Set Theory Proof Help needed. I have a question in my exercises that I am lost on. It is a set theory question, simple enough. However, it has a constraint on it that has made it dubiously more difficult for me. I will write it out in its entirety below.
Let $R,S,T$ be sets. Without using Theorem $6.24$, prove that
$(R \cap S)∪(R \cap T) \subseteq R \cap (S \cup T)$.
Theorem $6.24$ (as per my textbook) is Distributivity for Sets, the property that allows me to simplify and prove this quickly. Without this tool, I'm helpless with this proof. Any and all help is appreciated! Stay safe guys.
 A: Let $x\in(R∩S)∪(R∩T)$. 
Thus, $x\in R$.
Also, $x\in S$ or $x\in T$.
Thus, $x\in S∪T$ and from here $x\in  R∩(S∪T).$
A: You need to show that if $x\in (R \cap S)\cup (R \cap T)$ then $x\in R\cap (S\cup T)$. Try doing this using the definition of $\cap$ and $\cup$.
I'll help you get started - if $x$ belongs to the LHS, then $x$ either $x$ is in both $R$ and $S$, or $x$ is in both $R$ and $T$. What does this tell you about $x$? How can you show it belongs to the set on the RHS using this information?
If you need more help let me know.
A: Element-chasing:
Let $x \in (R\cap S)\cup (R \cap T)$;
Then $x \in (R \cap S)$ or  $\in (R \cap T)$.
1) $x \in (R \cap S):$ 
Then $x \in R$ and  $x \in S$, and  
since $S\subset S \cup T,$  we have $x \in S\cup T$;
Summing up:  $x \in (R \cap S)$ then
$x \in R$ and $x \in S \cup T,$ i.e 
$x \in R \cap (S \cup T)$
2) $x \in (R \cap T).$ Can you finish?
A: Below a proof , with an emphasise on logical structure 
Dsjunctive syllogism is the following inference rule : 
from (X OR Y) and ( not-X) , infer : Y

