Is Goedel term (in incomleteness theorem) both true and unproveable?

Question

In Goedel incompleteness theorem is Goedel term both true and unproveable, or just unproveable and truth neutral?

Can we add Goedel term to the theory as axiom and get new theory?

Can we add Goedel term negation to the theory and get a new theory? What can be said abot this last one theory: will it be inconsistent but this inconsistency will never manifestate?

Answer 1

I am interpreting your question as being about the usual Gödel sentence $\theta$ which is interpreted as saying "there is no proof of $\theta$ from $\text{PA}$," or perhaps more formally as "$\neg ( \exists n ) ( \text{Proof}(n,\ulcorner \theta \urcorner) )$," where $\text{Proof}(x,y)$ is the predicate asserting that $x$ codes a proof (from $\text{PA}$) of the sentence coded by $y$. Note that $\theta$ has the property that if $\text{PA}$ is ($\omega$-)consistent, then $\text{PA} \not\vdash \theta$ and $\text{PA} \not\vdash \neg \theta$.

(As a disclaimer of sorts, whenever I speak of "proof" (in English) below, I mean a "proof from $\text{PA}$.")

In a certain sense, $\theta$ is neither of the options you list... at least not without some extra assumptions. As $\text{PA}$ is either consistent or inconsistent, let's look at these cases separately:

• If $\text{PA}$ is consistent (which it is probably safe to say most mathematicians believe), then as $\text{PA} \not\vdash \theta$ it follows that $\theta$ speaks the truth about itself (as there is no proof of $\theta$ from $\text{PA}$ no natural number can encode a proof of $\theta$ from $\text{PA}$).
• If $\text{PA}$ is inconsistent, then $\text{PA}$ proves everything, and in particular $\text{PA} \vdash \theta$, and so the assertion that "$\theta$ has no proof in $\text{PA}$" is false, i.e., $\theta$ is false. (We can find a proof of $\theta$ from $\text{PA}$, and convert this proof into a number witnessing $( \mathbb{N} , ... ) \models \neg \theta$.)

To summarise: if $\text{PA}$ is consistent, then $\theta$ is an unprovable sentence which is true; if $\text{PA}$ is inconsistent, then $\theta$ is a provable sentence which is false.

The first case is the interesting one for the remainder of your questions. (If $\text{PA}$ is inconsistent, then so are $\text{PA} + \theta$ and $\text{PA} + \neg \theta$.) Recall that if $T$ is any theory and $\phi$ is any sentence, then $T + \phi$ is consistent iff $T \not\vdash \neg \phi$. So if $\text{PA}$ is consistent we have both $\text{PA} \not\vdash \theta$ and $\text{PA} \not\vdash \neg\theta$, and so both $\text{PA} + \neg \theta$ and $\text{PA} + \theta$ are consistent.

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Perhaps the more interesting sentence regarding the second and third questions is $\text{Con} ( \text{PA} )$ which expresses the consistency of $\text{PA}$; something to the effect of $\neg ( \exists n ) ( \text{Proof} ( n , \ulcorner 0 = 0 \wedge \neg 0 = 0 \urcorner )$. This is another sentence known to be independent of $\text{PA}$, provided that $\text{PA}$ is consistent. Following similar reasoning to the above, if $\text{PA}$ is consistent, then $\text{Con} ( \text{PA} )$ is a true (unprovable) sentence, and if $\text{PA}$ is inconsistent, then $\text{Con} ( \text{PA} )$ is a false (provable) sentence. Again, in the former case both $\text{Con} ( \text{PA} )$ and $\neg \text{Con} ( \text{PA} )$ may be appended to $\text{PA}$ to yield a consistent theory.

Looking at the consistency of $\text{PA} + \neg \text{Con} ( \text{PA} )$, recall that this just means, via Gödel's Completeness Theorem, that it has some model $\mathcal{M}$. As $\mathcal{M} \models \neg \text{Con} ( \text{PA} )$, then there is some $a \in \mathcal{M}$ such that $\mathcal{M} \models \text{Proof} ( a , \ulcorner 0=0 \wedge \neg 0=0 \urcorner )$, i.e., $\mathcal{M}$ "thinks" that $a$ codes a proof of $0=0 \wedge \neg 0=0$. However this $a$ will not correspond to any real natural number, so we cannot translate this object into a real proof of $0=0 \wedge \neg 0=0$. ($\mathcal{M}$ will be a nonstandard model of $\text{PA}$, and will contain objects which you can think of as "infinitely big natural numbers;" $a$ will be one of these.)

Answer 2

I am not an expert on these matters, but I think I can answer all of your questions, so here goes.

1. We assume the theory $T$ is strong enough to prove arithmetical truths; the technical qualifications are not that important here. The Goedel term $G$ for a theory $T$ informally states that '$G$ cannot be proved within $T$'. If $T$ is consistent (which we assume, otherwise $T$ would not be very interesting) then $T$ cannot prove $G$ (we'd get a contradiction). Therefore the term $G$ is actually true $-$ it really can't be proved within $T$.

2. We can add $G$ to $T$ as an axiom to obtain new theory $T'$. There is no problem doing this. We knew $G$ was true, so we are just enlarging $T$, so that it its more powerful and can prove more true statements about the world. $T'$ can now prove $G$ (trivially) but it is still not complete since Goedel's theorem produces a new true term $G'$ that $T'$ can't prove.

3. We can add $\neg G$ to $T$ as an axiom to obtain new theory $T'$. This new theory will still be consistent if $T$ was because $T$ itself can't prove $G$ and obviously $\lnot G$ won't make proving $G$ any easier. Nevertheless, we know that $G$ was true. Therefore this new theory $T'$ is not sound $-$ it is proving theorems that are not true.