Proof using.
induction ys; intros; simpl in *.
- break_match; intuition auto using subseq_refl.
- repeat break_match.
+ auto.
+ discriminate.
+ simpl in *. subst. right. auto using subseq_app_r.
+ simpl in *. find_inversion. intuition.
rewrite app_comm_cons. auto.
Qed.

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