proposition: Lemma equiv_union: forall L1 L2 L3 L4, Equiv L1 L3 -> Equiv L2 L4 -> Equiv (Union L1 L2) (Union L3 L4).
proof: Proof. split; intros. - eapply union_in_equiv; eauto. - eapply union_in_equiv; eauto. + apply equiv_sym. assumption. + apply equiv_sym. assumption. Qed.
imports: Coq.Lists.List