Simplify Boolean Expression Help

Question

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?

Answer 1

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.

Answer 2

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

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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 $(x+y'+z')(x+y+z)$.