# Simplify the boolean expression #2

• larry21
In summary, the task at hand is to simplify an expression using boolean algebra postulates, laws, and theorems. The expression is (\overline{\overline{x + \overline{y})(xz + \overline{y}}) + x (\overline{yz}}). The simplified expression is xyz + x'y'. The steps taken to simplify the expression are as follows: 1. Applying De Morgan's Law, 2. Distributive Property, 3. Distributive Property again, 4. Factoring, 5. Distributive Property, 6. Distributive Property, 7. Simplifying using the identity law, 8. Simplifying using the identity law again, 9. Applying
larry21

## 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.

Last edited:
larry21 said:

## 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.

## 1. What is a boolean expression?

A boolean expression is a mathematical expression that can only have one of two values: true or false. It is often used in computer programming and logic to make decisions based on certain conditions.

## 2. Why is it important to simplify a boolean expression?

Simplifying a boolean expression makes it easier to understand and evaluate. It can also help reduce the number of steps needed to solve a problem or make a decision based on the expression.

## 3. How do you simplify a boolean expression?

To simplify a boolean expression, you can use algebraic laws and rules, such as the distributive property, De Morgan's laws, and the laws of negation. You can also use truth tables or Karnaugh maps for more complex expressions.

## 4. What is the benefit of simplifying a boolean expression using algebraic laws?

Using algebraic laws to simplify a boolean expression can help reduce the expression to its simplest form, making it easier to understand and evaluate. It can also help identify any errors or inconsistencies in the expression.

## 5. Can a boolean expression be simplified further if it contains both AND and OR operators?

Yes, a boolean expression can be simplified further if it contains both AND and OR operators. In this case, you can use the distributive property to expand the expression and then apply other simplification rules to reduce it to its simplest form.

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