How do you selectively simplify arguments to each time a function is called, without evaluating the function itself?

Question

I'm using Coq 8.5pl1.

To make a contrived but illustrative example,

(* fix so simpl will automatically unfold. *)
Definition double := fix f n := 2*n.

Theorem contrived n : double (2 + n) = 2 + double (1 + n).

Now, I only want to simplify the arguments to double, and not any part outside of it. (For example, because the rest has already carefully been put into the correct form.)

simpl.
   S (S (n + S (S (n + 0)))) = S (S (S (n + S (n + 0))))

This converted the outside (2 + ...) to (S (S ...)) as well as unfolding double.

I can match one of them by doing:

match goal with | |- (double ?A) = _ => simpl A end.
   double (S (S n)) = 2 + double (1 + n)

How do I say that I want to simplify all of them? That is, I want to get

   double (S (S n)) = 2 + double (S n)

without having to put a separate pattern for each call to double.

I can simplify except for double itself with

remember double as x; simpl; subst x.
   double (S (S n)) = S (S (double (S n)))

Answer 1

You may find the ssreflect pattern selection language and rewrite mechanism useful here. For instance, you can have a finer grained control with patterns + the simpl operator /=:

From mathcomp Require Import ssreflect.
Definition double := fix f n := 2*n.
Theorem contrived n : double (2 + n) = 2 + double (1 + n).
rewrite ![_+n]/=.

Will display:

n : nat
============================
double (S (S n)) = 2 + double (S n)

You can also use anonymous rewrite rules:

rewrite (_ : double (2+n) = 2 + double (1+n)) //.

I would personally factor the rewrite in smaller lemmas:

Lemma doubleE n : double n = n + n.
Proof. by elim: n => //= n ihn; rewrite -!plus_n_Sm -plus_n_O. Qed.

Lemma doubleS n : double (1 + n) = 2 + double n.
Proof. by rewrite !doubleE /= -plus_n_Sm. Qed.

Theorem contrived n : double (1+n) = 2 + double n.
rewrite doubleS.