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Re: Lots of Idiotic Silly Braces?

From: Brian Selzer <brian_at_selzer-software.com>
Date: Sun, 22 Jul 2007 16:57:37 GMT
Message-ID: <50Moi.40361$Um6.18249@newssvr12.news.prodigy.net>

"paul c" <toledobythesea_at_oohay.ac> wrote in message news:biuoi.135995$1i1.14227_at_pd7urf3no...
> paul c wrote:

>> Brian Selzer wrote:
>>
>>> "paul c" <toledobythesea_at_oohay.ac> wrote in message 
>>> news:Lf4oi.131821$xq1.73460_at_pd7urf1no...
>>>
>>>> Brian Selzer wrote:
>>>> ...
>>>>
>>>>> Can rva's be keys?  A relation value being the extension of a 
>>>>> predicate, the set of tuples in a relation value represents a set of 
>>>>> positive atomic formulae, and under the closed world assumption, that 
>>>>> set implies the negation of each atomic formula that conforms to the 
>>>>> schema but is not represented by a tuple.  How, then, can a relation 
>>>>> valued attribute be a key?  Consider, the schema R{S{A, B}}, and the 
>>>>> following relation value, r:
>>>>>
>>>>> r = {{S={{A=3, B=4}, {A=3, B=5}}}, {S={A=3, B=4}}}
>>>>>
>>>>> Now, suppose that P(A, B) is the predicate of S.  The first tuple of r 
>>>>> asserts that P(3, 5) is true, but the second tuple implies that P(3, 
>>>>> 5) is false. ...
>>>>
>>>>
>>>> I think "S" here is an attribute name, not a relation. S refers to two 
>>>> different relation values which happen to appear as tuples in r. The 
>>>> first tuple of of r asserts that the ("first", relation-valued) tuple 
>>>> that combines (P(3,4) AND P(3,5)) is true in one relation value. The 
>>>> second tuple implies that (P (3,5)) is false in the "second" relation 
>>>> value.
>>>>
>>>
>>>
>>> But that's the point.  How can they both be true?  And if they can, then 
>>> how can it be consistent since there isn't also a tuple {S={A=3, B=5}}?
>>> ...
>>
>>
>> Those are the kind of questions that people who don't believe in Santa 
>> Claus ask.  We are talking about the behaviour of a machine here, not 
>> mythology!
>>
>

> That's maybe not a very useful reply, let me try again. If P(A,B) is the
> predicate of S, let's say P2(P(A,B)) is the predicate of r. You say the
> second tuple of r implies that P(3,5) is false. I say the second tuple of
> r implies that P2(P(3,5)) is false and that r has nothing to say about the
> truth or falseness of proposition P(3,5).
>

> UNGROUP (at least the D&D version which is the only one I'm familiar with)
> seems to have the effect of discarding the P2 predicate. I think you are
> mentally ungrouping r and pretending that the result has the same
> predicate as r, which I think is trying to have your cake and eat it too.
>

I think I see where you're going with this: correct me if I'm wrong, but aren't you saying that P and P2 are independent? I'm not so sure that they are. P2 ranges over relations--the possible extensions of P. It is my understanding that under a given interpretation, there can only be one independent extension of P. The truth value for r is the conjunction of the truth values for its constituent tuples:

  P2(P(3,4) /\ P(3,5)) /\ P2(P(3,4) /\ ~P(3,5))

P(3,5) cannot be both true and false, leaving,

  P2(T) /\ P2(F)

which is true if and only if P2(T) = P2(F).

UNGROUP does not discard the predicate, it transforms a second-order predicate into one that is first-order. Something like,

From X --> Q(X) and Q(X) = (E /\ F /\ G), you get,

X --> (E /\ F /\ G) which is logically equivalent to

(X --> E) /\ (X --> F) /\ (X --> G)

It follows then that the second-order predicate for r should be logically equivalent to the first order predicate for r UNGROUP S, but isn't, since there is only one attribute in r. There are no attributes in r whose values determine the values for S; therefore, there can be at most one value for S, and thus only one tuple for r.

So, for base relations, an rva can be a determinant if and only if it is also a dependent. In other words, an rva cannot be the only key on a base relation.

I guess I was wrong, then. An rva can be a key, as long as it isn't the only key.

> p
Received on Sun Jul 22 2007 - 11:57:37 CDT

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