[@Tobias](https://forum.azimuthproject.org/profile/2235/Tobias%20Fritz), I also attempted to quotient Anindya's counterexample to that three-element monoid, but I think it loses something. Here's a picture of what we're dealing with, and a multiplication table:

![](https://i.imgur.com/VzVPhYj.png)


|Iab
-+---
I|Iab
a|aaa
b|bbb


Here, we have that \\(b \le a\\), as well as \\(b = b \otimes b = b \otimes a = b\\), so the counterexample fails. This is because we've identified all of the non-A words into a single class, thereby making them comparable when they weren't before. We might be able to get away with this if we added another element, \\(b'\\), so that \\(b \otimes a = b'\\).