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Is the polynomial time hierarchy regarded today as metamathematics? … This example therefore illustrates my point that there is really no coherent distinction into mathematics/metamathematics/meta-metamathematics. …

Set theory provides a good example. It is often convenient in set theory to work with the concept of "classes" and treat them as mathematical objects of their own kind. The standard axiomatization of …

Although your question is vague in certain ways, one robust answer to it is provided by the subject known as Reverse Mathematics. The nature of this answer is different from what you had suggested or …

You must read the charming essay lampooning this notation, while also giving a thorough logical analysis of it, by Adrian Mathias. Adrian Mathias, A Term of Length 4,523,659,424,929, Synthese 133 (20 …

The answer is yes, provided these theories are consistent. For example, PA proves that there is a number $n$, such that if there is no proof of a contradiction in PA of size at most $n$, then there is …

Cantor's Attic, which I founded with Victoria Gitman, is a compendium of information on the consistency strength hierarchy in set theory, spanning the range from ZFC (which in this context is viewed a …

Vitali famously constructed a set of reals that is not Lebesgue measurable by using the Axiom of Choice. Most people expect that it is not possible to carry out such a construction without the Axiom o …

The research area known as Reverse Mathematics is concerned with finding out the weakest theory that suffices to prove a given mathematical statement over a very weak base theory. The project has now …

The use of the Boolean-valued model approach to forcing allows one to avoid any consideration of countable transitive models in the proof that Con(ZFC) implies Con(ZFC+$\neg$CH), and gives in my opini …

There can be no such theory $T$, even if you weaken the requirement to $T$ being merely arithmetically definable, rather than insisting it must be effective. To see this, consider the theory T+TA, wh …

The assertion that there is (or is not) a truth predicate is expressible in the second-order language of set theory, but assuming consistency, not by any first-order assertion. Second-order. In the …

There are several subtle issues with your post. It is not in general possible to express the notion of "definable", because it leads to contradictions. For example, the class $D$ is not definable in …

In this answer, let me assume as you indicated in the comments that you are working in a second-order set theory with a truth-predicate for first-order truth. Such a theory goes strictly beyond ZFC in …

When one studies a set with structures that do not interact with each other, then really all that matters about the set is its cardinality. And so the situation of your question can be viewed as arisi …

If ZFC is consistent, then no, it does not prove the assertion "CH is not provable in ZFC", since the non-provability of any assertion in a theory implies the consistency of that theory, and so this w …