User blog comment:Wythagoras/Extension to Friedmann's stuff/@comment-11227630-20150501095933/@comment-1605058-20150501120743

It's better to understand this "collapse" in terms of whole subgraphs. It just means that instead of treating subgraph as group of vertices, we treat it like one "supervertex", which is connected to other "supervertex" which there is an edge connecting two subgraphs.

When we collapse a subgraph of \(G_2\), it doesn't have any specific label. It can be thought to have all of the labels of subgraph we collapse, and we only require one of these labels to be \(\geq\) than corresponding label of vertex in \(G_1\).

You are right - just being the well-quasi-order doesn't mean that there always exists a maximal length of a sequence. But it exists as soon as we put any kind of constraint which makes it so that at every place we have only finitely many possible elements (the proof uses Konig's lemma, and I can write it down if you are interested). Here, if we restrict ourselves to finitely many labels, and say that n-th graph has at most n+i vertices, then we obviously have only finitely many graphs which can possibly be the n-th graph. It follows that the longest sequence must exist in this case.