How to simplify the Boolean function $A'B'C + A'BC' + ABC + AB'C'$? So the question I have asks to implement the circuit with $XOR$ gates.
So I am 3/4 through the problem when I am having problems simplifying the Boolean expressions below:
$$A'B'C + A'BC' + ABC + AB'C'$$
According to the professor this can be simplified to $A \,\,XOR \,\,B\,\, XOR \,\,C$.
The next one:   $$A'B'C'D + A'B'CD' + A'BC'D' + A'BCD + ABC'D + ABCD' + AB'C'D' + AB'CD $$
Can be simplified to: $A\,\,XOR \,\,B \,\,XOR\,\,C \,\,XOR\,\,D $.
Again I have no idea how to simplify it to that.
I have not the slightest clue how to even get to that. I have tried many MANY methods I must be looking at this the wrong way. Can anyone help?
 A: $$A′B′C+A′BC′+ABC+AB′C′$$
$$=A^{'}(B^{'}C + BC^{'}) + A(BC + B^{'}C^{'})$$
Now 
$X \oplus Y = XY^{'} + X^{'}Y$
$(X \oplus Y)^{'}= XY + X^{'}Y^{'}$
Thus
$$=A^{'}(B \oplus C) + A(B \oplus C)^{'}$$
$$=A \oplus B \oplus C$$
Note that $(\oplus)^{'}$ is the logical expression XNOR
A: This is very easy! If exclusive or is $A\oplus B=\bar AB+A\bar B,~$ then coincidence is $A\odot B$ 
$=AB+\bar A\bar B.~$ At the same time, the two operations are obviously opposite to one 
another; i.e., if A is the same as B, then they coincide, and their XOR yields false; 
and vice-versa. In other words, $A\odot B=\overline{A\oplus B}.~$ Now, factoring as you already 
have, $~\bar A\Big(B\oplus C\Big)+A\Big(B\odot C\Big)=\bar A\Big(B\oplus C\Big)+A\cdot\overline{B\oplus C}=A\oplus\Big(B\oplus C\Big)=$ 
$=A\oplus B\oplus C$.
A: Let's denote $\bf{XOR}$ by $\oplus$ as usual. It is in fact addition $\pmod 2$.
You can ask the question: what is the Disjunctive Normal Form of the Boolean function $(A,B,C)\mapsto A \oplus B \oplus C$. We can look at the table of values of the function
$$
\begin{array}{c|c|c|c}
 A & B & C & A\oplus B\oplus C
\\ \hline
0 & 0 & 0 & 0 \\\hline
0 & 0 & 1 & 1 \\\hline
0 & 1 & 0 & 1 \\\hline
0 & 1 & 1 & 0 \\\hline
1 & 0 & 0 & 1 \\\hline
1 & 0 & 1 & 0 \\\hline
1 & 1 & 0 & 0 \\\hline
1 & 1 & 1 & 1 \\\hline
\end{array}
$$
Now we can build the Disjunctive Normal Form of the function $A\oplus B \oplus C$. Focus on the values $(A,B,C)$ where the functions takes value $1$. For each such value, say $(0,1,0)$, consider the conjunction $A'B C'$.  Now take the disjunction ($+$) of all of these conjunctions. You get your formula.
In general, $A_1 \oplus A_2 \oplus \cdots A_n$ takes value $1$ if and only if an odd number of $A_i$ take value $1$. Therefore its disjunctive normal form is
$$\sum_{\epsilon \in \{ \  , '\}^n} \prod_{i=1}^n A_i^{\epsilon_i} $$
where the sum is over all those $\epsilon$ with an odd number of $\epsilon_i = \ $ ( the empty symbol) ( that is, for an odd number of $A_i$ we considered the variable $A_i$ and not its complement $A_i'$).
