Fredrick wrote:

> Do we know that \\( \mathcal{X} \\) will be a preorder?

No. In in comment #41 wrote:

> It might been wiser to call the things we're studying so far "enriched preorders". A plain old preorder has a _truth value_ for any pair of objects - that is, an element of \\(\textbf{Bool}\\). We are "enriching" this idea - in the sense of making it more exciting - by replacing \\(\textbf{Bool}\\) by an arbitrary symmetric monoidal preorder \\(\mathcal{V}\\).

> However, even the term "enriched preorder" could confuse you into thinking that every enriched preorder is a preorder. _It's not!_ After all, there's no reason that having an element of \\(\mathcal{V}\\) gives you an element of \\(\textbf{Bool}\\) in any interesting way.