# Reflection principles

Let con(ZFC) be a sentence in ZFC asserting that ZFC has an omega-model M. Let $A_{M}$ be an wff over M. Let S be the theory ZFC+con(ZFC). Is the reflection for S: $Bew_{S}(A_{M}) \implies A_{M}$ is satisfied? I asking also for an explanation of the paradox in the link

http://cs.nyu.edu/pipermail/fom/2007-October/012035.html

of the case when ZFC is replaced on S=ZFC+(ZFC has omega-model)?

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Could you explain what does $Bew_S(A_M)$ mean? Also, perhaps you could re-word your final question somehow; I don't really understand it as it is written. – Joel David Hamkins Mar 6 '13 at 22:20
I don't know what Bew_S(A) means here. Are you asking for an explanation of the paradox in the link you mention? – Joel David Hamkins Mar 6 '13 at 22:35
Joel, I think Bew_S(A_M) is supposed to be (a formalization of) the statement that A_M is provable in S. ("Bew" was, I believe, used by Gödel to abbreviate "beweisbar".) – Andreas Blass Mar 6 '13 at 22:41
Of course Bew_S(X)--->X is true for any X, because all the axioms of S are true. But that argument uses information that goes beyond ZFC, so presumably the question should be whether Bew_S(A_M)--->A_M is provable in some (yet to be specified) formal system. It should also be explained what is meant by a wff being "over M" and in particular why such a wff is in the language of S so that Bew_S(A_M) makes sense. – Andreas Blass Mar 6 '13 at 22:45

I suppose the "paradox" you're asking about is the passage marked with >> at the link you gave, but with "$\omega$-model" in place of "model" and with "has an $\omega$-model" in place of "is consistent". But then there is no longer any justification for the statement (on lines 9 & 10) that there's a proof in ZFC of the negation of con(ZFC) (which now becomes the negation of "ZFC has an $\omega$-model"). What you have is rather that this negation holds in all $\omega$-models of ZFC, but that doesn't immediately translate into a syntactic fact about existence of a proof, which you could then translate into English.