haskell - 这总是正确的: fmap (foldr f z) .sequenceA=foldr(liftA2f)(purez)

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)

我强烈怀疑表达式 left 和 right 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

如果你能证明fmap toList 。序列A = 序列A 。 toList，并且您最初的声明适用于 t = [] 您应该可以开始了。