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Concurrent Systems/notes/3.md

@@ -163,4 +163,19 @@ If $p_i$ is in the CS, it has terminated its first wait for $j$
 	
 W.r.t. the execution of $p_j$, it can be that
 - this read overlaps with `FLAG[j] <- up` by Lemma 1, `MY_TURN[i] < MY_TURN[j]` and ok
-- this read is contained within the computation of `MY_TURN[j]`, but it's not possible, since `MY_TURN` is computed with the 
+- this read is contained within the computation of `MY_TURN[j]`, but it's not possible, since `MY_TURN` is computed with the `FLAG` up
+- this read overlaps with `FLAG[j] <- down` or this read happens when $p_j$ is in the bakery:
+	- `MY_TURN[j] has been ...
+continuare
+
+**MUTEX:** $p_i$ and $p_j$ cannot simultaneously be in the CS:
+*Proof:*
+By contradiction, by Lemma 2 applied twice, we would have:
+$$⟨MY\_TURN[i] , i⟩ < ⟨MY\_TURN[j] , j⟩ \land ⟨MY\_TURN[j] , j⟩ < ⟨MY\_TURN[i] , i⟩$$
+
+#### Deadlock freedom proof
+By contradiction, assume that there is a lock but nobody enters its CS
+- All processes in the bakery (we will call this set Q) are blocked in their wait
+- The first wait cannot block forever
+	- All $p_i \in Q$ have their FLAG down
+	- All $p_i \not \in Q$ have their FLAG down