Proof using.
induction ys; intros; simpl in *; break_match; intuition.
subst. simpl in *. specialize (IHys l). concludes. auto with *.
Qed.

Lemma subseq_subseq_eq :
forall (xs ys : list A),
subseq xs ys ->
subseq ys xs ->
xs = ys.