proposition: Lemma rpo_2_2 : forall ta1 ta2 tb1 tb2 , rpo ta1 ta2 -> rpo tb1 (Term ord_psi (ta2:: tb2::nil)) -> rpo (Term ord_psi (ta1:: tb1 ::nil)) (Term ord_psi (ta2:: tb2 ::nil)).
proof: Proof. intros; apply Top_eq_lex. - cbn; auto with T2. - left. + auto with T2. + auto with T2. - inversion_clear 1; try subst s'. + apply rpo_trans with ta2;auto with T2. eapply Subterm; [| eleft]. now left. + destruct H2. * subst s'; auto with T2. * case H1. Qed.
imports: Coq(Arith List Lia Compare_dec Relations Wellfounded RelationClasses) hydras(Epsilon0) hydras(More_Arith Restriction) hydras(ON_Generic) hydras(rpo.term rpo.rpo) hydras(T2) hydras(Compat815)