[Haskell-cafe] For class Monoid; better names than mempty & mappend might have been: mid (mident) & mbinop

[wren ng thornton]

On 7/24/11 10:09 PM, Sebastian Fischer wrote:
>>
>> because list is a (the?) free monoid.
>
>   Yes, all free monoids are isomorphic (to lists).
>
> Sebastian

For completeness...

The free monoid over a set S is the set of all finite sequences of
elements drawn from S. Often this is written with the Kleene star: S^*.
Finite lists are a particular representation for the free monoid (namely
with all operations associated to the right), though this representation
may not be "isomorphic enough" for certain analyses (e.g., algorithmic
complexity).

Note that the free semigroup over a set S is the set of all finite
non-empty sequences of elements drawn from S. We can name this by S^+.
Other than the fact that S^+ does not contain the empty sequence, the
other important difference is that, while both S^* and S^+ are both closed
under the operation, only S^* is co-closed under the operation (i.e., not
all elements of S^+ can be decomposed into two elements of S^+ joined by
the operator).

-- 
Live well,
~wren