proposition: Lemma cons_nf : forall a n a' n' b, lt a' a -> nf a -> nf(cons a' n' b)-> nf(cons a n (cons a' n' b)).
proof: Proof. unfold nf. simpl. intros a n a' n' b' Hlta Ha. apply (bool_decide_eq_true _) in Hlta. destruct b'. - intro Hnfa'. now rewrite Ha, Hnfa'. - rewrite Ha. simpl. destruct b'2; intro H; now rewrite H. 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)