Re: Relational symmetric difference is well defined

Vadim Tropashko <vadimtro_invalid_at_yahoo.com> wrote in news:1181944396.435969.12660_at_j4g2000prf.googlegroups.com:

> On May 31, 6:40 pm, "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. 

"Natural" join or relation composition is a first order expression allright if written properly:

A(x,y) join B(y,z) def. {(x,z)| exists(y)( A(x,y)&B(y,z) ) }

>
> So having the second order definition is not necessarily that bad.
>

Sure, second order lang. is more expressive, but its expressivity does not come for free !

> The situation may be analogous to the fundamental theorem of algebra,
> which states that the complex numbers field is algebraically closed.
> The most common proof of the theorem (by Gauss) depends on the
> analytic structure of C and uses second order arguments. However, one
> can give a first order, purely algebraic proof of more general result
> that if R is any real closed field, then C=R[i] is algebraically
> closed.
>
>
Received on Sat Jun 16 2007 - 01:21:25 CEST