reserve f,g,h for Function,
  A for set;
reserve F for Function,
  B,x,y,y1,y2,z for set;
reserve x,z for object;
reserve X for non empty set,
  Y for set,
  F for BinOp of X,
  f,g,h for Function of Y,X,
  x,x1,x2 for Element of X;
reserve Y for non empty set,
  F for BinOp of X,
  f,g,h for Function of Y,X,
  x,x1,x2 for Element of X;
reserve Y for set,
  F for BinOp of X,
  f,g,h for Function of Y,X,
  x,x1,x2 for Element of X;
reserve Y for non empty set,
  F for BinOp of X,
  f,g,h for Function of Y,X,
  x,x1,x2 for Element of X;
reserve Y for set,
  F for BinOp of X,
  f,g,h for Function of Y,X,
  x,x1,x2 for Element of X;
reserve Y for non empty set,
  F for BinOp of X,
  f,g,h for Function of Y,X,
  x,x1,x2 for Element of X;
reserve Y for set,
  F for BinOp of X,
  f,g,h for Function
  of Y,X,
  x,x1,x2 for Element of X;

theorem
  for X,Y,Z being non empty set for f being Function of X, [:Y,Z:] holds
  rng (f~) = (rng f)~
proof
  let X,Y,Z be non empty set;
  let f be Function of X, [:Y,Z:];
  let x,y be object;
  thus [x,y] in rng (f~) implies [x,y] in (rng f)~
  proof
    assume [x,y] in rng (f~);
    then consider z being object such that
A1: z in dom (f~) and
A2: [x,y] = f~.z by FUNCT_1:def 3;
A3: z in dom f by A1,Def1;
    f.z = f~~.z
      .= [y,x] by A1,A2,Def1;
    then [y,x] in rng f by A3,FUNCT_1:def 3;
    hence thesis by RELAT_1:def 7;
  end;
  assume [x,y] in (rng f)~;
  then [y,x] in rng f by RELAT_1:def 7;
  then consider z being object such that
A4: z in dom f & [y,x] = f.z by FUNCT_1:def 3;
  z in dom (f~) & f~.z = [x,y] by A4,Def1;
  hence thesis by FUNCT_1:def 3;
end;
