Set Difference with Union

Theorem

Let $R, S, T$ be sets.

Then:

$R \setminus \paren {S \cup T} = \paren {R \cup T} \setminus \paren {S \cup T} = \paren {R \setminus S} \setminus T = \paren {R \setminus T} \setminus S$

where:

$R \setminus S$ denotes set difference

$R \cup T$ denotes set union.

Illustration by Venn Diagram

Proof

Consider $R, S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the universal set.

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

$=$

$\ds \paren {R \cup T} \cap \overline {\paren {S \cup T} }$

Set Difference as Intersection with Complement

$\ds $

$=$

$\ds \paren {R \cup T} \cap \paren {\overline S \cap \overline T}$

De Morgan's Laws: Complement of Union

$\ds $

$=$

$\ds \paren {\paren {R \cup T} \cap \overline T} \cap \overline S$

Intersection is Associative and Intersection is Commutative

$\ds $

$=$

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

Set Difference as Intersection with Complement

$\ds $

$=$

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

Set Difference with Union is Set Difference

$\Box$

Then:

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

$=$

$\ds R \cap \overline {\paren {S \cup T} }$

Set Difference as Intersection with Complement

$\ds $

$=$

$\ds R \cap \paren {\overline S \cap \overline T}$

De Morgan's Laws: Complement of Union

$\ds $

$=$

$\ds \paren {R \cap \overline S} \cap \overline T$

Intersection is Associative

$\ds $

$=$

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

Set Difference as Intersection with Complement

$\Box$

Then:

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

$=$

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

Union is Commutative

$\ds $

$=$

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

from above

$\blacksquare$

Sources

1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Exercise $3.4 \ \text{(c)}$
2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $\S 2$: Problem $1 \ \text{(ii)}$