Re: Relational symmetric difference is well defined
From: V.J. Kumar <vjkmail_at_gmail.com>
Date: Fri, 1 Jun 2007 03:40:22 +0200 (CEST)
Message-ID: <Xns9941DCB7E1855vdghher_at_194.177.96.26>
>
> 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".
Date: Fri, 1 Jun 2007 03:40:22 +0200 (CEST)
Message-ID: <Xns9941DCB7E1855vdghher_at_194.177.96.26>
Vadim Tropashko <vadimtro_invalid_at_yahoo.com> wrote in news:1180628927.976321.267880_at_a26g2000pre.googlegroups.com:
> On May 30, 8:52 pm, Marshall <marshall.spi..._at_gmail.com> wrote:
[Quoted] [Quoted] >> 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. Received on Fri Jun 01 2007 - 03:40:22 CEST