Project euler problem 21 solution in haskell with point-free amicable definition

Amicable2.hs

module Amicable2 where

d :: Int -> Int
d x = sum [n | n <- [1 .. (x `div` 2)], x `mod` n == 0]

(===) :: (Eq b) => (a -> b) -> (a -> b) -> a -> Bool
(f === g) x = f x == g x

(/==) :: (Eq b) => (a -> b) -> (a -> b) -> a -> Bool
(f /== g) x = f x /= g x      

(&&&) :: (a -> Bool) -> (a -> Bool) -> a -> Bool
(f &&& g) x = f x && g x

amicable :: Int -> Bool 
amicable = d /== id &&& ((d . d) === id)

main = print $ sum $ filter amicable [1 .. 9999]