# Re: foreign key constraint versus referential integrity constraint

Date: Mon, 26 Oct 2009 15:01:43 -0400

Message-ID: <0K6dnb447sWKb3jXnZ2dnUVZ_qKdnZ2d_at_giganews.com>

"Tegiri Nenashi" <tegirinenashi_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. Received on Mon Oct 26 2009 - 20:01:43 CET