diff --git a/Concurrent Systems/notes/6 - Atomicity.md b/Concurrent Systems/notes/6 - Atomicity.md
index 86f3ea4..d54f470 100644
--- a/Concurrent Systems/notes/6 - Atomicity.md	
+++ b/Concurrent Systems/notes/6 - Atomicity.md	
@@ -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