Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 25 2 ii Solution Created 2026-10-03 Updated 2026-10-06
For finite rooted trees write when there is an injective function of vertices preserving greatest common ancestors, equivalently an infimum-preserving rooted-tree homeomorphic embedding. No linear ordering of siblings is imposed. Kruskal's tree theorem says that every infinite sequence has an earlier tree embedding into a later tree: this embedding relation is a well-quasi-ordering.
The finite form uses a parameter to restrict the growth of the trees. It asserts the existence of a length for whichHere is the number of vertices. The parameter is universally quantified; changing the starting index just shifts the parameter. This is Friedman's finite form of Kruskal's theorem.
Fix and suppose there were arbitrarily long sequences violating the conclusion. Form a tree whose nodes are their finite prefixes, including the empty prefix. Choose one representative of each finite rooted-tree isomorphism type. At position there are only finitely many choices, since the tree has at most vertices. Hence this finite bad-sequence tree is finitely branching. It has nodes at arbitrarily large heights by the supposed failure of the finite form. König infinity lemma supplies an infinite branch. The branch is an infinite sequence satisfying the size bounds and with no pair having , contradicting Kruskal's tree theorem. Therefore the finite length exists for every .
The significance is that this is a true statement about finite objects and natural numbers which is not provable in Peano arithmetic; in fact it is not provable in the stronger arithmetical transfinite recursion theory in second-order arithmetic, . It is a natural combinatorial instance of incompleteness. For fixed , the condition can be checked by finite search over trees and injective vertex maps, so its uniform arithmetic form is with a computable, indeed primitive recursive, predicate . Thus its least sufficient length is a total computable function, but its totality cannot be proved in those theories. Each fixed numerical instance can nevertheless be proved in Peano arithmetic by verifying some particular finite witness. The obstruction concerns the uniform statement for all parameters, rather than the decidability of any one finite search.