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 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;
