Search Results
| 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 lo.logic
Search options not deleted
user 74918
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.
0
votes
1
answer
184
views
Combinatorially defined effectively closed set
Is there a combinatorially defined, nonempty effectively closed set $Q\subseteq 2^\omega$ such that all members of $Q$ are incomputable?
Combinatorially defined means that the definition of $Q$ does …
- 909
6
votes
1
answer
193
views
Complexity of a combinatorial constraint
For two $k$-partitions $X,Y\in k^\omega$ of $\omega$
(seen as functions $\omega\rightarrow k$),
we say $X,Y$ are almost disjoint
iff $X^{-1}(i)\cap Y^{-1}(i)$ is finite
for all $i<k$.
Question: Does …
- 909
8
votes
2
answers
484
views
Comprehension axiom that helps in the opposite direction
Usually, having more comprehension axiom means the more you can prove. We wonder if the converse can be the case.
Is there a natural problem $\mathsf{P}$ so that $\mathsf{P}+\neg(\Gamma-\mathsf{Compre …
- 909