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"Vadim Tropashko" <vadimtro_invalid_at_yahoo.com> wrote in message
news:1182964456.876858.28340_at_g37g2000prf.googlegroups.com...
> On Jun 26, 7:03 pm, Jan Hidders <hidd..._at_gmail.com> wrote:
>> I suspect that the condition
>> for r + (s * t) = (r + s) * (r + t)
>> should be A(r) * A(s) = A(r) * A(t) = A(s) * A(t).
>> So if two have an attribute, the third must also have it.
>
> This is more powerful condition than in the "First steps..." paper.
> Indeed, the correct criteria is:
>
> A(x,t) \/ (B(y,t) /\ C(z,t)) = (A(x,t) \/ B(y,t)) /\ (A(x,t) \/
> C(z,t))
>
> The proof is by writing down both sides in the set notation. Assuming
> general relation headers A(x,u,w,t), B(y,u,v,t), C(z,w,v,t) the left
> hand side works out to be
>
> exists xyzv : A \/ (B /\ C)
>
> The right hand side evaluates to
>
> exists zy : (exists xwv: A \/ B) /\ (exists xuv: A \/ C)
>
> In predicate logic we have
>
> exists x: (P /\ Q) <-> exists x: P /\ exists x: Q
>
While it is true that
exists x: (P \/ Q) implies exists x: P \/ exists x: Q
and
exists x: P \/ exists x: Q implies exists x: (P \/ Q),
isn't it true that
exists x: (P /\ Q) implies exists x: P /\ exists x: Q
but
exists x: P /\ exists x: Q does not imply exists x: (P /\ Q)?
Consequently,
exists x: (P /\ Q) is not equivalent to exists x: P /\ exists x: Q
> so that attribute x could be moved outside.
>
> So, your proof of (A /\ 00) \/ (A /\ 11) = A is correct, then. Yet,
> the earlier proof of the equality theorem is not. (So I keep my
> notation until at least one more spectacular derivation:-)
>
> I agree with your other post. It seems like advancing "the theory of
> relations with known attributes" is easier at this point than the
> "theory of general relations".
>
>
Received on Wed Jun 27 2007 - 16:17:11 CDT