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Roberto Vescarelli wrote:
> Dear Owen,
>
> Thanks for your reply,
>
> >IMO, there cannot be a distinction between being and existence.
> >Could you further clarify what you mean?
>
> I think that logic of common language must be different from logic of first
> order languages because
>
> (a) there are in common language such words like 'Father Christmas' used to
> denote non-existing objects
I don't agree that 'Santa Clause' or 'Father Christmas' denote
anything at all.
They are referred to only by the predicates that describe them. They
have no objectual reference.
There are no non-denoting objects, only their descriptions exist.
See: Frege's Sense and Reference.
>
> (b) there is, consequently, the possibility of doing right reasoning about
> non-existing objects like 'If Father Cristmas brings, sponte sua, a present
> to every good children, he must be a good person'
Again, I do not agree. Father Cristmas cannot (and does not) bring
presents to anyone.
He cannot be a good person, indeed he cannot be a person at all.
>
> Why are there similar phaenomena?
>
> The natural answer is that there are things that don't exist. So we must
> distinguish beetwen being and existence (like Meinong did).
There cannot be things which do not exist. Meinong is wrong, See
Russell.
>
>
> >Existence, is argueably not a property of things, but rather, it is the
> >logical sum of properties of things
>
> >E!x =df EF(Fx).
>
> If you have a property you have the being, but not necessarly you exists. I
> don't think the Descartes argument is valid. The round square is not
> existent, because it is a contraddictory concept, but if the side of the
> round square is 1, then the diagonal is the square root of 2. Therefore we
> can speak of some properties of inexistent objects.
I do believe that Descartes' argument is valid. i.e. to have the
property of thinking entails that you exist. I think therefore I am,
is tautologous.
The side of the round square cannot exist. There are no properties that
non-existent things have, including the property of non-existence.
To say that the round square has the property of non-existence is a
contradiction.
~(the round square exists) is not equivalent to (the round square has
non-existence).
The former is true while the latter is false.
>
> > 'Father Christmas has a white beard' is false.
>
> And what you think about 'Father Cristmass has not a white beard'?
"Father Cristmass has not a white beard' is ambiguious: does it mean,
1. It is not the case that (Father Christmas has a white beard), in
which case it is true, or,
2. Father Christmas has a non-white beard, in which case it is false.
~(Fx) <-> (~F)x, if and only if, x exists.
>
> > There is no physical property that the described thing 'Father
> > Christmas' has.
>
> Even the property of not haveing properties?.
Yes. What is it that can have the property of 'not having properties'?
Surely, that is a contradiction in terms.
>
> > It is not a member of any class.
>
> It is a member of the class of things that have the being but don't exist.
Nonsense.
In general I think we have to agree to disagree.
>
>
> >>
> >> As another istance consider a therorem of firs order logic that is
> invalid
> >> in our logic:
> >>
> >> (4) Pa -> exist an x such that (Px)
> >>
> >> It is easy to see that (4) is true in M but is false in -M. In fact if I
> >> exist, you can deduce from ?I have a white beard? that ?there exist white
> >> beared things?, but if Father Christmas doesn't exist, you can not deduce
> >> that ?there exist white beared things? from ?Father Christmas has a white
> >> beard?.
Pa -> ExPx is valid, iff a is a value of the variable x. That is to say
'a' must exist.
eg, ~(the present king of France = the present king of France) ->
Ex~(x=x), is a contradiction. The premise is true but the conclusion is
false.
Again, there is no set -M, non-existent things are not members of any
set.
The empty set does not contain non-existent things, rather it does not
contain any members at all.
>
> >??
>
> In first order logic if Pa is true, you can deduce that exist an x such that
> Px. But for the pourposes of free logic, you can not correctly extend this
> reasoning to the case in with 'a' is an inexistent object. If a is an
> inexistent object you can not deduce 'exist an x such that Px' from 'Pa'.
>
> Greetings,
> --
> Roberto Vescarelli
> http://www.faberbox.com/roby/
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