User blog:B1mb0w/Mapping D(l,0,1) to epsilon nought

From a previous blog post it is asserted that each iteration of \(D(2,0,n)\), e.g.:

\(D(2,0,0) = D(1,D(1,2,2),D(1,2,2))\) where \(D(1,2,2) >> f_{\omega+1}(3)\)

\(>> f_{\omega+1}^2(3)\)

\(D(2,0,1) = D(1,D(2,0,0),D(2,0,0)) >> f_{\omega+2}(3)\)\)

will progressively increase the ordinal strength of a lower bounded \(f\) function. This blog will match this asserted progression against the usual omega to epsilon nought ordinal hierarchy.

Iterating thru the \(f\) exponent (function nesting)
\(D(2,0,n)\) iterates the exponent value of a lower bounded \(f\) function. Here are the first few examples:

\(D(2,0,0) >> f_{\omega+1}^2(3)\)

\(D(2,0,1) >> f_{\omega+2}(3)\)

\(D(2,0,2) >> f_{\omega+2}^2(3)\)

\(D(2,0,3) >> f_{\omega.2}(3)\)

From here we can start mapping \(D(2,0,n)\) to \(f_{X(n/2)(3)} functions based on a X function to be defined, we can ignore exponents and focus on the index of the \(f\) function only.

Iterating thru to \(f_{\omega^{\omega}}\)
1. add  \(\omega\)  9 times to get  \(\omega^{\omega}\)

2. add  \(\omega^{\omega}\)  3 times to get  \(\omega^{\omega+1}\)

3. add  \(\omega^{\omega+1}\)  3 times to get  \(\omega^{\omega+2}\)

4. add  \(\omega^{\omega+2}\)  3 times to get  \(\omega^{\omega.2}\)

5. add  \(\omega^{\omega.2}\)  3 times to get  \(\omega^{\omega.2+1}\)

6. add  \(\omega^{\omega.2+1}\)  3 times to get  \(\omega^{\omega.2+2}\)

7. add  \(\omega^{\omega.2+2}\)  3 times to get  \(\omega^{\omega^2}\)

8. add  \(\omega^{\omega^2}\)  3 times to get  \(\omega^{\omega^2+1}\)

9. add  \(\omega^{\omega^2+1}\)  3 times to get  \(\omega^{\omega^2+2}\)

10. add  \(\omega^{\omega^2+2}\)  3 times to get  \(\omega^{\omega^2+\omega}\)

11. add  \(\omega^{\omega^2+\omega}\)  3 times to get  \(\omega^{\omega^2+\omega+1}\)

12. add  \(\omega^{\omega^2+\omega+1}\)  3 times to get  \(\omega^{\omega^2+\omega+2}\)

13. add  \(\omega^{\omega^2+\omega+2}\)  3 times to get  \(\omega^{\omega^2+\omega.2}\)

14. add  \(\omega^{\omega^2+\omega.2}\)  3 times to get  \(\omega^{\omega^2+\omega.2+1}\)

15. add  \(\omega^{\omega^2+\omega.2+1}\)  3 times to get  \(\omega^{\omega^2+\omega.2+2}\)

16. add  \(\omega^{\omega^2+\omega.2+2}\)  3 times to get  \(\omega^{\omega^2.\omega.2+\omega}\)

17. add  \(\omega^{\omega^2.2}\)  3 times to get  \(\omega^{\omega^2.2+1}\)

18. add  \(\omega^{\omega^2.2+1}\) 3 times to get  \(\omega^{\omega^2.2+2}\)

19. add  \(\omega^{\omega^2.2+2}\) 3 times to get  \(\omega^{\omega^2.2+\omega}\)

20. add  \(\omega^{\omega^2.2+\omega}\) 3 times to get  \(\omega^{\omega^\omega+1}\)

21. add  \(\omega^{\omega^2.2+\omega+1}\) 3 times to get  \(\omega^{\omega^\omega+2}\)

22. add  \(\omega^{\omega^2.2+\omega+2}\) 3 times to get  \(\omega^{\omega^\omega.2}\)

23. add  \(\omega^{\omega^2.2+\omega.2}\)  3 times to get  \(\omega^{\omega^\omega.2+1}\)

24. add  \(\omega^{\omega^2.2+\omega.2+1}\)  3 times to get  \(\omega^{\omega^\omega.2+2}\)

25. add  \(\omega^{\omega^2.2+\omega.2+2}\)  3 times to get  \(\epsilon_0\)

Calculating the total number of steps
All of the above steps are recursive. For example, Step 8 requires adding \(\omega^{\omega^2}\) 3 times. But adding \(\omega^{\omega^2}\) once requires all of Step 7 (adding \(\omega^{\omega.2+2}\) 3 times) and so on. This continues all the way to Step 1 and adding \(\omega\) 9 times.

Therefore the number of steps in total is \(3^24.9 = 3^{26} = 3^{3^3-1}\)

The general rule for any n other than 3 is \(n^{n\uparrow\uparrow(n-1)-1}\)