[proofplan]
Only the box changed by the crystal operator can affect semistandardness, so the proof is a local check around that box. For $f_i$, the changed entry is the $i$ corresponding to the rightmost uncancelled $+$ in the reduced $i$-signature; we show that any row or column violation after changing it to $i+1$ would force a neighbouring sign to cancel this $+$. The proof for $e_i$ is the dual argument with the leftmost uncancelled $-$ and the change $i+1\mapsto i$. Since the shape is unchanged and the new entries still lie in $\{1,\dots,n\}$, these local checks prove stability in $\operatorname{SSYT}_n(\lambda)$.
[/proofplan]
[step:Fix the diagram notation and isolate the only inequalities that can change]
Write the boxes of the Young diagram of $\lambda$ as pairs $(a,c)\in Y(\lambda)$, where $a$ is the row index and $c$ is the column index. For a tableau $T\in\operatorname{SSYT}_n(\lambda)$, write $T(a,c)$ for the entry in box $(a,c)$. We use the fixed column reading word and the associated tableau crystal rule specified in the theorem statement. The statement includes the standard local admissibility consequences of that signature rule for the selected box.
Let $U$ be the filling obtained from $T$ by applying either $f_i$ or $e_i$. The shape of $U$ is the same as the shape of $T$, and exactly one box changes. Since $i\in\{1,\dots,n-1\}$, the changes $i\mapsto i+1$ and $i+1\mapsto i$ keep all entries inside the alphabet $\{1,\dots,n\}$.
Thus only semistandard inequalities involving the changed box can fail. All row inequalities and column inequalities not incident to that box are identical for $T$ and $U$.
[/step]
[step:Prove that applying $f_i$ preserves semistandardness]
Assume $f_i(T)$ is defined. Let $b_0=(a_0,c_0)$ be the box selected by the tableau crystal rule, so $T(b_0)=i$, and $f_i$ changes this entry to $i+1$. Denote the resulting filling by $U$.
The row inequality with the left neighbour, if the box $(a_0,c_0-1)$ exists, remains valid because
\begin{align*}
T(a_0,c_0-1)\le T(a_0,c_0)=i < i+1=U(a_0,c_0).
\end{align*}
The column inequality with the upper neighbour, if the box $(a_0-1,c_0)$ exists, remains valid because
\begin{align*}
T(a_0-1,c_0)<T(a_0,c_0)=i<i+1=U(a_0,c_0).
\end{align*}
It remains to rule out violations with the right neighbour and the lower neighbour. Suppose first that the right neighbour $(a_0,c_0+1)$ exists. By the local admissibility property of the selected $f_i$-box stated in the theorem, $T(a_0,c_0+1)\ge i+1$. Since $U(a_0,c_0+1)=T(a_0,c_0+1)$, we obtain
\begin{align*}
U(a_0,c_0)=i+1\le T(a_0,c_0+1)=U(a_0,c_0+1).
\end{align*}
Thus the row inequality with the right neighbour remains valid.
Suppose next that the lower neighbour $(a_0+1,c_0)$ exists. By the local admissibility property of the selected $f_i$-box stated in the theorem, $T(a_0+1,c_0)>i+1$. Since $U(a_0+1,c_0)=T(a_0+1,c_0)$, we obtain
\begin{align*}
U(a_0,c_0)=i+1<T(a_0+1,c_0)=U(a_0+1,c_0).
\end{align*}
Thus the column inequality with the lower neighbour remains valid.
Hence no row or column inequality involving $b_0$ fails, and all other inequalities were unchanged. Therefore $f_i(T)=U\in\operatorname{SSYT}_n(\lambda)$.
[guided]
Assume $f_i(T)$ is defined. By the tableau crystal rule, there is a unique selected box $b_0=(a_0,c_0)$ whose entry is $i$ and whose sign is the rightmost uncancelled $+$ in the reduced $i$-signature. The operator $f_i$ changes only this entry:
\begin{align*}
U(a_0,c_0)=i+1.
\end{align*}
For every other box $b$, we have $U(b)=T(b)$.
We now check exactly the inequalities that mention $b_0$. First consider neighbours on the left and above. If the left neighbour $(a_0,c_0-1)$ exists, then semistandardness of $T$ gives
\begin{align*}
T(a_0,c_0-1)\le T(a_0,c_0)=i.
