I recently read the ordinal-based proof of Goodstein's Theorem, saying that all Goodstein sequences do terminate. However, I did not see why the crucial part worked.
To my knowledge, the proof involves the construction of a parallel ordinal sequence to each Goodstein sequence $G(m)(n)$, $P(m)(n)$, so that $\exists G(m)(n)\Longleftrightarrow\exists P(m)(n)$. Each $P(m)(n+1)$ is constructed by replacing each occurrence of $n+1$ in $G(m)(n)$ with $\omega$, so $2^2+2\rightarrow \omega^\omega+\omega$, for example.
The proof then says that since the base-changing aspect of the process going from $G(m)(n)$ doesn't affect the omegas in the corresponding $P(m)(n)$ and then we subtract one from both, the $P(m)(n)$ must become strictly smaller each step. But why is this the case?
If I have some $G(m)(n)$ which is equivalent to $2^2$, with a $P(m)(n)$ thus equal to $\omega^2$, although it is true that $3^3-1<3^3$, is it true that $\omega^2-1<\omega^2$? Is the expression "$\omega^2-1$" even well defined?
Without this strictly decreasing aspect of the $P(m)(n)$ sequence, it is not clear to me how the proof works...
Starting with $4=2^2,$ with base $2,$ the next number is obtained by changing the $2$'s to $3$'s and subtracting $1,$ giving $3^3-1,$ with base $3$. But the rules require this to be re-written in base $3$ as $3^2\cdot 2+3^1\cdot 2+2$ before going further. The next number is $4^2\cdot 2+4^1\cdot 2 +1,$ with base $4.$ The corresponding ordinals are $\omega^{\omega},\; \omega^2\cdot 2+\omega\cdot 2+2,\;\omega^2\cdot 2+\omega\cdot 2+1.$