Row insertion and reverse deletion are inverse

(First direction, row $s$.) In $U$ the appended letter is $x_s$ at position $r_s={\lambda }_s+1$ of row $s$, so row $s$ of $U$ equals row $s$ of $T$ followed by $x_s$, and the largest index $j_s$ with $U(s,j_s)<+\mathrm{\infty }$ is $r_s$. Reverse deletion therefore sets $x_s=U(s,r_s)$, empties that cell (so that $V$ has shape $[\mathrm{shape}(T)]$), and continues upward.

(Converse direction, insertion route.) If $s=1$, deletion removes the final entry $x_1$ of row1 and reinsertion appends it, so the converse is immediate. For $s\ge 2$, let $(V,x)=U-b$ with deletion positions $j_1,\dots ,j_s$, so $j_s=t$, row $i$ of $V$ agrees with row $i$ of $U$ off the single cell $(i,j_i)$ and carries $x_{i+1}$ there for $i<s$ (with the cell $b=(s,j_s)$ absent), and the expelled letter is $x=x_1$. In row $1$ of $V$, the entries left of $j_1$ are the entries of $U$ left of $(1,j_1)$, hence smaller than $x_1$, and the entries right of $j_1$ are the entries of $U$ right of it, hence larger than $x_1$; the entry at $j_1$ is $x_2$. So inserting $x_1$ replaces position $j_1$ and bumps $x_2$.

(First direction, induction upward.) Suppose reverse deletion carries $x_{i+1}$ into row $i<s$, after restoring the lower rows. Row $i$ is still the row of $U$: its entry at $r_i$ is $x_i<x_{i+1}$, its entries left of $r_i$ are smaller than $x_i$, and its entries right of $r_i$ are unchanged entries of $T$ strictly greater than the old displaced value $T(i,r_i)=x_{i+1}$. Thus the rightmost entry smaller than $x_{i+1}$ is exactly $r_i$. Reverse deletion carries $x_i$ upward and restores $T(i,r_i)=x_{i+1}$. Inducting from the final-box deletion in step 1.1 restores every row of $T$.

(Converse direction, induction downward.) At row $i<s$, deletion removed $x_i$ at $j_i$ and replaced it by $x_{i+1}>x_i$. The entries of $V$ left of $j_i$ are unchanged entries of $U$ smaller than $x_i$, and those to its right are unchanged entries larger than $x_{i+1}$, since $j_i$ was the rightmost entry smaller than $x_{i+1}$ and the entries are distinct. Therefore, when reinsertion carries $x_i$ into row $i$, it chooses exactly $j_i$, restores $x_i$, and bumps $x_{i+1}$ into row $i+1$. Starting at row1 with $x=x_1$, this induction reconstructs all replaced rows. In the final row $s$, deletion removed its row-end value $x_s$, so the remaining entries are smaller than $x_s$ and reinsertion appends it precisely in $b=(s,j_s)$.

(First direction, conclusion.) By step 1.1 and downward induction in step 2.1, reverse deletion visits the rows $s,s-1,\dots ,1$, restores in each row $i$ the entry of $T$ at $(i,r_i)$, expels $x_1=x$, and leaves the filling $T$ of shape $[\mathrm{shape}(T)]$; that is, $U-b=(T,x)$.

(Converse direction, conclusion.) By steps 1.2 and 2.2 the insertion of $x$ into $V$ follows the positions $j_1,\dots ,j_s$, rewrites the same entries as $U$ and appends at $b$; hence $V\leftarrow x=U$ with new box $b$. Finally $x\mathrm{\notin }V$, because by [L2] the entries of $V$ are the entries of $U$ with $x_1$ deleted, and $U$ has distinct entries.