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I'm trying to understand the statement of Gödel's second incompleteness theorem. For a set of axioms $\Phi$ containing $\Phi_{PA}$, the author claims the following sentence 'express' the consistency of $\Phi$: $$Con(\Phi) := \neg Prov_\Phi(0\equiv 1)$$

My question is: what is the precise relationship between consistency of $\Phi$ and the sentence $Con(\Phi)?$

My guess is

  1. Maybe he means $\Phi$ is consistent if and only if $Con(\Phi)$ is true/correct. But if so, what does 'true/correct' mean? There seems to be no model fixed.

  2. Maybe $Con(\Phi)$ is a purely arithmetically statement in the standard natural number model $\mathbb N$? So we can discuss the correctness of $Con(\Phi)$ in $\mathbb N$ in the usual sense.

Thomas Andrews
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Hydrogen
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    You should probably post a copy of the passage and a citation if you want someone to interpret an author’s claims. Otherwise we’re just guessing too. That said, my guess is that they just mean it’s a formal arithmetical sentence that’s true in the standard model iff $\Phi$ is consistent. – spaceisdarkgreen Sep 19 '24 at 20:06
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    If a first order logic has a proof of a contradiction, the system is inconsistent, and then it can prove anything, including $1=0.$ If you can prove $1=0,$ since you can also prove $1\neq0,$ you do not have consistency. – Thomas Andrews Sep 19 '24 at 20:17
  • So really, you can think of "$\Phi$ is inconsistent iff you can prove $1=0$ in $\Phi.$" – Thomas Andrews Sep 19 '24 at 20:19
  • @ThomasAndrews I think I understand '$\Phi$ is consistence iff I can prove $1=0$ in $\Phi$'. My question is, does this mean '$\Phi$ is consistent' is equivalent to '$Con(\Phi)$ is 'true''? – Hydrogen Sep 19 '24 at 21:29
  • What is $\Phi'?$ $\Phi$ is consistent iff you cannot prove $1=0$ in it. – Thomas Andrews Sep 19 '24 at 21:42
  • @ThomasAndrews I'm sorry I think it is a typo. Does the correctness of the sentence $Con(\Phi)$ itself make sense? – Hydrogen Sep 19 '24 at 22:41
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    Your guess number 2 is correct. In guess 1, "true" means "true in the standard model of arithmetic". – Andrés E. Caicedo Sep 20 '24 at 03:16
  • Gödel's second incomplete theorem: "No sufficiently strong formal theory can pr" – Misha Lavrov Oct 03 '24 at 14:08

2 Answers2

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In this field, 'true' is short for 'true in the standard model of arithmetic' $\mathbb{N}$. The relation between the consistency of a theory extending $\sf PA$ and a statement expressing it is quite tricky.

Usually, to define an arithmetized consistency statement for $\sf T$, you first need to define (among others such as the coding) a provability predicate and the $\sf T$-axiom set. Such choices do have an impact in the relation. Given the standard provability predicate $\sf Prov$, the standard $\sf PA$-axiom set $\sf A$, and your (standard) definition of consistency as $\mathsf{Con(A) := \neg Prov_A (\ulcorner \bot \urcorner)} $, you indeed have:

$\sf PA$ is consistent (i.e. $\mathsf{PA}\nvdash \bot$) iff $\sf Con(A)$ is true in $\mathbb{N}$ ($\mathbb{N}\models \neg \mathsf{Prov_A}(\ulcorner \bot \urcorner)$).

Any consistency statement should satisfy such equivalence. (Could we even call it a consistency statement otherwise?) In that sense, your $\sf Con(A)$ expresses the consistency of $\sf PA$, and similarly for extension of $\sf PA$ as well as weaker theories of arithmetic.

However, what we don't have in all cases is: $\sf PA$ is consistent iff $\sf PA \nvdash Con(A)$. This depends on your particular choices (among others) of the provability predicate and axiom set. For example, by using Rosser's provability predicate or Feferman's definition of a $\sf PA$-axiom set $\sf B$, the equivalence fails (provided that $\sf PA$ is consistent). See for example, here, here or directly in his Arithmetization of metamathematics in a general setting, Fundamenta mathematicae, vol. 49 no. 1 (1960), pp. 35–92.

In particular, $\sf Con(B)$ does express (in the above sense) consistency but only provided that $\sf PA$ is consistent. For these reasons, it is not so clear that $\sf Con(B)$ really expresses (in an intuitive sense) a consistency statement although for what we know it might express (in the above sense) a consistency statement. All these questions fall under the name of intensionality of metamathematics.

Léon Probst
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  • Now, 'true in the standard model of arithmetic $\mathbb N$ ' seems mysterious to me. I know the formal definition of this, i.e. when plugging in functions and elements, the statement is true, which is different from provable. But philosophically, does that mean it is true in some transcendental sense? How could we say a statement is true without any way to check it? – Hydrogen Oct 07 '24 at 00:07
  • True in $\mathbb{N}$ is simply true about the natural numbers. In the same way, you (perhaps) think that the axioms of $\sf PA$ are truths about the natural numbers. Answering your second question is quite difficult, and maybe more suitable for philosophy stackexchange. You might find interesting discussion here

    I also suggest to, in the context of Gödel's incompletness, to check Smith's An Introduction to Gödel's Theorems, especially § 23.2 on the truth of Gödel sentence.

     – Léon Probst Oct 07 '24 at 13:08
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This is essentially a legitimate statement for the consistency predicate; I shall outline an account of it. I borrow some expressions from Peter Smith's An Introduction to Gödel's Theorems simplifying a bit in order to focus on the underlying ideas. I use $\leftrightarrow$ instead of $=_{def}$ to be more in line with the language of logic. $\ulcorner\,\urcorner$ denotes Gödel numbering as usual.

We take $\Phi_{T}$ is a set of axioms for a theory $T$ such that the set of axioms of Peano Arithmetic $\Phi_{\mathrm{PA}}$ is a subset of it.

We define the numerical relation $Prf(m, n)$ which holds when $m$ is the Gödel number of a $\mathrm{PA}$ derivation of the closed formula coded by the Gödel number $n$ (viz., $n$ enumerates a $\mathrm{PA}$ theorem).

Let there be a formula $\psi(x, y)$ that arithmetically defines the relation $Prf(m, n)$; hence, for all natural numbers $m$ and $n$, we have $Prf(m, n)$ if and only if $\psi(m, n)$ holds.

Then, we define the provability predicate in the usual way we know from first-order predicate logic:

$$Prov(n)\leftrightarrow\exists vPrf(v, n)$$

Thus, we can define the consistency predicate:

$$Con(\Phi_{T})\leftrightarrow\neg Prov_{T}(\ulcorner 0=1\urcorner)$$

Therefore,

$$Con(\Phi_{T})\leftrightarrow\neg \exists vPrf_{T}(v, \ulcorner 0=1\urcorner)$$

which expresses that there is no natural number $v$ that would be the Gödel number for the proof of the closed formula $0=1$ for consistency to hold.

Consequently, we see that the consistency predicate is an arithmetical statement like the provability predicate.

Finally, I should remark that the consistency predicate could be construed in alternative ways and the outline given here can be formalised with additional details (e.g., requiring the variable $x$ of $Con(x)$ to be the Gödel number of the conjunction of the axioms involved).

Tankut Beygu
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