Does $(A\cup B)\cap(C\cup D)=(A\cap C)\cup(B\cap D)$? Let $A$, $B$, $C$, and $D$ be nonempty sets. 
Does $(A\cup B)\cap(C\cup D)=(A\cap C)\cup(B\cap D)$?
It seems to be true by looking at Venn diagram, but I'm getting mixed up with the proof.
 A: You could fill in this whole truth table.  If there is one row in which the last two columns disagree, then that answers your question.
$$
\begin{array}{|c|c|c|c|c|c|} \hline
x\in A & x\in B & x\in C & x\in D & x\in(A\cup B)\cap(C\cup D) & x\in(A\cap C)\cup(B\cap D)  \\
\hline
T & T & T & T & \cdots & \cdots \\
T & T & T & f & \cdots & \cdots \\
T & T & f & T & \cdots & \cdots \\
T & f & T & T & \cdots & \cdots \\
f & T & T & T & \cdots & \cdots \\
T & T & f & f & \cdots & \cdots \\
T & f & T & f & \cdots & \cdots \\
f & T & T & f & T & f \\
T & f & f & T & \cdots & \cdots \\
f & T & f & T & \cdots & \cdots \\
f & f & T & T & \cdots & \cdots \\
T & f & f & f & \cdots & \cdots \\
f & T & f & f & \cdots & \cdots \\
f & f & T & f & \cdots & \cdots \\
f & f & f & T & \cdots & \cdots \\
f & f & f & f & \cdots & \cdots \\
\hline
\end{array}
$$
A: No. Suppose $A=D=\emptyset$. Then $(A\cup B)\cap (C\cup D)=B\cap C$, but $(A\cap C)\cup (B\cap D)=\emptyset$.
This is still false if you require the sets to be nonempty. Let $A=D=\{0\}$. Then the first set is $(B\cap C)\cup \{0\}$ and the second is at most $\{0\}$, which are not equal in general.
