
theorem Th14:
for X,Y be non empty set, f be Function of X,Y st f is bijective holds
 ex g be Function of Y,X st g is bijective & g = f" & .:g = (.:f)"
proof
    let X,Y be non empty set, f be Function of X,Y;
    assume
A1: f is bijective; then
A2: rng f = dom(f") & dom f = rng(f") by FUNCT_1:33;
A3: dom f = X & rng f = Y by A1,FUNCT_2:def 1,def 3; then
    reconsider g = f" as Function of Y,X by A2,FUNCT_2:1;

    take g;
A4: g is onto by A2,A3,FUNCT_2:def 3;
    hence g is bijective by A1;
    thus g = f";

    .:f is bijective by A1,Th1; then
A5: dom (.:f) = bool X & rng (.:f) = bool Y by FUNCT_2:def 1,def 3;

    .:g is bijective by A1,A4,Th1; then
A6:dom (.:g) = bool Y & rng (.:g) = bool X by FUNCT_2:def 1,def 3;

    for x,y be object st x in dom (.:f) & y in dom (.:g) holds
     (.:f).x = y iff (.:g).y = x
    proof
     let x,y be object;
     assume
A7: x in dom (.:f) & y in dom (.:g); then
     reconsider s = x as Subset of X;
     reconsider t = y as Subset of Y by A7;
     now assume (.:f).x = y; then
      y = f.:s by A1,Th1; then
      (.:g).y = g.:(f.:s) by A1,A4,Th1; then
      (.:g).y =f"(f.:s) by FUNCT_1:85,A1;
      hence (.:g).y =x by A1,A3,FUNCT_1:94;
     end;
     hence (.:f).x = y implies (.:g).y = x;

     assume (.:g).y = x; then
     x = g.:t by A1,A4,Th1; then
     (.:f).x = f.:(g.:t) by A1,Th1; then
     (.:f).x =(g").:(g.:t) by FUNCT_1:43,A1; then
     (.:f).x =g"(g.:t) by A1,FUNCT_1:85;
     hence (.:f).x = y by A1,A2,A3,FUNCT_1:94;
    end;
    hence .:g = (.:f)" by A1,A5,A6,FUNCT_1:38;
end;
