Is ZFC+Con(ZFC) powerful enough to show there isn't any standard model of ZFC? What you think about it?



Standard model of  ZFC
Many set theorists define that a model $\langle M,E\rangle$ of set
theory is *standard* to mean that the set membership relation $E$
of the model is the actual set membership relation $\in$,
restricted to objects in $M$.


The existence of such a standard model of ZFC is equivalent to the
existence of a well-founded model of ZFC, since every well-founded
model is isomorphic to its transitive collapse, via the Mostowski
collapse. The hypothesis that there is such a transitive model of
ZFC should be considered a very weak large cardinal axiom. For
example, the [transitive model of ZFC](http://cantorsattic.info/Transitive_ZFC_model) hypothesis appears near the bottom of the
large cardinal hierarchy in [Cantor's Attic](http://cantorsattic.info).


In particular, nearly all the usual large cardinal axioms imply
the existence of a standard model of ZFC, and so very few set
theorists want or expect ZFC to rule them out. Since we think that
large cardinals are consistent with ZFC, we also expect that it is
consistent with ZFC that there are standard models of ZFC. 


For example, if $\kappa$ is an inaccessible cardinal, then the set $H\_\kappa$ of all sets having hereditary size less than $\kappa$ forms a transitive model of ZFC. Under the Grothendieck Universe Axiom, there will be a proper class of such inaccessible cardinals, and so the entire set-theoretic universe will be the union of such standard set models of ZFC.


Indeed, I would say much more: it is a fundamental part of the
reflection paradigm, the view that truths in the whole
set-theoretic universe are increasingly found in proper initial
segments of it, that we should expect many initial segments of the
universe to be ZFC models, and these will be standard models. This
is the kind of philosophical justification that one often hears
for various large cardinal hypotheses.


Meanwhile, it is easy to see that if ZFC is consistent, then it is consistent with ZFC that there is no standard model of ZFC. To prove this, take any model of ZFC. If it has no transitive models of ZFC, then we're done. If it does have a transitive model, then it will have an $\in$-minimal transitive model of ZFC, and this will be a model of ZFC having no transitive models of ZFC.


Furthermore, since every transitive model of ZFC has exactly the same arithmetic truths as the ambient universe, it follows that if there is a transitive model of ZFC, then there is one having no transitive model of ZFC, in which Con(ZFC) still holds. In particular, ZFC+Con(ZFC), if consistent, cannot prove the existence of a standard model of ZFC. One may iterate this to the [consistency hierarchy](http://cantorsattic.info/Transitive_ZFC_model#Consistency_hierarchy) of models of ZFC. Basically, no amount of iterating the Con operator will get you to the existence of a standard model of ZFC, and this seems relevant for your revised question.


Finally, let me point out that if ZFC is inconsistent, then of course it is strong
enough to prove any assertion, including the one you ask about,
and so we cannot say for sure that ZFC doesn't prove your 
assertion without going beyond ZFC in our axioms.


