Help to find X (logic) I need help to find X,if:

$(A \cap X)= B $ 

and

$(A \cup X)= C $ 

for $B \subseteq A\subseteq C$. 
what is X?thanks!
 A: A Venn diagram is helpful:

On the one hand $A\cap X=B$, so the only part of $X$ that’s inside the middle oval is the orange part, $B$. On the other hand, $A\cup X=C$, which is everything inside the outermost oval, so $X$ must include all of the blue. In other words, $X$ includes the orange and the blue but is disjoint from $A\setminus B$, the white band in the middle. There are several ways to describe that set in terms of $A,B$, and $C$; can you find at least one of them?
A: Logic allows us to construct a nice solution in this case, where we don't need the assumption $\;B \subseteq A \subseteq C\;$.
Starting with first equation, 
\begin{align}
& A \cap X \;=\; B \\
= & \;\;\;\;\;\text{"set extensionality; definition of $\;\cap\;$"} \\
& \langle \forall z :: z \in A \land z \in X \;\equiv\; z \in B \rangle \\
= & \;\;\;\;\;\text{"apply (what Dijkstra et al. call) the golden rule} \\
& \;\;\;\;\;\phantom"\text{-- this allows to separate out $\;X\;$ and also introduces $\;A \cup X\;$"} \\
& \langle \forall z :: z \in A \;\equiv\; z \in X \;\equiv\; z \in A \lor z \in X \;\equiv\; z \in B \rangle \\
= & \;\;\;\;\;\text{"use second equation and definition of $\;\cup\;$"} \\
& \langle \forall z :: z \in A \;\equiv\; z \in X \;\equiv\; z \in C \;\equiv\; z \in B \rangle \\
= & \;\;\;\;\;\text{"$\;\equiv\;$ is symmetric, twice -- the rightmost one is not essential"} \\
& \langle \forall z :: z \in X \;\equiv\; z \in A \;\equiv\; z \in B \;\equiv\; z \in C \rangle \\
= & \;\;\;\;\;\text{"double negation -- so that we can introduce $\;\triangle\;$ next"} \\
& \langle \forall z :: z \in X \;\equiv\; z \in A \;\not\equiv\; z \in B \;\not\equiv\; z \in C \rangle \\
= & \;\;\;\;\;\text{"definition of symmetric difference $\;\triangle\;$, twice; set extensionality"} \\
& X \;=\; A \triangle B \triangle C
\end{align}
(Here I use a not-so-well-known definition of symmetric set difference $\;\triangle\;$; see another answer of mine for more details.  I also use the fact that $\;\equiv,\not\equiv,\triangle\;$ are associative, and that $\;\equiv,\not\equiv\;$ are mutually associative.)
So regardless of $\;A,B,C\;$ the answer is $\;A \triangle B \triangle C\;$.

Update: Note that $\;A \cap X \;=\; B\;$ implies $\;B \subseteq A\;$, and $\;A \cup X \;=\; C\;$ implies $\;A \subseteq C\;$, so the assumption $\;B \subseteq A \subseteq C\;$ is already implicit in the two equations.
