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

@@ -130,18 +130,13 @@ $$A' \mid B \xrightarrow{\tau} A \mid B'$$
 
 ##### ▶️ Step 3: Transition from B'
 From the definition:
-- $BB' \triangleq \bar{c}.B$
-	- So: B′→cˉBB' \xrightarrow{\bar{c}} B
+- $B' \triangleq \bar{c}.B$
+	- So: $B' \xrightarrow{\bar{c}} B$
 
 Now use the **right parallel rule**:
-
-B′→cˉBA∣B′→cˉA∣B\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:**
-
-A∣B′→cˉA∣BA \mid B' \xrightarrow{\bar{c}} A \mid B
-
----
+$$A \mid B' \xrightarrow{\bar{c}} A \mid B$$
 
 ## 🔄 Full Transition Path