# Proving Algebraic Datatypes are “Algebraic”

Several programming languages allow programmers to define (potentially recursive) custom types, by composing together existing ones. For instance, in OCaml, one can define lists as follows:

```
type 'a list =
| Cons of 'a * 'a list
| Nil
```

This translates in Haskell as

```
data List a =
Cons a (List a)
| Nil
```

In Rust as

```
enum List<A> {
Cons(A, Box<List<a>>),
Nil,
}
```

Or in Coq as

```
Inductive list a :=
| cons : a -> list a -> list a
| nil
```

And so forth.

Each language will have its own specific constructions, and the type systems
of OCaml, Haskell, Rust and Coq —to only cite them— are far from being
equivalent. That being said, they often share a common “base formalism,”
usually (and sometimes abusively) referred to as *algebraic datatypes*. This
expression is used because under the hood any datatype can be encoded as a
composition of types using two operators: sum ($+$) and product ($*$) for
types.

- $a + b$ is the disjoint union of types $a$ and $b$. Any term of $a$ can be injected into $a + b$, and the same goes for $b$. Conversely, a term of $a + b$ can be projected into either $a$ or $b$.
- $a * b$ is the Cartesian product of types $a$ and $b$. Any term of $a * b$ is made of one term of $a$ and one term of $b$ (think tuples).

For an algebraic datatype, one constructor allows for defining “named tuples,” that is ad hoc product types. Besides, constructors are mutually exclusive: you cannot define the same term using two different constructors. Therefore, a datatype with several constructors is reminiscent of a disjoint union. Coming back to the $\mathrm{list}$ type, under the syntactic sugar of algebraic datatypes, we can define it as

$\mathrm{list}_\alpha \equiv \mathrm{unit} + \alpha * \mathrm{list}_\alpha$

where $\mathrm{unit}$ models the `nil`

case, and $α * \mathrm{list}_\alpha$
models the `cons`

case.

The set of types which can be defined in a language together with $+$ and $*$ form an “algebraic structure” in the mathematical sense, hence the name. It means the definitions of $+$ and $*$ have to satisfy properties such as commutativity or the existence of neutral elements. In this article, we will prove some of them in Coq. More precisely,

- $+$ is commutative, that is $\forall (x, y),\ x + y = y + x$
- $+$ is associative, that is $\forall (x, y, z),\ (x + y) + z = x + (y + z)$
- $+$ has a neutral element, that is $\exists e_s,\ \forall x,\ x + e_s = x$
- $*$ is commutative, that is $\forall (x, y),\ x * y = y * x$
- $*$ is associative, that is $\forall (x, y, z),\ (x * y) * z = x * (y * z)$
- $*$ has a neutral element, that is $\exists e_p,\ \forall x,\ x * e_p = x$
- The distributivity of $+$ and $*$, that is $\forall (x, y, z),\ x * (y + z) = x * y + x * z$
- $*$ has an absorbing element, that is $\exists e_a,\ \forall x, \ x * e_a = e_a$

For the record, the `sum`

($+$) and `prod`

($*$) types are defined
in Coq as follows:

```
Inductive sum (A B : Type) : Type :=
| inl : A -> sum A B
| inr : B -> sum A B
Inductive prod (A B : Type) : Type :=
| pair : A -> B -> prod A B
```

## An Equivalence for `Type`

Algebraic structures come with *equations* expected to be true. This means
there is an implicit dependency which is —to my opinion— too easily overlooked:
the definition of $=$. In Coq, `=`

is a built-in relation that states
that two terms are “equal” if they can be reduced to the same “hierarchy” of
constructors. This is too strong in the general case, and in particular for our
study of algebraic structures of `Type`

. It is clear that, to Coq’s
opinion, $\alpha + \beta$ is not structurally *equal* to $\beta + \alpha$, yet
we will have to prove they are “equivalent.”

