I'm not sure whether this is the best place to ask this, but is the XOR binary operator a combination of AND+NOT operators?
Is XOR a combination of AND and NOT operators?
6 Answers
Yes. In fact, any logical operation can be built from the NAND operator, where
A NAND B = NOT(A AND B)
See for instance http://en.wikipedia.org/wiki/NAND_logic, which gives
A XOR B = (A NAND (A NAND B)) NAND (B NAND (A NAND B))
Digression: There is a story about certain military devices being designed using only NAND gates, so that only one part needs to be certified, stocked as spares, etc. I don't know whether it's actually true.
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3AKA, [Sheffer stroke](http://en.wikipedia.org/wiki/Sheffer_stroke). – 2011-05-11
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3Of course, that can be simplified to (NOT(A AND B)) AND (NOT(NOT A AND NOT B)) :p – 2011-05-11
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0Which, of course, actually gives you an obvious pattern to convert an arbitrary binary operator into a combination of ANDs and NOTs, without having to go through NAND. – 2011-05-11
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2Another way to determine A XOR B is A != B. – 2011-05-11
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3Most flash drives are made entirely from NAND gates: http://www.tech-faq.com/nand-drive.html – 2011-05-11
Meanwhile, let me add the fact that XOR is not expressible by any expression whose operations use only AND, OR, IMPLIES, ConverseImplies and IFF.
To see this, observe merely that True XOR True is False, but the other operations always take True input to True output, a feature that would be inductively preserved in complex expressions.
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2Or, "you cannot NOT have `NOT`." – 2011-05-12
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2I was tempted to say that that I couldn't disagree less! But actually, you can not have NOT, since it has already been observed that NAND suffices, as does NOR. – 2011-05-12
Here are two similar but different versions derived using a OR b = NOT(NOT a AND NOT b)
, though there are others
p XOR q = ( p AND NOT q ) OR ( NOT p AND q ) = NOT ( NOT ( p AND NOT q ) AND NOT ( NOT p AND q ) )
or
p XOR q = ( p OR q ) AND NOT ( p AND q ) = NOT ( NOT p AND NOT q ) AND NOT ( p AND q )
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0one too many NOTs for my liking.. But thanks for the detailed reply – 2011-05-11
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0@Dark Star1: The first has the same number of NOTs as Nate Eldredge has NANDs, while the second has the same number of NOTs as IainM has NOTs. – 2011-05-11
$\{AND, NOT\}$ is a complete set of operators, i.e. you can express any boolean function $f:\{0,1\}^n \to \{0,1\}$ using them:
$$f(x_1,\cdots,x_n) = \lnot \bigwedge_{w\in f^{-1}(1)} \lnot \big(\bigwedge_{w_i=1}x_i \land \bigwedge_{w_i=0}\lnot x_i \big)$$