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I realize from the answer to this post that the fallacy in my "proof" of "ZF is inconsistent" was that I was not considering that there are models with non-standard integers. However now I think I developed an actual deduction of $T \vdash \text{Cons} T$ for any sufficiently powerful theory $T$ thus implying by Godel's Second Incompleteness Theorem that $T$ is inconsistent.

As before, let $\text{Prb}_T \sigma$ represent $T \vdash \sigma$ and $\text{Cons} T$ be the sentence $\neg \text{Prb}_T (0=1)$. By the fixed-point lemma we have the existence of a sentence $\sigma$ such that:

$$T \vdash (\sigma \leftrightarrow (\text{Prb}_T \sigma \rightarrow \text{Cons} T))$$

By reflection we have:

$(1) \; T \vdash \text{Prb}_T(\sigma \rightarrow (\text{Prb}_T \sigma \rightarrow \text{Cons} T))$

By formalized modus ponens we have:

$(2) \; T \vdash (\text{Prb}_T \sigma \rightarrow \text{Prb}_T(\text{Prb}_T \sigma \rightarrow \text{Cons} T))$

By formalized modus ponens again we have:

$(3) \; T \vdash (\text{Prb}_T \sigma \rightarrow (\text{Prb}_T \text{Prb}_T \sigma \rightarrow \text{Prb}_T \text{Cons} T))$

Now formalized reflection is $T \vdash (\text{Prb}_T \sigma \rightarrow \text{Prb}_T \text{Prb}_T \sigma)$ so from the last step and sentential logic we have:

$(4) \; T \vdash (\text{Prb}_T \sigma \rightarrow \text{Prb}_T \text{Cons} T))$

Now, Godel's Second Incompleteness Theorem formalized is: $T \vdash (\text{Prb}_T \text{Cons} T \rightarrow \neg \text{Cons} T)$. Since anything follows from a contradiction, we have $T \vdash (\neg \text{Cons} T \rightarrow \tau)$. Replacing $\tau$ with $\text{Cons} T$ and following this chain of implications, line (4) implies:

$(5) \; T \vdash (\text{Prb}_T \sigma \rightarrow \text{Cons} T)$

By our choice of $\sigma$ we now have $T \vdash \sigma$ which by reflection yields $T \vdash \text{Prb}_T \sigma$. From (5) therefore we have $T \vdash \text{Cons} T$.

Where am I going wrong here? The only thing I can think of is that I formalized Godel's Second Incompleteness Theorem incorrectly, but then how is it formalized?

Ari
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2 Answers2

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Your mistake is the sentence

Since anything follows from a contradiction, we have $T\vdash(\neg Cons(T)\rightarrow\tau)$ [for any $\tau$].

This is not the case. $T$ does prove that, if $T$ is consistent, then $T$ proves everything; that is, $$T\vdash \neg Cons(T)\rightarrow Pr_T(\tau)$$ for all $\tau$, but this is a far cry from what you claim.

Indeed, think about it this way: in a model $M$ of $T$ in which $T$ appears inconsistent, the sentence $$\mbox{"$0=1$"}$$ will definitely not be true (since $M\models T$ and $T\vdash 0\not=1$. So such a model will satisfy $$\mbox{"$\neg Cons(T)\wedge \neg(0=1)$."}$$ But this means that "$\neg Cons(T)\rightarrow (0=1)$" is not true in every model of $T$! So, by Soundness, $T\not\vdash \neg Cons(T)\rightarrow (0=1)$.


Another take on the same point: There are two senses in which "anything follows from a contradiction": the external version e.g. $$T\vdash 0=1\rightarrow \tau,$$ and the internal version e.g. $$T\vdash Pr_T(0=1)\rightarrow Pr_T(\tau).$$ The former, stronger case applies if the hypothesis is a contradiction, that is, a statement $T$ disproves; the latter applies if the hypothesis is merely the assertion that a contradiction occurs.

The issue with claiming $$T\vdash Pr_T(0=1)\rightarrow \tau,$$ however, is that $Pr_T(0=1)$ is not a contradiction! This is the whole thrust of Godel's incompleteness theorem.

