Lemma 2

Let $s = (s_1, \ldots, s_q)$, $\Phi(s) = (t_1, \ldots, t_q)$ and $w = s_1 \ldots s_q$. Let $T_w$ be the set of elements of $T$ such that $\eta(w, t) = -1$. Then $s$ is a reduced decomposition of $w$ if and only if the $t_i$ are distinct, and in that case $T_w = \{ t_1, \ldots, t_q \}$ and $\mathrm{Card}(T_w) = l(w)$.

Clearly $T_w \subset \{ t_1, \ldots, t_q \}$. Taking $s$ to be reduced, it follows that $\mathrm{Card}(T_w) \leq l(w)$. Moreover, if the $t_i$ are distinct, then $n(s, t)$ is equal to 1 or 0 according to whether $t$ does or does not belong to $\{ t_1, \ldots, t_q \}$. It follows that $T_w = \{ t_1, \ldots, t_q \}$ and that $q = \mathrm{Card}(T_w) \leq l(w)$, which implies that $s$ is reduced. Suppose finally that $t_i = t_j$ with $i < j$. This gives $s_i = us_j u^{-1}$, with $u = s_{i+1} \ldots s_{j-1}$, hence

This shows that $s$ is not a reduced decomposition of $w$.