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\\).