + p by A9,XREAL_1:8;
      hence thesis by A10,COMPLEX1:60;
    end;
    hence contradiction;
  end;
end;

theorem Th9:
  (ex g st for n being Element of NAT holds s.n = g) implies s is convergent
proof
  given g such that
A1: for n being Element of NAT holds s.n = g;
  take g;
  now
    let p be Real such that
A2: 0<p;
    take k = 0;
    let n such that
    k<=n;
    s.n = g by A1;
    hence |.s.n - g.| < p by A2,COMPLEX1:44;
  end;
  hence thesis;
end;

theorem Th10:
  for g st for n being Element of NAT holds s.n = g holds lim s = g
proof
  let g;
  assume
A1: for n being Element of NAT holds s.n = g;
A2: now
    let p be Real such that
A3: 0<p;
    take k = 0;
    let n such that
    k<=n;
    s.n = g by A1;
    hence |.s.n - g.| < p by A3,COMPLEX1:44;
  end;
  s is convergent by A1,Th9;
  hence thesis by A2,Def6;
end;

Lm1: for n holds sc.n = 0c by FUNCOP_1:7;

registration
  cluster convergent for Complex_Sequence;
  existence by Lm1,Th9;
end;

registration
  let s be convergent Complex_Sequence;
  cluster s*' -> convergent for Complex_Sequence;
  coherence
  proof let t be Complex_Sequence such that
A1:  t = s*';
    consider g such that
A2: for p being Real st 0<p ex n st for m st n<=m holds |.s.m-g.|<p by Def5;
     reconsider z = g*' as Element of COMPLEX;
    take r = g*';
    let p be Real;
    assume 0<p;
    then consider n such that
A3: for m st n<=m holds |.s.m-g.|<p by A2;
    take n;
    let m such that
A4: n<=m;
    |.(s*').m - r.| = |.(s.m)*' - g*'.| by Def2
      .= |.(s.m - g)*'.| by COMPLEX1:34
      .= |.s.m-g.| by COMPLEX1:53;
    hence thesis by A3,A4,A1;
  end;
end;

canceled;

theorem Th12:
  s is convergent implies lim(s*') = (lim s)*'
proof
  set g = lim s;
  assume
A1: s is convergent;
  then reconsider s1 = s as convergent Complex_Sequence;
A2: now
    let p be Real;
    assume 0<p;
    then consider n such that
A3: for m st n<=m holds |.s.m-g.|<p by A1,Def6;
    take n;
    let m such that
A4: n<=m;
    |.(s*').m - g*'.| = |.(s.m)*' - g*'.| by Def2
      .= |.(s.m - g)*'.| by COMPLEX1:34
      .= |.s.m - g.| by COMPLEX1:53;
    hence |.(s*').m - (lim s)*'.| < p by A3,A4;
  end;
  s1*' is convergent;
  hence thesis by A2,Def6;
end;

begin 

registration
  let s1,s2 be convergent Complex_Sequence;
  cluster s1 + s2 -> convergent for Complex_Sequence;
  coherence
proof let s be Complex_Sequence such that
A1: s = s1 + s2;
  consider g such that
A2: for p be Real st 0<p ex n st for m st n<=m holds |.s1.m-g.|<p by Def5;
  consider g9 such that
A3: for p be Real st 0<p ex n st for m st n<=m holds |.s2.m-g9.|<p by Def5;
  take g1 = g + g9;
  let p be Real;
  assume
A4: p>0;
  then consider n1 such that
A5: for m st n1<=m holds |.s1.m - g.|<p/2 by A2;
  consider n2 such that
A6: for m st n2<=m holds |.s2.m - g9.|<p/2 by A3,A4;
  reconsider n = max(n1,n2) as Element of NAT by FINSEQ_2:1;
  take n;
  let m such that
A7: n<=m;
  n2<=n by XXREAL_0:25;
  then n + n2 <= n + m by A7,XREAL_1:7;
  then n2<=m by XREAL_1:6;
  then
A8: |.s2.m - g9.|<p/2 by A6;
A9: |.(s1 + s2).m - g1.| = |.s1.m + s2.m - (g + g9).| by VALUED_1:1
    .= |.(s1.m - g) + (s2.m - g9).|;
  n1<=n by XXREAL_0:25;
  then n + n1 <= n + m by A7,XREAL_1:7;
  then n1<=m by XREAL_1:6;
  then |.s1.m - g.|<p/2 by A5;
  then |.s1.m - g.| + |.s2.m - g9.| < p/2 + p/2 by A8,XREAL_1:8;
  then |.s1.m - g.| + |.s2.m - g9.| + |.(s1 + s2).m - g1.|
     < p + (|.s1.m - g.| + |.s2.m - g9.|) by A9,COMPLEX1:56,XREAL_1:8;
  hence thesis by A1,XREAL_1:6;
end;
end;

canceled;

theorem Th14:
 for s,s9 being convergent Complex_Sequence
  holds lim (s + s9)=(lim s)+ (lim s9)
proof
 let s,s9 be convergent Complex_Sequence;
  for p be Real st 0<p ex n st for m st n<=m holds |.(s + s9).m-((lim s)+(
  lim s9)).|<p
  proof
    let p be Real;
    assume 0<p;
    then
A1: 0<p/2;
    then consider n1 such that
A2: for m st n1<=m holds |.s.m-lim s.|<p/2 by Def6;
    consider n2 such that
A3: for m st n2<=m holds |.s9.m-lim s9.|<p/2 by A1,Def6;
    reconsider n = max(n1,n2) as Element of NAT by FINSEQ_2:1;
    take n;
    let m such that
A4: n<=m;
    n2<=n by XXREAL_0:25;
    then n + n2 <= n + m by A4,XREAL_1:7;
    then n2<=m by XREAL_1:6;
    then
A5: |.s9.m-lim s9.|<p/2 by A3;
A6: |.(s + s9).m-((lim s)+(lim s9)).| = |.s.m + s9.m - ((lim s)+(lim s9))
    .| by VALUED_1:1
      .=|.(s.m -lim s) + (s9.m - lim s9).|;
    n1<=n by XXREAL_0:25;
    then n + n1 <= n + m by A4,XREAL_1:7;
    then n1<=m by XREAL_1:6;
    then |.s.m-lim s.|<p/2 by A2;
    then |.s.m-lim s.| + |.s9.m-lim s9.| < p/2 + p/2 by A5,XREAL_1:8;
    then |.s.m-lim s.| + |.s9.m -lim s9.| + |.(s + s9).m-((lim s)+(lim s9)).|
    < p + (|.s.m-lim s.| + |.s9.m -lim s9.|) by A6,COMPLEX1:56,XREAL_1:8;
    hence thesis by XREAL_1:6;
  end;
  hence thesis by Def6;
end;

canceled;

theorem
 for s,s9 being convergent Complex_Sequence
  holds lim (s + s9)*' = (lim s)*' + (lim s9)*'
proof
 let s,s9 be convergent Complex_Sequence;
  thus lim (s + s9)*' = (lim (s + s9))*' by Th12
    .= ((lim s) + (lim s9))*' by Th14
    .= (lim s)*' + (lim s9)*' by COMPLEX1:32;
end;

registration
  let s be convergent Complex_Sequence, c be complex number;
  cluster c(#)s -> convergent for Complex_Sequence;
  coherence
  proof
A1: now
      let c be Element of COMPLEX;
      per cases;
      suppose
A2:     c =0c;
        now
          let n;
          thus (c(#)s).n = 0c*s.n by A2,VALUED_1:6
            .= 0c;
        end;
        hence c(#)s is convergent by Th9;
      end;
      suppose
A3:     c <> 0c;
        thus c(#)s is convergent
        proof
          consider g such that
A4:       for p be Real st 0<p ex n st for m st n<=m holds |.s.m-g.|<
          p by Def5;
          take g9= c*g;
          let p be Real such that
A5:       0<p;
A6:       |.c.|>0 by A3,COMPLEX1:47;
          consider n such that
A7:       for m st n<=m holds |.s.m-g.|<p/|.c.| by A6,A4,A5;
          take n;
          let m;
          assume n<=m;
          then |.s.m-g.|<p/|.c.| by A7;
          then |.s.m-g.|*|.c.|<p / |.c.|*|.c.| by A6,XREAL_1:68;
          then |.s.m-g.|*|.c.|<p * (|.c.|)"*|.c.| by XCMPLX_0:def 9;
          then |.s.m-g.|*|.c.|<p * ((|.c.|)"*|.c.|);
          then
A8:       |.s.m-g.|*|.c.|<p * 1 by A6,XCMPLX_0:def 7;
          |.(c(#)s).m-g9.| = |.c*s.m-c*g.| by VALUED_1:6
            .= |.c*(s.m-g).|
            .= |.s.m-g.|*|.c.| by COMPLEX1:65;
          hence thesis by A8;
        end;
      end;
    end;
    c in COMPLEX by XCMPLX_0:def 2;
    hence thesis by A1;
  end;
end;

canceled;

theorem Th18:
  for s being convergent Complex_Sequence, r being complex number
   holds lim(r(#)s)=r*(lim s)
proof
  let s being convergent Complex_Sequence, r be complex number;
  reconsider r as Element of COMPLEX by XCMPLX_0:def 2;
  per cases;
   suppose
A1: r<>0c;
    for p be Real st p>0 holds ex n st for m st n<=m holds |.(r(#)s)
    .m-r*(lim s).| <p
    proof
      let p be Real such that
A2:   p>0;
A3:   |.r.|>0 by A1,COMPLEX1:47;
      p / |.r.| > 0 by A3,A2;
      then consider n such that
A4:   for m st n<=m holds |.s.m-(lim s).|< p/|.r.| by Def6;
      take n;
      let m;
      assume n<=m;
      then |.s.m-(lim s).|<p/|.r.| by A4;
      then |.s.m-(lim s).|*|.r.|<p / |.r.|*|.r.| by A3,XREAL_1:68;
      then |.s.m-(lim s).|*|.r.|< p * (|.r.|)"*|.r.| by XCMPLX_0:def 9;
      then |.s.m-(lim s).|*|.r.|< p * ((|.r.|)"*|.r.|);
      then
A5:   |.s.m-(lim s).|*|.r.|< p * 1 by A3,XCMPLX_0:def 7;
      |.(r(#)s).m - r*(lim s).|=|.r*s.m - r*(lim s).| by VALUED_1:6
        .= |. r *(s.m-(lim s)).|
        .= |.s.m-(lim s).|*|.r.| by COMPLEX1:65;
      hence thesis by A5;
    end;
    hence thesis by Def6;
   end;
  suppose
A6: r=0c;
    now
      let n;
      thus (r(#)s).n = 0c*s.n by A6,VALUED_1:6
        .= 0c;
    end;
    hence thesis by A6,Th10;
   end;
end;

canceled;

theorem
  for s being convergent Complex_Sequence
  holds lim (r(#)s)*' = (r*')*(lim s)*'
proof
  let s being convergent Complex_Sequence;
  thus lim (r(#)s)*' = (lim (r(#)s))*' by Th12
    .= (r*(lim s))*' by Th18
    .= (r*')*(lim s)*' by COMPLEX1:35;
end;

registration
  let s be convergent Complex_Sequence;
  cluster -s -> convergent for Complex_Sequence;
  coherence;
end;

canceled;

theorem Th22:
  for s being convergent Complex_Sequence
  holds lim(-s)=-(lim s)
proof
  let s being convergent Complex_Sequence;
   lim(-s) = (-1)*(lim s) by Th18
    .= - 1r*(lim s) by COMPLEX1:def 4;
  hence thesis by COMPLEX1:def 4;
end;

canceled;

theorem
 for s being convergent Complex_Sequence
  holds lim (-s)*' = -(lim s)*'
proof
  let s being convergent Complex_Sequence;
  thus lim (-s)*' = (lim (-s))*' by Th12
    .= (-(lim s))*' by Th22
    .= -(lim s)*' by COMPLEX1:33;
end;

registration
  let s1,s2 be convergent Complex_Sequence;
  cluster s1 - s2 -> convergent for Complex_Sequence;
  coherence
proof
  - s2 is convergent;
  hence thesis;
end;
end;

canceled;

theorem Th26:
  for s,s9 being convergent Complex_Sequence
  holds lim(s - s9)=(lim s)-( lim s9)
proof
  let s,s9 be convergent Complex_Sequence;
  lim(s - s9) = (lim s) + lim(-s9) by Th14
    .= (lim s) + -(lim s9) by Th22;
  hence thesis;
end;

canceled;

theorem
  for s,s9 being convergent Complex_Sequence
  holds lim (s - s9)*' = (lim s)*' - (lim s9)*'
proof
  let s,s9 be convergent Complex_Sequence;
  thus lim (s - s9)*' = (lim (s - s9))*' by Th12
    .= ((lim s) - (lim s9))*' by Th26
    .= (lim s)*' - (lim s9)*' by COMPLEX1:34;
end;

registration
  cluster convergent -> bounded for Complex_Sequence;
  coherence
  proof
    let s;
    assume s is convergent;
    then consider g such that
A1: for p be Real st 0<p ex n st for m st n<=m holds |.s.m-g.|<p by Def5;
    consider n1 such that
A2: for m st n1<=m holds |.s.m-g.|<1 by A1;
    now
      take R=|.g.| +1;
      0+0 < R by COMPLEX1:46,XREAL_1:8;
      hence 0<R;
      let m;
A3:   |.s.m.|- |.g.|+|.g.|=|.s.m.|;
      assume n1<=m;
      then
A4:   |.s.m-g.|<1 by A2;
      |.s.m.| - |.g.| <= |.s.m-g.| by COMPLEX1:59;
      then |.s.m.| - |.g.| < 1 by A4,XXREAL_0:2;
      hence |.s.m.|<R by A3,XREAL_1:6;
    end;
    then consider R1 be real number such that
A5: 0<R1 and
A6: for m st n1<=m holds |.s.m.|< R1;
    consider R2 such that
A7: 0<R2 and
A8: for m st m<=n1 holds |.s.m.|< R2 by Th2;
    take R=R1+R2;
    let m;
A9: R2+0 < R by A5,XREAL_1:8;
A10: now
      assume m<=n1;
      then |.s.m.|< R2 by A8;
      hence thesis by A9,XXREAL_0:2;
    end;
A11: R1+0 < R by A7,XREAL_1:8;
    now
      assume n1<=m;
      then |.s.m.|< R1 by A6;
      hence thesis by A11,XXREAL_0:2;
    end;
    hence thesis by A10;
  end;
end;

registration
  cluster non bounded -> non convergent for Complex_Sequence;
  coherence;
end;

registration
  let s1,s2 be convergent Complex_Sequence;
  cluster s1 (#) s2 -> convergent for Complex_Sequence;
  coherence
proof let s be Complex_Sequence such that
A1: s = s1 (#) s2;
  consider g1 such that
A2: for p being Real st 0<p ex n st for m st n<=m holds |.s1.m-g1.|<p by Def5;
  consider g2 such that
A3: for p being Real st 0<p ex n st for m st n<=m holds |.s2.m-g2.|<p by Def5;
  take g=g1*g2;
  let p be Real;
  consider R such that
A4: 0<R and
A5: for n holds |.s1.n.|<R by Th8;
A6: 0+0<|.g2.|+R by A4,COMPLEX1:46,XREAL_1:8;
  assume
A7: 0<p;
  then consider n1 such that
A8: for m st n1<=m holds |.s1.m-g1.|<p/(|.g2.|+R) by A2,A6;
  consider n2 such that
A9: for m st n2<=m holds |.s2.m-g2.|<p/(|.g2.|+R) by A3,A6,A7;
  take n=n1+n2;
  let m such that
A10: n<=m;
  n1<=n1+n2 by NAT_1:12;
  then n1<=m by A10,XXREAL_0:2;
  then
A11: |.s1.m-g1.|<=p/(|.g2.|+R) by A8;
  0<=|.g2.| & |.(s1.m-g1)*g2.|=|.g2.|*|.s1.m-g1.| by COMPLEX1:46,65;
  then
A12: |.(s1.m-g1)*g2.|<=|.g2.|*(p/(|.g2.|+R)) by A11,XREAL_1:64;
  |.((s1(#)s2).m)-g.|=|.(s1.m*s2.m) -s1.m*g2+s1.m*g2-g1*g2.| by VALUED_1:5
    .=|.s1.m*(s2.m-g2)+(s1.m-g1)*g2.|;
  then
A13: |.((s1(#)s2).m)-g.|<= |.s1.m*(s2.m-g2).|+|.(s1.m-g1)*g2.| by COMPLEX1:56;
  n2<=n by NAT_1:12;
  then n2<=m by A10,XXREAL_0:2;
  then
A14: |.s2.m-g2.|<p/(|.g2.|+R) by A9;
A15: 0<=|.s1.m.| & 0<=|.s2.m-g2.| by COMPLEX1:46;
  |.s1.m.|<R by A5;
  then |.s1.m.|*|.s2.m-g2.|<R*(p/(|.g2.|+R)) by A15,A14,XREAL_1:96;
  then
A16: |.s1.m*(s2.m-g2).|<R*(p/(|.g2.|+R)) by COMPLEX1:65;
  R*(p/(|.g2.|+R))+|.g2.|*(p/(|.g2.|+R)) =(p/(|.g2.|+R))*(|.g2.|+R)
    .=p by A6,XCMPLX_1:87;
  then |.s1.m*(s2.m-g2).|+|.(s1.m-g1)*g2.|<p by A16,A12,XREAL_1:8;
  hence thesis by A1,A13,XXREAL_0:2;
end;
end;

canceled;

theorem Th30:
 for s,s9 being convergent Complex_Sequence
  holds lim(s(#)s9)=(lim s)*(lim s9)
proof
  let s,s9 be convergent Complex_Sequence;
  consider R such that
A1: 0<R and
A2: for n holds |.s.n.|<R by Th8;
A3: 0+0<|.(lim s9).|+R by A1,COMPLEX1:46,XREAL_1:8;
A4: 0<=|.(lim s9).| by COMPLEX1:46;
  now
    let p be Real;
    assume 0<p;
    then
A5: 0<p/(|.(lim s9).|+R) by A3;
    then consider n1 such that
A6: for m st n1<=m holds |.s.m-(lim s).|<p/(|.(lim s9).|+R) by Def6;
    consider n2 such that
A7: for m st n2<=m holds |.s9.m-(lim s9).|<p/(|.(lim s9).|+R) by A5,Def6;
    take n=n1+n2;
    let m such that
A8: n<=m;
    n1<=n1+n2 by NAT_1:12;
    then n1<=m by A8,XXREAL_0:2;
    then
A9: |.s.m-(lim s).|<=p/(|.(lim s9).|+R) by A6;
    |.(s.m-(lim s))*(lim s9).| =|.(lim s9).|*|.s.m-(lim s).| by COMPLEX1:65;
    then
A10: |.(s.m-(lim s))*(lim s9).|<=|.(lim s9).|* (p/(|.(lim s9).|+R)) by A4,A9,
XREAL_1:64;
A11: 0<=|.s.m.| & 0<=|.s9.m-(lim s9).| by COMPLEX1:46;
    n2<=n by NAT_1:12;
    then n2<=m by A8,XXREAL_0:2;
    then
A12: |.s9.m-(lim s9).|<p/(|.(lim s9).|+R) by A7;
    |.((s(#)s9).m)-(lim s)*(lim s9).| =|.(s.m*s9.m-s.m*(lim s9)+s.m*(lim
    s9))- (lim s)*(lim s9).| by VALUED_1:5
      .=|.s.m*(s9.m-(lim s9))+ (s.m-(lim s))*(lim s9).|;
    then
A13: |.((s(#)s9).m)-(lim s)*(lim s9).|<= |.s.m*(s9.m-(lim s9)).|+|.(s.m-(
    lim s))*(lim s9).| by COMPLEX1:56;
    |.s.m.|<R by A2;
    then |.s.m.|*|.s9.m-(lim s9).|<R*(p/(|.(lim s9).|+R)) by A11,A12,XREAL_1:96
;
    then
A14: |.s.m*(s9.m-(lim s9)).|<R*(p/(|.(lim s9).|+R)) by COMPLEX1:65;
    R*(p/(|.(lim s9).|+R))+|.(lim s9).|*(p/(|.(lim s9).|+R)) =(p/(|.(lim
    s9).|+R))*(|.(lim s9).|+R)
      .=p by A3,XCMPLX_1:87;
    then |.s.m*(s9.m-(lim s9)).|+|.(s.m-(lim s))* (lim s9).|<p by A14,A10,
XREAL_1:8;
    hence |.((s(#)s9).m)-(lim s)*(lim s9).|<p by A13,XXREAL_0:2;
  end;
  hence thesis by Def6;
end;

canceled;

theorem
  for s,s9 being convergent Complex_Sequence
  holds lim (s(#)s9)*' = (lim s)*' * (lim s9)*'
proof
  let s,s9 be convergent Complex_Sequence;
  thus lim (s(#)s9)*' = (lim (s(#)s9))*' by Th12
    .= ((lim s) * (lim s9))*' by Th30
    .= (lim s)*' * (lim s9)*' by COMPLEX1:35;
end;

theorem Th33:
  for s being convergent Complex_Sequence st lim s <> 0c
   ex n st for m st n <=m holds |.(lim s).|/2<|.s.m.|
proof
  let s be convergent Complex_Sequence such that
A1: (lim s)<>0c;
  0<|.(lim s).| by A1,COMPLEX1:47;
  then 0<|.(lim s).|/2;
  then consider n1 such that
A2: for m st n1<=m holds |.s.m-(lim s).|<|.(lim s).|/2 by Def6;
  take n=n1;
  let m;
  assume n<=m;
  then
A3: |.s.m-(lim s).|<|.(lim s).|/2 by A2;
A4: |.(lim s)-s.m.|=|.-(s.m-(lim s)).| .=|.s.m-(lim s).| by COMPLEX1:52;
A5: |.(lim s).|/2+(|.s.m.|- |.(lim s).|/2) =|.s.m.| & |.(lim s).|- |.s.m.|+
  (|.s. m.|- |.(lim s).|/2) =|.(lim s).|/2;
  |.(lim s).|- |.s.m.|<=|.(lim s)-s.m.| by COMPLEX1:59;
  then |.(lim s).|- |.s.m.|<|.(lim s).|/2 by A3,A4,XXREAL_0:2;
  hence thesis by A5,XREAL_1:6;
end;

theorem Th34:
  for s being convergent Complex_Sequence st lim s <> 0c & s is non-zero
   holds s" is convergent
proof
  let s be convergent Complex_Sequence;
  assume that
A1: (lim s)<>0c and
A2: s is non-zero;
  consider n1 such that
A3: for m st n1<=m holds |.(lim s).|/2<|.s.m.| by A1,Th33;
  take(lim s)";
  let p be Real;
  assume
A4: 0<p;
A5: 0<|.(lim s).| by A1,COMPLEX1:47;
  then 0*0<|.(lim s).|*|.(lim s).|;
  then 0<(|.(lim s).|*|.(lim s).|)/2;
  then 0*0<p*((|.(lim s).|*|.(lim s).|)/2) by A4;
  then consider n2 such that
A6: for m st n2<=m holds |.s.m-(lim s).|<p*((|.(lim s).|*|.(lim s).|)/2
  ) by Def6;
  take n=n1+n2;
  let m such that
A7: n<=m;
  n1<=n1+n2 by NAT_1:12;
  then n1<=m by A7,XXREAL_0:2;
  then
A8: |.(lim s).|/2<|.s.m.| by A3;
A9: 0<|.(lim s).|/2 by A5;
  then 0*0<p*(|.(lim s).|/2) by A4;
  then
A10: (p*(|.(lim s).|/2))/|.s.m.|< (p*(|.(lim s).|/2))/(|.(lim s).|/2) by A8,A9
,XREAL_1:76;
A11: 0<>|.(lim s).|/2 by A1,COMPLEX1:47;
A12: (p*(|.(lim s).|/2))/(|.(lim s).|/2 ) =(p*(|.(lim s).|/2))*(|.(lim s).|/
  2 )" by XCMPLX_0:def 9
    .=p*((|.(lim s).|/2)*(|.(lim s).|/2 )")
    .=p*1 by A11,XCMPLX_0:def 7
    .=p;
A13: 0<>|.(lim s).| by A1,COMPLEX1:47;
A14: (p*((|.(lim s).|*|.(lim s).|)/2))/(|.s.m.|*|.(lim s).|) =(p*(2"*(|.(lim
  s).|*|.(lim s).|)))* (|.s.m.|*|.(lim s).|)" by XCMPLX_0:def 9
    .=p*2"*((|.(lim s).|*|.(lim s).|)*(|.(lim s).|*|.s.m.|)")
    .=p*2"*((|.(lim s).|*|.(lim s).|)* (|.(lim s).|"*|.s.m.|")) by XCMPLX_1:204
    .=p*2"*(|.(lim s).|*(|.(lim s).|*|.(lim s).|")*|.s.m.|")
    .=p*2"*(|.(lim s).|*1*|.s.m.|") by A13,XCMPLX_0:def 7
    .=p*(|.(lim s).|/2)*|.s.m.|"
    .=(p*(|.(lim s).|/2))/|.s.m.| by XCMPLX_0:def 9;
  s.m*(lim s)<>0c by A1,A2;
  then 0<|.s.m*(lim s).| by COMPLEX1:47;
  then
A15: 0<|.s.m.|*|.(lim s).| by COMPLEX1:65;
  n2<=n by NAT_1:12;
  then n2<=m by A7,XXREAL_0:2;
  then |.s.m-(lim s).|<p*((|.(lim s).|*|.(lim s).|)/2) by A6;
  then
A16: |.s.m-(lim s).|/(|.s.m.|*|.(lim s).|)< (p*((|.(lim s).|*|.(lim s).|)/2)
  )/(|.s.m.|*|.(lim s).|) by A15,XREAL_1:74;
  |.(s").m-(lim s)".|=|.(s.m)"-(lim s)".| by VALUED_1:10
    .=|.s.m-(lim s).|/(|.s.m.|*|.(lim s).|) by A1,A2,Th1;
  hence thesis by A16,A14,A10,A12,XXREAL_0:2;
end;

theorem Th35:
  s is convergent & (lim s)<>0c & s is non-zero implies lim s"=( lim s)"
proof
  assume that
A1: s is convergent and
A2: (lim s)<>0c and
A3: s is non-zero;
  consider n1 such that
A4: for m st n1<=m holds |.(lim s).|/2<|.s.m.| by A1,A2,Th33;
A5: 0<|.(lim s).| by A2,COMPLEX1:47;
  then 0*0<|.(lim s).|*|.(lim s).|;
  then
A6: 0<(|.(lim s).|*|.(lim s).|)/2;
A7: 0<>|.(lim s).| by A2,COMPLEX1:47;
A8: now
    let p be Real;
A9: 0<>|.(lim s).|/2 by A2,COMPLEX1:47;
A10: (p*(|.(lim s).|/2))/(|.(lim s).|/2 ) =(p*(|.(lim s).|/2))*(|.(lim s)
    .|/2 )" by XCMPLX_0:def 9
      .=p*((|.(lim s).|/2)*(|.(lim s).|/2 )")
      .=p*1 by A9,XCMPLX_0:def 7
      .=p;
    assume
A11: 0<p;
    then 0*0<p*((|.(lim s).|*|.(lim s).|)/2) by A6;
    then consider n2 such that
A12: for m st n2<=m holds |.s.m-(lim s).|<p*((|.(lim s).|*|.(lim s).|)
    /2) by A1,Def6;
    take n=n1+n2;
    let m such that
A13: n<=m;
    n1<=n1+n2 by NAT_1:12;
    then n1<=m by A13,XXREAL_0:2;
    then
A14: |.(lim s).|/2<|.s.m.| by A4;
A15: 0<|.(lim s).|/2 by A5;
    then 0*0<p*(|.(lim s).|/2) by A11;
    then
A16: (p*(|.(lim s).|/2))/|.s.m.|< (p*(|.(lim s).|/2))/(|.(lim s).|/2) by A14
,A15,XREAL_1:76;
A17: (p*((|.(lim s).|*|.(lim s).|)/2))/(|.s.m.|*|.(lim s).|) =(p*(2"*(|.(
    lim s).|*|.(lim s).|)))* (|.s.m.|*|.(lim s).|)" by XCMPLX_0:def 9
      .=p*2"*((|.(lim s).|*|.(lim s).|)*(|.(lim s).|*|.s.m.|)")
      .=p*2"*((|.(lim s).|*|.(lim s).|)* (|.(lim s).|"*|.s.m.|")) by
XCMPLX_1:204
      .=p*2"*(|.(lim s).|*(|.(lim s).|*|.(lim s).|")*|.s.m.|")
      .=p*2"*(|.(lim s).|*1*|.s.m.|") by A7,XCMPLX_0:def 7
      .=p*(|.(lim s).|/2)*|.s.m.|"
      .=(p*(|.(lim s).|/2))/|.s.m.| by XCMPLX_0:def 9;
    s.m*(lim s)<>0c by A2,A3;
    then 0<|.s.m*(lim s).| by COMPLEX1:47;
    then
A18: 0<|.s.m.|*|.(lim s).| by COMPLEX1:65;
    n2<=n by NAT_1:12;
    then n2<=m by A13,XXREAL_0:2;
    then |.s.m-(lim s).|<p*((|.(lim s).|*|.(lim s).|)/2) by A12;
    then
A19: |.s.m-(lim s).|/(|.s.m.|*|.(lim s).|)< (p*((|.(lim s).|*|.(lim s).|)/
    2))/(|.s.m.|*|.(lim s).|) by A18,XREAL_1:74;
    |.(s").m-(lim s)".|=|.(s.m)"-(lim s)".| by VALUED_1:10
      .=|.s.m-(lim s).|/(|.s.m.|*|.(lim s).|) by A2,A3,Th1;
    hence |.(s").m-(lim s)".|<p by A19,A17,A16,A10,XXREAL_0:2;
  end;
  s" is convergent by A1,A2,A3,Th34;
  hence thesis by A8,Def6;
end;

canceled;

theorem
  s is convergent & (lim s)<>0c & s is non-zero implies lim (s")*' = ((
  lim s)*')"
proof
  assume
A1: s is convergent & (lim s)<>0c & s is non-zero;
  then s" is convergent by Th34;
  hence lim (s")*' = (lim s")*' by Th12
    .= ((lim s)")*' by A1,Th35
    .= ((lim s)*')" by COMPLEX1:36;
end;

theorem Th38:
  s9 is convergent & s is convergent & (lim s)<>0c & s is non-zero
  implies s9/"s is convergent
proof
  assume that
A1: s9 is convergent and
A2: s is convergent & (lim s)<>0c & s is non-zero;
  s" is convergent by A2,Th34;
  hence thesis by A1;
end;

theorem Th39:
  s9 is convergent & s is convergent & (lim s)<>0c & s is non-zero
  implies lim(s9/"s)=(lim s9)/(lim s)
proof
  assume that
A1: s9 is convergent and
A2: s is convergent & (lim s)<>0c & s is non-zero;
  s" is convergent by A2,Th34;
  then lim (s9(#)(s"))=(lim s9)*(lim s") by A1,Th30
    .=(lim s9)*(lim s)" by A2,Th35
    .=(lim s9)/(lim s) by XCMPLX_0:def 9;
  hence thesis;
end;

canceled;

theorem
  s9 is convergent & s is convergent & (lim s)<>0c & s is non-zero
  implies lim (s9/"s)*' = ((lim s9)*')/((lim s)*')
proof
  assume
A1: s9 is convergent & s is convergent & (lim s)<>0c & s is non-zero;
  then s9/"s is convergent by Th38;
  hence lim (s9/"s)*' = (lim (s9/"s))*' by Th12
    .= ((lim s9)/(lim s))*' by A1,Th39
    .= ((lim s9)*')/((lim s)*') by COMPLEX1:37;
end;

theorem Th42:
  s is convergent & s1 is bounded & (lim s)=0c implies s(#)s1 is convergent
proof
  assume that
A1: s is convergent and
A2: s1 is bounded and
A3: (lim s) = 0c;
  take g=0c;
  consider R such that
A4: 0<R and
A5: for m holds |.s1.m.|<R by A2,Th8;
  let p be Real such that
A6: 0<p;
A7: 0<p/R by A6,A4;
  then consider n1 such that
A8: for m st n1<=m holds |.s.m-0c.|<p/R by A1,A3,Def6;
  take n=n1;
  let m such that
A9: n<=m;
A10: |.((s(#)s1).m)-g.|=|.s.m*s1.m.| by VALUED_1:5
    .=|.s.m.|*|.s1.m.| by COMPLEX1:65;
  |.s.m.|=|.s.m-0c.|;
  then
A11: |.s.m.|<p/R by A8,A9;
  now
    (p/R)*R=p*R"*R by XCMPLX_0:def 9
      .=p*(R"*R)
      .=p*1 by A4,XCMPLX_0:def 7
      .=p;
    then
A12: (p/R)*|.s1.m.|<p by A5,A7,XREAL_1:68;
A13: 0<=|.s1.m.| by COMPLEX1:46;
    assume |.s1.m.|<>0;
    then |.((s(#)s1).m)-g.|<(p/R)*|.s1.m.| by A11,A10,A13,XREAL_1:68;
    hence thesis by A12,XXREAL_0:2;
  end;
  hence thesis by A6,A10;
end;

theorem Th43:
  s is convergent & s1 is bounded & (lim s)=0c implies lim(s(#)s1) =0c
proof
  assume that
A1: s is convergent and
A2: s1 is bounded and
A3: lim s=0c;
A4: now
    consider R such that
A5: 0<R and
A6: for m holds |.s1.m.|<R by A2,Th8;
    let p be Real such that
A7: 0<p;
A8: 0<p/R by A7,A5;
    then consider n1 such that
A9: for m st n1<=m holds |.s.m-0c.|<p/R by A1,A3,Def6;
    take n=n1;
    let m;
    (p/R)*R=p*R"*R by XCMPLX_0:def 9
      .=p*(R"*R)
      .=p*1 by A5,XCMPLX_0:def 7;
    then
A10: (p/R)*|.s1.m.|<p by A6,A8,XREAL_1:68;
    assume n<=m;
    then
A11: |.s.m-0c.|<p/R by A9;
A12: |.((s(#)s1).m)-0c.|=|.s.m*s1.m.| by VALUED_1:5
      .=|.s.m.|*|.s1.m.| by COMPLEX1:65;
A13: 0<=|.s1.m.| by COMPLEX1:46;
    now
      assume |.s1.m.|<>0;
      then |.((s(#)s1).m)-0c.|<(p/R)*|.s1.m.| by A11,A12,A13,XREAL_1:68;
      hence |.((s(#)s1).m)-0c.|<p by A10,XXREAL_0:2;
    end;
    hence |.((s(#)s1).m)-0c.|<p by A7,A12;
  end;
  s(#)s1 is convergent by A1,A2,A3,Th42;
  hence thesis by A4,Def6;
end;

canceled;

theorem
  s is convergent & s1 is bounded & (lim s)=0c implies lim (s(#)s1)*' = 0c
proof
  assume
A1: s is convergent & s1 is bounded & (lim s)=0c;
  then s(#)s1 is convergent by Th42;
  hence lim (s(#)s1)*' = (lim (s(#)s1))*' by Th12
    .= 0c by A1,Th43,COMPLEX1:28;
end;














begin

reserve n,n1,n2,m for Element of NAT,
  r,r1,r2,p,g1,g2,g for real number,
  seq,seq9,seq1 for Real_Sequence,
  y for set;

theorem Th1:
  -g<r & r<g iff abs(r)<g
proof
  thus -g<r & r<g implies abs(r)<g
  proof
    assume that
A1: -g<r and
A2: r<g;
A3: abs(r)<=g by A1,A2,ABSVALUE:5;
    now
      assume
A4:   abs(r)=g;
      now
        assume r<0;
        then g=-r by A4,ABSVALUE:def 1;
        hence contradiction by A1;
      end;
      hence contradiction by A2,A4,ABSVALUE:def 1;
    end;
    hence thesis by A3,XXREAL_0:1;
  end;
  assume
A5: abs(r)<g;
  then
A6: -g<=r by ABSVALUE:5;
A7: 0<=abs(r) by COMPLEX1:46;
A8: 0<g by A5,COMPLEX1:46;
A9: -g<-0 by A5,A7;
  now
    assume r=-g;
    then abs(r)=--g by A9,ABSVALUE:def 1
      .=g;
    hence contradiction by A5;
  end;
  hence -g<r by A6,XXREAL_0:1;
  thus thesis by A5,A8,ABSVALUE:def 1;
end;

theorem Th2:
  g<>0 & r<>0 implies abs(g"-r")=abs(g-r)/(abs(g)*abs(r))
proof
  assume that
A1: g<>0 and
A2: r<>0;
  thus abs(g"-r")=abs(1/g-r") by XCMPLX_1:215
    .=abs(1/g-1/r) by XCMPLX_1:215
    .=abs((1*r-1*g)/(g*r)) by A1,A2,XCMPLX_1:130
    .=abs(r-g)/abs(g*r) by COMPLEX1:67
    .=abs(-(g-r))/(abs(g)*abs(r)) by COMPLEX1:65
    .=abs(g-r)/(abs(g)*abs(r)) by COMPLEX1:52;
end;

definition
  let f be real-valued Function;
  attr f is bounded_above means
  :Def1:
  ex r st for y st y in dom f holds f.y<r;
  attr f is bounded_below means
  :Def2:
  ex r st for y st y in dom f holds r<f.y;
end;

definition
  let seq;
A1: dom seq = NAT by SEQ_1:1;
  redefine attr seq is bounded_above means
  :Def3:
  ex r st for n holds seq.n<r;
  compatibility
  proof
    thus seq is bounded_above implies ex r st for n holds seq.n<r
    proof
      given r such that
A2:   for y st y in dom seq holds seq.y<r;
      take r;
      let n;
      thus thesis by A1,A2;
    end;
    given r such that
A3: for n holds seq.n<r;
    take r;
    let y;
    assume y in dom seq;
    hence thesis by A1,A3;
  end;
  redefine attr seq is bounded_below means
  :Def4:
  ex r st for n holds r<seq.n;
  compatibility
  proof
    thus seq is bounded_below implies ex r st for n holds r < seq.n
    proof
      given r such that
A4:   for y st y in dom seq holds r < seq.y;
      take r;
      let n;
      thus thesis by A1,A4;
    end;
    given r such that
A5: for n holds r < seq.n;
    take r;
    let y;
    assume y in dom seq;
    hence thesis by A1,A5;
  end;
end;

registration
  cluster bounded -> bounded_above bounded_below for real-valued Function;
  coherence
  proof
    let f be real-valued Function;
    given r such that
A1: for y st y in dom f holds abs(f.y) <r;
    thus f is bounded_above
    proof
      take r;
      let y;
A2:   f.y <= abs(f.y) by ABSVALUE:4;
      assume y in dom f;
      hence thesis by A1,A2,XXREAL_0:2;
    end;
    take -abs r;
    let y;
A3: -abs(f.y) <= f.y by ABSVALUE:4;
    r <= abs r by ABSVALUE:4;
    then
A4: - abs r <= -r by XREAL_1:24;
    assume y in dom f;
    then abs(f.y) <r by A1;
    then -r < -abs(f.y) by XREAL_1:24;
    then -abs r < -abs(f.y) by A4,XXREAL_0:2;
    hence thesis by A3,XXREAL_0:2;
  end;
  cluster bounded_above bounded_below -> bounded for real-valued Function;
  coherence
  proof
    let f be real-valued Function;
    assume f is bounded_above;
    then consider r1 such that
A5: for y st y in dom f holds f.y<r1 by Def1;
    assume f is bounded_below;
    then consider r2 such that
A6: for y st y in dom f holds r2<f.y by Def2;
    take g=abs(r1)+abs(r2)+1;
A7: 0<=abs(r1) by COMPLEX1:46;
    let y such that
A8: y in dom f;
    r1<=abs(r1) by ABSVALUE:4;
    then f.y<abs(r1) by A5,A8,XXREAL_0:2;
    then f.y +0<abs(r1)+abs(r2) by COMPLEX1:46,XREAL_1:8;
    then
A9: f.y<g by XREAL_1:8;
    -abs(r2)<=r2 by ABSVALUE:4;
    then -abs(r2)<f.y by A6,A8,XXREAL_0:2;
    then -abs(r1)+-abs(r2)<0+f.y by A7,XREAL_1:8;
    then
A10: -abs(r1)-abs(r2)+-1<f.y+0 by XREAL_1:8;
    -abs(r1)-abs(r2)-1=-g;
    hence abs(f.y)<g by A9,A10,Th1;
  end;
end;

theorem Th3:
  seq is bounded iff ex r st (0<r & for n holds abs(seq.n)<r)
proof
  thus seq is bounded implies ex r st (0<r & for n holds abs(seq.n)<r)
  proof
    assume
A1: seq is bounded;
    then consider r1 such that
A2: for n holds seq.n<r1 by Def3;
    consider r2 such that
A3: for n holds r2<seq.n by A1,Def4;
    take g=abs(r1)+abs(r2)+1;
A4: 0<=abs(r1) by COMPLEX1:46;
    0<=abs(r2) by COMPLEX1:46;
    hence 0<g by A4;
    let n;
    r1<=abs(r1) by ABSVALUE:4;
    then seq.n<abs(r1) by A2,XXREAL_0:2;
    then seq.n +0<abs(r1)+abs(r2) by COMPLEX1:46,XREAL_1:8;
    then
A5: seq.n<g by XREAL_1:8;
    -abs(r2)<=r2 by ABSVALUE:4;
    then -abs(r2)<seq.n by A3,XXREAL_0:2;
    then -abs(r1)+-abs(r2)<0+seq.n by A4,XREAL_1:8;
    then
A6: -abs(r1)-abs(r2)+-1<seq.n+0 by XREAL_1:8;
    -abs(r1)-abs(r2)-1=-g;
    hence thesis by A5,A6,Th1;
  end;
  given r such that 0<r and
A7: for n holds abs(seq.n)<r;
  take r;
  let y;
  assume y in dom seq;
  then y is Element of NAT by FUNCT_2:def 1;
  hence thesis by A7;
end;

theorem Th4:
  for n ex r st (0<r & for m st m<=n holds abs(seq.m)<r)
proof
  defpred X[Nat] means ex r1 st 0<r1 & for m st m<=$1 holds abs(seq.m)<r1;
A1: X[0]
  proof
    reconsider r=abs(seq.0)+1 as real number;
    take r;
    0+0<abs(seq.0)+1 by COMPLEX1:46,XREAL_1:8;
    hence 0<r;
    let m;
    assume m<=0;
    then 0=m by NAT_1:2;
    then abs(seq.m)+0<r by XREAL_1:8;
    hence thesis;
  end;
A2: for n st X[n] holds X[n+1]
  proof
    let n;
    given r1 such that
A3: 0<r1 and
A4: for m st m<=n holds abs(seq.m)<r1;
A5: now
      assume
A6:   abs(seq.(n+1))<=r1;
      take r=r1+1;
      thus 0<r by A3;
      let m such that
A7:   m<=n+1;
A8:   now
        assume m<=n;
        then
A9:     abs(seq.m)<r1 by A4;
        r1+0<r by XREAL_1:8;
        hence abs(seq.m)<r by A9,XXREAL_0:2;
      end;
      now
        assume
A10:    m=n+1;
        r1+0<r by XREAL_1:8;
        hence abs(seq.m)<r by A6,A10,XXREAL_0:2;
      end;
      hence abs(seq.m)<r by A7,A8,NAT_1:8;
    end;
    now
      assume
A11:  r1<=abs(seq.(n+1));
      take r=abs(seq.(n+1))+1;
      0+0<r by COMPLEX1:46,XREAL_1:8;
      hence 0<r;
      let m such that
A12:  m<=n+1;
A13:  now
        assume m<=n;
        then abs(seq.m)<r1 by A4;
        then abs(seq.m)<abs(seq.(n+1)) by A11,XXREAL_0:2;
        then abs(seq.m)+0<r by XREAL_1:8;
        hence abs(seq.m)<r;
      end;
      now
        assume m=n+1;
        then abs(seq.m)+0<r by XREAL_1:8;
        hence abs(seq.m)<r;
      end;
      hence abs(seq.m)<r by A12,A13,NAT_1:8;
    end;
    hence thesis by A5;
  end;
  thus for n holds X[n] from NAT_1:sch 1(A1,A2);
end;






definition
 canceled;
  let seq;
  redefine attr seq is convergent means
  :Def6:
  ex g st for p st 0<p ex n st for m st n<=m holds abs (seq.m-g) < p;
  compatibility
   proof
    thus seq is convergent implies
     ex g st for p st 0<p ex n st for m st n<=m holds abs (seq.m-g) < p
     proof assume seq is convergent;
       then consider g being Element of COMPLEX such that
A1:     for p be Real st 0<p ex n st for m st n<=m holds |.seq.m-g.|<p
              by COMSEQ_2:def 5;
A2:     Re g+(Im g)*<i> = g by COMPLEX1:13;
       now assume
A3:       g is not real;
A4:       Im g <> 0 by A2,A3;
         set p = abs Im g;
A5:       p > 0 by A4,COMPLEX1:47;
         not ex n st for m st n<=m holds |.seq.m-g.|< p
          proof let n; take n; thus n<=n;
A6:          Im(seq.n) = 0 by COMPLEX1:def 2;
A7:          Im(seq.n-g) = Im(seq.n)-Im g by COMPLEX1:19;
A8:          |.seq.n-g.| = sqrt ((Re(seq.n-g))^2 + (Im g)^2) by A7,A6;
            (Re(seq.n-g))^2 >= 0 by XREAL_1:63;
            then sqrt((Re(seq.n-g))^2+|.Im g.|^2) >= |.Im g.| by SQUARE_1:58;
           hence |.seq.n-g.| >= |.Im g.| by A8,COMPLEX1:75;
          end;
        hence contradiction by A1,A5;
       end;
       then reconsider g as real number;
      take g;
      let p;
       reconsider p as Real by XREAL_0:def 1;
       0<p implies ex n st for m st n<=m holds abs (seq.m-g) < p by A1;
      hence thesis;
     end;
    given g such that
A9:   for p st 0<p ex n st for m st n<=m holds abs (seq.m-g) < p;
     g in REAL by XREAL_0:def 1;
     then reconsider g as Element of COMPLEX by NUMBERS:11;
    take g;
    thus thesis by A9;
   end;
end;

definition
  let seq;
  assume
A1: seq is convergent;
  func lim seq -> real number means
  :Def7:
  for p st 0<p ex n st for m st n<=m holds abs(seq.m-it)<p;
  existence by A1,Def6;
  uniqueness
  proof
    given g1,g2 such that
A2: for p st 0<p ex n st for m st n<=m holds abs(seq.m-g1)<p and
A3: for p st 0<p ex n st for m st n<=m holds abs(seq.m-g2)<p and
A4: g1<>g2;
A5: now
      assume abs(g1-g2)=0;
      then 0+g2=g1-g2+g2 by ABSVALUE:2;
      hence contradiction by A4;
    end;
A6: 0<=abs(g1-g2) by COMPLEX1:46;
    then consider n1 such that
A7: for m st n1<=m holds abs(seq.m-g1)<abs(g1-g2)/4 by A2,A5;
    consider n2 such that
A8: for m st n2<=m holds abs(seq.m-g2)<abs(g1-g2)/4 by A3,A5,A6;
A9: abs(seq.(n1+n2)-g1)<abs(g1-g2)/4 by A7,NAT_1:12;
    abs(seq.(n1+n2)-g2)<abs(g1-g2)/4 by A8,NAT_1:12;
    then
A10: abs(seq.(n1+n2)-g2)+abs(seq.(n1+n2)-g1)<abs(g1-g2)/4+abs(g1-g2)/4
    by A9,XREAL_1:8;
    abs(g1-g2)=abs(-(seq.(n1+n2)-g1)+(seq.(n1+n2)-g2));
    then abs(g1-g2)<=abs(-(seq.(n1+n2)-g1))+abs(seq.(n1+n2)-g2) by COMPLEX1:56
;
    then
A11: abs(g1-g2)<=abs(seq.(n1+n2)-g1)+abs(seq.(n1+n2)-g2) by COMPLEX1:52;
    abs(g1-g2)/2 <abs(g1-g2) by A5,A6,XREAL_1:216;
    hence contradiction by A10,A11,XXREAL_0:2;
  end;
end;

definition
  let f be real-valued Function;
  redefine attr f is bounded means
  f is bounded_above bounded_below;
  compatibility;
end;

definition
  let seq;
  redefine func lim seq -> Real;
  coherence by XREAL_0:def 1;
end;

registration
  cluster constant -> convergent for Real_Sequence;
  coherence
  proof
    let seq be Real_Sequence;
    assume seq is constant;
    then consider r being Real such that
A1: for n being Nat holds seq.n=r by VALUED_0:def 18;
    take g=r;
    let p such that
A2: 0<p;
    take n=0;
    let m such that
    n<=m;
    abs((seq.m)-g)=abs(r-g) by A1
      .=0 by ABSVALUE:2;
    hence thesis by A2;
  end;
end;

registration
  cluster constant for Real_Sequence;
  existence
  proof
    reconsider cs = NAT --> 0 as Real_Sequence by FUNCOP_1:45;
    take cs;
    thus thesis;
  end;
end;

theorem Th5:
  seq is convergent & seq9 is convergent implies seq + seq9 is convergent
proof
  assume that
A1: seq is convergent and
A2: seq9 is convergent;
  consider g1 such that
A3: for p st 0<p ex n st for m st n<=m holds abs(seq.m-g1)<p by A1,Def6;
  consider g2 such that
A4: for p st 0<p ex n st for m st n<=m holds abs(seq9.m-g2)<p by A2,Def6;
  take g=g1+g2;
  let p;
  assume
A5: 0<p;
  then consider n1 such that
A6: for m st n1<=m holds abs(seq.m-g1)<p/2 by A3;
  consider n2 such that
A7: for m st n2<=m holds abs(seq9.m-g2)<p/2 by A4,A5;
  take k=n1+n2;
  let m such that
A8: k<=m;
  n1<=n1+n2 by NAT_1:12;
  then n1<=m by A8,XXREAL_0:2;
  then
A9: abs(seq.m-g1)<p/2 by A6;
  n2<=k by NAT_1:12;
  then n2<=m by A8,XXREAL_0:2;
  then abs(seq9.m-g2)<p/2 by A7;
  then
A10: abs(seq.m-g1)+abs(seq9.m-g2)<p/2+p/2 by A9,XREAL_1:8;
A11: abs((seq+seq9).m-g)=abs(seq.m+seq9.m-(g1+g2)) by SEQ_1:7
    .=abs(seq.m-g1+(seq9.m-g2));
  abs(seq.m-g1+(seq9.m-g2))<=abs(seq.m-g1)+abs(seq9.m-g2) by COMPLEX1:56;
  hence thesis by A10,A11,XXREAL_0:2;
end;

registration
  let seq1,seq2 be convergent Real_Sequence;
  cluster seq1 + seq2 -> convergent for Real_Sequence;
  coherence by Th5;
end;

theorem Th6:
  seq is convergent & seq9 is convergent implies
  lim (seq + seq9)=(lim seq)+(lim seq9)
proof
  assume that
A1: seq is convergent and
A2: seq9 is convergent;
  now
    let p;
    assume 0<p;
    then
A3: 0<p/2;
    then consider n1 such that
A4: for m st n1<=m holds abs(seq.m-(lim seq))<p/2 by A1,Def7;
    consider n2 such that
A5: for m st n2<=m holds abs(seq9.m-(lim seq9))<p/2 by A2,A3,Def7;
    take k=n1+n2;
    let m such that
A6: k<=m;
    n1<=n1+n2 by NAT_1:12;
    then n1<=m by A6,XXREAL_0:2;
    then
A7: abs(seq.m-(lim seq))<p/2 by A4;
    n2<=k by NAT_1:12;
    then n2<=m by A6,XXREAL_0:2;
    then abs(seq9.m-(lim seq9))<p/2 by A5;
    then
A8: abs(seq.m-(lim seq))+abs(seq9.m-(lim seq9))<p/2+p/2 by A7,XREAL_1:8;
A9: abs(((seq+seq9).m)-((lim seq)+(lim seq9)))
    =abs(seq.m+seq9.m-((lim seq)+(lim seq9))) by SEQ_1:7
      .=abs(seq.m-(lim seq)+(seq9.m-(lim seq9)));
    abs(seq.m-(lim seq)+(seq9.m-(lim seq9)))<=
    abs(seq.m-(lim seq))+abs(seq9.m-(lim seq9)) by COMPLEX1:56;
    hence abs((seq+seq9).m-((lim seq)+(lim seq9)))<p by A8,A9,XXREAL_0:2;
  end;
  hence thesis by A1,A2,Def7;
end;

theorem Th7:
  seq is convergent implies r(#)seq is convergent
proof
  assume seq is convergent;
  then consider g1 such that
A1: for p st 0<p ex n st for m st n<=m holds abs(seq.m-g1)<p by Def6;
  take g=r*g1;
A2: now
    assume
A3: r=0;
    let p such that
A4: 0<p;
    take k=0;
    let m such that k<=m;
    abs(((r(#)seq).m)-g)=abs(0*seq.m-0) by A3,SEQ_1:9
      .=0 by ABSVALUE:2;
    hence abs(((r(#)seq).m)-g)<p by A4;
  end;
  now
    assume
A5: r<>0;
    then
A6: 0<abs(r) by COMPLEX1:47;
    let p such that
A7: 0<p;
A8: 0<>abs(r) by A5,COMPLEX1:47;
    consider n1 such that
A9: for m st n1<=m holds abs(seq.m-g1)<p/abs(r) by A1,A6,A7;
    take k=n1;
    let m;
    assume k<=m;
    then
A10: abs(seq.m-g1)<p/abs(r) by A9;
A11: abs(((r(#)seq).m)-g)=abs(r*seq.m-r*g1) by SEQ_1:9
      .=abs(r*(seq.m-g1))
      .=abs(r)*abs(seq.m-g1) by COMPLEX1:65;
    abs(r)*(p/abs(r))=abs(r)*((abs(r))"*p) by XCMPLX_0:def 9
      .=abs(r)*(abs(r))"*p
      .=1*p by A8,XCMPLX_0:def 7
      .=p;
    hence abs(((r(#)seq).m)-g)<p by A6,A10,A11,XREAL_1:68;
  end;
  hence thesis by A2;
end;

registration
  let r be real number;
  let seq be convergent Real_Sequence;
  cluster r(#)seq -> convergent for Real_Sequence;
  coherence by Th7;
end;

theorem Th8:
  seq is convergent implies lim(r(#)seq)=r*(lim seq)
proof
  assume
A1: seq is convergent;
A2: now
    assume
A3: r=0;
    let p such that
A4: 0<p;
    take k=0;
    let m such that k<=m;
    abs(((r(#)seq).m)-r*(lim seq))=abs(0*seq.m-0) by A3,SEQ_1:9
      .=0 by ABSVALUE:2;
    hence abs(((r(#)seq).m)-r*(lim seq))<p by A4;
  end;
  now
    assume
A5: r<>0;
    then
A6: 0<abs(r) by COMPLEX1:47;
    let p such that
A7: 0<p;
A8: 0<>abs(r) by A5,COMPLEX1:47;
    0<p/abs(r) by A6,A7;
    then consider n1 such that
A9: for m st n1<=m holds abs(seq.m-(lim seq))<p/abs(r) by A1,Def7;
    take k=n1;
    let m;
    assume k<=m;
    then
A10: abs(seq.m-(lim seq))<p/abs(r) by A9;
A11: abs(((r(#)seq).m)-r*(lim seq))=abs(r*seq.m-r*(lim seq)) by SEQ_1:9
      .=abs(r*(seq.m-(lim seq)))
      .=abs(r)*abs(seq.m-(lim seq)) by COMPLEX1:65;
    abs(r)*(p/abs(r))=abs(r)*((abs(r))"*p) by XCMPLX_0:def 9
      .=abs(r)*(abs(r))"*p
      .=1*p by A8,XCMPLX_0:def 7
      .=p;
    hence abs(((r(#)seq).m)-r*(lim seq))<p by A6,A10,A11,XREAL_1:68;
  end;
  hence thesis by A1,A2,Def7;
end;

theorem
  seq is convergent implies -seq is convergent;

registration
  let seq be convergent Real_Sequence;
  cluster -seq -> convergent for Real_Sequence;
  coherence;
end;

theorem Th10:
  seq is convergent implies lim(-seq) = -(lim seq)
proof
  assume seq is convergent;
  then lim ((-1)(#)seq)=(-1)*(lim seq) by Th8;
  hence thesis;
end;

theorem Th11:
  seq is convergent & seq9 is convergent implies seq - seq9 is convergent
proof
  assume that
A1: seq is convergent and
A2: seq9 is convergent;
  -seq9 is convergent by A2;
  hence thesis by A1;
end;

registration
  let seq1,seq2 be convergent Real_Sequence;
  cluster seq1 - seq2 -> convergent for Real_Sequence;
  coherence by Th11;
end;

theorem Th12:
  seq is convergent & seq9 is convergent implies
  lim(seq - seq9)=(lim seq)-(lim seq9)
proof
  assume that
A1: seq is convergent and
A2: seq9 is convergent;
  thus lim(seq - seq9)=(lim seq)+(lim(- seq9)) by A1,A2,Th6
    .=(lim seq)+-(lim seq9) by A2,Th10
    .=(lim seq)-(lim seq9);
end;

theorem Th13:
  seq is convergent implies seq is bounded
proof
  assume seq is convergent;
  then consider g such that
A1: for p st 0<p ex n st for m st n<=m holds abs(seq.m-g)<p by Def6;
  consider n1 such that
A2: for m st n1<=m holds abs(seq.m-g)<1 by A1;
A3: now
    take r=abs(g)+1;
    0+0<r by COMPLEX1:46,XREAL_1:8;
    hence 0<r;
    let m;
    assume n1<=m;
    then
A4: abs(seq.m-g)<1 by A2;
    abs(seq.m)-abs(g)<=abs(seq.m-g) by COMPLEX1:59;
    then
A5: abs(seq.m)-abs(g)<1 by A4,XXREAL_0:2;
    abs(seq.m)-abs(g)+abs(g)=abs(seq.m);
    hence abs(seq.m)<r by A5,XREAL_1:6;
  end;
  now consider r1 such that
A6: 0<r1 and
A7: for m st n1<=m holds abs(seq.m)<r1 by A3;
    consider r2 such that
A8: 0<r2 and
A9: for m st m<=n1 holds abs(seq.m)<r2 by Th4;
    take r=r1+r2;
    thus 0<r by A6,A8;
A10: r1+0<r by A8,XREAL_1:8;
A11: 0+r2<r by A6,XREAL_1:8;
    let m;
A12: now
      assume n1<=m;
      then abs(seq.m)<r1 by A7;
      hence abs(seq.m)<r by A10,XXREAL_0:2;
    end;
    now
      assume m<=n1;
      then abs(seq.m)<r2 by A9;
      hence abs(seq.m)<r by A11,XXREAL_0:2;
    end;
    hence abs(seq.m)<r by A12;
  end;
  hence thesis by Th3;
end;

registration
  cluster convergent -> bounded for Real_Sequence;
  coherence by Th13;
end;

theorem Th14:
  seq is convergent & seq9 is convergent implies seq (#) seq9 is convergent
proof
  assume that
A1: seq is convergent and
A2: seq9 is convergent;
  consider g1 such that
A3: for p st 0<p ex n st for m st n<=m holds abs(seq.m-g1)<p by A1,Def6;
  consider g2 such that
A4: for p st 0<p ex n st for m st n<=m holds abs(seq9.m-g2)<p by A2,Def6;
  take g=g1*g2;
  consider r such that
A5: 0<r and
A6: for n holds abs(seq.n)<r by A1,Th3;
A7: 0<=abs(g2) by COMPLEX1:46;
A8: 0+0<abs(g2)+r by A5,COMPLEX1:46,XREAL_1:8;
  let p;
  assume
A9: 0<p;
  then consider n1 such that
A10: for m st n1<=m holds abs(seq.m-g1)<p/(abs(g2)+r) by A3,A8;
  consider n2 such that
A11: for m st n2<=m holds abs(seq9.m-g2)<p/(abs(g2)+r) by A4,A8,A9;
  take n=n1+n2;
  let m such that
A12: n<=m;
A13: 0<=abs(seq.m) by COMPLEX1:46;
A14: 0<=abs(seq9.m-g2) by COMPLEX1:46;
  n2<=n by NAT_1:12;
  then n2<=m by A12,XXREAL_0:2;
  then
A15: abs(seq9.m-g2)<p/(abs(g2)+r) by A11;
  n1<=n1+n2 by NAT_1:12;
  then n1<=m by A12,XXREAL_0:2;
  then
A16: abs(seq.m-g1)<=p/(abs(g2)+r) by A10;
  abs(((seq(#)seq9).m)-g)=abs(seq.m*seq9.m-(seq.m*g2-seq.m*g2)-g1*g2) by
SEQ_1:8
    .=abs(seq.m*(seq9.m-g2)+(seq.m-g1)*g2);
  then
A17: abs(((seq(#)seq9).m)-g)<=
  abs(seq.m*(seq9.m-g2))+abs((seq.m-g1)*g2) by COMPLEX1:56;
  abs(seq.m)<r by A6;
  then abs(seq.m)*abs(seq9.m-g2)<r*(p/(abs(g2)+r)) by A13,A14,A15,XREAL_1:96;
  then
A18: abs(seq.m*(seq9.m-g2))<r*(p/(abs(g2)+r)) by COMPLEX1:65;
  abs((seq.m-g1)*g2)=abs(g2)*abs(seq.m-g1) by COMPLEX1:65;
  then
A19: abs((seq.m-g1)*g2)<=abs(g2)*(p/(abs(g2)+r)) by A7,A16,XREAL_1:64;
  r*(p/(abs(g2)+r))+abs(g2)*(p/(abs(g2)+r)) =(p/(abs(g2)+r))*(abs(g2)+r)
    .=p by A8,XCMPLX_1:87;
  then abs(seq.m*(seq9.m-g2))+abs((seq.m-g1)*g2)<p by A18,A19,XREAL_1:8;
  hence thesis by A17,XXREAL_0:2;
end;

registration
  let seq1,seq2 be convergent Real_Sequence;
  cluster seq1 (#) seq2 -> convergent for Real_Sequence;
  coherence by Th14;
end;

theorem Th15:
  seq is convergent & seq9 is convergent implies
  lim(seq(#)seq9)=(lim seq)*(lim seq9)
proof
  assume that
A1: seq is convergent and
A2: seq9 is convergent;
   consider r such that
A3: 0<r and
A4: for n holds abs(seq.n)<r by A1,Th3;
A5: 0<=abs((lim seq9)) by COMPLEX1:46;
A6: 0+0<abs((lim seq9))+r by A3,COMPLEX1:46,XREAL_1:8;
  now
    let p;
    assume 0<p;
    then
A7: 0<p/(abs((lim seq9))+r) by A6;
    then consider n1 such that
A8: for m st n1<=m holds abs(seq.m-(lim seq))<p/(abs((lim seq9))+r) by A1,Def7;
    consider n2 such that
A9: for m st n2<=m holds abs(seq9.m-(lim seq9))<p/(abs((lim seq9))+r)
    by A2,A7,Def7;
    take n=n1+n2;
    let m such that
A10: n<=m;
A11: 0<=abs(seq.m) by COMPLEX1:46;
A12: 0<=abs(seq9.m-(lim seq9)) by COMPLEX1:46;
    n2<=n by NAT_1:12;
    then n2<=m by A10,XXREAL_0:2;
    then
A13: abs(seq9.m-(lim seq9))<p/(abs((lim seq9))+r) by A9;
    n1<=n1+n2 by NAT_1:12;
    then n1<=m by A10,XXREAL_0:2;
    then
A14: abs(seq.m-(lim seq))<=p/(abs((lim seq9))+r) by A8;
    abs(((seq(#)seq9).m)-(lim seq)*(lim seq9))
    =abs(seq.m*seq9.m-(seq.m*(lim seq9)-seq.m*(lim seq9))-
    (lim seq)*(lim seq9)) by SEQ_1:8
      .=abs(seq.m*(seq9.m-(lim seq9))+
    (seq.m-(lim seq))*(lim seq9));
    then
A15: abs(((seq(#)seq9).m)-(lim seq)*(lim seq9))<=
    abs(seq.m*(seq9.m-(lim seq9)))+
    abs((seq.m-(lim seq))*(lim seq9)) by COMPLEX1:56;
    abs(seq.m)<r by A4;
    then abs(seq.m)*abs(seq9.m-(lim seq9))<r*(p/(abs((lim seq9))+r))
    by A11,A12,A13,XREAL_1:96;
    then
A16: abs(seq.m*(seq9.m-(lim seq9)))<r*(p/(abs((lim seq9))+r)) by COMPLEX1:65;
    abs((seq.m-(lim seq))*(lim seq9))
    =abs((lim seq9))*abs(seq.m-(lim seq)) by COMPLEX1:65;
    then
A17: abs((seq.m-(lim seq))*(lim seq9))<=abs((lim seq9))*
    (p/(abs((lim seq9))+r)) by A5,A14,XREAL_1:64;
    r*(p/(abs((lim seq9))+r))+abs((lim seq9))*(p/(abs((lim seq9))+r))
    =(p/(abs((lim seq9))+r))*(abs((lim seq9))+r)
      .=p by A6,XCMPLX_1:87;
    then abs(seq.m*(seq9.m-(lim seq9)))+abs((seq.m-(lim seq))*
    (lim seq9))<p by A16,A17,XREAL_1:8;
    hence abs(((seq(#)seq9).m)-(lim seq)*(lim seq9))<p by A15,XXREAL_0:2;
  end;
  hence thesis by A1,A2,Def7;
end;

theorem Th16:
  seq is convergent implies (lim seq<>0 implies
  ex n st for m st n<=m holds abs(lim seq)/2<abs(seq.m))
proof
  assume that
A1: seq is convergent and
A2: lim seq<>0;
  0<abs(lim seq) by A2,COMPLEX1:47;
  then 0<abs(lim seq)/2;
  then consider n1 such that
A3: for m st n1<=m holds abs(seq.m-(lim seq))<abs(lim seq)/2 by A1,Def7;
  take n=n1;
  let m;
  assume n<=m;
  then
A4: abs(seq.m-(lim seq))<abs(lim seq)/2 by A3;
A5: abs(lim seq)-abs(seq.m)<=abs((lim seq)-seq.m) by COMPLEX1:59;
  abs((lim seq)-seq.m)=abs(-(seq.m-(lim seq)))
    .=abs(seq.m-(lim seq)) by COMPLEX1:52;
  then
A6: abs(lim seq)-abs(seq.m)<abs(lim seq)/2 by A4,A5,XXREAL_0:2;
A7: abs(lim seq)/2+(abs(seq.m)-abs(lim seq)/2) = abs(seq.m);
  abs(lim seq)-abs(seq.m)+(abs(seq.m)-abs(lim seq)/2) = abs(lim seq)/2;
  hence thesis by A6,A7,XREAL_1:6;
end;

theorem Th17:
  seq is convergent & (for n holds 0<=seq.n) implies 0<=lim seq
proof
  assume that
A1: seq is convergent and
A2: for n holds 0<=seq.n and
A3: not 0<=(lim seq);
  0<-(lim seq) by A3;
  then consider n1 such that
A4: for m st n1<=m holds abs(seq.m-(lim seq))<-(lim seq) by A1,Def7;
  abs(seq.n1-(lim seq))<=-(lim seq) by A4;
  then seq.n1-(lim seq)<=-(lim seq) by ABSVALUE:5;
  then
A5: seq.n1-(lim seq)+(lim seq)<=-(lim seq)+(lim seq) by XREAL_1:6;
  now
    assume seq.n1=0;
    then abs(seq.n1-(lim seq))=-(lim seq) by A3,ABSVALUE:def 1;
    hence contradiction by A4;
  end;
  hence contradiction by A2,A5;
end;

theorem
  seq is convergent & seq9 is convergent &
  (for n holds seq.n<=(seq9.n)) implies (lim seq)<=(lim seq9)
proof
  assume that
A1: seq is convergent and
A2: seq9 is convergent and
A3: for n holds seq.n<=(seq9.n);
  now
    let n;
    seq.n<=seq9.n by A3;
    then
A4: seq.n-seq.n<=seq9.n-seq.n by XREAL_1:9;
    (seq9-seq).n=seq9.n+(-seq).n by SEQ_1:7
      .=seq9.n+-seq.n by SEQ_1:10
      .=seq9.n-seq.n;
    hence 0<=(seq9-seq).n by A4;
  end;
  then
A5: 0<=lim(seq9-seq) by A1,A2,Th17;
  lim (seq9-seq)=(lim seq9)- (lim seq) by A1,A2,Th12;
  then 0+(lim seq)<=(lim seq9)-(lim seq)+(lim seq) by A5,XREAL_1:6;
  hence thesis;
end;

theorem Th19:
  seq is convergent & seq9 is convergent &
  (for n holds seq.n<=(seq1.n) & seq1.n<=seq9.n) & (lim seq)=(lim seq9)
  implies seq1 is convergent
proof
  assume that
A1: seq is convergent and
A2: seq9 is convergent and
A3: for n holds seq.n<=(seq1.n) & seq1.n<=seq9.n and
A4: (lim seq)=(lim seq9);
  take lim seq;
  let p;
  assume
A5: 0<p;
  then consider n1 such that
A6: for m st n1<=m holds abs(seq.m-(lim seq))<p by A1,Def7;
  consider n2 such that
A7: for m st n2<=m holds abs(seq9.m-(lim seq))<p by A2,A4,A5,Def7;
  take n=n1+n2;
  let m such that
A8: n<=m;
  n2<=n by NAT_1:12;
  then n2<=m by A8,XXREAL_0:2;
  then abs(seq9.m-(lim seq))<p by A7;
  then
A9: seq9.m-(lim seq)<p by Th1;
  n1<=n1+n2 by NAT_1:12;
  then n1<=m by A8,XXREAL_0:2;
  then abs(seq.m-(lim seq))<p by A6;
  then
A10: -p<seq.m-(lim seq) by Th1;
  seq.m<=seq1.m by A3;
  then seq.m-(lim seq)<=seq1.m-(lim seq) by XREAL_1:9;
  then
A11: -p<seq1.m-(lim seq) by A10,XXREAL_0:2;
  seq1.m<=seq9.m by A3;
  then seq1.m-(lim seq)<=seq9.m-(lim seq) by XREAL_1:9;
  then seq1.m-(lim seq)<p by A9,XXREAL_0:2;
  hence thesis by A11,Th1;
end;

theorem
  seq is convergent & seq9 is convergent &
  (for n holds seq.n<=(seq1.n) & seq1.n<=seq9.n) &
  lim seq = lim seq9 implies lim seq1 = lim seq
proof
  assume that
A1: seq is convergent and
A2: seq9 is convergent and
A3: for n holds seq.n<=(seq1.n) & seq1.n<=seq9.n and
A4: lim seq = lim seq9;
A5: seq1 is convergent by A1,A2,A3,A4,Th19;
  now
    let p;
    assume
A6: 0<p;
    then consider n1 such that
A7: for m st n1<=m holds abs(seq.m-(lim seq))<p by A1,Def7;
    consider n2 such that
A8: for m st n2<=m holds abs(seq9.m-(lim seq))<p by A2,A4,A6,Def7;
    take n=n1+n2;
    let m such that
A9: n<=m;
    n2<=n by NAT_1:12;
    then n2<=m by A9,XXREAL_0:2;
    then abs(seq9.m-(lim seq))<p by A8;
    then
A10: seq9.m-(lim seq)<p by Th1;
    n1<=n1+n2 by NAT_1:12;
    then n1<=m by A9,XXREAL_0:2;
    then abs(seq.m-(lim seq))<p by A7;
    then
A11: -p<seq.m-(lim seq) by Th1;
    seq.m<=seq1.m by A3;
    then seq.m-(lim seq)<=seq1.m-(lim seq) by XREAL_1:9;
    then
A12: -p<seq1.m-(lim seq) by A11,XXREAL_0:2;
    seq1.m<=seq9.m by A3;
    then seq1.m-(lim seq)<=seq9.m-(lim seq) by XREAL_1:9;
    then seq1.m-(lim seq)<p by A10,XXREAL_0:2;
    hence abs(seq1.m-(lim seq))<p by A12,Th1;
  end;
  hence thesis by A5,Def7;
end;

theorem Th21:
  seq is convergent & lim seq <> 0 & seq is non-zero implies seq" is convergent
proof
  assume that
A1: seq is convergent and
A2: lim seq<>0 and
A3: seq is non-zero;
A4: 0<abs(lim seq) by A2,COMPLEX1:47;
A5: 0<>abs(lim seq) by A2,COMPLEX1:47;
  consider n1 such that
A6: for m st n1<=m holds abs(lim seq)/2<abs(seq.m) by A1,A2,Th16;
  0*0<abs(lim seq)*abs(lim seq) by A4;
  then
A7: 0<(abs(lim seq)*abs(lim seq))/2;
  take (lim seq)";
  let p;
  assume
A8: 0<p;
  then 0*0<p*((abs(lim seq)*abs(lim seq))/2) by A7;
  then consider n2 such that
A9: for m st n2<=m holds
  abs(seq.m-(lim seq))<p*((abs(lim seq)*abs(lim seq))/2) by A1,Def7;
  take n=n1+n2;
  let m such that
A10: n<=m;
  n2<=n by NAT_1:12;
  then n2<=m by A10,XXREAL_0:2;
  then
A11: abs(seq.m-(lim seq))<p*((abs(lim seq)*abs(lim seq))/2) by A9;
  n1<=n1+n2 by NAT_1:12;
  then n1<=m by A10,XXREAL_0:2;
  then
A12: abs(lim seq)/2<abs(seq.m) by A6;
A13: seq.m<>0 by A3,SEQ_1:5;
  then seq.m*(lim seq)<>0 by A2;
  then 0<abs(seq.m*(lim seq)) by COMPLEX1:47;
  then 0<abs(seq.m)*abs(lim seq) by COMPLEX1:65;
  then
A14: abs(seq.m-(lim seq))/(abs(seq.m)*abs(lim seq))<
  (p*((abs(lim seq)*abs(lim seq))/2))/(abs(seq.m)*abs(lim seq))
  by A11,XREAL_1:74;
A15: (p*((abs(lim seq)*abs(lim seq))/2))/(abs(seq.m)*abs(lim seq))
  =(p*(2"*(abs(lim seq)*abs(lim seq))))*
  (abs(seq.m)*abs(lim seq))" by XCMPLX_0:def 9
    .=p*2"*((abs(lim seq)*abs(lim seq))*(abs(lim seq)*abs(seq.m))")
    .=p*2"*((abs(lim seq)*abs(lim seq))*
  (abs(lim seq)"*(abs(seq.m))")) by XCMPLX_1:204
    .=p*2"*(abs(lim seq)*(abs(lim seq)*abs(lim seq)")*abs(seq.m)")
    .=p*2"*(abs(lim seq)*1*(abs(seq.m))") by A5,XCMPLX_0:def 7
    .=p*(abs(lim seq)/2)*abs(seq.m)"
    .=(p*(abs(lim seq)/2))/abs(seq.m) by XCMPLX_0:def 9;
A16: abs((seq").m-(lim seq)")=abs((seq.m)"-(lim seq)") by VALUED_1:10
    .=abs(seq.m-(lim seq))/(abs(seq.m)*abs(lim seq)) by A2,A13,Th2;
A17: 0<abs(lim seq)/2 by A4;
A18: 0<>abs(lim seq)/2 by A2,COMPLEX1:47;
  0*0<p*(abs(lim seq)/2) by A8,A17;
  then
A19: (p*(abs(lim seq)/2))/abs(seq.m)<
  (p*(abs(lim seq)/2))/(abs(lim seq)/2) by A12,A17,XREAL_1:76;
  (p*(abs(lim seq)/2))/(abs(lim seq)/2 )
  =(p*(abs(lim seq)/2))*(abs(lim seq)/2 )" by XCMPLX_0:def 9
    .=p*((abs(lim seq)/2)*(abs(lim seq)/2 )")
    .=p*1 by A18,XCMPLX_0:def 7
    .=p;
  hence thesis by A14,A15,A16,A19,XXREAL_0:2;
end;

theorem Th22:
  seq is convergent & lim seq <> 0 & seq is non-zero
  implies lim seq"=(lim seq)"
proof
  assume that
A1: seq is convergent and
A2: (lim seq)<>0 and
A3: seq is non-zero;
A4: seq" is convergent by A1,A2,A3,Th21;
A5: 0<abs(lim seq) by A2,COMPLEX1:47;
A6: 0<>abs(lim seq) by A2,COMPLEX1:47;
  consider n1 such that
A7: for m st n1<=m holds abs(lim seq)/2<abs(seq.m) by A1,A2,Th16;
  0*0<abs(lim seq)*abs(lim seq) by A5;
  then
A8: 0<(abs(lim seq)*abs(lim seq))/2;
  now
    let p;
    assume
A9: 0<p;
    then 0*0<p*((abs(lim seq)*abs(lim seq))/2) by A8;
    then consider n2 such that
A10: for m st n2<=m holds
    abs(seq.m-(lim seq))<p*((abs(lim seq)*abs(lim seq))/2) by A1,Def7;
    take n=n1+n2;
    let m such that
A11: n<=m;
    n2<=n by NAT_1:12;
    then n2<=m by A11,XXREAL_0:2;
    then
A12: abs(seq.m-(lim seq))<p*((abs(lim seq)*abs(lim seq))/2) by A10;
    n1<=n1+n2 by NAT_1:12;
    then n1<=m by A11,XXREAL_0:2;
    then
A13: abs(lim seq)/2<abs(seq.m) by A7;
A14: seq.m<>0 by A3,SEQ_1:5;
    then seq.m*(lim seq)<>0 by A2;
    then 0<abs(seq.m*(lim seq)) by COMPLEX1:47;
    then 0<abs(seq.m)*abs(lim seq) by COMPLEX1:65;
    then
A15: abs(seq.m-(lim seq))/(abs(seq.m)*abs(lim seq))<
    (p*((abs(lim seq)*abs(lim seq))/2))/(abs(seq.m)*abs(lim seq))
    by A12,XREAL_1:74;
A16: (p*((abs(lim seq)*abs(lim seq))/2))/(abs(seq.m)*abs(lim seq))
    =(p*(2"*(abs(lim seq)*abs(lim seq))))*
    (abs(seq.m)*abs(lim seq))" by XCMPLX_0:def 9
      .=p*2"*((abs(lim seq)*abs(lim seq))*(abs(lim seq)*abs(seq.m))")
      .=p*2"*((abs(lim seq)*abs(lim seq))*
    ((abs(lim seq))"*(abs(seq.m))")) by XCMPLX_1:204
      .=p*2"*(abs(lim seq)*(abs(lim seq)*abs(lim seq)")*abs(seq.m)")
      .=p*2"*(abs(lim seq)*1*abs(seq.m)") by A6,XCMPLX_0:def 7
      .=p*(abs(lim seq)/2)*abs(seq.m)"
      .=(p*(abs(lim seq)/2))/abs(seq.m) by XCMPLX_0:def 9;
A17: abs((seq").m-(lim seq)")=abs((seq.m)"-(lim seq)") by VALUED_1:10
      .=abs(seq.m-(lim seq))/(abs(seq.m)*abs(lim seq)) by A2,A14,Th2;
A18: 0<abs(lim seq)/2 by A5;
A19: 0<>abs(lim seq)/2 by A2,COMPLEX1:47;
    0*0<p*(abs(lim seq)/2) by A9,A18;
    then
A20: (p*(abs(lim seq)/2))/abs(seq.m)<
    (p*(abs(lim seq)/2))/(abs(lim seq)/2) by A13,A18,XREAL_1:76;
    (p*(abs(lim seq)/2))/(abs(lim seq)/2 )
    =(p*(abs(lim seq)/2))*(abs(lim seq)/2 )" by XCMPLX_0:def 9
      .=p*((abs(lim seq)/2)*(abs(lim seq)/2 )")
      .=p*1 by A19,XCMPLX_0:def 7
      .=p;
    hence abs((seq").m-(lim seq)")<p by A15,A16,A17,A20,XXREAL_0:2;
  end;
  hence thesis by A4,Def7;
end;

theorem
  seq9 is convergent & seq is convergent & lim seq <> 0
  & seq is non-zero implies seq9/"seq is convergent
proof
  assume that
A1: seq9 is convergent and
A2: seq is convergent and
A3: (lim seq)<>0 and
A4: seq is non-zero;
  seq" is convergent by A2,A3,A4,Th21;
  hence thesis by A1;
end;

theorem
  seq9 is convergent & seq is convergent & lim seq <> 0
  & seq is non-zero implies lim(seq9/"seq)=(lim seq9)/(lim seq)
proof
  assume that
A1: seq9 is convergent and
A2: seq is convergent and
A3: (lim seq)<>0 and
A4: seq is non-zero;
  seq" is convergent by A2,A3,A4,Th21;
  then lim (seq9(#)(seq"))=(lim seq9)*(lim seq") by A1,Th15
    .=(lim seq9)*(lim seq)" by A2,A3,A4,Th22
    .=(lim seq9)/(lim seq) by XCMPLX_0:def 9;
  hence thesis;
end;

theorem Th25:
  seq is convergent & seq1 is bounded & lim seq=0 implies
  seq(#)seq1 is convergent
proof
  assume that
A1: seq is convergent and
A2: seq1 is bounded and
A3: lim seq=0;
  reconsider g1=0 as Real;
  take g=g1;
  let p such that
A4: 0<p;
  consider r such that
A5: 0<r and
A6: for m holds abs(seq1.m)<r by A2,Th3;
A7: 0<p/r by A4,A5;
  then consider n1 such that
A8: for m st n1<=m holds abs(seq.m-0)<p/r by A1,A3,Def7;
  take n=n1;
  let m such that
A9: n<=m;
  abs(seq.m)=abs(seq.m-0);
  then
A10: abs(seq.m)<p/r by A8,A9;
A11: abs(((seq(#)seq1).m)-g)=abs(seq.m*seq1.m-0) by SEQ_1:8
    .=abs(seq.m)*abs(seq1.m) by COMPLEX1:65;
  now
    assume
A12: abs(seq1.m)<>0;
    (p/r)*r=p*r"*r by XCMPLX_0:def 9
      .=p*(r"*r)
      .=p*1 by A5,XCMPLX_0:def 7
      .=p;
    then
A13: (p/r)*abs(seq1.m)<p by A6,A7,XREAL_1:68;
    0<=abs(seq1.m) by COMPLEX1:46;
    then abs(((seq(#)seq1).m)-g)<(p/r)*abs(seq1.m) by A10,A11,A12,XREAL_1:68;
    hence thesis by A13,XXREAL_0:2;
  end;
  hence thesis by A4,A11;
end;

theorem
  seq is convergent & seq1 is bounded & lim seq=0 implies lim(seq(#)seq1)=0
proof
  assume that
A1: seq is convergent and
A2: seq1 is bounded and
A3: lim seq=0;
A4: seq(#)seq1 is convergent by A1,A2,A3,Th25;
  now
    let p such that
A5: 0<p;
    consider r such that
A6: 0<r and
A7: for m holds abs(seq1.m)<r by A2,Th3;
A8: 0<p/r by A5,A6;
    then consider n1 such that
A9: for m st n1<=m holds abs(seq.m-0)<p/r by A1,A3,Def7;
    take n=n1;
    let m;
    assume n<=m;
    then
A10: abs(seq.m-0)<p/r by A9;
A11: abs(((seq(#)seq1).m)-0)=abs(seq.m*seq1.m-0) by SEQ_1:8
      .=abs(seq.m)*abs(seq1.m) by COMPLEX1:65;
    now
      assume
A12:  abs(seq1.m)<>0;
      (p/r)*r=p*r"*r by XCMPLX_0:def 9
        .=p*(r"*r)
        .=p*1 by A6,XCMPLX_0:def 7;
      then
A13:  (p/r)*abs(seq1.m)<p by A7,A8,XREAL_1:68;
      0<=abs(seq1.m) by COMPLEX1:46;
      then abs(((seq(#)seq1).m)-0)<(p/r)*abs(seq1.m) by A10,A11,A12,XREAL_1:68;
      hence abs(((seq(#)seq1).m)-0)<p by A13,XXREAL_0:2;
    end;
    hence abs(((seq(#)seq1).m)-0)<p by A5,A11;
  end;
  hence thesis by A4,Def7;
end;



reserve g for Element of COMPLEX;

registration
  let s be convergent Complex_Sequence;
  cluster |.s.| -> convergent for Real_Sequence;
  coherence
  proof
    |.s.| is convergent
    proof
      consider g such that
A1:   for p be Real st 0<p ex n st for m st n<=m holds |.s.m-g.|<p
        by COMSEQ_2:def 5;
      take |.g.|;
      let p be real number;
      assume
A2:   0<p;
      p is Real by XREAL_0:def 1;
      then consider n such that
A3:   for m st n<=m holds |.s.m-g.|<p by A1,A2;
      take n;
      let m;
      assume n<=m;
      then |.s.m-g.|<p by A3;
      then abs(|.s.m.|- |.g.|) + |.s.m-g.| < p+|.s.m-g.| by COMPLEX1:64
,XREAL_1:8;
      then
      abs(|.s.m.|- |.g.|) + |.s.m-g.|- |.s.m-g.| < p+|.s.m-g.|- |.s.m-g.|
      by XREAL_1:9;
      hence thesis by VALUED_1:18;
    end;
    hence thesis;
  end;
end;

reserve s,s1,s9 for Complex_Sequence;

theorem Th27:
  s is convergent implies lim |.s.| = |.lim s.|
proof
  set g = lim s;
  assume
A1: s is convergent;
  then reconsider s1 = s as convergent Complex_Sequence;
  reconsider w = |.s1.| as Real_Sequence;
A2: w is convergent;
  reconsider w = |.lim s.| as Real;
  now
    let p be real number;
    assume
A3: 0<p;
    p is Real by XREAL_0:def 1;
    then consider n such that
A4: for m st n<=m holds |.s.m-g.|<p by A1,A3,COMSEQ_2:def 6;
    take n;
    let m;
    assume n<=m;
    then |.s.m-g.|<p by A4;
    then abs(|.s.m.|- |.g.|) + |.s.m-g.| < p+|.s.m-g.|
      by COMPLEX1:64,XREAL_1:8;
    then abs(|.s.m.|- |.g.|) + |.s.m-g.|- |.s.m-g.| < p+|.s.m-g.|- |.s.m-g.|
    by XREAL_1:9;
    hence abs(|.s.|.m - w) < p by VALUED_1:18;
  end;
  hence thesis by A2,Def7;
end;

theorem
 for s,s9 being convergent Complex_Sequence
  holds lim |.(s + s9).| = |.(lim s)+(lim s9).|
proof
 let s,s9 be convergent Complex_Sequence;
  thus lim |.(s + s9).| = |.lim (s + s9).| by Th27
    .= |.(lim s)+(lim s9).| by COMSEQ_2:14;
end;

theorem
  for s being convergent Complex_Sequence
   holds lim |.(r(#)s).| = |.r.|*|.(lim s).|
proof
  let s being convergent Complex_Sequence;
  thus lim |.(r(#)s).| = |.lim (r(#)s).| by Th27
    .= |.r*(lim s).| by COMSEQ_2:18
    .= |.r.|*|.(lim s).| by COMPLEX1:65;
end;

theorem
  for s being convergent Complex_Sequence
  holds lim |.-s.| = |.lim s.|
proof
  let s being convergent Complex_Sequence;
  thus lim |.-s.| = |.lim (-s).| by Th27
    .= |.-(lim s).| by COMSEQ_2:22
    .= |.lim s.| by COMPLEX1:52;
end;

theorem
    for s,s9 being convergent Complex_Sequence
  holds lim |.s - s9.| = |.(lim s) - (lim s9).|
proof
  let s,s9 be convergent Complex_Sequence;
  thus lim |.s - s9.| = |.lim (s - s9).| by Th27
    .= |.(lim s) - (lim s9).| by COMSEQ_2:26;
end;

theorem
  for s,s9 being convergent Complex_Sequence
  holds lim |.s(#)s9.| = |.lim s.|*|.lim s9.|
proof
  let s,s9 be convergent Complex_Sequence;
  thus lim |.s(#)s9.| = |.lim (s(#)s9).| by Th27
    .= |.(lim s)*(lim s9).| by COMSEQ_2:30
    .= |.lim s.|*|.lim s9.| by COMPLEX1:65;
end;

theorem
  s is convergent & (lim s)<>0c & s is non-zero implies lim |.s".| = (|.
  lim s.|)"
proof
  assume
A1: s is convergent & (lim s)<>0c & s is non-zero;
  then s" is convergent by COMSEQ_2:34;
  hence lim |.s".| = |.lim s".| by Th27
    .= |.(lim s)".| by A1,COMSEQ_2:35
    .= (|.lim s.|)" by COMPLEX1:66;
end;

theorem
  s9 is convergent & s is convergent & (lim s)<>0c & s is non-zero
  implies lim |.(s9/"s).|=|.(lim s9).|/|.(lim s).|
proof
  assume
A1: s9 is convergent & s is convergent & (lim s)<>0c & s is non-zero;
  then s9/"s is convergent by COMSEQ_2:38;
  hence lim |.s9/"s.| = |.lim (s9/"s).| by Th27
    .= |.(lim s9)/(lim s).| by A1,COMSEQ_2:39
    .= |.(lim s9).|/|.(lim s).| by COMPLEX1:67;
end;

theorem
  s is convergent & s1 is bounded & (lim s)=0c implies lim |.s(#)s1.| = 0
proof
  assume
A1: s is convergent & s1 is bounded & (lim s)=0c;
  then s(#)s1 is convergent by COMSEQ_2:42;
  hence lim |.s(#)s1.| = |.(lim (s(#)s1)).| by Th27
    .= 0 by A1,COMSEQ_2:43,COMPLEX1:44;
end;















begin

reserve x,y for set;
reserve C,C9,D,D9,E for non empty set;
reserve c for Element of C;
reserve c9 for Element of C9;
reserve d,d1,d2,d3,d4,e for Element of D;
reserve d9 for Element of D9;



theorem Th1:
  for f being Function holds <:{},f:> = {} & <:f,{}:> = {}
proof
  let f be Function;
  dom <:{},f:> = dom {} /\ dom f & dom <:f,{}:> = dom f /\ dom {} by
FUNCT_3:def 7;
  hence thesis;
end;

theorem
  for f being Function holds [:{},f:] = {} & [:f,{}:] = {}
proof
  let f be Function;
  dom [:{},f:] = [:dom {},dom f:] & dom [:f,{}:] = [:dom f,dom {}:] by
FUNCT_3:def 8;
  hence thesis by ZFMISC_1:90;
end;

theorem Th3:
  for F,f being Function holds F.:({},f) = {} & F.:(f,{}) = {}
proof
  let F,f be Function;
  F.:({},f) = F*{} by Th1;
  hence thesis by Th1;
end;

theorem
  for F being Function holds F[:]({},x) = {}
proof
  let F be Function;
  F[:]({},x) = F.:({}, dom {} --> x);
  hence thesis by Th3;
end;

theorem
  for F being Function holds F[;](x,{}) = {}
proof
  let F be Function;
  F[;](x,{}) = F.:(dom {} --> x, {});
  hence thesis by Th3;
end;

theorem Th6:
  for X being set, x1,x2 being set holds <:X-->x1,X-->x2:> = X -->[ x1,x2]
proof
  let X be set, x1,x2 be set;
  set f = X-->x1, g = X-->x2;
  set fg = <:f,g:>;
  now
    per cases;
    suppose
A1:   X = {};
      thus thesis by A1;
    end;
    suppose
A2:   X <> {};
      rng fg c= [:{x1},{x2}:];
      then
A3:   rng fg c= {[x1,x2]} by ZFMISC_1:29;
      set x = the Element of X;
A4:   dom f = X & dom g = X by FUNCOP_1:13;
      then
A5:   dom fg = X by FUNCT_3:50;
      f.x = x1 & g.x = x2 by A2,FUNCOP_1:7;
      then fg.x = [x1,x2] by A2,A4,FUNCT_3:49;
      then [x1,x2] in rng fg by A2,A5,FUNCT_1:def 3;
      then {[x1,x2]} c= rng fg by ZFMISC_1:31;
      then rng fg = {[x1,x2]} by A3,XBOOLE_0:def 10;
      hence thesis by A5,FUNCOP_1:9;
    end;
  end;
  hence thesis;
end;

theorem Th7:
  for F being Function,X being set, x1,x2 being set st [x1,x2] in
  dom F holds F.:(X-->x1,X-->x2) = X --> F.(x1,x2)
proof
  let F be Function, X be set, x1,x2 be set such that
A1: [x1,x2] in dom F;
  set f = X-->x1, g = X-->x2;
  thus F.:(f,g) = F*(X-->[x1,x2]) by Th6
    .= X --> F.(x1,x2) by A1,FUNCOP_1:17;
end;

reserve i,j for Nat;
reserve F for Function of [:D,D9:],E;
reserve p,q for FinSequence of D,
  p9,q9 for FinSequence of D9;

definition
  let D,D9,E,F,p,p9;
  redefine func F.:(p,p9) -> FinSequence of E;
  coherence by FINSEQ_2:70;
end;

definition
  let D,D9,E,F,p,d9;
  redefine func F[:](p,d9) -> FinSequence of E;
  coherence by FINSEQ_2:83;
end;

definition
  let D,D9,E,F,d,p9;
  redefine func F[;](d,p9) -> FinSequence of E;
  coherence by FINSEQ_2:77;
end;

reserve f,f9 for Function of C,D,
  h for Function of D,E;

definition
  let D,E be set, p be FinSequence of D, h be Function of D,E;
  redefine func h*p -> FinSequence of E;
  coherence by FINSEQ_2:32;
end;

theorem Th8:
  h*(p^<*d*>) = (h*p)^<*h.d*>
proof
  set q = h*(p^<*d*>);
  set r = (h*p)^<*h.d*>;
  set i = len p + 1;
A1: len q = len(p^<*d*>) by FINSEQ_2:33;
A2: len(h*p) = len p by FINSEQ_2:33;
  len(p^<*d*>) = i by FINSEQ_2:16;
  then
A3: dom q = Seg i by A1,FINSEQ_1:def 3;
A4: now
    let j be Nat;
A5: Seg len p = dom p by FINSEQ_1:def 3;
A6: Seg len(h*p) = dom(h*p) by FINSEQ_1:def 3;
    assume
A7: j in dom q;
    now
      per cases by A3,A7,A5,FINSEQ_2:7;
      suppose
A8:     j in dom p;
        hence h.((p^<*d*>).j) = h.(p.j) by FINSEQ_1:def 7
          .= (h*p).j by A2,A5,A6,A8,FUNCT_1:12
          .= r.j by A2,A5,A6,A8,FINSEQ_1:def 7;
      end;
      suppose
A9:     j = i;
        hence h.((p^<*d*>).j) = h.d by FINSEQ_1:42
          .= r.j by A2,A9,FINSEQ_1:42;
      end;
    end;
    hence q.j = r.j by A7,FUNCT_1:12;
  end;
  len r = len(h*p)+1 by FINSEQ_2:16;
  hence thesis by A1,A2,A4,FINSEQ_2:9,16;
end;

theorem
  h*(p^q) = (h*p)^(h*q)
proof
  defpred P[FinSequence of D] means h*(p^$1) = (h*p)^(h*$1);
A1: for q,d st P[q] holds P[q^<*d*>]
  proof
    let q,d such that
A2: h*(p^q) = (h*p)^(h*q);
    thus h*(p^(q^<*d*>)) = h*((p^q)^<*d*>) by FINSEQ_1:32
      .= (h*(p^q))^<*h.d*> by Th8
      .= (h*p)^((h*q)^<*h.d*>) by A2,FINSEQ_1:32
      .= (h*p)^(h*(q^<*d*>)) by Th8;
  end;
  h*(p^<*>D) = h*p by FINSEQ_1:34
    .= (h*p)^(h*(<*>D)) by FINSEQ_1:34;
  then
A3: P[<*>D];
  for q holds P[q] from FINSEQ_2:sch 2(A3,A1);
  hence thesis;
end;

reserve T,T1,T2,T3 for Tuple of i,D;
reserve T9 for Tuple of i, D9;
reserve S for Tuple of j, D;
reserve S9 for Tuple of j, D9;

Lm1: for T being Tuple of 0, D holds F.:(T,T9) = <*>E
proof
  let T be Tuple of 0, D;
  T = <*>D;
  hence thesis by FINSEQ_2:73;
end;

Lm2: for T9 being Tuple of 0, D9 holds F[;](d,T9) = <*>E
proof
  let T9 be Tuple of 0, D9;
  T9 = <*>D9;
  hence thesis by FINSEQ_2:79;
end;

Lm3: for T being Tuple of 0, D holds F[:](T,d9) = <*>E
proof
  let T be Tuple of 0, D;
  T = <*>D;
  hence thesis by FINSEQ_2:85;
end;

theorem Th10:
  F.:(T^<*d*>,T9^<*d9*>) = (F.:(T,T9))^<*F.(d,d9)*>
proof
  set p = T^<*d*>, q = T9^<*d9*>, pq = F.:(T,T9);
  set r = F.:(p,q), s = pq^<*F.(d,d9)*>;
A1: len T9 = i by CARD_1:def 7;
  then
A2: len q = i+1 by FINSEQ_2:16;
A3: len T = i by CARD_1:def 7;
  then
A4: len pq = i by A1,FINSEQ_2:72;
  len p = i+1 by A3,FINSEQ_2:16;
  then
A5: len r = i+1 by A2,FINSEQ_2:72;
  then
A6: dom r = Seg(i+1) by FINSEQ_1:def 3;
A7: now
    let j be Nat;
    assume
A8: j in dom r;
    now
      per cases by A6,A8,FINSEQ_2:7;
      suppose
A9:     j in Seg i;
        Seg len T = dom T by FINSEQ_1:def 3;
        then
A10:    p.j = T.j by A3,A9,FINSEQ_1:def 7;
A11:    Seg len pq = dom pq by FINSEQ_1:def 3;
        Seg len T9 = dom T9 by FINSEQ_1:def 3;
        then
A12:    q.j = T9.j by A1,A9,FINSEQ_1:def 7;
        j in dom pq by A4,A9,FINSEQ_1:def 3;
        hence F.(p.j,q.j) = pq.j by A10,A12,FUNCOP_1:22
          .= s.j by A4,A9,A11,FINSEQ_1:def 7;
      end;
      suppose
A13:    j = i+1;
        then p.j = d & q.j = d9 by A3,A1,FINSEQ_1:42;
        hence F.(p.j,q.j) = s.j by A4,A13,FINSEQ_1:42;
      end;
    end;
    hence r.j = s.j by A8,FUNCOP_1:22;
  end;
  len s = len pq + 1 by FINSEQ_2:16;
  hence thesis by A5,A4,A7,FINSEQ_2:9;
end;

theorem
  F.:(T^S,T9^S9) = (F.:(T,T9))^(F.:(S,S9))
proof
  defpred P[Nat] means for S being Tuple of $1, D, S9 being Tuple
  of $1, D9 holds F.:(T^S,T9^S9) = (F.:(T,T9))^(F.:(S,S9));
  now
    let j such that
A1: for S,S9 holds F.:(T^S,T9^S9) = (F.:(T,T9))^(F.:(S,S9));
    let S be Tuple of j+1, D;
    let S9 be Tuple of j+1, D9;
    consider S1 being Element of j-tuples_on D, d such that
A2: S = S1^<*d*> by FINSEQ_2:117;
A3: T^S = T^S1^<*d*> by A2,FINSEQ_1:32;
    reconsider S1 as Tuple of j, D;
    consider S19 being Element of j-tuples_on D9, d9 such that
A4: S9 = S19^<*d9*> by FINSEQ_2:117;
A5: T9^S9 = T9^S19^<*d9*> by A4,FINSEQ_1:32;
    reconsider S19 as Tuple of j, D9;
    T^S1 is Tuple of i+j, D & T9^S19 is Tuple of i+j, D9 by FINSEQ_2:107;
    hence F.:(T^S,T9^S9) = (F.:(T^S1,T9^S19))^<*F.(d,d9)*> by A3,A5,Th10
      .= (F.:(T,T9))^(F.:(S1,S19))^<*F.(d,d9)*> by A1
      .= (F.:(T,T9))^((F.:(S1,S19))^<*F.(d,d9)*>) by FINSEQ_1:32
      .= (F.:(T,T9))^(F.:(S,S9)) by A2,A4,Th10;
  end;
  then
A6: for j st P[j] holds P[j+1];
  now
    let S be Tuple of 0, D;
    let S9 be Tuple of 0, D9;
    S = <*>D;
    then
A7: F.:(S,S9) = <*>E by FINSEQ_2:73;
    T^S = T & T9^S9 = T9 by FINSEQ_2:95;
    hence F.:(T^S,T9^S9) = (F.:(T,T9))^(F.:(S,S9)) by A7,FINSEQ_1:34;
  end;
  then
A8: P[0];
  for j holds P[j] from NAT_1:sch 2(A8,A6);
  hence thesis;
end;

theorem Th12:
  F[;](d,p9^<*d9*>) = (F[;](d,p9))^<*F.(d,d9)*>
proof
  set pd = p9^<*d9*>, q = F[;](d,p9);
  set r = F[;](d,pd), s = q^<*F.(d,d9)*>;
  set i = len p9;
A1: len q = i by FINSEQ_2:78;
  len pd = i + 1 by FINSEQ_2:16;
  then
A2: len r = i + 1 by FINSEQ_2:78;
  then
A3: dom r = Seg(i+1) by FINSEQ_1:def 3;
A4: now
    let j be Nat;
    assume
A5: j in dom r;
    now
      per cases by A3,A5,FINSEQ_2:7;
      suppose
A6:     j in Seg i;
        then
A7:     j in dom q by A1,FINSEQ_1:def 3;
A8:     Seg len q = dom q by FINSEQ_1:def 3;
        Seg len p9 = dom p9 by FINSEQ_1:def 3;
        hence F.(d,pd.j) = F.(d,p9.j) by A6,FINSEQ_1:def 7
          .= q.j by A7,FUNCOP_1:32
          .= s.j by A1,A6,A8,FINSEQ_1:def 7;
      end;
      suppose
A9:     j = i+1;
        hence F.(d,pd.j) = F.(d,d9) by FINSEQ_1:42
          .= s.j by A1,A9,FINSEQ_1:42;
      end;
    end;
    hence r.j = s.j by A5,FUNCOP_1:32;
  end;
  len s = len q + 1 by FINSEQ_2:16;
  hence thesis by A1,A2,A4,FINSEQ_2:9;
end;

theorem
  F[;](d,p9^q9) = (F[;](d,p9))^(F[;](d,q9))
proof
  defpred P[FinSequence of D9] means F[;](d,p9^($1)) = (F[;](d,p9))^(F[;](d,$1
  ));
A1: for q9,d9 st P[q9] holds P[q9^<*d9*>]
  proof
    let q9,d9 such that
A2: F[;](d,p9^q9) = (F[;](d,p9))^(F[;](d,q9));
    thus F[;](d,p9^(q9^<*d9*>)) = F[;](d,(p9^q9)^<*d9*>) by FINSEQ_1:32
      .= (F[;](d,p9^q9))^<*F.(d,d9)*> by Th12
      .= (F[;](d,p9))^((F[;](d,q9))^<*F.(d,d9)*>) by A2,FINSEQ_1:32
      .= (F[;](d,p9))^(F[;](d,q9^<*d9*>)) by Th12;
  end;
  F[;](d,p9^(<*>D9)) = F[;](d,p9) by FINSEQ_1:34
    .= (F[;](d,p9))^(<*>E) by FINSEQ_1:34
    .= (F[;](d,p9))^(F[;](d,<*>D9)) by FINSEQ_2:79;
  then
A3: P[<*>D9];
  for q9 holds P[q9] from FINSEQ_2:sch 2(A3,A1);
  hence thesis;
end;

theorem Th14:
  F[:](p^<*d*>,d9) = (F[:](p,d9))^<*F.(d,d9)*>
proof
  set pd = p^<*d*>, q = F[:](p,d9);
  set r = F[:](pd,d9), s = q^<*F.(d,d9)*>;
  set i = len p;
A1: len q = i by FINSEQ_2:84;
  len pd = i + 1 by FINSEQ_2:16;
  then
A2: len r = i + 1 by FINSEQ_2:84;
  then
A3: dom r = Seg(i+1) by FINSEQ_1:def 3;
A4: now
    let j be Nat;
    assume
A5: j in dom r;
    now
      per cases by A3,A5,FINSEQ_2:7;
      suppose
A6:     j in Seg i;
        then
A7:     j in dom q by A1,FINSEQ_1:def 3;
        Seg len p = dom p by FINSEQ_1:def 3;
        hence F.(pd.j,d9) = F.(p.j,d9) by A6,FINSEQ_1:def 7
          .= q.j by A7,FUNCOP_1:27
          .= s.j by A7,FINSEQ_1:def 7;
      end;
      suppose
A8:     j = i+1;
        hence F.(pd.j,d9) = F.(d,d9) by FINSEQ_1:42
          .= s.j by A1,A8,FINSEQ_1:42;
      end;
    end;
    hence r.j = s.j by A5,FUNCOP_1:27;
  end;
  len s = len q + 1 by FINSEQ_2:16;
  hence thesis by A1,A2,A4,FINSEQ_2:9;
end;

theorem
  F[:](p^q,d9) = (F[:](p,d9))^(F[:](q,d9))
proof
  defpred P[FinSequence of D] means F[:](p^($1),d9) = (F[:](p,d9))^(F[:]($1,d9
  ));
A1: for q,d st P[q] holds P[q^<*d*>]
  proof
    let q,d such that
A2: F[:](p^q,d9) = (F[:](p,d9))^(F[:](q,d9));
    thus F[:](p^(q^<*d*>),d9) = F[:]((p^q)^<*d*>,d9) by FINSEQ_1:32
      .= (F[:](p^q,d9))^<*F.(d,d9)*> by Th14
      .= (F[:](p,d9))^((F[:](q,d9))^<*F.(d,d9)*>) by A2,FINSEQ_1:32
      .= (F[:](p,d9))^(F[:](q^<*d*>,d9)) by Th14;
  end;
  F[:](p^(<*>D),d9) = F[:](p,d9) by FINSEQ_1:34
    .= (F[:](p,d9))^(<*>E) by FINSEQ_1:34
    .= (F[:](p,d9))^(F[:](<*>D,d9)) by FINSEQ_2:85;
  then
A3: P[<*>D];
  for q holds P[q] from FINSEQ_2:sch 2(A3,A1);
  hence thesis;
end;

theorem
  for h being Function of D,E holds h*(i|->d) = i|->(h.d)
proof
  let h be Function of D,E;
  d in D;
  then d in dom h by FUNCT_2:def 1;
  hence thesis by FUNCOP_1:17;
end;

theorem Th17:
  F.:(i|->d,i|->d9) = i |-> (F.(d,d9))
proof
  [d,d9] in [:D,D9:] by ZFMISC_1:def 2;
  then [d,d9] in dom F by FUNCT_2:def 1;
  hence thesis by Th7;
end;

theorem
  F[;](d,i|->d9) = i |-> (F.(d,d9))
proof
  thus F[;](d,i|->d9) = F.:(i |->d,i|->d9) by FUNCOP_1:13
    .= i |-> (F.(d,d9)) by Th17;
end;

theorem
  F[:](i|->d,d9) = i |-> (F.(d,d9))
proof
  thus F[:](i|->d,d9) = F.:(i |->d,i|->d9) by FUNCOP_1:13
    .= i |-> (F.(d,d9)) by Th17;
end;

theorem
  F.:(i|->d,T9) = F[;](d,T9)
proof
  dom T9 = Seg len T9 by FINSEQ_1:def 3
    .= Seg i by CARD_1:def 7;
  hence thesis;
end;

theorem
  F.:(T,i|->d) = F[:](T,d)
proof
  dom T = Seg len T by FINSEQ_1:def 3
    .= Seg i by CARD_1:def 7;
  hence thesis;
end;

theorem
  F[;](d,T9) = F[;](d,id D9)*T9
proof
  rng T9 c= D9;
  hence F[;](d,T9) = F[;](d,(id D9)*T9) by RELAT_1:53
    .= F[;](d,(id D9))*T9 by FUNCOP_1:34;
end;

theorem
  F[:](T,d) = F[:](id D,d)*T
proof
  rng T c= D;
  hence F[:](T,d) = F[:]((id D)*T,d) by RELAT_1:53
    .= F[:]((id D),d)*T by FUNCOP_1:29;
end;



reserve F,G for BinOp of D;
reserve u for UnOp of D;
reserve H for BinOp of E;

Lm4: T is Function of Seg i,D
proof
  len T = i by CARD_1:def 7;
  then Seg i = dom T by FINSEQ_1:def 3;
  hence thesis by FINSEQ_2:26;
end;

theorem Th24:
  F is associative implies F[;](d,id D)*(F.:(f,f9)) = F.:(F[;](d, id D)*f,f9)
proof
  assume
A1: F is associative;
  now
    let c;
    thus (F[;](d,id D)*(F.:(f,f9))).c = (F[;](d,id D)).((F.:(f,f9)).c) by
FUNCT_2:15
      .= (F[;](d,id D)).(F.(f.c,f9.c)) by FUNCOP_1:37
      .= F.(d,(id D).(F.(f.c,f9.c))) by FUNCOP_1:53
      .= F.(d,F.(f.c,f9.c)) by FUNCT_1:18
      .= F.(F.(d,f.c),f9.c) by A1,BINOP_1:def 3
      .= F.((F[;](d,f)).c,f9.c) by FUNCOP_1:53
      .= F.(((F[;](d,id D))*f).c,f9.c) by FUNCOP_1:55
      .= (F.:(F[;](d,id D)*f,f9)).c by FUNCOP_1:37;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem Th25:
  F is associative implies F[:](id D,d)*(F.:(f,f9)) = F.:(f,F[:]( id D,d)*f9)
proof
  assume
A1: F is associative;
  now
    let c;
    thus (F[:](id D,d)*(F.:(f,f9))).c = (F[:](id D,d)).((F.:(f,f9)).c) by
FUNCT_2:15
      .= (F[:](id D,d)).(F.(f.c,f9.c)) by FUNCOP_1:37
      .= F.((id D).(F.(f.c,f9.c)),d) by FUNCOP_1:48
      .= F.(F.(f.c,f9.c),d) by FUNCT_1:18
      .= F.(f.c,F.(f9.c,d)) by A1,BINOP_1:def 3
      .= F.(f.c,(F[:](f9,d)).c) by FUNCOP_1:48
      .= F.(f.c,((F[:](id D,d))*f9).c) by FUNCOP_1:50
      .= (F.:(f,F[:](id D,d)*f9)).c by FUNCOP_1:37;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  F is associative implies F[;](d,id D)*(F.:(T1,T2)) = F.:(F[;](d,id D)*
  T1,T2 )
proof
  assume
A1: F is associative;
  per cases;
  suppose
A2: i = 0;
    then F.:(T1,T2) = <*>D by Lm1;
    then
A3: F[;](d,id D)*(F.:(T1,T2)) = <*>D;
    T2 = <*>D by A2;
    hence thesis by A3,FINSEQ_2:73;
  end;
  suppose
    i <> 0;
    then reconsider C = Seg i as non empty set;
    T1 is Function of C,D & T2 is Function of C,D by Lm4;
    hence thesis by A1,Th24;
  end;
end;

theorem
  F is associative implies F[:](id D,d)*(F.:(T1,T2)) = F.:(T1,F[:](id D,
  d)*T2 )
proof
  assume
A1: F is associative;
  per cases;
  suppose
A2: i = 0;
    then F.:(T1,T2) = <*>D by Lm1;
    then
A3: F[:](id D,d)*(F.:(T1,T2)) = <*>D;
    T1 = <*>D by A2;
    hence thesis by A3,FINSEQ_2:73;
  end;
  suppose
    i <> 0;
    then reconsider C = Seg i as non empty set;
    T1 is Function of C,D & T2 is Function of C,D by Lm4;
    hence thesis by A1,Th25;
  end;
end;

theorem
  F is associative implies F.:(F.:(T1,T2),T3) = F.:(T1,F.:(T2,T3))
proof
  assume
A1: F is associative;
  per cases;
  suppose
A2: i = 0;
    then F.:(T1,T2) = <*>D by Lm1;
    then
A3: F.:(F.:(T1,T2),T3) = <*>D by FINSEQ_2:73;
    F.:(T2,T3) = <*>D by A2,Lm1;
    hence thesis by A3,FINSEQ_2:73;
  end;
  suppose
    i <> 0;
    then reconsider C = Seg i as non empty set;
A4: T3 is Function of C,D by Lm4;
    T1 is Function of C,D & T2 is Function of C,D by Lm4;
    hence thesis by A1,A4,FUNCOP_1:61;
  end;
end;

theorem
  F is associative implies F[:](F[;](d1,T),d2) = F[;](d1,F[:](T,d2))
proof
  assume
A1: F is associative;
  per cases;
  suppose
A2: i = 0;
    then F[;](d1,T) = <*>D by Lm2;
    then
A3: F[:](F[;](d1,T),d2) = <*>D by FINSEQ_2:85;
    F[:](T,d2) = <*>D by A2,Lm3;
    hence thesis by A3,FINSEQ_2:79;
  end;
  suppose
    i <> 0;
    then reconsider C = Seg i as non empty set;
    T is Function of C,D by Lm4;
    hence thesis by A1,FUNCOP_1:59;
  end;
end;

theorem
  F is associative implies F.:(F[:](T1,d),T2) = F.:(T1,F[;](d,T2))
proof
  assume
A1: F is associative;
  per cases;
  suppose
A2: i = 0;
    then F[:](T1,d) = <*>D by Lm3;
    then
A3: F.:(F[:](T1,d),T2) = <*>D by FINSEQ_2:73;
    F[;](d,T2) = <*>D by A2,Lm2;
    hence thesis by A3,FINSEQ_2:73;
  end;
  suppose
    i <> 0;
    then reconsider C = Seg i as non empty set;
    T1 is Function of C,D & T2 is Function of C,D by Lm4;
    hence thesis by A1,FUNCOP_1:60;
  end;
end;

theorem
  F is associative implies F[;](F.(d1,d2),T) = F[;](d1,F[;](d2,T))
proof
  assume
A1: F is associative;
  per cases;
  suppose
    i = 0;
    then T = <*>D & F[;](d2,T) = <*>D by Lm2;
    hence thesis;
  end;
  suppose
    i <> 0;
    then reconsider C = Seg i as non empty set;
    T is Function of C,D by Lm4;
    hence thesis by A1,FUNCOP_1:62;
  end;
end;

theorem
  F is associative implies F[:](T,F.(d1,d2)) = F[:](F[:](T,d1),d2)
proof
  assume
A1: F is associative;
  per cases;
  suppose
    i = 0;
    then T = <*>D & F[:](T,d1) = <*>D by Lm3;
    hence thesis;
  end;
  suppose
    i <> 0;
    then reconsider C = Seg i as non empty set;
    T is Function of C,D by Lm4;
    hence thesis by A1,FUNCOP_1:63;
  end;
end;

theorem
  F is commutative implies F.:(T1,T2) = F.:(T2,T1)
proof
  assume
A1: F is commutative;
  per cases;
  suppose
A2: i = 0;
    then F.:(T1,T2) = <*>D by Lm1;
    hence thesis by A2,Lm1;
  end;
  suppose
    i <> 0;
    then reconsider C = Seg i as non empty set;
    T1 is Function of C,D & T2 is Function of C,D by Lm4;
    hence thesis by A1,FUNCOP_1:65;
  end;
end;

theorem
  F is commutative implies F[;](d,T) = F[:](T,d)
proof
  assume
A1: F is commutative;
  per cases;
  suppose
A2: i = 0;
    then F[;](d,T) = <*>D by Lm2;
    hence thesis by A2,Lm3;
  end;
  suppose
    i <> 0;
    then reconsider C = Seg i as non empty set;
    T is Function of C,D by Lm4;
    hence thesis by A1,FUNCOP_1:64;
  end;
end;

theorem Th35:
  F is_distributive_wrt G implies F[;](G.(d1,d2),f) = G.:(F[;](d1,
  f),F[;] (d2,f))
proof
  assume
A1: F is_distributive_wrt G;
  now
    let c;
    thus (F[;](G.(d1,d2),f)).c = F.(G.(d1,d2),f.c) by FUNCOP_1:53
      .= G.(F.(d1,f.c),F.(d2,f.c)) by A1,BINOP_1:11
      .= G.(F[;](d1,f).c,F.(d2,f.c)) by FUNCOP_1:53
      .= G.((F[;](d1,f)).c,(F[;](d2,f)).c) by FUNCOP_1:53
      .= (G.:(F[;](d1,f),F[;](d2,f))).c by FUNCOP_1:37;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem Th36:
  F is_distributive_wrt G implies F[:](f,G.(d1,d2)) = G.:(F[:](f,
  d1),F[:] (f,d2))
proof
  assume
A1: F is_distributive_wrt G;
  now
    let c;
    thus (F[:](f,G.(d1,d2))).c = F.(f.c,G.(d1,d2)) by FUNCOP_1:48
      .= G.(F.(f.c,d1),F.(f.c,d2)) by A1,BINOP_1:11
      .= G.(F[:](f,d1).c,F.(f.c,d2)) by FUNCOP_1:48
      .= G.((F[:](f,d1)).c,(F[:](f,d2)).c) by FUNCOP_1:48
      .= (G.:(F[:](f,d1),F[:](f,d2))).c by FUNCOP_1:37;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem Th37:
  (for d1,d2 holds h.(F.(d1,d2)) = H.(h.d1,h.d2)) implies h*(F.:(f
  ,f9)) = H.:(h*f,h*f9)
proof
  assume
A1: for d1,d2 holds h.(F.(d1,d2)) = H.(h.d1,h.d2);
  now
    let c;
    thus (h*(F.:(f,f9))).c = h.((F.:(f,f9)).c) by FUNCT_2:15
      .= h.(F.(f.c,f9.c)) by FUNCOP_1:37
      .= H.(h.(f.c),h.(f9.c)) by A1
      .= H.((h*f).c,h.(f9.c)) by FUNCT_2:15
      .= H.((h*f).c,(h*f9).c) by FUNCT_2:15
      .= (H.:(h*f,h*f9)).c by FUNCOP_1:37;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem Th38:
  (for d1,d2 holds h.(F.(d1,d2)) = H.(h.d1,h.d2)) implies h*(F[;](
  d,f)) = H[;](h.d,h*f)
proof
  assume
A1: for d1,d2 holds h.(F.(d1,d2)) = H.(h.d1,h.d2);
  reconsider g = C --> d as Function of C,D;
A2: dom h = D & dom(h*f) = C by FUNCT_2:def 1;
  thus h*(F[;](d,f)) = h*(F.:(g,f)) by FUNCT_2:def 1
    .= H.:(h*g,h*f) by A1,Th37
    .= H[;](h.d,h*f) by A2,FUNCOP_1:17;
end;

theorem Th39:
  (for d1,d2 holds h.(F.(d1,d2)) = H.(h.d1,h.d2)) implies h*(F[:](
  f,d)) = H[:](h*f,h.d)
proof
  assume
A1: for d1,d2 holds h.(F.(d1,d2)) = H.(h.d1,h.d2);
  reconsider g = C --> d as Function of C,D;
A2: dom h = D & dom(h*f) = C by FUNCT_2:def 1;
  thus h*(F[:](f,d)) = h*(F.:(f,g)) by FUNCT_2:def 1
    .= H.:(h*f,h*g) by A1,Th37
    .= H[:](h*f,h.d) by A2,FUNCOP_1:17;
end;

theorem
  u is_distributive_wrt F implies u*(F.:(f,f9)) = F.:(u*f,u*f9)
proof
  assume for d1,d2 holds u.(F.(d1,d2)) = F.(u.d1,u.d2);
  hence thesis by Th37;
end;

theorem
  u is_distributive_wrt F implies u*(F[;](d,f)) = F[;](u.d,u*f)
proof
  assume for d1,d2 holds u.(F.(d1,d2)) = F.(u.d1,u.d2);
  hence thesis by Th38;
end;

theorem
  u is_distributive_wrt F implies u*(F[:](f,d)) = F[:](u*f,u.d)
proof
  assume for d1,d2 holds u.(F.(d1,d2)) = F.(u.d1,u.d2);
  hence thesis by Th39;
end;

theorem Th43:
  F is having_a_unity implies F.:(C-->the_unity_wrt F,f) = f & F.:
  (f,C-->the_unity_wrt F) = f
proof
  set e = the_unity_wrt F;
  reconsider g = C-->e as Function of C,D;
  assume
A1: F is having_a_unity;
  now
    let c;
    thus (F.:(g,f)).c = F.(g.c,f.c) by FUNCOP_1:37
      .= F.(e,f.c) by FUNCOP_1:7
      .= f.c by A1,SETWISEO:15;
  end;
  hence F.:(C-->e,f) = f by FUNCT_2:63;
  now
    let c;
    thus (F.:(f,g)).c = F.(f.c,g.c) by FUNCOP_1:37
      .= F.(f.c,e) by FUNCOP_1:7
      .= f.c by A1,SETWISEO:15;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem Th44:
  F is having_a_unity implies F[;](the_unity_wrt F,f) = f
proof
  set e = the_unity_wrt F;
  assume
A1: F is having_a_unity;
  now
    let c;
    thus (F[;](e,f)).c = F.(e,f.c) by FUNCOP_1:53
      .= f.c by A1,SETWISEO:15;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem Th45:
  F is having_a_unity implies F[:](f,the_unity_wrt F) = f
proof
  set e = the_unity_wrt F;
  assume
A1: F is having_a_unity;
  now
    let c;
    thus (F[:](f,e)).c = F.(f.c,e) by FUNCOP_1:48
      .= f.c by A1,SETWISEO:15;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  F is_distributive_wrt G implies F[;](G.(d1,d2),T) = G.:(F[;](d1,T),F
  [;] (d2,T))
proof
  assume
A1: F is_distributive_wrt G;
  per cases;
  suppose
    i = 0;
    then F[;](d1,T) = <*>D & F[;](G.(d1,d2),T) = <*>D by Lm2;
    hence thesis by FINSEQ_2:73;
  end;
  suppose
    i <> 0;
    then reconsider C = Seg i as non empty set;
    T is Function of C,D by Lm4;
    hence thesis by A1,Th35;
  end;
end;

theorem
  F is_distributive_wrt G implies F[:](T,G.(d1,d2)) = G.:(F[:](T,d1),F
  [:] (T,d2))
proof
  assume
A1: F is_distributive_wrt G;
  per cases;
  suppose
    i = 0;
    then F[:](T,d1) = <*>D & F[:](T,G.(d1,d2)) = <*>D by Lm3;
    hence thesis by FINSEQ_2:73;
  end;
  suppose
    i <> 0;
    then reconsider C = Seg i as non empty set;
    T is Function of C,D by Lm4;
    hence thesis by A1,Th36;
  end;
end;

theorem Th48:
  (for d1,d2 holds h.(F.(d1,d2)) = H.(h.d1,h.d2)) implies h*(F.:(
  T1,T2)) = H.:(h*T1,h*T2)
proof
  assume
A1: for d1,d2 holds h.(F.(d1,d2)) = H.(h.d1,h.d2);
  per cases;
  suppose
A2: i = 0;
    then F.:(T1,T2) = <*>D by Lm1;
    then
A3: h*(F.:(T1,T2)) = <*>E;
    h*T1 = <*>E by A2;
    hence thesis by A3,FINSEQ_2:73;
  end;
  suppose
    i <> 0;
    then reconsider C = Seg i as non empty set;
    T1 is Function of C,D & T2 is Function of C,D by Lm4;
    hence thesis by A1,Th37;
  end;
end;

theorem Th49:
  (for d1,d2 holds h.(F.(d1,d2)) = H.(h.d1,h.d2)) implies h*(F[;](
  d,T)) = H[;](h.d,h*T)
proof
  assume
A1: for d1,d2 holds h.(F.(d1,d2)) = H.(h.d1,h.d2);
  per cases;
  suppose
A2: i = 0;
    then F[;](d,T) = <*>D by Lm2;
    then
A3: h*(F[;](d,T)) = <*>E;
    h*T = <*>E by A2;
    hence thesis by A3,FINSEQ_2:79;
  end;
  suppose
    i <> 0;
    then reconsider C = Seg i as non empty set;
    T is Function of C,D by Lm4;
    hence thesis by A1,Th38;
  end;
end;

theorem Th50:
  (for d1,d2 holds h.(F.(d1,d2)) = H.(h.d1,h.d2)) implies h*(F[:](
  T,d)) = H[:](h*T,h.d)
proof
  assume
A1: for d1,d2 holds h.(F.(d1,d2)) = H.(h.d1,h.d2);
  per cases;
  suppose
A2: i = 0;
    then F[:](T,d) = <*>D by Lm3;
    then
A3: h*(F[:](T,d)) = <*>E;
    h*T = <*>E by A2;
    hence thesis by A3,FINSEQ_2:85;
  end;
  suppose
    i <> 0;
    then reconsider C = Seg i as non empty set;
    T is Function of C,D by Lm4;
    hence thesis by A1,Th39;
  end;
end;

theorem
  u is_distributive_wrt F implies u*(F.:(T1,T2)) = F.:(u*T1,u*T2)
proof
  assume for d1,d2 holds u.(F.(d1,d2)) = F.(u.d1,u.d2);
  hence thesis by Th48;
end;

theorem
  u is_distributive_wrt F implies u*(F[;](d,T)) = F[;](u.d,u*T)
proof
  assume for d1,d2 holds u.(F.(d1,d2)) = F.(u.d1,u.d2);
  hence thesis by Th49;
end;

theorem
  u is_distributive_wrt F implies u*(F[:](T,d)) = F[:](u*T,u.d)
proof
  assume for d1,d2 holds u.(F.(d1,d2)) = F.(u.d1,u.d2);
  hence thesis by Th50;
end;

theorem
  G is_distributive_wrt F & u = G[;](d,id D) implies u is_distributive_wrt F
proof
  assume that
A1: G is_distributive_wrt F and
A2: u = G[;](d,id D);
  let d1,d2;
  thus u.(F.(d1,d2)) = G.(d,(id D).(F.(d1,d2))) by A2,FUNCOP_1:53
    .= G.(d,F.(d1,d2)) by FUNCT_1:18
    .= F.(G.(d,d1),G.(d,d2)) by A1,BINOP_1:11
    .= F.(G.(d,(id D).d1),G.(d,d2)) by FUNCT_1:18
    .= F.(G.(d,(id D).d1),G.(d,(id D).d2)) by FUNCT_1:18
    .= F.(u.d1,G.(d,(id D).d2)) by A2,FUNCOP_1:53
    .= F.(u.d1,u.d2) by A2,FUNCOP_1:53;
end;

theorem
  G is_distributive_wrt F & u = G[:](id D,d) implies u is_distributive_wrt F
proof
  assume that
A1: G is_distributive_wrt F and
A2: u = G[:](id D,d);
  let d1,d2;
  thus u.(F.(d1,d2)) = G.((id D).(F.(d1,d2)),d) by A2,FUNCOP_1:48
    .= G.(F.(d1,d2),d) by FUNCT_1:18
    .= F.(G.(d1,d),G.(d2,d)) by A1,BINOP_1:11
    .= F.(G.((id D).d1,d),G.(d2,d)) by FUNCT_1:18
    .= F.(G.((id D).d1,d),G.((id D).d2,d)) by FUNCT_1:18
    .= F.(u.d1,G.((id D).d2,d)) by A2,FUNCOP_1:48
    .= F.(u.d1,u.d2) by A2,FUNCOP_1:48;
end;

theorem
  F is having_a_unity implies F.:(i|->the_unity_wrt F,T) = T & F.:(T,i
  |->the_unity_wrt F) = T
proof
  assume
A1: F is having_a_unity;
  per cases;
  suppose
A2: i = 0;
    then T = <*>D;
    hence thesis by A2,Lm1;
  end;
  suppose
    i <> 0;
    then reconsider C = Seg i as non empty set;
    T is Function of C,D by Lm4;
    hence thesis by A1,Th43;
  end;
end;

theorem
  F is having_a_unity implies F[;](the_unity_wrt F,T) = T
proof
  assume
A1: F is having_a_unity;
  per cases;
  suppose
 i = 0;
    then T = <*>D;
    hence thesis by Lm2;
  end;
  suppose
    i <> 0;
    then reconsider C = Seg i as non empty set;
    T is Function of C,D by Lm4;
    hence thesis by A1,Th44;
  end;
end;

theorem
  F is having_a_unity implies F[:](T,the_unity_wrt F) = T
proof
  assume
A1: F is having_a_unity;
  per cases;
  suppose
 i = 0;
    then T = <*>D;
    hence thesis by Lm3;
  end;
  suppose
    i <> 0;
    then reconsider C = Seg i as non empty set;
    T is Function of C,D by Lm4;
    hence thesis by A1,Th45;
  end;
end;

definition
  let D,u,F;
  pred u is_an_inverseOp_wrt F means
  :Def1:
  for d holds F.(d,u.d) = the_unity_wrt F & F.(u.d,d) = the_unity_wrt F;
end;

definition
  let D,F;
  attr F is having_an_inverseOp means
  :Def2:
  ex u st u is_an_inverseOp_wrt F;
end;

definition
  let D,F;
  assume that
A1: F is having_a_unity and
A2: F is associative and
A3: F is having_an_inverseOp;
  func the_inverseOp_wrt F -> UnOp of D means
  :Def3:
  it is_an_inverseOp_wrt F;
  existence by A3,Def2;
  uniqueness
  proof
    set e = the_unity_wrt F;
    let u1,u2 be UnOp of D such that
A4: for d holds F.(d,u1.d) = e & F.(u1.d,d) = e and
A5: for d holds F.(d,u2.d) = e & F.(u2.d,d) = e;
    now
      let d;
      set d1 = u1.d, d2 = u2.d;
      thus u1.d = F.(d1,e) by A1,SETWISEO:15
        .= F.(d1,F.(d,d2)) by A5
        .= F.(F.(d1,d),d2) by A2,BINOP_1:def 3
        .= F.(e,d2) by A4
        .= u2.d by A1,SETWISEO:15;
    end;
    hence thesis by FUNCT_2:63;
  end;
end;

theorem Th59:
  F is having_a_unity & F is associative & F is
having_an_inverseOp implies F.((the_inverseOp_wrt F).d,d) = the_unity_wrt F & F
  .(d,(the_inverseOp_wrt F).d) = the_unity_wrt F
proof
  assume F is having_a_unity & F is associative & F is having_an_inverseOp;
  then the_inverseOp_wrt F is_an_inverseOp_wrt F by Def3;
  hence thesis by Def1;
end;

theorem Th60:
  F is having_a_unity & F is associative & F is
  having_an_inverseOp & F.(d1,d2) = the_unity_wrt F implies d1 = (
  the_inverseOp_wrt F).d2 & (the_inverseOp_wrt F).d1 = d2
proof
  assume that
A1: F is having_a_unity and
A2: F is associative and
A3: F is having_an_inverseOp and
A4: F.(d1,d2) = the_unity_wrt F;
  set e = the_unity_wrt F, d3 = (the_inverseOp_wrt F).d2;
  F.(F.(d1,d2),d3) = d3 by A1,A4,SETWISEO:15;
  then F.(d1,F.(d2,d3)) = d3 by A2,BINOP_1:def 3;
  then F.(d1,e) = d3 by A1,A2,A3,Th59;
  hence d1 = d3 by A1,SETWISEO:15;
  set d3 = (the_inverseOp_wrt F).d1;
  F.(d3,F.(d1,d2)) = d3 by A1,A4,SETWISEO:15;
  then F.(F.(d3,d1),d2) = d3 by A2,BINOP_1:def 3;
  then F.(e,d2) = d3 by A1,A2,A3,Th59;
  hence thesis by A1,SETWISEO:15;
end;

theorem Th61:
  F is having_a_unity & F is associative & F is
  having_an_inverseOp implies (the_inverseOp_wrt F).(the_unity_wrt F) =
  the_unity_wrt F
proof
  assume that
A1: F is having_a_unity and
A2: F is associative & F is having_an_inverseOp;
  set e = the_unity_wrt F;
  F.(e,e) = e by A1,SETWISEO:15;
  hence thesis by A1,A2,Th60;
end;

theorem Th62:
  F is having_a_unity & F is associative & F is
having_an_inverseOp implies (the_inverseOp_wrt F).((the_inverseOp_wrt F).d) = d
proof
  assume
A1: F is having_a_unity & F is associative & F is having_an_inverseOp;
  then F.(d,(the_inverseOp_wrt F).d) = the_unity_wrt F by Th59;
  hence thesis by A1,Th60;
end;

theorem Th63:
  F is having_a_unity & F is associative & F is commutative & F is
  having_an_inverseOp implies (the_inverseOp_wrt F) is_distributive_wrt F
proof
  assume that
A1: F is having_a_unity and
A2: F is associative and
A3: F is commutative and
A4: F is having_an_inverseOp;
  let d1,d2;
  set e = the_unity_wrt F, u = the_inverseOp_wrt F;
  F.(F.(u.d1,u.d2),F.(d1,d2)) = F.(u.d1,F.(u.d2,F.(d1,d2))) by A2,BINOP_1:def 3
    .= F.(u.d1,F.(F.(u.d2,d1),d2)) by A2,BINOP_1:def 3
    .= F.(u.d1,F.(F.(d1,u.d2),d2)) by A3,BINOP_1:def 2
    .= F.(u.d1,F.(d1,F.(u.d2,d2))) by A2,BINOP_1:def 3
    .= F.(u.d1,F.(d1,e)) by A1,A2,A4,Th59
    .= F.(F.(u.d1,d1),e) by A2,BINOP_1:def 3
    .= F.(e,e) by A1,A2,A4,Th59
    .= e by A1,SETWISEO:15;
  hence u.(F.(d1,d2)) = F.(u.d1,u.d2) by A1,A2,A4,Th60;
end;

theorem Th64:
  F is having_a_unity & F is associative & F is
having_an_inverseOp & (F.(d,d1) = F.(d,d2) or F.(d1,d) = F.(d2,d)) implies d1 =
  d2
proof
  assume that
A1: F is having_a_unity and
A2: F is associative and
A3: F is having_an_inverseOp and
A4: F.(d,d1) = F.(d,d2) or F.(d1,d) = F.(d2,d);
  set e = the_unity_wrt F, u = the_inverseOp_wrt F;
  per cases by A4;
  suppose
A5: F.(d,d1) = F.(d,d2);
    thus d1 = F.(e,d1) by A1,SETWISEO:15
      .= F.(F.(u.d,d),d1) by A1,A2,A3,Th59
      .= F.(u.d,F.(d,d2)) by A2,A5,BINOP_1:def 3
      .= F.(F.(u.d,d),d2) by A2,BINOP_1:def 3
      .= F.(e,d2) by A1,A2,A3,Th59
      .= d2 by A1,SETWISEO:15;
  end;
  suppose
A6: F.(d1,d) = F.(d2,d);
    thus d1 = F.(d1,e) by A1,SETWISEO:15
      .= F.(d1,F.(d,u.d)) by A1,A2,A3,Th59
      .= F.(F.(d2,d),u.d) by A2,A6,BINOP_1:def 3
      .= F.(d2,F.(d,u.d)) by A2,BINOP_1:def 3
      .= F.(d2,e) by A1,A2,A3,Th59
      .= d2 by A1,SETWISEO:15;
  end;
end;

theorem Th65:
  F is having_a_unity & F is associative & F is
  having_an_inverseOp & (F.(d1,d2) = d2 or F.(d2,d1) = d2) implies d1 =
  the_unity_wrt F
proof
  assume that
A1: F is having_a_unity and
A2: F is associative & F is having_an_inverseOp;
  set e = the_unity_wrt F;
  assume F.(d1,d2) = d2 or F.(d2,d1) = d2;
  then F.(d1,d2) = F.(e,d2) or F.(d2,d1) = F.(d2,e) by A1,SETWISEO:15;
  hence thesis by A1,A2,Th64;
end;

theorem Th66:
  F is associative & F is having_a_unity & F is
having_an_inverseOp & G is_distributive_wrt F & e = the_unity_wrt F implies for
  d holds G.(e,d) = e & G.(d,e) = e
proof
  assume that
A1: F is associative and
A2: F is having_a_unity and
A3: F is having_an_inverseOp and
A4: G is_distributive_wrt F and
A5: e = the_unity_wrt F;
  let d;
  set ed = G.(e,d);
  F.(ed,ed) = G.(F.(e,e),d) by A4,BINOP_1:11
    .= ed by A2,A5,SETWISEO:15;
  hence ed = e by A1,A2,A3,A5,Th65;
  set de = G.(d,e);
  F.(de,de) = G.(d,F.(e,e)) by A4,BINOP_1:11
    .= de by A2,A5,SETWISEO:15;
  hence thesis by A1,A2,A3,A5,Th65;
end;

theorem Th67:
  F is having_a_unity & F is associative & F is
having_an_inverseOp & u = the_inverseOp_wrt F & G is_distributive_wrt F implies
  u.(G.(d1,d2)) = G.(u.d1,d2) & u.(G.(d1,d2)) = G.(d1,u.d2)
proof
  assume that
A1: F is having_a_unity & F is associative & F is having_an_inverseOp and
A2: u = the_inverseOp_wrt F and
A3: G is_distributive_wrt F;
  set e = the_unity_wrt F;
  F.(G.(d1,d2),G.(u.d1,d2)) = G.(F.(d1,u.d1),d2) by A3,BINOP_1:11
    .= G.(e,d2) by A1,A2,Th59
    .= e by A1,A3,Th66;
  hence u.(G.(d1,d2)) = G.(u.d1,d2) by A1,A2,Th60;
  F.(G.(d1,d2),G.(d1,u.d2)) = G.(d1,F.(d2,u.d2)) by A3,BINOP_1:11
    .= G.(d1,e) by A1,A2,Th59
    .= e by A1,A3,Th66;
  hence thesis by A1,A2,Th60;
end;

theorem
  F is having_a_unity & F is associative & F is having_an_inverseOp & u
= the_inverseOp_wrt F & G is_distributive_wrt F & G is having_a_unity implies G
  [;](u.(the_unity_wrt G),id D) = u
proof
  assume that
A1: F is having_a_unity & F is associative & F is having_an_inverseOp &
  u = the_inverseOp_wrt F & G is_distributive_wrt F and
A2: G is having_a_unity;
  set o = the_unity_wrt G;
  for d holds (G[;](u.o,id D)).d = u.d
  proof
    let d;
    thus (G[;](u.o,id D)).d = G.(u.o,(id D).d) by FUNCOP_1:53
      .= G.(u.o,d) by FUNCT_1:18
      .= u.(G.(o,d)) by A1,Th67
      .= u.d by A2,SETWISEO:15;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  F is associative & F is having_a_unity & F is having_an_inverseOp & G
  is_distributive_wrt F implies G[;](d,id D).the_unity_wrt F = the_unity_wrt F
proof
  assume that
A1: F is associative and
A2: F is having_a_unity and
A3: F is having_an_inverseOp and
A4: G is_distributive_wrt F;
  set e = the_unity_wrt F, i = the_inverseOp_wrt F;
  G.(d,e) = G.(d,F.(e,e)) by A2,SETWISEO:15
    .= F.(G.(d,e),G.(d,e)) by A4,BINOP_1:11;
  then e = F.(F.(G.(d,e),G.(d,e)),i.(G.(d,e))) by A1,A2,A3,Th59;
  then e = F.(G.(d,e),F.(G.(d,e),i.(G.(d,e)))) by A1,BINOP_1:def 3;
  then e = F.(G.(d,e),e) by A1,A2,A3,Th59;
  then e = G.(d,e) by A2,SETWISEO:15;
  then G.(d,(id D).e) = e by FUNCT_1:18;
  hence thesis by FUNCOP_1:53;
end;

theorem
  F is associative & F is having_a_unity & F is having_an_inverseOp & G
  is_distributive_wrt F implies G[:](id D,d).the_unity_wrt F = the_unity_wrt F
proof
  assume that
A1: F is associative and
A2: F is having_a_unity and
A3: F is having_an_inverseOp and
A4: G is_distributive_wrt F;
  set e = the_unity_wrt F, i = the_inverseOp_wrt F;
  G.(e,d) = G.(F.(e,e),d) by A2,SETWISEO:15
    .= F.(G.(e,d),G.(e,d)) by A4,BINOP_1:11;
  then e = F.(F.(G.(e,d),G.(e,d)),i.(G.(e,d))) by A1,A2,A3,Th59;
  then e = F.(G.(e,d),F.(G.(e,d),i.(G.(e,d)))) by A1,BINOP_1:def 3;
  then e = F.(G.(e,d),e) by A1,A2,A3,Th59;
  then e = G.(e,d) by A2,SETWISEO:15;
  then G.((id D).e,d) = e by FUNCT_1:18;
  hence thesis by FUNCOP_1:48;
end;

theorem Th71:
  F is having_a_unity & F is associative & F is
having_an_inverseOp implies F.:(f,(the_inverseOp_wrt F)*f) = C-->the_unity_wrt
  F & F.:((the_inverseOp_wrt F)*f,f) = C-->the_unity_wrt F
proof
  set u = the_inverseOp_wrt F;
  reconsider g = C-->the_unity_wrt F as Function of C,D;
  assume
A1: F is having_a_unity & F is associative & F is having_an_inverseOp;
  now
    let c;
    thus (F.:(f,u*f)).c = F.(f.c,(u*f).c) by FUNCOP_1:37
      .= F.(f.c,u.(f.c)) by FUNCT_2:15
      .= the_unity_wrt F by A1,Th59
      .= g.c by FUNCOP_1:7;
  end;
  hence F.:(f,(the_inverseOp_wrt F)*f) = C-->the_unity_wrt F by FUNCT_2:63;
  now
    let c;
    thus (F.:(u*f,f)).c = F.((u*f).c,f.c) by FUNCOP_1:37
      .= F.(u.(f.c),f.c) by FUNCT_2:15
      .= the_unity_wrt F by A1,Th59
      .= g.c by FUNCOP_1:7;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem Th72:
  F is associative & F is having_an_inverseOp & F is
having_a_unity & F.:(f,f9) = C-->the_unity_wrt F implies f = (the_inverseOp_wrt
  F)*f9 & (the_inverseOp_wrt F)*f = f9
proof
  assume that
A1: F is associative & F is having_an_inverseOp & F is having_a_unity and
A2: F.:(f,f9) = C-->the_unity_wrt F;
  set u = the_inverseOp_wrt F;
  set e = the_unity_wrt F;
  reconsider g = C-->e as Function of C,D;
  now
    let c;
    F.(f.c,f9.c) = g.c by A2,FUNCOP_1:37
      .= e by FUNCOP_1:7;
    hence f.c = u.(f9.c) by A1,Th60
      .= (u*f9).c by FUNCT_2:15;
  end;
  hence f = (the_inverseOp_wrt F)*f9 by FUNCT_2:63;
  now
    let c;
    F.(f.c,f9.c) = g.c by A2,FUNCOP_1:37
      .= e by FUNCOP_1:7;
    then f9.c = u.(f.c) by A1,Th60;
    hence (u*f).c = f9.c by FUNCT_2:15;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  F is having_a_unity & F is associative & F is having_an_inverseOp
  implies F.:(T,(the_inverseOp_wrt F)*T) = i|->the_unity_wrt F & F.:((
  the_inverseOp_wrt F)*T,T) = i|->the_unity_wrt F
proof
  assume
A1: F is having_a_unity & F is associative & F is having_an_inverseOp;
  reconsider uT = the_inverseOp_wrt F*T as Element of i-tuples_on D by
FINSEQ_2:113;
  per cases;
  suppose
A2: i = 0;
    then F.:(T,uT) = <*>D & F.:(uT,T) = <*>D by Lm1;
    hence thesis by A2;
  end;
  suppose
    i <> 0;
    then reconsider C = Seg i as non empty set;
    T is Function of C,D by Lm4;
    hence thesis by A1,Th71;
  end;
end;

theorem
  F is associative & F is having_an_inverseOp & F is having_a_unity & F
  .:(T1,T2) = i|->the_unity_wrt F implies T1 = (the_inverseOp_wrt F)*T2 & (
  the_inverseOp_wrt F)*T1 = T2
proof
  assume
A1: F is associative & F is having_an_inverseOp & F is having_a_unity &
  F .:(T1,T2) = i|->the_unity_wrt F;
  per cases;
  suppose
    i = 0;
    then T1 = <*>D & T2 = <*>D;
    hence thesis;
  end;
  suppose
    i <> 0;
    then reconsider C = Seg i as non empty set;
    T1 is Function of C,D & T2 is Function of C,D by Lm4;
    hence thesis by A1,Th72;
  end;
end;

theorem Th75:
  F is associative & F is having_a_unity & e = the_unity_wrt F & F
  is having_an_inverseOp & G is_distributive_wrt F implies G[;](e,f) = C-->e
proof
  reconsider g = C-->e as Function of C,D;
  assume
A1: F is associative & F is having_a_unity & e = the_unity_wrt F & F is
  having_an_inverseOp & G is_distributive_wrt F;
  now
    let c;
    thus (G[;](e,f)).c = G.(e,f.c) by FUNCOP_1:53
      .= e by A1,Th66
      .= g.c by FUNCOP_1:7;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  F is associative & F is having_a_unity & e = the_unity_wrt F & F is
  having_an_inverseOp & G is_distributive_wrt F implies G[;](e,T) = i|->e
proof
  assume
A1: F is associative & F is having_a_unity & e = the_unity_wrt F & F is
  having_an_inverseOp & G is_distributive_wrt F;
  per cases;
  suppose
A2: i = 0;
    then G[;](e,T) = <*>D by Lm2;
    hence thesis by A2;
  end;
  suppose
    i <> 0;
    then reconsider C = Seg i as non empty set;
    T is Function of C,D by Lm4;
    hence thesis by A1,Th75;
  end;
end;

definition
  let F,f,g be Function;
  func F*(f,g) -> Function equals
  F*[:f,g:];
  correctness;
end;

theorem Th77:
  for F,f,g being Function st [x,y] in dom(F*(f,g)) holds (F*(f,g)
  ).(x,y) = F.(f.x,g.y)
proof
  let F,f,g be Function such that
A1: [x,y] in dom(F*(f,g));
  [x,y] in dom [:f,g:] by A1,FUNCT_1:11;
  then [x,y] in [:dom f,dom g:] by FUNCT_3:def 8;
  then [:f,g:].(x,y) = [f.x,g.y] by FUNCT_3:65;
  hence thesis by A1,FUNCT_1:12;
end;

theorem
  for F,f,g being Function st [x,y] in dom(F*(f,g)) holds (F*(f,g)).(x,y
  ) = F.(f.x,g.y) by Th77;

theorem Th79:
  for F being Function of [:D,D9:],E, f being Function of C,D, g
  being Function of C9,D9 holds F*(f,g) is Function of [:C,C9:],E
proof
  let F be Function of [:D,D9:],E, f be Function of C,D, g be Function of C9,
  D9;
  F*(f,g) = F*[:f,g:];
  hence thesis;
end;

theorem
  for u,u9 being Function of D,D holds F*(u,u9) is BinOp of D by Th79;

definition
  let D,F;
  let f,f9 be Function of D,D;
  redefine func F*(f,f9) -> BinOp of D;
  coherence by Th79;
end;

theorem Th81:
  for F being Function of [:D,D9:],E, f being Function of C,D, g
  being Function of C9,D9 holds (F*(f,g)).(c,c9) = F.(f.c,g.c9)
proof
  let F be Function of [:D,D9:],E, f be Function of C,D, g be Function of C9,
  D9;
  set H = F*(f,g);
  H is Function of [:C,C9:],E by Th79;
  then dom H = [:C,C9:] by FUNCT_2:def 1;
  then [c,c9] in dom H by ZFMISC_1:def 2;
  hence thesis by Th77;
end;

theorem Th82:
  for u being Function of D,D holds (F*(id D,u)).(d1,d2) = F.(d1,u
  .d2) & (F*(u,id D)).(d1,d2) = F.(u.d1,d2)
proof
  let u be Function of D,D;
  (id D).d1 = d1 & (id D).d2 = d2 by FUNCT_1:18;
  hence thesis by Th81;
end;

theorem Th83:
  (F*(id D,u)).:(f,f9) = F.:(f,u*f9)
proof
  now
    let c;
    thus ((F*(id D,u)).:(f,f9)).c = (F*(id D,u)).(f.c,f9.c) by FUNCOP_1:37
      .= F.(f.c,u.(f9.c)) by Th82
      .= F.(f.c,(u*f9).c) by FUNCT_2:15
      .= (F.:(f,u*f9)).c by FUNCOP_1:37;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  (F*(id D,u)).:(T1,T2) = F.:(T1,u*T2)
proof
  now
    per cases;
    suppose
      i = 0;
      then (F*(id D,u)).:(T1,T2) = <*>D & u*T2 = <*>D by Lm1;
      hence thesis by FINSEQ_2:73;
    end;
    suppose
      i <> 0;
      then reconsider C = Seg i as non empty set;
      T1 is Function of C,D & T2 is Function of C,D by Lm4;
      hence thesis by Th83;
    end;
  end;
  hence thesis;
end;

theorem
  F is associative & F is having_a_unity & F is commutative & F is
having_an_inverseOp & u = the_inverseOp_wrt F implies u.(F*(id D,u).(d1,d2)) =
  F*(u,id D).(d1,d2) & F*(id D,u).(d1,d2) = u.(F*(u,id D).(d1,d2))
proof
  assume that
A1: F is associative & F is having_a_unity and
A2: F is commutative and
A3: F is having_an_inverseOp & u = the_inverseOp_wrt F;
A4: u is_distributive_wrt F by A1,A2,A3,Th63;
  thus u.(F*(id D,u).(d1,d2)) = u.(F.(d1,u.d2)) by Th82
    .= F.(u.d1,u.(u.d2)) by A4,BINOP_1:def 12
    .= F.(u.d1,d2) by A1,A3,Th62
    .= F*(u,id D).(d1,d2) by Th82;
  hence thesis by A1,A3,Th62;
end;

theorem
  F is associative & F is having_a_unity & F is having_an_inverseOp
  implies (F*(id D,the_inverseOp_wrt F)).(d,d) = the_unity_wrt F
proof
  assume
A1: F is associative & F is having_a_unity & F is having_an_inverseOp;
  set u = the_inverseOp_wrt F;
  thus F*(id D,u).(d,d) = F.(d,u.d) by Th82
    .= the_unity_wrt F by A1,Th59;
end;

theorem
  F is associative & F is having_a_unity & F is having_an_inverseOp
  implies (F*(id D,the_inverseOp_wrt F)).(d,the_unity_wrt F) = d
proof
  assume that
A1: F is associative and
A2: F is having_a_unity and
A3: F is having_an_inverseOp;
  set u = the_inverseOp_wrt F, e = the_unity_wrt F;
  thus (F*(id D,u)).(d,e) = F.(d,u.e) by Th82
    .= F.(d,e) by A1,A2,A3,Th61
    .= d by A2,SETWISEO:15;
end;

theorem
  F is having_a_unity implies (F*(id D,u)).(the_unity_wrt F,d) = u.d
proof
  assume
A1: F is having_a_unity;
  set e = the_unity_wrt F;
  thus (F*(id D,u)).(e,d) = F.(e,u.d) by Th82
    .= u.d by A1,SETWISEO:15;
end;

theorem
  F is commutative & F is associative & F is having_a_unity & F is
having_an_inverseOp & G = F*(id D,the_inverseOp_wrt F) implies for d1,d2,d3,d4
  holds F.(G.(d1,d2),G.(d3,d4))= G.(F.(d1,d3),F.(d2,d4))
proof
  assume that
A1: F is commutative and
A2: F is associative and
A3: F is having_a_unity & F is having_an_inverseOp and
A4: G = F*(id D,the_inverseOp_wrt F);
  set u = the_inverseOp_wrt F;
A5: u is_distributive_wrt F by A1,A2,A3,Th63;
  let d1,d2,d3,d4;
  thus F.(G.(d1,d2),G.(d3,d4)) = F.(F.(d1,u.d2),G.(d3,d4)) by A4,Th82
    .= F.(F.(d1,u.d2),F.(d3,u.d4)) by A4,Th82
    .= F.(d1,F.(u.d2,F.(d3,u.d4))) by A2,BINOP_1:def 3
    .= F.(d1,F.(F.(u.d2,d3),u.d4)) by A2,BINOP_1:def 3
    .= F.(d1,F.(F.(d3,u.d2),u.d4)) by A1,BINOP_1:def 2
    .= F.(d1,F.(d3,F.(u.d2,u.d4))) by A2,BINOP_1:def 3
    .= F.(F.(d1,d3),F.(u.d2,u.d4)) by A2,BINOP_1:def 3
    .= F.(F.(d1,d3),u.(F.(d2,d4))) by A5,BINOP_1:def 12
    .= G.(F.(d1,d3),F.(d2,d4)) by A4,Th82;
end;














begin

reserve p,q,r for FinSequence;
reserve u,v,x,y,y1,y2,z,A,D,X,Y for set;
reserve i,j,k,l,m,n for Nat;

theorem Th1:
  Seg 3 = {1,2,3}
proof
  thus Seg 3 = {1,2} \/ {2 + 1} by FINSEQ_1:2,9
    .= {1,2,3} by ENUMSET1:3;
end;

theorem Th2:
  Seg 4 = {1,2,3,4}
proof
  thus Seg 4 = {1,2,3} \/ {3 + 1} by Th1,FINSEQ_1:9
    .= {1,2,3,4} by ENUMSET1:6;
end;

theorem Th3:
  Seg 5 = {1,2,3,4,5}
proof
  thus Seg 5 = {1,2,3,4} \/ {4 + 1} by Th2,FINSEQ_1:9
    .= {1,2,3,4,5} by ENUMSET1:10;
end;

theorem Th4:
  Seg 6 = {1,2,3,4,5,6}
proof
  thus Seg 6 = {1,2,3,4,5} \/ {5 + 1} by Th3,FINSEQ_1:9
    .= {1,2,3,4,5,6} by ENUMSET1:15;
end;

theorem Th5:
  Seg 7 = {1,2,3,4,5,6,7}
proof
  thus Seg 7 = {1,2,3,4,5,6} \/ {6 + 1} by Th4,FINSEQ_1:9
    .= {1,2,3,4,5,6,7} by ENUMSET1:21;
end;

theorem
  Seg 8 = {1,2,3,4,5,6,7,8}
proof
  thus Seg 8 = {1,2,3,4,5,6,7} \/ {7 + 1} by Th5,FINSEQ_1:9
    .= {1,2,3,4,5,6,7,8} by ENUMSET1:28;
end;

theorem Th7:
  Seg k = {} iff not k in Seg k
   proof
    thus Seg k = {} implies not k in Seg k;
    assume not k in Seg k;
    then k = 0 by FINSEQ_1:3;
    hence Seg k = {};
   end;

theorem Th8:
  not k + 1 in Seg k
proof
  assume k + 1 in Seg k;
  then
A1: k + 1 <= k by FINSEQ_1:1;
  k <= k + 1 by NAT_1:12;
  then k + 0 = k + 1 by A1,XXREAL_0:1;
  hence thesis;
end;

theorem
  k <> 0 implies k in Seg(k + n)
proof
  assume k <> 0;
  then
A1: 0 + 1 <= k by NAT_1:13;
  k <= k + n by NAT_1:12;
  hence thesis by A1,FINSEQ_1:1;
end;

theorem Th10:
  k + n in Seg k implies n = 0
proof
  assume k + n in Seg k;
  then k + n <= k + 0 by FINSEQ_1:1;
  hence thesis by XREAL_1:6;
end;

theorem
  k < n implies k + 1 in Seg n
proof
  assume k < n;
  then
A1: k + 1 <= n by NAT_1:13;
  1 <= k + 1 by NAT_1:12;
  hence thesis by A1,FINSEQ_1:1;
end;

theorem Th12:
  k in Seg n & m < k implies k - m in Seg n
proof
  assume that
A1: k in Seg n and
A2: m < k;
  consider i being Nat such that
A3: k = m + i by A2,NAT_1:10;
  reconsider x = k - m as Element of NAT by A3,ORDINAL1:def 12;
A4: now
    assume not 1 <= x;
    then x = 0 by NAT_1:14;
    hence contradiction by A2;
  end;
A5: k <= n by A1,FINSEQ_1:1;
  i <= k by A3,NAT_1:12;
  then x <= n by A3,A5,XXREAL_0:2;
  hence thesis by A4,FINSEQ_1:1;
end;

theorem
  k - n in Seg k iff n < k
proof
  thus k - n in Seg k implies n < k
  proof
    assume
A1: k - n in Seg k;
    then reconsider x = k - n as Element of NAT;
    assume not n < k;
    then k - n <= n - n by XREAL_1:9;
    then x = 0;
    hence contradiction by A1,FINSEQ_1:1;
  end;
  thus thesis by Th12,FINSEQ_1:3;
end;

theorem Th14:
  Seg k misses {k + 1}
proof
  set x = the Element of Seg k /\ {k + 1};
  assume not thesis;
  then
A1: Seg k /\ {k + 1} <> {} by XBOOLE_0:def 7;
  then
A2: x in Seg k by XBOOLE_0:def 4;
  then reconsider x as Element of NAT;
  x in {k + 1} by A1,XBOOLE_0:def 4;
  then
A3: x = k + 1 by TARSKI:def 1;
  x <= k by A2,FINSEQ_1:1;
  hence thesis by A3,XREAL_1:29;
end;

theorem
  Seg(k + 1) \ Seg k = {k + 1}
proof
A1: Seg(k + 1) = Seg k \/ {k + 1} by FINSEQ_1:9;
  Seg k misses {k + 1} by Th14;
  hence thesis by A1,XBOOLE_1:88;
end;




theorem
  Seg k <> Seg(k + 1) by Th8,FINSEQ_1:4;

theorem
  Seg k = Seg(k + n) implies n = 0 by Th10,FINSEQ_1:3;

theorem Th18:
  Seg k c= Seg(k + n)
proof
  k <= k + n by NAT_1:12;
  hence thesis by FINSEQ_1:5;
end;

theorem Th19:
  Seg k, Seg n are_c=-comparable
proof
  n <= k or k <= n;
  then Seg n c= Seg k or Seg k c= Seg n by FINSEQ_1:5;
  hence thesis by XBOOLE_0:def 9;
end;

theorem Th20:
  Seg k = {y} implies k = 1 & y = 1
proof
  assume
A1: Seg k = {y};
  now
    per cases;
    suppose
      k = 0;
      hence thesis by A1;
    end;
    suppose
      k <> 0;
      then
A2:   k in Seg k by FINSEQ_1:3;
      then 1 <= k by FINSEQ_1:1;
      then Seg 1 c= Seg k by FINSEQ_1:5;
      then Seg 1 = {y} by A1,ZFMISC_1:33;
      hence thesis by A1,A2,FINSEQ_1:2,TARSKI:def 1,ZFMISC_1:3;
    end;
  end;
  hence thesis;
end;

theorem
  Seg k = {x,y} & x <> y implies k = 2 & {x,y} = {1,2}
proof
  assume that
A1: Seg k = {x,y} and
A2: x <> y;
  now
    per cases;
    suppose
      k = 0;
      hence thesis by A1;
    end;
    suppose
A3:   k <> 0;
      now
        per cases;
        suppose
          k = 1;
          hence thesis by A1,A2,FINSEQ_1:2,ZFMISC_1:5;
        end;
        suppose
A4:       k <> 1;
          1 <= k by A3,NAT_1:14;
          then 1 < k by A4,XXREAL_0:1;
          then
A5:       1 + 1 <= k by NAT_1:13;
          then Seg 2 c= Seg k by FINSEQ_1:5;
          then
A6:       1 = x & 2 = x or 1 = x & 2 = y or 2 = x & 1 = y or 1 = y & 2 =
          y by A1,FINSEQ_1:2,ZFMISC_1:22;
          now
            k in Seg k by A1,Th7;
            then
A7:         k = 1 or k = 2 by A1,A6,TARSKI:def 2;
            assume not k <= 2;
            hence contradiction by A7;
          end;
          hence thesis by A1,A5,FINSEQ_1:2,XXREAL_0:1;
        end;
      end;
      hence thesis;
    end;
  end;
  hence thesis;
end;

theorem Th22:
  x in dom p implies x in dom(p ^ q)
proof
  dom p c= dom(p ^ q) by FINSEQ_1:26;
  hence thesis;
end;

theorem
  x in dom p implies x is Element of NAT;

theorem Th24:
  x in dom p implies x <> 0
proof
  assume x in dom p;
  then x in Seg(len p) by FINSEQ_1:def 3;
  hence thesis by FINSEQ_1:1;
end;

theorem Th25:
  n in dom p iff 1 <= n & n <= len p
proof
  thus n in dom p implies 1 <= n & n <= len p
  proof
    assume n in dom p;
    then n in Seg(len p) by FINSEQ_1:def 3;
    hence thesis by FINSEQ_1:1;
  end;
  assume that
A1: 1 <= n and
A2: n <= len p;
  n in Seg(len p) by A1,A2,FINSEQ_1:1;
  hence thesis by FINSEQ_1:def 3;
end;

theorem
  n in dom p iff n - 1 is Element of NAT & len p - n is Element of NAT
proof
  thus n in dom p implies n - 1 is Element of NAT & len p - n is Element of NAT
  proof
    assume
A1: n in dom p;
    then
A2: n <= len p by Th25;
    1 <= n by A1,Th25;
    hence thesis by A2,INT_1:5;
  end;
  assume that
A3: n - 1 is Element of NAT and
A4: len p - n is Element of NAT;
A5: 0 + n <= len p by A4,XREAL_1:19;
  0 + 1 <= n by A3,XREAL_1:19;
  hence thesis by A5,Th25;
end;

canceled 2;

theorem Th29:
  len p = len q iff dom p = dom q
proof
  dom p = Seg(len p) by FINSEQ_1:def 3;
  hence thesis by FINSEQ_1:def 3;
end;

theorem Th30:
  len p <= len q iff dom p c= dom q
proof
A1: dom q = Seg(len q) by FINSEQ_1:def 3;
  dom p = Seg(len p) by FINSEQ_1:def 3;
  hence thesis by A1,FINSEQ_1:5;
end;

theorem Th31:
  x in rng p implies 1 in dom p
proof
  assume x in rng p;
  then p <> {};
  then
A1: 1 <= len p by NAT_1:14;
  dom p = Seg(len p) by FINSEQ_1:def 3;
  hence thesis by A1,FINSEQ_1:1;
end;

theorem
  rng p <> {} implies 1 in dom p
proof
  set y = the Element of rng p;
  assume rng p <> {};
  then y in rng p;
  hence thesis by Th31;
end;

theorem
  {} <> <* x,y *>;

theorem
  {} <> <* x,y,z *>;

theorem Th35:
  <* x *> <> <* y,z *>
proof
  len<* x *> = 1 by FINSEQ_1:40;
  hence thesis by FINSEQ_1:44;
end;

theorem
  <* u *> <> <* x,y,z *>
proof
  len<* u *> = 1 by FINSEQ_1:40;
  hence thesis by FINSEQ_1:45;
end;

theorem
  <* u,v *> <> <* x,y,z *>
proof
  len<* u,v *> = 2 by FINSEQ_1:44;
  hence thesis by FINSEQ_1:45;
end;

theorem Th38:
  len r = len p + len q & (for k being Element of NAT st k in dom
p holds r.k = p.k) & (for k being Element of NAT st k in dom q holds r.(len p +
  k) = q.k) implies r = p ^ q
proof
  assume len r = len p + len q;
  then
A1: dom r = Seg(len p + len q) by FINSEQ_1:def 3;
  assume that
A2: for k being Element of NAT st k in dom p holds r.k = p.k and
A3: for k being Element of NAT st k in dom q holds r.(len p + k) = q.k;
A4: for k being Nat st k in dom q holds r.(len p + k) = q.k by A3;
  for k being Nat st k in dom p holds r.k = p.k by A2;
  hence thesis by A1,A4,FINSEQ_1:def 7;
end;

Lm1: A c= Seg k implies Sgm A is one-to-one
proof
  assume
A1: A c= Seg k;
  then
A2: rng(Sgm A) = A by FINSEQ_1:def 13;
  let x,y;
  assume that
A3: x in dom(Sgm A) and
A4: y in dom(Sgm A) and
A5: Sgm(A).x = Sgm(A).y and
A6: x <> y;
  Sgm(A).y in rng(Sgm A) by A4,FUNCT_1:def 3;
  then
A7: Sgm(A).y in Seg k by A1,A2;
  Sgm(A).x in rng(Sgm A) by A3,FUNCT_1:def 3;
  then Sgm(A).x in Seg k by A1,A2;
  then reconsider m = Sgm(A).x, n = Sgm(A).y as Element of NAT by A7;
  reconsider i = x, j = y as Element of NAT by A3,A4;
A8: dom(Sgm A) = Seg(len(Sgm A)) by FINSEQ_1:def 3;
  now
    per cases by A6,XXREAL_0:1;
    suppose
A9:   i < j;
A10:  j <= len(Sgm A) by A4,A8,FINSEQ_1:1;
      1 <= i by A3,A8,FINSEQ_1:1;
      then m < n by A1,A9,A10,FINSEQ_1:def 13;
      hence contradiction by A5;
    end;
    suppose
A11:  j < i;
A12:  i <= len(Sgm A) by A3,A8,FINSEQ_1:1;
      1 <= j by A4,A8,FINSEQ_1:1;
      then n < m by A1,A11,A12,FINSEQ_1:def 13;
      hence contradiction by A5;
    end;
  end;
  hence thesis;
end;

theorem Th39:
  for A being finite set st A c= Seg k holds len(Sgm A) = card A
proof
  let A be finite set;
A1: dom(Sgm A) = Seg(len(Sgm A)) by FINSEQ_1:def 3;
A2: card(Seg(len(Sgm A))) = len(Sgm A) by FINSEQ_1:57;
  assume
A3: A c= Seg k;
  then
A4: rng(Sgm A) = A by FINSEQ_1:def 13;
  Sgm A is one-to-one by A3,Lm1;
  then Seg(len(Sgm A)),A are_equipotent by A1,A4,WELLORD2:def 4;
  hence thesis by A2,CARD_1:5;
end;

theorem
  for A being finite set st A c= Seg k holds dom(Sgm A) = Seg(card A)
proof
  let A be finite set;
  assume A c= Seg k;
  then len(Sgm A) = card A by Th39;
  hence thesis by FINSEQ_1:def 3;
end;

theorem Th41:
  X c= Seg i & k < l & 1 <= n & m <= len(Sgm X) & Sgm(X).m = k &
  Sgm(X).n = l implies m < n
proof
  assume that
A1: X c= Seg i and
A2: k < l and
A3: 1 <= n and
A4: m <= len(Sgm X) and
A5: Sgm(X).m = k and
A6: Sgm(X).n = l and
A7: not m < n;
  n < m by A2,A5,A6,A7,XXREAL_0:1;
  hence thesis by A1,A2,A3,A4,A5,A6,FINSEQ_1:def 13;
end;

theorem Th42:
  X c= Seg i & Y c= Seg j implies ((for m,n being Element of NAT
  st m in X & n in Y holds m < n) iff Sgm(X \/ Y) = Sgm(X) ^ Sgm(Y))
proof
  assume that
A1: X c= Seg i and
A2: Y c= Seg j;
  set r = Sgm(X \/ Y);
  set q = Sgm Y;
  set p = Sgm X;
  Seg i, Seg j are_c=-comparable by Th19;
  then Seg i c= Seg j or Seg j c= Seg i by XBOOLE_0:def 9;
  then
A3: X c= Seg j or Y c= Seg i by A1,A2,XBOOLE_1:1;
  then
A4: X \/ Y c= Seg i or X \/ Y c= Seg j by A1,A2,XBOOLE_1:8;
  thus (for m,n being Element of NAT st m in X & n in Y holds m < n) implies
  Sgm(X \/ Y) = Sgm(X) ^ Sgm(Y)
  proof
    reconsider X1 = X, Y1 = Y as finite set by A1,A2;
    defpred P[Element of NAT] means $1 in dom p implies r.$1 = p.$1;
    assume
A5: for m,n being Element of NAT st m in X & n in Y holds m < n;
    X /\ Y = {}
    proof
      set x = the Element of X /\ Y;
      X = rng p by A1,FINSEQ_1:def 13;
      then
A6:   X c= NAT;
      assume
A7:   not thesis;
      then x in X by XBOOLE_0:def 4;
      then reconsider m = x as Element of NAT by A6;
A8:   m in Y by A7,XBOOLE_0:def 4;
      m in X by A7,XBOOLE_0:def 4;
      hence thesis by A5,A8;
    end;
    then
A9: X misses Y by XBOOLE_0:def 7;
A10: len r = card(X1 \/ Y1) by A1,A2,A3,Th39,XBOOLE_1:8
      .= card X1 + card Y1 by A9,CARD_2:40
      .= len p + card Y1 by A1,Th39
      .= len p + len q by A2,Th39;
A11: now
      let k be Element of NAT;
      assume
A12:  P[k];
      thus P[k+1]
      proof
        assume
A13:    k + 1 in dom p;
        then
A14:    p.(k + 1) in rng p by FUNCT_1:def 3;
        then reconsider n = p.(k + 1) as Element of NAT;
A15:    n in X by A1,A14,FINSEQ_1:def 13;
        len p <= len r by A10,NAT_1:12;
        then Seg(len p) c= Seg(len r) by FINSEQ_1:5;
        then dom p c= Seg(len r) by FINSEQ_1:def 3;
        then
A16:    dom p c= dom r by FINSEQ_1:def 3;
        then
A17:    r.(k + 1) in rng r by A13,FUNCT_1:def 3;
        then reconsider m = r.(k + 1) as Element of NAT;
A18:    m in X \/ Y by A4,A17,FINSEQ_1:def 13;
        assume
A19:    r.(k + 1) <> p.(k + 1);
A20:    k + 1 in dom r by A13,A16;
        now
          per cases;
          suppose
A21:        k in dom p;
            then p.k in rng p by FUNCT_1:def 3;
            then reconsider n1 = p.k as Element of NAT;
            r.k in rng r by A16,A21,FUNCT_1:def 3;
            then reconsider m1 = r.k as Element of NAT;
            now
              per cases by A19,XXREAL_0:1;
              suppose
A22:            m < n;
A23:            1 <= k + 1 by A13,Th25;
                not m in Y by A5,A15,A22;
                then m in X by A18,XBOOLE_0:def 3;
                then m in rng p by A1,FINSEQ_1:def 13;
                then consider x such that
A24:            x in dom p and
A25:            p.x = m by FUNCT_1:def 3;
                reconsider x as Element of NAT by A24;
                x <= len p by A24,Th25;
                then
A26:            x < k + 1 by A1,A22,A25,A23,Th41;
                k + 1 in Seg(len r) by A20,FINSEQ_1:def 3;
                then
A27:            k + 1 <= len r by FINSEQ_1:1;
                k in Seg(len p) by A21,FINSEQ_1:def 3;
                then
A28:            1 <= k by FINSEQ_1:1;
                k < k + 1 by XREAL_1:29;
                then
A29:            n1 < m by A4,A12,A21,A28,A27,FINSEQ_1:def 13;
A30:            k <= len p by A21,Th25;
                1 <= x by A24,Th25;
                then k < x by A1,A25,A29,A30,Th41;
                hence contradiction by A26,NAT_1:13;
              end;
              suppose
A31:            n < m;
A32:            1 <= k + 1 by A13,Th25;
                n in X \/ Y by A15,XBOOLE_0:def 3;
                then n in rng r by A4,FINSEQ_1:def 13;
                then consider x such that
A33:            x in dom r and
A34:            r.x = n by FUNCT_1:def 3;
                reconsider x as Element of NAT by A33;
                x <= len r by A33,Th25;
                then
A35:            x < k + 1 by A1,A2,A3,A31,A34,A32,Th41,XBOOLE_1:8;
A36:            k + 1 <= len p by A13,Th25;
A37:            k < k + 1 by XREAL_1:29;
                1 <= k by A21,Th25;
                then
A38:            m1 < n by A1,A12,A21,A37,A36,FINSEQ_1:def 13;
A39:            k <= len r by A16,A21,Th25;
                1 <= x by A33,Th25;
                then k < x by A1,A2,A3,A34,A38,A39,Th41,XBOOLE_1:8;
                hence contradiction by A35,NAT_1:13;
              end;
            end;
            hence contradiction;
          end;
          suppose
A40:        not k in dom p;
A41:        k < k + 1 by XREAL_1:29;
            k < 1 or len p < k by A40,Th25;
            then
A42:        k = 0 or len p < k + 1 & k + 1 <= len p by A13,A41,Th25,NAT_1:14
,XXREAL_0:2;
            now
              per cases by A19,XXREAL_0:1;
              suppose
A43:            m < n;
                then not m in Y by A5,A15;
                then m in X by A18,XBOOLE_0:def 3;
                then m in rng p by A1,FINSEQ_1:def 13;
                then consider x such that
A44:            x in dom p and
A45:            p.x = m by FUNCT_1:def 3;
A46:            1 <= k + 1 by A13,Th25;
                reconsider x as Element of NAT by A44;
                x <= len p by A44,Th25;
                then x < k + 1 by A1,A43,A45,A46,Th41;
                hence contradiction by A42,A44,Th24,NAT_1:14;
              end;
              suppose
A47:            n < m;
A48:            1 <= k + 1 by A13,Th25;
                n in X \/ Y by A15,XBOOLE_0:def 3;
                then n in rng r by A4,FINSEQ_1:def 13;
                then consider x such that
A49:            x in dom r and
A50:            r.x = n by FUNCT_1:def 3;
                reconsider x as Element of NAT by A49;
                x <= len r by A49,Th25;
                then x < k + 1 by A1,A2,A3,A47,A50,A48,Th41,XBOOLE_1:8;
                hence contradiction by A42,A49,Th24,NAT_1:14;
              end;
            end;
            hence contradiction;
          end;
        end;
        hence contradiction;
      end;
    end;
A51: P[0] by Th24;
A52: for k being Element of NAT holds P[k] from NAT_1:sch 1(A51,A11);
    defpred P[Element of NAT] means $1 in dom q implies r.(len p + $1) = q.$1;
A53: now
      let k be Element of NAT;
      assume
A54:  P[k];
      thus P[k+1]
      proof
        set a = len p + (k + 1);
        assume
A55:    k + 1 in dom q;
        then k + 1 <= len q by Th25;
        then
A56:    a <= len r by A10,XREAL_1:7;
A57:    1 <= k + 1 by NAT_1:12;
        then 1 <= a by NAT_1:12;
        then
A58:    a in dom r by A56,Th25;
        then
A59:    r.a in rng r by FUNCT_1:def 3;
        reconsider m = r.a as Element of NAT by A59;
A60:    now
          assume m in X;
          then m in rng p by A1,FINSEQ_1:def 13;
          then consider x such that
A61:      x in dom p and
A62:      p.x = m by FUNCT_1:def 3;
          reconsider x as Element of NAT by A61;
          x <= len p by A61,Th25;
          then
A63:      x <= len r by A10,NAT_1:12;
A64:      r is one-to-one by A1,A2,A3,Lm1,XBOOLE_1:8;
          1 <= x by A61,Th25;
          then
A65:      x in dom r by A63,Th25;
          r.x = r.a by A52,A61,A62;
          then x = a by A58,A64,A65,FUNCT_1:def 4;
          then len p + (k + 1) <= len p + 0 by A61,Th25;
          hence contradiction by XREAL_1:29;
        end;
A66:    q.(k + 1) in rng q by A55,FUNCT_1:def 3;
        then reconsider n = q.(k + 1) as Element of NAT;
A67:    n in Y by A2,A66,FINSEQ_1:def 13;
        assume
A68:    r.(len p + (k + 1)) <> q.(k + 1);
A69:    m in X \/ Y by A4,A59,FINSEQ_1:def 13;
        now
          per cases;
          suppose
A70:        k in dom q;
            then q.k in rng q by FUNCT_1:def 3;
            then reconsider n1 = q.k as Element of NAT;
            1 <= k by A70,Th25;
            then
A71:        1 <= len p + k by NAT_1:12;
A72:        k <= len q by A70,Th25;
            then len p + k <= len r by A10,XREAL_1:7;
            then len p + k in dom r by A71,Th25;
            then r.(len p + k) in rng r by FUNCT_1:def 3;
            then reconsider m1 = r.(len p + k) as Element of NAT;
            now
              per cases by A68,XXREAL_0:1;
              suppose
A73:            m < n;
A74:            1 <= k + 1 by A55,Th25;
                m in Y by A69,A60,XBOOLE_0:def 3;
                then m in rng q by A2,FINSEQ_1:def 13;
                then consider x such that
A75:            x in dom q and
A76:            q.x = m by FUNCT_1:def 3;
                reconsider x as Element of NAT by A75;
                x <= len q by A75,Th25;
                then
A77:            x < k + 1 by A2,A73,A76,A74,Th41;
                len p + k < len p + k + 1 by XREAL_1:29;
                then
A78:            n1 < m by A4,A54,A56,A70,A71,FINSEQ_1:def 13;
A79:            k <= len q by A70,Th25;
                1 <= x by A75,Th25;
                then k < x by A2,A76,A78,A79,Th41;
                hence contradiction by A77,NAT_1:13;
              end;
              suppose
A80:            n < m;
A81:            1 <= a by A57,NAT_1:12;
                n in X \/ Y by A67,XBOOLE_0:def 3;
                then n in rng r by A4,FINSEQ_1:def 13;
                then consider x such that
A82:            x in dom r and
A83:            r.x = n by FUNCT_1:def 3;
                reconsider x as Element of NAT by A82;
                x <= len r by A82,Th25;
                then
A84:            x < len p + k + 1 by A1,A2,A3,A80,A83,A81,Th41,XBOOLE_1:8;
A85:            k + 1 <= len q by A55,Th25;
A86:            k < k + 1 by XREAL_1:29;
                1 <= k by A70,Th25;
                then
A87:            m1 < n by A2,A54,A70,A86,A85,FINSEQ_1:def 13;
A88:            len p + k <= len r by A10,A72,XREAL_1:7;
                1 <= x by A82,Th25;
                then len p + k < x by A1,A2,A3,A83,A87,A88,Th41,XBOOLE_1:8;
                hence contradiction by A84,NAT_1:13;
              end;
            end;
            hence contradiction;
          end;
          suppose
A89:        not k in dom q;
A90:        k < k + 1 by XREAL_1:29;
            k < 1 or len q < k by A89,Th25;
            then
A91:        k = 0 or len q < k + 1 & k + 1 <= len q by A55,A90,Th25,NAT_1:14
,XXREAL_0:2;
            now
              per cases by A68,XXREAL_0:1;
              suppose
A92:            m < n;
A93:            1 <= k + 1 by A55,Th25;
                m in Y by A69,A60,XBOOLE_0:def 3;
                then m in rng q by A2,FINSEQ_1:def 13;
                then consider x such that
A94:            x in dom q and
A95:            q.x = m by FUNCT_1:def 3;
                reconsider x as Element of NAT by A94;
                x <= len q by A94,Th25;
                then x < k + 1 by A2,A92,A95,A93,Th41;
                hence contradiction by A91,A94,Th24,NAT_1:14;
              end;
              suppose
A96:           n < m;
A97:           1 <= len p + 1 by NAT_1:12;
                n in X \/ Y by A67,XBOOLE_0:def 3;
                then n in rng r by A4,FINSEQ_1:def 13;
                then consider x such that
A98:           x in dom r and
A99:           r.x = n by FUNCT_1:def 3;
                reconsider x as Element of NAT by A98;
                x <= len r by A98,Th25;
                then x < len p + 1 by A1,A2,A3,A91,A96,A99,A97,Th41,XBOOLE_1:8;
                then
A100:           x <= len p by NAT_1:13;
                1 <= x by A98,Th25;
                then
A101:           x in dom p by A100,Th25;
                then
A102:           p.x in rng p by FUNCT_1:def 3;
                n = p.x by A52,A99,A101;
                then n in X by A1,A102,FINSEQ_1:def 13;
                hence contradiction by A5,A67;
              end;
            end;
            hence contradiction;
          end;
        end;
        hence contradiction;
      end;
    end;
A103: P[0] by Th24;
    for k being Element of NAT holds P[k] from NAT_1:sch 1(A103,A53);
    hence thesis by A10,A52,Th38;
  end;
  assume
A104: Sgm(X \/ Y) = Sgm(X) ^ Sgm(Y);
  let m,n be Element of NAT;
  assume that
A105: m in X and
A106: n in Y;
  n in rng q by A2,A106,FINSEQ_1:def 13;
  then consider y such that
A107: y in dom q and
A108: q.y = n by FUNCT_1:def 3;
  reconsider y as Element of NAT by A107;
A109: n = r.(len p + y) by A104,A107,A108,FINSEQ_1:def 7;
  y <= len q by A107,Th25;
  then len p + y <= len p + len q by XREAL_1:7;
  then
A110: len p + y <= len r by A104,FINSEQ_1:22;
  m in rng(Sgm X) by A1,A105,FINSEQ_1:def 13;
  then consider x such that
A111: x in dom p and
A112: p.x = m by FUNCT_1:def 3;
  reconsider x as Element of NAT by A111;
A113: x in Seg(len p) by A111,FINSEQ_1:def 3;
  then
A114: 1 <= x by FINSEQ_1:1;
A115: x <= len p by A113,FINSEQ_1:1;
  y <> 0 by A107,Th24;
  then
A116: x < len p + y by A115,NAT_1:16,XXREAL_0:2;
  m = r.x by A104,A111,A112,FINSEQ_1:def 7;
  hence thesis by A4,A109,A114,A116,A110,FINSEQ_1:def 13;
end;

theorem Th43:
  Sgm {} = {}
proof
  {} c= Seg 0;
  hence thesis by FINSEQ_1:51;
end;



theorem
  0 <> n implies Sgm{n} = <* n *>
proof
  assume 0 <> n;
  then n in Seg n by FINSEQ_1:3;
  then
A1: {n} c= Seg n by ZFMISC_1:31;
  then len(Sgm{n}) = card{n} by Th39;
  then
A2: len(Sgm{n}) = 1 by CARD_1:30;
  rng(Sgm{n}) = {n} by A1,FINSEQ_1:def 13;
  hence thesis by A2,FINSEQ_1:39;
end;

theorem
  0 < n & n < m implies Sgm{n,m} = <* n,m *>
proof
  assume that
A1: 0 < n and
A2: n < m;
A3: m in Seg m by A2,FINSEQ_1:3;
A4: n in Seg n by A1,FINSEQ_1:3;
  Seg n c= Seg m by A2,FINSEQ_1:5;
  then
A5: {n,m} c= Seg m by A3,A4,ZFMISC_1:32;
  then
A6: Sgm{n,m} is one-to-one by Lm1;
A7: len(Sgm{n,m}) = card{n,m} by A5,Th39
    .= 2 by A2,CARD_2:57;
  then
A8: dom(Sgm{n,m}) = {1,2} by FINSEQ_1:2,def 3;
  then
A9: 1 in dom(Sgm{n,m}) by TARSKI:def 2;
A10: 2 in dom(Sgm{n,m}) by A8,TARSKI:def 2;
A11: rng(Sgm{n,m}) = {n,m} by A5,FINSEQ_1:def 13;
  then
A12: Sgm{n,m}.2 in {n,m} by A10,FUNCT_1:def 3;
A13: Sgm{n,m}.1 in {n,m} by A9,A11,FUNCT_1:def 3;
  now
    per cases by A13,A12,TARSKI:def 2;
    suppose
      Sgm{n,m}.1 = n & Sgm{n,m}.2 = n;
      hence Sgm{n,m}.1 = n & Sgm{n,m}.2 = m by A9,A10,A6,FUNCT_1:def 4;
    end;
    suppose
      Sgm{n,m}.1 = m & Sgm{n,m}.2 = m;
      hence Sgm{n,m}.1 = n & Sgm{n,m}.2 = m by A9,A10,A6,FUNCT_1:def 4;
    end;
    suppose
      Sgm{n,m}.1 = n & Sgm{n,m}.2 = m;
      hence Sgm{n,m}.1 = n & Sgm{n,m}.2 = m;
    end;
    suppose
      Sgm{n,m}.1 = m & Sgm{n,m}.2 = n;
      hence Sgm{n,m}.1 = n & Sgm{n,m}.2 = m by A2,A5,A7,FINSEQ_1:def 13;
    end;
  end;
  hence thesis by A7,FINSEQ_1:44;
end;

theorem Th46:
  len(Sgm(Seg k)) = k
proof
  card(Seg k) = k by FINSEQ_1:57;
  hence thesis by Th39;
end;

Lm2: Sgm(Seg(k + 0)) | Seg k = Sgm(Seg k)
proof
  card(Seg k) = k by FINSEQ_1:57;
  then len(Sgm(Seg k)) = k by Th39;
  then dom(Sgm(Seg k)) = Seg k by FINSEQ_1:def 3;
  hence thesis by RELAT_1:68;
end;

Lm3: now
  let n;
  assume
A1: for k holds Sgm(Seg(k + n)) | Seg k = Sgm(Seg k);
  let k;
  set X = Sgm(Seg(k + (n + 1)));
  set Y = Sgm(Seg(k + 1));
A2: Y | Seg k = Sgm(Seg k)
  proof
    reconsider p = Y | Seg k as FinSequence of NAT by FINSEQ_1:18;
A3: len Y = k + 1 by Th46;
    then
A4: dom Y = Seg(k + 1) by FINSEQ_1:def 3;
A5: k <= k + 1 by NAT_1:12;
    then
A6: dom p = Seg k by A3,FINSEQ_1:17;
A7: rng Y = Seg(k + 1) by FINSEQ_1:def 13;
A8: Y.(k + 1) = k + 1
    proof
      k + 1 in dom Y by A4,FINSEQ_1:4;
      then
A9:   Y.(k + 1) in Seg(k + 1) by A7,FUNCT_1:def 3;
      then reconsider n = Y.(k + 1) as Element of NAT;
A10:  k < k + 1 by XREAL_1:29;
      k + 1 in rng Y by A7,FINSEQ_1:4;
      then consider x such that
A11:  x in dom Y and
A12:  Y.x = k + 1 by FUNCT_1:def 3;
      reconsider x as Element of NAT by A11;
      assume not thesis;
      then not Y.(k + 1) in {k + 1} by TARSKI:def 1;
      then Y.(k + 1) in Seg(k + 1) \ {k + 1} by A9,XBOOLE_0:def 5;
      then
A13:  Y.(k + 1) in Seg k by FINSEQ_1:10;
A14:  now
        assume
A15:    x = k + 1;
        n <= k by A13,FINSEQ_1:1;
        hence contradiction by A12,A15,XREAL_1:29;
      end;
      x <= k + 1 by A3,A11,Th25;
      then
A16:  x < k + 1 by A14,XXREAL_0:1;
      1 <= x by A11,Th25;
      then
A17:  k + 1 < n by A3,A12,A16,FINSEQ_1:def 13;
      n <= k by A13,FINSEQ_1:1;
      hence contradiction by A17,A10,XXREAL_0:2;
    end;
A18: Y is one-to-one by Lm1;
    rng p = rng Y \ {Y.(k + 1)}
    proof
      thus rng p c= rng Y \ {Y.(k + 1)}
      proof
        let x;
        assume
A19:    x in rng p;
A20:    now
          assume x in {k + 1};
          then
A21:      x = k + 1 by TARSKI:def 1;
A22:      k < k + 1 by XREAL_1:29;
A23:      k + 1 in dom Y by A4,FINSEQ_1:4;
A24:      Seg k c= Seg(k + 1) by Th18;
          consider y such that
A25:      y in dom p and
A26:      p.y = x by A19,FUNCT_1:def 3;
          reconsider y as Element of NAT by A25;
A27:      Y.y = p.y by A25,FUNCT_1:47;
          y <= k by A6,A25,FINSEQ_1:1;
          hence contradiction by A18,A4,A6,A8,A25,A26,A24,A27,A21,A23,A22,
FUNCT_1:def 4;
        end;
        rng p c= rng Y by RELAT_1:70;
        hence thesis by A8,A19,A20,XBOOLE_0:def 5;
      end;
      let x;
      assume
A28:  x in rng Y \ {Y.(k + 1)};
      then x in rng Y by XBOOLE_0:def 5;
      then consider y such that
A29:  y in dom Y and
A30:  Y.y = x by FUNCT_1:def 3;
      now
        assume y in {k + 1};
        then
A31:    x = k + 1 by A8,A30,TARSKI:def 1;
        not x in {k + 1} by A8,A28,XBOOLE_0:def 5;
        hence contradiction by A31,TARSKI:def 1;
      end;
      then y in Seg(k + 1) \ {k + 1} by A4,A29,XBOOLE_0:def 5;
      then
A32:  y in dom p by A6,FINSEQ_1:10;
      then p.y = Y.y by FUNCT_1:47;
      hence thesis by A30,A32,FUNCT_1:def 3;
    end;
    then
A33: rng p = Seg k by A7,A8,FINSEQ_1:10;
    now
A34:  len p = k by A3,A5,FINSEQ_1:17;
      let i,j,l,m;
      assume that
A35:  1 <= i and
A36:  i < j and
A37:  j <= len p and
A38:  l = p.i and
A39:  m = p.j;
      i <= len p by A36,A37,XXREAL_0:2;
      then i in dom p by A6,A35,A34,FINSEQ_1:1;
      then
A40:  p.i = Y.i by FUNCT_1:47;
      1 <= j by A35,A36,XXREAL_0:2;
      then j in dom p by A6,A37,A34,FINSEQ_1:1;
      then
A41:  p.j = Y.j by FUNCT_1:47;
      len Y = k + 1 by Th46;
      then j <= len Y by A37,A34,NAT_1:12;
      hence l < m by A35,A36,A38,A39,A40,A41,FINSEQ_1:def 13;
    end;
    hence thesis by A33,FINSEQ_1:def 13;
  end;
  X = Sgm(Seg(k + 1 + n));
  then X | Seg(k + 1) = Y by A1;
  then Sgm(Seg k) = X | (Seg(k + 1) /\ Seg k) by A2,RELAT_1:71;
  hence Sgm(Seg(k + (n + 1))) | Seg k = Sgm(Seg k) by FINSEQ_1:7,NAT_1:12;
end;

Lm4: for k holds Sgm(Seg(k + n)) | Seg k = Sgm(Seg k)
proof
  defpred P[Nat] means for k holds Sgm(Seg(k + $1)) | Seg k = Sgm(
  Seg k);
A1: for k st P[k] holds P[k+1] by Lm3;
A2: P[0] by Lm2;
  for n holds P[n] from NAT_1:sch 2(A2,A1);
  hence thesis;
end;

theorem
  Sgm(Seg(k + n)) | Seg k = Sgm(Seg k) by Lm4;

Lm5: now
  let k be Nat;
  assume
A1: Sgm(Seg k) = idseq k;
A2: len(idseq(k + 1)) = k + 1 by CARD_1:def 7;
  then
A3: len(Sgm(Seg(k + 1))) = len(idseq(k + 1)) by Th46;
  then
A4: dom Sgm(Seg(k + 1)) = Seg(k+1) by A2,FINSEQ_1:def 3;
  now
    let j be Nat;
    assume
A5: j in dom Sgm(Seg(k + 1));
    then
A6: j in Seg k \/ {k + 1} by A4,FINSEQ_1:9;
    now
      per cases by A6,XBOOLE_0:def 3;
      suppose
A7:     j in Seg k;
A8:     idseq(k + 1).j = j by A4,A5,FUNCT_1:18;
A9:     Sgm(Seg(k + 1)) | Seg k = Sgm(Seg k) by Lm4;
        Sgm(Seg k).j = j by A1,A7,FUNCT_1:18;
        hence Sgm(Seg(k + 1)).j = idseq(k + 1).j by A7,A8,A9,FUNCT_1:49;
      end;
      suppose
A10:    j in {k + 1};
        set Y = Sgm(Seg k);
        set X = Sgm(Seg(k + 1));
A11:    j = k + 1 by A10,TARSKI:def 1;
        then
A12:    j in Seg(k + 1) by FINSEQ_1:4;
        now
          rng X = Seg(k + 1) by FINSEQ_1:def 13;
          then
A13:      X.j in Seg(k + 1) by A5,FUNCT_1:def 3;
          then reconsider n = X.j as Element of NAT;
          assume X.j <> j;
          then not X.j in {j} by TARSKI:def 1;
          then X.j in Seg(k + 1) \ {k + 1} by A11,A13,XBOOLE_0:def 5;
          then
A14:      X.j in Seg k by FINSEQ_1:10;
A15:      X is one-to-one by Lm1;
A16:      dom X = Seg(k + 1) by A2,A3,FINSEQ_1:def 3;
A17:      k < k + 1 by XREAL_1:29;
          X | Seg k = Y by Lm4;
          then
A18:      X.n = Y.n by A14,FUNCT_1:49
            .= n by A1,A14,FUNCT_1:18;
          n <= k by A14,FINSEQ_1:1;
          hence contradiction by A11,A12,A16,A13,A15,A18,A17,FUNCT_1:def 4;
        end;
        hence Sgm(Seg(k + 1)).j = idseq(k + 1).j by A11,FINSEQ_1:4,FUNCT_1:18;
      end;
    end;
    hence Sgm(Seg(k + 1)).j = idseq(k + 1).j;
  end;
  hence Sgm(Seg(k + 1)) = idseq(k + 1) by A3,FINSEQ_2:9;
end;

theorem Th48:
  Sgm(Seg k) = idseq k
proof
  defpred P[Nat] means Sgm(Seg $1) = idseq $1;
A1: for k being Nat st P[k] holds P[k+1] by Lm5;
A2: P[0] by Th43;
  for k being Nat holds P[k] from NAT_1:sch 2(A2,A1);
  hence thesis;
end;

theorem Th49:
  p | Seg n = p iff len p <= n
proof
  thus p | Seg n = p implies len p <= n by FINSEQ_1:86;
  assume len p <= n;
  then Seg(len p) c= Seg n by FINSEQ_1:5;
  then dom p c= Seg n by FINSEQ_1:def 3;
  then
A1: dom p = dom p /\ Seg n by XBOOLE_1:28;
  for x st x in dom p holds p.x = p.x;
  hence thesis by A1,FUNCT_1:46;
end;

theorem Th50:
  idseq(n + k) | Seg n = idseq n
proof
A1: Sgm(Seg n) = idseq n by Th48;
  Sgm(Seg(n + k)) = idseq(n + k) by Th48;
  hence thesis by A1,Lm4;
end;

theorem
  idseq n | Seg m = idseq m iff m <= n
proof
  thus idseq n | Seg m = idseq m implies m <= n
  proof
    assume idseq n | Seg m = idseq m;
    then len(idseq m) <= len(idseq n) by FINSEQ_1:79;
    then m <= len(idseq n) by CARD_1:def 7;
    hence thesis by CARD_1:def 7;
  end;
  assume m <= n;
  then ex j being Nat st n = m + j by NAT_1:10;
  hence thesis by Th50;
end;

theorem
  idseq n | Seg m = idseq n iff n <= m
proof
  len(idseq n) = n by CARD_1:def 7;
  hence thesis by Th49;
end;

theorem Th53:
  len p = k + l & q = p | Seg k implies len q = k
proof
  assume that
A1: len p = k + l and
A2: q = p | Seg k;
  k <= len p by A1,NAT_1:12;
  hence thesis by A2,FINSEQ_1:17;
end;

theorem
  len p = k + l & q = p | Seg k implies dom q = Seg k
proof
  assume that
A1: len p = k + l and
A2: q = p | Seg k;
  len q = k by A1,A2,Th53;
  hence thesis by FINSEQ_1:def 3;
end;

theorem Th55:
  len p = k + 1 & q = p | Seg k implies p = q ^ <* p.(k + 1) *>
proof
  assume that
A1: len p = k + 1 and
A2: q = p | Seg k;
A3: for l being Element of NAT holds l in dom q implies p.l = q.l by A2,
FUNCT_1:47;
  set r = <* p.(k + 1) *>;
A4: now
    let l be Element of NAT;
    assume l in dom r;
    then l in {1} by FINSEQ_1:2,38;
    then
A5: l = 1 by TARSKI:def 1;
    hence p.(len q + l) = p.(k + 1) by A1,A2,Th53
      .= r.l by A5,FINSEQ_1:40;
  end;
  len p = len q + 1 by A1,A2,Th53
    .= len q + len<* p.(k + 1) *> by FINSEQ_1:39;
  hence thesis by A3,A4,Th38;
end;

theorem
  p | X is FinSequence iff ex k being Element of NAT st X /\ dom p = Seg k
proof
  thus p | X is FinSequence implies ex k being Element of NAT st X /\ dom p =
  Seg k
  proof
    assume p | X is FinSequence;
    then consider k being Nat such that
A1: dom(p | X) = Seg k by FINSEQ_1:def 2;
    reconsider k as Element of NAT by ORDINAL1:def 12;
    take k;
    thus thesis by A1,RELAT_1:61;
  end;
  given k being Element of NAT such that
A2: X /\ dom p = Seg k;
  dom(p | X) = Seg k by A2,RELAT_1:61;
  hence thesis by FINSEQ_1:def 2;
end;

theorem Th57:
  card((p ^ q) " A) = card(p " A) + card(q " A)
proof
  set X = (p ^ q) " A;
  set B = {len p + n where n is Element of NAT : n in q " A};
  defpred P[set,set] means ex i st $1 = i & $2 = len p + i;
A1: X = p " A \/ B
  proof
    thus X c= p " A \/ B
    proof
      let x;
      assume
A2:   x in X;
      then
A3:   x in dom(p ^ q) by FUNCT_1:def 7;
      then reconsider k = x as Element of NAT;
      now
        per cases by A3,FINSEQ_1:25;
        suppose
A4:       k in dom p;
          then (p ^ q).k = p.k by FINSEQ_1:def 7;
          then p.k in A by A2,FUNCT_1:def 7;
          then k in p " A by A4,FUNCT_1:def 7;
          hence thesis by XBOOLE_0:def 3;
        end;
        suppose
          ex m being Nat st m in dom q & k = len p + m;
          then consider m being Nat such that
A5:       m in dom q and
A6:       k = len p + m;
          q.m = (p ^ q).k by A5,A6,FINSEQ_1:def 7;
          then q.m in A by A2,FUNCT_1:def 7;
          then
A7:       m in q " A by A5,FUNCT_1:def 7;
          m in NAT by ORDINAL1:def 12;
          then k in B by A6,A7;
          hence thesis by XBOOLE_0:def 3;
        end;
      end;
      hence thesis;
    end;
    let x;
    assume
A8: x in p " A \/ B;
    now
      per cases by A8,XBOOLE_0:def 3;
      suppose
A9:     x in p " A;
        then
A10:    x in dom p by FUNCT_1:def 7;
        then reconsider k = x as Element of NAT;
        (p ^ q).k = p.k by A10,FINSEQ_1:def 7;
        then
A11:    (p ^ q).x in A by A9,FUNCT_1:def 7;
        x in dom(p ^ q) by A10,Th22;
        hence thesis by A11,FUNCT_1:def 7;
      end;
      suppose
        x in B;
        then consider n being Element of NAT such that
A12:    x = len p + n and
A13:    n in q " A;
A14:    n in dom q by A13,FUNCT_1:def 7;
        then (p ^ q).(len p + n) = q.n by FINSEQ_1:def 7;
        then
A15:    (p ^ q).x in A by A12,A13,FUNCT_1:def 7;
        x in dom(p ^ q) by A12,A14,FINSEQ_1:28;
        hence thesis by A15,FUNCT_1:def 7;
      end;
    end;
    hence thesis;
  end;
  p " A /\ B = {}
  proof
    set x = the Element of p " A /\ B;
    assume
A16: not thesis;
    then x in B by XBOOLE_0:def 4;
    then consider n being Element of NAT such that
A17: x = len p + n and
A18: n in q " A;
    len p + n in p " A by A16,A17,XBOOLE_0:def 4;
    then len p + n in dom p by FUNCT_1:def 7;
    then len p + n <= len p + 0 by Th25;
    then
A19: n = 0 by XREAL_1:6;
    n in dom q by A18,FUNCT_1:def 7;
    hence thesis by A19,Th24;
  end;
  then
A20: p " A misses B by XBOOLE_0:def 7;
  reconsider B as finite set by A1,FINSET_1:1,XBOOLE_1:7;
A21: card X = card(p " A) + card B by A1,A20,CARD_2:40;
A22: for x st x in q " A ex y st P[x,y]
  proof
    let x;
    assume x in q " A;
    then x in dom q by FUNCT_1:def 7;
    then reconsider i = x as Element of NAT;
    reconsider y = len p + i as set;
    take y;
    take i;
    thus thesis;
  end;
  consider f being Function such that
A23: dom f = q " A and
A24: for x st x in q " A holds P[x,f.x] from CLASSES1:sch 1(A22);
A25: rng f = B
  proof
    thus rng f c= B
    proof
      let x;
      assume x in rng f;
      then consider y such that
A26:  y in dom f and
A27:  f.y = x by FUNCT_1:def 3;
      consider i such that
A28:  y = i and
A29:  f.y = len p + i by A23,A24,A26;
      i is Element of NAT by ORDINAL1:def 12;
      hence thesis by A23,A26,A27,A28,A29;
    end;
    let x;
    assume x in B;
    then consider n being Element of NAT such that
A30: x = len p + n and
A31: n in q " A;
    ex i st n = i & f.n = len p + i by A24,A31;
    hence thesis by A23,A30,A31,FUNCT_1:def 3;
  end;
  f is one-to-one
  proof
    let x,y;
    assume that
A32: x in dom f and
A33: y in dom f and
A34: f.x = f.y;
A35: ex j st y = j & f.y = len p + j by A23,A24,A33;
    ex i st x = i & f.x = len p + i by A23,A24,A32;
    hence thesis by A34,A35;
  end;
  then q " A,B are_equipotent by A23,A25,WELLORD2:def 4;
  hence thesis by A21,CARD_1:5;
end;

theorem Th58:
  p " A c= (p ^ q) " A
proof
  let x;
A1: dom p c= dom(p ^ q) by FINSEQ_1:26;
  assume
A2: x in p " A;
  then
A3: x in dom p by FUNCT_1:def 7;
  then reconsider k = x as Element of NAT;
A4: p.k in A by A2,FUNCT_1:def 7;
  p.k = (p ^ q).k by A3,FINSEQ_1:def 7;
  hence thesis by A3,A1,A4,FUNCT_1:def 7;
end;

definition
  let p,A;
  func p - A -> FinSequence equals
  p * Sgm ((dom p) \ p " A);
  coherence
  proof
    now
      assume p <> {};
      then reconsider D = Seg(len p) as non empty Subset of NAT;
      Seg(len p) \ p " A c= Seg(len p) by XBOOLE_1:36;
      then rng Sgm(Seg(len p) \ p " A) c= D by FINSEQ_1:def 13;
      then reconsider q = Sgm(Seg(len p) \ p " A) as FinSequence of D by
FINSEQ_1:def 4;
      reconsider r = p * q as FinSequence by FINSEQ_2:30;
      take rr = r;
      thus rr = p * Sgm((dom p) \ p " A) by FINSEQ_1:def 3;
    end;
    hence thesis;
  end;
end;

theorem Th59:
  len(p - A) = len p - card(p " A)
proof
  set q = Sgm(Seg(len p) \ p " A);
A1: Seg len p = dom p by FINSEQ_1:def 3;
  Seg(len p) \ p " A c= Seg(len p) by XBOOLE_1:36;
  then rng q c= dom p by A1,FINSEQ_1:def 13;
  then
A2: dom q = dom(p - A) by A1,RELAT_1:27;
A3: dom q = Seg(len q) by FINSEQ_1:def 3;
A4: p " A c= Seg(len p)
  proof
    let x;
A5: p " A c= dom p by RELAT_1:132;
A6: dom p = Seg(len p) by FINSEQ_1:def 3;
    assume x in p " A;
    hence thesis by A5,A6;
  end;
  len q = card(Seg(len p) \ p " A) by Th39,XBOOLE_1:36;
  then len(p - A) = card(Seg(len p) \ p " A) by A2,A3,FINSEQ_1:def 3;
  hence len(p - A) = card(Seg(len p)) - card(p " A) by A4,CARD_2:44
    .= len p - card(p " A) by FINSEQ_1:57;
end;

theorem Th60:
  len(p - A) <= len p
proof
  len(p - A) = len p - card(p " A) by Th59;
  hence thesis by XREAL_1:43;
end;

theorem Th61:
  len(p - A) = len p implies A misses rng p
proof
  assume
A1: len(p - A) = len p;
A2: len(p - A) = len p - card(p " A) by Th59;
  assume A meets rng p;
  then p " A <> {} by RELAT_1:138;
  hence thesis by A1,A2;
end;

theorem
  n = len p - card(p " A) implies dom(p - A) = Seg n
proof
  assume n = len p - card(p " A);
  then len(p - A) = n by Th59;
  hence thesis by FINSEQ_1:def 3;
end;

theorem
  dom(p - A) c= dom p
proof
A1: dom(p - A) = Seg(len(p - A)) by FINSEQ_1:def 3;
A2: dom p = Seg(len p) by FINSEQ_1:def 3;
  len(p - A) <= len p by Th60;
  hence thesis by A1,A2,FINSEQ_1:5;
end;

theorem
  dom(p - A) = dom p implies A misses rng p
proof
A1: dom(p - A) = Seg(len(p - A)) by FINSEQ_1:def 3;
A2: dom p = Seg(len p) by FINSEQ_1:def 3;
  assume dom(p - A) = dom p;
  hence thesis by A1,A2,Th61,FINSEQ_1:6;
end;

theorem Th65:
  rng(p - A) = rng p \ A
proof
  set q = Sgm(Seg(len p) \ p " A);
A1: dom p = Seg(len p) by FINSEQ_1:def 3;
A2: dom p = Seg len p by FINSEQ_1:def 3;
  thus rng(p - A) c= rng p \ A
  proof
    let x;
A3: rng(p * q) c= rng p by RELAT_1:26;
    assume
A4: x in rng(p - A);
A5: now
A6:   Seg(len p) \ p " A c= Seg(len p) by XBOOLE_1:36;
      assume
A7:   x in A;
      consider y such that
A8:   y in dom(p - A) and
A9:   (p - A).y = x by A4,FUNCT_1:def 3;
      set z = q.y;
A10:  y in dom q by A2,A8,FUNCT_1:11;
      then
A11:  (p - A).y = p.z by A2,FUNCT_1:13;
      z in rng q by A10,FUNCT_1:def 3;
      then z in Seg(len p) \ p " A by A6,FINSEQ_1:def 13;
      then
A12:  not z in p " A by XBOOLE_0:def 5;
      z in dom p by A2,A8,FUNCT_1:11;
      hence contradiction by A7,A9,A11,A12,FUNCT_1:def 7;
    end;
    x in rng(p * q) by A4,FINSEQ_1:def 3;
    hence thesis by A3,A5,XBOOLE_0:def 5;
  end;
  let x;
  assume
A13: x in rng p \ A;
  then consider y such that
A14: y in dom p and
A15: p.y = x by FUNCT_1:def 3;
  Seg(len p) \ p " A c= Seg(len p) by XBOOLE_1:36;
  then
A16: rng q = Seg(len p) \ p " A by FINSEQ_1:def 13;
  not p.y in A by A13,A15,XBOOLE_0:def 5;
  then not y in p " A by FUNCT_1:def 7;
  then y in rng q by A14,A1,A16,XBOOLE_0:def 5;
  then consider z such that
A17: z in dom q and
A18: q.z = y by FUNCT_1:def 3;
A19: (p - A).z = x by A2,A15,A17,A18,FUNCT_1:13;
  z in dom(p - A) by A2,A14,A17,A18,FUNCT_1:11;
  hence thesis by A19,FUNCT_1:def 3;
end;

theorem
  rng(p - A) c= rng p
proof
  rng(p - A) = rng p \ A by Th65;
  hence thesis;
end;

theorem
  rng(p - A) = rng p implies A misses rng p
proof
  assume rng(p - A) = rng p;
  then rng p \ A = rng p by Th65;
  hence thesis by XBOOLE_1:83;
end;

theorem Th68:
  p - A = {} iff rng p c= A
proof
  thus p - A = {} implies rng p c= A
  proof
    assume that
A1: p - A = {} and
A2: not rng p c= A;
    rng p \ A <> {} by A2,XBOOLE_1:37;
    then rng(p - A) <> {} by Th65;
    hence thesis by A1;
  end;
  assume
A3: rng p c= A;
  rng(p - A) = rng p \ A by Th65;
  hence thesis by A3,XBOOLE_1:37;
end;

theorem Th69:
  p - A = p iff A misses rng p
proof
  thus p - A = p implies A misses rng p
  proof
    assume that
A1: p - A = p and
A2: not A misses rng p;
    len(p - A) <> len p by A2,Th61;
    hence contradiction by A1;
  end;
  assume A misses rng p;
  then p " A = {} by RELAT_1:138;
  then Sgm(Seg(len p) \ p " A) = idseq(len p) by Th48;
  then p * Sgm(Seg(len p) \ p " A) = p by FINSEQ_2:54;
  hence thesis by FINSEQ_1:def 3;
end;

theorem
  p - {x} = p iff not x in rng p
proof
  thus p - {x} = p implies not x in rng p
  proof
    assume p - {x} = p;
    then
A1: {x} misses rng p by Th69;
    x in {x} by TARSKI:def 1;
    hence thesis by A1,XBOOLE_0:3;
  end;
  assume
A2: not x in rng p;
  {x} misses rng p
  proof
    assume {x} meets rng p;
    then ex y st y in {x} & y in rng p by XBOOLE_0:3;
    hence thesis by A2,TARSKI:def 1;
  end;
  hence thesis by Th69;
end;

theorem
  p - {} = p
proof
  {} misses rng p by XBOOLE_1:65;
  hence thesis by Th69;
end;

theorem
  p - rng p = {} by Th68;

Lm6: <* x *> - A = <* x *> iff not x in A
proof
  thus <* x *> - A = <* x *> implies not x in A
  proof
    assume <* x *> - A = <* x *>;
    then
A1: rng<* x *> misses A by Th69;
A2: rng<* x *> = {x} by FINSEQ_1:39;
    x in {x} by TARSKI:def 1;
    hence thesis by A2,A1,XBOOLE_0:3;
  end;
  assume
A3: not x in A;
  rng<* x *> misses A
  proof
    assume rng<* x *> meets A;
    then consider y such that
A4: y in rng<* x *> and
A5: y in A by XBOOLE_0:3;
    y in {x} by A4,FINSEQ_1:39;
    hence thesis by A3,A5,TARSKI:def 1;
  end;
  hence thesis by Th69;
end;

Lm7: <* x *> - A = {} iff x in A
proof
A1: rng<* x *> = {x} by FINSEQ_1:39;
  thus <* x *> - A = {} implies x in A
  proof
    assume <* x *> - A = {};
    then rng<* x *> c= A by Th68;
    then {x} c= A by FINSEQ_1:39;
    hence thesis by ZFMISC_1:31;
  end;
  assume x in A;
  then rng<* x *> c= A by A1,ZFMISC_1:31;
  hence thesis by Th68;
end;

Lm8: (p ^ {}) - A = (p - A) ^ ({} - A)
proof
  thus (p ^ {}) - A = p - A by FINSEQ_1:34
    .= (p - A) ^ ({} - A) by FINSEQ_1:34;
end;

Lm9: (p ^ <* x *>) - A = (p - A) ^ (<* x *> - A)
proof
  set r = (p ^ <* x *>) - A;
  set q = <* x *> - A;
  set t = p ^ <* x *>;
A1: len t = len p + len<* x *> by FINSEQ_1:22
    .= len p + 1 by FINSEQ_1:40;
A2: now
    t " A c= dom t by RELAT_1:132;
    then
A3: t " A c= Seg(len t) by FINSEQ_1:def 3;
    set S = Seg(len(p ^ <* x *>)) \ (p ^ <* x *>) " A;
    set s = Sgm(Seg(len(p ^ <* x *>)) \ (p ^ <* x *>) " A);
    let k be Element of NAT;
A4: S c= Seg(len t) by XBOOLE_1:36;
    assume
A5: k in dom q;
    then
A6: not x in A by Lm7,RELAT_1:38;
    then
A7: q = <* x *> by Lm6;
    then
A8: dom q = {1} by FINSEQ_1:2,def 8;
    then
A9: k = 1 by A5,TARSKI:def 1;
A10: p " A = (p ^ <* x *>) " A
    proof
      thus p " A c= (p ^ <* x *>) " A
      proof
        let y;
        assume
A11:    y in p " A;
        then
A12:    y in dom p by FUNCT_1:def 7;
        then reconsider z = y as Element of NAT;
        p.y in A by A11,FUNCT_1:def 7;
        then
A13:    t.z in A by A12,FINSEQ_1:def 7;
        y in dom t by A12,Th22;
        hence thesis by A13,FUNCT_1:def 7;
      end;
      let y;
      assume
A14:  y in (p ^ <* x *>) " A;
      then
A15:  y in dom t by FUNCT_1:def 7;
      then reconsider z = y as Element of NAT;
A16:  t.y in A by A14,FUNCT_1:def 7;
A17:  now
        given n such that
A18:    n in dom q and
A19:    z = len p + n;
        n = 1 by A8,A18,TARSKI:def 1;
        hence contradiction by A6,A16,A19,FINSEQ_1:42;
      end;
A20:  z in dom p or ex n being Nat st n in dom q & z = len p + n by A7,A15,
FINSEQ_1:25;
      then t.z = p.z by A17,FINSEQ_1:def 7;
      hence thesis by A16,A20,A17,FUNCT_1:def 7;
    end;
A21: now
      assume
A22:  len t in t " A;
      t.(len t) = x by A1,FINSEQ_1:42;
      hence contradiction by A6,A22,FUNCT_1:def 7;
    end;
A23: dom s = Seg(len s) by FINSEQ_1:def 3;
    len t in Seg(len t) by A1,FINSEQ_1:4;
    then len t in S by A21,XBOOLE_0:def 5;
    then len t in rng s by A4,FINSEQ_1:def 13;
    then consider y such that
A24: y in dom s and
A25: s.y = len t by FUNCT_1:def 3;
    reconsider y as Element of NAT by A24;
A26: len(p - A) + k = len p - card(p " A) + k by Th59
      .= len p + 1 - card((p ^ <* x *>) " A) by A9,A10
      .= len p + len<* x *> - card((p ^ <* x *>) " A) by FINSEQ_1:39
      .= len(p ^ <* x *>) - card((p ^ <* x *>) " A) by FINSEQ_1:22
      .= len r by Th59;
A27: len s = card S by Th39,XBOOLE_1:36
      .= card(Seg(len t)) - card(t " A) by A3,CARD_2:44
      .= len t - card(p " A) by A10,FINSEQ_1:57
      .= len p - card(p " A) + k by A1,A9
      .= len r by A26,Th59;
    then
A28: len s = len(p - A) + 1 by A5,A8,A26,TARSKI:def 1;
    then
A29: len r in dom s by A27,A23,FINSEQ_1:4;
A30: now
A31:  s.(len s) in rng s by A27,A29,FUNCT_1:def 3;
      reconsider w = s.(len s) as Element of NAT by A31;
      w in S by A4,A31,FINSEQ_1:def 13;
      then
A32:  w in Seg(len t) by XBOOLE_0:def 5;
      assume
A33:  y <> len s;
A34:  y in Seg(len s) by A24,FINSEQ_1:def 3;
      then y <= len s by FINSEQ_1:1;
      then
A35:  y < len s by A33,XXREAL_0:1;
      1 <= y by A34,FINSEQ_1:1;
      then len t < w by A4,A25,A35,FINSEQ_1:def 13;
      hence contradiction by A32,FINSEQ_1:1;
    end;
    dom t = Seg len t by FINSEQ_1:def 3;
    hence r.(len(p - A) + k) = t.(len t) by A26,A27,A28,A23,A25,A30,FINSEQ_1:4
,FUNCT_1:13
      .= x by A1,FINSEQ_1:42
      .= q.k by A7,A9,FINSEQ_1:def 8;
  end;
A36: now
A37: x in A implies t " A = p " A \/ {len p + 1}
    proof
      assume
A38:  x in A;
      thus t " A c= p " A \/ {len p + 1}
      proof
        let y;
        assume
A39:    y in t " A;
        then y in dom t by FUNCT_1:def 7;
        then y in Seg(len p + 1) by A1,FINSEQ_1:def 3;
        then
A40:    y in Seg(len p) \/ {len p + 1} by FINSEQ_1:9;
        now
          per cases by A40,XBOOLE_0:def 3;
          suppose
A41:        y in Seg(len p);
            then reconsider j = y as Element of NAT;
A42:        t.y in A by A39,FUNCT_1:def 7;
A43:        y in dom p by A41,FINSEQ_1:def 3;
            then p.j = t.j by FINSEQ_1:def 7;
            then y in p " A by A43,A42,FUNCT_1:def 7;
            hence thesis by XBOOLE_0:def 3;
          end;
          suppose
            y in {len p + 1};
            hence thesis by XBOOLE_0:def 3;
          end;
        end;
        hence thesis;
      end;
      let y;
      assume
A44:  y in p " A \/ {len p + 1};
      now
        per cases by A44,XBOOLE_0:def 3;
        suppose
A45:      y in p " A;
          p " A c= t " A by Th58;
          hence thesis by A45;
        end;
        suppose
          y in {len p + 1};
          then
A46:      y = len p + 1 by TARSKI:def 1;
          then y in Seg(len t) by A1,FINSEQ_1:4;
          then
A47:      y in dom t by FINSEQ_1:def 3;
          t.y in A by A38,A46,FINSEQ_1:42;
          hence thesis by A47,FUNCT_1:def 7;
        end;
      end;
      hence thesis;
    end;
    t " A c= dom t by RELAT_1:132;
    then
A48: t " A c= Seg(len t) by FINSEQ_1:def 3;
A49: not x in A implies t " A = p " A
    proof
      assume
A50:  not x in A;
      thus t " A c= p " A
      proof
        let y;
        assume
A51:    y in t " A;
        then
A52:    y in dom t by FUNCT_1:def 7;
        then reconsider l = y as Element of NAT;
A53:    now
          assume l = len p + 1;
          then t.l = x by FINSEQ_1:42;
          hence contradiction by A50,A51,FUNCT_1:def 7;
        end;
A54:    y in Seg(len t) by A52,FINSEQ_1:def 3;
        then l <= len p + 1 by A1,FINSEQ_1:1;
        then l < len p + 1 by A53,XXREAL_0:1;
        then
A55:    l <= len p by NAT_1:13;
        1 <= l by A54,FINSEQ_1:1;
        then l in Seg(len p) by A55,FINSEQ_1:1;
        then
A56:    y in dom p by FINSEQ_1:def 3;
        t.l in A by A51,FUNCT_1:def 7;
        then p.l in A by A56,FINSEQ_1:def 7;
        hence thesis by A56,FUNCT_1:def 7;
      end;
      thus thesis by Th58;
    end;
    set s2 = Sgm(Seg(len p) \ p " A);
    set s1 = Sgm(Seg(len t) \ t " A);
    let k be Element of NAT;
A57: dom p = Seg len p by FINSEQ_1:def 3;
A58: now
      {len p + 1} /\ p " A = {}
      proof
        set z = the Element of {len p + 1} /\ p " A;
A59:    p " A c= dom p by RELAT_1:132;
        assume
A60:    not thesis;
        then z in {len p + 1} by XBOOLE_0:def 4;
        then
A61:    z = len p + 1 by TARSKI:def 1;
A62:    dom p = Seg(len p) by FINSEQ_1:def 3;
        z in p " A by A60,XBOOLE_0:def 4;
        then len p + 1 <= len p by A61,A59,A62,FINSEQ_1:1;
        hence thesis by XREAL_1:29;
      end;
      then
A63:  {len p + 1} misses p " A by XBOOLE_0:def 7;
      assume t " A = p " A;
      hence Seg(len t) \ t " A = (Seg(len p) \/ {len p + 1}) \ p " A by A1,
FINSEQ_1:9
        .= (Seg(len p) \ p " A) \/ ({len p + 1} \ p " A) by XBOOLE_1:42
        .= (Seg(len p) \ p " A) \/ {len p + 1} by A63,XBOOLE_1:83;
    end;
    Seg(len p) /\ {len p + 1} = {}
    proof
      set z = the Element of Seg(len p) /\ {len p + 1};
      assume
A64:  not thesis;
      then
A65:  z in Seg(len p) by XBOOLE_0:def 4;
      then reconsider f = z as Element of NAT;
      f in {len p + 1} by A64,XBOOLE_0:def 4;
      then
A66:  f = len p + 1 by TARSKI:def 1;
      f <= len p by A65,FINSEQ_1:1;
      hence thesis by A66,XREAL_1:29;
    end;
    then
A67: Seg(len p) misses {len p + 1} by XBOOLE_0:def 7;
A68: now
      assume t " A = p " A \/ {len p + 1};
      hence
      Seg(len t) \ t " A = (Seg(len p) \/ {len p + 1}) \ (p " A \/ {len p
      + 1}) by A1,FINSEQ_1:9
        .= (Seg(len p) \ (p " A \/ {len p + 1}) \/ ({len p + 1} \ (p " A \/
      {len p + 1}))) by XBOOLE_1:42
        .= (Seg(len p) \ (p " A \/ {len p + 1})) \/ {} by XBOOLE_1:46
        .= (Seg(len p) \ p " A) /\ (Seg(len p) \ {len p + 1}) by XBOOLE_1:53
        .= (Seg(len p) \ p " A) /\ Seg(len p) by A67,XBOOLE_1:83
        .= (Seg(len p) \ p " A) by XBOOLE_1:28,36;
    end;
    p " A c= dom p by RELAT_1:132;
    then
A69: p " A c= Seg(len p) by FINSEQ_1:def 3;
    len s2 = card(Seg(len p) \ p " A) by Th39,XBOOLE_1:36
      .= card(Seg(len p)) - card(p " A) by A69,CARD_2:44;
    then
A70: len s2 = len p - card(p " A) by FINSEQ_1:57;
    Seg(len t) \ t " A c= Seg(len t) by XBOOLE_1:36;
    then
A71: rng s1 c= Seg(len t) by FINSEQ_1:def 13;
A72: Seg(len p) \ p " A c= Seg(len p) by XBOOLE_1:36;
    then
A73: rng s2 c= Seg(len p) by FINSEQ_1:def 13;
    assume k in dom(p - A);
    then
A74: k in dom s2 by A57,FUNCT_1:11;
    then s2.k in rng s2 by FUNCT_1:def 3;
    then s2.k in Seg(len p) by A73;
    then
A75: s2.k in dom p by FINSEQ_1:def 3;
    len s1 = card(Seg(len t) \ t " A) by Th39,XBOOLE_1:36
      .= card(Seg(len t)) - card(t " A) by A48,CARD_2:44;
    then
A76: len s1 = len t - card(t " A) by FINSEQ_1:57;
A77: dom s2 c= dom s1
    proof
      <* x *> " A c= dom<* x *> by RELAT_1:132;
      then <* x *> " A c= {1} by FINSEQ_1:2,def 8;
      then
A78:  <* x *> " A = {} or <* x *> " A = {1} by ZFMISC_1:33;
      let y;
A79:  card(p " A) <= card(p " A) + 1 by NAT_1:12;
      assume
A80:  y in dom s2;
      then reconsider l = y as Element of NAT;
A81:  y in Seg(len s2) by A80,FINSEQ_1:def 3;
      then l <= len p - card(p " A) by A70,FINSEQ_1:1;
      then l + card(p " A) <= len p by XREAL_1:19;
      then
A82:  l + card(p " A) + 1 <= len t by A1,XREAL_1:7;
      card(t " A) = card(p " A) + card(<* x *> " A) by Th57;
      then l + card(t " A) <= l + (card(p " A) + 1) by A78,A79,CARD_1:27,30
,XREAL_1:7;
      then l + card(t " A) <= len t by A82,XXREAL_0:2;
      then
A83:  l <= len s1 by A76,XREAL_1:19;
      1 <= l by A81,FINSEQ_1:1;
      then l in Seg(len s1) by A83,FINSEQ_1:1;
      hence thesis by FINSEQ_1:def 3;
    end;
    then s1.k in rng s1 by A74,FUNCT_1:def 3;
    then s1.k in Seg(len t) by A71;
    then reconsider l = s1.k as Element of NAT;
    dom t = Seg len t by FINSEQ_1:def 3;
    then
A84: r.k = t.l by A74,A77,FUNCT_1:13;
    len p + 1 in Seg(len p + 1) by FINSEQ_1:4;
    then
A85: {len p + 1} c= Seg(len p + 1) by ZFMISC_1:31;
A86: now
      per cases by A37,A49,A58,A68;
      suppose
        Seg(len p) \ p " A = Seg(len t) \ t " A;
        hence s1.k = s2.k;
      end;
      suppose
A87:    Seg(len t) \ t " A = (Seg(len p) \ p " A) \/ {len p + 1};
        now
          let m,n be Element of NAT;
          assume that
A88:      m in Seg(len p) \ p " A and
A89:      n in {len p + 1};
          m in Seg(len p) by A88,XBOOLE_0:def 5;
          then
A90:      m <= len p by FINSEQ_1:1;
A91:      len p < len p + 1 by XREAL_1:29;
          n = len p + 1 by A89,TARSKI:def 1;
          hence m < n by A90,A91,XXREAL_0:2;
        end;
        then s1 = s2 ^ Sgm{len p + 1} by A72,A85,A87,Th42;
        hence s1.k = s2.k by A74,FINSEQ_1:def 7;
      end;
    end;
    (p - A).k = p.(s2.k) by A57,A74,FUNCT_1:13;
    hence r.k = (p - A).k by A75,A86,A84,FINSEQ_1:def 7;
  end;
  len r = len(p ^ <* x *>) - card(t " A) by Th59
    .= len p + len<* x *> - card(t " A) by FINSEQ_1:22
    .= (len p + len<* x *>) - (card(p " A) + card(<* x *> " A)) by Th57
    .= len p - card(p " A) + len<* x *> + (- card(<* x *> " A))
    .= len(p - A) + len<* x *> + (- card(<* x *> " A)) by Th59
    .= len(p - A) + (len<* x *> - card(<* x *> " A))
    .= len(p - A) + len q by Th59;
  hence thesis by A36,A2,Th38;
end;

Lm10: now
  let q,x;
  assume
A1: for p,A holds (p ^ q) - A = (p - A) ^ (q - A);
  let p,A;
  thus (p ^ (q ^ <* x *>)) - A = (p ^ q ^ <* x *>) - A by FINSEQ_1:32
    .= ((p ^ q) - A) ^ (<* x *> - A) by Lm9
    .= (p - A) ^ (q - A) ^ (<* x *> - A) by A1
    .= (p - A) ^ ((q - A) ^ (<* x *> - A)) by FINSEQ_1:32
    .= (p - A) ^ ((q ^ <* x *>) - A) by Lm9;
end;

Lm11: for q,p,A holds (p ^ q) - A = (p - A) ^ (q - A)
proof
  defpred P[FinSequence] means (p ^ $1) - A = (p - A) ^ ($1 - A);
A1: for q,x st P[q] holds P[q^<*x*>] by Lm10;
A2: P[{}] by Lm8;
  for q holds P[q] from FINSEQ_1:sch 3(A2,A1);
  hence thesis;
end;

theorem
  (p ^ q) - A = (p - A) ^ (q - A) by Lm11;

theorem
  {} - A = {};

theorem
  <* x *> - A = <* x *> iff not x in A by Lm6;

theorem
  <* x *> - A = {} iff x in A by Lm7;

theorem Th77:
  <* x,y *> - A = {} iff x in A & y in A
proof
A1: <* x,y *> = <* x *> ^ <* y *> by FINSEQ_1:def 9;
  thus <* x,y *> - A = {} implies x in A & y in A
  proof
    assume <* x,y *> - A = {};
    then rng<* x,y *> c= A by Th68;
    then {x,y} c= A by FINSEQ_2:127;
    hence thesis by ZFMISC_1:32;
  end;
  assume that
A2: x in A and
A3: y in A;
A4: <* y *> - A = {} by A3,Lm7;
   card {} +card {} = card({} ^ {}) by FINSEQ_1:22;
   then
A5: card({} ^ {}) = 0;
  <* x *> - A = {} by A2,Lm7;
  hence <* x,y *> - A = {} ^ {} by A4,A1,Lm11
    .= {} by A5;
end;

theorem Th78:
  x in A & not y in A implies <* x,y *> - A = <* y *>
proof
  assume that
A1: x in A and
A2: not y in A;
A3: <* y *> - A = <* y *> by A2,Lm6;
A4: <* x,y *> = <* x *> ^ <* y *> by FINSEQ_1:def 9;
  <* x *> - A = {} by A1,Lm7;
  hence <* x,y *> - A = {} ^ <* y *> by A3,A4,Lm11
    .= <* y *> by FINSEQ_1:34;
end;

theorem
  <* x,y *> - A = <* y *> & x <> y implies x in A & not y in A
proof
  assume that
A1: <* x,y *> - A = <* y *> and
A2: x <> y;
  assume
A3: not thesis;
A4: y in A implies <* y *> - A = {} by Lm7;
A5: not x in A implies <* x *> - A = <* x *> by Lm6;
A6: not y in A implies <* y *> - A = <* y *> by Lm6;
A7: x in A implies <* x *> - A = {} by Lm7;
A8: <* x *>.1 = x by FINSEQ_1:40;
  <* y *> = (<* x *> ^ <* y *>) - A by A1,FINSEQ_1:def 9
    .= (<* x *> - A) ^ (<* y *> - A) by Lm11;
  then <* y *> = {} or <* x *> = <* y *> or <* y *> = <* x,y *> by A7,A5,A4,A6
,A3,FINSEQ_1:34,def 9;
  hence thesis by A2,A8,Th35,FINSEQ_1:40;
end;

theorem Th80:
  not x in A & y in A implies <* x,y *> - A = <* x *>
proof
  assume that
A1: not x in A and
A2: y in A;
A3: <* y *> - A = {} by A2,Lm7;
A4: <* x,y *> = <* x *> ^ <* y *> by FINSEQ_1:def 9;
  <* x *> - A = <* x *> by A1,Lm6;
  hence <* x,y *> - A = <* x *> ^ {} by A3,A4,Lm11
    .= <* x *> by FINSEQ_1:34;
end;

theorem
  <* x,y *> - A = <* x *> & x <> y implies not x in A & y in A
proof
  assume that
A1: <* x,y *> - A = <* x *> and
A2: x <> y;
  assume
A3: not thesis;
A4: y in A implies <* y *> - A = {} by Lm7;
A5: not x in A implies <* x *> - A = <* x *> by Lm6;
A6: not y in A implies <* y *> - A = <* y *> by Lm6;
A7: x in A implies <* x *> - A = {} by Lm7;
A8: <* x *>.1 = x by FINSEQ_1:40;
  <* x *> = (<* x *> ^ <* y *>) - A by A1,FINSEQ_1:def 9
    .= (<* x *> - A) ^ (<* y *> - A) by Lm11;
  then <* x *> = {} or <* x *> = <* y *> or <* x *> = <* x,y *> by A7,A5,A4,A6
,A3,FINSEQ_1:34,def 9;
  hence thesis by A2,A8,Th35,FINSEQ_1:40;
end;

theorem
  <* x,y *> - A = <* x,y *> iff not x in A & not y in A
proof
A1: <* x,y *> = <* x *> ^ <* y *> by FINSEQ_1:def 9;
  thus <* x,y *> - A = <* x,y *> implies not x in A & not y in A
  proof
    assume
A2: <* x,y *> - A = <* x,y *>;
    assume not thesis;
    then x in A & y in A or not x in A & y in A or x in A & not y in A;
    then
    <* x,y *> - A = {} or <* x,y *> - A = <* x *> or <* x,y *> - A = <* y
    *> by Th77,Th78,Th80;
    hence thesis by A2,Th35;
  end;
  assume that
A3: not x in A and
A4: not y in A;
A5: <* y *> - A = <* y *> by A4,Lm6;
  <* x *> - A = <* x *> by A3,Lm6;
  hence thesis by A5,A1,Lm11;
end;

theorem Th83:
  len p = k + 1 & q = p | Seg k implies (p.(k + 1) in A iff p - A = q - A)
proof
  assume that
A1: len p = k + 1 and
A2: q = p | Seg k;
  thus p.(k + 1) in A implies p - A = q - A
  proof
    assume
A3: p.(k + 1) in A;
    thus p - A = (q ^ <* p.(k + 1) *>) - A by A1,A2,Th55
      .= (q - A) ^ (<* p.(k + 1) *> - A) by Lm11
      .= (q - A) ^ {} by A3,Lm7
      .= q - A by FINSEQ_1:34;
  end;
  assume that
A4: p - A = q - A and
A5: not p.(k + 1) in A;
  q - A = (q ^ <* p.(k + 1) *>) - A by A1,A2,A4,Th55
    .= (q - A) ^ (<* p.(k + 1) *> - A) by Lm11
    .= (q - A) ^ <* p.(k + 1) *> by A5,Lm6;
  hence thesis by FINSEQ_1:87;
end;

theorem Th84:
  len p = k + 1 & q = p | Seg k implies (not p.(k + 1) in A iff p
  - A = (q - A) ^ <* p.(k + 1) *>)
proof
  assume that
A1: len p = k + 1 and
A2: q = p | Seg k;
  thus not p.(k + 1) in A implies p - A = (q - A) ^ <* p.(k + 1) *>
  proof
    assume
A3: not p.(k + 1) in A;
    thus p - A = (q ^ <* p.(k + 1) *>) - A by A1,A2,Th55
      .= (q - A) ^ (<* p.(k + 1) *> - A) by Lm11
      .= (q - A) ^ <* p.(k + 1) *> by A3,Lm6;
  end;
  assume
A4: p - A = (q - A) ^ <* p.(k + 1) *>;
  assume p.(k + 1) in A;
  then q - A = (q - A) ^ <* p.(k + 1) *> by A1,A2,A4,Th83;
  hence contradiction by FINSEQ_1:87;
end;

Lm12: for l st (for p,A st len p = l for n st n in dom p for B being finite
set st B = {k where k is Element of NAT : k in dom p & k <= n & p.k in A} holds
p.n in A or (p - A).(n - card B) = p.n) holds for p,A st len p = l + 1 for n st
n in dom p for C being finite set st C = {m where m is Element of NAT : m in
dom p & m <= n & p.m in A} holds p.n in A or (p - A).(n - card C) = p.n
proof
  let l;
  assume
A1: for p,A st len p = l for n st n in dom p for B being finite set st B
= {k where k is Element of NAT : k in dom p & k <= n & p.k in A} holds p.n in A
  or (p - A).(n - card B) = p.n;
  let p,A;
  reconsider q = p | Seg l as FinSequence by FINSEQ_1:15;
  assume
A2: len p = l + 1;
  then
A3: len q = l by Th53;
  let n;
  set B = {k where k is Element of NAT: k in dom q & k <= n & q.k in A};
  B c= dom q
  proof
    let x be set;
    assume x in B;
    then
    ex k being Element of NAT st x = k & k in dom q & k <= n & q.k in A;
    hence thesis;
  end;
  then reconsider B as finite set;
  assume
A4: n in dom p;
  let C be finite set such that
A5: C = {m where m is Element of NAT: m in dom p & m <= n & p.m in A};
  assume
A6: not p.n in A;
  now
    per cases;
    suppose
A7:   p.(l + 1) in A;
A8:   n in Seg(l + 1) by A2,A4,FINSEQ_1:def 3;
      not n in {l + 1} by A6,A7,TARSKI:def 1;
      then n in Seg(l + 1) \ {l + 1} by A8,XBOOLE_0:def 5;
      then
A9:   n in Seg l by FINSEQ_1:10;
A10:  B = C
      proof
        thus B c= C
        proof
          let x;
A11:      dom q c= dom p by RELAT_1:60;
          assume x in B;
          then consider k being Element of NAT such that
A12:      x = k and
A13:      k in dom q and
A14:      k <= n and
A15:      q.k in A;
          p.k in A by A13,A15,FUNCT_1:47;
          hence thesis by A5,A12,A13,A14,A11;
        end;
        let x;
        assume x in C;
        then consider m being Element of NAT such that
A16:    x = m and
A17:    m in dom p and
A18:    m <= n and
A19:    p.m in A by A5;
        m in Seg(len p) by A17,FINSEQ_1:def 3;
        then
A20:    1 <= m by FINSEQ_1:1;
        n <= l by A9,FINSEQ_1:1;
        then m <= l by A18,XXREAL_0:2;
        then m in Seg l by A20,FINSEQ_1:1;
        then
A21:    m in dom q by A3,FINSEQ_1:def 3;
        then q.m in A by A19,FUNCT_1:47;
        hence thesis by A16,A18,A21;
      end;
A22:  q - A = p - A by A2,A7,Th83;
A23:  n in dom q by A3,A9,FINSEQ_1:def 3;
      then p.n = q.n by FUNCT_1:47;
      hence thesis by A1,A6,A3,A23,A22,A10;
    end;
    suppose
      not p.(l + 1) in A;
      then
A24:  p - A = (q - A) ^ <* p.(l + 1) *> by A2,Th84;
      now
        per cases;
        suppose
A25:      n = l + 1;
          p " A = C
          proof
            thus p " A c= C
            proof
              let x;
              assume
A26:          x in p " A;
              then
A27:          x in dom p by FUNCT_1:def 7;
              then reconsider z = x as Element of NAT;
A28:          p.x in A by A26,FUNCT_1:def 7;
              z in Seg n by A2,A25,A27,FINSEQ_1:def 3;
              then z <= n by FINSEQ_1:1;
              hence thesis by A5,A27,A28;
            end;
            let x;
            assume x in C;
            then
            ex m being Element of NAT st x = m & m in dom p & m <= n & p.
            m in A by A5;
            hence thesis by FUNCT_1:def 7;
          end;
          then
A29:      len(p - A) = n - card C by A2,A25,Th59;
          len<* p.(l + 1) *> = 1 by FINSEQ_1:39;
          then (p - A).(n - card C) = (p - A).(len(q - A) + 1) by A24,A29,
FINSEQ_1:22
            .= p.(l + 1) by A24,FINSEQ_1:42;
          hence thesis by A25;
        end;
        suppose
          n <> l + 1;
          then
A30:      not n in {l + 1} by TARSKI:def 1;
          n in Seg(l + 1) by A2,A4,FINSEQ_1:def 3;
          then n in Seg(l + 1) \ {l + 1} by A30,XBOOLE_0:def 5;
          then
A31:      n in Seg l by FINSEQ_1:10;
          then 1 <= n by FINSEQ_1:1;
          then n in Seg n by FINSEQ_1:1;
          then
A32:      {n} c= Seg n by ZFMISC_1:31;
A33:      Seg n \ B c= Seg l \ q " A
          proof
            let x;
            assume
A34:        x in Seg n \ B;
            then reconsider z = x as Element of NAT;
A35:        x in Seg n by A34,XBOOLE_0:def 5;
            then
A36:        1 <= z by FINSEQ_1:1;
A37:        z <= n by A35,FINSEQ_1:1;
A38:        now
              assume
A39:          z in q " A;
              then
A40:          z in dom q by FUNCT_1:def 7;
              q.z in A by A39,FUNCT_1:def 7;
              then z in B by A37,A40;
              hence contradiction by A34,XBOOLE_0:def 5;
            end;
            n <= l by A31,FINSEQ_1:1;
            then z <= l by A37,XXREAL_0:2;
            then z in Seg l by A36,FINSEQ_1:1;
            hence thesis by A38,XBOOLE_0:def 5;
          end;
A41:      B = C
          proof
            thus B c= C
            proof
              let x;
A42:          dom q c= dom p by RELAT_1:60;
              assume x in B;
              then consider k being Element of NAT such that
A43:          x = k and
A44:          k in dom q and
A45:          k <= n and
A46:          q.k in A;
              p.k in A by A44,A46,FUNCT_1:47;
              hence thesis by A5,A43,A44,A45,A42;
            end;
            let x;
            assume x in C;
            then consider m being Element of NAT such that
A47:        x = m and
A48:        m in dom p and
A49:        m <= n and
A50:        p.m in A by A5;
            m in Seg(len p) by A48,FINSEQ_1:def 3;
            then
A51:        1 <= m by FINSEQ_1:1;
            n <= l by A31,FINSEQ_1:1;
            then m <= l by A49,XXREAL_0:2;
            then m in Seg l by A51,FINSEQ_1:1;
            then
A52:        m in dom q by A3,FINSEQ_1:def 3;
            then q.m in A by A50,FUNCT_1:47;
            hence thesis by A47,A49,A52;
          end;
          q " A c= dom q by RELAT_1:132;
          then
A53:      q " A c= Seg l by A3,FINSEQ_1:def 3;
          q " A \/ (Seg l \ q " A) = q " A \/ Seg l by XBOOLE_1:39
            .= Seg l by A53,XBOOLE_1:12;
          then card(Seg l) = card(q " A) + card(Seg l \ q " A) by CARD_2:40
,XBOOLE_1:79;
          then
A54:      len q = card(q " A) + card(Seg(len q) \ q " A) by A3,FINSEQ_1:57;
          set b = card B;
          set a = n - card B;
A55:      card(Seg n) = n by FINSEQ_1:57;
A56:      n in dom q by A3,A31,FINSEQ_1:def 3;
          then
A57:      not q.n in A by A6,FUNCT_1:47;
A58:      B c= (Seg n) \ {n}
          proof
            let x;
            assume x in B;
            then consider k being Element of NAT such that
A59:        x = k and
A60:        k in dom q and
A61:        k <= n and
A62:        q.k in A;
            k in Seg(len q) by A60,FINSEQ_1:def 3;
            then 1 <= k by FINSEQ_1:1;
            then
A63:        k in Seg n by A61,FINSEQ_1:1;
            not k in {n} by A57,A62,TARSKI:def 1;
            hence thesis by A59,A63,XBOOLE_0:def 5;
          end;
          then card B <= card((Seg n) \ {n}) by NAT_1:43;
          then card B <= card(Seg n) - card{n} by A32,CARD_2:44;
          then b <= card(Seg n) - 1 by CARD_1:30;
          then b <= n - 1 by FINSEQ_1:57;
          then b + 1 <= n by XREAL_1:19;
          then
A64:      0 + b < a + b by NAT_1:13;
          then 0 <= a by XREAL_1:6;
          then reconsider a as Element of NAT by INT_1:3;
          Seg n \ {n} c= Seg n by XBOOLE_1:36;
          then a = card(Seg n \ B) by A58,A55,CARD_2:44,XBOOLE_1:1;
          then a <= len q - card(q " A) by A3,A54,A33,NAT_1:43;
          then
A65:      a <= len(q - A) by Th59;
          1 <= a by A64,NAT_1:14;
          then a in Seg(len(q - A)) by A65,FINSEQ_1:1;
          then
A66:      a in dom(q - A) by FINSEQ_1:def 3;
          p.n = q.n by A56,FUNCT_1:47;
          then (q - A).(n - card B) = p.n by A1,A6,A3,A56;
          hence thesis by A24,A41,A66,FINSEQ_1:def 7;
        end;
      end;
      hence thesis;
    end;
  end;
  hence thesis;
end;

