# Distinctness I: Injectivity

[injectivity]

Type checkers spend much of their time verifying that pairs of types are equal, or at least compatible, checking for instance that the type of the argument to a function matches the type it expects.

Occasionally, they need to do the opposite, and verify that two types are distinct. One common case of this is allowing the programmer to omit those cases of a pattern match which are impossible because they would imply that two distinct types are equal. For instance, given a value which is either a string, or evidence that $\n{Int} = \n{String}$, then matching on this value need only handle the first case as the second is impossible.

type (_,_) eq = Refl : ('a, 'a) eq
type ('a, 'b) either = Left of 'a | Right of 'b

let f (x : (string, (int,string) eq) either) =
match x with
| Left s -> s
(* no case for Right, and yet the match is exhaustive *)

-- Agda can reason about distinctness of more than just types
open import Agda.Builtin.String
data Eq (a : String) : String -> Set where
Refl : Eq a a

data Sum (A B : Set) : Set where
Left : A -> Sum A B
Right : B -> Sum A B

f : Sum String (Eq "foo" "bar") -> String
f (Left s) = s
f (Right ()) -- "()" is the absurd pattern


Of course, this is unsound if the type system is wrong about two types being distinct. One common way that this can occur is when the type system assumes that all type constructors are injective.

Injectivity of a type constructor $F$ means that if $F[A] = F[B]$, then the types $A$ and $B$ are equal. Most common type constructors (List, Array and so on) are in fact injective, so this is a natural property to expect. However, if it is possible to define a type constructor $F$ by $F[A] = \n{Int}$, then injectivity fails. This issue occurred in Dotty1, a research Scala compiler, as well as in Agda2 (with the injective-type-constructors option). Curiously, Haskell makes the same assumption, but without unsoundness.

// Counterexample by Aleksander Boruch-Gruszecki
object Test {
sealed trait EQ[A, B]
final case class Refl[T]() extends EQ[T, T]

def absurd[F[_], X, Y](eq: EQ[F[X], F[Y]], x: X): Y = eq match {
case Refl() => x
}

var ex: Exception = _
try {
type Unsoundness[X] = Int
val s: String = absurd[Unsoundness, Int, String](Refl(), 0)
} catch {
case e: ClassCastException => ex = e
}

def main(args: Array[String]) =
assert(ex != null)
}

-- Counterexample by Andreas Abel
{-# OPTIONS --injective-type-constructors #-}

open import Common.Prelude
open import Common.Equality

abstract
f : Bool → Bool
f x = true

same : f true ≡ f false
same = refl

not-same : f true ≡ f false → ⊥
not-same ()

absurd : ⊥
absurd = not-same same

data Equ a b where
Refl : Equ a a

-- not in fact unsound! see below
everythingIsInjective :: Equ (f a) (f b) -> Equ a b
everythingIsInjective Refl = Refl


The reason that it is sound for Haskell to assume an arbitrary type constructor $F$ is injective is that it draws a distinction between type constructors and type-level functions. The following two declarations are distinct:

type family Foo :: * -> *
type family Bar a :: *


Here, Foo is defined as a type constructor, something of kind * -> *. This kind only contains injective type constructors (like lists, etc.). By contrast, Bar is defined as a type-level function. Bar is not necessarily injective, as one can define:

type instance Bar a = Int


However, Bar is not of kind * -> *, and cannot be passed as the f parameter to everythingIsInjective:

eqFoo :: Equ (Foo a) (Foo b) -> Equ a b
eqFoo = everythingIsInjective -- works

eqBar :: Equ (Bar a) (Bar b) -> Equ a b
eqBar = everythingIsInjective -- error


In fact, in Haskell there is currently no way to quantify over things like Bar: Haskell's higher-kinded types only allow passing around injective type constructors. A recent paper by Kiss et al.3 proposes adding parameterisation by type functions to Haskell, by adding a new kind * => * which does not carry injectiveness assumptions.

3

Higher-order Type-level Programming in Haskell (ICFP '19), Csongor Kiss, Susan Eisenbach, Tony Field, Simon Peyton Jones (2019)