TITLE: What's wrong with this proof of ZF being inconsistent?
QUESTION [6 upvotes]: I'm studying (independently) mathematical logic and in investigating
self-referential statements I developed a result which I don't know
how to interpret.  I'll use the notation from Enderton's "A
mathematical introduction to logic" (p 266-267):
Let $T$ be any sufficiently strong recursively axiomizable theory (PA
or ZF will do) and let $\sf Prb_T \sigma$ (Here $\sf Prb_T$
abbreviates "provable in $T$") express that $\sigma$ has a deductive
proof in $T$. (i.e. $\sf Prb_T \sigma$ means $T \vdash \sigma$).  Let
$\sf Cons T$ be a sentence expressing that $T$ is consistent (such as
$\neg \sf Prb_T (0=1))$.  Then using the fixed-point lemma we can get a
sentence $\sigma$ such that:
$$T \vdash (\sigma \leftrightarrow \sf Prb_T (\sf Cons T \rightarrow
\neg \sf Prb_T \sigma))$$
Now  either $T \vdash \sigma$ or $T \nvdash \sigma$.  I will show that
we cannot have either assuming $T$ is consistent.
If $T \vdash \sigma$ then by our definition of $\sigma$ we have that
$T \vdash \sf Prb_T (\sf Cons T \rightarrow \neg \sf Prb_T \sigma)$.  This implies that $T \vdash (\sf Cons T \rightarrow \neg \sf Prb_T \sigma)$
But this means that we have a proof of $T$ being consistent implies $T \nvdash \sigma$ which
contradicts this initial hypothesis of $T \vdash \sigma$. Therefore,
assuming $T$ is consistent, we must have $T \nvdash \sigma$.  But this
too also leads to a contradiction, for by our construction of $\sigma$
we now have $T \nvdash \sf Prb_T (\sf Cons T \rightarrow \neg \sf
Prb_T \sigma)$.  This means that there is no proof of $T$ being
consistent implying that $T \nvdash \sigma$.  However, we just
provided a proof (albeit meta-mathematical in nature) testifying to
this fact.
So what should I logically conclude from this?  I figure that this
implies that $T$ is inconsistent.  I guess there is another
possibility:  That this whole result/proof cannot be formalized thus
removing the contradiction (or maybe not because then we might have
a contradiction with the Completeness Theorem for first-order logic).  I know that by Godel's Second Incompleteness Theorem assuming $T$ is consistent then $T \nvdash \sf Cons T$.  However $T$ can still prove some facts about implications involving $\sf Cons T$ (In fact, Enderton gives one such example on the bottom of page 267).  My understanding is that most meta-mathematical results such
as Godel's incompleteness theorem and Tarski's undefinability theorem
can be represented and proven in the deductive calculus. Am I wrong on
this?  If I'm not then what makes this result any different?

REPLY [6 votes]: The problem with the argument, indeed with most of the claims of this sort is that the internalization of logic and proofs are only as good as your standard integers, but become uncontrollable when we move forward from them.
What do I mean by that? We know that every real, meta-theoretic, proof corresponds to a specific natural number. And we know that $T$ itself has a predicate that is interpreted as exactly the codes for the axioms of $T$.
But this only happens in models where the integers are standard. If you have non-standard integers, you invariably add new axioms to $T$ and you add new inference rules to your logic, and so you added new ways of proving that $0=1$.
So the way to reconcile this is to notice that indeed a model of $\sf ZFC+\lnot\operatorname{Con}(ZFC)$ must disagree with its meta-theory about the integers (namely, it must have non-standard integers which will witness this).