the introduction of an additional clock, $t$, that measures absolute time. Now, instead of comparing directly the language of two TAs, a problem which is in general undecidable, we can compare their discretized TAs and have the following (see Theorem \[th:inclusion\_gap\]): if ${\mathfrak{L}}({{A}_d})$, the language of the discretized TA of $A$, is not included in ${\mathfrak{L}}({{B}_d})$, the language of the discretized TA of $B$, then the same holds for ${\mathfrak{L}}(A)$ with respect to ${\mathfrak{L}}(B)$. If, however, ${\mathfrak{L}}({{A}_d}) \subseteq {\mathfrak{L}}({{B}_d})$ then ${\mathfrak{L}}(A)$ is included in the topological closure of ${\mathfrak{L}}(B)$.

The next natural question, in case ${\mathfrak{L}}(A) \nsubseteq {\mathfrak{L}}(B)$, is how far away is a timed trace of ${\mathfrak{L}}(A)$ from all timed traces of ${\mathfrak{L}}(B)$, that is, what is the conformance distance $c({\mathfrak{L}}(A), {\mathfrak{L}}(B))$, the distance of ${\mathfrak{L}}(A)$ from being conformed with ${\mathfrak{L}}(B))$. When an untimed word of ${\mathfrak{L}}(A)$ is not in ${\mathfrak{L}}(B)$ or when a transition in $A$ which is not bound in time is not met with a similar transition in ${\mathfrak{L}}(B)$ of the same action label then $c({\mathfrak{L}}(A), {\mathfrak{L}}(B)) = \infty$ and the existence of these cases is decidable. A more challenging question is whether there is a sequence of timed traces of ${\mathfrak{L}}(A)$ which tend to diverge 