proposition: Lemma beq_rule_ok : forall a1 a2, beq_rule a1 a2 = true <-> a1 = a2.
proof: Proof. destruct a1 as [l1 r1]. destruct a2 as [l2 r2]. simpl. intuition. rewrite andb_eq, !beq_string_ok in H. destruct H. subst. refl. inversion H. subst. rewrite !(beq_refl beq_string_ok). refl. Qed.
imports: CoLoR(LogicUtil RelUtil ListUtil BoolUtil EqUtil ListDec)