vault backup: 2025-03-18 15:50:04

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Marco Realacci 2025-03-18 15:50:04 +01:00
parent 4bc441c75d
commit 36a06a3c35

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@ -32,6 +32,7 @@ $\hat{H}$ is linearizable if $\hat{H}|_{X}$ is linearizable, for all X in H
For all X, let $\hat{S}_{X}$ be a linearization of $\hat{H}_{X}$
- $\hat{S}_{X}$ defines a total order on the operations on X (call it $\to_{X}$)
- a union of relations is a union of pairs!
Let $\to$ denote $\to_{H} \cup \bigcup_{X \in H} \to _{X}$
@ -43,7 +44,8 @@ We now show that $\to$ is acyclic.
- 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 b**e $op \to_X op' \to_H op$ (or vice versa)
- then, $op' \to op$ means that $r$
- then, $op' \to op$ means that $res(op') <_H inv(op)$
- Since $\hat{S}_X$ is a linearization of $\hat{H}|_X$ and op/op' are on X, this implies $res(op') <_X inv(op)$, which means that $op' \to_X op$, and so $\to_X$ would be cyclic.
> [!PDF|red] class 6, p.6> we would have a cycle of length