[Haskell-cafe] Confusion on the third monad law when using lambda abstractions

[Jake McArthur]

Jake McArthur wrote:
> Generally, you can transform anything of the form:
> 
>     baz x1 = a =<< b =<< ... =<< z x1
> 
> into:
> 
>     baz = a <=< b <=< ... <=< z

I was just looking through the source for the recently announced Hyena 
library and decided to give a more concrete example from a real-world 
project. Consider this function from the project's Data.Enumerator 
module[1]:

     compose enum1 enum2 f initSeed = enum1 f1 (Right initSeed) >>= k
         where
           f1 (Right seed) bs = ...
           k (Right seed) = ...

First, I would flip the `(>>=)` into a `(=<<)` (and I will ignore the 
`where` portion of the function from now on):

     compose enum1 enum2 f initSeed = k =<< enum1 f1 (Right initSeed)

Next, transform the `(=<<)` into a `(<=<)`:

     compose enum1 enum2 f initSeed = k <=< enum1 f1 $ Right initSeed

We can "move" the `($)` to the right by using `(.)`:

     compose enum1 enum2 f initSeed = k <=< enum1 f1 . Right $ initSeed

Finally, we can drop the `initSeed` from both sides:

     compose enum1 enum2 f = k <=< enum1 f1 . Right

I didn't test that my transformation preserved the semantics of the 
function or even that the type is still the same, but even if it's wrong 
it should give you the idea.

- Jake

[1] 
http://github.com/tibbe/hyena/blob/9655e9e6473af1e069d22d3ee75537ad3b88a732/Data/Enumerator.hs#L117