Set Intersection Distributes over Set Difference

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Theorem

Set intersection is distributive over set difference.


Let $R, S, T$ be sets.


Then:

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

where:

$R \setminus S$ denotes set difference
$R \cap T$ denotes set intersection.


Proof

\(\ds \paren {R \cap T} \setminus \paren {S \cap T}\) \(=\) \(\ds \paren {\paren {R \cap T} \setminus S} \cup \paren {\paren {R \cap T} \setminus T}\) De Morgan's Laws: Difference with Intersection
\(\ds \) \(=\) \(\ds \paren {\paren {R \cap T} \setminus S} \cup \O\) Set Difference of Intersection with Set is Empty Set
\(\ds \) \(=\) \(\ds \paren {R \cap T} \setminus S\) Union with Empty Set
\(\ds \) \(=\) \(\ds \paren {R \setminus S} \cap T\) Intersection with Set Difference is Set Difference with Intersection

$\Box$


Then:

\(\ds R \cap \paren {S \setminus T}\) \(=\) \(\ds \paren {S \setminus T} \cap R\) Intersection is Commutative
\(\ds \) \(=\) \(\ds \paren {S \cap R} \setminus \paren {T \cap R}\) from above
\(\ds \) \(=\) \(\ds \paren {R \cap S} \setminus \paren {R \cap T}\) Intersection is Commutative

$\blacksquare$


Sources

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