COQ 中子序列的传递性

Question

我一直在研究逻辑基础，并且在子序列练习的传递性上非常卡住。

(** **** 练习：2 星，高级（子序列）

A list is a _subsequence_ of another list if all of the elements
in the first list occur in the same order in the second list,
possibly with some extra elements in between. For example,

- (Optional, harder) Prove [subseq_trans] that subsequence is
  transitive -- that is, if [l1] is a subsequence of [l2] and [l2]
  is a subsequence of [l3], then [l1] is a subsequence of [l3].
  Hint: choose your induction carefully! *)

Inductive subseq : list nat -> list nat -> Prop :=
  | sseq_e (l2 : list nat) : subseq ([]) (l2)
  | sseq_m (l1 l2 : list nat) (n : nat) (H: subseq l1 l2) : subseq (n::l1) (n::l2)
  | sseq_nm (l1 l2 : list nat) (n : nat) (H: subseq l1 l2) : subseq (l1) (n::l2) 
  .

Theorem subseq_trans : forall (l1  l2 l3 : list nat),
  subseq l1 l2 ->
  subseq l2 l3 ->
  subseq l1 l3.
Proof.
intros l1 l2 l3 H H0.
induction H.
  -  apply sseq_e.
  - induction l3. 
  -- inversion H0. 
  -- inversion H0. 
  ---   apply sseq_m.

在尝试了几种不同的方法后，我无法获得正确的归纳假设。我尝试了多种方法并最终遇到这样一种情况，在我的假设中，我有类似 subseq l2 x::l3 的东西，但随后我需要证明 subseq l2 l3 这似乎是一个死胡同。任何正确方向的指示将不胜感激。

Answer 1

I have something like subseq l2 x::l3 but then I need to prove subseq l2 l3 which seems like a dead end.

该经验建议将归纳假设推广到 l3。

(* near the beginning of the proof *)
generalize dependent l3.
induction H.

COQ 中子序列的传递性

Transitivity of subsequence in COQ

coq