$\frac{1}{2}{\mathbb{N}_0}$ because it can be shown that for any other solution the timed traces can be shifted so that we are nearer $\frac{1}{2}{\mathbb{N}_0}$. In fact, it is quite clear that the solution should be looked for when considering the integral end-points of the event intervals. The solution is, in general, in $\frac{1}{2}{\mathbb{N}_0}$ and not in ${\mathbb{N}_0}$ as can be seen from the following example. Suppose that an event of $\tau^A$ occurs at time $0 < t < 1$ where the corresponding event in $B$ can occur at time $0$ or at time $1$. Then, the maximal time difference, namely $\frac{1}{2}$, occurs when we choose the event of $\tau^A$ to be at time $t=\frac{1}{2}$.

By the way they are defined, the untimed runs on the augmented region automaton ${\mathfrak{R}^t}(A)$, as well as those on the discretized timed automaton ${{A}_d}$, are identical to the untimed runs on $A$. The runs differ in the exact time on which each event occurs. When the absolute time of occurrence of an event is $t_0 \in {\mathbb{N}_0}$ then ${{A}_d}$ agrees with $A$. When $t_0 = n+\varepsilon$, $n \in {\mathbb{N}_0}$, $0 < \varepsilon <1$ then the time of the event on ${{A}_d}$ is set to be $n 