For any two sets A and B,

(c) $(A \cup B)-(A \cap B)$

$(A-B) \cup(B-A)=\left(A \cap B^{\prime}\right) \cup\left(B \cap A^{\prime}\right)$

$=\left[A \cup\left(B \cap A^{\prime}\right)\right] \cap\left[B^{\prime} \cup\left(B \cap A^{\prime}\right)\right]$ [Using distribution law]

$=\left[(A \cup B) \cap\left(A \cup A^{\prime}\right)\right] \cap\left[\left(B^{\prime} \cup B\right) \cap\left(B^{\prime} \cup A^{\prime}\right)\right] \quad[$ Using distribution law $]$

$=[(A \cup B) \cap(U)] \cap\left[(U) \cap\left(B^{\prime} \cup A^{\prime}\right)\right] \quad\left[A \cup A^{\prime}=U=B^{\prime} \cup B\right]$

$=[A \cup B] \cap\left[B^{\prime} \cup A^{\prime}\right]$

$\left[(A \cup B) \cap(U)=(A \cup B)\right.$ and $\left.(U) \cap\left(B^{\prime} \cup A^{\prime}\right)=\left(B^{\prime} \cup A^{\prime}\right)\right]$

$=[A \cup B] \cap\left[(A \cap B)^{\prime}\right] \quad\left[(A \cap B)^{\prime}=B^{\prime} \cup A^{\prime}\right]$

$=[A \cup B] \cap[(A \cup B)-(A \cap B)]$

$=[(A \cup B)-(A \cap B)]$