proposition: Lemma remove_app : forall x l1 l2, remove x (l1 ++ l2) = remove x l1 ++ remove x l2.
proof: Proof. intros x l1; induction l1 as [|a l1 IHl1]; intros l2; simpl. - reflexivity. - destruct (eq_dec x a). + apply IHl1. + rewrite <- app_comm_cons; f_equal. apply IHl1. Qed.
imports: PeanoNat