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.

 

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

 

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

 

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

Complement law - (Y + Y') = 1
Therefore it follows from Identites that
(Y' + X)(Y' + Y)
= (Y' + X)(1)
= Y' + X

 

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)

 

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)

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...

 

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!