Formal letters act by mutually inverse permutations on the set of reduced words

For $r\in\mathcal R(X)$, define $\lambda_a(r)$ by deleting the first letter when $r$ begins with $a^{-1}$, and by prepending $a$ otherwise; in the second case the only new seam is not an inverse pair, so the output is reduced, while deletion from a reduced word also leaves a reduced word.

If $r=a^{-1}s$ is reduced, then $s$ does not begin with $a$, so $\lambda_a(r)=s$ and $\lambda_{a^{-1}}(s)=a^{-1}s=r$. If $r$ does not begin with $a^{-1}$, then $\lambda_a(r)=ar$ begins with $a$, so $\lambda_{a^{-1}}(ar)=r$. Thus $\lambda_{a^{-1}}\circ\lambda_a=\operatorname{id}$, and replacing $a$ by $a^{-1}$ gives $\lambda_a\circ\lambda_{a^{-1}}=\operatorname{id}$.

Hence each $\lambda_a$ is a bijection of $\mathcal R(X)$, so it is a permutation by [F2], and $\lambda_{a^{-1}}=\lambda_a^{-1}$.

For a word $w=a_1\cdots a_n$, construct $\Lambda_w=\lambda_{a_1}\circ\cdots\circ\lambda_{a_n}$, with the empty composite equal to the identity. Composition acts from right to left. If the suffix $a_{k+1}\cdots a_n$ of a reduced word has already been obtained from $\varepsilon$, then it does not begin with $a_k^{-1}$, so $\lambda_{a_k}$ prepends $a_k$. Induction on the suffix length using [L1] therefore gives $\Lambda_r(\varepsilon)=r$ for every reduced $r$, including $r=\varepsilon$.

Inserting or deleting an adjacent pair $aa^{-1}$ inserts or deletes the adjacent composite $\lambda_a\circ\lambda_{a^{-1}}=\operatorname{id}$ inside $\Lambda_w$; therefore one elementary move leaves $\Lambda_w$ unchanged, and so does any finite sequence of such moves.