Question

我试图证明一些关于奇数和偶数自然数的公理。我在我的证明中使用了三种定义的数据类型。

data Nat = Z | S Nat

data Even (a :: Nat) :: * where
  ZeroEven :: Even Z
  NextEven :: Even n -> Even (S (S n))

data Odd (a :: Nat) :: * where
  OneOdd :: Odd (S Z)
  NextOdd :: Odd n -> Odd (S (S n))

我还为添加和乘法定义了以下类型系列。

type family   Add (n :: Nat) (m :: Nat) :: Nat
type instance Add Z m = m
type instance Add (S n) m = S (Add n m)

type family   Mult (n :: Nat) (m :: Nat) :: Nat
type instance Mult Z m = Z
type instance Mult (S n) m = Add (Mult n m) n

我有定义的功能，证明两个平均值的总和是均匀的，两个平均值的乘积是偶数。

evenPlusEven :: Even n -> Even m -> Even (Add n m)
evenPlusEven ZeroEven m = m
evenPlusEven (NextEven n) m = NextEven (evenPlusEven n m)

evenTimesEven :: Even n -> Even m -> Even (Mult n m)
evenTimesEven ZeroEven m = ZeroEven
evenTimesEven (NextEven n) m = evenPlusEven (EvenTimesEven n m) n

我使用的是GADTs，DataKinds，TypeFamilies和UndecidableInstances语言扩展程序以及GHC版本7.10.3。运行evenPlusEven给出了我期望的结果，但是当我包含evenTimesEven时，我收到了编译错误。错误是：

Could not deduce (Add (Add (Mult n1 m) n1) ('S n1)
                  ~ Add (Mult n1 m) n1)
from the context (n ~ 'S ('S n1))
  bound by a pattern with constructor
             NextEven :: forall (n :: Nat). Even n -> Even ('S ('S n)),
           in an equation for `evenTimesEven'
  at OddsAndEvens.hs:71:16-25
NB: `Add' is a type function, and may not be injective
Expected type: Even (Mult n m)
  Actual type: Even (Add (Mult n1 m) n1)
Relevant bindings include
  m :: Even m
    (bound at OddsAndEvens.hs:71:28)
  n :: Even n1
    (bound at OddsAndEvens.hs:71:25)
  evenTimesEven :: Even n -> Even m -> Even (Mult n m)
    (bound at OddsAndEvens.hs:70:1)
In the expression: evenPlusEven (evenTimesEven n m) n
In an equation for `evenTimesEven':
    evenTimesEven (NextEven n) m = evenPlusEven (evenTimesEven n m) n

Mult的类型族实例编译正常，如果我用错误抛出替换evenTimesEven的最后一行，我可以编译代码，并且函数运行良好，输入为{{1} }这让我觉得我的ZeroEven实例是正确的，我Mult的实现是问题所在，但我不确定原因。

不应该evenTimesEven和Even (Mult n m)有同类吗？

Answer 1

下面，我会滥用常用的数学符号。

由此，我们得到from the context (n ~ 'S ('S n1))。

n = 2+n1

我们需要证明Expected type: Even (Mult n m)偶数，即n*m偶数。

(2+n1)*m

我们已证明Actual type: Even (Add (Mult n1 m) n1)。这不一样。附加条款应为(n1*m)+n1，而非m，并且应添加两次。

类型函数可能不是单射的

1 个答案: