Jonathan wrote in [#9](
>[David wrote](
>> 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.