(******************************************  Copied From Coq.Logic.JMeq.  No longer involving JMeq_eq, which is potentially inconsistenty with homotopy type theory.  TODO: let reflexivity, symmetry and transitivity work again.  ******************************************)
Definition JMeq {A:Type} (x:A) {B:Type} (y: B): Prop :=
{H: @eq Type A B | @eq_rect Type A (fun T: Type => T) x B H = y}.
Hint Resolve JMeq_refl.
Hint Immediate JMeq_sym.
(** [JMeq] is equivalent to [eq_dep Type (fun X => X)] *)
(** [eq_dep U P p x q y] is strictly finer than [JMeq (P p) x (P q) y] *)
(** However, when the dependencies are equal, [JMeq (P p) x (P q) y] is as strong as [eq_dep U P p x q y] (this uses [JMeq_eq]) *)

Lemma JMeq_eq_dep :
forall (U:Type) (P:U->Prop) p q (x:P p) (y:P q),
p = q -> JMeq x y -> eq_dep U P p x q y.