[Haskell-cafe] Re: what is inverse of mzero and return?

Keean Schupke k.schupke at imperial.ac.uk
Sun Jan 23 07:21:43 EST 2005

Ashley Yakeley wrote:

> I think it would be helpful if all these classes came with their laws
>prominently attached in their Haddock documentation or wherever. The 
>trouble with MonadPlus is that the precise set of associated laws is 
>either unspecified or not the most useful (I assume there's a paper on 
>the class somewhere). I think everyone can agree on  these:
>  mplus mzero a = a
>  mplus a mzero = a
>  mplus (mplus a b) c = mplus a (mplus b c)
I think it would be even better if the classes came with assertions
that 'enforced the laws'... I think this requires dependant types

>  mzero >>= a = mzero
>But what about this?
>  a >> mzero = mzero
Well it was in the list I saw... I we consider the familar arithmetic 
a * b * 0 -- it is clear that an mzero anywhere in a sequence should result
in mzero:

a >> b >> mzero >> c >> d = mzero

But that says nothing about the co-product... where mzero should be the
identity IE:

a `mplus` mzero = a
mzero `mplus` a = a

But I am not sure there are any requirements on commutivity or associativity
on the co-product operation?

>It's satisfied by [] and Maybe, but not IO (for instance, when a is 
>'putStrLn "Hello"'), but IO has been declared an instance of MonadPlus. 
>And then there are the two I gave:
>  (mplus a b) >>= c = mplus (a >>= c) (b >>= c)
This was not on the list I saw - guess it could either be an omission,
or it has nothing to do with "MonadPlus" ... monads with identity and

>...which is satisfied by [], but not Maybe or IO.
>  mplus (return a) b = return a
>...which is satisfied by Maybe and IO, but not [], although your 
>alternative declaration would make [] satisfy this and not the previous 
But one could make up any arbitrary law that is satisfied by some
definition of a Monad and not others. Presumably there has to be
some sound category-theoretic reason for including the law?


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