\cdots > a_{p}$ and $b_1 > \cdots > b_{q}$. As $w$ is frank, it is the reading word of a Young tableau of skew shape $\rot{(q,p)^{t}}$. Thus, $a_i\leq b_i$ for $1\leq i\leq p$. \looseness-1 Instead of computing $\iota(w)$ by rectifying the appropriate two-columned tableau, one may perform successive Schensted column insertions of the numbers $a_p$ down to $a_1$ starting from the single-columned tableau with column word $w^{(2)}$. See Figure~\ref{fig:iota via column insertion} for an example. Compare with Figure~\ref{fig:jdt on two columns} which established the same fact using jeu-de-taquin. \begin{figure}[htb]\centering \ytableausetup{mathmode,boxsize=1em} \begin{ytableau} 7\\6\\4\\3\\2 \end{ytableau} $\longrightarrow $ \begin{ytableau} 7\\6\\4\\3\\*(blue!50)1 & *(red!30)2 \end{ytableau} $ \longrightarrow$ \begin{ytableau} 7\\6\\4\\*(blue!50)2 & *(red!30)3\\1 & 2 \end{ytableau} $ \longrightarrow$ \begin{ytableau} 7\\*(blue!50)6\\4 & *(red!30)6\\2 & 3\\1 & 2 \end{ytableau} \caption{Establishing that $\iota(\framebox{621} \hspace{1mm} \framebox{76432})=\framebox{76421}\hspace{1mm}\framebox{632}$ by column-inserting 1,2, and 6 into the single-columned tableau with reading word $76432$. Blue boxes show the entries being inserted whereas red boxes contain the entries bumped.}\label{fig:iota via column insertion} \end{figure} In each intermediate step of this column insertion procedure, the number $a_i$ being inserted into the current tableau bumps a distinct element from $\{b_1,\dots, b_q\}$. Furthermore this bumped entry is guaranteed to be strictly greater than the entries in the second column in the current tableau. Therefore, the insertion tableau is completely determined by the entries that get bumped. More precisely, for $i$ from $p$ down to $1$, define the integer ${\sf m}(i)$ recursively as follows. We define ${\sf m}(p)$ to be the largest integer $j$ such that $a_p\leq b_{j}$. Subsequently, for $i=p-1,\dots,1$, define ${\sf m}(i)$ to be the largest integer $j$ such that $j<{\sf m}(i+1) $ and $a_i\leq b_{j}$. Observe that in our Schensted column-insertion procedure, the entry $a_i$ bumps $b_{{\sf m}(i)}$. Therefore, the set of entries that get bumped is $\{b_{{\sf m}(i)}\suchthat 1\leq i\leq p\}$. For the example in Figure~\ref{fig:iota via column insertion}, we have ${\sf m}(3)=5$, ${\sf m}(2)=4$ and ${\sf m}(1)=2$. Therefore the set of entries that get bumped is $\{b_5,b_4,b_2\}=\{2,3,6\}$. \looseness-1 Consider $\crw{\phi(w)}=u_1\cdots u_n$. The word $v\coloneqq v_1\cdots v_n$ obtained by recording the column to which each $u_i$ belongs gives us the weakly increasing arrangement of letters in $w$. Furthermore, for $1\leq i\leq p$ (respectively $1\leq i\leq q$) the letter in $v$ corresponding to the $i$th $2$ (respectively 1) from the left in $\crw{\phi(w)}$ is equal to $a_i$ (respectively $b_i$). Recall that the crystal reflection operator $s_1$ acting on $\phi(w)$ begins by pairing each $2$ in $\crw{\phi(w)}$ to the closest unpaired $1$ to its right. Equivalently, in our current context, a $2$ corresponding to $a_i$ for some $1\leq i\leq p$ gets paired with the $1$ in $\crw{\phi(w)}$ corresponding to $b_{{\sf m}(i)}$. We infer that the unpaired $1$s in $\crw{\phi(w)}$ correspond to those $b_j$ that are not bumped. These are precisely the $b_j$ that determine which $1$s in $\phi(w)$ turn into $2$s in computing $s_1(\phi(w))$. Thus we infer that $s_1(\phi(w))=\phi(\iota(w))$. This establishes the claim in the case $p