Proof using.
intros x s P al. case (always_Cons x s P al); trivial.
Qed.

(* Specifications: *)
Lemma always_now' :
forall (P : infseq T -> Prop) ex,
always P ex ->
P ex.
Lemma always_tl :
forall (s: infseq T) P, always P s -> always P (tl s).