If we assume that $S_{k} = R_{k} \cup i_{A}$, then: $$\begin{align} S_{1} \cap S_{2} &= (R_{1} \cup i_{A}) \cap (R_{2} \cup i_{A}) \\&= [(R_{1} \cup i_{A}) \cap R_{2}] \cup [(R_{1} \cup i_{A}) \cap i_{A}] \\&= (R_{1} \cap R_{2}) \cup (R_{2} \cap i_{A}) \cup (R_{1} \cap i_{A})\cup(i_{A} \cap i_{A}) \\&= (R_{1} \cap R_{2}) \cup (R_{2} \cap i_{A}) \cup (R_{1} \cap i_{A})\cup i_{A} \\&= [(R_{1} \cap R_{2}) \cup i_{A}] \cup (R_{1} \cap i_{A}) \cup (R_{2} \cap i_{A}) \\&= (R \cup i_{A}) \cup (...) \\&= S \cup (...) \end{align}$$
So $S \subset S_{1} \cap S_{2} $
I checked a couple of solutions online: One proved that $S_{1} \cap S_{2} = S$ with very large jumps in logic, so I couldn't quite follow. Another one did the procedure I showed and concluded that $S_{1} \cap S_{2} \subset S$. I ended up concluding that $S \subset S_{1} \cap S_{2}$ thus $S = S_{1} \cap S_{2}$. But I'm having trouble believing it, because if $A = B \cup (...)$ then how can $A$ and $B$ be equal?
Here's my attempt at checking that $S_{1} \cap S_{2} \subset S $:
Suppose $(x,y) \in S_{1} \cap S_{2}$. Then $(x,y)\in S_{1}$ and $(x,y)\in S_{2}$.
Take $(x,y) \in S_{1}$. Either $(x,y) \in R_{1}$ or $(x,y) \notin R_{1}$.
Suppose $(x,y) \notin R_{1}$. Either $x = y$ or $x \neq y$. If $x = y$, then $(x,y) \in S$, because $S$ is a reflexive closure.
If $x \neq y$, then it can be shown that $T = S - {(x,y)}$ is the reflexive closure of $R_{1}$, which is a contradiction.
The case for $R_{2}$ is analogous. Since we have proven that $(x,y) \in S$ whenever $(x,y) \notin R_{1}$ or $(x,y) \notin R_{2}$, all that's left is to prove that $(x,y) \in S$ whenever $(x,y) \in R_{1}$ and $(x,y) \in R_{2}$.
Suppose $(x,y) \in R_{1}$ and $(x,y) \in R_{2}$. Then $(x,y) \in R_{1} \cap R_{2} = R$. Since $S$ is the reflexive closure of $R$, $R \subset S$. So $(x,y) \in S$.
Thus, $S_{1} \cap S_{2} \subset S$.
Is this right? I can't see where I could have gone wrong, but the fact that $S_{1} \cap S_{2} = S \cup (...)$ is throwing me off.
Thanks.
For immediate results use distributivity:
(A $\cup$ D) $\cap$ (B $\cup$ D) = (A $\cap$ B) $\cup$ D.