Proof through consistency

Take first-order Peano Arithmetic PA. We know that Gentzen proved PA consistent. Now, if one sets for example $\varphi$ to represent Fermat's theorem in FO, would proving PA+$\varphi$ consistent be accepted as a proof for $\varphi$ to "hold"?

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Consistency of $PA+\phi$ is not a proof of $\phi$. Consider axioms of euclidean geometry with $X$ meaning the first four and $\phi$ denoting the fith. We know that $X+\phi$ is consistent, i.e. there exists a model where all the five axioms are true. However, there are other models where $X$ is true, but $\phi$ is not.

Similarly if you would take $\phi_i$ to denote $x = i$ then of course there is some model of $PA+\phi_2$, but this does not force $\phi_2$, actually the number $2$ isn't special here.

On the other hand if you would take $\phi$ to denote $(a+b)\cdot c = a \cdot c + b\cdot c$ then there will be model of $PA+\phi$, but you won't find any model of $PA+\neg \phi$. Still, it doesn't follow from consistency of $PA+\phi$, but directly from $PA$.

Edit: To answer the comment, the key part is whether there exists any models of $PA+\neg\phi$. If they do, then $\phi$ cannot be regarded as true for all models of PA. You can build math on it, but what you build will be true only in those models where $\phi$ is true, that is not all, and so your new theory will not be true for all models of PA as well.

In fact, to prove $\phi$ it is enough to show that $PA+\neg\phi$ is not consistent. But I think you already know this technique: this the well known proof by contradiction. To give even more intuition: to prove something it is not enough to show an example, but it is sufficient to show that no counter-examples exists.

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@rank See the first edit, comment was too short. –  dtldarek Apr 20 '12 at 9:49
@rank You are right, "universally true" is an unfortunate term, I changed it to "true in all models of PA". –  dtldarek Apr 20 '12 at 10:42