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