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

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- Jun 9th 2010, 09:00 AMwkn0524Simplify 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 AMundefined
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 AMAckbeet
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 AMAckbeet
Incidentally, this question really ought to be in the Discrete Mathematics, Set Theory and Logic forum, not in Calculus.

- Jun 9th 2010, 09:38 AMwkn0524
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 AMAckbeet
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 AMwkn0524
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 AMundefined
- Jun 9th 2010, 10:00 AMAckbeet
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 AMp0oint
$\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'