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Why is there something rather than nothing?

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Brian Holtz

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Aug 9, 2001, 7:18:26 PM8/9/01
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"Paul Filseth" <pg...@lsil.com> wrote [in alt.atheism.moderated] :

> > > "Why is there something instead of nothing?" cannot be explained
> > > correctly, because any answer you might offer qualifies as
> > > "something" and is therefore part of what you were supposed to
> > > explain.
> >
> > Only if you assume that the explanatory relation can never
> > be reflexive (A explains A) or symmetric (A explains B and
> > B explains A).
>
> That's not an assumption; it follows from what it means to
> explain something.

Not necessarily. Nozick writes that "a small literature exists
that attempts to formulate precise conditions whereby circular
explanations are excluded" and cites work by Hempel and others.

> > In his _Philosophical Investigations_ Nozick explores the alternative
> > idea of "explanatory self-subsumption" and gives a hypothetical
> > example:
> > P says: any lawlike statement have characteristic C is true.
> > Let us imagine this is our deepest law[...] Next we face the
> > question of why P holds true, and we notice that P itself has
> > characteristics C. [..] Our question is not whether such
> > self-subsumption as an instance of itself can constitute
> > a proof, but whether it can constitute an explanation. [..]
> > <snip> If a brute fact is something that cannot be explained by
> > anything, then a self-subsumable principle isn't a brute fact; <snip>
>
> To me, that looks an awful lot like Nozick just stuck in his
> conclusion as a premise.

He asked whether self-subsumption could reduce the
brute-fact quality of a deepest law, and considered two possible
senses of "brute fact". Under one sense (which you quoted) it
does reduce it, while under the other it doesn't. Still,
non-self-subsuming deepest laws would be considered brute under
both senses, and so self-subsumption indeed seems to reduce (but
perhaps not eliminate) this brute-fact quality.

> > But I'm
> > not so ready to completely dismiss the whole concept of necessary
> > existence. In particular, I wonder if logical possibility itself
> > exists necessarily? If nothing existed instead something, would
> > it therefore be true that nothing can even possibly exist?
> >
> > If the answer is no, then there may be an answer to the Big Why
> > ("why is there something rather than nothing?"). The answer might be:
> > Nothing exists except logical possibility, which necessarily exists,
> > and our perception of material existence is an epiphenomenon of our
> > being logical subcomponents of a logically possible universe.
>
> Well, let's start by not being "deceived by grammar". I don't
> know what it means to say logical possibility exists. The formal way
> to talk about existence is to phrase it as quantification over some
> set of properties. You don't say "God exists."; you say "There exists
> an x such that ((x created the universe) and (x is intelligent) and
> (For all y, if (y created the universe) then (y = x)))." Existence
> statements that can't be rephrased that way are category errors. Can
> you put "Logical possibility exists necessarily" into that form? What
> are the properties of logical possibility?

So how would we say nothing exists in 1st-order PC? I guess in
2nd-order PC it might be: (for all x)(for all P)(! Px).

But to deal with possibility, don't we need modal logic and its
operators M (it-is-possible-that) and L (it-is-necessary-that)?
The modal logical system S5 is built on the extra assumption that
Mp -> LMp: if a proposition is possible, its being possible is a
necessary truth. This assumption sounds like what I meant above,
but it is an axiom rather than a theorem.

Hmm, I guess I need to learn more about modal logic:
http://plato.stanford.edu/entries/logic-modal/

--
Brian...@sun.com
Knowledge is dangerous. Take a risk:
http://humanknowledge.net

gr...@ling.lll.hawaii.edu

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Aug 9, 2001, 9:28:38 PM8/9/01
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Brian Holtz <Brian...@sun.com> wrote:
...

> So how would we say nothing exists in 1st-order PC?

How about: ~(Ex)(x=x) ?

...
--
Greg Lee <l...@hawaii.edu>

Bill Taylor

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Aug 11, 2001, 12:07:33 AM8/11/01
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gr...@ling.lll.hawaii.edu writes:

|> > So how would we say nothing exists in 1st-order PC?
|>
|> How about: ~(Ex)(x=x) ?

Or, of course, if you're a true mathie and don't mind empty whatnots, just

~(Ex)()

...and you might even be able to do without the empty bracket pair!

_
_| |___ ==================================
_| _| _|_ Bill Taylor UV = V U% = %
|___|_ | _|___ ==================================
|_| |_ _| W.Ta...@math.canterbury.ac.nz ^V = % ^% = V (!)
|___| | ==================================
W F C T |_|

Virgil

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Aug 11, 2001, 12:37:36 AM8/11/01
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If there were nothing, we wouldn't be here to wonder about it.

"Cogito, ...."

George Dance

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Aug 12, 2001, 11:38:50 AM8/12/01
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> Brian Holtz <Brian...@sun.com> wrote:
> ...
> > So how would we say nothing exists in 1st-order PC?
>
gr...@ling.lll.hawaii.edu wrote in message news:<9kvdc6$o8k$1...@news.hawaii.edu>...
> How about: ~(Ex)(x=x) ?
>

Because that would be equivalent to saying: (x)~(x=x), which is
clearly contradictory.

> ...

George Dance

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Aug 12, 2001, 11:44:32 AM8/12/01
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mat...@math.canterbury.ac.nz (Bill Taylor) wrote in message news:<9l2b25$bbr$2...@cantuc.canterbury.ac.nz>...


But those assertions would be equivalent to asserting: (x)~() and
(x)~, both of which are meaningless.

George Dance

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Aug 12, 2001, 11:49:36 AM8/12/01
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Virgil <vmh...@home.com> wrote in message news:<vmhjr2-AF3457....@news1.denver1.co.home.com>...

> If there were nothing, we wouldn't be here to wonder about it.
>
> "Cogito, ...."

Well, Descartes was wrong about the "Cogito." - he was assuming that
he was thinking, when all he knew for sure was that there was an
experience of thinking.

But the proof that there is something rather than nothing follows even
from this weakened premise. For if it is certainly the case that
there is an experience of thinking, then it certainly not the case
that there is no experience of thinking. Therefore there is something
(an experience of thinking) rather than nothing (no experience of
thinking). QED

gr...@ling.lll.hawaii.edu

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Aug 12, 2001, 1:46:58 PM8/12/01
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That is not clear to me. If there were an individual a, then together
with (x)(x=x), it would imply both a=a and ~(a=a), which is a
contradiction. But all that shows is that there is no individual
a, which is what I meant to say.

--
Greg Lee <l...@hawaii.edu>

Virgil

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Aug 12, 2001, 5:10:48 PM8/12/01
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In article <e5361400.01081...@posting.google.com>,
georg...@my-deja.com (George Dance) wrote:

If you are going that route, the best you can say is that you
believe that there is something.

