(* The converse seems to require much more work *)

Definition fstAB := fst (A:=A) (B:=B).

Lemma inf_often_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), inf_often P s -> inf_often Q (map f s).