0

I'm taking Computer Architecture and we were given a list of simple Boolean Rules. I'm having trouble finding out where to start.

This is my work so far

F = y'z' + xy'z + xz' + x'yz'

F = y'((1)z'+x(1)z)+z'(x+x'y(1))

F = y'(z'+xz)+z'(x+x'y)

F = y'(xz'+zz')+z'(xx'+xy)

F = y'(xz'+0)+z'(0+xy)

F= y'(xz')+z'(xy)

F= y'(xz')+z'(xy(1))

F= y'(xz')+z'(xy(x+x'))

At this point I feel like I have too many terms.

Zoe Brig
  • 3

1 Answers1

0

As a general approach, you should see which variable or negation thereof the terms have in common. In fact, see which one(s) is being shared the most. In this case, I see $3$ terms that include $z'$, so I would start with that:

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

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

Now let's see what we can do with the $y'+x+x'y$ term:

Well, as a very useful principle we have:

Reduction

$p+p'q=p+q$ (in the context of $p$, the $p'q$ term reduces to just $q$

If Reduction is not on your list, too bad, since it is really super handy. We can derive it from more basic principles, which is why it is not always given to you:

$p+p'q=(p+p')(p+q)=(1)(p+q)=p+q$

Applying Reduction:

$y'+x+x'y=y'+y=1$

Ah, super! So this means that your function can be simplified to:

$z'(y'+x+x'y)+xy'z=z'(1)+xy'z=z'+xy'z$

And now we can do one more application of Reduction to get:

$z'+xy'$

Bram28
  • 100,612
  • 6
  • 70
  • 118
  • Thank you so much!! I understand now. I'm going to practice more so I can get better. We were also not taught reduction, but it will make my homework go much faster. – Zoe Brig Feb 02 '20 at 19:59
  • I'm still a little confused on the last step, how the reduction worked whenever you reduced xy'z to xy' – Zoe Brig Feb 02 '20 at 20:50
  • @ZoeWhite It is because you also have z'. That is: z'+xy'z becomes z'+xy' ... this is another instance of Reduction: p'+pq=p+q – Bram28 Feb 02 '20 at 21:22