Set Difference Equals First Set iff Empty Intersection

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Theorem

$S \setminus T = S \iff S \cap T = \O$


Proof

Assume $S, T \subseteq \Bbb U$ where $\Bbb U$ is a universal set.

\(\displaystyle S \setminus T\) \(=\) \(\displaystyle S\) $\quad$ $\quad$
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle S \cap \map \complement T\) \(=\) \(\displaystyle S\) $\quad$ Set Difference as Intersection with Complement $\quad$
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle S\) \(\subseteq\) \(\displaystyle \map \complement T\) $\quad$ Intersection with Subset is Subset‎ $\quad$
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle S \cap \map \complement {\map \complement T}\) \(=\) \(\displaystyle \O\) $\quad$ Intersection with Complement is Empty iff Subset $\quad$
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle S \cap T\) \(=\) \(\displaystyle \O\) $\quad$ Complement of Complement $\quad$

$\blacksquare$


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