Re: Relational symmetric difference is well defined
From: V.J. Kumar <vjkmail_at_gmail.com>
Date: Tue, 19 Jun 2007 22:37:42 +0200 (CEST)
Message-ID: <Xns9954A927FC658vdghher_at_194.177.96.26>
>> which is a different query altogether since forall y: A(x,y) <->
>> B(y,z) evaluates to true only in trivial cases and does not give a
>> set valued join which was specified originally with set builder
>> notation ! E.g. if you have
>>
>> A:
>>
>> 1 2
>> 1 3
>> 2 5
>> 2 6
>>
>> B:
>>
>> 2 a
>> 3 a
>> 5 b
>> 6 b
>>
>> then your formula will produce an empty set instead of {(1 a), (2 b)}
>> !
Date: Tue, 19 Jun 2007 22:37:42 +0200 (CEST)
Message-ID: <Xns9954A927FC658vdghher_at_194.177.96.26>
Jan Hidders <hidders_at_gmail.com> wrote in news:1182264646.065242.294970_at_e9g2000prf.googlegroups.com:
>>
>> {(x,z)| (Forall y : A(x,y) <-> B(y,z) }
> > Ok. >
>> which is a different query altogether since forall y: A(x,y) <->
>> B(y,z) evaluates to true only in trivial cases and does not give a
>> set valued join which was specified originally with set builder
>> notation ! E.g. if you have
>>
>> A:
>>
>> 1 2
>> 1 3
>> 2 5
>> 2 6
>>
>> B:
>>
>> 2 a
>> 3 a
>> 5 b
>> 6 b
>>
>> then your formula will produce an empty set instead of {(1 a), (2 b)}
>> !
> > Really? For x="1", z="a" the formula says "forall y : A(1,y) <-> > B(y,a)". For y=2 and y=3 the propositions A(1,y) and B(y,a) are both > true. For all other values for y both are false. So I would think the > formula holds for x=1, z=a.
I was wrong by being pessimistic about the formula, but you are not right either. You formula is overly optimistic. Assuming quantification domains X = 1..100 and Z = 'a'..'z' and according to your predicate, the following pairs would be legit:
(3, c), (4, d), (3, d),... etc., i.e. every pair such that A(x,y) and B (y,z) evaluate to false for each y. If you go ahead and try to set domain boundaries, you'd still need to deal with gaps.
> > -- Jan Hidders > >Received on Tue Jun 19 2007 - 22:37:42 CEST