Lm13: for l for p,A st len p = l for n st n in dom p holds for B being finite
set st B = {k where k is Element of NAT: k in dom p & k <= n & p.k in A} holds
p.n in A or (p - A).(n - card B) = p.n
proof
  defpred P[Element of NAT] means for p,A st len p = $1 for n st n in dom p
holds for B being finite set st B = {k where k is Element of NAT : k in dom p &
  k <= n & p.k in A} holds p.n in A or (p - A).(n - card B) = p.n;
A1: P[0]
  proof
    let p,A;
    assume len p = 0;
    then p = {};
    hence thesis;
  end;
A2: for k being Element of NAT st P[k] holds P[k+1] by Lm12;
  for n being Element of NAT holds P[n] from NAT_1:sch 1(A1,A2);
  hence thesis;
end;

theorem
  n in dom p implies for B being finite set st B = {k where k is Element
of NAT : k in dom p & k <= n & p.k in A} holds p.n in A or (p - A).(n - card B)
  = p.n
proof
  len p = len p;
  hence thesis by Lm13;
end;

theorem
  p is FinSequence of D implies p - A is FinSequence of D
proof
  assume p is FinSequence of D;
  then
A1: rng p c= D by FINSEQ_1:def 4;
  rng(p - A) = rng p \ A by Th65;
  then rng(p - A) c= D by A1,XBOOLE_1:1;
  hence thesis by FINSEQ_1:def 4;
end;

theorem
  p is one-to-one implies p - A is one-to-one
proof
  assume
A1: p is one-to-one;
A2: p - A = p * Sgm(Seg(len p) \ p " A) by FINSEQ_1:def 3;
  Sgm(Seg(len p) \ p " A) is one-to-one by Lm1,XBOOLE_1:36;
  hence thesis by A1,A2;
end;

theorem Th88:
  p is one-to-one implies len(p - A) = len p - card(A /\ rng p)
proof
A1: p " A c= dom p by RELAT_1:132;
  assume
A2: p is one-to-one;
  p " A,A /\ rng p are_equipotent
  proof
    deffunc F(set) = p.$1;
    consider f being Function such that
A3: dom f = p " A and
A4: for x st x in p " A holds f.x = F(x) from FUNCT_1:sch 3;
    take f;
    thus f is one-to-one
    proof
      let x,y;
      assume that
A5:   x in dom f and
A6:   y in dom f and
A7:   f.x = f.y;
A8:   f.y = p.y by A3,A4,A6;
      p.x = f.x by A3,A4,A5;
      hence thesis by A2,A1,A3,A5,A6,A7,A8,FUNCT_1:def 4;
    end;
    thus dom f = p " A by A3;
    thus rng f c= A /\ rng p
    proof
      let x;
      assume x in rng f;
      then consider y such that
A9:   y in dom f and
A10:  f.y = x by FUNCT_1:def 3;
A11:  p.y in A by A3,A9,FUNCT_1:def 7;
      y in dom p by A3,A9,FUNCT_1:def 7;
      then
A12:  p.y in rng p by FUNCT_1:def 3;
      p.y = f.y by A3,A4,A9;
      hence thesis by A10,A11,A12,XBOOLE_0:def 4;
    end;
    let x;
    assume
A13: x in A /\ rng p;
    then x in rng p by XBOOLE_0:def 4;
    then consider y such that
A14: y in dom p and
A15: p.y = x by FUNCT_1:def 3;
    p.y in A by A13,A15,XBOOLE_0:def 4;
    then
A16: y in p " A by A14,FUNCT_1:def 7;
    then f.y = x by A4,A15;
    hence thesis by A3,A16,FUNCT_1:def 3;
  end;
  then card(p " A) = card(A /\ rng p) by CARD_1:5;
  hence thesis by Th59;
end;

theorem Th89:
  for A being finite set st p is one-to-one & A c= rng p holds len
  (p - A) = len p - card A
proof
  let A be finite set;
  assume that
A1: p is one-to-one and
A2: A c= rng p;
  A /\ rng p = A by A2,XBOOLE_1:28;
  hence thesis by A1,Th88;
end;

theorem
  p is one-to-one & x in rng p implies len(p - {x}) = len p - 1
proof
  assume that
A1: p is one-to-one and
A2: x in rng p;
  {x} c= rng p by A2,ZFMISC_1:31;
  then len(p - {x}) = len p - card{x} by A1,Th89;
  hence thesis by CARD_1:30;
end;

theorem Th91:
  rng p misses rng q & p is one-to-one & q is one-to-one iff p ^ q
  is one-to-one
proof
  thus rng p misses rng q & p is one-to-one & q is one-to-one implies p ^ q is
  one-to-one
  proof
    assume that
A1: rng p misses rng q and
A2: p is one-to-one and
A3: q is one-to-one;
    let x,y;
    assume that
A4: x in dom(p ^ q) and
A5: y in dom(p ^ q) and
A6: (p ^ q).x = (p ^ q).y;
    reconsider k1 = x, k2 = y as Element of NAT by A4,A5;
    now
      per cases by A4,A5,FINSEQ_1:25;
      suppose
A7:     k1 in dom p & k2 in dom p;
        then
A8:     (p ^ q).k2 = p.k2 by FINSEQ_1:def 7;
        (p ^ q).k1 = p.k1 by A7,FINSEQ_1:def 7;
        hence thesis by A2,A6,A7,A8,FUNCT_1:def 4;
      end;
      suppose
A9:     k1 in dom p & ex n being Nat st n in dom q & k2 = len p + n;
        then consider n being Nat such that
A10:    n in dom q and
A11:    k2 = len p + n;
A12:    q.n in rng q by A10,FUNCT_1:def 3;
A13:    (p ^ q).k1 = p.k1 by A9,FINSEQ_1:def 7;
        (p ^ q).k2 = q.n by A9,A11,FINSEQ_1:def 7;
        then q.n in rng p by A6,A9,A13,FUNCT_1:def 3;
        hence thesis by A1,A12,XBOOLE_0:3;
      end;
      suppose
A14:    k2 in dom p & ex n being Nat st n in dom q & k1 = len p + n;
        then consider n being Nat such that
A15:    n in dom q and
A16:    k1 = len p + n;
A17:    q.n in rng q by A15,FUNCT_1:def 3;
A18:    (p ^ q).k2 = p.k2 by A14,FINSEQ_1:def 7;
        (p ^ q).k1 = q.n by A14,A16,FINSEQ_1:def 7;
        then q.n in rng p by A6,A14,A18,FUNCT_1:def 3;
        hence thesis by A1,A17,XBOOLE_0:3;
      end;
      suppose
A19:    (ex n being Nat st n in dom q & k1 = len p + n) & ex n being
        Nat st n in dom q & k2 = len p + n;
        then consider n2 being Nat such that
A20:    n2 in dom q and
A21:    k2 = len p + n2;
A22:    (p ^ q).k2 = q.n2 by A20,A21,FINSEQ_1:def 7;
        consider n1 being Nat such that
A23:    n1 in dom q and
A24:    k1 = len p + n1 by A19;
        (p ^ q).k1 = q.n1 by A23,A24,FINSEQ_1:def 7;
        hence thesis by A3,A6,A23,A24,A20,A21,A22,FUNCT_1:def 4;
      end;
    end;
    hence thesis;
  end;
  assume
A25: p ^ q is one-to-one;
  thus rng p misses rng q
  proof
    assume not rng p misses rng q;
    then consider x such that
A26: x in rng p and
A27: x in rng q by XBOOLE_0:3;
    consider y1 such that
A28: y1 in dom p and
A29: p.y1 = x by A26,FUNCT_1:def 3;
A30: y1 in Seg(len p) by A28,FINSEQ_1:def 3;
    consider y2 such that
A31: y2 in dom q and
A32: q.y2 = x by A27,FUNCT_1:def 3;
A33: y2 in Seg(len q) by A31,FINSEQ_1:def 3;
    reconsider y1,y2 as Element of NAT by A28,A31;
A34: len p + y2 in dom(p ^ q) by A31,FINSEQ_1:28;
A35: (p ^ q).y1 = p.y1 by A28,FINSEQ_1:def 7;
A36: (p ^ q).(len p + y2) = q.y2 by A31,FINSEQ_1:def 7;
    y1 in dom(p ^ q ) by A28,Th22;
    then
A37: y1 = len p + y2 by A25,A29,A32,A34,A35,A36,FUNCT_1:def 4;
A38: y1 = y1 + 0;
A39: len p <= len p + y2 by NAT_1:12;
    y1 <= len p by A30,FINSEQ_1:1;
    then y1 = len p by A37,A39,XXREAL_0:1;
    hence thesis by A33,A37,A38,FINSEQ_1:1;
  end;
  thus p is one-to-one
  proof
    let x,y;
    assume that
A40: x in dom p and
A41: y in dom p and
A42: p.x = p.y;
    reconsider k = x, l = y as Element of NAT by A40,A41;
A43: (p ^ q).k = p.k by A40,FINSEQ_1:def 7;
A44: (p ^ q).l = p.l by A41,FINSEQ_1:def 7;
A45: l in dom(p ^q) by A41,Th22;
    k in dom(p ^ q) by A40,Th22;
    hence thesis by A25,A42,A43,A44,A45,FUNCT_1:def 4;
  end;
  let x,y;
  assume that
A46: x in dom q and
A47: y in dom q and
A48: q.x = q.y;
  consider l being Nat such that
A49: y = l and
A50: len p + l in dom(p ^ q) by A47,FINSEQ_1:27;
A51: (p ^ q).(len p + l) = q.l by A47,A49,FINSEQ_1:def 7;
  consider k being Nat such that
A52: x = k and
A53: len p + k in dom(p ^ q) by A46,FINSEQ_1:27;
  (p ^ q).(len p + k) = q.k by A46,A52,FINSEQ_1:def 7;
  then len p + k = len p + l by A25,A48,A52,A53,A49,A50,A51,FUNCT_1:def 4;
  hence thesis by A52,A49;
end;

theorem
  A c= Seg k implies Sgm A is one-to-one by Lm1;

theorem Th93:
  <* x *> is one-to-one
proof
  let y1,y2;
  assume that
A1: y1 in dom<* x *> and
A2: y2 in dom<* x *> and
  <* x *>.y1 = <* x *>.y2;
  y1 in {1} by A1,FINSEQ_1:2,def 8;
  then
A3: y1 = 1 by TARSKI:def 1;
  y2 in {1} by A2,FINSEQ_1:2,def 8;
  hence thesis by A3,TARSKI:def 1;
end;

theorem Th94:
  x <> y iff <* x,y *> is one-to-one
proof
A1: <* x,y *>.2 = y by FINSEQ_1:44;
  2 in {1,2} by TARSKI:def 2;
  then
A2: 2 in dom<* x,y *> by FINSEQ_1:2,89;
  thus x <> y implies <* x,y *> is one-to-one
  proof
    assume
A3: x <> y;
    let y1,y2;
    assume that
A4: y1 in dom<* x,y *> and
A5: y2 in dom<* x,y *> and
A6: <* x,y *>.y1 = <* x,y *>.y2;
A7: y2 in {1,2} by A5,FINSEQ_1:2,89;
A8: y1 in {1,2} by A4,FINSEQ_1:2,89;
    now
      per cases by A8,A7,TARSKI:def 2;
      suppose
        y1 = 1 & y2 = 1 or y1 = 2 & y2 = 2;
        hence thesis;
      end;
      suppose
A9:     y1 = 1 & y2 = 2;
        then <* x,y *>.y1 = x by FINSEQ_1:44;
        hence thesis by A3,A6,A9,FINSEQ_1:44;
      end;
      suppose
A10:    y1 = 2 & y2 = 1;
        then <* x,y *>.y1 = y by FINSEQ_1:44;
        hence thesis by A3,A6,A10,FINSEQ_1:44;
      end;
    end;
    hence thesis;
  end;
  assume that
A11: <* x,y *> is one-to-one and
A12: x = y;
  1 in {1,2} by TARSKI:def 2;
  then
A13: 1 in dom<* x,y *> by FINSEQ_1:2,89;
  <* x,y *>.1 = x by FINSEQ_1:44;
  hence thesis by A11,A12,A13,A2,A1,FUNCT_1:def 4;
end;

theorem Th95:
  x <> y & y <> z & z <> x iff <* x,y,z *> is one-to-one
proof
  set p = <* x,y,z *>;
A1: p.1 = x by FINSEQ_1:45;
A2: p.3 = z by FINSEQ_1:45;
  thus x <> y & y <> z & z <> x implies <* x,y,z *> is one-to-one
  proof
    assume that
A3: x <> y and
A4: y <> z and
A5: z <> x;
    {x,y} /\ {z} = {}
    proof
      set y1 = the Element of {x,y} /\ {z};
      assume
A6:   not thesis;
      then y1 in {x,y} by XBOOLE_0:def 4;
      then
A7:   y1 = x or y1 = y by TARSKI:def 2;
      y1 in {z} by A6,XBOOLE_0:def 4;
      hence thesis by A4,A5,A7,TARSKI:def 1;
    end;
    then {} = rng<* x,y *> /\ {z} by FINSEQ_2:127
      .= rng<* x,y *> /\ rng<* z *> by FINSEQ_1:38;
    then
A8: rng<* x,y *> misses rng<* z *> by XBOOLE_0:def 7;
A9: <* z *> is one-to-one by Th93;
    <* x,y *> is one-to-one by A3,Th94;
    then <* x,y *> ^ <* z *> is one-to-one by A8,A9,Th91;
    hence thesis by FINSEQ_1:43;
  end;
A10: p.2 = y by FINSEQ_1:45;
  1 in {1,2,3} by ENUMSET1:def 1;
  then
A11: 1 in dom p by Th1,FINSEQ_1:89;
  3 in {1,2,3} by ENUMSET1:def 1;
  then
A12: 3 in dom p by Th1,FINSEQ_1:89;
  2 in {1,2,3} by ENUMSET1:def 1;
  then
A13: 2 in dom p by Th1,FINSEQ_1:89;
  assume <* x,y,z *> is one-to-one;
  hence thesis by A11,A13,A12,A1,A10,A2,FUNCT_1:def 4;
end;

theorem Th96:
  p is one-to-one & rng p = {x} implies len p = 1
proof
  assume that
A1: p is one-to-one and
A2: rng p = {x};
A3: now
    given y1,y2 such that
A4: y1 in dom p and
A5: y2 in dom p and
A6: y1 <> y2;
    p.y2 in rng p by A5,FUNCT_1:def 3;
    then
A7: p.y2 = x by A2,TARSKI:def 1;
    p.y1 in rng p by A4,FUNCT_1:def 3;
    then p.y1 = x by A2,TARSKI:def 1;
    hence contradiction by A1,A4,A5,A6,A7,FUNCT_1:def 4;
  end;
  set y = the Element of dom p;
A8: dom p <> {} by A2,RELAT_1:42;
A9: dom p = {y}
  proof
    thus dom p c= {y}
    proof
      let x;
      assume x in dom p;
      then x = y by A3;
      hence thesis by TARSKI:def 1;
    end;
    let x;
    y in dom p by A8;
    hence thesis by TARSKI:def 1;
  end;
  dom p = Seg(len p) by FINSEQ_1:def 3;
  hence thesis by A9,Th20;
end;

theorem
  p is one-to-one & rng p = {x} implies p = <* x *>
proof
  assume that
A1: p is one-to-one and
A2: rng p = {x};
  len p = 1 by A1,A2,Th96;
  hence thesis by A2,FINSEQ_1:39;
end;

theorem Th98:
  p is one-to-one & rng p = {x,y} & x <> y implies len p = 2
proof
  assume that
A1: p is one-to-one and
A2: rng p = {x,y} and
A3: x <> y;
  set q = <* x,y *>;
A4: rng q = {x,y} by FINSEQ_2:127;
A5: len q = 2 by FINSEQ_1:44;
  q is one-to-one by A3,Th94;
  hence thesis by A1,A2,A4,A5,FINSEQ_1:48;
end;

theorem
  p is one-to-one & rng p = {x,y} & x <> y implies p = <* x,y *> or p =
  <* y,x *>
proof
  assume that
A1: p is one-to-one and
A2: rng p = {x,y} and
A3: x <> y;
A4: len p = 2 by A1,A2,A3,Th98;
  then
A5: dom p = {1,2} by FINSEQ_1:2,def 3;
  then
A6: 2 in dom p by TARSKI:def 2;
  then
A7: p.2 in rng p by FUNCT_1:def 3;
A8: 1 in dom p by A5,TARSKI:def 2;
  then p.1 in rng p by FUNCT_1:def 3;
  then p.1 = x & p.2 = x or p.1 = x & p.2 = y or p.1 = y & p.2 = x or p.1 = y
  & p.2 = y by A2,A7,TARSKI:def 2;
  hence thesis by A1,A4,A8,A6,FINSEQ_1:44,FUNCT_1:def 4;
end;

theorem Th100:
  p is one-to-one & rng p = {x,y,z} & <* x,y,z *> is one-to-one
  implies len p = 3
proof
A1: len <* x,y,z *> = 3 by FINSEQ_1:45;
  rng <* x,y,z *> = {x,y,z} by FINSEQ_2:128;
  hence thesis by A1,FINSEQ_1:48;
end;

theorem
  p is one-to-one & rng p = {x,y,z} & x <> y & y <> z & x <> z implies
  len p = 3
proof
  assume that
A1: p is one-to-one and
A2: rng p = {x,y,z} and
A3: x <> y and
A4: y <> z and
A5: x <> z;
  <* x,y,z *> is one-to-one by A3,A4,A5,Th95;
  hence thesis by A1,A2,Th100;
end;

begin 


theorem
  for D being non empty set, df being FinSequence of D holds df is non
empty implies ex d being Element of D, df1 being FinSequence of D st d = df.1 &
  df = <*d*>^df1
proof
  let D be non empty set, df be FinSequence of D;
  deffunc F(Nat) = df.($1+1);
  assume
A1: df is non empty;
  then reconsider lend1 = len df - 1 as Element of NAT by INT_1:5,NAT_1:14;
  1 <= len df by A1,NAT_1:14;
  then 1 in dom df by Th25;
  then reconsider d = df.1 as Element of D by FINSEQ_2:11;
  take d;
  consider dta being FinSequence such that
A2: len dta = lend1 and
A3: for j being Nat st j in dom dta holds dta.j=F(j) from FINSEQ_1:sch 2;
  now
    let j be Nat;
    assume
A4: j in dom dta;
    then j in Seg len dta by FINSEQ_1:def 3;
    then j + 1 in Seg (len dta + 1) by FINSEQ_1:60;
    then j+1 in dom df by A2,FINSEQ_1:def 3;
    then df.(j+1) in D by FINSEQ_2:11;
    hence dta.j in D by A3,A4;
  end;
  then reconsider dta as FinSequence of D by FINSEQ_2:12;
  take dta;
  thus d = df.1;
  now
    thus
A5: len (<*d*>^dta) = len <*d*> + len dta by FINSEQ_1:22
      .= 1 + (len df - 1) by A2,FINSEQ_1:40
      .= len df;
    let i be Nat;
A6: dom df = Seg len df by FINSEQ_1:def 3;
    assume
A7: i in dom df;
    then
A8: 1 <= i by A6,FINSEQ_1:1;
A9: i <= len df by A6,A7,FINSEQ_1:1;
    per cases by A8,XXREAL_0:1;
    suppose
      i = 1;
      hence df.i = (<*d*>^dta).i by FINSEQ_1:41;
    end;
    suppose
A10:  i > 1;
      then consider j being Element of NAT such that
A11:  j = i-1 and
A12:  1 <= j by INT_1:51;
      i - 1 <= lend1 by A9,XREAL_1:9;
      then
A13:  j in dom dta by A2,A11,A12,Th25;
      len <*d*> = 1 by FINSEQ_1:40;
      then (<*d*>^dta).i = dta.j by A5,A9,A10,A11,FINSEQ_1:24
        .= df.(j+1) by A3,A13
        .= df.i by A11;
      hence df.i = (<*d*>^dta).i;
    end;
  end;
  hence thesis by FINSEQ_2:9;
end;

theorem
  for df being FinSequence, d being set holds i in dom df implies (<*d*>
  ^df).(i+1) = df.i
proof
  let df be FinSequence, d be set;
A1: len (<*d*>^df) = len <*d*> + len df by FINSEQ_1:22
    .= 1 + len df by FINSEQ_1:40;
  assume
A2: i in dom df;
  then i in Seg len df by FINSEQ_1:def 3;
  then i+1 in Seg len (<*d*>^df) by A1,FINSEQ_1:60;
  then i+1 in dom (<*d*>^df) by FINSEQ_1:def 3;
  then
A3: i+1 <= len (<*d*>^df) by Th25;
A4: len <*d*> = 1 by FINSEQ_1:40;
  1 <= i by A2,Th25;
  then 1 < i+1 by NAT_1:13;
  hence (<*d*>^df).(i+1) = df.(i + 1 - len <*d*>) by A4,A3,FINSEQ_1:24
    .= df.i by A4;
end;



definition
  let i;
  let p be FinSequence;
  func Del(p,i) -> FinSequence equals
  p * Sgm ((dom p) \ {i});
  coherence
  proof
    set q = Sgm((dom p) \ {i});
A1: (Seg len p) \ {i} c= Seg len p by XBOOLE_1:36;
    dom p = Seg len p by FINSEQ_1:def 3;
    then rng q c= dom p by A1,FINSEQ_1:def 13;
    then
A2: dom (p*q) = dom q by RELAT_1:27;
    dom q = Seg len q by FINSEQ_1:def 3;
    hence thesis by A2,FINSEQ_1:def 2;
  end;
end;

theorem Th104:
  for p being FinSequence holds
  (i in dom p implies ex m being Nat st len p = m + 1 & len Del(p,i) = m) &
  (not i in dom p implies Del(p,i) = p)
proof
  let p be FinSequence;
  hereby
    not i in (Seg(len p) \ {i})
    proof
      assume i in (Seg(len p) \ {i});
      then not i in {i} by XBOOLE_0:def 5;
      hence thesis by TARSKI:def 1;
    end;
    then
A1: card ((Seg(len p) \ {i}) \/ {i})=card (Seg(len p) \ {i}) +1 by CARD_2:41;
    assume
A2: i in dom p;
    then reconsider D9 = dom p as non empty set;
    reconsider D = rng p as non empty set by A2,FUNCT_1:3;
    reconsider r = p as Function of D9,D by FUNCT_2:1;
A3: dom p = Seg len p by FINSEQ_1:def 3;
    for x holds x in (Seg(len p) \ {i}) implies x in Seg(len p) by
XBOOLE_0:def 5;
    then
A4: (Seg(len p) \ {i}) c= Seg len p by TARSKI:def 3;
    then rng Sgm(Seg(len p) \ {i}) c= Seg len p by FINSEQ_1:def 13;
    then reconsider q = Sgm((dom p) \ {i}) as FinSequence of D9 by A3,
FINSEQ_1:def 4;
    p <> {} by A2;
    then consider m being Nat such that
A5: len p = m+1 by NAT_1:6;
    take m;
A6: len(r*q) = len q by FINSEQ_2:33;
    i in Seg len p by A2,FINSEQ_1:def 3;
    then for x holds x in {i} implies x in Seg(len p) by TARSKI:def 1;
    then {i} c= Seg len p by TARSKI:def 3;
    then card Seg(len p)=card (Seg(len p) \ {i}) +1 by A1,XBOOLE_1:45;
    then card (Seg(len p) \ {i}) +1 = m+1 by A5,FINSEQ_1:57;
    then len (Sgm(Seg(len p) \ {i}))=m by A4,Th39;
    hence len p=m + 1 & len Del(p,i) = m by A5,A6,FINSEQ_1:def 3;
  end;
  assume not i in dom p;
  then for x st x in {i} holds not x in dom p by TARSKI:def 1;
  then {i} misses dom p by XBOOLE_0:3;
  hence Del(p,i) = p * Sgm(dom p) by XBOOLE_1:83
    .= p * Sgm(Seg len p) by FINSEQ_1:def 3
    .= p * (idseq len p) by Th48
    .= p|(Seg len p) by RELAT_1:65
    .= p|(dom p) by FINSEQ_1:def 3
    .= p;
end;

theorem
  for D being non empty set for p being FinSequence of D holds
  Del(p,i) is FinSequence of D
proof
  let D be non empty set, p be FinSequence of D;
  per cases;
  suppose
    i in dom p;
    then reconsider D9=Seg(len p) as non empty set by FINSEQ_1:def 3;
    for x holds x in (Seg(len p) \ {i}) implies x in Seg(len p) by
XBOOLE_0:def 5;
    then Seg(len p) \ {i} c= Seg(len p) by TARSKI:def 3;
    then rng Sgm(Seg(len p) \ {i}) c= Seg(len p) by FINSEQ_1:def 13;
    then reconsider q=Sgm(Seg(len p) \ {i}) as FinSequence of D9 by
FINSEQ_1:def 4;
    p * q = Del(p,i) by FINSEQ_1:def 3;
    hence thesis by FINSEQ_2:31;
  end;
  suppose
    not i in dom p;
    hence thesis by Th104;
  end;
end;



theorem
  for p be FinSequence holds rng Del(p,i) c= rng p
proof
  let p be FinSequence;
  let x;
  assume x in rng Del(p,i);
  hence thesis by FUNCT_1:14;
end;



theorem Th107:
  n = m + 1 & i in Seg n implies len Sgm(Seg n \ {i}) = m
proof
  assume that
A1: n = m + 1 and
A2: i in Seg n;
  set X = Seg n \ {i};
  i in {i} by TARSKI:def 1;
  then not i in X by XBOOLE_0:def 5;
  then card X + 1 = card (X \/ {i}) by CARD_2:41
    .= card ((Seg n) \/ {i}) by XBOOLE_1:39
    .= card (Seg n) by A2,ZFMISC_1:40
    .= m + 1 by A1,FINSEQ_1:57;
  hence thesis by Th39,XBOOLE_1:36;
end;

reserve J for Element of NAT;

theorem Th108:
  for i,k,m,n being Nat st n=m+1 & k in Seg n & i in Seg m holds
  (1<=i & i<k implies Sgm(Seg n \ {k}).i = i) &
  (k<=i & i<=m implies Sgm(Seg n \ {k}).i = i+1)
proof
  defpred P[Nat] means for n,k,i being Nat st n = $1 + 1
& k in Seg n & i in Seg $1 holds (1<=i & i<k implies Sgm(Seg n \ {k}).i = i) &
  (k<=i & i<=$1implies Sgm(Seg n \ {k}).i = i+1);
  let i,k,m,n be Nat;
  assume that
A1: n = m + 1 and
A2: k in Seg n and
A3: i in Seg m;
A4: for k being Nat st P[k] holds P[k+1]
  proof
    let k being Nat such that
A5: P[k];
    let g,i,n be Nat;
    assume that
A6: g = k+1 + 1 and
A7: i in Seg g and
A8: n in Seg(k+1);
A9: 1<=i by A7,FINSEQ_1:1;
    set X = Seg g \ {i};
A10: len Sgm(X)=k+1 by A6,A7,Th107;
A11: 1<=n by A8,FINSEQ_1:1;
A12: i<=g by A7,FINSEQ_1:1;
A13: X c= Seg g by XBOOLE_1:36;
    then
A14: rng Sgm(X) = X by FINSEQ_1:def 13;
A15: n<=k+1 by A8,FINSEQ_1:1;
    now
      per cases;
      suppose
A16:    i=g;
        then
A17:    k+1<i by A6,NAT_1:13;
        X = Seg(k+1)
        proof
          thus X c= Seg(k+1)
          proof
            let x;
A18:        Seg g = {J: 1<=J & J<= g} by FINSEQ_1:def 1;
            assume
A19:        x in X;
            then x in Seg g by XBOOLE_0:def 5;
            then consider J such that
A20:        x=J and
A21:        1<=J and
A22:        J<=g by A18;
            not x in {i} by A19,XBOOLE_0:def 5;
            then x<>i by TARSKI:def 1;
            then J<g by A16,A20,A22,XXREAL_0:1;
            then J<=k+1 by A6,NAT_1:13;
            hence thesis by A20,A21,FINSEQ_1:1;
          end;
          let x;
          assume x in Seg(k+1);
          then x in {J: 1<=J & J<=k+1} by FINSEQ_1:def 1;
          then consider J such that
A23:      x=J and
A24:      1<=J and
A25:      J<=k+1;
          k+1<=g by A6,NAT_1:11;
          then J<=g by A25,XXREAL_0:2;
          then
A26:      J in Seg g by A24,FINSEQ_1:1;
          not J in {i} by A17,A25,TARSKI:def 1;
          hence thesis by A23,A26,XBOOLE_0:def 5;
        end;
        then Sgm X = idseq (k+1) by Th48;
        hence 1<=n & n<i implies Sgm(X).n = n by A8,FUNCT_1:17;
        thus i<=n & n<=k+1 implies Sgm(Seg g \ {i}).n = n+1 by A6,A16,NAT_1:13;
      end;
      suppose
A27:    i<>g;
        set A = {l where l is Element of NAT : 1<=l & l<=g & l<>i};
A28:    X = A
        proof
          thus X c= A
          proof
            let x;
A29:        Seg g= {J: 1<=J & J<=g } by FINSEQ_1:def 1;
            assume
A30:        x in X;
            then x in Seg g by XBOOLE_0:def 5;
            then consider m being Element of NAT such that
A31:        x=m and
A32:        1<=m and
A33:        m<=g by A29;
            not x in {i} by A30,XBOOLE_0:def 5;
            then m<>i by A31,TARSKI:def 1;
            hence thesis by A31,A32,A33;
          end;
          let x;
          assume x in A;
          then consider m being Element of NAT such that
A34:      x=m and
A35:      1<=m and
A36:      m<=g and
A37:      m<>i;
A38:      not m in {i} by A37,TARSKI:def 1;
          m in Seg g by A35,A36,FINSEQ_1:1;
          hence thesis by A34,A38,XBOOLE_0:def 5;
        end;
        1<=k+1 by A11,A15,XXREAL_0:2;
        then k+1 in dom Sgm(X) by A10,Th25;
        then Sgm(X).(k+1) in X by A14,FUNCT_1:def 3;
        then
A39:    ex J st Sgm(X).(k+1) = J & 1<=J & J<=g & J<>i by A28;
A40:     g in NAT by ORDINAL1:def 12;
        1<=g by A9,A12,XXREAL_0:2;
        then g in rng Sgm(X) by A14,A27,A28,A40;
        then consider x be Nat such that
A41:    x in dom Sgm(X) and
A42:    Sgm(X).x=g by FINSEQ_2:10;
        1<=x by A41,Th25;
        then
A43:    k+1<=x by A13,A10,A39,A42,FINSEQ_1:def 13;
A44:    i<g by A12,A27,XXREAL_0:1;
        then
A45:    i<=k+1 by A6,NAT_1:13;
A46:    x<=k+1 by A10,A41,Th25;
        now
          per cases by A15,XXREAL_0:1;
          suppose
A47:        n=k+1;
            hence 1<=n & n<i implies Sgm(X).n = n by A6,A44,NAT_1:13;
            thus i<=n & n<=k+1 implies Sgm(Seg g \ {i}).n=n+1 by A6,A42,A46,A43
,A47,XXREAL_0:1;
          end;
          suppose
A48:        n<k+1;
            set Y = Seg(k+1) \ {i};
A49:        now
              let j1,j be Element of NAT;
              assume that
A50:          j1 in Y and
A51:          j in {g};
              j1 in Seg(k+1) by A50,XBOOLE_0:def 5;
              then
A52:          j1<=k+1 by FINSEQ_1:1;
              j=g by A51,TARSKI:def 1;
              hence j1<j by A6,A52,NAT_1:13;
            end;
A53:        X = Y \/ {g}
            proof
              thus X c= Y \/ {g}
              proof
                let x;
                assume x in X;
                then consider J such that
A54:            x=J and
A55:            1<=J and
A56:            J<=g and
A57:            J<>i by A28;
                now
                  per cases by A56,XXREAL_0:1;
                  suppose
                    J=g;
                    then x in {g} by A54,TARSKI:def 1;
                    hence thesis by XBOOLE_0:def 3;
                  end;
                  suppose
                    J<g;
                    then J<=k+1 by A6,NAT_1:13;
                    then
A58:                J in Seg(k+1) by A55,FINSEQ_1:1;
                    not J in {i} by A57,TARSKI:def 1;
                    then x in Y by A54,A58,XBOOLE_0:def 5;
                    hence thesis by XBOOLE_0:def 3;
                  end;
                end;
                hence thesis;
              end;
              let x such that
A59:          x in Y \/ {g};
              now
                per cases by A59,XBOOLE_0:def 3;
                suppose
A60:              x in Y;
A61:              Seg(k+1)={ s where s is Element of NAT :1<=s & s<=k+1}
                  by FINSEQ_1:def 1;
                  x in Seg(k+1) by A60,XBOOLE_0:def 5;
                  then consider s be Element of NAT such that
A62:              x=s and
A63:              1<=s and
A64:              s<=k+1 by A61;
                  not x in {i} by A60,XBOOLE_0:def 5;
                  then
A65:              s<>i by A62,TARSKI:def 1;
                  k+1<=g by A6,NAT_1:11;
                  then s<=g by A64,XXREAL_0:2;
                  hence thesis by A28,A62,A63,A65;
                end;
                suppose
                  x in {g};
                  then
A66:              x=g by TARSKI:def 1;
                  1<=g by A9,A12,XXREAL_0:2;
                  hence thesis by A27,A28,A66,A40;
                end;
              end;
              hence thesis;
            end;
            then {g} c= X by XBOOLE_1:7;
            then
A67:        {g} c= Seg g by A13,XBOOLE_1:1;
            Y c=X by A53,XBOOLE_1:7;
            then Y c= Seg g by A13,XBOOLE_1:1;
            then
A68:        Sgm(X) = Sgm(Y)^Sgm({g}) by A53,A49,A67,Th42;
            n<=k by A48,NAT_1:13;
            then
A69:        n in Seg k by A11,FINSEQ_1:1;
A70:        i in Seg(k+1) by A9,A45,FINSEQ_1:1;
            then len Sgm(Y) = k by Th107;
            then
A71:        n in dom Sgm(Y) by A69,FINSEQ_1:def 3;
            n<=k by A48,NAT_1:13;
            then
A72:        n in Seg k by A11,FINSEQ_1:1;
            then
A73:        1<=n & n<i implies Sgm(Seg(k+1) \ {i}).n = n by A5,A70;
            now
              assume that
A74:          1<=n and
A75:          n<i;
              n in Seg len Sgm(Y) by A72,A70,Th107;
              then n in dom Sgm(Y) by FINSEQ_1:def 3;
              hence Sgm(X).n = n by A73,A68,A74,A75,FINSEQ_1:def 7;
            end;
            hence 1<=n & n<i implies Sgm(X).n = n;
            assume that
A76:        i<=n and
            n<=k+1;
            i<=n & n<=k implies Sgm(Seg(k+1) \ {i}).n = n+1 by A5,A72,A70;
            hence Sgm(X).n = n+1 by A48,A68,A76,A71,FINSEQ_1:def 7,NAT_1:13;
          end;
        end;
        hence thesis;
      end;
    end;
    hence thesis;
  end;
A77: P[0];
  for k being Nat holds P[k] from NAT_1:sch 2(A77,A4);
  hence thesis by A1,A2,A3;
end;

theorem Th109:
  for f being FinSequence st len f = m+1 & n in dom f holds len Del(f,n) = m
proof
  let f be FinSequence such that
A1: len f = m+1 and
A2: n in dom f;
  ex k being Nat st len f = k+1 & len Del(f,n) = k by A2,Th104;
  hence thesis by A1;
end;

theorem
  for f being FinSequence st k < n holds Del(f,n).k = f.k
proof
  let f be FinSequence;
  assume
A1: k<n;
  per cases;
  suppose that
A2: n in dom f and
A3: f <> {};
    consider m being Nat such that
A4: len f = m+1 by A3,NAT_1:6;
    now
      per cases;
      suppose
A5:     1 <= k;
        set X = dom f \ {n};
A6:     dom Sgm(X)=Seg len Sgm(X) by FINSEQ_1:def 3;
A7:     dom f=Seg len f by FINSEQ_1:def 3;
        then
A8:     len Sgm X = m by A4,A2,Th107;
        X c= Seg len f by A7,XBOOLE_1:36;
        then rng Sgm(X) = X by FINSEQ_1:def 13;
        then
A9:     dom (f*Sgm X) = dom Sgm X by RELAT_1:27,XBOOLE_1:36;
        n<=m+1 by A4,A2,Th25;
        then k<m+1 by A1,XXREAL_0:2;
        then k<=m by NAT_1:13;
        then
A10:     k in Seg m by A5,FINSEQ_1:1;
        then 1<=k & k<n implies Sgm(X).k = k by A4,A2,A7,Th108;
        hence thesis by A1,A5,A10,A6,A9,A8,FUNCT_1:12;
      end;
      suppose
A11:    not 1 <= k;
        Seg len Del (f,n) = Seg m by A4,A2,Th109;
        then dom Del (f,n) = Seg m by FINSEQ_1:def 3;
        then
A12:    not k in dom Del (f,n) by A11,FINSEQ_1:1;
        not k in dom f by A11,Th25;
        then f.k = {} by FUNCT_1:def 2;
        hence thesis by A12,FUNCT_1:def 2;
      end;
    end;
    hence thesis;
  end;
  suppose
    not n in dom f or f = {};
    hence thesis by Th104;
  end;
end;

theorem
  for f being FinSequence st len f = m+1 & n in dom f & n<=k & k<=m
  holds Del(f,n).k = f.(k+1)
proof
  let f be FinSequence;
  assume that
A1: len f = m+1 and
A2: n in dom f and
A3: n<=k and
A4: k<=m;
  set X = dom f \ {n};
A5: dom f=Seg len f by FINSEQ_1:def 3;
  then X c= Seg len f by XBOOLE_1:36;
  then rng Sgm(X) = X by FINSEQ_1:def 13;
  then
A6: dom (f*Sgm(X)) = dom Sgm(X) by RELAT_1:27,XBOOLE_1:36;
A7: len Sgm(X) = m by A1,A2,A5,Th107;
A8: dom Sgm(X)=Seg len Sgm(X) by FINSEQ_1:def 3;
  1<=n by A2,Th25;
  then 1<=k by A3,XXREAL_0:2;
  then
A9: k in Seg m by A4,FINSEQ_1:1;
  then n<=k & k<=m implies Sgm(X).k = k+1 by A1,A2,A5,Th108;
  hence thesis by A3,A4,A9,A8,A6,A7,FUNCT_1:12;
end;



theorem Th112:
  for f being FinSequence st k<=n holds (f|n).k=f.k
proof
  let f be FinSequence;
  assume
A1: k<=n;
  per cases by NAT_1:14;
  suppose
A2: k = 0;
    then
A3: not k in dom f by Th25;
    not k in dom(f|n) by A2,Th25;
    hence (f|n).k = {} by FUNCT_1:def 2
      .= f.k by A3,FUNCT_1:def 2;
  end;
  suppose
    1 <= k;
    then k in Seg n by A1,FINSEQ_1:1;
    then (f|Seg n).k=f.k by FUNCT_1:49;
    hence thesis by FINSEQ_1:def 15;
  end;
end;



theorem
  for p,q being FinSequence st p c= q holds q|len p = p
proof
  let p,q be FinSequence such that
A1: p c= q;
A2: for k being Nat st k in dom p holds p.k = (q|len p).k
  proof
    let k be Nat;
    assume
A3: k in dom p;
    then
A4: k <= len p by Th25;
    thus p.k = q.k by A1,A3,GRFUNC_1:2
      .= (q|len p).k by A4,Th112;
  end;
  len p <= len q by A1,FINSEQ_1:63;
  then len(q|len p) = len p by FINSEQ_1:59;
  then dom(q|len p) = dom p by Th29;
  hence thesis by A2,FINSEQ_1:13;
end;



theorem Th114:
  for A being set, F being FinSequence holds Sgm(F"A)^Sgm(F"(rng
  F \ A)) is Permutation of dom F
proof
  let A be set;
  let F be FinSequence;
A1: dom F = Seg len F by FINSEQ_1:def 3;
  set p = Sgm(F"A)^Sgm(F"(rng F \ A));
  A misses (rng F \ A) by XBOOLE_1:79;
  then
A2: A /\ (rng F \ A) = {} by XBOOLE_0:def 7;
A3: F"A /\ F"(rng F \ A) = F"(A /\ (rng F \ A)) by FUNCT_1:68
    .= {} by A2;
  then
A4: F"A misses F"(rng F \ A) by XBOOLE_0:def 7;
A5: F"(rng F \ A) c= dom F by RELAT_1:132;
  then
A6: F"(rng F \ A) c= Seg len F by FINSEQ_1:def 3;
  then
A7: Sgm(F"(rng F \ A)) is one-to-one by Lm1;
A8: F"A c= dom F by RELAT_1:132;
  then
A9: F"A c= Seg len F by FINSEQ_1:def 3;
  then rng Sgm(F"A) /\ rng Sgm(F"(rng F \ A)) = F"A /\ rng Sgm(F"(rng F \ A))
  by FINSEQ_1:def 13
    .= {} by A6,A3,FINSEQ_1:def 13;
  then
A10: rng Sgm(F"A) misses rng Sgm(F"(rng F \ A)) by XBOOLE_0:def 7;
A11: rng p = rng Sgm(F"A) \/ rng Sgm(F"(rng F \ A)) by FINSEQ_1:31
    .= F"A \/ rng Sgm(F"(rng F \ A)) by A9,FINSEQ_1:def 13
    .= F"A \/ F"(rng F \ A) by A6,FINSEQ_1:def 13
    .= F"(A \/ (rng F \ A)) by RELAT_1:140
    .= F"((rng F) \/ A) by XBOOLE_1:39
    .= F"(rng F) \/ F"A by RELAT_1:140
    .= (dom F) \/ F"A by RELAT_1:134
    .= dom F by RELAT_1:132,XBOOLE_1:12;
  Sgm(F"A) is one-to-one by A9,Lm1;
  then
A12: p is one-to-one by A7,A10,Th91;
  len p = len Sgm(F"A) + len Sgm(F"(rng F \ A)) by FINSEQ_1:22
    .= card (F"A) + len Sgm(F"(rng F \ A)) by A9,Th39
    .= card (F"A) + card (F"(rng F \ A)) by A6,Th39
    .= card (F"A \/ F"(rng F \ A)) by A4,CARD_2:40
    .= len Sgm (F"A \/ F"(rng F \ A)) by A1,A8,A5,Th39,XBOOLE_1:8
    .= len Sgm (F"(A \/ (rng F \ A))) by RELAT_1:140
    .= len Sgm (F"((rng F) \/ A)) by XBOOLE_1:39
    .= len Sgm (F"(rng F) \/ F"A) by RELAT_1:140
    .= len Sgm (dom F \/ F"A) by RELAT_1:134
    .= len Sgm dom F by RELAT_1:132,XBOOLE_1:12
    .= card Seg len F by A1,Th39
    .= len F by FINSEQ_1:57;
  then dom p = dom F by Th29;
  then p is Function of dom F, dom F by A11,FUNCT_2:1;
  hence thesis by A12,A11,FUNCT_2:57;
end;

theorem
  for F being FinSequence, A be Subset of rng F
  ex p being Permutation of dom F st (F-A`)^(F-A) = F*p
proof
  let F be FinSequence;
  let A be Subset of rng F;
  set F1 = F-A`, F2 = F-A;
  A` = (rng F) \ A by SUBSET_1:def 4;
  then F"A` misses F"A by FUNCT_1:71,XBOOLE_1:79;
  then card (F"A`) + card (F"A) =  card ((F"A`) \/ (F"A)) by CARD_2:40
                               .=card (F"(A`\/A)) by RELAT_1:140
                               .= card (F"[#]rng F) by SUBSET_1:10
                               .= card (F"rng F) by SUBSET_1:def 3
                               .=card dom F by RELAT_1:134
                               .=len F by CARD_1:62;
  then A1:len F = len F -card (F"A`) + (len F -card (F"A))
               .= len F1 + (len F -card (F"A)) by Th59
               .= len F1 + len F2 by Th59;
 (rng F) \ (rng F1) = (rng F) \ (rng F \ A`) by Th65
    .= rng F \ (rng F \ (rng F \ A)) by SUBSET_1:def 4
    .= rng F \ (rng F /\ A) by XBOOLE_1:48
    .= rng F \ A by XBOOLE_1:47
    .= rng F2 by Th65;
  then reconsider
  p = Sgm(F"(rng F1))^Sgm(F"(rng F2)) as Permutation of dom F by Th114;
  reconsider F3 = F * p as FinSequence by FINSEQ_2:40;
  take p;
A2: now
A3: F"A` = F"(rng F \ A) by SUBSET_1:def 4
      .= F"(rng F) \ F"A by FUNCT_1:69
      .= dom F \ F"A by RELAT_1:134;
A4: F"(rng F) = dom F by RELAT_1:134;
    then
A5: card(F"(rng F) \ F"A`) = card (F"(rng F)) - card (F"A`) by A3,CARD_2:44
,XBOOLE_1:36
      .= card Seg len F - card (F"A`) by A4,FINSEQ_1:def 3
      .= len F - card (F"A`) by FINSEQ_1:57
      .= len (F - A`) by Th59;
    F"(rng F1) c= dom F by RELAT_1:132;
    then
A6: F"(rng F1) c= Seg len F by FINSEQ_1:def 3;
    let k be Element of NAT such that
A7: k in dom F1;
A8: dom Sgm(F"(rng F1)) = Seg len Sgm(F"(rng F1)) by FINSEQ_1:def 3
      .= Seg card(F"(rng F1)) by A6,Th39
      .= Seg card(F"(rng F \ A`)) by Th65
      .= Seg len (F - A`) by A5,FUNCT_1:69
      .= dom F1 by FINSEQ_1:def 3;
    dom Sgm(F"(rng F1)) c= dom p by FINSEQ_1:26;
    hence (F*p).k = F.(p.k) by A7,A8,FUNCT_1:13
      .= F.(Sgm(F"(rng F1)).k) by A7,A8,FINSEQ_1:def 7
      .= (F*Sgm(F"(rng F1))).k by A7,A8,FUNCT_1:13
      .= (F*Sgm(F"(rng F \ A`))).k by Th65
      .= (F*Sgm(F"(rng F \ (rng F \ A)))).k by SUBSET_1:def 4
      .= (F*Sgm(F"(rng F /\ A))).k by XBOOLE_1:48
      .= (F*Sgm(F"(rng F) /\ F"A)).k by FUNCT_1:68
      .= (F*Sgm(dom F /\ F"A)).k by RELAT_1:134
      .= F1.k by A3,XBOOLE_1:48;
  end;
A9: now
A10: F"(rng F) = dom F by RELAT_1:134;
    F"A c= dom F by RELAT_1:132;
    then F"A c= F"(rng F) by RELAT_1:134;
    then
A11: card(F"(rng F) \ F"A) = card (F"(rng F)) - card (F"A) by CARD_2:44
      .= card Seg len F - card (F"A) by A10,FINSEQ_1:def 3
      .= len F - card (F"A) by FINSEQ_1:57
      .= len (F - A) by Th59;
    dom F = Seg len F by FINSEQ_1:def 3;
    then len Sgm(F"(rng F2)) = card(F"(rng F2)) by Th39,RELAT_1:132
      .= card(F"(rng F \ A)) by Th65
      .= len (F - A) by A11,FUNCT_1:69;
    then
A12: dom Sgm(F"(rng F2)) = dom F2 by Th29;
A13: F"(rng F) = dom F by RELAT_1:134;
    F"A` = F"(rng F \ A) by SUBSET_1:def 4
      .= F"(rng F) \ F"A by FUNCT_1:69
      .= dom F \ F"A by RELAT_1:134;
    then
A14: card(F"(rng F) \ F"A`) = card (F"(rng F)) - card (F"A`) by A13,CARD_2:44
,XBOOLE_1:36
      .= card Seg len F - card (F"A`) by A13,FINSEQ_1:def 3
      .= len F - card (F"A`) by FINSEQ_1:57
      .= len (F - A`) by Th59;
    let k be Element of NAT such that
A15: k in dom F2;
    F"(rng F1) c= dom F by RELAT_1:132;
    then F"(rng F1) c= Seg len F by FINSEQ_1:def 3;
    then
A16: len Sgm(F"(rng F1)) = card(F"(rng F1)) by Th39
      .= card(F"(rng F \ A`)) by Th65
      .= len (F - A`) by A14,FUNCT_1:69;
    then len F1 + k in dom p by A15,A12,FINSEQ_1:28;
    hence (F*p).(len F1 + k) = F.(p.(len F1 + k)) by FUNCT_1:13
      .= F.(Sgm(F"(rng F2)).k) by A15,A12,A16,FINSEQ_1:def 7
      .= (F*Sgm(F"(rng F2))).k by A15,A12,FUNCT_1:13
      .= (F*Sgm(F"(rng F \ A))).k by Th65
      .= (F*Sgm(F"(rng F) \ F"A)).k by FUNCT_1:69
      .= F2.k by RELAT_1:134;
  end;
  len F3 = len F1 + len F2 by A1,FINSEQ_2:44;
  hence thesis by A2,A9,Th38;
end;



theorem
  for f being FinSubsequence holds f is FinSequence implies Seq f = f
proof
  let f be FinSubsequence;
  assume f is FinSequence;
  then reconsider ss9=f as FinSequence;
  consider k be Nat such that
A1: dom ss9 = Seg k by FINSEQ_1:def 2;
  k in NAT by ORDINAL1:def 12;
  then
A2: len ss9 <= k by A1,FINSEQ_1:def 3;
  Seq f = f* Sgm Seg k by A1,FINSEQ_1:def 14
    .= f* idseq k by Th48
    .= f by A2,FINSEQ_2:54;
  hence thesis;
end;



theorem
  for t be FinSequence of INT holds t is FinSequence of REAL
proof
  let t be FinSequence of INT;
  now
    let i be Nat;
    assume i in dom t;
    then
A1: t.i in rng t by FUNCT_1:def 3;
    t.i in INT by A1;
    hence t.i in REAL by NUMBERS:15;
  end;
  hence thesis by FINSEQ_2:12;
end;



theorem
  len p < len q iff dom p c< dom q
proof
A1: len p = len q iff dom p = dom q by Th29;
  len p <= len q iff dom p c= dom q by Th30;
  hence thesis by A1,XBOOLE_0:def 8,XXREAL_0:1;
end;



theorem Th119:
  for A being set, i being Nat holds i <> 0 & A = {} iff i-tuples_on A = {}
proof
  let A be set, i be Nat;
  hereby
    assume i <> 0;
    then
A1: 0+1 <= i by NAT_1:13;
    assume that
A2: A = {} and
A3: i-tuples_on A <> {};
    reconsider B = i-tuples_on A as non empty FinSequenceSet of A by A3;
    set p = the Element of B;
    B = {s where s is Element of A*: len s = i} & p in B;
    then ex s being Element of A* st p = s & len s = i;
    then 1 in dom p by A1,Th25;
    then
A4:    rng p <> {} by RELAT_1:42;
    thus contradiction by A2,A4;
  end;
  assume that
A5: i-tuples_on A = {} and
A6: i = 0 or A <> {};
  len (<*>A) = 0;
  then reconsider A as non empty set by A5,A6,FINSEQ_2:133;
  i-tuples_on A is non empty;
  hence contradiction by A5;
end;



registration
  let i be Nat, D be set;
  cluster i-tuples_on D -> with_common_domain;
  coherence
  proof
    set S = i-tuples_on D;
    let f, g be Function such that
A1: f in S and
A2: g in S;
A3: ex s being Element of D* st f = s & len s = i by A1;
    ex s being Element of D* st g = s & len s = i by A2;
    hence thesis by A3,Th29;
  end;
end;

registration
  let i be Nat, D be set;
  cluster i-tuples_on D -> product-like;
  coherence
  proof
    set S = i-tuples_on D;
    per cases;
    suppose D = {} & i = 0;
      then S = { <*>D } by FINSEQ_2:94
        .= {{}};
      hence thesis by CARD_3:10;
    end;
    suppose D = {} & i <> 0;
      then
A1:   S = {} by Th119;
      take f = 0 .--> {};
      rng f = {{}} by FUNCOP_1:8;
      then {} in rng f by TARSKI:def 1;
      hence thesis by A1,CARD_3:26;
    end;
    suppose D <> {};
      then reconsider D as non empty set;
      set S = i-tuples_on D;
      take product" S;
      S = product product" S
      proof
        thus S c= product product" S by CARD_3:77;
        let x be set;
        assume x in product product" S;
        then consider g being Function such that
A2:     x = g and
A3:     dom g = dom product" S and
A4:     for z being set st z in dom product" S holds g.z in (product" S).z
        by CARD_3:def 5;
        set s = the Element of S;
      s in S;
        then consider t being Element of D* such that
A5:     s = t and
A6:     len t = i;
A7:     dom g = DOM S by A3,CARD_3:def 12
          .= dom s by CARD_3:108;
        dom s = Seg len t by A5,FINSEQ_1:def 3;
        then
A8:     g is FinSequence by A7,FINSEQ_1:def 2;
        rng g c= D
        proof
          let y be set;
          assume y in rng g;
          then consider a being set such that
A9:      a in dom g and
A10:      g.a = y by FUNCT_1:def 3;
          g.a in (product" S).a by A3,A4,A9;
          then g.a in pi(S,a) by A3,A9,CARD_3:def 12;
          then consider f being Function such that
A11:      f in S and
A12:      g.a = f.a by CARD_3:def 6;
          consider w being Element of D* such that
A13:      f = w and len w = i by A11;
          dom g = dom w by A7,A11,A13,CARD_3:def 10;
          then
A14:          w.a in rng w by A9,FUNCT_1:def 3;
          thus thesis by A10,A12,A13,A14;
        end;
        then reconsider g as FinSequence of D by A8,FINSEQ_1:def 4;
A15:    g in D* by FINSEQ_1:def 11;
        len g = i by A5,A6,A7,Th29;
        hence thesis by A2,A15;
      end;
      hence thesis;
    end;
  end;
end;

begin 

reserve
  n for Element of NAT;

theorem
  for D1,D2 be non empty set, p be FinSequence of D1, f be Function
  of D1,D2 holds dom(f*p) = dom p & len (f*p) = len p & for n being Nat st n in
  dom (f*p) holds (f*p).n = f.(p.n)
proof
  let D1,D2 be non empty set, p be FinSequence of D1, f be Function of D1,D2;
A1: rng p c= D1 & dom f = D1 by FUNCT_2:def 1;
  hence dom(f*p) = dom p by RELAT_1:27;
  dom(f*p) = dom p by A1,RELAT_1:27;
  hence len(f*p) = len p by Th29;
  let n be Nat;
  assume n in dom (f*p);
  hence thesis by FUNCT_1:12;
end;

definition
  let D be non empty set, R be Relation of D;
  func ExtendRel(R) -> Relation of D* means
  :Def3:
  for x,y be FinSequence of D
holds [x,y] in it iff len x = len y & for n st n in dom x holds [x.n,y.n] in R;
  existence
  proof
    defpred P[set,set] means for a,b be FinSequence of D st a = $1 & b = $2
    holds len a = len b & for n st n in dom a holds [a.n,b.n] in R;
    consider P be Relation of D*,D* such that
A1: for x,y be set holds [x,y] in P iff x in D* & y in D* & P[x,y]
    from RELSET_1:sch 1;
    take P;
    let a,b be FinSequence of D;
    thus [a,b] in P implies len a = len b & for n st n in dom a holds [a.n,b.n
    ] in R by A1;
    assume len a = len b & for n st n in dom a holds [a.n,b.n] in R;
    then
A2: P[a,b];
    a in D* & b in D* by FINSEQ_1:def 11;
    hence thesis by A1,A2;
  end;
  uniqueness
  proof
    let P,Q be Relation of D*;
    assume that
A3: for x,y be FinSequence of D holds [x,y] in P iff len x = len y &
    for n st n in dom x holds [x.n,y.n] in R and
A4: for x,y be FinSequence of D holds [x,y] in Q iff len x = len y &
    for n st n in dom x holds [x.n,y.n] in R;
    for a,b be set holds [a,b] in P iff [a,b] in Q
    proof
      let a,b be set;
      thus [a,b] in P implies [a,b] in Q
      proof
        assume
A5:     [a,b] in P;
        then reconsider a1 = a, b1 = b as Element of D* by ZFMISC_1:87;
        len a1 = len b1 & for n st n in dom a1 holds [a1.n,b1.n] in R by A3,A5;
        hence thesis by A4;
      end;
      assume
A6:   [a,b] in Q;
      then reconsider a1 = a, b1 = b as Element of D* by ZFMISC_1:87;
      len a1 = len b1 & for n st n in dom a1 holds [a1.n,b1.n] in R by A4,A6;
      hence thesis by A3;
    end;
    hence thesis by RELAT_1:def 2;
  end;
end;

theorem
  for D be non empty set holds ExtendRel(id D) = id (D*)
proof
  let D be non empty set;
  set P = ExtendRel(id D), Q = id(D*);
  for a,b be set holds [a,b] in P iff [a,b] in Q
  proof
    let a,b be set;
    thus [a,b] in P implies [a,b] in Q
    proof
      assume
A1:   [a,b] in P;
      then reconsider a1 = a, b1 = b as Element of D* by ZFMISC_1:87;
A2:   now
        let n be Nat;
        assume n in dom a1;
        then [a1.n,b1.n] in id D by A1,Def3;
        hence a1.n = b1.n by RELAT_1:def 10;
      end;
      len a1 = len b1 by A1,Def3;
      then a1 = b1 by A2,FINSEQ_2:9;
      hence thesis by RELAT_1:def 10;
    end;
    assume
A3: [a,b] in Q;
    then reconsider a1 = a, b1 = b as Element of D* by ZFMISC_1:87;
A4: a1 = b1 by A3,RELAT_1:def 10;
A5: for n st n in dom a1 holds [a1.n,b1.n] in id D
    proof
      let n;
      assume n in dom a1;
      then a1.n in D by FINSEQ_2:11;
      hence thesis by A4,RELAT_1:def 10;
    end;
    len a1 = len b1 by A3,RELAT_1:def 10;
    hence thesis by A5,Def3;
  end;
  hence thesis by RELAT_1:def 2;
end;

definition
  let D be non empty set, R be Equivalence_Relation of D;
  let y be FinSequence of Class(R), x be FinSequence of D;
  pred x is_representatives_FS y means
  len x = len y & for n st n in dom x holds Class(R,x.n) = y.n;
end;

theorem
  for D be non empty set, R be Equivalence_Relation of D, y be
  FinSequence of Class(R) ex x be FinSequence of D st x is_representatives_FS y
proof
  let D be non empty set, R be Equivalence_Relation of D, y be FinSequence of
  Class(R);
  defpred P[set,set] means for u be Element of D st $2 = u holds Class(R,u) =
  y.$1;
A1: for e be set st e in dom y ex u be set st u in D & P[e,u]
  proof
    let e be set;
    assume e in dom y;
    then y.e in rng y by FUNCT_1:def 3;
    then consider a be Element of D such that
A2: y.e = Class(R,a) by EQREL_1:36;
    reconsider b = a as set;
    take b;
    thus b in D;
    let u be Element of D;
    assume b = u;
    hence thesis by A2;
  end;
  consider f being Function such that
A3: dom f = dom y & rng f c= D & for e be set st e in dom y holds P[e,f.
  e] from FUNCT_1:sch 5(A1);
  dom f = Seg len y by A3,FINSEQ_1:def 3;
  then reconsider f as FinSequence by FINSEQ_1:def 2;
  reconsider f as FinSequence of D by A3,FINSEQ_1:def 4;
  take f;
  thus len f = len y by A3,Th29;
  let n;
  assume
A4: n in dom f;
  then f.n in rng f by FUNCT_1:def 3;
  then reconsider u = f.n as Element of D;
  Class(R,u) = y.n by A3,A4;
  hence thesis;
end;



reserve x,y,y1,y2,z,a,b,X,Y,Z,V1,V2 for set,
  f,g,h,h9,f1,f2 for Function,
  i for Nat,
  P for Permutation of X,
  D,D1,D2,D3 for non empty set,
  d1 for Element of D1,
  d2 for Element of D2,
  d3 for Element of D3;

theorem Th123:
  x in product <*X*> iff ex y st y in X & x = <*y*>
proof
A1: dom <*X*> = Seg 1 by FINSEQ_1:def 8;
A2: <*X*>.1 = X by FINSEQ_1:def 8;
A3: 1 in Seg 1 by FINSEQ_1:2,TARSKI:def 1;
  thus x in product <*X*> implies ex y st y in X & x = <*y*>
  proof
    assume x in product <*X*>;
    then consider f such that
A4: x = f and
A5: dom f = dom <*X*> and
A6: for x st x in dom <*X*> holds f.x in <*X*>.x by CARD_3:def 5;
    reconsider f as FinSequence by A1,A5,FINSEQ_1:def 2;
    take f.1;
    thus thesis by A1,A3,A2,A4,A5,A6,FINSEQ_1:def 8;
  end;
  given y such that
A7: y in X and
A8: x = <*y*>;
A9: now
    let a;
    assume a in Seg 1;
    then a = 1 by FINSEQ_1:2,TARSKI:def 1;
    hence <*y*>.a in <*X*>.a by A2,A7,FINSEQ_1:def 8;
  end;
  dom <*y*> = Seg 1 by FINSEQ_1:def 8;
  hence thesis by A1,A8,A9,CARD_3:def 5;
end;

theorem Th124:
  z in product <*X,Y*> iff ex x,y st x in X & y in Y & z = <*x,y*>
proof
A1: <*X,Y*>.1 = X & <*X,Y*>.2 = Y by FINSEQ_1:44;
  len <*X,Y*> = 2 by FINSEQ_1:44;
  then
A2: dom <*X,Y*> = Seg 2 by FINSEQ_1:def 3;
A3: 1 in Seg 2 & 2 in Seg 2 by FINSEQ_1:2,TARSKI:def 2;
  thus z in product <*X,Y*> implies ex x,y st x in X & y in Y & z = <*x,y*>
  proof
    assume z in product <*X,Y*>;
    then consider f such that
A4: z = f and
A5: dom f = dom <*X,Y*> and
A6: for x st x in dom <*X,Y*> holds f.x in <*X,Y*>.x by CARD_3:def 5;
    reconsider f as FinSequence by A2,A5,FINSEQ_1:def 2;
    take f.1, f.2;
    len f = 2 by A2,A5,FINSEQ_1:def 3;
    hence thesis by A2,A3,A1,A4,A6,FINSEQ_1:44;
  end;
  given x,y such that
A7: x in X & y in Y and
A8: z = <*x,y*>;
A9: now
    let a;
    assume a in Seg 2;
    then a = 1 or a = 2 by FINSEQ_1:2,TARSKI:def 2;
    hence <*x,y*>.a in <*X,Y*>.a by A1,A7,FINSEQ_1:44;
  end;
  len <*x,y*> = 2 by FINSEQ_1:44;
  then dom <*x,y*> = Seg 2 by FINSEQ_1:def 3;
  hence thesis by A2,A8,A9,CARD_3:def 5;
end;

theorem Th125:
  a in product <*X,Y,Z*> iff ex x,y,z st x in X & y in Y & z in Z &
  a = <*x,y,z*>
proof
A1: 3 in Seg 3 by Th1,ENUMSET1:def 1;
A2: <*X,Y,Z*>.1 = X & <*X,Y,Z*>.2 = Y by FINSEQ_1:45;
A3: <*X,Y,Z*>.3 = Z by FINSEQ_1:45;
  len <*X,Y,Z*> = 3 by FINSEQ_1:45;
  then
A4: dom <*X,Y,Z*> = Seg 3 by FINSEQ_1:def 3;
A5: 1 in Seg 3 & 2 in Seg 3 by Th1,ENUMSET1:def 1;
  thus a in product <*X,Y,Z*> implies ex x,y,z st x in X & y in Y & z in Z & a
  = <*x,y,z*>
  proof
    assume a in product <*X,Y,Z*>;
    then consider f such that
A6: a = f and
A7: dom f = dom <*X,Y,Z*> and
A8: for x st x in dom <*X,Y,Z*> holds f.x in <*X,Y,Z*>.x by CARD_3:def 5;
    reconsider f as FinSequence by A4,A7,FINSEQ_1:def 2;
    take f.1, f.2, f.3;
    len f = 3 by A4,A7,FINSEQ_1:def 3;
    hence thesis by A4,A5,A1,A2,A3,A6,A8,FINSEQ_1:45;
  end;
  given x,y,z such that
A9: x in X & y in Y & z in Z and
A10: a = <*x,y,z*>;
A11: now
    let a;
    assume a in Seg 3;
    then a = 1 or a = 2 or a = 3 by Th1,ENUMSET1:def 1;
    hence <*x,y,z*>.a in <*X,Y,Z*>.a by A2,A3,A9,FINSEQ_1:45;
  end;
  len <*x,y,z*> = 3 by FINSEQ_1:45;
  then dom <*x,y,z*> = Seg 3 by FINSEQ_1:def 3;
  hence thesis by A4,A10,A11,CARD_3:def 5;
end;

theorem
  product <*D*> = 1-tuples_on D
proof
  <*D*> = 1 |-> D by FINSEQ_2:59
    .= Seg 1 --> D;
  then product <*D*> = Funcs(Seg 1,D) by CARD_3:11;
  hence thesis by FINSEQ_2:93;
end;

theorem Th127:
  product <*D1,D2*> = { <*d1,d2*>: not contradiction }
proof
  thus product <*D1,D2*> c= { <*d1,d2*>: not contradiction }
  proof
    let a;
    assume a in product <*D1,D2*>;
    then ex x,y st x in D1 & y in D2 & a = <*x,y*> by Th124;
    hence thesis;
  end;
  let a;
  assume a in { <*d1,d2*>: not contradiction };
  then ex d1,d2 st a = <*d1,d2*>;
  hence thesis by Th124;
end;

theorem
  product <*D,D*> = 2-tuples_on D
proof
  thus product <*D,D*> = {<*d1,d2*> where d1 is Element of D, d2 is Element of
D: not contradiction} by Th127
    .= 2-tuples_on D by FINSEQ_2:99;
end;

theorem Th129:
  product <*D1,D2,D3*> = { <*d1,d2,d3*>: not contradiction }
proof
  thus product <*D1,D2,D3*> c= { <*d1,d2,d3*>: not contradiction }
  proof
    let a;
    assume a in product <*D1,D2,D3*>;
    then ex x,y,z st x in D1 & y in D2 & z in D3 & a = <*x,y,z*> by Th125;
    hence thesis;
  end;
  let a;
  assume a in { <*d1,d2,d3*>: not contradiction };
  then ex d1,d2,d3 st a = <*d1,d2,d3*>;
  hence thesis by Th125;
end;

theorem
  product <*D,D,D*> = 3-tuples_on D
proof
  thus product <*D,D,D*> = {<*d1,d2,d3*> where d1 is Element of D, d2 is
  Element of D, d3 is Element of D: not contradiction} by Th129
    .= 3-tuples_on D by FINSEQ_2:102;
end;

theorem Th131:
 for D being set holds product (i |-> D) = i-tuples_on D
proof let D be set;
  thus product (i |-> D) = product (Seg i --> D)
    .= Funcs(Seg i,D) by CARD_3:11
    .= i-tuples_on D by FINSEQ_2:93;
end;

theorem Th132:
  doms <*f*> = <*dom f*> & rngs <*f*> = <*rng f*>
proof
A1: rng <*f*> = {f} & <*f*>"rng <*f*> = dom <*f*> by FINSEQ_1:38,RELAT_1:134;
A2: dom doms <*f*> = <*f*>"SubFuncs rng <*f*> & dom <*dom f*> = Seg 1
 by FINSEQ_1:38,FUNCT_6:def 2;
A3: <*f*>.1 = f by FINSEQ_1:40;
A4: dom <*f*> = Seg 1 & SubFuncs {f} = {f} by FINSEQ_1:38,FUNCT_6:20;
A5: <*rng f*>.1 = rng f by FINSEQ_1:40;
A6: now
    let x;
    assume
A7: x in {1};
    then x = 1 by TARSKI:def 1;
    hence (rngs <*f*>).x = <*rng f*>.x
     by A1,A3,A5,A4,A7,FINSEQ_1:2,FUNCT_6:def 3;
  end;
A8: <*dom f*>.1 = dom f by FINSEQ_1:40;
  now
    let x;
    assume
A9: x in {1};
    then x = 1 by TARSKI:def 1;
    hence (doms <*f*>).x = <*dom f*>.x
       by A1,A3,A8,A4,A9,FINSEQ_1:2,FUNCT_6:def 2;
  end;
  hence doms <*f*> = <*dom f*> by A1,A2,A4,FINSEQ_1:2,FUNCT_1:2;
  dom rngs <*f*> = <*f*>"SubFuncs rng <*f*> & dom <*rng f*> = Seg 1
   by FINSEQ_1:38,FUNCT_6:def 3;
  hence thesis by A1,A4,A6,FINSEQ_1:2,FUNCT_1:2;
end;

theorem Th133:
  doms <*f,g*> = <*dom f, dom g*> & rngs <*f,g*> = <*rng f, rng g *>
proof
A1: rng <*f,g*> = {f,g} & <*f,g*>"rng <*f,g*> = dom <*f,g*> by FINSEQ_2:127
,RELAT_1:134;
A2: dom doms <*f,g*> = <*f,g*>"SubFuncs rng <*f,g*> & dom <*dom f, dom g*> =
  Seg 2 by FINSEQ_1:89,FUNCT_6:def 2;
A3: <*f,g*>.1 = f & <*f,g*>.2 = g by FINSEQ_1:44;
A4: dom <*f,g*> = Seg 2 & SubFuncs {f,g} = {f,g} by FINSEQ_1:89,FUNCT_6:20;
A5: <*rng f, rng g*>.1 = rng f & <*rng f, rng g*>.2 = rng g by FINSEQ_1:44;
A6: now
    let x;
    assume
A7: x in {1,2};
    then x = 1 or x = 2 by TARSKI:def 2;
    hence (rngs <*f,g*>).x = <*rng f, rng g*>.x
     by A1,A3,A5,A4,A7,FINSEQ_1:2,FUNCT_6:def 3;
  end;
A8: <*dom f, dom g*>.1 = dom f & <*dom f, dom g*>.2 = dom g by FINSEQ_1:44;
  now
    let x;
    assume
A9: x in {1,2};
    then x = 1 or x = 2 by TARSKI:def 2;
    hence (doms <*f,g*>).x = <*dom f, dom g*>.x
     by A1,A3,A8,A4,A9,FINSEQ_1:2,FUNCT_6:def 2;
  end;
  hence doms <*f,g*> = <*dom f, dom g*> by A1,A2,A4,FINSEQ_1:2,FUNCT_1:2;
  dom rngs <*f,g*> = <*f,g*>"SubFuncs rng <*f,g*> & dom <*rng f, rng g*>
  = Seg 2 by FINSEQ_1:89,FUNCT_6:def 3;
  hence thesis by A1,A4,A6,FINSEQ_1:2,FUNCT_1:2;
end;

theorem
  doms <*f,g,h*> = <*dom f, dom g, dom h*> & rngs <*f,g,h*> = <*rng f,
  rng g, rng h*>
proof
A1: rng <*f,g,h*> = {f,g,h} & <*f,g,h*>"rng <*f,g,h*> = dom <*f,g,h*> by
FINSEQ_2:128,RELAT_1:134;
A2: <*rng f, rng g, rng h*>.3 = rng h by FINSEQ_1:45;
A3: <*f,g,h*>.1 = f & <*f,g,h*>.2 = g by FINSEQ_1:45;
A4: <*f,g,h*>.3 = h by FINSEQ_1:45;
A5: dom <*f,g,h*> = Seg 3 & SubFuncs {f,g,h} = {f,g,h}
        by FINSEQ_1:89,FUNCT_6:20;
A6: <*rng f, rng g, rng h*>.1 = rng f & <*rng f, rng g, rng h*>.2 = rng g
  by FINSEQ_1:45;
A7: now
    let x;
    assume
A8: x in {1,2,3};
    then x = 1 or x = 2 or x = 3 by ENUMSET1:def 1;
    hence (rngs <*f,g,h*>).x = <*rng f, rng g, rng h*>.x by A1,A3,A4,A6,A2,A5
,A8,Th1,FUNCT_6:def 3;
  end;
A9: dom doms <*f,g,h*> = <*f,g,h*>"SubFuncs rng <*f,g,h*> & dom <*dom f, dom
  g, dom h*> = Seg 3 by FINSEQ_1:89,FUNCT_6:def 2;
A10: <*dom f, dom g, dom h*>.3 = dom h by FINSEQ_1:45;
A11: <*dom f, dom g, dom h*>.1 = dom f & <*dom f, dom g, dom h*>.2 = dom g
  by FINSEQ_1:45;
  now
    let x;
    assume
A12: x in {1,2,3};
    then x = 1 or x = 2 or x = 3 by ENUMSET1:def 1;
    hence (doms <*f,g,h*>).x = <*dom f, dom g, dom h*>.x by A1,A3,A4,A11,A10,A5
,A12,Th1,FUNCT_6:def 2;
  end;
  hence doms <*f,g,h*> = <*dom f, dom g, dom h*> by A1,A9,A5,Th1,FUNCT_1:2;
  dom rngs <*f,g,h*> = <*f,g,h*>"SubFuncs rng <*f,g,h*> & dom <*rng f,
  rng g, rng h*> = Seg 3 by FINSEQ_1:89,FUNCT_6:def 3;
  hence thesis by A1,A5,A7,Th1,FUNCT_1:2;
end;

theorem Th135:
  Union <*X*> = X & meet <*X*> = X
proof
  thus Union <*X*> = union rng <*X*>
    .= union {X} by FINSEQ_1:38
    .= X by ZFMISC_1:25;
  thus meet <*X*> = meet {X} by FINSEQ_1:38
    .= X by SETFAM_1:10;
end;

theorem Th136:
  Union <*X,Y*> = X \/ Y & meet <*X,Y*> = X /\ Y
proof
  thus Union <*X,Y*> = union rng <*X,Y*>
    .= union {X,Y} by FINSEQ_2:127
    .= X \/ Y by ZFMISC_1:75;
  thus meet <*X,Y*> = meet {X,Y} by FINSEQ_2:127
    .= X /\ Y by SETFAM_1:11;
end;

theorem
  Union <*X,Y,Z*> = X \/ Y \/ Z & meet <*X,Y,Z*> = X /\ Y /\ Z
proof
A1: union ({X,Y} \/ {Z}) = union {X,Y} \/ union {Z} & union {X,Y} = X \/ Y
  by ZFMISC_1:75,78;
A2: union {Z} = Z by ZFMISC_1:25;
A3: {X,Y} \/ {Z} = {X,Y,Z} by ENUMSET1:3;
  thus Union <*X,Y,Z*> = union rng <*X,Y,Z*>
    .= X \/ Y \/ Z by A1,A2,A3,FINSEQ_2:128;
A4: meet {Z} = Z by SETFAM_1:10;
  meet ({X,Y} \/ {Z}) = meet {X,Y} /\ meet {Z} & meet {X,Y} = X /\ Y by
SETFAM_1:9,11;
  hence thesis by A4,A3,FINSEQ_2:128;
end;

theorem
  x in dom f implies <*f*>..(1,x) = f.x & <*f,g*>..(1,x) = f.x & <*f,g,h
  *>..(1,x) = f.x
proof
A1: <*f,g,h*>.1 = f & 1 in Seg 1 by FINSEQ_1:2,45,TARSKI:def 1;
A2: dom <*f*> = Seg 1 & dom <*f,g*> = Seg 2 by FINSEQ_1:89;
A3: 1 in Seg 2 & 1 in Seg 3 by Th1,ENUMSET1:def 1,FINSEQ_1:2,TARSKI:def 2;
A4: dom <*f,g,h*> = Seg 3 by FINSEQ_1:89;
  <*f*>.1 = f & <*f,g*>.1 = f by FINSEQ_1:44,def 8;
  hence thesis by A1,A3,A2,A4,FUNCT_5:38;
end;

theorem
  x in dom g implies <*f,g*>..(2,x) = g.x & <*f,g,h*>..(2,x) = g.x
proof
A1: 2 in Seg 2 & 2 in Seg 3 by Th1,ENUMSET1:def 1,FINSEQ_1:2,TARSKI:def 2;
A2: dom <*f,g*> = Seg 2 & dom <*f,g,h*> = Seg 3 by FINSEQ_1:89;
  <*f,g*>.2 = g & <*f,g,h*>.2 = g by FINSEQ_1:44,45;
  hence thesis by A1,A2,FUNCT_5:38;
end;

theorem
  x in dom h implies <*f,g,h*>..(3,x) = h.x
proof
A1: dom <*f,g,h*> = Seg 3 by FINSEQ_1:89;
  <*f,g,h*>.3 = h & 3 in Seg 3 by Th1,ENUMSET1:def 1,FINSEQ_1:45;
  hence thesis by A1,FUNCT_5:38;
end;

theorem
  dom <:<*h*>:> = dom h & for x st x in dom h holds <:<*h*>:>.x = <*h.x *>
proof
A1: 1 in Seg 1 by FINSEQ_1:2,TARSKI:def 1;
A2: rng <:<*h*>:> c= product rngs <*h*> & rngs <*h*> = <*rng h*>
 by Th132,FUNCT_6:29;
  thus
A3: dom <:<*h*>:> = meet doms <*h*> by FUNCT_6:29
    .= meet <*dom h*> by Th132
    .= dom h by Th135;
  let x;
  assume
A4: x in dom h;
  then <:<*h*>:>.x in rng <:<*h*>:> by A3,FUNCT_1:def 3;
  then consider g such that
A5: <:<*h*>:>.x = g and
A6: dom g = dom <*rng h*> and
  for y st y in dom <*rng h*> holds g.y in <*rng h*>.y by A2,CARD_3:def 5;
A7: dom g = Seg 1 by A6,FINSEQ_1:38;
  dom <*h*> = Seg 1 & <*h*>.1 = h by FINSEQ_1:38,40;
  then
A8: (uncurry <*h*>).(1,x) = h.x by A4,A1,FUNCT_5:38;
  reconsider g as FinSequence by A7,FINSEQ_1:def 2;
A9: len g = 1 by A7,FINSEQ_1:def 3;
  g.1 = (uncurry <*h*>).(1,x) by A3,A4,A5,A7,A1,FUNCT_6:31;
  hence thesis by A5,A8,A9,FINSEQ_1:40;
end;

theorem Th142:
  dom <:<*f1,f2*>:> = dom f1 /\ dom f2 & for x st x in dom f1 /\
  dom f2 holds <:<*f1,f2*>:>.x = <*f1.x,f2.x*>
proof
A1: dom <*f1,f2*> = Seg 2 by FINSEQ_1:89;
A2: rng <:<*f1,f2*>:> c= product rngs <*f1,f2*> & rngs <*f1,f2*> = <*rng f1,
  rng f2*> by Th133,FUNCT_6:29;
  thus
A3: dom <:<*f1,f2*>:> = meet doms <*f1,f2*> by FUNCT_6:29
    .= meet <*dom f1, dom f2*> by Th133
    .= dom f1 /\ dom f2 by Th136;
  let x;
  assume
A4: x in dom f1 /\ dom f2;
  then <:<*f1,f2*>:>.x in rng <:<*f1,f2*>:> by A3,FUNCT_1:def 3;
  then consider g such that
A5: <:<*f1,f2*>:>.x = g and
A6: dom g = dom <*rng f1, rng f2*> and
  for y st y in dom <*rng f1, rng f2*> holds g.y in <*rng f1, rng f2*> .y
  by A2,CARD_3:def 5;
A7: dom g = Seg 2 by A6,FINSEQ_1:89;
A8: 1 in Seg 2 by FINSEQ_1:2,TARSKI:def 2;
  reconsider g as FinSequence by A7,FINSEQ_1:def 2;
A9: 2 in Seg 2 by FINSEQ_1:2,TARSKI:def 2;
  then
A10: g.2 = (uncurry <*f1,f2*>).(2,x) by A3,A4,A5,A7,FUNCT_6:31;
  <*f1,f2*>.2 = f2 & x in dom f2 by A4,FINSEQ_1:44,XBOOLE_0:def 4;
  then
A11: (uncurry <*f1,f2*>).(2,x) = f2.x by A1,A9,FUNCT_5:38;
A12: len g = 2 by A7,FINSEQ_1:def 3;
  <*f1,f2*>.1 = f1 & x in dom f1 by A4,FINSEQ_1:44,XBOOLE_0:def 4;
  then
A13: (uncurry <*f1,f2*>).(1,x) = f1.x by A1,A8,FUNCT_5:38;
  g.1 = (uncurry <*f1,f2*>).(1,x) by A3,A4,A5,A7,A8,FUNCT_6:31;
  hence thesis by A5,A13,A10,A11,A12,FINSEQ_1:44;
end;

theorem
  dom Frege<*h*> = product <*dom h*> & rng Frege<*h*> = product <*rng h
  *> & for x st x in dom h holds (Frege<*h*>).<*x*> = <*h.x*>
proof
A1: <*h*>.1 = h by FINSEQ_1:def 8;
A2: rngs <*h*> = <*rng h*> by Th132;
A3: doms <*h*> = <*dom h*> by Th132;
  hence
  dom Frege<*h*> = product <*dom h*> & rng Frege<*h*> = product <*rng h*>
  by A2,FUNCT_6:38,def 7;
  let x;
  assume x in dom h;
  then
A4: <*x*> in product doms <*h*> by A3,Th123;
  then consider f such that
A5: (Frege<*h*>).<*x*> = f and
  dom f = <*h*>"SubFuncs rng <*h*> and
  for y st y in dom f holds f.y = (uncurry <*h*>).(y,<*x*>.y) by FUNCT_6:def 7;
A6: dom <*h*> = Seg 1 & 1 in Seg 1 by FINSEQ_1:2,def 8,TARSKI:def 1;
  then dom <*rng h*> = Seg 1 & f in product rngs <*h*> by A1,A4,A5,
FINSEQ_1:def 8,FUNCT_6:37;
  then
A7: dom f = Seg 1 by A2,CARD_3:9;
A8: <*x*>.1 = x by FINSEQ_1:def 8;
  reconsider f as FinSequence by A7,FINSEQ_1:def 2;
  f.1 = h.(<*x*>.1) by A6,A1,A4,A5,FUNCT_6:37;
  hence thesis by A5,A7,A8,FINSEQ_1:def 8;
end;

theorem Th144:
  dom Frege<*f1,f2*> = product <*dom f1, dom f2*> & rng Frege<*f1,
f2*> = product <*rng f1, rng f2*> & for x,y st x in dom f1 & y in dom f2 holds
  (Frege<*f1,f2*>).<*x,y*> = <*f1.x, f2.y*>
proof
A1: rngs <*f1,f2*> = <*rng f1, rng f2*> by Th133;
  len <*rng f1, rng f2*> = 2 by FINSEQ_1:44;
  then
A2: dom <*rng f1, rng f2*> = Seg 2 by FINSEQ_1:def 3;
  len <*f1,f2*> = 2 by FINSEQ_1:44;
  then
A3: dom <*f1,f2*> = Seg 2 by FINSEQ_1:def 3;
A4: doms <*f1,f2*> = <*dom f1, dom f2*> by Th133;
  hence dom Frege<*f1,f2*> = product <*dom f1, dom f2*> & rng Frege<*f1,f2*> =
  product <*rng f1, rng f2*> by A1,FUNCT_6:38,def 7;
  let x,y;
  assume x in dom f1 & y in dom f2;
  then
A5: <*x,y*> in product doms <*f1,f2*> by A4,Th124;
  then consider f such that
A6: (Frege<*f1,f2*>).<*x,y*> = f and
  dom f = <*f1,f2*>"SubFuncs rng <*f1,f2*> and
  for z st z in dom f holds f.z = (uncurry <*f1,f2*>).(z,<*x,y*>.z)
   by FUNCT_6:def 7;
A7: <*x,y*>.1 = x & <*x,y*>.2 = y by FINSEQ_1:44;
A8: 1 in Seg 2 & <*f1,f2*>.1 = f1 by FINSEQ_1:2,44,TARSKI:def 2;
  then f in product rngs <*f1,f2*> by A3,A5,A6,FUNCT_6:37;
  then
A9: dom f = Seg 2 by A1,A2,CARD_3:9;
  then reconsider f as FinSequence by FINSEQ_1:def 2;
  2 in Seg 2 & <*f1,f2*>.2 = f2 by FINSEQ_1:2,44,TARSKI:def 2;
  then
A10: f.2 = f2.(<*x,y*>.2) by A3,A5,A6,FUNCT_6:37;
A11: len f = 2 by A9,FINSEQ_1:def 3;
  f.1 = f1.(<*x,y*>.1) by A3,A8,A5,A6,FUNCT_6:37;
  hence thesis by A6,A10,A11,A7,FINSEQ_1:44;
end;

theorem
  x in dom f1 & x in dom f2 implies for y1,y2 holds <:f1,f2:>.x = [y1,y2
  ] iff <:<*f1,f2*>:>.x = <*y1,y2*>
proof
A1: <*f1.x,f2.x*>.1 = f1.x & <*f1.x,f2.x*>.2 = f2.x by FINSEQ_1:44;
  assume x in dom f1 & x in dom f2;
  then
A2: x in dom f1 /\ dom f2 by XBOOLE_0:def 4;
  let y1,y2;
A3: <*y1,y2*>.1 = y1 & <*y1,y2*>.2 = y2 by FINSEQ_1:44;
  [f1.x,f2.x] = [y1,y2] iff f1.x = y1 & f2.x = y2 by XTUPLE_0:1;
  hence thesis by A2,A1,A3,Th142,FUNCT_3:48;
end;

theorem
  x in dom f1 & y in dom f2 implies for y1,y2 holds [:f1,f2:].(x,y) = [
  y1,y2] iff (Frege<*f1,f2*>).<*x,y*> = <*y1,y2*>
proof
  assume
A1: x in dom f1 & y in dom f2;
  let y1,y2;
A2: <*f1.x,f2.y*>.1 = f1.x & <*f1.x,f2.y*>.2 = f2.y by FINSEQ_1:44;
A3: <*y1,y2*>.1 = y1 & <*y1,y2*>.2 = y2 by FINSEQ_1:44;
  [f1.x,f2.y] = [y1,y2] iff f1.x = y1 & f2.y = y2 by XTUPLE_0:1;
  hence thesis by A1,A2,A3,Th144,FUNCT_3:def 8;
end;

theorem
  Funcs(<*X*>,Y) = <*Funcs(X,Y)*>
proof
A1: dom <*X*> = Seg 1 by FINSEQ_1:def 8;
A2: dom Funcs(<*X*>,Y) = dom <*X*> by FUNCT_6:def 8;
  then reconsider p = Funcs(<*X*>,Y) as FinSequence by A1,FINSEQ_1:def 2;
  <*X*>.1 = X & 1 in Seg 1 by FINSEQ_1:2,def 8,TARSKI:def 1;
  then p.1 = Funcs(X,Y) by A1,FUNCT_6:def 8;
  hence thesis by A2,A1,FINSEQ_1:def 8;
end;

theorem
  Funcs(<*X,Y*>,Z) = <*Funcs(X,Z), Funcs(Y,Z)*>
proof
A1: Seg len <*X,Y*> = dom <*X,Y*> by FINSEQ_1:def 3;
A2: dom Funcs(<*X,Y*>,Z) = dom <*X,Y*> by FUNCT_6:def 8;
  then reconsider p = Funcs(<*X,Y*>,Z) as FinSequence by A1,FINSEQ_1:def 2;
A3: len <*X,Y*> = 2 by FINSEQ_1:44;
  then
A4: len p = 2 by A2,A1,FINSEQ_1:def 3;
  <*X,Y*>.2 = Y & 2 in Seg 2 by FINSEQ_1:2,44,TARSKI:def 2;
  then
A5: p.2 = Funcs(Y,Z) by A3,A1,FUNCT_6:def 8;
  <*X,Y*>.1 = X & 1 in Seg 2 by FINSEQ_1:2,44,TARSKI:def 2;
  then p.1 = Funcs(X,Z) by A3,A1,FUNCT_6:def 8;
  hence thesis by A4,A5,FINSEQ_1:44;
end;

theorem
  Funcs(X,<*Y*>) = <*Funcs(X,Y)*>
proof
A1: dom <*Y*> = Seg 1 by FINSEQ_1:def 8;
A2: dom Funcs(X,<*Y*>) = dom <*Y*> by FUNCT_6:def 9;
  then reconsider p = Funcs(X,<*Y*>) as FinSequence by A1,FINSEQ_1:def 2;
  <*Y*>.1 = Y & 1 in Seg 1 by FINSEQ_1:2,def 8,TARSKI:def 1;
  then p.1 = Funcs(X,Y) by A1,FUNCT_6:def 9;
  hence thesis by A2,A1,FINSEQ_1:def 8;
end;

theorem
  Funcs(X,<*Y,Z*>) = <*Funcs(X,Y), Funcs(X,Z)*>
proof
A1: Seg len <*Y,Z*> = dom <*Y,Z*> by FINSEQ_1:def 3;
A2: dom Funcs(X,<*Y,Z*>) = dom <*Y,Z*> by FUNCT_6:def 9;
  then reconsider p = Funcs(X,<*Y,Z*>) as FinSequence by A1,FINSEQ_1:def 2;
A3: len <*Y,Z*> = 2 by FINSEQ_1:44;
  then
A4: len p = 2 by A2,A1,FINSEQ_1:def 3;
  <*Y,Z*>.2 = Z & 2 in Seg 2 by FINSEQ_1:2,44,TARSKI:def 2;
  then
A5: p.2 = Funcs(X,Z) by A3,A1,FUNCT_6:def 9;
  <*Y,Z*>.1 = Y & 1 in Seg 2 by FINSEQ_1:2,44,TARSKI:def 2;
  then p.1 = Funcs(X,Y) by A3,A1,FUNCT_6:def 9;
  hence thesis by A4,A5,FINSEQ_1:44;
end;



theorem
  for f being FinSequence st rng f={x,y} & len f=2 holds f.1=x & f.
  2=y or f.1=y & f.2=x
proof
  let f be FinSequence;
  assume that
A1: rng f={x,y} and
A2: len f=2;
  2 in dom f by A2,Th25;
  then
A3: f.2 in rng f by FUNCT_1:def 3;
A4: now
    x in rng f by A1,TARSKI:def 2;
    then consider z such that
A5: z in dom f and
A6: x=f.z by FUNCT_1:def 3;
    reconsider nz=z as Element of NAT by A5;
A7: 1<=nz & nz<=len f by A5,Th25;
    assume that
A8: f.1=y and
A9: f.2=y;
    per cases by A2,A7,NAT_1:9;
    suppose
      nz=1;
      hence f.1=x & f.2=y by A9,A6;
    end;
    suppose
      nz=1+1;
      hence f.1=x & f.2=y by A8,A9,A6;
    end;
  end;
A10: now
    y in rng f by A1,TARSKI:def 2;
    then consider z such that
A11: z in dom f and
A12: y=f.z by FUNCT_1:def 3;
    reconsider nz=z as Element of NAT by A11;
A13: 1<=nz & nz<=len f by A11,Th25;
    assume that
A14: f.1=x and
A15: f.2=x;
    per cases by A2,A13,NAT_1:9;
    suppose
      nz=1;
      hence f.1=x & f.2=y by A14,A15,A12;
    end;
    suppose
      nz=1+1;
      hence f.1=x & f.2=y by A14,A12;
    end;
  end;
  1 in dom f by A2,Th25;
  then f.1 in rng f by FUNCT_1:def 3;
  hence thesis by A1,A3,A10,A4,TARSKI:def 2;
end;



theorem
  for X being set, k being Element of NAT st X c= Seg k holds for m
  ,n being Element of NAT st m in dom (Sgm X) & n = (Sgm X).m holds m <= n
proof
  let X be set, k be Element of NAT;
  defpred P[Nat] means ($1 in dom (Sgm X) & (ex n being Element of NAT st n=(
  Sgm X).$1 & $1 <= n)) or (not $1 in dom (Sgm X));
  assume
A1: X c= Seg k;
  now
    let x be non empty Nat;
    assume
A2: P[x];
    now
      per cases by A2;
      suppose
A3:     x in dom (Sgm X) & ex n being Element of NAT st n = (Sgm X).x & x <= n;
A4:     x+0 < x+1 by XREAL_1:8;
        consider n being Element of NAT such that
A5:     n = Sgm(X).x and
A6:     x <= n by A3;
A7:     1 <= x by A3,Th25;
        now
          set n1 = (Sgm X).(x+1);
          assume
A8:       x+1 in dom (Sgm X);
          then (Sgm X).(x+1) in rng (Sgm X) by FUNCT_1:3;
          then reconsider n1 as Element of NAT;
          take n1;
          thus n1 = (Sgm X).(x+1);
          x+1 <= len Sgm X by A8,Th25;
          then n < n1 by A1,A7,A4,A5,FINSEQ_1:def 13;
          then x < n1 by A6,XXREAL_0:2;
          hence x+1 <= n1 by NAT_1:13;
        end;
        hence P[x+1];
      end;
      suppose
        not x in dom (Sgm X);
        then x < 0+1 or x > len Sgm X by Th25;
        then x+1 > len Sgm X + 0 by NAT_1:13;
        hence P[x+1] by Th25;
      end;
    end;
    hence P[x+1];
  end;
  then
A9: for x being non empty Nat st P[x] holds P[x+1];
  let m, n be Element of NAT;
  assume that
A10: m in dom (Sgm X) and
A11: n = (Sgm X).m;
  reconsider m9=m as non empty Element of NAT by A10,Th25;
  now
    set n = (Sgm X).1;
A12: m <= len (Sgm X) by A10,Th25;
    1 <= m by A10,Th25;
    then 1 <= len (Sgm X) by A12,XXREAL_0:2;
    hence 1 in dom (Sgm X) by Th25;
    then
A13: (Sgm X).1 in rng (Sgm X) by FUNCT_1:3;
    then reconsider n as Element of NAT;
    take n;
    thus n = (Sgm X).1;
    rng (Sgm X) = X by A1,FINSEQ_1:def 13;
    hence 1 <= n by A1,A13,FINSEQ_1:1;
  end;
  then
A14: P[1];
  for x being non empty Nat holds P[x] from NAT_1:sch 10(A14,A9);
  then P[m9];
  hence thesis by A10,A11;
end;

registration
  let i be Nat; let D be finite set;
  cluster i-tuples_on D -> finite;
 coherence
  proof
    now
      let x be set;
      assume x in dom (i |-> D);
      then x in Seg i by FUNCOP_1:13;
      hence (i |-> D).x is finite by FUNCOP_1:7;
    end;
    then
A1: i |-> D is finite-yielding by FINSET_1:def 4;
    product (i |-> D) = i-tuples_on D by Th131;
    hence thesis by A1;
  end;
end;














begin

reserve x,z for set;
reserve k for Element of NAT;
reserve D for non empty set;

definition
  let B be non empty set, A be set, b be Element of B;
  redefine func A --> b -> Function of A,B;
  coherence
  proof
    set f = A --> b;
A1: dom f = A by FUNCOP_1:13;
    thus thesis by A1;
  end;
end;

definition
  let IT be FinSequence-membered set;
  redefine attr IT is with_common_domain means
  :Def1:
   for a,b being FinSequence st a in IT & b in IT holds len a = len b;
  compatibility
   proof
    thus IT is with_common_domain implies
   for a,b being FinSequence st a in IT & b in IT holds len a = len b
     proof
      assume
A1:     IT is with_common_domain;
      let a,b being FinSequence;
      assume a in IT & b in IT;
       then dom a = dom b by A1,CARD_3:def 10;
      hence len a = len b by FINSEQ_3:29;
     end;
    assume
A2:  for a,b being FinSequence st a in IT & b in IT holds len a = len b;
    let f,g be Function;
    assume
A3:    f in IT & g in IT;
     then reconsider f,g as FinSequence;
     len f = len g by A2,A3;
    hence thesis by FINSEQ_3:29;
   end;
end;

registration
  cluster FinSequence-membered with_common_domain for set;
  existence
  proof
    take {};
    thus (for x being set st x in {} holds x is FinSequence) & for a,b being
    FinSequence st a in {} & b in {} holds len a = len b;
  end;
end;

definition
  mode relation is FinSequence-membered with_common_domain set;
end;

reserve X for set;
reserve p,r for relation;
reserve a,a1,a2,b for FinSequence;

theorem
  X c= p implies X is relation;

theorem
  {a} is relation
proof
  for z st z in {a} holds z is FinSequence by TARSKI:def 1;
  then reconsider X ={a} as FinSequence-membered set by FINSEQ_1:def 18;
  X is with_common_domain;
 hence thesis;
end;

scheme
  relexist{A() -> set, P[FinSequence]}: ex r st for a holds a in r iff a in A(
  ) & P[a]
provided
A1: for a,b st P[a] & P[b] holds len a = len b
proof
  defpred P1[set] means ex a st P[a] & $1 = a;
  consider X such that
A2: x in X iff x in A() & P1[x] from XBOOLE_0:sch 1;
   X is FinSequence-membered
  proof
    let x be set;
    assume x in X;
    then ex a st P[a] & x = a by A2;
    hence thesis;
  end;
   then reconsider X as FinSequence-membered set;
    X is with_common_domain
  proof
    let a,b;
    assume that
A3: a in X and
A4: b in X;
A5: ex d being FinSequence st P[d] & b = d by A2,A4;
    ex c being FinSequence st P[c] & a = c by A2,A3;
    hence thesis by A1,A5;
  end;
  then reconsider r = X as relation;
  for a holds a in r iff a in A() & P[a]
  proof
    let a;
    now
      assume
A6:   a in r;
      then ex c being FinSequence st P[c] & a = c by A2;
      hence a in A() & P[a] by A2,A6;
    end;
    hence thesis by A2;
  end;
  hence thesis;
end;

definition
  let p,r;
  redefine pred p = r means
  for a holds a in p iff a in r;
  compatibility
  proof
    thus p = r implies for a holds a in p iff a in r;
    thus (for a holds a in p iff a in r) implies p = r
    proof
      assume for a holds a in p iff a in r;
      then for x holds x in p iff x in r;
      hence thesis by TARSKI:1;
    end;
  end;
end;

registration
  cluster empty -> with_common_domain for set;
  coherence
  proof
    let X be set;
    assume
A1:   X is empty;
    then
    for a,b being FinSequence st a in X & b in X holds len a = len b;
   hence thesis by Def1,A1;
  end;
end;

theorem Th3:
  for p st for a holds not a in p holds p = {}
proof
  let p such that
A1: for a holds not a in p;
  assume p <> {};
  then ex x being set st x in p by XBOOLE_0:def 1;
  hence contradiction by A1;
end;

definition
  let p;
  assume
A1: p <> {};

  func the_arity_of p -> Element of NAT means
  for a st a in p holds it = len a;
  existence
  proof
    consider c being FinSequence such that
A2: c in p by A1,Th3;
    for a st a in p holds len c = len a by A2,Def1;
    hence thesis;
  end;
  uniqueness
  proof
    let n1,n2 be Element of NAT;
    assume that
A3: for a st a in p holds n1 = len a and
A4: for a st a in p holds n2 = len a;
    consider a such that
A5: a in p by A1,Th3;
    len a = n1 by A3,A5;
    hence thesis by A4,A5;
  end;
end;

definition
  let k;
  mode relation_length of k -> relation means
    for a st a in it holds len a = k;
  existence
  proof
    take {};
    thus thesis;
  end;
end;

definition
  let X be set;
  mode relation of X -> relation means
    for a st a in it holds rng a c= X;
  existence
  proof
    take {};
    thus thesis;
  end;
end;

theorem Th4:
  {} is relation of X
proof
  thus a in {} implies rng a c= X;
end;

theorem Th5:
  {} is relation_length of k
proof
  thus a in {} implies len a = k;
end;

definition
  let X, k;
  mode relation of X,k -> relation means
    it is relation of X & it is relation_length of k;
  existence
  proof
    take {};
    thus thesis by Th4,Th5;
  end;
end;

definition
  let D;
  func relations_on D -> set means
  :Def7:
  for X holds X in it iff X c= D* &
  for a,b being FinSequence of D st a in X & b in X holds len a = len b;
  existence
  proof
    defpred P[set] means ex Y being set st Y = $1 & Y c= D* & for a,b being
    FinSequence of D st a in Y & b in Y holds len a = len b;
    consider A being set such that
A1: for x holds x in A iff x in bool(D*) & P[x] from XBOOLE_0:sch 1;
    take A;
    for X being set holds X in A iff X c= D* & for a,b being FinSequence
    of D st a in X & b in X holds len a = len b
    proof
      let X be set;
      thus X in A implies X c= D* & for a,b being FinSequence of D st a in X &
      b in X holds len a = len b
      proof
        assume X in A;
        then
        ex Y being set st Y = X & Y c= D* & for a,b being FinSequence of D
        st a in Y & b in Y holds len a=len b by A1;
        hence thesis;
      end;
      thus thesis by A1;
    end;
    hence thesis;
  end;
  uniqueness
  proof
    let A1,A2 be set;
    assume that
A2: for X being set holds X in A1 iff X c= D* & for a,b being
    FinSequence of D st a in X & b in X holds len a=len b and
A3: for X being set holds X in A2 iff X c= D* & for a,b being
    FinSequence of D st a in X & b in X holds len a=len b;
    for x being set holds x in A1 iff x in A2
    proof
      let x;
      thus x in A1 implies x in A2
      proof
        assume
A4:     x in A1;
        then
A5:     for a,b being FinSequence of D st a in x & b in x holds len a=len
        b by A2;
        x c= D* by A2,A4;
        hence thesis by A3,A5;
      end;
      thus x in A2 implies x in A1
      proof
        assume
A6:     x in A2;
        then
A7:     for a,b being FinSequence of D st a in x & b in x holds len a=len
        b by A3;
        x c= D* by A3,A6;
        hence thesis by A2,A7;
      end;
    end;
    hence thesis by TARSKI:1;
  end;
end;

registration
  let D;
  cluster relations_on D -> non empty;
  coherence
  proof
A1: for a,b being FinSequence of D st a in {} & b in {} holds len a = len b;
    defpred P[set] means ex Y being set st Y = $1 & Y c= D* & for a,b being
    FinSequence of D st a in Y & b in Y holds len a = len b;
    consider XX being set such that
A2: for x holds x in XX iff x in bool(D*) & P[x] from XBOOLE_0:sch 1;
    {} c= D* by XBOOLE_1:2;
    then reconsider A = XX as non empty set by A2,A1;
    for X being set holds X in A iff X c= D* & for a,b being FinSequence
    of D st a in X & b in X holds len a = len b
    proof
      let X be set;
      thus X in A implies X c= D* & for a,b being FinSequence of D st a in X &
      b in X holds len a = len b
      proof
        assume X in A;
        then
        ex Y being set st Y = X & Y c= D* & for a,b being FinSequence of D
        st a in Y & b in Y holds len a=len b by A2;
        hence thesis;
      end;
      thus thesis by A2;
    end;
    hence thesis by Def7;
  end;
end;

definition
  let D be non empty set;
  mode relation of D is Element of relations_on D;
end;

reserve a,b for FinSequence of D;
reserve p,r for Element of relations_on D;

theorem
  X c= r implies X is Element of relations_on D
proof
  assume
A1: X c= r;
  then
A2: for a,b st a in X & b in X holds len a = len b by Def7;
  r c= D* by Def7;
  then X c= D* by A1,XBOOLE_1:1;
  hence thesis by A2,Def7;
end;

theorem
  {a} is Element of relations_on D
proof
A1: for a1,a2 being FinSequence of D st a1 in {a} & a2 in {a} holds len a1 =
  len a2
  proof
    let a1,a2 be FinSequence of D;
    assume that
A2: a1 in {a} and
A3: a2 in {a};
    a1 = a by A2,TARSKI:def 1;
    hence thesis by A3,TARSKI:def 1;
  end;
  a in D* by FINSEQ_1:def 11;
  then {a} c= D* by ZFMISC_1:31;
  hence thesis by A1,Def7;
end;

theorem
  for x,y being Element of D holds {<*x,y*>} is Element of relations_on D
proof
  let x,y be Element of D;
A1: for a1,a2 being FinSequence of D st a1 in {<*x,y*>} & a2 in {<*x,y*>}
  holds len a1 = len a2
  proof
    let a1,a2 be FinSequence of D;
    assume that
A2: a1 in {<*x,y*>} and
A3: a2 in {<*x,y*>};
    a1 = <*x,y*> by A2,TARSKI:def 1;
    hence thesis by A3,TARSKI:def 1;
  end;
  <*x*>^<*y*> is FinSequence of D;
  then <*x,y*> in D* by FINSEQ_1:def 11;
  then {<*x,y*>} c= D* by ZFMISC_1:31;
  hence thesis by A1,Def7;
end;

definition
  let D,p,r;
  redefine pred p = r means
  :Def8:
  for a holds a in p iff a in r;
  compatibility
  proof
    thus p = r implies for a holds a in p iff a in r;
    thus (for a holds a in p iff a in r) implies p = r
    proof
      assume
A1:   for a holds a in p iff a in r;
      now
        let x;
A2:     now
          assume
A3:       x in r;
          r is Subset of D* by Def7;
          then x is FinSequence of D by A3,FINSEQ_1:def 11;
          hence x in p by A1,A3;
        end;
        now
          assume
A4:       x in p;
          p is Subset of D* by Def7;
          then x is FinSequence of D by A4,FINSEQ_1:def 11;
          hence x in r by A1,A4;
        end;
        hence x in p iff x in r by A2;
      end;
      hence thesis by TARSKI:1;
    end;
  end;
end;

scheme
  relDexist{D() -> non empty set, P[FinSequence of D()]}: ex r being Element
  of relations_on D() st for a being FinSequence of D() holds a in r iff P[a]
provided
A1: for a,b being FinSequence of D() st P[a] & P[b] holds len a = len b
proof
  defpred P1[set] means ex a being FinSequence of D() st P[a] & $1 = a;
  consider X being set such that
A2: x in X iff x in D()* & P1[x] from XBOOLE_0:sch 1;
A3: for a,b being FinSequence of D() st a in X & b in X holds len a = len b
  proof
    let a,b be FinSequence of D();
    assume that
A4: a in X and
A5: b in X;
A6: ex d being FinSequence of D() st P[d] & b = d by A2,A5;
    ex c being FinSequence of D() st P[c] & a = c by A2,A4;
    hence thesis by A1,A6;
  end;
  for x holds x in X implies x in D()* by A2;
  then X c= D()* by TARSKI:def 3;
  then reconsider r = X as Element of relations_on D() by A3,Def7;
  for a being FinSequence of D() holds a in r iff P[a]
  proof
    let a be FinSequence of D();
A7: now
A8:   a in D()* by FINSEQ_1:def 11;
      assume P[a];
      hence a in r by A2,A8;
    end;
    now
      assume a in r;
      then ex c being FinSequence of D() st P[c] & a = c by A2;
      hence P[a];
    end;
    hence thesis by A7;
  end;
  hence thesis;
end;

definition
  let D;
  func empty_rel(D) -> Element of relations_on D means
  :Def9:
  not a in it;
  existence
  proof
    defpred P[FinSequence of D] means contradiction;
A1: P[a] & P[b] implies len a = len b;
    consider r such that
A2: for a holds a in r iff P[a] from relDexist(A1);
    take r;
    thus thesis by A2;
  end;
  uniqueness
  proof
    let r1,r2 be Element of relations_on D;
    assume that
A3: not a in r1 and
A4: not a in r2;
    for a holds ( a in r1 iff a in r2) by A3,A4;
    hence r1 = r2 by Def8;
  end;
end;

theorem
  empty_rel(D) = {}
proof
  assume
A1: not thesis;
  set x = the Element of empty_rel(D);
  empty_rel(D) is Subset of D* by Def7;
  then x in D* by A1,TARSKI:def 3;
  then reconsider a = x as FinSequence of D by FINSEQ_1:def 11;
  a in empty_rel(D) by A1;
  hence contradiction by Def9;
end;

definition
  let D,p;
  assume
A1: p <> empty_rel(D);
  func the_arity_of p -> Element of NAT means
  a in p implies it = len a;
  existence
  proof
    consider c being FinSequence of D such that
A2: c in p by A1,Def9;
    a in p implies len c = len a by A2,Def7;
    hence thesis;
  end;
  uniqueness
  proof
    let n1,n2 be Element of NAT;
    assume that
A3: a in p implies n1 = len a and
A4: a in p implies n2 = len a;
    consider a such that
A5: a in p by A1,Def9;
    len a = n1 by A3,A5;
    hence thesis by A4,A5;
  end;
end;

scheme
  relDexist2{D() -> non empty set, k() -> Element of NAT, P[FinSequence of D()
]}: ex r being Element of relations_on D() st for a being FinSequence of D() st
  len a = k() holds a in r iff P[a] proof
  defpred P1[set] means ex a being FinSequence of D() st len a = k() & P[a] &
  $1 = a;
  consider X being set such that
A1: x in X iff x in D()* & P1[x] from XBOOLE_0:sch 1;
A2: for a,b being FinSequence of D() st a in X & b in X holds len a = len b
  proof
    let a,b be FinSequence of D();
    assume that
A3: a in X and
A4: b in X;
A5: ex d being FinSequence of D() st len d = k() & P[d] & b = d by A1,A4;
    ex c being FinSequence of D() st len c = k() & P[c] & a = c by A1,A3;
    hence thesis by A5;
  end;
  for x holds x in X implies x in D()* by A1;
  then X c= D()* by TARSKI:def 3;
  then reconsider r = X as Element of relations_on D() by A2,Def7;
  for a being FinSequence of D() st len a = k() holds a in r iff P[a]
  proof
    let a be FinSequence of D() such that
A6: len a = k();
A7: now
A8:   a in D()* by FINSEQ_1:def 11;
      assume P[a];
      hence a in r by A1,A6,A8;
    end;
    now
      assume a in r;
      then
      ex c being FinSequence of D() st len c = k() & P[c] & a = c by A1;
      hence P[a];
    end;
    hence thesis by A7;
  end;
  hence thesis;
end;

definition
  func BOOLEAN -> set equals
  {0,1};
  coherence;
end;

registration
  cluster BOOLEAN -> non empty;
  coherence;
end;

definition
  redefine func FALSE -> Element of BOOLEAN;
  coherence by TARSKI:def 2;
  redefine func TRUE -> Element of BOOLEAN;
  coherence by TARSKI:def 2;
end;

definition
  let x be set;
  redefine attr x is boolean means
  :Def12:
  x in BOOLEAN;
  compatibility
  proof
    hereby
      assume x is boolean;
      then x = FALSE or x = TRUE by XBOOLEAN:def 3;
      hence x in BOOLEAN;
    end;
    assume x in BOOLEAN;
    hence x = FALSE or x = TRUE by TARSKI:def 2;
  end;
end;

registration
  cluster -> boolean for Element of BOOLEAN;
  coherence by Def12;
end;

reserve u,v,w for boolean set;

definition
  let v be boolean set;
  redefine func 'not' v equals
  TRUE if v = FALSE otherwise FALSE;
  compatibility
  proof
    let w;
    thus v = FALSE implies (w = 'not' v iff w = TRUE);
    assume v <> FALSE;
    then v = TRUE by XBOOLEAN:def 3;
    hence thesis;
  end;
  consistency;
  let w be boolean set;
  redefine func v '&' w equals
  TRUE if v = TRUE & w =TRUE otherwise FALSE;
  compatibility
  proof
    let u be set;
    thus v = TRUE & w =TRUE implies (u = v '&' w iff u = TRUE);
    assume v <> TRUE or w <>TRUE;
    then v = FALSE or w = FALSE by XBOOLEAN:def 3;
    hence thesis;
  end;
  consistency;
end;

definition
  let v be Element of BOOLEAN;
  redefine func 'not' v -> Element of BOOLEAN;
  correctness by Def12;
  let w be Element of BOOLEAN;
  redefine func v '&' w -> Element of BOOLEAN;
  correctness by Def12;
end;

canceled;

theorem
  (v = FALSE iff 'not' v = TRUE) & (v = TRUE iff 'not' v = FALSE);

theorem
  (v '&' w = TRUE iff v = TRUE & w = TRUE) & (v '&' w = FALSE iff v =
  FALSE or w = FALSE) by XBOOLEAN:101,140;

theorem
  FALSE '&' v = FALSE;

theorem
  TRUE '&' v = v;

theorem
  v '&' v = FALSE implies v = FALSE;

theorem
  v '&' (w '&' u) = (v '&' w) '&' u;

definition
  let X;
  func ALL(X) equals
  :Def15:
  TRUE if not FALSE in X otherwise FALSE;
  correctness;
end;

registration
  let X;
  cluster ALL X -> boolean;
  correctness by Def15;
end;

definition
  let X;
  redefine func ALL X -> Element of BOOLEAN;
  correctness by Def12;
end;

theorem
  (not FALSE in X iff ALL(X) = TRUE) & (FALSE in X iff ALL(X) = FALSE)
  by Def15;

begin 


definition
  let f be Relation;
  attr f is boolean-valued means
  :Def16:
  rng f c= BOOLEAN;
end;

registration
  cluster boolean-valued for Function;
  existence
  proof
    take {};
    thus rng {} c= BOOLEAN by XBOOLE_1:2;
  end;
end;

registration
  let f be boolean-valued Function, x be set;
  cluster f.x -> boolean;
  coherence
  proof
    per cases;
    suppose
      not x in dom f;
      then f.x = FALSE by FUNCT_1:def 2;
      hence f.x in BOOLEAN;
    end;
    suppose
A1:   x in dom f;
A2:   rng f c= BOOLEAN by Def16;
      f.x in rng f by A1,FUNCT_1:def 3;
      hence f.x in BOOLEAN by A2;
    end;
  end;
end;

definition
  let p be boolean-valued Function;
  func 'not' p -> boolean-valued Function means
  :Def17:
  dom it = dom p & for x being set st x in dom p holds it.x = 'not'(p.x);
  existence
  proof
    deffunc F(set) = 'not'(p.$1);
    consider q being Function such that
A1: dom q = dom p and
A2: for x being set st x in dom p holds q.x = F(x) from FUNCT_1:sch 3;
    q is boolean-valued
    proof
      let x be set;
      assume x in rng q;
      then consider y being set such that
A3:   y in dom q and
A4:   x = q.y by FUNCT_1:def 3;
      x = 'not'(p.y) by A1,A2,A3,A4;
      then x = FALSE or x = TRUE by XBOOLEAN:def 3;
      hence thesis;
    end;
    hence thesis by A1,A2;
  end;
  uniqueness
  proof
    let q1,q2 be boolean-valued Function such that
A5: dom q1 = dom p and
A6: for x being set st x in dom p holds q1.x = 'not'(p.x) and
A7: dom q2 = dom p and
A8: for x being set st x in dom p holds q2.x = 'not'(p.x);
    for x being set st x in dom p holds q1.x = q2.x
    proof
      let x be set;
      assume
A9:   x in dom p;
      then q1.x = 'not'(p.x) by A6;
      hence thesis by A8,A9;
    end;
    hence thesis by A5,A7,FUNCT_1:2;
  end;
  involutiveness
  proof
    let q,p be boolean-valued Function;
    assume that
A10: dom q = dom p and
A11: for x being set st x in dom p holds q.x = 'not'(p.x);
    thus dom p = dom q by A10;
    let x be set;
    assume
A12: x in dom q;
    thus p.x = 'not' 'not'(p.x) .= 'not'(q.x) by A10,A11,A12;
  end;
  let q be boolean-valued Function;
  func p '&' q -> boolean-valued Function means
  :Def18:
  dom it = dom p /\ dom
  q & for x being set st x in dom it holds it.x = (p.x) '&' (q.x);
  existence
  proof
    deffunc F(set) = (p.$1) '&' (q.$1);
    consider s being Function such that
A13: dom s = dom p /\ dom q and
A14: for x being set st x in dom p /\ dom q holds s.x = F(x) from
    FUNCT_1:sch 3;
    s is boolean-valued
    proof
      let x be set;
      assume x in rng s;
      then consider y being set such that
A15:  y in dom s and
A16:  x = s.y by FUNCT_1:def 3;
      x = (p.y) '&' (q.y) by A13,A14,A15,A16;
      then x = FALSE or x = TRUE by XBOOLEAN:def 3;
      hence thesis;
    end;
    hence thesis by A13,A14;
  end;
  uniqueness
  proof
    let s1,s2 be boolean-valued Function such that
A17: dom s1 = dom p /\ dom q and
A18: for x being set st x in dom s1 holds s1.x = (p.x) '&' (q.x) and
A19: dom s2 = dom p /\ dom q and
A20: for x being set st x in dom s2 holds s2.x = (p.x) '&' (q.x);
    for x being set st x in dom s1 holds s1.x = s2.x
    proof
      let x be set;
      assume
A21:  x in dom s1;
      then s1.x = (p.x) '&' (q.x) by A18;
      hence thesis by A17,A19,A20,A21;
    end;
    hence thesis by A17,A19,FUNCT_1:2;
  end;
  commutativity;
  idempotence;
end;

registration
  let A be set;
  cluster -> boolean-valued for Function of A,BOOLEAN;
  coherence
  proof
   let F be Function of A,BOOLEAN;
   thus rng F c= BOOLEAN by RELAT_1:def 19;
  end;
end;

definition
  let A be non empty set;
  let p be Function of A,BOOLEAN;
  redefine func 'not' p -> Function of A,BOOLEAN means
  for x being Element of A holds it.x = 'not'(p.x);
  coherence
  proof
A1: dom 'not' p = dom p by Def17
      .= A by PARTFUN1:def 2;
    rng 'not' p c= BOOLEAN by Def16;
    hence thesis by A1,FUNCT_2:2;
  end;
  compatibility
  proof
    let IT be Function of A,BOOLEAN;
A2: dom IT = A by FUNCT_2:def 1;
    hereby
      assume
A3:   IT = 'not' p;
      let x be Element of A;
      x in A;
      then x in dom p by FUNCT_2:def 1;
      hence IT.x = 'not'(p.x) by A3,Def17;
    end;
A4: dom p = A by FUNCT_2:def 1;
    assume for x being Element of A holds IT.x = 'not'(p.x);
    then for x being set st x in dom p holds IT.x = 'not'(p.x) by A4;
    hence thesis by A2,A4,Def17;
  end;
  let q be Function of A,BOOLEAN;
  redefine func p '&' q -> Function of A,BOOLEAN means
  for x being Element of A holds it.x = (p.x) '&' (q.x);
  coherence
  proof
A5: rng(p '&' q) c= BOOLEAN by Def16;
    dom p = A & dom q = A by PARTFUN1:def 2;
    then dom(p '&' q) = A /\ A by Def18
      .= A;
    hence thesis by A5,FUNCT_2:2;
  end;
  compatibility
  proof
    let IT be Function of A,BOOLEAN;
A6: dom IT = A by FUNCT_2:def 1;
    hereby
      assume
A7:   IT = p '&' q;
      let x be Element of A;
A8:   dom q = A by FUNCT_2:def 1;
      dom p = A by FUNCT_2:def 1;
      then dom(p '&' q) = A /\ A by A8,Def18
        .= A;
      hence IT.x = (p.x) '&' (q.x) by A7,Def18;
    end;
A9: dom q = A by FUNCT_2:def 1;
A10: dom IT = A /\ A by FUNCT_2:def 1
      .= dom p /\ dom q by A9,FUNCT_2:def 1;
    assume for x being Element of A holds IT.x = (p.x) '&' (q.x);
    then for x being set st x in dom IT holds IT.x = (p.x) '&' (q.x) by A6;
    hence thesis by A10,Def18;
  end;
end;

begin 

reserve A,z for set,
  x,y for FinSequence of A,
  h for PartFunc of A*,A,
  n,m for Nat;

definition
  let IT be Relation;
  attr IT is homogeneous means
:Def21: dom IT is with_common_domain;
end;

definition
  let A;
  let IT be PartFunc of A*,A;
  attr IT is quasi_total means
:Def22: for x,y st len x = len y & x in dom IT holds y in dom IT;
end;

registration
  let f be Relation;
  let X be with_common_domain set;
  cluster f|X -> homogeneous;
  coherence
  proof
    dom(f|X) c= X by RELAT_1:58;
   hence dom(f|X) is with_common_domain;
  end;
end;

registration
  let A be non empty set, f be PartFunc of A*,A;
  cluster dom f -> FinSequence-membered;
  coherence
   proof
     dom f c= A* by RELAT_1:def 18;
    hence thesis;
   end;
end;

registration
  let A be non empty set;
  cluster homogeneous quasi_total non empty for PartFunc of A*,A;
  existence
  proof
    set a = the Element of A;
    set f = <*>A .-->a;
A1: dom f = {<*>A} by FUNCOP_1:13;
A2: dom f c= A*
    proof
      let z;
      assume z in dom f;
      then z = <*>A by A1,TARSKI:def 1;
      hence thesis by FINSEQ_1:def 11;
    end;
A3: rng f = {a} by FUNCOP_1:8;
    rng f c= A
    proof
      let z;
      assume z in rng f;
      then z = a by A3,TARSKI:def 1;
      hence thesis;
    end;
    then reconsider f as PartFunc of A*,A by A2,RELSET_1:4;
A4: f is quasi_total
    proof
      let x,y be FinSequence of A;
      assume that
A5:   len x = len y and
A6:   x in dom f;
      x = <*>A by A1,A6,TARSKI:def 1;
      then len x = 0;
      then y = <*>A by A5;
      hence thesis by A1,TARSKI:def 1;
    end;
    f is homogeneous
    proof
      let x,y be FinSequence;
      assume that
A7:   x in dom f and
A8:   y in dom f;
      x = <*>A by A1,A7,TARSKI:def 1;
      hence thesis by A1,A8,TARSKI:def 1;
    end;
    hence thesis by A4;
  end;
end;

registration
  cluster homogeneous non empty for Function;
  existence
  proof
    set f = the homogeneous non empty PartFunc of {{}}*,{{}};
    take f;
    thus thesis;
  end;
end;

registration let R be homogeneous Relation;
 cluster dom R -> with_common_domain;
 coherence by Def21;
end;

theorem Th18:
  for A being non empty set, a being Element of A holds <*>A .-->a
  is homogeneous quasi_total non empty PartFunc of A*,A
proof
  let A be non empty set, a be Element of A;
  set f = <*>A .-->a;
A1: dom f = {<*>A} by FUNCOP_1:13;
A2: dom f c= A*
  proof
    let z;
    assume z in dom f;
    then z = <*>A by A1,TARSKI:def 1;
    hence thesis by FINSEQ_1:def 11;
  end;
A3: rng f = {a} by FUNCOP_1:8;
  rng f c= A
  proof
    let z;
    assume z in rng f;
    then z = a by A3,TARSKI:def 1;
    hence thesis;
  end;
  then reconsider f as PartFunc of A*,A by A2,RELSET_1:4;
A4: f is quasi_total
  proof
    let x,y be FinSequence of A;
    assume that
A5: len x = len y and
A6: x in dom f;
    x = <*>A by A1,A6,TARSKI:def 1;
    then len x = 0;
    then y = <*>A by A5;
    hence thesis by A1,TARSKI:def 1;
  end;
  f is homogeneous
  proof
    let x,y be FinSequence;
    assume that
A7: x in dom f and
A8: y in dom f;
    x = <*>A by A1,A7,TARSKI:def 1;
    hence thesis by A1,A8,TARSKI:def 1;
  end;
  hence thesis by A4;
end;

theorem
  for A being non empty set, a being Element of A holds <*>A .-->a
  is Element of PFuncs(A*,A)
proof
  let A be non empty set, a be Element of A;
  set f = <*>A .-->a;
A1: dom f = {<*>A} by FUNCOP_1:13;
A2: dom f c= A*
  proof
    let z;
    assume z in dom f;
    then z = <*>A by A1,TARSKI:def 1;
    hence thesis by FINSEQ_1:def 11;
  end;
A3: rng f = {a} by FUNCOP_1:8;
  rng f c= A
  proof
    let z;
    assume z in rng f;
    then z = a by A3,TARSKI:def 1;
    hence thesis;
  end;
  then reconsider f as PartFunc of A*,A by A2,RELSET_1:4;
  f in PFuncs(A*,A) by PARTFUN1:45;
  hence thesis;
end;

definition
  let A;
  mode PFuncFinSequence of A is FinSequence of PFuncs(A*,A);
end;

definition
  let A;
  let IT be PFuncFinSequence of A;
  attr IT is homogeneous means
  :Def23:
  for n,h st n in dom IT & h = IT.n holds h is homogeneous;
end;

definition
  let A;
  let IT be PFuncFinSequence of A;
  attr IT is quasi_total means
  :Def24:
  for n,h st n in dom IT & h = IT.n holds h is quasi_total;
end;

definition
  let A be non empty set;
  let x be Element of PFuncs(A*,A);
  redefine func <*x*> -> PFuncFinSequence of A;
  coherence
  proof
    <*x*> is FinSequence of PFuncs(A*,A);
    hence thesis;
  end;
end;

registration
  let A be non empty set;
  cluster homogeneous quasi_total non-empty for PFuncFinSequence of A;
  existence
  proof
    set a = the Element of A;
    reconsider f = <*>A .-->a as PartFunc of A*,A by Th18;
    reconsider f as Element of PFuncs(A*,A) by PARTFUN1:45;
    take <*f*>;
    thus <*f*> is homogeneous
    proof
      let n;
      let h be PartFunc of A*,A;
      assume that
A1:   n in dom <*f*> and
A2:   h =<*f*>.n;
      n in {1} by A1,FINSEQ_1:2,def 8;
      then h = <*f*>.1 by A2,TARSKI:def 1;
      then h = f by FINSEQ_1:def 8;
      hence thesis by Th18;
    end;
    thus <*f*> is quasi_total
    proof
      let n;
      let h be PartFunc of A*,A;
      assume that
A3:   n in dom <*f*> and
A4:   h =<*f*>.n;
      n in {1} by A3,FINSEQ_1:2,def 8;
      then h = <*f*>.1 by A4,TARSKI:def 1;
      then h = f by FINSEQ_1:def 8;
      hence thesis by Th18;
    end;
    thus <*f*> is non-empty
    proof
      let n be set;
      assume
A5:   n in dom <*f*>;
      then reconsider n as Element of NAT;
      n in {1} by A5,FINSEQ_1:2,def 8;
      then n = 1 by TARSKI:def 1;
      hence thesis by FINSEQ_1:def 8;
    end;
  end;
end;

registration
  let A be non empty set;
  let f be homogeneous PFuncFinSequence of A;
  let i be set;
  cluster f.i -> homogeneous;
  coherence
  proof
    per cases;
    suppose
A1:   i in dom f;
A2:   rng f c= PFuncs(A*,A) by RELAT_1:def 19;
      f.i in rng f by A1,FUNCT_1:3;
      then f.i is PartFunc of A*,A by A2,PARTFUN1:46;
      hence thesis by A1,Def23;
    end;
    suppose
A3:   not i in dom f;
      let x be Function;
      thus thesis by A3,FUNCT_1:def 2,RELAT_1:38;
    end;
  end;
end;

reserve A for non empty set,
  h for PartFunc of A*,A,

  a for Element of A;

theorem
  for x be Element of PFuncs(A*,A) st x = <*>A .--> a holds <*x*>
  is homogeneous quasi_total non-empty
proof
  let x be Element of PFuncs(A*,A) such that
A1: x = <*>A .--> a;
A2: for n,h st n in dom <*x*> & h = <*x*>.n holds h is homogeneous
  proof
    let n,h;
    assume that
A3: n in dom <*x*> and
A4: h =<*x*>.n;
    n in {1} by A3,FINSEQ_1:2,def 8;
    then h = <*x*>.1 by A4,TARSKI:def 1;
    then h = x by FINSEQ_1:def 8;
    hence thesis by A1,Th18;
  end;
A5: for n,h st n in dom <*x*> & h = <*x*>.n holds h is quasi_total
  proof
    let n,h;
    assume that
A6: n in dom <*x*> and
A7: h =<*x*>.n;
    n in {1} by A6,FINSEQ_1:2,def 8;
    then h = <*x*>.1 by A7,TARSKI:def 1;
    then h = x by FINSEQ_1:def 8;
    hence thesis by A1,Th18;
  end;
  for n being set st n in dom <*x*> holds <*x*>.n is non empty
  proof
    let n be set;
    assume n in dom <*x*>;
    then n in {1} by FINSEQ_1:2,def 8;
    then <*x*>.n = <*x*>.1 by TARSKI:def 1;
    hence thesis by A1,FINSEQ_1:def 8;
  end;
  hence thesis by A2,A5,Def23,Def24,FUNCT_1:def 9;
end;

definition
  let f be homogeneous Relation;
  func arity(f) -> Nat means
:Def25:  for x being FinSequence st x in dom f holds it =
  len x if ex x being FinSequence st x in dom f otherwise it = 0;
  consistency;
  existence
  proof
    thus (ex x being FinSequence st x in dom f) implies ex n st for x being
    FinSequence st x in dom f holds n = len x
    proof
      given x being FinSequence such that
A1:   x in dom f;
      take len x;
      let y be FinSequence;
      assume y in dom f;
       then dom x = dom y by A1,CARD_3:def 10;
      hence thesis by FINSEQ_3:29;
    end;
    thus thesis;
  end;
  uniqueness
  proof
    let n,m;
    hereby
      given x0 being FinSequence such that
A2:   x0 in dom f;
      assume that
A3:   for x being FinSequence st x in dom f holds n = len x and
A4:   for x being FinSequence st x in dom f holds m = len x;
      n = len x0 by A2,A3;
      hence n = m by A2,A4;
    end;
    thus thesis;
  end;
end;

definition
  let f be homogeneous Function;
  redefine func arity(f) -> Element of NAT;
  coherence by ORDINAL1:def 12;
end;

begin 

theorem
  for n be Nat, D be non empty set, D1 be non empty Subset of D
  holds n-tuples_on D /\ n-tuples_on D1 = n-tuples_on D1
proof
  let n be Nat,D be non empty set, D1 be non empty Subset of D;
  n-tuples_on D1 c= n-tuples_on D
  proof
    let z be set;
    assume z in n-tuples_on D1;
    then z is Tuple of n,D1 by FINSEQ_2:131;
    then z is Element of n-tuples_on D by FINSEQ_2:109;
    hence thesis;
  end;
  hence thesis by XBOOLE_1:28;
end;

theorem
  for D being non empty set for h being homogeneous quasi_total non
  empty PartFunc of D*,D holds dom h = (arity(h))-tuples_on D
proof
  let D be non empty set;
  let f be homogeneous quasi_total non empty PartFunc of D*,D;
  set y = the Element of dom f;
A1: dom f c= D* by RELAT_1:def 18;
  then
A2: y in D* by TARSKI:def 3;
  thus dom f c= (arity(f))-tuples_on D
  proof
    let x be set;
    assume
A3: x in dom f;
    then reconsider x9 = x as FinSequence of D by A1,FINSEQ_1:def 11;
    len x9 = arity f by A3,Def25;
    then x9 is Element of (arity(f))-tuples_on D by FINSEQ_2:92;
    hence thesis;
  end;
  reconsider y as FinSequence of D by A2,FINSEQ_1:def 11;
  let x be set;
  assume x in (arity(f))-tuples_on D;
  then x in {s where s is Element of D* : len s = arity(f)} by FINSEQ_2:def 4;
  then
A4: ex s being Element of D* st x = s & len s = arity(f);
  len y = arity f by Def25;
  hence thesis by A4,Def22;
end;

definition
  let D be non empty set;
  mode PFuncsDomHQN of D -> non empty set means
    :Def26:
    for x be Element of it
    holds x is homogeneous quasi_total non empty PartFunc of D*,D;
  existence
  proof
    set a = the Element of D;
    reconsider A = {{<*>D} -->a} as non empty set;
    take A;
    let x be Element of A;
    x= <*>D .-->a by TARSKI:def 1;
    hence thesis by Th18;
  end;
end;

definition
  let D be non empty set, P be PFuncsDomHQN of D;
  redefine mode Element of P -> homogeneous quasi_total non empty PartFunc of
    D*,D;
  coherence by Def26;
end;














begin

reserve X,Y,Z,x,y,z for set;

registration
  cluster empty -> reflexive irreflexive symmetric antisymmetric asymmetric
    connected strongly_connected transitive for Relation;
  coherence
  proof
    let R be Relation such that
A1: R is empty;
    x in {} implies [x,x] in {};
    then {} is_reflexive_in field {} by RELAT_2:def 1;
    hence R is reflexive by A1,RELAT_2:def 9;
    {} is_irreflexive_in field {}
    proof
      let x;
      assume x in field {};
      thus thesis;
    end;
    hence R is irreflexive by A1,RELAT_2:def 10;
    thus R is symmetric
    proof
      let x,y;
      assume that
      x in field R and
      y in field R and
A2:   [x,y] in R;
      thus thesis by A1,A2;
    end;
    {} is_antisymmetric_in field {}
    proof
      let x,y;
      assume that
A3:   x in field {} and
      y in field {} and
      [x,y] in {} and
      [y,x] in {};
      thus thesis by A3;
    end;
    hence R is antisymmetric by A1,RELAT_2:def 12;
    {} is_asymmetric_in field {}
    proof
      let x,y;
      assume that
      x in field {} and
      y in field {} and
      [x,y] in {};
      thus thesis;
    end;
    hence R is asymmetric by A1,RELAT_2:def 13;
    {} is_connected_in field {}
    proof
      let x,y;
      assume that
A4:   x in field {} and
      y in field {} and
      x <> y;
      thus thesis by A4;
    end;
    hence R is connected by A1,RELAT_2:def 14;
    {} is_strongly_connected_in field {}
    proof
      let x,y;
      assume that
A5:   x in field {} and
      y in field {};
      thus thesis by A5;
    end;
    hence R is strongly_connected by A1,RELAT_2:def 15;
    {} is_transitive_in field {}
    proof
      let x,y,z;
      assume that
A6:   x in field {} and
      y in field {} and
      z in field {} and
      [x,y] in {} and
      [y,z] in {};
      thus thesis by A6;
    end;
    hence R is transitive by A1,RELAT_2:def 16;
  end;
end;



notation
  let X;
  synonym Total X for nabla X;
end;

definition
  let R be Relation, Y be set;
  redefine func R |_2 Y -> Relation of Y,Y;
  coherence by XBOOLE_1:17;
end;

theorem
  rng Total X = X
proof
  for x holds x in X iff ex y st [y,x] in Total X
  proof
    let x;
    thus x in X implies ex y st [y,x] in Total X
    proof
      assume
A1:   x in X;
      take x;
      [x,x] in [:X,X:] by A1,ZFMISC_1:87;
      hence thesis by EQREL_1:def 1;
    end;
    given y such that
A2: [y,x] in Total X;
    thus thesis by A2,ZFMISC_1:87;
  end;
  hence thesis by XTUPLE_0:def 13;
end;

theorem Th2:
  for x,y st x in X & y in X holds [x,y] in Total X
proof
  let x,y;
  assume x in X & y in X;
  then [x,y] in [:X,X:] by ZFMISC_1:87;
  hence thesis by EQREL_1:def 1;
end;

theorem
  for x,y st x in field Total X & y in field Total X holds [x,y] in Total X
proof
  let x,y;
  assume x in field Total X & y in field Total X;
  then x in X & y in X by ORDERS_1:12;
  hence thesis by Th2;
end;

theorem
  Total X is strongly_connected
proof
  let x,y;
  assume x in field Total X & y in field Total X;
  then x in X & y in X by ORDERS_1:12;
  then [x,y] in [:X,X:] by ZFMISC_1:87;
  hence thesis by EQREL_1:def 1;
end;

theorem
  Total X is connected
proof
  let x,y;
  assume that
A1: x in field Total X & y in field Total X and
  x<>y;
  x in X & y in X by A1,ORDERS_1:12;
  then [x,y] in [:X,X:] by ZFMISC_1:87;
  hence thesis by EQREL_1:def 1;
end;



reserve T,R for Tolerance of X;

theorem
  for T being Tolerance of X holds rng T = X
proof
  let T be Tolerance of X;
  for x holds x in rng T iff x in X
  proof
    let x;
    x in X implies x in rng T
    proof
      field T = X by ORDERS_1:12;
      then
A1:   T is_reflexive_in X by RELAT_2:def 9;
      assume x in X;
      then [x,x] in T by A1,RELAT_2:def 1;
      hence thesis by XTUPLE_0:def 13;
    end;
    hence thesis;
  end;
  hence thesis by TARSKI:1;
end;

theorem Th7:
  for T being total reflexive Relation of X holds x in X iff [x,x] in T
proof
  let T be total reflexive Relation of X;
  thus x in X implies [x,x] in T by EQREL_1:5;
  assume
A1: [x,x] in T;
  field T = X by ORDERS_1:12;
  hence thesis by A1,RELAT_1:15;
end;

theorem
  for T being Tolerance of X holds T is_reflexive_in X
proof
  let T be Tolerance of X;
  field T = X by ORDERS_1:12;
  hence thesis by RELAT_2:def 9;
end;

theorem
  for T being Tolerance of X holds T is_symmetric_in X
proof
  let T be Tolerance of X;
  field T = X by ORDERS_1:12;
  hence thesis by RELAT_2:def 11;
end;

theorem Th10:
  for R be Relation of X,Y st R is symmetric holds R |_2 Z is symmetric
proof
  let R be Relation of X,Y;
  assume R is symmetric;
  then
A1: R is_symmetric_in field R by RELAT_2:def 11;
  now
    let x,y;
    assume that
A2: x in field(R|_2 Z) & y in field(R|_2 Z) and
A3: [x,y] in R|_2 Z;
A4: [x,y] in R by A3,XBOOLE_0:def 4;
A5: [y,x] in [:Z,Z:] by A3,ZFMISC_1:88;
    x in field R & y in field R by A2,WELLORD1:12;
    then [y,x] in R by A1,A4,RELAT_2:def 3;
    hence [y,x] in R|_2 Z by A5,XBOOLE_0:def 4;
  end;
  then R|_2 Z is_symmetric_in field(R|_2 Z) by RELAT_2:def 3;
  hence thesis by RELAT_2:def 11;
end;

definition
  let X,T;
  let Y be Subset of X;
  redefine func T |_2 Y -> Tolerance of Y;
  coherence
  proof
    now
      let x;
      assume x in Y;
      then [x,x] in [:Y,Y:] & [x,x] in T by Th7,ZFMISC_1:87;
      then [x,x] in T |_2 Y by XBOOLE_0:def 4;
      hence x in dom (T |_2 Y) by XTUPLE_0:def 12;
    end;
    then Y c= dom(T |_2 Y) by TARSKI:def 3;
    then dom(T |_2 Y) = Y by XBOOLE_0:def 10;
    hence T |_2 Y is Tolerance of Y by Th10,PARTFUN1:def 2,WELLORD1:15;
  end;
end;

theorem
  Y c= X implies T|_2 Y is Tolerance of Y
proof
  assume Y c= X;
  then reconsider Z = Y as Subset of X;
  T |_2 Z is Tolerance of Z;
  hence thesis;
end;



definition
  let X;
  let T be Tolerance of X;

  mode TolSet of T -> set means
    :Def1:
    for x,y st x in it & y in it holds [x,y ] in T;
  existence
  proof
    take {};
    let x,y;
    assume that
A1: x in {} and
    y in {};
    thus thesis by A1;
  end;
end;

theorem Th12:
  {} is TolSet of T
proof
  let x,y;
  assume that
A1: x in {} and
  y in {};
  thus thesis by A1;
end;

definition
  let X;
  let T be Tolerance of X;
  let IT be TolSet of T;
  attr IT is TolClass-like means
  :Def2:
  for x st not x in IT & x in X ex y st y in IT & not [x,y] in T;
end;

registration
  let X;
  let T be Tolerance of X;
  cluster TolClass-like for TolSet of T;
  existence
  proof
    defpred X[set] means $1 is TolSet of T;
    consider TS being set such that
A1: for x holds x in TS iff x in bool X & X[x] from XBOOLE_0:sch 1;
A2: TS c= bool X
    proof
      let x;
      assume x in TS;
      hence thesis by A1;
    end;
A3: for Z st Z c= TS & Z is c=-linear ex Y st Y in TS & for X1 being set
    st X1 in Z holds X1 c= Y
    proof
      let Z such that
A4:   Z c= TS and
A5:   Z is c=-linear;
      for x,y st x in union Z & y in union Z holds [x,y] in T
      proof
        let x,y;
        assume that
A6:     x in union Z and
A7:     y in union Z;
        consider Zy being set such that
A8:     y in Zy and
A9:     Zy in Z by A7,TARSKI:def 4;
        reconsider Zy as TolSet of T by A1,A4,A9;
        consider Zx being set such that
A10:    x in Zx and
A11:    Zx in Z by A6,TARSKI:def 4;
        reconsider Zx as TolSet of T by A1,A4,A11;
        Zx,Zy are_c=-comparable by A5,A11,A9,ORDINAL1:def 8;
        then Zx c= Zy or Zy c= Zx by XBOOLE_0:def 9;
        hence thesis by A10,A8,Def1;
      end;
      then
A12:  union Z is TolSet of T by Def1;
      take union Z;
      Z c= bool X by A2,A4,XBOOLE_1:1;
      then union Z c= union bool X by ZFMISC_1:77;
      then union Z c= X by ZFMISC_1:81;
      hence union Z in TS by A1,A12;
      let X1 be set;
      assume X1 in Z;
      hence thesis by ZFMISC_1:74;
    end;
    {} c= X & {} is TolSet of T by Th12,XBOOLE_1:2;
    then TS <> {} by A1;
    then consider Y such that
A13: Y in TS and
A14: for Z st Z in TS & Z <> Y holds not Y c= Z by A3,ORDERS_1:65;
    reconsider Y as TolSet of T by A1,A13;
    take Y;
    let x such that
A15: not x in Y and
A16: x in X;
    set Y1 = Y \/ {x};
A17: {x} c= X by A16,ZFMISC_1:31;
    assume
A18: for y st y in Y holds [x,y] in T;
    for y,z st y in Y1 & z in Y1 holds [y,z] in T
    proof
      let y,z;
      assume that
A19:  y in Y1 and
A20:  z in Y1;
      y in Y or y in {x} by A19,XBOOLE_0:def 3;
      then
A21:  y in Y or y = x by TARSKI:def 1;
      z in Y or z in {x} by A20,XBOOLE_0:def 3;
      then
A22:  z in Y or z = x by TARSKI:def 1;
      assume
A23:  not [y,z] in T;
      then not [z,y] in T by EQREL_1:6;
      hence contradiction by A16,A18,A21,A22,A23,Def1,Th7;
    end;
    then
A24: Y1 is TolSet of T by Def1;
A25: Y1 <> Y
    proof
A26:  x in {x} by TARSKI:def 1;
      assume Y1 = Y;
      hence contradiction by A15,A26,XBOOLE_0:def 3;
    end;
    Y in bool X by A1,A13;
    then Y1 c= X by A17,XBOOLE_1:8;
    then Y1 in TS by A1,A24;
    hence contradiction by A14,A25,XBOOLE_1:7;
  end;
end;

definition
  let X;
  let T be Tolerance of X;
  mode TolClass of T is TolClass-like TolSet of T;
end;

theorem
  for T being Tolerance of X st {} is TolClass of T holds T={}
proof
  let T be Tolerance of X;
  assume {} is TolClass of T;
  then reconsider 00 = {} as TolClass of T;
  assume T <> {};
  then consider x,y such that
A1: [x,y] in T by RELAT_1:37;
  x in X by A1,ZFMISC_1:87;
  then ex z st z in 00 & not [x,z] in T by Def2;
  hence contradiction;
end;

theorem
  {} is Tolerance of {} by RELSET_1:12;

theorem Th15:
  for x,y st [x,y] in T holds {x,y} is TolSet of T
proof
  let x,y;
  assume
A1: [x,y] in T;
  then
A2: x in X & y in X by ZFMISC_1:87;
  for a,b being set st a in {x,y} & b in {x,y} holds [a,b] in T
  proof
    let a,b be set;
    assume that
A3: a in {x,y} and
A4: b in {x,y};
A5: b = x or b = y by A4,TARSKI:def 2;
    a = x or a = y by A3,TARSKI:def 2;
    hence thesis by A1,A2,A5,Th7,EQREL_1:6;
  end;
  hence thesis by Def1;
end;

theorem
  for x st x in X holds {x} is TolSet of T
proof
  let x;
  assume x in X;
  then [x,x] in T by Th7;
  then {x,x} is TolSet of T by Th15;
  hence thesis by ENUMSET1:29;
end;

theorem
  for Y,Z st Y is TolSet of T holds Y /\ Z is TolSet of T
proof
  let Y,Z such that
A1: Y is TolSet of T;
  let x,y;
  assume x in Y /\ Z & y in Y /\ Z;
  then x in Y & y in Y by XBOOLE_0:def 4;
  hence thesis by A1,Def1;
end;

theorem Th18:
  Y is TolSet of T implies Y c= X
proof
  assume
A1: Y is TolSet of T;
  let x;
  assume x in Y;
  then [x,x] in T by A1,Def1;
  hence thesis by Th7;
end;

theorem Th19:
  for Y being TolSet of T ex Z being TolClass of T st Y c= Z
proof
  let Y be TolSet of T;
  defpred X[set] means $1 is TolSet of T & ex Z st $1=Z & Y c= Z;
  consider TS being set such that
A1: for x holds x in TS iff x in bool X & X[x] from XBOOLE_0:sch 1;
A2: for x being set holds x in TS iff x in bool X & x is TolSet of T & Y c= x
  proof
    let x be set;
    thus x in TS implies x in bool X & x is TolSet of T & Y c= x
    proof
      assume
A3:   x in TS;
      hence x in bool X & x is TolSet of T by A1;
      ex Z st x=Z & Y c= Z by A1,A3;
      hence thesis;
    end;
    thus thesis by A1;
  end;
  Y c= X by Th18;
  then
A4: TS <> {} by A2;
A5: TS c= bool X
  proof
    let x;
    assume x in TS;
    hence thesis by A1;
  end;
  for Z st Z c= TS & Z is c=-linear ex Y st Y in TS & for X1 being set st
  X1 in Z holds X1 c= Y
  proof
    let Z such that
A6: Z c= TS and
A7: Z is c=-linear;
A8: for x,y st x in union Z & y in union Z holds [x,y] in T
    proof
      let x,y;
      assume that
A9:   x in union Z and
A10:  y in union Z;
      consider Zy being set such that
A11:  y in Zy and
A12:  Zy in Z by A10,TARSKI:def 4;
      reconsider Zy as TolSet of T by A1,A6,A12;
      consider Zx being set such that
A13:  x in Zx and
A14:  Zx in Z by A9,TARSKI:def 4;
      reconsider Zx as TolSet of T by A1,A6,A14;
      Zx, Zy are_c=-comparable by A7,A14,A12,ORDINAL1:def 8;
      then Zx c= Zy or Zy c= Zx by XBOOLE_0:def 9;
      hence thesis by A13,A11,Def1;
    end;
A15: Z <> {} implies thesis
    proof
      assume
A16:  Z <> {};
A17:  Y c= union Z
      proof
        set y = the Element of Z;
        y in TS by A6,A16,TARSKI:def 3;
        then
A18:    Y c= y by A2;
        let x;
        assume x in Y;
        hence thesis by A16,A18,TARSKI:def 4;
      end;
      Z c= bool X by A5,A6,XBOOLE_1:1;
      then union Z c= union bool X by ZFMISC_1:77;
      then
A19:  union Z c= X by ZFMISC_1:81;
      take union Z;
      union Z is TolSet of T by A8,Def1;
      hence union Z in TS by A2,A19,A17;
      let X1 be set;
      assume X1 in Z;
      hence thesis by ZFMISC_1:74;
    end;
    Z = {} implies thesis
    proof
      set Y = the Element of TS;
      assume
A20:  Z = {};
      take Y;
      thus Y in TS by A4;
      let X1 be set;
      assume X1 in Z;
      hence thesis by A20;
    end;
    hence thesis by A15;
  end;
  then consider Z1 being set such that
A21: Z1 in TS and
A22: for Z st Z in TS & Z<>Z1 holds not Z1 c= Z by A4,ORDERS_1:65;
  reconsider Z1 as TolSet of T by A1,A21;
  Z1 is TolClass of T
  proof
    assume not thesis;
    then consider x such that
A23: not x in Z1 and
A24: x in X and
A25: for y st y in Z1 holds [x,y] in T by Def2;
    set Y1 = Z1 \/ {x};
A26: {x} c= X by A24,ZFMISC_1:31;
    for y,z st y in Y1 & z in Y1 holds [y,z] in T
    proof
      let y,z;
      assume that
A27:  y in Y1 and
A28:  z in Y1;
      y in Z1 or y in {x} by A27,XBOOLE_0:def 3;
      then
A29:  y in Z1 or y = x by TARSKI:def 1;
      z in Z1 or z in {x} by A28,XBOOLE_0:def 3;
      then
A30:  z in Z1 or z = x by TARSKI:def 1;
      assume
A31:  not [y,z] in T;
      then not [z,y] in T by EQREL_1:6;
      hence contradiction by A24,A25,A29,A30,A31,Def1,Th7;
    end;
    then
A32: Y1 is TolSet of T by Def1;
    Y c= Z1 & Z1 c= Y1 by A2,A21,XBOOLE_1:7;
    then
A33: Y c= Y1 by XBOOLE_1:1;
A34: Y1 <> Z1
    proof
A35:  x in {x} by TARSKI:def 1;
      assume Y1 = Z1;
      hence contradiction by A23,A35,XBOOLE_0:def 3;
    end;
    Z1 in bool X by A1,A21;
    then Y1 c= X by A26,XBOOLE_1:8;
    then Y1 in TS by A2,A32,A33;
    hence contradiction by A22,A34,XBOOLE_1:7;
  end;
  then reconsider Z1 as TolClass of T;
  take Z1;
  thus thesis by A2,A21;
end;

theorem Th20:
  for x,y st [x,y] in T ex Z being TolClass of T st x in Z & y in Z
proof
  let x,y;
  assume
A1: [x,y] in T;
  then
A2: x in X & y in X by ZFMISC_1:87;
  for a,b being set st a in {x,y} & b in {x,y} holds [a,b] in T
  proof
    let a,b be set;
    assume that
A3: a in {x,y} and
A4: b in {x,y};
A5: b = x or b = y by A4,TARSKI:def 2;
    a = x or a = y by A3,TARSKI:def 2;
    hence thesis by A1,A2,A5,Th7,EQREL_1:6;
  end;
  then reconsider PS = {x,y} as TolSet of T by Def1;
  consider Z being TolClass of T such that
A6: PS c= Z by Th19;
  take Z;
  x in {x,y} by TARSKI:def 2;
  hence x in Z by A6;
  y in {x,y} by TARSKI:def 2;
  hence thesis by A6;
end;

theorem Th21:
  for x st x in X ex Z being TolClass of T st x in Z
proof
  let x;
  assume x in X;
  then [x,x] in T by Th7;
  then ex Z being TolClass of T st x in Z & x in Z by Th20;
  hence thesis;
end;

theorem
  T c= Total X
proof
  let x,y;
  assume [x,y] in T;
  then [x,y] in [:X,X:];
  hence thesis by EQREL_1:def 1;
end;

theorem
  id X c= T
proof
  let x,y;
  assume [x,y] in id X;
  then x in X & x = y by RELAT_1:def 10;
  hence thesis by Th7;
end;

scheme
  ToleranceEx{A() -> set,P[set,set]}: ex T being Tolerance of A() st for x,y
  st x in A() & y in A() holds [x,y] in T iff P[x,y]
provided
A1: for x st x in A() holds P[x,x] and
A2: for x,y st x in A() & y in A() & P[x,y] holds P[y,x]
proof
  defpred X[set] means ex y,z st $1 = [y,z] & P[y,z];
  consider T being set such that
A3: for x holds x in T iff x in [:A(),A():] & X[x] from XBOOLE_0:sch 1;
  for x st x in T holds x in [:A(),A():] by A3;
  then reconsider T as Relation of A(),A() by TARSKI:def 3;
A4: field T c= A() \/ A() by RELSET_1:8;
  for x st x in field T holds [x,x] in T
  proof
    let x;
    assume x in field T;
    then [x,x] in [:A(),A():] & P[x,x] by A1,A4,ZFMISC_1:87;
    hence thesis by A3;
  end;
  then
A5: T is_reflexive_in field T by RELAT_2:def 1;
  for x,y st x in field T & y in field T & [x,y] in T holds [y,x] in T
  proof
    let x,y such that
A6: x in field T & y in field T and
A7: [x,y] in T;
    x in A() & y in A() & P[x,y]
    proof
      thus x in A() & y in A() by A4,A6;
      consider a,b being set such that
A8:   [x,y] = [a,b] and
A9:   P[a,b] by A3,A7;
      x = a by A8,XTUPLE_0:1;
      hence thesis by A8,A9,XTUPLE_0:1;
    end;
    then [y,x] in [:A(),A():] & P[y,x] by A2,ZFMISC_1:87;
    hence thesis by A3;
  end;
  then
A10: T is_symmetric_in field T by RELAT_2:def 3;
  for x st x in A() holds x in dom T
  proof
    let x;
    assume x in A();
    then [x,x] in [:A(),A():] & P[x,x] by A1,ZFMISC_1:87;
    then [x,x] in T by A3;
    hence thesis by XTUPLE_0:def 12;
  end;
  then A() c= dom T by TARSKI:def 3;
  then dom T = A() by XBOOLE_0:def 10;
  then reconsider T as Tolerance of A() by A5,A10,PARTFUN1:def 2,RELAT_2:def 9
,def 11;
  take T;
  let x,y;
  assume
A11: x in A() & y in A();
  thus [x,y] in T implies P[x,y]
  proof
    assume [x,y] in T;
    then consider a,b being set such that
A12: [x,y] = [a,b] and
A13: P[a,b] by A3;
    x = a by A12,XTUPLE_0:1;
    hence thesis by A12,A13,XTUPLE_0:1;
  end;
  assume
A14: P[x,y];
  [x,y] in [:A(),A():] by A11,ZFMISC_1:87;
  hence thesis by A3,A14;
end;

theorem
  for Y ex T being Tolerance of union Y st for Z st Z in Y holds Z is
  TolSet of T
proof
  let Y;
  defpred X[set,set] means ex Z st Z in Y & $1 in Z & $2 in Z;
A1: for x st x in union Y holds X[x,x]
  proof
    let x;
    assume x in union Y;
    then ex Z st x in Z & Z in Y by TARSKI:def 4;
    hence thesis;
  end;
A2: for x,y st x in union Y & y in union Y & X[x,y] holds X[y,x];
  consider T being Tolerance of union Y such that
A3: for x,y st x in union Y & y in union Y holds [x,y] in T iff X[x,y]
  from ToleranceEx(A1,A2);
  take T;
  let Z such that
A4: Z in Y;
  for x,y st x in Z & y in Z holds [x,y] in T
  proof
    let x,y;
    assume
A5: x in Z & y in Z;
    then x in union Y & y in union Y by A4,TARSKI:def 4;
    hence thesis by A3,A4,A5;
  end;
  hence thesis by Def1;
end;

theorem
  for Y being set for T,R being Tolerance of union Y st (for x,y holds [
x,y] in T iff ex Z st Z in Y & x in Z & y in Z) & (for x,y holds [x,y] in R iff
  ex Z st Z in Y & x in Z & y in Z) holds T = R
proof
  let Y be set;
  let T,R be Tolerance of union Y such that
A1: for x,y holds [x,y] in T iff ex Z st Z in Y & x in Z & y in Z and
A2: for x,y holds [x,y] in R iff ex Z st Z in Y & x in Z & y in Z;
  for x,y holds [x,y] in T iff [x,y] in R
  proof
    let x,y;
    thus [x,y] in T implies [x,y] in R
    proof
      assume [x,y] in T;
      then ex Z st Z in Y & x in Z & y in Z by A1;
      hence thesis by A2;
    end;
    assume [x,y] in R;
    then ex Z st Z in Y & x in Z & y in Z by A2;
    hence thesis by A1;
  end;
  hence thesis by RELAT_1:def 2;
end;

theorem Th26:
  for T,R being Tolerance of X st for Z holds Z is TolClass of T
  iff Z is TolClass of R holds T = R
proof
  let T,R be Tolerance of X;
  assume
A1: for Z holds Z is TolClass of T iff Z is TolClass of R;
  for x,y holds [x,y] in T iff [x,y] in R
  proof
    let x,y;
    thus [x,y] in T implies [x,y] in R
    proof
      assume [x,y] in T;
      then consider Z being TolClass of T such that
A2:   x in Z & y in Z by Th20;
      reconsider Z as TolClass of R by A1;
      Z is TolSet of R;
      hence thesis by A2,Def1;
    end;
    assume [x,y] in R;
    then consider Z being TolClass of R such that
A3: x in Z & y in Z by Th20;
    reconsider Z as TolClass of T by A1;
    Z is TolSet of T;
    hence thesis by A3,Def1;
  end;
  hence thesis by RELAT_1:def 2;
end;



notation
  let X, Y;
  let T be Relation of X, Y;
  let x;
  synonym neighbourhood (x, T) for Class (T,x);
end;

theorem Th27:
  for y being set holds y in neighbourhood(x,T) iff [x,y] in T
proof
  let y be set;
  hereby
    assume y in neighbourhood(x,T);
    then [y,x] in T by EQREL_1:19;
    hence [x,y] in T by EQREL_1:6;
  end;
  assume [x,y] in T;
  then [y,x] in T by EQREL_1:6;
  hence thesis by EQREL_1:19;
end;

theorem
  for Y st for Z being set holds Z in Y iff x in Z & Z is TolClass of T
  holds neighbourhood(x,T) = union Y
proof
  let Y such that
A1: for Z being set holds Z in Y iff x in Z & Z is TolClass of T;
  for y holds y in neighbourhood(x,T) iff y in union Y
  proof
    let y;
    thus y in neighbourhood(x,T) implies y in union Y
    proof
      assume y in neighbourhood(x,T);
      then [x,y] in T by Th27;
      then consider Z being TolClass of T such that
A2:   x in Z and
A3:   y in Z by Th20;
      Z in Y by A1,A2;
      hence thesis by A3,TARSKI:def 4;
    end;
    assume y in union Y;
    then consider Z such that
A4: y in Z and
A5: Z in Y by TARSKI:def 4;
    reconsider Z as TolClass of T by A1,A5;
    x in Z by A1,A5;
    then [x,y] in T by A4,Def1;
    hence thesis by Th27;
  end;
  hence thesis by TARSKI:1;
end;

theorem
  for Y st for Z holds Z in Y iff x in Z & Z is TolSet of T holds
  neighbourhood(x,T) = union Y
proof
  let Y such that
A1: for Z holds Z in Y iff x in Z & Z is TolSet of T;
  for y holds y in neighbourhood(x,T) iff y in union Y
  proof
    let y;
    thus y in neighbourhood(x,T) implies y in union Y
    proof
      assume y in neighbourhood(x,T);
      then [x,y] in T by Th27;
      then consider Z being TolClass of T such that
A2:   x in Z and
A3:   y in Z by Th20;
      Z in Y by A1,A2;
      hence thesis by A3,TARSKI:def 4;
    end;
    assume y in union Y;
    then consider Z such that
A4: y in Z and
A5: Z in Y by TARSKI:def 4;
    reconsider Z as TolSet of T by A1,A5;
    x in Z by A1,A5;
    then [x,y] in T by A4,Def1;
    hence thesis by Th27;
  end;
  hence thesis by TARSKI:1;
end;



definition
  let X;
  let T be Tolerance of X;

  func TolSets T -> set means
  :Def3:
  for Y holds Y in it iff Y is TolSet of T;
  existence
  proof
    defpred X[set] means $1 is TolSet of T;
    consider Z being set such that
A1: x in Z iff x in bool X & X[x] from XBOOLE_0:sch 1;
    take Z;
    let Y;
    thus Y in Z implies Y is TolSet of T by A1;
    assume
A2: Y is TolSet of T;
    for x holds x in Y implies x in X
    proof
      let x;
      assume x in Y;
      then [x,x] in T by A2,Def1;
      hence thesis by ZFMISC_1:87;
    end;
    then Y c= X by TARSKI:def 3;
    hence thesis by A1,A2;
  end;
  uniqueness
  proof
    defpred P[set] means $1 is TolSet of T;
    let Z1,Z2 be set such that
A3: for Y holds Y in Z1 iff P[Y] and
A4: for Y holds Y in Z2 iff P[Y];
    Z1 = Z2 from XBOOLE_0:sch 2 (A3, A4);
    hence thesis;
  end;
  func TolClasses T -> set means
  :Def4:
  for Y holds Y in it iff Y is TolClass of T;
  existence
  proof
    defpred X[set] means $1 is TolClass of T;
    consider Z being set such that
A5: x in Z iff x in bool X & X[x] from XBOOLE_0:sch 1;
    take Z;
    let Y;
    thus Y in Z implies Y is TolClass of T by A5;
    assume
A6: Y is TolClass of T;
    for x holds x in Y implies x in X
    proof
      let x;
      assume x in Y;
      then [x,x] in T by A6,Def1;
      hence thesis by ZFMISC_1:87;
    end;
    then Y c= X by TARSKI:def 3;
    hence thesis by A5,A6;
  end;
  uniqueness
  proof
    defpred P[set] means $1 is TolClass of T;
    let C1,C2 be set such that
A7: for Y holds Y in C1 iff P[Y] and
A8: for Y holds Y in C2 iff P[Y];
    C1 = C2 from XBOOLE_0:sch 2 (A7, A8);
    hence thesis;
  end;
end;

theorem
  TolClasses R c= TolClasses T implies R c= T
proof
  assume
A1: TolClasses R c= TolClasses T;
  let x,y;
  assume [x,y] in R;
  then consider Z being TolClass of R such that
A2: x in Z & y in Z by Th20;
  Z in TolClasses R by Def4;
  then Z is TolSet of T by A1,Def4;
  hence thesis by A2,Def1;
end;

theorem
  for T,R being Tolerance of X st TolClasses T = TolClasses R holds T = R
proof
  let T,R be Tolerance of X;
  assume
A1: TolClasses T = TolClasses R;
  for Z holds Z is TolClass of T iff Z is TolClass of R
  proof
    let Z;
    Z is TolClass of T iff Z in TolClasses R by A1,Def4;
    hence thesis by Def4;
  end;
  hence thesis by Th26;
end;

theorem
  union TolClasses T = X
proof
  for x holds x in union TolClasses T iff x in X
  proof
    let x;
    thus x in union TolClasses T implies x in X
    proof
      assume x in union TolClasses T;
      then consider Z such that
A1:   x in Z and
A2:   Z in TolClasses T by TARSKI:def 4;
      Z is TolSet of T by A2,Def4;
      then Z c= X by Th18;
      hence thesis by A1;
    end;
    assume x in X;
    then consider Z being TolClass of T such that
A3: x in Z by Th21;
    Z in TolClasses T by Def4;
    hence thesis by A3,TARSKI:def 4;
  end;
  hence thesis by TARSKI:1;
end;

theorem
  union TolSets T = X
proof
  for x holds x in union TolSets T iff x in X
  proof
    let x;
    thus x in union TolSets T implies x in X
    proof
      assume x in union TolSets T;
      then consider Z such that
A1:   x in Z and
A2:   Z in TolSets T by TARSKI:def 4;
      Z is TolSet of T by A2,Def3;
      then Z c= X by Th18;
      hence thesis by A1;
    end;
    assume x in X;
    then consider Z being TolClass of T such that
A3: x in Z by Th21;
    Z in TolSets T by Def3;
    hence thesis by A3,TARSKI:def 4;
  end;
  hence thesis by TARSKI:1;
end;

theorem
  (for x st x in X holds neighbourhood(x,T) is TolSet of T) implies T is
  transitive
proof
  assume
A1: for x st x in X holds neighbourhood(x,T) is TolSet of T;
A2: field T = X by ORDERS_1:12;
  for x,y,z st x in field T & y in field T & z in field T & [x,y] in T & [
  y,z] in T holds [x,z] in T
  proof
    let x,y,z;
    assume that
    x in field T and
A3: y in field T and
    z in field T and
A4: [x,y] in T and
A5: [y,z] in T;
    reconsider N = neighbourhood(y,T) as TolSet of T by A2,A1,A3;
    [y,x] in T by A4,EQREL_1:6;
    then
A6: x in N by Th27;
    z in N by A5,Th27;
    hence thesis by A6,Def1;
  end;
  then T is_transitive_in field T by RELAT_2:def 8;
  hence thesis by RELAT_2:def 16;
end;

theorem
  T is transitive implies for x st x in X holds neighbourhood(x,T) is
  TolClass of T
proof
  assume
A1: T is transitive;
  let x;
  assume
A2: x in X;
  set N = Class(T,x);
  field T = X by ORDERS_1:12;
  then
A3: T is_transitive_in X by A1,RELAT_2:def 16;
  for y,z st y in N & z in N holds [y,z] in T
  proof
    let y,z such that
A4: y in N and
A5: z in N;
    [x,y] in T by A4,Th27;
    then
A6: [y,x] in T by EQREL_1:6;
    [x,z] in T by A5,Th27;
    hence thesis by A3,A2,A4,A5,A6,RELAT_2:def 8;
  end;
  then reconsider Z = N as TolSet of T by Def1;
  for x st not x in Z & x in X ex y st y in Z & not [x,y] in T
  proof
    let y such that
A7: not y in Z and
    y in X;
A8: x in Z by A2,EQREL_1:20;
    assume for z st z in Z holds [y,z] in T;
    then [y,x] in T by A8;
    then [x,y] in T by EQREL_1:6;
    hence contradiction by A7,Th27;
  end;
  hence thesis by Def2;
end;

theorem
  for x for Y being TolClass of T st x in Y holds Y c= neighbourhood(x,T )
proof
  let x;
  let Y be TolClass of T such that
A1: x in Y;
  for y st y in Y holds y in neighbourhood(x,T)
  proof
    let y;
    assume y in Y;
    then [x,y] in T by A1,Def1;
    hence thesis by Th27;
  end;
  hence thesis by TARSKI:def 3;
end;

theorem
  TolSets R c= TolSets T iff R c= T
proof
  thus TolSets R c= TolSets T implies R c= T
  proof
    assume
A1: TolSets R c= TolSets T;
    let x,y;
    assume [x,y] in R;
    then consider Z being TolClass of R such that
A2: x in Z & y in Z by Th20;
    Z in TolSets R by Def3;
    then Z is TolSet of T by A1,Def3;
    hence thesis by A2,Def1;
  end;
  assume
A3: R c= T;
  let x;
  assume x in TolSets R;
  then reconsider Z = x as TolSet of R by Def3;
  for x,y st x in Z & y in Z holds [x,y] in T
  proof
    let x,y;
    assume x in Z & y in Z;
    then [x,y] in R by Def1;
    hence thesis by A3;
  end;
  then Z is TolSet of T by Def1;
  hence thesis by Def3;
end;

theorem
  TolClasses T c= TolSets T
proof
  let x;
  assume x in TolClasses T;
  then x is TolSet of T by Def4;
  hence thesis by Def3;
end;

theorem
  (for x st x in X holds neighbourhood(x,R) c= neighbourhood(x,T))
  implies R c= T
proof
  assume
A1: for x st x in X holds neighbourhood(x,R) c= neighbourhood(x,T);
  let x,y;
  assume
A2: [x,y] in R;
  then x in X by ZFMISC_1:87;
  then
A3: neighbourhood(x,R) c= neighbourhood(x,T) by A1;
  y in neighbourhood(x,R) by A2,Th27;
  hence thesis by A3,Th27;
end;

theorem
  T c= T*T
proof
  let x,y;
  assume
A1: [x,y] in T;
  then y in X by ZFMISC_1:87;
  then [y,y] in T by Th7;
  hence thesis by A1,RELAT_1:def 8;
end;














begin

reserve
  X,x,y,z for set,
  k,n,m for Element of NAT,
  f for Function,
  p,q,r for FinSequence of NAT;






notation
  let p,q be FinSequence;
  synonym p is_a_prefix_of q for p c= q;
end;

definition
  let p,q be FinSequence;
  redefine pred p is_a_prefix_of q means
  :Def1:
  ex n st p = q|Seg n;
  compatibility
  proof
    thus p c= q implies ex n st p = q|Seg n
    proof
      assume
A1:   p c= q;
      consider n be Nat such that
A2:   dom p = Seg n by FINSEQ_1:def 2;
      reconsider n as Element of NAT by ORDINAL1:def 12;
      take n;
      thus thesis by A1,A2,GRFUNC_1:23;
    end;
    thus thesis by RELAT_1:59;
  end;
end;

theorem Th1:
  for p,q being FinSequence holds p is_a_prefix_of q iff
  ex r being FinSequence st q = p^r
proof
  let p,q be FinSequence;
  thus p is_a_prefix_of q implies ex r being FinSequence st q = p^r
  proof
    given n such that
A1: p = q|Seg n;
    thus thesis by A1,FINSEQ_1:80;
  end;
  given r being FinSequence such that
A2: q = p^r;
 dom p = Seg len p & p = q|dom p by A2,FINSEQ_1:21,def 3;
  hence thesis by Def1;
end;

theorem Th2:
  for p,q being FinSequence st p is_a_prefix_of q & len p = len q holds p = q
proof
  let p,q be FinSequence;
  assume that
A1: p is_a_prefix_of q and
A2: len p = len q;
  consider r being FinSequence such that
A3: q = p^r by A1,Th1;
 len p = len p + len r by A2,A3,FINSEQ_1:22;
then  r = {};
  hence thesis by A3,FINSEQ_1:34;
end;

theorem Th3:
  <*x*> is_a_prefix_of <*y*> iff x = y
proof
  thus <*x*> is_a_prefix_of <*y*> implies x = y
  proof
    assume
A1: <*x*> is_a_prefix_of <*y*>;
 len <*x*> = 1 & len <*y*> = 1 by FINSEQ_1:40;
then A2: <*x*> = <*y*> by A1,Th2;
 <*x*>.1 = x by FINSEQ_1:40;
    hence thesis by A2,FINSEQ_1:40;
  end;
  thus thesis;
end;

notation
  let p,q be FinSequence;
  synonym p is_a_proper_prefix_of q for p c< q;
end;

Lm1: for A,B being finite set st A c= B & card A = card B holds A = B
proof
  let A,B be finite set such that
A1: A c= B and
A2: card A = card B;
 card(B \ A) = card B - card A by A1,CARD_2:44
    .= 0 by A2;
then  B \ A = {};
then  B c= A by XBOOLE_1:37;
  hence thesis by A1,XBOOLE_0:def 10;
end;

theorem Th4:
  for p,q being finite set st p,q are_c=-comparable & card p = card q holds
  p = q
proof
  let p,q be finite set such that
A1: p c= q or q c= p and
A2: card p = card q;
  per cases by A1;
  suppose
 p c= q;
    hence thesis by A2,Lm1;
  end;
  suppose
 q c= p;
    hence thesis by A2,Lm1;
  end;
end;

reserve p1,p2,p3 for FinSequence;

theorem Th5:
  <*x*>,<*y*> are_c=-comparable implies x = y
proof
    assume
A1: <*x*>,<*y*> are_c=-comparable;
 len <*x*> = 1 & len <*y*> = 1 by FINSEQ_1:40;
then A2: <*x*> = <*y*> by A1,Th4;
    <*x*>.1 = x by FINSEQ_1:40;
    hence thesis by A2,FINSEQ_1:40;
end;

theorem Th6:
  for p,q being finite set st p c< q holds card p < card q
proof
  let p,q be finite set;
  assume that
A1: p c= q and
A2: p <> q;
A3: card p <= card q by A1,NAT_1:43;
  p,q are_c=-comparable by A1,XBOOLE_0:def 9;
  hence thesis by A3,A2,Th4,XXREAL_0:1;
end;

theorem Th7:
  p1^<*x*> is_a_prefix_of p2 implies p1 is_a_proper_prefix_of p2
proof
  assume p1^<*x*> is_a_prefix_of p2;
  then consider p3 such that
A1: p2 = p1^<*x*>^p3 by Th1;
 p2 = p1^(<*x*>^p3) by A1,FINSEQ_1:32;
  hence p1 is_a_prefix_of p2 by Th1;
  assume p1 = p2;
then  len p1 = len(p1^<*x*>) + len p3 by A1,FINSEQ_1:22
    .= len p1 + len <*x*> + len p3 by FINSEQ_1:22
    .= len p1 + 1 + len p3 by FINSEQ_1:39;
then  len p1 + 1 <= len p1 by NAT_1:11;
  hence contradiction by NAT_1:13;
end;

theorem Th8:
  p1 is_a_prefix_of p2 implies p1 is_a_proper_prefix_of p2^<*x*>
proof
  assume p1 is_a_prefix_of p2;
  then consider p3 such that
A1: p2 = p1^p3 by Th1;
 p2^<*x*> = p1^(p3^<*x*>) by A1,FINSEQ_1:32;
  hence p1 is_a_prefix_of p2^<*x*> by Th1;
  assume p1 = p2^<*x*>;
then  len p2 = len(p2^<*x*>) + len p3 by A1,FINSEQ_1:22
    .= len p2 + len <*x*> + len p3 by FINSEQ_1:22
    .= len p2 + 1 + len p3 by FINSEQ_1:39;
then  len p2 + 1 <= len p2 by NAT_1:11;
  hence contradiction by NAT_1:13;
end;

theorem Th9:
  p1 is_a_proper_prefix_of p2^<*x*> implies p1 is_a_prefix_of p2
proof
  assume that
A1: p1 is_a_prefix_of p2^<*x*> and
A2: p1 <> p2^<*x*>;
A3: ex p3 st p2^<*x*> = p1^p3 by A1,Th1;
A4: len p1 <= len(p2^<*x*>) by A1,NAT_1:43;
   len(p2^<*x*>) = len p2 + len <*x*> & len <*x*> = 1 by FINSEQ_1:22,39;
then  len p1 < len p2 + 1 by A1,A2,A4,Th2,XXREAL_0:1;
then  len p1 <= len p2 by NAT_1:13;
then  ex p3 st p1^p3 = p2 by A3,FINSEQ_1:47;
  hence thesis by Th1;
end;

theorem
  {} is_a_proper_prefix_of p2 or {} <> p2 implies
  p1 is_a_proper_prefix_of p1^p2
proof
  assume
A1: {} is_a_proper_prefix_of p2 or {} <> p2;
  thus p1 is_a_prefix_of p1^p2 by Th1;
  assume p1 = p1^p2;
then  len p1 = len p1 + len p2 by FINSEQ_1:22;
then  p2 = {};
  hence contradiction by A1;
end;





definition
  let p be FinSequence;

  func ProperPrefixes p -> set means
  :Def2:
  x in it iff ex q being FinSequence st x = q & q is_a_proper_prefix_of p;
  existence
  proof
    set E = rng p;
    defpred X[set] means ex q being FinSequence st $1 = q &
    q is_a_proper_prefix_of p;
    consider X such that
A1: x in X iff x in E* & X[x] from XBOOLE_0:sch 1;
    take X;
    let x;
    thus x in X implies ex q being FinSequence st x = q &
    q is_a_proper_prefix_of p by A1;
    given q be FinSequence such that
A2: x = q and
A3: q is_a_proper_prefix_of p;
    q is_a_prefix_of p by A3,XBOOLE_0:def 8;
    then ex n st q = p|Seg n by Def1;
    then rng q c= E by RELAT_1:70;
    then reconsider q as FinSequence of E by FINSEQ_1:def 4;
    q in E* by FINSEQ_1:def 11;
    hence thesis by A1,A2,A3;
  end;
  uniqueness
  proof
    let S1,S2 be set such that
A4: x in
    S1 iff ex q being FinSequence st x = q & q is_a_proper_prefix_of p and
A5: x in S2 iff ex q being FinSequence st x = q & q is_a_proper_prefix_of p;
    defpred X[set] means
    ex q being FinSequence st $1 = q & q is_a_proper_prefix_of p;
A6: x in S1 iff X[x] by A4;
A7: x in S2 iff X[x] by A5;
    thus thesis from XBOOLE_0:sch 2(A6,A7);
  end;
end;

theorem Th11:
  for p being FinSequence st x in ProperPrefixes p holds x is FinSequence
proof
  let p be FinSequence;
  assume x in ProperPrefixes p;
  then
  ex q being FinSequence st x = q & q is_a_proper_prefix_of p by Def2;
  hence thesis;
end;

theorem Th12:
  for p,q being FinSequence holds
  p in ProperPrefixes q iff p is_a_proper_prefix_of q
proof
  let p,q be FinSequence;
  thus p in ProperPrefixes q implies p is_a_proper_prefix_of q
  proof
    assume p in ProperPrefixes q;
then
 ex r being FinSequence st p = r & r is_a_proper_prefix_of q by Def2;
    hence thesis;
  end;
  thus thesis by Def2;
end;

theorem Th13:
  for p,q being FinSequence st p in ProperPrefixes q holds len p < len q
proof
  let p,q be FinSequence;
  assume p in ProperPrefixes q;
  then p is_a_proper_prefix_of q by Th12;
  hence thesis by Th6;
end;

theorem
  for p,q,r being FinSequence st q^r in ProperPrefixes p holds
  q in ProperPrefixes p
proof
  let p,q,r be FinSequence;
  assume
A1: q^r in ProperPrefixes p;
A2: q is_a_prefix_of q^r by Th1;
 q^r is_a_proper_prefix_of p by A1,Th12;
 then q is_a_proper_prefix_of p by A2,XBOOLE_1:59;
  hence thesis by Th12;
end;

theorem Th15:
  ProperPrefixes {} = {} proof set x = the Element of ProperPrefixes {};
  assume
A1: not thesis;
  then reconsider x as FinSequence by Th11;
 x is_a_proper_prefix_of {} by A1,Th12;
  hence contradiction by XBOOLE_1:115;
end;

set D = { {} };

theorem Th16:
  ProperPrefixes <*x*> = { {} }
proof
  thus ProperPrefixes <*x*> c= D
  proof
    let y;
    assume y in ProperPrefixes <*x*>;
    then consider p being FinSequence such that
A1: y = p and
A2: p is_a_proper_prefix_of <*x*> by Def2;
A3: len p < len <*x*> by A2,Th6;
 len <*x*> = 1 & 1 = 0+1 by FINSEQ_1:39;
then  p = {} by A3,NAT_1:13;
    hence thesis by A1,TARSKI:def 1;
  end;
  let y;
  assume y in D;
then A4: y = {} by TARSKI:def 1;
 {} is_a_prefix_of <*x*> by XBOOLE_1:2;
then  {} is_a_proper_prefix_of <*x*> by XBOOLE_0:def 8;
  hence thesis by A4,Def2;
end;

theorem
  for p,q being FinSequence st p is_a_prefix_of q holds
  ProperPrefixes p c= ProperPrefixes q
proof
  let p,q be FinSequence such that
A1: p is_a_prefix_of q;
  let x;
  assume
A2: x in ProperPrefixes p;
  then reconsider r = x as FinSequence by Th11;
 r is_a_proper_prefix_of p by A2,Th12;
then  r is_a_proper_prefix_of q by A1,XBOOLE_1:58;
  hence thesis by Th12;
end;

theorem Th18:
  for p,q,r being FinSequence st
  q in ProperPrefixes p & r in ProperPrefixes p holds q,r are_c=-comparable
proof
  let p,q,r be FinSequence;
  assume q in ProperPrefixes p;
then
A1: ex q1 being FinSequence st q = q1 & q1 is_a_proper_prefix_of p by Def2;
  assume r in ProperPrefixes p;
then
A2: ex r1 being FinSequence st r = r1 & r1 is_a_proper_prefix_of p by Def2;
 q is_a_prefix_of p by A1,XBOOLE_0:def 8;
  then consider n such that
A3: q = p|Seg n by Def1;
 r is_a_prefix_of p by A2,XBOOLE_0:def 8;
  then consider k such that
A4: r = p|Seg k by Def1;
A5: n <= k implies thesis
  proof
    assume n <= k;
then  Seg n c= Seg k by FINSEQ_1:5;
then  q = r|Seg n by A3,A4,FUNCT_1:51;
then  q is_a_prefix_of r by Def1;
    hence thesis by XBOOLE_0:def 9;
  end;
 k <= n implies thesis
  proof
    assume k <= n;
then  Seg k c= Seg n by FINSEQ_1:5;
then  r = q|Seg k by A3,A4,FUNCT_1:51;
then  r is_a_prefix_of q by Def1;
    hence thesis by XBOOLE_0:def 9;
  end;
  hence thesis by A5;
end;





definition
  let X;
  attr X is Tree-like means
  :Def3:
  X c= NAT* & (for p st p in X holds ProperPrefixes p c= X) &
  for p,k,n st p^<*k*> in X & n <= k holds p^<*n*> in X;
end;

registration
  cluster { {} } -> Tree-like;
  coherence
  proof
    thus D c= NAT*
    proof
      let x;
      assume x in D;
      then x = {} by TARSKI:def 1;
      hence thesis by FINSEQ_1:49;
    end;
    thus for p st p in D holds ProperPrefixes p c= D
    proof
      let p;
      assume p in D;
      then p = {} by TARSKI:def 1;
      hence thesis by Th15,XBOOLE_1:2;
    end;
    thus thesis by TARSKI:def 1;
  end;
end;

registration
  cluster non empty Tree-like for set;
  existence
  proof
    take D = { <*>(NAT) };
    thus thesis;
  end;
end;

definition
  mode Tree is Tree-like non empty set;
end;

reserve T,T1 for Tree;

theorem Th19:
  x in T implies x is FinSequence of NAT
proof
  T c= NAT* by Def3;
  hence thesis by FINSEQ_1:def 11;
end;

definition
  let T;
  redefine mode Element of T -> FinSequence of NAT;
  coherence by Th19;
end;

theorem Th20:
  for p,q being FinSequence st p in T & q is_a_prefix_of p holds q in T
proof
  let p,q be FinSequence;
  assume that
A1: p in T and
A2: q is_a_prefix_of p;
  reconsider r = p as Element of T by A1;
A3: ProperPrefixes r c= T by Def3;
 q is_a_proper_prefix_of p or q = p by A2,XBOOLE_0:def 8;
then  q in ProperPrefixes p or p = q by Th12;
  hence thesis by A3;
end;

theorem Th21:
  for r being FinSequence st q^r in T holds q in T
proof
  let r be FinSequence;
  assume
A1: q^r in T;
  then reconsider p = q^r as FinSequence of NAT by Th19;
  reconsider s = p|Seg len q as FinSequence of NAT by FINSEQ_1:18;
 len p = len q + len r by FINSEQ_1:22;
then  len q <= len p by NAT_1:11;
then A2: len s = len q by FINSEQ_1:17;
 now
    let n be Nat;
    assume 1 <= n & n <= len q;
then A3: n in Seg len q by FINSEQ_1:1;
 Seg len q = dom q by FINSEQ_1:def 3;
then  p.n = q.n by A3,FINSEQ_1:def 7;
    hence q.n = s.n by A3,FUNCT_1:49;
  end;
then  q = s by A2,FINSEQ_1:14;
then A4: q is_a_prefix_of p by Def1;
 now
    assume q <> p;
then  q is_a_proper_prefix_of p by A4,XBOOLE_0:def 8;
then A5: q in ProperPrefixes p by Def2;
 ProperPrefixes p c= T by A1,Def3;
    hence thesis by A5;
  end;
  hence thesis by A1;
end;

theorem Th22:
  {} in T & <*> NAT in T
  proof
    reconsider x = the Element of T as FinSequence of NAT;
    x in T;
    then
A1: {}^x in T by FINSEQ_1:34;
    {} = <*> NAT;
    hence thesis by A1,Th21;
  end;

theorem
  { {} } is Tree;

registration
  let T,T1;
  cluster T \/ T1 -> Tree-like;
  coherence
  proof
    set D = T \/ T1;
    T c= NAT* & T1 c= NAT* by Def3;
    hence D c= NAT* by XBOOLE_1:8;
    thus for p st p in D holds ProperPrefixes p c= D
    proof
      let p;
      assume p in D;
      then p in T or p in T1 by XBOOLE_0:def 3;
      then
A1:   ProperPrefixes p c= T or ProperPrefixes p c= T1 by Def3;
      T c= D & T1 c= D by XBOOLE_1:7;
      hence thesis by A1,XBOOLE_1:1;
    end;
    let p,k,n;
    assume that
A2: p^<*k*> in D and
A3: n <= k;
    p^<*k*> in T or p^<*k*> in T1 by A2,XBOOLE_0:def 3;
    then  p^<*n*> in T or p^<*n*> in T1 by A3,Def3;
    hence thesis by XBOOLE_0:def 3;
  end;

  cluster T /\ T1 -> Tree-like non empty;
  coherence
  proof
    set D = T /\ T1;
    thus D is Tree-like
    proof
      T c= NAT* & T /\ T1 c= T by Def3,XBOOLE_1:17;
      hence D c= NAT* by XBOOLE_1:1;
      thus for p st p in D holds ProperPrefixes p c= D
      proof
        let p;
        assume
A4:     p in D;
        then
A5:     p in T by XBOOLE_0:def 4;
A6:     p in T1 by A4,XBOOLE_0:def 4;
A7:     ProperPrefixes p c= T by A5,Def3;
        ProperPrefixes p c= T1 by A6,Def3;
        hence thesis by A7,XBOOLE_1:19;
      end;
      let p,k,n;
      assume that
A8:   p^<*k*> in D and
A9:   n <= k;
A10:   p^<*k*> in T by A8,XBOOLE_0:def 4;
A11:   p^<*k*> in T1 by A8,XBOOLE_0:def 4;
A12:   p^<*n*> in T by A9,A10,Def3;
      p^<*n*> in T1 by A9,A11,Def3;
      hence thesis by A12,XBOOLE_0:def 4;
    end;
    {} in T & {} in T1 by Th22;
    hence thesis by XBOOLE_0:def 4;
  end;
end;

theorem
  T \/ T1 is Tree;

theorem
  T /\ T1 is Tree;





registration
  cluster finite for Tree;
  existence
  proof
    take D;
    thus thesis;
  end;
end;

reserve fT,fT1 for finite Tree;

theorem
  fT \/ fT1 is finite Tree;

theorem
  fT /\ T is finite Tree;





definition
  let n;

  func elementary_tree n -> Tree equals
  { <*k*> : k < n } \/ { {} };
  correctness
  proof
    set IT = { <*k*> : k < n } \/ D;
    IT is Tree-like
    proof
      thus IT c= NAT*
      proof
        let x;
        assume x in IT;
        then
A1:     x in { <*k*> : k < n } or x in D by XBOOLE_0:def 3;
A2:     {} in NAT* by FINSEQ_1:49;
        now
          assume x in { <*k*> : k < n };
          then  ex k st x = <*k*> & k < n;
          hence thesis by FINSEQ_1:def 11;
        end;
        hence thesis by A1,A2,TARSKI:def 1;
      end;
      thus for p st p in IT holds ProperPrefixes p c= IT
      proof
        let p;
        assume p in IT;
        then
A3:     p in { <*k*> : k < n } or p in D by XBOOLE_0:def 3;
A4:     {} c= IT by XBOOLE_1:2;
        now
          assume p in { <*k*> : k < n };
          then ex k st p = <*k*> & k < n;
          then ProperPrefixes p = D by Th16;
          hence thesis by XBOOLE_1:7;
        end;
        hence thesis by A3,A4,Th15,TARSKI:def 1;
      end;
      let p,k,m;
      assume p^<*k*> in IT;
      then p^<*k*> in
      { <*j*> where j is Element of NAT : j < n } or p^<*k*> in D
      by XBOOLE_0:def 3;
      then consider l being Element of NAT such that
A5:   p^<*k*> = <*l*> and
A6:   l < n by TARSKI:def 1;
      len p + len <*k*> = len <*l*> by A5,FINSEQ_1:22
      .= 1 by FINSEQ_1:39;
      then   len p + 1 = 0 + 1 by FINSEQ_1:39;
      then A7:  p = {};
      then A8:  <*k*> = <*l*> by A5,FINSEQ_1:34;
      <*k*>.1 = k by FINSEQ_1:def 8;
      then A9:  k = l by A8,FINSEQ_1:def 8;
      assume
A10:  m <= k;
A11:  p^<*m*> = <*m*> by A7,FINSEQ_1:34;
      m < n by A6,A9,A10,XXREAL_0:2;
      then p^<*m*> in { <*j*> where j is Element of NAT : j < n } by A11;
      hence thesis by XBOOLE_0:def 3;
    end;
    hence thesis;
  end;
end;

registration
  let n;
  cluster elementary_tree n -> finite;
  coherence
  proof
    set IT = elementary_tree n;
    IT,Seg(n+1) are_equipotent
    proof
      defpred F[set,set] means
      $1 = {} & $2 = 1 or ex n st $1 = <*n*> & $2 = n+2;
A1:  x in IT & F[x,y] & F[x,z] implies y = z
      proof
        assume that x in IT and
A2:    (x = {} & y = 1 or ex n st x = <*n*> & y = n+2)&( x = {} & z = 1 or
        ex n st x = <*n*> & z = n+2);
        now
          given n1 being Element of NAT such that
A3:      x = <*n1*> & y = n1+2;
          given n2 being Element of NAT such that
A4:      x = <*n2*> & z = n2+2;
          <*n1*>.1 = n1 by FINSEQ_1:def 8;
          hence thesis by A3,A4,FINSEQ_1:def 8;
        end;
        hence thesis by A2;
      end;
A5:  x in IT implies ex y st F[x,y]
      proof
        assume
A6:    x in IT;
A7:    now
          assume x in { <*k*> : k < n };
          then consider k such that
A8:      x = <*k*> and k < n;
          reconsider y = k+2 as set;
          take y;
          thus F[x,y] by A8;
        end;
        now
          assume
A9:      x in D;
          reconsider y = 1 as set;
          take y;
          thus F[x,y] by A9,TARSKI:def 1;
        end;
        hence thesis by A6,A7,XBOOLE_0:def 3;
      end;
      consider f such that
A10:  dom f = IT & for x st x in IT holds F[x,f.x] from CLASSES1:sch 1(A5);
      take f;
      thus f is one-to-one
      proof
        let x,y;
        assume that
A11:    x in dom f and
A12:    y in dom f and
A13:    f.x = f.y;
A14:    x = {} & f.x = 1 or ex n st x = <*n*> & f.x = n+2 by A10,A11;
A15:    y = {} & f.y = 1 or ex n st y = <*n*> & f.y = n+2 by A10,A12;
A16:    now
          assume that
          x = {} and
A17:      f.x = 1;
          given n such that
          y = <*n*> and
A18:      f.y = n+2;
      0+1 = n+1+1 by A13,A17,A18;
          hence contradiction;
        end;
    now
          assume that
          y = {} and
A19:      f.y = 1;
          given n such that
          x = <*n*> and
A20:      f.x = n+2;
      0+1 = n+1+1 by A13,A19,A20;
          hence contradiction;
        end;
        hence thesis by A13,A14,A15,A16;
      end;
      thus dom f = IT by A10;
      thus rng f c= Seg(n+1)
      proof
        let x;
        assume x in rng f;
        then consider y such that
A21:    y in dom f and
A22:    x = f.y by FUNCT_1:def 3;
    1 <= 1+n by NAT_1:11;
then A23:    1 in Seg(n+1) by FINSEQ_1:1;
    now
          given k such that
A24:      y = <*k*> and
A25:      x = k+2;
      y in { <*j*> where j is Element of NAT : j < n } or y in D
          by A10,A21,XBOOLE_0:def 3;
          then consider l being Element of NAT such that
A26:      y = <*l*> & l < n by A24,TARSKI:def 1;
      <*k*>.1 = k & <*l*>.1 = l by FINSEQ_1:def 8;
then       k+1 <= n by A24,A26,NAT_1:13;
then       1+0 <= k+1+1 & k+1+1 <= n+1 by XREAL_1:7;
          hence thesis by A25,FINSEQ_1:1;
        end;
        hence thesis by A10,A21,A22,A23;
      end;
      let x;
      assume
A27:  x in Seg(n+1);
      then reconsider k = x as Element of NAT;
A28:  1 <= k by A27,FINSEQ_1:1;
A29:  k <= n+1 by A27,FINSEQ_1:1;
      {} in D by TARSKI:def 1;
then A30:  {} in IT by XBOOLE_0:def 3;
then   F[{},f.{}] by A10;
then A31:  1 in rng f by A10,A30,FUNCT_1:def 3;
  now
        assume 1 < k;
then     1+1 <= k by NAT_1:13;
        then consider m be Nat such that
A32:    k = 2+m by NAT_1:10;
        reconsider m as Element of NAT by ORDINAL1:def 12;
    1+1+m = 1+(1+m);
then     1+m <= n by A29,A32,XREAL_1:6;
then     m < n by NAT_1:13;
then     <*m*> in { <*j*> where j is Element of NAT : j < n };
then A33:    <*m*> in IT by XBOOLE_0:def 3;
then     F[<*m*>,f.<*m*>] by A10;
then     k = f.<*m*> by A1,A32,A33;
        hence k in rng f by A10,A33,FUNCT_1:def 3;
      end;
      hence thesis by A28,A31,XXREAL_0:1;
    end;
    hence thesis by CARD_1:38;
  end;
end;

theorem Th28:
  k < n implies <*k*> in elementary_tree n
proof
  assume k < n;
  then <*k*> in { <*j*> where j is Element of NAT : j < n };
  hence thesis by XBOOLE_0:def 3;
end;

theorem Th29:
  elementary_tree 0 = { {} }
proof
 now
    set x = the Element of { <*j*> where j is Element of NAT : j < 0 };
    assume { <*j*> where j is Element of NAT : j < 0 } <> {};
then  x in { <*j*> where j is Element of NAT : j < 0 };
then  ex k st x = <*k*> & k < 0;
    hence contradiction;
  end;
  hence thesis;
end;

theorem
  p in elementary_tree n implies p = {} or ex k st k < n & p = <*k*>
proof
  assume p in elementary_tree n;
then A1: p in D or p in { <*k*> : k < n } by XBOOLE_0:def 3;
 p in { <*k*> : k < n } implies ex k st p = <*k*> & k < n;
  hence thesis by A1,TARSKI:def 1;
end;





definition
  let T;
  func Leaves T -> Subset of T means
  :Def5:
  p in it iff p in T & not ex q st q in T & p is_a_proper_prefix_of q;
  existence
  proof
    defpred X[set] means
    for p st $1 = p for q st q in T holds not p is_a_proper_prefix_of q;
    consider X such that
A1: x in X iff x in T & X[x] from XBOOLE_0:sch 1;
 X c= T
    proof
      let x;
      thus thesis by A1;
    end;
    then reconsider X as Subset of T;
    take X;
    let p;
    thus
    p in X implies p in T & not ex q st q in T & p is_a_proper_prefix_of
    q by A1;
    assume that
A2: p in T and
A3: not ex q st q in T & p is_a_proper_prefix_of q;
 for r being FinSequence of NAT st p = r
    for q st q in T holds not r is_a_proper_prefix_of q by A3;
    hence thesis by A1,A2;
  end;
  uniqueness
  proof
    let L1,L2 be Subset of T such that
A4: p in L1 iff p in T & not ex q st q in T & p is_a_proper_prefix_of q and
A5: p in L2 iff p in T & not ex q st q in T & p is_a_proper_prefix_of q;
A6: T c= NAT* by Def3;
then A7: L1 c= NAT* by XBOOLE_1:1;
A8: L2 c= NAT* by A6,XBOOLE_1:1;
 now
      let x;
      thus x in L1 implies x in L2
      proof
        assume
A9:    x in L1;
        then reconsider p = x as FinSequence of NAT by A7,FINSEQ_1:def 11;
    not ex q st q in T & p is_a_proper_prefix_of q by A4,A9;
        hence thesis by A5,A9;
      end;
      assume
A10:  x in L2;
      then reconsider p = x as FinSequence of NAT by A8,FINSEQ_1:def 11;
  not ex q st q in T & p is_a_proper_prefix_of q by A5,A10;
      hence x in L1 by A4,A10;
    end;
    hence thesis by TARSKI:1;
  end;
  let p such that
B11: p in T;
  func T|p -> Tree means  
  :Def6:
  q in it iff p^q in T;
  existence
  proof
    defpred X[set] means for q st $1 = q holds p^q in T;
    consider X such that
A11: x in X iff x in NAT* & X[x] from XBOOLE_0:sch 1;
     <*>
 NAT in NAT* & for q st <*> NAT = q holds p^q in T by B11,FINSEQ_1:34,def
11;
    then reconsider X as non empty set by A11;
A12: X c= NAT*
    proof
      let x;
      thus thesis by A11;
    end;
A13: now
      let q such that
A14:  q in X;
      thus ProperPrefixes q c= X
      proof
        let x;
        assume x in ProperPrefixes q;
        then consider r being FinSequence such that
A15:    x = r and
A16:    r is_a_proper_prefix_of q by Def2;
    r is_a_prefix_of q by A16,XBOOLE_0:def 8;
then A17:    ex n st r = q|Seg n by Def1;
        then reconsider r as FinSequence of NAT by FINSEQ_1:18;
        consider s being FinSequence such that
A18:    q = r^s by A17,FINSEQ_1:80;
A19:    p^q in T by A11,A14;
    p^q = p^r^s by A18,FINSEQ_1:32;
        then     r
 in NAT* & for q st r = q holds p^q in T by A19,Th21,FINSEQ_1:def 11;
        hence thesis by A11,A15;
      end;
    end;
 now
      let q,k,n;
      assume that
A20:  q^<*k*> in X and
A21:  n <= k;
  p^(q^<*k*>) in T by A11,A20;
then   p^q^<*k*> in T by FINSEQ_1:32;
then   p^q^<*n*> in T by A21,Def3;
      then   q
^<*n*> in NAT* & for r st r = q^<*n*> holds p^r in T by FINSEQ_1:32,def 11;
      hence q^<*n*> in X by A11;
    end;
    then reconsider X as Tree by A12,A13,Def3;
    take X;
    let q;
    thus q in X implies p^q in T by A11;
    assume p^q in T;
    then  q
 in NAT* & for r be FinSequence of NAT st q = r holds p^r in T by
FINSEQ_1:def 11;
    hence thesis by A11;
  end;
  uniqueness
  proof
    let T1,T2 be Tree such that
A22: q in T1 iff p^q in T and
A23: q in T2 iff p^q in T;
 now
      let x;
      thus x in T1 implies x in T2
      proof
        assume
A24:    x in T1;
        then reconsider q = x as FinSequence of NAT by Th19;
    p^q in T by A22,A24;
        hence thesis by A23;
      end;
      assume
A25:  x in T2;
      then reconsider q = x as FinSequence of NAT by Th19;
  p^q in T by A23,A25;
      hence x in T1 by A22;
    end;
    hence thesis by TARSKI:1;
  end;
end;

theorem
  T|(<*> NAT) = T
proof
A1: <*> NAT in T by Th22;
  thus T|(<*> NAT) c= T
  proof
    let x;
    assume x in T|(<*> NAT);
    then reconsider p = x as Element of T|(<*> NAT);
 {}^p = p by FINSEQ_1:34;
    hence thesis by A1,Def6;
  end;
  let x;
  assume x in T;
  then reconsider p = x as Element of T;
 {}^p = p by FINSEQ_1:34;
  hence thesis by A1,Def6;
end;

registration
  let T be finite Tree;
  let p be Element of T;
  cluster T|p -> finite;
  coherence
  proof
    consider t being Function such that
A1: rng t = T and
A2: dom t in omega by FINSET_1:def 1;
    defpred P[set,set] means ex q st t.$1 = q & ((ex r st $2 = r & q = p^r) or
    (for r holds q <> p^r) & $2 = <*> NAT);
A3: for x st x in dom t ex y st P[x,y]
    proof
      let x;
      assume x in dom t;
then    t.x in T by A1,FUNCT_1:def 3;
      then reconsider q = t.x as FinSequence of NAT by Th19;
   (ex r st q = p^r) implies thesis;
      hence thesis;
    end;
    consider f being Function such that
A4: dom f = dom t & for x st x in dom t holds P[x,f.x] from CLASSES1:sch 1(
    A3);
 T|p is finite
    proof
      take f;
      thus rng f c= T|p
      proof
        let x;
        assume x in rng f;
        then consider y such that
A5:    y in dom f and
A6:    x = f.y by FUNCT_1:def 3;
        consider q such that
A7:    t.y = q and
A8:    (ex r st x = r & q = p^r) or (for r holds q <> p^r) & x = <*> NAT
        by A4,A5,A6;
        assume
A9:    not x in T|p;
            p
^(<*> NAT) = p & q in T by A1,A4,A5,A7,FINSEQ_1:34,FUNCT_1:def 3;
        hence contradiction by A8,A9,Def6;
      end;
      thus T|p c= rng f
      proof
        let x;
        assume
A10:    x in T|p;
        then reconsider q = x as FinSequence of NAT by Th19;
    p^q in T by A10,Def6;
        then consider y such that
A11:    y in dom t and
A12:    p^q = t.y by A1,FUNCT_1:def 3;
    P[y,f.y] by A4,A11;
then     x = f.y by A12,FINSEQ_1:33;
        hence thesis by A4,A11,FUNCT_1:def 3;
      end;
      thus thesis by A2,A4;
    end;
    hence thesis;
  end;
end;

definition
  let T;
  assume
A1: Leaves T <> {};
  mode Leaf of T -> Element of T means
  it in Leaves T;
  existence
  proof set x = the Element of Leaves T;
    reconsider x as Element of T by A1,TARSKI:def 3;
    take x;
    thus thesis by A1;
  end;
end;

definition
  let T;
  mode Subtree of T -> Tree means
    ex p being Element of T st it = T|p;
  existence
  proof set p = the Element of T;
    take T|p, p;
    thus thesis;
  end;
end;

reserve t for Element of T;

definition
  let T,p,T1;
  assume
A1: p in T;
  func T with-replacement (p,T1) -> Tree means
  :Def9:
  q in it iff q in T & not p is_a_proper_prefix_of q or
  ex r st r in T1 & q = p^r;
  existence
  proof
    reconsider p9 = p as Element of T by A1;
 not p is_a_proper_prefix_of p9;
then  p in { t : not p is_a_proper_prefix_of t };
    then reconsider X = { t : not p is_a_proper_prefix_of t } \/
    { p^s where s is Element of T1 : s = s } as non empty set;
A2: x in { t : not p is_a_proper_prefix_of t } implies
    x is FinSequence of NAT & x in NAT* & x in T
    proof
      assume x in { t : not p is_a_proper_prefix_of t };
then A3:   ex t st x = t & not p is_a_proper_prefix_of t;
      hence x is FinSequence of NAT;
      thus thesis by A3,FINSEQ_1:def 11;
    end;
 X is Tree-like
    proof
      thus X c= NAT*
      proof
        let x;
        assume x in X;
then A4:     x in { t : not p is_a_proper_prefix_of t } or
        x in { p^s where s is Element of T1 : s = s } by XBOOLE_0:def 3;
    now
          assume x in { p^s where s is Element of T1 : s = s };
then       ex s being Element of T1 st x = p^s & s = s;
          hence x is FinSequence of NAT;
        end;
        hence thesis by A2,A4,FINSEQ_1:def 11;
      end;
      thus for q st q in X holds ProperPrefixes q c= X
      proof
        let q such that
A5:    q in X;
A6:    now
          assume q in { t : not p is_a_proper_prefix_of t };
then A7:      ex t st q = t & not p is_a_proper_prefix_of t;
then A8:      ProperPrefixes q c= T by Def3;
          thus ProperPrefixes q c= X
          proof
            let x;
            assume
A9:        x in ProperPrefixes q;
            then consider p1 such that
A10:        x = p1 and
A11:        p1 is_a_proper_prefix_of q by Def2;
            reconsider p1 as Element of T by A8,A9,A10;
        not p is_a_proper_prefix_of p1 by A7,A11,XBOOLE_1:56;
then         x in { s1 where s1 is Element of T :
            not p is_a_proper_prefix_of s1 } by A10;
            hence thesis by XBOOLE_0:def 3;
          end;
        end;
    now
          assume q in { p^s where s is Element of T1 : s = s };
          then consider s being Element of T1 such that
A12:      q = p^s and s = s;
          thus ProperPrefixes q c= X
          proof
            let x;
            assume
A13:        x in ProperPrefixes q;
            then reconsider r = x as FinSequence by Th11;
        r is_a_proper_prefix_of p^s by A12,A13,Th12;
then         r is_a_prefix_of p^s by XBOOLE_0:def 8;
            then consider r1 being FinSequence such that
A14:        p^s = r^r1 by Th1;
A15:        now
              assume len p <= len r;
              then consider r2 being FinSequence such that
A16:          p^r2 = r by A14,FINSEQ_1:47;
          p^s = p^(r2^r1) by A14,A16,FINSEQ_1:32;
then           s = r2^r1 by FINSEQ_1:33;
              then A17:          dom
 r2 = Seg len r2 & s|dom r2 = r2 by FINSEQ_1:21,def 3;
              then reconsider r2 as FinSequence of NAT by FINSEQ_1:18;
          r2 is_a_prefix_of s by A17,Def1;
              then reconsider r2 as Element of T1 by Th20;
          r = p^r2 by A16;
then           r in { p^v where v is Element of T1 : v = v };
              hence r in X by XBOOLE_0:def 3;
            end;
        now
              assume len r <= len p;
then           ex r2 being FinSequence st r^r2 = p by A14,FINSEQ_1:47;
then A18:          p|dom r = r by FINSEQ_1:21;
A19:          dom r = Seg len r by FINSEQ_1:def 3;
              then
reconsider r3 = r as FinSequence of NAT by A18,FINSEQ_1:18;
A20:          r3 is_a_prefix_of p by A18,A19,Def1;
then A21:          not p is_a_proper_prefix_of r3 by XBOOLE_0:def 8;
              reconsider r3 as Element of T by A1,A20,Th20;
          r3 in { t : not p is_a_proper_prefix_of t } by A21;
              hence r in X by XBOOLE_0:def 3;
            end;
            hence thesis by A15;
          end;
        end;
        hence thesis by A5,A6,XBOOLE_0:def 3;
      end;
      let q,k,n such that
A22:  q^<*k*> in X and
A23:  n <= k;
A24:  now
        assume q^<*k*> in { t : not p is_a_proper_prefix_of t };
then A25:    ex s being Element of T st
        q^<*k*> = s & not p is_a_proper_prefix_of s;
        then reconsider u = q^<*n*> as Element of T by A23,Def3;
    now
          assume p is_a_proper_prefix_of u;
then       p is_a_prefix_of q by Th9;
          hence contradiction by A25,Th8;
        end;
then     q^<*n*> in { t : not p is_a_proper_prefix_of t };
        hence thesis by XBOOLE_0:def 3;
      end;
  now
        assume q^<*k*> in { p^s where s is Element of T1 : s = s };
        then consider s being Element of T1 such that
A26:    q^<*k*> = p^s and s = s;
A27:    now
          assume len q <= len p;
          then consider r being FinSequence such that
A28:      q^r = p by A26,FINSEQ_1:47;
      q^<*k*> = q^(r^s) by A26,A28,FINSEQ_1:32;
then A29:      <*k*> = r^s by FINSEQ_1:33;
A30:      now
            assume
A31:        r = <*k*>;
            then reconsider s = q^<*n*> as Element of T by A1,A23,A28,Def3;
        now
              assume
A32:          p is_a_proper_prefix_of s;
then A33:          p is_a_prefix_of s by XBOOLE_0:def 8;
          len p = len q + len <*k*> by A28,A31,FINSEQ_1:22
                .= len q + 1 by FINSEQ_1:40
                .= len q + len <*n*> by FINSEQ_1:40
                .= len s by FINSEQ_1:22;
              hence contradiction by A32,A33,Th2;
            end;
            hence q^<*n*> in { t : not p is_a_proper_prefix_of t };
          end;
      now
            assume that
A34:        s = <*k*> and
A35:        r = {};
        s = {}^s & {} = <*> NAT by FINSEQ_1:34;
then         {}^<*n*> in T1 by A23,A34,Def3;
            then reconsider t = <*n*> as Element of T1 by FINSEQ_1:34;
        q^<*n*> = p^t by A28,A35,FINSEQ_1:34;
            hence
q^<*n*> in { p^v where v is Element of T1 : v = v };
          end;
          hence thesis by A29,A30,FINSEQ_1:88,XBOOLE_0:def 3;
        end;
    now
          assume len p <= len q;
          then consider r being FinSequence such that
A36:      p^r = q by A26,FINSEQ_1:47;
      p^(r^<*k*>) = p^s by A26,A36,FINSEQ_1:32;
then A37:      r^<*k*> = s by FINSEQ_1:33;
          then
      dom r = Seg len r & s|dom r = r by FINSEQ_1:21,def 3;
          then reconsider r as FinSequence of NAT by FINSEQ_1:18;
          reconsider t = r^<*n*> as Element of T1 by A23,A37,Def3;
      q^<*n*> = p^t by A36,FINSEQ_1:32;
then       q^<*n*> in { p^v where v is Element of T1 : v = v };
          hence thesis by XBOOLE_0:def 3;
        end;
        hence thesis by A27;
      end;
      hence thesis by A22,A24,XBOOLE_0:def 3;
    end;
    then reconsider X as Tree;
    take X;
    let q;
    thus q in X implies q in T & not p is_a_proper_prefix_of q or
    ex r st r in T1 & q = p^r
    proof
      assume
A38:  q in X;
A39:  now
        assume q in { t : not p is_a_proper_prefix_of t };
then     ex s being Element of T st q = s & not p is_a_proper_prefix_of s;
        hence thesis;
      end;
  now
        assume q in { p^s where s is Element of T1 : s = s };
then     ex s being Element of T1 st q = p^s & s = s;
        hence ex r st r in T1 & q = p^r;
      end;
      hence thesis by A38,A39,XBOOLE_0:def 3;
    end;
    assume
A40: q in T & not p is_a_proper_prefix_of q or ex r st r in T1 & q = p^r;
A41: q in T & not p is_a_proper_prefix_of q implies
    q in { t : not p is_a_proper_prefix_of t };
 (ex r st r in T1 & q = p^r)
    implies q in { p^v where v is Element of T1 : v = v };
    hence thesis by A40,A41,XBOOLE_0:def 3;
  end;
  uniqueness
  proof
    let S1,S2 be Tree such that
A42: q in S1 iff q in T & not p is_a_proper_prefix_of q or
    ex r st r in T1 & q = p^r and
A43: q in S2 iff q in T & not p is_a_proper_prefix_of q or
    ex r st r in T1 & q = p^r;
    thus S1 c= S2
    proof
      let x;
      assume
A44:  x in S1;
      then reconsider q = x as FinSequence of NAT by Th19;
        q
 in T & not p is_a_proper_prefix_of q or ex r st r in T1 & q = p^ r
      by A42,A44;
      hence thesis by A43;
    end;
    let x;
    assume
A45: x in S2;
    then reconsider q = x as FinSequence of NAT by Th19;
 q in T & not p is_a_proper_prefix_of q or ex r st r in T1 & q = p^r
    by A43,A45;
    hence thesis by A42;
  end;
end;

theorem Th32:
  p in T implies T with-replacement (p,T1) =
  { t1 where t1 is Element of T : not p is_a_proper_prefix_of t1 } \/
  { p^s where s is Element of T1 : s = s }
proof
  assume
A1: p in T;
  thus
  T with-replacement (p,T1) c= { t : not p is_a_proper_prefix_of t } \/
  { p^s where s is Element of T1 : s = s }
  proof
    let x;
    assume
A2: x in T with-replacement (p,T1);
    then reconsider q = x as FinSequence of NAT by Th19;
A3: (ex r st r in T1 & q = p^r) implies
    x in { p^s where s is Element of T1 : s = s };
 q in T & not p is_a_proper_prefix_of q implies
    x in { t : not p is_a_proper_prefix_of t };
    hence thesis by A1,A2,A3,Def9,XBOOLE_0:def 3;
  end;
  let x such that
A4: x in { t : not p is_a_proper_prefix_of t } \/
  { p^s where s is Element of T1 : s = s };
A5: now
    assume x in { p^s where s is Element of T1 : s = s };
then  ex s being Element of T1 st x = p^s & s = s;
    hence thesis by A1,Def9;
  end;
 now
    assume x in { t : not p is_a_proper_prefix_of t };
then  ex t st x = t & not p is_a_proper_prefix_of t;
    hence thesis by A1,Def9;
  end;
  hence thesis by A4,A5,XBOOLE_0:def 3;
end;

theorem
  p in T implies T1 = T with-replacement (p,T1)|p
proof
  assume
A1: p in T;
then A2: p in T with-replacement (p,T1) by Def9;
  thus T1 c= T with-replacement (p,T1)|p
  proof
    let x;
    assume
A3: x in T1;
    then reconsider q = x as FinSequence of NAT by Th19;
 p^q in T with-replacement (p,T1) by A1,A3,Def9;
    hence thesis by A2,Def6;
  end;
  let x;
  assume
A4: x in T with-replacement (p,T1)|p;
  then reconsider q = x as FinSequence of NAT by Th19;
A5: p^q in T with-replacement (p,T1) by A2,A4,Def6;
A6: now
    assume that
    p^q in T and
A7: not p is_a_proper_prefix_of p^q;
 p is_a_prefix_of p^q by Th1;
then  p^q = p by A7,XBOOLE_0:def 8
      .= p^{} by FINSEQ_1:34;
then  q = {} by FINSEQ_1:33;
    hence q in T1 by Th22;
  end;
 (ex r st r in T1 & p^q = p^r) implies q in T1 by FINSEQ_1:33;
  hence thesis by A1,A5,A6,Def9;
end;

registration
  let T be finite Tree, t be Element of T;
  let T1 be finite Tree;
  cluster T with-replacement (t,T1) -> finite;
  coherence
  proof
A1: { s where s is Element of T : not t is_a_proper_prefix_of s } c= T
    proof
      let x;
      assume
   x in { s where s is Element of T : not t is_a_proper_prefix_of s };
then
   ex s being Element of T st x = s & not t is_a_proper_prefix_of s;
      hence thesis;
    end;
 T1,{ t^s where s is Element of T1 : s = s } are_equipotent
    proof
      defpred P[set,set] means ex q st $1 = q & $2 = t^q;
A2:   x in T1 implies ex y st P[x,y]
      proof
        assume x in T1;
        then reconsider q = x as FinSequence of NAT by Th19;
     t^q = t^q;
        hence thesis;
      end;
      consider f such that
A3:   dom f = T1 & for x st x in T1 holds P[x,f.x] from CLASSES1:sch 1(A2);
      take f;
      thus f is one-to-one
      proof
        let x,y;
        assume that
A4:     x in dom f & y in dom f and
A5:    f.x = f.y;
            (
 ex q st x = q & f.x = t^q)& ex r st y = r & f.y = t^r by A3,A4;
        hence thesis by A5,FINSEQ_1:33;
      end;
      thus dom f = T1 by A3;
      thus rng f c= { t^s where s is Element of T1 : s = s }
      proof
        let x;
        assume x in rng f;
        then consider y such that
A6:    y in dom f and
A7:    x = f.y by FUNCT_1:def 3;
        consider q such that
A8:    y = q and
A9:    f.y = t^q by A3,A6;
        reconsider q as Element of T1 by A3,A6,A8;
    x = t^q by A7,A9;
        hence thesis;
      end;
      let x;
      assume x in { t^s where s is Element of T1 : s = s };
      then consider s being Element of T1 such that
A10:  x = t^s and s = s;
  P[s,f.s] by A3;
      hence thesis by A3,A10,FUNCT_1:def 3;
    end;
then  { t^s where s is Element of T1 : s = s } is finite by CARD_1:38;
then
 { v where v is Element of T : not t is_a_proper_prefix_of v } \/ { t^s
    where s is Element of T1 : s = s } is finite by A1;
    hence thesis by Th32;
  end;
end;

reserve w for FinSequence;

theorem Th34:
  for p being FinSequence holds ProperPrefixes p,dom p are_equipotent
proof
  let p be FinSequence;
  defpred P[set,set] means ex w st $1 = w & $2 = (len w)+1;
A1: for x st x in ProperPrefixes p ex y st P[x,y]
  proof
    let x;
    assume x in ProperPrefixes p;
    then consider q being FinSequence such that
A2: x = q and q is_a_proper_prefix_of p by Def2;
    reconsider y = (len q)+1 as set;
    take y,q;
    thus thesis by A2;
  end;
  consider f such that
A3: dom f = ProperPrefixes p and
A4: for x st x in ProperPrefixes p holds P[x,f.x] from CLASSES1:sch 1(A1);
  take f;
  thus f is one-to-one
  proof
    let x,y;
    assume that
A5: x in dom f & y in dom f and
A6: f.x = f.y;
     (
ex q being FinSequence st x = q & f.x = (len q)+1 )& ex r being FinSequence
    st y = r & f.x = (len r)+1 by A3,A4,A5,A6;
    hence thesis by A3,A5,Th4,Th18;
  end;
  thus dom f = ProperPrefixes p by A3;
  thus rng f c= dom p
  proof
    let x;
    assume x in rng f;
    then consider y such that
A7: y in dom f and
A8: x = f.y by FUNCT_1:def 3;
    consider q being FinSequence such that
A9: y = q and
A10: x = (len q)+1 by A3,A4,A7,A8;
 len q < len p by A3,A7,A9,Th13;
then  0+1 <= (len q)+1 & (len q)+1 <= len p by NAT_1:13;
then  x in Seg len p by A10,FINSEQ_1:1;
    hence thesis by FINSEQ_1:def 3;
  end;
  let x;
  assume
A11: x in dom p;
then A12: x in Seg len p by FINSEQ_1:def 3;
  reconsider n = x as Element of NAT by A11;
A13: 1 <= n by A12,FINSEQ_1:1;
A14: n <= len p by A12,FINSEQ_1:1;
  consider m be Nat such that
A15: n = 1+m by A13,NAT_1:10;
  reconsider m as Element of NAT by ORDINAL1:def 12;
  reconsider q = p|Seg m as FinSequence by FINSEQ_1:15;
A16: m <= len p by A14,A15,NAT_1:13;
A17: m <> len p by A14,A15,NAT_1:13;
A18: len q = m by A16,FINSEQ_1:17;
A19: q is_a_prefix_of p by Def1;
   len q = m by A16,FINSEQ_1:17;
then  q is_a_proper_prefix_of p by A17,A19,XBOOLE_0:def 8;
then A20: q in ProperPrefixes p by Th12;
then  ex r being FinSequence st q = r & f.q = (len r)+1 by A4;
  hence thesis by A3,A15,A18,A20,FUNCT_1:def 3;
end;

registration
  let p be FinSequence;
  cluster ProperPrefixes p -> finite;
  coherence
  proof
 ProperPrefixes p,dom p are_equipotent by Th34;
    hence thesis by CARD_1:38;
  end;
end;

theorem
  for p being FinSequence holds card ProperPrefixes p = len p
proof
  let p be FinSequence;
A1: dom p = Seg len p by FINSEQ_1:def 3;
A2: ProperPrefixes p,dom p are_equipotent by Th34;
 card dom p = card len p by A1,FINSEQ_1:55;
  hence thesis by A2,CARD_1:5;
end;





definition
  let IT be set;
  attr IT is AntiChain_of_Prefixes-like means
  :Def10:
  (for x st x in IT holds x is FinSequence) &
  for p1,p2 st p1 in IT & p2 in IT & p1 <> p2 holds
  not p1,p2 are_c=-comparable;
end;

registration
  cluster AntiChain_of_Prefixes-like for set;
  existence
  proof
    take {};
    thus for x st x in {} holds x is FinSequence;
    let p1,p2;
    thus thesis;
  end;
end;

definition
  mode AntiChain_of_Prefixes is AntiChain_of_Prefixes-like set;
end;

theorem Th36:
  { w } is AntiChain_of_Prefixes-like
proof
  thus for x st x in { w } holds x is FinSequence by TARSKI:def 1;
  let p1,p2;
  assume that
A1: p1 in { w } and
A2: p2 in { w };
 p1 = w by A1,TARSKI:def 1;
  hence thesis by A2,TARSKI:def 1;
end;

theorem Th37:
  not p1,p2 are_c=-comparable implies { p1,p2 } is AntiChain_of_Prefixes-like
proof
  assume
A1: not p1,p2 are_c=-comparable;
  thus for x st x in { p1,p2 } holds x is FinSequence by TARSKI:def 2;
  let q1,q2 be FinSequence;
  assume that
A2: q1 in { p1,p2 } and
A3: q2 in { p1,p2 };
A4: q1 = p1 or q1 = p2 by A2,TARSKI:def 2;
 q2 = p1 or q2 = p2 by A3,TARSKI:def 2;
  hence thesis by A1,A4;
end;

definition
  let T;
  mode AntiChain_of_Prefixes of T -> AntiChain_of_Prefixes means
  :Def11:
  it c= T;
  existence
  proof set t = the Element of T;
    reconsider S = { t } as AntiChain_of_Prefixes by Th36;
    take S;
    let x;
    assume x in S;
    then x = t by TARSKI:def 1;
    hence thesis;
  end;
end;

reserve t1,t2 for Element of T;

theorem Th38:
  {} is AntiChain_of_Prefixes of T & { {} } is AntiChain_of_Prefixes of T
proof
 {} is AntiChain_of_Prefixes-like
  proof
    thus for x st x in {} holds x is FinSequence;
    let p1,p2;
    thus thesis;
  end;
  then reconsider S = {} as AntiChain_of_Prefixes;
 S c= T by XBOOLE_1:2;
  hence {} is AntiChain_of_Prefixes of T by Def11;
  reconsider S = D as AntiChain_of_Prefixes by Th36;
 S is AntiChain_of_Prefixes of T
  proof
    let x;
    assume x in S;
then  x = {} by TARSKI:def 1;
    hence thesis by Th22;
  end;
  hence thesis;
end;

theorem
  { t } is AntiChain_of_Prefixes of T
proof reconsider S = { t } as AntiChain_of_Prefixes by Th36;
 S is AntiChain_of_Prefixes of T
  proof
    let x;
    assume x in S;
then  x = t by TARSKI:def 1;
    hence thesis;
  end;
  hence thesis;
end;

theorem
  not t1,t2 are_c=-comparable implies { t1,t2 } is AntiChain_of_Prefixes of T
proof
  assume not t1,t2 are_c=-comparable;
  then reconsider A = { t1,t2 } as AntiChain_of_Prefixes by Th37;
 A is AntiChain_of_Prefixes of T
  proof
    let x;
    assume x in A;
then  x = t1 or x = t2 by TARSKI:def 2;
    hence thesis;
  end;
  hence thesis;
end;

registration
  let T be finite Tree;
  cluster -> finite for AntiChain_of_Prefixes of T;
  coherence
  proof
    let X be AntiChain_of_Prefixes of T;
 X c= T by Def11;
    hence thesis;
  end;
end;

definition
  let T be finite Tree;
  func height T -> Element of NAT means
  :Def12:
  (ex p st p in T & len p = it) & for p st p in T holds len p <= it;
  existence
  proof
    consider n be Nat such that
A1: T,Seg n are_equipotent by FINSEQ_1:56;
    defpred X[Nat] means for p st p in T holds len p <= $1;
A2: ex n be Nat st X[n]
    proof
   now
        given p such that
A3:     p in T and
A4:     not len p <= n;
A5:     ProperPrefixes p c= T by A3,Def3;
A6:     ProperPrefixes p,dom p are_equipotent by Th34;
A7:     card ProperPrefixes p c= card T by A5,CARD_1:11;
A8:     card ProperPrefixes p = card dom p by A6,CARD_1:5;
        A9:    card
 T = card Seg n & dom p = Seg len p by A1,CARD_1:5,FINSEQ_1:def 3;
    Seg n c= Seg len p by A4,FINSEQ_1:5;
then A10:    card Seg n c= card Seg len p by CARD_1:11;
            card
 Seg n = card n & card Seg len p = card len p by FINSEQ_1:55;
then     card n = card len p by A7,A8,A9,A10,XBOOLE_0:def 10;
        hence contradiction by A4,CARD_1:40;
      end;
      then consider n be Nat such that
A11:  X[n];
      reconsider n as Element of NAT by ORDINAL1:def 12;
      take n;
      thus thesis by A11;
    end;
    consider n being Nat such that
A12: X[n] and
A13: for m be Nat st X[m] holds n <= m from NAT_1:sch 5(A2);
    set x = the Element of T;
    reconsider n as Element of NAT by ORDINAL1:def 12;
    take n;
    thus ex p st p in T & len p = n
    proof
      assume
A14:  for p st p in T holds len p <> n;
      reconsider x as FinSequence of NAT;
  len x <= n by A12;
then   len x < n by A14,XXREAL_0:1;
      then consider k be Nat such that
A15:  n = k+1 by NAT_1:6;
      reconsider k as Element of NAT by ORDINAL1:def 12;
  for p st p in T holds len p <= k
      proof
        let p;
        assume
A16:    p in T;
then     len p <= n by A12;
then     len p < k+1 by A14,A15,A16,XXREAL_0:1;
        hence thesis by NAT_1:13;
      end;
then   n <= k by A13;
      hence contradiction by A15,NAT_1:13;
    end;
    let p;
    assume p in T;
    hence thesis by A12;
  end;
  uniqueness
  proof
    let l1,l2 be Element of NAT;
    given p1 being FinSequence of NAT such that
A17: p1 in T & len p1 = l1;
    assume
A18: for p st p in T holds len p <= l1;
    given p2 being FinSequence of NAT such that
A19: p2 in T & len p2 = l2;
    assume for p st p in T holds len p <= l2;
then A20: l1 <= l2 by A17;
 l2 <= l1 by A18,A19;
    hence thesis by A20,XXREAL_0:1;
  end;
  func width T -> Element of NAT means
  :Def13:
  ex X being AntiChain_of_Prefixes of T st it = card X &
  for Y being AntiChain_of_Prefixes of T holds card Y <= card X;
  existence
  proof
    defpred X[Nat] means ex X being finite set st $1 = card X & X c= T &
    (for p,q st p in X & q in X & p <> q holds not p,q are_c=-comparable);
     0 = card {} & for p,q st p in {} & q in {} & p <> q holds not p,q
    are_c=-comparable;
then A21: ex n be Nat st X[n] by XBOOLE_1:2;
A22: for n be Nat st X[n] holds n <= card T
    proof
      let n be Nat;
      given X being finite set such that
A23:  n = card X & X c= T and
      for p,q st p in X & q in X & p <> q holds not p,q are_c=-comparable;
A24:  card X c= card T & card X = card n by A23,CARD_1:11;
  card T = card card T;
      hence thesis by A24,NAT_1:40;
    end;
    consider n being Nat such that
A25: X[n] and
A26: for m be Nat st X[m] holds m <= n from NAT_1:sch 6(A22,A21);
    consider X being finite set such that
A27: n = card X and
A28: X c= T and
A29: for p,q st p in X & q in X & p <> q holds not p,q are_c=-comparable
    by A25;
 X is AntiChain_of_Prefixes-like
    proof
      thus for x st x in X holds x is FinSequence
      proof
        let x;
        assume
A30:    x in X;
        T c= NAT* by Def3;
        hence thesis by A30,A28;
      end;
      let p1,p2;
      assume
A31:  p1 in X & p2 in X;
      then reconsider q1 = p1, q2 = p2 as Element of T by A28;
  p1 = q1 & p2 = q2;
      hence thesis by A29,A31;
    end;
    then reconsider X as AntiChain_of_Prefixes;
    reconsider X as AntiChain_of_Prefixes of T by A28,Def11;
    reconsider n as Element of NAT by ORDINAL1:def 12;
    take n,X;
    thus n = card X by A27;
    let Y be AntiChain_of_Prefixes of T;
     Y
 c= T & for p,q st p in Y & q in Y & p <> q holds not p,q are_c=-comparable
    by Def10,Def11;
    hence thesis by A26,A27;
  end;
  uniqueness
  proof
    let n1,n2 be Element of NAT;
    given X1 being AntiChain_of_Prefixes of T such that
A32: n1 = card X1 and
A33: for Y being AntiChain_of_Prefixes of T holds card Y <= card X1;
    given X2 being AntiChain_of_Prefixes of T such that
A34: n2 = card X2 and
A35: for Y being AntiChain_of_Prefixes of T holds card Y <= card X2;
A36: card X1 <= card X2 by A35;
 card X2 <= card X1 by A33;
    hence thesis by A32,A34,A36,XXREAL_0:1;
  end;
end;

theorem
  1 <= width fT
proof
   (
 ex X being AntiChain_of_Prefixes of fT st width fT = card X & for Y being
  AntiChain_of_Prefixes of fT holds card Y <= card X)&
  D is AntiChain_of_Prefixes of fT by Def13,Th38;
  then card D <= width fT;
  hence thesis by CARD_1:30;
end;

theorem
  height elementary_tree 0 = 0
proof
 now
    thus ex p st p in elementary_tree 0 & len p = 0
    proof
      take <*> NAT;
      thus thesis by Th29,TARSKI:def 1;
    end;
    let p;
    assume p in elementary_tree 0;
    then  p = {} by Th29,TARSKI:def 1;
    hence len p <= 0;
  end;
  hence thesis by Def12;
end;

theorem
  height fT = 0 implies fT = elementary_tree 0
proof
  assume
A1: height fT = 0;
  thus fT c= elementary_tree 0
  proof
    let x;
    assume x in fT;
    then reconsider t = x as Element of fT;
 len t = 0 by A1,Def12;
then  x = {};
    hence thesis by Th22;
  end;
  let x;
  assume x in elementary_tree 0;
then  x = {} by Th29,TARSKI:def 1;
  hence thesis by Th22;
end;

theorem
  height elementary_tree(n+1) = 1
proof
  set T = elementary_tree(n+1);
 now
    thus ex p st p in T & len p = 1
    proof
      take p = <*0*>;
      thus p in T by Th28;
      thus thesis by FINSEQ_1:40;
    end;
    let p such that
A1: p in T;
A2: p in D implies p = {} by TARSKI:def 1;
 now
      assume p in { <*k*> : k < n+1 };
then    ex k st p = <*k*> & k < n+1;
      hence len p = 1 by FINSEQ_1:40;
    end;
    hence len p <= 1 by A1,A2,XBOOLE_0:def 3;
  end;
  hence thesis by Def12;
end;

theorem
  width elementary_tree 0 = 1
proof
  set T = elementary_tree 0;
 now reconsider X = D as AntiChain_of_Prefixes of T by Th38;
    take X;
    thus 1 = card X by CARD_1:30;
    let Y be AntiChain_of_Prefixes of T;
 Y c= X by Def11,Th29;
    hence card Y <= card X by NAT_1:43;
  end;
  hence thesis by Def13;
end;

theorem
  width elementary_tree(n+1) = n+1
proof
  set T = elementary_tree(n+1);
 now
 { <*k*> : k < n+1 } is AntiChain_of_Prefixes-like
    proof
      thus x in { <*k*> : k < n+1 } implies x is FinSequence
      proof
        assume x in { <*k*> : k < n+1 };
then      ex k st x = <*k*> & k < n+1;
        hence thesis;
      end;
      let p1,p2;
      assume p1 in { <*k*> : k < n+1 } & p2 in { <*m*> : m < n+1 };
      then    (
 ex k st p1 = <*k*> & k < n+1)& ex m st p2 = <*m*> & m < n+1;
      hence thesis by Th5;
    end;
    then reconsider X = { <*k*> : k < n+1 } as AntiChain_of_Prefixes;
 X c= T by XBOOLE_1:7;
    then reconsider X as AntiChain_of_Prefixes of T by Def11;
    take X;
 X,Seg(n+1) are_equipotent
    proof
      defpred P[set,set] means ex n st $1 = <*n*> & $2 = n+1;
A1:   x in X & P[x,y] & P[x,z] implies y = z
      proof
        assume x in X;
        given n1 being Element of NAT such that
A2:    x = <*n1*> & y = n1+1;
        given n2 being Element of NAT such that
A3:    x = <*n2*> & z = n2+1;
    <*n1*>.1 = n1 by FINSEQ_1:def 8;
        hence thesis by A2,A3,FINSEQ_1:def 8;
      end;
A4:  x in X implies ex y st P[x,y]
      proof
        assume x in X;
        then consider k such that
A5:    x = <*k*> and k < n+1;
        reconsider y = k+1 as set;
        take y;
        thus thesis by A5;
      end;
      consider f such that
A6:  dom f = X & for x st x in X holds P[x,f.x] from CLASSES1:sch 1(A4);
      take f;
      thus f is one-to-one
      proof
        let x,y;
        assume that
A7:    x in dom f & y in dom f and
A8:    f.x = f.y;
            (
 ex k1 being Element of NAT st x = <*k1*> & f.x = k1+1)& ex k2 being Element
        of NAT st y = <*k2*> & f.y = k2+1 by A6,A7;
        hence thesis by A8;
      end;
      thus dom f = X by A6;
      thus rng f c= Seg(n+1)
      proof
        let x;
        assume x in rng f;
        then consider y such that
A9:    y in dom f and
A10:    x = f.y by FUNCT_1:def 3;
        consider k such that
A11:    y = <*k*> and
A12:    x = k+1 by A6,A9,A10;
        consider m such that
A13:    y = <*m*> & m < n+1 by A6,A9;
    <*k*>.1 = k & <*m*>.1 = m by FINSEQ_1:def 8;
then     0+1 <= k+1 & k+1 <= n+1 by A11,A13,NAT_1:13;
        hence thesis by A12,FINSEQ_1:1;
      end;
      let x;
      assume
A14:  x in Seg(n+1);
      then reconsider k = x as Element of NAT;
A15:  1 <= k by A14,FINSEQ_1:1;
A16:  k <= n+1 by A14,FINSEQ_1:1;
      consider m be Nat such that
A17:  k = 1+m by A15,NAT_1:10;
      reconsider m as Element of NAT by ORDINAL1:def 12;
  m < n+1 by A16,A17,NAT_1:13;
then A18:  <*m*> in X;
then   P[<*m*>,f.<*m*>] by A6;
then   x = f.<*m*> by A1,A17,A18;
      hence thesis by A6,A18,FUNCT_1:def 3;
    end;
then A19: card Seg(n+1) = card X by CARD_1:5;
    hence n+1 = card X by FINSEQ_1:57;
    let Y be AntiChain_of_Prefixes of T;
A20: Y c= T by Def11;
A21: {} in Y implies Y = D
    proof
      assume that
A22:  {} in Y and
A23:  Y <> D;
      consider x such that
A24:  not (x in Y iff x in D) by A23,TARSKI:1;
A25:  {} <> x by A22,A24,TARSKI:def 1;
      reconsider x as FinSequence of NAT by A20,A24,Th19;
  {} is_a_prefix_of x by XBOOLE_1:2;
then   {},x are_c=-comparable by XBOOLE_0:def 9;
      hence contradiction by A22,A24,A25,Def10,TARSKI:def 1;
    end;
A26: card D = 1 & 1 <= 1+n by CARD_1:30,NAT_1:11;
 now
      assume
A27:  not {} in Y;
  Y c= X
      proof
        let x;
        assume
A28:    x in Y;
then     x in { <*k*> : k < n+1 } or x in D by A20,XBOOLE_0:def 3;
        hence thesis by A27,A28,TARSKI:def 1;
      end;
      hence card Y <= card X by NAT_1:43;
    end;
    hence card Y <= card X by A19,A21,A26,FINSEQ_1:57;
  end;
  hence thesis by Def13;
end;

theorem
  for t being Element of fT holds height(fT|t) <= height fT
proof
  let t be Element of fT;
  consider p such that
A1: p in fT|t and
A2: len p = height(fT|t) by Def12;
 t^p in fT by A1,Def6;
then A3: len(t^p) <= height fT by Def12;
   len(t^p) = len t + len p & len p <= len p + len t by FINSEQ_1:22,NAT_1:11
  ;
  hence thesis by A2,A3,XXREAL_0:2;
end;

theorem Th48:
  for t being Element of fT st t <> {} holds height(fT|t) < height fT
proof
  let t be Element of fT;
  assume t <> {};
then A1: 0+1 <= len t by NAT_1:13;
  consider p such that
A2: p in fT|t and
A3: len p = height(fT|t) by Def12;
 t^p in fT by A2,Def6;
then A4: len(t^p) <= height fT by Def12;
   len(t^p) = len t + len p & len p + 1 <= len t + len p by A1,FINSEQ_1:22
,XREAL_1:7;
then  height(fT|t)+1 <= height fT by A3,A4,XXREAL_0:2;
  hence thesis by NAT_1:13;
end;

scheme TreeInd { P[Tree] }: for fT holds P[fT]
provided
A1: for fT st for n st <*n*> in fT holds P[fT|<*n*>] holds P[fT]
proof
  defpred X[set] means for fT holds height fT = $1 implies P[fT];
A2: for n being Nat st for k being Nat st k < n holds X[k] holds X[n]
  proof
    let n be Nat such that
A3: for k being Nat st k < n for fT st height fT = k holds P[fT];
    let fT such that
A4: height fT = n;
 now
      let k;
      assume <*k*> in fT;
      then reconsider k9 = <*k*> as Element of fT;
   height(fT|k9) < height fT by Th48;
      hence P[fT|<*k*>] by A3,A4;
    end;
    hence thesis by A1;
  end;
A5: for n be Nat holds X[n] from NAT_1:sch 4(A2);
  let fT;
 height fT = height fT;
  hence thesis by A5;
end;

begin 


reserve s,t for FinSequence;

theorem
  w^t is_a_proper_prefix_of w^s implies t is_a_proper_prefix_of s
proof
  assume
A1: w^t is_a_proper_prefix_of w^s;
then  w^t is_a_prefix_of w^s by XBOOLE_0:def 8;
  then consider a be FinSequence such that
A2: w^s = w^t^a by Th1;
 w^t^a = w^(t^a) by FINSEQ_1:32;
then  s= t^a by A2,FINSEQ_1:33;
then  t is_a_prefix_of s by Th1;
  hence thesis by A1,XBOOLE_0:def 8;
end;

theorem
  n <> m implies not <*n*> is_a_prefix_of <*m*>^s
proof
  assume
A1: n <> m;
  assume <*n*> is_a_prefix_of <*m*>^s;
then A2: ex a be FinSequence st <*m*>^s = <*n*>^a by Th1;
 m = (<*m*>^s).1 by FINSEQ_1:41
    .= n by A2,FINSEQ_1:41;
  hence contradiction by A1;
end;

theorem
  elementary_tree 1 = {{},<*0*>}
proof
 now
    let x;
    thus x in {{},<*0*>} implies x in { <*n*> : n < 1 } or x in D
    proof
      assume x in {{},<*0*>};
then    x = {} or x = <*0*> by TARSKI:def 2;
      hence thesis by TARSKI:def 1;
    end;
    assume
A1: x in { <*n*> : n < 1 } or x in D;
 now per cases by A1;
      suppose
     x in { <*n*> : n < 1 };
        then consider n such that
A2:     x = <*n*> and
A3:     n < 1;
     n = 0 by A3,NAT_1:25;
        hence x in {{},<*0*>} by A2,TARSKI:def 2;
      end;
      suppose x in D;
then     x = {} by TARSKI:def 1;
        hence x in {{},<*0*>} by TARSKI:def 2;
      end;
    end;
    hence x in {{},<*0*>};
  end;
  hence thesis by XBOOLE_0:def 3;
end;

theorem
  not <*n*> is_a_proper_prefix_of <*m*>
proof
  assume
A1: not thesis;
then  <*n*> is_a_prefix_of <*m*> by XBOOLE_0:def 8;
  hence contradiction by A1,Th3;
end;

theorem
  elementary_tree 2 = {{},<*0*>,<*1*>}
proof
 now
    let x;
    thus
    x in {{},<*0*>,<*1*>} implies x in { <*n*> : n < 2 } or x in D
    proof
      assume x in {{},<*0*>,<*1*>};
then    x = {} or x = <*0*> or x = <*1*> by ENUMSET1:def 1;
      hence thesis by TARSKI:def 1;
    end;
    assume
A1: x in { <*n*> : n < 2 } or x in D;
 now per cases by A1;
      suppose
     x in { <*n*> : n < 2 };
        then consider n such that
A2:     x = <*n*> and
A3:     n < 2;
     n = 0 or n = 1 by A3,NAT_1:26;
        hence x in {{},<*0*>,<*1*>} by A2,ENUMSET1:def 1;
      end;
      suppose x in D;
then     x = {} by TARSKI:def 1;
        hence x in {{},<*0*>,<*1*>} by ENUMSET1:def 1;
      end;
    end;
    hence x in {{},<*0*>,<*1*>};
  end;
  hence thesis by XBOOLE_0:def 3;
end;



theorem
  for T being Tree, t being Element of T holds
  t in Leaves T iff not t^<*0*> in T
proof
  let T be Tree, t be Element of T;
  hereby
    assume
A1: t in Leaves T;
 t is_a_proper_prefix_of t^<*0*> by Th7;
    hence not t^<*0*> in T by A1,Def5;
  end;
  assume that
A2: not t^<*0*> in T and
A3: not t in Leaves T;
  consider q being FinSequence of NAT such that
A4: q in T and
A5: t is_a_proper_prefix_of q by A3,Def5;
 t is_a_prefix_of q by A5,XBOOLE_0:def 8;
  then consider r being FinSequence such that
A6: q = t^r by Th1;
  reconsider r as FinSequence of NAT by A6,FINSEQ_1:36;
 len q = len t+len r by A6,FINSEQ_1:22;
then len r <> 0 by A5,Th6;
then r <> {};
  then consider p being FinSequence of NAT,
   x being Element of NAT such that
A7: r = <*x*>^p by FINSEQ_2:130;
  reconsider x as Element of NAT;
 q = t^<*x*>^p by A6,A7,FINSEQ_1:32;
then  t^<*x*> in T by A4,Th21;
  hence contradiction by A2,Def3;
end;

theorem
  for T being Tree, t being Element of T holds
  t in Leaves T iff not ex n being Element of NAT st t^<*n*> in T
proof
  let T be Tree, t be Element of T;
  hereby
    assume
A1: t in Leaves T;
    given n being Element of NAT such that
A2: t^<*n*> in T;
A3: not t is_a_proper_prefix_of t^<*n*> by A1,A2,Def5;
 t is_a_prefix_of t^<*n*> by Th1;
then A4: t = t^<*n*> by A3,XBOOLE_0:def 8;
 t = t^{} by FINSEQ_1:34;
    hence contradiction by A4,FINSEQ_1:33;
  end;
  assume that
A5: not(ex n being Element of NAT st t^<*n*> in T) and
A6: not t in Leaves T;
  consider q being FinSequence of NAT such that
A7: q in T and
A8: t is_a_proper_prefix_of q by A6,Def5;
 t is_a_prefix_of q by A8,XBOOLE_0:def 8;
  then consider r being FinSequence such that
A9: q = t^r by Th1;
  reconsider r as FinSequence of NAT by A9,FINSEQ_1:36;
 len q = len t+len r by A9,FINSEQ_1:22;
then  len r <> 0 by A8,Th6;
then  r <> {};
  then consider p being FinSequence of NAT,
  x being Element of NAT such that
A10: r = <*x*>^p by FINSEQ_2:130;
  reconsider x as Element of NAT;
  q = t^<*x*>^p by A9,A10,FINSEQ_1:32;
  hence contradiction by A5,A7,Th21;
end;

definition
  func TrivialInfiniteTree equals
  { k |-> 0 where k is Element of NAT: not contradiction };
  coherence;
end;

registration
  cluster TrivialInfiniteTree -> non empty Tree-like;
  coherence
  proof
    set X = TrivialInfiniteTree;
    0 |-> 0 in X;
    hence X is non empty;
    thus X c= NAT*
    proof
      let x be set;
      assume x in X;
      then ex k being Element of NAT st x = k |-> 0;
      hence thesis by FINSEQ_1:def 11;
    end;
    thus for p being FinSequence of NAT st p in X holds ProperPrefixes p c= X
    proof
      let p be FinSequence of NAT;
      assume p in X;
      then consider m being Element of NAT such that
A1:   p = m |-> 0;
      let x be set;
      assume
A2:   x in ProperPrefixes p;
      then reconsider x as FinSequence by Th11;
A3:   for k being Nat st 1 <= k & k <= len x holds x.k = (len x |-> 0).k
      proof
        x is_a_proper_prefix_of p by A2,Th12;
        then
A4:     x c= p by XBOOLE_0:def 8;
        let k be Nat;
        assume 1 <= k & k <= len x;
        then
A5:     k in dom x by FINSEQ_3:25;
        thus (len x |-> 0).k = 0
          .= p.k by A1
          .= x.k by A5,A4,GRFUNC_1:2;
      end;
      len x = len (len x |-> 0) by CARD_1:def 7;
      then x = len x |-> 0 by A3,FINSEQ_1:14;
      hence thesis;
    end;
    let p be FinSequence of NAT, m, n be Element of NAT;
    assume p^<*m*> in X;
    then consider k being Element of NAT such that
A7: p^<*m*> = k |-> 0;
    assume
A8: n <= m;
    len (p^<*m*>) = len p + 1 by FINSEQ_2:16;
    then (p^<*m*>).len (p^<*m*>) = m by FINSEQ_1:42;
    then
A11: m = 0 by A7;
A12: for z being Nat st 1 <= z & z <= len (p^<*n*>) holds (len (p^<*n*>)
    |-> 0).z = (p^<*n*>).z
    proof
      let z be Nat;
      assume 1 <= z & z <= len (p^<*n*>);
      thus (len (p^<*n*>) |-> 0).z = 0
        .= (p^<*m*>).z by A7
        .= (p^<*n*>).z by A8,A11;
    end;
    len (p^<*n*>) = len (len (p^<*n*>) |-> 0) by CARD_1:def 7;
    then len (p^<*n*>) |-> 0 = p^<*n*> by A12,FINSEQ_1:14;
    hence thesis;
  end;
end;

theorem Th56:
  NAT,TrivialInfiniteTree are_equipotent
proof
  defpred P[Element of NAT,set] means $2 = $1 |-> 0;
A1: for x being Element of NAT ex y being Element of TrivialInfiniteTree st
  P[x,y]
  proof
    let x be Element of NAT;
    x |-> 0 in TrivialInfiniteTree;
    then reconsider y = x |-> 0 as Element of TrivialInfiniteTree;
    take y;
    thus thesis;
  end;
  consider f being Function of NAT,TrivialInfiniteTree such that
A2: for x being Element of NAT holds P[x,f.x] from FUNCT_2:sch 3(A1);
  take f;
  thus f is one-to-one
  proof
    let x,y be set;
    assume
A3: x in dom f & y in dom f;
    assume
A4: f.x = f.y;
    reconsider x, y as Element of NAT by A3,FUNCT_2:def 1;
    ( P[x,f.x])& P[y,f.y] by A2;
    hence thesis by A4,FINSEQ_2:143;
  end;
  thus
A5: dom f = NAT by FUNCT_2:def 1;
  thus rng f c= TrivialInfiniteTree by RELAT_1:def 19;
  let a be set;
  assume a in TrivialInfiniteTree;
  then consider k being Element of NAT such that
A6: a = k |-> 0;
  f.k = a by A2,A6;
  hence thesis by A5,FUNCT_1:def 3;
end;

registration
  cluster TrivialInfiniteTree -> infinite;
  coherence by Th56,CARD_1:38;
end;














begin

reserve f for Function;
reserve p,q for FinSequence;
reserve A,B,C,x,x1,x2,y,z for set;
reserve k,l,m,n for Nat;
reserve a for Nat;

definition
  let f,x;
  pred f is_one-to-one_at x means
  :Def1:
  f"Im(f,x) = {x};
end;

theorem Th1:
  f is_one-to-one_at x implies x in dom f
proof
  assume f is_one-to-one_at x;
  then f"Im(f,x) = {x} by Def1;
  then x in f"Im(f,x) by TARSKI:def 1;
  hence thesis by FUNCT_1:def 7;
end;

theorem Th2:
  f is_one-to-one_at x iff x in dom f & f " {f.x} = {x}
proof
  thus f is_one-to-one_at x implies x in dom f & f " {f.x} = {x}
  proof
    assume
A1: f is_one-to-one_at x;
    hence
A2: x in dom f by Th1;
    f"Im(f,x) = {x} by A1,Def1;
    hence thesis by A2,FUNCT_1:59;
  end;
  assume x in dom f & f " {f.x} = {x};
  hence f"Im(f,x) = {x} by FUNCT_1:59;
end;

theorem Th3:
  f is_one-to-one_at x iff x in dom f & for z st z in dom f & x <>
  z holds f.x <> f.z
proof
  thus f is_one-to-one_at x implies x in dom f & for z st z in dom f & x <> z
  holds f.x <> f.z
  proof
    assume
A1: f is_one-to-one_at x;
    hence x in dom f by Th1;
    let z;
    assume that
A2: z in dom f and
A3: x <> z and
A4: f.x = f.z;
    f.x in {f.x} by TARSKI:def 1;
    then z in f " {f.x} by A2,A4,FUNCT_1:def 7;
    then z in {x} by A1,Th2;
    hence thesis by A3,TARSKI:def 1;
  end;
  assume that
A5: x in dom f and
A6: for z st z in dom f & x <> z holds f.x <> f.z and
A7: not f is_one-to-one_at x;
  f " {f.x} <> {x} by A5,A7,Th2;
  then consider y such that
A8: y in f " {f.x} & not y in {x} or y in {x} & not y in f " {f.x} by TARSKI:1;
  f.x in {f.x} by TARSKI:def 1;
  then
A9: x in f " {f.x} by A5,FUNCT_1:def 7;
  now
    per cases by A8;
    suppose
A10:  y in f " {f.x} & not y in {x};
      then f.y in {f.x} by FUNCT_1:def 7;
      then
A11:  f.y = f.x by TARSKI:def 1;
      y in dom f & x <> y by A10,FUNCT_1:def 7,TARSKI:def 1;
      hence thesis by A6,A11;
    end;
    suppose
      not y in f " {f.x} & y in {x};
      hence thesis by A9,TARSKI:def 1;
    end;
  end;
  hence thesis;
end;

theorem
  (for x st x in dom f holds f is_one-to-one_at x) iff f is one-to-one
proof
  thus (for x st x in dom f holds f is_one-to-one_at x) implies f is one-to-one
  proof
    assume
A1: for x st x in dom f holds f is_one-to-one_at x;
    let x1,x2;
    assume that
A2: x1 in dom f and
A3: x2 in dom f & f.x1 = f.x2;
    f is_one-to-one_at x1 by A1,A2;
    hence thesis by A3,Th3;
  end;
  assume
A4: f is one-to-one;
  let x;
  assume
A5: x in dom f;
  then for z holds z in dom f & x <> z implies f.x <> f.z by A4,FUNCT_1:def 4;
  hence thesis by A5,Th3;
end;

definition
  let R be Relation, y be set;
  pred R just_once_values y means
  :Def2:
  card Coim(R,y) = 1;
end;

theorem Th5:
  f just_once_values y implies y in rng f
proof
  assume f just_once_values y;
  then card Coim(f,y) = 1 by Def2;
  then rng f meets {y} by CARD_1:27,RELAT_1:138;
  then consider x being set such that
A1: x in rng f /\ {y} by XBOOLE_0:4;
  x in {y} by A1,XBOOLE_0:def 4;
  then y = x by TARSKI:def 1;
  hence thesis by A1,XBOOLE_0:def 4;
end;

theorem Th6:
  f just_once_values y iff ex x st {x} = f " {y}
proof
  thus f just_once_values y implies ex x st {x} = f " {y}
  proof
    assume
A1: f just_once_values y;
    then y in rng f by Th5;
    then consider x such that
A2: x in dom f and
A3: f.x = y by FUNCT_1:def 3;
    take x;
    card Coim(f,y) = 1 by A1,Def2;
    then consider z such that
A4: f " {y} = {z} by CARD_2:42;
    f.x in {y} by A3,TARSKI:def 1;
    then x in {z} by A2,A4,FUNCT_1:def 7;
    hence thesis by A4,TARSKI:def 1;
  end;
  assume ex x st {x} = f " {y};
  hence card Coim(f,y) = 1 by CARD_1:30;
end;

theorem Th7:
  f just_once_values y iff ex x st x in dom f & y = f.x & for z st
  z in dom f & z <> x holds f.z <> y
proof
  thus f just_once_values y implies ex x st x in dom f & y = f.x & for z st z
  in dom f & z <> x holds f.z <> y
  proof
    assume
A1: f just_once_values y;
    then
A2: card Coim(f,y) = 1 by Def2;
    y in rng f by A1,Th5;
    then consider x1 such that
A3: x1 in dom f and
A4: f.x1 = y by FUNCT_1:def 3;
    f.x1 in {y} by A4,TARSKI:def 1;
    then
A5: x1 in f " {y} by A3,FUNCT_1:def 7;
    take x1;
    thus x1 in dom f & y = f.x1 by A3,A4;
    let z;
    assume that
A6: z in dom f and
A7: z <> x1 and
A8: f.z = y;
A9: f"{y} is finite by A2;
    f.z in {y} by A8,TARSKI:def 1;
    then z in f " {y} by A6,FUNCT_1:def 7;
    then {z,x1} c= f " {y} by A5,ZFMISC_1:32;
    then card{z,x1} <= 1 by A9,A2,NAT_1:43;
    then 2 <= 1 by A7,CARD_2:57;
    hence thesis;
  end;
  given x such that
A10: x in dom f and
A11: y = f.x and
A12: for z st z in dom f & z <> x holds f.z <> y;
A13: {x} = f " {y}
  proof
    thus {x} c= f " {y}
    proof
      let x1;
      assume x1 in {x};
      then
A14:  x1 = x by TARSKI:def 1;
      f.x in {y} by A11,TARSKI:def 1;
      hence thesis by A10,A14,FUNCT_1:def 7;
    end;
    let x1;
    assume
A15: x1 in f " {y};
    then f.x1 in {y} by FUNCT_1:def 7;
    then
A16: f.x1 = y by TARSKI:def 1;
    x1 in dom f by A15,FUNCT_1:def 7;
    then x1 = x by A12,A16;
    hence thesis by TARSKI:def 1;
  end;
  card Coim(f,y) = 1 by A13,CARD_1:30;
  hence thesis by Def2;
end;

theorem Th8:
  f is one-to-one iff for y st y in rng f holds f just_once_values y
proof
  thus f is one-to-one implies for y st y in rng f holds f just_once_values y
  proof
    assume
A1: f is one-to-one;
    let y;
    assume y in rng f;
    then consider x such that
A2: x in dom f & f.x = y by FUNCT_1:def 3;
    for z holds z in dom f & z <> x implies f.z <> y by A1,A2,FUNCT_1:def 4;
    hence thesis by A2,Th7;
  end;
  assume
A3: for y st y in rng f holds f just_once_values y;
  let x,y;
  assume that
A4: x in dom f and
A5: y in dom f & f.x = f.y;
  f.x in rng f by A4,FUNCT_1:def 3;
  then f just_once_values f.x by A3;
  then consider x1 such that
A6: {x1} = f " {f.x} by Th6;
A7: f.x in {f.x} by TARSKI:def 1;
  then
A8: y in f "{f.x} by A5,FUNCT_1:def 7;
  x in f " {f.x} by A4,A7,FUNCT_1:def 7;
  then x = x1 by A6,TARSKI:def 1;
  hence thesis by A6,A8,TARSKI:def 1;
end;

theorem Th9:
  f is_one-to-one_at x iff x in dom f & f just_once_values f.x
proof
  thus f is_one-to-one_at x implies x in dom f & f just_once_values f.x
  proof
    assume
A1: f is_one-to-one_at x;
    hence x in dom f by Th1;
    {x} = f " {f.x} by A1,Th2;
    hence thesis by Th6;
  end;
  assume that
A2: x in dom f and
A3: f just_once_values f.x;
  consider z such that
A4: f " {f.x} = {z} by A3,Th6;
  f.x in {f.x} by TARSKI:def 1;
  then x in {z} by A2,A4,FUNCT_1:def 7;
  then x = z by TARSKI:def 1;
  hence thesis by A2,A4,Th2;
end;

definition
  let f,y;
  assume
A1: f just_once_values y;
  func f <- y -> set means
  :Def3:
  it in dom f & f.it = y;
  existence
  proof
    y in rng f by A1,Th5;
    then consider x such that
A2: x in dom f & f.x = y by FUNCT_1:def 3;
    take x;
    thus thesis by A2;
  end;
  uniqueness
  proof
    let x1,x2;
    assume that
A3: x1 in dom f & f.x1 = y and
A4: x2 in dom f & f.x2 = y;
    consider x such that
    x in dom f and
    f.x = y and
A5: for z st z in dom f & z <> x holds f.z <> y by A1,Th7;
    x = x1 by A3,A5;
    hence thesis by A4,A5;
  end;
end;

theorem
  f just_once_values y implies Im(f, f<-y) = {y}
proof
  assume
A1: f just_once_values y;
  then f <- y in dom f by Def3;
  hence Im(f,f <- y) = {f.(f <- y)} by FUNCT_1:59
    .= {y} by A1,Def3;
end;

theorem Th11:
  f just_once_values y implies f " {y} = {f <- y}
proof
  assume
A1: f just_once_values y;
  then consider x such that
A2: {x} = f " {y} by Th6;
A3: x in f " {y} by A2,ZFMISC_1:31;
  then f.x in {y} by FUNCT_1:def 7;
  then
A4: f.x = y by TARSKI:def 1;
  x in dom f by A3,FUNCT_1:def 7;
  hence thesis by A1,A2,A4,Def3;
end;

theorem
  f is one-to-one & y in rng f implies (f").y = f <- y
proof
  assume that
A1: f is one-to-one and
A2: y in rng f;
  consider x such that
A3: x in dom f & f.x = y by A2,FUNCT_1:def 3;
  f just_once_values y by A1,A2,Th8;
  then x = f <- y by A3,Def3;
  hence thesis by A1,A3,FUNCT_1:32;
end;

theorem
  f is_one-to-one_at x implies f <- (f.x) = x
proof
  assume f is_one-to-one_at x;
  then x in dom f & f just_once_values f.x by Th9;
  hence thesis by Def3;
end;

theorem
  f just_once_values y implies f is_one-to-one_at f <- y
proof
  assume
A1: f just_once_values y;
A2: now
    let x;
    assume x in dom f & x <> f <- y;
    then f.x <> y by A1,Def3;
    hence f.x <> f.(f <- y) by A1,Def3;
  end;
  f <- y in dom f by A1,Def3;
  hence thesis by A2,Th3;
end;

reserve D for non empty set;
reserve d,d1,d2,d3 for Element of D;

definition
  let D;
  let d1,d2;
  redefine func <* d1,d2 *> -> FinSequence of D;
  coherence by FINSEQ_2:13;
end;

definition
  let D;
  let d1,d2,d3;
  redefine func <* d1,d2,d3 *> -> FinSequence of D;
  coherence by FINSEQ_2:14;
end;

theorem
  for i being Nat for D being set, P being FinSequence of D st 1 <= i &
  i <= len P holds P/.i = P.i
proof
  let i be Nat;
  let D be set, P be FinSequence of D;
  assume 1 <= i & i <= len P;
  then i in dom P by FINSEQ_3:25;
  hence thesis by PARTFUN1:def 6;
end;

theorem
  <* d *>/.1 = d
proof
A1: 1 in {1} by FINSEQ_1:2;
  dom<* d *> = {1} & <* d *>.1 = d by FINSEQ_1:2,def 8;
  hence thesis by A1,PARTFUN1:def 6;
end;

theorem
  <* d1,d2 *>/.1 = d1 & <* d1,d2 *>/.2 = d2
proof
  set s = <* d1,d2 *>;
A1: s.2 = d2 & 1 in {1,2} by FINSEQ_1:2,44;
A2: 2 in {1,2} by FINSEQ_1:2;
  dom s = {1,2} & s.1 = d1 by FINSEQ_1:2,44,89;
  hence thesis by A1,A2,PARTFUN1:def 6;
end;

theorem
  <* d1,d2,d3 *>/.1 = d1 & <* d1,d2,d3 *>/.2 = d2 & <* d1,d2,d3 *>/.3 = d3
proof
  set s = <* d1,d2,d3 *>;
A1: s.2 = d2 & s.3 = d3 by FINSEQ_1:45;
A2: 1 in {1,2,3} & 2 in {1,2,3} by FINSEQ_3:1;
A3: 3 in {1,2,3} by FINSEQ_3:1;
  dom s = {1,2,3} & s.1 = d1 by FINSEQ_1:45,89,FINSEQ_3:1;
  hence thesis by A1,A2,A3,PARTFUN1:def 6;
end;

definition
  let p,x;
  func x..p -> Element of NAT equals
  Sgm(p " {x}).1;
  coherence
  proof
    set q = Sgm(p " {x});
    per cases;
    suppose not 1 in dom q;
      then q.1 = 0 by FUNCT_1:def 2;
      hence thesis;
    end;
    suppose
A1:   1 in dom q;
A2:   rng q c= NAT by FINSEQ_1:def 4;
      q.1 in rng q by A1,FUNCT_1:def 3;
      hence thesis by A2;
    end;
  end;
end;

theorem Th19:
  x in rng p implies p.(x..p) = x
proof
  set q = Sgm(p " {x});
A1: p " {x} c= dom p & dom p = Seg(len p) by FINSEQ_1:def 3,RELAT_1:132;
  assume x in rng p;
  then p " {x} <> {} by FUNCT_1:72;
  then rng q <> {} by A1,FINSEQ_1:def 13;
  then 1 in dom q by FINSEQ_3:32;
  then
A2: q.1 in rng q by FUNCT_1:def 3;
  rng q = p " {x} by A1,FINSEQ_1:def 13;
  then p.(x..p) in {x} by A2,FUNCT_1:def 7;
  hence thesis by TARSKI:def 1;
end;

theorem Th20:
  x in rng p implies x..p in dom p
proof
A1: p " {x} c= dom p & dom p = Seg(len p) by FINSEQ_1:def 3,RELAT_1:132;
  assume x in rng p;
  then p " {x} <> {} by FUNCT_1:72;
  then rng(Sgm(p " {x})) <> {} by A1,FINSEQ_1:def 13;
  then 1 in dom(Sgm(p " {x})) by FINSEQ_3:32;
  then x..p in rng(Sgm(p " {x})) by FUNCT_1:def 3;
  then x..p in p " {x} by A1,FINSEQ_1:def 13;
  hence thesis by FUNCT_1:def 7;
end;

theorem Th21:
  x in rng p implies 1 <= x..p & x..p <= len p
proof
  assume x in rng p;
  then x..p in dom p by Th20;
  hence thesis by FINSEQ_3:25;
end;

theorem Th22:
  x in rng p implies x..p - 1 is Element of NAT & len p - x..p is
  Element of NAT
proof
  assume x in rng p;
  then 1 <= x..p & x..p <= len p by Th21;
  hence thesis by INT_1:5;
end;

theorem Th23:
  x in rng p implies x..p in p " {x}
proof
  assume
A1: x in rng p;
  then p.(x..p) = x by Th19;
  then
A2: p.(x..p) in {x} by TARSKI:def 1;
  x..p in dom p by A1,Th20;
  hence thesis by A2,FUNCT_1:def 7;
end;

theorem Th24:
  for k st k in dom p & k < x..p holds p.k <> x
proof
  let k;
  set q = Sgm(p " {x});
  assume that
A1: k in dom p and
A2: k < x..p and
A3: p.k = x;
A4: x in {x} by TARSKI:def 1;
A5: p " {x} c= dom p & dom p = Seg(len p) by FINSEQ_1:def 3,RELAT_1:132;
  then rng q = p " {x} by FINSEQ_1:def 13;
  then k in rng q by A1,A3,A4,FUNCT_1:def 7;
  then consider y such that
A6: y in dom q and
A7: q.y = k by FUNCT_1:def 3;
  reconsider y as Element of NAT by A6;
A8: now
    assume not 1 < y;
    then 1 = y or y < 1 by XXREAL_0:1;
    hence contradiction by A2,A6,A7,FINSEQ_3:24,NAT_1:14;
  end;
  dom q = Seg(len q) by FINSEQ_1:def 3;
  then y <= len q by A6,FINSEQ_1:1;
  hence contradiction by A2,A5,A7,A8,FINSEQ_1:def 13;
end;

theorem Th25:
  p just_once_values x implies p <- x = x..p
proof
  assume
A1: p just_once_values x;
  then x in rng p by Th5;
  then x..p in dom p & p.(x..p) = x by Th19,Th20;
  hence thesis by A1,Def3;
end;

theorem Th26:
  p just_once_values x implies for k st k in dom p & k <> x..p holds p.k <> x
proof
  assume
A1: p just_once_values x;
  let k;
  assume
A2: k in dom p & k <> x..p & p.k = x;
  p <- x = x..p by A1,Th25;
  hence thesis by A1,A2,Def3;
end;

theorem Th27:
  x in rng p & (for k st k in dom p & k <> x..p holds p.k <> x)
  implies p just_once_values x
proof
  assume that
A1: x in rng p and
A2: for k st k in dom p & k <> x..p holds p.k <> x;
A3: for z st z in dom p & z <> x..p holds p.z <> x by A2;
  p.(x..p) = x & x..p in dom p by A1,Th19,Th20;
  hence thesis by A3,Th7;
end;

theorem Th28:
  p just_once_values x iff x in rng p & {x..p} = p " {x}
proof
  thus p just_once_values x implies x in rng p & {x..p} = p " {x}
  proof
    assume
A1: p just_once_values x;
    then x..p = p <- x by Th25;
    hence thesis by A1,Th5,Th11;
  end;
  assume that
A2: x in rng p and
A3: {x..p} = p " {x};
A4: now
    let z;
    assume that
A5: z in dom p and
A6: z <> x..p and
A7: p.z = x;
    p.z in {x} by A7,TARSKI:def 1;
    then z in p " {x} by A5,FUNCT_1:def 7;
    hence contradiction by A3,A6,TARSKI:def 1;
  end;
  p.(x..p) = x & x..p in dom p by A2,Th19,Th20;
  hence thesis by A4,Th7;
end;

theorem
  p is one-to-one & x in rng p implies {x..p} = p " {x}
proof
  assume that
A1: p is one-to-one and
A2: x in rng p;
  thus {x..p} c= p " {x}
  proof
    let y;
    assume y in {x..p};
    then y = x..p by TARSKI:def 1;
    hence thesis by A2,Th23;
  end;
  let y;
  assume
A3: y in p " {x};
  then
A4: y in dom p by FUNCT_1:def 7;
  p.y in {x} by A3,FUNCT_1:def 7;
  then
A5: p.y = x by TARSKI:def 1;
  p.(x..p) = x & x..p in dom p by A2,Th19,Th20;
  then x..p = y by A1,A4,A5,FUNCT_1:def 4;
  hence thesis by TARSKI:def 1;
end;

theorem Th30:
  p just_once_values x iff len(p - {x}) = len p - 1
proof
  thus p just_once_values x implies len(p - {x}) = len p - 1
  proof
    assume p just_once_values x;
    then len(p - {x}) = len p - card(p " {x}) & p " {x} = {x..p} by Th28,
FINSEQ_3:59;
    hence thesis by CARD_1:30;
  end;
A1: p"{x} = Coim(p,x);
  assume len(p - {x}) = len p - 1;
  then len p + -1 = len p - card(p " {x}) by FINSEQ_3:59
    .= len p + - (card (p"{x}));
  hence thesis by A1,Def2;
end;

reserve L,M for Element of NAT;

theorem Th31:
  p just_once_values x implies for k st k in dom(p - {x}) holds (k
< x..p implies (p - {x}).k = p.k) & (x..p <= k implies (p - {x}).k = p.(k + 1))
proof
  assume
A1: p just_once_values x;
  set q = p - {x};
  let k;
  assume
A2: k in dom(p - {x});
  set A = {L : L in dom p & L <= k & p.L in {x}};
A3: dom(p - {x}) c= dom p by FINSEQ_3:63;
  thus k < x..p implies (p - {x}).k = p.k
  proof
    assume
A4: k < x..p;
    A c= {}
    proof
      let y;
      assume y in A;
      then consider L such that
      y = L and
A5:   L in dom p & L <= k and
A6:   p.L in {x};
      p.L <> x by A1,A4,A5,Th26;
      hence thesis by A6,TARSKI:def 1;
    end;
    then
A7: A = {};
    p.k <> x by A1,A2,A3,A4,Th26;
    then not p.k in {x} by TARSKI:def 1;
    then q.(k - 0) = p.k by A2,A3,A7,CARD_1:27,FINSEQ_3:85;
    hence thesis;
  end;
  set B = {M : M in dom p & M <= k + 1 & p.M in {x}};
  assume
A8: x..p <= k;
A9: x in rng p by A1,Th5;
A10: B = {x..p}
  proof
    thus B c= {x..p}
    proof
      let y;
      assume y in B;
      then consider M such that
A11:  M = y and
A12:  M in dom p and
      M <= k + 1 and
A13:  p.M in {x};
      p.M = x by A13,TARSKI:def 1;
      then M = x..p by A1,A12,Th26;
      hence thesis by A11,TARSKI:def 1;
    end;
    let y;
    assume y in {x..p};
    then
A14: y = x..p by TARSKI:def 1;
    p.(x..p) = x by A9,Th19;
    then
A15: p.(x..p) in {x} by TARSKI:def 1;
    x..p in dom p & x..p <= k + 1 by A9,A8,Th20,NAT_1:12;
    hence thesis by A14,A15;
  end;
  then reconsider B as finite set;
  dom q = Seg(len q) & len(p - {x}) = len p - 1 by A1,Th30,FINSEQ_1:def 3;
  then k <= len p - 1 by A2,FINSEQ_1:1;
  then 1 <= k + 1 & k + 1 <= len p by NAT_1:12,XREAL_1:19;
  then k + 1 in Seg(len p);
  then
A16: k + 1 in dom p by FINSEQ_1:def 3;
  now
    x..p <> k + 1 by A8,XREAL_1:29;
    then
A17: p.(k + 1) <> x by A1,A16,Th26;
    assume p.(k + 1) in {x};
    hence contradiction by A17,TARSKI:def 1;
  end;
  then (p - {x}).((k + 1) - card B) = p.(k + 1) by A16,FINSEQ_3:85;
  then q.((k + 1) - 1) = p.(k + 1) by A10,CARD_1:30;
  hence thesis;
end;

theorem
  p is one-to-one & x in rng p implies for k st k in dom(p - {x}) holds
  ((p - {x}).k = p.k iff k < x..p) & ((p - {x}).k = p.(k + 1) iff x..p <= k)
proof
  assume that
A1: p is one-to-one and
A2: x in rng p;
  set q = p - {x};
  let k;
  assume
A3: k in dom(p - {x});
A4: p just_once_values x by A1,A2,Th8;
  then dom q = Seg(len q) & len(p - {x}) = len p - 1 by Th30,FINSEQ_1:def 3;
  then k <= len p - 1 by A3,FINSEQ_1:1;
  then 1 <= k + 1 & k + 1 <= len p by NAT_1:12,XREAL_1:19;
  then k + 1 in Seg(len p);
  then
A5: dom(p - {x}) c= dom p & k + 1 in dom p by FINSEQ_1:def 3,FINSEQ_3:63;
  thus (p - {x}).k = p.k implies k < x..p
  proof
    assume that
A6: (p - {x}).k = p.k and
A7: not k < x..p;
    q.k = p.(k + 1) by A4,A3,A7,Th31;
    then k + 0 = k + 1 by A1,A3,A5,A6,FUNCT_1:def 4;
    hence thesis;
  end;
  thus k < x..p implies (p - {x}).k = p.k by A4,A3,Th31;
  thus (p - {x}).k = p.(k + 1) implies x..p <= k
  proof
    assume
A8: (p - {x}).k = p.(k + 1);
    assume not x..p <= k;
    then p.(k + 1) = p.k by A4,A3,A8,Th31;
    then k + 0 = k + 1 by A1,A3,A5,FUNCT_1:def 4;
    hence thesis;
  end;
  thus thesis by A4,A3,Th31;
end;

definition
  let p;
  let x;
  assume
A1: x in rng p;
  func p -| x -> FinSequence means
  :Def5:
  ex n st n = x..p - 1 & it = p | Seg n;
  existence
  proof
    reconsider n = x..p - 1 as Element of NAT by A1,Th22;
    reconsider q = p | Seg n as FinSequence by FINSEQ_1:15;
    take q;
    thus thesis;
  end;
  uniqueness;
end;

theorem Th33:
  x in rng p & n = x..p - 1 implies p | Seg n = p -| x
proof
  assume x in rng p;
  then ex m st m = x..p - 1 & p | Seg m = p -| x by Def5;
  hence thesis;
end;

theorem Th34:
  x in rng p implies len(p -| x) = x..p - 1
proof
  assume
A1: x in rng p;
  then consider n such that
A2: n = x..p - 1 and
A3: p | Seg n = p -| x by Def5;
A4: n <= n + 1 by NAT_1:12;
  n + 1 <= len p by A1,A2,Th21;
  then n <= len p by A4,XXREAL_0:2;
  hence thesis by A2,A3,FINSEQ_1:17;
end;

theorem Th35:
  x in rng p & n = x..p - 1 implies dom(p -| x) = Seg n
proof
  assume x in rng p;
  then len(p -| x) = x..p - 1 by Th34;
  hence thesis by FINSEQ_1:def 3;
end;

theorem Th36:
  x in rng p & k in dom(p -| x) implies p.k = (p -| x).k
proof
  assume that
A1: x in rng p and
A2: k in dom(p -| x);
  ex n st n = x..p - 1 & p | Seg n = p -| x by A1,Def5;
  hence thesis by A2,FUNCT_1:47;
end;

theorem Th37:
  x in rng p implies not x in rng(p -| x)
proof
  assume that
A1: x in rng p and
A2: x in rng(p -| x);
  reconsider n = x..p - 1 as Element of NAT by A1,Th22;
  set r = p | Seg n;
A3: r = p -| x by A1,Th33;
  then consider y such that
A4: y in dom r and
A5: r.y = x by A2,FUNCT_1:def 3;
A6: dom r = Seg n by A1,A3,Th35;
  then reconsider y as Element of NAT by A4;
  y <= n by A4,A6,FINSEQ_1:1;
  then
A7: y + 1 <= x..p by XREAL_1:19;
  y < y + 1 by XREAL_1:29;
  then dom r c= dom p & y < x..p by A7,RELAT_1:60,XXREAL_0:2;
  then p.y <> x by A4,Th24;
  hence thesis by A4,A5,FUNCT_1:47;
end;

theorem
  x in rng p implies rng(p -| x) misses {x}
proof
  assume x in rng p;
  then not x in rng(p -| x) by Th37;
  then for y st y in rng(p -| x) holds not y in {x} by TARSKI:def 1;
  hence thesis by XBOOLE_0:3;
end;

theorem
  x in rng p implies rng(p -| x) c= rng p
proof
  assume x in rng p;
  then ex n st n = x..p - 1 & p | Seg n = p -| x by Def5;
  hence thesis by RELAT_1:70;
end;

theorem
  x in rng p implies (x..p = 1 iff p -| x = {})
proof
  assume
A1: x in rng p;
  thus x..p = 1 implies p -| x = {}
  proof
    assume
A2: x..p = 1;
    len(p -| x) = x..p - 1 by A1,Th34
      .= 0 by A2;
    hence thesis;
  end;
  assume p -| x = {};
  then
A3: len(p -| x) = 0;
  len(p -| x) = x..p - 1 by A1,Th34;
  hence thesis by A3;
end;

theorem
  x in rng p & p is FinSequence of D implies p -| x is FinSequence of D
proof
  assume x in rng p;
  then ex n st n = x..p - 1 & p | Seg n = p -| x by Def5;
  hence thesis by FINSEQ_1:18;
end;

definition
  let p;
  let x;
  assume
A1: x in rng p;
  func p |-- x -> FinSequence means
  :Def6:
  len it = len p - x..p & for k st k in dom it holds it.k = p.(k + x..p);
  existence
  proof
    deffunc F(Nat) = p.($1 + x..p);
    reconsider n = len p - x..p as Element of NAT by A1,Th22;
    consider q such that
A2: len q = n & for k be Nat st k in dom q holds q.k = F(k) from
    FINSEQ_1:sch 2;
    take q;
    thus thesis by A2;
  end;
  uniqueness
  proof
    let q,r be FinSequence;
    assume that
A3: len q = len p - x..p and
A4: for k st k in dom q holds q.k = p.(k + x..p);
    assume that
A5: len r = len p - x..p and
A6: for k st k in dom r holds r.k = p.(k + x..p);
    now
      let k be Nat;
A7:   dom q = Seg(len q) & dom r = Seg(len r) by FINSEQ_1:def 3;
      assume
A8:   k in dom q;
      then q.k = p.(k + x..p) by A4;
      hence q.k = r.k by A3,A5,A6,A8,A7;
    end;
    hence thesis by A3,A5,FINSEQ_2:9;
  end;
end;

theorem Th42:
  x in rng p & n = len p - x..p implies dom(p |-- x) = Seg n
proof
  assume x in rng p;
  then len(p |-- x) = len p - x..p by Def6;
  hence thesis by FINSEQ_1:def 3;
end;

theorem Th43:
  x in rng p & n in dom(p |-- x) implies n + x..p in dom p
proof
  assume that
A1: x in rng p and
A2: n in dom(p |-- x);
  reconsider m = len p - x..p as Element of NAT by A1,Th22;
  n in Seg m by A1,A2,Th42;
  then n <= len p - x..p by FINSEQ_1:1;
  then
A3: n + x..p <= len p by XREAL_1:19;
  1 <= n by A2,FINSEQ_3:25;
  then 1 <= n + x..p by NAT_1:12;
  hence thesis by A3,FINSEQ_3:25;
end;

theorem
  x in rng p implies rng(p |-- x) c= rng p
proof
  assume
A1: x in rng p;
  let y;
  assume y in rng(p |-- x);
  then consider z such that
A2: z in dom(p |-- x) and
A3: (p |-- x).z = y by FUNCT_1:def 3;
  reconsider z as Element of NAT by A2;
  y = p.(z + x..p) & z + x..p in dom p by A1,A2,A3,Def6,Th43;
  hence thesis by FUNCT_1:def 3;
end;

theorem Th45:
  p just_once_values x iff x in rng p & not x in rng(p |-- x)
proof
  thus p just_once_values x implies x in rng p & not x in rng(p |-- x)
  proof
    assume
A1: p just_once_values x;
    hence
A2: x in rng p by Th5;
    assume x in rng(p |-- x);
    then consider z such that
A3: z in dom(p |-- x) and
A4: (p |-- x).z = x by FUNCT_1:def 3;
    reconsider z as Element of NAT by A3;
    (p |-- x).z = p.(z + x..p) & z + x..p in dom p by A2,A3,Def6,Th43;
    then z + x..p = 0 + x..p by A1,A4,Th26;
    hence thesis by A3,FINSEQ_3:24;
  end;
  assume that
A5: x in rng p and
A6: not x in rng(p |-- x);
  now
    let k;
    assume that
A7: k in dom p and
A8: k <> x..p and
A9: p.k = x;
    now
      per cases by A8,XXREAL_0:1;
      suppose
        k < x..p;
        then k + 1 <= x..p by NAT_1:13;
        then k <= x..p - 1 by XREAL_1:19;
        then
A10:    k <= len(p -| x) by A5,Th34;
        1 <= k by A7,FINSEQ_3:25;
        then
A11:    k in dom(p -| x) by A10,FINSEQ_3:25;
        then x = (p -| x).k by A5,A9,Th36;
        then x in rng(p -| x) by A11,FUNCT_1:def 3;
        hence contradiction by A5,Th37;
      end;
      suppose
A12:    x..p < k;
        then consider m be Nat such that
A13:    k = x..p + m by NAT_1:10;
        x..p + 0 < x..p + m by A12,A13;
        then 0 < m;
        then
A14:    0 + 1 <= m by NAT_1:13;
        m + x..p <= len p by A7,A13,FINSEQ_3:25;
        then m <= len p - x..p by XREAL_1:19;
        then m <= len(p |-- x) by A5,Def6;
        then
A15:    m in dom(p |-- x) by A14,FINSEQ_3:25;
        then (p |-- x).m = p.k by A5,A13,Def6;
        hence contradiction by A6,A9,A15,FUNCT_1:def 3;
      end;
    end;
    hence contradiction;
  end;
  hence thesis by A5,Th27;
end;

theorem Th46:
  x in rng p & p is one-to-one implies not x in rng(p |-- x)
proof
  assume that
A1: x in rng p and
A2: p is one-to-one and
A3: x in rng(p |-- x);
A4: len(p |-- x) = len p - x..p by A1,Def6;
  consider y such that
A5: y in dom(p |-- x) and
A6: (p |-- x).y = x by A3,FUNCT_1:def 3;
  reconsider y as Element of NAT by A5;
A7: 1 <= y + x..p by A1,Th21,NAT_1:12;
A8: y in Seg(len(p |-- x)) by A5,FINSEQ_1:def 3;
  then y <= len(p |-- x) by FINSEQ_1:1;
  then y + x..p <= len p by A4,XREAL_1:19;
  then y + x..p in Seg(len p) by A7;
  then
A9: y + x..p in dom p by FINSEQ_1:def 3;
A10: x..p in dom p & p.(x..p) = x by A1,Th19,Th20;
  (p |-- x).y = p.(y + x..p) by A1,A5,Def6;
  then 0 + x..p = y + x..p by A2,A6,A10,A9,FUNCT_1:def 4;
  hence thesis by A8,FINSEQ_1:1;
end;

theorem
  p just_once_values x iff x in rng p & rng(p |-- x) misses {x}
proof
  thus p just_once_values x implies x in rng p & rng(p |-- x) misses {x}
  proof
    assume
A1: p just_once_values x;
    hence x in rng p by Th45;
    assume not rng(p |-- x) misses {x};
    then
A2: ex y st y in rng(p |-- x) & y in {x} by XBOOLE_0:3;
    not x in rng(p |-- x) by A1,Th45;
    hence thesis by A2,TARSKI:def 1;
  end;
  assume that
A3: x in rng p and
A4: rng(p |-- x) misses {x};
  now
A5: x in {x} by TARSKI:def 1;
    assume x in rng(p |-- x);
    hence contradiction by A4,A5,XBOOLE_0:3;
  end;
  hence thesis by A3,Th45;
end;

theorem
  x in rng p & p is one-to-one implies rng(p |-- x) misses {x}
proof
  assume x in rng p & p is one-to-one;
  then not x in rng(p |-- x) by Th46;
  then for y st y in rng(p |-- x) holds not y in {x} by TARSKI:def 1;
  hence thesis by XBOOLE_0:3;
end;

theorem
  x in rng p implies (x..p = len p iff p |-- x = {})
proof
  assume
A1: x in rng p;
  thus x..p = len p implies p |-- x = {}
  proof
    assume
A2: x..p = len p;
    len(p |-- x) = len p - x..p by A1,Def6
      .= 0 by A2;
    hence thesis;
  end;
  assume p |-- x = {};
  then
A3: len(p |-- x) = 0;
  len(p |-- x) = len p - x..p by A1,Def6;
  hence thesis by A3;
end;

theorem
  x in rng p & p is FinSequence of D implies p |-- x is FinSequence of D
proof
  assume that
A1: x in rng p and
A2: p is FinSequence of D;
  rng(p |-- x) c= D
  proof
A3: len(p |-- x) = len p - x..p by A1,Def6;
    let y;
    assume y in rng(p |-- x);
    then consider z such that
A4: z in dom(p |-- x) and
A5: (p |-- x).z = y by FUNCT_1:def 3;
    reconsider z as Element of NAT by A4;
    dom(p |-- x) = Seg(len(p |-- x)) by FINSEQ_1:def 3;
    then z <= len(p |-- x) by A4,FINSEQ_1:1;
    then
A6: z + x..p <= len p by A3,XREAL_1:19;
    1 <= z + x..p by A1,Th21,NAT_1:12;
    then z + x..p in Seg(len p) by A6;
    then
A7: z + x..p in dom p by FINSEQ_1:def 3;
    y = p.(z + x..p) by A1,A4,A5,Def6;
    then
A8: y in rng p by A7,FUNCT_1:def 3;
    rng p c= D by A2,FINSEQ_1:def 4;
    hence thesis by A8;
  end;
  hence thesis by FINSEQ_1:def 4;
end;

theorem Th51:
  x in rng p implies p = (p -| x) ^ <* x *> ^ (p |-- x)
proof
  set q1 = p -| x;
  set q2 = p |-- x;
  set r = q1 ^ <* x *>;
  assume
A1: x in rng p;
A2: now
    let k be Element of NAT;
    assume k in dom q2;
    then q2.k = p.(x..p - 1 + 1 + k) by A1,Def6
      .= p.(len q1 + 1 + k) by A1,Th34
      .= p.(len q1 + len<* x *> + k) by FINSEQ_1:40
      .= p.(len r + k) by FINSEQ_1:22;
    hence p.(len r + k) = q2.k;
  end;
A3: now
    let k be Element of NAT;
    assume
A4: k in dom r;
    now
      per cases by A4,FINSEQ_1:25;
      suppose
A5:     k in dom q1;
        hence r.k = q1.k by FINSEQ_1:def 7
          .= p.k by A1,A5,Th36;
      end;
      suppose
        ex n be Nat st n in dom<* x *> & k = len q1 + n;
        then consider n be Nat such that
A6:     n in dom <* x *> and
A7:     k = len q1 + n;
        n in {1} by A6,FINSEQ_1:2,def 8;
        then
A8:     n = 1 by TARSKI:def 1;
        hence r.k = <* x *>.1 by A6,A7,FINSEQ_1:def 7
          .= x by FINSEQ_1:def 8
          .= p.(x..p - 1 + 1) by A1,Th19
          .= p.k by A1,A7,A8,Th34;
      end;
    end;
    hence p.k = r.k;
  end;
  len p = len p - x..p + x..p .= x..p - 1 + 1 + len q2 by A1,Def6
    .= len q1 + 1 + len q2 by A1,Th34
    .= len q1 + len<* x *> + len q2 by FINSEQ_1:40
    .= len r + len q2 by FINSEQ_1:22;
  hence thesis by A3,A2,FINSEQ_3:38;
end;

theorem
  x in rng p & p is one-to-one implies p -| x is one-to-one
proof
  assume x in rng p;
  then p = (p -| x) ^ <* x *> ^ (p |-- x) by Th51
    .= (p -| x) ^ (<* x *> ^ (p |-- x)) by FINSEQ_1:32;
  hence thesis by FINSEQ_3:91;
end;

theorem
  x in rng p & p is one-to-one implies p |-- x is one-to-one
proof
  assume x in rng p;
  then p = (p -| x) ^ <* x *> ^ (p |-- x) by Th51;
  hence thesis by FINSEQ_3:91;
end;

theorem Th54:
  p just_once_values x iff x in rng p & p - {x} = (p -| x) ^ (p |-- x)
proof
  set q = p - {x};
  set r = p -| x;
  set s = p |-- x;
  thus p just_once_values x implies x in rng p & p - {x} = (p -| x) ^ (p |-- x
  )
  proof
    assume
A1: p just_once_values x;
    hence
A2: x in rng p by Th5;
A3: now
      x..p <= len p by A2,Th21;
      then x..p - 1 <= len p - 1 by XREAL_1:9;
      then len r <= len p - 1 by A2,Th34;
      then len r <= len q by A1,Th30;
      then
A4:   Seg(len r) c= Seg(len q) by FINSEQ_1:5;
      let k be Nat;
A5:   Seg(len r) = dom r & Seg(len q) = dom q by FINSEQ_1:def 3;
      assume
A6:   k in dom r;
      then k in Seg(len r) by FINSEQ_1:def 3;
      then k <= len r by FINSEQ_1:1;
      then k <= x..p - 1 by A2,Th34;
      then
A7:   k + 1 <= x..p by XREAL_1:19;
      k < k + 1 by XREAL_1:29;
      then
A8:   k < x..p by A7,XXREAL_0:2;
      r.k = p.k by A2,A6,Th36;
      hence q.k = r.k by A1,A6,A4,A5,A8,Th31;
    end;
A9: now
      reconsider m = x..p - 1 as Element of NAT by A2,Th22;
      let k be Nat;
      set z = k + m;
      assume
A10:  k in dom s;
      then
A11:  s.k = p.(k + x..p) by A2,Def6;
A12:  dom s = Seg(len s) by FINSEQ_1:def 3;
      then
A13:  1 <= k by A10,FINSEQ_1:1;
      then x..p + 1 <= k + x..p by XREAL_1:7;
      then
A14:  x..p <= k + x..p - 1 by XREAL_1:19;
      k <= len s by A10,A12,FINSEQ_1:1;
      then k <= len p - x..p by A2,Def6;
      then k + x..p <= len p by XREAL_1:19;
      then k + x..p - 1 <= len p - 1 by XREAL_1:9;
      then
A15:  k + x..p - 1 <= len q by A1,Th30;
      1 <= x..p by A2,Th21;
      then 1 + 1 <= k + x..p by A13,XREAL_1:7;
      then
A16:  1 <= k + x..p - 1 by XREAL_1:19;
      dom q = Seg(len q) by FINSEQ_1:def 3;
      then z in dom q by A16,A15;
      then q.z = p.(z + 1) by A1,A14,Th31
        .= p.(k + x..p);
      hence q.(len r + k) = s.k by A2,A11,Th34;
    end;
    len r + len s = (x..p - 1) + len s by A2,Th34
      .= (x..p - 1) + (len p - x..p) by A2,Def6
      .= len p - 1
      .= len q by A1,Th30;
    then dom q = Seg(len r + len s) by FINSEQ_1:def 3;
    hence thesis by A3,A9,FINSEQ_1:def 7;
  end;
  assume
A17: x in rng p;
  assume
A18: p - {x} = (p -| x) ^ (p |-- x);
  now
    let k;
    assume that
A19: k in dom p and
A20: k <> x..p and
A21: p.k = x;
    {x..p,k} c= p " {x}
    proof
      let y;
      assume
A22:  y in {x..p,k};
A23:  x in {x} by TARSKI:def 1;
      x..p in dom p & p.(x..p) = x by A17,Th19,Th20;
      then
A24:  x..p in p " {x} by A23,FUNCT_1:def 7;
      k in p " {x} by A19,A21,A23,FUNCT_1:def 7;
      hence thesis by A22,A24,TARSKI:def 2;
    end;
    then card{x..p,k} <= card(p " {x}) by NAT_1:43;
    then
A25: 2 <= card(p " {x}) by A20,CARD_2:57;
A26: len q = len p - card(p " {x}) by FINSEQ_3:59
      .= len p + (- card(p " {x}));
    len q = len r + len s by A18,FINSEQ_1:22
      .= (x..p - 1) + len s by A17,Th34
      .= (x..p - 1) + (len p - x..p) by A17,Def6
      .= len p + (- 1);
    hence contradiction by A26,A25;
  end;
  hence thesis by A17,Th27;
end;

theorem
  x in rng p & p is one-to-one implies p - {x} = (p -| x) ^ (p |-- x)
proof
  assume x in rng p & p is one-to-one;
  then p just_once_values x by Th8;
  hence thesis by Th54;
end;

theorem Th56:
  x in rng p & p - {x} is one-to-one & p - {x} = (p -| x) ^ (p |--
  x) implies p is one-to-one
proof
  assume that
A1: x in rng p and
A2: p - {x} is one-to-one and
A3: p - {x} = (p -| x) ^ (p |-- x);
  set q = p - {x};
  let x1,x2;
  assume that
A4: x1 in dom p and
A5: x2 in dom p and
A6: p.x1 = p.x2;
  reconsider k1 = x1, k2 = x2 as Element of NAT by A4,A5;
A7: p just_once_values x by A1,A3,Th54;
  now
    per cases by XXREAL_0:1;
    suppose
      x1 = x..p & x2 = x..p;
      hence thesis;
    end;
    suppose
      x1 = x..p & x..p < k2;
      hence thesis by A1,A5,A6,A7,Th19,Th26;
    end;
    suppose
      x1 = x..p & k2 < x..p;
      hence thesis by A1,A5,A6,A7,Th19,Th26;
    end;
    suppose
      k1 < x..p & x..p = x2;
      hence thesis by A1,A4,A6,A7,Th19,Th26;
    end;
    suppose
A8:   k1 < x..p & x..p < k2;
      x..p <= len p by A1,Th21;
      then k1 < len p by A8,XXREAL_0:2;
      then k1 + 1 <= len p by NAT_1:13;
      then k1 <= len p - 1 by XREAL_1:19;
      then
A9:   k1 <= len q by A7,Th30;
      k1 in Seg(len p) by A4,FINSEQ_1:def 3;
      then 1 <= k1 by FINSEQ_1:1;
      then k1 in Seg(len q) by A9;
      then
A10:  k1 in dom q by FINSEQ_1:def 3;
      then
A11:  q.k1 = p.k1 by A7,A8,Th31;
      consider m2 be Nat such that
A12:  k2 = m2 + 1 by A8,NAT_1:6;
      reconsider m2 as Element of NAT by ORDINAL1:def 12;
A13:  x..p <= m2 by A8,A12,NAT_1:13;
      k2 in Seg(len p) by A5,FINSEQ_1:def 3;
      then k2 <= len p by FINSEQ_1:1;
      then m2 <= len p - 1 by A12,XREAL_1:19;
      then
A14:  m2 <= len q by A7,Th30;
      1 <= x..p by A1,Th21;
      then 1 <= m2 by A13,XXREAL_0:2;
      then m2 in Seg(len q) by A14;
      then
A15:  m2 in dom(p - {x}) by FINSEQ_1:def 3;
      then q.m2 = p.k2 by A7,A12,A13,Th31;
      hence thesis by A2,A6,A8,A10,A13,A15,A11,FUNCT_1:def 4;
    end;
    suppose
A16:  k1 < x..p & k2 < x..p;
A17:  x..p <= len p by A1,Th21;
      then k2 < len p by A16,XXREAL_0:2;
      then k2 + 1 <= len p by NAT_1:13;
      then k2 <= len p - 1 by XREAL_1:19;
      then
A18:  k2 <= len q by A7,Th30;
      k2 in Seg(len p) by A5,FINSEQ_1:def 3;
      then 1 <= k2 by FINSEQ_1:1;
      then k2 in Seg(len q) by A18;
      then
A19:  k2 in dom q by FINSEQ_1:def 3;
      then
A20:  p.k2 = (p - {x}).k2 by A7,A16,Th31;
      k1 < len p by A16,A17,XXREAL_0:2;
      then k1 + 1 <= len p by NAT_1:13;
      then k1 <= len p - 1 by XREAL_1:19;
      then
A21:  k1 <= len q by A7,Th30;
      k1 in Seg(len p) by A4,FINSEQ_1:def 3;
      then 1 <= k1 by FINSEQ_1:1;
      then k1 in Seg(len q) by A21;
      then
A22:  k1 in dom q by FINSEQ_1:def 3;
      then p.k1 = (p - {x}).k1 by A7,A16,Th31;
      hence thesis by A2,A6,A22,A19,A20,FUNCT_1:def 4;
    end;
    suppose
A23:  x..p < k1 & x..p < k2;
      then consider m2 be Nat such that
A24:  k2 = m2 + 1 by NAT_1:6;
      consider m1 be Nat such that
A25:  k1 = m1 + 1 by A23,NAT_1:6;
      reconsider m1, m2 as Element of NAT by ORDINAL1:def 12;
      k2 in Seg(len p) by A5,FINSEQ_1:def 3;
      then k2 <= len p by FINSEQ_1:1;
      then m2 <= len p - 1 by A24,XREAL_1:19;
      then
A26:  m2 <= len q by A7,Th30;
A27:  1 <= x..p by A1,Th21;
A28:  x..p <= m2 by A23,A24,NAT_1:13;
      then 1 <= m2 by A27,XXREAL_0:2;
      then m2 in Seg(len q) by A26;
      then
A29:  m2 in dom(p - {x}) by FINSEQ_1:def 3;
      then
A30:  p.k2 = (p - {x}).m2 by A7,A24,A28,Th31;
      k1 in Seg(len p) by A4,FINSEQ_1:def 3;
      then k1 <= len p by FINSEQ_1:1;
      then m1 <= len p - 1 by A25,XREAL_1:19;
      then
A31:  m1 <= len q by A7,Th30;
A32:  x..p <= m1 by A23,A25,NAT_1:13;
      then 1 <= m1 by A27,XXREAL_0:2;
      then m1 in Seg(len q) by A31;
      then
A33:  m1 in dom(p - {x}) by FINSEQ_1:def 3;
      then p.k1 = (p - {x}).m1 by A7,A25,A32,Th31;
      hence thesis by A2,A6,A25,A24,A33,A29,A30,FUNCT_1:def 4;
    end;
    suppose
      x..p < k1 & x..p = x2;
      hence thesis by A1,A4,A6,A7,Th19,Th26;
    end;
    suppose
A34:  x..p < k1 & k2 < x..p;
      x..p <= len p by A1,Th21;
      then k2 < len p by A34,XXREAL_0:2;
      then k2 + 1 <= len p by NAT_1:13;
      then k2 <= len p - 1 by XREAL_1:19;
      then
A35:  k2 <= len q by A7,Th30;
      k2 in Seg(len p) by A5,FINSEQ_1:def 3;
      then 1 <= k2 by FINSEQ_1:1;
      then k2 in Seg(len q) by A35;
      then
A36:  k2 in dom q by FINSEQ_1:def 3;
      then
A37:  q.k2 = p.k2 by A7,A34,Th31;
      consider m2 be Nat such that
A38:  k1 = m2 + 1 by A34,NAT_1:6;
      reconsider m2 as Element of NAT by ORDINAL1:def 12;
A39:  x..p <= m2 by A34,A38,NAT_1:13;
      k1 in Seg len p by A4,FINSEQ_1:def 3;
      then k1 <= len p by FINSEQ_1:1;
      then m2 <= len p - 1 by A38,XREAL_1:19;
      then
A40:  m2 <= len q by A7,Th30;
      1 <= x..p by A1,Th21;
      then 1 <= m2 by A39,XXREAL_0:2;
      then m2 in Seg len q by A40;
      then
A41:  m2 in dom(p - {x}) by FINSEQ_1:def 3;
      then q.m2 = p.k1 by A7,A38,A39,Th31;
      hence thesis by A2,A6,A34,A36,A39,A41,A37,FUNCT_1:def 4;
    end;
  end;
  hence thesis;
end;

theorem
  x in rng p & p is one-to-one implies rng(p -| x) misses rng(p |-- x)
proof
  assume that
A1: x in rng p and
A2: p is one-to-one;
  p = (p -| x) ^ <* x *> ^ (p |-- x) by A1,Th51;
  then rng(p |-- x) misses rng((p -| x) ^ <* x *>) by A2,FINSEQ_3:91;
  hence thesis by FINSEQ_1:29,XBOOLE_1:63;
end;

theorem Th58:
  A is finite implies ex p st rng p = A & p is one-to-one
proof
  defpred P[set] means ex p st rng p = $1 & p is one-to-one;
  rng {} = {};
  then
A1: P[{}];
  now
    let B,C;
    assume that
    B in A and
    C c= A;
    given p such that
A2: rng p = C and
A3: p is one-to-one;
A4: now
      assume
A5:   not B in C;
      take q = p ^ <* B *>;
      thus rng q = rng p \/ rng<* B *> by FINSEQ_1:31
        .= C \/ {B} by A2,FINSEQ_1:38;
      thus q is one-to-one
      proof
        let x,y;
        assume that
A6:     x in dom q & y in dom q and
A7:     q.x = q.y;
        reconsider k = x, l = y as Element of NAT by A6;
A8:     now
          assume
A9:       k in dom p;
          given n be Nat such that
A10:      n in dom<* B *> and
A11:      l = len p + n;
          n in {1} by A10,FINSEQ_1:2,38;
          then
A12:      n = 1 by TARSKI:def 1;
          <* B *>.n = q.k by A7,A10,A11,FINSEQ_1:def 7
            .= p.k by A9,FINSEQ_1:def 7;
          then B = p.k by A12,FINSEQ_1:def 8;
          hence thesis by A2,A5,A9,FUNCT_1:def 3;
        end;
A13:    now
          assume
A14:      l in dom p;
          given n be Nat such that
A15:      n in dom<* B *> and
A16:      k = len p + n;
          n in {1} by A15,FINSEQ_1:2,38;
          then
A17:      n = 1 by TARSKI:def 1;
          <* B *>.n = q.l by A7,A15,A16,FINSEQ_1:def 7
            .= p.l by A14,FINSEQ_1:def 7;
          then B = p.l by A17,FINSEQ_1:def 8;
          hence thesis by A2,A5,A14,FUNCT_1:def 3;
        end;
A18:    now
          given m1 being Nat such that
A19:      m1 in dom<* B *> and
A20:      k = len p + m1;
          m1 in {1} by A19,FINSEQ_1:2,def 8;
          then
A21:      m1 = 1 by TARSKI:def 1;
          given m2 being Nat such that
A22:      m2 in dom<* B *> and
A23:      l = len p + m2;
          m2 in {1} by A22,FINSEQ_1:2,def 8;
          hence thesis by A20,A23,A21,TARSKI:def 1;
        end;
        now
          assume
A24:      k in dom p & l in dom p;
          then q.k = p.k & q.l = p.l by FINSEQ_1:def 7;
          hence thesis by A3,A7,A24,FUNCT_1:def 4;
        end;
        hence thesis by A6,A8,A13,A18,FINSEQ_1:25;
      end;
    end;
    now
      assume
A25:  B in C;
      take q = p;
      thus rng q = C \/ {B} & q is one-to-one by A2,A3,A25,ZFMISC_1:40;
    end;
    hence ex p st rng p = C \/ {B} & p is one-to-one by A4;
  end;
  then
A26: for B,C being set st B in A & C c= A & P[C] holds P[C \/ {B}];
  assume
A27: A is finite;
  thus P[A] from FINSET_1:sch 2(A27,A1,A26);
end;

theorem Th59:
  rng p c= dom p & p is one-to-one implies rng p = dom p
proof
  defpred P[Nat] means for q st len q = $1 & rng q c= dom q & q is one-to-one
  holds rng q = dom q;
A1: len p = len p;
  now
    let k;
    assume
A2: for q st len q = k & rng q c= dom q & q is one-to-one holds rng q = dom q;
    let q;
    assume that
A3: len q = k + 1 and
A4: rng q c= dom q and
A5: q is one-to-one;
A6: dom q = Seg(k + 1) by A3,FINSEQ_1:def 3;
    dom q c= rng q
    proof
      let x;
      assume
A7:   x in dom q;
      then reconsider n = x as Element of NAT;
      per cases;
      suppose
A8:     k + 1 in rng q;
        now
          per cases;
          suppose
            n = k + 1;
            hence thesis by A8;
          end;
          suppose
            n <> k + 1;
            then not x in {k + 1} by TARSKI:def 1;
            then x in Seg(k + 1) \ {k + 1} by A6,A7,XBOOLE_0:def 5;
            then
A9:         x in Seg k by FINSEQ_1:10;
            set r = q - {k + 1};
A10:        len r = (k + 1) - 1 by A3,A5,A8,FINSEQ_3:90;
            then
A11:        dom r = Seg k by FINSEQ_1:def 3;
A12:        rng r = rng q \ {k + 1} by FINSEQ_3:65;
            then rng r c= Seg(k + 1) \ {k + 1} by A4,A6,XBOOLE_1:33;
            then rng r c= dom r by A11,FINSEQ_1:10;
            then rng r = dom r by A2,A5,A10,FINSEQ_3:87;
            hence thesis by A12,A11,A9;
          end;
        end;
        hence thesis;
      end;
      suppose
A13:    not k + 1 in rng q;
A14:    rng q c= Seg k
        proof
          let x;
          assume
A15:      x in rng q;
          then not x in {k + 1} by A13,TARSKI:def 1;
          then x in Seg(k + 1) \ {k + 1} by A4,A6,A15,XBOOLE_0:def 5;
          hence thesis by FINSEQ_1:10;
        end;
A16:    k + 1 in Seg(k + 1) by FINSEQ_1:4;
        then
A17:    q.(k + 1) in rng q by A6,FUNCT_1:def 3;
        reconsider r = q | Seg k as FinSequence by FINSEQ_1:15;
A18:    dom r c= dom q & k < k + 1 by RELAT_1:60,XREAL_1:29;
A19:    len r = k by A3,FINSEQ_3:53;
        then
A20:    dom r = Seg k by FINSEQ_1:def 3;
        rng r c= rng q & r is one-to-one by A5,FUNCT_1:52,RELAT_1:70;
        then rng r = dom r by A2,A19,A20,A14,XBOOLE_1:1;
        then consider x such that
A21:    x in dom r and
A22:    r.x = q.(k + 1) by A20,A14,A17,FUNCT_1:def 3;
        reconsider n = x as Element of NAT by A21;
        r.x = q.x & n <= k by A20,A21,FINSEQ_1:1,FUNCT_1:49;
        hence thesis by A5,A6,A16,A21,A22,A18,FUNCT_1:def 4;
      end;
    end;
    hence rng q = dom q by A4,XBOOLE_0:def 10;
  end;
  then
A23: for k st P[k] holds P[k+1];
  now
    let q;
    assume
A24: len q = 0;
    assume that
    rng q c= dom q and
    q is one-to-one;
    q = {} by A24;
    hence rng q = dom q by RELAT_1:38;
  end;
  then
A25: P[0];
  for k holds P[k] from NAT_1:sch 2(A25,A23);
  hence thesis by A1;
end;

theorem Th60:
  rng p = dom p implies p is one-to-one
proof
  defpred P[Nat] means for p st len p = $1 & rng p = dom p holds p is
  one-to-one;
A1: len p = len p;
A2: now
    let k;
    assume
A3: P[k];
    thus P[k+1]
    proof
      set x = k + 1;
      let p;
      set q = p - {k + 1};
      assume that
A4:   len p = k + 1 and
A5:   rng p = dom p;
A6:   dom p = Seg(k + 1) by A4,FINSEQ_1:def 3;
      then
A7:   k + 1 in rng p by A5,FINSEQ_1:4;
      now
        rng q = Seg(k + 1) \ {k + 1} by A5,A6,FINSEQ_3:65
          .= Seg k by FINSEQ_1:10;
        then card(rng q) = k by FINSEQ_1:57;
        then
A8:     card k = card(rng q);
        p.(x..p) = x by A5,A6,Th19,FINSEQ_1:4;
        then
A9:     p.(x..p) in {k + 1} by TARSKI:def 1;
        let l;
        assume that
A10:    l in dom p and
A11:    l <> (k + 1)..p and
A12:    p.l = k + 1;
A13:    card{x..p,l} = 2 by A11,CARD_2:57;
        p.l in {k + 1} by A12,TARSKI:def 1;
        then
A14:    l in p " {x} by A10,FUNCT_1:def 7;
        x..p in dom p by A5,A6,Th20,FINSEQ_1:4;
        then x..p in p " {x} by A9,FUNCT_1:def 7;
        then {x..p,l} c= p " {k + 1} by A14,ZFMISC_1:32;
        then
A15:    2 <= card(p " {k + 1}) by A13,NAT_1:43;
        len q = (k + 1) - card(p " {k + 1}) by A4,FINSEQ_3:59;
        then
        2 + len q <= card(p " {x}) + ((k + 1) - card(p " {k + 1})) by A15,
XREAL_1:6;
        then len q + 1 + 1 <= k + 1;
        then len q + 1 <= k by XREAL_1:6;
        then
A16:    len q <= k - 1 by XREAL_1:19;
        dom q = Seg(len q) by FINSEQ_1:def 3;
        then card(rng q) c= card(dom q) & card(len q) = card(dom q) by
CARD_1:12,FINSEQ_1:57;
        then k <= len q by A8,NAT_1:40;
        then k <= k - 1 by A16,XXREAL_0:2;
        then k + 1 <= k + 0 by XREAL_1:19;
        hence contradiction by XREAL_1:6;
      end;
      then
A17:  p just_once_values k + 1 by A7,Th27;
      then
A18:  len q = (k + 1) - 1 by A4,Th30
        .= k;
A19:  q = (p -| (k + 1)) ^ (p |-- (k + 1)) by A17,Th54;
      rng q = Seg(k + 1) \ {k + 1} by A5,A6,FINSEQ_3:65
        .= Seg k by FINSEQ_1:10;
      then dom q = rng q by A18,FINSEQ_1:def 3;
      hence thesis by A3,A7,A18,A19,Th56;
    end;
  end;
A20: P[0]
  proof
    let p;
    assume len p = 0;
    then p = {};
    hence thesis;
  end;
  for k holds P[k] from NAT_1:sch 2(A20,A2);
  hence thesis by A1;
end;

theorem
  rng p = rng q & len p = len q & q is one-to-one implies p is one-to-one
proof
  assume that
A1: rng p = rng q and
A2: len p = len q and
A3: q is one-to-one;
A4: rng p = dom(q") by A1,A3,FUNCT_1:33;
  then
A5: dom (q" * p) = dom p by RELAT_1:27
    .= Seg(len p) by FINSEQ_1:def 3;
  then reconsider r = q" * p as FinSequence by FINSEQ_1:def 2;
  rng r = rng(q") by A4,RELAT_1:28
    .= dom q by A3,FUNCT_1:33
    .= Seg(len q) by FINSEQ_1:def 3;
  then r is one-to-one by A2,A5,Th60;
  hence thesis by A4,FUNCT_1:26;
end;

