Question

我有以下使用SnocList

的功能

import Data.List.Views

takeLast : (n : Nat) -> List a -> List a
takeLast n xs with (snocList xs)
  takeLast Z [] | Empty = []
  takeLast Z (ys ++ [x]) | (Snoc rec) = []
  takeLast (S k) [] | Empty = []
  takeLast (S k) (ys ++ [x]) | (Snoc rec) = takeLast k ys ++ [x] | rec

它需要给定List中的最后n个元素。我希望它等同于以下函数。

takeLast' : (n : Nat) -> (xs : List a) -> List a
takeLast' n xs = reverse $ take n $ reverse xs

当我尝试以下示例证明

时

takeLastsAreEquivalent : takeLast' 2 [1, 2, 3] = takeLast 2 [1, 2, 3]
takeLastsAreEquivalent = Refl

我从类型检查器中收到以下错误。

When checking right hand side of takeLastsAreEquivalent with expected type
        takeLast' 2 [1, 2, 3] = takeLast 2 [1, 2, 3]

Type mismatch between
        with block in Main.takeLast Integer 2 [1, 2, 3] (snocList [1, 2, 3]) =
        with block in Main.takeLast Integer 2 [1, 2, 3] (snocList [1, 2, 3]) (Type of Refl)
and
        [2, 3] = with block in Main.takeLast Integer 2 [1, 2, 3] (snocList [1, 2, 3]) (Expected type)

Specifically:
        Type mismatch between
                with block in Main.takeLast Integer 2 [1, 2, 3] (snocList [1, 2, 3])
        and
                [2, 3]
Holes: Main.takeLastsAreEquivalent

当我在REPL中运行它们时，一切似乎都很好。

*/Example> takeLast' 2 [1, 2, 3]
[2, 3] : List Integer
*/Example> :t takeLast' 2 [1, 2, 3]
takeLast' 2 [1, 2, 3] : List Integer
*/Example> takeLast 2 [1, 2, 3]
[2, 3] : List Integer
*/Example> :t takeLast 2 [1, 2, 3]
takeLast 2 [1, 2, 3] : List Integer

我知道这不是一个真实的证明，但我只想说明这些是如何工作的，我正在研究一个例子。

不能给Idris证明使用SnocList的功能

0 个答案: