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Chapter 1: Problem 1

Show the distributive laws $$ A \cap(B \cup C)=(A \cap B) \cup(A \cap C), \quad A \cup(B \cap C)=(A \cup B) \cap(A \cup C) $$

Short Answer

Expert verified
The distributive laws of set theory \( A \cap(B \cup C)=(A \cap B) \cup(A \cap C) \) and \( A \cup(B \cap C)=(A \cup B) \cap(A \cup C) \) are indeed valid.

Step by step solution

01

Prove the first distributive law

To prove the first distributive law \( A \cap(B \cup C)=(A \cap B) \cup(A \cap C) \), we consider an arbitrary element \( x \). For \( x \) to be in \( A \cap(B \cup C) \), \( x \) must be in both \( A \) and \( B \cup C \). If \( x \) is in \( B \cup C \), then \( x \) must be either in \( B \) or in \( C \), or in both. If \( x \) is in \( A \) and \( B \), then \( x \) is in \( A \cap B \). If \( x \) is in \( A \) and \( C \), then \( x \) is in \( A \cap C \). Hence, in both cases, \( x \) is in \( (A \cap B) \cup(A \cap C) \). So \( A \cap(B \cup C)=(A \cap B) \cup(A \cap C) \).
02

Prove the second distributive law

To prove the second distributive law \( A \cup(B \cap C)=(A \cup B) \cap(A \cup C) \), we again consider an arbitrary element \( x \). For \( x \) to be in \( A \cup(B \cap C) \), \( x \) must be either in \( A \), or in \( B \cap C \), or in both. If \( x \) is in \( B \cap C \), then \( x \) must be in both \( B \) and \( C \). If \( x \) is in \( A \) or \( B \), then \( x \) is in \( A \cup B \). If \( x \) is in \( A \) or \( C \), then \( x \) is in \( A \cup C \). Hence, in both cases, \( x \) is in \( (A \cup B) \cap(A \cup C) \). So \( A \cup(B \cap C)=(A \cup B) \cap(A \cup C) \).

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