vault backup: 2025-04-14 15:25:00

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Marco Realacci 2025-04-14 15:25:00 +02:00
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@ -56,14 +56,14 @@ To let p0 be simulated by q0, we should have that p1 is simulated by q1 or q2.
If S contained one among (p1,q1) or (p1,q2), then it would not be a simulation: indeed, p1 can perform both a c (whereas q1 cannot) and a b (whereas q2 cannot). If S contained one among (p1,q1) or (p1,q2), then it would not be a simulation: indeed, p1 can perform both a c (whereas q1 cannot) and a b (whereas q2 cannot).
**Remark:** for proving equivalence, it is NOT enough to find a simulation of p by q and a simulation of q by p **Remark:** for proving equivalence, it is NOT enough to find a simulation of p by q and a simulation of q by p
![](../../Pasted%20image%2020250414082824.png) ![](images/Pasted%20image%2020250414082824.png)
p0 is simulated by q0: $$S = {(p0, q0), (p1, q2), (p2, q3)}$$ p0 is simulated by q0: $$S = {(p0, q0), (p1, q2), (p2, q3)}$$
q0 is simulated by p0: $$S ={(q0,p0),(q1,p1),(q2,p1),(q3,p2)}$$ q0 is simulated by p0: $$S ={(q0,p0),(q1,p1),(q2,p1),(q3,p2)}$$
However, p0 and q0 are not bisimilar: the transition q0 -> a -> q1 is not bisimulable by any transition from p0 (ndeed, p0 a> p1 does not suffice, because p1 can perform a b and so cannot be bisimilar to q1). However, p0 and q0 are not bisimilar: the transition q0 -> a -> q1 is not bisimulable by any transition from p0 (ndeed, p0 a> p1 does not suffice, because p1 can perform a b and so cannot be bisimilar to q1).
#### Bisimulation is NOT isomorfism #### Bisimulation is NOT isomorfism
![](../../Pasted%20image%2020250414084615.png) ![](images/Pasted%20image%2020250414084615.png)
$S = \{(p0,q0), (p1,q1), (p2,q1), (p0,q2)\}$ $S = \{(p0,q0), (p1,q1), (p2,q1), (p0,q2)\}$
$S^{1} = \{(q0,p0), (q1,p1), (q1,p2), (q2,p0)\}$ $S^{1} = \{(q0,p0), (q1,p1), (q1,p2), (q2,p0)\}$
@ -101,10 +101,10 @@ So, (p', q') ∈
Let (p,q) ∈ S. Then, there exists a bisimulation that contains the pair (p, q); thus, (p, q) ∈ . Let (p,q) ∈ S. Then, there exists a bisimulation that contains the pair (p, q); thus, (p, q) ∈ .
## A syntax for LTSs ## A syntax for LTSs
![](../../Pasted%20image%2020250414091521.png) ![](images/Pasted%20image%2020250414091521.png)
![](../../Pasted%20image%2020250414091528.png) ![](images/Pasted%20image%2020250414091528.png)
The only ingredients we used to write down an LTS are: The only ingredients we used to write down an LTS are:
- sequential composition (of an action and a process) - sequential composition (of an action and a process)
@ -117,7 +117,7 @@ For every identifier (denoted with capital letters A,B,..), we shall assume a un
Let us denote with P{b1/x1 . . . bn/xn} the process obtained from P by replacing name xi with name bi, for every i = 1,..., n. Let us denote with P{b1/x1 . . . bn/xn} the process obtained from P by replacing name xi with name bi, for every i = 1,..., n.
![](../../Pasted%20image%2020250414092202.png) ![](images/Pasted%20image%2020250414092202.png)
>[!def] >[!def]
>The set of non-deterministic processes is given by the following grammar: >The set of non-deterministic processes is given by the following grammar:
@ -133,7 +133,9 @@ We shall usually omit tail occurrences of .0 and, for example, simply writ
From the syntax to the LTS We have shown how it is possible, starting from an LTS, to generate a corresponding process From the syntax to the LTS We have shown how it is possible, starting from an LTS, to generate a corresponding process
It is also possible the inverse translation and then the two formalisms do coincide; the rules that have to be used in this translation are: It is also possible the inverse translation and then the two formalisms do coincide; the rules that have to be used in this translation are:
![](../../Pasted%20image%2020250414093017.png) ![](images/Pasted%20image%2020250414093017.png)
#### Examples
![](images/Pasted%20image%2020250414152221.png)
examples ![](images/Pasted%20image%2020250414152234.png)
![](images/Pasted%20image%2020250414152247.png)
![](images/Pasted%20image%2020250414152300.png)

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@ -15,7 +15,7 @@ Given a set of names N (that denote events)
When two processes synchronize, an external observer has no way of understanding what is happening in the system When two processes synchronize, an external observer has no way of understanding what is happening in the system
- synchronization is not observable from the outside; it produces a special silent action, that we denote with τ - synchronization is not observable from the outside; it produces a special silent action, that we denote with τ
The set of actions we shall consider is: ![150](../../Pasted%20image%2020250414102733.png) The set of actions we shall consider is: ![150](images/Pasted%20image%2020250414102733.png)
It is also useful to force some processes of the system to synchronize between them (without the possibility of showing to the outside some actions) It is also useful to force some processes of the system to synchronize between them (without the possibility of showing to the outside some actions)
@ -23,8 +23,8 @@ The restriction operator P\a restricts the scope of name a to process P (a is vi
This is similar to local variables in a procedure of an imperative program This is similar to local variables in a procedure of an imperative program
![](../../Pasted%20image%2020250414103800.png) ![](images/Pasted%20image%2020250414103800.png)
![](../../Pasted%20image%2020250414104010.png) ![](images/Pasted%20image%2020250414104010.png)
fino alla 7 compresa... fino alla 7 compresa...

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