# [Discrete math] Proving the form of a function

fawk3s
I wasnt really sure where I was supposed to post this problem, so I figured this place is as good as any.

## Homework Statement

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

## The Attempt at a Solution

Well I tried replacing x2 with x2, but I don't 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 doesn't make much sense to me.

Last edited:

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

## Homework Statement

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

## The Attempt at a Solution

Well I tried replacing x2 with x2, but I don't 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 doesn't make much sense to me.

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.

fawk3s
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.

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 doesn't 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.

Mentor

(The word you want is "latter".)

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

## Homework Statement

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

## The Attempt at a Solution

Well I tried replacing x2 with x2, but I don't 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 doesn't make much sense to me.

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

fawk3s