Jonathan wrote in [#9](https://forum.azimuthproject.org/discussion/comment/20134/#Comment_20134)
>[David wrote](https://forum.azimuthproject.org/discussion/comment/20132/#Comment_20132):
>> Answer to **Puzzle 194**: \$$S\multimap T=\bigvee \lbrace A : \; S \otimes A \le T \rbrace=\bigvee \lbrace A : \; S \cup A \subseteq T\rbrace\$$, so this is the set with the most elements such that its union with \$$S\$$ is \$$T\$$. This set is \$$S^c\cap T\$$ (where \$$S^c\$$ denotes the set complement operation), since the addition of any more elements of \$$X\$$ would make its union with \$$S\$$ a proper superset of \$$T\$$.

>Couldn't it be \$$S \cup (S^c \cap T)\$$ as well, since union is idempotent? If \$$S \cup A \subseteq T\$$ is true, then \$$S \cup (S \cup A) \subseteq T\$$ is also, and vice versa.

>EDIT: And this just simplifies to \$$S \cup T\$$.

Yes, you're right, but it is not true that \$$S\cup T\subseteq T\$$ (unless \$$S\subseteq T\$$). In fact, even if we try with \$$S\multimap T=\emptyset\$$, we would have to have \$$S\subseteq T\$$ in order for \$$S\cup\emptyset\subseteq T\$$ to hold. So, I was wrong, this monoidal poset is not closed since \$$S\multimap T\$$ does not exist, in general.