theorem Th115:
  rng f c= A implies f+~(x,y) = (id A+*(x,y))*f
proof
  assume
A1: rng f c= A;
  per cases;
  suppose
    x in A;
    then
A2: x in dom id A;
    thus f+~(x,y) = (id A)*f+*((x.-->y)*f) by A1,RELAT_1:53
      .= (id A+*(x.-->y))*f by Th10
      .= (id A+*(x,y))*f by A2,Def2;
  end;
  suppose
A3: not x in A;
    then
A4: not x in dom id A;
    not x in rng f by A1,A3;
    hence f+~(x,y) = f by FUNCT_4:103
      .= (id A)*f by A1,RELAT_1:53
      .= (id A+*(x,y))*f by A4,Def2;
  end;
end;
