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:P), merely negating "fundamental" axioms does not yield strong in-system consequences. …

No. Take for instance a sequence of two-element sets $(A_i)_{i\in\omega}$ such that every infinite subsequence has empty product ("Russellian socks"), and let $B_i=A_i\sqcup\{\{i\}\}$. Then the produc …

Your ordinal $\beta_\mathcal{L}$ is perfectly well-defined: in my opinion it's more easily thought of as $$\sup\{\alpha: \forall \beta<\alpha(V_\beta\not\equiv V_\alpha)\},$$ and this definition shoul …

EDIT: I've rewritten for clarity. First, re: your claim "It appears that what Boolos is saying is that: when we extend the rough iterative conception of set with a ranking function, then we get Rep …

Unless I'm missing something, this scheme is inconsistent. Take $$F(x)=\bigcup_{y\in x}(\mathcal{P}(y)).$$ Then for all $X$ with more than one element and any $a$ with $a<X$ we have $X$ ACC$^F$ $a$: …

The following is an easy recipe for building a model of ZFA+AC with one atom, from a model $M$ of ZFC (it's trivial to modify this construction to include arbitrarily large sets of atoms, and even a p …

Certainly if $\{x\in\mathbb{N}: P(x)\}$ is incomputable, then - for any computable and true set of axioms $T$ - there will be some (in fact, infinitely many) $n$ such that $T$ neither proves nor disproves … Diophantine equations with zeroes - is incomputable, since otherwise we could form a silly set of axioms which is computable and decides each instance of the problem! …

A stark demonstration of why precisely defining how you form $PA_{\lambda +1}$ for $\lambda$ a limit ordinal: in 1939 Turing showed that if $\varphi$ is a true $\Pi^0_1$ statement, there is a notation …