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.

Proof

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

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

$\blacksquare$