Set Difference Subset

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Theorem

$S \setminus T \subseteq S$

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

$\blacksquare$

The result follows from the definition of subset.

	$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle S \setminus T$	$=$	$\displaystyle S \cap \complement_S \left({T}\right)$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Set Difference Relative Complement
	$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$\subseteq$	$\displaystyle S$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Intersection Subset

$\blacksquare$

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

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