Error while simplifying boolean expression Expression $((A \oplus B) \land \lnot C) \lor (\lnot(A \oplus B) \land C)$ simplifies to $A \oplus B \oplus C$.
This is my attempt at simplification:
1. $((A \oplus B) \land \lnot C) \lor (\lnot(A \oplus B) \land C)$
2. $(((A \land \lnot B) \lor (\lnot A \land B)) \land \lnot C) \lor (\lnot ((A \land \lnot B) \lor (\lnot A \land B)) \land C)$
3. $((\lnot C \land A \land \lnot B) \lor (\lnot C \land \lnot A \land B)) \lor (C \land \lnot A \land B) \lor (C \land A \land \lnot B)$
4. $(\lnot A \land ((\lnot C \land B) \lor (C \land B))) \lor (\lnot B \land ((\lnot C \land A) \lor (C \land A))$
5. $(\lnot A \land (B \land (\lnot C \lor C))) \lor (\lnot B \land (A \land (\lnot C \lor C)))$
6. $(\lnot A \land B \land \top) \lor (\lnot B \land A \land \top)$
7. $(\lnot A \land B) \lor (\lnot B \land A)$
8. $A \oplus B$
What am I doing wrong? It looks to me like everything is correct, but my book says that $A \oplus B \oplus C$ is the simplest form of this expression. If you know the correct way to simplify this, please write it in step by step form.
 A: Well.. Since you tried it's probably OK to give you the answer.
((A⊕B)∧¬C)∨(¬(A⊕B)∧C)
(D∧¬C)∨(¬D∧C)
D⊕C - Reverse XOR
A⊕B⊕C - OK because XOR is associative
A: Seeing how you go from line 1 to line 2, you must have the equivalence rule:
$$P \oplus Q \Leftrightarrow (P \land \neg Q) \lor (\neg P \land Q)$$
But if you have that, then $((A \oplus B) \land \lnot C) \lor (\lnot(A \oplus B) \land C) \Leftrightarrow A \oplus B \oplus C$ is just an instance of this very equivalence.
OK, but your question was: what did you do wrong?
Well, you made a mistake in going from line 2 to line 3.  The second conjunct of line 2 is:
$$(\lnot ((A \land \lnot B) \lor (\lnot A \land B)) \land C)$$
and on line 3 you rewrote that as:
$$(C \land \lnot A \land B) \lor (C \land A \land \lnot B)$$
but that is not correct. Instead you get:
$$(\lnot ((A \land \lnot B) \lor (\lnot A \land B)) \land C) \Leftrightarrow$$
$$(\lnot (A \land \lnot B) \land \lnot (\lnot A \land B) \land C) \Leftrightarrow$$
$$((\lnot A \lor B) \land (A \lor \lnot B) \land C)$$
So basically from line 3 on everything you had was wrong.
A: Do not try to jump over two hurdles at once.   Just take it one step at a time.
Also, keep the end goal in sight. 
$\begin{split} & (A\oplus B)\oplus C
\\
=~&(~(A\oplus B)\land\lnot C~)\lor(\lnot(A\oplus B)\land C)
\\
=~&(~(~(A\land\lnot  B)\lor(\lnot A\land B)~)\land\lnot C~) ~\lor~(~\lnot(A\oplus B)\land C~)
\\
=~&{(A\land\lnot B\land\lnot C)\lor(\lnot A\land B\land\lnot C) ~\lor~(~\lnot(A\oplus B)\land C~)}
\\
\vdots~~&
\\
=~&(A\land\lnot B\land\lnot C)\lor(\lnot A\land B\land\lnot C) ~\lor~ (\lnot A\land\lnot B\land C)\lor(A\land B\land C)
\\
=~&A\oplus B\oplus C
\\ \hline
\therefore \quad &(A\oplus B)\oplus C = A\oplus B\oplus C
\end{split}\\ $
