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 computability-theory
Search options not deleted
user 3091
computable sets and functions, Turing degrees, c.e. degrees, models of computability, primitive recursion, oracle computation, models of computability, decision problems, undecidability, Turing jump, halting problem, notions of computable randomness, computable model theory, computable equivalence relation theory, arithmetic and hyperarithmetic hierarchy, infinitary computability, $\alpha$-recursion, complexity theory.
0
votes
Can you prove equivalence without being able to calculate it?
Isn't this done all the time in basic set theory with the Schroeder-Bernstein Theorem, where we deduce equivalence (in this context, the existence of a bijection) without presenting one?