Set Difference is Anticommutative

Theorem

Set difference is an anticommutative operation:

$S = T \iff S \setminus T = T \setminus S = \O$

Proof

From Set Difference with Superset is Empty Set‎ we have:

$S \subseteq T \iff S \setminus T = \O$

$T \subseteq S \iff T \setminus S = \O$

The result follows from definition of set equality:

$S = T \iff \paren {S \subseteq T} \land \paren {T \subseteq S}$

$\blacksquare$

Also see

• Union is Commutative
• Intersection is Commutative
• Symmetric Difference is Commutative

Sources

• 1964: W.E. Deskins: Abstract Algebra ... (previous) ... (next): $\S 1.1$: Theorem $1.7$
• 1964: W.E. Deskins: Abstract Algebra ... (previous) ... (next): Exercise $1.1: \ 7$