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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
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0 1   1
----------
1 0   1
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1 1   0
----------

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

$$A⊕B⊕A.B= A+B$$

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If this is homework, please add the homework tag. –  Ben Crowell Oct 14 '12 at 22:50
    
Okay thanks Ben. –  Leo Oct 15 '12 at 16:54

1 Answer 1

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.

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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

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