@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.