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以下是结构归纳的定义吗?

foldr f a (xs::ys) = foldr f (foldr f a ys) xs

有人可以给我一个 Haskell 结构归纳的例子吗?

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1 回答 1

24

您没有指定它,但我假设::是指 list concanation 和 use ++,因为那是 Haskell 中使用的运算符。为了证明这一点,我们将对 进行归纳xs。首先,我们证明该陈述适用于基本情况(即xs = []

foldr f a (xs ++ ys) 
{- By definition of xs -}
= foldr f a ([] ++ ys)
{- By definition of ++ -}
= foldr f a ys

foldr f (foldr f a ys) xs
{- By definition of xs -}
= foldr f (foldr f a ys) []
{- By definition of foldr -}
= foldr f a ys

现在,我们假设归纳假设foldr f a (xs ++ ys) = foldr f (foldr f a ys) xs成立xs并表明它也将成立于列表 x:xs

foldr f a (x:xs ++ ys)
{- By definition of ++ -}
= foldr f a (x:(xs ++ ys))
{- By definition of foldr -}
= x `f` foldr f a (xs ++ ys)
         ^------------------ call this k1
= x `f` k1

foldr f (foldr f a ys) (x:xs)
{- By definition of foldr -}
= x `f` foldr f (foldr f a ys) xs
         ^----------------------- call this k2
= x `f` k2

现在,根据我们的归纳假设,我们知道k1并且k2是相等的,因此

x `f` k1 =  x `f` k2

从而证明了我们的假设。

于 2013-02-19T15:01:39.320 回答