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Dear Owen,
Thanks for your reply,
>IMO, there cannot be a distinction between being and existence.
>Could you further claify 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
(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'
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).
>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.
> 'Father Christmas has a white beard' is false.
And what you think about 'Father Cristmass has not a white beard'?
> There is no physical property that the described thing 'Father
> Christmas' has.
Even the property of not haveing properties?.
> 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.
>>
>> 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?.
>??
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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