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The content is given two relationships: R₁ and R₂ prove that s(R₁ ∩ R₂)=s(R₁) ∩ s(R₂)

My teacher has taught us the UNION versions in class, and I figure it's easy. Also I have already finished the other two intersection versions( transitive and reflexive closure), but the symmetric practice stucks me a lot,I found if expand one side directly ,it seems impossible to testify that they are equal.

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closed as unclear what you're asking by Jean-Claude Arbaut, Mike Earnest, Joshua Mundinger, Lord Shark the Unknown, Lee David Chung Lin Apr 20 at 1:17

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Let’s prove, that $S(R_1 \cap R_2) \subset S(R_1) \cap S(R_2)$:

If $(a, b) \in S(R_1 \cap R_2)$, that means that either $(a, b) \in R_1 \cap R_2$ or $(b, a) \in R_1 \cap R_2$. That means that either $(a, b)$ or $(b, a)$ is in both $R_1$ and $R_2$ at the same time. And that can be rephrased as $(a, b)$ being in both $S(R_1)$ and $S(R_2)$ at the same time, which is exactly $(a, b) \in S(R_1) \cap S(R_2)$.

However, as pointed in the comments, the converse does not hold: If $R_1 = \{(a, b)\}$ and $R_2 = \{(b, a)\}$, then $S(R_1) \cap S(R_2) = \{(a, b), (b, a)\}$, but $S(R_1 \cap R_2) = \emptyset$

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  • $\begingroup$ I guess I understand, so I should testify that they contain each other? $\endgroup$ – heiheihei hahaha Apr 19 at 10:27
  • $\begingroup$ Hmm. I thought the other containment is the difficult one. In fact, isn't it false? If $R_1=\{(a,b)\}$ and $R_2=\{(b,a)\}$ then $S(R_1)\cap S(R_2)=\{(a,b),(b,a)\}$ but $S(R_1\cap R_2)$ is empty unless I misunderstood something. $\endgroup$ – Jyrki Lahtonen Apr 19 at 19:12
  • $\begingroup$ @JyrkiLahtonen, actually, it was me, who made the mistake. My proof works only in one direction (the only one in which OP’s statement seems to be true). $\endgroup$ – Yanior Weg Apr 19 at 20:38

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