proposition: Theorem app_assoc : forall X (lst1 lst2 lst3 : list X), lst1 ++ lst2 ++ lst3 = (lst1 ++ lst2) ++ lst3.
proof: Proof. intros X lst1 lst2 lst3. induction lst1 as [| h1 t1]. - simpl. reflexivity. - simpl. rewrite IHt1. reflexivity. Qed.