Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $\Sigma$ denotes the busy beaver function, how can I then show, that there is an $t\in \mathbb{N}$ such that for all $x\geqslant t$ we have $\Sigma(x)>f(x)$, where $f$ is an arbitrary partial recursive function?

share|cite|improve this question

The (even original) proof of the theorem can be found here on page 880.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.