Haskell的HSpec和Delicious入门？

我是新来的Haskell，我试图让hspec工作与美味（使用可口-hspec）与堆栈。我已经看到了将Haste与HUnit一起使用的示例，如下所示：

import Test.Tasty
import Test.Tasty.SmallCheck as SC
import Test.Tasty.QuickCheck as QC
import Test.Tasty.HUnit

import Data.List
import Data.Ord

main = defaultMain tests

tests :: TestTree
tests = testGroup "Tests" [properties, unitTests]

properties :: TestTree
properties = testGroup "Properties" [scProps, qcProps]

scProps = testGroup "(checked by SmallCheck)"
  [ SC.testProperty "sort == sort . reverse" $
      \list -> sort (list :: [Int]) == sort (reverse list)
  , SC.testProperty "Fermat's little theorem" $
      \x -> ((x :: Integer)^7 - x) `mod` 7 == 0
  -- the following property does not hold
  , SC.testProperty "Fermat's last theorem" $
      \x y z n ->
        (n :: Integer) >= 3 SC.==> x^n + y^n /= (z^n :: Integer)
  ]

qcProps = testGroup "(checked by QuickCheck)"
  [ QC.testProperty "sort == sort . reverse" $
      \list -> sort (list :: [Int]) == sort (reverse list)
  , QC.testProperty "Fermat's little theorem" $
      \x -> ((x :: Integer)^7 - x) `mod` 7 == 0
  -- the following property does not hold
  , QC.testProperty "Fermat's last theorem" $
      \x y z n ->
        (n :: Integer) >= 3 QC.==> x^n + y^n /= (z^n :: Integer)
  ]

unitTests = testGroup "Unit tests"
  [ testCase "List comparison (different length)" $
      [1, 2, 3] `compare` [1,2] @?= GT

  -- the following test does not hold
  , testCase "List comparison (same length)" $
      [1, 2, 3] `compare` [1,2,2] @?= LT
  ]

但是我不想使用hunit，smallcheck或quickcheck，而是要使用hspec。我尝试了以下方法：

module Spec where

import Game

import Test.Tasty
import Test.Tasty.Hspec

spec_beats :: Spec
spec_beats = do

  it "hello" $
    Rock `beats` Scissors `shouldBe` True

tests :: TestTree
tests = testGroup "Tests" [spec_beats]

main = defaultMain tests

但这不能编译：

    • Couldn't match type ‘hspec-core-2.7.1:Test.Hspec.Core.Spec.Monad.SpecM
                             () ()’
                     with ‘TestTree’
      Expected type: TestTree
        Actual type: Spec
    • In the expression: spec_beats
      In the second argument of ‘testGroup’, namely ‘[spec_beats]’
      In the expression: testGroup "Tests" [spec_beats]
   |        
15 | tests = testGroup "Tests" [spec_beats]
   |                            ^^^^^^^^^^

所以我的问题是，我如何使我的hspec示例工作好吃？