功能证明 (Haskell)

我没能读懂《RWH》；我命令没有人放弃Haskell：函数式编程的技巧。现在我对第 146 页上的这些功能证明很好奇。具体来说，我试图证明 8.5.1sum (reverse xs) = sum xs。我可以做一些归纳证明，但后来我陷入困境。

HYP:

sum ( reverse xs ) = sum xs

BASE:

sum ( reverse [] ) = sum []

Left  = sum ( [] ) (reverse.1)
      = 0          (sum.1)

Right = 0          (sum.1)

就职：

sum ( reverse (x:xs) ) = sum (x:xs) 

Left = sum ( reverse xs ++ [x] )    (reverse.2)

Right = sum (x:xs)   
      = x + sum xs                  (sum.2)

所以现在我只是想证明这一点Left sum ( reverse xs ++ [x] )等于Right x + sum xs，但这离我开始的地方并不太远sum ( reverse (x:xs) ) = sum (x:xs).

我不太清楚为什么需要证明这一点，使用符号证明似乎是完全合理的reverse x:y:z = z:y:x(by defn)，并且因为 + 是可交换的 (arth) 那么reverse 1+2+3 = 3+2+1,

sum (reverse [])     = sum []                     -- def reverse
sum (reverse (x:xs)) = sum (reverse xs ++ [x])    -- def reverse
                     = sum (reverse xs) + sum [x] -- sum lemma below
                     = sum (reverse xs) + x       -- def sum
                     = x + sum (reverse xs)       -- commutativity assumption!
                     = x + sum xs                 -- inductive hypothesis
                     = sum (x:xs)                 -- definition of sum

然而，结合性和交换性的基本假设并没有得到严格保证，并且这对于许多数字类型（例如Float and Double违反这些假设的地方。

Lemma: sum (xs ++ ys) == sum xs + sum ys给定关联性(+)

Proof:

sum ([] ++ ys)     = sum ys           -- def (++)
                   = 0 + sum ys       -- identity of addition
                   = sum [] ++ sum ys -- def sum

sum ((x:xs) ++ ys) = sum (x : (xs ++ ys))  -- def (++)
                   = x + sum (xs ++ ys)    -- def sum 
                   = x + (sum xs + sum ys) -- inductive hypothesis
                   = (x + sum xs) + sum ys -- associativity assumption!
                   = sum (x:xs) + sum ys   -- def sum