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(** * 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:
<<
enum List<A> {
Cons(A, Box< List<a> >),
Nil,
}
>>
In Coq:
<<
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] (remember tuples?).
For an algebraic datatype, one constructor allows for defining “named
tuples”, that is adhoc product types. Besides, constructors are mutually
exclusive: you cannot define the same term using two different constructors.
Therefore, a datatype with several constructors is reminescent of a disjoint
union. Coming back to the [list] type, under the syntactic sugar of
algebraic datatypes, the [list α] type is equivalent to [unit + α * list α],
where [unit] models the [nil] case, and [α * list α] 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 #<span class="imath">#\forall (x, y),\ x + y
= y + x#</span>#
 [+] is associative, that is #<span class="imath">#\forall (x, y, z),\ (x
+ y) + z = x + (y + z)#</span>#
 [+] has a neutral element, that is #<span class="imath">#\exists e_s,
\ \forall x,\ x + e_s = x#</span>#
 [*] is commutative, that is #<span class="imath">#\forall (x, y),\ x * y
= y * x#</span>#
 [*] is associative, that is #<span class="imath">#\forall (x, y, z),\ (x
* y) * z = x * (y * z)#</span>#
 [*] has a neutral element, that is #<span class="imath">#\exists e_p,
\ \forall x,\ x * e_p = x#</span>#
 The distributivity of [+] and [*], that is #<span class="imath">#\forall
(x, y, z),\ x * (y + z) = x * y + x * z#</span>#
 [*] has an absorbing element, that is #<span class="imath">#\exists e_a,
\ \forall x, \ x * e_a = e_a#</span>#
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
>>
#<div id="generatetoc"></div>#
#<div id="history">site/posts/AlgebraicDatatypes.v</div># *)
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 builtin 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 counterpart 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:
#<span class="dmath">#\forall (x : α),\ x = g(f(x))#</span>#
#<span class="dmath">#\forall (y : β),\ y = f(g(y))#</span>#
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. *)
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 *)
(** [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. *)
Lemma type_equiv_refl (α : Type) : α == α.
(** This proof is straightforward. A type [α] is equivalent to itself because:
#<span class="imath">#\forall (x : α),\ x = id(id(x))#</span># *)
Proof.
now equiv with (@id α) and (@id α).
Qed.
(** [type_equiv] is symmetric. *)
Lemma type_equiv_sym {α β} (equ : α == β) : β == α.
(** 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 *)
Lemma type_equiv_trans {α β γ} (equ1 : α == β) (equ2 : β == γ)
: α == γ.
(** 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]: *)
Infix ">>>" := (fun f g x => g (f x)) (at level 70).
(** 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.
