{\mathfrak{L}}_2) + c(L_2, L_3).$$

\[th:integer\_gap\] Let $A,B \in {\mathrm{NTA}}$. Then $c({\mathfrak{L}}(A), {\mathfrak{L}}(B)) \in \frac{1}{2}{\mathbb{N}_0}\cup \{\infty\}$.

Clearly, the conformance distance $c({\mathfrak{L}}(A), {\mathfrak{L}}(B))$ can be $\infty$, for example, when the untimed language of $A$ contains a word that is not in the untimed language of $B$. Suppose now that $\delta = c({\mathfrak{L}}(A), {\mathfrak{L}}(B)) < \infty$. It suffices to show the following. Given a path $\gamma^A$ in $A$ and another path $\gamma^B$ in $B$, where both define the same untimed trace (identical sequence of actions), let $T^A$ ($T^B$) be the set of all timed traces along $\gamma^A$ ($\gamma^B$). We need to show that $$\label{eq:dist_path}
     \sup_{\tau^A \in T^A} \inf_{\tau^B \in T^B} d(\tau^A, \tau^B) \in \frac{1}{2}{\mathbb{N}_0}.$$ By [@R19], the timestamp of each of the events along $\gamma^A$ and $\gamma^B$ is an interval of the form $(m,n)$, $(m,n]$, $[m,n)$ or $[m,n]$, where $m \leq n$ and $m \in {\mathbb{N}_0}$, $n \in {\mathbb{N}_0}\cup \{\infty\}$. This can be shown by writing equalities and inequalities over the integers and variables $z_i$, where $z_i$ represents the time of the $i$-th event along the path. Then becomes an optimization problem over the integers and variables for the events along $\gamma^A$ as well as for those along $\gamma^B$. The solution lies in 