proposition: Lemma plus_cons_cons_eqn a n b a' n' b': (cons a n b) + (cons a' n' b') = match compare a a' with | Eq => cons a (S (n + n')) b' | Lt => cons a' n' b' | Gt => cons a n (T1add b (cons a' n' b')) end.
proof: Proof. reflexivity. Qed.
imports: Coq(Arith Max Bool Lia Compare_dec Relations Ensembles Wellfounded Bool RelationClasses Operators_Properties ArithRing Logic.Eqdep_dec) hydras(More_Arith Restriction DecPreOrder) hydras(OrdNotations)