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Let $\mathbb{N}$ be the standard model of the natural numbers. For any statement in the language of arithmetic, we can translate into a statement in the language of set theory by asking if it is true of $\mathbb{N}$.

Let's say that a statement in arithmetic is "extraneous" if it is independent of PA. For example, ZFC proves Con(PA), which is extraneous.

My question is, is there a set of statements $S$ (in the language of set theory), such that $S$ proves no extraneous statements, and $S+PA=ZFC$ (or perhaps $S+PA \vdash ZFC$).

Edit: We can also consider the same question, but with PA replaced with the set of arithmetical statements provable in ZFC.

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    $\begingroup$ I think this question is a bit muddled in its language. What does 'independent of PA' mean in a non-arithmetic context? For instance, the statement 'For any two sets $x,y$ there exists the set $\{x,y\}$ is 'independent of $PA$'. Is that not innately an extraneous statement? $\endgroup$
    – Steven Stadnicki
    Commented Sep 5, 2017 at 23:24
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    $\begingroup$ What do you mean by $S+PA=ZFC$? $\mathsf{PA} $ and $\mathsf{ZFC}$ are stated in different languages. Or do you mean the translation of $\mathsf{PA}$ to the language of set theory rather than $\mathsf{PA}$ itself? What precisely is that translation? A straightforward version replaces each $\phi $ with the claim that $\mathbb N $ satisfies $\phi $, but then the translation of $\mathsf{PA}$ witnesses the consistency of $\mathsf{PA}$. $\endgroup$
    – Andrés E. Caicedo
    Commented Sep 6, 2017 at 0:39
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    $\begingroup$ @JoelDavidHamkins I didn't mean for it to have a negative connotation. I just wanted to pick some statements of arithmetic, and PA was the first thing I thought of. The set of tautologies or set of arithmetical theorems of ZFC would also be interesting. $\endgroup$
    – Christopher King
    Commented Sep 6, 2017 at 3:25
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    $\begingroup$ @Andres: We can translate each claim about natural numbers into a claim about finite ordinals, and thus turn sentences in the language of arithmetic into sentences in the language of set theory, to turn PA into a theory in the language of set theory which is strictly weaker than ZFC and in particular which does not itself prove Con(PA) in any standard sense. $\endgroup$
    – Sridhar Ramesh
    Commented Sep 6, 2017 at 4:59
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    $\begingroup$ We could carry out this translation even in the context of a set theory with no axiom of infinity, which would fail to prove Con(PA). $\endgroup$
    – Sridhar Ramesh
    Commented Sep 6, 2017 at 5:24

2 Answers 2

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If $S$ proves no "extraneous" statements, then $S$ cannot prove $X \rightarrow Con(PA)$ for any arithmetic statement $X$ which $PA$ proves. It follows that $S + PA$ cannot prove $Con(PA)$, and therefore $S + PA$ cannot entail all of $ZFC$.

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    $\begingroup$ I would suggest that perhaps more in the spirit of what I would've thought this question would be would be to search for some S and some T such that S proves no extraneous statements, T consists only of arithmetic statements, and S + T = ZFC. Without loss of generality, we may even take T to consist of all arithmetic consequences of ZFC. I suspect no suitable S can be found, but do not know a proof at the moment. $\endgroup$
    – Sridhar Ramesh
    Commented Sep 6, 2017 at 0:26
  • $\begingroup$ actually yeah, I like that question better. $\endgroup$
    – Christopher King
    Commented Sep 6, 2017 at 1:28
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I think you are asking for ZFC without the Axiom of Infinity, leaving the heriditarily finite sets. This proves the same arithmetic theorems as PA, if I understand it right.

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