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.



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