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

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 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:

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Tony Morris writes:

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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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