# R E S P E C T - Find Out What It Means To The Coq Standard Library

Tags: coqMay 15 2020

By

**Lucas Silver**

## An Introduction To Rewriting In Coq

This blog post is a companion to the following code. Please follow along with that file.

Prerequisites:

Basic Coq Knowledge: at least through

`Relations.v`

from Software FoundationsBasic Typeclasses Knowledge: background from any language should be fine

## What is Rewriting?

In this blog post, I will introduce the technique of term *rewriting*, implemented by the `rewrite`

tactic in Coq. Informally, rewriting is the process of replacing a term in a mathematical proposition in a way that preserves the correctness of that statement. For instance, equal terms can be used interchangeably in any context, and integers equivalent modulo `k`

can be used interchangeable if the context cares only about their remainders when divided by `k`

and not any of their other features.

In informal mathematics, rewriting is so ubiquitous that it is not really even explicitly taught. We do it all the time without bothering to justify it. However, in a formal logical system like Coq, we cannot do anything without justification. In order to construct justifications for rewriting, we must first understand when exactly these rewrites are justified.

Suppose, we have some relation `R : A -> A -> Prop`

, two values `x y : A`

such that `R x y`

. We have some context `K`

. Our goal is `K[x]`

and we would prefer our goal to be `K[y]`

. To do this we want to rewrite `x`

into `y`

under `K`

. In order for this to be justified, we need to know that as long as we know `R x y`

, we also know that `K[y]`

implies `K[x]`

. To introduce some other vocabularly, to say `R`

is proper with respect to `K`

is the same as the previous sentence.

## Rewriting With Coq Equality

The `=`

relation in Coq is strong enough that it is proper under any relation. This means that, if we know `x = y`

, then for any context `K`

, we can transform a goal of `K[x]`

into one of `K[y]`

These notions of respect and proper are implemented by typeclasses in the Coq standard library. We will go over the exact instances we will need to provide later. For now, just keep in mind that Coq has provided a series of tactics that, when given certain type class instances, will do all of the justifications and transform your goal for you.

Let’s go through some concrete examples showing the strength of rewriting. Consider the following proposition and proof script.

```
Goal forall (a b c d : Z), a = b -> b = c -> c = d -> a = d.
Proof.
intros a b c d Hab Hbc Hcd. rewrite Hab. rewrite Hbc. auto.
Qed.
```

That is essentially the transitivity property of equality extended to 4 elements. Note how we prove this using two rewrites instead of two applications of the transitivity property. This instance of rewriting happens when the context is exactly the relation we are rewriting with. This simplifies the whole rewriting process.

Now let’s look at a slightly more complicated example. The following proposition is trivially simple with rewriting.

```
Goal forall (a b c x y z : Z), a = x -> b = y -> c = z ->
a + b + c = x + y + z.
Proof.
intros a b c x y z Hax Hby Hcz. rewrite Hax. rewrite Hby. rewrite Hcz.
reflexivity.
Qed.
```

Unlike the previous example, it is not immediately obvious how we would prove this proposition without rewriting. In this case we are rewriting integers under the context of addition. In fact, we can rewrite using `=`

under any context. This is a very useful property of the `=`

relation that is, unsurprisingly, not true for arbitrary relations. Now, we will move on to discuss how to generalized rewriting with arbitrary relations.

## Rewriting Under Arbitrary Equivalences

Suppose we have an integer `k`

and consider modular arithmetic modulo `k`

. So our relation `x ≡ y`

means that `x`

is equivalent to `y`

mod `k`

. The definition given in the file uses Bezout coefficients, which is equivalent to a division based definition. However, this is not really important for our purposes. Note that `≡`

is the notation we will use for the relation defined by `equiv`

in the file.

To prove that a relation is an equivalence relation, we must show that it is reflexive, symmetry and transitive. In isolation, proving each of these properties enables useful tactics. `reflexivity`

lets you discharge goals of the form `x ≡ x`

, and `symmetry`

lets you change goals and hypotheses from `x ≡ y`

to `y ≡ x`

. In order to do the most basic rewriting, all that you need is the transitive property.

We can declare the typeclass instance with the following code.

The proof is unrelated to the subject of this post, so it has been elided. With the `Transitive`

instance defined, we can now demonstrate some simple rewriting.

```
Goal forall (a b c : Z), a ≡ b -> b ≡ c -> a ≡ c.
Proof.
intros. rewrite H. Fail (rewrite <- H).
auto.
Qed.
```

This is just a statement of the transitive property for the relation. Run the code for yourself, and note that we can rewrite `a`

into `b`

, but when we try to rewrite `b`

back into `a`

it fails. Because we only proved this relation to be `Transitive`

, we can only rewrite in the goal by replacing the value on the left of the relation with the value on the right (and vice-versa when rewriting in hypotheses).

