Prove $(A \cap B) \cup (A \cap B^c) = A$ I need to prove the following statement: 
$$(A \cup B) \cap (A \cup B^c) = A$$
I did the following steps:
\begin{align}
&(A \cup B) \cap (A \cup B^c) = A \\
&A \cup B \cap (A \cup A) = A \\
&A \cup B \cap A = A \quad \text{via absorption law} \\
&A = A
\end{align}
Am I correct in my reasoning? 
Thanks!
 A: No. Using distribution law$$(A \cup B) \cap (A \cup B^c) = A \cup( B \cap  B^c) = A \cup\emptyset=A$$
A: I can't follow how you got from the first step to the second step without some justification, but here is one approach:
\begin{align}
(A\cup B)\cap (A\cup B^c) & \text{Given} \\
((A\cup B)\cap A)\cup ((A\cup B)\cap B^c) & \text{Using Distributive Law} \\
((A\cap A)\cup (A\cap B)) \cup ((A\cap B^c)\cup (B\cap B^c)) & \text{Using Distributivity Again} \\
(A\cup (A\cap B)) \cup(A\cap B^c) & \text{Using $A\cup A=A$, $B\cap B^c=\emptyset$, and $\emptyset \cup C=C$} \\
A \cup (A\cap B^c) & \text{Using Absorption Law} \\
A & \text{Using Absorption Law one last time}
\end{align}
A: First some feedback.  Most importantly, I have no idea how you got from the first line to the second line.
Also, please try to avoid writing $\;X \cup Y \cap Z\;$: always write parentheses in such a case, even if common convention would allow skipping a few.  Economy of symbols becomes harmful when there is potential confusion.  And more importantly, $\;\cup\;$ and $\;\cap\;$ are perfect duals, so any precedence convention destroys symmetry.
Now, I would write down the proof Semsem's answer in a different format.  That format helps to make clear the intention of what you are doing.
We're trying to simplify $\;(A \cup B) \cap (A \cup B^c)\;$ to $\;A\;$.  Therefore, we calculate
\begin{align}
& (A \cup B) \cap (A \cup B^c) \\
= & \qquad \text{"simplify: distributive law in reverse -- bringing the $\;B\;$'s together"} \\
& A \cup (B \cap B^c) \\
= & \qquad \text{"simplify: set and complement have nothing in common"} \\
& A \cup \emptyset \\
= & \qquad \text{"simplify: adding empty set is a no-op"} \\
& A \\
\end{align}
This proves the original statement.
An advantage of this format (see EWD1300 under "The proof format") is there is explicit room to explain each of the steps.  As Dijkstra says, "it is absolutely essential that at least one line is available for each hint, so as to discourage economizing on them."
A: Your second step is invalid. $A \cup B^c \neq A \cup A$
A better option may be to distribute the intersection over the union as follow
$((A \cup B) \cap A) \cup ((A \cup B) \cap B^c)$ and work from there
