Show Functor List or [] that satisfied two Functor laws:

    class Functor f where
        fmap:: (a -> b) -> f a -> f b

    instance Functor [] where
        fmap g [] = []   -- g = (a -> b),  [] = f a
        fmap g (x:cx) = (g x):(fmap g cx)

    Two Functor laws:
    1. fmap id = id
    2. fmap (f . g) = (fmap f).(fmap g)

    Prove:
    1. fmap id = id
       fmap id [] = [] (from def. of fmap)
       id []      = [] (from def. of identity function)
    => fmap id [] = id []
    => fmap id    = id

       fmap id (x:cx) = (id x):(fmap id cx)
               (x:cx) = (x:cx)

    => fmap id = id

    2. fmap(f . g) = (fmap f).(fmap g) 

        fmap(f . g) (x:cx) = (fmap f) . (fmap g) (x:cx)
                             (fmap f)(fmap g x:cx)
                             (fmap f)((g x:fmap g cx))
                              (f g x): (fmap f (fmap g cx))
                              (f (g x)):(fmap f (fmpa g cx))
                              fmap f (g x : fmap g cx) = (f g x): (fmap f (fmap g cx))
                              fmap f (fmap g x:cx)
                              (fmap f) . (fmap g) (c:cx)
    => fmap (f . g) = (fmap f).(fmap g)

    1. 2. => List or [] is Functor