### Introducing `type_equiv`

Since `=`

for `Type`

is not suitable for reasoning about algebraic
datatypes, we introduce our own equivalence relation, denoted `==`

. We
say two types $\alpha$ and $\beta$ are equivalent up to an isomorphism
when for any term of type $\alpha$, there exists a counterpart term of type
$\beta$ and vice versa. In other words, $\alpha$ and $\beta$ are equivalent if
we can exhibit two functions $f$ and $g$ such that:

$\forall (x : α),\ x = g(f(x))$

$\forall (y : β),\ y = f(g(y))$

This translates into the following inductive type.

```
Reserved Notation "x == y" (at level 72).
Inductive type_equiv (α β : Type) : Prop :=
| mk_type_equiv (f : α -> β) (g : β -> α)
(equ1 : forall (x : α), x = g (f x))
(equ2 : forall (y : β), y = f (g y))
: α == β
where "x == y" := (type_equiv x y).
```

As mentioned earlier, we prove two types are equivalent by exhibiting
two functions, and proving these functions satisfy two properties. We
introduce a `Ltac`

notation to that end.

```
Tactic Notation "equiv" "with" uconstr(f) "and" uconstr(g)
:= apply (mk_type_equiv f g).
```

The tactic `equiv with f and g`

will turn a goal of the form `α == β`

into two subgoals to prove `f`

and `g`

form an isomorphism.

`type_equiv`

is an Equivalence

We can prove it by demonstrating it is

- Reflexive,
- Symmetric, and
- Transitive.

`type_equiv`

is reflexive

This proof is straightforward. A type $\alpha$ is equivalent to itself because:

$\forall (x : α),\ x = id(id(x))$

```
Lemma type_equiv_refl (α : Type) : α == α.
Proof.
now equiv with (@id α) and (@id α).
Qed.
```

`type_equiv`

is symmetric

If $\alpha = \beta$, then we know there exists two functions $f$ and $g$ which satisfy the expected properties. We can “swap” them to prove that $\beta == \alpha$.

```
Lemma type_equiv_sym {α β} (equ : α == β) : β == α.
Proof.
destruct equ as [f g equ1 equ2].
now equiv with g and f.
Qed.
```

`type_equiv`

is transitive

If $\alpha = \beta$, we know there exists two functions $f_\alpha$ and
$g_\beta$ which satisfy the expected properties of `type_equiv`

.
Similarly, because $\beta = \gamma$,
we know there exists two additional functions $f_\beta$ and $g_\gamma$. We can
compose these functions together to prove $\alpha = \gamma$.

As a reminder, composing two functions $f$ and $g$ (denoted by `f >>> g`

thereafter) consists in using the result of $f$ as the input of $g$.

```
Infix ">>>" := (fun f g x => g (f x)) (at level 70).
Lemma type_equiv_trans {α β γ} (equ1 : α == β) (equ2 : β == γ)
: α == γ.
Proof.
destruct equ1 as [fα gβ equαβ equβα],
equ2 as [fβ gγ equβγ equγβ].
equiv with (fα >>> fβ) and (gγ >>> gβ).
+ intros x.
rewrite <- equβγ.
now rewrite <- equαβ.
+ intros x.
rewrite <- equβα.
now rewrite <- equγβ.
Qed.
```

#### Conclusion

The Coq standard library introduces the `Equivalence`

type class. We can
provide an instance of this type class for `type_equiv`

, using the three
lemmas we have proven in this section.

```
#[refine]
Instance type_equiv_Equivalence : Equivalence type_equiv :=
{}.
Proof.
+ intros x.
apply type_equiv_refl.
+ intros x y.
apply type_equiv_sym.
+ intros x y z.
apply type_equiv_trans.
Qed.
```

### Examples

`list`

’s canonical form

We now come back to our initial example, given in the introduction of this
write-up. We can prove our assertion, that is `list α == unit + α * list α`

.

```
Lemma list_equiv (α : Type)
: list α == unit + α * list α.
Proof.
equiv with (fun x => match x with
| [] => inl tt
| x :: rst => inr (x, rst)
end)
and (fun x => match x with
| inl _ => []
| inr (x, rst) => x :: rst
end).
+ now intros [| x rst].
+ now intros [[] | [x rst]].
Qed.
```

