Two questions on the Gödel sentence

Question

1. If we add the Gödel sentence $G$ as an axiom of our theory, then does it merely become false? - since now there's a proof of it. And then what, the theory becomes inconsistent? Is there more to it than just that?

2. From one point of view, $G$ is unprovable merely because it is not implied by the axioms. Is there another way of given a set of axioms, constructing a sentence not implied by the axioms thus unprovable or is there something special about $G$?

Answer 1

The other answers are quite correct, but let me say a bit more regarding your second question.

First, "unprovable" and "not implied by the axioms" mean exactly the same thing. (Note that "undecidable" means something slightly different - unprovable and undisprovable.) So the answer to your question as phrased is, no.

However, I take it that you mean that the Godel sentence is undecidable because it refers to provability-from-the-axioms in an explicit way, and you're asking whether there are sentences which are undecidable without going this route. If this is the case, the answer is yes, in a very strong sense:

Since no computable consistent extension of $PA$ is complete, the set of $PA$-undecidable sentences is not computable. This shows that there is no "form" that $PA$-undecidable sentences always must have; there will always be surprising examples (and even examples that are $PA$-undecidable, but which we cannot prove are $PA$-undecidable even assuming $Con(PA)$!). The same goes for other reasonable theories which are strong enough to represent arithmetic, like $ZFC$ (set theory). Some interesting examples include:

• Goodstein's theorem is undecidable in $PA$ (see also the Paris-Harrington principle).

• The Continuum Hypothesis is undecidable in $ZFC$.

• "Every vector space has a basis" is undecidable in $ZF$ (= set theory without choice).

And there are many others. In terms of what methods may be used for establishing unprovability/undecidability over $PA$, see https://www.cs.umd.edu/~gasarch/TOPICS/largeramsey/bovINTRO.pdf.

Also, purely as a curiosity, check out Kripke's model-theoretic proof of the incompleteness of $PA$!

Answer 2

If adding a sentence made the theory inconsistent, then it must be the case that the theory implies its negation.

Indeed, you are making one basic mistake when you assume that adding $G$ as an axiom makes it false. $G$'s construction is such that $T\vdash G\leftrightarrow \neg\square_{T}G$; that is, there is no proof of $G$ in the original theory $T$.

Thus if we consider $T+G$, the theory to which we add $G$ as an axiom, then $T+G\vdash \neg\square_{T} G$, and that is perfectly consistent.

Same if we consider $T+\neg G$ instead. Only that such a theory will imply that $T+\neg G\vdash \square_{T} G$, which is arithmetically false, and thus $T+\neg G$ is $\omega$-inconsistent.

Note that we could then construct $G'$ in a similar fashion such that $T+G\vdash G'\leftrightarrow \square_{T+G}G'$, the corresponding Gödel sentence for the theory $T+G$, so we cannot complete the theory that way.

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Regarding your second question, there are plenty more undecidable sentences in the literature. For example, Rosser's sentence.

$G$ is special because it is equivalent to the consistency of the theory it is derived from, in the sense that $T\vdash G\leftrightarrow \neg \square_{T}\bot$. This can be seen using the fixed point machinery of provability logic.

Answer 3

(1) Asking if $G$ becomes false is meaningless; truth or falsity is only meaningful relative to some particular model. What you can ask meaningfully in this situation is whether it's provable from the new theory or not. That's easy, though — if you've added it as an axiom, it's now provable in the new theory.

The new theory will be consistent. The whole point of the sentence $G$ is that both $G$ and $\lnot G$ are consistent with the original theory.

(2) There are other ways of finding a sentence $\varphi$ that is neither provable or disprovable from a theory $T.$ The most straightforward approach is to find a model of $T\cup\lbrace\varphi\rbrace,$ and then also a model of $T\cup\lbrace\lnot\varphi\rbrace.$

A simple example of this is to let $T$ be the set of axioms for a group, and let $\varphi$ be the sentence $(\forall x)(\forall y)(xy=yx).$ To prove that $\varphi$ is independent of $T,$ you just have to exhibit an Abelian group and a non-Abelian group.

More complicated examples of this can be found in set theory, in the proofs of the independence of AC from ZF, the independence of CH from ZFC, etc.