Proof using.
intros f P Q PQ.
apply always_map; apply eventually_map; assumption.
Qed.

Lemma inf_often_map_conv :
forall (f: A->B) (P: infseq A->Prop) (Q: infseq B->Prop),
extensional P -> extensional Q ->
(forall s, Q (map f s) -> P s) ->
forall (s: infseq A), inf_often Q (map f s) -> inf_often P s.