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Minimize the following boolean equation.

shamieh

Active member
Sep 13, 2013
539
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:
\(\displaystyle =(wx + w\bar{x})(wz + xy)
=w(x + \bar{x})
=w * 1 = w(wz + xy) = wz + xy\)
 
Last edited:

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
$f${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

Active member
Sep 13, 2013
539
So what's the solution? Because I'm lost.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
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

Active member
Sep 13, 2013
539
Wow ach, thank you so much!