proposition: Theorem zip_map_eq : forall (A B C : Set)(ls : list A)(f1 : A -> B)(f2 : A -> C), zip (map f1 ls) (map f2 ls) = map (fun a => (f1 a, f2 a)) ls.
proof: Proof. induction ls; intuition; simpl in *. f_equal. eauto. Qed.
imports: FCF.FCF Permutation