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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 |
#2
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Re: Question about DISCRETE MATH!
Not enough information.
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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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