Re: Relational symmetric difference is well defined

Vadim Tropashko <vadimtro_invalid_at_yahoo.com> wrote in news:1182287660.790811.172340_at_q19g2000prn.googlegroups.com:

> On Jun 19, 1:37 pm, "V.J. Kumar" <vjkm..._at_gmail.com> wrote:

>> Jan Hidders <hidd..._at_gmail.com> wrote > > 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.

This would not work 'cause you cannot quantify twice over the same variable (y).

>
> or
>
> {(x,z)| (exists y: A(x,y) & exists y: B(y,z) -> forall y : A(x,y) <->
> B(y,z) }

Ditto.

>
>
>
Received on Tue Jun 19 2007 - 23:58:18 CEST