This is Buchholz's hydra labeled with ordinals that can be larger than omega.
- If \(a\) has label 0, we proceed as in Kirby-Paris' game. Call the node's parent \(b\), and its grandparent \(c\) (if it exists). First we delete \(a\). If \(c\) exists (i.e. \(b\) is not the root), we make \(n\) copies of \(b\) and all its children and attach them to \(c\).
- If \(a\) has successor label \(\alpha + 1\), we go down the tree looking for a node \(b\) with label \(v \leq \alpha\) (which is guaranteed to exist, a every child of the root node has label 0). Consider the subtree rooted at \(b\) — call it \(S\). Create a copy of \(S\), call it \(S'\). Within \(S'\), we relabel \(b\) with \(\alpha\) and relabel \(a\) with \(0\). Back in the original tree, replace \(a\) with \(S'\).
- If \(a\) has transfinite label \(\beta\), we simply relabel it with \(\beta[n + 1]\).