haskell - 这是否总是正确的： fmap (foldr fz) 。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式相等，但是如何证明呢？

Answer 1

至少这是一个开始：

\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

如果您能证明这一点fmap toList . sequenceA = sequenceA . toList，并且您最初的主张对您有效，t = []那么您应该可以继续使用。