# [Discrete math] Proving the form of a function

1. Sep 26, 2012

### fawk3s

I wasnt really sure where I was supposed to post this problem, so I figured this place is as good as any.

1. The problem statement, all variables and given/known data

x2 - the inversion of x2. (Yes, I was too dumb to figure out how to get "_" on it.)

We are given that
f(x1,x2,x3) = x1x2 v x2x3 v x1x3 = x1&x2 v x2&x3 v x1&x3

Prove, that

f(x1,x2,x3) = x2

2. Relevant equations

3. The attempt at a solution

Well I tried replacing x2 with x2, but I dont really know how that changes anything. I have no idea how you are supposed to end up with x2. And even when trying to replace the x'es with 0's and 1's, I DO NOT get that

x1x2 v x2x3 v x1x3=x2

that is

f(x1,x2,x3) = x2

Am I doing this whole thing wrong? Because it doesnt make much sense to me.

Last edited: Sep 26, 2012
2. Sep 26, 2012

### Staff: Mentor

Just to make typing simpler, you could dispense with the subscripts and use A, B, and C. Also, there are other notations for the negation of something, such as ~A or A'.

So your function can be written as f(A, B, C) = AB + BC + AC, and you are to show that f(A, ~B, C) = B. (Here + means "or" and a product means "and".)

I made columns for A, ~B, and C and filled in the eight rows of the truth table. I added columns for A(~B), ~BC, and AC, and then one more column for A(~B) + ~BC + AC. The truth values in the last column are different from those of B, so I'm wondering if there's an error in the problem or that you might have copied it down wrong.

3. Sep 26, 2012

### fawk3s

No, its copied down right. After trying to crack it and simplify it for a while (which I now realize was stupid because it's in its simplest form...), I made the truth table as well and saw that B doesnt match the new function's values. After that I was sure that either I was doing something totally wrong or there had to be an error in it. So I guess it was the ladder this time.

Thank you.

4. Sep 26, 2012

### Staff: Mentor

(The word you want is "latter".)

5. Sep 26, 2012

### Ray Vickson

The result you are asked to show is incorrect. We have
$$f(0,x_2,0) = 0 \vee 0 \vee 0 = 0, \text{ for any value of }x_2,\\ f(1,x_2,1) = x_2 \vee x_2 \vee 1 = 1 \text{ for any value of }x_2.$$

RGV

6. Sep 26, 2012

### fawk3s

I think it's too late to edit now