occupy the same collection of columns as those of $M$ and so no two rooks are in the same column. Finally, $T\cap f_2(M) \neq \emptyset$ since the rook in $S$ got sent to a rook in $T$. Thus $f_2(M)$ is a matching in $A_T$.

Using $f_1$ and $f_2$ we know where to send all matchings in $B^+$ that have a rook in $P'$ and not $P$. Moreover, the images of $f_1$ and $f_2$ are disjoint and thus there is an injection from matchings in $B^+$ with a rook in $P'$ and not $P$ to matchings in $B^+$ that have a rook in $P$ and not $P'$. Since this injection preserves the size of the matching, $m_k(B') \leq m_k(B)$.

It remains to prove that $B'$ has strictly fewer matchings than $B$. Suppose that there are no rooks of $M$ that are in $E$ and fix an injection $s: S^*(M)\to T^*(M)$ Since $\abs{S^*(M)}<\abs{T^*(M)}$ there exists $Q\in T^*(M)$ such that $Q\neq s(R)$ for any $R\in S^*(M)$. Then $\{Q,P\}\in A_T$ and there does not exist $M\in A_S$ such that $f_2(M) = \{Q,P\}$. Thus, $m(B')<m(B)$.

Given a Young diagram $B$ with an out-corner $P$ and an in-corner $P'$ with $s(P)<s(P')$ we call the move that 