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I‘m not very versed in Gödel‘s incompleteness theorem but in a naive way: If a statement of existence is not provable, there you cannot find an example which fulfills the statement (otherwise the statement would be provable with this example). But when there is no element which fulfills the statement, doesn’t that imply that the statement ist false?

I thought about that one in the context of the measure problem - because the statement

$$\exists \text{ measure function } \mu: 2^{\mathbb R} \to [0,\infty] \, \forall I = [a,b] \subseteq \bar{\mathbb R}: \mu(I) = b - a$$

is neither provable nor refutable. But if I cannot prove there is a measure function, I cannot find a $\mu$ for which the statement is true. Because finding such a $\mu$ would prove the statement. But when there is no such $\mu$, the statement of existence is false, isn‘t it? Where is my mistake in thinking?

ATW
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  • In classical logic, a statement is considered false, if you can give a proof of its negation. – Simon Marynissen Aug 28 '20 at 08:47
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    Not always can we be sure that the existence of an example is provable, but often this is the case. If Goldbach's conjecture is false, there must be a proof that it is false. But if the Collatz conjecture has a counterexample, this need not be provable because we might not be able to prove a divergent sequence to be actually divergent. – Peter Aug 28 '20 at 09:01
  • If we could prove that Goldbach's conjecture is not disprovable within PA (for example in ZFC), we would have proven it. This fact could however not be proven within PA, because then , PA would have shown that there cannot be a counterexample. – Peter Aug 28 '20 at 09:06
  • @Peter Can you express the Collatz Conjecture as a first-order sentence in the language of Peano arithmetic? – Robert Shore Aug 28 '20 at 09:23
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    I am pretty sure this is possible, this is even possible for the Goodstein-sequences where PA cannot prove that they all terminate. But PA can "understand" the Goodstein-sequences. – Peter Aug 28 '20 at 09:29
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    @SimonMarynissen Not in every case of a false statement, we will be able to prove the negation. "false" is not the same as "disprovable" – Peter Aug 28 '20 at 09:31
  • I meant to say, if a statement is known to be false, then we have a proof of its negation. – Simon Marynissen Aug 28 '20 at 10:20

1 Answers1

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It's not correct that if a statement of existence isn't provable, then you can't find an example that fulfills the statement. It means the theory has at least one model in which you can't find an example to fulfill the statement. But for the statement to be false, you need to be able to say that you can't find an example to fulfill the statement in any model of the theory.

To take a simple example, in the theory of fields with characteristic $0$, the statement $\exists x~(x^2+1 =0)$ is neither provable nor refutable. That just means in some models (fields) it's false (e.g., $\Bbb R$) but in others (e.g., $\Bbb C$), it's true.

If a sentence $\sigma$ is neither provable nor refutable from a theory $T$, that means the theory has at least one model in which the sentence is true and it also has at least one model in which it's false. That's because both $T \cup \{ \sigma \}$ and $T \cup \{ \lnot \sigma \}$ are consistent so by the Completeness Theorem, they both must have models.

Robert Shore
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  • Why the downvote? – Robert Shore Aug 28 '20 at 08:54
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    I'm the up-voter, not the down-voter, but I think I think the last sentence is a bit too strong. This isn't the only case where an existential statement is neither provable nor refutable is it? I'm thinking of the same Collatz example Peter gave in his comment n the OP. – saulspatz Aug 28 '20 at 09:06
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    What I wrote is a consequence of the Completeness Theorem. If $\sigma$ is neither provable nor refutable from $T$, then both $T \cup { \sigma }$ and $T \cup { \lnot \sigma }$ are consistent, which means they each must have models. – Robert Shore Aug 28 '20 at 09:09