# Minimize the following boolean equation.

#### shamieh

##### Active member
Minimize the following Boolean Equation. Write out the simplified formula (SOP FORM).

Need someone to check this. I got the answer wrong somehow.

$f${w,x,y,z) = $$\displaystyle w*x + w * \bar{x} + w * z + x * y$$

$$\displaystyle =(wx + w\bar{x})(wz + xy) =w(x + \bar{x}) =w * 1 = w(wz + xy) = wz + xy$$

Last edited:

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
$f${w,x,y,z) = $$\displaystyle w*x + w * \bar{x} + w * z + x * y$$

$$\displaystyle =(wx + w\bar{x})(wz + xy)$$
I don't understand why you replaced + between $(wx+w\bar{x})$ and $(wz+xy)$ in the original expression with * in the second expression. Which operation is in the problem statement?

$$\displaystyle =w(x + \bar{x}) =w * 1$$
What happened to $(wz+xy)$? When you write =, it should indeed mean "equal", not "I will work on one subexpression and later return to the other one".

$$\displaystyle w(wz + xy) = wz + xy$$
Here you return the second factor $(wz+xy)$.

#### shamieh

##### Active member
So what's the solution? Because I'm lost.

#### Ackbach

##### Indicium Physicus
Staff member
I would do
\begin{align*}
f&=wx + w \bar{x} + wz + xy \qquad \text{(original expression)} \\
&=w(x+ \bar{x})+wz+xy \qquad \text{(distribution law)} \\
&=w \cdot 1+wz+xy \qquad \text{(complementation or excluded middle)} \\
&=w+wz+xy \qquad \text{(identity for $\cdot$)} \\