Proof using.
intros P invP.
cofix c.
intros [x s] Pxs; apply Always; trivial.
apply c; apply invP in Pxs.
assumption.
Qed.

Lemma always_Cons :
forall (x: T) (s: infseq T) P,
always P (Cons x s) -> P (Cons x s) /\ always P s.