sieb :: [a -> Bool] -> [a] -> [[a]]
sieb (p:ps) subjects = (filter p subjects)

{- PROBLEM:
    Couldn't match type `a' with `[a]'
      `a' is a rigid type variable bound by
          the type signature for sieb :: [a -> Bool] -> [a] -> [[a]]
          at u4.hs:17:1
    Expected type: [a] -> Bool
      Actual type: a -> Bool
    In the first argument of `filter', namely `p'
    In the expression: (filter p subjects)
-}