Proof (eqmod_neg modulus).
Hint Resolve eqm_neg: ints.

(* Specifications: *)
Lemma eqm_sub:
forall a b c d, eqm a b -> eqm c d -> eqm (a - c) (b - d).
Lemma eqm_sub:
forall a b c d, eqm a b -> eqm c d -> eqm (a - c) (b - d).