construction, the position of $l$ in $\Delta_1$ is at least one unit to the right of the positions of $\pi$ and $l$ in $\Gamma_0$. It follows that the point at which $l$ is placed in $\Delta_1$ lies in the part $\Pi_{\Gamma_0}$ of $\mathcal R_{\Gamma_0}$ that is delimited by the representation of the edge $(\pi,l)$ in $\Gamma_0$ from the left and by the horizontal lines through the positions of $\pi$ and $l$ in $\Gamma_0$ from above and from below, respectively. The convexity of $\Pi_{\Gamma_0}$ implies that the edge $(\pi,l)$ lies in $\mathcal R_{\Gamma_0}$ throughout $\langle \Gamma_0,\Delta_1 \rangle$.

The proof that the morph $\langle \Delta_k,\Gamma_1 \rangle$ is planar is symmetric.

We now prove that the drawing $\Delta_i$ is planar, for each $i=0,\dots,k$; refer to \[fig:upward-linearsteps-intermediate\]. Since $\Delta'_i$ is planar, by induction, any crossing in $\Delta_i$ involves the edge $(\pi,l)$. Suppose, for a contradiction, that an edge $(u,v)$ crosses $(\pi,l)$. If $(u,v)$ is an edge of $R_{T}\setminus \{(\pi,l)\}$, then $(u,v)$ and $(\pi,l)$ are separated by the horizontal line through $\pi$ and thus do not cross. Assume hence that $(u,v)$ is not an edge of $R_{T}\setminus \{(\pi,l)\}$. The intersection between $(u,v)$ and $(\pi,l)$ in $\Delta_i$ has to happen at an interior point $c$ of 