Set Difference Intersection with First Set is Set Difference

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Theorem

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


Let $S, T$ be sets.


Then:

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


Proof 1

\(\displaystyle \left({S \setminus T}\right)\) \(\subseteq\) \(\displaystyle S\) Set Difference is Subset
\(\displaystyle \implies \ \ \) \(\displaystyle \left({S \setminus T}\right) \cap S\) \(=\) \(\displaystyle S \setminus T\) Intersection with Subset is Subset‎

$\blacksquare$


Proof 2

\(\displaystyle \left({S \setminus T}\right) \cap S\) \(=\) \(\displaystyle \left({S \cap S}\right) \setminus T\) Intersection with Set Difference is Set Difference with Intersection
\(\displaystyle \) \(=\) \(\displaystyle S \setminus T\) Intersection is Idempotent

$\blacksquare$