Let $n\in\mathbb N$ with $n\ge 2$, let $\lambda=(\lambda_1\ge\cdots\ge\lambda_r>0)$ be a partition with Young diagram $Y(\lambda)=\{(a,c):1\le a\le r,\ 1\le c\le\lambda_a\}$, and let $\operatorname{SSYT}_n(\lambda)$ denote the set of maps $T:Y(\lambda)\to\{1,\dots,n\}$ such that $T(a,c)\le T(a,c+1)$ whenever both boxes exist and $T(a,c)<T(a+1,c)$ whenever both boxes exist. Fix $i\in\{1,\dots,n-1\}$. Equip $\operatorname{SSYT}_n(\lambda)$ with the type $A_{n-1}$ tableau crystal rule obtained from the fixed column reading word used in this theorem, the $i$-signature obtained by writing $+$ for each entry $i$ and $-$ for each entry $i+1$, and cancellation of adjacent pairs $-+$. Assume this tableau rule has the standard local admissibility property: if $f_i$ selects a box $b=(a,c)$ with $T(b)=i$, then any right neighbour satisfies $T(a,c+1)\ge i+1$ and any lower neighbour satisfies $T(a+1,c)>i+1$; if $e_i$ selects a box $b=(a,c)$ with $T(b)=i+1$, then any left neighbour satisfies $T(a,c-1)\le i$ and any upper neighbour satisfies $T(a-1,c)<i$. If $T\in\operatorname{SSYT}_n(\lambda)$ and either $f_i(T)$ or $e_i(T)$ is defined as a tableau rather than the null value $0$, then the resulting filling is again an element of $\operatorname{SSYT}_n(\lambda)$.