@John

> $$
\begin{array}{c|c}
\text{monoid} & \text{effect of b) on $\mathcal{X}(x,y)$} \\\\
\hline
(\oplus, \tt{false}) & \text{If $\mathcal{X}(x,y)$, then for all objects $z$, either $\mathcal{X}(x,z)$ or $\mathcal{X}(z,x)$ or both} \\\\
(\wedge, \tt{true}) & \text{If $\mathcal{X}(x,y)$ and $\mathcal{X}(y,z)$, then $\mathcal{X}(x,z)$} \\\\
(\leftrightarrow, \tt{true}) & \text{If $\mathcal{X}(x,y) = \mathcal{X}(y,z)$, then $\mathcal{X}(x,z)$} \\\\
(\vee, \tt{false}) & \text{Either $\mathcal{X}(x,y) = \tt{false}$, or $\mathcal{X}(x,y) = \tt{true}$, for all objects $x,y$.} \\\\
\end{array}$$
>
> In [Lecture 30](https://forum.azimuthproject.org/discussion/2124/lecture-30-chapter-1-preorders-as-enriched-categories) I consider the second row in detail: with this particular monoidal structure on \\(\mathbf{Bool}\\), a \\(\mathbf{Bool}\\)-enriched category is just a preorder, since
>
> If \\(\mathcal{X}(x,y) = \mathcal{X}(y,z)\\), then \\(\mathcal{X}(x,z)\\)
>
> is the good old transitive law. Let me think a bit about the other three: I haven't really studied them, but they must be good for something!

Hopefully I can recover a tiny bit from my screw-ups here.

**Theorem**. We can always define a preorder over \\(\mathrm{Obj}(\mathcal{X})\\) from any **Bool**-enriched category using the following rule:

$$
x \preceq_{\mathcal{X}} y \iff \mathcal{X}(x,y) = \mathcal{X}(x,x)
$$

This generic construction works for the case considered by you in [Lecture 30](https://forum.azimuthproject.org/discussion/2124/lecture-30-chapter-1-preorders-as-enriched-categories), as well as the other cases Dan and I have been considering.

**Proof**.

The proof goes by considering each preorder over **Bool**. Reflexivity is trivial to see by the rule, so the only thing we need to show is transitivity.

In the case of the trivial preorder, since \\(\mathcal{X}(a,b) = \mathcal{X}(c,d)\\) for all \\(a,b,c,d \in \mathrm{Obj}(\mathcal{X})\\). So the preorder generated by the rule is trivial preorder over \\(\mathrm{Obj}(\mathcal{X})\\).

In the case of the discrete poset, we know that \\(\mathcal{X}(x,x) = I\\) for all \\(x\\) (see my post [# 3](https://forum.azimuthproject.org/discussion/comment/18372/#Comment_18372) in this thread). Suppose that \\(a \preceq_{\mathcal{X}} b\\) and \\(b \preceq_{\mathcal{X}} c\\). Then it must be that \\(\mathcal{X}(a,b) = \mathcal{X}(b,c) = I\\). By the rule (b) of **Bool**-enriched categories we have:

$$
\mathcal{X}(a,b) \otimes \mathcal{X}(b,c) \leq \mathcal{X}(a,c)
$$

Since \\(\mathcal{X}(a,b) \otimes \mathcal{X}(b,c) = I\\), it must be that \\(I \leq \mathcal{X}(a,c)\\). Because we are consider the discrete poset, that means that \\(\mathcal{X}(a,c) = I = \mathcal{X}(a,a)\\). Since \\(\mathcal{X}(a,b) = \mathcal{X}(b,a)\\), this is an equivalence relation.

The next case is for the lattice \\(\mathtt{false} \leq \mathtt{true}\\). Either \\(I = \mathtt{true}\\) or \\(I = \mathtt{false}\\).

Assume that \\(I = \mathtt{true}\\). It must be that \\(\otimes = \wedge\\). We can figure this out by just going through the truth table for \\(\wedge\\). We know that \\(\mathtt{true} \otimes \mathtt{true} = \mathtt{true}\\), and \\(\mathtt{true} \otimes \mathtt{false} = \mathtt{false} \otimes \mathtt{true} = \mathtt{false}\\). Finally, since \\(\mathtt{false} \leq \mathtt{true}\\) it must be \\(\mathtt{false} \otimes \mathtt{false} \leq \mathtt{false} \otimes \mathtt{true}\\), hence \\(\mathtt{false} \otimes \mathtt{false} = \mathtt{false}\\). The rule follows from the proof John writes in [Lecture 30](https://forum.azimuthproject.org/discussion/2124/lecture-30-chapter-1-preorders-as-enriched-categories) along with the fact that \\(\mathcal{X}(a,a) = I.\forall a\\).

Finally, consider the case where \\(I = \mathtt{false}\\). Following logic similar to the above argument, it must be that \\(\otimes = \vee\\). Next, we have a kind of collapse argument: \\(\forall a,b,c,d. \mathcal{X}(a,b) = \mathcal{X}(c,d)\\). For suppose that some pair \\(\mathcal{X}(a,b) = \mathtt{true}\\), then since \\(\mathcal{X}(c,a) \otimes \mathcal{X}(a,b) \leq \mathcal{X}(c,b)\\), it must be \\(\mathcal{X}(a,c) = \mathtt{true}\\) since \\(\otimes = \vee\\). However, we also have \\(\mathcal{X}(c,b) \otimes \mathcal{X}(b,d) \leq \mathcal{X}(c,d)\\), hence \\(\mathcal{X}(c,d) = \mathtt{true}\\). Since every element is comparable to every other element, this is just a trivial preorder.

\\(\Box\\)

As an aside, the case considered in Lecture 30 is the only *interesting* case in the above proof.