# Extending a Model of $T + \operatorname {Con} ( T )$ to a model of $T + \neg \operatorname {Con} ( T )$

Let $$T$$ be a recursively axiomatizable extension of $$\mathsf {PA}$$ and $$\mathfrak M$$ be a model of $$T + \operatorname {Con} ( T )$$. Is it true that there must exist a model $$\mathfrak N$$ such that $$\mathfrak M \subset \mathfrak N$$ and $$\mathfrak N \models T + \neg \operatorname {Con} ( T )$$?

My intuition is that the answer should be affirmitive. I think since $$\neg \operatorname {Con} ( T )$$ is $$\Sigma _ 1$$, we may be able add a new element to $$| \mathfrak M |$$ coding a (nonstandard) proof of contradiction in $$T$$. I suppose some sort of argument similar to that of Gödel's incompleteness theorems might suffice for proving the existence of such extension; but I don't know if that's true.

What I tried to do was taking the positive diagram $$\Delta$$ of $$\mathfrak M$$, and trying to prove that $$T \cup \Delta + \neg \operatorname {Con} ( T )$$ is consistent. As my first attempt, I tried to find theorems from model theory, like Robinson's joint consistency theorem, that would do the job. After failing to find such theorems, I tried to prove the consistency of the mentioned theory via an argument by contradiction: if it's not consistent, finitely many members of $$\Delta$$, say $$\delta _ 0 , \dots , \delta _ { n - 1 }$$, would suffice to derive a contradiction. Letting $$\delta = \bigwedge _ { i < n } \delta _ i$$, we would then have $$T + \delta \vdash \operatorname {Con} ( T )$$. As the parameters from $$\mathfrak M$$ do not appear in $$T$$ or $$\operatorname {Con} ( T )$$, we can replace the parameters appearing in $$\delta$$ with variables and take the existential closure of the resulting formula to get a $$\Sigma _ 1$$ sentence $$\gamma$$ in the language of arithmetic such that $$T + \gamma \vdash \operatorname {Con} ( T )$$. Now, since $$\operatorname {Con} ( T )$$ is $$\Pi _ 1$$, I doubt that the existence of such $$\gamma$$ would be possible, but I couldn't go further and prove this fact.

• What is meant by $\mathfrak{M} \subset \mathfrak{N}$? I assume you mean more than mere subset. Do you mean that the inclusion preserves $+, \cdot, 0, 1, S$? Do you mean something more? Jan 29 at 4:43
• This is a great question. Note that there are mutually inconsistent (over $\mathsf{PA}$, say) $\Sigma_1$ sentences; for example, with a bit of care you can whip up a pair of r.e. theories $T,S$ such that "There is a $\perp$-proof in $T$ shorter than any $\perp$-proof in $S$" and "There is a $\perp$-proof in $S$ shorter than any $\perp$-proof in $T$" are each consistent with $\mathsf{PA}$ (and via internalized MRDP, you can even get purely-existential examples). So something more than mere existential-ness of $\neg Con(\mathsf{PA})$ will be needed for a positive answer here. Jan 29 at 4:48
• @MarkSaving At least in the models-of-arithmetic context it's fairly standard in my experience to use "$\subset$" for "substructure." Jan 29 at 4:48

Suppose $$T$$ is as above, $$\mathfrak{M}\models T$$, and $$\mathfrak{M}$$ has no extension to a model of $$T+\neg Con(T)$$. Then by compactness, we can extract an existential sentence $$\varphi$$ such that $$T\vdash\varphi\rightarrow Con(T)$$ - basically, $$\varphi$$ corresponds to a finite subset of the atomic diagram of $$\mathfrak{M}$$ which together with $$T$$ prevents $$\mathfrak{M}$$ from being expanded to a model of $$\neg Con(T)$$.
However, we now can use the internal $$\Sigma_1$$ completeness of $$\mathsf{PA}$$ (see here). Consider the theory $$S=T+\varphi$$. Since $$T$$, and hence $$S$$, contains $$\mathsf{PA}$$, we have that $$S$$ proves "$$T$$ proves every true $$\Sigma_1$$ sentence." In particular, $$S$$ proves "If $$\varphi$$ is true then $$T$$ proves $$\varphi$$," and so a fortiori "If $$T$$ is consistent then $$T+\varphi$$ is consistent." By Godel's second incompleteness theorem, $$S$$ cannot prove its own consistency, so $$S$$ cannot prove $$T$$'s consistency either. But this is exactly saying that $$T+\varphi+\neg Con(T)$$ has a model, contradicting our assumption on $$\varphi$$ above.
Note that this leaves open the situation for extremely weak theories such as $$\mathsf{EA}$$, in the context of which extensions by true $$\Sigma_1$$ sentences can be quite odd; see Visser's paper Oracle bites theory.