@Tobias – I think I might have an answer to your **Puzzle TF1**... (it's an adaptation of @Jonathan's attempt at #18)

Let's consider the set of words on an alphabet (including the empty word). This is a monoid by string concatenation.

We will define an A-word to be any word beginning with the letter A.

Consider the following rather ungainly preorder. Every word is \$$\leq\$$ itself. But we also have the rule that any A-word is \$$\geq\$$ any other word (whether it is an A-word or not). So the preorder is mostly discrete, except with all the A-words lumped together and sitting on top, so to speak.

Now we can check that \$$x\leq x' \Rightarrow x\otimes y\leq x'\otimes y\$$ – the only non-trivial case is when \$$x'\$$ is an A-word, in which case \$$x'\otimes y\$$ is an A-word too.

But we do *not* have \$$y\leq y' \Rightarrow x\otimes y\leq x\otimes y'\$$ – consider the case when \$$y'\$$ is an A-word, but \$$x\$$ is not (eg B \$$\leq\$$ A but CB \$$\nleq\$$ CA).

So your "one-sided" condition is *not* enough to obtain a monoidal preorder in general.