Haskell: Given the following code, prove for finite lists by using structural induction, given:

append:: [a] => [a] => [a] append ys = ys append (x:xs) y = x: (append xs ys) filter p [] = 0 filter p (a:as) pa = a: (filter p as) I otherwise = filter p as that filter p (append xs ys) append (filter p xs) (filter pys) append:: [a] => [a] => [a] append ys = ys append (x:xs) y = x: (append xs ys) filter p [] = 0 filter p (a:as) pa = a: (filter p as) I otherwise = filter p as that filter p (append xs ys) append (filter p xs) (filter pys)