# Clarification on models, interpretations, undefinability

I am an absolute beginner in this area and this is my third attempt to ask this question as my past posts have some inaccurate notations and I deleted them. Any help is much appreciated ....

From (Jech, 2006 p. 163), by Godel's 2nd incompleteness theorem, we cannot prove consistency of $$ZF$$ using $$ZF$$ alone. If given some axiom $$A$$ and we want to show that $$ZF + A$$ is consistent, the way to go is to use models $$M$$ s.t. $$M \models \phi^M$$ for all $$ZF$$ axioms $$\phi$$ as well as $$M \models A$$.

Now, in pp. 165-166, Jech showed that $$V \models \phi^V$$ for $$ZF$$ axioms $$\phi$$ as well as $$ZF \models \text{axiom of regularity}$$. The same is done in p. 176 for $$L$$.

This is where I am getting confused a bit. We are showing that $$V$$ and $$L$$ are models of $$ZF$$ (such as p. 176, Th. 13.3), but if we have shown that $$V$$ and $$L$$ are models of $$ZF$$ then we have shown using $$ZF$$ that $$ZF$$ is consistent. One answer to this in my past post holds that these are class-models and are not set-sized models so there's no contradiction...

I am trying to go deeper on this idea and would like to know if my understanding below is correct.

Define $$\models_n^M \phi$$ as a relation that a formula $$\phi$$ of $$ZF$$ of length $$n$$ is true in $$M$$ (where $$M$$ can be a proper class). In showing that $$V$$ is a model of $$ZF$$ (i.e. pp. 165-166), we are proving that for any fixed $$\phi$$ of length $$n$$, the ff is provable - perhaps using transfinite induction:

$$\vdash_{ZF} (V \models^V_n \phi^V) \quad{(1)}$$

This is different from saying that "$$V$$ is a model of $$ZF$$" and proving it in $$ZF$$ which is:

$$\vdash_{ZF} \forall_{n \in \mathbb{N}}(V \models^V_n \phi^V) \quad{(2)}$$

Given that $$\forall_{n \in \mathbb{N}}(V \models^V_n \phi^V)$$ is not formally definable from Tarski's undefinability theorem (Th 12.7 of Jech), much less provable from Godel's 2nd incompleteness theorem.

But why is there a need at all to show that $$(1)$$ holds for all axioms $$\phi$$ of $$ZF$$ ? ... My guess is that even though we cannot prove in $$ZF$$ that "$$V$$ is a model of $$ZF$$" in the sense of $$(2)$$, by showing that $$V$$ is a model of $$ZF$$ in the sense of $$(1)$$, we are actually referring to $$V$$ as an interpretation as defined in Schoenfield, where $$I$$ is an interpretation of a theory $$T$$ if for any formula $$\phi$$, we have $$\vdash_T \phi$$ implies $$\vdash_I \phi^I$$. We are then using the interpretation theorem in Schoenfield that: If $$I$$ is an interpretation of $$T$$ and $$I$$ is consistent, then $$T$$ is consistent.

So by proving $$(1)$$ for any $$\phi$$ we have shown that $$V$$ is an interpretation of $$ZF$$ so that for any additional axiom $$A$$, if we can prove (using $$ZF$$, no problem at all) that $$A$$ holds in $$V$$, then $$V$$ is a model of $$A$$ (i.e. $$A$$ is consistent in $$V$$ and there's no problem here given that $$A$$ is assumed to belong to some fixed level in the Levy heirarchy so that $$V \models_n A$$ is provable in $$ZF$$ under some fixed $$n$$) such that if $$ZF$$ is consistent, by the interpretation theorem, then $$ZF+A$$ is consistent.

• Maybe the proof of $M \vDash \phi ^M$ is not "in" $\mathsf {ZFC}$ but in the meta-theory... Commented May 20 at 9:03
• And yes, if we prove that $M$ is a model, than this implies that the theory is consistent. BUT if the proof is not done "in" $\mathsf {ZFC}$, this does not contradict GIT2. Commented May 20 at 9:04
• See page 161: "We shall also consider models of set theory that are proper classes. However, due to Godel’s Second Incompleteness Theorem, we have to be careful how the generalization is formulated." And see page 163. Commented May 20 at 9:07
• Commented May 20 at 9:23
• Commented May 20 at 9:27