# Boolean Algebra simplification problem

• delta59
In summary, the conversation discusses the simplification of a Boolean function using DeMorgan's law and whether it works in all cases or only specific ones. One person suggests trying all possible combinations of variables to see if the simplification is correct, while another mentions that it is not always correct according to DeMorgan's law. The original poster is unsure and asks for clarification.

## Homework Statement

I not sure if this is the right place for this, so move me if you need.

Ok let's say I have the function xyz+xy'z' - can I pull a x out giving x(yz+y'z') and say that yz=a giving x(a+a') which is x(1) or x. If this is true that means a=yz so a'=(yz)'. Can you distribute the "not" so a'=y'z' or is this considered wrong.

I have a book question I just did and the simplification I did above worked and gave the same answer when I simplified using another method.

So my question is, does this simplification work all the time or just in specific cases. Or does it not work at all and the planets are just aligned so the answers came out the same. When I ask my professor he just smiles and says you will find out later. I would post the problem but there are 4 inputs and 8 min terms so its kind of long, but if you really want I can

thanks

delta59 said:

## Homework Statement

I not sure if this is the right place for this, so move me if you need.

Ok let's say I have the function xyz+xy'z' - can I pull a x out giving x(yz+y'z') and say that yz=a giving x(a+a') which is x(1) or x. If this is true that means a=yz so a'=(yz)'. Can you distribute the "not" so a'=y'z' or is this considered wrong.

I have a book question I just did and the simplification I did above worked and gave the same answer when I simplified using another method.

So my question is, does this simplification work all the time or just in specific cases. Or does it not work at all and the planets are just aligned so the answers came out the same. When I ask my professor he just smiles and says you will find out later. I would post the problem but there are 4 inputs and 8 min terms so its kind of long, but if you really want I can

thanks
It's easy enough to check. Just try all four possibilities of y and z. You'll find what you did isn't correct. If a=yz, then a'=(yz)' = y'+z'. This is one of DeMorgan's laws.

I know its not right by Demorgan's law, but the cryptic answer my prof gave and the fact that the problem worked out. Leads me to believe there is something to this. I was just wondering if anyone knows.

There isn't. Just try all four combinations that you can have for y and z. The two expressions are not equal.

## What is Boolean Algebra simplification problem?

Boolean Algebra simplification problem is a mathematical method used to simplify complex logical expressions in Boolean Algebra. It involves using various laws and rules to reduce a given logical expression to its simplest form.

## Why is Boolean Algebra simplification important?

Boolean Algebra simplification is important because it helps to reduce complex logical expressions into simpler forms, making them easier to understand and use in digital logic circuits. It also helps to identify redundant or unnecessary components in a logical expression, leading to more efficient and optimized circuit designs.

## What are the basic laws and rules used in Boolean Algebra simplification?

The basic laws and rules used in Boolean Algebra simplification include the commutative law, associative law, distributive law, De Morgan's laws, and the identity and complement laws. These laws help in rearranging logical expressions and simplifying them into a more manageable form.

## What is the process of simplifying a Boolean Algebra expression?

The process of simplifying a Boolean Algebra expression involves the following steps:1. Apply the commutative, associative, and distributive laws to rearrange the expression.2. Use De Morgan's laws to simplify expressions with negated variables.3. Combine like terms and eliminate redundant terms.4. Use the identity and complement laws to reduce the expression to its simplest form.

## Can Boolean Algebra simplification be applied to any logical expression?

Yes, Boolean Algebra simplification can be applied to any logical expression, as long as it follows the basic laws and rules of Boolean Algebra. However, the process may become more complex for larger and more complex expressions, and may require the use of advanced techniques such as Karnaugh maps or Quine-McCluskey method.

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