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Let's say a nonconstructive proof was given in ZFC that ZFC was inconsistent.

Note that this doesn't automatically make ZFC inconsistent. Given a consistent theory $X$, $X + \neg \text{Con}(X)$ is consistent given Godel's second incompleteness theorem. But $X + \neg \text{Con}(X)$ proves there is a contradiction in $X$, and therefore $X + \neg \text{Con}(X)$. So we have consistent and possibly very usable theory that proves its own inconsistency!

Yet, ZFC proving its own inconsistency would still surely have interesting consequences! In particular:

  • What other systems would ZFC prove inconsistent.
  • What systems would be inconsistent.
    • ZFC + Con(ZFC) would be inconsistent. I'm not sure what the consequences of this would be.
  • What other consequences would there be? Would ZFC still be usable, or would too many problematic consequences pop up?

Note: The non-constructive part isn't important. I would like to use the assumption that ZFC is consistient though, unless ZFC proving ZFC inconsistient somehow leads to ZFC being inconsistient.

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closed as unclear what you're asking by Andrés E. Caicedo, YoTengoUnLCD, Watson, user91500, Math1000 Jun 28 '16 at 13:09

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Constructiveness is just a red herring here, isn't it? $\endgroup$ – Steven Stadnicki Jun 28 '16 at 1:33
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    $\begingroup$ What precisely do you mean by "nonconstructive proof"? $\endgroup$ – Andrés E. Caicedo Jun 28 '16 at 1:43
  • $\begingroup$ @StevenStadnicki basically, the proof doesn't give any specific contradiction. If it did that would show ZFC is inconsistent. $\endgroup$ – PyRulez Jun 28 '16 at 2:07
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    $\begingroup$ That doesn't seem to make sense. Could you clarify a bit more? $\endgroup$ – Andrés E. Caicedo Jun 28 '16 at 3:10
  • $\begingroup$ @AndrésE.Caicedo A constructive proof would mean exhibiting a specific contradiction, meaning ZFC is inconsistent. A nonconstructive proof could be something like using the axiom of choice or using a proof by contradiction: it shows that ZFC proves that ZFC is inconsistent, but won't necessarily entail a contradiction in ZFC externally. $\endgroup$ – PyRulez Jun 28 '16 at 3:17
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Well, it is a little unclear what you mean by nonconstructive proof still. But you can think of what models of $ZFC+\neg{Con(ZFC)}$ looks like when you consider $ZFC+\neg{Con(ZFC)}$ as a formal theory, which seems to be part of the question you are asking.

First of all a model of the sort of theory you are looking at would have non-standard natural numbers else the inconsistency would be witnessed making $ZFC$ itself inconsistent. Also to make matters even more confusing, there maybe models of what it believes to be $ZFC$ in there! (This is weird nesting sort of argument because we can develop set theory inside of set theory inside of set theory inside of set theory........)

Again, note that what I'm talking about is a little tricky. I'm assuming that you are working within set theory and formalize, a separate copy if you will, of ZFC in there. Now you can do the proof of the incompleteness theorem, which gives you that (under the assumption that $ZFC$ is consistent) $ZFC+\neg{Con(ZFC)}$ is consistent. So you can look at those models of that theory.

Also it should be noted: considered as a formal theory, the incompleteness theorem applies to $ZFC+\neg{Con(ZFC)}$ too. So as a result we have that $ZFC+\neg{Con(ZFC)}+{Con(ZFC+\neg{Con(ZFC)})}$ and $ZFC+\neg{Con(ZFC)}+\neg{Con(ZFC+\neg{Con(ZFC)})}$. In fact this idea can be used to show that $ZFC+\neg{Con(ZFC)}$ if consistent has continuum many different completions to complete theories. So even at a countable level there are a maximum number of non-isomorphic models. As can be seen already, it will be a fairly complicated thing to study, as what exactly the model thinks about certain things will be dependent on what exactly it believes about statements like ${Con(ZFC+\neg{Con(ZFC)})}$ etc.

If you are interested; the study of non-standard Peano Arithmetic is well developed. As far as I know their interest is different from what you are asking about and I don't know how much of the ideas would carry over, but it seems as good a starting point as any.

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Take any consistent first-order theory $T$ that is nice enough (has decidable proof validity). Let $T' = T + \neg Con(T)$. Then $T'$ is consistent and $T' \vdash \neg Con(T)$ and hence $T' \vdash \neg Con(T')$. So $T'$ would prove 'its own inconsistency', but interpreting this correctly is slightly subtle (see this post). $\def\prov{\square}$

Now it is better to use provability logic as described in this post. Then what we have here is that $T \vdash \prov \bot$. An immediate consequence is that $T \vdash \prov φ$ for any sentence $φ$. So if by non-constructive proof of inconsistency you mean that $T$ proves the existence of a proof of "$\bot$" over $T$, then this also implies a non-constructive proof that every sentence is true, which is of course ridiculous and hence bad enough for us to reject $T$ as being a useful foundational system, even though $T$ is 'not as bad' as an outright inconsistent system.

The crucial point is that as long as $T$ is $Σ_1$-sound ($T$ interprets $PA$ via a translation $i$, and for every $Σ_1$-sentence $φ$ over $PA$, if $T \vdash i(φ)$ then $\mathbb{N} \vDash φ$), then $T \vdash \prov φ$ implies $T \vdash φ$ for any sentence $φ$ over $T$, and so consistency of $T$ implies that $T$ cannot prove "$\prov \bot$". So the finer answer would be that if you believe that $ZFC$ is $Σ_1$-sound, then there cannot be a non-constructive proof that $ZFC$ is inconsistent (in the above sense).

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