# Decidability of Gödel sentences.

Letting $$\text{Pr}(x)$$ be a formula that weakly represents the set $$\{x:x\text{ is a Gödel code of a provable sentence in PA}\}$$, where PA means Peano Arithmetic.

Gödel's first incompleteness theorem shows that there is a sentence $$\phi$$ such that $$\text{PA}\vdash (\phi \leftrightarrow \neg \text{Pr}(\ulcorner \phi \urcorner)).$$ The theorem shows that $$\phi$$ is true but not provable in PA.

Now consider the following two sentences $$\alpha$$ and $$\beta$$ such that: $$\text{PA}\vdash(\alpha \leftrightarrow \text{Pr}(\ulcorner \alpha\urcorner)\vee \text{Pr}(\ulcorner\text{Pr}(\ulcorner \alpha\urcorner) \urcorner)) \quad\text{and}\quad \text{PA}\vdash(\beta \leftrightarrow \text{Pr}(\ulcorner \beta\urcorner)\wedge \text{Pr}(\ulcorner\text{Pr}(\ulcorner \beta\urcorner) \urcorner)).$$ Is $$\alpha$$ (resp. $$\beta$$) decidable in PA? Is $$\alpha$$ (resp. $$\beta$$) true?

Generally, positive self-referentiality with the provability relation results in PA theorems (so there's a bit of asymmetry here). Specifically, Lob's theorem says that whenever PA proves $$Pr(\ulcorner\varphi\urcorner)\rightarrow\varphi$$ then PA already proves $$\varphi$$.
This immediately gives that PA proves $$\alpha$$. As for $$\beta$$, PA proves that PA is $$\Sigma_1$$-complete, so in particular PA proves $$Pr(\ulcorner\varphi\urcorner)\rightarrow Pr(\ulcorner Pr(\ulcorner\varphi\urcorner)\urcorner)$$ for each $$\varphi$$; so PA proves $$Pr(\ulcorner\beta\urcorner)\rightarrow\beta$$, and we may again use Lob's theorem to conclude that PA proves $$\beta$$.