Set Theory/Examples/(A cap B) cup (C cap A) cup Complement (Complement A cup Complement B)

Example in Set Theory

The expression:

$\paren {A \cap B} \cup \paren {C \cap A} \cup \relcomp {\Bbb U} {\relcomp {\Bbb U} A \cup \relcomp {\Bbb U} B}$

can be simplified to:

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

Proof

\(\ds \)

\(\)

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

\(\ds \)

\(=\)

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

De Morgan's Laws: Complement of Intersection

\(\ds \)

\(=\)

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

Complement of Complement

\(\ds \)

\(=\)

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

Union is Commutative, Union is Associative, Intersection is Commutative

\(\ds \)

\(=\)

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

Set Union is Idempotent

\(\ds \)

\(=\)

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

Intersection Distributes over Union

$\blacksquare$

Sources

• 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $12$