TITLE: Generalizing a result of Kreisel on $\omega$-consistency
QUESTION [6 upvotes]: In  (reference)The following result is attributed to Kreisel:
Lemma1(Kreisel) If $T$ is an $\omega$-consistent theory in the language of arithmetic and $\pi$ is a true $\Pi_1$ sentence, then $T+\pi$ is also $\omega$-consistent.
My question is:
Question: If $T$ is an $\omega$-consistent theory in the language of arithmetic, is $T+Th_{\Pi_1}(\mathbb{N})$ also $\omega$-consistent?  ($Th_{\Pi_1}(\mathbb{N})$ is the set of all true $\Pi_1$ sentences).
The best i could do is the following result:
lemma2) If $T\supset I\Sigma_1$ is an $\omega$-consistent theory in the language of arithmetic and $A$ an r.e. subset of $Th_{\Pi_1}(\mathbb{N})$, then $T+A$ is also
$\omega$-consistent.
proof:
if $T_1=T+A$  was $\omega$-inconsistent, then there was a sentence $\exists x \alpha(x)$ such that $T+A\vdash \exists x \alpha(x)$ and also $T+A\vdash \neg \alpha(\overline{n})$ (for all $n$).
Let $A=\{\pi_{i}\}_{i \in \mathbb{N}}$. By the compactness theorem, there are indexes $j_{1},\ldots,j_{m}$ such that:
$T+\{ \pi_{j_{1}},\ldots,\pi_{j_{m}}\}\vdash \exists x \alpha(x)$.
It follows from conservation theorem(reference, theorem 5.2.1) that  $T+con(T_{1})\vdash  \pi_{i}$ (for every $i\in \mathbb{N}$). Then
$T+con(T_{1})\vdash \exists x \alpha(x)$,
and by similar reason:
$T+con(T_{1})\vdash \neg \alpha(\overline{n})$  (for every $n\in \mathbb{N}$).
Then $T_1=T+con(T_{1})$ is an $\omega$-inconsistent theory and it contradicts the lemma1.

REPLY [9 votes]: The property does not hold in general.
First, I’ll recall some basic properties of $\omega$-consistency. Let $T\vdash_1\phi$ denote the relation that $\phi$ is derivable from $T$ using rules of first-order logic, and unnested instances of the $\omega$-rule.

If $T\vdash_1\phi$, then $\phi$ is derivable from $T$ using a single instance of the $\omega$-rule. In particular, $T$ is $\omega$-inconsistent if and only if $T\vdash_1\bot$.

If $T$ is r.e., then $T\vdash_1\phi$ is a $\Sigma_3$ property of $\phi$. Thus, the $\omega$-consistency of $T$ is a $\Pi_3$ statement.

On the other hand, $Q\vdash_1\phi$ for every true $\Sigma_3$ sentence $\phi$: write $\phi=\exists x\,\forall y\,\psi(x,y)$, where $\psi\in\Sigma_1$, and fix $n\in\omega$ such that $\mathbb N\models\forall y\,\psi(\bar n,y)$. Then $Q\vdash\psi(\bar n,\bar m)$ for every $m$ by $\Sigma_1$-completeness of $Q$, hence $Q\vdash_1\forall y\,\psi(\bar n,y)$ using the $\omega$-rule.

Similarly, $Q+\mathrm{Th}_{\Pi_1}(\mathbb N)\vdash_1\phi$ for every true $\Sigma_4$ sentence $\phi$: this follows by the same argument as above, with $\psi\in\Sigma_2$.

If $T$ is an r.e. extension of $I\Delta_0+\mathit{EXP}$ (this can be negotiated down with a bit of care), then $T\vdash_1\phi$ satisfies the Bernays–Löb derivability conditions: that is, if we write $\Box_{T,1}\phi$ for the natural arithmetization of $T\vdash_1\phi$, we have

$T\vdash_1\phi\implies T\vdash_1\Box_{T,1}\phi$: this is a consequence of 2 and 3.

$T\vdash\Box_{T,1}(\phi\to\psi)\to(\Box_{T,1}\phi\to\Box_{T,1}\psi)$: we can concatenate two proofs.

$T\vdash\Box_{T,1}\phi\to\Box_{T,1}\Box_{T,1}\phi$; more generally, if $\psi\in\Sigma_3$, then $T\vdash\psi\to\Box_{T,1}\psi$. This follows by formalizing the argument in 3, using the ordinary formalized $\Sigma_1$-completeness of $Q$.


