证明GHC的类型不等式

Question

出于教育目的,我一直在尝试通过使用各种语言扩展和单例类型来重构Haskell中的"使用Idris进行类型驱动开发"(即RemoveElem.idr)一书中的示例.它的要点是一个从非空向量中移除元素的函数,给出了元素实际上在向量中的证明.以下代码是自包含的(GHC 8.4或更高版本).问题出现在最后:

{-# LANGUAGE EmptyCase #-}
{-# LANGUAGE DataKinds #-}
{-# LANGUAGE GADTs #-}
{-# LANGUAGE LambdaCase #-}
{-# LANGUAGE RankNTypes #-}
{-# LANGUAGE StandaloneDeriving #-}
{-# LANGUAGE TypeFamilies #-}
{-# LANGUAGE TypeOperators #-}
{-# LANGUAGE TypeInType #-}

import Data.Kind
import Data.Type.Equality
import Data.Void

-- | Inductively defined natural numbers.
data Nat = Z | S Nat deriving (Eq, Show)

-- | Singleton types for natural numbers.
data SNat :: Nat -> Type where
    SZ :: SNat 'Z
    SS :: SNat n -> SNat ('S n)

deriving instance Show (SNat n)

-- | "Demote" a singleton-typed natural number to an ordinary 'Nat'.
fromSNat :: SNat n -> Nat
fromSNat SZ = Z
fromSNat (SS n) = S (fromSNat n)

-- | A decidable proposition.
data Dec a = Yes a | No (a -> Void)

-- | Propositional equality of natural numbers.
eqSNat :: SNat a -> SNat b -> Dec (a :~: b)
eqSNat  SZ     SZ    = Yes Refl
eqSNat  SZ    (SS _) = No (\case {})
eqSNat (SS _)  SZ    = No (\case {})
eqSNat (SS a) (SS b) = case eqSNat a b of
    No  f    -> No (\case Refl -> f Refl)
    Yes Refl -> Yes Refl

-- | A length-indexed list (aka vector).
data Vect :: Nat -> Type -> Type where
    Nil   :: Vect 'Z a
    (:::) :: a -> Vect n a -> Vect ('S n) a

infixr 5 :::

deriving instance Show a => Show (Vect n a)

-- | @Elem a v@ is the proposition that an element of type @a@
-- is contained in a vector of type @v@. To be useful, @a@ and @v@
-- need to refer to singleton types.
data Elem :: forall a n. a -> Vect n a -> Type where
    Here  :: Elem x (x '::: xs)
    There :: Elem x xs -> Elem x (y '::: xs)

deriving instance Show a => Show (Elem a v)

------------------------------------------------------------------------
-- From here on, to simplify things, only vectors of natural
-- numbers are considered.

-- | Singleton types for vectors of 'Nat's.
data SNatVect :: forall n. Nat -> Vect n Nat -> Type where
    SNatNil  :: SNatVect 'Z 'Nil
    SNatCons :: SNat a -> SNatVect n v -> SNatVect ('S n) (a '::: v)

deriving instance Show (SNatVect n v)

-- | "Demote" a singleton-typed vector of 'SNat's to an
-- ordinary vector of 'Nat's.
fromSNatVect :: SNatVect n v -> Vect n Nat
fromSNatVect SNatNil = Nil
fromSNatVect (SNatCons a v) = fromSNat a ::: fromSNatVect v

-- | Decide whether a value is in a vector.
isElem :: SNat a -> SNatVect n v -> Dec (Elem a v)
isElem _  SNatNil        = No (\case {})
isElem a (SNatCons b as) = case eqSNat a b of
    Yes Refl   -> Yes Here
    No notHere -> case isElem a as of
        Yes there   -> Yes (There there)
        No notThere -> No $ \case
            Here        -> notHere Refl
            There there -> notThere there

type family RemoveElem (a :: Nat) (v :: Vect ('S n) Nat) :: Vect n Nat where
    RemoveElem a (a '::: as) = as
    RemoveElem a (b '::: as) = b '::: RemoveElem a as

