1.(a)
f :: forall a. [a] -> [a]
Prove: for all AL, AR,
               h :: AL -> AR,
               xs :: [AL] :
            map h (f xs) = f (map h xs)
 



1.(b)
e :: forall b. (Int -> b -> b) -> b -> b
Prove: for all BR,
               opR :: Int -> BR -> BR
               zR :: BR
         foldr opR zR (e (:) []) = e opR zR



1.(c)

<T> LinkedList<T> bad(T x) {

}

bad :: Show a => a -> [a]
bad x =