Aside

Suppose you’re a Humean best-system theorist about laws of nature. For S to be a law of nature is for S to feature (as an axiom? As a theorem?) in the best theory that predicts local matters of fact.

Suppose we have some candidate theory T—mentioning perhaps material objects, their physical properties, and locations in space-time. T is interpreted—-perhaps via interpretation I with domain D. Now apply the Loewenheim-Skolem theorem to this, to produce a countable domain D* with pretty much the same interpretation I. (That is: any names have the same referents, and the extension assigned to predicates is as on the original except intersected with the new domain. If you thought of the interpretation as first assigning properties, whose extension relative to a domain was just the set of things within that domain that instantiate the property, they we can leave the interpretation untouched). D*, I will make true exactly the same sentences in the language of T as D,I does. And in particular, it makes true the sentences that figure in T. So let T continue to stand for the interpreted theory, whose quantifiers range over D, and T* stand for the distinct interpreted theory, whose quantifiers range over D*. T and T* are syntactically identical (hence have the same syntactic consequences). And since their interpretations differ only wrt the quantifiers, if we concentrate on their predictions for “local matters of fact” (which I’ll assume for now can be expressed using quantifier-free vocabulary) we won’t find any differences in how these consequences are interpreted. So it looks like they “fit” the data to exactly the same extent.

If T is our best theory, then the laws will be exceptionless regularities. If T* is the best theory, then the laws will be restricted. After all, the domain of T*, and hence the scope of the quantifiers featuring in its generalizations, is countable, and so only ranges over e.g. countably many regions of space-time.  Datum: the scientific laws are those that T projects, not the cunning skolemite restrictions that T* talks about. How does the Humean secure this?

In Lewis’s story, we get this via simplicity considerations. Simplicity is measured for Lewis by economy of expression in a canonical language. And (on the way I read Lewis) the canonical language will contain an unrestricted quantifier (or at least, something unrestricted enough to range over all physically relevant stuff). But there’s a weirdness here. The principled fix that Lewis gives us on the canonical language concerns the predicates it contains (those corresponding to natural properties, rather than artificial stuff like “being such that T obtains”). But it’s really not clear what the principled reason is for selecting the rest of the canonical language—-why it should contain these connectives, those quantifiers, or whatever. Ted Sider notes this sort of lacuna, and proposes extending the Lewisian natural/non-natural distinction so it covers all categories of entity. That would do the trick here, if with Sider you think that the unrestricted quantifier is perfectly natural. But this is a considerable metaphysical commitment, both in extending the naturalness ideology and in endorsing the particular thesis about its application to unrestricted quantification (that’s partly Sider’s point). Instead of going Sider’s way, perhaps the line will be just that the canonical language contains unrestricted quantification, and there’s *no* deeper explanation for why. That’s just the way it has to be for the Humean proposal to return the extensionally correct results. Ok: but a bit disappointing if that’s the end of the story.

Another thing to focus on here is whether T* is as informative as T. After all, telling us only about how countably many things behave *sounds* less informative than telling us about how *all* things behave. But there’s an issue with how we unpack this idea. First, as noted, syntactically they have the same consequences (and in the quantifier-free case there’s no relevant difference of interpretation).

But there’s a more semantic way of understanding “informativeness”. It could be measured by, e.g. the volume of worlds at which the theory is true (smaller volume=more informative). In general, modal approaches to informativeness (whether framed in terms of volume, or the shape of the region defined in modal space, or anything else) will turn on whether the theory is true in this world vs. that world. To get a grip on this, we need a theory equipped not merely with an extensional interpretation, but with an intensional one. We can run the skolemite trick to build up something like this too (walk round from world to world, and at each world w, apply the skolemite construction to the extensions induced at w by the intended unrestricted interpretation. You now have countable domains for each w, such that holding the rest of the interpretation untouched, you get exactly the same L-sentences coming out true at w on the restricted as on the unrestricted interpretation. Now you can piece together a property P, whose extension at w is the skolemite domain you constructed at w. And by keeping the original intended intensional interpretation, except restricting the quantifier to P, we guarantee that that the resulting intensional interpretation will assign the same sentence-intensions to L-sentences as the intended one did. Net result: if we take T* to be the interpreted theory with quantifiers restricted to P, we have the same informativeness as T by any modal criterion. And of course, at every world, T* ranges over only countably many things.

So simplicity might knock T* out, but at least on Lewis’s proposal, only because we built unrestricted quantification rather than skolemite quantification into the canonical language from which simplicity is assessed from the get-go. Informativeness I don’t think would discriminate T from T*, if we extend the skolemite construction as indicated (this takes a bit of thinking through—at first it seems *obvious* that restricted claims are less informative than unrestricted ones—it’s well worth thinking through exactly how the skolemite construction works to undercut that natural thought). Is there some more satisfying way for the Humean to establish that laws of nature are truly general and unrestricted?

Here’s one idea : in order to derive *any* predictions about local matters of fact, it looks like we need in the language of the theory singular terms for local circumstances (the *local* prediction isn’t that any scenario with such-and-such qualitative characteristics will have thus-and-so features; but that *this* scenario has thus-and-so features, *because* it has such-and-such characteristics). And there’s a question about what singular terms do feature in the  language of the theory. The pure language of science (or Lewis’s canonical language) may not contain demonstratives—but to account for applications then it’s natural to supplement it with just such resources. If we just supplement the canonical language with just that stuff we actually use in applications, then presumably there’ll only be countably many such singular terms, and the argument can run as previously. If we supplement it with terms for each scenario whatsoever, then the argument is blocked—-for there’ll be more than countably-many local scenarios, and each such named scenario will have to feature in the domain.

(Thanks to Jason Turner for discussion of some of these ideas; though he shouldn’t be blamed for deficiencies in the above. Literature question: are there places out there where this issue is discussed?)

## Meaningless identities?

I run across the idea from time to time that some identity statements are “category mistakes” or “meaningless”—even when the terms flanking the identity have referents. In the philosophy of mathematics, it’s pretty common to see it suggested that “Julius Caesar=7” and “the natural number 2=the third von Neumann ordinal” are meaningless. Agustin Rayo’s recent book “The construction of logical space”, for example, advertises it as an advantage of his favoured approach to mathematical language that it allows for this phenomenon. Huw Price’s collection of papers on Naturalism and Representationalism suggests something similar for terms taken from different “discourses”. It’s an intriguing suggestion, but I can’t see how to make it work—for reasons independent of the “big picture” theories of ontology and meaning that it is supposed to articulate/challenge.

Here’s an argument against meaningless identities between referential singular terms. (The cogniscienti will recognize it as a variant of Evans’ famous argument against indeterminate identity; and in the light of that, I wouldn’t be surprised to learn that the point is already in the literature. References please!)

To start with, suppose that “A=B” is our candidate for a meaningless identity, and “A” and “B” have referents (thus, I’ll feel free to use those terms, rather than simply mention them). I offer two premises:

(1) A is an x such that there’s some term N referring to x, such that “N=A” is meaningful and true.
(2) B is not an x such that there’s some term N referring to x, such that “N=A” is meaningful and true.

By Leibniz’s law, since A possesses a property that B lacks, it follows:

(3) A is not identical to B.

I say that anyone who takes “A=B” to be meaningless is committed to (1) and (2), but must not commit herself to (3). Since (1) and (2) entail (3), this leaves her in an incoherent position.

The commitment to (1) is the easiest to defend. Just let the term N be “A”. I wouldn’t recommend resisting at this point.

The need to avoid commitment to (3) is also pretty obvious, so long as nothing meaningless can be true.

The most involved case to make is that our target theorist is committed to (2).  Here’s the argument for that: Suppose that there was an N such that B possessed the property mentioned. Then, ex hypothesi “N=A” would be meaningful and true, where “A” refers to A and “N” refers to B. Hence B=A. So possessing that property would entail the truth of something that’s meaningless. Just as in the previous paragraph, that’s impossible. So B must lack the property, just as (2) reports.

