Functor, Applicative, Monad, Foldable, Traversable instances for (, , ) a b

[Benno Fünfstück]

Correct me I I'm wrong but I believe Tony Morris is saying the following:

* not every operation defined for the integers "makes sense" for all
integers. Example: '/' does not make sense in the case of 'x / 0'
* yet, we still include '0' in the integers even though not every operation
makes sense for '0'
* -> not every operation in Foldable makes sense for `(,) a`
* -> but we also include `0` in the integers, so "not every operation makes
sense" is not an argument to exclude `(,) a` from being foldable,
  just as `0` is not excluded from the integers.

Kind regards,
Benno

Tony Morris <tonymorris at gmail.com> schrieb am So., 9. Apr. 2017 um
14:26 Uhr:

> These two things are true:
>
> * 0 is in the set of integers
> * ∀ a. ((,) a) is Foldable, and as one of many consequences, the length
> of all values in the set ∀ a. ((,) a) is 1.
>
> There are four possible positions to take on these claims:
>
> 1. Both are true.
> 2. Both are false.
> 3. The first true and second false.
> 4. The second true and the first false.
>
> I respect arguments 1 and 2. If I chose 1 and you chose 2, I'd say "well
> rightio then mate and cheers to that", we'd clink glasses and move on.
> Same if it were vice versa.
>
> I do not have the same respect for positions 3 and 4.
>
> On 09/04/17 19:48, Jon Fairbairn wrote:
> > Tony Morris <tonymorris at gmail.com> writes:
> >
> >> I don't think it is the "appropriate" set. It's an example. 0 is in the
> >> set of integers. The value 0 is in many sets.
> > OK, so I clearly do not understand your argument. The
> > implication I took from “and 0 is not an integer” is that the
> > foldable instance for ((,) a) should be present because it is
> > the zero case of something that has integers as its domain, and
> > I wanted to know what that something is.  If this was not the
> > intention of your argument, what was?
> >
>
>
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