Re: "Armstrong's axioms" augmentation - help plz
Date: Wed, 09 Mar 2005 09:26:58 GMT
Message-ID: <CjzXd.60321$uc.8848_at_trnddc03>
"Jan Hidders" <jan.hidders_at_REMOVETHIS.pandora.be> wrote in message news:OfyXd.33305$zl5.3358827_at_phobos.telenet-ops.be...
> Dan wrote:
>> "Jan Hidders" <jan.hidders_at_REMOVETHIS.pandora.be> wrote in message
>> news:np4Xd.32216$ub7.3221532_at_phobos.telenet-ops.be...
>>
>>>love boat via DBMonster.com wrote:
>>>
>>>>I understand the Augmentation rule:
>>>>{ X -> Y } |= XZ -> YZ
>>>>
>>>>but I don't understand why the rule can also be stated as:
>>>>
>>>>{ X -> Y } |= XZ -> Y
>>>>
>>>>Why is this?
>>>
>>>It cannot.
>>
>> A simple proof by contradiction should prove that it an equivalent
>> statement of the augmentation rule.
>>
>> Suppose that X->Y holds, but XZ -> Y does not. Thus, for any two
>> arbitrary tuples in a relation we assume,
>> 1) t1[X] = t2[X];
>> 2) t1[Y] = t2[Y];
>> 3) t1[XZ] = t2[XZ];
>> 4) but by our supposition, t1[Y] <> t2[Y].
>>
>> However we see that this (4) cannot be true, since by 2) we see that
>> t1[Y] does equal t2[Y].
> > ?? You prove that the rule is correct, but that is not what is disputed. >
Hello Jan,
What is being disputed here? I thought the question from the OP was whether
the augmentation rule could be expressed as either:
1) {X->Y} |= XZ->YZ
By augmentation of the first form of the rule, we get {XZ->Y}|= {XZ->YZ}.
2) {X->Y} |= XZ->Y
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.
>>> If you replace the first rule with the second you will not
>>> derive all FDs that hold.
>>
>> 1) XZ->Y (given)
>> 2) X->XZ (reflixivity)
>> 3) X->Y (transitivity)
>> 4) XZ-YZ (augmentation)
> > Your step 2 is incorrect (reflexivity says that XZ->X but not that X->XZ) > and in step 4 you assume the rule that you are trying to derive.
Yes. My only defence is dyslexia (in visualizing the subset operator) and
stupidity for including it in the first place, since it really seems
irrevalent.
>
> -- Jan Hidders
Received on Wed Mar 09 2005 - 10:26:58 CET