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I was reading @Noah Schweber's response to the following post and it got me wondering how do we actually prove $T \vdash \neg Consistent(T) \rightarrow Prv_T(0 = 1)$ where $T$ is as described in the asker's post (that is, a consistent axiomatizable theory extending the theory of $A_E$ (slide 5 here--using the asker's link from the linked post)?

Rex
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    Where does Noah say $T\vdash \lnot Con(T)\to 0=1$? – spaceisdarkgreen Jun 13 '18 at 03:03
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    We have $T\vdash 0\ne 1,$ so if we had $T\vdash 0 \ne 1 \to Con(T)$ we would have $T\vdash Con(T).$ I think you must mean $T\vdash \lnot Con(T)\to Prov_T(\ulcorner 0=1 \urcorner).$ – spaceisdarkgreen Jun 13 '18 at 03:07
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    @spaceisdarkgreen You and Noah are both right and it is my mistake -- I will edit the question to say how does $T$ prove $\neg Consistent(T) \rightarrow Prov_T(\ulcorner \alpha \urcorner)$ -- one can substitute $\alpha$ with anything – Rex Jun 13 '18 at 03:31
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    @Rex If someone said "${\rm Prov}_T(p)$ is a $\Sigma_1$ formula, would that make sense to you? – DanielV Jun 13 '18 at 03:32
  • @DanielV sure in terms of hierarchy -- why is that relevant? – Rex Jun 13 '18 at 03:51
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    @Rex Because your question is a special case of the axiom $\square A \to \square (A \to B) \to \square B$, which generally uses the witnesses for the first two $\Sigma$ to prove the final $\Sigma$. – DanielV Jun 13 '18 at 03:58
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    Not applicable to your question, but a lovely constructive proof of ex falso is possible in the world of Heyting arithmetic if we define $\bot := 0 = 1$. Then an inhabitant of this type can be used to prove by induction that all integers are equal, and hence that all arithmetic statements are true. – Patrick Stevens Jun 13 '18 at 06:14

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As I commented, if $T$ is consistent (and is an axiomatizable extension of Robinson arithmetic or somesuch) then $T\not\vdash \lnot \operatorname{Con}(T)\to 0=1.$ What I think you are possibly asking about (edit: as I came back to finish writing, I see now you have confirmed this) is the phrase "T proves inconsistent theories prove anything" which would be more appropriately phrased $$ T\vdash\lnot\operatorname{Con}(T) \to \operatorname{Prov}_T(\ulcorner 0=1\urcorner).$$ Well, that's usually how $\operatorname{Con}(T)$ is defined... so I guess the nontrivial thing to prove would be $$T\vdash\lnot\operatorname{Con}(T) \to \operatorname{Prov}_T(\ulcorner \phi\urcorner) $$ for any sentence $\phi.$ Well, we can write this, per our definition of $\operatorname{Con}(T)$, as $$T\vdash\operatorname{Prov}_T(\ulcorner 0=1 \urcorner) \to \operatorname{Prov}_T(\ulcorner \phi\urcorner).$$

One property of $\operatorname{Prov}_T$ that we need is that if $T\vdash \psi$ then $T\vdash\operatorname{Prov}_T(\ulcorner\psi\urcorner)$. Furthermore, since $T\vdash 0\ne 1,$ we have $T\vdash (0=1) \to \phi,$ so we are done if we can show $$ T \vdash \psi \implies T\vdash\operatorname{Prov}_T(\ulcorner\psi\urcorner)\\T\vdash \operatorname{Prov}_T(\ulcorner A\to B\urcorner) \implies T\vdash \operatorname{Prov}_T(\ulcorner A \urcorner) \to \operatorname{Prov}_T(\ulcorner B\urcorner).$$

Although these seem plausible, it's tedious to actually show explicitly since we'd need to get into the ugly nuts and bolts of $\operatorname{Prov}_T$. The gist is that proving $\operatorname{Prov}_T(n)$ involves finding a number $m$ that represents a proof of the sentence $n$ represents. If $T\vdash \psi,$ then we can formalize that proof and turn it into a number. Then to prove $\operatorname{Prov}_T(\psi)$ in $T,$ we check that $m$ is actually a proof of $n.$ Checking a proof is a mechanical process, which is formalizable in $T.$ As for the second notion, it's mostly the same idea: we need to chain a couple of proofs together and formalize that in $T.$

spaceisdarkgreen
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Actually, that's not what I said; the difference may seem small at first, but it's actually quite importnat. (Nor is the claim attributed to me true: for "reasonable" $T$, if $T$ proves $\alpha\rightarrow 0=1$ then by contrapositive $T$ proves $\neg\alpha$, so we would have $T$ proving its own consistency.)

