# Thread: Boolean Simplify : (x+y)'(x'+y') = ?

1. ## Boolean Simplify : (x+y)'(x'+y') = ?

Hello,

I have a question about simplifying this expression:

Here is my progress, but i cannot seem to get the answer --> x'y'

(x+y)'(x'+y')

= (x'y') (x'+y')
= (x'y' * x') + (x'y' * y')
= (x''y') + (x'y'')
= (xy') + (x'y)
= ........................

Where have i gone wrong ?

Thanks.

2. ## Re: Boolean Simplify : (x+y)'(x'+y') = ?

x'x' does not equal x

x'x' = x'

3. ## Re: Boolean Simplify : (x+y)'(x'+y') = ?

Originally Posted by anpk05
I have a question about simplifying this expression:
Here is my progress, but i cannot seem to get the answer --> x'y'
(x+y)'(x'+y') = (x'y') (x'+y')
= (x'y' * x') + (x'y' * y')
= (x''y') + (x'y'')
= (xy') + (x'y)
\begin{align*}(x+y)'(x'+y')&=(x'y')(x'+y')\\&=(x'y 'x'+x'y'y')\\&=(x'x'y'+x'y'y')\\&=(x'y'+x'y')\\&=( x'y')\end{align*}

4. ## Re: Boolean Simplify : (x+y)'(x'+y') = ?

Originally Posted by abender
x'x' does not equal x

x'x' = x'
Thanks for that, i don't know how i came up with that. Since i knew x * x = x.

5. ## Re: Boolean Simplify : (x+y)'(x'+y') = ?

Originally Posted by Plato
\begin{align*}(x+y)'(x'+y')&=(x'y')(x'+y')\\&=(x'y 'x'+x'y'y')\\&=(x'x'y'+x'y'y')\\&=(x'y'+x'y')\\&=( x'y')\end{align*}
How does (x'y' + x'y' ) = (x'y') ?

Thanks

6. ## Re: Boolean Simplify : (x+y)'(x'+y') = ?

The property that i know is, (x+y)' = (x'y')

But how can (x'y' + x'y') = (x'y') ? hmm......

7. ## Re: Boolean Simplify : (x+y)'(x'+y') = ?

Originally Posted by anpk05
The property that i know is, (x+y)' = (x'y')

But how can (x'y' + x'y') = (x'y') ? hmm......
Suppose z is (x'y').

(x'y' + x'y') = (z + z) = z = (x'y')

(x' AND y') OR (x' AND y') = (x' AND y')