To prove that \((A \cap C)' = A' \cup C'\), we can use the definition of the complement of a set and properties of set operations.
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Definitions:
- The complement of a set \(X\), denoted \(X'\), is defined as the set of all elements not in \(X\): \[ X' = { x \mid x \notin X } \]
- The intersection of two sets \(A\) and \(C\) is defined as: \[ A \cap C = { x \mid x \in A \text{ and } x \in C } \]
- The union of two sets \(A'\) and \(C'\) is defined as: \[ A' \cup C' = { x \mid x \in A' \text{ or } x \in C' } \]
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Left-hand Side:
- We start with the left-hand side \((A \cap C)'\): \[ (A \cap C)' = { x \mid x \notin (A \cap C) } \] By the definition of intersection, this means: \[ = { x \mid x \notin A \text{ or } x \notin C } \] This statement uses De Morgan's Laws, which allow us to express the complement of an intersection in terms of the union of complements.
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Right-hand Side:
- Next, we consider the right-hand side \(A' \cup C'\): \[ A' \cup C' = { x \mid x \in A' \text{ or } x \in C' } \] By the definition of the complement, this expands to: \[ = { x \mid x \notin A \text{ or } x \notin C } \]
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Conclusion:
- Now we can see that both sides are equal: \[ (A \cap C)' = { x \mid x \notin A \text{ or } x \notin C } = A' \cup C' \]
- Therefore, we have proved that: \[ (A \cap C)' = A' \cup C' \]
Thus, the proof is complete.