Gödel’s second incompleteness theorem tells us that no (sufficiently powerful) consistent theory can prove the statement of its own consistency. But did you know that such a theory *can* prove the statement of its own *inconsistency*? A consistent theory that claims to be inconsistent is what I’ll call a self-hating theory.

My convention in what follows: ℕ refers to the *real*, *true* natural numbers, the set consisting of {0, 1, 2, 3, …} and nothing else. ω refers to the *formal object *that exists in a theory of arithmetic that is “supposed to be” ℕ, but (in first-order logic) cannot be guaranteed to be so.

When I write ℕ ⊨ ψ, I am saying that the sentence ψ is true of the natural numbers. When I write T ⊢ ψ (resp. T ⊬ ψ), I am saying that the sentence ψ can be (resp. can’t be) proven from the axioms of the theory T. And when I write T ⊨ ψ, I am saying that the axioms of T *semantically entail* the truth of ψ (or in other words, that ψ comes out true in all models of T). The next two paragraphs will give some necessary background on Gödel’s encoding, and then we’ll explore the tantalizing claims I started with.

Gödel’s breathtakingly awesome insight was that within any language that is expressive enough to talk about natural number arithmetic, one can encode sentences as numbers and talk about syntactic features of these sentences as *properties of numbers*. When a number n encodes a sentence ψ, we write n = ⟦ψ⟧. Then Gödel showed that you can have sentences talking about the *provability* of other sentences. (The next step, of course, was showing that you can have sentences talking about *their own* provability – sneaking in self-reference through the back door of any structure attempting to describe arithmetic.)

In particular, in any theory of natural number arithmetic T, one can write a sentence that on its surface appears to just be a statement about the properties of natural numbers, but when looked at through the lens of Gödel’s encoding, ends up actually encoding the sentence “T ⊢ ψ”. And this sentence is itself encoded as some natural number! So there’s a natural number n such that n = ⟦T ⊢ ψ⟧. It’s a short step from here to generating a sentence that encodes the statement of T’s own consistency. We merely need to encode the sentence “¬∃n (n = ⟦T ⊢ 0=1⟧)”, or in English, there’s no number n such that n encodes a proof of “0=1” from the axioms of T. In even plainer English, no number encodes a proof of contradiction from T (from which it follows that there IS no proof of contradiction from T, as any proof of contradiction would be represented by some number). We write this sentence as Con(T).

Okay, now we’re in a position to write the original claim of this post more formally. If a theory T is consistent, then ℕ ⊨ Con(T). And Gödel’s second incompleteness theorem tells us that if ℕ ⊨ Con(T), then T ⊬ Con(T). But if T doesn’t prove the sentence Con(T), then no contradiction can be derived by adding ¬Con(T) as an axiom! So (T + ¬Con(T)) is itself a consistent theory, i.e. ℕ ⊨ Con(T + ¬Con(T)). But hold on! (T + ¬Con(T)) can prove its own inconsistency! Why? Because (T + ¬Con(T)) ⊢ ¬Con(T), i.e. it proves that a contradiction can be derived from the axioms of T, and it also has as axioms every one of the axioms of T! So the same number that encodes a proof of the inconsistency of T, also counts as a proof of the inconsistency of (T + ¬Con(T))!

Summarizing this all:

ℕ ⊨ Con(T)

⇒

T ⊬ Con(T)

⇒

ℕ ⊨ Con(T + ¬Con(T)),

but

(T + ¬Con(T)) ⊢ ¬Con(T + ¬Con(T))

There we have it, a theory that is consistent but proves its own inconsistency!

Expressed another way:

T ⊢ ∃n (n = ⟦T ⊢ 0=1⟧),

but

T ⊬ 0=1

Ok, so believe it or not, a lot of the strangeness of this can be explained away by thinking about the implications of nonstandard models of arithmetic. One easy way to see this is to reflect on the fact that, as we saw above, “T is consistent” becomes in Gödel’s encoding, “There is no natural number n such that n encodes a proof of T’s inconsistency.” Or more precisely, “T is consistent” becomes “There is no natural number n such that n = ⟦T ⊢ 0=1⟧.”

Now, no first-order theory can pin down the natural numbers.

