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I was thinking about the incompleteness theorem. In a book it said that say RH might be unprovable and that a Mathematician could be working on a problem that is unprovable. But, was wondering is it even worse than that.

Are there problems such that it's unprovable, but cannot be proven unprovable. So the problem wouldn't be able to be solved by a Mathematician and the Mathematician will never know if it can be solved.

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    The word "unprovable" makes it easy to confuse oneself, because it invites thinking that "unprovable" is a property of a particular formula _in itself_. Really, however, it is a _relation_ between a formula and a particular set of axioms/assumptions. So, you want a sentence $\phi$ such that $T\not\vdash \phi$ but $S\not\vdash (T\not\vdash \phi)$ for some theories $S$ and $T$? Do you require that $S$ and $T$ are the same theory?2012-04-05

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Let $\gamma_{PA}$ be the Gödel sentence for $PA$, and let $\mathbf{Proof}_{PA}(\theta,y)$ be the predicate the says $y$ encodes a proof for $\theta$ from $PA$. Then we have

$ \mathbb{N} \models \gamma_{PA} \longleftrightarrow (\forall y) \neg \mathbf{Proof}_{PA}(\theta,y)$

From the proof of the first incompleteness theorem, we know that

$ PA \nvdash \gamma_{PA}, ~~~ PA \nvdash \neg \gamma_{PA}$

Thus $\mathbb{N} \models \gamma_{PA}$, so that the Gödel is sentence is "true," albeit unprovable in $PA$.

Now, let $\square_{PA}(\theta)$ be the predicate stating that $\theta$ is provable in $PA$. Suppose that

$PA \vdash \neg \square_{PA}(\gamma_{PA})$

Then $PA$ would be able to prove that $PA$ cannot prove something, and hence

$PA \vdash \mathbf{Const}_{PA}$

which contradicts the Second Incompleteness Theorem.

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    I do not understand the answer. Is the answer to the OP yes or no?2015-10-27