如何让 ghc 相信类型级别的加法是可交换的(以实现依赖类型的反向)?

How to convince ghc that type level addition is commutative (to implement dependently typed reverse)?

这不会编译,因为 ghc 告诉我 Add 不是单射的。我如何告诉编译器 Add 确实是可交换的(也许通过告诉它 Add 是单射的)?从hasochism paper看来,必须以某种方式提供代理。

{-# LANGUAGE DataKinds             #-}
{-# LANGUAGE TypeOperators         #-}
{-# LANGUAGE KindSignatures        #-}
{-# LANGUAGE GADTs                 #-}
{-# LANGUAGE RankNTypes            #-}
{-# LANGUAGE ScopedTypeVariables   #-}
{-# LANGUAGE TypeFamilies          #-}
{-# LANGUAGE UndecidableInstances  #-}

data Nat = Z | S Nat

type family Add a b where
  Add  Z    n = n
  Add  n    Z = n
  Add (S n) k = S (Add n k)

data VecList n a where
  Nil  :: VecList Z a
  Cons :: a -> VecList n a -> VecList (S n) a

safeRev :: forall a n . VecList n a -> VecList n a
safeRev xs = safeRevAux Nil xs
  where
    safeRevAux :: VecList p a -> VecList q a -> VecList (Add p q) a
    safeRevAux acc Nil = acc
    safeRevAux acc (Cons y ys) = safeRevAux (Cons y acc) ys

一个人可以做到这一点,但我觉得幕后发生的事情太多了。

{-# LANGUAGE TypeOperators         #-}
{-# LANGUAGE GADTs                 #-}
{-# LANGUAGE RankNTypes            #-}
{-# LANGUAGE ScopedTypeVariables   #-}
{-# LANGUAGE TypeFamilies          #-}

import Data.Proxy
import Data.Type.Equality

data Nat = Z | S Nat

type family n1 + n2 where
  Z + n2 = n2
  (S n1) + n2 = S (n1 + n2)

-- singleton for Nat
data SNat :: Nat -> * where
  SZero :: SNat Z
  SSucc :: SNat n -> SNat (S n)

-- inductive proof of right-identity of +
plus_id_r :: SNat n -> ((n + Z) :~: n)
plus_id_r SZero = Refl
plus_id_r (SSucc n) = gcastWith (plus_id_r n) Refl

-- inductive proof of simplification on the rhs of +
plus_succ_r :: SNat n1 -> Proxy n2 -> ((n1 + (S n2)) :~: (S (n1 + n2)))
plus_succ_r SZero _ = Refl
plus_succ_r (SSucc n1) proxy_n2 = gcastWith (plus_succ_r n1 proxy_n2) Refl

data VecList n a where
  V0  :: VecList Z a
  Cons :: a -> VecList n a -> VecList (S n) a

reverseList :: VecList n a -> VecList n a
reverseList V0 = V0
reverseList list = go SZero V0 list
  where
    go :: SNat n1 -> VecList n1  a-> VecList n2 a -> VecList (n1 + n2) a
    go snat acc V0 = gcastWith (plus_id_r snat) acc
    go snat acc (Cons h (t :: VecList n3 a)) =
      gcastWith (plus_succ_r snat (Proxy :: Proxy n3))
              (go (SSucc snat) (Cons h acc) t)

safeHead :: VecList (S n) a -> a
safeHead (Cons x _) = x

test = safeHead $ reverseList (Cons 'a' (Cons 'b' V0))

请参阅 https://www.haskell.org/pipermail/haskell-cafe/2014-September/115919.html 了解原始想法。

编辑:

@user3237465 这很有趣,更符合我的想法 (虽然经过反思我的问题可能不是很好 制定)。

我好像有"axioms"

type family n1 :+ n2 where
  Z :+ n2 = n2
  (S n1) :+ n2 = S (n1 + n2)

因此可以产生像

这样的证明
plus_id_r :: SNat n -> ((n + Z) :~: n)
plus_id_r SZero = Refl
plus_id_r (SSucc n) = gcastWith (plus_id_r n) Refl

我觉得这很简洁。我通常会这样推理

对于你的解决方案,你给出 "axioms"

type family n :+ m where
    Z   :+ m = m
    S n :+ m = n :+ S m

有了这些,(n + Z) :~: n 的证明将不起作用。

我可以从新的 "axiom" 产生原始秒的证明 second "axiom"(所以我的 second "axiom" 现在是引理?)。

succ_plus_id :: SNat n1 -> SNat n2 -> (((S n1) :+ n2) :~: (S (n1 :+ n2)))
succ_plus_id SZero _ = Refl
succ_plus_id (SSucc n) m = gcastWith (succ_plus_id n (SSucc m)) Refl

所以现在我应该能够让原始校样发挥作用,但我 不确定目前如何。

到目前为止我的推理是否正确?

