Okay, I tried recovering the attempt in my previous post and failed to get anywhere. I think there's some kind of fundamental issue between the two properties we're supposed to obey for an enriched category. Transitivity of divisibility gives us that "if x divides y and y divides z, then x divides z"; we want this to translate into the second property, so \$$\mathcal{X}(x, y)\$$ should assign true iff x divides y, and \$$\otimes\$$ should be "and", \$$\land\$$. The identity of \$$\land\$$ is \$$\mathrm{true}\$$, so everything we assign should be bounded below by true. If we pick the order \$$\mathrm{false} \le \mathrm{true}\$$, we have to assign everything \$$\mathrm{true}\$$, which is _not_ what we want. And if we pick the order \$$\mathrm{true} \le \mathrm{false}\$$, we violate the translation of the second property -- it would say that if x divides y and y divides z, it's okay if x doesn't divide z.

So that leaves us with two options: the discrete preorder and the codiscrete preorder. The discrete preorder won't do either, since it's okay if x doesn't divide y or y doesn't divide z for x to still divide z. But the codiscrete preorder isn't strong enough to guarantee transitivity (or much of anything, really). I guess it just trivially satisfies both properties. But this feels _really_ unsatisfying, and I can't imagine this is what was intended.

So what do we do? I feel like this argument has exhausted all options for representing the poset of divisibility on the positive integers as a **Bool**-enriched category. Am I just way off track?