Proof using.
intros f P Q PQ [x s]; apply PQ.
Qed.

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