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

(My partial answer)

There are three posets over **Bool** - the trivial preorder, the discrete poset, and the lattice from **Puzzle 86**.

In the case of the trivial preorder and discrete poset there is only one monoid up to isomorphism.

The trivial preorder corresponds to the complete graph, and the monoidal operation can effectively be anything since all objects are equivalent. This is trivially a symmetric preoder.

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.