[Discrete math] Proving the form of a function

AI Thread Summary
The discussion centers on a problem in discrete mathematics involving the function f(x1, x2, x3) = x1x2 v x2x3 v x1x3, with the goal of proving that f(x1, x2, x3) equals x2. Participants express confusion over how to manipulate the function to arrive at the desired result and question whether the problem statement is accurate. Truth tables are referenced as a method to verify the function's outputs, but discrepancies lead to doubts about the correctness of the original claim. Ultimately, it is suggested that the statement to be proven is indeed incorrect, as demonstrated by specific evaluations of the function.
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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
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.<br />

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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