proposition: Lemma ssubrPlus1 : forall r s0 s1, nPlus r -> ssubr (Plus s0 s1) r = ssubr s0 r || ssubr s1 r.
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)