# 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? Thanks in advance

• Here's something that might help: $\neg(A \text{ xor } B) = A \text{ xnor } B = AB + \neg A \neg B$. – Cameron Williams Sep 30 '14 at 4:12
• A Wolfram Alpha widget suggests it doesn't simplify nicely... – Cameron Williams Sep 30 '14 at 4:18

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.

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 $A \text{ xor } (B \text { or } C)$.