Friday, September 07, 2012

You can follow this conversation by subscribing to the comment feed for this post.

(1) is I think false because names in MPL must denote existing objects. Hence, in the possible worlds in which Venus exists '~(∃x)(x = Venus)' is false and in the possible worlds in which it does not exist '~(∃x)(x = Venus)' is nonsense, ie. not a proposition at all.

You have argued before that *Venus might have not existed* cannot be adequately parsed in MPL. I think you're right. At any rate, (1) is not it.

I don't see how the first could possibly be true, in classical MPL (but I am not an expert).

I agree with Jan and Ed. I would have to reject (1) on the grounds that I don't know how to marry the semantics of first order logic (with equality) to the semantics of the modal operators. In particular, logical constants present a real problem. For there are possible worlds in which the referent of 'Venus' is not at all well-defined, and as you have pointed out, this defeats the conditions for applying a first order language with such a logical constant to that world. Section 3.2 of the SEP article Modern origins of modal logic seems partly to address this issue, but it's not clear to me what conclusion is arrived at. I'd very much appreciate enlightenment on this from an expert.

Jan and Ed,

Are you guys familiar with the New Testament phrase, "Straining at a gnat while swallowing a camel"? It seems apropos in this context.

First of all, would you stumble over

'Venus does not exist' is possibly true?

I hope not. Venus exists. But it might not have existed, i. e., it is possible that Venus not exist. (1) is simply the Quinean translation of this.

Now put on your thinking caps and try again!

And just now I find that David has joined Jan and Ed.

David,

You agree that 'Venus does not exist' is logically equivalent to '~(∃x)(x = Venus).'

Do you agree that it is possible that Venus not exist? I hope so.

Well then, what is the problem with expressing that in terms of (1)?

Dr Vallicella,

I assumed your argument is given in a particular formal system --- MPL extended with modal operators. If this is the case, I challenge you to solve the triad without denying (1). In this formal system the falsity of (1) is not straining at a gnat but a matter of fact (as far as I know the facts).

On the other hand the argument may not be intended to be read in a formal system and the logical expressions are just shorthand for natural language expressions. It should then be read as:

1*. It might have been the case that nothing that existed was identical to Venus.
---
3*. It might have been the case that something was not identical to anything that existed. (by existential generalisation).

But (3*) is not self contradictory because the "somethings" compared belong to different possible worlds. Indeed, there is something in the actual world that is not identical to anything in some other possible world. The straightforward reading of this commits one to non-existing objects, which is another can of worms. The immediate contradiction has however been resolved (or so I say).

"[W]ould you stumble over

'Venus does not exist' is possibly true?"

Not at all, though I'd prefer to say 'Venus might have not existed' (this expression makes it clear this is in natural language and not in a formal system). This is obviously true.

Do I agree that 'Venus does not exist' is logically equivalent to '~(∃x)(x = Venus).' No, I do not. I'm sure the Quinean formulation in terms of identity is wrong. I don't think 'Venus does not exist' denotes a fact. Rather it's a dialectical move which rescinds an earlier implicit existential assertion by which the proper name 'Venus' acquires a meaning. More here.

David,

Surely 'Venus exists' is log. equivalent to 'Something is identical to Venus.' In MPL notation: '(Ex)(x = Venus).'

Of course, there is no fact corresponding to 'Venus does not exist.' But there is the proposition expressed by that sentence, and surely it is log. equivalent to the prop. expressed by '(Ex)(x = Venus.'

If you don't agree with this, then we have precious little common ground, and we cannot even begin to get on to the interesting questions.

Jan,

Will try to respond to you tomorrow.

Bill,

Let us suppose that '~(∃x)(x = Venus)' is true. Then '(∀x)~(x=Venus)' is true. Hence, by UI, '~(Venus=Venus)'. But this is false. Ergo, our assumption, '~(∃x)(x = Venus)', is false. And this argument appears to work in any world, so we have shown that '~(∃x)(x = Venus)' is necessarily false. Hence it cannot be a good regimentation of/be logically equivalent to 'Venus does not exist', for there are worlds in which we would deem the latter true. So the Quinean thin theory looks decidedly dodgy, which I took to be your theme in this series of posts. So we agree to that extent.

If we look a little deeper into why this goes wrong I think we arrive at Jan's objection, viz, in MPL, '(x=Venus)' is applicable/makes sense only in worlds in which Venus exists. So MPL lacks the resources to express these negative singular existentials.

I'm interested to see where you are taking this line of thought, but I'm interested in these problems of formalisation too.

Jan,

Of the three propositions, the second is the most dubious and ought to be rejected. The inference is from

~(Ex)(x = Venus)
to
(Ey)~(Ex)(x = y).

If the premise is true, then 'Venus' is an empty name. But then one cannot existentially generalize on it to arrive at the conclusion. If it is not the case that something is identical to Venus, it does not follow that there is something y such that y is not identical to anything.

Excellent comment, David, which I will try to respond to later in the day. But now, on to less ethereal pursuits.

David,

"And this argument appears to work in any world, so we have shown that '~(∃x)(x = Venus)' is necessarily false."

To be precise, you have shown that '~(∃x)(x = Venus)' is false in every world in which it is meaningful. If that is what one means by 'necessarily false', so be it. This is also my position.

Dr Vallicella,

"If the premise is true, then 'Venus' is an empty name"

Empty names are not allowed in MPL. Could you please clarify if your argument is to be read in MPL, some other formal system or not in a formal system at all?

Addendum: you have shown that '~(∃x)(x = Venus)' is false in every world in which it is meaningful *under MPL*. Some other logic, or natural language reasoning, may fare better. This simply shows that '~(∃x)(x = Venus)' is not an adequate parsing of 'Venus does not exist' in MPL. If I'm not mistaken, Quine implicitly acknowledges this by parsing 'Pegasus does not exist' as 'nothing pegasises' (∀x ~Px) and not as 'nothing is identical to Pegasus' (~(∃x)(x = Pegasus)).

