无法将类型A与A1匹配,但类型看起来相同

我在学习haskell,编译器给了我一个我不太明白的错误。它说预期类型和实际类型不匹配,但对我来说,它们看起来是相同的。有人能帮我理解这个错误想说什么吗?

/app/app/Main.hs:75:11: error:
    • Couldn't match type ‘a’ with ‘a1’
      ‘a’ is a rigid type variable bound by
        the type signature for:
          app :: forall a.
                 (ReaderT (W.Options -> String -> IO (Response BL.ByteString)) IO a
                  -> IO a)
                 -> IO Middleware
        at /app/app/Main.hs:68:1-99
      ‘a1’ is a rigid type variable bound by
        a type expected by the context:
          forall a1.
          ReaderT (W.Options -> String -> IO (Response BL.ByteString)) IO a1
          -> IO a1
        at /app/app/Main.hs:75:3-12
      Expected type: ReaderT
                       (W.Options -> String -> IO (Response BL.ByteString)) IO a1
                     -> IO a1
        Actual type: ReaderT
                       (W.Options -> String -> IO (Response BL.ByteString)) IO a
                     -> IO a
    • In the first argument of ‘spockT’, namely ‘(r)’
      In the expression: spockT (r)
      In a stmt of a 'do' block: spockT (r) $ routes apiKey template
    • Relevant bindings include
        r :: ReaderT
               (W.Options -> String -> IO (Response BL.ByteString)) IO a
             -> IO a
          (bound at /app/app/Main.hs:69:5)
        app :: (ReaderT
                  (W.Options -> String -> IO (Response BL.ByteString)) IO a
                -> IO a)
               -> IO Middleware
          (bound at /app/app/Main.hs:69:1)
   |
75 |   spockT (r) $ routes apiKey template
   |           ^

更让我困惑的是,我所做的只是将一个函数从被直接调用变成一个参数。

失败的代码在下面(简化),但是如果我内联 runner 也就是说,将其作为 app 一切正常。

main :: IO ()
main = do
  runSpock 8080 $ app runner

runner :: ReaderT (W.Options -> String -> IO (Response BL.ByteString)) m a -> m a
runner r = runReaderT r W.getWith

app :: (ReaderT (W.Options -> String -> IO (Response BL.ByteString)) IO a -> IO a) -> IO Middleware
app runner = do
  spockT (runner) $ routes

routes :: MonadIO m => SpockCtxT ctx (ReaderT (W.Options -> String -> IO (Response BL.ByteString)) m) ()
routes = do
  get "healthz" $ text "ok"