Union as Symmetric Difference with Intersection

Theorem

Let $A$ and $B$ be sets.

Then:

$A \cup B = \paren {A \symdif B} \symdif \paren {A \cap B}$

where:

$A \cup B$ denotes set union

$A \cap B$ denotes set intersection

$A \symdif B$ denotes set symmetric difference

Proof

From the definition of symmetric difference:

$\paren {A \symdif B} \symdif \paren {A \cap B} = \paren {\paren {A \symdif B} \cup \paren {A \cap B} } \setminus \paren {\paren {A \symdif B} \cap \paren {A \cap B} }$

Also from the definition of symmetric difference:

$\paren {A \symdif B} \cap \paren {A \cap B} = \paren {\paren {A \cup B} \setminus \paren {A \cap B} } \cap \paren {A \cup B}$

From Set Difference Intersection with Second Set is Empty Set:

$\paren {S \setminus T} \cap T = \O$

Hence:

$\paren {\paren {A \cup B} \setminus \paren {A \cup B} } \cap \paren {A \cup B} = \O$

This leaves:

\(\ds \)

\(\)

\(\ds \paren {\paren {A \symdif B} \cup \paren {A \cap B} } \setminus \paren {\paren {A \symdif B} \cap \paren {A \cap B} }\)

\(\ds \)

\(=\)

\(\ds \paren {\paren {A \symdif B} \cup \paren {A \cap B} } \setminus \O\)

\(\ds \)

\(=\)

\(\ds \paren {A \symdif B} \cup \paren {A \cap B}\)

Set Difference with Empty Set is Self

Then:

\(\ds \)

\(\)

\(\ds \paren {A \symdif B} \cup \paren {A \cap B}\)

\(\ds \)

\(=\)

\(\ds \paren {A \setminus B} \cup \paren {B \setminus A} \cup \paren {A \cap B}\)

Definition 1 of Symmetric Difference

\(\ds \)

\(=\)

\(\ds \paren {\paren {A \setminus B} \cup \paren {A \cap B} } \cup \paren {\paren {B \setminus A} \cup \paren {A \cap B} }\)

Set Union is Idempotent, Union is Commutative and Union is Associative

\(\ds \)

\(=\)

\(\ds A \cup B\)

Set Difference Union Intersection

Hence the result.

$\blacksquare$