# Thread: Boole Algebra: DNF help

1. ## Boole Algebra: DNF help

Find the DNF for:

$f(x,y,z) = y\overline{z} + (y+z)(\overline{z+\overline{x}})$

so by using the axioms and stuff I found:

$f(x,y,z)= xy\overline{z}+ \overline{x}y\overline{z}$

but by putting everything into a table and filling in 1 for the 3 $x,y,z,\overline{x},\overline{y},\overline{z}$ that give $f = 1$ I found:

$f(x,y,z) = \overline{x}y\overline{z} + x\overline{y}z + xy\overline{z} + xyz$

So what did I do wrong or did I overlook something?

2. Hello shinhidora
Originally Posted by shinhidora
Find the DNF for:

$f(x,y,z) = y\overline{z} + (y+z)(\overline{z+\overline{x}})$

so by using the axioms and stuff I found:

$f(x,y,z)= xy\overline{z}+ \overline{x}y\overline{z}$

but by putting everything into a table and filling in 1 for the 3 $x,y,z,\overline{x},\overline{y},\overline{z}$ that give $f = 1$ I found:

$f(x,y,z) = \overline{x}y\overline{z} + x\overline{y}z + xy\overline{z} + xyz$

So what did I do wrong or did I overlook something?
If the original expression was:
$f(x,y,z) = \bar{x}y\overline{z} + (y+z)(\overline{z+\overline{x}})$?
then I agree with your first answer:
$f(x,y,z)= xy\overline{z}+ \overline{x}y\overline{z}$
except that you can then simplify it further:
$f(x,y,z)= xy\overline{z}+ \overline{x}y\overline{z}$
$= (x+\bar{x})y\bar{z}$

$=y\bar{z}$
But if the original expression was correct, then I should write:
$f(x,y,z) = y\overline{z} + (y+z)(\overline{z+\overline{x}})$
$= y\overline{z} + (y+z)(\overline{z}x)$

$=y\bar{z}+xy\bar{z} +xz\bar{z}$

$=y\bar{z}(1+x)$

$=y\bar{z}$
which turns out to be the same anyway!

3. Originally Posted by Grandad
But if the original expression was correct, then I should write:
$f(x,y,z) = y\overline{z} + (y+z)(\overline{z+\overline{x}})$
$= y\overline{z} + (y+z)(\overline{z}x)$

$=y\bar{z}+xy\bar{z} +xz\bar{z}$

$=y\bar{z}(1+x)$

$=y\bar{z}$
which turns out to be the same anyway!

The original was correct...

And I found the first answer in my Mathbook $xy\overline{z}+\overline{x}y\overline{z}$ so I assume it's correct?

While the second answer I found by putting everything into a table:

$x y z | \overline{x} \overline{y} \overline{z} | f = y\overline{z}+(y+z)(\overline{z+\overline{x}})$
000|111| 0
001|110| 0
010|101| 1
100|011| 0
011|100| 0
101|010| 1
110|001| 1
111|000| 1

so I get $f = \overline{x}y\overline{z} + x\overline{y}z + xy\overline{z} + xyz$

4. I know I'm not allowed to bump, but I have exam in 9 hours... :\$

and I'm really stuck :s

5. Originally Posted by shinhidora
The original was correct...

And I found the first answer in my Mathbook $xy\overline{z}+\overline{x}y\overline{z}$ so I assume it's correct?

While the second answer I found by putting everything into a table:

$x y z | \overline{x} \overline{y} \overline{z} | f = y\overline{z}+(y+z)(\overline{z+\overline{x}})$
000|111| 0
001|110| 0
010|101| 1
100|011| 0
011|100| 0
101|010| 1
110|001| 1
111|000| 1

so I get $f = \overline{x}y\overline{z} + x\overline{y}z + xy\overline{z} + xyz$
If I'm not mistaking, the 2 red 1's are actually 0's. I didn't check the 0's.
I think it's because you missed the line that overlaps $z+\overline{x}$

Good luck for your exam

6. I fail at life... Thx Moo

7. Hello shinhidora

Sorry I wasn't able to get back to you last night. This may be a bit late for your immediate need - your exam - but let me see if I can clear this up.

The answer $f = xy\bar{z} + \bar{x}y\bar{z}$ is in full disjunctive normal form, which requires each variable to appear in every clause. However, this will clearly simplify to:
$f = xy\bar{z} + \bar{x}y\bar{z}$
$=(x+\bar{x})y\bar{z}$

$=1y\bar{z}$

$=y\bar{z}$
which is also, of course, in DNF, but not full DNF because $x$ has been eliminated.

As far as your table is concerned, Moo is right: you have made an error in your working (although I'm not sure where, since you don't show it). Look at the attached diagram, where I have shown all the working. (Note: I have used $x'$, etc to denote $\bar{x}$ as it's easier to type.)

You'll see that this does give the answer $f = xy\bar{z} + \bar{x}y\bar{z}$.