As mentioned, this is a variant of Evans’ argument. And various—but not all—the responses to Evans might have traction here. For example: a nonclassical logic might help resist certain of the moves; or endorse certain versions of Leibniz’s law but resist the one relied on above. On the other hand, what I think of as the most widely accepted caveat/objection to Evans’ argument—-inapplicability in cases of referential indeterminacy in the terms involved—doesn’t seem to me have any interesting analogue here.

There’s certainly new material in here that could be pinpointed as the problem. For example, the argument shoots up and down the semantic staircase, between the metalinguistic claim that “A=B” is meaningless, and the object language claims: A=B/~A=B. These are involved in (3) and (in more than one place) in the defence of (2). So questioning those moves, once they’re formalized in a fully spelled out form of the argument sketched, is one way forward our advocate of meaningless identities might favour.

However, though I see some options in the table, I’ve no idea exactly which way the friends of these “meaningless identity” view would jump in response. Ideas welcome!

## Substitutions and Models, Part 3: Boolos (and Gödel)

Last time, we saw Quine’s argument that, in languages rich enough for elementary number theory,

(EQ-R) $|\!\!\models P$ iff $\models P$

And in Part 1, we saw a compelling looking argument — relying on deduction and compactness theorems for substitutional and model-theoretic consequence — that (EQ-R) will entail

(EQ) $\Delta |\!\!\models P$ iff $\Delta \models P$

Today, we’ll see Boolos’s argument that (EQ) is false in languages rich enough for elementary number theory.

Section 1: Consequence and Consistency

Boolos’s argument is not direct. It goes, rather, via the notion of consistency. Let’s remind ourselves of the connection between the two.

(Boolos in fact talks of satisfiability; like many logicians (and for decent reasons)  he reserves the term “consistency” for a purely syntactic notion. I’m not going to follow suit, though. “Consistency” here will be a catch-all term for concepts in the general neighborhood, and we’ll define “satisfiability” in a moment. This shouldn’t cause any confusion, since the syntactic notion Boolos would use “consistency” won’t show up in today’s post.)

Let’s first remind ourselves of the connection between consistency and consequence.

The basic idea is this: if a set of sentences is consistent, all members of the set can be true together — “can” in the same sense that, if an argument is valid, the conclusion can’t be false if the premises are true.

We say that a set $\Delta$ is substitutionally consistent if and only if there is a substitution scheme S where $S(\Delta)$ is true, and we say that it is model-theoretically consistent, or satisfiable, if and only if there is a model on which all the members of $\Delta$ are true.

The crucial connection we want to preserve with these definitions is that, if $\Delta \Rightarrow P$, then the set $\Delta \cup \{\neg P\}$ is not consistent. Let’s verify that this holds for our two different notions.

First, model-theoretically: if $\Delta \models P$, then every model of $\Delta$ is a model of P; since no model is a model of both P and $\neg P$, no model is a model of $\Delta \cup \{\neg P\}$, and so that set is unsatisfiable. Conversely, if $\Delta \not\models P$, then there is some model M of $\Delta$ that is not also a model of P; it will be a model of $\neg P$, and so $\Delta \cup \{\neg P\}$ will be satisfiable.

Now, substitutionally: if $\Delta |\!\!\models P$, then every substitution scheme that makes $\Delta$ true makes P true, too. Since no substitution scheme makes both P and $\neg P$ true, every substitution scheme that makes $\Delta$ true does not make $\neg P$ true, and so $\Delta \cup \{\neg P\}$ is not substitutionally consistent. Conversely, suppose $\Delta |\!\!\not\models P$. Then some substitution scheme S makes all of $\Delta$ true but does not make P true. Since S(P) has to be a sentence, and every sentence is either true or false, S(P) must be false. So $\neg S(P)$ is true; but $\neg S(P) = S(\neg P)$ (since S has to preserve truth-functional connectives), so S is a substitution scheme that makes all of $\Delta \cup \{\neg P\}$ true.

This gets us two biconditionals:

(TransM) $\Delta \models P$ iff $\Delta \cup \{\neg P\}$ is not satisfiable.

(TransS) $\Delta \!\!\models P$ iff $\Delta \cup \{\neg P\}$ is not substitutionally consistent.

Putting these biconditionals together with EQ and negating both sides gets us

(ConEQ*) For any set $\Delta$ and sentence P, $\Delta \cup \{\neg P\}$ is satisfiable iff it is substitutionally consistent.

But notice that, if ConEQ* holds, so does

(ConEQ) For any set $\Gamma$, $\Gamma$ is satisfiable iff it is substitutionally consistent.

Why is that? Well, either one of the sentences in $\Gamma$ is of the form $\neg P$, or not. If so, then $\Gamma = \Gamma \cup \{\neg P\}$, and so we get ConEQ by plugging this identity into ConEQ*. So suppose no sentence in $\Gamma$ is of the form $\neg P$; then they will all be of the form P. Choose one such sentence, and notice that (a) every model of P is also a model of $\neg \neg P$, and (b) for every substitution scheme S, S(P) is true iff S($\neg\neg P$) is true. So $\Gamma$ will be satisfiable/substitutionally consistent if and only if $\Gamma \cup \{\neg\neg P\}$ is. But by ConEQ*, $\Gamma \cup \{\neg\neg P\}$ is satisfiable iff it is substitutionally consistent, and the result follows.

(The limiting case not covered here is the empty set. We can make an arbitrary choice as to whether to count the empty set as consistent or not; so long as we choose the same way on both the model-theoretic and substitutional side of things, ConEQ will hold.)

So if EQ is true, so is ConEQ. Or, more precisely (since these theses have to be implicitly indexed to languages), if EQ holds of a language, so does ConEQ. Boolos, however, will show us that ConEQ fails for first-order languages rich enough for arithmetic. Thus, so does EQ.

Section 2: Gödel and Tarski: An Aside

In the next section, I’ll give Boolos’s counterexample to ConEQ. But first we’ll briefly touch on some much-needed background on Gödel’s famous incompleteness theorem and a related result by Tarski. (I brushed up against Gödel’s results in the last post, but they’ll be more central to Boolos’s argument here so I want to go through them in a bit more detail.)

Gödel proved that first-order arithmetic was incomplete, in (roughly) the following sense: every finitely specifiable axiomatization of arithmetic will fail to prove some true arithmetical statement. The way he proved it was very clever. He noticed that syntactic entities, such as finite strings of symbols, etc., could each be “coded” by a unique natural number. He then showed it was possible to define arithmetical predicates that applied to natural numbers if and only if those numbers coded syntactic strings with various properties (i.e., properties like being a sentence, being a proof, and so on). And using these resources he showed how to build a sentence that essentially said of itself that it was unprovable. If it was false, it would be provable (and so true); so it must be true, and therefore unprovable.

Let’s get a bit more precise. We suppose we have a language LA strong enough for simple arithmetic, and PA is a true axiomatization of Peano Arithmetic in LA. Then using a Gödel numbering scheme, we can assign each finite string of LA symbols a unique code number (or “Gödel number”). There are actually many different, equally good Gödel numbering scheme; but let’s suppose we’ve fixed on one in particular.

If A is a string of symbols of LA, let $\ulcorner A\urcorner$ denote its Gödel code number. Gödel also showed it is possible to define predicates $SENT$ and $PRF_{T}$ where:

(1a) If A is a sentence, PA $\vdash SENT(\ulcorner A \urcorner)$, and
(1b) If A is not a sentence, PA $\vdash \neg SENT(\ulcorner A \urcorner)$.

