（（->）r）类型的适用法律

Question

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)

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

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.

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)

Answer 1

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 -> uxy .

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.