George Dance

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Aug 13, 2001, 6:36:36 AM8/13/01
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Virgil <vmh...@home.com> wrote in message news:<vmhjr2-3BE9E7....@news1.denver1.co.home.com>...

I don't think so. I'm having the experience; that is a brute fact.
Because I'm having the experience, I can infer that there are other
things existing besides the experience. Eg, there is an experience
that I'm drinking coffee and typing on my computer keyboard, early in
the morning, in my apartment. So I infer that I, the coffee, the
keyboard, the apartment, the sun, and all the rest exist. And I'm
justified in doing so, so long as none of the inferences entail any
oppositions (contradictories or contraries).

David C. Ullrich

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Aug 13, 2001, 11:12:38 AM8/13/01
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In _standard_ fol the "models" are all assumed to be non-empty,
so that (x)~(x=x) is indeed contradictory.

Not to imply that standard fol is what you were referring to,
and yes, this bit of standard fol is a part that bugs more
than one person. But we want (x), Ex, and ~ to work the way
they do, and if we allow empty universes they don't.

>--
>Greg Lee <l...@hawaii.edu>


David C. Ullrich

George Dance

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Aug 14, 2001, 6:58:12 PM8/14/01
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> >> Brian Holtz <Brian...@sun.com> wrote:
> >> ...
> >> > So how would we say nothing exists in 1st-order PC?
> >>
> gr...@ling.lll.hawaii.edu wrote in message news:<9kvdc6$o8k$1...@news.hawaii.edu>...
> >> How about: ~(Ex)(x=x) ?
>
> George Dance <georg...@my-deja.com> wrote:
> > Because that would be equivalent to saying: (x)~(x=x), which is
> > clearly contradictory.
>
gr...@ling.lll.hawaii.edu wrote in message news:<9l6fei$cbp$1...@news.hawaii.edu>...

> That is not clear to me. If there were an individual a, then together
> with (x)(x=x), it would imply both a=a and ~(a=a), which is a
> contradiction. But all that shows is that there is no individual
> a, which is what I meant to say.

I don't think it is necessary to instantiate a constant, in order to
show the contradiction. It can be done by instantiating x as a free
variable as well.

Richard L. Epstein defines '=' as the predicate '__is identical to__',
which he interprets informally as "__is the same # as__', where # is a
noun or non phrase for the objects of the universe, and formally as:
"For any open wff A and any assignment of references %,%|=
x=y->(A(x)<->A(y/x), where y replaces some but not necessarily all
occurrences of x in A for which it is free." (*Predicate Logic,*
Wadsworth 2001, 210; symbols substituted).

Following that interpretation, and discussing a universe or domain of
possible objects, means reading ~Ex(x=x) as "There is no possible
object that is the same possible object as itself," and (x)~(x=x) as
"Every possible object is not the same possible object as itself."
Since possible objects can be conceived that are the same objects as
themselves, the assertion is contradictory.

gr...@ling.lll.hawaii.edu

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Aug 14, 2001, 8:30:04 PM8/14/01
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George Dance <georg...@my-deja.com> wrote:
...
>> George Dance <georg...@my-deja.com> wrote:
>> > Because that would be equivalent to saying: (x)~(x=x), which is
>> > clearly contradictory.
>>
> gr...@ling.lll.hawaii.edu wrote in message news:<9l6fei$cbp$1...@news.hawaii.edu>...
>> That is not clear to me. If there were an individual a, then together
...

> I don't think it is necessary to instantiate a constant, in order to
> show the contradiction. It can be done by instantiating x as a free
> variable as well.

I don't understand what you mean by "instantiating x as a free variable".
If you mean passing from (x)f(x) to f(x), if you justify that by
the rule of universal instantiation, then f(x) is an instance, and
x is an individual. That's what instantiation means. Often a
different style letter is used, instead of an x, to make this
clear.

Your use of "free variables" has the potential to become confusing.

> Richard L. Epstein defines '=' as the predicate '__is identical to__',
> which he interprets informally as "__is the same # as__', where # is a
> noun or non phrase for the objects of the universe, and formally as:
> "For any open wff A and any assignment of references %,%|=
> x=y->(A(x)<->A(y/x), where y replaces some but not necessarily all
> occurrences of x in A for which it is free." (*Predicate Logic,*
> Wadsworth 2001, 210; symbols substituted).

Note he says "any assignment of references". In the empty universe,
there are no references to assign.

> Following that interpretation, and discussing a universe or domain of
> possible objects, means reading ~Ex(x=x) as "There is no possible
> object that is the same possible object as itself," and (x)~(x=x) as
> "Every possible object is not the same possible object as itself."
> Since possible objects can be conceived that are the same objects as
> themselves, the assertion is contradictory.

What's the difference between a "possible object" and an "object"?
If you're saying there's a _possible_ non-empty universe, I guess I'd
go along with that. Tentatively.

In my understanding, you show a sentence or set of them to be contradictory
by showing that one can infer both a sentence and its negation.

--
Greg Lee <l...@hawaii.edu>

Owen Holden

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Aug 15, 2001, 7:12:59 AM8/15/01
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gr...@ling.lll.hawaii.edu wrote in message news:<9kvdc6$o8k$1...@news.hawaii.edu>...

> Brian Holtz <Brian...@sun.com> wrote:
> ...
> > So how would we say nothing exists in 1st-order PC?
>
> How about: ~(Ex)(x=x) ?
>
> ...


IMO, If we define 'x exists' as:

E!x, defined, Ey(x=y), then

1. Some x exists, something exists, ExE!x, means, ExEy(x=y)

2. Everything exists AxE!x, means, AxEy(x=y)

3. Nothing exists ~ExE!x, means, ~ExEy(x=y).

gr...@ling.lll.hawaii.edu

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Aug 15, 2001, 12:45:06 PM8/15/01
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> E!x, defined, Ey(x=y), then

But isn't this last equivalent to ~Ex(x=x) ? (I can prove it if
for individuals u, v, I can go from u=v to u=u.)
--
Greg Lee <l...@hawaii.edu>

Owen Holden

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Aug 15, 2001, 6:19:47 PM8/15/01
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gr...@ling.lll.hawaii.edu wrote in message <9le8ui$qc2$1...@news.hawaii.edu>...


I agree that they are equivalent. they are both theorems.

E!x<->x=x.

Curiously we can also say []AxE!x, and Ax[]E!x.

That everything is necessarily existent, or that it is necessary
that everything exists, denies all contingent existences.

Do you agree that this is not a suitable consequence of FOPL?

