Why do people speak about truth value of undecidable propositions?

Question

I have heard people say things like "If the Goldbach conjecture was proven to be independent of PA, then it would follow it's true." The reasoning behind this is that, if it was false, we could explicitly provide the counterexample, thus proving it false. Hence, it's true.

Now, this reasoning does make sense to me. What doesn't really make sense is that it doesn't make sense to ask whether an undecidable proposition is true or not before fixing a model. If the Goldbach conjecture was in fact independent of PA, there would be a model of PA in which it's true and a model where it's false. So to say that it would follow it's true doesn't really make sense to me.

Maybe what is meant is that it would follow that GC is true in the standard model of PA?

But nonetheless, I have read discussions about the truth of the axiom of choice. It's independent of ZF, so, if you accept that ZF is consistent, you accept the existence of a model of it in which AC is true. How can you argue that "it's false"?

Answer 1

The Goldbach conjecture is a statement about the natural numbers. When we say that it is false or true we mean just that there is or there isn't an even number greater than $2$ that is not the sum of two primes. When we say that if it is not refutable in $\sf PA$, then it is true we mean that if the formal statement of the negation of the Goldbach conjecture isn't a theorem of $\sf PA$, then every even number greater than $2$ is the sum of two primes.

In fact, we can prove a formalization of this conditional statement in $\sf PA$: if $\varphi$ is the formal statement of the Goldbach conjecture, then $\sf PA \vdash \lnot Prov_{PA}(\ulcorner \lnot\varphi\urcorner)\to \varphi.$ The same can be said for any $\Pi_1^0$ arithmetical statement, and holds equally well if we replace $\sf PA$ with any reasonable foundational system. If $\sf PA$ actually can't prove $\varphi$ or $\lnot \varphi$ there is no contradiction here, since by the incompleteness theorem, $\sf PA$ can't prove any statement of the form $\lnot \sf Prov_{PA}(\ulcorner\psi\urcorner),$ as this implies $\sf Con_{PA}.$

Say we could show in $\sf ZFC$ that the Goldbach conjecture is independent $\sf PA$. Then by the above theorem, we would have a proof in $\sf ZFC$ of the Goldbach conjecture. We could also then prove that there is nonstandard model of $\sf PA$ where the Goldbach conjecture $\varphi$ is false, and by the above theorem, $\sf Prov_{PA}(\ulcorner\lnot \varphi\urcorner)$ would be true in the model and so the model would think there was a proof in $\sf PA$ of the negation of the Goldbach conjecture. This is par for the course: nonstandard models of $\sf PA$ often disagree with the metatheory about things. (The model also disagrees about the Goldbach conjecture in this case, of course.)

We don't need to take a philosophical stance on absolute truth even about natural numbers in the above discussion: everything I said are just formal consequences of certain theories. Maybe we're pluralists of some sort and we think Golbachian and anti-Goldbachian mathematics are both interesting. When we're doing anti-Goldbachian math, we necessarily assume the Goldbach conjecture is refutable in any reasonable theory (though we don't have a concrete refutation any more than we have a concrete counterexample to Goldbach), as that's a straightforward consequence of the negation of the Goldbach conjecture.

Of course, neither Goldbachian nor anti-Goldbachian mathematics presently has a consistency proof, and if one day the conjecture is proven or refuted, one side will either have a bunch of useless results, or perhaps can retreat to studying weaker systems where the conjecture is still independent. But this is just a highly formalistic way of framing ordinary mathematics: people study consequences of both sides of undecided conjectures all the time.

On the other hand we could also think the natural numbers are a real thing and the Goldbach conjecture is a sharp statement about them that may be true or false, and if it's false nearly any reasonable formal system will prove this. (That's how I actually think about it.)

On the axiom of choice, people could mean a lot of different things when they say it's false. They could mean it's false in the model they're working in right now, they could mean they are assuming its negation (in the background theory they're working in, or informally), or they could mean they really believe there is such a thing as abstract sets and that they believe the axiom of choice is not a correct statement about sets.

Some people might not even mean its false in any sense, just that they choose not to assume it when they do mathematics. This could be because they are realists and aren't sure if it's true. Or maybe they're quite certain it is true, but they're topos theorists and they only want to use set theoretical reasoning that is valid in any topos (and avoid excluded middle and some other stuff too). Maybe they're traditional logicians and are interested in consequences of $\sf ZF$ without the $\sf C$ at the moment, their philosophical positions notwithstanding.

