Models and Inconsistency.

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.

-
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

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 $\varphi$.

-
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.

-
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