| Oracle FAQ | Your Portal to the Oracle Knowledge Grid | |
Home -> Community -> Usenet -> comp.databases.theory -> Re: Relational symmetric difference is well defined
Vadim Tropashko <vadimtro_invalid_at_yahoo.com> wrote in
news:1181951119.615693.101990_at_d30g2000prg.googlegroups.com:
> On Jun 15, 4:21 pm, "V.J. Kumar" <vjkm..._at_gmail.com> wrote:
>> Sure, second order lang. is more expressive, but its expressivity
does
>> not come for free !
Of course it is, e.g.(project def. project on some attributes A not in R2):
R1 divide R2 = project(R1) - project(project(R1) times R2) - R1).
Converting the above to a first-order formula is left as an exercise;)
>
>
Received on Sat Jun 16 2007 - 09:16:46 CDT
![]() |
![]() |