# 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 | NilThis translates in Haskell as

data List a = Cons a (List a) | NilIn Rust:

enum List<A> { Cons(A, Box< List<a> >), Nil, }In Coq:

Inductive list a := | cons : a -> list a -> list a | nilAnd 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 (remember tuples?).

- + 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 $\exists e_a, \ \forall x, \ x * e_a = e_a$

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
- The sum Operator
- The prod Operator
- prod has an Absorbing Element
- prod and sum Distributivity
- Bonus: Algebraic Datatypes and Metaprogramming

## Revisions

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You can consult the source of this file in its current version here.

2020-07-12 | More spellchecking and typos | 48a9b49 |

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From Coq Require Import

Basics Setoid Equivalence Morphisms

List FunctionalExtensionality.

Import ListNotations.

Set Implicit Arguments.

## 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, α + β is not structurally

*equal*to β + α, 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 α and β are equivalent up to an isomorphism (denoted by α == β) when for any term of type α, there exists a counter-part term of type β and vice versa. In other words, α and β are equivalent if we can exhibit two functions f and g such that: $\forall (x : α),\ x = g(f(x))$ $\forall (y : β),\ y = f(g(y))$ In Coq, this translates into the following inductive types.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.
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

type_equiv is an equivalence, and we can prove it by demonstrating it is (1) reflexive, (2) symmetric, and (3) transitive. type_equiv is reflexive.
This proof is straightforward. A type α is equivalent to itself because:
$\forall (x : α),\ x = id(id(x))$

type_equiv is symmetric.

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

Proof.

destruct equ as [f g equ1 equ2].

now equiv with g and f.

Qed.

type_equiv is transitive

If α == β, we know there exists two functions fα and gβ which satisfy
the expected properties of type_equiv. Similarly, because β == γ, we
know there exists two additional functions fβ and gγ. We can compose
these functions together to prove α == γ.
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:

Then comes the proof.

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.

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 prod_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 prod_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

This means that if α == α' and β == β', then α + β == α' + β'. To prove this, we compose together the functions whose existence is implied by α == α' and β == β'. 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.

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 perspective readers; if 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 α + e == α for any type α (similarly to $x~+~0~=~x$ for any natural number $x$ that is). 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 syntax:
Note that the following definition is erroneous.

Inductive empty.Using Print on such a type illustrates the issue.

Inductive empty : Prop := Build_empty { }That is, when the := is omitted, Coq defines an inductive type with one constructor. Coming back to empty being the neutral element of sum. From a high-level perspective, this 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

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.

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.

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.

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 number is $0$ (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).

Here is the definition of fold_list_type, as outputed by Print.

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