reserve i,j,k,n,m for Nat,
  x,y,z,y1,y2 for object, X,Y,D for set,
  p,q for XFinSequence;
reserve k1,k2 for Nat;

theorem Th31:
  for X,Y being finite natural-membered set st X <N< Y
  & i< len (Sgm0 Y) holds (Sgm0 Y).i = (Sgm0 (X \/ Y)).(i+len (Sgm0 X))
proof
  let X,Y be finite natural-membered set;
  assume that
A1: X <N< Y and
A2: i< len (Sgm0 Y);
  consider m being Nat such that
A3: Y c= Segm m by Th2;
  reconsider h=(Sgm0 (X \/ Y))/^(len (Sgm0 X)) as XFinSequence of NAT;
A4: len (Sgm0 X)=card X by Th20;
A5: len (Sgm0 Y)=card Y by Th20;
  then
A6: h.i=(Sgm0 (X \/ Y)).(i+card X) by A1,A2,A4,Th30;
A7: len (Sgm0 (X \/ Y))=card (X \/Y) by Th20;
  X/\Y=(X/\(Y/\NAT)) by A3,XBOOLE_1:1,28
    .= (X/\Y/\NAT) by XBOOLE_1:16
    .={} by A1,Th23;
  then X misses Y;
  then
A8: card Y +card X=card (X\/Y) by CARD_2:40;
  len h=len ((Sgm0 (X \/ Y))) -' len (Sgm0 X) by Def2
    .= card (X) + card Y -' card X by A8,A7,Th20
    .= card Y by NAT_D:34
    .= len (Sgm0 Y) by Th20;
  then
A9: len h=card Y by Th20;
A10: for l,m,k1,k2 being Nat st l < m & m < len h & k1=h.l & k2=h.m holds k1
  < k2
  proof
    let l,m,k1,k2 be Nat;
    assume that
A11: l < m and
A12: m < len h and
A13: k1=h.l and
A14: k2=h.m;
A15: m+card X <len (Sgm0 (X \/ Y)) by A8,A7,A9,A12,XREAL_1:6;
    set m3=m+card X;
    set l3=l+card X;
A16: l3<m3 by A11,XREAL_1:6;
    l < card Y by A9,A11,A12,XXREAL_0:2;
    then
A17: h.l= (Sgm0 (X \/ Y)).(l+card X) by A1,A4,Th30;
    h.m=(Sgm0 (X \/ Y)).(m+card X) by A1,A4,A9,A12,Th30;
    hence thesis by A13,A14,A17,A15,A16,Def4;
  end;
  rng h=Y by A1,A2,A4,A5,Th30;
  hence thesis by A4,A10,A6,Def4;
end;
