1. **Stating the problem:** We need to prove or disprove the set theory statement $A \cap B = A \cup B$. 2. **Recall definitions:** - The intersection $A \cap B$ is the set of all elements that are in both $A$ and $B$. - The union $A \cup B$ is the set of all elements that are in $A$, or in $B$, or in both. 3. **Important rule:** For any sets $A$ and $B$, generally $A \cap B \subseteq A \cup B$, but $A \cap B$ is not equal to $A \cup B$ unless $A = B$. 4. **Check if $A \cap B = A \cup B$ can hold:** - If $A = B$, then $A \cap B = A = B$ and $A \cup B = A = B$, so equality holds. - If $A \neq B$, then there exists an element in $A$ or $B$ not in the other, so $A \cup B$ contains more elements than $A \cap B$. 5. **Conclusion:** The statement $A \cap B = A \cup B$ is generally false unless $A = B$. Therefore, the equality $A \cap B = A \cup B$ holds if and only if $A = B$. **Final answer:** $$A \cap B = A \cup B \iff A = B$$