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 not deleted user 33842

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.

6 votes

Computability complexity of the first-order theory of arithmetic?

The theory of arithmetic is $\Delta^1_1$, thus it cannot be $\Pi^1_1$-complete (since $\Delta^1_1$ is a strict subset of $\Pi^1_1$ and these classes are closed under recursive preimages).
The User's user avatar
  • 2,442
5 votes

Basic results with three or more hypotheses

The HSP-theorem from universal algebra: If a class of algebraic structures (over a given signature) is closed under homomorphic images, substructures and products, then it is defined by a set of equat …
4 votes
3 answers
819 views

Impact of the axiom of replacement on finite sets

The axiom of replacement is usually used to prove the existence of large sets, to provide a reflection principle, for transfinite recursion… However, I am wondering how it affects finite sets. Let me …
The User's user avatar
  • 2,442
2 votes

What would be some major consequences of the inconsistency of ZFC?

Notice that you can drop the axiom of replacement or replace it by a weaker reflection principle. Without this axiom you have less consistency strength—it might still be consistent even if ZFC is not …
The User's user avatar
  • 2,442
2 votes

Existential quantification over regular predicates

That is a central point about automatic structures: By projection (“existential quantification”) you get another regular predicate, and regular predicates are also closed under intersection and comple …
The User's user avatar
  • 2,442
2 votes
Accepted

How many subsets of $\mathbb{R}$ are order isomorphic to $\mathbb{Q}$?

There are continuum many countable subsets of the continuum (because $\mathfrak{c}^{\aleph_0}=2^{\aleph_0}$). Thus the answer is $\mathfrak{c}$. See this question.
The User's user avatar
  • 2,442