PS: ghc 同意我关于为什么正确身份存在的证明不起作用的推理

Could not deduce ((n1 :+ 'S 'Z) ~ 'S n1)
...
or from ((n1 :+ 'Z) ~ n1)

你可以稍微简化一下reverse的定义:

{-# LANGUAGE GADTs, KindSignatures, DataKinds    #-}
{-# LANGUAGE TypeFamilies, UndecidableInstances  #-}
{-# LANGUAGE TypeOperators                       #-}

data Nat = Z | S Nat

data Vec a n where
    Nil   :: Vec a Z
    (:::) :: a -> Vec a n -> Vec a (S n)

type family n :+ m where
    Z   :+ m = m
    S n :+ m = n :+ S m

elim0 :: Vec a (n :+ Z) -> Vec a n
elim0 = undefined

accrev :: Vec a n -> Vec a n
accrev = elim0 . go Nil where
    go :: Vec a m -> Vec a n -> Vec a (n :+ m)
    go acc  Nil       = acc
    go acc (x ::: xs) = go (x ::: acc) xs

(:+) 运算符是根据 (:::) 运算符定义的。 (:::) 案例中的统一过程如下:

x ::: xs 导致 n 成为 S n。所以结果的类型变成 Vec a (S n :+ m) 或者,在 beta-reduction 之后,变成 Vec a (n :+ S m)。而

x ::: acc         :: Vec a (S m)
xs                :: Vec a  n
go (x ::: acc) xs :: Vec a (n :+ S m)

所以我们有一场比赛。但是现在您需要定义 elim0 :: Vec a (n :+ Z) -> Vec a n,这需要您的问题的两个证明。

Agda 中的全部代码:http://lpaste.net/117679


顺便说一句,这不是真的,无论如何你都需要证据。以下是 reverse 在 Agda 标准库中的定义:

foldl : ∀ {a b} {A : Set a} (B : ℕ → Set b) {m} →
        (∀ {n} → B n → A → B (suc n)) →
        B zero →
        Vec A m → B m
foldl b _⊕_ n []       = n
foldl b _⊕_ n (x ∷ xs) = foldl (λ n → b (suc n)) _⊕_ (n ⊕ x) xs

reverse : ∀ {a n} {A : Set a} → Vec A n → Vec A n
reverse {A = A} = foldl (Vec A) (λ rev x → x ∷ rev) []

那是因为 foldl 携带了关于 _⊕_ 行为的额外类型信息,所以你在每一步都满足类型检查器的要求,不需要证明。

{-# LANGUAGE GADTs                #-}
{-# LANGUAGE KindSignatures       #-}
{-# LANGUAGE DataKinds            #-}
{-# LANGUAGE TypeFamilies         #-}
{-# LANGUAGE UndecidableInstances #-}
{-# LANGUAGE ExplicitForAll #-}

import Data.Type.Equality

data Nat = Z | S Nat

type family (n :: Nat) :+ (m :: Nat) :: Nat where
    Z   :+ m = m
    S n :+ m = n :+ S m

-- Singleton for Nat
data SNat :: Nat -> * where
  SZero :: SNat Z
  SSucc :: SNat n -> SNat (S n)

succ_plus_id :: SNat n1 -> SNat n2 -> (((S n1) :+ n2) :~: (S (n1 :+ n2)))
succ_plus_id SZero _ = Refl
succ_plus_id (SSucc n) m = gcastWith (succ_plus_id n (SSucc m)) Refl

plus_id_r :: SNat n -> ((n :+ Z) :~: n)
plus_id_r SZero = Refl
plus_id_r (SSucc x) = gcastWith (plus_id_r x) (succ_plus_id x SZero)

data Vec a n where
    Nil   :: Vec a Z
    (:::) :: a -> Vec a n -> Vec a (S n)

size :: Vec a n -> SNat n
size Nil         = SZero
size (_ ::: xs)  = SSucc $ size xs

elim0 :: SNat n -> (Vec a (n :+ Z) -> Vec a n)
elim0 n x = gcastWith (plus_id_r n) x

accrev :: Vec a n -> Vec a n
accrev x = elim0 (size x) $ go Nil x where
    go :: Vec a m -> Vec a n -> Vec a (n :+ m)
    go acc  Nil       = acc
    go acc (x ::: xs) = go (x ::: acc) xs

safeHead :: Vec a (S n) -> a
safeHead (x ::: _) = x