Re: "Armstrong's axioms" augmentation - help plz

From: Dan <guntermann_at_verizon.com>
Date: Wed, 09 Mar 2005 19:44:59 GMT
Message-ID: <%mIXd.75247$uc.51780_at_trnddc08>


"Jan Hidders" <jan.hidders_at_REMOVETHIS.pandora.be> wrote in message news:y2HXd.33648$rG.3316803_at_phobos.telenet-ops.be...
> Dan wrote:
>>
>> Hello Jan,
>
> Hi Dan,
>
>> 1) {X->Y} |= XZ->YZ
>> 2) {X->Y} |= XZ->Y
>>
>> By augmentation of the first form of the rule, we get {XZ->Y}|= {XZ->YZ}.
>> That is the extent of the OP's question, as far as I can tell.
>
> Really? Are you saying that OPs "can also be stated as" should be
> interpreted as "allows me to derive that"? That doesn't make much sense to
> me. I'm sorry Dan, I'm having a really hard time trying to follow you
> here.

Sorry, Jan. I was having a hard time following as well.
>
>> I interpreted your answer of *it cannot* to mean that the second
>> inference rule expression was invalid (re: "I don't understand why the
>> second rule can also be stated by..."). But now that I read closer, I
>> think you are asserting something different.
>
> Indeed I am. Armstrong's rules are an inference mechanism. Given a set of
> FDs they allow you to derive more FDs, and, as you know, the nice thing
> about them is that they are sound and complete. Now, what *I* meant was
> that if in the usual set of three rules you replace the augmentation rule
> with the other rule then your inference mechanisme will still be sound
> (the rule is correct) but no longer complete (it is weaker).

Ok. I see your point, but just make sure its crystal clear: the first three rules form the basis for the rest of the rules (including the alternative for augmentation that the OP stated). Altering the augmentation rule in the form the OP presented would undercut a degree of completeness (as the basis for all remaining inferences are altered).

> I really don't see what other meaningful interpretation of "the
> augmentation rule can also be stated as ..." there could be. Note by the
> way that in this sense the augmentation rule is equivalent with the other
> rule plus the union rule:

I didn't realize that this argument was talking about *replacing* the augmentation rule and thus didn't see a problem with a *shortcut*. But now I understand where you are coming from. Someone should probably notify Elmasri and Navathe that their wording could be misleading.
>
> 3) {X->Y, X->Z } |= X->YZ

Yes. This is what I was thinking from the very beginning (I left this part off above) and therefore couldn't understand why you'd have a problem with the restatement. Now I do. Thanks for taking the time to explain.
>
>
> -- Jan Hidders
Received on Wed Mar 09 2005 - 20:44:59 CET

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