`list`

is a morphism

This means that if `α == β`

, then `list α == list β`

. We prove this
by defining an instance of the `Proper`

type class.

```
Instance list_Proper
: Proper (type_equiv ==> type_equiv) list.
Proof.
add_morphism_tactic.
intros α β [f g equαβ equβα].
equiv with (map f) and (map g).
all: setoid_rewrite map_map; intros l.
+ replace (fun x : α => g (f x))
with (@id α).
++ symmetry; apply map_id.
++ apply functional_extensionality.
apply equαβ.
+ replace (fun x : β => f (g x))
with (@id β).
++ symmetry; apply map_id.
++ apply functional_extensionality.
apply equβα.
Qed.
```

The use of the `Proper`

type class allows for leveraging hypotheses of
the form `α == β`

with the `rewrite`

tactic. I personally consider
providing instances of `Proper`

whenever it is possible to be a good
practice, and would encourage any Coq programmers to do so.

`nat`

is a special purpose `list`

Did you notice? Now, using `type_equiv`

, we can prove it!

```
Lemma nat_and_list : nat == list unit.
Proof.
equiv with (fix to_list n :=
match n with
| S m => tt :: to_list m
| _ => []
end)
and (fix of_list l :=
match l with
| _ :: rst => S (of_list rst)
| _ => 0
end).
+ induction x; auto.
+ induction y; auto.
rewrite <- IHy.
now destruct a.
Qed.
```

#### Non-empty lists

We can introduce a variant of `list`

which contains at least one element
by modifying the `nil`

constructor so that it takes one argument instead
of none.

```
Inductive non_empty_list (α : Type) :=
| ne_cons : α -> non_empty_list α -> non_empty_list α
| ne_singleton : α -> non_empty_list α.
```

We can demonstrate the relation between `list`

and
`non_empty_list`

, which reveals an alternative implementation of
`non_empty_list`

. More precisely, we can prove that `forall (α : Type), non_empty_list α == α * list α`

. It is a bit more cumbersome, but not that
much. We first define the conversion functions, then prove they satisfy the
properties expected by `type_equiv`

.

```
Fixpoint non_empty_list_of_list {α} (x : α) (l : list α)
: non_empty_list α :=
match l with
| y :: rst => ne_cons x (non_empty_list_of_list y rst)
| [] => ne_singleton x
end.
#[local]
Fixpoint list_of_non_empty_list {α} (l : non_empty_list α)
: list α :=
match l with
| ne_cons x rst => x :: list_of_non_empty_list rst
| ne_singleton x => [x]
end.
Definition list_of_non_empty_list {α} (l : non_empty_list α)
: α * list α :=
match l with
| ne_singleton x => (x, [])
| ne_cons x rst => (x, list_of_non_empty_list rst)
end.
Lemma ne_list_list_equiv (α : Type)
: non_empty_list α == α * list α.
Proof.
equiv with list_of_non_empty_list
and (prod_curry non_empty_list_of_list).
+ intros [x rst|x]; auto.
cbn.
revert x.
induction rst; intros x; auto.
cbn; now rewrite IHrst.
+ intros [x rst].
cbn.
destruct rst; auto.
change (non_empty_list_of_list x (α0 :: rst))
with (ne_cons x (non_empty_list_of_list α0 rst)).
replace (α0 :: rst)
with (list_of_non_empty_list
(non_empty_list_of_list α0 rst)); auto.
revert α0.
induction rst; intros y; [ reflexivity | cbn ].
now rewrite IHrst.
Qed.
```

## The `sum`

Operator

`sum`

Is a Morphism

To prove this, we compose together the functions whose existence is implied by
$\alpha = \alpha'$ and $\beta = \beta'$. To that end, we introduce the
auxiliary function `lr_map`

.

```
Definition lr_map_sum {α β α' β'} (f : α -> α') (g : β -> β')
(x : α + β)
: α' + β' :=
match x with
| inl x => inl (f x)
| inr y => inr (g y)
end.
```

