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For all set A,B and C is (A cap B)cupC =...

For all set A,B and C is `(A cap B)cupC = A cap (B cup C)`?Justify your answer.

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To determine whether the expression \((A \cap B) \cup C = A \cap (B \cup C)\) holds for all sets \(A\), \(B\), and \(C\), we will analyze both sides of the equation step by step. ### Step 1: Understand the left-hand side The left-hand side of the equation is \((A \cap B) \cup C\). 1. **Calculate \(A \cap B\)**: This is the intersection of sets \(A\) and \(B\), which includes all elements that are common to both sets. 2. **Union with \(C\)**: After finding \(A \cap B\), we take the union of this result with set \(C\). The union includes all elements from both sets, without duplicates. ### Step 2: Understand the right-hand side The right-hand side of the equation is \(A \cap (B \cup C)\). 1. **Calculate \(B \cup C\)**: This is the union of sets \(B\) and \(C\), which includes all elements that are in either set \(B\) or set \(C\). 2. **Intersection with \(A\)**: After finding \(B \cup C\), we take the intersection of this result with set \(A\). This will include all elements that are common to both \(A\) and the union of \(B\) and \(C\). ### Step 3: Provide a counterexample To show that these two expressions are not equal for all sets, we can provide a specific example: - Let \(A = \{1, 2\}\) - Let \(B = \{2, 3\}\) - Let \(C = \{3, 4\}\) Now, we calculate both sides: **Left-hand side:** 1. \(A \cap B = \{2\}\) (common element) 2. \((A \cap B) \cup C = \{2\} \cup \{3, 4\} = \{2, 3, 4\}\) **Right-hand side:** 1. \(B \cup C = \{2, 3\} \cup \{3, 4\} = \{2, 3, 4\}\) 2. \(A \cap (B \cup C) = \{1, 2\} \cap \{2, 3, 4\} = \{2\}\) ### Step 4: Compare both sides - Left-hand side: \((A \cap B) \cup C = \{2, 3, 4\}\) - Right-hand side: \(A \cap (B \cup C) = \{2\}\) Since \(\{2, 3, 4\} \neq \{2\}\), we conclude that the two sides are not equal. ### Conclusion Thus, we have shown that \((A \cap B) \cup C \neq A \cap (B \cup C)\) for the given sets \(A\), \(B\), and \(C\). Therefore, the statement is false for all sets.
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