Understanding ST's quantification and phantom type

Question

I am trying to write monadic test code using Test.QuickCheck.Monadic.

The module provides a harness

monadicST :: Testable a => (forall s. PropertyM (ST s) a) -> Property

to execute ST-based tests, and a function

run :: Monad m => m a -> PropertyM m a

to lift any action into a test monad.

In my code I have a polymorphic monadic action:

act :: Monad m => a -> m ()
act = undefined

The following function

st :: a -> Property
st = monadicST . run . act

fails to compile:

• Couldn't match type ‘PropertyM m0 ()’
                 with ‘forall s. PropertyM (ST s) a0’
  Expected type: PropertyM m0 () -> Property
    Actual type: (forall s. PropertyM (ST s) a0) -> Property
• In the first argument of ‘(.)’, namely ‘monadicST’
  In the expression: monadicST . run . act
  In an equation for ‘st’: st = monadicST . run . act

but this one

st' :: a -> Property
st' a = monadicST $ run $ act a

compiles without a problem.

Can anyone explain?