Re: equivalence of functional dependencies

From: shannon <shannon_at_nolunchmeat.com>
Date: Tue, 06 Jan 2004 22:48:17 +0100
Message-ID: <btfadu$86r$04$1@news.t-online.com>

I have tried an example from the elmasri book, perhaps somebody can pass judgement on my logic,

two sets of functional dependencies F= {A > C, AC > D, E > AD, E > H} and G = {A > CD, E > AH}. Check whether or not they are equivalent.

here I make conclusion that they are not equivalent,

E > AD, E > H will as union give me E > ADH

therefore E > AH not deductible

A > C, AC > D will not help to derive A > CD either,

thanks

Shannon

Jonathan Leffler wrote:
> shannon wrote:
>

>> FD has been bugging me for a month now,
>>
>> F = {A  -> BC, A  -> D, CD -> E}
>> G = {A -> BCE, A -> ABD, CD -> E}
>>
>> I have been told that the sets of functional dependencies above are 
>> equivalent, can anybody explain to me how I can come to this 
>> conclusion step by step,
>>
>> I understand that armstrong's axioms are used to come to the 
>> conclusion, I have seen these axioms written down,
>>
>> please use another example if you are in fear of doing 'homework'

>
>
> In F, A -> BC and A -> D means A -> BCD. Since CD -> E, transitivity
> means that A -> CD => A -> E. Hence, the first term of G is deducible.
> The second term in G is similarly deducible from the trivial A -> A (not
> cited; it is a tautology), plus A -> BC plus A -> D. The third term is
> identical in F and G.
>
>
Received on Tue Jan 06 2004 - 15:48:17 CST