Intersection with Set Difference is Set Difference with Intersection

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Theorem

Let $R, S, T$ be sets.


Then:

$\left({R \setminus S}\right) \cap T = \left({R \cap T}\right) \setminus S$

where:

$R \setminus S$ denotes set difference
$R \cap T$ denotes set intersection.


Proof 1

Consider $R, S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the universe.


Then:

\(\displaystyle \left({R \setminus S}\right) \cap T\) \(=\) \(\displaystyle \left({R \cap \complement \left({S}\right)}\right) \cap T\) Set Difference as Intersection with Complement
\(\displaystyle \) \(=\) \(\displaystyle \left({R \cap T}\right) \cap \complement \left({S}\right)\) Intersection is Commutative and Intersection is Associative
\(\displaystyle \) \(=\) \(\displaystyle \left({R \cap T}\right) \setminus S\) Set Difference as Intersection with Complement

$\blacksquare$


Proof 2

\(\displaystyle \) \(\) \(\displaystyle x \in \paren {R \setminus S} \cap T\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \) \(\) \(\displaystyle \paren {x \in R \land x \notin S} \land x \in T\) Definition of Set Intersection and Definition of Set Difference
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \) \(\) \(\displaystyle \paren {x \in R \land x \in T} \land x \notin S\) Rule of Commutation and Rule of Association
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \) \(\) \(\displaystyle x \in \paren {R \cap T} \setminus S\) Definition of Set Intersection and Definition of Set Difference

$\blacksquare$