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.
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.
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.
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} $