Proof (eqmod_sub modulus).
Hint Resolve eqm_sub: ints.

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