[FOM] Finite axiomatizations extending ZFC

Robert M. Solovay solovay at Math.Berkeley.EDU
Fri May 9 04:11:38 EDT 2008

There is no such example. (I believe this result is due to Richard 
Montague.) Here is a sketch of the proof.

Let Phi be an axiom (in the language of ZFC) such that Phi implies (in 
first order logic) all the axioms of ZFC. Then not Phi is logically valid.

Proof: By the reflection theorem, ZFC proves "Phi implies Con(Phi)" for 
any sentence Phi. It follows that our particular Phi has Con(Phi) as a 
consequence. Godel's incompleteness easily applies to Phi since Phi has 
all the axioms of ZFC as a consequence. So Phi is inconsistent.

Precisely this argument appears in Kunen's text on set theory as 
Corrollary IV 7. 7 on page 138.

 	--Bob Solovay

On Thu, 8 May 2008, pax0 at seznam.cz wrote:

> Can someone give an example of a *finite* set of sentences T (in the language { \epsilon })
> with consequences (necessarily strictly) containing all instances of axioms of ZFC ?
> No schemas are allowed, and nothing like 0=1, of course.
> The lesser the total number of symbols in T is, the better.
> Thank you, Jan
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