[John wrote](https://forum.azimuthproject.org/discussion/comment/20130/#Comment_20130):
> 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.

[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\$$.