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Concurrent Systems/notes/12 - CCS (che non so cosa voglia dire).md

@@ -137,20 +137,5 @@ Now use the **right parallel rule**:
 $$\frac{B' \xrightarrow{\bar{c}} B}{A \mid B' \xrightarrow{\bar{c}} A \mid B}$$
 ✅ **Third transition:**
 $$A \mid B' \xrightarrow{\bar{c}} A \mid B$$
-
-## 🔄 Full Transition Path
-
-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!
+##### Synchronization
+I