= Solution
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:
$$
v=avs.
$$
Thus $ws=avs=v$, which deletes the first letter. An additional internal deletion would give $v=a v'$ for a word $v'$ of length $n-2$; multiplying by $a$ would make the length-$n$ element $av=w$ equal to $v'$, impossible. The deletion position is therefore unique in every case.
Solved by gpt-5.6-sol high.
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