Question

我已经向这个证据提出了一个不同的问题，所以我很抱歉重复。只是没有人会回答，因为另一个人有答案。

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)

这些是foldr的定义：

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xs的基本情况。我有这个，但我不知道它是否正确。

foldr f st ([]++ys) = f (foldr f st []) (foldr f st ys)
LHS:
    foldr f st ([]++ys) 
  = foldr f st ys                                              by (++) and by (foldr.1)
RHS:
    f (foldr f st []) (foldr f st ys) = 
  = f st (foldr f st ys)                                       by (foldr.1)
  = foldr f st ys                                              by def of st = 0 and f = (+)

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)

LHS:
    foldr f st (x:xs ++ ys)  
  = f x (foldr f st (xs++yx))                                      (by foldr.2)
  = f x (f (foldr f st xs) (foldr f st ys) )                       (by ind. hyp)
  = f (f x (foldr f st xs)) (foldr f st ys)                        (by assosiativity of f)

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)

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

在您开始证明某事之前，请查看一些示例，例如

xs = ys = [] ; st = True ; f _ _ = False

Answer 2

更新：问题被多次编辑。以下问题在问题revision 1中有意义。

这是不可能证明的。您需要f成为幺半群：f a (f b c) == f (f a b) c f a st == a和f st a == a。

可证明的陈述是foldr f st (xs++ys) == foldr f (foldr f st ys) xs

假设f st定义了一个monoid，我们得到以下语句：

foldr f st ([]++ys) ==
     -- by definition of neutral element of list monoid, []++ys == ys
                    == foldr f st ys
     -- by definition of neutral element of `f` monoid, f st a == a
                    == f st (foldr f st ys)
     -- by definition of foldr, (foldr f st []) == st
                    == f (foldr f st []) (foldr f st ys)

然后，

foldr f st ((x:xs)++ys) ==
     -- by associativity of list monoid, (x:xs)++ys == x:(xs++ys)
                    == foldr f st (x:(xs++ys))
     -- by definition of foldr, foldr f st (x:t) == f x (foldr f st t)
                    == f x (foldr f st (xs++ys))
     -- by induction, foldr f st (xs++ys) == f (foldr f st xs) (foldr f st ys)
                    == f x (f (foldr f st xs) (foldr f st ys))
     -- by associativity of monoid `f`, f x (f y z) == f (f x y) z
                    == f (f x (foldr f st xs)) (foldr f st ys)
     -- by definition of foldr, f x (foldr f st t) == foldr f st (x:t)
                    == f (foldr f st (x:xs)) (foldr f st ys)

实质上，这是幺半群同态的证明。列表是一个自由的幺半群，同态存在，并且正是你正在寻找的形式 - 它是f的变形。

这是我写的正确的证据，如果没有，那是什么问题？

2 个答案: