Set Difference is Right Distributive over Set Intersection/Proof 1

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Theorem

$\paren {A \cap B} \setminus C = \paren {A \setminus C} \cap \paren {B \setminus C}$


Proof

Consider $A, B, C \subseteq \mathbb U$, where $\mathbb U$ is considered as the universe.


\(\ds \paren {A \cap B} \setminus C\) \(=\) \(\ds \paren {A \cap B} \cap \map \complement C\) Set Difference as Intersection with Complement
\(\ds \) \(=\) \(\ds \paren {A \cap B} \cap \paren {\map \complement C \cap \map \complement C}\) Set Intersection is Idempotent
\(\ds \) \(=\) \(\ds \paren {A \cap \map \complement C} \cap \paren {B \cap \map \complement C}\) Intersection is Associative and Intersection is Commutative
\(\ds \) \(=\) \(\ds \paren {A \setminus C} \cap \paren {B \setminus C}\) Set Difference as Intersection with Complement

$\blacksquare$