Can GADTs be used to prove type inequalities in GHC?

Question

So, in my ongoing attempts to half-understand Curry-Howard through small Haskell exercises, I've gotten stuck at this point:

{-# LANGUAGE GADTs #-}

import Data.Void

type Not a = a -> Void

-- | The type of type equality proofs, which can only be instantiated if a = b.
data Equal a b where
    Refl :: Equal a a

-- | Derive a contradiction from a putative proof of @Equal Int Char@.
intIsNotChar :: Not (Equal Int Char)
intIsNotChar intIsChar = ???

Clearly the type Equal Int Char has no (non-bottom) inhabitants, and thus semantically there ought to be an absurdEquality :: Equal Int Char -> a function... but for the life of me I can't figure out any way to write one other than using undefined.

So either:

1. I'm missing something, or
2. There is some limitation of the language that makes this an impossible task, and I haven't managed to understand what it is.

I suspect the answer is something like this: the compiler is unable to exploit the fact that there are no Equal constructors that don't have a = b. But if that is so, what makes it true?

Corney

See: typesandkinds.wordpress.com/2012/12/01/… – Carboloy

@dbaupp: Not a duplicate--that question isn't trying to do anything with a proof of inequality. – Slog

@C.A.McCann, ah yeah, I guess I should read more closely before I start throwing around wild accusations. :) – Grapery

Answer 1

Here's a shorter version of Philip JF's solution, which is the way dependent type theorists have been refuting equations for years.

type family Discriminate x
type instance Discriminate Int  = ()
type instance Discriminate Char = Void

transport :: Equal a b -> Discriminate a -> Discriminate b
transport Refl d = d

refute :: Equal Int Char -> Void
refute q = transport q ()

In order to show that things are different, you have to catch them behaving differently by providing a computational context which results in distinct observations. Discriminate provides exactly such a context: a type-level program which treats the two types differently.

It is not necessary to resort to undefined to solve this problem. Total programming sometimes involves rejecting impossible inputs. Even where undefined is available, I would recommend not using it where a total method suffices: the total method explains why something is impossible and the typechecker confirms; undefined merely documents your promise. Indeed, this method of refutation is how Epigram dispenses with "impossible cases" whilst ensuring that a case analysis covers its domain.

As for computational behaviour, note that refute, via transport is necessarily strict in q and that q cannot compute to head normal form in the empty context, simply because no such head normal form exists (and because computation preserves type, of course). In a total setting, we'd be sure that refute would never be invoked at run time. In Haskell, we're at least certain that its argument will diverge or throw an exception before we're obliged to respond to it. A lazy version, such as

absurdEquality e = error "you have a type error likely to cause big problems"

will ignore the toxicity of e and tell you that you have a type error when you don't. I prefer

absurdEquality e = e `seq` error "sue me if this happens"

if the honest refutation is too much like hard work.

Tobacconist

Note that seq e (error "sue me if this happens") can evaluate the second argument before the first one, if it is an issue you can use pseq (hackage.haskell.org/package/base-4.9.1.0/docs/…). – Tanker

Answer 2

I don't understand the problem with using undefined every type is inhabited by bottom in Haskell. Our language is not strongly normalizing... You are looking for the wrong thing. Equal Int Char leads to type errors not nice well kept exceptions. See

{-# LANGUAGE GADTs, TypeFamilies #-}

data Equal a b where
    Refl :: Equal a a

type family Pick cond a b
type instance Pick Char a b = a
type instance Pick Int a b = b

newtype Picker cond a b = Picker (Pick cond a b)

pick :: b -> Picker Int a b
pick = Picker

unpick :: Picker Char a b -> a
unpick (Picker x) = x

samePicker :: Equal t1 t2 -> Picker t1 a b -> Picker t2 a b
samePicker Refl x = x

absurdCoerce :: Equal Int Char -> a -> b
absurdCoerce e x = unpick (samePicker e (pick x))

you could use this to create the function you want

absurdEquality e = absurdCoerce e ()

but that will produce undefined behavior as its computation rule. false should cause programs to abort, or at the very least run for ever. Aborting is the computation rule that is akin to turning minimal logic into intiutionistic logic by adding not. The correct definition is

absurdEquality e = error "you have a type error likely to cause big problems"

as to the question in the title: essentially no. To the best of my knowledge, type inequality is not representable in a practical way in current Haskell. Coming changes to the type system may lead to this getting nicer, but as of right now, we have equalities but not inequalites.

Deme