Jan,

'Venus' is not an empty name like 'Vulcan.' So why isn't '~(∃x)(x = Venus)' kosher in MPL? Surely the negation of that formula is a wff (well-formed formula). So why isn't it a wff?

My argument is to be read in MPL. So (1) is to be read as

Poss ~(∃x)(x = Venus)

where 'Poss' stands in for the diamond as that symbol is used in standard modal propositional logic.

Jan,

What I have been examining is van Inwagen's Quinean version of the thin theory according to which there is an eschewal of the notion that 'exists' is a 2nd level predicate.

But of course one could take '= Venus' to pick out a property in '~(∃x)(x = Venus)' in which case the formula says that identity-with-Venus is not instantiated. But that brings us back to the 2nd level analysis which has its own problems.

Is it clear what I am saying?

David,

Your argument is similar to one I already gave. But then you need to deal with Andrew's objection.

You say, >>And this argument appears to work in any world, so we have shown that '~(∃x)(x = Venus)' is necessarily false.<<

Now your argument works in the actual world. But in worlds in which Venus does not exist, it won't be true that (x)~(x = Venus).

I had argued that from the logical necessity of '(x)(x = x)' one can validly infer the logical necessity of 'Venus = Venus' Andrew balked at that move. His point was that Venus = Venus only in worlds in which it exists, so that Venus' self-identity is as contingent as its existence.

Dr Vallicella,

"Is it clear what I am saying?"
You are saying (correct me if I am wrong) that according to Q-vI '=Venus' does not induce a property, therefore one cannot perform existential generalisation on it, therefore (2) is ungrounded. Clear. However, when Quine says 'nothing Pegasises', he seems to me explicitly committed to properties of the aforementioned kind. Is this a tension (or even a contradiction) in Quine's theory or my misunderstanding?

"So why isn't '~(∃x)(x = Venus)' kosher in MPL? Surely the negation of that formula is a wff (well-formed formula). So why isn't it a wff?"

It is a wff in the actual world. It is not a wff in a world in which Venus doesn't exist, because is such a world 'Venus' is an empty name.

"Now [David's] argument works in the actual world. But in worlds in which Venus does not exist, it won't be true that (x)~(x = Venus)."

Yes. What is important is that it won't be false either because it is not well formed. Hence '~(∃x)(x = Venus)' is true in no possible world, hence (1) is false.

Jan,

I would have thought that whether a formula is a wff depends on purely syntactical considerations, i.e., rules for concatenating uninterpreted signs, and not on any semantic considerations.

You seem to be saying that '~(∃x)(x = Venus)' is a wff in the actual world because 'Venus' has a referent in the actual world, but that in possible worlds in which Venus does not exist, it is not a wff.

Is that what you are saying?

There is also this consideration. One might argue that all that's required is that the formula be a wff in the actual world. We can evaluate it relative to some merely possible world without it having to exist there.

Jan,

I finally get what you are saying.

In the actual world, '~(∃x)(x = Venus)' is false, and in every other world it is not well-formulated, hence neither true nor false. Therefore, the formula is not true in any world!

The objection to this is that well-formedness is purely syntactical, hence world-invariant: it cannot vary from world to world.

"In the actual world, '~(∃x)(x = Venus)' is false, and in every other world it is not well-formulated, hence neither true nor false. Therefore, the formula is not true in any world!"

With the modification that the formula is not well defined in any world in which in Venus does not exist and not in any non-actual world simplicitier, this has indeed been my main point since the first comment.

"The objection to this is that well-formedness is purely syntactical, hence world-invariant"

It is a not well-formed expression due to semantic and not syntactical considerations. Of course the syntax is fine as it is world-invariant. Semantics however isn't; the universe of discourse changes from world to world. In a world in which Venus does not exist 'Venus' is an empty name and all expressions containing the name are not well formed. I will offer an argument.

Denote Socrates by s and Let P be any essential property of Socrates, say 'is a rational animal'. Let W be a world in which no animals exist (I will be using 'exist' tenselessly). Hence, Socrates does not exist in W.
1. Ps is well defined in W (for reductio).
2. Ps is true in W (from (1) and because P is an essential property).
3. ∃x: Px in W (from (2) and existential generalisation)
4. There exists a rational animal in W. Contradiction with the definition of W.
(Remember that each possible world comes with its own domain of discourse and the operator '∃' quantifies over the domain of discourse of W and not of the actual world).

I don't understand why you affirm (2).

Do you agree that (1) implies that Ps has a truth value?

Bill,

I think Jan and I are pretty much on the same page with this, except that, rather than say it's a question of well-formedness (this being a purely syntactic issue, as you point out) I'd say that applying a first order language to a world in such a way as to leave a logical constant like 'Venus' without a referent defeats the guarantee that the first order rules of inference are truth-preserving. We should perhaps see the theory of first order languages as a piece of applicable mathematics. The mathematics shows that if certain conditions are met then the rules of inference are truth-preserving. If these conditions, one of which is that all logical constants must have referents, are not met then there are no guarantees. This is just as it always is when applying a piece of mathematics to the real world. If we want to allow logical constants to lack references then we have to move to a different logic (some species of free logic perhaps) which will offer guarantees for a somewhat different set of rules of inference.

The comments to this entry are closed.

Other Maverick Philosopher Sites

WWW
Member since 10/2008

May 2013

Sun Mon Tue Wed Thu Fri Sat
1 2 3 4
5 6 7 8 9 10 11
12 13 14 15 16 17 18
19 20 21 22 23 24 25
26 27 28 29 30 31