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 |
$\blacksquare$