Can we have a consistent and effective (fulfilling Godel's criteria) first order theory $T$, that is both arithmetically unsound and $\omega$-inconsistent, and yet doesn't prove its own inconsistency ( i.e. $T \not \vdash \neg \operatorname {Con}(T)$)?
Consistent Unsound Omega-Inconsistent Effective Theory – Logic
lo.logic
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Surprisingly, the answer is yes! Well, let me say that the answer is yes for what I find to be a reasonable way to understand what you've asked.
Specifically, what I claim is that if PA is consistent, then there is a consistent theory $T$ in the language of arithmetic with the following properties:
- The axioms of $T$ are definable in the language of arithmetic.
- PA proves, of every particular axiom of PA, that it satisfies the defining property of $T$, and so $T$ extends PA.
- $T$ proves that the set of axioms satisfying that definition forms a consistent theory. In other words, $T$ proves that $T$ is consistent.
In this sense, the theory $T$ is a positive instance of what you request.
But actually, a bit more is true about the theory $T$ I have in mind, and it may lead you to think a little about what exactly you want.
- Actually, PA proves that $T$ is consistent.
- Furthermore, the theory $T$ has exactly the same axioms as PA.
I believe that this was observed first by Feferman: S. Feferman, Arithmetization of metamathematics in a general setting, Fund. Math. 49 (1960-1961), 35--92. (Thanks to Andreas Blass for pointing out the precise reference.)
The idea of the proof is simple. We shall simply describe the axioms of PA in a different way, rather than enumerating them in the usual way. Specifically, let $T$ consist of the usual axioms of PA, added one at a time, except that we add the next axiom only so long as the resulting theory remains consistent.
Since we assumed that PA is consistent, it follows that actually all the axioms of PA will actually satisfy the defining property of $T$, and so PA will be contained in $T$. Furthermore, since PA proves of any particular finite number of axioms of PA that they are consistent, it follows that PA proves that any particular axiom of PA will be in $T$.
Because of how we defined it, however, it is clear that PA and hence also $T$ proves that $T$ is consistent, since if it weren't, there would be a first stage where the inconsistency arises, and then we wouldn't have added the axiom making it inconsistent. Almost by definition, $T$ is consistent, and PA can prove that. So $T$ proves that $T$, as defined by the definition we gave for it, is consistent. So this theory $T$ actually proves its own consistency!
Meanwhile, let me point out that if one makes slightly stronger requirements on what is wanted, then the question has a negative answer, essentially by the usual proof of the second incompleteness theorem:
Theorem. Suppose that $T$ is a arithmetically definable theory extending PA, such that if $\sigma$ is an axiom of $T$, then $T$ proves that $\sigma$ is an axiom of $T$ and furthermore PA proves these things about $T$. If $T$ is consistent, then it does not prove its own consistency.
Proof. By the Gödel fixed-point lemma, let $\psi$ be a sentence for which PA proves $\psi\leftrightarrow\ \not\vdash_T\psi$. Thus, PA proves that $\psi$ asserts its own non-provability in $T$.
I claim, first, that $T$ does not prove $\psi$, since if it did, then since $T$ proves that its actual axioms are indeed axioms, it follows that $T$ would prove that that proof is indeed a proof, and so $T$ would prove that $\psi$ is provable in $T$, a statement which PA and hence $T$ proves is equivalent to $\neg\psi$, and so $T$ would also prove $\neg\psi$, contrary to consistency. So $T$ does not prove $\psi$. And this is precisely what $\psi$ asserts, so $\psi$ is true.
In the previous paragraph, we argued that if $T$ is consistent, then $\psi$ is true. By formalizing that argument in arithmetic, then since we assumed that PA proved our hypotheses on $T$, we see that PA proves that $\text{Con}(T)\to\psi$. So if $T$ were to prove $\text{Con}(T)$, then it would prove $\psi$, contradicting our earlier observation. So $T$ does not prove $\text{Con}(T)$. QED
To my way of thinking, the principal case of interest with regard to the first incompleteness theorem, the case carrying almost fully the philosophical interest and fascination of the theorem, is the case of true arithmetic theories. In this formulation, the theorem states that there is no computable axiomatization of a theory $T$ that proves all and only the true statements of arithmetic, that is, the statements that are true in the intended standard model $\langle\mathbb{N},+,\cdot,0,1,<\rangle$. In short, no computably axiomatized true theory of arithmetic is complete. We cannot provide a computable list of axioms for the true theory of arithmetic.
