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Godel's first incompleteness theorem states that any consistent formal system $F$ within which a certain amount of elementary arithmetic can be carried out is incomplete. I know we can prove Continuum Hypothesis to be independent of ZFC.

Now I wonder if there could be some statement in some formal system $F$ within which a certain amount of elementary arithmetic can be carried out such that the statement can neither be proved to be true nor false nor prove it's independent of the formal system.

We cannot prove $F$ to be consistent within $F$ itself. I thought if such a statement exists, maybe we can add the statement for $F$, maybe this new formal system can prove the statement to be independent? If not, it seems to be very sad that we will just never know something.

Logic is not my focus, and I'm not very familiar with it, just asking out of curiosity, so please forgive my ignorance.

wsz_fantasy
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  • How about $1\div 0 = 0$? By the usual definition of division on R, its truth value cannot be determined since the definition is not applicable for a divisor of zero. – Dan Christensen Jan 30 '24 at 17:13
  • @DanChristensen 1/0 is undefined, but we may define it arbitrarily when we define the division. For example, when we define division as a total function, we simply set $a/0=0$. The definition works with no problem, and it only fails to satisfy properties of division. – Hanul Jeon Jan 30 '24 at 17:16
  • Have you checked the statement of Gödel's second incompleteness theorem? – Hanul Jeon Jan 30 '24 at 17:17
  • @HanulJeon It is important to know the domain of definition for any given function. Outside of that domain, that function is said to be undefined. – Dan Christensen Jan 30 '24 at 17:24
  • @DanChristensen I mean, your example does not constitute a valid example for this question. I know all of what you said, but any reasonably strong systems can disprove $1/0=0$ under your terminology since they can prove $(1,0)$ is not in the domain. – Hanul Jeon Jan 30 '24 at 17:26
  • @HanulJeon The usual definition of division on $R$ is something like: $\forall x, y, z \in R: [y\neq 0 \implies [ x\div y=z \iff x=y\cdot z]].$ This definition is not applicable for $y=0$. – Dan Christensen Jan 30 '24 at 17:32
  • @DanChristensen I should acknowledge my first comment might be out of point, but your reply is not actually against my last comment. The OP asked if there is a statement that is neither provable nor disprovable nor proved to be independent. Your statement $1/0=0$ is disprovable. – Hanul Jeon Jan 30 '24 at 17:37
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    The answer might be different if we interpret $1/0=0$ differently. I understand it as an shorthand for "$1/0$ is defined and its value is $0$." Then it is disprovable since $1/0$ is undefined, which is provable weak fragment of arithmetic (like $\mathsf{PRA}$.) Or the statement $1/0=0$ may even not make sense. – Hanul Jeon Jan 30 '24 at 17:38
  • This 1/0 business doesn't seem to be in the spirit of the question, anyway. – Karl Jan 30 '24 at 17:42
  • @HanulJeon I'm just saying that the above definition cannot be used to prove $1\div 0 = 0$. Likewise, it cannot be used to prove $1\div 0 \neq 0$. – Dan Christensen Jan 30 '24 at 17:55
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    This is actually the case for any undecidable statement $G$ in a sufficiently powerful theory. What the system can actually prove internally is 'if I am consistent then there is no proof or disproof of $G$'. But since the system also can't prove its own consistency, the 'core' statement 'there is no proof or disproof of $G$' is undecidable. – Steven Stadnicki Jan 30 '24 at 18:16

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Nonconstructively, such statements should exist, at least for “good” foundational theories. In particular, fix a recursive $\Sigma_1$-sound extension of arithmetic $T$ (in particular, ZFC is such an extension under relatively modest assumptions). Suppose that for all $\Sigma_1$ sentences $\phi$, either $T \vdash \phi$ or $T \vdash ((T \vdash \phi) \to T \vdash \bot)$. In other words, suppose that for all such $\phi$, either $T$ proves $\phi$ or $T$ proves that $\neg \phi$ is relatively consistent with $T$.

In this case, we have an algorithm for computing whether $\phi$ is true. For if $\phi$ is true, then $T$ proves it. If it were also the case that $T$ proves $((T \vdash \phi) \to T \vdash \bot)$, then we would have $T \vdash (T \vdash \bot)$, which contradicts that $T$ is $\Sigma_1$-sound. By contrast, if $\phi$ is false, then $T$ can’t prove $\phi$ by $\Sigma_1$-soundness. Thus, for all such $\phi$, exactly one of the following holds: (1) $T \vdash \phi$ and $\phi$ is true, and (2) $((T \vdash \phi) \to T \vdash \bot)$ and $\phi$ is false. By simply enumerating proofs until we figure out which holds, we can algorithmically decide the truth of $\phi$. However, the problem of determining the truth of a $\Sigma_1$-sentence is algorithmically equivalent to the Halting Problem and is thus undecidable.

I hypothesise this argument can be extended to work with any consistent extension of arithmetic and that it can be made constructive.

Mark Saving
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