
theorem Th17:
  for A,B,x being set holds not x in A implies ((id A)+*(B --> x)) "({x}) = B
proof
  let A,B,x be set;
  set f = (id A)+*(B --> x);
  assume
A1: not x in A;
  thus f"({x}) c= B
  proof
    let y be object;
    assume
A2: y in f"({x});
    then
A3: y in dom f by FUNCT_1:def 7;
A4: f.y in {x} by A2,FUNCT_1:def 7;
    per cases;
    suppose
      y in dom (B --> x);
      hence thesis;
    end;
    suppose
A5:   not y in dom (B --> x);
      then
A6:   f.y = (id A).y by FUNCT_4:11;
A7:   y in dom (B --> x) or y in dom (id A) by A3,FUNCT_4:12;
      then (id A).y = y by A5,FUNCT_1:18;
      then y = x by A4,A6,TARSKI:def 1;
      hence thesis by A1,A7;
    end;
  end;
  let y be object;
  assume
A8: y in B;
  then
A9: y in dom (B-->x) by FUNCOP_1:13;
  then f.y = (B-->x).y by FUNCT_4:13;
  then f.y = x by A8,FUNCOP_1:7;
  then
A10: f.y in {x} by TARSKI:def 1;
  y in dom f by A9,FUNCT_4:12;
  hence thesis by A10,FUNCT_1:def 7;
end;
