
theorem Th13:
  for A,B,C being set, f being Function of [:A,B:],C st ~f is onto
  holds f is onto
proof
  let A,B,C be set, f be Function of [:A,B:],C;
A1: rng~f c= rng f by FUNCT_4:41;
  assume ~f is onto;
  then rng~f = C;
  hence rng f = C by A1;
end;
