marking in ${[\![E]\!]}$, $d<1-\max({{\it frac}(M)})$ and $M{\longrightarrow_{d}}M'$. The assumption on $d$ implies that for every token $(p,c) \in M$ we have ${{\it int}(c)}={{\it int}(c+d)}$. In other words, the integral part of the token age remained the same. Therefore $(p,{{\it int}(c)}) = (p,{{\it int}(c+d)}) \in M'$. Also from the assumption on $d$ we get that $${{\it frac}(M')}= \{x+d \mid x \in {{\it frac}(M)}\}$$ Recall that ${\mathit{abs}_{}(M)}={\mathit{abs}_{S}(M)}$ and ${\mathit{abs}_{}(M')}={\mathit{abs}_{S'}(M')}$ for the sets $S{{\overset{\text{\tinydef}}{=}}}\{0\}\cup {{\it frac}(M)}$ and $S'{{\overset{\text{\tinydef}}{=}}}\{0\}\cup {{\it frac}(M')}$. Clearly, $0\notin {{\it frac}(M')}$. There are two cases:

1.  $0\in{{\it frac}(M)}$. Then ${\mathit{abs}_{}(M')}=\emptyset{\mathit{abs}_{}(M)}\in {{\mathcal{L}}{\mathopen{}\mathclose\bgroup\originalleft}(\emptyset E{\aftergroup\egroup\originalright})}$, and consequently, $M'\in{[\![\emptyset E]\!]}$.

2.  $0\notin{{\it frac}(M)}$. Then ${\mathit{abs}_{}(M')}={\mathit{abs}_{}(M)}=\emptyset w\in {{\mathcal{L}}{\mathopen{}\mathclose\bgroup\originalleft}(E{\aftergroup\egroup\originalright})}$. Suppose that $E=x_0\alpha$, i.e., $E$ has $x_0\in\Sigma$ as its leftmost symbol, and $w\in{{\mathcal{L}}{\mathopen{}\mathclose\bgroup\originalleft}(\alpha{\aftergroup\egroup\originalright})}$. If $x_0=\emptyset$ then ${[\![E]\!]}={[\![\emptyset E]\!]}$ and thus ${\mathit{abs}_{}(M')}\in{[\![\emptyset E]\!]}$. Otherwise, if $x_0\neq\emptyset$ then $x_0w\in{{\mathcal{L}}{\mathopen{}\mathclose\bgroup\originalleft}(E{\aftergroup\egroup\originalright})}$ and $x_0w={\mathit{abs}_{}(M'')}$ for some marking $M''\ge M'$. So again, $M'\in{[\![\emptyset E]\!]}$.

*“$\subseteq$”*: W.l.o.g., pick a non-empty marking $M'\in {[\![\emptyset E]\!]}$. If $E$ has $\emptyset$ as its leftmost symbol, then ${[\![\emptyset E]\!]}={[\![E]\!]}$ and the claim follows using $d=0$, since then $M'\in{[\![E]\!]}$. So suppose that $E$ does not start with $\emptyset$. Note that by \[oneclock:abstr\], there are no tokens in the marking $M'$ whose clocks have fractional value zero. Let $$d{{\overset{\text{\tinydef}}{=}}}\min({{\it frac}(M')})$$ 