proposition: Lemma union_assoc_in_1: forall w L1 L2 L3, In w (Union L1 (Union L2 L3)) -> In w (Union (Union L1 L2) L3).
proof: Proof. intros. destruct H as [H1|[H2|H3]]. - left. left. assumption. - left. right. assumption. - right. assumption. Qed.
imports: Coq.Lists.List