theorem Th19:
  G1 * idt F1 = idt (G1*F1)
proof
  now
    let a be Object of A;
    thus (G1*(idt F1))!a = G1.((idt F1)!a) by Th11
      .= G1.(idm (F1.a)) by FUNCTOR2:4
      .= idm (G1.(F1.a)) by FUNCTOR2:1
      .= idm ((G1*F1).a) by FUNCTOR0:33
      .= (idt (G1*F1))!a by FUNCTOR2:4;
  end;
  hence thesis by FUNCTOR2:3;
end;