Then, we prove `sum`

is a morphism by defining a `Proper`

instance.

```
Instance sum_Proper
: Proper (type_equiv ==> type_equiv ==> type_equiv) sum.
Proof.
add_morphism_tactic.
intros α α' [fα gα' equαα' equα'α]
β β' [fβ gβ' equββ' equβ'β].
equiv with (lr_map_sum fα fβ)
and (lr_map_sum gα' gβ').
+ intros [x|y]; cbn.
++ now rewrite <- equαα'.
++ now rewrite <- equββ'.
+ intros [x|y]; cbn.
++ now rewrite <- equα'α.
++ now rewrite <- equβ'β.
Qed.
```

`sum`

Is Commutative

```
Definition sum_invert {α β} (x : α + β) : β + α :=
match x with
| inl x => inr x
| inr x => inl x
end.
Lemma sum_com {α β} : α + β == β + α.
Proof.
equiv with sum_invert and sum_invert;
now intros [x|x].
Qed.
```

`sum`

Is Associative

The associativity of `sum`

is straightforward to prove, and should not
pose a particular challenge to prospective readersIf we assume that this article is well written, that is.
.

```
Lemma sum_assoc {α β γ} : α + β + γ == α + (β + γ).
Proof.
equiv with (fun x =>
match x with
| inl (inl x) => inl x
| inl (inr x) => inr (inl x)
| inr x => inr (inr x)
end)
and (fun x =>
match x with
| inl x => inl (inl x)
| inr (inl x) => inl (inr x)
| inr (inr x) => inr x
end).
+ now intros [[x|x]|x].
+ now intros [x|[x|x]].
Qed.
```

`sum`

Has A Neutral Element

We need to find a type $e$ such that $\alpha + e = \alpha$ for any type $\alpha$ (similarly to $x + 0 = x$ for any natural number $x$).

Any empty type (that is, a type with no term such as `False`

) can act as
the natural element of `Type`

. As a reminder, empty types in Coq are
defined with the following syntaxNote that `Inductive empty.`

is erroneous.
When the `:=`

is omitted, Coq defines an inductive type with one
constructor, making such a type equivalent to `unit`

, not
`False`

.
:

```
Inductive empty := .
```

From a high-level perspective, `empty`

being the neutral element of
`sum`

makes sense. Because we cannot construct a term of type `empty`

,
then `α + empty`

contains exactly the same numbers of terms as `α`

.
This is the intuition. Now, how can we convince Coq that our intuition is
correct? Just like before, by providing two functions of types:

`α -> α + empty`

`α + empty -> α`

The first function is `inl`

, that is one of the constructors of
`sum`

.

The second function is trickier to write in Coq, because it comes down to
writing a function of type is `empty -> α`

.

```
Definition from_empty {α} : empty -> α :=
fun x => match x with end.
```

It is the exact same trick that allows Coq to encode proofs by contradiction.

If we combine `from_empty`

with the generic function

```
Definition unwrap_left_or {α β}
(f : β -> α) (x : α + β)
: α :=
match x with
| inl x => x
| inr x => f x
end.
```

Then, we have everything to prove that `α == α + empty`

.

```
Lemma sum_neutral (α : Type) : α == α + empty.
Proof.
equiv with inl and (unwrap_left_or from_empty);
auto.
now intros [x|x].
Qed.
```

## The `prod`

Operator

This is very similar to what we have just proven for `sum`

, so expect
less text for this section.

`prod`

Is A Morphism

```
Definition lr_map_prod {α α' β β'}
(f : α -> α') (g : β -> β')
: α * β -> α' * β' :=
fun x => match x with (x, y) => (f x, g y) end.
Instance prod_Proper
: Proper (type_equiv ==> type_equiv ==> type_equiv) prod.
Proof.
add_morphism_tactic.
intros α α' [fα gα' equαα' equα'α]
β β' [fβ gβ' equββ' equβ'β].
equiv with (lr_map_prod fα fβ)
and (lr_map_prod gα' gβ').
+ intros [x y]; cbn.
rewrite <- equαα'.
now rewrite <- equββ'.
+ intros [x y]; cbn.
rewrite <- equα'α.
now rewrite <- equβ'β.
Qed.
```

