# Boolean Algebra simplification problem

## Homework Statement

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

Ok lets 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

## Answers and Replies

vela
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Science Advisor
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## Homework Statement

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

Ok lets 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.

vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
There isn't. Just try all four combinations that you can have for y and z. The two expressions are not equal.