Re: Relational symmetric difference is well defined
From: V.J. Kumar <vjkmail_at_gmail.com>
Date: Tue, 19 Jun 2007 15:28:56 +0200 (CEST)
Message-ID: <Xns9954607668228vdghher_at_194.177.96.26>
>> 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 !
Date: Tue, 19 Jun 2007 15:28:56 +0200 (CEST)
Message-ID: <Xns9954607668228vdghher_at_194.177.96.26>
Jan Hidders <hidders_at_gmail.com> wrote in news: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 ? You cannot do 'forall' over 'y' because your predicate will be false except in some trivial cases. What then ?
>and > hence this query is expressible in the relational algebra. That makes > it a first order expression by the usual definition of the term. What > other definition did you have in mind? > > -- Jan Hidders > >Received on Tue Jun 19 2007 - 15:28:56 CEST