Union minus Symmetric Difference equals Intersection
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Theorem
- $\paren {A \cup B} \setminus \paren {A \symdif B} = A \cap B$
Proof
| \(\ds \paren {A \cup B} \setminus \paren {A \symdif B}\) | \(=\) | \(\ds \paren {A \cup B} \setminus \paren {\paren {A \cup B} \setminus \paren {A \cap B} }\) | Definition 2 of Symmetric Difference | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {A \cup B} \cap \paren {A \cap B}\) | Set Difference with Set Difference | |||||||||||
| \(\ds \) | \(=\) | \(\ds A \cap B\) | Intersection of Union with Intersection |
$\blacksquare$