我在idris中有一个函数计数,定义为:
count : Eq a => a -> Vect n a -> Nat
count x [] = Z
count x (y::ys) = with (x == y)
| True = S (count x ys)
| False = count x ys
最大值计数的证明可以返回:
countLTELen : Eq a => (x : a) -> (l : Vect n a) -> LTE (count x l) n
countLTELen x [] = lteRefl
countLteLen x (y::ys) with (x == y)
| True = LTESucc (countLTELen x ys)
| False = lteSuccRight (countLTELen x ys)
这一切都很好。我现在想编写一个从列表中删除所有元素的函数,removeAll:
removeAll : Eq a => (x : a) -> (l : Vect n a) -> Vect (n - (count x l)) a
removeAll x [] = []
removeAll x (y::ys) with (x == y)
| True = removeAll x ys
| False = x :: removeAll x ys
但是这个定义给出了一个错误:
|
56 | removeAll : Eq a => (x : a) -> (l : Vect n a) -> Vect (n - (count x l)) a
| ^
When checking type of Proof.removeAll:
When checking argument smaller to function Prelude.Nat.-:
Can't find a value of type
LTE (count a n constraint x l) n
如何使用我的证据告知Idris此类型签名是否正确?
答案 0 :(得分:4)
目前,Idris无法找到{auto smaller : LTE n m}
的证据(-)
。
所以要么你需要明确:
removeAll : Eq a => (x : a) -> (l : Vect n a) ->
Vect ((-) {smaller=countLTELen x l} n (count x l) ) a
或者,因为smaller
是auto
- 参数,您可以将编译器提示给您的校对函数。然后,当auto
- 为LTE (count x l) n
找到值时,将尝试此功能。
%hint
countLTELen : Eq a => (x : a) -> (l : Vect n a) -> LTE (count x l) n