Union of Symmetric Differences

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

$\blacksquare$