(2a) If Pr is a proof (from the axioms of a well-behaved theory T) of A, and Pr’s Gödel code is n, then PA $\vdash PRF_{T}(n,\ulcorner A\urcorner)$; otherwise
(2b) PA $\vdash \neg PRF_{T}(n,\ulcorner A\urcorner)$

Now, there are a number of ways to proceed from here. But one way of proving Gödel’s result (which is not quite the way Gödel himself originally did it) is via the diagonalization lemma:

DIAG: If a theory T includes PA, and if F(x) is a formula open in x, then there is a sentence A where $T \vdash A \leftrightarrow F(\ulcorner A\urcorner)$.*

Now take F(x) as $\neg\exists y(PRF_{PA}(y,x))$. By DIAG, we have that there is a sentence $A_{G}$ where

PA $\vdash A_{G} \leftrightarrow \neg\exists y(PRF_{PA}(y,\ulcorner A_{G}\urcorner))$

($PRF_{PA}(n,m)$, recall, means that there is a proof from the axioms of PA with Gödel number n, and the theorem it proves has Gödel number m.) Call this sentence $DIAG_{G}$. Since PA proves it, PA is true, and our proof theory is sound, $DIAG_{G}$ must be true.

Now suppose $A_{G}$ is false. But since $DIAG_{G}$ is true, this means that $\exists y(PRF_{PA}(y,\ulcorner A_{G}\urcorner)$. So there is a PA-proof of $A_{G}$. But PA is true and its proof system is sound, so whatever it proves is true, so $A_{G}$ is true. Contradiction.

This means $A_{G}$ must be true. But if it is true, $DIAG_{G}$ tells us it can’t be proven in PA. So there is a true arithmetical statement — namely, $A_{G}$ — that cannot be proven in PA. This is known as PA’s Gödel sentence.

(Could you add $A_{G}$ to PA’s axioms to get a new theory, PA+ that could prove $A_{G}$? Yes. But now you could re-run the argument from DIAG using the open sentence $\neg\exists y(PRF_{PA+}(y,x))$, where $PRF_{PA+}$ codes for provability-in-PA+. This gives us a new sentence, $A_{G+}$, which by the above reasoning must be true but not provable in PA+. And so it goes.)

As interesting as all this is, what we need to understand Boolos’s counterxample isn’t Gödel’s proof itself, but rather a related corollary, proven by Tarski, that no sufficiently rich language can define its own truth predicate. More accurately: if L is a language with all the arithmetical vocabulary needed for arithmetic, and if PA (as formulated in L) is true, then there is no arithmetical formula T of L such that T(n) is true if and only if n is the Gödel number of a true sentence of L.

This proof also goes via the diagonalization lemma. Suppose we have a language L, and the axioms of PA are among the truths of L. Suppose L also has a formula $\neg T(x)$ which is true only of Gödel numbers of sentences that are not truths of L. Then the diagonalization lemma tells us that there is a sentence LIAR such that

$PA \vdash \textrm{LIAR} \leftrightarrow \neg T(\ulcorner\textrm{LIAR}\urcorner)$

Since PA is true and has a sound proof system, this sentence must be true as well. But if LIAR is true, then $\neg T(\ulcorner\textrm{LIAR}\urcorner)$ is false, in which case LIAR isn’t a true sentence of L after all. And if LIAR is false, then $\neg T(\ulcorner\textrm{LIAR}\urcorner)$ is false, in which case $\ulcorner\textrm{LIAR}\urcorner$ numbers a true sentence of L, in which case LIAR is true. Either way leads to contradiction; this is a reductio of the premise that L contains a predicate T true of all and only the Gödel numbers of truths of L.

With Gödel’s proof, we could enrich PA to be able to prove $A_{G}$ (but were then faced with a new unprovable truth, $A_{G+}$). Similarly, with Tarski’s proof, we can enrich L to a new language, L+, and in L+ it may be possible to define an arithmetical formula T such that T(n) is true if and only if n numbers a true sentence of L. But T cannot get things right for all the sentences that are in L+ but not L, too, in just the same way that PA+ had a new Gödel sentence it could not prove.

Section 3: The Counterexample

The Setup

OK, so here’s the deal. Let LA be a language that is rich enough for simple arithmetic, in the very strongest sense: it has expressions for the various numerical operators, and these mean their associated numerical operations. To fix ideas, let’s suppose LA has no constants at all, and the following predicates:

 Predicate Means $Z(x)$ x = 0 $S(x,y)$ y is the successor of x (y = x + 1) $A(x,y,z)$ x + y = z $P(x,y,z)$ x $\times$ y = z

We’ll also suppose LA has one further predicate, $G(x)$, which means x is a number’. (We also have the quantifiers of LA implicitly restricted to natural numbers, so $\forall x G(x)$ is a truth of LA.) Let’s denote the set of all truths of LA, “Tr(LA)”.

Before going on, let’s notice something about how this language will talk about numbers. Suppose we wanted to say that 1+2=3. We usually do this using a functor “+”, numerals (“1”, “2”, etc.), and an equality sign “=”. Many standard formulations of PA work like this, except that they only have one (constant) numeral, “0”, and then a “successor function” “S”, where “Sx” is the number one greater than x. Here, we would represent “1+2=3” as something like “S0+SS0=SSS0”. In such formulations of PA, the expressions consisting of “S” applied n times to “0” is called a standard numeral, and (more precisely) the standard numeral for n. Sometimes it can be useful to use some sort of typographical shorthand, e.g. a boldfaced convention where “3” is understood to stand for 3’s standard numeral, “SSS0”. Here, we can take an ordinary numerical statement, say “1+2=3”, bold each of its terms, to get “1+2=3“, and then unpack the boldfaced convention to get “S0+SS0=SSS0”.

But our language LA isn’t like this at all: we don’t have identity, and we don’t have functors. So if we use boldfaced numerals to stand for some conventional expression of a certain number in a certain language, we can’t trade out a boldfaced numeral for one cooked up out of “0” and a successor function. We must do something else.

What we do instead is describe the number. For instance, we know that three is something which succeeds something which succeeds something which succeeds zero. (And “succeeding zero” is the same as “succeeding something which zeroes” — that is, succeeds an x where $Z(x)$.) So, formally, our boldfaced convention will work as follows: when we get a sentence, say

3 …,

we understand that as shorthand for

$\exists x_{0}\exists x_{1}\exists x_{2}\exists x_{3}(Z(x_{0}) \wedge S(x_{0}, x_{1}) \wedge S(x_{1}, x_{2}) \wedge S(x_{2}, x_{3}) \wedge \ldots x_{3} \ldots)$

Of course, that’s pretty long, but it does the trick. (And it should be clear how to extend this for arbitrary “numerals”.) There’s another, equivalent way to describe what we’re doing. Instead of having standard numerals, we’ll have standard numeral-predicates. Thus, for instance, 3(x) will be a shorthand for the open sentence

$\exists x_{0}\exists x_{1}\exists x_{2}\exists x_{3}(Z(x_{0}) \wedge S(x_{0}, x_{1}) \wedge S(x_{1}, x_{2}) \wedge S(x_{2}, x))$

3

as

$\exists$x(3(x) $\wedge$ … x … )

This predicate-numeral convention will be useful later.

(Why the convoluted strategy? Boolos does this to stack the deck in Quine’s favor; Quine’s favored language was free of functors, which he replaced with a description trick like this.)

OK, enough of LA. Now consider a language LB. Syntactically, it is exactly the same. Semantically, it is almost the same — but not quite. The only difference is that, in LB, $G(x)$ is true of all and only the numbers that are Gödel numbers of sentences in Tr(LA). Following earlier usage, we’ll denote the set of all truths of LB, “Tr(LB)”. And we’ll use the same boldfaced numeral (and numeral-predicate) convention for LB as we used for LA.

Notice that the terms of the each languages are syntactically the same. This means that they will share Gödel coding schemes. More precisely, since Gödel schemes simply assign numbers to strings on the basis of their syntactic properties — semantic ones never enter into it — then any sensible Gödel coding scheme will assign syntactically identical strings of LA and LB (whether they be sentences, proofs, whatever) the very same Gödel codes.

Despite this syntactic similarity, it will be useful for us to keep tabs on which language we’re talkig about, though. We’ll want to distinguish, e.g., $G$‘ as it appears as a term of LA, and as it appears as a term of LB. So we’ll add subscripts to each of the terms as appropriate, writing the (non-logical) terms of LA as

$Z_{A}, S_{A}, A_{A}, P_{A}$, and $G_{A}$,

and the (non-logical) terms of LB as

$Z_{B}, S_{B}, A_{B}, P_{B}$, and $G_{B}$.

