set theory - Definition of ZFC2 (second order logic) - Mathematics ...
Nov 28, 2021 · ZFC with 1st order logic is known. My question may be formulated either way. What is the definition of ZFC2 (ZFC+second order logic)? (I assume that formulas also have quantifications of two kind over object and over predicates. Then what are some examples of statements in ZFC2 that cannot be proved in ZFC1?)
Why is ZFC recursively axiomatizable? - Mathematics Stack Exchange
Oct 16, 2018 · If it is, $\theta$ is a ZFC axiom. Otherwise, check if $\theta$ is a Replacement instance. If it is, then $\theta$ is a ZFC axiom. Otherwise, we finally conclude that $\theta$ is not a ZFC axiom. Now the above is very specific; various related sets are not recursive, namely: The set of sentences which are equivalent (over ZFC) to one of the ...
logic - Why did we settle for ZFC? - Mathematics Stack Exchange
Jan 10, 2018 · $\begingroup$ ZFC is named after Zermelo, but it is not like he went up to the mountain alone and came back with stone tablets that have been used unchanged ever since. He was one link in a multiple-person effort to construct a theory that (a) didn't allow the paradoxes to be derived, and (b) allowed all of ordinary mathematics to be formalized ...
What would happen if ZFC were found to be inconsistent?
IMHO, most likely — if ZFC turns out to be inconsistent — it would be possible to slightly restrict the replacement axiom and get a “consistent” axiom system in which all mathematics can be formalized (of course, one cannot be sure that the new axiom system is consistent). $\endgroup$
ZFC + "There exists an inaccessible cardinal" proves Con (ZFC)
This relies exactly on looking at the way we're defining ZFC in the object language. (Note that I said "axiom of ZFC," not "theorem of ZFC.") We can indeed whip up a code for a theory ZFC' such that ZFC=ZFC' under reasonable assumptions but ZFC can't prove "$\forall x,y(x=y)$ is not a ZFC'-axiom." $\endgroup$ –
ZFC and axiom of power set - Mathematics Stack Exchange
Jul 7, 2019 · The independence of the axiom of powerset from the remaining axioms can be shown by constructing a model of $\mathsf{ZFC}^-$, that is $\mathsf{ZFC}$ without the axiom of powerset, which is not also a model of $\mathsf{ZFC}$.
ZFC replacement schema in FOL - Mathematics Stack Exchange
Sep 19, 2024 · The axioms aren’t like a “recipe” for “constructing” a domain of sets. Rather, they’re like a set of “requirements” that any model of ZFC set theory must satisfy. I can take $\mathfrak{M}$, a model of ZFC, and add more sets to it (e.g. by forcing extension) to yield a new, bigger model $\mathfrak{M}_{2}$ which is also a model of ...
Are category-theory and set-theory on the equal foundational …
That said, choice of foundations actually matters here. The category of (ZFC) sets does not exist in ZFC. Typically, to formulate category theory as used into something like ZFC requires axiomatically adding inaccessible cardinals or even Grothendieck universes. On the other hand, the way category theory is typically used already assumes set ...
Basic question about encoding ZFC into PA
However this is formalized, $\mathsf{ZFC}$ should be more than enough to carry out this assignment. It follows that, since $\mathsf{ZFC}$ proves the consistency of $\mathsf{PA}+A$, then the latter cannot interpret $\mathsf{ZFC}$: Otherwise, $\mathsf{ZFC}$ would prove its own consistency, against the incompleteness theorem.
What is wrong with ZFC? - Mathematics Stack Exchange
Sep 7, 2015 · Apparently ZFC (let's assume with Universes) is a headache for category theorists. Something about ZFC makes category theory harder than it would be in, say, homotopy type theoretic foundations. So far all I've seen is hand waving - can anyone give an example of where ZFC(+U) truly makes life so much more difficult for category theorists?