Proof using.
intros f J P K Q JK PQ s un.
induction un.
- apply U0, PQ. assumption.
- rewrite map_Cons.
apply U_next.
* rewrite <- map_Cons.
apply JK.
assumption.
* assumption.
Qed.

Lemma release_map :
forall (f: A->B) (J P: infseq A->Prop) (K Q: infseq B->Prop),
(forall s, J s -> K (map f s)) ->
(forall s, P s -> Q (map f s)) ->
forall (s: infseq A),
release J P s -> release K Q (map f s).