(** In fact, [fold] on empty sets is more than equivalent to the initial element, it is Leibniz-equal to it. *)

(* Specifications: *)
Lemma fold_1b :
forall s (A : Type)(i : A) (f : elt -> A -> A),
Empty s -> (fold f s i) = i.
Lemma fold_1b :
forall s (A : Type)(i : A) (f : elt -> A -> A),
Empty s -> (fold f s i) = i.