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the table of logic and ($ \wedge $) is

0 0 0 
0 1 0
1 0 0
1 1 1

I can build a "logic and" operation with additions and subtractions?

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I don't think so, but you can do it with addition, subtraction, and logical or. – Dan Brumleve Jan 11 '12 at 3:52

If you mean truncated subtraction, so that negative results are truncated to zero, that is, $x\dot-y=0$ when $x\leq y$, then since $\neg x=1\dot-x$, it follows that de Morgan's law amounts to $$x\wedge y=1\dot-\Bigl((1\dot-x)+(1\dot- y)\Bigr),$$ which is what you want.

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this is the same of $(x+y)\dot- 1$ ? – juanpablo Jan 11 '12 at 21:21
Yes, indeed, it is the same. – JDH Jan 11 '12 at 21:45

If I understand you correctly, then yes you can. You could have one rule for (0, 0) and one rule for the other cases as follows: If (x, y)=(0, 0), then (x^y)=(x+y). Otherwise (x^y)=((x+y)-1).

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mmm I think in this solution, but I need the if ... I only have additions and subtractions – juanpablo Jan 11 '12 at 13:41

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