Proof using.
unfold impl_tl; intuition.
Qed.

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