Re: Relational symmetric difference is well defined

On 19 jun, 22:37, "V.J. Kumar" <vjkm..._at_gmail.com> wrote:
> Jan Hidders <hidd..._at_gmail.com> wrote innews: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.

They should. My claim was that it is equivalent with the set equality join as defined by Vadim:

 A(x,y)/=B(y,z) is {(x,z)| {y|A(x,y)}={y|B(y,z)} }

and this also has these pairs in its result. Of course this is not a completely correct definition of the set equality join but, as already indicated by Vadim, it's very easy to see how this can be fixed in the definition and in the formula while staying in first-order logic (just add "and (exists y : A(x,y)) and (exists y : B(y, z))"). In fact, the biggest problem seems to be to get the brackets right. (Sorry Vadim! ;-))

• Jan Hidders