1

The Gödel sentence G means "G and ¬G are independent of Peano arithmetic". Couldn't it be false? Couldn't there be actually a way to get to ¬G just from Peano axioms?

Otakar Molnár López
  • 21
  • Exactly; the Gödel sentence $G$ is undecidable in (first-order) Peano arithmetic, i.e. neither $\mathsf {PA} \vdash G$ nor $\mathsf {PA} \vdash \lnot G$ – Mauro ALLEGRANZA Apr 08 '21 at 12:07
  • But $G$ is an arithmetical formula; thus, it expresses a fact about natural numbers that it must hold or not. But the construction of $G$ is such that $G$ - when read through" the encoding - "reads" as a formula that asserts that $G$ itself is underivable in $\mathsf {PA}$, and so it is. Thus, $G$ is TRUE in the "common sense" view of the term. – Mauro ALLEGRANZA Apr 08 '21 at 12:10
  • How are you so sure? Maybe there is a very long proof within PA that gets to ¬. – Otakar Molnár López Apr 08 '21 at 12:10
  • No, if $\mathsf {PA}$ is consistent, there is no proof of $G$ in $\mathsf {PA}$. How can we be sure? Exactly as about Pythagoras Theorem: w ehave proved it. – Mauro ALLEGRANZA Apr 08 '21 at 12:11
  • If I'm correct, Con(PA) => G is derived from the acceptance that G is true and not backwards. Please, tell me if I'm wrong. Then, it might follow that ¬G is reacheable within PA and if a proof is found, Con(PA) => ¬G. – Otakar Molnár López Apr 08 '21 at 12:17
  • 1
    The Gödel sentence $G$ is a sentence of PA asserting that $G$ is not provable in PA. That is not the same as "$G$ and $\lnot G$ are independent of PA" (which is clearly true of $\lnot G$ if it is true of $G$). – Rob Arthan Apr 08 '21 at 12:52
  • We could if PA is incoherent or inconsistent. Any incoherent theory can prove anything. This is very unlikely though. Edward Nelson thought he found a proof but it was shown to be wrong. – Giorgio Genovesi Apr 08 '21 at 12:57
  • @RobArthan and Mauro. I can see my mistake now. Have a nice day. – Otakar Molnár López Apr 08 '21 at 12:58
  • Let me see if the question with G = "G is not provable in PA" makes sense. – Otakar Molnár López Apr 08 '21 at 13:04
  • It doesn't. Thanks a lot. – Otakar Molnár López Apr 08 '21 at 13:09

2 Answers2

1

As RobArthan said in his comment, the Gödel sentence is a sentence of PA asserting that is not provable in PA. That is not the same as " and ¬ are independent of PA".

Otakar Molnár López
  • 21
0

As noted, the sentence in your OP isn't actually a Godel sentence. However, we can still ask what its status is with repect to (dis)provability in $\mathsf{PA}$. In particular, given its "Godelish flavor," it may seem plausible that it too should be independent of $\mathsf{PA}$. The situation is in fact rather surprising:

Suppose $A$ is a sentence of the type in the OP. Within $\mathsf{PA}$, we can argue that if $A$ were $\mathsf{PA}$-disprovable then $A$ would be false; that is, $\mathsf{PA}\vdash Prov_{\mathsf{PA}}(\neg A)\rightarrow \neg A$. This is a direct consequence of the OP-property of $A$.

Now we apply Lob's theorem to $\neg A$: since $\mathsf{PA}$ proves "If $\neg A$ is $\mathsf{PA}$-provable, then it is true," we in fact get an outright $\mathsf{PA}$-proof of $\neg A$! Snappily:

"I am $\mathsf{PA}$-undecidable" is $\mathsf{PA}$-disprovable.

(A more common application of Lob's theorem which may be easier to think about is that "I am $\mathsf{PA}$-provable" is $\mathsf{PA}$-provable.) See here for some discussion of the intuition behind Lob's theorem; the general topic of provability logic may be of interest as well.

Noah Schweber
  • 260,658