# Absoluteness of completeness

Suppose $V_0, V_1$ are (not necessarily well-founded) models of ZFC and suppose $\varphi$ is a first order sentence in a finite language $L$ (in our background model of set theory). Because every true finite set is an element of the well-founded part of both $V_0$ and $V_1$ we can consider $\varphi$ as an element of both $V_0$ and $V_1$.

Is it the case that:

$V_0 \models$ The deductive closure of $\varphi$ is a complete $L$-theory

if and only if

$V_1 \models$ The deductive closure of $\varphi$ is a complete $L$-theory?

Thanks.

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The deductive closure of $\phi$ could be complete in one model and incomplete in the other.

Here's one way to see that (perhaps not the most elegant). Take any reasonable finitely axiomatized theory $T$ in a finite language $L$ capable of sufficient coding to carry out the proof of Gödel's Theorem and not implying the consistency of $ZFC$. (My favorite choice is $I\Sigma_1$, but others might prefer something in the language of set theory.) Fix your favorite encoding of $Con(ZFC)$ in this theory.

Let $T^-$ be some other finitely axiomatized theory on the same language which is complete and so that $T^-\vdash\neg Con(ZFC)$. It need not have any relation to $T$, so it could just interpret all functions as projections onto the first coordinate and all relations as empty, assuming that deduces $\neg Con(ZFC)$. (Note that $\neg Con(ZFC)$ has no meaning in this theory---it's just some arbitrary existential sentence.)

Now consider the sentence $\phi$ which is $$(Con(ZFC)\rightarrow T)\wedge(\neg Con(ZFC)\rightarrow T^-)$$ (Interpreting $T$ and $T^-$ as the sentences which are the conjunctions of all their axioms.)

In any model of $ZFC$ where $Con(ZFC)$ holds, this is incomplete because $T$ is incomplete. (And also because it's basically a disjunction of two theories and can't tell which case holds.)

In a model $V_1$ of $ZFC$ where $Con(ZFC)$ fails, $T$ can actually prove $\neg Con(ZFC)$, because this is a $\Sigma_1$ formula. Therefore the $T$ branch of the disjunction is self-invalidating: $V_1\vDash T\vdash\neg Con(ZFC)$, so $V_1\vDash \phi\vdash Con(ZFC)\rightarrow\neg Con(ZFC)$, so $V_1\vDash\phi\vdash T^-$. And (the deductive closure of) $T^-$ is complete.

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