proposition: Lemma ssublPlus1 : forall r s0 s1, nPlus r -> ssubl r (Plus s0 s1) = ssubl r s0 || ssubl r s1.
proof: Proof. by case. Qed.
imports: Coq(RelationClasses Setoid Morphisms Permutation) mathcomp(ssreflect ssrfun ssrbool eqtype ssrnat seq fintype) mathcomp(bigop path) RegLang(languages) RegexpBrzozowski(regexp finite_der equiv)