Let $T\_0$ be the set theory axiomatized by $ZFC^-$ (that is $ZFC$ without powerset) + every set is countable + $\mathbb{V}=\mathbb{L}$. 


**Question 1:** Suppose $\phi$ is a sentence of set theory. Must there be a large cardinal axiom $A$ such that $\phi$ is decided by $T\_0$ + "there are transitive set models of $A$ of arbitrarily high ordinal height?" 


**Update:** no.


**Question 2:** Is there a set $T$ of $\Pi\_2$ sentences, such that $ZFC^- \cup T$ is complete? 


**Update:** no, essentially. See my answer below.


**Questions 3 and 4 after comments.**


**Comments.**


* Question 1 is related to, but different from several previous questions on Mathoverflow, e.g. [Nice Algebraic Statements Independent from ZF + V=L (constructibility)](https://mathoverflow.net/questions/118081/nice-algebraic-statements-independent-from-zf-v-l-constructibility) or [On statements independent of ZFC + V=L](https://mathoverflow.net/questions/11480/on-statements-independent-of-zfc-v-l?noredirect=1&lq=1) or [Natural statements independent from true $\Pi^0\_2$ sentences](https://mathoverflow.net/questions/81190/natural-statements-independent-from-true-pi0-2-sentences?rq=1). Regarding the first two---Hamkins gives a long list of examples of things independent of $\mathbb{V}= \mathbb{L}$, but it seems that my schema takes care of all of them. (Of course in those questions $ZFC$ was assumed.)
* Moreover, Question 1 was partly motivated by Dorais' answer to the second question above, which references a question of Shelah from [The Future of Set Theory](http://shelah.logic.at/E16/E16.html).
* I'm leaving "large cardinal axiom" undefined here (so Question 1 can't be formalized), but obviously we should exclude inconsistent axioms, or things like $ZFC + $ "there are no transitive set models of ZFC."
* If we pick a definition of ``large cardinal axiom", then "every set is countable" + $\mathbb{V}=\mathbb{L}$ + "there are transitive set models of $A$ of arbitrarily high ordinal height" (for each large cardinal axiom $A$) is a set of $\Pi\_2$ sentences. So a negative answer to Question 2 implies a negative answer to Question 1.
* We can ignore (recursive) large cardinal axiom schemas $\Gamma$ because we can just replace them by $ZFC\_0$+"there is a transitive set model of $\Gamma$" where $ZFC\_0$ is some large enough finite fragment of $ZFC$. In particular we don't have to write "$ZFC + A$".
* $T\_0$ + this axiom schema (loosely speaking) is of personal significance to me: in fact I believe it to be true. (Namely, given whatever universe of sets $\mathbb{V}$ in which we are working, it seems reasonable to suppose there is a larger universe of sets $\mathbb{W} \models \mathbb{V}=\mathbb{L}$ in which $\mathbb{V}$ is countable, or such that $\mathbb{W}$ is a model of a given large cardinal axiom $A$. Ergo,...)


Let $\mathcal{L}\_{\mbox{set}}$ be the language of set theory $\{\in\}$ and let $\mathcal{L}\_1$ be $\mathcal{L}\_{\mbox{set}} \cup \{P\}$, $P$ a new unary relation symbol. Let $T\_1$ be $T\_0$ + the axioms asserting that $P \subseteq \mbox{ON}$ is stationary (for $\in$-definable classes) and for every $\alpha \in P$, $(\mathbb{V}\_\alpha, \in) \preceq (\mathbb{V}, \in)$. We insist that large cardinal axioms $A$ be sentences of $\mathcal{L}\_{\mbox{set}}$.


**Question 3:** Suppose $\phi$ is a sentence of set theory (i.e. of $\mathcal{L}\_{\mbox{set}}$). Must there be a large cardinal axiom $A$ such that $\phi$ is decided by $T\_1$ + "there are transitive set models of $A$ of arbitrarily high ordinal height?" 


**Question 4:** Is there a set $T$ of $\Pi\_2$ sentences of set theory, such that $T\_1 \cup T$ decides every sentence of set theory?



A "Completion" of $ZFC^-$
Without a definition of “large cardinal axiom”, I’m going to ignore Q1 and Q3.


The answers to Q2 and Q4 are negative by the following general principle. (For Q4, we take $T\_0$ to be the set of $\mathcal L\_{\mathrm{set}}$-consequences of $T\_1$.)



