Suppose $U$ is a set aand each $A$ and $B$ is a subset of $U$. Prove $(A \cap B)^c = A^c \cup B^c$.
\begin{enumerate}
\item $A \in B \Longrightarrow B^c \in A^c$
By remark 1, we may assume $B^c$ is not empty. Let $x \in B^c$, so $x \not\in B$. Since $A \in B$, $x \not\in B$ implies $x \not\in A$. So $x \in A^c$.
\item $B^c \in A^c \Longrightarrow A \in B$
Since $B^c \in A^c$, then by part 1. $(A^c)^c \in (B^c)^c$. By theorem 6a, $A \in B$.
\therefore $(A \cap B)^c = A^c \cup B^c$.
