Proof using.
intros P Q inv is_inv PeQ s invs ev. induction ev as [s Ps | x s ev IHev].
- apply PeQ; assumption.
- constructor 2. apply IHev. apply is_inv with x; assumption.
Qed.

Lemma not_eventually :
forall (P : infseq T -> Prop),
forall x s, ~ eventually P (Cons x s) -> ~ eventually P s.