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Let our first-order signature be $(N;+,\cdot,<,0,1)$. Let $S$ be the set of true first-order sentences of that sentences that have no variables, not even bound to a quantifier. For example, $((1>1 \vee 1<(1+1)) \land 1=1)$ belongs to $S$. Let $G(S)$ be the set of godel numbers of $S$, for an arbitrary Gödel numbering $G$. Is $G(S)$ definable in Peano Arithmetic (PA)?

leftaroundabout
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user107952
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1 Answers1

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$G(S)$ is computable, since comparison of terms in the language of arithmetic is computable, and so $G(S)$ is a fortiori arithmetically definable. In particular, it is $\Delta^0_1$.

However, there is an important point re: language to make here. It's not really proper to talk about a theory defining something. Definability takes place in structures/languages; theories prove things. Hence my use of the term "arithmetically definable" above. The connection with a specific theory comes in when we ask what is needed to prove basic facts about (some fixed definition of) the set in question. Here, for instance, the equivalence between computability and representability over (say) Robinson arithmetic tells us that $\mathsf{PA}$ is massive overkill for answering individual membership questions about $G(S)$.

(Re: the italicized parenthetical, using a pathological definition of $G(S)$ could change the picture. For example, letting $\gamma$ be some fixed "reasonable" definition of $G(S)$, consider the formula $\theta(x)\equiv$ "$\gamma(x)\leftrightarrow Con(\mathsf{PA})$." This $\theta$ also defines $G(S)$, but $\mathsf{PA}$ can't prove the equivalence between $\theta$ and $\gamma$; from $\mathsf{PA}$'s perspective, it's also possible that $\theta$ is equivalent to $\neg\gamma$ instead! See also this old answer of mine for a related issue regarding pathological expressions, this time in the context of Lob's theorem.)

Noah Schweber
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