Re: Declaring super types
From: Vadim Tropashko <vadimtro_at_gmail.com>
Date: Mon, 19 Apr 2010 13:57:04 -0700 (PDT)
Message-ID: <99d22d3d-26bb-4b68-8513-597f6f4c4696_at_w20g2000prm.googlegroups.com>
Date: Mon, 19 Apr 2010 13:57:04 -0700 (PDT)
Message-ID: <99d22d3d-26bb-4b68-8513-597f6f4c4696_at_w20g2000prm.googlegroups.com>
On Apr 19, 1:06 pm, r..._at_raampje.lan (Reinier Post) wrote:
> ... "is a" does have a simple, useful and consistent
> definition for the relational model: for relvars R, S, we can write
>
> R is a S
>
> as a shorthand for:
>
> + all attributes of S are attributes of R;
> + those attributes are a (possibly super)key of R, and
> + R projected on those attributes is always a subset of S.
> ...
Are you saying
"R is a S"
is eqivalent to
"R join S = R"? Received on Mon Apr 19 2010 - 22:57:04 CEST