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 $S$:

$A \setminus B = A \cap \relcomp S B$


Proof

\(\displaystyle A \setminus B\) \(=\) \(\displaystyle \set {x: x \in A \land x \notin B}\) $\quad$ Definition of Set Difference $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \set {x: \paren {x \in A \land x \in X} \land x \notin B}\) $\quad$ Definition of Subset, Modus Ponens and Rule of Conjunction $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \set {x: x \in A \land \paren {x \in X \land x \notin B} }\) $\quad$ Conjunction is Associative $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \set {x: x \in A \land x \in \relcomp S B}\) $\quad$ Definition of Relative Complement $\quad$
\(\displaystyle \) \(=\) \(\displaystyle A \cap \relcomp S B\) $\quad$ Definition of Set Intersection $\quad$

$\blacksquare$


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