Symmetric Difference of Unions

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

$\blacksquare$