Lm1: for A,B being finite set, f being Function of A,B holds card A = card B &
rng f = B implies f is one-to-one
proof
  let A,B be finite set, f be Function of A,B;
  assume that
A1: card A = card B and
A2: rng f = B;
A3: A,B are_equipotent by A1,CARD_1:5;
  consider p such that
A4: rng p = A and
A5: p is one-to-one by Th58;
  dom p,p .: (dom p) are_equipotent by A5,CARD_1:33;
  then dom p,A are_equipotent by A4,RELAT_1:113;
  then
A6: dom p,B are_equipotent by A3,WELLORD2:15;
  reconsider X = dom p as finite set;
  consider q such that
A7: rng q = B and
A8: q is one-to-one by Th58;
A9: dom q = Seg(len q) by FINSEQ_1:def 3;
  dom q,q .: (dom q) are_equipotent by A8,CARD_1:33;
  then dom q,B are_equipotent by A7,RELAT_1:113;
  then dom p,dom q are_equipotent by A6,WELLORD2:15;
  then card X = card(Seg(len q)) by A9,CARD_1:5
    .= len q by FINSEQ_1:57;
  then
A10: len q = card(Seg(len p)) by FINSEQ_1:def 3
    .= len p by FINSEQ_1:57;
  now
    per cases;
    suppose
      B = {};
      hence thesis;
    end;
    suppose
A11:  B <> {};
      then rng p = dom f by A4,FUNCT_2:def 1;
      then
A12:  rng(f * p) = B by A2,RELAT_1:28
        .= dom(q") by A7,A8,FUNCT_1:33;
      dom(q") = rng q by A8,FUNCT_1:33;
      then rng f c= dom(q") by A7,RELAT_1:def 19;
      then dom(q" * f) = dom f by RELAT_1:27;
      then rng p = dom(q" * f) by A4,A11,FUNCT_2:def 1;
      then
A13:  dom(q" * f * p) = dom p by RELAT_1:27
        .= Seg(len p) by FINSEQ_1:def 3;
      then reconsider g = q" * f * p as FinSequence by FINSEQ_1:def 2;
      g = q" * (f * p) by RELAT_1:36;
      then rng g = rng(q") by A12,RELAT_1:28
        .= dom q by A8,FUNCT_1:33
        .= Seg(len q) by FINSEQ_1:def 3;
      then
A14:  g is one-to-one by A10,A13,Th60;
      q * g = q * (q " * (f * p)) by RELAT_1:36
        .= q * q" * (f * p) by RELAT_1:36
        .= id(rng q) * (f * p) by A8,FUNCT_1:39
        .= id(rng q) * f * p by RELAT_1:36
        .= f * p by A2,A7,RELAT_1:54;
      then q * g * p" = f * (p * p") by RELAT_1:36
        .= f * id(rng p) by A5,FUNCT_1:39
        .= f * id(dom f) by A4,A11,FUNCT_2:def 1
        .= f by RELAT_1:52;
      hence thesis by A5,A8,A14;
    end;
  end;
  hence thesis;
end;

theorem Th62:
  p is one-to-one iff card(rng p) = len p
proof
  thus p is one-to-one implies card(rng p) = len p
  proof
    assume p is one-to-one;
    then dom p,p .: (dom p) are_equipotent by CARD_1:33;
    then dom p,rng p are_equipotent by RELAT_1:113;
    then Seg(len p),rng p are_equipotent by FINSEQ_1:def 3;
    then card(Seg(len p)) = card(rng p) by CARD_1:5;
    hence thesis by FINSEQ_1:57;
  end;
  reconsider f = p as Function of dom p, rng p by FUNCT_2:1;
  reconsider B = dom p as finite set;
  assume card(rng p) = len p;
  then card(rng p) = card(Seg(len p)) by FINSEQ_1:57;
  then
A1: card(rng p) = card B by FINSEQ_1:def 3;
  dom f is finite;
  hence thesis by A1,Lm1;
end;





reserve f for Function of A,B;

theorem
  for A,B being finite set, f being Function of A,B holds card A = card
  B & f is one-to-one implies rng f = B
proof
  let A,B be finite set;
  let f be Function of A,B;
  assume that
A1: card A = card B and
A2: f is one-to-one;
A3: A,B are_equipotent by A1,CARD_1:5;
  consider p such that
A4: rng p = A and
A5: p is one-to-one by Th58;
  dom p,p .: (dom p) are_equipotent by A5,CARD_1:33;
  then dom p,A are_equipotent by A4,RELAT_1:113;
  then
A6: dom p,B are_equipotent by A3,WELLORD2:15;
  consider q such that
A7: rng q = B and
A8: q is one-to-one by Th58;
A9: dom q = Seg(len q) by FINSEQ_1:def 3;
  dom q,q .: (dom q) are_equipotent by A8,CARD_1:33;
  then dom q,B are_equipotent by A7,RELAT_1:113;
  then dom p,dom q are_equipotent by A6,WELLORD2:15;
  then card dom p = card(Seg(len q)) by A9,CARD_1:5
    .= len q by FINSEQ_1:57;
  then
A10: len q = card(Seg(len p)) by FINSEQ_1:def 3
    .= len p by FINSEQ_1:57;
  per cases;
  suppose
    B = {};
    hence thesis;
  end;
  suppose
A11: B <> {};
    dom(q") = rng q by A8,FUNCT_1:33;
    then rng f c= dom(q") by A7,RELAT_1:def 19;
    then dom(q" * f) = dom f by RELAT_1:27;
    then rng p = dom(q" * f) by A4,A11,FUNCT_2:def 1;
    then
A12: dom(q" * f * p) = dom p by RELAT_1:27
      .= Seg(len p) by FINSEQ_1:def 3;
    then reconsider g = q" * f * p as FinSequence by FINSEQ_1:def 2;
    g = q" * (f * p) by RELAT_1:36;
    then rng g c= rng(q") by RELAT_1:26;
    then rng g c= dom q by A8,FUNCT_1:33;
    then rng g c= dom g by A10,A12,FINSEQ_1:def 3;
    then rng g = dom g by A2,A5,A8,Th59;
    then rng g = dom q by A10,A12,FINSEQ_1:def 3;
    then
A13: rng(q * g) = B by A7,RELAT_1:28;
A14: rng f c= rng q by A7,RELAT_1:def 19;
A15: rng p = dom f by A4,A11,FUNCT_2:def 1;
    q * g = q * (q " * (f * p)) by RELAT_1:36
      .= q * q" * (f * p) by RELAT_1:36
      .= id(rng q) * (f * p) by A8,FUNCT_1:39
      .= id(rng q) * f * p by RELAT_1:36
      .= f * p by A14,RELAT_1:53;
    hence thesis by A13,A15,RELAT_1:28;
  end;
end;

theorem
  for A,B being finite set, f being Function of A,B st card A = card B &
  rng f = B holds f is one-to-one by Lm1;



theorem
  card B in card A & B <> {} implies ex x,y st x in A & y in A & x <> y
  & f.x = f.y
proof
  assume that
A1: card B in card A and
A2: B <> {} and
A3: for x,y st x in A & y in A & x <> y holds f.x <> f.y;
A4: dom f = A by A2,FUNCT_2:def 1;
  then for x,y holds x in dom f & y in dom f & f.x = f.y implies x = y by A3;
  then f is one-to-one by FUNCT_1:def 4;
  then dom f,f .: (dom f) are_equipotent by CARD_1:33;
  then dom f,rng f are_equipotent by RELAT_1:113;
  then
A5: card A = card(rng f) by A4,CARD_1:5;
  rng f c= B by RELAT_1:def 19;
  then card A c= card B by A5,CARD_1:11;
  hence contradiction by A1,CARD_1:4;
end;

theorem Th66:
  card A in card B implies ex x st x in B & for y st y in A holds f.y <> x
proof
  assume that
A1: card A in card B and
A2: for x st x in B ex y st y in A & f.y = x;
A3: dom f = A by A1,CARD_1:27,FUNCT_2:def 1;
  rng f = B
  proof
    thus rng f c= B by RELAT_1:def 19;
    let x;
    assume x in B;
    then ex y st y in A & f.y = x by A2;
    hence thesis by A3,FUNCT_1:def 3;
  end;
  then card B c= card A by A3,CARD_1:12;
  hence thesis by A1,CARD_1:4;
end;

begin 


theorem
  for D being non empty set, f being FinSequence of D, p being Element
  of D holds (f^<*p*>)/.(len f + 1) = p
proof
  let D be non empty set, f be FinSequence of D, p be Element of D;
  len(f^<*p*>) = len f + len<*p*> by FINSEQ_1:22;
  then 1 <= len f + 1 & len f + 1 <= len(f^<*p*>) by FINSEQ_1:39,NAT_1:11;
  then len f + 1 in dom(f^<*p*>) by FINSEQ_3:25;
  hence (f^<*p*>)/.(len f + 1) = (f^<*p*>).(len f + 1) by PARTFUN1:def 6
    .= p by FINSEQ_1:42;
end;



theorem
  for E being non empty set, p,q being FinSequence of E st k in dom p
  holds (p ^ q)/.k = p/.k
proof
  let E be non empty set, p,q be FinSequence of E;
  assume
A1: k in dom p;
  then k in dom(p ^ q) by FINSEQ_3:22;
  hence (p ^ q)/.k = (p ^ q).k by PARTFUN1:def 6
    .= p.k by A1,FINSEQ_1:def 7
    .= p/.k by A1,PARTFUN1:def 6;
end;

theorem
  for E being non empty set, p,q being FinSequence of E st k in dom q
  holds (p ^ q)/.(len p + k) = q/.k
proof
  let E be non empty set, p,q be FinSequence of E;
  assume
A1: k in dom q;
  then len p + k in dom(p ^ q) by FINSEQ_1:28;
  hence (p ^ q)/.(len p + k) = (p ^ q).(len p + k) by PARTFUN1:def 6
    .= q.k by A1,FINSEQ_1:def 7
    .= q/.k by A1,PARTFUN1:def 6;
end;



theorem
  for m being Nat for D being set, p being FinSequence of D st a in dom(
  p|m) holds (p|m qua FinSequence of D)/.a = p/.a
proof
  let m be Nat;
  let D be set, p being FinSequence of D;
  assume
A1: a in dom(p|m);
  then a in dom p /\ (Seg m) by RELAT_1:61;
  then
A2: a in dom p by XBOOLE_0:def 4;
  thus (p|m)/.a = (p|Seg m).a by A1,PARTFUN1:def 6
    .= p.a by A1,FUNCT_1:47
    .= p/.a by A2,PARTFUN1:def 6;
end;



theorem
  for D being set, f being FinSequence of D, n,m st n in dom f & m in
  Seg n holds m in dom f & (f|n)/.m = f/.m
proof
  let D be set, f be FinSequence of D, n,m;
  assume that
A1: n in dom f and
A2: m in Seg n;
  dom f = Seg len f & n<=len f by A1,FINSEQ_1:def 3,FINSEQ_3:25;
  then
A3: Seg n c= dom f by FINSEQ_1:5;
  hence m in dom f by A2;
A4: Seg n = dom f /\ Seg n by A3,XBOOLE_1:28
    .= dom(f|n) by RELAT_1:61;
  hence (f|n)/.m = (f|n).m by A2,PARTFUN1:def 6
    .= f.m by A2,A4,FUNCT_1:47
    .= f/.m by A2,A3,PARTFUN1:def 6;
end;



theorem
  for X being finite set st n <= card X holds ex A being finite Subset
  of X st card A = n
proof
  let X be finite set such that
A1: n <= card X;
  consider p being FinSequence such that
A2: rng p = X and
A3: p is one-to-one by Th58;
  reconsider q = p|Seg n as FinSequence by FINSEQ_1:15;
  n <= len p by A1,A2,A3,Th62;
  then
A4: len q = n by FINSEQ_1:17;
  reconsider A = rng q as Subset of X by A2,RELAT_1:70;
  q is one-to-one by A3,FUNCT_1:52;
  then card A = n by A4,Th62;
  hence thesis;
end;

reserve f for Function;

theorem
  f is one-to-one & x in rng f implies card Coim (f,x) = 1
proof
  assume f is one-to-one & x in rng f;
  then f just_once_values x by Th8;
  hence thesis by Def2;
end;



definition
  let x1,x2,x3,x4 be set;
  func <*x1,x2,x3,x4*> -> set equals
  <*x1,x2,x3*>^<*x4*>;
  correctness;
  let x5 be set;
  func <*x1,x2,x3,x4,x5*> -> set equals
  <*x1,x2,x3*>^<*x4,x5*>;
  correctness;
end;

registration
  let x1,x2,x3,x4 be set;
  cluster <*x1,x2,x3,x4*> -> non empty Function-like Relation-like;
  coherence;
  let x5 be set;
  cluster <*x1,x2,x3,x4,x5*> -> non empty Function-like Relation-like;
  coherence;
end;

registration
  let x1,x2,x3,x4 be set;
  cluster <*x1,x2,x3,x4*> -> FinSequence-like;
  coherence;
  let x5 be set;
  cluster <*x1,x2,x3,x4,x5*> -> FinSequence-like;
  coherence;
end;

definition
  let D be non empty set,x1,x2,x3,x4 be Element of D;
  redefine func <* x1,x2,x3,x4*> -> FinSequence of D;
  coherence
  proof
    <*x1,x2,x3,x4*>=<* x1,x2,x3*>^<*x4*>;
    hence thesis;
  end;
end;

definition
  let D be non empty set,x1,x2,x3,x4,x5 be Element of D;
  redefine func <*x1,x2,x3,x4,x5*> -> FinSequence of D;
  coherence
  proof
    <*x1,x2,x3,x4,x5*>=<* x1,x2,x3*>^<*x4,x5*>;
    hence thesis;
  end;
end;

reserve x1,x2,x3,x4,x5 for set;

theorem Th74:
  <*x1,x2,x3,x4*>=<*x1,x2,x3*>^<*x4*> & <*x1,x2,x3,x4*>=<*x1,x2*>^
<*x3,x4*> & <*x1,x2,x3,x4*>=<*x1*>^<*x2,x3,x4*> & <*x1,x2,x3,x4*>=<*x1*>^<*x2*>
  ^<*x3*>^<*x4*>
proof
  thus <*x1,x2,x3,x4*>=<*x1,x2,x3*>^<*x4*>;
  thus <*x1,x2,x3,x4*>=<*x1,x2*>^<*x3,x4*> by FINSEQ_1:32;
  hence <*x1,x2,x3,x4*>=<*x1*>^(<*x2*>^<*x3,x4*>) by FINSEQ_1:32
    .=<*x1*>^<*x2,x3,x4*> by FINSEQ_1:43;
  thus thesis;
end;

theorem Th75:
  <*x1,x2,x3,x4,x5*>=<*x1,x2,x3*>^<*x4,x5*> & <*x1,x2,x3,x4,x5*>=
<*x1,x2,x3,x4*>^<*x5*> & <*x1,x2,x3,x4,x5*>=<*x1*>^<*x2*>^<*x3*>^<*x4*>^<*x5*>
& <*x1,x2,x3,x4,x5*>=<*x1,x2*>^<*x3,x4,x5*> & <*x1,x2,x3,x4,x5*>=<*x1*>^<*x2,x3
  ,x4,x5*>
proof
  thus <*x1,x2,x3,x4,x5*>=<*x1,x2,x3*>^<*x4,x5*>;
  thus
A1: <*x1,x2,x3,x4,x5*>=<*x1,x2,x3,x4*>^<*x5*> by FINSEQ_1:32;
  thus <*x1,x2,x3,x4,x5*>=<*x1*>^<*x2*>^<*x3*>^<*x4*>^<*x5*> by FINSEQ_1:32;
  thus <*x1,x2,x3,x4,x5*>=<*x1,x2*>^(<*x3*>^<*x4*>)^<*x5*> by A1,FINSEQ_1:32
    .=<*x1,x2*>^<*x3,x4,x5*> by FINSEQ_1:32;
  hence <*x1,x2,x3,x4,x5*>=<*x1*>^(<*x2*>^<*x3,x4,x5*>) by FINSEQ_1:32
    .=<*x1*>^<*x2,x3,x4,x5*> by Th74;
end;

reserve p for FinSequence;

theorem Th76:
  p = <*x1,x2,x3,x4*> iff len p = 4 & p.1 = x1 & p.2 = x2 & p.3=x3 & p.4= x4
proof
  thus p = <*x1,x2,x3,x4*> implies len p = 4 & p.1 = x1 & p.2 = x2 & p.3 = x3
  & p.4 = x4
  proof
    set p3=<*x1,x2,x3*>;
    assume
A1: p =<*x1,x2,x3,x4*>;
    hence len p =len <*x1,x2,x3*> + len <*x4*> by FINSEQ_1:22
      .=3 + len <*x4*> by FINSEQ_1:45
      .=3+1 by FINSEQ_1:40
      .=4;
A2: dom p3 ={1,2,3} by FINSEQ_3:1,FINSEQ_1:89;
    then 1 in dom p3 by ENUMSET1:def 1;
    hence p.1 = p3.1 by A1,FINSEQ_1:def 7
      .=x1 by FINSEQ_1:45;
    2 in dom p3 by A2,ENUMSET1:def 1;
    hence p.2 = p3.2 by A1,FINSEQ_1:def 7
      .=x2 by FINSEQ_1:45;
    3 in dom p3 by A2,ENUMSET1:def 1;
    hence p.3 = p3.3 by A1,FINSEQ_1:def 7
      .=x3 by FINSEQ_1:45;
    1 in {1} by TARSKI:def 1;
    then
A3: 1 in dom <*x4*> by FINSEQ_1:2,def 8;
    thus p.4 = (p3^<*x4*>).(3+1) by A1
      .=(p3^<*x4*>).(len p3 + 1) by FINSEQ_1:45
      .= <*x4*>.1 by A3,FINSEQ_1:def 7
      .= x4 by FINSEQ_1:def 8;
  end;
  assume that
A4: len p = 4 and
A5: p.1 = x1 and
A6: p.2 = x2 and
A7: p.3 = x3 and
A8: p.4=x4;
A9: for k be Nat st k in dom <*x1,x2,x3*> holds p.k=<*x1,x2,x3*>.k
  proof
A10: len <*x1,x2,x3*> = 3 by FINSEQ_1:45;
    let k be Nat;
    assume k in dom <*x1,x2,x3*>;
    then
A11: k in {1,2,3} by A10,FINSEQ_1:def 3,FINSEQ_3:1;
    per cases by A11,ENUMSET1:def 1;
    suppose
      k=1;
      hence thesis by A5,FINSEQ_1:45;
    end;
    suppose
      k=2;
      hence thesis by A6,FINSEQ_1:45;
    end;
    suppose
      k=3;
      hence thesis by A7,FINSEQ_1:45;
    end;
  end;
A12: for k be Nat st k in dom <*x4*> holds p.( (len <*x1,x2,x3*>) + k) = <*
  x4*>.k
  proof
    let k be Nat;
    assume k in dom <*x4*>;
    then k in {1} by FINSEQ_1:2,def 8;
    then
A13: k = 1 by TARSKI:def 1;
    hence p.( (len <*x1,x2,x3*>) + k) = p.(3+1) by FINSEQ_1:45
      .=<*x4*>.k by A8,A13,FINSEQ_1:def 8;
  end;
  dom p = Seg(3+1) by A4,FINSEQ_1:def 3
    .= Seg((len <*x1,x2,x3*>) + 1) by FINSEQ_1:45
    .= Seg((len <*x1,x2,x3*>) + len <*x4*>) by FINSEQ_1:39;
  hence thesis by A9,A12,FINSEQ_1:def 7;
end;

canceled;

theorem Th78:
  p = <*x1,x2,x3,x4,x5*> iff len p = 5 & p.1 = x1 & p.2 = x2 & p.3
  =x3 & p.4= x4 & p.5= x5
proof
  set p4=<*x1,x2,x3,x4*>;
  thus p = <*x1,x2,x3,x4,x5*> implies len p = 5 & p.1 = x1 & p.2 = x2 & p.3 =
  x3 & p.4 = x4 & p.5 = x5
  proof
    set p4=<*x1,x2,x3,x4*>;
    1 in {1} by TARSKI:def 1;
    then
A1: 1 in dom <*x5*> by FINSEQ_1:2,def 8;
    assume
A2: p =<*x1,x2,x3,x4,x5*>;
    then
A3: p = (p4^<*x5*>) by Th75;
    thus len p = len (<*x1,x2,x3,x4*>^<*x5*>) by A2,Th75
      .=len <*x1,x2,x3,x4*> + len <*x5*> by FINSEQ_1:22
      .=4 + len <*x5*> by Th76
      .=4+1 by FINSEQ_1:40
      .=5;
A4: dom p4 ={1,2,3,4} by FINSEQ_1:89,FINSEQ_3:2;
    then 1 in dom p4 by ENUMSET1:def 2;
    hence p.1 = p4.1 by A3,FINSEQ_1:def 7
      .=x1 by Th76;
    2 in dom p4 by A4,ENUMSET1:def 2;
    hence p.2 = p4.2 by A3,FINSEQ_1:def 7
      .=x2 by Th76;
    3 in dom p4 by A4,ENUMSET1:def 2;
    hence p.3 = p4.3 by A3,FINSEQ_1:def 7
      .=x3 by Th76;
    4 in dom p4 by A4,ENUMSET1:def 2;
    hence p.4 = p4.4 by A3,FINSEQ_1:def 7
      .=x4 by Th76;
    thus p.5 = (p4^<*x5*>).(4+1) by A2,Th75
      .=(p4^<*x5*>).(len p4+ 1) by Th76
      .= <*x5*>.1 by A1,FINSEQ_1:def 7
      .= x5 by FINSEQ_1:def 8;
  end;
  assume that
A5: len p = 5 and
A6: p.1 = x1 and
A7: p.2 = x2 and
A8: p.3 = x3 and
A9: p.4=x4 and
A10: p.5=x5;
A11: for k be Nat st k in dom p4 holds p.k=p4.k
  proof
A12: len p4 = 4 by Th76;
    let k be Nat;
    assume k in dom p4;
    then
A13: k in {1,2,3,4} by A12,FINSEQ_1:def 3,FINSEQ_3:2;
    per cases by A13,ENUMSET1:def 2;
    suppose
      k=1;
      hence thesis by A6,Th76;
    end;
    suppose
      k=2;
      hence thesis by A7,Th76;
    end;
    suppose
      k=3;
      hence thesis by A8,Th76;
    end;
    suppose
      k=4;
      hence thesis by A9,Th76;
    end;
  end;
A14: for k be Nat st k in dom <*x5*> holds p.( len p4 + k) = <*x5*>.k
  proof
    let k be Nat;
    assume k in dom <*x5*>;
    then k in {1} by FINSEQ_1:2,def 8;
    then
A15: k = 1 by TARSKI:def 1;
    hence p.( len p4 + k) = p.(4+1) by Th76
      .=<*x5*>.k by A10,A15,FINSEQ_1:def 8;
  end;
  dom p = Seg(4+1) by A5,FINSEQ_1:def 3
    .= Seg(len p4 + 1) by Th76
    .= Seg(len p4 + len <*x5*>) by FINSEQ_1:39;
  hence p=p4^<*x5*> by A11,A14,FINSEQ_1:def 7
    .=<*x1,x2,x3,x4,x5*> by Th75;
end;

reserve ND for non empty set;
reserve y1,y2,y3,y4,y5 for Element of ND;

canceled;

theorem
  <*y1,y2,y3,y4*>/.1 = y1 & <*y1,y2,y3,y4*>/.2 = y2 & <*y1,y2,y3,y4*>/.3
  = y3 & <*y1,y2,y3,y4*>/.4=y4
proof
  set s = <* y1,y2,y3,y4 *>;
A1: 2 in {1,2,3,4} & 3 in {1,2,3,4} by FINSEQ_3:2;
A2: s.2 = y2 & s.3 = y3 by Th76;
A3: 4 in {1,2,3,4} by FINSEQ_3:2;
A4: s.4 = y4 & 1 in {1,2,3,4} by Th76,FINSEQ_3:2;
  dom s = {1,2,3,4} & s.1 = y1 by Th76,FINSEQ_1:89,FINSEQ_3:2;
  hence thesis by A2,A4,A1,A3,PARTFUN1:def 6;
end;

theorem
  <*y1,y2,y3,y4,y5*>/.1 = y1 & <*y1,y2,y3,y4,y5*>/.2 = y2 & <*y1,y2,y3,
  y4,y5*>/.3 = y3 & <*y1,y2,y3,y4,y5*>/.4=y4 & <*y1,y2,y3,y4,y5*>/.5=y5
proof
  set s = <* y1,y2,y3,y4,y5 *>, i5={1,2,3,4,5};
A1: 1 in i5 & 2 in i5 by FINSEQ_3:3;
A2: 3 in i5 & 4 in i5 by FINSEQ_3:3;
A3: s.4 = y4 & s.5=y5 by Th78;
A4: s.2 = y2 & s.3 = y3 by Th78;
A5: 5 in i5 by FINSEQ_3:3;
  dom s =i5 & s.1 = y1 by Th78,FINSEQ_1:89,FINSEQ_3:3;
  hence thesis by A4,A3,A1,A2,A5,PARTFUN1:def 6;
end;



scheme
  Sch1{D()->non empty set, N()->Nat, P[set,set]}: ex f being FinSequence of D(
  ) st len f = N() & for n st n in Seg N() holds P[n,f/.n]
provided
A1: for n st n in Seg N() ex d being Element of D() st P[n,d]
proof
  defpred P1[set,set] means P[$1,$2] & $2 in D();
  reconsider X=D() as set;
A2: now
    let x be set;
    assume
A3: x in Seg N();
    then reconsider x9=x as Element of NAT;
    consider d be Element of D() such that
A4: P[x9,d] by A1,A3;
    reconsider y=d as set;
    take y;
    thus y in X & P1[x,y] by A4;
  end;
  consider f be Function such that
A5: dom f = Seg N() & rng f c= X & for x be set st x in Seg N() holds P1
  [x,f.x] from FUNCT_1:sch 5(A2);
  reconsider f as FinSequence by A5,FINSEQ_1:def 2;
  reconsider f as FinSequence of D() by A5,FINSEQ_1:def 4;
  take f;
  N() in NAT by ORDINAL1:def 12;
  hence len f = N() by A5,FINSEQ_1:def 3;
  let n;
  assume
A6: n in Seg N();
  then P[n,f.n] by A5;
  hence thesis by A5,A6,PARTFUN1:def 6;
end;



theorem Th82:
  for D being non empty set, p,q being FinSequence of D st p c= q
  holds ex p9 being FinSequence of D st p ^ p9 = q
proof
  let D be non empty set, p,q be FinSequence of D;
  assume
A1: p c= q;
  dom p = Seg len p & dom q = Seg len q by FINSEQ_1:def 3;
  then Seg len p c= Seg len q by A1,GRFUNC_1:2;
  then len p <= len q by FINSEQ_1:5;
  then reconsider N = len q - len p as Element of NAT by INT_1:3,XREAL_1:48;
  defpred P[Nat,set] means q/.(len p + $1) = $2;
A2: for n being Nat st n in Seg N ex d being Element of D st P[n,d];
  consider f being FinSequence of D such that
A3: len f = N and
A4: for n being Nat st n in Seg N holds P[n,f/.n] from Sch1(A2);
  take f;
A5: len (p ^ f) = len p + N by A3,FINSEQ_1:22
    .= len q;
  now
    let k be Nat;
    assume that
A6: 1 <= k and
A7: k <= len (p ^ f);
A8: k in NAT by ORDINAL1:def 12;
    then k in Seg len q by A5,A6,A7;
    then
A9: k in dom q by FINSEQ_1:def 3;
    per cases;
    suppose
      k <= len p;
      then k in Seg len p by A6,A8;
      then
A10:  k in dom p by FINSEQ_1:def 3;
      hence (p ^ f).k = p.k by FINSEQ_1:def 7
        .= q.k by A1,A10,GRFUNC_1:2;
    end;
    suppose
A11:  len p < k;
      then k - len p > 0 by XREAL_1:50;
      then reconsider kk = k - len p as Element of NAT by INT_1:3;
      k <= len p + len f by A7,FINSEQ_1:22;
      then
A12:  kk <= len p + len f - len p by XREAL_1:9;
      k - len p >= 0 + 1 by A11,INT_1:7,XREAL_1:50;
      then
A13:  kk in Seg len f by A12;
      then
A14:  kk in dom f by FINSEQ_1:def 3;
      thus (p ^ f).k = f.kk by A7,A11,FINSEQ_1:24
        .= f/.kk by A14,PARTFUN1:def 6
        .= q/.(len p + kk) by A3,A4,A13
        .= q.k by A9,PARTFUN1:def 6;
    end;
  end;
  hence thesis by A5,FINSEQ_1:14;
end;

theorem
  for D being non empty set, p,q being FinSequence of D, i being Element
  of NAT st p c= q & 1 <= i & i <= len p holds q.i = p.i
proof
  let D be non empty set, p,q be FinSequence of D, i be Element of NAT;
  assume p c= q;
  then
A1: ex p9 being FinSequence of D st p ^ p9 = q by Th82;
  assume 1 <= i & i <= len p;
  hence thesis by A1,FINSEQ_1:64;
end;



scheme
  PiLambdaD{D()-> non empty set, l()->Nat, F(set)-> Element of D()}: ex g
being FinSequence of D() st len g=l() & for n being Nat st n in dom g holds g/.
  n=F(n) proof
  consider g being FinSequence of D() such that
A1: len g=l() & for n be Nat st n in dom g holds g.n=F(n) from FINSEQ_2:
  sch 1;
  take g;
  thus len g=l() by A1;
  let n be Nat;
  assume
A2: n in dom g;
  then g.n = F(n) by A1;
  hence thesis by A2,PARTFUN1:def 6;
end;



registration
  let x1,x2,x3,x4 be set;
  cluster <*x1,x2,x3,x4*> -> 4-element;
  coherence;
  let x5 be set;
  cluster <*x1,x2,x3,x4,x5*> -> 5-element;
  coherence;
end;

begin 

theorem
  for m, n being Nat holds m < n implies ex p being Element of NAT
  st n = m+p & 1 <= p
proof
  let m, n be Nat;
  assume
A1: m < n;
  then consider p being Nat such that
A2: n=m+p by NAT_1:10;
  reconsider p as Element of NAT by ORDINAL1:def 12;
  take p;
  thus n = m+p by A2;
  assume p < 1;
  then p < 0+1;
  then p = 0 by NAT_1:13;
  hence contradiction by A1,A2;
end;

theorem
  for S being set, D1, D2 being non empty set, f1 being Function of
  S, D1, f2 being Function of D1, D2 holds f1 is bijective & f2 is bijective
  implies f2*f1 is bijective
proof
  let S be set, D1, D2 be non empty set, f1 be Function of S, D1, f2 be
  Function of D1, D2;
  set f3 = f2*f1;
A1: dom f2 = D1 by FUNCT_2:def 1;
  assume
A2: f1 is bijective & f2 is bijective;
  then rng f2 = D2 & rng f1 = D1 by FUNCT_2:def 3;
  then rng f3 = D2 by A1,RELAT_1:28;
  then f3 is one-to-one onto by A2,FUNCT_2:def 3;
  hence thesis;
end;



theorem
  for Y being set, E1, E2 being Equivalence_Relation of Y holds
  Class E1 = Class E2 implies E1 = E2
proof
  let Y be set, E1, E2 be Equivalence_Relation of Y such that
A1: Class E1 = Class E2;
  now
    let x be set;
    hereby
      assume
A2:   x in E1;
      then consider a, b being set such that
A3:   x = [a, b] and
A4:   a in Y and
A5:   b in Y by RELSET_1:2;
      reconsider a,b as Element of Y by A4,A5;
      Class (E1, b) in Class E2 by A1,A4,EQREL_1:def 3;
      then consider c being set such that
      c in Y and
A6:   Class (E1, b) = Class (E2, c) by EQREL_1:def 3;
      b in Class (E1, b) by A4,EQREL_1:20;
      then [b, c] in E2 by A6,EQREL_1:19;
      then
A7:   [c, b] in E2 by EQREL_1:6;
      a in Class (E1, b) by A2,A3,EQREL_1:19;
      then [a, c] in E2 by A6,EQREL_1:19;
      hence x in E2 by A3,A7,EQREL_1:7;
    end;
    assume
A8: x in E2;
    then consider a, b being set such that
A9: x = [a, b] and
A10: a in Y and
A11: b in Y by RELSET_1:2;
    reconsider a, b as Element of Y by A10,A11;
    Class (E2, b) in Class E1 by A1,A10,EQREL_1:def 3;
    then consider c being set such that
    c in Y and
A12: Class (E2, b) = Class (E1, c) by EQREL_1:def 3;
    b in Class (E2, b) by A10,EQREL_1:20;
    then [b, c] in E1 by A12,EQREL_1:19;
    then
A13: [c, b] in E1 by EQREL_1:6;
    a in Class (E2, b) by A8,A9,EQREL_1:19;
    then [a, c] in E1 by A12,EQREL_1:19;
    hence x in E1 by A9,A13,EQREL_1:7;
  end;
  hence thesis by TARSKI:1;
end;

registration
  let Z be finite set;
  cluster -> finite for a_partition of Z;
  coherence;
end;

registration
  let X be non empty finite set;
  cluster non empty finite for a_partition of X;
  existence
  proof
    set p = the a_partition of X;
    take p;
    thus thesis;
  end;
end;

reserve X, A for non empty finite set,
  PX for a_partition of X,
  PA1, PA2 for a_partition of A;

theorem Th87:
  for X being non empty set, PX being a_partition of X for Pi
  being set st Pi in PX ex x being Element of X st x in Pi
proof
  let X be non empty set, PX be a_partition of X;
  let Pi be set;
  assume Pi in PX;
  then reconsider Pi as Element of PX;
  set q = the Element of Pi;
A1: Pi <> {} by EQREL_1:def 4;
  then q in Pi;
  then reconsider q as Element of X;
  take q;
  thus thesis by A1;
end;

theorem
  card PX <= card X
proof
  assume card PX > card X;
  then card card X in card card PX by NAT_1:41;
  then consider Pi being set such that
A1: Pi in PX and
A2: for x being set st x in X holds (proj PX).x <> Pi by Th66;
  reconsider Pi as Element of PX by A1;
  consider q being Element of X such that
A3: q in Pi by Th87;
  reconsider Pq = (proj PX).q as Element of PX;
A4: Pq = Pi or Pq misses Pi by EQREL_1:def 4;
  q in Pq by EQREL_1:def 9;
  hence contradiction by A2,A3,A4,XBOOLE_0:3;
end;

theorem
  PA1 is_finer_than PA2 implies card PA2 <= card PA1
proof
  defpred P[set,set] means $1 c= $2;
  assume PA1 is_finer_than PA2;
  then
A1: for e being set st e in PA1 ex u being set st u in PA2 & P[e,u] by
SETFAM_1:def 2;
  consider f being Function of PA1, PA2 such that
A2: for e being set st e in PA1 holds P[e,f.e] from FUNCT_2:sch 1 (A1);
  assume card PA1 < card PA2;
  then card card PA1 in card card PA2 by NAT_1:41;
  then consider p2i being set such that
A3: p2i in PA2 and
A4: for x being set st x in PA1 holds f.x <> p2i by Th66;
  reconsider p2i as Element of PA2 by A3;
  consider q being Element of A such that
A5: q in p2i by Th87;
  reconsider p2q = f.((proj PA1).q) as Element of PA2;
A6: p2q=p2i or p2q misses p2i by EQREL_1:def 4;
  q in (proj PA1).q & (proj PA1).q c= p2q by A2,EQREL_1:def 9;
  hence contradiction by A4,A5,A6,XBOOLE_0:3;
end;

theorem Th90:
  PA1 is_finer_than PA2 implies for p2 being Element of PA2 ex p1
  being Element of PA1 st p1 c=p2
proof
  assume
A1: PA1 is_finer_than PA2;
  let p2 be Element of PA2;
  consider E1 being Equivalence_Relation of A such that
A2: PA1 = Class E1 by EQREL_1:34;
  reconsider p29 = p2 as Subset of A;
  consider E2 being Equivalence_Relation of A such that
A3: PA2 = Class E2 by EQREL_1:34;
  consider a being set such that
A4: a in A and
A5: p29 = Class (E2, a) by A3,EQREL_1:def 3;
A6: a in Class (E1, a) by A4,EQREL_1:20;
  reconsider E1a = Class (E1, a) as Element of PA1 by A2,A4,EQREL_1:def 3;
  consider p22 being set such that
A7: p22 in PA2 and
A8: E1a c= p22 by A1,SETFAM_1:def 2;
  reconsider p229 = p22 as Subset of A by A7;
  take E1a;
  ex b being set st b in A & p229 = Class (E2, b) by A3,A7,EQREL_1:def 3;
  hence thesis by A5,A8,A6,EQREL_1:23;
end;

theorem
  PA1 is_finer_than PA2 & card PA1 = card PA2 implies PA1 = PA2
proof
  defpred P[set,set] means $1 c= $2;
  assume that
A1: PA1 is_finer_than PA2 and
A2: card PA1 = card PA2;
A3: for e being set st e in PA1 ex u being set st u in PA2 & P[e,u] by A1,
SETFAM_1:def 2;
  consider f being Function of PA1, PA2 such that
A4: for e being set st e in PA1 holds P[e,f.e] from FUNCT_2:sch 1 (A3);
A5: dom f = PA1 by FUNCT_2:def 1;
A6: PA2 c= rng f
  proof
    let p2 be set;
    assume p2 in PA2;
    then reconsider p29 = p2 as Element of PA2;
    consider p1 being Element of PA1 such that
A7: p1 c= p29 by A1,Th90;
    reconsider fp1 = f.p1 as Subset of A by TARSKI:def 3;
A8: p1 c= f.p1 by A4;
    p29 meets f.p1
    proof
      ex x being Element of A st x in p1 by Th87;
      hence thesis by A7,A8,XBOOLE_0:3;
    end;
    then p29 = fp1 by EQREL_1:def 4;
    hence thesis by A5,FUNCT_1:def 3;
  end;
  rng f c= PA2 by RELAT_1:def 19;
  then rng f = PA2 by A6,XBOOLE_0:def 10;
  then
A9: f is one-to-one by A2,Lm1;
A10: for x being Element of PA1 holds x = f.x
  proof
    let x be Element of PA1;
    reconsider fx = f.x as Subset of A by TARSKI:def 3;
    consider E1 being Equivalence_Relation of A such that
A11: PA1 = Class E1 by EQREL_1:34;
    assume x <> f.x;
    then consider a being set such that
A12: a in x & not a in f.x or a in f.x & not a in x by TARSKI:1;
A13: fx c= A;
    then
A14: a in Class (E1, a) by A12,EQREL_1:20;
    reconsider CE1a = Class (E1, a) as Element of PA1 by A13,A12,A11,
EQREL_1:def 3;
    reconsider fCE1a = f.CE1a as Subset of A by TARSKI:def 3;
A15: x c= f.x by A4;
    CE1a c= f.CE1a by A4;
    then fx meets fCE1a by A15,A12,A14,XBOOLE_0:3;
    then fx = fCE1a by EQREL_1:def 4;
    hence contradiction by A5,A9,A15,A12,A14,FUNCT_1:def 4;
  end;
  now
    let x be set;
    hereby
      assume x in PA1;
      then reconsider x9 = x as Element of PA1;
      set fx = f.x9;
      fx in PA2;
      hence x in PA2 by A10;
    end;
    assume x in PA2;
    then consider y being set such that
A16: y in dom f and
A17: x = f.y by A6,FUNCT_1:def 3;
    reconsider y9 = y as Element of PA1 by A16,FUNCT_2:def 1;
    y9 = f.y9 by A10;
    hence x in PA1 by A17;
  end;
  hence thesis by TARSKI:1;
end;

registration
  let D be set, M be FinSequence of D*;
  let k;
  cluster M/.k  -> Function-like Relation-like; 
  coherence;
  
end;

registration
  let D be set, M be FinSequence of D*;
  let k;
  cluster M/.k -> FinSequence-like D-valued for Function;
  coherence by FINSEQ_1:def 11;
end;















begin

reserve x,y,y1,y2 for set,
  D for non empty set,
  d,d1,d2,d3 for Element of D,
  F ,G,H,H1,H2 for FinSequence of D,
  f,f1,f2 for Function of NAT,D,
  g for BinOp of D,
  k,n,i,l for Element of NAT,
  P for Permutation of dom F;

definition
  let D,F,g;
  assume
A1: g is having_a_unity or len F >= 1;
  func g "**" F -> Element of D means
  :Def1:
  it = the_unity_wrt g if g is
having_a_unity & len F = 0 otherwise ex f st f.1 = F.1 & (for n st 0 <> n & n <
  len F holds f.(n + 1) = g.(f.n,F.(n + 1))) & it = f.(len F);
  existence
  proof
    now
      per cases;
      suppose
        len F = 0;
        hence thesis by A1;
      end;
      suppose
A2:     len F <> 0;
        defpred P[Element of NAT] means for F st len F = $1 & len F <> 0 ex d,
f st f.1 = F.1 & (for n st 0 <> n & n < len F holds f.(n + 1) = g.(f.n,F.(n + 1
        ))) & d = f.(len F);
A3:     for k st P[k] holds P[k + 1]
        proof
          let k;
          assume
A4:       P[k];
          let F;
          assume that
A5:       len F = k + 1 and
          len F <> 0;
          reconsider G = F | Seg k as FinSequence of D by FINSEQ_1:18;
A6:       len G = k by A5,FINSEQ_3:53;
          now
            per cases;
            suppose
A7:           len G = 0;
              set f = NAT --> F.1;
A8:           rng f c= D
              proof
                let x;
                assume x in rng f;
                then ex y st y in dom f & f.y = x by FUNCT_1:def 3;
                then
A9:             x = F.1 by FUNCOP_1:7;
                1 in dom F by A5,A6,A7,FINSEQ_3:25;
                then
A10:            x in rng F by A9,FUNCT_1:def 3;
                rng F c= D by FINSEQ_1:def 4;
                hence thesis by A10;
              end;
              dom f = NAT by FUNCOP_1:13;
              then reconsider f as Function of NAT,D by A8,RELSET_1:4;
              take d = f.1,f;
              thus f.1 = F.1 by FUNCOP_1:7;
              thus for n st 0 <> n & n < len F holds f.(n + 1) = g.(f.n,F.(n +
              1)) by A5,A6,A7,NAT_1:14;
              thus d = f.(len F) by A5,A6,A7;
            end;
            suppose
A11:          len G <> 0;
              reconsider j = len F as Element of NAT;
              1 <= len F by A5,NAT_1:12;
              then len F in dom F by FINSEQ_3:25;
              then
A12:          F.(len F) in rng F by FUNCT_1:def 3;
              rng F c= D by FINSEQ_1:def 4;
              then reconsider t = F.(len F) as Element of D by A12;
              len G >= 1 by A11,NAT_1:14;
              then
A13:          1 in dom G by FINSEQ_3:25;
              consider d,f such that
A14:          f.1 = G.1 and
A15:          for n st 0 <> n & n < len G holds f.(n + 1) = g.(f.n,G.
              (n + 1)) and
A16:          d = f.(len G) by A4,A5,A11,FINSEQ_3:53;
              deffunc F(Element of NAT) = f.$1;
              consider h being Function of NAT,D such that
A17:          h.j = g.(d,t) and
A18:          for n being Element of NAT st n <> j holds h.n = F(n)
              from FUNCT_2:sch 6;
              take a = h.j,h;
              1 <> j by A5,A11,FINSEQ_3:53;
              hence h.1 = G.1 by A14,A18
                .= F.1 by A13,FUNCT_1:47;
              thus for n st 0 <> n & n < len F holds h.(n + 1) = g.(h.n,F.(n +
              1))
              proof
                let n;
                assume that
A19:            n <> 0 and
A20:            n < len F;
                now
                  per cases;
                  suppose
A21:                n + 1 = len F;
                    len G <> len F by A5,A6;
                    hence thesis by A5,A6,A16,A17,A18,A21;
                  end;
                  suppose
A22:                n + 1 <> len F;
                    n + 1 <= len F by A20,NAT_1:13;
                    then
A23:                n + 1 < len F by A22,XXREAL_0:1;
                    then
A24:                n < len G by A5,A6,XREAL_1:6;
                    1 <= n + 1 & n + 1 <= len G by A5,A6,A23,NAT_1:12,13;
                    then
A25:                n + 1 in dom G by FINSEQ_3:25;
                    h.(n + 1) = f.(n + 1) by A18,A22
                      .= g.(f.n,G.(n + 1)) by A15,A19,A24
                      .= g.(f.n,F.(n + 1)) by A25,FUNCT_1:47
                      .= g.(h.n,F.(n + 1)) by A18,A20;
                    hence thesis;
                  end;
                end;
                hence thesis;
              end;
              thus a = h.(len F);
            end;
          end;
          hence thesis;
        end;
A26:    P[0];
        for k holds P[k] from NAT_1:sch 1(A26,A3);
        hence thesis by A2;
      end;
    end;
    hence thesis;
  end;
  uniqueness
  proof
    let d1,d2;
    thus g is having_a_unity & len F = 0 & d1 = the_unity_wrt g & d2 =
    the_unity_wrt g implies d1 = d2;
    assume
A27: not g is having_a_unity or len F <> 0;
    given f1 such that
A28: f1.1 = F.1 and
A29: for n st 0 <> n & n < len F holds f1.(n + 1) = g.(f1.n,F.(n + 1)) and
A30: d1 = f1.(len F);
    given f2 such that
A31: f2.1 = F.1 and
A32: for n st 0 <> n & n < len F holds f2.(n + 1) = g.(f2.n,F.(n + 1)) and
A33: d2 = f2.(len F);
    defpred P[Element of NAT] means $1 <> 0 & $1 <= len F implies f1.$1 = f2.
    $1;
A34: for n st P[n] holds P[n + 1]
    proof
      let n;
      assume
A35:  n <> 0 & n <= len F implies f1.n = f2.n;
      assume that
      n + 1 <> 0 and
A36:  n + 1 <= len F;
      now
        per cases;
        suppose
          n = 0;
          hence thesis by A28,A31;
        end;
        suppose
A37:      n <> 0;
A38:      n < len F by A36,NAT_1:13;
          then f1.(n + 1) = g.(f1.n,F.(n + 1)) by A29,A37;
          hence thesis by A32,A35,A37,A38;
        end;
      end;
      hence thesis;
    end;
A39: P[0];
    for n holds P[n] from NAT_1:sch 1(A39,A34);
    hence thesis by A1,A27,A30,A33;
  end;
  consistency;
end;

theorem
  len F >= 1 implies ex f st f.1 = F.1 & (for n st 0 <> n & n < len F
  holds f.(n + 1) = g.(f.n,F.(n + 1))) & g "**" F = f.(len F) by Def1;

theorem
  len F >= 1 & (ex f st f.1 = F.1 & (for n st 0 <> n & n < len F holds f
  .(n + 1) = g.(f.n,F.(n + 1))) & d = f.(len F)) implies d = g "**" F by Def1;

definition
  let A be non empty set, F be Function of NAT,A, p be FinSequence of A;
  redefine func F +* p -> Function of NAT,A;
  coherence
  proof
A1: dom F = NAT by FUNCT_2:def 1;
    dom(F +* p) = dom F \/ dom p by FUNCT_4:def 1
      .= NAT by A1,XBOOLE_1:12;
    hence thesis by FUNCT_2:def 1;
  end;
end;

notation
  let f be FinSequence;
  synonym findom f for dom f;
end;

definition
  let f be FinSequence;
  redefine func findom f -> Element of Fin NAT;
  coherence
  proof
    dom f = Seg len f by FINSEQ_1:def 3;
    hence thesis by FINSUB_1:def 5;
  end;
end;

Lm1: len F >= 1 & g is associative & g is commutative implies g "**" F = g $$(
findom F,(NAT-->the_unity_wrt g)+*F)
proof
  assume that
A1: len F >= 1 and
A2: g is associative & g is commutative;
  set A = findom F;
  set h = (NAT-->the_unity_wrt g)+*F;
A3: dom F = Seg len F by FINSEQ_1:def 3;
  then consider G being Function of Fin NAT,D such that
A4: g $$ (A,h) = G.A and
  for d st d is_a_unity_wrt g holds G.{} = d and
A5: for n holds G.{n} = h.n and
A6: for B being Element of Fin NAT st B c= A & B <> {} for n st n in A
  \ B holds G.(B \/ {n}) = g.(G.B,h.n) by A1,A2,SETWISEO:def 3;
  consider f such that
A7: f.1 = F.1 and
A8: for n st 0 <> n & n < len F holds f.(n + 1) = g.(f.n,F.(n + 1)) and
A9: g "**" F = f.(len F) by A1,Def1;
  defpred P[Element of NAT] means $1 <> 0 & $1 <= len F implies f.$1 = G.(Seg
  $1);
A10: for n st P[n] holds P[n + 1]
  proof
    let n;
    assume
A11: n <> 0 & n <= len F implies f.n = G.(Seg n);
    assume that
    n + 1 <> 0 and
A12: n + 1 <= len F;
    now
      per cases;
      suppose
A13:    n = 0;
        1 in dom F by A1,A3,FINSEQ_1:1;
        then h.1 = F.1 by FUNCT_4:13;
        hence thesis by A7,A5,A13,FINSEQ_1:2;
      end;
      suppose
A14:    n <> 0;
        reconsider B = Seg n as Element of Fin NAT by FINSUB_1:def 5;
        n + 1 >= 1 by NAT_1:12;
        then
A15:    n + 1 in dom F by A12,FINSEQ_3:25;
A16:    n < len F by A12,NAT_1:13;
        then
A17:    f.(n + 1) = g.(f.n,F.(n + 1)) by A8,A14;
        not n + 1 in Seg n by FINSEQ_3:10;
        then
A18:    n + 1 in A \ Seg n by A15,XBOOLE_0:def 5;
A19:    Seg n c= A by A3,A16,FINSEQ_1:5;
        G.(Seg(n + 1)) = G.(Seg n \/ {n + 1}) by FINSEQ_1:9
          .= g.(G.B,h.(n + 1)) by A6,A14,A19,A18;
        hence thesis by A11,A12,A14,A17,A15,FUNCT_4:13,NAT_1:13;
      end;
    end;
    hence thesis;
  end;
A20: P[0];
  for n holds P[n] from NAT_1:sch 1(A20,A10);
  hence thesis by A1,A9,A3,A4;
end;

Lm2: len F = 0 & g is having_a_unity & g is associative & g is commutative
implies g "**" F = g $$(findom F,(NAT-->the_unity_wrt g)+*F)
proof
  assume that
A1: len F = 0 and
A2: g is having_a_unity and
A3: g is associative & g is commutative;
  F = {} by A1;
  then
A4: dom F = {}.NAT;
  thus g "**" F = the_unity_wrt g by A1,A2,Def1
    .= g $$(findom F,(NAT-->the_unity_wrt g)+*F) by A2,A3,A4,SETWISEO:31;
end;

theorem
  (g is having_a_unity or len F >= 1) & g is associative & g is
  commutative implies g "**" F = g $$(findom F,(NAT-->the_unity_wrt g)+*F)
proof
  len F = 0 or len F >= 1 by NAT_1:14;
  hence thesis by Lm1,Lm2;
end;

Lm3: g is having_a_unity implies g "**" <*>D = the_unity_wrt g
proof
A1: len <*>D = 0;
  assume g is having_a_unity;
  hence thesis by A1,Def1;
end;

Lm4: g "**" <* d *> = d
proof
A1: len<* d *> = 1 by FINSEQ_1:39;
  then
  ex f st f.1 = <* d *>.1 & (for n st 0 <> n & n < len<* d *> holds f.(n +
  1) = g.(f.n,<* d *>.(n + 1))) & g "**" <* d *> = f.len<* d *> by Def1;
  hence thesis by A1,FINSEQ_1:def 8;
end;

Lm5: len F >= 1 implies g "**" (F ^ <* d *>) = g.(g "**" F,d)
proof
  set G = F ^ <* d *>;
A1: G.(len F + 1) = d by FINSEQ_1:42;
A2: len G = len F + len<* d *> by FINSEQ_1:22
    .= len F + 1 by FINSEQ_1:39;
  then 1 <= len G by NAT_1:12;
  then consider f1 such that
A3: f1.1 = G.1 and
A4: for n st 0 <> n & n < len G holds f1.(n + 1) = g.(f1.n,G.(n + 1)) and
A5: g "**" G = f1.(len G) by Def1;
  assume
A6: len F >= 1;
  then consider f such that
A7: f.1 = F.1 and
A8: for n st 0 <> n & n < len F holds f.(n + 1) = g.(f.n,F.(n + 1)) and
A9: g "**" F = f.(len F) by Def1;
  defpred P[Element of NAT] means 0 <> $1 & $1 < len G implies f.$1 = f1.$1;
A10: for n st P[n] holds P[n + 1]
  proof
    let n;
    assume
A11: P[n];
    assume that
A12: 0 <> n + 1 and
A13: n + 1 < len G;
A14: n + 1 >= 1 by A12,NAT_1:14;
    now
      per cases by A14,XXREAL_0:1;
      suppose
A15:    n + 1 = 1;
        1 in dom F by A6,FINSEQ_3:25;
        hence thesis by A7,A3,A15,FINSEQ_1:def 7;
      end;
      suppose
A16:    n + 1 > 1;
        then n <> 0;
        then
A17:    f1.(n + 1) = g.(f1.n,G.(n + 1)) by A4,A13,NAT_1:12;
A18:    n + 1 <= len F by A2,A13,NAT_1:13;
        then
A19:    n < len F by NAT_1:13;
        1 <= n + 1 by NAT_1:12;
        then
A20:    n + 1 in dom F by A18,FINSEQ_3:25;
        n + 1 > 0 + 1 by A16;
        then f.(n + 1) = g.(f.n,F.(n + 1)) by A8,A19;
        hence thesis by A11,A13,A16,A17,A20,FINSEQ_1:def 7,NAT_1:12;
      end;
    end;
    hence thesis;
  end;
A21: P[0];
A22: for n holds P[n] from NAT_1:sch 1(A21,A10);
  g "**" G = g.(f1.(len F),G.(len F + 1)) by A6,A2,A4,A5,XREAL_1:29;
  hence thesis by A6,A9,A2,A1,A22,XREAL_1:29;
end;

Lm6: g is having_a_unity & len F = 0 implies g "**" (F ^ <* d *>) = g.(g "**"
F,d)
proof
  assume that
A1: g is having_a_unity and
A2: len F = 0;
  F = <*>D by A2;
  hence g "**" (F ^ <* d *>) = g "**" <* d *> by FINSEQ_1:34
    .= d by Lm4
    .= g.(the_unity_wrt g,d) by A1,SETWISEO:15
    .= g.(g "**" F,d) by A1,A2,Def1;
end;

theorem Th4:
  g is having_a_unity or len F >= 1 implies g "**" (F ^ <* d *>) =
  g.(g "**" F,d)
proof
  len F >= 1 or len F = 0 by NAT_1:14;
  hence thesis by Lm5,Lm6;
end;

theorem Th5:
  g is associative & (g is having_a_unity or len F >= 1 & len G >=
  1) implies g "**" (F ^ G) = g.(g "**" F,g "**" G)
proof
  defpred P[FinSequence of D] means for F,g st g is associative & (g is
having_a_unity or len F >= 1 & len $1 >= 1) holds g "**" (F ^ $1) = g.(g "**" F
  ,g "**" $1);
A1: for G,d st P[G] holds P[G ^ <* d *>]
  proof
    let G,d;
    assume
A2: P[G];
    let F,g;
    assume that
A3: g is associative and
A4: g is having_a_unity or len F >= 1 & len(G ^ <* d *>) >= 1;
A5: now
A6:   len(F ^ G) = len F + len G by FINSEQ_1:22;
      assume not g is having_a_unity;
      hence len(F ^ G) >= 1 by A4,A6,NAT_1:12;
    end;
A7: g "**" (F ^ (G ^ <* d *>)) = g "**" (F ^ G ^ <* d *>) by FINSEQ_1:32
      .= g.(g "**" (F ^ G),d) by A5,Th4;
    now
      per cases;
      suppose
A8:     len G <> 0;
        then len G >= 1 by NAT_1:14;
        hence
        g "**" (F ^ (G ^ <* d *>)) = g.(g.(g "**" F,g "**" G),d) by A2,A3,A4,A7
          .= g.(g "**" F,g.(g "**" G,d)) by A3,BINOP_1:def 3
          .= g.(g "**" F,g "**" (G ^ <* d *>)) by A8,Th4,NAT_1:14;
      end;
      suppose
        len G = 0;
        then
A9:     G = {};
        hence g "**" (F ^ (G ^ <* d *>)) = g "**" (F ^ <* d *>) by FINSEQ_1:34
          .= g.(g "**" F,d) by A4,Th4
          .= g.(g "**" F,g "**" <* d *>) by Lm4
          .= g.(g "**" F,g "**" (G ^ <* d *>)) by A9,FINSEQ_1:34;
      end;
    end;
    hence thesis;
  end;
A10: P[<*>D]
  proof
    let F,g;
    assume that
    g is associative and
A11: g is having_a_unity or len F >= 1 & len <*>D >= 1;
    thus g "**" (F ^ <*>D) = g "**" F by FINSEQ_1:34
      .= g.(g "**" F,the_unity_wrt g) by A11,SETWISEO:15
      .= g.(g "**" F,g "**" <*>D) by A11,Lm3;
  end;
  for G holds P[G] from FINSEQ_2:sch 2(A10,A1);
  hence thesis;
end;

theorem Th6:
  g is associative & (g is having_a_unity or len F >= 1) implies g
  "**" (<* d *> ^ F) = g.(d,g "**" F)
proof
A1: len<* d *> = 1 by FINSEQ_1:39;
  assume g is associative &( g is having_a_unity or len F >= 1);
  hence g "**" (<* d *> ^ F) = g.(g "**" <* d *>,g "**" F) by A1,Th5
    .= g.(d,g "**" F) by Lm4;
end;

Lm7: g is associative & g is commutative implies for f being Permutation of
dom F st len F >= 1 & len F = len G & (for i st i in dom G holds G.i = F.(f.i))
holds g "**" F = g "**" G
proof
  assume that
A1: g is associative and
A2: g is commutative;
  defpred P[Element of NAT] means for H1,H2 st len H1 >= 1 & len H1 = $1 & len
H1 = len H2 for f being Permutation of dom H1 st (for i st i in dom H2 holds H2
  .i = H1.(f.i)) holds g "**" H1 = g "**" H2;
A3: now
    let k;
    assume
A4: P[k];
    thus P[k+1]
    proof
      let H1,H2;
      assume that
      len H1 >= 1 and
A5:   len H1 = k + 1 and
A6:   len H1 = len H2;
      reconsider p = H2 | (Seg k) as FinSequence of D by FINSEQ_1:18;
      let f be Permutation of dom H1;
A7:   dom H1 = Seg(k + 1) by A5,FINSEQ_1:def 3;
      then
A8:   rng f = Seg(k + 1) by FUNCT_2:def 3;
A9:   now
        let n;
        assume n in dom f;
        then f.n in Seg(k + 1) by A8,FUNCT_1:def 3;
        hence f.n is Element of NAT;
      end;
A10:  rng H2 c= D by FINSEQ_1:def 4;
      Seg(k + 1) = {} implies Seg(k + 1) = {};
      then
A11:  dom f = Seg(k + 1) by A7,FUNCT_2:def 1;
A12:  k + 1 in Seg(k + 1) by FINSEQ_1:4;
      then
A13:  f.(k + 1) in Seg(k + 1) by A11,A8,FUNCT_1:def 3;
      then reconsider n = f.(k + 1) as Element of NAT;
A14:  dom H2 = Seg(k + 1) by A5,A6,FINSEQ_1:def 3;
      then H2.(k + 1) in rng H2 by A12,FUNCT_1:def 3;
      then reconsider d = H2.(k + 1) as Element of D by A10;
A15:  n <= k + 1 by A13,FINSEQ_1:1;
      then consider m2 being Nat such that
A16:  n + m2 = k + 1 by NAT_1:10;
      defpred P[Nat,set] means $2 = H1.(n + $1);
      1 <= n by A13,FINSEQ_1:1;
      then consider m1 being Nat such that
A17:  1 + m1 = n by NAT_1:10;
      reconsider m1, m2 as Element of NAT by ORDINAL1:def 12;
A18:  for j be Nat st j in Seg m2 ex x st P[j,x];
      consider q2 being FinSequence such that
A19:  dom q2 = Seg m2 and
A20:  for k be Nat st k in Seg m2 holds P[k,q2.k] from FINSEQ_1:sch 1
      (A18);
      rng q2 c= D
      proof
        let x;
        assume x in rng q2;
        then consider y such that
A21:    y in findom q2 and
A22:    x = q2.y by FUNCT_1:def 3;
        reconsider y as Element of NAT by A21,SETWISEO:9;
        1 <= y by A19,A21,FINSEQ_1:1;
        then
A23:    1 <= n + y by NAT_1:12;
        y <= m2 by A19,A21,FINSEQ_1:1;
        then n + y <= len H1 by A5,A16,XREAL_1:7;
        then n + y in Seg(len H1) by A23,FINSEQ_1:1;
        then n + y in dom H1 by FINSEQ_1:def 3;
        then
A24:    H1.(n + y) in rng H1 by FUNCT_1:def 3;
        rng H1 c= D by FINSEQ_1:def 4;
        then reconsider xx = H1.(n + y) as Element of D by A24;
        xx in D;
        hence thesis by A19,A20,A21,A22;
      end;
      then reconsider q2 as FinSequence of D by FINSEQ_1:def 4;
      reconsider q1 = H1 | (Seg m1) as FinSequence of D by FINSEQ_1:18;
      defpred P[Nat,set] means (f.$1 in dom q1 implies $2 = f.$1) & (not f.$1
      in dom q1 implies for l st l = f.$1 holds $2 = l - 1);
A25:  k <= k + 1 by NAT_1:12;
      then
A26:  Seg k c= Seg(k + 1) by FINSEQ_1:5;
A27:  for i be Nat st i in Seg k ex y st P[i,y]
      proof
        let i be Nat;
        assume
A28:    i in Seg k;
        now
          f.i in Seg(k + 1) by A11,A8,A26,A28,FUNCT_1:def 3;
          then reconsider j = f.i as Element of NAT;
          assume
A29:      not f.i in dom q1;
          take y = j - 1;
          thus f.i in dom q1 implies y = f.i by A29;
          assume not f.i in dom q1;
          let t be Element of NAT;
          assume t = f.i;
          hence y = t - 1;
        end;
        hence thesis;
      end;
      consider gg being FinSequence such that
A30:  dom gg = Seg k and
A31:  for i be Nat st i in Seg k holds P[i,gg.i] from FINSEQ_1:sch 1(
      A27);
A32:  dom p = Seg k by A5,A6,A25,FINSEQ_1:17;
      m1 <= n by A17,NAT_1:11;
      then
A33:  m1 <= k + 1 by A15,XXREAL_0:2;
      then
A34:  dom q1 = Seg m1 by A5,FINSEQ_1:17;
A35:  now
        let i,l;
        assume that
A36:    l = f.i and
A37:    not f.i in dom q1 and
A38:    i in dom gg;
A39:    l < 1 or m1 < l by A34,A36,A37,FINSEQ_1:1;
A40:    now
          assume m1 + 1 = l;
          then k + 1 = i by A12,A11,A17,A26,A30,A36,A38,FUNCT_1:def 4;
          then k + 1 <= k + 0 by A30,A38,FINSEQ_1:1;
          hence contradiction by XREAL_1:6;
        end;
        f.i in rng f by A11,A26,A30,A38,FUNCT_1:def 3;
        then m1 + 1 <= l by A8,A36,A39,FINSEQ_1:1,NAT_1:13;
        then m1 + 1 < l by A40,XXREAL_0:1;
        then m1 + 1 + 1 <= l by NAT_1:13;
        hence m1 + 2 <= l;
      end;
      1 + k = 1 + (m1 + m2) by A17,A16;
      then
A41:  m1 <= k by NAT_1:11;
A42:  rng gg c= dom p
      proof
        let y;
        assume y in rng gg;
        then consider x such that
A43:    x in findom gg and
A44:    gg.x = y by FUNCT_1:def 3;
        reconsider x as Element of NAT by A43,SETWISEO:9;
        now
          per cases;
          suppose
A45:        f.x in dom q1;
A46:        dom q1 c= dom p by A41,A34,A32,FINSEQ_1:5;
            f.x = gg.x by A30,A31,A43,A45;
            hence thesis by A44,A45,A46;
          end;
          suppose
A47:        not f.x in dom q1;
            reconsider j = f.x as Element of NAT by A11,A26,A9,A30,A43;
A48:        f.x in Seg(k + 1) by A11,A8,A26,A30,A43,FUNCT_1:def 3;
            j < 1 or m1 < j by A34,A47,FINSEQ_1:1;
            then reconsider l = j - 1 as Element of NAT by A48,FINSEQ_1:1
,NAT_1:20;
            j <= k + 1 by A48,FINSEQ_1:1;
            then
A49:        l <= (k + 1) - 1 by XREAL_1:9;
            m1 + 2 <= j by A35,A43,A47;
            then
A50:        m1 + 2 - 1 <= l by XREAL_1:9;
            1 <= m1 + 1 by NAT_1:12;
            then
A51:        1 <= l by A50,XXREAL_0:2;
            gg.x = j - 1 by A30,A31,A43,A47;
            hence thesis by A32,A44,A51,A49,FINSEQ_1:1;
          end;
        end;
        hence thesis;
      end;
A52:  len q1 = m1 by A5,A33,FINSEQ_1:17;
A53:  now
        let j be Nat;
        assume
A54:    j in dom q2;
        len(q1 ^ <* d *>) = m1 + len<* d *> by A52,FINSEQ_1:22
          .= n by A17,FINSEQ_1:39;
        hence H1.(len(q1 ^ <* d *>) + j) = q2.j by A19,A20,A54;
      end;
      set q = q1 ^ q2;
A55:  len q2 = m2 by A19,FINSEQ_1:def 3;
      then
A56:  len q = m1 + m2 by A52,FINSEQ_1:22;
      then
A57:  dom q = Seg k by A17,A16,FINSEQ_1:def 3;
      then reconsider gg as Function of dom q, dom q by A32,A30,A42,
FUNCT_2:2;
A58:  len p = k by A5,A6,A25,FINSEQ_1:17;
A59:  rng gg = dom q
      proof
        thus rng gg c= dom q by A17,A16,A56,A32,A42,FINSEQ_1:def 3;
        let y;
        assume
A60:    y in dom q;
        then reconsider j = y as Element of NAT;
        consider x such that
A61:    x in dom f and
A62:    f.x = y by A8,A26,A57,A60,FUNCT_1:def 3;
        reconsider x as Element of NAT by A11,A61;
        now
          per cases;
          suppose
A63:        x in dom gg;
            now
              per cases;
              suppose
                f.x in dom q1;
                then gg.x = f.x by A30,A31,A63;
                hence thesis by A62,A63,FUNCT_1:def 3;
              end;
              suppose
A64:            not f.x in dom q1;
                j <= k by A57,A60,FINSEQ_1:1;
                then 1 <= j + 1 & j + 1 <= k + 1 by NAT_1:12,XREAL_1:7;
                then j + 1 in rng f by A8,FINSEQ_1:1;
                then consider x1 being set such that
A65:            x1 in dom f and
A66:            f.x1 = j + 1 by FUNCT_1:def 3;
A67:            now
                  assume not x1 in dom gg;
                  then x1 in Seg(k + 1) \ Seg k by A7,A30,A65,XBOOLE_0:def 5;
                  then x1 in {k + 1} by FINSEQ_3:15;
                  then
A68:              j + 1 = m1 +1 by A17,A66,TARSKI:def 1;
                  1 <= j by A57,A60,FINSEQ_1:1;
                  hence contradiction by A34,A62,A64,A68,FINSEQ_1:1;
                end;
                j < 1 or m1 < j by A34,A62,A64,FINSEQ_1:1;
                then not j + 1 <= m1 by A57,A60,FINSEQ_1:1,NAT_1:13;
                then not f.x1 in dom q1 by A34,A66,FINSEQ_1:1;
                then gg.x1 = j + 1 - 1 by A30,A31,A66,A67
                  .= y;
                hence thesis by A67,FUNCT_1:def 3;
              end;
            end;
            hence thesis;
          end;
          suppose
A69:        not x in dom gg;
            j <= k by A57,A60,FINSEQ_1:1;
            then 1 <= j + 1 & j + 1 <= k + 1 by NAT_1:12,XREAL_1:7;
            then j + 1 in rng f by A8,FINSEQ_1:1;
            then consider x1 being set such that
A70:        x1 in dom f and
A71:        f.x1 = j + 1 by FUNCT_1:def 3;
            x in Seg(k + 1) \ Seg k by A7,A30,A61,A69,XBOOLE_0:def 5;
            then x in {k + 1} by FINSEQ_3:15;
            then
A72:        x = k + 1 by TARSKI:def 1;
A73:        now
              assume not x1 in dom gg;
              then x1 in Seg(k + 1) \ Seg k by A7,A30,A70,XBOOLE_0:def 5;
              then x1 in {k + 1} by FINSEQ_3:15;
              then j + 1 = j + 0 by A62,A72,A71,TARSKI:def 1;
              hence contradiction;
            end;
            m1 <= j by A17,A62,A72,XREAL_1:29;
            then not j + 1 <= m1 by NAT_1:13;
            then not f.x1 in dom q1 by A34,A71,FINSEQ_1:1;
            then gg.x1 = j + 1 - 1 by A30,A31,A71,A73
              .= y;
            hence thesis by A73,FUNCT_1:def 3;
          end;
        end;
        hence thesis;
      end;
      assume
A74:  for i st i in dom H2 holds H2.i = H1.(f.i);
      then
A75:  H2.(k + 1) = H1.(f.(k + 1)) by A14,FINSEQ_1:4;
A76:  now
        let j be Nat;
        assume
A77:    j in dom(q1 ^ <* d *>);
A78:    now
          assume j in Seg m1;
          then
A79:      j in dom q1 by A5,A33,FINSEQ_1:17;
          then q1.j = H1.j by FUNCT_1:47;
          hence H1.j = (q1 ^ <* d *>).j by A79,FINSEQ_1:def 7;
        end;
A80:    now
          1 in Seg 1 & len<* d *> = 1 by FINSEQ_1:1,39;
          then 1 in dom <* d *> by FINSEQ_1:def 3;
          then
A81:      (q1 ^ <* d *>).(len q1 + 1) = <* d *>.1 by FINSEQ_1:def 7;
          assume j in {n};
          then j = n by TARSKI:def 1;
          hence (q1 ^ <* d *>).j = H1.j by A75,A17,A52,A81,FINSEQ_1:40;
        end;
        len(q1 ^ <* d *>) = m1 + len <* d *> by A52,FINSEQ_1:22
          .= m1 + 1 by FINSEQ_1:40;
        then j in Seg(m1 + 1) by A77,FINSEQ_1:def 3;
        then j in Seg m1 \/ {n} by A17,FINSEQ_1:9;
        hence H1.j = (q1 ^ <* d *>).j by A78,A80,XBOOLE_0:def 3;
      end;
      gg is one-to-one
      proof
        let y1,y2;
        assume that
A82:    y1 in dom gg and
A83:    y2 in dom gg and
A84:    gg.y1 = gg.y2;
        reconsider j1 = y1, j2 = y2 as Element of NAT by A30,A82,A83;
A85:    f.y2 in Seg(k + 1) by A11,A8,A26,A30,A83,FUNCT_1:def 3;
A86:    f.y1 in Seg(k + 1) by A11,A8,A26,A30,A82,FUNCT_1:def 3;
        then reconsider a = f.y1, b = f.y2 as Element of NAT by A85;
        now
          per cases;
          suppose
            f.y1 in dom q1 & f.y2 in dom q1;
            then gg.j1 = f.y1 & gg.j2 = f.y2 by A30,A31,A82,A83;
            hence thesis by A11,A26,A30,A82,A83,A84,FUNCT_1:def 4;
          end;
          suppose
A87:        f.y1 in dom q1 & not f.y2 in dom q1;
            then
A88:        a <= m1 by A34,FINSEQ_1:1;
            gg.j1 = a & gg.j2 = b - 1 by A30,A31,A82,A83,A87;
            then
A89:        (b - 1) + 1 <= m1 + 1 by A84,A88,XREAL_1:6;
            1 <= b by A85,FINSEQ_1:1;
            then
A90:        b in Seg(m1 + 1) by A89,FINSEQ_1:1;
            not b in Seg m1 by A5,A33,A87,FINSEQ_1:17;
            then b in Seg(m1 + 1) \ Seg m1 by A90,XBOOLE_0:def 5;
            then b in {m1 + 1} by FINSEQ_3:15;
            then b = m1 + 1 by TARSKI:def 1;
            then y2 = k + 1 by A12,A11,A17,A26,A30,A83,FUNCT_1:def 4;
            hence thesis by A30,A83,FINSEQ_3:8;
          end;
          suppose
A91:        not f.y1 in dom q1 & f.y2 in dom q1;
            then
A92:        b <= m1 by A34,FINSEQ_1:1;
            gg.j1 = a - 1 & gg.j2 = b by A30,A31,A82,A83,A91;
            then
A93:        (a - 1) + 1 <= m1 + 1 by A84,A92,XREAL_1:6;
            1 <= a by A86,FINSEQ_1:1;
            then
A94:        a in Seg(m1 + 1) by A93,FINSEQ_1:1;
            not a in Seg m1 by A5,A33,A91,FINSEQ_1:17;
            then a in Seg(m1 + 1) \ Seg m1 by A94,XBOOLE_0:def 5;
            then a in {m1 + 1} by FINSEQ_3:15;
            then a = m1 + 1 by TARSKI:def 1;
            then y1 = k + 1 by A12,A11,A17,A26,A30,A82,FUNCT_1:def 4;
            hence thesis by A30,A82,FINSEQ_3:8;
          end;
          suppose
A95:        not f.y1 in dom q1 & not f.y2 in dom q1;
            then gg.j2 = b - 1 by A30,A31,A83;
            then
A96:        gg.y2 = b + (- 1);
            gg.j1 = a - 1 by A30,A31,A82,A95;
            then gg.j1 = a + (- 1);
            then a = b by A84,A96,XCMPLX_1:2;
            hence thesis by A11,A26,A30,A82,A83,FUNCT_1:def 4;
          end;
        end;
        hence thesis;
      end;
      then reconsider gg as Permutation of dom q by A59,FUNCT_2:57;
      len(q1 ^ <* d *>) + len q2 = len q1 + len<* d *> + m2 by A55,FINSEQ_1:22
        .= k + 1 by A17,A16,A52,FINSEQ_1:40;
      then dom H1 = Seg(len(q1 ^ <* d *>) + len q2) by A5,FINSEQ_1:def 3;
      then
A97:  H1 = q1 ^ <* d *> ^ q2 by A76,A53,FINSEQ_1:def 7;
A98:  now
        let i;
        assume
A99:    i in dom p;
        then f.i in rng f by A11,A26,A32,FUNCT_1:def 3;
        then reconsider j = f.i as Element of NAT by A8;
        now
          per cases;
          suppose
A100:       f.i in dom q1;
            then
A101:       f.i = gg.i & H1.(j) = q1.(j) by A32,A31,A99,FUNCT_1:47;
            H2.i = p.i & H2.i = H1.(f.i) by A74,A14,A26,A32,A99,FUNCT_1:47;
            hence p.i = q.(gg.i) by A100,A101,FINSEQ_1:def 7;
          end;
          suppose
A102:       not f.i in dom q1;
            then m1 + 2 <= j by A32,A30,A35,A99;
            then
A103:       m1 + 2 - 1 <= j - 1 by XREAL_1:9;
            m1 < m1 + 1 by XREAL_1:29;
            then
A104:       m1 < j - 1 by A103,XXREAL_0:2;
            then m1 < j by XREAL_1:146,XXREAL_0:2;
            then reconsider j1 = j - 1 as Element of NAT by NAT_1:20;
A105:       not j1 in dom q1 by A34,A104,FINSEQ_1:1;
A106:       gg.i = j - 1 by A32,A31,A99,A102;
            then j - 1 in dom q by A32,A30,A59,A99,FUNCT_1:def 3;
            then consider a being Nat such that
A107:       a in dom q2 and
A108:       j1 = len q1 + a by A105,FINSEQ_1:25;
A109:       len<* d *> = 1 by FINSEQ_1:39;
A110:       H2.i = p.i & H2.i = H1.(f.i) by A74,A14,A26,A32,A99,FUNCT_1:47;
A111:       H1 = q1 ^ (<* d *> ^ q2) by A97,FINSEQ_1:32;
            j in dom H1 by A7,A11,A8,A26,A32,A99,FUNCT_1:def 3;
            then consider b being Nat such that
A112:       b in dom(<* d *> ^ q2) and
A113:       j = len q1 + b by A102,A111,FINSEQ_1:25;
A114:       H1.j = (<* d *> ^ q2).b by A111,A112,A113,FINSEQ_1:def 7;
A115:       b = 1 + a by A108,A113;
            q.(j - 1) = q2.a by A107,A108,FINSEQ_1:def 7;
            hence p.i = q.(gg.i) by A106,A110,A107,A114,A115,A109,
FINSEQ_1:def 7;
          end;
        end;
        hence p.i = q.(gg.i);
      end;
      now
        per cases;
        suppose
A116:     len p <> 0;
A117:     H2 = p ^ <* d *> by A5,A6,FINSEQ_3:55;
          g "**" p = g "**" q by A4,A17,A16,A58,A56,A98,A116,NAT_1:14;
          then
A118:     g "**" H2 = g.(g "**" q,d) by A116,A117,Th4,NAT_1:14;
          now
            per cases;
            suppose
A119:         len q1 <> 0 & len q2 <> 0;
              len(<* d *> ^ q2) = len<* d *> + len q2 & len<* d *> = 1
              by FINSEQ_1:22,40;
              then
A120:         len(<* d *> ^ q2) >= 1 by NAT_1:12;
A121:         len q1 >= 1 by A119,NAT_1:14;
              len q2 >= 1 by A119,NAT_1:14;
              then g "**" H2 = g.(g.(g "**" q1,g "**" q2),d) by A1,A118,A121
,Th5
                .= g.(g "**" q1,g.(g "**" q2,d)) by A1,BINOP_1:def 3
                .= g.(g "**" q1,g.(d,g "**" q2)) by A2,BINOP_1:def 2
                .= g.(g "**" q1,g "**" (<* d *> ^ q2)) by A1,A119,Th6,NAT_1:14
                .= g "**" (q1 ^ (<* d *> ^ q2)) by A1,A121,A120,Th5
                .= g "**" H1 by A97,FINSEQ_1:32;
              hence thesis;
            end;
            suppose
              len q1 = 0 & len q2 <> 0;
              then
A122:         q1 = {};
              then
A123:         H1 = <* d *> ^ q2 by A97,FINSEQ_1:34
                .= <* d *> ^ q by A122,FINSEQ_1:34;
              g "**" H2 = g.(d,g "**" q) by A2,A118,BINOP_1:def 2
                .= g "**" (<* d *> ^ q) by A1,A17,A16,A58,A56,A116,Th6,NAT_1:14
;
              hence thesis by A123;
            end;
            suppose
              len q1 <> 0 & len q2 = 0;
              then
A124:         q2 = {};
              then H1 = q1 ^ <* d *> by A97,FINSEQ_1:34
                .= q ^ <* d *> by A124,FINSEQ_1:34;
              hence thesis by A17,A16,A58,A56,A116,A118,Th4,NAT_1:14;
            end;
            suppose
              len q1 = 0 & len q2 = 0;
              then len q = 0 + 0 by FINSEQ_1:22;
              hence thesis by A5,A6,A17,A16,A56,A116,FINSEQ_1:17;
            end;
          end;
          hence thesis;
        end;
        suppose
A125:     len p = 0;
          then dom H1 = {1} by A5,A58,FINSEQ_1:2,def 3;
          then
A126:     dom f = {1} & rng f = {1} by FUNCT_2:def 1,def 3;
          1 in {1} by TARSKI:def 1;
          then f.1 in {1} by A126,FUNCT_1:def 3;
          then H1.1 = H2.1 by A75,A58,A125,TARSKI:def 1;
          then H1 = <* d *> by A5,A58,A125,FINSEQ_1:40;
          hence thesis by A5,A6,A58,A125,FINSEQ_1:40;
        end;
      end;
      hence thesis;
    end;
  end;
  let f be Permutation of dom F;
A127: P[0];
  for k holds P[k] from NAT_1:sch 1(A127,A3);
  hence thesis;
end;

Lm8: g is having_a_unity & len F = 0 & G = F * P implies g "**" F = g "**" G
proof
  assume that
A1: g is having_a_unity and
A2: len F = 0;
  assume
A3: G = F * P;
  F = {} by A2;
  then G = {} by A3,RELAT_1:39;
  then
A4: len G = 0;
  thus g "**" F = the_unity_wrt g by A1,A2,Def1
    .= g "**" G by A1,A4,Def1;
end;

theorem Th7:
  g is commutative associative & (g is having_a_unity or len F >= 1
  ) & G = F * P implies g "**" F = g "**" G
proof
  assume that
A1: g is commutative associative and
A2: g is having_a_unity or len F >= 1;
  assume
A3: G = F * P;
  now
    per cases;
    suppose
      len F = 0;
      hence thesis by A2,A3,Lm8;
    end;
    suppose
A4:   len F <> 0;
      len F = len G & for i st i in dom G holds G.i = F.(P.i) by A3,FINSEQ_2:44
,FUNCT_1:12;
      hence thesis by A1,A4,Lm7,NAT_1:14;
    end;
  end;
  hence thesis;
end;

Lm9: g is associative commutative & F is one-to-one & G is one-to-one & rng F
= rng G & len F >= 1 implies g "**" F = g "**" G
proof
  assume that
A1: g is associative commutative and
A2: F is one-to-one and
A3: G is one-to-one and
A4: rng F = rng G and
A5: len F >= 1;
A6: len F = len G by A2,A3,A4,FINSEQ_1:48;
  set P = F" * G;
A7: dom(F") = rng F by A2,FUNCT_1:33;
  then
A8: dom P = dom G by A4,RELAT_1:27
    .= Seg len F by A6,FINSEQ_1:def 3
    .= dom F by FINSEQ_1:def 3;
  rng(F") = dom F by A2,FUNCT_1:33;
  then
A9: rng P c= dom F by RELAT_1:26;
  dom F = Seg len F by FINSEQ_1:def 3;
  then reconsider P as Function of dom F, dom F by A5,A8,A9,FUNCT_2:def 1
,RELSET_1:4;
  rng P = rng(F") by A4,A7,RELAT_1:28
    .= dom F by A2,FUNCT_1:33;
  then reconsider P as Permutation of dom F by A2,A3,FUNCT_2:57;
  F * P = (F * F") * G by RELAT_1:36
    .= id(rng G) * G by A2,A4,FUNCT_1:39
    .= G by RELAT_1:54;
  hence thesis by A1,A5,Th7;
end;

Lm10: len F = 0 & g is having_a_unity & F is one-to-one & G is one-to-one &
rng F = rng G implies g "**" F = g "**" G
proof
  assume that
A1: len F = 0 & g is having_a_unity and
A2: F is one-to-one & G is one-to-one & rng F = rng G;
  len G = len F & g "**" F = the_unity_wrt g by A1,A2,Def1,FINSEQ_1:48;
  hence thesis by A1,Def1;
end;

theorem
  (g is having_a_unity or len F >= 1) & g is associative commutative & F
  is one-to-one & G is one-to-one & rng F = rng G implies g "**" F = g "**" G
proof
  len F >= 1 or len F = 0 by NAT_1:14;
  hence thesis by Lm9,Lm10;
end;

Lm11: len F = 1 implies g "**" F = F.1
proof
  assume
A1: len F = 1;
  then F = <* F.1 *> by FINSEQ_1:40
    .= <* F/.1 *> by A1,FINSEQ_4:15;
  hence g "**" F = F/.1 by Lm4
    .= F.1 by A1,FINSEQ_4:15;
end;

Lm12: g is associative & g is commutative & len F >= 1 & len F = len G & len F
= len H & (for k st k in dom F holds H.k = g.(F.k,G.k)) implies g "**" H = g.(g
"**" F,g "**" G)
proof
  assume that
A1: g is associative and
A2: g is commutative;
  defpred P[Element of NAT] means for F,G,H st len F >= 1 & len F = $1 & len F
= len G & len F = len H & (for k st k in dom F holds H.k = g.(F.k,G.k)) holds g
  "**" H = g.(g "**" F,g "**" G);
A3: now
    let k;
    assume
A4: P[k];
    thus P[k+1]
    proof
      let F,G,H;
      assume that
      len F >= 1 and
A5:   len F = k + 1 and
A6:   len F = len G and
A7:   len F = len H and
A8:   for k st k in dom F holds H.k = g.(F.k,G.k);
      reconsider f = F | Seg k,gg = G | Seg k,h = H | Seg k as FinSequence of
      D by FINSEQ_1:18;
A9:   len h = k by A5,A7,FINSEQ_3:53;
A10:  len f = k by A5,FINSEQ_3:53;
A11:  len gg = k by A5,A6,FINSEQ_3:53;
A12:  now
        k <= k + 1 by NAT_1:12;
        then Seg len f c= Seg len F by A5,A10,FINSEQ_1:5;
        then Seg len f c= dom F by FINSEQ_1:def 3;
        then
A13:    dom f c= dom F by FINSEQ_1:def 3;
        let i;
        assume
A14:    i in dom f;
        then i in Seg len gg by A10,A11,FINSEQ_1:def 3;
        then i in dom gg by FINSEQ_1:def 3;
        then
A15:    G.i = gg.i by FUNCT_1:47;
        i in Seg len h by A10,A9,A14,FINSEQ_1:def 3;
        then i in dom h by FINSEQ_1:def 3;
        then
A16:    H.i = h.i by FUNCT_1:47;
        F.i = f.i by A14,FUNCT_1:47;
        hence h.i = g.(f.i,gg.i) by A8,A14,A15,A16,A13;
      end;
      now
        per cases by NAT_1:14;
        suppose
A17:      len f >= 1;
A18:      rng H c= D by FINSEQ_1:def 4;
A19:      rng F c= D & rng G c= D by FINSEQ_1:def 4;
A20:      k + 1 in Seg(k + 1) by FINSEQ_1:4;
          then k + 1 in dom G by A5,A6,FINSEQ_1:def 3;
          then
A21:      (G.(k + 1)) in rng G by FUNCT_1:def 3;
          k + 1 in dom H by A5,A7,A20,FINSEQ_1:def 3;
          then
A22:      (H.(k + 1)) in rng H by FUNCT_1:def 3;
A23:      k + 1 in dom F by A5,A20,FINSEQ_1:def 3;
          then (F.(k + 1)) in rng F by FUNCT_1:def 3;
          then reconsider
          d = F.(k + 1),d1 = G.(k + 1),d2 = H.(k + 1) as Element of
          D by A21,A22,A19,A18;
A24:      d2 = g.(F.(k + 1),G.(k + 1)) by A8,A23;
          F = f ^ <* d *> by A5,FINSEQ_3:55;
          then
A25:      g "**" F = g.(g "**" f,d) by A17,Th4;
          G = gg ^ <* d1 *> by A5,A6,FINSEQ_3:55;
          then
A26:      g "**" G = g.(g "**" gg,d1) by A10,A11,A17,Th4;
A27:      H = h ^ <* d2 *> by A5,A7,FINSEQ_3:55;
          g "**" h = g.(g "**" f,g "**" gg) by A4,A10,A11,A9,A12,A17;
          hence g "**" H = g.(g.(g "**" f,g "**" gg),g.(d,d1)) by A10,A9,A17
,A27,A24,Th4
            .= g.(g.(g.(g "**" f,g "**" gg),d),d1) by A1,BINOP_1:def 3
            .= g.(g.(g "**" f,g.(g "**" gg,d)),d1) by A1,BINOP_1:def 3
            .= g.(g.(g "**" f,g.(d,g "**" gg)),d1) by A2,BINOP_1:def 2
            .= g.(g.(g "**" F,g "**" gg),d1) by A1,A25,BINOP_1:def 3
            .= g.(g "**" F,g "**" G) by A1,A26,BINOP_1:def 3;
        end;
        suppose
A28:      len f = 0;
          then
A29:      g "**" G = G.1 & 1 in dom F by A5,A6,A10,Lm11,FINSEQ_3:25;
          g "**" H = H.1 & g "**" F = F.1 by A5,A7,A10,A28,Lm11;
          hence thesis by A8,A29;
        end;
      end;
      hence thesis;
    end;
  end;
  assume
A30: len F >= 1 & len F = len G & len F = len H;
A31: P[0];
  for k holds P[k] from NAT_1:sch 1(A31,A3);
  hence thesis by A30;
end;

Lm13: g is having_a_unity & len F = 0 & len F = len G & len F = len H implies
g "**" F = g.(g "**" G,g "**" H)
proof
  assume that
A1: g is having_a_unity and
A2: len F = 0 and
A3: len F = len G and
A4: len F = len H;
  thus g "**" F = the_unity_wrt g by A1,A2,Def1
    .= g.(the_unity_wrt g,the_unity_wrt g) by A1,SETWISEO:15
    .= g.(g "**" G,the_unity_wrt g) by A1,A2,A3,Def1
    .= g.(g "**" G,g "**" H) by A1,A2,A4,Def1;
end;

theorem
  g is associative commutative & (g is having_a_unity or len F >= 1) &
  len F = len G & len F = len H & (for k st k in dom F holds F.k = g.(G.k,H.k))
  implies g "**" F = g.(g "**" G,g "**" H)
proof
A1: dom F = Seg len F & dom G = Seg len G by FINSEQ_1:def 3;
A2: len F = 0 or len F >= 1 by NAT_1:14;
  assume g is associative commutative &( g is having_a_unity or len F >= 1);
  hence thesis by A1,A2,Lm12,Lm13;
end;

theorem
  g is having_a_unity implies g "**" <*>D = the_unity_wrt g by Lm3;

theorem
  g "**" <* d *> = d by Lm4;

theorem Th12:
  g "**" <* d1,d2 *> = g.(d1,d2)
proof
A1: len<* d1 *> = 1 by FINSEQ_1:39;
  thus g "**" <* d1,d2 *> = g "**" (<* d1 *> ^ <* d2 *>) by FINSEQ_1:def 9
    .= g.(g "**" <* d1 *>,d2) by A1,Th4
    .= g.(d1,d2) by Lm4;
end;

theorem
  g is commutative implies g "**" <* d1,d2 *> = g "**" <* d2,d1 *>
proof
  assume
A1: g is commutative;
  thus g "**" <* d1,d2 *> = g.(d1,d2) by Th12
    .= g.(d2,d1) by A1,BINOP_1:def 2
    .= g "**" <* d2,d1 *> by Th12;
end;

theorem Th14:
  g "**" <* d1,d2,d3 *> = g.(g.(d1,d2),d3)
proof
A1: len<* d1,d2 *> = 2 by FINSEQ_1:44;
  thus g "**" <* d1,d2,d3 *> = g "**" (<* d1,d2 *> ^ <* d3 *>) by FINSEQ_1:43
    .= g.(g "**" <* d1,d2 *>,d3) by A1,Th4
    .= g.(g.(d1,d2),d3) by Th12;
end;

theorem
  g is commutative implies g "**" <* d1,d2,d3 *> = g "**" <* d2,d1,d3 *>
proof
  assume
A1: g is commutative;
  thus g "**" <* d1,d2,d3 *> = g.(g.(d1,d2),d3) by Th14
    .= g.(g.(d2,d1),d3) by A1,BINOP_1:def 2
    .= g "**" <* d2,d1,d3 *> by Th14;
end;

theorem Th16:
  g "**" (1 |-> d) = d
proof
  thus g "**" (1 |-> d) = g "**" <* d *> by FINSEQ_2:59
    .= d by Lm4;
end;

theorem
  g "**" (2 |-> d) = g.(d,d)
proof
  thus g "**" (2 |-> d) = g "**" <* d,d *> by FINSEQ_2:61
    .= g.(d,d) by Th12;
end;

theorem Th18:
  g is associative & (g is having_a_unity or k <> 0 & l <> 0)
  implies g "**" ((k + l) |-> d) = g.(g "**" (k |-> d),g "**" (l |-> d))
proof
  k <> 0 & l <> 0 implies len(k |-> d) <> 0 & len(l |-> d) <> 0 by CARD_1:def 7
;
  then
A1: k <> 0 & l <> 0 implies len(k |-> d) >= 1 & len(l |-> d) >= 1 by NAT_1:14;
  (k + l) |-> d = (k |-> d) ^ (l |-> d) by FINSEQ_2:123;
  hence thesis by A1,Th5;
end;

theorem
  g is associative & (g is having_a_unity or k <> 0 & l <> 0) implies g
  "**" (k * l |-> d) = g "**" (l |-> (g "**" (k |-> d)))
proof
  defpred P[Element of NAT] means for g,k,d st g is associative & (g is
  having_a_unity or k <> 0 & $1 <> 0) holds g "**" (k * $1 |-> d) = g "**" ($1
  |-> (g "**" (k |-> d)));
A1: for l st P[l] holds P[l + 1]
  proof
    let l;
    assume
A2: P[l];
    let g,k,d;
    assume that
A3: g is associative and
A4: g is having_a_unity or k <> 0 & l + 1 <> 0;
    now
      per cases;
      suppose
        l = 0;
        hence thesis by Th16;
      end;
      suppose
A5:     l <> 0;
        then
A6:     k <> 0 implies k * l <> 0 by XCMPLX_1:6;
        g "**" (k * (l + 1) |-> d) = g "**" ((k * l + k * 1) |-> d)
          .= g.(g "**" (k * l |-> d),g "**" (k |-> d)) by A3,A4,A6,Th18
          .= g.(g "**" (l |-> (g "**" (k |-> d))),g "**" (k |-> d)) by A2,A3,A4
,A5
          .= g.(g "**" (l |-> (g "**" (k |-> d))),g "**" (1 |-> (g "**" (k
        |-> d)))) by Th16;
        hence thesis by A3,A5,Th18;
      end;
    end;
    hence thesis;
  end;
A7: P[0];
  for l holds P[l] from NAT_1:sch 1(A7,A1);
  hence thesis;
end;

theorem
  len F = 1 implies g "**" F = F.1 by Lm11;

theorem
  len F = 2 implies g "**" F = g.(F.1,F.2)
proof
  assume
A1: len F = 2;
  then F = <* F.1,F.2 *> by FINSEQ_1:44
    .= <* F/.1,F.2 *> by A1,FINSEQ_4:15
    .= <* F/.1,F/.2 *> by A1,FINSEQ_4:15;
  hence g "**" F = g.(F/.1,F/.2) by Th12
    .= g.(F.1,F/.2) by A1,FINSEQ_4:15
    .= g.(F.1,F.2) by A1,FINSEQ_4:15;
end;














begin

reserve x,y for set;
reserve C,C9,D,E for non empty set;
reserve c,c9,c1,c2,c3 for Element of C;
reserve B,B9,B1,B2 for Element of Fin C;
reserve A for Element of Fin C9;
reserve d,d1,d2,d3,d4,e for Element of D;
reserve F,G for BinOp of D;
reserve u for UnOp of D;
reserve f,f9 for Function of C,D;
reserve g for Function of C9,D;
reserve H for BinOp of E;
reserve h for Function of D,E;
reserve i,j for Nat;
reserve s for Function;
reserve p,q for FinSequence of D;
reserve T1,T2 for Element of i-tuples_on D;

Lm1: for i be Nat holds Seg i is Element of Fin NAT by FINSUB_1:def 5;

Lm2: now
  let i;
  assume i+1 in Seg i;
  then i+1 <= i by FINSEQ_1:1;
  hence contradiction by XREAL_1:29;
end;

theorem Th1:
  F is commutative associative & c1 <> c2 implies F $$ ({.c1,c2.},f
  ) = F.(f.c1, f.c2)
proof
  assume that
A1: F is commutative associative and
A2: c1 <> c2;
  consider g being Function of Fin C, D such that
A3: F $$ ({.c1,c2.},f) = g.{c1,c2} and
  for e st e is_a_unity_wrt F holds g.{} = e and
A4: for c holds g.{c} = f.c and
A5: for B st B c= { c1,c2 } & B <> {} for c st c in { c1,c2 } \ B holds
  g.(B \/ {c}) = F.(g.B,f.c) by A1,SETWISEO:def 3;
  c1 in {c1} & not c2 in {c1} by A2,TARSKI:def 1;
  then {c1,c2} \ {c1} = {c2} by ZFMISC_1:62;
  then
A6: c2 in {c1,c2} \ {c1} by TARSKI:def 1;
  thus F $$ ({.c1,c2.},f) = g.({.c1.} \/ {.c2.}) by A3,ENUMSET1:1
    .= F.(g.{c1}, f.c2) by A5,A6,ZFMISC_1:7
    .= F.(f.c1, f.c2) by A4;
end;

theorem Th2:
  F is commutative associative & (B <> {} or F is having_a_unity) &
  not c in B implies F $$(B \/ {.c.}, f) = F.(F $$(B,f),f.c)
proof
  assume that
A1: F is commutative associative and
A2: B <> {} or F is having_a_unity and
A3: not c in B;
  per cases;
  suppose
A4: B = {};
    then B = {}.C;
    then F $$(B,f) = the_unity_wrt F by A1,A2,SETWISEO:31;
    hence F.(F $$(B,f),f.c) = f.c by A2,A4,SETWISEO:15
      .= F $$(B \/ {.c.}, f) by A1,A4,SETWISEO:17;
  end;
  suppose
A5: B <> {};
    consider g9 being Function of Fin C, D such that
A6: F $$ (B,f) = g9.B and
    for e st e is_a_unity_wrt F holds g9.{} = e and
A7: for c9 holds g9.{c9} = f.c9 and
A8: for B9 st B9 c= B & B9 <> {} for c9 st c9 in B \ B9 holds g9.(B9
    \/ {c9}) = F.(g9.B9,f.c9) by A1,A2,SETWISEO:def 3;
    consider g being Function of Fin C, D such that
A9: F $$ (B \/ {.c.},f) = g.(B \/ {c}) and
    for e st e is_a_unity_wrt F holds g.{} = e and
A10: for c9 holds g.{c9} = f.c9 and
A11: for B9 st B9 c= B \/ {c} & B9 <> {} for c9 st c9 in (B \/ {c}) \
    B9 holds g.(B9 \/ {c9}) = F.(g.B9,f.c9) by A1,SETWISEO:def 3;
    defpred X[set] means $1 <> {} & $1 c= B implies g.($1) = g9.($1);
A12: for B9 being Element of Fin C, b being Element of C holds X[B9] & not
    b in B9 implies X[B9 \/ {b}]
    proof
A13:  B c= B \/ {c} by XBOOLE_1:7;
      let B9, c9 such that
A14:  B9 <> {} & B9 c= B implies g.B9 = g9.B9 and
A15:  not c9 in B9 and
      B9 \/ {c9} <> {} and
A16:  B9 \/ {c9} c= B;
A17:  c9 in B by A16,ZFMISC_1:137;
      then
A18:  c9 in B \ B9 by A15,XBOOLE_0:def 5;
      c9 in (B \/ {c}) by A17,XBOOLE_0:def 3;
      then
A19:  c9 in (B \/ {c}) \ B9 by A15,XBOOLE_0:def 5;
      B9 c= B by A16,XBOOLE_1:11;
      then
A20:  B9 c= B \/ {c} by A13,XBOOLE_1:1;
      per cases;
      suppose
A21:    B9 = {};
        then g.(B9 \/ {c9}) = f.c9 by A10;
        hence thesis by A7,A21;
      end;
      suppose
A22:    B9 <> {};
        hence g.(B9 \/ {c9}) = F.(g9.B9, f.c9) by A11,A14,A16,A20,A19,
XBOOLE_1:11
          .= g9.(B9 \/ {c9}) by A8,A16,A18,A22,XBOOLE_1:11;
      end;
    end;
A23: X[{}.C];
A24: for B9 holds X[B9] from SETWISEO:sch 2(A23,A12);
    c in B \/ {c} by ZFMISC_1:136;
    then c in (B \/ {c}) \ B by A3,XBOOLE_0:def 5;
    hence F $$(B \/ {.c.}, f) = F.(g.B, f.c) by A5,A9,A11,XBOOLE_1:7
      .= F.(F $$(B,f),f.c) by A5,A6,A24;
  end;
end;

theorem
  F is commutative associative & c1 <> c2 & c1 <> c3 & c2 <> c3 implies
  F $$ ({.c1,c2,c3.},f) = F.(F.(f.c1, f.c2),f.c3)
proof
  assume that
A1: F is commutative associative and
A2: c1 <> c2;
  assume c1 <> c3 & c2 <> c3;
  then
A3: not c3 in {c1,c2} by TARSKI:def 2;
  thus F $$ ({.c1,c2,c3.},f) = F $$ ({.c1,c2.} \/ {.c3.},f) by ENUMSET1:3
    .= F.(F $$ ({.c1,c2.},f),f.c3) by A1,A3,Th2
    .= F.(F.(f.c1, f.c2),f.c3) by A1,A2,Th1;
end;

theorem
  F is commutative associative & (B1 <> {} & B2 <> {} or F is
having_a_unity) & B1 misses B2 implies F $$(B1 \/ B2, f) = F.(F $$(B1,f),F $$(
  B2,f))
proof
  assume that
A1: F is commutative associative and
A2: B1 <> {} & B2 <> {} or F is having_a_unity and
A3: B1 /\ B2 = {};
  now
    per cases;
    suppose
A4:   B1 = {} or B2 = {};
      now
        per cases by A4;
        suppose
A5:       B1 = {};
          hence F $$(B1 \/ B2, f) = F.(the_unity_wrt F,F $$(B2,f)) by A2,
SETWISEO:15
            .= F.(F $$({}.C,f),F $$(B2,f)) by A1,A2,A4,SETWISEO:31
            .= F.(F $$(B1,f),F $$(B2,f)) by A5;
        end;
        suppose
A6:       B2 = {};
          hence F $$(B1 \/ B2, f) = F.(F $$(B1,f),the_unity_wrt F) by A2,
SETWISEO:15
            .= F.(F $$(B1,f),F $$({}.C,f)) by A1,A2,A4,SETWISEO:31
            .= F.(F $$(B1,f),F $$(B2,f)) by A6;
        end;
      end;
      hence thesis;
    end;
    suppose
A7:   B1 <> {} & B2 <> {};
      defpred X[Element of Fin C] means $1 <> {} & B1 /\ $1 = {} implies F $$(
      B1 \/ $1,f) = F.(F $$(B1,f),F $$($1,f));
A8:   for B9 being Element of Fin C, b being Element of C holds X[B9] &
      not b in B9 implies X[B9 \/ {.b.}]
      proof
        let B,c such that
A9:     B <> {} & B1 /\ B = {} implies F $$(B1 \/ B,f) = F.(F $$ (B1,
        f),F $$(B,f)) and
A10:    not c in B and
        B \/ {c} <> {};
        assume
A11:    B1 /\ (B \/ {c}) = {};
        then
A12:    B1 misses (B \/ {c}) by XBOOLE_0:def 7;
        c in B \/ {c} by ZFMISC_1:136;
        then
A13:    not c in B1 by A11,XBOOLE_0:def 4;
A14:    B <> {} & B1 misses B implies F $$(B1 \/ B,f) = F.(F $$ (B1,f),F
        $$(B,f)) by A9,XBOOLE_0:def 7;
        now
          per cases;
          suppose
A15:        B = {};
            hence
            F $$ (B1 \/ (B \/ {.c.}),f) = F.(F $$ (B1,f), f.c) by A1,A7,A13,Th2
              .= F.(F $$ (B1,f),F $$(B \/ {.c.},f)) by A1,A15,SETWISEO:17;
          end;
          suppose
A16:        B <> {};
A17:        not c in B1 \/ B by A10,A13,XBOOLE_0:def 3;
            thus F $$ (B1 \/ (B \/ {.c.}),f) = F $$ (B1 \/ B \/ {.c.},f) by
XBOOLE_1:4
              .= F.(F.(F $$ (B1,f),F $$(B,f)), f.c) by A1,A14,A12,A16,A17,Th2,
XBOOLE_1:7,63
              .= F.(F $$ (B1,f),F.(F $$(B,f), f.c)) by A1,BINOP_1:def 3
              .= F.(F $$(B1,f),F $$(B \/ {.c.},f)) by A1,A10,A16,Th2;
          end;
        end;
        hence thesis;
      end;
A18:  X[{}.C];
      for B2 holds X[B2] from SETWISEO:sch 2(A18,A8);
      hence thesis by A3,A7;
    end;
  end;
  hence thesis;
end;

theorem Th5:
  F is commutative associative & (A <> {} or F is having_a_unity) &
(ex s st dom s = A & rng s = B & s is one-to-one & g|A = f*s) implies F $$(A,g)
  = F $$(B,f)
proof
  defpred X[Element of Fin C9] means $1 <> {} or F is having_a_unity implies
for B st ex s st dom s = $1 & rng s = B & s is one-to-one & g|$1 = f*s holds F
  $$($1,g) = F $$(B,f);
  assume
A1: F is commutative associative;
A2: for B9 being Element of Fin C9, b being Element of C9 holds X[B9] & not
  b in B9 implies X[B9 \/ {.b.}]
  proof
    let A9 be Element of Fin C9,a be Element of C9 such that
A3: A9 <> {} or F is having_a_unity implies for B st ex s st dom s =
    A9 & rng s = B & s is one-to-one & g|A9 = f*s holds F $$(A9,g) = F $$(B,f)
    and
A4: not a in A9;
    assume A9 \/ {a} <> {} or F is having_a_unity;
    let B;
    set A = A9 \/ {.a.};
    given s such that
A5: dom s = A and
A6: rng s = B and
A7: s is one-to-one and
A8: g|A = f*s;
A9: a in A by ZFMISC_1:136;
    then
A10: s.a in B by A5,A6,FUNCT_1:def 3;
    B c= C by FINSUB_1:def 5;
    then reconsider c = s.a as Element of C by A10;
    set B9 = B \ {.c.};
    set s9 = s|A9;
A11: s9 is one-to-one by A7,FUNCT_1:52;
    now
      let y;
      thus y in rng s9 implies y in B9
      proof
        assume y in rng s9;
        then consider x such that
A12:    x in dom s9 and
A13:    y = s9.x by FUNCT_1:def 3;
A14:    s9.x = s.x by A12,FUNCT_1:47;
A15:    x in dom s /\ A9 by A12,RELAT_1:61;
        then x in dom s & x <> a by A4,XBOOLE_0:def 4;
        then s.x <> c by A5,A7,A9,FUNCT_1:def 4;
        then
A16:    not y in {c} by A13,A14,TARSKI:def 1;
        x in dom s by A15,XBOOLE_0:def 4;
        then y in B by A6,A13,A14,FUNCT_1:def 3;
        hence thesis by A16,XBOOLE_0:def 5;
      end;
      assume
A17:  y in B9;
      then y in B by XBOOLE_0:def 5;
      then consider x such that
A18:  x in dom s and
A19:  y = s.x by A6,FUNCT_1:def 3;
A20:  x in A9 or x in {a} by A5,A18,XBOOLE_0:def 3;
      not y in {c} by A17,XBOOLE_0:def 5;
      then x <> a by A19,TARSKI:def 1;
      then x in dom s /\ A9 by A18,A20,TARSKI:def 1,XBOOLE_0:def 4;
      then x in dom s9 & s9.x = s.x by FUNCT_1:48,RELAT_1:61;
      hence y in rng s9 by A19,FUNCT_1:def 3;
    end;
    then
A21: rng s9 = B9 by TARSKI:1;
    now
      let x;
      thus x in dom(g|A9) implies x in dom(f*s9)
      proof
        assume x in dom(g|A9);
        then
A22:    x in dom g /\ A9 by RELAT_1:61;
        then
A23:    x in A9 by XBOOLE_0:def 4;
A24:    x in dom g by A22,XBOOLE_0:def 4;
        x in A by A23,ZFMISC_1:136;
        then x in dom g /\ A by A24,XBOOLE_0:def 4;
        then
A25:    x in dom(f*s) by A8,RELAT_1:61;
        then
A26:    s.x in dom f by FUNCT_1:11;
        x in dom s by A25,FUNCT_1:11;
        then x in dom s /\ A9 by A23,XBOOLE_0:def 4;
        then
A27:    x in dom s9 by RELAT_1:61;
        then s9.x = s.x by FUNCT_1:47;
        hence thesis by A26,A27,FUNCT_1:11;
      end;
      assume
A28:  x in dom(f*s9);
      then
A29:  x in dom s9 by FUNCT_1:11;
      then
A30:  x in dom s /\ A9 by RELAT_1:61;
      then
A31:  x in dom s by XBOOLE_0:def 4;
      s9.x in dom f by A28,FUNCT_1:11;
      then s.x in dom f by A29,FUNCT_1:47;
      then x in dom(g|A) by A8,A31,FUNCT_1:11;
      then x in dom g /\ A by RELAT_1:61;
      then
A32:  x in dom g by XBOOLE_0:def 4;
      x in A9 by A30,XBOOLE_0:def 4;
      then x in dom g /\ A9 by A32,XBOOLE_0:def 4;
      hence x in dom(g|A9) by RELAT_1:61;
    end;
    then
A33: dom(g|A9) = dom(f*s9) by TARSKI:1;
    a in C9;
    then a in dom g by FUNCT_2:def 1;
    then a in dom g /\ A by A9,XBOOLE_0:def 4;
    then a in dom(f*s) & g.a = (f*s).a by A8,FUNCT_1:48,RELAT_1:61;
    then
A34: g.a = f.c by FUNCT_1:12;
    B9 \/ {c} = B \/ {c} by XBOOLE_1:39;
    then
A35: B = B9 \/ {c} by A10,ZFMISC_1:40;
A36: dom s9 = A9 by A5,RELAT_1:62,XBOOLE_1:7;
A37: for x st x in dom(g|A9) holds (g|A9).x = (f*s9).x
    proof
      let x;
      assume x in dom(g|A9);
      then
A38:  x in dom g /\ A9 by RELAT_1:61;
      then
A39:  x in A9 by XBOOLE_0:def 4;
      then
A40:  x in A by ZFMISC_1:136;
      x in dom g by A38,XBOOLE_0:def 4;
      then x in dom g /\ A by A40,XBOOLE_0:def 4;
      then x in dom(f*s) by A8,RELAT_1:61;
      then
A41:  x in dom s by FUNCT_1:11;
      then x in dom s /\ A9 by A39,XBOOLE_0:def 4;
      then
A42:  x in dom s9 by RELAT_1:61;
      then
A43:  s9.x = s.x by FUNCT_1:47;
      thus (g|A9).x = g.x by A39,FUNCT_1:49
        .= (f*s).x by A8,A40,FUNCT_1:49
        .= f.(s.x) by A41,FUNCT_1:13
        .= (f*s9).x by A42,A43,FUNCT_1:13;
    end;
    then
A44: g|A9 = f*s9 by A33,FUNCT_1:2;
    now
      let y;
      thus y in g.:A9 implies y in f.:B9
      proof
        assume y in g.:A9;
        then consider x such that
A45:    x in dom g and
A46:    x in A9 and
A47:    y = g.x by FUNCT_1:def 6;
        x in dom g /\ A9 by A45,A46,XBOOLE_0:def 4;
        then
A48:    x in dom(g|A9) by RELAT_1:61;
        then x in dom s9 by A33,FUNCT_1:11;
        then
A49:    s9.x in B9 by A21,FUNCT_1:def 3;
        y = (f*s9).x by A44,A46,A47,FUNCT_1:49;
        then
A50:    y = f.(s9.x) by A33,A48,FUNCT_1:12;
        s9.x in dom f by A33,A48,FUNCT_1:11;
        hence thesis by A50,A49,FUNCT_1:def 6;
      end;
      assume y in f.:B9;
      then consider x such that
      x in dom f and
A51:  x in B9 and
A52:  y = f.x by FUNCT_1:def 6;
      set x9 = s9".x;
A53:  x9 in A9 by A11,A36,A21,A51,FUNCT_1:32;
      A9 c= C9 by FINSUB_1:def 5;
      then x9 in C9 by A53;
      then
A54:  x9 in dom g by FUNCT_2:def 1;
      s9.x9 = x by A11,A21,A51,FUNCT_1:35;
      then y = (f*s9).x9 by A36,A52,A53,FUNCT_1:13
        .= g.x9 by A44,A53,FUNCT_1:49;
      hence y in g.:A9 by A53,A54,FUNCT_1:def 6;
    end;
    then
A55: f.:B9 = g.:A9 by TARSKI:1;
A56: not c in B9 by ZFMISC_1:56;
    now
      per cases;
      suppose
A57:    A9 = {};
        B9 c= C by FINSUB_1:def 5;
        then B9 c= dom f by FUNCT_2:def 1;
        then
A58:    B9 = {} by A55,A57;
        thus F $$(A,g) = f.c by A1,A34,A57,SETWISEO:17
          .= F $$(B,f) by A1,A35,A58,SETWISEO:17;
      end;
      suppose
A59:    A9 <> {};
        A9 c= C9 by FINSUB_1:def 5;
        then A9 c= dom g by FUNCT_2:def 1;
        then
A60:    B9 <> {} by A55,A59;
        thus F $$(A,g) = F.(F $$(A9,g),g.a) by A1,A4,A59,Th2
          .= F.(F $$(B9,f),f.c) by A3,A34,A11,A36,A21,A33,A37,A59,FUNCT_1:2
          .= F $$(B,f) by A1,A35,A56,A60,Th2;
      end;
    end;
    hence thesis;
  end;
A61: X[{}.C9]
  proof
    assume
A62: {}.C9 <> {} or F is having_a_unity;
    let B;
    given s such that
A63: dom s = {}.C9 & rng s = B & s is one-to-one and
    g|({}.C9) = f*s;
    B,{} are_equipotent by A63,WELLORD2:def 4;
    then
A64: B = {}.C by CARD_1:26;
    F $$({}.C9,g) = the_unity_wrt F by A1,A62,SETWISEO:31;
    hence thesis by A1,A62,A64,SETWISEO:31;
  end;
  for A holds X[A] from SETWISEO:sch 2(A61,A2);
  hence thesis;
end;

theorem
  H is commutative associative & (B <> {} or H is having_a_unity) & f is
  one-to-one implies H $$(f.:B,h) = H $$(B,h*f)
proof
  assume that
A1: H is commutative associative &( B <> {} or H is having_a_unity) and
A2: f is one-to-one;
  set s = f|B;
A3: rng s = f.:B & (h*f)|B = h*s by RELAT_1:83,115;
  B c= C by FINSUB_1:def 5;
  then B c= dom f by FUNCT_2:def 1;
  then
A4: dom s = B by RELAT_1:62;
  s is one-to-one by A2,FUNCT_1:52;
  hence thesis by A1,A4,A3,Th5;
end;

theorem
  F is commutative associative & (B <> {} or F is having_a_unity) & f|B
  = f9|B implies F $$(B,f) = F $$(B,f9)
proof
  assume
A1: F is commutative associative &( B <> {} or F is having_a_unity);
  set s = id B;
A2: dom s = B & rng s = B;
  assume f|B = f9|B;
  then f|B = f9*s by RELAT_1:65;
  hence thesis by A1,A2,Th5;
end;

theorem
  F is commutative associative & F is having_a_unity & e = the_unity_wrt
  F & f.:B = {e} implies F$$(B,f) = e
proof
  assume that
A1: F is commutative associative and
A2: F is having_a_unity and
A3: e = the_unity_wrt F;
  defpred X[Element of Fin C] means f.:($1) = {e} implies F$$($1,f) = e;
A4: for B9 being Element of Fin C, b being Element of C holds X[B9] & not b
  in B9 implies X[B9 \/ {.b.}]
  proof
    let B9,c such that
A5: f.:(B9) = {e} implies F$$(B9,f) = e and
A6: not c in B9 and
A7: f.:(B9 \/ {c}) = {e};
A8: now
      per cases;
      suppose
        B9 = {};
        then
A9:     B9 = {}.C;
        thus F$$(B9 \/ {.c.},f) = F.(F $$(B9,f),f.c) by A1,A2,A6,Th2
          .= F.(e,f.c) by A1,A2,A3,A9,SETWISEO:31;
      end;
      suppose
A10:    B9 <> {};
        B9 c= C by FINSUB_1:def 5;
        then
A11:    B9 c= dom f by FUNCT_2:def 1;
        f.:B9 c= {e} by A7,RELAT_1:123,XBOOLE_1:7;
        hence F$$(B9 \/ {.c.},f) = F.(e,f.c) by A1,A5,A6,A10,A11,Th2,
ZFMISC_1:33;
      end;
    end;
    {.c.} c= C by FINSUB_1:def 5;
    then
A12: {c} c= dom f by FUNCT_2:def 1;
    then
A13: c in dom f by ZFMISC_1:31;
    Im(f,c) c= {e} by A7,RELAT_1:123,XBOOLE_1:7;
    then Im(f,c) = {e} by A12,ZFMISC_1:33;
    then {e} = {f.c} by A13,FUNCT_1:59;
    then f.c = e by ZFMISC_1:3;
    hence thesis by A2,A3,A8,SETWISEO:15;
  end;
A14: X[{}.C];
  for B holds X[B] from SETWISEO:sch 2(A14,A4);
  hence thesis;
end;

theorem Th9:
  F is commutative associative & F is having_a_unity & e =
the_unity_wrt F & G.(e,e) = e & (for d1,d2,d3,d4 holds F.(G.(d1,d2),G.(d3,d4))=
  G.(F.(d1,d3),F.(d2,d4))) implies G.(F$$(B,f),F$$(B,f9)) = F $$(B,G.:(f,f9))
proof
  assume that
A1: F is commutative associative & F is having_a_unity and
A2: e = the_unity_wrt F and
A3: G.(e,e) = e and
A4: for d1,d2,d3,d4 holds F.(G.(d1,d2),G.(d3,d4))= G.(F.(d1,d3),F.(d2,d4 ));
  defpred X[Element of Fin C] means G.(F$$($1,f),F$$($1,f9)) = F $$($1,G.:(f,
  f9));
A5: for B9 being Element of Fin C, b being Element of C holds X[B9] & not b
  in B9 implies X[B9 \/ {.b.}]
  proof
    let B,c such that
A6: G.(F$$(B,f),F$$(B,f9)) = F $$(B,G.:(f,f9)) and
A7: not c in B;
    set s9 = F$$(B,f9);
    set s = F$$(B,f);
    F$$(B \/ {.c.},f) = F.(s,f.c) & F$$(B \/ {.c.},f9) = F.(s9,f9.c) by A1,A7
,Th2;
    hence G.(F$$(B \/ {.c.},f),F$$(B \/ {.c.},f9)) = F.(G.(s,s9),G.(f.c,f9.c))
    by A4
      .= F.(G.(s,s9),G.:(f,f9).c) by FUNCOP_1:37
      .= F $$(B \/ {.c.},G.:(f,f9)) by A1,A6,A7,Th2;
  end;
  F$$({}.C,f) = e & F$$({}.C,f9) = e by A1,A2,SETWISEO:31;
  then
A8: X[{}.C] by A1,A2,A3,SETWISEO:31;
  for B holds X[B] from SETWISEO:sch 2(A8,A5);
  hence thesis;
end;

Lm3: F is commutative associative implies for d1,d2,d3,d4 holds F.(F.(d1,d2),F
.(d3,d4))= F.(F.(d1,d3),F.(d2,d4))
proof
  assume that
A1: F is commutative and
A2: F is associative;
  let d1,d2,d3,d4;
  thus F.(F.(d1,d2),F.(d3,d4)) = F.(d1,F.(d2,F.(d3,d4))) by A2,BINOP_1:def 3
    .= F.(d1,F.(F.(d2,d3),d4)) by A2,BINOP_1:def 3
    .= F.(d1,F.(F.(d3,d2),d4)) by A1,BINOP_1:def 2
    .= F.(d1,F.(d3,F.(d2,d4))) by A2,BINOP_1:def 3
    .= F.(F.(d1,d3),F.(d2,d4)) by A2,BINOP_1:def 3;
end;

theorem
  F is commutative associative & F is having_a_unity implies F.(F$$(B,f)
  ,F$$(B,f9)) = F $$(B,F.:(f,f9))
proof
  set e = the_unity_wrt F;
  assume
A1: F is commutative & F is associative & F is having_a_unity;
  then
  F.(e,e) = e & for d1,d2,d3,d4 holds F.(F.(d1,d2),F.(d3,d4))= F.(F.(d1,d3
  ),F. (d2,d4)) by Lm3,SETWISEO:15;
  hence thesis by A1,Th9;
end;

theorem
  F is commutative associative & F is having_a_unity & F is
having_an_inverseOp & G = F*(id D,the_inverseOp_wrt F) implies G.(F$$(B,f),F$$(
  B,f9)) = F $$(B,G.:(f,f9))
proof
  assume that
A1: F is commutative associative & F is having_a_unity and
A2: F is having_an_inverseOp & G = F*(id D,the_inverseOp_wrt F);
  set e = the_unity_wrt F;
  G.(e,e) = e & for d1,d2,d3,d4 holds F.(G.(d1,d2),G.(d3,d4))= G.(F.(d1,d3
  ),F. (d2,d4)) by A1,A2,FINSEQOP:86,89;
  hence thesis by A1,Th9;
end;

theorem Th12:
  F is commutative associative & F is having_a_unity & e =
the_unity_wrt F & G is_distributive_wrt F & G.(d,e) = e implies G.(d,F$$(B,f))
  = F $$(B,G[;](d,f))
proof
  assume that
A1: F is commutative associative & F is having_a_unity and
A2: e = the_unity_wrt F and
A3: G is_distributive_wrt F;
  defpred X[Element of Fin C] means G.(d,F$$($1,f)) = F $$($1,G[;](d,f));
A4: for B9 being Element of Fin C, b being Element of C holds X[B9] & not b
  in B9 implies X[B9 \/ {.b.}]
  proof
    let B9,c such that
A5: G.(d,F$$(B9,f)) = F $$(B9,G[;](d,f)) and
A6: not c in B9;
    thus G.(d,F$$(B9 \/ {.c.},f)) = G.(d,F.(F$$(B9,f),f.c)) by A1,A6,Th2
      .= F.(G.(d,F$$(B9,f)),G.(d,f.c)) by A3,BINOP_1:11
      .= F.(F $$(B9,G[;](d,f)),(G[;](d,f)).c) by A5,FUNCOP_1:53
      .= F $$(B9 \/ {.c.},G[;](d,f)) by A1,A6,Th2;
  end;
  assume G.(d,e) = e;
  then G.(d,F$$({}.C,f)) = e by A1,A2,SETWISEO:31
    .= F $$({}.C,G[;](d,f)) by A1,A2,SETWISEO:31;
  then
A7: X[{}.C];
  for B holds X[B] from SETWISEO:sch 2(A7,A4);
  hence thesis;
end;

theorem Th13:
  F is commutative associative & F is having_a_unity & e =
the_unity_wrt F & G is_distributive_wrt F & G.(e,d) = e implies G.(F$$(B,f),d)
  = F $$(B,G[:](f,d))
proof
  assume that
A1: F is commutative associative & F is having_a_unity and
A2: e = the_unity_wrt F and
A3: G is_distributive_wrt F;
  defpred X[Element of Fin C] means G.(F$$($1,f),d) = F $$($1,G[:](f,d));
A4: for B9 being (Element of Fin C), b being Element of C holds X[B9] & not
  b in B9 implies X[B9 \/ {.b.}]
  proof
    let B9,c such that
A5: G.(F$$(B9,f),d) = F $$(B9,G[:](f,d)) and
A6: not c in B9;
    thus G.(F$$(B9 \/ {.c.},f),d) = G.(F.(F$$(B9,f),f.c),d) by A1,A6,Th2
      .= F.(G.(F$$(B9,f),d),G.(f.c,d)) by A3,BINOP_1:11
      .= F.(F $$(B9,G[:](f,d)),(G[:](f,d)).c) by A5,FUNCOP_1:48
      .= F $$(B9 \/ {.c.},G[:](f,d)) by A1,A6,Th2;
  end;
  assume G.(e,d) = e;
  then G.(F$$({}.C,f),d) = e by A1,A2,SETWISEO:31
    .= F $$({}.C,G[:](f,d)) by A1,A2,SETWISEO:31;
  then
A7: X[{}.C];
  for B holds X[B] from SETWISEO:sch 2(A7,A4);
  hence thesis;
end;

theorem
  F is commutative associative & F is having_a_unity & F is
having_an_inverseOp & G is_distributive_wrt F implies G.(d,F$$(B,f)) = F $$(B,G
  [;](d,f))
proof
  assume that
A1: F is commutative associative & F is having_a_unity and
A2: F is having_an_inverseOp and
A3: G is_distributive_wrt F;
  set e = the_unity_wrt F;
  G.(d,e) = e by A1,A2,A3,FINSEQOP:66;
  hence thesis by A1,A3,Th12;
end;

theorem
  F is commutative associative & F is having_a_unity & F is
having_an_inverseOp & G is_distributive_wrt F implies G.(F$$(B,f),d) = F $$(B,G
  [:](f,d))
proof
  assume that
A1: F is commutative associative & F is having_a_unity and
A2: F is having_an_inverseOp and
A3: G is_distributive_wrt F;
  set e = the_unity_wrt F;
  G.(e,d) = e by A1,A2,A3,FINSEQOP:66;
  hence thesis by A1,A3,Th13;
end;

theorem Th16:
  F is commutative associative & F is having_a_unity & H is
  commutative associative & H is having_a_unity & h.the_unity_wrt F =
the_unity_wrt H & (for d1,d2 holds h.(F.(d1,d2)) = H.(h.d1,h.d2)) implies h.(F
  $$(B,f)) = H $$(B,h*f)
proof
  assume that
A1: F is commutative associative & F is having_a_unity and
A2: H is commutative associative & H is having_a_unity and
A3: h.the_unity_wrt F = the_unity_wrt H and
A4: for d1,d2 holds h.(F.(d1,d2)) = H.(h.d1,h.d2);
  defpred X[Element of Fin C] means h.(F$$($1,f)) = H $$($1,h*f);
A5: for B9 being (Element of Fin C), b being Element of C holds X[B9] & not
  b in B9 implies X[B9 \/ {.b.}]
  proof
    let B,c such that
A6: h.(F$$(B,f)) = H $$(B,h*f) and
A7: not c in B;
    thus h.(F$$(B \/ {.c.},f)) = h.(F.(F$$(B,f),f.c)) by A1,A7,Th2
      .= H.(H $$(B,h*f),h.(f.c)) by A4,A6
      .= H.(H $$(B,h*f),(h*f).c) by FUNCT_2:15
      .= H $$(B \/ {.c.},h*f) by A2,A7,Th2;
  end;
  h.(F$$({}.C,f)) = the_unity_wrt H by A1,A3,SETWISEO:31
    .= H $$({}.C,h*f) by A2,SETWISEO:31;
  then
A8: X[{}.C];
  for B holds X[B] from SETWISEO:sch 2(A8,A5);
  hence thesis;
end;

theorem
  F is commutative associative & F is having_a_unity & u.the_unity_wrt F
= the_unity_wrt F & u is_distributive_wrt F implies u.(F$$(B,f)) = F $$(B,u*f)
proof
  assume
A1: F is commutative associative & F is having_a_unity & u.the_unity_wrt
  F = the_unity_wrt F;
  assume for d1,d2 holds u.(F.(d1,d2)) = F.(u.d1,u.d2);
  hence thesis by A1,Th16;
end;

theorem
  F is commutative associative & F is having_a_unity & F is
having_an_inverseOp & G is_distributive_wrt F implies G[;](d,id D).(F$$(B,f)) =
  F $$(B,G[;](d,id D)*f)
proof
  assume that
A1: F is commutative associative & F is having_a_unity and
A2: F is having_an_inverseOp and
A3: G is_distributive_wrt F;
  set e = the_unity_wrt F;
  set u = G[;](d,id D);
  u is_distributive_wrt F by A3,FINSEQOP:54;
  then
A4: for d1,d2 holds u.(F.(d1,d2)) = F.(u.d1,u.d2) by BINOP_1:def 12;
  G[;](d,id D).e = e by A1,A2,A3,FINSEQOP:69;
  hence thesis by A1,A4,Th16;
end;

theorem
  F is commutative associative & F is having_a_unity & F is
  having_an_inverseOp implies (the_inverseOp_wrt F).(F$$(B,f)) = F $$(B,(
  the_inverseOp_wrt F)*f)
proof
  assume that
A1: F is commutative associative & F is having_a_unity and
A2: F is having_an_inverseOp;
  set e = the_unity_wrt F, u = the_inverseOp_wrt F;
  u is_distributive_wrt F by A1,A2,FINSEQOP:63;
  then
A3: for d1,d2 holds u.(F.(d1,d2)) = F.(u.d1,u.d2) by BINOP_1:def 12;
  u.e = e by A1,A2,FINSEQOP:61;
  hence thesis by A1,A3,Th16;
end;

definition
  let D,p,d;
  func [#](p,d) -> Function of NAT,D equals
  (NAT --> d) +* p;
  coherence;
end;

theorem Th20:
  (i in dom p implies [#](p,d).i = p.i) & (not i in dom p implies
  [#](p,d).i = d)
proof
  thus i in dom p implies [#](p,d).i = p.i by FUNCT_4:13;
A1: i in NAT by ORDINAL1:def 12;
  assume not i in dom p;
  hence [#](p,d).i = (NAT --> d).i by FUNCT_4:11
    .= d by A1,FUNCOP_1:7;
end;

theorem
  [#](p,d)|(dom p) = p
proof
  set k = len p, f = [#](p,d);
  Seg k c= NAT;
  then Seg k c= dom f by FUNCT_2:def 1;
  then
A1: dom (f|Seg k) = Seg k by RELAT_1:62;
A2: dom p = Seg k by FINSEQ_1:def 3;
  now
    let x;
    assume
A3: x in Seg k;
    then (f|Seg k).x = f.x by A1,FUNCT_1:47;
    hence (f|Seg k).x = p.x by A2,A3,Th20;
  end;
  hence thesis by A1,A2,FUNCT_1:2;
end;

theorem
  [#](p^q,d)|(dom p) = p
proof
  set k = len p, f = [#](p^q,d);
  Seg k c= NAT;
  then Seg k c= dom f by FUNCT_2:def 1;
  then
A1: dom (f|Seg k) = Seg k by RELAT_1:62;
A2: dom p = Seg k by FINSEQ_1:def 3;
  now
    let x;
    k <= k + len q by NAT_1:12;
    then k <= len (p^q) by FINSEQ_1:22;
    then
A3: Seg(len(p^q)) = dom (p^q) & Seg k c= Seg len(p^q) by FINSEQ_1:5,def 3;
    assume
A4: x in Seg k;
    then (f|Seg k).x = f.x by A1,FUNCT_1:47;
    hence (f|Seg k).x = (p^q).x by A4,A3,Th20
      .= p.x by A2,A4,FINSEQ_1:def 7;
  end;
  hence thesis by A1,A2,FUNCT_1:2;
end;

theorem
  rng [#](p,d) = rng p \/ {d}
proof
  now
    let y;
    thus y in rng [#](p,d) implies y in rng p \/ {d}
    proof
      assume y in rng [#](p,d);
      then consider x such that
A1:   x in dom [#](p,d) and
A2:   y = [#](p,d).x by FUNCT_1:def 3;
      reconsider i = x as Element of NAT by A1;
      now
        per cases;
        case
A3:       i in dom p;
          then y = p.i by A2,Th20;
          hence y in rng p by A3,FUNCT_1:def 3;
        end;
        case
          not i in dom p;
          then y = d by A2,Th20;
          hence y in {d} by TARSKI:def 1;
        end;
      end;
      hence thesis by XBOOLE_0:def 3;
    end;
    assume
A4: y in rng p \/ {d};
    now
      per cases by A4,XBOOLE_0:def 3;
      suppose
        y in rng p;
        then consider i being Nat such that
A5:     i in dom p and
A6:     y = p.i by FINSEQ_2:10;
        y = [#](p,d).i by A5,A6,Th20;
        hence y in rng [#](p,d) by A5,FUNCT_2:4;
      end;
      suppose
A7:     y in {d};
        dom p = Seg len p by FINSEQ_1:def 3;
        then
A8:     not len p + 1 in dom p by Lm2;
        y = d by A7,TARSKI:def 1;
        then y = [#](p,d).(len p + 1) by A8,Th20;
        hence y in rng [#](p,d) by FUNCT_2:4;
      end;
    end;
    hence y in rng [#](p,d);
  end;
  hence thesis by TARSKI:1;
end;

theorem
  h*[#](p,d) = [#](h*p,h.d)
proof
  now
    let i be Element of NAT;
A1: dom(h*p)=Seg len(h*p) by FINSEQ_1:def 3;
A2: len(h*p) = len p & Seg len p = dom p by FINSEQ_1:def 3,FINSEQ_2:33;
    now
      per cases;
      suppose
A3:     i in dom p;
        hence h.([#](p,d).i) = h.(p.i) by Th20
          .= (h*p).i by A3,FUNCT_1:13
          .= [#](h*p,h.d).i by A2,A1,A3,Th20;
      end;
      suppose
A4:     not i in dom p;
        hence h.([#](p,d).i) = h.d by Th20
          .= [#](h*p,h.d).i by A2,A1,A4,Th20;
      end;
    end;
    hence (h*[#](p,d)).i = [#](h*p,h.d).i by FUNCT_2:15;
  end;
  hence thesis by FUNCT_2:63;
end;

Lm4: len p = len q & F.(e,e) = e implies F.:([#](p,e),[#](q,e)) = [#](F.:(p,q)
,e)
proof
  assume that
A1: len p = len q and
A2: F.(e,e) = e;
  set r = F.:(p,q);
A3: len r = len p by A1,FINSEQ_2:72;
A4: dom r = Seg len r by FINSEQ_1:def 3;
A5: dom p = Seg len p by FINSEQ_1:def 3;
A6: dom q = Seg len q by FINSEQ_1:def 3;
  now
    let i be Element of NAT;
    now
      per cases;
      suppose
A7:     i in dom p;
        hence F.([#](p,e).i,[#](q,e).i) = F.(p.i,[#](q,e).i) by Th20
          .= F.(p.i,q.i) by A1,A5,A6,A7,Th20
          .= r.i by A3,A5,A4,A7,FUNCOP_1:22
          .= [#](r,e).i by A3,A5,A4,A7,Th20;
      end;
      suppose
A8:     not i in dom p;
        hence F.([#](p,e).i,[#](q,e).i) = F.(e,[#](q,e).i) by Th20
          .= e by A1,A2,A5,A6,A8,Th20
          .= [#](r,e).i by A3,A5,A4,A8,Th20;
      end;
    end;
    hence F.:([#](p,e),[#](q,e)).i = [#](r,e).i by FUNCOP_1:37;
  end;
  hence thesis by FUNCT_2:63;
end;

Lm5: F.(e,d) = e implies F[:]([#](p,e),d) = [#](F[:](p,d),e)
proof
  assume
A1: F.(e,d) = e;
  now
    let i be Element of NAT;
A2: dom p = Seg len p by FINSEQ_1:def 3;
A3: len(F[:](p,d)) = len p & dom(F[:](p,d)) = Seg len(F[:](p,d)) by
FINSEQ_1:def 3,FINSEQ_2:84;
    now
      per cases;
      suppose
A4:     i in dom p;
        hence F.([#](p,e).i,d) = F.(p.i,d) by Th20
          .= (F[:](p,d)).i by A3,A2,A4,FUNCOP_1:27
          .= ([#](F[:](p,d),e)).i by A3,A2,A4,Th20;
      end;
      suppose
A5:     not i in dom p;
        hence F.([#](p,e).i,d) = F.(e,d) by Th20
          .= ([#](F[:](p,d),e)).i by A1,A3,A2,A5,Th20;
      end;
    end;
    hence (F[:]([#](p,e),d)).i = ([#](F[:](p,d),e)).i by FUNCOP_1:48;
  end;
  hence thesis by FUNCT_2:63;
end;

Lm6: F.(d,e) = e implies F[;](d,[#](p,e)) = [#](F[;](d,p),e)
proof
  assume
A1: F.(d,e) = e;
  now
    let i be Element of NAT;
A2: dom p = Seg len p by FINSEQ_1:def 3;
A3: len(F[;](d,p)) = len p & dom(F[;](d,p)) = Seg len(F[;](d,p)) by
FINSEQ_1:def 3,FINSEQ_2:78;
    now
      per cases;
      suppose
A4:     i in dom p;
        hence F.(d,[#](p,e).i) = F.(d,p.i) by Th20
          .= (F[;](d,p)).i by A3,A2,A4,FUNCOP_1:32
          .= ([#](F[;](d,p),e)).i by A3,A2,A4,Th20;
      end;
      suppose
A5:     not i in dom p;
        hence F.(d,[#](p,e).i) = F.(d,e) by Th20
          .= ([#](F[;](d,p),e)).i by A1,A3,A2,A5,Th20;
      end;
    end;
    hence (F[;](d,[#](p,e))).i = ([#](F[;](d,p),e)).i by FUNCOP_1:53;
  end;
  hence thesis by FUNCT_2:63;
end;

notation
  let i be Nat;
  synonym finSeg i for Seg i;
end;

definition
  let i be Nat;
  redefine func finSeg i -> Element of Fin NAT;
  coherence by Lm1;
end;

notation
  let D,p,F;
  synonym F $$ p for F "**" p;
end;

definition
  let D,p,F;
  assume
A1: ( F is having_a_unity or len p >= 1)& F is associative commutative;
  redefine func F $$ p equals
  :Def2:
  F $$(findom p,[#](p,the_unity_wrt F));
  compatibility by A1,FINSOP_1:3;
end;

theorem Th25:
  F is having_a_unity implies F"**"(i|->the_unity_wrt F) = the_unity_wrt F
proof
  set e = the_unity_wrt F;
  defpred X[Nat] means F"**"($1|->e) = e;
  assume
A1: F is having_a_unity;
A2: for i st X[i] holds X[i+1]
  proof
    let i;
    assume
A3: F"**"(i|->e) = e;
    thus F"**"((i+1)|->e) = F"**"((i|->e)^<*e*>) by FINSEQ_2:60
      .= F.(F"**"(i|->e),e) by A1,FINSOP_1:4
      .= e by A1,A3,SETWISEO:15;
  end;
  F"**"(0|->e) = F"**" <*>D .= e by A1,FINSOP_1:10;
  then
A4: X[0];
  for i holds X[i] from NAT_1:sch 2(A4,A2);
  hence thesis;
end;

theorem Th26:
  F is associative & (i>=1 & j>=1 or F is having_a_unity) implies
  F"**"((i+j)|->d) = F.(F"**"(i|->d),F"**"(j|->d))
proof
  assume
A1: F is associative;
  set p1 = (i|->d),p2 = (j|->d);
  assume i>=1 & j>=1 or F is having_a_unity;
  then len p1 >= 1 & len p2 >= 1 or F is having_a_unity by CARD_1:def 7;
  then F "**"(p1^p2) = F.(F"**"p1,F"**"p2) by A1,FINSOP_1:5;
  hence thesis by FINSEQ_2:123;
end;

theorem
  F is commutative associative & (i>=1 & j>=1 or F is having_a_unity)
  implies F"**"((i*j)|->d) = F"**"(j|->(F"**"(i|->d)))
proof
  assume that
A1: F is commutative associative and
A2: i>=1 & j>=1 or F is having_a_unity;
  per cases by NAT_1:3;
  suppose
A3: i = 0 or j = 0;
    set e = the_unity_wrt F;
A4: now
      per cases by A3;
      suppose
        i = 0;
        then i|->d = <*>D;
        then F"**"(i|->d) = e by A2,A3,FINSOP_1:10;
        hence F"**"(j|->(F"**"(i|->d))) = e by A2,A3,Th25;
      end;
      suppose
        j = 0;
        then j|->(F"**"(i|->d)) = <*>D;
        hence F"**"(j|->(F"**"(i|->d))) = e by A2,A3,FINSOP_1:10;
      end;
    end;
    (i*j)|->d = <*>D by A3;
    hence thesis by A2,A3,A4,FINSOP_1:10;
  end;
  suppose
A5: i > 0 & j > 0;
    defpred X[Nat] means $1 <> 0 implies F"**"((i*$1)|->d) = F"**"($1|->(F"**"
    (i|->d)));
A6: for j st X[j] holds X[j+1]
    proof
      let j such that
A7:   j <> 0 implies F"**"((i*j)|->d) = F"**"(j|->(F"**"(i|->d)));
      now
        per cases by NAT_1:14;
        suppose
A8:       j = 0;
          1|->(F"**"(i|->d)) = <*F"**"(i|->d)*> by FINSEQ_2:59;
          hence thesis by A8,FINSOP_1:11;
        end;
        suppose
A9:       j >= 1+0;
          then j > 0 by NAT_1:13;
          then i*j > i*0 by A5,XREAL_1:68;
          then
A10:      i*j >= 1+0 by NAT_1:13;
          F"**"((i*(j+1))|->d) = F"**"((i*j+i*1)|->d)
            .= F.(F"**"((i*j)|->d),F"**"(i|->d)) by A1,A2,A10,Th26
            .= F.(F"**"((i*j)|->d),F"**"(1|->(F"**"(i|->d)))) by FINSOP_1:16
            .= F"**"((j+1)|->(F"**"(i|->d))) by A1,A7,A9,Th26;
          hence thesis;
        end;
      end;
      hence thesis;
    end;
A11: X[0];
    for j holds X[j] from NAT_1:sch 2(A11,A6);
    hence thesis by A5;
  end;
end;

theorem Th28:
  F is having_a_unity & H is having_a_unity & h.the_unity_wrt F =
the_unity_wrt H & (for d1,d2 holds h.(F.(d1,d2)) = H.(h.d1,h.d2)) implies h.(F
  "**"p) = H "**"(h*p)
proof
  assume that
A1: F is having_a_unity and
A2: H is having_a_unity and
A3: h.the_unity_wrt F = the_unity_wrt H and
A4: for d1,d2 holds h.(F.(d1,d2)) = H.(h.d1,h.d2);
  defpred X[FinSequence of D] means h.(F"**"$1) = H "**"(h*$1);
A5: for q,d st X[q] holds X[q^<*d*>]
  proof
    let q,d such that
A6: h.(F"**"q) = H "**"(h*q);
    thus h.(F"**"(q^<*d*>)) = h.(F.(F"**"q,d)) by A1,FINSOP_1:4
      .= H.(H "**"(h*q),h.d) by A4,A6
      .= H "**" ((h*q)^<*h.d*>) by A2,FINSOP_1:4
      .= H "**"(h*(q^<*d*>)) by FINSEQOP:8;
  end;
  h.(F"**"<*>D) = h.(the_unity_wrt F) by A1,FINSOP_1:10
    .= H "**"(<*>E) by A2,A3,FINSOP_1:10
    .= H "**"(h*<*>D);
  then
A7: X[<*>D];
  X[q] from FINSEQ_2:sch 2(A7,A5);
  hence thesis;
end;

theorem
  F is having_a_unity & u.the_unity_wrt F = the_unity_wrt F & u
  is_distributive_wrt F implies u.(F"**"p) = F "**"(u*p)
proof
  assume
A1: F is having_a_unity & u.the_unity_wrt F = the_unity_wrt F;
  assume for d1,d2 holds u.(F.(d1,d2)) = F.(u.d1,u.d2);
  hence thesis by A1,Th28;
end;

theorem
  F is associative & F is having_a_unity & F is having_an_inverseOp & G
  is_distributive_wrt F implies G[;](d,id D).(F"**"p) = F "**"(G[;](d,id D)*p)
proof
  assume that
A1: F is associative and
A2: F is having_a_unity and
A3: F is having_an_inverseOp and
A4: G is_distributive_wrt F;
  set e = the_unity_wrt F;
  set u = G[;](d,id D);
  u is_distributive_wrt F by A4,FINSEQOP:54;
  then
A5: for d1,d2 holds u.(F.(d1,d2)) = F.(u.d1,u.d2) by BINOP_1:def 12;
  G[;](d,id D).e = e by A1,A2,A3,A4,FINSEQOP:69;
  hence thesis by A2,A5,Th28;
end;

theorem
  F is commutative associative & F is having_a_unity & F is
  having_an_inverseOp implies (the_inverseOp_wrt F).(F"**"p) = F "**"((
  the_inverseOp_wrt F)*p)
proof
  assume that
A1: F is commutative associative and
A2: F is having_a_unity and
A3: F is having_an_inverseOp;
  set e = the_unity_wrt F, u = the_inverseOp_wrt F;
  u is_distributive_wrt F by A1,A2,A3,FINSEQOP:63;
  then
A4: for d1,d2 holds u.(F.(d1,d2)) = F.(u.d1,u.d2) by BINOP_1:def 12;
  u.e = e by A1,A2,A3,FINSEQOP:61;
  hence thesis by A2,A4,Th28;
end;

theorem Th32:
  F is commutative associative & F is having_a_unity & e =
the_unity_wrt F & G.(e,e) = e & (for d1,d2,d3,d4 holds F.(G.(d1,d2),G.(d3,d4))=
G.(F.(d1,d3),F.(d2,d4))) & len p = len q implies G.(F"**"p,F"**"q) = F "**"(G.:
  (p,q))
proof
  assume that
A1: F is commutative & F is associative & F is having_a_unity & e =
  the_unity_wrt F and
A2: G.(e,e) = e and
A3: F.(G.(d1,d2),G.(d3,d4))= G.(F.(d1,d3),F.(d2,d4)) and
A4: len p = len q;
A5: len p = len(G.:(p,q)) by A4,FINSEQ_2:72;
A6: dom(G.:(p,q)) = Seg len(G.:(p,q)) by FINSEQ_1:def 3;
A7: dom q = Seg len q by FINSEQ_1:def 3;
A8: dom p = Seg len p by FINSEQ_1:def 3;
  thus G.(F "**"p,F "**"q) = G.(F $$(findom p,[#](p,e)),F "**"q) by A1,Def2
    .= G.(F $$(findom p,[#](p,e)),F $$(findom q,[#](q,e))) by A1,Def2
    .= F $$(findom p,G.:([#](p,e),[#](q,e))) by A1,A2,A3,A4,A8,A7,Th9
    .= F $$(findom(G.:(p,q)),[#](G.:(p,q),e)) by A2,A4,A5,A8,A6,Lm4
    .= F "**"(G.:(p,q)) by A1,Def2;
end;

theorem Th33:
  F is commutative associative & F is having_a_unity & e =
the_unity_wrt F & G.(e,e) = e & (for d1,d2,d3,d4 holds F.(G.(d1,d2),G.(d3,d4))=
  G.(F.(d1,d3),F.(d2,d4))) implies G.(F"**"T1,F"**"T2) = F "**"(G.:(T1,T2))
proof
  len T1 = i & len T2 = i by CARD_1:def 7;
  hence thesis by Th32;
end;

theorem Th34:
  F is commutative associative & F is having_a_unity & len p = len
  q implies F.(F"**"p,F"**"q) = F "**"(F.:(p,q))
proof
  set e = the_unity_wrt F;
  assume
A1: F is commutative & F is associative & F is having_a_unity;
  then
  F.(e,e) = e & for d1,d2,d3,d4 holds F.(F.(d1,d2),F.(d3,d4))= F.(F.(d1,d3
  ),F. (d2,d4)) by Lm3,SETWISEO:15;
  hence thesis by A1,Th32;
end;

theorem Th35:
  F is commutative associative & F is having_a_unity implies
  F.(F"**"T1,F"**"T2) = F "**"(F.:(T1,T2))
proof
  len T1 = i & len T2 = i by CARD_1:def 7;
  hence thesis by Th34;
end;

theorem
  F is commutative associative & F is having_a_unity implies
  F"**"(i|->(F.(d1,d2))) = F.(F"**"(i|->d1),F"**"(i|->d2))
proof
  reconsider T1 = i|->d1, T2 = i|->d2 as Element of i-tuples_on D;
  i|->(F.(d1,d2)) = F.:(T1,T2) by FINSEQOP:17;
  hence thesis by Th35;
end;

theorem
  F is commutative associative & F is having_a_unity & F is
having_an_inverseOp & G = F*(id D,the_inverseOp_wrt F) implies G.(F"**"T1,F"**"
  T2) = F "**"(G.:(T1,T2))
proof
  assume that
A1: F is commutative associative & F is having_a_unity and
A2: F is having_an_inverseOp & G = F*(id D,the_inverseOp_wrt F);
  set e = the_unity_wrt F;
  G.(e,e) = e & for d1,d2,d3,d4 holds F.(G.(d1,d2),G.(d3,d4))= G.(F.(d1,d3
  ),F. (d2,d4)) by A1,A2,FINSEQOP:86,89;
  hence thesis by A1,Th33;
end;

theorem Th38:
  F is commutative associative & F is having_a_unity & e =
the_unity_wrt F & G is_distributive_wrt F & G.(d,e) = e implies G.(d,F"**"p) =
  F "**"(G[;](d,p))
proof
  assume that
A1: F is commutative & F is associative & F is having_a_unity & e =
  the_unity_wrt F and
A2: G is_distributive_wrt F and
A3: G.(d,e) = e;
A4: len p = len(G[;](d,p)) & Seg len p = dom p by FINSEQ_1:def 3,FINSEQ_2:78;
A5: Seg len(G[;](d,p)) = dom(G[;](d,p)) by FINSEQ_1:def 3;
  thus G.(d,F"**"p) = G.(d,F$$(findom p,[#](p,e))) by A1,Def2
    .= F $$(findom p,G[;](d,[#](p,e))) by A1,A2,A3,Th12
    .= F $$(findom p,[#](G[;](d,p),e)) by A3,Lm6
    .= F "**"(G[;](d,p)) by A1,A4,A5,Def2;
end;

theorem Th39:
  F is commutative associative & F is having_a_unity & e =
the_unity_wrt F & G is_distributive_wrt F & G.(e,d) = e implies G.(F"**"p,d) =
  F "**"(G[:](p,d))
proof
  assume that
A1: F is commutative associative & F is having_a_unity & e =
  the_unity_wrt F and
A2: G is_distributive_wrt F and
A3: G.(e,d) = e;
A4: len p = len(G[:](p,d)) & Seg len p = dom p by FINSEQ_1:def 3,FINSEQ_2:84;
A5: Seg len(G[:](p,d)) = dom(G[:](p,d)) by FINSEQ_1:def 3;
  thus G.(F"**"p,d) = G.(F$$(findom p,[#](p,e)),d) by A1,Def2
    .= F $$(findom p,G[:]([#](p,e),d)) by A1,A2,A3,Th13
    .= F $$(findom p,[#](G[:](p,d),e)) by A3,Lm5
    .= F "**"(G[:](p,d)) by A1,A4,A5,Def2;
end;

theorem
  F is commutative associative & F is having_a_unity & F is
  having_an_inverseOp & G is_distributive_wrt F implies G.(d,F"**"p) = F "**"(G
  [;](d,p))
proof
  assume that
A1: F is commutative associative & F is having_a_unity and
A2: F is having_an_inverseOp and
A3: G is_distributive_wrt F;
  set e = the_unity_wrt F;
  G.(d,e) = e by A1,A2,A3,FINSEQOP:66;
  hence thesis by A1,A3,Th38;
end;

theorem
  F is commutative associative & F is having_a_unity & F is
  having_an_inverseOp & G is_distributive_wrt F implies G.(F"**"p,d) = F "**"(G
  [:](p,d))
proof
  assume that
A1: F is commutative associative & F is having_a_unity and
A2: F is having_an_inverseOp and
A3: G is_distributive_wrt F;
  set e = the_unity_wrt F;
  G.(e,d) = e by A1,A2,A3,FINSEQOP:66;
  hence thesis by A1,A3,Th39;
end;














begin

reserve x,X,Y for set;
reserve C for non empty set;
reserve c for Element of C;
reserve f,f1,f2,f3,g,g1 for complex-valued Function;
reserve r,p for complex number;

Lm1: (-1)"=-1;





definition
  let f1,f2 be complex-valued Function;

  func f1/f2 -> Function means
  :Def1:
  dom it = dom f1 /\ (dom f2 \ f2"{0}) &
  for c being set st c in dom it holds it.c = f1.c * (f2.c)";
  existence
  proof
    deffunc F(set) = f1.$1 * (f2.$1)";
    ex f being Function st dom f = dom f1 /\ (dom f2 \ f2"{0}) & for x st
    x in dom f1 /\ (dom f2 \ f2"{0}) holds f.x = F(x) from FUNCT_1:sch 3;
    hence thesis;
  end;
  uniqueness
  proof
    let f, g be Function such that
A1: dom f = dom f1 /\ (dom f2 \ f2"{0}) and
A2: for c being set st c in dom f holds f.c = f1.c * (f2.c)" and
A3: dom g = dom f1 /\ (dom f2 \ f2"{0}) and
A4: for c being set st c in dom g holds g.c = f1.c * (f2.c)";
    now
      let x;
      assume
A5:   x in dom f;
      hence f.x = f1.x * (f2.x)" by A2
        .= g.x by A1,A3,A4,A5;
    end;
    hence thesis by A1,A3,FUNCT_1:2;
  end;
end;

registration
  let f1, f2 be complex-valued Function;
  cluster f1/f2 -> complex-valued;
  coherence
  proof
    let x be set;
    set F = f1/f2;
    assume x in dom F;
    then F.x = f1.x * (f2.x)" by Def1;
    hence thesis;
  end;
end;

registration
  let f1, f2 be real-valued Function;
  cluster f1/f2 -> real-valued;
  coherence
  proof
    let x be set;
    set F = f1/f2;
    assume x in dom F;
    then F.x = f1.x * (f2.x)" by Def1;
    hence thesis;
  end;
end;

definition
  let C be set, D be complex-membered set;
  let f1, f2 be PartFunc of C,D;
  redefine func f1/f2 -> PartFunc of C,COMPLEX;
  coherence
  proof
    set F = f1/f2;
A1: rng F c= COMPLEX by VALUED_0:def 1;
    dom F = dom f1 /\ (dom f2 \ f2"{0}) by Def1;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let C be set, D be real-membered set;
  let f1, f2 be PartFunc of C,D;
  redefine func f1/f2 -> PartFunc of C,REAL;
  coherence
  proof
    set F = f1/f2;
    rng F c= REAL by VALUED_0:def 3;
    then F is PartFunc of dom F, REAL by RELSET_1:4;
    hence thesis by RELSET_1:5;
  end;
end;

definition
  let f be complex-valued Function;

  func f^ -> Function means
  :Def2:
  dom it = dom f \ f"{0} & for c being set st c in dom it holds it.c = (f.c)";
  existence
  proof
    deffunc F(set) = (f.$1)";
    ex h being Function st dom h = dom f \ f"{0} & for x st x in dom f \ f
    "{0} holds h.x = F(x) from FUNCT_1:sch 3;
    hence thesis;
  end;
  uniqueness
  proof
    let h, g be Function such that
A1: dom h = dom f \ f"{0} and
A2: for c being set st c in dom h holds h.c = (f.c)" and
A3: dom g = dom f \ f"{0} and
A4: for c being set st c in dom g holds g.c = (f.c)";
    now
      let x;
      assume
A5:   x in dom h;
      hence h.x = (f.x)" by A2
        .= g.x by A1,A3,A4,A5;
    end;
    hence thesis by A1,A3,FUNCT_1:2;
  end;
end;

registration
  let f be complex-valued Function;
  cluster f^ -> complex-valued;
  coherence
  proof
    let x be set;
    set F = f^;
    assume x in dom F;
    then F.x = (f.x)" by Def2;
    hence thesis;
  end;
end;

registration
  let f be real-valued Function;
  cluster f^ -> real-valued;
  coherence
  proof
    let x be set;
    set F = f^;
    assume x in dom F;
    then F.x = (f.x)" by Def2;
    hence thesis;
  end;
end;

definition
  let C be set, D be complex-membered set, f be PartFunc of C,D;
  redefine func f^ -> PartFunc of C,COMPLEX;
  coherence
  proof
    set F = f^;
A1: rng F c= COMPLEX by VALUED_0:def 1;
    dom F = dom f \ f"{0} by Def2;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let C be set, D be real-membered set, f be PartFunc of C,D;
  redefine func f^ -> PartFunc of C,REAL;
  coherence
  proof
    set F = f^;
    rng F c= REAL by VALUED_0:def 3;
    then F is PartFunc of dom F, REAL by RELSET_1:4;
    hence thesis by RELSET_1:5;
  end;
end;

theorem Th1:
  dom (g^) c= dom g & dom g /\ (dom g \ g"{0}) = dom g \ g"{0}
proof
  dom (g^) = dom g \ g"{0} by Def2;
  hence dom (g^) c= dom g;
  thus thesis by XBOOLE_1:28;
end;

theorem Th2:
  dom (f1(#)f2) \ (f1(#)f2)"{0} = (dom f1 \ f1"{0}) /\ (dom f2 \ f2"{0})
proof
  thus dom (f1(#)f2) \ (f1(#)f2)"{0} c= (dom f1 \ (f1)"{0}) /\ (dom f2 \ (f2)"
  {0})
  proof
    let x;
    assume
A1: x in dom (f1(#)f2) \ (f1(#)f2)"{0};
    then not x in (f1(#)f2)"{0} by XBOOLE_0:def 5;
    then not (f1(#)f2).x in {0} by A1,FUNCT_1:def 7;
    then (f1(#)f2).x <> 0 by TARSKI:def 1;
    then
A2: f1.x * f2.x <> 0 by VALUED_1:5;
    then f2.x <> 0;
    then not f2.x in {0} by TARSKI:def 1;
    then
A3: not x in (f2)"{0} by FUNCT_1:def 7;
    x in dom (f1(#)f2) by A1;
    then
A4: x in dom f1 /\ dom f2 by VALUED_1:def 4;
    then x in dom f2 by XBOOLE_0:def 4;
    then
A5: x in dom f2 \ (f2)"{0} by A3,XBOOLE_0:def 5;
    f1.x <> 0 by A2;
    then not f1.x in {0} by TARSKI:def 1;
    then
A6: not x in (f1)"{0} by FUNCT_1:def 7;
    x in dom f1 by A4,XBOOLE_0:def 4;
    then x in dom f1 \ (f1)"{0} by A6,XBOOLE_0:def 5;
    hence thesis by A5,XBOOLE_0:def 4;
  end;
  thus (dom f1 \ (f1)"{0}) /\ (dom f2 \ (f2)"{0}) c= dom (f1(#)f2) \ (f1(#)f2)
  "{0}
  proof
    let x;
    assume
A7: x in (dom f1 \ (f1)"{0}) /\ (dom f2 \ (f2)"{0});
    then x in dom f2 \ (f2)"{0} by XBOOLE_0:def 4;
    then not x in (f2)"{0} by XBOOLE_0:def 5;
    then not f2.x in {0} by A7,FUNCT_1:def 7;
    then
A8: f2.x <> 0 by TARSKI:def 1;
A9: x in dom f1 \ (f1)"{0} by A7,XBOOLE_0:def 4;
    then not x in (f1)"{0} by XBOOLE_0:def 5;
    then not f1.x in {0} by A9,FUNCT_1:def 7;
    then f1.x <> 0 by TARSKI:def 1;
    then f1.x * f2.x <>0 by A8;
    then (f1(#)f2).x <> 0 by VALUED_1:5;
    then not (f1(#)f2).x in {0} by TARSKI:def 1;
    then
A10: not x in (f1(#)f2)"{0} by FUNCT_1:def 7;
    x in dom f1 /\ dom f2 by A7,A9,XBOOLE_0:def 4;
    then x in dom (f1(#)f2) by VALUED_1:def 4;
    hence thesis by A10,XBOOLE_0:def 5;
  end;
end;

theorem Th3:
  c in dom (f^) implies f.c <> 0
proof
  assume that
A1: c in dom (f^) and
A2: f.c = 0;
A3: c in dom f \ f"{0} by A1,Def2;
  then
A4: not c in f"{0} by XBOOLE_0:def 5;
  now
    per cases by A4,FUNCT_1:def 7;
    suppose
      not c in dom f;
      hence contradiction by A3;
    end;
    suppose
      not f.c in {0};
      hence contradiction by A2,TARSKI:def 1;
    end;
  end;
  hence contradiction;
end;

theorem Th4:
  (f^)"{0} = {}
proof
  set x = the Element of (f^)"{0};
  assume
A1: (f^)"{0} <> {};
  then
A2: x in dom (f^) by FUNCT_1:def 7;
  then
A3: x in dom f \ f"{0} by Def2;
  then not x in f"{0} by XBOOLE_0:def 5;
  then
A4: not f.x in {0} by A3,FUNCT_1:def 7;
  ((f^) qua Function).x in {0} by A1,FUNCT_1:def 7;
  then (f^).x = 0 by TARSKI:def 1;
  then (f.x)" = 0 by A2,Def2;
  hence contradiction by A4,TARSKI:def 1,XCMPLX_1:202;
end;

theorem Th5:
  (abs(f))"{0} = f"{0} & (-f)"{0} = f"{0}
proof
  now
    let c be set;
    thus c in (abs(f))"{0} implies c in f"{0}
    proof
      assume
A1:   c in (abs(f))"{0};
      then (abs(f)).c in {0} by FUNCT_1:def 7;
      then (abs(f)).c = 0 by TARSKI:def 1;
      then abs(f.c) = 0 by VALUED_1:18;
      then f.c = 0 by COMPLEX1:45;
      then
A2:   f.c in {0} by TARSKI:def 1;
      c in dom (abs(f)) by A1,FUNCT_1:def 7;
      then c in dom f by VALUED_1:def 11;
      hence thesis by A2,FUNCT_1:def 7;
    end;
    assume
A3: c in (f)"{0};
    then f.c in {0} by FUNCT_1:def 7;
    then f.c = 0 by TARSKI:def 1;
    then abs(f.c) = 0 by ABSVALUE:2;
    then (abs(f)).c = 0 by VALUED_1:18;
    then
A4: (abs(f)).c in {0} by TARSKI:def 1;
    c in dom f by A3,FUNCT_1:def 7;
    then c in dom (abs(f)) by VALUED_1:def 11;
    hence c in (abs(f))"{0} by A4,FUNCT_1:def 7;
  end;
  hence (abs(f))"{0} = f"{0} by TARSKI:1;
  now
    let c be set;
    thus c in (-f)"{0} implies c in f"{0}
    proof
      assume
A5:   c in (-f)"{0};
      then (-f).c in {0} by FUNCT_1:def 7;
      then (-f).c = 0 by TARSKI:def 1;
      then --(f.c) = -0 by VALUED_1:8;
      then
A6:   f.c in {0} by TARSKI:def 1;
      c in dom (-f) by A5,FUNCT_1:def 7;
      then c in dom f by VALUED_1:8;
      hence thesis by A6,FUNCT_1:def 7;
    end;
    assume
A7: c in (f)"{0};
    then f.c in {0} by FUNCT_1:def 7;
    then f.c = 0 by TARSKI:def 1;
    then (-f).c = -0 by VALUED_1:8;
    then
A8: (-f).c in {0} by TARSKI:def 1;
    c in dom f by A7,FUNCT_1:def 7;
    then c in dom (-f) by VALUED_1:8;
    hence c in (-f)"{0} by A8,FUNCT_1:def 7;
  end;
  hence thesis by TARSKI:1;
end;

theorem Th6:
  dom (f^^) = dom (f|(dom (f^)))
proof
A1: dom (f^) = dom f \ f"{0} by Def2;
  thus dom (f^^) = dom (f^) \(f^)"{0} by Def2
    .= dom (f^) \ {} by Th4
    .= dom f /\ dom (f^) by A1,XBOOLE_1:28
    .= dom (f|(dom (f^))) by RELAT_1:61;
end;

theorem Th7:
  r<>0 implies (r(#)f)"{0} = f"{0}
proof
  assume
A1: r<>0;
  now
    let c be set;
    thus c in (r(#)f)"{0} implies c in f"{0}
    proof
      assume
A2:   c in (r(#)f)"{0};
      then
A3:   c in dom (r(#)f) by FUNCT_1:def 7;
      (r(#)f).c in {0} by A2,FUNCT_1:def 7;
      then (r(#)f).c = 0 by TARSKI:def 1;
      then r*f.c = 0 by A3,VALUED_1:def 5;
      then f.c = 0 by A1;
      then
A4:   f.c in {0} by TARSKI:def 1;
      c in dom f by A3,VALUED_1:def 5;
      hence thesis by A4,FUNCT_1:def 7;
    end;
    assume
A5: c in (f)"{0};
    then f.c in {0} by FUNCT_1:def 7;
    then f.c = 0 by TARSKI:def 1;
    then
A6: r*f.c = 0;
A7: c in dom f by A5,FUNCT_1:def 7;
    then c in dom (r(#)f) by VALUED_1:def 5;
    then (r(#)f).c = 0 by A6,VALUED_1:def 5;
    then
A8: (r(#)f).c in {0} by TARSKI:def 1;
    c in dom (r(#)f) by A7,VALUED_1:def 5;
    hence c in (r(#)f)"{0} by A8,FUNCT_1:def 7;
  end;
  hence thesis by TARSKI:1;
end;





theorem
  (f1 + f2) + f3 = f1 + (f2 + f3)
proof
A1: dom (f1 + f2 + f3) = dom (f1 + f2) /\ dom f3 by VALUED_1:def 1
    .= dom f1 /\ dom f2 /\ dom f3 by VALUED_1:def 1
    .= dom f1 /\ (dom f2 /\ dom f3) by XBOOLE_1:16
    .= dom f1 /\ dom (f2 + f3) by VALUED_1:def 1
    .= dom (f1 + (f2 + f3)) by VALUED_1:def 1;
  now
    let c be set;
    assume
A2: c in dom (f1 + f2 + f3);
    then c in dom (f1 + f2) /\ dom f3 by VALUED_1:def 1;
    then
A3: c in dom (f1 + f2) by XBOOLE_0:def 4;
    c in dom f1 /\ dom (f2 + f3) by A1,A2,VALUED_1:def 1;
    then
A4: c in dom (f2 + f3) by XBOOLE_0:def 4;
    thus (f1 + f2 + f3).c = (f1 + f2).c + f3.c by A2,VALUED_1:def 1
      .= f1.c + f2.c + f3.c by A3,VALUED_1:def 1
      .= f1.c + (f2.c + f3.c)
      .= f1.c + (f2 + f3).c by A4,VALUED_1:def 1
      .= (f1 + (f2 + f3)).c by A1,A2,VALUED_1:def 1;
  end;
  hence thesis by A1,FUNCT_1:2;
end;

theorem Th9:
  (f1 (#) f2) (#) f3 = f1 (#) (f2 (#) f3)
proof
A1: now
    let c be set;
    assume c in dom (f1(#)f2(#)f3);
    thus (f1 (#) f2 (#) f3).c = (f1 (#) f2).c * f3.c by VALUED_1:5
      .= f1.c * f2.c * f3.c by VALUED_1:5
      .= f1.c * (f2.c * f3.c)
      .= f1.c * (f2 (#) f3).c by VALUED_1:5
      .= (f1 (#) (f2 (#) f3)).c by VALUED_1:5;
  end;
  dom (f1 (#) f2 (#) f3) = dom (f1 (#) f2) /\ dom f3 by VALUED_1:def 4
    .= dom f1 /\ dom f2 /\ dom f3 by VALUED_1:def 4
    .= dom f1 /\ (dom f2 /\ dom f3) by XBOOLE_1:16
    .= dom f1 /\ dom (f2 (#) f3) by VALUED_1:def 4
    .= dom (f1 (#) (f2 (#) f3)) by VALUED_1:def 4;
  hence thesis by A1,FUNCT_1:2;
end;

theorem Th10:
  (f1 + f2) (#) f3=f1 (#) f3 + f2 (#) f3
proof
A1: dom ((f1 + f2) (#) f3) = dom (f1 + f2) /\ dom f3 by VALUED_1:def 4
    .= dom f1 /\ dom f2 /\ (dom f3 /\ dom f3) by VALUED_1:def 1
    .= dom f1 /\ dom f2 /\ dom f3 /\ dom f3 by XBOOLE_1:16
    .= dom f1 /\ dom f3 /\ dom f2 /\ dom f3 by XBOOLE_1:16
    .= dom f1 /\ dom f3 /\ (dom f2 /\ dom f3) by XBOOLE_1:16
    .= dom (f1 (#) f3) /\ (dom f2 /\ dom f3) by VALUED_1:def 4
    .= dom (f1 (#) f3) /\ dom (f2 (#) f3) by VALUED_1:def 4
    .= dom (f1 (#) f3 + f2 (#) f3) by VALUED_1:def 1;
  now
    let c be set;
    assume
A2: c in dom ((f1 + f2)(#)f3);
    then c in dom (f1 + f2) /\ dom f3 by VALUED_1:def 4;
    then
A3: c in dom (f1 + f2) by XBOOLE_0:def 4;
    thus ((f1 + f2) (#) f3).c = (f1 + f2).c * f3.c by VALUED_1:5
      .= (f1.c + f2.c) * f3.c by A3,VALUED_1:def 1
      .= f1.c * f3.c + f2.c * f3.c
      .= (f1 (#) f3).c + f2.c* f3.c by VALUED_1:5
      .= (f1 (#) f3).c + (f2 (#) f3).c by VALUED_1:5
      .=((f1 (#) f3) + (f2 (#) f3)).c by A1,A2,VALUED_1:def 1;
  end;
  hence thesis by A1,FUNCT_1:2;
end;

theorem
  f3 (#) (f1 + f2)=f3(#)f1 + f3(#)f2 by Th10;

theorem Th12:
  r(#)(f1(#)f2)=r(#)f1(#)f2
proof
A1: dom (r(#)(f1 (#) f2)) = dom (f1 (#) f2) by VALUED_1:def 5
    .= dom f1 /\ dom f2 by VALUED_1:def 4
    .= dom (r(#)f1) /\ dom f2 by VALUED_1:def 5
    .= dom (r(#)f1(#)f2) by VALUED_1:def 4;
  now
    let c be set;
    assume
A2: c in dom (r(#)(f1(#)f2));
    then c in dom (r(#)f1) /\ dom f2 by A1,VALUED_1:def 4;
    then
A3: c in dom (r(#)f1) by XBOOLE_0:def 4;
    thus (r(#)(f1(#)f2)).c = r * (f1(#)f2).c by A2,VALUED_1:def 5
      .= r*(f1.c * f2.c) by VALUED_1:5
      .= (r*f1.c) * f2.c
      .= (r(#)f1).c * f2.c by A3,VALUED_1:def 5
      .= (r(#)f1 (#) f2).c by VALUED_1:5;
  end;
  hence thesis by A1,FUNCT_1:2;
end;

theorem Th13:
  r(#)(f1(#)f2)=f1(#)(r(#)f2)
proof
A1: dom (r(#)(f1 (#) f2)) = dom (f1 (#) f2) by VALUED_1:def 5
    .= dom f1 /\ dom f2 by VALUED_1:def 4
    .= dom f1 /\ dom (r(#)f2) by VALUED_1:def 5
    .= dom (f1(#)(r(#)f2)) by VALUED_1:def 4;
  now
    let c be set;
    assume
A2: c in dom (r(#)(f1(#)f2));
    then c in dom f1 /\ dom (r(#)f2) by A1,VALUED_1:def 4;
    then
A3: c in dom (r(#)f2) by XBOOLE_0:def 4;
    thus (r(#)(f1(#)f2)).c = r * (f1(#)f2).c by A2,VALUED_1:def 5
      .= r * (f1.c * f2.c) by VALUED_1:5
      .= f1.c * (r * f2.c)
      .= f1.c * (r(#)f2).c by A3,VALUED_1:def 5
      .= (f1(#)(r(#)f2)).c by VALUED_1:5;
  end;
  hence thesis by A1,FUNCT_1:2;
end;

theorem Th14:
  (f1 - f2)(#)f3=f1(#)f3 - f2(#)f3
proof
A1: dom ((f1 - f2) (#) f3) = dom (f1 - f2) /\ dom f3 by VALUED_1:def 4
    .= dom f1 /\ dom f2 /\ (dom f3 /\ dom f3) by VALUED_1:12
    .= dom f1 /\ dom f2 /\ dom f3 /\ dom f3 by XBOOLE_1:16
    .= dom f1 /\ dom f3 /\ dom f2 /\ dom f3 by XBOOLE_1:16
    .= dom f1 /\ dom f3 /\ (dom f2 /\ dom f3) by XBOOLE_1:16
    .= dom (f1 (#) f3) /\ (dom f2 /\ dom f3) by VALUED_1:def 4
    .= dom (f1 (#) f3) /\ dom (f2 (#) f3) by VALUED_1:def 4
    .= dom (f1 (#) f3 - f2 (#) f3) by VALUED_1:12;
  now
    let c be set;
    assume
A2: c in dom ((f1 - f2)(#)f3);
    then c in dom (f1 - f2) /\ dom f3 by VALUED_1:def 4;
    then
A3: c in dom (f1 - f2) by XBOOLE_0:def 4;
    thus ((f1 - f2) (#) f3).c = (f1 - f2).c * f3.c by VALUED_1:5
      .= (f1.c - f2.c) * f3.c by A3,VALUED_1:13
      .= f1.c * f3.c - f2.c * f3.c
      .= (f1 (#) f3).c - f2.c * f3.c by VALUED_1:5
      .= (f1 (#) f3).c - (f2 (#) f3).c by VALUED_1:5
      .=((f1 (#) f3) - (f2 (#) f3)).c by A1,A2,VALUED_1:13;
  end;
  hence thesis by A1,FUNCT_1:2;
end;

theorem
  f3(#)f1 - f3(#)f2 = f3(#)(f1 - f2) by Th14;

theorem
  r(#)(f1 + f2) = r(#)f1 + r(#)f2
proof
A1: dom (r(#)(f1 + f2)) = dom (f1 + f2) by VALUED_1:def 5
    .= dom f1 /\ dom f2 by VALUED_1:def 1
    .= dom f1 /\ dom (r(#)f2) by VALUED_1:def 5
    .= dom (r(#)f1) /\ dom (r(#)f2) by VALUED_1:def 5
    .= dom (r(#)f1 + r(#)f2) by VALUED_1:def 1;
  now
    let c be set;
    assume
A2: c in dom (r(#)(f1 + f2));
    then
A3: c in dom (f1 + f2) by VALUED_1:def 5;
A4: c in dom (r(#)f1) /\ dom (r(#)f2) by A1,A2,VALUED_1:def 1;
    then
A5: c in dom (r(#)f1) by XBOOLE_0:def 4;
A6: c in dom (r(#)f2) by A4,XBOOLE_0:def 4;
    thus (r(#)(f1 + f2)).c = r * (f1 + f2).c by A2,VALUED_1:def 5
      .= r * (f1.c + f2.c) by A3,VALUED_1:def 1
      .= r * f1.c + r * f2.c
      .= (r(#)f1).c + r * f2.c by A5,VALUED_1:def 5
      .= (r(#)f1).c + (r(#)f2).c by A6,VALUED_1:def 5
      .= (r(#)f1 + r(#)f2).c by A1,A2,VALUED_1:def 1;
  end;
  hence thesis by A1,FUNCT_1:2;
end;

theorem
  (r*p)(#)f = r(#)(p(#)f)
proof
A1: dom ((r*p) (#) f) = dom f by VALUED_1:def 5
    .= dom (p(#)f) by VALUED_1:def 5
    .= dom (r(#)(p(#)f)) by VALUED_1:def 5;
  now
    let c be set;
    assume
A2: c in dom ((r*p)(#)f);
    then
A3: c in dom (p(#)f) by A1,VALUED_1:def 5;
    thus ((r*p)(#)f).c = r*p * f.c by A2,VALUED_1:def 5
      .= r*(p * f.c)
      .= r * (p(#)f).c by A3,VALUED_1:def 5
      .= (r(#)(p(#)f)).c by A1,A2,VALUED_1:def 5;
  end;
  hence thesis by A1,FUNCT_1:2;
end;

theorem
  r(#)(f1 - f2) = r(#)f1 - r(#)f2
proof
A1: dom (r(#)(f1 - f2)) = dom (f1 - f2) by VALUED_1:def 5
    .= dom f1 /\ dom f2 by VALUED_1:12
    .= dom f1 /\ dom (r(#)f2) by VALUED_1:def 5
    .= dom (r(#)f1) /\ dom (r(#)f2) by VALUED_1:def 5
    .= dom (r(#)f1 - r(#)f2) by VALUED_1:12;
  now
    let c be set;
    assume
A2: c in dom (r(#)(f1 - f2));
    then
A3: c in dom (f1 - f2) by VALUED_1:def 5;
A4: c in dom (r(#)f1) /\ dom (r(#)f2) by A1,A2,VALUED_1:12;
    then
A5: c in dom (r(#)f1) by XBOOLE_0:def 4;
A6: c in dom (r(#)f2) by A4,XBOOLE_0:def 4;
    thus (r(#)(f1 - f2)).c = r * (f1 - f2).c by A2,VALUED_1:def 5
      .= r * (f1.c - f2.c) by A3,VALUED_1:13
      .= r * f1.c - r * f2.c
      .= (r(#)f1).c - r * f2.c by A5,VALUED_1:def 5
      .= (r(#)f1).c - (r(#)f2).c by A6,VALUED_1:def 5
      .= (r(#)f1 - r(#)f2).c by A1,A2,VALUED_1:13;
  end;
  hence thesis by A1,FUNCT_1:2;
end;

theorem
  f1-f2 = (-1)(#)(f2-f1)
proof
A1: dom (f1 - f2) = dom f2 /\ dom f1 by VALUED_1:12
    .= dom (f2 - f1) by VALUED_1:12
    .= dom ((-1)(#)(f2 - f1)) by VALUED_1:def 5;
  now
A2: dom (f1 - f2) = dom f2 /\ dom f1 by VALUED_1:12
      .= dom (f2 - f1) by VALUED_1:12;
    let c be set such that
A3: c in dom (f1-f2);
    thus (f1 - f2).c = f1.c - f2.c by A3,VALUED_1:13
      .= (-1)*(f2.c - f1.c)
      .= (-1)*((f2 - f1).c) by A3,A2,VALUED_1:13
      .= ((-1)(#)(f2 - f1)).c by A1,A3,VALUED_1:def 5;
  end;
  hence thesis by A1,FUNCT_1:2;
end;

theorem
  f1 - (f2 + f3) = f1 - f2 - f3
proof
A1: dom (f1 - (f2 + f3)) = dom f1 /\ dom (f2 + f3) by VALUED_1:12
    .= dom f1 /\ (dom f2 /\ dom f3) by VALUED_1:def 1
    .= dom f1 /\ dom f2 /\ dom f3 by XBOOLE_1:16
    .= dom (f1 - f2) /\ dom f3 by VALUED_1:12
    .= dom (f1 - f2 - f3) by VALUED_1:12;
  now
    let c be set;
    assume
A2: c in dom (f1 - (f2 + f3));
    then c in dom f1 /\ dom (f2 + f3) by VALUED_1:12;
    then
A3: c in dom (f2 + f3) by XBOOLE_0:def 4;
    c in dom (f1 - f2) /\ dom f3 by A1,A2,VALUED_1:12;
    then
A4: c in dom (f1 - f2) by XBOOLE_0:def 4;
    thus (f1 - (f2 + f3)).c = f1.c - (f2 + f3).c by A2,VALUED_1:13
      .= f1.c - (f2.c + f3.c) by A3,VALUED_1:def 1
      .= f1.c - f2.c - f3.c
      .= (f1 - f2).c - f3.c by A4,VALUED_1:13
      .= (f1 - f2 - f3).c by A1,A2,VALUED_1:13;
  end;
  hence thesis by A1,FUNCT_1:2;
end;

theorem
  1(#)f = f
proof
A1: now
    let c be set;
    assume c in dom (1(#)f);
    hence (1(#)f).c = 1 * f.c by VALUED_1:def 5
      .= f.c;
  end;
  dom (1(#)f) = dom f by VALUED_1:def 5;
  hence thesis by A1,FUNCT_1:2;
end;

theorem
  f1 - (f2 - f3) = f1 - f2 + f3
proof
A1: dom (f1 - (f2 - f3)) = dom f1 /\ dom (f2 - f3) by VALUED_1:12
    .= dom f1 /\ (dom f2 /\ dom f3) by VALUED_1:12
    .= dom f1 /\ dom f2 /\ dom f3 by XBOOLE_1:16
    .= dom (f1 - f2) /\ dom f3 by VALUED_1:12
    .= dom (f1 - f2 + f3) by VALUED_1:def 1;
  now
    let c be set;
    assume
A2: c in dom (f1 - (f2 - f3));
    then c in dom f1 /\ dom (f2 - f3) by VALUED_1:12;
    then
A3: c in dom (f2 - f3) by XBOOLE_0:def 4;
    c in dom (f1 - f2) /\ dom f3 by A1,A2,VALUED_1:def 1;
    then
A4: c in dom (f1 - f2) by XBOOLE_0:def 4;
    thus (f1 - (f2 - f3)).c = f1.c - (f2 - f3).c by A2,VALUED_1:13
      .= f1.c - (f2.c - f3.c) by A3,VALUED_1:13
      .= f1.c - f2.c + f3.c
      .= (f1 - f2).c + f3.c by A4,VALUED_1:13
      .= (f1 - f2 + f3).c by A1,A2,VALUED_1:def 1;
  end;
  hence thesis by A1,FUNCT_1:2;
end;

theorem
  f1 + (f2 - f3) = f1 + f2 - f3
proof
A1: dom (f1 + (f2 - f3)) = dom f1 /\ dom (f2 - f3) by VALUED_1:def 1
    .= dom f1 /\ (dom f2 /\ dom f3) by VALUED_1:12
    .= dom f1 /\ dom f2 /\ dom f3 by XBOOLE_1:16
    .= dom (f1 + f2) /\ dom f3 by VALUED_1:def 1
    .= dom (f1 + f2 - f3) by VALUED_1:12;
  now
    let c be set;
    assume
A2: c in dom (f1 + (f2 - f3));
    then c in dom f1 /\ dom (f2 - f3) by VALUED_1:def 1;
    then
A3: c in dom (f2 - f3) by XBOOLE_0:def 4;
    c in dom (f1 + f2) /\ dom f3 by A1,A2,VALUED_1:12;
    then
A4: c in dom (f1 + f2) by XBOOLE_0:def 4;
    thus (f1 + (f2 - f3)).c = f1.c + (f2 - f3).c by A2,VALUED_1:def 1
      .= f1.c + (f2.c - f3.c) by A3,VALUED_1:13
      .= f1.c + f2.c - f3.c
      .= (f1 + f2).c - f3.c by A4,VALUED_1:def 1
      .= (f1 + f2 - f3).c by A1,A2,VALUED_1:13;
  end;
  hence thesis by A1,FUNCT_1:2;
end;

theorem Th24:
  abs(f1(#)f2) = abs(f1)(#)abs(f2)
proof
A1: now
    let c be set;
    assume c in dom (abs(f1 (#) f2));
    thus (abs(f1(#)f2)).c = abs((f1(#)f2).c) by VALUED_1:18
      .= abs(f1.c * f2.c) by VALUED_1:5
      .= abs(f1.c) * abs(f2.c) by COMPLEX1:65
      .= ((abs(f1)).c) * abs(f2.c) by VALUED_1:18
      .= ((abs(f1)).c) * (abs(f2)).c by VALUED_1:18
      .= (abs(f1)(#)abs(f2)).c by VALUED_1:5;
  end;
  dom (abs(f1 (#) f2)) = dom (f1 (#) f2) by VALUED_1:def 11
    .= dom f1 /\ dom f2 by VALUED_1:def 4
    .= dom f1 /\ dom (abs(f2)) by VALUED_1:def 11
    .= dom (abs(f1)) /\ dom (abs(f2)) by VALUED_1:def 11
    .= dom (abs(f1)(#)abs(f2)) by VALUED_1:def 4;
  hence thesis by A1,FUNCT_1:2;
end;

theorem
  abs(r(#)f) = abs(r)(#)abs(f)
proof
A1: dom (abs(r(#)f)) = dom (r(#)f) by VALUED_1:def 11
    .= dom f by VALUED_1:def 5
    .= dom (abs(f)) by VALUED_1:def 11
    .= dom (abs(r)(#)abs(f)) by VALUED_1:def 5;
  now
    let c be set;
    assume
A2: c in dom (abs(r(#)f));
    then
A3: c in dom (r(#)f) by VALUED_1:def 11;
    thus (abs(r(#)f)).c = abs((r(#)f).c) by VALUED_1:18
      .=abs(r*(f.c)) by A3,VALUED_1:def 5
      .=abs(r)*abs(f.c) by COMPLEX1:65
      .=abs(r)*(abs(f)).c by VALUED_1:18
      .=(abs(r)(#)abs(f)).c by A1,A2,VALUED_1:def 5;
  end;
  hence thesis by A1,FUNCT_1:2;
end;

theorem Th26:
  f^^ = f|(dom (f^))
proof
A1: dom (f^^) = dom (f|(dom (f^))) by Th6;
  now
    let c be set;
    assume
A2: c in dom (f^^);
    then c in dom f /\ dom (f^) by A1,RELAT_1:61;
    then
A3: c in dom (f^) by XBOOLE_0:def 4;
    thus (f^^).c = ((f^).c)" by A2,Def2
      .= ((f.c)")" by A3,Def2
      .= (f|(dom (f^))).c by A1,A2,FUNCT_1:47;
  end;
  hence thesis by A1,FUNCT_1:2;
end;

theorem Th27:
  (f1(#)f2)^ = (f1^)(#)(f2^)
proof
A1: dom ((f1(#)f2)^) = dom (f1(#)f2) \ (f1(#)f2)"{0} by Def2
    .= (dom f1 \ f1"{0}) /\ (dom f2 \ (f2)"{0}) by Th2
    .= dom (f1^) /\ (dom f2 \ (f2)"{0}) by Def2
    .= dom (f1^) /\ dom (f2^) by Def2
    .= dom ((f1^) (#) (f2^)) by VALUED_1:def 4;
  now
    let c be set;
    assume
A2: c in dom ((f1(#)f2)^);
    then
A3: c in dom (f1^) /\ dom (f2^) by A1,VALUED_1:def 4;
    then
A4: c in dom (f1^) by XBOOLE_0:def 4;
A5: c in dom (f2^) by A3,XBOOLE_0:def 4;
    thus ((f1(#)f2)^).c = ((f1(#)f2).c)" by A2,Def2
      .= (f1.c * f2.c)" by VALUED_1:5
      .= (f1.c)"* (f2.c)" by XCMPLX_1:204
      .= ((f1^).c)*(f2.c)" by A4,Def2
      .= ((f1^).c) *((f2^).c) by A5,Def2
      .= ((f1^) (#) (f2^)).c by VALUED_1:5;
  end;
  hence thesis by A1,FUNCT_1:2;
end;

theorem Th28:
  r<>0 implies (r(#)f)^ = r" (#) (f^)
proof
  assume
A1: r<>0;
A2: dom ((r(#)f)^) = dom (r(#)f) \ (r(#)f)"{0} by Def2
    .= dom (r(#)f) \ f"{0} by A1,Th7
    .= dom f \ f"{0} by VALUED_1:def 5
    .= dom (f^) by Def2
    .= dom (r"(#)(f^)) by VALUED_1:def 5;
  now
    let c be set;
    assume
A3: c in dom ((r(#)f)^);
    then
A4: c in dom (r(#)f) \ (r(#)f)"{0} by Def2;
A5: c in dom (f^) by A2,A3,VALUED_1:def 5;
    thus ((r(#)f)^).c = ((r(#)f).c)" by A3,Def2
      .= (r*(f.c))" by A4,VALUED_1:def 5
      .= r"* (f.c)" by XCMPLX_1:204
      .= r"* ((f^).c) by A5,Def2
      .= (r" (#) (f^)).c by A2,A3,VALUED_1:def 5;
  end;
  hence thesis by A2,FUNCT_1:2;
end;

theorem
  (-f)^ = (-1)(#)(f^) by Lm1,Th28;

theorem Th30:
  (abs(f))^ = abs(f^)
proof
A1: dom ((abs(f))^) = dom (abs(f)) \ (abs(f))"{0} by Def2
    .= dom f \ (abs(f))"{0} by VALUED_1:def 11
    .= dom f \ f"{0} by Th5
    .= dom (f^) by Def2
    .= dom (abs((f^))) by VALUED_1:def 11;
  now
    let c be set;
    assume
A2: c in dom ((abs(f))^);
    then
A3: c in dom (f^) by A1,VALUED_1:def 11;
    thus ((abs(f))^).c = ((abs(f)).c)" by A2,Def2
      .= (abs(f.c))" by VALUED_1:18
      .= 1/abs(f.c) by XCMPLX_1:215
      .= abs(1/f.c) by COMPLEX1:80
      .= abs((f.c)") by XCMPLX_1:215
      .= abs(f^.c) by A3,Def2
      .= (abs(f^)).c by VALUED_1:18;
  end;
  hence thesis by A1,FUNCT_1:2;
end;

theorem Th31:
  f/g = f(#) (g^)
proof
A1: now
    let c be set;
    assume
A2: c in dom (f/g);
    then c in dom f /\ (dom g \ g"{0}) by Def1;
    then c in dom f /\ dom (g^) by Def2;
    then
A3: c in dom (g^) by XBOOLE_0:def 4;
    thus (f/g).c = f.c * (g.c)" by A2,Def1
      .= f.c * (g^).c by A3,Def2
      .= (f (#) (g^)).c by VALUED_1:5;
  end;
  dom (f/g) = dom f /\ (dom g \ g"{0}) by Def1
    .= dom f /\ dom (g^) by Def2
    .= dom (f(#)(g^)) by VALUED_1:def 4;
  hence thesis by A1,FUNCT_1:2;
end;

theorem Th32:
  r(#)(g/f) = (r(#)g)/f
proof
  thus r(#)(g/f) = r(#)(g(#)(f^)) by Th31
    .= (r(#)g)(#)(f^) by Th12
    .= (r(#)g)/f by Th31;
end;

theorem
  (f/g)(#)g = (f|dom(g^))
proof
A1: dom ((f/g)(#)g) = dom (f/g) /\ dom g by VALUED_1:def 4
    .= dom f /\ (dom g \ g"{0}) /\ dom g by Def1
    .= dom f /\ ((dom g \ g"{0}) /\ dom g) by XBOOLE_1:16
    .= dom f /\ (dom (g^) /\ dom g) by Def2
    .= dom f /\ dom (g^) by Th1,XBOOLE_1:28
    .= dom (f|(dom (g^))) by RELAT_1:61;
  now
    let c be set;
    assume
A2: c in dom ((f/g)(#)g);
    then c in dom f /\ dom (g^) by A1,RELAT_1:61;
    then
A3: c in dom (g^) by XBOOLE_0:def 4;
    then
A4: g.c <> 0 by Th3;
    thus ((f/g)(#)g).c = ((f/g).c) * g.c by VALUED_1:5
      .= (f(#)(g^)).c * g.c by Th31
      .= (f.c) *((g^).c) * g.c by VALUED_1:5
      .= (f.c)*(g.c)"*g.c by A3,Def2
      .= (f.c)*((g.c)" * (g.c))
      .= (f.c)*1 by A4,XCMPLX_0:def 7
      .= (f|(dom (g^))).c by A1,A2,FUNCT_1:47;
  end;
  hence thesis by A1,FUNCT_1:2;
end;

theorem Th34:
  (f/g)(#)(f1/g1) = (f(#)f1)/(g(#)g1)
proof
A1: now
    let c be set;
    assume c in dom ((f/g)(#)(f1/g1));
    thus ((f/g)(#)(f1/g1)).c = ((f/g).c)* (f1/g1).c by VALUED_1:5
      .= ((f(#)(g^)).c) * (f1/g1).c by Th31
      .= ((f(#)(g^)).c) * (f1(#)(g1^)).c by Th31
      .= (f.c) * ((g^).c) * (f1(#)(g1^)).c by VALUED_1:5
      .= (f.c) * ((g^).c) * ((f1.c)* (g1^).c) by VALUED_1:5
      .= (f.c) * ((f1.c) * (((g^).c) * (g1^).c))
      .= (f.c) * ((f1.c) * ((g^)(#)(g1^)).c) by VALUED_1:5
      .= (f.c) * (f1.c) * ((g^)(#)(g1^)).c
      .= (f.c) * (f1.c) * ((g(#)g1)^).c by Th27
      .= ((f(#)f1).c) * ((g(#)g1)^).c by VALUED_1:5
      .= ((f(#)f1)(#)((g(#)g1)^)).c by VALUED_1:5
      .= ((f(#)f1)/(g(#)g1)).c by Th31;
  end;
  dom ((f/g)(#)(f1/g1)) = dom (f/g) /\ dom (f1/g1) by VALUED_1:def 4
    .= dom f /\ (dom g \ g"{0}) /\ dom (f1/g1) by Def1
    .= dom f /\ (dom g \ g"{0}) /\ (dom f1 /\ (dom g1 \ g1"{0})) by Def1
    .= dom f /\ ((dom g \ g"{0}) /\ (dom f1 /\ (dom g1 \ g1"{0}))) by
XBOOLE_1:16
    .= dom f /\ (dom f1 /\ ((dom g \ g"{0}) /\ (dom g1 \ g1"{0}))) by
XBOOLE_1:16
    .= dom f /\ dom f1 /\ ((dom g \ g"{0}) /\ (dom g1 \ g1"{0})) by XBOOLE_1:16
    .= dom (f(#)f1) /\ ((dom g \ g"{0}) /\ (dom g1 \ g1"{0})) by VALUED_1:def 4
    .= dom (f(#)f1) /\ (dom (g(#)g1) \ (g(#)g1)"{0}) by Th2
    .= dom ((f(#)f1)/(g(#)g1)) by Def1;
  hence thesis by A1,FUNCT_1:2;
end;

theorem Th35:
  (f1/f2)^ = (f2|dom(f2^))/f1
proof
  thus (f1/f2)^ = (f1(#)(f2^))^ by Th31
    .= (f1^)(#)(f2^^) by Th27
    .= (f2|dom(f2^))(#)(f1^) by Th26
    .= (f2|dom(f2^))/f1 by Th31;
end;

theorem Th36:
  g (#) (f1/f2) = (g (#) f1)/f2
proof
  thus g (#) (f1/f2) = g (#) (f1 (#) (f2^)) by Th31
    .= g (#) f1 (#) (f2^) by Th9
    .= (g (#) f1)/f2 by Th31;
end;

theorem
  g/(f1/f2) = (g(#)(f2|dom(f2^)))/f1
proof
  thus g/(f1/f2) = g (#) ((f1/f2)^) by Th31
    .= g (#) ((f2|dom(f2^))/f1) by Th35
    .= (g (#) (f2|dom(f2^)))/f1 by Th36;
end;

theorem
  -f/g = (-f)/g & f/(-g) = -f/g
proof
  thus -f/g = (-f)/g by Th32;
  thus f/(-g) = f (#) ((-g)^) by Th31
    .= f (#) ((-1) (#) (g^)) by Lm1,Th28
    .= -(f (#) (g^)) by Th13
    .= -(f/g) by Th31;
end;

theorem
  f1/f + f2/f = (f1 + f2)/f & f1/f - f2/f = (f1 - f2)/f
proof
  thus f1/f + f2/f = f1(#)(f^) +f2/f by Th31
    .= f1(#)(f^) + f2(#)(f^) by Th31
    .= (f1+f2) (#) (f^) by Th10
    .= (f1+f2)/f by Th31;
  thus f1/f - f2/f = f1(#)(f^) - f2/f by Th31
    .= f1(#)(f^) -f2(#)(f^) by Th31
    .= (f1-f2)(#)(f^) by Th14
    .= (f1-f2)/f by Th31;
end;

theorem Th40:
  f1/f + g1/g = (f1(#)g + g1(#)f)/(f(#)g)
proof
A1: now
    let c be set;
A2: dom (g^) c= dom g by Th1;
    assume
A3: c in dom ((f1/f) + (g1/g));
    then
A4: c in dom (f1/f) /\ dom (g1/g) by VALUED_1:def 1;
    then
A5: c in dom (f1/f) by XBOOLE_0:def 4;
A6: c in dom (g1/g) by A4,XBOOLE_0:def 4;
A7: dom (f^) c= dom f by Th1;
A8: c in dom (f1 (#)(f^)) /\ dom (g1/g) by A4,Th31;
    then c in dom (f1 (#)(f^)) by XBOOLE_0:def 4;
    then
A9: c in dom f1 /\ dom(f^) by VALUED_1:def 4;
    then
A10: c in dom(f^) by XBOOLE_0:def 4;
    then
A11: f.c <> 0 by Th3;
    c in dom (f1 (#)(f^)) /\ dom (g1(#)(g^)) by A8,Th31;
    then c in dom (g1(#)(g^)) by XBOOLE_0:def 4;
    then
A12: c in dom g1 /\ dom(g^) by VALUED_1:def 4;
    then
A13: c in dom(g^) by XBOOLE_0:def 4;
    then
A14: g.c <> 0 by Th3;
    c in dom g1 by A12,XBOOLE_0:def 4;
    then c in dom g1 /\ dom f by A10,A7,XBOOLE_0:def 4;
    then
A15: c in dom (g1(#)f) by VALUED_1:def 4;
    c in dom f1 by A9,XBOOLE_0:def 4;
    then c in dom f1 /\ dom g by A13,A2,XBOOLE_0:def 4;
    then c in dom (f1(#)g) by VALUED_1:def 4;
    then c in dom (f1(#)g) /\ dom (g1(#)f) by A15,XBOOLE_0:def 4;
    then
A16: c in dom (f1(#)g + g1(#)f) by VALUED_1:def 1;
    c in dom (f^) /\ dom (g^) by A10,A13,XBOOLE_0:def 4;
    then c in dom ((f^)(#)(g^)) by VALUED_1:def 4;
    then c in dom ((f(#)g)^) by Th27;
    then c in dom (f1(#)g + g1(#)f) /\ dom ((f(#)g)^) by A16,XBOOLE_0:def 4;
    then c in dom ((f1(#)g + g1(#)f)(#)((f(#)g)^)) by VALUED_1:def 4;
    then
A17: c in dom ((f1(#)g + g1(#)f)/(f(#)g)) by Th31;
    thus (f1/f + g1/g).c = (f1/f).c + (g1/g).c by A3,VALUED_1:def 1
      .= (f1.c) * (f.c)" + (g1/g).c by A5,Def1
      .= (f1.c) *(1*(f.c)") + (g1.c) * 1 * (g.c)" by A6,Def1
      .= (f1.c) *((g.c) *(g.c)"* (f.c)") + (g1.c) * (1 * (g.c)") by A14,
XCMPLX_0:def 7
      .= (f1.c) *(g.c *((g.c)"* (f.c)")) + (g1.c) * ((f.c) *(f.c)" * (g.c)")
    by A11,XCMPLX_0:def 7
      .= (f1.c) *((g.c) *((g.c * f.c)")) + (g1.c) * ((f.c) *((f.c)" * (g.c)"
    )) by XCMPLX_1:204
      .= (f1.c) *((g.c) *((f.c* g.c)")) + (g1.c) * ((f.c) *((f.c * g.c)"))
    by XCMPLX_1:204
      .= (f1.c) *((g.c) * ((f(#)g).c)") + (g1.c) * ((f.c) *((f.c * g.c)"))
    by VALUED_1:5
      .= (f1.c) *(g.c) * ((f(#)g).c)" + (g1.c) * ((f.c) * ((f(#) g).c)") by
VALUED_1:5
      .= (f1(#)g).c * ((f(#)g).c)" + (g1.c) *f.c *((f(#)g).c)" by VALUED_1:5
      .= (f1(#)g).c * ((f(#)g).c)" + (g1(#)f).c *((f(#)g).c)" by VALUED_1:5
      .= ((f1(#)g).c + (g1(#)f).c) *((f(#)g).c)"
      .= (f1(#)g + g1(#)f).c *((f(#)g).c)" by A16,VALUED_1:def 1
      .= ((f1(#)g + g1(#)f)/(f(#)g)).c by A17,Def1;
  end;
  dom ((f1/f) + (g1/g)) = dom (f1/f) /\ dom (g1/g) by VALUED_1:def 1
    .= dom f1 /\ (dom f \ f"{0}) /\ dom (g1/g) by Def1
    .= dom f1 /\ (dom f \ f"{0}) /\ (dom g1 /\ (dom g \ g"{0})) by Def1
    .= dom f1 /\ (dom f /\ (dom f \ f"{0})) /\ (dom g1 /\ (dom g \ g"{0}))
  by Th1
    .= dom f /\ (dom f \ f"{0}) /\ dom f1 /\ (dom g /\ (dom g \ g"{0}) /\
  dom g1) by Th1
    .= dom f /\ (dom f \ f"{0}) /\ (dom f1 /\ (dom g /\ (dom g \ g"{0}) /\
  dom g1)) by XBOOLE_1:16
    .= dom f /\ (dom f \ f"{0}) /\ (dom f1 /\ (dom g /\ (dom g \ g"{0})) /\
  dom g1) by XBOOLE_1:16
    .= dom f /\ (dom f \ f"{0}) /\ (dom f1 /\ dom g /\ (dom g \ g"{0}) /\
  dom g1) by XBOOLE_1:16
    .= dom f /\ (dom f \ f"{0}) /\ (dom (f1(#)g) /\ (dom g \ g"{0}) /\ dom
  g1) by VALUED_1:def 4
    .= dom f /\ (dom f \ f"{0}) /\ (dom (f1(#)g) /\ (dom g1 /\ (dom g \ g"{0
  }))) by XBOOLE_1:16
    .= dom (f1(#)g) /\ ((dom f \ f"{0}) /\ dom f /\ (dom g1 /\ (dom g \ g"{0
  }))) by XBOOLE_1:16
    .= dom (f1(#)g) /\ ((dom f \ f"{0}) /\ (dom f /\ (dom g1 /\ (dom g \ g"{
  0})))) by XBOOLE_1:16
    .= dom (f1(#)g) /\ ((dom f \ f"{0}) /\ (dom g1 /\ dom f /\ (dom g \ g"{0
  }))) by XBOOLE_1:16
    .= dom (f1(#)g) /\ ((dom f \ f"{0}) /\ (dom (g1(#)f) /\ (dom g \ g"{0}))
  ) by VALUED_1:def 4
    .= dom (f1(#)g) /\ (dom (g1(#)f) /\ ((dom f \ f"{0}) /\ (dom g \ g"{0}))
  ) by XBOOLE_1:16
    .= dom (f1(#)g) /\ dom (g1(#) f) /\ ((dom f \ f"{0}) /\ (dom g \ g"{0}))
  by XBOOLE_1:16
    .= dom (f1(#)g + g1(#)f) /\ ((dom f \ f"{0}) /\ (dom g \ g"{0})) by
VALUED_1:def 1
    .= dom (f1(#)g + g1(#)f) /\ (dom (f(#)g) \ (f(#)g)"{0}) by Th2
    .= dom ((f1(#)g + g1(#)f)/(f(#)g)) by Def1;
  hence thesis by A1,FUNCT_1:2;
end;

theorem
  (f/g)/(f1/g1) = (f(#)(g1|dom(g1^)))/(g(#)f1)
proof
  thus (f/g)/(f1/g1) = (f/g)(#)((f1/g1)^) by Th31
    .= (f/g)(#)(((g1|dom(g1^)))/f1) by Th35
    .= (f(#)(g1|dom(g1^)))/(g(#)f1) by Th34;
end;

theorem
  f1/f - g1/g = (f1(#)g - g1(#)f)/(f(#)g)
proof
  thus f1/f - g1/g = f1/f + ((-1)(#) g1)/g by Th32
    .= (f1(#)g + (-1)(#) g1(#)f)/(f(#)g) by Th40
    .= (f1(#)g - (g1(#)f))/(f(#)g) by Th12;
end;

theorem
  abs(f1/f2) = abs(f1)/abs(f2)
proof
  thus abs(f1/f2) = abs(f1(#)(f2^)) by Th31
    .= abs(f1)(#)abs((f2^)) by Th24
    .= abs(f1)(#)((abs(f2))^) by Th30
    .= abs(f1)/abs(f2) by Th31;
end;

theorem Th44:
  (f1+f2)|X = f1|X + f2|X & (f1+f2)|X = f1|X + f2 & (f1+f2)|X = f1 + f2|X
proof
A1: now
    let c be set;
    assume
A2: c in dom ((f1+f2)|X);
    then
A3: c in dom (f1+f2) /\ X by RELAT_1:61;
    then
A4: c in X by XBOOLE_0:def 4;
A5: c in dom (f1+f2) by A3,XBOOLE_0:def 4;
    then
A6: c in dom f1 /\ dom f2 by VALUED_1:def 1;
    then c in dom f2 by XBOOLE_0:def 4;
    then c in dom f2 /\ X by A4,XBOOLE_0:def 4;
    then
A7: c in dom (f2|X) by RELAT_1:61;
    c in dom f1 by A6,XBOOLE_0:def 4;
    then c in dom f1 /\ X by A4,XBOOLE_0:def 4;
    then
A8: c in dom (f1|X) by RELAT_1:61;
    then c in dom (f1|X) /\ dom (f2|X) by A7,XBOOLE_0:def 4;
    then
A9: c in dom ((f1|X) + (f2|X)) by VALUED_1:def 1;
    thus ((f1+f2)|X).c = (f1+f2).c by A2,FUNCT_1:47
      .= (f1.c) + (f2.c) by A5,VALUED_1:def 1
      .= ((f1|X).c) + (f2.c) by A8,FUNCT_1:47
      .= ((f1|X).c) + ((f2|X).c) by A7,FUNCT_1:47
      .= ((f1|X)+(f2|X)).c by A9,VALUED_1:def 1;
  end;
  dom ((f1+f2)|X) = dom (f1+f2) /\ X by RELAT_1:61
    .= dom f1 /\ dom f2 /\ (X /\ X) by VALUED_1:def 1
    .= dom f1 /\ (dom f2 /\ (X /\ X)) by XBOOLE_1:16
    .= dom f1 /\ (dom f2 /\ X /\ X) by XBOOLE_1:16
    .= dom f1 /\ (X /\ dom (f2|X)) by RELAT_1:61
    .= dom f1 /\ X /\ dom (f2|X) by XBOOLE_1:16
    .= dom (f1|X) /\ dom (f2|X) by RELAT_1:61
    .= dom ((f1|X)+(f2|X)) by VALUED_1:def 1;
  hence (f1+f2)|X = f1|X + f2|X by A1,FUNCT_1:2;
A10: now
    let c be set;
    assume
A11: c in dom ((f1+f2)|X);
    then
A12: c in dom (f1+f2) /\ X by RELAT_1:61;
    then
A13: c in X by XBOOLE_0:def 4;
A14: c in dom (f1+f2) by A12,XBOOLE_0:def 4;
    then
A15: c in dom f1 /\ dom f2 by VALUED_1:def 1;
    then c in dom f1 by XBOOLE_0:def 4;
    then c in dom f1 /\ X by A13,XBOOLE_0:def 4;
    then
A16: c in dom (f1|X) by RELAT_1:61;
    c in dom f2 by A15,XBOOLE_0:def 4;
    then c in dom (f1|X) /\ dom f2 by A16,XBOOLE_0:def 4;
    then
A17: c in dom ((f1|X) + f2) by VALUED_1:def 1;
    thus ((f1+f2)|X).c = (f1+f2).c by A11,FUNCT_1:47
      .= (f1.c) +(f2.c) by A14,VALUED_1:def 1
      .= ((f1|X).c) +(f2.c) by A16,FUNCT_1:47
      .= ((f1|X)+f2).c by A17,VALUED_1:def 1;
  end;
  dom ((f1+f2)|X) = dom (f1+f2) /\ X by RELAT_1:61
    .= dom f1 /\ dom f2 /\ X by VALUED_1:def 1
    .= dom f1 /\ X /\ dom f2 by XBOOLE_1:16
    .= dom (f1|X) /\ dom f2 by RELAT_1:61
    .= dom ((f1|X)+ f2) by VALUED_1:def 1;
  hence (f1+f2)|X = f1|X + f2 by A10,FUNCT_1:2;
A18: now
    let c be set;
    assume
A19: c in dom ((f1+f2)|X);
    then
A20: c in dom (f1+f2) /\ X by RELAT_1:61;
    then
A21: c in X by XBOOLE_0:def 4;
A22: c in dom (f1+f2) by A20,XBOOLE_0:def 4;
    then
A23: c in dom f1 /\ dom f2 by VALUED_1:def 1;
    then c in dom f2 by XBOOLE_0:def 4;
    then c in dom f2 /\ X by A21,XBOOLE_0:def 4;
    then
A24: c in dom (f2|X) by RELAT_1:61;
    c in dom f1 by A23,XBOOLE_0:def 4;
    then c in dom f1 /\ dom (f2|X) by A24,XBOOLE_0:def 4;
    then
A25: c in dom (f1 + (f2|X)) by VALUED_1:def 1;
    thus ((f1+f2)|X).c = (f1+f2).c by A19,FUNCT_1:47
      .= (f1.c) +(f2.c) by A22,VALUED_1:def 1
      .= (f1.c) +((f2|X).c) by A24,FUNCT_1:47
      .= (f1+(f2|X)).c by A25,VALUED_1:def 1;
  end;
  dom ((f1+f2)|X) = dom (f1+f2) /\ X by RELAT_1:61
    .= dom f1 /\ dom f2 /\ X by VALUED_1:def 1
    .= dom f1 /\ (dom f2 /\ X) by XBOOLE_1:16
    .= dom f1 /\ dom (f2|X) by RELAT_1:61
    .= dom (f1 + (f2|X)) by VALUED_1:def 1;
  hence thesis by A18,FUNCT_1:2;
end;

theorem Th45:
  (f1(#)f2)|X = f1|X (#) f2|X & (f1(#)f2)|X = f1|X (#) f2 & (f1(#)
  f2)|X = f1 (#) f2|X
proof
A1: now
    let c be set;
    assume
A2: c in dom ((f1(#)f2)|X);
    then
A3: c in dom (f1(#)f2) /\ X by RELAT_1:61;
    then
A4: c in X by XBOOLE_0:def 4;
    c in dom (f1(#)f2) by A3,XBOOLE_0:def 4;
    then
A5: c in dom f1 /\ dom f2 by VALUED_1:def 4;
    then c in dom f1 by XBOOLE_0:def 4;
    then c in dom f1 /\ X by A4,XBOOLE_0:def 4;
    then
A6: c in dom (f1|X) by RELAT_1:61;
    c in dom f2 by A5,XBOOLE_0:def 4;
    then c in dom f2 /\ X by A4,XBOOLE_0:def 4;
    then
A7: c in dom (f2|X) by RELAT_1:61;
    thus ((f1(#)f2)|X).c = (f1(#)f2).c by A2,FUNCT_1:47
      .= (f1.c) *(f2.c) by VALUED_1:5
      .= ((f1|X).c) *(f2.c) by A6,FUNCT_1:47
      .= ((f1|X).c) *((f2|X).c) by A7,FUNCT_1:47
      .= ((f1|X)(#)(f2|X)).c by VALUED_1:5;
  end;
  dom ((f1(#)f2)|X) = dom (f1(#)f2) /\ X by RELAT_1:61
    .= dom f1 /\ dom f2 /\ (X /\ X) by VALUED_1:def 4
    .= dom f1 /\ (dom f2 /\ (X /\ X)) by XBOOLE_1:16
    .= dom f1 /\ (dom f2 /\ X /\ X) by XBOOLE_1:16
    .= dom f1 /\ (X /\ dom (f2|X)) by RELAT_1:61
    .= dom f1 /\ X /\ dom (f2|X) by XBOOLE_1:16
    .= dom (f1|X) /\ dom (f2|X) by RELAT_1:61
    .= dom ((f1|X)(#)(f2|X)) by VALUED_1:def 4;
  hence (f1(#)f2)|X = f1|X (#) f2|X by A1,FUNCT_1:2;
A8: now
    let c be set;
    assume
A9: c in dom ((f1(#)f2)|X);
    then
A10: c in dom (f1(#)f2) /\ X by RELAT_1:61;
    then c in dom (f1(#)f2) by XBOOLE_0:def 4;
    then c in dom f1 /\ dom f2 by VALUED_1:def 4;
    then
A11: c in dom f1 by XBOOLE_0:def 4;
    c in X by A10,XBOOLE_0:def 4;
    then c in dom f1 /\ X by A11,XBOOLE_0:def 4;
    then
A12: c in dom (f1|X) by RELAT_1:61;
    thus ((f1(#)f2)|X).c = (f1(#)f2).c by A9,FUNCT_1:47
      .= (f1.c) *(f2.c) by VALUED_1:5
      .= ((f1|X).c) *(f2.c) by A12,FUNCT_1:47
      .= ((f1|X)(#)f2).c by VALUED_1:5;
  end;
  dom ((f1(#)f2)|X) = dom (f1(#)f2) /\ X by RELAT_1:61
    .= dom f1 /\ dom f2 /\ X by VALUED_1:def 4
    .= dom f1 /\ X /\ dom f2 by XBOOLE_1:16
    .= dom (f1|X) /\ dom f2 by RELAT_1:61
    .= dom ((f1|X)(#) f2) by VALUED_1:def 4;
  hence (f1(#)f2)|X = f1|X (#) f2 by A8,FUNCT_1:2;
A13: now
    let c be set;
    assume
A14: c in dom ((f1(#)f2)|X);
    then
A15: c in dom (f1(#)f2) /\ X by RELAT_1:61;
    then c in dom (f1(#)f2) by XBOOLE_0:def 4;
    then c in dom f1 /\ dom f2 by VALUED_1:def 4;
    then
A16: c in dom f2 by XBOOLE_0:def 4;
    c in X by A15,XBOOLE_0:def 4;
    then c in dom f2 /\ X by A16,XBOOLE_0:def 4;
    then
A17: c in dom (f2|X) by RELAT_1:61;
    thus ((f1(#)f2)|X).c = (f1(#)f2).c by A14,FUNCT_1:47
      .= (f1.c) *(f2.c) by VALUED_1:5
      .= (f1.c) *((f2|X).c) by A17,FUNCT_1:47
      .= (f1(#)(f2|X)).c by VALUED_1:5;
  end;
  dom ((f1(#)f2)|X) = dom (f1(#)f2) /\ X by RELAT_1:61
    .= dom f1 /\ dom f2 /\ X by VALUED_1:def 4
    .= dom f1 /\ (dom f2 /\ X) by XBOOLE_1:16
    .= dom f1 /\ dom (f2|X) by RELAT_1:61
    .= dom (f1 (#) (f2|X)) by VALUED_1:def 4;
  hence thesis by A13,FUNCT_1:2;
end;

theorem Th46:
  (-f)|X = -(f|X) & (f^)|X = (f|X)^ & (abs(f))|X = abs((f|X))
proof
A1: now
    let c be set;
    assume
A2: c in dom ((-f)|X);
    then
A3: c in dom (-f) /\ X by RELAT_1:61;
    then c in dom (-f) by XBOOLE_0:def 4;
    then
A4: c in dom f by VALUED_1:8;
    c in X by A3,XBOOLE_0:def 4;
    then c in dom f /\ X by A4,XBOOLE_0:def 4;
    then
A5: c in dom (f|X) by RELAT_1:61;
    thus ((-f)|X).c = (-f).c by A2,FUNCT_1:47
      .= -(f.c) by VALUED_1:8
      .= -((f|X).c) by A5,FUNCT_1:47
      .= (-(f|X)).c by VALUED_1:8;
  end;
  dom ((-f)|X) = dom (-f) /\ X by RELAT_1:61
    .= dom f /\ X by VALUED_1:8
    .= dom (f|X) by RELAT_1:61
    .= dom (-(f|X)) by VALUED_1:8;
  hence (-f)|X = -(f|X) by A1,FUNCT_1:2;
A6: dom ((f^)|X) = dom (f^) /\ X by RELAT_1:61
    .= (dom f \ f"{0}) /\ X by Def2
    .= dom f /\ X \ f"{0} /\ X by XBOOLE_1:50
    .= dom (f|X) \ X /\ f"{0} by RELAT_1:61
    .= dom (f|X) \ (f|X)"{0} by FUNCT_1:70
    .= dom ((f|X)^) by Def2;
A7: dom ((f|X)^) c= dom (f|X) by Th1;
  now
    let c be set;
    assume
A8: c in dom ((f^)|X);
    then c in dom (f^) /\ X by RELAT_1:61;
    then
A9: c in dom (f^) by XBOOLE_0:def 4;
    thus ((f^)|X).c = (f^).c by A8,FUNCT_1:47
      .= (f.c)" by A9,Def2
      .= ((f|X).c)" by A7,A6,A8,FUNCT_1:47
      .= ((f|X)^).c by A6,A8,Def2;
  end;
  hence (f^)|X = (f|X)^ by A6,FUNCT_1:2;
A10: dom ((abs(f))|X) = dom (abs(f)) /\ X by RELAT_1:61
    .= dom f /\ X by VALUED_1:def 11
    .= dom (f|X) by RELAT_1:61
    .= dom (abs((f|X))) by VALUED_1:def 11;
  now
    let c be set;
    assume
A11: c in dom ((abs(f))|X);
    then
A12: c in dom (f|X) by A10,VALUED_1:def 11;
    thus ((abs(f))|X).c = (abs(f)).c by A11,FUNCT_1:47
      .= abs(f.c) by VALUED_1:18
      .= abs((f|X).c) by A12,FUNCT_1:47
      .= (abs((f|X))).c by VALUED_1:18;
  end;
  hence thesis by A10,FUNCT_1:2;
end;

theorem
  (f1-f2)|X = f1|X - f2|X & (f1-f2)|X = f1|X - f2 & (f1-f2)|X = f1 - f2| X
proof
  thus (f1-f2)|X = (f1|X)+ (-f2)|X by Th44
    .= (f1|X) - (f2|X) by Th46;
  thus (f1-f2)|X = (f1|X) - f2 by Th44;
  thus (f1-f2)|X = f1+ (-f2)|X by Th44
    .= f1 - (f2|X) by Th46;
end;

theorem
  (f1/f2)|X = f1|X / f2|X & (f1/f2)|X = f1|X / f2 &(f1/f2)|X = f1 / f2|X
proof
  thus (f1/f2)|X = (f1(#)(f2^))|X by Th31
    .= (f1|X) (#) (f2^|X) by Th45
    .= (f1|X) (#) ((f2|X)^) by Th46
    .= (f1|X)/(f2|X) by Th31;
  thus (f1/f2)|X = (f1(#)(f2^))|X by Th31
    .= (f1|X) (#) (f2^) by Th45
    .= (f1|X)/f2 by Th31;
  thus (f1/f2)|X = (f1(#)(f2^))|X by Th31
    .= f1 (#) (f2^)|X by Th45
    .= f1 (#) ((f2|X)^) by Th46
    .= f1/(f2|X) by Th31;
end;

theorem Th49:
  (r(#)f)|X = r(#)(f|X)
proof
A1: now
    let c be set;
    assume
A2: c in dom ((r(#)f)|X);
    then
A3: c in dom (r(#)f) /\ X by RELAT_1:61;
    then
A4: c in X by XBOOLE_0:def 4;
A5: c in dom (r(#)f) by A3,XBOOLE_0:def 4;
    then c in dom f by VALUED_1:def 5;
    then c in dom f /\ X by A4,XBOOLE_0:def 4;
    then
A6: c in dom (f|X) by RELAT_1:61;
    then
A7: c in dom (r(#)(f|X)) by VALUED_1:def 5;
    thus ((r(#)f)|X).c = (r(#)f).c by A2,FUNCT_1:47
      .= r*(f.c) by A5,VALUED_1:def 5
      .= r*(f|X).c by A6,FUNCT_1:47
      .= (r(#)(f|X)).c by A7,VALUED_1:def 5;
  end;
  dom ((r(#)f)|X) = dom (r(#)f) /\ X by RELAT_1:61
    .= dom f /\ X by VALUED_1:def 5
    .= dom (f|X) by RELAT_1:61
    .= dom (r(#)(f|X)) by VALUED_1:def 5;
  hence thesis by A1,FUNCT_1:2;
end;





reserve r,r1,r2,p for real number;
reserve f,f1,f2 for PartFunc of C,REAL;

theorem Th50:
  (f1 is total & f2 is total iff f1+f2 is total) & (f1 is total &
  f2 is total iff f1-f2 is total) & (f1 is total & f2 is total iff f1(#)f2 is
  total)
proof
  thus f1 is total & f2 is total iff f1+f2 is total
  proof
    thus f1 is total & f2 is total implies f1+f2 is total;
    assume f1+f2 is total;
    then dom (f1+f2) = C by PARTFUN1:def 2;
    then
A4: dom f1 /\ dom f2 = C by VALUED_1:def 1;
    then
A5: C c= dom f2 by XBOOLE_1:17;
    C c= dom f1 by A4,XBOOLE_1:17;
    hence dom f1 = C & dom f2 = C by A5,XBOOLE_0:def 10;
  end;
  thus f1 is total & f2 is total iff f1-f2 is total
  proof
    thus f1 is total & f2 is total implies f1-f2 is total;
    assume f1-f2 is total;
    then dom (f1-f2) = C by PARTFUN1:def 2;
    then
A9: dom f1 /\ dom f2 = C by VALUED_1:12;
    then
A10: C c= dom f2 by XBOOLE_1:17;
    C c= dom f1 by A9,XBOOLE_1:17;
    hence dom f1 = C & dom f2 = C by A10,XBOOLE_0:def 10;
  end;
  thus f1 is total & f2 is total implies f1(#)f2 is total;
  assume f1(#)f2 is total;
  then dom (f1(#)f2) = C by PARTFUN1:def 2;
  then
A14: dom f1 /\ dom f2 = C by VALUED_1:def 4;
  then
A15: C c= dom f2 by XBOOLE_1:17;
  C c= dom f1 by A14,XBOOLE_1:17;
  hence dom f1 = C & dom f2 = C by A15,XBOOLE_0:def 10;
end;

theorem Th51:
  f is total iff r(#)f is total
proof
  thus f is total implies r(#)f is total;
  assume r(#)f is total;
  then dom (r(#)f) = C by PARTFUN1:def 2;
  hence dom f = C by VALUED_1:def 5;
end;

theorem
  f is total iff -f is total by Th51;

theorem
  f is total iff abs(f) is total
proof
  thus f is total implies abs(f) is total;
  assume abs(f) is total;
  then dom (abs(f)) = C by PARTFUN1:def 2;
  hence dom f = C by VALUED_1:def 11;
end;

theorem Th54:
  f^ is total iff f"{0} = {} & f is total
proof
  thus f^ is total implies f"{0} = {} & f is total
  proof
    assume f^ is total;
    then
A1: dom (f^) = C by PARTFUN1:def 2;
    f"{0} c= C;
    then f"{0} c= dom f \ f"{0} by A1,Def2;
    hence f"{0} = {} by XBOOLE_1:38;
    then C = dom f \ {} by A1,Def2;
    hence dom f = C;
  end;
  assume that
A2: f"{0} = {} and
A3: f is total;
  thus dom (f^) = dom f \ f"{0} by Def2
    .= C by A2,A3,PARTFUN1:def 2;
end;

theorem Th55:
  f1 is total & f2"{0} = {} & f2 is total iff f1/f2 is total
proof
  thus f1 is total & f2"{0} = {} & f2 is total implies f1/f2 is total
  proof
    assume that
A1: f1 is total and
A2: f2"{0} = {} and
A3: f2 is total;
    f2^ is total by A2,A3,Th54;
    then f1(#)(f2^) is total by A1;
    hence thesis by Th31;
  end;
  assume f1/f2 is total;
  then
A4: f1(#)(f2^) is total by Th31;
  hence f1 is total by Th50;
  f2^ is total by A4,Th50;
  hence thesis by Th54;
end;

theorem
  f1 is total & f2 is total implies (f1+f2).c = f1.c + f2.c & (f1-f2).c
  = f1.c - f2.c & (f1(#)f2).c = f1.c * f2.c
proof
  assume that
A1: f1 is total and
A2: f2 is total;
  f1+f2 is total by A1,A2;
  then dom (f1+f2) = C by PARTFUN1:def 2;
  hence (f1+f2).c = f1.c + f2.c by VALUED_1:def 1;
  f1-f2 is total by A1,A2;
  then dom (f1-f2) = C by PARTFUN1:def 2;
  hence (f1-f2).c = f1.c - f2.c by VALUED_1:13;
  thus thesis by VALUED_1:5;
end;

theorem
  f is total implies (r(#)f).c = r * (f.c)
proof
  assume f is total;
  then r(#)f is total;
  then dom (r(#)f) = C by PARTFUN1:def 2;
  hence thesis by VALUED_1:def 5;
end;

theorem
  f is total implies (-f).c = - f.c & (abs(f)).c = abs( f.c ) by VALUED_1:8,18;

theorem
  f^ is total implies (f^).c = (f.c)"
proof
  assume f^ is total;
  then dom (f^) = C by PARTFUN1:def 2;
  hence thesis by Def2;
end;

theorem
  f1 is total & f2^ is total implies (f1/f2).c = f1.c *(f2.c)"
proof
  assume that
A1: f1 is total and
A2: f2^ is total;
A3: f2 is total by A2,Th54;
  f2"{0} = {} by A2,Th54;
  then f1/f2 is total by A1,A3,Th55;
  then dom (f1/f2) = C by PARTFUN1:def 2;
  hence thesis by Def1;
end;





definition
  let X,C be set;
  redefine func chi(X,C) -> PartFunc of C,REAL;
  coherence
  proof
    now
      let x;
      assume x in rng chi(X,C);
      then x = 0 or x = 1 by TARSKI:def 2;
      hence x in REAL;
    end;
    then
A1: rng chi(X,C) c= REAL by TARSKI:def 3;
    dom chi(X,C) =C by FUNCT_3:def 3;
    hence thesis by A1,RELSET_1:4;
  end;
end;

registration
  let X,C be set;
  cluster chi(X,C) -> total for PartFunc of C,REAL;
  coherence
  proof
    dom chi(X,C) =C by FUNCT_3:def 3;
    hence thesis by PARTFUN1:def 2;
  end;
end;

theorem Th61:
  f= chi(X,C) iff ( dom f = C & for c holds (c in X implies f.c =
  1) & (not c in X implies f.c = 0))
proof
  thus f= chi(X,C) implies (dom f = C & for c holds (c in X implies f.c = 1) &
  (not c in X implies f.c = 0)) by FUNCT_3:def 3;
  assume that
A1: dom f = C and
A2: for c holds (c in X implies f.c = 1) & (not c in X implies f.c = 0);
  for x st x in C holds (x in X implies (f qua Function).x = 1) & (not x
  in X implies (f qua Function).x = 0) by A2;
  hence thesis by A1,FUNCT_3:def 3;
end;

theorem
  chi(X,C) is total;

theorem
  c in X iff chi(X,C).c = 1 by Th61;

theorem
  not c in X iff chi(X,C).c = 0 by Th61;

theorem
  c in C \ X iff chi(X,C).c = 0
proof
  thus c in C \ X implies chi(X,C).c = 0
  proof
    assume c in C \ X;
    then not c in X by XBOOLE_0:def 5;
    hence thesis by Th61;
  end;
  assume chi(X,C).c = 0;
  then not c in X by Th61;
  hence thesis by XBOOLE_0:def 5;
end;

theorem
  chi(C,C).c = 1 by Th61;

theorem Th67:
  chi(X,C).c <> 1 iff chi(X,C).c = 0
proof
  thus chi(X,C).c <> 1 implies chi(X,C).c = 0
  proof
    c in C;
    then c in dom chi(X,C) by Th61;
    then
A1: chi(X,C).c in rng chi(X,C) by FUNCT_1:def 3;
A2: rng chi(X,C) c= {0,1} by FUNCT_3:39;
    assume chi(X,C).c <> 1;
    hence thesis by A1,A2,TARSKI:def 2;
  end;
  assume that
A3: chi(X,C).c = 0 and
A4: chi(X,C).c = 1;
  thus contradiction by A3,A4;
end;

theorem
  X misses Y implies chi(X,C) + chi(Y,C) = chi(X \/ Y,C)
proof
  assume
A1: X /\ Y = {};
A2: now
    let c such that
A3: c in dom (chi(X,C) + chi(Y,C));
    now
      per cases;
      suppose
A4:     c in X;
        then not c in Y by A1,XBOOLE_0:def 4;
        then
A5:     chi(Y,C).c = 0 by Th61;
A6:     c in X \/ Y by A4,XBOOLE_0:def 3;
        chi(X,C).c = 1 by A4,Th61;
        hence chi(X,C).c + chi(Y,C).c = chi(X \/ Y,C).c by A5,A6,Th61;
      end;
      suppose
A7:     not c in X;
        then
A8:     chi(X,C).c = 0 by Th61;
        now
          per cases;
          suppose
A9:         c in Y;
            then
A10:        c in X \/ Y by XBOOLE_0:def 3;
            chi(Y,C).c = 1 by A9,Th61;
            hence chi(X,C).c + chi(Y,C).c = chi(X \/ Y,C).c by A8,A10,Th61;
          end;
          suppose
A11:        not c in Y;
            then
A12:        not c in X \/ Y by A7,XBOOLE_0:def 3;
            chi(Y,C).c = 0 by A11,Th61;
            hence chi(X,C).c + chi(Y,C).c = chi(X \/ Y,C).c by A8,A12,Th61;
          end;
        end;
        hence chi(X,C).c + chi(Y,C).c = chi(X \/ Y,C).c;
      end;
    end;
    hence (chi(X,C) + chi(Y,C)).c = chi(X \/ Y,C).c by A3,VALUED_1:def 1;
  end;
  dom (chi(X,C) + chi(Y,C)) = dom chi(X,C) /\ dom chi(Y,C) by VALUED_1:def 1
    .= C /\ dom chi(Y,C) by Th61
    .= C /\ C by Th61
    .= dom chi(X \/ Y,C) by Th61;
  hence thesis by A2,PARTFUN1:5;
end;

theorem
  chi(X,C) (#) chi(Y,C) = chi(X /\ Y,C)
proof
A1: now
    let c such that
    c in dom (chi(X,C) (#) chi(Y,C));
    now
      per cases;
      suppose
A2:     chi(X,C).c * chi(Y,C).c = 0;
        now
          per cases by A2;
          suppose
            chi(X,C).c = 0;
            then not c in X by Th61;
            then not c in X /\ Y by XBOOLE_0:def 4;
            hence chi(X,C).c * chi(Y,C).c = chi(X /\ Y,C).c by A2,Th61;
          end;
          suppose
            chi(Y,C).c = 0;
            then not c in Y by Th61;
            then not c in X /\ Y by XBOOLE_0:def 4;
            hence chi(X,C).c * chi(Y,C).c = chi(X /\ Y,C).c by A2,Th61;
          end;
        end;
        hence chi(X,C).c * chi(Y,C).c = chi(X /\ Y,C).c;
      end;
      suppose
A3:     chi(X,C).c * chi(Y,C).c <> 0;
        then
A4:     chi(Y,C).c <> 0;
        then
A5:     chi(Y,C).c = 1 by Th67;
A6:     c in Y by A4,Th61;
A7:     chi(X,C).c <> 0 by A3;
        then c in X by Th61;
        then
A8:     c in X /\ Y by A6,XBOOLE_0:def 4;
        chi(X,C).c = 1 by A7,Th67;
        hence chi(X,C).c * chi(Y,C).c = chi(X /\ Y,C).c by A5,A8,Th61;
      end;
    end;
    hence (chi(X,C)(#)chi(Y,C)).c = chi(X /\ Y,C).c by VALUED_1:5;
  end;
  dom (chi(X,C) (#) chi(Y,C)) = dom chi(X,C) /\ dom chi(Y,C) by VALUED_1:def 4
    .= C /\ dom chi(Y,C) by Th61
    .= C /\ C by Th61
    .= dom chi(X /\ Y,C) by Th61;
  hence thesis by A1,PARTFUN1:5;
end;





reserve f for real-valued Function;

theorem Th70:
  f|Y is bounded_above iff ex r st for c being set st c in Y /\
  dom f holds f.c <= r
proof
  thus f|Y is bounded_above implies ex r st for c being set st c in Y /\ dom f
  holds f.c <= r
  proof
    given r being real number such that
A1: for p being set st p in dom(f|Y) holds (f|Y).p<r;
    take r;
    let c be set;
    assume c in Y /\ dom f;
    then
A2: c in dom(f|Y) by RELAT_1:61;
    then (f|Y).c < r by A1;
    hence thesis by A2,FUNCT_1:47;
  end;
  given r such that
A3: for c being set st c in Y /\ dom f holds f.c <= r;
  reconsider r1 = r+1 as real number;
  take r1;
  let p be set;
  assume
A4: p in dom(f|Y);
  then p in Y /\ dom f by RELAT_1:61;
  then f.p <= r by A3;
  then
A5: (f|Y).p <= r by A4,FUNCT_1:47;
  r < r1 by XREAL_1:29;
  hence thesis by A5,XXREAL_0:2;
end;

theorem Th71:
  f|Y is bounded_below iff ex r st for c being set st c in Y /\
  dom f holds r <= f.c
proof
  thus f|Y is bounded_below implies ex r st for c being set st c in Y /\ dom f
  holds r <= f.c
  proof
    given r being real number such that
A1: for p being set st p in dom(f|Y) holds r < (f|Y).p;
    take r;
    let c be set;
    assume c in Y /\ dom f;
    then
A2: c in dom(f|Y) by RELAT_1:61;
    then r < (f|Y).c by A1;
    hence thesis by A2,FUNCT_1:47;
  end;
  given r such that
A3: for c being set st c in Y /\ dom f holds r <= f.c;
  reconsider r1 = r-1 as real number;
  take r1;
  let p be set;
  assume
A4: p in dom(f|Y);
  then p in Y /\ dom f by RELAT_1:61;
  then r <= f.p by A3;
  then
A5: r <= (f|Y).p by A4,FUNCT_1:47;
  r1 < r by XREAL_1:44;
  hence thesis by A5,XXREAL_0:2;
end;

theorem Th72:
  f is bounded iff ex r st for c being set st c in dom f holds abs (f.c)<=r
proof
  thus f is bounded implies ex r st for c being set st c in dom f holds abs(f.
  c)<=r
  proof
    assume
A1: f is bounded;
    then consider r1 such that
A2: for c being set st c in dom f holds f.c < r1 by SEQ_2:def 1;
    consider r2 such that
A3: for c being set st c in dom f holds r2 < f.c by A1,SEQ_2:def 2;
    take g=abs(r1)+abs(r2);
    let c be set such that
A4: c in dom f;
A5: 0 <= abs(r2) by COMPLEX1:46;
    r1 <= abs(r1) by ABSVALUE:4;
    then f.c <= abs(r1) by A2,A4,XXREAL_0:2;
    then
A6: f.c + 0 <= abs(r1)+abs(r2) by A5,XREAL_1:7;
A7: 0 <= abs(r1) by COMPLEX1:46;
    -abs(r2) <= r2 by ABSVALUE:4;
    then -abs(r2) <= f.c by A3,A4,XXREAL_0:2;
    then
A8: -abs(r1)+-abs(r2) <= 0 + f.c by A7,XREAL_1:7;
    -abs(r1)+-abs(r2) = -g;
    hence thesis by A6,A8,ABSVALUE:5;
  end;
  given r such that
A9: for c being set st c in dom f holds abs(f.c) <= r;
  thus f is bounded_above
  proof
    take r+1;
    let c be set;
    assume c in dom f;
    then
A10: abs(f.c) < r+1 by A9,XREAL_1:39;
    f.c <= abs(f.c) by ABSVALUE:4;
    hence thesis by A10,XXREAL_0:2;
  end;
  take -(r+1);
  let c be set;
  assume c in dom f;
  then abs(f.c) < r+1 by A9,XREAL_1:39;
  then
A11: -(r+1) < -abs(f.c) by XREAL_1:24;
  -abs(f.c) <= f.c by ABSVALUE:4;
  hence thesis by A11,XXREAL_0:2;
end;

theorem
  f|Y is bounded iff ex r st for c being set st c in Y /\ dom f holds
  abs(f.c)<=r
proof
  thus f|Y is bounded implies ex r st for c being set st c in Y /\ dom f holds
  abs(f.c)<=r
  proof
    assume
A1: f|Y is bounded;
    then consider r1 such that
A2: for c being set st c in Y /\ dom f holds f.c <= r1 by Th70;
    consider r2 such that
A3: for c being set st c in Y /\ dom f holds r2 <= f.c by A1,Th71;
    take g=abs(r1)+abs(r2);
A4: r1 <= abs(r1) by ABSVALUE:4;
    let c be set such that
A5: c in Y /\ dom f;
    f.c <= r1 by A2,A5;
    then
A6: f.c <= abs(r1) by A4,XXREAL_0:2;
A7: -abs(r2) <= r2 by ABSVALUE:4;
    r2 <= f.c by A3,A5;
    then
A8: -abs(r2) <= f.c by A7,XXREAL_0:2;
    0 <= abs(r1) by COMPLEX1:46;
    then
A9: -abs(r1)+-abs(r2) <= 0 + f.c by A8,XREAL_1:7;
    0 <= abs(r2) by COMPLEX1:46;
    then
A10: f.c + 0 <= abs(r1)+abs(r2) by A6,XREAL_1:7;
    -abs(r1)+-abs(r2) = -g;
    hence thesis by A10,A9,ABSVALUE:5;
  end;
  given r such that
A11: for c being set st c in Y /\ dom f holds abs(f.c) <= r;
  now
    let c be set;
    assume c in Y /\ dom f;
    then abs(f.c) <= r by A11;
    then
A12: -r <= -abs(f.c) by XREAL_1:24;
    -abs(f.c) <= f.c by ABSVALUE:4;
    hence -r <= f.c by A12,XXREAL_0:2;
  end;
  then
A13: f|Y is bounded_below by Th71;
  now
    let c be set;
    assume c in Y /\ dom f;
    then
A14: abs(f.c) <= r by A11;
    f.c <= abs(f.c) by ABSVALUE:4;
    hence f.c <= r by A14,XXREAL_0:2;
  end;
  then f|Y is bounded_above by Th70;
  hence thesis by A13;
end;

theorem
  (Y c= X & f|X is bounded_above implies f|Y is bounded_above) & (Y c= X
& f|X is bounded_below implies f|Y is bounded_below) & (Y c= X & f|X is bounded
  implies f|Y is bounded)
proof
  thus
A1: Y c= X & f|X is bounded_above implies f|Y is bounded_above
  proof
    assume that
A2: Y c= X and
A3: f|X is bounded_above;
    consider r such that
A4: for c being set st c in X /\ dom f holds f.c <= r by A3,Th70;
    now
      take r;
      let c be set;
      assume
A5:   c in Y /\ dom f;
      then
A6:   c in dom f by XBOOLE_0:def 4;
      c in Y by A5,XBOOLE_0:def 4;
      then c in X /\ dom f by A2,A6,XBOOLE_0:def 4;
      hence f.c <= r by A4;
    end;
    hence thesis by Th70;
  end;
  thus
A7: Y c= X & f|X is bounded_below implies f|Y is bounded_below
  proof
    assume that
A8: Y c= X and
A9: f|X is bounded_below;
    consider r such that
A10: for c being set st c in X /\ dom f holds r <= f.c by A9,Th71;
    now
      take r;
      let c be set;
      assume
A11:  c in Y /\ dom f;
      then
A12:  c in dom f by XBOOLE_0:def 4;
      c in Y by A11,XBOOLE_0:def 4;
      then c in X /\ dom f by A8,A12,XBOOLE_0:def 4;
      hence r <= f.c by A10;
    end;
    hence thesis by Th71;
  end;
  assume that
A13: Y c= X and
A14: f|X is bounded;
  thus thesis by A1,A7,A13,A14;
end;

theorem
  f|X is bounded_above & f|Y is bounded_below implies f|(X /\ Y) is bounded
proof
  assume that
A1: f|X is bounded_above and
A2: f|Y is bounded_below;
  consider r1 such that
A3: for c being set st c in X /\ dom f holds f.c <= r1 by A1,Th70;
  now
    let c be set;
    assume
A4: c in X /\ Y /\ dom f;
    then c in X /\ Y by XBOOLE_0:def 4;
    then
A5: c in X by XBOOLE_0:def 4;
    c in dom f by A4,XBOOLE_0:def 4;
    then c in X /\ dom f by A5,XBOOLE_0:def 4;
    hence f.c <= r1 by A3;
  end;
  hence f|(X /\ Y) is bounded_above by Th70;
  consider r2 such that
A6: for c being set st c in Y /\ dom f holds r2 <= f.c by A2,Th71;
  now
    let c be set;
    assume
A7: c in X /\ Y /\ dom f;
    then c in X /\ Y by XBOOLE_0:def 4;
    then
A8: c in Y by XBOOLE_0:def 4;
    c in dom f by A7,XBOOLE_0:def 4;
    then c in Y /\ dom f by A8,XBOOLE_0:def 4;
    hence r2 <= f.c by A6;
  end;
  hence thesis by Th71;
end;

registration
  cluster constant -> bounded for real-valued Function;
  coherence
  proof
    let f be real-valued Function;
    per cases;
    suppose
A1:   f is empty;
      reconsider z=0 as real number;
      assume f is constant;
      thus f is bounded_above
      proof
        take z;
        let x be set;
        thus thesis by A1;
      end;
      take z;
      let x be set;
      thus thesis by A1;
    end;
    suppose
A2:   f is not empty;
      assume f is constant;
      then consider r such that
A3:   for c being set st c in dom f holds f.c = r by A2,VALUED_0:15;
      thus f is bounded_above
      proof
        take r+1;
        let y be set;
        assume y in dom f;
        then f.y = r by A3;
        hence thesis by XREAL_1:39;
      end;
      take r-1;
      let y be set;
      assume y in dom f;
      then f.y = r by A3;
      hence thesis by XREAL_1:231;
    end;
  end;
end;

theorem
  X misses dom f implies f|X is bounded
proof
  assume X /\ dom f = {};
  then dom(f|X) = {} by RELAT_1:61;
  then f|X = {};
  hence thesis;
end;

theorem
  (0(#)f)|Y is bounded
proof
  reconsider p1 = 0 as Real;
  set r = 0;
  now
    take p=p1;
    let c be set;
    assume c in Y /\ dom (r(#)f);
    then
A1: c in dom (r(#)f) by XBOOLE_0:def 4;
    r*f.c <= 0;
    hence (r(#)f).c <= p by A1,VALUED_1:def 5;
  end;
  hence (r(#)f)|Y is bounded_above by Th70;
  now
    take p=p1;
    let c be set;
    assume c in Y /\ dom (r(#)f);
    then
A2: c in dom (r(#)f) by XBOOLE_0:def 4;
    0 <= r*f.c;
    hence p <= (r(#)f).c by A2,VALUED_1:def 5;
  end;
  hence thesis by Th71;
end;

registration
  let f be bounded_above real-valued Function, X be set;
  cluster f|X -> bounded_above for real-valued Function;
  coherence
  proof
    consider r such that
A1: for y being set st y in dom f holds f.y<r by SEQ_2:def 1;
    f|X is bounded_above
    proof
      take r;
      let y be set;
      assume
A2:   y in dom(f|X);
      then y in dom f by RELAT_1:57;
      then f.y < r by A1;
      hence thesis by A2,FUNCT_1:47;
    end;
    hence thesis;
  end;
end;

registration
  let f be bounded_below real-valued Function, X be set;
  cluster f|X -> bounded_below for real-valued Function;
  coherence
  proof
    consider r such that
A1: for y being set st y in dom f holds r < f.y by SEQ_2:def 2;
    f|X is bounded_below
    proof
      take r;
      let y be set;
      assume
A2:   y in dom(f|X);
      then y in dom f by RELAT_1:57;
      then r < f.y by A1;
      hence thesis by A2,FUNCT_1:47;
    end;
    hence thesis;
  end;
end;

registration
  let f be bounded_above real-valued Function;
  let r be non negative real number;
  cluster r(#)f -> bounded_above for real-valued Function;
  coherence
  proof
    consider r1 such that
A1: for c being set st c in dom f holds f.c < r1 by SEQ_2:def 1;
    now
      take p = r*abs(r1)+1;
      let c be set;
A2:   r1 <= abs(r1) by ABSVALUE:4;
      assume
A3:   c in dom (r(#)f);
      then c in dom f by VALUED_1:def 5;
      then f.c < abs(r1) by A1,A2,XXREAL_0:2;
      then r * (f.c) <= abs(r1)*r by XREAL_1:64;
      then (r(#)f).c <= abs(r1)*r by A3,VALUED_1:def 5;
      hence (r(#)f).c < p by XREAL_1:39;
    end;
    hence thesis by SEQ_2:def 1;
  end;
end;

registration
  let f be bounded_below real-valued Function;
  let r be non negative real number;
  cluster r(#)f -> bounded_below for real-valued Function;
  coherence
  proof
    consider r1 such that
A1: for c being set st c in dom f holds r1 < f.c by SEQ_2:def 2;
    now
      take p = r*(-(abs r1))-1;
      let c be set;
A2:   -abs(r1) <= r1 by ABSVALUE:4;
      assume
A3:   c in dom (r(#)f);
      then c in dom f by VALUED_1:def 5;
      then f.c >= -abs(r1) by A1,A2,XXREAL_0:2;
      then r * (f.c) >= (-abs(r1))*r by XREAL_1:64;
      then (r(#)f).c >= (-abs(r1))*r by A3,VALUED_1:def 5;
      hence (r(#)f).c > p by XREAL_1:231;
    end;
    hence thesis by SEQ_2:def 2;
  end;
end;

registration
  let f be bounded_above real-valued Function;
  let r be non positive real number;
  cluster r(#)f -> bounded_below for real-valued Function;
  coherence
  proof
    consider r1 such that
A1: for c being set st c in dom f holds f.c < r1 by SEQ_2:def 1;
    now
      take p=r*abs(r1)-1;
      let c be set;
A2:   r1<=abs(r1) by ABSVALUE:4;
      assume
A3:   c in dom (r(#)f);
      then c in dom f by VALUED_1:def 5;
      then f.c <= abs(r1) by A1,A2,XXREAL_0:2;
      then r*abs(r1) <= r * f.c by XREAL_1:65;
      then r*abs(r1) <= (r(#)f).c by A3,VALUED_1:def 5;
      hence p < (r(#)f).c by XREAL_1:231;
    end;
    hence thesis by SEQ_2:def 2;
  end;
end;

registration
  let f be bounded_below real-valued Function;
  let r be non positive real number;
  cluster r(#)f -> bounded_above for real-valued Function;
  coherence
  proof
    consider r1 such that
A1: for c being set st c in dom f holds r1 < f.c by SEQ_2:def 2;
    now
      take p=r*(-abs(r1))+1;
      let c be set;
A2:   -abs(r1) <= r1 by ABSVALUE:4;
      assume
A3:   c in dom (r(#)f);
      then c in dom f by VALUED_1:def 5;
      then -abs(r1) <= f.c by A1,A2,XXREAL_0:2;
      then r * f.c <= r*(-abs(r1)) by XREAL_1:65;
      then (r(#)f).c <= r*(-abs(r1)) by A3,VALUED_1:def 5;
      hence (r(#)f).c < p by XREAL_1:39;
    end;
    hence thesis by SEQ_2:def 1;
  end;
end;

theorem Th78:
  (f|Y is bounded_above & 0<=r implies (r(#)f)|Y is bounded_above)
  & (f|Y is bounded_above & r<=0 implies (r(#)f)|Y is bounded_below)
proof
  (r(#)f)|Y = r(#)(f|Y) by Th49;
  hence thesis;
end;

theorem Th79:
  (f|Y is bounded_below & 0<=r implies (r(#)f)|Y is bounded_below)
  & (f|Y is bounded_below & r<=0 implies (r(#)f)|Y is bounded_above)
proof
  (r(#)f)|Y = r(#)(f|Y) by Th49;
  hence thesis;
end;

theorem Th80:
  f|Y is bounded implies (r(#)f)|Y is bounded
proof
  assume
A1: f|Y is bounded;
  per cases;
  suppose
A2: 0 <= r;
    hence (r(#)f)|Y is bounded_above by A1,Th78;
    thus thesis by A1,A2,Th79;
  end;
  suppose
A3: r <= 0;
    hence (r(#)f)|Y is bounded_above by A1,Th79;
    thus thesis by A1,A3,Th78;
  end;
end;

registration
  let f;
  cluster abs f -> bounded_below;
  coherence
  proof
    now
      reconsider p = -1 as real number;
      take p;
      let c be set;
      assume c in dom (abs(f));
      0 <= abs(f.c) by COMPLEX1:46;
      hence p < (abs(f)).c by VALUED_1:18;
    end;
    hence thesis by SEQ_2:def 2;
  end;
end;

theorem
  (abs f)|X is bounded_below;

registration
  let f be bounded real-valued Function;
  cluster abs f -> bounded for real-valued Function;
  coherence
  proof
    consider r1 such that
A1: for c being set st c in dom f holds abs(f.c) <= r1 by Th72;
    now
      take r1;
      let c be set;
      assume c in dom abs f;
      then c in dom f by VALUED_1:def 11;
      then abs abs(f.c) <= r1 by A1;
      hence abs((abs f).c) <= r1 by VALUED_1:18;
    end;
    hence thesis by Th72;
  end;
end;

registration
  let f be bounded_above real-valued Function;
  cluster -f -> bounded_below for real-valued Function;
  coherence;
end;

theorem Th82:
  f|Y is bounded implies (abs f)|Y is bounded & (-f)|Y is bounded
proof
  assume
A1: f|Y is bounded;
  (abs f)|Y = abs(f|Y) by Th46;
  hence (abs f)|Y is bounded by A1;
  thus thesis by A1,Th80;
end;

reserve f1,f2 for real-valued Function;

registration
  let f1,f2 be bounded_above real-valued Function;
  cluster f1+f2 -> bounded_above for real-valued Function;
  coherence
  proof
    consider r2 such that
A1: for c being set st c in dom f2 holds f2.c < r2 by SEQ_2:def 1;
    consider r1 such that
A2: for c being set st c in dom f1 holds f1.c < r1 by SEQ_2:def 1;
    now
      take r=r1+r2;
      let c be set;
      assume
A3:   c in dom (f1+f2);
      then
A4:   c in dom f1 /\ dom f2 by VALUED_1:def 1;
      then c in dom f1 by XBOOLE_0:def 4;
      then
A5:   f1.c < r1 by A2;
      c in dom f2 by A4,XBOOLE_0:def 4;
      then f1.c + f2.c < r by A1,A5,XREAL_1:8;
      hence (f1+f2).c < r by A3,VALUED_1:def 1;
    end;
    hence thesis by SEQ_2:def 1;
  end;
end;

registration
  let f1,f2 be bounded_below real-valued Function;
  cluster f1+f2 -> bounded_below for real-valued Function;
  coherence
  proof
    consider r2 such that
A1: for c being set st c in dom f2 holds f2.c > r2 by SEQ_2:def 2;
    consider r1 such that
A2: for c being set st c in dom f1 holds f1.c > r1 by SEQ_2:def 2;
    now
      take r=r1+r2;
      let c be set;
      assume
A3:   c in dom (f1+f2);
      then
A4:   c in dom f1 /\ dom f2 by VALUED_1:def 1;
      then c in dom f1 by XBOOLE_0:def 4;
      then
A5:   f1.c > r1 by A2;
      c in dom f2 by A4,XBOOLE_0:def 4;
      then f1.c + f2.c > r by A1,A5,XREAL_1:8;
      hence (f1+f2).c > r by A3,VALUED_1:def 1;
    end;
    hence thesis by SEQ_2:def 2;
  end;
end;

theorem Th83:
  (f1|X is bounded_above & f2|Y is bounded_above implies (f1+f2)|
  (X /\ Y) is bounded_above) & (f1|X is bounded_below & f2|Y is bounded_below
implies (f1+f2)|(X /\ Y) is bounded_below) & (f1|X is bounded & f2|Y is bounded
  implies (f1+f2)|(X /\ Y) is bounded)
proof
  (f1+f2)|(X /\ Y) = f1|(X /\ Y)+f2|(X /\ Y) by Th44
    .= f1|(X /\ Y)+f2|Y|X by RELAT_1:71
    .= f1|X|Y+f2|Y|X by RELAT_1:71;
  hence thesis;
end;

registration
  let f1,f2 be bounded real-valued Function;
  cluster f1(#)f2 -> bounded for real-valued Function;
  coherence
  proof
    consider r2 such that
A1: for c being set st c in dom f2 holds abs(f2.c) <= r2 by Th72;
    consider r1 such that
A2: for c being set st c in dom f1 holds abs(f1.c) <= r1 by Th72;
    now
      take r=r1*r2;
      let c be set;
A3:   0<=abs(f2.c) by COMPLEX1:46;
      assume c in dom (f1(#)f2);
      then
A4:   c in dom f1 /\ dom f2 by VALUED_1:def 4;
      then c in dom f1 by XBOOLE_0:def 4;
      then
A5:   abs(f1.c) <= r1 by A2;
      c in dom f2 by A4,XBOOLE_0:def 4;
      then
A6:   abs(f2.c) <= r2 by A1;
      0<=abs(f1.c) by COMPLEX1:46;
      then abs(f1.c)*abs(f2.c) <= r by A5,A6,A3,XREAL_1:66;
      then abs(f1.c * f2.c) <= r by COMPLEX1:65;
      hence abs((f1(#)f2).c) <= r by VALUED_1:5;
    end;
    hence thesis by Th72;
  end;
end;

theorem Th84:
  f1|X is bounded & f2|Y is bounded implies (f1(#)f2)|(X /\ Y) is
  bounded & (f1-f2)|(X /\ Y) is bounded
proof
  assume that
A1: f1|X is bounded and
A2: f2|Y is bounded;
  (f1(#)f2)|(X /\ Y) = f1|(X /\ Y)(#)f2|(X /\ Y) by Th45
    .= f1|(X /\ Y)(#)f2|Y|X by RELAT_1:71
    .= f1|X|Y(#)f2|Y|X by RELAT_1:71;
  hence (f1(#)f2)|(X /\ Y) is bounded by A1,A2;
  (-f2)|Y is bounded by A2,Th82;
  hence thesis by A1,Th83;
end;

theorem Th85:
  f|X is bounded_above & f|Y is bounded_above implies f|(X \/ Y)
  is bounded_above
proof
  assume that
A1: f|X is bounded_above and
A2: f|Y is bounded_above;
  consider r1 such that
A3: for c being set st c in X /\ dom f holds f.c <= r1 by A1,Th70;
  consider r2 such that
A4: for c being set st c in Y /\ dom f holds f.c <= r2 by A2,Th70;
  now
    take r = abs(r1) + abs(r2);
    let c be set;
    assume
A5: c in (X \/ Y) /\ dom f;
    then
A6: c in dom f by XBOOLE_0:def 4;
A7: c in X \/ Y by A5,XBOOLE_0:def 4;
    now
      per cases by A7,XBOOLE_0:def 3;
      suppose
        c in X;
        then c in X /\ dom f by A6,XBOOLE_0:def 4;
        then
A8:     f.c <= r1 by A3;
A9:     0 <= abs(r2) by COMPLEX1:46;
        r1 <= abs(r1) by ABSVALUE:4;
        then f.c <= abs(r1) by A8,XXREAL_0:2;
        then f.c + 0 <= r by A9,XREAL_1:7;
        hence f.c <= r;
      end;
      suppose
        c in Y;
        then c in Y /\ dom f by A6,XBOOLE_0:def 4;
        then
A10:    f.c <= r2 by A4;
A11:    0 <= abs(r1) by COMPLEX1:46;
        r2 <= abs(r2) by ABSVALUE:4;
        then f.c <= abs(r2) by A10,XXREAL_0:2;
        then 0 + f.c <= r by A11,XREAL_1:7;
        hence f.c <= r;
      end;
    end;
    hence f.c <= r;
  end;
  hence thesis by Th70;
end;

theorem Th86:
  f|X is bounded_below & f|Y is bounded_below implies f|(X \/ Y)
  is bounded_below
proof
  assume that
A1: f|X is bounded_below and
A2: f|Y is bounded_below;
  consider r1 such that
A3: for c being set st c in X /\ dom f holds r1 <= f.c by A1,Th71;
  consider r2 such that
A4: for c being set st c in Y /\ dom f holds r2 <= f.c by A2,Th71;
  now
    take r = -abs(r1) - abs(r2);
    let c be set;
    assume
A5: c in (X \/ Y) /\ dom f;
    then
A6: c in dom f by XBOOLE_0:def 4;
A7: c in X \/ Y by A5,XBOOLE_0:def 4;
    now
      per cases by A7,XBOOLE_0:def 3;
      suppose
        c in X;
        then c in X /\ dom f by A6,XBOOLE_0:def 4;
        then
A8:     r1 <= f.c by A3;
A9:     0 <= abs(r2) by COMPLEX1:46;
        -abs(r1) <= r1 by ABSVALUE:4;
        then -abs(r1) <= f.c by A8,XXREAL_0:2;
        then r <= f.c - 0 by A9,XREAL_1:13;
        hence r <= f.c;
      end;
      suppose
        c in Y;
        then c in Y /\ dom f by A6,XBOOLE_0:def 4;
        then
A10:    r2 <= f.c by A4;
A11:    0 <= abs(r1) by COMPLEX1:46;
        -abs(r2) <= r2 by ABSVALUE:4;
        then -abs(r2) <= f.c by A10,XXREAL_0:2;
        then -abs(r2) - abs(r1) <= f.c - 0 by A11,XREAL_1:13;
        hence r <= f.c;
      end;
    end;
    hence r <= f.c;
  end;
  hence thesis by Th71;
end;

theorem
  f|X is bounded & f|Y is bounded implies f|(X \/ Y) is bounded
proof
  assume that
A1: f|X is bounded and
A2: f|Y is bounded;
  thus f|(X \/ Y) is bounded_above by A1,A2,Th85;
  thus thesis by A1,A2,Th86;
end;

reserve f,f1,f2 for PartFunc of C,REAL;

registration
  let C;
  let f1,f2 be constant PartFunc of C,REAL;
  cluster f1+f2 -> constant;
  coherence
  proof
    consider r2 being Real such that
A1: for c st c in dom f2 holds f2.c = r2 by PARTFUN2:def 1;
    consider r1 being Real such that
A2: for c st c in dom f1 holds f1.c = r1 by PARTFUN2:def 1;
    now
      let c;
      assume
A3:   c in dom(f1+f2);
      then
A4:   c in dom f1 /\ dom f2 by VALUED_1:def 1;
      then
A5:   c in dom f1 by XBOOLE_0:def 4;
A6:   c in dom f2 by A4,XBOOLE_0:def 4;
      thus (f1+f2).c = f1.c + f2.c by A3,VALUED_1:def 1
        .= f1.c + r2 by A1,A6
        .= r1+r2 by A2,A5;
    end;
    hence thesis by PARTFUN2:def 1;
  end;
  cluster f1-f2 -> constant;
  coherence
  proof
    consider r2 being Real such that
A7: for c st c in dom f2 holds f2.c = r2 by PARTFUN2:def 1;
    consider r1 being Real such that
A8: for c st c in dom f1 holds f1.c = r1 by PARTFUN2:def 1;
    now
      let c;
      assume
A9:   c in dom(f1-f2);
      then
A10:  c in dom f1 /\ dom f2 by VALUED_1:12;
      then
A11:  c in dom f1 by XBOOLE_0:def 4;
A12:  c in dom f2 by A10,XBOOLE_0:def 4;
      thus (f1-f2).c = f1.c - f2.c by A9,VALUED_1:13
        .= f1.c - r2 by A7,A12
        .= r1-r2 by A8,A11;
    end;
    hence thesis by PARTFUN2:def 1;
  end;
  cluster f1(#)f2 -> constant;
  coherence
  proof
    consider r2 being Real such that
A13: for c st c in dom f2 holds f2.c = r2 by PARTFUN2:def 1;
    consider r1 being Real such that
A14: for c st c in dom f1 holds f1.c = r1 by PARTFUN2:def 1;
    now
      let c;
      assume
A15:  c in dom(f1(#)f2);
      then
A16:  c in dom f1 /\ dom f2 by VALUED_1:def 4;
      then
A17:  c in dom f1 by XBOOLE_0:def 4;
A18:  c in dom f2 by A16,XBOOLE_0:def 4;
      thus (f1(#)f2).c = f1.c * f2.c by A15,VALUED_1:def 4
        .= f1.c * r2 by A13,A18
        .= r1*r2 by A14,A17;
    end;
    hence thesis by PARTFUN2:def 1;
  end;
end;

theorem
  f1|X is constant & f2|Y is constant implies (f1+f2)|(X /\ Y) is
  constant & (f1-f2)|(X /\ Y) is constant & (f1(#)f2)|(X /\ Y) is constant
proof
  assume that
A1: f1|X is constant and
A2: f2|Y is constant;
  consider r1 be Real such that
A3: for c st c in X /\ dom f1 holds f1.c = r1 by A1,PARTFUN2:57;
  consider r2 be Real such that
A4: for c st c in Y /\ dom f2 holds f2.c = r2 by A2,PARTFUN2:57;
  now
    let c;
    assume
A5: c in X /\ Y /\ dom (f1+f2);
    then
A6: c in X /\ Y by XBOOLE_0:def 4;
    then
A7: c in X by XBOOLE_0:def 4;
A8: c in dom (f1+f2) by A5,XBOOLE_0:def 4;
    then
A9: c in (dom f1 /\ dom f2) by VALUED_1:def 1;
    then c in dom f1 by XBOOLE_0:def 4;
    then
A10: c in X /\ dom f1 by A7,XBOOLE_0:def 4;
A11: c in Y by A6,XBOOLE_0:def 4;
    c in dom f2 by A9,XBOOLE_0:def 4;
    then
A12: c in Y /\ dom f2 by A11,XBOOLE_0:def 4;
    thus (f1+f2).c =f1.c + f2.c by A8,VALUED_1:def 1
      .= r1 + f2.c by A3,A10
      .= r1 + r2 by A4,A12;
  end;
  hence (f1+f2)|(X /\ Y) is constant by PARTFUN2:57;
  now
    let c;
    assume
A13: c in X /\ Y /\ dom (f1-f2);
    then
A14: c in X /\ Y by XBOOLE_0:def 4;
    then
A15: c in X by XBOOLE_0:def 4;
A16: c in dom (f1-f2) by A13,XBOOLE_0:def 4;
    then
A17: c in (dom f1 /\ dom f2) by VALUED_1:12;
    then c in dom f1 by XBOOLE_0:def 4;
    then
A18: c in X /\ dom f1 by A15,XBOOLE_0:def 4;
A19: c in Y by A14,XBOOLE_0:def 4;
    c in dom f2 by A17,XBOOLE_0:def 4;
    then
A20: c in Y /\ dom f2 by A19,XBOOLE_0:def 4;
    thus (f1-f2).c = f1.c - f2.c by A16,VALUED_1:13
      .= r1 - f2.c by A3,A18
      .= r1 - r2 by A4,A20;
  end;
  hence (f1-f2)|(X /\ Y) is constant by PARTFUN2:57;
  now
    let c;
    assume
A21: c in X /\ Y /\ dom (f1(#)f2);
    then
A22: c in X /\ Y by XBOOLE_0:def 4;
    c in dom (f1(#)f2) by A21,XBOOLE_0:def 4;
    then
A23: c in (dom f1 /\ dom f2) by VALUED_1:def 4;
    then
A24: c in dom f1 by XBOOLE_0:def 4;
A25: c in dom f2 by A23,XBOOLE_0:def 4;
    c in Y by A22,XBOOLE_0:def 4;
    then
A26: c in Y /\ dom f2 by A25,XBOOLE_0:def 4;
    c in X by A22,XBOOLE_0:def 4;
    then
A27: c in X /\ dom f1 by A24,XBOOLE_0:def 4;
    thus (f1(#)f2).c =f1.c * f2.c by VALUED_1:5
      .= r1 * f2.c by A3,A27
      .= r1 * r2 by A4,A26;
  end;
  hence thesis by PARTFUN2:57;
end;

registration
  let C;
  let f be constant PartFunc of C,REAL;
  cluster -f -> constant;
  coherence
  proof
    consider r be Real such that
A1: for c be Element of C st c in dom f holds f.c = r by PARTFUN2:def 1;
    now
      take p=-r;
      let c be Element of C;
      assume c in dom (-f);
      then c in dom f by VALUED_1:8;
      then -f.c = p by A1;
      hence (-f).c = p by VALUED_1:8;
    end;
    hence thesis by PARTFUN2:def 1;
  end;
  cluster abs f -> constant;
  coherence
  proof
    consider r be Real such that
A2: for c be Element of C st c in dom f holds f.c = r by PARTFUN2:def 1;
    now
      take p= abs r;
      let c be Element of C;
      assume c in dom (abs f);
      then c in dom f by VALUED_1:def 11;
      then abs(f.c) = p by A2;
      hence (abs f).c = p by VALUED_1:18;
    end;
    hence thesis by PARTFUN2:def 1;
  end;
  let p;
  cluster p(#)f -> constant;
  coherence
  proof
    consider r be Real such that
A3: for c be Element of C st c in dom f holds f.c = r by PARTFUN2:def 1;
    now
      take r1= p*r;
      let c be Element of C;
      assume c in dom (p(#)f);
      then c in dom f by VALUED_1:def 5;
      then p*(f.c) = r1 by A3;
      hence (p(#)f).c = r1 by VALUED_1:6;
    end;
    hence thesis by PARTFUN2:def 1;
  end;
end;

theorem Th89:
  f|Y is constant implies (p(#)f)|Y is constant
proof
  (p(#)f)|Y = p(#)(f|Y) by Th49;
  hence thesis;
end;

theorem Th90:
  f|Y is constant implies (-f)|Y is constant
proof
  (-f)|Y = -(f|Y) by Th46;
  hence thesis;
end;

theorem Th91:
  f|Y is constant implies (abs f)|Y is constant
proof
  (abs f)|Y = abs(f|Y) by Th46;
  hence thesis;
end;

theorem
  f|Y is constant implies (for r holds (r(#)f)|Y is bounded) & (-f)|Y is
  bounded & (abs f)|Y is bounded
proof
  assume
A1: f|Y is constant;
  hereby
    let r;
    (r(#)f)|Y is constant by A1,Th89;
    hence (r(#)f)|Y is bounded;
  end;
  (-f)|Y is constant by A1,Th90;
  hence (-f)|Y is bounded;
  (abs f)|Y is constant by A1,Th91;
  hence thesis;
end;

theorem
  (f1|X is bounded_above & f2|Y is constant implies (f1+f2)|(X /\ Y) is
  bounded_above) & (f1|X is bounded_below & f2|Y is constant implies (f1+f2)|(X
/\ Y) is bounded_below) & (f1|X is bounded & f2|Y is constant implies (f1+f2)|(
  X /\ Y) is bounded) by Th83;

theorem
  (f1|X is bounded_above & f2|Y is constant implies (f1-f2)|(X /\ Y) is
  bounded_above) & (f1|X is bounded_below & f2|Y is constant implies (f1-f2)|(X
/\ Y) is bounded_below) & (f1|X is bounded & f2|Y is constant implies (f1-f2)|(
X /\ Y) is bounded & (f2-f1)|(X /\ Y) is bounded& (f1(#)f2)|(X /\ Y) is bounded
  )
proof
  thus f1|X is bounded_above & f2|Y is constant implies (f1-f2)|(X /\ Y) is
  bounded_above
  proof
    assume that
A1: f1|X is bounded_above and
A2: f2|Y is constant;
    (-f2)|Y is constant by A2,Th90;
    hence thesis by A1,Th83;
  end;
  thus f1|X is bounded_below & f2|Y is constant implies (f1-f2)|(X /\ Y) is
  bounded_below
  proof
    assume that
A3: f1|X is bounded_below and
A4: f2|Y is constant;
    (-f2)|Y is constant by A4,Th90;
    hence thesis by A3,Th83;
  end;
  assume that
A5: f1|X is bounded and
A6: f2|Y is constant;
  (-f2)|Y is constant by A6,Th90;
  hence (f1-f2)|(X /\ Y) is bounded by A5,Th83;
  thus (f2-f1)|(X /\ Y) is bounded by A5,A6,Th84;
  thus thesis by A5,A6,Th84;
end;














begin 

registration
  let n be Nat;
  cluster n-element natural-valued for FinSequence;
  existence
  proof
    take n|->0;
    thus thesis;
  end;
end;

registration
  cluster real-valued for FinSequence;
  existence
  proof
    the FinSequence of REAL is real-valued;
    hence thesis;
  end;
end;

definition
  let F be real-valued Relation;
  redefine func rng F -> Subset of REAL;
  coherence by VALUED_0:def 3;
end;

registration
  let D be non empty set;
  let F be Function of REAL,D;
  let F1 be real-valued FinSequence;
  cluster F*F1 -> FinSequence-like;
  coherence
  proof
    consider n1 being Nat such that
A1: dom F1 = Seg n1 by FINSEQ_1:def 2;
    take n1;
    dom F = REAL & rng F1 c= REAL by FUNCT_2:def 1;
    hence thesis by A1,RELAT_1:27;
  end;
end;

registration
  let r be real number;
  cluster <*r*> -> real-valued;
  coherence
  proof
    reconsider p = r as Element of REAL by XREAL_0:def 1;
    <*p*> is FinSequence-like;
    hence thesis;
  end;
end;

registration
  let r1, r2 be real number;
  cluster <*r1, r2*> -> real-valued;
  coherence
  proof
    reconsider p1 = r1, p2 = r2 as Element of REAL by XREAL_0:def 1;
    <*p1,p2*> is FinSequence-like;
    hence thesis;
  end;
end;

registration
  let r1, r2, r3 be real number;
  cluster <*r1, r2, r3*> -> real-valued;
  coherence
  proof
    reconsider p1 = r1, p2 = r2, p3 = r3 as Element of REAL by XREAL_0:def 1;
    <*p1,p2,p3*> is FinSequence-like;
    hence thesis;
  end;
end;

registration
  let r1, r2, r3, r4 be real number;
  cluster <*r1, r2, r3, r4*> -> real-valued;
  coherence
  proof
    reconsider p1 = r1, p2 = r2, p3 = r3, p4 = r4 as Element of REAL
     by XREAL_0:def 1;
    <*p1,p2,p3,p4*> is FinSequence-like;
    hence thesis;
  end;
end;

registration
  let j be Nat, r be real number;
  cluster j |-> r -> real-valued;
  coherence;
end;

registration
  let f, g be real-valued FinSequence;
  cluster f ^ g -> real-valued;
  coherence
  proof
    rng (f ^ g) = rng f \/ rng g by FINSEQ_1:31;
    then rng (f ^ g) c= REAL by XBOOLE_1:8;
    hence thesis by VALUED_0:def 3;
  end;
end;

reserve s for set,
  i,j,k for Nat,
  x,x1,x2,x3 for Element of REAL,
  r,r1,r2,r3,r4 for real number,
  F,F1,F2,F3 for real-valued FinSequence,
  R,R1,R2 for Element of i-tuples_on REAL;

theorem
  0 is_a_unity_wrt addreal by BINOP_2:2,SETWISEO:14;

definition
  redefine func diffreal equals
  addreal*(id REAL,compreal);
  compatibility
  proof
    let b be BinOp of REAL;
    now
      let r1,r2 be Element of REAL;
      thus diffreal.(r1,r2) = r1 - r2 by BINOP_2:def 10
        .= r1 + - r2
        .= addreal.(r1,- r2) by BINOP_2:def 9
        .= addreal.(r1,compreal.r2) by BINOP_2:def 7
        .= (addreal*(id REAL,compreal)).(r1,r2) by FINSEQOP:82;
    end;
    hence thesis by BINOP_1:2;
  end;
  correctness;
end;

definition
  func sqrreal -> UnOp of REAL means
  :Def2:
  for r holds it.r = r^2;
  existence
  proof
    deffunc F(Element of REAL) = $1^2;
    consider G being Function of REAL,REAL such that
A1: for x being Element of REAL holds G.x = F(x) from FUNCT_2:sch 4;
    take G;
    let r;
    r in REAL by XREAL_0:def 1;
    hence thesis by A1;
  end;
  uniqueness
  proof
    let G1,G2 be UnOp of REAL such that
A2: for r holds G1.r = r^2 and
A3: for r holds G2.r = r^2;
    now
      let x;
      G1.(x) = (x)^2 by A2;
      hence G1.x = G2.x by A3;
    end;
    hence thesis by FUNCT_2:63;
  end;
end;

theorem
  1 is_a_unity_wrt multreal by BINOP_2:7,SETWISEO:14;

theorem Th3:
  multreal is_distributive_wrt addreal
proof
  now
    let x1,x2,x3;
    thus multreal.(x1,addreal.(x2,x3)) = multreal.(x1,x2+x3) by BINOP_2:def 9
      .= x1*(x2+x3) by BINOP_2:def 11
      .= x1*x2+x1*x3
      .= addreal.(x1*x2,x1*x3) by BINOP_2:def 9
      .= addreal.(multreal.(x1,x2),x1*x3) by BINOP_2:def 11
      .= addreal.(multreal.(x1,x2),multreal.(x1,x3)) by BINOP_2:def 11;
    thus multreal.(addreal.(x1,x2),x3) = multreal.(x1+x2,x3) by BINOP_2:def 9
      .= (x1+x2)*x3 by BINOP_2:def 11
      .= x1*x3+x2*x3
      .= addreal.(x1*x3,x2*x3) by BINOP_2:def 9
      .= addreal.(multreal.(x1,x3),x2*x3) by BINOP_2:def 11
      .= addreal.(multreal.(x1,x3),multreal.(x2,x3)) by BINOP_2:def 11;
  end;
  hence thesis by BINOP_1:11;
end;

theorem
  sqrreal is_distributive_wrt multreal
proof
  let x1,x2;
  thus sqrreal.(multreal.(x1,x2)) = sqrreal.(x1*x2) by BINOP_2:def 11
    .= (x1*x2)^2 by Def2
    .= x1^2*x2^2
    .= multreal.(x1^2,x2^2) by BINOP_2:def 11
    .= multreal.(sqrreal.x1,x2^2) by Def2
    .= multreal.(sqrreal.x1,sqrreal.x2) by Def2;
end;

definition
  let x be real number;
  func x multreal -> UnOp of REAL equals
  multreal[;](x,id REAL);
  coherence
  proof
    reconsider y = x as Real by XREAL_0:def 1;
    multreal[;](y,id REAL) is UnOp of REAL;
    hence thesis;
  end;
end;

