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[英]Proving equivalence of sequence definitions from Applicative and Monad
[英]Proving equality of streams
我有一個數據類型
data N a = N a [N a]
玫瑰樹和應用實例
instance Applicative N where
pure a = N a (repeat (pure a))
(N f xs) <*> (N a ys) = N (f a) (zipWith (<*>) xs ys)
並需要證明適用法律。 然而, 純粹創造無限深,無限分枝的樹木。 因此,例如,在證明同態定律
pure f <*> pure a = pure (f a)
我認為這證明了平等
zipWith (<*>) (repeat (pure f)) (repeat (pure a)) = repeat (pure (f a))
通過近似(或采取)引理將起作用。 然而,我的嘗試導致歸納步驟中的“惡性循環”。 特別是減少
approx (n + 1) (zipWith (<*>) (repeat (pure f)) (repeat (pure a))
給
(pure f <*> pure a) : approx n (repeat (pure (f a)))
其中approx是近似函數。 如果沒有明確的共同證據,我怎樣才能證明這種平等?
以下是我認為有用並且保持在程序語法和等式推理層面的一些草圖。
基本的直覺是, repeat x
推理要比推理流(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)
你為什么需要共同誘導? 只是歸納。
pure f <*> pure a = pure (f a)
也可以寫(你需要證明左右相等)
N f [(pure f)] <*> N a [(pure a)] = N (f a) [(pure (f a))]
這允許你一次關閉一個學期。 這會給你感應。
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