true if $n=1$ and it is true by induction if $n>1$ since $\mathcal M$ has $2$ steps more than $\mathcal M'$.

For each $i=1,2,\dots,k$, the drawing $\Delta_i$ is straight-line by construction; further, $\Delta_i$ is strictly-upward since $\Delta'_i$ is strictly-upward and since $l$ lies one unit below $\pi$ in $\Delta_i$. By \[obs:upward-stays-upward\], the morph $\mathcal M$ is upward. Since $\ell_{\Gamma_0}, \ell_{\Delta'_2},\ell_{\Delta'_3},\dots,\ell_{\Delta'_{k-1}},\ell_{\Gamma_1}$ are grid columns and since $\pi$ is placed at a grid point in each of $\Delta'_1,\Delta'_2,\dots,\Delta'_k$, we have that $l$ is placed at a grid point in each of $\Delta_1,\Delta_2,\dots,\Delta_k$, hence each of such drawings is a grid drawing.

We now analyze $w(\mathcal M)$. If $n=1$, then $\mathcal M=\langle \Gamma_0,\Gamma_1\rangle$, hence $w(\mathcal M)=\max\{w(\Gamma_0),w(\Gamma_1)\} +n-1$. Assume next that $n\geq 2$. Consider any $i\in \{2,3,\dots,k-1\}$. By construction $\Delta_i$ occupies one more grid column than $\Delta'_i$. By induction $w(\Delta'_i)\leq \max\{w(\Gamma'_0),w(\Gamma'_1)\} +n-2$; further, $w(\Gamma'_0)\leq w(\Gamma_0)$ and $w(\Gamma'_1)\leq w(\Gamma_1)$. Hence, $w(\Delta_i)\leq  \max\{w(\Gamma_0),w(\Gamma_1)\} +n-1$, as required. In order to bound $w(\Delta_1)$, the argument is the same as the one above if $\ell_{\Gamma_0}$ contains a node of $\Delta'_1$. Otherwise, $\ell_{\Gamma_0}$ contains $l$ and no other node of $\Gamma_0$; then, by construction, we have $w(\Delta_1)= w(\Gamma_0)+1\leq \max\{w(\Gamma_0),w(\Gamma_1)\} +n-1$, as required. The proof that $w(\Delta_k)\leq \max\{w(\Gamma_0),w(\Gamma_1)\} +n-1$ is analogous.

