Re: completeness of the relational lattice

On 28 jun, 17:52, Vadim Tropashko <vadimtro_inva..._at_yahoo.com> wrote:
> On Jun 28, 2:30 am, Jan Hidders <hidd..._at_gmail.com> wrote:
>
>
>
> > 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.
>
> At the "distribution" step:
>
> { (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))) }
>
> you applied one more rule
> exists x: (P /\ Q) --> exists x: P /\ exists x: Q
> after distributivity.

?? After distributivity I apply once more "P or ( Exists x : Q(x) ) <=> Exists x : ( P and Q(x) )". I can do this because each variable appears only on one side of the "and".

Anyway, I've made a little progress with checking my proof, but I could not avoid adding new axioms. I also noticed that dealing with W / 10 / 11 now caused agaiin a lot of extra axioms, so for now I'm leaving them out again. The current list of axioms (and that I conjecture to be complete) is as follows:

We consider the following algebra:

E ::= R | E * E | E + E | [H] | <H>.

Where:

- R : a relation with name R (and known header)
- E * E : natural join
- E + E : generalized union
- [H] : the empty relation with header H (possibly empty)
- <H> : the equality relation { (x, ..., x) | x in Domain } over

- A(R) = H where H is the header of R
- A(r * s) = A(r) + A(s)
- A(r + s) = A(r) * A(s)
- A([H]) = H
- A(<H>) = H

(1)    r * r = r
(2)    r * s = s * r
(3)    r * (s * t) = (r * s) * t

(4)    r + r = r

(10)   R = R + [H]                                 if H is the header

(11)   [H] * [S] = [H + S]
(12)   [H] + [S] = [H * S]
(22)   R * [] = [H]                                if H is the header

(27)   <H> * <S> = <H + S>       if H * S nonempty
(28)   R * <x> = R               if x a single attribute in header H

(26)   <H> + [S] = <H * S>
(29)   [H] * <S> = [H + S]
(31)   ((r * <S>) + [H]) * <S> = ((r * <S>) + [H + S]) * <S>       if

I invite you to challenge me to show that it can prove an equation that holds. Of course you should also check if all these equations can be derived by you.

• Jan Hidders