And we’ll adopt the further convention that for any syntactic string S (belonging to both languages), we’ll use $S_{A}$ to denote that string as a string of LA, and $S_{B}$ to denote it as one of LB. Notice that these subscripts are not an official part of the languages LA and LB; rather, they’re just our own metalinguistic commentary on elements of those languages. So these subscripts don’t figure into the Gödel codes strings get assigned. Notice, however, that we can now rewrite our earlier observation as follows: if $\ulcorner\cdot\urcorner$ is any well-defined system of Gödel coding, then for any string S, $\ulcorner S_{A}\urcorner = \ulcorner S_{B}\urcorner$.

Notice also: $G_{B}$ is a truth-predicate for LA. By definition, for any number n, $G_{B}(n)$ is true if and only if n is the Gödel number of a truth of LA. And so it must also be a truth-predicate for the fragment of LA that doesn’t have `$G_{A}$‘ in it.

3.a: A First Pass

OK, now let’s look at Boolos’s argument. Consider the set $Tr(LB)_{A}$ — the set of all the sentences of LA that are syntactically just like the ones in Tr(LB). This set is satisfiable: we know full well it has a model. We know this because the intended interpretation of LB provides just such a model. More precisely, let M be a model of LA that assigns each predicate $P_{A}$ the extension that $P_{B}$ in fact has; since Tr(LB) Is in fact true, and M makes LA’s terms mean just what LB’s already do, M must be a model of $Tr(LB)_{A}$. The question, then, is whether it is substitutionally consistent — that is, whether there is a substitution scheme T where $T(Tr(LB)_{A})$ is true, too.

Let’s start by noting that, if there is, then $T(Tr(LB)_{A}) \subseteq Tr(LA)$. This is because we’re working in the language LA here; if $T(Tr(LB)_{A})$ are all true, then there some (if not all) of the truths of LA, and so some of the members of Tr(LA).

Now one thing we may have noticed is that $Tr(LB)_{A}$ and Tr(LA) have a lot of overlap. In fact, any sentence that does not have “G” in it will be in one of these if and only if it is in the other. So we might have thought that an easy way to cook up a translation scheme T that makes $T(Tr(LB)_{A})$ all true would be to have it give an interesting substitution for “G” and leave the other predicates alone. But this won’t work, and it’s worth taking a minute to see why.

Suppose n is a Gödel number of a true sentence of LA. Then $G_{B}$(n) is in Tr(LB), and so $G_{A}$(n) is in Tr(A). And this, remember, just is the sentence

$\exists x_{0}, x_{1}, \ldots, x_{n}(Z_{A}(x_{0}) \wedge S_{A}(x_{0},x_{1}) \wedge \ldots \wedge S_{A}(x_{n-1}, x_{n}) \wedge G_{A}(x_{n}))$

Now, when we take the substitution instance of this that T gives us, that simply swaps out all the predicates here for their values under T. But T, by hypothesis, leaves everything but “G” alone. So the result of applying T to this sentence gives us

$\exists x_{0}, x_{1}, \ldots, x_{n}(Z_{A}(x_{0}) \wedge S_{A}(x_{0},x_{1}) \wedge \ldots \wedge S_{A}(x_{n-1}, x_{n}) \wedge T(G_{A})(x_{n}))$

In other words, it gives us the sentence T($G_{A}$)(n). Likewise, if n is not a Gödel number of a true sentence of LA, $\neg G_{B}$(n) is in Tr(LB), and a similar argument shows that $\neg T(G_{A})$(n).

We’re supposing that T takes us to all and only truths. But (since the satisfier of the numeral-predicate n just is the number n!) this means that $T(G_{A})$ is a formula of LA that is true of all and only the Gödel numbers of truths of LA. And this is what Tarski’s theorem told us we couldn’t have. Reductio.

3.b: For More General Cases

But a reductio of what? Of the assumption that there was a substitution scheme T that made $Tr(LB)_{A}$ true and didn’t fiddle with any of the predicates other than $G_{A}$. What happens if we relax that second assumption? Basically, the same thing. Only it takes more effort to see why. That’s what I’ll argue in this section.

OK, so suppose we have a substitution scheme T, where $T[Tr(LB)_{A}]$ are all true.

Here’s the first thing to notice. $G_{B}(\textbf{n})$ will be true if and only if n is the Gödel number of a true sentence of LA. So $T[G_{A}(\textbf{n})]$ will be true if and only if n is the Gödel number of a true sentence of LA, too.

But what is $G_{A}(\textbf{n})$? Recall our numeral-predicate strategy: it is the sentence

$\exists x(\textbf{n}(x) \wedge G_{A}(x))$

So the substitution instance of this is

(1) $\exists x(T[\textbf{n}](x) \wedge T[G_{A}(x)])$

And notice: (1) is true if and only if n is the Gödel number of a true sentence of LA. But this doesn’t give us quite what we want: an open formula that is true of and only of the true Gödel numbers of LA sentences. So we’re not quite done.

What we need is a formula of LA, R, where is true of n and m if and only if m satisfies T[n(x)]. If we had that, then (1) would be equivalent to

(2) $\exists x(R(\textbf{n},x) \wedge T[G_{A}(x)])$

and would be true if and only if n was the Gödel number of a truth of LA. Moreover, the open formula

(3) $\exists xR(y,x) \wedge T[G_{A}(x)]$

would be a ‘truth-formula’ for LA: it would be true of and only of the Gödel numbers of LA truths. Since Tarski’s theorem tells us there can’t be any such formula, our reductio would be complete.

But how do we show that there is such a formula R? Our answer comes in two parts. The first requires a bit of background. Start with the notion of a (total) recursive function. A function f is recursive if and only if (more or less, assuming the Church-Turing thesis) there is an effective algorithm for determining, for each $\vec{x}$ and y, whether f($\vec{x}$) = y. From this, we can also define the notion of a recursive relation: an n-place relation R’ is recursive if and only if there is an n-place recursive function f where f($\vec{x}$) = 1 iff $\vec{x}$ stand in R’, and 0 otherwise. And finally, we have the following useful result: if R’ is any n-placed recursive relation, then there is a formula R* of LA where

PA $\vdash$ R*($\vec{\textbf{x}}$) if $\vec{x}$ stand in R’, and
PA $\vdash \neg$ R*($\vec{\textbf{x}}$) otherwise.

OK, so far, so good. Now notice that, given any number n, there is a recursive function that will deliver the Gödel number of the formula “T[n](x)”. We can see this by imagining a purely mechanical process for taking n and generating “T[n](x)”‘s Gödel number.

The first thing we need to do is figure out just what formula T[n](x) is. But this will clearly be a simple algorithmic process: it’s purely mechanical to write down $Z_{A}$, n instances of $S_{A}$, and then conjoin them and fill in and bind the right variables in the right places. This gets us the formula n(x). And it’s also purely mechanical to get from this to T[n](x): just look up each predicate’s assignment in T’s table and write it in the right spot. But (given a fixed Gödel coding) it’s also purely mechanical to take any formula and find its Gödel number. So there is a binary recursive relation GN* that n bears to x if and only if m is the Gödel number of “T[n](x)”. And since there is such a relation, there is also a predicate GN where GN(n,x) is satisfied when and only when x is the Gödel number of “T[n](x)”.

Second, there will be a satisfaction predicate SAT for PA that applies to “T[n](x)” — that is, a predicate SAT where SAT(x,m) is true if and only if m satisfies the formula with Gödel code x.** So now consider the formula

$\exists x(GN*(n,x) \wedge SAT(x,m))$

This is a formula that is satisfied by n and m if and only if m satisfies T[n]. But (putting this all together), this means that the open formula

$\exists y[\exists x(GN*(n,x) \wedge SAT(x,y)) \wedge T[G_{A}](y)]$

is a truth-formula for LA. For n satisfies this formula if and only if there is a number m that satisfies both $T[G_{A}]$ and T[n] if and only if the sentence $T[G_{A}(\textbf{n})]$ is true if and only if the sentence $G_{B}(\textbf{n})$ is true if and only if n is the Gödel number of a truth of LA. Boolos’s reductio is complete.