Noah Schweber
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  • Do you mean above that "T does prove that if T is inconsistent then T proves everything" since that is what is reflected in the notation. – Ari Jan 20 '16 at 16:11
  • @Ari Yes. Informally, what's going on is this: T does prove "If I am inconsistent, then I prove everything." But T does not prove "If I am inconsistent, then everything is true", because $T$ does not prove "The things I prove are true". Indeed, $T$ can't even express "the things I prove are true" by Tarski's theorem on the undefinability of truth. (And, by the way, this weakness on the part of $T$ is why we should expect reflection principles to sometimes add strength to a theory; and there's a lot of work in proof theory on expressing theories as reflection plus weaker theories.) – Noah Schweber Jan 20 '16 at 16:13
  • But can't this whole problem be avoided if instead I started with $T \vdash (\sigma \leftrightarrow (Prb_T \sigma \rightarrow Prb_T Cons T))$. Then we would have $T \vdash (Prb_T \sigma \rightarrow Prb_T (Prb_T Cons T)$ Now by Godel's Second Incompleteness Theorem we should have $T \vdash (Prb_T (Prb_T Cons T) \rightarrow Prb_T (\neg Cons T)$ (i.e. "If I have a proof that I have a proof of me being consistent, I can prove that I am inconsistent") By the internal version you cited above we would have $T \vdash (Prb_T(\neg Cons T) \rightarrow Prb_T(\tau))$. – Ari Jan 20 '16 at 17:06
  • Then if we take $\tau$ to be $Cons T$ we can proceed as before. – Ari Jan 20 '16 at 17:07
  • @Ari Your sentence "Now by Godel's Second Incompleteness Theorem we should have $T\vdash (Prb_T(Prb_T(Cons(T))))\rightarrow Prb_T(\neg Cons(T))$" is incorrect. Within $T$, just because $T$ proves that $T$ proves its own consistency does not mean that $T$ proves that $T$ is inconsistent - this is because (again, within $T$) just because $T$ proves that $T$ proves $\tau$ doesn't mean $T$ proves $\tau$. – Noah Schweber Jan 20 '16 at 17:22
  • I agree that in general we do not have $T \vdash (Prb_T Prb_T \tau \rightarrow Prb_T \tau)$ due to the existence of non-standard models. However, how is Godel's Incompleteness Theorem formalized within $T$ if not by $T \vdash (Prb_T Cons T \rightarrow \neg Cons T)$? If this is so then we have by reflection and formalized modus ponens what I had above. Unless, you are saying that we would still not have $T \vdash (Prb_T(Prb_T(0=1)) \rightarrow \neg Cons T)$, which along with the internalized $T \vdash (\neg Cons T \rightarrow Prb_T(\tau))$ would yield with the chain of implications: – Ari Jan 21 '16 at 16:42
  • $T \vdash (Prb_T \sigma \rightarrow Prb_T Cons T)$. But why can't we have that but we can have the internalization of Godel's incompleteness theorem (assuming I got it correct)? – Ari Jan 21 '16 at 16:44
  • @Ari I am indeed saying we do not have $T\vdash (Prb_T(Prb_T(0=1)))\implies \neg ConsT$. To see why we don't have it, try proving it - you'll want the reflection principle "If $T$ proves $\exists x P(x)$, then $T$ proves $P(t)$ for some term $t$" ($\Sigma^0_1$-Soundness) to be provable inside $T$, but it's not. – Noah Schweber Jan 21 '16 at 19:06
  • But is claiming $T \vdash (Prb_T(Prb_T(0=1))) \implies \neg Cons T$ the same as claiming $T \vdash (Prb_T(Prb_T(0=1)) \rightarrow \neg Cons T)$ (which I had) and then using the principle of modus ponens in the chain of implications to get the desired result – Ari Jan 21 '16 at 20:36
  • @Ari Neither "$T\vdash [(Prb_T(Prb_T(0=1)))\rightarrow \neg ConsT]$" nor "$[T\vdash Prb_T(Prb_T(0=1))]\implies \neg ConsT$" is true. So, yes, they're equivalent, but false. – Noah Schweber Jan 21 '16 at 20:41
  • But just so I know, my formulation of Godel's Second Incompleteness Theorem within $T$ was correct? – Ari Jan 21 '16 at 20:43
  • @Ari Yes, that looks right to me. – Noah Schweber Jan 21 '16 at 20:46
  • I'm confused still but I guess I'll just have to settle for the fact that we can have $T \vdash (Prb_T(\neg Prb_T(0=1)) \rightarrow \neg Cons T)$ but we cannot have $T \vdash (Prb_T(Prb(0=1)) \rightarrow \neg Cons T)$ since it goes against the soundness theorem. I just thought that if I'm making an argument for $T \vdash Cons T$ for all sufficiently powerful theory $T$ then all arguments against it have to come from a syntactic level not a semantic one since our mathematical thinking is trapped inside some powerful theory $T$. – Ari Jan 21 '16 at 21:11
  • I guess that's just a philosophical point though that I have to get over that permeates all of meta-mathematics – Ari Jan 21 '16 at 21:12
  • @Ari There's no need to resort to semantics, certainly - we can work on a purely syntactic level. It's just that I find that's usually less illuminating. (And yes, it is very weird that proving "I don't prove $0=1$" is worse than proving "I do prove $0=1$"!) – Noah Schweber Jan 22 '16 at 00:33
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Just to sum up Noah's answer, the lengthy comment exchange which ensued (from which you must have learned a lot), as well as briefly touch on the philosophical point you raised in your final comment, I'll say the following:

