Как присвоить выражениям общие типы в предложении «где»?

_{(извините за ужасно надуманный пример)}

Я хочу указать типы в предложении where -:

somemap :: (a -> b) -> [a] -> [b]
somemap f xs = ys
  where
    some = take 5 xs :: [a]
    ys = map f some :: [b]

Но это вызывает ошибку :

*Main> :load file.hs 
[1 of 1] Compiling Main             ( file.hs, interpreted )

fil.hs:15:18:
    Couldn't match expected type `a1' against inferred type `a'
      `a1' is a rigid type variable bound by
           an expression type signature at file.hs:15:25
      `a' is a rigid type variable bound by
          the type signature for `somemap' at file.hs:12:12
      Expected type: [a1]
      Inferred type: [a]
    In the second argument of `take', namely `xs'
    In the expression: take 5 xs :: [a]

file.hs:16:13:
    Couldn't match expected type `b1' against inferred type `b'
      `b1' is a rigid type variable bound by
           an expression type signature at file.hs:16:24
      `b' is a rigid type variable bound by
          the type signature for `somemap' at file.hs:12:17
    In the first argument of `map', namely `f'
    In the expression: map f some :: [b]
    In the definition of `ys': ys = map f some :: [b]
Failed, modules loaded: none.

. Принимая во внимание, что если я просто укажу конкретные типы, заменив Intна aи Boolна b, нет проблем :

somemap :: (Int -> Bool) -> [Int] -> [Bool]
somemap f xs = ys
  where
    some = take 5 xs :: [Int]
    ys = map f some :: [Bool]

. Итак, мой вопрос::Как мне указать общий типы и ограничения типов в предложении where -?