Proving `forall x xs ys, subseq (x :: xs) ys -> subseq xs ys` in Coq

Question

I have the following definition

Inductive subseq : list nat -> list nat -> Prop :=
| empty_subseq : subseq [] []
| add_right : forall y xs ys, subseq xs ys -> subseq xs (y::ys)
| add_both : forall x y xs ys, subseq xs ys -> subseq (x::xs) (y::ys)
.

Using this, I wish to prove the following lemma

Lemma del_l_preserves_subseq : forall x xs ys, subseq (x :: xs) ys -> subseq xs ys.

So, I tried looking at the proof of subseq (x :: xs) ys by doing destruct H.

Proof.
  intros. induction H.

3 subgoals (ID 209)

  x : nat
  xs : list nat
  ============================
  subseq xs [ ]

subgoal 2 (ID 216) is:
 subseq xs (y :: ys)
subgoal 3 (ID 222) is:
 subseq xs (y :: ys)

Why does the first subgoal ask me to prove subseq xs []? Shouldn't the destruct tactic know that the proof cannot be of the form empty_subseq since the type contains x :: xs and not []?

In general how do I prove the lemma that I am trying to prove?

score 5 · Accepted Answer · answered Jan 26 '19 at 12:55

Shouldn't the destruct tactic know that the proof cannot be of the form empty_subseq since the type contains x :: xs and not []?

In fact, destruct doesn't know that much. It just replaces x :: xs and xs with [] and [] in the empty_subseq case. In particular, this frequently leads to lost information in the context. Better alternatives:

Use inversion instead of destruct.
Use remember to ensure both type indices of subseq are variables before destruct. (remember (x :: xs) as xxs in H.) This more explicit goal management also works well with induction.

score 0 · Answer 2 · answered Jan 27 '19 at 06:45

Li-yao's answer was actually useful. This is a proof of the lemma.

Lemma del_l_preserves_subseq : forall x xs ys, subseq (x :: xs) ys -> subseq xs ys.
Proof.
  intros x xs ys.
  induction ys as [|y ys'].
  - intros. inversion H. (* Inversion will detect that no constructor matches the type of H *)
  - intros. inversion H. (* Inversion will automatically discharge the first case *)
    + (* When [subseq (x :: xs) ys'] holds *)
      apply IHys' in H2. now apply add_right.
    + (* When [subseq xs ys'] holds *)
      now apply add_right.
Qed

Proving `forall x xs ys, subseq (x :: xs) ys -> subseq xs ys` in Coq

2 Answers2