Set Difference is Subset/Proof 1

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Theorem

$S \setminus T \subseteq S$


Proof

\(\displaystyle x \in S \setminus T\) \(\implies\) \(\displaystyle x \in S \land x \notin T\) $\quad$ Definition of Set Difference $\quad$
\(\displaystyle \) \(\implies\) \(\displaystyle x \in S\) $\quad$ Rule of Simplification $\quad$

The result follows from the definition of subset.

$\blacksquare$


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