# Boolean algebra help

• Nov 19th 2009, 06:11 AM
Jones
Boolean algebra help
Hi,

I have a boolean function that im not sure how to solve.

$\displaystyle (x+z')'*(y+z')' + xy' +xz'$

im supposed to write this on disjunctive normal form.

I tried multiplying the $\displaystyle xy'$ and the $\displaystyle xz'$ term
by (z+z') and (y+y') but that makes the whole expression really long.

Any ideas?

Jones
• Nov 19th 2009, 08:41 PM
qmech
A start to rewriting this
I assume (x+z')' means not ( x or (not z)). If so, then doesn't this simplify to
x'z? Then your expression simplifies to:

= x'zy'z + xy' + xz'
= x'y'z + xy' + xz'
• Nov 20th 2009, 09:53 AM
Jones
No, thats incorrect.

The correct answer should be $\displaystyle x'y'z + xy'z' + xy'z + xyz'$
• Nov 20th 2009, 09:59 AM
qmech
A little more manipulation
Note:
xy' = xy'z + xy'z'
because z + z'=1
• Nov 21st 2009, 12:50 PM
Jones
So,

Simplifying the first two paranthesis
$\displaystyle \overline{x+\overline{z}} \ast \overline{y+\overline{z}}$

Should simplify down to just $\displaystyle \overline{x}z \ast \overline{y}z$

Applying Morgans law gives:

$\displaystyle \overline{\overline{x}z \ast \overline{y}z} \rightarrow x+\overline{z} + y+\overline{z}$

Using de morgans again gives:$\displaystyle \overline{x}z+\overline{y}z$

The whole expression now becomes $\displaystyle \overline{x}z+\overline{y}z+x\overline{y}+x\overli ne{z}$

Is this correct?
• Nov 21st 2009, 01:43 PM
mosta86
you need my advice it is a 3 literals function so it is easy to build the truth table and see where the function set to 1 then build the karnough map and you will get the most simplified form
• Nov 21st 2009, 01:50 PM
mosta86
the simplified form is z'x+zy'

truth table :
X Y Z F
0 0 0 0
0 0 1 1
0 1 0 0
0 1 1 0
1 0 0 1
1 0 1 1
1 1 0 1
1 1 1 0

K-map :
xy 00 01 11 10
z 0 0 0 (1 1)
1 1) 0 0 (1
=> z'x+zy'