Set Difference is Right Distributive over Set Intersection/General Case/Proof
Jump to navigation
Jump to search
Theorem
Let $U$ be a collection of sets.
Let $T$ be a set.
Then:
- $\ds \bigcap_{X \mathop \in U} \paren {X \setminus T} = \paren {\bigcap_{X \mathop \in U} X} \setminus T$
That is, the difference with an intersection equals the intersection of the differences.
Proof
\(\ds x\) | \(\in\) | \(\ds \bigcap_{X \mathop \in U} \paren {X \setminus T}\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \forall X \in U: \, \) | \(\ds x\) | \(\in\) | \(\ds X \setminus T\) | Definition of Set Intersection | |||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \forall X \in U: \, \) | \(\ds x\) | \(\in\) | \(\ds X\) | Definition of Set Difference | |||||||||
\(\, \ds \land \, \) | \(\ds x\) | \(\not \in\) | \(\ds T\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds x\) | \(\in\) | \(\ds \bigcap_{X \mathop \in U}\) | Definition of Set Intersection | ||||||||||
\(\, \ds \land \, \) | \(\ds x\) | \(\not \in\) | \(\ds T\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds x\) | \(\in\) | \(\ds \paren {\bigcap_{X \mathop \in U} X} \setminus T\) | Definition of Set Difference |
$\blacksquare$