Union of Intersections of 2 from 3 equals Intersection of Unions of 2 from 3

Theorem

Let $A$, $B$ and $C$ be sets.

Then:

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

Proof

\(\ds \paren {A \cap B} \cup \paren {B \cap C} \cup \paren {C \cap A}\)

\(=\)

\(\ds \paren {B \cap \paren {A \cup C} } \cup \paren {C \cap A}\)

Intersection Distributes over Union

\(\ds \)

\(=\)

\(\ds \paren {B \cup \paren {C \cap A} } \cap \paren {\paren {A \cup C} \cup \paren {C \cap A} }\)

Union Distributes over Intersection

\(\ds \)

\(=\)

\(\ds \paren {B \cup \paren {C \cap A} } \cap \paren {C \cup A}\)

Intersection is Subset of Union, Union with Superset is Superset

\(\ds \)

\(=\)

\(\ds \paren {B \cup C} \cap \paren {B \cup A} \cap \paren {C \cup A}\)

Union Distributes over Intersection

$\blacksquare$

Sources

• 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $13$