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

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 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?