Can Boolean Algebra Be Simplified to Just XY + Z?

[Nyasha]

Homework Statement

F(x,y,z)=X'Y'Z + X'YZ+ XY'Z + XYZ' + XYZ

The Attempt at a Solution

F(x,y,z) = X'Y'Z + X'YZ+ XY'Z + XY (Z'+Z)

F(x,y,z) = X'Y'Z + X'YZ+ XY'Z + XY

I am stuck here l don't know how to further simplify it so that l can end up with F= XY + Z

 

[sylas]

Homework Statement

F(x,y,z)=X'Y'Z + X'YZ+ XY'Z + XYZ' + XYZ

The Attempt at a Solution

F(x,y,z) = X'Y'Z + X'YZ+ XY'Z + XY (Z'+Z)

F(x,y,z) = X'Y'Z + X'YZ+ XY'Z + XY

I am stuck here l don't know how to further simplify it so that l can end up with F= XY + Z

Your first step is fine. You brought out a common factor XY from two terms, and then simplified what was left.

You can keep doing that. Look for the common factors.

However, since you already know the answer, there is something you can do that will help. You would like to have as many terms as possible with Z as a factor, which will give you the best chance of getting it all reduced to Z. So consider ALL terms involving Z... including the XYZ term that you've reduced to XY.

Cheers -- sylas

 

[Nyasha]

Your first step is fine. You brought out a common factor XY from two terms, and then simplified what was left.

You can keep doing that. Look for the common factors.

However, since you already know the answer, there is something you can do that will help. You would like to have as many terms as possible with Z as a factor, which will give you the best chance of getting it all reduced to Z. So consider ALL terms involving Z... including the XYZ term that you've reduced to XY.

Cheers -- sylas

X'Y'Z + X'YZ+ XY'Z + XY

X'Y'Z + Z (X'Y + XY') + XY

From here l can not think of any boolean property which can be used to simplfy this expression

 

[sylas]

X'Y'Z + X'YZ+ XY'Z + XY

X'Y'Z + Z (X'Y + XY') + XY

From here l can not think of any boolean property which can be used to simplfy this expression

Your first term there also has Z as a common factor. Bring it out of this term as well, just like you did for the middle two terms.

----

Your last term XY is equivalent to XY + XYZ

So you can get ANOTHER term XYZ with Z as a common factor. It was also in the original expression explicitly. So use it also. Get all the terms with Z as a common factor. That was the point of my previous hint.

Cheers -- sylas

 

[Nyasha]

=Z(X'Y'+X'Y+XY') + XY

= Z(X'(Y'+Y))+XY') + XY

=Z(X'+XY')+XY

Man l don't know where am l going wrong on this thing ?

 

[sylas]

=Z(X'Y'+X'Y+XY') + XY

= Z(X'(Y'+Y))+XY') + XY

=Z(X'+XY')+XY

Man l don't know where am l going wrong on this thing ?

You are not going wrong. You are doing well. The next step is the hint I mentioned.

You know the general rule in boolean algebra

AB + A = A​

Well, this means you can proceed from the end of your sequence above as follows, by adding a term XYZ back that will be useful.

Z(X'+XY')+XY = Z(X'+XY')+XY+XYZ

Note that you are adding back the XYZ term, which was there at the start. It turns out this will be useful; because you know you want to end up with XY+Z, and so the more Z terms you can involve the better.

From this, now do the common factor step with Z, yet again...

Cheers -- sylas