That something exists, Ex(x=x) or ExE!x, seems OK but, can we say
AxE!x or Ax(x=x), are theorems?

Do we affirm Quines dictum, 'to be is to be a value of a variable',
necessarily?

[](Greg Lee = Greg Lee), []E!(Greg Lee), seem to me very doubtful.


>Greg Lee <l...@hawaii.edu>


Owen Holden

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Aug 15, 2001, 6:46:54 PM8/15/01
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David C. Ullrich wrote in message <3b77edd3...@nntp.sprynet.com>...

Could you show, please, why the empty universe denies expected theorems,
given that 'Fa' is not meaningless when 'a' does not exist.

gr...@ling.lll.hawaii.edu

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Aug 15, 2001, 10:03:12 PM8/15/01
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Owen Holden <owenh...@yahoo.com> wrote:
...

> That everything is necessarily existent, or that it is necessary
> that everything exists, denies all contingent existences.

> Do you agree that this is not a suitable consequence of FOPL?

Yes. I've seen, somewhere or other, some discussions of logics free
of existence presuppositions. Perhaps by Nicholas Rescher.

David C. Ullrich

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Aug 16, 2001, 10:33:06 AM8/16/01
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??? I didn't say anything about "meaningless". The formula

Ax f(x) -> Ex f(x)

is a theorem of standard first-order logic. So it had better
be true in every interpretation, or soundness&completeness
goes away. If we allow an empty universe it is false in that
structure.

Find a book on logic.

>>>--
>>>Greg Lee <l...@hawaii.edu>
>>
>>
>>David C. Ullrich
>
>


David C. Ullrich

Owen Holden

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Aug 16, 2001, 11:22:35 AM8/16/01
to

David C. Ullrich wrote in message <3b7bd90d...@nntp.sprynet.com>...


You are wrong, if f(x) is false for non-existent x's then Axf(x) is false.
If f(x) is true for non-existent x's then Exf(x) is true.
As long as f(x) is meaningful, true or false, then Axf(x)->Exf(x) is valid,
empty universe or not!

>Find a book on logic.

Thanks for your advice?!

David C. Ullrich

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Aug 16, 2001, 5:43:24 PM8/16/01
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On Thu, 16 Aug 2001 15:22:35 GMT, "Owen Holden"
<owenh...@yahoo.com> wrote:

Very funny. Now note that I've included the word "standard"
everywhere, find a book and look up the standard definitions.
The definitions in standard first-order logic are not the
same as in Holden logic.

>>Find a book on logic.
>
>Thanks for your advice?!

No charge.

>>
>>>>>--
>>>>>Greg Lee <l...@hawaii.edu>
>>>>
>>>>
>>>>David C. Ullrich
>>>
>>>
>>
>>
>>David C. Ullrich
>
>


David C. Ullrich

George Dance

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Aug 19, 2001, 12:00:04 PM8/19/01
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gr...@ling.lll.hawaii.edu wrote in message news:<9lcfqc$rv0$1...@news.hawaii.edu>...

> George Dance <georg...@my-deja.com> wrote:
> ...
> >> George Dance <georg...@my-deja.com> wrote:
> >> > Because that would be equivalent to saying: (x)~(x=x), which is
> >> > clearly contradictory.
> >>
> gr...@ling.lll.hawaii.edu wrote in message news:<9l6fei$cbp$1...@news.hawaii.edu>...
> >> That is not clear to me. If there were an individual a, then together
> ...
> > I don't think it is necessary to instantiate a constant, in order to
> > show the contradiction. It can be done by instantiating x as a free
> > variable as well.
>
> I don't understand what you mean by "instantiating x as a free variable".
> If you mean passing from (x)f(x) to f(x), if you justify that by
> the rule of universal instantiation, then f(x) is an instance, and
> x is an individual. That's what instantiation means. Often a
> different style letter is used, instead of an x, to make this
> clear.
>
> Your use of "free variables" has the potential to become confusing.

I don't think I have a non-standard idea of free variables; but it
seems very possible that I'm inferring and saying non-standard things
about them. Which is one reason I'm pursuing this topic; to check
what I'm saying against what else is being said, including what is
standardly said.

In this case, I think you can instantiate (x)Fx in two ways: as a
constant, Fa
which means to me "(at least) one thing in particular is an F;" or as
a free variable, Fy, which means to me "any thing, whatever thing one
can pick out, is an F." Existential instantiation Fa is out in this
case, as it begs the question: it refutes the claim that nothing
exists by assuming that something exists; whereas universal
instantiation, as it does not assume that something exists, is
non-question-begging.

> > Richard L. Epstein defines '=' as the predicate '__is identical to__',
> > which he interprets informally as "__is the same # as__', where # is a
> > noun or non phrase for the objects of the universe, and formally as:
> > "For any open wff A and any assignment of references %,%|=
> > x=y->(A(x)<->A(y/x), where y replaces some but not necessarily all
> > occurrences of x in A for which it is free." (*Predicate Logic,*
> > Wadsworth 2001, 210; symbols substituted).
>
> Note he says "any assignment of references". In the empty universe,
> there are no references to assign.
>
> > Following that interpretation, and discussing a universe or domain of
> > possible objects, means reading ~Ex(x=x) as "There is no possible
> > object that is the same possible object as itself," and (x)~(x=x) as
> > "Every possible object is not the same possible object as itself."
> > Since possible objects can be conceived that are the same objects as
> > themselves, the assertion is contradictory.
>
> What's the difference between a "possible object" and an "object"?
> If you're saying there's a _possible_ non-empty universe, I guess I'd
> go along with that. Tentatively.

That wasn't what I'd meant to say. What I meant was more like: If
we're going to analyze a claim that "nothing exists," we cannot assume
that any objects exist (that's the question-begging I noted above.)
All that there is, is experience or thought - and since we can't talk
about that non-self-referentially in fol, we can't even say in fol
that the experience or thought "exists". What we can say, though, is
that we can experience or think that some objects exist, without that
entailing a contradiction. That means it's possible for such objects
to exist, meaning only that their non-existence (if indeed they do not
exist - we can't assume that, either) is purely contingent. If it's
possible for one such object to exist, then that object exists as a
"possible object," and the same with all the others.

> In my understanding, you show a sentence or set of them to be contradictory
> by showing that one can infer both a sentence and its negation.

You're right. I was wrong. While I was claiming that (x)~(x=x) is
self-contradictory, that wasn't what I was actually thinking: I was
actually assuming (x)(x=x), and thinking that (x)~(x=x) contradicted
with that.