See [1] for more information about the provability logic of $\vdash_1$ and related provability predicates.


Now, let $T_0$ be an $\omega$-consistent r.e. extension of $I\Delta_0+\mathit{EXP}$ (such as $I\Sigma_1$ or $\mathit{PA}$), and put
$$T=T_0+\Box_{T_0,1}\bot$$
(that is, $T_0$ + its own formalized $\omega$-inconsistency). The standard proof of the second Gödel incompleteness theorem using the derivability conditions shows that $T$ is $\omega$-consistent. On the other hand, $\neg\Box_{T_0,1}\bot$ is a true $\Pi_3$ sentence, hence
$$T_0+\mathrm{Th}_{\Pi_1}(\mathbb N)\vdash_1\neg\Box_{T_0,1}\bot$$
by 4, thus $T+\mathrm{Th}_{\Pi_1}(\mathbb N)$ is $\omega$-inconsistent.

Kreisel’s lemma can be strengthened in a different direction, namely it holds for a larger class of formulas than $\Pi_1$:

Proposition 1: If $T\supseteq Q$ is $\omega$-consistent, and $\phi$ is a true $\Sigma_3$ sentence, then $T+\phi$ is $\omega$-consistent.

Proof: Otherwise $T+\phi\vdash_1\bot$, hence $T\vdash_1\neg\phi$. On the other hand, $Q\vdash_1\phi$ by property 3 above, hence $T\vdash_1\bot$, i.e., $T$ is $\omega$-inconsistent.

Corollary 2: If $T\supseteq Q$ is $\omega$-consistent, and $A$ is an r.e. set of true $\Pi_2$ sentences, then $T+A$ is $\omega$-consistent.

Proof: We may assume $T\supseteq I\Delta_0+\mathit{EXP}$ by Proposition 2, as $I\Delta_0+\mathit{EXP}$ is a finitely axiomatizable true $\Pi_2$ theory. It suffices to find a true $\Pi_2$ (or $\Sigma_3$) sentence $\phi$ such that $T+\phi\vdash A$. So, let $\alpha(x)$ be a $\Sigma_1$ definition of $A$ in $\mathbb N$, and $\mathrm{Tr}_{\Pi_2}(x)$ a universal $\Pi_2$ formula. Then we can take $\phi=\forall x\,(\alpha(x)\to\mathrm{Tr}_{\Pi_2}(x))$.
Notice that both statements are optimal with respect to arithmetic complexity:

The Proposition may fail for true $\Pi_3$ sentences $\phi$: for example, take the $\omega$-consistent theory $T=T_0+\Box_{T_0,1}\bot$ considered above, and $\phi=\neg\Box_{T_0,1}\bot$, which is a true $\Pi_3$ sentence. Then $T+\phi$ is inconsistent.

The Corollary may fail for r.e. sets $A$ of true $\Sigma_2$ sentences: continuing the previous example, write $\phi=\forall x\,\psi(x)$ with $\psi\in\Sigma_2$, and put $A=\{\psi(\bar n):n\in\omega\}$. Then $T+A\vdash_1\phi$, while $\neg\phi\in T$, hence $T+A$ is $\omega$-inconsistent.



For yet another extension of Kreisel’s lemma, I recall that Smoryński proved that for any r.e. theory $T\supseteq Q$,
$$T\vdash_1\phi\iff\mathrm{Tr}_{\Sigma_3}(\mathbb N)+\mathrm{RFN}_T\vdash\phi,$$
where $\mathrm{RFN}_T$ is the uniform reflection principle for $T$: the schema
$$\forall x\:\bigl(\Box_T\psi(\dot x)\to\psi(x)\bigr)$$
for all formulas $\psi(x)$, where $\Box_T$ denotes the usual provability predicate for $T$. It is also known that if $T\supseteq Q$ is finite, then $Q+\mathrm{RFN}_T\equiv T+\mathrm{PA}$. Thus:

Proposition 3: If $T\supseteq Q$ is $\omega$-consistent, then $T+I\Sigma_n$ is $\omega$-consistent for every $n\in\omega$.
Moreover, if $S\subseteq T$ is r.e., and $\phi(x)$ is any formula, then $T+\forall x\,\bigl(\Box_S\phi(\dot x)\to\phi(x)\bigr)$ is $\omega$-consistent.


References:
[1] George Boolos, The Logic of Provability, Cambridge University Press 1993.
[2] Craig Smoryński, Self-reference and modal logic, Springer, 1985.