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This is related to my previous question. The thing was to show :

$ZF \vdash \neg Con(ZF + AF) \longrightarrow \neg Con(ZF)$

If $ZF + AF$ is inconsistant then there is a finite number of $ZF + AF$ axioms, $\psi_1,\dots,\psi_n$ such as $\psi_1 \wedge \dots \wedge \psi_n \longrightarrow \phi \wedge \neg \phi$

We know that we have a model $WF$ such as for all $ZF+AF$ axioms, $F$, we can write

$ZF \vdash F^{WF}$

So $ZF \vdash \psi_1^{WF} \wedge \dots \wedge \psi_n^{WF}$

So $ZF \vdash \phi^{WF} \wedge \neg \phi^{WF}$

So $ZF \vdash \neg Con(ZF)$

My problem is that I wonder if somehow, we didn't do something weaker than $ZF \vdash \neg Con(ZF + AF) \longrightarrow \neg Con(ZF)$.

Indeed, if $ZF \vdash \phi \wedge \neg \phi$ then $ZF \vdash \neg Con(ZF)$ (in fact even $Peano \vdash \neg Con(ZF)$).

But why $ZF \vdash \neg Con(ZF+AF)$ would imply that $ZF+AF \vdash \phi \wedge \neg \phi$ ?

Couldn't the demonstration be non standard ?

Thanks in advance to anyone who has some clue...

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I am not sure I understand the question. The way you phrase it is a bit confusing. Nevertheless this is the answer to what I think that you are asking:

It can be proved that if we have a recursively enumerable set of natural numbers $X$ then there exists a formula $\phi$ of the language of Peano arithmetic such that:

$PA\vdash\phi(\bar{m}) \iff m\in X$

This is essentially what is used to prove the first incompleteness theorem since the set of the consequences of PA is recursively enumerable (here I used $\bar{m}$ to denote the element $m$ in the language of PA).

Of course we can prove something similar in a theory stronger than PA like ZF or ZF+AF (or any other theory that extends PA as long as it's axiomatic). For such a theory (let's call it $T$) since its consequences are recursively enumerable we have that:

$T\vdash\phi\iff T\vdash Pr(\ulcorner\phi\urcorner)$

If we can prove something in ZF we can of course prove it in ZF+AF. Since the consistency is a statement about the provability of a contradiction it is immediate that the contradiction is provable.

What this says essentially is that we cannot prove the existence of a proof of a contradiction without being able to prove the contradiction. Or we can say that a non-standard demonstration is not demonstratable.

P.S.: If I misinterpreted the question please let me know what it is you are looking for.

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    Yes, if fact, 'what you say (existence of a non standard proof implies the existence of another standard proof)' + 'the normal proof we can see of relative consistency' would show that $ZF \vdash \dots$. Unfortunatly I cannot see that anywhere (and I looked at A LOT of books). Everybody skip that like it's trivial... But I guess this is another question, I ll think about it and if I don't find anything during the week, I ll ask it here. Anyway, thanks a lot for all your answers and your time ! You helped me a lot.2011-03-24