Why is the Gödel sentence accepted as true?

Question

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?

Answer 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".

Answer 2

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.