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

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

$$A⊕B⊕A.B= A+B$$

1 Answers 1

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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. Scott2012-10-15
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    $$A⊕B⊕A.B= A+B$$ $$\begin{array}{c|c|c} A&B&A\oplus B&A\cdot B&(A\oplus B)\oplus A\cdot B&A+B\\ \hline 0&0&0&0&0\oplus 0=0&0\\ 0&1&1&0&1\oplus 0=1&0\\ 1&0&1&0&0\oplus 1=1&0\\ 1&1&0&1&1\oplus 1=0&1 \end{array}$$2012-10-15
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    @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 $0$.2012-10-15
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    Oh, I see what I did wrong. I get it thanks!2012-10-15
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    @Leo: Great! (Just as a final check, your last two columns should both end up being $0,1,1,1$, verifying the identity.)2012-10-15