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The incompleteness theorem states that one cannot prove whether ZF or ZFC is consistent, but what about ZFC withouth Axiom of infinity? (Assuming the empty set exists)

Furthermore, let $M$ be a consistent model not invoking infinity and $A,B$ be statements invoking infinity such that $A$ contradicts $B$. Then, let's assume both $M+A$ and $M+B$ are consistent. If statements $\phi_A$ and $\phi_B$ invoking infinity are provable in $M+A$ and $M+B$ relatively, then are finite pieces of $\phi_A$ and $\phi_B$ both provable in $M$?

Asaf Karagila
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Jj-
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    The second paragraph doesn't make any sense. – Asaf Karagila Apr 04 '13 at 23:05
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    Did you perhaps mean that $M$ is a theory? What does "invoking" mean in this context? – Asaf Karagila Apr 04 '13 at 23:08
  • @Asad I meant "a statement which gurantees existence of any infinite object". And yes i know the incompleteness theorem states any axiomatic system in which arithmetic of natural number is well-defined cannot prove its own consistency. – Jj- Apr 04 '13 at 23:13
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    Removing an axiom cannot make a consistent theory inconsistent. – user76284 Feb 28 '20 at 05:08

3 Answers3

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The corollary from the incompleteness theorems is that you cannot prove the consistency of $\sf ZFC$ from $\sf ZFC$ itself. You have to have a stronger theory.

For example, in $\sf ZFC+\text{There exists an inaccessible cardinal}$, you can in fact prove the consistency of $\sf ZFC$ because this is a stronger theory.

Similarly this is the case of $\sf ZF_{fin}$ ($\sf ZF$ without infinity). The theory itself cannot prove its own consistency. However $\sf ZFC$ is a strictly stronger theory, and it proves the consistency of $\sf ZF_{fin}$. It does so by exhibiting a set which is a model of the theory, $V_\omega$ - the set of the hereditarily finite sets.

Large cardinal axioms are often called "strong infinity axioms" because they mimic the axiom of infinity, in the sense that they make a stronger theory by describing that a certain set of ordinals exists.

Asaf Karagila
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    I had this strong suspicion, just a hunch, mind you, that you'd be drawn like a magnet to answering this question, go wonder? ;-) +1 – amWhy Apr 04 '13 at 23:12
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    I just happened to come across it on my way to bed (which include two more cruises down the front page, and a third from my iPhone before shutting my eyes)... ;-) – Asaf Karagila Apr 04 '13 at 23:18
  • Sleep tight! ;-) – amWhy Apr 04 '13 at 23:20
  • I hardly think so, but thanks anyway. – Asaf Karagila Apr 04 '13 at 23:22
  • @Asaf Thank you, but since i wasn't specific, i think your answer does not answer my second question. (I don't know whether this is an appropriate example) For example, say $\phi_A$ is provable in ZFC and $\phi_B$ is provable in ZF+AD, then are these two statement both provable for finite objects in ZF? I'm curious to know how infinity affects finite universe – Jj- Apr 04 '13 at 23:30
  • Sorry, I'm unfamiliar with this sort of thing. I thought we couldn't demonstrate the consistency of ZFC at all. Is there a resource I can look at that demonstrates the consistency of ZFC through your 'stronger theory' approach? – Josh Keneda Apr 04 '13 at 23:35
  • @Josh: If there exists an inaccessible cardinal $\kappa$ then the set of those with rank $<\kappa$ is a model of $\sf ZFC$, which prove its consistency (through the completeness theorem). See this for the details on the fact this is a model. But to drive the point home, the incompleteness theorem tells us that we cannot prove the consistency of $\sf PA$ from $\sf PA$; but we can prove the consistency of $\sf PA$ from $\sf ZFC$. Why? Because $\sf ZFC$ is a stronger theory. – Asaf Karagila Apr 04 '13 at 23:39
  • @Josh: If you want any references, probably any modern set theory book would suffice. Kanamori The Higher Infinite might be a good start for this (in particular for large cardinal axioms). You may also want to read a bit in Cantor's Attic about the assumption of consistency (see Con ZFC). – Asaf Karagila Apr 04 '13 at 23:41
  • @Katlus: Note that both $\sf ZFC$ and $\sf ZF+AD$ include the axiom of infinity. But what may interest you is the fact that if we fix a universe of $\sf ZF$, then every model (set or class alike) of $\sf ZF_{fin}$ (including those of the full theory) will have a copy of the "real" $V_\omega$. So in some sense the truly finite universe is sort of absolute (when we fix one particular meta-universe, of course). – Asaf Karagila Apr 04 '13 at 23:44
  • @Josh: Let me add one last point, and that the whole topic of relative consistency and the consistency of $\sf ZFC$ is really full of delicate and fine points. It took me a year into my masters before someone pointed out that adding large cardinal axioms is strengthening the theory, and can therefore prove the consistency of $\sf ZFC$. If you're not used to wading through logic related arguments; or at least familiar with the various caveats of provability vs. truth and so on - it can get very confusing and full of false beliefs. – Asaf Karagila Apr 04 '13 at 23:48
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ZF without the Axiom of Infinity is still strong enough to derive Peano's axioms PA. So we cannot prove its consistency without proving the consistency of PA.

Andrés E. Caicedo
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Usally "ZFC without Axiom of Infinity" is considered consistent, since $\sf ZFC \setminus AOI + \neg AOI$ has the natural numbers as a model via Ackermann Coding. If $\sf ZFC \setminus AOI + \neg AOI$ has a model, so does $\sf ZFC \setminus AOI$. This is a model like the Hereditary Finite Sets:

Hereditarily finite set
https://en.wikipedia.org/wiki/Hereditarily_finite_set

The Ackermann Coding was shown here:

Ackermann, Wilhelm (1937), "Die Widerspruchsfreiheit
der allgemeinen Mengenlehre", Mathematische Annalen,
114 (1): 305–315

On the other hand, to formally prove it, "ZFC with Axiom of Infinity, the usual ZFC" should do, it would show that $V_\omega$ is a model of $\sf ZFC \setminus AOI + \neg AOI$. So you would need to first accept $\sf ZFC$.

Remark: For the rest of your question concerning $M+A$, I have no answer. How do you combine a model $M$ and a formula $A$?