I am currently on the road, so please excuse my poor formatting.

This theorem does not fix \$$\mathcal{V}\$$. I have been reading your blog article from a few years ago and it’s references to acquaint myself with enriched categories, Simon. I see now that not fixing \$$\mathcal{V}\$$ is unorthodox. I am sorry, I didn’t intend to defy convention.

Consider some \$$\mathcal{X}\$$ enriched in some \$$\mathcal{V}\$$. Since I am not fixing a \$$\mathcal{V}\$$, I can’t talk about a particular preorder \$$\le\$$ when comparing \$$\mathcal{X}(a,b)\$$ and \$$\mathcal{X}(c,d)\$$. Because of this, I thought it would make since to clarify the order associated with \$$\mathcal{X}\$$ as \$$\le_\mathcal{X}\$$.

A ‘monoidal preorder over a discrete poset’ is a monoidal preorder where the underlying preorder is a discrete poset. That is not a great definition. I will attempt to clarify.

In a discrete poset, \$$a \le b\$$ is true if and only if \$$a = b\$$.

Every monoid can be made into a monoidal preorder over a discrete poset. Just construct the trivial poset over the domain of the monoid.

A simple example would be \$$\mathbb{Z}/(2)\$$. Since discrete posets are equivalence relations, we can represent them with their equivalence classes. In this case the equivalence class is {{0},{1}}.

I will try to go into how to construct an enriched category for the discretely ordered \$$\mathbb{Z}/(2)\$$ group or any other discretely ordered group later.