You can build up more intuition for this by practicing rewriting with relations that are only transitive, and not symmentric, like the `<=`

relation.

In order to rewrite in both directions, we will need to show that our relation is `Symmetric`

in addition to being `Transitive`

.

Now that we have this instance, we can return to the previous example.

```
Goal forall (a b c : Z), a ≡ b -> b ≡ c -> a ≡ c.
Proof.
intros. rewrite H. rewrite <- H. rewrite <- H0.
rewrite H0. rewrite H. auto.
Qed.
```

Note that some of these rewrites would not have worked previously.

In general, you can declare the relation to be an `Equivalence`

relation in order to get all the benefits of `Transitive`

, `Symmetric`

, and `Reflexive`

properties without declaring them separately.

Now we can try something similar to our first example. Once again we will be able to complete the proof only using rewrites, without specifically apply any lemmas or knowing anything about the structure of our relation.

```
Goal forall (a b c d : Z), a ≡ b -> b ≡ c -> c ≡ d -> a ≡ d.
Proof.
intros a b c d Hab Hcb Hcd. rewrite Hab. rewrite Hcb. auto.
Qed.
```

## Rewriting Under Arbitrary Contexts

Now let us proceed to a more complicated example. We can start by observing and proving that `k ≡ 0`

.

We might want to prove some basic propositions by rewriting `k`

into `0`

. But follow in the companion code what happens with the following proof script.

So it didn’t work. The terrifying error message you should have seen is related to typeclass resolution in Coq and is not obviously helpful.

Even without useful error messages to guide us, we can still figure out what went wrong. Recall what I mentioned earlier about rewriting under different contexts. We need to prove that `≡`

respects itself under the context of addition. It makes sense that we should need to prove this, because we can easily construct functions that are not proper. For instance, consider

which clearly distinguishes between `0`

and `k`

in general. For example, let `k = 5`

. `f 0 0 = 0`

and `f 5 5 = 1`

.

In the scary error message you may have noticed the term `Proper`

and the notation `==>`

. So we can start our search there.

`Proper`

is the function that takes a relation `R`

, a single element `a`

and uses `a`

for both of `R`

’s inputs. We provide some more basic examples in the attached code. We haven’t gotten very far with this definition. So let’s try to learn more about the `==>`

. Note that `==>`

is notation for the function `respectful`

.

```
Definition respectful {A B : Type} (R : A -> A -> Prop)
(R' : B -> B -> Prop) : (A -> B) -> (A -> B) -> Prop :=
fun f g => forall x y, R x y -> R' (f x) (g y)
```

Now we have found an interesting looking definition! `respectful`

is a function that lifts relations over inputs and outputs to relations over functions. `respectful R R' f g`

is true if, given any inputs `x,y`

that satisfy `R`

, `f x, g y`

must satisfy `R'`

.

The specific typeclass we are interested in instantiating is the application of `Proper`

to `respectful R R'`

. This is equivalent to applying `respectful R R'`

to the same context `K`

twice. This gives us the proposition

This should start to look like the informal definitions from the beginning of this post. Consider the following proposition

After introducing the first 3 variables, we have, in our context, `x,y : Z`

and `H : x ≡ y`

, and we have, as our goal, `x + 2 ≡ y + 2`

. For a general rule consider the following proposition

It means that given two sequences of variables `x1 ... xn`

and `y1 .. yn`

such that given any `i`

, `Ri xi yi`

,

So all of the relations to the left of a `==>`

relate arguments and the rightmost relation relates the outputs.

We have everything in place to begin some rewriting. Declaring the `Proper`

instance has given Coq all of the tools it needs to look under the context of addition and replace the equivalent terms when we invoke the `rewrite`

tactic. Now we can return to our modular arithmetic example. First let us prove that addition respects `≡`

in its second argument. Note that `equiv`

is the same relation as `≡`

.

Now we can rewrite in the second argument of addition. Let us put that to the test in a simple example.

```
Goal forall x, x + k ≡ x.
Proof.
intros. rewrite k_equiv_0.
rewrite Z.add_comm. simpl. reflexivity.
Qed.
```

Note that we still can’t rewrite in the left argument.