> 
> $\DeclareMathOperator\Tr{Tr}\DeclareMathOperator\Wit{Wit}\let\eq\leftrightarrow\def\gonu#1{\ulcorner#1\urcorner}\let\ob\overline$**Proposition:** Let $T\_0$ be an r.e. theory interpreting Robinson’s arithmetic, and $\Gamma$ a set of sentences for which $T\_0$ has a truth predicate $\Tr\_\Gamma(x)$, that is,
>  $$\tag{$\*$}T\_0\vdash\phi\eq\Tr\_\Gamma(\ob{\gonu\phi})$$
>  for all $\phi\in\Gamma$. Then no extension of $T\_0$ by a set of $\Gamma$-sentences is a consistent complete theory.
> 
> 
> 


**Proof:** Let $S\subseteq\Gamma$, and assume for contradiction that $T=T\_0+S$ is consistent and complete. The basic idea of the proof is that we can define in $T$ a truth predicate $\Tr(x)$ for *all* sentences as “$x$ is $T\_0$-provable from a set of true $\Gamma$-sentences”, contradicting Tarski’s theorem on the undefinability of truth.


In more detail, we fix an interpretation of, say, $S^1\_2$ in $T\_0$ so that we have basic coding of sequences of integers, and a (polynomial-time) proof predicate for $T\_0$. We define $\Wit(w,x)$ to be the formula


“the sequence $w$ is a $T\_0$-proof of a sentence $x$ from extra axioms, each of which is a sentence $a\in\Gamma$ such that $\Tr\_\Gamma(a)$,”


and we put


$$\Tr(x)\eq\exists w\,(\Wit(w,x)\land\forall w'<w\,\neg\Wit(w',\gonu{\neg x})).$$



> 
> **Claim:** Whenever $T$ proves a sentence $\phi$, it also proves $\Tr(\ob{\gonu\phi})$. Whenever $T$ proves $\neg\phi$, it also proves $\neg\Tr(\ob{\gonu\phi})$.
> 
> 
> 


Using the Claim, we can easily finish the proof of the Proposition: by Gödel’s diagonal lemma, there is a sentence $\alpha$ such that


$$T\_0\vdash\alpha\eq\neg\Tr(\ob{\gonu\alpha}).$$


Since $T$ is complete, it proves $\alpha$ or $\neg\alpha$. If $T\vdash\alpha$, then $T$ proves $\neg\Tr(\ob{\gonu\alpha})$ by the definition of $\alpha$, and $\Tr(\ob{\gonu\alpha})$ by the Claim, hence $T$ is inconsistent, contrary to our assumptions. The case $T\vdash\neg\alpha$ is similar.


Now, to prove the Claim, assume $T\vdash\phi$. We can fix a $T\_0$-proof of $\phi$ from some $\psi\_1,\dots,\psi\_k\in S$, which has a standard Gödel number $n$. 


By $\Sigma^0\_1$-completeness, $T$ proves that $\ob n$ is a $T\_0$-proof of $\ob{\gonu\phi}$ from $\ob{\gonu{\psi\_1}},\dots,\ob{\gonu{\psi\_k}}$. Moreover, $T$ proves each $\psi\_i$, hence also $\Tr\_\Gamma(\ob{\gonu{\psi\_i}})$ by $(\*)$. Thus,


$$T\vdash\Wit(\ob n,\ob{\gonu{\phi}}).$$


On the other hand, let $m<n$, we will show


$$T\vdash\neg\Wit(\ob m,\ob{\gonu{\neg\phi}}).$$


This again follows by $\Sigma^0\_1$-completeness unless $m$ is an actual Gödel number of an actual $T\_0$-proof of $\neg\phi$ from some sentences $\chi\_1,\dots,\chi\_l\in\Gamma$. Since $T$ also proves $\phi$, this means


$$T\vdash\neg\chi\_1\lor\dots\lor\neg\chi\_l,$$


hence


$$T\vdash\neg\Tr\_\Gamma(\ob{\gonu{\chi\_1}})\lor\dots\lor\neg\Tr\_\Gamma(\ob{\gonu{\chi\_l}})$$


by $(\*)$, hence $T\vdash\neg\Wit(\ob m,\ob{\gonu{\neg\phi}})$ as needed.


Since $T$ knows that the only numbers below $\ob n$ are $\ob0,\dots,\ob{n-1}$, we have established $T\vdash\Tr(\ob{\gonu\phi})$.


The second part of the Claim is similar: assuming $T\vdash\neg\phi$, we fix its standard proof with Gödel number $n$, and we show


$$T\vdash\Wit(\ob n,\ob{\gonu{\neg\phi}})$$


and


$$T\vdash\neg\Wit(\ob m,\ob{\gonu\phi})$$


for all $m\le n$, which implies $T\vdash\neg\Tr(\ob{\gonu\phi})$.


