proposition: Lemma comp_hom_compatible : forall (a b c : C) (x x' : Hom b c) (y y' : Hom a b), Equal x x' -> Equal y y' -> Equal (comp_hom x y) (comp_hom x' y').
proof: Proof. intros a b c x x' y y' H' H'0; try assumption. unfold comp_hom in |- *; auto with algebra. Qed.
imports: Cartesian