Simplify Boolean Expression Help I have a question: is this true the way I solve it?
$$\begin{align*}
xyz&+xyz'+xy'z+xy'z'+x'y'z+x'yz'\\
&= xy+xy'+x'y'z+x'yz'\\
&= x(y+y')+x'y'z+x'yz'\\
&= x+x'y'z+x'yz'\\
&= x+x'z'z\\
&=x
\end{align*}$$
Am I doing something wrong here?
 A: It clearly can’t be right, because the original expression is true when $x$ and $y$ are false and $z$ is true, and the final expression is not. 
You were doing fine up through $x+x'y'z+x'yz'$, but the next step doesn’t work. However,
$$\begin{align*}
x+x'y'z+x'yz'&=x+x'(y'z+yz')\\
&=\big(x+x(y'z+yz')\big)+x'(y'z+yz')\\
&=x+\big(x(y'z+yz')+x'(y'z+yz')\big)\\
&=x+y'z+yz'\;,
\end{align*}$$
and you’re not going to get it any simpler than that.
A: The following truth-table indicates that you are indeed doing something wrong:
 x | y | z | xyz | xyz' | xy'z | xy'z' | x'y'z | x'yz' | result | equal to x
---|---|---|-----|------|------|-------|-------|-------|--------|------------
 0 | 0 | 0 |  0  |  0   |  0   |  0    |   0   |   0   |   0    |     yes
 0 | 0 | 1 |  0  |  0   |  0   |  0    |   1   |   0   |   1    |     no
 0 | 1 | 0 |  0  |  0   |  0   |  0    |   0   |   1   |   1    |     no
 0 | 1 | 1 |  0  |  0   |  0   |  0    |   0   |   0   |   0    |     yes
 1 | 0 | 0 |  0  |  0   |  0   |  1    |   0   |   0   |   1    |     yes
 1 | 0 | 1 |  0  |  0   |  1   |  0    |   0   |   0   |   1    |     yes
 1 | 1 | 0 |  0  |  1   |  0   |  0    |   0   |   0   |   1    |     yes
 1 | 1 | 1 |  1  |  0   |  0   |  0    |   0   |   0   |   1    |     yes


An alternative solution:
This expression is an OR of $6$ out of $8$ possible ANDs.
So you can take the negation of an OR of the remaining $2$ ANDs.
The remaining $2$ ANDs are $x'yz$ and $x'y'z'$.
The negation of their OR is $(x'yz+x'y'z')'$.
Using de-morgan law, it is equivalent to $(x+y'+z')(x+y+z)$.
