這總是如此：fmap（foldr fz）。 sequenceA = foldr（liftA2 f）（純z）

Question

import Prelude hiding (foldr)

import Control.Applicative
import Data.Foldable
import Data.Traversable

left, right :: (Applicative f, Traversable t) => (a -> b -> b) -> b -> t (f a) -> f b
left f z = fmap (foldr f z) . sequenceA
right f z = foldr (liftA2 f) (pure z)

我強烈懷疑左右表達是否相等，但如何證明呢？

Answer 1

這至少是一個開始：

\f z -> fmap (foldr f z) . sequenceA
== (definition of Foldable foldr)
\f z -> fmap (foldr f z . toList) . sequenceA
== (distributivity of fmap)
\f z -> fmap (foldr f z) . fmap toList . sequenceA
== (need to prove this step, but it seems intuitive to me)
\f z -> fmap (foldr f z) . sequenceA . toList

\f z -> foldr (liftA2 f) (pure z)
== (definition of Foldable foldr)
\f z -> foldr (liftA2 f) (pure z) . toList

如果你能證明fmap toList . sequenceA = sequenceA . toList fmap toList . sequenceA = sequenceA . toList fmap toList . sequenceA = sequenceA . toList ，並且你的原始聲明適用於t = []你應該很高興去。