Moreover, `m a` is 'a' plus some terminal element , for example
Nothing, [], Left _ etc, So a morphism (a - m a) contains all the
morphisms of (m a - m a).
2012/9/5 Alberto G. Corona agocor...@gmail.com:
Alexander,
In my post (excuses for my dyslexia) I try to demonstrate that the
Thanks, Kristopher
2012/9/4 Kristopher Micinski krismicin...@gmail.com:
Your post feels similar to another one posted recently...
http://web.jaguarpaw.co.uk/~tom/blog/2012/09/02/what-is-a-monad-really.html
just fyi, :-),
kris
On Tue, Sep 4, 2012 at 6:39 AM, Alberto G. Corona
Alexander,
In my post (excuses for my dyslexia) I try to demonstrate that the
codomain (m a), from the point of view of C. Theory, can be seen as
'a' plus, optionally, some additional element, so a monadic morphism
(a - m a) is part of a endofunctor in (m a - m a)
When considering the
Monads are monoids in the category of endofunctors
This Monoid instance for the endofunctors of the set of all elements
of (m a) typematch in Haskell with FlexibleInstances:
instance Monad m = Monoid (a - m a) where
mappend = (=) -- kleisly operator
mempty = return
The article can
Your post feels similar to another one posted recently...
http://web.jaguarpaw.co.uk/~tom/blog/2012/09/02/what-is-a-monad-really.html
just fyi, :-),
kris
On Tue, Sep 4, 2012 at 6:39 AM, Alberto G. Corona agocor...@gmail.com wrote:
Monads are monoids in the category of endofunctors
This
Not to mention the ugly formatting ;)
2012/9/5 Richard O'Keefe o...@cs.otago.ac.nz:
On 4/09/2012, at 10:39 PM, Alberto G. Corona wrote:
Monads are monoids in the category of endofunctors
This Monoid instance for the endofunctors of the set of all elements
of (m a) typematch in Haskell
On Tue, Sep 4, 2012 at 3:39 AM, Alberto G. Corona agocor...@gmail.comwrote:
Monads are monoids in the category of endofunctors
This Monoid instance for the endofunctors of the set of all elements
of (m a) typematch in Haskell with FlexibleInstances:
instance Monad m = Monoid (a - m a)
On Tue, Sep 4, 2012 at 4:21 PM, Alexander Solla alex.so...@gmail.comwrote:
On Tue, Sep 4, 2012 at 3:39 AM, Alberto G. Corona agocor...@gmail.comwrote:
Monads are monoids in the category of endofunctors
This Monoid instance for the endofunctors of the set of all elements
of (m a)
