Search Results

The metatheory involved in proving "if ZFC is $\omega$-consistent then it doesn't prove its own inconsistency" is some tiny fragment of PA. I'd expect primitive recursive arithmetic (PRA) to be more t …

Saul Kripke gave a proof of incompleteness using nonstandard models and a notion of "fulfillability". Roughly speaking, a sequence fulfills a (prenex) formula if the formula is true when its successi …

As indicated by Feferman, the key distinction is between two sorts of arithmetical definitions of certain concepts (like "being the Gödel number of a theorem of a given theory"). Suppose I have some …

Hindman's theorem is an example where (slightly?) higher infinity simplifies a proof greatly. The theorem asserts that, if you partition the set of positive integers into finitely many subsets, then …

I agree with the previous answers, but it seems worthwhile to separate two points. (1) You are right to use "Let $x\in A$" even when $A$ might be empty. (2) Even if you had been wrong, such matters …

Gödel's second incompleteness theorem leads to a curious example as follows. The diagonalization method (as in the proof of the first incompleteness theorem) produces, for any reasonable theory T tha …

Both Carl Mummert and I answered your previous question, in comments, but it seems you haven't understood what we wrote. The problem is with induction, not with the deduction theorem. Your argument …

You seem to use "independent of $T$" to mean "unprovable in $T$", so I'll interpret the question that way (not as "neither provable nor refutable in $T$). If $ZF\vdash\phi$ is true, then it can be p …

Let me address the part of the question about "what the linkages back to the tau mean, what the boxes mean." The usual notation for using Hilbert's epsilon symbol is that one writes $(\varepsilon x)\ …

A very familiar example is given by the different ways to express the completeness property of the real line --- Cauchy sequences converge, bounded nonempty sets have suprema, etc.

$\DeclareMathOperator\BB{BB}$Philosophical issues, like acceptance (or non-acceptance) of large cardinals, won't affect $\BB(n)$, because the busy beaver function is defined arithmetically and so depe …