Lm1: (multreal[;](r,id REAL)).r1 = r*r1
proof
  reconsider a=r, s=r1 as Element of REAL by XREAL_0:def 1;
  thus (multreal[;](r,id REAL)).r1 = multreal.(a,(id REAL).s) by FUNCOP_1:53
    .= multreal.(a,s) by FUNCT_1:18
    .= r*r1 by BINOP_2:def 11;
end;

theorem
  (r multreal).r1 = r*r1 by Lm1;

theorem
  r multreal is_distributive_wrt addreal
proof
  r in REAL by XREAL_0:def 1;
  hence thesis by Th3,FINSEQOP:54;
end;

theorem Th7:
  compreal is_an_inverseOp_wrt addreal
proof
  let x;
  thus addreal.(x,compreal.x) = x+compreal.x by BINOP_2:def 9
    .= x+-x by BINOP_2:def 7
    .= the_unity_wrt addreal by BINOP_2:2;
  thus addreal.(compreal.x,x) = compreal.x+x by BINOP_2:def 9
    .= -x+x by BINOP_2:def 7
    .= the_unity_wrt addreal by BINOP_2:2;
end;

theorem Th8:
  addreal is having_an_inverseOp by Th7,FINSEQOP:def 2;

theorem Th9:
  the_inverseOp_wrt addreal = compreal by Th7,Th8,FINSEQOP:def 3;

theorem
  compreal is_distributive_wrt addreal by Th8,Th9,FINSEQOP:63;



Lm2: F is FinSequence of REAL
proof
  thus rng F c= REAL;
end;

definition
  let F1,F2;
  redefine func F1 + F2 -> FinSequence of REAL equals
  addreal.:(F1,F2);
  coherence by Lm2;
  compatibility
  proof
    reconsider F3=F1,F4=F2 as FinSequence of REAL by Lm2;
    let F be FinSequence of REAL;
    dom addreal = [:REAL,REAL:] by FUNCT_2:def 1;
    then [:rng F3, rng F4:] c= dom addreal by ZFMISC_1:96;
    then
A1: dom(addreal.:(F1,F2)) = dom F1 /\ dom F2 by FUNCOP_1:69;
A2: dom (F1+F2) = dom F1 /\ dom F2 by VALUED_1:def 1;
    thus F = F1+F2 implies F = addreal.:(F1,F2)
    proof
      assume
A3:   F = F1+F2;
      for z being set st z in dom(addreal.:(F1,F2)) holds F.z = addreal.(
      F1.z,F2.z)
      proof
        let z be set;
        assume z in dom(addreal.:(F1,F2));
        hence F.z = F1.z + F2.z by A2,A1,A3,VALUED_1:def 1
          .= addreal.(F1.z,F2.z) by BINOP_2:def 9;
      end;
      hence thesis by A2,A1,A3,FUNCOP_1:21;
    end;
    assume
A4: F = addreal.:(F1,F2);
    now
      let c be set;
      assume c in dom F;
      hence F.c = addreal.(F1.c,F2.c) by A4,FUNCOP_1:22
        .= F1.c + F2.c by BINOP_2:def 9;
    end;
    hence thesis by A1,A4,VALUED_1:def 1;
  end;
  commutativity
  proof
    let F be FinSequence of REAL;
    let F1,F2;
    assume
A5: F = addreal.:(F1,F2);
    reconsider F1,F2 as FinSequence of REAL by Lm2;
A6: dom addreal = [:REAL,REAL:] by FUNCT_2:def 1;
    then
A7: [:rng F2, rng F1:] c= dom addreal by ZFMISC_1:96;
    [:rng F1, rng F2:] c= dom addreal by A6,ZFMISC_1:96;
    then
A8: dom(addreal.:(F1,F2)) = dom F1 /\ dom F2 by FUNCOP_1:69
      .= dom(addreal.:(F2,F1)) by A7,FUNCOP_1:69;
    for z being set st z in dom(addreal.:(F2,F1)) holds F.z = addreal.(F2
    .z,F1.z)
    proof
      let z be set;
      assume z in dom(addreal.:(F2,F1));
      hence F.z = addreal.(F1.z,F2.z) by A5,A8,FUNCOP_1:22
        .= F1.z + F2.z by BINOP_2:def 9
        .= addreal.(F2.z,F1.z) by BINOP_2:def 9;
    end;
    hence thesis by A5,A8,FUNCOP_1:21;
  end;
end;

definition
  let i,R1,R2;
  redefine func R1 + R2 -> Element of i-tuples_on REAL;
  coherence by FINSEQ_2:120;
end;

theorem
  (R1+R2).s = R1.s + R2.s
proof
  per cases;
  suppose
A1: not s in Seg i;
    then
A2: not s in dom R2 by FINSEQ_2:124;
A3: not s in dom R1 by A1,FINSEQ_2:124;
    not s in dom(R1+R2) by A1,FINSEQ_2:124;
    hence (R1+R2).s = (0 qua Element of NAT)+(0 qua Element of NAT) by
FUNCT_1:def 2
      .= R1.s + (0 qua Element of NAT) by A3,FUNCT_1:def 2
      .= R1.s + R2.s by A2,FUNCT_1:def 2;
  end;
  suppose
    s in Seg i;
    then s in dom (R1 + R2) by FINSEQ_2:124;
    hence thesis by VALUED_1:def 1;
  end;
end;

theorem
  <*>REAL + F = <*>REAL
proof
  F is FinSequence of REAL by Lm2;
  hence thesis by FINSEQ_2:73;
end;

theorem
  <*r1*> + <*r2*> = <*r1+r2*>
proof
  reconsider s1=r1, s2=r2 as Element of REAL by XREAL_0:def 1;
  <*s1*> + <*s2*> = <*addreal.(s1,s2)*> by FINSEQ_2:74
    .= <*r1+r2*> by BINOP_2:def 9;
  hence thesis;
end;

theorem
  (i|->r1) + (i|->r2) = i|->(r1+r2)
proof
  reconsider s1=r1, s2=r2 as Element of REAL by XREAL_0:def 1;
  (i|->s1) + (i|->s2) = i|->(addreal.(s1,s2)) by FINSEQOP:17
    .= i|->(r1+r2) by BINOP_2:def 9;
  hence thesis;
end;

theorem
  F1 + (F2 + F3) = F1 + F2 + F3 by RFUNCT_1:8;

theorem
  R + (i|->(0 qua Real)) = R by BINOP_2:2,FINSEQOP:56;

theorem
  (-F).s = -F.s by VALUED_1:8;

definition
  let F;
  redefine func -F -> FinSequence of REAL equals
  compreal*F;
  coherence by Lm2;
  compatibility
  proof
    let F1 be FinSequence of REAL;
A1: dom(-F) = dom F by VALUED_1:8;
A2: rng F c= REAL & dom compreal = REAL by FUNCT_2:def 1;
    then
A3: dom(compreal*F) = dom F by RELAT_1:27;
    thus F1 = -F implies F1 = compreal*F
    proof
      assume
A4:   F1 = -F;
      now
        let c be set;
        assume
A5:     c in dom F1;
        thus F1.c = -(F.c) by A4,VALUED_1:8
          .= compreal.(F.c) by BINOP_2:def 7
          .= (compreal*F).c by A1,A4,A5,FUNCT_1:13;
      end;
      hence thesis by A3,A4,FUNCT_1:2,VALUED_1:8;
    end;
    assume
A6: F1 = compreal*F;
    now
      let c be set;
      assume
A7:   c in dom F1;
      thus (-F).c = -(F.c) by VALUED_1:8
        .= compreal.(F.c) by BINOP_2:def 7
        .= (compreal*F).c by A6,A7,FUNCT_1:12;
    end;
    hence thesis by A1,A2,A6,FUNCT_1:2,RELAT_1:27;
  end;
end;

definition
  let i,R;
  redefine func -R -> Element of i-tuples_on REAL;
  coherence by FINSEQ_2:113;
end;

theorem
  (-F).s = -(F.s) by VALUED_1:8;

theorem
  -(<*>REAL) = <*>REAL;

theorem
  -<*r*> = <*-r*>
proof
  reconsider s=r as Element of REAL by XREAL_0:def 1;
  -<*s*> = <*compreal.s*> by FINSEQ_2:35
    .= <*-r*> by BINOP_2:def 7;
  hence thesis;
end;

theorem Th21:
  -(i|->r) = i|->-r
proof
  reconsider s=r as Element of REAL by XREAL_0:def 1;
  -(i|->s) = i|->(compreal.s) by FINSEQOP:16
    .= i|->-r by BINOP_2:def 7;
  hence thesis;
end;

theorem
  R + -R = i|->0 by Th8,Th9,BINOP_2:2,FINSEQOP:73;

theorem
  R1 + R2 = i|->0 implies R1 = -R2 by Th8,Th9,BINOP_2:2,FINSEQOP:74;

theorem
  for R1, R2 being complex-valued Function st -R1 = -R2 holds R1 = R2
proof
  let R1, R2 be complex-valued Function;
  assume -R1 = -R2;
  hence R1 = --R2 .= R2;
end;

theorem
  R1 + R = R2 + R implies R1 = R2
proof
  assume R1 + R = R2 + R;
  then R1 + (R + -R)= (R2 + R)+-R by FINSEQOP:28;
  then
A1: R1 + (R + -R)= R2 + (R + -R) by FINSEQOP:28;
  R + -R = i|->0 by Th8,Th9,BINOP_2:2,FINSEQOP:73;
  then R1 = R2 + (i|->(0 qua Real)) by A1,BINOP_2:2,FINSEQOP:56;
  hence thesis by BINOP_2:2,FINSEQOP:56;
end;

theorem Th26:
  -(F1 + F2) = -F1 + -F2
proof
A1: dom -(F1 + F2) = dom(F1+F2) by VALUED_1:8;
A2: dom(F1+F2) = dom F1 /\ dom F2 by VALUED_1:def 1;
A3: dom (-F1 + -F2) = dom (-F1) /\ dom -F2 by VALUED_1:def 1
    .= dom F1 /\ dom -F2 by VALUED_1:8
    .= dom F1 /\ dom F2 by VALUED_1:8;
  now
    let i;
    assume
A4: i in dom -(F1+F2);
    thus (-(F1 + F2)).i = -((F1+F2).i) by VALUED_1:8
    .= -(F1.i+F2.i) by A1,A4,VALUED_1:def 1
    .= -(F1.i)+-(F2.i)
    .= -(F1.i) + (-F2).i by VALUED_1:8
    .= (-F1).i + (-F2).i by VALUED_1:8
    .= (-F1 + -F2).i by A1,A2,A3,A4,VALUED_1:def 1;
  end;
  hence thesis by A2,A3,FINSEQ_1:13,VALUED_1:8;
end;

definition
  let F1,F2;
  redefine func F1 - F2 -> FinSequence of REAL equals
  diffreal.:(F1,F2);
  coherence by Lm2;
  compatibility
  proof
    reconsider F3=F1,F4=F2 as FinSequence of REAL by Lm2;
    let F be FinSequence of REAL;
A1: dom (F1-F2) = dom F1 /\ dom F2 by VALUED_1:12;
    dom diffreal = [:REAL,REAL:] by FUNCT_2:def 1;
    then
A2: [:rng F3, rng F4:] c= dom diffreal by ZFMISC_1:96;
    then
A3: dom(diffreal.:(F1,F2)) = dom F1 /\ dom F2 by FUNCOP_1:69;
    thus F = F1-F2 implies F = diffreal.:(F1,F2)
    proof
      assume
A4:   F = F1-F2;
      for z being set st z in dom(diffreal.:(F1,F2)) holds F.z = diffreal.
      (F1.z,F2.z)
      proof
        let z be set;
        assume z in dom(diffreal.:(F1,F2));
        hence F.z = F1.z - F2.z by A1,A3,A4,VALUED_1:13
          .= diffreal.(F1.z,F2.z) by BINOP_2:def 10;
      end;
      hence thesis by A1,A3,A4,FUNCOP_1:21;
    end;
    assume
A5: F = diffreal.:(F1,F2);
    now
      let c be set;
      assume c in dom F;
      hence F.c = diffreal.(F1.c,F2.c) by A5,FUNCOP_1:22
        .= F1.c - F2.c by BINOP_2:def 10;
    end;
    hence thesis by A1,A2,A5,FUNCOP_1:69,VALUED_1:14;
  end;
end;

definition
  let i,R1,R2;
  redefine func R1 - R2 -> Element of i-tuples_on REAL;
  coherence by FINSEQ_2:120;
end;

theorem
  (R1-R2).s = R1.s - R2.s
proof
  per cases;
  suppose
A1: not s in Seg i;
    then
A2: not s in dom R2 by FINSEQ_2:124;
A3: not s in dom R1 by A1,FINSEQ_2:124;
    not s in dom(R1-R2) by A1,FINSEQ_2:124;
    hence (R1-R2).s = (0 qua Element of NAT)-(0 qua Element of NAT) by
FUNCT_1:def 2
      .= R1.s - (0 qua Element of NAT) by A3,FUNCT_1:def 2
      .= R1.s - R2.s by A2,FUNCT_1:def 2;
  end;
  suppose s in Seg i;
    then s in dom (R1 - R2) by FINSEQ_2:124;
    hence thesis by VALUED_1:13;
  end;
end;

theorem
  <*>REAL - F = <*>REAL & F - <*>REAL = <*>REAL
proof
  F is FinSequence of REAL by Lm2;
  hence thesis by FINSEQ_2:73;
end;

theorem
  <*r1*> - <*r2*> = <*r1-r2*>
proof
  reconsider s1=r1, s2=r2 as Element of REAL by XREAL_0:def 1;
  <*s1*> - <*s2*> = <*diffreal.(s1,s2)*> by FINSEQ_2:74
    .= <*r1-r2*> by BINOP_2:def 10;
  hence thesis;
end;

theorem
  (i|->r1) - (i|->r2) = i|->(r1-r2)
proof
  reconsider s1=r1, s2=r2 as Element of REAL by XREAL_0:def 1;
  (i|->s1) - (i|->s2) = i|->(diffreal.(s1,s2)) by FINSEQOP:17
    .= i|->(r1-r2) by BINOP_2:def 10;
  hence thesis;
end;

theorem
  F1 - F2 = F1 + - F2;

theorem
  R - (i|->(0 qua Real)) = R
proof
  thus R - (i|->(0 qua Real)) = R + (i|->(-(0 qua Element of NAT))) by Th21
    .= R by BINOP_2:2,FINSEQOP:56;
end;

theorem
  (i|->(0 qua Real)) - R = -R by BINOP_2:2,FINSEQOP:56;

theorem
  F1 - -F2 = F1 + F2;

theorem
  -(F1 - F2) = F2 - F1
proof
  thus -(F1 - F2) = -F1 + --F2 by Th26
    .= F2 - F1;
end;

theorem
  -(F1 - F2) = -F1 + F2
proof
  thus -(F1 - F2) = -F1 +--F2 by Th26
    .= -F1 + F2;
end;

theorem
  R - R = i|->0 by Th8,Th9,BINOP_2:2,FINSEQOP:73;

theorem
  R1 - R2 = i|->0 implies R1 = R2
proof
  assume R1 - R2 = i|->0;
  then R1 = --R2 by Th8,Th9,BINOP_2:2,FINSEQOP:74;
  hence thesis;
end;

theorem
  F1 - F2 - F3 = F1 - (F2 + F3) by RFUNCT_1:20;

theorem
  F1 + (F2 - F3) = F1 + F2 - F3 by RFUNCT_1:23;

theorem
  F1 - (F2 - F3) = F1 - F2 + F3 by RFUNCT_1:22;

theorem
  R1 = R1 + R - R
proof
  thus R1 = R1 + (i|->(0 qua Real)) by BINOP_2:2,FINSEQOP:56
    .= R1 + (R - R) by Th8,Th9,BINOP_2:2,FINSEQOP:73
    .= R1 + R - R by RFUNCT_1:23;
end;

theorem
  R1 = R1 - R + R
proof
  thus R1 = R1 + (i|->(0 qua Real)) by BINOP_2:2,FINSEQOP:56
    .= R1 + (-R + R) by Th8,Th9,BINOP_2:2,FINSEQOP:73
    .= R1 - R + R by FINSEQOP:28;
end;

notation
  let F,r;
  synonym r*F for r(#)F;
end;

theorem
  (r*F).s = r*(F.s) by VALUED_1:6;

definition
  let F,r;
  redefine func r*F -> FinSequence of REAL equals
  (r multreal)*F;
  coherence by Lm2;
  compatibility
  proof
    let F1 be FinSequence of REAL;
A1: dom(r*F) = dom F by VALUED_1:def 5;
A2: rng F c= REAL & dom(r multreal) = REAL by FUNCT_2:def 1;
    then
A3: dom((r multreal)*F) = dom F by RELAT_1:27;
    thus F1 = r*F implies F1 = (r multreal)*F
    proof
      assume
A4:   F1 = r*F;
      now
        let c be set;
        assume
A5:     c in dom F1;
        hence F1.c = r*(F.c) by A4,VALUED_1:def 5
          .= (r multreal).(F.c) by Lm1
          .= ((r multreal)*F).c by A1,A4,A5,FUNCT_1:13;
      end;
      hence thesis by A1,A3,A4,FUNCT_1:2;
    end;
    assume
A6: F1 = (r multreal)*F;
    now
      let c be set;
      assume
A7:   c in dom F1;
      thus (r*F).c = r*(F.c) by VALUED_1:6
        .= (r multreal).(F.c) by Lm1
        .= ((r multreal)*F).c by A6,A7,FUNCT_1:12;
    end;
    hence thesis by A1,A2,A6,FUNCT_1:2,RELAT_1:27;
  end;
end;

definition
  let i,R,r;
  redefine func r*R -> Element of i-tuples_on REAL;
  coherence by FINSEQ_2:113;
end;

theorem
  (r*F).s = r*(F.s) by VALUED_1:6;

theorem
  r*(<*>REAL) = <*>REAL;

theorem
  r*<*r1*> = <*r*r1*>
proof
  reconsider s=r, s1=r1 as Element of REAL by XREAL_0:def 1;
  s*<*s1*> = <*(multreal[;](s,id REAL)).s1*> by FINSEQ_2:35
    .= <*r*r1*> by Lm1;
  hence thesis;
end;

theorem Th48:
  r1*(i|->r2) = i|->(r1*r2)
proof
  reconsider s1=r1, s2=r2 as Element of REAL by XREAL_0:def 1;
  s1*(i|->s2) = i|->((multreal[;](s1,id REAL)).s2) by FINSEQOP:16
    .= i|->(r1*r2) by Lm1;
  hence thesis;
end;

theorem
  (r1*r2)*F = r1*(r2*F) by RFUNCT_1:17;

theorem
  (r1 + r2)*F = r1*F + r2*F
proof
A1: dom ((r1 + r2)*F) = dom F by VALUED_1:def 5;
A2: dom (r1*F + r2*F) = dom (r1*F) /\ dom (r2*F) by VALUED_1:def 1;
A3: dom (r1*F) = dom F by VALUED_1:def 5;
A4: dom (r2*F) = dom F by VALUED_1:def 5;
    now
      let i;
      assume
A5:   i in dom ((r1+r2)*F);
      thus ((r1+r2)*F).i = (r1+r2)*(F.i) by VALUED_1:6
      .= r1*(F.i) + r2*(F.i)
      .= r1*(F.i) + (r2*F).i by VALUED_1:6
      .= (r1*F).i + (r2*F).i by VALUED_1:6
      .= (r1*F+r2*F).i by A1,A2,A3,A4,A5,VALUED_1:def 1;
    end;
    hence thesis by A1,A2,A3,A4,FINSEQ_1:13;
end;

theorem
  r*(F1+F2) = r*F1 + r*F2 by RFUNCT_1:16;

theorem
  1*F = F by RFUNCT_1:21;

theorem
  0*R = i|->0
proof
A1: rng R c= REAL;
  thus 0*R = multreal[;](0,(id REAL)*R) by FUNCOP_1:34
    .= multreal[;](0,R) by A1,RELAT_1:53
    .= i|->0 by Th3,Th8,BINOP_2:2,FINSEQOP:76;
end;

theorem
  (-1)*F = -F;

notation
  let F;
  synonym sqr F for F^2;
end;

definition
  let F;
  redefine func sqr F -> FinSequence of REAL equals
  sqrreal*F;
  compatibility
  proof
    let f be FinSequence of REAL;
    sqr F = sqrreal*F
    proof
      dom sqrreal = REAL by FUNCT_2:def 1;
      then
A1:   rng F c= dom sqrreal;
A2:   dom sqr F = dom F by VALUED_1:11
        .= dom(sqrreal*F) by A1,RELAT_1:27;
      hence len sqr F = len(sqrreal*F) by FINSEQ_3:29;
      let k;
      assume 1 <= k & k <= len sqr F;
      then
A3:   k in dom sqr F by FINSEQ_3:25;
      thus (sqr F).k = (F.k)^2 by VALUED_1:11
        .= sqrreal.(F.k) by Def2
        .= (sqrreal*F).k by A2,A3,FUNCT_1:12;
    end;
    hence thesis;
  end;
  coherence by Lm2;
end;

definition
  let i,R;
  redefine func sqr R -> Element of i-tuples_on REAL;
  coherence by FINSEQ_2:113;
end;

theorem
  sqr <*r*> = <*r^2*>
proof
  reconsider s=r as Element of REAL by XREAL_0:def 1;
  sqr <*s*> = <*sqrreal.s*> by FINSEQ_2:35
    .= <*r^2*> by Def2;
  hence thesis;
end;

theorem
  sqr(i |-> r) = i |-> r^2
proof
  reconsider s=r as Element of REAL by XREAL_0:def 1;
  sqr(i |-> s) = i |-> (sqrreal.s) by FINSEQOP:16
    .= i |-> r^2 by Def2;
  hence thesis;
end;

theorem Th57:
  sqr -F = sqr F
proof
A1: dom sqr -F = dom -F by VALUED_1:11
    .= dom F by VALUED_1:8;
A2: dom sqr F = dom F by VALUED_1:11;
  now
    let j;
    assume j in dom sqr -F;
    set r = F.j, r9 = (-F).j;
    thus (sqr -F).j = (r9)^2 by VALUED_1:11
      .= (-r)^2 by VALUED_1:8
      .= r^2
      .= (sqr F).j by VALUED_1:11;
  end;
  hence thesis by A1,A2,FINSEQ_1:13;
end;

theorem Th58:
  sqr (r*F) = r^2 * sqr F
proof
A1: dom (r*F) = dom F by VALUED_1:def 5;
A2: dom (r^2 * sqr F) = dom sqr F by VALUED_1:def 5;
A3: dom sqr F = dom F by VALUED_1:11;
    now
      let i;
      assume i in dom sqr (r*F);
      thus (sqr (r*F)).i = ((r*F).i)^2 by VALUED_1:11
      .= (r*(F.i))^2 by VALUED_1:6
      .= r^2 * (F.i)^2
      .= r^2 * (sqr F).i by VALUED_1:11
      .= (r^2 * sqr F).i by VALUED_1:6;
    end;
    hence thesis by A1,A2,A3,FINSEQ_1:13,VALUED_1:11;
  end;

notation
  let F1,F2;
  synonym mlt(F1,F2) for F1(#)F2;
end;

definition
  let F1,F2;
  redefine func mlt(F1,F2) -> FinSequence of REAL equals
  multreal.:(F1,F2);
  coherence by Lm2;
  compatibility
  proof
    reconsider F3=F1,F4=F2 as FinSequence of REAL by Lm2;
    let F be FinSequence of REAL;
    dom multreal = [:REAL,REAL:] by FUNCT_2:def 1;
    then [:rng F3, rng F4:] c= dom multreal by ZFMISC_1:96;
    then
A1: dom(multreal.:(F1,F2)) = dom F1 /\ dom F2 by FUNCOP_1:69;
A2: dom mlt(F1,F2) = dom F1 /\ dom F2 by VALUED_1:def 4;
    thus F = mlt(F1,F2) implies F = multreal.:(F1,F2)
    proof
      assume
A3:   F = mlt(F1,F2);
      for z being set st z in dom(multreal.:(F1,F2)) holds F.z = multreal.
      (F1.z,F2.z)
      proof
        let z be set;
        assume z in dom(multreal.:(F1,F2));
        thus F.z = F1.z * F2.z by A3,VALUED_1:5
          .= multreal.(F1.z,F2.z) by BINOP_2:def 11;
      end;
      hence thesis by A2,A1,A3,FUNCOP_1:21;
    end;
    assume
A4: F = multreal.:(F1,F2);
    now
      let c be set;
      assume c in dom F;
      hence F.c = multreal.(F1.c,F2.c) by A4,FUNCOP_1:22
        .= F1.c * F2.c by BINOP_2:def 11;
    end;
    hence thesis by A1,A4,VALUED_1:def 4;
  end;
  commutativity
  proof
    let F be FinSequence of REAL;
    let F1,F2;
    reconsider F3=F1,F4=F2 as FinSequence of REAL by Lm2;
A5: dom multreal = [:REAL,REAL:] by FUNCT_2:def 1;
    then
A6: [:rng F4, rng F3:] c= dom multreal by ZFMISC_1:96;
    [:rng F3, rng F4:] c= dom multreal by A5,ZFMISC_1:96;
    then
A7: dom(multreal.:(F1,F2)) = dom F1 /\ dom F2 by FUNCOP_1:69
      .= dom(multreal.:(F2,F1)) by A6,FUNCOP_1:69;
    assume
A8: F = multreal.:(F1,F2);
    for z being set st z in dom(multreal.:(F2,F1)) holds F.z = multreal.(
    F2.z,F1.z)
    proof
      let z be set such that
A9:   z in dom(multreal.:(F2,F1));
      set F1z = F1.z, F2z =F2.z;
      thus F.z = multreal.(F1.z,F2.z) by A8,A7,A9,FUNCOP_1:22
        .= F1z * F2z by BINOP_2:def 11
        .= multreal.(F2.z,F1.z) by BINOP_2:def 11;
    end;
    hence thesis by A8,A7,FUNCOP_1:21;
  end;
end;

theorem
  mlt(F1,F2).s = F1.s * F2.s by VALUED_1:5;

definition
  let i,R1,R2;
  redefine func mlt(R1,R2) -> Element of i-tuples_on REAL;
  coherence by FINSEQ_2:120;
end;

theorem 
  mlt(F1,F2).s = F1.s * F2.s by VALUED_1:5;

theorem
  mlt(<*>REAL,F) = <*>REAL
proof
  F is FinSequence of REAL by Lm2;
  hence thesis by FINSEQ_2:73;
end;

theorem
  mlt(<*r1*>,<*r2*>) = <*r1*r2*>
proof
  reconsider s1=r1, s2=r2 as Element of REAL by XREAL_0:def 1;
  mlt(<*s1*>,<*s2*>) = <*multreal.(s1,s2)*> by FINSEQ_2:74
    .= <*r1*r2*> by BINOP_2:def 11;
  hence thesis;
end;

theorem Th63:
  mlt(i|->r,R) = r*R
proof
  reconsider s=r as Element of REAL by XREAL_0:def 1;
  mlt(i|->s,R) = multreal[;](s,R) by FINSEQOP:20
    .= r*R by FINSEQOP:22;
  hence thesis;
end;

theorem
  mlt(i|->r1,i|->r2) = i|->(r1*r2)
proof
  reconsider s1=r1, s2=r2 as Element of REAL by XREAL_0:def 1;
  mlt(i|->s1,i|->s2) = s1*(i|->s2) by Th63
    .= i|->(r1*r2) by Th48;
  hence thesis;
end;

theorem
  r*mlt(F1,F2) = mlt(r*F1,F2) by RFUNCT_1:12;

theorem 
  r*R = mlt(i|->r,R) by Th63;

theorem
  sqr(F) = mlt(F,F);

theorem Th68:
  sqr(F1 + F2) = sqr F1 + 2*mlt(F1,F2) + sqr F2
proof
A1: dom sqr(F1 + F2) = dom (F1+F2) by VALUED_1:11;
A2: dom(F1+F2) = dom F1 /\ dom F2 by VALUED_1:def 1;
A3: dom(sqr F1 + 2*mlt(F1,F2))
    = dom(sqr F1) /\ dom (2*mlt(F1,F2)) by VALUED_1:def 1
    .= dom F1 /\ dom (2*mlt(F1,F2)) by VALUED_1:11
    .= dom F1 /\ dom (mlt(F1,F2)) by VALUED_1:def 5
    .= dom F1 /\ (dom F1 /\ dom F2) by VALUED_1:def 4
    .= dom F1 /\ dom F1 /\ dom F2 by XBOOLE_1:16
    .= dom F1 /\ dom F2;
then A4: dom(sqr F1 + 2*mlt(F1,F2) + sqr F2)
    = dom F1 /\ dom F2 /\ dom sqr F2 by VALUED_1:def 1
    .= dom F1 /\ dom F2 /\ dom F2 by VALUED_1:11
    .= dom F1 /\ (dom F2 /\ dom F2) by XBOOLE_1:16;
  now
    let j;
    assume
A5: j in dom sqr(F1+F2);
    set r1r2 = (F1 + F2).j, r1 = F1.j, r2 = F2.j, r192 = (sqr F1).j,
    r292 = (sqr F2).j, r1r2a = mlt(F1,F2).j, 2r1r2 = (2*mlt(F1,F2)).j,
    r1922r1r2 = (sqr F1 + 2*mlt(F1,F2)).j;
    thus (sqr(F1 + F2)).j = r1r2^2 by VALUED_1:11
      .= (r1 + r2)^2 by A1,A5,VALUED_1:def 1
      .= r1^2+2*r1*r2+r2^2
      .= r192+2*(r1*r2)+r2^2 by VALUED_1:11
      .= r192+2*(r1*r2)+r292 by VALUED_1:11
      .= r192+2*(r1r2a)+r292 by VALUED_1:5
      .= r192+2r1r2+r292 by VALUED_1:6
      .= r1922r1r2+r292 by A1,A2,A3,A5,VALUED_1:def 1
      .= (sqr F1 + 2*mlt(F1,F2) + sqr F2).j by A1,A2,A4,A5,VALUED_1:def 1;
  end;
  hence thesis by A2,A4,FINSEQ_1:13,VALUED_1:11;
end;

theorem Th69:
  sqr(F1 - F2) = sqr F1 - 2*mlt(F1,F2) + sqr F2
proof
  thus sqr(F1 - F2) = sqr F1 + 2*mlt(F1,-F2) + sqr -F2 by Th68
    .= sqr F1 + 2*mlt(F1,-F2) + sqr F2 by Th57
    .= sqr F1 + 2*((-1)*mlt(F1,F2)) + sqr F2 by RFUNCT_1:12
    .= sqr F1 + ((-1)*2)*mlt(F1,F2) + sqr F2 by RFUNCT_1:17
    .= sqr F1 - 2*mlt(F1,F2) + sqr F2 by RFUNCT_1:17;
end;

theorem
  sqr mlt(F1,F2) = mlt(sqr F1,sqr F2)
proof
A1: dom mlt(F1,F2) = dom F1 /\ dom F2 by VALUED_1:def 4;
A2: dom mlt(sqr F1,sqr F2) = dom sqr F1 /\ dom sqr F2 by VALUED_1:def 4;
A3: dom sqr F1 = dom F1 by VALUED_1:11;
A4: dom sqr F2 = dom F2 by VALUED_1:11;
    now
      let i;
      assume i in dom sqr mlt(F1,F2);
      thus (sqr mlt(F1,F2)).i = (mlt(F1,F2).i)^2 by VALUED_1:11
      .= (F1.i*F2.i)^2 by VALUED_1:5
      .= (F1.i)^2*(F2.i)^2
      .= (sqr F1).i*(F2.i)^2 by VALUED_1:11
      .= (sqr F1).i * (sqr F2).i by VALUED_1:11
      .= (mlt(sqr F1,sqr F2)).i by VALUED_1:5;
    end;
    hence thesis by A1,A2,A3,A4,FINSEQ_1:13,VALUED_1:11;
  end;



notation
  let F be Relation;
  synonym F is complex-yielding for F is complex-valued;
end;

registration
  cluster -> complex-valued for FinSequence of COMPLEX;
  coherence;
  cluster real-valued complex-valued for FinSequence;
  existence
  proof
    <*>REAL is real-valued;
    hence thesis;
  end;
end;

definition
  let F be complex-valued FinSequence;

  func Sum F -> complex number means
  :Def10:
  ex f being FinSequence of COMPLEX st f = F & it = addcomplex $$ f;
  existence
  proof
    rng F c= COMPLEX by VALUED_0:def 1;
    then reconsider f = F as FinSequence of COMPLEX by FINSEQ_1:def 4;
    take addcomplex $$ f;
    thus thesis;
  end;
  uniqueness;
end;

registration
  let F be real-valued FinSequence;
  cluster Sum F -> real;
  coherence
  proof
    set mc = addcomplex;
    consider f being FinSequence of COMPLEX such that
A1: f = F and
A2: Sum F = addcomplex $$ f by Def10;
    set g = [#](f,the_unity_wrt mc);
    defpred P[Nat] means mc $$ (finSeg $1,[#](f,the_unity_wrt mc)) is real;
A3: for k being Nat st P[k] holds P[k+1]
    proof
      let k be Nat;
      assume
A4:   P[k];
      reconsider k as Element of NAT by ORDINAL1:def 12;
      g.(k+1) is real
      proof
        per cases;
        suppose
          k+1 in dom f;
          then g.(k+1) = f.(k+1) by SETWOP_2:20;
          hence thesis by A1;
        end;
        suppose
          not k+1 in dom f;
          hence thesis by BINOP_2:1,SETWOP_2:20;
        end;
      end;
      then reconsider a = g.(k+1), b = mc $$(finSeg k,g) as real number by A4;
      not (k + 1) in Seg k by FINSEQ_3:8;
      then mc $$ (finSeg k \/ {.k+1.},g) = mc.(mc $$(finSeg k,g),g.(k+1)) by
SETWOP_2:2
        .= b + a by BINOP_2:def 3;
      hence thesis by FINSEQ_1:9;
    end;
A5: mc $$ f = mc $$ (findom f,[#](f,the_unity_wrt mc)) & ex n being Nat st
    dom f = Seg n by FINSEQ_1:def 2,SETWOP_2:def 2;
    Seg 0 = {}.NAT;
    then
A6: P[0] by BINOP_2:1,SETWISEO:31;
    for n being Nat holds P[n] from NAT_1:sch 2(A6,A3);
    hence thesis by A2,A5;
  end;
end;

theorem Th71:
  for F being FinSequence of REAL holds Sum F = addreal $$ F
proof
  set g = addreal, h = addcomplex;
  let F be FinSequence of REAL;
  rng F c= COMPLEX by NUMBERS:11,XBOOLE_1:1;
  then reconsider f = F as FinSequence of COMPLEX by FINSEQ_1:def 4;
  defpred P[Nat] means g $$ (finSeg $1,[#](F,the_unity_wrt g)) = h $$ (finSeg
  $1,[#](f,the_unity_wrt h));
  consider n being Nat such that
A1: dom f = Seg n by FINSEQ_1:def 2;
A2: g $$ F = g $$ (finSeg n,[#](F,the_unity_wrt g)) & h $$ f = h $$ (finSeg
  n, [#](f,the_unity_wrt h)) by A1,SETWOP_2:def 2;
A3: for k being Nat st P[k] holds P[k+1]
  proof
    set j = [#](f,the_unity_wrt h);
    set i = [#](F,the_unity_wrt g);
    let k be Nat;
    assume
A4: P[k];
    reconsider k as Element of NAT by ORDINAL1:def 12;
A5: i.(k+1) = j.(k+1)
    proof
      per cases;
      suppose
A6:     k+1 in dom f;
        then j.(k+1) = F.(k+1) by SETWOP_2:20
          .= i.(k+1) by A6,SETWOP_2:20;
        hence thesis;
      end;
      suppose
A7:     not k+1 in dom f;
        then j.(k+1) = the_unity_wrt h by SETWOP_2:20
          .= i.(k+1) by A7,BINOP_2:1,2,SETWOP_2:20;
        hence thesis;
      end;
    end;
A8: not (k + 1) in Seg k by FINSEQ_3:8;
    g $$ (finSeg (k+1),i) = g $$ (finSeg k \/ {.k+1.},i) by FINSEQ_1:9
      .= g.(g $$(finSeg k,i),i.(k+1)) by A8,SETWOP_2:2
      .= g $$(finSeg k,i) + (i.(k+1)) by BINOP_2:def 9
      .= h.(h $$(finSeg k,j),j.(k+1)) by A4,A5,BINOP_2:def 3
      .= h $$ (finSeg k \/ {.k+1.},j) by A8,SETWOP_2:2
      .= h $$ (finSeg (k+1),j) by FINSEQ_1:9;
    hence thesis;
  end;
A9: Seg 0 = {}.NAT;
  then g $$ (finSeg 0,[#](F,the_unity_wrt g)) = the_unity_wrt h by BINOP_2:1,2
,SETWISEO:31
    .= h $$ (finSeg 0,[#](f,the_unity_wrt h)) by A9,SETWISEO:31;
  then
A10: P[0];
  for k being Nat holds P[k] from NAT_1:sch 2(A10,A3);
  then g $$ F = h $$ f by A2;
  hence thesis by Def10;
end;

definition
  let F be FinSequence of COMPLEX;
  redefine func Sum F -> Element of COMPLEX equals
  addcomplex $$ F;
  coherence by XCMPLX_0:def 2;
  compatibility by Def10;
end;

definition
  let F be FinSequence of REAL;
  redefine func Sum F -> Element of REAL equals
  addreal $$ F;
  coherence by XREAL_0:def 1;
  compatibility by Th71;
end;

theorem Th72:
  Sum(<*> REAL) = 0 by BINOP_2:2,FINSOP_1:10;

theorem
  Sum <*r*> = r
proof
  reconsider r as Element of REAL by XREAL_0:def 1;
  Sum <*r*> = r by FINSOP_1:11;
  hence thesis;
end;

theorem Th74:
  Sum(F^<*r*>) = Sum F + r
proof
  reconsider s=r as Element of REAL by XREAL_0:def 1;
  reconsider F1=F as FinSequence of REAL by Lm2;
  thus Sum(F^<*r*>) = Sum(F1^<*s*>) .= addreal.(addreal $$ F1,s) by FINSOP_1:4
    .= Sum F1 + r by BINOP_2:def 9
    .= Sum F + r;
end;

theorem Th75:
  Sum(F1^F2) = Sum F1 + Sum F2
proof
  reconsider F3=F1,F4=F2 as FinSequence of REAL by Lm2;
  thus Sum(F1^F2) = Sum(F3^F4)
    .= addreal.(addreal $$ F3,addreal $$ F4) by FINSOP_1:5
    .= Sum F3 + Sum F4 by BINOP_2:def 9
    .= Sum F1 + Sum F2;
end;

theorem
  Sum(<*r*>^F) = r + Sum F
proof
  reconsider s=r as Element of REAL by XREAL_0:def 1;
  thus Sum(<*r*>^F) = Sum <*s*> + Sum F by Th75
    .= r + Sum F by FINSOP_1:11;
end;

theorem Th77:
  Sum<*r1,r2*> = r1 + r2
proof
  reconsider s1=r1 as Element of REAL by XREAL_0:def 1;
  thus Sum<*r1,r2*> = Sum<*s1*> + r2 by Th74
    .= r1 + r2 by FINSOP_1:11;
end;

theorem Th78:
  Sum<*r1,r2,r3*> = r1 + r2 + r3
proof
  thus Sum<*r1,r2,r3*> = Sum<*r1,r2*> + r3 by Th74
    .= r1 + r2 + r3 by Th77;
end;

theorem
  for R being Element of 0-tuples_on REAL holds Sum R = 0 by Th72;

theorem Th80:
  Sum(i |-> r) = i*r
proof
  reconsider r as Element of REAL by XREAL_0:def 1;
  defpred P[Nat] means Sum($1 |->r) = $1*r;
A1: for i st P[i] holds P[(i+1)]
  proof
    let i such that
A2: Sum(i |-> r) = i*r;
    thus Sum((i+1) |-> r) = Sum((i |-> r)^<*r*>) by FINSEQ_2:60
      .= i*r + 1*r by A2,Th74
      .= (i+1)*r;
  end;
A3: P[0] by Th72;
  for i holds P[i] from NAT_1:sch 2(A3,A1);
  hence thesis;
end;

theorem Th81:
  Sum(i |-> (0 qua Real)) = 0
proof
  thus Sum(i |-> (0 qua Real)) = i*(0 qua Element of NAT) by Th80
    .= 0;
end;

theorem Th82:
  (for j st j in Seg i holds R1.j <= R2.j) implies Sum R1 <= Sum R2
proof
  defpred P[Nat] means for R1,R2 being Element of $1-tuples_on REAL st for j
  st j in Seg $1 holds R1.j <= R2.j holds Sum R1 <= Sum R2;
A1: for i st P[i] holds P[i+1]
  proof
    let i such that
A2: for R1,R2 being Element of i-tuples_on REAL st for j st j in Seg i
    holds R1.j <= R2.j holds Sum R1 <= Sum R2;
    set n = i+1;
    let R1,R2 be Element of (i+1)-tuples_on REAL such that
A3: for j st j in Seg (i+1) holds R1.j <= R2.j;
    consider R19 being (Element of i-tuples_on REAL), x1 such that
A4: R1 = R19^<*x1*> by FINSEQ_2:117;
    consider R29 being (Element of i-tuples_on REAL), x2 such that
A5: R2 = R29^<*x2*> by FINSEQ_2:117;
    for j st j in Seg i holds R19.j <= R29.j
    proof
      let j such that
A6:   j in Seg i;
      Seg len R29 = dom R29 & len R29 = i by CARD_1:def 7,FINSEQ_1:def 3;
      then
A7:   R29.j = R2.j by A5,A6,FINSEQ_1:def 7;
      Seg len R19 = dom R19 & len R19 = i by CARD_1:def 7,FINSEQ_1:def 3;
      then R19.j = R1.j by A4,A6,FINSEQ_1:def 7;
      hence thesis by A3,A6,A7,FINSEQ_2:8;
    end;
    then
A8: Sum R19 <= Sum R29 by A2;
A9: R2.n = x2 by A5,FINSEQ_2:116;
    R1.n = x1 by A4,FINSEQ_2:116;
    then
A10: x1 <= x2 by A3,A9,FINSEQ_1:4;
A11: Sum R2 = Sum R29 + x2 by A5,Th74;
    Sum R1 = Sum R19 + x1 by A4,Th74;
    hence thesis by A10,A8,A11,XREAL_1:7;
  end;
A12: P[0];
  for i holds P[i] from NAT_1:sch 2(A12,A1);
  hence thesis;
end;

theorem Th83:
  (for j st j in Seg i holds R1.j <= R2.j) & (ex j st j in Seg i
  & R1.j < R2.j) implies Sum R1 < Sum R2
proof
  defpred P[Nat] means for R1,R2 be Element of $1-tuples_on REAL st (for j st
j in Seg $1 holds R1.j <= R2.j) & (ex j st j in Seg $1 & R1.j < R2.j) holds Sum
  R1 < Sum R2;
  now
    let i such that
A1: P[i];
    let R1,R2 be Element of (i+1)-tuples_on REAL such that
A2: for j st j in Seg (i+1) holds R1.j <= R2.j;
    given j such that
A3: j in Seg (i+1) and
A4: R1.j < R2.j;
    consider R19 being (Element of i-tuples_on REAL), x1 such that
A5: R1 = R19^<*x1*> by FINSEQ_2:117;
    consider R29 being (Element of i-tuples_on REAL), x2 such that
A6: R2 = R29^<*x2*> by FINSEQ_2:117;
A7: for j st j in Seg i holds R19.j <= R29.j
    proof
      let j such that
A8:   j in Seg i;
      Seg len R29 = dom R29 & len R29 = i by CARD_1:def 7,FINSEQ_1:def 3;
      then
A9:   R29.j = R2.j by A6,A8,FINSEQ_1:def 7;
      Seg len R19 = dom R19 & len R19 = i by CARD_1:def 7,FINSEQ_1:def 3;
      then R19.j = R1.j by A5,A8,FINSEQ_1:def 7;
      hence thesis by A2,A8,A9,FINSEQ_2:8;
    end;
A10: R2.(i+1) = x2 by A6,FINSEQ_2:116;
A11: R1.(i+1) = x1 by A5,FINSEQ_2:116;
    now
      per cases by A3,FINSEQ_2:7;
      suppose
A12:    j in Seg i;
        Seg len R29 = dom R29 & len R29 = i by CARD_1:def 7,FINSEQ_1:def 3;
        then
A13:    R29.j = R2.j by A6,A12,FINSEQ_1:def 7;
A14:    Sum R1 = Sum R19 + x1 & Sum R2 = Sum R29 + x2 by A5,A6,Th74;
A15:    x1 <= x2 by A2,A11,A10,FINSEQ_1:4;
        Seg len R19 = dom R19 & len R19 = i by CARD_1:def 7,FINSEQ_1:def 3;
        then R19.j = R1.j by A5,A12,FINSEQ_1:def 7;
        then Sum R19 < Sum R29 by A1,A4,A7,A12,A13;
        hence Sum R1 < Sum R2 by A14,A15,XREAL_1:8;
      end;
      suppose
A16:    j = i+1;
A17:    Sum R2 = Sum R29 + x2 by A6,Th74;
        Sum R19 <= Sum R29 & Sum R1 = Sum R19 + x1 by A5,A7,Th74,Th82;
        hence Sum R1 < Sum R2 by A4,A11,A10,A16,A17,XREAL_1:8;
      end;
    end;
    hence Sum R1 < Sum R2;
  end;
  then
A18: for i st P[i] holds P[i+1];
A19: P[0];
  for i holds P[i] from NAT_1:sch 2(A19,A18);
  hence thesis;
end;

theorem Th84:
  (for i st i in dom F holds 0 <= F.i) implies 0 <= Sum F
proof
  reconsider F1 = F as FinSequence of REAL by Lm2;
  set i = len F;
  set R1 = i|->(0 qua Real);
  reconsider R2 = F1 as Element of i-tuples_on REAL by FINSEQ_2:92;
A1: Seg len F = dom F by FINSEQ_1:def 3;
  assume
A2: for i st i in dom F holds 0 <= F.i;
  for j st j in Seg i holds R1.j <= R2.j by A2,A1;
  then Sum R1 <= Sum R2 by Th82;
  hence thesis by Th81;
end;

theorem Th85:
  (for i st i in dom F holds 0 <= F.i) & (ex i st i in dom F & 0
  < F.i) implies 0 < Sum F
proof
  reconsider F1 = F as FinSequence of REAL by Lm2;
  set i = len F, R1 = i|->(0 qua Real);
  reconsider R2 = F1 as Element of i-tuples_on REAL by FINSEQ_2:92;
A1: Seg len F = dom F by FINSEQ_1:def 3;
  assume
A2: for i st i in dom F holds 0 <= F.i;
A3: for j st j in Seg i holds R1.j <= R2.j by A2,A1;
  given j such that
A5: j in dom F and
A6: 0 < F.j;
  R1.j = 0;
  then Sum R1 < Sum R2 by A1,A3,A5,A6,Th83;
  hence thesis by Th81;
end;

theorem Th86:
  0 <= Sum sqr F
proof
  now
    let i such that i in dom sqr F;
    0 <= (F.i)^2 by XREAL_1:63;
    hence 0 <= (sqr F).i by VALUED_1:11;
  end;
  hence thesis by Th84;
end;

theorem Th87:
  Sum(r*F) = r*(Sum F)
proof
  reconsider F1=F as FinSequence of REAL by Lm2;
  reconsider s=r as Element of REAL by XREAL_0:def 1;
  set rM = multreal[;](s,id REAL);
  thus Sum (r*F) = rM.(addreal $$ F1) by Th3,Th8,SETWOP_2:30
    .= r*(Sum F1) by Lm1
    .= r*(Sum F);
end;

theorem Th88:
  Sum -F = -(Sum F)
proof
  reconsider F1=F as FinSequence of REAL by Lm2;
  thus Sum -F = compreal.(addreal $$ F1) by Th8,Th9,SETWOP_2:31
    .= -(Sum F1) by BINOP_2:def 7
    .= -(Sum F);
end;

theorem Th89:
  Sum(R1 + R2) = Sum R1 + Sum R2
proof
  thus Sum(R1 + R2) = addreal.(addreal$$R1,addreal$$R2) by SETWOP_2:35
    .= Sum R1 + Sum R2 by BINOP_2:def 9;
end;

theorem Th90:
  Sum(R1 - R2) = Sum R1 - Sum R2
proof
  thus Sum(R1 - R2)= Sum R1 + Sum -R2 by Th89
  .= Sum R1 +- Sum R2 by Th88
  .= Sum R1 - Sum R2;
end;

theorem
  Sum sqr R = 0 implies R = i |-> 0
proof
  assume
A1: Sum sqr R = 0;
A2: len R = i by CARD_1:def 7;
A3: len(i |-> 0) = i by CARD_1:def 7;
  assume R <> i |-> 0;
  then consider j such that
A4: j in dom R and
A5: R.j <> (i |-> 0).j by A2,A3,FINSEQ_2:9;
  set x = R.j,x9 = (sqr R).j;
A6: dom R = Seg len R by FINSEQ_1:def 3;
  x <> 0 by A5;
  then 0 < x^2 by SQUARE_1:12;
  then
A7: 0 < x9 by VALUED_1:11;
A8: now
    let k such that
    k in dom sqr R;
    set r = (sqr R).k;
    set x = R.k;
    0 <= x^2 by XREAL_1:63;
    hence 0 <= r by VALUED_1:11;
  end;
  dom sqr R = Seg len sqr R by FINSEQ_1:def 3;
  then j in dom sqr R by A4,A6,FINSEQ_2:33;
  hence thesis by A1,A8,A7,Th85;
end;

theorem
  (Sum mlt(R1,R2))^2 <= (Sum sqr R1)*(Sum sqr R2)
proof
  defpred P[Nat] means for R1,R2 being Element of $1-tuples_on REAL holds (Sum
  mlt(R1,R2))^2 <= (Sum sqr R1)*(Sum sqr R2);
A1: for i st P[i] holds P[i+1]
  proof
    let i such that
A2: for R1,R2 being Element of i-tuples_on REAL holds (Sum mlt(R1,R2))
    ^2 <= (Sum sqr R1)*(Sum sqr R2);
    let R1,R2 be Element of (i+1)-tuples_on REAL;
    consider R19 being (Element of i-tuples_on REAL), x1 such that
A3: R1 = R19^<*x1*> by FINSEQ_2:117;
    consider R29 being (Element of i-tuples_on REAL), x2 such that
A4: R2 = R29^<*x2*> by FINSEQ_2:117;
A5: Sum sqr (R^<*r*>) = Sum sqr R + r^2
    proof
      reconsider s=r as Element of REAL by XREAL_0:def 1;
      sqr(R^<*s*>) = (sqrreal*R)^<*sqrreal.s*> by FINSEQOP:8
        .= (sqr R)^<*r^2*> by Def2;
      hence thesis by Th74;
    end;
    then
A6: (Sum sqr R29 + x2^2) = Sum sqr R2 by A4;
    (Sum mlt(R19,R29))^2 + (0 qua Element of NAT) <= (Sum sqr R19)*(Sum
    sqr R29) by A2;
    then
A7: 0 <= (Sum sqr R19)*(Sum sqr R29) - (Sum mlt(R19,R29))^2 by XREAL_1:19;
    mlt(R19^<*x1*>,R29^<*x2*>) = (multreal.:(R19,R29))^<*multreal.(x1,x2)
    *> by FINSEQOP:10
      .= (mlt(R19,R29))^<*x1*x2*> by BINOP_2:def 11;
    then
A8: Sum mlt(R19^<*x1*>,R29^<*x2*>) = Sum mlt(R19,R29) + x1*x2 by Th74;
A9: 2*(x1*x2)*Sum mlt(R19,R29) = 2*((x1*x2)*Sum mlt(R19,R29))
      .= 2*Sum((x1*x2)*mlt(R19,R29)) by Th87
      .= 2*Sum(x1*(x2*mlt(R19,R29))) by RFUNCT_1:17
      .= 2*Sum(x1*mlt(R29,x2*R19)) by RFUNCT_1:12
      .= 2*Sum(mlt(x1*R29,x2*R19)) by FINSEQOP:26;
A10: -(Sum mlt(R19,R29)+x1*x2)^2 = -(x1*x2)^2+-(2*(x1*x2)*Sum mlt(R19,R29)
    +(Sum mlt(R19,R29))^2)
      .= -x1^2*x2^2+(-(Sum mlt(R19,R29))^2+ -2*Sum(mlt(x1*R29,x2*R19))) by A9;
A11: 0 <= Sum sqr (x1*R29-x2*R19) by Th86;
A12: (Sum sqr R19 + x1^2)*(Sum sqr R29 + x2^2) = (Sum sqr R19)*(Sum sqr R29
    )+(x1^2*(Sum sqr R29)+(Sum sqr R19)*x2^2)+x1^2*x2^2
      .= (Sum sqr R19)*(Sum sqr R29)+(Sum(x1^2*sqr R29)+(Sum sqr R19)*x2^2)+
    x1^2*x2^2 by Th87
      .= (Sum sqr R19)*(Sum sqr R29)+(Sum sqr(x1*R29)+x2^2*(Sum sqr R19))+x1
    ^2*x2^2 by Th58
      .= (Sum sqr R19)*(Sum sqr R29)+(Sum sqr(x1*R29)+Sum (x2^2*sqr R19))+x1
    ^2*x2^2 by Th87
      .= (Sum sqr R19)*(Sum sqr R29)+(Sum sqr(x1*R29)+Sum sqr(x2*R19))+x1^2*
    x2^2 by Th58;
A13: Sum sqr(x1*R29)+Sum sqr(x2*R19) + -2*Sum(mlt(x1*R29,x2*R19)) = Sum
    sqr(x1*R29)-2*Sum(mlt(x1*R29,x2*R19))+Sum sqr(x2*R19)
      .= Sum sqr(x1*R29)-Sum(2*mlt(x1*R29,x2*R19))+Sum sqr(x2*R19) by Th87
      .= Sum(sqr(x1*R29)-2*mlt(x1*R29,x2*R19))+Sum sqr(x2*R19) by Th90
      .= Sum(sqr (x1*R29)-2*mlt(x1*R29,x2*R19)+sqr(x2*R19)) by Th89;
    (Sum sqr R19 + x1^2) = Sum sqr R1 by A3,A5;
    then
    (Sum sqr R1)*(Sum sqr R2) - (Sum mlt(R1,R2))^2 = (Sum sqr R19 + x1^2)
    *(Sum sqr R29 + x2^2) + -(Sum mlt(R19,R29)+x1*x2)^2 by A3,A4,A8,A6
      .= (Sum sqr R19)*(Sum sqr R29)+-(Sum mlt(R19,R29))^2 + ((Sum sqr(x1*
    R29)+Sum sqr(x2*R19)) +- 2*Sum(mlt(x1*R29,x2*R19))) by A12,A10
      .= (Sum sqr R19)*(Sum sqr R29)-(Sum mlt(R19,R29))^2 + Sum sqr (x1*R29-
    x2*R19) by A13,Th69;
    then (Sum mlt(R1,R2))^2 + (0 qua Element of NAT) <= (Sum sqr R1)*(Sum sqr
    R2) by A7,A11,XREAL_1:19;
    hence thesis;
  end;
A14: P[0];
  for i holds P[i] from NAT_1:sch 2(A14,A1);
  hence thesis;
end;



definition
  let F be complex-valued FinSequence;
  func Product F -> complex number means
  :Def13:
  ex f being FinSequence of COMPLEX st f = F & it = multcomplex $$ f;
  existence
  proof
    rng F c= COMPLEX by VALUED_0:def 1;
    then reconsider f = F as FinSequence of COMPLEX by FINSEQ_1:def 4;
    take multcomplex $$ f;
    thus thesis;
  end;
  uniqueness;
end;

registration
  let F be real-valued FinSequence;
  cluster Product F -> real;
  coherence
  proof
    set mc = multcomplex;
    consider f being FinSequence of COMPLEX such that
A1: f = F and
A2: Product F = multcomplex $$ f by Def13;
    set g = [#](f,the_unity_wrt mc);
    defpred P[Nat] means mc $$ (finSeg $1,[#](f,the_unity_wrt mc)) is real;
A3: for k being Nat st P[k] holds P[k+1]
    proof
      let k be Nat;
      assume
A4:   P[k];
      reconsider k as Element of NAT by ORDINAL1:def 12;
      g.(k+1) is real
      proof
        per cases;
        suppose
          k+1 in dom f;
          then g.(k+1) = f.(k+1) by SETWOP_2:20;
          hence thesis by A1;
        end;
        suppose
          not k+1 in dom f;
          hence thesis by BINOP_2:6,SETWOP_2:20;
        end;
      end;
      then reconsider a = g.(k+1), b = mc $$(finSeg k,g) as real number by A4;
      not (k + 1) in Seg k by FINSEQ_3:8;
      then mc $$ (finSeg k \/ {.k+1.},g) = mc.(mc $$(finSeg k,g),g.(k+1)) by
SETWOP_2:2
        .= b * a by BINOP_2:def 5;
      hence thesis by FINSEQ_1:9;
    end;
A5: mc $$ f = mc $$ (findom f,[#](f,the_unity_wrt mc)) & ex n being Nat st
    dom f = Seg n by FINSEQ_1:def 2,SETWOP_2:def 2;
    Seg 0 = {}.NAT;
    then
A6: P[0] by BINOP_2:6,SETWISEO:31;
    for n being Nat holds P[n] from NAT_1:sch 2(A6,A3);
    hence thesis by A2,A5;
  end;
end;

theorem Th93:
  for F being FinSequence of REAL holds Product F = multreal $$ F
proof
  set g = multreal, h = multcomplex;
  let F be FinSequence of REAL;
  rng F c= COMPLEX by NUMBERS:11,XBOOLE_1:1;
  then reconsider f = F as FinSequence of COMPLEX by FINSEQ_1:def 4;
  defpred P[Nat] means g $$ (finSeg $1,[#](F,the_unity_wrt g)) = h $$ (finSeg
  $1,[#](f,the_unity_wrt h));
  consider n being Nat such that
A1: dom f = Seg n by FINSEQ_1:def 2;
A2: g $$ F = g $$ (finSeg n,[#](F,the_unity_wrt g)) & h $$ f = h $$ (finSeg
  n, [#](f,the_unity_wrt h)) by A1,SETWOP_2:def 2;
A3: for k being Nat st P[k] holds P[k+1]
  proof
    set j = [#](f,the_unity_wrt h);
    set i = [#](F,the_unity_wrt g);
    let k be Nat;
    assume
A4: P[k];
A5: i.(k+1) = j.(k+1)
    proof
      per cases;
      suppose
A6:     k+1 in dom f;
        then j.(k+1) = F.(k+1) by SETWOP_2:20
          .= i.(k+1) by A6,SETWOP_2:20;
        hence thesis;
      end;
      suppose
A7:     not k+1 in dom f;
        then j.(k+1) = the_unity_wrt h by SETWOP_2:20
          .= i.(k+1) by A7,BINOP_2:6,7,SETWOP_2:20;
        hence thesis;
      end;
    end;
A8: not (k + 1) in Seg k by FINSEQ_3:8;
    reconsider k as Element of NAT by ORDINAL1:def 12;
    g $$ (finSeg (k+1),i) = g $$ (finSeg k \/ {.k+1.},i) by FINSEQ_1:9
      .= g.(g $$(finSeg k,i),i.(k+1)) by A8,SETWOP_2:2
      .= g $$(finSeg k,i) * (i.(k+1)) by BINOP_2:def 11
      .= h.(h $$(finSeg k,j),j.(k+1)) by A4,A5,BINOP_2:def 5
      .= h $$ (finSeg k \/ {.k+1.},j) by A8,SETWOP_2:2
      .= h $$ (finSeg (k+1),j) by FINSEQ_1:9;
    hence thesis;
  end;
A9: Seg 0 = {}.NAT;
  then g $$ (finSeg 0,[#](F,the_unity_wrt g)) = the_unity_wrt h by BINOP_2:6,7
,SETWISEO:31
    .= h $$ (finSeg 0,[#](f,the_unity_wrt h)) by A9,SETWISEO:31;
  then
A10: P[0];
  for k being Nat holds P[k] from NAT_1:sch 2(A10,A3);
  then g $$ F = h $$ f by A2;
  hence thesis by Def13;
end;

definition
  let F be FinSequence of COMPLEX;
  redefine func Product F -> Element of COMPLEX equals
  multcomplex $$ F;
  coherence by XCMPLX_0:def 2;
  compatibility by Def13;
end;

definition
  let F be FinSequence of REAL;
  redefine func Product F -> Element of REAL equals
  multreal $$ F;
  coherence by XREAL_0:def 1;
  compatibility by Th93;
end;

Lm3: for F being empty FinSequence holds Product F = 1
proof
  Product <*>REAL = 1 by BINOP_2:7,FINSOP_1:10;
  hence thesis;
end;

theorem Th94:
  Product <*>REAL = 1 by Lm3;

registration
  let r be complex number;
  cluster <*r*> -> complex-valued;
  coherence
  proof
    reconsider p = r as Element of COMPLEX by XCMPLX_0:def 2;
    reconsider f = <*p*> as FinSequence of COMPLEX;
    f is FinSequence-like;
    hence thesis;
  end;
end;

registration
  let r1, r2 be complex number;
  cluster <*r1, r2*> -> complex-valued;
  coherence
  proof
    reconsider p1 = r1, p2 = r2 as Element of COMPLEX by XCMPLX_0:def 2;
    reconsider f = <*p1,p2*> as FinSequence of COMPLEX;
    f is FinSequence-like;
    hence thesis;
  end;
end;

registration
  let r1, r2, r3 be complex number;
  cluster <*r1, r2, r3*> -> complex-valued;
  coherence
  proof
    reconsider p1 = r1, p2 = r2, p3 = r3 as Element of COMPLEX by
XCMPLX_0:def 2;
    reconsider f = <*p1,p2,p3*> as FinSequence of COMPLEX;
    f is FinSequence-like;
    hence thesis;
  end;
end;

theorem Th95:
  for r being complex number holds Product <*r*> = r
proof
  let r be complex number;
  reconsider r9 = r as Element of COMPLEX by XCMPLX_0:def 2;
  reconsider F = <*r9*> as FinSequence of COMPLEX;
  multcomplex $$ F = r by FINSOP_1:11;
  hence thesis by Def13;
end;

registration
  let f, g be complex-valued FinSequence;
  cluster f ^ g -> complex-valued;
  coherence
  proof
A1: rng (f ^ g) = rng f \/ rng g by FINSEQ_1:31;
    rng f c= COMPLEX & rng g c= COMPLEX by VALUED_0:def 1;
    then rng (f ^ g) c= COMPLEX by A1,XBOOLE_1:8;
    hence thesis by VALUED_0:def 1;
  end;
end;

theorem Th96:
  for F being complex-valued FinSequence, r being complex
  number holds Product (F^<*r*>) = Product F * r
proof
  let F be complex-valued FinSequence, r be complex number;
  reconsider p = r as Element of COMPLEX by XCMPLX_0:def 2;
  rng F c= COMPLEX & rng (F^<*p*>) c= COMPLEX by VALUED_0:def 1;
  then reconsider Fr = F^<*r*>, Ff = F as FinSequence of COMPLEX
  by FINSEQ_1:def 4;
  thus Product (F^<*r*>) = multcomplex $$ Fr by Def13
    .= multcomplex.(Product Ff,p) by FINSOP_1:4
    .= Product F * r by BINOP_2:def 5;
end;

theorem Th97:
  for F1, F2 being complex-valued FinSequence holds
  Product (F1^F2) = Product F1 * Product F2
proof
  let F1, F2 be complex-valued FinSequence;
A1: rng (F1^F2) c= COMPLEX by VALUED_0:def 1;
  rng F1 c= COMPLEX & rng F2 c= COMPLEX by VALUED_0:def 1;
  then reconsider FF = F1^F2, f1 = F1, f2 = F2 as FinSequence of COMPLEX
  by A1,FINSEQ_1:def 4;
  thus Product (F1^F2) = multcomplex $$ FF by Def13
    .= multcomplex.(Product f1,Product f2) by FINSOP_1:5
    .= Product F1 * Product F2 by BINOP_2:def 5;
end;

theorem
  Product (<*r*>^F) = r * Product F
proof
  thus Product (<*r*>^F) = Product <*r*> * Product F by Th97
    .= r * Product F by Th95;
end;

theorem Th99:
  for r1, r2 being complex number holds Product <*r1,r2*> = r1 * r2
proof
  let r1, r2 be complex number;
  thus Product <*r1,r2*> = Product <*r1*> * r2 by Th96
    .= r1 * r2 by Th95;
end;

theorem
  for r1, r2, r3 being complex number holds Product <*r1,r2,r3*> = r1 * r2 * r3
proof
  let r1, r2, r3 be complex number;
  thus Product <*r1,r2,r3*> = Product <*r1,r2*> * r3 by Th96
    .= r1 * r2 * r3 by Th99;
end;

theorem
  for R being Element of 0-tuples_on REAL holds Product R = 1 by Lm3;

theorem
  Product (i|->1) = 1
proof
  Product (i |-> the_unity_wrt multreal) = the_unity_wrt multreal by
SETWOP_2:25;
  hence thesis by BINOP_2:7;
end;

Lm4: for p being complex-valued FinSequence st len p <> 0 ex q being
complex-valued FinSequence, d being complex number st p = q^<*d*>
proof
  let p be complex-valued FinSequence;
  assume len p <> 0;
  then consider q being FinSequence,d being set such that
A1: p = q^<*d*> by CARD_1:27,FINSEQ_1:46;
  rng p c= COMPLEX by VALUED_0:def 1;
  then
A2: p is FinSequence of COMPLEX by FINSEQ_1:def 4;
  for i st i in dom q holds q.i in COMPLEX
  proof
    let i;
    assume i in dom q;
    then p.i = q.i & i in dom p by A1,FINSEQ_1:def 7,FINSEQ_2:15;
    hence thesis by A2,FINSEQ_2:11;
  end;
  then
A3: q is FinSequence of COMPLEX by FINSEQ_2:12;
  len p = len q + 1 by A1,FINSEQ_2:16;
  then p.(len p) = d by A1,FINSEQ_1:42;
  hence thesis by A1,A3;
end;

theorem
  for F being complex-valued FinSequence holds
  (ex k st k in dom F & F.k = 0) iff Product F = 0
proof
  defpred P[Nat] means for F being complex-valued FinSequence st len F = $1
  holds (ex k st k in Seg $1 & F.k = 0) iff Product F = 0;
  let F be complex-valued FinSequence;
A1: Seg len F = dom F by FINSEQ_1:def 3;
A2: for i st P[i] holds P[i+1]
  proof
    let i such that
A3: for F being complex-valued FinSequence st len F = i holds (ex k
    st k in Seg i & F.k = 0) iff Product F = 0;
    let F be complex-valued FinSequence;
    assume
A4: len F = i+1;
    then consider
    F9 being complex-valued FinSequence, x being complex number
    such that
A5: F = F9^<*x*> by Lm4;
A6: len F = len F9 + 1 by A5,FINSEQ_2:16;
A7: Product F = Product F9 * x by A5,Th96;
    thus (ex k st k in Seg (i+1) & F.k = 0) implies Product F = 0
    proof
      given k such that
A8:   k in Seg (i+1) and
A9:   F.k = 0;
      now
        per cases by A8,FINSEQ_2:7;
        suppose
A10:      k in Seg i;
          Seg len F9 = dom F9 by FINSEQ_1:def 3;
          then F9.k = F.k by A4,A5,A6,A10,FINSEQ_1:def 7;
          then Product F9 = 0 by A3,A4,A6,A9,A10;
          hence thesis by A7;
        end;
        suppose
          k = i+1;
          then x = 0 by A4,A5,A6,A9,FINSEQ_1:42;
          hence thesis by A7;
        end;
      end;
      hence thesis;
    end;
    assume
A11: Product F = 0;
    per cases by A7,A11;
    suppose
      Product F9 = 0;
      then consider k such that
A12:  k in Seg i and
A13:  F9.k = 0 by A3,A4,A6;
      Seg len F9 = dom F9 by FINSEQ_1:def 3;
      then F.k = 0 by A4,A5,A6,A12,A13,FINSEQ_1:def 7;
      hence thesis by A12,FINSEQ_2:8;
    end;
    suppose
      x = 0;
      then F.(i+1) = 0 by A4,A5,A6,FINSEQ_1:42;
      hence thesis by FINSEQ_1:4;
    end;
  end;
A14: P[0] proof let F be complex-valued FinSequence;
   assume len F = 0;
    then F = <*>COMPLEX;
   hence thesis by Th94;
  end;
  for i holds P[i] from NAT_1:sch 2(A14,A2);
  hence thesis by A1;
end;

theorem
  Product ((i+j) |->r ) = (Product (i|->r))*(Product (j|->r))
proof
  reconsider s=r as Element of REAL by XREAL_0:def 1;
  Product ((i+j) |->s) = multreal.(multreal$$(i|->s),multreal$$(j|->s)) by
SETWOP_2:26
    .= (Product (i|->s))*(Product (j|->s)) by BINOP_2:def 11;
  hence thesis;
end;

theorem
  Product ((i*j) |->r) = Product (j |-> (Product (i|->r)))
proof
  reconsider r as Element of REAL by XREAL_0:def 1;
  Product ((i*j) |->r) = Product (j |-> (Product (i|->r))) by SETWOP_2:27;
  hence thesis;
end;

theorem
  Product (i|->(r1*r2)) = (Product (i|->r1))*(Product (i|->r2))
proof
  reconsider s1=r1, s2=r2 as Element of REAL by XREAL_0:def 1;
  Product (i|->(s1*s2)) = multreal$$(i|->multreal.(s1,s2)) by BINOP_2:def 11
    .= multreal.(multreal$$(i|->s1),multreal$$(i|->s2)) by SETWOP_2:36
    .= (Product (i|->s1))*(Product (i|->s2)) by BINOP_2:def 11;
  hence thesis;
end;

theorem Th107:
  Product mlt(R1,R2) = Product R1 * Product R2
proof
  thus Product (mlt(R1,R2)) = multreal.(multreal$$R1,multreal$$R2)
  by SETWOP_2:35
    .= Product R1 * Product R2 by BINOP_2:def 11;
end;

theorem
  Product (r*R) = Product (i|->r) * Product R
proof
  reconsider s=r as Element of REAL by XREAL_0:def 1;
  thus Product (r*R) = Product mlt(i|->s,R) by Th63
    .= Product (i|->r) * Product R by Th107;
end;

theorem
  Product sqr R = (Product R)^2 by Th107;

begin 


reserve z,z1,z2 for Element of COMPLEX;

theorem
  for F being FinSequence of COMPLEX holds Product F = multcomplex $$ F;

theorem
  Product ((i+j) |->z) = (Product (i|->z))*(Product (j|->z))
proof
  thus Product ((i+j) |->z) = multcomplex.(multcomplex$$(i|->z),multcomplex$$(j
  |->z)) by SETWOP_2:26
    .= (Product (i|->z))*(Product (j|->z)) by BINOP_2:def 5;
end;

theorem
  Product ((i*j) |->z) = Product (j |-> (Product (i|->z))) by SETWOP_2:27;

theorem
  Product (i|->(z1*z2)) = (Product (i|->z1))*(Product (i|->z2))
proof
  thus Product (i|->(z1*z2)) = multcomplex$$(i|->multcomplex.(z1,z2)) by
BINOP_2:def 5
    .= multcomplex.(multcomplex$$(i|->z1),multcomplex$$(i|->z2)) by SETWOP_2:36
    .= (Product (i|->z1))*(Product (i|->z2)) by BINOP_2:def 5;
end;

begin 

reserve  n for Nat,
  x, y, a for real number,
  p, p1, p2, p3, q, q1, q2 for Element of n-tuples_on REAL;

theorem Th114:
  for x being real-valued FinSequence holds len (-x)=len x
proof
  let x be real-valued FinSequence;
  dom (-x)=dom x by VALUED_1:8;
  hence thesis by FINSEQ_3:29;
end;

theorem Th115:
  for x1,x2 being real-valued FinSequence st len x1=len x2 holds
  len (x1+x2)=len x1
proof
  let x1,x2 be real-valued FinSequence;
  set n=len x1;
A1: x2 is FinSequence of REAL by Lm2;
  x1 is FinSequence of REAL by Lm2;
  then reconsider z1=x1 as Element of (len x1)-tuples_on REAL by FINSEQ_2:92;
  assume len x1=len x2;
  then reconsider z2=x2 as Element of n-tuples_on REAL by A1,FINSEQ_2:92;
  dom (z1+z2)=Seg n by FINSEQ_2:124;
  hence thesis by FINSEQ_1:def 3;
end;

theorem Th116:
  for x1,x2 being real-valued FinSequence st len x1=len x2 holds
  len (x1-x2)=len x1
proof
  let x1,x2 be real-valued FinSequence;
  set n=len x1;
A1: x2 is FinSequence of REAL by Lm2;
  x1 is FinSequence of REAL by Lm2;
  then reconsider z1=x1 as Element of (len x1)-tuples_on REAL by FINSEQ_2:92;
  assume len x1=len x2;
  then reconsider z2=x2 as Element of n-tuples_on REAL by A1,FINSEQ_2:92;
  dom (z1-z2)=Seg n by FINSEQ_2:124;
  hence thesis by FINSEQ_1:def 3;
end;

theorem Th117:
  for a being real number, x being real-valued FinSequence
   holds len (a*x)=len x
proof
  let a be real number, x be real-valued FinSequence;
  set n=len x;
  x is FinSequence of REAL by Lm2;
  then reconsider z=x as Element of n-tuples_on REAL by FINSEQ_2:92;
  len (a*z)=n by CARD_1:def 7;
  hence thesis;
end;

theorem Th118:
  for x,y,z being real-valued FinSequence st len x=len y & len y=len z
  holds mlt(x+y,z) = mlt(x,z)+mlt(y,z)
proof
  let x,y,z be real-valued FinSequence;
A1:x is FinSequence of REAL & y is FinSequence of REAL &
  z is FinSequence of REAL by Lm2;
  assume len x=len y & len y=len z;
  then reconsider x2=x, y2=y, z2=z as Element of (len x)-tuples_on REAL by A1,
FINSEQ_2:92;
A2: dom (mlt(x+y,z))=Seg len(mlt(x2+y2,z2)) by FINSEQ_1:def 3
    .=Seg len x by CARD_1:def 7
    .=Seg len (mlt(x2,z2)+mlt(y2,z2)) by CARD_1:def 7
    .= dom (mlt(x2,z2)+mlt(y2,z2)) by FINSEQ_1:def 3;
A3: dom (mlt(x,z))=Seg len(mlt(x2,z2)) by FINSEQ_1:def 3
    .=Seg len x by CARD_1:def 7
    .=Seg len (mlt(x2,z2)+mlt(y2,z2)) by CARD_1:def 7
    .= dom (mlt(x2,z2)+mlt(y2,z2)) by FINSEQ_1:def 3;
  for i being Nat st i in dom (mlt(x+y,z)) holds mlt(x+y,z).i=(mlt(x,z)+
  mlt(y,z)).i
  proof
    let i be Nat;
    assume
A4: i in dom mlt(x+y,z);
    set a=x.i, b=y.i, d=(x+y).i, e=z.i;
    len (x2+y2)=len x by CARD_1:def 7;
    then dom (x2+y2)=Seg len x by FINSEQ_1:def 3
      .=Seg len(mlt(x2,z2)) by CARD_1:def 7
      .=dom (mlt(x,z)) by FINSEQ_1:def 3;
    then
A5: d=a+b by A2,A3,A4,VALUED_1:def 1;
    thus mlt(x+y,z).i=d*e by VALUED_1:5
      .=a*e+b*e by A5
      .=mlt(x,z).i +b*e by VALUED_1:5
      .=mlt(x,z).i +mlt(y,z).i by VALUED_1:5
      .=(mlt(x,z)+mlt(y,z)).i by A2,A4,VALUED_1:def 1;
  end;
  hence thesis by A2,FINSEQ_1:13;
end;

begin 

definition
  let x1,x2 be real-valued FinSequence;
  func |( x1,x2 )| -> real number equals
  Sum mlt(x1,x2);
  correctness;
  commutativity;
end;

definition
  let x1,x2 be real-valued FinSequence;
  redefine func |( x1,x2 )| -> Element of REAL;
  correctness;
  commutativity;
end;

theorem Th119:
  for x being real-valued FinSequence holds |(x,x)| >= 0
proof
  let x be real-valued FinSequence;
  set n=len x;
  x is FinSequence of REAL by Lm2;
  then reconsider w = x as Element of n-tuples_on REAL by FINSEQ_2:92;
  |(x, x)| = Sum sqr(w);
  hence thesis by Th86;
end;

theorem Th120:
  for x,y,z being real-valued FinSequence st len x=len y & len y=len z
  holds |((x+y),z)| = |(x,z)| + |(y,z)|
proof
  let x,y,z be real-valued FinSequence;
A1:x is FinSequence of REAL & y is FinSequence of REAL &
  z is FinSequence of REAL by Lm2;
  assume
A2: len x=len y & len y=len z;
  then reconsider x2=x,y2=y,z2=z as Element of (len x)-tuples_on REAL by A1,
FINSEQ_2:92;
  |((x+y),z)|= Sum(mlt(x,z)+mlt(y,z)) by A2,Th118
    .= Sum(mlt(x2,z2))+Sum(mlt(y2,z2)) by Th89
    .= |(x,z)| + |(y,z)|;
  hence thesis;
end;

theorem Th121:
  for x,y being real-valued FinSequence,a being real number st len x=
  len y holds |(a*x,y)| = a*|(x,y)|
proof
  let x,y be real-valued FinSequence,a be real number;
  reconsider a2=a as Element of REAL by XREAL_0:def 1;
A1:x is FinSequence of REAL & y is FinSequence of REAL by Lm2;
  assume len x=len y;
  then reconsider x2=x, y2 = y as Element of (len x)-tuples_on REAL by A1,
FINSEQ_2:92;
  |(a*x,y)| =Sum(a2*mlt(x2,y2)) by FINSEQOP:26
    .=a*|(x,y)| by Th87;
  hence thesis;
end;

theorem Th122:
  for x1,x2 being real-valued FinSequence st len x1=len x2 holds
  |(-x1, x2)| = -|(x1, x2)|
proof
  let x1,x2 be real-valued FinSequence;
  assume len x1=len x2;
  then |(-x1, x2)| = (-1)*|(x1, x2)| by Th121;
  hence thesis;
end;

theorem
  for x1,x2 being real-valued FinSequence st len x1=len x2 holds
  |(-x1, -x2)| = |(x1, x2)|
proof
  let x1,x2 be real-valued FinSequence;
  assume
A1: len x1=len x2;
  then len (-x2)=len x1 by Th114;
  then |(-x1, -x2)| = -|(x1, -x2)| by Th122
    .= --|(x1, x2)| by A1,Th122;
  hence thesis;
end;

theorem Th124:
  for x1,x2,x3 being real-valued FinSequence st len x1=len x2 & len x2
  =len x3 holds |(x1-x2, x3)| = |(x1, x3)| - |(x2, x3)|
proof
  let x1,x2,x3 be real-valued FinSequence;
  assume that
A1: len x1=len x2 and
A2: len x2=len x3;
  len (-x2)=len x2 by Th114;
  then |(x1 - x2, x3)| = |(x1, x3)| + |(-x2, x3)| by A1,A2,Th120
    .= |(x1, x3)| + - |(x2, x3)| by A2,Th122;
  hence thesis;
end;

theorem
  for x,y being real number,x1,x2,x3 being real-valued FinSequence st len x1
  =len x2 & len x2=len x3 holds |((x*x1+y*x2), x3)| = x*|(x1,x3)| + y*|(x2,x3)|
proof
  let x,y be real number,x1,x2,x3 be real-valued FinSequence;
  assume that
A1: len x1=len x2 and
A2: len x2=len x3;
  len (x*x1)=len x1 & len (y*x2)=len x2 by Th117;
  then |(x*x1 + y*x2, x3)| = |(x*x1, x3)| + |(y*x2, x3)| by A1,A2,Th120
    .= x*|(x1,x3)| + |(y*x2,x3)| by A1,A2,Th121
    .= x*|(x1,x3)| + y*|(x2,x3)| by A2,Th121;
  hence thesis;
end;

theorem Th126:
  for x1,x2,y1,y2 being real-valued FinSequence st len x1=len x2 & len
x2=len y1 & len y1=len y2 holds |(x1+x2, y1+y2)| = |(x1, y1)| + |(x1, y2)| + |(
  x2, y1)| + |(x2, y2)|
proof
  let x1,x2,y1,y2 be real-valued FinSequence;
  assume that
A1: len x1=len x2 and
A2: len x2=len y1 and
A3: len y1=len y2;
  |(x1+x2, y2)| = |(x1, y2)| + |(x2, y2)| by A1,A2,A3,Th120;
  then
A4: |(x1+x2, y1)| +|(x1+x2, y2)| = (|(x1, y1)|+|(x2, y1)|) + (|(x1, y2)| +
  |(x2, y2)|) by A1,A2,Th120;
  len (x1+x2)=len x1 by A1,Th115;
  hence thesis by A1,A2,A3,A4,Th120;
end;

theorem Th127:
  for x1,x2,y1,y2 being real-valued FinSequence st len x1=len x2 & len
x2=len y1 & len y1=len y2 holds |(x1-x2, y1-y2)| = |(x1, y1)| - |(x1, y2)| - |(
  x2, y1)| + |(x2, y2)|
proof
  let x1,x2,y1,y2 be real-valued FinSequence;
  assume that
A1: len x1=len x2 and
A2: len x2=len y1 and
A3: len y1=len y2;
  |(x1,y1-y2)| = |(x1,y1)| - |(x1,y2)| by A1,A2,A3,Th124;
  then
A4: |(x1,y1-y2)| - |(x2,y1-y2)| = (|(x1,y1)|-|(x1,y2)|)-(|(x2,y1)|-|(x2,y2)|
  ) by A2,A3,Th124;
  len (y1 - y2)=len y1 by A3,Th116;
  hence thesis by A1,A2,A4,Th124;
end;

theorem
  for x,y being real-valued FinSequence st len x=len y holds |(x+y, x+
  y)| = |(x, x)| + 2*|(x, y)| + |(y, y)|
proof
  let x,y be real-valued FinSequence;
A1: |(x, x)| + |(x, y)| + |(x, y)| = |(x, x)| + 2*|(x, y)|;
  assume len x=len y;
  hence thesis by A1,Th126;
end;

theorem
  for x,y being real-valued FinSequence st len x=len y holds |(x-y, x-
  y)| = |(x, x)| - 2*|(x, y)| + |(y, y)|
proof
  let x,y be real-valued FinSequence;
  assume len x=len y;
  then |(x-y, x-y)| = |(x,x)| - |(x,y)| - |(y,x)| + |(y, y)| by Th127
    .= |(x,x)| - 2*|(x,y)| + |(y, y)|;
  hence thesis;
end;

theorem Th130:
  |(p1 + p2, p3)| = |(p1, p3)| + |(p2, p3)|
proof
  reconsider f1=p1, f2=p2, f3=p3 as real-valued FinSequence;
  len f2=n by CARD_1:def 7;
  then len f1=len f2 & len f2=len f3 by CARD_1:def 7;
  hence thesis by Th120;
end;

theorem Th131:
  for x being real number holds |(x*p1, p2)| = x*|(p1, p2)|
proof
  let x be real number;
  reconsider f1=p1,f2=p2 as real-valued FinSequence;
  len f1=n & len f2=n by CARD_1:def 7;
  hence thesis by Th121;
end;

theorem Th132:
  |(-p1, p2)| = -|(p1, p2)|
proof
  |(-p1, p2)| = |((-1)*p1, p2)| .= (-1)*|(p1, p2)| by Th131;
  hence thesis;
end;

theorem
  |(-p1, -p2)| = |(p1, p2)|
proof
  |(-p1, -p2)| = -|(p1, -p2)| by Th132
    .= --|(p1, p2)| by Th132;
  hence thesis;
end;

theorem Th134:
  |(p1-p2, p3)| = |(p1, p3)| - |(p2, p3)|
proof
  |(p1 - p2, p3)| = |(p1, p3)| + |(-p2, p3)| by Th130
    .= |(p1, p3)| + - |(p2, p3)| by Th132;
  hence thesis;
end;

theorem
  |((x*p1+y*p2), p3)| = x*|(p1,p3)| + y*|(p2,p3)|
proof
  |(x*p1 + y*p2, p3)| = |(x*p1, p3)| + |(y*p2, p3)| by Th130
    .= x*|(p1,p3)| + |(y*p2,p3)| by Th131
    .= x*|(p1,p3)| + y*|(p2,p3)| by Th131;
  hence thesis;
end;

theorem Th136:
  |(p1+p2, q1+q2)| = |(p1, q1)| + |(p1, q2)| + |(p2, q1)| + |(p2, q2)|
proof
A1: |(p1+p2, q1)| = |(p1, q1)| + |(p2, q1)| & |(p1+p2, q2)| = |(p1, q2)| +
  |(p2, q2)| by Th130;
  |(p1+p2, q1+q2)| = |(p1+p2, q1)| + |(p1+p2, q2)| by Th130
    .= |(p1, q1)|+|(p1, q2)|+|(p2, q1)|+|(p2, q2)| by A1;
  hence thesis;
end;

theorem Th137:
  |(p1-p2, q1-q2)| = |(p1, q1)| - |(p1, q2)| - |(p2, q1)| + |(p2, q2)|
proof
A1: |(p1,q1-q2)| = |(p1,q1)| - |(p1,q2)| & |(p2,q1-q2)| = |(p2,q1)| - |(p2,
  q2)| by Th134;
  |(p1-p2, q1-q2)| = |(p1, q1-q2)| - |(p2, q1-q2)| by Th134
    .= |(p1,q1)|-|(p1,q2)|-|(p2,q1)|+|(p2,q2)| by A1;
  hence thesis;
end;

theorem
  |(p+q, p+q)| = |(p, p)| + 2*|(p, q)| + |(q, q)|
proof
  |(p, p)| + |(p, q)| + |(p, q)| = |(p, p)| + 2*|(p, q)|;
  hence thesis by Th136;
end;

theorem Th139:
  |(p-q, p-q)| = |(p, p)| - 2*|(p, q)| + |(q, q)|
proof
  |(p-q, p-q)| = |(p,p)| - |(p,q)| - |(p,q)| + |(q, q)| by Th137
    .= |(p,p)| - 2*|(p,q)| + |(q, q)|;
  hence thesis;
end;

theorem
  |(p,q)| <= |(p,p)| + |(q,q)|
proof
  0 <= |(p,p)| & 0 <= |(q,q)| by Th119;
  then
A1: 0/2 <= (|(p,p)| + |(q,q)|)/2;
  |(p-q, p-q)| = |(p,p)| - 2*|(p,q)| + |(q,q)| by Th139
    .= |(p,p)| + |(q,q)| - 2*|(p,q)|;
  then 2*|(p,q)| <= |(p,p)| + |(q,q)| - 0 by Th119,XREAL_1:11;
  then (2*|(p,q)|)/2 <= (|(p,p)| + |(q,q)|)/2 by XREAL_1:72;
  then (0 qua Element of NAT) + |(p,q)|
   <= (|(p,p)| + |(q,q)|)/2 + (|(p,p)| + |(q,q)|)/2 by A1,XREAL_1:7;
  hence thesis;
end;

definition
  let p, q be real-valued FinSequence;
  pred p,q are_orthogonal means
  :Def17:
  |(p,q)| = 0;
  symmetry;
end;

theorem
  p, q are_orthogonal implies a*p,q are_orthogonal
proof
  assume p, q are_orthogonal;
  then |(p,q)|=0 by Def17;
  then a*(|(p,q)|)=0;
  then |(a*p,q)|=0 by Th131;
  hence thesis by Def17;
end;



theorem
  Sum<*r1,r2,r3,r4*> = r1 + r2 + r3 + r4
proof
  thus Sum<*r1,r2,r3,r4*> = Sum<*r1,r2,r3*> + r4 by Th74
    .= r1 + r2 + r3 + r4 by Th78;
end;



reserve f,g for FinSequence of REAL;

theorem Th143:
  len f = len sqr f & dom f = dom sqr f
proof
  rng f c= REAL & dom sqrreal = REAL by FUNCT_2:def 1;
  hence len f = len (sqrreal*f) by FINSEQ_2:29
    .= len sqr f;
  hence thesis by FINSEQ_3:29;
end;

theorem
  sqr (f^g) = sqr f ^ sqr g
proof
A1: len f = len sqr f by Th143;
A2: len sqr (f^g) = len (f^g) by Th143;
A3: len g = len sqr g & len(f^g) = len f + len g by Th143,FINSEQ_1:22;
A4: for i be Nat st 1 <= i & i <= len sqr (f^g) holds sqr (f^g).i = (sqr f ^
  sqr g).i
  proof
    let i be Nat;
    assume that
A5: 1 <= i and
A6: i <= len sqr (f^g);
A7: i in dom (f^g) by A2,A5,A6,FINSEQ_3:25;
    per cases;
    suppose
A8:   i in dom f;
      then
A9:   i in dom sqr f by Th143;
      thus sqr (f^g).i = (sqrreal*(f^g)).i
        .= sqrreal.((f^g).i) by A7,FUNCT_1:13
        .= sqrreal.(f.i) by A8,FINSEQ_1:def 7
        .= (f.i)^2 by Def2
        .= (sqr f).i by VALUED_1:11
        .= (sqr f ^ sqr g).i by A9,FINSEQ_1:def 7;
    end;
    suppose
      not i in dom f;
      then
A10:  len f < i by A5,FINSEQ_3:25;
      then reconsider j = i - len f as Element of NAT by INT_1:5;
A11:  i <= len(f^g) by A6,Th143;
A12:  i <= len(sqr f ^ sqr g) by A1,A3,A2,A6,FINSEQ_1:22;
      thus sqr (f^g).i = (sqrreal*(f^g)).i
        .= sqrreal.((f^g).i) by A7,FUNCT_1:13
        .= sqrreal.(g.j) by A10,A11,FINSEQ_1:24
        .= (g.j)^2 by Def2
        .= (sqr g).j by VALUED_1:11
        .= (sqr f ^ sqr g).i by A1,A10,A12,FINSEQ_1:24;
    end;
  end;
  len (sqr (f^g)) = len (sqr f ^ sqr g) by A1,A3,A2,FINSEQ_1:22;
  hence thesis by A4,FINSEQ_1:14;
end;

theorem
  for F being real-valued FinSequence holds F is FinSequence of REAL
proof
  let F be real-valued FinSequence;
  thus rng F c= REAL;
end;














begin

reserve i,j,k,n,m,l,s,t,r for Nat;
reserve a,b for real number;
reserve F,G,H for FinSequence of REAL;

theorem Th1:
  n>=1 implies Seg n = {1} \/ {k where k is Element of NAT: 1<k & k <n} \/ {n}
proof
  assume
A1: n>=1;
A2: {1} \/ {k where k is Element of NAT: 1<k & k<n} \/ {n} c= Seg n
  proof
    let d be set;
A3: {k where k is Element of NAT: 1<k & k<n} c= Seg n
    proof
      let d be set;
      assume d in {k where k is Element of NAT: 1<k & k<n};
      then ex k being Element of NAT st d = k & 1<k & k<n;
      hence thesis;
    end;
    1 in Seg n by A1; then
    {1} c= Seg n by ZFMISC_1:31; then
A4: {1} \/ {k where k is Element of NAT: 1<k & k<n} c= Seg n by A3,XBOOLE_1:8;
    n in Seg n by A1,FINSEQ_1:3; then
    {n} c= Seg n by ZFMISC_1:31; then
    {1} \/ {k where k is Element of NAT: 1<k & k<n} \/ {n} c= Seg n
      by A4,XBOOLE_1:8;
    hence thesis;
  end;
  Seg n c= {1} \/ {k where k is Element of NAT: 1<k & k<n} \/ {n}
  proof
    let d be set;
    assume
A5: d in Seg n;
    per cases by A1,XXREAL_0:1;
    suppose
A6:   n>1;
      reconsider l = d as Element of NAT by A5;
A7:   l <= n by A5,FINSEQ_1:1;
      1 <= l by A5,FINSEQ_1:1;
      then 1 = l or 1<l & l<n or l = n or 1 = n by A7,XXREAL_0:1;
      then
      d in {1} or d in {k where k is Element of NAT: 1<k & k<n} or d in {n
      } by A6,TARSKI:def 1;
      then
      d in {1} \/ {k where k is Element of NAT: 1<k & k<n} or d in {n} by
XBOOLE_0:def 3;
      hence thesis by XBOOLE_0:def 3;
    end;
    suppose
      n = 1;
      hence thesis by A5,FINSEQ_1:2,XBOOLE_0:def 3;
    end;
  end;
  hence thesis by A2,XBOOLE_0:def 10;
end;

theorem Th2:
  len (a*F) = len F
proof
  reconsider G = F as Element of (len F)-tuples_on REAL by FINSEQ_2:92;
  len (a*G) = len G by CARD_1:def 7;
  hence thesis;
end;

theorem
  dom G = dom (a*G) by VALUED_1:def 5;



registration
  let x be complex number, n be Nat;
  cluster n |-> x -> complex-yielding;
  coherence;
end;

definition
  let x be complex number, n be Nat;
  func x |^ n equals
  Product (n |-> x);
  coherence;
end;

registration
  let x be real number, n be Nat;
  cluster x |^ n -> real;
  coherence;
end;

definition
  let x be Real, n be Nat;
  redefine func x |^ n -> Real;
  coherence by XREAL_0:def 1;
end;

reserve z for complex number;

registration
  let z be complex number, n be Nat;
  cluster z|^n -> complex;
  coherence;
end;

theorem
  z|^0 = 1 by RVSUM_1:94;

theorem Th5:
  z|^1 = z
proof
  thus z|^1 = Product(<*z*>) by FINSEQ_2:59
    .= z by RVSUM_1:95;
end;

theorem Th6:
  z|^(s+1) = z|^s * z
proof
  thus z|^(s+1) = Product((s |-> z) ^ <*z*>) by FINSEQ_2:60
    .= z|^s*z by RVSUM_1:96;
end;

registration
  let x, n be Nat;
  cluster x|^n -> natural;
  coherence
  proof
    defpred P[Nat] means x|^$1 is natural;
A1: for a being Nat st P[a] holds P[a+1]
    proof
      let a be Nat;
      assume P[a];
      then reconsider b = x|^a as Nat;
      x|^(a+1) = b*x by Th6;
      hence thesis;
    end;
A2: P[0] by RVSUM_1:94;
    for a being Nat holds P[a] from NAT_1:sch 2(A2,A1);
    hence thesis;
  end;
end;

reserve x,y for complex number;

theorem
  (x*y) |^ s = x|^s * y|^s
proof
  reconsider x, y as Element of COMPLEX by XCMPLX_0:def 2;
  x|^s*y|^s = Product(s |-> (x*y)) by RVSUM_1:113;
  hence thesis;
end;

theorem Th8:
  x|^(s+t) = x|^s * x|^t
proof
  reconsider x as Element of COMPLEX by XCMPLX_0:def 2;
  x|^s*x|^t = Product((s+t) |-> x) by RVSUM_1:111;
  hence thesis;
end;

theorem
  (x|^s) |^ t = x|^(s*t)
proof
  reconsider x as Element of COMPLEX by XCMPLX_0:def 2;
  x|^(s*t) = Product(t |-> x|^s) by RVSUM_1:112;
  hence thesis;
end;

theorem Th10:
  1|^s = 1
proof
  defpred P[Nat] means 1|^$1 = 1;
A1: now
    let s;
    assume P[s];
    then 1|^(s+1) = 1 * 1 by Th6;
    hence P[s+1];
  end;
A2: P[0] by RVSUM_1:94;
  for s holds P[s] from NAT_1:sch 2(A2,A1);
  hence thesis;
end;

theorem Th11:
  s >= 1 implies 0|^s = 0
proof
  defpred P[Nat] means 0|^$1 = 0;
A1: now
    let n be Nat;
    assume n>=1 & P[n];
    0|^(n+1) = 0|^n * 0 by Th6
      .= 0;
    hence P[n+1];
  end;
A2: P[1] by Th5;
  for n be Nat st n >=1 holds P[n] from NAT_1:sch 8(A2,A1);
  hence thesis;
end;



registration
  let n be Nat;
  cluster idseq n -> natural-valued;
  coherence;
end;

definition
  let n be Nat;
  func n! -> Element of REAL equals
  Product idseq n;
  coherence by XREAL_0:def 1;
end;

registration
  let n be Nat;
  cluster n! -> real;
  coherence;
end;

theorem Th12:
  0! = 1 by RVSUM_1:94;

theorem
  1! = 1 by FINSEQ_2:50,RVSUM_1:95;

theorem
  2! = 2
proof
  thus 2! = 1*2 by FINSEQ_2:52,RVSUM_1:99
  .= 2;
end;

theorem Th15:
  (s+1)! = (s!) * (s+1)
proof
  idseq (s+1) = idseq s ^ <* s+1 *> by FINSEQ_2:51;
  hence thesis by RVSUM_1:96;
end;

theorem Th16:
  s! is Element of NAT
proof
  defpred P[Nat] means $1! is Element of NAT;
A1: now
    let s;
    assume P[s];
    then reconsider k=s! as Element of NAT;
    (s+1)! = (s+1) * k by Th15;
    hence P[s+1];
  end;
A2: P[0] by RVSUM_1:94;
  for s holds P[s] from NAT_1:sch 2(A2,A1);
  hence thesis;
end;

registration
  let n;
  cluster n! -> natural;
  coherence by Th16;
end;

theorem Th17:
  s! > 0
proof
  defpred P[Nat] means $1!>0;
A1: now
    let s;
    assume P[s];
    then (s+1) * (s!)>(s+1) * 0;
    hence P[s+1] by Th15;
  end;
A2: P[0] by RVSUM_1:94;
  for s holds P[s] from NAT_1:sch 2(A2,A1);
  hence thesis;
end;

registration
  let n;
  cluster n! -> positive;
  coherence by Th17;
end;

theorem
  (s!) * (t!) <> 0;



definition
  let k,n be Nat;
  func n choose k means
  :Def3:
  for l be Nat st l = n-k holds it = (n!)/((k!) * (l!)) if n >= k
  otherwise it = 0;
  consistency;
  existence
  proof
    thus n >= k implies ex r1 be set st for l being Nat st l = n-k holds r1 =
    (n!)/((k!) * (l!))
    proof
      assume n >= k;
      then reconsider m = n-k as Element of NAT by INT_1:5;
      take (n!)/((k!)*(m!));
      thus thesis;
    end;
    thus thesis;
  end;
  uniqueness
  proof
    let r1,r2 be set;
    thus n>=k & (for l being Nat st l = n-k holds r1 = (n!)/((k!) * (l!))) & (
    for l being Nat st l = n-k holds r2 = (n!)/((k!) * (l!))) implies r1 = r2
    proof
      assume n>=k;
      then reconsider m = n-k as Element of NAT by INT_1:5;
      assume that
A1:   for l being Nat st l = n-k holds r1 = (n!)/((k!) * (l!)) and
A2:   for l being Nat st l = n-k holds r2 = (n!)/((k!) * (l!));
      r1 = (n!)/((k!) * (m!)) by A1;
      hence thesis by A2;
    end;
    thus thesis;
  end;
end;

registration
  let k,n be Nat;
  cluster n choose k -> real;
  coherence
  proof
    per cases;
    suppose
A1:   n >= k;
      then reconsider l = n-k as Element of NAT by INT_1:5;
      n choose k = (n!)/((k!) * (l!)) by A1,Def3;
      hence thesis;
    end;
    suppose n < k;
      hence thesis by Def3;
    end;
  end;
end;

definition
  let k,n be Nat;
  redefine func n choose k -> Real;
  coherence by XREAL_0:def 1;
end;

theorem Th19:
  s choose 0 = 1
proof
  reconsider m = s-0 as Element of NAT by ORDINAL1:def 12;
  m = s;
  then (s choose 0) = (s!)/(1 * (s!)) by Def3,Th12
    .= 1 by XCMPLX_1:60;
  hence thesis;
end;

theorem Th20:
  s >= t & r = s-t implies s choose t = s choose r
proof
  assume
A1: s>=t;
  t-s>=0-s by XREAL_1:9;
  then
A2: --s>=-(t-s) by XREAL_1:24;
  assume
A3: r = s-t;
  then t = s-r;
  then s choose r = (s!)/((r!) * (t!)) by A2,Def3;
  hence thesis by A1,A3,Def3;
end;

theorem Th21:
  s choose s = 1
proof
  reconsider m = s-s as Element of NAT by INT_1:5;
  m = 0;
  then s choose s = s choose 0 by Th20;
  hence thesis by Th19;
end;

theorem Th22:
  (t+1) choose (s+1) = (t choose (s+1)) + (t choose s)
proof
  per cases by XXREAL_0:1;
  suppose
A1: s<t;
    thus ((t+1) choose (s+1)) = (t choose (s+1)) + (t choose s)
    proof
      reconsider m1 = t-s as Element of NAT by A1,INT_1:5;
A2:   s+1<=t by A1,NAT_1:13;
      then reconsider m2 = t-(s+1) as Element of NAT by INT_1:5;
A3:   s+1<=t+1 by A1,XREAL_1:6;
      then reconsider m = (t+1)-(s+1) as Element of NAT by INT_1:5;
      (t choose (s+1)) = (t!)/(((s+1)!) * (m2!)) by A2,Def3;
      then
A4:   (t choose (s+1)) + (t choose s) = (t!)/(((s+1)!) * (m2!)) + (t!)/((s
      !) * (m1!)) by A1,Def3
        .= ((t!)*((s!)*(m1!))+(t!)*(((s+1)!)*(m2!)))/((((s+1)!)*(m2!))*((s!)
      *(m1!))) by XCMPLX_1:116
        .= (t!*((s!)*(m1!))+(t!)*(s!*(s+1)*(m2!)))/((((s+1)!)*(m2!))*((s!)*(
      m1!))) by Th15
        .= (s!*((t!)*((m1!)+(s+1)*(m2!))))/(s!*((m1!)*((s+1)!*(m2!))))
        .=(t!*(((m2+1)!)+(s+1)*(m2!)))/(m1!*((s+1)!*(m2!))) by XCMPLX_1:91
        .= (t!*((m2!)*(m2+1)+(m2!)*(s+1)))/(m1!*((s+1)!*(m2!))) by Th15
        .= (m2!*((t!)*((m2+1)+(s+1))))/(m2!*(((s+1)!)*(m1!)))
        .= ((t!)*(t-(s-(s+1))))/(((s+1)!)*(m1!)) by XCMPLX_1:91
        .= ((t+1)!)/(((s+1)!)*(m1!)) by Th15;
      m = m1;
      hence thesis by A4,A3,Def3;
    end;
  end;
  suppose
A5: s=t;
    then s<t+1 by NAT_1:13;
    then
A6: t choose (s+1)=0 by A5,Def3;
    (t+1) choose (s+1)=1 by A5,Th21;
    hence thesis by A5,A6,Th21;
  end;
  suppose
A7: s>t;
    then s+1 >t+1 by XREAL_1:8;
    then
A8: (t+1) choose (s+1)=0 by Def3;
A9: s+1>t+0 by A7,XREAL_1:8;
    t choose s =0 by A7,Def3;
    hence thesis by A9,A8,Def3;
  end;
end;

theorem Th23:
  s >= 1 implies s choose 1 = s
proof
  defpred P[Nat] means $1 choose 1 = $1;
A1: now
    let n be Nat;
    assume that
    n>=1 and
A2: P[n];
    (n+1) choose 1 = (n+1) choose (0+1) .= n + n choose 0 by A2,Th22
      .= n+1 by Th19;
    hence P[n+1];
  end;
A3: P[1] by Th21;
  for n be Nat st n >= 1 holds P[n] from NAT_1:sch 8(A3,A1);
  hence thesis;
end;

theorem
  s>=1 & t = s-1 implies s choose t = s
proof
  assume that
A1: s>=1 and
A2: t = s-1;
  s choose t = s choose 1 by A1,A2,Th20;
  hence thesis by A1,Th23;
end;

theorem Th25:
  s choose r is Element of NAT
proof
  defpred P[Nat] means for r holds ($1 choose r) is Element of NAT;
A1: for s st P[s] holds P[s+1]
  proof
    let s;
    assume
A2: P[s];
A3: for r st r <> 0 holds ((s+1) choose r) is Element of NAT
    proof
      let r;
      assume
A4:   r <> 0;
      ((s+1) choose r) is Element of NAT
      proof
        consider t being Nat such that
A5:     r = t+1 by A4,NAT_1:6;
        reconsider t as Element of NAT by ORDINAL1:def 12;
        reconsider m1 = (s choose t), m2 = (s choose (t+1)) as Element of NAT
        by A2;
        m1 + m2 =(s choose t) + (s choose (t+1));
        hence thesis by A5,Th22;
      end;
      hence thesis;
    end;
    let r;
    r = 0 or r<>0;
    hence thesis by A3,Th19;
  end;
A6: P[0]
  proof
    let r;
    r = 0 or r > 0;
    hence thesis by Def3,Th19;
  end;
  for s holds P[s] from NAT_1:sch 2(A6,A1);
  hence thesis;
end;

theorem
  for m,F st m <> 0 & len F = m & (for i,l st i in dom F & l = n+i-1
  holds F.i = l choose n) holds Sum F = (n+m) choose (n+1)
proof
  defpred P[Nat] means for F st $1 <> 0 & len F = $1 & (for i,l st i in dom F
  & l = n+i-1 holds F.i = l choose n) holds Sum F = (n+$1) choose (n+1);
A1: for m st P[m] holds P[m+1]
  proof
    let m;
    assume
A2: P[m];
    let F;
    assume that
    m+1 <> 0 and
A3: len F = m+1 and
A4: for i,l st i in dom F & l = n+i-1 holds F.i = l choose n;
    consider G be FinSequence of REAL,x being Real such that
A5: F = G^<*x*> by A3,FINSEQ_2:19;
A6: m+1 = len G +1 by A3,A5,FINSEQ_2:16;
    per cases;
    suppose
A7:   m = 0;
A8:   n = n+1-1;
A9:   dom F = Seg 1 by A3,A7,FINSEQ_1:def 3;
      then 1 in dom F;
      then F.1 = n choose n by A4,A8;
      hence Sum F = Sum<*n choose n*> by A9,FINSEQ_1:def 8
        .= n choose n by FINSOP_1:11
        .= 1 by Th21
        .= (n+(m+1)) choose (n+1) by A7,Th21;
    end;
    suppose
A10:  m <> 0;
A11:  n+m = n +(m+1)-1;
A12:  for i,l st i in dom G & l = n+i-1 holds G.i = l choose n
      proof
A13:    dom G c= dom F by A5,FINSEQ_1:26;
        let i,l;
        assume that
A14:    i in dom G and
A15:    l = n+i-1;
        G.i = F.i by A5,A14,FINSEQ_1:def 7;
        hence thesis by A4,A14,A15,A13;
      end;
      dom F = Seg (m+1) by A3,FINSEQ_1:def 3;
      then
A16:  (n+m) choose n = (G^<*x*>).(len G +1) by A4,A5,A6,A11,FINSEQ_1:4
        .= x by FINSEQ_1:42;
      thus Sum F = Sum G + x by A5,RVSUM_1:74
        .= (n+m) choose (n+1) + (n+m) choose n by A2,A6,A10,A12,A16
        .= (n+m+1) choose (n+1) by Th22
        .= (n+(m+1)) choose (n+1);
    end;
  end;
A17: P[0];
  thus for m holds P[m] from NAT_1:sch 2(A17,A1);
end;

registration
  let k,n be Nat;
  cluster n choose k -> natural;
  coherence by Th25;
end;

definition
  let k,n be Nat;
  redefine func n choose k -> Element of NAT;
  coherence by ORDINAL1:def 12;
end;

definition
  let a,b be real number;
  let n be Nat;
  func (a,b) In_Power n -> FinSequence of REAL means
  :Def4:
  len it = n+1 & for i,l,m being Nat st i in dom it & m = i - 1 & l = n-m holds
    it.i = (n choose m) * a|^l * b|^m;
  existence
  proof
    reconsider n as Element of NAT by ORDINAL1:def 12;
    defpred P[Nat,set] means for l,m be Nat st m = $1 - 1 & l = n-m holds $2 =
    (n choose m)* a|^l * b|^m;
A1: for k be Nat st k in Seg(n+1) ex x being set st P[k,x]
    proof
      let k be Nat;
      assume
A2:   k in Seg (n+1);
      then k >= 1 by FINSEQ_1:1;
      then reconsider m1 = k-1 as Element of NAT by INT_1:5;
      k <= n+1 by A2,FINSEQ_1:1;
      then k-1<=n+1-1 by XREAL_1:9;
      then reconsider l1 = n-m1 as Element of NAT by INT_1:5;
      take (n choose m1)*a|^l1*b|^m1;
      thus thesis;
    end;
    consider p being FinSequence such that
A3: dom p = Seg(n+1) and
A4: for k be Nat st k in Seg(n+1) holds P[k,p.k] from FINSEQ_1:sch 1(A1);
    rng p c= REAL
    proof
      let x be set;
      assume x in rng p;
      then consider y be set such that
A5:   y in dom p and
A6:   x = p.y by FUNCT_1:def 3;
      reconsider y as Element of NAT by A5;
      y >= 1 by A3,A5,FINSEQ_1:1;
      then reconsider m1 = y-1 as Element of NAT by INT_1:5;
      y <= n+1 by A3,A5,FINSEQ_1:1;
      then y-1<=n+1-1 by XREAL_1:9;
      then reconsider l1 = n-m1 as Element of NAT by INT_1:5;
      p.y = (n choose m1) *a|^l1*b|^m1 by A3,A4,A5;
      then reconsider x as Real by A6;
      x in REAL;
      hence thesis;
    end;
    then reconsider p as FinSequence of REAL by FINSEQ_1:def 4;
    take p;
    thus thesis by A3,A4,FINSEQ_1:def 3;
  end;
  uniqueness
  proof
    let G,H;
    assume that
A7: len G = n+1 and
A8: for i,l,m be Nat st i in dom G & m = i - 1 & l = n-m holds G.i =
    (n choose m)* a|^l * b|^m;
    assume that
A9: len H = n+1 and
A10: for i,l,m be Nat st i in dom H & m = i - 1 & l = n-m holds H.i =
    (n choose m)* a|^l * b|^m;
    for i st i in dom G holds G.i = H.i
    proof
      let i;
      assume
A11:  i in dom G;
      then
A12:  i in Seg (n+1) by A7,FINSEQ_1:def 3;
      then i>=1 by FINSEQ_1:1;
      then reconsider m = i-1 as Element of NAT by INT_1:5;
      i<=n+1 by A12,FINSEQ_1:1;
      then i-1<=n+1-1 by XREAL_1:9;
      then reconsider l = n-m as Element of NAT by INT_1:5;
      i in dom H by A9,A12,FINSEQ_1:def 3;
      then H.i = (n choose m)* a|^l *b|^m by A10;
      hence thesis by A8,A11;
    end;
    hence thesis by A7,A9,FINSEQ_2:9;
  end;
end;

theorem Th27:
  (a,b) In_Power 0 = <*1*>
proof
  set F = (a,b) In_Power 0;
A1: len F = 0+1 by Def4
    .= 1;
  then
A2: dom F = {1} by FINSEQ_1:2,def 3;
  then 1 in dom F by TARSKI:def 1;
  then consider i be set such that
A3: i in dom F;
  reconsider i as Element of NAT by A3;
A4: i = 1 by A2,A3,TARSKI:def 1;
  then reconsider m1 = i-1 as Element of NAT by INT_1:5;
  reconsider l1 = 0-m1 as Element of NAT by A4;
  1 in dom (a,b) In_Power 0 by A2,TARSKI:def 1;
  then F.1 = (0 choose 0)*a|^l1*b|^m1 by A4,Def4
    .= 1*a|^0*b|^0 by A4,Th21
    .= 1 by RVSUM_1:94;
  hence thesis by A1,FINSEQ_1:40;
end;

theorem Th28:
  ((a,b) In_Power s).1 = a|^s
proof
  reconsider m1 = 1-1 as Element of NAT by INT_1:5;
  reconsider l1 = s-m1 as Nat;
  len ((a,b) In_Power s) = s+1 by Def4;
  then
A1: dom ((a,b) In_Power s) = Seg (s+1) by FINSEQ_1:def 3;
  s+1>=0+1 by XREAL_1:6;
  then 1 in dom ((a,b) In_Power s) by A1;
  then ((a,b) In_Power s).1 = (s choose 0)*a|^l1*b|^m1 by Def4
    .= 1*a|^l1*b|^m1 by Th19
    .= a|^s by RVSUM_1:94;
  hence thesis;
end;

theorem Th29:
  ((a,b) In_Power s).(s+1) = b|^s
proof
  reconsider m1 = s+1-1 as Element of NAT by INT_1:5,NAT_1:12;
  reconsider l1 = s-m1 as Element of NAT by INT_1:5;
  len ((a,b) In_Power s) = s+1 by Def4;
  then dom ((a,b) In_Power s) = Seg (s+1) by FINSEQ_1:def 3;
  then s+1 in dom ((a,b) In_Power s) by FINSEQ_1:4;
  then ((a,b) In_Power s).(s+1) = (s choose s)*a|^l1*b|^m1 by Def4
    .= 1*a|^l1*b|^m1 by Th21
    .= 1*b|^m1 by RVSUM_1:94
    .= b|^s;
  hence thesis;
end;

theorem Th30:
  (a+b) |^ s = Sum((a,b) In_Power s)
proof
  defpred P[Nat] means (a+b) |^ $1 = Sum((a,b) In_Power $1);
A1: for n st P[n] holds P[n+1]
  proof
    reconsider a, b as Real by XREAL_0:def 1;
    let n;
    reconsider G1 = ((a*((a,b) In_Power n))^<*0 qua Real*>) as FinSequence of
    REAL;
    set G2 = (<*0 qua Real*>^(b*((a,b) In_Power n)));
    assume P[n];
    then
A2: (a+b) |^ (n+1) = (a+b)*Sum((a,b) In_Power n) by Th6
      .= a*Sum((a,b) In_Power n) + b*Sum((a,b) In_Power n)
      .= Sum(a*((a,b) In_Power n)) + b*Sum((a,b) In_Power n) by RVSUM_1:87
      .= Sum(a*((a,b) In_Power n)) + Sum(b*((a,b) In_Power n)) by RVSUM_1:87;
    len G2 = len <*0*> + len (b*((a,b) In_Power n)) by FINSEQ_1:22
      .= 1+ len (b*((a,b) In_Power n)) by FINSEQ_1:40
      .= 1+ len ((a,b) In_Power n) by Th2
      .= n+1+1 by Def4;
    then reconsider F2 = G2 as Element of (n+1+1)-tuples_on REAL by FINSEQ_2:92
;
    len G1 = len (a*((a,b) In_Power n)) + len <*0*> by FINSEQ_1:22
      .= len (a*((a,b) In_Power n)) +1 by FINSEQ_1:40
      .= len ((a,b) In_Power n) +1 by Th2
      .= n+1+1 by Def4;
    then reconsider F1 = G1 as Element of (n+1+1)-tuples_on REAL by FINSEQ_2:92
;
A3: Sum F2 = 0+ Sum(b*((a,b) In_Power n)) by RVSUM_1:76
      .= Sum(b*((a,b) In_Power n));
    Sum F1 = Sum(a*((a,b) In_Power n)) +0 by RVSUM_1:74
      .= Sum(a*((a,b) In_Power n));
    then
A4: Sum(G1+G2) = Sum(a*((a,b) In_Power n))+ Sum (b*((a,b) In_Power n)) by A3,
RVSUM_1:89;
    set F = F1 + F2;
A5: len F2 = n+1+1 by CARD_1:def 7;
A6: len F = n+1+1 by CARD_1:def 7;
A7: len F1 = n+1+1 by CARD_1:def 7;
A8: for i st i in dom ((a,b) In_Power (n+1)) holds F.i = ((a,b) In_Power
    (n+1)).i
    proof
      let i;
      assume
A9:   i in dom ((a,b) In_Power (n+1));
      set r2 = F2/.i;
      set r1 = F1/.i;
      set r = ((a,b) In_Power n)/.i;
A10:  len ((a,b) In_Power (n+1)) = n+1+1 by Def4;
      then
A11:  i in Seg (n+1+1) by A9,FINSEQ_1:def 3;
      then
A12:  i in dom F1 by A7,FINSEQ_1:def 3;
A13:  i in dom F2 by A5,A11,FINSEQ_1:def 3;
A14:  i in {1} implies F.i = ((a,b) In_Power (n+1)).i
      proof
        n+1>=0+1 by XREAL_1:6;
        then 1 in Seg (n+1);
        then 1 in Seg len (((a,b) In_Power n)) by Def4;
        then
A15:    1 in dom ((a,b) In_Power n) by FINSEQ_1:def 3;
        then
A16:    1 in dom (a*((a,b) In_Power n)) by VALUED_1:def 5;
        assume i in {1};
        then
A17:    i = 1 by TARSKI:def 1;
        then
A18:    r = ((a,b) In_Power n).1 by A15,PARTFUN1:def 6;
        r = ((a,b) In_Power n).i by A17,A15,PARTFUN1:def 6;
        then
A19:    r = a|^n by A17,Th28;
A20:    r1 = ((a*((a,b) In_Power n))^<*0*>).1 by A12,A17,PARTFUN1:def 6
          .= (a*((a,b) In_Power n)).1 by A16,FINSEQ_1:def 7
          .= a *a|^n by A18,A19,RVSUM_1:44
          .= a|^(n+1) by Th6;
A21:    r2 = F2.i by A13,PARTFUN1:def 6;
A22:    r1 = F1.i by A12,PARTFUN1:def 6;
        r2 = (<*0*>^(b*((a,b) In_Power n))).1 by A13,A17,PARTFUN1:def 6
          .= 0 by FINSEQ_1:41;
        then F.i = r1+0 by A22,A21,RVSUM_1:11
          .= ((a,b) In_Power (n+1)).i by A17,A20,Th28;
        hence thesis;
      end;
      i>=1 by A11,FINSEQ_1:1;
      then reconsider j = i-1 as Element of NAT by INT_1:5;
      set x = ((a,b) In_Power n)/.j;
A23:  i = j+1;
A24:  i in dom F by A6,A11,FINSEQ_1:def 3;
A25:  i in {k where k is Element of NAT: k>1 & k<n+1+1} implies F.i = ((a
      ,b) In_Power (n+1)).i
      proof
        assume i in {k where k is Element of NAT: 1<k & k<n+1+1};
        then
A26:    ex k being Element of NAT st k=i & 1<k & k<n+1+1;
        then reconsider m1 = i-1 as Element of NAT by INT_1:5;
A27:    r1 = ((a*((a,b) In_Power n))^<*0*>).i by A12,PARTFUN1:def 6;
        1<=j by A23,A26,NAT_1:13;
        then reconsider m2 = j-1 as Element of NAT by INT_1:5;
A28:    j<=n+1 by A23,A26,XREAL_1:6;
        then j-1<=n+1-1 by XREAL_1:9;
        then reconsider l2 = n-m2 as Element of NAT by INT_1:5;
        1<=j by A23,A26,NAT_1:13;
        then j in Seg (n+1) by A28;
        then j in Seg len ((a,b) In_Power n) by Def4;
        then
A29:    j in dom ((a,b) In_Power n) by FINSEQ_1:def 3;
        then
A30:    x = ((a,b) In_Power n).j by PARTFUN1:def 6;
A31:    i<=n+1 by A26,NAT_1:13;
        then i in Seg (n+1) by A26;
        then i in Seg len ((a,b) In_Power n) by Def4;
        then
A32:    i in dom ((a,b) In_Power n) by FINSEQ_1:def 3;
        then
A33:    r = ((a,b) In_Power n).i by PARTFUN1:def 6;
        i in dom (a*((a,b) In_Power n)) by A32,VALUED_1:def 5;
        then
A34:    r1 = (a*((a,b) In_Power n)).i by A27,FINSEQ_1:def 7
          .= a*r by A33,RVSUM_1:44;
        i-1<=n+1-1 by A31,XREAL_1:9;
        then reconsider l1 = n-m1 as Element of NAT by INT_1:5;
A35:    l1+1 = n+1-(m2+1);
A36:    j in dom (b*((a,b) In_Power n)) by A29,VALUED_1:def 5;
A37:    r2 = (<*0*>^(b*((a,b) In_Power n))).i by A13,PARTFUN1:def 6;
        then r2 = (<*0*>^(b*((a,b) In_Power n))).(len <*0*> +j) by A23,
FINSEQ_1:40
          .= (b*((a,b) In_Power n)).j by A36,FINSEQ_1:def 7
          .= b*x by A30,RVSUM_1:44;
        then F.i = a*r + b*x by A24,A27,A37,A34,VALUED_1:def 1
          .= a*(a|^l1*(n choose m1)*b|^m1) + b*x by A32,A33,Def4
          .= a*a|^l1*((n choose m1)*b|^m1) + b*x
          .= a|^(l1+1)*((n choose m1)*b|^m1) + b*x by Th6
          .= a|^(l1+1)*((n choose m1)*b|^m1) + b*(b|^m2*((n choose m2)*a|^l2
        )) by A29,A30,Def4
          .= a|^(l1+1)*((n choose m1)*b|^m1) + b*b|^m2*((n choose m2)*a|^l2)
          .= a|^(l1+1)*((n choose (m2+1))*b|^(m2+1)) + b|^(m2+1)*((n choose
        m2)*a|^l2) by Th6
          .= ((n choose (m2+1)) + (n choose m2))*a|^(l1+1)*b|^(m2+1)
          .= ((n+1) choose (m2+1))*a|^(l1+1)*b|^(m2+1) by Th22
          .= ((a,b) In_Power (n+1)).i by A9,A35,Def4;
        hence thesis;
      end;
A38:  i in {n+1+1} implies F.i = ((a,b) In_Power (n+1)).i
      proof
        assume
A39:    i in {n+1+1};
        then
A40:    i = n+1+1 by TARSKI:def 1;
A41:    j = n+1+1-1 by A39,TARSKI:def 1
          .= n+1;
        n+1 in Seg (n+1) by FINSEQ_1:4;
        then j in Seg len (((a,b) In_Power n)) by A41,Def4;
        then
A42:    j in dom ((a,b) In_Power n) by FINSEQ_1:def 3;
        then
A43:    x = ((a,b) In_Power n).(n+1) by A41,PARTFUN1:def 6
          .= b|^n by Th29;
A44:    x = ((a,b) In_Power n).j by A42,PARTFUN1:def 6;
A45:    j in dom (b*((a,b) In_Power n)) by A42,VALUED_1:def 5;
A46:    r2 = (<*0*>^(b*((a,b) In_Power n))).(1+n+1) by A13,A40,PARTFUN1:def 6
          .= (<*0*>^(b*((a,b) In_Power n))).(len <*0*> +j) by A41,FINSEQ_1:39
          .= (b*((a,b) In_Power n)).j by A45,FINSEQ_1:def 7
          .= b*b|^n by A44,A43,RVSUM_1:44
          .= b|^(n+1) by Th6;
A47:    r2 = F2.i by A13,PARTFUN1:def 6;
        n+1 = len ((a,b) In_Power n) by Def4
          .= len (a*((a,b) In_Power n)) by Th2;
        then
A48:    r1 = ((a*((a,b) In_Power n))^<*0*>).(len (a*((a,b) In_Power n))+1
        ) by A12,A40,PARTFUN1:def 6
          .= 0 by FINSEQ_1:42;
        r1 = F1.i by A12,PARTFUN1:def 6;
        then F.i = 0+r2 by A48,A47,RVSUM_1:11
          .= ((a,b) In_Power (n+1)).i by A40,A46,Th29;
        hence thesis;
      end;
A49:  now
        assume
        i in {1} \/ {k where k is Element of NAT: 1<k & k<n+1+1} \/ { n+1+1};
        then i in {1} \/ {k where k is Element of NAT: 1<k & k<n+1+1} or i in
        {n+1+1} by XBOOLE_0:def 3;
        hence thesis by A14,A25,A38,XBOOLE_0:def 3;
      end;
      dom ((a,b) In_Power (n+1)) = Seg (n+1+1) by A10,FINSEQ_1:def 3;
      hence thesis by A9,A49,Th1,NAT_1:12;
    end;
    len ((a,b) In_Power (n+1)) = len F by A6,Def4;
    hence thesis by A2,A4,A8,FINSEQ_2:9;
  end;
  (a+b) |^ 0 = Sum<*1 qua Real*> by FINSOP_1:11,RVSUM_1:94
    .= Sum((a,b) In_Power 0) by Th27;
  then
A50: P[0];
  for n holds P[n] from NAT_1:sch 2(A50,A1);
  hence thesis;
end;

definition
  let n be Nat;
  func Newton_Coeff n -> FinSequence of REAL means
  :Def5:
  len it = n+1 & for i,k be Nat st i in dom it & k = i-1 holds
  it.i = n choose k;
  existence
  proof
    reconsider n as Element of NAT by ORDINAL1:def 12;
    defpred P[Nat,set] means for s st s = $1-1 holds $2 = n choose s;
A1: for l be Nat st l in Seg (n+1) ex x being set st P[l,x]
    proof
      let l be Nat;
      assume l in Seg (n+1);
      then l>=1 by FINSEQ_1:1;
      then reconsider k = l-1 as Element of NAT by INT_1:5;
      set x = n choose k;
      take x;
      thus thesis;
    end;
    consider F being FinSequence such that
A2: dom F = Seg (n+1) & for l be Nat st l in Seg (n+1) holds P[l,F.l]
    from FINSEQ_1:sch 1(A1);
    rng F c= REAL
    proof
      let x be set;
      assume x in rng F;
      then consider y be set such that
A3:   y in dom F and
A4:   x = F.y by FUNCT_1:def 3;
      reconsider y as Element of NAT by A3;
      y >= 1 by A2,A3,FINSEQ_1:1;
      then reconsider m1 = y-1 as Element of NAT by INT_1:5;
      F.y = n choose m1 by A2,A3;
      then reconsider x as Real by A4;
      x in REAL;
      hence thesis;
    end;
    then reconsider F as FinSequence of REAL by FINSEQ_1:def 4;
    take F;
    thus thesis by A2,FINSEQ_1:def 3;
  end;
  uniqueness
  proof
    let G,H;
    assume that
A5: len G = n+1 and
A6: for i,m be Nat st i in dom G & m = i - 1 holds G.i = n choose m;
    assume that
A7: len H = n+1 and
A8: for i,m be Nat st i in dom H & m = i - 1 holds H.i = n choose m;
    for i st i in dom G holds G.i = H.i
    proof
      let i;
      assume
A9:   i in dom G;
      then
A10:  i in Seg (n+1) by A5,FINSEQ_1:def 3;
      then i>=1 by FINSEQ_1:1;
      then reconsider m = i-1 as Element of NAT by INT_1:5;
      i in dom H by A7,A10,FINSEQ_1:def 3;
      then H.i = n choose m by A8;
      hence thesis by A6,A9;
    end;
    hence thesis by A5,A7,FINSEQ_2:9;
  end;
end;

theorem Th31:
  Newton_Coeff s = (1,1) In_Power s
proof
A1: for i st i in dom (Newton_Coeff s) holds (Newton_Coeff s).i = ((1,1)
  In_Power s).i
  proof
    let i;
    assume
A2: i in dom (Newton_Coeff s);
A3: dom (Newton_Coeff s) = Seg len (Newton_Coeff s) by FINSEQ_1:def 3
      .= Seg (s+1) by Def5;
    then i>=1 by A2,FINSEQ_1:1;
    then reconsider m1 = i-1 as Element of NAT by INT_1:5;
    i<=(s+1) by A2,A3,FINSEQ_1:1;
    then s+1-1>=i-1 by XREAL_1:9;
    then reconsider l1 = s-m1 as Element of NAT by INT_1:5;
    dom (Newton_Coeff s) = Seg len ((1,1) In_Power s) by A3,Def4
      .= dom ((1,1) In_Power s) by FINSEQ_1:def 3;
    then ((1,1) In_Power s).i = (s choose m1)*1|^l1*1|^m1 by A2,Def4
      .= (s choose m1)*1*1|^m1 by Th10
      .= (s choose m1)*1 by Th10
      .= (Newton_Coeff s).i by A2,Def5;
    hence thesis;
  end;
  len (Newton_Coeff s) = s+1 by Def5
    .= len ((1,1) In_Power s) by Def4;
  hence thesis by A1,FINSEQ_2:9;
end;

theorem
  2|^s = Sum(Newton_Coeff s)
proof
  2|^s = (1+1) |^ s .= Sum((1,1) In_Power s) by Th30
    .= Sum(Newton_Coeff s) by Th31;
  hence thesis;
end;

begin 


theorem Th33:
  l >= 1 implies k*l >= k
proof
  assume
A1: l>=1;
  for k holds k*l>=k
  proof
    defpred P[Nat] means $1*l>=$1;
A2: for k st P[k] holds P[k+1]
    proof
      let k;
A3:   (k + 1)*l = k * l + 1*l;
A4:   k+l>=k+1 by A1,XREAL_1:7;
      assume k*l>=k;
      then (k+1)*l>=k+l by A3,XREAL_1:7;
      hence (k+1)*l>=k+1 by A4,XXREAL_0:2;
    end;
A5: P[0];
    thus for k holds P[k] from NAT_1:sch 2(A5,A2);
  end;
  hence thesis;
end;

theorem Th34:
  l >= 1 & n >= k*l implies n >= k
proof
  assume that
A1: l>=1 and
A2: n>=k*l;
  k*l>=k by A1,Th33;
  hence thesis by A2,XXREAL_0:2;
end;

definition
  let n;
  redefine func n! -> Element of NAT;
  coherence by Th16;
end;

theorem Th35:
  l <> 0 implies l divides l!
proof
  assume l<>0;
  then consider n being Nat such that
A1: l = n+1 by NAT_1:6;
  reconsider n as Element of NAT by ORDINAL1:def 12;
  (n+1)! = (n+1) * (n!) by Th15;
  hence thesis by A1,NAT_D:def 3;
end;

theorem
  n <> 0 implies (n+1)/n > 1
proof
  assume
A1: n<>0;
  (n+1)/n=n/n+1/n by XCMPLX_1:62
    .=1+1/n by A1,XCMPLX_1:60;
  hence thesis by A1,XREAL_1:29;
end;

theorem Th37:
  k/(k+1) < 1
proof
  m < m+1 by XREAL_1:29;
  hence thesis by XREAL_1:189;
end;

theorem Th38:
  l! >= l
proof
  defpred P[Nat] means $1! >= $1;
A1: for l st P[l] holds P[l+1]
  proof
    let l;
    assume
A2: l!>=l;
A3: l!*(l+1)=(l+1)! by Th15;
    l=0 & (l+1)!>=(l+1) or l>=1 & (l+1)!>=(l+1)
    proof
      per cases by NAT_1:14;
      case
        l=0;
        hence thesis by FINSEQ_2:50,RVSUM_1:95;
      end;
      case
A4:     l>=1;
        (l+1)!>=(l+1)*l by A2,A3,XREAL_1:64;
        hence thesis by A4,Th34;
      end;
    end;
    hence thesis;
  end;
A5: P[0];
  for l being Nat holds P[l] from NAT_1:sch 2(A5,A1);
  hence thesis;
end;

theorem Th39:
  m<>1 & m divides n implies not m divides (n+1)
proof
  assume that
A1: m<>1 and
A2: m divides n and
A3: m divides (n+1);
  consider t being Nat such that
A4: n = m * t by A2,NAT_D:def 3;
  consider s being Nat such that
A5: n+1 = m * s by A3,NAT_D:def 3;
  t <= s
  proof
    (n+1)*t=m*s*t by A5
      .= n*s by A4;
    then
A6: t=n*s/(n+1) by XCMPLX_1:89
      .=s*(n/(n+1)) by XCMPLX_1:74;
    assume
A7: t > s;
    s>0 by A5;
    hence contradiction by A7,A6,Th37,XREAL_1:157;
  end;
  then reconsider r =s-t as Element of NAT by INT_1:5;
  1=m*s-m*t by A4,A5;
  then 1=m*r;
  hence contradiction by A1,NAT_1:15;
end;

theorem Th40:
  j<>0 implies j divides (j+k)!
proof
  defpred P[Nat] means for j st j<>0 holds j divides (j+$1)!;
A1: for k st P[k] holds P[k+1]
  proof
    let k;
    assume
A2: for j st j<>0 holds j divides (j+k)!;
    let j;
    assume j<>0;
    then j divides ((j+k)!)*((j+k)+1) by A2,NAT_D:9;
    hence thesis by Th15;
  end;
A3: P[0] by Th35;
  for k holds P[k] from NAT_1:sch 2(A3,A1);
  hence thesis;
end;

theorem Th41:
  j<=l & j<>0 implies j divides l!
proof
  assume that
A1: j<=l and
A2: j<>0;
  ex k being Nat st l=j+k by A1,NAT_1:10;
  hence thesis by A2,Th40;
end;

theorem
  j<>1 & j<>0 & j divides (l!+1) implies j>l by Th39,Th41;



theorem
  m lcm (n lcm k) = (m lcm n) lcm k
proof
  set M = n lcm k;
  set K = m lcm M;
  set N = m lcm n;
  set L = N lcm k;
A1: m divides K by NAT_D:def 4;
A2: M divides K by NAT_D:def 4;
  n divides M by NAT_D:def 4;
  then n divides K by A2,NAT_D:4;
  then
A3: N divides K by A1,NAT_D:def 4;
  k divides M by NAT_D:def 4;
  then k divides K by A2,NAT_D:4;
  then
A4: L divides K by A3,NAT_D:def 4;
A5: N divides L by NAT_D:def 4;
A6: k divides L by NAT_D:def 4;
  n divides N by NAT_D:def 4;
  then n divides L by A5,NAT_D:4;
  then
A7: M divides L by A6,NAT_D:def 4;
  m divides N by NAT_D:def 4;
  then m divides L by A5,NAT_D:4;
  then K divides L by A7,NAT_D:def 4;
  hence thesis by A4,NAT_D:5;
end;

theorem Th44:
  m divides n iff m lcm n = n
proof
  thus m divides n implies m lcm n = n
  proof
    set M = m lcm n;
    assume m divides n;
    then
A1: M divides n by NAT_D:def 4;
    n divides M by NAT_D:def 4;
    hence thesis by A1,NAT_D:5;
  end;
  thus thesis by NAT_D:def 4;
end;

theorem
  n divides m & k divides m iff n lcm k divides m
proof
  n lcm k divides m implies n divides m & k divides m
  proof
    set M = n lcm k;
A1: n divides M by NAT_D:def 4;
A2: k divides M by NAT_D:def 4;
    assume n lcm k divides m;
    hence thesis by A1,A2,NAT_D:4;
  end;
  hence thesis by NAT_D:def 4;
end;

theorem
  m lcm 1 = m
proof
  set M = m lcm 1;
  1 divides m by NAT_D:6;
  then
A1: M divides m by NAT_D:def 4;
  m divides M by NAT_D:def 4;
  hence thesis by A1,NAT_D:5;
end;

theorem Th47:
  m lcm n divides m*n
proof
  set s=m*n;
A1: n divides s by NAT_D:9;
  m divides s by NAT_D:9;
  hence thesis by A1,NAT_D:def 4;
end;

theorem
  m gcd (n gcd k) = (m gcd n) gcd k
proof
  set M = n gcd k;
  set K = m gcd M;
  set N = m gcd n;
  set L = N gcd k;
A1: K divides M by NAT_D:def 5;
A2: K divides m by NAT_D:def 5;
  M divides n by NAT_D:def 5;
  then K divides n by A1,NAT_D:4;
  then
A3: K divides N by A2,NAT_D:def 5;
A4: L divides N by NAT_D:def 5;
A5: L divides k by NAT_D:def 5;
  N divides n by NAT_D:def 5;
  then L divides n by A4,NAT_D:4;
  then
A6: L divides M by A5,NAT_D:def 5;
  N divides m by NAT_D:def 5;
  then L divides m by A4,NAT_D:4;
  then
A7: L divides K by A6,NAT_D:def 5;
  M divides k by NAT_D:def 5;
  then K divides k by A1,NAT_D:4;
  then K divides L by A3,NAT_D:def 5;
  hence thesis by A7,NAT_D:5;
end;

theorem Th49:
  n divides m implies n gcd m = n
proof
  set N = n gcd m;
  assume n divides m;
  then
A1: n divides N by NAT_D:def 5;
  N divides n by NAT_D:def 5;
  hence thesis by A1,NAT_D:5;
end;

theorem
  m divides n & m divides k iff m divides n gcd k
proof
  m divides n gcd k implies m divides n & m divides k
  proof
    set M = n gcd k;
A1: M divides n by NAT_D:def 5;
A2: M divides k by NAT_D:def 5;
    assume m divides n gcd k;
    hence thesis by A1,A2,NAT_D:4;
  end;
  hence thesis by NAT_D:def 5;
end;

theorem
  m gcd 1 = 1
proof
  set M = m gcd 1;
A1: 1 divides M by NAT_D:6;
  M divides 1 by NAT_D:def 5;
  hence thesis by A1,NAT_D:5;
end;

theorem
  m gcd 0 = m
proof
  set M = m gcd 0;
  m divides 0 by NAT_D:6;
  then
A1: m divides M by NAT_D:def 5;
  M divides m by NAT_D:def 5;
  hence thesis by A1,NAT_D:5;
end;

theorem Th53:
  (m gcd n) lcm n = n
proof
  set M = m gcd n;
  M divides n by NAT_D:def 5;
  hence thesis by Th44;
end;

theorem Th54:
  m gcd (m lcm n) = m
proof
  set M = m lcm n;
  m divides M by NAT_D:def 4;
  hence thesis by Th49;
end;

theorem
  m gcd (m lcm n) = (n gcd m) lcm m
proof
  m gcd (m lcm n) = m by Th54;
  hence thesis by Th53;
end;

theorem
  m divides n implies m gcd k divides n gcd k
proof
  set M = m gcd k;
A1: M divides k by NAT_D:def 5;
  assume
A2: m divides n;
  M divides m by NAT_D:def 5;
  then M divides n by A2,NAT_D:4;
  hence thesis by A1,NAT_D:def 5;
end;

theorem
  m divides n implies k gcd m divides k gcd n
proof
  set M = k gcd m;
A1: M divides k by NAT_D:def 5;
  assume
A2: m divides n;
  M divides m by NAT_D:def 5;
  then M divides n by A2,NAT_D:4;
  hence thesis by A1,NAT_D:def 5;
end;

theorem Th58:
  n > 0 implies n gcd m > 0
proof
  assume that
A1: n>0 and
A2: n gcd m <= 0;
A3: (n gcd m) divides n by NAT_D:def 5;
  n gcd m = 0 or n gcd m < 0 by A2;
  then ex r being Nat st n = 0*r by A3,NAT_D:def 3;
  hence contradiction by A1;
end;

theorem
  m > 0 & n > 0 implies m lcm n > 0
proof
  assume that
A1: m>0 and
A2: n>0 and
A3: m lcm n <= 0;
A4: (m lcm n) divides m*n by Th47;
  m lcm n = 0 or m lcm n < 0 by A3;
  then ex r being Nat st m*n = 0*r by A4,NAT_D:def 3;
  hence contradiction by A1,A2;
end;

theorem
  (n gcd m) lcm (n gcd k) divides n gcd (m lcm k)
proof
  set M = m lcm k;
  set N = n gcd k;
  set P = n gcd m;
  set L = P lcm N;
A1: N divides k by NAT_D:def 5;
  k divides M by NAT_D:def 4;
  then
A2: N divides M by A1,NAT_D:4;
A3: P divides m by NAT_D:def 5;
  m divides M by NAT_D:def 4;
  then P divides M by A3,NAT_D:4;
  then
A4: L divides M by A2,NAT_D:def 4;
A5: P divides n by NAT_D:def 5;
  N divides n by NAT_D:def 5;
  then L divides n by A5,NAT_D:def 4;
  hence thesis by A4,NAT_D:def 5;
end;

theorem
  m divides l implies m lcm (n gcd l) divides (m lcm n) gcd l
proof
  set M = m lcm n;
  set K = M gcd l;
  set N = n gcd l;
A1: m divides M by NAT_D:def 4;
A2: N divides n by NAT_D:def 5;
A3: N divides l by NAT_D:def 5;
  n divides M by NAT_D:def 4;
  then N divides M by A2,NAT_D:4;
  then
A4: N divides K by A3,NAT_D:def 5;
  assume m divides l;
  then m divides K by A1,NAT_D:def 5;
  hence thesis by A4,NAT_D:def 4;
end;

theorem
  n gcd m divides n lcm m
proof
A1: n divides n lcm m by NAT_D:def 4;
  n gcd m divides n by NAT_D:def 5;
  hence thesis by A1,NAT_D:4;
end;



reserve p,q for Nat;
reserve i0,i,i1,i2,i4 for Integer;

theorem
  0 < m implies n mod m = n - m * (n div m)
proof
  reconsider z1=m * (n div m),z2=(n mod m) as Integer;
  assume m > 0;
  then n - z1 = z1 + z2 -z1 by NAT_D:2;
  hence thesis;
end;

theorem
  i2 >= 0 implies i1 mod i2 >= 0 by INT_1:57;

theorem
  i2 > 0 implies i1 mod i2 < i2 by INT_1:58;

theorem
  i2 <> 0 implies i1 = (i1 div i2) * i2 + (i1 mod i2) by INT_1:59;


theorem
  m>0 or n>0 implies ex i,i1 st i*m + i1*n = m gcd n
proof
  defpred P[Nat] means $1>0 & ex i,i1 st i*m+i1*n=$1;
  assume
A1: m>0 or n>0;
A2: ex p being Nat st P[p]
  proof
    per cases by A1;
    suppose
A3:   m>0;
      1*m + 0*n=m;
      hence thesis by A3;
    end;
    suppose
A4:   n>0;
      0*m + 1*n=n;
      hence thesis by A4;
    end;
  end;
  ex p be Nat st P[p] & for n be Nat st P[n] holds p <= n from NAT_1:sch 5
  (A2);
  then consider p being Nat such that
A5: p>0 and
A6: ex i,i0 st i*m+i0*n=p and
A7: for k be Nat st (k>0 & ex i1,i2 st i1*m+i2*n=k) holds p <= k;
  consider i,i0 such that
A8: i*m+i0*n=p by A6;
A9: for k st ex i1,i2 st i1*m+i2*n=k holds p divides k
  proof
    let k;
    given i1,i2 such that
A10: i1*m+i2*n=k;
    consider l,s being Nat such that
A11: k=p*l+s and
A12: s<p by A5,NAT_1:17;
    reconsider l,s as Element of NAT by ORDINAL1:def 12;
    s=0
    proof
      s = k - p*l by A11;
      then
A13:  s=(i1-i*l)*m+(i2-i0*l)*n by A8,A10;
      assume s<>0;
      hence contradiction by A7,A12,A13;
    end;
    hence thesis by A11,NAT_D:def 3;
  end;
A14: for s be Nat st s divides m & s divides n holds s divides p
  proof
    let s be Nat;
    assume that
A15: s divides m and
A16: s divides n;
    consider r being Nat such that
A17: n = s*r by A16,NAT_D:def 3;
    reconsider p9=p,s9=s as Integer;
    consider l being Nat such that
A18: m = s*l by A15,NAT_D:def 3;
    ex i4 st p9=s9*i4
    proof
      take i*l+i0*r;
      thus thesis by A8,A18,A17;
    end;
    hence thesis by INT_1:def 3;
  end;
  0*m + 1*n=n;
  then
A19: p divides n by A9;
  1*m + 0*n=m;
  then p divides m by A9;
  then p= m gcd n by A19,A14,NAT_D:def 5;
  hence thesis by A6;
end;



definition
  func SetPrimes -> Subset of NAT means
  :Def6:
  for n being Nat holds n in it iff n is prime;
  existence
  proof
    defpred P[Nat] means $1 is prime;
    consider X being Subset of NAT such that
A1: for n being Element of NAT holds n in X iff P[n] from SUBSET_1:sch 3;
    take X;
    let n be Nat;
    thus n in X implies n is prime by A1;
    assume
A2: n is prime;
    n in NAT by ORDINAL1:def 12;
    hence thesis by A1,A2;
  end;
  uniqueness
  proof
    let A, B be Subset of NAT such that
A3: for n being Nat holds n in A iff n is prime and
A4: for n being Nat holds n in B iff n is prime;
    thus A c= B
    proof
      let x be set;
      assume
A5:   x in A;
      then reconsider x as Element of NAT;
      x is prime by A3,A5;
      hence thesis by A4;
    end;
    let x be set;
    assume
A6: x in B;
    then reconsider x as Element of NAT;
    x is prime by A4,A6;
    hence thesis by A3;
  end;
end;

registration
  cluster prime for Element of NAT;
  existence by INT_2:28;
  cluster prime for Nat;
  existence by INT_2:28;
end;

definition
  mode Prime is prime Nat;
end;

reserve x for set;

definition
  let p be Nat;
  func SetPrimenumber p -> Subset of NAT means
  :Def7:
  for q being Nat holds q in it iff q < p & q is prime;
  existence
  proof
    defpred P[Nat] means $1 < p & $1 is prime;
    consider X being Subset of NAT such that
A1: for q being Element of NAT holds q in X iff P[q] from SUBSET_1:sch 3;
    take X;
    let q be Nat;
    thus q in X implies q < p & q is prime by A1;
    assume that
A2: q < p and
A3: q is prime;
    q in NAT by ORDINAL1:def 12;
    hence thesis by A1,A2,A3;
  end;
  uniqueness
  proof
    let A, B be Subset of NAT such that
A4: for q being Nat holds q in A iff q < p & q is prime and
A5: for q being Nat holds q in B iff q < p & q is prime;
    thus A c= B
    proof
      let x be set;
      assume
A6:   x in A;
      then reconsider x as Element of NAT;
A7:   x is prime by A4,A6;
      x < p by A4,A6;
      hence thesis by A5,A7;
    end;
    let x be set;
    assume
A8: x in B;
    then reconsider x as Element of NAT;
A9: x is prime by A5,A8;
    x < p by A5,A8;
    hence thesis by A4,A9;
  end;
end;

theorem Th68:
  SetPrimenumber p c= SetPrimes
proof
  let x;
  assume
A1: x in SetPrimenumber p;
  then reconsider x9 = x as Element of NAT;
  x9 is prime by A1,Def7;
  hence thesis by Def6;
end;

theorem
  p <= q implies SetPrimenumber p c= SetPrimenumber q
proof
  assume
A1: p <= q;
  let x;
  assume
A2: x in SetPrimenumber p;
  then reconsider x9 = x as Element of NAT;
  x9 < p by A2,Def7;
  then
A3: x9 < q by A1,XXREAL_0:2;
  x9 is prime by A2,Def7;
  hence thesis by A3,Def7;
end;

theorem Th70:
  SetPrimenumber p c= Seg p
proof
  let x be set;
  assume
A1: x in SetPrimenumber p;
  then reconsider q = x as Element of NAT;
  q is prime by A1,Def7;
  then
A2: 1 <= q by INT_2:def 4;
  q < p by A1,Def7;
  hence thesis by A2;
end;

theorem Th71:
  SetPrimenumber p is finite
proof
  SetPrimenumber p c= Seg p by Th70;
  hence thesis;
end;

registration
  let n be Nat;
  cluster SetPrimenumber n -> finite;
  coherence by Th71;
end;

reserve p for Prime;

theorem Th72:
  ex p st p is prime & p>l
proof
  reconsider two = 2 as Prime by INT_2:28;
  reconsider l as Element of NAT by ORDINAL1:def 12;
  l=0 & (ex p st p is prime & p>l) or l=1 & (ex p st p is prime & p>l) or
  2<=l & ex p st p is prime & p>l
  proof
    per cases by NAT_1:26;
    case
A1:   l=0;
      take two;
      thus thesis by A1;
    end;
    case
A2:   l=1;
      take two;
      thus thesis by A2;
    end;
    case
A3:   2<=l;
      l<=l! by Th38;
      then 2<=l! by A3,XXREAL_0:2;
      then (2+0) <= (l!+1) by XREAL_1:7;
      then consider j being Element of NAT such that
A4:   j is prime and
A5:   j divides (l!+1) by INT_2:31;
      reconsider j9=j as Prime by A4;
      take j9;
A6:   j<>0 by A4,INT_2:def 4;
      j<>1 by A4,INT_2:def 4;
      hence thesis by A5,A6,Th39,Th41;
    end;
  end;
  hence thesis;
end;

Lm2:
  SetPrimenumber 2 = {}
proof
  assume SetPrimenumber 2<>{};
  then consider x being set such that
A1: x in SetPrimenumber 2 by XBOOLE_0:def 1;
  reconsider x as Nat by A1;
A2: x is prime by A1,Def7;
  x<2 by A1,Def7;
  then 0 is prime or 1 is prime by A2,NAT_1:26;
  hence contradiction by INT_2:def 4;
end;

registration
  cluster SetPrimes -> non empty;
  coherence by Def6,INT_2:28;
end;

registration
  cluster SetPrimenumber 2 -> empty;
  coherence by Lm2;
end;

theorem
  SetPrimenumber m c= Seg m
proof
  let x be set;
  assume
A1: x in SetPrimenumber m;
  then reconsider x as Element of NAT;
  x is prime by A1,Def7;
  then
A2: 1 <= x by INT_2:def 4;
  x < m by A1,Def7;
  hence thesis by A2;
end;

theorem
  k>=m implies not k in SetPrimenumber m by Def7;

theorem
  SetPrimenumber n \/ {n} is finite;

theorem Th76:
  for f being Prime, g being Nat st f < g holds SetPrimenumber f
  \/ {f} c= SetPrimenumber g
proof
  let f be Prime,g be Nat;
  assume
A1: f<g;
  let x be set;
  assume
A2: x in SetPrimenumber f\/{f};
  then reconsider x as Nat;
  per cases by A2,XBOOLE_0:def 3;
  suppose
A3: x in SetPrimenumber f;
    then x < f by Def7;
    then
A4: x < g by A1,XXREAL_0:2;
    x is prime by A3,Def7;
    hence thesis by A4,Def7;
  end;
  suppose
    x in {f};
    then x = f by TARSKI:def 1;
    hence thesis by A1,Def7;
  end;
end;

theorem
  k>=m implies not k in SetPrimenumber m by Def7;

Lm3: SetPrimenumber n = {k where k is Element of NAT:k<n & k is prime}
proof
  thus SetPrimenumber n c= {k where k is Element of NAT:k<n & k is prime}
  proof
    let x be set;
    assume
A1: x in SetPrimenumber n;
    then reconsider x as Element of NAT;
A2: x is prime by A1,Def7;
    x < n by A1,Def7;
    hence thesis by A2;
  end;
  let x be set;
  assume x in {k where k is Element of NAT:k<n & k is prime};
  then ex k be Element of NAT st x=k & k<n & k is prime;
  hence thesis by Def7;
end;

definition
  let n be Nat;
  func primenumber n -> prime Element of NAT means
  :Def8:
  n = card SetPrimenumber it;
  existence
  proof
    defpred P1[Nat] means ex m st $1=card SetPrimenumber m & m is Prime;
A1: for n st P1[n] holds P1[n+1]
    proof
      let n;
      given m such that
A2:   n=card SetPrimenumber m and
A3:   m is Prime;
      defpred P[Nat] means $1 is prime & $1>m;
A4:   not m in SetPrimenumber m by Def7;
A5:   ex k be Nat st P[k]
      proof
        ex p st p is prime & p>m by Th72;
        hence thesis;
      end;
      consider k be Nat such that
A6:   P[k] & for l be Nat st P[l] holds k <= l from NAT_1:sch 5(A5);
      take k;
A7:   SetPrimenumber k c= SetPrimenumber m\/{m}
      proof
        let s be set;
        assume
A8:     s in SetPrimenumber k;
        then reconsider s as Element of NAT;
A9:     s is prime by A8,Def7;
        s<k by A8,Def7;
        then
A10:    s<=m by A6,A9;
        per cases by A10,XXREAL_0:1;
        suppose
          s=m;
          then s in {m} by TARSKI:def 1;
          hence thesis by XBOOLE_0:def 3;
        end;
        suppose
A11:      s<m;
A12:      SetPrimenumber m c= SetPrimenumber m\/{m} by XBOOLE_1:7;
          s in SetPrimenumber m by A9,A11,Def7;
          hence thesis by A12;
        end;
      end;
      SetPrimenumber m\/{m} c= SetPrimenumber k by A3,A6,Th76;
      then SetPrimenumber k=SetPrimenumber m\/{m} by A7,XBOOLE_0:def 10;
      hence n + 1 = card SetPrimenumber k by A2,A4,CARD_2:41;
      thus thesis by A6;
    end;
    SetPrimenumber 2 = {k where k is Element of NAT:k<2 & k is prime} by Lm3;
    then
A13: P1[0] by CARD_1:27,INT_2:28;
    for n holds P1[n] from NAT_1:sch 2(A13,A1);
    then consider m such that
A14: n=card SetPrimenumber m and
A15: m is Prime;
    m in NAT by ORDINAL1:def 12;
    hence thesis by A14,A15;
  end;
  uniqueness
  proof
    let f,g be prime Element of NAT;
    assume that
A16: n = card SetPrimenumber f and
A17: n = card SetPrimenumber g;
    assume
A18: f<>g;
    f<g & n+1 <= n or g<f & n+1 <= n
    proof
      per cases by A18,XXREAL_0:1;
      case
A19:    f<g;
A20:    not f in SetPrimenumber f by Def7;
        card (SetPrimenumber f\/{f}) <= card SetPrimenumber g by A19,Th76,
NAT_1:43;
        hence thesis by A16,A17,A20,CARD_2:41;
      end;
      case
A21:    g<f;
A22:    not g in SetPrimenumber g by Def7;
        card (SetPrimenumber g\/{g}) <= card SetPrimenumber f by A21,Th76,
NAT_1:43;
        hence thesis by A16,A17,A22,CARD_2:41;
      end;
    end;
    hence contradiction by NAT_1:13;
  end;
end;

theorem
  SetPrimenumber n = {k where k is Element of NAT: k<n & k is prime} by Lm3;


theorem Th79:
  SetPrimes is infinite
proof
  assume
A1: SetPrimes is finite;
  then reconsider SP = SetPrimes as finite set;
  consider n be Nat such that
A2: SetPrimes,Seg n are_equipotent by A1,FINSEQ_1:56;
  card SetPrimes = card Seg n by A2,CARD_1:5;
  then card SetPrimes = card n by FINSEQ_1:55;
  then
A3: card SP = n by CARD_1:def 2;
  set p=primenumber (n+1);
  set SPp = SetPrimenumber p;
A4: n+1 = card SetPrimenumber p by Def8;
  card SPp <= card SP by Th68,NAT_1:43;
  hence contradiction by A3,A4,NAT_1:13;
end;

registration
  cluster SetPrimes -> non empty infinite;
  coherence by Th79;
end;

reserve i for Nat;



theorem
  i is prime implies for m,n being Nat holds i divides m * n implies
  i divides m or i divides n
proof
  defpred P[Nat] means $1 is prime & ex m,n being Nat st $1 divides m * n &
  not $1 divides m & not $1 divides n;
  assume
A1: i is prime;
  given m,n being Nat such that
A2: i divides m * n and
A3: not( (i divides m or i divides n));
A4: ex i be Nat st P[i] by A1,A2,A3;
  consider p9 being Nat such that
A5: P[p9] and
A6: for p1 being Nat st P[p1] holds p9 <= p1 from NAT_1:sch 5(A4);
  defpred Q[Nat] means ex n being Nat st p9 divides $1 * n & not p9 divides $1
  & not p9 divides n;
A7: ex i be Nat st Q[i] by A5;
  consider m be Nat such that
A8: Q[m] and
A9: for m1 being Nat st Q[m1] holds m <= m1 from NAT_1:sch 5(A7);
A10: m > 0 by A8;
  consider n being Nat such that
A11: p9 divides m * n and
A12: not p9 divides m and
A13: not p9 divides n by A8;
A14: m < p9
  proof
    set m1 = m mod p9;
    assume
A15: m >= p9;
A16: p9 > 0 by A5,INT_2:def 4;
    then m = p9* (m div p9) + m1 by NAT_D:2;
    then
A17: m * n = p9 * (m div p9) * n + m1 * n;
    m1 < p9 by A16,NAT_D:1;
    then
A18: m1 < m by A15,XXREAL_0:2;
A19: p9 divides p9 * (m div p9) by NAT_D:def 3;
    p9 divides p9 * ((m div p9) * n) by NAT_D:def 3;
    then
A20: p9 divides m1 * n by A11,A17,NAT_D:10;
    m = p9* (m div p9) + m1 by A16,NAT_D:2;
    then not p9 divides m1 by A12,A19,NAT_D:8;
    hence contradiction by A9,A13,A18,A20;
  end;
A21: m < 2
  proof
    assume m >= 2;
    then consider p1 being Element of NAT such that
A22: p1 is prime and
A23: p1 divides m by INT_2:31;
A24: p1 > 1 by A22,INT_2:def 4;
A25: not p1 divides p9
    proof
      assume p1 divides p9;
      then p1 = 1 or p1 = p9 by A5,INT_2:def 4;
      hence contradiction by A8,A22,A23,INT_2:def 4;
    end;
    p1 <= m by A10,A23,NAT_D:7;
    then
A26: not p9 <= p1 by A14,XXREAL_0:2;
A27: p1 <> 0 by A22,INT_2:def 4;
    consider k being Nat such that
A28: m * n = p9 * k by A11,NAT_D:def 3;
    p1 divides p9 * k by A23,A28,NAT_D:9;
    then p1 divides k by A6,A22,A26,A25;
    then consider k1 being Nat such that
A29: k = p1 * k1 by NAT_D:def 3;
    consider m1 being Nat such that
A30: m = p1 * m1 by A23,NAT_D:def 3;
A31: m1 > 0 by A8,A30;
A32: m1 <> m
    proof
      assume m1 = m;
      then m = 1 * m1;
      hence contradiction by A24,A30,A31,XREAL_1:68;
    end;
    m1 divides m by A30,NAT_D:def 3;
    then m1 <= m by A10,NAT_D:7;
    then
A33: m1 < m by A32,XXREAL_0:1;
    (p9 * k1) * p1 = p1 *(m1 * n) by A28,A29,A30;
    then m1 * n = p9 * k1 by A27,XCMPLX_1:5;
    then
A34: p9 divides m1 * n by NAT_D:def 3;
    not p9 divides m1 by A12,A30,NAT_D:9;
    hence contradiction by A9,A13,A33,A34;
  end;
  m >= 2
  proof
A35: m >= 0 + 1 by A10,NAT_1:13;
    assume
A36: m < 2;
    then m <= 1 + 1;
    then m = 1 by A36,A35,NAT_1:9;
    hence contradiction by A11,A13;
  end;
  hence thesis by A21;
end;

theorem
  for x being complex number holds x |^ 2 = x * x & x^2 = x |^ 2
proof
  let x be complex number;
  thus x * x = (x |^ 1) * x by Th5
    .= x |^ (1+1) by Th6
    .= x |^ 2;
  hence thesis;
end;



theorem
  m qua Integer div n = m div n & m qua Integer mod n = m mod n;



reserve x for real number;

theorem
  x > 0 implies x|^k > 0
proof
  defpred P[Nat] means for x st x > 0 holds x|^$1 > 0;
A1: for k holds P[k] implies P[k+1]
  proof
    let k such that
A2: for x st x > 0 holds x|^k > 0;
    let x;
A3: x|^(k+1) = x * x|^k by Th6;
    assume
A4: x > 0;
    then x|^k > 0 by A2;
    hence thesis by A4,A3;
  end;
A5: P[0] by RVSUM_1:94;
  for k holds P[k] from NAT_1:sch 2(A5,A1);
  hence thesis;
end;



theorem
  n > 0 implies 0 |^ n = 0
proof
  assume n>0;
  then n>=1+0 by NAT_1:13;
  hence thesis by Th11;
end;



definition
  let m,n be Element of NAT;
  redefine func m|^n -> Element of NAT;
  coherence by ORDINAL1:def 12;
end;



defpred P[Nat] means 2|^$1 >= $1 + 1;

Lm4: P[0] by RVSUM_1:94;

Lm5: for n st P[n] holds P[n+1]
proof
  let n;
  assume 2|^n >= n+1;
  then 2|^n - n >= 1 by XREAL_1:19;
  then
A1: 2*(2|^n - n) >= 2*1 by XREAL_1:64;
  2|^(n+1) - (n + 1 + 1) = 2*2|^n - (2*n - n + 2) by Th6
    .= 2*(2|^n - n) + (n - 2);
  then 2|^(n+1) - (n + 1 + 1) >= 2 + (n - 2) by A1,XREAL_1:6;
  then 2|^(n+1) >= 0 + (n + 1 + 1) by XREAL_1:19;
  hence 2|^(n+1) >= n + 1 + 1;
end;

theorem Th85:
  2|^n >= n + 1
proof
  for n holds P[n] from NAT_1:sch 2(Lm4,Lm5);
  hence thesis;
end;

theorem
  2|^n > n
proof
  2|^n >= n + 1 by Th85;
  hence thesis by NAT_1:13;
end;



reserve k for Element of NAT;

scheme
  Euklides9 { F(Nat) -> Element of NAT, G(Nat) -> Element of NAT,
    a() -> Element of NAT, b() -> Element of NAT } :
    ex k st F(k) = a() gcd b() & G(k) = 0
provided
A1: 0 < b() and
A2: b() < a() and
A3: F(0) = a() and
A4: G(0) = b() and
A5: for k st G(k) > 0 holds F(k+1) = G(k) & G(k+1) = F(k) mod G(k)
proof
  deffunc F(Nat,set) = IFEQ($2,0,0,G($1));
  consider q being Function of NAT,NAT such that
A6: q.0 = a() and
A7: for i being Nat holds q.(i+1) = F(i,q.i) from NAT_1:sch 12;
  deffunc Q(Element of NAT) = q.$1;
  q.(0+1) = IFEQ(q.0,0,0,G(0)) by A7
    .= G(0) by A2,A6,FUNCOP_1:def 8;
  then
A8: Q(0) = a() & Q(1 qua Element of NAT) = b() by A4,A6;
A9: for k st q.k qua Element of NAT > 0 holds q.k = F(k)
  proof
    let k such that
A10: q.k qua Element of NAT > 0;
    now
      per cases;
      case
        k = 0;
        hence thesis by A3,A6;
      end;
      case
        k <> 0;
        then consider i being Nat such that
A11:    k = i+1 by NAT_1:6;
        reconsider i as Element of NAT by ORDINAL1:def 12;
A12:    now
          assume
A13:      q.i qua Element of NAT = 0;
          q.k = IFEQ(q.i,0,0,G(i)) by A7,A11
            .= 0 by A13,FUNCOP_1:def 8;
          hence contradiction by A10;
        end;
        q.k = IFEQ(q.i,0,0,G(i)) by A7,A11
          .= G(i) by A12,FUNCOP_1:def 8;
        hence thesis by A5,A10,A11;
      end;
    end;
    hence thesis;
  end;
A14: for k holds Q(k+2) = Q(k) mod Q(k+1)
  proof
    let k;
    now
      per cases;
      case
A15:    q.(k+1) <> 0;
A16:    now
A17:      q.k = 0 or q.k <> 0;
          assume
A18:      G(k) = 0;
          q.(k+1) = IFEQ(q.k,0,0,G(k)) by A7
            .= 0 by A18,A17,FUNCOP_1:def 8;
          hence contradiction by A15;
        end;
A19:    q.(k+(1+1)) = q.((k+1)+1) .= IFEQ(q.(k+1),0,0,G(k+1)) by A7
          .= G(k+1) by A15,FUNCOP_1:def 8
          .= F(k) mod G(k) by A5,A16;
A20:    now
          assume
A21:      q.k = 0;
          q.(k+1) = IFEQ(q.k,0,0,G(k)) by A7
            .= 0 by A21,FUNCOP_1:def 8;
          hence contradiction by A15;
        end;
        q.(k+1) = IFEQ(q.k,0,0,G(k)) by A7
          .= G(k) by A20,FUNCOP_1:def 8;
        hence thesis by A9,A19,A20;
      end;
      case
A22:    q.(k+1) = 0;
        thus q.(k+2) = q.((k+1)+1) .= IFEQ(q.(k+1),0,0,G(k+1)) by A7
          .= 0 by A22,FUNCOP_1:def 8
          .= q.(k) mod q.(k+1) by A22,NAT_D:def 2;
      end;
    end;
    hence thesis;
  end;
A23: 0 < b() & b() < a() by A1,A2;
  consider k such that
A24: Q(k) = a() gcd b() & Q(k + 1) = 0 from NAT_D:sch 1(A23,A8,A14);
  take k;
  thus F(k) = a() gcd b() by A1,A9,A24,Th58;
  a() gcd b() > 0 by A1,Th58;
  hence G(k) = IFEQ(q.k,0,0,G(k)) by A24,FUNCOP_1:def 8
    .= 0 by A7,A24;
end;



reserve k,n,n1,n2,m1,m2 for Nat;

theorem Th87:
  r <> 0 or n = 0 iff r|^n <> 0
proof
  defpred P[Nat] means r <> 0 or $1 = 0 iff r|^$1 <> 0;
A1: P[k] implies P[k+1]
  proof
    assume
A2: P[k];
    r|^(k+1) = r|^k*r by Th6;
    hence thesis by A2;
  end;
A3: P[0] by RVSUM_1:94;
  P[k] from NAT_1:sch 2(A3,A1);
  hence thesis;
end;

Lm6: (2|^n1)*(2*m1+1) = (2|^n2)*(2*m2+1) implies n1 <= n2
proof
  assume
A1: (2|^n1)*(2*m1+1) = (2|^n2)*(2*m2+1);
  assume
A2: n1 > n2;
  then consider n being Nat such that
A3: n1 = n2+n by NAT_1:10;
  n <> 0 by A2,A3;
  then consider k being Nat such that
A4: n = k+1 by NAT_1:6;
A5: 2|^n2 <> 0 by Th87;
  reconsider k as Element of NAT by ORDINAL1:def 12;
  2|^n1 = (2|^n2)*(2|^(k+1)) by A3,A4,Th8;
  then (2|^n1)*(2*m1+1) = 2|^n2*(2|^(k+1)*(2*m1+1));
  then 2|^(k+1)*(2*m1+1) = 2*m2+1 by A1,A5,XCMPLX_1:5;
  then 2*m2+1 = 2|^k*(2|^1)*(2*m1+1) by Th8
    .= 2*(2|^k)*(2*m1+1) by Th5
    .= 2*((2|^k)*(2*m1+1));
  then
A6: 2 divides 2*m2+1 by NAT_D:def 3;
  2 divides 2*m2 by NAT_D:def 3;
  hence contradiction by A6,NAT_D:7,10;
end;

theorem
  (2|^n1)*(2*m1+1) = (2|^n2)*(2*m2+1) implies n1 = n2 & m1 = m2
proof
A1: 2|^n1 <> 0 by Th87;
  assume
A2: (2|^n1)*(2*m1+1) = (2|^n2)*(2*m2+1);
  then
A3: n2 <= n1 by Lm6;
  n1 <= n2 by A2,Lm6;
  hence n1 = n2 by A3,XXREAL_0:1;
  then 2*m1+1 = 2*m2+1 by A2,A1,XCMPLX_1:5;
  hence thesis;
end;

theorem
  k <= n implies m |^ k divides m |^ n
proof
  assume k <= n;
  then consider t being Nat such that
A1: n = k + t by NAT_1:10;
  reconsider t as Element of NAT by ORDINAL1:def 12;
  m |^ n = (m |^ k)*(m |^ t) by A1,Th8;
  hence thesis by NAT_D:def 3;
end;















begin

reserve X,Y,Z,x,y,y1,y2 for set,
  D for non empty set,
  k,n,n1,n2,m2,m1 for Nat,

  L,K,M,N for Cardinal,
  f,g for Function;

theorem
  X is finite implies X is countable;

theorem
  omega is countable;

reserve r for Real;

theorem Th3:
  r <> 0 or n = 0 iff r|^n <> 0
proof
  defpred P[Nat] means r <> 0 or $1 = 0 iff r|^$1 <> 0;
A1: P[k] implies P[k+1]
  proof
A2: r|^(k+1) = r|^k*r by NEWTON:6;
    assume P[k];
    hence thesis by A2;
  end;
A3: P[0] by NEWTON:4;
  P[k] from NAT_1:sch 2(A3,A1);
  hence thesis;
end;

Lm1: (2|^n1)*(2*m1+1) = (2|^n2)*(2*m2+1) implies n1 <= n2
proof
  assume
A1: (2|^n1)*(2*m1+1) = (2|^n2)*(2*m2+1);
  assume
A2: n1 > n2;
  then consider n being Nat such that
A3: n1 = n2+n by NAT_1:10;
  n <> 0 by A2,A3;
  then consider k being Nat such that
A4: n = k+1 by NAT_1:6;
A5: 2|^n2 <> 0 by Th3;
  reconsider k as Element of NAT by ORDINAL1:def 12;
  2|^n1 = (2|^n2)*(2|^(k+1)) by A3,A4,NEWTON:8;
  then (2|^n1)*(2*m1+1) = 2|^n2*(2|^(k+1)*(2*m1+1));
  then 2|^(k+1)*(2*m1+1) = 2*m2+1 by A1,A5,XCMPLX_1:5;
  then 2*m2+1 = 2|^k*(2|^1)*(2*m1+1) by NEWTON:8
    .= 2*(2|^k)*(2*m1+1) by NEWTON:5
    .= 2*((2|^k)*(2*m1+1));
  then
A6: 2 divides 2*m2+1 by NAT_D:def 3;
  2 divides 2*m2 by NAT_D:def 3;
  hence contradiction by A6,NAT_D:7,10;
end;

theorem Th4:
  (2|^n1)*(2*m1+1) = (2|^n2)*(2*m2+1) implies n1 = n2 & m1 = m2
proof
A1: 2|^n1 <> 0 by Th3;
  assume
A2: (2|^n1)*(2*m1+1) = (2|^n2)*(2*m2+1);
  then n1 <= n2 & n2 <= n1 by Lm1;
  hence n1 = n2 by XXREAL_0:1;
  then 2*m1+1 = 2*m2+1 by A2,A1,XCMPLX_1:5;
  hence thesis;
end;

Lm2: x in [:NAT,NAT:] implies ex n1,n2 being Element of NAT st x = [n1,n2]
proof
  assume
A1: x in [:NAT,NAT:];
  then reconsider n1 = x`1, n2 = x`2 as Element of NAT by MCART_1:10;
  take n1,n2;
  thus thesis by A1,MCART_1:21;
end;

theorem Th5:
  [:NAT,NAT:],NAT are_equipotent & card NAT = card [:NAT,NAT:]
proof
  [:NAT,{0}:] c= [:NAT,NAT:] by ZFMISC_1:95;
  then card [:NAT,{0}:] c= card [:NAT,NAT:] by CARD_1:11;
  then
A1: card NAT c= card [:NAT,NAT:] by CARD_1:69;
  deffunc f(Element of NAT,Element of NAT) = (2|^$1)*(2*$2+1);
  consider f being Function of [:NAT,NAT:],NAT such that
A2: for n,m being Element of NAT holds f.(n,m) = f(n,m) from BINOP_1:sch
  4;
A3: f is one-to-one
  proof
    let x,y;
    assume x in dom f;
    then consider n1,m1 being Element of NAT such that
A4: x = [n1,m1] by Lm2;
    assume y in dom f;
    then consider n2,m2 being Element of NAT such that
A5: y = [n2,m2] by Lm2;
    assume
A6: f.x = f.y;
A7: (2|^n1)*(2*m1+1) = f.(n1,m1) by A2
      .= f.(n2,m2) by A4,A5,A6
      .= (2|^n2)*(2*m2+1) by A2;
    then n1 = n2 by Th4;
    hence thesis by A4,A5,A7,Th4;
  end;
  dom f = [:NAT,NAT:] & rng f c= NAT by FUNCT_2:def 1;
  then card [:NAT,NAT:] c= card NAT by A3,CARD_1:10;
  then card NAT = card [:NAT,NAT:] by A1,XBOOLE_0:def 10;
  hence thesis by CARD_1:5;
end;

theorem Th6:
  (omega)*`(omega) = omega by Th5,CARD_1:47,CARD_2:def 2;

theorem Th7:
  X is countable & Y is countable implies [:X,Y:] is countable
proof
  assume card X c= omega & card Y c= omega;
  then [:card X,card Y:] c= [:omega,omega:] by ZFMISC_1:96;
  then card [:card X,card Y:] c= card [:omega,omega:] by CARD_1:11;
  then card [:card X,card Y:] c= (omega)*`omega by CARD_2:def 2;
  hence card [:X,Y:] c= omega by Th6,CARD_2:7;
end;

theorem Th8:
  1-tuples_on D,D are_equipotent & card (1-tuples_on D) = card D
proof
  deffunc f(set) = <*$1*>;
  consider f such that
A1: dom f = D & for x st x in D holds f.x = f(x) from FUNCT_1:sch 3;
  D,1-tuples_on D are_equipotent
  proof
    take f;
    thus f is one-to-one
    proof
      let x,y;
      assume x in dom f & y in dom f;
      then
A2:   f.x = <*x*> & f.y = <*y*> by A1;
      <*x*>.1 = x by FINSEQ_1:def 8;
      hence thesis by A2,FINSEQ_1:def 8;
    end;
    thus dom f = D by A1;
    thus rng f c= 1-tuples_on D
    proof
      let x;
      assume x in rng f;
      then consider y such that
A3:   y in dom f and
A4:   x = f.y by FUNCT_1:def 3;
      reconsider y as Element of D by A1,A3;
      x = <*y*> by A1,A4;
      then x in { <*d*> where d is Element of D : not contradiction };
      hence thesis by FINSEQ_2:96;
    end;
    let x;
    assume x in 1-tuples_on D;
    then reconsider y = x as Element of 1-tuples_on D;
    consider z being Element of D such that
A5: y = <*z*> by FINSEQ_2:97;
    x = f.z by A1,A5;
    hence thesis by A1,FUNCT_1:def 3;
  end;
  hence thesis by CARD_1:5;
end;

reserve p,q for FinSequence,
  k,m,n,n1,n2,n3 for Element of NAT;

theorem Th9:
  [:n-tuples_on D, m-tuples_on D:],(n+m)-tuples_on D
are_equipotent & card [:n-tuples_on D, m-tuples_on D:] = card ((n+m)-tuples_on
  D)
proof
  defpred P[set,set] means ex p be Element of n-tuples_on D,q be Element of m
  -tuples_on D st $1 = [p,q] & $2 = p^q;
  set A = [:n-tuples_on D, m-tuples_on D:];
  set B = (n+m)-tuples_on D;
A1: for x st x in A ex y st P[x,y]
  proof
    let x;
    assume
A2: x in A;
    then reconsider p = x`1 as Element of n-tuples_on D by MCART_1:10;
    reconsider q = x`2 as Element of m-tuples_on D by A2,MCART_1:10;
    reconsider y = p^q as set;
    take y;
    x = [x`1,x`2] by A2,MCART_1:21;
    hence thesis;
  end;
  consider f such that
A3: dom f = A & for x st x in A holds P[x,f.x] from CLASSES1:sch 1(A1);
  thus [:n-tuples_on D, m-tuples_on D:],(n+m)-tuples_on D are_equipotent
  proof
    take f;
    thus f is one-to-one
    proof
      let x,y;
      assume x in dom f;
      then consider
      p1 be Element of n-tuples_on D, q1 be Element of m-tuples_on D
      such that
A4:   x = [p1,q1] and
A5:   f.x = p1^q1 by A3;
      assume y in dom f;
      then consider
      p2 be Element of n-tuples_on D, q2 be Element of m-tuples_on D
      such that
A6:   y = [p2,q2] and
A7:   f.y = p2^q2 by A3;
      assume
A8:   f.x = f.y;
A9:   len p1 = n & len p2 = n by CARD_1:def 7;
      then consider p such that
A10:  p1^p = p2 by A5,A7,A8,FINSEQ_1:47;
      consider q such that
A11:  p2^q = p1 by A5,A7,A8,A9,FINSEQ_1:47;
      len p1+0 = len(p1^p)+len q by A10,A11,FINSEQ_1:22
        .= len p1+len p+len q by FINSEQ_1:22
        .= len p1+(len p+len q);
      then p = {};
      then p1 = p2 by A10,FINSEQ_1:34;
      hence thesis by A4,A5,A6,A7,A8,FINSEQ_1:33;
    end;
    thus dom f = A by A3;
    thus rng f c= B
    proof
      let x;
      assume x in rng f;
      then consider y such that
A12:  y in dom f & x = f.y by FUNCT_1:def 3;
      ex p being Element of n-tuples_on D, q being Element of m-tuples_on
      D st y = [p,q] & x = p^q by A3,A12;
      then x is Tuple of n+m, D;
      then x is Element of (n+m)-tuples_on D by FINSEQ_2:131;
      hence thesis;
    end;
    let x;
    assume x in B;
    then reconsider x as Element of B;
    consider p being Element of n-tuples_on D, q being Element of m-tuples_on
    D such that
A13: x = p^q by FINSEQ_2:106;
    consider p1 being Element of n-tuples_on D, q1 being Element of m
    -tuples_on D such that
A14: [p,q] = [p1,q1] and
A15: f.[p,q] = p1^q1 by A3;
    p1 = p & q1 = q by A14,XTUPLE_0:1;
    hence thesis by A3,A13,A15,FUNCT_1:def 3;
  end;
  hence thesis by CARD_1:5;
end;

theorem Th10:
  D is countable implies n-tuples_on D is countable
proof
  defpred P[Nat] means $1-tuples_on D is countable;
  assume card D c= omega;
  then card (1-tuples_on D) c= omega by Th8;
  then
A1: 1-tuples_on D is countable by CARD_3:def 14;
A2: for k being Nat holds P[k] implies P[k+1]
  proof
    let k be Nat;
    assume P[k];
    then [:k-tuples_on D, 1-tuples_on D:] is countable by A1,Th7;
    then
A3: card [:k-tuples_on D, 1-tuples_on D:] c= omega by CARD_3:def 14;
    k in NAT by ORDINAL1:def 12;
    then card [:k-tuples_on D, 1-tuples_on D:] = card ((k+1)-tuples_on D) by
Th9;
    hence thesis by A3,CARD_3:def 14;
  end;
  { <*>D } is countable;
  then
A4: P[0] by FINSEQ_2:94;
  for k being Nat holds P[k] from NAT_1:sch 2(A4,A2);
  hence thesis;
end;

theorem Th11:
  for f st dom f is countable & for x st x in dom f holds f.x is
  countable holds Union f is countable
proof
  let f such that
A1: card dom f c= omega and
A2: for x st x in dom f holds f.x is countable;
  now
    let x;
    assume x in dom f;
    then f.x is countable by A2;
    hence card (f.x) c= omega by CARD_3:def 14;
  end;
  hence card Union f c= omega by A1,Th6,CARD_2:86;
end;

theorem
  (X is countable & for Y st Y in X holds Y is countable) implies union
  X is countable
proof
  assume that
A1: card X c= omega and
A2: for Y st Y in X holds Y is countable;
  now
    let Y;
    assume Y in X;
    then Y is countable by A2;
    hence card Y c= omega by CARD_3:def 14;
  end;
  hence card union X c= omega by A1,Th6,CARD_2:87;
end;

theorem
  D is countable implies D* is countable
proof
  defpred P[set,set] means ex n st $1 = n & $2 = n-tuples_on D;
A1: for x st x in NAT ex y st P[x,y]
  proof
    let x;
    assume x in NAT;
    then reconsider n = x as Element of NAT;
    reconsider y = n-tuples_on D as set;
    take y,n;
    thus thesis;
  end;
  consider f such that
A2: dom f = NAT & for x st x in NAT holds P[x,f.x] from CLASSES1:sch 1(
  A1);
A3: D* = union { k-tuples_on D : not contradiction } by FINSEQ_2:108;
A4: Union f = D*
  proof
    thus Union f c= D*
    proof
      let x;
      assume x in Union f;
      then consider X such that
A5:   x in X and
A6:   X in rng f by TARSKI:def 4;
      consider y such that
A7:   y in dom f and
A8:   X = f.y by A6,FUNCT_1:def 3;
      reconsider y as Element of NAT by A2,A7;
      ex n st y = n & X = n-tuples_on D by A2,A8;
      then X in { k-tuples_on D : not contradiction };
      hence thesis by A3,A5,TARSKI:def 4;
    end;
    let x;
    assume x in D*;
    then consider X such that
A9: x in X & X in { k-tuples_on D : not contradiction } by A3,TARSKI:def 4;
    consider n such that
A10: X = n-tuples_on D by A9;
    ex k st n = k & f.n = k-tuples_on D by A2;
    then X in rng f by A2,A10,FUNCT_1:def 3;
    hence thesis by A9,TARSKI:def 4;
  end;
  assume
A11: D is countable;
  now
    let x;
    assume x in dom f;
    then ex n st x = n & f.x = n-tuples_on D by A2;
    hence f.x is countable by A11,Th10;
  end;
  hence thesis by A2,A4,Th11;
end;

theorem
  omega c= card (D*)
proof
  defpred P[set,set] means ex p st $1 = p & $2 = len p;
A1: {} in D* & len {} = 0 by FINSEQ_1:49;
A2: for x st x in D* ex y st P[x,y]
  proof
    let x;
    assume x in D*;
    then reconsider p = x as Element of (D*);
    reconsider p as FinSequence;
    reconsider y = len p as set;
    take y,p;
    thus thesis;
  end;
  consider f such that
A3: dom f = D* & for x st x in D* holds P[x,f.x] from CLASSES1:sch 1(A2);
  defpred P[Nat] means $1 in f.:(D*);
A4: for n being Nat holds P[n] implies P[n+1]
  proof
    set y = the Element of D;
    let n be Nat;
    assume n in f.:(D*);
    then consider x such that
A5: x in dom f and
A6: x in D* and
A7: n = f.x by FUNCT_1:def 6;
    consider p such that
A8: x = p and
A9: n = len p by A3,A5,A7;
    reconsider p as FinSequence of D by A6,A8,FINSEQ_1:def 11;
A10: len (p^<*y*>) = n+len <*y*> by A9,FINSEQ_1:22
      .= n+1 by FINSEQ_1:40;
A11: p^<*y*> in D* by FINSEQ_1:def 11;
    then ex q being FinSequence st p^<*y*> = q & f.(p^<*y*>) = len q by A3;
    hence thesis by A3,A11,A10,FUNCT_1:def 6;
  end;
  ex p st {} = p & f.{} = len p by A3,FINSEQ_1:49;
  then
A12: P[0] by A3,A1,FUNCT_1:def 6;
A13: for n being Nat holds P[n] from NAT_1:sch 2(A12,A4);
  NAT c= f.:(D*)
  proof
    let x;
    assume x in NAT;
    then reconsider n = x as Element of NAT;
    n in f.:(D*) by A13;
    hence thesis;
  end;
  hence thesis by CARD_1:47,66;
end;

scheme
  FraenCoun1 { f(set)->set, P[set] } : { f(n) : P[n] } is countable proof
  consider f such that
A1: dom f = NAT & for x st x in NAT holds f.x = f(x) from FUNCT_1:sch 3;
  { f(n) : P[n] } c= rng f
  proof
    let x;
    assume x in { f(n) : P[n] };
    then consider n such that
A2: x = f(n) and
    P[n];
    f.n = x by A1,A2;
    hence thesis by A1,FUNCT_1:def 3;
  end;
  hence thesis by A1,CARD_3:93;
end;

scheme
  FraenCoun2 { f(set,set)->set, P[set,set] } : { f(n1,n2) : P[n1,n2] } is
  countable proof
  consider N being Function such that
  N is one-to-one and
A1: dom N = NAT and
A2: rng N = [:NAT,NAT:] by Th5,WELLORD2:def 4;
  deffunc g(set) = f((N.$1)`1,(N.$1)`2);
  consider f such that
A3: dom f = NAT & for x st x in NAT holds f.x = g(x) from FUNCT_1:sch 3;
  { f(n1,n2) : P[n1,n2] } c= rng f
  proof
    let x;
    assume x in { f(n1,n2) : P[n1,n2] };
    then consider n1,n2 such that
A4: x = f(n1,n2) and
    P[n1,n2];
    consider y such that
A5: y in dom N and
A6: [n1,n2] = N.y by A2,FUNCT_1:def 3;
    [n1,n2]`1 = n1 & [n1,n2]`2 = n2;
    then x = f.y by A1,A3,A4,A5,A6;
    hence thesis by A1,A3,A5,FUNCT_1:def 3;
  end;
  hence thesis by A3,CARD_3:93;
end;

scheme
  FraenCoun3 { f(set,set,set)->set, P[set,set,set] } : { f(n1,n2,n3) : P[n1,n2
  ,n3] } is countable proof
  [:NAT,NAT:],[:[:NAT,NAT:],NAT:] are_equipotent by Th5,CARD_2:8;
  then
A1: NAT,[:[:NAT,NAT:],NAT:] are_equipotent by Th5,WELLORD2:15;
  [:[:NAT,NAT:],NAT:] = [:NAT,NAT,NAT:] by ZFMISC_1:def 3;
  then consider N being Function such that
  N is one-to-one and
A2: dom N = NAT and
A3: rng N = [:NAT,NAT,NAT:] by A1,WELLORD2:def 4;
  deffunc g(set) = f((N.$1)`1`1,(N.$1)`1`2,(N.$1)`2);
  consider f such that
A4: dom f = NAT & for x st x in NAT holds f.x = g(x) from FUNCT_1:sch 3;
  { f(n1,n2,n3) : P[n1,n2,n3] } c= rng f
  proof
    reconsider NAT9 = NAT as non empty set;
    let x;
    assume x in { f(n1,n2,n3) : P[n1,n2,n3] };
    then consider n1,n2,n3 such that
A5: x = f(n1,n2,n3) and
    P[n1,n2,n3];
    reconsider n1,n2,n3 as Element of NAT9;
A6: [n1,n2,n3]`3_3 = n3 & [n1,n2,n3]`1_3 = [n1,n2,n3]`1`1;
    consider y such that
A8: y in dom N and
A9: [n1,n2,n3] = N.y by A3,FUNCT_1:def 3;
    [n1,n2,n3]`1_3 = n1 & [n1,n2,n3]`2_3 = n2;
    then x = f.y by A2,A4,A5,A8,A9,A6;
    hence thesis by A2,A4,A8,FUNCT_1:def 3;
  end;
  hence thesis by A4,CARD_3:93;
end;

reserve f,f1,f2 for Function,
  X1,X2 for set;


theorem Th15:
  not M is finite implies M*`M = M
proof
  defpred P[set] means ex f st f = $1 & f is one-to-one & dom f = [:rng f,rng
  f:];
  consider X such that
A1: x in X iff x in PFuncs([:M,M:],M) & P[x] from XBOOLE_0:sch 1;
A2: x in X implies x is Function
  proof
    assume x in X;
    then ex f st f = x & f is one-to-one & dom f = [:rng f,rng f:] by A1;
    hence thesis;
  end;
A3: for Z st Z <> {} & Z c= X & Z is c=-linear holds union Z in X
  proof
    let Z;
    assume that
    Z <> {} and
A4: Z c= X and
A5: Z is c=-linear;
    union Z is Relation-like Function-like
    proof
      set F = union Z;
      thus for x st x in F ex y1,y2 st [y1,y2] = x
      proof
        let x;
        assume x in F;
        then consider Y such that
A6:     x in Y and
A7:     Y in Z by TARSKI:def 4;
        reconsider f = Y as Function by A2,A4,A7;
        for x st x in f ex y1,y2 st [y1,y2] = x by RELAT_1:def 1;
        hence thesis by A6;
      end;
      let x,y1,y2;
      assume [x,y1] in F;
      then consider X1 such that
A8:   [x,y1] in X1 and
A9:   X1 in Z by TARSKI:def 4;
      assume [x,y2] in F;
      then consider X2 such that
A10:  [x,y2] in X2 and
A11:  X2 in Z by TARSKI:def 4;
      reconsider f1 = X1, f2 = X2 as Function by A2,A4,A9,A11;
      X1,X2 are_c=-comparable by A5,A9,A11,ORDINAL1:def 8;
      then X1 c= X2 or X2 c= X1 by XBOOLE_0:def 9;
      then
      [x,y2] in X1 & (for x,y1,y2 st [x,y1] in f1 & [x,y2] in f1 holds y1
= y2) or [x,y1] in X2 & for x,y1,y2 st [x,y1] in f2 & [x,y2] in f2 holds y1 =
      y2 by A8,A10,FUNCT_1:def 1;
      hence thesis by A8,A10;
    end;
    then reconsider f = union Z as Function;
A12: f is one-to-one
    proof
      let x1,x2 be set;
      assume that
A13:  x1 in dom f and
A14:  x2 in dom f;
      [x1,f.x1] in f by A13,FUNCT_1:1;
      then consider X1 such that
A15:  [x1,f.x1] in X1 and
A16:  X1 in Z by TARSKI:def 4;
      [x2,f.x2] in f by A14,FUNCT_1:1;
      then consider X2 such that
A17:  [x2,f.x2] in X2 and
A18:  X2 in Z by TARSKI:def 4;
      consider f2 such that
A19:  f2 = X2 and
A20:  f2 is one-to-one and
      dom f2 = [:rng f2,rng f2:] by A1,A4,A18;
      consider f1 such that
A21:  f1 = X1 and
A22:  f1 is one-to-one and
      dom f1 = [:rng f1,rng f1:] by A1,A4,A16;
      X1, X2 are_c=-comparable by A5,A16,A18,ORDINAL1:def 8;
      then X1 c= X2 or X2 c= X1 by XBOOLE_0:def 9;
      then x1 in dom f1 & x2 in dom f1 & f.x1 = f1.x1 & f.x2 = f1.x2 or x1 in
      dom f2 & x2 in dom f2 & f.x1 = f2.x1 & f.x2 = f2.x2 by A15,A17,A21,A19,
FUNCT_1:1;
      hence thesis by A22,A20,FUNCT_1:def 4;
    end;
A23: dom f = [:rng f,rng f:]
    proof
      thus dom f c= [:rng f,rng f:]
      proof
        let x;
        assume x in dom f;
        then [x,f.x] in f by FUNCT_1:def 2;
        then consider Y such that
A24:    [x,f.x] in Y and
A25:    Y in Z by TARSKI:def 4;
        consider g being Function such that
A26:    g = Y and
        g is one-to-one and
A27:    dom g = [:rng g,rng g:] by A1,A4,A25;
        g c= f by A25,A26,ZFMISC_1:74;
        then rng g c= rng f by RELAT_1:11;
        then
A28:    dom g c= [:rng f,rng f:] by A27,ZFMISC_1:96;
        x in dom g by A24,A26,FUNCT_1:1;
        hence thesis by A28;
      end;
      let x1,x2 be set;
      assume
A29:  [x1,x2] in [:rng f,rng f:];
      [x1,x2]`1 in rng f by A29,MCART_1:10;
      then consider y1 such that
A31:  y1 in dom f & [x1,x2]`1 = f.y1 by FUNCT_1:def 3;
      [x1,x2] `2 in rng f by A29,MCART_1:10;
      then consider y2 such that
A32:  y2 in dom f & [x1,x2]`2 = f.y2 by FUNCT_1:def 3;
      [y2,[x1,x2]`2] in f by A32,FUNCT_1:1;
      then consider X2 such that
A33:  [y2,[x1,x2]`2] in X2 and
A34:  X2 in Z by TARSKI:def 4;
      consider f2 such that
A35:  f2 = X2 and
      f2 is one-to-one and
A36:  dom f2 = [:rng f2,rng f2:] by A1,A4,A34;
      f2 c= f by A34,A35,ZFMISC_1:74;
      then
A37:  dom f2 c= dom f by RELAT_1:11;
      [y1,[x1,x2]`1] in f by A31,FUNCT_1:1;
      then consider X1 such that
A38:  [y1,[x1,x2]`1] in X1 and
A39:  X1 in Z by TARSKI:def 4;
      consider f1 such that
A40:  f1 = X1 and
      f1 is one-to-one and
A41:  dom f1 = [:rng f1,rng f1:] by A1,A4,A39;
      X1, X2 are_c=-comparable by A5,A39,A34,ORDINAL1:def 8;
      then X1 c= X2 or X2 c= X1 by XBOOLE_0:def 9;
      then y1 in dom f1 & y2 in dom f1 & f1.y1 = [x1,x2]`1 & f1.y2 = [x1,x2]
`2 or y1 in dom f2 & y2 in dom f2 & f2.y1 = [x1,x2]`1 & f2.y2 = [x1,x2]`2 by
A38,A33,A40,A35,FUNCT_1:1;
      then [x1,x2]`1 in rng f1 & [x1,x2]`2 in rng f1 or [x1,x2]`1 in rng f2 &
      [x1,x2]`2 in rng f2 by FUNCT_1:def 3;
      then
A42:  [x1,x2] in dom f1 or [x1,x2] in dom f2 by A41,A36,ZFMISC_1:87;
      f1 c= f by A39,A40,ZFMISC_1:74;
      then dom f1 c= dom f by RELAT_1:11;
      hence thesis by A42,A37;
    end;
A43: rng f c= M
    proof
      let y;
      assume y in rng f;
      then consider x such that
A44:  x in dom f & y = f.x by FUNCT_1:def 3;
      [x,y] in union Z by A44,FUNCT_1:def 2;
      then consider Y such that
A45:  [x,y] in Y and
A46:  Y in Z by TARSKI:def 4;
      Y in PFuncs([:M,M:],M) by A1,A4,A46;
      then consider g being Function such that
A47:  Y = g and
      dom g c= [:M,M:] and
A48:  rng g c= M by PARTFUN1:def 3;
      x in dom g & g.x = y by A45,A47,FUNCT_1:1;
      then y in rng g by FUNCT_1:def 3;
      hence thesis by A48;
    end;
    dom f c= [:M,M:]
    proof
      let x;
      assume x in dom f;
      then [x,f.x] in union Z by FUNCT_1:def 2;
      then consider Y such that
A49:  [x,f.x] in Y and
A50:  Y in Z by TARSKI:def 4;
      Y in PFuncs([:M,M:],M) by A1,A4,A50;
      then consider g being Function such that
A51:  Y = g and
A52:  dom g c= [:M,M:] and
      rng g c= M by PARTFUN1:def 3;
      x in dom g by A49,A51,FUNCT_1:1;
      hence thesis by A52;
    end;
    then f in PFuncs([:M,M:],M) by A43,PARTFUN1:def 3;
    hence thesis by A1,A12,A23;
  end;
  consider f such that
A53: f is one-to-one and
A54: dom f = [:omega,omega:] & rng f = omega by Th5,WELLORD2:def 4;
  assume
A55: not M is finite;
  then not M in omega;
  then
A56: omega c= M by CARD_1:4;
  then [:omega,omega:] c= [:M,M:] by ZFMISC_1:96;
  then f in PFuncs([:M,M:],M) by A54,A56,PARTFUN1:def 3;
  then X <> {} by A1,A53,A54;
  then consider Y such that
A57: Y in X and
A58: for Z st Z in X & Z <> Y holds not Y c= Z by A3,ORDERS_1:67;
  consider f such that
A59: f = Y and
A60: f is one-to-one and
A61: dom f = [:rng f,rng f:] by A1,A57;
  set A = rng f;
A62: [:A,A:],A are_equipotent by A60,A61,WELLORD2:def 4;
  Y in PFuncs([:M,M:],M) by A1,A57;
  then
A63: ex f st Y = f & dom f c= [:M,M:] & rng f c= M by PARTFUN1:def 3;
  set N = card A;
A64: card M = M by CARD_1:def 2;
  then
A65: N c= M by A59,A63,CARD_1:11;
A66: now
    (omega \ A) misses A by XBOOLE_1:79;
    then
A67: (omega \ A) /\ A = {} by XBOOLE_0:def 7;
    then [:(omega \ A) /\ A,A /\ (omega \ A):] = {} by ZFMISC_1:90;
    then
A68: [:omega \ A,A:] /\ [:A,omega \ A:] = {} by ZFMISC_1:100;
    [:(omega \ A) /\ (omega \ A),(omega \ A) /\ A:] = {} by A67,ZFMISC_1:90;
    then [:omega \ A,omega \ A:] /\ [:omega \ A,A:] = {} by ZFMISC_1:100;
    then
A69: [:omega \ A,omega \ A:] misses [:omega \ A,A:] by XBOOLE_0:def 7;
    [:A /\ A,(omega \ A) /\ A:] = {} by A67,ZFMISC_1:90;
    then
A70: [:A,omega \ A:] /\ [:A,A:] = {} by ZFMISC_1:100;
    [:(omega \ A) /\ A,A /\ A:] = {} by A67,ZFMISC_1:90;
    then
A71: {} \/ {} = {} & [:omega \ A,A:] /\ [:A,A:] = {} by ZFMISC_1:100;
    [:(omega \ A) /\ A,(omega \ A) /\ A:] = {} by A67,ZFMISC_1:90;
    then [:omega \ A,omega \ A:] /\ [:A,A:] = {} by ZFMISC_1:100;
    then
A72: ([:omega \ A,omega \ A:] \/ [:omega \ A,A:]) /\ [:A,A:] = {} by A71,
XBOOLE_1:23;
A73: omega c= omega +` N & omega +` omega = omega by CARD_2:75,94;
    assume
A74: A is finite;
    then N in omega by CARD_1:47,CARD_3:42;
    then omega +` N c= omega +` omega by CARD_2:83;
    then
A75: omega = omega +` N by A73,XBOOLE_0:def 10;
    N = card card A;
    then (omega)*`N c= omega by A74,CARD_2:89;
    then
A76: omega +` (omega)*`N = omega by CARD_2:76;
    card omega = omega by CARD_1:def 2;
    then
A77: omega = card (omega \ A) by A74,A75,CARD_2:98,CARD_3:42;
    [:(omega \ A) /\ A,(omega \ A) /\ (omega \ A):] = {} by A67,ZFMISC_1:90;
    then [:omega \ A,omega \ A:] /\ [:A,omega \ A:] = {} by ZFMISC_1:100;
    then ([:omega \ A,omega \ A:] \/ [:omega \ A,A:]) /\ [:A,omega \ A:] = {}
    \/ {} by A68,XBOOLE_1:23
      .= {};
    then ([:omega \ A,omega \ A:] \/ [:omega \ A,A:]) misses [:A,omega \ A:]
    by XBOOLE_0:def 7;
    then card ([:omega \ A,omega \ A:] \/ [:omega \ A,A:] \/ [:A,omega \ A:])
= card ([:omega \ A,omega \ A:] \/ [:omega \ A,A:]) +` card [:A,omega \ A:] by
CARD_2:35
      .= card [:omega \ A,omega \ A:] +` card [:omega \ A,A:] +` card [:A,
    omega \ A:] by A69,CARD_2:35
      .= card [:omega \ A,omega \ A:] +` card [:omega,N:] +` card [:A,omega
    \ A:] by A77,CARD_2:7
      .= card [:omega,omega:] +` card [:omega,N:] +` card [:A,omega \ A:] by
A77,CARD_2:7
      .= omega +` card [:omega,N:] +` card [:N,omega:] by A77,Th5,CARD_2:7
      .= omega +` (omega)*`N +` card [:N,omega:] by CARD_2:def 2
      .= omega +` (omega)*`N +` N*`(omega) by CARD_2:def 2;
    then [:omega \ A,omega \ A:] \/ [:omega \ A,A:] \/ [:A,omega \ A:],omega
    \ A are_equipotent by A77,A76,CARD_1:5;
    then consider g being Function such that
A78: g is one-to-one and
A79: dom g = [:omega \ A,omega \ A:] \/ [:omega \ A,A:] \/ [:A,omega \ A:] and
A80: rng g = omega \ A by WELLORD2:def 4;
A81: dom (g+*f) = dom g \/ dom f by FUNCT_4:def 1;
    then
A82: dom (g+*f) = [:omega \ A,(omega \ A) \/ A:] \/ [:A,omega \ A:] \/ [:
    A,A:] by A61,A79,ZFMISC_1:97
      .= [:omega \ A,(omega \ A) \/ A:] \/ ([:A,omega \ A:] \/ [:A,A:]) by
XBOOLE_1:4
      .= [:omega \ A,(omega \ A) \/ A:] \/ [:A,(omega \ A) \/ A:] by
ZFMISC_1:97
      .= [:(omega \ A) \/ A,(omega \ A) \/ A:] by ZFMISC_1:97
      .= [:omega \/ A,(omega \ A) \/ A:] by XBOOLE_1:39
      .= [:omega \/ A,omega \/ A:] by XBOOLE_1:39;
    {} \/ {} = {};
    then dom g /\ dom f = {} by A61,A79,A72,A70,XBOOLE_1:23;
    then
A83: dom g misses dom f by XBOOLE_0:def 7;
    then g c= g+*f by FUNCT_4:32;
    then rng f c= rng (g+*f) & rng g c= rng (g+*f) by FUNCT_4:18,RELAT_1:11;
    then rng (g+*f) c= rng g \/ rng f & rng g \/ rng f c= rng (g+*f) by
FUNCT_4:17,XBOOLE_1:8;
    then
A84: rng (g+*f) = rng g \/ rng f by XBOOLE_0:def 10
      .= omega \/ A by A80,XBOOLE_1:39;
A85: g+*f is one-to-one
    proof
      rng f misses rng g by A80,XBOOLE_1:79;
      then
A86:  rng f /\ rng g = {} by XBOOLE_0:def 7;
      let x,y;
      assume that
A87:  x in dom (g+*f) and
A88:  y in dom (g+*f);
A89:  y in dom g or y in dom f by A81,A88,XBOOLE_0:def 3;
      x in dom f or x in dom g by A81,A87,XBOOLE_0:def 3;
      then (g+*f).x = f.x & (g+*f).y = f.y & (f.x = f.y implies x = y) or (g
+*f).x = g.x & (g+*f).y = g.y & (g.x = g.y implies x = y) or (g+*f).x = f.x & (
g+*f).y = g.y & f.x in rng f & g.y in rng g or (g+*f).x = g.x & (g+*f).y = f.y
      & g.x in rng g & f.y in rng f by A60,A78,A83,A89,FUNCT_1:def 3,def 4
,FUNCT_4:13,16;
      hence thesis by A86,XBOOLE_0:def 4;
    end;
    set x = the Element of omega \ A;
    omega \ A <> {} by A74,A75,CARD_1:68,CARD_3:42;
    then
A90: x in omega & not x in A by XBOOLE_0:def 5;
A91: omega \/ A c= M by A56,A59,A63,XBOOLE_1:8;
    then [:omega \/ A,omega \/ A:] c= [:M,M:] by ZFMISC_1:96;
    then g+*f in PFuncs([:M,M:],M) by A84,A82,A91,PARTFUN1:def 3;
    then g+*f in X by A1,A84,A82,A85;
    then g+*f = f by A58,A59,FUNCT_4:25;
    hence contradiction by A84,A90,XBOOLE_0:def 3;
  end;
A92: now
    N*`N = card [:N,N:] by CARD_2:def 2;
    then
A93: N*`N = card [:A,A:] by CARD_2:7;
    [:A,A:],A are_equipotent by A60,A61,WELLORD2:def 4;
    then
A94: N*`N = N by A93,CARD_1:5;
    assume N <> M;
    then
A95: N in M by A65,CARD_1:3;
    M+`N = M by A55,A65,CARD_2:76;
    then card (M \ A) = M by A64,A95,CARD_2:98;
    then consider h being Function such that
A96: h is one-to-one & dom h = A and
A97: rng h c= M \ A by A65,CARD_1:10;
    set B = rng h;
    A,B are_equipotent by A96,WELLORD2:def 4;
    then
A98: N = card B by CARD_1:5;
    A misses (M \ A) & A /\ B c= A /\ (M \ A) by A97,XBOOLE_1:26,79;
    then A /\ B c= {} by XBOOLE_0:def 7;
    then A /\ B = {};
    then
A99: A misses B by XBOOLE_0:def 7;
    (A \/ B) \ A = B \ A by XBOOLE_1:40
      .= B by A99,XBOOLE_1:83;
    then
A100: [:B,B:] c= [:(A \/ B) \ A,A \/ B:] by ZFMISC_1:96;
    [:(A \/ B) \ A,A \/ B:] c= [:(A \/ B) \ A,A \/ B:] \/ [:A \/ B,(A \/
B ) \ A :] & [:A \/ B,A \/ B:] \ [:A,A:] = [:(A \/ B) \ A,A \/ B:] \/ [:A \/ B,
    (A \/ B ) \ A:] by XBOOLE_1:7,ZFMISC_1:103;
    then
A101: [:B,B:] c= [:A \/ B,A \/ B:] \ [:A,A:] by A100,XBOOLE_1:1;
    N+`N = N by A66,CARD_2:75;
    then card (A \/ B) = N by A98,A99,CARD_2:35;
    then card [:A \/ B,A \/ B:] = card [:N,N:] by CARD_2:7
      .= N by A94,CARD_2:def 2;
    then
A102: card ([:A \/ B,A \/ B:] \ [:A,A:]) c= N by CARD_1:11;
    N = card [:N,N:] by A94,CARD_2:def 2;
    then N = card [:B,B:] by A98,CARD_2:7;
    then N c= card ([:A \/ B,A \/ B:] \ [:A,A:]) by A101,CARD_1:11;
    then card ([:A \/ B,A \/ B:] \ [:A,A:]) = N by A102,XBOOLE_0:def 10;
    then [:A \/ B,A \/ B:] \ [:A,A:],B are_equipotent by A98,CARD_1:5;
    then consider g such that
A103: g is one-to-one and
A104: dom g = [:A \/ B,A \/ B:] \ [:A,A:] and
A105: rng g = B by WELLORD2:def 4;
A106: dom (g+*f) = dom g \/ dom f by FUNCT_4:def 1;
    then A c= A \/ B & dom (g+*f) = [:A \/ B,A \/ B:] \/ [:A,A:] by A61,A104,
XBOOLE_1:7,39;
    then
A107: dom (g+*f) = [:A \/ B,A \/ B:] by XBOOLE_1:12,ZFMISC_1:96;
A108: ([:A \/ B,A \/ B:] \ [:A,A:]) misses [:rng f,rng f:] by XBOOLE_1:79;
A109: g+*f is one-to-one
    proof
      let x,y;
      assume that
A110: x in dom (g+*f) and
A111: y in dom (g+*f);
A112: y in dom g or y in dom f by A106,A111,XBOOLE_0:def 3;
      x in dom f or x in dom g by A106,A110,XBOOLE_0:def 3;
      then
A113: (g+*f).x = f.x & (g+*f).y = f.y & (f.x = f.y implies x = y) or (g
+*f).x = g.x & (g+*f).y = g.y & (g.x = g.y implies x = y) or (g+*f).x = f.x & (
g+*f).y = g.y & f.x in rng f & g.y in rng g or (g+*f).x = g.x & (g+*f).y = f.y
      & g.x in rng g & f.y in rng f by A60,A61,A103,A104,A108,A112,
FUNCT_1:def 3,def 4,FUNCT_4:13,16;
      A misses (M \ A) & A /\ B c= A /\ (M \ A) by A97,XBOOLE_1:26,79;
      then
A114: rng f /\ rng g c= {} by A105,XBOOLE_0:def 7;
      assume (g+*f).x = (g+*f).y;
      hence thesis by A114,A113,XBOOLE_0:def 4;
    end;
    set x = the Element of B;
A115: B <> {} by A66,A98;
    then x in M \ A by A97,TARSKI:def 3;
    then
A116: not x in rng f by XBOOLE_0:def 5;
    g c= g+*f by A61,A104,FUNCT_4:32,XBOOLE_1:79;
    then rng f c= rng (g+*f) & rng g c= rng (g+*f) by FUNCT_4:18,RELAT_1:11;
    then
A117: rng g \/ rng f c= rng(g+*f) by XBOOLE_1:8;
    rng(g+*f) c= rng g \/ rng f by FUNCT_4:17;
    then
A118: rng (g+*f) = rng g \/ rng f by A117,XBOOLE_0:def 10;
    B c= M by A97,XBOOLE_1:1;
    then
A119: A \/ B c= M by A59,A63,XBOOLE_1:8;
    then [:A \/ B,A \/ B:] c= [:M,M:] by ZFMISC_1:96;
    then g+*f in PFuncs([:M,M:],M) by A105,A118,A107,A119,PARTFUN1:def 3;
    then
A120: g+*f in X by A1,A105,A118,A107,A109;
    x in rng (g+*f) by A105,A118,A115,XBOOLE_0:def 3;
    hence contradiction by A58,A59,A120,A116,FUNCT_4:25;
  end;
  then M*`M = card [:N,N:] by CARD_2:def 2
    .= card [:A,A:] by CARD_2:7;
  hence thesis by A92,A62,CARD_1:5;
end;

theorem Th16:
  not M is finite & 0 in N & (N c= M or N in M) implies M*`N = M & N*`M = M
proof
A1: 1*`M = M by CARD_2:21;
  assume not M is finite;
  then
A2: M*`M = M by Th15;
  assume 0 in N;
  then 1 c= N by CARD_2:68;
  then
A3: 1*`M c= N*`M by CARD_2:90;
  assume N c= M or N in M;
  then N*`M c= M*`M by CARD_2:90;
  hence thesis by A2,A3,A1,XBOOLE_0:def 10;
end;

theorem Th17:
  not M is finite & (N c= M or N in M) implies M*`N c= M & N*`M c= M
proof
  assume not M is finite & (N c= M or N in M);
  then M*`N = M or not 0 in N by Th16;
  then M*`N c= M or N = 0 & M*`0 = 0 & 0 c= M
  by CARD_2:20,ORDINAL3:8,XBOOLE_1:2;
  hence thesis;
end;

theorem
  not X is finite implies [:X,X:],X are_equipotent & card [:X,X:] = card X
proof
  assume not X is finite;
  then (card X)*`(card X) = card X by Th15;
  then card [:card X,card X:] = card X by CARD_2:def 2;
  then card [:X,X:] = card X by CARD_2:7;
  hence thesis by CARD_1:5;
end;

theorem
  not X is finite & Y is finite & Y <> {} implies [:X,Y:],X
  are_equipotent & card [:X,Y:] = card X
proof
  assume that
A1: not X is finite and
A2: Y is finite & Y <> {};
  card Y c= card X & 0 in card Y by A1,A2,ORDINAL3:8;
  then (card X)*`(card Y) = card X by A1,Th16;
  then card [:card X,card Y:] = card X by CARD_2:def 2;
  then card [:X,Y:] = card X by CARD_2:7;
  hence thesis by CARD_1:5;
end;

theorem
  K in L & M in N implies K*`M in L*`N & M*`K in L*`N
proof
A1: for K,L,M,N st K in L & M in N & L c= N holds K*`M in L*`N
  proof
    let K,L,M,N;
    assume that
A2: K in L and
A3: M in N and
A4: L c= N;
A5: now
      assume
A6:   N is finite;
      then reconsider N as finite Cardinal;
      reconsider L,M,K as finite Cardinal by A2,A3,A4,A6,CARD_3:92;
A7:   card N = N by CARD_1:def 2;
A8:   card M = M by CARD_1:def 2;
      then card M < card N by A3,A7,NAT_1:41;
      then
A9:   card K * card M <= card K * card N by XREAL_1:64;
A10:  card L = L by CARD_1:def 2;
      then
A11:  L*`N = card (card L * card N) by A7,CARD_2:39;
A12:  card K = K by CARD_1:def 2;
      then card K < card L by A2,A10,NAT_1:41;
      then card K * card N < card L * card N by A3,XREAL_1:68;
      then
A13:  card K * card M < card L * card N by A9,XXREAL_0:2;
      K*`M = card (card K * card M) by A12,A8,CARD_2:39;
      hence thesis by A11,A13,NAT_1:41;
    end;
A14: 0 in L by A2,ORDINAL3:8;
    now
      assume
A15:  not N is finite;
      then
A16:  L*`N = N by A4,A14,Th16;
A17:  omega c= N by A15,CARD_3:85;
A18:  now
        assume K is finite & M is finite;
        then reconsider K,M as finite Cardinal;
        card K = K & card M = M by CARD_1:def 2;
        then K*`M = card (card K * card M) by CARD_2:39
          .= (card K * card M) by CARD_1:def 2;
        hence thesis by A16,A17,TARSKI:def 3;
      end;
      K c= M & (M is finite or not M is finite) or M c= K & (K is finite
      or not K is finite);
      then
      K is finite & M is finite or K*`M c= M or K*`M c= K & K in N by A2,A4
,Th17;
      hence thesis by A3,A16,A18,ORDINAL1:12;
    end;
    hence thesis by A5;
  end;
  L c= N or N c= L;
  hence thesis by A1;
end;

theorem Th21:
  not X is finite implies card X = (omega)*`card X
proof
  assume
A1: not X is finite;
  then omega c= card X by CARD_3:85;
  hence thesis by A1,Th16;
end;

theorem
  X <> {} & X is finite & not Y is finite implies card Y *` card X = card Y
proof
  assume that
A1: X <> {} & X is finite and
A2: not Y is finite;
  card X in card Y & 0 in card X by A1,A2,CARD_3:86,ORDINAL3:8;
  hence thesis by A2,Th16;
end;

theorem Th23:
  not D is finite & n <> 0 implies n-tuples_on D,D are_equipotent
  & card (n-tuples_on D) = card D
proof
  assume that
A1: not D is finite and
A2: n <> 0;
  defpred P[Nat] means $1 <> 0 implies card ($1-tuples_on D) = card D;
A3: for k being Nat holds P[k] implies P[k+1]
  proof
    let k be Nat;
    0-tuples_on D = { <*>D } by FINSEQ_2:94;
    then
A4: 0 in card (0-tuples_on D) & card (0-tuples_on D) c= card D by A1,ORDINAL3:8
;
    k in NAT by ORDINAL1:def 12;
    then
A5: card ((k+1)-tuples_on D) = card [:(k-tuples_on D),1-tuples_on D:] by Th9
      .= card [:card (k-tuples_on D),card (1-tuples_on D):] by CARD_2:7
      .= card [:card (k-tuples_on D),card D:] by Th8
      .= card (k-tuples_on D) *` card D by CARD_2:def 2;
    assume P[k];
    hence thesis by A1,A5,A4,Th16;
  end;
A6: P[0];
  for k being Nat holds P[k] from NAT_1:sch 2(A6,A3);
  then card (n-tuples_on D) = card D by A2;
  hence thesis by CARD_1:5;
end;

theorem
  not D is finite implies card D = card (D*)
proof
  defpred P[set] means not contradiction;
  deffunc f(Element of NAT) = $1-tuples_on D;
A1: D* = union {k-tuples_on D: not contradiction} by FINSEQ_2:108;
  assume
A2: not D is finite;
A3: for X st X in {k-tuples_on D: not contradiction} holds card X c= card D
  proof
    let X;
    assume X in {k-tuples_on D: not contradiction};
    then consider k such that
A4: X = k-tuples_on D;
    0-tuples_on D = { <*>D } by FINSEQ_2:94;
    then card (0-tuples_on D) c= card D & k = 0 or k <> 0 by A2;
    hence thesis by A2,A4,Th23;
  end;
  1-tuples_on D in {k-tuples_on D: not contradiction};
  then card (1-tuples_on D) c= card (D*) by A1,CARD_1:11,ZFMISC_1:74;
  then
A5: card D c= card (D*) by Th8;
  {f(k): P[k]} is countable from FraenCoun1;
  then card {k-tuples_on D: not contradiction} c= omega by CARD_3:def 14;
  then card union {k-tuples_on D: not contradiction} c= omega *` card D by A3,
CARD_2:87;
  then card (D*) c= card D by A2,A1,Th21;
  hence thesis by A5,XBOOLE_0:def 10;
end;














begin 

reserve k,n,m for Element of NAT,
  A,B,C for Ordinal,
  X for set,
  x,y,z for set;
Lm1: 1 = card 1 by CARD_1:def 2;
Lm2: 2 = card 2 by CARD_1:def 2;

begin 

reserve f,g,h,fx for Function,
  K,M,N for Cardinal,
  phi,psi for
  Ordinal-Sequence;

theorem Th1:
  nextcard card X = nextcard X
proof
  card card X in nextcard X & for M st card card X in M holds nextcard X
  c= M by CARD_1:def 3;
  hence thesis by CARD_1:def 3;
end;

theorem Th2:
  y in Union f iff ex x st x in dom f & y in f.x
proof
  thus y in Union f implies ex x st x in dom f & y in f.x
  proof
    assume y in Union f;
    then consider X such that
A1: y in X and
A2: X in rng f by TARSKI:def 4;
    consider x such that
A3: x in dom f & X = f.x by A2,FUNCT_1:def 3;
    take x;
    thus thesis by A1,A3;
  end;
  given x such that
A4: x in dom f and
A5: y in f.x;
  f.x in rng f by A4,FUNCT_1:def 3;
  hence thesis by A5,TARSKI:def 4;
end;

theorem Th3:
  alef A is infinite
proof
  {} c= A by XBOOLE_1:2;
  then omega c= alef A by CARD_1:23,46;
  hence thesis;
end;

theorem Th4:
  M is infinite implies ex A st M = alef A
proof
  defpred P[set] means $1 is infinite implies ex A st $1 = alef A;
A1: P[K] implies P[nextcard K]
  proof
    assume that
A2: P[K] and
A3: not nextcard K is finite;
    now
      assume K is finite;
      then reconsider K9 = K as finite set;
      card K = K by CARD_1:def 2;
      then nextcard K = card (card K9 + 1) by NAT_1:42;
      hence contradiction by A3;
    end;
    then consider A such that
A4: K = alef A by A2;
    take succ A;
    thus thesis by A4,CARD_1:19;
  end;
A5: K <> {} & K is limit_cardinal & (for N st N in K holds P[N]) implies P[
  K]
  proof
    deffunc a(Ordinal) = alef $1;
    defpred R[set,set] means ex A st $1 = alef A & $2 = A;
    assume that
    K <> {} and
A6: K is limit_cardinal and
A7: for N st N in K holds P[N] and
A8: not K is finite;
    defpred P[set] means ex N st N = $1 & not N is finite;
    consider X such that
A9: x in X iff x in K & P[x] from XBOOLE_0:sch 1;
A10: for x st x in X ex y st R[x,y]
    proof
      let x;
      assume
A11:  x in X;
      then consider N such that
A12:  N = x and
A13:  not N is finite by A9;
      N in K by A9,A11,A12;
      then ex A st x = alef A by A7,A12,A13;
      hence thesis;
    end;
    consider f such that
A14: dom f = X & for x st x in X holds R[x,f.x] from CLASSES1:sch 1(
    A10);
    now
      let x;
      assume x in rng f;
      then consider y such that
A15:  y in X and
A16:  x = f.y by A14,FUNCT_1:def 3;
      consider A such that
A17:  y = alef A and
A18:  x = A by A14,A15,A16;
      thus x is Ordinal by A18;
      thus x c= rng f
      proof
        let z;
        assume
A19:    z in x;
        then reconsider z9 = z as Ordinal by A18;
A20:    alef z9 in alef A by A18,A19,CARD_1:21;
        alef A in K by A9,A15,A17;
        then
A21:    alef z9 in K by A20,ORDINAL1:10;
        not alef z9 is finite by Th3;
        then
A22:    alef z9 in X by A9,A21;
        then ex A st alef z9 = alef A & f.(alef z9) = A by A14;
        then z9 = f.(alef z9) by CARD_1:22;
        hence thesis by A14,A22,FUNCT_1:def 3;
      end;
    end;
    then reconsider A = rng f as Ordinal by ORDINAL1:19;
    consider L being T-Sequence such that
A23: dom L = A & for B st B in A holds L.B = a(B) from ORDINAL2:sch 2;
    now
      let B;
      assume B in A;
      then consider x such that
A24:  x in X and
A25:  B = f.x by A14,FUNCT_1:def 3;
      consider C such that
A26:  x = alef C and
A27:  B = C by A14,A24,A25;
A28:  alef succ C = nextcard alef C by CARD_1:19;
      alef C in K by A9,A24,A26;
      then
A29:  alef succ C c= K by A28,CARD_3:90;
      alef succ C <> K by A6,A28,CARD_1:def 4;
      then
A30:  alef succ C in K by A29,CARD_1:3;
      not alef succ C is finite by Th3;
      then
A31:  alef succ C in X by A9,A30;
      then consider D being Ordinal such that
A32:  alef succ C = alef D and
A33:  f.(alef succ C) = D by A14;
      succ C = D by A32,CARD_1:22;
      hence succ B in A by A14,A27,A31,A33,FUNCT_1:def 3;
    end;
    then A is limit_ordinal by ORDINAL1:28;
    then
A34: A = {} or alef A = card sup L by A23,CARD_1:20;
    take A;
    sup L c= K
    proof
      let x;
      assume
A35:  x in sup L;
      then reconsider x9 = x as Ordinal;
      consider C such that
A36:  C in rng L and
A37:  x9 c= C by A35,ORDINAL2:21;
      consider y such that
A38:  y in dom L and
A39:  C = L.y by A36,FUNCT_1:def 3;
      reconsider y as Ordinal by A38;
A40:  C = alef y by A23,A38,A39;
      consider z such that
A41:  z in dom f and
A42:  y = f.z by A23,A38,FUNCT_1:def 3;
      ex D being Ordinal st z = alef D & y = D by A14,A41,A42;
      then C in K by A9,A14,A40,A41;
      hence thesis by A37,ORDINAL1:12;
    end;
    then card sup L c= card K by CARD_1:11;
    then
A43: card sup L c= K by CARD_1:def 2;
    now
      per cases;
      case
        A = {};
        then not omega in X by A14,RELAT_1:42;
        then not omega in K by A9;
        then
A44:    K c= omega by CARD_1:4;
        omega c= K by A8,CARD_3:85;
        hence K = omega by A44,XBOOLE_0:def 10;
      end;
      case
A45:    A <> {};
        assume K <> alef A;
        then
A46:    card sup L in K by A34,A43,A45,CARD_1:3;
        not alef A is finite by Th3;
        then
A47:    card sup L in X by A9,A34,A45,A46;
        then consider B such that
A48:    card sup L = alef B and
A49:    f.(card sup L) = B by A14;
        A = B by A34,A45,A48,CARD_1:22;
        then A in A by A14,A47,A49,FUNCT_1:def 3;
        hence contradiction;
      end;
    end;
    hence thesis by CARD_1:46;
  end;
A50: P[{}];
  for M holds P[M] from CARD_1:sch 1(A50,A1,A5);
  hence thesis;
end;

registration
  let phi;
  cluster Union phi -> ordinal;
  coherence
  proof
    ex A st rng phi c= A by ORDINAL2:def 4;
    then On rng phi = rng phi by ORDINAL3:6;
    hence thesis by ORDINAL3:5;
  end;
end;

theorem Th5:
  X c= A implies ex phi st phi = canonical_isomorphism_of(RelIncl
  order_type_of RelIncl X, RelIncl X) & phi is increasing & dom phi =
  order_type_of RelIncl X & rng phi = X
proof
  set R = RelIncl X;
  set B = order_type_of R;
  set phi = canonical_isomorphism_of (RelIncl B,R);
  assume
A1: X c= A;
  then R is well-ordering by WELLORD2:8;
  then R, RelIncl B are_isomorphic by WELLORD2:def 2;
  then RelIncl B is well-ordering & RelIncl B, R are_isomorphic by WELLORD1:40
,WELLORD2:8;
  then
A2: phi is_isomorphism_of RelIncl B,R by WELLORD1:def 9;
  then
A3: phi is one-to-one by WELLORD1:def 7;
A4: field RelIncl B = B by WELLORD2:def 1;
  then
A5: dom phi = B by A2,WELLORD1:def 7;
A6: field R = X by WELLORD2:def 1;
  then
A7: rng phi = X by A2,WELLORD1:def 7;
  reconsider phi as T-Sequence by A5,ORDINAL1:def 7;
  reconsider phi as Ordinal-Sequence by A1,A7,ORDINAL2:def 4;
  take phi;
  thus phi = canonical_isomorphism_of (RelIncl order_type_of RelIncl X,
  RelIncl X);
  thus phi is increasing
  proof
    let a,b be Ordinal;
    assume that
A8: a in b and
A9: b in dom phi;
A10: a in dom phi by A8,A9,ORDINAL1:10;
    reconsider a9 = phi.a, b9 = phi.b as Ordinal;
A11: b9 in X by A7,A9,FUNCT_1:def 3;
    a c= b by A8,ORDINAL1:def 2;
    then [a,b] in RelIncl B by A5,A9,A10,WELLORD2:def 1;
    then
A12: [a9,b9] in R by A2,WELLORD1:def 7;
    a9 in X by A7,A10,FUNCT_1:def 3;
    then
A13: a9 c= b9 by A12,A11,WELLORD2:def 1;
    a <> b by A8;
    then a9 <> b9 by A3,A9,A10,FUNCT_1:def 4;
    then a9 c< b9 by A13,XBOOLE_0:def 8;
    hence thesis by ORDINAL1:11;
  end;
  thus thesis by A2,A4,A6,WELLORD1:def 7;
end;

theorem Th6:
  X c= A implies sup X is_cofinal_with order_type_of RelIncl X
proof
  assume
A1: X c= A;
  then consider phi such that
  phi = canonical_isomorphism_of (RelIncl order_type_of RelIncl X, RelIncl
  X) and
A2: phi is increasing & dom phi = order_type_of RelIncl X & rng phi = X
  by Th5;
  take phi;
  On X = X by A1,ORDINAL3:6;
  hence thesis by A2,ORDINAL2:def 3;
end;

theorem Th7:
  X c= A implies card X = card order_type_of RelIncl X
proof
  set R = RelIncl X;
  set B = order_type_of R;
  set phi = canonical_isomorphism_of (RelIncl B,R);
  assume X c= A;
  then R is well-ordering by WELLORD2:8;
  then R, RelIncl B are_isomorphic by WELLORD2:def 2;
  then RelIncl B is well-ordering & RelIncl B, R are_isomorphic by WELLORD1:40
,WELLORD2:8;
  then
A1: phi is_isomorphism_of RelIncl B,R by WELLORD1:def 9;
  field R = X by WELLORD2:def 1;
  then
A2: rng phi = X by A1,WELLORD1:def 7;
  field RelIncl B = B by WELLORD2:def 1;
  then
A3: dom phi = B by A1,WELLORD1:def 7;
  phi is one-to-one by A1,WELLORD1:def 7;
  then B,X are_equipotent by A3,A2,WELLORD2:def 4;
  hence thesis by CARD_1:5;
end;

theorem Th8:
  ex B st B c= card A & A is_cofinal_with B
proof
  set M = card A;
  M,A are_equipotent by CARD_1:def 2;
  then consider f such that
A1: f is one-to-one and
A2: dom f = M and
A3: rng f = A by WELLORD2:def 4;
  defpred P[set] means not f.$1 in Union (f|$1);
  reconsider f as T-Sequence by A2,ORDINAL1:def 7;
  reconsider f as Ordinal-Sequence by A3,ORDINAL2:def 4;
  consider X such that
A4: x in X iff x in M & P[x] from XBOOLE_0:sch 1;
  set R = RelIncl X;
  set B = order_type_of R;
  take B;
A5: X c= M
  proof
    let x;
    thus thesis by A4;
  end;
  hence B c= card A by WELLORD2:14;
  set phi = canonical_isomorphism_of (RelIncl B,R);
  R is well-ordering by A5,WELLORD2:8;
  then R, RelIncl B are_isomorphic by WELLORD2:def 2;
  then RelIncl B is well-ordering & RelIncl B, R are_isomorphic by WELLORD1:40
,WELLORD2:8;
  then
A6: phi is_isomorphism_of RelIncl B,R by WELLORD1:def 9;
  then
A7: phi is one-to-one by WELLORD1:def 7;
  field RelIncl B = B by WELLORD2:def 1;
  then
A8: dom phi = B by A6,WELLORD1:def 7;
  field R = X by WELLORD2:def 1;
  then
A9: rng phi = X by A6,WELLORD1:def 7;
  reconsider phi as T-Sequence by A8,ORDINAL1:def 7;
  reconsider phi as Ordinal-Sequence by A5,A9,ORDINAL2:def 4;
A10: dom (f*phi) = B by A2,A5,A8,A9,RELAT_1:27;
  then reconsider xi = f*phi as T-Sequence by ORDINAL1:def 7;
  rng (f*phi) c= A by A3,RELAT_1:26;
  then reconsider xi as Ordinal-Sequence by ORDINAL2:def 4;
  take xi;
  thus dom xi = B & rng xi c= A by A2,A3,A5,A8,A9,RELAT_1:26,27;
  thus xi is increasing
  proof
    let a,b be Ordinal;
    assume that
A11: a in b and
A12: b in dom xi;
A13: a in dom xi by A11,A12,ORDINAL1:10;
    then
A14: phi.a in X by A8,A9,A10,FUNCT_1:def 3;
    a <> b by A11;
    then
A15: phi.a <> phi.b by A8,A7,A10,A12,A13,FUNCT_1:def 4;
    reconsider a9 = phi.a, b9 = phi.b as Ordinal;
    reconsider a99 = f.a9, b99 = f.b9 as Ordinal;
A16: phi.b in X by A8,A9,A10,A12,FUNCT_1:def 3;
    then not b99 in Union (f|b9) by A4;
    then
A17: Union (f|b9) c= b99 by ORDINAL1:16;
    a c= b by A11,ORDINAL1:def 2;
    then [a,b] in RelIncl B by A10,A12,A13,WELLORD2:def 1;
    then [phi.a,phi.b] in R by A6,WELLORD1:def 7;
    then a9 c= b9 by A14,A16,WELLORD2:def 1;
    then a9 c< b9 by A15,XBOOLE_0:def 8;
    then a9 in b9 by ORDINAL1:11;
    then a99 c= Union (f|b9) by A2,A5,A14,FUNCT_1:50,ZFMISC_1:74;
    then
A18: a99 c= b99 by A17,XBOOLE_1:1;
    a99 <> b99 by A1,A2,A5,A15,A14,A16,FUNCT_1:def 4;
    then
A19: a99 c< b99 by A18,XBOOLE_0:def 8;
    a99 = xi.a & b99 = xi.b by A11,A12,FUNCT_1:12,ORDINAL1:10;
    hence thesis by A19,ORDINAL1:11;
  end;
  thus A c= sup xi
  proof
    let x;
    assume x in A;
    then consider y such that
A20: y in dom f and
A21: x = f.y by A3,FUNCT_1:def 3;
    reconsider x9 = x, y as Ordinal by A20,A21;
    now
      per cases;
      suppose
        not f.y in Union (f|y);
        then y in X by A2,A4,A20;
        then consider z such that
A22:    z in B & y = phi.z by A8,A9,FUNCT_1:def 3;
        x9 = xi.z & xi.z in rng xi by A10,A21,A22,FUNCT_1:12,def 3;
        hence thesis by ORDINAL2:19;
      end;
      suppose
A23:    f.y in Union (f|y);
        defpred P[Ordinal] means $1 in y & f.y in f.$1;
        consider z such that
A24:    z in dom (f|y) and
A25:    f.y in (f|y).z by A23,Th2;
        reconsider z as Ordinal by A24;
A26:    (f|y).z = f.z by A24,FUNCT_1:47;
        dom (f|y) = dom f /\ y by RELAT_1:61;
        then z in y by A24,XBOOLE_0:def 4;
        then
A27:    ex C st P[C] by A25,A26;
        consider C such that
A28:    P[C] & for B st P[B] holds C c= B from ORDINAL1:sch 1(A27);
        now
          thus C in M by A2,A20,A28,ORDINAL1:10;
          assume f.C in Union (f|C);
          then consider a be set such that
A29:      a in dom (f|C) and
A30:      f.C in (f|C).a by Th2;
          reconsider a as Ordinal by A29;
          reconsider fa = (f|C).a, fy = f.y as Ordinal;
          f.a = fa by A29,FUNCT_1:47;
          then
A31:      fy in f.a by A28,A30,ORDINAL1:10;
          dom (f|C) = dom f /\ C by RELAT_1:61;
          then
A32:      a in C by A29,XBOOLE_0:def 4;
          then a in y by A28,ORDINAL1:10;
          hence contradiction by A28,A32,A31,ORDINAL1:5;
        end;
        then C in X by A4;
        then consider z such that
A33:    z in B and
A34:    C = phi.z by A8,A9,FUNCT_1:def 3;
        reconsider z as Ordinal by A33;
        reconsider xz = xi.z as Ordinal;
        xz in rng xi by A10,A33,FUNCT_1:def 3;
        then
A35:    xz in sup xi by ORDINAL2:19;
        x9 in xz by A10,A21,A28,A33,A34,FUNCT_1:12;
        hence thesis by A35,ORDINAL1:10;
      end;
    end;
    hence thesis;
  end;
  sup A = A by ORDINAL2:18;
  hence thesis by A3,ORDINAL2:22,RELAT_1:26;
end;

theorem Th9:
  ex M st M c= card A & A is_cofinal_with M & for B st A
  is_cofinal_with B holds M c= B
proof
  defpred P[Ordinal] means $1 c= card A & A is_cofinal_with $1;
A1: ex B st P[B] by Th8;
  consider C such that
A2: P[C] and
A3: for B st P[B] holds C c= B from ORDINAL1:sch 1(A1);
  take M = card C;
  consider B such that
A4: B c= M and
A5: C is_cofinal_with B by Th8;
A6: M c= C by CARD_1:8;
  then
A7: B c= C by A4,XBOOLE_1:1;
  then B c= card A by A2,XBOOLE_1:1;
  then C c= B by A2,A3,A5,ORDINAL4:37;
  then
A8: C = B by A7,XBOOLE_0:def 10;
  hence M c= card A & A is_cofinal_with M by A2,A4,A6,XBOOLE_0:def 10;
  let B;
  assume that
A9: A is_cofinal_with B and
A10: not M c= B;
A11: C = M by A4,A6,A8,XBOOLE_0:def 10;
  then not B c= card A by A3,A9,A10;
  hence contradiction by A2,A11,A10,XBOOLE_1:1;
end;

Lm3: rng phi = rng psi & phi is increasing & psi is increasing implies for A
st A in dom phi holds A in dom psi & phi.A = psi.A

proof
  assume that
A1: rng phi = rng psi and
A2: phi is increasing;

  defpred P[Ordinal] means $1 in dom phi implies $1 in dom psi & phi.$1 = psi.
  $1;

  assume
A3: for A,B st A in B & B in dom psi holds psi.A in psi.B;
A4: for A st for B st B in A holds P[B] holds P[A]
  proof
    let A;
    assume that
A5: for B st B in A & B in dom phi holds B in dom psi & phi.B = psi.B and
A6: A in dom phi;
    phi.A in rng phi by A6,FUNCT_1:def 3;
    then consider x such that
A7: x in dom psi and
A8: phi.A = psi.x by A1,FUNCT_1:def 3;
    reconsider x as Ordinal by A7;
A9: now
      assume
A10:  A in x;
      then
A11:  A in dom psi by A7,ORDINAL1:10;
      then psi.A in rng psi by FUNCT_1:def 3;
      then consider y such that
A12:  y in dom phi and
A13:  psi.A = phi.y by A1,FUNCT_1:def 3;
      reconsider y as Ordinal by A12;
      psi.A in psi.x by A3,A7,A10;
      then not A c= y by A2,A8,A12,A13,ORDINAL1:5,ORDINAL4:9;
      then
A14:  y in A by ORDINAL1:16;
      then
A15:  psi.y = phi.y by A5,A12;
      psi.y in psi.A by A3,A11,A14;
      hence contradiction by A13,A15;
    end;
    now
      assume
A16:  x in A;

      then phi.x in phi.A & x in dom phi by A2,A6,ORDINAL1:10,ORDINAL2:def 12;

      then phi.A in phi.A by A5,A8,A16;
      hence contradiction;
    end;
    hence thesis by A7,A8,A9,ORDINAL1:14;
  end;
  thus P[A] from ORDINAL1:sch 2(A4);
end;

theorem
  rng phi = rng psi & phi is increasing & psi is increasing implies phi
  = psi
proof
  assume
A1: rng phi = rng psi & phi is increasing & psi is increasing;
A2: dom phi = dom psi
  proof
    thus dom phi c= dom psi
    proof
      let x;
      assume x in dom phi;
      hence thesis by A1,Lm3;
    end;
    let x;
    assume x in dom psi;
    hence thesis by A1,Lm3;
  end;
  for x st x in dom phi holds phi.x = psi.x by A1,Lm3;
  hence thesis by A2,FUNCT_1:2;
end;

theorem Th11:
  phi is increasing implies phi is one-to-one
proof
  assume
A1: for A,B st A in B & B in dom phi holds phi.A in phi.B;
  let x,y;
  assume that
A2: x in dom phi & y in dom phi and
A3: phi.x = phi.y;
  reconsider A = x, B = y as Ordinal by A2;
A4: A in B or A = B or B in A by ORDINAL1:14;
  not phi.A in phi.B by A3;
  hence thesis by A1,A2,A3,A4;
end;

theorem Th12:
  (phi^psi)|(dom phi) = phi
proof
  dom (phi^psi) = (dom phi)+^(dom psi) by ORDINAL4:def 1;
  then dom phi c= dom (phi^psi) by ORDINAL3:24;
  then
A1: dom phi = (dom (phi^psi))/\(dom phi) by XBOOLE_1:28;
  for x st x in dom phi holds phi.x = (phi^psi).x by ORDINAL4:def 1;
  hence thesis by A1,FUNCT_1:46;
end;

theorem
  X <> {} implies card { Y where Y is Subset of X: card Y in M } c= M*`
  exp(card X,M)
proof
  set Z = { Y where Y is Subset of X: card Y in M };
  X,card X are_equipotent by CARD_1:def 2;
  then consider f such that
A1: f is one-to-one and
A2: dom f = X and
A3: rng f = card X by WELLORD2:def 4;
  defpred P[set,set] means ex A be Ordinal, phi be Ordinal-Sequence st $2 = [A
,phi] & dom phi = M & phi|A is increasing & rng (phi|A) = f.:$1 & for B st A c=
  B & B in M holds phi.B = {};
A4: for x st x in Z ex y st P[x,y]
  proof
    deffunc f(set) = {};
    let x;
    set A = order_type_of RelIncl (f.:x);
    consider xi2 being Ordinal-Sequence such that
A5: dom xi2 = M -^ A & for B st B in M -^ A holds xi2.B = f(B) from
    ORDINAL2:sch 3;
    assume x in Z;
    then
A6: ex Y being Subset of X st x = Y & card Y in M;
    consider xi1 being Ordinal-Sequence such that
    xi1 = canonical_isomorphism_of (RelIncl A, RelIncl (f.:x)) and
A7: xi1 is increasing and
A8: dom xi1 = A and
A9: rng xi1 = f.:x by A3,Th5,RELAT_1:111;
    set phi = xi1^xi2;
    take y = [A,phi], A, phi;
    card (f.:x) = card A by A3,Th7,RELAT_1:111;
    then card A in M by A6,CARD_1:67,ORDINAL1:12;
    then A in M by CARD_3:44;
    then A c= M by ORDINAL1:def 2;
    then A+^( M -^ A) = M by ORDINAL3:def 5;
    hence y = [A,phi] & dom phi = M & phi|A is increasing & rng (phi|A) = f.:x
    by A7,A8,A9,A5,Th12,ORDINAL4:def 1;
    let B;
    assume that
A10: A c= B and
A11: B in M;
A12: B-^A in M-^A by A10,A11,ORDINAL3:53;
    B = A+^(B-^A) by A10,ORDINAL3:def 5;
    then phi.B = xi2.(B-^A) by A8,A5,A12,ORDINAL4:def 1;
    hence thesis by A5,A12;
  end;
  consider g such that
A13: dom g = Z & for x st x in Z holds P[x,g.x] from CLASSES1:sch 1(A4);
  assume
A14: X <> {};
  rng g c= [:M,Funcs(M,card X):]
  proof
    let x;
    assume x in rng g;
    then consider y such that
A15: y in dom g and
A16: x = g.y by FUNCT_1:def 3;
    consider A,phi such that
A17: x = [A,phi] and
A18: dom phi = M and
A19: phi|A is increasing and
A20: rng (phi|A) = f.:y and
A21: for B st A c= B & B in M holds phi.B = {} by A13,A15,A16;
A22: ex Y being Subset of X st y = Y & card Y in M by A13,A15;
    rng phi c= card X
    proof
      let x;
      assume x in rng phi;
      then consider z such that
A23:  z in dom phi and
A24:  x = phi.z by FUNCT_1:def 3;
      reconsider z as Ordinal by A23;
      z in A or A c= z by ORDINAL1:16;
      then x in f.:y & f.:y c= card X or x = {} by A3,A18,A20,A21,A23,A24,
FUNCT_1:50,RELAT_1:111;
      hence thesis by A14,ORDINAL3:8;
    end;
    then
A25: phi in Funcs(M,card X) by A18,FUNCT_2:def 2;
A26: A c= M or M c= A;
    phi|A is one-to-one by A19,Th11;
    then dom (phi|A),f.:y are_equipotent by A20,WELLORD2:def 4;
    then card dom (phi|A) = card (f.:y) by CARD_1:5;
    then card dom (phi|A) in M by A22,CARD_1:67,ORDINAL1:12;
    then
A27: dom (phi|A) in M by CARD_3:44;
    then dom (phi|A) <> M;
    then A in M by A18,A27,A26,RELAT_1:62,68;
    hence thesis by A17,A25,ZFMISC_1:87;
  end;
  then
A28: card rng g c= card [:M,Funcs(M,card X):] by CARD_1:11;
  g is one-to-one
  proof
    let x,y;
    assume that
A29: x in dom g and
A30: y in dom g and
A31: g.x = g.y;
    consider A2 be Ordinal, phi2 be Ordinal-Sequence such that
A32: g.y = [A2,phi2] and
    dom phi2 = M and
    phi2|A2 is increasing and
A33: rng (phi2|A2) = f.:y and
    for B st A2 c= B & B in M holds phi2.B = {} by A13,A30;
    consider A1 be Ordinal, phi1 be Ordinal-Sequence such that
A34: g.x = [A1,phi1] and
    dom phi1 = M and
    phi1|A1 is increasing and
A35: rng (phi1|A1) = f.:x and
    for B st A1 c= B & B in M holds phi1.B = {} by A13,A29;
A36: A1 = A2 & phi1 = phi2 by A31,A34,A32,XTUPLE_0:1;
A37: ex Y being Subset of X st x = Y & card Y in M by A13,A29;
    thus x c= y
    proof
      let z;
      assume
A38:  z in x;
      then f.z in f.:x by A2,A37,FUNCT_1:def 6;
      then ex x1 being set st x1 in dom f & x1 in y & f.z = f.x1 by A35,A33,A36
,FUNCT_1:def 6;
      hence thesis by A1,A2,A37,A38,FUNCT_1:def 4;
    end;
    let z;
A39: ex Y being Subset of X st y = Y & card Y in M by A13,A30;
    assume
A40: z in y;
    then f.z in f.:y by A2,A39,FUNCT_1:def 6;
    then
    ex x1 being set st x1 in dom f & x1 in x & f.z = f.x1 by A35,A33,A36,
FUNCT_1:def 6;
    hence thesis by A1,A2,A39,A40,FUNCT_1:def 4;
  end;
  then
A41: Z,rng g are_equipotent by A13,WELLORD2:def 4;
  card [:M,Funcs(M,card X):] = card [:M,card Funcs(M,card X):] by CARD_2:7
    .= M*`card Funcs(M,card X) by CARD_2:def 2
    .= M*`exp(card X,M) by CARD_2:def 3;
  hence thesis by A41,A28,CARD_1:5;
end;

theorem Th14:
  M in exp(2,M)
proof
  card bool M = exp(2,card M) & card M = M by CARD_1:def 2,CARD_2:31;
  hence thesis by CARD_1:14;
end;

registration
  cluster infinite for Cardinal;
  existence
  proof
    take omega;
    thus thesis;
  end;
end;

registration
  cluster infinite -> non empty for set;
  coherence;
end;

definition
  mode Aleph is infinite Cardinal;
  let M;

  func cf M -> Cardinal means
  :Def1:
  M is_cofinal_with it & for N st M
  is_cofinal_with N holds it c= N;
  existence
  proof
    defpred P[Ordinal] means M is_cofinal_with $1 & $1 is Cardinal;
A1: ex A st P[A];
    consider A such that
A2: P[A] & for B st P[B] holds A c= B from ORDINAL1:sch 1(A1);
    reconsider K = A as Cardinal by A2;
    take K;
    thus M is_cofinal_with K by A2;
    let N;
    assume M is_cofinal_with N;
    hence thesis by A2;
  end;
  uniqueness
  proof
    let K1,K2 be Cardinal;
    assume M is_cofinal_with K1 & ( for N st M is_cofinal_with N holds K1 c=
    N)& M is_cofinal_with K2 & for N st M is_cofinal_with N holds K2 c= N;
    then K1 c= K2 & K2 c= K1;
    hence thesis by XBOOLE_0:def 10;
  end;
  let N;
  func N-powerfunc_of M -> Cardinal-Function means
  : Def2:
  (for x holds x in
  dom it iff x in M & x is Cardinal) & for K st K in M holds it.K = exp(K,N);
  existence
  proof
    deffunc f(set) = exp(card $1,N);
    defpred P[set] means $1 is Cardinal;
    consider X such that
A3: x in X iff x in M & P[x] from XBOOLE_0:sch 1;
    consider f being Cardinal-Function such that
A4: dom f = X & for x st x in X holds f.x = f(x) from CARD_3:sch 1;
    take f;
    thus x in dom f iff x in M & x is Cardinal by A3,A4;
    let K;
    assume K in M;
    then K = card K & K in X by A3,CARD_1:def 2;
    hence thesis by A4;
  end;
  uniqueness
  proof
    defpred P[set] means $1 in M & $1 is Cardinal;
    let f1,f2 be Cardinal-Function;
    assume that
A5: for x holds x in dom f1 iff P[x] and
A6: for K st K in M holds f1.K = exp(K,N) and
A7: for x holds x in dom f2 iff P[x] and
A8: for K st K in M holds f2.K = exp(K,N);
A9: now
      let x;
      assume
A10:  x in dom f1;
      then reconsider K = x as Cardinal by A5;
A11:  K in M by A5,A10;
      hence f1.x = exp(K,N) by A6
        .= f2.x by A8,A11;
    end;
    dom f1 = dom f2 from XBOOLE_0:sch 2(A5,A7);
    hence thesis by A9,FUNCT_1:2;
  end;
end;

registration
  let A;
  cluster alef A -> infinite;
  coherence by Th3;
end;

begin 

reserve a,b for Aleph;

theorem
  ex A st a = alef A by Th4;

theorem Th16:
  a <> 0 & a <> 1 & a <> 2 & a <> card n & card n in a & omega c= a
proof
  omega c= a & card n in omega by CARD_3:85;
  hence thesis;
end;

theorem Th17:
  a c= M or a in M implies M is Aleph
proof
  assume
A1: a c= M or a in M;
  omega c= a by Th16;
  then omega c= M by A1,XBOOLE_1:1;
  hence thesis;
end;

theorem Th18:
  a c= M or a in M implies a +` M = M & M +` a = M & a *` M = M &
  M *` a = M
proof
A1: card 0 in a by Th16;
  assume
A2: a c= M or a in M;
  then M is infinite by Th17;
  hence thesis by A2,A1,CARD_2:76,CARD_4:16;
end;

theorem
  a +` a = a & a *` a = a by CARD_2:75,CARD_4:15;

theorem Th20:
  M c= exp(M,a)
proof
  1 in a by Lm1,Th16;
  then M = 0 & {} c= exp(M,a) or exp(M,1) c= exp(M,a) & exp(M,1) = M by
CARD_2:27,93,XBOOLE_1:2;
  hence thesis;
end;

theorem
  union a = a by ORDINAL1:def 6;

registration
  let a,M;
  cluster a +` M -> infinite;
  coherence
  proof
    a c= M or M c= a;
    then a +` M = M & M is Aleph or a +` M = a by Th18,CARD_2:76;
    hence thesis;
  end;
end;

registration
  let M,a;
  cluster M +` a -> infinite;
  coherence;
end;

registration
  let a,b;
  cluster a *` b -> infinite;
  coherence
  proof
    a c= b or b c= a;
    hence thesis by Th18;
  end;
  cluster exp(a,b) -> infinite;
  coherence
  proof
    1 in b by Lm1,Th16;
    then
A1: exp(a,1) c= exp(a,b) by CARD_2:92;
    exp(a,1) = a & omega c= a by Th16,CARD_2:27;
    then omega c= exp(a,b) by A1,XBOOLE_1:1;
    hence thesis;
  end;
end;

begin 

definition
  let IT be Aleph;
  attr IT is regular means
  cf IT = IT;
end;

notation
  let IT be Aleph;
  antonym IT is irregular for IT is regular;
end;

registration
  let a;
  cluster nextcard a -> infinite;
  coherence
  proof
    a in nextcard a by CARD_1:18;
    then omega c= nextcard a;
    hence thesis;
  end;
end;

theorem Th22:
  cf omega = omega
proof
  defpred P[set,set] means $2 c= $1;
  assume
A1: cf omega <> omega;
  cf omega c= omega by Def1;
  then cf omega in omega by A1,CARD_1:3;
  then reconsider B = cf omega as finite set;
  set n = card B;
A2: for x,y st P[x,y] & P[y,x] holds x = y by XBOOLE_0:def 10;
A3: for x,y,z st P[x,y] & P[y,z] holds P[x,z] by XBOOLE_1:1;
  card cf omega = cf omega by CARD_1:def 2;
  then omega is_cofinal_with n by Def1;
  then consider xi being Ordinal-Sequence such that
A4: dom xi = n and
A5: rng xi c= omega and
  xi is increasing and
A6: omega = sup xi by ORDINAL2:def 17;
  reconsider rxi = rng xi as finite set by A4,FINSET_1:8;
A7: rxi <> {} by A6,ORDINAL2:18;
  consider x such that
A8: x in rxi & for y st y in rxi & y <> x holds not P[y,x]
    from CARD_2:sch 1(A7,A2,A3);
  reconsider x as Ordinal by A5,A8;
  now
    let A;
    assume A in rng xi;
    then A c= x or not x c= A by A8;
    hence A in succ x by ORDINAL1:22;
  end;
  then
A9: omega c= succ x by A6,ORDINAL2:20;
  succ x in omega by A5,A8,ORDINAL1:28;
  hence contradiction by A9;
end;

Lm4: 1 = card 1 by CARD_1:def 2;

theorem Th23:
  cf nextcard a = nextcard a
proof
  nextcard a is_cofinal_with cf nextcard a by Def1;
  then consider xi being Ordinal-Sequence such that
A1: dom xi = cf nextcard a and
A2: rng xi c= nextcard a and
  xi is increasing and
A3: nextcard a = sup xi by ORDINAL2:def 17;
A4: card Union xi c= Sum Card xi & Sum (cf nextcard a --> a) = (cf nextcard
  a)*` a by CARD_2:65,CARD_3:39;
A5: card nextcard a = nextcard a & succ Union xi = (Union xi) +^ 1 by
CARD_1:def 2,ORDINAL2:31;
A6: now
    let x;
    assume
A7: x in cf nextcard a;
    xi.x in rng xi by A1,A7,FUNCT_1:def 3;
    then
A8: card (xi.x) in nextcard a by A2,CARD_1:8,ORDINAL1:12;
    (Card xi).x = card (xi.x) & (cf nextcard a --> a).x = a by A1,A7,
CARD_3:def 2,FUNCOP_1:7;
    hence (Card xi).x c= (cf nextcard a --> a).x by A8,CARD_3:91;
  end;
  dom Card xi = dom xi & dom (cf nextcard a --> a) = cf nextcard a by
CARD_3:def 2,FUNCOP_1:13;
  then Sum Card xi c= Sum (cf nextcard a --> a) by A1,A6,CARD_3:30;
  then card Union xi c= (cf nextcard a)*`a by A4,XBOOLE_1:1;
  then
A9: (card Union xi) +` 1 c= (cf nextcard a)*`a +` 1 by CARD_2:84;
A10: card ((Union xi) +^ 1) = (card Union xi) +` card 1 by CARD_2:13;
  ex A st rng xi c= A by ORDINAL2:def 4;
  then On rng xi = rng xi by ORDINAL3:6;
  then
A11: card nextcard a c= card succ Union xi by A3,CARD_1:11,ORDINAL3:72;
A12: cf nextcard a c= nextcard a by Def1;
  now
    per cases;
    suppose
      cf nextcard a = 0;
      then
A13:  (cf nextcard a)*`a = 0 by CARD_2:20;
      0+`1 = 1 & 1 in nextcard a by Lm1,Th16,CARD_2:18;
      hence thesis by A11,A5,A9,A10,A13,Lm4;
    end;
    suppose
A14:  cf nextcard a <> 0;
      0 c= cf nextcard a by XBOOLE_1:2;
      then
A15:  0 in cf nextcard a by A14,CARD_1:3;
A16:  cf nextcard a c= a or a c= cf nextcard a;
      1 in a by Lm1,Th16;
      then
A17:  (cf nextcard a)*`a = a & a+`1 = a & a in nextcard a or (cf nextcard
a)*`a = cf nextcard a & cf nextcard a is Aleph by A15,A16,Th18,CARD_1:18
,CARD_2:76,CARD_4:16;
      then
      nextcard a c= (cf nextcard a) +` 1 & 1 in cf nextcard a by A11,A5,A9,A10
,Lm1,Th16,CARD_1:4,XBOOLE_1:1;
      then nextcard a c= cf nextcard a by A11,A5,A9,A10,A17,Lm1,CARD_1:4
,CARD_2:76;
      hence thesis by A12,XBOOLE_0:def 10;
    end;
  end;
  hence thesis;
end;

theorem Th24:
  omega c= cf a
proof
A1: a is_cofinal_with cf a by Def1;
  then cf a <> {} by ORDINAL2:50;
  then
A2: {} in cf a by ORDINAL3:8;
  cf a is limit_ordinal by A1,ORDINAL4:38;
  hence thesis by A2,ORDINAL1:def 11;
end;

theorem
  cf 0 = 0 & cf card (n+1) = 1
proof
A1: succ n is_cofinal_with 1 by ORDINAL3:73;
  card (n+1) = n+1 & n+1 = succ n by CARD_1:def 2,NAT_1:38;
  then cf card (n+1) c= 1 by A1,Def1;
  then
A2: cf card (n+1) = 1 or cf card (n+1) = 0 & {} c= n & n in succ n by
ORDINAL1:6,ORDINAL3:16,XBOOLE_1:2;
  cf 0 c= 0 & 0 c= cf 0 by Def1,XBOOLE_1:2;
  hence cf 0 = 0 by XBOOLE_0:def 10;
  card (n+1) is_cofinal_with cf card (n+1) by Def1;
  hence thesis by A2,ORDINAL2:50;
end;

theorem Th26:
  X c= M & card X in cf M implies sup X in M & union X in M
proof
  assume that
A1: X c= M and
A2: card X in cf M;
  set A = order_type_of (RelIncl X);
A3: card A = card X by A1,Th7;
  consider N such that
A4: N c= card A and
A5: A is_cofinal_with N and
  for C st A is_cofinal_with C holds N c= C by Th9;
  sup X is_cofinal_with A by A1,Th6;
  then
A6: sup X is_cofinal_with N by A5,ORDINAL4:37;
A7: now
    assume sup X = M;
    then cf M c= N by A6,Def1;
    hence contradiction by A2,A3,A4,CARD_1:4;
  end;
  for x st x in X holds x is Ordinal by A1;
  then reconsider A = union X as Ordinal by ORDINAL1:23;
A8: sup M = M by ORDINAL2:18;
  sup X c= sup M by A1,ORDINAL2:22;
  then
A9: sup X c< M by A8,A7,XBOOLE_0:def 8;
  hence sup X in M by ORDINAL1:11;
  A c= sup X
  proof
    let x;
    assume x in A;
    then consider Y being set such that
A10: x in Y and
A11: Y in X by TARSKI:def 4;
    reconsider Y as Ordinal by A1,A11;
    Y in sup X by A11,ORDINAL2:19;
    then Y c= sup X by ORDINAL1:def 2;
    hence thesis by A10;
  end;
  hence thesis by A9,ORDINAL1:11,12;
end;

theorem Th27:
  dom phi = M & rng phi c= N & M in cf N implies sup phi in N &
  Union phi in N
proof
  assume that
A1: dom phi = M and
A2: rng phi c= N and
A3: M in cf N;
A4: card M = M by CARD_1:def 2;
  card rng phi c= card M by A1,CARD_1:12;
  then card rng phi in cf N by A3,A4,ORDINAL1:12;
  hence thesis by A2,Th26;
end;

registration
  let a;
  cluster cf a -> infinite;
  coherence by Th17,Th24;
end;

theorem Th28:
  cf a in a implies a is limit_cardinal
proof
  assume
A1: cf a in a;
  given M such that
A2: a = nextcard M;
  cf a c= M by A1,A2,CARD_3:91;
  then reconsider M as Aleph;
  nextcard M in nextcard M by A1,A2,Th23;
  hence contradiction;
end;

theorem Th29:
  cf a in a implies ex xi being Ordinal-Sequence st dom xi = cf a
& rng xi c= a & xi is increasing & a = sup xi & xi is Cardinal-Function & not 0
  in rng xi
proof
  a is_cofinal_with cf a by Def1;
  then consider xi being Ordinal-Sequence such that
A1: dom xi = cf a and
A2: rng xi c= a and
  xi is increasing and
A3: a = sup xi by ORDINAL2:def 17;
  deffunc f(T-Sequence) = (nextcard (xi.dom $1)) \/ nextcard sup $1;
  consider phi being T-Sequence such that
A4: dom phi = cf a & for A for psi being T-Sequence st A in cf a & psi =
  phi|A holds phi.A = f(psi) from ORDINAL1:sch 4;
  phi is Ordinal-yielding
  proof
    take sup rng phi;
    let x;
    assume
A5: x in rng phi;
    then consider y such that
A6: y in dom phi and
A7: x = phi.y by FUNCT_1:def 3;
    reconsider y as Ordinal by A6;
    y c= dom phi by A6,ORDINAL1:def 2;
    then dom (phi|y) = y by RELAT_1:62;
    then x = ( nextcard (xi.y)) \/ nextcard sup (phi|y) by A4,A6,A7;
    hence thesis by A5,ORDINAL2:19;
  end;
  then reconsider phi as Ordinal-Sequence;
  defpred P[Ordinal] means $1 in cf a implies phi.$1 in a;
  assume cf a in a;
  then
A8: a is limit_cardinal by Th28;
A9: now
    let A such that
A10: for B st B in A holds P[B];
    thus P[A]
    proof
      assume
A11:  A in cf a;
      then
A12:  card A in cf a by CARD_1:9;
      A c= dom phi by A4,A11,ORDINAL1:def 2;
      then
A13:  dom (phi|A) = A by RELAT_1:62;
      then phi.A = (nextcard (xi.A)) \/ nextcard sup (phi|A) by A4,A11;
      then
A14:  phi.A = nextcard (xi.A) or phi.A = nextcard sup (phi|A) by ORDINAL3:12;
A15:  rng (phi|A) c= a
      proof
        let x;
        assume x in rng (phi|A);
        then consider y such that
A16:    y in dom (phi|A) and
A17:    x = (phi|A).y by FUNCT_1:def 3;
        reconsider y as Ordinal by A16;
        x = phi.y & y in cf a by A11,A13,A16,A17,FUNCT_1:47,ORDINAL1:10;
        hence thesis by A10,A13,A16;
      end;
      (phi|A).:A = rng (phi|A) by A13,RELAT_1:113;
      then card rng (phi|A) in cf a by A12,CARD_1:67,ORDINAL1:12;
      then sup rng (phi|A) in a by A15,Th26;
      then card sup (phi|A) in a by CARD_1:9;
      then
A18:  nextcard card sup (phi|A) c= a by CARD_3:90;
      xi.A in rng xi by A1,A11,FUNCT_1:def 3;
      then card (xi.A) in a by A2,CARD_1:9;
      then
A19:  nextcard card (xi.A) c= a by CARD_3:90;
A20:  nextcard card sup (phi|A) <> a & nextcard card sup (phi|A) =
      nextcard sup ( phi|A) by A8,Th1,CARD_1:def 4;
      a <> nextcard card (xi.A) & nextcard card (xi.A) = nextcard (xi.A)
      by A8,Th1,CARD_1:def 4;
      hence thesis by A19,A18,A20,A14,CARD_1:3;
    end;
  end;
A21: for A holds P[A] from ORDINAL1:sch 2(A9);
A22: rng phi c= a
  proof
    let x;
    assume x in rng phi;
    then ex y st y in dom phi & x = phi.y by FUNCT_1:def 3;
    hence thesis by A4,A21;
  end;
  take phi;
  thus dom phi = cf a by A4;
  thus rng phi c= a
  proof
    let x;
    assume x in rng phi;
    then ex y st y in dom phi & x = phi.y by FUNCT_1:def 3;
    hence thesis by A4,A21;
  end;
  thus phi is increasing
  proof
    let A,B;
    assume that
A23: A in B and
A24: B in dom phi;
    reconsider C = phi.A as Ordinal;
    A in dom phi by A23,A24,ORDINAL1:10;
    then C in rng (phi|B) by A23,FUNCT_1:50;
    then
A25: C in sup (phi|B) by ORDINAL2:19;
    sup (phi|B) in nextcard sup (phi|B) by CARD_1:18;
    then
A26: C in nextcard sup (phi|B) by A25,ORDINAL1:10;
    phi.B = ( nextcard (xi.dom (phi|B))) \/ nextcard sup (phi|B) by A4,A24;
    hence thesis by A26,XBOOLE_0:def 3;
  end;
  thus a c= sup phi
  proof
    let x;
    assume x in a;
    then reconsider x as Element of a;
    consider A such that
A27: A in rng xi and
A28: x c= A by A3,ORDINAL2:21;
    consider y such that
A29: y in dom xi and
A30: A = xi.y by A27,FUNCT_1:def 3;
    reconsider y as Ordinal by A29;
    y c= cf a by A1,A29,ORDINAL1:def 2;
    then dom (phi|y) = y by A4,RELAT_1:62;
    then
    A in nextcard A & phi.y = ( nextcard A) \/ nextcard sup (phi|y) by A1,A4
,A29,A30,CARD_1:18;
    then A in phi.y by XBOOLE_0:def 3;
    then
A31: A c= phi.y by ORDINAL1:def 2;
    phi.y in rng phi by A1,A4,A29,FUNCT_1:def 3;
    then phi.y in sup phi by ORDINAL2:19;
    hence thesis by A28,A31,ORDINAL1:12,XBOOLE_1:1;
  end;
  sup a = a by ORDINAL2:18;
  hence sup phi c= a by A22,ORDINAL2:22;
  phi is Cardinal-yielding
  proof
    let y;
    assume
A32: y in dom phi;
    then reconsider y as Ordinal;
    y c= dom phi by A32,ORDINAL1:def 2;
    then dom (phi|y) = y by RELAT_1:62;
    then phi.y = ( nextcard (xi.y)) \/ nextcard sup (phi|y) by A4,A32;
    hence thesis by ORDINAL3:12;
  end;
  hence phi is Cardinal-Function;
  assume 0 in rng phi;
  then consider x such that
A33: x in dom phi and
A34: 0 = phi.x by FUNCT_1:def 3;
  reconsider x as Ordinal by A33;
  x c= dom phi by A33,ORDINAL1:def 2;
  then dom (phi|x) = x by RELAT_1:62;
  then 0 = ( nextcard (xi.x)) \/ nextcard sup (phi|x) by A4,A33,A34;
  then 0 = nextcard (xi.x) or 0 = nextcard sup (phi|x);
  hence contradiction by CARD_1:15;
end;

theorem
  omega is regular & nextcard a is regular
proof
  thus cf omega = omega by Th22;
  thus cf nextcard a = nextcard a by Th23;
end;

begin 

reserve a,b for Aleph;

theorem Th31:
  a c= b implies exp(a,b) = exp(2,b)
proof
A1: card bool a = exp(2,card a) by CARD_2:31;
  card a = a & card a in card bool a by CARD_1:14,def 2;
  then
A2: exp(a,b) c= exp(exp(2,a),b) by A1,CARD_2:92;
  assume a c= b;
  then
A3: exp(exp(2,a),b) = exp(2,a*`b) & a*`b = b by Th18,CARD_2:30;
  2 in a by Lm2,Th16;
  then exp(2,b) c= exp(a,b) by CARD_2:92;
  hence thesis by A2,A3,XBOOLE_0:def 10;
end;

theorem
  exp(nextcard a,b) = exp(a,b) *` nextcard a
proof
  now
    per cases by CARD_1:4;
    suppose
A1:   a in b;
      then a c= b by CARD_1:3;
      then
A2:   exp(a,b) = exp(2,b) by Th31;
      nextcard a c= b by A1,CARD_3:90;
      then exp(nextcard a,b) = exp(2,b) & nextcard a in exp(2,b) by Th14,Th31,
ORDINAL1:12;
      hence thesis by A2,Th18;
    end;
    suppose
A3:   b c= a;
      deffunc f(Ordinal) = Funcs(b,$1);
      consider phi being T-Sequence such that
A4:   dom phi = nextcard a & for A st A in nextcard a holds phi.A = f
      (A) from ORDINAL2:sch 2;
A5:   cf nextcard a = nextcard a by Th23;
A6:   b in nextcard a by A3,CARD_3:91;
      Funcs(b,nextcard a) c= Union phi
      proof
        let x;
        assume x in Funcs(b,nextcard a);
        then consider f be Function such that
A7:     x = f and
A8:     dom f = b and
A9:     rng f c= nextcard a by FUNCT_2:def 2;
        reconsider f as T-Sequence by A8,ORDINAL1:def 7;
        reconsider f as Ordinal-Sequence by A9,ORDINAL2:def 4;
        sup f in nextcard a by A6,A5,A8,A9,Th27;
        then
A10:    phi.sup f in rng phi by A4,FUNCT_1:def 3;
        rng f c= sup f by ORDINAL2:49;
        then
A11:    f in Funcs(b,sup f) by A8,FUNCT_2:def 2;
        phi.sup f = Funcs(b,sup f) by A6,A5,A4,A8,A9,Th27;
        hence thesis by A7,A11,A10,TARSKI:def 4;
      end;
      then
A12:  card Funcs(b,nextcard a) c= card Union phi by CARD_1:11;
      card Funcs(b,nextcard a) = exp(nextcard a,b) & card Union phi c=
      Sum Card phi by CARD_2:def 3,CARD_3:39;
      then
A13:  exp(nextcard a,b) c= Sum Card phi by A12,XBOOLE_1:1;
      a in nextcard a by CARD_1:18;
      then
A14:  exp(nextcard a,b) *` exp(nextcard a,b) = exp(nextcard a,b) & exp(a,
      b) c= exp (nextcard a,b) by CARD_2:92,CARD_4:15;
      exp(nextcard a,1) = nextcard a & 1 in b by Lm1,Th16,CARD_2:27;
      then nextcard a c= exp(nextcard a,b) by CARD_2:92;
      then
A15:  exp(a,b)*`nextcard a c= exp(nextcard a,b) by A14,CARD_2:90;
A16:  now
        let x;
A17:    card card x = card x & card b = card b;
        assume
A18:    x in nextcard a;
        then reconsider x9 = x as Ordinal;
A19:    phi.x9 = Funcs(b,x9) by A4,A18;
        card x in nextcard a by A18,CARD_1:8,ORDINAL1:12;
        then card x c= a by CARD_3:91;
        then Funcs(b,card x) c= Funcs(b,a) by FUNCT_5:56;
        then
A20:    card Funcs(b,card x) c= card Funcs(b,a) by CARD_1:11;
A21:    card Funcs(b,a) = exp(a,b) by CARD_2:def 3;
        (nextcard a --> exp(a,b)).x = exp(a,b) & Card phi.x = card (phi.x
        ) by A4,A18,CARD_3:def 2,FUNCOP_1:7;
        hence Card phi.x c= (nextcard a --> exp(a,b)).x by A19,A17,A20,A21,
FUNCT_5:61;
      end;
      dom Card phi = dom phi & dom (nextcard a --> exp(a,b)) = nextcard a
      by CARD_3:def 2,FUNCOP_1:13;
      then
A22:  Sum Card phi c= Sum (nextcard a --> exp(a,b)) by A4,A16,CARD_3:30;
      Sum (nextcard a --> exp(a,b)) = (nextcard a)*`exp(a,b) by CARD_2:65;
      then exp(nextcard a,b) c= exp(a,b)*`nextcard a by A13,A22,XBOOLE_1:1;
      hence thesis by A15,XBOOLE_0:def 10;
    end;
  end;
  hence thesis;
end;

theorem Th33:
  Sum (b-powerfunc_of a) c= exp(a,b)
proof
  set X = { c where c is Element of a: c is Cardinal};
  set f = X --> exp(a,b);
  X c= a
  proof
    let x;
    assume x in X;
    then ex c being Element of a st x = c & c is Cardinal;
    hence thesis;
  end;
  then
A1: f c= a --> exp(a,b) by FUNCT_4:4;
  Sum (a --> exp(a,b)) = a*`exp( a,b) by CARD_2:65;
  then
A2: Sum f c= a*`exp(a,b) by A1,CARD_3:33;
A3: dom f = X & dom (b-powerfunc_of a) = X
  proof
    thus dom f = X by FUNCOP_1:13;
    thus dom (b-powerfunc_of a) c= X
    proof
      let x;
      assume x in dom (b-powerfunc_of a);
      then x in a & x is Cardinal by Def2;
      hence thesis;
    end;
    let x;
    assume x in X;
    then ex c being Element of a st x = c & c is Cardinal;
    hence thesis by Def2;
  end;
  1 in b & exp(a,1) = a by Lm1,Th16,CARD_2:27;
  then a c= exp(a,b) by CARD_2:93;
  then
A4: a*`exp(a,b) = exp(a,b) by Th18;
  now
    let x;
    assume
A5: x in X;
    then consider c being Element of a such that
A6: x = c and
A7: c is Cardinal;
    reconsider c as Cardinal by A7;
A8: f.x = exp(a,b) by A5,FUNCOP_1:7;
    (b-powerfunc_of a).x = exp(c,b) by A6,Def2;
    hence (b-powerfunc_of a).x c= f.x by A8,CARD_2:93;
  end;
  then Sum (b-powerfunc_of a) c= Sum f by A3,CARD_3:30;
  hence thesis by A2,A4,XBOOLE_1:1;
end;

theorem
  a is limit_cardinal & b in cf a implies exp(a,b) = Sum (b-powerfunc_of
  a)
proof
  assume that
A1: a is limit_cardinal and
A2: b in cf a;
  deffunc f(Ordinal) = Funcs(b,$1);
  consider L being T-Sequence such that
A3: dom L = a & for A st A in a holds L.A = f(A) from ORDINAL2:sch 2;
A4: now
    let x;
A5: card Union (b-powerfunc_of a) c= Sum (b-powerfunc_of a) by CARD_3:40;
    assume
A6: x in a;
    then reconsider x9 = x as Ordinal;
    set m = card x9;
A7: m in a by A6,CARD_1:8,ORDINAL1:12;
    then m in dom (b-powerfunc_of a) by Def2;
    then
A8: (b-powerfunc_of a).(card x) in rng (b-powerfunc_of a) by FUNCT_1:def 3;
    x9,m are_equipotent by CARD_1:def 2;
    then
A9: card Funcs(b,x9) = card Funcs(b,card x9) by FUNCT_5:60;
    L.x = Funcs(b,x9) & (Card L).x = card (L.x) by A3,A6,CARD_3:def 2;
    then
A10: (Card L).x = exp(m,b) by A9,CARD_2:def 3;
    (b-powerfunc_of a).(card x) = exp(card x,b) by A7,Def2;
    then card exp(card x,b) c= card Union (b-powerfunc_of a) by A8,CARD_1:11
,ZFMISC_1:74;
    then
A11: card exp(card x,b) c= Sum (b-powerfunc_of a) by A5,XBOOLE_1:1;
    card exp(card x,b) = exp(card x,b) by CARD_1:def 2;
    hence (Card L).x c= (a --> Sum (b-powerfunc_of a)).x by A6,A10,A11,
FUNCOP_1:7;
  end;
  dom (a --> Sum (b-powerfunc_of a)) = a & dom Card L = dom L by CARD_3:def 2
,FUNCOP_1:13;
  then
A12: Sum Card L c= Sum (a --> Sum (b-powerfunc_of a)) by A3,A4,CARD_3:30;
  a c= Sum (b-powerfunc_of a)
  proof
    let x;
    assume
A13: x in a;
    then reconsider x9 = x as Ordinal;
    set m = card x9;
    m in a by A13,CARD_1:8,ORDINAL1:12;
    then
A14: nextcard m c= a by CARD_3:90;
    nextcard m <> a by A1,CARD_1:def 4;
    then
A15: nextcard m in a by A14,CARD_1:3;
    then nextcard m in dom (b-powerfunc_of a) by Def2;
    then
A16: (b-powerfunc_of a).(nextcard m) in rng (b-powerfunc_of a) by FUNCT_1:def 3
;
    (b-powerfunc_of a).(nextcard m) = exp(nextcard m,b) by A15,Def2;
    then
A17: card exp(nextcard m,b) c= card Union (b-powerfunc_of a) by A16,CARD_1:11
,ZFMISC_1:74;
A18: nextcard m c= exp(nextcard m,b) by Th20;
    card x = card card x;
    then
A19: x9 in nextcard x9 & nextcard card x = nextcard x by CARD_1:18,CARD_3:88;
    card exp(nextcard m,b) = exp(nextcard m,b) & card Union (b
    -powerfunc_of a) c= Sum (b-powerfunc_of a) by CARD_1:def 2,CARD_3:40;
    then exp(nextcard m,b) c= Sum (b-powerfunc_of a) by A17,XBOOLE_1:1;
    then nextcard card x c= Sum (b-powerfunc_of a) by A18,XBOOLE_1:1;
    hence thesis by A19;
  end;
  then
A20: a*`Sum (b-powerfunc_of a) = Sum (b-powerfunc_of a) by Th18;
  Funcs(b,a) c= Union L
  proof
    let x;
    assume x in Funcs(b,a);
    then consider f such that
A21: x = f and
A22: dom f = b and
A23: rng f c= a by FUNCT_2:def 2;
    reconsider f as T-Sequence by A22,ORDINAL1:def 7;
    reconsider f as Ordinal-Sequence by A23,ORDINAL2:def 4;
    rng f c= sup f by ORDINAL2:49;
    then
A24: x in Funcs(b,sup f) by A21,A22,FUNCT_2:def 2;
    sup f in a by A2,A22,A23,Th27;
    then
A25: L.sup f in rng L by A3,FUNCT_1:def 3;
    L.sup f = Funcs(b,sup f) by A2,A3,A22,A23,Th27;
    hence thesis by A24,A25,TARSKI:def 4;
  end;
  then
A26: card Funcs(b,a) c= card Union L by CARD_1:11;
  card Union L c= Sum Card L by CARD_3:39;
  then card Funcs(b,a) c= Sum Card L by A26,XBOOLE_1:1;
  then
A27: exp(a,b) c= Sum Card L by CARD_2:def 3;
A28: Sum (b-powerfunc_of a) c= exp(a,b) by Th33;
  Sum (a --> Sum (b-powerfunc_of a)) = a*`Sum (b-powerfunc_of a) by CARD_2:65;
  then exp(a,b) c= a*`Sum (b-powerfunc_of a) by A27,A12,XBOOLE_1:1;
  hence thesis by A28,A20,XBOOLE_0:def 10;
end;

theorem
  cf a c= b & b in a implies exp(a,b) = exp(Sum (b-powerfunc_of a), cf a
  )
proof
  assume that
A1: cf a c= b and
A2: b in a;
  consider phi such that
A3: dom phi = cf a and
A4: rng phi c= a and
  phi is increasing and
A5: a = sup phi and
A6: phi is Cardinal-Function and
A7: not 0 in rng phi by A1,A2,Th29,ORDINAL1:12;
  defpred R[set,set] means ex g,h st g = $1 & h = $2 & dom g = b & rng g c= a
  & dom h = cf a & for y st y in cf a ex fx st h.y = [fx,phi.y] & dom fx = b &
  for z st z in b holds (g.z in phi.y implies fx.z = g.z) & (not g.z in phi.y
  implies fx.z = 0);
A8: for x st x in Funcs(b,a) ex x1 being set st R[x,x1]
  proof
    let x;
    assume x in Funcs(b,a);
    then consider g such that
A9: x = g & dom g = b & rng g c= a by FUNCT_2:def 2;
    defpred P[set,set] means ex fx st $2 = [fx,phi.$1] & dom fx = b & for z st
z in b holds (g.z in phi.$1 implies fx.z = g.z) & (not g.z in phi.$1 implies fx
    .z = 0);
A10: for y st y in cf a ex x2 being set st P[y,x2]
    proof
      deffunc g(set) = 0;
      deffunc f(set) = g.$1;
      let y such that
      y in cf a;
      defpred C[set] means g.$1 in phi.y;
      consider fx such that
A11:  dom fx = b & for z st z in b holds (C[z] implies fx.z = f(z)) &
      (not C[z] implies fx.z = g(z)) from PARTFUN1:sch 1;
      take [fx,phi.y], fx;
      thus thesis by A11;
    end;
    consider h such that
A12: dom h = cf a & for y st y in cf a holds P[y,h.y] from CLASSES1:
    sch 1 ( A10);
    take h, g, h;
    thus thesis by A9,A12;
  end;
  consider f such that
A13: dom f = Funcs(b,a) & for x st x in Funcs(b,a) holds R[x,f.x] from
  CLASSES1:sch 1(A8);
  deffunc f(set) = Funcs(b,$1);
  consider F being Function such that
A14: dom F = dom (b-powerfunc_of a) & for x st x in dom (b-powerfunc_of
  a) holds F.x = f(x) from FUNCT_1:sch 3;
A15: rng f c= Funcs(cf a, Union disjoin F)
  proof
    let y;
    assume y in rng f;
    then consider x such that
A16: x in dom f and
A17: y = f.x by FUNCT_1:def 3;
    consider g,h such that
    g = x and
A18: h = f.x and
    dom g = b and
    rng g c= a and
A19: dom h = cf a and
A20: for y st y in cf a ex fx st h.y = [fx,phi.y] & dom fx = b & for z
st z in b holds (g.z in phi.y implies fx.z = g.z) & (not g.z in phi.y implies
    fx.z = 0) by A13,A16;
    rng h c= Union disjoin F
    proof
      let x1 be set;
      assume x1 in rng h;
      then consider x2 being set such that
A21:  x2 in dom h and
A22:  x1 = h.x2 by FUNCT_1:def 3;
      consider fx such that
A23:  x1 = [fx,phi.x2] and
A24:  dom fx = b and
A25:  for z st z in b holds (g.z in phi.x2 implies fx.z = g.z) & (not
      g.z in phi.x2 implies fx.z = 0) by A19,A20,A21,A22;
      rng fx c= phi.x2
      proof
        reconsider x2 as Ordinal by A19,A21;
        let z;
        reconsider A = phi.x2 as Ordinal;
        assume z in rng fx;
        then consider z9 being set such that
A26:    z9 in dom fx & z = fx.z9 by FUNCT_1:def 3;
A27:    z = g.z9 & g.z9 in phi.x2 or z = 0 by A24,A25,A26;
        A <> 0 by A3,A7,A19,A21,FUNCT_1:def 3;
        hence thesis by A27,ORDINAL3:8;
      end;
      then
A28:  fx in Funcs(b,phi.x2) by A24,FUNCT_2:def 2;
A29:  [fx,phi.x2]`1 = fx & [fx,phi.x2]`2 = phi.x2;
      phi.x2 in rng phi & phi.x2 is Cardinal by A3,A6,A19,A21,CARD_3:def 1
,FUNCT_1:def 3;
      then
A30:  phi.x2 in dom (b-powerfunc_of a) by A4,Def2;
      then F.(phi.x2) = Funcs(b,phi.x2) by A14;
      hence thesis by A14,A23,A28,A30,A29,CARD_3:22;
    end;
    hence thesis by A17,A18,A19,FUNCT_2:def 2;
  end;
  card card Union disjoin F = card Union disjoin F & card cf a = cf a by
CARD_1:def 2;
  then
A31: card Funcs(cf a, Union disjoin F) = card Funcs(cf a, card Union
  disjoin F) by FUNCT_5:61
    .= exp(card Union disjoin F, cf a) by CARD_2:def 3;
A32: dom Card F = dom F by CARD_3:def 2;
A33: f is one-to-one
  proof
    let x,y;
    assume that
A34: x in dom f and
A35: y in dom f and
A36: f.x = f.y;
    consider g1, h1 being Function such that
A37: g1 = x and
A38: h1 = f.x and
A39: dom g1 = b and
A40: rng g1 c= a and
    dom h1 = cf a and
A41: for x1 being set st x1 in cf a ex fx st h1.x1 = [fx,phi.x1] & dom
fx = b & for z st z in b holds (g1.z in phi.x1 implies fx.z = g1.z) & (not g1.z
    in phi.x1 implies fx.z = 0) by A13,A34;
    consider g2, h2 being Function such that
A42: g2 = y and
A43: h2 = f.y and
A44: dom g2 = b and
A45: rng g2 c= a and
    dom h2 = cf a and
A46: for x2 being set st x2 in cf a ex fx st h2.x2 = [fx,phi.x2] & dom
fx = b & for z st z in b holds (g2.z in phi.x2 implies fx.z = g2.z) & (not g2.z
    in phi.x2 implies fx.z = 0) by A13,A35;
    now
      let x1 be set;
      assume x1 in b;
      then reconsider X = x1 as Element of b;
      g1.X in rng g1 & g2.X in rng g2 by A39,A44,FUNCT_1:def 3;
      then reconsider A1 = g1.x1, A2 = g2.x1 as Element of a by A40,A45;
      set A = A1 \/ A2;
      A = A1 or A = A2 by ORDINAL3:12;
      then succ A in a by ORDINAL1:28;
      then consider B such that
A47:  B in rng phi and
A48:  succ A c= B by A5,ORDINAL2:21;
      consider x2 being set such that
A49:  x2 in dom phi and
A50:  B = phi.x2 by A47,FUNCT_1:def 3;
A51:  A in succ A by ORDINAL1:6;
      then
A52:  A2 in B by A48,ORDINAL1:12,XBOOLE_1:7;
      consider f1 being Function such that
A53:  h1.x2 = [f1,phi.x2] and
      dom f1 = b and
A54:  for z st z in b holds (g1.z in phi.x2 implies f1.z = g1.z) & (
      not g1.z in phi.x2 implies f1.z = 0) by A3,A41,A49;
      A1 in B by A48,A51,ORDINAL1:12,XBOOLE_1:7;
      then
A55:  f1.X = g1.x1 by A50,A54;
      consider f2 being Function such that
A56:  h2.x2 = [f2,phi.x2] and
      dom f2 = b and
A57:  for z st z in b holds (g2.z in phi.x2 implies f2.z = g2.z) & (
      not g2.z in phi.x2 implies f2.z = 0) by A3,A46,A49;
      f1 = f2 by A36,A38,A43,A53,A56,XTUPLE_0:1;
      hence g1.x1 = g2.x1 by A50,A57,A52,A55;
    end;
    hence thesis by A37,A39,A42,A44,FUNCT_1:2;
  end;
A58: dom disjoin F = dom F by CARD_3:def 3;
A59: now
    let x;
    assume
A60: x in dom F;
    then
A61: (disjoin F).x = [:F.x,{x}:] by CARD_3:def 3;
    (Card F).x = card (F.x) & (Card disjoin F).x = card ((disjoin F).x)
    by A58,A60,CARD_3:def 2;
    hence (Card disjoin F).x = (Card F).x by A61,CARD_1:69;
  end;
  now
    let x;
    assume
A62: x in dom F;
    then reconsider M = x as Cardinal by A14,Def2;
    M in a by A14,A62,Def2;
    then
A63: (b-powerfunc_of a).M = exp(M,b) by Def2;
    (Card F).x = card (F.x) & F.x = Funcs(b,x) by A14,A62,CARD_3:def 2;
    hence (Card F).x = (b-powerfunc_of a).x by A63,CARD_2:def 3;
  end;
  then
A64: Card F = b-powerfunc_of a by A14,A32,FUNCT_1:2;
  dom Card disjoin F = dom disjoin F by CARD_3:def 2;
  then Card F = Card disjoin F by A58,A32,A59,FUNCT_1:2;
  then card Union disjoin F c= Sum (b-powerfunc_of a) by A64,CARD_3:39;
  then
A65: exp(card Union disjoin F, cf a) c= exp(Sum (b-powerfunc_of a), cf a)
  by CARD_2:93;
  exp(a,b) = card Funcs(b,a) by CARD_2:def 3;
  then exp(a,b) c= card Funcs(cf a, Union disjoin F) by A13,A33,A15,CARD_1:10;
  then
A66: exp(exp(a,b), cf a) = exp(a,b*`cf a) & exp(a,b) c= exp(Sum (b
  -powerfunc_of a ), cf a) by A31,A65,CARD_2:30,XBOOLE_1:1;
  Sum (b-powerfunc_of a) c= exp(a,b) by Th33;
  then
A67: exp(Sum (b-powerfunc_of a), cf a) c= exp(exp(a,b), cf a) by CARD_2:93;
  b*`cf a = b by A1,Th18;
  hence thesis by A67,A66,XBOOLE_0:def 10;
end;

reserve O for Ordinal,
        F for Subset of omega;



