# Re: completeness of the relational lattice

Date: Wed, 27 Jun 2007 23:23:53 -0000

Message-ID: <1182986633.411188.36730_at_n60g2000hse.googlegroups.com>

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))

- Jan Hidders