Re: Undefinedness

On Nov 23, 9:44 pm, Cimode <cim..._at_hotmail.com> wrote:
> On 21 nov, 14:34, JOG <j..._at_cs.nott.ac.uk> wrote:
>
> > Word up CDT. How the devil are you all? Well, I return with a question
> > that as ever highlights my complete lack of formal mathematical
> > training, and in light of knowing no logicians in my daily life (funny
> > that), I was hoping that one of you kind folks might be able to
> > advise:
>
> > Say I had a set of 3 encoded propositions:
> > R := { {(Name, Tom), (Age, 42)}, {(Name, Dick), (Age, 16)}, {(Name,
> > Harry)} }
>
> > (note that Harry's Age is missing, so instead of adding a null, i've
> > intentionally just left the attribute out. Just ride with such oddness
> > for now if you would.)
>
> > What if I deigned to create a simple 'adults' subset of this set of
> > propositions, by creating a predicate that only returned the elements,
> > p, which contained an age attribute greater than 18. Could I state
> > this as (where E signifies set membership):
>
> > Adults := { p E R | EXISTSx ( x > 18 && (Age, x) E p ) }
>
> > My question obviously hinges around Harry's missing age attribute. In
> > this case would the EXISTSx (...) part of the set's intension simply
> > return a FALSE, or will I end up in the quagmire of 3VL with an
> > UNDEFINED? My instinct is that I am still in 2VL given there is no
> > null floating about, but since the recent, excellent discussions of
> > Jan's DEF operator, and having delved into beeson's logic of partial
> > terms, I am not at all confident.
>
> > Any comments are much appreciated, and regards to all, Jim.
>
> I do not understand how you can already go to any form of subtyping
> without a valid proposition allowing to establish relation R?

Well first, I'm not sure that I'd refer to specifying a subset of propositions as 'subtyping', second, all the propositions are valid as far as I can tell (their pretty simple ones after all), and third, R isn't a relation. Regards, J.

> I suggest decomposing R before attempting to constitute Adults such
> as...
>
> R := { {(Name, Tom), (Age, 42)}, {(Name, Dick), (Age, 16)}, {(Name,
> Harry)} }
> into ....(I assume Name as being a unique identifier)
> RName:= { {(Name, Tom)}, {(Name, Dick)}, {(Name,
> Harry)} } -->p1
> and
> RAges := { {(Name, Tom), (Age, 42)}, {(Name, Dick), (Age, 16)} } -->
> p2
> You may then constitute....
>
> Adults := { p1 E R1 | EXISTSx ( x > 18 && (Age, x) E p1 ) }
Received on Sat Nov 24 2007 - 02:03:26 CET