\leq n -1$. Furthermore, $\mu$ is linear in every cell.

We prove this by induction on $m + n$. The base cases ($m = 0$, $n = 0$, and $m = n = 1$) follow from Lemma \[lem:pointcurve\] and Lemma \[lem:linesegments\].

For induction, we assume that $m \geq 1$, $n \geq 1$, and $m + n > 2$. By Lemma \[lem:realizingset\], a realizing set $E_\text{r}$ exists for $P$ and $Q$, say at value $\varepsilon_\text{r}$. The set contains realizing events $e_1, \ldots, e_k$ ($k \geq 1$), numbered in lexicographic order. By definition, $\varepsilon_\text{r} \leq \varepsilon$ holds. Suppose that $E_\text{r}$ splits curve $P$ into $P_1, \ldots, P_{k+1}$ and curve $Q$ into $Q_1, \ldots, Q_{k+1}$, where $P_i$ has $m_i$ edges, $Q_i$ has $n_i$ edges. By definition of a realizing event, none of the events in $E_\text{r}$ occur on the right or top boundary of cell $(m,n)$. Hence, for any $i$ ($1 \leq i \leq k+1$), it holds that $m_i \leq m$, $n_i \leq n$, and $m_i < m$ or $n_i < n$. Since a path exists in the free space diagram at $\varepsilon_\text{r}$ through all events in $E_\text{r}$, the induction hypothesis implies that, for any $i$ ($1 \leq i \leq k+1$), a locally 