Question

Let A, B and C be sets. Prove that (A intersection C)'= A' Union C'

Answer 1

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.

1. **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:
\[ A' \cup C' = \{ x \mid x \in A' \text{ or } x \in C' \} \]

2. **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.

3. **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:
\[
= \{ x \mid x \notin A \text{ or } x \notin C \}
\]

4. **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.