from Haskell

6. (6 points) Give the following code prove, for finite lists by using structural induction, given the definitions: append:: [a] => [a] => [a] append lys ys append (x:xs) y = x: (append xs ys) filter p [] = [] filter p (a:as) | p a = a: (filter pas) | otherwise = filter pas that filter p (append xs ys) append (filter p xs) (filter pys) 6. (6 points) Give the following code prove, for finite lists by using structural induction, given the definitions: append:: [a] => [a] => [a] append lys ys append (x:xs) y = x: (append xs ys) filter p [] = [] filter p (a:as) | p a = a: (filter pas) | otherwise = filter pas that filter p (append xs ys) append (filter p xs) (filter pys)