proposition: Lemma cons_lt_epsilon0 : forall a b n c, cons a b n c < epsilon0 -> nf (cons a b n c) -> a = zero /\ b < epsilon0 /\ c < epsilon0.
proof: Proof. intros a b n c H. inversion H using lt_02. split. apply lt_one_inv;auto with T2. split. unfold epsilon0. exact H1. apply transitivity with (cons a b n c);auto with T2. apply lt_tail;auto with T2. inversion 1. inversion 2. inversion 1. inversion 1. Qed.
imports: EPSILON0 Arith List Omega Compare_dec Relations Wellfounded Max Tools More_nat AccP not_decreasing EPSILON0 More_nat Gamma0_prelude Gamma0_length term rpo