Set Difference over Subset

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Theorem

Let $A$, $B$, and $S$ be sets.

Let $A \subseteq B$.


Then:

$A \setminus S \subseteq B \setminus S$


Proof

\(\displaystyle A \setminus S\) \(=\) \(\displaystyle A \cap \map \complement S\) Set Difference as Intersection with Complement
\(\displaystyle \) \(\subseteq\) \(\displaystyle B \cap \map \complement S\) Corollary to Set Intersection Preserves Subsets
\(\displaystyle \) \(=\) \(\displaystyle B \setminus S\) Set Difference as Intersection with Complement

$\blacksquare$