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Re: completeness of the relational lattice

From: Vadim Tropashko <vadimtro_invalid_at_yahoo.com>
Date: Wed, 27 Jun 2007 14:55:10 -0700
Message-ID: <1182981310.688913.42760@z28g2000prd.googlegroups.com>


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 to
http://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... Received on Wed Jun 27 2007 - 16:55:10 CDT

Original text of this message

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