# Usefulness of Gödel's incompleteness theorem

Given a first-order language $$L$$ and a theory $$T$$ in that language (a set of formulas of $$L$$), if $$T$$ is strong enough to prove arithmetic, then Gödel's second incompleteness theorem tells us that $$T$$ cannot prove its own consitency, ie $$T\not\vdash \text{Cons}(T)$$.

Now, imagine for a second that Gödel's second incompleteness theorem is false, and that we do have a formal proof that $$T\vdash \text{Cons}(T)$$. What confidence does it give us in $$T$$ ? If $$T$$ is inconsistent, it proves all formulas and in particular $$T\vdash \text{Cons}(T)$$. That makes this imaginary proof of $$T\vdash \text{Cons}(T)$$ fundamentally useless.

So if I rephrase Gödel's theorem tongue-in-cheek, it reads : do not look for a useless proof of consistency, because apart from being useless, it is nonexistent.

Then what makes the incompleteness theorem so famous ? It halted Hilbert's program, but I fail to pinpoint exactly what Hilbert hoped to achieve.

• @MauroALLEGRANZA I know Gödel's theorem is true. I said "imagine" it is false :) – V. Semeria Sep 27 '18 at 11:03
• Hilbert hoped to achieve a "normal" mathematical proof of consistency of arithmetic (and then analysis). We have math proofs of consistency : prop calculus, first-order logic, etc. and they need mathematical "resources" of a quite elementary level. – Mauro ALLEGRANZA Sep 27 '18 at 11:58
• Hilbert project aimed at use "reasonable" mathematical resources (basically: Primitive Recursive Arithmetic) to develop a "combinatoric" proof (in a nutshell, a formal theory is a collection of finite sequences of finite strings) of e.g. Peano's arithmetic (formalized with predicate calculus) using "elementary" mathematical resources. – Mauro ALLEGRANZA Sep 27 '18 at 12:00

The second incompleteness theorem is useful because it points out an unprovable statement with a more natural meaning than the unprovable statement we get from the first incompleteness theorem.

This can be used as a part of independence proofs: One way to prove that theory $$T$$ cannot prove some interesting statement $$A$$ is to show that $$T$$ proves $$A\to\operatorname{Con}(T)$$ -- so if $$T$$ also proved $$A$$, it would prove its own consistency, which we know it doesn't.

As one example, this is a quick way to know that ZFC (if it is consistent) does not prove that there are inaccessible cardinals. This gives the positive knowledge that if ZFC is consistent, it has a model with no inaccessible cardinal in it, which may be further useful.

If you don't care at all about knowing that this-or-that theory cannot possibly prove $$A$$, then I suppose this would not count as useful for you. However, if you've banged your head into a wall for long enough trying to prove $$A$$ -- because that would certainly be useful for your research if only you knew how! -- then you might not dismiss this so quickly.

• Fair point. I actually knew how to use incompleteness to prove some independence results. But I think this is a clever indirect application of incompleteness that came afterwards, it is not what historically happened between Hilbert and Gödel. Hilbert wanted to prove that mathematics were correct and Gödel stopped him. I don't see what Hilbert wanted to do, since $T \vdash\text{Cons}(T)$ tells us nothing about the consistency of $T$. As you said, only about some other statements independent of $T$. – V. Semeria Sep 27 '18 at 11:00
• @V.Semeria: Hilbert's program was not just to find a theory that proved itself consistent. It was to bootstrap mathematics by starting with a simple theory $T_0$ that everyone could believe in on intuitive grounds (something like PA, for example), and then prove in $T_0$ that some stronger and richer theory $T_1$ (something like set theory) is also trustworthy. This is not inherently meaningless or useless. However, the second incompleteness theorem prevents this if $T_1$ proves everything $T_0$ does. It $T_0$ proves $T_1$ consistent, then $T_1$ would prove itself consistent. – Henning Makholm Sep 27 '18 at 11:14
• Even if what we got was only a relative consistency result ($T_0$ proves that if $T_0$ is consistent then so is $T_1$), we would still run into problems if $T_1$ is strong enough to prove $T_0$ consistent. This is the case for $T_0$ being $PA$ and $T_1$ being ZF set theory, a possibility that definitely was on the Hilbertians' minds. – Henning Makholm Sep 27 '18 at 11:17
• So do you say that the main impact of incompleteness was that $\text{PA}\not\vdash\text{Cons}(\text{ZF})$ ? (because if it did, then $\text{PA}\vdash\text{Cons}(\text{PA})$). In that case it looks obvious, even without proving the incompleteness theorem, that the bootstrapping collapses to the question of PA proving the consistency of PA, which is ill-founded. – V. Semeria Sep 27 '18 at 11:32
• @V.Semeria: I don't know what you mean by "which is ill-founded" there. Your argument in the question post is at best that PA proving its own consistency would be uninteresting; there's nothing there which implies that it should be expected to be impossible, except for the incompleteness theorem stating that it is. – Henning Makholm Sep 27 '18 at 12:12