Christopher wrote:

> So we need to change \$$\mathcal{V} \, \text{is commutative}\$$ to \$$\mathcal{V} \, \text{is symmetric (or braided? Not sure what those terms mean exactly})\$$.

It would be good to review [Lecture 22](https://forum.azimuthproject.org/discussion/2084/lecture-22-chapter-2-symmetric-monoidal-preorders/p1) then, where I explained "symmetric" and "commutative" monoidal preorders. Here's just a portion:

**Definition.** A monoidal preorder \$$(X, \le, \otimes, I) \$$ is **symmetric** if

$y \otimes x \le x \otimes y$

for all \$$x,y\in X\$$.

Since \$$x\$$ and \$$y\$$ are _anything_, we can switch their roles and conclude that

$y \otimes x \le x \otimes y \textrm{ and } x \otimes y \le y \otimes x$

for all \$$x,y\$$ in a symmetric monoidal preorder. You might guess that this implies \$$x \otimes y = y \otimes x\$$, but no! This will follow if our preorder is a poset, since a poset is a preorder where

$a \le b \textrm{ and } b \le a \textrm{ implies } a = b .$

But it's not true in general:

**Puzzle 67.** A monoidal preorder is **commutative** if \$$x \otimes y = y \otimes x\$$ for all \$$x,y\$$. Find a symmetric monoidal preorder that is not commutative.

In Chapter 4, for convenience, I am only working with monoidal posets - not more general monoidal preorders. From the above, a monoidal poset is symmetric iff it is commutative.

So, in Chapter 4, I only talk about _commutative_ monoidal posets!

When dealing with monoidal _categories_ we get yet another flavor of commutativity: braided. But that's irrelevant now.