The problem is to simplify: $$ xz+\bar{x}y+zy $$ I have an answer key that says the answer is: $$ xz + \bar{x}y $$ I have no idea how they got this expression, though. The first thing I tried was to find the dual. Setting the original expression equal to some variable c, I got $$ \bar{c} = \bar{x}\bar{y} + \bar{z}x + \bar{z}\bar{y} $$ This, however, doesn't simplify anything. Taking the dual of this only gets me the original expression again. How do I get the $ zy $ term to drop out?
Asked
Active
Viewed 342 times
2 Answers
2
You can get the $zy$ term to drop out by conjoining it with $\,1 = x + \bar x\,$ first, and then distributing $zy=yz$ over $x + \bar x$.
Then, by "reverse" distribution (twice) you collect common factors (there will be two terms with $xz$, and two terms with $\bar xy$). Doing this makes nice things happen:
$$\begin{align} xz+\bar{x}y+zy & = xz+\bar xy + (1)zy \\ \\ & = xz + \bar xy + (x + \bar x)yz\\ \\ & = xz + \bar xy + xyz + \bar x y z\\ \\ & = xz(1 + y) + \bar xy(1 + z)\\ \\ & = xz(1) + \bar xy(1)\\ \\ & = xz + \bar xy \end{align}$$
amWhy
- 209,954
-
Interesting. This doesn't seem like an obvious thing to do. I suppose something like this would just be a consequence of pattern recognition. Does this come up often? – David May 18 '14 at 18:30
-
It comes up here and there. You're right that it isn't, initially, intuitive. In some cases, using this "trick" is a bit more intuitive, but as with most things, it becomes more intuitive with practice. Once you see it a few times, and understand why it works, plus actively look for situations to apply it, it will stick with you. – amWhy May 18 '14 at 18:33
1
solution
$xz + x'y + yz = xz(1+y) + x'y(1+z) + zy\\ =xz + xzy + x'y + x'yz + zy\\ =xz + zy(x + x') + x'y + yz\\ =xz + 2yz + x'y$
on comparing LHS and RHS , we get $$\Rightarrow yz=0 \Rightarrow xz + x'y + yz = xz + x'y$$
Vladhagen
- 4,878
user3721155
- 11