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How to express existential types using higher rank (rank-N) type polymorphism?

We're used to having universally quantified types for polymorphic functions. Existentially quantified types are used much less often. How can we express existentially quantified types using universal type quantifiers?

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Petr Avatar asked Nov 30 '12 21:11

Petr


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2 Answers

It turns out that existential types are just a special case of Σ-types (sigma types). What are they?

Sigma types

Just as Π-types (pi types) generalise our ordinary function types, allowing the resulting type to depend on the value of its argument, Σ-types generalise pairs, allowing the type of second component to depend on the value of the first one.

In a made-up Haskell-like syntax, Σ-type would look like this:

data Sigma (a :: *) (b :: a -> *)
    = SigmaIntro
        { fst :: a
        , snd :: b fst
        }

-- special case is a non-dependent pair
type Pair a b = Sigma a (\_ -> b)

Assuming * :: * (i.e. the inconsistent Set : Set), we can define exists a. a as:

Sigma * (\a -> a)

The first component is a type and the second one is a value of that type. Some examples:

foo, bar :: Sigma * (\a -> a)
foo = SigmaIntro Int  4
bar = SigmaIntro Char 'a'

exists a. a is fairly useless - we have no idea what type is inside, so the only operations that can work with it are type-agnostic functions such as id or const. Let's extend it to exists a. F a or even exists a. Show a => F a. Given F :: * -> *, the first case is:

Sigma * F   -- or Sigma * (\a -> F a)

The second one is a bit trickier. We cannot just take a Show a type class instance and put it somewhere inside. However, if we are given a Show a dictionary (of type ShowDictionary a), we can pack it with the actual value:

Sigma * (\a -> (ShowDictionary a, F a))
-- inside is a pair of "F a" and "Show a" dictionary

This is a bit inconvenient to work with and assumes that we have a Show dictionary around, but it works. Packing the dictionary along is actually what GHC does when compiling existential types, so we could define a shortcut to have it more convenient, but that's another story. As we will learn soon enough, the encoding doesn't actually suffer from this problem.


Digression: thanks to constraint kinds, it's possible to reify the type class into concrete data type. First, we need some language pragmas and one import:

{-# LANGUAGE ConstraintKinds, GADTs, KindSignatures  #-}
import GHC.Exts -- for Constraint

GADTs already give us the option to pack a type class along with the constructor, for example:

data BST a where
    Nil  :: BST a
    Node :: Ord a => a -> BST a -> BST a -> BST a

However, we can go one step further:

data Dict :: Constraint -> * where
    D :: ctx => Dict ctx

It works much like the BST example above: pattern matching on D :: Dict ctx gives us access to the whole context ctx:

show' :: Dict (Show a) -> a -> String
show' D = show

(.+) :: Dict (Num a) -> a -> a -> a
(.+) D = (+)

We also get quite natural generalisation for existential types that quantify over more type variables, such as exists a b. F a b.

Sigma * (\a -> Sigma * (\b -> F a b))
-- or we could use Sigma just once
Sigma (*, *) (\(a, b) -> F a b)
-- though this looks a bit strange

The encoding

Now, the question is: can we encode Σ-types with just Π-types? If yes, then the existential type encoding is just a special case. In all glory, I present you the actual encoding:

newtype SigmaEncoded (a :: *) (b :: a -> *)
    = SigmaEncoded (forall r. ((x :: a) -> b x -> r) -> r)

There are some interesting parallels. Since dependent pairs represent existential quantification and from classical logic we know that:

(∃x)R(x) ⇔ ¬(∀x)¬R(x) ⇔ (∀x)(R(x) → ⊥) → ⊥

forall r. r is almost , so with a bit of rewriting we get:

(∀x)(R(x) → r) → r

And finally, representing universal quantification as a dependent function:

forall r. ((x :: a) -> R x -> r) -> r

Also, let's take a look at the type of Church-encoded pairs. We get a very similar looking type:

Pair a b  ~  forall r. (a -> b -> r) -> r

We just have to express the fact that b may depend on the value of a, which we can do by using dependent function. And again, we get the same type.

