If \( \mathrm{A}, \mathrm{B}, \mathrm{C} \) are any sets, prove that \( A-(B \cup C)=(A-B) \cap(A-C) \)

To prove the set equality \( A-(B \cup C)=(A-B) \cap(A-C) \), we will show that each side is a subset of the other. ### Left to Right: \( A-(B \cup C) \subseteq (A-B) \cap (A-C) \) Let \( x \in A-(B \cup C) \). This means that \( x \in A \) and \( x \notin (B \cup C) \). By the definition of the union, \( x \notin (B \cup C) \) implies that \( x \notin B \) and \( x \notin C \). Therefore, \( x \in A-B \) and \( x \in A-C \). Since \( x \in A-B \) and \( x \in A-C \), it follows that \( x \in (A-B) \cap (A-C) \). Hence, \( A-(B \cup C) \subseteq (A-B) \cap (A-C) \). ### Right to Left: \( (A-B) \cap (A-C) \subseteq A-(B \cup C) \) Let \( x \in (A-B) \cap (A-C) \). This means that \( x \in A-B \) and \( x \in A-C \). By the definition of set difference, \( x \in A-B \) implies that \( x \in A \) and \( x \notin B \). Similarly, \( x \in A-C \) implies that \( x \in A \) and \( x \notin C \). Since \( x \notin B \) and \( x \notin C \), it follows that \( x \notin (B \cup C) \). Therefore, \( x \in A \) and \( x \notin (B \cup C) \), which means \( x \in A-(B \cup C) \). Hence, \( (A-B) \cap (A-C) \subseteq A-(B \cup C) \). ### Conclusion Since we have shown that \( A-(B \cup C) \subseteq (A-B) \cap (A-C) \) and \( (A-B) \cap (A-C) \subseteq A-(B \cup C) \), we can conclude that \( A-(B \cup C) = (A-B) \cap (A-C) \).