Proof using.
induction zs; intros; simpl in *;
repeat break_match; subst; simpl in *; intuition; subst; eauto;
right; (eapply IHzs; [|eauto]); simpl; eauto.
Qed.

Lemma subseq_In :
forall (ys xs : list A) x,
subseq xs ys ->
In x xs ->
In x ys.