Proof using.
intros f P (x, (y, s)) nP; assumption.
Qed.

Lemma always_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), always P s -> always Q (map f s).