Can someone simplify $\Big(\big(A^c\cup(B\cup A^c)^c\big)^c\cap(A\cup B^c)\Big)^c$ ? I'd appreciate it if it was simplified in steps as well to help me understand the process.

$$\Big(\big(A^c\cup(B\cup A^c)^c\big)^c\cap(A\cup B^c)\Big)^c$$

I reached this point and couldn't go any further: 

 A: Well we have
$$\begin{array}{cc}
& \left(\left(A^c\cup(B\cup A^c)^c\right)^c\cap(A\cup B^c)\right)^c \\
= & \left(\left(A^c\cup(B^c\cap A)\right)^c\cap(A\cup B^c)\right)^c \\
= & \left(\left((A^c\cup B^c)\cap (A^c\cup A)\right)^c\cap(A\cup B^c)\right)^c \\
= & \left(\left(A^c\cup B^c\right)^c\cap(A\cup B^c)\right)^c \\
= & \left(\left(A\cap B\right)\cap(A\cup B^c)\right)^c \\
= & \left(\left(A\cap B\cap A\right)\cup\left(A\cap B\cap B^c\right)\right)^c \\
= & \left(\left(A\cap B\right)\cup\emptyset\right)^c \\
= & \left(A\cap B\right)^c \\
= & A^c\cup B^c
\end{array}$$
In your work, you most likely got stuck when you wrote $A\cup A^c=U$, where $U$ is the "set of everything". Therefore $(A^c\cup B^c)\subset U$ and $(A^c\cup B^c)\cap U=A^c\cup B^c$.
A: Note that I'm using the following properties here:  $(X \cup Y)^{c} = X^{c} \cap Y^{c}$ and $(X \cap Y)^{c} = X^{c} \cup Y^{c}$.
\begin{split} \Big(\big(A^c\cup(B\cup A^c)^c\big)^c\cap(A\cup B^c)\Big)^c &= \Big(\underbrace{\big(A^c\cup(B\cup A^c)^c\big)^c}_{A \cap (B \cup A^{c}) = A \cap B}\cap(A\cup B^c)\Big)^c \\ &= \Big((A \cap B)\cap(A\cup B^c)\Big)^c \\ &= (A \cap B)^{c} \cup (A \cup B^{c})^{c} \\ &= A^{c} \cup B^{c} \cup \underbrace{(A^{c} \cap B)}_{\text{subset of } A^{c}} \\ &= A^{c} \cup B^{c}\end{split}
