# Absorption Laws (Set Theory)/Union with Intersection

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## Contents

## Theorem

- $S \cup \paren {S \cap T} = S$

## Proof 1

\(\displaystyle \) | \(\) | \(\displaystyle \paren {S \cap T} \subseteq S\) | Intersection is Subset | ||||||||||

\(\displaystyle \) | \(\leadsto\) | \(\displaystyle S \cup \paren {S \cap T} = S\) | Union with Superset is Superset |

$\blacksquare$

## Proof 2

\(\displaystyle x\) | \(\in\) | \(\displaystyle S \cup \paren {S \cap T}\) | |||||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle x\) | \(\in\) | \(\displaystyle S \lor \paren {x \in S \land x \in T}\) | Definition of Set Intersection and Definition of Set Union | |||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle x\) | \(\in\) | \(\displaystyle S\) | Disjunction Absorbs Conjunction |

$\blacksquare$

## Also see

- Intersection Absorbs Union, where it is proved that $S \cap \paren {S \cup T} = S$

These two results together are known as the Absorption Laws, corresponding to the equivalent results in logic.

## Sources

- 1965: J.A. Green:
*Sets and Groups*... (previous) ... (next): Chapter $1$. Sets: Exercise $3$ - 1975: T.S. Blyth:
*Set Theory and Abstract Algebra*... (previous) ... (next): $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $3$