Re: equivalence of functional dependencies

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.

-- 
Jonathan Leffler                   #include <disclaimer.h>
Email: jleffler_at_earthlink.net, jleffler_at_us.ibm.com
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