Corollary 2

Assume that R is reduced. Let $w \in W$, let q be the length of $w$ with respect to S (Chap. IV, § 1, no. 1), and let $w = s_1 \ldots s_q$ be a reduced decomposition of $w$. Let $\alpha_1, \ldots, \alpha_q$ be the elements of $B(C)$ corresponding to $s_1, \ldots, s_q$. Put

The roots $\theta_i$ are $> 0$, pairwise distinct, $w(\theta_i) < 0$, and every root $\alpha > 0$ such that $w(\alpha) < 0$ is equal to one of the $\theta_i$.

Let X be the set of $\alpha > 0$ such that $w(\alpha) < 0$. By (ii),

On the other hand, if $\alpha \in X$ it is clear that there exists $i \in \{1, q\}$ such that

By Cor. 1, this implies that $s_{i+1} \ldots s_q(\alpha) = \alpha_i$ and hence that $\alpha = \theta_i$. The set X is thus contained in the set of $\theta_i$. Since $\operatorname{Card}(X) = q$, this is possible only if X is equal to the set of $\theta_i$ and these are pairwise distinct. Hence the corollary.