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

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.

• $\theta$ is not true, but $\lnot\theta$ is not provable. Nov 28, 2019 at 22:19
• @AndrésE.Caicedo Thank you for your response! I am trying to understand why this is the case, any suggestions on how to see it? Nov 28, 2019 at 22:26
• $N$ is sound and $\mathfrak N\models N$. Argue from there. Nov 28, 2019 at 22:28

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 $$N\vdash\psi$$.