Instead, what is true (and this is clarified a bit in the comments to my answer) is this (again, supposing $T$ is a "reasonable" theory):

For each sentence $\alpha$, $T$ proves "$\neg Con(T)\implies Prov_T(\ulcorner\alpha\urcorner)$."

Here "$Prov_T$" is the usual Godel provability predicate, and "$\ulcorner$$\cdot$$\urcorner$" is the usual Godel number operation.$^1$

In particular and colloquially:

  • $T$ does prove the statement "If $T$ is inconsistent, then $T$ proves $0=1$."

  • $T$ does not prove the statement "If $T$ is inconsistent, then $0=1$."

Intuitively, this is an example of the more general phenomenon that $T$ "lacks confidence:" it is not the case that $T$ proves "if I prove $\alpha$, then $\alpha$ is true" for general statements $\alpha$, and indeed by Lob's theorem the only way this will happen at all is if $T$ already proves $\alpha$.

$^1$And actually I should really write "$\neg Con(T)\implies Prov_T(\underline{\ulcorner\alpha\urcorner})$," where "$\underline{n}$" is the numeral corresponding to $n$ for a natural number $n$, but that's a minor issue here.

EDIT: While it's correct to say "go look at the proof of Godel's theorem first," that's maybe a bit discouraging. So here's a more concrete problem which will require the same techniques to solve, but is hopefully more approachable:

$T$ proves "$Prov_T(\ulcorner\alpha\urcorner)\wedge Prov_T(\ulcorner\beta\urcorner)\implies Prov_T(\ulcorner\alpha\wedge\beta\urcorner)$."

That is, $T$ proves "If $T$ proves $\alpha$ and $T$ proves $\beta$, then $T$ proves $\alpha\wedge\beta$." (Recall that "$\wedge$" means "and," and see the footnote above re: missing underlines.)

As you trace through Godel's argument, I think you'll see rather quickly how to establish this. And with that practice, it's not too hard to prove the result you care about.

Noah Schweber
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  • Thank you and I was mistaken in this assertion. I still think it is an interesting problem to consider how $T \vdash \neg Con(T) \rightarrow Prov_T(\ulcorner \alpha \urcorner)$ – Rex Jun 13 '18 at 03:32
  • I do not believe this is obvious that $T$ proves it – Rex Jun 13 '18 at 03:32
  • It is obvious that this is true but is this because it is a tautology from $T$'s perspective as we are told $T$ is consistent and $\neg Con(T)$ is never going to be true? – Rex Jun 13 '18 at 03:33
  • @Rex It depends on one's background. If you're familiar with the usual proof of Godel's theorem, then it should be - the point being that along the way to proving Godel's theorem you already have to show that $T$ can prove basic facts about provability. – Noah Schweber Jun 13 '18 at 03:33
  • The problem is that if one isn't familiar with the usual arithmetization of logic involved in proving Godel's theorems, then there's a lot of material to cover, more than I think is appropriate to cover here. I think the first thing to do is to understand the proof of Godel's incompleteness theorem on its own, and there are many texts which cover it (in my previous answer I linked to one by Peter Smith I quite like). Specific questions about the proof, or - following the proof - specific questions about how to use those techniques here, would be good MSE questions in my opinion. – Noah Schweber Jun 13 '18 at 03:37
  • Do you mean the fixed point lemma in Godel's first incompleteness theorem or are you referring to a different incompleteness theorem? – Rex Jun 13 '18 at 03:37
  • @Rex I do not mean the fixed point lemma. I mean the rest of the proof, actually - everything but the fixed point lemma. The development of logic inside arithmetic (e.g. how we even try to express "is provable," before getting to "this statement is provable"). – Noah Schweber Jun 13 '18 at 03:38
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    It's quite unfortunate, but in some presentations the fixed point lemma is highlighted as "the cool bit" and the whole arithmetization apparatus is largely ignored or brushed under the rug as "annoying technicality." This is terrible: the arithmetization of logic is the heart of the argument. So it's precisely in those annoying technical details that we find the tools to talk about questions like this. – Noah Schweber Jun 13 '18 at 03:40
  • So you recommend I review the proof of Godel's first incompleteness theorem -- is that correct? – Rex Jun 13 '18 at 03:43
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    @Rex Yes - specifically, the pesky formal details. And I recommend Smith's book, although of course it's merely one of many excellent sources. I do recommend an actual published source as opposed to random notes online, though, since while there are many excellent expositions online there are also many terrible expositions online and it can be hard to gauge quality at first. – Noah Schweber Jun 13 '18 at 03:45
  • Thank you very much for your thoughtful comments I will look into these sources! – Rex Jun 13 '18 at 03:49
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    @user170039 I think you're confused - I'm talking about the incompleteness theorem, not the completeness theorem. – Noah Schweber Jun 13 '18 at 05:14