Set Difference is Subset/Proof 2

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Theorem

$S \setminus T \subseteq S$


Proof

\(\ds S \setminus T\) \(=\) \(\ds S \cap \complement_S \left({T}\right)\) Set Difference as Intersection with Relative Complement
\(\ds \) \(\subseteq\) \(\ds S\) Intersection is Subset

$\blacksquare$