Boolean expression simplification - is my solution correct?

Question

I have this boolean expression:

(x′ ∧ y ∧ z′) ∨ (x′ ∧ z) ∨ (x ∧ y)

and I simplified it using K-maps to this:

(y ∧ z′) ∨ x

Is my solution correct? Thanks!

Answer 1

Is my solution correct? $(y ∧ z′) ∨ x$

No.   You have this boolean expression:

$$\begin{align}\quad&(x′ ∧ y ∧ z′) ∨ (x′ ∧ z) ∨ (x ∧ y)\\=~&(y ∧ x′ ∧ z′) ∨ (y ∧ x) ∨ (x′ ∧ z)&&\text{by commutation}\\=~&&&\text{by distributing out the common term}\\=~&&&\\=~&&&\\=~&y\lor(x'\land z)&&\end{align}$$

Fill in the missing details.

Answer 2

I used Stanford's truth table generator to get:

x   y   z   ((¬x ∧ (y ∧ ¬z)) ∨ ((¬x ∧ z) ∨ (x ∧ y)))
F   F   F   F
F   F   T   T
F   T   F   T
F   T   T   T
T   F   F   F
T   F   T   F
T   T   F   T
T   T   T   T

Now, we know the value from x, y, and z to the result, automatically generated--no mistakes.

From here, you could opt to draw a karnaugh map, or simply look at the results.

Here's a k-map:

  y'z' y'z yz yz'
x  0    0  1  1
x' 0    1  1  1

If you notice, wherever $y$ is $1$, we get $1$. So:

$$y$$

However, there is one exception, and that's $x'y'z$, so:

$$y + x'y'z$$

Let's double check our work...

x   y   z   (y ∨ (¬x ∧ (¬y ∧ z)))
F   F   F   F
F   F   T   T
F   T   F   T
F   T   T   T
T   F   F   F
T   F   T   F
T   T   F   T
T   T   T   T