9.4.1 - proof by induction over as

Induction base:
    Suppose as = []. Then we have:
    map f (as ++ bs) = map f ([] ++ bs) = map f bs = [] ++ (map f bs) = (map f []) ++ (map f bs) = (map f as) ++ (map f bs).

Induction step:
    Suppose map f (as ++ bs) = (map f as) ++ (map f bs) for certain as and any bs. Then we have:
    map f ([a:as] ++ bs) = map f [a:as ++ bs] = [f a : map f (as ++ bs)] = [f a : (map f as) ++ (map f bs)] = [f a : map f as] ++ (map f bs) = (map f [a:as]) ++ (map f bs).

By the principle of induction we have now proven that map f (as ++ bs) = (map f as) ++ (map f bs) for any finite lists as, bs.