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I’m trying to show that a first-order theory $ T $ is inconsistent if and only if $ T \vdash \varphi $ for every w.f.f. $ \varphi $.

I understand that there might be a need to use the axioms for $ \sf PA $, but I’m not sure if it will be too helpful here.

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Why do you think that you might need to use the PA axioms? Please, take the time to write better questions. – Carl Mummert Mar 12 '13 at 11:27
up vote 2 down vote accepted

This is the converse of Samuel’s answer.

Let $ T $ be a first-order theory in the language $ \mathcal{L} $.

  • Suppose that $ T $ is inconsistent.

  • By definition, $ T \vdash (\phi \land \neg \phi) $ for some $ \mathcal{L} $-formula $ \phi $.

  • For all $ \mathcal{L} $-formulas $ \varphi $, we have the tautology $ ((\phi \land \neg \phi) \to \varphi) $.

  • It follows that $ T \vdash ((\phi \land \neg \phi) \to \varphi) $ for all $ \mathcal{L} $-formulas $ \varphi $.

  • Therefore, by modus ponens, $ T \vdash \varphi $ for all $ \mathcal{L} $-formulas $ \varphi $.

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So you showed the reverse direction, and Samuel showed the forward direction? Still trying to understand the proof – Buddy Holly Mar 12 '13 at 5:30
@Buddy Holly: Yes, I proved the reverse direction. :) – Haskell Curry Mar 12 '13 at 5:33
OK, Thanks for your help – Buddy Holly Mar 12 '13 at 5:35

Since you are quantifying over all well-formed formulas in your language then surely you can have a substitution instance for $\varphi := A$ and $\varphi := \neg A$ and by your assumption $T \vdash \varphi$, $\forall \varphi$ and so $T \vdash A$ and $T \vdash \neg A$ for some $A$. Therefore, $T$ is inconsistent.

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Is this the idea of Russell's Paradox, in analogy? – Buddy Holly Mar 12 '13 at 5:28
@BuddyHolly: Russell's Paradox is a set-theoretic paradox which states that the set of all sets is a set which does not contain itself. That is a distinct idea from a model being inconsistent if it satisfies a statement and the negation of that statement. – Samuel Reid Mar 12 '13 at 5:31

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