# True, but unprovable statements and the Hydra problem. (Gödel's incompleteness)

Gödel's incompleteness theorem states that PA is negation-incomplete. One theorem that has been shown to be independent of the axioms of PA is the theorem that it's impossible to lose the Hydra game.

The theorem is nonetheless viewed as true, and it is indeed provable in set theory. However, since PA is independent of this theorem assuming PA is consistent, we should be able to augment the axioms of PA with an axiom that would guarantee that the negation of the theorem is provable.

At this point some confusion arises. In the new formal system, the theorem is false, but nonetheless it's considered "intuitively" true. Have we just added an axiom that is in fact "intuitively" false to the system? Considering also that in PA there is a model where the statement is false, shouldn't the PA axioms too be viewed with some suspicion?

• No, they shouldn't. This is just an obvious consequence of incompleteness. PA is sound, meaning that its axioms (and theorems) are true in the standard model (the natural numbers). That a sentence $\phi$ is independent of PA means that neither $\phi$ nor its negation is provable, which implies that both PA $+\{\phi\}$ and PA $+\{\lnot\phi\}$ are consistent. Obviously, exactly one of these two extensions is again sound, while the other is not. You could argue that the "right" extension is the sound one, but note that none of this says anything about issues with PA itself. Commented May 29, 2017 at 14:46
• As an aside, is "negation-incomplete" the terminology used in some specific reference you have consulted? It is not standard terminology. Commented May 29, 2017 at 14:49
• I think I saw a question/answer there where such formulation was used. I just wanted to distinguish from the system being semantically complete (which Gödel proved it is), see also the wikipedia article: en.wikipedia.org/wiki/Completeness_(logic) So reading your comment, the standard model is taken as "True", I think I understand now.
– Dole
Commented May 29, 2017 at 15:21
• @AndrésE.Caicedo Can I still ask for a clarification on what you mean by the standard model? Do you simply mean a model which has addition and multiplication functions and constants 0,1, and if I get this right, no axioms?
– Dole
Commented Jun 4, 2017 at 17:54

I think there are a couple confusions here. For simplicity, I'll adopt a Platonist viewpoint.

First of all, the "standard model" just refers to the structure $\mathcal{N}=(\mathbb{N}; +,\times)$. The point is that the PA axioms constitute a description of $\mathcal{N}$. In the context of arithmetic, we use the word "true" as shorthand for "true in the specific structure $\mathcal{N}$."

(Incidentally, I don't know what you mean when you ask whether the standard model is just "a model which has addition and multiplication functions and constants 0,1, and if I get this right, no axioms" - the fist part just describes any model of PA, not the standard one specifically, and models don't have axioms so I'm not sure what the second part means.)

We tend to be very confident that the PA axioms are in fact true. There is however no reason to believe that PA completely describes $\mathcal{N}$. You interpret this as a cause to doubt the PA axioms:

Considering also that in PA there is a model where the statement is false, shouldn't the PA axioms too be viewed with some suspicion?

But this has things backwards. Surely we are confident that addition of natural numbers is commutative; this means that $T=\{\forall x\forall y(x+y=y+x)\}$ is a set of axioms (well, axiom) which we are confident is true of $\mathcal{N}$. But there are lots of models of the theory $T$ where obviously true statements about $\mathcal{N}$ are false - e.g. $T$ has a model with exactly one element! So by your logic we should be skeptical of the truth of $T$ in $\mathcal{N}$.

The failure of a theory to prove every true statement about $\mathcal{N}$ should not negatively impact our confidence in the truth of the axioms of that theory. In fact, strength and plausibility really pull in opposite directions: if we're super confident that a theory $T_0$ is true of $\mathcal{N}$, and $T_1$ is a stronger theory than $T_0$, then we can't be more confident that $T_1$ is true of $\mathcal{N}$. Saying that the PA axioms are true of $\mathcal{N}$ just means that $\mathcal{N}$ is one of the models of PA; the presence of other models, even other nonequivalent models, is merely a comment on the strength of PA.

This is not to say that doubting PA is silly. Some serious mathematicians (granted, a tiny tiny minority) have indeed expressed such doubts. But the incompleteness of PA is in no way motivation for those doubts; indeed, doubting the truth of PA (or even stronger, believing that PA is inconsistent) is really the claim that PA is too strong - that it has axioms which shouldn't be there.