Intersection with Set Difference is Set Difference with Intersection

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:

\(\ds \paren {R \setminus S} \cap T\)

\(=\)

\(\ds \paren {R \cap \map \complement S} \cap T\)

Set Difference as Intersection with Complement

\(\ds \)

\(=\)

\(\ds \paren {R \cap T} \cap \map \complement S\)

Intersection is Commutative and Intersection is Associative

\(\ds \)

\(=\)

\(\ds \paren {R \cap T} \setminus S\)

Set Difference as Intersection with Complement

$\blacksquare$

Proof 2

\(\ds \)

\(\)

\(\ds x \in \paren {R \setminus S} \cap T\)

\(\ds \leadstoandfrom \ \ \)

\(\ds \)

\(\)

\(\ds \paren {x \in R \land x \notin S} \land x \in T\)

Definition of Set Intersection and Definition of Set Difference

\(\ds \leadstoandfrom \ \ \)

\(\ds \)

\(\)

\(\ds \paren {x \in R \land x \in T} \land x \notin S\)

Rule of Commutation and Rule of Association

\(\ds \leadstoandfrom \ \ \)

\(\ds \)

\(\)

\(\ds x \in \paren {R \cap T} \setminus S\)

Definition of Set Intersection and Definition of Set Difference

$\blacksquare$