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Re: completeness of the relational lattice

From: Jan Hidders <hidders_at_gmail.com>
Date: Thu, 28 Jun 2007 09:30:24 -0000
Message-ID: <1183023024.749003.318200@q75g2000hsh.googlegroups.com>


On 28 jun, 03:18, Vadim Tropashko <vadimtro_inva..._at_yahoo.com> wrote:
> On Jun 27, 4:23 pm, Jan Hidders <hidd..._at_gmail.com> wrote:
>
>
>
> > On 27 jun, 23:55, Vadim Tropashko <vadimtro_inva..._at_yahoo.com> wrote:
>
> > > On Jun 27, 2:17 pm, "Brian Selzer" <b..._at_selzer-software.com> wrote:
>
> > > > "Vadim Tropashko" <vadimtro_inva..._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
>
> > > That's right, I was looking tohttp://en.wikipedia.org/wiki/First-order_logic#Identities
> > > before posting, but apparently didn't notice this subtlety. So I don't
> > > see how we can justify attribute x in the distributivity law and fail
> > > to find a counter example either...
>
> > I think what you missed is rules like "P or ( Exists x : Q(x) ) <=>
> > Exists x : ( P and Q(x) )" where P has no free occurrence of x.
>
> > So you start with:
>
> > A(x, t) \/ (B(y, t) /\ C(z, t) which is formally:
>
> > { (t) | (Exists x : A(x, t)) \/
> > (Exists y, z : B(y, t) /\ C(z, t)) }
>
> > = /* pull y and z quantifiers to the front */
>
> > { (t) | Exists y, z :
> > (Exists x : A(x, t)) \/ (B(y, t) /\ C(z, t)) }
>
> > = /* distribution */
>
> > { (t) | Exists y, z :
> > ((Exists x : A(x, t)) \/ B(y, t)) /\
> > (Exists x : A(x, t)) \/ C(z, t))) }
>
> > = /* push y and z quantifiers back to their variables */
>
> > { (t) | ((Exists x : A(x, t)) \/ (Exists y : B(y, t))) /\
> > (Exists x : A(x, t)) \/ (Exists z : C(z, t)))) }
>
> > which is indeed the representation of
>
> > (A(x,t) \/ B(y,t)) /\ (A(x,t) \/ C(z,t))
>
> Right. If the rule
> exists x: (P /\ Q) --> exists x: P /\ exists x: Q
> don't allow to pull x to the outermost level, it could be leveraged to
> push the x back in front of A\/B and A\/C.

I'm not sure what you mean here. The rule I used for pushing is the same as the one I used for pulling, but applied in reverse.

Received on Thu Jun 28 2007 - 04:30:24 CDT

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