*Really for this theorem, T need not include PA but only include the weaker Robinson arithmetic; but since PA includes that, and since we’re going to be dealing explicitly with cases where our theories include PA, we can ignore this slight weakening.

**This isn’t quite right, and for reasons that have to do with Tarski’s theorem. Just as there can be no perfectly general truth predicate for LA, there can also be no perfectly general satisfaction predicate for LA. What we do have, though, is a hierarchy for satisfaction predicates that follows the arithmetical hierarchy. That is, for every complexity $\Pi_{n}$ or $\Sigma_{n}$ in the arithmetical hierarchy, there is a satisfaction predicate SAT$_{\Pi_{n}}$ or SAT$_{\Sigma_{n}}$ that applies to all formula of that complexity. That is, if P is (say) a $\Sigma_{n}$ formula, then SAT$_{\Sigma_{n}}(\vec{x},\ulcorner P\urcorner)$ if and only if $\vec{x}$ satisfies P. However, if P is a formula of greater complexity than $\Sigma_{n}$, SAT$_{\Sigma_{n}}$ can’t be trusted to ‘get the facts right’ about what satisfies it.

However, for any translation scheme T, there will be some particular level of complexity that all formulae of the form T[n(x)] will share. This is because all such formulae will be of the form

$\exists y_{0},\ldots,y_{n-1}(T(Z_{A})(y_{0}) \wedge T(S_{A})(y_{0},y_{1}) \wedge \ldots T(S_{A})(y_{n-1},y))$

That is, it will be a block of quantifiers followed by a conjunction of translations of primitive formulae of LA. We can think of this conjunction as having two main conjuncts:

$(T(Z_{A})(y_{0})) \wedge (T(S_{A})(y_{0},y_{1}) \wedge \ldots T(S_{A})(y_{n-1},y))$

Here’s an argument for one of the possible cases; it will be easy to reconstruct the other cases from it. Suppose $T[Z_{A}]$ is of complexity $\Delta_{i}$, and $T[S_{A}]$ is of complexity $\Delta_{j}$. Notice that, in this case, any conjunction of formulae all of the form of $T[S_{A}]$ is also $\Delta_{j}$. If i = j, then the complexity of this entire formula is $\Delta_{j}$. If i j, the complexity of this formula is $\Delta_{i}$. So no matter what there is a fixed k such that, no matter how many iterations of T(S_{A}) are added, this formula is of complexity $\Delta_{k}$, in which case the complexity of T[n(x)] is $\Sigma_{k+1}$, and we can use SAT$_{\Sigma_{k+1}}$ for the argument in the text.

## Substitutions and Models, Part 2: Quine

The second post in the series that started here. Apologies for the months-long delay! The third will go up later this week.

Quine claims that, for a certain class of languages, the substitutional and model-theoretic definitions of logical truth are equivalent:

(EQ-R) $|\!\!\models P$ iff $\models P$

Which languages? Ones that have two properties:

(P1) They are “rich enough for elementary number theory” (Philosophy of Logic, p. 53).
(P2) They are first-order. (In particular, completeness and a specific version of the Loewnheim-Skolem theorem hold for them.)

Here are today’s questions:

1. What counts as being “rich enough for number theory”?
2. How is this result consistent with the arguments of section 3 of the last post?
3. What, exactly, is Quine’s argument? How does it work, and on what assumptions does it depend?

I won’t tackle these in order, though. I’ll start with 2, and then move on to 3. 1 will be covered towards the end.

Section 1: Motivation

Before getting into the nitty gritty, it’s worth taking a step back to ask why we — or Quine — should care, either about EQ-R or the full EQ.

First, consider how we would expect Etchemendy to react. Since the EQs hold, at best, only for a restricted class of languages, he will likely be unimpressed. Etchemendy’s approach in The Concept of Logical Consequence shows that he wants a perfectly general analysis which, when thrown an arbitrary language, will give the right predictions. Since we know a substitutional account generates bad predictions for at least some languages, by Etchemendy’s lights there’s no use continuing to go on about it.

Not so for Quine. He has little truck with the notion of an analysis. (If we could analyze logical consequence, we would have an instance of synonymy, which he famously decries.) As a pragmatist, the important question for him is, “What well-behaved definition will suit us best for our purposes?” And if our purposes keep us well within the limited class of languages for which an EQ result holds, then this tells us that, if the model-theoretic account works for our purposes, so does the substitutional.

For Quine, though, the result also goes two steps deeper. For one thing, Quine thinks there are reasons to use the substitutional account when we can for — unlike the model-theoretic one — we don’t have to deploy sets for the substitutional account. The objection here isn’t the “there are no sets, so let’s be substitutionalists” one — Quine eventually accepted sets, after all, thinking them indispensable to science — but rather that we should do as much as we can with as little as we can get away with, so if we’re not forced to, we should avoid appealing to them in our definition of logical consequence.

Second, and in Boolos’s eyes most important, Quine thinks the EQ results give an argument for identifying “logic” with first-order logic. (There is, I think, a real issue as to just what this identification means — both for us and for Quine — but set that aside.) We choose a logic just as we choose other theories: by seeing which does best at the various theoretical virtues. Since the EQ result, as well as completeness, apply only to first-order theories, only if logic is first-order can we collapse proof-theoretic, model-theoretic, and substitutional accounts of consequence. Quine sees this as a virtue, and thus grist for his “logic-is-first-order” mill.

Section 2: Reconciliation

Ok, next: resolving the last post’s puzzle. There, we saw that a counting sentence like

(E2) $\exists x\exists y(x \ne y)$

will count as a substitutional, but not a model-theoretic, truth, and thus threaten a counterexample to (EQ-R). How is Quine not worried about this? The answer comes a few pages after his argument, where we learn “Truths of identity theory [such as $\forall x(x=x)$] do not count as logical truths under the contemplated definitions of logical truth, since they are falsifiable by substituting other predicates for $=$.” (p. 61). Quine does not think that all substitution schemes take (E2) to (E2), for — unlike what Etchemendy assumed — he does not reckon the identity predicate one of the logical constants.

That answers one side of the problem. But in standard textbook model-theoretic treatments of first-order logic with identity, $\forall x(x=x)$ does come out true on all models. If this is not to be a further counterexample to EQ-R, Quine had better have a not-quite-so textbook model theory. And indeed he does. How his differs from the textbook depends on just what the textbook says.

In some textbooks, the interpretation function of a model does not assign an extension to $=$. Rather, satisfaction-conditions for sentences of the form $\alpha = \beta$ are hard-wired into the definitions of truth and satisfaction on a model. In other textbooks, the definitions of truth and satisfaction on a model don’t treat identity predicates in any special way. But a set $\langle D, I\rangle$ only counts as a model if the interpretation function I assigns the set $\{\langle x, x\rangle : x \in D\}$ to I. Quine’s model theory differs from the first textbook’s by not having any special truth-or-satisfaction clause for identity statements, and differs from the second’s by allowing models to assign different extensions to $=$. So on Quine’s model theory, $\forall x(x=x)$ ends up not being a model-theoretic truth either, and the potential counterexamples are all blocked.

(For those keeping track: this will, clearly, affect Quine’s overall pro-first-order argument. If he wants to argue that first-order logic is best on the basis of its collapsing model-theoretic, proof-theoretic, and substitutional accounts of logical truth, he’ll have to argue that first-order logic without identity is best — because with identity, first-order logic doesn’t have this property.

Alternatively, as Günther pointed out in the comments to the last post, Quine might instead plump for the modified substitution scheme. Then he can keep identity in, so long as he is willing to argue (as he apparently is) that inclusive (i.e., empty-domain-allowing) first-order logic without names is best. Either way, Quine has to depart from the textbook in some direction or other. As a matter of fact, he departs in all three: he also eschews names and endorses (in Mathematical Logic, anyway) inclusion. So he has a fairly large menu of options here.)

