Well, this didn't pan out, but perhaps it will help someone else thinking about **TF1**. But I guess it counts for **Puzzle 63**!

Consider the lexicographical order on the free monoid over a ordered alphabet \\(\Sigma\\). (This is the set of all words, or finite sequences, that can be put together using elements of the alphabet.) This is not a commutative monoid, as \\(abc\\) is a different word from \\(cba\\). Prefixes dominate in the lexicographical order, so appending a common suffix will not change how two words are ordered -- thus we have right-multiplication. We _also_ have left-multiplication, since prepending a common prefix also has no effect on the ordering. So we can clearly have a monoidal preorder without requiring commutativity.

(It seems clear that requiring multiplication on both sides to preserve the order is equivalent to requiring \\((x, x') \le (y, y') \implies x \otimes y \le x' \otimes y'\\).) (**EDIT:** Got this slightly backwards; fixed in #20.)

Consider the lexicographical order on the free monoid over a ordered alphabet \\(\Sigma\\). (This is the set of all words, or finite sequences, that can be put together using elements of the alphabet.) This is not a commutative monoid, as \\(abc\\) is a different word from \\(cba\\). Prefixes dominate in the lexicographical order, so appending a common suffix will not change how two words are ordered -- thus we have right-multiplication. We _also_ have left-multiplication, since prepending a common prefix also has no effect on the ordering. So we can clearly have a monoidal preorder without requiring commutativity.

(It seems clear that requiring multiplication on both sides to preserve the order is equivalent to requiring \\((x, x') \le (y, y') \implies x \otimes y \le x' \otimes y'\\).) (**EDIT:** Got this slightly backwards; fixed in #20.)