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In response to a number of intractable metaphysical puzzles and issues in quantum mechanics some philosophers have proposed that it must be possible for identity to be indeterminate, i.e., for there to be an x and a y where it’s neither true nor false that x=y. Here’s an example of a metaphysical puzzle that might lead one to think this. Say there’s a boat made of some boards. As time passes, the boards are replaced one by one with new boards. Eventually, they’re all replaced. Call the original boat ‘B’ and the one with the replaced boards ‘B*.’ It seems consistent with everything said so far that B=B*. Yet now imagine another case where after having built B we disassemble it and then reassemble it. Call the reassembled boat ‘B**.’ Once again, it seems consistent with everything I’ve said about this case that B=B**. And this is so even if we imagine that it took a really long time to disassemble B. Say we go in one day and remove one board, months later we remove another board, and so on, until finally it has been disassembled completely, and then we reassemble it with the old boards. It doesn’t seem that the length of time it took to disassemble the old boat should prevent it from being the case that the original boat, B, is identical to the reassembled one, B**. But now imagine a third case. But really this isn’t a new case at all but just the first two combined into one. Remember that in the second case we would take a board out of the boat every few months until eventually we had removed all its boards. Now imagine that when I told you that I forgot to tell you something else. Whenever we removed the board, we replaced it with an old one. So when we had the boat completely "disassembled" we still had a boat sitting there in the shop, and this boat was B*. Similarly, when I told you about the first case I forgot to mention something. Whenever we replaced one of the boat’s boards, we left the old one lying around the shop. Then we took all the old boards and put them back together again, leaving us with a boat in its own right, namely B**. Filling in the details like this seems to create a problem, for B* and B** are definitely different boats. Before we concluded that B=B* and B=B**. And in discussing the first case, didn’t I tell everything that was relevant to the identity of B*? Why should facts about the origin and identity of some other boat (namely B**) have any relevance to the question of what boat B* is? Similarly, why should facts about the origin and identity of B* have any relevance to the issue of what boat B** is, since B*¹ B**?

The philosophical literature presents us with four main responses to identity puzzles such as the one just presented. First, one might say that composite entities such as boats don’t exist, strictly speaking, or second, one might say that when such objects undergo any changes the result is a new object. Third, one might say that it’s really B* and not B** that is identical to B (or vice-versa). And finally, one might claim that there is no fact of the matter as to whether it’s B* or B** that is identical to B. Sometimes the identity relation can be indeterminate or vague, and this is a case in point. Thus it’s neither true nor false that B=B* and neither true nor false that B=B**. The first option will seem unattractive to most philosophers because of the prima facie implausibility of the claim there are no (persisting) boats. Neither does it seem plausible that one can destroy an old boat and create a new one just by giving the old one a small scratch. The third option, however, is also problematic, because for many such identity puzzles it will always seem arbitrary to say that it’s one particular identity that holds rather than the other. Thus the apparent plausibility of the final option.

So the thesis that indeterminate identity is possible is well motivated. However, others have argued that indeterminate identity (II) must be impossible. The most well known objection to the II thesis is this. Say it was neither true nor false that x=y. It seems to follow that x is such that it’s indeterminate whether it is y. But it’s not the case that y is such that it’s indeterminate whether it is y, for it’s definitely true that y is y. So it looks like there is at least one way in which x and y are different. Yet it seems that if there is any way in which x and y are different, then x and y must be different things. So, if it’s indeterminate that x=y, then it’s false that x=y. Therefore, it can’t be indeterminate that x=y. There are well known responses the defender of II can make to this argument, and in this paper I will not be adding anything to the discussion of whether it’s sound. Instead, I will present another problem for the II thesis. More specifically, I will present a problem for any attempt to provide a consistent set theory that allows for sets containing an object that is indeterminately identical to something (a vague object). Parsons and Woodruff (1999) have provided the most worked out attempt at a framework for set theory that allows for sets containing vague objects. Yet I will show that given the commitments of their system, they must deny that vague objects can be members of singleton sets unless they deny at least one of two claims. The first is that for any x, it’s not the case that {x} has two or more members; the second is that a set A has two or more members just in case there are distinct objects that are both members of A. Yet both of these claims seem obviously true. But I shall argue that if vague objects cannot belong to singleton sets, then they cannot belong to any finite sets. So it seems that Parsons and Woodruff’s attempt to allow for sets with vague objects in them is a failure. Furthermore, it seems that if there could be vague objects, there also could be singleton sets with vague objects in them, or at least could be some finite sets with vague objects in them. So we have another problem for the II thesis.

