Basic row-column incidence lemma

Fix $r\ge 1$ and a column $j$ of $[\lambda ]$. The first $r$ rows of $s$ contribute at most one entry each to column $j$ of $t$, by the hypothesis, and column $j$ contains only ${\lambda }_j^{\mathrm{\prime }}$ boxes; so column $j$ of $t$ contains at most $\min (r,{\lambda }_j^{\mathrm{\prime }})$ entries drawn from the first $r$ rows of $s$. Double counting the nodes of $[\lambda ]$ lying in its first $r$ rows, column $j$ contributes exactly $\min (r,{\lambda }_j^{\mathrm{\prime }})$ of them, whence ${\sum }_j\min (r,{\lambda }_j^{\mathrm{\prime }})={\sum }_{i\le r}{\lambda }_i$.

Summing the bound of step 1.1 over all columns: the first $r$ rows of $s$ contain exactly ${\sum }_{i\le r}{\mu }_i$ entries by [L4], and each of them lies in exactly one column of $t$, so ${\sum }_{i\le r}{\mu }_i\le {\sum }_j\min (r,{\lambda }_j^{\mathrm{\prime }})={\sum }_{i\le r}{\lambda }_i$. Since $r\ge 1$ was arbitrary, $\lambda ⊵\mu$ by [L3], which is the first clause of the statement.

Assume now that $\lambda =\mu$. Then the outer terms of the inequality of step 2.1 are equal for every $r\ge 1$, so each of the ${\sum }_j$ many column bounds of step 1.1 is attained: for all $r\ge 1$ and all columns $j$, exactly $\min (r,{\lambda }_j^{\mathrm{\prime }})$ entries of the first $r$ rows of $s$ lie in column $j$ of $t$.

Let $A$ be the matrix with $A_{ij}=1$ when row $i$ of $s$ meets column $j$ of $t$ and $A_{ij}=0$ otherwise. Step 3.1 says ${\sum }_{i\le r}{A}_{ij}=\min (r,{\lambda }_j^{\mathrm{\prime }})$ for all $r\ge 1$ and all $j$; taking $r\ge$ (number of rows of $s$) shows ${\sum }_i{A}_{ij}={\lambda }_j^{\mathrm{\prime }}$, and comparing with general $r$ shows the ones in column $j$ of $A$ occur exactly in rows $i\le {\lambda }_j^{\mathrm{\prime }}$. Since ${\lambda }_j^{\mathrm{\prime }}\ge i$ holds exactly when ${\lambda }_i\ge j$ for the weakly decreasing sequence $\lambda$, row $i$ of $s$ meets column $j$ of $t$ precisely when $j\le {\lambda }_i$.

Define $v(i,j)$, for each node $(i,j)\in [\lambda ]$, as the unique entry of $s$ that lies in row $i$ of $s$ and in column $j$ of $t$; step 4.1 supplies existence and uniqueness for exactly the nodes of $[\lambda ]$, and the $n$ entries of $s$ are distributed bijectively over those nodes, so $v:[\lambda ]\to \{1,\dots ,n\}$ is a bijection, that is, a $\lambda$-tableau.

For every row $i$, the entries $v(i,j)$ with $1\le j\le {\lambda }_i$ are exactly the entries of row $i$ of $s$, rearranged. Define $\sigma \in S_n$ on row $i$ of $s$ by sending the entry in box $(i,j)$ of $s$ to $v(i,j)$; as $j$ runs over $1,\dots ,{\lambda }_i$ this is a permutation of the entries of row $i$ of $s$, so $\sigma$ preserves every row set of $s$ and $\sigma \cdot s=v$ holds by construction, whence $\sigma \in R_s$.

For every column $j$, the entries $v(i,j)$ with $1\le i\le {\lambda }_j^{\mathrm{\prime }}$ are ${\lambda }_j^{\mathrm{\prime }}$ distinct entries of the set $B_j$ of entries of column $j$ of $t$, hence they are exactly $B_j$. Define $\gamma \in S_n$ by $\gamma (t(i,j)):=v(i,j)$ for all nodes $(i,j)$; this is well defined because $t$ is a bijection, it maps $B_j$ bijectively onto itself for every column $j$, and it satisfies $\gamma \cdot t=v$, so $\gamma \in C_t$.

Steps 6.1 and 7.1 give $v=\sigma \cdot s=\gamma \cdot t$ with $\sigma \in R_s$ and $\gamma \in C_t$, the equality clause of the statement with $\rho :=\sigma$ and $\gamma$ as constructed, and step 2.1 proved the dominance clause; hence the lemma holds for every $n\ge 0$. ∎