Every base of a root system arises as indecomposable positive roots of a regular element? I'm confused about a line in the Theorem p48 in Humphrey's's book on Lie Algebras.
He's proving that every base $\Delta$ of a root system $\Phi$ arises as the set of $\Delta(\gamma)$ of indecomposable roots in $\Phi^+(\gamma)=\{\alpha\in\Phi:(\gamma,\alpha)>0\}$ for some regular element $\gamma$, regular meaning $(\gamma,\alpha)\neq 0$ for any $\alpha\in\Phi$.
In step 5, he proves $\Phi^+=\Phi^+(\gamma)$, where $\Phi^+$ is the set of roots of positive height with respect to $\Delta$, and then says clearly $\Delta$ consists of indecomposable elements. Why is it clear that $\Delta$ consists of indecompoable elements? If $\alpha\in\Delta$ can be expressed as $\alpha=\beta_1+\beta_2$ for $\beta_i\in\Phi^+$, where's the problem?
 A: Suppose that $\alpha=\beta_1+\beta_2$ for $\beta_1,\beta_2\in\Phi^+$. Observe that $\beta_2\neq\alpha$, since then $\beta_1=0$, but $0\notin\Phi$, and $\beta_2\neq-\alpha$ since then $(\gamma,\beta_2)=-(\gamma,\alpha)<0$, since $(\gamma,\alpha)>0$, a contradiction since $\beta_2\in\Phi^+(\gamma)$. 
Since $\beta_2$ is not a multiple of $\alpha$, it follows that when writing $\beta_2=\sum_{\delta\in\Delta}k_\delta\delta$, for $k_\delta\geq 0$, there must be a nonzero coefficient $k_\delta$, for some $\delta\neq\alpha$. Then
$$
\beta_1=\alpha-\beta_2=(1-k_\alpha)\alpha-\sum_{\delta\in\Delta\setminus\{\alpha\}}k_\delta\delta
$$
So the unique expression of $\beta_1$ in terms of the base $\Delta$ has a negative coefficient, contrary to $\beta_1$ being a positive root. 
Alternatively, and more or less equivalently, now writing $\beta_1=\sum_{\delta\in\Delta}k'_\delta\delta$ with $k'_\delta\geq 0$, we have
$$
\alpha=\sum_{\delta\in\Delta}(k'_\delta+k_\delta)\delta
$$
so the coefficient of that particular $\delta\neq\alpha$ is still nonzero, so
$$
0=-\alpha+\sum_{\delta\in\Delta}(k'_\delta+k_\delta)\delta
$$
is a nontrivial linear combination equal to $0$, contrary to linear independence of a base, axiom (B1) in Humphrey's. 
