Set Difference as Intersection with Relative Complement

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Theorem

Let $A, B \subseteq S$.


Then the set difference between $A$ and $B$ can be expressed as the intersection with the relative complement with respect to $S$:

$A \setminus B = A \cap \complement_S \left({B}\right)$


Proof

\(\displaystyle A \setminus B\) \(=\) \(\displaystyle \left\{ {x : x \in A \land x \notin B}\right\}\) $\quad$ Definition of Set Difference $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left\{ {x : \left(x \in A \land x \in X\right) \land x \notin B}\right\}\) $\quad$ Definition of Subset; Modus Ponens and Rule of Conjunction $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left\{ {x : x \in A \land \left(x \in X \land x \notin B\right)}\right\}\) $\quad$ Rule of Association for Conjunction $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left\{ {x : x \in A \land x \in \complement_S \left({B}\right)}\right\}\) $\quad$ Definition of Relative Complement $\quad$
\(\displaystyle \) \(=\) \(\displaystyle A \cap \complement_S \left({B}\right)\) $\quad$ Definition of Set Intersection $\quad$

$\blacksquare$


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