Informally, my argument against the existence of singleton sets with vague objects in them is this. Say it’s neither true nor false that x=y. It’s true that {x} doesn’t have two or more members, and so it’s true that it’s not the case that for some v and some w, v¹ w & vÎ {x} & wÎ {x}. Therefore, it’s true that for any v and any w, v=w Ú vÏ {x} Ú wÏ {x}; instantiating ‘v’ to ‘x’ and ‘w’ to ‘y’, we get that it’s true that x=y Ú xÏ {x} Ú yÏ {x}, and so either it’s true that x=y, it’s true that xÏ {x}, or it’s true that yÏ {x}. But none of these things can be true given our assumption that it’s neither true nor false that x=y. For by hypothesis it’s not true that x=y and it’s obviously not true that xÏ {x}, so it must be true that yÏ {x}. The problem is that x definitely is a member of {x}, and if it’s true that x is a member of some set while it’s also true that y isn’t a member of that same set, then x and y must be different things. So, on the assumption that it’s indeterminate that x=y and that there is a set {x}, it’s not indeterminate that x=y. It follows that vague objects cannot be members of singleton sets. (As we shall see, the above argument doesn’t assume the contrapositive of Leibniz’s Law or some other principle defenders of II typically deny.)

Before giving a more formal version of my argument, I need to define two logical connectives, and state some inference rules. As is usual, ‘Ñ ’ will be equivalent to ‘it is indeterminate that,’ so the proposition Ñ P will be true just in case it’s neither true nor false that P, and false otherwise. Following Parsons (2000), ‘!P’ will be true just in case it’s true that P, and false otherwise.

I will assume that an inference is valid just in case it’s truth-preserving. Given this understanding of validity, the following is a valid rule of inference:

In a three-valued logic, reductio and conditional proofs are different than in a bivalent logic. Just because one can derive a contradiction from P doesn’t mean that P is false, rather, though it does mean that P isn’t true. Still, if we can derive a contradiction from Ñ P, we may infer ~!Ñ P. But since Ñ P can only be true or false, if it isn’t true it’s false. Therefore, if we can derive a contradiction from Ñ P, we may validly infer ~Ñ P, as the following inference rule states:

IR2: Ñ P

. . .

Q

\ ~Ñ P

Finally, since neither true nor indeterminate propositions are false, IR3 and IR4 are valid, while IR5 is valid because indeterminate propositions aren’t true:

IR3: !P IR4: Ñ P IR5: Ñ P

\ ~!~P \ ~!~P \ ~!P

Now it’s time for my argument. It depends on four main assumptions. The first is that IR1 above is valid, which I will say more about later. Here are my other assumptions:

A1: For any set A, " x" y[!(xÎ A) & !(yÏ A) ® x¹ y]

A2: " x~({x} has 2 or more members)

D1: Set A has 2 or more members =_df $ x$ y[x¹ y & xÎ A & yÎ A]

A1 is explicitly accepted by Parsons and Woodruff (1999) in their attempt to allow for sets with vague objects in them, and rightly so. If it’s definitely true that x is a member of A and definitely false that y is, then how could x be y? But A2 seems equally obvious. It only says that singleton sets don’t have two or more members, which is the case since singleton sets have only a single member. And since for a set to have two or more members is for it to have a pair of distinct members, D1.

In the following argument assume that ‘a’ and ‘b’ denote objects, and that there is a set {a}.

1. Ñ (a=b) hypothesis

2. !~({a} has 2 or more members) A2, universal instantiation, IR1

3. !~$ x$ y[x¹ y & xÎ {a} & yÎ {a}] 2, D1

4. !(a=b Ú aÏ {a} Ú bÏ {a}) 3, quantification theory, DeM’s Laws,

5. !(a=b) Ú !(aÏ {a}) Ú !(bÏ {a}) 4, truth-table for ‘Ú ’

6. ~!(a=b) 1, IR5

7. !(aÎ {a}) from the definition of ‘{a}’

8. ~!(aÏ {a}) 7, IR3

9. !(bÏ {a}) 5, 6, disjunctive syllogism, 8, disjunctive

10. !(aÎ {a}) & !(bÏ {a}) ® a¹ b A1, universal instantiation (three times)

11. !(a¹ b) 7, 9, conjunction, 10, modus ponens, IR1

12. ~!(a¹ b) 1, IR4

13. ~Ñ (a=b) 1-12, IR2

We were able to infer 13 on the assumption that there was a set {a}. Thus if it’s true that this set exists, then 13. The same conclusion could have been inferred had we posited the existence of {b} instead of {a}, and so if it’s true that there is a set {a} or {b}, then 13. By universal generalization we may further infer that

