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Marco Realacci 2025-04-14 17:20:00 +02:00
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Concurrent Systems/notes/12 - CCS (che non so cosa voglia dire).md
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@ -137,20 +137,5 @@ Now use the **right parallel rule**:
$$\frac{B' \xrightarrow{\bar{c}} B}{A \mid B' \xrightarrow{\bar{c}} A \mid B}$$ $$\frac{B' \xrightarrow{\bar{c}} B}{A \mid B' \xrightarrow{\bar{c}} A \mid B}$$
✅ **Third transition:** ✅ **Third transition:**
$$A \mid B' \xrightarrow{\bar{c}} A \mid B$$ $$A \mid B' \xrightarrow{\bar{c}} A \mid B$$
##### Synchronization
## 🔄 Full Transition Path I
Putting it all together, the full trace is:
1. A∣B→aA′∣BA \mid B \xrightarrow{a} A' \mid B
2. A′∣B→τA∣B′A' \mid B \xrightarrow{\tau} A \mid B'
3. A∣B′→cˉA∣BA \mid B' \xrightarrow{\bar{c}} A \mid B
So the system loops back to the starting state!
---
Let me know if you want a **state diagram in LaTeX (TikZ)** for this or a different trace involving `B'` and `A'` again!