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How to show that the Peano Arithmetic theory is not scapegoat?

Note:

  • Peano Arithmetic is a consistent theory.

  • A theory T is scapegoat if for every formula $A$ with only one free variable there exist a closed term $s$ such that $T$ proves: $$(\exists x \; (\neg A(x)) )\Rightarrow \neg A(s)$$

("$\neg$" means "not"). I think we can start by assuming that the theory is scapegoat then we get a contradiction, but how!?

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    @Sandra: Is it supposed to be $(\exists x\neg A(x))\Rightarrow (\forall x\neg A(x))$, or $\exists x(\neg A(x)\Rightarrow \neg A(x))$? The latter seems a bit silly, so I suspect it's the former...2011-03-09
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    Perhaps you meant for the closed term s to appear in the specified theorem of T? As in "not A(s)"?2011-03-09
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    @ Answer#1: What about the theory itself??2011-03-09
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    @ Answer#2: I didn't get the idea! Could you be more clear!2011-03-09
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    How is the negation relevant, especially since we are working in Peano Arithmetic?2011-03-09

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