Creating a completely dependent concatenation

Question

A nice true fact about concatenation is that if I know any two variables in the equation:

a ++ b = c

Then I know the third.

I would like to capture this idea in my own concat so I use a functional dependency.

{-# Language DataKinds, GADTs, FlexibleContexts, FlexibleInstances, FunctionalDependencies, KindSignatures, PolyKinds, TypeOperators, UndecidableInstances #-}
import Data.Kind (Type)

class Concatable
   (m  :: k -> Type)
   (as :: k)
   (bs :: k)
   (cs :: k)
   | as bs -> cs
   , as cs -> bs
   , bs cs -> as
   where
     concat' :: m as -> m bs -> m cs

Now I conjure heterogeneous list like so:

data HList ( as :: [ Type ] ) where
  HEmpty :: HList '[]
  HCons  :: a -> HList as -> HList (a ': as)

But when I try to declare these as Concatable I have an issue

instance Concatable HList '[] bs bs where
  concat' HEmpty bs = bs
instance
  ( Concatable HList as bs cs
  )
    => Concatable HList (a ': as) bs (a ': cs)
  where
    concat' (HCons head tail) bs = HCons head (concat' tail bs)

I don't satisfy my third functional dependency. Or rather the compiler believes we do not. This is because the compiler believes that in our second instance it might be the case that bs ~ (a ': cs). And it could be the case if Concatable as (a ': cs) cs.

How can I adjust my instances so that all three of the dependencies are satisfied?

Answer 1

So, as comments suggest, GHC is not gonna figure it out on it's own, but we can help it with a bit of type level programming. Let's introduce some TypeFamilies. All of these functions are pretty straightforward translations of list manipulation lifted to the type level:

-- | This will produce the suffix of `cs` without `as`
type family DropPrefix (as :: [Type]) (cs :: [Type]) where
  DropPrefix '[] cs = cs
  DropPrefix (a ': as) (a ': cs) = DropPrefix as cs

-- Similar to the logic in the question itself: list concatenation. 
type family Concat (as :: [Type]) (bs :: [Type]) where
  Concat '[] bs = bs
  Concat (head ': tail) bs = head ': Concat tail bs

-- | Naive list reversal with help of concatenation.
type family Reverse (xs :: [Type]) where
  Reverse '[] = '[]
  Reverse (x ': xs) = Concat (Reverse xs) '[x]

-- | This will produce the prefix of `cs` without `bs`
type family DropSuffix (bs :: [Type]) (cs :: [Type]) where
  DropSuffix bs cs = Reverse (DropPrefix (Reverse bs) (Reverse cs))

-- | Same definition of `HList` as in the question
data HList (as :: [Type]) where
  HEmpty :: HList '[]
  HCons :: a -> HList as -> HList (a ': as)

-- | Definition of concatenation at the value level
concatHList :: (cs ~ Concat as bs) => HList as -> HList bs -> HList cs
concatHList HEmpty bs = bs
concatHList (HCons head tail) bs = HCons head (concatHList tail bs)

With this tools at our disposal we can actually get to hour goal, but first let's define a function with the desired properties:

• Ability to deduce cs from as and bs
• Ability to deduce as from bs and cs
• Ability to deduce bs from as and cs

Voila:

concatH ::
     (cs ~ Concat as bs, bs ~ DropPrefix as cs, as ~ DropSuffix bs cs)
  => HList as
  -> HList bs
  -> HList cs
concatH = concatHList

Let's test it:

foo :: HList '[Char, Bool]
foo = HCons 'a' (HCons True HEmpty)

bar :: HList '[Int]
bar = HCons (1 :: Int) HEmpty

λ> :t concatH foo bar
concatH foo bar :: HList '[Char, Bool, Int]
λ> :t concatH bar foo
concatH bar foo :: HList '[Int, Char, Bool]

And finally the end goal:

class Concatable (m :: k -> Type) (as :: k) (bs :: k) (cs :: k)
  | as bs -> cs
  , as cs -> bs
  , bs cs -> as
  where
  concat' :: m as -> m bs -> m cs

instance (cs ~ Concat as bs, bs ~ DropPrefix as cs, as ~ DropSuffix bs cs) =>
         Concatable HList as bs cs where
  concat' = concatH

λ> :t concat' HEmpty bar
concat' HEmpty bar :: HList '[Int]
λ> :t concat' foo bar
concat' foo bar :: HList '[Char, Bool, Int]
λ> :t concat' bar foo
concat' bar foo :: HList '[Int, Char, Bool]