Symmetric Difference is Subset of Union of Symmetric Differences

Theorem
Let $R, S, T$ be sets.

Then:
 * $R \symdif S \subseteq \paren {R \symdif T} \cup \paren {S \symdif T}$

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

Proof
From the definition of symmetric difference, we have:


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

Then from Set Difference is Subset of Union of Differences, we have:
 * $R \setminus S \subseteq \paren {R \setminus T} \cup \paren {T \setminus S}$


 * $S \setminus R \subseteq \paren {S \setminus T} \cup \paren {T \setminus R}$

Thus: