Finite fragments of ZFC This is a question regarding exercise (II.2.15) in Kunen's Set Theory (2011):
The exercise reads:
"We shall see later that $ZFC\vdash $Con$(\Gamma)$ whenever $\Gamma$ is a finite subset of $ZFC$. Use this to define explicitly, in $ZFC$, a binary relation $E$ on $\omega$ such that $ZFC\vdash\varphi^{\omega,E}$ for each axiom $\varphi$ of $ZFC$".
My understanding is that I am proving a scheme in the metatheory. ie given an axiom $\varphi$ of $ZFC$, I want to prove that $ZFC\vdash[(\omega;E)\models\varphi$]. 
I understand that this does not necessarily imply $ZFC\vdash[(\omega;E)\models ZFC]$ since $(\omega;E)\models ZFC$ is a sentence of $ZFC$ and if this were the case then we would have $ZFC\vdash$Con$(ZFC)$.
Is my understanding correct thus far?
Now the hint reads:
"List the axioms of $ZFC$ in some computable way, as {$\xi_i:i<\omega$}; let $ZFC_n=${$\xi_i:i<\omega$}. Working in $ZFC$, we can define $\Gamma$ to be $ZFC$ if $ $Con$(ZFC)$; if $\neg $Con$(ZFC)$, let $\Gamma=ZFC_n$ where $n$ is largest such that $ $Con$(ZFC_n)$. Then clearly $ $Con$(\Gamma)$, and observe that the proof of the Completeness Theorem yields an explicit $E$ such that $(\omega;E)\models \Gamma$".
Now my understanding is that for each axiom $\varphi$ of $ZFC$ I want to show:
1) $ZFC+$Con$(ZFC)\vdash[(\omega;E)\models \varphi]$
2) $ZFC+\neg $Con$(ZFC)\vdash[(\omega;E)\models \varphi]$
and after showing 1) and 2), it will then follow that $ZFC\vdash[(\omega;E)\models\varphi$].
1) seems straightforward, since arguing in $ZFC+ $Con$(ZFC)$, I get that $ZFC$ has a model by the Completeness theorem, and so I can get a countable structure $(\omega,E)$ to model $ZFC$ with a bit of work and by using the proof of the Completeness Theorem. So since I've shown that $ZFC+$Con$(ZFC)\vdash[(\omega;E)\models ZFC]$, then it follows that $ZFC+$Con$(ZFC)\vdash[(\omega;E)\models \varphi]$ for each axiom $\varphi$ of $ZFC$. 
For 2), I can argue in $ZFC+\neg $Con$(ZFC)$ that some finite fragment of $ZFC$ must be inconsistent by Compactness. So then I get:
$ZFC+\neg $Con$(ZFC)\vdash\exists n\in\omega [$Con$(ZFC_n)\wedge \forall m\in\omega[m>n \implies \neg $Con$(ZFC_m)]]$.
So $ZFC+\neg $Con$(ZFC)\vdash $Con$(ZFC_i)\wedge \forall m\in\omega[m>i \implies \neg $Con$(ZFC_m)]]$ for some $i<\omega$ 
Now I want to use the fact that $ZFC\vdash$Con$(\Gamma)$ whenever $\Gamma$ is a finite subset of $ZFC$, to say that $ZFC\vdash$Con$(ZFC_{i+1})$, which means my reasoning is definitely flawed.
Assuming I am approaching this exercise correctly, how do I go about showing 2)?
I apologize for the long length in explaining my question.
Any help is appreciated, thanks.
 A: We want to define $E$ over $\mathrm{ZFC}$ such that $\mathrm{ZFC}\vdash\varphi^{\omega,E}$ for every $\varphi\in\mathrm{ZFC}$.  If $\mathrm{ZFC}$ is inconsistent in the real world, then it proves everything and we are done.  So we may assume $\mathrm{ZFC}$ is consistent in the real world.  I think this is the key point of the argument.
Let $\varphi\in\mathrm{ZFC}$.  To show that $\mathrm{ZFC}\vdash\varphi^{\omega,E}$, we take an arbitrary model $V\models\mathrm{ZFC}$, and attempt to prove $V\models\varphi^{\omega,E}$.  If $V\models\mathrm{Con}(\mathrm{ZFC})$, then we are in the ‘straightforward’ case as you observed.  So we may assume $V\models\neg\mathrm{Con}(\mathrm{ZFC})$.
Now $V\models\mathrm{ZFC}+\neg\mathrm{Con}(\mathrm{ZFC})$.  Following Kunen's hint, let $n$ be the largest $n\in\omega^V$ such that $V\models\mathrm{Con}(\mathrm{ZFC}_n)$.  This $n$ must be bigger than all natural numbers $i$ in the real world because of what Kunen says at the beginning.  As a result, if we apply the completeness theorem in $V$ to find a (countable) model $M$ of $\mathrm{ZFC}_n$ in $V$, then $M\models\mathrm{ZFC}_i$ for every natural number $i$ in the real world, which is the same as saying $M\models\mathrm{ZFC}$ in the real world.  Notice $V\not\models(M\models\mathrm{ZFC})$ in this case.
Sorry for referring to the ‘real world’ so many times.  I cannot find a better phrase for it.  Suggestions are welcome.
