Proof using.
intros. induction xs; simpl in *; try break_if; intuition congruence.
Qed.

(* Specifications: *)
Lemma remove_app_comm :
forall a xs ys,
remove A_eq_dec a (xs ++ ys) = remove A_eq_dec a xs ++ remove A_eq_dec a ys.
Lemma filter_app : forall (f : A -> bool) xs ys,
filter f (xs ++ ys) = filter f xs ++ filter f ys.