9.4.2 - proof by induction over xs

Induction base:
    Suppose xs = []. Then we have:
    flatten (map (map f) xs) = flatten (map (map f) []) = flatten [] = [] = map f [] = map f (flatten []) = map f (flatten xs).

Induction step:
    Suppose flatten (map (map f) xs) = map f (flatten xs) for certain xs of finite length. Then we have:
    flatten (map (map f) [x:xs]) = flatten [map f x : map (map f) xs] = (map f x) ++ flatten (map (map f) xs) = (map f x) ++ (map f (flatten xs)) =(9.4.1) map f (x ++ (flatten xs)) = map f (flatten [x:xs]).

By the principle of induction we have now proven that flatten (map (map f) xs) = map f (flatten xs) for any list of finite length xs.