Anindya wrote:

> **Puzzle 67** – how about we take the set of finite "words" on an "alphabet" (including the empty word)

> • this is a monoid by string concatenation, eg CAT \\(\otimes\\) FISH \\(=\\) CATFISH

> • it is a preorder by word length, eg CAT \\(\leq\\) DOG, DOG \\(\leq\\) CAT, CAT \\(\leq\\) FISH, FISH \\(\nleq\\) DOG

> • the monoid structure preserves \\(\leq\\), so it is a monoidal preorder

> • it is symmetric, eg CATFISH \\(\leq\\) FISHCAT and FISHCAT \\(\le\\) CATFISH

> • but it is not commutative, since CATFISH \\(\neq\\) FISHCAT

I like that!

Here's another: take any monoid \\(X\\) that's not commutative, and give it a preorder such that \\(x \le y\\) for _all_ \\(x,y \in X\\). Then this is a symmetric monoidal preorder that's not commutative.

Sometimes in math it pays to be ruthless.