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Let $a, n \in \mathbb{Z}^{*}_{+}$. $a$ is defined inductively in the base $n$-recursive. We first write $a$ in the base $n$, e.g., as a sum of terms of the form $k_tn^t$, with $0 \le k_t < n$. For each exponent $t$, we write $t$ in the base $n$-recursive, until all the numbers in the representation are less than $n$. For instance,

$1309 = 3^6 + 2.3^5 + 1.3^4 + 1.3^2 + 1.3 + 1$

$ = 3^{2.3} + 2.3^{3+2} + 1.3^{3+1} + 1.3^2 + 1$

Let $x_1 \in \mathbb{Z}$ arbitrary. We define $x_n$ recursively, as following: if $x_{n-1} > 0$, we write $x_{n-1}$ in the base $n$-recursive and we replace all the numbers $n$ for $n+1$ (even the exponents!), so we obtain the successor of $x_n$. If $x_{n-1} = 0$, then $x_n = 0$.

Example:

$x_1 = 2^{2^{2} + 2 + 1} + 2^{2+1} + 2 + 1$

$\Rightarrow x_2 = 3^{3^{3} + 3 + 1} + 3^{3+1} + 3$

$\Rightarrow x_3 = 4^{4^{4} + 4 + 1} + 4^{4+1} + 3$

$\Rightarrow x_4 = 5^{5^{5} + 5 + 1} + 5^{5+1} + 2$

$\Rightarrow x_5 = 6^{6^{6} + 6 + 1} + 6^{6+1} + 1$

$\Rightarrow x_6 = 7^{7^{7} + 7 + 1} + 7^{7+1}$

$\Rightarrow x_7 = 8^{8^{8} + 8 + 1} + 7.8^8 + 7.8^7 + 7.8^6 + ... + 7$

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Prove that $\exists N : x_N = 0$.

These are known as Goodstein sequences, and proving that no matter the number we start with, the sequence ends with $0$ is quite a difficult task, a rigorous proof of it requiring the theory of transfinite ordinal numbers.

I think this statement forgot to say that, at the stage where all numbers are passed, we also subtract at the end (however, their examples include this subtraction); otherwise it is clear that the sequence will continue to grow ...

Anyway, I don't know how to prove it.

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    What is your question? You can find linkls to expositions and proofs of Goodstein's theorem in this answer. – Bill Dubuque Nov 17 '19 at 16:21
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    At the end of each step, $1$ is subtracted. At some point the sequence becomes stationary, and from some point on, it goes to $0$ in steps of $1$. I know a site (but it is in german) : Joerg Resag shows the proof idea in his site "Die Grenzen der Berechenbarkeit" – Peter Nov 17 '19 at 16:33

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