David wrote:

> Is \$$\leq\$$ properly defined in **Puzzle 193**? I think we should have \$$I\leq S\$$ for all \$$S\in P(X)\$$, but it is not true that \$$X\subseteq S\$$ (unless \$$X=S\$$).

Aargh, you're right: \$$(P(X), \subseteq, \cup, \emptyset)\$$ is not a monoidal poset.

We could use reverse inclusion instead, but that's too obviously just the same puzzle as the one before, re-expressed using complements. I'm gonna replace this puzzle with one about an old friend, the poset of partitions.