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Concurrent Systems/notes/6 - Atomicity.md
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@ -40,7 +40,8 @@ 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)
- 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 $\to_X$
> [!PDF|red] class 6, p.6> we would have a cycle of length