Proof fun _ P x0 xs H
=> let H0
:  forall x, In x (x0 :: xs) -> P x
:= proj1 (Forall_forall P (x0 :: xs)) H in
let H1
:  forall x, In x xs -> P x
:= fun x H2
=> H0 x (or_intror (x0 = x) H2) in
proj2 (Forall_forall P xs) H1.
Arguments Forall_tail {A} {P} x0 xs.

(** Accepts two predicates, [P] and [Q], and a list, [xs], and proves that, if [P -> Q], and there exists an element in [xs] for which [P] is true, then there exists an element in [xs] for which [Q] is true. *)
Theorem Exists_impl
:  forall (A : Type) (P Q : A -> Prop),
(forall x : A, P x -> Q x) ->
forall xs : list A,
Exists P xs ->
Exists Q xs.