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我正在尝试通过自然数计算奇偶校验和一半的下限：

data IsEven : Nat -> Nat -> Type where
    Times2 : (n : Nat) -> IsEven (n + n) n

data IsOdd : Nat -> Nat -> Type where
    Times2Plus1 : (n : Nat) -> IsOdd (S (n + n)) n

parity : (n : Nat) -> Either (Exists (IsEven n)) (Exists (IsOdd n))

我尝试了明显的实现parity：

parity Z = Left $ Evidence _ $ Times2 0
parity (S Z) = Right $ Evidence _ $ Times2Plus1 0
parity (S (S n)) with (parity n)
  parity (S (S (k + k))) | Left (Evidence _ (Times2 k)) =
      Left $ rewrite plusSuccRightSucc k k in Evidence _ $ Times2 (S k)
  parity (S (S (S ((k + k))))) | Right (Evidence _ (Times2Plus1 k)) =
      Right $ rewrite plusSuccRightSucc k k in Evidence _ $ Times2Plus1 (S k)

这种类型检查并按预期工作。但是，如果我尝试标记parity为total，伊德里斯开始抱怨：

 parity is possibly not total due to: with block in parity

with我看到的唯一块parity是从parity (S (S n))to递归调用的块parity n，但显然这是有根据的，因为n在结构上小于S (S n).

我如何说服 Idrisparity是完全的？

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1 回答 1

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对我来说它看起来像一个错误，因为以下基于case通过整体检查器的解决方案：

total
parity : (n : Nat) -> Either (Exists (IsEven n)) (Exists (IsOdd n))
parity Z = Left $ Evidence _ $ Times2 0
parity (S Z) = Right $ Evidence _ $ Times2Plus1 0
parity (S (S k)) =
  case (parity k) of
    Left (Evidence k (Times2 k)) =>
      Left $ rewrite plusSuccRightSucc k k in Evidence _ $ Times2 (S k)
    Right (Evidence k (Times2Plus1 k)) =>
      Right $ rewrite plusSuccRightSucc k k in Evidence _ $ Times2Plus1 (S k)
于 2017-10-04T10:58:40.383 回答