I have once been told that Schanuel's conjecture (which is a transcendence statement about tuples of complex numbers and their exponentials) is set theoretically absolute, which means that its truth (or falsity) is independent of the set theoretical model (or assumptions), because it is a $\Pi^1_2$ statement. Is this correct?

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    $\begingroup$ (The reference to $\Pi_2$ is imprecise. See my comments on one of the answers.) $\endgroup$ Jul 26, 2013 at 14:09
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    $\begingroup$ I edited the question, thanks. $\endgroup$
    – user48900
    Jul 26, 2013 at 14:32

2 Answers 2


You need to be a bit careful how you state it. Remember that Con(ZF) is a $\Pi_1^0$ statement, and that is independent of ZF.

I think what you're looking for is Shoenfield absoluteness. This implies that $\Pi_2^1$ statements are independent of certain axioms that can be added to ZF. Amongst these axioms are the axiom of choice and the generalised continuum hypothesis. Hence if you have a proof of Schanuel's conjecture that assumes choice and CH, then there must be a proof using only the axioms of ZF.

It's also the case that if a $\Pi_2^1$ statement holds, then it also holds in any transitive models of ZF.

  • $\begingroup$ Thanks for your answer. What about Solovay's model: if I have a proof of Schanuel in Solovay model, does it hold also in ZFC? $\endgroup$
    – user48900
    Jul 26, 2013 at 12:41
  • $\begingroup$ I don't really know about Solovay's model, but it looks like it's built as an inner model of a forcing extension, so I think everything should work there. $\endgroup$
    – aws
    Jul 26, 2013 at 13:07
  • $\begingroup$ You are mixing things here. $\mathrm{CON}(\mathsf{ZF})$ is $\Pi^0_1$, meaning it has the form $\forall n\phi(n)$, where $\phi$ is a recursive statement (and $n$ varies over natural numbers). On the other hand, a straightforward translation of the standard formulation of Schanuel's conjecture, is $\Pi^1_2$, meaning it has the form $\forall r\exists y\psi(r,y)$, where $r,y$ vary over real (or complex) numbers, and $\psi$ is an arithmetic statement (meaning, all its quantifiers vary over numbers). (Cont.) $\endgroup$ Jul 26, 2013 at 14:07
  • $\begingroup$ On the other hand, $\Pi_1$ and $\Pi_2$ mean yet something different from these two definitions, and Shoenfield's theorem does not apply to $\Pi_2$ statement. In fact, $\Pi_1$ statements are not absolute. $\endgroup$ Jul 26, 2013 at 14:08
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    $\begingroup$ Re the Solovay model thing, there is a different problem in this case: To build Solovay's model, you need resources beyond $\mathsf{ZFC}$, on consistency strength grounds. If Schanuel's conjecture is proved in Solovay's model, then it is true, but the proof may require the theory $\mathsf{ZFC}+$"There is an inaccessible cardinal". $\endgroup$ Jul 26, 2013 at 14:11

This is an extremely sloppy reference to the Shoenfield absoluteness theorem.

This theorem says that given an arithmetical statement about the natural numbers which is not "too complicated" (first-order statements are not too complicated, but not just them), then the statement is true in every two transitive models of $\sf ZFC$ which have the same ordinals.

This doesn't mean that the truth value is independent of $\sf ZFC$. It just means that we cannot prove independence using the two "usual" tools of set theory:

  1. Forcing, which does not change the ordinals of a model, nor destroy its transitivity; and
  2. Inner models, which are by definition a subclass of the model which is transitive and contains all the ordinals.

There might be other ways to prove independence, e.g. compactness arguments of some sort. But things like compactness usually destroy nice properties of models of $\sf ZFC$, and the absoluteness theorem no longer applies. On the other hand, we are less interested in not-nice models of $\sf ZFC$ in general, which is why set theorists focus on the two methods above, rather than other methods.

So the conclusion is that if we can force the truth value of the Schanuel conjecture, then we actually proved it.


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