
theorem thre:
for X being set ex Y being set st card X c= card Y & X /\ Y = {}
proof
let X be set;
set D = bool(X) /\ X;
set Y = bool(X) \ (bool(X) /\ X);
C: now assume H: Y /\ X <> {};
   set o = the Element of (Y /\ X);
   H1: o in Y & o in X by H,XBOOLE_0:def 4; then
   o in bool X & not o in D by XBOOLE_0:def 5;
   hence contradiction by H1,XBOOLE_0:def 4;
   end;
B: card bool X c= card Y  +` card X
   proof
   now let o be object;
     assume B1: o in bool X;
     per cases;
     suppose o in X;
       hence o in Y \/ X by XBOOLE_0:def 3;
       end;
     suppose not o in X;
       then not o in D by XBOOLE_0:def 4;
       then o in Y by B1,XBOOLE_0:def 5;
       hence o in Y \/ X by XBOOLE_0:def 3;
       end;
     end;
   then B1: bool X c= Y \/ X;
   card(Y \/ X) = card Y  +` card X by C,CARD_2:35,XBOOLE_0:def 7;
   hence thesis by B1,CARD_1:11;
   end;
per cases;
suppose AS: card X = 0;
  set u = the object;
  take Y = {u};
  thus card X c= card Y by AS;
  now assume A: X /\ Y <> {};
    set o = the Element of X /\ Y;
    o in X & o in Y by A,XBOOLE_0:def 4;
    hence contradiction by AS;
    end;
  hence thesis;
  end;
suppose AS: card X = 1;
  then consider o being object such that A: X = {o} by CARD_2:42;
  consider u being object such that B: not u in {o} by th1;
  take Y = {u};
  thus card X c= card Y by AS,CARD_2:42;
  now assume C: X /\ Y <> {};
    set x = the Element of X /\ Y;
    x in X & x in Y by C,XBOOLE_0:def 4;
    hence contradiction by A,B,TARSKI:def 1;
    end;
  hence thesis;
  end;
suppose AS: card X = 2;
  then consider x,y being object such that
  A: x <> y & X = {x,y} by CARD_2:60;
  consider u being object such that B: not u in {x,y} by th1;
  consider v being object such that C: not v in {x,y,u} by th1;
  take Y = {u,v};
  u <> v by C,ENUMSET1:def 1;
  hence card X c= card Y by AS,CARD_2:57;
  now assume H: X /\ Y <> {};
    set o = the Element of X /\ Y;
    E: o in X & o in Y by H,XBOOLE_0:def 4; then
    F: o = x or o = y by A,TARSKI:def 2;
    per cases by E,TARSKI:def 2;
    suppose o = u;
      hence contradiction by A,B,H,XBOOLE_0:def 4;
      end;
    suppose o = v;
      hence contradiction by F,C,ENUMSET1:def 1;
      end;
    end;
  hence thesis;
  end;
suppose AS: card X is finite & card X <> 0 & card X <> 1 & card X <> 2;
  reconsider n = card X as Nat by AS;
  D: now assume n < 2 + 1;
     then n <= 2 by NAT_1:13;
     then n = 0 or ... or n = 2;
     hence contradiction by AS;
     end;
  B0: card bool X = exp(2,n) by CARD_2:31;
  now assume card Y c= card X;
    then card Y +` n c= n +` n by CARD_2:83;
    then card bool X c= n +` n by B;
    hence contradiction by D,B0,th0b,CARD_1:4;
    end;
  hence thesis by C;
  end;
suppose AS: not card X is finite;
  now assume card Y c= card X;
    then card Y +` card X c= card X +` card X by CARD_2:83; then
    card Y +` card X c= card X by AS,CARD_2:75; then
    card bool X c= card X by B;
    hence contradiction by CARD_1:4,CARD_1:14;
    end;
  hence thesis by C;
  end;
end;
