theorem Th6:
  Union (X --> Y) c= Y
proof
  let x be object;
  assume x in Union (X --> Y);
  then consider Z such that
A1: x in Z and
A2: Z in rng (X --> Y) by TARSKI:def 4;
  ex z being object st z in dom (X --> Y) & Z = (X --> Y).z
by A2,FUNCT_1:def 3;
  hence thesis by A1,FUNCOP_1:7;
end;
