what follows from denying an axiom

you seem to be reasoning as follows:

If T proves A, then ~A proves ~con(T).

In the first place, this is not grammatical.
If we are talking about what T proves, then T is a THEORY.
~A is NOT a theory; it is just one sentence.
Since this was addressed to ME about how *I* seem to be
reasoning, let me invoke my state motto ("To be rather than to seem")
and say what I *factually* meant, which was that if T proves A then
T also proves ~A --> ~con(T)

where in your case, T = ZFC, A = the completeness theorem.


The cases can't possibly even matter.
What I am arguing is true all the time, or at least all the times
when "con(T)" ACTUALLY MEANS "T is consistent".

But that conclusion is *not* valid.

The conclusion that "~A proves" something is certainly not
valid, for grammatical reasons alone. WhatEVER I was trying
to say, it COULD NOT POSSIBLY have been THAT.

To see this, let T_0 be PA, and let G be the Godel
sentence for PA. Let T be T_0 + ~G.

Then T is consistent, and T proves ~G,

Exactly. Please note that your "+" here is adding just
more than ~G to the theory: a theory is a set that is closed
under consequence, so what your "+" actually means is
that you are adding "~G" AS AN AXIOM (to the axioms of T_0,
to get the axioms of T) and adding a lot of ITS consequences to T_0
to get T.
Among these consequences are ABSOLUTELY ALL conditionals whose
hypothesis is G.

but the statement

G -> ~con(T)

is not true.

It IS, however, PROVABLE IN T, which means IT IS true
IN ALL MODELS OF THE AXIOMS OF T.
This is ALL the "truth" that I personally HAVE EVER been on about.
It risks confusion to say "just plain" true ["simpliciter"] in this
context.
It is WORTH the trouble to BOTHER saying "true in the standard
model",
despite the fact that it is less than clear what the standard model of
ZFC is
(is the GCH true or false in it? what about large-cardinal axioms?).


As a matter of fact, the statement

G -> con(T)

is provable (within T, or within PA).

G -> whatEVERthe*** is provable in *T*, since ~G is an axiom of T.
In particular, ALL three of
G --> con(T),
G --> ~con(T), and
G --> ( (con(T)/\~con(T) ) are provable in T.

So G --> "T is not consistent" is a reasonable translation.
But that statement is merely provable in T; it is not necessarily
"true",
if "true truth" is what you are looking for. My point, though, is
simply
that where I come from, WE DON'T DO false axioms. The PURPOSE of the
axioms is to LIMIT the investigation to models in which they are true.

And while T proves this, it would of course still be the case the
G (as one sentence) fails to prove it, except by relying on T.

.

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