这应该是一个直截了当的事情要证明，但我一直陷入困境。非常感谢帮助...
Require Import Arith.
Fixpoint At n (l:list nat) :=
match n with
| 0 => match l with nil => None | cons x _ => Some x end
| S n' => match l with nil => None | cons _ l' => At n' l' end
end.
Definition ordered l :=
forall i j, i < j -> j < length l ->
exists x y, At i l = Some x /\ At j l = Some y /\ x <= y.
Lemma ordered_cons:
forall l a b, ordered (cons b l) -> a <= b -> ordered (cons a (cons b l)).
intros l a b H H0 i j Hij Hj.
destruct i,j;
(* solve three of the four cases, and leave one untouched *)
try (inversion Hij; fail);
try (apply H; simpl in *; omega).
编辑：这是完成证明的解决方案。我需要考虑j的下两个案例。
destruct j;
try (exists a,b; auto; fail).
destruct (H 0 (S j)) as [x [y [Ha [Hb Hc]]]]; auto with arith;
simpl in *; inversion Ha; exists a, y;
repeat split; try tauto; omega.
Qed.