Lemma 3

Let $w \in W$ and $s \in S$ be such that $l(sw) \leq l(w)$. For any sequence $s = (s_1, \ldots, s_q)$ of elements of $S$ with $w = s_1 \ldots s_q$, there exists an integer $j$ such that $1 \leq j \leq q$ and

Let $p$ be the length of $w$ and $w' = sw$. By the Remark of no. 3, $l(w') \equiv l(w) + 1$ mod. 2. The hypothesis $l(w') \leq l(w)$ and the relation

thus leads to $l(w') = p - 1$. Choose a reduced decomposition

of $w'$ and put $s = (s, s'_1, \ldots, s'_{p-1})$ and $\Phi(s') = (t'_1, \ldots, t'_p)$. It is clear that $s'$ is a reduced decomposition of $w$ and that $t'_1 = s$. The elements $t'_1, \ldots, t'_p$ being distinct by Lemma 2, we have $n(s', s) = 1$. Since $w$ is the product of the sequence $s$, we have $n(s, s) \equiv n(s', s)$ mod. 2 by Lemma 1, hence $n(s, s) \neq 0$. Consequently, $s$ is equal to one of the elements $t_j$ of the sequence $\Phi(s)$, hence the lemma.