[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

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.

 

[Mark44]

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.

 

[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.

 

[Mark44]

Just don't walk under that ladder - it's bad luck.

(The word you want is "latter".)

 

[Ray Vickson]

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

RGV

 

[fawk3s]

Just don't walk under that ladder - it's bad luck.

(The word you want is "latter".)

I think it's too late to edit now