> That's not so surprising, since that's what I was in the midst of proving.
Whoops. My main point was the relation to hypothetical syllogism, but I clearly had a dumb moment on the way there.
> 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\\).