element of $\Gamma$, $w$ can be expressed as $$\begin{aligned}
w&=&u_1 \hat \gamma_{i_1}^{\varepsilon_1} u_2 \hat \gamma_{i_2}^{\varepsilon_2} \cdots \hat \gamma_{i_s}^{\varepsilon_s} u_{s+1}\\
&=&v_1 \gamma_{i_1}^{\varepsilon_1} v_2 \gamma_{i_2}^{\varepsilon_2} \cdots \gamma_{i_s}^{\varepsilon_s} v_{s+1}\end{aligned}$$ where $v_\ell=v_\ell(\sigma, \tau)$ is a reduced word in $\{ \sigma^{\pm 1}, \tau^{\pm 1} \}$ for $\beta_{\ell-1} u_\ell \alpha_\ell$ (with convention $\alpha_{s+1}=\beta_0=1$) and $$\alpha_\ell=\begin{cases}
\sigma^{p_{i_\ell}} & \text{if $\varepsilon_\ell=+1$} \\
\tau^{p_{i_\ell}} & \text{if $\varepsilon_\ell=-1$}
\end{cases}
\;\;\;\;\;\;
\beta_\ell=\begin{cases}
\tau^{-p_{i_\ell}} & \text{if $\varepsilon_\ell=+1$} \\
\sigma^{-p_{i_\ell}} & \text{if $\varepsilon_\ell=-1$}
\end{cases}$$ The idea of the proof is to apply the ping-pong argument to $w$ to show that $w$ cannot represent the identity element of $\Gamma$. Provided with suitable initial points in $X$, the ping-pong argument applies easily to the strings $v_\ell$. What we need to show is that we can pass each $\gamma^{\varepsilon_\ell}_{i_\ell}$ in the ping-pong argument; in other words, we need to check that $\gamma^{\varepsilon_\ell}_{i_\ell}$ takes the terminal point from the ping-pong rally $v_{\ell+1}$ to a suitable initial point for the ping-pong rally $v_\ell$. We will see that our choice of $p_i$ prevents excessive cancellations, and we can indeed pass each $\gamma^{\varepsilon_\ell}_{i_\ell}$ under the conditions (2) and (3) in the statement of the proposition.

\[pass gamma\]For each $\ell$, $v_\ell$ is not an empty word. If $\varepsilon_\ell=+1$, then the last 