Re: "Armstrong's axioms" augmentation - help plz
From: Dan <guntermann_at_verizon.com>
Date: Wed, 09 Mar 2005 09:52:14 GMT
Message-ID: <iHzXd.81013$Dc.28962_at_trnddc06>
> The first rule implies the second as you pointed out, but the second
> cannot stand in for the first as the implication goes only one direction
> (from the first rule to the second and not from the second statement of a
> rule to the first).
> --dawn
>
Received on Wed Mar 09 2005 - 10:52:14 CET
Date: Wed, 09 Mar 2005 09:52:14 GMT
Message-ID: <iHzXd.81013$Dc.28962_at_trnddc06>
"Dawn M. Wolthuis" <dwolt_at_tincat-group.comREMOVE> wrote in message
news:d0j8fm$omh$1_at_news.netins.net...
> "paul c" <toledobythesea_at_oohay.moc> wrote in message
> news:n99Xd.599930$Xk.252349_at_pd7tw3no...
>> Jan Hidders wrote:
>>> 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. If you replace the first rule with the second you will not
>>> derive all FDs that hold.
>>>
>>> -- Jan Hidders
>>
>> The first 'rule' is X -> Y, and so is the second! What's the difference?
>>
>> p
>
> The first rule implies the second as you pointed out, but the second
> cannot stand in for the first as the implication goes only one direction
> (from the first rule to the second and not from the second statement of a
> rule to the first).
Now I am lost.
I think there might be confusion in terminology here. By the term "rule", are we talking about a term (i.e. a single FD), or are we talking in terms of Armstrong's rules (i.e. implications involving two FD's)?
By the first, you of course are right, but that has nothing to do with what the OP asked. The OP wasn't asking whether XZ->Y could replace X->Y.
By the second use of the term, the duality of Armstrong's 2nd and 4th rules means that the entailments are equivalent.
>
> --dawn
>
Received on Wed Mar 09 2005 - 10:52:14 CET