@[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)