Coq: Rewriting with 'forall' in hypothesis or goal

Question

I have proved 'correctness' of the reverse function on polymorphic Lists in Coq. The following proof works just fine, but I have a few questions about how the rewrite tactic works.

Here's the code:

Require Export Coq.Lists.List.
Import ListNotations.

Fixpoint rev {T:Type} (l:list T) : list T :=
  match l with
  | nil    => nil
  | h :: t => rev t ++ [h]
  end.

(* Prove rev_acc equal to above naive implementation. *)
Fixpoint rev_acc {T:Type} (l acc:list T) : list T :=
  match l with
  | nil => acc
  | h :: t => rev_acc t (h::acc)
  end.

Theorem app_assoc : forall  (T:Type) (l1 l2 l3 : list T),
  (l1 ++ l2) ++ l3 = l1 ++ (l2 ++ l3).
Proof.
Admitted.

Theorem rev_acc_correct : forall (T:Type) (l k :list T),
  rev l ++ k = rev_acc l k.
Proof.
  intros T l.
  induction l as [ | h l' IHl' ].
  - reflexivity.
  - simpl. 
    intro k.
    (* Why is "intro k" required for "rewrite -> app_assoc" *)
    (* But "rewrite -> IHl'" works regardless of "intro k".  *)
    (* generalize (rev l'), [h], k. *)
    rewrite -> app_assoc.
    simpl.
    rewrite -> IHl'.
    reflexivity.
Qed.

In the inductive step of the proof for rev_acc_correct if I skip intro k, then rewriting with app_assoc complains that it cannot find a matching subterm.

Found no subterm matching "(?M1058 ++ ?M1059) ++ ?M1060" in the current goal.

Here, I presume that the ? before the placeholder names denote that the terms are constrained, in this case to be of type List T for some type T; and since rev l' and [h] in the goal are instances of List T, one would expect a match in the goal.

On the other hand, rewriting with inductive hypothesis(rewrite -> IHl') instead of app_assoc goes through without needing an intro k before.

I find this behaviour of rewrite a bit confusing and the Coq manual doesn't provide any details. I don't want to have to read through the implementation but I need a good operational understanding of what the rewrite tactic does, especially with regards to how term matching works. Any answers/references in this direction are highly appreciated.

Your question is interesting! But since it provides an answer to a Software Foundations exercise, could you please delete it and post a minimal version that does not mention the exercise? — Arthur Azevedo De Amorim
I personally think the example is abstracted away enough from any particular SF exercise, the details are important to exposing these complications, and the question is more valuable than hiding this particular solution. Maybe you could just remove reference to Software Foundations in the question itself? — Tej Chajed
As far as I'm aware the only solution to an exercise in the above code is the implementation. I have made it as minimal as I can while still maintaining the context of the proof. Please feel free to edit if required. — Akay

Tej Chajed Tej Chajed · Accepted Answer · 2018-10-02T14:11:36

The complication with this rewrite is that there's a binder (the forall k), which can complicate things. If you just want things to work, use setoid_rewrite instead of rewrite and it will rewrite under binders.

rewrite IHl' looks like it happens under a binder, but the pattern being re-written doesn't actually involve the bound variable, so the binder isn't actually important. Here's what I mean: the goal is
```
forall k : list T, (rev l' ++ [h]) ++ k = rev_acc l' (h :: k)
```
which is the same thing as (that is, equal to):
```
 (fun l : list T => forall k : list T, l ++ k = rev_acc l' (h :: k)) (rev l' ++ [h])
```
which I got using pattern (rev l' ++ [h]) in Ltac. Now it's clear that you can just rewrite the part being applied to and ignore the binder. When you do rewrite IHl' Coq easily figures out that IHl should be specialized to [h] and the rewrite proceeds.
rewrite app_assoc, on the other hand, needs to be specialized to three lists, specifically rev l', [h], and k. It can't be specialized in the current context because the variable k is only bound underneath the forall. This is why the pattern (?x ++ ?y) ++ ?z doesn't appear in the goal.

So what do you actually do? You can of course introduce k so there is no binder, but there's a simpler and more general technique: Coq has generalized rewriting that can rewrite under binders, which you can use by instead calling setoid_rewrite (see Rewriting under binders in the Coq reference manual). The manual tells you you need to declare morphisms, but the relevant ones have all been implemented for you in this case for forall, so setoid_rewrite app_assoc will just work.

Note that while you can always introduce a forall to get rid of the binder, setoid_rewrite can be really handy when your goal is an exists. Rather than using eexists you can just rewrite under the binder.

Coq: Rewriting with 'forall' in hypothesis or goal

1 Answers