vault backup: 2025-03-18 16:20:04

Concurrent Systems/notes/6 - Atomicity.md

@@ -51,7 +51,9 @@ We now show that $\to$ is acyclic.
 	- by contradiction, consider a shortest cycle
 		- adjacent edges cannot belong to the same order (e.g. not both $\to_X$), otw. the cycle would be shortable, because of transitivity of the total order!
 		- adjacent edges cannot belong to orders on different objects
-			- this would mean that an operation is involved in both $\to_X$ and $\to_Y$
+			- this would mean that an operation is involved in both $\to_X$ and $\to_Y$ but it is not possible of course
+		- Hence, at least one $\to_X$ exists and it must be between two $\to_H$ i.e.: $$op1 \to_H op2 \to_X op3 \to_H op4$$
+		- can this be a cycle?
 
 
 > [!PDF|red] class 6, p.6>  we would have a cycle of length