[Haskell-cafe] A question about "monad laws"

[askyle]

ajb-2 wrote:
> 
> Define:
>      f >=> g = \x -> f x >>= g
> 

So you're either not taking (>=>) as primitive or you're stating the
additional
property that there exists (>>=) such that f >=> g  === (>>= g) . f
(from which you can easily show that (f . g) >=> h === (f >=> h) . g ).

A presentation of the monad laws based on (>=>) (I prefer (<=<) since it
meshes
better with (.) ) should (IMHO) regard (>=>) as primitive and state its
needed properties
whence one can derive all other formulations (Note that he Identity law then
requires the
existence of return as another primitive).

That done, you define (>>=), fmap and the rest in terms of (<=<) :

(>>=) = flip (<=< id)               -- I like to put it as (>>= g) = (g <=<
id)
fmap f = (return . f) <=< id


-----
Ariel J. Birnbaum
-- 
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