Re: "Armstrong's axioms" augmentation - help plz
Date: Wed, 09 Mar 2005 07:10:18 GMT
Message-ID: <ujxXd.74292$uc.57954_at_trnddc08>
"Jan Hidders" <jan.hidders_at_REMOVETHIS.pandora.be> wrote in message
news:np4Xd.32216$ub7.3221532_at_phobos.telenet-ops.be...
Suppose that X->Y holds, but XZ -> Y does not. Thus, for any two arbitrary
tuples in a relation we assume,
> 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.
However we see that this (4) cannot be true, since by 2) we see that t1[Y] does equal t2[Y].
Unless I'm missing something obvious, it seems to me that paul c. got this right.
*Note: my proof is a variant of a proof of augmentation given by Elmasri and Navathe, Fundamentals of Database Systems (3rd. ed) p. 479
Make sense?
If you replace the first rule with the second you will not
> derive all FDs that hold.
- XZ->Y (given)
- X->XZ (reflixivity)
- X->Y (transitivity)
- XZ-YZ (augmentation)
If we can get back to the first rule from the second, I'm not sure I understand how we will miss deriving (inferring) all FDs that would have held with just the first rule.
- Dan
> -- Jan Hidders
Received on Wed Mar 09 2005 - 08:10:18 CET