Re: more on delete from join

From: Mr. Scott <>
Date: Fri, 28 Aug 2009 02:16:44 -0400
Message-ID: <>

"paul c" <> wrote in message news:eimlm.40861$Db2.21494_at_edtnps83...

> Mr. Scott wrote:

>> "paul c" <> wrote in message
>> news:RYhlm.40820$Db2.17042_at_edtnps83...
>>> Mr. Scott wrote:
>>>> "paul c" <> wrote in message
>>> ...
>>>>> Insert to a base relvar represents a conclusion: R =: R UNION A 
>>>>> conludes R AND A.
>>>> It concludes R OR A.  R AND A would be just A, provided A is in R.
>>> I think you mean R <AND> A would be just A, provided A is in R and they 
>>> have the same heading.  Whether or not A is in R, after the assignment, 
>>> all propositions that A stands for and all propositions that R 
>>> previously stood for are true, eg., if r1,r2,...,rn and a1,a2,...,an are 
>>> the original propositions, r1 AND r2 AND ... a1 AND a2 is true as far as 
>>> R is concerned. .

>> I mean R OR A. Under the closed world interpretation, the disjunction of
>> the collection of facts represented by a table is interpreted as the
>> conjunction of those facts along with the negation of the facts
>> represented by the complement of the table, the assumption being that
>> what is stated is supposed to be true and what isn't isn't. I mean R OR
>> A because the disjunctions of the collections of facts represented in R
>> and in A are combined into the disjunction of a single collection of
>> facts prior to applying the closed world interpretation.
> I don't know if the plural 'disjunctions' is a typo'.  Do you mean that 
> the resulting value of R stands for "r1 OR r2, etc." is true?

I don't seem to be getting through.


The intersection of sets R and A contains those objects that are elements of both R AND A. The union of sets R and A contains those objects that are elements of either R OR A. A closed formula bound existentially evaluates to a set of propositions which may be either true or false. The truth value of the set of propositions is the disjunction of those propositions. "There is a ...." is only true when there is at least one .... Received on Fri Aug 28 2009 - 08:16:44 CEST

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