Can PA+Con(ZFC) prove every theorem of ZFC?
Can PA+Con(ZFC), prove everything ZFC can?
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set-theory
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0And all deduction rules and axioms would be, 'symbol i cant be followed by symbol j etc'. All deductions are then finite sequences of integers. – 2012-05-04
1 Answers
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No, since, as others have pointed out, ZFC is about sets, while PA is about integers. The language of ZFC can be translated into arithmetic, in the manner you suggest, but then PA is proving things about provability in ZFC, not proving the truth of the theorems themselves. For instance, if $ZFC\vdash \phi$ then already PA by itself (without Con(ZFC)), and even weaker theories, can prove the formula representing "$ZFC\vdash\phi$".
However it is true that if ZFC prove a $\Pi^0_1$ statement about natural numbers then PA+Con(ZFC) proves the same statement.
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2It gives us a proof directly: suppose ZFC proves $\forall x\psi(x)$. Then $PA$ proves $"ZFC\vdash\forall x\psi(x)"$. $PA$ also proves that $\exists x\neg\psi(x)\rightarrow "ZFC\vdash\exists x\neg\psi(x)"$. So $PA$ proves $\exists x\neg\psi(x)\rightarrow\neg Con(ZFC)$, and correspondingly $PA+Con(ZFC)$ proves $\forall x\psi(x)$. – 2012-05-04