Proof using.
induction xs; intros; simpl.
+ auto using subseq_refl.
+ break_match.
* auto.
* right. auto using subseq_nil.
Qed.

Lemma subseq_app_tail :
forall ys xs zs,
subseq (A:=A) xs ys ->
subseq (xs ++ zs) (ys ++ zs).