limit supremum and infimum question 
Question:
Show that
$\limsup A_n -\liminf A_n = \limsup(A_n A^c_{n+1}) =\limsup (A^c_n A_{n+1})$

the thing I understand from this queston is the following;
$$\bigcap_{n=1}^\infty \bigcup_{n=k}^\infty A_k  \setminus \bigcup_{n=1}^\infty\bigcap_{n=k}^\infty A_k =(\bigcap_{n=1}^\infty \bigcup_{n=k}^\infty A_k) \cap (\bigcup_{n=1}^\infty\bigcap_{n=k}^\infty A_k )^c = \bigcap_{n=1}^\infty \bigcup_{n=k}^\infty ( A_kA_k^c)  $$
and I know that $A\setminus B = A\cap B^c $
Probably this way is incorrect.I'm not sure. Please show me the correct solution way. Thank you for helping.
 A: It turns out that the claim in the question is incorrect. The equality is actually
$$\limsup_{n\to\infty} A_n \setminus\liminf_{n\to\infty} A_n = \limsup_{n\to\infty} (A_n\triangle A_{n+1}). $$
We have
$$\begin{align*}
\limsup_{n\to\infty} A_n \setminus\liminf_{n\to\infty} A_n 
&= \left(\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty A_k\right)\setminus\left(\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty A_k\right)\\
&= \left(\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty A_k\right)\cap\left(\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty A_k\right)^c\\
&= \left(\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty A_k\right)\cap\left(\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty A_k^c\right)\\
&= \bigcap_{n=1}^\infty\left(\bigcup_{k=n}^\infty A_k \cap \bigcup_{k=n}^\infty A_k^c \right)\\
&= \bigcap_{n=1}^\infty\left(\left(\bigcup_{k=n}^\infty A_k \cap \bigcup_{k=n}^\infty A_{k+1}^c\right) \cup \left(\bigcup_{k=n}^\infty A_k^c\cap\bigcup_{k=n}^\infty A_{k+1}\right) \right)\\
&= \bigcap_{n=1}^\infty\left(\bigcup_{k=n}^\infty (A_k\cap A_{k+1}^c) \cup \bigcup_{k=n}^\infty (A_k^c\cap A_{k+1}) \right)\\
&= \bigcap_{n=1}^\infty\left(\bigcup_{k=n}^\infty (A_k\cap A_{k+1}^c)\cup (A_k^c\cap A_{k+1}) \right)\\
&= \bigcap_{n=1}^\infty\left(\bigcup_{k=n}^\infty A_k\triangle A_{k+1}\right)\\
&= \limsup_{n\to\infty} (A_n\triangle A_{n+1}).
\end{align*}$$