Section 3: The Argument

Here is Quine’s basic argument for EQ-R:

(I). If P is true on all substitution schemes, P is true on all number-theoretic substitution schemes.
(II). If P is true on all number-theoretic substitution schemes, it is true on all models.
(III). If P is true on all models, it is true on all substitution schemes.
(C). Therefore, P is true on all substitution schemes iff P is true on all models.

But (C) is (EQ-R), the desired conclusion. This argument is clearly valid (it’s a standard proof technique to prove a biconditional P iff Q by proving a string of conditionals that start with P, go through Q, and end up back at P again.) But what of the premises?

Some of this, of course, will depend on what “number-theoretic substitution scheme” means. But whatever it means, so long as these are a kind of substitution scheme (and they are!), I is trivial. II and III not so much, though, so we’ll look at Quine’s arguments for them in detail.

3.a: The Argument for III.

Quine argues for this by appeal to

(COMPLETENESS) If $\models P$, then $\vdash P$

(where $\vdash P$ represents the provability of P in any standard first-order system). His reasoning is apparently something like the following. Where S is any (logical-term-preserving) substitution scheme whatsoever:

1. Suppose P is satisfied by every model.
2. Then, by COMPLETENESS, it is provable. So there is a proof Pr of P.
3. If Pr is the proof of P, then S(Pr) is a proof of S(P).
4. Thus, S(P) is provable.
5. Therefore, S(P) is true.

The first thing to note is that the move from 4 to 5 relies on the thought that falsehoods can’t be proven or, equivalently, that first-order deductive systems are intuitively sound. ‘Intuitive soundness’ is a notion that crops up in eg Kreisel’s squeezing argument.* The idea here (as Quine puts it, at any rate) is that, among the proof systems we can pick, there are some where we can tell, by inspection, that every individual move takes us from truths to truths. (Or from nothing at all to truths, as happens when e.g. we have an axiomatic system.) Furthermore, we can tell that stringing together series of moves each of which preserves truth will also preserve truth. So we can tell, just by inspection, that there will be no proof of a falsehood. So, since there is a proof of S(P) (line 4), it must be true (line 5).

The second thing to note is that this requires also that, if Pr is a proof, then S(Pr) (the result of taking the proof and replacing each sentence Q in it with S(Q)) is also a proof. This is premise 3. Call proof systems with this property preservational. Notice that not every proof system we might pick is preservational. Suppose, for instance, that we chose a system of semantic tableaux (or “truth-trees”) where only contradictions between literals (atomic predications or their negations) were allowed to close. A proof of ‘$R(a) \vee \neg R(a)$‘ will be a tree that starts with ‘$\neg(R(a) \vee \neg R(a))$‘ and ends with ‘$R(a)$‘ and ‘$\neg R(a)$‘ closing the (only) branch. Now consider the substitution scheme that takes $R(a)$ to $F(b) \wedge G(a)$. The tree that you get by taking the above and substituting with this scheme has ‘$F(b) \wedge G(a)$‘ and $\neg(F(b) \wedge G(a))$‘ closing, which is not a proper proof of $(F(b) \wedge G(a)) \vee \neg( F(b) \wedge G(a))$ in the imagined system, because it doesn’t have any contradiction between literals.

This isn’t a huge problem: there are preservational proof systems. And there are intuitively sound proof systems, too. In fact, there seem to be proof systems that are both — standard axiomatized systems seem to have both properties. (They are preservational because whether something counts as a proof is entirely schematic — depends entirely on whether lines of the proof fit certain schemas — and since the substitution schemes preserve logical constants, they can’t make a sentence stop matching a schema when it did before. They certainly seem to be intuitively sound, too, although this is perhaps a more controversial matter.) And good thing, too, because the argument for III goes through only if there are proof systems with both properties. Otherwise, we might have intuitively sound proofs turning into intuitively unsound ones under substitution (violating the move from 4 to 5), or instead intuitively sound proofs turning into things that aren’t even proofs (falsifying 3).

3.b: The Argument for II: Hilbert-Bernays Theorem.

OK, enough of III. It looks secure enough. On to II. Quine’s argument for it relies on a remarkable theorem, a strengthening of the Lowenheim-Skolem theorem, first shown by Hilbert and Bernays:

(HB) If P has a model, it has a constructible model.

A model is constructible iff its domain is a subset of the natural numbers, and each of its predicate extensions is given by a formula of simple arithmetic. That is to say: if F is an n-placed predicate and M a constructible model, then there is some formula A of simple arithmetic, open in n variables, where $\langle x_{1}, \ldots x_{n}\rangle$ is in F’s extension on M iff $x_{1}, \ldots x_{n}$ satisfy A. (Note: F is a predicate of the object-language here, but A is an arithmetical formula of the metalanguage.)

What does it mean to say that a formula is a ‘formula of simple arithmetic’? We’ll deal with this quite a bit more in Section 4. For now, just note that it will be a formula that uses the primitive vocabulary used in, say, Peano Arithmetic.

The argument for II goes as follows. Call a substitution scheme number-theoretic iff every predicate and term gets replaced for some formula or term of simple arithmetic. We first use (HB) to argue for

(HB*) If Q is true on some model, then for some number-theoretic substitution scheme S, S(Q) is true.

The very basic idea is that, if F is given an extension E on the number-theoretic model, and if A is the formula of simple arithmetic where $\langle x_{1}, \ldots x_{n}\rangle \in E$ iff $x_{1}, \ldots x_{n}$ satisfy A, then we can replace F for A. Notice what’s happening here: A appears both in the metalanguage and the object language, so we use the metalanguage characterization of the model (in terms of A) to cook up an object-language substitution scheme (using A). If we do this for all of the predicates and terms in Q, we end up with a sentence S(Q) that basically describes the constructible model we began with, and is true because S(Q) accurately describes that model. In other words: since P is true on the model, and S(Q) says, more or less, that P is true on the model, S(Q) is flat-out true.

Since (HB*) is a general principle, it holds even if Q is of the form ~P (for any sentence P). If we substitute in ~P, we get:

(HB~) If ~P is true on some model, then for some number-theoretic substitution scheme S, S(~P) is true.

Now take the contrapositive:

(HBC) If for every number-theoretic substitution scheme S, S(~P) is false, then ~P is false on every model.

But S(~P) = ~S(P), and ~S(P) is false if and only if S(P) is true. Likewise, ~P will be false on a model iff P is true on that model. So (HBC) becomes:

(II) If P is true on every number-theoretic substitution scheme, then P is true on every model.

This is Quine’s second premise. So we have it — if we can substantiate the move from HB to HB*.

4. “Rich Enough for Elementary Number Theory”.

Can we substantiate the move? We can if ‘is a formula of simple arithmetic’ means something like (i) ‘the formula uses Peano Arithmetic’s vocabulary’, (ii) ‘the vocabulary of Peano Arithmetic means arithmetic (‘+’ means plus, and so on), and (iii) Peano Arithmetic is true. In that case, HB ensures that there is a model made out of numbers where every formula’s satisfaction-conditions can be expressed in PA, and for any formula F we can, by uniform substitution, get a formula S(F) that explicitly states those numerical satisfaction-conditions.

4.a: Another Substitution Scheme?

Those are very strong assumptions, though. We might have hoped that we could get away with something considerably weaker. It’s tempting to think that we could have gotten away with mere provability from PA, in which case ‘rich enough for elementary number theory’ would only need to mean something like ‘has the syntactic resources of PA’. The temptation comes, in large part, because we want to think of HB as coding up a kind of logical connection: the premises of mathematics, so to speak, entail that if P has a model, S(P) is true.

Mere syntactic resources won’t be enough, but the tempting thought can be made substitutionally. We might reason like this. Suppose that  PA is false (formulated in our ‘vocabulary of number theory’), but has a true substitution instance under a scheme T. HB — the claim which tells us that every satisfiable sentence has a constructible model — tells us that $PA \vdash S(P)$. In this case, since our proof theory is preservational, $T(PA) \vdash T(S(P))$. But since T(PA) is true and our proof theory is intuitively sound, T(S(P)) is true. And T(S(P)) is a substitution instance of P; so the fact that P has a model is enough to show that it has a true substitution instance.

