Symmetric Difference of Unions
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Theorem
Let $R$, $S$ and $T$ be sets.
Then:
- $\paren {R \cup T} \symdif \paren {S \cup T} = \paren {R \symdif S} \setminus T$
where:
- $\symdif$ denotes the symmetric difference
- $\setminus$ denotes set difference
- $\cup$ denotes set union
Proof
\(\ds \paren {R \cup T} \symdif \paren {S \cup T}\) | \(=\) | \(\ds \paren {\paren {R \cup T} \setminus \paren {S \cup T} } \cup \paren {\paren {S \cup T} \setminus \paren {R \cup T} }\) | Definition 1 of Symmetric Difference | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\paren {R \setminus S} \setminus T} \cup \paren {\paren {S \setminus R} \setminus T}\) | Set Difference with Union | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\paren {R \setminus S} \cup \paren {S \setminus R} } \setminus T\) | Set Difference is Right Distributive over Union | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {R \symdif S} \setminus T\) | Definition 1 of Symmetric Difference |
$\blacksquare$