But I don't see how I cannot assume the truth of (x)(x=x), even for
"possible objects" (as defined above). For example, whatever object I
think of, at any time, has to be the same object as the one that I am
thinking of at that time; and to say that "The object that I am
thinking of now is not the same object as the one that I am thinking
of now" still _sounds_ to me like something that cannot be asserted
without contradiction.

George Dance

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Aug 19, 2001, 12:07:12 PM8/19/01
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"Brian Holtz" <Brian...@sun.com> wrote in message news:<6TEc7.47069$Kd7.27...@news1.rdc1.sfba.home.com>...

> So how would we say nothing exists in 1st-order PC? I guess in
> 2nd-order PC it might be: (for all x)(for all P)(! Px).

My (tentative) answer would be: Don't even try. Rather, I think the
best bet is to give "existence" the meaning it has in fol - something
"exists" if we can say something about it - and use a different
predicate, "__is real", to translate the standard non-logical meaning
of "__exists". That allows one to say things like Ex~Rx, which is
perfectly clear and sensible.

Owen Holden

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Aug 20, 2001, 5:47:11 AM8/20/01
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georg...@my-deja.com (George Dance) wrote in message news:<e5361400.01081...@posting.google.com>...


This 'meaning' of existence won't do. We can say many things about
non-existent things too. For example, It is not the case that the
present King of France is bald, is true, and says something about the
non-existent object..the present King of France.

'That which is not equal to itself is not equal to itself', is a true
statement about the non-existent object 'that which is not equal to
itself'.

'The whole number between 2 and 3 is greater than 0, is false' is true
and about the non-existent whole number between 2 and 3.

etc..

We cannot define 'X exists' as 'Some statement is true about X'!

George Dance

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Aug 22, 2001, 6:20:39 PM8/22/01
to
owenh...@home.com (Owen Holden) wrote in message news:<73412d9.01082...@posting.google.com>...

> georg...@my-deja.com (George Dance) wrote in message news:<e5361400.01081...@posting.google.com>...
> > "Brian Holtz" <Brian...@sun.com> wrote in message news:<6TEc7.47069$Kd7.27...@news1.rdc1.sfba.home.com>...
> >
> > > So how would we say nothing exists in 1st-order PC? I guess in
> > > 2nd-order PC it might be: (for all x)(for all P)(! Px).
> >
> > My (tentative) answer would be: Don't even try. Rather, I think the
> > best bet is to give "existence" the meaning it has in fol - something
> > "exists" if we can say something about it - and use a different
> > predicate, "__is real", to translate the standard non-logical meaning
> > of "__exists". That allows one to say things like Ex~Rx, which is
> > perfectly clear and sensible.
>
> "Rather, I think the
> best bet is to give "existence" the meaning it has in fol - something
> "exists" if we can say something about it -"

That was a sloppy way of putting it; should have been "if we can say
something true about it".


>
> This 'meaning' of existence won't do. We can say many things about
> non-existent things too. For example, It is not the case that the
> present King of France is bald, is true, and says something about the
> non-existent object..the present King of France.

~Ex(Fx^Bx) ["Fx=x is the King of France" "Bx=x is bald"] says nothing
about any such object; it's merely a way of saying (x)(~Fxv~Bx), which
is a statement about objects in general (that does assume that *at
least one thing exists* that either is not the present King of France
or is not bald).

> 'That which is not equal to itself is not equal to itself', is a true
> statement about the non-existent object 'that which is not equal to
> itself'.

(x)(~(x=x)->~(x=x)) sounds like a (weird and somewhat paradoxical) way
of saying (x)(x=x)->(x=x). It actually says nothing about that
non-existent object, either.

> 'The whole number between 2 and 3 is greater than 0, is false' is true
> and about the non-existent whole number between 2 and 3.

This is another "it is not the case..." sentence, as in your first
example.
Same comments apply.

> etc..
>
> We cannot define 'X exists' as 'Some statement is true about X'!

You actually haven't made any true statements about those objects.
Can you come up with any? Is it true, for example, that

1. The present King of France has hair on his scalp?; or that
2. The object that is not equal to itself is smaller (or larger, or
the same size - pick any one) than I am?; or that
3. The whole number between 2 and 3 is less than or equal to 0?

That is the only way to show that (x)Fx |= (Ex)Fx (which is not itself
a definition, but does follow from the definition of the quantifiers)
is invalid.

George Dance

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Aug 22, 2001, 9:45:12 PM8/22/01
to
Maybe I'm being overly sensitive; but I think these corrections should
be made by me rather than by a flamer:

georg...@my-deja.com (George Dance) wrote in message news:<e5361400.01082...@posting.google.com>...


> ~Ex(Fx^Bx) ["Fx=x is the King of France" "Bx=x is bald"] says nothing
> about any such object; it's merely a way of saying (x)(~Fxv~Bx), which
> is a statement about objects in general (that does

imply

> that *at least one thing exists* that either is not
> the present King of France
> or is not bald).

and

> (x)(~(x=x)->~(x=x)) sounds like a (weird and somewhat paradoxical) way
> of saying

(x)(x=x)->(x=x)).


> It actually says nothing about that
> non-existent object, either.

(Though that statement, too, implies that something exists.)

Owen Holden

unread,
Aug 23, 2001, 9:43:18 PM8/23/01
to

George Dance wrote in message ...

>owenh...@home.com (Owen Holden) wrote in message
news:<73412d9.01082...@posting.google.com>...
>> georg...@my-deja.com (George Dance) wrote in message
news:<e5361400.01081...@posting.google.com>...
>> > "Brian Holtz" <Brian...@sun.com> wrote in message
news:<6TEc7.47069$Kd7.27...@news1.rdc1.sfba.home.com>...
>> >
>> > > So how would we say nothing exists in 1st-order PC? I guess in
>> > > 2nd-order PC it might be: (for all x)(for all P)(! Px).
>> >
>> > My (tentative) answer would be: Don't even try. Rather, I think the
>> > best bet is to give "existence" the meaning it has in fol - something
>> > "exists" if we can say something about it - and use a different
>> > predicate, "__is real", to translate the standard non-logical meaning
>> > of "__exists". That allows one to say things like Ex~Rx, which is
>> > perfectly clear and sensible.
>>
>> "Rather, I think the
>> best bet is to give "existence" the meaning it has in fol - something
>> "exists" if we can say something about it -"
>
>That was a sloppy way of putting it; should have been "if we can say
>something true about it".

It is still not enough to say, (x exists)<->(there is some true statement
about x).
For example; (Fx v ~(Fx)), is true and true about x, whether x exists or
not.


