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Chapter 1: Problem 3

Prove the Distributive Laws: (a) \(A \cap(B \cup C)=(A \cap B) \cup(A \cap C)\), (b) \(A \cup(B \cap C)=(A \cup B) \cap(A \cup C)\).

Short Answer

Expert verified
Both the distributive laws of set theory, namely \(A \cap (B \cup C) = (A \cap B) \cup (A \cap C)\) and \(A \cup (B \cap C) = (A \cup B) \cap (A \cup C)\), hold true for all sets A, B, and C, as proved by applying the basic principles of logic to the definitions of set intersection and union.

Step by step solution

01

Proving the first law

For the first law \(A \cap (B \cup C) = (A \cap B) \cup (A \cap C)\), first assume an element \(x\) is in \(A \cap (B \cup C)\). That means that \(x\) is in \(A\) and \(x\) is in \(B \cup C\). Which means \(x\) is in \(A\) and that \(x\) is in either \(B\) or \(C\). So evidently, \(x\) is in \(A \cap B\) or in \(A \cap C\), which means \(x\) is in \((A \cap B) \cup (A \cap C)\). So, \(A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C)\). Reversing the argument shows that any element in \((A \cap B) \cup (A \cap C)\) is also in \(A \cap (B \cup C)\), hence \(A \cap (B \cup C) = (A \cap B) \cup (A \cap C)\).
02

Proving the second law

For the second law \(A \cup (B \cap C) = (A \cup B) \cap (A \cup C)\), assume an element \(x\) is in \(A \cup (B \cap C)\). This implies that \(x\) is in \(A\) or \(x\) is in \(B \cap C\), meaning \(x\) is in \(A\) or \(x\) is in both \(B\) and \(C\). evidently, \(x\) is in \(A \cup B\) and \(A \cup C\), which means \(x\) is in \((A \cup B) \cap (A \cup C)\). So, \(A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)\). Reversing the argument demonstrates that any element in \((A \cup B) \cap (A \cup C)\) is also in \(A \cup (B \cap C)\), hence \(A \cup (B \cap C) = (A \cup B) \cap (A \cup C)\).

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