Re: Relational symmetric difference is well defined

On 19 jun, 15:28, "V.J. Kumar" <vjkm..._at_gmail.com> wrote:
> Jan Hidders <hidd..._at_gmail.com> wrote innews:1182159542.073157.146030_at_g4g2000hsf.googlegroups.com:
>
>
>
> > On 18 jun, 03:11, "V.J. Kumar" <vjkm..._at_gmail.com> wrote:
> >> Jan Hidders <hidd..._at_gmail.com> wrote
> >> innews:1182119546.370762.75420_at_n2g2000hse.googlegroups.com:
>
> >> > On 16 jun, 15:13, "V.J. Kumar" <vjkm..._at_gmail.com> wrote:
> >> >> Jan Hidders <hidd..._at_gmail.com> wrote
> >> >> innews:1181948397.949921.43300_at_n2g2000hse.googlegroups.com:
>
> >> >> > On 15 jun, 23:17, "V.J. Kumar" <vjkm..._at_gmail.com> wrote:
> >> >> >> Jan Hidders <hidd..._at_gmail.com> wrote
> >> >> >> innews:1181919464.037821.176760_at_p77g2000hsh.googlegroups.com:
>
> >> >> >> > On 15 jun, 16:00, "V.J. Kumar" <vjkm..._at_gmail.com> wrote:
> >> >> >> >> Jan Hidders <hidd..._at_gmail.com> wrote
> >> >> >> >> innews:1181910061.495472.280190_at_c77g2000hse.googlegroups.com:
>
> >> >> >> >> > On 1 jun, 03:40, "V.J. Kumar" <vjkm..._at_gmail.com> wrote:
> >> >> >> >> >> Vadim Tropashko <vadimtro_inva..._at_yahoo.com> wrote
> >> >> >> >> >> innews:1180628927.976321.267880_at_a26g2000pre.googlegroups.c
> >> >> >> >> >> om:
>
> >> >> >> >> >> > On May 30, 8:52 pm, Marshall
> >> >> >> >> >> > <marshall.spi..._at_gmail.com> wrote:
> >> >> >> >> >> >> Can you clarify the difference between set containment
> >> >> >> >> >> >> join and set equality join? The inverse of join is
> >> >> >> >> >> >> much on my mind these days.
>
> >> >> >> >> >> > Set equality join
>
> >> >> >> >> >> > A(x,y)/=B(y,z) is {(x,z)| {y|A(x,y)}={y|A(y,z)} }
>
> >> >> >> >> >> > Set containment join
>
> >> >> >> >> >> > A(x,y)/=B(y,z) is {(x,z)| {y|A(x,y)}>{y|A(y,z)} }
>
> >> >> >> >> >> > where the ">" is "subset of".
>
> >> >> >> >> >> The above formulas obviously are no longer first-order
> >> >> >> >> >> expressions. Along with the increased expressive power
> >> >> >> >> >> (e.g. it's trivial to define a powerset), you will reap
> >> >> >> >> >> the usual drawbacks of the higher order logic.
>
> >> >> >> >> > This was perhaps already clear, but it is the
> >> >> >> >> > *formulation* of the semantics which is not first-order.
> >> >> >> >> > The semantics themselves are clearly first order since
> >> >> >> >> > they can be defined in first order logic or the flat
> >> >> >> >> > relational algebra.
>
> >> >> >> >> This is very intriguing !
>
> >> >> >> > Not really. It is pretty obvious that in the above
> >> >> >> > formulation of the semantics of the joins you can replace the
> >> >> >> > higher order expression with a first order formula.
>
> >> >> >> How ?
>
> >> >> > For example, the formula
>
> >> >> > {y|A(x,y)}={y|A(y,z)}
>
> >> >> > is equivalent with
>
> >> >> > (Forall y : A(x,y) -> A(y,z)) and (Forall y : A(y,z) ->
> >> >> > A(x,y))
>
> >> >> That's fine, but even with substituting a predicate expression
> >> >> for the set expression in the original expression you'd still
> >> >> quantify over predicates which is another way of saying 'over
> >> >> sets':
>
> >> >> {x| (setA = setB)} is no different from {x| (A <=>B)} because
> >> >> forall (setA=setB) is the same like forall(A<=B), so you are
> >> >> still dealing with the second order quantification.
>
> >> > Where in the formula that I gave did you see second order
> >> > quantifiers or quantification over predicates?
>
> >> [...snip..]
>
> >> Yes, your simple formulas expressing set equality are first order
> >> allright, but they don't help to make the formula describing a set
> >> valued join a first oder expression !
>
> > It makes the query that corresponds to the operator of the form
> > { (x,z) | \phi(x,z) } with \phi a formula in first order logic
>
> No, it does not make that. Look, the set valued join was defined as
>
> { (x,z)| Px <-> Pz } where Px def. A(x,y) for some fixed x and Pz def. B
> (y,z) for some fixed z. Px and Pz are predicate variables, that is
> second order variables. How do you intend to handle 'y' in \phi(x,z) def.
> A(x,y) <-> B(x,y) so that it would become a first order expression ?

?? Do you really want me to spell that out for you?

We start with

(1) {(x,z)| {y|A(x,y)}={y|A(y,z)} }

As already argued the proposition "{y|A(x,y)}={y|A(y,z)}" is equivalent with the proposition "(Forall y : A(x,y) -> A(y,z)) and (Forall y : A(y,z) -> A(x,y))". So simple substitution gives us:

(2) {(x,z)| (Forall y : A(x,y) -> A(y,z)) and (Forall y : A(y,z) -> A(x,y)) }

Voila, a query that corresponds to (1) and of the form { (x,z) | \phi(x,z) } with \phi a formula in first order logic.

• Jan Hidders