用foldr重写cp

cp = folder g [[]]
        where g xs xss = [x:ys|x<-xs,ys<-xss]

假设：（a）filter（all p）（cp xss）= cp（map（filter p）xs）= folder g （map（filter p）xs）
应证明等式（B）成立

(b)    filter (all p) (cp (xs:xss)) = cp (map (filter p) (xs:xss)) 

{# definition of map and cp #}
cp (map (filter p) (xs:xss)) = foldr g [[]] (map (filter p) (xs:xss))
{# definition of folder #}
g (filter p xs) (foldr g [[]] (map (filter p)) xss)

{#基于假设（a）#）g（filter p xs）（filter（all p）（cp xss））

{# definition of g #}
[x:ys|x<-filter p xs, ys<-filter (all p) (cp xss)]
{# definition of filter 
[x:ys|x<-[x' <- xs,p x'], ys <- [ys' <- (cp xss), (all p) ys']]

{# condiontial cp means cp after filter #}

filter (all p) (cp (xs:xss))

In conclusion,
cp (map (filter p) (xs:xss)) = filter (all p) (cp (xs:xss))