`prod`

Is Commutative

```
Definition prod_invert {α β} (x : α * β) : β * α :=
(snd x, fst x).
Lemma prod_com {α β} : α * β == β * α.
Proof.
equiv with prod_invert and prod_invert;
now intros [x y].
Qed.
```

`prod`

Is Associative

```
Lemma prod_assoc {α β γ}
: α * β * γ == α * (β * γ).
Proof.
equiv with (fun x =>
match x with
| ((x, y), z) => (x, (y, z))
end)
and (fun x =>
match x with
| (x, (y, z)) => ((x, y), z)
end).
+ now intros [[x y] z].
+ now intros [x [y z]].
Qed.
```

`prod`

Has A Neutral Element

```
Lemma prod_neutral (α : Type) : α * unit == α.
Proof.
equiv with fst and ((flip pair) tt).
+ now intros [x []].
+ now intros.
Qed.
```

`prod`

Has An Absorbing Element *)

And this absorbing element is `empty`

, just like the absorbing element of
the multiplication of natural numbers is $0$ (that is, the neutral element of
the addition).

```
Lemma prod_absord (α : Type) : α * empty == empty.
Proof.
equiv with (snd >>> from_empty)
and (from_empty).
+ intros [_ []].
+ intros [].
Qed.
```

`prod`

And `sum`

Distributivity

Finally, we can prove the distributivity property of `prod`

and
`sum`

using a similar approach to prove the associativity of `prod`

and `sum`

.

```
Lemma prod_sum_distr (α β γ : Type)
: α * (β + γ) == α * β + α * γ.
Proof.
equiv with (fun x => match x with
| (x, inr y) => inr (x, y)
| (x, inl y) => inl (x, y)
end)
and (fun x => match x with
| inr (x, y) => (x, inr y)
| inl (x, y) => (x, inl y)
end).
+ now intros [x [y | y]].
+ now intros [[x y] | [x y]].
Qed.
```

## Bonus: Algebraic Datatypes and Metaprogramming

Algebraic datatypes are very suitable for generating functions, as demonstrated
by the automatic deriving of typeclass in Haskell or trait in Rust. Because a
datatype can be expressed in terms of `sum`

and `prod`

, you just
have to know how to deal with these two constructions to start metaprogramming.

We can take the example of the `fold`

functions. A `fold`

function
is a function which takes a container as its argument, and iterates over the
values of that container in order to compute a result.

We introduce `fold_type INPUT CANON_FORM OUTPUT`

, a tactic to compute the
type of the fold function of the type `INPUT`

, whose “canonical form” (in terms
of `prod`

and `sum`

) is `CANON_FORM`

and whose result type is
`OUTPUT`

. Interested readers have to be familiar with `Ltac`

.

```
Ltac fold_args b a r :=
lazymatch a with
| unit =>
exact r
| b =>
exact (r -> r)
| (?c + ?d)%type =>
exact (ltac:(fold_args b c r) * ltac:(fold_args b d r))%type
| (b * ?c)%type =>
exact (r -> ltac:(fold_args b c r))
| (?c * ?d)%type =>
exact (c -> ltac:(fold_args b d r))
| ?a =>
exact (a -> r)
end.
Ltac currying a :=
match a with
| ?a * ?b -> ?c => exact (a -> ltac:(currying (b -> c)))
| ?a => exact a
end.
Ltac fold_type b a r :=
exact (ltac:(currying (ltac:(fold_args b a r) -> b -> r))).
```

We use it to compute the type of a `fold`

function for `list`

.

```
Definition fold_list_type (α β : Type) : Type :=
ltac:(fold_type (list α) (unit + α * list α)%type β).
```

```
fold_list_type =
fun α β : Type => β -> (α -> β -> β) -> list α -> β
: Type -> Type -> Type
```

It is exactly what you could have expected (as match the type of
`fold_right`

).

Generating the body of the function is possible in theory, but probably not in
`Ltac`

without modifying a bit `type_equiv`

. This could be a nice
use case for MetaCoq though.