There are two problems with this. The first is that any interesting mathematical consequences of PA will have to involve claims of identity. If Hilbert and Bernays had showed that, when P is satisfiable, $PA \vdash S(P)$, they surely would have relied on a proof theory that included identity. Since Quine doesn’t reckon identity among the logical constants, it’s not clear whether he can appeal to this claim. Perhaps he could finesse it by adding the axioms of identity theory to PA, but I’d want to see the reasoning spelled out in detail.

This first problem may not be a big deal. The second is: there’s no reason at all to think that PA $\vdash$ S(P), and at least some inductive reason to think that sometimes it doesn’t.

First, why is there no reason to think it does? Well, suppose that S(P) is in fact the Godel sentence for PA — the sentence G which says, in effect, that G — it itself — is not provable in PA. Since PA is (various kinds of) consistent, we know that PA $\not\vdash$ G. But G is going to be true on some constructible model. (It’s true on the standard model of PA, for instance, which is a constructible model.) And there’s just no guarantee that, when we applied S to P, G wasn’t what we ended up with.

Of course, this itself doesn’t tell us that there are any P for which S(P) is G; it just raises a spectre of doubt. Here’s another spectre of doubt. The result that Quine appeals to when discussing the Hilbert-Bernays theorem appears to be a corollary of Bernay’s Lemma. Where $Prf$ is the formula that captures (via Godel coding — if you’re not familiar, I’ll explain more in the next post) the provability predicate for the “empty theory” (that is, it captures theorem-of-first-order-logic-(without identity(??))-hood) this lemma says that, for any sentence P, there is a substitution scheme $S_{P}$ such that

(BL) $PA \vdash S_{P}(P) \rightarrow Prf(\ulcorner P \urcorner)$

(where $\ulcorner P \urcorner$ represents P’s Godel number). Since P will be a theorem if and only if $Prf(\ulcorner P \urcorner)$ is true, this means that the truth of $S_{P}(P)$ suffices for the theoremhood of P.

We can deduce HB* from this as follows:

1. $PA \vdash S_{\neg P}(\neg P) \rightarrow Prf(\ulcorner \neg P \urcorner)$ (Bernays lemma, with $\neg P$ substituted for P.)
2. $PA \vdash \neg Prf(\ulcorner \neg P \urcorner) \rightarrow \neg S_{\neg P}(\neg P)$ (Contraposing)
3. $PA \vdash \neg Prf(\ulcorner \neg P \urcorner) \rightarrow S_{\neg P}(P)$ (The internal negation commutes, and the two cancel)

If P is consistent, there is no proof of not-P, and so (assuming PA is true and our proof theory sound) $\neg Prf(\ulcorner \neg P \urcorner)$ will be true, in which case so will $S_{\neg P}(P)$. This means that $S_{\neg P}$ is the substitution scheme we need.

OK, so here’s that next spectre of doubt. The most natural way to show that $S_{\neg P}(P)$ will be provable (as opposed to just true) in PA for every P would be if $\neg Prf(\ulcorner \neg P\urcorner)$ was provable in PA for every consistent P. This latter sentence says, in effect, that P is first-order consistent. What we might hope to do is to show

(*) $PA \vdash \neg Prf(\ulcorner \neg P \urcorner)$

for all consistent P. Then we could use * and line 3 above to conclude (with appropriate subscripts) that PA $\vdash$ S(P).

But, alas, there’s no hope of this. Notice that we already have a procedure for telling whether any sentence Q is a theorem of first-order logic that’s guaranteed to terminate and give us a “yes” answer if Q in fact is a theorem. Moreover, we know that there is no “decision procedure” for determining Q’s theoremhood: there’s no procedure that, for any Q we throw at it, is guaranteed to eventually terminate and tell us “yes” if Q is a theorem and “no” if it’s not.

But if (*) were true for every consistent P, then we would have such a procedure. Just substitute $\neg Q$ for P. If Q is not a theorem, then there’s no proof of Q, and (since there is a proof of Q iff there is a proof of $\neg\neg Q$) $\neg Prf(\ulcorner \neg \neg Q \urcorner)$ is true. But if the truth of this latter claim guaranteed its derivability from PA (as (*) would have it do), then — since we have a procedure that will always deliver a “yes” verdict when there is a proof — we would have a procedure guaranteed to give us an “yes” verdict whenever we fed it an invalid Q.

But we could then stitch these procedures together to cook up a decision procedure for first-order theoremhood. First, spend five minutes on the procedure looking for a proof of Q. Then, if you haven’t gotten an answer yet,  spend five minutes looking for a proof of $\neg Prf(\ulcorner \neg \neg Q \urcorner)$. If you don’t find one in that time, go back to the first, and keep swapping every five minutes. Since no matter whether Q is a theorem or not we are guaranteed that at least one procedure will terminate, this procedure will eventually terminate. If it terminates with a “yes” answer on the “proof of Q” front, then you have the answer “theorem”; if it terminated with a “yes” on the “proof of $\neg Prf(\ulcorner \neg \neg Q \urcorner)$” front, you have the answer “not a theorem”. But since any such procedure is impossible, (*) must be false after all.

Now, none of this shows that PA can’t prove S(P) for every consistent P. It just shows that it can’t (always) do it via (*). But it is difficult to see what other considerations would lead us to think that PA could prove S(P) for every P.

I’d love to know whether PA’s proving power along these lines has been sussed out yet. I have a sneaking suspicion that this is an open problem (in the sense of nobody knowing the answer, rather than in the sense of people actively caring about and working on the answer). But the overall inductive weight of the evidence is pretty hefty against PA’s ability to prove S(P) for every P. I sure wouldn’t place much of a wager on it. And so, at the very least, it leaves the T-using argument above underdeveloped.

4.b: What Can We Say?

Let’s suppose that there are some consistent P for which PA can’t prove the relevant S(P). Then what happens to Quine’s result when PA is true but doesn’t mean arithmetic? Or when PA is false, regardless of whether it means numbers or not?

Suppose first that PA is true, and also means arithmetic. Then we’ll call its standard model $\mathbb{N}$, and following standard usage, let $Tr(\mathbb{N})$ be the set of all of the sentences of L that are true on $\mathbb{N}$. (Godel’s result shows us, in effect, that no recursive axiomatization can prove every sentence in $Tr(\mathbb{N})$.)  Then Quine’s argument at least shows us that, if P is consistent, S(P) is in $Tr(\mathbb{N})$.

Now suppose that PA is true but doesn’t mean arithmetic. Then we need to know whether the set of all the truths of our arithmetical language L is again $Tr(\mathbb{N})$. If PA is true because reality provides the (right sort of) non-standard model, then not all of $Tr(\mathbb{N})$ will be true. That’s just the result of PA not proving all of the sentences of $Tr(\mathbb{N})$. Suppose Q is a sentence of PA and PA $\not\vdash$ Q; then by the completeness theorem for first-order logic, there will be a model of PA that’s not also a model of Q. And if PA $\not\vdash$ S(P), then (if reality is making PA be true by being the wrong kind of model) it might be that S(P) is false.

So, if PA doesn’t mean arithmetic, we can say that Quine’s result holds so long as the set of truths of its language is $Tr(\mathbb{N})$. That might not seem very informative: it comes awfully close to saying something like “Quine’s result holds so long as, for every consistent P, it’s associated S(P) is true”. But we can do at least slightly better if we allow ourselves certain extensions of first-order logic in fixing the language. For instance, Hartry Field notes that if we have a “finitude” quantifier $\mathscr{F}$, where $\mathscr{F}$x(Gx) means “finitely many things are G”, then PA plus the axiom

(FIN) $\forall x\mathscr{F}y(y \leq x))$

are all true only on “standard” models, i.e., models isomorphic to $\mathbb{N}$, and therefore PA plus FIN guarantee the truth of $Tr(\mathbb{N})$.** Likewise, if the logic of PA is enriched with the omega rule, then we will again have PA $\vdash$ Q for every Q $\in Tr(\mathbb{N})$ (for our newly omega-enriched $\vdash$ relation), and so the arguments of the Quine post can be re-given.