-- | Remove a (singleton-typed) element from a (non-empty, singleton-typed)
-- vector, given a proof that the element is in the vector.
removeElem :: forall (a :: Nat) (v :: Vect ('S n) Nat)
    . SNat a
    -> Elem a v
    -> SNatVect ('S n) v
    -> SNatVect n (RemoveElem a v)
removeElem x prf (SNatCons y ys) = case prf of
    Here        -> ys
    There later -> case ys of
        SNatNil    -> case later of {}
        SNatCons{} -> SNatCons y (removeElem x later ys)
            -- ^ Could not deduce:
            --            RemoveElem a (y '::: (a2 '::: v2))
            --          ~ (y '::: RemoveElem a (a2 '::: v2))

显然,类型系统需要说服值的类型x并且y在代码的该分支中不可能相等,因此可以明确地使用类型族的第二个等式来根据需要减少返回类型.我不知道该怎么做.天真地,我希望构造函数There和模式匹配There later来携带/揭示GHC类型不等式的证明.

以下是一个明显冗余和部分的解决方案,它只是演示了GHC对递归调用进行类型检查所需的类型不等式:

        SNatCons{} -> case (x, y) of
            (SZ, SS _) -> SNatCons y (removeElem x later ys)
            (SS _, SZ) -> SNatCons y (removeElem x later ys)

现在,例如,这有效:

?> let vec = SNatCons SZ (SNatCons (SS SZ) (SNatCons SZ SNatNil))
?> :t vec
vec
  :: SNatVect ('S ('S ('S 'Z))) ('Z '::: ('S 'Z '::: ('Z '::: 'Nil)))
?> let Yes prf = isElem (SS SZ) vec
?> :t prf
prf :: Elem ('S 'Z) ('Z '::: ('S 'Z '::: ('Z '::: 'Nil)))
?> let vec' = removeElem (SS SZ) prf vec
?> :t vec'
vec' :: SNatVect ('S ('S 'Z)) ('Z '::: ('Z '::: 'Nil))
?> fromSNatVect vec'
Z ::: (Z ::: Nil)

解析度

正如在@ chi的评论中暗示并在HTNW的答案中详细阐述的那样,我试图通过removeElem使用上述类型签名和类型系列来解决错误的问题,如果我能够这样做,那么所得到的程序将是错误的类型.

以下是我根据HTNW的答案所做的更正(您可能需要在继续阅读之前阅读).

第一个错误,或不必要的并发症,是重复SNatVects类型的向量的长度.我觉得有必要写fromSNatVect,但肯定不是:

data SNatVect (v :: Vect n Nat) :: Type where
    SNatNil  :: SNatVect 'Nil
    SNatCons :: SNat a -> SNatVect v -> SNatVect (a '::: v)

deriving instance Show (SNatVect v)

fromSNatVect :: forall (v :: Vect n Nat). SNatVect v -> Vect n Nat
-- implementation unchanged

现在有两种写作方法removeElem.第一个接受a Elem,an SNatVect并返回Vect:

removeElem :: forall (a :: Nat) (n :: Nat) (v :: Vect ('S n) Nat)
    . Elem a v
    -> SNatVect v
    -> Vect n Nat
removeElem prf (SNatCons y ys) = case prf of
    Here        -> fromSNatVect ys
    There later -> case ys of
        SNatNil    -> case later of {}
        SNatCons{} -> fromSNat y ::: removeElem later ys

第二接受一个SElem,一个SNatVect并返回SNatVect,使用RemoveElem型家族,反映了值级函数:

data SElem (e :: Elem a (v :: Vect n k)) where
    SHere  :: forall x xs. SElem ('Here :: Elem x (x '::: xs))
    SThere :: forall x y xs (e :: Elem x xs). SElem e -> SElem ('There e :: Elem x (y '::: xs))

type family RemoveElem (xs :: Vect ('S n) a) (e :: Elem x xs) :: Vect n a where
    RemoveElem (x '::: xs) 'Here = xs
    RemoveElem (x '::: xs) ('There later) = x '::: RemoveElem xs later

sRemoveElem :: forall (xs :: Vect ('S n) Nat) (e :: Elem x xs)
    . SElem e
    -> SNatVect xs
    -> SNatVect (RemoveElem xs e)
sRemoveElem prf (SNatCons y ys) = case prf of
    SHere        -> ys
    SThere later -> case ys of
        SNatNil    -> case later of {}
        SNatCons{} -> SNatCons y (sRemoveElem later ys)

