Proof
fun x y z H H0
=> ex_ind
(fun u H1
=>  op_intro_r (x + z) (y + z) u H0
|| a = (y + z) + u @a by op_is_assoc x z u
|| x + (z + u) = a @a by op_is_assoc y z u
|| x + a = y + a   @a by <- H1
|| a = y + 0       @a by <- op_id_r x
|| x = a           @a by <- op_id_r y)
H.

(** Proves that an element's left inverse is unique. *)
Theorem op_inv_l_uniq
:  forall x : M, has_inv_r x -> forall y z : M, op_is_inv_l x y -> op_is_inv_l x z -> z = y.