The Weyl group is finite and faithful

Each generator $s_{\alpha }$ lies in $O(E)$ and satisfies $s_{\alpha }(\mathrm{\Phi })=\mathrm{\Phi }$ by [L1]. Consequently every $w\in W(\mathrm{\Phi })$, being a finite product of generators and their inverses, restricts to a bijection $w:\mathrm{\Phi }\to \mathrm{\Phi }$.

The assignment $\rho :W(\mathrm{\Phi })\to \mathrm{Sym}(\mathrm{\Phi })$, $\rho (w)=w{\mathrm{\mid }}_{\mathrm{\Phi }}$, is a group homomorphism: the restriction of a composition of linear maps is the composition of the restrictions. Hence $\rho (W(\mathrm{\Phi }))$ is a subgroup of the finite group $\mathrm{Sym}(\mathrm{\Phi })$.

If $\rho (w)=\mathrm{i}\mathrm{d}$ then $w(\alpha )=\alpha$ for every $\alpha \in \mathrm{\Phi }$; since $\mathrm{\Phi }$ spans $E$ and $w$ is linear, $w={\mathrm{i}\mathrm{d}}_E$. Thus $\ker \rho =\{1\}$: the restriction action on the finite set $\mathrm{\Phi }$ is faithful, and $\rho$ identifies $W(\mathrm{\Phi })$ with the finite subgroup $\rho (W(\mathrm{\Phi }))\le \mathrm{Sym}(\mathrm{\Phi })$. In particular $W(\mathrm{\Phi })$ is finite and $\mathrm{\mid }W(\mathrm{\Phi })\mathrm{\mid }=\mathrm{\mid }\rho (W(\mathrm{\Phi }))\mathrm{\mid }\le \mathrm{\mid }\mathrm{\Phi }\mathrm{\mid }!$.