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Can PA+Con(ZFC) prove every theorem of ZFC?

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Obviously not. PA+Con(ZFC) is not even in the same language as ZFC. –  Andres Caicedo May 4 '12 at 22:04
PA+Con(ZFC) can't even state the axioms of ZFC, let alone prove every theorem of ZFC. –  Arturo Magidin May 4 '12 at 22:05
Cant the symbols of the language be put in bijection with some finite set of integers? –  user1708 May 4 '12 at 22:08
And all deduction rules and axioms would be, 'symbol i cant be followed by symbol j etc'. All deductions are then finite sequences of integers. –  user1708 May 4 '12 at 22:14

1 Answer 1

up vote 6 down vote accepted

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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Thanks, this clarified it. Does knowing ZFC proves a $\pi_0^1$ statement necessarily help us in proving the same statement in PA+Con(ZFC)? Is there a way to 'use' the Con(ZFC) statement in speeding up the proving? –  user1708 May 4 '12 at 22:31
It 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 $\forall x\psi(x)$. –  Henry May 4 '12 at 22:37

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