Proof using.
unfold and_tl; intuition.
Qed.

Lemma and_tl_map_conv :
forall (f: A->B) (P P': infseq A->Prop) (Q Q': infseq B->Prop),
(forall s, Q (map f s) -> P s) ->
(forall s, Q' (map f s) -> P' s) ->
forall (s: infseq A),
(Q /\_ Q') (map f s) -> (P /\_ P') s.