letter of $v_\ell$ is $\sigma^{\pm 1}$ and the first letter of $v_{\ell+1}$ is $\tau^{\pm 1}$. If $\varepsilon_\ell=-1$, then the last letter of $v_\ell$ is $\tau^{\pm 1}$ and the first letter of $v_{\ell+1}$ is $\sigma^{\pm 1}$.

Let us show that, if $\varepsilon_\ell=+1$, then $v_\ell$ is a non-empty word ending with $\sigma^{\pm1}$. Since $\varepsilon_\ell=+1$, we have $\alpha_\ell= \sigma^{p_{i_\ell}}$ and $\beta_\ell= \tau^{-p_{i_\ell}}$. There are three cases to consider: (i) $u_\ell$ is an empty word; (ii) the last letter of $u_\ell$ is $\tau^{\pm 1}$; or (iii) the last letter of $u_\ell$ is $\sigma^{\pm 1}$.

Case (i): If $u_\ell$ is empty, then $v_\ell$ is the reduced word for $\beta_{\ell-1} \alpha_\ell$, and thus we have $$v_\ell=\begin{cases}
\sigma^{p_{i_\ell}} & \text{if $\ell=1$} \\
\tau^{-p_{i_{\ell-1}}} \sigma^{p_{i_\ell}} & \text{if $\ell \neq 1$ and $\varepsilon_{\ell-1}=+1$} \\
\sigma^{-p_{i_{\ell-1}}+p_{i_\ell}} & \text{if $\ell \neq 1$ and $\varepsilon_{\ell-1}=-1$} \\
\end{cases}$$ In the last subcase, since $\varepsilon_\ell=+1 \neq -1=\varepsilon_{\ell-1}$, we must have $i_\ell \neq i_{\ell-1}$ as noted before. Thus, we must have $|p_{i_\ell}-p_{i_{\ell-1}}|>M$, and it follows that $v_\ell$ is a nontrivial power of $\sigma$. In all subcases, $v_\ell$ is indeed a non-empty word ending with $\sigma^{\pm 1}$.

Case (ii): If the last letter of $u_\ell$ is $\tau^{\pm 1}$, then there is no cancellation between $u_\ell$ and 