Re: Newbie question about db normalization theory: redundant keys OK?

From: Hugo Kornelis <>
Date: Sun, 16 Dec 2007 23:55:07 +0100
Message-ID: <>

On Sun, 16 Dec 2007 08:48:25 -0000, David Portas wrote:

>"Hugo Kornelis" <> wrote in message
>> On Sat, 15 Dec 2007 09:54:11 -0400, Bob Badour wrote:
>>>6NF has at most one non-key attribute.
>> Hi Bob,
>> Isn't the latter statement an oversimplification? Unless I am truly
>> misunderstanding 6NF, I'd say that the scema below is 6NF even though R1
>> has two non-key attributes:
>> R1(a*,b,c)
>> R2(b*,c*,d)
>> Best, Hugo
>Your R1 is not in 6NF because it can be further decomposed as I already
>demonstrated in my example.
>In R2 does "b*,c*" represent one candidate key or two? If two, then it can
>be further decomposed and it isn't in 6NF.

Hi David,

Sorry for the unclarity. The notation b*,c* was meant to represent a single candidate key consisting of two attributes. Decomposing R1 would wreck the foreign key constraint between R1 and R2, that's why I wrote that it can't be nonloss decomposed.

Best, Hugo Received on Sun Dec 16 2007 - 23:55:07 CET

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