Proof using.
unfold not_tl; intuition.
Qed.

Lemma not_tl_map_conv :
forall (f: A->B) (P: infseq A->Prop) (Q: infseq B->Prop),
(forall s, P s -> Q (map f s)) ->
forall (s: infseq A), (~_ Q) (map f s) -> (~_ P) s.