proposition: Lemma ssubrVoid_ok : forall b, ssubr (Void char) b -> forall x, (x \in b) -> (x \in (Void char)).
proof: Proof. elim => //. move => r1 hr1 r2 hr2. rewrite ssubrPlus2; case/andP => h1 h2 x. rewrite -topredE /=; case/orP => hu. - by apply: hr1. by apply: hr2. 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)