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!