Inverse of FizzBuzz implementation in Haskell

BuzzFizz.hs

isFizz i = mod i 3 == 0
isBuzz i = mod i 5 == 0
isFizzBuzz i = isFizz i && isBuzz i
isFizzOrBuzz i = isFizz i || isBuzz i
isNeither i = (not.isBuzz) i && (not.isFizz) i

toFizzBuzz i
    | isFizzBuzz i = "FizzBuzz"
    | isFizz i = "Fizz"
    | isBuzz i = "Buzz"

fizzBuzz numbers = map toFizzBuzz $ filter isFizzOrBuzz numbers

testFizzBuzz s
    | s == "FizzBuzz" = (\i -> isFizzBuzz i)
    | s == "Fizz" = (\i -> isFizz i && (not.isBuzz) i)
    | s == "Buzz" = (\i -> isBuzz i && (not.isFizz) i)

buzzFizz :: [String] -> [Int]
buzzFizz (v:alues) =
    shortest results
  where
    shortest (x:xs) = foldl (\a b -> if length a > length b then b else a) x xs
    shortest [] = []
    results = filter (\r -> last r > 0) $ map (buzzFizz' (v:alues)) startNumbers
    startNumbers = filter (testFizzBuzz v) [1..100]

buzzFizz' :: [String] -> Int -> [Int]
buzzFizz' (value:values) i =
    if testFizzBuzz value i
        then [i] ++ buzzFizz' values (i+1)
        else if isNeither i
           then [i] ++ (buzzFizz' (value:values) (i+1))
           else [-1]

buzzFizz' [] i = []