Minimize the following boolean equation.

[shamieh]

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) = \(\displaystyle w*x + w * \bar{x} + w * z + x * y\)

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

 

[Evgeny.Makarov]

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

My answer:
\(\displaystyle =(wx + w\bar{x})(wz + xy)\)

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]

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

 

[Ackbach]

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 $\cdot$)} \\
&=w+xy \qquad \text{(absorption law)}.
\end{align*}

 

[shamieh]

Wow ach, thank you so much!