Skip to main content
Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Code code:"if (foo != bar)"
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Saves in:saves
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with
Search options answers only not deleted user 4600

first-order and higher-order logic, model theory, set theory, proof theory, computability theory, formal languages, definability, interplay of syntax and semantics, constructive logic, intuitionism, philosophical logic, modal logic, completeness, Gödel incompleteness, decidability, undecidability, theories of truth, truth revision, consistency.

2 votes
Accepted

What is ($\Pi^1_1$-CA)${}+{}$BI? And what is ID${}_\omega$?

BI = Bar Induction. See https://www.jstor.org/stable/2270902 for the abbreviation and https://en.wikipedia.org/wiki/Bar_induction for the definition.
Bjørn Kjos-Hanssen's user avatar
3 votes

Can we have a theory $T$ that is complete for simple sentences in the language of $T$ that a...

Partial answer: If there is such a theory $T$, it would mean that every simple sentence that's independent of $T$, must decide Con($T$). Note that every $\Pi^0_1$ sentence is equivalent to a simple …
Bjørn Kjos-Hanssen's user avatar
32 votes
Accepted

On sentences true in all finite groups

The answer is Yes for the second question, about $(\exists x)(\forall y)w=1$. Following Christian Remling's idea: If a sentence like $$\exists x(\forall y)(yxy^{-1}x^2y^{-9}\dots=1)$$ holds in all fin …
Bjørn Kjos-Hanssen's user avatar
7 votes
Accepted

Is DNC/DNR stronger than "prompt" non-computability?

The graph of the course-of-values variant $$\{(x,(f(0),\dots,f(x))): x\in \mathbb N\}$$ of such a function would be effectively immune. Namely, if we enumerate a subset of this graph then there is an …
Bjørn Kjos-Hanssen's user avatar
1 vote

Turing degree of finding independent formulas

From your use of "$\downarrow$" it seems you do not require $f$ to give an answer on non-$\omega$-consistent theories. At least we can get a function $f$ partial computable in $0^{(3)}$ (perhaps you …
Bjørn Kjos-Hanssen's user avatar
4 votes
Accepted

On fast-growing hierarchy

Let $\varphi_a $ be the $ a $ th partial computable function in a standard way. Given a recursively enumerable set $ W $, let $ f (n) $ be the maximum of $\varphi_a (b)$ over $ b\le n $ and $ a $ amo …
Bjørn Kjos-Hanssen's user avatar
5 votes

Relation between Turing degrees and functions computable with them

Your condition is equivalent to $A''\le_T B'$, that is $B$ is high above $A$. Here $'$ is the Turing jump operator, i.e., the relativized halting problem operator. In the case $A=0$, $B$ is high. The …
Bjørn Kjos-Hanssen's user avatar
1 vote

Is there a good list of nomenclature for modal axioms?

Apparently $\Box\Box\alpha\rightarrow\Box\alpha$ is known as C4 (Converse of (4)). Source: Stanford Encyclopedia of Philosophy.
Bjørn Kjos-Hanssen's user avatar
5 votes

${\frak b}$ and ${\frak d}$ defined with $\leq$ instead of $\leq^*$

${\frak b}'=\aleph_0$ since you can take $S$ to be the collection of constant functions. ${\frak d}' = {\frak d} + \aleph_0$ since you can take any family $S$ realizing $\frak d$ and close it under f …
Bjørn Kjos-Hanssen's user avatar
3 votes

Literature on Kripke models

Brian F. Chellas: Modal Logic: An Introduction, 1980. Starts very basic but covers in detail the beautiful completeness theorem proofs for the basic systems like $S5$, $S4$, $K$.
Bjørn Kjos-Hanssen's user avatar
18 votes

In what sense is GCD an extension of boolean OR?

In the ordering $\preceq$ of nonnegative integers by divisibility, 1 is the least element and 0 is the greatest, and we have for instance $$ 1\preceq 2\preceq 6\preceq 12\preceq\dots\preceq 0.$$ In th …
Bjørn Kjos-Hanssen's user avatar
1 vote
Accepted

Logic Alphabet for more than Two Variables

Well, when the number of Boolean variables is $n=3$, since the number of connectives is $2^{2^n}=(2^{2^2})^{n-1}$, we can use infix notation like $pxqyr$, where $p$, $q$, $r$ are Boolean variables and …
Bjørn Kjos-Hanssen's user avatar
3 votes
Accepted

Jump of strongly hyperhyperimmune degrees and DNR relative to 0'

Carl Jockusch and Frank Stephan. A cohesive set which is not high. Mathematical Logic Quarterly, 39:515-530, 1993. A corrective note (keeping the main results intact) appeared in the same journal, 43: …
Bjørn Kjos-Hanssen's user avatar
3 votes

Models for the given FOL statement

Yes. If there is a model $\mathcal M$ of size $n$ of any sentence $\phi$ that does not use = then you can take any element $a$ of $\mathcal M$ (using the fact that $n>0$) and let $\mathcal N$ be $\mat …
Bjørn Kjos-Hanssen's user avatar
21 votes
Accepted

Question arising from Voevodsky's talk on inconsistency

Let $S$ be a first order definable Martin-Löf random set such as Chaitin's $\Omega$. If Peano Arithmetic, or ZFC, or any other theory with a computable set of axioms, proves infinitely many facts of t …
Bjørn Kjos-Hanssen's user avatar

1
2 3 4 5 6
15 30 50 per page