Lm5: for X being finite set st X c= O holds order_type_of RelIncl X is finite
proof
  let X be finite set;
  assume X c= O;
  then RelIncl X is well-ordering by WELLORD2:8;

  then RelIncl X,RelIncl order_type_of RelIncl X are_isomorphic by
WELLORD2:def 2;

  hence thesis by CARD_3:105;
end;

theorem Th36:
  for X being finite set st X c= O holds order_type_of RelIncl X =
  card X
proof
  let X be finite set;
  assume
A1: X c= O;
  then order_type_of RelIncl X is finite by Lm5;
  then reconsider o = order_type_of RelIncl X as Nat;
  card X = card order_type_of RelIncl X by A1,Th7;
  then o = card X by CARD_1:def 2;
  hence thesis;
end;

theorem Th37:
  {x} c= O implies order_type_of RelIncl {x} = 1
proof
  card {x} = 1 by CARD_2:42;
  hence thesis by Th36;
end;

theorem
  {x} c= O implies canonical_isomorphism_of (RelIncl order_type_of
  RelIncl {x}, RelIncl {x}) = 0 .--> x
proof
  set X = {x};
  set R = RelIncl X;
  set C = canonical_isomorphism_of (RelIncl order_type_of R,R);
A1: RelIncl {0} = {[0,0]} by WELLORD2:22;
  assume
A2: X c= O;
  then
A3: order_type_of R = {0} by Th37,CARD_1:49;
  R is well-ordering by A2,WELLORD2:8;
  then R, RelIncl order_type_of R are_isomorphic by WELLORD2:def 2;
  then
A4: RelIncl order_type_of R, R are_isomorphic by WELLORD1:40;
  RelIncl order_type_of R is well-ordering by WELLORD2:8;
  then
A5: C is_isomorphism_of RelIncl order_type_of R, R by A4,WELLORD1:def 9;
  then
A6: rng canonical_isomorphism_of(RelIncl {0}, R) = field R by A3,WELLORD1:def 7
    .= X by WELLORD2:def 1;
  dom canonical_isomorphism_of(RelIncl {0}, R) = field RelIncl {0} by A5,A3,
WELLORD1:def 7
    .= {0} by A1,RELAT_1:173;
  hence thesis by A3,A6,FUNCT_4:112;
end;

registration
  let O be Ordinal, X be Subset of O, n be set;
  cluster canonical_isomorphism_of (RelIncl order_type_of RelIncl X,RelIncl X)
    .n -> ordinal;
  coherence
  proof
    consider phi being Ordinal-Sequence such that
A1: phi = canonical_isomorphism_of(RelIncl order_type_of RelIncl X,
    RelIncl X) and
    phi is increasing and
    dom phi = order_type_of RelIncl X and
A2: rng phi = X by Th5;
    per cases;
    suppose
      n in dom phi;
      then phi.n in rng phi by FUNCT_1:def 3;
      hence thesis by A1,A2;
    end;
    suppose
      not n in dom phi;
      hence thesis by A1,FUNCT_1:def 2;
    end;
  end;
end;

registration
  let X be natural-membered set, n be set;
  cluster canonical_isomorphism_of
   (RelIncl order_type_of RelIncl X,RelIncl X).n -> natural;
  coherence
  proof
    X c= NAT
    proof
      let x be set;
      assume x in X;
      hence thesis by ORDINAL1:def 12;
    end;
    then reconsider X as Subset of NAT;
    consider phi being Ordinal-Sequence such that
A1: phi = canonical_isomorphism_of(RelIncl order_type_of RelIncl X,
    RelIncl X) and
    phi is increasing and
    dom phi = order_type_of RelIncl X and
A2: rng phi = X by Th5;
    per cases;
    suppose
A3:   n in dom phi;
      phi.n in rng phi by A3,FUNCT_1:def 3;
      hence thesis by A1,A2;
    end;
    suppose
      not n in dom phi;
      hence thesis by A1,FUNCT_1:def 2;
    end;
  end;
end;

theorem
  card F c= order_type_of RelIncl F
proof
  card F = card order_type_of RelIncl F by Th7;
  hence thesis by CARD_1:8;
end;















begin

reserve x,y,z,a,b,c,X,X1,X2,Y,Z for set,
  W,W1,W2 for Tree,
  w,w9 for Element of W,
  f for Function,
  D,D9 for non empty set,
  i,k,k1,k2,l,m,n for Element of NAT,
  v,v1,v2 for FinSequence,
  p,q,r,r1,r2 for FinSequence of NAT;

theorem Th1:
  for v1,v2,v st v1 is_a_prefix_of v & v2 is_a_prefix_of v
  holds v1,v2 are_c=-comparable
proof
  let p,q,r be FinSequence;
  assume p is_a_prefix_of r;
then A1: ex p9 being FinSequence st r = p^p9 by TREES_1:1;
  assume q is_a_prefix_of r;
then A2: ex q9 being FinSequence st r = q^q9 by TREES_1:1;
 len p <= len q or len q <= len p;
then  (ex t being FinSequence st p^t = q) or
  ex t being FinSequence st q^t = p by A1,A2,FINSEQ_1:47;
  hence p is_a_prefix_of q or q is_a_prefix_of p by TREES_1:1;
end;

theorem Th2:
  for v1,v2,v st v1 is_a_proper_prefix_of v & v2 is_a_prefix_of v holds
  v1,v2 are_c=-comparable
proof
  let p,q,r be FinSequence;
  assume p is_a_proper_prefix_of r;
then  p is_a_prefix_of r by XBOOLE_0:def 8;
  hence thesis by Th1;
end;

theorem Th3:
  len v1 = k+1 implies ex v2,x st v1 = v2^<*x*> & len v2 = k
proof
  assume
A1: len v1 = k+1;
  reconsider v2 = v1|Seg k as FinSequence by FINSEQ_1:15;
 v2 is_a_prefix_of v1 by TREES_1:def 1;
  then consider v such that
A2: v1 = v2^v by TREES_1:1;
  take v2, v.1;
A3: k <= k+1 by NAT_1:11;
then  len v2 = k by A1,FINSEQ_1:17;
then  k + len v = k+1 by A1,A2,FINSEQ_1:22;
  hence thesis by A1,A2,A3,FINSEQ_1:17,40;
end;

theorem Th4:
  ProperPrefixes (v^<*x*>) = ProperPrefixes v \/ {v}
proof
  thus ProperPrefixes (v^<*x*>) c= ProperPrefixes v \/ {v}
  proof
    let y;
    assume y in ProperPrefixes (v^<*x*>);
    then consider v1 such that
A1: y = v1 and
A2: v1 is_a_proper_prefix_of v^<*x*> by TREES_1:def 2;
 v1 is_a_prefix_of v & v1 <> v or v1 = v by A2,TREES_1:9;
then
 v1 is_a_proper_prefix_of v or v1 in {v} by TARSKI:def 1,XBOOLE_0:def 8;
then  y in ProperPrefixes v or y in {v} by A1,TREES_1:def 2;
    hence thesis by XBOOLE_0:def 3;
  end;
  let y;
  assume y in ProperPrefixes v \/ {v};
then A3: y in ProperPrefixes v or y in {v} by XBOOLE_0:def 3;
A4: now
    assume y in ProperPrefixes v;
    then consider v1 such that
A5: y = v1 and
A6: v1 is_a_proper_prefix_of v by TREES_1:def 2;
 v is_a_prefix_of v^<*x*> by TREES_1:1;
then  v1 is_a_proper_prefix_of v^<*x*> by A6,XBOOLE_1:58;
    hence thesis by A5,TREES_1:def 2;
  end;
 v^{} = v by FINSEQ_1:34;
  then
 v is_a_prefix_of v^<*x*> & v <> v^<*x*> by FINSEQ_1:33,TREES_1:1;
then  v is_a_proper_prefix_of v^<*x*> by XBOOLE_0:def 8;
then  y in ProperPrefixes v or y = v & v in ProperPrefixes (v^<*x*>)
  by A3,TARSKI:def 1,TREES_1:def 2;
  hence thesis by A4;
end;

scheme TreeStructInd { T()->Tree, P[set] }:
  for t being Element of T() holds P[t]
provided
A1: P[{}] and
A2: for t being Element of T(), n st P[t] & t^<*n*> in T() holds P[t^<*n*>]
proof
  defpred X[set] means for t being Element of T() st len t = $1 holds P[t];
 card x = 0 implies x = {};
then A3: X[0] by A1;
A4: X[k] implies X[k+1]
  proof
    assume
A5: for t being Element of T() st len t = k holds P[t];
    let t be Element of T();
    assume len t = k+1;
    then consider v, x such that
A6: t = v^<*x*> and
A7: len v = k by Th3;
    reconsider v as FinSequence of NAT by A6,FINSEQ_1:36;
    reconsider v as Element of T() by A6,TREES_1:21;
 rng <*x*> c= rng t by A6,FINSEQ_1:30;
then  rng <*x*> = {x} & rng <*x*> c= NAT by FINSEQ_1:39,XBOOLE_1:1;
    then reconsider x as Element of NAT by ZFMISC_1:31;
A8: P[v] by A5,A7;
 x = x;
    hence thesis by A2,A6,A8;
  end;
A9: X[k] from NAT_1:sch 1(A3,A4);
  let t be Element of T();
 len t = len t;
  hence thesis by A9;
end;

theorem Th5:
  (for p holds p in W1 iff p in W2) implies W1 = W2
proof
  assume
A1: for p holds p in W1 iff p in W2;
  thus W1 c= W2
  proof
    let x;
    assume x in W1;
    then reconsider p = x as Element of W1;
 p in W2 by A1;
    hence thesis;
  end;
  let x;
  assume x in W2;
  then reconsider p = x as Element of W2;
 p in W1 by A1;
  hence thesis;
end;

definition
  let W1,W2;
  redefine pred W1 = W2 means
  for p holds p in W1 iff p in W2;
  compatibility by Th5;
end;

theorem
  p in W implies W = W with-replacement (p,W|p)
proof
  assume
A1: p in W;
 now
    let q;
    thus q in W implies q in W with-replacement (p,W|p)
    proof
      assume
A2:   q in W;
   now
        assume p is_a_proper_prefix_of q;
then      p is_a_prefix_of q by XBOOLE_0:def 8;
        then consider r being FinSequence such that
A3:     q = p^r by TREES_1:1;
     rng r c= rng q by A3,FINSEQ_1:30;
then      rng r c= NAT by XBOOLE_1:1;
        then reconsider r as FinSequence of NAT by FINSEQ_1:def 4;
        take r;
        thus r in W|p & q = p^r by A1,A2,A3,TREES_1:def 6;
      end;
      hence thesis by A1,A2,TREES_1:def 9;
    end;
    assume that
A4: q in W with-replacement (p,W|p) and
A5: not q in W;
 ex r st r in W|p & q = p^r by A1,A4,A5,TREES_1:def 9;
    hence contradiction by A1,A5,TREES_1:def 6;
  end;
  hence thesis by Th5;
end;

theorem Th7:
  p in W & q in W & not p is_a_prefix_of q implies
  q in W with-replacement (p,W1)
proof
 not p is_a_prefix_of q implies not p is_a_proper_prefix_of q
  by XBOOLE_0:def 8;
  hence thesis by TREES_1:def 9;
end;

theorem
  p in W & q in W & not p,q are_c=-comparable implies
  (W with-replacement (p,W1)) with-replacement (q,W2) =
  (W with-replacement (q,W2)) with-replacement (p,W1)
proof
  assume that
A1: p in W and
A2: q in W and
A3: not p,q are_c=-comparable;
A4: not p is_a_prefix_of q by A3,XBOOLE_0:def 9;
 not q is_a_prefix_of p by A3,XBOOLE_0:def 9;
then A5: p in W with-replacement (q,W2) by A1,A2,Th7;
A6: q in W with-replacement (p,W1) by A1,A2,A4,Th7;
  let r;
  thus r in (W with-replacement (p,W1)) with-replacement (q,W2) implies
  r in (W with-replacement (q,W2)) with-replacement (p,W1)
  proof
    assume r in (W with-replacement (p,W1)) with-replacement (q,W2);
then  r in W with-replacement (p,W1) & not q is_a_proper_prefix_of r or
    ex r1 st r1 in W2 & r = q^r1 by A6,TREES_1:def 9;
then
 r in W & not p is_a_proper_prefix_of r & not q is_a_proper_prefix_of r or
    (ex r2 st r2 in W1 & r = p^r2) & not q is_a_proper_prefix_of r or
    q is_a_prefix_of r & ex r1 st r1 in W2 & r = q^r1 by A1,TREES_1:1,def 9
    ;
then  r in W with-replacement (q,W2) & not p is_a_proper_prefix_of r or
    ex r1 st r1 in W1 & r = p^r1 by A2,A3,Th2,TREES_1:def 9;
    hence thesis by A5,TREES_1:def 9;
  end;
  assume r in (W with-replacement (q,W2)) with-replacement (p,W1);
then  r in W with-replacement (q,W2) & not p is_a_proper_prefix_of r or
  ex r1 st r1 in W1 & r = p^r1 by A5,TREES_1:def 9;
then
 r in W & not q is_a_proper_prefix_of r & not p is_a_proper_prefix_of r or
  (ex r2 st r2 in W2 & r = q^r2) & not p is_a_proper_prefix_of r or
  p is_a_prefix_of r & ex r1 st r1 in W1 & r = p^r1 by A2,TREES_1:1,def 9;
then  r in W with-replacement (p,W1) & not q is_a_proper_prefix_of r or
  ex r1 st r1 in W2 & r = q^r1 by A1,A3,Th2,TREES_1:def 9;
  hence thesis by A6,TREES_1:def 9;
end;

definition
  let IT be Tree;
  attr IT is finite-order means
  ex n st for t being Element of IT holds not t^<*n*> in IT;
end;

registration
  cluster finite-order for Tree;
  existence
  proof reconsider T = {{}} as Tree;
    take T,0;
    let t be Element of T;
    thus thesis by TARSKI:def 1;
  end;
end;

definition
  let W;
  mode Chain of W -> Subset of W means
    :
    Def3: for p,q st p in it & q in it holds p,q are_c=-comparable;
  existence
  proof reconsider S = {} as Subset of W by XBOOLE_1:2;
    take S;
    let p,q;
    thus thesis;
  end;
  mode Level of W -> Subset of W means
    : Def4:
    ex n being Nat st it = { w: len w = n};
  existence
  proof
 {} in W by TREES_1:22;
    then reconsider L = {{}} as Subset of W by ZFMISC_1:31;
    take L,0;
    thus L c= { w: len w = 0}
    proof
      let x;
      assume
A1:   x in L;
then A2:   x = {} by TARSKI:def 1;
   len {} = 0;
      hence thesis by A1,A2;
    end;
    let x;
    assume x in { w: len w = 0};
then  ex w st x = w & len w = 0;
then  x = {};
    hence thesis by TARSKI:def 1;
  end;
  let w;
  func succ w -> Subset of W equals
  { w^<*n*>: w^<*n*> in W };
  coherence
  proof
 { w^<*n*>: w^<*n*> in W } c= W
    proof
      let x;
      assume x in { w^<*n*>: w^<*n*> in W };
then   ex n st x = w^<*n*> & w^<*n*> in W;
      hence thesis;
    end;
    hence thesis;
  end;
end;

theorem
  for L being Level of W holds L is AntiChain_of_Prefixes of W
proof
  let L be Level of W;
  consider n being Nat such that
A1: L = { w: len w = n } by Def4;
 L is AntiChain_of_Prefixes-like
  proof
    thus for x st x in L holds x is FinSequence
    proof
      let x;
      assume x in L;
then    x is Element of W;
      hence thesis;
    end;
    let v1,v2;
    assume v1 in L;
then A2: ex w1 be Element of W st v1 = w1 & len w1 = n by A1;
    assume v2 in L;
then  ex w2 be Element of W st v2 = w2 & len w2 = n by A1;
    hence thesis by A2,TREES_1:4;
  end;
  hence thesis by TREES_1:def 11;
end;

theorem
  succ w is AntiChain_of_Prefixes of W
proof
 succ w is AntiChain_of_Prefixes-like
  proof
    thus for x st x in succ w holds x is FinSequence
    proof
      let x;
      assume x in succ w;
then    ex i st x = w^<*i*> & w^<*i*> in W;
      hence thesis;
    end;
    let v1,v2;
    assume v1 in succ w;
then A1: ex k1 st v1 = w^<*k1*> & w^<*k1*> in W;
    assume v2 in succ w;
then A2: ex k2 st v2 = w^<*k2*> & w^<*k2*> in W;
A3: len v1 = len w + 1 by A1,FINSEQ_2:16;
 len v2 = len w + 1 by A2,FINSEQ_2:16;
    hence thesis by A3,TREES_1:4;
  end;
  hence thesis by TREES_1:def 11;
end;

theorem
  for A being AntiChain_of_Prefixes of W, C being Chain of W
  ex w st A /\ C c= {w}
proof
  let A be AntiChain_of_Prefixes of W, C be Chain of W;
A1: now
    let p,q;
    assume
A2: p in A /\ C & q in A /\ C;
then A3: p in A & q in A by XBOOLE_0:def 4;
 p in C & q in C by A2,XBOOLE_0:def 4;
then  p,q are_c=-comparable by Def3;
    hence p = q by A3,TREES_1:def 10;
  end;
  set w = the Element of W;
 now per cases;
    suppose
   A /\ C = {};
then    A /\ C c= {w} by XBOOLE_1:2;
      hence thesis;
    end;
    suppose
A4:   A /\ C <> {};
      set x = the Element of A /\ C;
  x in C by A4,XBOOLE_0:def 4;
      then reconsider x as Element of W;
      take x;
      thus A /\ C c= {x}
      proof
        let y;
        assume
A5:    y in A /\ C;
then     y is Element of W;
        then reconsider y9 = y as FinSequence of NAT;
    x = y9 by A1,A5;
        hence thesis by TARSKI:def 1;
      end;
    end;
  end;
  hence thesis;
end;

definition
  let W;
  let n be Nat;
  func W-level n -> Level of W equals
  { w: len w = n };
  coherence
  proof
    defpred X[Element of W] means len $1 = n;
 { w: X[w] } is Subset of W from DOMAIN_1:sch 7;
    hence thesis by Def4;
  end;
end;

theorem
  w^<*n*> in succ w iff w^<*n*> in W;

theorem
  w = {} implies W-level 1 = succ w
proof
  assume
A1: w = {};
  thus W-level 1 c= succ w
  proof
    let x;
    assume x in W-level 1;
    then consider w9 such that
A2: x = w9 and
A3: len w9 = 1;
A4: w9 = <*w9.1*> by A3,FINSEQ_1:40;
then  rng w9 = {w9.1} by FINSEQ_1:39;
    then reconsider n = w9.1 as Element of NAT by ZFMISC_1:31;
 w9 = w^<*n*> by A1,A4,FINSEQ_1:34;
    hence thesis by A2;
  end;
  let x;
  assume x in succ w;
  then consider i such that
A5: x = w^<*i*> and
A6: w^<*i*> in W;
  reconsider w9 = w^<*i*> as Element of W by A6;
 len <*i*> = 1 & w9 = <*i*> by A1,FINSEQ_1:34,39;
  hence thesis by A5;
end;

theorem Th14:
  W = union { W-level n: not contradiction }
proof
  thus W c= union { W-level n: not contradiction }
  proof
    let x;
    assume x in W;
    then reconsider w = x as Element of W;
A1: x in W-level len w;
 W-level len w in { W-level n: not contradiction };
    hence thesis by A1,TARSKI:def 4;
  end;
  let x;
  assume x in union { W-level n: not contradiction };
  then consider X such that
A2: x in X & X in { W-level n: not contradiction } by TARSKI:def 4;
 ex n st X = W-level n by A2;
  hence thesis by A2;
end;

theorem
  for W being finite Tree holds W = union { W-level n: n <= height W }
proof
  let W be finite Tree;
  thus W c= union { W-level n: n <= height W }
  proof
    let x;
    assume x in W;
    then reconsider w = x as Element of W;
A1: len w <= height W by TREES_1:def 12;
A2: w in W-level len w;
 W-level len w in { W-level n: n <= height W } by A1;
    hence thesis by A2,TARSKI:def 4;
  end;
  let x;
  assume x in union { W-level n: n <= height W };
  then consider X such that
A3: x in X & X in { W-level n: n <= height W } by TARSKI:def 4;
 ex n st X = W-level n & n <= height W by A3;
  hence thesis by A3;
end;

theorem
  for L being Level of W ex n st L = W-level n
proof
  let L be Level of W;
  consider n being Nat such that
A1: L = { w: len w = n} by Def4;
  reconsider n as Element of NAT by ORDINAL1:def 12;
  take n;
  thus thesis by A1;
end;

scheme FraenkelCard { A() ->non empty set, X() -> set, F(set) -> set }:
  card { F(w) where w is Element of A(): w in X() } c= card X()
proof
  consider f such that
A1: dom f = X() & for x st x in X() holds f.x = F(x) from FUNCT_1:sch 3;
 { F(w) where w is Element of A(): w in X() } c= rng f
  proof
    let x;
    assume x in { F(w) where w is Element of A(): w in X() };
    then consider w being Element of A() such that
A2: x = F(w) and
A3: w in X();
 f.w = x by A1,A2,A3;
    hence thesis by A1,A3,FUNCT_1:def 3;
  end;
  hence thesis by A1,CARD_1:12;
end;

scheme FraenkelFinCard { A() ->non empty set,
  X,Y() -> finite set, F(set) -> set }: card Y() <= card X()
provided
A1: Y() = { F(w) where w is Element of A(): w in X() }
proof
 card { F(w) where w is Element of A(): w in X() } c= card X()
  from FraenkelCard;
  hence thesis by A1,NAT_1:39;
end;

theorem Th17:
  W is finite-order implies ex n st for w holds
  ex B being finite set st B = succ w & card B <= n
proof
  given n such that
A1: for w holds not w^<*n*> in W;
A2: Seg n is finite;
  take n;
  let w;
  deffunc U(Real) = w^<*$1-1*>;
A3: { U(r) where r is Real: r in Seg n } is finite from FRAENKEL:sch 21(A2
  );
A4: succ w c= { w^<*r-1*> where r is Real: r in Seg n }
  proof
    let x;
    assume x in succ w;
    then consider k such that
A5: x = w^<*k*> and
A6: w^<*k*> in W;
 not w^<*n*> in W by A1;
then  k < n by A6,TREES_1:def 3;
then A7: k+1 <= n by NAT_1:13;
 1 <= k+1 by NAT_1:11;
then A8: k+1 in Seg n by A7,FINSEQ_1:1;
 (k+1)-1 = k;
    hence thesis by A5,A8;
  end;
  then reconsider B = succ w as finite set by A3;
  take B;
  thus B = succ w;
  set C = { U(r) where r is Real: r in Seg n };
 C is finite from FRAENKEL:sch 21(A2);
  then reconsider C as finite set;
A9: C = { U(r) where r is Real: r in Seg n };
 card C <= card Seg n from FraenkelFinCard(A9);
then A10: card C <= n by FINSEQ_1:57;
 card B <= card C by A4,NAT_1:43;
  hence thesis by A10,XXREAL_0:2;
end;

theorem Th18:
  W is finite-order implies succ w is finite
proof
  assume W is finite-order;
  then consider n such that
A1: for w holds ex B being finite set st B = succ w & card B <= n by Th17;
 ex B being finite set st B = succ w & card B <= n by A1;
  hence thesis;
end;

registration
  let W be finite-order Tree;
  let w be Element of W;
  cluster succ w -> finite;
  coherence by Th18;
end;

theorem Th19:
  {} is Chain of W proof reconsider S = {} as Subset of W by XBOOLE_1:2;
 S is Chain of W
  proof
    let p,q;
    thus thesis;
  end;
  hence thesis;
end;

theorem Th20:
  {{}} is Chain of W
proof
 {} in W by TREES_1:22;
  then reconsider S = {{}} as Subset of W by ZFMISC_1:31;
 S is Chain of W
  proof
    let p,q;
    assume that
A1: p in S and
A2: q in S;
 p = {} by A1,TARSKI:def 1;
    hence thesis by A2,TARSKI:def 1;
  end;
  hence thesis;
end;

registration
  let W;
  cluster non empty for Chain of W;
  existence
  proof
 {{}} is Chain of W by Th20;
    hence thesis;
  end;
end;

definition
  let W;
  let IT be Chain of W;
  attr IT is Branch-like means
  :
  Def7: (for p st p in IT holds ProperPrefixes p c= IT) &
  not ex p st p in W & for q st q in IT holds q is_a_proper_prefix_of p;
end;

registration
  let W;
  cluster Branch-like for Chain of W;
  existence
  proof
    defpred X[set] means $1 is Chain of W &
    for p st p in $1 holds ProperPrefixes p c= $1;
    consider X such that
A1: Y in X iff Y in bool W & X[Y] from XBOOLE_0:sch 1;
     {}
 is Chain of W & for p st p in {} holds ProperPrefixes p c= {} by Th19;
then A2: X <> {} by A1;
 now
      let Z;
      assume that
      Z <> {} and
A3:   Z c= X and
A4:   Z is c=-linear;
   union Z c= W
      proof
        let x;
        assume x in union Z;
        then consider Y such that
A5:     x in Y and
A6:    Y in Z by TARSKI:def 4;
    Y in bool W by A1,A3,A6;
        hence thesis by A5;
      end;
      then reconsider Z9 = union Z as Subset of W;
A7:  Z9 is Chain of W
      proof
        let p,q;
        assume p in Z9;
        then consider X1 such that
A8:    p in X1 and
A9:    X1 in Z by TARSKI:def 4;
        assume q in Z9;
        then consider X2 such that
A10:    q in X2 and
A11:    X2 in Z by TARSKI:def 4;
    X1,X2 are_c=-comparable by A4,A9,A11,ORDINAL1:def 8;
then A12:    X1 c= X2 or X2 c= X1 by XBOOLE_0:def 9;
A13:    X1 is Chain of W by A1,A3,A9;
    X2 is Chain of W by A1,A3,A11;
        hence thesis by A8,A10,A12,A13,Def3;
      end;
  now
        let p;
        assume p in union Z;
        then consider X1 such that
A14:    p in X1 & X1 in Z by TARSKI:def 4;
            ProperPrefixes
 p c= X1 & X1 c= union Z by A1,A3,A14,ZFMISC_1:74;
        hence ProperPrefixes p c= union Z by XBOOLE_1:1;
      end;
      hence union Z in X by A1,A7;
    end;
    then consider Y such that
A15: Y in X and
A16: Z in X & Z <> Y implies not Y c= Z by A2,ORDERS_1:67;
    reconsider Y as Chain of W by A1,A15;
    take Y;
    thus for p st p in Y holds ProperPrefixes p c= Y by A1,A15;
    given p such that
A17: p in W and
A18: for q st q in Y holds q is_a_proper_prefix_of p;
    set Z = (ProperPrefixes p) \/ {p};
     ProperPrefixes p c= W & {p} c= W by A17,TREES_1:def 3,ZFMISC_1:31;
    then reconsider Z9 = Z as Subset of W by XBOOLE_1:8;
A19: Z9 is Chain of W
    proof
      let q,r;
      assume that
A20:  q in Z9 and
A21:  r in Z9;
A22:  q in ProperPrefixes p or q in {p} by A20,XBOOLE_0:def 3;
A23:  r in ProperPrefixes p or r in {p} by A21,XBOOLE_0:def 3;
A24:  q is_a_proper_prefix_of p or q = p by A22,TARSKI:def 1,TREES_1:12;
A25:  r is_a_proper_prefix_of p or r = p by A23,TARSKI:def 1,TREES_1:12;
A26:  q is_a_prefix_of p by A24,XBOOLE_0:def 8;
  r is_a_prefix_of p by A25,XBOOLE_0:def 8;
      hence thesis by A26,Th1;
    end;
 now
      let q;
      assume q in Z;
then   q in ProperPrefixes p or q in {p} by XBOOLE_0:def 3;
then   q is_a_proper_prefix_of p or q = p by TARSKI:def 1,TREES_1:12;
then   q is_a_prefix_of p by XBOOLE_0:def 8;
then A27:  ProperPrefixes q c= ProperPrefixes p by TREES_1:17;
  ProperPrefixes p c= Z by XBOOLE_1:7;
      hence ProperPrefixes q c= Z by A27,XBOOLE_1:1;
    end;
then A28: Z in X by A1,A19;
A29: p in {p} by TARSKI:def 1;
A30: not p in Y by A18;
A31: p in Z by A29,XBOOLE_0:def 3;
 Y c= Z
    proof
      let x;
      assume
A32:  x in Y;
      then reconsider t = x as Element of W;
  t is_a_proper_prefix_of p by A18,A32;
then   t in ProperPrefixes p by TREES_1:12;
      hence thesis by XBOOLE_0:def 3;
    end;
    hence thesis by A16,A28,A30,A31;
  end;
end;

definition
  let W;
  mode Branch of W is Branch-like Chain of W;
end;

registration
  let W;
  cluster Branch-like -> non empty for Chain of W;
  coherence
  proof
    let B be Chain of W such that
A1: B is Branch-like empty;
    set t = the Element of W;
 for q st q in B holds q is_a_proper_prefix_of t by A1;
    hence contradiction by A1,Def7;
  end;
end;

reserve C for Chain of W,
  B for Branch of W;

theorem Th21:
  v1 in C & v2 in C implies v1 in ProperPrefixes v2 or v2 is_a_prefix_of v1
proof
  assume
A1: v1 in C & v2 in C;
  then reconsider p = v1, q = v2 as Element of W;
  assume
A2: not v1 in ProperPrefixes v2;
A3: p,q are_c=-comparable by A1,Def3;
A4: not v1 is_a_proper_prefix_of v2 by A2,TREES_1:12;
 v1 is_a_prefix_of v2 or v2 is_a_prefix_of v1 by A3,XBOOLE_0:def 9;
  hence thesis by A4,XBOOLE_0:def 8;
end;

theorem Th22:
  v1 in C & v2 in C & v is_a_prefix_of v2 implies
  v1 in ProperPrefixes v or v is_a_prefix_of v1
proof
  assume that
A1: v1 in C & v2 in C and
A2: v is_a_prefix_of v2;
 v1 in ProperPrefixes v2 or v2 is_a_prefix_of v1 by A1,Th21;
then  v1 is_a_proper_prefix_of v2 or v is_a_prefix_of v1
  by A2,TREES_1:12,XBOOLE_1:1;
then  v,v1 are_c=-comparable by A2,Th2,XBOOLE_0:def 9;
then  v is_a_proper_prefix_of v1 or v = v1 or v1 is_a_proper_prefix_of v
  by XBOOLE_1:104;
  hence thesis by TREES_1:12,XBOOLE_0:def 8;
end;

registration
  let W;
  cluster finite for Chain of W;
  existence
  proof reconsider C = {} as Chain of W by Th19;
    take C;
    thus thesis;
  end;
end;

theorem Th23:
  for C being finite Chain of W st card C > n ex p st p in C & len p >= n
proof
  let C be finite Chain of W;
  defpred X[Element of NAT] means
  $1 < card C implies ex p st p in C & len p >= $1;
A1: X[0]
  proof
    assume
A2: 0 < card C;
then A3: C <> {};
    set x = the Element of C;
    reconsider x as Element of W by A2,CARD_1:27,TARSKI:def 3;
    reconsider x as FinSequence of NAT;
    take x;
    thus thesis by A3;
  end;
A4: X[k] implies X[k+1]
  proof
    assume that
A5: k < card C implies ex p st p in C & len p >= k and
A6: k+1 < card C;
A7: k <= k+1 by NAT_1:11;
then A8: k < card C by A6,XXREAL_0:2;
    consider p such that
A9: p in C and
A10: len p >= k by A5,A6,A7,XXREAL_0:2;
    reconsider q = p|Seg k as FinSequence by FINSEQ_1:15;
A11: len q = k by A10,FINSEQ_1:17;
then A12: card ProperPrefixes q = k by TREES_1:35;
then  card ProperPrefixes q in card C by A8,NAT_1:44;
then A13: C \ ProperPrefixes q <> {} by CARD_1:68;
    set x = the Element of C \ ProperPrefixes q;
A14: x in C by A13,XBOOLE_0:def 5;
A15: not x in ProperPrefixes q by A13,XBOOLE_0:def 5;
    reconsider x as Element of W by A14;
 card (ProperPrefixes q \/ {x}) = k+1 by A12,A15,CARD_2:41;
then  card (ProperPrefixes q \/ {x}) in card C by A6,NAT_1:44;
then A16: C \ (ProperPrefixes q \/ {x}) <> {} by CARD_1:68;
    set y = the Element of C \ (ProperPrefixes q \/ {x});
A17: y in C by A16,XBOOLE_0:def 5;
A18: not y in ProperPrefixes q \/ {x} by A16,XBOOLE_0:def 5;
    reconsider y as Element of W by A17;
A19: not y in ProperPrefixes q by A18,XBOOLE_0:def 3;
A20: not y in {x} by A18,XBOOLE_0:def 3;
A21: q is_a_prefix_of p by TREES_1:def 1;
then A22: q is_a_prefix_of x by A9,A14,A15,Th22;
A23: q is_a_prefix_of y by A9,A17,A19,A21,Th22;
A24: x <> y by A20,TARSKI:def 1;
 x = q or x <> q;
then  q is_a_proper_prefix_of y or q is_a_proper_prefix_of x
    by A22,A23,A24,XBOOLE_0:def 8;
then  k < len x or k < len y by A11,TREES_1:6;
then  k+1 <= len x or k+1 <= len y by NAT_1:13;
    hence thesis by A14,A17;
  end;
 X[k] from NAT_1:sch 1(A1,A4);
  hence thesis;
end;

theorem Th24:
  for C holds { w: ex p st p in C & w is_a_prefix_of p } is Chain of W
proof
  let C;
 { w: ex p st p in C & w is_a_prefix_of p } c= W
  proof
    let x;
    assume x in { w: ex p st p in C & w is_a_prefix_of p };
then  ex w st x = w & ex p st p in C & w is_a_prefix_of p;
    hence thesis;
  end;
  then reconsider Z = { w: ex p st p in C & w is_a_prefix_of p } as Subset of
  W;
 Z is Chain of W
  proof
    let p,q;
    assume p in Z;
then  ex w st p = w & ex p st p in C & w is_a_prefix_of p;
    then consider r1 such that
A1: r1 in C and
A2: p is_a_prefix_of r1;
    assume q in Z;
then  ex w9 st q = w9 & ex p st p in C & w9 is_a_prefix_of p;
    then consider r2 such that
A3: r2 in C and
A4: q is_a_prefix_of r2;
 r1,r2 are_c=-comparable by A1,A3,Def3;
then  r1 is_a_prefix_of r2 or r2 is_a_prefix_of r1 by XBOOLE_0:def 9;
then  p is_a_prefix_of r2 or q is_a_prefix_of r1 by A2,A4,XBOOLE_1:1;
    hence thesis by A2,A4,Th1;
  end;
  hence thesis;
end;

theorem Th25:
  p is_a_prefix_of q & q in B implies p in B
proof
  assume p is_a_prefix_of q;
then  p is_a_proper_prefix_of q or p = q by XBOOLE_0:def 8;
then A1: p in ProperPrefixes q or p = q by TREES_1:def 2;
  assume
A2: q in B;
then  ProperPrefixes q c= B by Def7;
  hence thesis by A1,A2;
end;

theorem Th26:
  {} in B
proof
  set x = the Element of B;
  reconsider x as Element of W;
 <*> NAT is_a_prefix_of x by XBOOLE_1:2;
  hence thesis by Th25;
end;

theorem Th27:
  p in C & q in C & len p <= len q implies p is_a_prefix_of q
proof
  assume p in C & q in C & len p <= len q & not p is_a_prefix_of q;
  then  q
 in ProperPrefixes p & not q is_a_proper_prefix_of p by Th21,TREES_1:6;
  hence contradiction by TREES_1:12;
end;

theorem Th28:
  ex B st C c= B
proof
  defpred X[set] means $1 is Chain of W & C c= $1 &
  for p st p in $1 holds ProperPrefixes p c= $1;
  consider X such that
A1: Y in X iff Y in bool W & X[Y] from XBOOLE_0:sch 1;
  set Z = { w: ex p st p in C & w is_a_prefix_of p };
A2: Z is Chain of W by Th24;
A3: C c= Z
  proof
    let x;
    assume
A4: x in C;
    then reconsider w = x as Element of W;
 w is_a_prefix_of w;
    hence thesis by A4;
  end;
 now
    let p;
    assume p in Z;
then A5: ex w st p = w & ex p st p in C & w is_a_prefix_of p;
    then consider q such that
A6: q in C and
A7: p is_a_prefix_of q;
    thus ProperPrefixes p c= Z
    proof
      let x;
      assume x in ProperPrefixes p;
      then consider r being FinSequence such that
A8:  x = r and
A9:  r is_a_proper_prefix_of p by TREES_1:def 2;
  r is_a_prefix_of p by A9,XBOOLE_0:def 8;
then   r is_a_prefix_of q & r in W by A5,A7,TREES_1:20,XBOOLE_1:1;
      hence thesis by A6,A8;
    end;
  end;
then A10: X <> {} by A1,A2,A3;
 now
    let Z;
    assume that
A11: Z <> {} and
A12: Z c= X and
A13: Z is c=-linear;
 union Z c= W
    proof
      let x;
      assume x in union Z;
      then consider Y such that
A14:  x in Y and
A15:  Y in Z by TARSKI:def 4;
  Y in bool W by A1,A12,A15;
      hence thesis by A14;
    end;
    then reconsider Z9 = union Z as Subset of W;
A16: Z9 is Chain of W
    proof
      let p,q;
      assume p in Z9;
      then consider X1 such that
A17:  p in X1 and
A18:  X1 in Z by TARSKI:def 4;
      assume q in Z9;
      then consider X2 such that
A19:  q in X2 and
A20:  X2 in Z by TARSKI:def 4;
  X1,X2 are_c=-comparable by A13,A18,A20,ORDINAL1:def 8;
then A21:  X1 c= X2 or X2 c= X1 by XBOOLE_0:def 9;
A22:  X1 is Chain of W by A1,A12,A18;
  X2 is Chain of W by A1,A12,A20;
      hence thesis by A17,A19,A21,A22,Def3;
    end;
A23: now
      let p;
      assume p in union Z;
      then consider X1 such that
A24:  p in X1 & X1 in Z by TARSKI:def 4;
        ProperPrefixes p c= X1 & X1 c= union Z by A1,A12,A24,ZFMISC_1:74;
      hence ProperPrefixes p c= union Z by XBOOLE_1:1;
    end;
    set x = the Element of Z;
 x in X by A11,A12,TARSKI:def 3;
then A25: C c= x by A1;
 x c= union Z by A11,ZFMISC_1:74;
then  C c= union Z by A25,XBOOLE_1:1;
    hence union Z in X by A1,A16,A23;
  end;
  then consider Y such that
A26: Y in X and
A27: for Z st Z in X & Z <> Y holds not Y c= Z by A10,ORDERS_1:67;
  reconsider Y as Chain of W by A1,A26;
 now
    thus for p st p in Y holds ProperPrefixes p c= Y by A1,A26;
    given p such that
A28: p in W and
A29: for q st q in Y holds q is_a_proper_prefix_of p;
    set Z = (ProperPrefixes p) \/ {p};
     ProperPrefixes p c= W & {p} c= W by A28,TREES_1:def 3,ZFMISC_1:31;
    then reconsider Z9 = Z as Subset of W by XBOOLE_1:8;
A30: Z9 is Chain of W
    proof
      let q,r;
      assume that
A31:  q in Z9 and
A32:  r in Z9;
A33:  q in ProperPrefixes p or q in {p} by A31,XBOOLE_0:def 3;
A34:  r in ProperPrefixes p or r in {p} by A32,XBOOLE_0:def 3;
A35:  q is_a_proper_prefix_of p or q = p by A33,TARSKI:def 1,TREES_1:12;
A36:  r is_a_proper_prefix_of p or r = p by A34,TARSKI:def 1,TREES_1:12;
A37:  q is_a_prefix_of p by A35,XBOOLE_0:def 8;
  r is_a_prefix_of p by A36,XBOOLE_0:def 8;
      hence thesis by A37,Th1;
    end;
A38: now
      let q;
      assume q in Z;
then   q in ProperPrefixes p or q in {p} by XBOOLE_0:def 3;
then   q is_a_proper_prefix_of p or q = p by TARSKI:def 1,TREES_1:12;
then   q is_a_prefix_of p by XBOOLE_0:def 8;
then A39:  ProperPrefixes q c= ProperPrefixes p by TREES_1:17;
  ProperPrefixes p c= Z by XBOOLE_1:7;
      hence ProperPrefixes q c= Z by A39,XBOOLE_1:1;
    end;
A40: Y c= Z
    proof
      let x;
      assume
A41:  x in Y;
      then reconsider t = x as Element of W;
  t is_a_proper_prefix_of p by A29,A41;
then   t in ProperPrefixes p by TREES_1:12;
      hence thesis by XBOOLE_0:def 3;
    end;
 C c= Y by A1,A26;
then  C c= Z by A40,XBOOLE_1:1;
then A42: Z in X by A1,A30,A38;
A43: p in {p} by TARSKI:def 1;
A44: not p in Y by A29;
 p in Z by A43,XBOOLE_0:def 3;
    hence contradiction by A27,A40,A42,A44;
  end;
  then reconsider Y as Branch of W by Def7;
  take Y;
  thus thesis by A1,A26;
end;

scheme FuncExOfMinNat { P[set,Nat], X()->set }:
  ex f st dom f = X() & for x st x in X() ex n st f.x = n & P[x,n] &
  for m st P[x,m] holds n <= m
provided
A1: for x st x in X() ex n st P[x,n]
proof
  defpred Q[set,set] means
  ex n st $2 = n & P[$1,n] & for m st P[$1,m] holds n <= m;
A2: for x st x in X() ex y st Q[x,y]
  proof
    let x;
    defpred X[Nat] means P[x,$1];
    assume x in X();
then  ex n st X[n] by A1;
then A3: ex n be Nat st X[n];
    consider n be Nat such that
A4: X[n] & for m be Nat st X[m] holds n <= m from NAT_1:sch 5(A3);
    take n;
 n in NAT & for m st X[m] holds n <= m by A4,ORDINAL1:def 12;
    hence thesis by A4;
  end;
  thus ex f st dom f = X() & for x st x in X() holds Q[x,f.x]
  from CLASSES1:sch 1(A2);
end;

Lm1: X is finite implies ex n st for k st k in X holds k <= n
proof
  assume
A1: X is finite;
  defpred P[set,set] means
  ex k1,k2 being Element of NAT st $1 = k1 & $2 = k2 & k1 >= k2;
 now per cases;
    suppose
A2:   X /\ NAT = {};
      thus thesis
      proof
        take 0;
        let k;
        assume k in X;
        hence thesis by A2,XBOOLE_0:def 4;
      end;
    end;
    suppose
A3:   X /\ NAT <> {};
      reconsider XN = X /\ NAT as finite set by A1;
A4:   XN <> {} by A3;
A5:   for x,y holds P[x,y] & P[y,x] implies x = y by XXREAL_0:1;
A6:   for x,y,z st P[x,y] & P[y,z] holds P[x,z] by XXREAL_0:2;
      consider x such that
A7:   x in XN & for y st y in XN & y <> x holds not P[y,x]
        from CARD_2:sch 1(A4,A5,A6);
      reconsider n = x as Element of NAT by A7,XBOOLE_0:def 4;
      take n;
      let k;
      assume k in X;
then   k in X /\ NAT by XBOOLE_0:def 4;
      hence k <= n by A7;
    end;
  end;
  hence thesis;
end;

scheme InfiniteChain { X()->set, a()->set, Q[set], P[set,set] }:
  ex f st dom f = NAT & rng f c= X() & f.0 = a() &
  for k holds P[f.k,f.(k+1)] & Q[f.k]
provided
A1: a() in X() & Q[a()] and
A2: for x st x in X() & Q[x] ex y st y in X() & P[x,y] & Q[y]
proof
  consider Y such that
A3: x in Y iff x in X() & Q[x] from XBOOLE_0:sch 1;
  defpred Q[set,set] means $2 in Y & P[$1,$2];
A4: for x st x in Y ex y st Q[x,y]
  proof
    let x;
    assume x in Y;
then  x in X() & Q[x] by A3;
    then consider y such that
A5: y in X() & P[x,y] & Q[y] by A2;
    take y;
    thus thesis by A3,A5;
  end;
  consider g be Function such that
A6: dom g = Y & for x st x in Y holds Q[x,g.x] from CLASSES1:sch 1(A4);
  deffunc U(set,set) = g.$2;
  consider f such that
A7: dom f = NAT & f.0 = a() & for n being Nat holds f.(n+1) = U(n,f.n)
  from NAT_1:sch 11;
  take f;
  thus dom f = NAT by A7;
  defpred X[Element of NAT] means f.$1 in Y;
A8: X[0] by A1,A3,A7;
A9: X[k] implies X[k+1]
  proof
    assume f.k in Y;
then  g.(f.k) in Y by A6;
    hence thesis by A7;
  end;
A10: X[k] from NAT_1:sch 1(A8,A9);
  thus rng f c= X()
  proof
    let x;
    assume x in rng f;
    then consider y such that
A11: y in dom f and
A12: x = f.y by FUNCT_1:def 3;
    reconsider y as Element of NAT by A7,A11;
 f.y in Y by A10;
    hence thesis by A3,A12;
  end;
  thus f.0 = a() by A7;
  let k;
A13: f.k in Y by A10;
then  P[f.k,g.(f.k)] by A6;
  hence thesis by A3,A7,A13;
end;

theorem Th29:
  for T being Tree st (for n ex C being finite Chain of T st card C = n) &
  for t being Element of T holds succ t is finite
  ex B being Chain of T st not B is finite
proof
  let T be Tree;
  assume that
A1: for n ex X being finite Chain of T st card X = n and
A2: for t being Element of T holds succ t is finite;
  defpred P[FinSequence] means for n ex B being Branch of T st
  $1 in B & ex p st p in B & len p >= len $1 + n;
A3: for x being Element of T st P[x] ex n st x^<*n*> in T & P[x^<*n*>]
  proof
    let x be Element of T such that
A4: P[x] and
A5: for n st x^<*n*> in T holds not P[x^<*n*>];
A6: succ x is finite by A2;
    defpred X[set,Element of NAT] means for B being Branch of T,p,q st
    p = $1 & $1 in B & q in B holds len p + $2 > len q;
A7: for y st y in succ x ex n st X[y,n]
    proof
      let y;
      assume y in succ x;
      then consider k such that
A8:   y = x^<*k*> and
A9:  x^<*k*> in T;
      consider n such that
A10:  for B being Branch of T st x^<*k*> in B for p st p in B holds
      len p < len (x^<*k*>) + n by A5,A9;
      take n;
      thus thesis by A8,A10;
    end;
    consider f such that
A11: dom f = succ x and
A12: for y st y in succ x ex n st f.y = n & X[y,n] &
    for m st X[y,m] holds n <= m from FuncExOfMinNat(A7);
    consider k such that
A13: for m st m in rng f holds m <= k by A6,A11,Lm1,FINSET_1:8;
    consider B being Branch of T such that
A14: x in B and
A15: ex p st p in B & len p >= len x + (k+1) by A4;
    consider p such that
A16: p in B and
A17: len p >= len x + (k+1) by A15;
    reconsider r = p|Seg(len x + 1) as FinSequence of NAT by FINSEQ_1:18;
 len x + 1 <= len x + 1 + k by NAT_1:11;
then A18: len p >= len x + 1 by A17,XXREAL_0:2;
A19: r is_a_prefix_of p by TREES_1:def 1;
A20: len r = len x + 1 by A18,FINSEQ_1:17;
A21: r in B by A16,A19,Th25;
then  x is_a_prefix_of r by A14,A20,Th27,NAT_1:11;
    then consider j being FinSequence such that
A22: r = x^j by TREES_1:1;
 len x + len j = len r by A22,FINSEQ_1:22;
then A23: j = <*j.1*> by A20,FINSEQ_1:40;
A24: dom r = Seg len r & 1 <= 1+len x by FINSEQ_1:def 3,NAT_1:11;
 len x+1 <= len r by A18,FINSEQ_1:17;
    then  (
x^<*j.1*>).(len x+1) = j.1 & len x + 1 in dom r by A24,FINSEQ_1:1,42;
then  j.1 in rng r by A22,A23,FUNCT_1:def 3;
    then reconsider l = j.1 as Element of NAT;
A25: x^<*l*> in succ x by A21,A22,A23;
    then consider n such that
A26: f.(x^<*l*>) = n and
A27: for B being Branch of T,p,q st p = x^<*l*> & x^<*l*> in B & q in B
    holds len p + n > len q
    and for m st for B being Branch of T,p,q st p = x^<*l*> & x^<*l*> in B
    & q in B holds len p + m > len q holds n <= m
    by A12;
 n in rng f by A11,A25,A26,FUNCT_1:def 3;
then  n <= k by A13;
then  len r + n <= len x + 1 + k by A20,XREAL_1:7;
    hence contradiction by A16,A17,A21,A22,A23,A27,XXREAL_0:2;
  end;
  reconsider e = {} as Element of T by TREES_1:22;
A28: P[e]
  proof
    given n such that
A29: for B being Branch of T st e in B for p st p in B holds len p < len e + n;
    consider C being finite Chain of T such that
A30: card C = n+1 by A1;
    consider B being Branch of T such that
A31: C c= B by Th28;
 n+0 < n+1 by XREAL_1:6;
then A32: ex p st p in C & len p >= n by A30,Th23;
 e in B & len e = 0 by Th26;
    hence contradiction by A29,A31,A32;
  end;
  defpred QQ[set] means ex t being Element of T st t = $1 & P[t];
  defpred PP[set,set] means ex p,n st $1 = p & p^<*n*> in T & $2 = p^<*n*>;
A33: e in T & QQ[e] by A28;
A34: for x st x in T & QQ[x] ex y st y in T & PP[x,y] & QQ[y]
  proof
    let x such that x in T;
    given t being Element of T such that
A35: t = x and
A36: P[t];
    consider n such that
A37: t^<*n*> in T and
A38: P[t^<*n*>] by A3,A36;
    reconsider y = t^<*n*> as set;
    take y;
    thus y in T & PP[x,y] by A35,A37;
    reconsider t1 = t^<*n*> as Element of T by A37;
    take t1;
    thus thesis by A38;
  end;
 ex f st dom f = NAT & rng f c= T & f.0 = e &
  for k holds PP[f.k,f.(k+1)] & QQ[f.k] from InfiniteChain(A33,A34);
  then consider f such that
A39: dom f = NAT and
A40: rng f c= T and
A41: f.0 = e and
A42: for k holds (ex p,n st f.k = p & p^<*n*> in T & f.(k+1) = p^<*n*>) &
  ex t being Element of T st t = f.k & P[t];
  reconsider C = rng f as Subset of T by A40;
A43: now
    let k be Nat;
    defpred X[Nat] means
    for p,q st p = f.k & q = f.(k+$1) holds p is_a_prefix_of q;
A44: X[0];
A45: for n be Nat st X[n] holds X[n+1]
    proof
      let n be Nat;
      assume
A46:  for p,q st p = f.k & q = f.(k+n) holds p is_a_prefix_of q;
      let p,q such that
A47:  p = f.k and
A48:  q = f.(k+(n+1));
      reconsider k,n as Element of NAT by ORDINAL1:def 12;
      consider r,l such that
A49:  f.(k+n) = r and r^<*l*> in T and
A50:  f.((k+n)+1) = r^<*l*> by A42;
A51:  p is_a_prefix_of r by A46,A47,A49;
  r is_a_prefix_of r^<*l*> by TREES_1:1;
      hence thesis by A48,A50,A51,XBOOLE_1:1;
    end;
    thus for n be Nat holds X[n] from NAT_1:sch 2(A44,A45);
  end;
A52: now
    let k,l;
    assume k <= l;
then  ex n be Nat st l = k+n by NAT_1:10;
    hence for p,q st p = f.k & q = f.l holds p is_a_prefix_of q by A43;
  end;
 C is Chain of T
  proof
    let p,q;
    assume that
A53: p in C and
A54: q in C;
    consider x such that
A55: x in NAT and
A56: p = f.x by A39,A53,FUNCT_1:def 3;
    consider y such that
A57: y in NAT and
A58: q = f.y by A39,A54,FUNCT_1:def 3;
    reconsider x, y as Element of NAT by A55,A57;
 x <= y or y <= x;
    hence p is_a_prefix_of q or q is_a_prefix_of p by A52,A56,A58;
  end;
  then reconsider C as Chain of T;
  take C;
  defpred X[Element of NAT] means for p st p = f.$1 holds len p = $1;
A59: X[0] by A41;
A60: X[k] implies X[k+1]
  proof
    assume
A61: for p st p = f.k holds len p = k;
    let p such that
A62: p = f.(k+1);
    consider q,n such that
A63: f.k = q and q^<*n*> in T and
A64: f.(k+1) = q^<*n*> by A42;
    thus len p = len q + len <*n*> by A62,A64,FINSEQ_1:22
      .= len q + 1 by FINSEQ_1:39
      .= k+1 by A61,A63;
  end;
A65: X[k] from NAT_1:sch 1(A59,A60);
 f is one-to-one
  proof
    let x,y;
    assume x in dom f & y in dom f;
    then reconsider x9 = x, y9 = y as Element of NAT by A39;
    consider p,n such that
A66: f.x9 = p and p^<*n*> in T
    and f.(x9+1) = p^<*n*> by A42;
A67: ex q,k st f.y9 = q & q^<*k*> in T & f.(y9+1) = q^<*k*> by A42;
 len p = x9 by A65,A66;
    hence thesis by A65,A66,A67;
  end;
then  NAT,C are_equipotent by A39,WELLORD2:def 4;
  hence thesis by CARD_1:38;
end;


theorem
  for T being finite-order Tree st
  for n ex C being finite Chain of T st card C = n
  ex B being Chain of T st not B is finite
proof
  let T be finite-order Tree;
 for t being Element of T holds succ t is finite;
  hence thesis by Th29;
end;

definition
  let IT be Relation;
  attr IT is DecoratedTree-like means
  :Def8:
  dom IT is Tree;
end;

registration
  cluster DecoratedTree-like for Function;
  existence
  proof set W = the Tree;
    take f = W --> 0;
    thus dom f is Tree by FUNCOP_1:13;
  end;
end;

definition
  mode DecoratedTree is DecoratedTree-like Function;
end;

reserve T,T1,T2 for DecoratedTree;

registration
  let T;
  cluster dom T -> non empty Tree-like;
  coherence by Def8;
end;

registration
  let D;
  cluster DecoratedTree-like D-valued for Function;
  existence
  proof set W = the Tree;
    set d = the Element of D;
    set f = W --> d;
 dom f = W by FUNCOP_1:13;
    then reconsider f as DecoratedTree by Def8;
 f is D-valued;
    hence thesis;
  end;
end;

definition
  let D;
  mode DecoratedTree of D is D-valued DecoratedTree;
end;

definition
  let D be non empty set, T be DecoratedTree of D, t be Element of dom T;
  redefine func T.t -> Element of D;
  coherence
  proof
 T.t in rng T & rng T c= D by FUNCT_1:def 3;
    hence thesis;
  end;
end;

theorem Th31:
  dom T1 = dom T2 & (for p st p in dom T1 holds T1.p = T2.p) implies T1 = T2
proof
  assume that
A1: dom T1 = dom T2 and
A2: for p st p in dom T1 holds T1.p = T2.p;
 now
    let x;
    assume x in dom T1;
    then reconsider p = x as Element of dom T1;
 T1.p = T2.p by A2;
    hence T1.x = T2.x;
  end;
  hence thesis by A1,FUNCT_1:2;
end;

scheme DTreeEx { T() -> Tree, P[set,set] }:
  ex T st dom T = T() & for p st p in T() holds P[p,T.p]
provided
A1: for p st p in T() ex x st P[p,x]
proof
A2: for x st x in T() ex y st P[x,y]
  proof
    let x;
    assume x in T();
    then reconsider p = x as Element of T();
 ex y st P[p,y] by A1;
    hence thesis;
  end;
  consider f such that
A3: dom f = T() & for x st x in T() holds P[x,f.x] from CLASSES1:sch 1(A2);
  reconsider T = f as DecoratedTree by A3,Def8;
  take T;
  thus thesis by A3;
end;

scheme DTreeLambda { T() -> Tree, f(set) -> set }:
  ex T st dom T = T() & for p st p in T() holds T.p = f(p)
proof
  consider f such that
A1: dom f = T() & for x st x in T() holds f.x = f(x) from FUNCT_1:sch 3;
  reconsider T = f as DecoratedTree by A1,Def8;
  take T;
  thus thesis by A1;
end;

definition

  let T;
  func Leaves T -> set equals
  T.:Leaves dom T;
  correctness;
  let p;
  func T|p -> DecoratedTree means
  :
  Def10: dom it = (dom T)|p & for q st q in (dom T)|p holds it.q = T.(p^q);
  existence
  proof
    deffunc U(FinSequence) = T.(p^$1);
    thus ex t being DecoratedTree st
    dom t = (dom T)|p & for q st q in (dom T)|p holds t.q = U(q)
    from DTreeLambda;
  end;
  uniqueness
  proof
    let T1,T2 such that
A1: dom T1 = (dom T)|p and
A2: for q st q in (dom T)|p holds T1.q = T.(p^q) and
A3: dom T2 = (dom T)|p and
A4: for q st q in (dom T)|p holds T2.q = T.(p^q);
 now
      let q;
      assume
A5:   q in (dom T)|p;
then    T1.q = T.(p^q) by A2;
      hence T1.q = T2.q by A4,A5;
    end;
    hence thesis by A1,A3,Th31;
  end;
end;

theorem Th32:
  p in dom T implies rng (T|p) c= rng T
proof
  assume
A1: p in dom T;
  let x;
  assume x in rng (T|p);
  then consider y such that
A2: y in dom (T|p) and
A3: x = (T|p).y by FUNCT_1:def 3;
  reconsider y as Element of dom (T|p) by A2;
A4: dom (T|p) = (dom T)|p by Def10;
then A5: p^y in dom T by A1,TREES_1:def 6;
 x = T.(p^y) by A3,A4,Def10;
  hence thesis by A5,FUNCT_1:def 3;
end;

definition
  let D;
  let T be DecoratedTree of D;
  redefine func Leaves T -> Subset of D;
  coherence
  proof
 T.:Leaves dom T c= rng T & rng T c= D by RELAT_1:111;
    hence thesis by XBOOLE_1:1;
  end;
end;

registration
  let D;
  let T be DecoratedTree of D;
  let p be Element of dom T;
  cluster T|p -> D-valued;
  coherence
  proof
 rng (T|p) c= rng T & rng T c= D by Th32;
then  rng (T|p) c= D by XBOOLE_1:1;
    hence thesis by RELAT_1:def 19;
  end;
end;

definition
  let T,p,T1;
  assume
A1: p in dom T;
  func T with-replacement (p,T1) -> DecoratedTree means
  dom it = dom T with-replacement (p,dom T1) &
  for q st q in dom T with-replacement (p,dom T1) holds
  not p is_a_prefix_of q & it.q = T.q or
  ex r st r in dom T1 & q = p^r & it.q = T1.r;
  existence
  proof
    defpred X[FinSequence,set] means not p is_a_prefix_of $1 & $2 = T.$1 or
    ex r st r in dom T1 & $1 = p^r & $2 = T1.r;
A2: for q st q in dom T with-replacement (p,dom T1) ex x st X[q,x]
    proof
      let q such that
A3:   q in dom T with-replacement (p,dom T1);
   now per cases by XBOOLE_0:def 8;
        suppose
       p is_a_proper_prefix_of q;
          then consider r such that
A4:       r in dom T1 & q = p^r by A1,A3,TREES_1:def 9;
          thus thesis
          proof
            take T1.r;
            thus thesis by A4;
          end;
        end;
        suppose
A5:       p = q;
          thus thesis
          proof
            take T1.({} NAT);
         {} NAT in dom T1 & q = p^(<*> NAT) by A5,FINSEQ_1:34,TREES_1:22;
            hence thesis;
          end;
        end;
        suppose
       not p is_a_prefix_of q;
          hence thesis;
        end;
      end;
      hence thesis;
    end;
    thus ex TT being DecoratedTree st
    dom TT = dom T with-replacement (p,dom T1) &
    for q st q in dom T with-replacement (p,dom T1) holds X[q,TT.q]
    from DTreeEx(A2);
  end;
  uniqueness
  proof
    let D1,D2 be DecoratedTree such that
A6: dom D1 = dom T with-replacement (p,dom T1) and
A7: for q st q in dom T with-replacement (p,dom T1) holds
    not p is_a_prefix_of q & D1.q = T.q or
    ex r st r in dom T1 & q = p^r & D1.q = T1.r and
A8: dom D2 = dom T with-replacement (p,dom T1) and
A9: for q st q in dom T with-replacement (p,dom T1) holds
    not p is_a_prefix_of q & D2.q = T.q or
    ex r st r in dom T1 & q = p^r & D2.q = T1.r;
 now
      let q;
      assume that
A10:  q in dom D1 and
A11:  D1.q <> D2.q;
A12:  not p is_a_prefix_of q & D1.q = T.q or
      ex r st r in dom T1 & q = p^r & D1.q = T1.r by A6,A7,A10;
  not p is_a_prefix_of q & D2.q = T.q or
      ex r st r in dom T1 & q = p^r & D2.q = T1.r by A6,A9,A10;
      hence contradiction by A11,A12,FINSEQ_1:33,TREES_1:1;
    end;
    hence thesis by A6,A8,Th31;
  end;
end;

registration
  let W,x;
  cluster W --> x -> DecoratedTree-like;
  coherence
  proof
 dom (W --> x) = W by FUNCOP_1:13;
    hence thesis by Def8;
  end;
end;

theorem Th33:
  (for x st x in D holds x is Tree) implies union D is Tree
proof
  assume
A1: for x st x in D holds x is Tree;
  then reconsider x = the Element of D as Tree;

 x c= union D by ZFMISC_1:74;
  then reconsider T = union D as non empty set;
 T is Tree-like
  proof
 for X st X in D holds X c= NAT*
    proof
      let X;
      assume X in D;
then    X is Tree by A1;
      hence thesis by TREES_1:def 3;
    end;
    hence T c= NAT* by ZFMISC_1:76;
    thus for p st p in T holds ProperPrefixes p c= T
    proof
      let p;
      assume p in T;
      then consider X such that
A2:   p in X and
A3:   X in D by TARSKI:def 4;
      reconsider X as Tree by A1,A3;
  ProperPrefixes p c= X & X c= T by A2,A3,TREES_1:def 3,ZFMISC_1:74;
      hence thesis by XBOOLE_1:1;
    end;
    let p,k,n;
    assume that
A4: p^<*k*> in T and
A5: n <= k;
    consider X such that
A6: p^<*k*> in X and
A7: X in D by A4,TARSKI:def 4;
    reconsider X as Tree by A1,A7;
 p^<*n*> in X by A5,A6,TREES_1:def 3;
    hence thesis by A7,TARSKI:def 4;
  end;
  hence thesis;
end;

theorem Th34:
  (for x st x in X holds x is Function) & X is c=-linear implies
  union X is Relation-like Function-like
proof
  assume that
A1: for x st x in X holds x is Function and
A2: X is c=-linear;
  thus a in union X implies ex b,c st [b,c] = a
  proof
    assume a in union X;
    then consider x be set such that
A3: a in x and
A4: x in X by TARSKI:def 4;
    reconsider x as Function by A1,A4;
 x = x;
    hence thesis by A3,RELAT_1:def 1;
  end;
  let a,b,c;
  assume that
A5: [a,b] in union X and
A6: [a,c] in union X;
  consider x be set such that
A7: [a,b] in x and
A8: x in X by A5,TARSKI:def 4;
  consider y be set such that
A9: [a,c] in y and
A10: y in X by A6,TARSKI:def 4;
  reconsider x as Function by A1,A8;
  reconsider y as Function by A1,A10;
 x,y are_c=-comparable by A2,A8,A10,ORDINAL1:def 8;
then  x c= y or y c= x by XBOOLE_0:def 9;
  hence thesis by A7,A9,FUNCT_1:def 1;
end;

theorem Th35:
  (for x st x in D holds x is DecoratedTree) & D is c=-linear
  implies union D is DecoratedTree
proof
  assume that
A1: for x st x in D holds x is DecoratedTree and
A2: D is c=-linear;
 for x holds x in D implies x is Function by A1;
  then reconsider T = union D as Function by A2,Th34;
  defpred X[set,set] means ex T1 st $1 = T1 & dom T1 = $2;
A3: for x st x in D ex y st X[x,y]
  proof
    let x;
    assume x in D;
    then reconsider T1 = x as DecoratedTree by A1;
 dom T1 = dom T1;
    hence thesis;
  end;
  consider f such that
A4: dom f = D & for x st x in D holds X[x,f.x] from CLASSES1:sch 1(A3);
  reconsider E = rng f as non empty set by A4,RELAT_1:42;
 now
    let x;
    assume x in E;
    then consider y such that
A5: y in dom f & x = f.y by FUNCT_1:def 3;
 ex T1 st y = T1 & dom T1 = x by A4,A5;
    hence x is Tree;
  end;
then A6: union E is Tree by Th33;
 dom T = union E
  proof
    thus dom T c= union E
    proof
      let x;
      assume x in dom T;
      then consider y such that
A7:  [x,y] in T by XTUPLE_0:def 12;
      consider X such that
A8:  [x,y] in X and
A9:  X in D by A7,TARSKI:def 4;
      consider T1 such that
A10:  X = T1 and
A11:  dom T1 = f.X by A4,A9;
A12:  dom T1 in rng f by A4,A9,A11,FUNCT_1:def 3;
A13:  x in dom T1 by A8,A10,XTUPLE_0:def 12;
  dom T1 c= union E by A12,ZFMISC_1:74;
      hence thesis by A13;
    end;
    let x;
    assume x in union E;
    then consider X such that
A14: x in X and
A15: X in E by TARSKI:def 4;
    consider y such that
A16: y in dom f and
A17: X = f.y by A15,FUNCT_1:def 3;
    consider T1 such that
A18: y = T1 and
A19: dom T1 = X by A4,A16,A17;
 [x,T1.x] in T1 by A14,A19,FUNCT_1:1;
then  [x,T1.x] in union D by A4,A16,A18,TARSKI:def 4;
    hence thesis by XTUPLE_0:def 12;
  end;
  hence thesis by A6,Def8;
end;

theorem Th36:
  (for x st x in D9 holds x is DecoratedTree of D) & D9 is c=-linear implies
  union D9 is DecoratedTree of D
proof
  assume that
A1: for x st x in D9 holds x is DecoratedTree of D and
A2: D9 is c=-linear;
 for x st x in D9 holds x is DecoratedTree by A1;
  then reconsider T = union D9 as DecoratedTree by A2,Th35;
 rng T c= D
  proof
    let x;
    assume x in rng T;
    then consider y such that
A3: y in dom T & x = T.y by FUNCT_1:def 3;
 [y,x] in T by A3,FUNCT_1:1;
    then consider X such that
A4: [y,x] in X and
A5: X in D9 by TARSKI:def 4;
    reconsider X as DecoratedTree of D by A1,A5;
 y in dom X & x = X.y by A4,FUNCT_1:1;
then A6: x in rng X by FUNCT_1:def 3;
    thus thesis by A6;
  end;
  hence thesis by RELAT_1:def 19;
end;

scheme DTreeStructEx
  { D() -> non empty set, d() -> Element of D(), F(set) -> set,
  S() -> Function of [:D(),NAT:],D()}:
  ex T being DecoratedTree of D() st T.{} = d() &
  for t being Element of dom T holds succ t = { t^<*k*>: k in F(T.t)} &
  for n st n in F(T.t) holds T.(t^<*n*>) = S().(T.t,n)
provided
A1: for d being Element of D(), k1,k2 st k1 <= k2 & k2 in F(d) holds k1 in F(d)
proof
  defpred P[Element of NAT] means
  ex T being DecoratedTree of D() st T.{} = d() &
  for t being Element of dom T holds len t <= $1 &
  (len t < $1 implies succ t = { t^<*k*>: k in F(T.t)} &
  for n,x st x = T.t & n in F(x) holds T.(t^<*n*>) = S().(x,n));
A2: P[0] proof reconsider W = {{}} as Tree;
    take T = W --> d();
 {} in W by TREES_1:22;
    hence T.{} = d() by FUNCOP_1:7;
    let t be Element of dom T;
 dom T = W by FUNCOP_1:13;
then  t = {} by TARSKI:def 1;
    hence len t <= 0;
    assume len t < 0;
    hence thesis;
  end;
A3: P[k] implies P[k+1]
  proof
    given T be DecoratedTree of D() such that
A4: T.{} = d() & for t being Element of dom T holds len t <= k &
    (len t < k implies succ t = { t^<*m*>: m in F(T.t)} &
    for n,x st x = T.t & n in F(x) holds T.(t^<*n*>) = S().(x,n));
    reconsider M
    = { t^<*n*> where t is Element of dom T: n in F(T.t) } \/ dom T
    as non empty set;
 M is Tree-like
    proof
      thus M c= NAT*
      proof
        let x;
        assume x in M;
then A5:    x in { t^<*n*> where t is Element of dom T: n in F(T.t) } or
        x in dom T & dom T c= NAT* by TREES_1:def 3,XBOOLE_0:def 3;
        assume
A6:    not x in NAT*;
then     ex n be Element of NAT,t being Element of dom T
        st x = t^<*n*> & n in F (T.t) by A5;
        hence thesis by A6,FINSEQ_1:def 11;
      end;
      thus for p st p in M holds ProperPrefixes p c= M
      proof
        let p;
        assume p in M;
then A7:    p in { t^<*n*> where t is Element of dom T: n in F(T.t) } or
        p in dom T by XBOOLE_0:def 3;
    now
          assume p in { t^<*n*> where t is Element of dom T: n in F(T.t) };
          then consider n be Element of NAT, t be Element of dom T such that
A8:      p = t^<*n*> and n in F(T.t);
A9:      ProperPrefixes t c= dom T by TREES_1:def 3;
A10:      dom T c= M by XBOOLE_1:7;
A11:      t in dom T;
A12:      ProperPrefixes t c= M by A9,A10,XBOOLE_1:1;
A13:      {t} c= M by A10,A11,ZFMISC_1:31;
      ProperPrefixes p = ProperPrefixes t \/ {t} by A8,Th4;
          hence thesis by A12,A13,XBOOLE_1:8;
        end;
then     ProperPrefixes p c= M or ProperPrefixes p c= dom T & dom T c= M
        by A7,TREES_1:def 3,XBOOLE_1:7;
        hence thesis by XBOOLE_1:1;
      end;
      let p,m,n;
      assume p^<*m*> in M;
then A14:  p^<*m*> in { t^<*l*> where t is Element of dom T: l in F(T.t) } or
      p^<*m*> in dom T by XBOOLE_0:def 3;
      assume that
A15:  n <= m and
A16:  not p^<*n*> in M;
  not p^<*n*> in dom T by A16,XBOOLE_0:def 3;
      then consider l be Element of NAT, t be Element of dom T such that
A17:  p^<*m*> = t^<*l*> and
A18:  l in F(T.t) by A14,A15,TREES_1:def 3;
      A19:  len (p^<*m*>) = len p + len <*m*> & len <*m*> = 1 by FINSEQ_1:22,40
      ;
      A20:  len (t^<*l*>) = len t + len <*l*> & len <*l*> = 1 by FINSEQ_1:22,40
      ;
      A21:  (
p^<*m*>).(len p + 1) = m & (t^<*l*>).(len t + 1) = l by FINSEQ_1:42;
then A22:  p = t by A17,A19,A20,FINSEQ_1:33;
  n in F(T.t) by A1,A15,A17,A18,A19,A20,A21;
then   p^<*n*> in { s^<*i*> where s is Element of dom T: i in F(T.s) }
      by A22;
      hence thesis by A16,XBOOLE_0:def 3;
    end;
    then reconsider M as Tree;
    defpred P[FinSequence,set] means $1 in dom T & $2 = T.$1 or
    not $1 in dom T & ex n,q st $1 = q^<*n*> & $2 = S().(T.q,n);
A23: for p st p in M ex x st P[p,x]
    proof
      let p;
      assume p in M;
then A24:  p in { t^<*l*> where t is Element of dom T: l in F(T.t) } or
      p in dom T by XBOOLE_0:def 3;
  now
        assume
A25:    not p in dom T;
        then consider l be Element of NAT, t be Element of dom T such that
A26:    p = t^<*l*> and l in F(T.t) by A24;
        take x = S().(T.t,l);
        thus p in dom T & x = T.p or
        not p in dom T & ex n,q st p = q^<*n*> & x = S().(T.q,n) by A25,A26;
      end;
      hence thesis;
    end;
    consider T9 be DecoratedTree such that
A27: dom T9 = M & for p st p in M holds P[p,T9.p] from DTreeEx(A23);
 rng T9 c= D()
    proof
      let x;
      assume x in rng T9;
      then consider y such that
A28:  y in dom T9 and
A29:  x = T9.y by FUNCT_1:def 3;
      reconsider y as Element of dom T9 by A28;
A30:  now
        assume y in dom T;
        then reconsider t = y as Element of dom T;
    T.t in D();
        hence thesis by A27,A29;
      end;
  now
        assume
A31:    not y in dom T;
        then consider n,q such that
A32:    y = q^<*n*> and
A33:    T9.y = S().(T.q,n) by A27;
    y in { t^<*l*> where t is Element of dom T: l in F(T.t) }
        by A27,A31,XBOOLE_0:def 3;
        then consider l be Element of NAT, t be Element of dom T such that
A34:    y = t^<*l*> and l in F(T.t);
A35:    len <*n*> = 1 by FINSEQ_1:39;
A36:    len <*l*> = 1 by FINSEQ_1:39;
A37:    len (q^<*n*>) = len q + 1 by A35,FINSEQ_1:22;
A38:    len (t^<*l*>) = len t + 1 by A36,FINSEQ_1:22;
            (
q^<*n*>).(len q + 1) = n & (t^<*l*>).(len t + 1) = l by FINSEQ_1:42;
then A39:    q = t by A32,A34,A37,A38,FINSEQ_1:33;
    T.t in D();
then     [T.q,n] in [:D(),NAT:] by A39,ZFMISC_1:87;
        hence thesis by A29,A33,FUNCT_2:5;
      end;
      hence thesis by A30;
    end;
    then reconsider T9 as DecoratedTree of D() by RELAT_1:def 19;
    take T9;
 <*> NAT in M & <*> NAT in dom T by TREES_1:22;
    hence T9. {} = d() by A4,A27;
    let t be Element of dom T9;
A40: now
      assume t in { s^<*l*> where s is Element of dom T: l in F(T.s) };
      then consider l be Element of NAT, s being Element of dom T such that
A41:  t = s^<*l*> and l in F(T.s);
  len s <= k by A4;
then   len <*l*> = 1 & len s + 1 <= k+1 by FINSEQ_1:39,XREAL_1:7;
      hence len t <= k+1 by A41,FINSEQ_1:22;
    end;
 now
      assume t in dom T;
      then reconsider s = t as Element of dom T;
  len s <= k & k <= k+1 by A4,NAT_1:11;
      hence len t <= k+1 by XXREAL_0:2;
    end;
    hence len t <= k+1 by A27,A40,XBOOLE_0:def 3;
    assume
A42: len t < k+1;
A43: now
      assume
A44:  not t in dom T;
then   t in { s^<*l*> where s is Element of dom T: l in F(T.s) }
      by A27,XBOOLE_0:def 3;
      then consider l be Element of NAT, s be Element of dom T such that
A45:  t = s^<*l*> and
A46:  l in F(T.s);
A47:  len t = len s + len <*l*> by A45,FINSEQ_1:22;
  len <*l*> = 1 & len t <= k by A42,FINSEQ_1:39,NAT_1:13;
then   len s < k by A47,NAT_1:13;
then   succ s = { s^<*m*>: m in F(T.s)} by A4;
then   t in succ s by A45,A46;
      hence contradiction by A44;
    end;
then A48: T9.t = T.t by A27;
    reconsider t9 = t as Element of dom T by A43;
    thus succ t c= { t^<*i*>: i in F(T9.t)}
    proof
      let x;
      assume x in succ t;
      then consider n such that
A49:  x = t^<*n*> and
A50:  t^<*n*> in dom T9;
  now per cases;
        suppose
A51:      t^<*n*> in dom T;
          then
reconsider s = t^<*n*>, t9 = t as Element of dom T by TREES_1:21;
      len s <= k & len s = len t + 1 by A4,FINSEQ_2:16;
then       len t < k by NAT_1:13;
then       succ t9 = { t9^<*m*>: m in F(T.t9) } by A4;
          hence thesis by A48,A49,A51;
        end;
        suppose
      not t^<*n*> in dom T;
then       t^<*n*> in { s^<*l*> where s is Element of dom T: l in F(T.s
          ) } by A27,A50,XBOOLE_0:def 3;
          then consider l be Element of NAT, s be Element of dom T such that
A52:      t^<*n*> = s^<*l*> and
A53:      l in F(T.s);
     t = s by A52,FINSEQ_2:17;
          hence thesis by A48,A49,A52,A53;
        end;
      end;
      hence thesis;
    end;
    thus
A54: { t^<*i*>: i in F(T9.t)} c= succ t
    proof
      let x;
      assume x in { t^<*i*>: i in F(T9.t)};
      then consider n such that
A55: x = t^<*n*> and
A56: n in F(T9.t);
 x = t9^<*n*> by A55;
then  x in { s^<*l*> where s is Element of dom T: l in F(T.s) }
      by A48,A56;
then  x in dom T9 by A27,XBOOLE_0:def 3;
      hence thesis by A55;
    end;
    let n,x;
    assume that
A57: x = T9.t and
A58: n in F(x);
 t^<*n*> in { t^<*i*>: i in F(T9.t)} by A57,A58;
then A59: t^<*n*> in succ t by A54;
 now per cases;
      suppose
A60:   t^<*n*> in dom T;
        then reconsider s = t^<*n*> as Element of dom T;
   len s <= k & len s = len t + 1 by A4,FINSEQ_2:16;
then    len t9 < k by NAT_1:13;
then    T.(t9^<*n*>) = S().(x,n) by A4,A48,A57,A58;
        hence thesis by A27,A59,A60;
      end;
      suppose
   not t^<*n*> in dom T;
        then consider l,q such that
A61:   t^<*n*> = q^<*l*> and
A62:   T9.(t^<*n*>) = S().(T.q,l) by A27,A59;
   t = q & n = l by A61,FINSEQ_2:17;
        hence thesis by A27,A43,A57,A62;
      end;
    end;
    hence thesis;
  end;
A63: P[k] from NAT_1:sch 1(A2,A3);
  defpred P[set,set] means ex T being DecoratedTree of D(), k st
  $1 = k & $2 = T & T.{} = d() &
  for t being Element of dom T holds len t <= k &
  (len t < k implies succ t = { t^<*i*>: i in F(T.t)} &
  for n,x st x = T.t & n in F(x) holds T.(t^<*n*>) = S().(x,n));
A64: for x st x in NAT ex y st P[x,y]
  proof
    let x;
    assume x in NAT;
    then reconsider n = x as Element of NAT;
    consider T being DecoratedTree of D() such that
A65: T.{} = d() & for t being Element of dom T holds len t <= n &
    (len t < n implies succ t = { t^<*k*>: k in F(T.t)} &
    for n,x st x = T.t & n in F(x) holds T.(t^<*n*>) = S().(x,n)) by A63;
    reconsider y = T as set;
    take y,T,n;
    thus thesis by A65;
  end;
  consider f such that
A66: dom f = NAT & for x st x in NAT holds P[x,f.x] from CLASSES1:sch 1(A64);
  reconsider E = rng f as non empty set by A66,RELAT_1:42;
A67: for x st x in E holds x is DecoratedTree of D()
  proof
    let x;
    assume x in E;
    then consider y such that
A68: y in dom f and
A69: x = f.y by FUNCT_1:def 3;
 ex T being DecoratedTree of D(), k st y = k & f.y = T & T.{} = d() &
    for t being Element of dom T holds len t <= k &
    (len t < k implies succ t = { t^<*i*>: i in F(T.t)} &
    for n,x st x = T.t & n in F(x) holds T.(t^<*n*>) = S().(x,n)) by A66,A68;
    hence thesis by A69;
  end;
A70: for T1,T2,k1,k2 st T1 = f.k1 & T2 = f.k2 & k1 <= k2 holds T1 c= T2
  proof
    let T1,T2;
    let x,y be Element of NAT such that
A71: T1 = f.x and
A72: T2 = f.y and
A73: x <= y;
    consider T19 being DecoratedTree of D(), k1 such that
A74: x = k1 & f.x = T19 & T19.{} = d() &
    for t being Element of dom T19 holds len t <= k1 &
    (len t < k1 implies succ t = { t^<*i*>: i in F(T19.t)} &
    for n,x st x = T19.t & n in F(x) holds T19.(t^<*n*>) = S().(x,n)) by A66;
    consider T29 being DecoratedTree of D(), k2 such that
A75: y = k2 & f.y = T29 & T29.{} = d() &
    for t being Element of dom T29 holds len t <= k2 &
    (len t < k2 implies succ t = { t^<*i*>: i in F(T29.t)} &
    for n,x st x = T29.t & n in F(x) holds T29.(t^<*n*>) = S().(x,n)) by A66;
    defpred I[Element of NAT] means
    for t being Element of dom T1 st len t <= $1 holds
    t in dom T2 & T1.t = T2.t;
A76: I[0]
    proof
      let t be Element of dom T1 such that
A77: len t <= 0;
 t = {} by A77;
      hence thesis by A71,A72,A74,A75,TREES_1:22;
    end;
A78: I[k] implies I[k+1]
    proof
      assume
A79: for t being Element of dom T1 st len t <= k holds
      t in dom T2 & T1.t = T2.t;
      let t be Element of dom T1;
      assume len t <= k+1;
then A80: len t <= k or len t = k+1 by NAT_1:8;
 now
        assume
A81:   len t = k+1;
        reconsider p = t|Seg k as FinSequence of NAT by FINSEQ_1:18;
   p is_a_prefix_of t by TREES_1:def 1;
        then reconsider p as Element of dom T1 by TREES_1:20;
A82:   k <= k+1 by NAT_1:11;
A83:   k+1 <= k1 by A71,A74,A81;
A84:   len p = k by A81,A82,FINSEQ_1:17;
A85:   k < k1 by A83,NAT_1:13;
A86:   T1.p = T2.p by A79,A84;
        reconsider p9 = p as Element of dom T2 by A79,A84;
   t <> {} by A81;
        then consider q being FinSequence, x being set such that
A87:   t = q^<*x*> by FINSEQ_1:46;
        A88:   p
 is_a_prefix_of t & q is_a_prefix_of t by A87,TREES_1:1,def 1;
   k+1 = len q + 1 by A81,A87,FINSEQ_2:16;
then A89:   p = q by A84,A88,Th1,TREES_1:4;
   <*x*> is FinSequence of NAT by A87,FINSEQ_1:36;
then A90:   rng <*x*> c= NAT by FINSEQ_1:def 4;
   rng <*x*> = {x} & x in {x} by FINSEQ_1:38,TARSKI:def 1;
        then reconsider x as Element of NAT by A90;
A91:   p^<*x*> in succ p by A87,A89;
   succ p = { p^<*i*>: i in F(T1.p)} by A71,A74,A84,A85;
        then consider i such that
A92:   p^<*x*> = p^<*i*> and
A93:   i in F(T1.p) by A91;
A94:   k < k2 by A73,A74,A75,A85,XXREAL_0:2;
then A95:   succ p9 = { p9^<*l*>: l in F(T2.p9) } by A72,A75,A84;
A96:   x = i by A92,FINSEQ_2:17;
A97:   t in succ p9 by A86,A87,A89,A92,A93,A95;
   T19.t = S().(T19.p,x) by A71,A74,A84,A85,A87,A89,A93,A96;
        hence
thesis by A71,A72,A74,A75,A84,A86,A87,A89,A93,A94,A96,A97;
      end;
      hence thesis by A79,A80;
    end;
A98: I[k] from NAT_1:sch 1(A76,A78);
    let x;
    assume
A99: x in T1;
    then consider y,z such that
A100: [y,z] = x by RELAT_1:def 1;
A101: T1.y = z by A99,A100,FUNCT_1:1;
    reconsider y as Element of dom T1 by A99,A100,FUNCT_1:1;
 len y <= len y;
then  y in dom T2 & T1.y = T2.y by A98;
    hence thesis by A100,A101,FUNCT_1:1;
  end;
 E is c=-linear
  proof
    let T1,T2 be set;
    assume
A102: T1 in E;
    then consider x such that
A103: x in dom f and
A104: T1 = f.x by FUNCT_1:def 3;
    assume
A105: T2 in E;
    then consider y such that
A106: y in dom f and
A107: T2 = f.y by FUNCT_1:def 3;
A108: T1 is DecoratedTree by A67,A102;
A109: T2 is DecoratedTree by A67,A105;
    reconsider x,y as Element of NAT by A66,A103,A106;
 x <= y or y <= x;
    hence T1 c= T2 or T2 c= T1 by A70,A104,A107,A108,A109;
  end;
  then reconsider T = union rng f as DecoratedTree of D() by A67,Th36;
  take T;
  consider T9 being DecoratedTree of D(), k such that
A110: 0 = k & f.0 = T9 & T9.{} = d() &
  for t being Element of dom T9 holds len t <= k &
  (len t < k implies succ t = { t^<*i*>: i in F(T9.t)} &
  for n,x st x = T9.t & n in F(x) holds T9.(t^<*n*>) = S().(x,n)) by A66;
 {} in dom T9 by TREES_1:22;
then A111: [{},d()] in T9 by A110,FUNCT_1:1;
 T9 in rng f by A66,A110,FUNCT_1:def 3;
then  [{},d()] in T by A111,TARSKI:def 4;
  hence T.{} = d() by FUNCT_1:1;
A112: for T1,x st T1 in E & x in dom T1 holds x in dom T & T1.x = T.x
  proof
    let T1,x;
    assume that
A113: T1 in E and
A114: x in dom T1;
 [x,T1.x] in T1 by A114,FUNCT_1:1;
then  [x,T1.x] in T by A113,TARSKI:def 4;
    hence thesis by FUNCT_1:1;
  end;
  let t be Element of dom T;
  thus succ t c= { t^<*i*>: i in F(T.t)}
  proof
    let x;
    assume x in succ t;
    then consider l such that
A115: x = t^<*l*> and
A116: t^<*l*> in dom T;
 [x,T.x] in T by A115,A116,FUNCT_1:1;
    then consider X such that
A117: [x,T.x] in X and
A118: X in rng f by TARSKI:def 4;
    consider y such that
A119: y in NAT and
A120: X = f.y by A66,A118,FUNCT_1:def 3;
    consider T1 being DecoratedTree of D(), k1 such that
A121: y = k1 & f.y = T1 & T1.{} = d() &
    for t being Element of dom T1 holds len t <= k1 &
    (len t < k1 implies succ t = { t^<*i*>: i in F(T1.t)} &
    for n,x st x = T1.t & n in F(x) holds T1.(t^<*n*>) = S().(x,n))
    by A66,A119;
A122: t^<*l*> in dom T1 by A115,A117,A120,A121,FUNCT_1:1;
    then
reconsider t9 = t, p = t^<*l*> as Element of dom T1 by TREES_1:21;
 len p <= k1 by A121;
then  len t + 1 <= k1 by FINSEQ_2:16;
then A123: len t9 < k1 by NAT_1:13;
then A124: succ t9 = { t9^<*i*>: i in F(T1.t9)} by A121;
 T1.t = T.t by A112,A118,A120,A121,A123;
    hence thesis by A115,A122,A124;
  end;
 [t,T.t] in T by FUNCT_1:1;
  then consider X such that
A125: [t,T.t] in X and
A126: X in E by TARSKI:def 4;
  consider y such that
A127: y in NAT and
A128: X = f.y by A66,A126,FUNCT_1:def 3;
  reconsider y as Element of NAT by A127;
  consider T1 being DecoratedTree of D(), k1 such that
A129: y = k1 & f.y = T1 & T1.{} = d() &
  for t being Element of dom T1 holds len t <= k1 &
  (len t < k1 implies succ t = { t^<*i*>: i in F(T1.t)} &
  for n,x st x = T1.t & n in F(x) holds T1.(t^<*n*>) = S().(x,n)) by A66;
  consider T2 being DecoratedTree of D(), k2 such that
A130: y+1 = k2 & f.(y+1) = T2 & T2.{} = d() &
  for t being Element of dom T2 holds len t <= k2 &
  (len t < k2 implies succ t = { t^<*i*>: i in F(T2.t)} &
  for n,x st x = T2.t & n in F(x) holds T2.(t^<*n*>) = S().(x,n)) by A66;
 y <= y+1 by NAT_1:11;
then A131: T1 c= T2 by A70,A129,A130;
  reconsider t1 = t as Element of dom T1 by A125,A128,A129,FUNCT_1:1;
A132: len t1 <= y by A129;
A133: T2.t = T.t by A125,A128,A129,A131,FUNCT_1:1;
  reconsider t2 = t as Element of dom T2 by A125,A128,A129,A131,FUNCT_1:1;
A134: len t2 < y+1 by A132,NAT_1:13;
then A135: succ t2 = { t2^<*i*>: i in F(T2.t2)} by A130;
  thus { t^<*i*>: i in F(T.t)} c= succ t
  proof
    let x;
    assume
A136: x in { t^<*i*>: i in F(T.t)};
then A137: ex l st x = t^<*l*> & l in F(T.t);
A138: x in succ t2 by A130,A133,A134,A136;
 T2 in E by A66,A130,FUNCT_1:def 3;
then  x in dom T by A112,A138;
    hence thesis by A137;
  end;
  let n;
  assume
A139: n in F(T.t);
then A140: t^<*n*> in succ t2 by A133,A135;
 T2 in E by A66,A130,FUNCT_1:def 3;
then  T2.(t^<*n*>) = T.(t^<*n*>) by A112,A140;
  hence thesis by A130,A133,A134,A139;
end;

scheme DTreeStructFinEx
  { D() -> non empty set, d() -> Element of D(), F(set) -> Element of NAT,
  S() -> Function of [:D(),NAT:],D()}:
  ex T being DecoratedTree of D() st T.{} = d() &
  for t being Element of dom T holds succ t = { t^<*k*>: k < F(T.t)} &
  for n st n < F(T.t) holds T.(t^<*n*>) = S().(T.t,n)
proof
  deffunc FF(Element of NAT) = { i: i < $1};
  deffunc U(set) = FF(F($1));
A1: for d being Element of D(), k1,k2 st k1 <= k2 & k2 in U(d) holds k1 in U(d)
  proof
    let d be Element of D(), k1,k2;
    assume
A2: k1 <= k2 & k2 in { i: i < F(d)};
then  ex i st k2 = i & i < F(d);
then  k1 < F(d) by A2,XXREAL_0:2;
    hence thesis;
  end;
  consider T being DecoratedTree of D() such that
A3: T.{} = d() &
  for t being Element of dom T holds succ t = { t^<*k*>: k in U(T.t)} &
  for n st n in U(T.t) holds T.(t^<*n*>) = S().(T.t,n) from DTreeStructEx(A1
  );
  take T;
  thus T.{} = d() by A3;
  let t be Element of dom T;
A4: succ t = { t^<*k*>: k in FF(F(T.t))} by A3;
  thus succ t c= { t^<*i*>: i < F(T.t)}
  proof
    let x;
    assume x in succ t;
    then consider l such that
A5: x = t^<*l*> and
A6: l in FF(F(T.t)) by A4;
 ex i st l = i & i < F(T.t) by A6;
    hence thesis by A5;
  end;
  thus { t^<*i*>: i < F(T.t)} c= succ t
  proof
    let x;
    assume x in { t^<*i*>: i < F(T.t)};
    then consider l such that
A7: x = t^<*l*> and
A8: l < F(T.t);
 l in FF(F(T.t)) by A8;
    hence thesis by A4,A7;
  end;
  let n;
  assume n < F(T.t);
then  n in FF(F(T.t));
  hence thesis by A3;
end;

begin 


registration
  let Tr be finite Tree, v be Element of Tr;
  cluster succ v -> finite;
  coherence;
end;

definition
  let Tr be finite Tree, v be Element of Tr;
  func branchdeg v equals
  card succ v;
  correctness;
end;

registration
  cluster finite for DecoratedTree;
  existence
  proof
    reconsider T = { {} } as Tree;
    take T --> {};
    thus thesis;
  end;
end;

registration
  let D be non empty set;
  cluster finite for DecoratedTree of D;
  existence
  proof set d = the Element of D;
    reconsider T = { {} } as Tree;
    take T --> d;
    thus thesis;
  end;
end;

registration
  let a,b be non empty set;
  cluster non empty for Relation of a,b;
  existence
  proof
 [:a,b:] c= [:a,b:];
    hence thesis;
  end;
end;



reserve x1,x2 for set,
  w for FinSequence of NAT;

theorem
  for Z1,Z2 being Tree,p being FinSequence of NAT st p in Z1 holds
  for v being Element of Z1 with-replacement (p,Z2),w being Element of Z2
  st v = p^w holds succ v,succ w are_equipotent
proof
  let Z1,Z2 be Tree,p be FinSequence of NAT such that
A1: p in Z1;
  set T = Z1 with-replacement (p,Z2);
  let t be Element of Z1 with-replacement (p,Z2), t2 be Element of Z2;
  assume
A2: t = p^t2;
then A3: p is_a_prefix_of t by TREES_1:1;
A4: for n holds t^<*n*> in T iff t2^<*n*> in Z2
  proof
    let n;
A5: p is_a_proper_prefix_of t^<*n*> by A3,TREES_1:8;
A6: t^<*n*> = p^(t2^<*n*>) by A2,FINSEQ_1:32;
    thus t^<*n*> in T implies t2^<*n*> in Z2
    proof
      assume t^<*n*> in T;
then    ex w st w in Z2 & t^<*n*> = p^w by A1,A5,TREES_1:def 9;
      hence thesis by A6,FINSEQ_1:33;
    end;
    assume t2^<*n*> in Z2;
    hence thesis by A1,A6,TREES_1:def 9;
  end;
  defpred P[set,set] means for n st $1 = t^<*n*> holds $2 = t2^<*n*>;
A7: for x st x in succ t ex y st P[x,y]
  proof
    let x;
    assume x in succ t;
    then consider n such that
A8: x = t^<*n*> and t^<*n*> in T;
    take t2^<*n*>;
    let m;
    assume x = t^<*m*>;
    hence thesis by A8,FINSEQ_1:33;
  end;
  consider f being Function such that
A9: dom f = succ t & for x st x in
  succ t holds P[x,f.x] from CLASSES1:sch 1(A7);
 now
    let x;
    thus x in rng f implies x in succ t2
    proof
      assume x in rng f;
      then consider y such that
A10:  y in dom f and
A11:  x = f.y by FUNCT_1:def 3;
      consider n such that
A12:  y = t^<*n*> and
A13:  t^<*n*> in T by A9,A10;
A14:  x = t2^<*n*> by A9,A10,A11,A12;
  t2^<*n*> in Z2 by A4,A13;
      hence thesis by A14;
    end;
    assume x in succ t2;
    then consider n such that
A15: x = t2^<*n*> and
A16: t2^<*n*> in Z2;
 t^<*n*> in T by A4,A16;
then A17: t^<*n*> in dom f by A9;
then  f.(t^<*n*>) = x by A9,A15;
    hence x in rng f by A17,FUNCT_1:def 3;
  end;
then A18: rng f = succ t2 by TARSKI:1;
 f is one-to-one
  proof
    let x1,x2;
    assume that
A19: x1 in dom f and
A20: x2 in dom f and
A21: f.x1 = f.x2;
    consider m such that
A22: x1 = t^<*m*> and t^<*m*> in T by A9,A19;
    consider k such that
A23: x2 = t^<*k*> and t^<*k*> in T by A9,A20;
 t2^<*m*> = f.x1 by A9,A19,A22
      .= t2^<*k*> by A9,A20,A21,A23;
    hence thesis by A22,A23,FINSEQ_1:33;
  end;
  hence thesis by A9,A18,WELLORD2:def 4;
end;

scheme DTreeStructEx
  { D() -> non empty set, d() -> Element of D(), Q[set,set],
  S() -> Function of [:D(),NAT:],D()}:
  ex T being DecoratedTree of D() st T.{} = d() &
  for t being Element of dom T holds succ t = { t^<*k*>: Q[k,T.t]} &
  for n st Q[n,T.t] holds T.(t^<*n*>) = S().(T.t,n)
provided
A1: for d being Element of D(), k1,k2 st k1 <= k2 & Q[k2,d] holds Q[k1,d]
proof
  defpred P[Element of NAT] means
  ex T being DecoratedTree of D() st T.{} = d() &
  for t being Element of dom T holds len t <= $1 &
  (len t < $1 implies succ t = { t^<*k*>: Q[k,T.t]} &
  for n,x st x = T.t & Q[n,x] holds T.(t^<*n*>) = S().(x,n));
A2: P[0] proof reconsider W = {{}} as Tree;
    take T = W --> d();
 {} in W by TREES_1:22;
    hence T.{} = d() by FUNCOP_1:7;
    let t be Element of dom T;
 dom T = W by FUNCOP_1:13;
then  t = {} by TARSKI:def 1;
    hence len t <= 0;
    assume len t < 0;
    hence thesis;
  end;
A3: P[k] implies P[k+1]
  proof
    given T be DecoratedTree of D() such that
A4: T.{} = d() & for t being Element of dom T holds len t <= k &
    (len t < k implies succ t = { t^<*m*>: Q[m,T.t]} &
    for n,x st x = T.t & Q[n,x] holds T.(t^<*n*>) = S().(x,n));
    reconsider M = { t^<*n*> where t is Element of dom T: Q[n,T.t] } \/ dom T
    as non empty set;
 M is Tree-like
    proof
      thus M c= NAT*
      proof
        let x;
        assume x in M;
then A5:    x in { t^<*n*> where t is Element of dom T: Q[n,T.t] } or
        x in dom T & dom T c= NAT* by TREES_1:def 3,XBOOLE_0:def 3;
        assume
A6:    not x in NAT*;
then     ex n be Element of NAT,t being Element of dom T
        st x = t^<*n*> & Q[n,T.t] by A5;
        hence thesis by A6,FINSEQ_1:def 11;
      end;
      thus for p st p in M holds ProperPrefixes p c= M
      proof
        let p;
        assume p in M;
then A7:    p in { t^<*n*> where t is Element of dom T: Q[n,T.t] } or
        p in dom T by XBOOLE_0:def 3;
    now
          assume p in { t^<*n*> where t is Element of dom T: Q[n,T.t] };
          then consider n be Element of NAT, t be Element of dom T such that
A8:      p = t^<*n*> and Q[n,T.t];
A9:      ProperPrefixes t c= dom T by TREES_1:def 3;
A10:      dom T c= M by XBOOLE_1:7;
A11:      t in dom T;
A12:      ProperPrefixes t c= M by A9,A10,XBOOLE_1:1;
A13:      {t} c= M by A10,A11,ZFMISC_1:31;
      ProperPrefixes p = ProperPrefixes t \/ {t} by A8,Th4;
          hence thesis by A12,A13,XBOOLE_1:8;
        end;
then     ProperPrefixes p c= M or ProperPrefixes p c= dom T & dom T c= M
        by A7,TREES_1:def 3,XBOOLE_1:7;
        hence thesis by XBOOLE_1:1;
      end;
      let p,m,n;
      assume p^<*m*> in M;
then A14:  p^<*m*> in { t^<*l*> where t is Element of dom T: Q[l,T.t] } or
      p^<*m*> in dom T by XBOOLE_0:def 3;
      assume that
A15:  n <= m and
A16:  not p^<*n*> in M;
  not p^<*n*> in dom T by A16,XBOOLE_0:def 3;
      then consider l be Element of NAT, t be Element of dom T such that
A17:  p^<*m*> = t^<*l*> and
A18:  Q[l,T.t] by A14,A15,TREES_1:def 3;
      A19:  len (p^<*m*>) = len p + len <*m*> & len <*m*> = 1 by FINSEQ_1:22,40
      ;
      A20:  len (t^<*l*>) = len t + len <*l*> & len <*l*> = 1 by FINSEQ_1:22,40
      ;
      A21:  (
p^<*m*>).(len p + 1) = m & (t^<*l*>).(len t + 1) = l by FINSEQ_1:42;
then A22:  p = t by A17,A19,A20,FINSEQ_1:33;
  Q[n,T.t] by A1,A15,A17,A18,A19,A20,A21;
then
  p^<*n*> in { s^<*i*> where s is Element of dom T: Q[i,T.s] } by A22;
      hence thesis by A16,XBOOLE_0:def 3;
    end;
    then reconsider M as Tree;
    defpred P[FinSequence,set] means $1 in dom T & $2 = T.$1 or
    not $1 in dom T & ex n,q st $1 = q^<*n*> & $2 = S().(T.q,n);
A23: for p st p in M ex x st P[p,x]
    proof
      let p;
      assume p in M;
then A24:  p in { t^<*l*> where t is Element of dom T: Q[l,T.t] } or
      p in dom T by XBOOLE_0:def 3;
  now
        assume
A25:    not p in dom T;
        then consider l be Element of NAT, t be Element of dom T such that
A26:    p = t^<*l*> and Q[l,T.t] by A24;
        take x = S().(T.t,l);
        thus p in dom T & x = T.p or
        not p in dom T & ex n,q st p = q^<*n*> & x = S().(T.q,n) by A25,A26;
      end;
      hence thesis;
    end;
    consider T9 be DecoratedTree such that
A27: dom T9 = M & for p st p in M holds P[p,T9.p] from DTreeEx(A23);
 rng T9 c= D()
    proof
      let x;
      assume x in rng T9;
      then consider y such that
A28:  y in dom T9 and
A29:  x = T9.y by FUNCT_1:def 3;
      reconsider y as Element of dom T9 by A28;
A30:  now
        assume y in dom T;
        then reconsider t = y as Element of dom T;
    T.t in D();
        hence thesis by A27,A29;
      end;
  now
        assume
A31:    not y in dom T;
        then consider n,q such that
A32:    y = q^<*n*> and
A33:    T9.y = S().(T.q,n) by A27;
    y in { t^<*l*> where t is Element of dom T: Q[l,T.t] }
        by A27,A31,XBOOLE_0:def 3;
        then consider l be Element of NAT, t be Element of dom T such that
A34:    y = t^<*l*> and Q[l,T.t];
A35:    len <*n*> = 1 by FINSEQ_1:39;
A36:    len <*l*> = 1 by FINSEQ_1:39;
A37:    len (q^<*n*>) = len q + 1 by A35,FINSEQ_1:22;
A38:    len (t^<*l*>) = len t + 1 by A36,FINSEQ_1:22;
            (
q^<*n*>).(len q + 1) = n & (t^<*l*>).(len t + 1) = l by FINSEQ_1:42;
then A39:    q = t by A32,A34,A37,A38,FINSEQ_1:33;
    T.t in D();
then     [T.q,n] in [:D(),NAT:] by A39,ZFMISC_1:87;
        hence thesis by A29,A33,FUNCT_2:5;
      end;
      hence thesis by A30;
    end;
    then reconsider T9 as DecoratedTree of D() by RELAT_1:def 19;
    take T9;
 <*> NAT in M & <*> NAT in dom T by TREES_1:22;
    hence T9. {} = d() by A4,A27;
    let t be Element of dom T9;
A40: now
      assume t in { s^<*l*> where s is Element of dom T: Q[l,T.s] };
      then consider l be Element of NAT, s being Element of dom T such that
A41:  t = s^<*l*> and Q[l,T.s];
  len s <= k by A4;
then   len <*l*> = 1 & len s + 1 <= k+1 by FINSEQ_1:39,XREAL_1:7;
      hence len t <= k+1 by A41,FINSEQ_1:22;
    end;
 now
      assume t in dom T;
      then reconsider s = t as Element of dom T;
  len s <= k & k <= k+1 by A4,NAT_1:11;
      hence len t <= k+1 by XXREAL_0:2;
    end;
    hence len t <= k+1 by A27,A40,XBOOLE_0:def 3;
    assume
A42: len t < k+1;
A43: now
      assume
A44:  not t in dom T;
then   t in { s^<*l*> where s is Element of dom T: Q[l,T.s] }
      by A27,XBOOLE_0:def 3;
      then consider l be Element of NAT, s be Element of dom T such that
A45:  t = s^<*l*> and
A46:  Q[l,T.s];
A47:  len t = len s + len <*l*> by A45,FINSEQ_1:22;
  len <*l*> = 1 & len t <= k by A42,FINSEQ_1:39,NAT_1:13;
then   len s < k by A47,NAT_1:13;
then   succ s = { s^<*m*>: Q[m,T.s]} by A4;
then   t in succ s by A45,A46;
      hence contradiction by A44;
    end;
then A48: T9.t = T.t by A27;
    reconsider t9 = t as Element of dom T by A43;
    thus succ t c= { t^<*i*>: Q[i,T9.t]}
    proof
      let x;
      assume x in succ t;
      then consider n such that
A49:  x = t^<*n*> and
A50:  t^<*n*> in dom T9;
  now per cases;
        suppose
A51:      t^<*n*> in dom T;
          then
reconsider s = t^<*n*>, t9 = t as Element of dom T by TREES_1:21;
      len s <= k & len s = len t + 1 by A4,FINSEQ_2:16;
then       len t < k by NAT_1:13;
then       succ t9 = { t9^<*m*>: Q[m,T.t9] } by A4;
          hence thesis by A48,A49,A51;
        end;
        suppose
      not t^<*n*> in dom T;
then       t^<*n*> in { s^<*l*> where s is Element of dom T: Q[l,T.s] }
          by A27,A50,XBOOLE_0:def 3;
          then consider l be Element of NAT, s be Element of dom T such that
A52:      t^<*n*> = s^<*l*> and
A53:      Q[l,T.s];
     t = s by A52,FINSEQ_2:17;
          hence thesis by A48,A49,A52,A53;
        end;
      end;
      hence thesis;
    end;
    thus
A54: { t^<*i*>: Q[i,T9.t]} c= succ t
    proof
      let x;
      assume x in { t^<*i*>: Q[i,T9.t]};
      then consider n such that
A55: x = t^<*n*> and
A56: Q[n,T9.t];
 x = t9^<*n*> by A55;
then  x in { s^<*l*> where s is Element of dom T: Q[l,T.s] } by A48,A56
      ;
then  x in dom T9 by A27,XBOOLE_0:def 3;
      hence thesis by A55;
    end;
    let n,x;
    assume that
A57: x = T9.t and
A58: Q[n,x];
 t^<*n*> in { t^<*i*>: Q[i,T9.t]} by A57,A58;
then A59: t^<*n*> in succ t by A54;
 now per cases;
      suppose
A60:   t^<*n*> in dom T;
        then reconsider s = t^<*n*> as Element of dom T;
   len s <= k & len s = len t + 1 by A4,FINSEQ_2:16;
then    len t9 < k by NAT_1:13;
then    T.(t9^<*n*>) = S().(x,n) by A4,A48,A57,A58;
        hence thesis by A27,A59,A60;
      end;
      suppose
   not t^<*n*> in dom T;
        then consider l,q such that
A61:   t^<*n*> = q^<*l*> and
A62:   T9.(t^<*n*>) = S().(T.q,l) by A27,A59;
   t = q & n = l by A61,FINSEQ_2:17;
        hence thesis by A27,A43,A57,A62;
      end;
    end;
    hence thesis;
  end;
A63: P[k] from NAT_1:sch 1(A2,A3);
  defpred P[set,set] means ex T being DecoratedTree of D(), k st
  $1 = k & $2 = T & T.{} = d() &
  for t being Element of dom T holds len t <= k &
  (len t < k implies succ t = { t^<*i*>: Q[i,T.t]} &
  for n,x st x = T.t & Q[n,x] holds T.(t^<*n*>) = S().(x,n));
A64: for x st x in NAT ex y st P[x,y]
  proof
    let x;
    assume x in NAT;
    then reconsider n = x as Element of NAT;
    consider T being DecoratedTree of D() such that
A65: T.{} = d() & for t being Element of dom T holds len t <= n &
    (len t < n implies succ t = { t^<*k*>: Q[k,T.t]} &
    for n,x st x = T.t & Q[n,x] holds T.(t^<*n*>) = S().(x,n)) by A63;
    reconsider y = T as set;
    take y,T,n;
    thus thesis by A65;
  end;
  consider f such that
A66: dom f = NAT & for x st x in NAT holds P[x,f.x] from CLASSES1:sch 1(A64
  );
  reconsider E = rng f as non empty set by A66,RELAT_1:42;
A67: for x st x in E holds x is DecoratedTree of D()
  proof
    let x;
    assume x in E;
    then consider y such that
A68: y in dom f and
A69: x = f.y by FUNCT_1:def 3;
 ex T being DecoratedTree of D(), k st y = k & f.y = T & T.{} = d() &
    for t being Element of dom T holds len t <= k &
    (len t < k implies succ t = { t^<*i*>: Q[i,T.t]} &
    for n,x st x = T.t & Q[n,x] holds T.(t^<*n*>) = S().(x,n)) by A66,A68;
    hence thesis by A69;
  end;
A70: for T1,T2,k1,k2 st T1 = f.k1 & T2 = f.k2 & k1 <= k2 holds T1 c= T2
  proof
    let T1,T2;
    let x,y be Element of NAT such that
A71: T1 = f.x and
A72: T2 = f.y and
A73: x <= y;
    consider T19 being DecoratedTree of D(), k1 such that
A74: x = k1 & f.x = T19 & T19.{} = d() &
    for t being Element of dom T19 holds len t <= k1 &
    (len t < k1 implies succ t = { t^<*i*>: Q[i,T19.t]} &
    for n,x st x = T19.t & Q[n,x] holds T19.(t^<*n*>) = S().(x,n)) by A66;
    consider T29 being DecoratedTree of D(), k2 such that
A75: y = k2 & f.y = T29 & T29.{} = d() &
    for t being Element of dom T29 holds len t <= k2 &
    (len t < k2 implies succ t = { t^<*i*>: Q[i,T29.t]} &
    for n,x st x = T29.t & Q[n,x] holds T29.(t^<*n*>) = S().(x,n)) by A66;
    defpred I[Element of NAT] means
    for t being Element of dom T1 st len t <= $1 holds
    t in dom T2 & T1.t = T2.t;
A76: I[0]
    proof
      let t be Element of dom T1 such that
A77: len t <= 0;
 t = {} by A77;
      hence thesis by A71,A72,A74,A75,TREES_1:22;
    end;
A78: I[k] implies I[k+1]
    proof
      assume
A79: for t being Element of dom T1 st len t <= k holds
      t in dom T2 & T1.t = T2.t;
      let t be Element of dom T1;
      assume len t <= k+1;
then A80: len t <= k or len t = k+1 by NAT_1:8;
 now
        assume
A81:   len t = k+1;
        reconsider p = t|Seg k as FinSequence of NAT by FINSEQ_1:18;
   p is_a_prefix_of t by TREES_1:def 1;
        then reconsider p as Element of dom T1 by TREES_1:20;
A82:   k <= k+1 by NAT_1:11;
A83:   k+1 <= k1 by A71,A74,A81;
A84:   len p = k by A81,A82,FINSEQ_1:17;
A85:   k < k1 by A83,NAT_1:13;
A86:   T1.p = T2.p by A79,A84;
        reconsider p9 = p as Element of dom T2 by A79,A84;
   t <> {} by A81;
        then consider q being FinSequence, x being set such that
A87:   t = q^<*x*> by FINSEQ_1:46;
        A88:   p
 is_a_prefix_of t & q is_a_prefix_of t by A87,TREES_1:1,def 1;
   k+1 = len q + 1 by A81,A87,FINSEQ_2:16;
then A89:   p = q by A84,A88,Th1,TREES_1:4;
   <*x*> is FinSequence of NAT by A87,FINSEQ_1:36;
then A90:   rng <*x*> c= NAT by FINSEQ_1:def 4;
   rng <*x*> = {x} & x in {x} by FINSEQ_1:38,TARSKI:def 1;
        then reconsider x as Element of NAT by A90;
A91:   p^<*x*> in succ p by A87,A89;
   succ p = { p^<*i*>: Q[i,T1.p]} by A71,A74,A84,A85;
        then consider i such that
A92:   p^<*x*> = p^<*i*> and
A93:   Q[i,T1.p] by A91;
A94:   k < k2 by A73,A74,A75,A85,XXREAL_0:2;
then A95:   succ p9 = { p9^<*l*>: Q[l,T2.p9] } by A72,A75,A84;
A96:   x = i by A92,FINSEQ_2:17;
A97:   t in succ p9 by A86,A87,A89,A92,A93,A95;
   T19.t = S().(T19.p,x) by A71,A74,A84,A85,A87,A89,A93,A96;
        hence
thesis by A71,A72,A74,A75,A84,A86,A87,A89,A93,A94,A96,A97;
      end;
      hence thesis by A79,A80;
    end;
A98: I[k] from NAT_1:sch 1(A76,A78);
    let x;
    assume
A99: x in T1;
    then consider y,z such that
A100: [y,z] = x by RELAT_1:def 1;
A101: T1.y = z by A99,A100,FUNCT_1:1;
    reconsider y as Element of dom T1 by A99,A100,FUNCT_1:1;
 len y <= len y;
then  y in dom T2 & T1.y = T2.y by A98;
    hence thesis by A100,A101,FUNCT_1:1;
  end;
 E is c=-linear
  proof
    let T1,T2 be set;
    assume
A102: T1 in E;
    then consider x such that
A103: x in dom f and
A104: T1 = f.x by FUNCT_1:def 3;
    assume
A105: T2 in E;
    then consider y such that
A106: y in dom f and
A107: T2 = f.y by FUNCT_1:def 3;
A108: T1 is DecoratedTree by A67,A102;
A109: T2 is DecoratedTree by A67,A105;
    reconsider x,y as Element of NAT by A66,A103,A106;
 x <= y or y <= x;
    hence T1 c= T2 or T2 c= T1 by A70,A104,A107,A108,A109;
  end;
  then reconsider T = union rng f as DecoratedTree of D() by A67,Th36;
  take T;
  consider T9 being DecoratedTree of D(), k such that
A110: 0 = k & f.0 = T9 & T9.{} = d() &
  for t being Element of dom T9 holds len t <= k &
  (len t < k implies succ t = { t^<*i*>: Q[i,T9.t]} &
  for n,x st x = T9.t & Q[n,x] holds T9.(t^<*n*>) = S().(x,n)) by A66;
 {} in dom T9 by TREES_1:22;
then A111: [{},d()] in T9 by A110,FUNCT_1:1;
 T9 in rng f by A66,A110,FUNCT_1:def 3;
then  [{},d()] in T by A111,TARSKI:def 4;
  hence T.{} = d() by FUNCT_1:1;
A112: for T1,x st T1 in E & x in dom T1 holds x in dom T & T1.x = T.x
  proof
    let T1,x;
    assume that
A113: T1 in E and
A114: x in dom T1;
 [x,T1.x] in T1 by A114,FUNCT_1:1;
then  [x,T1.x] in T by A113,TARSKI:def 4;
    hence thesis by FUNCT_1:1;
  end;
  let t be Element of dom T;
  thus succ t c= { t^<*i*>: Q[i,T.t]}
  proof
    let x;
    assume x in succ t;
    then consider l such that
A115: x = t^<*l*> and
A116: t^<*l*> in dom T;
 [x,T.x] in T by A115,A116,FUNCT_1:1;
    then consider X such that
A117: [x,T.x] in X and
A118: X in rng f by TARSKI:def 4;
    consider y such that
A119: y in NAT and
A120: X = f.y by A66,A118,FUNCT_1:def 3;
    consider T1 being DecoratedTree of D(), k1 such that
A121: y = k1 & f.y = T1 & T1.{} = d() &
    for t being Element of dom T1 holds len t <= k1 &
    (len t < k1 implies succ t = { t^<*i*>: Q[i,T1.t]} &
    for n,x st x = T1.t & Q[n,x] holds T1.(t^<*n*>) = S().(x,n)) by A66,A119;
A122: t^<*l*> in dom T1 by A115,A117,A120,A121,FUNCT_1:1;
    then
reconsider t9 = t, p = t^<*l*> as Element of dom T1 by TREES_1:21;
 len p <= k1 by A121;
then  len t + 1 <= k1 by FINSEQ_2:16;
then A123: len t9 < k1 by NAT_1:13;
then A124: succ t9 = { t9^<*i*>: Q[i,T1.t9]} by A121;
 T1.t = T.t by A112,A118,A120,A121,A123;
    hence thesis by A115,A122,A124;
  end;
 [t,T.t] in T by FUNCT_1:1;
  then consider X such that
A125: [t,T.t] in X and
A126: X in E by TARSKI:def 4;
  consider y such that
A127: y in NAT and
A128: X = f.y by A66,A126,FUNCT_1:def 3;
  reconsider y as Element of NAT by A127;
  consider T1 being DecoratedTree of D(), k1 such that
A129: y = k1 & f.y = T1 & T1.{} = d() &
  for t being Element of dom T1 holds len t <= k1 &
  (len t < k1 implies succ t = { t^<*i*>: Q[i,T1.t]} &
  for n,x st x = T1.t & Q[n,x] holds T1.(t^<*n*>) = S().(x,n)) by A66;
  consider T2 being DecoratedTree of D(), k2 such that
A130: y+1 = k2 & f.(y+1) = T2 & T2.{} = d() &
  for t being Element of dom T2 holds len t <= k2 &
  (len t < k2 implies succ t = { t^<*i*>: Q[i,T2.t]} &
  for n,x st x = T2.t & Q[n,x] holds T2.(t^<*n*>) = S().(x,n)) by A66;
 y <= y+1 by NAT_1:11;
then A131: T1 c= T2 by A70,A129,A130;
  reconsider t1 = t as Element of dom T1 by A125,A128,A129,FUNCT_1:1;
A132: len t1 <= y by A129;
A133: T2.t = T.t by A125,A128,A129,A131,FUNCT_1:1;
  reconsider t2 = t as Element of dom T2 by A125,A128,A129,A131,FUNCT_1:1;
A134: len t2 < y+1 by A132,NAT_1:13;
then A135: succ t2 = { t2^<*i*>: Q[i,T2.t2]} by A130;
  thus { t^<*i*>: Q[i,T.t]} c= succ t
  proof
    let x;
    assume
A136: x in { t^<*i*>: Q[i,T.t]};
then A137: ex l st x = t^<*l*> & Q[l,T.t];
A138: x in succ t2 by A130,A133,A134,A136;
 T2 in E by A66,A130,FUNCT_1:def 3;
then  x in dom T by A112,A138;
    hence thesis by A137;
  end;
  let n;
  assume
A139: Q[n,T.t];
then A140: t^<*n*> in succ t2 by A133,A135;
 T2 in E by A66,A130,FUNCT_1:def 3;
then  T2.(t^<*n*>) = T.(t^<*n*>) by A112,A140;
  hence thesis by A130,A133,A134,A139;
end;



theorem
  for T1, T2 being Tree st
     for n being Element of NAT holds T1-level n = T2-level n
   holds T1 = T2
proof
  let T1, T2 be Tree such that
A1: for n being Element of NAT holds T1-level n = T2-level n;
  for p being FinSequence of NAT holds p in T1 iff p in T2
  proof
    let p be FinSequence of NAT;
A2: T1 = union { T1-level n where n is Element of NAT: not contradiction }
    by Th14;
    hereby
      assume p in T1;
      then consider Y being set such that
A3:   p in Y and
A4:   Y in { T1-level n where n is Element of NAT: not contradiction }
      by A2,TARSKI:def 4;
      consider n being Element of NAT such that
A5:   Y = T1-level n by A4;
      Y = T2-level n by A1,A5;
      hence p in T2 by A3;
    end;
    assume
A6: p in T2;
    T2 = union { T2-level n where n is Element of NAT: not contradiction }
    by Th14;
    then consider Y being set such that
A7: p in Y and
A8: Y in { T2-level n where n is Element of NAT: not contradiction }
    by A6,TARSKI:def 4;
    consider n being Element of NAT such that
A9: Y = T2-level n by A8;
    Y = T1-level n by A1,A9;
    hence thesis by A7;
  end;
  hence thesis by Th5;
end;

theorem
  for n being Element of NAT holds TrivialInfiniteTree-level n = { n |-> 0 }
proof
  set T = TrivialInfiniteTree;
  let n be Element of NAT;
  set L = { w where w is Element of T: len w = n };
  set f = n |-> 0;
  {f} = L
  proof
    hereby
      let a be set;
      assume a in {f};
      then
A1:   a = f by TARSKI:def 1;
      f in T & len f = n by CARD_1:def 7;
      hence a in L by A1;
    end;
    let a be set;
    assume a in L;
    then consider w being Element of T such that
A2: a = w & len w = n;
    w in T;
    then ex k being Element of NAT st w = k |-> 0;
    then a = f by A2,CARD_1:def 7;
    hence thesis by TARSKI:def 1;
  end;
  hence thesis;
end;

theorem
  for X,Y being set for B being c=-linear Subset of PFuncs(X,Y)
  holds union B in PFuncs(X,Y)
proof
  let X,Y be set;
  let B be c=-linear Subset of PFuncs(X,Y);
  for x be set st x in B holds x is Function;
  then reconsider f = union B as Function by Th34;
  per cases;
  suppose
    B <> {};
    then reconsider D = B as non empty functional set;
A1: now
      let x be set;
      assume x in { dom g where g is Element of D: not contradiction };
      then consider g being Element of D such that
A2:   x = dom g;
      g in PFuncs(X,Y) by TARSKI:def 3;
      then ex f being Function st g = f & dom f c= X & rng f c= Y by
PARTFUN1:def 3;
      hence x c= X by A2;
    end;
A3: now
      let x be set;
      assume x in { rng g where g is Element of D: not contradiction };
      then consider g being Element of D such that
A4:   x = rng g;
      g in PFuncs(X,Y) by TARSKI:def 3;
      then ex f being Function st g = f & dom f c= X & rng f c= Y by
PARTFUN1:def 3;
      hence x c= Y by A4;
    end;
    rng f = union { rng g where g is Element of D: not contradiction }
          by FUNCT_1:110;
    then
A5: rng f c= Y by A3,ZFMISC_1:76;
    dom f = union { dom g where g is Element of D: not contradiction }
     by FUNCT_1:110;
    then dom f c= X by A1,ZFMISC_1:76;
    hence thesis by A5,PARTFUN1:def 3;
  end;
  suppose
A6: B = {};
    {} is PartFunc of X, Y by RELSET_1:12;
    hence thesis by A6,PARTFUN1:45,ZFMISC_1:2;
  end;
end;














begin 

reserve x, X, X1, X2 for set;

Lm1: now
  let X1,X2,X3 be set;
  thus X1 /\ (X2 /\ X3) = X1 /\ X1 /\ X2 /\ X3 by XBOOLE_1:16
    .= X1 /\ (X1 /\ X2) /\ X3 by XBOOLE_1:16
    .= (X1 /\ X2) /\ (X1 /\ X3) by XBOOLE_1:16;
end;

definition
  let Y be functional set;
  func DOMS(Y) equals
  union {dom f where f is Element of Y: not contradiction};
  coherence;
end;

definition
  let X;
  attr X is complex-functions-membered means
  :Def2:
  x in X implies x is complex-valued Function;
end;

definition
  let X;
  attr X is ext-real-functions-membered means
  :Def3:
  x in X implies x is ext-real-valued Function;
end;

definition
  let X;
  attr X is real-functions-membered means
  :Def4:
  x in X implies x is real-valued Function;
end;

definition
  let X;
  attr X is rational-functions-membered means
  :Def5:
  x in X implies x is RAT-valued Function;
end;

definition
  let X;
  attr X is integer-functions-membered means
  :Def6:
  x in X implies x is INT-valued Function;
end;

definition
  let X;
  attr X is natural-functions-membered means
  :Def7:
  x in X implies x is natural-valued Function;
end;

registration
  cluster natural-functions-membered -> integer-functions-membered for set;
  coherence
  proof
    let X;
    assume
A1: for x being set st x in X holds x is natural-valued Function;
    let x;
    assume x in X;
    then x is natural-valued Function by A1;
    hence thesis;
  end;
  cluster integer-functions-membered -> rational-functions-membered for set;
  coherence
  proof
    let X;
    assume
A2: for x being set st x in X holds x is INT-valued Function;
    let x;
    assume x in X;
    then x is INT-valued Function by A2;
    hence thesis;
  end;
  cluster rational-functions-membered -> real-functions-membered for set;
  coherence
  proof
    let X;
    assume
A3: for x being set st x in X holds x is RAT-valued Function;
    let x;
    thus thesis by A3;
  end;
  cluster real-functions-membered -> complex-functions-membered for set;
  coherence
  proof
    let X;
    assume
A4: for x being set st x in X holds x is real-valued Function;
    let x;
    thus thesis by A4;
  end;
  cluster real-functions-membered -> ext-real-functions-membered for set;
  coherence
  proof
    let X;
    assume
A5: for x being set st x in X holds x is real-valued Function;
    let x;
    thus thesis by A5;
  end;
end;

registration
  cluster empty -> natural-functions-membered for set;
  coherence
  proof
    let X be set such that
A1: X is empty;
    let x;
    thus thesis by A1;
  end;
end;

registration
  let f be complex-valued Function;
  cluster {f} -> complex-functions-membered;
  coherence
  proof
    let x;
    thus thesis by TARSKI:def 1;
  end;
end;

registration
  cluster complex-functions-membered -> functional for set;
  coherence
  proof
    let X;
    assume
A1: X is complex-functions-membered;
    let x;
    thus thesis by A1,Def2;
  end;
  cluster ext-real-functions-membered -> functional for set;
  coherence
  proof
    let X;
    assume
A2: X is ext-real-functions-membered;
    let x;
    thus thesis by A2,Def3;
  end;
end;

set ff = the natural-valued Function;

registration
  cluster natural-functions-membered non empty for set;
  existence
  proof
    take {ff};
    thus for x being set st x in {ff} holds x is natural-valued Function by
TARSKI:def 1;
    thus thesis;
  end;
end;

registration
  let X be complex-functions-membered set;
  cluster -> complex-functions-membered for Subset of X;
  coherence
  proof
    let S be Subset of X;
    let x;
    thus thesis by Def2;
  end;
end;

registration
  let X be ext-real-functions-membered set;
  cluster -> ext-real-functions-membered for Subset of X;
  coherence
  proof
    let S be Subset of X;
    let x;
    thus thesis by Def3;
  end;
end;

registration
  let X be real-functions-membered set;
  cluster -> real-functions-membered for Subset of X;
  coherence
  proof
    let S be Subset of X;
    let x;
    thus thesis by Def4;
  end;
end;

registration
  let X be rational-functions-membered set;
  cluster -> rational-functions-membered for Subset of X;
  coherence
  proof
    let S be Subset of X;
    let x;
    thus thesis by Def5;
  end;
end;

registration
  let X be integer-functions-membered set;
  cluster -> integer-functions-membered for Subset of X;
  coherence
  proof
    let S be Subset of X;
    let x;
    thus thesis by Def6;
  end;
end;

registration
  let X be natural-functions-membered set;
  cluster -> natural-functions-membered for Subset of X;
  coherence
  proof
    let S be Subset of X;
    let x;
    thus thesis by Def7;
  end;
end;

definition
  set A = COMPLEX;
  let D be set;
  defpred P[set] means $1 is PartFunc of D,A;
  func C_PFuncs(D) -> set means
  :Def8:
  for f being set holds f in it iff f is PartFunc of D,COMPLEX;
  existence
  proof
    consider X being set such that
A1: for x being set holds x in X iff x in PFuncs(D,A) & P[x] from
    XBOOLE_0:sch 1;
    take X;
    let f be set;
    thus f in X implies f is PartFunc of D,A by A1;
    assume
A2: f is PartFunc of D,A;
    then f in PFuncs(D,A) by PARTFUN1:45;
    hence thesis by A1,A2;
  end;
  uniqueness
  proof
    let P, Q be set;
    assume for f being set holds f in P iff f is PartFunc of D,A;
    then
A3: for f being set holds f in P iff P[f];
    assume for f being set holds f in Q iff f is PartFunc of D,A;
    then
A4: for f being set holds f in Q iff P[f];
    thus P = Q from XBOOLE_0:sch 2(A3,A4);
  end;
end;

definition
  set A = COMPLEX;
  let D be set;
  defpred P[set] means $1 is Function of D,A;
  func C_Funcs(D) -> set means
  :Def9:
  for f being set holds f in it iff f is Function of D,COMPLEX;
  existence
  proof
    consider X being set such that
A1: for x being set holds x in X iff x in Funcs(D,A) & P[x] from
    XBOOLE_0:sch 1;
    take X;
    let f be set;
    thus f in X implies f is Function of D,A by A1;
    assume
A2: f is Function of D,A;
    then f in Funcs(D,A) by FUNCT_2:8;
    hence thesis by A1,A2;
  end;
  uniqueness
  proof
    let P, Q be set;
    assume for f being set holds f in P iff f is Function of D,A;
    then
A3: for f being set holds f in P iff P[f];
    assume for f being set holds f in Q iff f is Function of D,A;
    then
A4: for f being set holds f in Q iff P[f];
    thus P = Q from XBOOLE_0:sch 2(A3,A4);
  end;
end;

definition
  set A = ExtREAL;
  let D be set;
  defpred P[set] means $1 is PartFunc of D,A;
  func E_PFuncs(D) -> set means
  :Def10:
  for f being set holds f in it iff f is PartFunc of D,ExtREAL;
  existence
  proof
    consider X being set such that
A1: for x being set holds x in X iff x in PFuncs(D,A) & P[x] from
    XBOOLE_0:sch 1;
    take X;
    let f be set;
    thus f in X implies f is PartFunc of D,A by A1;
    assume
A2: f is PartFunc of D,A;
    then f in PFuncs(D,A) by PARTFUN1:45;
    hence thesis by A1,A2;
  end;
  uniqueness
  proof
    let P, Q be set;
    assume for f being set holds f in P iff f is PartFunc of D,A;
    then
A3: for f being set holds f in P iff P[f];
    assume for f being set holds f in Q iff f is PartFunc of D,A;
    then
A4: for f being set holds f in Q iff P[f];
    thus P = Q from XBOOLE_0:sch 2(A3,A4);
  end;
end;

definition
  set A = ExtREAL;
  let D be set;
  defpred P[set] means $1 is Function of D,A;
  func E_Funcs(D) -> set means
  :Def11:
  for f being set holds f in it iff f is Function of D,ExtREAL;
  existence
  proof
    consider X being set such that
A1: for x being set holds x in X iff x in Funcs(D,A) & P[x] from
    XBOOLE_0:sch 1;
    take X;
    let f be set;
    thus f in X implies f is Function of D,A by A1;
    assume
A2: f is Function of D,A;
    then f in Funcs(D,A) by FUNCT_2:8;
    hence thesis by A1,A2;
  end;
  uniqueness
  proof
    let P, Q be set;
    assume for f being set holds f in P iff f is Function of D,A;
    then
A3: for f being set holds f in P iff P[f];
    assume for f being set holds f in Q iff f is Function of D,A;
    then
A4: for f being set holds f in Q iff P[f];
    thus P = Q from XBOOLE_0:sch 2(A3,A4);
  end;
end;

definition
  set A = REAL;
  let D be set;
  defpred P[set] means $1 is PartFunc of D,A;
  func R_PFuncs(D) -> set means
  :Def12:
  for f being set holds f in it iff f is PartFunc of D,REAL;
  existence
  proof
    consider X being set such that
A1: for x being set holds x in X iff x in PFuncs(D,A) & P[x] from
    XBOOLE_0:sch 1;
    take X;
    let f be set;
    thus f in X implies f is PartFunc of D,A by A1;
    assume
A2: f is PartFunc of D,A;
    then f in PFuncs(D,A) by PARTFUN1:45;
    hence thesis by A1,A2;
  end;
  uniqueness
  proof
    let P, Q be set;
    assume for f being set holds f in P iff f is PartFunc of D,A;
    then
A3: for f being set holds f in P iff P[f];
    assume for f being set holds f in Q iff f is PartFunc of D,A;
    then
A4: for f being set holds f in Q iff P[f];
    thus P = Q from XBOOLE_0:sch 2(A3,A4);
  end;
end;

definition
  set A = REAL;
  let D be set;
  defpred P[set] means $1 is Function of D,A;
  func R_Funcs(D) -> set means
  :Def13:
  for f being set holds f in it iff f is Function of D,REAL;
  existence
  proof
    consider X being set such that
A1: for x being set holds x in X iff x in Funcs(D,A) & P[x] from
    XBOOLE_0:sch 1;
    take X;
    let f be set;
    thus f in X implies f is Function of D,A by A1;
    assume
A2: f is Function of D,A;
    then f in Funcs(D,A) by FUNCT_2:8;
    hence thesis by A1,A2;
  end;
  uniqueness
  proof
    let P, Q be set;
    assume for f being set holds f in P iff f is Function of D,A;
    then
A3: for f being set holds f in P iff P[f];
    assume for f being set holds f in Q iff f is Function of D,A;
    then
A4: for f being set holds f in Q iff P[f];
    thus P = Q from XBOOLE_0:sch 2(A3,A4);
  end;
end;

definition
  set A = RAT;
  let D be set;
  defpred P[set] means $1 is PartFunc of D,A;
  func Q_PFuncs(D) -> set means
  :Def14:
  for f being set holds f in it iff f is PartFunc of D,RAT;
  existence
  proof
    consider X being set such that
A1: for x being set holds x in X iff x in PFuncs(D,A) & P[x] from
    XBOOLE_0:sch 1;
    take X;
    let f be set;
    thus f in X implies f is PartFunc of D,A by A1;
    assume
A2: f is PartFunc of D,A;
    then f in PFuncs(D,A) by PARTFUN1:45;
    hence thesis by A1,A2;
  end;
  uniqueness
  proof
    let P, Q be set;
    assume for f being set holds f in P iff f is PartFunc of D,A;
    then
A3: for f being set holds f in P iff P[f];
    assume for f being set holds f in Q iff f is PartFunc of D,A;
    then
A4: for f being set holds f in Q iff P[f];
    thus P = Q from XBOOLE_0:sch 2(A3,A4);
  end;
end;

definition
  set A = RAT;
  let D be set;
  defpred P[set] means $1 is Function of D,A;
  func Q_Funcs(D) -> set means
  :Def15:
  for f being set holds f in it iff f is Function of D,RAT;
  existence
  proof
    consider X being set such that
A1: for x being set holds x in X iff x in Funcs(D,A) & P[x] from
    XBOOLE_0:sch 1;
    take X;
    let f be set;
    thus f in X implies f is Function of D,A by A1;
    assume
A2: f is Function of D,A;
    then f in Funcs(D,A) by FUNCT_2:8;
    hence thesis by A1,A2;
  end;
  uniqueness
  proof
    let P, Q be set;
    assume for f being set holds f in P iff f is Function of D,A;
    then
A3: for f being set holds f in P iff P[f];
    assume for f being set holds f in Q iff f is Function of D,A;
    then
A4: for f being set holds f in Q iff P[f];
    thus P = Q from XBOOLE_0:sch 2(A3,A4);
  end;
end;

definition
  set A = INT;
  let D be set;
  defpred P[set] means $1 is PartFunc of D,A;
  func I_PFuncs(D) -> set means
  :Def16:
  for f being set holds f in it iff f is PartFunc of D,INT;
  existence
  proof
    consider X being set such that
A1: for x being set holds x in X iff x in PFuncs(D,A) & P[x] from
    XBOOLE_0:sch 1;
    take X;
    let f be set;
    thus f in X implies f is PartFunc of D,A by A1;
    assume
A2: f is PartFunc of D,A;
    then f in PFuncs(D,A) by PARTFUN1:45;
    hence thesis by A1,A2;
  end;
  uniqueness
  proof
    let P, Q be set;
    assume for f being set holds f in P iff f is PartFunc of D,A;
    then
A3: for f being set holds f in P iff P[f];
    assume for f being set holds f in Q iff f is PartFunc of D,A;
    then
A4: for f being set holds f in Q iff P[f];
    thus P = Q from XBOOLE_0:sch 2(A3,A4);
  end;
end;

definition
  set A = INT;
  let D be set;
  defpred P[set] means $1 is Function of D,A;
  func I_Funcs(D) -> set means
  :Def17:
  for f being set holds f in it iff f is Function of D,INT;
  existence
  proof
    consider X being set such that
A1: for x being set holds x in X iff x in Funcs(D,A) & P[x] from
    XBOOLE_0:sch 1;
    take X;
    let f be set;
    thus f in X implies f is Function of D,A by A1;
    assume
A2: f is Function of D,A;
    then f in Funcs(D,A) by FUNCT_2:8;
    hence thesis by A1,A2;
  end;
  uniqueness
  proof
    let P, Q be set;
    assume for f being set holds f in P iff f is Function of D,A;
    then
A3: for f being set holds f in P iff P[f];
    assume for f being set holds f in Q iff f is Function of D,A;
    then
A4: for f being set holds f in Q iff P[f];
    thus P = Q from XBOOLE_0:sch 2(A3,A4);
  end;
end;

definition
  set A = NAT;
  let D be set;
  defpred P[set] means $1 is PartFunc of D,A;
  func N_PFuncs(D) -> set means
  :Def18:
  for f being set holds f in it iff f is PartFunc of D,NAT;
  existence
  proof
    consider X being set such that
A1: for x being set holds x in X iff x in PFuncs(D,A) & P[x] from
    XBOOLE_0:sch 1;
    take X;
    let f be set;
    thus f in X implies f is PartFunc of D,A by A1;
    assume
A2: f is PartFunc of D,A;
    then f in PFuncs(D,A) by PARTFUN1:45;
    hence thesis by A1,A2;
  end;
  uniqueness
  proof
    let P, Q be set;
    assume for f being set holds f in P iff f is PartFunc of D,A;
    then
A3: for f being set holds f in P iff P[f];
    assume for f being set holds f in Q iff f is PartFunc of D,A;
    then
A4: for f being set holds f in Q iff P[f];
    thus P = Q from XBOOLE_0:sch 2(A3,A4);
  end;
end;

definition
  set A = NAT;
  let D be set;
  defpred P[set] means $1 is Function of D,A;
  func N_Funcs(D) -> set means
  :Def19:
  for f being set holds f in it iff f is Function of D,NAT;
  existence
  proof
    consider X being set such that
A1: for x being set holds x in X iff x in Funcs(D,A) & P[x] from
    XBOOLE_0:sch 1;
    take X;
    let f be set;
    thus f in X implies f is Function of D,A by A1;
    assume
A2: f is Function of D,A;
    then f in Funcs(D,A) by FUNCT_2:8;
    hence thesis by A1,A2;
  end;
  uniqueness
  proof
    let P, Q be set;
    assume for f being set holds f in P iff f is Function of D,A;
    then
A3: for f being set holds f in P iff P[f];
    assume for f being set holds f in Q iff f is Function of D,A;
    then
A4: for f being set holds f in Q iff P[f];
    thus P = Q from XBOOLE_0:sch 2(A3,A4);
  end;
end;

theorem Th1:
  C_Funcs(X) is Subset of C_PFuncs(X)
proof
  C_Funcs(X) c= C_PFuncs(X)
  proof
    let x;
    assume x in C_Funcs(X);
    then x is Function of X,COMPLEX by Def9;
    hence thesis by Def8;
  end;
  hence thesis;
end;

theorem Th2:
  E_Funcs(X) is Subset of E_PFuncs(X)
proof
  E_Funcs(X) c= E_PFuncs(X)
  proof
    let x;
    assume x in E_Funcs(X);
    then x is Function of X,ExtREAL by Def11;
    hence thesis by Def10;
  end;
  hence thesis;
end;

theorem Th3:
  R_Funcs(X) is Subset of R_PFuncs(X)
proof
  R_Funcs(X) c= R_PFuncs(X)
  proof
    let x;
    assume x in R_Funcs(X);
    then x is Function of X,REAL by Def13;
    hence thesis by Def12;
  end;
  hence thesis;
end;

theorem Th4:
  Q_Funcs(X) is Subset of Q_PFuncs(X)
proof
  Q_Funcs(X) c= Q_PFuncs(X)
  proof
    let x;
    assume x in Q_Funcs(X);
    then x is Function of X,RAT by Def15;
    hence thesis by Def14;
  end;
  hence thesis;
end;

theorem Th5:
  I_Funcs(X) is Subset of I_PFuncs(X)
proof
  I_Funcs(X) c= I_PFuncs(X)
  proof
    let x;
    assume x in I_Funcs(X);
    then x is Function of X,INT by Def17;
    hence thesis by Def16;
  end;
  hence thesis;
end;

theorem Th6:
  N_Funcs(X) is Subset of N_PFuncs(X)
proof
  N_Funcs(X) c= N_PFuncs(X)
  proof
    let x;
    assume x in N_Funcs(X);
    then x is Function of X,NAT by Def19;
    hence thesis by Def18;
  end;
  hence thesis;
end;

registration
  let X;
  cluster C_PFuncs(X) -> complex-functions-membered;
  coherence
  proof
    let x;
    thus thesis by Def8;
  end;
  cluster C_Funcs(X) -> complex-functions-membered;
  coherence
  proof
    reconsider C = C_Funcs(X) as Subset of C_PFuncs(X) by Th1;
    C is complex-functions-membered;
    hence thesis;
  end;
  cluster E_PFuncs(X) -> ext-real-functions-membered;
  coherence
  proof
    let x;
    thus thesis by Def10;
  end;
  cluster E_Funcs(X) -> ext-real-functions-membered;
  coherence
  proof
    reconsider C = E_Funcs(X) as Subset of E_PFuncs(X) by Th2;
    C is ext-real-functions-membered;
    hence thesis;
  end;
  cluster R_PFuncs(X) -> real-functions-membered;
  coherence
  proof
    let x;
    thus thesis by Def12;
  end;
  cluster R_Funcs(X) -> real-functions-membered;
  coherence
  proof
    reconsider C = R_Funcs(X) as Subset of R_PFuncs(X) by Th3;
    C is real-functions-membered;
    hence thesis;
  end;
  cluster Q_PFuncs(X) -> rational-functions-membered;
  coherence
  proof
    let x;
    thus thesis by Def14;
  end;
  cluster Q_Funcs(X) -> rational-functions-membered;
  coherence
  proof
    reconsider C = Q_Funcs(X) as Subset of Q_PFuncs(X) by Th4;
    C is rational-functions-membered;
    hence thesis;
  end;
  cluster I_PFuncs(X) -> integer-functions-membered;
  coherence
  proof
    let x;
    thus thesis by Def16;
  end;
  cluster I_Funcs(X) -> integer-functions-membered;
  coherence
  proof
    reconsider C = I_Funcs(X) as Subset of I_PFuncs(X) by Th5;
    C is integer-functions-membered;
    hence thesis;
  end;
  cluster N_PFuncs(X) -> natural-functions-membered;
  coherence
  proof
    let x;
    thus thesis by Def18;
  end;
  cluster N_Funcs(X) -> natural-functions-membered;
  coherence
  proof
    reconsider C = N_Funcs(X) as Subset of N_PFuncs(X) by Th6;
    C is natural-functions-membered;
    hence thesis;
  end;
end;

registration
  let X be complex-functions-membered set;
  cluster -> complex-valued for Element of X;
  coherence
  proof
    X is empty or X is non empty;
    hence thesis by Def2,SUBSET_1:def 1;
  end;
end;

registration
  let X be ext-real-functions-membered set;
  cluster -> ext-real-valued for Element of X;
  coherence
  proof
    X is empty or X is non empty;
    hence thesis by Def3,SUBSET_1:def 1;
  end;
end;

registration
  let X be real-functions-membered set;
  cluster -> real-valued for Element of X;
  coherence
  proof
    X is empty or X is non empty;
    hence thesis by Def4,SUBSET_1:def 1;
  end;
end;

registration
  let X be rational-functions-membered set;
  cluster -> RAT-valued for Element of X;
  coherence
  proof
    X is empty or X is non empty;
    hence thesis by Def5,SUBSET_1:def 1;
  end;
end;

registration
  let X be integer-functions-membered set;
  cluster -> INT-valued for Element of X;
  coherence
  proof
    X is empty or X is non empty;
    hence thesis by Def6,SUBSET_1:def 1;
  end;
end;

registration
  let X be natural-functions-membered set;
  cluster -> natural-valued for Element of X;
  coherence
  proof
    X is empty or X is non empty;
    hence thesis by Def7,SUBSET_1:def 1;
  end;
end;

registration
  let X, x be set;
  let Y be complex-functions-membered set;
  let f be PartFunc of X,Y;
  cluster f.x -> Function-like Relation-like;
  coherence;
end;

registration
  let X, x be set;
  let Y be ext-real-functions-membered set;
  let f be PartFunc of X,Y;
  cluster f.x -> Function-like Relation-like;
  coherence;
end;

registration
  let X, x;
  let Y be complex-functions-membered set;
  let f be PartFunc of X,Y;
  cluster f.x -> complex-valued;
  coherence
  proof
    per cases;
    suppose
      x in dom f;
      then f.x in rng f by FUNCT_1:def 3;
      hence thesis;
    end;
    suppose
      not x in dom f;
      hence thesis by FUNCT_1:def 2;
    end;
  end;
end;

registration
  let X, x;
  let Y be ext-real-functions-membered set;
  let f be PartFunc of X,Y;
  cluster f.x -> ext-real-valued;
  coherence
  proof
    per cases;
    suppose
      x in dom f;
      then f.x in rng f by FUNCT_1:def 3;
      hence thesis;
    end;
    suppose
      not x in dom f;
      hence thesis by FUNCT_1:def 2;
    end;
  end;
end;

registration
  let X, x;
  let Y be real-functions-membered set;
  let f be PartFunc of X,Y;
  cluster f.x -> real-valued;
  coherence
  proof
    per cases;
    suppose
      x in dom f;
      then f.x in rng f by FUNCT_1:def 3;
      hence thesis;
    end;
    suppose
      not x in dom f;
      hence thesis by FUNCT_1:def 2;
    end;
  end;
end;

registration
  let X, x;
  let Y be rational-functions-membered set;
  let f be PartFunc of X,Y;
  cluster f.x -> RAT-valued;
  coherence
  proof
    per cases;
    suppose
      x in dom f;
      then f.x in rng f by FUNCT_1:def 3;
      hence thesis;
    end;
    suppose
      not x in dom f;
      hence thesis by FUNCT_1:def 2;
    end;
  end;
end;

registration
  let X, x;
  let Y be integer-functions-membered set;
  let f be PartFunc of X,Y;
  cluster f.x -> INT-valued;
  coherence
  proof
    per cases;
    suppose
      x in dom f;
      then f.x in rng f by FUNCT_1:def 3;
      hence thesis;
    end;
    suppose
      not x in dom f;
      hence thesis by FUNCT_1:def 2;
    end;
  end;
end;

registration
  let X, x;
  let Y be natural-functions-membered set;
  let f be PartFunc of X,Y;
  cluster f.x -> natural-valued;
  coherence
  proof
    per cases;
    suppose
      x in dom f;
      then f.x in rng f by FUNCT_1:def 3;
      hence thesis;
    end;
    suppose
      not x in dom f;
      hence thesis by FUNCT_1:def 2;
    end;
  end;
end;

registration
  let X;
  let Y be complex-membered set;
  cluster PFuncs(X,Y) -> complex-functions-membered;
  coherence
  proof
    let x;
    assume x in PFuncs(X,Y);
    then consider f being Function such that
A1: x = f and
A2: dom f c= X & rng f c= Y by PARTFUN1:def 3;
    reconsider f as PartFunc of X,Y by A2,RELSET_1:4;
    f is set;
    hence thesis by A1;
  end;
end;

registration
  let X;
  let Y be ext-real-membered set;
  cluster PFuncs(X,Y) -> ext-real-functions-membered;
  coherence
  proof
    let x;
    assume x in PFuncs(X,Y);
    then consider f being Function such that
A1: x = f and
A2: dom f c= X & rng f c= Y by PARTFUN1:def 3;
    reconsider f as PartFunc of X,Y by A2,RELSET_1:4;
    f is set;
    hence thesis by A1;
  end;
end;

registration
  let X;
  let Y be real-membered set;
  cluster PFuncs(X,Y) -> real-functions-membered;
  coherence
  proof
    let x;
    assume x in PFuncs(X,Y);
    then consider f being Function such that
A1: x = f and
A2: dom f c= X & rng f c= Y by PARTFUN1:def 3;
    reconsider f as PartFunc of X,Y by A2,RELSET_1:4;
    f is set;
    hence thesis by A1;
  end;
end;

registration
  let X;
  let Y be rational-membered set;
  cluster PFuncs(X,Y) -> rational-functions-membered;
  coherence
  proof
    let x;
    assume x in PFuncs(X,Y);
    then consider f being Function such that
A1: x = f and
A2: dom f c= X & rng f c= Y by PARTFUN1:def 3;
    reconsider f as PartFunc of X,Y by A2,RELSET_1:4;
    f is set;
    hence thesis by A1;
  end;
end;

registration
  let X;
  let Y be integer-membered set;
  cluster PFuncs(X,Y) -> integer-functions-membered;
  coherence
  proof
    let x;
    assume x in PFuncs(X,Y);
    then consider f being Function such that
A1: x = f and
A2: dom f c= X & rng f c= Y by PARTFUN1:def 3;
    reconsider f as PartFunc of X,Y by A2,RELSET_1:4;
    f is set;
    hence thesis by A1;
  end;
end;

registration
  let X;
  let Y be natural-membered set;
  cluster PFuncs(X,Y) -> natural-functions-membered;
  coherence
  proof
    let x;
    assume x in PFuncs(X,Y);
    then consider f being Function such that
A1: x = f and
A2: dom f c= X & rng f c= Y by PARTFUN1:def 3;
    reconsider f as PartFunc of X,Y by A2,RELSET_1:4;
    f is set;
    hence thesis by A1;
  end;
end;

registration
  let X;
  let Y be complex-membered set;
  cluster Funcs(X,Y) -> complex-functions-membered;
  coherence
  proof
    let x;
    assume x in Funcs(X,Y);
    then consider f being Function such that
A1: x = f and
A2: dom f = X & rng f c= Y by FUNCT_2:def 2;
    reconsider f as PartFunc of X,Y by A2,RELSET_1:4;
    f is set;
    hence thesis by A1;
  end;
end;

registration
  let X;
  let Y be ext-real-membered set;
  cluster Funcs(X,Y) -> ext-real-functions-membered;
  coherence
  proof
    let x;
    assume x in Funcs(X,Y);
    then consider f being Function such that
A1: x = f and
A2: dom f = X & rng f c= Y by FUNCT_2:def 2;
    reconsider f as PartFunc of X,Y by A2,RELSET_1:4;
    f is set;
    hence thesis by A1;
  end;
end;

registration
  let X;
  let Y be real-membered set;
  cluster Funcs(X,Y) -> real-functions-membered;
  coherence
  proof
    let x;
    assume x in Funcs(X,Y);
    then consider f being Function such that
A1: x = f and
A2: dom f = X & rng f c= Y by FUNCT_2:def 2;
    reconsider f as PartFunc of X,Y by A2,RELSET_1:4;
    f is set;
    hence thesis by A1;
  end;
end;

registration
  let X;
  let Y be rational-membered set;
  cluster Funcs(X,Y) -> rational-functions-membered;
  coherence
  proof
    let x;
    assume x in Funcs(X,Y);
    then consider f being Function such that
A1: x = f and
A2: dom f = X & rng f c= Y by FUNCT_2:def 2;
    reconsider f as PartFunc of X,Y by A2,RELSET_1:4;
    f is set;
    hence thesis by A1;
  end;
end;

registration
  let X;
  let Y be integer-membered set;
  cluster Funcs(X,Y) -> integer-functions-membered;
  coherence
  proof
    let x;
    assume x in Funcs(X,Y);
    then consider f being Function such that
A1: x = f and
A2: dom f = X & rng f c= Y by FUNCT_2:def 2;
    reconsider f as PartFunc of X,Y by A2,RELSET_1:4;
    f is set;
    hence thesis by A1;
  end;
end;

registration
  let X;
  let Y be natural-membered set;
  cluster Funcs(X,Y) -> natural-functions-membered;
  coherence
  proof
    let x;
    assume x in Funcs(X,Y);
    then consider f being Function such that
A1: x = f and
A2: dom f = X & rng f c= Y by FUNCT_2:def 2;
    reconsider f as PartFunc of X,Y by A2,RELSET_1:4;
    f is set;
    hence thesis by A1;
  end;
end;

definition
  let R be Relation;
  attr R is complex-functions-valued means
  :Def20:
  rng R is complex-functions-membered;
  attr R is ext-real-functions-valued means
  :Def21:
  rng R is ext-real-functions-membered;
  attr R is real-functions-valued means
  :Def22:
  rng R is real-functions-membered;
  attr R is rational-functions-valued means
  :Def23:
  rng R is rational-functions-membered;
  attr R is integer-functions-valued means
  :Def24:
  rng R is integer-functions-membered;
  attr R is natural-functions-valued means
  :Def25:
  rng R is natural-functions-membered;
end;

registration
 let Y be complex-functions-membered set;
 cluster -> complex-functions-valued for Y-valued Function;
 coherence
  proof let f be Y-valued Function;
   thus rng f is complex-functions-membered;
  end;
end;

definition
  let f be Function;
  redefine attr f is complex-functions-valued means
  for x being set st x in dom f holds f.x is complex-valued Function;
  compatibility
  proof
    thus f is complex-functions-valued implies for x being set st x in dom f
    holds f.x is complex-valued Function
    proof
      assume
A1:   rng f is complex-functions-membered;
      let x;
      assume x in dom f;
      then f.x in rng f by FUNCT_1:def 3;
      hence thesis by A1;
    end;
    assume
A2: for x being set st x in dom f holds f.x is complex-valued Function;
    let y be set;
    assume y in rng f;
    then ex x being set st x in dom f & f.x = y by FUNCT_1:def 3;
    hence thesis by A2;
  end;
  redefine attr f is ext-real-functions-valued means
  for x being set st x in dom f holds f.x is ext-real-valued Function;
  compatibility
  proof
    thus f is ext-real-functions-valued implies for x being set st x in dom f
    holds f.x is ext-real-valued Function
    proof
      assume
A3:   rng f is ext-real-functions-membered;
      let x;
      assume x in dom f;
      then f.x in rng f by FUNCT_1:def 3;
      hence thesis by A3;
    end;
    assume
A4: for x being set st x in dom f holds f.x is ext-real-valued Function;
    let y be set;
    assume y in rng f;
    then ex x being set st x in dom f & f.x = y by FUNCT_1:def 3;
    hence thesis by A4;
  end;
  redefine attr f is real-functions-valued means
  for x being set st x in dom f holds f.x is real-valued Function;
  compatibility
  proof
    thus f is real-functions-valued implies for x being set st x in dom f
    holds f.x is real-valued Function
    proof
      assume
A5:   rng f is real-functions-membered;
      let x;
      assume x in dom f;
      then f.x in rng f by FUNCT_1:def 3;
      hence thesis by A5;
    end;
    assume
A6: for x being set st x in dom f holds f.x is real-valued Function;
    let y be set;
    assume y in rng f;
    then ex x being set st x in dom f & f.x = y by FUNCT_1:def 3;
    hence thesis by A6;
  end;
  redefine attr f is rational-functions-valued means
  for x being set st x in dom f holds f.x is RAT-valued Function;
  compatibility
  proof
    thus f is rational-functions-valued implies for x being set st x in dom f
    holds f.x is RAT-valued Function
    proof
      assume
A7:   rng f is rational-functions-membered;
      let x;
      assume x in dom f;
      then f.x in rng f by FUNCT_1:def 3;
      hence thesis by A7;
    end;
    assume
A8: for x being set st x in dom f holds f.x is RAT-valued Function;
    let y be set;
    assume y in rng f;
    then ex x being set st x in dom f & f.x = y by FUNCT_1:def 3;
    hence thesis by A8;
  end;
  redefine attr f is integer-functions-valued means
  for x being set st x in dom f holds f.x is INT-valued Function;
  compatibility
  proof
    thus f is integer-functions-valued implies for x being set st x in dom f
    holds f.x is INT-valued Function
    proof
      assume
A9:   rng f is integer-functions-membered;
      let x;
      assume x in dom f;
      then f.x in rng f by FUNCT_1:def 3;
      hence thesis by A9;
    end;
    assume
A10: for x being set st x in dom f holds f.x is INT-valued Function;
    let y be set;
    assume y in rng f;
    then ex x being set st x in dom f & f.x = y by FUNCT_1:def 3;
    hence thesis by A10;
  end;
  redefine attr f is natural-functions-valued means
  for x being set st x in dom f holds f.x is natural-valued Function;
  compatibility
  proof
    thus f is natural-functions-valued implies for x being set st x in dom f
    holds f.x is natural-valued Function
    proof
      assume
A11:  rng f is natural-functions-membered;
      let x;
      assume x in dom f;
      then f.x in rng f by FUNCT_1:def 3;
      hence thesis by A11;
    end;
    assume
A12: for x being set st x in dom f holds f.x is natural-valued Function;
    let y be set;
    assume y in rng f;
    then ex x being set st x in dom f & f.x = y by FUNCT_1:def 3;
    hence thesis by A12;
  end;
end;

registration
  cluster natural-functions-valued -> integer-functions-valued for Relation;
  coherence
  proof
    let R be Relation such that
A1: rng R is natural-functions-membered;
    let y be set;
    thus thesis by A1;
  end;
  cluster integer-functions-valued -> rational-functions-valued for Relation;
  coherence
  proof
    let R be Relation such that
A2: rng R is integer-functions-membered;
    let y be set;
    thus thesis by A2;
  end;
  cluster rational-functions-valued -> real-functions-valued for Relation;
  coherence
  proof
    let R be Relation such that
A3: rng R is rational-functions-membered;
    let y be set;
    thus thesis by A3;
  end;
  cluster real-functions-valued -> ext-real-functions-valued for Relation;
  coherence
  proof
    let R be Relation such that
A4: rng R is real-functions-membered;
    let y be set;
    thus thesis by A4;
  end;
  cluster real-functions-valued -> complex-functions-valued for Relation;
  coherence
  proof
    let R be Relation such that
A5: rng R is real-functions-membered;
    let y be set;
    thus thesis by A5;
  end;
end;

registration
  cluster empty -> natural-functions-valued for Relation;
  coherence
  proof
    let X be Relation such that
A1: X is empty;
    let x;
    thus thesis by A1;
  end;
end;

registration
  cluster natural-functions-valued for Function;
  existence
  proof
    take {};
    thus thesis;
  end;
end;

registration
  let R be complex-functions-valued Relation;
  cluster rng R -> complex-functions-membered;
  coherence by Def20;
end;

registration
  let R be ext-real-functions-valued Relation;
  cluster rng R -> ext-real-functions-membered;
  coherence by Def21;
end;

registration
  let R be real-functions-valued Relation;
  cluster rng R -> real-functions-membered;
  coherence by Def22;
end;

registration
  let R be rational-functions-valued Relation;
  cluster rng R -> rational-functions-membered;
  coherence by Def23;
end;

registration
  let R be integer-functions-valued Relation;
  cluster rng R -> integer-functions-membered;
  coherence by Def24;
end;

registration
  let R be natural-functions-valued Relation;
  cluster rng R -> natural-functions-membered;
  coherence by Def25;
end;

registration
  let X;
  let Y be complex-functions-membered set;
  cluster -> complex-functions-valued for PartFunc of X,Y;
  coherence;
end;

registration
  let X;
  let Y be ext-real-functions-membered set;
  cluster -> ext-real-functions-valued for PartFunc of X,Y;
  coherence
  proof
    let f be PartFunc of X,Y;
    let x;
    thus thesis;
  end;
end;

registration
  let X;
  let Y be real-functions-membered set;
  cluster -> real-functions-valued for PartFunc of X,Y;
  coherence
  proof
    let f be PartFunc of X,Y;
    let x;
    thus thesis;
  end;
end;

registration
  let X;
  let Y be rational-functions-membered set;
  cluster -> rational-functions-valued for PartFunc of X,Y;
  coherence
  proof
    let f be PartFunc of X,Y;
    let x;
    thus thesis;
  end;
end;

registration
  let X;
  let Y be integer-functions-membered set;
  cluster -> integer-functions-valued for PartFunc of X,Y;
  coherence
  proof
    let f be PartFunc of X,Y;
    let x;
    thus thesis;
  end;
end;

registration
  let X;
  let Y be natural-functions-membered set;
  cluster -> natural-functions-valued for PartFunc of X,Y;
  coherence
  proof
    let f be PartFunc of X,Y;
    let x;
    thus thesis;
  end;
end;

registration
  let f be complex-functions-valued Function;
  let x;
  cluster f.x -> Function-like Relation-like;
  coherence
  proof
    per cases;
    suppose
      x in dom f;
      then f.x in rng f by FUNCT_1:def 3;
      hence thesis;
    end;
    suppose
      not x in dom f;
      hence thesis by FUNCT_1:def 2;
    end;
  end;
end;

registration
  let f be ext-real-functions-valued Function;
  let x;
  cluster f.x -> Function-like Relation-like;
  coherence
  proof
    per cases;
    suppose
      x in dom f;
      then f.x in rng f by FUNCT_1:def 3;
      hence thesis;
    end;
    suppose
      not x in dom f;
      hence thesis by FUNCT_1:def 2;
    end;
  end;
end;

registration
  let f be complex-functions-valued Function;
  let x;
  cluster f.x -> complex-valued;
  coherence
  proof
    per cases;
    suppose
      x in dom f;
      then f.x in rng f by FUNCT_1:def 3;
      hence thesis;
    end;
    suppose
      not x in dom f;
      hence thesis by FUNCT_1:def 2;
    end;
  end;
end;

registration
  let f be ext-real-functions-valued Function;
  let x;
  cluster f.x -> ext-real-valued;
  coherence
  proof
    per cases;
    suppose
      x in dom f;
      then f.x in rng f by FUNCT_1:def 3;
      hence thesis;
    end;
    suppose
      not x in dom f;
      hence thesis by FUNCT_1:def 2;
    end;
  end;
end;

registration
  let f be real-functions-valued Function;
  let x;
  cluster f.x -> real-valued;
  coherence
  proof
    per cases;
    suppose
      x in dom f;
      then f.x in rng f by FUNCT_1:def 3;
      hence thesis;
    end;
    suppose
      not x in dom f;
      hence thesis by FUNCT_1:def 2;
    end;
  end;
end;

registration
  let f be rational-functions-valued Function;
  let x;
  cluster f.x -> RAT-valued;
  coherence
  proof
    per cases;
    suppose
      x in dom f;
      then f.x in rng f by FUNCT_1:def 3;
      hence thesis;
    end;
    suppose
      not x in dom f;
      hence thesis by FUNCT_1:def 2;
    end;
  end;
end;

registration
  let f be integer-functions-valued Function;
  let x;
  cluster f.x -> INT-valued;
  coherence
  proof
    per cases;
    suppose
      x in dom f;
      then f.x in rng f by FUNCT_1:def 3;
      hence thesis;
    end;
    suppose
      not x in dom f;
      hence thesis by FUNCT_1:def 2;
    end;
  end;
end;

registration
  let f be natural-functions-valued Function;
  let x;
  cluster f.x -> natural-valued;
  coherence
  proof
    per cases;
    suppose
      x in dom f;
      then f.x in rng f by FUNCT_1:def 3;
      hence thesis;
    end;
    suppose
      not x in dom f;
      hence thesis by FUNCT_1:def 2;
    end;
  end;
end;

begin 

reserve Y, Y1, Y2 for complex-functions-membered set,
  c, c1, c2 for complex number,
  f for PartFunc of X,Y,
  f1 for PartFunc of X1,Y1,
  f2 for PartFunc of X2, Y2,
  g, h, k for complex-valued Function;

theorem Th7:
  g <> {} & g + c1 = g + c2 implies c1 = c2
proof
  assume that
A1: g <> {} and
A2: g+c1 = g+c2;
  consider x such that
A3: x in dom g by A1,XBOOLE_0:def 1;
  dom g = dom(g+c2) by VALUED_1:def 2;
  then
A4: (g+c2).x = g.x+c2 by A3,VALUED_1:def 2;
  dom g = dom(g+c1) by VALUED_1:def 2;
  then (g+c1).x = g.x+c1 by A3,VALUED_1:def 2;
  hence c1 = c2 by A2,A4;
end;

theorem Th8:
  g <> {} & g - c1 = g - c2 implies c1 = c2
proof
  assume that
A1: g <> {} and
A2: g-c1 = g-c2;
  consider x such that
A3: x in dom g by A1,XBOOLE_0:def 1;
  dom g = dom(g-c2) by VALUED_1:def 2;
  then
A4: (g-c2).x = g.x-c2 by A3,VALUED_1:def 2;
  dom g = dom(g-c1) by VALUED_1:def 2;
  then (g-c1).x = g.x-c1 by A3,VALUED_1:def 2;
  hence c1 = c2 by A2,A4;
end;

theorem Th9:
  g <> {} & g is non-empty & g (#) c1 = g (#) c2 implies c1 = c2
proof
  assume that
A1: g <> {} and
A2: g is non-empty and
A3: g(#)c1 = g(#)c2;
  consider x such that
A4: x in dom g by A1,XBOOLE_0:def 1;
  g.x in rng g by A4,FUNCT_1:def 3;
  then
A5: g.x <> {} by A2,RELAT_1:def 9;
  (g(#)c1).x = g.x*c1 & (g(#)c2).x = g.x*c2 by VALUED_1:6;
  hence c1 = c2 by A3,A5,XCMPLX_1:5;
end;

theorem Th10:
  - (g + c) = -g - c
proof
A1: dom -(g+c) = dom(g+c) by VALUED_1:8;
A2: dom(g+c) = dom g & dom(-g-c) = dom -g by VALUED_1:def 2;
  hence dom -(g+c) = dom(-g-c) by A1,VALUED_1:8;
  let x;
  assume
A3: x in dom -(g+c);
A4: dom -g = dom g by VALUED_1:8;
  thus (-(g+c)).x = -(g+c).x by VALUED_1:8
    .= -(g.x+c) by A1,A3,VALUED_1:def 2
    .= -g.x-c
    .= (-g).x-c by VALUED_1:8
    .= (-g-c).x by A2,A1,A4,A3,VALUED_1:def 2;
end;

theorem Th11:
  - (g - c) = -g + c
proof
A1: dom -(g-c) = dom(g-c) by VALUED_1:8;
A2: dom(g-c) = dom g & dom(-g+c) = dom -g by VALUED_1:def 2;
  hence dom -(g-c) = dom(-g+c) by A1,VALUED_1:8;
  let x;
  assume
A3: x in dom -(g-c);
A4: dom -g = dom g by VALUED_1:8;
  thus (-(g-c)).x = -(g-c).x by VALUED_1:8
    .= -(g.x-c) by A1,A3,VALUED_1:def 2
    .= -g.x+c
    .= (-g).x+c by VALUED_1:8
    .= (-g+c).x by A2,A1,A4,A3,VALUED_1:def 2;
end;

theorem Th12:
  (g + c1) + c2 = g + (c1 + c2)
proof
A1: dom((g+c1)+c2) = dom(g+c1) by VALUED_1:def 2;
A2: dom(g+c1) = dom g by VALUED_1:def 2;
  hence dom((g+c1)+c2) = dom(g+(c1+c2)) by A1,VALUED_1:def 2;
  let x;
A3: dom(g+(c1+c2)) = dom(g) by VALUED_1:def 2;
  assume
A4: x in dom((g+c1)+c2);
  hence ((g+c1)+c2).x = (g+c1).x+c2 by VALUED_1:def 2
    .= g.x+c1+c2 by A1,A4,VALUED_1:def 2
    .= g.x+(c1+c2)
    .= (g+(c1+c2)).x by A1,A2,A3,A4,VALUED_1:def 2;
end;

theorem Th13:
  (g + c1) - c2 = g + (c1 - c2)
proof
A1: dom((g+c1)-c2) = dom(g+c1) by VALUED_1:def 2;
A2: dom(g+c1) = dom g by VALUED_1:def 2;
  hence dom((g+c1)-c2) = dom(g+(c1-c2)) by A1,VALUED_1:def 2;
  let x;
A3: dom(g+(c1-c2)) = dom(g) by VALUED_1:def 2;
  assume
A4: x in dom((g+c1)-c2);
  hence ((g+c1)-c2).x = (g+c1).x-c2 by VALUED_1:def 2
    .= g.x+c1-c2 by A1,A4,VALUED_1:def 2
    .= g.x+(c1-c2)
    .= (g+(c1-c2)).x by A1,A2,A3,A4,VALUED_1:def 2;
end;

theorem Th14:
  (g - c1) + c2 = g - (c1 - c2)
proof
A1: dom((g-c1)+c2) = dom(g-c1) by VALUED_1:def 2;
A2: dom(g-c1) = dom g by VALUED_1:def 2;
  hence dom((g-c1)+c2) = dom(g-(c1-c2)) by A1,VALUED_1:def 2;
  let x;
A3: dom(g-(c1-c2)) = dom(g) by VALUED_1:def 2;
  assume
A4: x in dom((g-c1)+c2);
  hence ((g-c1)+c2).x = (g-c1).x+c2 by VALUED_1:def 2
    .= g.x-c1+c2 by A1,A4,VALUED_1:def 2
    .= g.x-(c1-c2)
    .= (g-(c1-c2)).x by A1,A2,A3,A4,VALUED_1:def 2;
end;

theorem Th15:
  (g - c1) - c2 = g - (c1 + c2)
proof
A1: dom((g-c1)-c2) = dom(g-c1) by VALUED_1:def 2;
A2: dom(g-c1) = dom g by VALUED_1:def 2;
  hence dom((g-c1)-c2) = dom(g-(c1+c2)) by A1,VALUED_1:def 2;
  let x;
A3: dom(g-(c1+c2)) = dom(g) by VALUED_1:def 2;
  assume
A4: x in dom((g-c1)-c2);
  hence ((g-c1)-c2).x = (g-c1).x-c2 by VALUED_1:def 2
    .= g.x-c1-c2 by A1,A4,VALUED_1:def 2
    .= g.x-(c1+c2)
    .= (g-(c1+c2)).x by A1,A2,A3,A4,VALUED_1:def 2;
end;

theorem Th16:
  g (#) c1 (#) c2 = g (#) (c1 * c2)
proof
  dom((g(#)c1)(#)c2) = dom(g(#)c1) & dom(g(#)c1) = dom g by VALUED_1:def 5;
  hence dom((g(#)c1)(#)c2) = dom(g(#)(c1*c2)) by VALUED_1:def 5;
  let x;
  assume x in dom((g(#)c1)(#)c2);
  thus ((g(#)c1)(#)c2).x = (g(#)c1).x*c2 by VALUED_1:6
    .= g.x*c1*c2 by VALUED_1:6
    .= g.x*(c1*c2)
    .= (g(#)(c1*c2)).x by VALUED_1:6;
end;

theorem Th17:
  - (g + h) = (-g) - h
proof
A1: dom -(g+h) = dom(g+h) by VALUED_1:8;
  dom(g+h) = dom g /\ dom h & dom(-g-h) = dom(-g) /\ dom h by VALUED_1:12,def 1
;
  hence
A2: dom -(g+h) = dom(-g-h) by A1,VALUED_1:8;
  let x;
  assume
A3: x in dom -(g+h);
  thus (-(g+h)).x = -(g+h).x by VALUED_1:8
    .= -(g.x+h.x) by A1,A3,VALUED_1:def 1
    .= -g.x-h.x
    .= (-g).x-h.x by VALUED_1:8
    .= (-g-h).x by A2,A3,VALUED_1:13;
end;

theorem Th18:
  g - h = - (h - g)
proof
A1: dom -(h-g) = dom(h-g) by VALUED_1:8;
  dom(g-h) = dom g /\ dom h by VALUED_1:12;
  hence
A2: dom(g-h) = dom -(h-g) by A1,VALUED_1:12;
  let x;
  assume
A3: x in dom(g-h);
  hence (g-h).x = g.x-h.x by VALUED_1:13
    .= -(h.x-g.x)
    .= -(h-g).x by A1,A2,A3,VALUED_1:13
    .= (-(h-g)).x by VALUED_1:8;
end;

theorem Th19:
  g (#) h /" k = g (#) (h /" k)
proof
A1: dom(g (#) (h /" k)) = dom g /\ dom(h /" k) & dom(g (#) h /" k) = dom(g
  (#) h ) /\ dom k by VALUED_1:16,def 4;
  dom(g (#) h) = dom g /\ dom h & dom(h /" k) = dom h /\ dom k by VALUED_1:16
,def 4;
  hence dom(g (#) h /" k) = dom(g (#) (h /" k)) by A1,XBOOLE_1:16;
  let x;
  assume x in dom(g (#) h /" k);
  thus (g (#) h /" k).x = (g (#) h).x / k.x by VALUED_1:17
    .= g.x * h.x / k.x by VALUED_1:5
    .= g.x * (h.x / k.x)
    .= g.x * (h /" k).x by VALUED_1:17
    .= (g (#) (h /" k)).x by VALUED_1:5;
end;

theorem Th20:
  g /" h (#) k = g (#) k /" h
proof
A1: dom(g /" h (#) k) = dom(g /" h) /\ dom k & dom(g (#) k /" h) = dom(g (#)
  k) /\ dom h by VALUED_1:16,def 4;
  dom(g /" h) = dom g /\ dom h & dom(g (#) k) = dom g /\ dom k by VALUED_1:16
,def 4;
  hence dom(g /" h (#) k) = dom(g (#) k /" h) by A1,XBOOLE_1:16;
  let x;
  assume x in dom(g /" h (#) k);
  thus (g /" h (#) k).x = (g /" h).x * k.x by VALUED_1:5
    .= g.x / h.x * k.x by VALUED_1:17
    .= g.x * k.x / h.x
    .= (g(#)k).x / h.x by VALUED_1:5
    .= (g (#) k /" h).x by VALUED_1:17;
end;

theorem Th21:
  g /" h /" k = g /" (h (#) k)
proof
A1: dom(g /" h /" k) = dom(g /" h) /\ dom k & dom(g /" (h (#) k)) = dom g /\
  dom (h (#) k) by VALUED_1:16;
  dom(g /" h) = dom g /\ dom h & dom(h (#) k) = dom h /\ dom k by VALUED_1:16
,def 4;
  hence dom(g /" h /" k) = dom(g /" (h (#) k)) by A1,XBOOLE_1:16;
  let x;
  assume x in dom(g /" h /" k);
  thus (g /" h /" k).x = (g /" h).x / k.x by VALUED_1:17
    .= g.x / h.x / k.x by VALUED_1:17
    .= g.x / (h.x * k.x) by XCMPLX_1:78
    .= g.x / (h (#) k).x by VALUED_1:5
    .= (g /" (h (#) k)).x by VALUED_1:17;
end;

theorem Th22:
  c(#)-g = (-c)(#)g
proof
  dom(c(#)-g) = dom -g by VALUED_1:def 5
    .= dom g by VALUED_1:8;
  hence dom(c(#)-g) = dom((-c)(#)g) by VALUED_1:def 5;
  let x;
  assume x in dom(c(#)-g);
  thus (c(#)-g).x = c*((-g).x) by VALUED_1:6
    .= c*(-g.x) by VALUED_1:8
    .= (-c)*g.x
    .= ((-c)(#)g).x by VALUED_1:6;
end;

theorem Th23:
  c(#)-g = -(c(#)g)
proof
A1: dom(-(c(#)g)) = dom(c(#)g) by VALUED_1:8
    .= dom g by VALUED_1:def 5;
  dom(c(#)-g) = dom -g by VALUED_1:def 5
    .= dom g by VALUED_1:8;
  hence dom(c(#)-g) = dom -(c(#)g) by A1;
  let x;
  assume x in dom(c(#)-g);
  thus (c(#)-g).x = c*((-g).x) by VALUED_1:6
    .= c*(-g.x) by VALUED_1:8
    .= -(c*g.x)
    .= -(c(#)g).x by VALUED_1:6
    .= (-(c(#)g)).x by VALUED_1:8;
end;

theorem Th24:
  (-c)(#)g = -(c(#)g)
proof
  thus (-c)(#)g = c(#)-g by Th22
    .= -(c(#)g) by Th23;
end;

theorem Th25:
  - (g (#) h) = (-g) (#) h
proof
A1: dom -(g(#)h) = dom(g(#)h) by VALUED_1:8;
  dom(g(#)h) = dom g /\ dom h & dom((-g)(#)h) = dom(-g) /\ dom h by
VALUED_1:def 4;
  hence dom -(g(#)h) = dom((-g)(#)h) by A1,VALUED_1:8;
  let x;
  assume x in dom -(g(#)h);
  thus (-(g(#)h)).x = -(g(#)h).x by VALUED_1:8
    .= -(g.x*h.x) by VALUED_1:5
    .= (-g.x)*h.x
    .= (-g).x*h.x by VALUED_1:8
    .= ((-g)(#)h).x by VALUED_1:5;
end;

theorem
  - (g /" h) = (-g) /" h
proof
A1: dom -(g/"h) = dom(g/"h) by VALUED_1:8;
  dom(g/"h) = dom g /\ dom h & dom((-g)/"h) = dom(-g) /\ dom h by VALUED_1:16;
  hence dom -(g/"h) = dom((-g)/"h) by A1,VALUED_1:8;
  let x;
  assume x in dom -(g/"h);
  thus (-(g/"h)).x = -(g/"h).x by VALUED_1:8
    .= -(g.x/h.x) by VALUED_1:17
    .= (-g.x)/h.x
    .= (-g).x/h.x by VALUED_1:8
    .= ((-g)/"h).x by VALUED_1:17;
end;

theorem Th27:
  - (g /" h) = g /" -h
proof
A1: dom -h = dom h by VALUED_1:8;
  dom(g/"h) = dom g /\ dom h & dom(g/"-h) = dom g /\ dom -h by VALUED_1:16;
  hence dom -(g/"h) = dom(g/"-h) by A1,VALUED_1:8;
  let x;
  assume x in dom -(g/"h);
  thus (-(g/"h)).x = -(g/"h).x by VALUED_1:8
    .= -(g.x/h.x) by VALUED_1:17
    .= g.x/-h.x by XCMPLX_1:188
    .= g.x/(-h).x by VALUED_1:8
    .= (g/"-h).x by VALUED_1:17;
end;

definition
  let f be complex-valued Function, c be complex number;
  func f (/) c -> Function equals
  (1/c) (#) f;
  coherence;
end;

registration
  let f be complex-valued Function, c be complex number;
  cluster f (/) c -> complex-valued;
  coherence;
end;

registration
  let f be real-valued Function, r be real number;
  cluster f (/) r -> real-valued;
  coherence;
end;

registration
  let f be RAT-valued Function, r be rational number;
  cluster f (/) r -> RAT-valued;
  coherence;
end;

registration
  let f be complex-valued FinSequence, c be complex number;
  cluster f (/) c -> FinSequence-like;
  coherence;
end;

theorem
  dom(g(/)c) = dom g by VALUED_1:def 5;

theorem
  (g(/)c).x = g.x / c by VALUED_1:6;

theorem Th30:
  (-g) (/) c = -(g(/)c)
proof
  thus (-g) (/) c = (-(1/c)) (#) g by Th22
    .= -(g(/)c) by Th24;
end;

theorem Th31:
  g (/) -c = -(g(/)c)
proof
  thus g (/) -c = (-1/c) (#) g by XCMPLX_1:188
    .= -(g(/)c) by Th24;
end;

theorem
  g (/) -c = (-g) (/) c
proof
  thus g (/) -c = - (g(/)c) by Th31
    .= (-g) (/) c by Th30;
end;

theorem Th33:
  g <> {} & g is non-empty & g (/) c1 = g (/) c2 implies c1 = c2
proof
  assume that
A1: g <> {} and
A2: g is non-empty and
A3: g(/)c1 = g(/)c2;
  consider x such that
A4: x in dom g by A1,XBOOLE_0:def 1;
  g.x in rng g by A4,FUNCT_1:def 3;
  then
A5: g.x <> {} by A2,RELAT_1:def 9;
  (g(/)c1).x = g.x/c1 & (g(/)c2).x = g.x/c2 by VALUED_1:6;
  then c1" = c2" by A3,A5,XCMPLX_1:5;
  hence c1 = c2 by XCMPLX_1:201;
end;

theorem
  g (#) c1 (/) c2 = g (#) (c1 / c2)
proof
  dom(g(#)c1) = dom g & dom(g(#)c1(/)c2) = dom(g(#)c1) by VALUED_1:def 5;
  hence dom(g(#)c1(/)c2) = dom(g(#)(c1/c2)) by VALUED_1:def 5;
  let x;
  assume x in dom(g(#)c1(/)c2);
  thus (g(#)c1(/)c2).x = (g(#)c1).x * c2" by VALUED_1:6
    .= g.x * c1 * c2" by VALUED_1:6
    .= g.x*(c1/c2)
    .= (g(#)(c1/c2)).x by VALUED_1:6;
end;

theorem
  g (/) c1 (#) c2 = g (#) c2 (/) c1
proof
A1: dom(g(/)c1(#)c2) = dom(g(/)c1) by VALUED_1:def 5;
  dom(g(/)c1) = dom g & dom(g(#)c2) = dom g by VALUED_1:def 5;
  hence dom(g(/)c1(#)c2) = dom(g(#)c2(/)c1) by A1,VALUED_1:def 5;
  let x;
  assume x in dom(g(/)c1(#)c2);
  thus (g(/)c1(#)c2).x = (g(/)c1).x * c2 by VALUED_1:6
    .= g.x * c1" * c2 by VALUED_1:6
    .= (g.x*c2)*c1"
    .= (g(#)c2).x*c1" by VALUED_1:6
    .= (g(#)c2(/)c1).x by VALUED_1:6;
end;

theorem
  g (/) c1 (/) c2 = g (/) (c1*c2)
proof
  dom(g(/)c1) = dom g & dom(g(/)(c1*c2)) = dom g by VALUED_1:def 5;
  hence dom(g(/)c1(/)c2) = dom(g(/)(c1*c2)) by VALUED_1:def 5;
  let x;
  assume x in dom(g(/)c1(/)c2);
  thus (g(/)c1(/)c2).x = (g(/)c1).x * c2" by VALUED_1:6
    .= g.x * c1" * c2" by VALUED_1:6
    .= g.x * (c1" * c2")
    .= g.x * (c1*c2)" by XCMPLX_1:204
    .= (g(/)(c1*c2)).x by VALUED_1:6;
end;

theorem
  (g+h) (/) c = g(/)c + h(/)c
proof
A1: dom((g+h)(/)c) = dom(g+h) by VALUED_1:def 5;
A2: dom(g+h) = dom g /\ dom h by VALUED_1:def 1;
  dom(g(/)c) = dom g & dom(h(/)c) = dom h by VALUED_1:def 5;
  hence
A3: dom((g+h)(/)c) = dom(g(/)c+h(/)c) by A1,A2,VALUED_1:def 1;
  let x;
  assume
A4: x in dom((g+h)(/)c);
  thus ((g+h)(/)c).x = (g+h).x * c" by VALUED_1:6
    .= (g.x+h.x)*c" by A1,A4,VALUED_1:def 1
    .= g.x*c" + h.x*c"
    .= (g(/)c).x + h.x*c" by VALUED_1:6
    .= (g(/)c).x + (h(/)c).x by VALUED_1:6
    .= (g(/)c+h(/)c).x by A3,A4,VALUED_1:def 1;
end;

theorem
  (g-h) (/) c = g(/)c - h(/)c
proof
A1: dom((g-h)(/)c) = dom(g-h) by VALUED_1:def 5;
A2: dom(g-h) = dom g /\ dom h by VALUED_1:12;
  dom(g(/)c) = dom g & dom(h(/)c) = dom h by VALUED_1:def 5;
  hence
A3: dom((g-h)(/)c) = dom(g(/)c-h(/)c) by A1,A2,VALUED_1:12;
  let x;
  assume
A4: x in dom((g-h)(/)c);
  thus ((g-h)(/)c).x = (g-h).x * c" by VALUED_1:6
    .= (g.x-h.x)*c" by A1,A4,VALUED_1:13
    .= g.x*c" - h.x*c"
    .= (g(/)c).x - h.x*c" by VALUED_1:6
    .= (g(/)c).x - (h(/)c).x by VALUED_1:6
    .= (g(/)c-h(/)c).x by A3,A4,VALUED_1:13;
end;

theorem
  (g(#)h) (/) c = g (#) (h(/)c)
proof
A1: dom((g(#)h)(/)c) = dom(g(#)h) by VALUED_1:def 5;
  dom(g(#)h) = dom g /\ dom h & dom(h(/)c) = dom h by VALUED_1:def 4,def 5;
  hence dom((g(#)h)(/)c) = dom(g(#)(h(/)c)) by A1,VALUED_1:def 4;
  let x;
  assume x in dom((g(#)h)(/)c);
  thus ((g(#)h)(/)c).x = (g(#)h).x * c" by VALUED_1:6
    .= g.x*h.x*c" by VALUED_1:5
    .= g.x*(h.x*c")
    .= g.x * (h(/)c).x by VALUED_1:6
    .= (g(#)(h(/)c)).x by VALUED_1:5;
end;

theorem
  (g/"h) (/) c = g /" (h(#)c)
proof
A1: dom((g/"h)(/)c) = dom(g/"h) by VALUED_1:def 5;
  dom(g/"h) = dom g /\ dom h & dom(h(#)c) = dom h by VALUED_1:16,def 5;
  hence dom((g/"h)(/)c) = dom(g/"(h(#)c)) by A1,VALUED_1:16;
  let x;
  assume x in dom((g/"h)(/)c);
  thus ((g/"h)(/)c).x = (g/"h).x * c" by VALUED_1:6
    .= g.x/h.x/c by VALUED_1:17
    .= g.x/(h.x*c) by XCMPLX_1:78
    .= g.x / (h(#)c).x by VALUED_1:6
    .= (g/"(h(#)c)).x by VALUED_1:17;
end;

definition



  let f be complex-functions-valued Function;
  deffunc F(set) = -f.$1;
  func <->f -> Function means
  :Def33:
  dom it = dom f & for x being set st x in dom it holds it.x = -f.x;
  existence
  proof
    ex F being Function st dom F = dom f & for x being set st x in dom f
    holds F.x = F(x) from FUNCT_1:sch 3;
    hence thesis;
  end;
  uniqueness
  proof
    let F,G be Function such that
A1: dom F = dom f and
A2: for x being set st x in dom F holds F.x = F(x) and
A3: dom G = dom f and
A4: for x being set st x in dom G holds G.x = F(x);
    thus dom F = dom G by A1,A3;
    let x;
    assume
A5: x in dom F;
    hence F.x = F(x) by A2
      .= G.x by A1,A3,A4,A5;
  end;
end;

definition
  let X;
  let Y be complex-functions-membered set;
  let f be PartFunc of X,Y;
  redefine func <->f -> PartFunc of X, C_PFuncs(DOMS(Y));
  coherence
  proof
    set h = <->f;
A1: dom h = dom f by Def33;
    rng h c= C_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
A4:   h.x = -f.x by A2,Def33;
      then reconsider y as Function by A3;
A5:   rng y c= COMPLEX
      proof
        let b be set;
        thus thesis by A3,A4,XCMPLX_0:def 2;
      end;
      f.x in Y by A1,A2,PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:8;
      then y is PartFunc of DOMS(Y),COMPLEX by A6,A5,RELSET_1:4;
      hence thesis by Def8;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X;
  let Y be real-functions-membered set;
  let f be PartFunc of X,Y;
  redefine func <->f -> PartFunc of X, R_PFuncs(DOMS(Y));
  coherence
  proof
    set h = <->f;
A1: dom h = dom f by Def33;
    rng h c= R_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = -f.x by A2,Def33;
A5:   rng y c= REAL
      proof
        let b be set;
        thus thesis by A3,A4,XREAL_0:def 1;
      end;
      f.x in Y by A1,A2,PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:8;
      then y is PartFunc of DOMS(Y),REAL by A6,A5,RELSET_1:4;
      hence thesis by Def12;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X;
  let Y be rational-functions-membered set;
  let f be PartFunc of X,Y;
  redefine func <->f -> PartFunc of X, Q_PFuncs(DOMS(Y));
  coherence
  proof
    set h = <->f;
A1: dom h = dom f by Def33;
    rng h c= Q_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = -f.x by A2,Def33;
A5:   rng y c= RAT
      proof
        let b be set;
        thus thesis by A3,A4,RAT_1:def 2;
      end;
      f.x in Y by A1,A2,PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:8;
      then y is PartFunc of DOMS(Y),RAT by A6,A5,RELSET_1:4;
      hence thesis by Def14;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X;
  let Y be integer-functions-membered set;
  let f be PartFunc of X,Y;
  redefine func <->f -> PartFunc of X, I_PFuncs(DOMS(Y));
  coherence
  proof
    set h = <->f;
A1: dom h = dom f by Def33;
    rng h c= I_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = -f.x by A2,Def33;
A5:   rng y c= INT
      proof
        let b be set;
        thus thesis by A3,A4,INT_1:def 2;
      end;
      f.x in Y by A1,A2,PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:8;
      then y is PartFunc of DOMS(Y),INT by A6,A5,RELSET_1:4;
      hence thesis by Def16;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

registration
  let Y be complex-functions-membered set;
  let f be FinSequence of Y;
  cluster <->f -> FinSequence-like;
  coherence
  proof
    dom<->f = dom f & ex n being Nat st dom f = Seg n by Def33,FINSEQ_1:def 2;
    hence thesis by FINSEQ_1:def 2;
  end;
end;

theorem
  <-><->f = f
proof
  set f1 = <->f;
A1: dom f1 = dom f by Def33;
  hence
A2: dom<->f1 = dom f by Def33;
  let x;
  assume
A3: x in dom<->f1;
  hence (<->f1).x = -f1.x by Def33
    .= -(-f.x) by A1,A2,A3,Def33
    .= f.x;
end;

theorem
  <->f1 = <->f2 implies f1 = f2
proof
A1: dom <->f1 = dom f1 by Def33;
  assume
A2: <->f1 = <->f2;
  hence dom f1 = dom f2 by A1,Def33;
  let x;
  assume
A3: x in dom f1;
  thus f1.x = --f1.x .= -(<->f1).x by A1,A3,Def33
    .= --f2.x by A2,A1,A3,Def33
    .= f2.x;
end;

definition
  let X be complex-functions-membered set;
  let Y be set;
  let f be PartFunc of X,Y;
  defpred P[set,set] means ex a being complex-valued Function st $1 = a & $2 =
  f.-a;
  func f(-) -> Function means
  dom it = dom f & for x being complex-valued
  Function st x in dom it holds it.x = f.-x;
  existence
  proof
A1: for x being set st x in dom f ex y being set st P[x,y]
    proof
      let x;
      assume x in dom f;
      then reconsider a = x as complex-valued Function;
      take f.-a, a;
      thus thesis;
    end;
    consider F being Function such that
A2: dom F = dom f and
A3: for x being set st x in dom f holds P[x,F.x] from CLASSES1:sch 1(
    A1);
    take F;
    thus dom F = dom f by A2;
    let x be complex-valued Function;
    assume x in dom F;
    then P[x,F.x] by A2,A3;
    hence thesis;
  end;
  uniqueness
  proof
    let F, G be Function such that
A4: dom F = dom f and
A5: for x being complex-valued Function st x in dom F holds F.x = f.- x and
A6: dom G = dom f and
A7: for x being complex-valued Function st x in dom G holds G.x = f.- x;
    thus dom F = dom G by A4,A6;
    let x;
    assume
A8: x in dom F;
    then reconsider y = x as complex-valued Function by A4;
    thus F.x = f.-y by A5,A8
      .= G.x by A4,A6,A7,A8;
  end;
end;

definition


  let f be complex-functions-valued Function;
  deffunc F(set) = (f.$1)";
  func </>f -> Function means
  :Def35:
  dom it = dom f & for x being set st x in dom it holds it.x = (f.x)";
  existence
  proof
    ex F being Function st dom F = dom f & for x being set st x in dom f
    holds F.x = F(x) from FUNCT_1:sch 3;
    hence thesis;
  end;
  uniqueness
  proof
    let F,G be Function such that
A1: dom F = dom f and
A2: for x being set st x in dom F holds F.x = F(x) and
A3: dom G = dom f and
A4: for x being set st x in dom G holds G.x = F(x);
    thus dom F = dom G by A1,A3;
    let x;
    assume
A5: x in dom F;
    hence F.x = F(x) by A2
      .= G.x by A1,A3,A4,A5;
  end;
end;

definition
  let X;
  let Y be complex-functions-membered set;
  let f be PartFunc of X,Y;
  redefine func </>f -> PartFunc of X, C_PFuncs(DOMS(Y));
  coherence
  proof
    set h = </>f;
A1: dom h = dom f by Def35;
    rng h c= C_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
A4:   h.x = (f.x)" by A2,Def35;
      then reconsider y as Function by A3;
A5:   rng y c= COMPLEX
      proof
        let b be set;
        thus thesis by A3,A4,XCMPLX_0:def 2;
      end;
      f.x in Y by A1,A2,PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:def 7;
      then y is PartFunc of DOMS(Y),COMPLEX by A6,A5,RELSET_1:4;
      hence thesis by Def8;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X;
  let Y be real-functions-membered set;
  let f be PartFunc of X,Y;
  redefine func </>f -> PartFunc of X, R_PFuncs(DOMS(Y));
  coherence
  proof
    set h = </>f;
A1: dom h = dom f by Def35;
    rng h c= R_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = (f.x)" by A2,Def35;
A5:   rng y c= REAL
      proof
        let b be set;
        thus thesis by A3,A4,XREAL_0:def 1;
      end;
      f.x in Y by A1,A2,PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:def 7;
      then y is PartFunc of DOMS(Y),REAL by A6,A5,RELSET_1:4;
      hence thesis by Def12;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X;
  let Y be rational-functions-membered set;
  let f be PartFunc of X,Y;
  redefine func </>f -> PartFunc of X, Q_PFuncs(DOMS(Y));
  coherence
  proof
    set h = </>f;
A1: dom h = dom f by Def35;
    rng h c= Q_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = (f.x)" by A2,Def35;
A5:   rng y c= RAT
      proof
        let b be set;
        thus thesis by A3,A4,RAT_1:def 2;
      end;
      f.x in Y by A1,A2,PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:def 7;
      then y is PartFunc of DOMS(Y),RAT by A6,A5,RELSET_1:4;
      hence thesis by Def14;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

registration
  let Y be complex-functions-membered set;
  let f be FinSequence of Y;
  cluster </>f -> FinSequence-like;
  coherence
  proof
    dom </>f = dom f & ex n being Nat st dom f = Seg n by Def35,FINSEQ_1:def 2;
    hence thesis by FINSEQ_1:def 2;
  end;
end;

theorem
  </></>f = f
proof
  set f1 = </>f;
A1: dom f1 = dom f by Def35;
  hence
A2: dom </>f1 = dom f by Def35;
  let x;
  assume
A3: x in dom </>f1;
  hence (</>f1).x = (f1.x)" by Def35
    .= (f.x)"" by A1,A2,A3,Def35
    .= f.x;
end;

definition


  let f be complex-functions-valued Function;
  deffunc F(set) = abs(f.$1);
  func abs(f) -> Function means
  :Def36:
  dom it = dom f & for x being set st x in dom it holds it.x = abs(f.x);
  existence
  proof
    ex F being Function st dom F = dom f & for x being set st x in dom f
    holds F.x = F(x) from FUNCT_1:sch 3;
    hence thesis;
  end;
  uniqueness
  proof
    let F,G be Function such that
A1: dom F = dom f and
A2: for x being set st x in dom F holds F.x = F(x) and
A3: dom G = dom f and
A4: for x being set st x in dom G holds G.x = F(x);
    thus dom F = dom G by A1,A3;
    let x;
    assume
A5: x in dom F;
    hence F.x = F(x) by A2
      .= G.x by A1,A3,A4,A5;
  end;
end;

definition
  let X;
  let Y be complex-functions-membered set;
  let f be PartFunc of X,Y;
  redefine func abs(f) -> PartFunc of X, C_PFuncs(DOMS(Y));
  coherence
  proof
    set h = abs(f);
A1: dom h = dom f by Def36;
    rng h c= C_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
A4:   h.x = abs(f.x) by A2,Def36;
      then reconsider y as Function by A3;
A5:   rng y c= COMPLEX
      proof
        let b be set;
        thus thesis by A3,A4,XCMPLX_0:def 2;
      end;
      f.x in Y by A1,A2,PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:def 11;
      then y is PartFunc of DOMS(Y),COMPLEX by A6,A5,RELSET_1:4;
      hence thesis by Def8;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X;
  let Y be real-functions-membered set;
  let f be PartFunc of X,Y;
  redefine func abs(f) -> PartFunc of X, R_PFuncs(DOMS(Y));
  coherence
  proof
    set h = abs(f);
A1: dom h = dom f by Def36;
    rng h c= R_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = abs(f.x) by A2,Def36;
A5:   rng y c= REAL
      proof
        let b be set;
        thus thesis by A3,A4,XREAL_0:def 1;
      end;
      f.x in Y by A1,A2,PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:def 11;
      then y is PartFunc of DOMS(Y),REAL by A6,A5,RELSET_1:4;
      hence thesis by Def12;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X;
  let Y be rational-functions-membered set;
  let f be PartFunc of X,Y;
  redefine func abs(f) -> PartFunc of X, Q_PFuncs(DOMS(Y));
  coherence
  proof
    set h = abs(f);
A1: dom h = dom f by Def36;
    rng h c= Q_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = abs(f.x) by A2,Def36;
A5:   rng y c= RAT
      proof
        let b be set;
        thus thesis by A3,A4,RAT_1:def 2;
      end;
      f.x in Y by A1,A2,PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:def 11;
      then y is PartFunc of DOMS(Y),RAT by A6,A5,RELSET_1:4;
      hence thesis by Def14;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X;
  let Y be integer-functions-membered set;
  let f be PartFunc of X,Y;
  redefine func abs(f) -> PartFunc of X, N_PFuncs(DOMS(Y));
  coherence
  proof
    set h = abs(f);
A1: dom h = dom f by Def36;
    rng h c= N_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = abs(f.x) by A2,Def36;
A5:   rng y c= NAT
      proof
        let b be set;
        thus thesis by A3,A4,ORDINAL1:def 12;
      end;
      f.x in Y by A1,A2,PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:def 11;
      then y is PartFunc of DOMS(Y),NAT by A6,A5,RELSET_1:4;
      hence thesis by Def18;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

registration
  let Y be complex-functions-membered set;
  let f be FinSequence of Y;
  cluster abs(f) -> FinSequence-like;
  coherence
  proof
    dom abs(f) = dom f & ex n being Nat st dom f = Seg n by Def36,
FINSEQ_1:def 2;
    hence thesis by FINSEQ_1:def 2;
  end;
end;

theorem
  abs abs f = abs f
proof
  set f1 = abs f;
  thus
A1: dom abs f1 = dom abs f by Def36;
  let x;
  assume
A2: x in dom abs f1;
  hence (abs f1).x = abs(f1.x) by Def36
    .= abs(abs(f.x)) by A1,A2,Def36
    .= (abs f).x by A1,A2,Def36;
end;

definition

  let Y be complex-functions-membered set;
  let f be Y-valued Function;
  let c be complex number;
  deffunc F(set) = c+(f.$1);
  func f[+]c -> Function means
  :Def37:
  dom it = dom f & for x being set st x in dom it holds it.x = c + f.x;
  existence
  proof
    ex F being Function st dom F = dom f & for x being set st x in dom f
    holds F.x = F(x) from FUNCT_1:sch 3;
    hence thesis;
  end;
  uniqueness
  proof
    let F,G be Function such that
A1: dom F = dom f and
A2: for x being set st x in dom F holds F.x = F(x) and
A3: dom G = dom f and
A4: for x being set st x in dom G holds G.x = F(x);
    thus dom F = dom G by A1,A3;
    let x;
    assume
A5: x in dom F;
    hence F.x = F(x) by A2
      .= G.x by A1,A3,A4,A5;
  end;
end;

definition
  let X;
  let Y be complex-functions-membered set;
  let f be PartFunc of X,Y;
  let c be complex number;
  redefine func f[+]c -> PartFunc of X, C_PFuncs(DOMS(Y));
  coherence
  proof
    set h = f[+]c;
A1: dom h = dom f by Def37;
    rng h c= C_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
A4:   h.x = f.x + c by A2,Def37;
      then reconsider y as Function by A3;
A5:   rng y c= COMPLEX
      proof
        let b be set;
        thus thesis by A3,A4,XCMPLX_0:def 2;
      end;
      f.x in Y by A1,A2,PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:def 2;
      then y is PartFunc of DOMS(Y),COMPLEX by A6,A5,RELSET_1:4;
      hence thesis by Def8;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X;
  let Y be real-functions-membered set;
  let f be PartFunc of X,Y;
  let c be real number;
  redefine func f[+]c -> PartFunc of X, R_PFuncs(DOMS(Y));
  coherence
  proof
    set h = f[+]c;
A1: dom h = dom f by Def37;
    rng h c= R_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = f.x + c by A2,Def37;
A5:   rng y c= REAL
      proof
        let b be set;
        thus thesis by A3,A4,XREAL_0:def 1;
      end;
      f.x in Y by A1,A2,PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:def 2;
      then y is PartFunc of DOMS(Y),REAL by A6,A5,RELSET_1:4;
      hence thesis by Def12;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X;
  let Y be rational-functions-membered set;
  let f be PartFunc of X,Y;
  let c be rational number;
  redefine func f[+]c -> PartFunc of X, Q_PFuncs(DOMS(Y));
  coherence
  proof
    set h = f[+]c;
A1: dom h = dom f by Def37;
    rng h c= Q_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = f.x + c by A2,Def37;
A5:   rng y c= RAT
      proof
        let b be set;
        thus thesis by A3,A4,RAT_1:def 2;
      end;
      f.x in Y by A1,A2,PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:def 2;
      then y is PartFunc of DOMS(Y),RAT by A6,A5,RELSET_1:4;
      hence thesis by Def14;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X;
  let Y be integer-functions-membered set;
  let f be PartFunc of X,Y;
  let c be integer number;
  redefine func f[+]c -> PartFunc of X, I_PFuncs(DOMS(Y));
  coherence
  proof
    set h = f[+]c;
A1: dom h = dom f by Def37;
    rng h c= I_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = f.x + c by A2,Def37;
A5:   rng y c= INT
      proof
        let b be set;
        thus thesis by A3,A4,INT_1:def 2;
      end;
      f.x in Y by A1,A2,PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:def 2;
      then y is PartFunc of DOMS(Y),INT by A6,A5,RELSET_1:4;
      hence thesis by Def16;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X;
  let Y be natural-functions-membered set;
  let f be PartFunc of X,Y;
  let c be Nat;
  redefine func f[+]c -> PartFunc of X, N_PFuncs(DOMS(Y));
  coherence
  proof
    set h = f[+]c;
A1: dom h = dom f by Def37;
    rng h c= N_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = f.x + c by A2,Def37;
A5:   rng y c= NAT
      proof
        let b be set;
        thus thesis by A3,A4,ORDINAL1:def 12;
      end;
      f.x in Y by A1,A2,PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:def 2;
      then y is PartFunc of DOMS(Y),NAT by A6,A5,RELSET_1:4;
      hence thesis by Def18;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

theorem
  f [+] c1 [+] c2 = f [+] (c1+c2)
proof
  set f1 = f[+]c1;
A1: dom(f1[+]c2) = dom f1 by Def37;
  dom f1 = dom f by Def37;
  hence
A2: dom(f1[+]c2) = dom(f[+](c1+c2)) by A1,Def37;
  let x;
  assume
A3: x in dom(f1[+]c2);
  hence (f1[+]c2).x = f1.x + c2 by Def37
    .= f.x + c1 + c2 by A1,A3,Def37
    .= f.x + (c1 + c2) by Th12
    .= (f[+](c1+c2)).x by A2,A3,Def37;
end;

theorem
  f <> {} & f is non-empty & f [+] c1 = f [+] c2 implies c1 = c2
proof
  assume that
A1: f <> {} and
A2: f is non-empty and
A3: f[+]c1 = f[+]c2;
  consider x such that
A4: x in dom f by A1,XBOOLE_0:def 1;
  f.x in rng f by A4,FUNCT_1:def 3;
  then
A5: f.x <> {} by A2,RELAT_1:def 9;
  dom f = dom(f[+]c2) by Def37;
  then
A6: (f[+]c2).x = f.x+c2 by A4,Def37;
  dom f = dom(f[+]c1) by Def37;
  then (f[+]c1).x = f.x+c1 by A4,Def37;
  hence c1 = c2 by A3,A5,A6,Th7;
end;

definition

  let Y be complex-functions-membered set;
  let f be Y-valued Function;
  let c be complex number;
  func f[-]c -> Function equals
  f [+] -c;
  coherence;
end;

theorem
  dom(f[-]c) = dom f by Def37;

theorem
  x in dom(f[-]c) implies (f[-]c).x = f.x - c by Def37;

definition
  let X;
  let Y be complex-functions-membered set;
  let f be PartFunc of X,Y;
  let c be complex number;
  redefine func f[-]c -> PartFunc of X, C_PFuncs(DOMS(Y));
  coherence
  proof
    f[-]c = f [+] -c;
    hence thesis;
  end;
end;

definition
  let X;
  let Y be real-functions-membered set;
  let f be PartFunc of X,Y;
  let c be real number;
  redefine func f[-]c -> PartFunc of X, R_PFuncs(DOMS(Y));
  coherence
  proof
    f[-]c = f [+] -c;
    hence thesis;
  end;
end;

definition
  let X;
  let Y be rational-functions-membered set;
  let f be PartFunc of X,Y;
  let c be rational number;
  redefine func f[-]c -> PartFunc of X, Q_PFuncs(DOMS(Y));
  coherence
  proof
    f[-]c = f [+] -c;
    hence thesis;
  end;
end;

definition
  let X;
  let Y be integer-functions-membered set;
  let f be PartFunc of X,Y;
  let c be integer number;
  redefine func f[-]c -> PartFunc of X, I_PFuncs(DOMS(Y));
  coherence
  proof
    f[-]c = f [+] -c;
    hence thesis;
  end;
end;

theorem
  f <> {} & f is non-empty & f [-] c1 = f [-] c2 implies c1 = c2
proof
  assume that
A1: f <> {} and
A2: f is non-empty and
A3: f[-]c1 = f[-]c2;
  consider x such that
A4: x in dom f by A1,XBOOLE_0:def 1;
  f.x in rng f by A4,FUNCT_1:def 3;
  then
A5: f.x <> {} by A2,RELAT_1:def 9;
  dom f = dom(f[-]c2) by Def37;
  then
A6: (f[-]c2).x = f.x-c2 by A4,Def37;
  dom f = dom(f[-]c1) by Def37;
  then (f[-]c1).x = f.x-c1 by A4,Def37;
  hence c1 = c2 by A3,A5,A6,Th8;
end;

theorem
  f [+] c1 [-] c2 = f [+] (c1-c2)
proof
  set f1 = f[+]c1;
A1: dom(f1[-]c2) = dom f1 by Def37;
  dom f1 = dom f by Def37;
  hence
A2: dom(f1[-]c2) = dom(f[+](c1-c2)) by A1,Def37;
  let x;
  assume
A3: x in dom(f1[-]c2);
  hence (f1[-]c2).x = f1.x - c2 by Def37
    .= f.x + c1 - c2 by A1,A3,Def37
    .= f.x + (c1 - c2) by Th12
    .= (f[+](c1-c2)).x by A2,A3,Def37;
end;

theorem
  f [-] c1 [+] c2 = f [-] (c1-c2)
proof
  set f1 = f[-]c1;
A1: dom(f1[+]c2) = dom f1 by Def37;
  dom f1 = dom f by Def37;
  hence
A2: dom(f1[+]c2) = dom(f[-](c1-c2)) by A1,Def37;
  let x;
  assume
A3: x in dom(f1[+]c2);
  hence (f1[+]c2).x = f1.x + c2 by Def37
    .= f.x - c1 + c2 by A1,A3,Def37
    .= f.x - (c1 - c2) by Th14
    .= (f[-](c1-c2)).x by A2,A3,Def37;
end;

theorem
  f [-] c1 [-] c2 = f [-] (c1+c2)
proof
  set f1 = f[-]c1;
A1: dom(f1[-]c2) = dom f1 by Def37;
  dom f1 = dom f by Def37;
  hence
A2: dom(f1[-]c2) = dom(f[-](c1+c2)) by A1,Def37;
  let x;
  assume
A3: x in dom(f1[-]c2);
  hence (f1[-]c2).x = f1.x - c2 by Def37
    .= f.x - c1 - c2 by A1,A3,Def37
    .= f.x - (c1 + c2) by Th15
    .= (f[-](c1+c2)).x by A2,A3,Def37;
end;

definition

  let Y be complex-functions-membered set;
  let f be Y-valued Function;
  let c be complex number;
  deffunc F(set) = c(#)(f.$1);
  func f[#]c -> Function means
  :Def39:
  dom it = dom f & for x being set st x in dom it holds it.x = c (#) (f.x);
  existence
  proof
    ex F being Function st dom F = dom f & for x being set st x in dom f
    holds F.x = F(x) from FUNCT_1:sch 3;
    hence thesis;
  end;
  uniqueness
  proof
    let F,G be Function such that
A1: dom F = dom f and
A2: for x being set st x in dom F holds F.x = F(x) and
A3: dom G = dom f and
A4: for x being set st x in dom G holds G.x = F(x);
    thus dom F = dom G by A1,A3;
    let x;
    assume
A5: x in dom F;
    hence F.x = F(x) by A2
      .= G.x by A1,A3,A4,A5;
  end;
end;

definition
  let X;
  let Y be complex-functions-membered set;
  let f be PartFunc of X,Y;
  let c be complex number;
  redefine func f[#]c -> PartFunc of X, C_PFuncs(DOMS(Y));
  coherence
  proof
    set h = f[#]c;
A1: dom h = dom f by Def39;
    rng h c= C_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
A4:   h.x = c (#) f.x by A2,Def39;
      then reconsider y as Function by A3;
A5:   rng y c= COMPLEX
      proof
        let b be set;
        thus thesis by A3,A4,XCMPLX_0:def 2;
      end;
      f.x in Y by A1,A2,PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:def 5;
      then y is PartFunc of DOMS(Y),COMPLEX by A6,A5,RELSET_1:4;
      hence thesis by Def8;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X;
  let Y be real-functions-membered set;
  let f be PartFunc of X,Y;
  let c be real number;
  redefine func f[#]c -> PartFunc of X, R_PFuncs(DOMS(Y));
  coherence
  proof
    set h = f[#]c;
A1: dom h = dom f by Def39;
    rng h c= R_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = c (#) f.x by A2,Def39;
A5:   rng y c= REAL
      proof
        let b be set;
        thus thesis by A3,A4,XREAL_0:def 1;
      end;
      f.x in Y by A1,A2,PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:def 5;
      then y is PartFunc of DOMS(Y),REAL by A6,A5,RELSET_1:4;
      hence thesis by Def12;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X;
  let Y be rational-functions-membered set;
  let f be PartFunc of X,Y;
  let c be rational number;
  redefine func f[#]c -> PartFunc of X, Q_PFuncs(DOMS(Y));
  coherence
  proof
    set h = f[#]c;
A1: dom h = dom f by Def39;
    rng h c= Q_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = c (#) f.x by A2,Def39;
A5:   rng y c= RAT
      proof
        let b be set;
        thus thesis by A3,A4,RAT_1:def 2;
      end;
      f.x in Y by A1,A2,PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:def 5;
      then y is PartFunc of DOMS(Y),RAT by A6,A5,RELSET_1:4;
      hence thesis by Def14;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X;
  let Y be integer-functions-membered set;
  let f be PartFunc of X,Y;
  let c be integer number;
  redefine func f[#]c -> PartFunc of X, I_PFuncs(DOMS(Y));
  coherence
  proof
    set h = f[#]c;
A1: dom h = dom f by Def39;
    rng h c= I_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = c (#) f.x by A2,Def39;
A5:   rng y c= INT
      proof
        let b be set;
        thus thesis by A3,A4,INT_1:def 2;
      end;
      f.x in Y by A1,A2,PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:def 5;
      then y is PartFunc of DOMS(Y),INT by A6,A5,RELSET_1:4;
      hence thesis by Def16;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X;
  let Y be natural-functions-membered set;
  let f be PartFunc of X,Y;
  let c be Nat;
  redefine func f[#]c -> PartFunc of X, N_PFuncs(DOMS(Y));
  coherence
  proof
    set h = f[#]c;
A1: dom h = dom f by Def39;
    rng h c= N_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = c (#) f.x by A2,Def39;
A5:   rng y c= NAT
      proof
        let b be set;
        thus thesis by A3,A4,ORDINAL1:def 12;
      end;
      f.x in Y by A1,A2,PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:def 5;
      then y is PartFunc of DOMS(Y),NAT by A6,A5,RELSET_1:4;
      hence thesis by Def18;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

theorem
  f [#] c1 [#] c2 = f [#] (c1*c2)
proof
  set f1 = f[#]c1;
A1: dom(f1[#]c2) = dom f1 by Def39;
  dom f1 = dom f by Def39;
  hence
A2: dom(f1[#]c2) = dom(f[#](c1*c2)) by A1,Def39;
  let x;
  assume
A3: x in dom(f1[#]c2);
  hence (f1[#]c2).x = f1.x (#) c2 by Def39
    .= f.x (#) c1 (#) c2 by A1,A3,Def39
    .= f.x (#) (c1 * c2) by Th16
    .= (f[#](c1*c2)).x by A2,A3,Def39;
end;

theorem
  f <> {} & f is non-empty & (for x st x in dom f holds f.x is non-empty
  ) & f [#] c1 = f [#] c2 implies c1 = c2
proof
  assume that
A1: f <> {} and
A2: f is non-empty and
A3: for x st x in dom f holds f.x is non-empty and
A4: f[#]c1 = f[#]c2;
  consider x such that
A5: x in dom f by A1,XBOOLE_0:def 1;
  dom f = dom(f[#]c2) by Def39;
  then
A6: (f[#]c2).x = f.x(#)c2 by A5,Def39;
  dom f = dom(f[#]c1) by Def39;
  then
A7: (f[#]c1).x = f.x(#)c1 by A5,Def39;
  f.x in rng f by A5,FUNCT_1:def 3;
  then
A8: f.x <> {} by A2,RELAT_1:def 9;
  f.x is non-empty by A3,A5;
  hence c1 = c2 by A4,A8,A7,A6,Th9;
end;

definition

  let Y be complex-functions-membered set;
  let f be Y-valued Function;
  let c be complex number;
  func f[/]c -> Function equals
  f [#] (c");
  coherence;
end;

theorem
  dom(f[/]c) = dom f by Def39;

theorem
  x in dom(f[/]c) implies (f[/]c).x = c" (#) f.x by Def39;

definition
  let X;
  let Y be complex-functions-membered set;
  let f be PartFunc of X,Y;
  let c be complex number;
  redefine func f[/]c -> PartFunc of X, C_PFuncs(DOMS(Y));
  coherence
  proof
    f[/]c = f [#] (c");
    hence thesis;
  end;
end;

definition
  let X;
  let Y be real-functions-membered set;
  let f be PartFunc of X,Y;
  let c be real number;
  redefine func f[/]c -> PartFunc of X, R_PFuncs(DOMS(Y));
  coherence
  proof
    f[/]c = f [#] (c");
    hence thesis;
  end;
end;

definition
  let X;
  let Y be rational-functions-membered set;
  let f be PartFunc of X,Y;
  let c be rational number;
  redefine func f[/]c -> PartFunc of X, Q_PFuncs(DOMS(Y));
  coherence
  proof
    f[/]c = f [#] (c");
    hence thesis;
  end;
end;

theorem
  f [/] c1 [/] c2 = f [/] (c1*c2)
proof
  set f1 = f[/]c1;
A1: dom(f1[/]c2) = dom f1 by Def39;
  dom f1 = dom f by Def39;
  hence
A2: dom(f1[/]c2) = dom(f[/](c1*c2)) by A1,Def39;
  let x;
  assume
A3: x in dom(f1[/]c2);
  hence (f1[/]c2).x = f1.x (#) c2" by Def39
    .= f.x (#) c1" (#) c2" by A1,A3,Def39
    .= f.x (#) (c1" * c2") by Th16
    .= f.x (#) (c1*c2)" by XCMPLX_1:204
    .= (f[/](c1*c2)).x by A2,A3,Def39;
end;

theorem
  f <> {} & f is non-empty & (for x st x in dom f holds f.x is non-empty
  ) & f [/] c1 = f [/] c2 implies c1 = c2
proof
  assume that
A1: f <> {} and
A2: f is non-empty and
A3: for x st x in dom f holds f.x is non-empty and
A4: f[/]c1 = f[/]c2;
  consider x such that
A5: x in dom f by A1,XBOOLE_0:def 1;
  dom f = dom(f[/]c2) by Def39;
  then
A6: (f[/]c2).x = f.x(/)c2 by A5,Def39;
  dom f = dom(f[/]c1) by Def39;
  then
A7: (f[/]c1).x = f.x(/)c1 by A5,Def39;
  f.x in rng f by A5,FUNCT_1:def 3;
  then
A8: f.x <> {} by A2,RELAT_1:def 9;
  f.x is non-empty by A3,A5;
  hence c1 = c2 by A4,A8,A7,A6,Th33;
end;

definition

  let Y be complex-functions-membered set;
  let f be Y-valued Function;
  let g be complex-valued Function;
  deffunc F(set) = f.$1+g.$1;
  func f<+>g -> Function means
  :Def41:
  dom it = dom f /\ dom g & for x being
  set st x in dom it holds it.x = f.x + g.x;
  existence
  proof
    ex F being Function st dom F = dom f /\ dom g & for x being set st x
    in dom f /\ dom g holds F.x = F(x) from FUNCT_1:sch 3;
    hence thesis;
  end;
  uniqueness
  proof
    let F,G be Function such that
A1: dom F = dom f /\ dom g and
A2: for x being set st x in dom F holds F.x = F(x) and
A3: dom G = dom f /\ dom g and
A4: for x being set st x in dom G holds G.x = F(x);
    thus dom F = dom G by A1,A3;
    let x;
    assume
A5: x in dom F;
    hence F.x = F(x) by A2
      .= G.x by A1,A3,A4,A5;
  end;
end;

definition
  let X;
  let Y be complex-functions-membered set;
  let f be PartFunc of X,Y;
  let g be complex-valued Function;
  redefine func f<+>g -> PartFunc of X /\ dom g, C_PFuncs(DOMS(Y));
  coherence
  proof
    set h = f<+>g;
A1: dom h = dom f /\ dom g by Def41;
    rng h c= C_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
A4:   h.x = f.x + g.x by A2,Def41;
      then reconsider y as Function by A3;
A5:   rng y c= COMPLEX
      proof
        let b be set;
        thus thesis by A3,A4,XCMPLX_0:def 2;
      end;
      x in dom f by A1,A2,XBOOLE_0:def 4;
      then f.x in Y by PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:def 2;
      then y is PartFunc of DOMS(Y),COMPLEX by A6,A5,RELSET_1:4;
      hence thesis by Def8;
    end;
    hence thesis by A1,RELSET_1:4,XBOOLE_1:27;
  end;
end;

definition
  let X;
  let Y be real-functions-membered set;
  let f be PartFunc of X,Y;
  let g be real-valued Function;
  redefine func f<+>g -> PartFunc of X /\ dom g, R_PFuncs(DOMS(Y));
  coherence
  proof
    set h = f<+>g;
A1: dom h = dom f /\ dom g by Def41;
    rng h c= R_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = f.x + g.x by A2,Def41;
A5:   rng y c= REAL
      proof
        let b be set;
        thus thesis by A3,A4,XREAL_0:def 1;
      end;
      x in dom f by A1,A2,XBOOLE_0:def 4;
      then f.x in Y by PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:def 2;
      then y is PartFunc of DOMS(Y),REAL by A6,A5,RELSET_1:4;
      hence thesis by Def12;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X;
  let Y be rational-functions-membered set;
  let f be PartFunc of X,Y;
  let g be RAT-valued Function;
  redefine func f<+>g -> PartFunc of X /\ dom g, Q_PFuncs(DOMS(Y));
  coherence
  proof
    set h = f<+>g;
A1: dom h = dom f /\ dom g by Def41;
    rng h c= Q_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = f.x + g.x by A2,Def41;
A5:   rng y c= RAT
      proof
        let b be set;
        thus thesis by A3,A4,RAT_1:def 2;
      end;
      x in dom f by A1,A2,XBOOLE_0:def 4;
      then f.x in Y by PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:def 2;
      then y is PartFunc of DOMS(Y),RAT by A6,A5,RELSET_1:4;
      hence thesis by Def14;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X;
  let Y be integer-functions-membered set;
  let f be PartFunc of X,Y;
  let g be INT-valued Function;
  redefine func f<+>g -> PartFunc of X /\ dom g, I_PFuncs(DOMS(Y));
  coherence
  proof
    set h = f<+>g;
A1: dom h = dom f /\ dom g by Def41;
    rng h c= I_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = f.x + g.x by A2,Def41;
A5:   rng y c= INT
      proof
        let b be set;
        thus thesis by A3,A4,INT_1:def 2;
      end;
      x in dom f by A1,A2,XBOOLE_0:def 4;
      then f.x in Y by PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:def 2;
      then y is PartFunc of DOMS(Y),INT by A6,A5,RELSET_1:4;
      hence thesis by Def16;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X;
  let Y be natural-functions-membered set;
  let f be PartFunc of X,Y;
  let g be natural-valued Function;
  redefine func f<+>g -> PartFunc of X /\ dom g, N_PFuncs(DOMS(Y));
  coherence
  proof
    set h = f<+>g;
A1: dom h = dom f /\ dom g by Def41;
    rng h c= N_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = f.x + g.x by A2,Def41;
A5:   rng y c= NAT
      proof
        let b be set;
        thus thesis by A3,A4,ORDINAL1:def 12;
      end;
      x in dom f by A1,A2,XBOOLE_0:def 4;
      then f.x in Y by PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:def 2;
      then y is PartFunc of DOMS(Y),NAT by A6,A5,RELSET_1:4;
      hence thesis by Def18;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

theorem
  f <+> g <+> h = f <+> (g+h)
proof
  set f1 = f<+>g;
A1: dom(g+h) = dom g /\ dom h by VALUED_1:def 1;
A2: dom(f1<+>h) = dom f1 /\ dom h by Def41;
  dom f1 = dom f /\ dom g & dom(f<+>(g+h)) = dom f /\ dom(g+h) by Def41;
  hence
A3: dom(f1<+>h) = dom(f<+>(g+h)) by A2,A1,XBOOLE_1:16;
  let x;
  assume
A4: x in dom(f1<+>h);
  then
A5: x in dom f1 by A2,XBOOLE_0:def 4;
A6: x in dom(g+h) by A3,A4,XBOOLE_0:def 4;
  thus (f1<+>h).x = f1.x + h.x by A4,Def41
    .= f.x + g.x + h.x by A5,Def41
    .= f.x + (g.x + h.x) by Th12
    .= f.x + ((g+h).x) by A6,VALUED_1:def 1
    .= (f<+>(g+h)).x by A3,A4,Def41;
end;

theorem
  <->(f<+>g) = (<->f) <+> -g
proof
  set f1 = f<+>g, f2 = <->f;
A1: dom <->f1 = dom f1 by Def33;
A2: dom f1 = dom f /\ dom g & dom f2 = dom f by Def33,Def41;
  dom -g = dom g by VALUED_1:8;
  hence
A3: dom <->f1 = dom(f2<+>-g) by A1,A2,Def41;
  let x;
  assume
A4: x in dom <->f1;
  then
A5: x in dom f2 by A1,A2,XBOOLE_0:def 4;
  thus (<->f1).x = -f1.x by A4,Def33
    .= -(f.x+g.x) by A1,A4,Def41
    .= -f.x-g.x by Th10
    .= -f.x+(-g).x by VALUED_1:8
    .= f2.x+(-g).x by A5,Def33
    .= (f2<+>-g).x by A3,A4,Def41;
end;

definition

  let Y be complex-functions-membered set;
  let f be Y-valued Function;
  let g be complex-valued Function;
  func f<->g -> Function equals
  f <+> -g;
  coherence;
end;

theorem Th61:
  dom(f<->g) = dom f /\ dom g
proof
  thus dom(f<->g) = dom f /\ dom -g by Def41
    .= dom f /\ dom g by VALUED_1:8;
end;

theorem Th62:
  x in dom(f<->g) implies (f<->g).x = f.x - g.x
proof
  assume x in dom(f<->g);
  hence (f<->g).x = f.x + (-g).x by Def41
    .= f.x - g.x by VALUED_1:8;
end;

definition
  let X;
  let Y be complex-functions-membered set;
  let f be PartFunc of X,Y;
  let g be complex-valued Function;
  redefine func f<->g -> PartFunc of X /\ dom g, C_PFuncs(DOMS(Y));
  coherence
  proof
    set h = f<->g;
A1: dom h = dom f /\ dom g by Th61;
    rng h c= C_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
A4:   h.x = f.x - g.x by A2,Th62;
      then reconsider y as Function by A3;
A5:   rng y c= COMPLEX
      proof
        let b be set;
        thus thesis by A3,A4,XCMPLX_0:def 2;
      end;
      x in dom f by A1,A2,XBOOLE_0:def 4;
      then f.x in Y by PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      h.x = f.x - g.x by A2,Th62;
      then dom y = dom(f.x) by A3,VALUED_1:def 2;
      then y is PartFunc of DOMS(Y),COMPLEX by A6,A5,RELSET_1:4;
      hence thesis by Def8;
    end;
    hence thesis by A1,RELSET_1:4,XBOOLE_1:27;
  end;
end;

definition
  let X;
  let Y be real-functions-membered set;
  let f be PartFunc of X,Y;
  let g be real-valued Function;
  redefine func f<->g -> PartFunc of X /\ dom g, R_PFuncs(DOMS(Y));
  coherence
  proof
    set h = f<->g;
A1: dom h = dom f /\ dom g by Th61;
    rng h c= R_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = f.x - g.x by A2,Th62;
A5:   rng y c= REAL
      proof
        let b be set;
        thus thesis by A3,A4,XREAL_0:def 1;
      end;
      x in dom f by A1,A2,XBOOLE_0:def 4;
      then f.x in Y by PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:def 2;
      then y is PartFunc of DOMS(Y),REAL by A6,A5,RELSET_1:4;
      hence thesis by Def12;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X;
  let Y be rational-functions-membered set;
  let f be PartFunc of X,Y;
  let g be RAT-valued Function;
  redefine func f<->g -> PartFunc of X /\ dom g, Q_PFuncs(DOMS(Y));
  coherence
  proof
    set h = f<->g;
A1: dom h = dom f /\ dom g by Th61;
    rng h c= Q_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = f.x - g.x by A2,Th62;
A5:   rng y c= RAT
      proof
        let b be set;
        thus thesis by A3,A4,RAT_1:def 2;
      end;
      x in dom f by A1,A2,XBOOLE_0:def 4;
      then f.x in Y by PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:def 2;
      then y is PartFunc of DOMS(Y),RAT by A6,A5,RELSET_1:4;
      hence thesis by Def14;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X;
  let Y be integer-functions-membered set;
  let f be PartFunc of X,Y;
  let g be INT-valued Function;
  redefine func f<->g -> PartFunc of X /\ dom g, I_PFuncs(DOMS(Y));
  coherence
  proof
    set h = f<->g;
A1: dom h = dom f /\ dom g by Th61;
    rng h c= I_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = f.x - g.x by A2,Th62;
A5:   rng y c= INT
      proof
        let b be set;
        thus thesis by A3,A4,INT_1:def 2;
      end;
      x in dom f by A1,A2,XBOOLE_0:def 4;
      then f.x in Y by PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:def 2;
      then y is PartFunc of DOMS(Y),INT by A6,A5,RELSET_1:4;
      hence thesis by Def16;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

theorem
  f <-> -g = f <+> g;

theorem
  <->(f<->g) = (<->f) <+> g
proof
  set f1 = f<->g, f2 = <->f;
A1: dom <->f1 = dom f1 by Def33;
A2: dom f1 = dom f /\ dom g & dom f2 = dom f by Def33,Th61;
  hence
A3: dom <->f1 = dom(f2<+>g) by A1,Def41;
  let x;
  assume
A4: x in dom <->f1;
  then
A5: x in dom f2 by A1,A2,XBOOLE_0:def 4;
  thus (<->f1).x = -f1.x by A4,Def33
    .= -(f.x-g.x) by A1,A4,Th62
    .= -f.x+g.x by Th11
    .= f2.x+g.x by A5,Def33
    .= (f2<+>g).x by A3,A4,Def41;
end;

theorem
  f <+> g <-> h = f <+> (g-h)
proof
  set f1 = f<+>g;
A1: dom(g-h) = dom g /\ dom h by VALUED_1:12;
A2: dom(f1<->h) = dom f1 /\ dom h by Th61;
  dom f1 = dom f /\ dom g & dom(f<+>(g-h)) = dom f /\ dom(g-h) by Def41;
  hence
A3: dom(f1<->h) = dom(f<+>(g-h)) by A2,A1,XBOOLE_1:16;
  let x;
  assume
A4: x in dom(f1<->h);
  then
A5: x in dom f1 by A2,XBOOLE_0:def 4;
A6: x in dom(g-h) by A3,A4,XBOOLE_0:def 4;
  thus (f1<->h).x = f1.x - h.x by A4,Th62
    .= f.x + g.x - h.x by A5,Def41
    .= f.x + (g.x - h.x) by Th13
    .= f.x + ((g-h).x) by A6,VALUED_1:13
    .= (f<+>(g-h)).x by A3,A4,Def41;
end;

theorem
  f <-> g <+> h = f <-> (g-h)
proof
  set f1 = f<->g;
A1: dom(g-h) = dom g /\ dom h by VALUED_1:12;
A2: dom(f1<+>h) = dom f1 /\ dom h by Def41;
  dom f1 = dom f /\ dom g & dom(f<->(g-h)) = dom f /\ dom(g-h) by Th61;
  hence
A3: dom(f1<+>h) = dom(f<->(g-h)) by A2,A1,XBOOLE_1:16;
  let x;
  assume
A4: x in dom(f1<+>h);
  then
A5: x in dom f1 by A2,XBOOLE_0:def 4;
A6: x in dom(g-h) by A3,A4,XBOOLE_0:def 4;
  thus (f1<+>h).x = f1.x + h.x by A4,Def41
    .= f.x - g.x + h.x by A5,Th62
    .= f.x - (g.x - h.x) by Th14
    .= f.x - ((g-h).x) by A6,VALUED_1:13
    .= (f<->(g-h)).x by A3,A4,Th62;
end;

theorem
  f <-> g <-> h = f <-> (g+h)
proof
  set f1 = f<->g;
A1: dom(g+h) = dom g /\ dom h by VALUED_1:def 1;
A2: dom(f1<->h) = dom f1 /\ dom h by Th61;
  dom f1 = dom f /\ dom g & dom(f<->(g+h)) = dom f /\ dom(g+h) by Th61;
  hence
A3: dom(f1<->h) = dom(f<->(g+h)) by A2,A1,XBOOLE_1:16;
  let x;
  assume
A4: x in dom(f1<->h);
  then
A5: x in dom f1 by A2,XBOOLE_0:def 4;
A6: x in dom(g+h) by A3,A4,XBOOLE_0:def 4;
  thus (f1<->h).x = f1.x - h.x by A4,Th62
    .= f.x - g.x - h.x by A5,Th62
    .= f.x - (g.x + h.x) by Th15
    .= f.x - ((g+h).x) by A6,VALUED_1:def 1
    .= (f<->(g+h)).x by A3,A4,Th62;
end;

definition

  let Y be complex-functions-membered set;
  let f be Y-valued Function;
  let g be complex-valued Function;
  deffunc F(set) = f.$1(#)g.$1;
  func f<#>g -> Function means
  :Def43:
  dom it = dom f /\ dom g & for x being
  set st x in dom it holds it.x = f.x (#) g.x;
  existence
  proof
    ex F being Function st dom F = dom f /\ dom g & for x being set st x
    in dom f /\ dom g holds F.x = F(x) from FUNCT_1:sch 3;
    hence thesis;
  end;
  uniqueness
  proof
    let F,G be Function such that
A1: dom F = dom f /\ dom g and
A2: for x being set st x in dom F holds F.x = F(x) and
A3: dom G = dom f /\ dom g and
A4: for x being set st x in dom G holds G.x = F(x);
    thus dom F = dom G by A1,A3;
    let x;
    assume
A5: x in dom F;
    hence F.x = F(x) by A2
      .= G.x by A1,A3,A4,A5;
  end;
end;

definition
  let X;
  let Y be complex-functions-membered set;
  let f be PartFunc of X,Y;
  let g be complex-valued Function;
  redefine func f<#>g -> PartFunc of X /\ dom g, C_PFuncs(DOMS(Y));
  coherence
  proof
    set h = f<#>g;
A1: dom h = dom f /\ dom g by Def43;
    rng h c= C_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
A4:   h.x = f.x (#) g.x by A2,Def43;
      then reconsider y as Function by A3;
A5:   rng y c= COMPLEX
      proof
        let b be set;
        thus thesis by A3,A4,XCMPLX_0:def 2;
      end;
      x in dom f by A1,A2,XBOOLE_0:def 4;
      then f.x in Y by PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:def 5;
      then y is PartFunc of DOMS(Y),COMPLEX by A6,A5,RELSET_1:4;
      hence thesis by Def8;
    end;
    hence thesis by A1,RELSET_1:4,XBOOLE_1:27;
  end;
end;

definition
  let X;
  let Y be real-functions-membered set;
  let f be PartFunc of X,Y;
  let g be real-valued Function;
  redefine func f<#>g -> PartFunc of X /\ dom g, R_PFuncs(DOMS(Y));
  coherence
  proof
    set h = f<#>g;
A1: dom h = dom f /\ dom g by Def43;
    rng h c= R_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = f.x (#) g.x by A2,Def43;
A5:   rng y c= REAL
      proof
        let b be set;
        thus thesis by A3,A4,XREAL_0:def 1;
      end;
      x in dom f by A1,A2,XBOOLE_0:def 4;
      then f.x in Y by PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:def 5;
      then y is PartFunc of DOMS(Y),REAL by A6,A5,RELSET_1:4;
      hence thesis by Def12;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X;
  let Y be rational-functions-membered set;
  let f be PartFunc of X,Y;
  let g be RAT-valued Function;
  redefine func f<#>g -> PartFunc of X /\ dom g, Q_PFuncs(DOMS(Y));
  coherence
  proof
    set h = f<#>g;
A1: dom h = dom f /\ dom g by Def43;
    rng h c= Q_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = f.x (#) g.x by A2,Def43;
A5:   rng y c= RAT
      proof
        let b be set;
        thus thesis by A3,A4,RAT_1:def 2;
      end;
      x in dom f by A1,A2,XBOOLE_0:def 4;
      then f.x in Y by PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:def 5;
      then y is PartFunc of DOMS(Y),RAT by A6,A5,RELSET_1:4;
      hence thesis by Def14;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X;
  let Y be integer-functions-membered set;
  let f be PartFunc of X,Y;
  let g be INT-valued Function;
  redefine func f<#>g -> PartFunc of X /\ dom g, I_PFuncs(DOMS(Y));
  coherence
  proof
    set h = f<#>g;
A1: dom h = dom f /\ dom g by Def43;
    rng h c= I_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = f.x (#) g.x by A2,Def43;
A5:   rng y c= INT
      proof
        let b be set;
        thus thesis by A3,A4,INT_1:def 2;
      end;
      x in dom f by A1,A2,XBOOLE_0:def 4;
      then f.x in Y by PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:def 5;
      then y is PartFunc of DOMS(Y),INT by A6,A5,RELSET_1:4;
      hence thesis by Def16;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X;
  let Y be natural-functions-membered set;
  let f be PartFunc of X,Y;
  let g be natural-valued Function;
  redefine func f<#>g -> PartFunc of X /\ dom g, N_PFuncs(DOMS(Y));
  coherence
  proof
    set h = f<#>g;
A1: dom h = dom f /\ dom g by Def43;
    rng h c= N_PFuncs(DOMS(Y))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = f.x (#) g.x by A2,Def43;
A5:   rng y c= NAT
      proof
        let b be set;
        thus thesis by A3,A4,ORDINAL1:def 12;
      end;
      x in dom f by A1,A2,XBOOLE_0:def 4;
      then f.x in Y by PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y: not contradiction};
      then
A6:   dom(f.x) c= DOMS(Y) by ZFMISC_1:74;
      dom y = dom(f.x) by A3,A4,VALUED_1:def 5;
      then y is PartFunc of DOMS(Y),NAT by A6,A5,RELSET_1:4;
      hence thesis by Def18;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

theorem
  f <#> -g = (<->f) <#> g
proof
  set f1 = <->f;
A1: dom f1 = dom f & dom(f<#>-g) = dom f /\ dom -g by Def33,Def43;
  dom(f1<#>g) = dom f1 /\ dom g by Def43;
  hence
A2: dom(f<#>-g) = dom(f1<#>g) by A1,VALUED_1:8;
  let x;
  assume
A3: x in dom(f<#>-g);
  then
A4: x in dom f1 by A1,XBOOLE_0:def 4;
  thus (f<#>-g).x = f.x (#) (-g).x by A3,Def43
    .= f.x(#)-g.x by VALUED_1:8
    .= (-f.x)(#)g.x by Th22
    .= f1.x(#)g.x by A4,Def33
    .= (f1<#>g).x by A2,A3,Def43;
end;

theorem
  f <#> -g = <-> (f <#> g)
proof
  set f1 = f<#>g;
A1: dom(<->f1) = dom f1 by Def33;
  dom f1 = dom f /\ dom g & dom(f<#>-g) = dom f /\ dom -g by Def43;
  hence
A2: dom(f<#>-g) = dom(<->f1) by A1,VALUED_1:8;
  let x;
  assume
A3: x in dom(f<#>-g);
  hence (f<#>-g).x = f.x (#) (-g).x by Def43
    .= f.x(#)-g.x by VALUED_1:8
    .= -(f.x(#)g.x) by Th24
    .= -f1.x by A1,A2,A3,Def43
    .= (<->f1).x by A2,A3,Def33;
end;

theorem
  f <#> g <#> h = f <#> (g(#)h)
proof
  set f1 = f<#>g;
A1: dom(g(#)h) = dom g /\ dom h by VALUED_1:def 4;
A2: dom(f1<#>h) = dom f1 /\ dom h by Def43;
  dom f1 = dom f /\ dom g & dom(f<#>(g(#)h)) = dom f /\ dom(g(#)h) by Def43;
  hence
A3: dom(f1<#>h) = dom(f<#>(g(#)h)) by A2,A1,XBOOLE_1:16;
  let x;
  assume
A4: x in dom(f1<#>h);
  then
A5: x in dom f1 by A2,XBOOLE_0:def 4;
A6: x in dom(g(#)h) by A3,A4,XBOOLE_0:def 4;
  thus (f1<#>h).x = f1.x (#) h.x by A4,Def43
    .= f.x (#) g.x (#) h.x by A5,Def43
    .= f.x (#) (g.x * h.x) by Th16
    .= f.x (#) ((g(#)h).x) by A6,VALUED_1:def 4
    .= (f<#>(g(#)h)).x by A3,A4,Def43;
end;

definition

  let Y be complex-functions-membered set;
  let f be Y-valued Function;
  let g be complex-valued Function;
  func f</>g -> Function equals
  f <#> (g");
  coherence;
end;

theorem Th71:
  dom(f</>g) = dom f /\ dom g
proof
  thus dom(f</>g) = dom f /\ dom(g") by Def43
    .= dom f /\ dom g by VALUED_1:def 7;
end;

theorem Th72:
  x in dom(f</>g) implies (f</>g).x = f.x (/) g.x
proof
  assume x in dom(f</>g);
  hence (f</>g).x = f.x (#) (g").x by Def43
    .= f.x (/) g.x by VALUED_1:10;
end;

definition
  let X;
  let Y be complex-functions-membered set;
  let f be PartFunc of X,Y;
  let g be complex-valued Function;
  redefine func f</>g -> PartFunc of X /\ dom g, C_PFuncs(DOMS(Y));
  coherence
  proof
    f</>g = f <#> (g");
    hence thesis by VALUED_1:def 7;
  end;
end;

definition
  let X;
  let Y be real-functions-membered set;
  let f be PartFunc of X,Y;
  let g be real-valued Function;
  redefine func f</>g -> PartFunc of X /\ dom g, R_PFuncs(DOMS(Y));
  coherence
  proof
    f</>g = f <#> (g");
    hence thesis by VALUED_1:def 7;
  end;
end;

definition
  let X;
  let Y be rational-functions-membered set;
  let f be PartFunc of X,Y;
  let g be RAT-valued Function;
  redefine func f</>g -> PartFunc of X /\ dom g, Q_PFuncs(DOMS(Y));
  coherence
  proof
    f</>g = f <#> (g");
    hence thesis by VALUED_1:def 7;
  end;
end;

theorem
  f <#> g </> h = f <#> (g/"h)
proof
  set f1 = f<#>g;
A1: dom(g/"h) = dom g /\ dom h by VALUED_1:16;
A2: dom(f1</>h) = dom f1 /\ dom h by Th71;
  dom f1 = dom f /\ dom g & dom(f<#>(g/"h)) = dom f /\ dom(g/"h) by Def43;
  hence
A3: dom(f1</>h) = dom(f<#>(g/"h)) by A2,A1,XBOOLE_1:16;
  let x;
  assume
A4: x in dom(f1</>h);
  then
A5: x in dom f1 by A2,XBOOLE_0:def 4;
  thus (f1</>h).x = f1.x (/) h.x by A4,Th72
    .= f.x (#) g.x (/) h.x by A5,Def43
    .= f.x (#) (g.x / h.x) by Th16
    .= f.x (#) ((g/"h).x) by VALUED_1:17
    .= (f<#>(g/"h)).x by A3,A4,Def43;
end;

definition

  let Y1, Y2 be complex-functions-membered set;
  let f be Y1-valued Function;
  let g be Y2-valued Function;
  deffunc F(set) = f.$1+g.$1;
  func f<++>g -> Function means
  :Def45:
  dom it = dom f /\ dom g & for x being
  set st x in dom it holds it.x = f.x + g.x;
  existence
  proof
    ex F being Function st dom F = dom f /\ dom g & for x being set st x
    in dom f /\ dom g holds F.x = F(x) from FUNCT_1:sch 3;
    hence thesis;
  end;
  uniqueness
  proof
    let F,G be Function such that
A1: dom F = dom f /\ dom g and
A2: for x being set st x in dom F holds F.x = F(x) and
A3: dom G = dom f /\ dom g and
A4: for x being set st x in dom G holds G.x = F(x);
    thus dom F = dom G by A1,A3;
    let x;
    assume
A5: x in dom F;
    hence F.x = F(x) by A2
      .= G.x by A1,A3,A4,A5;
  end;
end;

definition
  let X1, X2 be set;
  let Y1, Y2 be complex-functions-membered set;
  let f be PartFunc of X1,Y1;
  let g be PartFunc of X2,Y2;
  redefine func f<++>g -> PartFunc of X1 /\ X2, C_PFuncs(DOMS(Y1)/\DOMS(Y2));
  coherence
  proof
    set h = f<++>g;
A1: dom h = dom f /\ dom g by Def45;
    rng h c= C_PFuncs(DOMS(Y1)/\DOMS(Y2))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
A4:   h.x = f.x + g.x by A2,Def45;
      then reconsider y as Function by A3;
A5:   rng y c= COMPLEX
      proof
        let b be set;
        thus thesis by A3,A4,XCMPLX_0:def 2;
      end;
      x in dom g by A1,A2,XBOOLE_0:def 4;
      then g.x in Y2 by PARTFUN1:4;
      then dom(g.x) in {dom m where m is Element of Y2: not contradiction};
      then
A6:   dom(g.x) c= DOMS(Y2) by ZFMISC_1:74;
      x in dom f by A1,A2,XBOOLE_0:def 4;
      then f.x in Y1 by PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y1: not contradiction};
      then
A7:   dom(f.x) c= DOMS(Y1) by ZFMISC_1:74;
      dom y = dom(f.x) /\ dom(g.x) by A3,A4,VALUED_1:def 1;
      then y is PartFunc of DOMS(Y1)/\DOMS(Y2),COMPLEX by A7,A6,A5,RELSET_1:4
,XBOOLE_1:27;
      hence thesis by Def8;
    end;
    hence thesis by A1,RELSET_1:4,XBOOLE_1:27;
  end;
end;

definition
  let X1, X2 be set;
  let Y1, Y2 be real-functions-membered set;
  let f be PartFunc of X1,Y1;
  let g be PartFunc of X2,Y2;
  redefine func f<++>g -> PartFunc of X1 /\ X2, R_PFuncs(DOMS(Y1)/\DOMS(Y2));
  coherence
  proof
    set h = f<++>g;
A1: dom h = dom f /\ dom g by Def45;
    rng h c= R_PFuncs(DOMS(Y1)/\DOMS(Y2))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = f.x + g.x by A2,Def45;
A5:   rng y c= REAL
      proof
        let b be set;
        thus thesis by A3,A4,XREAL_0:def 1;
      end;
      x in dom g by A1,A2,XBOOLE_0:def 4;
      then g.x in Y2 by PARTFUN1:4;
      then dom(g.x) in {dom m where m is Element of Y2: not contradiction};
      then
A6:   dom(g.x) c= DOMS(Y2) by ZFMISC_1:74;
      x in dom f by A1,A2,XBOOLE_0:def 4;
      then f.x in Y1 by PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y1: not contradiction};
      then
A7:   dom(f.x) c= DOMS(Y1) by ZFMISC_1:74;
      dom y = dom(f.x) /\ dom(g.x) by A3,A4,VALUED_1:def 1;
      then y is PartFunc of DOMS(Y1)/\DOMS(Y2),REAL by A7,A6,A5,RELSET_1:4
,XBOOLE_1:27;
      hence thesis by Def12;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X1, X2 be set;
  let Y1, Y2 be rational-functions-membered set;
  let f be PartFunc of X1,Y1;
  let g be PartFunc of X2,Y2;
  redefine func f<++>g -> PartFunc of X1 /\ X2, Q_PFuncs(DOMS(Y1)/\DOMS(Y2));
  coherence
  proof
    set h = f<++>g;
A1: dom h = dom f /\ dom g by Def45;
    rng h c= Q_PFuncs(DOMS(Y1)/\DOMS(Y2))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = f.x + g.x by A2,Def45;
A5:   rng y c= RAT
      proof
        let b be set;
        thus thesis by A3,A4,RAT_1:def 2;
      end;
      x in dom g by A1,A2,XBOOLE_0:def 4;
      then g.x in Y2 by PARTFUN1:4;
      then dom(g.x) in {dom m where m is Element of Y2: not contradiction};
      then
A6:   dom(g.x) c= DOMS(Y2) by ZFMISC_1:74;
      x in dom f by A1,A2,XBOOLE_0:def 4;
      then f.x in Y1 by PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y1: not contradiction};
      then
A7:   dom(f.x) c= DOMS(Y1) by ZFMISC_1:74;
      dom y = dom(f.x) /\ dom(g.x) by A3,A4,VALUED_1:def 1;
      then y is PartFunc of DOMS(Y1)/\DOMS(Y2),RAT by A7,A6,A5,RELSET_1:4
,XBOOLE_1:27;
      hence thesis by Def14;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X1, X2 be set;
  let Y1, Y2 be integer-functions-membered set;
  let f be PartFunc of X1,Y1;
  let g be PartFunc of X2,Y2;
  redefine func f<++>g -> PartFunc of X1 /\ X2, I_PFuncs(DOMS(Y1)/\DOMS(Y2));
  coherence
  proof
    set h = f<++>g;
A1: dom h = dom f /\ dom g by Def45;
    rng h c= I_PFuncs(DOMS(Y1)/\DOMS(Y2))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = f.x + g.x by A2,Def45;
A5:   rng y c= INT
      proof
        let b be set;
        thus thesis by A3,A4,INT_1:def 2;
      end;
      x in dom g by A1,A2,XBOOLE_0:def 4;
      then g.x in Y2 by PARTFUN1:4;
      then dom(g.x) in {dom m where m is Element of Y2: not contradiction};
      then
A6:   dom(g.x) c= DOMS(Y2) by ZFMISC_1:74;
      x in dom f by A1,A2,XBOOLE_0:def 4;
      then f.x in Y1 by PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y1: not contradiction};
      then
A7:   dom(f.x) c= DOMS(Y1) by ZFMISC_1:74;
      dom y = dom(f.x) /\ dom(g.x) by A3,A4,VALUED_1:def 1;
      then y is PartFunc of DOMS(Y1)/\DOMS(Y2),INT by A7,A6,A5,RELSET_1:4
,XBOOLE_1:27;
      hence thesis by Def16;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X1, X2 be set;
  let Y1, Y2 be natural-functions-membered set;
  let f be PartFunc of X1,Y1;
  let g be PartFunc of X2,Y2;
  redefine func f<++>g -> PartFunc of X1 /\ X2, N_PFuncs(DOMS(Y1)/\DOMS(Y2));
  coherence
  proof
    set h = f<++>g;
A1: dom h = dom f /\ dom g by Def45;
    rng h c= N_PFuncs(DOMS(Y1)/\DOMS(Y2))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = f.x + g.x by A2,Def45;
A5:   rng y c= NAT
      proof
        let b be set;
        thus thesis by A3,A4,ORDINAL1:def 12;
      end;
      x in dom g by A1,A2,XBOOLE_0:def 4;
      then g.x in Y2 by PARTFUN1:4;
      then dom(g.x) in {dom m where m is Element of Y2: not contradiction};
      then
A6:   dom(g.x) c= DOMS(Y2) by ZFMISC_1:74;
      x in dom f by A1,A2,XBOOLE_0:def 4;
      then f.x in Y1 by PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y1: not contradiction};
      then
A7:   dom(f.x) c= DOMS(Y1) by ZFMISC_1:74;
      dom y = dom(f.x) /\ dom(g.x) by A3,A4,VALUED_1:def 1;
      then y is PartFunc of DOMS(Y1)/\DOMS(Y2),NAT by A7,A6,A5,RELSET_1:4
,XBOOLE_1:27;
      hence thesis by Def18;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

theorem
  f1 <++> f2 = f2 <++> f1
proof
  dom(f1<++>f2) = dom f1 /\ dom f2 by Def45;
  hence
A1: dom(f1<++>f2) = dom(f2<++>f1) by Def45;
  let x;
  assume
A2: x in dom(f1<++>f2);
  hence (f1<++>f2).x = f1.x + f2.x by Def45
    .= (f2<++>f1).x by A1,A2,Def45;
end;

theorem
  f <++> f1 <++> f2 = f <++> (f1 <++> f2)
proof
  set f3 = f<++>f1, f4 = f1<++>f2;
A1: dom(f3<++>f2) = dom f3 /\ dom f2 by Def45;
A2: dom(f<++>f4) = dom f /\ dom f4 by Def45;
  dom f3 = dom f /\ dom f1 & dom f4 = dom f1 /\ dom f2 by Def45;
  hence
A3: dom(f3<++>f2) = dom(f<++>f4) by A1,A2,XBOOLE_1:16;
  let x;
  assume
A4: x in dom(f3<++>f2);
  then
A5: x in dom f4 by A2,A3,XBOOLE_0:def 4;
A6: x in dom f3 by A1,A4,XBOOLE_0:def 4;
  thus (f3<++>f2).x = f3.x + f2.x by A4,Def45
    .= f.x + f1.x + f2.x by A6,Def45
    .= f.x + (f1.x + f2.x) by RFUNCT_1:8
    .= f.x + f4.x by A5,Def45
    .= (f<++>f4).x by A3,A4,Def45;
end;

theorem
  <-> (f1 <++> f2) = (<->f1) <++> (<->f2)
proof
  set f3 = f1 <++> f2, f4 = <->f1, f5 = <->f2;
A1: dom(f1 <++> f2) = dom f1 /\ dom f2 by Def45;
A2: dom <->f2 = dom f2 by Def33;
A3: dom <->f3 = dom f3 by Def33;
A4: dom <->f1 = dom f1 by Def33;
  hence
A5: dom(<->f3) = dom(f4<++>f5) by A1,A2,A3,Def45;
  let x;
  assume
A6: x in dom(<->f3);
  then
A7: x in dom f4 by A1,A4,A3,XBOOLE_0:def 4;
A8: x in dom f5 by A1,A2,A3,A6,XBOOLE_0:def 4;
  thus (<->f3).x = -f3.x by A6,Def33
    .= -(f1.x+f2.x) by A3,A6,Def45
    .= -f1.x-f2.x by Th17
    .= f4.x + -f2.x by A7,Def33
    .= f4.x + f5.x by A8,Def33
    .= (f4<++>f5).x by A5,A6,Def45;
end;

definition

  let Y1, Y2 be complex-functions-membered set;
  let f be Y1-valued Function;
  let g be Y2-valued Function;
  deffunc F(set) = f.$1-g.$1;
  func f<-->g -> Function means
  :Def46:
  dom it = dom f /\ dom g & for x being
  set st x in dom it holds it.x = f.x - g.x;
  existence
  proof
    ex F being Function st dom F = dom f /\ dom g & for x being set st x
    in dom f /\ dom g holds F.x = F(x) from FUNCT_1:sch 3;
    hence thesis;
  end;
  uniqueness
  proof
    let F,G be Function such that
A1: dom F = dom f /\ dom g and
A2: for x being set st x in dom F holds F.x = F(x) and
A3: dom G = dom f /\ dom g and
A4: for x being set st x in dom G holds G.x = F(x);
    thus dom F = dom G by A1,A3;
    let x;
    assume
A5: x in dom F;
    hence F.x = F(x) by A2
      .= G.x by A1,A3,A4,A5;
  end;
end;

definition
  let X1, X2 be set;
  let Y1, Y2 be complex-functions-membered set;
  let f be PartFunc of X1,Y1;
  let g be PartFunc of X2,Y2;
  redefine func f<-->g -> PartFunc of X1 /\ X2, C_PFuncs(DOMS(Y1)/\DOMS(Y2));
  coherence
  proof
    set h = f<-->g;
A1: dom h = dom f /\ dom g by Def46;
    rng h c= C_PFuncs(DOMS(Y1)/\DOMS(Y2))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
A4:   h.x = f.x - g.x by A2,Def46;
      then reconsider y as Function by A3;
A5:   rng y c= COMPLEX
      proof
        let b be set;
        thus thesis by A3,A4,XCMPLX_0:def 2;
      end;
      x in dom g by A1,A2,XBOOLE_0:def 4;
      then g.x in Y2 by PARTFUN1:4;
      then dom(g.x) in {dom m where m is Element of Y2: not contradiction};
      then
A6:   dom(g.x) c= DOMS(Y2) by ZFMISC_1:74;
      x in dom f by A1,A2,XBOOLE_0:def 4;
      then f.x in Y1 by PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y1: not contradiction};
      then
A7:   dom(f.x) c= DOMS(Y1) by ZFMISC_1:74;
      dom y = dom(f.x) /\ dom(g.x) by A3,A4,VALUED_1:12;
      then y is PartFunc of DOMS(Y1)/\DOMS(Y2),COMPLEX by A7,A6,A5,RELSET_1:4
,XBOOLE_1:27;
      hence thesis by Def8;
    end;
    hence thesis by A1,RELSET_1:4,XBOOLE_1:27;
  end;
end;

definition
  let X1, X2 be set;
  let Y1, Y2 be real-functions-membered set;
  let f be PartFunc of X1,Y1;
  let g be PartFunc of X2,Y2;
  redefine func f<-->g -> PartFunc of X1 /\ X2, R_PFuncs(DOMS(Y1)/\DOMS(Y2));
  coherence
  proof
    set h = f<-->g;
A1: dom h = dom f /\ dom g by Def46;
    rng h c= R_PFuncs(DOMS(Y1)/\DOMS(Y2))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = f.x - g.x by A2,Def46;
A5:   rng y c= REAL
      proof
        let b be set;
        thus thesis by A3,A4,XREAL_0:def 1;
      end;
      x in dom g by A1,A2,XBOOLE_0:def 4;
      then g.x in Y2 by PARTFUN1:4;
      then dom(g.x) in {dom m where m is Element of Y2: not contradiction};
      then
A6:   dom(g.x) c= DOMS(Y2) by ZFMISC_1:74;
      x in dom f by A1,A2,XBOOLE_0:def 4;
      then f.x in Y1 by PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y1: not contradiction};
      then
A7:   dom(f.x) c= DOMS(Y1) by ZFMISC_1:74;
      dom y = dom(f.x) /\ dom(g.x) by A3,A4,VALUED_1:12;
      then y is PartFunc of DOMS(Y1)/\DOMS(Y2),REAL by A7,A6,A5,RELSET_1:4
,XBOOLE_1:27;
      hence thesis by Def12;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X1, X2 be set;
  let Y1, Y2 be rational-functions-membered set;
  let f be PartFunc of X1,Y1;
  let g be PartFunc of X2,Y2;
  redefine func f<-->g -> PartFunc of X1 /\ X2, Q_PFuncs(DOMS(Y1)/\DOMS(Y2));
  coherence
  proof
    set h = f<-->g;
A1: dom h = dom f /\ dom g by Def46;
    rng h c= Q_PFuncs(DOMS(Y1)/\DOMS(Y2))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = f.x - g.x by A2,Def46;
A5:   rng y c= RAT
      proof
        let b be set;
        thus thesis by A3,A4,RAT_1:def 2;
      end;
      x in dom g by A1,A2,XBOOLE_0:def 4;
      then g.x in Y2 by PARTFUN1:4;
      then dom(g.x) in {dom m where m is Element of Y2: not contradiction};
      then
A6:   dom(g.x) c= DOMS(Y2) by ZFMISC_1:74;
      x in dom f by A1,A2,XBOOLE_0:def 4;
      then f.x in Y1 by PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y1: not contradiction};
      then
A7:   dom(f.x) c= DOMS(Y1) by ZFMISC_1:74;
      dom y = dom(f.x) /\ dom(g.x) by A3,A4,VALUED_1:12;
      then y is PartFunc of DOMS(Y1)/\DOMS(Y2),RAT by A7,A6,A5,RELSET_1:4
,XBOOLE_1:27;
      hence thesis by Def14;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X1, X2 be set;
  let Y1, Y2 be integer-functions-membered set;
  let f be PartFunc of X1,Y1;
  let g be PartFunc of X2,Y2;
  redefine func f<-->g -> PartFunc of X1 /\ X2, I_PFuncs(DOMS(Y1)/\DOMS(Y2));
  coherence
  proof
    set h = f<-->g;
A1: dom h = dom f /\ dom g by Def46;
    rng h c= I_PFuncs(DOMS(Y1)/\DOMS(Y2))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = f.x - g.x by A2,Def46;
A5:   rng y c= INT
      proof
        let b be set;
        thus thesis by A3,A4,INT_1:def 2;
      end;
      x in dom g by A1,A2,XBOOLE_0:def 4;
      then g.x in Y2 by PARTFUN1:4;
      then dom(g.x) in {dom m where m is Element of Y2: not contradiction};
      then
A6:   dom(g.x) c= DOMS(Y2) by ZFMISC_1:74;
      x in dom f by A1,A2,XBOOLE_0:def 4;
      then f.x in Y1 by PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y1: not contradiction};
      then
A7:   dom(f.x) c= DOMS(Y1) by ZFMISC_1:74;
      dom y = dom(f.x) /\ dom(g.x) by A3,A4,VALUED_1:12;
      then y is PartFunc of DOMS(Y1)/\DOMS(Y2),INT by A7,A6,A5,RELSET_1:4
,XBOOLE_1:27;
      hence thesis by Def16;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

theorem
  f1 <--> f2 = <-> (f2 <--> f1)
proof
  set f = f2<-->f1;
A1: dom(f1<-->f2) = dom f1 /\ dom f2 & dom(f2<-->f1) = dom f2 /\ dom f1 by
Def46;
  hence
A2: dom(f1<-->f2) = dom<->f by Def33;
  let x;
  assume
A3: x in dom(f1<-->f2);
  hence (f1<-->f2).x = f1.x-f2.x by Def46
    .= -(f2.x-f1.x) by Th18
    .= -f.x by A1,A3,Def46
    .= (<->f).x by A2,A3,Def33;
end;

theorem
  <-> (f1 <--> f2) = (<->f1) <++> f2
proof
  set f3 = f1 <--> f2, f4 = <->f1;
A1: dom <->f3 = dom f3 by Def33;
A2: dom(f1 <--> f2) = dom f1 /\ dom f2 & dom <->f1 = dom f1 by Def33,Def46;
  hence
A3: dom(<->f3) = dom(f4<++>f2) by A1,Def45;
  let x;
  assume
A4: x in dom(<->f3);
  then
A5: x in dom f4 by A2,A1,XBOOLE_0:def 4;
  thus (<->f3).x = -f3.x by A4,Def33
    .= -(f1.x-f2.x) by A1,A4,Def46
    .= -f1.x--f2.x by Th17
    .= f4.x + f2.x by A5,Def33
    .= (f4<++>f2).x by A3,A4,Def45;
end;

theorem
  f <++> f1 <--> f2 = f <++> (f1<-->f2)
proof
  set f3 = f<++>f1, f4 = f1<-->f2;
A1: dom(f3<-->f2) = dom f3 /\ dom f2 by Def46;
A2: dom(f<++>f4) = dom f /\ dom f4 by Def45;
  dom f3 = dom f /\ dom f1 & dom f4 = dom f1 /\ dom f2 by Def45,Def46;
  hence
A3: dom(f3<-->f2) = dom(f<++>f4) by A1,A2,XBOOLE_1:16;
  let x;
  assume
A4: x in dom(f3<-->f2);
  then
A5: x in dom f4 by A2,A3,XBOOLE_0:def 4;
A6: x in dom f3 by A1,A4,XBOOLE_0:def 4;
  thus (f3<-->f2).x = f3.x - f2.x by A4,Def46
    .= f.x + f1.x - f2.x by A6,Def45
    .= f.x + (f1.x - f2.x) by RFUNCT_1:8
    .= f.x + f4.x by A5,Def46
    .= (f<++>f4).x by A3,A4,Def45;
end;

theorem
  f <--> f1 <++> f2 = f <--> (f1 <--> f2)
proof
  set f3 = f<-->f1, f4 = f1<-->f2;
A1: dom(f3<++>f2) = dom f3 /\ dom f2 by Def45;
A2: dom(f<-->f4) = dom f /\ dom f4 by Def46;
  dom f3 = dom f /\ dom f1 & dom f4 = dom f1 /\ dom f2 by Def46;
  hence
A3: dom(f3<++>f2) = dom(f<-->f4) by A1,A2,XBOOLE_1:16;
  let x;
  assume
A4: x in dom(f3<++>f2);
  then
A5: x in dom f4 by A2,A3,XBOOLE_0:def 4;
A6: x in dom f3 by A1,A4,XBOOLE_0:def 4;
  thus (f3<++>f2).x = f3.x + f2.x by A4,Def45
    .= f.x - f1.x + f2.x by A6,Def46
    .= f.x - (f1.x - f2.x) by RFUNCT_1:22
    .= f.x - f4.x by A5,Def46
    .= (f<-->f4).x by A3,A4,Def46;
end;

theorem
  f <--> f1 <--> f2 = f <--> (f1 <++> f2)
proof
  set f3 = f<-->f1, f4 = f1<++>f2;
A1: dom(f3<-->f2) = dom f3 /\ dom f2 by Def46;
A2: dom(f<-->f4) = dom f /\ dom f4 by Def46;
  dom f3 = dom f /\ dom f1 & dom f4 = dom f1 /\ dom f2 by Def45,Def46;
  hence
A3: dom(f3<-->f2) = dom(f<-->f4) by A1,A2,XBOOLE_1:16;
  let x;
  assume
A4: x in dom(f3<-->f2);
  then
A5: x in dom f4 by A2,A3,XBOOLE_0:def 4;
A6: x in dom f3 by A1,A4,XBOOLE_0:def 4;
  thus (f3<-->f2).x = f3.x - f2.x by A4,Def46
    .= f.x - f1.x - f2.x by A6,Def46
    .= f.x - (f1.x + f2.x) by RFUNCT_1:20
    .= f.x - f4.x by A5,Def45
    .= (f<-->f4).x by A3,A4,Def46;
end;

theorem
  f <--> f1 <--> f2 = f <--> f2 <--> f1
proof
  set f3 = f<-->f1, f4 = f<-->f2;
A1: dom(f3<-->f2) = dom f3 /\ dom f2 by Def46;
A2: dom(f4<-->f1) = dom f4 /\ dom f1 by Def46;
  dom f3 = dom f /\ dom f1 & dom f4 = dom f /\ dom f2 by Def46;
  hence
A3: dom(f3<-->f2) = dom(f4<-->f1) by A1,A2,XBOOLE_1:16;
  let x;
  assume
A4: x in dom(f3<-->f2);
  then
A5: x in dom f4 by A2,A3,XBOOLE_0:def 4;
A6: x in dom f3 by A1,A4,XBOOLE_0:def 4;
  thus (f3<-->f2).x = f3.x - f2.x by A4,Def46
    .= f.x - f1.x - f2.x by A6,Def46
    .= f.x - f2.x - f1.x by RFUNCT_1:23
    .= f4.x - f1.x by A5,Def46
    .= (f4<-->f1).x by A3,A4,Def46;
end;

definition

  let Y1, Y2 be complex-functions-membered set;
  let f be Y1-valued Function;
  let g be Y2-valued Function;
  deffunc F(set) = f.$1(#)g.$1;
  func f<##>g -> Function means
  :Def47:
  dom it = dom f /\ dom g & for x being
  set st x in dom it holds it.x = f.x (#) g.x;
  existence
  proof
    ex F being Function st dom F = dom f /\ dom g & for x being set st x
    in dom f /\ dom g holds F.x = F(x) from FUNCT_1:sch 3;
    hence thesis;
  end;
  uniqueness
  proof
    let F,G be Function such that
A1: dom F = dom f /\ dom g and
A2: for x being set st x in dom F holds F.x = F(x) and
A3: dom G = dom f /\ dom g and
A4: for x being set st x in dom G holds G.x = F(x);
    thus dom F = dom G by A1,A3;
    let x;
    assume
A5: x in dom F;
    hence F.x = F(x) by A2
      .= G.x by A1,A3,A4,A5;
  end;
end;

definition
  let X1, X2 be set;
  let Y1, Y2 be complex-functions-membered set;
  let f be PartFunc of X1,Y1;
  let g be PartFunc of X2,Y2;
  redefine func f<##>g -> PartFunc of X1 /\ X2, C_PFuncs(DOMS(Y1)/\DOMS(Y2));
  coherence
  proof
    set h = f<##>g;
A1: dom h = dom f /\ dom g by Def47;
    rng h c= C_PFuncs(DOMS(Y1)/\DOMS(Y2))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
A4:   h.x = f.x (#) g.x by A2,Def47;
      then reconsider y as Function by A3;
A5:   rng y c= COMPLEX
      proof
        let b be set;
        thus thesis by A3,A4,XCMPLX_0:def 2;
      end;
      x in dom g by A1,A2,XBOOLE_0:def 4;
      then g.x in Y2 by PARTFUN1:4;
      then dom(g.x) in {dom m where m is Element of Y2: not contradiction};
      then
A6:   dom(g.x) c= DOMS(Y2) by ZFMISC_1:74;
      x in dom f by A1,A2,XBOOLE_0:def 4;
      then f.x in Y1 by PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y1: not contradiction};
      then
A7:   dom(f.x) c= DOMS(Y1) by ZFMISC_1:74;
      dom y = dom(f.x) /\ dom(g.x) by A3,A4,VALUED_1:def 4;
      then y is PartFunc of DOMS(Y1)/\DOMS(Y2),COMPLEX by A7,A6,A5,RELSET_1:4
,XBOOLE_1:27;
      hence thesis by Def8;
    end;
    hence thesis by A1,RELSET_1:4,XBOOLE_1:27;
  end;
end;

definition
  let X1, X2 be set;
  let Y1, Y2 be real-functions-membered set;
  let f be PartFunc of X1,Y1;
  let g be PartFunc of X2,Y2;
  redefine func f<##>g -> PartFunc of X1 /\ X2, R_PFuncs(DOMS(Y1)/\DOMS(Y2));
  coherence
  proof
    set h = f<##>g;
A1: dom h = dom f /\ dom g by Def47;
    rng h c= R_PFuncs(DOMS(Y1)/\DOMS(Y2))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = f.x (#) g.x by A2,Def47;
A5:   rng y c= REAL
      proof
        let b be set;
        thus thesis by A3,A4,XREAL_0:def 1;
      end;
      x in dom g by A1,A2,XBOOLE_0:def 4;
      then g.x in Y2 by PARTFUN1:4;
      then dom(g.x) in {dom m where m is Element of Y2: not contradiction};
      then
A6:   dom(g.x) c= DOMS(Y2) by ZFMISC_1:74;
      x in dom f by A1,A2,XBOOLE_0:def 4;
      then f.x in Y1 by PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y1: not contradiction};
      then
A7:   dom(f.x) c= DOMS(Y1) by ZFMISC_1:74;
      dom y = dom(f.x) /\ dom(g.x) by A3,A4,VALUED_1:def 4;
      then y is PartFunc of DOMS(Y1)/\DOMS(Y2),REAL by A7,A6,A5,RELSET_1:4
,XBOOLE_1:27;
      hence thesis by Def12;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X1, X2 be set;
  let Y1, Y2 be rational-functions-membered set;
  let f be PartFunc of X1,Y1;
  let g be PartFunc of X2,Y2;
  redefine func f<##>g -> PartFunc of X1 /\ X2, Q_PFuncs(DOMS(Y1)/\DOMS(Y2));
  coherence
  proof
    set h = f<##>g;
A1: dom h = dom f /\ dom g by Def47;
    rng h c= Q_PFuncs(DOMS(Y1)/\DOMS(Y2))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = f.x (#) g.x by A2,Def47;
A5:   rng y c= RAT
      proof
        let b be set;
        thus thesis by A3,A4,RAT_1:def 2;
      end;
      x in dom g by A1,A2,XBOOLE_0:def 4;
      then g.x in Y2 by PARTFUN1:4;
      then dom(g.x) in {dom m where m is Element of Y2: not contradiction};
      then
A6:   dom(g.x) c= DOMS(Y2) by ZFMISC_1:74;
      x in dom f by A1,A2,XBOOLE_0:def 4;
      then f.x in Y1 by PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y1: not contradiction};
      then
A7:   dom(f.x) c= DOMS(Y1) by ZFMISC_1:74;
      dom y = dom(f.x) /\ dom(g.x) by A3,A4,VALUED_1:def 4;
      then y is PartFunc of DOMS(Y1)/\DOMS(Y2),RAT by A7,A6,A5,RELSET_1:4
,XBOOLE_1:27;
      hence thesis by Def14;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X1, X2 be set;
  let Y1, Y2 be integer-functions-membered set;
  let f be PartFunc of X1,Y1;
  let g be PartFunc of X2,Y2;
  redefine func f<##>g -> PartFunc of X1 /\ X2, I_PFuncs(DOMS(Y1)/\DOMS(Y2));
  coherence
  proof
    set h = f<##>g;
A1: dom h = dom f /\ dom g by Def47;
    rng h c= I_PFuncs(DOMS(Y1)/\DOMS(Y2))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = f.x (#) g.x by A2,Def47;
A5:   rng y c= INT
      proof
        let b be set;
        thus thesis by A3,A4,INT_1:def 2;
      end;
      x in dom g by A1,A2,XBOOLE_0:def 4;
      then g.x in Y2 by PARTFUN1:4;
      then dom(g.x) in {dom m where m is Element of Y2: not contradiction};
      then
A6:   dom(g.x) c= DOMS(Y2) by ZFMISC_1:74;
      x in dom f by A1,A2,XBOOLE_0:def 4;
      then f.x in Y1 by PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y1: not contradiction};
      then
A7:   dom(f.x) c= DOMS(Y1) by ZFMISC_1:74;
      dom y = dom(f.x) /\ dom(g.x) by A3,A4,VALUED_1:def 4;
      then y is PartFunc of DOMS(Y1)/\DOMS(Y2),INT by A7,A6,A5,RELSET_1:4
,XBOOLE_1:27;
      hence thesis by Def16;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X1, X2 be set;
  let Y1, Y2 be natural-functions-membered set;
  let f be PartFunc of X1,Y1;
  let g be PartFunc of X2,Y2;
  redefine func f<##>g -> PartFunc of X1 /\ X2, N_PFuncs(DOMS(Y1)/\DOMS(Y2));
  coherence
  proof
    set h = f<##>g;
A1: dom h = dom f /\ dom g by Def47;
    rng h c= N_PFuncs(DOMS(Y1)/\DOMS(Y2))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = f.x (#) g.x by A2,Def47;
A5:   rng y c= NAT
      proof
        let b be set;
        thus thesis by A3,A4,ORDINAL1:def 12;
      end;
      x in dom g by A1,A2,XBOOLE_0:def 4;
      then g.x in Y2 by PARTFUN1:4;
      then dom(g.x) in {dom m where m is Element of Y2: not contradiction};
      then
A6:   dom(g.x) c= DOMS(Y2) by ZFMISC_1:74;
      x in dom f by A1,A2,XBOOLE_0:def 4;
      then f.x in Y1 by PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y1: not contradiction};
      then
A7:   dom(f.x) c= DOMS(Y1) by ZFMISC_1:74;
      dom y = dom(f.x) /\ dom(g.x) by A3,A4,VALUED_1:def 4;
      then y is PartFunc of DOMS(Y1)/\DOMS(Y2),NAT by A7,A6,A5,RELSET_1:4
,XBOOLE_1:27;
      hence thesis by Def18;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

theorem Th83:
  f1 <##> f2 = f2 <##> f1
proof
  dom(f1<##>f2) = dom f1 /\ dom f2 by Def47;
  hence
A1: dom(f1<##>f2) = dom(f2<##>f1) by Def47;
  let x;
  assume
A2: x in dom(f1<##>f2);
  hence (f1<##>f2).x = f1.x (#) f2.x by Def47
    .= (f2<##>f1).x by A1,A2,Def47;
end;

theorem
  (f <##> f1) <##> f2 = f <##> (f1 <##> f2)
proof
  set f3 = f<##>f1, f4 = f1<##>f2;
A1: dom(f3<##>f2) = dom f3 /\ dom f2 by Def47;
A2: dom(f<##>f4) = dom f /\ dom f4 by Def47;
  dom f3 = dom f /\ dom f1 & dom f4 = dom f1 /\ dom f2 by Def47;
  hence
A3: dom(f3<##>f2) = dom(f<##>f4) by A1,A2,XBOOLE_1:16;
  let x;
  assume
A4: x in dom(f3<##>f2);
  then
A5: x in dom f4 by A2,A3,XBOOLE_0:def 4;
A6: x in dom f3 by A1,A4,XBOOLE_0:def 4;
  thus (f3<##>f2).x = f3.x (#) f2.x by A4,Def47
    .= f.x (#) f1.x (#) f2.x by A6,Def47
    .= f.x (#) (f1.x (#) f2.x) by RFUNCT_1:9
    .= f.x (#) f4.x by A5,Def47
    .= (f<##>f4).x by A3,A4,Def47;
end;

theorem
  (<->f1) <##> f2 = <-> (f1<##>f2)
proof
  set f3 = f1<##>f2, f4 = <->f1;
A1: dom f3 = dom f1 /\ dom f2 & dom f4 = dom f1 by Def33,Def47;
  dom(f4<##>f2) = dom f4 /\ dom f2 by Def47;
  hence
A2: dom(f4<##>f2) = dom(<->f3) by A1,Def33;
  let x;
  assume
A3: x in dom(f4<##>f2);
  then
A4: x in dom f3 by A1,Def47;
  then
A5: x in dom <->f1 by A1,XBOOLE_0:def 4;
  thus (f4<##>f2).x = f4.x (#) f2.x by A3,Def47
    .= (-(f1.x)) (#) f2.x by A5,Def33
    .= -(f1.x) (#) f2.x by Th25
    .= -f3.x by A4,Def47
    .= (<->f3).x by A2,A3,Def33;
end;

theorem
  f1 <##> <->f2 = <-> (f1<##>f2)
proof
  set f3 = f1<##>f2, f4 = <->f2;
A1: dom f3 = dom f1 /\ dom f2 & dom f4 = dom f2 by Def33,Def47;
  dom(f1<##>f4) = dom f1 /\ dom f4 by Def47;
  hence
A2: dom(f1<##>f4) = dom(<->f3) by A1,Def33;
  let x;
  assume
A3: x in dom(f1<##>f4);
  then
A4: x in dom f3 by A1,Def47;
  then
A5: x in dom <->f2 by A1,XBOOLE_0:def 4;
  thus (f1<##>f4).x = f1.x (#) f4.x by A3,Def47
    .= f1.x (#) -f2.x by A5,Def33
    .= -(f1.x) (#) f2.x by Th25
    .= -f3.x by A4,Def47
    .= (<->f3).x by A2,A3,Def33;
end;

theorem Th87:
  f <##> (f1<++>f2) = (f<##>f1) <++> (f<##>f2)
proof
  set f3 = f<##>f1, f4 = f<##>f2, f5 = f1<++>f2;
A1: dom(f<##>f5) = dom f /\ dom f5 by Def47;
A2: dom f5 = dom f1 /\ dom f2 by Def45;
A3: dom(f3<++>f4) = dom f3 /\ dom f4 by Def45;
  dom f3 = dom f /\ dom f1 & dom f4 = dom f /\ dom f2 by Def47;
  hence
A4: dom(f<##>f5) = dom(f3<++>f4) by A1,A3,A2,Lm1;
  let x;
  assume
A5: x in dom(f<##>f5);
  then
A6: x in dom f3 by A3,A4,XBOOLE_0:def 4;
A7: x in dom f5 by A1,A5,XBOOLE_0:def 4;
A8: x in dom f4 by A3,A4,A5,XBOOLE_0:def 4;
  thus (f<##>f5).x = f.x (#) f5.x by A5,Def47
    .= f.x (#) (f1.x + f2.x) by A7,Def45
    .= f.x (#) f1.x + f.x (#) f2.x by RFUNCT_1:10
    .= f3.x + f.x (#) f2.x by A6,Def47
    .= f3.x + f4.x by A8,Def47
    .= (f3<++>f4).x by A4,A5,Def45;
end;

theorem
  (f1<++>f2) <##> f = (f1<##>f) <++> (f2<##>f)
proof
  set f3 = f1<##>f, f4 = f2<##>f, f5 = f1<++>f2;
A1: f1<##>f = f<##>f1 & f2<##>f = f<##>f2 by Th83;
  thus f5 <##> f = f <##> f5 by Th83
    .= f3 <++> f4 by A1,Th87;
end;

theorem Th89:
  f <##> (f1<-->f2) = (f<##>f1) <--> (f<##>f2)
proof
  set f3 = f<##>f1, f4 = f<##>f2, f5 = f1<-->f2;
A1: dom(f<##>f5) = dom f /\ dom f5 by Def47;
A2: dom f5 = dom f1 /\ dom f2 by Def46;
A3: dom(f3<-->f4) = dom f3 /\ dom f4 by Def46;
  dom f3 = dom f /\ dom f1 & dom f4 = dom f /\ dom f2 by Def47;
  hence
A4: dom(f<##>f5) = dom(f3<-->f4) by A1,A3,A2,Lm1;
  let x;
  assume
A5: x in dom(f<##>f5);
  then
A6: x in dom f3 by A3,A4,XBOOLE_0:def 4;
A7: x in dom f5 by A1,A5,XBOOLE_0:def 4;
A8: x in dom f4 by A3,A4,A5,XBOOLE_0:def 4;
  thus (f<##>f5).x = f.x (#) f5.x by A5,Def47
    .= f.x (#) (f1.x - f2.x) by A7,Def46
    .= f.x (#) f1.x - f.x (#) f2.x by RFUNCT_1:15
    .= f3.x - f.x (#) f2.x by A6,Def47
    .= f3.x - f4.x by A8,Def47
    .= (f3<-->f4).x by A4,A5,Def46;
end;

theorem
  (f1<-->f2) <##> f = (f1<##>f) <--> (f2<##>f)
proof
  set f3 = f1<##>f, f4 = f2<##>f, f5 = f1<-->f2;
A1: f1<##>f = f<##>f1 & f2<##>f = f<##>f2 by Th83;
  thus f5 <##> f = f <##> f5 by Th83
    .= f3 <--> f4 by A1,Th89;
end;

definition

  let Y1, Y2 be complex-functions-membered set;
  let f be Y1-valued Function;
  let g be Y2-valued Function;
  deffunc F(set) = f.$1/"g.$1;
  func f<//>g -> Function means
  :Def48:
  dom it = dom f /\ dom g & for x being
  set st x in dom it holds it.x = f.x /" g.x;
  existence
  proof
    ex F being Function st dom F = dom f /\ dom g & for x being set st x
    in dom f /\ dom g holds F.x = F(x) from FUNCT_1:sch 3;
    hence thesis;
  end;
  uniqueness
  proof
    let F,G be Function such that
A1: dom F = dom f /\ dom g and
A2: for x being set st x in dom F holds F.x = F(x) and
A3: dom G = dom f /\ dom g and
A4: for x being set st x in dom G holds G.x = F(x);
    thus dom F = dom G by A1,A3;
    let x;
    assume
A5: x in dom F;
    hence F.x = F(x) by A2
      .= G.x by A1,A3,A4,A5;
  end;
end;

definition
  let X1, X2 be set;
  let Y1, Y2 be complex-functions-membered set;
  let f be PartFunc of X1,Y1;
  let g be PartFunc of X2,Y2;
  redefine func f<//>g -> PartFunc of X1 /\ X2, C_PFuncs(DOMS(Y1)/\DOMS(Y2));
  coherence
  proof
    set h = f<//>g;
A1: dom h = dom f /\ dom g by Def48;
    rng h c= C_PFuncs(DOMS(Y1)/\DOMS(Y2))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
A4:   h.x = f.x /" g.x by A2,Def48;
      then reconsider y as Function by A3;
A5:   rng y c= COMPLEX
      proof
        let b be set;
        thus thesis by A3,A4,XCMPLX_0:def 2;
      end;
      x in dom g by A1,A2,XBOOLE_0:def 4;
      then g.x in Y2 by PARTFUN1:4;
      then dom(g.x) in {dom m where m is Element of Y2: not contradiction};
      then
A6:   dom(g.x) c= DOMS(Y2) by ZFMISC_1:74;
      x in dom f by A1,A2,XBOOLE_0:def 4;
      then f.x in Y1 by PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y1: not contradiction};
      then
A7:   dom(f.x) c= DOMS(Y1) by ZFMISC_1:74;
      dom y = dom(f.x) /\ dom(g.x) by A3,A4,VALUED_1:16;
      then y is PartFunc of DOMS(Y1)/\DOMS(Y2),COMPLEX by A7,A6,A5,RELSET_1:4
,XBOOLE_1:27;
      hence thesis by Def8;
    end;
    hence thesis by A1,RELSET_1:4,XBOOLE_1:27;
  end;
end;

definition
  let X1, X2 be set;
  let Y1, Y2 be real-functions-membered set;
  let f be PartFunc of X1,Y1;
  let g be PartFunc of X2,Y2;
  redefine func f<//>g -> PartFunc of X1 /\ X2, R_PFuncs(DOMS(Y1)/\DOMS(Y2));
  coherence
  proof
    set h = f<//>g;
A1: dom h = dom f /\ dom g by Def48;
    rng h c= R_PFuncs(DOMS(Y1)/\DOMS(Y2))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = f.x /" g.x by A2,Def48;
A5:   rng y c= REAL
      proof
        let b be set;
        thus thesis by A3,A4,XREAL_0:def 1;
      end;
      x in dom g by A1,A2,XBOOLE_0:def 4;
      then g.x in Y2 by PARTFUN1:4;
      then dom(g.x) in {dom m where m is Element of Y2: not contradiction};
      then
A6:   dom(g.x) c= DOMS(Y2) by ZFMISC_1:74;
      x in dom f by A1,A2,XBOOLE_0:def 4;
      then f.x in Y1 by PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y1: not contradiction};
      then
A7:   dom(f.x) c= DOMS(Y1) by ZFMISC_1:74;
      dom y = dom(f.x) /\ dom(g.x) by A3,A4,VALUED_1:16;
      then y is PartFunc of DOMS(Y1)/\DOMS(Y2),REAL by A7,A6,A5,RELSET_1:4
,XBOOLE_1:27;
      hence thesis by Def12;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

definition
  let X1, X2 be set;
  let Y1, Y2 be rational-functions-membered set;
  let f be PartFunc of X1,Y1;
  let g be PartFunc of X2,Y2;
  redefine func f<//>g -> PartFunc of X1 /\ X2, Q_PFuncs(DOMS(Y1)/\DOMS(Y2));
  coherence
  proof
    set h = f<//>g;
A1: dom h = dom f /\ dom g by Def48;
    rng h c= Q_PFuncs(DOMS(Y1)/\DOMS(Y2))
    proof
      let y be set;
      assume y in rng h;
      then consider x such that
A2:   x in dom h and
A3:   h.x = y by FUNCT_1:def 3;
      reconsider y as Function by A3;
A4:   h.x = f.x /" g.x by A2,Def48;
A5:   rng y c= RAT
      proof
        let b be set;
        thus thesis by A3,A4,RAT_1:def 2;
      end;
      x in dom g by A1,A2,XBOOLE_0:def 4;
      then g.x in Y2 by PARTFUN1:4;
      then dom(g.x) in {dom m where m is Element of Y2: not contradiction};
      then
A6:   dom(g.x) c= DOMS(Y2) by ZFMISC_1:74;
      x in dom f by A1,A2,XBOOLE_0:def 4;
      then f.x in Y1 by PARTFUN1:4;
      then dom(f.x) in {dom m where m is Element of Y1: not contradiction};
      then
A7:   dom(f.x) c= DOMS(Y1) by ZFMISC_1:74;
      dom y = dom(f.x) /\ dom(g.x) by A3,A4,VALUED_1:16;
      then y is PartFunc of DOMS(Y1)/\DOMS(Y2),RAT by A7,A6,A5,RELSET_1:4
,XBOOLE_1:27;
      hence thesis by Def14;
    end;
    hence thesis by A1,RELSET_1:4;
  end;
end;

theorem
  (<->f1) <//> f2 = <-> (f1<//>f2)
proof
  set f3 = f1<//>f2, f4 = <->f1;
A1: dom f3 = dom f1 /\ dom f2 & dom f4 = dom f1 by Def33,Def48;
  dom(f4<//>f2) = dom f4 /\ dom f2 by Def48;
  hence
A2: dom(f4<//>f2) = dom(<->f3) by A1,Def33;
  let x;
  assume
A3: x in dom(f4<//>f2);
  then
A4: x in dom f3 by A1,Def48;
  then
A5: x in dom <->f1 by A1,XBOOLE_0:def 4;
  thus (f4<//>f2).x = f4.x /" f2.x by A3,Def48
    .= (-(f1.x)) /" f2.x by A5,Def33
    .= -(f1.x) /" f2.x by Th25
    .= -f3.x by A4,Def48
    .= (<->f3).x by A2,A3,Def33;
end;

theorem
  f1 <//> <->f2 = <-> (f1<//>f2)
proof
  set f3 = f1<//>f2, f4 = <->f2;
A1: dom f3 = dom f1 /\ dom f2 & dom f4 = dom f2 by Def33,Def48;
  dom(f1<//>f4) = dom f1 /\ dom f4 by Def48;
  hence
A2: dom(f1<//>f4) = dom(<->f3) by A1,Def33;
  let x;
  assume
A3: x in dom(f1<//>f4);
  then
A4: x in dom f3 by A1,Def48;
  then
A5: x in dom <->f2 by A1,XBOOLE_0:def 4;
  thus (f1<//>f4).x = f1.x /" f4.x by A3,Def48
    .= f1.x /" -f2.x by A5,Def33
    .= -(f1.x) /" f2.x by Th27
    .= -f3.x by A4,Def48
    .= (<->f3).x by A2,A3,Def33;
end;

theorem
  f <##> f1 <//> f2 = f <##> (f1<//>f2)
proof
  set f3 = f<##>f1, f4 = f1<//>f2;
A1: dom(f3<//>f2) = dom f3 /\ dom f2 by Def48;
A2: dom(f<##>f4) = dom f /\ dom f4 by Def47;
  dom f3 = dom f /\ dom f1 & dom f4 = dom f1 /\ dom f2 by Def47,Def48;
  hence
A3: dom(f3<//>f2) = dom(f<##>f4) by A1,A2,XBOOLE_1:16;
  let x;
  assume
A4: x in dom(f3<//>f2);
  then
A5: x in dom f4 by A2,A3,XBOOLE_0:def 4;
A6: x in dom f3 by A1,A4,XBOOLE_0:def 4;
  thus (f3<//>f2).x = f3.x /" f2.x by A4,Def48
    .= f.x (#) f1.x /" f2.x by A6,Def47
    .= f.x (#) (f1.x /" f2.x) by Th19
    .= f.x (#) f4.x by A5,Def48
    .= (f<##>f4).x by A3,A4,Def47;
end;

theorem
  f <//> f1 <##> f2 = f <##> f2 <//> f1
proof
  set f3 = f<//>f1, f4 = f<##>f2;
A1: dom(f3<##>f2) = dom f3 /\ dom f2 by Def47;
A2: dom(f4<//>f1) = dom f4 /\ dom f1 by Def48;
  dom f3 = dom f /\ dom f1 & dom f4 = dom f /\ dom f2 by Def47,Def48;
  hence
A3: dom(f3<##>f2) = dom(f4<//>f1) by A1,A2,XBOOLE_1:16;
  let x;
  assume
A4: x in dom(f3<##>f2);
  then
A5: x in dom f4 by A2,A3,XBOOLE_0:def 4;
A6: x in dom f3 by A1,A4,XBOOLE_0:def 4;
  thus (f3<##>f2).x = f3.x (#) f2.x by A4,Def47
    .= f.x /" f1.x (#) f2.x by A6,Def48
    .= f.x (#) f2.x /" f1.x by Th20
    .= f4.x /" f1.x by A5,Def47
    .= (f4<//>f1).x by A3,A4,Def48;
end;

theorem
  f <//> f1 <//> f2 = f <//> (f1 <##> f2)
proof
  set f3 = f<//>f1, f4 = f1<##>f2;
A1: dom(f3<//>f2) = dom f3 /\ dom f2 by Def48;
A2: dom(f<//>f4) = dom f /\ dom f4 by Def48;
  dom f3 = dom f /\ dom f1 & dom f4 = dom f1 /\ dom f2 by Def47,Def48;
  hence
A3: dom(f3<//>f2) = dom(f<//>f4) by A1,A2,XBOOLE_1:16;
  let x;
  assume
A4: x in dom(f3<//>f2);
  then
A5: x in dom f4 by A2,A3,XBOOLE_0:def 4;
A6: x in dom f3 by A1,A4,XBOOLE_0:def 4;
  thus (f3<//>f2).x = f3.x /" f2.x by A4,Def48
    .= f.x /" f1.x /" f2.x by A6,Def48
    .= f.x /" (f1.x (#) f2.x) by Th21
    .= f.x /" f4.x by A5,Def47
    .= (f<//>f4).x by A3,A4,Def48;
end;

theorem
  (f1<++>f2) <//> f = (f1<//>f) <++> (f2<//>f)
proof
  set f3 = f1<//>f, f4 = f2<//>f, f5 = f1<++>f2;
A1: dom(f5<//>f) = dom f /\ dom f5 by Def48;
A2: dom f5 = dom f1 /\ dom f2 by Def45;
A3: dom(f3<++>f4) = dom f3 /\ dom f4 by Def45;
  dom f3 = dom f1 /\ dom f & dom f4 = dom f2 /\ dom f by Def48;
  hence
A4: dom(f5<//>f) = dom(f3<++>f4) by A1,A3,A2,Lm1;
  let x;
  assume
A5: x in dom(f5<//>f);
  then
A6: x in dom f3 by A3,A4,XBOOLE_0:def 4;
A7: x in dom f5 by A1,A5,XBOOLE_0:def 4;
A8: x in dom f4 by A3,A4,A5,XBOOLE_0:def 4;
  thus (f5<//>f).x = f5.x /" f.x by A5,Def48
    .= (f1.x+f2.x) /" f.x by A7,Def45
    .= f1.x /" f.x + f2.x /" f.x by RFUNCT_1:10
    .= f3.x + f2.x /" f.x by A6,Def48
    .= f3.x + f4.x by A8,Def48
    .= (f3<++>f4).x by A4,A5,Def45;
end;

theorem
  (f1<-->f2) <//> f = (f1<//>f) <--> (f2<//>f)
proof
  set f3 = f1<//>f, f4 = f2<//>f, f5 = f1<-->f2;
A1: dom(f5<//>f) = dom f /\ dom f5 by Def48;
A2: dom f5 = dom f1 /\ dom f2 by Def46;
A3: dom(f3<-->f4) = dom f3 /\ dom f4 by Def46;
  dom f3 = dom f1 /\ dom f & dom f4 = dom f2 /\ dom f by Def48;
  hence
A4: dom(f5<//>f) = dom(f3<-->f4) by A1,A3,A2,Lm1;
  let x;
  assume
A5: x in dom(f5<//>f);
  then
A6: x in dom f3 by A3,A4,XBOOLE_0:def 4;
A7: x in dom f5 by A1,A5,XBOOLE_0:def 4;
A8: x in dom f4 by A3,A4,A5,XBOOLE_0:def 4;
  thus (f5<//>f).x = f5.x /" f.x by A5,Def48
    .= (f1.x-f2.x) /" f.x by A7,Def46
    .= f1.x /" f.x - f2.x /" f.x by RFUNCT_1:14
    .= f3.x - f2.x /" f.x by A6,Def48
    .= f3.x - f4.x by A8,Def48
    .= (f3<-->f4).x by A4,A5,Def46;
end;














begin

reserve n,m,k for Element of NAT;
reserve r,r1 for real number;
reserve f,seq,seq1 for Real_Sequence;
reserve x,y for set;

