the negation of $A \veebar B \veebar C $??

Question

I need to know the negation of $A \veebar B \veebar C $, with $\veebar$

thanks in advance!!

It is $\neg (A \veebar B \veebar C )$. – Git Gud Aug 22 '13 at 19:25 — Git Gud, Aug 22 '13 at 19:25

Git Gud · Accepted Answer · 2013-08-22T19:33:41.097

4

The truth table for $A\veebar B \veebar C$ is

$Truth table for $A\veebar B \veebar C$$ .

Therefore the truth table for its negation is

$Truth table for the negation of $A\veebar B \veebar C$$ .

Now you can construct the formula that yields the truth table above by inspecting it:

$$\underbrace{(A\land B\land \neg C)}_{\large \text{1st true line}} \lor \underbrace{(A\land \neg B\land C)}_{\large \text{2nd true line}}\lor \ldots$$

It's not pretty, but it works.

edited Aug 22 '13 at 19:33

answered Aug 22 '13 at 19:22

Git Gud

31,356

What have you used to create those tables? – dtldarek Aug 22 '13 at 19:33
@dtldarek Check the links in the answer to be surprised at how easy it is. (Thank you, amWhy). – Git Gud Aug 22 '13 at 19:34
Why that doesn't surprise me... Thanks ;-) – dtldarek Aug 22 '13 at 19:36
From terms page: "It is permitted to use and post individual, incidental results or small groups of results from Wolfram|Alpha on non-commercial websites and blogs, provided those results are properly credited to Wolfram|Alpha". I try to do something like this, and I think it is a good practice to attribute the source. Anyway, good job. – dtldarek Aug 22 '13 at 19:45
@dtldarek I sually try to mention it in my answer in a natural way whenever I use WA. I couldn't find one this time and since the links are there anyway, I kept it as it is. – Git Gud Aug 22 '13 at 20:05

score 3 · Answer 2 · answered Aug 22 '13 at 19:07

3

It means either none of them are true or exactly two of them are.

answered Aug 22 '13 at 19:07

Dan Brumleve

17,796

score 1 · Answer 3 · answered Aug 22 '13 at 19:20

Assuming that $\veebar$ means $\mathtt{xor}$, formula $$a_1 \veebar a_2 \veebar \ldots \veebar a_n$$

means that odd number of elements $a_i$-s is true. Interpreting truth as $1$ and falsehood as $0$ we arrive at

$$a_1 + a_2 + \ldots + a_n \equiv 1 \pmod{2}.$$

Naturally, its negation is that even number of elements is true,

$$a_1 + a_2 + \ldots + a_n \equiv 0 \pmod{2}$$

which can be transformed to

$$a_1 + a_2 + \ldots + a_n + 1 \equiv 1 \pmod{2},$$

that is

$$a_1 \veebar a_2 \veebar \ldots \veebar a_n \veebar \mathtt{true}.$$

In your case this is equivalent to $A \veebar B \veebar C \veebar \mathtt{true}$.

I hope this helps $\ddot\smile$

the negation of $A \veebar B \veebar C $??

3 Answers3