[Haskell] Re: Announce: revamped free theorems generator

[Heinrich Apfelmus]

Sebastian Fischer wrote:
> I am interested in the mentioned laws because I want to show the monad 
> laws for the definition
> 
>     instance Monad FreeMonoid where
>       return x = FreeMonoid ($x)
>       a >>= f  = a >>- f
> 
> This definition of `>>=` is *not* the usual one for continuation monads, 
> but if the mentioned properties hold, I think it also satisfies the 
> monad laws.

Yes, it does satisfy the monad laws because  FreeMonoid  is the "free 
algebra functor"  T  for monoids as a T-algebra. In other words, a type 
  A  is a monoid exactly when there is a map  T A -> A . I've learned 
this idea from Dan Piponi:

http://blog.sigfpe.com/2007/02/monads-for-vector-spaces-probability.html
http://blog.sigfpe.com/2007/06/monads-from-algebra-and-the-gray-code.html


Regards,
Heinrich Apfelmus

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