Lm1: n<m implies ex k st m=n+1+k
proof
  assume
A1: n<m;
  then consider k1 be Nat such that
A2: m=n+k1 by NAT_1:10;
  k1 <> 0 by A1,A2;
  then consider n1 be Nat such that
A3: k1=n1+1 by NAT_1:6;
  reconsider n1 as Element of NAT by ORDINAL1:def 12;
  take n1;
  thus thesis by A2,A3;
end;

Lm2: ((for n holds seq.n<seq.(n+1)) implies for n,k holds seq.n<seq.(n+1+k)) &
((for n,k holds seq.n<seq.(n+1+k)) implies for n,m st n<m holds seq.n<seq.m) &
((for n,m st n<m holds seq.n<seq.m) implies for n holds seq.n<seq.(n+1))
proof
  thus (for n holds seq.n<seq.(n+1)) implies for n,k holds seq.n<seq.(n+1+k)
  proof
    assume
A1: for n holds seq.n<seq.(n+1);
    let n;
    defpred X[Element of NAT] means seq.n<seq.(n+1+$1);
A2: now
      let k such that
A3:   X[k];
      seq.(n+1+k)<seq.(n+1+k+1) by A1;
      hence X[k+1] by A3,XXREAL_0:2;
    end;
A4: X[0] by A1;
    thus for k holds X[k] from NAT_1:sch 1(A4,A2);
  end;
  thus (for n,k holds seq.n<seq.(n+1+k)) implies for n,m st n<m holds seq.n<
  seq.m
  proof
    assume
A5: for n,k holds seq.n<seq.(n+1+k);
    let n,m;
    assume n<m;
    then ex k st m=n+1+k by Lm1;
    hence thesis by A5;
  end;
  assume
A6: for n,m st n<m holds seq.n<seq.m;
  let n;
  n<n+1 by NAT_1:13;
  hence thesis by A6;
end;

Lm3: ((for n holds seq.(n+1)<seq.n) implies for n,k holds seq.(n+1+k)<seq.n) &
((for n,k holds seq.(n+1+k)<seq.n) implies for n,m st n<m holds seq.m<seq.n) &
((for n,m st n<m holds seq.m<seq.n) implies for n holds seq.(n+1)<seq.n)
proof
  thus (for n holds seq.(n+1)<seq.n) implies for n,k holds seq.(n+1+k)<seq.n
  proof
    assume
A1: for n holds seq.(n+1)<seq.n;
    let n;
    defpred X[Element of NAT] means seq.(n+1+$1)<seq.n;
A2: now
      let k such that
A3:   X[k];
      seq.(n+1+k+1)<seq.(n+1+k) by A1;
      hence X[k+1] by A3,XXREAL_0:2;
    end;
A4: X[0] by A1;
    thus for k holds X[k] from NAT_1:sch 1(A4,A2);
  end;
  thus (for n,k holds seq.(n+1+k)<seq.n) implies for n,m st n<m holds seq.m<
  seq.n
  proof
    assume
A5: for n,k holds seq.(n+1+k)<seq.n;
    let n,m;
    assume n<m;
    then ex k st m=n+1+k by Lm1;
    hence thesis by A5;
  end;
  assume
A6: for n,m st n<m holds seq.m<seq.n;
  let n;
  n<n+1 by NAT_1:13;
  hence thesis by A6;
end;

Lm4: ((for n holds seq.n<=seq.(n+1)) implies for n,k holds seq.n<=seq.(n+k)) &
((for n,k holds seq.n<=seq.(n+k)) implies for n,m st n<=m holds seq.n<=seq.m) &
((for n,m st n<=m holds seq.n<=seq.m) implies for n holds seq.n<=seq.(n+1))
proof
  thus (for n holds seq.n<=seq.(n+1)) implies for n,k holds seq.n<=seq.(n+k)
  proof
    assume
A1: for n holds seq.n<=seq.(n+1);
    let n;
    defpred X[Element of NAT] means seq.n<=seq.(n+$1);
A2: now
      let k such that
A3:   X[k];
      seq.(n+k)<=seq.(n+k+1) by A1;
      hence X[k+1] by A3,XXREAL_0:2;
    end;
A4: X[0];
    thus for k holds X[k] from NAT_1:sch 1(A4,A2);
  end;
  thus (for n,k holds seq.n<=seq.(n+k)) implies for n,m st n<=m holds seq.n<=
  seq.m
  proof
    assume
A5: for n,k holds seq.n<=seq.(n+k);
    let n,m;
    assume n<=m;
    then consider k1 be Nat such that
A6: m=n+k1 by NAT_1:10;
    k1 in NAT by ORDINAL1:def 12;
    hence thesis by A5,A6;
  end;
  assume
A7: for n,m st n<=m holds seq.n<=seq.m;
  let n;
  n<=n+1 by NAT_1:13;
  hence thesis by A7;
end;

Lm5: ((for n holds seq.(n+1)<=seq.n) implies for n,k holds seq.(n+k)<=seq.n) &
((for n,k holds seq.(n+k)<=seq.n) implies for n,m st n<=m holds seq.m<=seq.n) &
((for n,m st n<=m holds seq.m<=seq.n) implies for n holds seq.(n+1)<=seq.n)
proof
  thus (for n holds seq.(n+1)<=seq.n) implies for n,k holds seq.(n+k)<=seq.n
  proof
    assume
A1: for n holds seq.(n+1)<=seq.n;
    let n;
    defpred X[Element of NAT] means seq.(n+$1)<=seq.n;
A2: now
      let k such that
A3:   X[k];
      seq.(n+k+1)<=seq.(n+k) by A1;
      hence X[k+1] by A3,XXREAL_0:2;
    end;
A4: X[0];
    thus for k holds X[k] from NAT_1:sch 1(A4,A2);
  end;
  thus (for n,k holds seq.(n+k)<=seq.n) implies for n,m st n<=m holds seq.m<=
  seq.n
  proof
    assume
A5: for n,k holds seq.(n+k)<=seq.n;
    let n,m;
    assume n<=m;
    then consider k1 be Nat such that
A6: m=n+k1 by NAT_1:10;
    k1 in NAT by ORDINAL1:def 12;
    hence thesis by A5,A6;
  end;
  assume
A7: for n,m st n<=m holds seq.m<=seq.n;
  let n;
  n<=n+1 by NAT_1:13;
  hence thesis by A7;
end;





reserve e1,e2 for ext-real number;

definition

  let f be NAT-defined REAL-valued Function;
  redefine attr f is increasing means
  :Def1:
  for m,n st m in dom f & n in dom f & m < n holds f.m < f.n;
  compatibility
  proof
    thus f is increasing implies for m,n st m in dom f & n in dom f & m < n
    holds f.m < f.n by VALUED_0:def 13;
    assume
A1: for m,n st m in dom f & n in dom f & m < n holds f.m < f.n;
    let e1,e2;
    dom f c= NAT by RELAT_1:def 18;
    hence thesis by A1;
  end;
  redefine attr f is decreasing means
  :Def2:
  for m,n st m in dom f & n in dom f & m < n holds f.m > f.n;
  compatibility
  proof
    thus f is decreasing implies for m,n st m in dom f & n in dom f & m < n
    holds f.m > f.n by VALUED_0:def 14;
    assume
A2: for m,n st m in dom f & n in dom f & m < n holds f.m > f.n;
    let e1,e2;
    dom f c= NAT by RELAT_1:def 18;
    hence thesis by A2;
  end;
  redefine attr f is non-decreasing means
  :Def3:
  for m,n st m in dom f & n in dom f & m <= n holds f.m <= f.n;
  compatibility
  proof
    thus f is non-decreasing implies for m,n st m in dom f & n in dom f & m <=
    n holds f.m <= f.n by VALUED_0:def 15;
    assume
A3: for m,n st m in dom f & n in dom f & m <= n holds f.m <= f.n;
    let e1,e2;
    dom f c= NAT by RELAT_1:def 18;
    hence thesis by A3;
  end;
  redefine attr f is non-increasing means
  :Def4:
  for m,n st m in dom f & n in dom f & m <= n holds f.m >= f.n;
  compatibility
  proof
    thus f is non-increasing implies for m,n st m in dom f & n in dom f & m <=
    n holds f.m >= f.n by VALUED_0:def 16;
    assume
A4: for m,n st m in dom f & n in dom f & m <= n holds f.m >= f.n;
    let e1,e2;
    dom f c= NAT by RELAT_1:def 18;
    hence thesis by A4;
  end;
end;

Lm6: f is increasing iff for n being Element of NAT holds f.n < f.(n+1)
proof
  thus f is increasing implies for n being Element of NAT holds f.n < f.(n+1 )
  proof
    assume
A1: f is increasing;
    let n be Element of NAT;
    dom f = NAT & n < n+1 by FUNCT_2:def 1,NAT_1:13;
    hence thesis by A1,Def1;
  end;
  assume
A2: for n being Element of NAT holds f.n < f.(n+1);
  let m,n;
  assume that
  m in dom f and
  n in dom f and
A3: m < n;
  ex k st n = m+1+k by A3,Lm1;
  hence thesis by A2,Lm2;
end;

Lm7: f is decreasing iff for n being Element of NAT holds f.n > f.(n+1)
proof
  thus f is decreasing implies for n being Element of NAT holds f.n > f.(n+1 )
  proof
    assume
A1: f is decreasing;
    let n be Element of NAT;
    dom f = NAT & n < n+1 by FUNCT_2:def 1,NAT_1:13;
    hence thesis by A1,Def2;
  end;
  assume
A2: for n being Element of NAT holds f.n > f.(n+1);
  let m,n;
  assume that
  m in dom f and
  n in dom f and
A3: m < n;
  ex k st n = m+1+k by A3,Lm1;
  hence thesis by A2,Lm3;
end;

Lm8: f is non-decreasing iff for n being Element of NAT holds f.n <= f.(n+1)
proof
  thus f is non-decreasing implies for n being Element of NAT holds f.n <= f.(
  n+1)
  proof
    assume
A1: f is non-decreasing;
    let n be Element of NAT;
    dom f = NAT & n < n+1 by FUNCT_2:def 1,NAT_1:13;
    hence thesis by A1,Def3;
  end;
  assume
A2: for n being Element of NAT holds f.n <= f.(n+1);
  let m,n;
  assume that
  m in dom f and
  n in dom f and
A3: m <= n;
  consider k be Nat such that
A4: n = m+k by A3,NAT_1:10;
  k in NAT by ORDINAL1:def 12;
  hence thesis by A2,A4,Lm4;
end;

Lm9: f is non-increasing iff for n being Element of NAT holds f.n >= f.(n+1)
proof
  thus f is non-increasing implies for n being Element of NAT holds f.n >= f.(
  n+1)
  proof
    assume
A1: f is non-increasing;
    let n be Element of NAT;
    dom f = NAT & n < n+1 by FUNCT_2:def 1,NAT_1:13;
    hence thesis by A1,Def4;
  end;
  assume
A2: for n being Element of NAT holds f.n >= f.(n+1);
  let m,n;
  assume that
  m in dom f and
  n in dom f and
A3: m <= n;
  consider k be Nat such that
A4: n = m+k by A3,NAT_1:10;
  k in NAT by ORDINAL1:def 12;
  hence thesis by A2,A4,Lm5;
end;

definition

  let seq;
  attr seq is monotone means
  :Def5:
  seq is non-decreasing or seq is non-increasing;
end;

theorem Th1:
  seq is increasing iff for n,m st n<m holds seq.n<seq.m
proof
  thus seq is increasing implies for n,m st n<m holds seq.n<seq.m
  proof
    assume seq is increasing;
    then for n holds seq.n<seq.(n+1) by Lm6;
    then for n,k holds seq.n<seq.(n+1+k) by Lm2;
    hence thesis by Lm2;
  end;
  assume
A1: for n,m st n<m holds seq.n<seq.m;
  let n,m;
  assume that
  n in dom seq and
  m in dom seq and
A2: n < m;
  thus thesis by A1,A2;
end;

theorem
  seq is increasing iff for n,k holds seq.n<seq.(n+1+k)
proof
  thus seq is increasing implies for n,k holds seq.n<seq.(n+1+k)
  proof
    assume seq is increasing;
    then for n holds seq.n<seq.(n+1) by Lm6;
    hence thesis by Lm2;
  end;
  assume for n,k holds seq.n<seq.(n+1+k);
  then for n,m st n<m holds seq.n<seq.m by Lm2;
  hence thesis by Th1;
end;

theorem Th3:
  seq is decreasing iff for n,k holds seq.(n+1+k)<seq.n
proof
  thus seq is decreasing implies for n,k holds seq.(n+1+k)<seq.n
  proof
    assume seq is decreasing;
    then for n holds seq.(n+1)<seq.n by Lm7;
    hence thesis by Lm3;
  end;
  assume
A1: for n,k holds seq.(n+1+k)<seq.n;
  let n,m;
  assume that
  n in dom seq and
  m in dom seq and
A2: n < m;
  thus thesis by A1,A2,Lm3;
end;

theorem Th4:
  seq is decreasing iff for n,m st n<m holds seq.m<seq.n
proof
  thus seq is decreasing implies for n,m st n<m holds seq.m<seq.n
  proof
    assume seq is decreasing;
    then for n,k holds seq.(n+1+k)<seq.n by Th3;
    hence thesis by Lm3;
  end;
  assume
A1: for n,m st n<m holds seq.m<seq.n;
  let n,m;
  assume that
  n in dom seq and
  m in dom seq and
A2: n < m;
  thus thesis by A1,A2;
end;

theorem Th5:
  seq is non-decreasing iff for n,k holds seq.n<=seq.(n+k)
proof
  thus seq is non-decreasing implies for n,k holds seq.n<=seq.(n+k)
  proof
    assume seq is non-decreasing;
    then for n holds seq.n<=seq.(n+1) by Lm8;
    hence thesis by Lm4;
  end;
  assume for n,k holds seq.n<=seq.(n+k);
  then for n holds seq.n<=seq.(n+1);
  hence thesis by Lm8;
end;

theorem Th6:
  seq is non-decreasing iff for n,m st n<=m holds seq.n<=seq.m
proof
  thus seq is non-decreasing implies for n,m st n<=m holds seq.n<=seq.m
  proof
    assume seq is non-decreasing;
    then for n,k holds seq.n<=seq.(n+k) by Th5;
    hence thesis by Lm4;
  end;
  assume
A1: for n,m st n<=m holds seq.n<=seq.m;
  let n,m;
  assume that
  n in dom seq and
  m in dom seq and
A2: n <= m;
  thus thesis by A1,A2;
end;

theorem Th7:
  seq is non-increasing iff for n,k holds seq.(n+k)<=seq.n
proof
  thus seq is non-increasing implies for n,k holds seq.(n+k)<=seq.n
  proof
    assume seq is non-increasing;
    then for n holds seq.(n+1)<=seq.n by Lm9;
    hence thesis by Lm5;
  end;
  assume for n,k holds seq.(n+k)<=seq.n;
  then for n holds seq.(n+1)<=seq.n;
  hence thesis by Lm9;
end;

theorem Th8:
  seq is non-increasing iff for n,m st n<=m holds seq.m<=seq.n
proof
  thus seq is non-increasing implies for n,m st n<=m holds seq.m<=seq.n
  proof
    assume seq is non-increasing;
    then for n,k holds seq.(n+k)<=seq.n by Th7;
    hence thesis by Lm5;
  end;
  assume
A1: for n,m st n<=m holds seq.m<=seq.n;
  let n,m;
  assume that
  n in dom seq and
  m in dom seq and
A2: n <= m;
  thus thesis by A1,A2;
end;





theorem
  seq is increasing implies for n st 0<n holds seq.0<seq.n by Th1;

theorem
  seq is decreasing implies for n st 0<n holds seq.n<seq.0 by Th4;

theorem Th11:
  seq is non-decreasing implies for n holds seq.0<=seq.n by Th6;

theorem Th12:
  seq is non-increasing implies for n holds seq.n<=seq.0 by Th8;

registration
  cluster constant -> non-decreasing non-increasing for PartFunc of NAT, REAL;
  coherence
  proof
    let f be PartFunc of NAT, REAL such that
A1: f is constant;
    thus f is non-decreasing
    proof
      let m,n;
      assume that
A2:   m in dom f & n in dom f and
      m <= n;
      thus thesis by A1,A2,FUNCT_1:def 10;
    end;
    let m,n;
    assume that
A3: m in dom f & n in dom f and
    m <= n;
    thus thesis by A1,A3,FUNCT_1:def 10;
  end;
  cluster non-decreasing non-increasing -> constant for PartFunc of NAT, REAL;
  coherence
  proof
    let f be PartFunc of NAT, REAL such that
A4: f is non-decreasing non-increasing;
    let x,y;
    assume
A5: x in dom f & y in dom f;
    dom f c= NAT by RELAT_1:def 18;
    then reconsider m = x, n = y as Element of NAT by A5;
    m <= n or n <= m;
    then f.m <= f.n & f.n <= f.m by A4,A5,Def3,Def4;
    hence thesis by XXREAL_0:1;
  end;
end;




Lm10: incl NAT is increasing natural-valued
proof
  set seq = incl NAT;
  now
    let n;
    seq.n=n & seq.(n+1)=n+1 by FUNCT_1:18;
    hence seq.n<seq.(n+1) by NAT_1:13;
  end;
  hence thesis by Lm6;
end;

Lm11: id NAT is increasing natural-valued;

registration
  cluster increasing natural-valued for Real_Sequence;
  existence by Lm10;
end;

registration
  cluster increasing for sequence of NAT;
  existence by Lm11;
end;

Lm12: for f being sequence of NAT holds f is increasing iff for n being
Element of NAT holds f.n < f.(n+1)
proof
  let f be sequence of NAT;
  reconsider f as Real_Sequence by FUNCT_2:7;
  f is increasing iff for n being Element of NAT holds f.n < f.(n+1) by Lm6;
  hence thesis;
end;

reserve Nseq for increasing sequence of NAT;

theorem
  seq is sequence of NAT iff for n holds seq.n is Element of NAT
proof
  thus seq is sequence of NAT implies for n holds seq.n is Element of NAT
  proof
    assume seq is sequence of NAT;
    then
A1: rng seq c= NAT by VALUED_0:def 6;
    let n;
    n in NAT;
    then n in dom seq by FUNCT_2:def 1;
    then seq.n in rng seq by FUNCT_1:def 3;
    hence thesis by A1;
  end;
  assume
A2: for n holds seq.n is Element of NAT;
  rng seq c= NAT
  proof
    let x be set;
    assume x in rng seq;
    then consider y such that
A3: y in dom seq and
A4: x = seq.y by FUNCT_1:def 3;
    reconsider y as Element of NAT by A3,FUNCT_2:def 1;
    x = seq.y by A4;
    then x is Element of NAT by A2;
    hence thesis;
  end;
  hence thesis by FUNCT_2:6;
end;

registration
  let Nseq,k;
  cluster Nseq ^\ k -> increasing natural-valued for ext-real-valued Function;
  coherence
  proof
    now
      let n;
      Nseq.(n+k)<Nseq.(n+k+1) by Lm12;
      then (Nseq ^\k).n<Nseq.(n+1+k) by NAT_1:def 3;
      hence (Nseq ^\k).n<(Nseq ^\k).(n+1) by NAT_1:def 3;
    end;
    hence thesis by Lm12;
  end;
end;

definition
  let f be Real_Sequence;
  redefine attr f is increasing means
  for n being Element of NAT holds f.n < f.(n+1);
  compatibility by Lm6;
  redefine attr f is decreasing means
  for n being Element of NAT holds f.n > f.(n+1);
  compatibility by Lm7;
  redefine attr f is non-decreasing means
  for n being Element of NAT holds f.n <= f.(n+1);
  compatibility by Lm8;
  redefine attr f is non-increasing means
  for n being Element of NAT holds f.n >= f.(n+1);
  compatibility by Lm9;
end;

theorem
  for n holds n<=Nseq.n
proof
  defpred X[Element of NAT] means $1<=Nseq.$1;
A1: now
    let k such that
A2: X[k];
    Nseq.k<Nseq.(k+1) by Lm12;
    then k<Nseq.(k+1) by A2,XXREAL_0:2;
    hence X[k+1] by NAT_1:13;
  end;
A3: X[0];
  thus for k holds X[k] from NAT_1:sch 1(A3,A1);
end;

registration
  let s be Real_Sequence, k be Nat;
  cluster s^\k ->real-valued;
  coherence;
end;

theorem Th15:
  (seq+seq1) ^\k=(seq ^\k) + (seq1 ^\k)
proof
  now
    let n;
    thus ((seq+seq1) ^\k).n=(seq+seq1).(n+k) by NAT_1:def 3
      .=seq.(n+k) + seq1.(n+k) by SEQ_1:7
      .=(seq ^\k).n +seq1.(n+k) by NAT_1:def 3
      .=(seq ^\k).n +(seq1 ^\k).n by NAT_1:def 3
      .=((seq ^\k) +(seq1 ^\k)).n by SEQ_1:7;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem Th16:
  (-seq) ^\k=-(seq ^\k)
proof
  now
    let n;
    thus ((-seq) ^\k).n=(-seq).(n+k) by NAT_1:def 3
      .=-(seq.(n+k)) by SEQ_1:10
      .=-((seq ^\ k).n) by NAT_1:def 3
      .=(-(seq ^\k)).n by SEQ_1:10;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  (seq-seq1) ^\k=(seq ^\k)-(seq1 ^\k)
proof
  thus (seq-seq1) ^\k=(seq ^\k)+((-seq1) ^\k) by Th15
    .=(seq ^\k)-(seq1 ^\k) by Th16;
end;

theorem Th18:
  (seq") ^\k=(seq ^\k)"
proof
  now
    let n;
    thus ((seq") ^\k).n=(seq").(n+k) by NAT_1:def 3
      .=(seq.(n+k))" by VALUED_1:10
      .=((seq ^\k).n)" by NAT_1:def 3
      .=((seq ^\k)").n by VALUED_1:10;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem Th19:
  (seq(#)seq1) ^\k=(seq ^\k)(#)(seq1 ^\k)
proof
  now
    let n;
    thus ((seq(#)seq1) ^\k).n=(seq(#)seq1).(n+k) by NAT_1:def 3
      .=seq.(n+k)*seq1.(n+k) by SEQ_1:8
      .=(seq ^\k).n*seq1.(n+k) by NAT_1:def 3
      .=(seq ^\k).n*(seq1 ^\k).n by NAT_1:def 3
      .=((seq ^\k)(#)(seq1 ^\k)).n by SEQ_1:8;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  (seq/"seq1) ^\k=(seq ^\k)/"(seq1 ^\k)
proof
  thus (seq/"seq1) ^\k=(seq ^\k)(#)((seq1") ^\k) by Th19
    .=(seq ^\k)/"(seq1 ^\k) by Th18;
end;

theorem
  (r(#)seq) ^\k=r(#)(seq ^\k)
proof
  now
    let n;
    thus ((r(#)seq) ^\k).n=(r(#)seq).(n+k) by NAT_1:def 3
      .=r*(seq.(n+k)) by SEQ_1:9
      .=r*((seq ^\k).n) by NAT_1:def 3
      .=(r(#)(seq ^\k)).n by SEQ_1:9;
  end;
  hence thesis by FUNCT_2:63;
end;






theorem
  seq is increasing & seq1 is subsequence of seq implies seq1 is increasing
proof
  assume that
A1: seq is increasing and
A2: seq1 is subsequence of seq;
  let n;
  consider Nseq such that
A3: seq1=seq*Nseq by A2,VALUED_0:def 17;
  Nseq.n<Nseq.(n+1) by Lm12;
  then (seq.(Nseq.n))<(seq.(Nseq.(n+1))) by A1,Th1;
  then (seq*Nseq).n<(seq.(Nseq.(n+1))) by FUNCT_2:15;
  hence thesis by A3,FUNCT_2:15;
end;

theorem
  seq is decreasing & seq1 is subsequence of seq implies seq1 is decreasing
proof
  assume that
A1: seq is decreasing and
A2: seq1 is subsequence of seq;
  let n;
  consider Nseq such that
A3: seq1=seq*Nseq by A2,VALUED_0:def 17;
  Nseq.n<Nseq.(n+1) by Lm12;
  then seq.(Nseq.(n+1))<seq.(Nseq.n) by A1,Th4;
  then (seq*Nseq).(n+1)<seq.(Nseq.n) by FUNCT_2:15;
  hence thesis by A3,FUNCT_2:15;
end;

theorem Th24:
  seq is non-decreasing & seq1 is subsequence of seq implies seq1
  is non-decreasing
proof
  assume that
A1: seq is non-decreasing and
A2: seq1 is subsequence of seq;
  let n;
  consider Nseq such that
A3: seq1=seq*Nseq by A2,VALUED_0:def 17;
  Nseq.n<=Nseq.(n+1) by Lm12;
  then (seq.(Nseq.n))<=(seq.(Nseq.(n+1))) by A1,Th6;
  then (seq*Nseq).n<=(seq.(Nseq.(n+1))) by FUNCT_2:15;
  hence thesis by A3,FUNCT_2:15;
end;

theorem Th25:
  seq is non-increasing & seq1 is subsequence of seq implies seq1
  is non-increasing
proof
  assume that
A1: seq is non-increasing and
A2: seq1 is subsequence of seq;
  let n;
  consider Nseq such that
A3: seq1=seq*Nseq by A2,VALUED_0:def 17;
  Nseq.n<=Nseq.(n+1) by Lm12;
  then (seq.(Nseq.(n+1)))<=(seq.(Nseq.n)) by A1,Th8;
  then (seq*Nseq).(n+1)<=(seq.(Nseq.n)) by FUNCT_2:15;
  hence thesis by A3,FUNCT_2:15;
end;

theorem
  seq is monotone & seq1 is subsequence of seq implies seq1 is monotone
proof
  assume that
A1: seq is monotone and
A2: seq1 is subsequence of seq;
A3: now
    assume seq is non-increasing;
    then seq1 is non-increasing by A2,Th25;
    hence thesis by Def5;
  end;
  now
    assume seq is non-decreasing;
    then seq1 is non-decreasing by A2,Th24;
    hence thesis by Def5;
  end;
  hence thesis by A1,A3,Def5;
end;

theorem Th27:
  seq is bounded_above & seq1 is subsequence of seq implies seq1
  is bounded_above
proof
  assume that
A1: seq is bounded_above and
A2: seq1 is subsequence of seq;
  consider r1 such that
A3: for n holds seq.n<r1 by A1,SEQ_2:def 3;
  consider Nseq such that
A4: seq1=seq*Nseq by A2,VALUED_0:def 17;
  take r=r1;
  let n be Element of NAT;
  seq1.n=seq.(Nseq.n) by A4,FUNCT_2:15;
  hence seq1.n<r by A3;
end;

theorem Th28:
  seq is bounded_below & seq1 is subsequence of seq implies seq1
  is bounded_below
proof
  assume that
A1: seq is bounded_below and
A2: seq1 is subsequence of seq;
  consider r1 such that
A3: for n holds r1<seq.n by A1,SEQ_2:def 4;
  consider Nseq such that
A4: seq1=seq*Nseq by A2,VALUED_0:def 17;
  take r=r1;
  let n be Element of NAT;
  seq1.n=seq.(Nseq.n) by A4,FUNCT_2:15;
  hence r<seq1.n by A3;
end;

theorem
  seq is bounded & seq1 is subsequence of seq implies seq1 is bounded
proof
  assume
A1: seq is bounded & seq1 is subsequence of seq;
  hence seq1 is bounded_above by Th27;
  thus thesis by A1,Th28;
end;





theorem
  (seq is increasing & 0<r implies r(#)seq is increasing) & (0=r implies
r(#)seq is constant) & (seq is increasing & r<0 implies r(#)seq is decreasing)
proof
  thus seq is increasing & 0<r implies r(#)seq is increasing
  proof
    assume that
A1: seq is increasing and
A2: 0<r;
    let n;
    seq.n<seq.(n+1) by A1,Lm6;
    then r*seq.n<r*seq.(n+1) by A2,XREAL_1:68;
    then (r(#)seq).n<r*seq.(n+1) by SEQ_1:9;
    hence thesis by SEQ_1:9;
  end;
  thus 0=r implies r(#)seq is constant
  proof
    assume
A3: 0=r;
    now
      let n be Nat;
      n in NAT & r*seq.n=r*seq.(n+1) by A3,ORDINAL1:def 12;
      then (r(#)seq).n=r*seq.(n+1) by SEQ_1:9;
      hence (r(#)seq).n=(r(#)seq).(n+1) by SEQ_1:9;
    end;
    hence thesis by VALUED_0:25;
  end;
  assume that
A4: seq is increasing and
A5: r<0;
  let n;
  seq.n<seq.(n+1) by A4,Lm6;
  then r*seq.(n+1)<r*seq.n by A5,XREAL_1:69;
  then (r(#)seq).(n+1)<r*seq.n by SEQ_1:9;
  hence thesis by SEQ_1:9;
end;

theorem
  (seq is decreasing & 0<r implies r(#)seq is decreasing) & (seq is
  decreasing & r<0 implies r(#)seq is increasing)
proof
  thus seq is decreasing & 0<r implies r(#)seq is decreasing
  proof
    assume that
A1: seq is decreasing and
A2: 0<r;
    let n;
    seq.(n+1)<seq.n by A1,Lm7;
    then r*seq.(n+1)<r*seq.n by A2,XREAL_1:68;
    then (r(#)seq).(n+1)<r*seq.n by SEQ_1:9;
    hence thesis by SEQ_1:9;
  end;
  assume that
A3: seq is decreasing and
A4: r<0;
  let n;
  seq.(n+1)<seq.n by A3,Lm7;
  then r*seq.n<r*seq.(n+1) by A4,XREAL_1:69;
  then (r(#)seq).n<r*seq.(n+1) by SEQ_1:9;
  hence thesis by SEQ_1:9;
end;

theorem
  (seq is non-decreasing & 0<=r implies r(#)seq is non-decreasing) & (
  seq is non-decreasing & r<=0 implies r(#)seq is non-increasing)
proof
  thus seq is non-decreasing & 0<=r implies r(#)seq is non-decreasing
  proof
    assume that
A1: seq is non-decreasing and
A2: 0<=r;
    let n;
    seq.n<=seq.(n+1) by A1,Lm8;
    then r*seq.n<=r*seq.(n+1) by A2,XREAL_1:64;
    then (r(#)seq).n<=r*seq.(n+1) by SEQ_1:9;
    hence thesis by SEQ_1:9;
  end;
  assume that
A3: seq is non-decreasing and
A4: r<=0;
  let n;
  seq.n<=seq.(n+1) by A3,Lm8;
  then r*seq.(n+1)<=r*seq.n by A4,XREAL_1:65;
  then (r(#)seq).(n+1)<=r*seq.n by SEQ_1:9;
  hence thesis by SEQ_1:9;
end;

theorem
  (seq is non-increasing & 0<=r implies r(#)seq is non-increasing) & (
  seq is non-increasing & r<=0 implies r(#)seq is non-decreasing)
proof
  thus seq is non-increasing & 0<=r implies r(#)seq is non-increasing
  proof
    assume that
A1: seq is non-increasing and
A2: 0<=r;
    let n;
    seq.(n+1)<=seq.n by A1,Lm9;
    then r*seq.(n+1)<=r*seq.n by A2,XREAL_1:64;
    then (r(#)seq).(n+1)<=r*seq.n by SEQ_1:9;
    hence thesis by SEQ_1:9;
  end;
  assume that
A3: seq is non-increasing and
A4: r<=0;
  let n;
  seq.(n+1)<=seq.n by A3,Lm9;
  then r*seq.n<=r*seq.(n+1) by A4,XREAL_1:65;
  then (r(#)seq).n<=r*seq.(n+1) by SEQ_1:9;
  hence thesis by SEQ_1:9;
end;

theorem Th34:
  (seq is increasing & seq1 is increasing implies seq+seq1 is
  increasing) & (seq is decreasing & seq1 is decreasing implies seq+seq1 is
decreasing) & (seq is non-decreasing & seq1 is non-decreasing implies seq+seq1
  is non-decreasing) & (seq is non-increasing & seq1 is non-increasing implies
  seq+seq1 is non-increasing)
proof
  thus seq is increasing & seq1 is increasing implies seq+seq1 is increasing
  proof
    assume
A1: seq is increasing & seq1 is increasing;
    let n;
    seq.n<seq.(n+1) & seq1.n<seq1.(n+1) by A1,Lm6;
    then seq.n+seq1.n<seq.(n+1)+seq1.(n+1) by XREAL_1:8;
    then (seq+seq1).n<seq.(n+1)+seq1.(n+1) by SEQ_1:7;
    hence thesis by SEQ_1:7;
  end;
  thus seq is decreasing & seq1 is decreasing implies seq+seq1 is decreasing
  proof
    assume
A2: seq is decreasing & seq1 is decreasing;
    let n;
    seq.(n+1)<seq.n & seq1.(n+1)<seq1.n by A2,Lm7;
    then seq.(n+1)+seq1.(n+1)<seq.n+seq1.n by XREAL_1:8;
    then (seq+seq1).(n+1)<seq.n+seq1.n by SEQ_1:7;
    hence thesis by SEQ_1:7;
  end;
  thus seq is non-decreasing & seq1 is non-decreasing implies seq+seq1 is
  non-decreasing
  proof
    assume
A3: seq is non-decreasing & seq1 is non-decreasing;
    let n;
    seq.n<=seq.(n+1) & seq1.n<=seq1.(n+1) by A3,Lm8;
    then seq.n+seq1.n<=seq.(n+1)+seq1.(n+1) by XREAL_1:7;
    then (seq+seq1).n<=seq.(n+1)+seq1.(n+1) by SEQ_1:7;
    hence thesis by SEQ_1:7;
  end;
  assume
A4: seq is non-increasing & seq1 is non-increasing;
  let n;
  seq.(n+1)<=seq.n & seq1.(n+1)<=seq1.n by A4,Lm9;
  then seq.(n+1)+seq1.(n+1)<=seq.n+seq1.n by XREAL_1:7;
  then (seq+seq1).(n+1)<=seq.n+seq1.n by SEQ_1:7;
  hence thesis by SEQ_1:7;
end;

theorem
  (seq is increasing & seq1 is constant implies seq+seq1 is increasing)
& (seq is decreasing & seq1 is constant implies seq+seq1 is decreasing) & (seq
is non-decreasing & seq1 is constant implies seq+seq1 is non-decreasing) & (seq
  is non-increasing & seq1 is constant implies seq+seq1 is non-increasing)
proof
  thus seq is increasing & seq1 is constant implies seq+seq1 is increasing
  proof
    assume that
A1: seq is increasing and
A2: seq1 is constant;
    let n;
    consider r1 being Real such that
A3: for n being Nat holds seq1.n=r1 by A2,VALUED_0:def 18;
    seq.n<seq.(n+1) by A1,Lm6;
    then seq.n+r1<seq.(n+1)+r1 by XREAL_1:6;
    then seq.n+seq1.n<seq.(n+1)+r1 by A3;
    then seq.n+seq1.n<seq.(n+1)+seq1.(n+1) by A3;
    then (seq+seq1).n<seq.(n+1)+seq1.(n+1) by SEQ_1:7;
    hence thesis by SEQ_1:7;
  end;
  thus seq is decreasing & seq1 is constant implies seq+seq1 is decreasing
  proof
    assume that
A4: seq is decreasing and
A5: seq1 is constant;
    let n;
    consider r1 being Real such that
A6: for n being Nat holds seq1.n=r1 by A5,VALUED_0:def 18;
    seq.(n+1)<seq.n by A4,Lm7;
    then seq.(n+1)+r1<seq.n+r1 by XREAL_1:6;
    then seq.(n+1)+seq1.(n+1)<seq.n+r1 by A6;
    then seq.(n+1)+seq1.(n+1)<seq.n+seq1.n by A6;
    then (seq+seq1).(n+1)<seq.n+seq1.n by SEQ_1:7;
    hence thesis by SEQ_1:7;
  end;
  thus seq is non-decreasing & seq1 is constant implies seq+seq1 is
  non-decreasing by Th34;
  assume seq is non-increasing & seq1 is constant;
  hence thesis by Th34;
end;





theorem Th36:
  (seq is bounded_above & 0<r implies r(#)seq is bounded_above) &
(0=r implies r(#)seq is bounded) & (seq is bounded_above & r<0 implies r(#)seq
  is bounded_below)
proof
  thus seq is bounded_above & 0<r implies r(#)seq is bounded_above;
  thus 0=r implies r(#)seq is bounded
  proof
    assume
A1: 0=r;
    now
      let n;
      (r(#)seq).n = 0*seq.n by A1,SEQ_1:9;
      hence (r(#)seq).n < 1;
    end;
    hence r(#)seq is bounded_above by SEQ_2:def 3;
    take p=-1;
    let n be Element of NAT;
    -1<0 & r*seq.n=0 by A1;
    hence p<(r(#)seq).n by SEQ_1:9;
  end;
  assume that
A2: seq is bounded_above and
A3: r<0;
  consider r1 such that
A4: for n holds seq.n<r1 by A2,SEQ_2:def 3;
  take p=r*abs r1;
  let n be Element of NAT;
  r1<=abs r1 by ABSVALUE:4;
  then seq.n<abs r1 by A4,XXREAL_0:2;
  then r*abs r1<r*seq.n by A3,XREAL_1:69;
  hence p<(r(#)seq).n by SEQ_1:9;
end;

theorem
  (seq is bounded implies for r holds r(#)seq is bounded) & (seq is
  bounded implies -seq is bounded) & (seq is bounded iff abs seq is bounded)
proof
  thus seq is bounded implies for r holds r(#)seq is bounded
  proof
    assume
A1: seq is bounded;
    let r;
    per cases;
    suppose
      0<r;
      hence thesis by A1;
    end;
    suppose
      0=r;
      hence thesis by Th36;
    end;
    suppose
      r<0;
      hence thesis by A1;
    end;
  end;
  thus seq is bounded implies -seq is bounded;
  thus seq is bounded implies abs seq is bounded;
  assume abs seq is bounded;
  then consider r be real number such that
A2: 0<r and
A3: for n be Element of NAT holds abs(abs seq.n)<r by SEQ_2:3;
  now
    let n be Element of NAT;
    abs(abs seq .n)=abs(abs(seq.n)) by SEQ_1:12
      .=abs(seq.n);
    hence abs(seq.n) < r by A3;
  end;
  hence thesis by A2,SEQ_2:3;
end;





theorem
  seq is bounded_above & seq1 is non-increasing implies seq+seq1 is
  bounded_above
proof
  assume that
A1: seq is bounded_above and
A2: seq1 is non-increasing;
  consider r1 such that
A3: for n holds seq.n<r1 by A1,SEQ_2:def 3;
  take r=r1+seq1.0;
  let n be Element of NAT;
  seq1.n<=seq1.0 by A2,Th12;
  then seq.n+seq1.n<r1+seq1.0 by A3,XREAL_1:8;
  hence (seq+seq1).n<r by SEQ_1:7;
end;

theorem
  seq is bounded_below & seq1 is non-decreasing implies seq+seq1 is
  bounded_below
proof
  assume that
A1: seq is bounded_below and
A2: seq1 is non-decreasing;
  consider r1 such that
A3: for n holds r1<seq.n by A1,SEQ_2:def 4;
  take r=r1+seq1.0;
  let n be Element of NAT;
  seq1.0<=seq1.n by A2,Th11;
  then r1+seq1.0<seq.n+seq1.n by A3,XREAL_1:8;
  hence r<(seq+seq1).n by SEQ_1:7;
end;

theorem
  incl NAT is increasing natural-valued by Lm10;

registration
  cluster -> natural-valued for FinSequence of NAT;
  coherence;
end;

begin 

reserve v for FinSequence of REAL,
  r,s for Real,
  n,m,i,j,k for Element of NAT;

theorem Th41:
  abs(r-s)=1 iff r>s & r=s+1 or r<s & s=r+1
proof
  thus abs(r-s)=1 implies r>s & r=s+1 or r<s & s=r+1
  proof
    assume
A1: abs(r-s) = 1;
    now
      per cases by XXREAL_0:1;
      case
A2:     r>s;
        then 0<r-s by XREAL_1:50;
        then r-s=1 by A1,ABSVALUE:def 1;
        hence thesis by A2;
      end;
      case
        r=s;
        hence contradiction by A1,ABSVALUE:2;
      end;
      case
A3:     r<s;
        then
A4:     0<s-r by XREAL_1:50;
        1=abs(-(r-s)) by A1,COMPLEX1:52
          .=s-r by A4,ABSVALUE:def 1;
        hence thesis by A3;
      end;
    end;
    hence thesis;
  end;
  assume
A5: r>s & r=s+1 or r<s & s=r+1;
  per cases by A5;
  suppose
    r>s & r=s+1;
    hence thesis by ABSVALUE:def 1;
  end;
  suppose
A6: s>r & s=r+1;
    thus abs(r-s)=abs(-(r-s)) by COMPLEX1:52
      .= 1 by A6,ABSVALUE:def 1;
  end;
end;

theorem
  abs(i-j)+abs(n-m)=1 iff abs(i-j)=1 & n=m or abs(n-m)=1 & i=j
proof
  thus abs(i-j)+abs(n-m)=1 implies abs(i-j)=1 & n=m or abs(n-m)=1 & i=j
  proof
    assume
A1: abs(i-j)+abs(n-m)=1;
    now
      per cases;
      suppose
A2:     n=m;
        then 1=abs(i-j)+0 by A1,ABSVALUE:2
          .= abs(i-j);
        hence thesis by A2;
      end;
      suppose
A3:     n<>m;
        now
          per cases by A3,XXREAL_0:1;
          suppose
A4:         n<m;
            then reconsider l=m-n as Element of NAT by INT_1:5;
            0<l by A4,XREAL_1:50;
            then
A5:         0+1<=l by NAT_1:13;
A6:         abs(-(m-n))=abs(l) by COMPLEX1:52;
            then
A7:         abs(n-m)=abs(l);
A8:         abs(l)=l by ABSVALUE:def 1;
            then
A9:         l<=1 by A1,A6,COMPLEX1:46,XREAL_1:31;
            then l=1 by A5,XXREAL_0:1;
            then i-j=0 by A1,A8,A7,ABSVALUE:2;
            hence thesis by A6,A5,A8,A9,XXREAL_0:1;
          end;
          suppose
A10:        n>m;
            then reconsider l=n-m as Element of NAT by INT_1:5;
            0<l by A10,XREAL_1:50;
            then
A11:        0+1<=l by NAT_1:13;
A12:        abs(n-m)=l by ABSVALUE:def 1;
            then
A13:        l<=1 by A1,COMPLEX1:46,XREAL_1:31;
            then l=1 by A11,XXREAL_0:1;
            then i-j=0 by A1,A12,ABSVALUE:2;
            hence thesis by A11,A12,A13,XXREAL_0:1;
          end;
        end;
        hence thesis;
      end;
    end;
    hence thesis;
  end;
  assume
A14: abs(i-j)=1 & n=m or abs(n-m)=1 & i=j;
  now
    per cases by A14;
    suppose
A15:  abs(i-j)=1 & n=m;
      then abs(n-m)=0 by ABSVALUE:2;
      hence thesis by A15;
    end;
    suppose
A16:  abs(n-m)=1 & i=j;
      then abs(i-j)=0 by ABSVALUE:2;
      hence thesis by A16;
    end;
  end;
  hence thesis;
end;

theorem
  n>1 iff ex m st n=m+1 & m>0
proof
  thus n>1 implies ex m st n=m+1 & m>0
  proof
    assume
A1: 1<n;
    then consider m being Nat such that
A2: n = m+1 by NAT_1:6;
    reconsider m as Element of NAT by ORDINAL1:def 12;
    take m;
    m <> 0 by A1,A2;
    hence thesis by A2;
  end;
  given m such that
A3: n=m+1 & m>0;
  0+1<n by A3,XREAL_1:6;
  hence thesis;
end;

theorem
  for f being FinSequence,n,m,k st len f = m+1 & n in dom f & k in
  Seg m holds Del(f,n).k = f.k or Del(f,n).k = f.(k+1)
proof
  let f be FinSequence,n,m,k such that
A1: len f = m+1 & n in dom f and
A2: k in Seg m;
  set X = dom f \ {n};
A3: dom f=Seg len f by FINSEQ_1:def 3;
  then
A4: n<=k & k<=m implies Sgm(X).k=k+1 by A1,A2,FINSEQ_3:108;
  X c= Seg len f by A3,XBOOLE_1:36;
  then rng Sgm(X) = X by FINSEQ_1:def 13;
  then
A5: dom (f*Sgm(X)) = dom Sgm(X) by RELAT_1:27,XBOOLE_1:36;
A6: dom Sgm(X)=Seg len Sgm(X) & Del(f,n) = f*Sgm(X) by FINSEQ_1:def 3
,FINSEQ_3:def 2;
A7: len Sgm(X) = m by A1,A3,FINSEQ_3:107;
  1<=k & k<n implies Sgm(X).k=k by A1,A2,A3,FINSEQ_3:108;
  hence thesis by A2,A6,A5,A7,A4,FINSEQ_3:25,FUNCT_1:12;
end;

definition
  let f be FinSequence;
  redefine attr f is constant means
  for n,m st n in dom f & m in dom f holds f.n=f.m;
  compatibility
  proof
    thus f is constant implies for n,m st n in dom f & m in dom f holds f.n=f.
    m by FUNCT_1:def 10;
    assume
A1: for n,m st n in dom f & m in dom f holds f.n=f.m;
    let n,m be set;
    assume n in dom f & m in dom f;
    hence thesis by A1;
  end;
end;

registration
  cluster -> real-valued for FinSequence of REAL;
  coherence;
end;

registration
  cluster non empty increasing for FinSequence of REAL;
  existence
  proof
    take v = <* 0 qua Real *>;
    thus v is non empty;
    let n,m;
    assume that
A1: n in dom v and
A2: m in dom v;
A3: dom<* 0 *> = { 1 } by FINSEQ_1:2,38;
    then n = 1 by A1,TARSKI:def 1;
    hence thesis by A3,A2,TARSKI:def 1;
  end;
end;

registration
  cluster constant for FinSequence of REAL;
  existence
  proof
    take v = <*>REAL;
    let n,m;
    assume that
 n in dom v and
    m in dom v;
    thus thesis;
  end;
end;

theorem Th45:
  v<>{} & rng v c= Seg n & v.(len v) = n & (for k st 1<=k & k<=len
v - 1 holds for r,s st r = v.k & s = v.(k+1) holds abs(r-s) = 1 or r=s) & i in
  Seg n & i+1 in Seg n & m in dom v & v.m = i & (for k st k in dom v & v.k = i
  holds k<=m) implies m+1 in dom v & v.(m+1)=i+1
proof
  assume that
A1: v<>{} and
A2: rng v c= Seg n and
A3: v.(len v) = n and
A4: for k st 1<=k & k<=len v - 1 holds for r,s st r = v.k & s = v.(k+1)
  holds abs(r-s) = 1 or r=s and
A5: i in Seg n and
A6: i+1 in Seg n and
A7: m in dom v and
A8: v.m = i and
A9: for k st k in dom v & v.k = i holds k<=m;
A10: m<=len v by A7,FINSEQ_3:25;
  0+1<=len v by A1,NAT_1:13;
  then
A11: len v in dom v by FINSEQ_3:25;
  reconsider r=v.(m+1) as Real;
A12: i<=n by A5,FINSEQ_1:1;
A13: 1<=m by A7,FINSEQ_3:25;
A14: i+1<=n by A6,FINSEQ_1:1;
A15: now
    assume not m+1 in dom v;
    then 1>m+1 or m+1>len v by FINSEQ_3:25;
    then
A16: len v <=m by NAT_1:11,13;
    i<n by A14,NAT_1:13;
    hence contradiction by A3,A8,A10,A16,XXREAL_0:1;
  end;
  then
A17: m+1<=len v by FINSEQ_3:25;
  then
A18: m<len v by NAT_1:13;
A19: m<=len v - 1 by A17,XREAL_1:19;
  now
    per cases by A4,A8,A13,A19;
    case
A20:  abs(i - r) = 1;
      now
        per cases by A20,Th41;
        case
A21:      i>r & i=r+1;
          defpred P[set] means for k,r st k=$1 & k>0 & m+k in dom v & r=v.(m+k
          ) holds r<i;
A22:      for k st P[k] holds P[k+1]
          proof
            let k such that
A23:        P[k];
            let j,s such that
A24:        j=k+1 and
A25:        j>0 and
A26:        m+j in dom v and
A27:        s=v.(m+j);
            now
              per cases;
              suppose
                k=0;
                hence thesis by A21,A24,A27;
              end;
              suppose
A28:            k<>0;
A29:            m+k<=m+k+1 by NAT_1:11;
                m<=m+k by NAT_1:11;
                then
A30:            1<=m+k by A13,XXREAL_0:2;
A31:            m+(k+1)<=len v by A24,A26,FINSEQ_3:25;
                then m+k<=len v by A29,XXREAL_0:2;
                then
A32:            m+k in dom v by A30,FINSEQ_3:25;
                then v.(m+k) in rng v by FUNCT_1:def 3;
                then v.(m+k) in Seg n by A2;
                then reconsider r1=v.(m+k) as Element of NAT;
A33:            r1<i by A23,A28,A32;
A34:            m+k<=len v - 1 by A29,A31,XREAL_1:19;
                now
                  per cases by A4,A24,A27,A29,A30,A34;
                  suppose
A35:                abs(r1-s)=1;
                    now
                      per cases by A35,Th41;
                      suppose
                        r1>s & r1=s+1;
                        hence thesis by A33,XXREAL_0:2;
                      end;
                      suppose
                        r1<s & s=r1+1;
                        then
A36:                    s<=i by A33,NAT_1:13;
                        now
                          per cases by A36,XXREAL_0:1;
                          case
                            s<i;
                            hence thesis;
                          end;
                          case
                            s=i;
                            then m+j<=m by A9,A26,A27;
                            then j<=m-m by XREAL_1:19;
                            hence contradiction by A25;
                          end;
                        end;
                        hence thesis;
                      end;
                    end;
                    hence thesis;
                  end;
                  suppose
                    r1=s;
                    hence thesis by A23,A28,A32;
                  end;
                end;
                hence thesis;
              end;
            end;
            hence thesis;
          end;
          reconsider l = len v - m as Element of NAT by A18,INT_1:5;
A37:      0< len v - m & m+l = len v by A17,XREAL_1:19;
A38:      P[0];
          for k holds P[k] from NAT_1:sch 1(A38,A22);
          hence contradiction by A3,A11,A12,A37;
        end;
        case
          i<r & r=i+1;
          hence thesis by A15;
        end;
      end;
      hence thesis;
    end;
    case
      i=r;
      then m+1<=m by A9,A15;
      then 1<=m-m by XREAL_1:19;
      then 1<=0;
      hence contradiction;
    end;
  end;
  hence thesis;
end;

theorem
  v<>{} & rng v c= Seg n & v.1 = 1 & v.(len v) = n & (for k st 1<=k & k
  <=len v - 1 holds for r,s st r = v.k & s = v.(k+1) holds abs(r-s) = 1 or r=s)
  implies (for i st i in Seg n ex k st k in dom v & v.k = i) & for m,k,i,r st m
  in dom v & v.m = i & (for j st j in dom v & v.j = i holds j<=m) & m<k & k in
  dom v & r = v.k
   holds i<r
proof
  assume that
A1: v<>{} and
A2: rng v c= Seg n and
A3: v.1 = 1 and
A4: v.(len v) = n and
A5: for k st 1<=k & k<=len v - 1 holds for r,s st r = v.k & s = v.(k+1)
  holds abs(r-s) = 1 or r=s;
  defpred P[Element of NAT] means $1 in Seg n implies ex k st k in dom v & v.k
  = $1;
A6: 0+1<=len v by A1,NAT_1:13;
  then
A7: len v in dom v by FINSEQ_3:25;
A8: for i st P[i] holds P[i+1]
  proof
    let i such that
A9: i in Seg n implies ex k st k in dom v & v.k = i;
    assume
A10: i+1 in Seg n;
    now
      per cases;
      suppose
A11:    i = 0;
        take k=1;
        thus k in dom v & v.k = i+1 by A3,A6,A11,FINSEQ_3:25;
      end;
      suppose
A12:    i<>0;
        defpred R[set] means $1 in dom v & v.$1 = i;
A13:    for k be Nat holds R[k] implies k<=len v by FINSEQ_3:25;
        i+1<=n by A10,FINSEQ_1:1;
        then
A14:    i<=n by NAT_1:13;
A15:    0+1<=i by A12,NAT_1:13;
        then
A16:    ex k be Nat st R[k] by A9,A14,FINSEQ_1:1;
        consider m be Nat such that
A17:    R[m] & for k be Nat st R[k] holds k<=m from NAT_1:sch 6(A13,
        A16);
A18:    for k st R[k] holds k<=m by A17;
        reconsider m as Element of NAT by ORDINAL1:def 12;
        take k = m+1;
        i in Seg n by A15,A14,FINSEQ_1:1;
        hence k in dom v & v.k = i+1 by A1,A2,A4,A5,A10,A17,A18,Th45;
      end;
    end;
    hence thesis;
  end;
A19: P[0] by FINSEQ_1:1;
  thus for i holds P[i] from NAT_1:sch 1(A19,A8);
  let m,k,i,r;
  assume that
A20: m in dom v and
A21: v.m = i and
A22: for j st j in dom v & v.j = i holds j<=m and
A23: m<k and
A24: k in dom v and
A25: r = v.k;
A26: m<=len v & k<=len v by A20,A24,FINSEQ_3:25;
A27: 1<=m by A20,FINSEQ_3:25;
A28: i in rng v by A20,A21,FUNCT_1:def 3;
  then
A29: i<=n by A2,FINSEQ_1:1;
  now
    per cases;
    suppose
      i=n;
      then len v<=m by A4,A7,A22;
      hence thesis by A23,A26,XXREAL_0:1;
    end;
    suppose
A30:  i<>n;
      defpred P[set] means for j,s st j=$1 & j>0 & m+j in dom v & s=v.(m+j)
      holds i<s;
      i<n by A29,A30,XXREAL_0:1;
      then 1<=i+1 & i+1<=n by NAT_1:11,13;
      then i+1 in Seg n by FINSEQ_1:1;
      then
A31:  v.(m+1) = i+1 by A1,A2,A4,A5,A20,A21,A22,A28,Th45;
A32:  for k st P[k] holds P[k+1]
      proof
        let k such that
A33:    P[k];
        let j,s such that
A34:    j=k+1 and
A35:    j>0 and
A36:    m+j in dom v and
A37:    s=v.(m+j);
        per cases;
        suppose
          k=0;
          hence thesis by A31,A34,A37,NAT_1:13;
        end;
        suppose
A38:      k<>0;
          m<=m+k by NAT_1:11;
          then
A39:      1<=m+k by A27,XXREAL_0:2;
          s in rng v by A36,A37,FUNCT_1:def 3;
          then s in Seg n by A2;
          then reconsider s1=s as Element of NAT;
A40:      m+(k+1)<=len v by A34,A36,FINSEQ_3:25;
          m+k<=m+k+1 by NAT_1:11;
          then m+k<=len v by A40,XXREAL_0:2;
          then
A41:      m+k in dom v by A39,FINSEQ_3:25;
          then v.(m+k) in rng v by FUNCT_1:def 3;
          then v.(m+k) in Seg n by A2;
          then reconsider r1=v.(m+k) as Element of NAT;
A42:      i<r1 by A33,A38,A41;
A43:      m+k+1 = m+(k+1);
          then
A44:      m+k<= len v - 1 by A40,XREAL_1:19;
          now
            per cases by A5,A34,A37,A43,A39,A44;
            suppose
A45:          abs(r1-s)=1;
              now
                per cases by A45,Th41;
                suppose
                  r1>s & r1=s+1;
                  then
A46:              i<=s1 by A42,NAT_1:13;
                  now
                    per cases by A46,XXREAL_0:1;
                    case
                      i<s;
                      hence thesis;
                    end;
                    case
                      s=i;
                      then m+j<=m by A22,A36,A37;
                      then j<=m-m by XREAL_1:19;
                      hence contradiction by A35;
                    end;
                  end;
                  hence thesis;
                end;
                suppose
                  r1<s & s=r1+1;
                  hence thesis by A42,XXREAL_0:2;
                end;
              end;
              hence thesis;
            end;
            suppose
              r1=s;
              hence thesis by A33,A38,A41;
            end;
          end;
          hence thesis;
        end;
      end;
      reconsider l = k - m as Element of NAT by A23,INT_1:5;
A47:  0< k - m & m+l = k by A23,XREAL_1:50;
A48:  P[0];
      for k holds P[k] from NAT_1:sch 1(A48,A32);
      hence thesis by A24,A25,A47;
    end;
  end;
  hence thesis;
end;














begin

reserve x,y for set,
  n,m for Element of NAT,
  r,s for Real;

registration
  let f be FinSequence, x be set;
  cluster Coim(f,x) -> finite;
  coherence;
end;

theorem Th1:
  for f,g,h be FinSequence holds f,g are_fiberwise_equipotent iff
  f^h, g^h are_fiberwise_equipotent
proof
  let f,g,h be FinSequence;
  thus f,g are_fiberwise_equipotent implies f^h, g^h are_fiberwise_equipotent
  proof
    assume
A1: f,g are_fiberwise_equipotent;
    now
      let y;
      card Coim(f,y) = card Coim(g,y) by A1,CLASSES1:def 9;
      hence card Coim(f^h,y) = card Coim(g,y) + card Coim(h,y) by FINSEQ_3:57
        .= card Coim(g^h,y) by FINSEQ_3:57;
    end;
    hence thesis by CLASSES1:def 9;
  end;
  assume
A2: f^h,g^h are_fiberwise_equipotent;
  now
    let x;
A3: card Coim(f^h,x) = card Coim(f,x)+card Coim(h,x) & card Coim(g^h,x) =
    card Coim(g,x)+card Coim(h,x) by FINSEQ_3:57;
    card Coim(f^h,x) = card Coim(g^h,x) by A2,CLASSES1:def 9;
    hence card Coim(f,x) = card Coim(g,x) by A3;
  end;
  hence thesis by CLASSES1:def 9;
end;

theorem
  for f,g be FinSequence holds f^g, g^f are_fiberwise_equipotent
proof
  let f,g be FinSequence;
  let y;
  thus card Coim(f^g,y) = card(g"{y})+ card(f"{y}) by FINSEQ_3:57
    .= card Coim(g^f,y) by FINSEQ_3:57;
end;

theorem Th3:
  for f,g be FinSequence st f,g are_fiberwise_equipotent holds len
  f = len g & dom f = dom g
proof
  let f,g be FinSequence;
A1: dom f = Seg len f by FINSEQ_1:def 3;
  assume f,g are_fiberwise_equipotent;
  then
A2: card(f"(rng f)) = card(g"(rng f)) & rng f = rng g by CLASSES1:75,78;
A3: Seg len g = dom g by FINSEQ_1:def 3;
  thus len f = card(Seg len f) by FINSEQ_1:57
    .= card(g"(rng g)) by A1,A2,RELAT_1:134
    .= card(Seg len g) by A3,RELAT_1:134
    .= len g by FINSEQ_1:57;
  hence thesis by A1,FINSEQ_1:def 3;
end;

theorem Th4:
  for f,g be FinSequence holds f,g are_fiberwise_equipotent iff ex
  P be Permutation of dom g st f = g*P
proof
  let f,g be FinSequence;
  thus f,g are_fiberwise_equipotent implies ex P be Permutation of dom g st f
  = g*P
  proof
    assume
A1: f,g are_fiberwise_equipotent;
    then dom f = dom g by Th3;
    hence thesis by A1,CLASSES1:80;
  end;
  given P be Permutation of dom g such that
A2: f = g*P;
  dom g = {} implies dom g = {};
  then rng P c= dom g & dom P = dom g by FUNCT_2:def 1,RELAT_1:def 19;
  then dom f = dom g by A2,RELAT_1:27;
  hence thesis by A2,CLASSES1:80;
end;

defpred P[Nat] means for X be finite set,F be Function st card(dom(F|X)) = $1
ex a be FinSequence st F|X, a are_fiberwise_equipotent;

Lm1: P[0]
proof
  let X be finite set, F be Function;
  assume
A1: card(dom(F|X)) = 0;
  take A = {};
  let x;
  dom(F|X) = {} by A1;
  hence card Coim(F|X,x) = card Coim(A,x) by RELAT_1:132,XBOOLE_1:3;
end;

Lm2: for n st P[n] holds P[n+1]
proof
  let n be Element of NAT;
  assume
A1: P[n];
  let X be finite set, F be Function;
  reconsider dx = dom(F|X) as finite set;
  set x = the Element of dx;
  set Y = X\{x}, dy = dom(F|Y);
A2: dy = dom F /\ Y by RELAT_1:61;
A3: dx = dom F /\ X by RELAT_1:61;
A4: dy = dx \ {x}
  proof
    thus dy c= dx \ {x}
    proof
      let y;
      assume
A5:   y in dy;
      then y in X \ {x} by A2,XBOOLE_0:def 4;
      then
A6:   not y in {x} by XBOOLE_0:def 5;
      y in dom F by A2,A5,XBOOLE_0:def 4;
      then y in dx by A2,A3,A5,XBOOLE_0:def 4;
      hence thesis by A6,XBOOLE_0:def 5;
    end;
    let y;
    assume
A7: y in dx \{x};
    then ( not y in {x})& y in X by A3,XBOOLE_0:def 4,def 5;
    then
A8: y in Y by XBOOLE_0:def 5;
    y in dom F by A3,A7,XBOOLE_0:def 4;
    hence thesis by A2,A8,XBOOLE_0:def 4;
  end;
  assume
A9: card dom(F|X) = n+1;
  then {x} c= dx by CARD_1:27,ZFMISC_1:31;
  then card(dom(F|Y)) = card dx - card {x} by A4,CARD_2:44
    .= n+1-1 by A9,CARD_1:30
    .= n;
  then consider a be FinSequence such that
A10: F|Y, a are_fiberwise_equipotent by A1;
  take A = a^<* F.x *>;
  dx <> {} by A9;
  then x in dx;
  then
A11: x in dom F /\ X by RELAT_1:61;
  then x in dom F by XBOOLE_0:def 4;
  then
A12: {x} c= dom F by ZFMISC_1:31;
  x in X by A11,XBOOLE_0:def 4;
  then
A13: {x} c= X by ZFMISC_1:31;
  now
    let y;
A14: Y misses {x} by XBOOLE_1:79;
A15: (F|Y)"{y} \/ (F|{x})"{y} = (Y /\ F"{y}) \/ (F|{x})"{y} by FUNCT_1:70
      .= (Y /\ F"{y}) \/ ({x} /\ F"{y}) by FUNCT_1:70
      .= (Y \/ {x}) /\ F"{y} by XBOOLE_1:23
      .= (X \/ {x}) /\ F"{y} by XBOOLE_1:39
      .= X /\ F"{y} by A13,XBOOLE_1:12
      .= (F|X)"{y} by FUNCT_1:70;
    reconsider FF = <*F.x*>"{y} as finite set;
A16: card Coim(F|Y,y) = card Coim(a,y) by A10,CLASSES1:def 9;
A17: dom(F|{x}) = {x} by A12,RELAT_1:62;
A18: dom<*F.x*> = {1} by FINSEQ_1:2,38;
A19: now
      per cases;
      case
A20:    y = F.x;
A21:    {x} c= (F|{x})"{y}
        proof
          let z be set;
A22:      y in {y} by TARSKI:def 1;
          assume
A23:      z in {x};
          then z = x by TARSKI:def 1;
          then y = (F|{x}).z by A17,A20,A23,FUNCT_1:47;
          hence thesis by A17,A23,A22,FUNCT_1:def 7;
        end;
        (F|{x})"{y} c= {x} by A17,RELAT_1:132;
        then (F|{x})"{y} = {x} by A21,XBOOLE_0:def 10;
        then
A24:    card((F|{x})"{y}) = 1 by CARD_1:30;
A25:    {1} c= <*F.x*>"{y}
        proof
          let z be set;
A26:      y in {y} by TARSKI:def 1;
          assume
A27:      z in {1};
          then z = 1 by TARSKI:def 1;
          then y = <*F.x*>.z by A20,FINSEQ_1:40;
          hence thesis by A18,A27,A26,FUNCT_1:def 7;
        end;
        <*F.x*>"{y} c= {1} by A18,RELAT_1:132;
        then <*F.x*>"{y} = {1} by A25,XBOOLE_0:def 10;
        hence card((F|{x})"{y}) = card FF by A24,CARD_1:30;
      end;
      case
A28:    y <> F.x;
A29:    now
          set z = the Element of <*F.x*>"{y};
          assume
A30:      <*F.x*>"{y} <> {};
          then <*F.x*>.z in {y} by FUNCT_1:def 7;
          then
A31:      <*F.x*>.z = y by TARSKI:def 1;
          z in {1} by A18,A30,FUNCT_1:def 7;
          then z = 1 by TARSKI:def 1;
          hence contradiction by A28,A31,FINSEQ_1:40;
        end;
        now
          set z = the Element of (F|{x})"{y};
          assume
A32:      (F|{x})"{y} <> {};
          then (F|{x}).z in {y} by FUNCT_1:def 7;
          then
A33:      (F|{x}).z = y by TARSKI:def 1;
A34:      z in {x} by A17,A32,FUNCT_1:def 7;
          then z = x by TARSKI:def 1;
          hence contradiction by A17,A28,A34,A33,FUNCT_1:47;
        end;
        hence card((F|{x})"{y}) = card FF by A29;
      end;
    end;
    (F|Y)"{y} /\ (F|{x})"{y} = Y /\ (F"{y}) /\ (F|{x})"{y} by FUNCT_1:70
      .= Y /\ F"{y} /\ ({x} /\ F"{y}) by FUNCT_1:70
      .= F"{y} /\ Y /\ {x} /\ F"{y} by XBOOLE_1:16
      .= F"{y} /\ (Y /\ {x}) /\ F"{y} by XBOOLE_1:16
      .= {} /\ F"{y} by A14,XBOOLE_0:def 7
      .= {};
    hence card Coim(F|X,y) = card((F|Y)"{y}) + card FF - card {} by A15,A19,
CARD_2:45
      .= card Coim(A,y) by A16,FINSEQ_3:57;
  end;
  hence thesis by CLASSES1:def 9;
end;

theorem
  for F be Function, X be finite set ex f be FinSequence st F|X, f
  are_fiberwise_equipotent
proof
  let F be Function, X be finite set;
A1: card(dom(F|X)) = card(dom(F|X));
  for n holds P[n] from NAT_1:sch 1(Lm1,Lm2);
  hence thesis by A1;
end;

definition
  let n be Nat, f be FinSequence;
  func f /^ n -> FinSequence means
  :Def1:
  len it = len f - n & for m be Nat st
  m in dom it holds it.m = f.(m+n) if n<=len f otherwise it = {};
  existence
  proof
    thus n<=len f implies ex p1 be FinSequence st len p1 = len f - n & for m
    be Nat st m in dom p1 holds p1.m = f.(m+n)
    proof
      assume n<=len f;
      then reconsider k = len f - n as Element of NAT by NAT_1:21;
      deffunc F(Nat)=f.($1+n);
      consider p be FinSequence such that
A1:   len p = k & for m be Nat st m in dom p holds p.m = F(m) from
      FINSEQ_1:sch 2;
      take p;
      thus len p = len f - n by A1;
      let m be Nat;
      assume m in dom p;
      hence thesis by A1;
    end;
    thus thesis;
  end;
  uniqueness
  proof
    let p1,p2 be FinSequence;
    thus n<=len f & (len p1 = len f - n & for m be Nat st m in dom p1 holds p1
.m = f.(m+n)) & (len p2 = len f - n & for m be Nat st m in dom p2 holds p2.m =
    f.(m+n)) implies p1 = p2
    proof
      assume that
      n<=len f and
A2:   len p1 = len f - n and
A3:   for m be Nat st m in dom p1 holds p1.m = f.(m+n) and
A4:   len p2 = len f - n and
A5:   for m be Nat st m in dom p2 holds p2.m = f.(m+n);
A6:   dom p1 = Seg len p1 & Seg len p2 = dom p2 by FINSEQ_1:def 3;
      now
        let m be Nat;
        assume
A7:     m in dom p1;
        then p1.m = f.(m+n) by A3;
        hence p1.m = p2.m by A2,A4,A5,A6,A7;
      end;
      hence thesis by A2,A4,FINSEQ_2:9;
    end;
    thus thesis;
  end;
  consistency;
end;

definition
  let D be set, n be Nat, f be FinSequence of D;
  redefine func f /^ n -> FinSequence of D;
  coherence
  proof
    set p = f /^ n;
    per cases;
    suppose
A1:   n<=len f;
      then reconsider k = len f - n as Element of NAT by NAT_1:21;
A2:   len p = k by A1,Def1;
      rng p c= rng f
      proof
        let x;
        assume x in rng p;
        then consider m being Nat such that
A3:     m in dom p and
A4:     p.m = x by FINSEQ_2:10;
        m<=len p by A3,FINSEQ_3:25;
        then
A5:     m+n<=len f by A2,XREAL_1:19;
A6:     m<=m+n by NAT_1:11;
        1<=m by A3,FINSEQ_3:25;
        then 1<=m+n by A6,XXREAL_0:2;
        then
A7:     m+n in dom f by A5,FINSEQ_3:25;
        p.m = f.(m+n) by A1,A3,Def1;
        hence thesis by A4,A7,FUNCT_1:def 3;
      end;
      then rng p c= D by XBOOLE_1:1;
      hence thesis by FINSEQ_1:def 4;
    end;
    suppose len f < n;
      then f /^ n = <*>D by Def1;
      hence thesis;
    end;
  end;
end;

Lm3: for n be Nat for D be non empty set, f be FinSequence of D st len f<=n
holds (f|n) = f
proof
  let n be Nat;
  let D be non empty set,f be FinSequence of D;
A1: dom f = Seg len f by FINSEQ_1:def 3;
  assume len f<=n;
  then f|n = f|Seg n & dom f c= Seg n by A1,FINSEQ_1:5,def 15;
  hence thesis by RELAT_1:68;
end;

theorem Th6:
  for D be non empty set, f be FinSequence of D, n,m be Nat holds
  n in dom f & m in Seg n implies (f|n).m = f.m & m in dom f
proof
  let D be non empty set,f be FinSequence of D, n,m be Nat;
  assume that
A1: n in dom f and
A2: m in Seg n;
A3: f|n = f|Seg n by FINSEQ_1:def 15;
  dom f = Seg len f & n<=len f by A1,FINSEQ_1:def 3,FINSEQ_3:25;
  then
A4: Seg n c= dom f by FINSEQ_1:5;
  then Seg n = dom f /\ Seg n by XBOOLE_1:28
    .= dom(f|n) by A3,RELAT_1:61;
  hence thesis by A2,A3,A4,FUNCT_1:47;
end;

theorem Th7:
  for D be non empty set, f be FinSequence of D, n be Nat, x be
  set st len f = n+1 & x = f.(n+1) holds f = (f|n) ^ <*x*>
proof
  let D be non empty set, f be FinSequence of D, n be Nat,x;
  assume that
A1: len f = n+1 and
A2: x = f.(n+1);
  set fn = f|n;
A3: len fn = n by A1,FINSEQ_1:59,NAT_1:11;
A4: dom f = Seg len f by FINSEQ_1:def 3;
A5: n<=n+1 by NAT_1:11;
A6: now
    let m be Nat;
    assume
A7: m in dom f;
    then
A8: 1<=m by A4,FINSEQ_1:1;
A9: m<=len f by A4,A7,FINSEQ_1:1;
    now
      per cases;
      case
        m = len f;
        hence f.m = (fn^<*x*>).m by A1,A2,A3,FINSEQ_1:42;
      end;
      case
        m <> len f;
        then m<n+1 by A1,A9,XXREAL_0:1;
        then
A10:    m<=n by NAT_1:13;
        then 1<=n by A8,XXREAL_0:2;
        then
A11:    n in dom f by A1,A5,FINSEQ_3:25;
A12:    Seg len fn = dom fn by FINSEQ_1:def 3;
A13:    m in dom fn by A3,A8,A10,FINSEQ_3:25;
        hence (fn^<*x*>).m = fn.m by FINSEQ_1:def 7
          .= f.m by A3,A13,A11,A12,Th6;
      end;
    end;
    hence f.m = (fn^<*x*>).m;
  end;
  len (fn^<*x*>) = n + len <*x*> by A3,FINSEQ_1:22
    .= len f by A1,FINSEQ_1:40;
  hence thesis by A6,FINSEQ_2:9;
end;

theorem Th8:
  for D be non empty set, f be FinSequence of D, n be Nat holds
  (f|n)^(f/^n) = f
proof
  let D be non empty set, f be FinSequence of D,n be Nat;
  set fn = f/^n;
  now
    per cases;
    case
      len f<n;
      then f/^n = <*>D & f|n = f by Def1,Lm3;
      hence thesis by FINSEQ_1:34;
    end;
    case
A1:   n<=len f;
A2:   dom f = Seg len f by FINSEQ_1:def 3;
A3:   len (f|n) = n by A1,FINSEQ_1:59;
A4:   len fn = len f - n by A1,Def1;
      then
A5:   len ((f|n)^(f/^n)) = n+(len f - n) by A3,FINSEQ_1:22
        .= len f;
A6:   dom(f|n) = Seg n by A3,FINSEQ_1:def 3;
      now
        let m be Nat;
        assume
A7:     m in dom f;
        then
A8:     m<=len f by A2,FINSEQ_1:1;
A9:     1<=m by A2,A7,FINSEQ_1:1;
        now
          per cases;
          case
A10:        m<=n;
            then 1<=n by A9,XXREAL_0:2;
            then
A11:        n in dom f by A1,FINSEQ_3:25;
A12:        m in Seg n by A9,A10,FINSEQ_1:1;
            hence ((f|n)^(f/^n)).m = (f|n).m by A6,FINSEQ_1:def 7
              .= f.m by A12,A11,Th6;
          end;
          case
A13:        n<m;
            then max(0,m-n) = m-n by FINSEQ_2:4;
            then reconsider k = m-n as Element of NAT by FINSEQ_2:5;
            n+1<=m by A13,NAT_1:13;
            then
A14:        1<=k by XREAL_1:19;
            k<=len fn by A4,A8,XREAL_1:9;
            then
A15:        k in dom fn by A14,FINSEQ_3:25;
            thus ((f|n) ^ (f/^n)).m = fn.k by A3,A5,A8,A13,FINSEQ_1:24
              .= f.(k+n) by A1,A15,Def1
              .= f.m;
          end;
        end;
        hence ((f|n) ^ (f/^n)).m = f.m;
      end;
      hence thesis by A5,FINSEQ_2:9;
    end;
  end;
  hence thesis;
end;

theorem
  for R1,R2 be FinSequence of REAL st R1,R2 are_fiberwise_equipotent
  holds Sum R1 = Sum R2
proof
  let R1,R2 be FinSequence of REAL;
  defpred P[Nat] means for f,g be FinSequence of REAL st f,g
  are_fiberwise_equipotent & len f = $1 holds Sum f = Sum g;
  assume
A1: R1,R2 are_fiberwise_equipotent;
A2: len R1 = len R1;
A3: for n st P[n] holds P[n+1]
  proof
    let n;
    assume
A4: P[n];
    let f,g be FinSequence of REAL;
    assume that
A5: f,g are_fiberwise_equipotent and
A6: len f = n+1;
    set a = f.(n+1);
A7: rng f = rng g by A5,CLASSES1:75;
    0<n+1 by NAT_1:3;
    then 0 qua Nat+1<=n+1 by NAT_1:13;
    then n+1 in dom f by A6,FINSEQ_3:25;
    then a in rng g by A7,FUNCT_1:def 3;
    then consider m being Nat such that
A8: m in dom g and
A9: g.m = a by FINSEQ_2:10;
    set gg = g/^m, gm = g|m;
    m<=len g by A8,FINSEQ_3:25;
    then
A10: len gm = m by FINSEQ_1:59;
A11: 1<=m by A8,FINSEQ_3:25;
    then max(0,m-1) = m-1 by FINSEQ_2:4;
    then reconsider m1 = m-1 as Element of NAT by FINSEQ_2:5;
A12: m = m1+1;
    then m1<=m by NAT_1:11;
    then
A13: Seg m1 c= Seg m by FINSEQ_1:5;
    m in Seg m by A11,FINSEQ_1:1;
    then gm.m = a by A8,A9,Th6;
    then
A14: gm = (gm|m1)^<*a*> by A10,A12,Th7;
    set fn = f|n;
A15: g = (g|m)^(g/^m) by Th8;
A16: gm|m1 = gm|(Seg m1) by FINSEQ_1:def 15
      .= (g|(Seg m))|(Seg m1) by FINSEQ_1:def 15
      .= g|((Seg m)/\(Seg m1)) by RELAT_1:71
      .= g|(Seg m1) by A13,XBOOLE_1:28
      .= g|m1 by FINSEQ_1:def 15;
A17: f = fn ^ <*a*> by A6,Th7;
    now
      let x;
      card Coim(f,x) = card Coim(g,x) by A5,CLASSES1:def 9;
      then
      card(fn"{x}) + card(<*a*>"{x}) = card(((g|m1)^<*a*>^(g/^m))"{x}) by A15
,A14,A16,A17,FINSEQ_3:57
        .= card(((g|m1)^<*a*>)"{x}) + card((g/^m)"{x}) by FINSEQ_3:57
        .= card((g|m1)"{x})+ card(<*a*>"{x}) + card((g/^m)"{x}) by FINSEQ_3:57
        .= card((g|m1)"{x}) + card((g/^m)"{x})+ card(<*a*>"{x})
        .= card(((g|m1)^(g/^m))"{x})+ card(<*a*>"{x}) by FINSEQ_3:57;
      hence card Coim(fn,x) = card Coim((g|m1)^(g/^m),x);
    end;
    then
A18: fn, (g|m1)^(g/^m) are_fiberwise_equipotent by CLASSES1:def 9;
    len fn = n by A6,FINSEQ_1:59,NAT_1:11;
    then Sum fn = Sum((g|m1)^gg) by A4,A18;
    hence Sum f = Sum((g|m1)^gg) + Sum <*a*> by A17,RVSUM_1:75
      .= Sum(g|m1) + Sum gg+ Sum <*a*> by RVSUM_1:75
      .= Sum(g|m1)+ Sum <*a*> + Sum gg
      .= Sum gm + Sum gg by A14,A16,RVSUM_1:75
      .= Sum g by A15,RVSUM_1:75;
  end;
A19: P[0]
  proof
    let f,g be FinSequence of REAL;
    assume f,g are_fiberwise_equipotent & len f = 0;
    then
A20:   len g = 0 & f = <*>REAL by Th3;
     then g = <*>REAL;
    hence thesis by A20;
  end;
  for n holds P[n] from NAT_1:sch 1(A19,A3);
  hence thesis by A1,A2;
end;

definition
  let R be FinSequence of REAL;
  func MIM(R) -> FinSequence of REAL means
  :Def2:
  len it = len R & it.(len it) = R.(len R) &
  for n be Nat st 1 <= n & n <= len it - 1 holds it.n = (R.n) - (R.(n+1));
  existence
  proof
    per cases;
    suppose
A1:   len R = 0;
      take a = R;
      thus len a = len R & a.(len a) = R.(len R);
      let n be Nat;
      assume 1<=n & n<=len a - 1;
      hence thesis by A1,XXREAL_0:2;
    end;
    suppose
      len R <> 0;
      then 0<len R by NAT_1:3;
      then 0 qua Nat+1<=len R by NAT_1:13;
      then max(0,len R -1) = len R - 1 by FINSEQ_2:4;
      then reconsider l = len R -1 as Element of NAT by FINSEQ_2:5;
      defpred P[Nat,set] means $2 = R.$1 - R.($1+1);
      set t = R.(len R);
A2:   for n be Nat st n in Seg l ex x be Real st P[n,x];
      consider p be FinSequence of REAL such that
A3:   dom p = Seg l & for n be Nat st n in Seg l holds P[n,p.n] from
      FINSEQ_1:sch 5(A2);
      take a = p^<*t*>;
      thus
A4:   len a = len p + len <*t*> by FINSEQ_1:22
        .= l+len <*t*> by A3,FINSEQ_1:def 3
        .= l+1 by FINSEQ_1:39
        .= len R;
      hence a.(len a) = a.(l+1) .= a.(len p +1) by A3,FINSEQ_1:def 3
        .= R.(len R) by FINSEQ_1:42;
      let n be Nat;
      assume 1<=n & n<=len a - 1;
      then
A5:   n in Seg l by A4,FINSEQ_1:1;
      then p.n = R.n - R.(n+1) by A3;
      hence thesis by A3,A5,FINSEQ_1:def 7;
    end;
  end;
  uniqueness
  proof
    let p1,p2 be FinSequence of REAL;
    assume that
A6: len p1 = len R and
A7: p1.(len p1) = R.(len R) and
A8: for n be Nat st 1<=n & n<=len p1 - 1 holds p1.n = R.n - R.(n+1) and
A9: len p2 = len R and
A10: p2.(len p2) = R.(len R) and
A11: for n be Nat st 1<=n & n<=len p2 - 1 holds p2.n = R.n - R.(n+1);
A12: dom p1 = Seg len R by A6,FINSEQ_1:def 3;
    now
      let n be Nat;
      set r = R.n;
      assume
A13:  n in dom p1;
      then
A14:  1<=n by A12,FINSEQ_1:1;
A15:  n<=len R by A12,A13,FINSEQ_1:1;
      now
        per cases;
        case
          n = len R;
          hence p1.n = p2.n by A6,A7,A9,A10;
        end;
        case
A16:      n <> len R;
          set s = R.(n+1);
          n<len R by A15,A16,XXREAL_0:1;
          then n+1<=len R by NAT_1:13;
          then
A17:      n<=len R - 1 by XREAL_1:19;
          then p1.n = r-s by A6,A8,A14;
          hence p1.n = p2.n by A9,A11,A14,A17;
        end;
      end;
      hence p1.n = p2.n;
    end;
    hence thesis by A6,A9,FINSEQ_2:9;
  end;
end;

theorem Th10:
  for R be FinSequence of REAL, r be Real, n be Element of NAT st
  len R = n+2 & R.(n+1) = r holds MIM(R|(n+1)) = MIM(R)|n ^ <* r *>
proof
  let R be FinSequence of REAL, s,n;
  assume that
A1: len R = n+2 and
A2: R.(n+1) = s;
  set f1 = R|(n+1), m1 = MIM(f1), mf = MIM(R), fn = mf|n;
  0<n+1 by NAT_1:3;
  then
A3: 0 qua Nat+1<=n+1 by NAT_1:13;
A4: n+1+1 = n+(1+1);
  then n+1<=n+2 by NAT_1:11;
  then
A5: Seg len R = dom R & n+1 in Seg(n+2) by A3,FINSEQ_1:1,def 3;
A6: len f1 = n+1 by A1,A4,FINSEQ_1:59,NAT_1:11;
  then
A7: len MIM(f1) = n+1 by Def2;
  then
A8: dom m1 = Seg(n+1) by FINSEQ_1:def 3;
  n+1 in Seg(n+1) by A3,FINSEQ_1:1;
  then f1.(n+1) = s by A1,A2,A5,Th6;
  then
A9: m1.(n+1) = s by A6,A7,Def2;
A10: Seg len fn = dom fn by FINSEQ_1:def 3;
A11: Seg len mf = dom mf by FINSEQ_1:def 3;
A12: len mf = n+2 by A1,Def2;
  then
A13: len fn = n by FINSEQ_1:59,NAT_1:11;
A14: n<=n+2 by NAT_1:11;
A15: now
    let m be Nat;
    assume
A16: m in dom m1;
    then
A17: 1<=m by A8,FINSEQ_1:1;
A18: m<=n+1 by A8,A16,FINSEQ_1:1;
    now
      per cases;
      case
        m = n+1;
        hence m1.m = (fn^<*s*>).m by A13,A9,FINSEQ_1:42;
      end;
      case
        m <> n+1;
        then
A19:    m<n+1 by A18,XXREAL_0:1;
        then
A20:    m<=n by NAT_1:13;
        then
A21:    m in Seg n by A17,FINSEQ_1:1;
        set r2 = R.(m+1);
        set r1 = R.m;
A22:    len mf -1 = n+1 by A12;
        1<=n by A17,A20,XXREAL_0:2;
        then n in Seg(n+2) by A14,FINSEQ_1:1;
        then fn.m = mf.m by A12,A11,A21,Th6;
        then
A23:    r1 - r2 = fn.m by A17,A18,A22,Def2
          .= (fn^<*s*>).m by A13,A10,A21,FINSEQ_1:def 7;
        1<=m+1 & m+1<=n+1 by A19,NAT_1:11,13;
        then m+1 in Seg(n+1) by FINSEQ_1:1;
        then
A24:    f1.(m+1) = r2 by A1,A5,Th6;
        len m1 -1 = n & f1.m = r1 by A1,A7,A5,A8,A16,Th6;
        hence m1.m = (fn^<*s*>).m by A17,A20,A23,A24,Def2;
      end;
    end;
    hence m1.m = (fn^<*s*>).m;
  end;
  len(fn^<*s*>) = n + len <*s*> by A13,FINSEQ_1:22
    .= n+1 by FINSEQ_1:40;
  hence thesis by A7,A15,FINSEQ_2:9;
end;

theorem Th11:
  for R be FinSequence of REAL, r,s be Real, n be Element of NAT
  st len R = n+2 & R.(n+1) = r & R.(n+2) = s holds
  MIM(R) = MIM(R)|n ^ <* r-s,s *>
proof
  let R be FinSequence of REAL, r,s,n;
  set mf = MIM(R), nf = mf|n;
  assume that
A1: len R = n+2 and
A2: R.(n+1) = r and
A3: R.(n+2) = s;
A4: len mf = n+2 by A1,Def2;
  then
A5: dom mf = Seg(n+2) by FINSEQ_1:def 3;
A6: n+(1+1) = n+1+1;
  then n+1<=n+2 by NAT_1:11;
  then
A7: n<n+2 by NAT_1:13;
A8: len nf = n by A4,FINSEQ_1:59,NAT_1:11;
  then len (nf^<* r-s,s *>) = n + len <* r-s,s *> by FINSEQ_1:22;
  then
A9: len(nf^<* r-s,s *>) = n+2 by FINSEQ_1:44;
A10: n<=n+2 by NAT_1:11;
  now
    let m be Nat;
    assume
A11: m in dom mf;
    then
A12: 1<=m by A5,FINSEQ_1:1;
A13: m<=n+2 by A5,A11,FINSEQ_1:1;
    now
      per cases;
      case
A14:    m = n+2;
        hence mf.m = s by A1,A3,A4,Def2
          .= <* r-s,s *>.(n+2-n) by FINSEQ_1:44
          .= (nf ^ <* r-s,s *>).m by A8,A9,A7,A14,FINSEQ_1:24;
      end;
      case
        m <> n+2;
        then m<n+2 by A13,XXREAL_0:1;
        then
A15:    m<=n+1 by A6,NAT_1:13;
A16:    len mf - 1 = n+1 by A4;
        now
          per cases;
          case
A17:        m = n+1;
            then
A18:        n<m by NAT_1:13;
            thus mf.m = r - s by A2,A3,A6,A12,A16,A17,Def2
              .= <* r-s,s *>.(n+1-n) by FINSEQ_1:44
              .= (nf ^ <* r-s,s *>).m by A8,A9,A13,A17,A18,FINSEQ_1:24;
          end;
          case
            m <> n+1;
            then m<n+1 by A15,XXREAL_0:1;
            then
A19:        m<=n by NAT_1:13;
            then
A20:        m in Seg n by A12,FINSEQ_1:1;
            1<=n by A12,A19,XXREAL_0:2;
            then
A21:        n in Seg(n+2) by A10,FINSEQ_1:1;
A22:        dom nf = Seg len nf by FINSEQ_1:def 3;
            dom mf = Seg len mf by FINSEQ_1:def 3;
            hence mf.m = nf.m by A4,A20,A21,Th6
              .= (nf ^ <* r-s,s *>).m by A8,A20,A22,FINSEQ_1:def 7;
          end;
        end;
        hence mf.m = (nf ^ <* r-s,s *>).m;
      end;
    end;
    hence mf.m = (nf ^ <* r-s,s *>).m;
  end;
  hence thesis by A4,A9,FINSEQ_2:9;
end;

theorem Th12:
  MIM( <*>REAL ) = <*>REAL
proof
  set o = <*>REAL, mo = MIM(o);
  len mo = len o by Def2;
  hence thesis;
end;

theorem Th13:
  for r be Real holds MIM(<*r*>) = <*r*>
proof
  let r be Real;
  set f = <*r*>;
A1: len f = 1 by FINSEQ_1:40;
  then
A2: len MIM(f) = 1 by Def2;
  then
A3: dom MIM(f) = Seg 1 by FINSEQ_1:def 3;
  now
    let n be Nat;
    assume n in dom MIM(f);
    then n = 1 by A3,FINSEQ_1:2,TARSKI:def 1;
    hence MIM(f).n = f.n by A1,A2,Def2;
  end;
  hence thesis by A1,A2,FINSEQ_2:9;
end;

theorem
  for r,s be Real holds MIM(<*r,s*>) = <*r-s,s*>
proof
  let r,s be Real;
  set f = <*r,s*>;
A1: len f = 2 & f.1 = r by FINSEQ_1:44;
A2: f.2 = s by FINSEQ_1:44;
  0 qua Nat+2 = 2 & 0 qua Nat +1 = 1;
  then MIM(f) = MIM(f)|0 ^ <*r-s,s*> by A1,A2,Th11;
  hence thesis by FINSEQ_1:34;
end;

theorem
  for R be FinSequence of REAL, n be Element of NAT holds MIM(R) /^ n =
  MIM(R/^n)
proof
  let R be FinSequence of REAL,n;
  set mf = MIM(R), fn = R/^n, mn = MIM(fn);
A1: len mf = len R by Def2;
A2: len mn = len fn by Def2;
  now
    per cases;
    case
A3:   len R<n;
      then mf /^ n = <*>REAL by A1,Def1;
      hence thesis by A3,Def1,Th12;
    end;
    case
A4:   n<=len R;
      then
A5:   len(mf/^n) = len R - n by A1,Def1;
A6:   len mn = len fn by Def2;
      then
A7:   dom mn = Seg len fn by FINSEQ_1:def 3;
A8:   Seg len(mf/^n) = dom(mf/^n) by FINSEQ_1:def 3;
A9:   len fn = len R - n by A4,Def1;
A10:  Seg len fn = dom fn by FINSEQ_1:def 3;
      now
        let m be Nat;
        set r1 = R.(m+n);
        assume
A11:    m in dom mn;
        then
A12:    1<=m by A7,FINSEQ_1:1;
A13:    m<=len fn by A7,A11,FINSEQ_1:1;
A14:    fn.m = r1 by A4,A10,A7,A11,Def1;
        m<=m+n by NAT_1:11;
        then
A15:    1<=m+n by A12,XXREAL_0:2;
        now
          per cases;
          case
A16:        m = len fn;
            thus (mf/^n).m = mf.(m+n) by A1,A4,A5,A9,A8,A7,A11,Def1
              .= r1 by A1,A9,A16,Def2
              .= mn.m by A6,A14,A16,Def2;
          end;
          case
            m <> len fn;
            then m<len fn by A13,XXREAL_0:1;
            then
A17:        m+1<=len fn by NAT_1:13;
            then
A18:        m<=len fn - 1 by XREAL_1:19;
            set r2 = R.(m+1+n);
A19:        m+1+n = m+n+1;
            1<=m+1 by NAT_1:11;
            then m+1 in dom fn by A17,FINSEQ_3:25;
            then
A20:        fn.(m+1) = r2 by A4,Def1;
            m+1+n<=len R by A9,A17,XREAL_1:19;
            then
A21:        m+n<=len R - 1 by A19,XREAL_1:19;
            thus (mf/^n).m = mf.(m+n) by A1,A4,A5,A9,A8,A7,A11,Def1
              .= r1-r2 by A1,A15,A19,A21,Def2
              .= mn.m by A2,A12,A14,A18,A20,Def2;
          end;
        end;
        hence (mf/^n).m = mn.m;
      end;
      hence thesis by A5,A9,A6,FINSEQ_2:9;
    end;
  end;
  hence thesis;
end;

theorem Th16:
  for R be FinSequence of REAL st len R <> 0 holds Sum MIM(R) = R. 1
proof
  defpred P[Nat] means for R be FinSequence of REAL st len R <> 0 & len R = $1
  holds Sum MIM(R) = R.1;
  let R be FinSequence of REAL;
  assume
A1: len R <> 0;
A2: for n st P[n] holds P[n+1]
  proof
    let n;
    assume
A3: P[n];
    let R be FinSequence of REAL;
    assume that
    len R <> 0 and
A4: len R = n+1;
    now
      per cases;
      case
        n = 0;
        then
A5:     R = <*R.1*> by A4,FINSEQ_1:40;
        then MIM(R) = R by Th13;
        hence thesis by A5,FINSOP_1:11;
      end;
      case
        n <> 0;
        then 0<n by NAT_1:3;
        then 0 qua Nat+1<=n by NAT_1:13;
        then max(0,n-1) = n-1 by FINSEQ_2:4;
        then reconsider n1 = n-1 as Element of NAT by FINSEQ_2:5;
A6:     0 qua Nat+1<=n1+1 by NAT_1:11;
        then
A7:     Seg len R = dom R & 1 in Seg(n1+1) by FINSEQ_1:1,def 3;
        n1+2 = n1+1+1;
        then n1+1<=n1+2 by NAT_1:11;
        then
A8:     n1+1 in Seg(n1+2) by A6,FINSEQ_1:1;
        set f1 = R|(n1+1);
        set s = R.(n1+2);
        set r = R.(n1+1);
A9:     n1+2 = len R by A4;
A10:    len f1 = n1+1 by A4,FINSEQ_1:59,NAT_1:11;
        thus Sum MIM(R) = Sum((MIM(R)|n1) ^ <* r-s,s *>) by A4,Th11
          .= Sum(MIM(R)|n1) + Sum(<* r-s,s *>) by RVSUM_1:75
          .= Sum(MIM(R)|n1) + (r-s + s) by RVSUM_1:77
          .= Sum((MIM(R)|n1) ^<*r*>) by RVSUM_1:74
          .= Sum MIM(f1) by A9,Th10
          .= f1.1 by A3,A10
          .= R.1 by A4,A8,A7,Th6;
      end;
    end;
    hence thesis;
  end;
A11: P[0];
  for n holds P[n] from NAT_1:sch 1(A11,A2);
  hence thesis by A1;
end;

theorem
  for R be FinSequence of REAL, n be Element of NAT st n<len R holds Sum
  MIM(R/^n) = R.(n+1)
proof
  let R be FinSequence of REAL,n;
  assume
A1: n<len R;
  then
A2: len(R/^n) = len R - n by Def1;
  n+1<=len R by A1,NAT_1:13;
  then 1<=len R - n by XREAL_1:19;
  then
A3: 1 in dom(R/^n) by A2,FINSEQ_3:25;
  len(R/^n) <> 0 by A1,A2;
  hence Sum MIM(R/^n) = (R/^n).1 by Th16
    .= R.(n+1) by A1,A3,Def1;
end;

definition
  let IT be FinSequence of REAL;
  redefine attr IT is non-increasing means
  :Def3:
  for n be Element of NAT st n in dom IT & n+1 in dom IT holds
  IT.n >= IT.(n+1);
  compatibility
  proof
    thus IT is non-increasing implies for n be Element of NAT st n in dom IT &
    n+1 in dom IT holds IT.n >= IT.(n+1)
    proof
      assume
A1:   IT is non-increasing;
      let n be Element of NAT such that
A2:   n in dom IT & n+1 in dom IT;
      n+(0 qua Nat) <= n+1 by XREAL_1:6;
      hence thesis by A1,A2,SEQM_3:def 4;
    end;
    assume
A3: for n be Element of NAT st n in dom IT & n+1 in dom IT holds IT.n
    >= IT.(n+1);
    let m, n such that
A4: m in dom IT and
A5: n in dom IT & m <= n;
    defpred P[Nat] means $1 in dom IT & m <= $1 implies IT.m >= IT.$1;
A6: for k being Element of NAT st P[k] holds P[k+1]
    proof
      let k be Element of NAT such that
A7:   P[k] and
A8:   k+1 in dom IT;
      assume m <= k+1;
      then
A9:   m < k+1 or m = k+1 by XXREAL_0:1;
      per cases by A9,NAT_1:13;
      suppose
A10:    m <= k;
        k+(0 qua Nat) <= k+1 & k+1 <= len IT by A8,FINSEQ_3:25,XREAL_1:6;
        then
A11:    k <= len IT by XXREAL_0:2;
        1 <= m by A4,FINSEQ_3:25;
        then
A12:    1 <= k by A10,XXREAL_0:2;
        then k in dom IT by A11,FINSEQ_3:25;
        then IT.k >= IT.(k+1) by A3,A8;
        hence thesis by A7,A10,A12,A11,FINSEQ_3:25,XXREAL_0:2;
      end;
      suppose
        m = k+1;
        hence thesis;
      end;
    end;
A13: P[0] by FINSEQ_3:24;
    for k being Element of NAT holds P[k] from NAT_1:sch 1(A13,A6);
    hence thesis by A5;
  end;
end;

registration
  cluster non-increasing for FinSequence of REAL;
  existence
  proof
    take <*>REAL;
    let n;
    thus thesis;
  end;
end;

theorem Th18:
  for R be FinSequence of REAL st len R = 0 or len R = 1 holds
  R is non-increasing
proof
  let R be FinSequence of REAL;
  assume
A1: len R = 0 or len R = 1;
  now
    per cases by A1;
    case
      len R = 0;
      then R = <*>REAL;
      then for n st n in dom R & n+1 in dom R holds R.n>=R.(n+1);
      hence thesis by Def3;
    end;
    case
      len R = 1;
      then
A2:   dom R = {1} by FINSEQ_1:2,def 3;
      now
        let n;
        assume that
A3:     n in dom R and
A4:     n+1 in dom R;
        n = 1 by A2,A3,TARSKI:def 1;
        hence R.n>=R.(n+1) by A2,A4,TARSKI:def 1;
      end;
      hence thesis by Def3;
    end;
  end;
  hence thesis;
end;

theorem Th19:
  for R be FinSequence of REAL holds R is non-increasing iff for n
  ,m be Element of NAT st n in dom R & m in dom R & n<m holds R.n>=R.m
proof
  let R be FinSequence of REAL;
  thus R is non-increasing implies for n,m st n in dom R & m in dom R & n<m
  holds R.n>=R.m
  proof
    defpred P[Nat] means $1 in dom R implies for n st n in dom R & n<$1 holds
    R.n>=R.$1;
    assume
A1: R is non-increasing;
A2: for m st P[m] holds P[m+1]
    proof
      let m;
      assume
A3:   P[m];
      assume
A4:   m+1 in dom R;
      then m+1<=len R by FINSEQ_3:25;
      then
A5:   m<=len R by NAT_1:13;
      let n;
      assume that
A6:   n in dom R and
A7:   n<m+1;
      set t = R.m, r = R.n, s = R.(m+1);
A8:   n<=m by A7,NAT_1:13;
      1<=n by A6,FINSEQ_3:25;
      then
A9:   1<=m by A8,XXREAL_0:2;
      then
A10:  m in dom R by A5,FINSEQ_3:25;
      now
        per cases;
        case
          n = m;
          hence thesis by A1,A4,A6,Def3;
        end;
        case
          n <> m;
          then n<m by A8,XXREAL_0:1;
          then
A11:      r>=t by A3,A6,A9,A5,FINSEQ_3:25;
          t>=s by A1,A4,A10,Def3;
          hence thesis by A11,XXREAL_0:2;
        end;
      end;
      hence thesis;
    end;
    Seg len R = dom R by FINSEQ_1:def 3;
    then
A12: P[0] by FINSEQ_1:1;
    for m holds P[m] from NAT_1:sch 1(A12,A2);
    hence thesis;
  end;
  assume
A13: for n,m st n in dom R & m in dom R & n<m holds R.n>=R.m;
  let n;
A14: n<n+1 by NAT_1:13;
  assume n in dom R & n+1 in dom R;
  hence thesis by A13,A14;
end;

theorem Th20:
  for R be non-increasing FinSequence of REAL, n be Element of NAT
  holds R|n is non-increasing FinSequence of REAL
proof
  let f be non-increasing FinSequence of REAL, n;
  set fn = f|n;
  now
    per cases;
    case
      n = 0;
      then len fn = 0;
      hence thesis by Th18;
    end;
    case
      n <> 0;
      then 0<n by NAT_1:3;
      then
A1:   0 qua Nat+1<=n by NAT_1:13;
      now
        per cases;
        case
          len f<=n;
          hence thesis by Lm3;
        end;
        case
          n<len f;
          then
A2:       n in dom f & len fn = n by A1,FINSEQ_1:59,FINSEQ_3:25;
          now
            let m;
A3:         dom fn = Seg len fn by FINSEQ_1:def 3;
            assume
A4:         m in dom fn & m+1 in dom fn;
            then
A5:         m in dom f & m+1 in dom f by A2,A3,Th6;
            f.m = fn.m & f.(m+1) = fn.(m+1) by A2,A4,A3,Th6;
            hence fn.m>=fn.(m+1) by A5,Def3;
          end;
          hence thesis by Def3;
        end;
      end;
      hence thesis;
    end;
  end;
  hence thesis;
end;

theorem
  for R be non-increasing FinSequence of REAL, n be Element of NAT holds
  R /^ n is non-increasing
proof
  let f be non-increasing FinSequence of REAL, n;
  set fn = f /^ n;
  now
    let m;
    assume that
A1: m in dom fn and
A2: m+1 in dom fn;
A3: m<=m+n by NAT_1:11;
    1<=m by A1,FINSEQ_3:25;
    then
A4: 1<=m+n by A3,XXREAL_0:2;
A5: 1<=m+n+1 by NAT_1:11;
A6: m+1<=len fn by A2,FINSEQ_3:25;
    set r = fn.m, s = fn.(m+1);
A7: m+1+n = m+n+1;
A8: m<=len fn by A1,FINSEQ_3:25;
    now
      per cases;
      case
A9:     n<=len f;
        then
A10:    len fn = len f - n by Def1;
        then m+n<=len f by A8,XREAL_1:19;
        then
A11:    m+n in dom f by A4,FINSEQ_3:25;
        m+n+1<=len f by A6,A7,A10,XREAL_1:19;
        then
A12:    m+n+1 in dom f by A5,FINSEQ_3:25;
        r = f.(m+n) & s = f.(m+n+1) by A1,A2,A7,A9,Def1;
        hence r>=s by A11,A12,Def3;
      end;
      case
        len f<n;
        then fn = <*>REAL by Def1;
        hence r>=s;
      end;
    end;
    hence r>=s;
  end;
  hence thesis by Def3;
end;

Lm4: for f,g be non-increasing FinSequence of REAL, n st len f = n+1 & len f =
len g & f,g are_fiberwise_equipotent holds f.(len f) = g.(len g) & f|n, g|n
are_fiberwise_equipotent
proof
  let f,g be non-increasing FinSequence of REAL, n;
  assume that
A1: len f = n+1 and
A2: len f = len g and
A3: f,g are_fiberwise_equipotent;
  set r = f.(n+1), s = g.(n+1);
  0<n+1 by NAT_1:3;
  then
A4: 0 qua Nat+1<=n+1 by NAT_1:13;
  then
A5: n+1 in dom f by A1,FINSEQ_3:25;
A6: f = (f|n)^<*r*> by A1,Th7;
A7: n+1 in dom g by A1,A2,A4,FINSEQ_3:25;
A8: now
A9: rng f = rng g by A3,CLASSES1:75;
    assume
A10: r <> s;
    now
      per cases by A10,XXREAL_0:1;
      case
A11:    r>s;
        s in rng f by A7,A9,FUNCT_1:def 3;
        then consider m be Nat such that
A12:    m in dom f and
A13:    f.m = s by FINSEQ_2:10;
A14:    m<=len f by A12,FINSEQ_3:25;
        now
          per cases;
          case
            m = len f;
            hence contradiction by A1,A11,A13;
          end;
          case
            m <> len f;
            then m<n+1 by A1,A14,XXREAL_0:1;
            hence contradiction by A5,A11,A12,A13,Th19;
          end;
        end;
        hence contradiction;
      end;
      case
A15:    r<s;
        r in rng g by A5,A9,FUNCT_1:def 3;
        then consider m be Nat such that
A16:    m in dom g and
A17:    g.m = r by FINSEQ_2:10;
A18:    m<=len g by A16,FINSEQ_3:25;
        now
          per cases;
          case
            m = len g;
            hence contradiction by A1,A2,A15,A17;
          end;
          case
            m <> len g;
            then m<n+1 by A1,A2,A18,XXREAL_0:1;
            hence contradiction by A7,A15,A16,A17,Th19;
          end;
        end;
        hence contradiction;
      end;
    end;
    hence contradiction;
  end;
  hence f.(len f) = g.(len g) by A1,A2;
A19: g = (g|n)^<*r*> by A1,A2,A8,Th7;
  now
    let x;
    card Coim(f|n,x) + card(<*r*>"{x}) = card Coim(f,x) by A6,FINSEQ_3:57
      .= card Coim(g,x) by A3,CLASSES1:def 9
      .= card Coim(g|n,x) + card(<*r*>"{x}) by A19,FINSEQ_3:57;
    hence card Coim(f|n,x) = card Coim(g|n,x);
  end;
  hence thesis by CLASSES1:def 9;
end;

defpred P[Nat] means for R be FinSequence of REAL st $1 = len R ex b be
non-increasing FinSequence of REAL st R,b are_fiberwise_equipotent;

Lm5: P[0]
proof
  let R be FinSequence of REAL;
  assume len R = 0;
  then reconsider a = R as non-increasing FinSequence of REAL by Th18;
  take a;
  thus thesis;
end;

Lm6: for n st P[n] holds P[n+1]
proof
  let n;
  assume
A1: P[n];
  let R be FinSequence of REAL;
  set fn = R|(Seg n);
A2: fn = R|n by FINSEQ_1:def 15;
  set q = R.(n+1);
  reconsider fn as FinSequence by FINSEQ_1:15;
  rng fn c= rng R by RELAT_1:70;
  then rng fn c= REAL by XBOOLE_1:1;
  then reconsider fn as FinSequence of REAL by FINSEQ_1:def 4;
  n<=n+1 by NAT_1:11;
  then
A3: dom fn = dom R /\ Seg n & Seg n c= Seg(n+1) by FINSEQ_1:5,RELAT_1:61;
  assume
A4: n+1 = len R;
  then dom R = Seg(n+1) by FINSEQ_1:def 3;
  then dom fn = Seg n by A3,XBOOLE_1:28;
  then
A5: len fn = n by FINSEQ_1:def 3;
  then consider a be non-increasing FinSequence of REAL such that
A6: fn,a are_fiberwise_equipotent by A1;
A7: len fn = len a by A6,Th3;
A8: Seg len a = dom a by FINSEQ_1:def 3;
  now
    per cases;
    case
A9:   for t be Real st t in rng a holds q<=t;
      set b = a^<*q*>;
A10:  len b = n + len <*q*> by A5,A7,FINSEQ_1:22
        .= n+1 by FINSEQ_1:39;
      now
        let m;
        assume that
A11:    m in dom b and
A12:    m+1 in dom b;
A13:    1<=m+1 by A12,FINSEQ_3:25;
        set r = b.m, s = b.(m+1);
A14:    1<=m by A11,FINSEQ_3:25;
A15:    m+1<=len b by A12,FINSEQ_3:25;
        then m<=n+1-1 by A10,XREAL_1:19;
        then m in Seg n by A14,FINSEQ_1:1;
        then
A16:    m in dom a by A5,A7,FINSEQ_1:def 3;
        then
A17:    r = a.m by FINSEQ_1:def 7;
A18:    a.m in rng a by A16,FUNCT_1:def 3;
        now
          per cases;
          case
            m+1 = len b;
            then s = q by A5,A7,A10,FINSEQ_1:42;
            hence r>=s by A9,A17,A18;
          end;
          case
            m+1 <> len b;
            then m+1<len b by A15,XXREAL_0:1;
            then m+1<=n+1-1 by A10,NAT_1:13;
            then m+1 in Seg n by A13,FINSEQ_1:1;
            then
A19:        m+1 in dom a by A5,A7,FINSEQ_1:def 3;
            then a.(m+1) = s by FINSEQ_1:def 7;
            hence r>=s by A16,A17,A19,Def3;
          end;
        end;
        hence r>=s;
      end;
      then reconsider b as non-increasing FinSequence of REAL by Def3;
      take b;
      fn^<*q*> = R by A4,A2,Th7;
      hence R,b are_fiberwise_equipotent by A6,Th1;
    end;
    case
A20:  ex t be Real st t in rng a & not q<=t;
      defpred Q[Nat] means $1 in dom a & for r st r = a.$1 holds r<q;
A21:  ex k be Nat st Q[k]
      proof
        consider t be Real such that
A22:    t in rng a and
A23:    t<q by A20;
        consider k be Nat such that
A24:    k in dom a and
A25:    a.k = t by A22,FINSEQ_2:10;
        reconsider k as Element of NAT by ORDINAL1:def 12;
        take k;
        thus k in dom a by A24;
        let r;
        assume r = a.k;
        hence thesis by A23,A25;
      end;
      consider mi be Nat such that
A26:  Q[mi] & for m be Nat st Q[m] holds mi <= m from NAT_1:sch 5(A21
      );
      1<=mi by A26,FINSEQ_3:25;
      then max(0,mi-1) = mi-1 by FINSEQ_2:4;
      then reconsider k = mi-1 as Element of NAT by FINSEQ_2:5;
      mi<mi+1 by NAT_1:13;
      then
A27:  k<=mi by XREAL_1:19;
A28:  mi<=len a by A26,FINSEQ_3:25;
      then
A29:  k<=len a by A27,XXREAL_0:2;
      then
A30:  len(a/^k) = len a - k by Def1;
A31:  dom((a|k)^<*q*>) = Seg len((a|k)^<*q*>) by FINSEQ_1:def 3;
A32:  dom (a|k) c= dom((a|k)^<*q*>) by FINSEQ_1:26;
      set ak = a/^k, b = (a|k)^<*q*>^ak;
A33:  dom (a|k) = Seg len (a|k) by FINSEQ_1:def 3;
A34:  len(a|k) = k by A28,A27,FINSEQ_1:59,XXREAL_0:2;
      then
A35:  len((a|k)^<*q*>) = k+len <*q*> by FINSEQ_1:22
        .= k+1 by FINSEQ_1:39;
      then
A36:  len b = k+1 + len(a/^k) by FINSEQ_1:22;
      k+1<=len a by A26,FINSEQ_3:25;
      then
A37:  1<=len(a/^k) by A30,XREAL_1:19;
      now
        let m;
        assume that
A38:    m in dom b and
A39:    m+1 in dom b;
A40:    1<=m+1 by A39,FINSEQ_3:25;
A41:    m+1<=len b by A39,FINSEQ_3:25;
        set r = b.m, s = b.(m+1);
A42:    1<=m by A38,FINSEQ_3:25;
A43:    m<=len b by A38,FINSEQ_3:25;
        now
          per cases;
          case
A44:        m+1<=k;
            dom(a|k) c= dom((a|k)^(a/^k)) by FINSEQ_1:26;
            then
A45:        dom(a|k) c= dom a by Th8;
A46:        dom a = Seg len a by FINSEQ_1:def 3;
            m<=k by A44,NAT_1:13;
            then
A47:        m in Seg k by A42,FINSEQ_1:1;
            1<=k by A40,A44,XXREAL_0:2;
            then
A48:        k in dom a by A29,A46,FINSEQ_1:1;
            then
A49:        a.m = (a|k).m by A47,Th6;
A50:        m+1 in Seg k by A40,A44,FINSEQ_1:1;
            then
A51:        a.(m+1) = (a|k).(m+1) by A48,Th6;
A52:        b.(m+1) = ((a|k)^<*q*>).(m+1) by A34,A33,A32,A50,FINSEQ_1:def 7
              .= a.(m+1) by A34,A33,A50,A51,FINSEQ_1:def 7;
            b.m = ((a|k)^<*q*>).m by A34,A33,A32,A47,FINSEQ_1:def 7
              .= a.m by A34,A33,A47,A49,FINSEQ_1:def 7;
            hence r>=s by A34,A33,A47,A50,A52,A45,Def3;
          end;
          case
            k<m+1;
            then
A53:        k<=m by NAT_1:13;
            now
              per cases;
              case
A54:            k = m;
                then m+1 in dom((a|k)^<*q*>) by A35,A31,FINSEQ_1:4;
                then
A55:            b.(m+1) = ((a|k)^<*q*>).(k+1) by A54,FINSEQ_1:def 7
                  .= q by A34,FINSEQ_1:42;
A56:            m in Seg k by A42,A54,FINSEQ_1:1;
A57:            k in dom a by A8,A29,A42,A54,FINSEQ_1:1;
                then
A58:            a.m = (a|k).m by A56,Th6;
A59:            b.m = ((a|k)^<*q*>).m by A34,A33,A32,A56,FINSEQ_1:def 7
                  .= a.m by A34,A33,A56,A58,FINSEQ_1:def 7;
                now
                  assume s>r;
                  then for t be Real st t = a.k holds t<q by A54,A59,A55;
                  then mi<=k by A26,A57;
                  hence contradiction by XREAL_1:44;
                end;
                hence r>=s;
              end;
              case
                k <> m;
                then k<m by A53,XXREAL_0:1;
                then
A60:            k+1<=m by NAT_1:13;
                then
A61:            k+1<m+1 by NAT_1:13;
                max(0,m-(k+1)) = m-(k+1) by A60,FINSEQ_2:4;
                then reconsider l = m-(k+1) as Element of NAT by FINSEQ_2:5;
A62:            l+1 = m+1-(k+1);
                then
A63:            l+1<=len b -(k+1) by A41,XREAL_1:13;
                l<=l+1 by NAT_1:11;
                then
A64:            l<=len(a/^k) by A36,A63,XXREAL_0:2;
                1<=l+1 by NAT_1:11;
                then
A65:            l+1 in dom(a/^k) by A36,A63,FINSEQ_3:25;
                1<=l+1+k by A38,FINSEQ_3:25;
                then
A66:            k+l+1<=len a by A30,A36,A63,XREAL_1:19;
                then
A67:            k+l+1 in dom a by A42,FINSEQ_3:25;
                k+l<=k+l+1 by NAT_1:11;
                then
A68:            k+l<=len a by A66,XXREAL_0:2;
A69:            k+(l+1) <=len a by A30,A36,A63,XREAL_1:19;
                now
                  per cases;
                  case
A70:                k+1 = m;
                    then m in Seg (k+1) by A42,FINSEQ_1:1;
                    then
A71:                b.m = ((a|k)^<*q*>).(k+1) by A35,A31,A70,FINSEQ_1:def 7
                      .= q by A34,FINSEQ_1:42;
A72:                1 in dom(a/^k) by A37,FINSEQ_3:25;
                    b.(m+1) = (a/^k).(k+1+1 - (k+1)) by A35,A41,A61,A70,
FINSEQ_1:24
                      .= a.mi by A29,A72,Def1;
                    hence r>=s by A26,A71;
                  end;
                  case
                    k+1 <> m;
                    then
A73:                k+1<m by A60,XXREAL_0:1;
                    then k+1+1<=m by NAT_1:13;
                    then
A74:                1<=l by XREAL_1:19;
                    then
A75:                l in dom(a/^k) by A64,FINSEQ_3:25;
                    l<=k+l by NAT_1:11;
                    then 1<=k+l by A74,XXREAL_0:2;
                    then
A76:                k+l in dom a by A68,FINSEQ_3:25;
A77:                b.m = (a/^k).l by A35,A43,A73,FINSEQ_1:24
                      .= a.(k+l) by A29,A75,Def1;
                    b.(m+1) = (a/^k).(l+1) by A35,A41,A61,A62,FINSEQ_1:24
                      .= a.(k+l+1) by A29,A65,A69,Def1;
                    hence r>=s by A67,A77,A76,Def3;
                  end;
                end;
                hence r>=s;
              end;
            end;
            hence r>=s;
          end;
        end;
        hence r>=s;
      end;
      then reconsider b as non-increasing FinSequence of REAL by Def3;
      take b;
      now
        let x;
A78:    card Coim(fn,x) = card Coim(a,x) by A6,CLASSES1:def 9;
        thus card Coim(b,x) = card (((a|k)^<*q*>)"{x}) + card(ak"{x}) by
FINSEQ_3:57
          .= card((a|k)"{x}) + card(<*q*>"{x}) + card(ak"{x}) by FINSEQ_3:57
          .= card((a|k)"{x}) + card(ak"{x}) + card(<*q*>"{x})
          .= card(((a|k)^ak)"{x}) + card(<*q*>"{x}) by FINSEQ_3:57
          .= card(fn"{x}) + card(<*q*>"{x}) by A78,Th8
          .= card((fn^<*q*>)"{x}) by FINSEQ_3:57
          .= card Coim(R,x) by A4,A2,Th7;
      end;
      hence R,b are_fiberwise_equipotent by CLASSES1:def 9;
    end;
  end;
  hence thesis;
end;

theorem Th22:
  for R be FinSequence of REAL ex R1 be non-increasing FinSequence of
  REAL st R,R1 are_fiberwise_equipotent
proof
  let R be FinSequence of REAL;
A1: len R = len R;
  for n holds P[n] from NAT_1:sch 1(Lm5,Lm6);
  hence thesis by A1;
end;

Lm7: for n holds for g1,g2 be non-increasing FinSequence of REAL st n = len g1
& g1,g2 are_fiberwise_equipotent holds g1 = g2
proof
  defpred P[Nat] means for g1,g2 be non-increasing FinSequence of REAL st $1 =
  len g1 & g1,g2 are_fiberwise_equipotent holds g1 = g2;
A1: for n st P[n] holds P[n+1]
  proof
    let n;
    assume
A2: P[n];
    let g1,g2 be non-increasing FinSequence of REAL;
    set r = g1.(n+1);
    reconsider g1n = g1|n, g2n = g2|n as non-increasing FinSequence of REAL by
Th20;
    assume that
A3: len g1 = n+1 and
A4: g1,g2 are_fiberwise_equipotent;
A5: len g2 = len g1 by A4,Th3;
    then
A6: g1.(len g1) = g2.(len g2) by A3,A4,Lm4;
A7: (g1|n)^<*r*> = g1 by A3,Th7;
    len(g1|n) = n by A3,FINSEQ_1:59,NAT_1:11;
    then g1n = g2n by A2,A3,A4,A5,Lm4;
    hence thesis by A3,A5,A6,A7,Th7;
  end;
A8: P[0]
  proof
    let g1,g2 be non-increasing FinSequence of REAL;
    assume len g1 = 0 & g1,g2 are_fiberwise_equipotent;
    then len g2 = len g1 & g1 = <*>REAL by Th3;
    hence thesis;
  end;
  thus for n holds P[n] from NAT_1:sch 1(A8,A1);
end;

theorem
  for R1,R2 be non-increasing FinSequence of REAL st R1,R2
  are_fiberwise_equipotent holds R1 = R2
proof
  let g1,g2 be non-increasing FinSequence of REAL;
A1: len g1 = len g1;
  assume g1,g2 are_fiberwise_equipotent;
  hence thesis by A1,Lm7;
end;

theorem
  for R be FinSequence of REAL, r,s be Real st r <> 0 holds R"{s/r} = (r
  *R)"{ s }
proof
  let R be FinSequence of REAL, r,s;
A1: Seg len R = dom R & dom(r*R) = Seg len(r*R) by FINSEQ_1:def 3;
  assume
A2: r <> 0;
  thus R"{s/r} c= (r*R)"{s}
  proof
    let x;
    assume
A3: x in R"{s/r};
    then reconsider i = x as Element of NAT;
    R.i in {s/r} by A3,FUNCT_1:def 7;
    then R.i = s/r by TARSKI:def 1;
    then r*(R.i) = s by A2,XCMPLX_1:87;
    then (r*R).i = s by RVSUM_1:44;
    then
A4: (r*R).i in {s} by TARSKI:def 1;
    i in dom R by A3,FUNCT_1:def 7;
    then i in dom (r*R) by A1,FINSEQ_2:33;
    hence thesis by A4,FUNCT_1:def 7;
  end;
  let x;
  assume
A5: x in (r*R)"{s};
  then reconsider i = x as Element of NAT;
  (r*R).i in {s} by A5,FUNCT_1:def 7;
  then (r*R).i = s by TARSKI:def 1;
  then r*R.i = s by RVSUM_1:44;
  then R.i = s/r by A2,XCMPLX_1:89;
  then
A6: R.i in {s/r} by TARSKI:def 1;
  i in dom(r*R) by A5,FUNCT_1:def 7;
  then i in dom R by A1,FINSEQ_2:33;
  hence thesis by A6,FUNCT_1:def 7;
end;

theorem
  for R be FinSequence of REAL holds (0*R)"{0} = dom R
proof
  let R be FinSequence of REAL;
A1: Seg len(0*R) = dom(0*R) by FINSEQ_1:def 3;
A2: len(0*R) = len R & dom R = Seg len R by FINSEQ_1:def 3,FINSEQ_2:33;
  hence (0*R)"{0} c= dom R by A1,RELAT_1:132;
  let x;
  assume
A3: x in dom R;
  then reconsider i = x as Element of NAT;
  (0*R).i = 0*R.i by RVSUM_1:44
    .= 0;
  then (0*R).i in {0} by TARSKI:def 1;
  hence thesis by A2,A1,A3,FUNCT_1:def 7;
end;

begin 


reserve f, g for Function;

theorem
  for f, g st rng f = rng g & f is one-to-one & g is one-to-one holds
  f,g are_fiberwise_equipotent
proof
  let f, g be Function such that
A1: rng f = rng g and
A2: f is one-to-one and
A3: g is one-to-one;
  let x;
  per cases;
  suppose
A4: x in rng f;
    then card Coim(f,x) = 1 by A2,FINSEQ_4:73;
    hence thesis by A1,A3,A4,FINSEQ_4:73;
  end;
  suppose
A5: not x in rng f;
    then card(f"{x}) = 0 by CARD_1:27,FUNCT_1:72;
    hence thesis by A1,A5,CARD_1:27,FUNCT_1:72;
  end;
end;



theorem
  for D being set, f being FinSequence of D holds f /^ len f = {}
proof
  let D be set, f be FinSequence of D;
  len (f /^ len f) = len f - len f by Def1
    .= 0;
  hence thesis;
end;



theorem
  for f,g be Function,m,n be set st f.m=g.n & f.n=g.m & m in dom f & n
in dom f & dom f = dom g & (for k be set st k<>m & k<>n & k in dom f holds f.k=
  g.k) holds f,g are_fiberwise_equipotent
proof
  let f,g be Function,m,n be set;
  assume that
A1: f.m=g.n and
A2: f.n=g.m and
A3: m in dom f and
A4: n in dom f and
A5: dom f = dom g and
A6: for k be set st k<>m & k<>n & k in dom f holds f.k=g.k;
  set t=id dom f, nm=n .--> m,mn=m .--> n, p=t +* nm +* mn;
A7: dom nm ={ n } by FUNCOP_1:13;
A8: dom t = dom f;
A9: rng t = dom (t") by FUNCT_1:33
    .= dom f by A8,FUNCT_1:45;
  dom mn ={ m } by FUNCOP_1:13;
  then
A10: dom p = dom (t +* nm) \/ {m} by FUNCT_4:def 1
    .= dom t \/ {n} \/ {m} by A7,FUNCT_4:def 1
    .= dom f \/ {n} \/ {m}
    .= dom f \/ {m} by A4,ZFMISC_1:40
    .= dom f by A3,ZFMISC_1:40;
A11: now
    let x be set;
    assume
A12: x in dom f;
    then
A13: (p*p).x=p.(p.x) by A10,FUNCT_1:13;
    per cases;
    suppose
A14:  x=m;
      hence (p*p).x=p.n by A13,FUNCT_4:89
        .=x by A14,FUNCT_4:90;
    end;
    suppose
A15:  x<>m;
      now
        per cases;
        suppose
A16:      x=n;
          hence (p*p).x=p.m by A13,FUNCT_4:90
            .=x by A16,FUNCT_4:89;
        end;
        suppose
A17:      x<>n;
          hence (p*p).x=p.(t.x) by A13,A15,FUNCT_4:91
            .=p.x by A12,FUNCT_1:17
            .=t.x by A15,A17,FUNCT_4:91
            .=x by A12,FUNCT_1:17;
        end;
      end;
      hence (p*p).x=x;
    end;
  end;
  rng nm ={m} by FUNCOP_1:8;
  then rng t \/ rng nm =dom f by A3,ZFMISC_1:40;
  then
A18: rng(t +* nm) \/ rng mn c= dom f \/ rng mn by FUNCT_4:17,XBOOLE_1:9;
  for z be set st z in rng(p*p) holds z in rng p by FUNCT_1:14;
  then
A19: rng(p*p) c= rng p by TARSKI:def 3;
A20: rng p c= rng(t +* nm) \/ rng mn by FUNCT_4:17;
A21: rng mn ={n} by FUNCOP_1:8;
  then dom f \/ rng mn =dom p by A4,A10,ZFMISC_1:40;
  then
A22: dom (p*p) =dom f by A10,A18,A20,RELAT_1:27,XBOOLE_1:1;
  then p*p=t by A11,FUNCT_1:17;
  then
A23: p is one-to-one by A10,FUNCT_1:31;
  rng p c= dom f \/ rng mn by A18,A20,XBOOLE_1:1;
  then
A24: rng p c= dom p by A4,A10,A21,ZFMISC_1:40;
  then
A25: dom (g*p) =dom f by A5,A10,RELAT_1:27;
  now
    let x be set;
    assume
A26: x in dom f;
    then
A27: (g*p).x=g.(p.x) by A25,FUNCT_1:12;
    per cases;
    suppose
      x=m;
      hence (g*p).x=f.x by A1,A27,FUNCT_4:89;
    end;
    suppose
A28:  x<>m;
      now
        per cases;
        suppose
          x=n;
          hence (g*p).x=f.x by A2,A27,FUNCT_4:90;
        end;
        suppose
A29:      x<>n;
          hence (g*p).x=g.(t.x) by A27,A28,FUNCT_4:91
            .=g.x by A26,FUNCT_1:17
            .=f.x by A6,A26,A28,A29;
        end;
      end;
      hence (g*p).x=f.x;
    end;
  end;
  then
A30: f=g*p by A25,FUNCT_1:2;
  rng(p*p)=dom f by A9,A22,A11,FUNCT_1:17;
  then rng p = dom g by A5,A10,A24,A19,XBOOLE_0:def 10;
  hence thesis by A10,A23,A30,CLASSES1:77;
end;



theorem
  for D being non empty set, f being FinSequence of D, k being Nat holds
  len (f/^k)=len f-'k
proof
  let D be non empty set, f be FinSequence of D,k be Nat;
  per cases;
  suppose
A1: k<=len f;
    then len f-'k=len f-k by XREAL_1:233;
    hence thesis by A1,Def1;
  end;
  suppose
A2: k>len f;
    then (f/^k)=<*>D by Def1;
    then
A3: len (f/^k)=0;
    len f-k<0 by A2,XREAL_1:49;
    hence thesis by A3,XREAL_0:def 2;
  end;
end;



theorem Th30: 
  for f,g be FinSequence,x be set st x in dom g & f,g
  are_fiberwise_equipotent holds ex y be set st y in dom g & f.x=g.y
proof
  let f,g be FinSequence,x be set;
  assume that
A1: x in dom g and
A2: f,g are_fiberwise_equipotent;
  consider P be Permutation of dom g such that
A3: f = g*P by A2,Th4;
  take y=P.x;
  thus y in dom g by A1,FUNCT_2:5;
  thus thesis by A1,A3,FUNCT_2:15;
end;

theorem Th31:
  for f,g,h be FinSequence holds f,g are_fiberwise_equipotent iff
  h^f, h^g are_fiberwise_equipotent
proof
  let f,g,h be FinSequence;
  thus f,g are_fiberwise_equipotent implies h^f, h^g are_fiberwise_equipotent
  proof
    assume
A1: f,g are_fiberwise_equipotent;
    now
      let y be set;
      card Coim(f,y) = card Coim(g,y) by A1,CLASSES1:def 9;
      hence card Coim(h^f,y) = card(g"{y}) + card(h"{y}) by FINSEQ_3:57
        .= card Coim(h^g,y) by FINSEQ_3:57;
    end;
    hence thesis by CLASSES1:def 9;
  end;
  assume
A2: h^f,h^g are_fiberwise_equipotent;
  now
    let x be set;
A3: card Coim(h^f,x) = card Coim(f,x)+card(h"{x}) & card((h^g)"{x}) =
    card(g"{x} )+card(h"{x}) by FINSEQ_3:57;
    card Coim(h^f,x) = card Coim(h^g,x) by A2,CLASSES1:def 9;
    hence card Coim(f,x) = card Coim(g,x) by A3;
  end;
  hence thesis by CLASSES1:def 9;
end;

theorem
  for f,g be FinSequence,m,n,j be Element of NAT st f,g
are_fiberwise_equipotent & m<=n & n <= len f & (for i be Element of NAT st 1<=i
& i<=m holds f.i=g.i) & (for i be Element of NAT st n<i & i<=len f holds f.i=g.
  i) & m<j & j<=n holds ex k be Element of NAT st m<k & k<=n & f.j=g.k
proof
  let f,g be FinSequence,m,n,j be Element of NAT;
  assume
A1: f,g are_fiberwise_equipotent;
  assume that
A2: m<=n and
A3: n <= len f;
  assume
A4: for i be Element of NAT st 1<=i & i<=m holds f.i=g.i;
  reconsider m3=len f-n as Element of NAT by A3,INT_1:3,XREAL_1:48;
  len g=n+m3 by A1,Th3;
  then consider s2,r2 be FinSequence such that
A5: len s2 = n and
A6: len r2 = m3 and
A7: g = s2^r2 by FINSEQ_2:22;
A8: len f=n+m3;
  then consider s1,r1 be FinSequence such that
A9: len s1 = n and
A10: len r1 = m3 and
A11: f = s1^r1 by FINSEQ_2:22;
A12: dom r1 = Seg m3 by A10,FINSEQ_1:def 3;
  assume
A13: for i be Element of NAT st n<i & i<=len f holds f.i=g.i;
  now
    let i be Nat;
    reconsider a = i as Element of NAT by ORDINAL1:def 12;
A14: n < n+1 by XREAL_1:29;
    assume
A15: i in dom r1;
    then
A16: i in dom r2 by A6,A12,FINSEQ_1:def 3;
    1 <= i by A12,A15,FINSEQ_1:1;
    then n+1 <= n+i by XREAL_1:6;
    then
A17: n< i+n by A14,XXREAL_0:2;
    i <= m3 by A12,A15,FINSEQ_1:1;
    then
A18: len s1+i <= len f by A8,A9,XREAL_1:6;
    thus r1.i=f.(len s1+i) by A11,A15,FINSEQ_1:def 7
      .=g.(len s2+a) by A13,A9,A5,A17,A18
      .=r2.i by A7,A16,FINSEQ_1:def 7;
  end;
  then r1=r2 by A10,A6,FINSEQ_2:9;
  then
A19: s1,s2 are_fiberwise_equipotent by A1,A11,A7,Th1;
  reconsider m2=n-m as Element of NAT by A2,INT_1:3,XREAL_1:48;
A20: m+1 > m by XREAL_1:29;
  len s2=m+m2 by A5;
  then consider p2,q2 be FinSequence such that
A21: len p2 = m and
A22: len q2 = m2 and
A23: s2 = p2^q2 by FINSEQ_2:22;
A24: Seg m = dom p2 by A21,FINSEQ_1:def 3;
  len s1=m+m2 by A9;
  then consider p1,q1 be FinSequence such that
A25: len p1 = m and
A26: len q1 = m2 and
A27: s1 = p1^q1 by FINSEQ_2:22;
A28: f=p1^(q1^r1) by A11,A27,FINSEQ_1:32;
A29: dom p1 = Seg m by A25,FINSEQ_1:def 3;
A30: g=p2^(q2^r2) by A7,A23,FINSEQ_1:32;
  now
    let i be Nat;
    reconsider a = i as Element of NAT by ORDINAL1:def 12;
    assume
A31: i in dom p1;
    then
A32: 1<= i & i <= m by A29,FINSEQ_1:1;
    thus p1.i=f.i by A28,A31,FINSEQ_1:def 7
      .=g.a by A4,A32
      .=p2.i by A30,A24,A29,A31,FINSEQ_1:def 7;
  end;
  then p1=p2 by A25,A21,FINSEQ_2:9;
  then
A33: q1,q2 are_fiberwise_equipotent by A27,A23,A19,Th31;
  assume that
A34: m<j and
A35: j<=n;
  j-len p1 > 0 by A34,A25,XREAL_1:50;
  then reconsider x=j-len p1 as Element of NAT by INT_1:3;
A36: x <= n-len p1 by A35,XREAL_1:9;
A37: Seg m2 = dom q2 by A22,FINSEQ_1:def 3;
A38: 1+ 0<= x by A34,A25,INT_1:7,XREAL_1:50;
  then x in dom q2 by A25,A37,A36,FINSEQ_1:1;
  then consider y be set such that
A39: y in dom q2 and
A40: q1.x=q2.y by A33,Th30;
  reconsider y as Element of NAT by A39;
A41: len p2 + y in dom s2 by A23,A39,FINSEQ_1:28;
  reconsider k=len p2+y as Element of NAT by ORDINAL1:def 12;
  take k;
  1<=y by A37,A39,FINSEQ_1:1;
  then k>=len p2+1 by XREAL_1:6;
  hence m<k by A21,A20,XXREAL_0:2;
  y<=m2 by A37,A39,FINSEQ_1:1;
  then k <= m2+len p2 by XREAL_1:6;
  hence k<=n by A21;
  Seg m2 = dom q1 by A26,FINSEQ_1:def 3;
  then
A42: x in dom q1 by A25,A38,A36,FINSEQ_1:1;
  then len p1 + x in dom s1 by A27,FINSEQ_1:28;
  hence f.j=s1.(len p1+x) by A11,FINSEQ_1:def 7
    .=q2.y by A27,A40,A42,FINSEQ_1:def 7
    .=s2.(len p2+y) by A23,A39,FINSEQ_1:def 7
    .=g.k by A7,A41,FINSEQ_1:def 7;
end;

theorem
  for t being FinSequence of INT holds ex u being FinSequence of REAL
  st t,u are_fiberwise_equipotent & u is FinSequence of INT
  & u is non-increasing
proof
  let t be FinSequence of INT;
  t is FinSequence of REAL by FINSEQ_3:117;
  then consider u be non-increasing FinSequence of REAL such that
A1: t,u are_fiberwise_equipotent by Th22;
  take u;
  thus t,u are_fiberwise_equipotent by A1;
  rng t = rng u by A1,CLASSES1:75;
  hence u is FinSequence of INT by FINSEQ_1:def 4;
  thus thesis;
end;















begin

reserve n,k,k1,m,m1,n1,n2,l for Element of NAT;
reserve r,r1,r2,p,p1,g,g1,g2,s,s1,s2,t for real number;
reserve seq,seq1,seq2 for Real_Sequence;
reserve Nseq for increasing sequence of NAT;
reserve x for set;
reserve X,Y for Subset of REAL;

theorem Th1:
  for X,Y st for r,p st r in X & p in Y holds r<p ex g st for r,p
  st r in X & p in Y holds r<=g & g<=p
proof
  let X,Y;
  assume for r,p st r in X & p in Y holds r<p;
  then for r,p st r in X & p in Y holds r<=p;
  then consider g such that
A1: for r,p st r in X & p in Y holds r <= g & g <= p by AXIOMS:1;
  reconsider g as Real by XREAL_0:def 1;
  take g;
  thus thesis by A1;
end;

theorem Th2:
  0<p & (ex r st r in X) & (for r st r in X holds r+p in X) implies
  for g ex r st r in X & g<r
proof
  assume that
A1: 0<p and
A2: ex r st r in X and
A3: for r st r in X holds r+p in X and
A4: ex g st for r st r in X holds not g<r;
  defpred X[Real] means for r st r in X holds not $1<r;
  consider Y such that
A5: for p1 being Real holds p1 in Y iff X[p1] from SUBSET_1:sch 3;
  now
    let r,p1 such that
A6: r in X and
A7: p1 in Y;
    r+p in X by A3,A6;
    then
A8: r+p<=p1 by A5,A7;
    r+(0 qua Nat)<r+p by A1,XREAL_1:8;
    hence r<p1 by A8,XXREAL_0:2;
  end;
  then consider g2 such that
A9: for r,p1 st r in X & p1 in Y holds r<=g2 & g2<=p1 by Th1;
  consider g1 such that
A10: for r st r in X holds not g1<r by A4;
  g1 is Real by XREAL_0:def 1;
  then
A11: g1 in Y by A10,A5;
A12: now
    assume not g2-p in Y;
    then consider r1 such that
A13: r1 in X & g2-p<r1 by A5;
    g2-p+p<r1+p & r1+p in X by A3,A13,XREAL_1:6;
    hence contradiction by A11,A9;
  end;
  -p<-(0 qua Nat) by A1,XREAL_1:24;
  then g2+-p<g2+(0 qua Nat) by XREAL_1:6;
  hence contradiction by A2,A9,A12;
end;

theorem Th3:
  for r ex n st r<n
proof
  let r;
  for r st r in NAT holds r+1 in NAT by AXIOMS:2;
  then consider p such that
A1: p in NAT and
A2: r<p by Th2;
  consider n1 such that
A3: n1=p by A1;
  take n1;
  thus thesis by A2,A3;
end;

theorem
  X is real-bounded iff ex s st 0<s & for r st r in X holds abs(r)<s
proof
  thus X is real-bounded implies ex s st 0<s & for r st r in X holds abs(r)<s
  proof
    assume
A1: X is real-bounded;
    then consider s1 such that
A2:  s1 is UpperBound of X by XXREAL_2:def 10;
A3: for r st r in X holds r<=s1 by A2,XXREAL_2:def 1;
    consider s2 such that
A4: s2 is LowerBound of X by A1,XXREAL_2:def 9;
A5: for r st r in X holds s2<=r by A4,XXREAL_2:def 2;
    take s=abs(s1)+abs(s2)+1;
A6: 0<=abs(s1) by COMPLEX1:46;
A7: 0<=abs(s2) by COMPLEX1:46;
    hence 0<s by A6;
    let r such that
A8: r in X;
    s1<=abs(s1) & r<=s1 by A3,A8,ABSVALUE:4;
    then r<=abs(s1) by XXREAL_0:2;
    then r+(0 qua Nat)<=abs(s1)+abs(s2) by A7,XREAL_1:7;
    then
A9: r<s by XREAL_1:8;
    -abs(s2)<=s2 & s2<=r by A5,A8,ABSVALUE:4;
    then -abs(s2)<=r by XXREAL_0:2;
    then -abs(s1)+-abs(s2)<=(0 qua Nat)+r by A6,XREAL_1:7;
    then
A10: -abs(s1)-abs(s2)+-1<r+(0 qua Nat) by XREAL_1:8;
    -abs(s1)-abs(s2)-1=-(abs(s1)+abs(s2)+1);
    hence thesis by A9,A10,SEQ_2:1;
  end;
  given s such that
  0<s and
A11: for r st r in X holds abs(r)<s;
  thus X is bounded_below
  proof
    take -s;
    let r be ext-real number;
    assume
A12:   r in X;
     then reconsider r as real number;
    abs(r)<s by A11,A12;
    then
A13: -s<-abs(r) by XREAL_1:24;
    -abs(r)<=r by ABSVALUE:4;
    hence thesis by A13,XXREAL_0:2;
  end;
    take s;
    let r be ext-real number;
    assume
A14:   r in X;
     then reconsider r as real number;
     r<=abs(r) by ABSVALUE:4;
    hence thesis by A11,A14,XXREAL_0:2;
end;

definition

  let r;
  redefine func {r} -> Subset of REAL;
  coherence
  proof
    {r} c= REAL
    proof
      let x be set;
      assume x in {r};
      hence thesis by XREAL_0:def 1;
    end;
    hence thesis;
  end;
end;

theorem Th5:
  for X being real-membered set holds X is non empty bounded_above
implies ex g st (for r st r in X holds r<=g) & for s st 0<s ex r st r in X & g-
  s<r
proof
  let X be real-membered set;
  assume that
A1: X is non empty and
A2: X is bounded_above;
  consider p1 such that
A3: p1 is UpperBound of X by A2,XXREAL_2:def 10;
A4: for r st r in X holds r<=p1 by A3,XXREAL_2:def 1;
  defpred X[Real] means for r st r in X holds r<=$1;
  consider Y such that
A5: for p be Real holds p in Y iff X[p] from SUBSET_1:sch 3;
  X is Subset of REAL & for r,p st r in X & p in Y holds r<=p by A5,MEMBERED:3;
  then consider g1 such that
A6: for r,p st r in X & p in Y holds r<=g1 & g1<=p by AXIOMS:1;
  reconsider g1 as Real by XREAL_0:def 1;
  take g=g1;
A7: ex r1 being real number st r1 in X by A1,MEMBERED:9;
A8: now
    given s1 such that
A9: 0<s1 and
A10: for r st r in X holds not g-s1<r;
    g-s1 in Y by A5,A10;
    then g<=g-s1 by A7,A6;
    then g-(g-s1)<=g-s1-(g-s1) by XREAL_1:9;
    hence contradiction by A9;
  end;
  p1 is Real by XREAL_0:def 1;
  then p1 in Y by A4,A5;
  hence thesis by A6,A8;
end;

theorem Th6:
  for X being real-membered set holds (for r st r in X holds r<=g1
) & (for s st 0<s ex r st (r in X & g1-s<r)) & (for r st r in X holds r<=g2) &
  (for s st 0<s ex r st (r in X & g2-s<r)) implies g1=g2
proof
  let X be real-membered set;
  assume that
A1: for r st r in X holds r<=g1 and
A2: for s st 0<s ex r st r in X & g1-s<r and
A3: for r st r in X holds r<=g2 and
A4: for s st 0<s ex r st r in X & g2-s<r;
A5: now
    assume g2<g1;
    then ex r1 st r1 in X & g1-(g1-g2)<r1 by A2,XREAL_1:50;
    hence contradiction by A3;
  end;
  now
    assume g1<g2;
    then ex r1 st r1 in X & g2-(g2-g1)<r1 by A4,XREAL_1:50;
    hence contradiction by A1;
  end;
  hence thesis by A5,XXREAL_0:1;
end;

theorem Th7:
  for X being real-membered set holds X is non empty bounded_below
implies ex g st (for r st r in X holds g<=r) & for s st 0<s ex r st r in X & r<
  g+s
proof
  let X be real-membered set;
  assume that
A1: X is non empty and
A2: X is bounded_below;
A3: ex r1 being real number st r1 in X by A1,MEMBERED:9;
  consider p1 such that
A4: p1 is LowerBound of X by A2,XXREAL_2:def 9;
A5: for r st r in X holds p1<=r by A4,XXREAL_2:def 2;
  reconsider X as Subset of REAL by MEMBERED:3;
  defpred X[Real] means for r st r in X holds $1<=r;
  consider Y such that
A6: for p be Real holds p in Y iff X[p] from SUBSET_1:sch 3;
  for p,r st p in Y & r in X holds p<=r by A6;
  then consider g1 such that
A7: for p,r st p in Y & r in X holds p<=g1 & g1<=r by AXIOMS:1;
  reconsider g1 as Real by XREAL_0:def 1;
  take g=g1;
A8: now
    given s1 such that
A9: 0<s1 and
A10: for r st r in X holds not r<g+s1;
    g+s1 in Y by A6,A10;
    then g+s1<=g by A3,A7;
    then g+s1-g<=g-g by XREAL_1:9;
    hence contradiction by A9;
  end;
  p1 is Real by XREAL_0:def 1;
  then p1 in Y by A5,A6;
  hence thesis by A7,A8;
end;

theorem Th8:
  for X being real-membered set holds (for r st r in X holds g1<=r
) & (for s st 0<s ex r st (r in X & r<g1+s)) & (for r st r in X holds g2<=r) &
  (for s st 0<s ex r st (r in X & r<g2+s)) implies g1=g2
proof
  let X be real-membered set;
  assume that
A1: for r st r in X holds g1<=r and
A2: for s st 0<s ex r st r in X & r<g1+s and
A3: for r st r in X holds g2<=r and
A4: for s st 0<s ex r st r in X & r<g2+s;
A5: now
    assume g2<g1;
    then ex r1 st r1 in X & r1<g2+(g1-g2) by A4,XREAL_1:50;
    hence contradiction by A1;
  end;
  now
    assume g1<g2;
    then ex r1 st r1 in X & r1<g1+(g2-g1) by A2,XREAL_1:50;
    hence contradiction by A3;
  end;
  hence thesis by A5,XXREAL_0:1;
end;

definition

  let X be real-membered set;
  assume
A1: X is non empty bounded_above;
  func upper_bound X -> real number means
  :Def1:
  (for r st r in X holds r<=it) & for s st 0<s ex r st r in X & it-s<r;
  existence by A1,Th5;
  uniqueness by Th6;
end;

definition
  let X be real-membered set;
  assume
A1: X is non empty bounded_below;
  func lower_bound X -> real number means
  :Def2:
  (for r st r in X holds it<=r) & for s st 0<s ex r st r in X & r<it+s;
  existence by A1,Th7;
  uniqueness by Th8;
end;

Lm1: for X being non empty real-membered set st (for s st s in X holds s >= r)
& for t st for s st s in X holds s >= t holds r >= t holds r = lower_bound X
proof
  let X be non empty real-membered set;
  assume that
A1: for s st s in X holds s >= r and
A2: for t st for s st s in X holds s >= t holds r >= t;
A3: now
    let s be real number such that
A4: 0 < s;
    assume for t be real number st t in X holds t >= r+s;
    then r >= r+s by A2;
    hence contradiction by A4,XREAL_1:29;
  end;
  X is bounded_below
   proof
    take r;
    let s be ext-real number;
    assume s in X;
    hence thesis by A1;
   end;
  hence thesis by A1,A3,Def2;
end;

Lm2: for X being non empty real-membered set, r st (for s st s in X holds s <=
r) & for t st for s st s in X holds s <= t holds r <= t holds r = upper_bound X
proof
  let X be non empty real-membered set, r;
  assume that
A1: for s st s in X holds s <= r and
A2: for t st for s st s in X holds s <= t holds r <= t;
A3: now
    let s be real number such that
A4: 0 < s;
    assume for t be real number st t in X holds r-s >= t;
    then r <= r-s by A2;
    then r+s <= r by XREAL_1:19;
    hence contradiction by A4,XREAL_1:29;
  end;
  X is bounded_above
   proof
    take r;
    let s be ext-real number;
    assume s in X;
    hence thesis by A1;
   end;
  hence thesis by A1,A3,Def1;
end;

registration
  let X be non empty bounded_below real-membered set;
  identify lower_bound X with inf X;
  compatibility
  proof
A1: now
      let t;
      assume for s st s in X holds s >= t;
      then for x being ext-real number st x in X holds t <= x;
      then t is LowerBound of X by XXREAL_2:def 2;
      hence inf X >= t by XXREAL_2:def 4;
    end;
    for s st s in X holds s >= inf X by XXREAL_2:3;
    hence thesis by A1,Lm1;
  end;
end;

registration
  let X be non empty bounded_above real-membered set;
  identify upper_bound X with sup X;
  compatibility
  proof
A1: now
      let t;
      assume for s st s in X holds t >= s;
      then for x being ext-real number st x in X holds x <= t;
      then t is UpperBound of X by XXREAL_2:def 1;
      hence t >= sup X by XXREAL_2:def 3;
    end;
    for s st s in X holds s <= sup X by XXREAL_2:4;
    hence thesis by A1,Lm2;
  end;
end;

definition
  let X;
  redefine func upper_bound X -> Real;
  coherence by XREAL_0:def 1;
  redefine func lower_bound X -> Real;
  coherence by XREAL_0:def 1;
end;

theorem Th9:
  lower_bound {r} = r & upper_bound {r} = r by XXREAL_2:11,13;

theorem Th10:
  lower_bound {r} = upper_bound {r}
proof
  lower_bound {r}=r by XXREAL_2:13;
  hence thesis by XXREAL_2:11;
end;

theorem
  X is real-bounded non empty implies lower_bound X <= upper_bound X
proof
  assume X is real-bounded non empty;
  then reconsider X as real-bounded non empty real-membered set;
  lower_bound X <= upper_bound X by XXREAL_2:40;
  hence thesis;
end;

theorem
  X is real-bounded non empty implies ((ex r,p st r in X & p in X & p<>r) iff
  lower_bound X < upper_bound X)
proof
  assume that
A1: X is real-bounded and
A2: X is non empty;
  thus (ex r,p st r in X & p in X & p<>r) implies lower_bound X<upper_bound X
  proof
    given r,p such that
A3: r in X and
A4: p in X and
A5: p<>r;
A6: now
      assume
A7:   r<p;
      lower_bound X<=r by A1,A3,Def2;
      then
A8:   lower_bound X<p by A7,XXREAL_0:2;
      p<=upper_bound X by A1,A4,Def1;
      hence thesis by A8,XXREAL_0:2;
    end;
    now
      assume
A9:   p<r;
      lower_bound X<=p by A1,A4,Def2;
      then
A10:  lower_bound X<r by A9,XXREAL_0:2;
      r<=upper_bound X by A1,A3,Def1;
      hence thesis by A10,XXREAL_0:2;
    end;
    hence thesis by A5,A6,XXREAL_0:1;
  end;
  consider r being Element of REAL such that
A11: r in X by A2,SUBSET_1:4;
  assume that
A12: lower_bound X<upper_bound X and
A13: for r,p st r in X & p in X holds p=r;
  x in X iff x=r by A13,A11;
  then X={r} by TARSKI:def 1;
  hence contradiction by A12,Th10;
end;





theorem Th13:
  seq is convergent implies abs seq is convergent
proof
  assume seq is convergent;
  then consider g1 such that
A1: for p st 0<p ex n st for m st n<=m holds abs((seq.m)-g1)<p by SEQ_2:def 6;
  reconsider g1 as Real by XREAL_0:def 1;
  take g=abs(g1);
  let p;
  assume 0<p;
  then consider n1 such that
A2: for m st n1<=m holds abs((seq.m)-g1)<p by A1;
  take n=n1;
  let m;
  abs(g1-seq.m)=abs(-((seq.m)-g1)) .=abs((seq.m)-g1) by COMPLEX1:52;
  then g-abs(seq.m)<=abs((seq.m)-g1) by COMPLEX1:59;
  then abs(seq.m)-g<=abs((seq.m)-g1) & -abs((seq.m)-g1)<= -(g-abs(seq.m)) by
COMPLEX1:59,XREAL_1:24;
  then abs(abs(seq.m)-g)<=abs((seq.m)-g1) by ABSVALUE:5;
  then
A3: abs(((abs seq).m)-g) <=abs((seq.m)-g1) by SEQ_1:12;
  assume n<=m;
  then abs((seq.m)-g1)<p by A2;
  hence thesis by A3,XXREAL_0:2;
end;

registration
  let seq be convergent Real_Sequence;
  cluster abs seq -> convergent for Real_Sequence;
  coherence by Th13;
end;

theorem
  seq is convergent implies lim abs seq = abs lim seq
proof
  assume
A1: seq is convergent;
  now
    let p;
    assume 0<p;
    then consider n1 such that
A2: for m st n1<=m holds abs((seq.m)-(lim seq))<p by A1,SEQ_2:def 7;
    take n=n1;
    let m;
    abs((lim seq)-seq.m)=abs(-((seq.m)-(lim seq)))
      .=abs((seq.m)-(lim seq)) by COMPLEX1:52;
    then abs(lim seq)-abs(seq.m)<=abs((seq.m)-(lim seq)) by COMPLEX1:59;
    then abs(seq.m)-abs(lim seq)<=abs((seq.m)-(lim seq)) & -abs((seq.m)-(lim
    seq))<=- (abs(lim seq)-abs(seq.m)) by COMPLEX1:59,XREAL_1:24;
    then abs(abs(seq.m)-abs(lim seq))<=abs((seq.m)-(lim seq)) by ABSVALUE:5;
    then
A3: abs(((abs seq).m)-abs(lim seq)) <=abs((seq.m)-(lim seq)) by SEQ_1:12;
    assume n<=m;
    then abs((seq.m)-(lim seq))<p by A2;
    hence abs(((abs seq).m)-abs(lim seq))<p by A3,XXREAL_0:2;
  end;
  hence thesis by A1,SEQ_2:def 7;
end;

theorem
  abs seq is convergent & lim abs seq=0 implies seq is convergent & lim seq=0
proof
  assume that
A1: abs seq is convergent and
A2: lim abs seq=0;
A3: now
    let n;
    -abs(seq.n)<=seq.n by ABSVALUE:4;
    then
A4: -((abs seq).n)<=seq.n by SEQ_1:12;
    seq.n<=abs(seq.n) by ABSVALUE:4;
    hence (-abs seq).n<=seq.n & seq.n<=(abs seq).n by A4,SEQ_1:10,12;
  end;
A5: -abs seq is convergent & lim(-abs seq)=-lim abs seq by A1,SEQ_2:10;
  hence seq is convergent by A1,A2,A3,SEQ_2:19;
  thus thesis by A1,A2,A5,A3,SEQ_2:20;
end;

theorem Th16:
  seq1 is subsequence of seq & seq is convergent implies seq1 is convergent
proof
  assume that
A1: seq1 is subsequence of seq and
A2: seq is convergent;
  consider g1 such that
A3: for p st 0<p ex n st for m st n<=m holds abs((seq.m)-g1)<p by A2,
SEQ_2:def 6;
  take g1;
  let p;
  assume 0<p;
  then consider n1 such that
A4: for m st n1<=m holds abs((seq.m)-g1)<p by A3;
  take n=n1;
  let m such that
A5: n<=m;
  consider Nseq such that
A6: seq1=seq*Nseq by A1,VALUED_0:def 17;
  m<=Nseq.m by SEQM_3:14;
  then
A7: n<=Nseq.m by A5,XXREAL_0:2;
  seq1.m=seq.(Nseq.m) by A6,FUNCT_2:15;
  hence thesis by A4,A7;
end;

theorem Th17:
  seq1 is subsequence of seq & seq is convergent implies lim seq1= lim seq
proof
  assume that
A1: seq1 is subsequence of seq and
A2: seq is convergent;
  consider Nseq such that
A3: seq1=seq*Nseq by A1,VALUED_0:def 17;
A4: now
    let p;
    assume 0<p;
    then consider n1 such that
A5: for m st n1<=m holds abs((seq.m)-(lim seq))<p by A2,SEQ_2:def 7;
    take n=n1;
    let m such that
A6: n<=m;
    m<=Nseq.m by SEQM_3:14;
    then
A7: n<=Nseq.m by A6,XXREAL_0:2;
    seq1.m=seq.(Nseq.m) by A3,FUNCT_2:15;
    hence abs((seq1.m)-(lim seq))<p by A5,A7;
  end;
  seq1 is convergent by A1,A2,Th16;
  hence thesis by A4,SEQ_2:def 7;
end;

theorem Th18:
  seq is convergent & (ex k st for n st k<=n holds seq1.n=seq.n)
  implies seq1 is convergent
proof
  assume that
A1: seq is convergent and
A2: ex k st for n st k<=n holds seq1.n=seq.n;
  consider g1 such that
A3: for p st 0<p ex n st for m st n<=m holds abs((seq.m)-g1)<p by A1,
SEQ_2:def 6;
  consider k such that
A4: for n st k<=n holds seq1.n=seq.n by A2;
  take g=g1;
  let p;
  assume 0<p;
  then consider n1 such that
A5: for m st n1<=m holds abs((seq.m)-g1)<p by A3;
A6: now
    assume
A7: n1<=k;
    take n=k;
    let m;
    assume
A8: n<=m;
    then n1<=m by A7,XXREAL_0:2;
    then abs((seq.m)-g1)<p by A5;
    hence abs((seq1.m)-g)<p by A4,A8;
  end;
  now
    assume
A9: k<=n1;
    take n=n1;
    let m;
    assume
A10: n<=m;
    then seq1.m=seq.m by A4,A9,XXREAL_0:2;
    hence abs((seq1.m)-g)<p by A5,A10;
  end;
  hence thesis by A6;
end;

theorem
  seq is convergent & (ex k st for n st k<=n holds seq1.n=seq.n) implies
  lim seq=lim seq1
proof
  assume that
A1: seq is convergent and
A2: ex k st for n st k<=n holds seq1.n=seq.n;
  consider k such that
A3: for n st k<=n holds seq1.n=seq.n by A2;
A4: now
    let p;
    assume 0<p;
    then consider n1 such that
A5: for m st n1<=m holds abs((seq.m)-(lim seq))<p by A1,SEQ_2:def 7;
A6: now
      assume
A7:   n1<=k;
      take n=k;
      let m;
      assume
A8:   n<=m;
      then n1<=m by A7,XXREAL_0:2;
      then abs((seq.m)-(lim seq))<p by A5;
      hence abs((seq1.m)-(lim seq))<p by A3,A8;
    end;
    now
      assume
A9:   k<=n1;
      take n=n1;
      let m;
      assume
A10:  n<=m;
      then seq1.m=seq.m by A3,A9,XXREAL_0:2;
      hence abs((seq1.m)-(lim seq))<p by A5,A10;
    end;
    hence ex n st for m st n<=m holds abs(seq1.m-(lim seq))<p by A6;
  end;
  seq1 is convergent by A1,A2,Th18;
  hence thesis by A4,SEQ_2:def 7;
end;

registration
  cluster constant -> convergent for Real_Sequence;
  coherence;
end;

registration
  cluster constant for Real_Sequence;
  existence
  proof
    take the constant Real_Sequence;
    thus thesis;
  end;
end;

registration
  let seq be convergent Real_Sequence;
  let k;
  cluster seq ^\ k -> convergent for Real_Sequence;
  coherence by Th16;
end;

theorem
  seq is convergent implies lim (seq^\k)=lim seq by Th17;

theorem Th21:
  seq^\k is convergent implies seq is convergent
proof
  assume seq^\k is convergent;
  then consider g1 such that
A1: for p st 0<p ex n st for m st n<=m holds abs((seq^\k).m-g1)<p by
SEQ_2:def 6;
  take g1;
  let p;
  assume 0<p;
  then consider n1 such that
A2: for m st n1<=m holds abs((seq^\k).m-g1)<p by A1;
  take n=n1+k;
  let m;
  assume
A3: n<=m;
  then consider l be Nat such that
A4: m=n1+k+l by NAT_1:10;
  reconsider l as Element of NAT by ORDINAL1:def 12;
  m-k=n1+l+k+-k by A4;
  then consider m1 be Element of NAT such that
A5: m1=m-k;
  now
    assume not n1<=m1;
    then m1+k<n1+k by XREAL_1:6;
    hence contradiction by A3,A5;
  end;
  then
A6: abs((seq^\k).m1-g1)<p by A2;
  m1+k=m by A5;
  hence thesis by A6,NAT_1:def 3;
end;

theorem
  seq^\k is convergent implies lim seq = lim (seq^\k)
proof
  assume
A1: seq^\k is convergent;
A2: now
    let p;
    assume 0<p;
    then consider n1 such that
A3: for m st n1<=m holds abs((seq^\k).m-(lim (seq^\k)))<p by A1,SEQ_2:def 7;
    take n=n1+k;
    let m;
    assume
A4: n<=m;
    then consider l be Nat such that
A5: m=n1+k+l by NAT_1:10;
    reconsider l as Element of NAT by ORDINAL1:def 12;
    m-k=n1+l+k+-k by A5;
    then consider m1 such that
A6: m1=m-k;
    now
      assume not n1<=m1;
      then m1+k<n1+k by XREAL_1:6;
      hence contradiction by A4,A6;
    end;
    then
A7: abs((seq^\k).m1-(lim(seq^\k)))<p by A3;
    m1+k=m by A6;
    hence abs(seq.m-(lim(seq^\k)))<p by A7,NAT_1:def 3;
  end;
  seq is convergent by A1,Th21;
  hence thesis by A2,SEQ_2:def 7;
end;

theorem Th23:
  seq is convergent & lim seq<>0 implies ex k st (seq ^\k) is non-zero
proof
  assume that
A1: seq is convergent and
A2: lim seq<>0;
  consider n1 such that
A3: for m st n1<=m holds abs(lim seq)/2<abs((seq.m)) by A1,A2,SEQ_2:16;
  take k=n1;
  now
    let n;
    (0 qua Nat)+k<=n+k by XREAL_1:7;
    then abs(lim seq)/2<abs((seq.(n+k))) by A3;
    then
A4: abs(lim seq)/2<abs(((seq ^\k).n)) by NAT_1:def 3;
    0<abs(lim seq) by A2,COMPLEX1:47;
    then 0/2<abs(lim seq)/2 by XREAL_1:74;
    hence (seq ^\k).n<>0 by A4,ABSVALUE:2;
  end;
  hence thesis by SEQ_1:5;
end;

theorem
  seq is convergent & lim seq<>0 implies ex seq1 st seq1 is subsequence
  of seq & seq1 is non-zero
proof
  assume seq is convergent & lim seq <>0;
  then consider k such that
A1: seq ^\k is non-zero by Th23;
  take seq ^\k;
  thus thesis by A1;
end;

theorem Th25:
 seq is constant & (r in rng seq or ex n st seq.n=r ) implies lim seq=r
proof
 assume
A1: seq is constant;
   then consider r1 being Real such that
A2: rng seq={r1} by FUNCT_2:111;
A3: now
    assume that
A4: r in rng seq;
    consider r2 being Real such that
A5: for n being Nat holds seq.n=r2 by A1,VALUED_0:def 18;
A6: r=r1 by A4,A2,TARSKI:def 1;
    now
      let p such that
A7:   0<p;
      take n=0;
      let m such that
      n<=m;
      m in NAT;
      then m in dom seq by FUNCT_2:def 1;
      then seq.m in rng seq by FUNCT_1:def 3;
      then r2 in rng seq by A5;
      then r2=r by A2,A6,TARSKI:def 1;
      then seq.m=r by A5;
      hence abs((seq.m)-r)<p by A7,ABSVALUE:2;
    end;
    hence thesis by A1,SEQ_2:def 7;
  end;
A8: now
    assume that
A9: ex n st seq.n=r;
    consider n such that
A10: seq.n=r by A9;
    n in NAT;
    then n in dom seq by FUNCT_2:def 1;
    hence thesis by A3,A10,FUNCT_1:def 3;
  end;
  assume r in rng seq or ex n st seq.n=r;
  hence thesis by A3,A8;
end;

theorem
  seq is constant implies for n holds lim seq=seq.n by Th25;

theorem
  seq is convergent & lim seq<>0 implies for seq1 st seq1 is subsequence
  of seq & seq1 is non-zero holds lim (seq1")=(lim seq)"
proof
  assume that
A1: seq is convergent and
A2: lim seq<>0;
  let seq1 such that
A3: seq1 is subsequence of seq and
A4: seq1 is non-zero;
  lim seq1=lim seq by A1,A3,Th17;
  hence thesis by A1,A2,A3,A4,Th16,SEQ_2:22;
end;

theorem Th28:
  0<=r & (for n holds seq.n=1/(n+r)) implies seq is convergent
proof
  assume that
A1: 0<=r and
A2: for n holds seq.n=1/(n+r);
  take 0;
  let p;
  assume
A3: 0<p;
  consider k1 such that
A4: p"<k1 by Th3;
  p"+(0 qua Nat)<k1+r by A1,A4,XREAL_1:8;
  then 1/(k1+r)<1/p" by A3,XREAL_1:76;
  then
A5: 1/(k1+r)<1*p" " by XCMPLX_0:def 9;
  take n=k1;
  let m;
  assume n<=m;
  then
A6: n+r<=m+r by XREAL_1:6;
  0<p" by A3;
  then 1/(m+r)<=1/(n+r) by A1,A4,A6,XREAL_1:118;
  then
A7: 1/(m+r)<p by A5,XXREAL_0:2;
  seq.m=1/(m+r) by A2;
  hence thesis by A1,A7,ABSVALUE:def 1;
end;

theorem Th29:
  0<=r & (for n holds seq.n=1/(n+r)) implies lim seq=0
proof
  assume that
A1: 0<=r and
A2: for n holds seq.n=1/(n+r);
A3: now
    let p;
    assume
A4: 0<p;
    consider k1 such that
A5: p"<k1 by Th3;
    p"+(0 qua Nat)<k1+r by A1,A5,XREAL_1:8;
    then 1/(k1+r)<1/p" by A4,XREAL_1:76;
    then
A6: 1/(k1+r)<1*p" " by XCMPLX_0:def 9;
    take n=k1;
    let m;
    assume n<=m;
    then
A7: n+r<=m+r by XREAL_1:6;
    0<p" by A4;
    then 1/(m+r)<=1/(n+r) by A1,A5,A7,XREAL_1:118;
    then
A8: 1/(m+r)<p by A6,XXREAL_0:2;
    seq.m=1/(m+r) & 0<=m by A2;
    hence abs(seq.m-0)<p by A1,A8,ABSVALUE:def 1;
  end;
  seq is convergent by A1,A2,Th28;
  hence thesis by A3,SEQ_2:def 7;
end;

theorem
  (for n holds seq.n=1/(n+1)) implies seq is convergent & lim seq=0 by Th28
,Th29;

theorem
  0<=r & (for n holds seq.n=g/(n+r)) implies seq is convergent & lim seq =0
proof
  assume that
A1: 0<=r and
A2: for n holds seq.n=g/(n+r);
  reconsider r1 = r as Real by XREAL_0:def 1;
  deffunc U(Element of NAT) = 1/($1+r1);
  consider seq1 such that
A3: for n holds seq1.n=U(n) from SEQ_1:sch 1;
A4: now
    let n;
    thus (g(#)seq1).n=g*(seq1.n) by SEQ_1:9
      .=g*(1/(n+r)) by A3
      .=g*(1*(n+r)") by XCMPLX_0:def 9
      .=g/(n+r) by XCMPLX_0:def 9
      .=seq.n by A2;
  end;
A5: g(#)seq1 is convergent by A1,A3,Th28,SEQ_2:7;
  lim (g(#)seq1)=g*(lim seq1) by A1,A3,Th28,SEQ_2:8
    .=g*(0 qua Nat) by A1,A3,Th29
    .=0;
  hence thesis by A5,A4,FUNCT_2:63;
end;

theorem Th32:
  0<=r & (for n holds seq.n=1/(n*n+r)) implies seq is convergent
proof
  assume that
A1: 0<=r and
A2: for n holds seq.n=1/(n*n+r);
  take 0;
  let p;
  consider k1 such that
A3: p"<k1 by Th3;
  assume
A4: 0<p;
  then
A5: k1>0 by A3;
  then k1>=1+(0 qua Nat) by NAT_1:13;
  then k1<=k1*k1 by XREAL_1:151;
  then
A6: k1+r<=k1*k1+r by XREAL_1:6;
  take n=k1;
  let m such that
A7: n<=m;
  n*n<=m*m by A7,XREAL_1:66;
  then
A8: n*n+r<=m*m+r by XREAL_1:6;
  p"+(0 qua Nat)<k1+r by A1,A3,XREAL_1:8;
  then p"+(0 qua Nat)<k1*k1+r by A6,XXREAL_0:2;
  then 1/(k1*k1+r)<1/p" by A4,XREAL_1:76;
  then
A9: 1/(k1*k1+r)<1*p" " by XCMPLX_0:def 9;
  0<n^2 by A5,SQUARE_1:12;
  then 1/(m*m+r)<=1/(n*n+r) by A1,A8,XREAL_1:118;
  then
A10: 1/(m*m+r)<p by A9,XXREAL_0:2;
  seq.m=1/(m*m+r) by A2;
  hence thesis by A1,A10,ABSVALUE:def 1;
end;

theorem Th33:
  0<=r & (for n holds seq.n=1/(n*n+r)) implies lim seq=0
proof
  assume that
A1: 0<=r and
A2: for n holds seq.n=1/(n*n+r);
A3: now
    let p;
    consider k1 such that
A4: p"<k1 by Th3;
    assume
A5: 0<p;
    then
A6: k1>0 by A4;
    then k1>=1+(0 qua Nat) by NAT_1:13;
    then k1<=k1*k1 by XREAL_1:151;
    then
A7: k1+r<=k1*k1+r by XREAL_1:6;
    take n=k1;
    let m;
    assume n<=m;
    then n*n<=m*m by XREAL_1:66;
    then
A8: n*n+r<=m*m+r by XREAL_1:6;
    p"+(0 qua Nat)<k1+r by A1,A4,XREAL_1:8;
    then p"+(0 qua Nat)<k1*k1+r by A7,XXREAL_0:2;
    then 1/(k1*k1+r)<1/p" by A5,XREAL_1:76;
    then
A9: 1/(k1*k1+r)<1*p" " by XCMPLX_0:def 9;
    0<n^2 by A6,SQUARE_1:12;
    then 1/(m*m+r)<=1/(n*n+r) by A1,A8,XREAL_1:118;
    then
A10: 1/(m*m+r)<p by A9,XXREAL_0:2;
    seq.m=1/(m*m+r) & 0<=m*m by A2;
    hence abs(seq.m-0)<p by A1,A10,ABSVALUE:def 1;
  end;
  seq is convergent by A1,A2,Th32;
  hence thesis by A3,SEQ_2:def 7;
end;

theorem
  (for n holds seq.n=1/(n*n+1)) implies seq is convergent & lim seq=0 by Th32
,Th33;

theorem
  0<=r & (for n holds seq.n=g/(n*n+r)) implies seq is convergent & lim seq=0
proof
  assume that
A1: 0<=r and
A2: for n holds seq.n=g/(n*n+r);
  reconsider r1 = r as Real by XREAL_0:def 1;
  deffunc U(Element of NAT) = 1/($1*$1+r1);
  consider seq1 such that
A3: for n holds seq1.n=U(n) from SEQ_1:sch 1;
A4: now
    let n;
    thus (g(#)seq1).n=g*(seq1.n) by SEQ_1:9
      .=g*(1/(n*n+r)) by A3
      .=g*(1*(n*n+r)") by XCMPLX_0:def 9
      .=g/(n*n+r) by XCMPLX_0:def 9
      .=seq.n by A2;
  end;
A5: g(#)seq1 is convergent by A1,A3,Th32,SEQ_2:7;
  lim (g(#)seq1)=g*(lim seq1) by A1,A3,Th32,SEQ_2:8
    .=g*(0 qua Nat) by A1,A3,Th33
    .=0;
  hence thesis by A5,A4,FUNCT_2:63;
end;

registration
  cluster non-decreasing bounded_above -> convergent for Real_Sequence;
  coherence
  proof
    let seq be Real_Sequence;
    assume that
A1: seq is non-decreasing and
A2: seq is bounded_above;
    consider r2 such that
A3: for n holds seq.n<r2 by A2,SEQ_2:def 3;
    defpred X[Real] means ex n st $1=seq.n;
    consider X such that
A4: for p be Real holds p in X iff X[p] from SUBSET_1:sch 3;
A5: now
      take r=r2;
      let p;
      assume p in X;
      then ex n1 st p=seq.n1 by A4;
      hence p<=r by A3;
    end;
A6: (ex r st for p being real number st p in X holds p<=r)
     implies X is bounded_above
     proof
      given r such that
A7:    for p being real number st p in X holds p<=r;
      take r; let p be ext-real number;
      thus thesis by A7;
     end;
    take g=upper_bound X;
    let s;
    assume
A8: 0<s;
    seq.0 in X by A4;
    then consider p1 such that
A9: p1 in X and
A10: upper_bound X-s<p1 by A6,A8,Def1;
    consider n1 such that
A11: p1=seq.n1 by A4,A9;
    take n=n1;
    let m;
    assume n<=m;
    then seq.n<=seq.m by A1,SEQM_3:6;
    then g+ -s<seq.m by A10,A11,XXREAL_0:2;
    then
A12: -s<seq.m -g by XREAL_1:20;
    seq.m in X by A4;
    then seq.m<=g by A6,A5,Def1;
    then seq.m +(0 qua Nat)<g+s by A8,XREAL_1:8;
    then seq.m -g<s by XREAL_1:19;
    hence thesis by A12,SEQ_2:1;
  end;
end;

registration
  cluster non-increasing bounded_below -> convergent for Real_Sequence;
  coherence
  proof
    let seq be Real_Sequence;
    assume that
A1: seq is non-increasing and
A2: seq is bounded_below;
    defpred X[Real] means ex n st $1=seq.n;
    consider X such that
A3: for p be Real holds p in X iff X[p] from SUBSET_1:sch 3;
    take g=lower_bound X;
    let s;
    assume
A4: 0<s;
A5: (ex r st for p being real number st p in X holds r<=p)
     implies X is bounded_below
     proof
      given r such that
A6:    for p being real number st p in X holds r<=p;
      take r; let p be ext-real number;
      thus thesis by A6;
     end;
    seq.0 in X by A3;
    then consider p1 such that
A7: p1 in X and
A8: p1<lower_bound X+s by A5,A4,Def2;
    consider n1 such that
A9: p1=seq.n1 by A3,A7;
    take n=n1;
    let m;
    consider r1 such that
A10: for n holds r1< seq.n by A2,SEQ_2:def 4;
A11: now
      take r=r1;
      let p;
      assume p in X;
      then ex n1 st p=seq.n1 by A3;
      hence r<=p by A10;
    end;
    seq.m in X by A3;
    then (0 qua Nat)+g<=seq.m by A5,A11,Def2;
    then
A12: 0<=seq.m-g by XREAL_1:19;
    assume n<=m;
    then seq.m<=seq.n by A1,SEQM_3:8;
    then seq.m <g+s by A8,A9,XXREAL_0:2;
    then
A13: seq.m -g<s by XREAL_1:19;
    -s<-(0 qua Nat) by A4,XREAL_1:24;
    hence thesis by A13,A12,SEQ_2:1;
  end;
end;

registration
  cluster monotone bounded -> convergent for Real_Sequence;
  coherence
  proof
    let seq be Real_Sequence;
    assume that
A1: seq is monotone and
A2: seq is bounded;
A3: seq is non-increasing implies thesis by A2;
    seq is non-decreasing implies thesis by A2;
    hence thesis by A1,A3,SEQM_3:def 5;
  end;
end;

theorem Th36:
  seq is monotone & seq is bounded implies seq is convergent;

theorem
  seq is bounded_above & seq is non-decreasing implies for n holds seq.n
  <=lim seq
proof
  assume that
A1: seq is bounded_above and
A2: seq is non-decreasing;
  let m;
  set seq1 = NAT --> seq.m;
  deffunc U(Nat) = seq.($1+m);
  consider seq2 such that
A3: for n holds seq2.n=U(n) from SEQ_1:sch 1;
A4: now
    let n;
    seq1.n=seq.m & seq2.n=seq.(m+n) by A3,FUNCOP_1:7;
    hence seq1.n<=seq2.n by A2,SEQM_3:5;
  end;
  seq1.0=seq.m by FUNCOP_1:7;
  then
A5: lim seq1=seq.m by Th25;
  now
    let n be Nat;
    n in NAT by ORDINAL1:def 12;
    hence seq2.n=U(n) by A3;
  end;
  then
A6: seq2=seq^\m by NAT_1:def 3;
  then lim seq2=lim seq by A1,A2,Th17;
  hence thesis by A1,A2,A5,A6,A4,SEQ_2:18;
end;

theorem
  seq is bounded_below & seq is non-increasing implies for n holds lim
  seq <= seq.n
proof
  assume that
A1: seq is bounded_below and
A2: seq is non-increasing;
  let m;
  set seq1 = NAT --> seq.m;
  deffunc U(Nat) = seq.($1+m);
  consider seq2 such that
A3: for n holds seq2.n=U(n) from SEQ_1:sch 1;
A4: now
    let n;
    seq1.n=seq.m & seq2.n=seq.(m+n) by A3,FUNCOP_1:7;
    hence seq2.n<=seq1.n by A2,SEQM_3:7;
  end;
  seq1.0=seq.m by FUNCOP_1:7;
  then
A5: lim seq1=seq.m by Th25;
  now
    let n be Nat;
    n in NAT by ORDINAL1:def 12;
    hence seq2.n=U(n) by A3;
  end;
  then
A6: seq2=seq^\m by NAT_1:def 3;
  then lim seq2=lim seq by A1,A2,Th17;
  hence thesis by A1,A2,A5,A6,A4,SEQ_2:18;
end;

theorem Th39:
  for seq ex Nseq st seq*Nseq is monotone
proof
  let seq;
  defpred X[Element of NAT] means for m st $1<m holds seq.$1<seq.m;
  consider XN being Subset of NAT such that
A1: for n holds n in XN iff X[n] from SUBSET_1:sch 3;
A2: now
    given k1 such that
A3: for n st n in XN holds n<=k1;
    set seq1=seq ^\(1+k1);
    defpred P[set,set,set] means for k,l st k=$2 & l=$3 holds k<l & seq1.l<=
    seq1.k & for m st k<m & seq1.m<=seq1.k holds l<=m;
A4: now
      let k;
      assume k1<k;
      then not k in XN by A3;
      then consider m such that
A5:   k<m and
A6:   not seq.k<seq.m by A1;
      take n=m;
      thus k<n by A5;
      thus seq.n<=seq.k by A6;
    end;
A7: now
      let k;
      (0 qua Nat)+k1<k+1+k1 by XREAL_1:8;
      then consider n such that
A8:   k+1+k1<n and
A9:   seq.n<=seq.(k+1+k1) by A4;
      consider m be Nat such that
A10:  n=k+1+k1+m by A8,NAT_1:10;
      reconsider m as Element of NAT by ORDINAL1:def 12;
      n-(1+k1)=k+(0 qua Nat)+m by A10;
      then consider n1 be Element of NAT such that
A11:  n1=n-(1+k1);
      seq.n<=seq.(k+(1+k1)) by A9;
      then
A12:  seq.n<=seq1.k by NAT_1:def 3;
      take l=n1;
      now
        assume not k<l;
        then n-(1+k1)+(1+k1)<=k+(1+k1) by A11,XREAL_1:6;
        hence contradiction by A8;
      end;
      hence k<l;
      n=l+(1+k1) by A11;
      hence seq1.l<=seq1.k by A12,NAT_1:def 3;
    end;
A13: for n being Element of NAT for x being Element of NAT ex y being
    Element of NAT st P[n,x,y]
    proof
      let n be Element of NAT,x be Element of NAT;
      defpred X[Nat] means x<$1 & seq1.$1<=seq1.x;
      ex n be Element of NAT st X[n] by A7;
      then
A14:  ex n be Nat st X[n];
      ex l be Nat st X[l] & for m be Nat st X[m] holds l <= m from NAT_1:
      sch 5(A14);
      then consider l be Nat such that
A15:  x<l & seq1.l<=seq1.x & for m be Nat st x<m & seq1.m<=seq1.x holds l <= m;
      take l;
      l in NAT by ORDINAL1:def 12;
      hence thesis by A15;
    end;
    consider f being Function of NAT,NAT such that
    f.0 = 0 and
A16: for n being Element of NAT holds P[n,f.n,f.(n+1)] from RECDEF_1:
    sch 2 (A13);
A17: dom f =NAT by FUNCT_2:def 1;
A18: rng f c= NAT by RELAT_1:def 19;
    rng f c= REAL;
    then reconsider f as Real_Sequence by A17,RELSET_1:4;
A19: now
      let n;
      f.n in rng f by A17,FUNCT_1:def 3;
      hence f.n is Element of NAT by A18;
    end;
    now
      let n;
      f.n is Element of NAT & f.(n+1) is Element of NAT by A19;
      hence f.n<f.(n+1) by A16;
    end;
    then reconsider f as increasing sequence of NAT by SEQM_3:def 6;
    consider Nseq such that
A20: seq1=seq*Nseq by VALUED_0:def 17;
    reconsider Nseq1=Nseq*f as increasing sequence of NAT;
    take Nseq1;
    now
      let n;
      seq1.(f.(n+1))<=seq1.(f.n) by A16;
      then (seq1*f).(n+1)<=seq1.(f.n) by FUNCT_2:15;
      then (seq*Nseq*f).(n+1)<=(seq1*f).n by A20,FUNCT_2:15;
      then (seq*Nseq1).(n+1)<=(seq*Nseq*f).n by A20,RELAT_1:36;
      hence (seq*Nseq1).(n+1)<=(seq*Nseq1).n by RELAT_1:36;
    end;
    then seq*Nseq1 is non-increasing by SEQM_3:def 9;
    hence seq*Nseq1 is monotone by SEQM_3:def 5;
  end;
  now
    defpred P[set,set,set] means for k,l being Element of NAT st k=$2 & l=$3
    holds l in XN & k<l & for m st m in XN & k<m holds l<=m;
    assume
A21: for l ex n st n in XN & not n<=l;
    then consider n1 such that
A22: n1 in XN and
    not n1<=0;
    reconsider 09=n1 as Element of NAT qua non empty set;
A23: for n being Element of NAT for x being Element of NAT ex y being
    Element of NAT st P[n,x,y]
    proof
      let n be Element of NAT,x be Element of NAT;
      defpred X[Nat] means $1 in XN & x<$1;
      ex n be Element of NAT st X[n] by A21;
      then
A24:  ex n be Nat st X[n];
      ex l be Nat st X[l] & for m be Nat st X[m] holds l <= m from NAT_1:
      sch 5(A24);
      then consider l be Nat such that
A25:  l in XN & x<l & for m be Nat st m in XN & x<m holds l <= m;
      reconsider l as Element of NAT by ORDINAL1:def 12;
      take l;
      thus thesis by A25;
    end;
    consider f being Function of NAT,NAT such that
A26: f.0 = 09 and
A27: for n being Element of NAT holds P[n,f.n,f.(n+1)] from RECDEF_1:
    sch 2 (A23);
A28: dom f =NAT by FUNCT_2:def 1;
A29: rng f c= NAT by RELAT_1:def 19;
    rng f c= REAL;
    then reconsider f as Real_Sequence by A28,RELSET_1:4;
A30: now
      let n;
      thus n is Element of NAT;
A31:  f.n in rng f by A28,FUNCT_1:def 3;
      hence f.n is Element of NAT by A29;
      thus f.n is Element of NAT by A29,A31;
    end;
A32: now
      let n;
      f.n is Element of NAT & f.(n+1) is Element of NAT by A30;
      hence f.n<f.(n+1) by A27;
    end;
A33: now
      defpred X[Element of NAT] means f.$1 in XN;
      let n;
A34:  now
        let k such that
        X[k];
        f.k is Element of NAT & f.(k+1) is Element of NAT by A30;
        hence X[k+1] by A27;
      end;
A35:  X[0] by A22,A26;
      thus for n holds X[n] from NAT_1:sch 1(A35,A34);
    end;
    reconsider f as increasing sequence of NAT by A32,SEQM_3:def 6;
    take Nseq=f;
    now
      let n;
      Nseq.n in XN & Nseq.n<Nseq.(n+1) by A33,A32;
      then seq.(Nseq.n)<seq.(Nseq.(n+1)) by A1;
      then (seq*Nseq).n<seq.(Nseq.(n+1)) by FUNCT_2:15;
      hence (seq*Nseq).n<(seq*Nseq).(n+1) by FUNCT_2:15;
    end;
    then seq*Nseq is increasing by SEQM_3:def 6;
    hence seq*Nseq is monotone by SEQM_3:def 5;
  end;
  hence thesis by A2;
end;


theorem Th40:
  seq is bounded implies ex seq1 st seq1 is subsequence of seq &
  seq1 is convergent
proof
  assume
A1: seq is bounded;
  consider Nseq such that
A2: seq*Nseq is monotone by Th39;
  take seq1=seq*Nseq;
  thus seq1 is subsequence of seq;
  thus thesis by A1,A2,Th36,SEQM_3:29;
end;


theorem 
  seq is convergent iff for s st 0<s ex n st for m st n<=m holds abs(seq
  .m -seq.n)<s
proof
  thus seq is convergent implies for s st 0<s ex n st for m st n<=m holds abs(
  seq.m -seq.n)<s
  proof
    assume seq is convergent;
    then consider g1 such that
A1: for s st 0<s ex n st for m st n<=m holds abs(seq.m -g1)<s by SEQ_2:def 6;
    let s;
    assume 0<s;
    then consider n1 such that
A2: for m st n1<=m holds abs(seq.m -g1)<s/2 by A1,XREAL_1:215;
    take n=n1;
    let m;
    assume n<=m;
    then
A3: abs(seq.m -g1)<s/2 by A2;
A4: abs(seq.m -g1+(g1-seq.n))<=abs(seq.m -g1)+abs(g1-seq.n) by COMPLEX1:56;
    abs(seq.n -g1)<s/2 by A2;
    then abs(-(g1-seq.n))<s/2;
    then abs(g1-seq.n)<s/2 by COMPLEX1:52;
    then abs(seq.m -g1)+abs(g1-seq.n)<s/2+s/2 by A3,XREAL_1:8;
    hence thesis by A4,XXREAL_0:2;
  end;
  assume
A5: for s st 0<s ex n st for m st n<=m holds abs(seq.m -seq.n)<s;
  then consider n1 such that
A6: for m st n1<=m holds abs(seq.m -seq.n1)<1;
  consider r1 such that
A7: 0<r1 and
A8: for m st m<=n1 holds abs(seq.m)<r1 by SEQ_2:4;
  now
    take r=r1+abs(seq.n1)+1;
    (0 qua Nat)+(0 qua Nat)<r1+abs(seq.n1) by A7,COMPLEX1:46,XREAL_1:8;
    hence 0<r;
    let m;
A9: now
      assume n1<=m;
      then
A10:  abs(seq.m -seq.n1)<1 by A6;
      abs(seq.m)-abs(seq.n1)<=abs(seq.m -seq.n1) by COMPLEX1:59;
      then abs(seq.m)-abs(seq.n1)<1 by A10,XXREAL_0:2;
      then abs(seq.m)-abs(seq.n1)+abs(seq.n1)<1+abs(seq.n1) by XREAL_1:6;
      then (0 qua Nat)+abs(seq.m)<r1+(abs(seq.n1)+1) by A7,XREAL_1:8;
      hence abs(seq.m)<r;
    end;
    now
      assume m<=n1;
      then abs(seq.m)<r1 by A8;
      then abs(seq.m)+(0 qua Nat)<r1+abs(seq.n1) by COMPLEX1:46,XREAL_1:8;
      hence abs(seq.m)<r by XREAL_1:8;
    end;
    hence abs(seq.m)<r by A9;
  end;
  then seq is bounded by SEQ_2:3;
  then consider seq1 such that
A11: seq1 is subsequence of seq and
A12: seq1 is convergent by Th40;
  consider g1 such that
A13: for s st 0<s ex n st for m st n<=m holds abs(seq1.m -g1)<s by A12,
SEQ_2:def 6;
  take g1;
  let s;
  assume
A14: 0<s;
  then consider n1 such that
A15: for m st n1<=m holds abs(seq1.m -g1)<s/3 by A13,XREAL_1:222;
  consider n2 such that
A16: for m st n2<=m holds abs(seq.m -seq.n2)<s/3 by A5,A14,XREAL_1:222;
  take n=n1+n2;
  let m such that
A17: n<=m;
  consider Nseq such that
A18: seq1=seq*Nseq by A11,VALUED_0:def 17;
  n1<=n by NAT_1:12;
  then n1<=m by A17,XXREAL_0:2;
  then abs((seq*Nseq).m -g1)<s/3 by A18,A15;
  then
A19: abs(seq.(Nseq.m) -g1)<s/3 by FUNCT_2:15;
  n2<=n by NAT_1:12; then
A20: n2<=m by A17,XXREAL_0:2;
  m<=Nseq.m by SEQM_3:14;
  then n2<=Nseq.m by A20,XXREAL_0:2;
  then abs(seq.(Nseq.m) -seq.n2)<s/3 by A16;
  then abs(-(seq.n2-seq.(Nseq.m)))<s/3;
  then
A21: abs(seq.n2-seq.(Nseq.m))<s/3 by COMPLEX1:52;
  abs(seq.m -seq.n2)<s/3 by A16,A20;
  then
A22: abs(seq.m -seq.n2)+abs(seq.n2-seq.(Nseq.m))<s/3+s/3 by A21,XREAL_1:8;
  abs(seq.m -seq.n2+(seq.n2-seq.(Nseq.m)))<= abs(seq.m -seq.n2)+abs(seq.
  n2-seq.(Nseq.m)) by COMPLEX1:56;
  then abs(seq.m -seq.(Nseq.m))<s/3+s/3 by A22,XXREAL_0:2;
  then
A23: abs(seq.m -seq.(Nseq.m))+abs(seq.(Nseq.m) -g1)<s/3+s/3+s/3 by A19,
XREAL_1:8;
  abs(seq.m -seq.(Nseq.m)+(seq.(Nseq.m) -g1))<= abs(seq.m -seq.(Nseq.m))+
  abs(seq.(Nseq.m) -g1) by COMPLEX1:56;
  hence thesis by A23,XXREAL_0:2;
end;

theorem
  seq is constant & seq1 is convergent implies lim (seq+seq1) =(seq.0) +
lim seq1 & lim (seq-seq1) =(seq.0) - lim seq1 & lim (seq1-seq) =(lim seq1) -seq
  .0 & lim (seq(#)seq1) =(seq.0) * (lim seq1)
proof
  assume that
A1: seq is constant and
A2: seq1 is convergent;
  thus lim (seq+seq1)=(lim seq)+(lim seq1) by A1,A2,SEQ_2:6
    .=(seq.0)+(lim seq1) by A1,Th25;
  thus lim (seq-seq1)=(lim seq)-(lim seq1) by A1,A2,SEQ_2:12
    .=(seq.0)-(lim seq1) by A1,Th25;
  thus lim (seq1-seq)=(lim seq1)-(lim seq) by A1,A2,SEQ_2:12
    .=(lim seq1)-(seq.0) by A1,Th25;
  thus lim (seq(#)seq1)=(lim seq)*(lim seq1) by A1,A2,SEQ_2:15
    .=(seq.0)*(lim seq1) by A1,Th25;
end;

begin 


theorem Th43:
  for X being non empty real-membered set for t st for s st s in X
  holds s >= t holds lower_bound X >= t
proof
  let X be non empty real-membered set;
  set r = lower_bound X;
  let t;
  assume
A1: for s st s in X holds s >= t;
  set s = t-r;
  assume r < t; then
A2: s > 0 by XREAL_1:50;
  X is bounded_below proof take t; let s be ext-real number;
    thus thesis by A1;
   end;
  then ex t9 be real number st t9 in X & t9 < r+s by A2,Def2;
  hence contradiction by A1;
end;

theorem Th44:
  for X being non empty real-membered set st (for s st s in X holds s >=
r) & for t st for s st s in X holds s >= t holds r >= t holds r = lower_bound X
  by Lm1;

theorem Th45:
  for X being non empty real-membered set, r for t st for s st s
  in X holds s <= t holds upper_bound X <= t
proof
  let X be non empty real-membered set, r;
  set r = upper_bound X;
  let t;
  assume
A1: for s st s in X holds s <= t;
  set s = r-t;
  assume r > t; then
A2: s > 0 by XREAL_1:50;
  X is bounded_above proof take t;
    let s be ext-real number;
    thus thesis by A1;
   end;
  then ex t9 be real number st t9 in X & r-s < t9 by A2,Def1;
  hence contradiction by A1;
end;

theorem Th46:
  for X being non empty real-membered set, r st (for s st s in X holds s
  <= r) & for t st for s st s in X holds s <= t holds r <= t holds r =
  upper_bound X by Lm2;

theorem
  for X being non empty real-membered set, Y being real-membered set st
  X c= Y & Y is bounded_below holds lower_bound Y <= lower_bound X
proof
  let X be non empty real-membered set, Y be real-membered set;
  assume X c= Y & Y is bounded_below;
  then t in X implies t >= lower_bound Y by Def2;
  hence thesis by Th43;
end;

theorem
  for X being non empty real-membered set, Y being real-membered set st
  X c= Y & Y is bounded_above holds upper_bound X <= upper_bound Y
proof
  let X be non empty real-membered set, Y be real-membered set;
  assume X c= Y & Y is bounded_above;
  then t in X implies t <= upper_bound Y by Def1;
  hence thesis by Th45;
end;



definition
  let A be non empty natural-membered set;
  redefine func min A -> Element of NAT;
  coherence by ORDINAL1:def 12;
end;

begin 

reserve k,n for Element of NAT,
  r,r9,r1,r2 for Real,
  c,c9,c1,c2,c3 for Element of COMPLEX;

theorem
  0c is_a_unity_wrt addcomplex by BINOP_2:1,SETWISEO:14;

theorem Th50:
  compcomplex is_an_inverseOp_wrt addcomplex
proof
  let c;
  thus addcomplex.(c,compcomplex.c) = c+compcomplex.c by BINOP_2:def 3
    .= c+-c by BINOP_2:def 1
    .= the_unity_wrt addcomplex by BINOP_2:1;
  thus addcomplex.(compcomplex.c,c) = compcomplex.c+c by BINOP_2:def 3
    .= -c+c by BINOP_2:def 1
    .= the_unity_wrt addcomplex by BINOP_2:1;
end;

theorem Th51:
  addcomplex is having_an_inverseOp by Th50,FINSEQOP:def 2;

theorem Th52:
  the_inverseOp_wrt addcomplex = compcomplex by Th50,Th51,FINSEQOP:def 3;

definition
  redefine func diffcomplex equals
  addcomplex*(id COMPLEX,compcomplex);
  compatibility
  proof
    let b be BinOp of COMPLEX;
    now
      let c1,c2;
      thus diffcomplex.(c1,c2) = c1 - c2 by BINOP_2:def 4
        .= c1 + - c2
        .= addcomplex.(c1,- c2) by BINOP_2:def 3
        .= addcomplex.(c1,compcomplex.c2) by BINOP_2:def 1
        .= (addcomplex*(id COMPLEX,compcomplex)).(c1,c2) by FINSEQOP:82;
    end;
    hence thesis by BINOP_1:2;
  end;
end;

theorem
  1r is_a_unity_wrt multcomplex by BINOP_2:6,SETWISEO:14;

theorem Th54:
  multcomplex is_distributive_wrt addcomplex
proof
  now
    let c1,c2,c3;
    thus multcomplex.(c1,addcomplex.(c2,c3)) = multcomplex.(c1,c2+c3) by
BINOP_2:def 3
      .= c1*(c2+c3) by BINOP_2:def 5
      .= c1*c2+c1*c3
      .= addcomplex.(c1*c2,c1*c3) by BINOP_2:def 3
      .= addcomplex.(multcomplex.(c1,c2),c1*c3) by BINOP_2:def 5
      .= addcomplex.(multcomplex.(c1,c2),multcomplex.(c1,c3)) by BINOP_2:def 5;
    thus multcomplex.(addcomplex.(c1,c2),c3) = multcomplex.(c1+c2,c3) by
BINOP_2:def 3
      .= (c1+c2)*c3 by BINOP_2:def 5
      .= c1*c3+c2*c3
      .= addcomplex.(c1*c3,c2*c3) by BINOP_2:def 3
      .= addcomplex.(multcomplex.(c1,c3),c2*c3) by BINOP_2:def 5
      .= addcomplex.(multcomplex.(c1,c3),multcomplex.(c2,c3)) by BINOP_2:def 5;
  end;
  hence thesis by BINOP_1:11;
end;

definition let c be complex number;
  func c multcomplex -> UnOp of COMPLEX equals
  multcomplex[;](c,id COMPLEX);
  coherence
  proof
    reconsider c9 = c as Element of COMPLEX by XCMPLX_0:def 2;
    multcomplex[;](c9,id COMPLEX) is UnOp of COMPLEX;
    hence thesis;
  end;
end;

Lm3:(multcomplex[;](c,id COMPLEX)).c9 = c*c9
proof
  thus (multcomplex[;](c,id COMPLEX)).c9 = multcomplex.(c,(id COMPLEX).c9) by
FUNCOP_1:53
    .= multcomplex.(c,c9) by FUNCT_1:18
    .= c*c9 by BINOP_2:def 5;
end;

theorem
  (c multcomplex).c9 = c*c9 by Lm3;

theorem
  c multcomplex is_distributive_wrt addcomplex by Th54,FINSEQOP:54;

definition
  func abscomplex -> Function of COMPLEX,REAL means :Def5:
  for c holds it.c = |.c.|;
  existence from FUNCT_2:sch 4;
  uniqueness from BINOP_2:sch 1;
end;

reserve z,z1,z2 for FinSequence of COMPLEX;

definition let z1,z2;
  redefine func z1 + z2 -> FinSequence of COMPLEX equals
  addcomplex.:(z1,z2);
  coherence
  proof
    thus rng (z1+z2) c= COMPLEX;
  end;
  compatibility
  proof
    set g = addcomplex.:(z1,z2);
    dom addcomplex = [:COMPLEX,COMPLEX:] by FUNCT_2:def 1;
    then [:rng z1, rng z2:] c= dom addcomplex by ZFMISC_1:96;
    then
A1: dom(z1+z2) = dom z1 /\ dom z2 & dom g = dom z1 /\ dom z2 by FUNCOP_1:69
,VALUED_1:def 1;
    now
      let x be set;
      assume
A2:   x in dom (z1+z2);
      hence (z1+z2).x = z1.x + z2.x by VALUED_1:def 1
        .= addcomplex.(z1.x,z2.x) by BINOP_2:def 3
        .= g.x by A1,A2,FUNCOP_1:22;
    end;
    hence thesis by A1,FUNCT_1:2;
  end;
  redefine func z1 - z2 -> FinSequence of COMPLEX equals
  diffcomplex.:(z1,z2);
  coherence
  proof
    thus rng (z1-z2) c= COMPLEX;
  end;
  compatibility
  proof
    set g = diffcomplex.:(z1,z2);
    dom diffcomplex = [:COMPLEX,COMPLEX:] by FUNCT_2:def 1;
    then [:rng z1, rng z2:] c= dom diffcomplex by ZFMISC_1:96;
    then
A3: dom g = dom z1 /\ dom z2 by FUNCOP_1:69;
A4: dom(z1-z2) = dom z1 /\ dom z2 by VALUED_1:12;
    now
      let x be set;
      assume
A5:   x in dom (z1-z2);
      hence (z1-z2).x = z1.x - z2.x by VALUED_1:13
        .= diffcomplex.(z1.x,z2.x) by BINOP_2:def 4
        .= g.x by A4,A3,A5,FUNCOP_1:22;
    end;
    hence thesis by A3,FUNCT_1:2,VALUED_1:12;
  end;
end;

definition let z;
  redefine func -z -> FinSequence of COMPLEX equals
  compcomplex*z;
  coherence
  proof
    thus rng -z c= COMPLEX;
  end;
  compatibility
  proof
    set g = compcomplex*z;
    dom compcomplex = COMPLEX by FUNCT_2:def 1;
    then rng z c= dom compcomplex;
    then
A1: dom g = dom z by RELAT_1:27;
A2: dom -z = dom z by VALUED_1:8;
    now
      let x be set;
      assume
A3:   x in dom -z;
      thus (-z).x = -z.x by VALUED_1:8
        .= compcomplex.(z.x) by BINOP_2:def 1
        .= g.x by A2,A1,A3,FUNCT_1:12;
    end;
    hence thesis by A1,FUNCT_1:2,VALUED_1:8;
  end;
end;

notation let z;
  let c be complex number;
  synonym c*z for c(#)z;
end;

definition let z;
  let c be complex number;
  redefine func c*z -> FinSequence of COMPLEX equals
  (c multcomplex)*z;
  coherence
  proof
    thus rng (c(#)z) c= COMPLEX;
  end;
  compatibility
  proof
    set g = (c multcomplex)*z;
    dom (c multcomplex) = COMPLEX by FUNCT_2:def 1;
    then rng z c= dom (c multcomplex);
    then
A1: dom (c(#)z) = dom z & dom g = dom z by RELAT_1:27,VALUED_1:def 5;
    now
      let x be set;
      assume
A2:   x in dom (c(#)z);
A3:   c is Element of COMPLEX & z.x is Element of COMPLEX by XCMPLX_0:def 2;
      thus (c(#)z).x = c*z.x by VALUED_1:6
        .= (c multcomplex).(z.x) by A3,Lm3
        .= g.x by A1,A2,FUNCT_1:12;
    end;
    hence thesis by A1,FUNCT_1:2;
  end;
end;

definition let z;
  redefine func abs z -> FinSequence of REAL equals
  abscomplex*z;
  coherence
  proof
    thus rng abs z c= REAL;
  end;
  compatibility
  proof
    set g = abscomplex*z;
    dom abscomplex = COMPLEX by FUNCT_2:def 1;
    then rng z c= dom abscomplex;
    then
A1: dom abs z = dom z & dom g = dom z by RELAT_1:27,VALUED_1:def 11;
    now
      let x be set;
      assume
A2:   x in dom abs z;
A3:   z.x is Element of COMPLEX by XCMPLX_0:def 2;
      thus (abs z).x = abs(z.x) by VALUED_1:18
        .= abscomplex.(z.x) by A3,Def5
        .= g.x by A1,A2,FUNCT_1:12;
    end;
    hence thesis by A1,FUNCT_1:2;
  end;
end;

definition let n;
  func COMPLEX n -> FinSequenceSet of COMPLEX equals
  n-tuples_on COMPLEX;
  correctness;
end;

registration let n;
  cluster COMPLEX n -> non empty;
  coherence;
end;

reserve x,z,z1,z2,z3 for Element of COMPLEX n,
  A,B for Subset of COMPLEX n;

Lm4: dom z = Seg n
proof
  len z = n by CARD_1:def 7;
  hence thesis by FINSEQ_1:def 3;
end;

theorem Th57:
  k in Seg n implies z.k in COMPLEX
proof
  len z = n by CARD_1:def 7;
  then Seg n = dom z by FINSEQ_1:def 3;
  hence thesis by FINSEQ_2:11;
end;

definition let n,z1,z2;
  redefine func z1 + z2 -> Element of COMPLEX n;
  coherence by FINSEQ_2:120;
end;

theorem Th58:
  k in Seg n & c1 = z1.k & c2 = z2.k implies (z1 + z2).k = c1 + c2
proof
  assume that
A1: k in Seg n and
A2: c1 = z1.k & c2 = z2.k;
  k in dom(z1+z2) by A1,Lm4;
  hence (z1 + z2).k = addcomplex.(c1,c2) by A2,FUNCOP_1:22
    .= c1 + c2 by BINOP_2:def 3;
end;

definition let n;
  func 0c n -> FinSequence of COMPLEX equals
  n |-> 0c;
  correctness;
end;

definition let n;
  redefine func 0c n -> Element of COMPLEX n;
  coherence;
end;

theorem
  z + 0c n = z & z = 0c n + z by BINOP_2:1,FINSEQOP:56;

definition let n,z;
  redefine func -z -> Element of COMPLEX n;
  coherence by FINSEQ_2:113;
end;

theorem Th60:
  k in Seg n & c = z.k implies (-z).k = -c
proof
  assume that
A1: k in Seg n and
A2: c = z.k;
  k in dom(-z) by A1,Lm4;
  hence (-z).k = compcomplex.c by A2,FUNCT_1:12
    .= -c by BINOP_2:def 1;
end;

Lm5: -0c n = 0c n
proof
  thus -0c n = n|->(compcomplex.0c) by FINSEQOP:16
    .= n|->-(0c qua complex number) by BINOP_2:def 1
    .= 0c n;
end;

theorem
  z + -z = 0c n & -z + z = 0c n by Th51,Th52,BINOP_2:1,FINSEQOP:73;

theorem
  z1 + z2 = 0c n implies z1 = -z2 & z2 = -z1 by Th51,Th52,BINOP_2:1,FINSEQOP:74
;

canceled;

theorem
  -z1 = -z2 implies z1 = z2
proof
  assume -z1 = -z2;
  hence z1 = -(-z2)  .= z2;
end;

Lm6: z1 + z = z2 + z implies z1 = z2
proof
  assume z1 + z = z2 + z;
  then z1 + (z + -z)= (z2 + z) + -z by FINSEQOP:28;
  then
A1: z1 + (z + -z)= z2 + (z + -z) by FINSEQOP:28;
  z + -z = 0c n by Th51,Th52,BINOP_2:1,FINSEQOP:73;
  then z1 = z2 + 0c n by A1,BINOP_2:1,FINSEQOP:56;
  hence thesis by BINOP_2:1,FINSEQOP:56;
end;

theorem
  z1 + z = z2 + z or z1 + z = z + z2 implies z1 = z2 by Lm6;

theorem Th66:
  -(z1 + z2) = -z1 + -z2
proof
  (z1 + z2) + (-z1 + -z2) = z2 + z1 + -z1 + -z2 by FINSEQOP:28
    .= z2 + (z1 + -z1) + -z2 by FINSEQOP:28
    .= z2 + 0c n + -z2 by Th51,Th52,BINOP_2:1,FINSEQOP:73
    .= z2 + -z2 by BINOP_2:1,FINSEQOP:56
    .= 0c n by Th51,Th52,BINOP_2:1,FINSEQOP:73;
  hence thesis by Th51,Th52,BINOP_2:1,FINSEQOP:74;
end;

definition let n,z1,z2;
  redefine func z1 - z2 -> Element of COMPLEX n;
  coherence by FINSEQ_2:120;
end;

theorem
  k in Seg n implies (z1 - z2).k = z1.k - z2.k
proof
  assume that
A1: k in Seg n;
  set c1 = z1.k, c2 = z2.k;
  k in dom(z1 - z2) by A1,Lm4;
  hence (z1 - z2).k = diffcomplex.(c1,c2) by FUNCOP_1:22
    .= c1 - c2 by BINOP_2:def 4;
end;

theorem
  z - 0c n = z
proof
  thus z - 0c n = z + 0c n by Lm5
    .= z by BINOP_2:1,FINSEQOP:56;
end;

theorem
  0c n - z = -z
proof
  thus 0c n - z = 0c n + -z .= -z by BINOP_2:1,FINSEQOP:56;
end;

theorem
  z1 - -z2 = z1 + z2;

theorem Th71:
  -(z1 - z2) = z2 - z1
proof
  thus -(z1 - z2) = -z1 + -(-z2) by Th66
    .= z2 - z1;
end;

theorem Th72:
  -(z1 - z2) = -z1 + z2
proof
  thus -(z1 - z2) = -z1 + -(-z2) by Th66
    .= -z1 + z2;
end;

theorem Th73:
  z - z = 0c n
proof
  thus z - z = z + -z .= 0c n by Th51,Th52,BINOP_2:1,FINSEQOP:73;
end;

theorem Th74:
  z1 - z2 = 0c n implies z1 = z2
proof
  assume z1 - z2 = 0c n;
  then z1 + - z2 = 0c n;
  then z1 = -(-z2) by Th51,Th52,BINOP_2:1,FINSEQOP:74;
  hence thesis;
end;

theorem Th75:
  z1 - z2 - z3 = z1 - (z2 + z3)
proof
  thus z1 - z2 - z3 = z1 + -z2 + -z3 .= z1 + (- z2 + -z3) by FINSEQOP:28
    .= z1 - (z2 + z3) by Th66;
end;

theorem Th76:
  z1 + (z2 - z3) = z1 + z2 - z3
proof
  thus z1 + (z2 - z3) = z1 + (z2 + -z3) .= z1 + z2 + -z3 by FINSEQOP:28
    .= z1 + z2 - z3;
end;

theorem
  z1 - (z2 - z3) = z1 - z2 + z3
proof
  thus z1 - (z2 - z3) = z1 + (-z2 + z3) by Th72
    .= z1 + -z2 + z3 by FINSEQOP:28
    .= z1 - z2 + z3;
end;

theorem
  z1 - z2 + z3 = z1 + z3 - z2 by Th76;

theorem Th79:
  z1 = z1 + z - z
proof
  thus z1 = z1 + 0c n by BINOP_2:1,FINSEQOP:56
    .= z1 + (z - z) by Th73
    .= z1 + z - z by Th76;
end;

theorem Th80:
  z1 + (z2 - z1) = z2
proof
  thus z1 + (z2 - z1) = z2 + -z1 + z1 .= z2 + (-z1 + z1) by FINSEQOP:28
    .= z2 + 0c n by Th51,Th52,BINOP_2:1,FINSEQOP:73
    .= z2 by BINOP_2:1,FINSEQOP:56;
end;

theorem Th81:
  z1 = z1 - z + z
proof
  thus z1 = z1 + 0c n by BINOP_2:1,FINSEQOP:56
    .= z1 + (-z + z) by Th51,Th52,BINOP_2:1,FINSEQOP:73
    .= z1 + -z + z by FINSEQOP:28
    .= z1 - z + z;
end;

definition let n,z,c;
  redefine func c*z -> Element of COMPLEX n;
  coherence by FINSEQ_2:113;
end;

theorem Th82:
  k in Seg n & c9 = z.k implies (c*z).k = c*c9
proof
  assume that
A1: k in Seg n and
A2: c9 = z.k;
  k in dom(c*z) by A1,Lm4;
  hence (c*z).k = (c multcomplex).c9 by A2,FUNCT_1:12
    .= c*c9 by Lm3;
end;

theorem
  c1*(c2*z) = (c1*c2)*z
proof
  thus (c1*c2)*z = multcomplex[;](multcomplex.(c1,c2),id COMPLEX)*z by
BINOP_2:def 5
    .= multcomplex[;](c1,multcomplex[;](c2,id COMPLEX))*z by FUNCOP_1:62
    .= (multcomplex[;](c1,id COMPLEX)*multcomplex[;](c2,id COMPLEX))*z by
FUNCOP_1:55
    .= c1*(c2*z) by RELAT_1:36;
end;

theorem
  (c1 + c2)*z = c1*z + c2*z
proof
  set c1M = multcomplex[;](c1,id COMPLEX), c2M = multcomplex[;](c2,id COMPLEX);
  thus (c1 + c2)*z = multcomplex[;](addcomplex.(c1,c2),id COMPLEX)*z by
BINOP_2:def 3
    .= addcomplex.:(c1M,c2M)*z by Th54,FINSEQOP:35
    .= c1*z + c2*z by FUNCOP_1:25;
end;

theorem
  c*(z1+z2) = c*z1 + c*z2 by Th54,FINSEQOP:51,54;

theorem
  1r*z = z
proof
A1: rng z c= COMPLEX;
  thus 1r*z = (id COMPLEX)*z by BINOP_2:6,FINSEQOP:44
    .= z by A1,RELAT_1:53;
end;

theorem
  0c*z = 0c n
proof
A1: rng z c= COMPLEX;
  thus 0c*z = multcomplex[;](0c,(id COMPLEX)*z) by FUNCOP_1:34
    .= multcomplex[;](0c,z) by A1,RELAT_1:53
    .= 0c n by Th51,Th54,BINOP_2:1,FINSEQOP:76;
end;

theorem
  (-1r)*z = -z;

definition let n,z;
  redefine func abs z -> Element of n-tuples_on REAL;
  correctness by FINSEQ_2:113;
end;

theorem Th89:
  k in Seg n & c = z.k implies (abs z).k = |.c.|
proof
  assume that
A1: k in Seg n and
A2: c = z.k;
  len abs z = n by CARD_1:def 7;
  then k in dom abs z by A1,FINSEQ_1:def 3;
  hence (abs z).k = abscomplex.c by A2,FUNCT_1:12
    .= |.c.| by Def5;
end;

theorem Th90:
  abs 0c n = n |-> 0
proof
  thus abs 0c n = n |-> abscomplex.0c by FINSEQOP:16
    .= n |-> 0 by Def5,COMPLEX1:44;
end;

theorem Th91:
  abs -z = abs z
proof
  now
    let j be Nat;
    assume
A1: j in Seg n;
    then reconsider c = z.j, c9 = (-z).j as Element of COMPLEX by Th57;
    thus (abs -z).j = |.c9.| by A1,Th89
      .= |.-c.| by A1,Th60
      .= |.c.| by COMPLEX1:52
      .= (abs z).j by A1,Th89;
  end;
  hence thesis by FINSEQ_2:119;
end;

theorem Th92:
  abs(c*z) = |.c.|*(abs z)
proof
  now
    let j be Nat;
    reconsider w = j as Element of NAT by ORDINAL1:def 12;
    assume
A1: j in Seg n;
    then reconsider c9 = z.j, cc = (c*z).j as Element of COMPLEX by Th57;
    reconsider ac = (abs z).w as Real;
    thus (abs(c*z)).j = |.cc.| by A1,Th89
      .= |.c*c9.| by A1,Th82
      .= |.c.|*|.c9.| by COMPLEX1:65
      .= |.c.|*ac by A1,Th89
      .= (|.c.|*(abs z)).j by RVSUM_1:45;
  end;
  hence thesis by FINSEQ_2:119;
end;

definition
  let z be FinSequence of COMPLEX;
  func |.z.| -> Real equals
  sqrt Sum sqr abs z;
  correctness;
end;

theorem Th93:
  |.0c n.| = 0
proof
  thus |.0c n .| = sqrt Sum sqr (n |-> (0 qua Real)) by Th90
    .= sqrt Sum (n |-> (0 qua Real)^2) by RVSUM_1:56
    .= sqrt (n*(0 qua Nat)) by RVSUM_1:80
    .= 0 by SQUARE_1:17;
end;

theorem Th94:
  |.z.| = 0 implies z = 0c n
proof
  assume
A1: |.z.| = 0;
  now
    let j be Nat;
    assume
A2: j in Seg n;
    then reconsider c = z.j as Element of COMPLEX by Th57;
    0 <= Sum sqr abs z by RVSUM_1:86;
    then (abs z).j = (n|->0).j by A1,RVSUM_1:91,SQUARE_1:24;
    then |.c.| = (n|-> 0).j by A2,Th89;
    then c = 0c by COMPLEX1:45;
    hence z.j = (n|->0c).j;
  end;
  hence thesis by FINSEQ_2:119;
end;

theorem Th95:
  0 <= |.z.|
proof
  0 <= Sum sqr abs z by RVSUM_1:86;
  hence thesis by SQUARE_1:def 2;
end;

theorem
  |.-z.| = |.z.| by Th91;

theorem
  |.c*z.| = |.c.|*|.z.|
proof
A1: 0 <= |.c.|^2 & 0 <= Sum sqr abs z by RVSUM_1:86,XREAL_1:63;
  thus |.c*z.| = sqrt Sum sqr (|.c.|*abs z) by Th92
    .= sqrt Sum (|.c.|^2 * sqr abs z) by RVSUM_1:58
    .= sqrt (|.c.|^2 * Sum sqr abs z) by RVSUM_1:87
    .= sqrt |.c.|^2 * sqrt Sum sqr abs z by A1,SQUARE_1:29
    .= |.c.|*|.z.| by COMPLEX1:46,SQUARE_1:22;
end;

theorem Th98:
  |.z1 + z2.| <= |.z1.| + |.z2.|
proof
A1: 0 <= Sum sqr abs (z1 + z2) by RVSUM_1:86;
A2: 0 <= Sum sqr abs z1 by RVSUM_1:86;
  then
A3: 0 <= sqrt Sum sqr abs z1 by SQUARE_1:def 2;
A4: for k be Nat holds k in Seg n implies 0 <= (mlt(abs z1,abs z2)).k
  proof
    let k be Nat;
    set r = (mlt(abs z1,abs z2)).k;
    assume
A5: k in Seg n;
    then reconsider c1 = z1.k, c2 = z2.k as Element of COMPLEX by Th57;
    (abs z1).k = |.c1.| & (abs z2).k = |.c2.| by A5,Th89;
    then
A6: r = |.c1.|*|.c2.| by RVSUM_1:60;
    0 <= |.c1.| & 0 <= |.c2.| by COMPLEX1:46;
    hence thesis by A6;
  end;
  0 <= (Sum mlt(abs z1,abs z2))^2 by XREAL_1:63;
  then
A7: sqrt(Sum mlt(abs z1,abs z2))^2 <= sqrt((Sum sqr abs z1)*(Sum sqr abs z2
  )) by RVSUM_1:92,SQUARE_1:26;
  len mlt(abs z1,abs z2) = n by CARD_1:def 7;
  then dom mlt(abs z1,abs z2) = Seg n by FINSEQ_1:def 3;
  then Sum mlt(abs z1,abs z2) <= sqrt((Sum sqr abs z1)*(Sum sqr abs z2)) by A4
,A7,RVSUM_1:84,SQUARE_1:22;
  then 2*Sum mlt(abs z1,abs z2) <= 2*sqrt((Sum sqr abs z1)*(Sum sqr abs z2))
  by XREAL_1:64;
  then
  Sum sqr abs z1+(2*Sum mlt(abs z1,abs z2)) <= Sum sqr abs z1+2*sqrt((Sum
  sqr abs z1)*(Sum sqr abs z2)) by XREAL_1:7;
  then
A8: Sum sqr abs z1+(2*Sum mlt(abs z1,abs z2)) + Sum sqr abs z2 <= Sum sqr
  abs z1+2*sqrt((Sum sqr abs z1)*(Sum sqr abs z2)) + Sum sqr abs z2 by
XREAL_1:7;
A9: for k be Nat holds k in Seg n implies (sqr abs (z1 + z2)).k <= (sqr (abs
  z1 + abs z2)).k
  proof
    let k be Nat;
    set r2 = (sqr (abs z1 + abs z2)).k;
    len(abs z1 + abs z2) = n by CARD_1:def 7;
    then
A10: dom (abs z1 + abs z2) = Seg n by FINSEQ_1:def 3;
    assume
A11: k in Seg n;
    then reconsider c12 = (z1 + z2).k as Element of COMPLEX by Th57;
    reconsider abs912 = (abs (z1 + z2)).k as Real;
    0 <= |.c12.| by COMPLEX1:46;
    then
A12: 0 <= abs912 by A11,Th89;
    reconsider abs1 = (abs z1).k, abs2 = (abs z2).k as Real;
    reconsider c1 = z1.k, c2 = z2.k as Element of COMPLEX by A11,Th57;
    reconsider abs12 = (abs z1 + abs z2).k as Real;
    |.c1 + c2.| <= |.c1.| + |.c2.| by COMPLEX1:56;
    then |.c12.| <= |.c1.| + |.c2.| by A11,Th58;
    then |.c12.| <= |.c1.| + abs2 by A11,Th89;
    then |.c12.| <= abs1 + abs2 by A11,Th89;
    then |.c12.| <= abs12 by A11,A10,VALUED_1:def 1;
    then abs912 <= abs12 by A11,Th89;
    then (abs912)^2 <= (abs12)^2 by A12,SQUARE_1:15;
    then (abs912)^2 <= r2 by VALUED_1:11;
    hence thesis by VALUED_1:11;
  end;
A13: 0 <= Sum sqr abs z2 by RVSUM_1:86;
  then
A14: 0 <= sqrt Sum sqr abs z2 by SQUARE_1:def 2;
A15: Sum sqr abs z1 = (sqrt Sum sqr abs z1)^2 by A2,SQUARE_1:def 2;
A16: (sqrt Sum sqr abs z2)^2 = Sum sqr abs z2 by A13,SQUARE_1:def 2;
  Sum sqr (abs z1 + abs z2) = Sum (sqr abs z1 + 2*mlt(abs z1,abs z2) + sqr
  abs z2) by RVSUM_1:68
    .= Sum(sqr abs z1 + 2*mlt(abs z1,abs z2)) + Sum sqr abs z2 by RVSUM_1:89
    .= Sum sqr abs z1 + Sum(2*mlt(abs z1,abs z2)) + Sum sqr abs z2 by
RVSUM_1:89
    .= Sum sqr abs z1 + (2*Sum mlt(abs z1,abs z2))+Sum sqr abs z2 by RVSUM_1:87
;
  then Sum sqr abs (z1 + z2) <= Sum sqr abs z1 + (2*Sum mlt(abs z1,abs z2))+
  Sum sqr abs z2 by A9,RVSUM_1:82;
  then Sum sqr abs (z1 + z2) <= Sum sqr abs z1+2*sqrt((Sum sqr abs z1)*(Sum
  sqr abs z2)) + Sum sqr abs z2 by A8,XXREAL_0:2;
  then Sum sqr abs (z1 + z2) <= Sum sqr abs z1+2*((sqrt Sum sqr abs z1)*(sqrt
  Sum sqr abs z2)) + Sum sqr abs z2 by A2,A13,SQUARE_1:29;
  then sqrt Sum sqr abs (z1 + z2) <= sqrt(((sqrt Sum sqr abs z1) + (sqrt Sum
  sqr abs z2))^2) by A15,A16,A1,SQUARE_1:26;
  hence thesis by A3,A14,SQUARE_1:22;
end;

theorem Th99:
  |.z1 - z2.| <= |.z1.| + |.z2.|
proof
  |.z1 - z2.| <= |.z1.| + |.-z2.| by Th98;
  hence thesis by Th91;
end;

theorem
  |.z1.| - |.z2.| <= |.z1 + z2.|
proof
  z1 = z1 + z2 - z2 by Th79;
  then |.z1.| <= |.z1 + z2.| + |.z2.| by Th99;
  hence thesis by XREAL_1:20;
end;

theorem
  |.z1.| - |.z2.| <= |.z1 - z2.|
proof
  z1 = z1 - z2 + z2 by Th81;
  then |.z1.| <= |.z1 - z2.| + |.z2.| by Th98;
  hence thesis by XREAL_1:20;
end;

theorem Th102:
  |.z1 - z2.| = 0 iff z1 = z2
proof
  thus |.z1 - z2.| = 0 implies z1 = z2
  proof
    assume |.z1 - z2.| = 0;
    then z1 - z2 = 0c n by Th94;
    hence thesis by Th74;
  end;
  assume z1 = z2;
  then z1 - z2 = 0c n by Th73;
  hence thesis by Th93;
end;

theorem
  z1 <> z2 implies 0 < |.z1 - z2.|
proof
  assume z1 <> z2;
  then 0 <> |.z1 - z2.| by Th102;
  hence thesis by Th95;
end;

theorem Th104:
  |.z1 - z2.| = |.z2 - z1.|
proof
  thus |.z1 - z2.| = |.-(z2 - z1).| by Th71
    .= |.z2 - z1.| by Th91;
end;

theorem Th105:
  |.z1 - z2.| <= |.z1 - z.| + |.z - z2.|
proof
  |.z1 - z2.| = |.z1 - z + z - z2.| by Th81
    .= |.(z1 - z) + (z - z2).| by Th76;
  hence thesis by Th98;
end;

definition
  let n;
  let A be Subset of COMPLEX n;
  attr A is open means
  :Def14:
  for x st x in A ex r st 0 < r & for z st |.z.| < r holds x + z in A;
end;

definition let n;
  let A be Subset of COMPLEX n;
  attr A is closed means
  for x st for r st r > 0 ex z st |.z.| < r & x + z in A holds x in A;
end;

theorem
  for A being Subset of COMPLEX n st A = {} holds A is open
proof
  let A be Subset of COMPLEX n;
  assume
A1: A = {};
  let x;
  thus thesis by A1;
end;

theorem
  for A being Subset of COMPLEX n st A = COMPLEX n holds A is open
proof
  let A be Subset of COMPLEX n;
  assume
A1: A = COMPLEX n;
  let x such that
  x in A;
  take j=1;
  thus 0 < j;
  let z such that
  |.z.| < j;
  thus thesis by A1;
end;

theorem
  for AA being Subset-Family of COMPLEX n st for A being Subset of
  COMPLEX n st A in AA holds A is open for A being Subset of COMPLEX n st A =
  union AA holds A is open
proof
  let AA be Subset-Family of COMPLEX n such that
A1: for A being Subset of COMPLEX n st A in AA holds A is open;
  let A be Subset of COMPLEX n such that
A2: A = union AA;
  let x;
  assume x in A;
  then consider B being set such that
A3: x in B and
A4: B in AA by A2,TARSKI:def 4;
  reconsider B as Subset of COMPLEX n by A4;
  B is open by A1,A4;
  then consider r such that
A5: 0 < r and
A6: for z st |.z.| < r holds x + z in B by A3,Def14;
  take r;
  thus 0 < r by A5;
  let z;
  assume |.z.| < r;
  then x + z in B by A6;
  hence thesis by A2,A4,TARSKI:def 4;
end;

theorem
  for A,B being Subset of COMPLEX n st A is open & B is open for C
  being Subset of COMPLEX n st C = A /\ B holds C is open
proof
  let A,B be Subset of COMPLEX n such that
A1: A is open and
A2: B is open;
  let C be Subset of COMPLEX n such that
A3: C = A /\ B;
  let x;
  assume
A4: x in C;
  then x in A by A3,XBOOLE_0:def 4;
  then consider r1 such that
A5: 0 < r1 and
A6: for z st |.z.| < r1 holds x + z in A by A1,Def14;
  x in B by A3,A4,XBOOLE_0:def 4;
  then consider r2 such that
A7: 0 < r2 and
A8: for z st |.z.| < r2 holds x + z in B by A2,Def14;
  take min(r1,r2);
  thus 0 < min(r1,r2) by A5,A7,XXREAL_0:15;
  let z;
  assume
A9: |.z.| < min(r1,r2);
  min(r1,r2) <= r2 by XXREAL_0:17;
  then |.z.| < r2 by A9,XXREAL_0:2;
  then
A10: x + z in B by A8;
  min(r1,r2) <= r1 by XXREAL_0:17;
  then |.z.| < r1 by A9,XXREAL_0:2;
  then x + z in A by A6;
  hence thesis by A3,A10,XBOOLE_0:def 4;
end;

definition
  let n,x,r;
  func Ball(x,r) -> Subset of COMPLEX n equals
  { z : |.z - x.| < r };
  coherence
  proof
    defpred P[FinSequence of COMPLEX] means |.$1 - x.| < r;
    { z : P[z] } c= COMPLEX n from FRAENKEL:sch 10;
    hence thesis;
  end;
end;

theorem Th110:
  z in Ball(x,r) iff |.x - z.| < r
proof
A1: z in { z2 : |.z2 - x.| < r } iff ex z1 st z = z1 & |.z1 - x.| < r;
  |.z - x.| = |.x - z.| by Th104;
  hence thesis by A1;
end;

theorem
  0 < r implies x in Ball(x,r)
proof
  assume
A1: 0 < r;
  |.x - x.| = 0 by Th102;
  hence thesis by A1;
end;

theorem
  Ball(z1,r1) is open
proof
  let x;
  assume x in Ball(z1,r1); then
A1: |.z1 - x.| < r1 by Th110;
  take r = r1 - |.z1 - x.|;
  thus 0 < r by A1,XREAL_1:50;
  let z;
  assume |.z.| < r; then
A2: |.z.| + |.z1 - x.| < r + |.z1 - x.| by XREAL_1:6;
  z1 - x - z = z1 - (x + z) by Th75;
  then |.z1 - (x + z).| <= |.z.| + |.z1 - x.| by Th99;
  then |.z1 - (x + z).| < r + |.z1 - x.| by A2,XXREAL_0:2;
  hence thesis by Th110;
end;

definition let n,x,A;
  func dist(x,A) -> Real means
  :Def17:
  for X being Subset of REAL st X = {|.x
  - z.| : z in A} holds it = lower_bound X;
  existence
  proof
    deffunc f(Element of COMPLEX n) = |.x - $1.|;
    defpred P[set] means $1 in A;
    reconsider X = {f(z) : P[z]} as Subset of REAL from DOMAIN_1:sch 8;
    take lower_bound X;
    thus thesis;
  end;
  uniqueness
  proof
    deffunc f(Element of COMPLEX n) = |.x - $1.|;
    defpred P[set] means $1 in A;
    reconsider X = {f(z) : P[z]} as Subset of REAL from DOMAIN_1:sch 8;
    let r1,r2 be Real such that
A1: for X being Subset of REAL st X = {|.x - z.| : z in A} holds r1 =
    lower_bound X and
A2: for X being Subset of REAL st X = {|.x - z.| : z in A} holds r2 =
    lower_bound X;
    r1 = lower_bound X by A1;
    hence thesis by A2;
  end;
end;

definition let n,A,r;
  func Ball(A,r) -> Subset of COMPLEX n equals
  { z : dist(z,A) < r };
  coherence
  proof
    defpred P[Element of COMPLEX n] means dist($1,A) < r;
    { z : P[z] } c= COMPLEX n from FRAENKEL:sch 10;
    hence thesis;
  end;
end;

theorem Th113:
  for X being Subset of REAL, r st X <> {} & for r9 st r9 in X
  holds r <= r9 holds lower_bound X >= r
proof
  let X be Subset of REAL, r such that
A1: X <> {} and
A2: for r9 st r9 in X holds r <= r9;
  for r9 be ext-real number st r9 in X holds r <= r9 by A2;
  then r is LowerBound of X by XXREAL_2:def 2;
  then
A3: X is bounded_below by XXREAL_2:def 9;
  now
    let r9 be real number;
    assume r9 > 0;
    then consider r1 be real number such that
A4: r1 in X and
A5: r1 < lower_bound X + r9 by A1,A3,Def2;
    r <= r1 by A2,A4;
    hence lower_bound X + r9 >= r by A5,XXREAL_0:2;
  end;
  hence thesis by XREAL_1:41;
end;

theorem Th114:
  A <> {} implies dist(x,A) >= 0
proof
  defpred P[set] means $1 in A;
  deffunc f(Element of COMPLEX n) = |.x - $1.|;
  reconsider X = {f(z) : P[z]} as Subset of REAL from DOMAIN_1:sch 8;
  assume A <> {};
  then consider z1 such that
A1: z1 in A by SUBSET_1:4;
A2: |.x - z1.| in X by A1;
A3: now
    let r9;
    assume r9 in X;
    then ex z st r9 = |.x - z.| & z in A;
    hence r9>= 0 by Th95;
  end;
  dist(x,A) = lower_bound X by Def17;
  hence thesis by A2,A3,Th113;
end;

theorem Th115:
  A <> {} implies dist(x + z,A) <= dist(x,A) + |.z.|
proof
  defpred P[set] means $1 in A;
  deffunc g(Element of COMPLEX n) = |.x + z - $1.|;
  reconsider Y = {g(z1) : P[z1]} as Subset of REAL from DOMAIN_1:sch 8;
  deffunc f(Element of COMPLEX n) = |.x - $1.|;
A1: Y is bounded_below
  proof
    take 0;
    let r be ext-real number;
    assume r in Y;
    then ex z1 st r = |.x + z - z1 .| & z1 in A;
    hence thesis by Th95;
  end;
  reconsider X = {f(z1) : P[z1]} as Subset of REAL from DOMAIN_1:sch 8;
  assume A <> {};
  then consider z2 such that
A2: z2 in A by SUBSET_1:4;
A3: dist(x+z,A) = lower_bound Y by Def17;
A4: now let r9;
    assume r9 in X;
    then consider z3 such that
A5: r9 = |.x - z3.| and
A6: z3 in A;
    |.x + z - z3.| = |.x - z3 + z.| by Th76; then
A7: |.x + z - z3.| <= r9 + |.z.| by A5,Th98;
    |.x + z - z3.| in Y by A6;
    then |.x + z - z3.| >= dist(x + z,A) by A3,A1,Def2;
    then r9+ |.z.| >= dist(x + z,A) by A7,XXREAL_0:2;
    hence r9 >= dist(x + z,A) - |.z.| by XREAL_1:20;
  end;
A8: |.x - z2.| in X by A2;
  dist(x,A) = lower_bound X by Def17;
  then dist(x + z,A) - |.z.| <= dist(x,A) by A8,A4,Th113;
  hence thesis by XREAL_1:20;
end;

theorem Th116:
  x in A implies dist(x,A) = 0
proof
  defpred P[set] means $1 in A;
  deffunc f(Element of COMPLEX n) = |.x - $1.|;
  reconsider X = {f(z): P[z]} as Subset of REAL from DOMAIN_1:sch 8;
  assume
A1: x in A;
  then
A2: |.x - x.| in X;
A3: now
    reconsider r = |.x - x.| as real number;
    let r1 be real number such that
A4: 0<r1;
    take r;
    thus r in X by A1;
    thus r<(0 qua Nat)+r1 by A4,Th102;
  end;
A5: now
    let r be real number;
    assume r in X;
    then ex z st r = |.x - z .| & z in A;
    hence 0<=r by Th95;
  end;
A6: X is bounded_below
  proof
    take 0;
    let x be ext-real number;
    thus thesis by A5;
  end;
  thus dist(x,A) = lower_bound X by Def17
    .= 0 by A2,A5,A6,A3,Def2;
end;

theorem
  not x in A & A <> {} & A is closed implies dist(x,A) > 0
proof
  assume that
A1: not x in A and
A2: A <> {} and
A3: for x st for r st r > 0 ex z st |.z.| < r & x + z in A holds x in A and
A4: dist(x,A) <= 0;
A5: dist(x,A) = 0 by A2,A4,Th114;
  now
    deffunc f(Element of COMPLEX n) = |.x - $1.|;
    defpred P[set] means $1 in A;
    reconsider X = {f(z) : P[z]} as Subset of REAL from DOMAIN_1:sch 8;
    let r such that
A6: r > 0;
    consider z such that
A7: z in A by A2,SUBSET_1:4;
A8: X is bounded_below
    proof
      take 0;
      let r be ext-real number;
      assume r in X;
      then ex z st r = |.x - z .| & z in A;
      hence thesis by Th95;
    end;
A9: |.x - z.| in X by A7;
    dist(x,A) = lower_bound X & (0 qua Nat)+r = r by Def17;
    then consider r9 be real number such that
A10: r9 in X and
A11: r9< r by A5,A6,A9,A8,Def2;
    consider z1 such that
A12: r9 = |.x - z1.| & z1 in A by A10;
    take z = z1 - x;
    thus |.z.| < r & x + z in A by A11,A12,Th80,Th104;
  end;
  hence contradiction by A1,A3;
end;

theorem
  A <> {} implies |.z1 - x.| + dist(x,A) >= dist(z1,A)
proof
  x + (z1 - x) = z1 by Th80;
  hence thesis by Th115;
end;

theorem Th119:
  z in Ball(A,r) iff dist(z,A) < r
proof
  z in { z2 : dist(z2,A) < r } iff ex z1 st z = z1 & dist(z1,A) < r;
  hence thesis;
end;

theorem Th120:
  0 < r & x in A implies x in Ball(A,r)
proof
  assume that
A1: 0 < r and
A2: x in A;
  dist(x,A) = 0 by A2,Th116;
  hence thesis by A1;
end;

theorem
  0 < r implies A c= Ball(A,r)
proof
  assume
A1: r > 0;
  let x be set;
  assume x in A;
  hence thesis by A1,Th120;
end;

theorem
  A <> {} implies Ball(A,r1) is open
proof
  assume
A1: A <> {};
  let x;
  assume x in Ball(A,r1); then
A2: dist(x,A) < r1 by Th119;
  take r = r1 - dist(x,A);
  thus 0 < r by A2,XREAL_1:50;
  let z;
  assume |.z.| < r; then
A3: |.z.| + dist(x,A) < r + dist(x,A) by XREAL_1:6;
  dist(x + z,A) <= |.z.| + dist(x,A) by A1,Th115;
  then dist(x + z,A) < r + dist(x,A) by A3,XXREAL_0:2;
  hence thesis;
end;

definition
  let n,A,B;
  func dist(A,B) -> Real means
  :Def19:
  for X being Subset of REAL st X = {|.x
  - z.| : x in A & z in B} holds it = lower_bound X;
  existence
  proof
    deffunc f(Element of COMPLEX n,Element of COMPLEX n) = |.$1 - $2.|;
    defpred P[set,set] means $1 in A & $2 in B;
    reconsider X = {f(x,z) : P[x,z]} as Subset of REAL from DOMAIN_1:sch 9;
    take lower_bound X;
    thus thesis;
  end;
  uniqueness
  proof
    deffunc f(Element of COMPLEX n,Element of COMPLEX n) = |.$1 - $2.|;
    defpred P[set,set] means $1 in A & $2 in B;
    reconsider X = {f(x,z): P[x,z]} as Subset of REAL from DOMAIN_1:sch 9;
    let r1,r2 be Real such that
A1: for X being Subset of REAL st X = {|.x - z.| : x in A & z in B}
    holds r1 = lower_bound X and
A2: for X being Subset of REAL st X = {|.x - z.| : x in A & z in B}
    holds r2 = lower_bound X;
    r1 = lower_bound X by A1;
    hence thesis by A2;
  end;
end;

theorem
  for X,Y being Subset of REAL holds X <> {} & Y <> {} implies X ++ Y <> {};

theorem Th124:
  for X,Y being Subset of REAL holds X is bounded_below &
    Y is bounded_below implies X++Y is bounded_below
proof
  let X,Y be Subset of REAL;
  assume X is bounded_below;
  then consider r1 be real number such that
A1: r1 is LowerBound of X by XXREAL_2:def 9;
A2: for r be real number st r in X holds r1<=r by A1,XXREAL_2:def 2;
  assume Y is bounded_below;
  then consider r2 be real number such that
A3: r2 is LowerBound of Y by XXREAL_2:def 9;
A4: for r be real number st r in Y holds r2<=r by A3,XXREAL_2:def 2;
  take r1 + r2;
  let r be ext-real number;
  assume r in X++Y; then
  r in {r22+r12 where r22 is Element of COMPLEX,
    r12 is Element of COMPLEX : r22 in X & r12 in Y} by MEMBER_1:def 6;
  then consider r22,r12 being Element of COMPLEX such that
A5: r = r22 + r12 and
A6: r22 in X and
A7: r12 in Y;
    reconsider r9 = r22, r99 = r12 as real number by A6,A7;
A8: r2 <= r99 by A4,A7;
  r1 <= r9 by A2,A6;
  hence thesis by A5,A8,XREAL_1:7;
end;

theorem Th125:
  for X,Y being Subset of REAL st
  X <> {} & X is bounded_below & Y <> {} & Y is bounded_below holds
  lower_bound (X ++ Y) = lower_bound X + lower_bound Y
proof
  let X,Y be Subset of REAL such that
A1: X <> {} and
A2: X is bounded_below and
A3: Y <> {} and
A4: Y is bounded_below;
A5: now
    let r9 be real number;
    assume 0<r9;
    then
A6: r9/2 > 0 by XREAL_1:215;
    then consider r1 be real number such that
A7: r1 in X and
A8: r1<lower_bound X+r9/2 by A1,A2,Def2;
    consider r2 be real number such that
A9: r2 in Y and
A10: r2<lower_bound Y+r9/2 by A3,A4,A6,Def2;
     reconsider r = r1 + r2 as real number;
    take r;
    thus r in X++Y by A7,A9,MEMBER_1:46;
    lower_bound X + r9/2 + (lower_bound Y + r9/2) = lower_bound X +
    lower_bound Y + r9;
    hence r<lower_bound X + lower_bound Y+r9 by A8,A10,XREAL_1:8;
  end;
A11: now
    let r be real number;
    assume r in X++Y; then
    r in {r22+r12 where r22 is Element of COMPLEX,
      r12 is Element of COMPLEX : r22 in X & r12 in Y} by MEMBER_1:def 6;
    then consider r11,r22 being Element of COMPLEX such that
A12: r = r11 + r22 and
A13: r11 in X & r22 in Y;
    reconsider r1 = r11, r2 = r22 as real number by A13;
    r1 >= lower_bound X & r2 >= lower_bound Y by A2,A4,A13,Def2;
    hence lower_bound X + lower_bound Y<=r by A12,XREAL_1:7;
  end;
  X ++ Y <> {} & X ++ Y is bounded_below by A1,A2,A3,A4,Th124;
  hence thesis by A11,A5,Def2;
end;

theorem Th126:
  for X,Y being Subset of REAL st Y is bounded_below & X <> {} &
  for r st r in X ex r1 st r1 in Y & r1 <= r holds
    lower_bound X >= lower_bound Y
proof
  let X,Y be Subset of REAL such that
A1: Y is bounded_below and
A2: X <> {} and
A3: for r st r in X ex r1 st r1 in Y & r1 <= r;
  now
    let r1;
    assume r1 in X;
    then consider r2 such that
A4: r2 in Y and
A5: r2 <= r1 by A3;
    lower_bound Y <= r2 by A1,A4,Def2;
    hence r1 >= lower_bound Y by A5,XXREAL_0:2;
  end;
  hence thesis by A2,Th113;
end;

theorem Th127:
  A <> {} & B <> {} implies dist(A,B) >= 0
proof
  defpred P[set,set] means $1 in A & $2 in B;
  deffunc f(Element of COMPLEX n,Element of COMPLEX n) = |.$1 - $2.|;
  reconsider Z = {f(z1,z): P[z1,z]} as Subset of REAL from DOMAIN_1:sch 9;
  assume that
A1: A <> {} and
A2: B <> {};
  consider z1 such that
A3: z1 in A by A1,SUBSET_1:4;
A4: now
    let r9;
    assume r9 in Z;
    then ex z1,z st r9 = |.z1 - z.| & z1 in A & z in B;
    hence r9>= 0 by Th95;
  end;
  consider z2 such that
A5: z2 in B by A2,SUBSET_1:4;
A6: dist(A,B) = lower_bound Z by Def19;
  |.z1 - z2.| in Z by A3,A5;
  hence thesis by A6,A4,Th113;
end;

theorem
  dist(A,B) = dist(B,A)
proof
  defpred R[set,set] means $1 in B & $2 in A;
  deffunc f(Element of COMPLEX n,Element of COMPLEX n) = |.$1 - $2.|;
  reconsider Y = {f(z,z1) : R[z,z1]} as Subset of REAL from DOMAIN_1:sch 9;
  defpred P[set,set] means $1 in A & $2 in B;
  reconsider X = {f(z1,z) : P[z1,z]} as Subset of REAL from DOMAIN_1:sch 9;
A1: now
    let r;
A2: now
      given z,z1 such that
A3:   r =|.z - z1.| & z in B & z1 in A;
      take z1,z;
      thus r =|.z1 - z.| & z1 in A & z in B by A3,Th104;
    end;
    now
      given z1,z such that
A4:   r =|.z1 - z.| & z1 in A & z in B;
      take z,z1;
      thus r =|.z - z1.| & z in B & z1 in A by A4,Th104;
    end;
    hence r in X iff r in Y by A2;
  end;
  dist(A,B) = lower_bound X & dist(B,A) = lower_bound Y by Def19;
  hence thesis by A1,SUBSET_1:3;
end;

theorem Th129:
  A <> {} & B <> {} implies dist(x,A) + dist(x,B) >= dist(A,B)
proof
  defpred Y[set] means $1 in B;
  deffunc g(Element of COMPLEX n) = |.x - $1.|;
  reconsider Y = {g(z) : Y[z]} as Subset of REAL from DOMAIN_1:sch 8;
  defpred P[set,set] means $1 in A & $2 in B;
  defpred X[set] means $1 in A;
  deffunc f(Element of COMPLEX n,Element of COMPLEX n) = |.$1 - $2.|;
  reconsider X = {g(z) : X[z]} as Subset of REAL from DOMAIN_1:sch 8;
  assume that
A1: A <> {} and
A2: B <> {};
  consider z2 such that
A3: z2 in B by A2,SUBSET_1:4;
A4: Y <> {} & Y is bounded_below
  proof
    |.x - z2.| in Y by A3;
    hence Y <> {};
    take 0;
    let r be ext-real number;
    assume r in Y;
    then ex z1 st r = |.x - z1 .| & z1 in B;
    hence thesis by Th95;
  end;
A5: lower_bound X = dist(x,A) & lower_bound Y = dist(x,B) by Def17;
A6: |.x - z2 .| in Y by A3;
  reconsider Z = {f(z1,z) : P[z1,z]} as Subset of REAL from DOMAIN_1:sch 9;
  consider z1 such that
A7: z1 in A by A1,SUBSET_1:4;
A8: now
    let r;
    assume r in X ++ Y; then
    r in {r2+r1 where r2 is Element of COMPLEX,
      r1 is Element of COMPLEX : r2 in X & r1 in Y} by MEMBER_1:def 6;
    then consider r2,r1 being Element of COMPLEX such that
A9: r = r2 + r1 and
A10: r2 in X and
A11: r1 in Y;
    consider z2 such that
A12: r1 = |.x - z2.| & z2 in B by A11;
    consider z1 such that
A13: r2 = |.x - z1.| and
A14: z1 in A by A10;
    take r3 = |.z1 - z2.|;
    r2 = |.z1 - x.| by A13,Th104;
    hence r3 in Z & r3 <= r by A9,A14,A12,Th105;
  end;
A15: Z is bounded_below
  proof
    take 0;
    let r be ext-real number;
    assume r in Z;
    then ex z1,z st r = |.z1 - z .| & z1 in A & z in B;
    hence thesis by Th95;
  end;
A16: X <> {} & X is bounded_below
  proof
    |.x - z1.| in X by A7;
    hence X <> {};
    take 0;
    let r be ext-real number;
    assume r in X;
    then ex z st r = |.x - z .| & z in A;
    hence thesis by Th95;
  end;
A17: lower_bound Z = dist(A,B) by Def19;
A18:  X++Y c= REAL by MEMBERED:3;
  |.x - z1.| in X by A7;
  then |.x - z1.| + |.x - z2 .| in X ++ Y by A6,MEMBER_1:46;
  then lower_bound (X ++ Y) >= lower_bound Z by A15,A8,Th126,A18;
  hence thesis by A16,A4,A5,A17,Th125;
end;

theorem
  A meets B implies dist(A,B) = 0
proof
  assume A meets B;
  then consider z being set such that
A1: z in A and
A2: z in B by XBOOLE_0:3;
  reconsider z as Element of COMPLEX n by A1;
  dist(z,A) = 0 & dist(z,B) = 0 by A1,A2,Th116;
  then (0 qua Nat) + (0 qua Nat) >= dist(A,B) by A1,A2,Th129;
  hence thesis by A1,A2,Th127;
end;

definition
  let n;
  func ComplexOpenSets(n) -> Subset-Family of COMPLEX n equals
  {A where A is
  Subset of COMPLEX n: A is open};
  coherence
  proof
    set S = {A where A is Subset of COMPLEX n: A is open};
    S c= bool COMPLEX n
    proof
      let x be set;
      assume x in S;
      then ex A being Subset of COMPLEX n st x = A & A is open;
      hence thesis;
    end;
    hence thesis;
  end;
end;

theorem
  for A being Subset of COMPLEX n holds A in ComplexOpenSets n iff A is open
proof
  let B be Subset of COMPLEX n;
  B in {A where A is Subset of COMPLEX n: A is open} iff ex C being Subset
  of COMPLEX n st B = C & C is open;
  hence thesis;
end;

theorem
  for A being Subset of COMPLEX n holds A is closed iff A` is open
proof
  let A be Subset of COMPLEX n;
  thus A is closed implies A` is open
  proof
    assume
A1: for x st for r st r > 0 ex z st |.z.| < r & x + z in A holds x in A;
    let x;
    assume x in A`;
    then not x in A by XBOOLE_0:def 5;
    then consider r such that
A2: r > 0 and
A3: for z st |.z.| < r holds not x + z in A by A1;
    take r;
    thus 0 < r by A2;
    let z;
    assume |.z.| < r;
    hence thesis by A3,SUBSET_1:29;
  end;
  assume
A4: for x st x in A` ex r st 0 < r & for z st |.z.| < r holds x + z in A`;
  let x such that
A5: for r st r > 0 ex z st |.z.| < r & x + z in A;
  now
    let r;
    assume r > 0;
    then consider z such that
A6: |.z.| < r & x + z in A by A5;
    take z;
    thus |.z.| < r & not x + z in A` by A6,XBOOLE_0:def 5;
  end;
  then not x in A` by A4;
  hence thesis by SUBSET_1:29;
end;

begin 

reserve
  v,v1,v2 for FinSequence of REAL,
  n,m,k for Element of NAT,
  x for set;

defpred P[Nat] means for R being finite Subset of REAL st R <> {} & card R =
$1 holds R is bounded_above & upper_bound(R) in R & R is bounded_below &
lower_bound(R) in R;

