Once again, let's consider the divisibility poset on positive integers. We can construct this as a **Bool**-enriched category as follows.

First, let's consider **Bool** to be \\(\langle \\{\mathrm{true}, \mathrm{false}\\}, \le, \land, \mathrm{true} \rangle\\), where the order is given by \\(false \le true\\). This is clearly a poset; it's monoidal because \\(\land\\) is a meet under this order; and it's symmetric because \\(\land\\) is commutative.

Now let's render our divisiblity poset as an enriched category. Define \\(\mathrm{Ob}(\mathcal{X}) = \mathbb{Z}^+\\). This is the set of objects in our **Bool**-enriched category; 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\\).

Now we can check the two conditions. First, since \\(x \le x\\) is always true in our preorder, we trivially have that \\(I = \mathcal{X}(x, x)\\), satisfying the first condition. And we have that \\(\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z)\\) because divisibility is transitive: this property translates to "if x divides y and y divides z, then x divides z" under our definitions. (If one or the other does not hold, then the left side is \\(\mathrm{false}\\), leaving the right side entirely unconstrained.)

So \\(\langle \mathbb{Z}^+, \mid \rangle\\) is just a **Bool**-enriched category!