In order to rewrite in the left argument, we need to show that addition is proper in all of its arguments. This proof follows from our existing proper instance and the fact that addition is commutative.

```
Instance add_proper : Proper (equiv ==> equiv ==> equiv) Z.add.
Proof.
intros x y Hxy z w Hzw.
rewrite <- Hzw. Fail rewrite Hxy.
rewrite Z.add_comm. rewrite Hxy. rewrite Z.add_comm.
reflexivity.
Qed.
```

Now we can stress test our rewriting with the following example.

```
Goal forall x : Z,
k + (x + ( k + k ) + k) ≡
k + k + (k + k) + ( (k + k) + (k + k) ) + x.
Proof.
intros. rewrite k_equiv_0. simpl. rewrite Z.add_comm. simpl.
rewrite Z.add_comm. reflexivity.
Qed.
```

Note that there is only one `x`

on each side of the equivalence, so once we replace all `k`

’s with `0`

, the proof is vastly simpler.

## Rewriting Equivalent Notions of Computation

To wrap up I will introduce another example where rewriting is incredibly useful. Suppose we want to model stateful computations in Coq, where there are no stateful operations. We can do that with the following type (where our type of states is `S`

).

This captures the notion that all stateful computations require an input state and return an output state as well as a result. We can give this type a commonly used interface and specification by making it a monad. First we define the `ret`

and `bind`

functions.

```
Definition ret {A} (a : A) := fun s : S => (s,a).
Definition bind {A B} (m : State A) (f : A -> State B) :=
fun s => let '(s',a) := m s in f a s'.
```

`ret`

captures the notion of a pure computation returning `a`

, and therefore leaves its input state alone.

`bind`

captures the notion of concatenating two stateful computations by running the first one and threading the output state as input to the next one. It actually generalizes that notion a bit, taking as its argument a function `f : A -> State B`

, which can be views as a family of stateful computations indexed over `A`

. As is customary, we denote `bind`

with the infix operator `>>=`

. To learn more about the State monad, check out this page from the Haskell Wiki.

We can create a notion of stateful computation equivalence with the following function.

We can denote this equivalence with `≈`

, and provide an instance of the `Equivalence`

type class.

To follow the monad specification, these functions must satisfy the following equations.

```
Lemma bind_ret : forall (A B : Type) (a : A) (f : A -> State B),
ret a >>= f ≈ f a.
Proof.
...
Qed.
Lemma ret_bind : forall (A : Type) (m : State A),
m >>= ret ≈ m.
Proof.
...
Qed.
Lemma bind_bind : forall (A B C : Type) (m : State A)
(f : A -> State B) (g : B -> State C),
(m >>= f) >>= g ≈ (m >>= (fun a => f a >>= g)).
Proof.
...
Qed.
```

Now suppose we want to rewrite these equations. We currently can’t, because we have no `Proper`

instances for `bind`

or `ret`

with respect to state equivalence. This would be a nice thing to have because these equations require no knowledge of the definitions of `ret`

and `bind`

.

First, we can lift our equivalence over `State B`

to an equivalence over `A -> State B`

using the `pointwise_relation`

function.

```
Definition pointwise_relation {A B : Type} (R : B -> B -> Prop) :
(A -> B) -> (A -> B) -> Prop :=
fun f g => forall a, R (f a) (g a)
```

This captures the notion that given equal inputs, `f`

and `g`

produce related outputs.

Now we want to show that `bind`

respects the stateful equivalence relation given arguments that are similarly equivalent.

```
Instance proper_monad {A B: Type} : Proper
(@state_eq A ==> pointwise_relation A state_eq ==> @state_eq B)
(bind).
Proof.
...
Qed.
```

Now we can rewrite the monad laws under bind. Consider the following example.

```
Goal forall (A B :Type) (a : A) (f : A -> State B),
ret a >>= (fun a' => f a') >>= ret ≈ f a.
Proof.
intros. rewrite bind_bind. rewrite bind_ret. rewrite ret_bind.
reflexivity.
Qed.
```

## Conclusion

In addition to the `rewrite`

tactic discussed in this post, Coq has a similar, but more powerful `setoid_rewrite`

tactic. This tactic is enabled by the same typeclasses as `rewrite`

, but is capable of rewriting underneath contexts like universal or existential quantifiers. As a rule of thumb, if you feel like a rewrite should work, try to use `setoid_rewrite`

. To learn more about `setoid_rewrite`

, checkout the relevant sections of the Coq documentation here.

Hopefully, you found this to be a useful introduction to rewriting in Coq. There is much more to learn, but you should have a strong enough foundation to learn it on your own. Good luck proving things!