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 |
$\blacksquare$