[Discrete math] Proving the form of a function

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Homework Help Overview

The discussion revolves around a problem in discrete mathematics concerning the proof of a function's form. The function is defined as f(x1, x2, x3) = x1x2 v x2x3 v x1x3, and the task is to prove that f(x1, x2, x3) = x2.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss attempts to manipulate the function by substituting values and constructing truth tables. There is uncertainty about how to arrive at the conclusion that f(x1, x2, x3) equals x2. Some participants express confusion regarding the correctness of the problem statement itself.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some have provided guidance on constructing truth tables, while others question the validity of the original statement. There is no explicit consensus on the correctness of the problem or the approaches taken.

Contextual Notes

Participants note potential errors in the problem statement and discuss the implications of their findings based on truth table results. There is mention of notation preferences and the challenge of representing negation clearly.

fawk3s
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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

Homework Equations


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:
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fawk3s said:
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


Homework Equations





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.
 
Mark44 said:
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.
 
Just don't walk under that ladder - it's bad luck.:-p

(The word you want is "latter".)
 
fawk3s said:
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


Homework Equations





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
[tex]f(0,x_2,0) = 0 \vee 0 \vee 0 = 0, \text{ for any value of }x_2,\\<br /> f(1,x_2,1) = x_2 \vee x_2 \vee 1 = 1 \text{ for any value of }x_2.[/tex]

RGV
 
Mark44 said:
Just don't walk under that ladder - it's bad luck.:-p

(The word you want is "latter".)

I think it's too late to edit now :biggrin:
 

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