[proofplan] We isolate the signs in the $i$-signature and use a canonical matching description of the cancellation process. Under cancellation of adjacent $-+$ pairs, the uncancelled signs have normal form: all uncancelled $+$ signs precede all uncancelled $-$ signs. Changing the rightmost uncancelled $+$ to $-$ therefore turns it into the leftmost uncancelled $-$ in the new reduced signature, so $e_i$ reverses $f_i$. The converse is the same argument with $+$ and $-$ interchanged in the selected position. [/proofplan] [step:Encode the cancellation process by a canonical matching] Fix $w=w_1\cdots w_r\in B$. Let \begin{align*} J(w):=\{j\in\{1,\dots,r\}:w_j\in\{i,i+1\}\} \end{align*} with the inherited order from $\{1,\dots,r\}$. Define the sign map \begin{align*} \sigma_w:J(w)\to\{+,-\} \end{align*} by $\sigma_w(j)=+$ if $w_j=i$ and $\sigma_w(j)=-$ if $w_j=i+1$. We use the following deterministic matching rule. Read the signs in increasing order of positions. Maintain a list $M$ of currently unmatched minus positions. When the current sign is $-$, append its position to $M$. When the current sign is $+$ and $M$ is nonempty, match this plus position with the largest position in $M$ and delete that minus position from $M$. When the current sign is $+$ and $M$ is empty, declare this plus position unmatched. At the end, the unmatched minus positions are the entries remaining in $M$. This matching realizes the same uncancelled positions as repeated cancellation of adjacent $-+$ pairs. Indeed, matching a plus with the nearest unmatched minus to its left is exactly the local operation of removing one pair $-+$ after all signs between them that can be cancelled have already been removed. Thus the reduced signature is obtained from the unmatched signs in their original order. [guided] The only point that needs care is that the phrase “repeatedly cancelling adjacent $-+$ pairs” could seem to depend on the order of cancellations. We avoid that ambiguity by giving an equivalent canonical procedure. For the fixed word $w=w_1\cdots w_r$, we first ignore all letters except $i$ and $i+1$. The set \begin{align*} J(w):=\{j\in\{1,\dots,r\}:w_j\in\{i,i+1\}\} \end{align*} records the positions that survive this deletion. The map \begin{align*} \sigma_w:J(w)\to\{+,-\} \end{align*} is defined by $\sigma_w(j)=+$ for the letter $i$ and $\sigma_w(j)=-$ for the letter $i+1$. Now read these signs from left to right. Each time we see a minus, we store its position as a minus that is available to be cancelled. Each time we see a plus, there are two possibilities. If an available minus lies to its left, we cancel the plus with the nearest such minus, namely the largest stored position. If no available minus lies to its left, this plus cannot be part of a $-+$ cancellation, so it remains uncancelled. This is the stack description of adjacent $-+$ cancellation. A plus can only cancel a minus that occurs before it, and choosing the nearest currently unmatched minus exactly corresponds to performing all cancellations inside the intervening interval first. Therefore the unmatched positions produced by this deterministic procedure are precisely the positions that remain after all possible adjacent $-+$ cancellations have been performed. [/guided] [/step] [step:Put the reduced signature in normal form] Let $P(w)$ be the ordered list of unmatched plus positions produced by the matching rule, and let $N(w)$ be the ordered list of unmatched minus positions produced by the same rule. Then every position in $P(w)$ is smaller than every position in $N(w)$. To prove this, suppose $a\in N(w)$ and $b\in P(w)$ with $a<b$. When the algorithm reaches $b$, the minus at $a$ is still unmatched, because $a\in N(w)$ at the end. Hence the list $M$ is nonempty at position $b$, so the plus at $b$ would be matched with some unmatched minus rather than placed in $P(w)$. This contradicts $b\in P(w)$. Therefore no such pair $a<b$ exists. Consequently the reduced $i$-signature has the form \begin{align*} +\cdots+\, -\cdots- \end{align*} where the plus positions are exactly $P(w)$ and the minus positions are exactly $N(w)$. [/step] [step:Show that $e_i$ reverses a nonzero application of $f_i$] Assume $f_i(w)=w'$ for some $w'\in B$. Since $w'\in B$, the value $f_i(w)$ is nonzero, so $P(w)$ is nonempty. Let $p\in P(w)$ be the largest position in $P(w)$. By definition of $f_i$, the word $w'$ is obtained from $w$ by changing $w_p=i$ to $w'_p=i+1$ and leaving all other letters unchanged. We compare the canonical matchings for $w$ and $w'$. All signs strictly before $p$ are unchanged. Since $p$ is an unmatched plus for $w$, the matching process for $w$ has no unmatched minus available when it reaches $p$; otherwise the plus at $p$ would be matched. Thus, when processing $w'$, the new minus at $p$ is appended to the list of unmatched minus positions. Because $p$ is the largest unmatched plus position of $w$, every plus position after $p$ is matched in the matching for $w$. Those later pluses are matched using minus positions after $p$: no minus before or at $p$ is available to them in $w'$, since the new minus at $p$ is the first unmatched minus after all signs up to $p$ have been processed, and a later plus would match the nearest available unmatched minus only if it were needed. The same pairings after $p$ therefore remain valid, and the new minus at $p$ survives unmatched. By the normal form proved above, all unmatched plus positions in $w'$ lie before all unmatched minus positions in $w'$. The unmatched plus positions are precisely $P(w)\setminus\{p\}$, and the unmatched minus positions are precisely $\{p\}\cup N(w)$. Hence $p$ is the leftmost unmatched minus position in the reduced $i$-signature of $w'$. By the definition of $e_i$, the word $e_i(w')$ is obtained by changing $w'_p=i+1$ back to $i$ and leaving all other letters unchanged. Therefore $e_i(w')=w$. [/step] [step:Show that $f_i$ reverses a nonzero application of $e_i$] Assume $e_i(w')=w$ for some $w,w'\in B$. Since $w\in B$, the value $e_i(w')$ is nonzero, so $N(w')$ is nonempty. Let $p\in N(w')$ be the smallest position in $N(w')$. By definition of $e_i$, the word $w$ is obtained from $w'$ by changing $w'_p=i+1$ to $w_p=i$ and leaving all other letters unchanged. Apply the same comparison as in the previous step, with the changed sign viewed in reverse. In the reduced signature of $w'$, the position $p$ is the leftmost uncancelled minus, and all uncancelled plus positions lie strictly to its left. Changing this sign from $-$ to $+$ removes that leftmost uncancelled minus and creates a new uncancelled plus after all other uncancelled pluses. Thus the unmatched plus positions for $w$ are $P(w')\cup\{p\}$, the unmatched minus positions for $w$ are $N(w')\setminus\{p\}$, and $p$ is the rightmost unmatched plus position in the reduced $i$-signature of $w$. By the definition of $f_i$, the word $f_i(w)$ is obtained by changing the letter at this rightmost unmatched plus position $p$ from $i$ back to $i+1$ and leaving all other letters unchanged. Hence $f_i(w)=w'$. [/step] [step:Conclude the equivalence] The previous two steps prove both implications. If $f_i(w)=w'$, then $e_i(w')=w$. Conversely, if $e_i(w')=w$, then $f_i(w)=w'$. Therefore, for all $w,w'\in B$ and all $i\in\{1,\dots,n-1\}$, \begin{align*} f_i(w)=w' \iff e_i(w')=w. \end{align*} This is the desired inverse property of the Kashiwara operators on words. [/step]