How Do You Simplify This Boolean Expression Using DeMorgan's Law?

[genu]

Homework Statement

I'm trying to understand an example in the book. They're simplifying the following expression:

<br /> S=z \oplus (x \oplus y)
=z&#039;(xy&#039;+x&#039;y)+z(xy&#039;+x&#039;y)&#039;
I don't get how they go to this from the above
=z&#039;(xy&#039;+x&#039;y)+z(xy+x&#039;y&#039;)
=xy&#039;z&#039;+x&#039;yz&#039;+xyz+x&#039;y&#039;z

Homework Equations

DeMorgan law: (x+y)' = x'y'

The Attempt at a Solution

By DeMorgan's law, negating that last term should give z(z'y'*xy)

 

[Zryn]

Draw a truth table for X XOR Y, and then put an extra output that is NOT X XOR Y, and you should see that anything that isn't x'y + xy' is x'y' + xy.

*Edit: An explanation

DeMorgan law: (x+y)' = x'y'

You need to use both of DeMorgans laws in this case:

1. (x+y)' = x'y'
2. (xy)' = x' + y'

You have [xy' + x'y]'. You need to think of it as [xy']' + [x'y]'. This becomes [x'+y''] + [x''+y'] by 2. You need to think of as [x+y']' + [x'+y]' and then as [(x+y') + (x'+y)]'. This becomes (x+y')'(x'+y)' by 1. Simplify a bit and it becomes (x'+y)(x+y') which becomes x'x + x'y' + xy + yy' = 0 + x'y' + xy + 0 = x'y' + xy.

The truth table is a quick way (in this instance) of verifying the math.