# Restriction and re-labeling on CCS

In a process like $$R \stackrel{def}=((a.\bar{b}.0)\setminus\{b\})[a\to b]\mid(\bar{b}.b.0)+\bar{b}.c.0$$

$b$ is restricted to perform on the inner process of RHS $(a.\bar{b}.0)\setminus\{b\}$, thereafter the $a$ is relabeled to $b$. What I am confused is in drawing the LTS of it, the $a$ action occur at first process (we can have a transition with label $a$ in the LTS) and it is the next process that renaming takes place? can the following LTS be valid as first actions transitions? (the bar is replaced by underline) I am not completely sure about what is causing your confusion, but perhaps this can help:

$a.\bar{b}.0$ can only perform $a$.

$(a.\bar{b}.0)\setminus\{b\}$ can only perform $a$.

$((a.\bar{b}.0)\setminus\{b\})[a\to b]$ can only perform $b$ (i.e. $a$ after substitution).

The full process $((a.\bar{b}.0)\setminus\{b\})[a\to b]\mid(\bar{b}.b.0)+\bar{b}.c.0$ can perform $b$, $\bar b$ (two distinct ways), and $\tau$ (two distinct ways).

Also note that the unrelated process $((a.\bar{b}.0)[a\to b])\setminus\{b\}$ would instead be stuck.

• Thanks, I have edited the question and added LTS. so as you have mentioned there are two distinct Tau actions. my confusion is in that. where do they happen? – Amir-Mousavi May 7 '18 at 20:57
• @Amir-Mousavi There should be no $a$ there on the left arrow, only $b$ because of the renaming. The $\tau$s are caused by synchronization, where one side of the parallel does $b$ and the other does $\bar b$. – chi May 7 '18 at 21:00