Re: Relational symmetric difference is well defined
From: V.J. Kumar <vjkmail_at_gmail.com>
Date: Fri, 1 Jun 2007 12:58:11 +0200 (CEST)
Message-ID: <Xns994247275E9F0vdghher_at_194.177.96.26>
>
> Can you expand on that?
>
> Although it's clear the above operator(s) are different in
> kind from join and union, it's not obvious (to me anyway)
> that this difference translates into higher-orderness.
Date: Fri, 1 Jun 2007 12:58:11 +0200 (CEST)
Message-ID: <Xns994247275E9F0vdghher_at_194.177.96.26>
Marshall <marshall.spight_at_gmail.com> wrote in news:1180663055.923481.55990_at_r19g2000prf.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.
>
> Can you expand on that?
>
> Although it's clear the above operator(s) are different in
> kind from join and union, it's not obvious (to me anyway)
> that this difference translates into higher-orderness.
{y|A(x,y)} is a set variable that can range over arbitrary sets of domain elements rather than just the domain elements. Similarly, {r|r<=R}, a powerset of R, where R is some relation and <= is the set containment symbol, is also a second order formula.
>
>
> Marshall
>
>
Received on Fri Jun 01 2007 - 12:58:11 CEST