re **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