Set Difference is Right Distributive over Union

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Theorem

Set difference is right distributive over union.


Let $R, S, T$ be sets.


Then:

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

where:

$R \setminus S$ denotes set difference
$R \cup T$ denotes set union.


Proof

\(\displaystyle \paren {R \cup S} \setminus T\) \(=\) \(\displaystyle \paren {R \cup S} \cap \overline T\) Set Difference as Intersection with Complement
\(\displaystyle \) \(=\) \(\displaystyle \paren {R \cap \overline T} \cup \paren {S \cap \overline T}\) Intersection Distributes over Union
\(\displaystyle \) \(=\) \(\displaystyle \paren {R \setminus T} \cup \paren {S \setminus T}\) Set Difference as Intersection with Complement

$\blacksquare$