Stuck in Boolean Algebra equation

Question

I have this equation in Boolean Algebra:

$x*y*z+x'*y*z+x*y'*z+x*y*z' = y*z+x*z+x*y$

I got this:

$= yz(x+x')+xy'z+xyz'$

$= yz+xy'z+xyz'$

And from here I tried multiple things but it goes wrong like:

$yz+x(y'z+yz')$

$yz + x$

Since y'*y and z'*z =0 they disappear so this seemed to be the only answer for me what am I doing wrong?

That must be where I'm going wrong then.. I'm trying to go on from yz+x(y'z+yz') now — Sinan Samet, Jun 01 '16 at 12:58

score 1 · Answer 1 · answered Jun 01 '16 at 12:53

1

Hint: You could add in a couple of $xyz$ terms into the left side of the original equation.

answered Jun 01 '16 at 12:53

paw88789

So I can use the first xyz term without complements a couple times? – Sinan Samet Jun 01 '16 at 12:55
Yes because $xyz=xyz+xyz$ – paw88789 Jun 01 '16 at 12:58
yes it's evaluating to the result! So it's good to say here that we might get different expressions while simplifying, but the expression on the left and right would evaluate to same value when we substitute x,y and z in the equation. – Jasser Jun 01 '16 at 13:00
Alright I will try to answer my own question using this then – Sinan Samet Jun 01 '16 at 13:05

score 0 · Answer 2 · edited Jun 07 '16 at 08:24

0

So using the hint of @paw88789 I got this:

$xyz+x'yz+xy'z+xyz'$

$=yz(x+x')+xz(y+y')+xy(z+z')$

$=yz + xz + xy$

Is this right?

If it is, can I have a little more explanation on why I can use xyz a couple times and if I can do this with all the other terms too?

edited Jun 07 '16 at 08:24

Hailey

answered Jun 01 '16 at 13:10

Sinan Samet

2 Answers2