Question

我需要证明

f (g xs) == g (f xs)

当xs是Ints的有限列表时。

假设f和g都是[Int] - ＆gt; [Int]

类型

Answer 1

反例反驳：

f xs = []
g xs = [1]

如果您希望保留此属性，则需要对f和g的更多具体限制。

你可能在想法律

(map f . map g) == map (f . g)

确实可以证明。

Answer 2

从某个地方找到这个

Theorem: (map f . map g) xs = map (f . g) xs
       - Proof is by induction on xs
    *** Base Case, xs = []
        - Left side: (map f . map g) [] = map f (map g []) = map f [] = []
        - Right side: map (f . g) [] = []
    *** Inductive Case, xs = k:ks
        - Inductive Hypothesis: (map f . map g) ks = map (f . g) ks
        - Left Side
          + (map f . map g) xs
          + map f (map g (k:ks))
          + map f ((g k) : (map g ks))
          + (f (g k)) : (map f (map g ks))
          + ((f . g) k) : ((map f . map g) ks)  i.e. change from bracket form back to point form
          + ((f . g) k) : (map (f . g) ks) by inductive hypothesis
          + map (f . g) (k:ks) by definition of map
          + map (f . g) xs

如何在haskell中设置归纳证明？

2 个答案: