[proofplan] We prove that the content sequence determines a standard Young tableau by induction on the number of boxes. Removing the box containing the largest entry $n$ leaves standard Young tableaux of size $n-1$, and the hypotheses give the same content sequence for the remaining entries. The induction hypothesis identifies the smaller tableaux, and then the last box is determined because distinct addable boxes of a fixed Young diagram have distinct contents. [/proofplan] [step:Start the induction with the one-box tableau] We prove the statement by induction on $n \in \mathbb{N}$. If $n=1$, a standard Young tableau with one box has shape $(1)$ and contains the single entry $1$, so there is only one such tableau. Hence $S=T$ in the base case. [/step] [step:Delete the boxes containing the largest entries] Assume $n\ge 2$ and assume the theorem holds for standard Young tableaux with $n-1$ boxes. Let $b_S=(i_S,j_S)$ be the box containing $n$ in $S$, and let $b_T=(i_T,j_T)$ be the box containing $n$ in $T$. The box $b_S$ is removable from the Young diagram of $S$. Indeed, if the box immediately to the right of $b_S$ or the box immediately below $b_S$ belonged to the diagram, then the row-increasing or column-increasing condition would force that neighbouring entry to be strictly larger than $n$, impossible in a tableau filled by $\{1,\dots,n\}$. The same argument shows that $b_T$ is removable from the Young diagram of $T$. Define $S'$ to be the tableau obtained from $S$ by deleting the box $b_S$, and define $T'$ to be the tableau obtained from $T$ by deleting the box $b_T$. Since deletion removes only the largest entry and the deleted boxes are removable, $S'$ and $T'$ are standard Young tableaux with entries $\{1,\dots,n-1\}$. [/step] [step:Apply the induction hypothesis to the deleted tableaux] For each $k \in \{1,\dots,n-1\}$, the box containing $k$ is unchanged when passing from $S$ to $S'$ and from $T$ to $T'$. Therefore \begin{align*} c_k(S')=c_k(S)=c_k(T)=c_k(T'). \end{align*} By the induction hypothesis applied to the standard Young tableaux $S'$ and $T'$ of size $n-1$, we obtain \begin{align*} S'=T'. \end{align*} Let $\mu$ denote their common Young diagram. [/step] [step:Use distinct contents of addable boxes to identify the last box] The tableaux $S$ and $T$ are obtained from the common tableau $S'=T'$ of shape $\mu$ by adding one addable box and filling it with $n$. We show that two addable boxes of $\mu$ cannot have the same content. Write $\mu=(\mu_1,\dots,\mu_r)$ with $\mu_1\ge \cdots \ge \mu_r\ge 1$. An addable box in row $a\le r$ has the form $(a,\mu_a+1)$ and can occur only when $a=1$ or $\mu_{a-1}>\mu_a$; there is also possibly the new bottom-row addable box $(r+1,1)$. If $(a,\mu_a+1)$ and $(b,\mu_b+1)$ are two addable boxes with $1\le a<b\le r$, equality of contents would give \begin{align*} \mu_a+1-a=\mu_b+1-b. \end{align*} Thus \begin{align*} \mu_a-\mu_b=a-b<0, \end{align*} contradicting $\mu_a\ge \mu_b$. If $(a,\mu_a+1)$ with $a\le r$ had the same content as $(r+1,1)$, then \begin{align*} \mu_a+1-a=1-(r+1), \end{align*} so \begin{align*} \mu_a=a-r-1<0, \end{align*} contradicting $\mu_a\ge 1$. Hence all addable boxes of $\mu$ have distinct contents. Since the added box in $S$ has content $c_n(S)$ and the added box in $T$ has content $c_n(T)$, the hypothesis $c_n(S)=c_n(T)$ forces the two added boxes to be the same addable box of $\mu$. [guided] At this point the smaller tableaux have already been identified: $S'=T'$, and their common shape is a Young diagram $\mu$. The only remaining question is whether the entry $n$ could be placed in two different addable boxes of $\mu$ while having the same content. We prove that this cannot happen. Write the common Young diagram as a partition $\mu=(\mu_1,\dots,\mu_r)$, where row $a$ has $\mu_a$ boxes and \begin{align*} \mu_1\ge \mu_2\ge \cdots \ge \mu_r\ge 1. \end{align*} An addable box in an existing row $a$ has coordinates $(a,\mu_a+1)$, because it is placed immediately after the last box in that row. The new bottom-row addable box, if used, has coordinates $(r+1,1)$. The content of a box $(i,j)$ is $j-i$. Take two addable boxes in existing rows, say $(a,\mu_a+1)$ and $(b,\mu_b+1)$ with $1\le a<b\le r$. If their contents were equal, then \begin{align*} \mu_a+1-a=\mu_b+1-b. \end{align*} Rearranging gives \begin{align*} \mu_a-\mu_b=a-b. \end{align*} The right-hand side is negative because $a<b$, while the left-hand side is nonnegative because $\mu$ is a partition and hence $\mu_a\ge \mu_b$. This contradiction shows that no two addable boxes in existing rows have the same content. It remains to compare an existing-row addable box with the new bottom-row addable box. Suppose $(a,\mu_a+1)$ with $a\le r$ had the same content as $(r+1,1)$. Then \begin{align*} \mu_a+1-a=1-(r+1). \end{align*} Rearranging gives \begin{align*} \mu_a=a-r-1. \end{align*} Since $a\le r$, the right-hand side is at most $-1$, contradicting $\mu_a\ge 1$. Therefore every addable box of $\mu$ has a distinct content. Now the added box in $S$ is an addable box of $\mu$ whose content is $c_n(S)$, and the added box in $T$ is an addable box of $\mu$ whose content is $c_n(T)$. The hypothesis gives $c_n(S)=c_n(T)$. Since addable boxes of $\mu$ have distinct contents, these two added boxes must be the same box. [/guided] [/step] [step:Conclude that all entries occupy the same boxes] The equality $S'=T'$ shows that every entry $k\in\{1,\dots,n-1\}$ occupies the same box in $S$ and in $T$. The preceding step shows that the entry $n$ also occupies the same box in $S$ and in $T$. Hence every entry of $\{1,\dots,n\}$ occupies the same box in both tableaux, so \begin{align*} S=T. \end{align*} This completes the induction and proves the theorem. [/step]