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What does it mean to be a "standard" theory? By any account, the theory ZFC + V=HOD already is one of the "standard" theories. The axiom V=HOD is intensely studied by set theorists; it appears as a hy …

In my blog post Transfinite recursion as a fundamental principle in set theory, I prove that the principle of transfinite recursion is equivalent to the replacement axiom.

No. Consider the model $\langle V_{\omega_1},\in\rangle$. This is a model of Zermelo's theory, but all ordinals in it are countable. Thus, it satisfies the ordinal-replacement axiom, since the image o …

If I understand you correctly, the answer is yes, providing that you don't just add those axioms directly to $T$, but instead add the assertion $\phi^D$ for each axiom of $T$. …

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 …

If you make your requirement only for regular cardinals $\kappa$, then we can easily get an equiconsistency. Theorem. The following theories are equiconsistent over ZFC: There are unboundedly many …

Your transfer principle contradicts the axiom of foundation. To see this, observe that under foundation, every nonempty finite set $x$ has an $\in$-maximal element, a set $z\in x$ with $z\notin u$ fo …

I think that you haven't quite asked the question you intended to ask, since $L$ is not a set, but rather a proper class, and of course by the incompleteness theorem no subtheory of ZFC, if consistent …

Now, by the usual Zermelo axioms, we can project onto the second coordinate (this does not need replacement), and get the desired instance of replacement. Update. …

These axioms thus have independent interest, and yet fulfill the predicted tower of consistency strength. …

In weak set theories, using classical logic and interpreting "small subclass" as "set", this principle amounts to an alternative formulation of the collection axiom. For example, in Zermelo set theory …

We require only that $T$ has a computably enumerable list of axioms, that $T$ is strong enough to express the decision problem $A$, and that $T$ is sufficiently sound. … This argument works even as you strengthen your axioms, and so it doesn't matter whether you use PA or ZFC or ZFC plus large cardinals or what have you. …

Assuming that ZFC is consistent, then your system will not settle the question of whether 1cof is an element of 2cof or vice versa. One can see this by observing that from any model of ZFC we can make …

There is an ambiguity in this kind of question, where you refer to ZFC-A for an axiom A, since what the theory is that you refer to depends on the particular way that you axiomatize ZFC. For example, …

The ZFC axioms of set theory are first order, but have a second-order analogue ZFC2 which is obtained in much the same way as your move from PA to second-order PA, namely, we replace the first-order schemes …