Re: Relational symmetric difference is well defined

On 19 jun, 16:30, "V.J. Kumar" <vjkm..._at_gmail.com> wrote:
> Jan Hidders <hidd..._at_gmail.com> wrote innews:1182262078.054309.135620_at_n2g2000hse.googlegroups.com:
>
>
>
> > 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.c
> >> >> >> >> >> om:
>
> >> >> >> >> >> > 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.googlegroup
> >> >> >> >> >> >> s.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)} }
>
> The original posting contains a typo I think. It should ahve been:
>
> {(x,z)| {y|A(x,y)}={y|B(y,z)} }

Yes, indeed. Sorry for copying that typo.

> > 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)) }
>
> The substitution gives:
>
> {(x,z)| (Forall y : A(x,y) <-> B(y,z) }

Ok.

> which is a different query altogether since forall y: A(x,y) <-> B(y,z)
> evaluates to true only in trivial cases and does not give a set valued
> join which was specified originally with set builder notation ! E.g. if
> you have
>
> A:
>
> 1 2
> 1 3
> 2 5
> 2 6
>
> B:
>
> 2 a
> 3 a
> 5 b
> 6 b
>
> then your formula will produce an empty set instead of {(1 a), (2 b)} !

Really? For x="1", z="a" the formula says "forall y : A(1,y) <-> B(y,a)". For y=2 and y=3 the propositions A(1,y) and B(y,a) are both true. For all other values for y both are false. So I would think the formula holds for x=1, z=a.

• Jan Hidders