User blog comment:LittlePeng9/FORMAL comparison of Hydra battles and Hardy hierarchy/@comment-10429372-20141017061807

This is about the thing I was writing in my paper, so basically I can throw that away now. (The BH and formal comparisons didn't go that well together). However, I found an upper bound on the modified hierarchy using transfinite recursion: H'α(n) ≤ Hα(n+1).

Base case: H’0(n) = n = H0(n) = H0(H0(n))

Successor case: H’α+1(n) = H’α(n+1) ≤ Hα(n+2) = Hα+1(n+1) Limit case: H’α(n) = H’α[n](n+1) ≤ Hα[n](n+2) < Hα[n]+1(n+1) ≤ Hα[n+1](n+1) = H’α(n+1)

So $$Hydra(4) = H'_{\omega^{\omega^\omega}}(2) = H'_{\omega^{\omega^2}}(3) =$$

$$H'_{\omega^{\omega3}}(4) = H'_{\omega^{\omega2+4}}(5) = H'_{\omega^{\omega2+3}5}(6) =$$

$$H'_{\omega^{\omega2+3}5}(6) = H'_{\omega^{\omega2+3}4+\omega^{\omega2+2}6}(7) =$$

$$ H'_{\omega^{\omega2+3}4+\omega^{\omega2+2}5+\omega^{\omega2+1}7}(8) =$$

$$ H'_{\omega^{\omega2+3}4+\omega^{\omega2+2}5+\omega^{\omega2+1}6+\omega^{\omega2}8}(9) ≤$$

$$ H_{\omega^{\omega2+3}4+\omega^{\omega2+2}5+\omega^{\omega2+1}6+\omega^{\omega2}8}(10) =$$

$$ f^4_{omega2+3}(f^5_{\omega2+2}(f^6_{\omega2+1}(f^8_{\omega2}(10))))$$