Question

我有数据类型

data N a = N a [N a]

玫瑰树的

和应用实例

instance Applicative N where
 pure a = N a (repeat (pure a))
 (N f xs) <*> (N a ys) = N (f a) (zipWith (<*>) xs ys)

需要为其证明适用法律。但是， pure 会创建无限深度，无限分支的树。因此，例如，在证明同态定律

pure f <*> pure a = pure (f a)

我认为这证明了平等

zipWith (<*>) (repeat (pure f)) (repeat (pure a)) = repeat (pure (f a))

通过近似（或取）引理可行。然而，我的尝试导致归纳步骤中的“恶性循环”。特别是，减少

approx (n + 1) (zipWith (<*>) (repeat (pure f)) (repeat (pure a))

给出

(pure f <*> pure a) : approx n (repeat (pure (f a)))

其中 approx 是近似函数。 如何在没有明确的共同证明证明的情况下证明平等？

Answer 1

我会使用展开的通用属性（因为重复和适当的不合格的拉链都展开）。有一个相关的讨论on my blog。但您可能也喜欢Ralf Hinze关于独特修复点ICFP2008（以及随后的JFP论文）的论文。

（只是检查一下：你所有的玫瑰树都是无限宽而且无限深？我猜这些法则不会另有规定。）

Answer 2

以下是我认为有效并且仍然处于程序化语法和等式推理水平的草图。

基本的直觉是，推理repeat x比推理流（更糟糕的是，列表）更容易。

uncons (repeat x) = (x, repeat x)

zipWithAp xss yss = 
    let (x,xs) = uncons xss
        (y,ys) = uncons yss
    in (x <*> y) : zipWithAp xs ys

-- provide arguments to zipWithAp
zipWithAp (repeat x) (repeat y) = 
    let (x',xs) = uncons (repeat x)
        (y',ys) = uncons (repeat y)
    in (x' <*> y') : zipWithAp xs ys

-- substitute definition of uncons...
zipWithAp (repeat x) (repeat y) = 
    let (x,repeat x) = uncons (repeat x)
        (y,repeat y) = uncons (repeat y)
    in (x <*> y) : zipWithAp (repeat x) (repeat y)

-- remove now extraneous let clause
zipWithAp (repeat x) (repeat y) = (x <*> y) : zipWithAp (repeat x) (repeat y)

-- unfold identity by one step
zipWithAp (repeat x) (repeat y) = (x <*> y) : (x <*> y) : zipWithAp (repeat x) (repeat y)

-- (co-)inductive step
zipWithAp (repeat x) (repeat y) = repeat (x <*> y)

Answer 3

为什么你需要共同诱导？只是归纳。

pure f <*> pure a = pure (f a)

也可以写（你需要证明左右相等）

N f [(pure f)] <*> N a [(pure a)] = N (f a) [(pure (f a))]

允许您一次关闭一个词。这会给你感应。

证明流的平等

3 个答案: