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Concurrent Systems/notes/6 - Atomicity.md

@@ -41,7 +41,8 @@ We now show that $\to$ is acyclic.
 2. it cannot have cycles with 2 edges:
 	- let's assume that $op \to op' \to op$
 	- both arrows cannot be $\to_H$ nor $\to_X$ (for some X), otw. it won't be a total order (and would be cyclic)
-	- it cannot be that one is $\to_X$ 
+	- it cannot be that one is $\to_X$ and the other $\to_Y$ (for some $X \neq Y$), otherwise op/op' would be on 2 different objects.
+	- So it must be $op \to_X op' \to_H op$
 
 
 > [!PDF|red] class 6, p.6>  we would have a cycle of length