'idiomatic' Haskell type inequality

Question

(edited from previous question where I thought code below doesn\'t work)

I wish to implement a haskell function f that has a restriction such that its 2 parameters m

Answer 1

With new features being added to GHC, you'll be able to write:

{-# LANGUAGE DataKinds, PolyKinds, TypeFamilies #-}

type family Equal (a :: k) (b :: k) :: Bool
type instance where
   Equal a a = True
   Equal a b = False

Answer 2

This is how it is done in the HList library

{-# LANGUAGE FlexibleInstances, 
    MultiParamTypeClasses, 
    FunctionalDependencies, 
    UndecidableInstances ,
    IncoherentInstances
 #-}

data HTrue; data HFalse;

class TypeCast a b | a -> b
instance TypeCast a a

class TypeEq a b c | a b -> c
instance TypeEq x x HTrue
instance (TypeCast HFalse b) => TypeEq x y b
-- instance TypeEq x y HFalse -- would violate functional dependency

You can fully infer type equality now:

typeEq :: TypeEq a b c => a -> b -> c
typeEq _ _ = undefined

Note that typeEq 0 1 == HFalse since 0 :: Num a => a and 1 :: Num b => b.