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 ${\lambda }_{a^{-1}}\circ {\lambda }_a=\mathrm{id}$, and replacing $a$ by $a^{-1}$ gives ${\lambda }_a\circ {\lambda }_{a^{-1}}=\mathrm{id}$.

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 ${\mathrm{\Lambda }}_w={\lambda }_{a_1}\circ \cdots \circ {\lambda }_{a_n}$, 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 $\epsilon$, 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 ${\mathrm{\Lambda }}_r(\epsilon )=r$ for every reduced $r$, including $r=\epsilon$.

Inserting or deleting an adjacent pair $a{a}^{-1}$ inserts or deletes the adjacent composite ${\lambda }_a\circ {\lambda }_{a^{-1}}=\mathrm{id}$ inside ${\mathrm{\Lambda }}_w$; therefore one elementary move leaves ${\mathrm{\Lambda }}_w$ unchanged, and so does any finite sequence of such moves.