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.
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 $S(R_1 \cap R_2) = \emptyset$