proposition: Lemma compare_cons : forall a n b a' n' b', compare (cons a n b) (cons a' n' b') = match compare a a' with | Lt => Lt | Gt => Gt | Eq => (match n ?= n' with | Eq => compare b b' | comp => comp end) end.
proof: Proof. intros; 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)