4 ms·
Yup, 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!
by lsb 16y ago
Yup,
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!