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Expand the Column antisymmetrizer of a Young tableau:If two entries in one column of $v$ lie in the same row of $w$, their transposition belongs to both $C(v)$ and the row stabilizer of $w$, so the terms cancel in pairs. The assumption $b_v\{w\}\ne 0$ therefore says that every row of $w$ meets every column of $v$ in at most one entry.

The first row of $w$ has ${\lambda }_1$ entries, while $v$ has exactly ${\lambda }_1$ nonempty columns. It must consequently contain exactly one entry from each column of $v$. Permuting within each column puts these entries in the first row positions of $v$. Delete the matched first rows and repeat the argument on the remaining Young diagram. The product of the resulting column permutations is an element $h\in C(v)$ for which the row sets of $hv$ are those of $w$. ThusThis is the nonzero column antisymmetrizer criterion.

Write the transposition as $\tau =rc$ with $r\in R(t)$ and $c\in C(t)$. If $x$ is not one of the two entries moved by $\tau$, then $r(c(x))=x$. The entries $x$ and $c(x)$ lie in one column of $t$, while $c(x)$ and $r(c(x))=x$ lie in one row. A Young diagram has only one cell at the intersection of a specified row and column, so $c(x)=x$ and then $r(x)=x$.

Consequently both $r$ and $c$ fix every entry outside the support of $\tau$. On the two remaining entries each is either the identity or their transposition. They cannot both transpose them, since then $rc=1$, and they cannot both be the identity. Exactly one of $r,c$ is therefore $\tau$, proving that $\tau$ lies in exactly one of $R(t)$ and $C(t)$.

A removable node is a cell whose deletion leaves a Young diagram. The set ${\lambda }^{-}$ consists of the distinct partitions obtained by deleting one removable node from $\lambda$.