>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)
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,删除所有:
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
如何使用我的证明通知伊德里斯此类型签名是正确的?
现在,伊德里斯找不到(-)
的证明{auto smaller : LTE n m}
。
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,由于smaller
是一个auto
参数,因此可以将编译器提示到证明函数。然后,当auto
-查找LTE (count x l) n
的值时,将尝试此函数。
%hint
countLTELen : Eq a => (x : a) -> (l : Vect n a) -> LTE (count x l) n