Your approach as presented in your question itself is flawed because as Noah pointed out we do not have $T \vdash (\neg Cons T \rightarrow \tau)$, but rather $T \vdash (\neg Cons T \rightarrow Prb_T \tau)$.

In addition, even your approach as outlined in the comments is wrong. While Godel's Second Incompleteness Theorem (i.e. $T \vdash Cons T \implies \text{ "T is inconsistent"}$) and its representation within $T$ (i.e. $T \vdash (Prb_T Cons T \rightarrow \neg Cons T)$) are true, it is not the case that $T \vdash (Prb_T (\neg Cons T) \rightarrow \neg Cons T)$ and therefore we also do not have the external claim of $T \vdash \neg Cons T \implies \text{"T is inconsistent"}$.

This is due to the simple fact as pointed out in this answer to another one of your questions that when $T$ is consistent we have $T \nvdash Cons T$ so we can add $\neg Cons T$ to $T$ without affecting the consistency of the resulting theory. This new theory $R$ can prove $R \vdash Prb_T \tau \rightarrow Prb_R \tau$, so we can get $R \vdash Prb_R(0=1)$.

As to the philosophical issue about your tension over the use of semantic vs. syntactic proofs (which should be dispelled by the Soundness and Completeness Theorems) as well as your anxiety over all mathematical reasoning "being trapped inside a powerful theory $T$" (which by Godel's Second Incompleteness Theorem would provide itself with proofs that it is thinking consistently precisely when it is not) it occurred to me that you may want to try to think about it this way:

Mathematical reasoning, like all human thought, is based upon certain assumptions and value judgements (e.g. "Humans ought to be able to think deductively", "Mathematicians should avoid circular reasoning and contradictions", etc.) so it is more closely related to modal logic rather than first-order logic. Thus, mathematicians first make the assumption that they are capable of thinking deductively and then move on from there to prove theorems (e.g. the Soundness Theorem, the Completeness Theorem, etc.) The amazing ability of mathematics to help us comprehend the physical world in which we live as well as its deep aesthetic value (at least for those not afflicted with "mathphobia") testify to its worthiness of being pursued. As such, there is no reason to have an "existential crisis" over the possibility of a syntactic proof of $T \vdash Cons T$ for a powerful theory $T$ since this can be ruled out by the fact that we have models for these theories. (although you would undoubtedly counter that by completeness we have $\text{(Mathematical Reasoning)} \vdash \text{(Soundness Theorem)}$ so if there is somehow "a deep flaw" in mathematical reasoning itself it is only a matter of time until we develop a proof of $\neg \text{(Soundness Theorem)}$, ...). Life's simply too short to worry about such things.

Ari
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  • Wow, a self-referential answer to a question involving self-reference. I love it. (It's almost as good as a self-referential comment). If it's any solace, you can see now from what you just learned that even powerful consistent $T$ are "insecure" about their ability to prove theorems in the following sense: If $T\vdash \neg Cons T$, then we have $T \vdash Prb_T \tau$ as before. Replace $\tau$ with $\neg Prb_T(0\neq 1)$ and we get that $T \vdash Prb_T (\neg Pbr_T(0\neq 1))$ (i.e. "$T$ proves that it can prove that it cannot prove that which it can, namely $( 0\neq 1)$"). – Ari Jan 25 '16 at 16:52