Let $\Phi$ be a root system of type ADE, $\Lambda$ be the lattice in the Euclidean space spanned by $\Phi$. If $\Lambda$ is spanned by $\{v_i\}\subset \Phi$ (as a $\mathbb Z$-module), and let $\sigma_i$ be the reflection along hyperplane associated to $v_i$. Is it true that $\{\sigma_i\}$ generate the Weyl group $W(\Phi)$?
My approach
I know there is a proposition says reflections along vectors in a base (of a root system, or in some book called simple system) generate the full $W(\Phi)$. So I tried to show $\{v_i\}$ contains a base, but have no idea how to do this. More precisely, I don't know how to use the condition $\{v_i\}$ generate $\Lambda$.
If this too general, what about the case $\Phi=E_6$? If $6$ vectors $v_1,\ldots,v_6$ in the root system generate the lattice, is it true that after possibally replacing $v_i$ by $-v_i$ for some $i$, they will form a base of $E_6$?
Let $\Phi'$ be the root subsystem of $\Phi$ generated by $v_1,\ldots,v_n$ where $n$ is the rank of $\Phi$. Then, $\Phi'$ itself also has rank $n$ (because the $v_1,\ldots,v_n$ are linearly independent). Because $\Phi$ is simply laced, it follows that $\Phi=\Phi'$. Indeed, the Dynkin diagram of $\Phi'$ is a subdiagram of the one of $\Phi$ (each time with respect to some base) and both have $n$ vertices. Since there are only simple edges, we conclude that the diagrams are equal. It follows that $v_1,\ldots,v_n$ form a base up to sign. By replacing $v_i$ with $-v_i$ if necessary, which does not change the reflection along $v_i$, we may assume that $v_1,\ldots,v_n$ form a base of $\Phi$. Hence, $s_{v_1},\ldots,s_{v_n}$ generate $W$.