Let $U = \{1, 2, 3, 4, 5, 6\}$,$A = \{2, 3\}$,and $B = \{3, 4, 5\}$. Find $A'$,$B'$,$A' \cap B'$,$A \cup B$,and hence show that $(A \cup B)' = A' \cap B'$.

(A) Given $U = \{1, 2, 3, 4, 5, 6\}$,$A = \{2, 3\}$,and $B = \{3, 4, 5\}$.
$A' = U - A = \{1, 4, 5, 6\}$.
$B' = U - B = \{1, 2, 6\}$.
$A' \cap B' = \{1, 4, 5, 6\} \cap \{1, 2, 6\} = \{1, 6\}$.
$A \cup B = \{2, 3, 4, 5\}$.
$(A \cup B)' = U - (A \cup B) = \{1, 2, 3, 4, 5, 6\} - \{2, 3, 4, 5\} = \{1, 6\}$.
Since $(A \cup B)' = \{1, 6\}$ and $A' \cap B' = \{1, 6\}$,we have $(A \cup B)' = A' \cap B'$.
This verifies De Morgan's Law.