Symmetric Difference of Complements
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Theorem
The symmetric difference of two sets equals the symmetric difference of their complements:
- $\map \complement S \symdif \map \complement T = S \symdif T$
Proof
| \(\ds \map \complement S \symdif \map \complement T\) | \(=\) | \(\ds \paren {\map \complement S \setminus \map \complement T} \cup \paren {\map \complement T \setminus \map \complement S}\) | Definition of Symmetric Difference | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {T \setminus S} \cup \paren {S \setminus T}\) | Set Difference of Complements | |||||||||||
| \(\ds \) | \(=\) | \(\ds S \symdif T\) | Definition of Symmetric Difference |
$\blacksquare$
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $1$: The Notation and Terminology of Set Theory: $\S 8 \beta$