grid (-1.2,-1); (-1,-1.1) – (-.8,-1.1); (-.2,-1.2) rectangle (0,-1); (-.6,-1.2) grid (0,-1);

Again we need to show that $r_2(a,b) \in B^+$ for all $(a,b)\in B^+$. If $(a,b)\in S$ then $(i',s(a))\in B^+$ as $(i',s(a))\in T$. If the rook $(a,b)$ is in row $i$ then $r_2(a,b) = (a_0,b)\in B^+$ as $a_0 <i$ and $(i,b)\in B^+$. Finally, if $(a,b) = (a,s(a_0))$, then $r_2(a,b) = (a,j') \in B$ since $a<i'$ and $(i',j')$ is in $B^+$.

For $M\in A_S$, define $$f_2(M):=\{r_2(a,b): (a,b)\in M\}.$$ We claim that if $M$ is a matching in $A_S$ then $f_2(M)$ is a matching in $A_T$. First we will show that no two rooks are in the same row. There are only three rooks that change rows when we apply $r_2$. These rooks originally have rows $i'$, $i$ and $a_0$. After applying $r_2$ these rooks are in rows $i$, $a_0$ and $i'$ respectively. Thus the rooks of $f_2(M)$ occupy the same collection of rows as those of $M$. Now we must show that no two rooks occupy the same column. Similarly, there are only three rooks that change columns. These rooks originally occupy columns $j',j,$ and $s(a_0)$. After applying $r_2$, these rooks occupy $j,s(a_0),$ and $j'$ respectively. So the rooks of $f_2(M)$ 