Given $\varepsilon>0$ we apply Theorem 5.1 to $\varphi+\varepsilon$ and $\psi$. This gives $\MA(\max\{\varphi+\varepsilon,\psi\})=\MA(\varphi)$ on $G\subseteq\{\varphi+\varepsilon>\psi\}$. Letting $\varepsilon\to 0$ and using Theorem 3.1 we get $\MA(\max\{\varphi,\psi\})=\MA(\varphi)$ on $G$. Exchanging the roles of $\varphi$ and $\psi$ shows that $\MA(\varphi)=\MA(\psi)$ on $G$. ∎