Proof (eqmod_refl2 modulus).
Hint Resolve eqm_refl2: ints.

(* Specifications: *)
Lemma eqm_sym: forall x y, eqm x y -> eqm y x.
Proof (eqmod_sym modulus).
Lemma eqm_sym: forall x y, eqm x y -> eqm y x.
Proof (eqmod_sym modulus).