((->)r)类型的适用法律 [英] Applicative Laws for the ((->) r) type

问题描述

我正在尝试检查适用法律是否适用于函数类型((->) r)，这是我到目前为止所拥有的:

I'm trying to check that the Applicative laws hold for the function type ((->) r), and here's what I have so far:

-- Identiy
pure (id) <*> v = v 
-- Starting with the LHS
pure (id) <*> v
const id <*> v
(\x -> const id x (g x))
(\x -> id (g x))
(\x -> g x)
g x
v


-- Homomorphism
pure f <*> pure x = pure (f x)
-- Starting with the LHS
pure f <*> pure x
const f <*> const x
(\y -> const f y (const x y))
(\y -> f (x))
(\_ -> f x)
pure (f x)

我是否正确执行了前两个定律的步骤?

Did I perform the steps for the first two laws correctly?

我正在努力与&组成法.对于交换，到目前为止，我有以下内容:

I'm struggling with the interchange & composition laws. For interchange, so far I have the following:

-- Interchange
u <*> pure y = pure ($y) <*> u
-- Starting with the LHS
u <*> pure y
u <*> const y
(\x -> g x (const y x))
(\x -> g x y)
-- I'm not sure how to proceed beyond this point.

我希望为验证Interchange& amp;的步骤提供任何帮助. ((->) r)类型的成分适用法律.供参考，《组成法》的适用法律如下:

I would appreciate any help for the steps to verify the Interchange & Composition applicative laws for the ((->) r) type. For reference, the Composition applicative law is as follows:

pure (.) <*> u <*> v <*> w = u <*> (v <*> w)

推荐答案

我认为，在您的身份"证明中，您应该在任何地方都用v替换g(否则，g以及它来自何处) ?).同样，到目前为止，在交换"证明中，一切看起来还不错，但是神奇出现的g应该只是u.要继续证明，您可以开始降低RHS并确认它还会产生\x -> u x y.

I think in your "Identity" proof, you should replace g with v everywhere (otherwise what is g and where did it come from?). Similarly, in your "Interchange" proof, things look okay so far, but the g that magically appears should just be u. To continue that proof, you could start reducing the RHS and verify that it also produces \x -> u x y.

组成大致相同:在两侧插入pure和(<*>)的定义，然后在两侧开始计算.您很快就会来一些容易证明是等效的裸lambda.

Composition is more of the same: plug in the definitions of pure and (<*>) on both sides, then start calculating on both sides. You'll soon come to some bare lambdas that will be easy to prove equivalent.

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