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Although the other answers point out correctly that the exact value of $\text{BB}(n)$ is independent of ZF for large enough and even moderately sized values of $n$, nevertheless I should like to point …

Peter Koepke and Martin Koerwien developed the theory of sets of ordinals as a foundation of mathematics, showing senses in which it is equivalent to ZFC as a foundation. Peter Koeopke and Martin Koe …

It remains open whether the won-position problem of infinite chess is decidable, the problem of determining whether a given finite position in infinite chess is winning for white or not. See Richard S …

The quotient of one language $L$ by another $R$ is the set of strings $x$ such that $xy\in L$ for some $y\in R$. If both $L$ and $R$ are computably enumerable (what you call RE), then the quotient i …

Allow me to mention that since the players in effect adopt the roles of the quantifiers $\forall$ and $\exists$, as Bob has a winning strategy just in case for every move for Alice, there is a reply b …

The answer is yes. Suppose a function $g$ is computable by a procedure $p$ whose computation running time is bounded by a function $h\in\newcommand\PR{\text{PR}}\PR(f)$. I claim that $g\in\PR(f)$. T …

The ordinals of the form $\omega_D^{CK}$, as you denote it, are exactly the countable admissible ordinals, and these ordinals are intensely studied in the context of admissible set theory and fine str …

Let $b_k$ be the assertion that the busy beaver function at $k$ has the value that it actually has, that is, the value it has in the standard natural numbers of the meta-theory. We know that not all o …

I'm not sure what you mean by $\oplus$, but here is a counterexample. Let each $\phi_i(\vec x)$ be a tautology. So $S(\phi_i)=2$. But $\phi_1\oplus\cdots\phi_x$ is also trivial (depending on what you …

A Büchi automaton is a finite automaton that one runs on infinitely long strings (length $\omega$), with the proviso that the string is accepted if infinitely often the machine had visited an acceptin …

It's simple. If the halting problem is undecidable, then PA is not complete, since otherwise, you could solve the halting problem by searching for proofs in PA. And the same argument works for any sou …

Here is one way of interpreting your question. In my joint paper: Joel David Hamkins and Alexei Miasnikov, The halting problem is decidable on a set of asymptotic probability one, Notre Dame J. Form …

As I mentioned in the comments, it is a consequence of the MRDP theorem that the computably enumerable sets of natural numbers are precisely the projections of the natural-number zero sets of the mult …

The answer seems to be no, this is not decidable. You seem to have a concept in mind of what it means for a program $p$ to be UTP, and it involves the idea that pieces of any given computation histor …

Let $TA$ be the theory of true arithmetic, that is, the set of all truths of the usual standard model of arithmetic $\langle\mathbb{N},+,\cdot,0,1,<\rangle$. It is a theorem of ZFC that TA is consiste …