$\Delta \subset \Phi$ is a base in a root system imples $\Delta^\vee \subset \Phi^\vee$ is a base in a root system

(the notation here is compatible with J.E. Humphrey's "Introduction to Lie Algebras and Representation Theory")

Let $\Phi \subset E$ be a root system. Let $\Delta \subset \Phi$ be a base. I already know that $\Phi^\vee \subset E$ is a root system. I'd like to show that $\Delta^\vee \subset \Phi^\vee$ is a base.
(the notation $\alpha^\vee$ means $\frac{2\alpha}{(\alpha,\alpha)}$)

I tried 2 approaches and was stuck with about the same problem.

Here's one attempt: I'll denote $\Delta=\Delta(\gamma)$ for some regular $\gamma \in E$. I'd like to show that $\Delta(\gamma)^\vee=\Delta(\gamma^\vee)$. I can easily show that $\Phi^+(\gamma)^\vee=(\Phi^\vee)^+(\gamma^\vee)$. But then, how do I show that the indecomposable elements in $(\Phi^\vee)^+(\gamma^\vee)$ are exactly the duals of the indecomposable elements in $\Phi^+(\gamma)$? (If this is true, than it goes the other way around too, so I'll try to show that the dual of a decomposable element is decomposable). Denote $\alpha=\beta_1+\beta_2$ for some $\alpha,\beta_1,\beta_2 \in \Phi^+(\gamma)$. I'd like to show that $\alpha^\vee$ is decomposable too (in $(\Phi^+(\gamma))^\vee$). It is not true that $\alpha^\vee=\beta_1^\vee+\beta_2^\vee$ as I've seen by checking 2-dimensional examples. That's where I'm stuck.

(the other attempt was to go straight from the definition of a base, and there I had a problem with proving the integrality of the coefficients)

Another idea was to use the correspondence between Weyl chambers and bases. Both $\Phi$ and $\Phi^\vee$ define the same Weyl chambers. This gives a correspondence between the bases of $\Phi$ and the bases of $\Phi^\vee$. The correspondence is probably given by taking the dual of a base, but that's what I'm stuck at showing.

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