Symmetric Difference is Commutative

Theorem

Symmetric difference is commutative:

$S \symdif T = T \symdif S$

$\ds S \symdif T$

$=$

$\ds \paren {S \setminus T} \cup \paren {T \setminus S}$

Definition of Symmetric Difference

$\ds $

$=$

$\ds \paren {T \setminus S} \cup \paren {S \setminus T}$

$\ds $

$=$

$\ds T \symdif S$

Definition of Symmetric Difference

$\blacksquare$

1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 5$: Complements and Powers
1965: J.A. Green: Sets and Groups ... (previous) ... (next): Chapter $1$. Sets: Exercise $7 \ \text{(i)}$
1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $14$
2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): symmetric difference: $\text {(iii)}$
2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): algebra of sets: $\text {(ii)}$