Sum of Products (Boolean Algebra)

Question

I am having real trouble getting to the corrects answers when asked to simply Sum of products expressions. For instance:

Determine whether the left and right hand sides represent the same function:

a) $x_1\bar{x}_3+x_2x_3+\bar{x}_2\bar{x}_3 = (x_1+\bar{x}_2+x_3)(x_1+x_2+\bar{x}_3)(\bar{x}_1+x_2+\bar{x}_3)$

They answer is that they are equivalent. Here is my logic for solving the left hand side, I first expand so each term has 3 variables:

$$=x_1x_2\bar{x}_3+x_1\bar{x}_2\bar{x}_3+x_1x_2x_3+\bar{x}_1x_2x_3+x_1\bar{x}_2\bar{x}_3+\bar{x}_1\bar{x}_2\bar{x}_3$$

combining terms

$$\begin{align*} &=x_1x_2(\bar{x}_3+x_3)+(x_1+\bar{x}_1)\bar{x}_2\bar{x}_3+x_1\bar{x}_2\bar{x}_3\\ &=x_1x_2+\bar{x}_2\bar{x}_3+x_1\bar{x}_2\bar{x}_3 \end{align*}$$

Then if I apply De Morgan's law in order to get it in Product of sum form, I get nothing close to the equivalent of the right hand side.

Answer 1

After expanding RHS using distributive law and reducing it you should get:

$x_1\bar{x}_3+\bar{x}_2\bar{x}_3+x_2x_3+x_1x_2 $

Now, make Karnaugh map in order to get minimal disjunctive form:

If you group ones as on picture above you will get next expression:

$x_1\bar{x}_3+\bar{x}_2\bar{x}_3+x_2x_3 $

Answer 2

Just as when proving trig identities, in general it’s better to start by trying to simplify the more complicated side, which in this case is the right-hand side. If you expand the RHS by brute force and use the identities $xx=x$, $x\bar{x}=0$, $x+\bar{x}=1$, and $x+xy=x$ repeatedly, you shouldn’t have much trouble reducing it to $$x_1x_2+x_1\bar{x}_3+x_2x_3+\bar{x}_2\bar{x}_3,$$ which is almost what you want. Now write $$x_1x_2 = x_1x_2 1 = x_1x_2 (x_3 + \bar{x}_3),$$ expand, and use the absorption identity again to get rid of the extra terms.