Proposition 15

Let B be a basis of R and $\alpha$ an indivisible root. There exist $\beta \in B$ and $w \in W(R)$ such that $\alpha = w(\beta)$.

Let C be the chamber such that $B = B(C)$. The hyperplane $L_\alpha$ is a wall of a chamber $C'$ of R, and there exists an element of $W(R)$ that transforms $C'$ to C. We are therefore reduced to the case where $L_\alpha$ is a wall of C. Then $\alpha$ is proportional to an element $\beta$ of R. Since $\alpha$ and $\beta$ are indivisible, $\alpha = \pm \beta$. If $\alpha = -\beta$, then $\alpha = s_\beta(\beta)$, hence the proposition.