(I’ve written about this here and here.) I.e. no first order theory can express a quantification like “there is no natural number N such that …”. You can try, for sure, by defining some object ω and adding axioms to restrict its structure to look more and more like ℕ, but no matter how hard you try, no matter how many axioms you add, there will always be models of the theory in which ω ≠ ℕ. In particular, ω will be a strict superset of ℕ in all of these nonstandard models (ℕ ⊂ ω), so that ω contains all the naturals but also additional nonstandard numbers.

So now consider what happens when we try to quantify over the naturals by saying “∀x ∈ ω”. This quantifier inevitably ranges over ALL of the elements of ω in each model, so it also touches the nonstandard numbers in the nonstandard models. This means that the theory only semantically entails quantified statements that are true of all possible nonstandard numbers! (Remember, T ⊨ ψ means that ψ is true in ALL models of T.)

One nice consequence of this is that if T has a model in which ω = ℕ then in this model “∀x∈ω Φ(x)” is true only if Φ(x) is true of all natural numbers. By the completeness of first-order logic, this means that T can’t prove “∀x∈ω Φ(x)” unless it’s true of ℕ. This is reassuring; if T ⊢ ∀x∈ω Φ(x) and T has a model in which ω = ℕ, then ℕ ⊨ ∀x∈ω Φ(x).

But the implication doesn’t go the other way! ℕ ⊨ ∀x∈ω Φ(x) does not guarantee us that T ⊢ ∀x∈ω Φ(x), because T can only prove that which is true in *EVERY* model. So T can only prove “∀x∈ω Φ(x)” if Φ(x) is true of all the naturals and every nonstandard number in every model of T!

This is the reason that we don’t know for sure that if Goldbach’s conjecture is true of ℕ, then it’s provable in Peano arithmetic. On the face of it, this looks quite puzzling; Goldbach’s conjecture can be written as a first-order sentence and first-order logic is complete, so if it’s true then how could we possibly not prove it? The answer is hopefully clear enough now: Goldbach’s conjecture might be true of all of ℕ but false of some nonstandard models of Peano arithmetic (henceforth PA).

You might be thinking “Well if so, then we can just add Goldbach’s conjecture as an axiom to PA and get rid of those nonstandard models!” And you’re right, you will get rid of *those* nonstandard models. But you won’t get rid of all the nonstandard models in which Goldbach’s conjecture is true! You can keep adding as axioms statements that are true of ℕ but false of some nonstandard model, and as you do this you rule out more and more nonstandard models. At the end of this process (once your theory consists of literally all the first-order sentences that are true of ℕ), you will have created what is known as “True Arithmetic”: {ψ | ℕ ⊨ ψ}.

But guess what! At this point, have you finally ruled out all the nonstandard models? No! There’s still many many more (infinitely many, in fact! Nonstandard models of every cardinality! So many models that *no cardinality is large enough to describe how many!*) Pretty depressing, right? There are all these models that agree with ℕ on every first order sentence! But they are still not ℕ (most obviously because they contain numbers larger than 0, 1, 2, and all the rest of ℕ).

The nonstandard models of True Arithmetic are the models that are *truly irremovable* in any first-order theory of arithmetic. Any axiom you add to try to remove them will also remove ℕ as a model. And when you remove ℕ as a model, some pretty wacky stuff begins to happen.

Fully armed now with new knowledge of nonstandard numbers, let’s return to the statement I started with at the top of this post: there are consistent theories that prove their own inconsistency. The crucial point, the thing that explains this apparent paradox, is that *all such theories lack ℕ* *as a model*.

If you think about this for a minute, it should make sense why this must be the case. If a theory T is consistent, then the sentence “∀x∈ω (x ≠ ⟦T ⊢ 0 = 1⟧)” is true in a model where ω = ℕ. So if T has such a model, then T simply *can’t* prove its own inconsistency, as it’s actually *not* inconsistent and the model where ω = ℕ will be able to see that! And once more, T can only prove what’s true in all of its models.

