Haskel Currying merging functions

Question

Quick question I'm unable to understand how and why the following type definition gets formulated

:t const
const :: a -> b -> a

:t id
id :: a -> a

:t (const id)
(const id) :: b -> a -> a

Doesn't the following actually resolve too a -> a -> b -> a where a -> a is feed inside of a -> b -> a AKA (const(id _))

anyone mind explaining how this is actually being transformed?

Answer 1

const x :: b -> ??, where ?? is the type of x. In this case, x is the function id, which has type a -> a. Thus, const id :: b -> (a -> a), and you can drop the parentheses because -> is right-associative.

Answer 2

const :: a -> b -> a
id :: c -> c

const before applied with id

const :: (c->c) -> b -> (c->c)

const after applied id

const id :: b -> (c -> c)
const id :: b -> c -> c

No matter what you apply to const id you get c -> c