4 ms·
Don't you mean forall f g xs. map f (map g xs) = map (f.g) xs ? As for filter, I presume something like forall f g xs. filter f (filter g xs) = filter (
by uros643 16y ago
Don't you mean
forall f g xs. map f (map g xs) = map (f.g) xs
?
As for filter, I presume something like
forall f g xs. filter f (filter g xs) = filter (\x. g x && f x) xs
would work.
- lsb 16y agoYup, forall f g xs . filter f (filter g xs) = filter (\x -> g x && f x) xs was exactly what I meant, thanks for the catch!