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

[Tony Morris]

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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