# Boolean algebra simplification

Thread Starter

#### digesh9870

Joined Nov 8, 2016
11
Can someone help me. I am still a beginner and hence this minor doubt. We know the basic boolean algebra rule that 1 + X = 1. Does this also hold good for 1 + XY and 1 + XY' ? Is the following simplification right ?
XY' (Z + YZ') + Z
= XY'Z + XY'YZ' + Z
= XY'Z + Z
= Z (XY' + 1)

#### wayneh

Joined Sep 9, 2010
17,073
We know the basic boolean algebra rule that 1 + X = 1.
Huh?

Is the following simplification right ?
XY' (Z + YZ') + Z
= XY'Z + XY'YZ' + Z
= XY'Z + Z
What happened to XY'YZ' ?

#### WBahn

Joined Mar 31, 2012
26,398
Can someone help me. I am still a beginner and hence this minor doubt. We know the basic boolean algebra rule that 1 + X = 1. Does this also hold good for 1 + XY and 1 + XY' ?
Assumptions: 1 = Logical True; + = Boolean OR operator

Yes, (True) OR'ed with (Anything) is (True).

Is the following simplification right ?
XY' (Z + YZ') + Z
= XY'Z + XY'YZ' + Z
= XY'Z + Z
= Z (XY' + 1)
Yes, as far as you've taken it. Now you just need to finish it off.

You want to start verifying the correctness of your work from the result you get. This is almost always possible in most fields of engineering.

In this case you have:

XY' (Z + YZ') + Z

And your end result (once you finish) is that this is equivalent to Z. So that means that this expression must be True whenever Z is True and it must be False whenever Z if False. So check those two conditions and see if they hold.

#### WBahn

Joined Mar 31, 2012
26,398
Huh?

What happened to XY'YZ' ?
Look at the title (plus he mentions it in the first post) -- he's talking Boolean algebra.

• digesh9870
Thread Starter

#### digesh9870

Joined Nov 8, 2016
11
Assumptions: 1 = Logical True; + = Boolean OR operator

Yes, (True) OR'ed with (Anything) is (True).

Yes, as far as you've taken it. Now you just need to finish it off.

You want to start verifying the correctness of your work from the result you get. This is almost always possible in most fields of engineering.

In this case you have:

XY' (Z + YZ') + Z

And your end result (once you finish) is that this is equivalent to Z. So that means that this expression must be True whenever Z is True and it must be False whenever Z if False. So check those two conditions and see if they hold.
Thank you . Yes, I am talking about boolean algebra . I was curious to know that 1 + X = 1 also holds good for 1 + XY = 1 & 1 + XY' = 1. So I tried solving the aforementioned expression. As you said, the final answer (simplified form) would just be Z. I checked this using truth table and XY'(Z+YZ') + Z is nothing but Z only. Please correct me if I am wrong . Thank you both for your time.

#### wayneh

Joined Sep 9, 2010
17,073
Look at the title (plus he mentions it in the first post) -- he's talking Boolean algebra.
Oops! *facepalm*

#### WBahn

Joined Mar 31, 2012
26,398
Thank you . Yes, I am talking about boolean algebra . I was curious to know that 1 + X = 1 also holds good for 1 + XY = 1 & 1 + XY' = 1. So I tried solving the aforementioned expression. As you said, the final answer (simplified form) would just be Z. I checked this using truth table and XY'(Z+YZ') + Z is nothing but Z only. Please correct me if I am wrong . Thank you both for your time.
You are correct. Clearly if Z is True then the expression is True. Now we just need to see if there is any way for the expression to be True if Z is False. If Z is False the expression collapses to XYY' and regardless of the value of Y, either Y or Y' is False.

• digesh9870

#### WBahn

Joined Mar 31, 2012
26,398
Thread Starter

#### digesh9870

Joined Nov 8, 2016
11
Once again , thank you guys . This cleared a lot of confusion for me.

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