$w$ in the same left-to-right order (since $\Gamma_0$ and $\Gamma_1$ are order-preserving). Now since $q_0$ is to the left of $\ell_0^v$ and $q_1$ is to the left of $\ell_1^v$, Corollary 7.2 from [@DBLP:journals/siamcomp/AlamdariABCLBFH17] implies that $q_{t}$ is to the left of the line that passes through $v_{t}$ and $v'_{t}$, while $q_{t}$ is supposed to be a point of $\overline{v_{t} v'_{t}}$. This contradiction proves the lemma.

Upward planar morphs between strictly-upward drawings of rooted ordered trees maintain the drawing order-preserving at all times, as proved in the following.

\[le:upwardplanar-implies-order\] Let $\Gamma_0$ and $\Gamma_1$ be two order-preserving strictly-upward straight-line planar drawings of a rooted ordered tree $T$. Let $\mathcal M$ be any upward planar morph between $\Gamma_0$ and $\Gamma_1$. Then any intermediate drawing of $\mathcal M$ is order-preserving.

Assume that the morph $\mathcal M$ happens between the time instants $t=0$ and $t=1$. For any $t\in [0,1]$, denote by $\Gamma_t$ the drawing of $T$ in $\mathcal M$ at time $t$. Since $\mathcal M$ is upward, the drawing $\Gamma_t$ is strictly-upward, for any $t\in [0,1]$. Hence, it suffices to prove that, for any internal node $v$ of $T$, the edges from $v$ to its children enter $v$ in $\Gamma_t$ in the left-to-right order associated 