proposition: Lemma rpo_3_2 : forall ta1 tb1 tb2 , rpo tb1 tb2 -> rpo (Term ord_psi (ta1:: tb1 ::nil)) (Term ord_psi (ta1:: tb2 ::nil)).
proof: Proof. intros; apply Top_eq_lex. - cbn; auto with T2. - right; left; auto with T2. - inversion_clear 1; try subst s'. + eapply Subterm. * eleft; reflexivity. * left; destruct H1; unfold In in H; try subst s'. + eapply rpo_trans with tb2;auto with T2. * inversion_clear H1; subst; auto. inversion H0. * eapply Subterm; [| eleft]. right; now left. 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)