Okay, so now supposing T is consistent (i.e. ℕ ⊨ Con(T)), by Gödel’s second incompleteness theorem, T cannot prove its own consistency. This means that (T + ¬Con(T)) is a consistent theory! But (T + ¬Con(T)) *no longer has ℕ as a model*. Why? Because ℕ ⊨ Con(T) and (T + ¬Con(T)) ⊨ ¬Con(T). So for any consistent theory T, (T + ¬Con(T)) only has nonstandard models. What does this mean about the things that T + ¬Con(T) proves? It means that they no longer have to be true of ℕ. So for instance, even though ℕ ⊨ Con(T + ¬Con(T)), (T + ¬Con(T)) might end up proving ¬Con(T + ¬Con(T)). And in fact, it does prove this! As we saw up at the top of this post, a moment’s examination will show that (T + ¬Con(T)) asserts as an axiom that a contradiction can be derived from the axioms of T, but also contains all the axioms of T! So by monotonicity, (T + ¬Con(T)) proves ¬Con(T + ¬Con(T)).

What do we say of this purported proof of contradiction from (T + ¬Con(T))? Well, we know for sure that it’s not a standard proof, one that would be accepted by a mathematician. I.e., it asserts that there’s some n in ω that encodes a proof of contradiction from (T + ¬Con(T)). But this n is not actually a natural number, it’s a nonstandard number. And nonstandards encode proofs only in the syntactical sense; a nonstandard proof is a proof according to Gödel’s arithmetic encoding, but Gödel’s arithmetic encoding only applies to natural numbers. So if we attempted to translate n, we’d find that the “proof” it encoded was actually nonsense all along: a fake proof that passes as acceptable by wearing the arithmetic guise of a real proof, but in actuality proves nothing whatsoever.

Summarizing:

In first order logic, every theory of arithmetic has nonstandard models that foil our attempts to prove all the truths of ℕ. Theories of arithmetic with ONLY nonstandard models and no standard model can prove things that don’t actually hold true of ℕ. In particular, since theories of arithmetic can encode statements about their own consistency, theories that don’t have ℕ as a model can prove their own inconsistency, even if they really are consistent.

So much for first order logic. What about

Second Order Logic?

As you might already know, second order logic is capable of ruling out all nonstandard models. There are second order theories that are categorical for ℕ. But there’s a large price tag for this achievement: second order logic has no sound and complete proof system!

Sigh. People sometimes talk about nature being tricky, trying to hide aspects of itself from us. Often you hear this in the context of discussions about quantum mechanics and black holes. But I think that the ultimate trickster is logic itself! Want a logic that’s sound and complete? Ok, but you’ll have to give up the expressive power to allow yourself to talk categorically about ℕ. Want to have a logic with the expressive power to talk about ℕ? Ok, but you’ll have to give up the possibility of a sound and complete proof system. The ultimate structure of ℕ remains shifty, slipping from our view as soon as we try to look closely enough at it.

Suppose that T is a second order theory that is categorical for ℕ. Then for every second-order sentence ψ that is true of ℕ, T ⊨ ψ. But we can’t make the leap from T ⊨ ψ to T ⊢ ψ without a complete proof system! So there will be semantic implications of T that cannot actually be proven from T.

In particular, suppose T is consistent. Then T ⊨ Con(T), but T ⊬ Con(T), by Gödel’s second. And since T ⊬ Con(T), (T + ¬Con(T)) is consistent. But since T ⊨ Con(T), (T + ¬Con(T)) ⊨ Con(T). So (T + ¬Con(T)) ⊨ Con(T) ∧ ¬Con(T)!

In other words, T + ¬Con(T) actually has no model! But it’s consistent! There are consistent second-order theories that are *actually not logically possible* – that semantically entail a contradiction and have no models. How’s that for trickiness?

Great post! There is a spirited discussion of this topic on Scott Aronson’s blog.

Thank you! Link to the discussion?

https://www.scottaaronson.com/blog/?p=4974

However, there is a subtle point here, because second-order Categorical Arithmetic is not effectively axiomatizable, the incompleteness Theorems cannot be applied to it directly. So it seems to need a different proof of the syntactic incompleteness of second-order Categorical Arithmetic.

I disagree that the incompleteness theorems don’t apply in second order logic. Any consistent effectively axiomatized theory of arithmetic stronger than Robinson arithmetic Q will be incomplete, which means that any complete & consistent theory of arithmetic stronger than Q will not be effectively axiomatizable. The fact that second order logic allows a categorical theory of arithmetic (Peano arithmetic with the second-order induction axiom) tells us that it cannot be effectively axiomatizable.

The issue here is addressed in “A Proof Of Syntactic Incompleteness Of The Second-Order Categorical Arithmetic, DOI: 10.4236/oalib.1103969”.