@Matthew Doty, @Anindya Bhattacharyya: beautiful! One can make your argument even slightly more elegant by noting that it's enough to consider the case \$$I=d\$$; the other cases are not only similar, but reduce to this one by symmetry! You've already noted this for \$$I=c\$$, but it also applies to \$$I=a\$$ and \$$I=b\$$ if you turn the poset upside down: if the given poset had a monoidal structure with \$$a\$$ as the unit, then the same monoidal structure would work for the upside-down poset. But now one of the two maximal elements would be the unit.

Here's one more nut to crack:

**Puzzle TF1.** Suppose that \$$X\$$ is a monoid equipped with a preorder \$$\leq\$$ obeying:

$$x\leq x' \textrm{ implies } x\otimes y\leq x'\otimes y.$$

Does this make \$$X\$$ into a monoidal preorder?