Haskell将多个参数缩写为单个变量

所以我有一个函数 必须 有某种类型。我的实现类似于以下内容:

f :: Int -> Int -> Int -> Int -> Int -> Int -> Int -> Int -> Int -> Int
f t1 t2 t3 t4 t5 t6 t7 t8 t9
    = filterFirst checkFunc p
    where
        p = findAll [1..9]
        checkFunc = validate t1 t2 t3 t4 t5 t6 t7 t8 t9

现在有没有办法把t值缩写为 或者类似的事情:

f :: Int -> Int -> Int -> Int -> Int -> Int -> Int -> Int -> Int -> Int
f ts
    = filterFirst checkFunc p
    where
        p = findAll [1..9]
        checkFunc = validate ts

一个让它看起来更干净的方法将会是惊人的。

编辑: 更多细节

validate :: Int -> Int -> Int -> Int -> Int -> Int -> Int -> Int -> Int -> [Int] -> Bool
validate t1 t2 t3 t4 t5 t6 t7 t8 t9 is =
    [t1, t2, t3, t4, t5, t6, t7, t8, t9] == sums is

-- Calculates sums from specific indexes in list
sums :: [Int] -> [Int]

-- from https://stackoverflow.com/a/28904773/1218369
filterFirst :: (a -> Bool) -> [a] -> [a]

-- Find all possible permutations
findAll :: [a] -> [[a]]
-- Basically Data.List (permutations)