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  #1  
Old 09-26-2007, 06:32 PM
anto anto is offline
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Join Date: Feb 2007
Posts: 19
Default Question about DISCRETE MATH!

Can anyone please help me answer this question?

We define the operator "*" as the following : p * q = p -> !q (negation of q).

Without using any other operators EXCEPT the * (you cant use !, ^, v etc.), and using only the propositions p and q, express the following :

(a) !p
(b) p ^ q
(c) p v q
(d) p -> q
(e) p <-> q

Any help would be appreciated.
Thanks in advance
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  #2  
Old 09-26-2007, 07:48 PM
tshort tshort is offline
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Default Re: Question about DISCRETE MATH!

Not enough information.
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  #3  
Old 09-26-2007, 09:02 PM
Siegmund Siegmund is offline
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Posts: 1,850
Default Re: Question about DISCRETE MATH!

"P implies !Q" is the same thing as "!(P and Q)". Maybe this is enough of a hint to get OP started?

In other words, this is the classic computer-science result that any logic circuit can be built from NAND gates.
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