14. " x" y[!(there is a set {x}) Ú !(there is a set {y}) ® ~Ñ (x=y)]

It’s a controversial question what the truth table for the conditional should be in a three-valued logic, and it’s also controversial whether one may infer from ~Q ® ~P from P ® Q in a three valued logic. In particular, defenders of II typically deny the validity of contraposition when P or Q might be indeterminate. Yet in 14 the propositions in the antecedent and in the consequent can only be true or false, and so we can set aside such concerns and infer 15 from it, from which we may infer 16:

15. " x" y[Ñ (x=y) ® ~(!(there is a set {x}) Ú !(there is a set {y}))]

16. " x" y[Ñ (x=y) ® ~!(there is a set {x}) & ~!(there is a set {y})]

So whatever is a vague object cannot be truly a member of a singleton set.

Now take an arbitrary finite non-empty set { . . . , x, . . . }. In Zermelo-Fränkel Set theory, " x_1" x₂ . . . " x_n{x₁, x₂, . . . , x_n}={x₁}È {x₂}È . . . È {x_n}, and so assuming the validity of IR1 we have !({ . . . , x, . . . )= . . . È {x}È . . . ). Presumably it follows from this that the sentence ‘{ . . . , x, . . . )= . . . È {x}È . . . ’ is true. So even if x is a vague object, ‘{ . . . , x, . . . )= . . . È {x}È . . . ’ is a true sentence. But this sentence cannot be true if x is a vague object given 16, for then it won’t be true that there is a set {x}. But if there isn’t truly such a set, then ‘{x}’ will be a non-referring term, making it the case that ‘{ . . . , x, . . . )= . . . È {x}È . . . ’ is neither true nor false. Therefore, if we assume that x is a vague object and that it’s the member of a set { . . . , x, . . . }, it follows that a certain sentence both is and isn’t true. So either it isn’t true after all that x is a vague object or else it isn’t true that there is a set { . . . , x, . . . }. Therefore, given 16 vague objects cannot truly be members of finite sets.

How might Parsons and Woodruff respond to the above argument? It has been shown that 16 is a consequence of the validity of IR1 – IR5, A1, A2, and D1. The validity of IR2 – IR5 follows from the meaning of the logical connectives ‘!’ and ‘Ñ ,’ which they use themselves. Since they affirm that "valid inferences are ones that preserve truth" (Parsons and Woodruff (1999: 474)), they’re also committed to IR1, and I’ve already mentioned their endorsement of A1. So unless they deny A2 or D1, they must accept 16. But to deny either is absurd. So they cannot block my argument for 16. But as we have seen, it then seems that they cannot allow for finite sets having vague objects in them as they thought they could.

If we want to allow for sets having vague objects in them, it seems to me that the most promising strategy for blocking my argument is to deny the validity of IR1. As far as I can tell, that IR1 is valid is the only assumption of my argument that has been denied by someone (by van Inwagen (1988) and Cowles (1994)). One will take IR1 to be valid if one thinks that valid inferences are the ones that preserve truth, but according to van Inwagen valid inferences have to preserve at least as high a truth value in the conclusion as in the premises. So, for instance, if a form of inference is such that it guarantees that true premise(s) lead to a true conclusion, but also such that it allows indeterminate premise(s) to lead to a false conclusion, the inference won’t be valid. This is the sort of inference IR1 is. If P is indeterminate, then !P will be false, and therefore in inferring !P from P one will go from something that is indeterminate to something that is false. So according to van Inwagen’s understanding of validity, IR1 isn’t valid.

However, it seems to me that this way of understanding validity is mistaken. If we have it that P, it just seems to me that it should follow that it’s true that P. Imagine, for instance, that someone asserts that P, and we then say to him, "So, it’s true that P?" Now imagine he says, "Hold on, it doesn’t follow from what I said that it’s true that P, all I’m saying is that P!" I definitely have the intuition that this guy is mistaken, and that the inference he rejects is actually valid. However, I also realize that some philosophers have very strong intuitions that indeterminate identity must be possible, and most of these philosophers will want to allow for sets with vague objects in them. But in the very least I think I have shown that such philosophers must make a difficult decision. They must either reject the II thesis, reject the validity of IR1, or else claim that while there can be vague objects, such objects cannot be members of finite sets. Thus any set theory which presupposes the truth preserving understanding of validity as Parsons and Woodruff’s does won’t also be able to allow for sets containing vague objects.

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