Set problem trying to reduce a messy set equation Prove that any set equation involving one unknown set X whose RHS is ∅ can be
rewritten as (A ∩ X) ∪ (B ∩ X') = ∅ where neither A nor B involve X . (Assume that the equation is written using only ∩ , ∪ , and complement. Notice also by the second proof in part a that this means that this equation is equivalent to two
simultaneous equations A ∩ X = ∅ and B ∩ X' = ∅.)
I have attempted many solutions. But, I just don't seem to find out how to start with this here. This is partly because it could be any set equation! It is obvious that A is disjunct to X and B is disjunct with X'. However, I do not know how to start off with a completely unknown equation and lead to something, which looks like (A ∩ X) ∪ (B ∩ X') = ∅. I have asked around, but no one seems to be able to find a solution for this.
Any suggestions?
 A: You need to do a proof by induction on the term $\phi(X)$. Start with the atomic terms:
\begin{eqnarray}
X \\
C 
\end{eqnarray}
where $C$ is a known (i.e. a constant - independent of $X$) set.
Next, suppose that $\psi(X)$ and $\theta(X)$ can be both expressed in the desired form.  All you have to do is prove that $\phi(X)$ can also be expressed in this form when 


*

*$\phi(X)=(\psi(X))^{c}$

*$\phi(X)=\psi(X)\cap\theta(X)$

*$\phi(X)=\psi(X)\cup\theta(X)$ 



Edited to include example:
For example, for the atomic term $X$, put $A=\mathcal{U}$ (the universe) and $B=\emptyset$.  By doing this, we have found $A$ and $B$ such that $X=(A\cap X) \cup (B\cap X^{c})$. The equation can be rewritten:
$$(\mathcal{U}\cap X)\cup (\emptyset\cap X^{c})=\emptyset$$
which is clearly equivalent to $X=\emptyset$, and so we are done.
For the atomic term $C$, put $A=B=C$.  Then we have
$$(C\cap X)\cup (C\cap X^{c})=C$$
so $$(C\cap X)\cup (C\cap X^{c})=\emptyset \iff  C=\emptyset$$ as required.
Now, for condition 1:
Suppose that we have $A_{1}$ and $B_{1}$ such that $\psi(X)$ is equivalent to $$(A_{1}\cap X)\cup (B_{1}\cap X^{c}).$$
Now consider the term 
\begin{eqnarray}
\psi(X)^{c}&=&(A_{1}\cap X)^{c}\cap (B_{1}\cap X^{c})^{c}\\
&=&(A_{1}^{c}\cup X^{c})\cap (B_{1}^{c}\cup X) 
\end{eqnarray}
If you do some distributions, you should be able to get this into the desired form, which is  $$(A_{2}\cap X)\cup(B_{2}\cap X^{c}).$$ Note that $A_{2}$ and $B_{2}$ will depend on $A_{1}$ and $B_{1}$. This is OK, since by the induction hypothesis, these terms don't depend on $X$. 
A: I think you have to solve it by parts. Since the set is empty, that implies that (A ∩ X) = ∅ and (B ∩ X) = ∅. Since there is a union operator that is balancing both operations, that implies that both need to be equal to ∅ for this to be True. 
Hope this helps.
