Within CSS we talk about binary interactions. Does this prevent synchronizing multiple processes in one step? Example:

$P_1 ::= a.D$

$P_2 ::= \overline{a}.D$

$P_3 ::= a.D$

Let's consider the following LTS: $(P_1 | P_2 | P_3)\restriction_a$

Do I get $P_1, P_2, P_3$ $\tau$-synchronized simultaneously if an action $\tau$ happens or only two of them get synchronyzed and so I get $D|D|P_3$?


From $P \stackrel{def}{=} (P_1 | P_2 | P_3)\restriction_a$ you have exactly two outgoing transitions:

$$ P \stackrel{\tau}{\rightarrow} (D | D | P_3)\restriction_a $$


$$ P \stackrel{\tau}{\rightarrow} (P_1 | D | D)\restriction_a $$

CCS does not have ternary synchronisation, only binary. Indeed it can be shown that ternary synchronisation is strictly more expressive than binary synchronisation, in the sense that there cannot be a "nice" encoding of ternary synchronisation into calculi with only binary synchronisation.


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