Single Symmetric Formulation of Generalized DeMorgan's Formula? Sample exercise/riddle
2.9 from an online text, Elementary Topology: Problem Textbook, on Topology on page 13 suggests that there is a single symmetric formulation that combines both generalized formulations of De Morgan's law.

"These formulas are nonsymmetric cases of a single formulation, which contains in a symmetric way sets and their complements, unions, and intersections. 
2.9. Riddle.Find such a formulation."

Anyone have any thoughts?
Thanks in advance.
 A: From
$$X\setminus\bigcup_{A\in \Gamma} A=\bigcap_{A\in \Gamma} \left(X\setminus A\right)\quad (1)$$
$$X\setminus\bigcap_{A\in \Gamma} A=\bigcup_{A\in \Gamma} \left(X\setminus A\right)\quad (2)$$
we can get
$$\begin{align}
\left(X\setminus\bigcap_{A\in \Gamma} A\right)\setminus \left(X\setminus\bigcup_{A\in \Gamma} A\right) &= \left(\bigcup_{A\in \Gamma} \left(X\setminus A\right)\right)\setminus \left(\bigcap_{A\in \Gamma} \left(X\setminus A\right)\right)\\
\left(\bigcup_{A\in \Gamma} A\right)\setminus \left(\bigcap_{A\in \Gamma} A\right)&= \left(\bigcup_{A\in \Gamma} \left(X\setminus A\right)\right)\setminus \left(\bigcap_{A\in \Gamma} \left(X\setminus A\right)\right) \quad (3)
\end{align}$$
Notice that both $(1)$ and $(2)$ are special cases of $(3)$.

Proof of $(3)\Rightarrow (2)$: Let $\Gamma'=\Gamma\cup\{X\}$. Then applies $(3)$.
Proof of $(3)\Rightarrow (1)$: Let $\Gamma''=\Gamma\cup\{\emptyset\}$. After appling $(3)$, we get $\displaystyle \bigcup_{A\in \Gamma} A=X\setminus \left(\bigcap_{A\in \Gamma}(X\setminus A)\right)$. Taking complements of both sides give us $(1)$.

Summary: $$\bbox[5px,border:2px solid red]
{\left(\bigcup_{A\in \Gamma} A\right)\setminus \left(\bigcap_{A\in \Gamma} A\right)= \left(\bigcup_{A\in \Gamma} A^c\right)\setminus \left(\bigcap_{A\in \Gamma} A^c\right)}$$
is the desired symmetric formulation.
