us elaborate on that. When $\delta$ is exactly $k$ and not $3k^{+}$ or $3k^{-}$ then it means that it is achieved on specific timed traces and not as a limit. That is, it refers to an even occurring at time $t^A$ on a run on $A$ and and event occurring at time $t^B$ on a run on $B$, with $| t^A - t^B | = 3k$. Since the fractional parts of $t^A$ and $t^B$ are identical, the discretization in the corresponding runs on ${{A}_d}$ and ${{B}_d}$ are identical so that they occur at times $t^{{{A}_d}}$ and $t^{{{B}_d}}$ with $| t^{{{A}_d}} - t^{{{B}_d}} | = 3k$. The same applies when $\delta$ is exactly $3k+\frac{3}{2}$ since we are working with a resolution of $\frac{1}{2}$.

When $\delta = 3k^{+}$ or $\delta = 3k^{-}$ then it is achieved as a limit of timed traces. If $\delta = 3k^{+}$ then $\delta_d$ can be $3k + \frac{1}{2}$, for example, when $t^A=3$ and $t^B = 3 + \varepsilon$, $\varepsilon>0$. Then the discretized traces will occur at times $t^{{{A}_d}}=3$ and $t^{{{B}_d}}=3\frac{1}{2}$. Then by choosing a sequence of timed traces of ${\mathfrak{L}}(B)$ the time difference can tend to $0$ while in the discretized automata it will remain $\frac{1}{2}$.

The 