# Peano and consistency, how to understand it rightly.

I'm struggling with the notion of consistency, and a few cases :

I'm writing in the following $Con(T)$ to denote the arithmetic formula which expresses the consistency of $T$, for $T$ a consistent and recursive theory extending Peano (denoted $PA$).

Firstly, note that $PA$ isn't a complete theory, since Gödel's first incompleteness theorem states that a recursive extension of $Rob.$ cannot be both complete and consistent.

Now, my problem arises when I'm asking myself if this does hold : $$PA \cup \{Con(PA) \} \vdash \neg Con(PA)$$

And how about $$PA \cup \{\neg Con(PA) \} \vdash Con(PA)$$

Here are my thoughts : I know that from second incompleteness theorem, $PA \nvdash Con(PA)$, so it means that there is at least a model of $PA$ in which $\neg Con(PA)$ is true, for otherwise, it would be false in all models and so $Con(PA)$ would be true in the theory, which it cannot be.

And I would like to say that since the standard model is such that $PA$ is consistent, it is a model of $PA \cup \{Con(PA)\}$ too and it cannot proves $\neg Con(PA)$ otherwise, it would be inconsistent and so it may prove anything, including its own consistency, a blatant violation of Gödel's second theorem.

Hence to answer my first question, it doesn't hold.

Next, for my second problem, I know from above that there is a model of $PA$ in which there is $\neg Con(PA)$, so in this model, if I were to proves $Con(PA)$, I would have inconsistency, hence be able to prove anything, including the consistency of $PA \cup \{\neg Con(PA) \}$, which violates again the second theorem.

So answer to both my questions would be "No". Is that right?

On another note : is there any inconsistent and recursive theory extending Peano? I'd say that if there were one it would have to be complete by Gödel's first theorem, so I may just pick an axiom of Peano and add its negation to Peano and get such a monster, couldn't I? But now, if I were to consider a decidable and complete extension of Peano, I couldn't possibly have a recursive extension, but how should I build such an extension, how can I get Peano to "become" decidable?

Any other useful consideration to be sure to understand Gödel's theorem and implications is welcome.

For the bonus question: Yes, there are plenty of inconsistent recursive theories that extend PA. One simple example would be PA${}\cup\{1=0\}$.
• Wait. If it is no contradiction, then I don't have any argument about why it cannot prove $\neg Con(PA)$... So my arguments are empty. What am I missing? – Lery Jun 5 '15 at 22:21