
theorem Th2:
  for X being set, Y being Subset of X holds rng ((id X)|Y) = Y
proof
  let X be set, Y be Subset of X;
  set f = id X;
A1: f|Y is Function of Y,X by FUNCT_2:32;
  hereby
    let y be object;
    assume y in rng (f|Y);
    then consider x being object such that
A2: x in dom (f|Y) and
A3: y = (f|Y).x by FUNCT_1:def 3;
    (f|Y).x = f.x by A2,FUNCT_1:47
      .= x by A2,FUNCT_1:17;
    hence y in Y by A1,A2,A3,FUNCT_2:def 1;
  end;
  let y be object;
  X = {} implies Y = {};
  then
A4: dom (f|Y) = Y by A1,FUNCT_2:def 1;
  assume
A5: y in Y;
  then (f|Y).y = f.y by FUNCT_1:49
    .= y by A5,FUNCT_1:18;
  hence thesis by A4,A5,FUNCT_1:def 3;
end;
