I've attempted Mathew Doty's puzzle and I've made the same mistake as Keith E. Peterson – I've plugged in John's formula from [Lecture 6](https://forum.azimuthproject.org/discussion/1901/lecture-6-chapter-1-computing-adjoints#latest).

However, if we assume a complete lattice, is the following reasoning correct?

$$

\begin{eqnarray}

r(x,y) &=& \bigvee \\{a \in A : \; \Delta(a) \leq_{A\times A} (x,y) \\} \\\\

&=& \bigvee \\{a \in A : \; (a,a) \leq_{A\times A} (x,y) \\} \\\\

&=& \bigvee \\{a \in A : \; a \leq_A x, a \leq_A y \\} \\\\

&=& x \vee y

\end{eqnarray}

$$

Edit: I think the last step is wrong: initially I thought that \\(x\\) and \\(y\\) are in the set \\(R = \\{a \in A : \; a \leq_A x, a \leq_A y \\} \\), but that's not true. The set \\(R\\) might contain one of them if there is a relation between \\(x\\) and \\(y\\) (either \\(x \le y\\) or \\(y \le x\\)), but generally there isn't (the set is not totally ordered).

However, if we assume a complete lattice, is the following reasoning correct?

$$

\begin{eqnarray}

r(x,y) &=& \bigvee \\{a \in A : \; \Delta(a) \leq_{A\times A} (x,y) \\} \\\\

&=& \bigvee \\{a \in A : \; (a,a) \leq_{A\times A} (x,y) \\} \\\\

&=& \bigvee \\{a \in A : \; a \leq_A x, a \leq_A y \\} \\\\

&=& x \vee y

\end{eqnarray}

$$

Edit: I think the last step is wrong: initially I thought that \\(x\\) and \\(y\\) are in the set \\(R = \\{a \in A : \; a \leq_A x, a \leq_A y \\} \\), but that's not true. The set \\(R\\) might contain one of them if there is a relation between \\(x\\) and \\(y\\) (either \\(x \le y\\) or \\(y \le x\\)), but generally there isn't (the set is not totally ordered).