$T \vdash Prov_T(\sigma)$ but $T \nvdash \sigma$

Question

If we have a consistent and axiomatizable extension of PA, which we'll call $T$, I'm wondering if it's possible to have a sentence $\phi$ such that $T \vdash Prov_T(\phi)$ but $T \nvdash \phi$ where $Prov_T(x)$ expresses "$x$ is provable in $T$".

Note very importantly that we assume $T$ contains "sufficient induction" such that it satisfies the following derivability conditions:

(1) "reflection" which is if $T \vdash \sigma$ then $T \vdash Prov_T(\sigma)$ and (2) a "formal" version of modus ponens such that $T \vdash Prov_T(a) \land Prov_T(a \rightarrow b) \rightarrow Prov_T(b)$ for any sentences $a$ and $b$.

My intuition is that there is such a sentence $\phi$ since @Noah Schweber points out in this post that soundness might not hold.

On the other hand, I also think Lob's theorem is relevant and some adaptation of it might actually prove that $T \vdash Prov_T(\phi)$ then $T \vdash \phi$ if we use the derivability to adapt Lob's theorem. Any hints?

Answer 1

Take $T$ to be Peano arithmetic (PA) + the sentence "PA is inconsistent" (for instance $Prov_{PA}(0=1)$)

Then $T$ is recursively axiomatizable; and by Gödel's incompleteness theorem, $T$ is consistent (otherwise $PA \vdash Con(PA)$, and thus PA is inconsistent, which is absurd; unless PA is inconsistent, but then the whole discussion is rendered useless because there is no consistent extension of PA - anyway if we're working in ZF, PA is clearly consistent so we are content with that)

Since $T$ is consistent, and $PA \vdash \neg (0=1)$, then $T\nvdash 0=1$.

However, by very definition of $T$, $T\vdash Prov_{PA}(0=1)$ and so $T\vdash Prov_T(0=1)$ because for any recursively axiomatisable extension of PA $S$ and every $\phi$, we have $PA\vdash (Prov_{PA}(\phi) \implies Prov_S(\phi))$.

Hence we have found such a $T$.

This first part (before edit) answered the "is it possible ?" question; now if you were wondering about an arbitrary $T$ (can we find a $\phi$ for all $T$ ?), the answer is no, the best example being PA. Indeed, if $PA\vdash Prov_{PA}(\sigma)$, then $\mathbb{N}\models Prov_{PA}(\sigma)$ and so there is an actual proof in PA of $\sigma$: $PA\vdash \sigma$. In fact this argument shows that any recursively axiomatizable extension of PA that is contained in true arithmetic (the theory of $\mathbb{N}$) will have the property that there is no such $\phi$: indeed we have a soundness condition: if $\mathbb{N}\models Prov_T(\sigma)$, then $T\vdash \sigma$

Answer 2

Building off of Max's excellent answer, what we're dealing with here is that there are many different "niceness" properties a theory might have, and they yield different conclusions.

• Being consistent. 'Nuff said.

• Being recursively axiomatizable. This is necessary in order to even talk about provability from a theory in the language of arithmetic. (OK that's not quite true, in general all we need is that the theory be definable, but working with arithmetically defined theories can lead to weird results so it's reasonable to restrict attention to recursive axiomatizations here.)

• Being sufficiently strong. A theory extending PA (indeed, even much less) has the nice property that it is $\Sigma_1$ complete: every true $\Sigma_1$ sentence is provable in the theory. This in particular implies that if $T$ is sufficiently strong (and recursively axiomatizable) then $T\vdash\phi$ implies $T\vdash Prov_T(\phi)$ for each $\phi$.

• Being sufficiently correct. Consistent sufficiently strong recursively axiomatizable theories can still be quite weird - e.g. PA+$\neg$Con(PA). We have a separate interest in not-too-stupid theories - that is, theories which don't prove "blatantly false" sentences. Such theories are called "sound" - more specifically, if $T$ is a theory which proves only (not necessarily every!) true sentence of type $\Gamma$, then we say $T$ is $\Gamma$-sound. E.g. PA+$\neg$Con(PA) is not $\Sigma_1$ sound, since it proves the false $\Sigma_1$ sentence "$Prov_{PA}(\underline{\ulcorner 0=1\urcorner})$." Correctness is particularly relevant here, and somewhat subtle, so let me give three further comments on it:

  • First, note that correctness is in tension with completeness. To make a theory "more complete" we would strengthen it, while to make a theory "more correct" we would weaken it (by removing offending axioms).

  • Second, $\Gamma$-soundess principles aren't the only kinds of correctness which are of interest. For example, consider the following property: $$\mbox{If $T$ proves $\neg \phi(\underline{n})$ for each $n\in\mathbb{N}$, then $T$ doesn't prove $\exists x\phi(x)$.}$$ (Note that the conclusion is not "then $T$ proves $\forall x\neg\phi(x)$"!) This property is called $\omega$-consistency. Historically this was an assumption Godel needed in his proof, which was later removed by Rosser. The term "consistency," rather than "soundness," in the name of the principle reflects the fact that it really is a consistency property *with respect to a stronger-than-usual proof system. Indeed, soundness principles in general can be viewed as "higher consistency principles."

  • Finally, it's worth noting that on the face of it soundness principles can require some kind of "Platonic commitment" - to make sense of the phrase "$\Sigma_{17}$-sound," we need to (on the face of it) commit to the notion that $\Sigma_{17}$ statements have definite truth values. If we believe that $\mathbb{N}$ "exists" (???), then this is unproblematic, but the more "mathematical skepticism" we adopt can rapidly make us suspicious of even the notion of soundness (whereas consistency, being $\Pi_1$, is somewhat more palatable). This takes us well into foundational issues, and is really orthogonal to your question here, so I don't want to go into it in more detail - in particular, in my opinion it's best when learning the basics of this subject to at least temporarily adopt the position that $\mathbb{N}$ "really exists" and all arithmetic statements (at least) have definite truth values - but it's nonetheless worth mentioning.

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The connection with your question is the following:

Suppose $T$ is a consistent, recursively axiomatizable, sufficiently strong theory. Then it is possible that there is some $\phi$ such that $T\vdash Prov_T(\phi)$ but $T\not\vdash\phi$; for example, take $T$ to be PA+$\neg$Con(PA) and $\phi$ to be "$0=1$." However, this is ruled out if $T$ is additionally assumed to be $\omega$-consistent$^1$ or even $\Sigma_1$-sound$^2$.

$^1$This was how Godel used $\omega$-consistency in his original argument, although he could have made due with the weaker$^3$ assumption of $\Sigma_1$-soundness without changing the argument at all; I don't know when this was first observed.

$^2$Note that $\Sigma_1$-soundness and $\Sigma_1$-completeness - the latter of which follows from the assumed strength of $T$ - together imply that the $\Sigma_1$ theorems of $T$ are exactly the true $\Sigma_1$ statements, but do not imply that $T$ proves every true $\Pi_1$ statement, and indeed since the set of true $\Sigma_1$ statements isn't recursive we know this can't happen since $T$ is recursively axiomatizable.

$^3$Yes, $\omega$-consistency is strictly stronger than $\Sigma_1$ soundness! The point is that $\omega$-consistency applies to all $\phi$, not just to $\phi$ of a certain complexity.