[proofplan] We fix one index $i$ and translate the statement into a purely combinatorial assertion about a finite word in the two signs $+$ and $-$. Under the convention that adjacent pairs $-+$ are cancelled, an uncancelled $-$ remains exactly when some suffix has more $-$ signs than $+$ signs. Since $+$ records the letter $i$ and $-$ records the letter $i+1$, this is exactly the suffix ballot condition. Finally, a word is highest weight precisely when every raising operator $e_i$ is undefined, equivalently when no reduced $i$-signature contains an uncancelled $-$. [/proofplan] [step:Reduce the highest weight condition to uncancelled signs in each $i$-signature] Fix $i\in\{1,\dots,n-1\}$. Define the sign word \begin{align*} \sigma_i(w)\in\{+,-\}^{m_i} \end{align*} to be the word obtained from $w$ by deleting all letters other than $i$ and $i+1$, replacing each remaining letter $i$ by $+$ and each remaining letter $i+1$ by $-$. Here $m_i$ denotes the number of letters of $w$ equal to either $i$ or $i+1$. By the definition of the type $A$ word-crystal raising operator with this signature convention, $e_i(w)$ is undefined if and only if the reduced word obtained from $\sigma_i(w)$ by repeatedly cancelling adjacent pairs $-+$ contains no uncancelled $-$. Therefore $w$ is highest weight if and only if this reduced $i$-signature contains no uncancelled $-$ for every $i\in\{1,\dots,n-1\}$. [/step] [step:Characterize absence of uncancelled minus signs by suffix balances] We prove the following elementary cancellation criterion. [claim:Suffix cancellation criterion] Let $\sigma=s_1\cdots s_m$ be a finite word in the alphabet $\{+,-\}$. Repeatedly cancel adjacent pairs $-+$ until no such adjacent pair remains. Then the final reduced word contains no $-$ if and only if every suffix $\tau$ of $\sigma$ satisfies \begin{align*} \#\{k : \tau_k=+\}\ge \#\{k : \tau_k=-\}. \end{align*} [/claim] [proof] Process $\sigma$ from right to left by the following pairing algorithm. Let $A_j$ denote the number of currently available unpaired $+$ signs after the suffix $s_j\cdots s_m$ has been processed. Start with $A_{m+1}=0$. When $s_j=+$, set $A_j=A_{j+1}+1$. When $s_j=-$ and $A_{j+1}>0$, pair this $-$ with one later available $+$ and set $A_j=A_{j+1}-1$. When $s_j=-$ and $A_{j+1}=0$, this $-$ is unpaired. This right-to-left pairing is exactly the matching produced by cancelling adjacent pairs $-+$. Indeed, deleting a pair $-+$ pairs a $-$ with a later $+$, and repeated adjacent deletions can only create further adjacent pairs between signs that were separated by already paired signs. Conversely, if the right-to-left algorithm pairs a $-$ at position $a$ with a $+$ at position $b>a$, then every sign between positions $a$ and $b$ is paired internally before this pair is removed, so after those internal cancellations the two signs become adjacent and form a cancellable pair $-+$. Now suppose every suffix of $\sigma$ has at least as many $+$ signs as $-$ signs. If the algorithm left some $-$ at position $j$ unpaired, then at the moment $s_j$ was processed there was no available later $+$. Hence, in the suffix $s_j\cdots s_m$, the sign $s_j=-$ together with all later signs would contain more $-$ signs than $+$ signs after accounting for all possible later pairings. Equivalently, that suffix would have \begin{align*} \#\{k\ge j : s_k=-\}>\#\{k\ge j : s_k=+\}, \end{align*} contradicting the suffix hypothesis. Conversely, suppose some suffix $s_j\cdots s_m$ has more $-$ signs than $+$ signs. Each cancelled pair removes one $-$ and one later $+$. Thus the excess \begin{align*} \#\{k\ge j : s_k=-\}-\#\{k\ge j : s_k=+\} \end{align*} cannot be eliminated by pair cancellations inside that suffix. At least one $-$ from the suffix remains uncancelled. Therefore the final reduced word contains an uncancelled $-$. The two implications prove the criterion. [/proof] [guided] The key point is that the cancellation rule is oriented: the cancellable adjacent pair is $-+$, not $+ -$. Thus a $-$ can disappear only by being paired with a $+$ that occurs later in the sign word. Let $\sigma=s_1\cdots s_m$ be a word in $\{+,-\}$. We scan from right to left. At any stage, keep a count $A$ of available later $+$ signs that have not already been paired. When we meet a $+$, we add it to the available pool. When we meet a $-$, there are two cases. If $A>0$, we pair this $-$ with one of those later $+$ signs and reduce $A$ by $1$. If $A=0$, this $-$ cannot be paired with any later $+$, so it remains uncancelled. This is the same pairing as repeated adjacent cancellation of $-+$ pairs. A cancellation always pairs a $-$ with a later $+$. If there are already completely paired signs between them, those inner pairs can be removed first, after which the chosen $-$ and $+$ become adjacent and cancel as $-+$. Thus the right-to-left pairing algorithm records exactly which $-$ signs survive. Now assume every suffix of $\sigma$ has at least as many $+$ signs as $-$ signs. If a $-$ at position $j$ were left unpaired by the right-to-left algorithm, then among the suffix $s_j\cdots s_m$ there would be no enough later $+$ signs to match all the $-$ signs in that suffix. Hence \begin{align*} \#\{k\ge j : s_k=-\}>\#\{k\ge j : s_k=+\}, \end{align*} which contradicts the assumed suffix inequality. Conversely, assume there is a suffix $s_j\cdots s_m$ with more $-$ signs than $+$ signs. Each allowed cancellation removes one $-$ and one later $+$, so cancellations preserve the difference between the number of $-$ signs and the number of $+$ signs in that suffix except for deleting matched pairs. Since the suffix starts with a positive excess of $-$ signs, at least one $-$ from the suffix cannot be matched with a later $+$. Therefore the reduced sign word contains an uncancelled $-$. Thus the reduced sign word has no uncancelled $-$ exactly when every suffix has at least as many $+$ signs as $-$ signs. [/guided] [/step] [step:Translate the sign criterion back to letters of the word] Fix $i\in\{1,\dots,n-1\}$. For any suffix $u$ of $w$, deleting all letters other than $i$ and $i+1$ turns $u$ into a suffix of the sign word $\sigma_i(w)$ after replacing $i$ by $+$ and $i+1$ by $-$. Therefore the suffix inequality for signs becomes \begin{align*} \#\{j : u_j=i\}\ge \#\{j : u_j=i+1\}. \end{align*} By the suffix cancellation criterion, the reduced $i$-signature has no uncancelled $-$ if and only if this inequality holds for every suffix $u$ of $w$. Applying this equivalence for every $i\in\{1,\dots,n-1\}$, and using the definition of highest weight as the condition that every $e_i(w)$ is undefined, proves that $w$ is highest weight if and only if every suffix satisfies the stated ballot inequalities. This is the desired criterion. [/step]