The corresponding encoding/decoding functions are:

encode :: Sigma a b -> SigmaEncoded a b
encode (SigmaIntro a b) = SigmaEncoded (\f -> f a b)

decode :: SigmaEncoded a b -> Sigma a b
decode (SigmaEncoded f) = f SigmaIntro
-- recall that SigmaIntro is a constructor

The special case actually simplifies things enough that it becomes expressible in Haskell, let's take a look:

newtype ExistsEncoded (F :: * -> *)
    = ExistsEncoded (forall r. ((x :: *) -> (ShowDictionary x, F x) -> r) -> r)
    -- simplify a bit
    = ExistsEncoded (forall r. (forall x. (ShowDictionary x, F x) -> r) -> r)
    -- curry (ShowDictionary x, F x) -> r
    = ExistsEncoded (forall r. (forall x. ShowDictionary x -> F x -> r) -> r)
    -- and use the actual type class
    = ExistsEncoded (forall r. (forall x. Show x => F x -> r) -> r)

Note that we can view f :: (x :: *) -> x -> x as f :: forall x. x -> x. That is, a function with extra * argument behaves as a polymorphic function.

And some examples:

showEx :: ExistsEncoded [] -> String
showEx (ExistsEncoded f) = f show

someList :: ExistsEncoded []
someList = ExistsEncoded $ \f -> f [1]

showEx someList == "[1]"

Notice that someList is actually constructed via encode, but we dropped the a argument. That's because Haskell will infer what x in the forall x. part you actually mean.

From Π to Σ?

Strangely enough (although out of the scope of this question), you can encode Π-types via Σ-types and regular function types:

newtype PiEncoded (a :: *) (b :: a -> *)
    = PiEncoded (forall r. Sigma a (\x -> b x -> r) -> r)
-- \x -> is lambda introduction, b x -> r is a function type
-- a bit confusing, I know

encode :: ((x :: a) -> b x) -> PiEncoded a b
encode f = PiEncoded $ \sigma -> case sigma of
    SigmaIntro a bToR -> bToR (f a)

decode :: PiEncoded a b -> (x :: a) -> b x
decode (PiEncoded f) x = f (SigmaIntro x (\b -> b))
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Vitus Avatar answered Oct 21 '22 14:10

Vitus


I found an anwer in Proofs and Types by Jean-Yves Girard, Yves Lafont and Paul Taylor.

Imagine we have some one-argument type t :: * -> * and construct an existential type that holds t a for some a: exists a. t a. What can we do with such a type? In order to compute something out of it we need a function that can accept t a for arbitrary a, that means a function of type forall a. t a -> b. Knowing this, we can encode an existential type simply as a function that takes functions of type forall a. t a -> b, supplies the existential value to them and returns the result b:

{-# LANGUAGE RankNTypes #-}

newtype Exists t = Exists (forall b. (forall a. t a -> b) -> b)

Creating an existential value is now easy:

exists :: t a -> Exists t
exists x = Exists (\f -> f x)

And if we want to unpack the existential value, we just apply its content to a function that produces the result:

unexists :: (forall a. t a -> b) -> Exists t -> b
unexists f (Exists e) = e f

However, purely existential types are of very little use. We cannot do anything reasonable with a value we know nothing about. More often we need an existential type with a type class constraint. The procedure is just the same, we just add a type class constraint for a. For example:

newtype ExistsShow t = ExistsShow (forall b. (forall a. Show a => t a -> b) -> b)

existsShow :: Show a => t a -> ExistsShow t
existsShow x = ExistsShow (\f -> f x)

unexistsShow :: (forall a. Show a => t a -> b) -> ExistsShow t -> b
unexistsShow f (ExistsShow e) = e f

Note: Using existential quantification in functional programs is often considered a code-smell. It can indicate that we haven't liberated ourselves from OO thinking.

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Petr Avatar answered Oct 21 '22 14:10

Petr