这总是如此：fmap（foldr f z）。 sequenceA = foldr（liftA2 f）（纯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)

我强烈怀疑左右表达是否相等，但如何证明呢？

Answer 1

这至少是一个开始：

\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

如果您可以证明fmap toList . sequenceA = sequenceA . toList，并且您的原始声明适用于t = []，那么您应该很高兴。