Re: Relational symmetric difference is well defined

From: V.J. Kumar <vjkmail_at_gmail.com>
Date: Fri, 15 Jun 2007 16:00:30 +0200 (CEST)
Message-ID: <Xns995065B86DBvdghher@194.177.96.26>

Jan Hidders <hidders_at_gmail.com> wrote in news: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.com: 
>>
>> > 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 ! Mainly, 'cause by itself first order logic is meaningless, it is just a collection of well formed formulas or well formed strings of characters and some derivation rules to get new formulas by purely mechanical ways, there is no semantis without interpretation. It is, so, a syntactical system. Then, how do you define semantics in first order logic ?

>
> -- Jan Hidders
>
>
Received on Fri Jun 15 2007 - 09:00:30 CDT