Set Difference Intersection with First Set is Set Difference

Theorem

The intersection of the set difference with the first set is the set difference.

Let $S, T$ be sets.

Then:

$\paren {S \setminus T} \cap S = S \setminus T$

Proof 1

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

\(\subseteq\)

\(\ds S\)

Set Difference is Subset

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds S \setminus T\)

Intersection with Subset is Subset‎

$\blacksquare$

Proof 2

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

\(=\)

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

Intersection with Set Difference is Set Difference with Intersection

\(\ds \)

\(=\)

\(\ds S \setminus T\)

Set Intersection is Idempotent

$\blacksquare$