since $\Gamma$ is strictly-upward, it intersects the interior of $\mathcal R_{\Gamma}$ as well, a case which we already ruled out; we thus get a contradiction.

This concludes the proof.

We now exploit \[pr:empty-right\] to prove the planarity of $\mathcal M$. Since $\mathcal M'$ is planar, by induction, and since $\mathcal M$ coincides with $\mathcal M'$ when restricted to the nodes and edges of $T'$, it follows that $\mathcal M$ is planar, as long as the edge $(\pi,l)$ does not intersect any edge, other than at a common end-point, throughout $\mathcal M$. We now argue about the possible intersections of the edge $(\pi,l)$ in $\mathcal M$.

First, we deal with the morph $\langle \Gamma_0,\Delta_1 \rangle$, in which only the node $l$ moves. By \[pr:empty-right\] applied to $T$ and $\Gamma_0$, no node or edge of $T$, other than those of the righmost path $R_T$ of $T$, intersects $\mathcal R_{\Gamma_0}$. Hence, it suffices to prove that the edge $(\pi,l)$ lies in $\mathcal R_{\Gamma_0}$ throughout $\langle \Gamma_0,\Delta_1 \rangle$; refer to \[fig:upward-linearsteps-polyarea-planar\]. Since $\Gamma_0$ is a strictly-upward grid drawing, we have that $l$ lies at least one unit below $\pi$ in $\Gamma_0$; further, by construction, $l$ lies one unit below $\pi$ in $\Delta_1$. Moreover, by 