Simplification of boolean expression with xor I need to simplify the following boolean expression
¬(A xor B) xor (B + ¬C)
I know A xor B = ¬AB + A¬B
Then the expression will become
¬(¬AB + A¬B) xor (B + ¬C)
However, I stuck on it and I don't know how to simplify it further. Can someone give me a hint or push me in the right direction?
 A: If your expression is $¬(A \oplus B) \oplus (B \vee ¬C)$, then 
$¬(A \oplus B)=1\oplus A\oplus B$, $¬C=1\oplus C$ and $B\vee\neg C=B\oplus(\neg C)\oplus B(\neg C)$ so 
$¬(A \oplus B) \oplus (B \vee ¬C)=1\oplus A\oplus B\oplus B\oplus 1\oplus C\oplus B(1+C)=$
$A\oplus B\oplus C\oplus BC=$ 

$=A\oplus(B\vee C)$


Since $\oplus$ (XOR) is associative, and distributive over $\cdot$ (AND), and $X\oplus X=0$.
($0$ is FALSE and $1$ is TRUE)
The elements $\displaystyle\bigoplus_{i}A^{a_i}B^{b_i}C^{c_i}$, where $a_1,b_i,c_i=0,1$, is the elements of a Boolean ring and is very straightforward to manipulate the algebraic way.
A: Start by building a Karnaugh map:
$$\begin{array} {c|cc}
\text{AB \ C} & F & T  \\ \hline
           FF & F & T  \\
           FT & T & T  \\
           TT & F & F  \\
           TF & T & F   
\end{array}$$
Which looks like:
$$\begin{array} {c|cc}
\text{AB \ C} & T & T  \\ \hline
           FF & T & T  \\
           FT & T & T  \\
           TT &   &    \\
           TF &   &     
\end{array} \text{ xor }\begin{array} {c|cc}
\text{AB \ C} & F & T  \\ \hline
           FF & T &    \\
           FT &   &    \\
           TT &   &    \\
           TF & T &     
\end{array}$$
Which is $(\lnot A) \text{ xor } (\lnot B \text { and } \lnot C)$.  You can factor the $\lnot$ out to get $A \text{ xor } (B \text { or } C)$.
