Is $\theta$ a true sentence if the following holds in Robinson Arithmetic

Question

I already asked this but too many typos caused confusion and the editing was confusing me so I re-wrote it here:

If $\theta$ is a sentence in the language of number theory, and we know the following, with the corner braces being the Gödel number: $$N \vdash \theta \leftrightarrow Thm_N(\overline{\ulcorner \neg \theta \urcorner})$$ Note that $N$ is the acioms of Robinsons arithmetic and we have defined $Thm_N(f) \equiv (\exists c)Deduction(c,\,f)$ which basically says $f$ is a Gödel-number of a formula such that $c$ is a Gödel-number for a deduction of $f$.

The question is: is $\theta$ a true sentence (in $\mathfrak{N}$)?

My immediate thought is no, but Im worried I'm interpreting this wrong. Is $Thm_N(\overline{\ulcorner \neg \theta \urcorner})$ saying that there exists a deduction of $\neg \theta$? If so does that mean that $\theta$ must be false? I'm having trouble understanding.

Answer 1

Below I conflate sentences with their Godel numbers' numerals, and assume the soundness of $N$.

Pretty clearly $\theta$ can't be $N$-decidable, the key point being the very nice property of $\Sigma_1$-completeness - or rather, its particular instance that, for any sentence $\psi$, $N\vdash\psi$ implies $N\vdash Thm_N(\psi)$:

• If $N$ proved $\theta$, by $\Sigma_1$-completeness $N$ would also prove $Thm_N(\theta)$. By the assumption on $\theta$ this means that $N$ would prove $Thm_N(\theta)\wedge Thm_N(\neg\theta)$ - that is, $N$ would prove that $N$ is inconsistent. This contradicts the soundness of $N$.

• If $N$ proved $\neg\theta$, by $\Sigma_1$-completeness $N$ would also prove $Thm_N(\neg\theta)$, and so by assumption on $\theta$ $N$ would in fact prove $\theta$. This contradicts the consistency of $N$.

And this addresses the truth value of $\theta$ (in the standard model $\mathfrak{N}$) as well: the second bulletpoint above shows that $\theta$ is in fact false in $\mathfrak{N}$ since for all $\psi$ we have $\mathfrak{N}\models Thm_N(\psi)$ iff $N\vdash\psi$.