>>
>> This 'meaning' of existence won't do. We can say many things about
>> non-existent things too. For example, It is not the case that the
>> present King of France is bald, is true, and says something about the
>> non-existent object..the present King of France.
>
>~Ex(Fx^Bx) ["Fx=x is the King of France" "Bx=x is bald"] says nothing
>about any such object; it's merely a way of saying (x)(~Fxv~Bx), which
>is a statement about objects in general (that does assume that *at
>least one thing exists* that either is not the present King of France
>or is not bald).

You are confusing 'A king' with The King'.

Fx = x is King of France, Bx = x is bald,

A King of France is bald, means, Ex(Fx & Bx)

The King of France is bald, means, EyAx(x=y<->Fx .& Bx).
The definite article 'the' requires uniqueness, unlike the indefinite
article.

A King of France exists, means, ExFx.
The King of France exists, means EyAx(x=y<->Fx).

The King of France does not exist, ie. ~EyAx(x=y<->Fx).

Fx, is false for all x, because there is no Kingship in France at the
present time.

Therefore EyAx(x=y<->False)
<->EyAx~(x=y)
<->~AyEx(x=y)
But, AyEx(x=y) is a theorem of Identity, see *13.19 of Principia
Mathematica.

Therefore 'The present King of France exists' is false. That is, ~E!(the
present King of France).

That is, the described object 'the present King of France' does not exist.

The King of France is bald, is likewise false, that is, ~EyAx(x=y<->Fx .&
Bx).

The King of France is not bald, is also false but,
It is not the case that, the King of France is not bald, is true.

Ey(x=y<->Fx .& Bx) is false.
~(Ey(x=y<->Fx .& Bx) is true
EyAx(x=y<->Fx .& ~Bx) is false.
~(EyAx(x=y<->Fx .& ~Bx)) is true

>
>> 'That which is not equal to itself is not equal to itself', is a true
>> statement about the non-existent object 'that which is not equal to
>> itself'.
>
>(x)(~(x=x)->~(x=x)) sounds like a (weird and somewhat paradoxical) way
>of saying (x)(x=x)->(x=x). It actually says nothing about that
>non-existent object, either.
>

(that which is not equal to itself)=(that which is not equal to itself),
means,

(the x: ~(x=x))=(the x:~(x=x))<->EyAx(x=y<->~(x=x) .& y=y)
(the x: ~(x=x))=(the x:~(x=x))<->EyAx(x=y<->~(x=x)) since
y=y.
(the x: ~(x=x))=(the x:~(x=x))<->EyAx~(x=y)
(the x: ~(x=x))=(the x:~(x=x))<->~AyEx(x=y)
(the x: ~(x=x))=(the x:~(x=x))<->False
Therefore
~((the x: ~(x=x))=(the x:~(x=x))).

~E!(the x: ~(x=x)), is proven the same way.

It is false that (that which is not equal to itself) exists , is a theorem!

E!x<->x=x, if x exists or not, for names or descriptions.


>> 'The whole number between 2 and 3 is greater than 0, is false' is true
>> and about the non-existent whole number between 2 and 3.
>
>This is another "it is not the case..." sentence, as in your first
>example.
>Same comments apply.
>
>> etc..
>>
>> We cannot define 'X exists' as 'Some statement is true about X'!
>
>You actually haven't made any true statements about those objects.
>Can you come up with any? Is it true, for example, that
>

There are no primary predicates that are true of non-existent objects.

The definition of existence is: x exists, defined, EF(F!x), where !
indicates primary F.


>1. The present King of France has hair on his scalp?; or that
>2. The object that is not equal to itself is smaller (or larger, or
>the same size - pick any one) than I am?; or that
>3. The whole number between 2 and 3 is less than or equal to 0?
>
>That is the only way to show that (x)Fx |= (Ex)Fx (which is not itself
>a definition, but does follow from the definition of the quantifiers)
>is invalid.

It is not my intension to show that (x)Fx |= (Ex)Fx , is invalid, nor am I a
flamer.


George Dance

unread,
Aug 24, 2001, 6:39:05 AM8/24/01
to
"Owen Holden" <owenh...@yahoo.com> wrote in message news:<Wiih7.14624$n75.2...@news4.rdc1.on.home.com>...
snip

>
> It is not my intension to show that (x)Fx |= (Ex)Fx , is invalid, nor am I a
> flamer.

I can't even give your post the detailed study it deserves, let alone
the detailed reply I would like to make to it, right this minute. But
this last sentence of yours calls out for an immediate reply.

I think you are seriously debating a question of truth, because you
are interested in that question, and have given considerable thought
to it. I do not think that you are a "flamer," and my comment about
being flamers here in my brief correction was not a reference to you
or to anything you have written. I am sorry that it was
misunderstood, and hope that this corrects the misunderstanding.

George Dance

unread,
Aug 26, 2001, 8:58:02 AM8/26/01
to
"Owen Holden" <owenh...@yahoo.com> wrote in message news:<Wiih7.14624$n75.2...@news4.rdc1.on.home.com>...

> >> georg...@my-deja.com (George Dance) wrote in message
> news:<e5361400.01081...@posting.google.com>...
> >> > .... Rather, I think the

> >> > best bet is to give "existence" the meaning it has in fol - something
> >> > "exists" if we can say something about it - and use a different
> >> > predicate, "__is real", to translate the standard non-logical meaning
> >> > of "__exists". That allows one to say things like Ex~Rx, which is
> >> > perfectly clear and sensible.
> >>
> >owenh...@home.com (Owen Holden) wrote in message
> news:<73412d9.01082...@posting.google.com>...
> >> "Rather, I think the
> >> best bet is to give "existence" the meaning it has in fol - something
> >> "exists" if we can say something about it -"
> >
> George Dance wrote in message ...
> >That was a sloppy way of putting it; should have been "if we can say
> >something true about it".
>
> It is still not enough to say, (x exists)<->(there is some true statement
> about x).
> For example; (Fx v ~(Fx)), is true and true about x, whether x exists or
> not.

(Fx v ~(Fx)) is not a proposition, so it is not true. How could you
assert it it as a proposition? Asserting it as (x)(Fxv~Fx) entails
asserting (Ex)(Fxv~Fx); that the object(s) it is true of exists.
Asserting it as (Ex)(Fxv~Fx) is identical to asserting that the
object(s) it is true of exists.

> >> This 'meaning' of existence won't do. We can say many things about
> >> non-existent things too. For example, It is not the case that the
> >> present King of France is bald, is true, and says something about the
> >> non-existent object..the present King of France.
> >
> >~Ex(Fx^Bx) ["Fx=x is the King of France" "Bx=x is bald"] says nothing
> >about any such object; it's merely a way of saying (x)(~Fxv~Bx), which
> >is a statement about objects in general (that does assume that *at
> >least one thing exists* that either is not the present King of France
> >or is not bald).
>
> You are confusing 'A king' with The King'.

