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The Goldbach Conjecture asserts:

It is possible to write every even number greater that 2 as the sum of two primes.

Assume I can prove that the Goldbach Conjecture is unprovable from the Peano arithmetic. That would mean I could not prove the conjecture, nor the negation.

But wouldn't that mean that I can not find a counter-example? Because that would be a finite proof to show that its negation is true. And if I can't find an counter example, doesn't that imply that the conjecture is true?

In other words, wouldn't it be a proof to the conjecture if I can show that it is unprovable?

amWhy
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Adam
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    Is true that if you cannot find a counter-example, then there is no counter example? – Fimpellizzeri May 10 '18 at 19:25
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    If the Goldbach conjecture is false, then you can verify this, since you can run a computer long enough to find you a counterexample $2n$. So if the Goldbach Conjecture is false, then it's provable. – Alex R. May 10 '18 at 19:35
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    Do you believe you can prove it is unprovable in Peano arithmetic? – it's a hire car baby May 10 '18 at 19:48
  • Adam: Prove is a verb (to prove, the infinitive. We say, e.g., I prove, you prove, he proves, she proves, we prove, they prove ... a statement.) A Proof of a statement is the thing that we hope to arrive at after proving the statement. – amWhy May 10 '18 at 21:27
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    Take a different example such as Goodstein's theorem about certain fast-growing arithmetically defined sequences ending at $0$ - it cannot be proved in Peano arithmetic (that statement is the Kirby–Paris theorem) but it is true (and can be proved using techniques outside Peano arithmetic) so there are no counter-examples – Henry May 10 '18 at 22:29
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    @Fimpellizieri At least in the context of natural numbers, yes. Because not being able to find a counter example is equivalent to saying that a program that enumerates natural numbers and halts when it finds a counter example never terminates. – DanielV May 10 '18 at 22:30
  • @Henry does the Kirby-Paris theorem then already imply that the statement is true? Because if one knows that it is unprovable inside PA, then one can not construct a counter example? So the other proof with ordinal numbers is short and nice, but not necessary? – Adam May 11 '18 at 08:22
  • The Kirby-Paris theorem (that Goodstein's theorem cannot be proved in Peano arithmetic) also uses techniques outside Peano arithmetic, and is a harder proof that the use of ordinal numbers. So that means something special is necessary to prove Goodstein's theorem despite there being no counter-examples – Henry May 11 '18 at 09:18
  • @Henry if there are no counter-examples, how could the statement be possibly wrong? – Adam May 11 '18 at 09:41
  • A statement can be true, but may not be provable within the constraints of certain techniques. You can use ruler and compass to construct an irrational number such as $\sqrt{2}$. Similarly $\sqrt[3]{2}$ exists as a number since it is possible to double a cube (e.g. using origami), but it is not possible to prove this existence within the constraints of ruler and compass constructions – Henry May 11 '18 at 09:50
  • @Henry: Incidentally, Goodstein's theorem does not need ordinals to prove. Firstly, it suffices to prove that the hydra game terminates. The hydra game termination is indeed implied by the well-ordering of $ε_0$, but can also be proven in ACA (which is just PA plus conservative extension plus induction). In general, it seems that for some results one can obtain them either by a stronger induction principle or a stronger well-ordering principle. – user21820 May 11 '18 at 15:58

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Unprovable ≠ Undecidable. If PA can prove neither the conjecture nor its negation, it is undecidable in PA.

If you ever prove such a result, you certainly cannot be working within PA, because PA cannot prove that it cannot prove something, otherwise PA can prove that it cannot prove contradiction, which is impossible by Godel's second incompleteness theorem (assuming PA is consistent). Thus there is no paradox; your proof of unprovability over PA has to be a proof in some system other than PA.

So let's fix your foundational system MS as any reasonable formal system (at least proves existence of a model of PA), and let's reason within MS. If PA does not disprove Goldbach, then PA does not prove its negation, which is a $Σ_1$-sentence, and hence that negation cannot be true because PA is $Σ_1$-complete. So if PA does not disprove Goldbach, then Goldbach is actually true.

The caveat is that you are working within MS, so you should at least convince yourself that MS is consistent. Observe that if MS is inconsistent, you (working within MS) would be able to prove that PA does not prove Goldbach, but that would not mean anything.

Note that this argument does not necessarily apply to other open problems. For example the twin-prime conjecture can be written as a $Π_2$-sentence, and currently it is not known to be equivalent to a sentence of lower complexity, so the argument does not work, since PA is not $Σ_2$-complete.

user21820
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