# Help with boolean simplification

Discussion in 'Homework Help' started by exothesis, Aug 31, 2010.

1. ### exothesis Thread Starter New Member

Aug 31, 2010
3
0
Simplify XY'Z + X'Y'Z + XYZ

So far I have got:
Y'Z(X + X') + XYZ
--> Y'Z + XYZ

What do I do next? Any help would be appreciated.

2. ### exothesis Thread Starter New Member

Aug 31, 2010
3
0
Never mind, I managed to solve the problem

--> XY'Z + X'Y'Z + XYZ
--> Y'Z(X + X') + XYZ
--> Y'Z(1) + XYZ
--> Y'Z + XYZ
--> Z(Y' + XY)
--> Z((Y' + X)(Y' + Y))
--> Z((Y' + X)(1))
--> Z(Y' + X)
--> Y'Z + XZ

3. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
I' not quite sure this is correct. In step 6:
(Y'+X)(Y'+Y)=Y'Y'+Y'Y+XY'+XY=Y'+XY'+XY
wich is different from
Y'+XY

4. ### exothesis Thread Starter New Member

Aug 31, 2010
3
0
Complement law - (Y + Y') = 1
Therefore it follows from Identites that
(Y' + X)(Y' + Y)
= (Y' + X)(1)
= Y' + X

5. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
I don't argue about how (Y' + X)(Y' + Y) becomes (Y' + X).
What troubles me is how (Y' + XY) becomes (Y' + X)(Y' + Y)

6. ### AMIT_GOHEL Member

Jul 13, 2010
67
7
Hello Georacer,

You may have confusion between math's we everyday use nd the boolean maths..

What you have asked is Distributive low of Boolean algibra.

Regards...

7. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
You are correct indeed! It's been a while since I read my Boolean Algebra manual (Morris Mano). It seems in Boolean logic the * operation (and) is distributive for the + (or) operation
i.e. $x \cdot (y + z)= x \cdot y + x \cdot z$

but also the + operation is distributive for the * operation
i.e. $x+(y \cdot z)=(x+y)\cdot (x+z)$

There's todays minimum positive knowledge for me!