But those sorts of enrichments seem especially problematic for Quine. For Quine is trying to argue that substitutional and model-theoretic notions collapse, which in turn relies on the notions applying to the first-order case. If we enrich the logic by introducing new rules or operators, we would have to go back and re-start the argument from the beginning. And it’s entirely unclear how we would do this, since the Hilbert-Bernays theorem it relies on was proven for first-order consistencies, rather than sentences were which “consistent” in the extended sense that would come with an extra inference rule or quantifier. It is not at all clear that if we have an enriched proof relation, and an associated enriched notion of consistency, anything like the Hilbert-Bernays theorem could be proven. If we enrich our proof system in certain ways, the resulting notion of “proof” won’t be effectively decidable, and so there will be no arithmetically definable predicate $Prf$ that codes for it, and so there will be no Hilbert-Bernays theorem to appeal to, and so there will be no Quinean collapse argument.

(Perhaps the best Quinean strategy in this area is to go ahead and allow the finitude quantifier, deny it any status as “logic” (including any inferential rules coded into the proof theory), but then insist that we’re only considering interpretations where it means “finitely many”, and so only have to worry about models where it is true of some formula iff finitely many things satisfy the formula. The biggest worry with this approach, I think, would be that the notion of “consistency” would no longer look even reasonably plausible: sentences such as

$\mathscr{F}xGx \wedge \neg\exists x Gx$

would come out as consistent, when they seem flagrantly not so. If we can make substitutional and model-theoretic accounts of logic collapse only by introducing new resources and then asserting what look like gross implausibilities about what is and is not consistent using those resources, we might legitimately wonder whether the game’s still worth the candle.)

So perhaps we can do no better than saying something like, “Quine’s theorem will apply if PA is true and means arithmetic, or if $Tr(\mathbb{N})$ is true (regardless of whether or not it means arithmetic).

But what if PA is flat-out false?

Well, suppose PA is flat-out false. (Then $Tr(\mathbb{N})$ will be flat-out false, too, because it has PA as a part.) But suppose also that there is a substitution scheme U such that U(PA) is true. Then we can ask: is $U(Tr(\mathbb{N}))$ true, too? If the answer is yes, then U(S(P)) is also true. (As S(P) $\in Tr(\mathbb{N})$, U(S(P)) $\in U(Tr(\mathbb{N}))$.) And the question as to whether $U(Tr(\mathbb{N}))$ is true is (more or less) the question as to whether the chunk of reality that makes U(PA) true is like a standard model of PA, or like a non-standard one. Or, more precisely, it’s asking something like this: if you took each predicate P that appears in PA, looked at the formula F that U swapped P out for, and then built a model M by putting all and only the F-satisfiers in the domain of P, would the result be a standard model of PA? If yes, then $U(Tr(\mathbb{N}))$ will be true, and so Quine’s result will follow. Notice that if the answer is “no”, though, that doesn’t quite tell us that Quine’s result does not follow. For there are non-standard models of PA that are also models of $Tr(\mathbb{N})$; perhaps the model built is one of these. If so, then $U(Tr(\mathbb{N}))$ will again be true and Quine’s result will again follow. But if it’s the wrong sort of non-standard model of PA, S(P) may be false and we can’t rely on collapse.

So that, I submit, is what we can say with confidence: for Quine’s theorem to work, “rich enough for simple arithmetic” needs to mean that there is a substitution scheme U (which might be the identity scheme) for which $U(Tr(\mathbb{N}))$ is true. A sufficient (but not necessary) condition for this is for the model of PA reverse-engineered from the (true) U(PA) to be a standard model of PA.***

* Peter Smith argues that there is no genuinely “intuitive” notion of validity (and hence, I imagine, no “intuitive” notion of soundness); to get Kreisel’s argument off the ground, you need rather a semi-technical, quasi-intuitive notion first, and then the argument can show some fully technical apparatus sufficient to capture the semi-technical notion you’ve gotten toward. I think all of that can be accepted while still accepting that any quasi-intuitive notion of soundness you might end up with has this property: no proof that is quasi-intuitively sound has all true premises and a false conclusion. And that’s the strength of premise Quine needs at this stage of the argument.

** If I’m understanding aright, this constraint is strictly stronger than merely making all of $Tr(\mathbb{N})$ true, as there will be non-standard models of $Tr(\mathbb{N})$ that aren’t also models of PA+FIN.

*** An equivalent (and less messy-sounding) way to state the condition is that the domain of the so-constructed model (or the extension of “is a natural number” on the model, if the quantification isn’t restricted) is an $\omega$-sequence under “<“. But that will hold iff the domain of the quantifiers of L is an $\omega$-sequence under U(<) (or, if the domain isn’t restricted, that the extension of “U(natural number)” is an $\omega$-sequence under U(<)). So a different, not so model-theoretic-sounding way of phrasing the sufficient condition is simply “has as a domain (or as the extension of ‘U(natural number)’) an $\omega$-sequence under U(<)”.

## Thought Special Issue: Time and Modality

Thought is having a special issue on the metaphysics of time and modality.  Deadline is the end of May.  Call for papers follows:

Metaphysicians of modality argue over whether ontology extends beyond the actual just as metaphysicians of time argue over whether ontology extends beyond the present; and we might also ask whether it is a stable position to hold that reality includes the non-present but not the non-actual. There are modal analogues of McTaggart’s infamous argument for the unreality of time, and we can ask whether the modal and temporal arguments stand or fall together. We might wonder whether trans-world identity should be treated differently from identity across time, and whether if existence is contingent it must also be temporary, etc.

For this special issue of Thought we invite papers that make a contribution to either the metaphysics of time or of modality, or that illuminate the connections between them. Papers should correspond to the standard Thought guidelines and be no longer than 4500 words, including footnotes. Papers are to be submitted before 31st May 2013. When submitting please ensure you select article type as “The Metaphysics of Time and Modality Special Issue” to ensure your paper is reviewed via the special issue route.

## NIP early career conference (papers to be published in Thought)

University of Aberdeen, Scotland.
27-29 June 2013

KEYNOTE SPEAKERS
Professor Jennifer Saul (Sheffield)
Professor Brian Weatherson (Michigan)

CONFERENCE AIMS
Our ultimate aim is to showcase outstanding research by early career researchers. While there are a number of opportunities for graduate students and more senior philosophers to present and discuss their research, there are relatively few for researchers who have recently finished their PhDs and are building a career in philosophy. This series has been established as a step towards addressing this imbalance.

THOUGHT: A Journal of Philosophy EARLY CAREER SPECIAL ISSUE
The papers selected for the conference will be invited for inclusion in a special Early Career issue of the journal Thought.

The editors of the journal may request revisions which must be met to their satisfaction before publication of the paper will be guaranteed.

CRITERIA FOR SUBMISSIONS
We invite submissions from researchers who received their PhD within the last 5 years. Papers must fall under the research remit of the Northern Institute of Philosophy: metaphysics, epistemology and the philosophies of logic, language, mathematics and mind.

GUIDELINE FOR SUBMISSIONS
We invite papers suitable for a 40-minute presentation and of no longer than 4000 words. Papers should be accompanied by an abstract of no more than 150 words and must be suitable for blind refereeing. Please indicate at the top of the first page of your paper which of the following categories it falls under: Metaphysics, Epistemology, Philosophy of Logic, Philosophy of Language, Philosophy of Mathematics, Philosophy of Mind. Please include a separate cover sheet including name, title, institution, month and year of PhD award and contact details. The deadline for receipt of submissions is 12th April 2013. We aim to notify authors of the decision regarding their papers by 21st May 2013. Submissions must be in .doc or .pdf format and are to be submitted by e-mail to earlycareerconference@gmail.com

Meals (lunches and one social dinner) and accommodation will be provided for speakers.

Please direct any questions to the conference organisers on earlycareerconference@gmail.com