Set Difference over Subset

Theorem

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

Let $A \subseteq B$.


Then:

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


Proof

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

$\blacksquare$

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