Simplify the applicative instance of Free

[ianliu]

Is it possible to simplify the following two lines

free/src/Control/Monad/Free.hs

Lines 222 to 223 in a06c37f

	Pure a <*> Pure b = Pure $ a b
	Pure a <*> Free mb = Free $ fmap a <$> mb

to this?

  (Pure a) <*> b = fmap a b

I'm still learning Haskell and I was trying to implement the Free monad myself and came here to compare my implementation with yours; so I apologize in advance if this "issue" is just plain bogus.

Thanks,
Ian

[RyanGlScott]

Yes, those are equivalent as far as I can tell.

[Icelandjack]

Yes and a lot clearer, this part of the Applicative docs becomes relevant

As a consequence of these laws, the Functor instance for f will satisfy

fmap f x = pure f <*> x