It is this formulation of the theorem that addresses the Hilbert program, for example, aimed at providing a complete theory of the arithmetic truths. Gödel's theorem shows that we shall be unable to provide such a theory by a computable list of axioms.
A similar formulation of the theorem is the statement that every consistent computably axiomatizable theory $T$ (and containing a sufficent weak fragment of arithmetic) will admit true but unprovable statements.
These versions of the theorem can be proved by introducing the Gödel sentence $\sigma$, which asserts its own nonprovability in $T$. This statement is both true and not provable in the system. If the theory $T$ is true, then of course we also know that $T$ does not prove $\neg\sigma$, and so the theory is incomplete.
If we aim at a stronger version of the theorem, however, without the assumption that $T$ is true — but one must admit that this is a strange case in which to be interested! — then this argument doesn't quite seem to establish that the theory $T$ is incomplete, since perhaps $T$ would prove $\neg\sigma$. To address this, Gödel had recognized however that if we know the theory is $\omega$-consistent, then we may deduce also that $T$ does not prove $\neg\sigma$, and so $T$ is incomplete. So this was a technical condition, a weakening of truth to $\omega$-consistency, which sufficed to prove his desired conclusion in a more general context.
Soon after, however, Rosser pointed out that in fact no additional assumption is needed for the conclusion in the general case — one needs rather only to consider a slightly revised sentence, while using essentially similar methods. The Rosser sentence $\rho$ asserts that for every proof of $\rho$, there is a smaller proof of $\neg\rho$, smaller in terms of the Gödel code. One can then show, assuming only that $T$ is consistent and computably axiomatizable (and containing a sufficient weak fragment of arithmetic), that $T$ proves neither $\rho$ nor $\neg\rho$.
Thus, the Rosser argument shows that we can simply drop the $\omega$-consistency assumption from the theorem. It is irrelevant.
To my way of thinking, therefore, we frankly have little need to consider or discuss $\omega$-consistency at all in connection with the incompleteness theorem (except as a historical matter). It was a technical hypothesis considered by Gödel as a weakening of truth, but quickly seen to be unnecessary in light of Rosser's argument, which shows that essentially similar ideas as Gödel's prove the theorem without that hypothesis.
When I prove the incompleteness theorem, I start with the easy case of true arithmetic theories, the main case, a case also where the proof is slightly easier. For a stronger result, we go directly to the Gödel-Rosser theorem, which can be proved by essentially similar methods. There is no need or reason at all to consider $\omega$-consistency, and I typically mention it in passing only, in order to give historical context to Rosser's innovation.
In fact, there are many different proofs of the incompleteness theorem, and when I teach this topic, as I shall next semester, I prefer to give as many as possible. I am currently writing a a book entitled Ten proofs of Gödel incompleteness.
Best Answer
Sure. Consider something like $T=\mathsf{PA}+\neg\mathit{Con}(\mathsf{ZFC})$ (assuming $\mathsf{ZFC}$ is actually consistent of course).
$T$ is consistent but not $\omega$-consistent (it proves that each specific natural number fails to be a $\mathsf{ZFC}$-proof of $\perp$, but still proves that some number is a $\mathsf{ZFC}$-proof of $\perp$).
More interestingly, $T\not\vdash\neg \mathit{Con}(T)$. To see this, note that if it did we would have $\mathsf{PA}$-provably that $\neg\mathit{Con}(\mathsf{ZFC})\rightarrow\neg\mathit{Con}(T)$. Taking the contrapositive and shifting to the stronger theory $\mathsf{ZFC}$, we would get $$\mathsf{ZFC}\vdash\mathit{Con}(T)\rightarrow\mathit{Con}(\mathsf{ZFC}),$$ a contradiction since $\mathsf{ZFC}$ already proves the consistency of $T$. (Why does $\mathsf{ZFC}$ prove the consistency of $T$? If $T$ were inconsistent then $\mathsf{PA}$ would prove $\mathit{Con}(\mathsf{ZFC})$ and hence be inconsistent. So since $\mathsf{ZFC}$ proves that $\mathsf{PA}$ is consistent, $\mathsf{ZFC}$ also proves that $T$ is consistent.)