# Simplify the boolean expression #2

## Homework Statement

Simplify the expression using boolean algebra postulates, laws, theorems.

$(\overline{\overline{x + \overline{y})(xz + \overline{y}}) + x (\overline{yz}})$

## The Attempt at a Solution

$(\overline{\overline{x + \overline{y})(xz + \overline{y}}) + x (\overline{yz}})$

1. ((x + y')(xz + y'))'' (x(yz)')'
2. (x + y')(xz + y') (x' + (yz)'')
3. (x + y')(xz + y') (x' + yz)
4. (x + y')(x + y')(z + y')(x' + yz)
5. (x + y')(z + y')(x' + yz)
6. (xz + y')(x' + yz)
7. xzx' + xzyz + y'x' + y'yz
8. xx'z + xyzz + x'y' + y'yz
9. 0*z + xyz + x'y' + 0*z
10. 0 + xyz + x'y' + 0
11. xyz + x'y'

Am I on the right track? I keep thinking that I've done something wrong here... Thanks in advance for your help.

Edit: Added steps 4-11. I believe this is the most simplified. Any critique on my methods? Perhaps there is a more efficient way to simplify this? Thanks.

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berkeman
Mentor

## Homework Statement

Simplify the expression using boolean algebra postulates, laws, theorems.

$(\overline{\overline{x + \overline{y})(xz + \overline{y}}) + x (\overline{yz}})$

## The Attempt at a Solution

$(\overline{\overline{x + \overline{y})(xz + \overline{y}}) + x (\overline{yz}})$

1. ((x + y')(xz + y'))'' (x(yz)')'
2. (x + y')(xz + y') (x' + (yz)'')
3. (x + y')(xz + y') (x' + yz)
4. (x + y')(x + y')(z + y')(x' + yz)
5. (x + y')(z + y')(x' + yz)
6. (xz + y')(x' + yz)
7. xzx' + xzyz + y'x' + y'yz
8. xx'z + xyzz + x'y' + y'yz
9. 0*z + xyz + x'y' + 0*z
10. 0 + xyz + x'y' + 0
11. xyz + x'y'

Am I on the right track? I keep thinking that I've done something wrong here... Thanks in advance for your help.

Edit: Added steps 4-11. I believe this is the most simplified. Any critique on my methods? Perhaps there is a more efficient way to simplify this? Thanks.
What are the double nots " versus the single nots ' for? Sorry if I'm missing the obvious.

Can you post the Karnaugh maps for the original problem and your solution? That's how I check my answers on these types of problems. 