in $S$. Define $E:= \{(a,b): (a,b)\in B^+, a>i', b>j\}$. In Figure \[boardB+\] this is collection of blocks that are both to the right of $P$ and below $P'$. Let $S^*(M)\subset \{i'+1,\dots,i-1\}$ be the rows of blocks in $S$ that do not share a row with any rooks of $M$ that are in $E$. Similarly, let $T^*(M)\subset\{j+1,\dots,j'-1\}$ be the columns of blocks in $T$ that do not share a column with any rooks of $M$ that are in $E$. We claim that $\abs{S^*(M)}<\abs{T^*(M)}$. Note that $\abs{S} = i-i'-2$ and $\abs{T} = j'-j-2$. Since $i+j< i'+j'$ we have $i-i'< j-j'$ and so $\abs{S}< \abs{T}$. Letting $a$ be the number of rooks in $E$, then $\abs{S^*(M)} = \abs{S} - a$ and $\abs{T^*(M)} = \abs{T} - a$. So $\abs{S^*(M)}< \abs{T^*(M)}$. Thus, there is an injection $s: S^*(M)\to T^*(M)$. We fix some arbitrary injection $s$ and let $(a_0,j)$ be the location of the rook in $S$. Define $r_2: B^+ \to B^+$ by $$r_2(a,b):=
\begin{cases}
(i,j) &\text{ if } (a,b) = (i',j')\\
(i',s(a))  &\text{ if } (a,b)\in S\\
(a_0,b) &\text{ if } a=i\\
(a,j') &\text{ if } b=s(a_0)\\
(a,b) &\text{otherwise}
\end{cases}.$$

In Figure \[r2sketch\] the gray boxes are the images of the black boxes under the 