# Simplify Boolean Function

• Jun 9th 2010, 09:00 AM
wkn0524
Simplify Boolean Function
1. $\displaystyle p= x'y' + (xy')' (z(x + z'))'$

my problem was from here im stuck and does not know to using Boolean algebra or karnaugh maps for getting simplify this question. Thanks
• Jun 9th 2010, 09:10 AM
undefined
Quote:

Originally Posted by wkn0524
1. $\displaystyle p= x'y' + (xy')' (z(x + z'))'$

I'm pretty unfamiliar with the subject, so my reply may not be very helpful, but I'm wondering: is addition the same as XOR and multiplication the same as AND? This would lead to

0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 0

0 * 0 = 0
0 * 1 = 0
1 * 0 = 0
1 * 1 = 1

If so, then a (probably slow) way of solving is to construct an 8-row truth table and then find a simpler expression from that. I'm curious to see what people more experienced with these problems post.
• Jun 9th 2010, 09:31 AM
Ackbeet
Just to use slightly more standard syntax: you want to simplify

$\displaystyle p=\overline{x}\,\overline{y}+\overline{x\overline{ y}}\:\overline{z(x+\overline{z})}$, correct?

Are you wanting to put the result in disjunctive or conjunctive normal form, or are you just trying to minimize the number of logic gates required?

Note to undefined: in digital logic, at least, the plus symbol is usually the inclusive or, and you use something like $\displaystyle \otimes$ for exclusive or.

I would probably use Karnaugh maps or DeMorgan's rules to simplify.
• Jun 9th 2010, 09:33 AM
Ackbeet
Incidentally, this question really ought to be in the Discrete Mathematics, Set Theory and Logic forum, not in Calculus.
• Jun 9th 2010, 09:38 AM
wkn0524
Ya, that the method of using Boolean algebra, but i got compare to the Boolean algebra with Karnaugh maps, the result are not the same. :(
• Jun 9th 2010, 09:40 AM
Ackbeet
Show us what you did. I'm a fan of disjunctive normal form myself. What did you get for that?
• Jun 9th 2010, 09:53 AM
wkn0524
Quote:

Originally Posted by Ackbeet
Just to use slightly more standard syntax: you want to simplify

$\displaystyle p=\overline{x}\,\overline{y}+\overline{x\overline{ y}}\:\overline{z(x+\overline{z})}$, correct?

Are you wanting to put the result in disjunctive or conjunctive normal form, or are you just trying to minimize the number of logic gates required?

Note to undefined: in digital logic, at least, the plus symbol is usually the inclusive or, and you use something like $\displaystyle \otimes$ for exclusive or.

I would probably use Karnaugh maps or DeMorgan's rules to simplify.

Ya, at first De morgan's rules is needed.
There are two steps to simplify are using the rules of Boolean Algebra and karnaugh map.
its mean that i only need to simplify the question.

Here my step:
i got change the question.

$\displaystyle p=\overline{(\overline{x} + y)(x\overline{y}+z(x + \overline{z})}$
After DeMorgan rules:

$\displaystyle p= x\overline{y} + (\overline{x} + y)(\overline{z} + \overline{x}z)$

$\displaystyle p= x\overline{y} + (\overline{x}\,\overline{z} +\overline{x}z + y\overline{z} + \overline{x}yz)$

Maybe im stuck at here.
• Jun 9th 2010, 09:56 AM
undefined
Quote:

Originally Posted by Ackbeet
Just to use slightly more standard syntax: you want to simplify

$\displaystyle p=\overline{x}\,\overline{y}+\overline{x\overline{ y}}\:\overline{z(x+\overline{z})}$, correct?

Are you wanting to put the result in disjunctive or conjunctive normal form, or are you just trying to minimize the number of logic gates required?

Note to undefined: in digital logic, at least, the plus symbol is usually the inclusive or, and you use something like $\displaystyle \otimes$ for exclusive or.

I would probably use Karnaugh maps or DeMorgan's rules to simplify.

Thanks for the info. I guessed XOR because then we would have normal addition and multiplication in $\displaystyle \mathbb{Z}/2\mathbb{Z}$. Hopefully I'll get to studying the field soon.
• Jun 9th 2010, 10:00 AM
Ackbeet
For disjunctive normal form, I got

$\displaystyle p= \overline{x}\,\overline{y} +\overline{x}\,\overline{z} +\overline{x}\,z +y\,\overline{z} +\overline{x}\,y\,z$.

Is that what you got? If so, what did you do then?
• Jun 9th 2010, 11:59 AM
p0oint
$\displaystyle p= x'y' + (x'+y) (z'+(x + z')') =x'y'+(x'+y)(z'+x'z)=$
$\displaystyle =x'y'+(x'z'+x'x'z+yz'+yx'z)=x'y'+x'z'+x'z+yz'+x'yz$

Now here is the thing you need to do so you can create the karnogh map:

$\displaystyle x'y'(z+z')+x'z'(y+y')+x'z(y+y')+yz'(x+x')+x'yz=$
$\displaystyle x'y'z'+x'y'z+x'y'z+x'yz'+x'y'z+x'yz+xyz'+x'yz'+x'y z=$
Remove the duplicates:
$\displaystyle =x'y'z'+x'y'z+x'yz'+x'yz+xyz'$

Karnough map:
1 1 1 1
0 0 0 1

Take the 4 neighbour '1' (1st row of karnough map) and 2 neighbour '1' in the right angle.

The minimization is: yz'+x'