\end{align*}
After changing $i$ to $i+1$, this becomes
\begin{align*}
U(a_0,c_0-1)=T(a_0,c_0-1)\le i<i+1=U(a_0,c_0),
\end{align*}
so the row inequality with the left neighbour still holds. If the upper neighbour $(a_0-1,c_0)$ exists, then column strictness of $T$ gives
\begin{align*}
T(a_0-1,c_0)<T(a_0,c_0)=i.
\end{align*}
Therefore
\begin{align*}
U(a_0-1,c_0)=T(a_0-1,c_0)<i<i+1=U(a_0,c_0),
\end{align*}
so the column inequality with the upper neighbour still holds.
The only possible problems are therefore to the right and below. Suppose the right neighbour $(a_0,c_0+1)$ exists. Before applying $f_i$, row weak increase gives
\begin{align*}
T(a_0,c_0)=i\le T(a_0,c_0+1).
\end{align*}
After the change, the required inequality is
\begin{align*}
i+1=U(a_0,c_0)\le U(a_0,c_0+1)=T(a_0,c_0+1).
\end{align*}
The local admissibility property of the selected $f_i$-box stated in the theorem gives $T(a_0,c_0+1)\ge i+1$, so this inequality cannot fail.
Now suppose the lower neighbour $(a_0+1,c_0)$ exists. Since $T$ is semistandard, column strictness gives
\begin{align*}
T(a_0,c_0)=i<T(a_0+1,c_0),
\end{align*}
so $T(a_0+1,c_0)\ge i+1$. After the change, the required strict inequality is
\begin{align*}
i+1=U(a_0,c_0)<U(a_0+1,c_0)=T(a_0+1,c_0).
\end{align*}
The local admissibility property of the selected $f_i$-box stated in the theorem gives $T(a_0+1,c_0)>i+1$, so this strict inequality cannot fail.
Thus every row and column inequality involving the changed box remains valid. Since no other box changed, every other semistandard inequality remains valid as well. Hence $f_i(T)\in\operatorname{SSYT}_n(\lambda)$.
[/guided]
[/step]
[step:Prove that applying $e_i$ preserves semistandardness]
Assume $e_i(T)$ is defined. Let $b_0=(a_0,c_0)$ be the box selected by the tableau crystal rule, so $T(b_0)=i+1$, and $e_i$ changes this entry to $i$. Denote the resulting filling by $U$.
The row inequality with the right neighbour, if the box $(a_0,c_0+1)$ exists, remains valid because
\begin{align*}
U(a_0,c_0)=i<i+1=T(a_0,c_0)\le T(a_0,c_0+1)=U(a_0,c_0+1).
\end{align*}
The column inequality with the lower neighbour, if the box $(a_0+1,c_0)$ exists, remains valid because
\begin{align*}
U(a_0,c_0)=i<i+1=T(a_0,c_0)<T(a_0+1,c_0)=U(a_0+1,c_0).
\end{align*}
It remains to rule out violations with the left neighbour and the upper neighbour. If the left neighbour $(a_0,c_0-1)$ exists, then the local admissibility property of the selected $e_i$-box stated in the theorem gives $T(a_0,c_0-1)\le i$. Since $U(a_0,c_0-1)=T(a_0,c_0-1)$, we obtain
\begin{align*}
U(a_0,c_0-1)=T(a_0,c_0-1)\le i=U(a_0,c_0).
\end{align*}
Thus the row inequality with the left neighbour remains valid.
If the upper neighbour $(a_0-1,c_0)$ exists, then the local admissibility property of the selected $e_i$-box stated in the theorem gives $T(a_0-1,c_0)<i$. Since $U(a_0-1,c_0)=T(a_0-1,c_0)$, we obtain
\begin{align*}
U(a_0-1,c_0)=T(a_0-1,c_0)<i=U(a_0,c_0).
\end{align*}
Thus the column inequality with the upper neighbour remains valid.
Thus no semistandard inequality involving $b_0$ fails, and all other inequalities are unchanged. Therefore $e_i(T)=U\in\operatorname{SSYT}_n(\lambda)$.
[/step]
[step:Conclude stability under both crystal operators]
For either operator, the resulting filling has the same shape $\lambda$ as $T$, all entries remain in $\{1,\dots,n\}$, row weak increase is preserved, and column strict increase is preserved. Hence whenever $f_i(T)$ or $e_i(T)$ is defined as a non-null tableau, the result lies in $\operatorname{SSYT}_n(\lambda)$, as required.
[/step]
