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Hello.

Let $\operatorname{sat} X$ denote the satisfiability of a theory $X$.

From Gödel's second incompleteness theorem and his completeness theorem follows $$ZF \not\vdash \lceil \operatorname{sat} ZF \rceil$$ Assuming $\operatorname{sat} ZF$ in the metatheory, we can also say that $$ZF \not\vdash \lnot \lceil \operatorname{sat} ZF \rceil$$ that is, $\operatorname{sat} ZF$ is independent of $ZF$.

Most texts I have read about large cardinals assume the consistency of $ZF$ (and stronger systems created on top of $ZF$). I was wondering whether there is any research done about the system $ZF\cup \{ \lnot \lceil \operatorname{sat} ZF \rceil \}$, let us call it $ZF\bot$ in the following. Trivially $$ \operatorname{sat} ZF \Leftrightarrow \operatorname{sat} ZF\bot $$

I guess that a model $\mathbb{ZF}$ of $ZF\bot$ has another concept of "finite" sets, so i guess $$ \not\exists_{P(x)\in \operatorname{For}} \forall_{X\in \mathbb{ZF}} . P^{\mathbb{ZF}}(X) \Leftrightarrow |X|<\omega$$ because $\mathbb{ZF}$ must consider some infinite proof trees of $ZF\vdash\bot$ finite - but I have no formal argument for this. This paper seems to be somewhat related, as it deals with the length of inconsistency proofs.

However, except for this, I have not found anything related, but it would be interesting to know.

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  • $\begingroup$ I suggest looking at the answers to the following related question (in the context of $PA$). mathoverflow.net/questions/65837/… $\endgroup$
    – Ali Enayat
    Commented Mar 15, 2013 at 4:30
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    $\begingroup$ You need to assume sat ZF also for the first claim, since indeed, the nonprovability of any formula at all already implies consistency. For Goedel's proof, what you need for the second claim is the $\omega$-consistency of the theory, a stronger assumption than mere consistency. Rosser's formulation omits the need for this extra assumption. $\endgroup$
    – Joel David Hamkins
    Commented Mar 15, 2013 at 10:17

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A model of ZF$\perp$ must contain non-standard natural numbers (for instance, the Gödel number of the proof of a contradiction in ZF) and this implies that it cannot be well-founded (starting from a non-standard number, you can define a $\in$-decreasing sequence). Hence, if we are assuming the foundation axiom as a part of ZF, your model $\mathbb Z\mathbb F$ must be a pair $(M,R)$, where the relation $R\subset M\times M$ cannot be the true membership relation, since this is well-founded.

Hence, I assume that when you write $|X|$, you are talking in fact about the cardinality of the extension of $X$, i.e. $|\{x\in M\mid x\,R\,X\}|$. The true cardinality of $X$ as a set would have no direct relation with the model.

In this setting, such a formula $P(X)$ cannot exist. If so, you could define in $(M,R)$ the set $S=\{n\in\omega\mid P(n)\}$, which would be the set of all standard natural numbers. This would be an inductive set ($0\in S\land \forall n\in \omega (n\in S\rightarrow n+1\in S))$, but $S\neq \omega$. Hence the induction principle would fail in the model.

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    $\begingroup$ Note, though, that the converse fails: nonstandard models of set theory (or arithmetic; set theory isn't needed) don't necessarily 'think' that ZFC/arithmetic is inconsistent, and there are interesting results about models of arithmetic that disagree with the standard model about certain things, but believe that arithmetic is consistent. $\endgroup$
    – Noah Schweber
    Commented Jan 9, 2014 at 2:54
  • $\begingroup$ There are non-well-founded models of ZFC. Consider ZFC + "c is larger than 1" + "c is larger that 2" + ..., for some constant c, then this system implies ZFC and, by the compactness theorem, must be consistent. But c is not well-founded anymore, at least extrinsically. Intrinsically, it is, since there is no infinite descending chain inside this model. $\endgroup$
    – Christoph-Simon Senjak
    Commented Jan 9, 2014 at 21:07
  • $\begingroup$ Of course. The answer to you question is negative in a very general setting: in any model of set theory / arithmetic containing non-standard natural numbers, the sets that are externally finite cannot be characterized by a formula, since otherwise the induction principle would be violated inside the model. $\endgroup$
    – Carlos
    Commented Jan 9, 2014 at 23:08
  • $\begingroup$ One interesting observation: induction is the only reason that nonstandard models can't define a cut (=proper inductive initial segment of the model's natural numbers). As an example of what I mean, starting with a weak theory of arithmetic assuming only $\Sigma^0_n$-induction (so, $P^-+I\Sigma_n$), it is consistent that there is a $\Sigma^0_{n+1}$-definable cut. One thing I don't know: are there models of $P^-+I\Sigma_n$ in which the true natural numbers form a $\Sigma^0_{n+1}$-definable cut? $\endgroup$
    – Noah Schweber
    Commented Mar 10, 2014 at 3:43

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