Set Difference of Complements
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Theorem
- $\map \complement S \setminus \map \complement T = T \setminus S$
Proof
\(\ds \map \complement S \setminus \map \complement T\) | \(=\) | \(\ds \set {x: x \in \map \complement S \land x \notin \map \complement T}\) | Definition of Set Difference | |||||||||||
\(\ds \) | \(=\) | \(\ds \set {x: x \notin S \land x \in T}\) | Definition of Set Complement | |||||||||||
\(\ds \) | \(=\) | \(\ds \set {x: x \in T \land x \notin S}\) | Rule of Commutation | |||||||||||
\(\ds \) | \(=\) | \(\ds T \setminus S\) | Definition of Set Difference |
$\blacksquare$