Lm7: P[0];

Lm8: for n st P[n] holds P[n+1]
proof
  let n such that
A1: P[n];
  let R be finite Subset of REAL such that
A2: R <> {} and
A3: card R = n+1;
  now
    per cases;
    suppose
A4:   n=0;
      thus R is bounded_above;
      consider x such that
A5:   R={x} by A3,A4,CARD_2:42;
      x in R by A5,TARSKI:def 1;
      then reconsider r=x as Real;
      upper_bound R = r by A5,Th9;
      hence upper_bound R in R by A5,TARSKI:def 1;
      thus R is bounded_below;
      lower_bound R = r by A5,Th9;
      hence lower_bound R in R by A5,TARSKI:def 1;
    end;
    suppose
A6:   n<>0;
      set x = the Element of R;
      reconsider x as Real by A2,TARSKI:def 3;
      reconsider X = R \ {x} as finite Subset of REAL;
      set u = upper_bound X, m = max(x,u), l = lower_bound X, mi = min(x,l);
      {x} c= R by A2,ZFMISC_1:31;
      then card (R\{x}) = card R - card {x} by CARD_2:44;
      then
A7:   card X = n+1-1 by A3,CARD_1:30
        .=n;
      then
A8:   upper_bound X in X by A1,A6,CARD_1:27;
A9:   now
        reconsider s = m as real number;
        let r be real number such that
A10:    0<r;
        take s;
        now
          per cases by XXREAL_0:16;
          suppose
            m=x;
            hence s in R by A2;
            thus m-r<s by A10,XREAL_1:44;
          end;
          suppose
            m=u;
            hence s in R by A8,XBOOLE_0:def 5;
            thus m-r<s by A10,XREAL_1:44;
          end;
        end;
        hence s in R & m-r<s;
      end;
A11:  lower_bound X in X by A1,A6,A7,CARD_1:27;
A12:  now
        reconsider s = mi as real number;
        let r be real number such that
A13:    0<r;
        take s;
        now
          per cases by XXREAL_0:15;
          suppose
            mi=x;
            hence s in R by A2;
            thus s<mi+r by A13,XREAL_1:29;
          end;
          suppose
            mi=l;
            hence s in R by A11,XBOOLE_0:def 5;
            thus s<mi+r by A13,XREAL_1:29;
          end;
        end;
        hence s in R & s<mi+r;
      end;
      thus R is bounded_above;
A14:  u<=m by XXREAL_0:25;
A15:  now
        let s be real number such that
A16:    s in R;
        now
          per cases;
          suppose
            s=x;
            hence s<=m by XXREAL_0:25;
          end;
          suppose
            s<>x;
            then not s in {x} by TARSKI:def 1;
            then s in X by A16,XBOOLE_0:def 5;
            then s<=u by Def1;
            hence s<=m by A14,XXREAL_0:2;
          end;
        end;
        hence s<=m;
      end;
      now
        per cases by XXREAL_0:16;
        suppose
          m=x;
          hence m in R by A2;
        end;
        suppose
          m=u;
          hence m in R by A8,XBOOLE_0:def 5;
        end;
      end;
      hence upper_bound R in R by A15,A9,Def1;
      thus R is bounded_below;
A17:  mi<=l by XXREAL_0:17;
A18:  now
        let s be real number such that
A19:    s in R;
        now
          per cases;
          suppose
            s=x;
            hence mi<=s by XXREAL_0:17;
          end;
          suppose
            s<>x;
            then not s in {x} by TARSKI:def 1;
            then s in X by A19,XBOOLE_0:def 5;
            then l<=s by Def2;
            hence mi<=s by A17,XXREAL_0:2;
          end;
        end;
        hence mi<=s;
      end;
      now
        per cases by XXREAL_0:15;
        suppose
          mi=x;
          hence mi in R by A2;
        end;
        suppose
          mi=l;
          hence mi in R by A11,XBOOLE_0:def 5;
        end;
      end;
      hence lower_bound R in R by A18,A12,Def2;
    end;
  end;
  hence thesis;
end;

Lm9: for n holds P[n] from NAT_1:sch 1(Lm7,Lm8);

theorem Th133:
  for R being finite Subset of REAL holds R <> {} implies R is
bounded_above & upper_bound(R) in R & R is bounded_below & lower_bound(R) in R
proof
  let R be finite Subset of REAL;
  assume
A1: R <> {};
  P[card R] by Lm9;
  hence thesis by A1;
end;

theorem
  for n being Nat for f being FinSequence holds 1 <= n & n+1 <= len
  f iff n in dom f & n+1 in dom f
proof
  let n be Nat;
  let f be FinSequence;
  thus 1<=n & n+1 <= len f implies n in dom f & n+1 in dom f
  proof
    assume that
A1: 1<=n and
A2: n+1 <= len f;
    n<=n+1 by NAT_1:11;
    then 1<=n+1 & n<=len f by A2,NAT_1:11,XXREAL_0:2;
    hence thesis by A1,A2,FINSEQ_3:25;
  end;
  thus thesis by FINSEQ_3:25;
end;

theorem
  for n being Nat for f being FinSequence holds 1 <= n & n+2 <= len f
  iff n in dom f & n+1 in dom f & n+2 in dom f
proof
  let n be Nat;
  let f be FinSequence;
  thus 1<=n & n+2 <= len f implies n in dom f & n+1 in dom f & n+2 in dom f
  proof
    assume that
A1: 1<=n and
A2: n+2 <= len f;
    n+1<=n+1+1 by NAT_1:11;
    then
A3: 1<=n+1 & n+1<=len f by A2,NAT_1:11,XXREAL_0:2;
    n<=n+2 by NAT_1:11;
    then 1<=n+2 & n<=len f by A1,A2,XXREAL_0:2;
    hence thesis by A1,A2,A3,FINSEQ_3:25;
  end;
  assume that
A4: n in dom f and
  n+1 in dom f and
A5: n+2 in dom f;
  thus thesis by A4,A5,FINSEQ_3:25;
end;

theorem
  for D being non empty set, f1,f2 being FinSequence of D, n st 1
  <= n & n <= len f2 holds (f1^f2)/.(n + len f1) = f2/.n
proof
  let D be non empty set, f1,f2 be FinSequence of D, n such that
A1: 1 <= n and
A2: n <= len f2;
A3: len f1 < n + len f1 by A1,NAT_1:19;
  len(f1^f2) = len f1 + len f2 by FINSEQ_1:22;
  then
A4: n + len f1 <= len(f1^f2) by A2,XREAL_1:6;
  n + len f1 >= n by NAT_1:11;
  then n + len f1 >= 1 by A1,XXREAL_0:2;
  hence (f1^f2)/.(n + len f1) = (f1^f2).(n + len f1) by A4,FINSEQ_4:15
    .= f2.(n + len f1 - len f1) by A3,A4,FINSEQ_1:24
    .= f2/.n by A1,A2,FINSEQ_4:15;
end;

theorem
  v is increasing implies for n,m st n in dom v & m in dom v & n<=m
  holds v.n <= v.m
proof
  assume
A1: v is increasing;
  let n,m such that
A2: n in dom v & m in dom v and
A3: n<=m;
  now
    per cases;
    suppose
      n=m;
      hence thesis;
    end;
    suppose
      n<>m;
      then n<m by A3,XXREAL_0:1;
      hence thesis by A1,A2,SEQM_3:def 1;
    end;
  end;
  hence thesis;
end;

theorem Th138:
  v is increasing implies for n,m st n in dom v & m in dom v & n<>
  m holds v.n<>v.m
proof
  assume
A1: v is increasing;
  let n,m;
  assume that
A2: n in dom v & m in dom v and
A3: n<>m;
  now
    per cases by A3,XXREAL_0:1;
    suppose
      n<m;
      hence thesis by A1,A2,SEQM_3:def 1;
    end;
    suppose
      n>m;
      hence thesis by A1,A2,SEQM_3:def 1;
    end;
  end;
  hence thesis;
end;

theorem Th139:
  v is increasing & v1 = v|Seg n implies v1 is increasing
proof
  assume that
A1: v is increasing and
A2: v1 = v|Seg n;
  now
    per cases;
    suppose
      n<=len v;
      then Seg n c= Seg len v by FINSEQ_1:5;
      then
A3:   Seg n c= dom v by FINSEQ_1:def 3;
      then Seg n = dom v /\ Seg n by XBOOLE_1:28;
      then
A4:   dom v1 = Seg n by A2,RELAT_1:61;
      now
        let m,k such that
A5:     m in dom v1 & k in dom v1 and
A6:     m<k;
        set r=v1.m, s=v1.k;
        r=v.m & s=v.k by A2,A5,FUNCT_1:47;
        hence r<s by A1,A3,A4,A5,A6,SEQM_3:def 1;
      end;
      hence thesis by SEQM_3:def 1;
    end;
    suppose
      len v<=n;
      hence thesis by A1,A2,FINSEQ_2:20;
    end;
  end;
  hence thesis;
end;

defpred P1[Element of NAT] means for v st card rng v = $1 holds ex v1 st rng
v1 = rng v & len v1 = card rng v & v1 is increasing;

Lm10: P1[0]
proof
  let v;
  assume card rng v = 0;
  then
A1: rng v = {};
  take v1 = v;
  thus rng v1 = rng v;
  thus len v1 = card rng v by A1,RELAT_1:41;
  for n,m holds n in dom v1 & m in dom v1 & n<m implies v1.n < v1.m by A1,
RELAT_1:38,41;
  hence thesis by SEQM_3:def 1;
end;

Lm11: for n st P1[n] holds P1[n+1]
proof
  let n such that
A1: P1[n];
  let v such that
A2: card rng v = n+1;
  now
    reconsider R = rng v as finite Subset of REAL;
    set u = upper_bound(R), X = R \ {u};
    set f = X|`v;
    Seq f=f*Sgm(dom f) & rng Sgm(dom f)=dom f by FINSEQ_1:50;
    then
A3: rng Seq f = rng f by RELAT_1:28
      .= X by RELAT_1:89,XBOOLE_1:36;
    then reconsider f1 = Seq f as FinSequence of REAL by FINSEQ_1:def 4;
    reconsider X as finite set;
    upper_bound(R) in R by A2,Th133,CARD_1:27;
    then
A4: {u} c= R by ZFMISC_1:31;
    then
A5: card X = card R - card {u} by CARD_2:44;
A6: card {u}=1 by CARD_1:30;
    then consider v2 such that
A7: rng v2 = rng f1 and
A8: len v2 = card rng f1 and
A9: v2 is increasing by A1,A2,A3,A5;
    take v1 = v2 ^ <* u *>;
    thus rng v1 = X \/ rng <*u*> by A3,A7,FINSEQ_1:31
      .= (rng v \ {u}) \/ {u} by FINSEQ_1:39
      .= rng v \/ {u} by XBOOLE_1:39
      .= rng v by A4,XBOOLE_1:12;
    thus
A10: len v1 = n+ len <*u*> by A2,A3,A5,A6,A8,FINSEQ_1:22
      .= card rng v by A2,FINSEQ_1:39;
    now
      let k,m;
      assume that
A11:  k in dom v1 and
A12:  m in dom v1 and
A13:  k<m;
A14:  1<=m by A12,FINSEQ_3:25;
A15:  m<=len v1 by A12,FINSEQ_3:25;
      set r=v1.k, s=v1.m;
A16:  1<=k by A11,FINSEQ_3:25;
A17:  k<=len v1 by A11,FINSEQ_3:25;
      now
        per cases;
        suppose
A18:      m=len v1;
          k<len v1 by A13,A15,XXREAL_0:2;
          then k<=len v2 by A2,A3,A5,A6,A8,A10,NAT_1:13;
          then
A19:      k in dom v2 by A16,FINSEQ_3:25;
          then r=v2.k by FINSEQ_1:def 7;
          then
A20:      r in rng v2 by A19,FUNCT_1:def 3;
          then r in R by A3,A7,XBOOLE_0:def 5;
          then
A21:      r<=u by Def1;
          not r in {u} by A3,A7,A20,XBOOLE_0:def 5;
          then
A22:      r<>u by TARSKI:def 1;
          s=u by A2,A3,A5,A6,A8,A10,A18,FINSEQ_1:42;
          hence r<s by A21,A22,XXREAL_0:1;
        end;
        suppose
          m <>len v1;
          then m<len v1 by A15,XXREAL_0:1;
          then m<=len v2 by A2,A3,A5,A6,A8,A10,NAT_1:13;
          then
A23:      m in dom v2 by A14,FINSEQ_3:25;
          then
A24:      s=v2.m by FINSEQ_1:def 7;
          k<len v1 by A13,A17,A15,XXREAL_0:1;
          then k<=len v2 by A2,A3,A5,A6,A8,A10,NAT_1:13;
          then
A25:      k in dom v2 by A16,FINSEQ_3:25;
          then r=v2.k by FINSEQ_1:def 7;
          hence r<s by A9,A13,A25,A23,A24,SEQM_3:def 1;
        end;
      end;
      hence r<s;
    end;
    hence v1 is increasing by SEQM_3:def 1;
  end;
  hence thesis;
end;

Lm12: for n holds P1[n] from NAT_1:sch 1(Lm10,Lm11);

theorem
  for v holds ex v1 st rng v1 = rng v & len v1 = card rng v & v1 is
  increasing by Lm12;

defpred P3[Element of NAT] means for v1,v2 st len v1=$1 & len v2=$1 & rng v1=
rng v2 & v1 is increasing & v2 is increasing holds v1=v2;

Lm13: P3[0]
proof
  let v1,v2;
  assume that
A1: len v1=0 and
A2: len v2=0 and
  rng v1=rng v2 and
  v1 is increasing and
  v2 is increasing;
  v1={} by A1;
  hence thesis by A2;
end;

Lm14: for n st P3[n] holds P3[n+1]
proof
  let n such that
A1: P3[n];
  let v1,v2 such that
A2: len v1=n+1 and
A3: len v2=n+1 and
A4: rng v1=rng v2 and
A5: v1 is increasing and
A6: v2 is increasing;
  reconsider X = rng v1 as Subset of REAL;
  set u = upper_bound(X);
  X <> {} by A2,CARD_1:27,RELAT_1:41;
  then
A7: upper_bound(X) in X by Th133;
  then consider m being Nat such that
A8: m in dom v2 and
A9: v2.m=u by A4,FINSEQ_2:10;
  reconsider m as Element of NAT by ORDINAL1:def 12;
A10: m<=len v2 by A8,FINSEQ_3:25;
A11: n<=n+1 by NAT_1:11;
  then Seg n c= Seg(n+1) by FINSEQ_1:5;
  then
A12: Seg n = Seg(n+1) /\ Seg n by XBOOLE_1:28;
  dom v2=Seg len v2 by FINSEQ_1:def 3;
  then
A13: dom (v2|Seg n) =Seg n by A3,A12,RELAT_1:61;
  dom v1 = Seg len v1 by FINSEQ_1:def 3;
  then
A14: dom (v1|Seg n) = Seg n by A2,A12,RELAT_1:61;
  then reconsider f1 = v1|Seg n, f2 = v2|Seg n as FinSequence by A13,
FINSEQ_1:def 2;
  rng f2 c= rng v2 by RELAT_1:70; then
A15: rng f2 c= REAL by XBOOLE_1:1;
  rng f1 c= rng v1 by RELAT_1:70;
  then rng f1 c= REAL by XBOOLE_1:1;
  then reconsider f1, f2 as FinSequence of REAL by A15,FINSEQ_1:def 4;
  consider k being Nat such that
A16: k in dom v1 and
A17: v1.k=u by A7,FINSEQ_2:10;
  reconsider k as Element of NAT by ORDINAL1:def 12;
A18: k<=len v1 by A16,FINSEQ_3:25;
A19: 1<=k by A16,FINSEQ_3:25;
A20: now
    assume k<>len v1;
    then k<len v1 by A18,XXREAL_0:1;
    then
A21: k+1<=len v1 by NAT_1:13;
    reconsider s=v1.(k+1) as Real;
A22: k<k+1 by NAT_1:13;
    1<=k+1 by A19,NAT_1:13;
    then
A23: k+1 in dom v1 by A21,FINSEQ_3:25;
    then v1.(k+1) in rng v1 by FUNCT_1:def 3;
    then s<=u by Def1;
    hence contradiction by A5,A16,A17,A23,A22,SEQM_3:def 1;
  end;
A24: 1<=m by A8,FINSEQ_3:25;
A25: now
    assume m<>len v2;
    then m<len v2 by A10,XXREAL_0:1;
    then
A26: m+1<=len v2 by NAT_1:13;
    reconsider s=v2.(m+1) as Real;
A27: m<m+1 by NAT_1:13;
    1<=m+1 by A24,NAT_1:13;
    then
A28: m+1 in dom v2 by A26,FINSEQ_3:25;
    then v2.(m+1) in rng v2 by FUNCT_1:def 3;
    then s<=u by A4,Def1;
    hence contradiction by A6,A8,A9,A28,A27,SEQM_3:def 1;
  end;
A29: len f1 = n by A14,FINSEQ_1:def 3;
  then
A30: dom f1 c= dom v1 by A2,A11,FINSEQ_3:30;
A31: rng f1 = rng v1 \ {u}
  proof
    thus rng f1 c= rng v1 \ {u}
    proof
      let x;
      assume x in rng f1;
      then consider i being Nat such that
A32:  i in dom f1 and
A33:  f1.i=x by FINSEQ_2:10;
A34:  x=v1.i by A32,A33,FUNCT_1:47;
A35:  v1.i in rng v1 by A30,A32,FUNCT_1:def 3;
      i<=n by A14,A32,FINSEQ_1:1;
      then i<>n+1 by NAT_1:13;
      then x<>u by A2,A5,A16,A17,A20,A30,A32,A34,Th138;
      then not x in {u} by TARSKI:def 1;
      hence thesis by A34,A35,XBOOLE_0:def 5;
    end;
    let x;
    assume
A36: x in rng v1 \ {u};
    then x in rng v1 by XBOOLE_0:def 5;
    then consider i being Nat such that
A37: i in dom v1 and
A38: v1.i=x by FINSEQ_2:10;
A39: i<=len v1 by A37,FINSEQ_3:25;
    not x in {u} by A36,XBOOLE_0:def 5;
    then i <> len v1 by A17,A20,A38,TARSKI:def 1;
    then i<len v1 by A39,XXREAL_0:1;
    then
A40: i<=len f1 by A2,A29,NAT_1:13;
    1<=i by A37,FINSEQ_3:25;
    then
A41: i in dom f1 by A40,FINSEQ_3:25;
    then f1.i in rng f1 by FUNCT_1:def 3;
    hence thesis by A38,A41,FUNCT_1:47;
  end;
A42: dom v1 = Seg len v1 by FINSEQ_1:def 3;
A43: now
    let i be Nat;
    assume
A44: i in dom v1;
    then
A45: 1<=i by A42,FINSEQ_1:1;
A46: i<=len f1+1 by A2,A29,A42,A44,FINSEQ_1:1;
    now
      per cases;
      suppose
        i=len f1+1;
        hence (f1^<*u*>).i=v1.i by A2,A17,A20,A29,FINSEQ_1:42;
      end;
      suppose
        i<>len f1+1;
        then i < len f1+1 by A46,XXREAL_0:1;
        then i<=len f1 by NAT_1:13;
        then
A47:    i in dom f1 by A14,A29,A45,FINSEQ_1:1;
        hence (f1^<*u*>).i=f1.i by FINSEQ_1:def 7
          .= v1.i by A47,FUNCT_1:47;
      end;
    end;
    hence (f1^<*u*>).i=v1.i;
  end;
  len (f1^<*u*>) = n + len <*u*> by A29,FINSEQ_1:22
    .= len v1 by A2,FINSEQ_1:39;
  then
A48: v1=f1^<*u*> by A43,FINSEQ_2:9;
A49: len f2 = n by A13,FINSEQ_1:def 3;
  then
A50: len (f2^<*u*>) = n + len <*u*> by FINSEQ_1:22
    .= len v2 by A3,FINSEQ_1:39;
A51: dom f2 c= dom v2 by A3,A11,A49,FINSEQ_3:30;
A52: rng f2 = rng v2 \ {u}
  proof
    thus rng f2 c= rng v2 \ {u}
    proof
      let x;
      assume x in rng f2;
      then consider i being Nat such that
A53:  i in dom f2 and
A54:  f2.i=x by FINSEQ_2:10;
A55:  x=v2.i by A53,A54,FUNCT_1:47;
A56:  v2.i in rng v2 by A51,A53,FUNCT_1:def 3;
      i<=n by A13,A53,FINSEQ_1:1;
      then i<>n+1 by NAT_1:13;
      then x<>u by A3,A6,A8,A9,A25,A51,A53,A55,Th138;
      then not x in {u} by TARSKI:def 1;
      hence thesis by A55,A56,XBOOLE_0:def 5;
    end;
    let x;
    assume
A57: x in rng v2 \ {u};
    then x in rng v2 by XBOOLE_0:def 5;
    then consider i being Nat such that
A58: i in dom v2 and
A59: v2.i=x by FINSEQ_2:10;
A60: i<=len v2 by A58,FINSEQ_3:25;
    not x in {u} by A57,XBOOLE_0:def 5;
    then i <> len v2 by A9,A25,A59,TARSKI:def 1;
    then i<len v2 by A60,XXREAL_0:1;
    then
A61: i<=len f2 by A3,A49,NAT_1:13;
    1<=i by A58,FINSEQ_3:25;
    then
A62: i in dom f2 by A61,FINSEQ_3:25;
    then f2.i in rng f2 by FUNCT_1:def 3;
    hence thesis by A59,A62,FUNCT_1:47;
  end;
A63: dom v2 = Seg len v2 by FINSEQ_1:def 3;
A64: now
    let i be Nat;
    assume
A65: i in dom v2;
    then
A66: 1<=i by A63,FINSEQ_1:1;
A67: i<=len f2+1 by A3,A49,A63,A65,FINSEQ_1:1;
    now
      per cases;
      suppose
        i=len f2+1;
        hence (f2^<*u*>).i=v2.i by A3,A9,A25,A49,FINSEQ_1:42;
      end;
      suppose
        i<>len f2+1;
        then i < len f2+1 by A67,XXREAL_0:1;
        then i<=len f2 by NAT_1:13;
        then
A68:    i in dom f2 by A13,A49,A66,FINSEQ_1:1;
        hence (f2^<*u*>).i=f2.i by FINSEQ_1:def 7
          .= v2.i by A68,FUNCT_1:47;
      end;
    end;
    hence (f2^<*u*>).i=v2.i;
  end;
  f1 is increasing & f2 is increasing by A5,A6,Th139;
  then f1=f2 by A1,A4,A29,A49,A31,A52;
  hence thesis by A48,A50,A64,FINSEQ_2:9;
end;

Lm15: for n holds P3[n] from NAT_1:sch 1(Lm13,Lm14);

theorem
  for v1,v2 st len v1 = len v2 & rng v1 = rng v2 & v1 is increasing & v2
  is increasing holds v1 = v2 by Lm15;

definition
  let v;
  func Incr(v) ->increasing FinSequence of REAL means
  :Def21:
  rng it = rng v & len it = card rng v;
  existence
  proof
    consider v1 such that
A1: rng v1 = rng v & len v1 = card rng v and
A2: v1 is increasing by Lm12;
    reconsider v1 as increasing FinSequence of REAL by A2;
    take v1;
    thus thesis by A1;
  end;
  uniqueness by Lm15;
end;

registration
  let v be non empty FinSequence of REAL;
  cluster Incr v -> non empty;
  coherence
proof
A1: len Incr(v)=card rng v by Def21;
  assume Incr(v) is empty;
  then rng v = {} by A1;
  hence contradiction;
end;
end;



registration
  cluster non empty bounded_above bounded_below for Subset of REAL;
  existence
  proof
    reconsider A = {0} as Subset of REAL;
    take A;
    thus A is non empty;
    thus A is bounded_above;
    take 0;
    let t be ext-real number;
    assume t in A;
    hence thesis;
  end;
end;

theorem
  for A,B being non empty bounded_below Subset of REAL holds lower_bound(A
  \/ B) = min(lower_bound A,lower_bound B)
proof
  let A,B be non empty bounded_below Subset of REAL;
  set r = min(lower_bound A,lower_bound B);
A1: r <= lower_bound B by XXREAL_0:17;
A2: for r1 being real number st for t being real number st t in A \/ B holds
  t >= r1 holds r >= r1
  proof
    let r1 be real number;
    assume
A3: for t being real number st t in A \/ B holds t >= r1;
    now
      let t be real number;
      assume t in B;
      then t in A \/ B by XBOOLE_0:def 3;
      hence t >= r1 by A3;
    end;
    then
A4: lower_bound B >= r1 by Th43;
    now
      let t be real number;
      assume t in A;
      then t in A \/ B by XBOOLE_0:def 3;
      hence t >= r1 by A3;
    end;
    then lower_bound A >= r1 by Th43;
    hence thesis by A4,XXREAL_0:20;
  end;
A5: r <= lower_bound A by XXREAL_0:17;
  for t being real number st t in A \/ B holds t >= r
  proof
    let t be real number;
    assume t in A \/ B;
    then t in A or t in B by XBOOLE_0:def 3;
    then lower_bound A <= t or lower_bound B <= t by Def2;
    hence thesis by A5,A1,XXREAL_0:2;
  end;
  hence thesis by A2,Th44;
end;

theorem
  for A,B being non empty bounded_above Subset of REAL holds upper_bound(A
  \/ B) = max(upper_bound A,upper_bound B)
proof
  let A,B be non empty bounded_above Subset of REAL;
  set r = max(upper_bound A,upper_bound B);
A1: r >= upper_bound B by XXREAL_0:25;
A2: for r1 being real number st for t being real number st t in A \/ B holds
  t <= r1 holds r <= r1
  proof
    let r1 be real number;
    assume
A3: for t being real number st t in A \/ B holds t <= r1;
    now
      let t be real number;
      assume t in B;
      then t in A \/ B by XBOOLE_0:def 3;
      hence t <= r1 by A3;
    end;
    then
A4: upper_bound B <= r1 by Th45;
    now
      let t be real number;
      assume t in A;
      then t in A \/ B by XBOOLE_0:def 3;
      hence t <= r1 by A3;
    end;
    then upper_bound A <= r1 by Th45;
    hence thesis by A4,XXREAL_0:28;
  end;
A5: r >= upper_bound A by XXREAL_0:25;
  for t being real number st t in A \/ B holds t <= r
  proof
    let t be real number;
    assume t in A \/ B;
    then t in A or t in B by XBOOLE_0:def 3;
    then upper_bound A >= t or upper_bound B >= t by Def1;
    hence thesis by A5,A1,XXREAL_0:2;
  end;
  hence thesis by A2,Th46;
end;



theorem
  for R being non empty Subset of REAL,r0 being real number st for
  r being real number st r in R holds r <= r0 holds upper_bound R <= r0
proof
  let R be non empty Subset of REAL,r0 be real number;
  assume
A1: for r being real number st r in R holds r<=r0;
  then for r being ext-real number st r in R holds r<=r0;
  then r0 is UpperBound of R by XXREAL_2:def 1;
  then
A2: R is bounded_above by XXREAL_2:def 10;
  now
    set r1=(upper_bound R) -r0;
    assume upper_bound R >r0;
    then r1>0 by XREAL_1:50;
    then ex r being real number st r in R & (upper_bound R)-r1<r
     by A2,Def1;
    hence contradiction by A1;
  end;
  hence thesis;
end;















begin

reserve n,n1,m,k for Element of NAT;
reserve x,y for set;
reserve s,g,g1,g2,r,p,q,t for real number;
reserve s1,s2,s3 for Real_Sequence;
reserve Nseq for increasing sequence of NAT;
reserve X for Subset of REAL;

definition
  let g,s be real number;
  redefine func [. g,s .] -> Subset of REAL equals
  {r where r is Real: g<=r &
  r<=s };
  coherence
  proof
    now
      let x;
      assume x in [. g,s .];
      then
A1:   ex r being Element of ExtREAL st x = r & g <= r & r<=s;
      g in REAL & s in REAL by XREAL_0:def 1;
      hence x in REAL by A1,XXREAL_0:45;
    end;
    hence thesis by TARSKI:def 3;
  end;
  compatibility
  proof
    set Y = {r where r is Real: g<=r & r<=s };
    let X be Subset of REAL;
A2: [. g,s .] c= Y
    proof
      let x;
      assume x in [. g,s .];
      then consider r being Element of ExtREAL such that
A3:   x = r and
A4:   g <= r & r<=s;
      g in REAL & s in REAL by XREAL_0:def 1;
      then r in REAL by A4,XXREAL_0:45;
      hence thesis by A3,A4;
    end;
    Y c= [. g,s .]
    proof
      let x;
      assume x in Y;
      then consider r being Element of REAL such that
A5:   x = r & g <= r & r<=s;
      r in REAL;
      hence thesis by A5,NUMBERS:31;
    end;
    hence thesis by A2,XBOOLE_0:def 10;
  end;
end;

definition
  let g,s be ext-real number;
  redefine func ]. g,s .[ -> Subset of REAL equals
  {r where r is Real : g<r &
  r<s };
  coherence
  proof
    now
      let x;
      assume x in ]. g,s .[;
      then ex r being Element of ExtREAL st x = r & g < r & r<s;
      hence x in REAL by XXREAL_0:48;
    end;
    hence thesis by TARSKI:def 3;
  end;
  compatibility
  proof
    set Y = {r where r is Real: g<r & r<s };
    let X be Subset of REAL;
A1: Y c= ]. g,s .[
    proof
      let x;
      assume x in Y;
      then consider r being Element of REAL such that
A2:   x = r & g < r & r<s;
      r in REAL;
      hence thesis by A2,NUMBERS:31;
    end;
    ]. g,s .[ c= Y
    proof
      let x;
      assume x in ]. g,s .[;
      then consider r being Element of ExtREAL such that
A3:   x = r and
A4:   g < r & r<s;
      r in REAL by A4,XXREAL_0:48;
      hence thesis by A3,A4;
    end;
    hence thesis by A1,XBOOLE_0:def 10;
  end;
end;

theorem Th1:
  r in ].p-g,p+g.[ iff abs(r-p)<g
proof
  thus r in ].p-g,p+g.[ implies abs(r-p)<g
  proof
    assume r in ].p-g,p+g.[;
    then
A1: ex s be Real st r=s & p-g<s & s<p+g;
    then p+-g<r;
    then
A2: -g<r-p by XREAL_1:20;
    r-p<g by A1,XREAL_1:19;
    hence thesis by A2,SEQ_2:1;
  end;
  assume
A3: abs(r-p)<g;
  then r-p<g by SEQ_2:1;
  then
A4: r<p+g by XREAL_1:19;
  -g<r-p by A3,SEQ_2:1;
  then r is Real & p+-g<r by XREAL_0:def 1,XREAL_1:20;
  hence thesis by A4;
end;

theorem
  r in [.p,g.] iff abs(p+g-2*r)<=g-p
proof
  thus r in [.p,g.] implies abs(p+g-2*r)<=g-p
  proof
    assume r in [.p,g.];
    then
A1: ex s be Real st s=r & p<=s & s<=g;
    then 2*r<=2*g by XREAL_1:64;
    then -2*r>=-2*g by XREAL_1:24;
    then (p+g)+-2*r>=(p+g)+-2*g by XREAL_1:7;
    then
A2: p+g-2*r>=-(g-p);
    2*p<=2*r by A1,XREAL_1:64;
    then -2*p>=-2*r by XREAL_1:24;
    then (p+g)+-2*p>=(p+g)+-2*r by XREAL_1:7;
    hence thesis by A2,ABSVALUE:5;
  end;
  assume
A3: abs(p+g-2*r)<=g-p;
  then p+g-2*r>=-(g-p) by ABSVALUE:5;
  then p+g>=p-g+2*r by XREAL_1:19;
  then p+g-(p-g)>=2*r by XREAL_1:19;
  then
A4: 1/2*(2*g)>=1/2*(2*r) by XREAL_1:64;
  g-p>=p+g-2*r by A3,ABSVALUE:5;
  then 2*r+(g-p)>=p+g by XREAL_1:20;
  then 2*r>=p+g-(g-p) by XREAL_1:20;
  then 1/2*(2*r)>=1/2*(2*p) by XREAL_1:64;
  hence thesis by A4;
end;

theorem
  r in ].p,g.[ iff abs(p+g-2*r)<g-p
proof
  thus r in ].p,g.[ implies abs(p+g-2*r)<g-p
  proof
    assume r in ].p,g.[;
    then
A1: ex s be Real st s=r & p<s & s<g;
    then 2*r<2*g by XREAL_1:68;
    then -2*r>-2*g by XREAL_1:24;
    then (p+g)+-2*r>(p+g)+-2*g by XREAL_1:6;
    then
A2: p+g-2*r>-(g-p);
    2*p<2*r by A1,XREAL_1:68;
    then -2*p>-2*r by XREAL_1:24;
    then (p+g)+-2*p>(p+g)+-2*r by XREAL_1:6;
    hence thesis by A2,SEQ_2:1;
  end;
  assume
A3: abs(p+g-2*r)<g-p;
  then p+g-2*r>-(g-p) by SEQ_2:1;
  then p+g>p-g+2*r by XREAL_1:20;
  then p+g-(p-g)>2*r by XREAL_1:20;
  then
A4: 1/2*(2*g)>1/2*(2*r) by XREAL_1:68;
  g-p>p+g-2*r by A3,SEQ_2:1;
  then 2*r+(g-p)>p+g by XREAL_1:19;
  then 2*r>p+g-(g-p) by XREAL_1:19;
  then 1 /2*(2*r)>1/2*(2*p) by XREAL_1:68;
  hence thesis by A4;
end;

definition
  let X;
  attr X is compact means
  :Def3:
  for s1 st rng s1 c= X ex s2 st s2 is
  subsequence of s1 & s2 is convergent & lim s2 in X;
end;

definition
  let X;
  attr X is closed means
  :Def4:
  for s1 st rng s1 c= X & s1 is convergent holds lim s1 in X;
end;

definition
  let X;
  attr X is open means
  :Def5:
  X` is closed;
end;

theorem Th4:
  for s1 st rng s1 c= [.s,g.] holds s1 is bounded
proof
  let s1 such that
A1: rng s1 c= [.s,g.];
  thus s1 is bounded_above
  proof
    take r = g + 1;
A2: for t st t in rng s1 holds t < r
    proof
      let t;
      assume t in rng s1;
      then t in [.s,g.] by A1;
      then
A3:   ex p be Real st t = p & s <= p & p<=g;
      g < r by XREAL_1:29;
      hence thesis by A3,XXREAL_0:2;
    end;
    for n holds s1.n < r
    proof
      let n;
      n in NAT;
      then n in dom s1 by FUNCT_2:def 1;
      then s1.n in rng s1 by FUNCT_1:def 3;
      hence thesis by A2;
    end;
    hence thesis;
  end;
  thus s1 is bounded_below
  proof
    take r = s - 1;
A4: r + 1 = s - (1-1);
A5: for t st t in rng s1 holds r < t
    proof
      let t;
      assume t in rng s1;
      then t in [.s,g.] by A1;
      then
A6:   ex p be Real st t = p & s <= p & p<=g;
      r < s by A4,XREAL_1:29;
      hence thesis by A6,XXREAL_0:2;
    end;
    for n holds r < s1.n
    proof
      let n;
      n in NAT;
      then n in dom s1 by FUNCT_2:def 1;
      then s1.n in rng s1 by FUNCT_1:def 3;
      hence thesis by A5;
    end;
    hence thesis;
  end;
end;

theorem Th5:
  [.s,g.] is closed
proof
  for s1 st rng s1 c= [.s,g.] & s1 is convergent holds lim s1 in [.s,g.]
  proof
    let s1;
    assume that
A1: rng s1 c= [.s,g.] and
A2: s1 is convergent;
A3: s <= lim s1
    proof
      s in REAL by XREAL_0:def 1;
      then reconsider s2 = NAT --> s as Real_Sequence by FUNCOP_1:45;
A4:   now
        let n;
        n in NAT;
        then n in dom s1 by FUNCT_2:def 1;
        then s1.n in rng s1 by FUNCT_1:def 3;
        then s1.n in [.s,g.] by A1;
        then ex p be Real st s1.n = p & s <= p & p<=g;
        hence s2.n<=s1.n by FUNCOP_1:7;
      end;
      s2.0 = s by FUNCOP_1:7;
      then lim s2 = s by SEQ_4:25;
      hence thesis by A2,A4,SEQ_2:18;
    end;
    lim s1 <= g
    proof
      g in REAL by XREAL_0:def 1;
      then reconsider s2 = NAT --> g as Real_Sequence by FUNCOP_1:45;
A5:   now
        let n;
        n in NAT;
        then n in dom s1 by FUNCT_2:def 1;
        then s1.n in rng s1 by FUNCT_1:def 3;
        then s1.n in [.s,g.] by A1;
        then ex p be Real st s1.n = p & s <= p & p<=g;
        hence s1.n<=s2.n by FUNCOP_1:7;
      end;
      s2.0 = g by FUNCOP_1:7;
      then lim s2 = g by SEQ_4:25;
      hence thesis by A2,A5,SEQ_2:18;
    end;
    hence thesis by A3;
  end;
  hence thesis by Def4;
end;

theorem
  [.s,g.] is compact
proof
  for s1 st rng s1 c= [.s,g.] ex s2 st s2 is subsequence of s1 & s2 is
  convergent & lim s2 in [.s,g.]
  proof
    let s1;
A1: [.s,g.] is closed by Th5;
    assume
A2: (rng s1) c= [.s,g.];
    then consider s2 be Real_Sequence such that
A3: s2 is subsequence of s1 and
A4: s2 is convergent by Th4,SEQ_4:40;
    take s2;
    ex Nseq st s2 = s1*Nseq by A3,VALUED_0:def 17;
    then rng s2 c= rng s1 by RELAT_1:26;
    then rng s2 c= [.s,g.] by A2,XBOOLE_1:1;
    hence thesis by A3,A4,A1,Def4;
  end;
  hence thesis by Def3;
end;

theorem Th7:
  ].p,q.[ is open
proof
  defpred P[Real] means $1<=p or q<=$1;
  consider X such that
A1: for r be Real holds r in X iff P[r] from SUBSET_1:sch 3;
  now
    let s1 such that
A2: (rng s1) c= X and
A3: s1 is convergent;
    lim s1<=p or q<=lim s1
    proof
      assume
A4:   not thesis;
      then 0<lim s1 - p by XREAL_1:50;
      then consider n1 such that
A5:   for m st n1<=m holds abs(s1.m-lim s1)<lim s1-p by A3,SEQ_2:def 7;
      0<q - lim s1 by A4,XREAL_1:50;
      then consider n such that
A6:   for m st n<=m holds abs(s1.m-lim s1)<q-lim s1 by A3,SEQ_2:def 7;
      consider k such that
A7:   max(n,n1)<k by SEQ_4:3;
      n1<=max(n,n1) by XXREAL_0:25;
      then n1<=k by A7,XXREAL_0:2;
      then
A8:   abs(s1.k-lim s1)<lim s1 - p by A5;
      -abs(s1.k-lim s1) <= s1.k-lim s1 by ABSVALUE:4;
      then -(s1.k-lim s1) <= -(-abs(s1.k-lim s1)) by XREAL_1:24;
      then -(s1.k-lim s1)<lim s1 - p by A8,XXREAL_0:2;
      then -(lim s1 - p)< -(-(s1.k-lim s1)) by XREAL_1:24;
      then p - lim s1 < s1.k-lim s1;
      then
A9:   p < s1.k by XREAL_1:9;
      n<=max(n,n1) by XXREAL_0:25;
      then n<=k by A7,XXREAL_0:2;
      then
      s1.k-lim s1 <= abs(s1.k-lim s1) & abs(s1.k-lim s1)<q-lim s1 by A6,
ABSVALUE:4;
      then s1.k-lim s1 < q-lim s1 by XXREAL_0:2;
      then
A10:  s1.k < q by XREAL_1:9;
      dom s1 = NAT by FUNCT_2:def 1;
      then s1.k in rng s1 by FUNCT_1:def 3;
      hence contradiction by A1,A2,A10,A9;
    end;
    hence lim s1 in X by A1;
  end;
  then
A11: X is closed by Def4;
  now
    let r;
A12: r is Real by XREAL_0:def 1;
    assume r in ].p,q.[`;
    then not r in ].p,q.[ by XBOOLE_0:def 5;
    then r<=p or q<=r by A12;
    hence r in X by A1,A12;
  end;
  then
A13: ].p,q.[` c= X by MEMBERED:def 9;
  now
    let r;
    assume r in X;
    then not ex s be Real st s=r & p<s & s<q by A1;
    then r is Real & not r in ].p,q.[ by XREAL_0:def 1;
    hence r in ].p,q.[` by XBOOLE_0:def 5;
  end;
  then X c= ].p,q.[` by MEMBERED:def 9;
  then ].p,q.[` = X by A13,XBOOLE_0:def 10;
  hence thesis by A11,Def5;
end;

registration
  let p, q be real number;
  cluster ].p,q.[ -> open for Subset of REAL;
  coherence by Th7;
  cluster [.p,q.] -> closed for Subset of REAL;
  coherence by Th5;
end;

theorem Th8:
  X is compact implies X is closed
proof
  assume
A1: X is compact;
  assume not X is closed;
  then consider s1 such that
A2: rng s1 c= X and
A3: s1 is convergent & not lim s1 in X by Def4;
  ex s2 st s2 is subsequence of s1 & s2 is convergent & (lim s2) in X by A1,A2
,Def3;
  hence contradiction by A3,SEQ_4:17;
end;

registration
  cluster compact -> closed for Subset of REAL;
  coherence by Th8;
end;

theorem Th9:
  (for p st p in X ex r,n st 0<r & (for m st n<m holds r<abs(s1.m
- p))) implies for s2 st s2 is subsequence of s1 holds not (s2 is convergent &
  lim s2 in X)
proof
  assume that
A1: for p st p in X ex r,n st 0<r & for m st n<m holds r<abs(s1.m - p) and
A2: not for s2 st s2 is subsequence of s1 holds not (s2 is convergent &
  lim s2 in X);
  consider s2 such that
A3: s2 is subsequence of s1 and
A4: s2 is convergent and
A5: lim s2 in X by A2;
  consider r,n such that
A6: 0<r and
A7: for m st n<m holds r<abs(s1.m - lim s2) by A1,A5;
  consider n1 such that
A8: for m st n1<=m holds abs(s2.m-(lim s2))<r by A4,A6,SEQ_2:def 7;
  consider NS being increasing sequence of NAT such that
A9: s2 = s1*NS by A3,VALUED_0:def 17;
  set k = n + 1 + n1;
  abs((s1*NS).k-(lim s2))<r by A8,A9,NAT_1:11;
  then
A10: abs(s1.(NS.k)-(lim s2))<r by FUNCT_2:15;
  n + 1 <= k by NAT_1:11;
  then
A11: n < k by NAT_1:13;
  k<=NS.k by SEQM_3:14;
  then n<NS.k by A11,XXREAL_0:2;
  hence contradiction by A7,A10;
end;

theorem Th10:
  X is compact implies X is real-bounded
proof
  assume
A1: X is compact;
  assume
A2: not X is real-bounded;
    per cases by A2;
    suppose
A3:   not X is bounded_above;
      defpred P[Element of NAT,Element of REAL] means ex q st q =$2 & q in X &
      $1<q;
A4:   for n being Element of NAT ex p being Element of REAL st P[n,p]
      proof
        let n be Element of NAT;
        n is not UpperBound of X by A3,XXREAL_2:def 10;
        then consider t being ext-real number such that
A5:     t in X & n<t by XXREAL_2:def 1;
        take t;
        thus thesis by A5;
      end;
      consider f being Function of NAT,REAL such that
A6:   for n being Element of NAT holds P[n,f.n] from FUNCT_2:sch 3(A4);
A7:   now
        let n;
        ex q st q=f.n & q in X & n<q by A6;
        hence f.n in X & n<f.n;
      end;
A8:   for p st p in X ex r,n st 0<r & for m st n<m holds r<abs(f.m - p)
      proof
        let p such that
        p in X;
        consider q such that
A9:     q = 1;
        take r = q;
        consider k such that
A10:    p+1<k by SEQ_4:3;
        take n = k;
        thus 0 < r by A9;
        let m;
        assume n<m;
        then p+1 < m by A10,XXREAL_0:2;
        then p+1 < f.m by A7,XXREAL_0:2;
        then 1 < f.m - p by XREAL_1:20;
        hence thesis by A9,ABSVALUE:def 1;
      end;
      rng f c= X
      proof
        let x;
        assume x in rng f;
        then consider y such that
A11:    y in dom f and
A12:    x = f.y by FUNCT_1:def 3;
        reconsider y as Element of NAT by A11,FUNCT_2:def 1;
        f.y in X by A7;
        hence thesis by A12;
      end;
      then
      ex s2 st s2 is subsequence of f & s2 is convergent & lim s2 in X by A1
,Def3;
      hence contradiction by A8,Th9;
    end;
    suppose
A13:  not X is bounded_below;
      defpred P[Element of NAT,Element of REAL] means ex q st q=$2 & q in X &
      q<-$1;
A14:  for n being Element of NAT ex p being Element of REAL st P[n,p]
      proof
        let n be Element of NAT;
        -n is not LowerBound of X by A13,XXREAL_2:def 9;
        then consider t being ext-real number such that
A15:    t in X & t<-n by XXREAL_2:def 2;
        take t;
        thus thesis by A15;
      end;
      consider f being Function of NAT,REAL such that
A16:  for n being Element of NAT holds P[n,f.n] from FUNCT_2:sch 3(
      A14);
A17:  now
        let n;
        ex q st q=f.n & q in X & q<-n by A16;
        hence f.n in X & f.n<-n;
      end;
A18:  for p st p in X ex r,n st 0<r & for m st n<m holds r<abs(f.m - p)
      proof
        let p such that
        p in X;
        consider q such that
A19:    q = 1;
        take r = q;
        consider k such that
A20:    abs(p-1) <k by SEQ_4:3;
        take n = k;
        thus 0 < r by A19;
        let m;
        assume n<m;
        then
A21:    -m<-n by XREAL_1:24;
        -k <p-1 by A20,SEQ_2:1;
        then -m < p-1 by A21,XXREAL_0:2;
        then f.m < p-1 by A17,XXREAL_0:2;
        then f.m +1 < p by XREAL_1:20;
        then 1 < p - f.m by XREAL_1:20;
        then r < abs( -(f.m-p) ) by A19,ABSVALUE:def 1;
        hence thesis by COMPLEX1:52;
      end;
      rng f c= X
      proof
        let x;
        assume x in rng f;
        then consider y such that
A22:    y in dom f and
A23:    x = f.y by FUNCT_1:def 3;
        reconsider y as Element of NAT by A22,FUNCT_2:def 1;
        f.y in X by A17;
        hence thesis by A23;
      end;
      then
      ex s2 st s2 is subsequence of f & s2 is convergent & lim s2 in X by A1
,Def3;
      hence contradiction by A18,Th9;
    end;
end;

theorem
  X is real-bounded closed implies X is compact
proof
  assume that
A1: X is real-bounded and
A2: X is closed;
  now
    let s1 such that
A3: rng s1 c= X;
A4: s1 is bounded_below
    proof
      consider p such that
A5:    p is LowerBound of X by A1,XXREAL_2:def 9;
A6:   for q st q in X holds p<=q by A5,XXREAL_2:def 2;
      take r = p - 1;
A7:   r + 1 = p - (1-1);
A8:   for t st t in rng s1 holds r < t
      proof
        let t;
        assume t in rng s1;
        then
A9:     p<=t by A3,A6;
        r < p by A7,XREAL_1:29;
        hence thesis by A9,XXREAL_0:2;
      end;
      for n holds r < s1.n
      proof
        let n;
        n in NAT;
        then n in dom s1 by FUNCT_2:def 1;
        then s1.n in rng s1 by FUNCT_1:def 3;
        hence thesis by A8;
      end;
      hence thesis;
    end;
    s1 is bounded_above
    proof
      consider p such that
A10:    p is UpperBound of X by A1,XXREAL_2:def 10;
A11:   for q st q in X holds q<=p by A10,XXREAL_2:def 1;
      take r = p+1;
A12:  for t st t in rng s1 holds t < r
      proof
        let t;
        assume t in rng s1;
        then
A13:    t<=p by A3,A11;
        p < r by XREAL_1:29;
        hence thesis by A13,XXREAL_0:2;
      end;
      for n holds s1.n < r
      proof
        let n;
        n in NAT;
        then n in dom s1 by FUNCT_2:def 1;
        then s1.n in rng s1 by FUNCT_1:def 3;
        hence thesis by A12;
      end;
      hence thesis;
    end;
    then s1 is bounded by A4,SEQ_2:def 8;
    then consider s2 be Real_Sequence such that
A14: s2 is subsequence of s1 and
A15: s2 is convergent by SEQ_4:40;
    ex Nseq st s2 = s1*Nseq by A14,VALUED_0:def 17;
    then rng s2 c= rng s1 by RELAT_1:26;
    then rng s2 c= X by A3,XBOOLE_1:1;
    then lim s2 in X by A2,A15,Def4;
    hence ex s2 st s2 is subsequence of s1 & s2 is convergent & lim s2 in X by
A14,A15;
  end;
  hence thesis by Def3;
end;

theorem Th12:
  for X st X<>{} & X is closed & X is bounded_above holds upper_bound X in X
proof
  let X such that
A1: X<>{} and
A2: X is closed and
A3: X is bounded_above and
A4: not upper_bound X in X;
  set s1=upper_bound X;
  defpred P[Element of NAT,Element of REAL] means ex q st q=$2 & q in X & s1 -
  q < 1/($1+1);
A5: for n being Element of NAT ex p being Element of REAL st P[n,p]
  proof
    let n be Element of NAT;
    0<(n+1)";
    then 0<1/(n+1) by XCMPLX_1:215;
    then consider t such that
A6: t in X and
A7: s1 -1/(n+1) <t by A1,A3,SEQ_4:def 1;
    take t;
    s1 < t + 1/(n+1) by A7,XREAL_1:19;
    then s1 -t < 1/(n+1) by XREAL_1:19;
    hence thesis by A6;
  end;
  consider f being Function of NAT,REAL such that
A8: for n being Element of NAT holds P[n,f.n] from FUNCT_2:sch 3(A5);
A9: now
    let n;
    ex q st q = f.n & q in X & s1 -q < 1/(n+1) by A8;
    hence f.n in X & s1 - f.n<1/(n+1);
  end;
A10: rng f c= X
  proof
    let x;
    assume x in rng f;
    then consider y such that
A11: y in dom f and
A12: x = f.y by FUNCT_1:def 3;
    reconsider y as Element of NAT by A11,FUNCT_2:def 1;
    f.y in X by A9;
    hence thesis by A12;
  end;
A13: now
    let s;
    assume
A14: 0<s;
    consider n such that
A15: s"<n by SEQ_4:3;
    take k=n;
    let m;
    assume k<=m;
    then k+1<=m+1 by XREAL_1:6;
    then
A16: 1/(m+1)<=1/(k+1) by XREAL_1:118;
    f.m in X by A9;
    then f.m <= s1 by A3,SEQ_4:def 1;
    then
A17: 0 <= s1-f.m by XREAL_1:48;
    s"+0 <n+1 by A15,XREAL_1:8;
    then 1/(n+1)<1/s" by A14,XREAL_1:76;
    then 1/(n+1)<s by XCMPLX_1:216;
    then 1/(m+1)<s by A16,XXREAL_0:2;
    then s1 - f.m<s by A9,XXREAL_0:2;
    then abs(-(f.m - s1) )<s by A17,ABSVALUE:def 1;
    hence abs(f.m - s1)<s by COMPLEX1:52;
  end;
  then
A18: f is convergent by SEQ_2:def 6;
  then lim f=s1 by A13,SEQ_2:def 7;
  hence contradiction by A2,A4,A18,A10,Def4;
end;

theorem Th13:
  for X st X<>{} & X is closed & X is bounded_below holds lower_bound X in X
proof
  let X such that
A1: X<>{} and
A2: X is closed and
A3: X is bounded_below and
A4: not lower_bound X in X;
  set i1=lower_bound X;
  defpred P[Element of NAT,Element of REAL] means ex q st q=$2 & q in X & q-i1
  < 1/($1+1);
A5: for n being Element of NAT ex p being Element of REAL st P[n,p]
  proof
    let n be Element of NAT;
    0<(n+1)";
    then 0<1/(n+1) by XCMPLX_1:215;
    then consider t such that
A6: t in X and
A7: t<i1 + 1/(n+1) by A1,A3,SEQ_4:def 2;
    take t;
    t-i1 < 1/(n+1) by A7,XREAL_1:19;
    hence thesis by A6;
  end;
  consider f being Function of NAT,REAL such that
A8: for n being Element of NAT holds P[n,f.n] from FUNCT_2:sch 3(A5);
A9: now
    let n;
    ex q st q = f.n & q in X & q-i1 < 1/(n+1) by A8;
    hence f.n in X & f.n-i1<1/(n+1);
  end;
A10: rng f c= X
  proof
    let x;
    assume x in rng f;
    then consider y such that
A11: y in dom f and
A12: x = f.y by FUNCT_1:def 3;
    reconsider y as Element of NAT by A11,FUNCT_2:def 1;
    f.y in X by A9;
    hence thesis by A12;
  end;
A13: now
    let s;
    assume
A14: 0<s;
    consider n such that
A15: s"<n by SEQ_4:3;
    take k=n;
    let m;
    assume k<=m;
    then k+1<=m+1 by XREAL_1:6;
    then
A16: 1/(m+1)<=1/(k+1) by XREAL_1:118;
    f.m in X by A9;
    then i1<=f.m by A3,SEQ_4:def 2;
    then
A17: 0 <= f.m-i1 by XREAL_1:48;
    s"+0 <n+1 by A15,XREAL_1:8;
    then 1/(n+1)<1/s" by A14,XREAL_1:76;
    then 1/(n+1)<s by XCMPLX_1:216;
    then 1/(m+1)<s by A16,XXREAL_0:2;
    then f.m-i1<s by A9,XXREAL_0:2;
    hence abs( f.m-i1 )<s by A17,ABSVALUE:def 1;
  end;
  then
A18: f is convergent by SEQ_2:def 6;
  then lim f=i1 by A13,SEQ_2:def 7;
  hence contradiction by A2,A4,A18,A10,Def4;
end;

theorem
  for X st X<>{} & X is compact holds upper_bound X in X & lower_bound X in X
proof
  let X such that
A1: X<>{} and
A2: X is compact;
  X is real-bounded closed by A2,Th10;
  hence thesis by A1,Th12,Th13;
end;

theorem
  X is compact & (for g1,g2 st g1 in X & g2 in X holds [.g1,g2.] c= X)
  implies ex p,g st X = [.p,g.]
proof
  assume that
A1: X is compact and
A2: for g1,g2 st g1 in X & g2 in X holds [.g1,g2.] c= X;
  per cases;
  suppose
A3: X={};
    take 1;
    take 0;
    thus thesis by A3,XXREAL_1:29;
  end;
  suppose
A4: X<>{};
    take p=lower_bound X, g=upper_bound X;
A5: X is real-bounded closed by A1,Th10;
    now
      let r be Real;
      assume r in X;
      then r<=g & p<=r by A5,SEQ_4:def 1,def 2;
      hence r in [.p,g.];
    end;
    then
A6: X c= [.p,g.] by SUBSET_1:2;
    upper_bound X in X & lower_bound X in X by A4,A5,Th12,Th13;
    then [.p,g.] c= X by A2;
    hence thesis by A6,XBOOLE_0:def 10;
  end;
end;

registration
  cluster open for Subset of REAL;
  existence
  proof
    take ].0,1.[;
    thus thesis;
  end;
end;

definition
  let r be real number;

  mode Neighbourhood of r -> Subset of REAL means
    :Def6:
    ex g st 0<g & it = ]. r-g,r+g.[;
  existence
  proof
    take ].r-1,r+1.[;
    thus thesis;
  end;
end;

registration
  let r be real number;
  cluster -> open for Neighbourhood of r;
  coherence
  proof
    let A be Neighbourhood of r;
    ex g st 0<g & A = ].r-g,r+g.[ by Def6;
    hence thesis;
  end;
end;

theorem
  for N being Neighbourhood of r holds r in N
proof
  let N be Neighbourhood of r;
  ( ex g st 0<g & N = ].r-g,r+g.[)& abs(r-r) = 0 by Def6,ABSVALUE:2;
  hence thesis by Th1;
end;

theorem
  for r for N1,N2 being Neighbourhood of r ex N being Neighbourhood of r
  st N c= N1 & N c= N2
proof
  let r;
  let N1,N2 be Neighbourhood of r;
  consider g1 such that
A1: 0<g1 and
A2: ].r-g1,r+g1.[ = N1 by Def6;
  consider g2 such that
A3: 0<g2 and
A4: ].r-g2,r+g2.[ = N2 by Def6;
  set g = min(g1,g2);
  0<g by A1,A3,XXREAL_0:15;
  then reconsider N = ].r-g,r+g.[ as Neighbourhood of r by Def6;
  take N;
A5: g<=g1 by XXREAL_0:17;
  then
A6: r+g<=r+g1 by XREAL_1:7;
  -g1<=-g by A5,XREAL_1:24;
  then
A7: r+-g1<=r+-g by XREAL_1:7;
A8: g<=g2 by XXREAL_0:17;
  then -g2<=-g by XREAL_1:24;
  then
A9: r+-g2<=r+-g by XREAL_1:7;
A10: r+g<=r+g2 by A8,XREAL_1:7;
  now
    per cases;
    suppose
A11:  g1<=g2;
A12:  now
        let y;
        assume
A13:    y in ].r-g,r+g.[;
        then reconsider x1=y as Real;
        ex p2 be Real st y = p2 & r-g<p2 & p2<r+g by A13;
        then x1<r+g2 & r-g2<x1 by A10,A9,XXREAL_0:2;
        hence y in ].r-g2,r+g2.[;
      end;
      g1=g by A11,XXREAL_0:def 9;
      hence ].r-g,r+g.[ c= N1 & ].r-g,r+g.[ c= N2 by A2,A4,A12,TARSKI:def 3;
    end;
    suppose
A14:  g2<=g1;
A15:  now
        let y;
        assume
A16:    y in ].r-g,r+g.[;
        then reconsider x1=y as Real;
        ex p2 be Real st y = p2 & r-g<p2 & p2<r+g by A16;
        then x1<r+g1 & r-g1<x1 by A6,A7,XXREAL_0:2;
        hence y in ].r-g1,r+g1.[;
      end;
      g2=g by A14,XXREAL_0:def 9;
      hence ].r-g,r+g.[ c= N1 & ].r-g,r+g.[ c= N2 by A2,A4,A15,TARSKI:def 3;
    end;
  end;
  hence thesis;
end;

theorem Th18:
  for X being open Subset of REAL, r st r in X ex N being
  Neighbourhood of r st N c= X
proof
  let X be open Subset of REAL, r;
  assume that
A1: r in X and
A2: for N be Neighbourhood of r holds not N c= X;
  defpred P[Element of NAT,real number] means $2 in ].r - 1/($1+1),r + 1/($1+1
  ) .[ & $2 in X`;
A3: now
    let N be Neighbourhood of r;
    consider g such that
    0<g and
A4: N = ].r-g,r+g.[ by Def6;
    not N c= X by A2;
    then consider x such that
A5: x in N and
A6: not x in X by TARSKI:def 3;
    consider s be Real such that
A7: x=s and
    r-g<s and
    s<r+g by A5,A4;
    take s;
    thus s in N by A5,A7;
    thus s in X` by A6,A7,XBOOLE_0:def 5;
  end;
A8: for n ex s being Real st P[n,s]
  proof
    let n;
    0 < 1 * (n + 1)";
    then 0 < 1/(n + 1) by XCMPLX_0:def 9;
    then reconsider N=]. r - 1/(n+1), r + 1/(n+1) .[ as Neighbourhood of r by
Def6;
    ex s be Real st s in N & s in X` by A3;
    hence thesis;
  end;
  consider s3 such that
A9: for n holds P[n,s3.n] from FUNCT_2:sch 3(A8);
  deffunc G(Element of NAT)= r + 1/($1+1);
  deffunc F(Element of NAT)= r - 1/($1+1);
  consider s1 be Real_Sequence such that
A10: for n holds s1.n = F(n) from SEQ_1:sch 1;
  consider s2 be Real_Sequence such that
A11: for n holds s2.n = G(n) from SEQ_1:sch 1;
A12: for n holds s1.n <= s3.n & s3.n <= s2.n
  proof
    let n;
    s3.n in].r - 1/(n+1),r + 1/(n+1) .[ by A9;
    then
A13: ex s be Real st s = s3.n & r - 1/(n+1) <s & s<r + 1/(n+1);
    hence s1.n <= s3.n by A10;
    thus thesis by A11,A13;
  end;
A14: rng s3 c= X`
  proof
    let x;
    assume x in rng s3;
    then consider y such that
A15: y in dom s3 and
A16: s3.y=x by FUNCT_1:def 3;
    reconsider y as Element of NAT by A15,FUNCT_2:def 1;
    s3.y in X` by A9;
    hence thesis by A16;
  end;
A17: X` is closed by Def5;
A18: now
    let s;
    assume
A19: 0<s;
    consider n such that
A20: s"<n by SEQ_4:3;
    take n;
    let m;
    assume n<=m;
    then n + 1 <= m + 1 by XREAL_1:6;
    then
A21: 1/(m+1) <= 1/(n+1) by XREAL_1:118;
    s" + 0 < n + 1 by A20,XREAL_1:8;
    then 1/(n+1) < 1/s" by A19,XREAL_1:76;
    then
A22: 1/(n+1) < s by XCMPLX_1:216;
    abs( s2.m - r) = abs( r + 1/(m+1) - r) by A11
      .= 1/(m+1) by ABSVALUE:def 1;
    hence abs( s2.m - r) <s by A22,A21,XXREAL_0:2;
  end;
  then
A23: s2 is convergent by SEQ_2:def 6;
  then
A24: lim s2 = r by A18,SEQ_2:def 7;
A25: now
    let s;
    assume
A26: 0<s;
    consider n such that
A27: s"<n by SEQ_4:3;
    take n;
    let m;
    assume n<=m;
    then n + 1 <= m + 1 by XREAL_1:6;
    then
A28: 1/(m+1) <= 1/(n+1) by XREAL_1:118;
    s" + 0 < n + 1 by A27,XREAL_1:8;
    then 1/(n+1) < 1/s" by A26,XREAL_1:76;
    then
A29: 1/(n+1) < s by XCMPLX_1:216;
    abs( s1.m - r) = abs( r - 1/(m+1) - r) by A10
      .= abs( - 1/(m+1) )
      .= abs( 1/(m+1) ) by COMPLEX1:52
      .= 1/(m+1) by ABSVALUE:def 1;
    hence abs( s1.m - r) <s by A29,A28,XXREAL_0:2;
  end;
  then
A30: s1 is convergent by SEQ_2:def 6;
  then lim s1 = r by A25,SEQ_2:def 7;
  then s3 is convergent & lim s3 = r by A30,A23,A24,A12,SEQ_2:19,20;
  then r in X` by A14,A17,Def4;
  hence contradiction by A1,XBOOLE_0:def 5;
end;

theorem
  for X being open Subset of REAL, r st r in X ex g st 0<g & ].r-g,r+g.[ c= X
proof
  let X be open Subset of REAL, r;
  assume r in X;
  then consider N be Neighbourhood of r such that
A1: N c= X by Th18;
  consider g such that
A2: 0<g & N = ].r-g,r+g.[ by Def6;
  take g;
  thus thesis by A1,A2;
end;

theorem
  (for r st r in X holds ex N be Neighbourhood of r st N c= X) implies X
  is open
proof
  assume that
A1: for r st r in X holds ex N be Neighbourhood of r st N c= X and
A2: not X is open;
  not X` is closed by A2,Def5;
  then consider s1 such that
A3: rng s1 c= X` and
A4: s1 is convergent and
A5: not lim s1 in X` by Def4;
  consider N be Neighbourhood of (lim s1) such that
A6: N c= X by A1,A5,SUBSET_1:29;
  consider g such that
A7: 0<g and
A8: ]. (lim s1) - g, (lim s1) + g .[ = N by Def6;
  consider n such that
A9: for m st n<=m holds abs(s1.m - (lim s1) )<g by A4,A7,SEQ_2:def 7;
  n in NAT;
  then n in dom s1 by FUNCT_2:def 1;
  then
A10: s1.n in rng s1 by FUNCT_1:def 3;
A11: abs(s1.n - (lim s1) )<g by A9;
  then s1.n - (lim s1) <g by SEQ_2:1;
  then
A12: s1.n <(lim s1) + g by XREAL_1:19;
  -g < s1.n - (lim s1) by A11,SEQ_2:1;
  then (lim s1) +- g < (lim s1) + (s1.n - (lim s1)) by XREAL_1:6;
  then
  s1.n in {s where s is Real: (lim s1) - g < s & s <(lim s1) + g} by A12;
  hence contradiction by A3,A6,A8,A10,XBOOLE_0:def 5;
end;

theorem Th21:
  X is open bounded_above implies not upper_bound X in X
proof
  assume that
A1: X is open and
A2: X is bounded_above;
  assume upper_bound X in X;
  then consider N being Neighbourhood of upper_bound X such that
A3: N c= X by A1,Th18;
  consider t such that
A4: t>0 and
A5: N = ].upper_bound X-t,upper_bound X+t.[ by Def6;
A6: upper_bound X + t/2 > upper_bound X by A4,XREAL_1:29,215;
A7: upper_bound X + t/2 +t/2 > upper_bound X + t/2 by A4,XREAL_1:29,215;
  upper_bound X - t < upper_bound X by A4,XREAL_1:44;
  then upper_bound X - t < upper_bound X + t/2 by A6,XXREAL_0:2;
  then upper_bound X + t/2 in {s where s is Real: upper_bound X-t<s & s<
  upper_bound X+t} by A7;
  hence contradiction by A2,A3,A5,A6,SEQ_4:def 1;
end;

theorem Th22:
  X is open bounded_below implies not lower_bound X in X
proof
  assume that
A1: X is open and
A2: X is bounded_below;
  assume lower_bound X in X;
  then consider N being Neighbourhood of lower_bound X such that
A3: N c= X by A1,Th18;
  consider t such that
A4: t>0 and
A5: N = ].lower_bound X-t,lower_bound X+t.[ by Def6;
A6: lower_bound X - t/2 < lower_bound X by A4,XREAL_1:44,215;
A7: lower_bound X - t/2 - t/2 < lower_bound X - t/2 by A4,XREAL_1:44,215;
  lower_bound X < lower_bound X + t by A4,XREAL_1:29;
  then lower_bound X - t/2 < lower_bound X + t by A6,XXREAL_0:2;
  then lower_bound X - t/2 in {s where s is Real: lower_bound X-t<s & s<
  lower_bound X+t} by A7;
  hence contradiction by A2,A3,A5,A6,SEQ_4:def 2;
end;

theorem
  X is open real-bounded & (for g1,g2 st g1 in X & g2 in X holds [.g1,g2.] c=
  X) implies ex p,g st X = ].p,g.[
proof
  assume that
A1: X is open and
A2: X is real-bounded and
A3: for g1,g2 st g1 in X & g2 in X holds [.g1,g2.] c= X;
  per cases;
  suppose
A4: X={};
    take 1;
    take 0;
    thus thesis by A4,XXREAL_1:28;
  end;
  suppose
A5: X<>{};
    take p=lower_bound X, g=upper_bound X;
    now
      let r be Real;
      thus r in X implies r in ].p,g.[
      proof
        assume
A6:     r in X;
        then p<>r & p<=r by A1,A2,Th22,SEQ_4:def 2;
        then
A7:     p<r by XXREAL_0:1;
        g<>r & r<=g by A1,A2,A6,Th21,SEQ_4:def 1;
        then r<g by XXREAL_0:1;
        hence thesis by A7;
      end;
      assume r in ].p,g.[;
      then
A8:   ex s be Real st s=r & p<s & s<g;
      then g-r>0 by XREAL_1:50;
      then consider g2 such that
A9:   g2 in X & g-(g-r)<g2 by A2,A5,SEQ_4:def 1;
      r-p>0 by A8,XREAL_1:50;
      then consider g1 such that
A10:  g1 in X & g1<p+(r-p) by A2,A5,SEQ_4:def 2;
      reconsider g1, g2 as Real by XREAL_0:def 1;
      r in {s where s is Real: g1<=s & s<=g2} & [.g1,g2.] c= X by A3,A9,A10;
      hence r in X;
    end;
    hence thesis by SUBSET_1:3;
  end;
end;



definition
  let g be real number,s be ext-real number;
  redefine func [. g,s .[ -> Subset of REAL equals
  { r where r is Real : g<=r
  & r<s };
  coherence
  proof
    now
      let e be set;
      assume e in [. g,s .[;
      then
A1:   ex r being Element of ExtREAL st e = r & g <= r & r<s;
      g in REAL by XREAL_0:def 1;
      hence e in REAL by A1,XXREAL_0:46;
    end;
    hence thesis by TARSKI:def 3;
  end;
  compatibility
  proof
    set Y = {r where r is Real: g<=r & r<s };
    let X be Subset of REAL;
A2: [. g,s .[ c= Y
    proof
      let e be set;
      assume e in [. g,s .[;
      then consider r being Element of ExtREAL such that
A3:   e = r and
A4:   g <= r & r<s;
      g in REAL by XREAL_0:def 1;
      then r in REAL by A4,XXREAL_0:46;
      hence thesis by A3,A4;
    end;
    Y c= [. g,s .[
    proof
      let e be set;
      assume e in Y;
      then consider r being Element of REAL such that
A5:   e = r & g <= r & r<s;
      r in REAL;
      hence thesis by A5,NUMBERS:31;
    end;
    hence thesis by A2,XBOOLE_0:def 10;
  end;
end;

definition
  let g be ext-real number,s be real number;
  redefine func ]. g,s .] -> Subset of REAL equals
  { r where r is Real : g<r &
  r<=s };
  coherence
  proof
    now
      let e be set;
      assume e in ]. g,s .];
      then
A1:   ex r being Element of ExtREAL st e = r & g < r & r<=s;
      s in REAL by XREAL_0:def 1;
      hence e in REAL by A1,XXREAL_0:47;
    end;
    hence thesis by TARSKI:def 3;
  end;
  compatibility
  proof
    set Y = {r where r is Real: g<r & r<=s };
    let X be Subset of REAL;
A2: ]. g,s .] c= Y
    proof
      let e be set;
      assume e in ]. g,s .];
      then consider r being Element of ExtREAL such that
A3:   e = r and
A4:   g < r & r<=s;
      s in REAL by XREAL_0:def 1;
      then r in REAL by A4,XXREAL_0:47;
      hence thesis by A3,A4;
    end;
    Y c= ]. g,s .]
    proof
      let e be set;
      assume e in Y;
      then consider r being Element of REAL such that
A5:   e = r & g < r & r<=s;
      r in REAL;
      hence thesis by A5,NUMBERS:31;
    end;
    hence thesis by A2,XBOOLE_0:def 10;
  end;
end;














begin

reserve x,X,Y for set;
reserve g,r,r1,r2,p,p1,p2 for Element of REAL;
reserve R for Subset of REAL;
reserve seq,seq1,seq2,seq3 for Real_Sequence;
reserve Ns for increasing sequence of NAT;
reserve n for Element of NAT;
reserve W for non empty set;
reserve h,h1,h2 for PartFunc of W,REAL;





theorem Th1:
  seq1=seq2-seq3 iff for n holds seq1.n=seq2.n-seq3.n
proof
  thus seq1=seq2-seq3 implies for n holds seq1.n=seq2.n-seq3.n
  proof
    assume
A1: seq1=seq2-seq3;
    now
      let n;
      seq1.n=seq2.n+(-seq3).n by A1,SEQ_1:7;
      then seq1.n=seq2.n+-seq3.n by SEQ_1:10;
      hence seq1.n=seq2.n-seq3.n;
    end;
    hence thesis;
  end;
  assume
A2: for n holds seq1.n=seq2.n-seq3.n;
  now
    let n;
    thus seq1.n = seq2.n-seq3.n by A2
      .= seq2.n+-seq3.n
      .= seq2.n+(-seq3).n by SEQ_1:10;
  end;
  hence thesis by SEQ_1:7;
end;

theorem Th2:
  (seq1 + seq2)*Ns = (seq1*Ns) + (seq2*Ns) & (seq1 - seq2)*Ns = (
  seq1*Ns) - (seq2*Ns) & (seq1 (#) seq2)*Ns = (seq1*Ns) (#) (seq2*Ns)
proof
  now
    let n;
    thus ((seq1 + seq2)*Ns).n = (seq1 + seq2).(Ns.n) by FUNCT_2:15
      .= seq1.(Ns.n) + seq2.(Ns.n) by SEQ_1:7
      .= (seq1*Ns).n + seq2.(Ns.n) by FUNCT_2:15
      .= (seq1*Ns).n + (seq2*Ns).n by FUNCT_2:15
      .= (seq1*Ns + seq2*Ns).n by SEQ_1:7;
  end;
  hence (seq1 + seq2)*Ns = (seq1*Ns) + (seq2*Ns) by FUNCT_2:63;
  now
    let n;
    thus ((seq1 - seq2)*Ns).n = (seq1 - seq2).(Ns.n) by FUNCT_2:15
      .= seq1.(Ns.n) - seq2.(Ns.n) by Th1
      .= (seq1*Ns).n - seq2.(Ns.n) by FUNCT_2:15
      .= (seq1*Ns).n - (seq2*Ns).n by FUNCT_2:15
      .= (seq1*Ns - seq2*Ns).n by Th1;
  end;
  hence (seq1 - seq2)*Ns = (seq1*Ns) - (seq2*Ns) by FUNCT_2:63;
  now
    let n;
    thus ((seq1 (#) seq2)*Ns).n = (seq1 (#) seq2).(Ns.n) by FUNCT_2:15
      .= seq1.(Ns.n) * seq2.(Ns.n) by SEQ_1:8
      .= (seq1*Ns).n * seq2.(Ns.n) by FUNCT_2:15
      .= (seq1*Ns).n * (seq2*Ns).n by FUNCT_2:15
      .= ((seq1*Ns)(#)(seq2*Ns)).n by SEQ_1:8;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem Th3:
  (p(#)seq)*Ns = p(#)(seq*Ns)
proof
  now
    let n;
    thus ((p(#)seq)*Ns).n = (p(#)seq).(Ns.n) by FUNCT_2:15
      .= p*(seq.(Ns.n)) by SEQ_1:9
      .= p*((seq*Ns).n) by FUNCT_2:15
      .= (p(#)(seq*Ns)).n by SEQ_1:9;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  (-seq)*Ns = -(seq*Ns) & (abs(seq))*Ns = abs((seq*Ns))
proof
  thus (-seq)*Ns = ((-1)(#)seq)*Ns .= (-1)(#)(seq*Ns) by Th3
    .= -(seq*Ns);
  now
    let n;
    thus ((abs(seq))*Ns).n = (abs(seq)).(Ns.n) by FUNCT_2:15
      .= abs(seq.(Ns.n)) by SEQ_1:12
      .= abs((seq*Ns).n) by FUNCT_2:15
      .= (abs(seq*Ns)).n by SEQ_1:12;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem Th5:
  (seq*Ns)" = (seq")*Ns
proof
  now
    let n;
    thus ((seq*Ns)").n = ((seq*Ns).n)" by VALUED_1:10
      .= (seq.(Ns.n))" by FUNCT_2:15
      .= (seq").(Ns.n) by VALUED_1:10
      .= ((seq")*Ns).n by FUNCT_2:15;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  (seq1/"seq)*Ns = (seq1*Ns)/"(seq*Ns)
proof
  thus (seq1/"seq)*Ns = (seq1(#)(seq"))*Ns .= (seq1*Ns)(#)((seq")*Ns) by Th2
    .= (seq1*Ns)(#)((seq*Ns)") by Th5
    .= (seq1*Ns)/"(seq*Ns);
end;

theorem
  seq is convergent & (for n holds seq.n<=0) implies lim seq <= 0
proof
  assume that
A1: seq is convergent and
A2: for n holds seq.n<=0;
  set seq1 = -seq;
A3: now
    let n;
    seq1.n = -seq.n & seq.n<=0 by A2,SEQ_1:10;
    hence -(0 qua Real)<=seq1.n by XREAL_1:24;
  end;
  seq1 is convergent by A1,SEQ_2:9;
  then 0 <= lim seq1 by A3,SEQ_2:17;
  then -(0 qua Real) <= - lim seq by A1,SEQ_2:10;
  hence thesis by XREAL_1:24;
end;

theorem Th8:
  for h1,h2 being PartFunc of W,REAL, seq being sequence of W holds
  rng seq c= dom h1 /\ dom h2 implies (h1+h2)/*seq=h1/*seq+h2/*seq
  & (h1-h2)/*seq=h1/*seq-h2/*seq & (h1(#)h2)/*seq=(h1/*seq)(#)(h2/*seq)
proof
  let h1,h2 be PartFunc of W,REAL, seq be sequence of W;
A1: dom h1 /\ dom h2 c= dom h1 by XBOOLE_1:17;
A2: dom h1 /\ dom h2 c= dom h2 by XBOOLE_1:17;
  assume
A3: rng seq c= dom h1 /\ dom h2;
  then
A4: rng seq c= dom (h1 + h2) by VALUED_1:def 1;
  now
    let n;
A5: seq.n in rng seq by VALUED_0:28;
    thus ((h1+h2)/*seq).n = (h1+h2).(seq.n) by A4,FUNCT_2:108
      .= h1.(seq.n) + h2.(seq.n) by A4,A5,VALUED_1:def 1
      .= (h1/*seq).n + h2.(seq.n) by A3,A1,FUNCT_2:108,XBOOLE_1:1
      .= (h1/*seq).n + (h2/*seq).n by A3,A2,FUNCT_2:108,XBOOLE_1:1;
  end;
  hence (h1+h2)/*seq=h1/*seq+h2/*seq by SEQ_1:7;
A6: rng seq c= dom (h1 - h2) by A3,VALUED_1:12;
  now
    let n;
A7: seq.n in rng seq by VALUED_0:28;
    thus ((h1-h2)/*seq).n = (h1-h2).(seq.n) by A6,FUNCT_2:108
      .= h1.(seq.n) - h2.(seq.n) by A6,A7,VALUED_1:13
      .= (h1/*seq).n - h2.(seq.n) by A3,A1,FUNCT_2:108,XBOOLE_1:1
      .= (h1/*seq).n - (h2/*seq).n by A3,A2,FUNCT_2:108,XBOOLE_1:1;
  end;
  hence (h1-h2)/*seq=h1/*seq-h2/*seq by Th1;
A8: rng seq c= dom (h1 (#) h2) by A3,VALUED_1:def 4;
  now
    let n;
    thus ((h1(#)h2)/*seq).n = (h1(#)h2).(seq.n) by A8,FUNCT_2:108
      .= h1.(seq.n) * h2.(seq.n) by VALUED_1:5
      .= (h1/*seq).n * h2.(seq.n) by A3,A1,FUNCT_2:108,XBOOLE_1:1
      .= (h1/*seq).n * (h2/*seq).n by A3,A2,FUNCT_2:108,XBOOLE_1:1;
  end;
  hence thesis by SEQ_1:8;
end;

theorem Th9:
  for h being PartFunc of W,REAL, seq being sequence of W holds
  for r being real number holds rng seq c= dom h implies
  (r(#)h)/*seq = r(#) (h/*seq)
proof
  let h be PartFunc of W,REAL, seq be sequence of W;
  let r be real number;
  assume
A1: rng seq c= dom h;
  then
A2: rng seq c= dom (r(#)h) by VALUED_1:def 5;
  now
    let n;
A3: seq.n in rng seq by VALUED_0:28;
    thus ((r(#)h)/*seq).n = (r(#)h).(seq.n) by A2,FUNCT_2:108
      .= r * (h.(seq.n)) by A2,A3,VALUED_1:def 5
      .= r * (h/*seq).n by A1,FUNCT_2:108;
  end;
  hence thesis by SEQ_1:9;
end;

theorem
  for h being PartFunc of W,REAL, seq being sequence of W holds
  rng seq c= dom h implies abs(h/*seq) = (abs(h))/*seq & -(h/*seq) = (-h )/*seq
proof
  let h be PartFunc of W,REAL, seq be sequence of W;
  assume
A1: rng seq c= dom h;
  then
A2: rng seq c= dom abs(h) by VALUED_1:def 11;
  now
    let n;
    thus abs(h/*seq).n = abs((h/*seq).n) by SEQ_1:12
      .= abs(h.(seq.n)) by A1,FUNCT_2:108
      .= abs(h).(seq.n) by VALUED_1:18
      .= (abs(h)/*seq).n by A2,FUNCT_2:108;
  end;
  hence abs(h/*seq) = (abs(h))/*seq by FUNCT_2:63;
  thus -(h/*seq) = (-1)(#)(h/*seq) .= ((-1)(#)h)/*seq by A1,Th9
    .= (-h)/*seq;
end;

theorem
  for h being PartFunc of W,REAL, seq being sequence of W holds
  rng seq c= dom (h^) implies h/*seq is non-zero
proof
  let h be PartFunc of W,REAL, seq be sequence of W;
  assume
A1: rng seq c= dom (h^);
  then
A2: dom h \ h"{0} c= dom h & rng seq c= dom h \ h"{0} by RFUNCT_1:def 2
,XBOOLE_1:36;
  now
    given n such that
A3: (h/*seq).n=0;
    seq.n in rng seq by VALUED_0:28;
    then seq.n in dom (h^) by A1;
    then seq.n in dom h \ h"{0} by RFUNCT_1:def 2;
    then seq.n in dom h & not seq.n in h"{0} by XBOOLE_0:def 5;
    then
A4: not h.(seq.n) in {0} by FUNCT_1:def 7;
    h.(seq.n) =0 by A2,A3,FUNCT_2:108,XBOOLE_1:1;
    hence contradiction by A4,TARSKI:def 1;
  end;
  hence thesis by SEQ_1:5;
end;

theorem
  for h being PartFunc of W,REAL, seq being sequence of W holds
  rng seq c= dom (h^) implies (h^)/*seq =(h/*seq)"
proof
  let h be PartFunc of W,REAL, seq be sequence of W;
  assume
A1: rng seq c= dom (h^);
  then
A2: dom h \ h"{0} c= dom h & rng seq c= dom h \ h"{0} by RFUNCT_1:def 2
,XBOOLE_1:36;
  now
    let n;
A3: seq.n in rng seq by VALUED_0:28;
    thus ((h^)/*seq).n = (h^).(seq.n) by A1,FUNCT_2:108
      .= (h.(seq.n))" by A1,A3,RFUNCT_1:def 2
      .= ((h/*seq).n)" by A2,FUNCT_2:108,XBOOLE_1:1
      .= ((h/*seq)").n by VALUED_1:10;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  for h1,h2 being PartFunc of W,REAL, seq being sequence of W holds
  h1 is total & h2 is total implies
  (h1+h2)/*seq = h1/*seq + h2/*seq & (h1-h2)/*seq = h1/*seq - h2/*seq &
  (h1(#)h2)/*seq = (h1/*seq) (#) (h2/*seq)
proof
  let h1,h2 be PartFunc of W,REAL, seq be sequence of W;
  assume h1 is total & h2 is total;
  then dom (h1+h2) = W by PARTFUN1:def 2;
  then dom h1 /\ dom h2 = W by VALUED_1:def 1;
  then
A1: rng seq c= dom h1 /\ dom h2;
  hence (h1+h2)/*seq = h1/*seq + h2/*seq by Th8;
  thus (h1-h2)/*seq = h1/*seq - h2/*seq by A1,Th8;
  thus thesis by A1,Th8;
end;

theorem
  for h being PartFunc of W,REAL, seq being sequence of W holds
  h is total implies (r(#)h)/*seq = r(#)(h/*seq)
proof
  let h be PartFunc of W,REAL, seq be sequence of W;
  assume h is total;
  then dom h = W by PARTFUN1:def 2;
  then rng seq c= dom h;
  hence thesis by Th9;
end;

theorem
  for h being PartFunc of W,REAL, seq being sequence of W holds
  rng seq c= dom (h|X) implies abs((h|X)/*seq) = ((abs(h))|X)/*seq
proof
  let h be PartFunc of W,REAL, seq be sequence of W;
  assume
A1: rng seq c= dom (h|X);
A2: dom (h|X) = dom h /\ X by RELAT_1:61
    .= dom abs(h) /\ X by VALUED_1:def 11
    .= dom (abs(h)|X) by RELAT_1:61;
  now
    let n;
A3: seq.n in rng seq by VALUED_0:28;
    then seq.n in dom (h|X) by A1;
    then seq.n in dom h /\ X by RELAT_1:61;
    then
A4: seq.n in X by XBOOLE_0:def 4;
    thus (abs((h|X)/*seq)).n = abs( ((h|X)/*seq).n ) by SEQ_1:12
      .= abs( (h|X).(seq.n) ) by A1,FUNCT_2:108
      .= abs( h.(seq.n) ) by A1,A3,FUNCT_1:47
      .= (abs(h)).(seq.n) by VALUED_1:18
      .= ((abs(h))|X).(seq.n) by A4,FUNCT_1:49
      .= (((abs(h))|X)/*seq).n by A1,A2,FUNCT_2:108;
  end;
  hence thesis by FUNCT_2:63;
end;

theorem
  for h being PartFunc of W,REAL, seq being sequence of W holds
  rng seq c= dom (h|X) & h"{0}={} implies ((h^)|X)/*seq = ((h|X)/*seq)"
proof
  let h be PartFunc of W,REAL, seq be sequence of W;
  assume that
A1: rng seq c= dom (h|X) and
A2: h"{0}={};
  now
    let x;
    assume x in rng seq;
    then x in dom (h|X) by A1;
    then
A3: x in dom h /\ X by RELAT_1:61;
    then x in dom h \ h"{0} by A2,XBOOLE_0:def 4;
    then
A4: x in dom (h^) by RFUNCT_1:def 2;
    x in X by A3,XBOOLE_0:def 4;
    then x in dom (h^) /\ X by A4,XBOOLE_0:def 4;
    hence x in dom ((h^)|X) by RELAT_1:61;
  end;
  then
A5: rng seq c= dom ((h^)|X) by TARSKI:def 3;
  now
    let n;
A6: seq.n in rng seq by VALUED_0:28;
    then seq.n in dom (h|X) by A1;
    then
A7: seq.n in dom h /\ X by RELAT_1:61;
    then
A8: seq.n in X by XBOOLE_0:def 4;
    seq.n in dom h \ h"{0} by A2,A7,XBOOLE_0:def 4;
    then
A9: seq.n in dom (h^) by RFUNCT_1:def 2;
    thus (((h^)|X)/*seq).n = ((h^)|X).(seq.n) by A5,FUNCT_2:108
      .= (h^).(seq.n) by A8,FUNCT_1:49
      .= (h.(seq.n))" by A9,RFUNCT_1:def 2
      .= ((h|X).(seq.n))" by A1,A6,FUNCT_1:47
      .= (((h|X)/*seq).n)" by A1,FUNCT_2:108
      .= (((h|X)/*seq)").n by VALUED_1:10;
  end;
  hence thesis by FUNCT_2:63;
end;





registration
  let Z be set;
  let f be one-to-one Function;
  cluster f|Z -> one-to-one;
  coherence by FUNCT_1:52;
end;

theorem
  for h being one-to-one Function holds (h|X)" = (h")|(h.:X)
proof
  let h be one-to-one Function;
  reconsider hX = h|X as one-to-one Function;
  now
    let r be set;
    thus r in dom ((h|X)") implies r in dom ((h")|(h.:X))
    proof
      assume r in dom ((h|X)");
      then r in rng(hX) by FUNCT_1:33;
      then consider g being set such that
A1:   g in dom (h|X) and
A2:   (h|X).g=r by FUNCT_1:def 3;
A3:   h.g=r by A1,A2,FUNCT_1:47;
A4:   g in dom h /\ X by A1,RELAT_1:61;
      then g in dom h by XBOOLE_0:def 4;
      then r in rng h by A3,FUNCT_1:def 3;
      then
A5:   r in dom (h") by FUNCT_1:33;
      g in dom h & g in X by A4,XBOOLE_0:def 4;
      then r in h.:X by A3,FUNCT_1:def 6;
      then r in dom (h")/\(h.:X) by A5,XBOOLE_0:def 4;
      hence thesis by RELAT_1:61;
    end;
    assume r in dom ((h")|(h.:X));
    then r in dom(h") /\ (h.:X) by RELAT_1:61;
    then r in h.:X by XBOOLE_0:def 4;
    then consider t being set such that
A6: t in dom h and
A7: t in X and
A8: h.t=r by FUNCT_1:def 6;
    t in dom h/\X by A6,A7,XBOOLE_0:def 4;
    then
A9: t in dom(h|X) by RELAT_1:61;
    (h|X).t=r by A7,A8,FUNCT_1:49;
    then r in rng(hX) by A9,FUNCT_1:def 3;
    hence r in dom((h|X)") by FUNCT_1:33;
  end;
  then
A10: dom ((hX)")=dom ((h")|(h.:X)) by TARSKI:1;
  now
    given r being set such that
A11: r in dom((h|X)") and
A12: ((h|X)").r<>((h")|(h.:X)).r;
    r in dom(h")/\(h.:X) by A10,A11,RELAT_1:61;
    then r in h.:X by XBOOLE_0:def 4;
    then consider t be set such that
    t in dom h and
    t in X and
A13: h.t=r by FUNCT_1:def 6;
    r in rng (hX) by A11,FUNCT_1:33;
    then consider g be set such that
A14: g in dom(h|X) and
A15: (h|X).g=r by FUNCT_1:def 3;
A16: h.g=r by A14,A15,FUNCT_1:47;
    g in dom h /\ X by A14,RELAT_1:61;
    then
A17: g in dom h by XBOOLE_0:def 4;
    ((hX)").r<>(h").r by A10,A11,A12,FUNCT_1:47;
    then g<>(h").(h.t) by A14,A15,A13,FUNCT_1:34;
    hence contradiction by A13,A17,A16,FUNCT_1:34;
  end;
  hence thesis by A10,FUNCT_1:2;
end;

theorem Th18:
  for h being PartFunc of W,REAL holds
  rng h is real-bounded & upper_bound (rng h) = lower_bound (rng h)
  implies h is constant
proof
  let h be PartFunc of W,REAL;
  assume
A1: rng h is real-bounded & upper_bound (rng h) = lower_bound (rng h);
  assume not h is constant;
  then consider x1,x2 being set such that
A2: x1 in dom h & x2 in dom h and
A3: h.x1 <> h.x2 by FUNCT_1:def 10;
  h.x1 in rng h & h.x2 in rng h by A2,FUNCT_1:def 3;
  hence contradiction by A1,A3,SEQ_4:12;
end;

theorem
  for h being PartFunc of W,REAL holds
  h.:Y is real-bounded & upper_bound(h.:Y) = lower_bound(h.:Y) implies
  h|Y is constant
proof
  let h be PartFunc of W,REAL;
  rng (h|Y) = h.:Y by RELAT_1:115;
  hence thesis by Th18;
end;

reserve e1,e2 for ext-real number;
reserve h,h1,h2 for PartFunc of REAL,REAL;

definition
  let h;
  redefine attr h is increasing means
  :Def1:
  for r1,r2 st r1 in dom h & r2 in dom h & r1<r2 holds h.r1 < h.r2;
  compatibility
  proof
    thus h is increasing implies for r1,r2 st r1 in dom h & r2 in dom h & r1 <
    r2 holds h.r1 < h.r2 by VALUED_0:def 13;
    assume
A1: for r1,r2 st r1 in dom h & r2 in dom h & r1 < r2 holds h.r1 < h.r2;
    let e1,e2;
    thus thesis by A1;
  end;
  redefine attr h is decreasing means
  :Def2:
  for r1,r2 st r1 in dom h & r2 in dom h & r1<r2 holds h.r2 < h.r1;
  compatibility
  proof
    thus h is decreasing implies for r1,r2 st r1 in dom h & r2 in dom h & r1 <
    r2 holds h.r1 > h.r2 by VALUED_0:def 14;
    assume
A2: for r1,r2 st r1 in dom h & r2 in dom h & r1 < r2 holds h.r1 > h.r2;
    let e1,e2;
    thus thesis by A2;
  end;
  redefine attr h is non-decreasing means
  :Def3:
  for r1,r2 st r1 in dom h & r2 in dom h & r1<r2 holds h.r1 <= h.r2;
  compatibility
  proof
    thus h is non-decreasing implies for r1,r2 st r1 in dom h & r2 in dom h &
    r1 < r2 holds h.r1 <= h.r2 by VALUED_0:def 15;
    assume
A3: for r1,r2 st r1 in dom h & r2 in dom h & r1 < r2 holds h.r1 <= h. r2;
    let e1,e2 such that
A4: e1 in dom h & e2 in dom h;
    assume e1 <= e2;
    then e1 < e2 or e1 = e2 by XXREAL_0:1;
    hence thesis by A3,A4;
  end;
  redefine attr h is non-increasing means
  :Def4:
  for r1,r2 st r1 in dom h & r2 in dom h & r1<r2 holds h.r2 <= h.r1;
  compatibility
  proof
    thus h is non-increasing implies for r1,r2 st r1 in dom h & r2 in dom h &
    r1 < r2 holds h.r1 >= h.r2 by VALUED_0:def 16;
    assume
A5: for r1,r2 st r1 in dom h & r2 in dom h & r1 < r2 holds h.r1 >= h. r2;
    let e1,e2 such that
A6: e1 in dom h & e2 in dom h;
    assume e1 <= e2;
    then e1 < e2 or e1 = e2 by XXREAL_0:1;
    hence thesis by A5,A6;
  end;
end;

theorem Th20:
  h|Y is increasing iff for r1,r2 st r1 in Y /\ dom h & r2 in Y /\
  dom h & r1<r2 holds h.r1 < h.r2
proof
  thus h|Y is increasing implies for r1,r2 st r1 in Y /\ dom h & r2 in Y /\
  dom h & r1<r2 holds h.r2 > h.r1
  proof
    assume
A1: h|Y is increasing;
    let r1,r2 such that
A2: r1 in Y /\ dom h and
A3: r2 in Y /\ dom h and
A4: r1<r2;
A5: r2 in dom(h|Y) by A3,RELAT_1:61;
    then
A6: (h|Y).r2 = h.r2 by FUNCT_1:47;
A7: r1 in dom(h|Y) by A2,RELAT_1:61;
    then (h|Y).r1 = h.r1 by FUNCT_1:47;
    hence thesis by A1,A4,A7,A5,A6,Def1;
  end;
  assume
A8: for r1,r2 st r1 in Y /\ dom h & r2 in Y /\ dom h & r1<r2 holds h.r2 >h.r1;
  let r1,r2;
  assume that
A9: r1 in dom(h|Y) & r2 in dom(h|Y) and
A10: r1<r2;
A11: (h|Y).r1 = h.r1 & (h|Y).r2 = h.r2 by A9,FUNCT_1:47;
  r1 in Y /\ dom h & r2 in Y /\ dom h by A9,RELAT_1:61;
  hence thesis by A8,A10,A11;
end;

theorem Th21:
  h|Y is decreasing iff for r1,r2 st r1 in Y /\ dom h & r2 in Y /\
  dom h & r1<r2 holds h.r2 < h.r1
proof
  thus h|Y is decreasing implies for r1,r2 st r1 in Y /\ dom h & r2 in Y /\
  dom h & r1<r2 holds h.r2 < h.r1
  proof
    assume
A1: h|Y is decreasing;
    let r1,r2 such that
A2: r1 in Y /\ dom h and
A3: r2 in Y /\ dom h and
A4: r1<r2;
A5: r2 in dom(h|Y) by A3,RELAT_1:61;
    then
A6: (h|Y).r2 = h.r2 by FUNCT_1:47;
A7: r1 in dom(h|Y) by A2,RELAT_1:61;
    then (h|Y).r1 = h.r1 by FUNCT_1:47;
    hence thesis by A1,A4,A7,A5,A6,Def2;
  end;
  assume
A8: for r1,r2 st r1 in Y /\ dom h & r2 in Y /\ dom h & r1<r2 holds h.r2 <h.r1;
  let r1,r2;
  assume that
A9: r1 in dom(h|Y) & r2 in dom(h|Y) and
A10: r1<r2;
A11: (h|Y).r1 = h.r1 & (h|Y).r2 = h.r2 by A9,FUNCT_1:47;
  r1 in Y /\ dom h & r2 in Y /\ dom h by A9,RELAT_1:61;
  hence thesis by A8,A10,A11;
end;

theorem Th22:
  h|Y is non-decreasing iff for r1,r2 st r1 in Y /\ dom h & r2 in
  Y /\ dom h & r1<r2 holds h.r1 <= h.r2
proof
  thus h|Y is non-decreasing implies for r1,r2 st r1 in Y /\ dom h & r2 in Y
  /\ dom h & r1<r2 holds h.r1 <= h.r2
  proof
    assume
A1: h|Y is non-decreasing;
    let r1,r2 such that
A2: r1 in Y /\ dom h and
A3: r2 in Y /\ dom h and
A4: r1<r2;
A5: r2 in dom(h|Y) by A3,RELAT_1:61;
    then
A6: (h|Y).r2 = h.r2 by FUNCT_1:47;
A7: r1 in dom(h|Y) by A2,RELAT_1:61;
    then (h|Y).r1 = h.r1 by FUNCT_1:47;
    hence thesis by A1,A4,A7,A5,A6,Def3;
  end;
  assume
A8: for r1,r2 st r1 in Y /\ dom h & r2 in Y /\ dom h & r1<r2 holds h.r1
  <= h.r2;
  let r1,r2;
  assume that
A9: r1 in dom(h|Y) & r2 in dom(h|Y) and
A10: r1<r2;
A11: (h|Y).r1 = h.r1 & (h|Y).r2 = h.r2 by A9,FUNCT_1:47;
  r1 in Y /\ dom h & r2 in Y /\ dom h by A9,RELAT_1:61;
  hence thesis by A8,A10,A11;
end;

theorem Th23:
  h|Y is non-increasing iff for r1,r2 st r1 in Y /\ dom h & r2 in
  Y /\ dom h & r1<r2 holds h.r2 <= h.r1
proof
  thus h|Y is non-increasing implies for r1,r2 st r1 in Y /\ dom h & r2 in Y
  /\ dom h & r1<r2 holds h.r2 <= h.r1
  proof
    assume
A1: h|Y is non-increasing;
    let r1,r2 such that
A2: r1 in Y /\ dom h and
A3: r2 in Y /\ dom h and
A4: r1<r2;
A5: r2 in dom(h|Y) by A3,RELAT_1:61;
    then
A6: (h|Y).r2 = h.r2 by FUNCT_1:47;
A7: r1 in dom(h|Y) by A2,RELAT_1:61;
    then (h|Y).r1 = h.r1 by FUNCT_1:47;
    hence thesis by A1,A4,A7,A5,A6,Def4;
  end;
  assume
A8: for r1,r2 st r1 in Y /\ dom h & r2 in Y /\ dom h & r1<r2 holds h.r2 <=h.r1;
  let r1,r2;
  assume that
A9: r1 in dom(h|Y) & r2 in dom(h|Y) and
A10: r1<r2;
A11: (h|Y).r1 = h.r1 & (h|Y).r2 = h.r2 by A9,FUNCT_1:47;
  r1 in Y /\ dom h & r2 in Y /\ dom h by A9,RELAT_1:61;
  hence thesis by A8,A10,A11;
end;

definition
  let h;
  attr h is monotone means
  :Def5:
  h is non-decreasing or h is non-increasing;
end;

registration
  cluster non-decreasing -> monotone for PartFunc of REAL,REAL;
  coherence by Def5;
  cluster non-increasing -> monotone for PartFunc of REAL,REAL;
  coherence by Def5;
  cluster non monotone -> non non-decreasing non non-increasing for PartFunc of
    REAL,REAL;
  coherence;
end;

theorem Th24:
  h|Y is non-decreasing iff for r1,r2 st r1 in Y /\ dom h & r2 in
  Y /\ dom h & r1<=r2 holds h.r1 <= h.r2
proof
  thus h|Y is non-decreasing implies for r1,r2 st r1 in Y /\ dom h & r2 in Y
  /\ dom h & r1<=r2 holds h.r1 <= h.r2
  proof
    assume
A1: h|Y is non-decreasing;
    let r1,r2 such that
A2: r1 in Y /\ dom h & r2 in Y /\ dom h and
A3: r1<=r2;
    now
      per cases by A3,XXREAL_0:1;
      suppose
        r1 < r2;
        hence thesis by A1,A2,Th22;
      end;
      suppose
        r1 = r2;
        hence thesis;
      end;
    end;
    hence thesis;
  end;
  assume
A4: for r1,r2 st r1 in Y /\ dom h & r2 in Y /\ dom h & r1<=r2 holds h.
  r1 <= h.r2;
  let r1,r2;
  assume that
A5: r1 in dom(h|Y) & r2 in dom(h|Y) and
A6: r1<r2;
A7: (h|Y).r1 = h.r1 & (h|Y).r2 = h.r2 by A5,FUNCT_1:47;
  r1 in Y /\ dom h & r2 in Y /\ dom h by A5,RELAT_1:61;
  hence thesis by A4,A6,A7;
end;

theorem Th25:
  h|Y is non-increasing iff for r1,r2 st r1 in Y /\ dom h & r2 in
  Y /\ dom h & r1<=r2 holds h.r2 <= h.r1
proof
  thus h|Y is non-increasing implies for r1,r2 st r1 in Y /\ dom h & r2 in Y
  /\ dom h & r1<=r2 holds h.r2 <= h.r1
  proof
    assume
A1: h|Y is non-increasing;
    let r1,r2 such that
A2: r1 in Y /\ dom h & r2 in Y /\ dom h and
A3: r1<=r2;
    now
      per cases by A3,XXREAL_0:1;
      suppose
        r1 < r2;
        hence thesis by A1,A2,Th23;
      end;
      suppose
        r1 = r2;
        hence thesis;
      end;
    end;
    hence thesis;
  end;
  assume
A4: for r1,r2 st r1 in Y /\ dom h & r2 in Y /\ dom h & r1<=r2 holds h.
  r2<=h.r1;
  let r1,r2;
  assume that
A5: r1 in dom(h|Y) & r2 in dom(h|Y) and
A6: r1<r2;
A7: (h|Y).r1 = h.r1 & (h|Y).r2 = h.r2 by A5,FUNCT_1:47;
  r1 in Y /\ dom h & r2 in Y /\ dom h by A5,RELAT_1:61;
  hence thesis by A4,A6,A7;
end;

registration
  cluster non-decreasing non-increasing -> constant for PartFunc of REAL,REAL;
  coherence
  proof
    let h be PartFunc of REAL,REAL;
    assume
A1: h is non-decreasing non-increasing;
    let r1,r2 be set;
    assume
A2: r1 in dom h & r2 in dom h;
    then reconsider r1,r2 as Real;
    r1 < r2 or r1 = r2 or r2 < r1 by XXREAL_0:1;
    then h.r1 <= h.r2 & h.r2 <= h.r1 by A1,A2,Def3,Def4;
    hence thesis by XXREAL_0:1;
  end;
end;

registration
  cluster constant -> non-increasing non-decreasing for PartFunc of REAL,REAL;
  coherence
  proof
    let h be PartFunc of REAL,REAL;
    assume
A1: h is constant;
    thus h is non-increasing
    proof
      let r1,r2;
      assume r1 in dom h & r2 in dom h;
      hence thesis by A1,FUNCT_1:def 10;
    end;
    let r1,r2;
    assume r1 in dom h & r2 in dom h;
    hence thesis by A1,FUNCT_1:def 10;
  end;
end;

registration
  cluster trivial for PartFunc of REAL,REAL;
  existence
  proof
    reconsider f = {} as PartFunc of REAL,REAL by RELSET_1:12;
    take f;
    thus thesis;
  end;
end;

registration
  let h be increasing PartFunc of REAL,REAL, X be set;
  cluster h|X -> increasing for PartFunc of REAL,REAL;
  coherence
  proof
    now
      let r1,r2;
      assume
A1:   r1 in dom(h|X) & r2 in dom(h|X);
      then
A2:   h.r1 = (h|X).r1 & h.r2 = (h|X).r2 by FUNCT_1:47;
      assume
A3:   r1 < r2;
      r1 in dom h & r2 in dom h by A1,RELAT_1:57;
      hence (h|X).r1 < (h|X).r2 by A2,A3,Def1;
    end;
    hence thesis by Def1;
  end;
end;

registration
  let h be decreasing PartFunc of REAL,REAL, X be set;
  cluster h|X -> decreasing for PartFunc of REAL,REAL;
  coherence
  proof
    now
      let r1,r2;
      assume
A1:   r1 in dom(h|X) & r2 in dom(h|X);
      then
A2:   h.r1 = (h|X).r1 & h.r2 = (h|X).r2 by FUNCT_1:47;
      assume
A3:   r1 < r2;
      r1 in dom h & r2 in dom h by A1,RELAT_1:57;
      hence (h|X).r1 > (h|X).r2 by A2,A3,Def2;
    end;
    hence thesis by Def2;
  end;
end;

registration
  let h be non-decreasing PartFunc of REAL,REAL, X be set;
  cluster h|X -> non-decreasing for PartFunc of REAL,REAL;
  coherence
  proof
    now
      let r1,r2;
      assume
A1:   r1 in dom(h|X) & r2 in dom(h|X);
      then
A2:   h.r1 = (h|X).r1 & h.r2 = (h|X).r2 by FUNCT_1:47;
      assume
A3:   r1 < r2;
      r1 in dom h & r2 in dom h by A1,RELAT_1:57;
      hence (h|X).r1 <= (h|X).r2 by A2,A3,Def3;
    end;
    hence thesis by Def3;
  end;
end;

theorem
  Y misses dom h implies h|Y is increasing & h|Y is decreasing & h|Y is
  non-decreasing & h|Y is non-increasing & h|Y is monotone
proof
  assume
A1: Y /\ dom h = {};
  then
  for r1,r2 holds ( r1 in Y /\ dom h & r2 in Y /\ dom h & r1<r2 implies h.
  r1<h.r2);
  hence h|Y is increasing by Th20;
  for r1,r2 holds ( r1 in Y /\ dom h & r2 in Y /\ dom h & r1<r2 implies h.
  r2<h.r1) by A1;
  hence h|Y is decreasing by Th21;
  for r1,r2 holds ( r1 in Y /\ dom h & r2 in Y /\ dom h & r1<r2 implies h.
  r1<= h.r2) by A1;
  hence h|Y is non-decreasing by Th22;
A2: for r1,r2 holds ( r1 in Y /\ dom h & r2 in Y /\ dom h & r1<r2 implies h.
  r2<= h.r1) by A1;
  hence h|Y is non-increasing by Th23;
  thus thesis by A2,Th23;
end;

theorem
  h|Y is non-decreasing & h|X is non-increasing implies h|(Y /\ X) is constant
proof
  assume
A1: h|Y is non-decreasing & h|X is non-increasing;
  now
    per cases;
    suppose
      Y /\ X /\ dom h = {};
      then Y /\ X misses dom h by XBOOLE_0:def 7;
      hence thesis by PARTFUN2:39;
    end;
    suppose
A2:   Y /\ X /\ dom h <> {};
      set x = the Element of Y /\ X /\ dom h;
      x in dom h by A2,XBOOLE_0:def 4;
      then reconsider x as Real;
      now
        take r1 = h.x;
        now
          let p;
          assume
A3:       p in Y /\ X /\ dom h;
          then
A4:       p in Y /\ X by XBOOLE_0:def 4;
A5:       p in dom h by A3,XBOOLE_0:def 4;
          p in X by A4,XBOOLE_0:def 4;
          then
A6:       p in (X /\ dom h) by A5,XBOOLE_0:def 4;
A7:       x in dom h by A2,XBOOLE_0:def 4;
A8:       x in Y /\ X by A2,XBOOLE_0:def 4;
          then x in Y by XBOOLE_0:def 4;
          then
A9:       x in (Y /\ dom h) by A7,XBOOLE_0:def 4;
          x in X by A8,XBOOLE_0:def 4;
          then
A10:      x in (X /\ dom h) by A7,XBOOLE_0:def 4;
          p in Y by A4,XBOOLE_0:def 4;
          then
A11:      p in (Y /\ dom h) by A5,XBOOLE_0:def 4;
          now
            per cases;
            suppose
              x <= p;
              then h.x <= h.p & h.p <= h.x by A1,A11,A6,A9,A10,Th24,Th25;
              hence h.p = h.x by XXREAL_0:1;
            end;
            suppose
              p <= x;
              then h.p <= h.x & h.x <= h.p by A1,A11,A6,A9,A10,Th24,Th25;
              hence h.p = h.x by XXREAL_0:1;
            end;
          end;
          hence h.p = r1;
        end;
        hence for p st p in Y /\ X /\ dom h holds h.p = r1;
      end;
      hence thesis by PARTFUN2:57;
    end;
  end;
  hence thesis;
end;

theorem
  X c= Y & h|Y is increasing implies h|X is increasing
proof
  assume that
A1: X c= Y and
A2: h|Y is increasing;
  now
A3: X /\ dom h c= Y /\ dom h by A1,XBOOLE_1:26;
    let r1,r2;
    assume r1 in X /\ dom h & r2 in X /\ dom h & r1 < r2;
    hence h.r1 < h.r2 by A2,A3,Th20;
  end;
  hence thesis by Th20;
end;

theorem
  X c= Y & h|Y is decreasing implies h|X is decreasing
proof
  assume that
A1: X c= Y and
A2: h|Y is decreasing;
  now
A3: X /\ dom h c= Y /\ dom h by A1,XBOOLE_1:26;
    let r1,r2;
    assume r1 in X /\ dom h & r2 in X /\ dom h & r1 < r2;
    hence h.r1 > h.r2 by A2,A3,Th21;
  end;
  hence thesis by Th21;
end;

theorem
  X c= Y & h|Y is non-decreasing implies h|X is non-decreasing
proof
  assume that
A1: X c= Y and
A2: h|Y is non-decreasing;
  now
A3: X /\ dom h c= Y /\ dom h by A1,XBOOLE_1:26;
    let r1,r2;
    assume r1 in X /\ dom h & r2 in X /\ dom h & r1 < r2;
    hence h.r1 <= h.r2 by A2,A3,Th22;
  end;
  hence thesis by Th22;
end;

theorem
  X c= Y & h|Y is non-increasing implies h|X is non-increasing
proof
  assume that
A1: X c= Y and
A2: h|Y is non-increasing;
  now
A3: X /\ dom h c= Y /\ dom h by A1,XBOOLE_1:26;
    let r1,r2;
    assume r1 in X /\ dom h & r2 in X /\ dom h & r1 < r2;
    hence h.r1 >= h.r2 by A2,A3,Th23;
  end;
  hence thesis by Th23;
end;

theorem Th32:
  (h|Y is increasing & 0<r implies (r(#)h)|Y is increasing) & (r =
  0 implies (r(#)h)|Y is constant) & (h|Y is increasing & r<0 implies (r(#)h)|Y
  is decreasing)
proof
  thus h|Y is increasing & 0<r implies (r(#)h)|Y is increasing
  proof
    assume that
A1: h|Y is increasing and
A2: 0<r;
    now
      let r1,r2;
      assume that
A3:   r1 in Y /\ dom (r(#)h) and
A4:   r2 in Y /\ dom (r(#) h) and
A5:   r1<r2;
A6:   r2 in Y by A4,XBOOLE_0:def 4;
A7:   r2 in dom (r(#) h) by A4,XBOOLE_0:def 4;
      then r2 in dom h by VALUED_1:def 5;
      then
A8:   r2 in Y /\ dom h by A6,XBOOLE_0:def 4;
A9:   r1 in Y by A3,XBOOLE_0:def 4;
A10:  r1 in dom (r(#)h) by A3,XBOOLE_0:def 4;
      then r1 in dom h by VALUED_1:def 5;
      then r1 in Y /\ dom h by A9,XBOOLE_0:def 4;
      then h.r1 < h.r2 by A1,A5,A8,Th20;
      then r* h.r1 < r* h.r2 by A2,XREAL_1:68;
      then (r(#)h).r1 < r * h.r2 by A10,VALUED_1:def 5;
      hence (r(#)h).r1 < (r(#)h).r2 by A7,VALUED_1:def 5;
    end;
    hence thesis by Th20;
  end;
  thus r = 0 implies (r(#)h)|Y is constant
  proof
    assume
A11: r = 0;
    now
      let r1;
      assume r1 in Y /\ dom (r(#)h);
      then
A12:  r1 in dom (r(#)h) by XBOOLE_0:def 4;
      r* h.r1 = 0 by A11;
      hence (r(#)h).r1 = 0 by A12,VALUED_1:def 5;
    end;
    hence thesis by PARTFUN2:57;
  end;
  assume that
A13: h|Y is increasing and
A14: r<0;
  now
    let r1,r2;
    assume that
A15: r1 in Y /\ dom (r(#)h) and
A16: r2 in Y /\ dom (r(#) h) and
A17: r1<r2;
A18: r2 in Y by A16,XBOOLE_0:def 4;
A19: r2 in dom (r(#) h) by A16,XBOOLE_0:def 4;
    then r2 in dom h by VALUED_1:def 5;
    then
A20: r2 in Y /\ dom h by A18,XBOOLE_0:def 4;
A21: r1 in Y by A15,XBOOLE_0:def 4;
A22: r1 in dom (r(#)h) by A15,XBOOLE_0:def 4;
    then r1 in dom h by VALUED_1:def 5;
    then r1 in Y /\ dom h by A21,XBOOLE_0:def 4;
    then h.r1 < h.r2 by A13,A17,A20,Th20;
    then r* h.r2 < r* h.r1 by A14,XREAL_1:69;
    then (r(#)h).r2 < r * h.r1 by A19,VALUED_1:def 5;
    hence (r(#)h).r2 < (r(#)h).r1 by A22,VALUED_1:def 5;
  end;
  hence thesis by Th21;
end;

theorem
  (h|Y is decreasing & 0<r implies (r(#)h)|Y is decreasing) & (h|Y is
  decreasing & r<0 implies (r(#)h)|Y is increasing)
proof
  thus h|Y is decreasing & 0<r implies (r(#)h)|Y is decreasing
  proof
    assume that
A1: h|Y is decreasing and
A2: 0<r;
    now
      let r1,r2;
      assume that
A3:   r1 in Y /\ dom(r(#)h) and
A4:   r2 in Y /\ dom(r(#) h) and
A5:   r1<r2;
A6:   r2 in Y by A4,XBOOLE_0:def 4;
A7:   r2 in dom (r(#) h) by A4,XBOOLE_0:def 4;
      then r2 in dom h by VALUED_1:def 5;
      then
A8:   r2 in Y /\ dom h by A6,XBOOLE_0:def 4;
A9:   r1 in Y by A3,XBOOLE_0:def 4;
A10:  r1 in dom (r(#)h) by A3,XBOOLE_0:def 4;
      then r1 in dom h by VALUED_1:def 5;
      then r1 in Y /\ dom h by A9,XBOOLE_0:def 4;
      then h.r2 < h.r1 by A1,A5,A8,Th21;
      then r * h.r2 < r * h.r1 by A2,XREAL_1:68;
      then (r(#)h).r2 < r * h.r1 by A7,VALUED_1:def 5;
      hence (r(#)h).r2 < (r(#)h).r1 by A10,VALUED_1:def 5;
    end;
    hence thesis by Th21;
  end;
  assume that
A11: h|Y is decreasing and
A12: r<0;
  now
    let r1,r2;
    assume that
A13: r1 in Y /\ dom(r(#)h) and
A14: r2 in Y /\ dom(r(#) h) and
A15: r1<r2;
A16: r2 in Y by A14,XBOOLE_0:def 4;
A17: r2 in dom (r(#) h) by A14,XBOOLE_0:def 4;
    then r2 in dom h by VALUED_1:def 5;
    then
A18: r2 in Y /\ dom h by A16,XBOOLE_0:def 4;
A19: r1 in Y by A13,XBOOLE_0:def 4;
A20: r1 in dom (r(#)h) by A13,XBOOLE_0:def 4;
    then r1 in dom h by VALUED_1:def 5;
    then r1 in Y /\ dom h by A19,XBOOLE_0:def 4;
    then h.r2 < h.r1 by A11,A15,A18,Th21;
    then r* h.r1 < r* h.r2 by A12,XREAL_1:69;
    then (r(#) h).r1 < r* h.r2 by A20,VALUED_1:def 5;
    hence (r(#) h).r1 < (r(#) h).r2 by A17,VALUED_1:def 5;
  end;
  hence thesis by Th20;
end;

theorem Th34:
  (h|Y is non-decreasing & 0<=r implies (r(#)h)|Y is
  non-decreasing) & (h|Y is non-decreasing & r<=0 implies (r(#)h)|Y is
  non-increasing)
proof
  thus h|Y is non-decreasing & 0<=r implies (r(#)h)|Y is non-decreasing
  proof
    assume that
A1: h|Y is non-decreasing and
A2: 0<=r;
    now
      let r1,r2;
      assume that
A3:   r1 in Y /\ dom(r(#)h) and
A4:   r2 in Y /\ dom(r(#) h) and
A5:   r1<r2;
A6:   r2 in Y by A4,XBOOLE_0:def 4;
A7:   r2 in dom (r(#) h) by A4,XBOOLE_0:def 4;
      then r2 in dom h by VALUED_1:def 5;
      then
A8:   r2 in Y /\ dom h by A6,XBOOLE_0:def 4;
A9:   r1 in Y by A3,XBOOLE_0:def 4;
A10:  r1 in dom (r(#)h) by A3,XBOOLE_0:def 4;
      then r1 in dom h by VALUED_1:def 5;
      then r1 in Y /\ dom h by A9,XBOOLE_0:def 4;
      then h.r1 <= h.r2 by A1,A5,A8,Th22;
      then r* h.r1 <= (h.r2)*r by A2,XREAL_1:64;
      then (r(#)h).r1 <= r * h.r2 by A10,VALUED_1:def 5;
      hence (r(#)h).r1 <= (r(#)h).r2 by A7,VALUED_1:def 5;
    end;
    hence thesis by Th22;
  end;
  assume that
A11: h|Y is non-decreasing and
A12: r<=0;
  now
    let r1,r2;
    assume that
A13: r1 in Y /\ dom (r(#)h) and
A14: r2 in Y /\ dom (r(#) h) and
A15: r1<r2;
A16: r2 in Y by A14,XBOOLE_0:def 4;
A17: r2 in dom (r(#) h) by A14,XBOOLE_0:def 4;
    then r2 in dom h by VALUED_1:def 5;
    then
A18: r2 in Y /\ dom h by A16,XBOOLE_0:def 4;
A19: r1 in Y by A13,XBOOLE_0:def 4;
A20: r1 in dom (r(#)h) by A13,XBOOLE_0:def 4;
    then r1 in dom h by VALUED_1:def 5;
    then r1 in Y /\ dom h by A19,XBOOLE_0:def 4;
    then h.r1<=h.r2 by A11,A15,A18,Th22;
    then r*h.r2<=r*h.r1 by A12,XREAL_1:65;
    then (r(#)h).r2<=r*h.r1 by A17,VALUED_1:def 5;
    hence (r(#)h).r2<=(r(#)h).r1 by A20,VALUED_1:def 5;
  end;
  hence thesis by Th23;
end;

theorem
  (h|Y is non-increasing & 0<=r implies (r(#)h)|Y is non-increasing) & (
  h|Y is non-increasing & r<=0 implies (r(#)h)|Y is non-decreasing)
proof
  thus h|Y is non-increasing & 0<=r implies (r(#)h)|Y is non-increasing
  proof
    assume that
A1: h|Y is non-increasing and
A2: 0<=r;
    now
      let r1,r2;
      assume that
A3:   r1 in Y /\ dom(r(#)h) and
A4:   r2 in Y /\ dom(r(#) h) and
A5:   r1<r2;
A6:   r2 in Y by A4,XBOOLE_0:def 4;
A7:   r2 in dom (r(#) h) by A4,XBOOLE_0:def 4;
      then r2 in dom h by VALUED_1:def 5;
      then
A8:   r2 in Y /\ dom h by A6,XBOOLE_0:def 4;
A9:   r1 in Y by A3,XBOOLE_0:def 4;
A10:  r1 in dom (r(#)h) by A3,XBOOLE_0:def 4;
      then r1 in dom h by VALUED_1:def 5;
      then r1 in Y /\ dom h by A9,XBOOLE_0:def 4;
      then h.r2 <= h.r1 by A1,A5,A8,Th23;
      then r* h.r2 <= (h.r1)*r by A2,XREAL_1:64;
      then (r(#)h).r2 <= r * h.r1 by A7,VALUED_1:def 5;
      hence (r(#)h).r2 <= (r(#)h).r1 by A10,VALUED_1:def 5;
    end;
    hence thesis by Th23;
  end;
  assume that
A11: h|Y is non-increasing and
A12: r<=0;
  now
    let r1,r2;
    assume that
A13: r1 in Y /\ dom (r(#)h) and
A14: r2 in Y /\ dom (r(#)h) and
A15: r1<r2;
A16: r2 in Y by A14,XBOOLE_0:def 4;
A17: r2 in dom (r(#) h) by A14,XBOOLE_0:def 4;
    then r2 in dom h by VALUED_1:def 5;
    then
A18: r2 in Y /\ dom h by A16,XBOOLE_0:def 4;
A19: r1 in Y by A13,XBOOLE_0:def 4;
A20: r1 in dom (r(#)h) by A13,XBOOLE_0:def 4;
    then r1 in dom h by VALUED_1:def 5;
    then r1 in Y /\ dom h by A19,XBOOLE_0:def 4;
    then h.r2 <= h.r1 by A11,A15,A18,Th23;
    then r* h.r1 <= r* h.r2 by A12,XREAL_1:65;
    then (r(#)h).r1 <= r * h.r2 by A20,VALUED_1:def 5;
    hence (r(#)h).r1 <= (r(#)h).r2 by A17,VALUED_1:def 5;
  end;
  hence thesis by Th22;
end;

theorem Th36:
  r in X /\ Y /\ dom (h1+h2) implies r in X /\ dom h1 & r in Y /\ dom h2
proof
  assume
A1: r in X /\ Y /\ dom (h1+h2);
  then r in dom (h1+h2) by XBOOLE_0:def 4;
  then r in dom h1 /\ dom h2 by VALUED_1:def 1;
  then
A2: r in dom h1 & r in dom h2 by XBOOLE_0:def 4;
  r in X /\ Y by A1,XBOOLE_0:def 4;
  then r in X & r in Y by XBOOLE_0:def 4;
  hence thesis by A2,XBOOLE_0:def 4;
end;

theorem
  (h1|X is increasing & h2|Y is increasing implies (h1+h2)|(X /\ Y) is
increasing) & (h1|X is decreasing & h2|Y is decreasing implies (h1+h2)|(X /\ Y)
is decreasing) & (h1|X is non-decreasing & h2|Y is non-decreasing implies (h1+
  h2)|(X /\ Y) is non-decreasing) & (h1|X is non-increasing & h2|Y is
  non-increasing implies (h1+h2)|(X /\ Y) is non-increasing)
proof
  thus h1|X is increasing & h2|Y is increasing implies (h1+h2)|(X /\ Y) is
  increasing
  proof
    assume that
A1: h1|X is increasing and
A2: h2|Y is increasing;
    now
      let r1,r2;
      assume that
A3:   r1 in X /\ Y /\ dom (h1+h2) and
A4:   r2 in X /\ Y /\ dom (h1+h2) and
A5:   r1<r2;
      r1 in Y /\ dom h2 & r2 in Y /\ dom h2 by A3,A4,Th36;
      then
A6:   h2.r1 < h2.r2 by A2,A5,Th20;
A7:   r1 in dom (h1+h2) by A3,XBOOLE_0:def 4;
      r1 in X /\ dom h1 & r2 in X /\ dom h1 by A3,A4,Th36;
      then h1.r1 < h1.r2 by A1,A5,Th20;
      then h1.r1+h2.r1 < h1.r2+h2.r2 by A6,XREAL_1:8;
      then
A8:   (h1+h2).r1 < h1.r2 + h2.r2 by A7,VALUED_1:def 1;
      r2 in dom (h1+h2) by A4,XBOOLE_0:def 4;
      hence (h1+h2).r1 < (h1+h2).r2 by A8,VALUED_1:def 1;
    end;
    hence thesis by Th20;
  end;
  thus h1|X is decreasing & h2|Y is decreasing implies (h1+h2)|(X /\ Y) is
  decreasing
  proof
    assume that
A9: h1|X is decreasing and
A10: h2|Y is decreasing;
    now
      let r1,r2;
      assume that
A11:  r1 in X /\ Y /\ dom (h1+h2) and
A12:  r2 in X /\ Y /\ dom (h1+h2) and
A13:  r1<r2;
      r1 in Y /\ dom h2 & r2 in Y /\ dom h2 by A11,A12,Th36;
      then
A14:  h2.r2 < h2.r1 by A10,A13,Th21;
A15:  r2 in dom (h1+h2) by A12,XBOOLE_0:def 4;
      r1 in X /\ dom h1 & r2 in X /\ dom h1 by A11,A12,Th36;
      then h1.r2 < h1.r1 by A9,A13,Th21;
      then h1.r2+h2.r2 < h1.r1+h2.r1 by A14,XREAL_1:8;
      then
A16:  (h1+h2).r2 < h1.r1 + h2.r1 by A15,VALUED_1:def 1;
      r1 in dom (h1+h2) by A11,XBOOLE_0:def 4;
      hence (h1+h2).r2 < (h1+h2).r1 by A16,VALUED_1:def 1;
    end;
    hence thesis by Th21;
  end;
  thus h1|X is non-decreasing & h2|Y is non-decreasing implies (h1+h2)|(X /\ Y
  ) is non-decreasing
  proof
    assume that
A17: h1|X is non-decreasing and
A18: h2|Y is non-decreasing;
    now
      let r1,r2;
      assume that
A19:  r1 in X /\ Y /\ dom (h1+h2) and
A20:  r2 in X /\ Y /\ dom (h1+h2) and
A21:  r1<r2;
      r1 in Y /\ dom h2 & r2 in Y /\ dom h2 by A19,A20,Th36;
      then
A22:  h2.r1 <= h2.r2 by A18,A21,Th22;
A23:  r1 in dom (h1+h2) by A19,XBOOLE_0:def 4;
      r1 in X /\ dom h1 & r2 in X /\ dom h1 by A19,A20,Th36;
      then h1.r1 <= h1.r2 by A17,A21,Th22;
      then h1.r1+h2.r1 <= h1.r2+h2.r2 by A22,XREAL_1:7;
      then
A24:  (h1+h2).r1 <= h1.r2 + h2.r2 by A23,VALUED_1:def 1;
      r2 in dom (h1+h2) by A20,XBOOLE_0:def 4;
      hence (h1+h2).r1 <= (h1+h2).r2 by A24,VALUED_1:def 1;
    end;
    hence thesis by Th22;
  end;
  assume that
A25: h1|X is non-increasing and
A26: h2|Y is non-increasing;
  now
    let r1,r2;
    assume that
A27: r1 in X /\ Y /\ dom (h1+h2) and
A28: r2 in X /\ Y /\ dom (h1+h2) and
A29: r1<r2;
    r1 in Y /\ dom h2 & r2 in Y /\ dom h2 by A27,A28,Th36;
    then
A30: h2.r2 <= h2.r1 by A26,A29,Th23;
A31: r2 in dom (h1+h2) by A28,XBOOLE_0:def 4;
    r1 in X /\ dom h1 & r2 in X /\ dom h1 by A27,A28,Th36;
    then h1.r2 <= h1.r1 by A25,A29,Th23;
    then h1.r2+h2.r2 <= h1.r1+h2.r1 by A30,XREAL_1:7;
    then
A32: (h1+h2).r2 <= h1.r1 + h2.r1 by A31,VALUED_1:def 1;
    r1 in dom (h1+h2) by A27,XBOOLE_0:def 4;
    hence (h1+h2).r2 <= (h1+h2).r1 by A32,VALUED_1:def 1;
  end;
  hence thesis by Th23;
end;

theorem
  (h1|X is increasing & h2|Y is constant implies (h1+h2)|(X /\ Y) is
  increasing) & (h1|X is decreasing & h2|Y is constant implies (h1+h2)|(X /\ Y)
  is decreasing)
proof
  thus h1|X is increasing & h2|Y is constant implies (h1+h2)|(X /\ Y) is
  increasing
  proof
    assume that
A1: h1|X is increasing and
A2: h2|Y is constant;
    consider r such that
A3: for p st p in Y /\ dom h2 holds h2.p = r by A2,PARTFUN2:57;
    now
      let r1,r2;
      assume that
A4:   r1 in X /\ Y /\ dom (h1+h2) and
A5:   r2 in X /\ Y /\ dom (h1+h2) and
A6:   r1<r2;
      r1 in X /\ dom h1 & r2 in X /\ dom h1 by A4,A5,Th36;
      then h1.r1 < h1.r2 by A1,A6,Th20;
      then
A7:   h1.r1 + r < h1.r2 + r by XREAL_1:6;
      r1 in Y /\ dom h2 by A4,Th36;
      then
A8:   h1.r1 + h2.r1 < h1.r2 + r by A3,A7;
A9:   r1 in dom (h1+h2) by A4,XBOOLE_0:def 4;
      r2 in Y /\ dom h2 by A5,Th36;
      then h1.r1 + h2.r1 < h1.r2 + h2.r2 by A3,A8;
      then
A10:  (h1+h2).r1 < h1.r2 + h2.r2 by A9,VALUED_1:def 1;
      r2 in dom (h1+h2) by A5,XBOOLE_0:def 4;
      hence (h1+h2).r1 < (h1+h2).r2 by A10,VALUED_1:def 1;
    end;
    hence thesis by Th20;
  end;
  assume that
A11: h1|X is decreasing and
A12: h2|Y is constant;
  consider r such that
A13: for p st p in Y /\ dom h2 holds h2.p = r by A12,PARTFUN2:57;
  now
    let r1,r2;
    assume that
A14: r1 in X /\ Y /\ dom (h1+h2) and
A15: r2 in X /\ Y /\ dom (h1+h2) and
A16: r1<r2;
    r1 in X /\ dom h1 & r2 in X /\ dom h1 by A14,A15,Th36;
    then h1.r2 < h1.r1 by A11,A16,Th21;
    then
A17: h1.r2 + r < h1.r1 + r by XREAL_1:6;
    r2 in Y /\ dom h2 by A15,Th36;
    then
A18: h1.r2 + h2.r2 < h1.r1 + r by A13,A17;
A19: r2 in dom (h1+h2) by A15,XBOOLE_0:def 4;
    r1 in Y /\ dom h2 by A14,Th36;
    then h1.r2 + h2.r2 < h1.r1 + h2.r1 by A13,A18;
    then
A20: (h1+h2).r2 < h1.r1 + h2.r1 by A19,VALUED_1:def 1;
    r1 in dom (h1+h2) by A14,XBOOLE_0:def 4;
    hence (h1+h2).r2 < (h1+h2).r1 by A20,VALUED_1:def 1;
  end;
  hence thesis by Th21;
end;

theorem
  h1|X is increasing & h2|Y is non-decreasing implies (h1+h2)|(X /\ Y)
  is increasing
proof
  assume that
A1: h1|X is increasing and
A2: h2|Y is non-decreasing;
  now
    let r1,r2;
    assume that
A3: r1 in X /\ Y /\ dom (h1+h2) and
A4: r2 in X /\ Y /\ dom (h1+h2) and
A5: r1<r2;
A6: r2 in X /\ Y by A4,XBOOLE_0:def 4;
    then
A7: r2 in Y by XBOOLE_0:def 4;
A8: r2 in dom (h1+h2) by A4,XBOOLE_0:def 4;
    then
A9: r2 in dom h1 /\ dom h2 by VALUED_1:def 1;
    then r2 in dom h2 by XBOOLE_0:def 4;
    then
A10: r2 in Y /\ dom h2 by A7,XBOOLE_0:def 4;
A11: r1 in X /\ Y by A3,XBOOLE_0:def 4;
    then
A12: r1 in Y by XBOOLE_0:def 4;
A13: r1 in dom (h1+h2) by A3,XBOOLE_0:def 4;
    then
A14: r1 in dom h1 /\ dom h2 by VALUED_1:def 1;
    then r1 in dom h2 by XBOOLE_0:def 4;
    then r1 in Y /\ dom h2 by A12,XBOOLE_0:def 4;
    then
A15: h2.r1 <= h2.r2 by A2,A5,A10,Th22;
A16: r2 in X by A6,XBOOLE_0:def 4;
A17: r1 in X by A11,XBOOLE_0:def 4;
    r2 in dom h1 by A9,XBOOLE_0:def 4;
    then
A18: r2 in X /\ dom h1 by A16,XBOOLE_0:def 4;
    r1 in dom h1 by A14,XBOOLE_0:def 4;
    then r1 in X /\ dom h1 by A17,XBOOLE_0:def 4;
    then h1.r1 < h1.r2 by A1,A5,A18,Th20;
    then h1.r1 + h2.r1 < h1.r2 + h2.r2 by A15,XREAL_1:8;
    then (h1+h2).r1 < h1.r2 + h2.r2 by A13,VALUED_1:def 1;
    hence (h1+h2).r1 < (h1+h2).r2 by A8,VALUED_1:def 1;
  end;
  hence thesis by Th20;
end;

theorem
  h1|X is non-increasing & h2|Y is constant implies (h1+h2)|(X /\ Y) is
  non-increasing
proof
  assume that
A1: h1|X is non-increasing and
A2: h2|Y is constant;
  now
    let r1,r2;
    assume that
A3: r1 in X /\ Y /\ dom (h1+h2) and
A4: r2 in X /\ Y /\ dom (h1+h2) and
A5: r1<r2;
A6: r2 in X /\ Y by A4,XBOOLE_0:def 4;
    then
A7: r2 in X by XBOOLE_0:def 4;
A8: r2 in Y by A6,XBOOLE_0:def 4;
A9: r2 in dom (h1+h2) by A4,XBOOLE_0:def 4;
    then
A10: r2 in dom h1 /\ dom h2 by VALUED_1:def 1;
    then r2 in dom h2 by XBOOLE_0:def 4;
    then
A11: r2 in Y /\ dom h2 by A8,XBOOLE_0:def 4;
    r2 in dom h1 by A10,XBOOLE_0:def 4;
    then
A12: r2 in X /\ dom h1 by A7,XBOOLE_0:def 4;
A13: r1 in X /\ Y by A3,XBOOLE_0:def 4;
    then
A14: r1 in X by XBOOLE_0:def 4;
A15: r1 in Y by A13,XBOOLE_0:def 4;
A16: r1 in dom (h1+h2) by A3,XBOOLE_0:def 4;
    then
A17: r1 in dom h1 /\ dom h2 by VALUED_1:def 1;
    then r1 in dom h2 by XBOOLE_0:def 4;
    then r1 in Y /\ dom h2 by A15,XBOOLE_0:def 4;
    then
A18: h2.r2 = h2.r1 by A2,A11,PARTFUN2:58;
    r1 in dom h1 by A17,XBOOLE_0:def 4;
    then r1 in X /\ dom h1 by A14,XBOOLE_0:def 4;
    then h1.r2 <= h1.r1 by A1,A5,A12,Th23;
    then h1.r2 + h2.r2 <= h1.r1 + h2.r1 by A18,XREAL_1:6;
    then (h1+h2).r2 <= h1.r1 + h2.r1 by A9,VALUED_1:def 1;
    hence (h1+h2).r2 <= (h1+h2).r1 by A16,VALUED_1:def 1;
  end;
  hence thesis by Th23;
end;

theorem
  h1|X is decreasing & h2|Y is non-increasing implies (h1+h2)|(X /\ Y)
  is decreasing
proof
  assume that
A1: h1|X is decreasing and
A2: h2|Y is non-increasing;
  now
    let r1,r2;
    assume that
A3: r1 in X /\ Y /\ dom (h1+h2) and
A4: r2 in X /\ Y /\ dom (h1+h2) and
A5: r1<r2;
A6: r2 in X /\ Y by A4,XBOOLE_0:def 4;
    then
A7: r2 in Y by XBOOLE_0:def 4;
A8: r2 in dom (h1+h2) by A4,XBOOLE_0:def 4;
    then
A9: r2 in dom h1 /\ dom h2 by VALUED_1:def 1;
    then r2 in dom h2 by XBOOLE_0:def 4;
    then
A10: r2 in Y /\ dom h2 by A7,XBOOLE_0:def 4;
A11: r1 in X /\ Y by A3,XBOOLE_0:def 4;
    then
A12: r1 in Y by XBOOLE_0:def 4;
A13: r1 in dom (h1+h2) by A3,XBOOLE_0:def 4;
    then
A14: r1 in dom h1 /\ dom h2 by VALUED_1:def 1;
    then r1 in dom h2 by XBOOLE_0:def 4;
    then r1 in Y /\ dom h2 by A12,XBOOLE_0:def 4;
    then
A15: h2.r2 <= h2.r1 by A2,A5,A10,Th23;
A16: r2 in X by A6,XBOOLE_0:def 4;
A17: r1 in X by A11,XBOOLE_0:def 4;
    r2 in dom h1 by A9,XBOOLE_0:def 4;
    then
A18: r2 in X /\ dom h1 by A16,XBOOLE_0:def 4;
    r1 in dom h1 by A14,XBOOLE_0:def 4;
    then r1 in X /\ dom h1 by A17,XBOOLE_0:def 4;
    then h1.r2 < h1.r1 by A1,A5,A18,Th21;
    then h1.r2 + h2.r2 < h1.r1 + h2.r1 by A15,XREAL_1:8;
    then (h1+h2).r2 < h1.r1 + h2.r1 by A8,VALUED_1:def 1;
    hence (h1+h2).r2 < (h1+h2).r1 by A13,VALUED_1:def 1;
  end;
  hence thesis by Th21;
end;

theorem
  h1|X is non-decreasing & h2|Y is constant implies (h1+h2)|(X /\ Y) is
  non-decreasing
proof
  assume that
A1: h1|X is non-decreasing and
A2: h2|Y is constant;
  now
    let r1,r2;
    assume that
A3: r1 in X /\ Y /\ dom (h1+h2) and
A4: r2 in X /\ Y /\ dom (h1+h2) and
A5: r1<r2;
A6: r2 in X /\ Y by A4,XBOOLE_0:def 4;
    then
A7: r2 in X by XBOOLE_0:def 4;
A8: r2 in Y by A6,XBOOLE_0:def 4;
A9: r2 in dom (h1+h2) by A4,XBOOLE_0:def 4;
    then
A10: r2 in dom h1 /\ dom h2 by VALUED_1:def 1;
    then r2 in dom h2 by XBOOLE_0:def 4;
    then
A11: r2 in Y /\ dom h2 by A8,XBOOLE_0:def 4;
    r2 in dom h1 by A10,XBOOLE_0:def 4;
    then
A12: r2 in X /\ dom h1 by A7,XBOOLE_0:def 4;
A13: r1 in X /\ Y by A3,XBOOLE_0:def 4;
    then
A14: r1 in X by XBOOLE_0:def 4;
A15: r1 in Y by A13,XBOOLE_0:def 4;
A16: r1 in dom (h1+h2) by A3,XBOOLE_0:def 4;
    then
A17: r1 in dom h1 /\ dom h2 by VALUED_1:def 1;
    then r1 in dom h2 by XBOOLE_0:def 4;
    then r1 in Y /\ dom h2 by A15,XBOOLE_0:def 4;
    then
A18: h2.r2 = h2.r1 by A2,A11,PARTFUN2:58;
    r1 in dom h1 by A17,XBOOLE_0:def 4;
    then r1 in X /\ dom h1 by A14,XBOOLE_0:def 4;
    then h1.r1 <= h1.r2 by A1,A5,A12,Th22;
    then h1.r1 + h2.r1 <= h1.r2 + h2.r2 by A18,XREAL_1:6;
    then (h1+h2).r1 <= h1.r2 + h2.r2 by A16,VALUED_1:def 1;
    hence (h1+h2).r1 <= (h1+h2).r2 by A9,VALUED_1:def 1;
  end;
  hence thesis by Th22;
end;

theorem
  h|{x} is non-increasing
proof
  now
    let r1,r2;
    assume that
A1: r1 in {x} /\ dom h and
A2: r2 in {x} /\ dom h and
    r1<r2;
    r1 in {x} by A1,XBOOLE_0:def 4;
    then
A3: r1 = x by TARSKI:def 1;
    r2 in {x} by A2,XBOOLE_0:def 4;
    hence h.r1 >= h.r2 by A3,TARSKI:def 1;
  end;
  hence thesis by Th23;
end;

theorem
  h|{x} is decreasing
proof
  now
    let r1,r2;
    assume that
A1: r1 in {x} /\ dom h and
A2: r2 in {x} /\ dom h and
A3: r1<r2;
    r1 in {x} by A1,XBOOLE_0:def 4;
    then
A4: r1 = x by TARSKI:def 1;
    r2 in {x} by A2,XBOOLE_0:def 4;
    hence h.r1 > h.r2 by A3,A4,TARSKI:def 1;
  end;
  hence thesis by Th21;
end;

theorem
  h|{x} is non-decreasing
proof
  now
    let r1,r2;
    assume that
A1: r1 in {x} /\ dom h and
A2: r2 in {x} /\ dom h and
    r1<r2;
    r1 in {x} by A1,XBOOLE_0:def 4;
    then
A3: r1 = x by TARSKI:def 1;
    r2 in {x} by A2,XBOOLE_0:def 4;
    hence h.r1 <= h.r2 by A3,TARSKI:def 1;
  end;
  hence thesis by Th22;
end;

theorem
  h|{x} is non-increasing
proof
  now
    let r1,r2;
    assume that
A1: r1 in {x} /\ dom h and
A2: r2 in {x} /\ dom h and
    r1<r2;
    r1 in {x} by A1,XBOOLE_0:def 4;
    then
A3: r1 = x by TARSKI:def 1;
    r2 in {x} by A2,XBOOLE_0:def 4;
    hence h.r2 <= h.r1 by A3,TARSKI:def 1;
  end;
  hence thesis by Th23;
end;

theorem
  (id R)|R is increasing
proof
  now
    let r1,r2;
    assume that
A1: r1 in R /\ dom (id R) and
A2: r2 in R /\ dom (id R) and
A3: r1<r2;
    r1 in R by A1,XBOOLE_0:def 4;
    then
A4: (id R).r1 = r1 by FUNCT_1:18;
    r2 in R by A2,XBOOLE_0:def 4;
    hence (id R).r1 < (id R).r2 by A3,A4,FUNCT_1:18;
  end;
  hence thesis by Th20;
end;

theorem
  h|X is increasing implies (-h)|X is decreasing
proof
  assume h|X is increasing;
  then ((-1)(#)h)|X is decreasing by Th32;
  hence thesis;
end;

theorem
  h|X is non-decreasing implies (-h)|X is non-increasing
proof
  assume h|X is non-decreasing;
  then ((-1)(#)h)|X is non-increasing by Th34;
  hence thesis;
end;

theorem
  (h|[.p,g.] is increasing or h|[.p,g.] is decreasing) implies h|[.p,g.]
  is one-to-one
proof
  assume
A1: h|[.p,g.] is increasing or h|[.p,g.] is decreasing;
  now
    per cases by A1;
    suppose
A2:   h|[.p,g.] is increasing;
      now
        let p1,p2;
        assume that
A3:     p1 in dom(h|[.p,g.]) and
A4:     p2 in dom(h|[.p,g.]) and
A5:     (h|[.p,g.]).p1 = (h|[.p,g.]).p2;
A6:     p1 in [.p,g.] /\ dom h & p2 in [.p,g.] /\ dom h by A3,A4,RELAT_1:61;
A7:     h.p1=(h|[.p,g.]).p2 by A3,A5,FUNCT_1:47
          .=h.p2 by A4,FUNCT_1:47;
        thus p1=p2
        proof
          assume
A8:       p1<>p2;
          now
            per cases by A8,XXREAL_0:1;
            suppose
              p1<p2;
              hence contradiction by A2,A7,A6,Th20;
            end;
            suppose
              p2<p1;
              hence contradiction by A2,A7,A6,Th20;
            end;
          end;
          hence contradiction;
        end;
      end;
      hence thesis by PARTFUN1:8;
    end;
    suppose
A9:   h|[.p,g.] is decreasing;
      now
        let p1,p2;
        assume that
A10:    p1 in dom(h|[.p,g.]) and
A11:    p2 in dom(h|[.p,g.]) and
A12:    (h|[.p,g.]).p1 = (h|[.p,g.]).p2;
A13:    p1 in [.p,g.] /\ dom h & p2 in [.p,g.] /\ dom h by A10,A11,RELAT_1:61;
A14:    h.p1=(h|[.p,g.]).p2 by A10,A12,FUNCT_1:47
          .=h.p2 by A11,FUNCT_1:47;
        thus p1=p2
        proof
          assume
A15:      p1<>p2;
          now
            per cases by A15,XXREAL_0:1;
            suppose
              p1<p2;
              hence contradiction by A9,A14,A13,Th21;
            end;
            suppose
              p2<p1;
              hence contradiction by A9,A14,A13,Th21;
            end;
          end;
          hence contradiction;
        end;
      end;
      hence thesis by PARTFUN1:8;
    end;
  end;
  hence thesis;
end;

theorem
  for h being one-to-one PartFunc of REAL, REAL st h|[.p,g.] is
  increasing holds (h|[.p,g.])"|(h.:[.p,g.]) is increasing
proof
  let h be one-to-one PartFunc of REAL, REAL;
  assume that
A1: h|[.p,g.] is increasing and
A2: (h|[.p,g.])"|(h.:[.p,g.]) is not increasing;
  consider y1,y2 be Real such that
A3: y1 in h.:[.p,g.] /\ dom((h|[.p,g.])") and
A4: y2 in h.:[.p,g.] /\ dom((h|[.p,g.])") and
A5: y1<y2 and
A6: ((h|[.p,g.])").y1 >= ((h|[.p,g.])").y2 by A2,Th20;
  y1 in h.:[.p,g.] by A3,XBOOLE_0:def 4;
  then
A7: y1 in rng (h|[.p,g.]) by RELAT_1:115;
  y2 in h.:[.p,g.] by A4,XBOOLE_0:def 4;
  then
A8: y2 in rng (h|[.p,g.]) by RELAT_1:115;
A9: h|[.p,g.]|[.p,g.] is increasing by A1;
  now
    per cases;
    suppose
      ((h|[.p,g.])").y1 = ((h|[.p,g.])").y2;
      then y1 =(h|[.p,g.]).(((h|[.p,g.])").y2) by A7,FUNCT_1:35
        .=y2 by A8,FUNCT_1:35;
      hence contradiction by A5;
    end;
    suppose
A10:  ((h|[.p,g.])").y1 <> ((h|[.p,g.])").y2;
A11:  dom (h|[.p,g.])=dom ((h|[.p,g.])|[.p,g.]) by RELAT_1:72
        .=[.p,g.]/\dom(h|[.p,g.]) by RELAT_1:61;
A12:  ((h|[.p,g.])").y2 in dom (h|[.p,g.]) & ((h|[.p,g.])").y1 in dom (h|
      [.p,g.]) by A7,A8,PARTFUN2:60;
      ((h|[.p,g.])").y2 < ((h|[.p,g.])").y1 by A6,A10,XXREAL_0:1;
      then (h|[.p,g.]).(((h|[.p,g.])").y2) < (h|[.p,g.]).(((h|[.p,g.])").y1)
      by A9,A12,A11,Th20;
      then y2 < (h|[.p,g.]).(((h|[.p,g.])").y1) by A8,FUNCT_1:35;
      hence contradiction by A5,A7,FUNCT_1:35;
    end;
  end;
  hence contradiction;
end;

theorem
  for h being one-to-one PartFunc of REAL, REAL st h|[.p,g.] is
  decreasing holds (h|[.p,g.])"|(h.:[.p,g.]) is decreasing
proof
  let h be one-to-one PartFunc of REAL, REAL;
  assume that
A1: h|[.p,g.] is decreasing and
A2: not (h|[.p,g.])"|(h.:[.p,g.]) is decreasing;
  consider y1,y2 be Real such that
A3: y1 in h.:[.p,g.] /\ dom((h|[.p,g.])") and
A4: y2 in h.:[.p,g.] /\ dom((h|[.p,g.])") and
A5: y1<y2 and
A6: ((h|[.p,g.])").y2 >= ((h|[.p,g.])").y1 by A2,Th21;
  y1 in h.:[.p,g.] by A3,XBOOLE_0:def 4;
  then
A7: y1 in rng (h|[.p,g.]) by RELAT_1:115;
  y2 in h.:[.p,g.] by A4,XBOOLE_0:def 4;
  then
A8: y2 in rng (h|[.p,g.]) by RELAT_1:115;
A9: h|[.p,g.]|[.p,g.] is decreasing by A1;
  now
    per cases;
    suppose
      ((h|[.p,g.])").y1 = ((h|[.p,g.])").y2;
      then y1 =(h|[.p,g.]).(((h|[.p,g.])").y2) by A7,FUNCT_1:35
        .=y2 by A8,FUNCT_1:35;
      hence contradiction by A5;
    end;
    suppose
A10:  ((h|[.p,g.])").y1 <> ((h|[.p,g.])").y2;
A11:  dom (h|[.p,g.])=dom ((h|[.p,g.])|[.p,g.]) by RELAT_1:72
        .=[.p,g.]/\dom(h|[.p,g.]) by RELAT_1:61;
A12:  ((h|[.p,g.])").y2 in dom (h|[.p,g.]) & ((h|[.p,g.])").y1 in dom (h|
      [.p,g.]) by A7,A8,PARTFUN2:60;
      ((h|[.p,g.])").y2 > ((h|[.p,g.])").y1 by A6,A10,XXREAL_0:1;
      then (h|[.p,g.]).(((h|[.p,g.])").y2) < (h|[.p,g.]).(((h|[.p,g.])").y1)
      by A9,A12,A11,Th21;
      then y2 < (h|[.p,g.]).(((h|[.p,g.])").y1) by A8,FUNCT_1:35;
      hence contradiction by A5,A7,FUNCT_1:35;
    end;
  end;
  hence contradiction;
end;














begin

reserve x,y,X,Y for set;
reserve C for non empty set;
reserve c for Element of C;
reserve f,f1,f2,f3,g,g1 for PartFunc of C,COMPLEX;
reserve r1,r2,p1 for real number;
reserve r,q,cr1,cr2 for Element of COMPLEX;





definition
  mode Complex is complex number;
end;

definition
  let C,f1,f2;
  deffunc F(set) = f1/.$1 * (f2/.$1)";
  func f1/f2 -> PartFunc of C,COMPLEX means
  :Def1:
  dom it = dom f1 /\ (dom f2
  \ f2"{0}) & for c st c in dom it holds it/.c = f1/.c * (f2/.c)";
  existence
  proof
    defpred P[set] means $1 in dom f1 /\ (dom f2 \ f2"{0c});
    consider F being PartFunc of C,COMPLEX such that
A1: for c holds c in dom F iff P[c] and
A2: for c st c in dom F holds F/.c = F(c) from PARTFUN2:sch 2;
    take F;
    thus dom F = dom f1 /\ (dom f2 \ f2"{0}) by A1,SUBSET_1:3;
    thus thesis by A2;
  end;
  uniqueness
  proof
    set X = dom f1 /\ (dom f2 \ f2"{0});
    thus for f,g being PartFunc of C,COMPLEX st (dom f=X & for c be Element of
C st c in dom f holds f/.c = F(c)) & (dom g=X & for c be Element of C st c in
    dom g holds g/.c = F(c)) holds f = g from PARTFUN2:sch 3;
  end;
end;

definition
  let C,f;
  deffunc F(set) = (f/.$1)";
  func f^ -> PartFunc of C,COMPLEX means
  :Def2:
  dom it = dom f \ f"{0} & for c st c in dom it holds it/.c = (f/.c)";
  existence
  proof
    defpred P[set] means $1 in dom f \ f"{0c};
    consider F being PartFunc of C,COMPLEX such that
A1: for c holds c in dom F iff P[c] and
A2: for c st c in dom F holds F/.c = F(c) from PARTFUN2:sch 2;
    take F;
    thus dom F = dom f \ f"{0} by A1,SUBSET_1:3;
    let c;
    assume c in dom F;
    hence thesis by A2;
  end;
  uniqueness
  proof
    set X = dom f \ f"{0};
    thus for f,g being PartFunc of C,COMPLEX st (dom f=X & for c be Element of
C st c in dom f holds f/.c = F(c)) & (dom g=X & for c be Element of C st c in
    dom g holds g/.c = F(c)) holds f = g from PARTFUN2:sch 3;
  end;
end;

theorem Th1:
  dom (f1+f2) = dom f1 /\ dom f2 & for c st c in dom(f1+f2) holds (
  f1+f2)/.c = (f1/.c) + (f2/.c)
proof
  thus
A1: dom (f1+f2) = dom f1 /\ dom f2 by VALUED_1:def 1;
  now
    let c;
    assume
A2: c in dom (f1+f2);
    then c in dom f1 by A1,XBOOLE_0:def 4;
    then
A3: f1.c = f1/.c by PARTFUN1:def 6;
    c in dom f2 by A1,A2,XBOOLE_0:def 4;
    then
A4: f2.c = f2/.c by PARTFUN1:def 6;
    thus (f1+f2)/.c = (f1+f2).c by A2,PARTFUN1:def 6
      .= f1/.c+f2/.c by A2,A3,A4,VALUED_1:def 1;
  end;
  hence thesis;
end;

theorem Th2:
  dom (f1-f2) = dom f1 /\ dom f2 & for c st c in dom(f1-f2) holds (
  f1-f2)/.c = (f1/.c) - (f2/.c)
proof
A1: dom (f1-f2) = dom f1 /\ dom(-f2) by VALUED_1:def 1;
  hence dom (f1-f2) = dom f1 /\ dom f2 by VALUED_1:8;
  now
    let c;
    assume
A2: c in dom (f1-f2);
    then
A3: dom -f2 = dom f2 & c in dom -f2 by A1,VALUED_1:8,XBOOLE_0:def 4;
    c in dom f1 by A1,A2,XBOOLE_0:def 4;
    then
A4: f1/.c = f1.c by PARTFUN1:def 6;
    thus (f1-f2)/.c = (f1-f2).c by A2,PARTFUN1:def 6
      .= f1.c - f2.c by A2,VALUED_1:13
      .= f1/.c - f2/.c by A3,A4,PARTFUN1:def 6;
  end;
  hence thesis;
end;

theorem Th3:
  dom(f1(#)f2)=dom f1 /\ dom f2 & for c st c in dom(f1(#)f2) holds
  (f1(#)f2)/.c =(f1/.c) * (f2/.c)
proof
  thus
A1: dom (f1(#)f2) = dom f1 /\ dom f2 by VALUED_1:def 4;
  now
    let c;
    assume
A2: c in dom (f1(#)f2);
    then c in dom f1 by A1,XBOOLE_0:def 4;
    then
A3: f1.c = f1/.c by PARTFUN1:def 6;
    c in dom f2 by A1,A2,XBOOLE_0:def 4;
    then
A4: f2.c = f2/.c by PARTFUN1:def 6;
    thus (f1(#)f2)/.c = (f1(#)f2).c by A2,PARTFUN1:def 6
      .= f1/.c * f2/.c by A2,A3,A4,VALUED_1:def 4;
  end;
  hence thesis;
end;

theorem Th4:
  dom (r(#)f) = dom f & for c st c in dom (r(#)f) holds (r(#)f)/.c = r * (f/.c)
proof
  thus
A1: dom (r(#)f) = dom f by VALUED_1:def 5;
  now
    let c;
    assume
A2: c in dom (r(#)f);
    hence (r(#)f)/.c = (r(#)f).c by PARTFUN1:def 6
      .= r *(f.c) by VALUED_1:6
      .= r *(f/.c) by A1,A2,PARTFUN1:def 6;
  end;
  hence thesis;
end;

theorem Th5:
  dom (-f) = dom f & for c st c in dom (-f) holds (-f)/.c = -f/.c
proof
  thus
A1: dom (-f) = dom f by VALUED_1:8;
  now
    let c;
    assume
A2: c in dom (-f);
    hence (-f)/.c = (-f).c by PARTFUN1:def 6
      .=-f.c by VALUED_1:8
      .=-f/.c by A1,A2,PARTFUN1:def 6;
  end;
  hence thesis;
end;

Lm1: x in f"Y iff x in dom f & f/.x in Y by PARTFUN2:26;

theorem Th6:
  dom (g^) c= dom g & dom g /\ (dom g \ g"{0}) = dom g \ g"{0}
proof
  dom (g^) = dom g \ g"{0c} by Def2;
  hence dom (g^) c= dom g by XBOOLE_1:36;
  thus thesis by XBOOLE_1:28,36;
end;

theorem Th7:
  dom (f1(#)f2) \ (f1(#)f2)"{0} = (dom f1 \ (f1)"{0}) /\ (dom f2 \ (f2)"{0})
proof
  thus dom (f1(#)f2) \ (f1(#)f2)"{0} c= (dom f1 \ (f1)"{0}) /\ (dom f2 \ (f2)"
  {0})
  proof
    let x;
    assume
A1: x in dom (f1(#)f2) \ (f1(#)f2)"{0};
    then
A2: x in dom (f1(#)f2) by XBOOLE_0:def 5;
    reconsider x1=x as Element of C by A1;
    not x in (f1(#)f2)"{0c} by A1,XBOOLE_0:def 5;
    then not (f1(#)f2)/.x1 in {0c} by A2,PARTFUN2:26;
    then (f1(#)f2)/.x1 <> 0c by TARSKI:def 1;
    then
A3: (f1/.x1) * (f2/.x1) <> 0c by A2,Th3;
    then (f2/.x1) <> 0c;
    then not (f2/.x1) in {0c } by TARSKI:def 1;
    then
A4: not x1 in (f2)"{0c} by PARTFUN2:26;
A5: x1 in dom f1 /\ dom f2 by A2,Th3;
    then x1 in dom f2 by XBOOLE_0:def 4;
    then
A6: x in dom f2 \ (f2)"{0c} by A4,XBOOLE_0:def 5;
    (f1/.x1) <> 0c by A3;
    then not (f1/.x1) in {0c} by TARSKI:def 1;
    then
A7: not x1 in (f1)"{0c} by PARTFUN2:26;
    x1 in dom f1 by A5,XBOOLE_0:def 4;
    then x in dom f1 \ (f1)"{0c} by A7,XBOOLE_0:def 5;
    hence thesis by A6,XBOOLE_0:def 4;
  end;
  thus (dom f1 \ (f1)"{0}) /\ (dom f2 \ (f2)"{0}) c= dom (f1(#)f2) \ (f1(#)f2)
  "{0}
  proof
    let x;
    assume
A8: x in (dom f1 \ (f1)"{0}) /\ (dom f2 \ (f2)"{0});
    then reconsider x1=x as Element of C;
A9: x in dom f2 \ (f2)"{0c} by A8,XBOOLE_0:def 4;
    then
A10: x in dom f2 by XBOOLE_0:def 5;
    not x in (f2)"{0c } by A9,XBOOLE_0:def 5;
    then not (f2/.x1) in {0c} by A10,PARTFUN2:26;
    then
A11: (f2/.x1) <> 0c by TARSKI:def 1;
A12: x in dom f1 \ (f1)"{0c} by A8,XBOOLE_0:def 4;
    then
A13: x in dom f1 by XBOOLE_0:def 5;
    then x1 in dom f1 /\ dom f2 by A10,XBOOLE_0:def 4;
    then
A14: x1 in dom (f1(#)f2) by Th3;
    not x in (f1)"{0c} by A12,XBOOLE_0:def 5;
    then not (f1/.x1) in {0c} by A13,PARTFUN2:26;
    then (f1/.x1) <> 0c by TARSKI:def 1;
    then (f1/.x1) * (f2/.x1) <>0c by A11,XCMPLX_1:6;
    then (f1(#)f2)/.x1 <> 0c by A14,Th3;
    then not (f1(#)f2)/.x1 in {0c} by TARSKI:def 1;
    then not x in (f1(#)f2)"{0c} by PARTFUN2:26;
    hence thesis by A14,XBOOLE_0:def 5;
  end;
end;

theorem Th8:
  c in dom (f^) implies (f/.c) <> 0
proof
  assume that
A1: c in dom (f^) and
A2: (f/.c) = 0;
A3: c in dom f \ f"{0c} by A1,Def2;
  then
A4: not c in f"{0c} by XBOOLE_0:def 5;
  now
    per cases by A4,PARTFUN2:26;
    suppose
      not c in dom f;
      hence contradiction by A3,XBOOLE_0:def 5;
    end;
    suppose
      not (f/.c) in {0c};
      hence contradiction by A2,TARSKI:def 1;
    end;
  end;
  hence contradiction;
end;

theorem Th9:
  (f^)"{0} = {}
proof
  set x = the Element of (f^)"{0c};
  assume
A1: (f^)"{0} <> {};
  then
A2: x in dom (f^) by Lm1;
A3: (f^)/.x in {0c} by A1,Lm1;
  reconsider x as Element of C by A2;
  x in dom f \ f"{0c} by A2,Def2;
  then x in dom f & not x in f"{0c} by XBOOLE_0:def 5;
  then
A4: not f/.x in {0c} by PARTFUN2:26;
  (f^)/.x = 0c by A3,TARSKI:def 1;
  then (f/.x)" = 0c by A2,Def2;
  hence contradiction by A4,TARSKI:def 1,XCMPLX_1:202;
end;

theorem Th10:
  |.f.|"{0} = f"{0} & (-f)"{0} = f"{0}
proof
A1: dom |.f.| = dom f by VALUED_1:def 11;
  now
    let c;
    thus c in (|.f.|)"{0} implies c in f"{0c}
    proof
      assume
A2:   c in (|.f.|)"{0};
      then c in dom (|.f.|) by FUNCT_1:def 7;
      then
A3:   c in dom f by VALUED_1:def 11;
      (|.f.|).c in {0} by A2,FUNCT_1:def 7;
      then (|.f.|).c = 0 by TARSKI:def 1;
      then
A4:   |.(f.c).| = 0 by VALUED_1:18;
      c in dom (|.f.|) by A2,FUNCT_1:def 7;
      then f.c = f/.c by A1,PARTFUN1:def 6;
      then (f/.c) = 0c by A4,COMPLEX1:45;
      then (f/.c) in {0c} by TARSKI:def 1;
      hence thesis by A3,PARTFUN2:26;
    end;
    assume
A5: c in (f)"{0c};
    then (f/.c) in {0c} by PARTFUN2:26;
    then
A6: |.(f/.c).| = 0 by COMPLEX1:44,TARSKI:def 1;
A7: c in dom f by A5,PARTFUN2:26;
    then f.c = f/.c by PARTFUN1:def 6;
    then (|.f.|).c = 0 by A6,VALUED_1:18;
    then
A8: (|.f.|).c in {0} by TARSKI:def 1;
    c in dom (|.f.|) by A7,VALUED_1:def 11;
    hence c in (|.f.|)"{0} by A8,FUNCT_1:def 7;
  end;
  hence (|.f.|)"{0} = f"{0} by SUBSET_1:3;
  now
    let c;
    thus c in (-f)"{0c} implies c in f"{0c}
    proof
      assume
A9:   c in (-f)"{0c};
      then
A10:  c in dom (-f) by PARTFUN2:26;
      (-f)/.c in {0c} by A9,PARTFUN2:26;
      then (-f)/.c = 0c by TARSKI:def 1;
      then --((f/.c)) = -0c by A10,Th5;
      then
A11:  (f/.c) in {0c} by TARSKI:def 1;
      c in dom f by A10,Th5;
      hence thesis by A11,PARTFUN2:26;
    end;
    assume
A12: c in (f)"{0c};
    then (f/.c) in {0c} by PARTFUN2:26;
    then
A13: (f/.c) = 0c by TARSKI:def 1;
A14: c in dom f by A12,PARTFUN2:26;
    then c in dom (-f) by Th5;
    then (-f)/.c = -0c by A13,Th5;
    then
A15: (-f)/.c in {0c} by TARSKI:def 1;
    c in dom (-f) by A14,Th5;
    hence c in (-f)"{0c} by A15,PARTFUN2:26;
  end;
  hence thesis by SUBSET_1:3;
end;

theorem Th11:
  dom (f^^) = dom (f|(dom (f^)))
proof
A1: dom (f^) = dom f \ f"{0c} by Def2;
  thus dom (f^^) = dom (f^) \(f^)"{0c} by Def2
    .= dom (f^) \ {} by Th9
    .= dom f /\ dom (f^) by A1,XBOOLE_1:28,36
    .= dom (f|(dom (f^))) by RELAT_1:61;
end;

theorem Th12:
  r<>0 implies (r(#)f)"{0} = f"{0}
proof
  assume
A1: r<>0;
  now
    let c;
    thus c in (r(#)f)"{0c} implies c in f"{0c}
    proof
      assume
A2:   c in (r(#)f)"{0c};
      then
A3:   c in dom (r(#)f) by PARTFUN2:26;
      (r(#)f)/.c in {0c} by A2,PARTFUN2:26;
      then (r(#)f)/.c = 0c by TARSKI:def 1;
      then r*(f/.c) = 0c by A3,Th4;
      then (f/.c) = 0c by A1,XCMPLX_1:6;
      then
A4:   (f/.c) in {0c} by TARSKI:def 1;
      c in dom f by A3,Th4;
      hence thesis by A4,PARTFUN2:26;
    end;
    assume
A5: c in (f)"{0c};
    then (f/.c) in {0c} by PARTFUN2:26;
    then (f/.c) = 0c by TARSKI:def 1;
    then
A6: r*(f/.c) = 0c;
A7: c in dom f by A5,PARTFUN2:26;
    then c in dom (r(#)f) by Th4;
    then (r(#)f)/.c = 0c by A6,Th4;
    then
A8: (r(#)f)/.c in {0c} by TARSKI:def 1;
    c in dom (r(#)f) by A7,Th4;
    hence c in (r(#)f)"{0c} by A8,PARTFUN2:26;
  end;
  hence thesis by SUBSET_1:3;
end;

begin





theorem
  (f1 + f2) + f3 = f1 + (f2 + f3)
proof
A1: dom (f1 + f2 + f3) = dom (f1 + f2) /\ dom f3 by VALUED_1:def 1
    .= dom f1 /\ dom f2 /\ dom f3 by VALUED_1:def 1
    .= dom f1 /\ (dom f2 /\ dom f3) by XBOOLE_1:16
    .= dom f1 /\ dom (f2 + f3) by VALUED_1:def 1
    .= dom (f1 + (f2 + f3)) by VALUED_1:def 1;
  now
    let c;
    assume
A2: c in dom (f1 + f2 + f3);
    then c in dom (f1 + f2) /\ dom f3 by VALUED_1:def 1;
    then
A3: c in dom (f1 + f2) by XBOOLE_0:def 4;
    c in dom f1 /\ dom (f2 + f3) by A1,A2,VALUED_1:def 1;
    then
A4: c in dom (f2 + f3) by XBOOLE_0:def 4;
    thus (f1 + f2 + f3)/.c = (f1 + f2)/.c + (f3/.c) by A2,Th1
      .= ((f1/.c)) + ((f2/.c)) + (f3/.c) by A3,Th1
      .= ((f1/.c)) + (((f2/.c)) + (f3/.c))
      .= ((f1/.c)) + (f2 + f3)/.c by A4,Th1
      .= (f1 + (f2 + f3))/.c by A1,A2,Th1;
  end;
  hence thesis by A1,PARTFUN2:1;
end;

theorem Th14:
  (f1 (#) f2) (#) f3 = f1 (#) (f2 (#) f3)
proof
A1: dom (f1 (#) f2 (#) f3) = dom (f1 (#) f2) /\ dom f3 by Th3
    .= dom f1 /\ dom f2 /\ dom f3 by Th3
    .= dom f1 /\ (dom f2 /\ dom f3) by XBOOLE_1:16
    .= dom f1 /\ dom (f2 (#) f3) by Th3
    .= dom (f1 (#) (f2 (#) f3)) by Th3;
  now
    let c;
    assume
A2: c in dom (f1(#)f2(#)f3);
    then c in dom (f1 (#) f2) /\ dom f3 by Th3;
    then
A3: c in dom (f1 (#) f2) by XBOOLE_0:def 4;
    c in dom f1 /\ dom (f2(#)f3) by A1,A2,Th3;
    then
A4: c in dom (f2 (#) f3) by XBOOLE_0:def 4;
    thus (f1 (#) f2 (#) f3)/.c = (f1 (#) f2)/.c * (f3/.c) by A2,Th3
      .= ((f1/.c)) * ((f2/.c)) * (f3/.c) by A3,Th3
      .= ((f1/.c)) * (((f2/.c)) * (f3/.c))
      .= ((f1/.c)) * ((f2 (#) f3)/.c) by A4,Th3
      .= (f1 (#) (f2 (#) f3))/.c by A1,A2,Th3;
  end;
  hence thesis by A1,PARTFUN2:1;
end;

theorem Th15:
  (f1 + f2) (#) f3=f1 (#) f3 + f2 (#) f3
proof
A1: dom ((f1 + f2) (#) f3) = dom (f1 + f2) /\ dom f3 by Th3
    .= dom f1 /\ dom f2 /\ (dom f3 /\ dom f3) by VALUED_1:def 1
    .= dom f1 /\ dom f2 /\ dom f3 /\ dom f3 by XBOOLE_1:16
    .= dom f1 /\ dom f3 /\ dom f2 /\ dom f3 by XBOOLE_1:16
    .= dom f1 /\ dom f3 /\ (dom f2 /\ dom f3) by XBOOLE_1:16
    .= dom (f1 (#) f3) /\ (dom f2 /\ dom f3) by Th3
    .= dom (f1 (#) f3) /\ dom (f2 (#) f3) by Th3
    .= dom (f1 (#) f3 + f2 (#) f3) by VALUED_1:def 1;
  now
    let c;
    assume
A2: c in dom ((f1 + f2)(#)f3);
    then
A3: c in dom (f1(#)f3) /\ dom (f2(#)f3) by A1,VALUED_1:def 1;
    then
A4: c in dom (f1(#)f3) by XBOOLE_0:def 4;
    c in dom (f1 + f2) /\ dom f3 by A2,Th3;
    then
A5: c in dom (f1 + f2) by XBOOLE_0:def 4;
A6: c in dom (f2 (#) f3) by A3,XBOOLE_0:def 4;
    thus ((f1 + f2) (#) f3)/.c = (f1 + f2)/.c * (f3/.c) by A2,Th3
      .= (((f1/.c)) + ((f2/.c))) * (f3/.c) by A5,Th1
      .= ((f1/.c)) * (f3/.c) + ((f2/.c)) * (f3/.c)
      .= (f1 (#) f3)/.c + ((f2/.c))* (f3/.c) by A4,Th3
      .= (f1 (#) f3)/.c + (f2 (#) f3)/.c by A6,Th3
      .=((f1 (#) f3) + (f2 (#) f3))/.c by A1,A2,Th1;
  end;
  hence thesis by A1,PARTFUN2:1;
end;

theorem
  f3 (#) (f1 + f2)=f3(#)f1 + f3(#)f2 by Th15;

theorem Th17:
  r(#)(f1(#)f2)=r(#)f1(#)f2
proof
A1: dom (r(#)(f1 (#) f2)) = dom (f1 (#) f2) by Th4
    .= dom f1 /\ dom f2 by Th3
    .= dom (r(#)f1) /\ dom f2 by Th4
    .= dom (r(#)f1(#)f2) by Th3;
  now
    let c;
    assume
A2: c in dom (r(#)(f1(#)f2));
    then
A3: c in dom (f1(#)f2) by Th4;
    c in dom (r(#)f1) /\ dom f2 by A1,A2,Th3;
    then
A4: c in dom (r(#)f1) by XBOOLE_0:def 4;
    thus (r(#)(f1(#)f2))/.c = r * ((f1(#)f2)/.c) by A2,Th4
      .= r*(((f1/.c)) * ((f2/.c))) by A3,Th3
      .= (r*((f1/.c))) * ((f2/.c))
      .= (r(#)f1)/.c * ((f2/.c)) by A4,Th4
      .= (r(#)f1 (#) f2)/.c by A1,A2,Th3;
  end;
  hence thesis by A1,PARTFUN2:1;
end;

theorem Th18:
  r(#)(f1(#)f2)=f1(#)(r(#)f2)
proof
A1: dom (r(#)(f1 (#) f2)) = dom (f1 (#) f2) by Th4
    .= dom f1 /\ dom f2 by Th3
    .= dom f1 /\ dom (r(#)f2) by Th4
    .= dom (f1(#)(r(#)f2)) by Th3;
  now
    let c;
    assume
A2: c in dom (r(#)(f1(#)f2));
    then
A3: c in dom (f1(#)f2) by Th4;
    c in dom f1 /\ dom (r(#)f2) by A1,A2,Th3;
    then
A4: c in dom (r(#)f2) by XBOOLE_0:def 4;
    thus (r(#)(f1(#)f2))/.c = r * ((f1(#)f2)/.c) by A2,Th4
      .= r * (((f1/.c)) * ((f2/.c))) by A3,Th3
      .= ((f1/.c)) * (r * ((f2/.c)))
      .= ((f1/.c)) * ((r(#)f2)/.c) by A4,Th4
      .= (f1(#)(r(#)f2))/.c by A1,A2,Th3;
  end;
  hence thesis by A1,PARTFUN2:1;
end;

theorem Th19:
  (f1 - f2)(#)f3=f1(#)f3 - f2(#)f3
proof
A1: dom ((f1 - f2) (#) f3) = dom (f1 - f2) /\ dom f3 by Th3
    .= dom f1 /\ dom f2 /\ (dom f3 /\ dom f3) by Th2
    .= dom f1 /\ dom f2 /\ dom f3 /\ dom f3 by XBOOLE_1:16
    .= dom f1 /\ dom f3 /\ dom f2 /\ dom f3 by XBOOLE_1:16
    .= dom f1 /\ dom f3 /\ (dom f2 /\ dom f3) by XBOOLE_1:16
    .= dom (f1 (#) f3) /\ (dom f2 /\ dom f3) by Th3
    .= dom (f1 (#) f3) /\ dom (f2 (#) f3) by Th3
    .= dom (f1 (#) f3 - f2 (#) f3) by Th2;
  now
    let c;
    assume
A2: c in dom ((f1 - f2)(#)f3);
    then c in dom (f1 - f2) /\ dom f3 by Th3;
    then
A3: c in dom (f1 - f2) by XBOOLE_0:def 4;
A4: c in dom (f1(#)f3) /\ dom (f2(#)f3) by A1,A2,Th2;
    then
A5: c in dom (f1(#)f3) by XBOOLE_0:def 4;
A6: c in dom (f2 (#) f3) by A4,XBOOLE_0:def 4;
    thus ((f1 - f2) (#) f3)/.c = (f1 - f2)/.c * (f3/.c) by A2,Th3
      .= (((f1/.c)) - ((f2/.c))) * (f3/.c) by A3,Th2
      .= ((f1/.c)) * (f3/.c) - ((f2/.c)) * (f3/.c)
      .= (f1 (#) f3)/.c - ((f2/.c)) * (f3/.c) by A5,Th3
      .= (f1 (#) f3)/.c - (f2 (#) f3)/.c by A6,Th3
      .=((f1 (#) f3) - (f2 (#) f3))/.c by A1,A2,Th2;
  end;
  hence thesis by A1,PARTFUN2:1;
end;

theorem
  f3(#)f1 - f3(#)f2 = f3(#)(f1 - f2) by Th19;

theorem
  r(#)(f1 + f2) = r(#)f1 + r(#)f2
proof
A1: dom (r(#)(f1 + f2)) = dom (f1 + f2) by Th4
    .= dom f1 /\ dom f2 by VALUED_1:def 1
    .= dom f1 /\ dom (r(#)f2) by Th4
    .= dom (r(#)f1) /\ dom (r(#)f2) by Th4
    .= dom (r(#)f1 + r(#)f2) by VALUED_1:def 1;
  now
    let c;
    assume
A2: c in dom (r(#)(f1 + f2));
    then
A3: c in dom (f1 + f2) by Th4;
A4: c in dom (r(#)f1) /\ dom (r(#)f2) by A1,A2,VALUED_1:def 1;
    then
A5: c in dom (r(#)f1) by XBOOLE_0:def 4;
A6: c in dom (r(#)f2) by A4,XBOOLE_0:def 4;
    thus (r(#)(f1 + f2))/.c = r * ((f1 + f2)/.c) by A2,Th4
      .= r * (((f1/.c)) + ((f2/.c))) by A3,Th1
      .= r * ((f1/.c)) + r * ((f2/.c))
      .= (r(#)f1)/.c + r * ((f2/.c)) by A5,Th4
      .= (r(#)f1)/.c + (r(#)f2)/.c by A6,Th4
      .= (r(#)f1 + r(#)f2)/.c by A1,A2,Th1;
  end;
  hence thesis by A1,PARTFUN2:1;
end;

theorem
  (r*q)(#)f = r(#)(q(#)f)
proof
A1: dom ((r*q) (#) f) = dom f by Th4
    .= dom (q(#)f) by Th4
    .= dom (r(#)(q(#)f)) by Th4;
  now
    let c;
    assume
A2: c in dom ((r*q)(#)f);
    then
A3: c in dom (q(#)f) by A1,Th4;
    thus ((r*q)(#)f)/.c = r*q * (f/.c) by A2,Th4
      .= r*(q * (f/.c))
      .= r * ((q(#)f)/.c) by A3,Th4
      .= (r(#)(q(#)f))/.c by A1,A2,Th4;
  end;
  hence thesis by A1,PARTFUN2:1;
end;

theorem
  r(#)(f1 - f2) = r(#)f1 - r(#)f2
proof
A1: dom (r(#)(f1 - f2)) = dom (f1 - f2) by Th4
    .= dom f1 /\ dom f2 by Th2
    .= dom f1 /\ dom (r(#)f2) by Th4
    .= dom (r(#)f1) /\ dom (r(#)f2) by Th4
    .= dom (r(#)f1 - r(#)f2) by Th2;
  now
    let c;
    assume
A2: c in dom (r(#)(f1 - f2));
    then
A3: c in dom (f1 - f2) by Th4;
A4: c in dom (r(#)f1) /\ dom (r(#)f2) by A1,A2,Th2;
    then
A5: c in dom (r(#)f1) by XBOOLE_0:def 4;
A6: c in dom (r(#)f2) by A4,XBOOLE_0:def 4;
    thus (r(#)(f1 - f2))/.c = r * ((f1 - f2)/.c) by A2,Th4
      .= r * (((f1/.c)) - ((f2/.c))) by A3,Th2
      .= r * ((f1/.c)) - r * ((f2/.c))
      .= (r(#)f1)/.c - r * ((f2/.c)) by A5,Th4
      .= (r(#)f1)/.c - (r(#)f2)/.c by A6,Th4
      .= (r(#)f1 - r(#)f2)/.c by A1,A2,Th2;
  end;
  hence thesis by A1,PARTFUN2:1;
end;

theorem
  f1-f2 = (-1r)(#)(f2-f1)
proof
A1: dom (f1 - f2) = dom f2 /\ dom f1 by Th2
    .= dom (f2 - f1) by Th2
    .= dom ((-1r)(#)(f2 - f1)) by Th4;
  now
A2: dom (f1 - f2) = dom f2 /\ dom f1 by Th2
      .= dom (f2 - f1) by Th2;
    let c such that
A3: c in dom (f1-f2);
    thus (f1 - f2)/.c = ((f1/.c)) - ((f2/.c)) by A3,Th2
      .= (-1)*(((f2/.c)) - ((f1/.c)))
      .= (-1r)*((f2 - f1)/.c) by A3,A2,Th2,COMPLEX1:def 4
      .= ((-1r)(#)(f2 - f1))/.c by A1,A3,Th4;
  end;
  hence thesis by A1,PARTFUN2:1;
end;

theorem
  f1 - (f2 + f3) = f1 - f2 - f3
proof
A1: dom (f1 - (f2 + f3)) = dom f1 /\ dom (f2 + f3) by Th2
    .= dom f1 /\ (dom f2 /\ dom f3) by VALUED_1:def 1
    .= dom f1 /\ dom f2 /\ dom f3 by XBOOLE_1:16
    .= dom (f1 - f2) /\ dom f3 by Th2
    .= dom (f1 - f2 - f3) by Th2;
  now
    let c;
    assume
A2: c in dom (f1 - (f2 + f3));
    then c in dom f1 /\ dom (f2 + f3) by Th2;
    then
A3: c in dom (f2 + f3) by XBOOLE_0:def 4;
    c in dom (f1 - f2) /\ dom f3 by A1,A2,Th2;
    then
A4: c in dom (f1 - f2) by XBOOLE_0:def 4;
    thus (f1 - (f2 + f3))/.c = ((f1/.c)) - (f2 + f3)/.c by A2,Th2
      .= ((f1/.c)) - (((f2/.c)) + (f3/.c)) by A3,Th1
      .= ((f1/.c)) - ((f2/.c)) - (f3/.c)
      .= (f1 - f2)/.c - (f3/.c) by A4,Th2
      .= (f1 - f2 - f3)/.c by A1,A2,Th2;
  end;
  hence thesis by A1,PARTFUN2:1;
end;

theorem
  1r(#)f = f
proof
A1: now
    let c;
    assume c in dom (1r(#)f);
    hence (1r(#)f)/.c = 1r * (f/.c) by Th4
      .= (f/.c) by COMPLEX1:def 4;
  end;
  dom (1r(#)f) = dom f by Th4;
  hence thesis by A1,PARTFUN2:1;
end;

theorem
  f1 - (f2 - f3) = f1 - f2 + f3
proof
A1: dom (f1 - (f2 - f3)) = dom f1 /\ dom (f2 - f3) by Th2
    .= dom f1 /\ (dom f2 /\ dom f3) by Th2
    .= dom f1 /\ dom f2 /\ dom f3 by XBOOLE_1:16
    .= dom (f1 - f2) /\ dom f3 by Th2
    .= dom (f1 - f2 + f3) by VALUED_1:def 1;
  now
    let c;
    assume
A2: c in dom (f1 - (f2 - f3));
    then c in dom (f1 - f2) /\ dom f3 by A1,VALUED_1:def 1;
    then
A3: c in dom (f1 - f2) by XBOOLE_0:def 4;
    c in dom f1 /\ dom (f2 - f3) by A2,Th2;
    then
A4: c in dom (f2 - f3) by XBOOLE_0:def 4;
    thus (f1 - (f2 - f3))/.c = ((f1/.c)) - (f2 - f3)/.c by A2,Th2
      .= ((f1/.c)) - (((f2/.c)) - (f3/.c)) by A4,Th2
      .= ((f1/.c)) - ((f2/.c)) + (f3/.c)
      .= (f1 - f2)/.c + (f3/.c) by A3,Th2
      .= (f1 - f2 + f3)/.c by A1,A2,Th1;
  end;
  hence thesis by A1,PARTFUN2:1;
end;

theorem
  f1 + (f2 - f3) =f1 + f2 - f3
proof
A1: dom (f1 + (f2 - f3)) = dom f1 /\ dom (f2 - f3) by VALUED_1:def 1
    .= dom f1 /\ (dom f2 /\ dom f3) by Th2
    .= dom f1 /\ dom f2 /\ dom f3 by XBOOLE_1:16
    .= dom (f1 + f2) /\ dom f3 by VALUED_1:def 1
    .= dom (f1 + f2 - f3) by Th2;
  now
    let c;
    assume
A2: c in dom (f1 + (f2 - f3));
    then c in dom f1 /\ dom (f2 - f3) by VALUED_1:def 1;
    then
A3: c in dom (f2 - f3) by XBOOLE_0:def 4;
    c in dom (f1 + f2) /\ dom f3 by A1,A2,Th2;
    then
A4: c in dom (f1 + f2) by XBOOLE_0:def 4;
    thus (f1 + (f2 - f3))/.c = ((f1/.c)) + (f2 - f3)/.c by A2,Th1
      .= ((f1/.c)) + (((f2/.c)) - (f3/.c)) by A3,Th2
      .= ((f1/.c)) + ((f2/.c)) - (f3/.c)
      .= (f1 + f2)/.c - (f3/.c) by A4,Th1
      .= (f1 + f2 - f3)/.c by A1,A2,Th2;
  end;
  hence thesis by A1,PARTFUN2:1;
end;

theorem Th29:
  |.f1(#)f2.| = |.f1.|(#)|.f2.|
proof
A1: dom (|.f1 (#) f2.|) = dom (f1 (#) f2) by VALUED_1:def 11
    .= dom f1 /\ dom f2 by Th3
    .= dom f1 /\ dom (|.f2.|) by VALUED_1:def 11
    .= dom (|.f1.|) /\ dom (|.f2.|) by VALUED_1:def 11
    .= dom (|.f1.|(#)|.f2.|) by VALUED_1:def 4;
  now
A2: dom |.f2.| = dom f2 by VALUED_1:def 11;
    let c;
A3: dom |.f1.| = dom f1 by VALUED_1:def 11;
    assume
A4: c in dom (|.f1 (#) f2.|);
    then
A5: c in dom (f1 (#) f2) by VALUED_1:def 11;
A6: c in dom (|.f1.|) /\ dom (|.f2.|) by A1,A4,VALUED_1:def 4;
    then c in dom (|.f1.|) by XBOOLE_0:def 4;
    then
A7: f1.c = f1/.c by A3,PARTFUN1:def 6;
    c in dom (|.f2.|) by A6,XBOOLE_0:def 4;
    then
A8: f2/.c = f2.c by A2,PARTFUN1:def 6;
    thus (|.(f1(#)f2).|).c = |.(f1(#)f2).c.| by VALUED_1:18
      .= |.(f1(#)f2)/.c.| by A5,PARTFUN1:def 6
      .= |.((f1/.c)) * ((f2/.c)).| by A5,Th3
      .= |.((f1/.c)).| * |.((f2/.c)).| by COMPLEX1:65
      .= ((|.f1.|).c) *( |.((f2/.c)).|) by A7,VALUED_1:18
      .= ((|.f1.|).c) * ((|.f2.|).c) by A8,VALUED_1:18
      .= (|.f1.|(#)|.f2.|).c by VALUED_1:5;
  end;
  hence thesis by A1,PARTFUN1:5;
end;

theorem Th30:
  |.r(#)f.| = |.r.|(#)|.f.|
proof
A1: dom f = dom (|.f.|) by VALUED_1:def 11;
A2: dom (|.r(#)f.|) = dom (r(#)f) by VALUED_1:def 11
    .= dom f by Th4
    .= dom (|.r.|(#)|.f.|) by A1,VALUED_1:def 5;
  now
    let c;
    assume
A3: c in dom (|.r(#)f.|);
    then
A4: c in dom (r(#)f) by VALUED_1:def 11;
A5: c in dom (|.f.|) by A2,A3,VALUED_1:def 5;
    thus (|.r(#)f.|).c = |.(r(#)f).c.| by VALUED_1:18
      .=|.(r(#)f)/.c.| by A4,PARTFUN1:def 6
      .=|.r*((f/.c)).| by A4,Th4
      .=|.r.|*|.((f/.c)).| by COMPLEX1:65
      .=|.r.|*|.((f.c)).| by A1,A5,PARTFUN1:def 6
      .=|.r.|*((|.f.|).c) by VALUED_1:18
      .=(|.r.|(#)|.f.|).c by A2,A3,VALUED_1:def 5;
  end;
  hence thesis by A2,PARTFUN1:5;
end;

theorem
  -f = (-1r)(#)f by COMPLEX1:def 4;

canceled;

theorem
  f1 - (-f2) = f1 + f2;

theorem Th34:
  f^^ = f|(dom (f^))
proof
A1: dom (f^^) = dom (f|(dom (f^))) by Th11;
  now
    let c;
    assume
A2: c in dom (f^^);
    then c in dom f /\ dom (f^) by A1,RELAT_1:61;
    then
A3: c in dom (f^) by XBOOLE_0:def 4;
    thus (f^^)/.c = ((f^)/.c)" by A2,Def2
      .= (((f/.c))")" by A3,Def2
      .= (f|(dom (f^)))/.c by A1,A2,PARTFUN2:15;
  end;
  hence thesis by A1,PARTFUN2:1;
end;

theorem Th35:
  (f1(#)f2)^ = (f1^)(#)(f2^)
proof
A1: dom ((f1(#)f2)^) = dom (f1(#)f2) \ (f1(#)f2)"{0c} by Def2
    .= (dom f1 \ f1"{0c}) /\ (dom f2 \ (f2)"{0c}) by Th7
    .= dom (f1^) /\ (dom f2 \ (f2)"{0c}) by Def2
    .= dom (f1^) /\ dom (f2^) by Def2
    .= dom ((f1^) (#) (f2^)) by Th3;
  now
    let c;
    assume
A2: c in dom ((f1(#)f2)^);
    then c in dom (f1(#)f2) \ (f1(#)f2)"{0c} by Def2;
    then
A3: c in dom (f1(#)f2) by XBOOLE_0:def 5;
A4: c in dom (f1^) /\ dom (f2^) by A1,A2,Th3;
    then
A5: c in dom (f1^) by XBOOLE_0:def 4;
A6: c in dom (f2^) by A4,XBOOLE_0:def 4;
    thus ((f1(#)f2)^)/.c = ((f1(#)f2)/.c)" by A2,Def2
      .= (((f1/.c)) * ((f2/.c)))" by A3,Th3
      .= (((f1/.c)))"* (((f2/.c)))" by XCMPLX_1:204
      .= ((f1^)/.c)*(((f2/.c)))" by A5,Def2
      .= ((f1^)/.c) *((f2^)/.c) by A6,Def2
      .= ((f1^) (#) (f2^))/.c by A1,A2,Th3;
  end;
  hence thesis by A1,PARTFUN2:1;
end;

theorem Th36:
  r<>0 implies (r(#)f)^ = r" (#) (f^)
proof
  assume
A1: r<>0;
A2: dom ((r(#)f)^) = dom (r(#)f) \ (r(#)f)"{0c} by Def2
    .= dom (r(#)f) \ f"{0c} by A1,Th12
    .= dom f \ f"{0c} by Th4
    .= dom (f^) by Def2
    .= dom (r"(#)(f^)) by Th4;
  now
    let c;
    assume
A3: c in dom ((r(#)f)^);
    then
A4: c in dom (f^) by A2,Th4;
    c in dom (r(#)f) \ (r(#)f)"{0c} by A3,Def2;
    then
A5: c in dom (r(#)f) by XBOOLE_0:def 5;
    thus ((r(#)f)^)/.c = ((r(#)f)/.c)" by A3,Def2
      .= (r*((f/.c)))" by A5,Th4
      .= r"* ((f/.c))" by XCMPLX_1:204
      .= r"* ((f^)/.c) by A4,Def2
      .= (r" (#) (f^))/.c by A2,A3,Th4;
  end;
  hence thesis by A2,PARTFUN2:1;
end;

Lm2: (-1r)"=-1r by COMPLEX1:def 4;

theorem
  (-f)^ = (-1r)(#)(f^) by Lm2,Th36,COMPLEX1:def 4;

theorem Th38:
  |.f.|^ = |. f^ .|
proof
A1: dom ((|.f.|)^) = dom (|.f.|) \ (|.f.|)"{0} by RFUNCT_1:def 2
    .= dom f \ (|.f.|)"{0} by VALUED_1:def 11
    .= dom f \ f"{0c} by Th10
    .= dom (f^) by Def2
    .= dom (|.(f^).|) by VALUED_1:def 11;
  now
    let c;
A2: dom f = dom |.f.| by VALUED_1:def 11;
    assume
A3: c in dom ((|.f.|)^);
    then
A4: c in dom (f^) by A1,VALUED_1:def 11;
    c in dom (|.f.|) \ (|.f.|)"{0} by A3,RFUNCT_1:def 2;
    then
A5: c in dom (|.f.|) by XBOOLE_0:def 5;
    thus ((|.f.|)^).c = ((|.f.|).c)" by A3,RFUNCT_1:def 2
      .= (|.(f.c) .|)" by VALUED_1:18
      .= (|.(f/.c) .|)" by A5,A2,PARTFUN1:def 6
      .= |.1r.|/|.(f/.c) .| by COMPLEX1:48,XCMPLX_1:215
      .= |.1r/(f/.c) .| by COMPLEX1:67
      .= |.((f/.c))" .| by COMPLEX1:def 4,XCMPLX_1:215
      .= |.(f^)/.c .| by A4,Def2
      .= |.(f^).c .| by A4,PARTFUN1:def 6
      .= (|.(f^).|).c by VALUED_1:18;
  end;
  hence thesis by A1,PARTFUN1:5;
end;

theorem Th39:
  f/g = f(#) (g^)
proof
A1: now
    let c;
    assume
A2: c in dom (f/g);
    then c in dom f /\ (dom g \ g"{0c}) by Def1;
    then
A3: c in dom f /\ dom (g^) by Def2;
    then
A4: c in dom (g^) by XBOOLE_0:def 4;
A5: c in dom (f (#) (g^)) by A3,Th3;
    thus (f/g)/.c = (f/.c) * (g/.c)" by A2,Def1
      .= (f/.c) * ((g^)/.c) by A4,Def2
      .= (f (#) (g^))/.c by A5,Th3;
  end;
  dom (f/g) = dom f /\ (dom g \ g"{0c}) by Def1
    .= dom f /\ dom (g^) by Def2
    .= dom (f(#)(g^)) by Th3;
  hence thesis by A1,PARTFUN2:1;
end;

theorem Th40:
  r(#)(g/f) = (r(#)g)/f
proof
  thus r(#)(g/f) = r(#)(g(#)(f^)) by Th39
    .= (r(#)g)(#)(f^) by Th17
    .= (r(#)g)/f by Th39;
end;

theorem
  (f/g)(#)g = (f|dom(g^))
proof
A1: dom ((f/g)(#)g) = dom (f/g) /\ dom g by Th3
    .= dom f /\ (dom g \ g"{0c}) /\ dom g by Def1
    .= dom f /\ ((dom g \ g"{0c}) /\ dom g) by XBOOLE_1:16
    .= dom f /\ (dom (g^) /\ dom g) by Def2
    .= dom f /\ dom (g^) by Th6,XBOOLE_1:28
    .= dom (f|(dom (g^))) by RELAT_1:61;
  now
    let c;
    assume
A2: c in dom ((f/g)(#)g);
    then
A3: c in dom f /\ dom (g^) by A1,RELAT_1:61;
    then
A4: c in dom (f(#)(g^)) by Th3;
A5: c in dom (g^) by A3,XBOOLE_0:def 4;
    then
A6: g/.c <> 0c by Th8;
    thus ((f/g)(#)g)/.c = ((f/g)/.c) * (g/.c )by A2,Th3
      .= (f(#)(g^))/.c * (g/.c) by Th39
      .= ((f/.c)) *((g^)/.c) * (g/.c) by A4,Th3
      .= ((f/.c))*(g/.c)"*(g/.c) by A5,Def2
      .= ((f/.c))*((g/.c)" * (g/.c))
      .= ((f/.c))*1r by A6,COMPLEX1:def 4,XCMPLX_0:def 7
      .= (f|(dom (g^)))/.c by A1,A2,COMPLEX1:def 4,PARTFUN2:15;
  end;
  hence thesis by A1,PARTFUN2:1;
end;

theorem Th42:
  (f/g)(#)(f1/g1) = (f(#)f1)/(g(#)g1)
proof
A1: now
    let c;
    assume
A2: c in dom ((f/g)(#)(f1/g1));
    then c in dom (f/g) /\ dom (f1/g1) by Th3;
    then
A3: c in dom (f (#)(g^)) /\ dom (f1/g1) by Th39;
    then
A4: c in dom (f (#)(g^)) by XBOOLE_0:def 4;
    then
A5: c in dom f /\ dom(g^) by Th3;
    c in dom (f (#)(g^)) /\ dom (f1(#)(g1^)) by A3,Th39;
    then
A6: c in dom (f1(#)(g1^)) by XBOOLE_0:def 4;
    then
A7: c in dom f1 /\ dom(g1^) by Th3;
    then
A8: c in dom f1 by XBOOLE_0:def 4;
    c in dom f by A5,XBOOLE_0:def 4;
    then c in dom f /\ dom f1 by A8,XBOOLE_0:def 4;
    then
A9: c in dom (f(#)f1) by Th3;
A10: c in dom(g1^) by A7,XBOOLE_0:def 4;
    c in dom(g^) by A5,XBOOLE_0:def 4;
    then c in dom (g^) /\ dom (g1^) by A10,XBOOLE_0:def 4;
    then
A11: c in dom((g^)(#)(g1^)) by Th3;
    then c in dom((g(#)g1)^) by Th35;
    then c in dom (f(#)f1) /\ dom((g(#)g1)^) by A9,XBOOLE_0:def 4;
    then
A12: c in dom ((f(#)f1)(#)((g(#)g1)^)) by Th3;
    thus ((f/g)(#)(f1/g1))/.c = ((f/g)/.c)* ((f1/g1)/.c) by A2,Th3
      .= ((f(#)(g^))/.c) * ((f1/g1)/.c) by Th39
      .= ((f(#)(g^))/.c) * ((f1(#)(g1^))/.c) by Th39
      .= ((f/.c)) * ((g^)/.c) *((f1(#)(g1^))/.c) by A4,Th3
      .= ((f/.c)) * ((g^)/.c) * ((((f1/.c)))* ((g1^)/.c)) by A6,Th3
      .= ((f/.c)) * ((((f1/.c))) * (((g^)/.c) * ((g1^)/.c)))
      .= ((f/.c)) * ((((f1/.c))) * (((g^)(#)(g1^))/.c)) by A11,Th3
      .= ((f/.c)) * (((f1/.c))) * (((g^)(#)(g1^))/.c)
      .= ((f/.c)) * (((f1/.c))) * (((g(#)g1)^)/.c) by Th35
      .= ((f(#)f1)/.c) * (((g(#)g1)^)/.c) by A9,Th3
      .= ((f(#)f1)(#)((g(#)g1)^))/.c by A12,Th3
      .= ((f(#)f1)/(g(#)g1))/.c by Th39;
  end;
  dom ((f/g)(#)(f1/g1)) = dom (f/g) /\ dom (f1/g1) by Th3
    .= dom f /\ (dom g \ g"{0c}) /\ dom (f1/g1) by Def1
    .= dom f /\ (dom g \ g"{0c}) /\ (dom f1 /\ (dom g1 \ g1"{0c})) by Def1
    .= dom f /\ ((dom g \ g"{0c}) /\ (dom f1 /\ (dom g1 \ g1"{0c}))) by
XBOOLE_1:16
    .= dom f /\ (dom f1 /\ ((dom g \ g"{0c}) /\ (dom g1 \ g1"{0c}))) by
XBOOLE_1:16
    .= dom f /\ dom f1 /\ ((dom g \ g"{0c}) /\ (dom g1 \ g1"{0c})) by
XBOOLE_1:16
    .= dom (f(#)f1) /\ ((dom g \ g"{0c}) /\ (dom g1 \ g1"{0c})) by Th3
    .= dom (f(#)f1) /\ (dom (g(#)g1) \ (g(#)g1)"{0c}) by Th7
    .= dom ((f(#)f1)/(g(#)g1)) by Def1;
  hence thesis by A1,PARTFUN2:1;
end;

theorem Th43:
  (f1/f2)^ = (f2|dom(f2^))/f1
proof
  thus (f1/f2)^ = (f1(#)(f2^))^ by Th39
    .= (f1^)(#)(f2^^) by Th35
    .= (f2|dom(f2^))(#)(f1^) by Th34
    .= (f2|dom(f2^))/f1 by Th39;
end;

theorem Th44:
  g (#) (f1/f2) = (g (#) f1)/f2
proof
  thus g (#) (f1/f2) = g (#) (f1 (#) (f2^)) by Th39
    .= g (#) f1 (#) (f2^) by Th14
    .= (g (#) f1)/f2 by Th39;
end;

theorem
  g/(f1/f2) = (g(#)(f2|dom(f2^)))/f1
proof
  thus g/(f1/f2) = g (#) ((f1/f2)^) by Th39
    .= g (#) ((f2|dom(f2^))/f1) by Th43
    .= (g (#) (f2|dom(f2^)))/f1 by Th44;
end;

theorem
  -f/g = (-f)/g & f/(-g) = -f/g
proof
  thus -f/g = (-1r)(#)(f/g) by COMPLEX1:def 4
    .= (-f)/g by Th40,COMPLEX1:def 4;
  thus f/(-g) = f (#) ((-g)^) by Th39
    .= f (#) ((-1r) (#) (g^)) by Lm2,Th36,COMPLEX1:def 4
    .= -(f (#) (g^)) by Th18,COMPLEX1:def 4
    .= -(f/g) by Th39;
end;

theorem
  f1/f + f2/f = (f1 + f2)/f & f1/f - f2/f = (f1 - f2)/f
proof
  thus f1/f + f2/f = f1(#)(f^) +f2/f by Th39
    .= f1(#)(f^) + f2(#)(f^) by Th39
    .= (f1+f2) (#) (f^) by Th15
    .= (f1+f2)/f by Th39;
  thus f1/f - f2/f = f1(#)(f^) - f2/f by Th39
    .= f1(#)(f^) -f2(#)(f^) by Th39
    .= (f1-f2)(#)(f^) by Th19
    .= (f1-f2)/f by Th39;
end;

theorem Th48:
  f1/f + g1/g = (f1(#)g + g1(#)f)/(f(#)g)
proof
A1: now
    let c;
A2: dom (f^) c= dom f by Th6;
    assume
A3: c in dom ((f1/f) + (g1/g));
    then
A4: c in dom (f1/f) /\ dom (g1/g) by VALUED_1:def 1;
    then
A5: c in dom (f1/f) by XBOOLE_0:def 4;
A6: c in dom (g1/g) by A4,XBOOLE_0:def 4;
A7: c in dom (f1 (#)(f^)) /\ dom (g1/g) by A4,Th39;
    then c in dom (f1 (#)(f^)) by XBOOLE_0:def 4;
    then
A8: c in dom f1 /\ dom(f^) by Th3;
    then
A9: c in dom(f^) by XBOOLE_0:def 4;
    then
A10: (f/.c) <> 0c by Th8;
    c in dom (f1 (#)(f^)) /\ dom (g1(#)(g^)) by A7,Th39;
    then c in dom (g1(#)(g^)) by XBOOLE_0:def 4;
    then
A11: c in dom g1 /\ dom(g^) by Th3;
    then
A12: c in dom(g^) by XBOOLE_0:def 4;
    then
A13: g/.c <> 0c by Th8;
    c in dom g1 by A11,XBOOLE_0:def 4;
    then c in dom g1 /\ dom f by A9,A2,XBOOLE_0:def 4;
    then
A14: c in dom (g1(#)f) by Th3;
A15: dom (g^) c= dom g by Th6;
    then c in dom f /\ dom g by A9,A12,A2,XBOOLE_0:def 4;
    then
A16: c in dom (f(#)g) by Th3;
    c in dom f1 by A8,XBOOLE_0:def 4;
    then c in dom f1 /\ dom g by A12,A15,XBOOLE_0:def 4;
    then
A17: c in dom (f1(#)g) by Th3;
    then c in dom (f1(#)g) /\ dom (g1(#)f) by A14,XBOOLE_0:def 4;
    then
A18: c in dom (f1(#)g + g1(#)f) by VALUED_1:def 1;
    c in dom (f^) /\ dom (g^) by A9,A12,XBOOLE_0:def 4;
    then c in dom ((f^)(#)(g^)) by Th3;
    then c in dom ((f(#)g)^) by Th35;
    then c in dom (f1(#)g + g1(#)f) /\ dom ((f(#)g)^) by A18,XBOOLE_0:def 4;
    then c in dom ((f1(#)g + g1(#)f)(#)((f(#)g)^)) by Th3;
    then
A19: c in dom ((f1(#)g + g1(#)f)/(f(#)g)) by Th39;
    thus (f1/f + g1/g)/.c = (f1/f)/.c + (g1/g)/.c by A3,Th1
      .= (((f1/.c))) * ((f/.c))" + (g1/g)/.c by A5,Def1
      .= (((f1/.c))) *(1r*((f/.c))") + (g1/.c) * 1r * (g/.c)" by A6,Def1,
COMPLEX1:def 4
      .= (((f1/.c))) *((g/.c) *(g/.c)"* ((f/.c))") + (g1/.c) * (1r * (g/.c)"
    ) by A13,COMPLEX1:def 4,XCMPLX_0:def 7
      .= (((f1/.c))) *(g/.c *((g/.c)"* ((f/.c))")) + (g1/.c) * (((f/.c)) *((
    f/.c))" * (g/.c)") by A10,COMPLEX1:def 4,XCMPLX_0:def 7
      .= (((f1/.c))) *((g/.c) *((g/.c * (f/.c))")) + (g1/.c) * (((f/.c)) *((
    (f/.c))" * (g/.c)")) by XCMPLX_1:204
      .= (((f1/.c))) *((g/.c) *(((f/.c)* (g/.c))")) + (g1/.c) * (((f/.c)) *(
    ((f/.c) * (g/.c))")) by XCMPLX_1:204
      .= (((f1/.c))) *((g/.c) * ((f(#)g)/.c)") + (g1/.c) * (((f/.c)) *(((f/.
    c) * (g/.c))")) by A16,Th3
      .= (((f1/.c))) *(g/.c) * ((f(#)g)/.c)" + (g1/.c) * (((f/.c)) * ((f(#)
    g)/.c)") by A16,Th3
      .= (f1(#)g)/.c * ((f(#)g)/.c)" + (g1/.c) *(f/.c) *((f(#)g)/.c)" by A17
,Th3
      .= (f1(#)g)/.c * ((f(#)g)/.c)" + (g1(#)f)/.c *((f(#)g)/.c)" by A14,Th3
      .= ((f1(#)g)/.c + (g1(#)f)/.c) *((f(#)g)/.c)"
      .= (f1(#)g + g1(#)f)/.c *((f(#)g)/.c)" by A18,Th1
      .= ((f1(#)g + g1(#)f)/(f(#)g))/.c by A19,Def1;
  end;
  dom ((f1/f) + (g1/g)) = dom (f1/f) /\ dom (g1/g) by VALUED_1:def 1
    .= dom f1 /\ (dom f \ f"{0c}) /\ dom (g1/g) by Def1
    .= dom f1 /\ (dom f \ f"{0c}) /\ (dom g1 /\ (dom g \ g"{0c})) by Def1
    .= dom f1 /\ (dom f /\ (dom f \ f"{0c})) /\ (dom g1 /\ (dom g \ g"{0c}))
  by Th6
    .= dom f /\ (dom f \ f"{0c}) /\ dom f1 /\ (dom g /\ (dom g \ g"{0c}) /\
  dom g1) by Th6
    .= dom f /\ (dom f \ f"{0c}) /\ (dom f1 /\ (dom g /\ (dom g \ g"{0c}) /\
  dom g1)) by XBOOLE_1:16
    .= dom f /\ (dom f \ f"{0c}) /\ (dom f1 /\ (dom g /\ (dom g \ g"{0c}))
  /\ dom g1) by XBOOLE_1:16
    .= dom f /\ (dom f \ f"{0c}) /\ (dom f1 /\ dom g /\ (dom g \ g"{0c}) /\
  dom g1) by XBOOLE_1:16
    .= dom f /\ (dom f \ f"{0c}) /\ (dom (f1(#) g) /\ (dom g \ g"{0c}) /\
  dom g1) by Th3
    .= dom f /\ (dom f \ f"{0c}) /\ (dom (f1(#)g) /\ (dom g1 /\ (dom g \ g"{
  0c}))) by XBOOLE_1:16
    .= dom (f1(#)g) /\ ((dom f \ f"{0c}) /\ dom f /\ (dom g1 /\ (dom g \ g"{
  0c}))) by XBOOLE_1:16
    .= dom (f1(#)g) /\ ((dom f \ f"{0c}) /\ (dom f /\ (dom g1 /\ (dom g \ g"
  {0c})))) by XBOOLE_1:16
    .= dom (f1(#)g) /\ ((dom f \ f"{0c}) /\ (dom g1 /\ dom f /\ (dom g \ g"{
  0c}))) by XBOOLE_1:16
    .= dom (f1(#)g) /\ ((dom f \ f"{0c}) /\ (dom (g1(#)f) /\ (dom g \ g"{0c}
  ))) by Th3
    .= dom (f1(#)g) /\ (dom (g1(#)f) /\ ((dom f \ f"{0c}) /\ (dom g \ g"{0c}
  ))) by XBOOLE_1:16
    .= dom (f1(#)g) /\ dom (g1(#)f) /\ ((dom f \ f"{0c}) /\ (dom g \ g"{0c})
  ) by XBOOLE_1:16
    .= dom (f1(#)g + g1(#)f) /\ ((dom f \ f"{0c}) /\ (dom g \ g"{0c})) by
VALUED_1:def 1
    .= dom (f1(#)g + g1(#)f) /\ (dom (f(#)g) \ (f(#)g)"{0c}) by Th7
    .= dom ((f1(#)g + g1(#)f)/(f(#)g)) by Def1;
  hence thesis by A1,PARTFUN2:1;
end;

theorem
  (f/g)/(f1/g1) = (f(#)(g1|dom(g1^)))/(g(#)f1)
proof
  thus (f/g)/(f1/g1) = (f/g)(#)((f1/g1)^) by Th39
    .= (f/g)(#)(((g1|dom(g1^)))/f1) by Th43
    .= (f(#)(g1|dom(g1^)))/(g(#)f1) by Th42;
end;

theorem
  f1/f - g1/g = (f1(#)g - g1(#)f)/(f(#)g)
proof
  thus f1/f - g1/g = f1/f + ((-1r)(#) g1)/g by Th40,COMPLEX1:def 4
    .= (f1(#)g + (-1r)(#) g1(#)f)/(f(#)g) by Th48
    .= (f1(#)g - (g1(#)f))/(f(#)g) by Th17,COMPLEX1:def 4;
end;

theorem
  |.f1/f2.| = |.f1.|/|.f2.|
proof
  thus |.f1/f2.| = |.f1(#)(f2^).| by Th39
    .= |.f1.|(#)|.(f2^).| by Th29
    .= |.f1.|(#)((|.f2.|)^) by Th38
    .= |.f1.|/|.f2.| by RFUNCT_1:31;
end;

theorem Th52:
  (f1+f2)|X = f1|X + f2|X & (f1+f2)|X = f1|X + f2 & (f1+f2)|X = f1 + f2|X
proof
A1: now
    let c;
    assume
A2: c in dom ((f1+f2)|X);
    then
A3: c in dom (f1+f2) /\ X by RELAT_1:61;
    then
A4: c in X by XBOOLE_0:def 4;
A5: c in dom (f1+f2) by A3,XBOOLE_0:def 4;
    then
A6: c in dom f1 /\ dom f2 by VALUED_1:def 1;
    then c in dom f2 by XBOOLE_0:def 4;
    then c in dom f2 /\ X by A4,XBOOLE_0:def 4;
    then
A7: c in dom (f2|X) by RELAT_1:61;
    c in dom f1 by A6,XBOOLE_0:def 4;
    then c in dom f1 /\ X by A4,XBOOLE_0:def 4;
    then
A8: c in dom (f1|X) by RELAT_1:61;
    then c in dom (f1|X) /\ dom (f2|X) by A7,XBOOLE_0:def 4;
    then
A9: c in dom ((f1|X) + (f2|X)) by VALUED_1:def 1;
    thus ((f1+f2)|X)/.c = (f1+f2)/.c by A2,PARTFUN2:15
      .= (((f1/.c))) + (((f2/.c))) by A5,Th1
      .= ((f1|X)/.c) + (((f2/.c))) by A8,PARTFUN2:15
      .= ((f1|X)/.c) + ((f2|X)/.c) by A7,PARTFUN2:15
      .= ((f1|X)+(f2|X))/.c by A9,Th1;
  end;
  dom ((f1+f2)|X) = dom (f1+f2) /\ X by RELAT_1:61
    .= dom f1 /\ dom f2 /\ (X /\ X) by VALUED_1:def 1
    .= dom f1 /\ (dom f2 /\ (X /\ X)) by XBOOLE_1:16
    .= dom f1 /\ (dom f2 /\ X /\ X) by XBOOLE_1:16
    .= dom f1 /\ (X /\ dom (f2|X)) by RELAT_1:61
    .= dom f1 /\ X /\ dom (f2|X) by XBOOLE_1:16
    .= dom (f1|X) /\ dom (f2|X) by RELAT_1:61
    .= dom ((f1|X)+(f2|X)) by VALUED_1:def 1;
  hence (f1+f2)|X = f1|X + f2|X by A1,PARTFUN2:1;
A10: now
    let c;
    assume
A11: c in dom ((f1+f2)|X);
    then
A12: c in dom (f1+f2) /\ X by RELAT_1:61;
    then
A13: c in X by XBOOLE_0:def 4;
A14: c in dom (f1+f2) by A12,XBOOLE_0:def 4;
    then
A15: c in dom f1 /\ dom f2 by VALUED_1:def 1;
    then c in dom f1 by XBOOLE_0:def 4;
    then c in dom f1 /\ X by A13,XBOOLE_0:def 4;
    then
A16: c in dom (f1|X) by RELAT_1:61;
    c in dom f2 by A15,XBOOLE_0:def 4;
    then c in dom (f1|X) /\ dom f2 by A16,XBOOLE_0:def 4;
    then
A17: c in dom ((f1|X) + f2) by VALUED_1:def 1;
    thus ((f1+f2)|X)/.c = (f1+f2)/.c by A11,PARTFUN2:15
      .= (((f1/.c))) +(((f2/.c))) by A14,Th1
      .= ((f1|X)/.c) +(((f2/.c))) by A16,PARTFUN2:15
      .= ((f1|X)+f2)/.c by A17,Th1;
  end;
  dom ((f1+f2)|X) = dom (f1+f2) /\ X by RELAT_1:61
    .= dom f1 /\ dom f2 /\ X by VALUED_1:def 1
    .= dom f1 /\ X /\ dom f2 by XBOOLE_1:16
    .= dom (f1|X) /\ dom f2 by RELAT_1:61
    .= dom ((f1|X)+ f2) by VALUED_1:def 1;
  hence (f1+f2)|X = f1|X + f2 by A10,PARTFUN2:1;
A18: now
    let c;
    assume
A19: c in dom ((f1+f2)|X);
    then
A20: c in dom (f1+f2) /\ X by RELAT_1:61;
    then
A21: c in X by XBOOLE_0:def 4;
A22: c in dom (f1+f2) by A20,XBOOLE_0:def 4;
    then
A23: c in dom f1 /\ dom f2 by VALUED_1:def 1;
    then c in dom f2 by XBOOLE_0:def 4;
    then c in dom f2 /\ X by A21,XBOOLE_0:def 4;
    then
A24: c in dom (f2|X) by RELAT_1:61;
    c in dom f1 by A23,XBOOLE_0:def 4;
    then c in dom f1 /\ dom (f2|X) by A24,XBOOLE_0:def 4;
    then
A25: c in dom (f1 + (f2|X)) by VALUED_1:def 1;
    thus ((f1+f2)|X)/.c = (f1+f2)/.c by A19,PARTFUN2:15
      .= (((f1/.c))) +(((f2/.c))) by A22,Th1
      .= (((f1/.c))) +((f2|X)/.c) by A24,PARTFUN2:15
      .= (f1+(f2|X))/.c by A25,Th1;
  end;
  dom ((f1+f2)|X) = dom (f1+f2) /\ X by RELAT_1:61
    .= dom f1 /\ dom f2 /\ X by VALUED_1:def 1
    .= dom f1 /\ (dom f2 /\ X) by XBOOLE_1:16
    .= dom f1 /\ dom (f2|X) by RELAT_1:61
    .= dom (f1 + (f2|X)) by VALUED_1:def 1;
  hence thesis by A18,PARTFUN2:1;
end;

theorem Th53:
  (f1(#)f2)|X = f1|X (#) f2|X & (f1(#)f2)|X = f1|X (#) f2 & (f1(#)
  f2)|X = f1 (#) f2|X
proof
A1: now
    let c;
    assume
A2: c in dom ((f1(#)f2)|X);
    then
A3: c in dom (f1(#)f2) /\ X by RELAT_1:61;
    then
A4: c in X by XBOOLE_0:def 4;
A5: c in dom (f1(#)f2) by A3,XBOOLE_0:def 4;
    then
A6: c in dom f1 /\ dom f2 by Th3;
    then c in dom f2 by XBOOLE_0:def 4;
    then c in dom f2 /\ X by A4,XBOOLE_0:def 4;
    then
A7: c in dom (f2|X) by RELAT_1:61;
    c in dom f1 by A6,XBOOLE_0:def 4;
    then c in dom f1 /\ X by A4,XBOOLE_0:def 4;
    then
A8: c in dom (f1|X) by RELAT_1:61;
    then c in dom (f1|X) /\ dom (f2|X) by A7,XBOOLE_0:def 4;
    then
A9: c in dom ((f1|X) (#) (f2|X)) by Th3;
    thus ((f1(#)f2)|X)/.c = (f1(#)f2)/.c by A2,PARTFUN2:15
      .= (((f1/.c))) *(((f2/.c))) by A5,Th3
      .= ((f1|X)/.c) *(((f2/.c))) by A8,PARTFUN2:15
      .= ((f1|X)/.c) *((f2|X)/.c) by A7,PARTFUN2:15
      .= ((f1|X)(#)(f2|X))/.c by A9,Th3;
  end;
  dom ((f1(#)f2)|X) = dom (f1(#)f2) /\ X by RELAT_1:61
    .= dom f1 /\ dom f2 /\ (X /\ X) by Th3
    .= dom f1 /\ (dom f2 /\ (X /\ X)) by XBOOLE_1:16
    .= dom f1 /\ (dom f2 /\ X /\ X) by XBOOLE_1:16
    .= dom f1 /\ (X /\ dom (f2|X)) by RELAT_1:61
    .= dom f1 /\ X /\ dom (f2|X) by XBOOLE_1:16
    .= dom (f1|X) /\ dom (f2|X) by RELAT_1:61
    .= dom ((f1|X)(#)(f2|X)) by Th3;
  hence (f1(#)f2)|X = f1|X (#) f2|X by A1,PARTFUN2:1;
A10: now
    let c;
    assume
A11: c in dom ((f1(#)f2)|X);
    then
A12: c in dom (f1(#)f2) /\ X by RELAT_1:61;
    then
A13: c in X by XBOOLE_0:def 4;
A14: c in dom (f1(#)f2) by A12,XBOOLE_0:def 4;
    then
A15: c in dom f1 /\ dom f2 by Th3;
    then c in dom f1 by XBOOLE_0:def 4;
    then c in dom f1 /\ X by A13,XBOOLE_0:def 4;
    then
A16: c in dom (f1|X) by RELAT_1:61;
    c in dom f2 by A15,XBOOLE_0:def 4;
    then c in dom (f1|X) /\ dom f2 by A16,XBOOLE_0:def 4;
    then
A17: c in dom ((f1|X) (#) f2) by Th3;
    thus ((f1(#)f2)|X)/.c = (f1(#)f2)/.c by A11,PARTFUN2:15
      .= (((f1/.c))) *(((f2/.c))) by A14,Th3
      .= ((f1|X)/.c) *(((f2/.c))) by A16,PARTFUN2:15
      .= ((f1|X)(#)f2)/.c by A17,Th3;
  end;
  dom ((f1(#)f2)|X) = dom (f1(#)f2) /\ X by RELAT_1:61
    .= dom f1 /\ dom f2 /\ X by Th3
    .= dom f1 /\ X /\ dom f2 by XBOOLE_1:16
    .= dom (f1|X) /\ dom f2 by RELAT_1:61
    .= dom ((f1|X)(#) f2) by Th3;
  hence (f1(#)f2)|X = f1|X (#) f2 by A10,PARTFUN2:1;
A18: now
    let c;
    assume
A19: c in dom ((f1(#)f2)|X);
    then
A20: c in dom (f1(#)f2) /\ X by RELAT_1:61;
    then
A21: c in X by XBOOLE_0:def 4;
A22: c in dom (f1(#)f2) by A20,XBOOLE_0:def 4;
    then
A23: c in dom f1 /\ dom f2 by Th3;
    then c in dom f2 by XBOOLE_0:def 4;
    then c in dom f2 /\ X by A21,XBOOLE_0:def 4;
    then
A24: c in dom (f2|X) by RELAT_1:61;
    c in dom f1 by A23,XBOOLE_0:def 4;
    then c in dom f1 /\ dom (f2|X) by A24,XBOOLE_0:def 4;
    then
A25: c in dom (f1 (#) (f2|X)) by Th3;
    thus ((f1(#)f2)|X)/.c = (f1(#)f2)/.c by A19,PARTFUN2:15
      .= (((f1/.c))) *(((f2/.c))) by A22,Th3
      .= (((f1/.c))) *((f2|X)/.c) by A24,PARTFUN2:15
      .= (f1(#)(f2|X))/.c by A25,Th3;
  end;
  dom ((f1(#)f2)|X) = dom (f1(#)f2) /\ X by RELAT_1:61
    .= dom f1 /\ dom f2 /\ X by Th3
    .= dom f1 /\ (dom f2 /\ X) by XBOOLE_1:16
    .= dom f1 /\ dom (f2|X) by RELAT_1:61
    .= dom (f1 (#) (f2|X)) by Th3;
  hence thesis by A18,PARTFUN2:1;
end;

theorem Th54:
  (-f)|X = -(f|X) & (f^)|X = (f|X)^ & (|.f.|)|X = |.(f|X).|
proof
A1: dom ((f^)|X) = dom (f^) /\ X by RELAT_1:61
    .= (dom f \ f"{0c}) /\ X by Def2
    .= dom f /\ X \ f"{0c} /\ X by XBOOLE_1:50
    .= dom (f|X) \ X /\ f"{0c} by RELAT_1:61
    .= dom (f|X) \ (f|X)"{0c} by FUNCT_1:70
    .= dom ((f|X)^) by Def2;
A2: now
    let c;
    assume
A3: c in dom ((-f)|X);
    then
A4: c in dom (-f) /\ X by RELAT_1:61;
    then
A5: c in X by XBOOLE_0:def 4;
A6: c in dom (-f) by A4,XBOOLE_0:def 4;
    then c in dom f by Th5;
    then c in dom f /\ X by A5,XBOOLE_0:def 4;
    then
A7: c in dom (f|X) by RELAT_1:61;
    then
A8: c in dom (-(f|X)) by Th5;
    thus ((-f)|X)/.c = (-f)/.c by A3,PARTFUN2:15
      .= -((f/.c)) by A6,Th5
      .= -((f|X)/.c) by A7,PARTFUN2:15
      .= (-(f|X))/.c by A8,Th5;
  end;
  dom ((-f)|X) = dom (-f) /\ X by RELAT_1:61
    .= dom f /\ X by Th5
    .= dom (f|X) by RELAT_1:61
    .= dom (-(f|X)) by Th5;
  hence (-f)|X = -(f|X) by A2,PARTFUN2:1;
A9: dom ((f|X)^) c= dom (f|X) by Th6;
  now
    let c;
    assume
A10: c in dom ((f^)|X);
    then c in dom (f^) /\ X by RELAT_1:61;
    then
A11: c in dom (f^) by XBOOLE_0:def 4;
    thus ((f^)|X)/.c = (f^)/.c by A10,PARTFUN2:15
      .= ((f/.c))" by A11,Def2
      .= ((f|X)/.c)" by A9,A1,A10,PARTFUN2:15
      .= ((f|X)^)/.c by A1,A10,Def2;
  end;
  hence (f^)|X = (f|X)^ by A1,PARTFUN2:1;
A12: dom ((|.f.|)|X) = dom (|.f.|) /\ X by RELAT_1:61
    .= dom f /\ X by VALUED_1:def 11
    .= dom (f|X) by RELAT_1:61
    .= dom (|.(f|X).|) by VALUED_1:def 11;
  now
    let c;
A13: dom |.f.| = dom f by VALUED_1:def 11;
    assume
A14: c in dom ((|.f.|)|X);
    then
A15: c in dom (f|X) by A12,VALUED_1:def 11;
    c in dom (|.f.|) /\ X by A14,RELAT_1:61;
    then
A16: c in dom (|.f.|) by XBOOLE_0:def 4;
    thus ((|.f.|)|X).c = (|.f.|).c by A14,FUNCT_1:47
      .= |.(f.c).| by VALUED_1:18
      .= |.(f/.c).| by A16,A13,PARTFUN1:def 6
      .= |.(f|X)/.c.| by A15,PARTFUN2:15
      .= |.(f|X).c.| by A15,PARTFUN1:def 6
      .= (|.(f|X).|).c by VALUED_1:18;
  end;
  hence thesis by A12,PARTFUN1:5;
end;

theorem
  (f1-f2)|X = f1|X - f2|X & (f1-f2)|X = f1|X - f2 &(f1-f2)|X = f1 - f2|X
proof
  thus (f1-f2)|X = (f1|X)+ (-f2)|X by Th52
    .= (f1|X) - (f2|X) by Th54;
  thus (f1-f2)|X = (f1|X)+ -f2 by Th52
    .= (f1|X) - f2;
  thus (f1-f2)|X = f1+ (-f2)|X by Th52
    .= f1 - (f2|X) by Th54;
end;

theorem
  (f1/f2)|X = f1|X / f2|X & (f1/f2)|X = f1|X / f2 &(f1/f2)|X = f1 / f2|X
proof
  thus (f1/f2)|X = (f1(#)(f2^))|X by Th39
    .= (f1|X) (#) (f2^|X) by Th53
    .= (f1|X) (#) ((f2|X)^) by Th54
    .= (f1|X)/(f2|X) by Th39;
  thus (f1/f2)|X = (f1(#)(f2^))|X by Th39
    .= (f1|X) (#) (f2^) by Th53
    .= (f1|X)/f2 by Th39;
  thus (f1/f2)|X = (f1(#)(f2^))|X by Th39
    .= f1 (#) (f2^)|X by Th53
    .= f1 (#) ((f2|X)^) by Th54
    .= f1/(f2|X) by Th39;
end;

theorem
  (r(#)f)|X = r(#)(f|X)
proof
A1: now
    let c;
    assume
A2: c in dom ((r(#)f)|X);
    then
A3: c in dom (r(#)f) /\ X by RELAT_1:61;
    then
A4: c in X by XBOOLE_0:def 4;
A5: c in dom (r(#)f) by A3,XBOOLE_0:def 4;
    then c in dom f by Th4;
    then c in dom f /\ X by A4,XBOOLE_0:def 4;
    then
A6: c in dom (f|X) by RELAT_1:61;
    then
A7: c in dom (r(#)(f|X)) by Th4;
    thus ((r(#)f)|X)/.c = (r(#)f)/.c by A2,PARTFUN2:15
      .= r*((f/.c)) by A5,Th4
      .= r*((f|X)/.c) by A6,PARTFUN2:15
      .= (r(#)(f|X))/.c by A7,Th4;
  end;
  dom ((r(#)f)|X) = dom (r(#)f) /\ X by RELAT_1:61
    .= dom f /\ X by Th4
    .= dom (f|X) by RELAT_1:61
    .= dom (r(#)(f|X)) by Th4;
  hence thesis by A1,PARTFUN2:1;
end;

begin





theorem Th58:
  (f1 is total & f2 is total iff f1+f2 is total) & (f1 is total &
  f2 is total iff f1-f2 is total) & (f1 is total & f2 is total iff f1(#)f2 is
  total)
proof
  thus f1 is total & f2 is total iff f1+f2 is total
  proof
    thus f1 is total & f2 is total implies f1+f2 is total
    proof
      assume f1 is total & f2 is total;
      then dom f1 = C & dom f2 = C by PARTFUN1:def 2;
      hence dom (f1+f2) = C /\ C by VALUED_1:def 1
        .= C;
    end;
    assume f1+f2 is total;
    then dom (f1+f2) = C by PARTFUN1:def 2;
    then dom f1 /\ dom f2 = C by VALUED_1:def 1;
    then C c= dom f1 & C c= dom f2 by XBOOLE_1:17;
    hence dom f1 = C & dom f2 = C by XBOOLE_0:def 10;
  end;
  thus f1 is total & f2 is total iff f1-f2 is total
  proof
    thus f1 is total & f2 is total implies f1-f2 is total
    proof
      assume f1 is total & f2 is total;
      then dom f1 = C & dom f2 = C by PARTFUN1:def 2;
      hence dom (f1-f2) = C /\ C by Th2
        .= C;
    end;
    assume f1-f2 is total;
    then dom (f1-f2) = C by PARTFUN1:def 2;
    then dom f1 /\ dom f2 = C by Th2;
    then C c= dom f1 & C c= dom f2 by XBOOLE_1:17;
    hence dom f1 = C & dom f2 = C by XBOOLE_0:def 10;
  end;
  thus f1 is total & f2 is total implies f1(#)f2 is total
  proof
    assume f1 is total & f2 is total;
    then dom f1 = C & dom f2 = C by PARTFUN1:def 2;
    hence dom (f1(#)f2) = C /\ C by Th3
      .= C;
  end;
  assume f1(#)f2 is total;
  then dom (f1(#)f2) = C by PARTFUN1:def 2;
  then dom f1 /\ dom f2 = C by Th3;
  then C c= dom f1 & C c= dom f2 by XBOOLE_1:17;
  hence dom f1 = C & dom f2 = C by XBOOLE_0:def 10;
end;

theorem Th59:
  f is total iff r(#)f is total
proof
  thus f is total implies r(#)f is total
  proof
    assume f is total;
    then dom f = C by PARTFUN1:def 2;
    hence dom (r(#)f) = C by Th4;
  end;
  assume r(#)f is total;
  then dom (r(#)f) = C by PARTFUN1:def 2;
  hence dom f = C by Th4;
end;

theorem Th60:
  f is total iff -f is total
proof
A1: -f = (-1r)(#)f by COMPLEX1:def 4;
  thus f is total implies -f is total
  proof
A2: -f = (-1r)(#)f by COMPLEX1:def 4;
    assume f is total;
    hence thesis by A2,Th59;
  end;
  assume -f is total;
  hence thesis by A1,Th59;
end;

theorem Th61:
  f is total iff |.f.| is total
proof
  thus f is total implies |.f.| is total
  proof
    assume f is total;
    then dom f = C by PARTFUN1:def 2;
    hence dom (|.f.|) = C by VALUED_1:def 11;
  end;
  assume |.f.| is total;
  then dom (|.f.|) = C by PARTFUN1:def 2;
  hence dom f = C by VALUED_1:def 11;
end;

theorem Th62:
  f^ is total iff f"{0} = {} & f is total
proof
  thus f^ is total implies f"{0} = {} & f is total
  proof
    assume f^ is total;
    then
A1: dom (f^) = C by PARTFUN1:def 2;
    f"{0c} c= C;
    then f"{0c} c= dom f \ f"{0c} by A1,Def2;
    hence f"{0} = {} by XBOOLE_1:38;
    then C = dom f \ {} by A1,Def2;
    hence dom f = C;
  end;
  assume
A2: f"{0} = {} & f is total;
  thus dom (f^) = dom f \ f"{0c} by Def2
    .= C by A2,PARTFUN1:def 2;
end;

theorem Th63:
  f1 is total & f2"{0} = {} & f2 is total iff f1/f2 is total
proof
  thus f1 is total & f2"{0} = {} & f2 is total implies f1/f2 is total
  proof
    assume that
A1: f1 is total and
A2: f2"{0} = {} & f2 is total;
    f2^ is total by A2,Th62;
    then f1(#)(f2^) is total by A1,Th58;
    hence thesis by Th39;
  end;
  assume f1/f2 is total;
  then
A3: f1(#)(f2^) is total by Th39;
  hence f1 is total by Th58;
  f2^ is total by A3,Th58;
  hence thesis by Th62;
end;

theorem
  f1 is total & f2 is total implies (f1+f2)/.c = ((f1/.c)) + ((f2/.c)) &
  (f1-f2)/.c = ((f1/.c)) - ((f2/.c)) & (f1(#) f2)/.c = ((f1/.c)) * ((f2/.c))
proof
  assume
A1: f1 is total & f2 is total;
  then f1+f2 is total by Th58;
  then dom (f1+f2) = C by PARTFUN1:def 2;
  hence (f1+f2)/.c = ((f1/.c)) + ((f2/.c)) by Th1;
  f1-f2 is total by A1,Th58;
  then dom (f1-f2) = C by PARTFUN1:def 2;
  hence (f1-f2)/.c = ((f1/.c)) - ((f2/.c)) by Th2;
  f1(#)f2 is total by A1,Th58;
  then dom (f1(#)f2) = C by PARTFUN1:def 2;
  hence thesis by Th3;
end;

theorem
  f is total implies (r(#)f)/.c = r * ((f/.c))
proof
  assume f is total;
  then r(#)f is total by Th59;
  then dom (r(#)f) = C by PARTFUN1:def 2;
  hence thesis by Th4;
end;

theorem
  f is total implies (-f)/.c = - (f/.c) & (|.f.|).c = |. (f/.c) .|
proof
  assume
A1: f is total;
  then |.f.| is total by Th61;
  then
A2: dom |.f.| = dom f & dom (|.f.|) = C by PARTFUN1:def 2,VALUED_1:def 11;
  -f is total by A1,Th60;
  then dom (-f) = C by PARTFUN1:def 2;
  hence (-f)/.c = - (f/.c) by Th5;
  thus (|.f.|).c = |. (f.c) .| by VALUED_1:18
    .= |. (f/.c) .| by A2,PARTFUN1:def 6;
end;

theorem
  f^ is total implies (f^)/.c = ((f/.c))"
proof
  assume f^ is total;
  then dom (f^) = C by PARTFUN1:def 2;
  hence thesis by Def2;
end;

theorem
  f1 is total & f2^ is total implies (f1/f2)/.c = ((f1/.c)) *(((f2/.c))) "
proof
  assume that
A1: f1 is total and
A2: f2^ is total;
  f2"{0c} = {} & f2 is total by A2,Th62;
  then f1/f2 is total by A1,Th63;
  then dom (f1/f2) = C by PARTFUN1:def 2;
  hence thesis by Def1;
end;

begin





Lm3: |.f.| is bounded iff f is bounded
proof
A1: dom f = dom |.f.| by VALUED_1:def 11;
  thus |.f.| is bounded implies f is bounded
  proof
    assume |.f.| is bounded;
    then consider r1 being real number such that
A2: for y st y in dom |.f.| holds abs(|.f.|.y) < r1 by COMSEQ_2:def 3;
    take r1;
    let y;
    assume
A3: y in dom f;
    then abs(|.f.|.y) < r1 by A1,A2;
    then abs(|.f.y.|) < r1 by A1,A3,VALUED_1:def 11;
    hence thesis;
  end;
  given r1 such that
A4: for y st y in dom f holds abs(f.y) <r1;
  now
    take r1;
    let y;
    assume
A5: y in dom |.f.|;
    then abs(|.f.y.|) <r1 by A1,A4;
    hence abs(|.f.|.y) <r1 by A5,VALUED_1:def 11;
  end;
  hence thesis by COMSEQ_2:def 3;
end;

theorem Th69:
  f|Y is bounded iff ex p be real number st for c st c in Y /\ dom
  f holds |.(f/.c).|<= p
proof
A1: dom |.f.| = dom f by VALUED_1:def 11;
A2: dom(|.f.||Y) = Y /\ dom|.f.| by RELAT_1:61;
A3: |.f.||Y = |.f|Y.| by RFUNCT_1:46;
  hereby
    assume f|Y is bounded;
    then |.f.||Y is bounded by A3,Lm3;
    then consider p be real number such that
A4: for c being set st c in dom(|.f.||Y) holds abs((|.f.||Y).c) <= p
    by RFUNCT_1:72;
    now
      let c such that
A5:   c in Y /\ dom f;
      c in dom f by A5,XBOOLE_0:def 4;
      then
A6:   f.c = f/.c by PARTFUN1:def 6;
      abs((|.f.||Y).c) = abs(|.f.|.c) by A1,A2,A5,FUNCT_1:47
        .= abs(|.(f.c).|) by VALUED_1:18;
      hence |.(f/.c).| <= p by A1,A2,A4,A5,A6;
    end;
    hence
    ex p be real number st for c st c in Y /\ dom f holds |.(f/.c).| <= p;
  end;
  given p be real number such that
A7: for c st c in Y /\ dom f holds |.(f/.c).| <= p;
A8: dom |.f.| = dom f by VALUED_1:def 11;
  now
    let c be set such that
A9: c in dom(|.f.||Y);
A10: c in dom |.f.| by A9,RELAT_1:57;
    abs((|.f.||Y).c) = abs(|.f.|.c) by A9,FUNCT_1:47
      .= abs(|.(f.c).|) by VALUED_1:18
      .= abs(|.(f/.c).|) by A1,A10,PARTFUN1:def 6;
    hence abs((|.f.||Y).c) <= p by A2,A7,A8,A9;
  end;
  then |.f.||Y is bounded by RFUNCT_1:72;
  hence thesis by A3,Lm3;
end;

theorem
  Y c= X & f|X is bounded implies f|Y is bounded
proof
  assume that
A1: Y c= X and
A2: f|X is bounded;
  |.f.||X = |.f|X.| by RFUNCT_1:46;
  then |.f.||X is bounded by A2,Lm3;
  then |.f.||Y = |.f|Y.| & |.f.||Y is bounded by A1,RFUNCT_1:46,74;
  hence thesis by Lm3;
end;

theorem
  X misses dom f implies f|X is bounded
proof
  assume X /\ dom f = {};
  then for c holds c in X /\ dom f implies |.(f/.c).| <= 0;
  hence thesis by Th69;
end;

theorem Th72:
  f|Y is bounded implies (r(#)f)|Y is bounded
proof
  assume
A1: f|Y is bounded;
  |.f.||Y = |.f|Y.| by RFUNCT_1:46;
  then |.f.||Y is bounded by A1,Lm3;
  then (|.r.|(#)|.f.|)|Y is bounded by RFUNCT_1:80;
  then |.r(#)f.||Y = |.(r(#)f)|Y.| & |.r(#)f.||Y is bounded by Th30,RFUNCT_1:46
;
  hence thesis by Lm3;
end;

theorem
  |.f.||X is bounded_below
proof
  now
    take z=0;
    let c be set;
A1: z <= |. f/.c .| by COMPLEX1:46;
    assume c in X /\ dom (|.f.|);
    then
A2: c in dom (|.f.|) by XBOOLE_0:def 4;
    dom |.f.| = dom f by VALUED_1:def 11;
    then f.c = f/.c by A2,PARTFUN1:def 6;
    hence z <= (|.f.|).c by A1,VALUED_1:18;
  end;
  hence thesis by RFUNCT_1:71;
end;

theorem Th74:
  f|Y is bounded implies |.f.||Y is bounded & (-f)|Y is bounded
proof
  assume
A1: f|Y is bounded;
  |.f.||Y = |.f|Y.| by RFUNCT_1:46;
  hence |.f.||Y is bounded by A1,Lm3;
  (-1r)(#)(f|Y) = ((-1r)(#)f)|Y by RFUNCT_1:49;
  hence thesis by A1,Th72,COMPLEX1:def 4;
end;

theorem Th75:
  f1|X is bounded & f2|Y is bounded implies (f1+f2)|(X /\ Y) is bounded
proof
  assume that
A1: f1|X is bounded and
A2: f2|Y is bounded;
  consider r1 such that
A3: for c st c in X /\ dom f1 holds |.((f1/.c)).| <= r1 by A1,Th69;
  consider r2 such that
A4: for c st c in Y /\ dom f2 holds |.((f2/.c)).| <= r2 by A2,Th69;
  ex p1 st for c st c in X /\ Y /\ dom (f1+f2) holds |.(f1+f2)/.c.| <= p1
  proof
    take r0=r1+r2;
    let c;
A5: |.((f1/.c)) + ((f2/.c)).| <= |.((f1/.c)).|+|.((f2/.c)).| by COMPLEX1:56;
    assume
A6: c in X /\ Y /\ dom (f1+f2);
    then
A7: c in X /\ Y by XBOOLE_0:def 4;
    then
A8: c in X by XBOOLE_0:def 4;
A9: c in Y by A7,XBOOLE_0:def 4;
A10: c in dom (f1+f2) by A6,XBOOLE_0:def 4;
    then
A11: c in dom f1 /\ dom f2 by VALUED_1:def 1;
    then c in dom f2 by XBOOLE_0:def 4;
    then c in Y /\ dom f2 by A9,XBOOLE_0:def 4;
    then
A12: |.((f2/.c)).| <= r2 by A4;
    c in dom f1 by A11,XBOOLE_0:def 4;
    then c in X /\ dom f1 by A8,XBOOLE_0:def 4;
    then |.((f1/.c)).| <= r1 by A3;
    then |.((f1/.c)).|+|.((f2/.c)).| <= r0 by A12,XREAL_1:7;
    then |.((f1/.c)) + ((f2/.c)).| <= r0 by A5,XXREAL_0:2;
    hence thesis by A10,Th1;
  end;
  hence thesis by Th69;
end;

theorem Th76:
  f1|X is bounded & f2|Y is bounded implies (f1(#)f2)|(X /\ Y) is
  bounded & (f1-f2)|(X /\ Y) is bounded
proof
  assume that
A1: f1|X is bounded and
A2: f2|Y is bounded;
  consider r1 such that
A3: for c st c in X /\ dom f1 holds |.((f1/.c)).| <= r1 by A1,Th69;
  consider r2 such that
A4: for c st c in Y /\ dom f2 holds |.((f2/.c)).| <= r2 by A2,Th69;
  now
    take r=r1*r2;
    let c;
    assume
A5: c in X /\ Y /\ dom (f1(#)f2);
    then
A6: c in X /\ Y by XBOOLE_0:def 4;
    then
A7: c in X by XBOOLE_0:def 4;
A8: c in dom (f1(#)f2) by A5,XBOOLE_0:def 4;
    then
A9: c in dom f1 /\ dom f2 by Th3;
    then c in dom f1 by XBOOLE_0:def 4;
    then c in X /\ dom f1 by A7,XBOOLE_0:def 4;
    then
A10: |.((f1/.c)).| <= r1 by A3;
A11: c in Y by A6,XBOOLE_0:def 4;
    c in dom f2 by A9,XBOOLE_0:def 4;
    then c in Y /\ dom f2 by A11,XBOOLE_0:def 4;
    then
A12: |.((f2/.c)).| <= r2 by A4;
    0<=|.((f1/.c)).| & 0<=|.((f2/.c)).| by COMPLEX1:46;
    then |.((f1/.c)).|*|.((f2/.c)).| <= r by A10,A12,XREAL_1:66;
    then |.((f1/.c)) * ((f2/.c)).| <= r by COMPLEX1:65;
    hence |.(f1(#)f2)/.c.| <= r by A8,Th3;
  end;
  hence (f1(#)f2)|(X /\ Y) is bounded by Th69;
  (-f2)|Y is bounded by A2,Th74;
  hence thesis by A1,Th75;
end;

theorem
  f|X is bounded & f|Y is bounded implies f|(X \/ Y) is bounded
proof
  assume that
A1: f|X is bounded and
A2: f|Y is bounded;
  |.f.||Y = |.f|Y.| by RFUNCT_1:46;
  then
A3: |.f.||Y is bounded by A2,Lm3;
  |.f.||X = |.f|X.| by RFUNCT_1:46;
  then |.f.||X is bounded by A1,Lm3;
  then |.f.||(X \/ Y) = |.f|(X \/ Y).| & |.f.||(X \/ Y) is bounded by A3,
RFUNCT_1:46,87;
  hence thesis by Lm3;
end;

theorem
  f1|X is constant & f2|Y is constant implies (f1+f2)|(X /\ Y) is
  constant & (f1-f2)|(X /\ Y) is constant & (f1(#)f2)|(X /\ Y) is constant
proof
  assume that
A1: f1|X is constant and
A2: f2|Y is constant;
  consider cr1 such that
A3: for c st c in X /\ dom f1 holds (f1/.c) = cr1 by A1,PARTFUN2:35;
  consider cr2 such that
A4: for c st c in Y /\ dom f2 holds (f2/.c) = cr2 by A2,PARTFUN2:35;
  now
    let c;
    assume
A5: c in X /\ Y /\ dom (f1+f2);
    then
A6: c in X /\ Y by XBOOLE_0:def 4;
    then
A7: c in X by XBOOLE_0:def 4;
A8: c in dom (f1+f2) by A5,XBOOLE_0:def 4;
    then
A9: c in (dom f1 /\ dom f2) by VALUED_1:def 1;
    then c in dom f1 by XBOOLE_0:def 4;
    then
A10: c in X /\ dom f1 by A7,XBOOLE_0:def 4;
A11: c in Y by A6,XBOOLE_0:def 4;
    c in dom f2 by A9,XBOOLE_0:def 4;
    then
A12: c in Y /\ dom f2 by A11,XBOOLE_0:def 4;
    thus (f1+f2)/.c =((f1/.c)) + ((f2/.c)) by A8,Th1
      .= cr1 + ((f2/.c)) by A3,A10
      .= cr1 + cr2 by A4,A12;
  end;
  hence (f1+f2)|(X /\ Y) is constant by PARTFUN2:35;
  now
    let c;
    assume
A13: c in X /\ Y /\ dom (f1-f2);
    then
A14: c in X /\ Y by XBOOLE_0:def 4;
    then
A15: c in X by XBOOLE_0:def 4;
A16: c in dom (f1-f2) by A13,XBOOLE_0:def 4;
    then
A17: c in (dom f1 /\ dom f2) by Th2;
    then c in dom f1 by XBOOLE_0:def 4;
    then
A18: c in X /\ dom f1 by A15,XBOOLE_0:def 4;
A19: c in Y by A14,XBOOLE_0:def 4;
    c in dom f2 by A17,XBOOLE_0:def 4;
    then
A20: c in Y /\ dom f2 by A19,XBOOLE_0:def 4;
    thus (f1-f2)/.c = ((f1/.c)) - ((f2/.c)) by A16,Th2
      .= cr1 - ((f2/.c)) by A3,A18
      .= cr1 - cr2 by A4,A20;
  end;
  hence (f1-f2)|(X /\ Y) is constant by PARTFUN2:35;
  now
    let c;
    assume
A21: c in X /\ Y /\ dom (f1(#)f2);
    then
A22: c in X /\ Y by XBOOLE_0:def 4;
    then
A23: c in X by XBOOLE_0:def 4;
A24: c in dom (f1(#)f2) by A21,XBOOLE_0:def 4;
    then
A25: c in (dom f1 /\ dom f2) by Th3;
    then c in dom f1 by XBOOLE_0:def 4;
    then
A26: c in X /\ dom f1 by A23,XBOOLE_0:def 4;
A27: c in Y by A22,XBOOLE_0:def 4;
    c in dom f2 by A25,XBOOLE_0:def 4;
    then
A28: c in Y /\ dom f2 by A27,XBOOLE_0:def 4;
    thus (f1(#)f2)/.c =((f1/.c)) * ((f2/.c)) by A24,Th3
      .= cr1 * ((f2/.c)) by A3,A26
      .= cr1 * cr2 by A4,A28;
  end;
  hence thesis by PARTFUN2:35;
end;

theorem Th79:
  f|Y is constant implies (q(#)f)|Y is constant
proof
  assume f|Y is constant;
  then consider r such that
A1: for c st c in Y /\ dom f holds (f/.c) = r by PARTFUN2:35;
  now
    let c;
    assume
A2: c in Y /\ dom (q(#)f);
    then
A3: c in Y by XBOOLE_0:def 4;
A4: c in dom (q(#)f) by A2,XBOOLE_0:def 4;
    then c in dom f by Th4;
    then
A5: c in Y /\ dom f by A3,XBOOLE_0:def 4;
    thus (q(#)f)/.c = q * (f/.c) by A4,Th4
      .= q*r by A1,A5;
  end;
  hence thesis by PARTFUN2:35;
end;

theorem Th80:
  f|Y is constant implies |.f.||Y is constant & (-f)|Y is constant
proof
  assume f|Y is constant;
  then consider r such that
A1: for c st c in Y /\ dom f holds (f/.c) = r by PARTFUN2:35;
  now
    let c;
    assume
A2: c in Y /\ dom (|.f.|);
    then c in dom (|.f.|) by XBOOLE_0:def 4;
    then
A3: c in dom f by VALUED_1:def 11;
    c in Y by A2,XBOOLE_0:def 4;
    then
A4: c in Y /\ dom f by A3,XBOOLE_0:def 4;
    f.c = f/.c by A3,PARTFUN1:def 6;
    hence (|.f.|).c = |.(f/.c).| by VALUED_1:18
      .= |.r.| by A1,A4;
  end;
  hence |.f.||Y is constant by PARTFUN2:57;
  now
    take p=-r;
    let c;
    assume
A5: c in Y /\ dom (-f);
    then c in Y /\ dom f by Th5;
    then
A6: -(f/.c) = p by A1;
    c in dom (-f) by A5,XBOOLE_0:def 4;
    hence (-f)/.c = p by A6,Th5;
  end;
  hence thesis by PARTFUN2:35;
end;

theorem Th81:
  f|Y is constant implies f|Y is bounded
proof
  assume f|Y is constant;
  then consider r such that
A1: for c st c in Y /\ dom f holds (f/.c) = r by PARTFUN2:35;
  now
    take p=|.r.|;
    let c;
    assume c in Y /\ dom f;
    hence |.(f/.c).| <= p by A1;
  end;
  hence thesis by Th69;
end;

theorem
  f|Y is constant implies (for r holds (r(#)f)|Y is bounded) & (-f)|Y is
  bounded & |.f.||Y is bounded
proof
  assume
A1: f|Y is constant;
  hence for r holds (r(#)f)|Y is bounded by Th79,Th81;
  thus (-f)|Y is bounded by A1,Th80,Th81;
  |.f.||Y is constant by A1,Th80;
  hence thesis;
end;

theorem Th83:
  f1|X is bounded & f2|Y is constant implies (f1+f2)|(X /\ Y) is bounded
proof
  assume that
A1: f1|X is bounded and
A2: f2|Y is constant;
  f2|Y is bounded by A2,Th81;
  hence thesis by A1,Th75;
end;

theorem
  f1|X is bounded & f2|Y is constant implies (f1-f2)|(X /\ Y) is bounded
  & (f2-f1)|(X /\ Y) is bounded & (f1(#)f2)|(X /\ Y) is bounded
proof
  assume that
A1: f1|X is bounded and
A2: f2|Y is constant;
  (-f2)|Y is constant by A2,Th80;
  hence (f1-f2)|(X /\ Y) is bounded by A1,Th83;
A3: f2|Y is bounded by A2,Th81;
  hence (f2-f1)|(X /\ Y) is bounded by A1,Th76;
  thus thesis by A1,A3,Th76;
end;

theorem
  |.f.| is bounded iff f is bounded by Lm3;














begin

reserve n,m,k for Element of NAT;
reserve x, X,X1,Z,Z1 for set;
reserve s,g,r,p,x0,x1,x2 for real number;
reserve s1,s2,q1 for Real_Sequence;
reserve Y for Subset of REAL;
reserve f,f1,f2 for PartFunc of REAL,REAL;

definition
  let f,x0;
  pred f is_continuous_in x0 means
  :Def1:
  for s1 st rng s1 c= dom f & s1 is
  convergent & lim s1 = x0 holds f/*s1 is convergent & f.x0 = lim (f/*s1);
end;

theorem Th1:
  x0 in X & f is_continuous_in x0 implies f|X is_continuous_in x0
proof
  assume that
A1: x0 in X and
A2: f is_continuous_in x0;
  let s1 such that
A3: rng s1 c= dom(f|X) and
A4: s1 is convergent & lim s1 = x0;
  dom(f|X) = X /\ dom f by RELAT_1:61;
  then
A5: rng s1 c= dom f by A3,XBOOLE_1:18;
A6: (f|X)/*s1 = f/*s1 by A3,FUNCT_2:117;
  hence (f|X)/*s1 is convergent by A2,A4,A5,Def1;
  thus (f|X).x0 = f.x0 by A1,FUNCT_1:49
    .= lim ((f|X)/*s1) by A2,A4,A5,A6,Def1;
end;

theorem
  f is_continuous_in x0 iff for s1 st rng s1 c= dom f & s1 is convergent
& lim s1=x0 & (for n holds s1.n<>x0) holds f/*s1 is convergent & f.x0=lim(f/*s1
  )
proof
  thus f is_continuous_in x0 implies for s1 st rng s1 c= dom f & s1 is
convergent & lim s1=x0 & (for n holds s1.n<>x0) holds f/*s1 is convergent & f.
  x0=lim(f/*s1) by Def1;
  assume
A1: for s1 st rng s1 c=dom f & s1 is convergent & lim s1=x0 & (for n
  holds s1.n<>x0) holds f/*s1 is convergent & f.x0=lim(f/*s1);
  let s2 such that
A2: rng s2 c=dom f and
A3: s2 is convergent & lim s2=x0;
  now
    per cases;
    suppose
      ex n st for m st n<=m holds s2.m=x0;
      then consider N be Element of NAT such that
A4:   for m st N<=m holds s2.m=x0;
A5:   for n holds (s2^\N).n=x0
      proof
        let n;
        s2.(n+N)=x0 by A4,NAT_1:12;
        hence thesis by NAT_1:def 3;
      end;
A6:   f/*(s2^\N)=(f/*s2)^\N by A2,VALUED_0:27;
A7:   rng (s2^\N) c= rng s2 by VALUED_0:21;
A8:   now
        let p be real number such that
A9:     p>0;
        take n=0;
        let m such that
        n<=m;
        abs((f/*(s2^\N)).m-f.x0)=abs(f.((s2^\N).m)-f.x0) by A2,A7,FUNCT_2:108
,XBOOLE_1:1
          .=abs(f.x0-f.x0) by A5
          .=0 by ABSVALUE:2;
        hence abs((f/*(s2^\N)).m-f.x0)<p by A9;
      end;
      then
A10:  f/*(s2^\N) is convergent by SEQ_2:def 6;
      then f.x0=lim((f/*s2)^\N) by A8,A6,SEQ_2:def 7;
      hence thesis by A10,A6,SEQ_4:20,21;
    end;
    suppose
A11:  for n ex m st n<=m & s2.m<>x0;
      defpred P[Element of NAT,set,set] means for n,m st $2=n & $3=m holds n<m
      & s2.m<>x0 & for k st n<k & s2.k<>x0 holds m<=k;
      defpred P[set] means s2.$1<>x0;
      ex m1 be Element of NAT st 0<=m1 & s2.m1<>x0 by A11;
      then
A12:  ex m be Nat st P[m];
      consider M be Nat such that
A13:  P[M] & for n be Nat st P[n] holds M<=n from NAT_1:sch 5(A12);
      reconsider M9 = M as Element of NAT by ORDINAL1:def 12;
A14:  now
        let n;
        consider m such that
A15:    n+1<=m & s2.m<>x0 by A11;
        take m;
        thus n<m & s2.m<>x0 by A15,NAT_1:13;
      end;
A16:  for n for x be Element of NAT ex y be Element of NAT st P[n,x,y]
      proof
        let n;
        let x be Element of NAT;
        defpred P[Nat] means x<$1 & s2.$1<>x0;
        ex m st P[m] by A14;
        then
A17:    ex m be Nat st P[m];
        consider l be Nat such that
A18:    P[l] & for k be Nat st P[k] holds l<=k from NAT_1:sch 5(A17);
        take l;
        l in NAT by ORDINAL1:def 12;
        hence thesis by A18;
      end;
      consider F be Function of NAT,NAT such that
A19:  F.0=M9 & for n be Element of NAT holds P[n,F.n,F.(n+1)] from
      RECDEF_1:sch 2(A16);
A20:  rng F c= REAL by XBOOLE_1:1;
A21:  rng F c= NAT;
A22:  dom F=NAT by FUNCT_2:def 1;
      then reconsider F as Real_Sequence by A20,RELSET_1:4;
A23:  now
        let n;
        F.n in rng F by A22,FUNCT_1:def 3;
        hence F.n is Element of NAT by A21;
      end;
      now
        let n;
        F.n is Element of NAT & F.(n+1) is Element of NAT by A23;
        hence F.n<F.(n+1) by A19;
      end;
      then reconsider F as increasing sequence of NAT by SEQM_3:def 6;
A24:  s2*F is convergent & lim (s2*F)=x0 by A3,SEQ_4:16,17;
A25:  for n st s2.n<>x0 ex m st F.m=n
      proof
        defpred P[set] means s2.$1<>x0 & for m holds F.m<>$1;
        assume ex n st P[n];
        then
A26:    ex n be Nat st P[n];
        consider M1 be Nat such that
A27:    P[M1] & for n be Nat st P[n] holds M1<=n from NAT_1:sch 5(A26
        );
        defpred P[Nat] means $1<M1 & s2.$1<>x0 & ex m st F.m=$1;
A28:    ex n being Nat st P[n]
        proof
          take M;
          M<=M1 & M <> M1 by A13,A19,A27;
          hence M<M1 by XXREAL_0:1;
          thus s2.M<>x0 by A13;
          take 0;
          thus thesis by A19;
        end;
A29:    for n being Nat st P[n] holds n<=M1;
        consider MX be Nat such that
A30:    P[MX] & for n being Nat st P[n] holds n<=MX from NAT_1:sch 6(
        A29,A28);
A31:    for k st MX<k & k<M1 holds s2.k=x0
        proof
          given k such that
A32:      MX<k and
A33:      k<M1 & s2.k<>x0;
          now
            per cases;
            suppose
              ex m st F.m=k;
              hence contradiction by A30,A32,A33;
            end;
            suppose
              for m holds F.m<>k;
              hence contradiction by A27,A33;
            end;
          end;
          hence contradiction;
        end;
        consider m such that
A34:    F.m=MX by A30;
A35:    MX<F.(m+1) & s2.(F.(m+1))<>x0 by A19,A34;
        M1 in NAT by ORDINAL1:def 12;
        then
A36:    F.(m+1)<=M1 by A19,A27,A30,A34;
        now
          assume F.(m+1)<>M1;
          then F.(m+1)<M1 by A36,XXREAL_0:1;
          hence contradiction by A31,A35;
        end;
        hence contradiction by A27;
      end;
A37:  for n holds (s2*F).n<>x0
      proof
        defpred P[Element of NAT] means (s2*F).$1<>x0;
A38:    for k st P[k] holds P[k+1]
        proof
          let k such that
          (s2*F).k<>x0;
          P[k,F.k,F.(k+1)] by A19;
          then s2.(F.(k+1))<>x0;
          hence thesis by FUNCT_2:15;
        end;
A39:    P[0] by A13,A19,FUNCT_2:15;
        thus for n holds P[n] from NAT_1:sch 1(A39,A38);
      end;
A40:  rng (s2*F) c= rng s2 by VALUED_0:21;
      then rng (s2*F) c= dom f by A2,XBOOLE_1:1;
      then
A41:  f/*(s2*F) is convergent & f.x0=lim(f/*(s2*F)) by A1,A37,A24;
A42:  now
        let p be real number;
        assume
A43:    0<p;
        then consider n such that
A44:    for m st n<=m holds abs((f/*(s2*F)).m-f.x0)<p by A41,SEQ_2:def 7;
        take k=F.n;
        let m such that
A45:    k<=m;
        now
          per cases;
          suppose
            s2.m=x0;
            then abs((f/*s2).m-f.x0)= abs(f.x0-f.x0) by A2,FUNCT_2:108
              .=0 by ABSVALUE:2;
            hence abs((f/*s2).m-f.x0)<p by A43;
          end;
          suppose
            s2.m<>x0;
            then consider l be Element of NAT such that
A46:        m=F.l by A25;
            n<=l by A45,A46,SEQM_3:1;
            then abs((f/*(s2*F)).l-f.x0)<p by A44;
            then abs(f.((s2*F).l)-f.x0)<p by A2,A40,FUNCT_2:108,XBOOLE_1:1;
            then abs(f.(s2.m)-f.x0)<p by A46,FUNCT_2:15;
            hence abs((f/*s2).m-f.x0)<p by A2,FUNCT_2:108;
          end;
        end;
        hence abs((f/*s2).m-f.x0)<p;
      end;
      hence f/*s2 is convergent by SEQ_2:def 6;
      hence f.x0=lim(f/*s2) by A42,SEQ_2:def 7;
    end;
  end;
  hence thesis;
end;

theorem Th3:
  f is_continuous_in x0 iff for r st 0<r ex s st 0<s & for x1 st x1
  in dom f & abs(x1-x0)<s holds abs(f.x1-f.x0)<r
proof
  thus f is_continuous_in x0 implies for r st 0<r ex s st 0<s & for x1 st x1
  in dom f & abs(x1-x0)<s holds abs(f.x1-f.x0)<r
  proof
    assume
A1: f is_continuous_in x0;
    given r such that
A2: 0<r and
A3: for s holds not 0<s or ex x1 st x1 in dom f & abs(x1-x0)<s & not
    abs (f.x1-f.x0)<r;
    defpred P[Element of NAT,real number] means $2 in dom f & abs($2-x0) < 1/(
    $1+1) & not abs(f.$2-f.x0)<r;
A4: for n ex p being Real st P[n,p]
    proof
      let n;
      0<(n+1)";
      then 0<1/(n+1) by XCMPLX_1:215;
      then consider p such that
A5:   p in dom f & abs(p-x0) < 1/(n+1) & not abs(f.p-f.x0)<r by A3;
      take p;
      thus thesis by A5;
    end;
    consider s1 such that
A6: for n holds P[n,s1.n] from FUNCT_2:sch 3(A4);
A7: rng s1 c= dom f
    proof
      let x;
      assume x in rng s1;
      then ex n st x=s1.n by FUNCT_2:113;
      hence thesis by A6;
    end;
A8: now
      let n;
      not abs(f.(s1.n)-f.x0)<r by A6;
      hence not abs((f/*s1).n-f.x0)<r by A7,FUNCT_2:108;
    end;
A9: now
      let s be real number;
      assume
A10:  0<s;
      consider n such that
A11:  s"<n by SEQ_4:3;
      s"+(0 qua Nat) <n+1 by A11,XREAL_1:8;
      then 1/(n+1)<1/s" by A10,XREAL_1:76;
      then
A12:  1/(n+1)<s by XCMPLX_1:216;
      take k=n;
      let m;
      assume k<=m;
      then k+1<=m+1 by XREAL_1:6;
      then 1/(m+1)<=1/(k+1) by XREAL_1:118;
      then 1/(m+1)<s by A12,XXREAL_0:2;
      hence abs(s1.m-x0)<s by A6,XXREAL_0:2;
    end;
    then
A13: s1 is convergent by SEQ_2:def 6;
    then lim s1=x0 by A9,SEQ_2:def 7;
    then f/*s1 is convergent & f.x0=lim(f/*s1) by A1,A7,A13,Def1;
    then consider n such that
A14: for m st n<=m holds abs((f/*s1).m-f.x0)<r by A2,SEQ_2:def 7;
    abs((f/*s1).n-f.x0)<r by A14;
    hence contradiction by A8;
  end;
  assume
A15: for r st 0<r ex s st 0<s & for x1 st x1 in dom f & abs(x1-x0)<s
  holds abs(f.x1-f.x0)<r;
  now
    let s1 such that
A16: rng s1 c= dom f and
A17: s1 is convergent & lim s1 = x0;
A18: now
      let p be real number;
      assume 0<p;
      then consider s such that
A19:  0<s and
A20:  for x1 st x1 in dom f & abs(x1-x0)<s holds abs(f.x1-f.x0)<p by A15;
      consider n such that
A21:  for m st n<=m holds abs(s1.m-x0)<s by A17,A19,SEQ_2:def 7;
      take k=n;
      let m;
      assume k<=m;
      then s1.m in rng s1 & abs(s1.m-x0)<s by A21,VALUED_0:28;
      then abs(f.(s1.m)-f.x0)<p by A16,A20;
      hence abs((f/*s1).m - f.x0)<p by A16,FUNCT_2:108;
    end;
    then f/*s1 is convergent by SEQ_2:def 6;
    hence f/*s1 is convergent & f.x0 = lim (f/*s1) by A18,SEQ_2:def 7;
  end;
  hence thesis by Def1;
end;

theorem Th4:
  for f,x0 holds f is_continuous_in x0 iff for N1 being
Neighbourhood of f.x0 ex N being Neighbourhood of x0 st for x1 st x1 in dom f &
  x1 in N holds f.x1 in N1
proof
  let f,x0;
  thus f is_continuous_in x0 implies for N1 being Neighbourhood of f.x0 ex N
  being Neighbourhood of x0 st for x1 st x1 in dom f & x1 in N holds f.x1 in N1
  proof
    assume
A1: f is_continuous_in x0;
    let N1 be Neighbourhood of f.x0;
    consider r such that
A2: 0<r and
A3: N1 = ].f.x0-r,f.x0+r.[ by RCOMP_1:def 6;
    consider s such that
A4: 0<s and
A5: for x1 st x1 in dom f & abs(x1-x0)<s holds abs(f.x1-f.x0)<r by A1,A2,Th3;
    reconsider N=].x0-s,x0+s.[ as Neighbourhood of x0 by A4,RCOMP_1:def 6;
    take N;
    let x1;
    assume that
A6: x1 in dom f and
A7: x1 in N;
    abs(x1-x0)<s by A7,RCOMP_1:1;
    then abs(f.x1-f.x0)<r by A5,A6;
    hence thesis by A3,RCOMP_1:1;
  end;
  assume
A8: for N1 being Neighbourhood of f.x0 ex N being Neighbourhood of x0
  st for x1 st x1 in dom f & x1 in N holds f.x1 in N1;
  now
    let r;
    assume 0<r;
    then reconsider N1 = ].f.x0-r,f.x0+r.[ as Neighbourhood of f.x0 by
RCOMP_1:def 6;
    consider N2 being Neighbourhood of x0 such that
A9: for x1 st x1 in dom f & x1 in N2 holds f.x1 in N1 by A8;
    consider s such that
A10: 0<s and
A11: N2 = ].x0-s,x0+s.[ by RCOMP_1:def 6;
    take s;
    for x1 st x1 in dom f & abs(x1-x0)<s holds abs(f.x1-f.x0)<r
    proof
      let x1;
      assume that
A12:  x1 in dom f and
A13:  abs(x1-x0)<s;
      x1 in N2 by A11,A13,RCOMP_1:1;
      then f.x1 in N1 by A9,A12;
      hence thesis by RCOMP_1:1;
    end;
    hence 0<s & for x1 st x1 in dom f & abs(x1-x0)<s holds abs(f.x1-f.x0)<r by
A10;
  end;
  hence thesis by Th3;
end;

