Question

我试图通过结构归纳证明以下陈述：

foldr f st (xs++yx) = f (foldr f st xs) (foldr f st ys)        (foldr.3)

但是我甚至不确定如何定义foldr，所以我被困住了，因为没有为我提供任何定义。我现在相信foldr可以定义为

foldr f st [] = st                                             (foldr.1)
foldr f st x:xs = f x (foldr f st xs)                          (foldr.2)

现在我想开始处理将空列表传递给foldr的基本情况。我有这个，但我不认为这是正确的。

foldr f st ([]++[]) = f (foldr f st []) (foldr f st [])
LHS:
    foldr f st ([]++[]) = foldr f st []                        by (++)
    foldr f st [] = st                                         by (foldr.1)
RHS:
    f (foldr f st []) (foldr f st []) = f st st                by (foldr.1)
    = st                                                       by definition of identity, st = 0

LHS = RHS, therefore base case holds

现在这就是我的归纳步骤：

Assume that:
  foldr f st (xs ++ ys) = f (foldr f st xs) (foldr f st ys)        (ind. hyp)

Show that:
  foldr f st (x:xs ++ ys) = f (foldr f st x:xs) (foldr f st ys)    (inductive step)

LHS:
    foldr f st (x:xs ++ ys) = f x (foldr f st xs) (foldr f st ys)  (by foldr.2)
RHS:
    f (foldr f st x:xs) (foldr f st ys) = 
  = f f x (foldr f st xs) (foldr f st ys)                          (by foldr.2)
  = f x (foldr f st xs) (foldr f st ys)

LHS = RHS, therefore inductive step holds. End of proof.

我不确定此证明是否有效。我需要一些帮助来确定它是否正确，如果不正确 - 它的哪一部分不是。

Answer 1

首先：您可以通过API文档找到许多基本Haskell函数的定义，该文档可在Hackage上找到。 base的文档是here。 foldr导出foldr :: (a -> b -> b) -> b -> [a] -> b foldr k z = go where go [] = z go (y:ys) = y `k` go ys，其中包含指向Prelude的链接：

foldr f st [] = st
foldr f st (y:ys) = f y (foldr f st ys)

出于效率原因，它的定义是这样的;查找＆＃34;工人包装。＆＃34;它相当于

第二：在您所需的证明中，a -> a -> a的类型必须为a -> b -> b，这不如xs = ys = []一般。

让我们完成基本案例（foldr f st ([]++[]) = f (foldr f st []) (foldr f st []) -- Definition of ++ foldr f st [] = f (foldr f st []) (foldr f st []) -- Equation 1 of foldr st = f st st）。

st

这个等式一般不成立。要继续进行校对，您必须假设f是f的身份。

我相信，在非基本情况下，您还必须假设f是关联的。这两个假设相结合，表明st和{{1}}形成了一个幺半群。您是否想要证明its source code的某些内容？

证明foldr f st（xs ++ ys）= f（foldr f st xs）（foldr f st ys）

1 个答案: