> Puzzle 66. This is surprisingly hard to think through. I've got a weak hunch that the poset below cannot be made into a monoidal poset, but I'm still working through it. The poset can't be a lattice, per Puzzle 63, and this is the smallest non-semilattice non-trivial-order at my disposal.

I *think* that your poset can be made into a monoidal preorder by letting \$$\forall x. a \otimes x = x = x \otimes a\$$ and \$$\forall x \forall y. x \neq a \wedge y \neq a \implies x \otimes y = c\$$. Is that right?

In addition to the solution not being a meet-semi-lattice bounded below or a join-semi-lattice bounded above as you mention, I suspect that countable unbounded dense linear orders don't work.