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Arrange the lengths of all consecutive subwords of a word in a triangular grid. Under the folding condition, the boundary between length ascents and descents must contain a folding square. Its equal opposite vertices identify two letters that can be deleted. Induction proves the deletion condition for involutory generators.

Assume deletion. If $w=s_1\cdots s_m$ is reduced and $\ell (sw)<m$, then the word $s,s_1,\dots ,s_m$ is not reduced. Deletion removes two letters. It must remove the initial $s$: otherwise left cancellation by $s$ would give a word for $w$ shorter than $m$. Deleting $s$ and one $s_i$ gives the exchange condition.

Assume exchange, and suppose $sw$ and $wt$ are both ascents of $w$. Write $w=s_1\cdots s_m$ reduced. Then $s_1\cdots s_{m}t$ is reduced. If $swt$ is not a two-step ascent, left exchange deletes one letter from this expression. Deleting one of the $s_i$ would, after right cancellation by $t$, express $sw$ with at most $m-1$ letters, contradicting $\ell (sw)=m+1$. Thus exchange deletes the final $t$, and $swt=w$. This is the folding condition.

Finally assume folding. Consider a shortest counterexample to deletion and draw the array of lengths of all consecutive subwords $s_i\cdots s_j$. Adjacent entries differ by at most one because each generator is an involution. Starting at the first place where the full word fails to be geodesic and following the boundary between ascents and descents produces a square in which left and right multiplication are both ascents but the diagonal is not a two-step ascent. Folding identifies the opposite vertices of this square. Cancelling the common prefix and suffix says that two letters of the original word may be deleted. This contradicts the choice of a counterexample and proves deletion. This standard argument is the folding-grid proof of the deletion condition.

Hence

We prove the assertion by induction on $n$. Write $w=av$, where $a=s_{i_1}$ and $v=s_{i_2}\cdots s_{i_n}$; both displayed words are reduced. The first clause of the folding condition and part a imply that $\ell (vs)$ is either $n-2$ or $n$.

If $\ell (vs)=n-2$, the induction hypothesis deletes one unique letter from the reduced word for $v$. Prefixing $a$ gives the required deletion from $w$. If another deletion were possible, induction rules out another internal position, while deletion of $a$ would give $ws=v$ and hence $vs=av$, contradicting the lengths $n-2$ and $n$.

If $\ell (vs)=n$, both $av$ and $vs$ increase the length of $v$. Since $\ell (avs)=\ell (ws)=n-1$, the other alternative in the folding condition must hold:Thus $ws=avs=v$, which deletes the first letter. An additional internal deletion would give $v=a{v}^{\prime }$ for a word $v^{\prime }$ of length $n-2$; multiplying by $a$ would make the length-$n$ element $av=w$ equal to $v^{\prime }$, impossible. The deletion position is therefore unique in every case.