Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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"?

share|improve this question
add comment

1 Answer 1

up vote 3 down vote accepted

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.

share|improve this answer
    
So, to interpret your answer, you mean that if a statement $\varphi$ is true only in some non-standard models of PA (and possibly on some other models as well), then it wouldn't make sense to take $\varphi$ as a "true" mathematical statement in the sense that one could "build math on it"? –  rank Apr 20 '12 at 9:32
    
@rank See the first edit, comment was too short. –  dtldarek Apr 20 '12 at 9:49
    
Thank you for clarifying. But I'm afraid I'm still in doubt if I get this correctly. With "universally true", do you refer to a valid formula? (If so, I'm not sure if I understand how that is connected to the question.) –  rank Apr 20 '12 at 10:35
    
@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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.