Re: more closed-world chatter

From: Bob Badour <bbadour_at_pei.sympatico.ca>
Date: Fri, 11 May 2007 00:10:30 -0300
Message-ID: <4643de67$0$4062$9a566e8b@news.aliant.net>

Marshall wrote:

> On May 9, 2:21 am, Jon Heggland <jon.heggl..._at_idi.ntnu.no> wrote:
>

>>Marshall wrote:
>>
>>>On May 7, 4:15 am, Jon Heggland <jon.heggl..._at_idi.ntnu.no> wrote:
>>>
>>>>The crux of the matter in that prescription is the equivalence between
>>>>R1 & R2 and R1 - (R1 - R2). How do you handle that?
>>
>>>On further reflection, I don't think they're equivalent.
>>>We would say two expressions e1 and e2 on variables
>>>A and B were equivalent if
>>
>>>  forall A: forall B: e1(A) = e2(B)
>>
>>>But that's clearly not the case here. All we have here is
>>
>>>  exists A: exists B: e1(A) = e2(B)
>>
>>>And that's not a very strong claim. By those criteria,
>>>we could say that sqrt(x) and x/2 were equivalent,
>>>because, you know, 4. The fact that there is a *category*
>>>of A and B values for which the equation holds is
>>>a distraction, and not compelling.
>>
>>I'm not sure I understand. Are you disputing the equivalence
>>of the set intersection A INTERSECT B and the difference
>>A MINUS (A MINUS B)?

>
>
> Well, the earlier question was, given
>
> R1(a:t1)
> R2(a:t2)
>
> what is the type of a in R1 join R2?
>
> TTM says most-specific-supertype(t1, t2);
> I say least-specific-subtype(t1, t2).
>
> This to me is the primary question. There are other,
> related questions as well, but this is the most important
> one, as it affects join, which I consider to be the
> most important relational operator.

I see a major problem with what you desire. Suppose neither t1 is a supertype of t2 nor t2 is a supertype of t1. Suppose further that t1 intersect t2 is not empty.

No single data type might describe the intersection of t1 and t2. One might have to describe the intersection as the union of a set of data types.

Where does that leave your least specific subtype? Received on Thu May 10 2007 - 22:10:30 CDT