Proof
fun x y z H H0
=> op_inv_l_uniq x
(ex_intro (fun y => op_is_inv_r x y) y (proj2 H))
y z (proj1 H) (proj1 H0).

(** Proves that the identity element is its own left inverse. *)
Theorem op_inv_0_l
:  op_is_inv_l 0 0.