Proposition 4

Let $s \in S, w \in W$ and $s = (s_1, \ldots, s_q)$ be a reduced decomposition of $w$. Then one of the following must hold:

a) $l(sw) = l(w) + 1$ and $(s, s_1, \ldots, s_q)$ is a reduced decomposition of $sw$.

b) $l(sw) = l(w) - 1$ and there exists an integer $j$ such that $1 \leq j \leq q$, $(s_1, \ldots, s_{j-1}, s_{j+1}, \ldots, s_q)$ is a reduced decomposition of $sw$ and $(s, s_1, \ldots, s_{j-1}, s_{j+1}, \ldots, s_q)$ is a reduced decomposition of $w$.

Put $w' = sw$. By formula (3) of no. 1, we have

We distinguish two cases:

a) $l(w') > l(w)$. Then $l(w') = q + 1$ and $w' = ss_1 \ldots s_q$, so

is a reduced decomposition of $w'$.

b) $l(w') \leq l(w)$. By (E), there exists an integer $j$ such that $1 \leq j \leq q$ and (16) is satisfied. Then $w = ss_1 \ldots s_{j-1} s_{j+1} \ldots s_q$ and hence

Since $q - 1 \leq l(w') \leq q$, it follows that $l(w') = q - 1$ and that $(s_1, \ldots, s_{j-1}, s_{j+1}, \ldots, s_q)$ is a reduced decomposition of $w'$.