Sign Up

Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.

Have an account? Sign In

Have an account? Sign In Now

Sign In

Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.

Sign Up Here

Forgot Password?

Don't have account, Sign Up Here

Forgot Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Have an account? Sign In Now

You must login to ask a question.

Forgot Password?

Need An Account, Sign Up Here

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

Sign InSign Up

The Archive Base

The Archive Base Logo The Archive Base Logo

The Archive Base Navigation

  • SEARCH
  • Home
  • About Us
  • Blog
  • Contact Us
Search
Ask A Question

Mobile menu

Close
Ask a Question
  • Home
  • Add group
  • Groups page
  • Feed
  • User Profile
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Buy Points
  • Users
  • Help
  • Buy Theme
  • SEARCH
Home/ Questions/Q 6962933
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 27, 20262026-05-27T15:43:26+00:00 2026-05-27T15:43:26+00:00

I need to determine if L:={<M>|HP<=L(M)} is a recursive language, and if L is

  • 0

I need to determine if L:={<M>|HP<=L(M)} is a recursive language, and if L is a recursively enumerable language.

I think that Rice’s theorem can help prove L is not recursive but I don’t think that L is recursively enumerable…

  • 1 1 Answer
  • 0 Views
  • 0 Followers
  • 0
Share
  • Facebook
  • Report

Leave an answer
Cancel reply

You must login to add an answer.

Forgot Password?

Need An Account, Sign Up Here

1 Answer

  • Voted
  • Oldest
  • Recent
  • Random
  1. Editorial Team
    Editorial Team
    2026-05-27T15:43:27+00:00Added an answer on May 27, 2026 at 3:43 pm

    If L is not in RE then L is not in R either.

    You should try to reduce it to the halt problem. Lets say that X is a Turing machine that outputs false if L(X) is true and outputs true if L(X) is false.

    Is L(X) true? It is if and only if L(X) is false, which is a contradiction.

    Is L(X) false? It is if and only if L(X) is true, which is a contradiction too.

    The contradiction is in the implicit assumption that L is computable by a Turing machine. Hence L is not computable. The X Turing machine can’t exist. And Finally L is not in RE (and neither in R).

    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Related Questions

I'm not so strong with arrays but I need to determine how to count
I need to determine whether or not two sets contains exactly the same elements.
I need to determine if a URL entered by a site user is not
Consider that, for a Windows video game, I need to determine if the key
I need to determine if an input string input can be parsed by jodatimes
My winform application is launched by another application (the parent), I need determine the
I need to determine in 80% if a file is binary or text, is
I need to determine file type (i.e., MimeType) of stored data in the SQL
I need to determine if a given selection is in between a start line
I need to determine the clients .NET framework version in my web application. I'm

Explore

  • Home
  • Add group
  • Groups page
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Users
  • Help
  • SEARCH

Footer

© 2021 The Archive Base. All Rights Reserved
With Love by The Archive Base

Insert/edit link

Enter the destination URL

Or link to existing content

    No search term specified. Showing recent items. Search or use up and down arrow keys to select an item.