No, I was not. "It is not the case that the present King of France is
bald" can mean either "It is not the case that there is exactly one
King of France that is bald" or "It is not the case that there is any
King of France that is bald." As the first is very ambiguous (is it
true that there are two bald kings of France? 27 bald kings?), I chose
the second.


>
> Fx = x is King of France, Bx = x is bald,
>
> A King of France is bald, means, Ex(Fx & Bx)
>
> The King of France is bald, means, EyAx(x=y<->Fx .& Bx).
> The definite article 'the' requires uniqueness, unlike the indefinite
> article.

Not really. "I ride the bus to work every day" does not assert that I
ride a unique bus, and shouldn't be formalized as it it did. Rather, I
think you're making an assumption about kingship in present-day France
- that it requires one and only one king - which is false.

> A King of France exists, means, ExFx.
> The King of France exists, means EyAx(x=y<->Fx).

> The King of France does not exist, ie. ~EyAx(x=y<->Fx).

Which (I think) is equivalent to saying ~(Ex)Fx, which is equivalent
to asserting (Ax)(~(Fx)), which is what I said. Am I wrong about any
inference?


>
> Fx, is false for all x, because there is no Kingship in France at the
> present time.
>
> Therefore EyAx(x=y<->False)
> <->EyAx~(x=y)
> <->~AyEx(x=y)
> But, AyEx(x=y) is a theorem of Identity, see *13.19 of Principia
> Mathematica.
>
> Therefore 'The present King of France exists' is false. That is, ~E!(the
> present King of France).
>
> That is, the described object 'the present King of France' does not exist.

But you haven't described 'the present King of France' (or any of the
present Kings of France) at all.

> The King of France is bald, is likewise false, that is, ~EyAx(x=y<->Fx .&
> Bx).
> The King of France is not bald, is also false but,

But, you asserted earlier that Fxv~Fx (where F was not defined as
"__is the present King of France", but was meant to stand for any
possible predicate) was true of every object, including non-existent
objects. Let Fx stand for "x is bald." "The present King of France
is either bald or not bald" would then be a _true_ statement about the
present King of France. How can a disjunction be true, when both of
its disjuncts are false?

> It is not the case that, the King of France is not bald, is true.

> Ey(x=y<->Fx .& Bx) is false.
> ~(Ey(x=y<->Fx .& Bx) is true
> EyAx(x=y<->Fx .& ~Bx) is false.
> ~(EyAx(x=y<->Fx .& ~Bx)) is true

> >> 'That which is not equal to itself is not equal to itself', is a true
> >> statement about the non-existent object 'that which is not equal to
> >> itself'.
> >
> >(x)(~(x=x)->~(x=x)) sounds like a (weird and somewhat paradoxical) way
> >of saying (x)(x=x)->(x=x). It actually says nothing about that
> >non-existent object, either.

> (that which is not equal to itself)=(that which is not equal to itself),
> means,
>
> (the x: ~(x=x))=(the x:~(x=x))<->EyAx(x=y<->~(x=x) .& y=y)
> (the x: ~(x=x))=(the x:~(x=x))<->EyAx(x=y<->~(x=x)) since
> y=y.
> (the x: ~(x=x))=(the x:~(x=x))<->EyAx~(x=y)
> (the x: ~(x=x))=(the x:~(x=x))<->~AyEx(x=y)
> (the x: ~(x=x))=(the x:~(x=x))<->False
> Therefore
> ~((the x: ~(x=x))=(the x:~(x=x))).

Your conclusion then is that 'That which is not equal to itself is not
equal to itself', is a false statement (which is what the first tilde
says). So this is not a true statement about a non-existent object,
either, contrary to your earlier suggestion.

> ~E!(the x: ~(x=x)), is proven the same way.
>
> It is false that (that which is not equal to itself) exists , is a theorem!
>
> E!x<->x=x, if x exists or not, for names or descriptions.

Agreed.

> >> We cannot define 'X exists' as 'Some statement is true about X'!
> >
> >You actually haven't made any true statements about those objects.
> >Can you come up with any? Is it true, for example, that
> >
>
> There are no primary predicates that are true of non-existent objects.

> The definition of existence is: x exists, defined, EF(F!x), where !
> indicates primary F.

While this sounds like the definition I had in mind (though not
precisely the one I put on paper), it's not in the natural language,
nor am I sure how to put it there and keep its precision. 'There is
some primary predicate that is true of x'? or as 'There is some true
statement predicating something primary of x'? I would prefer not to
use the technical term 'primary' at all, if there were an equivalent
term with a more obvious meaning; but is there?

I'm interested in getting a true definition, not in defending a false
one (even if it's one that I previously made); and I would appreciate
your help in doing so.



> >1. The present King of France has hair on his scalp?; or that
> >2. The object that is not equal to itself is smaller (or larger, or
> >the same size - pick any one) than I am?; or that
> >3. The whole number between 2 and 3 is less than or equal to 0?
> >
> >That is the only way to show that (x)Fx |= (Ex)Fx (which is not itself
> >a definition, but does follow from the definition of the quantifiers)
> >is invalid.
>
> It is not my intension to show that (x)Fx |= (Ex)Fx , is invalid,

No, I can see that. If you believe AyEx(x=y) is a theorem, then you
obviously believe (Ax)Fx |= (Ex)Fx is valid as well.

I also see that you were not saying that my definition was completely
wrong, but only that it was inexact (which would still make it wrong).
I'm not completely convinced that it was inexact - that would depend
on your convincing me that "The King of France is either bald or not
bald" is a true statement - but I am willing to adopt a better
definition.

Alfred Einstead

unread,
Aug 30, 2001, 5:05:44 AM8/30/01
to
From Brian Holtz <Brian...@sun.com>:

> So how would we say nothing exists in 1st-order PC?

By formulating a theory that is self-contradictory.

So the ACTUAL question in the subject header is:

Why are the facts concerning the universe NOT self-contradictory?

A couple responses down the line:
> How about: ~(Ex)(x=x) ?

And a later response:


> 1. Some x exists, something exists, ExE!x, means, ExEy(x=y)

which is provably equivalent to Ex(x=x), using the axiom Ax(x=x).
This is also equivalent to Ex(0=0) or Ex(a->a) where "a" can be
replaced by any jumble of symbols which defines a closed
well-formed formula.

> 2. Everything exists AxE!x, means, AxEy(x=y)

A universal is considered true if vacuous. In that case, AxEy(x=y) is
true if nothing exists, which means that it can't be interpreted as

"Everything exists"
but rather as
"Everything that exists, exists".

