De Morgan's law gives $(A \cap B)'$ as:
A$A' \cap B'$, the intersection of complements directly here
B$A \cup B$, simply the union of the two original sets
C$A' \cup B'$, the union of the complements of $A$ and $B$
D$A - B$, the simple difference between A and B always
Answer & Solution
Correct answer: C. $A' \cup B'$, the union of the complements of $A$ and $B$
By De Morgan: $(A \cap B)' = A' \cup B'$.