Proof using.
intros f P (x, s) nP; assumption.
Qed.

Lemma next_map :
forall (f: A->B) (P: infseq A->Prop) (Q: infseq B->Prop),
(forall s, P s -> Q (map f s)) ->
forall (s: infseq A), next P s -> next Q (map f s).