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For the purposes of this question, $ZF$ $-$ Inf will denote the fragment $ZF$ with the axiom of infinity dropped and $ZF$ $-$ Inf + $\lnot$Inf will denote $ZF$ $-$ the axiom of infinity + the negation of the axiom of infinity. Inf, of course, is just ($\exists$$y$)($\emptyset$$\in$$y$ $\land$ ($\forall$$x$$\in $$y$)(($x$$\cup${x} $\in$$y$)), where $x$$\cup${$x$} is just the successor of $x$.

It is known that both $ZF$ $-$ Inf and $ZF$ $-$Inf + $\lnot$Inf are equiconsistent with $PA$ and also that $PRA$ + $TI({\epsilon_0})$ proves the consistency of both as well. It is also known that the universe of discourse for both $ZF$ $-$ Inf and $ZF$ $-$ Inf + $\lnot$Inf is $V_{\omega}$ (which would seem to suggest that, relative to $PA$, $\omega$ should be deemed a proper class).

Question: Do $ZF$ $-$ Inf and $ZF$ $-$ Inf + $\lnot$Inf prove the same theorems? Or does $ZF$ $-$ Inf + $\lnot$Inf prove theorems that $ZF$ $-$Inf can't?

More importantly, if $PRA$ + $TI({\epsilon_0})$ can prove the consistency of $ZF$ $-$ Inf, is there a way to consistently add Inf as a new axiom to $ZF$ $-$ Inf (say, by showing that Inf is independent of $ZF$ $-$ Inf--has this been done already)? If not, where do the problems with this 'bootstrapping' approach to the proof of the consistency of $ZF$ lie?

Thomas Benjamin
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    I might be misunderstanding, but: if full ZF is consistent, $\mathrm{ZF - Inf}$ does not prove $\neg\mathrm{Inf}$, since full $\mathrm{ZF}$ has a model. – Mees de Vries Jan 03 '17 at 11:21
  • @MeesdeVries: It doesn't have to--all I want to know is if $ZF$ $-$ Inf + $\lnot$ Inf proves theorems that $ZF$ $-$ Inf can't, and if there is a consistent way to add Inf to $ZF$ $-$ Inf. – Thomas Benjamin Jan 03 '17 at 11:49
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    @ThomasBenjamin: It certainly does; one such theorem is $\neg\rm Inf$ itself. – hmakholm left over Monica Jan 03 '17 at 12:50
  • We cannot show that con (ZF-Inf)\implies con(ZF) in any way that can be encoded in ZF-Inf as a sentence about Godel numbers in a Godel numbering system. You cannot refute "$\neg$ Inf is a theorem of ZF-Inf" without making extra assumptions that make your argument circular. This is a consequence of (1) Godel's incompleteness theorems,and (2) ZF proves that ZF-Inf+$\neg$ Inf has a set-model. – DanielWainfleet Jan 07 '17 at 05:46

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I think there is some confusion here.

ZF-Inf+$\neg$ Inf certainly does prove some theorems that ZF-Inf alone doesn't - namely, $\neg$ Inf! Keep in mind that every model of ZF is also a fortiriori a model of ZF-Inf.

Now, you write:

The universe of discourse for both [theories] is $V_\omega$.

I don't really know what you mean here. It is true that $V_\omega$ is the minimal model of each theory - however, that doesn't make it the universe of discourse, or even the intended model, of either.

I suspect you are asking about whether ZF-Inf and ZF-Inf+$\neg$Inf prove different things about arithmetic. The answer to this is no:

For any model $M$ of ZF-Inf, the susbtructure $(V_\omega)^M$ is a model of ZF-Inf+$\neg$Inf.

This is a straightforward verification.

You then ask:

More importantly, if $PRA$ + $TI({\epsilon_0})$ can prove the consistency of $ZF$ $-$ Inf, is there a way to consistently add Inf as a new axiom to $ZF$ $-$ Inf (say, by showing that Inf is independent of $ZF$ $-$ Inf--has this been done already)? If not, where do the problems with this 'bootstrapping' approach to the proof of the consistency of $ZF$ lie?

Well, "Inf is independent of ZF-Inf" is just a complicated way of saying "ZF is consistent"; and this is definitely provable from the right axioms e.g. ZF+large cardinals. I presume you're asking about whether $PRA$ + $TI({\epsilon_0})$ can prove this. In that case, the answer is of course "no": e.g. ZF proves that $PRA$ + $TI({\epsilon_0})$ is consistent! Indeed the consistency of ZF is galactically beyond the abilities of $PRA$ + $TI({\epsilon_0})$, so if $PRA$ + $TI({\epsilon_0})$ proved "ZF is consistent" then $PRA$ + $TI({\epsilon_0})$ would be inconsistent, by Goedel's Second.

(In detail: the key point is that $PRA$ + $TI({\epsilon_0})$ proves that ZF proves every true $\Sigma_1$ sentence, so in particular $PRA$ + $TI({\epsilon_0})$ proves that if $PRA$ + $TI({\epsilon_0})$ is inconsistent, then ZF proves that $PRA$ + $TI({\epsilon_0})$ is inconsistent; and moreover, since $PRA$ + $TI({\epsilon_0})$ proves every true $\Sigma_1$ sentence, $PRA$ + $TI({\epsilon_0})$ proves that ZF proves that $PRA$ + $TI({\epsilon_0})$ is consistent (since ZF does in fact prove this). So combining these two points, $PRA$ + $TI({\epsilon_0})$ proves "If ZF is consistent, then $PRA$ + $TI({\epsilon_0})$ is consistent" - so $PRA$ + $TI({\epsilon_0})$ better not prove that ZF is consistent! See this question for more analysis along these lines.)

I get the sense that you're trying to extend the Ackermann interpretation to get hold of more powerful theories than ZF-Inf+$\neg$Inf. As I explained in my answer to your other question, however, there is a serious obstacle to this: since the Ackermann interpretation satisfies $$(*)\quad\mbox{If $m$ codes a set containing $n$, then $m>n$},$$ the Ackermann structure associated to a model of PA will always satisfy $\neg$Inf. In particular, if you want a method for coding some models of ZF into models of PA which goes beyond just models of ZF-Inf+$\neg$Inf, then your coding method will need to be vastly more complicated.

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Noah Schweber
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  • It would be nice to have some references regarding such more complicated non-monotonic codings you refer to--surely there must be some survey articles regarding these.... By the way, the above answer was very helpful. Thanks. – Thomas Benjamin Jan 04 '17 at 05:31
  • @ThomasBenjamin I don't know of any off the top of my head. They would certainly have to be extremely complicated - for example, since PA doesn't prove Con(ZF), it would have to be the case that you could not prove that the interpretation worked in PA alone. – Noah Schweber Jan 04 '17 at 06:11