Haskell 如何说服ghc类型级加法是可交换的(实现依赖类型的反向)?
这不会编译,因为正如ghc告诉我的,Add不是内射的。我如何告诉编译器Add是真正可交换的(也许是告诉它Add是内射的)?从哈索希主义的论文看来,人们必须以某种方式提供一个代理Haskell 如何说服ghc类型级加法是可交换的(实现依赖类型的反向)?,haskell,dependent-type,Haskell,Dependent Type,这不会编译,因为正如ghc告诉我的,Add不是内射的。我如何告诉编译器Add是真正可交换的(也许是告诉它Add是内射的)?从哈索希主义的论文看来,人们必须以某种方式提供一个代理 {-# LANGUAGE DataKinds #-} {-# LANGUAGE TypeOperators #-} {-# LANGUAGE KindSignatures #-} {-# LANGUAGE GADTs #-} {-# L
{-# 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,但它感觉太多的是正在进行的封面下为我的口味
{-# 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))
请参阅以了解原始想法
编辑:
@user3237465这是非常有趣的,而且更符合我的想法
(尽管经过深思熟虑,我的问题可能不是很好
制定)
看来我有“公理”
这样就可以产生像这样的证据
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
- 在上面的最后一句中,我们有SSucc n::SNat(sk)so n::k
- 因此我们需要证明sk+Z:~:sk
- 通过第二个“公理”S k+Z=S(k+Z)
- 因此我们需要证明S(k+Z):~:sk
- plus_id_r n给出了(k+Z)~:k的“证明”
- Refl给出了m~n=>sm:~:sn的“证明”
- 因此,我们可以使用gcastWith统一这些证明,以给出所需的结果 结果
- 在最后一个子句中,我们再次得到了SSucc x的类型SNat(sk)
- 因此我们需要证明sk:+Z:~:sk
- 通过第二个新的“公理”,我们得到了sk+Z=k+sz
- 因此我们需要证明k+sz:~:sk
- 所以我们有更复杂的东西要证明:-(
Could not deduce ((n1 :+ 'S 'Z) ~ 'S n1)
...
or from ((n1 :+ 'Z) ~ n1)
您可以稍微简化
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。在(::)
案例中的统一过程如下:
x:::xs
将n
诱导为sn
。因此结果类型变为veca(sn:+m)
或者,在beta减少后,veca(n:+sm)
x ::: acc :: Vec a (S m)
xs :: Vec a n
go (x ::: acc) xs :: Vec a (n :+ S m)
所以我们有一个匹配项。但是现在你需要定义elim0::veca(n:+Z)->veca n
,这需要你的问题的两个证明
Agda中的整个代码:
顺便说一句,在任何情况下都不需要证据。以下是Agda标准库中如何定义
反向:
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) []
这是因为foldl
包含有关_⊕_代码>,因此你在每一步都满足类型检查器的要求,并且不需要任何证明。这个问题适用吗?在某种程度上,我无法获得用于反向累积版本的解决方案。编译器正确地认为Add不是内射的。你的第一个等式显示Add Z(s Z)~sz
但第一个和第二个表示添加(sz)Z~S(添加Z)~sz
。事实上,如果它是可交换的,它就不能是内射的。与您在示例中已经做过的相比,您希望得到什么呢?(+)
在任何情况下都必须在某个时候编写证明。例如,是否要删除gcast
-s?“我如何告诉编译器Add确实是可交换的?”正是您所做的。我会在Refl上进行模式匹配,而不是使用gcast,但这只是风格。即使在独立类型的语言中,使用相同的+和Nat定义,您仍然需要编写证明。非常感谢。我刚刚安装了Agda,但在以下任何位置都找不到模块函数的源:/yarr/repairs/Function.agda/yarr/repairs/Function.lagda/Library/Haskell/ghc-7.8.3/lib/agda-2.4.2.2/share/lib/prim/Function.agda/Library/Haskell/ghc-7.8.3/lib/agda-2.4.2.2/share/lib/prim/Function.lagda范围检查打开导入声明时Function@idontgetoutmuch,需要单独安装。您可以找到说明。另请参见下面的user3237465的实现:lpaste.net/118199和我的。我认为您应该接受您的答案,因为您的解决方案更好、更完整。我已经写了一篇博客文章总结了其中的大部分内容:
{-# 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
x ::: acc :: Vec a (S m)
xs :: Vec a n
go (x ::: acc) xs :: Vec a (n :+ S m)
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) []
{-# 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