known. In particular: $${\mathfrak{L}}({{A}_d}) \nsubseteq {\mathfrak{L}}({{B}_d}) \; \Rightarrow \; {\mathfrak{L}}(A) \nsubseteq {\mathfrak{L}}(B)$$ and $${\mathfrak{L}}({{A}_d}) \subseteq {\mathfrak{L}}({{B}_d}) \; \Rightarrow \; {\mathfrak{L}}(A) \subseteq \overline{{\mathfrak{L}}(B)},$$ so that the language inclusion problem between ${\mathfrak{L}}(A)$ and the topological closure of ${\mathfrak{L}}(B)$ is decidable.

$A$ and ${{A}_d}$ have the same untimed language. The timed languages ${\mathfrak{L}}(A)$ (with clocks running at triple speed) and ${\mathfrak{L}}({{A}_d})$ differ from one another in that every event of a run on $A$ that occurs at time $t$, with $t = n + \varepsilon$ and $0 < \varepsilon < 1$, occurs at the ‘rounded‘ time $n+\frac{1}{2}$ in the corresponding run on ${{A}_d}$. Similarly for $B$ with respect to ${{B}_d}$. It follows that $\delta = c({\mathfrak{L}}(A), {\mathfrak{L}}(B)) = \infty$ if and only if $\delta_d = c({\mathfrak{L}}({{A}_d}), {\mathfrak{L}}({{B}_d})) = \infty$.

Suppose now that $\delta < \infty$. We know that $\delta \in \frac{3}{2}{\mathbb{N}_0}$. Since the timed traces of ${\mathfrak{L}}({{A}_d})$ and ${\mathfrak{L}}({{B}_d})$ are discretized to the set $\frac{1}{2}{\mathbb{N}_0}$ then, when computing $\delta_d$ instead of $\delta$, we may have a difference of $\frac{1}{2}$ time units between the two. It follows that $$\label{eq:delta}
    \delta = \begin{cases}
    3k, &\quad \mathrm{if} \; \delta_d \in \{3k-\frac{1}{2}, 3k, 3k+\frac{1}{2}\}, \\
    3k+\frac{3}{2}, &\quad \mathrm{if} \; \delta_d \in \{3k+1, 3k+\frac{3}{2}, 3k+2 \}.
    \end{cases}$$

Let 