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Simplifying boolean expression

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Homework Statement



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

[tex]
S=z \oplus (x \oplus y)[/tex]
[tex]=z'(xy'+x'y)+z(xy'+x'y)'[/tex]
I don't get how they go to this from the above
[tex]=z'(xy'+x'y)+z(xy+x'y')[/tex]
[tex]=xy'z'+x'yz'+xyz+x'y'z[/tex]

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)
 
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Answers and Replies

  • #2
Zryn
Gold Member
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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.
 
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