Proof using.
cofix cf. intros (a, sA) (b, sB).
rewrite zip_Cons. rewrite map_Cons. constructor. apply cf.
Qed.

Lemma exteq_zip_map :
forall (f: A->B) (sA: infseq A) (sB: infseq B),
always (now (fun c: A*B => let (x, y) := c in y = f x)) (zip sA sB) ->
exteq sB (map f (map fstAB (zip sA (map f sA)))).