Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have to prove analytically to see if these equations are exclusively or.
$$A⊕ A=0$$

Do I solve this by using the truth table?

A A Output
0 0   0
0 1   1
1 0   1
1 1   0

How I am supposed to prove that this equation is a product of XOR?

$$A⊕B⊕A.B= A+B$$

share|cite|improve this question
up vote 1 down vote accepted

Yes, you can solve it by using the truth table. Consider $A\oplus A$, for instance; the truth table tells you that if $A=0$, then $A\oplus A=0\oplus 0=0$, and if $A=1$, then $A\oplus A=1\oplus 1=0$. These are the only possibilities, so it’s always true that $A\oplus A=0$.

You can do the same thing for the second problem; it just takes longer.

$$\begin{array}{c|c|c} A&B&A\oplus B&A\cdot B&(A\oplus B)\oplus A\cdot B&A+B\\ \hline 0&0&0&0&0\oplus 0=0&?\\ 0&1&1&0&1\oplus 0=1&?\\ 1&0&1&0&?&?\\ 1&1&0&1&?&? \end{array}$$

I’ll leave it to you to fill in the rest and decide whether the last two columns really are equal.

share|cite|improve this answer
Following what you demostrated is this what the table is suppose to look like?@Brian M. Scott – Leo Oct 15 '12 at 17:02
$$A⊕B⊕A.B= A+B$$ $$\begin{array}{c|c|c} A&B&A\oplus B&A\cdot B&(A\oplus B)\oplus A\cdot B&A+B\\ \hline 0&0&0&0&0\oplus 0=0&0\\ 0&1&1&0&1\oplus 0=1&0\\ 1&0&1&0&0\oplus 1=1&0\\ 1&1&0&1&1\oplus 1=0&1 \end{array}$$ – Leo Oct 15 '12 at 19:34
@Leo: I’m afraid not. In the fifth column the last two entries should be $1\oplus0=1$ and $0\oplus1=1$: the two numbers being XORed come from the $A\oplus B$ and $A\cdot B$ columns. For the sixth column you need to review the table for $+$: you’ve filled in the values for $A\cdot B$, not $A+B$. All cases of $A+B$ are $1$ except $0+0$, which is $0$. – Brian M. Scott Oct 15 '12 at 19:44
Oh, I see what I did wrong. I get it thanks! – Leo Oct 15 '12 at 20:11
@Leo: Great! (Just as a final check, your last two columns should both end up being $0,1,1,1$, verifying the identity.) – Brian M. Scott Oct 15 '12 at 20:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.