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For question in Proof Theory, where "proofs" themselves are the object of mathematical investigation. It is not to be used to request a proof of some result.
4
votes
What are the known large cardinal axioms for which weaker and stronger set theories "catch up"?
Every large cardinal property admits a formalization with the desired property. That is, every large cardinal property $\text{LC}$ admits a ZFC-provably equivalent formulation $A$ for which $\newcomma …
7
votes
Gödel coding and the function $z(x)$
Yes, for the reasons you mention, it is important to define your Gödel coding in such a way that the syntactic operations you want to undertake with assertions in the language are indeed expressible i …
7
votes
Accepted
Can we have consistent theories stating opposing provability statements that are non-standar...
This idea in play here is due to Rosser and is the main idea behind the Gödel-Rosser theorem.
Specifically, Rosser proposes to consider the sentence $\rho$ asserting that for every proof of $\rho$ in …
8
votes
Is the usual enumeration of $\mathsf{PA}$ "minimal for consistency strength"?
The cautious enumeration idea in my paper has some affinity with your suggestion.
Joel David Hamkins, Nonlinearity and illfoundedness in the hierarchy of large cardinal consistency strength, arxiv:22 …
7
votes
Computational complexity theoretic incompleteness: is that a thing?
These self-referential decision problems are already part of the subject of computational complexity. There are analogues of the halting problem, for example, for many of the various classes in the co …
34
votes
2
answers
2k
views
What is the logical status of the sentence combining the ideas of Löb and Rosser, "this sent...
Logicians are familiar with the variety of self-referential sentences expressible in the language of arithmetic:
The Gödel sentence, "this sentence is not provable", which indeed is not provable in w …
2
votes
Quantification over uncountable sets
There are several things one can say.
The theory of ZFC without powerset is often denoted by $\newcommand\ZFCm{\text{ZFC}^-}\ZFCm$. One has to be a little careful with what it means, since collection …
3
votes
Where did this presentation of Gödel's theorem appear?
This argument is essentially similar to the argument of Mel Fitting in his article, "Russell's paradox, Gödel's theorem" Chapter in book: Raymond Smullyan on self reference, 47–66, Outstanding Contri …
2
votes
Uniform incomparable consistency strengths
This does not answer your question, but I find it relevant. You ask for uniform incomparable statements $A_\tau$ and $B_\tau$, and then ask also for monotonicity. But I claim that if one asks for the …
3
votes
Interpreting proper elementarily equivalent end extensions?
There can be no such model.
The first observation is that if there is a model as you describe,
then I claim there will be an instance where $j$ is elementary. To
see this, suppose that $M$ is as in y …
12
votes
Infinite descending consistency chains
Here is perhaps a more relatable
example, which doesn't use self-reference. (I once heard a similar such example from W. Hugh Woodin.)$\newcommand\Con{\text{Con}}\newcommand\ZFC{\text{ZFC}}$
Let $\ps …
13
votes
Accepted
The Halting Problem and Church's Thesis
Let me point out that there are really a family of Church-Turing
theses assertions.
On the one hand, for what is sometimes described as the weak
Church-Turing thesis, one imagines an idealized human …
21
votes
Accepted
Is there a consistent arithmetically definable extension of PA that proves its own consistency?
Surprisingly, the answer is yes! Well, let me say that the answer
is yes for what I find to be a reasonable way to understand what
you've asked.
Specifically, what I claim is that if PA is consistent …
8
votes
Accepted
Does the notion of provably total function depend on the chosen representation?
Yes, this concept depends on how you represent the function.
For example, the constant zero function is provably
total, under that description, that is, using the formula
$\varphi(x,y)$ equal to "$y …
14
votes
Accepted
Peano arithmetic vs. fast-growing hierarchy with pathological fundamental sequences
The answer is no. Choose a fundamental sequence for $\epsilon_0$ itself in the usual way, which I think is $\epsilon_0[n]=\omega^{\omega^{{\vdots}^\omega}}$, and then modify the earlier fundamental se …