re **Puzzle 196.**

> Show that in any closed monoidal poset we have
>
> $x \multimap y = \bigvee \lbrace a : \; x \otimes a \le y \rbrace$

This just drops out of the rule for computing right adjoints, but there's no harm proving it directly.

Set \$$A = \bigvee \lbrace a : \; x \otimes a \le y \rbrace\$$.

Then \$$a\in A \implies x \otimes a \le y \implies a \le x \multimap y\$$, so \$$x \multimap y\$$ is an upper bound of \$$A\$$.

But our lemma tells us \$$x \otimes (x \multimap y) \le y\$$, ie \$$(x \multimap y)\in A\$$.

Hence \$$x \multimap y\$$ is the least upper bound of \$$A\$$.