@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?

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?