@[Michael](https://forum.azimuthproject.org/discussion/comment/18388/#Comment_18388): Let's take as example the divisibility poset on positive integers. We can construct this as a **Bool**-enriched category as follows.

Define \\(\mathrm{Ob}(\mathcal{X}) = \mathbb{Z}^+\\). This is the set of objects; in other words, every positive integer is an object.

Define \\(\mathcal{X}(x, y) = \mathrm{true}\\) iff \\(x \le y\\), and define \\(\mathcal{X}(x, y) = \mathrm{false}\\) iff \\(x \not\le y\\). This associates to every pair of objects -- natural numbers -- a value in **Bool**, corresponding to whether or not these objects are related by \\(\le\\). As an example, \\(\mathcal{X}(4, 8) = \mathrm{true}\\), since \\(4\\) divides \\(8\\).

Only thing is, I think we need to use \\(\lor\\) and \\(\mathrm{false}\\) for the monoidal operator and identity. By the chosen order, \\(\mathrm{true}\\) is the _greatest_ element of the preorder, so we're tacitly requiring that \\(\mathcal{X}(x, y) = \mathrm{true}\\) for all \\(x, y\\), since \\(I \le \mathcal{X}(x, y)\\) needs to hold. Is this a typo, John, or am I misreading things?

If we use \\(\lor\\) and \\(\mathrm{false}\\), then we can check the other conditions. First, \\(\mathrm{false}\\) is the least element of **Bool**, so the first property is trivially satisfied. And we have that \\(\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z)\\) because divisibility is transitive.

**EDIT:** But this actually fails the transitivity property once we switch to \\(\lor\\). I think we might actually want to use the order \\(\mathrm{true} \le \mathrm{false}\\), along with \\(\land\\) and \\(\mathrm{true}\\). Editing! (**EDIT 2:** nope, didn't work, see my next post below)