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.