> 3. Nothing exists ~ExE!x, means, ~ExEy(x=y).

or more simply the statement Ax(~(x=x)), since this can only be true if
vacuously true.

An additional comment about "personal existence":


> [](Greg Lee = Greg Lee), []E!(Greg Lee), seem to me very doubtful.

The identification of a train of space-time points as comprising the
"same object/entity" is not a well-defined concept and, at rock bottom,
not a physically meaningful concept at all.

I could just as easily (and arbitrarily) identify THIS the sequence of
space-time points comprising (using customary language for the sake
of argument here) as being the definition of a single entity X, which
the statement (X = X) is obviously true for (but not [](X = X); and the
same goes for [](Greg Lee = Greg Lee), which is equally arbitrary as
this entity whose description follows):

This train of space-time points/events defines the continuation of
a single entity:

A cosmic ray, up to the point it led to a train of particles upon
crashing into the atmosphere.

The sequence of particles that came off the cosmic ray, up to
the point the last particle in the sequence struck a certain
molecule in the air.

The sequence of molecules in the air, up to the point that
each came in contact with the next; finally to the point when the
last one hit a certain raindrop (I'm about to describe below).

The raindrop, up to the point it fell on a certain tree (that
I'm about to describe below).

The tree that a certain piece of lumber (I'm about to describe
below) came from, up to the time it was cut down, processed and
made into boards of lumber.

That piece of lumber which (I presume) went into making part of
the door on the house where your mother's father was raised.

The door on the house where your mother's father was raised,
up to the point when your mother's father touched it for the first
time.

Your mother's father, to the point when he had sex with your
mother's grandmother.

The sperm of your mother's father, at the point when it entered
her body.

The fusion of the sperm and egg that would eventually become your
mother

Your mother, up to the point of the fusion of the sperm and egg
that would become you

You, up to the point when you wrote this article.

Your article, up to the point when I saw it.

Me, from that point to the point when I wrote this article.

This article.

The entire USENET, a few minutes after the article was posted.

Owen Holden

unread,
Aug 31, 2001, 5:45:38 AM8/31/01
to

(Fx v ~(Fx)) is a universal propositional function, like x=x, Vx, etc..
It is true for any instance of x, a universal predicate.
It's true that FOPL interprets the existential quantifier (Ex)
as there is at least one 'existent' instance of the variable x.

For existent x's (Fx v ~Fx) is universal but, not so if x includes
descriptive values which do not exist. On the other hand
(Fx v ~(Fx)) is universal for existent and non-existent x's.

>
>> >> This 'meaning' of existence won't do. We can say many things about
>> >> non-existent things too. For example, It is not the case that the
>> >> present King of France is bald, is true, and says something about the
>> >> non-existent object..the present King of France.
>> >
>> >~Ex(Fx^Bx) ["Fx=x is the King of France" "Bx=x is bald"] says nothing
>> >about any such object; it's merely a way of saying (x)(~Fxv~Bx), which
>> >is a statement about objects in general (that does assume that *at
>> >least one thing exists* that either is not the present King of France
>> >or is not bald).
>>
>> You are confusing 'A king' with The King'.
>
>No, I was not. "It is not the case that the present King of France is
>bald" can mean either "It is not the case that there is exactly one
>King of France that is bald" or "It is not the case that there is any
>King of France that is bald." As the first is very ambiguous (is it
>true that there are two bald kings of France? 27 bald kings?), I chose
>the second.


1. "It is not the case that there is exactly one King of France that is
bald" , means
~(EyAx(x=y<->Fx .& Gy).

"The" requires uniqueness. I see no ambiguity in 1.

2. "It is not the case that there is any King of France that is bald."
means
~Ex(Fx & Gx)

This says 'It is not the case that 'a' present King of France is bald'.

It seem to me that you are confusing the definit article 'the' with the
indefinite
article 'a'.

>> Fx = x is King of France, Bx = x is bald,
>>
>> A King of France is bald, means, Ex(Fx & Bx)
>>
>> The King of France is bald, means, EyAx(x=y<->Fx .& Bx).
>> The definite article 'the' requires uniqueness, unlike the indefinite
>> article.
>
>Not really. "I ride the bus to work every day" does not assert that I
>ride a unique bus, and shouldn't be formalized as it it did.

How can you ride in a bus that is not particular, ie non-unique?

>Rather, I
>think you're making an assumption about kingship in present-day France
>- that it requires one and only one king - which is false.
>
>> A King of France exists, means, ExFx.
>> The King of France exists, means EyAx(x=y<->Fx).
>
>> The King of France does not exist, ie. ~EyAx(x=y<->Fx).
>
>Which (I think) is equivalent to saying ~(Ex)Fx, which is equivalent
>to asserting (Ax)(~(Fx)), which is what I said. Am I wrong about any
>inference?


Yes, you are wrong here IMO. ~ExFx says It is not the case that 'a King of
France exists'.
~EyAx(x=y<->Fx) says It is not the case that 'theKing of France exists'.

>> Fx, is false for all x, because there is no Kingship in France at the
>> present time.
>>
>> Therefore EyAx(x=y<->False)
>> <->EyAx~(x=y)
>> <->~AyEx(x=y)
>> But, AyEx(x=y) is a theorem of Identity, see *13.19 of Principia
>> Mathematica.
>>
>> Therefore 'The present King of France exists' is false. That is, ~E!(the
>> present King of France).
>>
>> That is, the described object 'the present King of France' does not
exist.
>
>But you haven't described 'the present King of France' (or any of the
>present Kings of France) at all.
>
>> The King of France is bald, is likewise false, that is, ~EyAx(x=y<->Fx .&
>> Bx).
>> The King of France is not bald, is also false but,
>
>But, you asserted earlier that Fxv~Fx (where F was not defined as
>"__is the present King of France", but was meant to stand for any
>possible predicate) was true of every object, including non-existent
>objects. Let Fx stand for "x is bald." "The present King of France
>is either bald or not bald" would then be a _true_ statement about the
>present King of France. How can a disjunction be true, when both of
>its disjuncts are false?

(The present King of France is bald, or, The present King of France is not
bald) is false.

(The present King of France is bald, or, It is not the case that The present
King of France is bald)
is tautologous.

