# Re: foreign key constraint versus referential integrity constraint

Date: Mon, 26 Oct 2009 12:51:46 -0700 (PDT)

Message-ID: <8e2f599d-bcb8-4f9a-81c5-5bbe9ff25d9c_at_g22g2000prf.googlegroups.com>

On Oct 26, 11:01 am, "Mr. Scott" <do_not_re..._at_noone.com> wrote:

> "Tegiri Nenashi" <tegirinena..._at_gmail.com> wrote in message

*>
**> news:84d21c7e-c6df-48a3-b5c9-f012caeded08_at_12g2000pri.googlegroups.com...
**>
**> > On Oct 25, 11:06 pm, "Mr. Scott" <do_not_re..._at_noone.com> wrote:
**> > > I think we should make the distinction, and formally.
**>
**> > > (p /\ q) -> r is not the same as (p -> r) /\ (q -> r)
**> > > but (p \/ q) -> r is the same as (p -> r) \/ (q -> r)
**> > I don't follow. If these are BA expressions with the "->" as material
**> > implication, then
**>
**> > (p v q) -> r = ~(p v q) v r = (~p ^ ~q) v r
**> > (p -> r) v (q -> r) = (~p v r) v (~q v r) = (~p v ~q) v r
**>
**> > If the "->" is interpreted as deduction symbol (that is partial
**> > boolean lattice order), then
**>
**> > (p v q) < r is not the same as (p < r) or (q < r)
**>
**> You're right.
**>
**> (p /\ q) -> r is not the same as (p -> r) /\ (q -> r)
**> but (p \/ q) -> r is.
*

(p ^ q) -> r = (p -> r) v (q -> r)

and, dually

(p v q) -> r = (p -> r) ^ (q -> r)

(Everything is perfectly symmetric in boolean algebra).

Now, before getting into constraint classification, what language do you suggest to express them in? Is it RC, RA, or something else? Received on Mon Oct 26 2009 - 20:51:46 CET