Proof using.
unfold and_tl; intuition.
Qed.

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