Let $\displaystyle A$ be a recursive set of sentences in a recursively numbered language with $\displaystyle \bold{0}$ and $\displaystyle \bold{S}$. Assume that every recursive relation is representable in the theory $\displaystyle \text{Cn }A$. Further assume that $\displaystyle A$ is $\displaystyle \omega$-consistent; i.e., there is no formula $\displaystyle \phi$ such that $\displaystyle A \vdash \exists x \phi(x)$ and for all $\displaystyle a \in \mathbb{N}$, $\displaystyle A \vdash \neg \phi(\bold{S}^a\bold{0})$. Construct a sentence $\displaystyle \sigma$ indirectly asserting that it is not a theorem of $\displaystyle A$, and show that neither $\displaystyle A \vdash \sigma$ nor $\displaystyle A \vdash \neg \sigma$.

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(Suggestion: See Section 3.0.)

Remark: This is a version of the incompleteness theorem that is closer to Gödel's original 1931 argument. Note that there is no requirement that the axioms $\displaystyle A$ be $\displaystyle true$ in $\displaystyle N=(\mathbb{N}; 0, S, <, +, \cdot, E)$. Nor is it required that $\displaystyle A$ include $\displaystyle A_E$; the fixed point argument can still be applied.

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If $\displaystyle \sigma$ says that it is not a theorem of $\displaystyle A$, then it is clear that $\displaystyle A \not\vdash \sigma$. How do I also show $\displaystyle A \not\vdash \neg\sigma$?

Another question is why $\displaystyle \omega$-consistent is considered here?

Thanks.