# Boolean Algebra Problem.

1. Sep 7, 2009

### Ithryndil

Ok, I am having a difficult time with the following problem. I am to show that the two sides are equal.

x1'x3 + x1x2x3' +x1'x2+x1x2' = x2'x3+x1x3'+x2x3'+x1'x2x3

For the LHS I can simplify it to:
x2(x1'+x1x3') + x1'x3 + x1x2'
x2x1' + x2x3' + x1'x3 + x1x2'
x1'x3 + x2x3' + x1x2'

For the RHS I can simplify to:
x2(x3' + x1'x3) + x2'x3 + x1x3'
x2x3' + x2x1' + x2'x3 + x1x3'
x1'x3 + x1'x2 + x2x3

At this point I am stuck. I know that both sides should be equal. I have to do this algebraically, so using a truth table is out of it.

2. Sep 8, 2009

### Elucidus

In order to show two Boolean expressions to be equal, it suffices to show they can be transformed into equivalent forms. The most rudimentary way is to find their complete-sum-of-products form (aka disjunctive normal form DNF).

When the number of variable is small, a Karnaugh map is also useful.

Removing the distraction of subscripts by letting x = x1, y = x2, and z = x3:

You need to show that x'z + xyz' + x'y + xy' = y'z + xz' + yz' + x'yz.

When finding the "completed" version of a fundamental product recall that

ab = ab(1) = ab(c + c') = abc + abc'

and then delete extra identical summands.

Or show that both expressions have the same Karnaugh map.

$$\begin{array}{c|c|c|c|c} & yz & yz' & y'z' & y'z \\ \hline x & & \checkmark & \checkmark & \checkmark \\ x' & \checkmark & \checkmark & & \checkmark \\ \end{array}$$

When the number of variables increases things get more difficult.

--Elucidus

3. Sep 12, 2009

### Kenneth Mann

I hope this isn't too late.
Also, I agree with ELUCIDUS about removing subscripts and making it easier, so I'll use his.

Hint:
You don't always have to minimize first. You can start out by expanding terms, such as:

X'Z = X'YZ + X'Y'Z
Then you can re-combine terms, then minimize. Try it!~

KM