有趣的是,两个版本都不会将元素作为单独的参数传递,因为该信息包含在Elem/ SElem值中.该value参数也可以从该函数的Idris版本中删除,但是removeElem_auto变体可能有点令人困惑,因为它只会将向量作为显式参数,如果隐式prf参数是,则删除向量的第一个元素未明确使用不同的证据.

Answer 1

考虑[1, 2, 1].RemoveElem 1 [1, 2, 1]是[2, 1].现在,调用removeElem 1 (There $ There $ Here) ([1, 2, 1] :: SNatVect 3 [1, 2, 1]) :: SNatVect 2 [2, 1],应该编译.这是错的.该Elem参数表示要删除的第三个元素,这将使[1, 2],但类型签名说,它必须是一个[2, 1].

首先,SNatVect有点破碎.它有两个Nat参数:

data SNatVect :: forall n. Nat -> Vect n a -> Type where ...

第一个是n,第二个是未命名的Nat.通过结构SNatVect,它们总是相等的.它允许一个SNatVect双重作为相等证明,但它可能不是那样的意图.你可能意味着

data SNatVect (n :: Nat) :: Vect n Nat -> Type where ...

只使用普通->语法就无法在源Haskell中编写此签名.但是,当GHC打印此类型时,您有时会得到

SNatVect :: forall (n :: Nat) -> Vect n Nat -> Type

但这是多余的.您可以将其Nat作为隐式forall参数,并从Vects类型推断:

data SNatVect (xs :: Vect n Nat) where
  SNatNil  :: SNatVect 'Nil
  SNatCons :: SNat x -> SNatVect xs -> SNatVect (x '::: xs)

这给了

SNatVect :: forall (n :: Nat). Vect n Nat -> Type

第二,尝试写作

removeElem :: forall (n :: Nat) (x :: Nat) (xs :: Vect (S n) Nat).
              Elem x xs -> SNatVect xs -> Vect n Nat

注意SNat参数是如何消失的,以及返回类型是如何简单的Vect.这个SNat论点使得这个类型"太大了",所以当函数没有意义时,你就会陷入困境.在SNatVect返回类型意味着你跳过一些步骤.粗略地说,每个函数都有三种形式:基本形式f :: a -> b -> c; 类型级别,type family F (x :: a) (y :: b) :: c; 和从属的,f :: forall (x :: a) (y :: b). Sing x -> Sing y -> Sing (F x y).每个都以"相同"的方式实现,但尝试实现一个而不实现其前任是一种让人感到困惑的可靠方法.

现在,你可以提升一点:

data SElem (e :: Elem x (xs :: Vect n k)) where
  SHere :: forall x xs. SElem ('Here :: Elem x (x '::: xs))
  SThere :: forall x y xs (e :: Elem x xs). SElem e -> SElem ('There e :: Elem x (y '::: xs))

type family RemoveElem (xs :: Vect (S n) a) (e :: Elem x xs) :: Vect n a

记下removeElem和的类型之间的关系RemoveElem.对参数的重新排序是因为类型e依赖于xs,因此需要相应地进行排序.或者:将xs参数从forall"d-and-implicitly-given" 提升为"明确给定",然后Sing xs参数被添加,因为它不包含任何信息,因为它是一个单例.

最后,你可以编写这个函数:

sRemoveElem :: forall (xs :: Vect (S n) Nat) (e :: Elem x xs).
               SElem e -> SNatVect xs -> SNatVect (RemoveElem xs e)