In ([reference](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0CCUQFjAA&url=http%3A%2F%2Fwww.philosophy.ox.ac.uk%2F__data%2Fassets%2Fpdf_file%2F0015%2F1194%2FSufficient_conditions.pdf&ei=7DAPVczxCoXfaqD_glg&usg=AFQjCNFrPLZT7E00XZ5-GjdmzvJ2Lwoqzw))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](http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0CB0QFjAA&url=http%3A%2F%2Fwww.compstat2004.cuni.cz%2F%7Ekrajicek%2Fsmorynski.pdf&ei=RjsPVYv_CY_3aor0gKgF&usg=AFQjCNF0y3C6S6X0z0_ViU6993GwAgn7Mw&sig2=JXr0dqT501lSJcC1foTefw), 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.



Generalizing a result of Kreisel on $\omega$-consistency
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.


1. 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 $T\vdash\_1\bot$.
2. 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.
3. 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.
4. 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 $\psi\in\Sigma\_2$.
5. 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.