In any event, the theorem that the axiom of choice is independent (assuming $\sf ZF$ is consistent) doesn't pose a challenge here. If we are realists about set theory who think the axiom is false, then we believe it's not true of actual sets. We don't care about what a model of $\sf ZFC$ (perhaps we should call it a nonstandard model of $\sf ZF$ in this case) thinks any more than we care what a nonstandard model of $\sf PA$ thinks about the Goldbach conjecture (in the unlikely event that it's independent of $\sf PA$).

Answer 2

It's true in the standard model.

Statement 1. If the Goldbach conjecture (GC) is independent of PA, then there is:

• a model $M_T$ of PA where GC is true, and
• a model $M_F$ of PA where GC is false.

In the above, $M_F$ cannot be the standard model (which we usually call "the natural numbers"), because the witness $n$ of the falsity of GC would also be a witness in $M_T$, so PA would be inconsistent.

Since $M_F$ cannot be the standard model, that means GC is definitely not false in the standard model, so it is true.

So we can rephrase Statement 1 as

Statement 2. If the Goldbach conjecture (GC) is independent of PA, then there is:

• a model $M_T$ of PA, namely the standard model, where GC is true,
• a model $M_F$ of PA, which is not the standard model, where GC is false.

So yes, by "it's true", people mean "it's true in the standard model".

Answer 3

As others have answered, saying that ZF is true or false either assumes you're talking about some specific model, or you subscribe to some form of mathematical realism and believe sets, or in the case of PA, the natural numbers, really exist in some form. (Or at the very least, that statements about sets or natural numbers are either true or false, irrespective of what can be proven about them.)

But, at least for the natural numbers, many people do believe they actually exist in at least some abstract way. Statements like "PA is consistent" or "ZF is consistent" are undecidable in their respective logical systems and thus independent, but many people still think there is a strong intuition that these statements have real truth values. (And most mathematicians believe these to be in fact true.)

Formulated another way, a computer program* (Turing machine) that attempts to prove the inconsistency of ZF by enumerating all provable ZF theorems will keep running forever (assuming ZF is in fact consistent). But that is not provable in ZF. Yet I think most people believe such a computer program would either keep running forever, or would stop with a result. (That question is known as the halting problem.) Do you think this question does not have a definite truth value? If you believe truth values only exist relative to models or logical theories, you should.

Computer programs can be defined in PA, so believing the halting problem of a certain program has a definite truth value only commits you to realism w.r.t. the natural numbers. Nonrealism regarding sets is a lot more common among mathematicians.

* a computer program with access to infinite memory, else it is not equivalent to a Turing machine.

Answer 4

My knowledge on the topic is quite limited, but here is how I see it.

When people speak about the natural numbers outside of a formal setting, they typically either think about the "actual" natural numbers (which presupposes a philosophical stance that such a structure really exists), or at least about a model of natural numbers within $\mathsf{ZF(C)}$. The first-order theory $\mathsf{PA}$ is just too weak to describe either of these, so statements about the natural numbers independent of $\mathsf{PA}$ but "actually true/false" can reasonably exist.

When Peano proposed his axioms of the natural numbers, he obviously wanted to describe a unique structure up to isomorphism (whether or not the notion existed at the time). And he essentially succeeded, but a second-order axiom of induction was used to account for all possible subsets of $\mathbb{N}$. Since $\mathsf{PA}$ as a first-order theory has to abide by the stricter rules, its axiom of induction is not completely equivalent. In fact, it is significantly weaker - it only accounts for the definable subsets of $\mathbb{N}$. And it must be split into infinitely many axioms rather than just one (i.e. it is an axiom schema). On top of that, the first-order Peano arithmetic is a recursively enumerable theory, so in the spirit of Gödel's theorems it is rather small - unlike the True Arithmetic $\text{Th}(\mathbb{N})$ within $\mathsf{ZFC}$. For these reasons $\mathsf{PA}$ is quite far from fully describing the natural numbers that everyone thinks about.

The argument about the axiom of choice being "actually" true or false, however, seems to me as a mainly philosophical discussion, which again presupposes the stance (much bolder than in case of numbers) that the universe of sets really exists.