> **Puzzle 87.** Figure out exactly what a \$$\mathbf{Bool}\$$-enriched category is, starting with the definition above.

On the other hand, in the case of the discrete poset we know that \$$\mathcal{X}(x,x) = I\$$ for all \$$x\$$ as per (a). Along with (b) we have \$$\mathcal{X}(x,y)\otimes\mathcal{X}(y,x) = I\$$ . This means that \$$\mathcal{X}(x,y) = \mathcal{X}(y,x)\$$ - if they were different then the identity law for monoids would be violated. If there are pairs \$$a, b, c\$$ and \$$d\$$ such that \$$\mathcal{X}(a,b) \neq \mathcal{X}(c,d) \$$ then \$$\mathtt{true} \otimes \mathtt{true} = \mathtt{false} \otimes \mathtt{false} = I\$$. The monoid \$$\langle\mathbf{Bool},\otimes, I\rangle\$$ must then be isomorphic to the [cyclic group \$$\mathbb{Z}/(2)\$$](https://en.wikipedia.org/wiki/Cyclic_group#Integer_and_modular_addition). Hence all of the monoids for \$$\mathbf{Bool}\$$-enriched category with discrete posets are symmetric and isomorphic.