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Concurrent Systems/notes/6 - Atomicity.md
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@ -42,7 +42,8 @@ We now show that $\to$ is acyclic.
- 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 $\to_X$ and the other $\to_Y$ (for some $X \neq Y$), otherwise op/op' would be on 2 different objects.
- So it must be $op \to_X op' \to_H op$
- **So it must b**e $op \to_X op' \to_H op$ (or vice versa)
- then, $op' \to op$ means that $r$
> [!PDF|red] class 6, p.6> we would have a cycle of length