# Can you remove all the extra arithmetic from ZFC (or other theories)?

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.

• 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? – Steven Stadnicki Sep 5 '17 at 23:24
• 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 $\mathsf{PA}$. – Andrés E. Caicedo Sep 6 '17 at 0:39
• @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. – PyRulez Sep 6 '17 at 3:25
• @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. – Sridhar Ramesh Sep 6 '17 at 4:59
• 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). – Sridhar Ramesh Sep 6 '17 at 5:24

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 $ZFC$.