>> It is not the case that, the King of France is not bald, is true.
>
>> Ey(x=y<->Fx .& Bx) is false.
>> ~(Ey(x=y<->Fx .& Bx) is true
>> EyAx(x=y<->Fx .& ~Bx) is false.
>> ~(EyAx(x=y<->Fx .& ~Bx)) is true
>
>> >> 'That which is not equal to itself is not equal to itself', is a true
>> >> statement about the non-existent object 'that which is not equal to
>> >> itself'.
>> >
>> >(x)(~(x=x)->~(x=x)) sounds like a (weird and somewhat paradoxical) way
>> >of saying (x)(x=x)->(x=x). It actually says nothing about that
>> >non-existent object, either.
>
>> (that which is not equal to itself)=(that which is not equal to itself),
>> means,
>>
>> (the x: ~(x=x))=(the x:~(x=x))<->EyAx(x=y<->~(x=x) .& y=y)
>> (the x: ~(x=x))=(the x:~(x=x))<->EyAx(x=y<->~(x=x)) since
>> y=y.
>> (the x: ~(x=x))=(the x:~(x=x))<->EyAx~(x=y)
>> (the x: ~(x=x))=(the x:~(x=x))<->~AyEx(x=y)
>> (the x: ~(x=x))=(the x:~(x=x))<->False
>> Therefore
>> ~((the x: ~(x=x))=(the x:~(x=x))).
>
>Your conclusion then is that 'That which is not equal to itself is not
>equal to itself', is a false statement (which is what the first tilde
>says).

Not so, It says, It is not the case that (that which is not equal to itself
is equal to itself).

>So this is not a true statement about a non-existent object,
>either, contrary to your earlier suggestion.
>
>> ~E!(the x: ~(x=x)), is proven the same way.
>>
>> It is false that (that which is not equal to itself) exists , is a
theorem!
>>
>> E!x<->x=x, if x exists or not, for names or descriptions.
>
>Agreed.
>
>> >> We cannot define 'X exists' as 'Some statement is true about X'!
>> >
>> >You actually haven't made any true statements about those objects.
>> >Can you come up with any? Is it true, for example, that
>> >
>>
>> There are no primary predicates that are true of non-existent objects.
>
>> The definition of existence is: x exists, defined, EF(F!x), where !
>> indicates primary F.
>
>While this sounds like the definition I had in mind (though not
>precisely the one I put on paper), it's not in the natural language,
>nor am I sure how to put it there and keep its precision. 'There is
>some primary predicate that is true of x'? or as 'There is some true
>statement predicating something primary of x'? I would prefer not to
>use the technical term 'primary' at all, if there were an equivalent
>term with a more obvious meaning; but is there?

Russell uses primary and secondary..see: Introduction to Mathematical
Philosophy
page 179.
Quine uses the term 'simple' predicate..see: Methods of Logic page 280.
See: Karl Lambert..Meinong and the Principle of Insdependence.

Owen

gr...@ling.lll.hawaii.edu

unread,
Aug 31, 2001, 2:37:58 PM8/31/01
to
Owen Holden <owenh...@yahoo.com> wrote:

> George Dance wrote in message ...

...


>>Not really. "I ride the bus to work every day" does not assert that I
>>ride a unique bus, and shouldn't be formalized as it it did.

> How can you ride in a bus that is not particular, ie non-unique?

He rides different buses every day. I do, in fact, ride the bus to
work every work day, and I transfer twice to a different bus going
to and from work, so the number of buses involved for me is six per
work day.

You seem to be taking Russell's theory as a serious description of how
"the" is used in English.

Owen Holden

unread,
Sep 1, 2001, 7:06:56 AM9/1/01
to

gr...@ling.lll.hawaii.edu wrote in message <9moli6$hvo$1...@news.hawaii.edu>...

>Owen Holden <owenh...@yahoo.com> wrote:
>
>> George Dance wrote in message ...
>...
>>>Not really. "I ride the bus to work every day" does not assert that I
>>>ride a unique bus, and shouldn't be formalized as it it did.
>
>> How can you ride in a bus that is not particular, ie non-unique?
>
>He rides different buses every day. I do, in fact, ride the bus to
>work every work day, and I transfer twice to a different bus going
>to and from work, so the number of buses involved for me is six per
>work day.

And each is unique.

>
>You seem to be taking Russell's theory as a serious description of how
>"the" is used in English.


Of course, what is the alternative?

>Greg Lee <l...@hawaii.edu>


gr...@ling.lll.hawaii.edu

unread,
Sep 1, 2001, 11:36:16 AM9/1/01
to
Owen Holden <owenh...@yahoo.com> wrote:

> gr...@ling.lll.hawaii.edu wrote in message <9moli6$hvo$1...@news.hawaii.edu>...
>>Owen Holden <owenh...@yahoo.com> wrote:
>>
>>> George Dance wrote in message ...
>>...
>>>>Not really. "I ride the bus to work every day" does not assert that I
>>>>ride a unique bus, and shouldn't be formalized as it it did.
>>
>>> How can you ride in a bus that is not particular, ie non-unique?
>>
>>He rides different buses every day. I do, in fact, ride the bus to
>>work every work day, and I transfer twice to a different bus going
>>to and from work, so the number of buses involved for me is six per
>>work day.

> And each is unique.

Each what? In reference to "the bus" I said I ride to work, no -- that's
the point of the example. In reference to the vehicles I referred to
by the phrase "number of buses", yes, in this example, each of those
is unique. (I can hardly give a relevant example without at some point
referring to two unique physical objects.)

>>
>>You seem to be taking Russell's theory as a serious description of how
>>"the" is used in English.


> Of course, what is the alternative?

What's the difference whether there is an alternative? If I don't have
an alternative, that doesn't convert Russell's theory from wrong to
right (considered as a theory of English "the").

_One_ alternative that works for some instances of "the" is to consider
it to be anaphoric -- it appears with the second of two noun phrases
with the same or corresponding reference. As in "If a wood house burns
down, you should rebuild the house using concrete." Or "My salamander
lost a leg, but eventually it grew the leg back again."

--
Greg Lee <l...@hawaii.edu>

Åke Persson

unread,
Sep 2, 2001, 4:36:30 PM9/2/01
to
Owen Holden wrote:

> For example; (Fx v ~(Fx)), is true and true about x, whether x exists or
> not.

'Not' means 'complement' or 'the other'. Only if Fx exists, there can exists
anything that contrasts to Fx. The disjunction (Fx v ~(Fx)) has only meaning
if the object x exists. Non-existing objects is only an abstraction. Indeed
there can not be such objects as non-existing and there are no reason to
concider the relevance of logic.

Brian Holtz wrote:

> So how would we say nothing exists in 1st-order PC?

We don't say it. Non-existing objects has no structure and follows no rules,
i.e. no logic.

Åke Persson

Stephane Ninin

unread,
Sep 5, 2001, 9:52:24 AM9/5/01
to
Alfred Einstead a ecrit:

>
> "Everything exists"
> but rather as
> "Everything that exists, exists".
>

copyright Douglas Adams. :)


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