How to apply a function once during simplification

2019-06-22 07:49发布

From what I understand, function calls in Coq are opaque. Sometimes, I need to use unfold to apply it and then fold to turn the function definition/body back to its name. This is often tedious. My question is, is there an easier way to let apply a specific instance of a function call?

As a minimal example, for a list l, to prove right-appending [] does not change l:

Theorem nil_right_app: forall {Y} (l: list Y), l ++ [] = l.
Proof.
  induction l. 
    reflexivity. 

This leaves:

1 subgoals
Y : Type
x : Y
l : list Y
IHl : l ++ [] = l
______________________________________(1/1)
(x :: l) ++ [] = x :: l

Now, I need to apply the definition of ++ (i.e. app) once (pretending there are other ++ in the goal which I don't want to apply/expand). Currently, the only way I know to implement this one time application is to first unfold ++ and then fold it:

    unfold app at 1. fold (app l []).

giving:

______________________________________(1/1)
x :: l ++ [] = x :: l

But this is inconvenient as I have to figure out the form of the term to use in fold. I did the computation, not Coq. My question boils down to:

Is there a simpler way to implement this one-time function application to the same effect?

1条回答
爷的心禁止访问
2楼-- · 2019-06-22 08:19

You can use simpl, compute or vm_compute if you want to ask Coq to perform some computation for you. If the definition of the function is Opaque, the above solution will fail, but you could first prove a rewriting lemma such as:

forall (A:Type) (a:A) (l1 l2: list A), (a :: l1) ++ l2 = a :: (l1 ++ l2).

using your technique, and then rewrite with it when necessary.

Here is an example using simpl:

Theorem nil_right_app: forall {Y} (l: list Y), l ++ nil = l.
Proof.
(* solve the first case directly *)
intros Y; induction l as [ | hd tl hi]; [reflexivity | ]. 
simpl app. (* or simply "simpl." *)
rewrite hi.
reflexivity.
Qed.

To answer your comment, I don't know how to tell cbv or compute to only compute a certain symbol. Note that in your case, they seem to compute too eagerly and simpl works better.

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