we already ruled out; we thus get a contradiction.

-   Suppose, for a contradiction, that a node of $S$ lies in the interior of $\mathcal R_{\Gamma}$. Since $S$ is connected, it follows that an edge of $S$ also intersects the interior of $\mathcal R_{\Gamma}$, a case which we already ruled out; we thus get a contradiction.

We now argue about possible intersections with the boundary of $\mathcal R_{\Gamma}$.

-   Suppose, for a contradiction, that a node $u$ not in $R_S$ lies on the boundary of $\mathcal R_{\Gamma}$. Since $\Gamma$ is planar, the node $u$ does not lie on $R_S$. Since $\Gamma$ is strictly-upward and the root $r(S)$ of $S$ is a node of $R_S$, the node $u$ does not lie on $h^1_{\Gamma}$. Finally, if $u$ lies on $h^2_{\Gamma}$, then since $\Gamma$ is strictly-upward, the edge from $u$ to its parent intersects the interior of $\mathcal R_{\Gamma}$, a case which we already ruled out; we thus get a contradiction.

-   Suppose, for a contradiction, that the interior of an edge $e$ not in $R_S$ intersects the boundary $\mathcal R_{\Gamma}$. Since $\Gamma$ is planar, the interior of $e$ does not intersect $R_S$. If the interior of $e$ intersects $h^1_{\Gamma}$ or $h^2_{\Gamma}$, then, 