theorem Th5:
  for f,x0 holds f is_continuous_in x0 iff for N1 being
  Neighbourhood of f.x0 ex N being Neighbourhood of x0 st f.:N c= N1
proof
  let f,x0;
  thus f is_continuous_in x0 implies for N1 being Neighbourhood of f.x0 ex N
  being Neighbourhood of x0 st f.:N c= N1
  proof
    assume
A1: f is_continuous_in x0;
    let N1 be Neighbourhood of f.x0;
    consider N being Neighbourhood of x0 such that
A2: for x1 st x1 in dom f & x1 in N holds f.x1 in N1 by A1,Th4;
    take N;
    now
      let r;
      assume r in f.:N;
      then ex x be Element of REAL st x in dom f & x in N & r=f.x by
PARTFUN2:59;
      hence r in N1 by A2;
    end;
    hence thesis by MEMBERED:def 9;
  end;
  assume
A3: for N1 being Neighbourhood of f.x0 ex N being Neighbourhood of x0 st
  f.:N c= N1;
  now
    let N1 be Neighbourhood of f.x0;
    consider N being Neighbourhood of x0 such that
A4: f.:N c= N1 by A3;
    take N;
    let x1;
    assume x1 in dom f & x1 in N;
    then f.x1 in f.:N by FUNCT_1:def 6;
    hence f.x1 in N1 by A4;
  end;
  hence thesis by Th4;
end;

theorem
  (ex N be Neighbourhood of x0 st dom f /\ N = {x0}) implies f
  is_continuous_in x0
proof
  given N be Neighbourhood of x0 such that
A1: dom f /\ N = {x0};
  x0 in dom f /\ N by A1,TARSKI:def 1;
  then
A2: x0 in dom f by XBOOLE_0:def 4;
  now
    let N1 be Neighbourhood of f.x0;
    take N;
A3: f.x0 in N1 by RCOMP_1:16;
    f.:N = Im(f,x0) by A1,RELAT_1:112
      .= {f.x0} by A2,FUNCT_1:59;
    hence f.:N c= N1 by A3,ZFMISC_1:31;
  end;
  hence thesis by Th5;
end;

theorem Th7:
  x0 in dom f1 /\ dom f2 & f1 is_continuous_in x0 & f2
is_continuous_in x0 implies f1+f2 is_continuous_in x0 & f1-f2 is_continuous_in
  x0 & f1(#)f2 is_continuous_in x0
proof
  assume
A1: x0 in dom f1 /\ dom f2;
  assume that
A2: f1 is_continuous_in x0 and
A3: f2 is_continuous_in x0;
  now
    let s1;
    assume that
A4: rng s1 c= dom(f1+f2) and
A5: s1 is convergent & lim s1=x0;
A6: rng s1 c= dom f1 /\ dom f2 by A4,VALUED_1:def 1;
    dom (f1+f2) = dom f1 /\ dom f2 by VALUED_1:def 1;
    then dom (f1+f2) c= dom f2 by XBOOLE_1:17;
    then
A7: rng s1 c= dom f2 by A4,XBOOLE_1:1;
    then
A8: f2/*s1 is convergent by A3,A5,Def1;
    dom (f1+f2) = dom f1 /\ dom f2 by VALUED_1:def 1;
    then dom (f1+f2) c= dom f1 by XBOOLE_1:17;
    then
A9: rng s1 c= dom f1 by A4,XBOOLE_1:1;
    then
A10: f1/*s1 is convergent by A2,A5,Def1;
    then f1/*s1+f2/*s1 is convergent by A8;
    hence (f1+f2)/*s1 is convergent by A6,RFUNCT_2:8;
A11: f1.x0 = lim (f1/*s1) by A2,A5,A9,Def1;
A12: f2.x0 = lim (f2/*s1) by A3,A5,A7,Def1;
    x0 in dom (f1+f2) by A1,VALUED_1:def 1;
    hence (f1+f2).x0 = f1.x0 + f2.x0 by VALUED_1:def 1
      .= lim (f1/*s1 + f2/*s1) by A10,A11,A8,A12,SEQ_2:6
      .= lim ((f1+f2)/*s1) by A6,RFUNCT_2:8;
  end;
  hence f1+f2 is_continuous_in x0 by Def1;
  now
    let s1;
    assume that
A13: rng s1 c= dom(f1-f2) and
A14: s1 is convergent & lim s1=x0;
A15: rng s1 c= dom f1 /\ dom f2 by A13,VALUED_1:12;
    dom (f1-f2) = dom f1 /\ dom f2 by VALUED_1:12;
    then dom (f1-f2) c= dom f2 by XBOOLE_1:17;
    then
A16: rng s1 c= dom f2 by A13,XBOOLE_1:1;
    then
A17: f2/*s1 is convergent by A3,A14,Def1;
    dom (f1-f2) = dom f1 /\ dom f2 by VALUED_1:12;
    then dom (f1-f2) c= dom f1 by XBOOLE_1:17;
    then
A18: rng s1 c= dom f1 by A13,XBOOLE_1:1;
    then
A19: f1/*s1 is convergent by A2,A14,Def1;
    then f1/*s1-f2/*s1 is convergent by A17;
    hence (f1-f2)/*s1 is convergent by A15,RFUNCT_2:8;
A20: f1.x0 = lim (f1/*s1) by A2,A14,A18,Def1;
A21: f2.x0 = lim (f2/*s1) by A3,A14,A16,Def1;
    x0 in dom (f1-f2) by A1,VALUED_1:12;
    hence (f1-f2).x0 = f1.x0 - f2.x0 by VALUED_1:13
      .= lim (f1/*s1 - f2/*s1) by A19,A20,A17,A21,SEQ_2:12
      .= lim ((f1-f2)/*s1) by A15,RFUNCT_2:8;
  end;
  hence f1-f2 is_continuous_in x0 by Def1;
  let s1;
  assume that
A22: rng s1 c= dom(f1(#) f2) and
A23: s1 is convergent & lim s1=x0;
  dom (f1(#)f2) = dom f1 /\ dom f2 by VALUED_1:def 4;
  then dom (f1(#)f2) c= dom f2 by XBOOLE_1:17;
  then
A24: rng s1 c= dom f2 by A22,XBOOLE_1:1;
  then
A25: f2/*s1 is convergent by A3,A23,Def1;
A26: rng s1 c= dom f1 /\ dom f2 by A22,VALUED_1:def 4;
  dom (f1(#)f2) = dom f1 /\ dom f2 by VALUED_1:def 4;
  then dom (f1(#)f2) c= dom f1 by XBOOLE_1:17;
  then
A27: rng s1 c= dom f1 by A22,XBOOLE_1:1;
  then
A28: f1/*s1 is convergent by A2,A23,Def1;
  then (f1/*s1)(#)(f2/*s1) is convergent by A25;
  hence (f1(#)f2)/*s1 is convergent by A26,RFUNCT_2:8;
A29: f1.x0 = lim (f1/*s1) by A2,A23,A27,Def1;
A30: f2.x0 = lim (f2/*s1) by A3,A23,A24,Def1;
  thus (f1(#)f2).x0 = f1.x0 * f2.x0 by VALUED_1:5
    .= lim ((f1/*s1)(#)(f2/*s1)) by A28,A29,A25,A30,SEQ_2:15
    .= lim ((f1(#)f2)/*s1) by A26,RFUNCT_2:8;
end;

theorem Th8:
  x0 in dom f & f is_continuous_in x0 implies r(#)f is_continuous_in x0
proof
  assume x0 in dom f;
  then
A1: x0 in dom (r(#)f) by VALUED_1:def 5;
  assume
A2: f is_continuous_in x0;
  let s1;
  assume that
A3: rng s1 c= dom(r(#)f) and
A4: s1 is convergent & lim s1=x0;
A5: rng s1 c= dom f by A3,VALUED_1:def 5;
  then
A6: f.x0 = lim (f/*s1) by A2,A4,Def1;
A7: f/*s1 is convergent by A2,A4,A5,Def1;
  then r(#)(f/*s1) is convergent;
  hence (r(#)f)/*s1 is convergent by A5,RFUNCT_2:9;
  thus (r(#)f).x0 = r*f.x0 by A1,VALUED_1:def 5
    .= lim (r(#)(f/*s1)) by A7,A6,SEQ_2:8
    .= lim ((r(#)f)/*s1) by A5,RFUNCT_2:9;
end;

theorem
  x0 in dom f & f is_continuous_in x0 implies abs(f) is_continuous_in x0
  & -f is_continuous_in x0
proof
  assume
A1: x0 in dom f;
  assume
A2: f is_continuous_in x0;
  now
    let s1;
    assume that
A3: rng s1 c= dom(abs(f)) and
A4: s1 is convergent & lim s1=x0;
A5: rng s1 c= dom f by A3,VALUED_1:def 11;
    then
A6: f.x0 = lim (f/*s1) by A2,A4,Def1;
A7: f/*s1 is convergent by A2,A4,A5,Def1;
    then abs(f/*s1) is convergent by SEQ_4:13;
    hence (abs(f))/*s1 is convergent by A5,RFUNCT_2:10;
    thus (abs(f)).x0 = abs(f.x0) by VALUED_1:18
      .= lim (abs(f/*s1)) by A7,A6,SEQ_4:14
      .= lim ((abs(f))/*s1) by A5,RFUNCT_2:10;
  end;
  hence abs(f) is_continuous_in x0 by Def1;
  thus thesis by A1,A2,Th8;
end;

theorem Th10:
  f is_continuous_in x0 & f.x0<>0 implies f^ is_continuous_in x0
proof
  assume that
A1: f is_continuous_in x0 and
A2: f.x0<>0;
  not f.x0 in {0} by A2,TARSKI:def 1;
  then
A3: not x0 in f"{0} by FUNCT_1:def 7;
  let s1;
  assume that
A4: rng s1 c= dom (f^) and
A5: s1 is convergent & lim s1=x0;
  dom f \ f"{0} c= dom f & rng s1 c= dom f \ f"{0} by A4,RFUNCT_1:def 2
,XBOOLE_1:36;
  then rng s1 c= dom f by XBOOLE_1:1;
  then
A6: f/*s1 is convergent & f.x0 = lim (f/*s1) by A1,A5,Def1;
  then (f/*s1)" is convergent by A2,A4,RFUNCT_2:11,SEQ_2:21;
  hence (f^)/*s1 is convergent by A4,RFUNCT_2:12;
  x0 in dom f by A2,FUNCT_1:def 2;
  then x0 in dom f \ f"{0} by A3,XBOOLE_0:def 5;
  then x0 in dom (f^) by RFUNCT_1:def 2;
  hence (f^).x0 = (f.x0)" by RFUNCT_1:def 2
    .= lim ((f/*s1)") by A2,A4,A6,RFUNCT_2:11,SEQ_2:22
    .= lim ((f^)/*s1) by A4,RFUNCT_2:12;
end;

theorem
  x0 in dom f2 & f1 is_continuous_in x0 & f1.x0<>0 & f2 is_continuous_in
  x0 implies f2/f1 is_continuous_in x0
proof
  assume
A1: x0 in dom f2;
  assume that
A2: f1 is_continuous_in x0 and
A3: f1.x0<>0 and
A4: f2 is_continuous_in x0;
  not f1.x0 in {0} by A3,TARSKI:def 1;
  then
A5: not x0 in f1"{0} by FUNCT_1:def 7;
  x0 in dom f1 by A3,FUNCT_1:def 2;
  then x0 in dom f1 \ f1"{0} by A5,XBOOLE_0:def 5;
  then x0 in dom(f1^) by RFUNCT_1:def 2;
  then
A6: x0 in dom(f1^) /\ dom f2 by A1,XBOOLE_0:def 4;
  f1^ is_continuous_in x0 by A2,A3,Th10;
  then f2(#)(f1^) is_continuous_in x0 by A4,A6,Th7;
  hence thesis by RFUNCT_1:31;
end;

theorem Th12:
  x0 in dom (f2*f1) & f1 is_continuous_in x0 & f2 is_continuous_in
  f1.x0 implies f2*f1 is_continuous_in x0
proof
  assume
A1: x0 in dom (f2*f1);
  assume that
A2: f1 is_continuous_in x0 and
A3: f2 is_continuous_in f1.x0;
  let s1 such that
A4: rng s1 c= dom (f2*f1) and
A5: s1 is convergent & lim s1 = x0;
A6: dom (f2*f1) c= dom f1 by RELAT_1:25;
  now
    let x;
    assume x in rng (f1/*s1);
    then consider n such that
A7: x=(f1/*s1).n by FUNCT_2:113;
    s1.n in rng s1 by VALUED_0:28;
    then f1.(s1.n) in dom f2 by A4,FUNCT_1:11;
    hence x in dom f2 by A4,A6,A7,FUNCT_2:108,XBOOLE_1:1;
  end;
  then
A8: rng (f1/*s1) c= dom f2 by TARSKI:def 3;
  now
    let n;
    s1.n in rng s1 by VALUED_0:28;
    then
A9: s1.n in dom f1 by A4,FUNCT_1:11;
    thus ((f2*f1)/*s1).n = (f2*f1).(s1.n) by A4,FUNCT_2:108
      .= f2.(f1.(s1.n)) by A9,FUNCT_1:13
      .= f2.((f1/*s1).n) by A4,A6,FUNCT_2:108,XBOOLE_1:1
      .= (f2/*(f1/*s1)).n by A8,FUNCT_2:108;
  end;
  then
A10: f2/*(f1/*s1) = (f2*f1)/*s1 by FUNCT_2:63;
  rng s1 c= dom f1 by A4,A6,XBOOLE_1:1;
  then
A11: f1/*s1 is convergent & f1.x0 = lim (f1/*s1) by A2,A5,Def1;
  then f2.(f1.x0) = lim (f2/*(f1/*s1)) by A3,A8,Def1;
  hence thesis by A1,A3,A11,A8,A10,Def1,FUNCT_1:12;
end;

definition
  let f;
  attr f is continuous means
  :Def2:
  for x0 st x0 in dom f holds f is_continuous_in x0;
end;

theorem Th13:
  for X,f st X c= dom f holds f|X is continuous iff for s1 st rng
s1 c= X & s1 is convergent & lim s1 in X holds f/*s1 is convergent & f.(lim s1)
  = lim (f/*s1)
proof
  let X,f such that
A1: X c= dom f;
  thus f|X is continuous implies for s1 st rng s1 c= X & s1 is convergent &
  lim s1 in X holds f/*s1 is convergent & f.(lim s1) = lim (f/*s1)
  proof
    assume
A2: f|X is continuous;
    now
      let s1 such that
A3:   rng s1 c= X and
A4:   s1 is convergent and
A5:   lim s1 in X;
A6:   dom (f|X) = dom f /\ X by RELAT_1:61
        .= X by A1,XBOOLE_1:28;
      then
A7:   f|X is_continuous_in (lim s1) by A2,A5,Def2;
      now
        let n;
A8:     s1.n in rng s1 by VALUED_0:28;
        thus ((f|X)/*s1).n = (f|X).(s1.n) by A3,A6,FUNCT_2:108
          .= f.(s1.n) by A3,A6,A8,FUNCT_1:47
          .= (f/*s1).n by A1,A3,FUNCT_2:108,XBOOLE_1:1;
      end;
      then
A9:   (f|X)/*s1 = f/*s1 by FUNCT_2:63;
      f.(lim s1) = (f|X).(lim s1) by A5,A6,FUNCT_1:47
        .= lim (f/*s1) by A3,A4,A6,A7,A9,Def1;
      hence
      f/*s1 is convergent & f.(lim s1) = lim (f/*s1) by A3,A4,A6,A7,A9,Def1;
    end;
    hence thesis;
  end;
  assume
A10: for s1 st rng s1 c= X & s1 is convergent & lim s1 in X holds f/*s1
  is convergent & f.(lim s1) = lim (f/*s1);
  now
A11: dom (f|X) = dom f /\ X by RELAT_1:61
      .= X by A1,XBOOLE_1:28;
    let x1 such that
A12: x1 in dom(f|X);
    now
      let s1 such that
A13:  rng s1 c= dom (f|X) and
A14:  s1 is convergent and
A15:  lim s1 = x1;
      now
        let n;
A16:    s1.n in rng s1 by VALUED_0:28;
        thus ((f|X)/*s1).n = (f|X).(s1.n) by A13,FUNCT_2:108
          .= f.(s1.n) by A13,A16,FUNCT_1:47
          .= (f/*s1).n by A1,A11,A13,FUNCT_2:108,XBOOLE_1:1;
      end;
      then
A17:  (f|X)/*s1 = f/*s1 by FUNCT_2:63;
      (f|X).(lim s1) = f.(lim s1) by A12,A15,FUNCT_1:47
        .= lim ((f|X)/*s1) by A10,A12,A11,A13,A14,A15,A17;
      hence
      (f|X)/*s1 is convergent & (f|X).x1 = lim ((f|X)/*s1) by A10,A12,A11,A13
,A14,A15,A17;
    end;
    hence f|X is_continuous_in x1 by Def1;
  end;
  hence thesis by Def2;
end;

theorem Th14:
  X c= dom f implies (f|X is continuous iff for x0,r st x0 in X &
0<r ex s st 0<s & for x1 st x1 in X & abs(x1-x0) < s holds abs(f.x1 - f.x0) < r
  )
proof
  assume
A1: X c= dom f;
  thus f|X is continuous implies for x0,r st x0 in X & 0<r ex s st 0<s & for
  x1 st x1 in X & abs(x1-x0) < s holds abs(f.x1 - f.x0) < r
  proof
    assume
A2: f|X is continuous;
    let x0,r;
    assume that
A3: x0 in X and
A4: 0<r;
    x0 in dom(f|X) by A1,A3,RELAT_1:62;
    then f|X is_continuous_in x0 by A2,Def2;
    then consider s such that
A5: 0<s and
A6: for x1 st x1 in dom(f|X) & abs(x1-x0)<s holds abs((f|X).x1-(f|X).
    x0) <r by A4,Th3;
    take s;
    thus 0<s by A5;
    let x1;
    assume that
A7: x1 in X and
A8: abs(x1-x0)<s;
A9: dom (f|X) = dom f /\ X by RELAT_1:61
      .= X by A1,XBOOLE_1:28;
    then abs(f.x1 - f.x0) = abs((f|X).x1 - f.x0) by A7,FUNCT_1:47
      .= abs((f|X).x1 - (f|X).x0) by A3,A9,FUNCT_1:47;
    hence thesis by A6,A9,A7,A8;
  end;
  assume
A10: for x0,r st x0 in X & 0<r ex s st 0<s & for x1 st x1 in X & abs(x1-
  x0) < s holds abs(f.x1 - f.x0) < r;
  now
    let x0 such that
A11: x0 in dom(f|X);
A12: x0 in X by A11;
    for r st 0<r ex s st 0<s & for x1 st x1 in dom(f|X) & abs(x1-x0)<s
    holds abs((f|X).x1-(f|X).x0)<r
    proof
      let r;
      assume 0<r;
      then consider s such that
A13:  0<s and
A14:  for x1 st x1 in X & abs(x1-x0) < s holds abs(f.x1 - f.x0) < r by A10,A12;
      take s;
      thus 0<s by A13;
      let x1 such that
A15:  x1 in dom(f|X) and
A16:  abs(x1-x0)<s;
      abs((f|X).x1-(f|X).x0) = abs((f|X).x1 - f.x0) by A11,FUNCT_1:47
        .= abs(f.x1 - f.x0) by A15,FUNCT_1:47;
      hence thesis by A14,A15,A16;
    end;
    hence f|X is_continuous_in x0 by Th3;
  end;
  hence thesis by Def2;
end;

registration
  cluster constant -> continuous for PartFunc of REAL,REAL;
  coherence
  proof
    let f be PartFunc of REAL,REAL;
    assume
A1: f is constant;
    now
      reconsider s = 1 as real number;
      let x0,r;
      assume that
A2:   x0 in dom f and
A3:   0<r;
      take s;
      thus 0<s;
      let x1;
      assume
A4:   x1 in dom f;
      assume abs(x1-x0) < s;
      f.x1 = f.x0 by A1,A2,A4,FUNCT_1:def 10;
      hence abs(f.x1 - f.x0) < r by A3,ABSVALUE:2;
    end;
    then f|dom f is continuous by Th14;
    hence thesis;
  end;
end;

registration
  cluster continuous for PartFunc of REAL,REAL;
  existence
  proof
    set f = the constant PartFunc of REAL,REAL;
    take f;
    thus thesis;
  end;
end;

registration
  let f be continuous PartFunc of REAL,REAL, X be set;
  cluster f|X -> continuous for PartFunc of REAL,REAL;
  coherence
  proof
    for x0 st x0 in dom(f|X) holds f|X is_continuous_in x0
    proof
      let x0;
      assume
A1:   x0 in dom(f|X);
      then x0 in dom f by RELAT_1:57;
      then
A2:   f is_continuous_in x0 by Def2;
      x0 in X by A1;
      hence thesis by A2,Th1;
    end;
    hence thesis by Def2;
  end;
end;

theorem
  f|X is continuous iff f|X|X is continuous;

theorem Th16:
  f|X is continuous & X1 c= X implies f|X1 is continuous
proof
  assume
A1: f|X is continuous;
  assume X1 c= X;
  then f|X1 = f|X|X1 by RELAT_1:74;
  hence thesis by A1;
end;

registration
  cluster empty -> continuous for PartFunc of REAL,REAL;
  coherence;
end;

registration
  let f;
  let X be trivial set;
  cluster f|X -> continuous for PartFunc of REAL,REAL;
  coherence
  proof
    per cases;
    suppose
      f|X is empty;
      hence thesis;
    end;
    suppose
      f|X is not empty;
      then consider x0 such that
A1:   x0 in dom(f|X qua PartFunc of REAL,REAL) by MEMBERED:9;
      x0 in X by A1,RELAT_1:57;
      then
A2:   X = {x0} by ZFMISC_1:132;
      now
        let p;
        assume p in dom(f|X);
        then
A3:     p in {x0} by A2;
        thus f|X is_continuous_in p
        proof
          let s1;
          assume that
A4:       rng s1 c= dom(f|X) and
          s1 is convergent and
          lim s1=p;
A5:       dom f /\ {x0} c= {x0} by XBOOLE_1:17;
          rng s1 c= dom f /\ {x0} by A2,A4,RELAT_1:61;
          then
A6:       rng s1 c= {x0} by A5,XBOOLE_1:1;
A7:       now
            let n;
            s1.n in rng s1 by VALUED_0:28;
            hence s1.n = x0 by A6,TARSKI:def 1;
          end;
A8:       p=x0 by A3,TARSKI:def 1;
A9:       now
            let g be real number such that
A10:        0<g;
            take n=0;
            let m such that
            n<=m;
            abs(((f|{x0})/*s1).m - (f|{x0}).p) = abs((f|{x0}).(s1.m) - (f
            |{x0}).x0) by A2,A8,A4,FUNCT_2:108
              .= abs((f|{x0}).x0 - (f|{x0}).x0) by A7
              .= 0 by ABSVALUE:2;
            hence abs(((f|{x0})/*s1).m - (f|{x0}).p) < g by A10;
          end;
          hence (f|X)/*s1 is convergent by A2,SEQ_2:def 6;
          hence thesis by A2,A9,SEQ_2:def 7;
        end;
      end;
      hence thesis by Def2;
    end;
  end;
end;

theorem
  f|{x0} is continuous;

registration
  let f1,f2 be continuous PartFunc of REAL,REAL;
  cluster f1+f2 -> continuous for PartFunc of REAL,REAL;
  coherence
  proof
    set X = dom(f1+f2);
    X c= dom f1 /\ dom f2 by VALUED_1:def 1;
    then
A1: X c= dom f1 & X c= dom f2 by XBOOLE_1:18;
A2: f1|X is continuous & f2|X is continuous;
    now
      let s1;
      assume that
A3:   rng s1 c= X and
A4:   s1 is convergent and
A5:   lim s1 in X;
A6:   f1/*s1 is convergent & f2/*s1 is convergent by A1,A2,A3,A4,A5,Th13;
      then
A7:   (f1/*s1)+(f2/*s1) is convergent;
A8:   rng s1 c= dom f1 /\ dom f2 by A3,VALUED_1:def 1;
      f1.(lim s1) = lim (f1/*s1) & f2.(lim s1) = lim (f2/*s1) by A1,A2,A3,A4,A5
,Th13;
      then (f1+f2).(lim s1) = lim (f1/*s1) + lim (f2/*s1) by A5,VALUED_1:def 1
        .= lim (f1/*s1 + f2/*s1) by A6,SEQ_2:6
        .= lim ((f1+f2)/*s1) by A8,RFUNCT_2:8;
      hence (f1+f2)/*s1 is convergent & (f1+f2).(lim s1)=lim((f1+f2)/*s1) by A8
,A7,RFUNCT_2:8;
    end;
    then (f1+f2)|X is continuous by Th13;
    hence thesis;
  end;
  cluster f1-f2 -> continuous for PartFunc of REAL,REAL;
  coherence
  proof
    set X = dom(f1-f2);
    X c= dom f1 /\ dom f2 by VALUED_1:12;
    then
A9: X c= dom f1 & X c= dom f2 by XBOOLE_1:18;
A10: f1|X is continuous & f2|X is continuous;
    now
      let s1;
      assume that
A11:  rng s1 c= X and
A12:  s1 is convergent and
A13:  lim s1 in X;
A14:  f1/*s1 is convergent & f2/*s1 is convergent by A9,A10,A11,A12,A13,Th13;
      then
A15:  (f1/*s1)-(f2/*s1) is convergent;
A16:  rng s1 c= dom f1 /\ dom f2 by A11,VALUED_1:12;
      f1.(lim s1) = lim (f1/*s1) & f2.(lim s1) = lim (f2/*s1) by A9,A10,A11,A12
,A13,Th13;
      then (f1-f2).(lim s1) = lim (f1/*s1) - lim (f2/*s1) by A13,VALUED_1:13
        .= lim (f1/*s1 - f2/*s1) by A14,SEQ_2:12
        .= lim ((f1-f2)/*s1) by A16,RFUNCT_2:8;
      hence (f1-f2)/*s1 is convergent & (f1-f2).(lim s1)=lim((f1-f2)/*s1) by
A16,A15,RFUNCT_2:8;
    end;
    then (f1-f2)|X is continuous by Th13;
    hence thesis;
  end;
  cluster f1(#)f2 -> continuous for PartFunc of REAL,REAL;
  coherence
  proof
    set X = dom(f1(#)f2);
    X c= dom f1 /\ dom f2 by VALUED_1:def 4;
    then
A17: X c= dom f1 & X c= dom f2 by XBOOLE_1:18;
A18: f1|X is continuous & f2|X is continuous;
    now
      let s1;
      assume that
A19:  rng s1 c= X and
A20:  s1 is convergent and
A21:  lim s1 in X;
A22:  f1/*s1 is convergent & f2/*s1 is convergent by A17,A18,A19,A20,A21,Th13;
      then
A23:  (f1/*s1)(#)(f2/*s1) is convergent;
A24:  rng s1 c= dom f1 /\ dom f2 by A19,VALUED_1:def 4;
      f1.(lim s1) = lim (f1/*s1) & f2.(lim s1) = lim (f2/*s1) by A17,A18,A19
,A20,A21,Th13;
      then (f1(#)f2).(lim s1) = lim (f1/*s1) * lim (f2/*s1) by A21,
VALUED_1:def 4
        .= lim ((f1/*s1)(#)(f2/*s1)) by A22,SEQ_2:15
        .= lim ((f1(#)f2)/*s1) by A24,RFUNCT_2:8;
      hence
      (f1(#)f2)/*s1 is convergent & (f1(#)f2).(lim s1)=lim((f1(#)f2)/*s1)
      by A24,A23,RFUNCT_2:8;
    end;
    then (f1(#)f2)|X is continuous by Th13;
    hence thesis;
  end;
end;

theorem Th18:
  for X,f1,f2 st X c= dom f1 /\ dom f2 & f1|X is continuous & f2|X
is continuous holds (f1+f2)|X is continuous & (f1-f2)|X is continuous & (f1(#)
  f2)|X is continuous
proof
  let X,f1,f2 such that
A1: X c= dom f1 /\ dom f2;
  assume
A2: f1|X is continuous & f2|X is continuous;
A3: X c= dom f1 & X c= dom f2 by A1,XBOOLE_1:18;
A4: now
    let s1;
    assume that
A5: rng s1 c= X and
A6: s1 is convergent & lim s1 in X;
A7: rng s1 c= dom f1 /\ dom f2 by A1,A5,XBOOLE_1:1;
A8: f1/*s1 is convergent & f2/*s1 is convergent by A3,A2,A5,A6,Th13;
    then
A9: (f1/*s1)(#)(f2/*s1) is convergent;
    f1.(lim s1) = lim (f1/*s1) & f2.(lim s1) = lim (f2/*s1) by A3,A2,A5,A6,Th13
;
    then (f1(#)f2).(lim s1) = lim (f1/*s1) * lim (f2/*s1) by VALUED_1:5
      .= lim ((f1/*s1) (#) (f2/*s1)) by A8,SEQ_2:15
      .= lim ((f1(#)f2)/*s1) by A7,RFUNCT_2:8;
    hence (f1(#)f2)/*s1 is convergent & (f1(#)f2).(lim s1)=lim((f1(#) f2)/*s1)
    by A7,A9,RFUNCT_2:8;
  end;
A10: X c= dom (f1+f2) by A1,VALUED_1:def 1;
  now
    let s1;
    assume that
A11: rng s1 c= X and
A12: s1 is convergent and
A13: lim s1 in X;
A14: f1/*s1 is convergent & f2/*s1 is convergent by A3,A2,A11,A12,A13,Th13;
    then
A15: (f1/*s1)+(f2/*s1) is convergent;
A16: rng s1 c= dom f1 /\ dom f2 by A1,A11,XBOOLE_1:1;
    f1.(lim s1) = lim (f1/*s1) & f2.(lim s1) = lim (f2/*s1) by A3,A2,A11,A12
,A13,Th13;
    then (f1+f2).(lim s1) = lim (f1/*s1) + lim (f2/*s1) by A10,A13,
VALUED_1:def 1
      .= lim (f1/*s1 + f2/*s1) by A14,SEQ_2:6
      .= lim ((f1+f2)/*s1) by A16,RFUNCT_2:8;
    hence
    (f1+f2)/*s1 is convergent & (f1+f2).(lim s1)=lim((f1+f2)/*s1) by A16,A15,
RFUNCT_2:8;
  end;
  hence (f1+f2)|X is continuous by A10,Th13;
A17: X c= dom (f1-f2) by A1,VALUED_1:12;
  now
    let s1;
    assume that
A18: rng s1 c= X and
A19: s1 is convergent and
A20: lim s1 in X;
A21: f1/*s1 is convergent & f2/*s1 is convergent by A3,A2,A18,A19,A20,Th13;
    then
A22: (f1/*s1)-(f2/*s1) is convergent;
A23: rng s1 c= dom f1 /\ dom f2 by A1,A18,XBOOLE_1:1;
    f1.(lim s1) = lim (f1/*s1) & f2.(lim s1) = lim (f2/*s1) by A3,A2,A18,A19
,A20,Th13;
    then (f1-f2).(lim s1) = lim (f1/*s1) - lim (f2/*s1) by A17,A20,VALUED_1:13
      .= lim (f1/*s1 - f2/*s1) by A21,SEQ_2:12
      .= lim ((f1-f2)/*s1) by A23,RFUNCT_2:8;
    hence
    (f1-f2)/*s1 is convergent & (f1-f2).(lim s1)=lim((f1-f2)/*s1) by A23,A22,
RFUNCT_2:8;
  end;
  hence (f1-f2)|X is continuous by A17,Th13;
  X c= dom (f1(#)f2) by A1,VALUED_1:def 4;
  hence thesis by A4,Th13;
end;

theorem
  for X,X1,f1,f2 st X c= dom f1 & X1 c= dom f2 & f1|X is continuous & f2
|X1 is continuous holds (f1+f2)|(X /\ X1) is continuous & (f1-f2)|(X /\ X1) is
  continuous & (f1(#)f2)|(X /\ X1) is continuous
proof
  let X,X1,f1,f2;
  assume X c= dom f1 & X1 c= dom f2;
  then
A1: X /\ X1 c= dom f1 /\ dom f2 by XBOOLE_1:27;
  assume f1|X is continuous & f2|X1 is continuous;
  then f1|(X /\ X1) is continuous & f2|(X /\ X1) is continuous by Th16,
XBOOLE_1:17;
  hence thesis by A1,Th18;
end;

registration
  let f be continuous PartFunc of REAL,REAL;
  let r;
  cluster r(#)f -> continuous for PartFunc of REAL,REAL;
  coherence
  proof
    set X = dom f;
A1: X c= dom(r(#)f) by VALUED_1:def 5;
A2: f|X is continuous;
A3: now
      let s1;
      assume that
A4:   rng s1 c= X and
A5:   s1 is convergent and
A6:   lim s1 in X;
A7:   f/*s1 is convergent by A2,A4,A5,A6,Th13;
      then
A8:   r(#)(f/*s1) is convergent;
      f.(lim s1) = lim (f/*s1) by A2,A4,A5,A6,Th13;
      then (r(#)f).(lim s1) = r * lim (f/*s1) by A1,A6,VALUED_1:def 5
        .= lim (r(#)(f/*s1)) by A7,SEQ_2:8
        .= lim ((r(#)f)/*s1) by A4,RFUNCT_2:9;
      hence
      (r(#)f)/*s1 is convergent & (r(#)f).(lim s1)=lim((r(#)f)/*s1) by A4,A8,
RFUNCT_2:9;
    end;
    dom(r(#)f) = X by VALUED_1:def 5;
    then (r(#)f)|X = r(#)f by RELAT_1:69;
    hence thesis by A1,A3,Th13;
  end;
end;

theorem Th20:
  for r,X,f st X c= dom f & f|X is continuous holds (r(#)f)|X is continuous
proof
  let r,X,f such that
A1: X c= dom f;
  assume
A2: f|X is continuous;
A3: X c= dom(r(#)f) by A1,VALUED_1:def 5;
  now
    let s1;
    assume that
A4: rng s1 c= X and
A5: s1 is convergent and
A6: lim s1 in X;
A7: f/*s1 is convergent by A1,A2,A4,A5,A6,Th13;
    then
A8: r(#)(f/*s1) is convergent;
    f.(lim s1) = lim (f/*s1) by A1,A2,A4,A5,A6,Th13;
    then (r(#)f).(lim s1) = r * lim (f/*s1) by A3,A6,VALUED_1:def 5
      .= lim (r(#)(f/*s1)) by A7,SEQ_2:8
      .= lim ((r(#)f)/*s1) by A1,A4,RFUNCT_2:9,XBOOLE_1:1;
    hence (r(#)f)/*s1 is convergent & (r(#)f).(lim s1)=lim((r(#)f)/*s1) by A1
,A4,A8,RFUNCT_2:9,XBOOLE_1:1;
  end;
  hence thesis by A3,Th13;
end;

theorem
  X c= dom f & f|X is continuous implies (abs f)|X is continuous & (-f)|
  X is continuous
proof
  assume
A1: X c= dom f;
  assume
A2: f|X is continuous;
  thus (abs f)|X is continuous
  proof
    let r;
    assume
A3: r in dom((abs f)|X);
    then r in dom abs f by RELAT_1:57;
    then
A4: r in dom f by VALUED_1:def 11;
    r in X by A3;
    then
A5: r in dom(f|X) by A4,RELAT_1:57;
    then
A6: f|X is_continuous_in r by A2,Def2;
    thus (abs(f))|X is_continuous_in r
    proof
      let s1;
      assume that
A7:   rng s1 c= dom ((abs(f))|X) and
A8:   s1 is convergent & lim s1 = r;
      rng s1 c= dom (abs(f)) /\ X by A7,RELAT_1:61;
      then rng s1 c= dom f /\ X by VALUED_1:def 11;
      then
A9:   rng s1 c= dom (f|X) by RELAT_1:61;
      now
        let n;
A10:    s1.n in rng s1 by VALUED_0:28;
        then s1.n in dom (f|X) by A9;
        then s1.n in dom f /\ X by RELAT_1:61;
        then
A11:    s1.n in X by XBOOLE_0:def 4;
        thus (abs((f|X)/*s1)).n = abs( ((f|X)/*s1).n ) by SEQ_1:12
          .=abs((f|X).(s1.n)) by A9,FUNCT_2:108
          .=abs(f.(s1.n)) by A9,A10,FUNCT_1:47
          .=(abs(f)).(s1.n) by VALUED_1:18
          .=((abs(f))|X).(s1.n) by A11,FUNCT_1:49
          .=(((abs(f))|X)/*s1).n by A7,FUNCT_2:108;
      end;
      then
A12:  abs((f|X)/*s1) = ((abs(f))|X)/*s1 by FUNCT_2:63;
A13:  abs((f|X).r) = abs(f.r) by A5,FUNCT_1:47
        .= (abs(f)).r by VALUED_1:18
        .= ((abs(f))|X).r by A3,FUNCT_1:47;
A14:  (f|X)/*s1 is convergent by A6,A8,A9,Def1;
      hence ((abs(f))|X)/*s1 is convergent by A12,SEQ_4:13;
      (f|X).r = lim ((f|X)/*s1) by A6,A8,A9,Def1;
      hence thesis by A14,A12,A13,SEQ_4:14;
    end;
  end;
  thus thesis by A1,A2,Th20;
end;

theorem Th22:
  f|X is continuous & f"{0} = {} implies f^|X is continuous
proof
  assume that
A1: f|X is continuous and
A2: f"{0} = {};
A3: dom(f^) = dom f \ {} by A2,RFUNCT_1:def 2
    .= dom f;
  let r;
  assume
A4: r in dom(f^|X);
  then
A5: r in dom(f^) by RELAT_1:57;
  r in X by A4;
  then
A6: r in dom(f|X) by A3,A5,RELAT_1:57;
  then
A7: f|X is_continuous_in r by A1,Def2;
  now
A8: now
      assume f.r = 0;
      then f.r in {0} by TARSKI:def 1;
      hence contradiction by A2,A3,A5,FUNCT_1:def 7;
    end;
    let s1;
    assume that
A9: rng s1 c= dom((f^)|X) and
A10: s1 is convergent & lim s1= r;
    rng s1 c= dom(f^) /\ X by A9,RELAT_1:61;
    then
A11: rng s1 c= dom(f|X) by A3,RELAT_1:61;
    then
A12: (f|X)/*s1 is convergent by A7,A10,Def1;
    now
      let n;
A13:  s1.n in rng s1 by VALUED_0:28;
      rng s1 c= dom f /\ X & dom f /\ X c= dom f by A3,A9,RELAT_1:61
,XBOOLE_1:17;
      then
A14:  rng s1 c= dom f by XBOOLE_1:1;
A15:  now
        assume f.(s1.n)=0;
        then f.(s1.n) in {0} by TARSKI:def 1;
        hence contradiction by A2,A14,A13,FUNCT_1:def 7;
      end;
      ((f|X)/*s1).n = (f|X).(s1.n) by A11,FUNCT_2:108
        .= f.(s1.n) by A11,A13,FUNCT_1:47;
      hence ((f|X)/*s1).n <>0 by A15;
    end;
    then
A16: (f|X)/*s1 is non-zero by SEQ_1:5;
    now
      let n;
A17:  s1.n in rng s1 by VALUED_0:28;
      then s1.n in dom((f^)|X) by A9;
      then s1.n in dom (f^) /\ X by RELAT_1:61;
      then
A18:  s1.n in dom (f^) by XBOOLE_0:def 4;
      thus (((f^)|X)/*s1).n = ((f^)|X).(s1.n) by A9,FUNCT_2:108
        .= (f^).(s1.n) by A9,A17,FUNCT_1:47
        .= (f.(s1.n))" by A18,RFUNCT_1:def 2
        .= ((f|X).(s1.n))" by A11,A17,FUNCT_1:47
        .= (((f|X)/*s1).n)" by A11,FUNCT_2:108
        .= (((f|X)/*s1)").n by VALUED_1:10;
    end;
    then
A19: ((f^)|X)/*s1 = ((f|X)/*s1)" by FUNCT_2:63;
A20: (f|X).r = f.r by A6,FUNCT_1:47;
    then lim ((f|X)/*s1) <> 0 by A7,A10,A11,A8,Def1;
    hence ((f^)|X)/*s1 is convergent by A12,A16,A19,SEQ_2:21;
    (f|X).r = lim ((f|X)/*s1) by A7,A10,A11,Def1;
    hence lim (((f^)|X)/*s1) = ((f|X).r)" by A12,A20,A8,A16,A19,SEQ_2:22
      .= (f.r)" by A6,FUNCT_1:47
      .= (f^).r by A5,RFUNCT_1:def 2
      .= ((f^)|X).r by A4,FUNCT_1:47;
  end;
  hence thesis by Def1;
end;

theorem
  f|X is continuous & (f|X)"{0} = {} implies f^|X is continuous
proof
  assume that
A1: f|X is continuous and
A2: (f|X)"{0} = {};
  f|X|X is continuous by A1;
  then (f|X)^|X is continuous by A2,Th22;
  then (f^)|X|X is continuous by RFUNCT_1:46;
  hence thesis;
end;

theorem
  X c= dom f1 /\ dom f2 & f1|X is continuous & f1"{0} = {} & f2|X is
  continuous implies (f2/f1)|X is continuous
proof
  assume
A1: X c= dom f1 /\ dom f2;
  assume that
A2: f1|X is continuous and
A3: f1"{0} = {} and
A4: f2|X is continuous;
A5: dom(f1^) = dom f1 \ {} by A3,RFUNCT_1:def 2
    .= dom f1;
  (f1^)|X is continuous by A2,A3,Th22;
  then (f2(#)(f1^))|X is continuous by A1,A4,A5,Th18;
  hence thesis by RFUNCT_1:31;
end;

registration
  let f1,f2 be continuous PartFunc of REAL,REAL;
  cluster f2*f1 -> continuous for PartFunc of REAL,REAL;
  coherence
  proof
    now
      let x0;
      assume
A1:   x0 in dom(f2*f1);
      then f1.x0 in dom f2 by FUNCT_1:11;
      then
A2:   f2 is_continuous_in f1.x0 by Def2;
      x0 in dom f1 by A1,FUNCT_1:11;
      then f1 is_continuous_in x0 by Def2;
      hence f2*f1 is_continuous_in x0 by A1,A2,Th12;
    end;
    hence thesis by Def2;
  end;
end;

theorem
  f1|X is continuous & f2|(f1.:X) is continuous implies (f2*f1)|X is continuous
proof
  (f2*f1)|X = (f2|(f1.:X))*(f1|X) by FUNCT_1:99;
  hence thesis;
end;

theorem
  f1|X is continuous & f2|X1 is continuous implies (f2*f1)|(X /\ (f1"X1)
  ) is continuous
proof
  (f2|X1)*(f1|X) = (f2*f1)|(X /\ (f1"X1)) by FUNCT_1:100;
  hence thesis;
end;

theorem
  f is total & (for x1,x2 holds f.(x1+x2) = f.x1 + f.x2) & (ex x0 st f
  is_continuous_in x0) implies f|REAL is continuous
proof
  assume that
A1: f is total and
A2: for x1,x2 holds f.(x1+x2) = f.x1 + f.x2;
A3: dom f = REAL by A1,PARTFUN1:def 2;
  given x0 such that
A4: f is_continuous_in x0;
A5: f.x0 + (0 qua Nat) = f.(x0+(0 qua Nat)) .= f.x0+f.0 by A2;
A6: now
    let x1;
    0 = f.(x1+-x1) by A5
      .= f.x1+f.(-x1) by A2;
    hence -(f.x1)=f.(-x1);
  end;
A7: now
    let x1,x2;
    thus f.(x1-x2)=f.(x1+-x2) .= f.x1 + f.(-x2) by A2
      .= f.x1 +- f.x2 by A6
      .= f.x1 - f.x2;
  end;
  now
    let x1,r;
    assume that
    x1 in REAL and
A8: r>0;
    set y=x1-x0;
    consider s such that
A9: 0<s and
A10: for x1 st x1 in dom f & abs(x1-x0)<s holds abs(f.x1-f.x0)<r by A4,A8,Th3;
    take s;
    thus s>0 by A9;
    let x2 such that
    x2 in REAL and
A11: abs(x2-x1)<s;
A12: x2-y is Real & abs(x2-y-x0)=abs(x2-x1) by XREAL_0:def 1;
    y+x0=x1;
    then abs(f.x2-f.x1) = abs(f.x2-(f.y+f.x0)) by A2
      .= abs(f.x2-f.y-f.x0)
      .= abs(f.(x2-y)-f.x0) by A7;
    hence abs(f.x2-f.x1)<r by A3,A10,A11,A12;
  end;
  hence thesis by A3,Th14;
end;

theorem Th28:
  for f st dom f is compact & f|dom f is continuous holds rng f is compact
proof
  let f;
  assume that
A1: dom f is compact and
A2: f|dom f is continuous;
  now
    let s1 such that
A3: rng s1 c= rng f;
    defpred P[set,set] means $2 in dom f & f.$2=s1.$1;
A4: for n ex p being Real st P[n,p]
    proof
      let n;
      s1.n in rng s1 by VALUED_0:28;
      then consider p being Element of REAL such that
A5:   p in dom f & s1.n=f.p by A3,PARTFUN1:3;
      take p;
      thus thesis by A5;
    end;
    consider q1 such that
A6: for n holds P[n,q1.n] from FUNCT_2:sch 3(A4);
    now
      let x;
      assume x in rng q1;
      then ex n st x = q1.n by FUNCT_2:113;
      hence x in dom f by A6;
    end;
    then
A7: rng q1 c= dom f by TARSKI:def 3;
    then consider s2 such that
A8: s2 is subsequence of q1 and
A9: s2 is convergent and
A10: (lim s2) in dom f by A1,RCOMP_1:def 3;
    now
      let n;
      f.(q1.n)=s1.n by A6;
      hence (f/*q1).n= s1.n by A7,FUNCT_2:108;
    end;
    then
A11: f/*q1=s1 by FUNCT_2:63;
    take q2 = f/*s2;
    lim s2 in dom(f|(dom f)) by A10;
    then f|(dom f) is_continuous_in (lim s2) by A2,Def2;
    then
A12: f is_continuous_in (lim s2);
    rng s2 c= rng q1 by A8,VALUED_0:21;
    then
A13: rng s2 c= dom f by A7,XBOOLE_1:1;
    then f.(lim s2) = lim (f/*s2) by A9,A12,Def1;
    hence
    q2 is subsequence of s1 & q2 is convergent & (lim q2) in rng f by A7,A11,A8
,A9,A10,A12,A13,Def1,FUNCT_1:def 3,VALUED_0:22;
  end;
  hence thesis by RCOMP_1:def 3;
end;

theorem
  Y c= dom f & Y is compact & f|Y is continuous implies (f.:Y) is compact
proof
  assume that
A1: Y c= dom f and
A2: Y is compact and
A3: f|Y is continuous;
A4: f|Y|Y is continuous by A3;
  dom (f|Y) = dom f /\ Y by RELAT_1:61
    .= Y by A1,XBOOLE_1:28;
  then rng (f|Y) is compact by A2,A4,Th28;
  hence thesis by RELAT_1:115;
end;

theorem Th30:
  for f st dom f<>{} & (dom f) is compact & f|dom f is continuous
  ex x1,x2 st x1 in dom f & x2 in dom f & f.x1 = upper_bound (rng f) & f.x2 =
  lower_bound (rng f)
proof
  let f;
  assume dom f <> {} & dom f is compact & f|dom f is continuous;
  then
A1: rng f <> {} & rng f is compact by Th28,RELAT_1:42;
  then consider x being Element of REAL such that
A2: x in dom f & upper_bound (rng f) = f.x by PARTFUN1:3,RCOMP_1:14;
  take x;
  consider y being Element of REAL such that
A3: y in dom f & lower_bound (rng f) = f.y by A1,PARTFUN1:3,RCOMP_1:14;
  take y;
  thus thesis by A2,A3;
end;


theorem
  for f,Y st Y<>{} & Y c= dom f & Y is compact & f|Y is continuous ex x1
,x2 st x1 in Y & x2 in Y & f.x1 = upper_bound (f.:Y) & f.x2 = lower_bound (f.:Y
  )
proof
  let f,Y such that
A1: Y <> {} and
A2: Y c= dom f and
A3: Y is compact and
A4: f|Y is continuous;
A5: dom (f|Y) = dom f /\ Y by RELAT_1:61
    .= Y by A2,XBOOLE_1:28;
  f|Y|Y is continuous by A4;
  then consider x1,x2 such that
A6: x1 in dom (f|Y) & x2 in dom (f|Y) and
A7: (f|Y).x1 = upper_bound (rng (f|Y)) & (f|Y).x2 = lower_bound (rng (f|
  Y)) by A1,A3,A5,Th30;
  take x1,x2;
  thus x1 in Y & x2 in Y by A6;
  f.x1=upper_bound(rng (f|Y)) & f.x2=lower_bound(rng (f|Y)) by A6,A7,FUNCT_1:47
;
  hence thesis by RELAT_1:115;
end;

definition
  let f;
  attr f is Lipschitzian means
  :Def3:
  ex r st 0<r & for x1,x2 st x1 in dom f &
  x2 in dom f holds abs(f.x1-f.x2)<=r*abs(x1-x2);
end;

theorem Th32:
  f|X is Lipschitzian iff ex r st 0<r & for x1,x2 st x1 in dom(f|X
  ) & x2 in dom(f|X) holds abs(f.x1-f.x2)<=r*abs(x1-x2)
proof
  thus f|X is Lipschitzian implies ex r st 0<r & for x1,x2 st x1 in dom(f|X) &
  x2 in dom(f|X) holds abs(f.x1-f.x2)<=r*abs(x1-x2)
  proof
    given r such that
A1: 0<r and
A2: for x1,x2 st x1 in dom(f|X) & x2 in dom(f|X) holds abs((f|X).x1-(f
    |X).x2)<=r*abs(x1-x2);
    take r;
    thus 0<r by A1;
    let x1,x2;
    assume
A3: x1 in dom(f|X) & x2 in dom(f|X);
    then (f|X).x1 = f.x1 & (f|X).x2 = f.x2 by FUNCT_1:47;
    hence thesis by A2,A3;
  end;
  given r such that
A4: 0<r and
A5: for x1,x2 st x1 in dom(f|X) & x2 in dom(f|X) holds abs(f.x1-f.x2)<=r
  *abs(x1-x2);
  take r;
  thus 0<r by A4;
  let x1,x2;
  assume
A6: x1 in dom(f|X) & x2 in dom(f|X);
  then (f|X).x1 = f.x1 & (f|X).x2 = f.x2 by FUNCT_1:47;
  hence thesis by A5,A6;
end;

registration
  cluster empty -> Lipschitzian for PartFunc of REAL,REAL;
  coherence
  proof
    let f be PartFunc of REAL,REAL;
    assume
A1: f is empty;
    take 1;
    thus thesis by A1;
  end;
end;

registration
  cluster empty for PartFunc of REAL,REAL;
  existence
  proof
    take the empty PartFunc of REAL,REAL;
    thus thesis;
  end;
end;

registration
  let f be Lipschitzian PartFunc of REAL,REAL, X be set;
  cluster f|X -> Lipschitzian for PartFunc of REAL,REAL;
  coherence
  proof
    consider r such that
A1: 0<r and
A2: for x1,x2 st x1 in dom f & x2 in dom f holds abs(f.x1-f.x2)<=r*abs
    (x1-x2) by Def3;
    now
      let x1,x2;
      assume x1 in dom(f|X) & x2 in dom(f|X);
      then x1 in dom f & x2 in dom f by RELAT_1:57;
      hence abs(f.x1-f.x2)<=r*abs(x1-x2) by A2;
    end;
    hence thesis by A1,Th32;
  end;
end;

theorem
  f|X is Lipschitzian & X1 c= X implies f|X1 is Lipschitzian
proof
  assume that
A1: f|X is Lipschitzian and
A2: X1 c= X;
  f|X1 = f|X|X1 by A2,RELAT_1:74;
  hence thesis by A1;
end;

registration
  let f1,f2 be Lipschitzian PartFunc of REAL,REAL;
  cluster f1+f2 -> Lipschitzian for PartFunc of REAL,REAL;
  coherence
  proof
    set X = dom f1, X1 = dom f2;
    consider s such that
A1: 0<s and
A2: for x1,x2 st x1 in dom(f1|(X /\ X1)) & x2 in dom(f1|(X /\ X1))
    holds abs(f1.x1-f1.x2)<=s*abs(x1-x2) by Th32;
    consider g such that
A3: 0<g and
A4: for x1,x2 st x1 in dom(f2|(X /\ X1)) & x2 in dom(f2|(X /\ X1))
    holds abs(f2.x1-f2.x2)<=g*abs(x1-x2) by Th32;
    now
      take p=s+g;
      thus 0<p by A1,A3;
      let x1,x2;
      assume that
A5:   x1 in dom(f1+f2) and
A6:   x2 in dom(f1+f2);
      abs((f1+f2).x1-(f1+f2).x2) = abs(f1.x1 + f2.x1-(f1+f2).x2) by A5,
VALUED_1:def 1
        .= abs(f1.x1 + f2.x1 - (f1.x2+f2.x2)) by A6,VALUED_1:def 1
        .= abs(f1.x1 - f1.x2 + (f2.x1-f2.x2));
      then
A7:   abs((f1+f2).x1-(f1+f2).x2)<=abs(f1.x1-f1.x2)+abs(f2.x1-f2.x2) by
COMPLEX1:56;
      dom(f2|(X /\ X1)) = dom f2 /\ (X /\ X1) by RELAT_1:61
        .= dom f2 /\ X1 /\ X by XBOOLE_1:16
        .= dom(f1+f2) by VALUED_1:def 1;
      then
A8:   abs(f2.x1-f2.x2)<=g*abs(x1-x2) by A4,A5,A6;
      dom(f1|(X /\ X1)) = dom f1 /\ (X /\ X1) by RELAT_1:61
        .= dom f1 /\ X /\ X1 by XBOOLE_1:16
        .= dom(f1+f2) by VALUED_1:def 1;
      then abs(f1.x1-f1.x2)<=s*abs(x1-x2) by A2,A5,A6;
      then abs(f1.x1-f1.x2)+abs(f2.x1-f2.x2)<= s*abs(x1-x2)+g*abs(x1-x2) by A8,
XREAL_1:7;
      hence abs((f1+f2).x1-(f1+f2).x2)<=p*abs(x1-x2) by A7,XXREAL_0:2;
    end;
    hence thesis by Def3;
  end;
  cluster f1-f2 -> Lipschitzian for PartFunc of REAL,REAL;
  coherence
  proof
    set X = dom f1, X1 = dom f2;
    consider s such that
A9: 0<s and
A10: for x1,x2 st x1 in dom(f1|(X /\ X1)) & x2 in dom(f1|(X /\ X1))
    holds abs(f1.x1-f1.x2)<=s*abs(x1-x2) by Th32;
    consider g such that
A11: 0<g and
A12: for x1,x2 st x1 in dom(f2|(X /\ X1)) & x2 in dom(f2|(X /\ X1))
    holds abs(f2.x1-f2.x2)<=g*abs(x1-x2) by Th32;
    now
      take p=s+g;
      thus 0<p by A9,A11;
      let x1,x2;
      assume that
A13:  x1 in dom(f1-f2) and
A14:  x2 in dom(f1-f2);
      abs((f1-f2).x1-(f1-f2).x2) = abs(f1.x1 - f2.x1-(f1-f2).x2) by A13,
VALUED_1:13
        .= abs(f1.x1 - f2.x1 - (f1.x2-f2.x2)) by A14,VALUED_1:13
        .= abs(f1.x1 - f1.x2 - (f2.x1-f2.x2));
      then
A15:  abs((f1-f2).x1-(f1-f2).x2)<=abs(f1.x1-f1.x2)+abs(f2.x1-f2.x2) by
COMPLEX1:57;
      dom(f2|(X /\ X1)) = dom f2 /\ (X /\ X1) by RELAT_1:61
        .= dom f2 /\ X1 /\ X by XBOOLE_1:16
        .= dom(f1-f2) by VALUED_1:12;
      then
A16:  abs(f2.x1-f2.x2)<=g*abs(x1-x2) by A12,A13,A14;
      dom(f1|(X /\ X1)) = dom f1 /\ (X /\ X1) by RELAT_1:61
        .= dom f1 /\ X /\ X1 by XBOOLE_1:16
        .= dom(f1-f2) by VALUED_1:12;
      then abs(f1.x1-f1.x2)<=s*abs(x1-x2) by A10,A13,A14;
      then abs(f1.x1-f1.x2)+abs(f2.x1-f2.x2)<= s*abs(x1-x2)+g*abs(x1-x2) by A16
,XREAL_1:7;
      hence abs((f1-f2).x1-(f1-f2).x2)<=p*abs(x1-x2) by A15,XXREAL_0:2;
    end;
    hence thesis by Def3;
  end;
end;

theorem
  f1|X is Lipschitzian & f2|X1 is Lipschitzian implies (f1+f2)|(X /\ X1)
  is Lipschitzian
proof
A1: f1|(X /\ X1) = f1|X|(X /\ X1) & f2|(X /\ X1) = f2|X1|(X /\ X1) by
RELAT_1:74,XBOOLE_1:17;
A2: (f1+f2)|(X /\ X1) = f1|(X /\ X1)+f2|(X /\ X1) by RFUNCT_1:44;
  assume f1|X is Lipschitzian & f2|X1 is Lipschitzian;
  hence thesis by A1,A2;
end;

theorem
  f1|X is Lipschitzian & f2|X1 is Lipschitzian implies (f1-f2)|(X /\ X1)
  is Lipschitzian
proof
A1: f1|(X /\ X1) = f1|X|(X /\ X1) & f2|(X /\ X1) = f2|X1|(X /\ X1) by
RELAT_1:74,XBOOLE_1:17;
A2: (f1-f2)|(X /\ X1) = f1|(X /\ X1)-f2|(X /\ X1) by RFUNCT_1:47;
  assume f1|X is Lipschitzian & f2|X1 is Lipschitzian;
  hence thesis by A1,A2;
end;

registration
  let f1,f2 be bounded Lipschitzian PartFunc of REAL,REAL;
  cluster f1 (#) f2 -> Lipschitzian for PartFunc of REAL,REAL;
  coherence
  proof
    set X = dom f1, X1 = dom f2;
    consider x1 such that
A1: for r be set st r in dom f1 holds abs(f1.r)<=x1 by RFUNCT_1:72;
    consider x2 such that
A2: for r be set st r in dom f2 holds abs(f2.r)<=x2 by RFUNCT_1:72;
    consider g such that
A3: 0<g and
A4: for x1,x2 st x1 in dom f2 & x2 in dom f2 holds abs(f2.x1-f2.x2)<=g
    * abs(x1-x2) by Def3;
    consider s such that
A5: 0<s and
A6: for x1,x2 st x1 in dom f1 & x2 in dom f1 holds abs(f1.x1-f1.x2)<=s
    * abs(x1-x2) by Def3;
A7: now
      let r;
      assume r in dom(f1(#)f2);
      then
A8:   r in dom f1 /\ dom f2 by VALUED_1:def 4;
      then r in dom f1 by XBOOLE_0:def 4;
      then
A9:   abs(f1.r)<=x1 by A1;
      r in dom f2 by A8,XBOOLE_0:def 4;
      then
A10:  abs(f2.r)<=x2 by A2;
      x1 <= abs(x1) by ABSVALUE:4;
      hence abs(f1.r) <= abs(x1) by A9,XXREAL_0:2;
      x2 <= abs(x2) by ABSVALUE:4;
      hence abs(f2.r) <= abs(x2) by A10,XXREAL_0:2;
    end;
    now
      take p = abs(x1)*g+abs(x2)*s+1;
A11:  0<=abs(x1) by COMPLEX1:46;
      0<=abs(x1) & 0<=abs(x2) by COMPLEX1:46;
      hence 0<p by A5,A3;
      let y1,y2 be real number;
      assume that
A12:  y1 in dom(f1(#)f2) and
A13:  y2 in dom(f1(#)f2);
A14:  y2 in X /\ X1 by A13,VALUED_1:def 4;
      then
A15:  y2 in X by XBOOLE_0:def 4;
      abs((f1.y1)*(f2.y1-f2.y2))=abs(f1.y1)*abs(f2.y1-f2.y2) & 0 <= abs(
      f2.y1-f2. y2) by COMPLEX1:46,65;
      then
A16:  abs((f1.y1)*(f2.y1-f2.y2))<=abs(x1)*abs(f2.y1-f2.y2) by A7,A12,XREAL_1:64
;
A17:  y2 in X1 by A14,XBOOLE_0:def 4;
A18:  y1 in X /\ X1 by A12,VALUED_1:def 4;
      then y1 in X1 by XBOOLE_0:def 4;
      then abs(x1)*abs(f2.y1-f2.y2)<=abs(x1)*(g*abs(y1-y2)) by A4,A17,A11,
XREAL_1:64;
      then
A19:  abs((f1.y1)*(f2.y1-f2.y2))<=abs(x1)*g*abs(y1-y2) by A16,XXREAL_0:2;
      0<=abs(y1-y2) by COMPLEX1:46;
      then
A20:  (abs(x1)*g+abs(x2)*s)*abs(y1-y2)+(0 qua Nat) <= (abs(x1)*g+abs(x2)*
      s)*abs(y1-y2)+1*abs(y1-y2) by XREAL_1:7;
      abs((f1(#)f2).y1-(f1(#)f2).y2) = abs((f1.y1)*(f2.y1) - (f1(#) f2).
      y2) by VALUED_1:5
        .= abs((f1.y1)*(f2.y1)+((f1.y1)*(f2.y2)-(f1.y1)*(f2.y2))-(f1.y2)*(f2
      .y2)) by VALUED_1:5
        .= abs((f1.y1)*((f2.y1)-(f2.y2))+((f1.y1)-(f1.y2))*(f2.y2));
      then
A21:  abs((f1(#)f2).y1-(f1(#)f2).y2)<=abs((f1.y1)*((f2.y1)-(f2.y2))) +
      abs(((f1.y1)-(f1.y2))*(f2.y2)) by COMPLEX1:56;
      abs(((f1.y1)-(f1.y2))*(f2.y2))=abs(f2.y2)*abs((f1.y1)-(f1.y2)) & 0
      <= abs(f1 .y1-f1.y2) by COMPLEX1:46,65;
      then
A22:  abs(((f1.y1)-(f1.y2))*(f2.y2))<=abs(x2)*abs(f1.y1-f1.y2) by A7,A13,
XREAL_1:64;
A23:  0<=abs(x2) by COMPLEX1:46;
      y1 in X by A18,XBOOLE_0:def 4;
      then abs(x2)*abs(f1.y1-f1.y2)<=abs(x2)*(s*abs(y1-y2)) by A6,A15,A23,
XREAL_1:64;
      then abs(((f1.y1)-(f1.y2))*(f2.y2))<=abs(x2)*(s*abs(y1-y2)) by A22,
XXREAL_0:2;
      then
      abs((f1.y1)*(f2.y1-f2.y2))+abs(((f1.y1)-(f1.y2))*(f2.y2)) <= abs(x1
      )*g*abs(y1-y2)+abs(x2)*s*abs(y1-y2) by A19,XREAL_1:7;
      then abs((f1(#)f2).y1-(f1(#)f2).y2)<= (abs(x1)*g+abs(x2)*s)*abs(y1-y2)
      by A21,XXREAL_0:2;
      hence abs((f1(#)f2).y1-(f1(#)f2).y2)<=p*abs(y1-y2) by A20,XXREAL_0:2;
    end;
    hence thesis by Def3;
  end;
end;

theorem
  f1|X is Lipschitzian & f2|X1 is Lipschitzian & f1|Z is bounded & f2|Z1
  is bounded implies (f1(#)f2)|(X /\ Z /\ X1 /\ Z1) is Lipschitzian
proof
A1: f1|(X /\ Z /\ X1 /\ Z1) = f1|(X1 /\ Z1 /\ (X /\ Z)) by XBOOLE_1:16
    .= f1|(X1 /\ Z1 /\ X /\ Z) by XBOOLE_1:16
    .= f1|Z|(X1 /\ Z1 /\ X) by RELAT_1:71;
A2: f1|(X /\ Z /\ X1 /\ Z1) = f1|(X1 /\ Z1 /\ (X /\ Z)) by XBOOLE_1:16
    .= f1|(X1 /\ Z1 /\ Z /\ X) by XBOOLE_1:16
    .= f1|X|(X1 /\ Z1 /\ Z) by RELAT_1:71;
A3: f2|(X /\ Z /\ X1 /\ Z1) = f2|(X /\ Z /\ Z1 /\ X1) by XBOOLE_1:16
    .= f2|X1|(Z /\ X /\ Z1) by RELAT_1:71;
A4: (f1(#)f2)|(X /\ Z /\ X1 /\ Z1) = f1|(X /\ Z /\ X1 /\ Z1)(#)f2|(X /\ Z /\
  X1 /\ Z1) & f2|(X /\ Z /\ X1 /\ Z1) = f2|Z1|(X /\ Z /\ X1) by RELAT_1:71
,RFUNCT_1:45;
  assume f1|X is Lipschitzian & f2|X1 is Lipschitzian & f1|Z is bounded & f2|
  Z1 is bounded;
  hence thesis by A1,A2,A4,A3;
end;

registration
  let f be Lipschitzian PartFunc of REAL, REAL;
  let p;
  cluster p(#)f -> Lipschitzian for PartFunc of REAL, REAL;
  coherence
  proof
    consider s such that
A1: 0<s and
A2: for x1,x2 st x1 in dom f & x2 in dom f holds abs(f.x1-f.x2)<=s*abs
    ( x1-x2) by Def3;
    per cases;
    suppose
A3:   p=0;
      now
        take s;
        thus 0<s by A1;
        let x1,x2;
        assume that
A4:     x1 in dom(p(#)f) and
A5:     x2 in dom(p(#)f);
A6:     0<=abs(x1-x2) by COMPLEX1:46;
        abs((p(#)f).x1-(p(#)f).x2) = abs(p*(f.x1)-(p(#)f).x2) by A4,
VALUED_1:def 5
          .= abs(0 qua Nat - p*(f.x2)) by A3,A5,VALUED_1:def 5
          .= 0 by A3,ABSVALUE:2;
        hence abs((p(#)f).x1-(p(#)f).x2)<=s*abs(x1-x2) by A1,A6;
      end;
      hence thesis by Def3;
    end;
    suppose
      p<>0;
      then 0<abs(p) by COMPLEX1:47;
      then
A7:   0*s<abs(p)*s by A1,XREAL_1:68;
      now
        take g = abs(p)*s;
A8:     0<=abs(p) by COMPLEX1:46;
        thus 0<g by A7;
        let x1,x2;
        assume that
A9:     x1 in dom(p(#)f) and
A10:    x2 in dom(p(#)f);
A11:    abs((p(#)f).x1-(p(#)f).x2) = abs(p*(f.x1)-(p(#)f).x2) by A9,
VALUED_1:def 5
          .= abs(p*(f.x1) - p*(f.x2)) by A10,VALUED_1:def 5
          .= abs(p*(f.x1 - f.x2))
          .= abs(p)*abs(f.x1 - f.x2) by COMPLEX1:65;
        x1 in dom f & x2 in dom f by A9,A10,VALUED_1:def 5;
        then abs(p)*abs(f.x1-f.x2)<=abs(p)*(s*abs(x1-x2)) by A2,A8,XREAL_1:64;
        hence abs((p(#)f).x1-(p(#)f).x2)<=g*abs(x1-x2) by A11;
      end;
      hence thesis by Def3;
    end;
  end;
end;

theorem
  f|X is Lipschitzian & X c= dom f implies (p(#)f)|X is Lipschitzian
proof
  (p(#)f)|X = p(#)(f|X) by RFUNCT_1:49;
  hence thesis;
end;

registration
  let f be Lipschitzian PartFunc of REAL, REAL;
  cluster abs f -> Lipschitzian for PartFunc of REAL, REAL;
  coherence
  proof
    consider s such that
A1: 0<s and
A2: for x1,x2 st x1 in dom f & x2 in dom f holds abs(f.x1-f.x2)<=s*abs
    ( x1-x2) by Def3;
    now
      take s;
      thus 0<s by A1;
      let x1,x2;
      assume x1 in dom abs f & x2 in dom abs f;
      then x1 in dom f & x2 in dom f by VALUED_1:def 11;
      then
A3:   abs(f.x1-f.x2)<=s*abs(x1-x2) by A2;
      abs((abs(f)).x1-(abs(f)).x2) = abs(abs(f.x1)-(abs(f)).x2) by VALUED_1:18
        .= abs(abs(f.x1)-abs(f.x2)) by VALUED_1:18;
      then abs((abs f).x1-(abs(f)).x2) <= abs(f.x1-f.x2) by COMPLEX1:64;
      hence abs((abs f).x1-(abs f).x2)<=s*abs(x1-x2) by A3,XXREAL_0:2;
    end;
    hence thesis by Def3;
  end;
end;

theorem
  f|X is Lipschitzian implies -f|X is Lipschitzian & (abs f)|X is Lipschitzian
proof
  assume
A1: f|X is Lipschitzian;
  hence -f|X is Lipschitzian;
  (abs f)|X = abs(f|X) by RFUNCT_1:46;
  hence thesis by A1;
end;

registration
  cluster constant -> Lipschitzian for PartFunc of REAL, REAL;
  coherence
  proof
    let f be PartFunc of REAL, REAL such that
A1: f is constant;
    now
      let x1,x2;
      assume x1 in dom f & x2 in dom f;
      then f.x1 = f.x2 by A1,FUNCT_1:def 10;
      then abs(f.x1-f.x2) = 0 by ABSVALUE:2;
      hence abs(f.x1-f.x2) <= 1*abs(x1-x2) by COMPLEX1:46;
    end;
    hence thesis by Def3;
  end;
end;

registration
  let Y;
  cluster id Y -> Lipschitzian for PartFunc of REAL,REAL;
  coherence
  proof
    reconsider r=1 as Real;
    id Y is Lipschitzian
    proof
      take r;
      thus r>0;
      let x1,x2;
      assume that
A1:   x1 in dom id Y and
A2:   x2 in dom id Y;
A3:   x2 in Y by A2;
      x1 in Y by A1;
      then abs((id Y).x1-(id Y).x2) = abs(x1-(id Y).x2) by FUNCT_1:18
        .= r*abs(x1-x2) by A3,FUNCT_1:18;
      hence thesis;
    end;
    hence thesis;
  end;
end;

registration

  cluster Lipschitzian -> continuous for PartFunc of REAL, REAL;
  coherence
  proof
    let f be PartFunc of REAL, REAL;
    set X = dom f;
    assume f is Lipschitzian;
    then consider r such that
A1: 0<r and
A2: for x1,x2 st x1 in X & x2 in X holds abs(f.x1-f.x2)<=r*abs(x1-x2) by Def3;
    now
      let x0 such that
A3:   x0 in X;
      for r st 0<r ex s st 0<s & for x1 st x1 in dom f & abs(x1-x0)<s
      holds abs(f.x1-f.x0)<r
      proof
        let g such that
A4:     0<g;
        set s=g/r;
        take s9=s;
A5:     now
          let x1;
          assume that
A6:       x1 in dom f and
A7:       abs(x1-x0)<s;
          r*abs(x1-x0)<(g/r)*r by A1,A7,XREAL_1:68;
          then
A8:       r*abs(x1-x0)<g by A1,XCMPLX_1:87;
          abs(f.x1-f.x0)<=r*abs(x1-x0) by A2,A3,A6;
          hence abs(f.x1-f.x0)<g by A8,XXREAL_0:2;
        end;
        s9=g*r" by XCMPLX_0:def 9;
        hence thesis by A1,A4,A5,XREAL_1:129;
      end;
      hence f is_continuous_in x0 by Th3;
    end;
    hence thesis by Def2;
  end;
end;

theorem
  for f st (ex r st rng f = {r}) holds f is continuous
proof
  let f;
  given r such that
A1: rng f = {r};
  now
    let x1,x2;
    assume that
A2: x1 in dom f and
A3: x2 in dom f;
    f.x2 in rng f by A3,FUNCT_1:def 3;
    then
A4: f.x2=r by A1,TARSKI:def 1;
    f.x1 in rng f by A2,FUNCT_1:def 3;
    then f.x1=r by A1,TARSKI:def 1;
    then abs(f.x1-f.x2) = 0 by A4,ABSVALUE:2;
    hence abs(f.x1-f.x2) <= 1*abs(x1-x2) by COMPLEX1:46;
  end;
  then f is Lipschitzian by Def3;
  hence thesis;
end;

theorem
  for f st (for x0 st x0 in dom f holds f.x0 = x0) holds f is continuous
proof
  let f such that
A1: for x0 st x0 in dom f holds f.x0 = x0;
  now
    let x1,x2;
    assume that
A2: x1 in dom f and
A3: x2 in dom f;
    f.x1 = x1 by A1,A2;
    hence abs(f.x1-f.x2) <= 1*abs(x1-x2) by A1,A3;
  end;
  then f is Lipschitzian by Def3;
  hence thesis;
end;

theorem Th41:
  (for x0 st x0 in X holds f.x0 = r*x0+p) implies f|X is continuous
proof
  assume
A1: for x0 st x0 in X holds f.x0 = r*x0+p;
A2: now
    let x1,x2;
    assume that
A3: x1 in dom(f|X) and
A4: x2 in dom(f|X);
    x2 in X by A4;
    then
A5: f.x2 = r*x2+p by A1;
A6: 0<=abs(x1-x2) by COMPLEX1:46;
    x1 in X by A3;
    then f.x1 = r*x1+p by A1;
    then abs(f.x1-f.x2) = abs(r*(x1-x2)) by A5
      .= abs(r)*abs(x1-x2) by COMPLEX1:65;
    then abs(f.x1-f.x2) + (0 qua Nat) <= abs(r)*abs(x1-x2) + 1*abs(x1-x2) by A6
,XREAL_1:7;
    hence abs(f.x1-f.x2) <= (abs(r)+1)*abs(x1-x2);
  end;
  (0 qua Nat)+(0 qua Nat)<abs(r)+1 by COMPLEX1:46,XREAL_1:8;
  then f|X is Lipschitzian by A2,Th32;
  hence thesis;
end;

theorem Th42:
  (for x0 st x0 in dom f holds f.x0 = x0^2) implies f|dom f is continuous
proof
  reconsider f1 = id dom f as PartFunc of REAL,REAL;
  assume
A1: for x0 st x0 in dom f holds f.x0 = x0^2;
A2: now
    let x0 be set;
    assume
A3: x0 in dom f;
    then reconsider x1 = x0 as Real;
    thus f.x0 = x1^2 by A1,A3
      .= f1.x1*x1 by A3,FUNCT_1:18
      .= f1.x0*f1.x0 by A3,FUNCT_1:18;
  end;
  dom f1 /\ dom f1 = dom f by RELAT_1:45;
  then f = f1(#)f1 by A2,VALUED_1:def 4;
  hence thesis;
end;

theorem
  X c= dom f & (for x0 st x0 in X holds f.x0 = x0^2) implies f|X is continuous
proof
  assume that
A1: X c= dom f and
A2: for x0 st x0 in X holds f.x0=x0^2;
  X=dom f /\ X by A1,XBOOLE_1:28;
  then
A3: X=dom(f|X) by RELAT_1:61;
  now
    let x0;
    assume
A4: x0 in dom(f|X);
    then f.x0=x0^2 by A2;
    hence (f|X).x0=x0^2 by A4,FUNCT_1:47;
  end;
  then f|X|X is continuous by A3,Th42;
  hence thesis;
end;

theorem Th44:
  (for x0 st x0 in dom f holds f.x0 = abs(x0)) implies f is continuous
proof
  assume
A1: for x0 st x0 in dom f holds f.x0 = abs(x0);
  now
    let x1,x2;
    assume x1 in dom f & x2 in dom f;
    then f.x1 = abs(x1) & f.x2 = abs(x2) by A1;
    hence abs(f.x1-f.x2) <= 1*abs(x1-x2) by COMPLEX1:64;
  end;
  then f is Lipschitzian by Def3;
  hence thesis;
end;

theorem
  (for x0 st x0 in X holds f.x0 = abs(x0)) implies f|X is continuous
proof
  assume that
A1: for x0 st x0 in X holds f.x0 = abs(x0);
  now
    let x0;
    assume
A2: x0 in dom(f|X);
    then f.x0=abs(x0) by A1;
    hence (f|X).x0=abs(x0) by A2,FUNCT_1:47;
  end;
  hence thesis by Th44;
end;

theorem Th46:
  f|X is monotone & (ex p,g st p<=g & f.:X=[.p,g.]) implies f|X is continuous
proof
  assume
A1: f|X is monotone;
  given p,g such that
A2: p<=g and
A3: f.:X=[.p,g.];
  reconsider p, g as Real by XREAL_0:def 1;
  now
    per cases by A2,XXREAL_0:1;
    suppose
      p=g;
      then f.:X = {p} by A3,XXREAL_1:17;
      then rng (f|X)= {p} by RELAT_1:115;
      then f|X is constant;
      hence thesis;

    end;
    suppose
A4:   p<g;
      now
        per cases by A1,RFUNCT_2:def 5;
        suppose
          f|X is non-decreasing;
          then
A5:       f|X|X is non-decreasing;
          for x0 st x0 in dom(f|X) holds f|X is_continuous_in x0
          proof
A6:         [.p,g.] = ].p,g.[ \/ {p,g} by A2,XXREAL_1:128;
            let x0;
A7:         (f|X).:X = f.:X by RELAT_1:129;
            assume
A8:         x0 in dom(f|X);
A9:        (f|X).x0 in (f|X).:X by A8,FUNCT_1:def 6;
            reconsider x0 as Real by XREAL_0:def 1;
            (f|X).x0 in [.p,g.] by A3,A9,RELAT_1:129;
            then
A10:        (f|X).x0 in ].p,g.[ or (f|X).x0 in {p,g} by A6,XBOOLE_0:def 3;
            now
              let N1 be Neighbourhood of (f|X).x0;
              now
                per cases by A10,TARSKI:def 2;
                suppose
                  (f|X).x0 in ].p,g.[;
                  then consider N2 being Neighbourhood of (f|X).x0 such that
A11:              N2 c= ].p,g.[ by RCOMP_1:18;
A12:              ].p,g.[ c= [.p,g.] by XXREAL_1:25;
                  consider N3 being Neighbourhood of (f|X).x0 such that
A13:              N3 c= N1 and
A14:              N3 c= N2 by RCOMP_1:17;
                  consider r such that
A15:              r>0 and
A16:              N3 = ].(f|X).x0-r,(f|X).x0+r.[ by RCOMP_1:def 6;
                  reconsider r as Real by XREAL_0:def 1;
A17:              (f|X).x0+r/2 < (f|X).x0+r/2+r/2 by A15,XREAL_1:29,215;
                  set M2 = (f|X).x0+r/2;
A18:              (f|X).x0 < (f|X).x0+r/2 by A15,XREAL_1:29,215;
A19:              (f|X).x0 < (f|X).x0+r by A15,XREAL_1:29;
                  then (f|X).x0-r < (f|X).x0+r-r by XREAL_1:9;
                  then (f|X).x0-r < (f|X).x0+r/2 by A18,XXREAL_0:2;
                  then
                  (f|X).x0+r/2 in {s where s is Real: (f|X).x0-r<s & s<(f
                  |X).x0+r} by A17;
                  then
A20:              M2 in ].(f|X).x0-r,(f|X).x0+r.[ by RCOMP_1:def 2;
                  then M2 in N2 by A14,A16;
                  then M2 in ].p,g.[ by A11;
                  then consider x2 be Real such that
A21:              x2 in dom (f|X) & x2 in X and
A22:              M2=(f|X).x2 by A3,A7,A12,PARTFUN2:59;
A23:              ].p,g.[ c= [.p,g.] by XXREAL_1:25;
                  set M1 = (f|X).x0-r/2;
A24:              (f|X).x0-r < (f|X).x0-r + r/2 by A15,XREAL_1:29,215;
                  (f|X).x0-r/2 < (f|X).x0 by A18,XREAL_1:19;
                  then (f|X).x0-r/2 < (f|X).x0+r by A19,XXREAL_0:2;
                  then
                  (f|X).x0-r/2 in {s where s is Real: (f|X).x0-r<s & s<(f
                  |X).x0+r} by A24;
                  then
A25:              M1 in ].(f|X).x0-r,(f|X).x0+r.[ by RCOMP_1:def 2;
                  then M1 in N2 by A14,A16;
                  then M1 in ].p,g.[ by A11;
                  then consider x1 be Real such that
A26:              x1 in dom (f|X) & x1 in X and
A27:              M1=(f|X).x1 by A3,A7,A23,PARTFUN2:59;
A28:              (f|X).x0 < (f|X).x0 + r/2 by A15,XREAL_1:29,215;
                  then
A29:              M1<(f|X).x0 by XREAL_1:19;
A30:              now
                    assume
A31:                x0<x1;
                    x0 in X /\ dom(f|X) & x1 in X /\ dom(f|X) by A8,A26,
XBOOLE_0:def 4;
                    hence contradiction by A5,A27,A29,A31,RFUNCT_2:22;
                  end;
A32:              M2>(f|X).x0 by A15,XREAL_1:29,215;
A33:              now
                    assume
A34:                x2<x0;
                    x0 in X /\ dom(f|X) & x2 in X /\ dom(f|X) by A8,A21,
XBOOLE_0:def 4;
                    hence contradiction by A5,A22,A32,A34,RFUNCT_2:22;
                  end;
                  x0<>x2 by A15,A22,XREAL_1:29,215;
                  then x0<x2 by A33,XXREAL_0:1;
                  then
A35:              x2-x0>0 by XREAL_1:50;
                  set R = min(x0-x1,x2-x0);
A36:              R<=x2-x0 by XXREAL_0:17;
                  x1<>x0 by A27,A28,XREAL_1:19;
                  then x1<x0 by A30,XXREAL_0:1;
                  then x0-x1>0 by XREAL_1:50;
                  then R>0 by A35,XXREAL_0:15;
                  then reconsider N=].x0-R,x0+R.[ as Neighbourhood of x0 by
RCOMP_1:def 6;
                  take N;
                  let x be Real;
                  assume that
A37:              x in dom(f|X) and
A38:              x in N;
A39:              x in X /\ dom(f|X) by A37,XBOOLE_1:28;
                  x in {s where s is Real:x0-R<s & s<x0+R} by A38,RCOMP_1:def 2
;
                  then
A40:              ex s be Real st s=x & x0-R<s & s<x0+R;
                  then x0<R+x by XREAL_1:19;
                  then
A41:              x0-x<R+x-x by XREAL_1:9;
                  R<=x0- x1 by XXREAL_0:17;
                  then x0-x<x0-x1 by A41,XXREAL_0:2;
                  then -(x0-x)>-(x0-x1) by XREAL_1:24;
                  then
A42:              x-x0+x0>x1-x0+x0 by XREAL_1:6;
                  x1 in X /\ dom(f|X) by A26,XBOOLE_0:def 4;
                  then
A43:              (f|X).x1 <= (f|X).x by A5,A42,A39,RFUNCT_2:22;
                  x-x0<R by A40,XREAL_1:19;
                  then x-x0<x2-x0 by A36,XXREAL_0:2;
                  then
A44:              x-x0+x0<x2-x0+x0 by XREAL_1:6;
                  x2 in X /\ dom(f|X) by A21,XBOOLE_0:def 4;
                  then (f|X).x <= (f|X).x2 by A5,A44,A39,RFUNCT_2:22;
                  then (f|X).x in {s where s is Real: M1<=s & s<=M2} by A27,A22
,A43;
                  then
A45:              (f|X).x in [.M1,M2.] by RCOMP_1:def 1;
                  [.M1,M2.] c= ].(f|X).x0-r,(f|X).x0+r.[ by A25,A20,
XXREAL_2:def 12;
                  then (f|X).x in N3 by A16,A45;
                  hence (f|X).x in N1 by A13;
                end;
                suppose
A46:              (f|X).x0 = p;
                  then consider r such that
A47:              r>0 and
A48:              N1 = ].p-r,p+r.[ by RCOMP_1:def 6;
                  reconsider r as Real by XREAL_0:def 1;
                  set R=min(r,g-p)/2;
                  g-p>0 by A4,XREAL_1:50;
                  then
A49:              min(r,g-p)>0 by A47,XXREAL_0:15;
                  then
A50:              R<min(r,g-p) by XREAL_1:216;
                  min(r,g-p) <= r by XXREAL_0:17;
                  then
A51:              R<r by A50,XXREAL_0:2;
                  then
A52:              p+R<p+r by XREAL_1:6;
A53:              p-R<p by A49,XREAL_1:44,215;
                  -r<-R by A51,XREAL_1:24;
                  then
A54:              p+-r<p+-R by XREAL_1:6;
                  p<p+r by A47,XREAL_1:29;
                  then p-R<p+r by A53,XXREAL_0:2;
                  then p-R in {s where s is Real: p-r<s & s<p+r} by A54;
                  then
A55:              p-R in ].p-r,p+r.[ by RCOMP_1:def 2;
A56:              ].p,g.[ c= [.p,g.] by XXREAL_1:25;
A57:              p<p+R by A49,XREAL_1:29,215;
                  min(r,g-p) <= g-p by XXREAL_0:17;
                  then R<g-p by A50,XXREAL_0:2;
                  then p+R<g by XREAL_1:20;
                  then p+R in {s where s is Real: p<s & s<g} by A57;
                  then p+R in ].p,g.[ by RCOMP_1:def 2;
                  then consider x1 be Real such that
A58:              x1 in dom (f|X) & x1 in X and
A59:              p+R=(f|X).x1 by A3,A7,A56,PARTFUN2:59;
A60:              x1 in X /\ dom(f|X) by A58,XBOOLE_0:def 4;
                  now
                    assume
A61:                x1<x0;
                    x0 in X /\ dom(f|X) & x1 in X /\ dom(f|X) by A8,A58,
XBOOLE_0:def 4;
                    hence contradiction by A5,A46,A57,A59,A61,RFUNCT_2:22;
                  end;
                  then x0<x1 by A46,A57,A59,XXREAL_0:1;
                  then reconsider
                  N=].x0-(x1-x0),x0+(x1-x0).[ as Neighbourhood of
                  x0 by RCOMP_1:def 6,XREAL_1:50;
                  take N;
                  let x be Real such that
A62:              x in dom(f|X) and
A63:              x in N;
                  x in {s where s is Real: x0-(x1-x0)<s & s<x0+(x1-x0)}
                  by A63,RCOMP_1:def 2;
                  then
A64:              ex s be Real st s=x & x0-(x1-x0)<s & s<x0+(x1-x0);
                  (f|X).x in [.p,g.] by A3,A7,A62,FUNCT_1:def 6;
                  then (f|X).x in {s where s is Real: p<=s & s<=g} by
RCOMP_1:def 1;
                  then ex s be Real st s=(f|X).x & p<=s & s<=g;
                  then
A65:              p-R<=(f|X).x by A53,XXREAL_0:2;
                  x in X /\ dom(f|X) by A62,XBOOLE_0:def 4;
                  then (f|X).x<=p+R by A5,A59,A60,A64,RFUNCT_2:22;
                  then (f|X).x in {s where s is Real: p-R<=s & s<= p+R} by A65;
                  then
A66:              (f|X).x in [.p-R,p+R.] by RCOMP_1:def 1;
                  p-r<p by A47,XREAL_1:44;
                  then p-r<p+R by A57,XXREAL_0:2;
                  then p+R in {s where s is Real: p-r<s & s<p+r} by A52;
                  then p+R in ].p-r,p+r.[ by RCOMP_1:def 2;
                  then [.p-R,p+R.] c= N1 by A48,A55,XXREAL_2:def 12;
                  hence (f|X).x in N1 by A66;
                end;
                suppose
A67:              (f|X).x0 = g;
A68:              ].p,g.[ c= [.p,g.] by XXREAL_1:25;
                  consider r such that
A69:              r>0 and
A70:              N1 = ].g-r,g+r.[ by A67,RCOMP_1:def 6;
                  reconsider r as Real by XREAL_0:def 1;
                  set R=min(r,g-p)/2;
                  g-p>0 by A4,XREAL_1:50;
                  then
A71:              min(r,g-p)>0 by A69,XXREAL_0:15;
                  then
A72:              R<min(r,g-p) by XREAL_1:216;
A73:              g-R<g by A71,XREAL_1:44,215;
                  min(r,g-p) <= g-p by XXREAL_0:17;
                  then R<g-p by A72,XXREAL_0:2;
                  then R+p<g by XREAL_1:20;
                  then g-R>p by XREAL_1:20;
                  then g-R in {s where s is Real: p<s & s<g} by A73;
                  then g-R in ].p,g.[ by RCOMP_1:def 2;
                  then consider x1 be Real such that
A74:              x1 in dom (f|X) & x1 in X and
A75:              g-R=(f|X).x1 by A3,A7,A68,PARTFUN2:59;
A76:              now
                    assume
A77:                x0<x1;
                    x0 in X /\ dom(f|X) & x1 in X /\ dom(f|X) by A8,A74,
XBOOLE_0:def 4;
                    hence contradiction by A5,A67,A73,A75,A77,RFUNCT_2:22;
                  end;
                  min(r,g-p) <= r by XXREAL_0:17;
                  then
A78:              R<r by A72,XXREAL_0:2;
                  then
A79:              g+R<g+r by XREAL_1:6;
                  -r<-R by A78,XREAL_1:24;
                  then
A80:              g+-r<g+-R by XREAL_1:6;
                  g<g+r by A69,XREAL_1:29;
                  then g-R<g+r by A73,XXREAL_0:2;
                  then g-R in {s where s is Real: g-r<s & s<g+r} by A80;
                  then
A81:              g-R in ].g-r,g+r.[ by RCOMP_1:def 2;
A82:              x1 in X /\ dom(f|X) by A74,XBOOLE_0:def 4;
A83:              g<g+R by A71,XREAL_1:29,215;
                  x1<>x0 by A67,A71,A75,XREAL_1:44,215;
                  then x1<x0 by A76,XXREAL_0:1;
                  then reconsider
                  N=].x0-(x0-x1),x0+(x0-x1).[ as Neighbourhood of
                  x0 by RCOMP_1:def 6,XREAL_1:50;
                  take N;
                  let x be Real such that
A84:              x in dom(f|X) and
A85:              x in N;
                  x in {s where s is Real: x0-(x0-x1)<s & s<x0+(x0-x1)}
                  by A85,RCOMP_1:def 2;
                  then
A86:              ex s be Real st s=x & x0-(x0-x1)<s & s<x0+(x0-x1);
                  (f|X).x in [.p,g.] by A3,A7,A84,FUNCT_1:def 6;
                  then (f|X).x in {s where s is Real: p<=s & s<=g} by
RCOMP_1:def 1;
                  then ex s be Real st s=(f|X).x & p<=s & s<=g;
                  then
A87:              (f|X).x<=g+R by A83,XXREAL_0:2;
                  x in X /\ dom(f|X) by A84,XBOOLE_0:def 4;
                  then g-R<=(f|X).x by A5,A75,A82,A86,RFUNCT_2:22;
                  then (f|X).x in {s where s is Real: g-R<=s & s<= g+R} by A87;
                  then
A88:              ( f|X).x in [.g-R,g+R.] by RCOMP_1:def 1;
                  g-r<g by A69,XREAL_1:44;
                  then g-r<g+R by A83,XXREAL_0:2;
                  then g+R in {s where s is Real: g-r<s & s<g+r} by A79;
                  then g+R in ].g-r,g+r.[ by RCOMP_1:def 2;
                  then [.g-R,g+R.] c= N1 by A70,A81,XXREAL_2:def 12;
                  hence (f|X).x in N1 by A88;
                end;
              end;
              then consider N being Neighbourhood of x0 such that
A89:          for x1 be Real st x1 in dom(f|X) & x1 in N holds (f|X) .x1 in N1;
              take N;
              thus for x1 be real number st x1 in dom(f|X) & x1 in N holds (f|
              X).x1 in N1 by A89;
            end;
            hence thesis by Th4;
          end;
          hence thesis by Def2;
        end;
        suppose
          f|X is non-increasing;
          then
A90:      f|X|X is non-increasing;
          for x0 st x0 in dom(f|X) holds f|X is_continuous_in x0
          proof
A91:        [.p,g.] = ].p,g.[ \/ {p,g} by A2,XXREAL_1:128;
            let x0;
A92:        (f|X).:X = f.:X by RELAT_1:129;
            assume
A93:        x0 in dom(f|X);
A94:        (f|X).x0 in (f|X).:X by A93,FUNCT_1:def 6;
            reconsider x0 as Real by XREAL_0:def 1;
            (f|X).x0 in [.p,g.] by A3,A94,RELAT_1:129;
            then
A95:        (f|X).x0 in ].p,g.[ or (f|X).x0 in {p,g} by A91,XBOOLE_0:def 3;
            now
              let N1 be Neighbourhood of (f|X).x0;
              now
                per cases by A95,TARSKI:def 2;
                suppose
                  (f|X).x0 in ].p,g.[;
                  then consider N2 being Neighbourhood of (f|X).x0 such that
A96:             N2 c= ].p,g.[ by RCOMP_1:18;
A97:             ].p,g.[ c= [.p,g.] by XXREAL_1:25;
                  consider N3 being Neighbourhood of (f|X).x0 such that
A98:             N3 c= N1 and
A99:             N3 c= N2 by RCOMP_1:17;
                  consider r such that
A100:             r>0 and
A101:             N3 = ].(f|X).x0-r,(f|X).x0+r.[ by RCOMP_1:def 6;
                  reconsider r as Real by XREAL_0:def 1;
A102:             (f|X).x0+r/2 < (f|X).x0+r/2+r/2 by A100,XREAL_1:29,215;
                  set M2 = (f|X).x0+r/2;
A103:             (f|X).x0 < (f|X).x0+r/2 by A100,XREAL_1:29,215;
A104:             (f|X).x0 < (f|X).x0+r by A100,XREAL_1:29;
                  then (f|X).x0-r < (f|X).x0+r-r by XREAL_1:9;
                  then (f|X).x0-r < (f|X).x0+r/2 by A103,XXREAL_0:2;
                  then (f|X).x0+r/2 in {s where s is Real: (f|X).x0-r<s & s<(
                  f|X).x0+r} by A102;
                  then
A105:             M2 in ].(f|X).x0-r,(f|X).x0+r.[ by RCOMP_1:def 2;
                  then M2 in N2 by A99,A101;
                  then M2 in ].p,g.[ by A96;
                  then consider x2 be Real such that
A106:             x2 in dom (f|X) & x2 in X and
A107:             M2=(f|X).x2 by A3,A92,A97,PARTFUN2:59;
A108:             ].p,g.[ c= [.p,g.] by XXREAL_1:25;
                  set M1 = (f|X).x0-r/2;
A109:             (f|X).x0-r < (f|X).x0-r + r/2 by A100,XREAL_1:29,215;
                  (f|X).x0-r/2 < (f|X).x0 by A103,XREAL_1:19;
                  then (f|X).x0-r/2 < (f|X).x0+r by A104,XXREAL_0:2;
                  then (f|X).x0-r/2 in {s where s is Real: (f|X).x0-r<s & s<(
                  f|X).x0+r} by A109;
                  then
A110:             M1 in ].(f|X).x0-r,(f|X).x0+r.[ by RCOMP_1:def 2;
                  then M1 in N2 by A99,A101;
                  then M1 in ].p,g.[ by A96;
                  then consider x1 be Real such that
A111:             x1 in dom (f|X) & x1 in X and
A112:             M1=(f|X).x1 by A3,A92,A108,PARTFUN2:59;
A113:             (f|X).x0 < (f|X).x0 + r/2 by A100,XREAL_1:29,215;
                  then
A114:             M1<(f|X).x0 by XREAL_1:19;
A115:             now
                    assume
A116:               x0>x1;
                    x0 in X /\ dom(f|X) & x1 in X /\ dom(f|X) by A93,A111,
XBOOLE_0:def 4;
                    hence contradiction by A90,A112,A114,A116,RFUNCT_2:23;
                  end;
A117:             M2>(f|X).x0 by A100,XREAL_1:29,215;
A118:             now
                    assume
A119:               x2>x0;
                    x0 in X /\ dom(f|X) & x2 in X /\ dom(f|X) by A93,A106,
XBOOLE_0:def 4;
                    hence contradiction by A90,A107,A117,A119,RFUNCT_2:23;
                  end;
                  x0<>x2 by A100,A107,XREAL_1:29,215;
                  then x0>x2 by A118,XXREAL_0:1;
                  then
A120:             x0-x2>0 by XREAL_1:50;
                  set R = min(x1-x0,x0-x2);
A121:             R<=x1-x0 by XXREAL_0:17;
                  x1<>x0 by A112,A113,XREAL_1:19;
                  then x1>x0 by A115,XXREAL_0:1;
                  then x1-x0>0 by XREAL_1:50;
                  then R>0 by A120,XXREAL_0:15;
                  then reconsider N=].x0-R,x0+R.[ as Neighbourhood of x0 by
RCOMP_1:def 6;
                  take N;
                  let x be Real;
                  assume that
A122:             x in dom(f|X) and
A123:             x in N;
A124:             x in X /\ dom(f|X) by A122,XBOOLE_1:28;
                  x in {s where s is Real:x0-R<s & s<x0+R} by A123,
RCOMP_1:def 2;
                  then
A125:             ex s be Real st s=x & x0-R<s & s<x0+R;
                  then x0<R+x by XREAL_1:19;
                  then
A126:             x0-x<R+x-x by XREAL_1:9;
                  x-x0<R by A125,XREAL_1:19;
                  then x-x0<x1-x0 by A121,XXREAL_0:2;
                  then
A127:             x-x0+x0<x1-x0+x0 by XREAL_1:6;
                  x1 in X /\ dom(f|X) by A111,XBOOLE_0:def 4;
                  then
A128:             (f|X).x1 <= (f|X).x by A90,A127,A124,RFUNCT_2:23;
                  R<=x0- x2 by XXREAL_0:17;
                  then x0-x<x0-x2 by A126,XXREAL_0:2;
                  then -(x0-x)>-(x0-x2) by XREAL_1:24;
                  then
A129:             x-x0+x0>x2-x0+x0 by XREAL_1:6;
                  x2 in X /\ dom(f|X) by A106,XBOOLE_0:def 4;
                  then (f|X).x <= (f|X).x2 by A90,A129,A124,RFUNCT_2:23;
                  then (f|X).x in {s where s is Real: M1<=s & s<=M2} by A112
,A107,A128;
                  then
A130:             (f|X).x in [.M1,M2.] by RCOMP_1:def 1;
                  [.M1,M2.] c= ].(f|X).x0-r,(f|X).x0+r.[ by A110,A105,
XXREAL_2:def 12;
                  then (f|X).x in N3 by A101,A130;
                  hence (f|X).x in N1 by A98;
                end;
                suppose
A131:             (f|X).x0 = p;
                  then consider r such that
A132:             r>0 and
A133:             N1 = ].p-r,p+r.[ by RCOMP_1:def 6;
                  reconsider r as Real by XREAL_0:def 1;
                  set R=min(r,g-p)/2;
                  g-p>0 by A4,XREAL_1:50;
                  then
A134:             min(r,g-p)>0 by A132,XXREAL_0:15;
                  then
A135:             R<min(r,g-p) by XREAL_1:216;
                  min(r,g-p) <= r by XXREAL_0:17;
                  then
A136:             R<r by A135,XXREAL_0:2;
                  then
A137:             p+R<p+r by XREAL_1:6;
A138:             p-R<p by A134,XREAL_1:44,215;
                  -r<-R by A136,XREAL_1:24;
                  then
A139:             p+-r<p+-R by XREAL_1:6;
                  p<p+r by A132,XREAL_1:29;
                  then p-R<p+r by A138,XXREAL_0:2;
                  then p-R in {s where s is Real: p-r<s & s<p+r} by A139;
                  then
A140:             p-R in ].p-r,p+r.[ by RCOMP_1:def 2;
A141:             ].p,g.[ c= [.p,g.] by XXREAL_1:25;
A142:             p<p+R by A134,XREAL_1:29,215;
                  min(r,g-p) <= g-p by XXREAL_0:17;
                  then R<g-p by A135,XXREAL_0:2;
                  then p+R<g by XREAL_1:20;
                  then p+R in {s where s is Real: p<s & s<g} by A142;
                  then p+R in ].p,g.[ by RCOMP_1:def 2;
                  then consider x1 be Real such that
A143:             x1 in dom (f|X) & x1 in X and
A144:             p+R=(f|X).x1 by A3,A92,A141,PARTFUN2:59;
A145:             x1 in X /\ dom(f|X) by A143,XBOOLE_0:def 4;
                  now
                    assume
A146:               x1>x0;
                    x0 in X /\ dom(f|X) & x1 in X /\ dom(f|X) by A93,A143,
XBOOLE_0:def 4;
                    hence contradiction by A90,A131,A142,A144,A146,RFUNCT_2:23;
                  end;
                  then x0>x1 by A131,A142,A144,XXREAL_0:1;
                  then reconsider
                  N=].x0-(x0-x1),x0+(x0-x1).[ as Neighbourhood of
                  x0 by RCOMP_1:def 6,XREAL_1:50;
                  take N;
                  let x be Real such that
A147:             x in dom(f|X) and
A148:             x in N;
                  x in {s where s is Real: x0-(x0-x1)<s & s<x0+(x0-x1)}
                  by A148,RCOMP_1:def 2;
                  then
A149:             ex s be Real st s=x & x0-(x0-x1)<s & s<x0+(x0-x1);
                  (f|X).x in [.p,g.] by A3,A92,A147,FUNCT_1:def 6;
                  then (f|X).x in {s where s is Real: p<=s & s<=g} by
RCOMP_1:def 1;
                  then ex s be Real st s=(f|X).x & p<=s & s<=g;
                  then
A150:             p-R<=(f|X).x by A138,XXREAL_0:2;
                  x in X /\ dom(f|X) by A147,XBOOLE_0:def 4;
                  then (f|X).x<=p+ R by A90,A144,A145,A149,RFUNCT_2:23;
                  then (f|X).x in {s where s is Real: p-R<=s & s<=p+R} by A150;
                  then
A151:             (f|X).x in [.p-R,p+R.] by RCOMP_1:def 1;
                  p-r<p by A132,XREAL_1:44;
                  then p-r<p+R by A142,XXREAL_0:2;
                  then p+R in {s where s is Real: p-r<s & s<p+r} by A137;
                  then p+R in ].p-r,p+r.[ by RCOMP_1:def 2;
                  then [.p-R,p+R.] c= N1 by A133,A140,XXREAL_2:def 12;
                  hence (f|X).x in N1 by A151;
                end;
                suppose
A152:             (f|X).x0 = g;
A153:             ].p,g.[ c= [.p,g.] by XXREAL_1:25;
                  consider r such that
A154:             r>0 and
A155:             N1 = ].g-r,g+r.[ by A152,RCOMP_1:def 6;
                  reconsider r as Real by XREAL_0:def 1;
                  set R=min(r,g-p)/2;
                  g-p>0 by A4,XREAL_1:50;
                  then
A156:             min(r,g-p)>0 by A154,XXREAL_0:15;
                  then
A157:             R<min(r,g-p) by XREAL_1:216;
A158:             g-R<g by A156,XREAL_1:44,215;
                  min(r,g-p) <= g-p by XXREAL_0:17;
                  then R<g-p by A157,XXREAL_0:2;
                  then R+p<g by XREAL_1:20;
                  then g-R>p by XREAL_1:20;
                  then g-R in {s where s is Real: p<s & s<g} by A158;
                  then g-R in ].p,g.[ by RCOMP_1:def 2;
                  then consider x1 be Real such that
A159:             x1 in dom (f|X) & x1 in X and
A160:             g-R=(f|X).x1 by A3,A92,A153,PARTFUN2:59;
A161:             now
                    assume
A162:               x0>x1;
                    x0 in X /\ dom(f|X) & x1 in X /\ dom(f|X) by A93,A159,
XBOOLE_0:def 4;
                    hence contradiction by A90,A152,A158,A160,A162,RFUNCT_2:23;
                  end;
                  min(r,g-p) <= r by XXREAL_0:17;
                  then
A163:             R<r by A157,XXREAL_0:2;
                  then
A164:             g+R<g+r by XREAL_1:6;
                  -r<-R by A163,XREAL_1:24;
                  then
A165:             g+-r<g+-R by XREAL_1:6;
                  g<g+r by A154,XREAL_1:29;
                  then g-R<g+r by A158,XXREAL_0:2;
                  then g-R in {s where s is Real: g-r<s & s<g+r} by A165;
                  then
A166:             g-R in ].g-r,g+r.[ by RCOMP_1:def 2;
A167:             x1 in X /\ dom(f|X) by A159,XBOOLE_0:def 4;
A168:             g<g+R by A156,XREAL_1:29,215;
                  x1<>x0 by A152,A156,A160,XREAL_1:44,215;
                  then x1>x0 by A161,XXREAL_0:1;
                  then reconsider
                  N=].x0-(x1-x0),x0+(x1-x0).[ as Neighbourhood of
                  x0 by RCOMP_1:def 6,XREAL_1:50;
                  take N;
                  let x be Real such that
A169:             x in dom(f|X) and
A170:             x in N;
                  x in {s where s is Real: x0-(x1-x0)<s & s<x0+(x1-x0)}
                  by A170,RCOMP_1:def 2;
                  then
A171:             ex s be Real st s=x & x0-(x1-x0)<s & s<x0+(x1-x0);
                  (f|X).x in [.p,g.] by A3,A92,A169,FUNCT_1:def 6;
                  then (f|X).x in {s where s is Real: p<=s & s<=g} by
RCOMP_1:def 1;
                  then ex s be Real st s=(f|X).x & p<=s & s<=g;
                  then
A172:             (f|X).x<=g+R by A168,XXREAL_0:2;
                  x in X /\ dom(f|X) by A169,XBOOLE_0:def 4;
                  then g-R<=(f|X). x by A90,A160,A167,A171,RFUNCT_2:23;
                  then (f|X).x in {s where s is Real: g-R<=s & s<=g+R} by A172;
                  then
A173:             (f|X).x in [.g-R,g+R.] by RCOMP_1:def 1;
                  g-r<g by A154,XREAL_1:44;
                  then g-r<g+R by A168,XXREAL_0:2;
                  then g+R in {s where s is Real: g-r<s & s<g+r} by A164;
                  then g+R in ].g-r,g+r.[ by RCOMP_1:def 2;
                  then [.g-R,g+R.] c= N1 by A155,A166,XXREAL_2:def 12;
                  hence (f|X).x in N1 by A173;
                end;
              end;
              then consider N being Neighbourhood of x0 such that
A174:         for x1 be Real st x1 in dom(f|X) & x1 in N holds (f|X) .x1 in N1;
              take N;
              thus for x1 st x1 in dom(f|X) & x1 in N holds (f|X).x1 in N1 by
A174;
            end;
            hence thesis by Th4;
          end;
          hence thesis by Def2;
        end;
      end;
      hence thesis;
    end;
  end;
  hence thesis;
end;

theorem
  for f being one-to-one PartFunc of REAL,REAL st p<=g & [.p,g.] c= dom
f & (f|[.p,g.] is increasing or f|[.p,g.] is decreasing) holds (f|[.p,g.])"|(f
  .:[.p,g.]) is continuous
proof
  let f be one-to-one PartFunc of REAL,REAL;
  assume that
A1: p<=g and
A2: [.p,g.] c= dom f and
A3: f|[.p,g.] is increasing or f|[.p,g.] is decreasing;
  reconsider p, g as Real by XREAL_0:def 1;
  now
    per cases by A3;
    suppose
A4:   f|[.p,g.] is increasing;
A5:   ((f|[.p,g.])").:(f.:[.p,g.]) = ((f|[.p,g.])").: (rng (f|[.p,g.])) by
RELAT_1:115
        .= ((f|[.p,g.])").:(dom ((f|[.p,g.])")) by FUNCT_1:33
        .= rng ((f|[.p,g.])") by RELAT_1:113
        .= dom (f|[.p,g.]) by FUNCT_1:33
        .= dom f /\ [.p,g.] by RELAT_1:61
        .= [.p,g.] by A2,XBOOLE_1:28;
      (f|[.p,g.])"|(f.:[.p,g.]) is increasing by A4,RFUNCT_2:51;
      hence thesis by A1,A5,Th46;
    end;
    suppose
A6:   f|[.p,g.] is decreasing;
A7:   ((f|[.p,g.])").:(f.:[.p,g.]) = ((f|[.p,g.])").: (rng (f|[.p,g.]))
      by RELAT_1:115
        .= ((f|[.p,g.])").:(dom ((f|[.p,g.])")) by FUNCT_1:33
        .= rng ((f|[.p,g.])") by RELAT_1:113
        .= dom (f|[.p,g.]) by FUNCT_1:33
        .= dom f /\ [.p,g.] by RELAT_1:61
        .= [.p,g.] by A2,XBOOLE_1:28;
      (f|[.p,g.])"|(f.:[.p,g.]) is decreasing by A6,RFUNCT_2:52;
      hence thesis by A1,A7,Th46;
    end;
  end;
  hence thesis;
end;



definition
  let a,b be real number;
  func AffineMap(a,b) -> Function of REAL, REAL means
  :Def4:
  for x being real number holds it.x = a*x + b;
  existence
  proof
    reconsider a9 = a, b9 = b as Element of REAL by XREAL_0:def 1;
    deffunc F(Real)=a9*$1+b9;
    consider f being Function of REAL, REAL such that
A1: for x being Element of REAL holds f.x =F(x) from FUNCT_2:sch 4;
    take f;
    let x be real number;
    reconsider x9 = x as Element of REAL by XREAL_0:def 1;
    f.x9 = a9*x+b9 by A1;
    hence thesis;
  end;
  uniqueness
  proof
    let f,f9 be Function of REAL, REAL such that
A2: for x being real number holds f.x = a*x + b and
A3: for x being real number holds f9.x = a*x + b;
    now
      let x be Element of REAL;
      thus f.x = a*x + b by A2
        .= f9.x by A3;
    end;
    hence f = f9 by FUNCT_2:63;
  end;
end;

registration
  let a,b be real number;
  cluster AffineMap(a,b) -> continuous;
  coherence
  proof
    set f = AffineMap(a,b);
    for x0 being real number st x0 in REAL holds f.x0 = a*x0+b by Def4;
    then REAL = dom f & AffineMap(a,b)|REAL is continuous by Th41,FUNCT_2:def 1
;
    hence thesis;
  end;
end;

registration
  cluster continuous for Function of REAL, REAL;
  existence
  proof
    take AffineMap(1,1);
    thus thesis;
  end;
end;

theorem Th48:
  for a,b being real number holds AffineMap(a,b).0 = b
proof
  let a,b be real number;
  thus AffineMap(a,b).0 = a*0 + b by Def4
    .= b;
end;

theorem Th49:
  for a,b being real number holds AffineMap(a,b).1 = a+b
proof
  let a,b be real number;
  thus AffineMap(a,b).1 = a*1 + b by Def4
    .= a + b;
end;

theorem Th50:
  for a,b being real number st a<> 0 holds AffineMap(a,b) is one-to-one
proof
  let a,b be real number such that
A1: a<> 0;
  let x1,x2 be set;
  set f = AffineMap(a,b);
  assume x1 in dom f;
  then reconsider r1 = x1 as real number;
  assume x2 in dom f;
  then reconsider r2 = x2 as real number;
  assume f.x1 = f.x2;
  then a*r1+b = f.x2 by Def4
    .= a*r2 +b by Def4;
  hence thesis by A1,XCMPLX_1:5;
end;

theorem
  for a,b,x,y being real number st a > 0 & x < y holds AffineMap(a,b).x
  < AffineMap(a,b).y
proof
  let a,b,x,y be real number;
  assume a > 0 & x < y;
  then
A1: a*x < a*y by XREAL_1:68;
  AffineMap(a,b).x = a*x + b & AffineMap(a,b).y = a*y + b by Def4;
  hence thesis by A1,XREAL_1:8;
end;

theorem
  for a,b,x,y being real number st a < 0 & x < y holds AffineMap(a,b).x
  > AffineMap(a,b).y
proof
  let a,b,x,y be real number;
  assume a < 0 & x < y;
  then
A1: a*x > a*y by XREAL_1:69;
  AffineMap(a,b).x = a*x + b & AffineMap(a,b).y = a*y + b by Def4;
  hence thesis by A1,XREAL_1:8;
end;

theorem Th53:
  for a,b,x,y being real number st a >= 0 & x <= y holds AffineMap
  (a,b).x <= AffineMap(a,b).y
proof
  let a,b,x,y be real number;
  assume a >= 0 & x <= y;
  then
A1: a*x <= a*y by XREAL_1:64;
  AffineMap(a,b).x = a*x + b & AffineMap(a,b).y = a*y + b by Def4;
  hence thesis by A1,XREAL_1:7;
end;

theorem
  for a,b,x,y being real number st a <= 0 & x <= y holds AffineMap(a,b).
  x >= AffineMap(a,b).y
proof
  let a,b,x,y be real number;
  assume a <= 0 & x <= y;
  then
A1: a*x >= a*y by XREAL_1:65;
  AffineMap(a,b).x = a*x + b & AffineMap(a,b).y = a*y + b by Def4;
  hence thesis by A1,XREAL_1:7;
end;

theorem Th55:
  for a,b being real number st a <> 0 holds rng AffineMap(a,b) = REAL
proof
  let a,b be real number such that
A1: a <> 0;
  thus rng AffineMap(a,b) c= REAL;
  let e be set;
  assume e in REAL;
  then reconsider r = e as Element of REAL;
  set s = (r - b)/a;
  AffineMap(a,b).s = a*s + b by Def4
    .= r - b + b by A1,XCMPLX_1:87
    .= r;
  hence thesis by FUNCT_2:4;
end;

theorem
  for a,b being real number st a <> 0 holds (AffineMap(a,b) qua
  Function)" = AffineMap(a",-b/a)
proof
  let a,b be real number such that
A1: a <> 0;
  for x being Element of REAL holds (AffineMap(a",-b/a)*AffineMap(a,b)).x
  = (id REAL).x
  proof
    let x being Element of REAL;
    thus (AffineMap(a",-b/a)*AffineMap(a,b)).x = AffineMap(a",-b/a).(AffineMap
    (a,b).x) by FUNCT_2:15
      .= AffineMap(a",-b/a).(a*x+b) by Def4
      .= a"*(a*x+b)+-b/a by Def4
      .= a"*a*x+a"*b +-b/a
      .= 1 *x+a"*b +-b/a by A1,XCMPLX_0:def 7
      .= x+a"*b -b/a
      .= x+ b/a -b/a by XCMPLX_0:def 9
      .= (id REAL).x by FUNCT_1:18;
  end;
  then
A2: AffineMap(a",-b/a)*AffineMap(a,b) = id REAL by FUNCT_2:63;
  rng AffineMap(a,b) = REAL by A1,Th55;
  hence thesis by A1,A2,Th50,FUNCT_2:30;
end;

theorem
  for a,b being real number st a > 0 holds AffineMap(a,b).:[.0,1.] = [.b,a+b.]
proof
  let a,b be real number such that
A1: a > 0;
  thus AffineMap(a,b).:[.0,1.] c= [.b,a+b.]
  proof
A2: AffineMap(a,b).1 = a+b by Th49;
    let u be set;
    assume u in AffineMap(a,b).:[.0,1.];
    then consider r being Element of REAL such that
A3: r in [.0,1.] and
A4: u = AffineMap(a,b).r by FUNCT_2:65;
    reconsider s = u as real number by A4;
    r <= 1 by A3,XXREAL_1:1;
    then
A5: s <= a+b by A1,A4,A2,Th53;
A6: AffineMap(a,b).0 = b by Th48;
    0 <= r by A3,XXREAL_1:1;
    then b <= s by A1,A4,A6,Th53;
    hence thesis by A5,XXREAL_1:1;
  end;
  let u be set;
  assume
A7: u in [.b,a+b.];
  then reconsider r = u as Element of REAL;
  set s = (r - b)/a;
A8: AffineMap(a,b).s = a*s + b by Def4
    .= r - b + b by A1,XCMPLX_1:87
    .= r;
  r <= a+b by A7,XXREAL_1:1;
  then r-b <= a by XREAL_1:20;
  then s <= a/a by A1,XREAL_1:72;
  then
A9: s <= 1 by A1,XCMPLX_1:60;
  b <= r by A7,XXREAL_1:1;
  then 0 <= r - b by XREAL_1:48;
  then s in [.0,1.] by A1,A9,XXREAL_1:1;
  hence thesis by A8,FUNCT_2:35;
end;















begin

reserve n,m for Element of NAT;
reserve x, X,X1,Z,Z1 for set;
reserve s,g,r,t,p,x0,x1,x2 for Real;
reserve s1,s2,q1 for Real_Sequence;
reserve Y for Subset of REAL;
reserve f,f1,f2 for PartFunc of REAL,REAL;

definition
  let f;
  attr f is uniformly_continuous means
  :Def1:
  for r st 0<r ex s st 0<s & for
x1,x2 st x1 in dom f & x2 in dom f & abs(x1 - x2) < s holds abs(f.x1 - f.x2) <
  r;
end;

theorem Th1:
  f|X is uniformly_continuous iff for r st 0<r ex s st 0<s & for x1
,x2 st x1 in dom(f|X) & x2 in dom(f|X) & abs(x1 - x2) < s holds abs(f.x1 - f.x2
  ) < r
proof
  thus f|X is uniformly_continuous implies for r st 0<r ex s st 0<s & for x1,
x2 st x1 in dom(f|X) & x2 in dom(f|X) & abs(x1 - x2) < s holds abs(f.x1 - f.x2)
  < r
  proof
    assume
A1: f|X is uniformly_continuous;
    let r;
    assume 0<r;
    then consider s such that
A2: 0<s and
A3: for x1,x2 st x1 in dom(f|X) & x2 in dom (f|X) & abs(x1 - x2) < s
    holds abs((f|X).x1 - (f|X).x2) < r by A1,Def1;
    take s;
    thus 0<s by A2;
    let x1,x2;
    assume that
A4: x1 in dom(f|X) and
A5: x2 in dom(f|X);
A6: (f|X).x2 = f.x2 by A5,FUNCT_1:47;
    (f|X).x1 = f.x1 by A4,FUNCT_1:47;
    hence thesis by A3,A4,A5,A6;
  end;
  assume
A7: for r st 0<r ex s st 0<s & for x1,x2 st x1 in dom(f|X) & x2 in dom(f
  |X) & abs(x1 - x2) < s holds abs(f.x1 - f.x2) < r;
  let r;
  assume 0<r;
  then consider s such that
A8: 0<s and
A9: for x1,x2 st x1 in dom(f|X) & x2 in dom(f|X) & abs(x1 - x2) < s
  holds abs(f.x1 - f.x2) < r by A7;
  take s;
  thus 0<s by A8;
  let x1,x2;
  assume that
A10: x1 in dom(f|X) and
A11: x2 in dom(f|X);
A12: (f|X).x2 = f.x2 by A11,FUNCT_1:47;
  (f|X).x1 = f.x1 by A10,FUNCT_1:47;
  hence thesis by A9,A10,A11,A12;
end;

theorem Th2:
  f|X is uniformly_continuous & X1 c= X implies f|X1 is uniformly_continuous
proof
  assume that
A1: f|X is uniformly_continuous and
A2: X1 c= X;
  now
    let r;
    assume 0<r;
    then consider s such that
A3: 0<s and
A4: for x1,x2 st x1 in dom(f|X) & x2 in dom(f|X) & abs(x1-x2)<s holds
    abs(f.x1-f.x2)<r by A1,Th1;
    take s;
    thus 0<s by A3;
    let x1,x2;
    assume that
A5: x1 in dom(f|X1) and
A6: x2 in dom(f|X1) and
A7: abs(x1-x2)<s;
    f|X1 c= f|X by A2,RELAT_1:75;
    then dom(f|X1) c= dom(f|X) by RELAT_1:11;
    hence abs(f.x1-f.x2)<r by A4,A5,A6,A7;
  end;
  hence thesis by Th1;
end;

theorem
  X c= dom f1 & X1 c= dom f2 & f1|X is uniformly_continuous & f2|X1 is
  uniformly_continuous implies (f1+f2)|(X/\X1) is uniformly_continuous
proof
  assume that
A1: X c= dom f1 and
A2: X1 c= dom f2;
A3: X /\ X1 c= dom f2 by A2,XBOOLE_1:108;
  X /\ X1 c= dom f1 by A1,XBOOLE_1:108;
  then X /\ X1 c= dom f1 /\ dom f2 by A3,XBOOLE_1:19;
  then
A4: X /\ X1 c= dom (f1+f2) by VALUED_1:def 1;
  assume that
A5: f1|X is uniformly_continuous and
A6: f2|X1 is uniformly_continuous;
A7: f2|(X /\ X1) is uniformly_continuous by A6,Th2,XBOOLE_1:17;
A8: f1|(X /\ X1) is uniformly_continuous by A5,Th2,XBOOLE_1:17;
  now
    let r;
A9: dom(f1+f2) = dom f1 /\ dom f2 by VALUED_1:def 1;
    assume 0<r;
    then
A10: 0<r/2 by XREAL_1:215;
    then consider s such that
A11: 0<s and
A12: for x1,x2 st x1 in dom(f1|(X/\X1)) & x2 in dom(f1|(X/\X1)) & abs(
    x1 -x2)<s holds abs(f1.x1-f1.x2)<r/2 by A8,Th1;
    consider g such that
A13: 0<g and
A14: for x1,x2 st x1 in dom(f2|(X/\X1)) & x2 in dom(f2|(X/\X1)) & abs(
    x1 -x2)<g holds abs(f2.x1-f2.x2)<r/2 by A7,A10,Th1;
    take p=min(s,g);
    thus 0<p by A11,A13,XXREAL_0:15;
    let x1,x2;
    assume that
A15: x1 in dom((f1+f2)|(X/\X1)) and
A16: x2 in dom((f1+f2)|(X/\X1)) and
A17: abs(x1-x2)<p;
A18: x2 in X/\X1 by A16,RELAT_1:57;
A19: x2 in dom(f1+f2) by A16,RELAT_1:57;
    then x2 in dom f2 by A9,XBOOLE_0:def 4;
    then
A20: x2 in dom(f2|(X/\X1)) by A18,RELAT_1:57;
    p <= g by XXREAL_0:17;
    then
A21: abs(x1-x2)<g by A17,XXREAL_0:2;
A22: x1 in X/\X1 by A15,RELAT_1:57;
    x2 in dom f1 by A9,A19,XBOOLE_0:def 4;
    then
A23: x2 in dom(f1|(X/\X1)) by A18,RELAT_1:57;
    p <= s by XXREAL_0:17;
    then
A24: abs(x1-x2)<s by A17,XXREAL_0:2;
A25: x1 in dom(f1+f2) by A15,RELAT_1:57;
    then x1 in dom f2 by A9,XBOOLE_0:def 4;
    then x1 in dom(f2|(X/\X1)) by A22,RELAT_1:57;
    then
A26: abs(f2.x1-f2.x2)<r/2 by A14,A21,A20;
    x1 in dom f1 by A9,A25,XBOOLE_0:def 4;
    then x1 in dom(f1|(X/\X1)) by A22,RELAT_1:57;
    then abs(f1.x1-f1.x2)<r/2 by A12,A24,A23;
    then
A27: abs(f1.x1-f1.x2)+abs(f2.x1-f2.x2)<r/2+r/2 by A26,XREAL_1:8;
A28: abs(f1.x1 - f1.x2 + (f2.x1-f2.x2)) <= abs(f1.x1-f1.x2) + abs(f2.x1-f2
    .x2) by COMPLEX1:56;
    abs((f1+f2).x1-(f1+f2).x2) = abs(f1.x1 + f2.x1-(f1+f2).x2) by A4,A22,
VALUED_1:def 1
      .= abs(f1.x1 + f2.x1 - (f1.x2+f2.x2)) by A4,A18,VALUED_1:def 1
      .= abs(f1.x1 - f1.x2 + (f2.x1-f2.x2));
    hence abs((f1+f2).x1 - (f1+f2).x2) < r by A27,A28,XXREAL_0:2;
  end;
  hence thesis by Th1;
end;

theorem
  X c= dom f1 & X1 c= dom f2 & f1|X is uniformly_continuous & f2|X1 is
  uniformly_continuous implies (f1-f2)|(X/\X1) is uniformly_continuous
proof
  assume that
A1: X c= dom f1 and
A2: X1 c= dom f2;
A3: X /\ X1 c= dom f2 by A2,XBOOLE_1:108;
  X /\ X1 c= dom f1 by A1,XBOOLE_1:108;
  then X /\ X1 c= dom f1 /\ dom f2 by A3,XBOOLE_1:19;
  then
A4: X /\ X1 c= dom (f1-f2) by VALUED_1:12;
  assume that
A5: f1|X is uniformly_continuous and
A6: f2|X1 is uniformly_continuous;
A7: f2|(X /\ X1) is uniformly_continuous by A6,Th2,XBOOLE_1:17;
A8: f1|(X /\ X1) is uniformly_continuous by A5,Th2,XBOOLE_1:17;
  now
    let r;
A9: dom(f1-f2) = dom f1 /\ dom f2 by VALUED_1:12;
    assume 0<r;
    then
A10: 0<r/2 by XREAL_1:215;
    then consider s such that
A11: 0<s and
A12: for x1,x2 st x1 in dom(f1|(X /\ X1)) & x2 in dom(f1|(X /\ X1)) &
    abs(x1-x2)<s holds abs(f1.x1-f1.x2)<r/2 by A8,Th1;
    consider g such that
A13: 0<g and
A14: for x1,x2 st x1 in dom(f2|(X /\ X1)) & x2 in dom(f2|(X /\ X1)) &
    abs(x1-x2)<g holds abs(f2.x1-f2.x2)<r/2 by A7,A10,Th1;
    take p=min(s,g);
    thus 0<p by A11,A13,XXREAL_0:15;
    let x1,x2;
    assume that
A15: x1 in dom((f1-f2)|(X/\X1)) and
A16: x2 in dom((f1-f2)|(X/\X1)) and
A17: abs(x1-x2)<p;
A18: x2 in X/\X1 by A16,RELAT_1:57;
A19: x2 in dom(f1-f2) by A16,RELAT_1:57;
    then x2 in dom f2 by A9,XBOOLE_0:def 4;
    then
A20: x2 in dom(f2|(X/\X1)) by A18,RELAT_1:57;
    p <= g by XXREAL_0:17;
    then
A21: abs(x1-x2)<g by A17,XXREAL_0:2;
A22: x1 in X/\X1 by A15,RELAT_1:57;
    x2 in dom f1 by A9,A19,XBOOLE_0:def 4;
    then
A23: x2 in dom(f1|(X/\X1)) by A18,RELAT_1:57;
    p <= s by XXREAL_0:17;
    then
A24: abs(x1-x2)<s by A17,XXREAL_0:2;
A25: x1 in dom(f1-f2) by A15,RELAT_1:57;
    then x1 in dom f2 by A9,XBOOLE_0:def 4;
    then x1 in dom(f2|(X/\X1)) by A22,RELAT_1:57;
    then
A26: abs(f2.x1-f2.x2)<r/2 by A14,A21,A20;
    x1 in dom f1 by A9,A25,XBOOLE_0:def 4;
    then x1 in dom(f1|(X/\X1)) by A22,RELAT_1:57;
    then abs(f1.x1-f1.x2)<r/2 by A12,A24,A23;
    then
A27: abs(f1.x1-f1.x2)+abs(f2.x1-f2.x2)<r/2+r/2 by A26,XREAL_1:8;
A28: abs(f1.x1 - f1.x2 - (f2.x1-f2.x2)) <= abs(f1.x1-f1.x2) + abs(f2.x1-f2
    .x2) by COMPLEX1:57;
    abs((f1-f2).x1-(f1-f2).x2) = abs(f1.x1 - f2.x1-(f1-f2).x2) by A4,A22,
VALUED_1:13
      .= abs(f1.x1 - f2.x1 - (f1.x2-f2.x2)) by A4,A18,VALUED_1:13
      .= abs(f1.x1 - f1.x2 - (f2.x1-f2.x2));
    hence abs((f1-f2).x1 - (f1-f2).x2) < r by A27,A28,XXREAL_0:2;
  end;
  hence thesis by Th1;
end;

theorem Th5:
  X c= dom f & f|X is uniformly_continuous implies (p(#)f)|X is
  uniformly_continuous
proof
  assume X c= dom f;
  then
A1: X c= dom (p(#)f) by VALUED_1:def 5;
  assume
A2: f|X is uniformly_continuous;
  per cases;
  suppose
A3: p=0;
    now
      let r;
      assume
A4:   0<r;
      then consider s such that
A5:   0<s and
      for x1,x2 st x1 in dom(f|X) & x2 in dom(f|X) & abs(x1-x2)<s holds
      abs(f.x1-f.x2)<r by A2,Th1;
      take s;
      thus 0<s by A5;
      let x1,x2;
      assume that
A6:   x1 in dom((p(#)f)|X) and
A7:   x2 in dom((p(#)f)|X) and
      abs(x1-x2)<s;
A8:   x2 in X by A7,RELAT_1:57;
      x1 in X by A6,RELAT_1:57;
      then abs((p(#)f).x1-(p(#)f).x2) = abs(p*(f.x1)-(p(#)f).x2) by A1,
VALUED_1:def 5
        .= abs(0 - p*(f.x2)) by A1,A3,A8,VALUED_1:def 5
        .= 0 by A3,ABSVALUE:2;
      hence abs((p(#)f).x1-(p(#)f).x2) <r by A4;
    end;
    hence thesis by Th1;
  end;
  suppose
A9: p<>0;
    then
A10: 0<abs(p) by COMPLEX1:47;
A11: 0 <> abs(p) by A9,COMPLEX1:47;
    now
      let r;
      assume 0<r;
      then 0 < r/abs(p) by A10,XREAL_1:139;
      then consider s such that
A12:  0<s and
A13:  for x1,x2 st x1 in dom(f|X) & x2 in dom(f|X) & abs(x1-x2)<s
      holds abs(f.x1-f.x2)<r/abs(p) by A2,Th1;
      take s;
      thus 0<s by A12;
      let x1,x2;
      assume that
A14:  x1 in dom((p(#)f)|X) and
A15:  x2 in dom((p(#)f)|X) and
A16:  abs(x1-x2)<s;
A17:  x2 in X by A15,RELAT_1:57;
A18:  x1 in X by A14,RELAT_1:57;
      then
A19:  abs((p(#)f).x1-(p(#)f).x2) = abs(p*(f.x1)-(p(#)f).x2) by A1,
VALUED_1:def 5
        .= abs(p*(f.x1) - p*(f.x2)) by A1,A17,VALUED_1:def 5
        .= abs(p*(f.x1 - f.x2))
        .= abs(p)*abs(f.x1 - f.x2) by COMPLEX1:65;
      x2 in dom(p(#)f) by A15,RELAT_1:57;
      then x2 in dom f by VALUED_1:def 5;
      then
A20:  x2 in dom(f|X) by A17,RELAT_1:57;
      x1 in dom(p(#)f) by A14,RELAT_1:57;
      then x1 in dom f by VALUED_1:def 5;
      then x1 in dom(f|X) by A18,RELAT_1:57;
      then abs(p)*abs(f.x1-f.x2)<r/abs(p)*abs(p) by A10,A13,A16,A20,XREAL_1:68;
      hence abs((p(#)f).x1-(p(#)f).x2) <r by A11,A19,XCMPLX_1:87;
    end;
    hence thesis by Th1;
  end;
end;

theorem
  X c= dom f & f|X is uniformly_continuous implies (-f)|X is
  uniformly_continuous
proof
  assume
A1: X c= dom f;
A2: -f = (-1)(#)f;
  assume f|X is uniformly_continuous;
  hence thesis by A1,A2,Th5;
end;

theorem
  f|X is uniformly_continuous implies (abs f)|X is uniformly_continuous
proof
  assume
A1: f|X is uniformly_continuous;
  now
    let r;
    assume 0<r;
    then consider s such that
A2: 0<s and
A3: for x1,x2 st x1 in dom(f|X) & x2 in dom(f|X) & abs(x1-x2)<s holds
    abs(f.x1-f.x2)<r by A1,Th1;
    take s;
    thus 0<s by A2;
    let x1,x2;
    assume that
A4: x1 in dom((abs f)|X) and
A5: x2 in dom((abs f)|X) and
A6: abs(x1-x2)<s;
    x2 in dom(abs f) by A5,RELAT_1:57;
    then
A7: x2 in dom f by VALUED_1:def 11;
    x2 in X by A5,RELAT_1:57;
    then
A8: x2 in dom(f|X) by A7,RELAT_1:57;
    abs((abs(f)).x1-(abs(f)).x2) = abs(abs(f.x1)-(abs(f)).x2) by VALUED_1:18
      .= abs(abs(f.x1)-abs(f.x2)) by VALUED_1:18;
    then
A9: abs((abs(f)).x1-(abs(f)).x2) <= abs(f.x1-f.x2) by COMPLEX1:64;
    x1 in dom(abs f) by A4,RELAT_1:57;
    then
A10: x1 in dom f by VALUED_1:def 11;
    x1 in X by A4,RELAT_1:57;
    then x1 in dom(f|X) by A10,RELAT_1:57;
    then abs(f.x1-f.x2)<r by A3,A6,A8;
    hence abs((abs f).x1 - (abs f).x2) < r by A9,XXREAL_0:2;
  end;
  hence thesis by Th1;
end;

theorem
  X c= dom f1 & X1 c= dom f2 & f1|X is uniformly_continuous & f2|X1 is
uniformly_continuous & f1|Z is bounded & f2|Z1 is bounded implies (f1(#)f2)|(X
  /\Z/\X1/\Z1) is uniformly_continuous
proof
  assume that
A1: X c= dom f1 and
A2: X1 c= dom f2;
  assume that
A3: f1|X is uniformly_continuous and
A4: f2|X1 is uniformly_continuous and
A5: f1|Z is bounded and
A6: f2|Z1 is bounded;
  consider x1 be real number such that
A7: for r being set st r in Z /\ dom f1 holds abs(f1.r)<=x1 by A5,RFUNCT_1:73;
  consider x2 be real number such that
A8: for r being set st r in Z1 /\ dom f2 holds abs(f2.r)<=x2 by A6,RFUNCT_1:73;
  reconsider x1, x2 as Real by XREAL_0:def 1;
  set M1=abs(x1)+1;
  set M2=abs(x2)+1;
  set M=max(M1,M2);
A9: now
    let r;
A10: M1 <= M by XXREAL_0:25;
    assume r in X /\ Z /\ X1 /\ Z1;
    then
A11: r in X /\ Z /\ (X1 /\ Z1) by XBOOLE_1:16;
    then
A12: r in X /\ Z by XBOOLE_0:def 4;
    then
A13: r in Z by XBOOLE_0:def 4;
    r in X by A12,XBOOLE_0:def 4;
    then r in Z /\ dom f1 by A1,A13,XBOOLE_0:def 4;
    then
A14: abs(f1.r)<=x1 by A7;
    x1+0 < M1 by ABSVALUE:4,XREAL_1:8;
    then abs(f1.r) < M1 by A14,XXREAL_0:2;
    hence abs(f1.r) < M by A10,XXREAL_0:2;
A15: M2 <= M by XXREAL_0:25;
A16: r in X1 /\ Z1 by A11,XBOOLE_0:def 4;
    then
A17: r in Z1 by XBOOLE_0:def 4;
    r in X1 by A16,XBOOLE_0:def 4;
    then r in Z1 /\ dom f2 by A2,A17,XBOOLE_0:def 4;
    then
A18: abs(f2.r)<=x2 by A8;
    x2+0 < M2 by ABSVALUE:4,XREAL_1:8;
    then abs(f2.r) < M2 by A18,XXREAL_0:2;
    hence abs(f2.r) < M by A15,XXREAL_0:2;
  end;
A19: 0+0 < abs(x2)+1 by COMPLEX1:46,XREAL_1:8;
  0+0 < abs(x1)+1 by COMPLEX1:46,XREAL_1:8;
  then
A20: 0 < M by A19,XXREAL_0:16;
  then
A21: 0<2*M by XREAL_1:129;
  for r st 0<r ex s st 0<s & for x1,x2 st x1 in dom((f1(#)f2)|(X/\Z/\X1/\
Z1)) & x2 in dom((f1(#)f2)|(X/\Z/\X1/\Z1)) & abs(x1 - x2) < s holds abs((f1(#)
  f2).x1 - (f1(#)f2).x2) < r
  proof
    let r;
    assume 0<r;
    then
A22: 0<r/(2*M) by A21,XREAL_1:139;
    then consider s such that
A23: 0<s and
A24: for x1,x2 st x1 in dom(f1|X) & x2 in dom(f1|X) & abs(x1-x2)<s
    holds abs(f1.x1-f1.x2)<r/(2*M) by A3,Th1;
    consider g such that
A25: 0<g and
A26: for x1,x2 st x1 in dom(f2|X1) & x2 in dom(f2|X1) & abs(x1-x2)<g
    holds abs(f2.x1-f2.x2)<r/(2*M) by A4,A22,Th1;
    take p=min(s,g);
    thus 0<p by A23,A25,XXREAL_0:15;
    let y1,y2 be Real;
    assume that
A27: y1 in dom((f1(#)f2)|(X/\Z/\X1/\Z1)) and
A28: y2 in dom((f1(#)f2)|(X/\Z/\X1/\Z1));
    assume
A29: abs(y1-y2)<p;
A30: y2 in X/\Z/\X1/\Z1 by A28,RELAT_1:57;
    then
A31: y2 in X /\ Z /\ (X1 /\ Z1) by XBOOLE_1:16;
    then y2 in X /\ Z by XBOOLE_0:def 4;
    then y2 in X by XBOOLE_0:def 4;
    then
A32: y2 in dom(f1|X) by A1,RELAT_1:62;
A33: y1 in X/\Z/\X1/\Z1 by A27,RELAT_1:57;
    then
A34: y1 in X /\ Z /\ (X1 /\ Z1) by XBOOLE_1:16;
    then y1 in X /\ Z by XBOOLE_0:def 4;
    then y1 in X by XBOOLE_0:def 4;
    then
A35: y1 in dom(f1|X) by A1,RELAT_1:62;
    y2 in X1 /\ Z1 by A31,XBOOLE_0:def 4;
    then y2 in X1 by XBOOLE_0:def 4;
    then
A36: y2 in dom(f2|X1) by A2,RELAT_1:62;
    y1 in X1 /\ Z1 by A34,XBOOLE_0:def 4;
    then y1 in X1 by XBOOLE_0:def 4;
    then
A37: y1 in dom(f2|X1) by A2,RELAT_1:62;
    p<=g by XXREAL_0:17;
    then abs(y1-y2)<g by A29,XXREAL_0:2;
    then
A38: abs(f2.y1-f2.y2)<r/(2*M) by A26,A37,A36;
    abs((f1(#)f2).y1-(f1(#)f2).y2) = abs((f1.y1)*(f2.y1) - (f1(#) f2).y2)
    by VALUED_1:5
      .= abs((f1.y1)*(f2.y1)+((f1.y1)*(f2.y2)-(f1.y1)*(f2.y2))-(f1.y2)*(f2.
    y2)) by VALUED_1:5
      .= abs((f1.y1)*((f2.y1)-(f2.y2))+((f1.y1)-(f1.y2))*(f2.y2));
    then
A39: abs((f1(#)f2).y1-(f1(#)f2).y2)<=abs((f1.y1)*((f2.y1)-(f2.y2))) + abs(
    ((f1.y1)-(f1.y2))*(f2.y2)) by COMPLEX1:56;
A40: abs((f1.y1)*((f2.y1)-(f2.y2)))=abs(f1.y1)*abs((f2.y1)-(f2.y2)) by
COMPLEX1:65;
A41: abs(f2.y2) < M by A9,A30;
A42: 0 <= abs(f2.y2) by COMPLEX1:46;
A43: 0 <= abs(f1.y1-f1.y2) by COMPLEX1:46;
A44: abs(((f1.y1)-(f1.y2))*(f2.y2))=abs(f2.y2)*abs((f1.y1)-(f1.y2)) by
COMPLEX1:65;
    p<=s by XXREAL_0:17;
    then abs(y1-y2)<s by A29,XXREAL_0:2;
    then abs(f1.y1-f1.y2)<r/(2*M) by A24,A35,A32;
    then
A45: abs(((f1.y1)-(f1.y2))*(f2.y2)) < M*(r/(2*M)) by A44,A41,A43,A42,XREAL_1:96
;
A46: 0 <= abs(f1.y1) by COMPLEX1:46;
A47: 0 <= abs(f2.y1-f2.y2) by COMPLEX1:46;
    abs(f1.y1) < M by A9,A33;
    then abs((f1.y1)*((f2.y1)-(f2.y2))) < M*(r/(2*M)) by A40,A38,A47,A46,
XREAL_1:96;
    then
A48: abs((f1.y1)*((f2.y1)-(f2.y2)))+abs(((f1.y1)-(f1.y2))*(f2.y2))< M*(r/(
    2*M))+M*(r/(2*M)) by A45,XREAL_1:8;
    M*(r/(2*M))+M*(r/(2*M)) = M*(r/(M*2)+r/(M*2)) .= r/M*M by XCMPLX_1:118
      .= r by A20,XCMPLX_1:87;
    hence thesis by A39,A48,XXREAL_0:2;
  end;
  hence thesis by Th1;
end;

theorem Th9:
  X c= dom f & f|X is uniformly_continuous implies f|X is continuous
proof
  assume
A1: X c= dom f;
  assume
A2: f|X is uniformly_continuous;
  now
    let x0,r be real number;
    assume that
A3: x0 in X and
A4: 0<r;
A5: x0 in dom(f|X) by A1,A3,RELAT_1:62;
    r is Real by XREAL_0:def 1;
    then consider s such that
A6: 0<s and
A7: for x1,x2 st x1 in dom(f|X) & x2 in dom(f|X) & abs(x1 - x2) < s
    holds abs(f.x1 - f.x2) < r by A2,A4,Th1;
    reconsider s as real number;
    take s;
    thus 0<s by A6;
    let x1 be real number;
    assume that
A8: x1 in X and
A9: abs(x1-x0) < s;
    x1 in dom(f|X) by A1,A8,RELAT_1:62;
    hence abs(f.x1 - f.x0) < r by A7,A9,A5;
  end;
  hence thesis by A1,FCONT_1:14;
end;

theorem Th10:
  f|X is Lipschitzian implies f|X is uniformly_continuous
proof
  assume f|X is Lipschitzian;
  then consider r be real number such that
A1: 0<r and
A2: for x1,x2 be real number st x1 in dom(f|X) & x2 in dom(f|X) holds
  abs(f.x1-f.x2)<=r*abs(x1-x2) by FCONT_1:32;
  now
    reconsider r as Real by XREAL_0:def 1;
    let p such that
A3: 0<p;
    take s=p/r;
    thus 0<s by A1,A3,XREAL_1:139;
    let x1,x2;
    assume that
A4: x1 in dom(f|X) and
A5: x2 in dom(f|X) and
A6: abs(x1 - x2) < s;
A7: r*abs(x1 - x2) < s*r by A1,A6,XREAL_1:68;
    abs(f.x1-f.x2)<=r*abs(x1-x2) by A2,A4,A5;
    then abs(f.x1-f.x2)<p/r*r by A7,XXREAL_0:2;
    hence abs(f.x1 - f.x2) < p by A1,XCMPLX_1:87;
  end;
  hence thesis by Th1;
end;

theorem Th11:
  for f,Y st Y c= dom f & Y is compact & f|Y is continuous holds f
  |Y is uniformly_continuous
proof
  let f,Y such that
A1: Y c= dom f and
A2: Y is compact and
A3: f|Y is continuous;
  assume not thesis;
  then consider r such that
A4: 0<r and
A5: for s st 0<s ex x1,x2 st x1 in dom(f|Y) & x2 in dom(f|Y) & abs(x1 -
  x2) < s & not abs(f.x1 - f.x2) < r by Th1;
  defpred P[Element of NAT,Real] means $2 in Y & ex x2 st x2 in Y & abs($2-x2)
  <1/($1+1) & not abs(f.($2)-f.x2) < r;
A6: now
    let n;
    consider x1 such that
A7: ex x2 st x1 in dom(f|Y) & x2 in dom(f|Y) & abs(x1 - x2) < 1/(n+1)
    & not abs(f.x1 - f.x2) < r by A5,XREAL_1:139;
    take x1;
    dom(f|Y) = Y by A1,RELAT_1:62;
    hence P[n,x1] by A7;
  end;
  consider s1 such that
A8: for n holds P[n,s1.n] from FUNCT_2:sch 3(A6);
  now
    let x;
    assume x in rng s1;
    then ex n st x=s1.n by FUNCT_2:113;
    hence x in Y by A8;
  end;
  then
A9: rng s1 c= Y by TARSKI:def 3;
  then consider q1 such that
A10: q1 is subsequence of s1 and
A11: q1 is convergent and
A12: (lim q1) in Y by A2,RCOMP_1:def 3;
  lim q1 in dom(f|Y) by A1,A12,RELAT_1:57;
  then
A13: f|Y is_continuous_in (lim q1) by A3,FCONT_1:def 2;
  rng q1 c= rng s1 by A10,VALUED_0:21;
  then
A14: rng q1 c= Y by A9,XBOOLE_1:1;
  then rng q1 c= dom f by A1,XBOOLE_1:1;
  then rng q1 c= dom f /\ Y by A14,XBOOLE_1:19;
  then
A15: rng q1 c= dom (f|Y) by RELAT_1:61;
  then
A16: (f|Y).(lim q1)=lim((f|Y)/*q1) by A11,A13,FCONT_1:def 1;
A17: (f|Y)/*q1 is convergent by A11,A13,A15,FCONT_1:def 1;
  defpred P[Element of NAT,real number] means $2 in Y & abs(s1.$1-$2) < 1/($1+
  1) & not abs(f.(s1.$1)-f.$2) < r;
A18: for n ex x2 st P[n,x2] by A8;
  consider s2 such that
A19: for n holds P[n,s2.n] from FUNCT_2:sch 3(A18);
  now
    let x;
    assume x in rng s2;
    then ex n st x=s2.n by FUNCT_2:113;
    hence x in Y by A19;
  end;
  then
A20: rng s2 c= Y by TARSKI:def 3;
  consider Ns1 be increasing sequence of NAT such that
A21: q1=s1*Ns1 by A10,VALUED_0:def 17;
  set q2 = q1 - (s1-s2)*Ns1;
A22: now
    let n;
    thus q2.n = (s1*Ns1).n - ((s1-s2)*Ns1).n by A21,RFUNCT_2:1
      .= s1.(Ns1.n) - ((s1-s2)*Ns1).n by FUNCT_2:15
      .= s1.(Ns1.n) - (s1-s2).(Ns1.n) by FUNCT_2:15
      .= s1.(Ns1.n) - (s1.(Ns1.n)-s2.(Ns1.n)) by RFUNCT_2:1
      .= (s2*Ns1).n by FUNCT_2:15;
  end;
  then
A23: q2 = s2*Ns1 by FUNCT_2:63;
  q2 is subsequence of s2 by A22,FUNCT_2:63,VALUED_0:def 17;
  then rng q2 c= rng s2 by VALUED_0:21;
  then
A24: rng q2 c= Y by A20,XBOOLE_1:1;
  then rng q2 c= dom f by A1,XBOOLE_1:1;
  then rng q2 c= dom f /\ Y by A24,XBOOLE_1:19;
  then
A25: rng q2 c= dom (f|Y) by RELAT_1:61;
A26: now
    let p be real number such that
A27: 0<p;
    consider k be Element of NAT such that
A28: p" < k by SEQ_4:3;
    take k;
    let m;
    assume k <= m;
    then k+1 <= m+1 by XREAL_1:6;
    then 1/(m+1) <= 1/(k+1) by XREAL_1:118;
    then
A29: abs(s1.m - s2.m) < 1/(k+1) by A19,XXREAL_0:2;
    k < k+1 by NAT_1:13;
    then p" < k+1 by A28,XXREAL_0:2;
    then 1/(k+1) < 1/p" by A27,XREAL_1:76;
    then
A30: 1/(k+1) < p by XCMPLX_1:216;
    abs((s1-s2).m - 0) = abs(s1.m - s2.m) by RFUNCT_2:1;
    hence abs((s1-s2).m - 0) < p by A30,A29,XXREAL_0:2;
  end;
  then
A31: s1-s2 is convergent by SEQ_2:def 6;
A32: (s1-s2)*Ns1 is subsequence of s1-s2 by VALUED_0:def 17;
  then
A33: (s1-s2)*Ns1 is convergent by A31,SEQ_4:16;
  lim (s1-s2) = 0 by A26,A31,SEQ_2:def 7;
  then lim ((s1-s2)*Ns1) = 0 by A31,A32,SEQ_4:17;
  then
A34: lim q2 = lim q1 - 0 by A11,A33,SEQ_2:12
    .= lim q1;
A35: q2 is convergent by A11,A33,SEQ_2:11;
  then
A36: (f|Y)/*q2 is convergent by A13,A34,A25,FCONT_1:def 1;
  then
A37: (f|Y)/*q1 - (f|Y)/*q2 is convergent by A17,SEQ_2:11;
  (f|Y).(lim q1) = lim ((f|Y)/*q2) by A13,A35,A34,A25,FCONT_1:def 1;
  then
A38: lim ((f|Y)/*q1 - (f|Y)/*q2) = (f|Y).(lim q1) - (f|Y).(lim q1) by A17,A16
,A36,SEQ_2:12
    .= 0;
  now
    let n;
    consider k be Element of NAT such that
A39: for m st k<=m holds abs(((f|Y)/*q1 - (f|Y)/*q2).m-0)<r by A4,A37,A38,
SEQ_2:def 7;
A40: q1.k in rng q1 by VALUED_0:28;
A41: q2.k in rng q2 by VALUED_0:28;
    abs(((f|Y)/*q1 - (f|Y)/*q2).k - 0) = abs(((f|Y)/*q1).k - ((f|Y)/*q2).
    k) by RFUNCT_2:1
      .= abs((f|Y).(q1.k) - ((f|Y)/*q2).k) by A15,FUNCT_2:108
      .= abs((f|Y).(q1.k) - (f|Y).(q2.k)) by A25,FUNCT_2:108
      .= abs(f.(q1.k) - (f|Y).(q2.k)) by A15,A40,FUNCT_1:47
      .= abs(f.(q1.k) - f.(q2.k)) by A25,A41,FUNCT_1:47
      .= abs(f.(s1.(Ns1.k)) - f.((s2*Ns1).k)) by A21,A23,FUNCT_2:15
      .= abs(f.(s1.(Ns1.k)) - f.(s2.(Ns1.k))) by FUNCT_2:15;
    hence contradiction by A19,A39;
  end;
  hence contradiction;
end;

theorem
  Y c= dom f & Y is compact & f|Y is uniformly_continuous implies f.:Y
  is compact by Th9,FCONT_1:29;

theorem
  for f,Y st Y <> {} & Y c= dom f & Y is compact & f|Y is
uniformly_continuous ex x1,x2 st x1 in Y & x2 in Y & f.x1 = upper_bound (f.:Y)
  & f.x2 = lower_bound (f.:Y)
proof
  let f,Y;
  assume that
A1: Y <> {} and
A2: Y c= dom f and
A3: Y is compact and
A4: f|Y is uniformly_continuous;
  ex x1,x2 be real number st x1 in Y & x2 in Y & f.x1 = upper_bound (f.:Y)
  & f.x2 = lower_bound (f.:Y) by A1,A2,A3,A4,Th9,FCONT_1:31;
  hence thesis;
end;

theorem
  X c= dom f & f|X is constant implies f|X is uniformly_continuous by Th10;

theorem Th15:
  p<= g & [.p,g.] c= dom f & f|[.p,g.] is continuous implies for r
  st r in [.f.p,f.g.] \/ [.f.g,f.p.] ex s st s in [.p,g.] & r = f.s
proof
  assume that
A1: p<=g and
A2: [.p,g.] c= dom f and
A3: f|[.p,g.] is continuous;
  let r such that
A4: r in [.f.p,f.g.] \/ [.f.g,f.p.];
A5: [.p,g.] is compact by RCOMP_1:6;
A6: f.p < f.g implies ex s st s in [.p,g.] & r = f.s
  proof
    reconsider f1 = [.p,g.] --> r as Function of [.p,g.], REAL by FUNCOP_1:45;
    assume that
A7: f.p < f.g and
A8: for s st s in [.p,g.] holds r <> f.s;
    [.f.p,f.g.] \/ [.f.g,f.p.] = [.f.p,f.g.] \/ {} by A7,XXREAL_1:29
      .= [.f.p,f.g.];
    then r in {s: f.p<=s & s<= f.g} by A4,RCOMP_1:def 1;
    then
A9: ex s st s=r & f.p<=s & s<=f.g;
    p in {s: p<=s & s<=g} by A1;
    then p in [.p,g.] by RCOMP_1:def 1;
    then
A10: r<> f.p by A8;
    reconsider f1 as PartFunc of REAL, REAL;
A11: dom f1 = [.p,g.] by FUNCOP_1:13;
    then
A12: [.p,g.] c= dom f1 /\ dom f by A2,XBOOLE_1:19;
    then
A13: [.p,g.] c= dom(f1-f) by VALUED_1:12;
A14: abs(f1-f)"{0} = {}
    proof
      assume abs(f1-f)"{0} <> {};
      then consider x2 such that
A15:  x2 in abs(f1-f)"{0} by SUBSET_1:4;
      x2 in dom (abs(f1-f)) by A15,FUNCT_1:def 7;
      then
A16:  x2 in dom (f1-f) by VALUED_1:def 11;
      then x2 in dom f1 /\ dom f by VALUED_1:12;
      then
A17:  x2 in [.p,g.] by A11,XBOOLE_0:def 4;
      (abs(f1-f)).x2 in {0} by A15,FUNCT_1:def 7;
      then (abs(f1-f)).x2 = 0 by TARSKI:def 1;
      then abs((f1-f).x2) = 0 by VALUED_1:18;
      then (f1-f).x2 = 0 by ABSVALUE:2;
      then f1.x2 -f.x2 = 0 by A16,VALUED_1:13;
      then r - f.x2 = 0 by A17,FUNCOP_1:7;
      hence contradiction by A8,A17;
    end;
A18: dom (abs(f1-f)^) = dom (abs(f1-f)) \ abs(f1-f)"{0} by RFUNCT_1:def 2
      .= dom (f1-f) by A14,VALUED_1:def 11
      .= [.p,g.] /\ dom f by A11,VALUED_1:12
      .= [.p,g.] by A2,XBOOLE_1:28;
    for x0 st x0 in [.p,g.] /\ dom f1 holds f1.x0=r by A11,FUNCOP_1:7;
    then f1|[.p,g.] is constant by PARTFUN2:57;
    then (f1-f)|[.p,g.] is continuous by A3,A12,FCONT_1:18;
    then abs(f1-f)|[.p,g.] is continuous by A13,FCONT_1:21;
    then abs(f1-f)^|[.p,g.] is continuous by A14,FCONT_1:22;
    then (abs(f1-f)^).:[.p,g.] is real-bounded by A5,A18,FCONT_1:29,RCOMP_1:10;
    then consider M being real number such that
A19: M is UpperBound of (abs(f1-f)^).:[.p,g.] by XXREAL_2:def 10;
A20: for x1 be real number st x1 in (abs(f1-f)^).:[.p,g.] holds x1<=M
        by A19,XXREAL_2:def 1;
    0+0<abs(M)+1 by COMPLEX1:46,XREAL_1:8;
    then 0<(abs(M)+1)";
    then
A21: 0<1/(abs(M)+1) by XCMPLX_1:215;
    f|[.p,g.] is uniformly_continuous by A2,A3,Th11,RCOMP_1:6;
    then consider s such that
A22: 0<s and
A23: for x1,x2 st x1 in dom(f|[.p,g.]) & x2 in dom(f|[.p,g .]) & abs(
    x1 - x2) < s holds abs(f.x1 - f.x2) < 1/(abs(M)+1) by A21,Th1;
A24: now
      let x1;
      assume
A25:  x1 in [.p,g.];
      then (abs(f1-f)^).x1 in (abs(f1-f)^).:[.p,g.] by A18,FUNCT_1:def 6;
      then (abs(f1-f)^).x1 <= M by A20;
      then ((abs(f1-f)).x1)" <= M by A18,A25,RFUNCT_1:def 2;
      then
A26:  abs((f1-f).x1)" <= M by VALUED_1:18;
      x1 in dom f1 /\ dom f by A2,A11,A25,XBOOLE_0:def 4;
      then x1 in dom (f1-f) by VALUED_1:12;
      then abs(f1.x1-f.x1)" <= M by A26,VALUED_1:13;
      then
A27:  abs(r-f.x1)" <= M by A25,FUNCOP_1:7;
      r-f.x1<>0 by A8,A25;
      then
A28:  0<abs(r-f.x1) by COMPLEX1:47;
      M+0<abs(M)+1 by ABSVALUE:4,XREAL_1:8;
      then abs(r-f.x1)" < abs(M)+1 by A27,XXREAL_0:2;
      then 1/(abs(M)+1)<1/abs(r-f.x1)" by A28,XREAL_1:76;
      hence 1/(abs(M)+1)<abs(r-f.x1) by XCMPLX_1:216;
    end;
    now
      per cases;
      suppose
        p=g;
        hence contradiction by A7;
      end;
      suppose
        p<>g;
        then p < g by A1,XXREAL_0:1;
        then
A29:    0 < g-p by XREAL_1:50;
        then
A30:    0 < (g-p)/s by A22,XREAL_1:139;
        consider k be Element of NAT such that
A31:    (g-p)/s < k by SEQ_4:3;
        (g-p)/s *s < s*k by A22,A31,XREAL_1:68;
        then (g-p) < s*k by A22,XCMPLX_1:87;
        then (g-p)/k < s*k/k by A31,A30,XREAL_1:74;
        then (g-p)/k < s*k*k" by XCMPLX_0:def 9;
        then (g-p)/k < s*(k*k");
        then
A32:    (g-p)/k < s*1 by A31,A30,XCMPLX_0:def 7;
        deffunc F(Element of NAT) = p + (g-p)/k*$1;
        consider s1 such that
A33:    for n holds s1.n = F(n) from SEQ_1:sch 1;
A34:    0<(g-p)/k by A29,A31,A30,XREAL_1:139;
A35:    now
          let n;
A36:      dom(f|[.p,g.]) = [.p,g.] by A2,RELAT_1:62;
          assume
A37:      n<k;
          then (g-p)/k*n < (g-p)/k*k by A34,XREAL_1:68;
          then (g-p)/k*n < (g-p) by A31,A30,XCMPLX_1:87;
          then p + (g-p)/k*n < p+(g-p) by XREAL_1:6;
          then
A38:      s1.n < p+(g-p) by A33;
          n+1<=k by A37,NAT_1:13;
          then (g-p)/k*(n+1) <= (g-p)/k*k by A29,XREAL_1:64;
          then (g-p)/k*(n+1) <= (g-p) by A31,A30,XCMPLX_1:87;
          then p + (g-p)/k*(n+1) <= p+(g-p) by XREAL_1:7;
          then
A39:      s1.(n+1) <= p+(g-p) by A33;
          p+0<=p+(g-p)/k*n by A29,XREAL_1:7;
          then p<=s1.n by A33;
          then s1.n in {x2: p<=x2 & x2<=g} by A38;
          hence
A40:      s1.n in [.p,g.] by RCOMP_1:def 1;
          p+0<=p+(g-p)/k*(n+1) by A29,XREAL_1:7;
          then p<=s1.(n+1) by A33;
          then s1.(n+1) in {x2: p<=x2 & x2<= g} by A39;
          hence
A41:      s1.(n+1) in [.p,g.] by RCOMP_1:def 1;
          abs(s1.(n+1) -s1.n) = abs(p + (g-p)/k*(n+1) - s1.n) by A33
            .= abs(p + (g-p)/k*(n+1) - (p + (g-p)/k*n)) by A33
            .= (g-p)/k by A29,ABSVALUE:def 1;
          hence abs(f.(s1.(n+1)) - f.(s1.n)) < 1/(abs(M)+1) by A23,A32,A40,A41
,A36;
        end;
        defpred P[Nat] means r<=f.(s1.$1);
A42:    s1.k = p + (g-p)/k*k by A33
          .= p + (g-p) by A31,A30,XCMPLX_1:87
          .= g;
        then
A43:    ex n be Nat st P[n] by A9;
        consider l be Nat such that
A44:    P[l] & for m be Nat st P[m] holds l<=m from NAT_1:sch 5(A43);
        s1.0 = p + (g-p)/k*0 by A33
          .= p;
        then l<>0 by A9,A10,A44,XXREAL_0:1;
        then consider l1 be Nat such that
A45:    l=l1+1 by NAT_1:6;
        reconsider l1 as Element of NAT by ORDINAL1:def 12;
A46:    r - f.(s1.l1) <= f.(s1.(l1+1)) - f.(s1.l1) by A44,A45,XREAL_1:9;
        l1+1 <= k by A9,A42,A44,A45;
        then
A47:    l1<k by NAT_1:13;
        then
A48:    abs(f.(s1.(l1+1)) - f.(s1.l1)) < 1/(abs(M)+1) by A35;
        f.(s1.l1) < r
        proof
          assume r <= f.(s1.l1);
          then l <= l1 by A44;
          then l+0 < l by A45,XREAL_1:8;
          hence contradiction;
        end;
        then
A49:    0<r-f.(s1.l1) by XREAL_1:50;
        then 0 < f.(s1.(l1+1)) - f.(s1.l1) by A44,A45,XREAL_1:9;
        then abs(f.(s1.(l1+1)) - f.(s1.l1)) = f.(s1.(l1+1)) - f.(s1.l1) by
ABSVALUE:def 1;
        then
A50:    abs(r - f.(s1.l1)) <= abs(f.(s1.(l1+1)) - f.(s1.l1)) by A49,A46,
ABSVALUE:def 1;
        s1.l1 in [.p,g.] by A35,A47;
        hence contradiction by A24,A50,A48,XXREAL_0:2;
      end;
    end;
    hence contradiction;
  end;
A51: f.g < f.p implies ex s st s in [.p,g.] & r = f.s
  proof
    reconsider f1 = [.p,g.] --> r as Function of [.p,g.], REAL by FUNCOP_1:45;
    assume that
A52: f.g < f.p and
A53: for s st s in [.p,g.] holds r <> f.s;
    [.f.p,f.g.] \/ [.f.g,f.p.] = {} \/ [.f.g,f.p.] by A52,XXREAL_1:29
      .= [.f.g,f.p.];
    then r in {s: f.g<=s & s<=f.p} by A4,RCOMP_1:def 1;
    then
A54: ex s st s=r & f.g<=s & s<=f.p;
    g in {s: p<=s & s<=g} by A1;
    then g in [.p,g.] by RCOMP_1:def 1;
    then
A55: r<> f.g by A53;
    reconsider f1 as PartFunc of REAL, REAL;
A56: dom f1 = [.p,g.] by FUNCOP_1:13;
    then
A57: [.p,g.] c= dom f1 /\ dom f by A2,XBOOLE_1:19;
    then
A58: [.p,g.] c= dom(f1-f) by VALUED_1:12;
A59: abs(f1-f)"{0} = {}
    proof
      assume abs(f1-f)"{0} <> {};
      then consider x2 such that
A60:  x2 in abs(f1-f)"{0} by SUBSET_1:4;
      x2 in dom (abs(f1-f)) by A60,FUNCT_1:def 7;
      then
A61:  x2 in dom (f1-f) by VALUED_1:def 11;
      then x2 in dom f1 /\ dom f by VALUED_1:12;
      then
A62:  x2 in [.p,g.] by A56,XBOOLE_0:def 4;
      (abs(f1-f)).x2 in {0} by A60,FUNCT_1:def 7;
      then (abs(f1-f)).x2 = 0 by TARSKI:def 1;
      then abs((f1-f).x2) = 0 by VALUED_1:18;
      then (f1-f).x2 = 0 by ABSVALUE:2;
      then f1.x2 -f.x2 = 0 by A61,VALUED_1:13;
      then r - f.x2 = 0 by A62,FUNCOP_1:7;
      hence contradiction by A53,A62;
    end;
A63: dom (abs(f1-f)^) = dom (abs(f1-f)) \ abs(f1-f)"{0} by RFUNCT_1:def 2
      .= dom (f1-f) by A59,VALUED_1:def 11
      .= [.p,g.] /\ dom f by A56,VALUED_1:12
      .= [.p,g.] by A2,XBOOLE_1:28;
    for x0 st x0 in [.p,g.] /\ dom f1 holds f1.x0=r by A56,FUNCOP_1:7;
    then f1|[.p,g.] is constant by PARTFUN2:57;
    then (f1-f)|[.p,g.] is continuous by A3,A57,FCONT_1:18;
    then abs(f1-f)|[.p,g.] is continuous by A58,FCONT_1:21;
    then abs(f1-f)^|[.p,g.] is continuous by A59,FCONT_1:22;
    then (abs(f1-f)^).:[.p,g.] is real-bounded by A5,A63,FCONT_1:29,RCOMP_1:10;
    then consider M being real number such that
A64: M is UpperBound of (abs(f1-f)^).:[.p,g.] by XXREAL_2:def 10;
A65: for x1 be real number st x1 in (abs(f1-f)^).:[.p,g.] holds x1<=M
          by A64,XXREAL_2:def 1;
    0+0<abs(M)+1 by COMPLEX1:46,XREAL_1:8;
    then 0<(abs(M)+1)";
    then
A66: 0<1/(abs(M)+1) by XCMPLX_1:215;
    f|[.p,g.] is uniformly_continuous by A2,A3,Th11,RCOMP_1:6;
    then consider s such that
A67: 0<s and
A68: for x1,x2 st x1 in dom(f|[.p,g.]) & x2 in dom(f|[.p,g .]) & abs(
    x1 - x2) < s holds abs(f.x1 - f.x2) < 1/(abs(M)+1) by A66,Th1;
A69: now
      let x1;
      assume
A70:  x1 in [.p,g.];
      then (abs(f1-f)^).x1 in (abs(f1-f)^).:[.p,g.] by A63,FUNCT_1:def 6;
      then (abs(f1-f)^).x1 <= M by A65;
      then ((abs(f1-f)).x1)" <= M by A63,A70,RFUNCT_1:def 2;
      then
A71:  abs((f1-f).x1)" <= M by VALUED_1:18;
      x1 in dom f1 /\ dom f by A2,A56,A70,XBOOLE_0:def 4;
      then x1 in dom (f1-f) by VALUED_1:12;
      then abs(f1.x1-f.x1)" <= M by A71,VALUED_1:13;
      then
A72:  abs(r-f.x1)" <= M by A70,FUNCOP_1:7;
      r-f.x1<>0 by A53,A70;
      then
A73:  0<abs(r-f.x1) by COMPLEX1:47;
      M+0<abs(M)+1 by ABSVALUE:4,XREAL_1:8;
      then abs(r-f.x1)" < abs(M)+1 by A72,XXREAL_0:2;
      then 1/(abs(M)+1)<1/abs(r-f.x1)" by A73,XREAL_1:76;
      hence 1/(abs(M)+1)<abs(r-f.x1) by XCMPLX_1:216;
    end;
    now
      per cases;
      suppose
        p=g;
        hence contradiction by A52;
      end;
      suppose
        p<>g;
        then p < g by A1,XXREAL_0:1;
        then
A74:    0 < g-p by XREAL_1:50;
        then
A75:    0 < (g-p)/s by A67,XREAL_1:139;
        consider k be Element of NAT such that
A76:    (g-p)/s < k by SEQ_4:3;
        (g-p)/s *s < s*k by A67,A76,XREAL_1:68;
        then (g-p) < s*k by A67,XCMPLX_1:87;
        then (g-p)/k < s*k/k by A76,A75,XREAL_1:74;
        then (g-p)/k < s*k*k" by XCMPLX_0:def 9;
        then (g-p)/k < s*(k*k");
        then
A77:    (g-p)/k < s*1 by A76,A75,XCMPLX_0:def 7;
        deffunc F(Element of NAT) = g - (g-p)/k*$1;
        consider s1 such that
A78:    for n holds s1.n = F(n) from SEQ_1:sch 1;
A79:    0<(g-p)/k by A74,A76,A75,XREAL_1:139;
A80:    now
          let n;
A81:      dom(f|[.p,g.]) = [.p,g.] by A2,RELAT_1:62;
          assume
A82:      n<k;
          then (g-p)/k*n < (g-p)/k*k by A79,XREAL_1:68;
          then (g-p)/k*n < (g-p) by A76,A75,XCMPLX_1:87;
          then g -(g-p) < g-(g-p)/k*n by XREAL_1:15;
          then
A83:      g-(g-p)< s1.n by A78;
          n+1<=k by A82,NAT_1:13;
          then (g-p)/k*(n+1) <= (g-p)/k*k by A74,XREAL_1:64;
          then (g-p)/k*(n+1) <= (g-p) by A76,A75,XCMPLX_1:87;
          then g -(g-p)<=g- (g-p)/k*(n+1) by XREAL_1:13;
          then
A84:      g-(g-p)<=s1.(n+1) by A78;
          g-(g-p)/k*n<=g-0 by A74,XREAL_1:13;
          then s1.n<=g by A78;
          then s1.n in {x2: p<=x2 & x2<=g} by A83;
          hence
A85:      s1.n in [.p,g.] by RCOMP_1:def 1;
          g-(g-p)/k*(n+1)<=g-0 by A74,XREAL_1:13;
          then s1.(n+1)<=g by A78;
          then s1.(n+1) in {x2: p<=x2 & x2<=g} by A84;
          hence
A86:      s1.(n+1) in [.p,g.] by RCOMP_1:def 1;
          abs(s1.(n+1) -s1.n) = abs(g- (g-p)/k*(n+1) - s1.n) by A78
            .= abs(g - (g-p)/k*(n+1) - (g - (g-p)/k*n)) by A78
            .= abs(-((g-p)/k*(n+1) - (g-p)/k*n))
            .= abs((g-p)/k*(n+1 - n)) by COMPLEX1:52
            .= (g-p)/k by A74,ABSVALUE:def 1;
          hence abs(f.(s1.(n+1)) - f.(s1.n)) < 1/(abs(M)+1) by A68,A77,A85,A86
,A81;
        end;
        defpred P[Nat] means r<=f.(s1.$1);
A87:    s1.k = g- (g-p)/k*k by A78
          .= g- (g-p) by A76,A75,XCMPLX_1:87
          .= p;
        then
A88:    ex n be Nat st P[n] by A54;
        consider l be Nat such that
A89:    P[l] & for m be Nat st P[m] holds l<=m from NAT_1:sch 5(A88
        );
        s1.0 = g- (g-p)/k*0 by A78
          .= g;
        then l<>0 by A54,A55,A89,XXREAL_0:1;
        then consider l1 be Nat such that
A90:    l=l1+1 by NAT_1:6;
        reconsider l1 as Element of NAT by ORDINAL1:def 12;
A91:    r - f.(s1.l1) <= f.(s1.(l1+1)) - f.(s1.l1) by A89,A90,XREAL_1:9;
        l1+1 <= k by A54,A87,A89,A90;
        then
A92:    l1<k by NAT_1:13;
        then
A93:    abs(f.(s1.(l1+1)) - f.(s1.l1)) < 1/(abs(M)+1) by A80;
        f.(s1.l1) < r
        proof
          assume r <= f.(s1.l1);
          then l <= l1 by A89;
          then l+0 < l by A90,XREAL_1:8;
          hence contradiction;
        end;
        then
A94:    0<r-f.(s1.l1) by XREAL_1:50;
        then 0 < f.(s1.(l1+1)) - f.(s1.l1) by A89,A90,XREAL_1:9;
        then abs(f.(s1.(l1+1)) - f.(s1.l1)) = f.(s1.(l1+1)) - f.(s1.l1) by
ABSVALUE:def 1;
        then
A95:    abs(r - f.(s1.l1)) <= abs(f.(s1.(l1+1)) - f.(s1.l1)) by A94,A91,
ABSVALUE:def 1;
        s1.l1 in [.p,g.] by A80,A92;
        hence contradiction by A69,A95,A93,XXREAL_0:2;
      end;
    end;
    hence contradiction;
  end;
  f.p = f.g implies ex s st s in [.p,g.] & r = f.s
  proof
    assume
A96: f.p = f.g;
    take p;
    thus p in [.p,g.] by A1,XXREAL_1:1;
    [.f.p,f.g.] \/ [.f.g,f.p.] = {f.p} by A96,XXREAL_1:17;
    hence thesis by A4,TARSKI:def 1;
  end;
  hence thesis by A6,A51,XXREAL_0:1;
end;

theorem Th16:
  p<=g & [.p,g.] c= dom f & f|[.p,g.] is continuous implies for r
st r in [.(lower_bound (f.:[.p,g.])),(upper_bound (f.:[.p,g.])).] ex s st s in
  [.p,g.] & r = f.s
proof
  assume that
A1: p<=g and
A2: [.p,g.] c= dom f and
A3: f|[.p,g.] is continuous;
  [.p,g.] is compact by RCOMP_1:6;
  then
A4: f.:[.p,g.] is real-bounded by A2,A3,FCONT_1:29,RCOMP_1:10;
  set ub=upper_bound (f.:[.p,g.]);
  set lb=lower_bound (f.:[.p,g.]);
  [.p,g.] <> {} by A1,XXREAL_1:1;
  then consider x2,x1 be real number such that
A5: x2 in [.p,g.] and
A6: x1 in [.p,g.] and
A7: f.x2=ub and
A8: f.x1=lb by A2,A3,FCONT_1:31,RCOMP_1:6;
  reconsider x1,x2 as Real by XREAL_0:def 1;
  let r such that
A9: r in [.lb,ub.];
  f.x1 in f.:[.p,g.] by A2,A6,FUNCT_1:def 6;
  then
A10: [.lb,ub.] = [.lb,ub.] \/ [.ub,lb.] by A4,SEQ_4:11,XXREAL_1:222;
  now
    per cases;
    suppose
A11:  x1 <= x2;
A12:  [.x1,x2.] c= [.p,g.] by A5,A6,XXREAL_2:def 12;
A13:  [.x1,x2.] c= [.p,g.] by A5,A6,XXREAL_2:def 12;
      then f|[.x1,x2.] is continuous by A3,FCONT_1:16;
      then consider s such that
A14:  s in [.x1,x2.] and
A15:  r=f.s by A2,A9,A7,A8,A10,A11,A12,Th15,XBOOLE_1:1;
      take s;
      thus s in [.p,g.] & r=f.s by A13,A14,A15;
    end;
    suppose
A16:  x2 <= x1;
A17:  [.x2,x1.] c= [.p,g.] by A5,A6,XXREAL_2:def 12;
A18:  [.x2,x1.] c= [.p,g.] by A5,A6,XXREAL_2:def 12;
      then f|[.x2,x1.] is continuous by A3,FCONT_1:16;
      then consider s such that
A19:  s in [.x2,x1.] and
A20:  r=f.s by A2,A9,A7,A8,A10,A16,A17,Th15,XBOOLE_1:1;
      take s;
      thus s in [.p,g.] & r=f.s by A18,A19,A20;
    end;
  end;
  hence thesis;
end;

theorem Th17:
  f is one-to-one & [.p,g.] c= dom f & p<=g & f|[.p,g.] is
  continuous implies f|[.p,g.] is increasing or f|[.p,g.] is decreasing
proof
  assume that
A1: f is one-to-one and
A2: [.p,g.] c= dom f and
A3: p<=g and
A4: f|[.p,g.] is continuous and
A5: not f|[.p,g.] is increasing and
A6: not f|[.p,g.] is decreasing;
  now
    per cases;
    suppose
      p=g;
      then [.p,g.]={p} by XXREAL_1:17;
      hence contradiction by A6,RFUNCT_2:44;
    end;
    suppose
A7:   p<>g;
A8:   g in [.p,g.] by A3,XXREAL_1:1;
A9:   p in [.p,g.] by A3,XXREAL_1:1;
      then
A10:  f.p<>f.g by A1,A2,A7,A8,FUNCT_1:def 4;
      now
        per cases by A10,XXREAL_0:1;
        suppose
A11:      f.p<f.g;
A12:      for x1 st p<=x1 & x1<=g holds f.p<=f.x1 & f.x1<=f.g
          proof
            let x1;
            assume that
A13:        p<=x1 and
A14:        x1<=g and
A15:        not (f.p<=f.x1 & f.x1 <=f.g);
            now
              per cases by A15;
              suppose
A16:            f.x1<f.p;
                then f.p in {r:f.x1<=r & r<=f.g} by A11;
                then f.p in [.f.x1,f.g.] by RCOMP_1:def 1;
                then
A17:            f.p in [.f.x1,f.g.] \/ [.f.g,f.x1.] by XBOOLE_0:def 3;
                x1 in {r: p<=r & r<=g} by A13,A14;
                then
A18:            x1 in [.p,g.] by RCOMP_1:def 1;
                g in [.p,g.] by A3,XXREAL_1:1;
                then
A19:            [.x1,g.] c= [.p,g.] by A18,XXREAL_2:def 12;
                then f|[.x1,g.] is continuous by A4,FCONT_1:16;
                then consider s such that
A20:            s in [.x1,g.] and
A21:            f.s=f.p by A2,A14,A19,A17,Th15,XBOOLE_1:1;
                s in {t: x1<=t & t<=g} by A20,RCOMP_1:def 1;
                then
A22:            ex r st r=s & x1<=r & r<=g;
A23:            x1>p by A13,A16,XXREAL_0:1;
                s in [.p,g.] by A19,A20;
                hence contradiction by A1,A2,A9,A23,A21,A22,FUNCT_1:def 4;
              end;
              suppose
A24:            f.g<f.x1;
                then f.g in {r:f.p<=r & r<=f.x1} by A11;
                then f.g in [.f.p,f.x1.] by RCOMP_1:def 1;
                then
A25:            f.g in [.f.p,f.x1.] \/ [.f.x1,f.p.] by XBOOLE_0:def 3;
                x1 in {r:p<=r & r<=g} by A13,A14;
                then
A26:            x1 in [.p,g.] by RCOMP_1:def 1;
                p in [.p,g.] by A3,XXREAL_1:1;
                then
A27:            [.p,x1.] c= [.p,g.] by A26,XXREAL_2:def 12;
                then f|[.p,x1.] is continuous by A4,FCONT_1:16;
                then consider s such that
A28:            s in [.p,x1.] and
A29:            f.s=f.g by A2,A13,A27,A25,Th15,XBOOLE_1:1;
                s in {t: p<=t & t<=x1} by A28,RCOMP_1:def 1;
                then
A30:            ex r st r=s & p<=r & r<=x1;
                s in [.p,g.] by A27,A28;
                then s=g by A1,A2,A8,A29,FUNCT_1:def 4;
                hence contradiction by A14,A24,A30,XXREAL_0:1;
              end;
            end;
            hence contradiction;
          end;
          consider x1,x2 such that
A31:      x1 in [.p,g.] /\ dom f and
A32:      x2 in [.p,g.] /\ dom f and
A33:      x1<x2 and
A34:      f.x2<=f.x1 by A5,RFUNCT_2:20;
A35:      x1 in [.p,g.] by A31,XBOOLE_0:def 4;
          then
A36:      [.p,x1.] c= [.p,g.] by A9,XXREAL_2:def 12;
A37:      x2 in [.p,g.] by A32,XBOOLE_0:def 4;
          then x2 in {r: p<=r & r<=g} by RCOMP_1:def 1;
          then ex r st r=x2 & p<=r & r<=g;
          then f.p <= f.x2 by A12;
          then f.x2 in {r: f.p<=r & r<=f.x1} by A34;
          then f.x2 in [.f.p,f.x1.] by RCOMP_1:def 1;
          then
A38:      f.x2 in [.f.p,f.x1.] \/ [.f.x1,f.p.] by XBOOLE_0:def 3;
          x1 in {t:p<=t & t<=g} by A35,RCOMP_1:def 1;
          then
A39:      ex r st r=x1 & p<=r & r<=g;
          p in [.p,g.] by A3,XXREAL_1:1;
          then
A40:      [.p,x1.] c= [.p,g.] by A35,XXREAL_2:def 12;
          then f|[.p,x1.] is continuous by A4,FCONT_1:16;
          then consider s such that
A41:      s in [.p,x1.] and
A42:      f.s=f.x2 by A2,A39,A38,A36,Th15,XBOOLE_1:1;
          s in {t: p<=t & t<=x1} by A41,RCOMP_1:def 1;
          then
A43:      ex r st r=s & p<=r & r<=x1;
          s in [.p,g.] by A40,A41;
          hence contradiction by A1,A2,A33,A37,A42,A43,FUNCT_1:def 4;
        end;
        suppose
A44:      f.p>f.g;
A45:      for x1 st p<=x1 & x1<=g holds f.g<=f.x1 & f.x1<=f.p
          proof
            let x1;
            assume that
A46:        p<=x1 and
A47:        x1<=g and
A48:        not (f.g<=f.x1 & f.x1 <=f.p);
            now
              per cases by A48;
              suppose
A49:            f.x1<f.g;
                now
                  per cases;
                  suppose
                    g=x1;
                    hence contradiction by A49;
                  end;
                  suppose
A50:                g<>x1;
                    x1 in {r: p<=r & r<=g} by A46,A47;
                    then
A51:                x1 in [.p,g.] by RCOMP_1:def 1;
                    f.g in {r:f.x1<=r & r<=f.p} by A44,A49;
                    then f.g in [.f.x1,f.p.] by RCOMP_1:def 1;
                    then
A52:                f.g in [.f.p,f.x1.] \/ [.f.x1,f.p.] by XBOOLE_0:def 3;
                    p in [.p,g.] by A3,XXREAL_1:1;
                    then
A53:                [.p,x1.] c= [.p,g.] by A51,XXREAL_2:def 12;
                    then f|[.p,x1.] is continuous by A4,FCONT_1:16;
                    then consider s such that
A54:                s in [.p,x1.] and
A55:                f.s=f.g by A2,A46,A53,A52,Th15,XBOOLE_1:1;
                    s in {t: p<=t & t<=x1} by A54,RCOMP_1:def 1;
                    then
A56:                ex r st r=s & p<=r & r<=x1;
                    s in [.p,g.] by A53,A54;
                    then s=g by A1,A2,A8,A55,FUNCT_1:def 4;
                    hence contradiction by A47,A50,A56,XXREAL_0:1;
                  end;
                end;
                hence contradiction;
              end;
              suppose
A57:            f.p<f.x1;
                now
                  per cases;
                  suppose
                    p=x1;
                    hence contradiction by A57;
                  end;
                  suppose
A58:                p<>x1;
                    x1 in {r:p<=r & r<=g} by A46,A47;
                    then
A59:                x1 in [.p,g.] by RCOMP_1:def 1;
                    f.p in {r:f.g<=r & r<=f.x1} by A44,A57;
                    then f.p in [.f.g,f.x1.] by RCOMP_1:def 1;
                    then
A60:                f.p in [.f.x1,f.g.] \/ [.f.g,f.x1.] by XBOOLE_0:def 3;
                    g in [.p,g.] by A3,XXREAL_1:1;
                    then
A61:                [.x1,g.] c= [.p,g.] by A59,XXREAL_2:def 12;
                    then f|[.x1,g.] is continuous by A4,FCONT_1:16;
                    then consider s such that
A62:                s in [.x1,g.] and
A63:                f.s=f.p by A2,A47,A61,A60,Th15,XBOOLE_1:1;
                    s in {t: x1<=t & t<=g} by A62,RCOMP_1:def 1;
                    then
A64:                ex r st r=s & x1<=r & r<=g;
                    s in [.p,g.] by A61,A62;
                    then s=p by A1,A2,A9,A63,FUNCT_1:def 4;
                    hence contradiction by A46,A58,A64,XXREAL_0:1;
                  end;
                end;
                hence contradiction;
              end;
            end;
            hence contradiction;
          end;
          consider x1,x2 such that
A65:      x1 in [.p,g.] /\ dom f and
A66:      x2 in [.p,g.] /\ dom f and
A67:      x1<x2 and
A68:      f.x1<=f.x2 by A6,RFUNCT_2:21;
A69:      x2 in [.p,g.] by A66,XBOOLE_0:def 4;
          then
A70:      [.x2,g.] c= [.p,g.] by A8,XXREAL_2:def 12;
A71:      x1 in [.p,g.] by A65,XBOOLE_0:def 4;
          then x1 in {t:p<=t & t<=g} by RCOMP_1:def 1;
          then ex r st r=x1 & p<=r & r<=g;
          then f.g <= f.x1 by A45;
          then f.x1 in {r: f.g<=r & r<=f.x2} by A68;
          then f.x1 in [.f.g,f.x2.] by RCOMP_1:def 1;
          then
A72:      f.x1 in [.f.x2,f.g.] \/ [.f.g,f.x2.] by XBOOLE_0:def 3;
          x2 in {r: p<=r & r<=g} by A69,RCOMP_1:def 1;
          then
A73:      ex r st r=x2 & p<=r & r<=g;
          g in [.p,g.] by A3,XXREAL_1:1;
          then
A74:      [.x2,g.] c= [.p,g.] by A69,XXREAL_2:def 12;
          then f|[.x2,g.] is continuous by A4,FCONT_1:16;
          then consider s such that
A75:      s in [.x2,g.] and
A76:      f.s=f.x1 by A2,A73,A72,A70,Th15,XBOOLE_1:1;
          s in {t: x2<=t & t<=g} by A75,RCOMP_1:def 1;
          then
A77:      ex r st r=s & x2<=r & r <=g;
          s in [.p,g.] by A74,A75;
          hence contradiction by A1,A2,A67,A71,A76,A77,FUNCT_1:def 4;
        end;
      end;
      hence contradiction;
    end;
  end;
  hence contradiction;
end;

theorem
  f is one-to-one & p<=g & [.p,g.] c= dom f & f|[.p,g.] is continuous
  implies lower_bound (f.:[.p,g.])=f.p & upper_bound (f.:[.p,g.])=f.g or
  lower_bound (f.:[.p,g.])=f.g & upper_bound (f.:[.p,g.])=f.p
proof
  assume that
A1: f is one-to-one and
A2: p<=g and
A3: [.p,g.] c= dom f and
A4: f|[.p,g.] is continuous;
A5: p in [.p,g.] by A2,XXREAL_1:1;
  then
A6: f.p in f.:[.p,g.] by A3,FUNCT_1:def 6;
A7: g in [.p,g.] by A2,XXREAL_1:1;
  then
A8: f.g in f.:[.p,g.] by A3,FUNCT_1:def 6;
A9: g in [.p,g.]/\dom f by A3,A7,XBOOLE_0:def 4;
A10: p in [.p,g.]/\dom f by A3,A5,XBOOLE_0:def 4;
  [.p,g.] <> {} by A2,XXREAL_1:1;
  then consider x1,x2 be real number such that
A11: x1 in [.p,g.] and
A12: x2 in [.p,g.] and
A13: f.x1=upper_bound (f.:[.p,g.]) and
A14: f.x2=lower_bound (f.:[.p,g.]) by A3,A4,FCONT_1:31,RCOMP_1:6;
A15: x2 in [.p,g.]/\dom f by A3,A12,XBOOLE_0:def 4;
  x2 in {t: p<=t & t<=g} by A12,RCOMP_1:def 1;
  then
A16: ex r st r=x2 & p<=r & r<=g;
  x1 in {r: p<=r & r<=g} by A11,RCOMP_1:def 1;
  then
A17: ex r st r=x1 & p<=r & r<=g;
  [.p,g.] is compact by RCOMP_1:6;
  then
A18: f.:[.p,g.] is real-bounded by A3,A4,FCONT_1:29,RCOMP_1:10;
A19: x1 in [.p,g.]/\dom f by A3,A11,XBOOLE_0:def 4;
  now
    per cases by A1,A2,A3,A4,Th17;
    suppose
A20:  f|[.p,g.] is increasing;
A21:  now
        assume x1<>g;
        then x1<g by A17,XXREAL_0:1;
        then f.x1<f.g by A9,A19,A20,RFUNCT_2:20;
        hence contradiction by A8,A18,A13,SEQ_4:def 1;
      end;
      now
        assume x2<>p;
        then p<x2 by A16,XXREAL_0:1;
        then f.p<f.x2 by A10,A15,A20,RFUNCT_2:20;
        hence contradiction by A6,A18,A14,SEQ_4:def 2;
      end;
      hence thesis by A13,A14,A21;
    end;
    suppose
A22:  f|[.p,g.] is decreasing;
A23:  now
        assume x2<>g;
        then x2<g by A16,XXREAL_0:1;
        then f.g<f.x2 by A9,A15,A22,RFUNCT_2:21;
        hence contradiction by A8,A18,A14,SEQ_4:def 2;
      end;
      now
        assume x1<>p;
        then p<x1 by A17,XXREAL_0:1;
        then f.x1<f.p by A10,A19,A22,RFUNCT_2:21;
        hence contradiction by A6,A18,A13,SEQ_4:def 1;
      end;
      hence thesis by A13,A14,A23;
    end;
  end;
  hence thesis;
end;


theorem Th19:
  p<=g & [.p,g.] c= dom f & f|[.p,g.] is continuous implies f.:[.p
  ,g.]=[.lower_bound (f.:[.p,g.]),upper_bound (f.:[.p,g.]).]
proof
  assume that
A1: p<=g and
A2: [.p,g.] c= dom f and
A3: f|[.p,g.] is continuous;
  [.p,g.] is compact by RCOMP_1:6;
  then
A4: f.:[.p,g.] is real-bounded by A2,A3,FCONT_1:29,RCOMP_1:10;
  now
    let y be Real;
    thus y in f.:[.p,g.] implies y in [.lower_bound(f.:[.p,g.]),upper_bound(f
    .:[.p,g.]).]
    proof
      assume
A5:   y in f.:[.p,g.];
      then
A6:   y>=lower_bound(f.:[.p,g.]) by A4,SEQ_4:def 2;
      y<=upper_bound(f.:[.p,g.]) by A4,A5,SEQ_4:def 1;
      then y in {s: lower_bound(f.:[.p,g.])<=s & s<=upper_bound(f.:[.p,g.])}
      by A6;
      hence thesis by RCOMP_1:def 1;
    end;
    assume y in [.lower_bound(f.:[.p,g.]),upper_bound(f.:[.p,g.]).];
    then ex x0 st x0 in [.p,g.] & y=f.x0 by A1,A2,A3,Th16;
    hence y in f.:[.p,g.] by A2,FUNCT_1:def 6;
  end;
  hence thesis by SUBSET_1:3;
end;

theorem
  for f be one-to-one PartFunc of REAL,REAL st p<=g & [.p,g.] c= dom f &
f|[.p,g.] is continuous holds f"|[.lower_bound (f.:[.p,g.]),upper_bound (f.:[.p
  ,g.]).] is continuous
proof
  let f be one-to-one PartFunc of REAL,REAL;
  assume that
A1: p<=g and
A2: [.p,g.] c= dom f and
A3: f|[.p,g.] is continuous;
  now
    per cases by A1,A2,A3,Th17;
    suppose
      f|[.p,g.] is increasing;
      then (f|[.p,g.])"|(f.:[.p,g.]) is increasing by RFUNCT_2:51;
      then f"|(f.:[.p,g.])|(f.:[.p,g.]) is increasing by RFUNCT_2:17;
      then f"|(f.:[.p,g.]) is monotone by RELAT_1:72;
      then
A4:   f"|[.lower_bound (f.:[.p,g.]),upper_bound (f.: [.p,g .]).] is
      monotone by A1,A2,A3,Th19;
      (f").:([.lower_bound (f.:[.p,g.]),upper_bound (f.:[.p,g.]).])=(f")
      .:(f .: [.p,g.]) by A1,A2,A3,Th19
        .=((f")|(f.:[.p,g.])).:(f.:[.p,g.]) by RELAT_1:129
        .=((f|[.p,g.])").:(f.:[.p,g.]) by RFUNCT_2:17
        .= ((f|[.p,g.])").:(rng (f|[.p,g.])) by RELAT_1:115
        .= ((f|[.p,g.])").:(dom ((f|[.p,g.])")) by FUNCT_1:33
        .= rng ((f|[.p,g.])") by RELAT_1:113
        .= dom (f|[.p,g.]) by FUNCT_1:33
        .= dom f /\ [.p,g.] by RELAT_1:61
        .= [.p,g.] by A2,XBOOLE_1:28;
      hence thesis by A1,A4,FCONT_1:46;
    end;
    suppose
      f|[.p,g.] is decreasing;
      then (f|[.p,g.])"|(f.:[.p,g.]) is decreasing by RFUNCT_2:52;
      then f"|(f.:[.p,g.])|(f.:[.p,g.]) is decreasing by RFUNCT_2:17;
      then f"|(f.:[.p,g.]) is monotone by RELAT_1:72;
      then
A5:   f"|[.lower_bound (f.:[.p,g.]),upper_bound (f.: [.p,g .]).] is
      monotone by A1,A2,A3,Th19;
      (f").:([.lower_bound (f.:[.p,g.]),upper_bound (f.:[.p,g.]).])=(f")
      .:(f .: [.p,g.]) by A1,A2,A3,Th19
        .=((f")|(f.:[.p,g.])).:(f.:[.p,g.]) by RELAT_1:129
        .=((f|[.p,g.])").:(f.:[.p,g.]) by RFUNCT_2:17
        .=((f|[.p,g.])").:(rng (f|[.p,g.])) by RELAT_1:115
        .=((f|[.p,g.])").:(dom ((f|[.p,g.])")) by FUNCT_1:33
        .=rng((f|[.p,g.])") by RELAT_1:113
        .=dom(f|[.p,g.]) by FUNCT_1:33
        .=dom f /\ [.p,g.] by RELAT_1:61
        .=[.p,g.] by A2,XBOOLE_1:28;
      hence thesis by A1,A5,FCONT_1:46;
    end;
  end;
  hence thesis;
end;














begin

reserve y,X for set;
reserve x,x0,x1,x2,g,g1,g2,r,r1,s,p,p1 for Real;
reserve n,m,k for Element of NAT;
reserve Y for Subset of REAL;
reserve Z for open Subset of REAL;
reserve s1,s3 for Real_Sequence;
reserve f,f1,f2 for PartFunc of REAL,REAL;

theorem Th1:
  (for r holds r in Y iff r in REAL) iff Y = REAL
proof
  thus (for r holds r in Y iff r in REAL) implies Y=REAL
  proof
    assume for r holds r in Y iff r in REAL;
    then for y holds y in Y iff y in REAL;
    hence thesis by TARSKI:1;
  end;
  assume
A1: Y=REAL;
  let r;
  thus r in Y implies r in REAL;
  thus thesis by A1;
end;

definition let x be real number;
  let IT be Real_Sequence;
  attr IT is x-convergent means  :Def1:
    IT is convergent & lim IT = x;
end;

registration
  cluster 0-convergent non-zero for Real_Sequence;
  existence
  proof
    deffunc F(Element of NAT) = 1/($1+1);
    consider s1 such that
A1: for n holds s1.n=F(n) from SEQ_1:sch 1;
    take s1;
    now
      let n;
      (n+1)" <> 0;
      then 1/(n+1) <> 0 by XCMPLX_1:215;
      hence s1.n <> 0 by A1;
    end; then
A2: s1 is non-zero by SEQ_1:5;
    lim s1 = 0 & s1 is convergent by A1,SEQ_4:30; then
    s1 is 0-convergent by Def1;
    hence thesis by A2;
  end;
end;

registration let f be 0-convergent Real_Sequence;
  cluster lim f -> empty;
  coherence
  proof
    thus thesis by Def1;
  end;
end;

registration
  cluster 0-convergent -> convergent for Real_Sequence;
  coherence
  proof
    let f be Real_Sequence;
    assume f is 0-convergent; then
    f is 0-convergent;
    hence thesis by Def1;
  end;
end;

reconsider cs = NAT --> 0 as Real_Sequence by FUNCOP_1:45;

reserve h for non-zero 0-convergent Real_Sequence;
reserve c for constant Real_Sequence;

definition
  let IT be PartFunc of REAL,REAL;
  attr IT is RestFunc-like means :Def2:
  IT is total & for h holds (h")(#)(IT/*h) is convergent &
  lim ((h")(#)(IT/*h)) = 0;
end;

reconsider cf = REAL --> 0 as Function of REAL, REAL by FUNCOP_1:45;

registration
  cluster RestFunc-like for PartFunc of REAL,REAL;
  existence
  proof
    take f = cf;
    thus f is total;
A1: dom f = REAL by FUNCOP_1:13;
    now
      let h;
      now
        let n be Nat;
A2:     rng h c= dom f by A1;
A3:     n in NAT by ORDINAL1:def 12;
        hence ((h")(#)(f/*h)).n = (h").n*((f/*h).n) by SEQ_1:8
          .= (h").n*(f.(h.n)) by A3,A2,FUNCT_2:108
          .= (h").n*0 by FUNCOP_1:7
          .= 0;
      end;
      then (h")(#)(f/*h) is constant & ((h")(#)(f/*h)).0 = 0 by VALUED_0:def 18
;
      hence (h")(#)(f/*h) is convergent & lim ((h")(#)(f/*h)) = 0 by SEQ_4:25;
    end;
    hence thesis;
  end;
end;

definition
  mode RestFunc is RestFunc-like PartFunc of REAL,REAL;
end;

definition
  let IT be PartFunc of REAL,REAL;
  attr IT is linear means  :Def3:
  IT is total & ex r st for p holds IT.p = r*p;
end;

registration
  cluster linear for PartFunc of REAL,REAL;
  existence
  proof
    deffunc F(Real) = 1*$1;
    defpred P[set] means $1 in REAL;
    consider f such that
A1: (for r holds r in dom f iff P[r]) & for r st r in dom f holds f.r=
    F(r) from SEQ_1:sch 3;
    take f;
    for y holds y in REAL implies y in dom f by A1;
    then REAL c= dom f by TARSKI:def 3;
    then
 dom f = REAL by XBOOLE_0:def 10;
    hence f is total by PARTFUN1:def 2;
    for p holds f.p = 1*p by A1;
    hence thesis;
  end;
end;

definition
  mode LinearFunc is linear PartFunc of REAL,REAL;
end;

reserve R,R1,R2 for RestFunc;
reserve L,L1,L2 for LinearFunc;

theorem Th2:
  L1+L2 is LinearFunc & L1-L2 is LinearFunc
proof
  consider g1 such that
A1: for p holds L1.p = g1*p by Def3;
  consider g2 such that
A2: for p holds L2.p = g2*p by Def3;
A3: L1 is total & L2 is total by Def3;
  now
    let p;
    thus (L1+L2).p = L1.p + L2.p by A3,RFUNCT_1:56
      .= g1*p + L2.p by A1
      .= g1*p + g2*p by A2
      .= (g1+g2)*p;
  end;
  hence L1+L2 is LinearFunc by A3,Def3;
  now
    let p;
    thus (L1-L2).p = L1.p - L2.p by A3,RFUNCT_1:56
      .= g1*p - L2.p by A1
      .= g1*p - g2*p by A2
      .= (g1-g2)*p;
  end;
  hence thesis by A3,Def3;
end;

theorem Th3:
  r(#)L is LinearFunc
proof
  consider g such that
A1: for p holds L.p = g*p by Def3;
A2: L is total by Def3;
  now
    let p;
    thus (r(#)L).p = r*L.p by A2,RFUNCT_1:57
      .= r*(g*p) by A1
      .= r*g*p;
  end;
  hence thesis by A2,Def3;
end;

theorem Th4:
  R1+R2 is RestFunc & R1-R2 is RestFunc & R1(#)R2 is RestFunc
proof
A1: R1 is total & R2 is total by Def2;
  now
    let h;
A2: (h")(#)((R1+R2)/*h) = (h")(#)((R1/*h)+(R2/*h)) by A1,RFUNCT_2:13
      .= ((h")(#)(R1/*h))+((h")(#)(R2/*h)) by SEQ_1:16;
A3: (h")(#)(R1/*h) is convergent & (h")(#)(R2/*h) is convergent by Def2;
    hence (h")(#)((R1+R2)/*h) is convergent by A2,SEQ_2:5;
    lim ((h")(#)(R1/*h)) = 0 & lim ((h")(#)(R2/*h)) = 0 by Def2;
    hence lim ((h")(#)((R1+R2)/*h)) = 0+0 by A3,A2,SEQ_2:6
      .= 0;
  end;
  hence R1+R2 is RestFunc by A1,Def2;
  now
    let h;
A4: (h")(#)((R1-R2)/*h) = (h")(#)((R1/*h)-(R2/*h)) by A1,RFUNCT_2:13
      .= ((h")(#)(R1/*h))-((h")(#)(R2/*h)) by SEQ_1:21;
A5: (h")(#)(R1/*h) is convergent & (h")(#)(R2/*h) is convergent by Def2;
    hence (h")(#)((R1-R2)/*h) is convergent by A4,SEQ_2:11;
    lim ((h")(#)(R1/*h)) = 0 & lim ((h")(#)(R2/*h)) = 0 by Def2;
    hence lim ((h")(#)((R1-R2)/*h)) = 0-0 by A5,A4,SEQ_2:12
      .= 0;
  end;
  hence R1-R2 is RestFunc by A1,Def2;
  now
    let h;
A6: (h")(#)(R2/*h) is convergent by Def2;
A7: h" is non-zero by SEQ_1:33;
A8: (h")(#)(R1/*h) is convergent & h is convergent by Def2;
    then
A9: h(#)((h")(#)(R1/*h)) is convergent by SEQ_2:14;
    lim ((h")(#)(R1/*h)) = 0 & lim h = 0 by Def2;
    then
A10: lim (h(#)((h")(#)(R1/*h))) = 0*0 by A8,SEQ_2:15
      .= 0;
A11: (h")(#)((R1(#)R2)/*h) = ((R1/*h)(#)(R2/*h))/"h by A1,RFUNCT_2:13
      .= ((R1/*h)(#)(R2/*h)(#)(h"))/"(h(#)(h")) by SEQ_1:43,A7
      .= ((R1/*h)(#)(R2/*h)(#)(h"))(#)((h"")(#)(h")) by SEQ_1:36
      .= h(#)(h")(#)((R1/*h)(#)((h")(#)(R2/*h))) by SEQ_1:14
      .= h(#)(h")(#)(R1/*h)(#)((h")(#)(R2/*h)) by SEQ_1:14
      .= h(#)((h")(#)(R1/*h))(#)((h")(#)(R2/*h)) by SEQ_1:14;
    hence (h")(#)((R1(#)R2)/*h) is convergent by A6,A9,SEQ_2:14;
    lim ((h")(#)(R2/*h)) = 0 by Def2;
    hence lim ((h")(#)((R1(#)R2)/*h)) = 0*0 by A6,A9,A10,A11,SEQ_2:15
      .= 0;
  end;
  hence thesis by A1,Def2;
end;

theorem Th5:
  r(#)R is RestFunc
proof
A1: R is total by Def2;
  now
    let h;
A2: (h")(#)((r(#)R)/*h) = (h")(#)(r(#)(R/*h)) by A1,RFUNCT_2:14
      .= r(#)((h")(#)(R/*h)) by SEQ_1:19;
A3: (h")(#)(R/*h) is convergent by Def2;
    hence (h")(#)((r(#)R)/*h) is convergent by A2,SEQ_2:7;
    lim ((h")(#)(R/*h)) = 0 by Def2;
    hence lim ((h")(#)((r(#)R)/*h)) = r*0 by A3,A2,SEQ_2:8
      .= 0;
  end;
  hence thesis by A1,Def2;
end;

theorem Th6:
  L1(#)L2 is RestFunc-like
proof
  consider x1 such that
A1: for p holds L1.p=x1*p by Def3;
A2: L1 is total & L2 is total by Def3;
  hence L1(#)L2 is total;
  consider x2 such that
A3: for p holds L2.p=x2*p by Def3;
  now
    let h;
    now
      let n;
A4:   h.n<>0 by SEQ_1:5;
      thus ((h")(#)((L1(#)L2)/*h)).n=(h").n *((L1(#)L2)/*h).n by SEQ_1:8
        .=(h").n*(L1(#)L2).(h.n) by A2,FUNCT_2:115
        .=(h").n*(L1.(h.n)*L2.(h.n)) by A2,RFUNCT_1:56
        .=(h").n*L1.(h.n)*L2.(h.n)
        .=((h.n)")*L1.(h.n)*L2.(h.n) by VALUED_1:10
        .=((h.n)")*((h.n)*x1)*L2.(h.n) by A1
        .=((h.n)")*(h.n)*x1*L2.(h.n)
        .=1*x1*L2.(h.n) by A4,XCMPLX_0:def 7
        .=x1*(x2*(h.n)) by A3
        .=x1*x2*(h.n)
        .=((x1*x2)(#)h).n by SEQ_1:9;
    end;
    then
A5: (h")(#)((L1(#)L2)/*h)=(x1*x2)(#)h by FUNCT_2:63;
    thus (h")(#)((L1(#)L2)/*h) is convergent by A5,SEQ_2:7;
    lim h=0;
    hence lim ((h")(#)((L1(#)L2)/*h)) = (x1*x2)*0 by A5,SEQ_2:8
      .=0;
  end;
  hence thesis;
end;

theorem Th7:
  R(#)L is RestFunc & L(#)R is RestFunc
proof
A1: L is total by Def3;
  consider x1 such that
A2: for p holds L.p=x1*p by Def3;
A3: R is total by Def2;
A4: now
    let h;
A5: (h")(#)(R/*h) is convergent by Def2;
    now
      let n;
      thus (L/*h).n=L.(h.n) by A1,FUNCT_2:115
        .=x1*(h.n) by A2
        .=(x1(#)h).n by SEQ_1:9;
    end;
    then
A6: (L/*h)=x1(#)h by FUNCT_2:63;
A7: L/*h is convergent by A6,SEQ_2:7;
    lim h=0;
    then
A8: lim (L/*h)=x1*0 by A6,SEQ_2:8
      .=0;
A9: (h")(#)((R(#)L)/*h)=(h")(#)((R/*h)(#)(L/*h)) by A3,A1,RFUNCT_2:13
      .=((h")(#)(R/*h))(#)(L/*h) by SEQ_1:14;
    hence (h")(#)((R(#)L)/*h) is convergent by A7,A5,SEQ_2:14;
    lim ((h")(#)(R/*h))=0 by Def2;
    hence lim ((h")(#)((R(#)L)/*h))=0*0 by A9,A7,A8,A5,SEQ_2:15
      .=0;
  end;
  hence R(#)L is RestFunc by A3,A1,Def2;
  thus thesis by A3,A1,A4,Def2;
end;

definition
  let f;
  let x0 be real number;
  pred f is_differentiable_in x0 means
  :Def4:
  ex N being Neighbourhood of x0
st N c= dom f & ex L,R st for x st x in N holds f.x - f.x0 = L.(x-x0) + R.(x-x0
  );
end;

definition
  let f;
  let x0 be real number;
  assume
A1: f is_differentiable_in x0;
  func diff(f,x0) -> Real means
  :Def5:
  ex N being Neighbourhood of x0 st N c= dom f &
  ex L,R st it=L.1 & for x st x in N holds f.x-f.x0 = L.(x-x0) + R.(x-x0);
  existence
  proof
    consider N being Neighbourhood of x0 such that
A2: N c= dom f and
A3: ex L,R st for x st x in N holds f.x - f.x0 = L.(x-x0) + R.(x-x0)
    by A1,Def4;
    consider L,R such that
A4: for x st x in N holds f.x - f.x0 = L.(x-x0) + R.(x-x0) by A3;
    consider r such that
A5: for p holds L.p = r*p by Def3;
    take r;
    L.1=r*1 by A5
      .=r;
    hence thesis by A2,A4;
  end;
  uniqueness
  proof
    let r,s;
    assume that
A6: ex N being Neighbourhood of x0 st N c= dom f & ex L,R st r=L.1 &
    for x st x in N holds f.x-f.x0 = L.(x-x0)+R.(x-x0) and
A7: ex N being Neighbourhood of x0 st N c= dom f & ex L,R st s=L.1 &
    for x st x in N holds f.x-f.x0 = L.(x-x0) + R.(x-x0);
    consider N being Neighbourhood of x0 such that
    N c= dom f and
A8: ex L,R st r=L.1 & for x st x in N holds f.x-f.x0 = L.(x-x0)+R.(x-
    x0) by A6;
    consider L,R such that
A9: r=L.1 and
A10: for x st x in N holds f.x-f.x0 = L.(x-x0) + R.(x-x0) by A8;
    consider r1 such that
A11: for p holds L.p = r1*p by Def3;
    consider N1 being Neighbourhood of x0 such that
    N1 c= dom f and
A12: ex L,R st s=L.1 & for x st x in N1 holds f.x-f.x0=L.(x-x0)+R.(x-
    x0) by A7;
    consider L1,R1 such that
A13: s =L1.1 and
A14: for x st x in N1 holds f.x-f.x0 = L1.(x-x0) + R1.(x-x0) by A12;
    consider p1 such that
A15: for p holds L1.p = p1*p by Def3;
    consider N0 be Neighbourhood of x0 such that
A16: N0 c= N & N0 c= N1 by RCOMP_1:17;
    consider g be real number such that
A17: 0<g and
A18: N0 = ].x0-g,x0+g.[ by RCOMP_1:def 6;
    deffunc F(Element of NAT) = g/($1+2);
    consider s1 such that
A19: for n holds s1.n = F(n) from SEQ_1:sch 1;
    now
      let n;
      0<>g/(n+2) by A17,XREAL_1:139;
      hence 0<>s1.n by A19;
    end;
    then
A20: s1 is non-zero by SEQ_1:5;
    s1 is convergent & lim s1 = 0 by A19,SEQ_4:31; then
    s1 is 0-convergent by Def1;
    then reconsider h = s1 as non-zero 0-convergent Real_Sequence
      by A20;
A21: for n ex x st x in N & x in N1 & h.n=x-x0
    proof
      let n;
      take x0+g/(n+2);
      0+1<n+1+1 by XREAL_1:6;
      then g/(n+2)<g/1 by A17,XREAL_1:76;
      then
A22:  x0 + g/(n+2)<x0 + g by XREAL_1:6;
      0<g/(n+2) by A17,XREAL_1:139;
      then x0+-g<x0+g/(n+2) by A17,XREAL_1:6;
      then x0+g/(n+2) in ].x0-g,x0+g.[ by A22;
      hence thesis by A16,A18,A19;
    end;
A23: s=p1*1 by A13,A15;
A24: r=r1*1 by A9,A11;
A25: now
      let x;
      assume that
A26:  x in N and
A27:  x in N1;
      f.x-f.x0 = L.(x-x0) + R.(x-x0) by A10,A26;
      then L.(x-x0) + R.(x-x0) = L1.(x-x0) + R1.(x-x0) by A14,A27;
      then r1*(x-x0) + R.(x-x0) = L1.(x-x0) + R1.(x-x0) by A11;
      hence r*(x-x0) + R.(x-x0) = s*(x-x0) + R1.(x-x0) by A15,A24,A23;
    end;
    now
      R1 is total by Def2;
      then dom R1 = REAL by PARTFUN1:def 2;
      then
A28:  rng h c= dom R1;
      let n be Nat;
      R is total by Def2;
      then dom R = REAL by PARTFUN1:def 2;
      then
A29:  rng h c= dom R;
A30:  n in NAT by ORDINAL1:def 12;
      then ex x st x in N & x in N1 & h.n=x-x0 by A21;
      then r*(h.n) + R.(h.n) = s*(h.n) + R1.(h.n) by A25;
      then
A31:  (r*(h.n))/(h.n)+(R.(h.n))/(h.n)=(s*(h.n)+R1.(h.n))/(h.n) by XCMPLX_1:62;
A32:  (R.(h.n))/(h.n) = (R.(h.n))*(h.n)" by XCMPLX_0:def 9
        .= (R.(h.n))*(h".n) by VALUED_1:10
        .= ((R/*h).n)*(h".n) by A30,A29,FUNCT_2:108
        .= ((h")(#)(R/*h)).n by A30,SEQ_1:8;
A33:  h.n <> 0 by A30,SEQ_1:5;
A34:  (R1.(h.n))/(h.n) = (R1.(h.n))*(h.n)" by XCMPLX_0:def 9
        .= (R1.(h.n))*(h".n) by VALUED_1:10
        .= ((R1/*h).n)*(h".n) by A30,A28,FUNCT_2:108
        .= ((h")(#)(R1/*h)).n by A30,SEQ_1:8;
A35:  (s*(h.n))/(h.n) = s*((h.n)/(h.n)) by XCMPLX_1:74
        .= s*1 by A33,XCMPLX_1:60
        .= s;
      (r*(h.n))/(h.n) = r*((h.n)/(h.n)) by XCMPLX_1:74
        .= r*1 by A33,XCMPLX_1:60
        .= r;
      then r + (R.(h.n))/(h.n) = s + (R1.(h.n))/(h.n) by A31,A35,XCMPLX_1:62;
      then r = s + (((h")(#)(R1/*h)).n - ((h")(#)(R/*h)).n) by A32,A34;
      hence r - s = (((h")(#)(R1/*h))-((h")(#)(R/*h))).n by A30,RFUNCT_2:1;
    end;
    then
    ((h")(#)(R1/*h))-((h")(#)(R/*h)) is constant & (((h")(#)(R1/*h))-((h"
    )(#)(R /*h))).1 = r-s by VALUED_0:def 18;
    then
A36: lim (((h")(#)(R1/*h))-((h")(#)(R/*h))) = r-s by SEQ_4:25;
A37: (h")(#)(R1/*h) is convergent & lim ((h")(#)(R1/*h)) = 0 by Def2;
    (h")(#)(R/*h) is convergent & lim ((h")(#)(R/*h)) = 0 by Def2;
    then r-s = 0-0 by A36,A37,SEQ_2:12;
    hence thesis;
  end;
end;

definition
  let f,X;
  pred f is_differentiable_on X means
  :Def6:
  X c= dom f & for x st x in X holds f|X is_differentiable_in x;
end;

theorem Th8:
  f is_differentiable_on X implies X is Subset of REAL
proof
  assume f is_differentiable_on X;
  then X c=dom f by Def6;
  hence thesis by XBOOLE_1:1;
end;

theorem Th9:
  f is_differentiable_on Z iff Z c= dom f & for x st x in Z holds
  f is_differentiable_in x
proof
  thus f is_differentiable_on Z implies Z c=dom f & for x st x in Z holds f
  is_differentiable_in x
  proof
    assume
A1: f is_differentiable_on Z;
    hence Z c=dom f by Def6;
    let x0;
    assume
A2: x0 in Z;
    then f|Z is_differentiable_in x0 by A1,Def6;
    then consider N being Neighbourhood of x0 such that
A3: N c= dom(f|Z) and
A4: ex L,R st for x st x in N holds (f|Z).x-(f|Z).x0=L.(x-x0)+R.(x-x0) by Def4;
    take N;
    dom(f|Z)=dom f/\Z by RELAT_1:61;
    then dom(f|Z) c=dom f by XBOOLE_1:17;
    hence N c= dom f by A3,XBOOLE_1:1;
    consider L,R such that
A5: for x st x in N holds (f|Z).x - (f|Z).x0 = L.(x-x0) + R.(x-x0) by A4;
    take L,R;
    let x;
    assume
A6: x in N;
    then (f|Z).x-(f|Z).x0=L.(x-x0)+R.(x-x0) by A5;
    then f.x-(f|Z).x0=L.(x-x0)+R.(x-x0) by A3,A6,FUNCT_1:47;
    hence thesis by A2,FUNCT_1:49;
  end;
  assume that
A7: Z c=dom f and
A8: for x st x in Z holds f is_differentiable_in x;
  thus Z c=dom f by A7;
  let x0;
  assume
A9: x0 in Z;
  then consider N1 being Neighbourhood of x0 such that
A10: N1 c= Z by RCOMP_1:18;
  f is_differentiable_in x0 by A8,A9;
  then consider N being Neighbourhood of x0 such that
A11: N c= dom f and
A12: ex L,R st for x st x in N holds f.x-f.x0=L.(x-x0)+R.(x-x0) by Def4;
  consider N2 being Neighbourhood of x0 such that
A13: N2 c= N1 and
A14: N2 c= N by RCOMP_1:17;
A15: N2 c= Z by A10,A13,XBOOLE_1:1;
  take N2;
  N2 c= dom f by A11,A14,XBOOLE_1:1;
  then N2 c= dom f/\Z by A15,XBOOLE_1:19;
  hence
A16: N2 c= dom(f|Z) by RELAT_1:61;
  consider L,R such that
A17: for x st x in N holds f.x-f.x0=L.(x-x0)+R.(x-x0) by A12;
A18: x0 in N2 by RCOMP_1:16;
  take L,R;
  let x;
  assume
A19: x in N2;
  then f.x-f.x0=L.(x-x0)+R.(x-x0) by A14,A17;
  then (f|Z).x-f.x0=L.(x-x0)+R.(x-x0) by A16,A19,FUNCT_1:47;
  hence thesis by A16,A18,FUNCT_1:47;
end;

theorem
  f is_differentiable_on Y implies Y is open
proof
  assume
A1: f is_differentiable_on Y;
  now
    let x0 be real number;
    assume x0 in Y;
    then f|Y is_differentiable_in x0 by A1,Def6;
    then consider N being Neighbourhood of x0 such that
A2: N c= dom(f|Y) and
    ex L,R st for x st x in N holds (f|Y).x-(f|Y).x0=L.(x-x0)+R.(x-x0) by Def4;
    take N;
    thus N c= Y by A2,XBOOLE_1:1;
  end;
  hence thesis by RCOMP_1:20;
end;

definition
  let f,X;
  assume
A1: f is_differentiable_on X;
  func f`|X -> PartFunc of REAL,REAL means
  :Def7:
  dom it = X & for x st x in X holds it.x = diff(f,x);
  existence
  proof
    deffunc F(Real) = diff(f,$1);
    defpred P[set] means $1 in X;
    consider F being PartFunc of REAL,REAL such that
A2: (for x holds x in dom F iff P[x]) & for x st x in dom F holds F.x
    = F(x) from SEQ_1:sch 3;
    take F;
    now
A3:   X is Subset of REAL by A1,Th8;
      let y;
      assume y in X;
      hence y in dom F by A2,A3;
    end;
    then
A4: X c= dom F by TARSKI:def 3;
    for y st y in dom F holds y in X by A2;
    then dom F c= X by TARSKI:def 3;
    hence dom F = X by A4,XBOOLE_0:def 10;
    now
      let x;
      assume x in X;
      then x in dom F by A2;
      hence F.x = diff(f,x) by A2;
    end;
    hence thesis;
  end;
  uniqueness
  proof
    let F,G be PartFunc of REAL,REAL;
    assume that
A5: dom F = X and
A6: for x st x in X holds F.x = diff(f,x) and
A7: dom G = X and
A8: for x st x in X holds G.x = diff(f,x);
    now
      let x;
      assume
A9:   x in dom F;
      then F.x = diff(f,x) by A5,A6;
      hence F.x=G.x by A5,A8,A9;
    end;
    hence thesis by A5,A7,PARTFUN1:5;
  end;
end;

theorem
  (Z c= dom f & ex r st rng f = {r}) implies f is_differentiable_on Z &
  for x st x in Z holds (f`|Z).x = 0
proof
  set R = cf;
A1: dom R = REAL by FUNCOP_1:13;
  now
    let h;
A2: now
      let n be Nat;
A3:   rng h c= dom R by A1;
A4:   n in NAT by ORDINAL1:def 12;
      hence ((h")(#)(R/*h)).n = (h".n)*((R/*h).n) by SEQ_1:8
        .= (h".n)*(R.(h.n)) by A4,A3,FUNCT_2:108
        .= (h".n)*0 by FUNCOP_1:7
        .= 0;
    end;
    then
A5: (h")(#)(R/*h) is constant by VALUED_0:def 18;
    hence (h")(#)(R/*h) is convergent;
    ((h")(#)(R/*h)).0 = 0 by A2;
    hence lim ((h")(#)(R/*h)) = 0 by A5,SEQ_4:25;
  end;
  then reconsider R as RestFunc by Def2;
  set L = cf;
  for p holds L.p = 0*p by FUNCOP_1:7;
  then reconsider L as LinearFunc by Def3;
  assume
A6: Z c= dom f;
  given r such that
A7: rng f = {r};
A8: now
    let x0;
    assume x0 in dom f;
    then f.x0 in {r} by A7,FUNCT_1:def 3;
    hence f.x0 = r by TARSKI:def 1;
  end;
A9: now
    let x0;
    assume
A10: x0 in Z;
    then consider N being Neighbourhood of x0 such that
A11: N c= Z by RCOMP_1:18;
A12: N c= dom f by A6,A11,XBOOLE_1:1;
    for x st x in N holds f.x - f.x0 = L.(x-x0) + R.(x-x0)
    proof
      let x;
      assume x in N;
      hence f.x - f.x0 = r - f.x0 by A8,A12
        .= r - r by A6,A8,A10
        .= L.(x-x0)+0 by FUNCOP_1:7
        .= L.(x-x0)+R.(x-x0) by FUNCOP_1:7;
    end;
    hence f is_differentiable_in x0 by A12,Def4;
  end;
  hence
A13: f is_differentiable_on Z by A6,Th9;
  let x0;
  assume
A14: x0 in Z;
  then
A15: f is_differentiable_in x0 by A9;
  then ex N being Neighbourhood of x0 st N c= dom f & ex L,R st for x st x in
  N holds f.x-f.x0=L.(x-x0)+R.(x-x0) by Def4;
  then consider N being Neighbourhood of x0 such that
A16: N c= dom f;
A17: for x st x in N holds f.x - f.x0 = L.(x-x0) + R.(x-x0)
  proof
    let x;
    assume x in N;
    hence f.x - f.x0 = r - f.x0 by A8,A16
      .=r - r by A6,A8,A14
      .=L.(x-x0) + 0 by FUNCOP_1:7
      .=L.(x-x0) + R.(x-x0) by FUNCOP_1:7;
  end;
  thus (f`|Z).x0 = diff(f,x0) by A13,A14,Def7
    .= L.1 by A15,A16,A17,Def5
    .=0 by FUNCOP_1:7;
end;

registration
  let h,n;
  cluster h^\n -> non-zero 0-convergent for Real_Sequence;
  coherence
  proof
    lim h = 0;
    then lim (h^\n) = 0 by SEQ_4:20; then
    h^\n is 0-convergent by Def1;
    hence thesis;
  end;
end;

theorem Th12:
  for x0 being real number for N being Neighbourhood of x0 st f
is_differentiable_in x0 & N c= dom f holds for h,c st rng c = {x0} & rng (h+c)
c= N holds h"(#)(f/*(h+c) - f/*c) is convergent & diff(f,x0) = lim (h"(#)(f/*(h
  +c) - f/*c))
proof
  let x0 be real number;
  let N be Neighbourhood of x0;
  assume that
A1: f is_differentiable_in x0 and
A2: N c= dom f;
  consider N1 be Neighbourhood of x0 such that
  N1 c= dom f and
A3: ex L,R st for x st x in N1 holds f.x - f.x0 = L.(x-x0) + R.(x-x0) by A1
,Def4;
  consider N2 be Neighbourhood of x0 such that
A4: N2 c= N and
A5: N2 c= N1 by RCOMP_1:17;
A6: N2 c= dom f by A2,A4,XBOOLE_1:1;
  let h,c such that
A7: rng c = {x0} and
A8: rng (h+c) c= N;
  consider g be real number such that
A9: 0<g and
A10: N2 = ].x0-g,x0+g.[ by RCOMP_1:def 6;
  x0 + 0 < x0 + g & x0 - g < x0 - 0 by A9,XREAL_1:8,15;
  then
A11: x0 in N2 by A10;
A12: rng c c= dom f
  proof
    let y;
    assume y in rng c;
    then y = x0 by A7,TARSKI:def 1;
    then y in N by A4,A11;
    hence thesis by A2;
  end;
  ex n st rng (c^\n) c= N2 & rng ((h+c)^\n) c= N2
  proof
    x0 in rng c by A7,TARSKI:def 1;
    then
A13: lim c = x0 by SEQ_4:25;
A14: h + c is convergent by SEQ_2:5;
    lim h = 0;
    then lim (h + c) = 0 + x0 by A13,SEQ_2:6
      .= x0;
    then consider n such that
A15: for m being Element of NAT st n<=m holds abs((h+c).m-x0)<g by A9,A14,
SEQ_2:def 7;
    take n;
A16: rng (c^\n) = {x0} by A7,VALUED_0:26;
    thus rng (c^\n) c= N2
    proof
      let y;
      assume y in rng (c^\n);
      hence thesis by A11,A16,TARSKI:def 1;
    end;
    let y;
    assume y in rng ((h+c)^\n);
    then consider m such that
A17: y = ((h+c)^\n).m by FUNCT_2:113;
    n + 0 <= n+m by XREAL_1:7;
    then
A18: abs((h+c).(n+m)-x0)<g by A15;
    then (h+c).(m+n) - x0 < g by SEQ_2:1;
    then ((h+c)^\n).m - x0 < g by NAT_1:def 3;
    then
A19: ((h+c)^\n).m < x0 + g by XREAL_1:19;
    -g < (h+c).(m+n) - x0 by A18,SEQ_2:1;
    then -g < ((h+c)^\n).m - x0 by NAT_1:def 3;
    then x0 +-g < ((h+c)^\n).m by XREAL_1:20;
    hence thesis by A10,A17,A19;
  end;
  then consider n such that
  rng (c^\n) c= N2 and
A20: rng ((h+c)^\n) c= N2;
  consider L,R such that
A21: for x st x in N1 holds f.x - f.x0 = L.(x-x0) + R.(x-x0) by A3;
A22: rng (c^\n) c= dom f
  proof
    let y;
    assume
A23: y in rng (c^\n);
    rng (c^\n) = rng c by VALUED_0:26;
    then y = x0 by A7,A23,TARSKI:def 1;
    then y in N by A4,A11;
    hence thesis by A2;
  end;
A24: L is total by Def3;
A25: (L/*(h^\n) + R/*(h^\n))(#)(h^\n)" is convergent & lim ((L/*(h^\n) + R/*
  (h^\n))(#)(h^\n)") = L.1
  proof
    deffunc F(Element of NAT) = L.1 + ((R/*(h^\n))(#)(h^\n)").$1;
    consider s1 such that
A26: for k holds s1.k = F(k) from SEQ_1:sch 1;
A27: now
A28:  ((h^\n)")(#)(R/*(h^\n)) is convergent & lim (((h^\n)")(#) (R/*(h^\n
      )))=0 by Def2;
      let r be real number;
      assume 0<r;
      then consider m such that
A29:  for k st m<=k holds abs((((h^\n)")(#) (R/*(h^\n))).k - 0)<r by A28,
SEQ_2:def 7;
      take n1=m;
      let k such that
A30:  n1<=k;
      abs(s1.k -L.1 ) = abs(L.1 + ((R/*(h^\n))(#)(h^\n)").k - L.1 ) by A26
        .= abs( (((h^\n)")(#)(R/*(h^\n))).k - 0 );
      hence abs(s1.k -L.1 )<r by A29,A30;
    end;
    consider s such that
A31: for p1 holds L.p1=s*p1 by Def3;
A32: L.1 = s*1 by A31
      .= s;
    now
      let m;
A33:  (h^\n).m <> 0 by SEQ_1:5;
      thus ((L/*(h^\n) + R/*(h^\n))(#)(h^\n)").m = ((L/*(h^\n) + R/*(h^\n)).m)
      *((h^\n)").m by SEQ_1:8
        .= ((L/*(h^\n)).m + (R/*(h^\n)).m) *((h^\n)").m by SEQ_1:7
        .= ((L/*(h^\n)).m)*((h^\n)").m+((R/*(h^\n)).m)*((h^\n)").m
        .= ((L/*(h^\n)).m)*((h^\n)").m + ((R/*(h^\n))(#)(h^\n)").m by SEQ_1:8
        .= ((L/*(h^\n)).m)*((h^\n).m)" + ((R/*(h^\n))(#)(h^\n)").m by
VALUED_1:10
        .= (L.((h^\n).m))*((h^\n).m)" + ((R/*(h^\n))(#) (h^\n)").m by A24,
FUNCT_2:115
        .= (s*((h^\n).m))*((h^\n).m)" + ((R/*(h^\n))(#)(h^\n)").m by A31
        .= s*(((h^\n).m)*((h^\n).m)") + ((R/*(h^\n))(#)(h^\n)").m
        .= s*1 + ((R/*(h^\n))(#)(h^\n)").m by A33,XCMPLX_0:def 7
        .= s1.m by A26,A32;
    end;
    then
A34: (L/*(h^\n) + R/*(h^\n))(#)(h^\n)" = s1 by FUNCT_2:63;
    hence (L/*(h^\n) + R/*(h^\n))(#)(h^\n)" is convergent by A27,SEQ_2:def 6;
    hence thesis by A34,A27,SEQ_2:def 7;
  end;
A35: rng ((h+c)^\n) c= dom f
  proof
    let y;
    assume y in rng ((h+c)^\n);
    then y in N2 by A20;
    then y in N by A4;
    hence thesis by A2;
  end;
A36: rng (h+c) c= dom f
  proof
    let y;
    assume y in rng (h+c);
    then y in N by A8;
    hence thesis by A2;
  end;
A37: for k holds f.(((h+c)^\n).k) - f.((c^\n).k) = L.((h^\n).k) + R.((h^\n). k)
  proof
    let k;
    ((h+c)^\n).k in rng ((h+c)^\n) by VALUED_0:28;
    then
A38: ((h+c)^\n).k in N2 by A20;
    (c^\n).k in rng (c^\n) & rng (c^\n) = rng c by VALUED_0:26,28;
    then
A39: (c^\n).k = x0 by A7,TARSKI:def 1;
    ((h+c)^\n).k - (c^\n).k = (h^\n + c^\n).k - (c^\n).k by SEQM_3:15
      .= (h^\n).k + (c^\n).k - (c^\n).k by SEQ_1:7
      .= (h^\n).k;
    hence thesis by A21,A5,A38,A39;
  end;
A40: R is total by Def2;
  now
    let k;
    thus (f/*((h+c)^\n)-f/*(c^\n)).k = (f/*((h+c)^\n)).k-(f/*(c^\n)).k by
RFUNCT_2:1
      .= f.(((h+c)^\n).k) - (f/*(c^\n)).k by A35,FUNCT_2:108
      .= f.(((h+c)^\n).k) - f.((c^\n).k) by A22,FUNCT_2:108
      .= L.((h^\n).k) + R.((h^\n).k) by A37
      .= (L/*(h^\n)).k + R.((h^\n).k) by A24,FUNCT_2:115
      .= (L/*(h^\n)).k + (R/*(h^\n)).k by A40,FUNCT_2:115
      .= (L/*(h^\n) + R/*(h^\n)).k by SEQ_1:7;
  end;
  then f/*((h+c)^\n) - f/*(c^\n) = L/*(h^\n) + R/*(h^\n) by FUNCT_2:63;
  then
A41: ((L/*(h^\n) + R/*(h^\n))(#)(h^\n)") = ((((f/*(h+c))^\n)-f/*(c^\n))(#) (
  h^\n)") by A36,VALUED_0:27
    .=((((f/*(h+c))^\n)-((f/*c)^\n))(#)(h^\n)") by A12,VALUED_0:27
    .=((((f/*(h+c))-(f/*c))^\n)(#)(h^\n)") by SEQM_3:17
    .=((((f/*(h+c))-(f/*c))^\n)(#)((h")^\n)) by SEQM_3:18
    .=((((f/*(h+c))-(f/*c))(#) h")^\n) by SEQM_3:19;
  then
A42: L.1 = lim ((h")(#)((f/*(h+c)) - (f/*c))) by A25,SEQ_4:22;
  thus h" (#) (f/*(h+c) - f/*c) is convergent by A25,A41,SEQ_4:21;
  for x st x in N2 holds f.x - f.x0 = L.(x-x0) + R.(x-x0) by A21,A5;
  hence thesis by A1,A6,A42,Def5;
end;

theorem Th13:
  f1 is_differentiable_in x0 & f2 is_differentiable_in x0 implies
  f1+f2 is_differentiable_in x0 & diff(f1+f2,x0)=diff(f1,x0)+diff(f2,x0)
proof
  assume that
A1: f1 is_differentiable_in x0 and
A2: f2 is_differentiable_in x0;
  consider N1 be Neighbourhood of x0 such that
A3: N1 c= dom f1 and
A4: ex L,R st for x st x in N1 holds f1.x - f1.x0 = L.(x-x0) + R.(x-x0)
  by A1,Def4;
  consider L1,R1 such that
A5: for x st x in N1 holds f1.x - f1.x0 = L1.(x-x0) + R1.(x-x0) by A4;
  consider N2 be Neighbourhood of x0 such that
A6: N2 c= dom f2 and
A7: ex L,R st for x st x in N2 holds f2.x - f2.x0 = L.(x-x0) + R.(x-x0)
  by A2,Def4;
  consider L2,R2 such that
A8: for x st x in N2 holds f2.x - f2.x0 = L2.(x-x0) + R2.(x-x0) by A7;
  reconsider R=R1+R2 as RestFunc by Th4;
  reconsider L=L1+L2 as LinearFunc by Th2;
A9: L1 is total & L2 is total by Def3;
  consider N be Neighbourhood of x0 such that
A10: N c= N1 and
A11: N c= N2 by RCOMP_1:17;
A12: N c= dom f2 by A6,A11,XBOOLE_1:1;
  N c= dom f1 by A3,A10,XBOOLE_1:1;
  then N /\ N c= dom f1 /\ dom f2 by A12,XBOOLE_1:27;
  then
A13: N c= dom (f1+f2) by VALUED_1:def 1;
A14: R1 is total & R2 is total by Def2;
A15: now
    let x;
A16: x0 in N by RCOMP_1:16;
    assume
A17: x in N;
    hence (f1+f2).x - (f1+f2).x0 = (f1.x+f2.x) - (f1+f2).x0 by A13,
VALUED_1:def 1
      .=f1.x+f2.x - (f1.x0+f2.x0) by A13,A16,VALUED_1:def 1
      .=(f1.x - f1.x0) + (f2.x - f2.x0)
      .=L1.(x-x0)+R1.(x-x0) + (f2.x - f2.x0) by A5,A10,A17
      .=L1.(x-x0)+R1.(x-x0) + (L2.(x-x0) + R2.(x-x0)) by A8,A11,A17
      .=(L1.(x-x0)+L2.(x-x0)) + (R1.(x-x0) + R2.(x-x0))
      .=L.(x-x0)+(R1.(x-x0) + R2.(x-x0)) by A9,RFUNCT_1:56
      .=L.(x-x0)+R.(x-x0) by A14,RFUNCT_1:56;
  end;
  hence f1+f2 is_differentiable_in x0 by A13,Def4;
  hence diff(f1+f2,x0)=L.1 by A13,A15,Def5
    .=L1.1 + L2.1 by A9,RFUNCT_1:56
    .=diff(f1,x0) + L2.1 by A1,A3,A5,Def5
    .=diff(f1,x0) + diff(f2,x0) by A2,A6,A8,Def5;
end;

theorem Th14:
  f1 is_differentiable_in x0 & f2 is_differentiable_in x0 implies
  f1-f2 is_differentiable_in x0 & diff(f1-f2,x0)=diff(f1,x0)-diff(f2,x0)
proof
  assume that
A1: f1 is_differentiable_in x0 and
A2: f2 is_differentiable_in x0;
  consider N1 be Neighbourhood of x0 such that
A3: N1 c= dom f1 and
A4: ex L,R st for x st x in N1 holds f1.x - f1.x0 = L.(x-x0) + R.(x-x0)
  by A1,Def4;
  consider L1,R1 such that
A5: for x st x in N1 holds f1.x - f1.x0 = L1.(x-x0) + R1.(x-x0) by A4;
  consider N2 be Neighbourhood of x0 such that
A6: N2 c= dom f2 and
A7: ex L,R st for x st x in N2 holds f2.x - f2.x0 = L.(x-x0) + R.(x-x0)
  by A2,Def4;
  consider L2,R2 such that
A8: for x st x in N2 holds f2.x - f2.x0 = L2.(x-x0) + R2.(x-x0) by A7;
  reconsider R=R1-R2 as RestFunc by Th4;
  reconsider L=L1-L2 as LinearFunc by Th2;
A9: L1 is total & L2 is total by Def3;
  consider N be Neighbourhood of x0 such that
A10: N c= N1 and
A11: N c= N2 by RCOMP_1:17;
A12: N c= dom f2 by A6,A11,XBOOLE_1:1;
  N c= dom f1 by A3,A10,XBOOLE_1:1;
  then N /\ N c= dom f1 /\ dom f2 by A12,XBOOLE_1:27;
  then
A13: N c= dom (f1-f2) by VALUED_1:12;
A14: R1 is total & R2 is total by Def2;
A15: now
    let x;
A16: x0 in N by RCOMP_1:16;
    assume
A17: x in N;
    hence (f1-f2).x - (f1-f2).x0 = (f1.x-f2.x) - (f1-f2).x0 by A13,VALUED_1:13
      .=f1.x - f2.x - (f1.x0-f2.x0) by A13,A16,VALUED_1:13
      .=f1.x - f1.x0 - (f2.x - f2.x0)
      .=L1.(x-x0) + R1.(x-x0) - (f2.x - f2.x0) by A5,A10,A17
      .=L1.(x-x0) + R1.(x-x0) - (L2.(x-x0) + R2.(x-x0)) by A8,A11,A17
      .=L1.(x-x0) - L2.(x-x0) + (R1.(x-x0) - R2.(x-x0))
      .=L.(x-x0) + (R1.(x-x0) - R2.(x-x0)) by A9,RFUNCT_1:56
      .=L.(x-x0) + R.(x-x0) by A14,RFUNCT_1:56;
  end;
  hence f1-f2 is_differentiable_in x0 by A13,Def4;
  hence diff(f1-f2,x0)=L.1 by A13,A15,Def5
    .=L1.1 - L2.1 by A9,RFUNCT_1:56
    .=diff(f1,x0) - L2.1 by A1,A3,A5,Def5
    .=diff(f1,x0) - diff(f2,x0) by A2,A6,A8,Def5;
end;

theorem Th15:
  f is_differentiable_in x0 implies r(#)f is_differentiable_in x0
  & diff((r(#)f),x0) = r*diff(f,x0)
proof
  assume
A1: f is_differentiable_in x0;
  then consider N1 be Neighbourhood of x0 such that
A2: N1 c= dom f and
A3: ex L,R st for x st x in N1 holds f.x - f.x0 = L.(x-x0) + R.(x-x0) by Def4;
  consider L1,R1 such that
A4: for x st x in N1 holds f.x - f.x0 = L1.(x-x0) + R1.(x-x0) by A3;
  reconsider R = r(#)R1 as RestFunc by Th5;
  reconsider L = r(#)L1 as LinearFunc by Th3;
A5: L1 is total by Def3;
A6: N1 c= dom(r(#)f) by A2,VALUED_1:def 5;
A7: R1 is total by Def2;
A8: now
    let x;
A9: x0 in N1 by RCOMP_1:16;
    assume
A10: x in N1;
    hence (r(#)f).x - (r(#)f).x0 = r*(f.x) - (r(#)f).x0 by A6,VALUED_1:def 5
      .= r*f.x - r*f.x0 by A6,A9,VALUED_1:def 5
      .= r*(f.x - f.x0)
      .= r*(L1.(x-x0) + R1.(x-x0)) by A4,A10
      .= r*L1.(x-x0) + r*R1.(x-x0)
      .= L.(x-x0) + r*R1.(x-x0) by A5,RFUNCT_1:57
      .= L.(x-x0) + R.(x-x0) by A7,RFUNCT_1:57;
  end;
  hence r(#)f is_differentiable_in x0 by A6,Def4;
  hence diff((r(#)f),x0) = L.1 by A6,A8,Def5
    .= r*L1.1 by A5,RFUNCT_1:57
    .= r*diff(f,x0) by A1,A2,A4,Def5;
end;

theorem Th16:
  f1 is_differentiable_in x0 & f2 is_differentiable_in x0 implies
f1(#)f2 is_differentiable_in x0 & diff(f1(#)f2,x0)=(f2.x0)*diff(f1,x0)+(f1.x0)*
  diff(f2,x0)
proof
  assume that
A1: f1 is_differentiable_in x0 and
A2: f2 is_differentiable_in x0;
  consider N1 be Neighbourhood of x0 such that
A3: N1 c= dom f1 and
A4: ex L,R st for x st x in N1 holds f1.x - f1.x0 = L.(x-x0) + R.(x-x0)
  by A1,Def4;
  consider L1,R1 such that
A5: for x st x in N1 holds f1.x - f1.x0 = L1.(x-x0) + R1.(x-x0) by A4;
  consider N2 be Neighbourhood of x0 such that
A6: N2 c= dom f2 and
A7: ex L,R st for x st x in N2 holds f2.x - f2.x0 = L.(x-x0) + R.(x-x0)
  by A2,Def4;
  consider L2,R2 such that
A8: for x st x in N2 holds f2.x - f2.x0 = L2.(x-x0) + R2.(x-x0) by A7;
  reconsider R18=R2(#)L1 as RestFunc by Th7;
  reconsider R17=R1(#)R2 as RestFunc by Th4;
A9: R18 is total by Def2;
  reconsider R16=R1(#)L2 as RestFunc by Th7;
  reconsider R14=L1(#)L2 as RestFunc by Th6;
  reconsider R19=R16+R17 as RestFunc by Th4;
  reconsider R20=R19+R18 as RestFunc by Th4;
A10: R14 is total by Def2;
  reconsider R12=(f1.x0)(#)R2 as RestFunc by Th5;
A11: R2 is total by Def2;
  reconsider L11=(f2.x0)(#)L1 as LinearFunc by Th3;
A12: L1 is total by Def3;
  reconsider R11=(f2.x0)(#)R1 as RestFunc by Th5;
A13: R1 is total by Def2;
  reconsider R13=R11+R12 as RestFunc by Th4;
  reconsider R15=R13+R14 as RestFunc by Th4;
  reconsider R=R15+R20 as RestFunc by Th4;
  consider N be Neighbourhood of x0 such that
A14: N c= N1 and
A15: N c= N2 by RCOMP_1:17;
A16: N c= dom f2 by A6,A15,XBOOLE_1:1;
  N c= dom f1 by A3,A14,XBOOLE_1:1;
  then N /\ N c= dom f1 /\ dom f2 by A16,XBOOLE_1:27;
  then
A17: N c= dom (f1(#)f2) by VALUED_1:def 4;
  reconsider L12=(f1.x0)(#)L2 as LinearFunc by Th3;
A18: L2 is total by Def3;
  reconsider L=L11+L12 as LinearFunc by Th2;
A19: R16 is total by Def2;
A20: L11 is total & L12 is total by Def3;
A21: now
    let x;
    assume
A22: x in N;
    then
A23: f1.x - f1.x0 + f1.x0 = L1.(x-x0) + R1.(x-x0) + f1.x0 by A5,A14;
    thus (f1(#)f2).x - (f1(#)f2).x0 = (f1.x) * (f2.x) - (f1(#) f2).x0 by
VALUED_1:5
      .=(f1.x)*(f2.x)+-(f1.x)*(f2.x0)+(f1.x)*(f2.x0)-(f1.x0)*(f2.x0) by
VALUED_1:5
      .=(f1.x)*((f2.x)-(f2.x0))+((f1.x)-(f1.x0))*(f2.x0)
      .=(f1.x)*((f2.x)-(f2.x0))+(L1.(x-x0)+R1.(x-x0))*(f2.x0) by A5,A14,A22
      .=(f1.x)*((f2.x)-(f2.x0))+((f2.x0)*L1.(x-x0)+R1.(x-x0)*(f2.x0))
      .=(f1.x)*((f2.x)-(f2.x0))+(L11.(x-x0)+(f2.x0)*R1.(x-x0)) by A12,
RFUNCT_1:57
      .=(L1.(x-x0) + R1.(x-x0) + f1.x0)*((f2.x)-(f2.x0))+(L11.(x-x0)+R11.(x-
    x0)) by A13,A23,RFUNCT_1:57
      .=(L1.(x-x0) + R1.(x-x0) + f1.x0)*(L2.(x-x0) + R2.(x-x0))+ (L11.(x-x0)
    +R11.(x-x0)) by A8,A15,A22
      .=(L1.(x-x0) + R1.(x-x0))*(L2.(x-x0) + R2.(x-x0))+ ((f1.x0)*L2.(x-x0)+
    (f1.x0)*R2.(x-x0)) + (L11.(x-x0)+R11.(x-x0))
      .=(L1.(x-x0) + R1.(x-x0))*(L2.(x-x0) + R2.(x-x0))+ (L12.(x-x0)+(f1.x0)
    *R2.(x-x0)) + (L11.(x-x0)+R11.(x-x0)) by A18,RFUNCT_1:57
      .=(L1.(x-x0) + R1.(x-x0))*(L2.(x-x0) + R2.(x-x0))+ (L12.(x-x0)+R12.(x-
    x0)) + (L11.(x-x0)+R11.(x-x0)) by A11,RFUNCT_1:57
      .=(L1.(x-x0) + R1.(x-x0))*(L2.(x-x0) + R2.(x-x0))+ (L12.(x-x0)+(L11.(x
    -x0)+(R11.(x-x0)+R12.(x-x0))))
      .=(L1.(x-x0) + R1.(x-x0))*(L2.(x-x0) + R2.(x-x0))+ (L12.(x-x0)+(L11.(x
    -x0)+R13.(x-x0))) by A13,A11,RFUNCT_1:56
      .=(L1.(x-x0) + R1.(x-x0))*(L2.(x-x0) + R2.(x-x0))+ (L11.(x-x0)+L12.(x-
    x0)+R13.(x-x0))
      .=(L1.(x-x0)*L2.(x-x0)+L1.(x-x0)*R2.(x-x0))+R1.(x-x0)*(L2.(x-x0)+R2.(x
    -x0))+ (L.(x-x0)+R13.(x-x0)) by A20,RFUNCT_1:56
      .=R14.(x-x0) + R2.(x-x0)*L1.(x-x0)+R1.(x-x0)*(L2.(x-x0)+R2.(x-x0))+ (L
    .(x-x0)+R13.(x-x0)) by A12,A18,RFUNCT_1:56
      .=R14.(x-x0) + R18.(x-x0)+(R1.(x-x0)*L2.(x-x0)+R1.(x-x0)*R2.(x-x0))+ (
    L.(x-x0)+R13.(x-x0)) by A12,A11,RFUNCT_1:56
      .=R14.(x-x0) + R18.(x-x0)+(R16.(x-x0)+R1.(x-x0)*R2.(x-x0))+ (L.(x-x0)+
    R13.(x-x0)) by A18,A13,RFUNCT_1:56
      .=R14.(x-x0) + R18.(x-x0)+(R16.(x-x0)+R17.(x-x0))+ (L.(x-x0)+R13.(x-x0
    )) by A13,A11,RFUNCT_1:56
      .=R14.(x-x0) + R18.(x-x0)+R19.(x-x0)+(L.(x-x0)+R13.(x-x0)) by A13,A11,A19
,RFUNCT_1:56
      .=R14.(x-x0) + (R19.(x-x0)+R18.(x-x0))+(L.(x-x0)+R13.(x-x0))
      .=L.(x-x0)+R13.(x-x0)+(R14.(x-x0) + R20.(x-x0)) by A13,A11,A19,A9,
RFUNCT_1:56
      .=L.(x-x0)+(R13.(x-x0)+R14.(x-x0) + R20.(x-x0))
      .=L.(x-x0)+(R15.(x-x0)+R20.(x-x0)) by A13,A11,A10,RFUNCT_1:56
      .=L.(x-x0)+R.(x-x0) by A13,A11,A10,A19,A9,RFUNCT_1:56;
  end;
  hence f1(#)f2 is_differentiable_in x0 by A17,Def4;
  hence diff(f1(#)f2,x0)=L.1 by A17,A21,Def5
    .= L11.1 + L12.1 by A20,RFUNCT_1:56
    .= f2.x0 * L1.1 + L12.1 by A12,RFUNCT_1:57
    .= f2.x0 * L1.1 + f1.x0 *L2.1 by A18,RFUNCT_1:57
    .= f2.x0 * diff(f1,x0) + f1.x0 * L2.1 by A1,A3,A5,Def5
    .= f2.x0 * diff(f1,x0) + f1.x0 * diff(f2,x0) by A2,A6,A8,Def5;
end;

theorem Th17:
  Z c= dom f & f|Z = id Z implies f is_differentiable_on Z & for x
  st x in Z holds (f`|Z).x = 1
proof
  set R = cf;
A1: dom R = REAL by FUNCOP_1:13;
  now
    let h;
A2: now
      let n be Nat;
A3:   rng h c= dom R by A1;
A4:   n in NAT by ORDINAL1:def 12;
      hence ((h")(#)(R/*h)).n = (h".n)*((R/*h).n) by SEQ_1:8
        .= (h".n)*(R.(h.n)) by A4,A3,FUNCT_2:108
        .= (h".n)*0 by FUNCOP_1:7
        .= 0;
    end;
    then
A5: (h")(#)(R/*h) is constant by VALUED_0:def 18;
    hence (h")(#)(R/*h) is convergent;
    ((h")(#)(R/*h)).0 = 0 by A2;
    hence lim ((h")(#)(R/*h)) = 0 by A5,SEQ_4:25;
  end;
  then reconsider R as RestFunc by Def2;
  reconsider L = id REAL as PartFunc of REAL,REAL;
  for p holds L.p = 1*p by FUNCT_1:18;
  then reconsider L as LinearFunc by Def3;
  assume that
A6: Z c= dom f and
A7: f|Z = id Z;
A8: now
    let x;
    assume
A9: x in Z;
    then f|Z.x = x by A7,FUNCT_1:18;
    hence f.x = x by A9,FUNCT_1:49;
  end;
A10: now
    let x0;
    assume
A11: x0 in Z;
    then consider N being Neighbourhood of x0 such that
A12: N c= Z by RCOMP_1:18;
A13: for x st x in N holds f.x - f.x0 = L.(x-x0) + R.(x-x0)
    proof
      let x;
      assume x in N;
      hence f.x - f.x0 = x - f.x0 by A8,A12
        .= x - x0 by A8,A11
        .= L.(x-x0)+0 by FUNCT_1:18
        .= L.(x-x0) + R.(x-x0) by FUNCOP_1:7;
    end;
    N c= dom f by A6,A12,XBOOLE_1:1;
    hence f is_differentiable_in x0 by A13,Def4;
  end;
  hence
A14: f is_differentiable_on Z by A6,Th9;
  let x0;
  assume
A15: x0 in Z;
  then consider N1 being Neighbourhood of x0 such that
A16: N1 c= Z by RCOMP_1:18;
A17: f is_differentiable_in x0 by A10,A15;
  then ex N being Neighbourhood of x0 st N c= dom f & ex L,R st for x st x in
  N holds f.x - f.x0 = L.(x-x0) + R.(x-x0) by Def4;
  then consider N being Neighbourhood of x0 such that
A18: N c= dom f;
  consider N2 being Neighbourhood of x0 such that
A19: N2 c= N1 and
A20: N2 c= N by RCOMP_1:17;
A21: N2 c= dom f by A18,A20,XBOOLE_1:1;
A22: for x st x in N2 holds f.x - f.x0 = L.(x-x0) + R.(x-x0)
  proof
    let x;
    assume x in N2;
    then x in N1 by A19;
    hence f.x - f.x0 = x - f.x0 by A8,A16
      .= x - x0 by A8,A15
      .= L.(x-x0)+0 by FUNCT_1:18
      .= L.(x-x0) + R.(x-x0) by FUNCOP_1:7;
  end;
  thus (f`|Z).x0 = diff(f,x0) by A14,A15,Def7
    .= L.1 by A17,A21,A22,Def5
    .= 1 by FUNCT_1:18;
end;

theorem
  Z c= dom (f1+f2) & f1 is_differentiable_on Z & f2 is_differentiable_on
Z implies f1+f2 is_differentiable_on Z & for x st x in Z holds ((f1+f2)`|Z).x =
  diff(f1,x) + diff(f2,x)
proof
  assume that
A1: Z c= dom (f1+f2) and
A2: f1 is_differentiable_on Z & f2 is_differentiable_on Z;
  now
    let x0;
    assume x0 in Z;
    then f1 is_differentiable_in x0 & f2 is_differentiable_in x0 by A2,Th9;
    hence f1+f2 is_differentiable_in x0 by Th13;
  end;
  hence
A3: f1+f2 is_differentiable_on Z by A1,Th9;
  now
    let x;
    assume
A4: x in Z;
    then
A5: f1 is_differentiable_in x & f2 is_differentiable_in x by A2,Th9;
    thus ((f1+f2)`|Z).x = diff((f1+f2),x) by A3,A4,Def7
      .= diff(f1,x) + diff(f2,x) by A5,Th13;
  end;
  hence thesis;
end;

theorem
  Z c= dom (f1-f2) & f1 is_differentiable_on Z & f2 is_differentiable_on
Z implies f1-f2 is_differentiable_on Z & for x st x in Z holds ((f1-f2)`|Z).x =
  diff(f1,x) - diff(f2,x)
proof
  assume that
A1: Z c= dom (f1-f2) and
A2: f1 is_differentiable_on Z & f2 is_differentiable_on Z;
  now
    let x0;
    assume x0 in Z;
    then f1 is_differentiable_in x0 & f2 is_differentiable_in x0 by A2,Th9;
    hence f1-f2 is_differentiable_in x0 by Th14;
  end;
  hence
A3: f1-f2 is_differentiable_on Z by A1,Th9;
  now
    let x;
    assume
A4: x in Z;
    then
A5: f1 is_differentiable_in x & f2 is_differentiable_in x by A2,Th9;
    thus ((f1-f2)`|Z).x = diff((f1-f2),x) by A3,A4,Def7
      .= diff(f1,x) - diff(f2,x) by A5,Th14;
  end;
  hence thesis;
end;

theorem
  Z c= dom (r(#)f) & f is_differentiable_on Z implies r(#)f
  is_differentiable_on Z & for x st x in Z holds ((r(#) f)`|Z).x =r*diff(f,x)
proof
  assume that
A1: Z c= dom (r(#)f) and
A2: f is_differentiable_on Z;
  now
    let x0;
    assume x0 in Z;
    then f is_differentiable_in x0 by A2,Th9;
    hence r(#)f is_differentiable_in x0 by Th15;
  end;
  hence
A3: r(#)f is_differentiable_on Z by A1,Th9;
  now
    let x;
    assume
A4: x in Z;
    then
A5: f is_differentiable_in x by A2,Th9;
    thus ((r(#)f)`|Z).x = diff((r(#)f),x) by A3,A4,Def7
      .= r*diff(f,x) by A5,Th15;
  end;
  hence thesis;
end;

theorem
  Z c= dom (f1(#)f2) & f1 is_differentiable_on Z & f2
is_differentiable_on Z implies f1(#)f2 is_differentiable_on Z & for x st x in Z
  holds ((f1(#)f2)`|Z).x = (f2.x)*diff(f1,x) + (f1.x)*diff(f2,x)
proof
  assume that
A1: Z c= dom (f1(#)f2) and
A2: f1 is_differentiable_on Z & f2 is_differentiable_on Z;
  now
    let x0;
    assume x0 in Z;
    then f1 is_differentiable_in x0 & f2 is_differentiable_in x0 by A2,Th9;
    hence f1(#)f2 is_differentiable_in x0 by Th16;
  end;
  hence
A3: f1(#)f2 is_differentiable_on Z by A1,Th9;
  now
    let x;
    assume
A4: x in Z;
    then
A5: f1 is_differentiable_in x & f2 is_differentiable_in x by A2,Th9;
    thus ((f1(#)f2)`|Z).x = diff((f1(#)f2),x) by A3,A4,Def7
      .= f2.x*diff(f1,x) + f1.x*diff(f2,x) by A5,Th16;
  end;
  hence thesis;
end;

theorem
  Z c= dom f & f|Z is constant implies f is_differentiable_on Z & for x
  st x in Z holds (f`|Z).x = 0
proof
  set R = cf;
A1: dom R = REAL by FUNCOP_1:13;
  now
    let h;
A2: now
      let n be Nat;
A3:   rng h c= dom R by A1;
A4:   n in NAT by ORDINAL1:def 12;
      hence ((h")(#)(R/*h)).n = (h".n)*((R/*h).n) by SEQ_1:8
        .=(h".n)*(R.(h.n)) by A4,A3,FUNCT_2:108
        .=(h".n)*0 by FUNCOP_1:7
        .=0;
    end;
    then
A5: (h")(#)(R/*h) is constant by VALUED_0:def 18;
    hence (h")(#)(R/*h) is convergent;
    ((h")(#)(R/*h)).0 = 0 by A2;
    hence lim ((h")(#)(R/*h)) = 0 by A5,SEQ_4:25;
  end;
  then reconsider R as RestFunc by Def2;
  set L = cf;
  for p holds L.p=0*p by FUNCOP_1:7;
  then reconsider L as LinearFunc by Def3;
  assume that
A6: Z c= dom f and
A7: f|Z is constant;
  consider r such that
A8: for x st x in Z/\dom f holds f.x=r by A7,PARTFUN2:57;
A9: now
    let x0;
    assume
A10: x0 in Z;
    then consider N being Neighbourhood of x0 such that
A11: N c= Z by RCOMP_1:18;
A12: N c= dom f by A6,A11,XBOOLE_1:1;
A13: x0 in Z/\dom f by A6,A10,XBOOLE_0:def 4;
    for x st x in N holds f.x-f.x0=L.(x-x0)+R.(x-x0)
    proof
      let x;
      assume x in N;
      then x in Z/\dom f by A11,A12,XBOOLE_0:def 4;
      hence f.x-f.x0=r-f.x0 by A8
        .=r - r by A8,A13
        .=L.(x-x0)+0 by FUNCOP_1:7
        .=L.(x-x0)+R.(x-x0) by FUNCOP_1:7;
    end;
    hence f is_differentiable_in x0 by A12,Def4;
  end;
  hence
A14: f is_differentiable_on Z by A6,Th9;
  let x0;
  assume
A15: x0 in Z;
  then consider N being Neighbourhood of x0 such that
A16: N c= Z by RCOMP_1:18;
A17: N c= dom f by A6,A16,XBOOLE_1:1;
A18: x0 in Z/\dom f by A6,A15,XBOOLE_0:def 4;
A19: for x st x in N holds f.x-f.x0=L.(x-x0)+R.(x-x0)
  proof
    let x;
    assume x in N;
    then x in Z/\dom f by A16,A17,XBOOLE_0:def 4;
    hence f.x - f.x0 = r - f.x0 by A8
      .=r - r by A8,A18
      .=L.(x-x0) + 0 by FUNCOP_1:7
      .=L.(x-x0) + R.(x-x0) by FUNCOP_1:7;
  end;
A20: f is_differentiable_in x0 by A9,A15;
  thus (f`|Z).x0 = diff(f,x0) by A14,A15,Def7
    .=L.1 by A20,A17,A19,Def5
    .=0 by FUNCOP_1:7;
end;

theorem
  Z c= dom f & (for x st x in Z holds f.x = r*x + p) implies f
  is_differentiable_on Z & for x st x in Z holds (f`|Z).x = r
proof
  set R = cf;
  defpred P[set] means $1 in REAL;
A1: dom R = REAL by FUNCOP_1:13;
  now
    let h;
A2: now
      let n be Nat;
A3:   rng h c= dom R by A1;
A4:   n in NAT by ORDINAL1:def 12;
      hence ((h")(#)(R/*h)).n = (h".n)*((R/*h).n) by SEQ_1:8
        .=(h".n)*(R.(h.n)) by A4,A3,FUNCT_2:108
        .=(h".n)*0 by FUNCOP_1:7
        .=0;
    end;
    then
A5: (h")(#)(R/*h) is constant by VALUED_0:def 18;
    hence (h")(#)(R/*h) is convergent;
    ((h")(#)(R/*h)).0 = 0 by A2;
    hence lim ((h")(#)(R/*h)) = 0 by A5,SEQ_4:25;
  end;
  then reconsider R as RestFunc by Def2;
  assume that
A6: Z c= dom f and
A7: for x st x in Z holds f.x = r*x + p;
  deffunc G(Real) = r*$1;
  consider L being PartFunc of REAL,REAL such that
A8: (for x holds x in dom L iff P[x]) & for x st x in dom L holds L.x=G(
  x) from SEQ_1:sch 3;
  dom L = REAL by A8,Th1;
  then
A9: L is total by PARTFUN1:def 2;
A10: now
    let x;
    thus L.x=r*x by A8;
  end;
  then reconsider L as LinearFunc by A9,Def3;
A11: now
    let x0;
    assume
A12: x0 in Z;
    then consider N being Neighbourhood of x0 such that
A13: N c= Z by RCOMP_1:18;
A14: for x st x in N holds f.x-f.x0=L.(x-x0)+R.(x-x0)
    proof
      let x;
      assume x in N;
      hence f.x-f.x0=r*x+p-f.x0 by A7,A13
        .=r*x+p - (r*x0+p) by A7,A12
        .=r*(x-x0)+0
        .=L.(x-x0)+0 by A10
        .=L.(x-x0)+R.(x-x0) by FUNCOP_1:7;
    end;
    N c= dom f by A6,A13,XBOOLE_1:1;
    hence f is_differentiable_in x0 by A14,Def4;
  end;
  hence
A15: f is_differentiable_on Z by A6,Th9;
  let x0;
  assume
A16: x0 in Z;
  then consider N being Neighbourhood of x0 such that
A17: N c= Z by RCOMP_1:18;
A18: for x st x in N holds f.x-f.x0=L.(x-x0)+R.(x-x0)
  proof
    let x;
    assume x in N;
    hence f.x - f.x0 = r*x+p - f.x0 by A7,A17
      .=r*x+p-(r*x0+p) by A7,A16
      .=r*(x-x0)+0
      .=L.(x-x0) + 0 by A10
      .=L.(x-x0) + R.(x-x0) by FUNCOP_1:7;
  end;
A19: N c= dom f by A6,A17,XBOOLE_1:1;
A20: f is_differentiable_in x0 by A11,A16;
  thus (f`|Z).x0 = diff(f,x0) by A15,A16,Def7
    .=L.1 by A20,A19,A18,Def5
    .=r*1 by A10
    .=r;
end;

theorem Th24:
  for x0 being real number holds f is_differentiable_in x0 implies
  f is_continuous_in x0
proof
  let x0 be real number;
  assume
A1: f is_differentiable_in x0;
  then consider N being Neighbourhood of x0 such that
A2: N c= dom f and
  ex L,R st for x st x in N holds f.x - f.x0 = L.(x-x0) + R.(x-x0) by Def4;
  now
    consider g be real number such that
A3: 0<g and
A4: N=].x0-g,x0+g.[ by RCOMP_1:def 6;
    x0 in REAL by XREAL_0:def 1;
    then reconsider s2 = NAT --> x0 as Real_Sequence by FUNCOP_1:45;
    let s1 such that
A5: rng s1 c= dom f and
A6: s1 is convergent and
A7: lim s1 = x0 and
A8: for n holds s1.n<>x0;
    consider l be Element of NAT such that
A9: for m st l<=m holds abs(s1.m-x0)<g by A6,A7,A3,SEQ_2:def 7;
    reconsider c =s2^\l as constant Real_Sequence;
    deffunc G(Real) = s1.$1-s2.$1;
    consider s3 such that
A10: for n holds s3.n=G(n) from SEQ_1:sch 1;
A11: s3=s1-s2 by A10,RFUNCT_2:1;
    then
A12: s3 is convergent by A6,SEQ_2:11;
A13: rng c = {x0}
    proof
      thus rng c c= {x0}
      proof
        let y;
        assume y in rng c;
        then consider n such that
A14:    y=(s2^\l).n by FUNCT_2:113;
        y=s2.(n+l) by A14,NAT_1:def 3;
        then y=x0 by FUNCOP_1:7;
        hence thesis by TARSKI:def 1;
      end;
      let y;
      assume y in {x0};
      then
A15:  y=x0 by TARSKI:def 1;
      c.0=s2.(0+l) by NAT_1:def 3
        .= y by A15,FUNCOP_1:7;
      hence thesis by VALUED_0:28;
    end;
A16: now
      let p be real number such that
A17:  0<p;
      take n=0;
      let m such that
      n<=m;
      x0 in N by RCOMP_1:16;
      then rng c c= dom f by A2,A13,ZFMISC_1:31;
      then abs((f/*c).m-f.x0)=abs(f.(c.m)-f.x0) by FUNCT_2:108
        .=abs(f.(s2.(m+l))-f.x0) by NAT_1:def 3
        .=abs(f.x0-f.x0) by FUNCOP_1:7
        .=0 by ABSVALUE:2;
      hence abs((f/*c).m-f.x0)<p by A17;
    end;
    then
A18: f/*c is convergent by SEQ_2:def 6;
    lim s2 = s2.0 by SEQ_4:26
      .=x0 by FUNCOP_1:7;
    then lim s3 =x0-x0 by A6,A7,A11,SEQ_2:12
      .=0;
    then
A19: lim(s3^\l)=0 by A12,SEQ_4:20;
A20: now
      given n such that
A21:  s3.n=0;
      s1.n-s2.n=0 by A10,A21;
      hence contradiction by A8,FUNCOP_1:7;
    end;
A22: now
      given n such that
A23:  (s3^\l).n=0;
      s3.(n+l)=0 by A23,NAT_1:def 3;
      hence contradiction by A20;
    end;
    s3^\l is 0-convergent by A12,A19,Def1;
    then reconsider h=s3^\l as 0-convergent non-zero Real_Sequence
      by A22,SEQ_1:5;
    now
      let n;
      thus (f/*(h+c)-f/*c+f/*c).n=(f/*(h+c)-f/*c).n+(f/*c).n by SEQ_1:7
        .=(f/*(h+c)).n-(f/*c).n+(f/*c).n by RFUNCT_2:1
        .=(f/*(h+c)).n;
    end;
    then
A24: f/*(h+c)-f/*c+(f/*c)=f/*(h+c) by FUNCT_2:63;
    now
      let n;
      thus (h+c).n=((s1-s2+s2)^\l).n by A11,SEQM_3:15
        .=(s1-s2+s2).(n+l) by NAT_1:def 3
        .=(s1-s2).(n+l)+s2.(n+l) by SEQ_1:7
        .=s1.(n+l)-s2.(n+l)+s2.(n+l) by RFUNCT_2:1
        .=(s1^\l).n by NAT_1:def 3;
    end;
    then
A25: f/*(h+c)-f/*c+(f/*c)=f/*(s1^\l) by A24,FUNCT_2:63
      .=(f/*s1)^\l by A5,VALUED_0:27;
    rng (h+c) c= N
    proof
      let y;
      assume y in rng(h+c);
      then consider n such that
A26:  y=(h+c).n by FUNCT_2:113;
      (h+c).n=((s1-s2+s2)^\l).n by A11,SEQM_3:15
        .=(s1-s2+s2).(n+l) by NAT_1:def 3
        .=(s1-s2).(n+l)+s2.(n+l) by SEQ_1:7
        .=s1.(n+l)-s2.(n+l)+s2.(n+l) by RFUNCT_2:1
        .=s1.(l+n);
      then abs((h+c).n-x0)<g by A9,NAT_1:12;
      hence thesis by A4,A26,RCOMP_1:1;
    end; then
A27: h"(#)(f/*(h+c) - f/*c) is convergent by A1,A2,A13,Th12;
    then
A28: lim (h(#)(h"(#)(f/*(h+c) - f/*c))) =0*lim(h"(#) (f/*(h+c) - f/*c)) by
A19,SEQ_2:15
      .=0;
    now
      let n;
A29:  h.n<>0 by A22;
      thus (h(#)(h"(#)(f/*(h+c) - f/*c))).n =h.n *(h"(#)(f/*(h+c) - f/*c)).n
      by SEQ_1:8
        .=h.n*((h").n*(f/*(h+c) - f/*c).n) by SEQ_1:8
        .=h.n*(((h.n)")*(f/*(h+c) - f/*c).n) by VALUED_1:10
        .=h.n*((h.n)")*(f/*(h+c) - f/*c).n
        .=1*(f/*(h+c) - f/*c).n by A29,XCMPLX_0:def 7
        .=(f/*(h+c) - f/*c).n;
    end;
    then
A30: h(#)(h"(#)(f/*(h+c) - f/*c))=f/*(h+c)-f/*c by FUNCT_2:63;
    then
A31: f/*(h+c)-f/*c is convergent by A27,SEQ_2:14;
    then
A32: f/*(h+c)-f/*c+f/*c is convergent by A18,SEQ_2:5;
    hence f/*s1 is convergent by A25,SEQ_4:21;
    lim(f/*c)=f.x0 by A16,A18,SEQ_2:def 7;
    then lim(f/*(h+c)-f/*c+f/*c)=0+f.x0 by A28,A30,A31,A18,SEQ_2:6
      .=f.x0;
    hence f.x0=lim(f/*s1) by A32,A25,SEQ_4:22;
  end;
  hence thesis by FCONT_1:2;
end;

theorem
  f is_differentiable_on X implies f|X is continuous
proof
  assume
A1: f is_differentiable_on X;
  let x be real number;
  assume x in dom(f|X);
  then x is Real & x in X by XREAL_0:def 1;
  then f|X is_differentiable_in x by A1,Def6;
  hence thesis by Th24;
end;

theorem Th26:
  f is_differentiable_on X & Z c= X implies f is_differentiable_on Z
proof
  assume that
A1: f is_differentiable_on X and
A2: Z c= X;
  X c= dom f by A1,Def6;
  hence Z c= dom f by A2,XBOOLE_1:1;
  let x0;
  assume
A3: x0 in Z;
  then f|X is_differentiable_in x0 by A1,A2,Def6;
  then consider N being Neighbourhood of x0 such that
A4: N c= dom(f|X) and
A5: ex L,R st for x st x in N holds (f|X).x-(f|X).x0=L.(x-x0)+R.(x-x0) by Def4;
  consider N1 being Neighbourhood of x0 such that
A6: N1 c= Z by A3,RCOMP_1:18;
  consider N2 being Neighbourhood of x0 such that
A7: N2 c= N and
A8: N2 c= N1 by RCOMP_1:17;
A9: N2 c= Z by A6,A8,XBOOLE_1:1;
  take N2;
  dom(f|X)=dom f/\X by RELAT_1:61;
  then dom(f|X) c=dom f by XBOOLE_1:17;
  then N c= dom f by A4,XBOOLE_1:1;
  then N2 c=dom f by A7,XBOOLE_1:1;
  then N2 c=dom f/\Z by A9,XBOOLE_1:19;
  hence
A10: N2 c=dom(f|Z) by RELAT_1:61;
  consider L,R such that
A11: for x st x in N holds (f|X).x-(f|X).x0=L.(x-x0)+R.(x-x0) by A5;
  take L,R;
  let x;
  assume
A12: x in N2;
  then (f|X).x-(f|X).x0=L.(x-x0)+R.(x-x0) by A7,A11;
  then
A13: (f|X).x-f.x0=L.(x-x0)+R.(x-x0) by A2,A3,FUNCT_1:49;
  x in N by A7,A12;
  then f.x-f.x0=L.(x-x0)+R.(x-x0) by A4,A13,FUNCT_1:47;
  then f.x-(f|Z).x0=L.(x-x0)+R.(x-x0) by A3,FUNCT_1:49;
  hence thesis by A10,A12,FUNCT_1:47;
end;

theorem 

ex R st R.0=0 & R is_continuous_in 0
proof
  A1: {} REAL is closed
  proof
    let a be Real_Sequence;
    assume rng a c= {} REAL & a is convergent;
    hence lim a in {}REAL by XBOOLE_1:3;
  end;
  ([#] REAL)` = {} REAL & REAL c= REAL & [#]REAL = REAL by XBOOLE_1:37;
  then
  reconsider Z = [#]REAL as open Subset of REAL by A1,RCOMP_1:def 5;
  set R = cf;
  reconsider f=R as PartFunc of REAL,REAL;
A2: dom R = REAL by FUNCOP_1:13;
  now
    let h;
A3: now
      let n be Nat;
A4:   rng h c= dom R by A2;
A5:   n in NAT by ORDINAL1:def 12;
      hence ((h")(#)(R/*h)).n = (h".n)*((R/*h).n) by SEQ_1:8
        .= (h".n)*(R.(h.n)) by A5,A4,FUNCT_2:108
        .= (h".n)*0 by FUNCOP_1:7
        .= 0;
    end;
    then
A6: (h")(#)(R/*h) is constant by VALUED_0:def 18;
    hence (h")(#)(R/*h) is convergent;
    ((h")(#)(R/*h)).0 = 0 by A3;
    hence lim ((h")(#)(R/*h)) = 0 by A6,SEQ_4:25;
  end;
  then reconsider R as RestFunc by Def2;
  set L = cf;
  for p holds L.p=0*p by FUNCOP_1:7;
  then reconsider L as LinearFunc by Def3;
  f|Z is constant;
  then consider r such that
A7: for x st x in Z/\dom f holds f.x=r by PARTFUN2:57;
A8:
now
    let x0;
    assume x0 in Z;
    set N = the Neighbourhood of x0;
A9: N c= dom f by A2;
A10: x0 in Z/\dom f by A2;
    for x st x in N holds f.x-f.x0=L.(x-x0)+R.(x-x0)
    proof
      let x;
      assume x in N;
      then x in Z/\dom f by A9,XBOOLE_0:def 4;
      hence f.x-f.x0=r-f.x0 by A7
        .=r - r by A7,A10
        .=L.(x-x0)+0 by FUNCOP_1:7
        .=L.(x-x0)+R.(x-x0) by FUNCOP_1:7;
    end;
    hence f is_differentiable_in x0 by A9,Def4;
  end;
  set x0 = the Real;
  f is_differentiable_in x0 by A8; then
  consider N being Neighbourhood of x0 such that
  N c= dom f and
A11: ex L,R st for x st x in N holds f.x - f.x0 = L.(x-x0) + R.(x-x0) by Def4;
  c