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

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

Lemma eventually_map_monotonic : forall (f: A->B) (P Q J: infseq A->Prop) (K : infseq B -> Prop), (forall x s, J (Cons x s) -> J s) -> (forall x s, K (map f (Cons x s)) -> K (map f s)) -> (forall s, J s -> K (map f s) -> Q s -> P s) -> forall s, J s -> K (map f s) -> eventually Q s -> eventually P s.