Set Difference of Intersection with Set is Empty Set

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Theorem

The set difference of the intersection of two sets with one of those sets is the empty set.


Let $S, T$ be sets.


Then:

$\left({S \cap T}\right) \setminus S = \varnothing$
$\left({S \cap T}\right) \setminus T = \varnothing$


Proof

From Set Difference is Right Distributive over Set Intersection we have:

$\left({R \cap S}\right) \setminus T = \left({R \setminus T}\right) \cap \left({S \setminus T}\right)$


Hence:

\(\displaystyle \left({S \cap T}\right) \setminus S\) \(=\) \(\displaystyle \left({S \setminus S}\right) \cap \left({T \setminus S}\right)\) Set Difference is Right Distributive over Set Intersection
\(\displaystyle \) \(=\) \(\displaystyle \varnothing \cap \left({T \setminus S}\right)\) Set Difference with Self is Empty Set
\(\displaystyle \) \(=\) \(\displaystyle \varnothing\) Intersection with Empty Set

$\Box$


\(\displaystyle \left({S \cap T}\right) \setminus T\) \(=\) \(\displaystyle \left({T \cap S}\right) \setminus T\) Intersection is Commutative
\(\displaystyle \) \(=\) \(\displaystyle \varnothing\) from above

$\blacksquare$