Union of Symmetric Differences

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Theorem

Let $R, S, T$ be sets.


Then:

$\paren {R \symdif S} \cup \paren {S \symdif T} = \paren {R \cup S \cup T} \setminus \paren {R \cap S \cap T}$

where $R \symdif S$ denotes the symmetric difference between $R$ and $S$.


Proof

From the definition of symmetric difference, we have:

$R \symdif S = \paren {R \setminus S} \cup \paren {S \setminus R}$


Thus, expanding:

\(\ds \paren {R \symdif S} \cup \paren {S \symdif T}\) \(=\) \(\ds \paren {R \setminus S} \cup \paren {S \setminus R} \cup \paren {S \setminus T} \cup \paren {T \setminus S}\)
\(\ds \) \(=\) \(\ds \paren {\paren {R \setminus S} \cup \paren {T \setminus S} } \cup \paren {\paren {S \setminus R} \cup \paren {S \setminus T} }\)
\(\ds \) \(=\) \(\ds \paren {\paren {R \cup T} \setminus S} \cup \paren {\paren {S \setminus R} \cup \paren {S \setminus T} }\) Set Difference is Right Distributive over Union
\(\ds \) \(=\) \(\ds \paren {\paren {R \cup T} \setminus S} \cup \paren {S \setminus \paren {R \cap T} }\) De Morgan's Laws: Difference with Intersection
\(\ds \) \(=\) \(\ds \paren {\paren {R \cup S \cup T} \setminus S} \cup \paren {S \setminus \paren {R \cap T} }\) Set Difference with Union is Set Difference
\(\ds \) \(=\) \(\ds \paren {R \cup S \cup T} \setminus \paren {R \cap S \cap T}\) De Morgan's Laws for Difference with Intersection: Corollary

$\blacksquare$


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