### Inline the definition of 'ap' in the Monad laws

The law as it is currently written is meaningless, because nowhere have we defined the implementation of 'ap'. The reader of the Control.Monad documentation is provided with only a type signature, > ap :: Monad m => m (a -> b) -> m a -> m b an informal description, > In many situations, the liftM operations can be replaced by uses of > ap, which promotes function application. and a relationship between 'ap' and the 'liftM' functions > return f `ap` x1 `ap` ... `ap` xn > is equivalent to > liftMn f x1 x2 ... xn Without knowing how 'ap' is defined, a law involving 'ap' cannot provide any guidance for how to write a lawful Monad instance, nor can we conclude anything from the law. I suspect that a reader equipped with the understanding that 'ap' was defined prior to the invention of the Applicative class could deduce that 'ap' must be defined in terms of (>>=), but nowhere as far as I can tell have we written this down explicitly for readers without the benefit of historical context. If the law is meant to express a relationship among (<*>), (>>=), and 'return', it seems that it is better off making this statement directly, sidestepping 'ap' altogether.

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