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An adjacency-preserving rooted-tree embedding is an injective vertex map preserving the root and every adjacency. Because each root-to-vertex path maps to a simple path of the same length, vertex depths are preserved. This is more restrictive than a homeomorphic embedding of a rooted tree, which may stretch edges into paths. The branching-depth antichain of rooted trees shows that the adjacency relation is not a well-quasi-ordering.

For rooted trees labelled in a preorder $Q$, a label-monotone tree embedding is a homeomorphic embedding of a rooted tree whose vertex map satisfies ${\ell }_T(v){\le }_Q{\ell }_S(h(v))$. The root need not map to the host root. For ordered child lists, require in addition that the map preserve the relative order of distinct branches.

A well-quasi-ordering is a reflexive transitive relation such that every infinite sequence $x_0,x_1,\dots$ has $i<j$ with $x_i\le x_j$. An infinite sequence without such a pair is a bad sequence. For a finite rooted tree, let $u\wedge v$ denote the lowest common ancestor of two vertices. A homeomorphic embedding of a rooted tree into another is an injective map $h$ withThus it preserves branching and sends each edge to a nonempty downward path, with different branches separated. The source root may map to a vertex below the target root. Kruskal's tree theorem says that finite rooted trees are well-quasi-ordered under these embeddings.

We first establish the word lemma used in the proof. The Higman lemma says that finite words over a well-quasi-ordering $Q$ are well-quasi-ordered by subsequence embedding with increased letters. Every infinite sequence in $Q$ has an infinite nondecreasing subsequence. To see this, some term must have infinitely many later terms above it: otherwise, repeatedly choosing past all finitely many successors of earlier chosen terms constructs a bad sequence. Apply this observation again inside that infinite upper cone, and repeat.

If the Higman lemma failed, choose a minimal bad sequence of words $w_0,w_1,\dots$, minimizing length at each position among choices admitting a bad continuation. None is empty. Write $w_i=u_i{a}_i$, with last letter $a_i$, and choose indices $i_0<i_1<\cdots$ on which the $a_i$ are nondecreasing. Thenis still a bad sequence. A comparison from the original prefix into some $u_{i_j}$ would give one into $w_{i_j}$. A comparison $u_{i_j}\le u_{i_k}$ would extend by $a_{i_j}\le a_{i_k}$ to $w_{i_j}\le w_{i_k}$. Both contradict the original badness. But the replacement at position $i_0$ is shorter than $w_{i_0}$, contradicting minimality. This proves the Higman lemma.

Now suppose Kruskal's tree theorem fails and choose a minimal bad sequence $T_0,T_1,\dots$, minimizing the number of vertices at each stage. Let $\mathcal{B}$ contain all proper descendant-rooted subtrees of all $T_i$. We claim that $\mathcal{B}$ is well-quasi-ordered by rooted-tree homeomorphic embeddings. Otherwise take a bad sequence $U_0,U_1,\dots$ from it. Each finite set of host trees supplies only finitely many subtrees, so after passing to a subsequence we may arrange that $U_j$ is a proper subtree of $T_{k_j}$ with $k_0<k_1<\cdots$. The sequenceis bad: a comparison from a prefix tree into $U_j$ would compose with its inclusion into $T_{k_j}$ and contradict the original badness; comparisons among the $U_j$ are excluded by construction. Yet $U_0$ is smaller than $T_{k_0}$, contradicting minimality. This proves the claim.

List the immediate-child subtrees of each $T_i$ in any fixed order. These are finite words over $\mathcal{B}$. By the Higman lemma, some earlier child list embeds into a later one with increased letters. The selected child subtrees embed into distinct child branches of the later tree. Map the source root to the target root and use these embeddings in the selected branches. The paths from the target root to the embedded child roots are separated because the target child branches are distinct. The resulting injection preserves lowest common ancestors, giving $T_i⪯T_j$, a contradiction. This proves Kruskal's tree theorem completely. It also gives the root-preserving homeomorphic version: once the weaker relation is well-quasi-ordered on all finite rooted trees, apply the Higman lemma to their child lists to obtain a comparison preserving the root.

For the proposed adjacency-preserving rooted-tree embedding, the answer is no: it is not a well-quasi-ordering. For every $m\ge 1$, form $T_m$ from a stem of $m$ edges starting at the root and then attach two leaves to its terminal vertex. The only vertex with two children is at depth $m$. A root-preserving adjacency injection preserves depths: the unique path of length $d$ from the root maps to a simple path of length $d$ from the target root. The image of a vertex with two children must still have two distinct children. Thus an embedding $T_m\to T_n$ must send the branch vertex at depth $m$ to the unique branch vertex at depth $n$, forcing $m=n$. ThereforeThis is the branching-depth antichain of rooted trees. Homeomorphic embeddings can stretch the stem and hence do not have this obstruction.

A well-quasi-ordering is a reflexive transitive relation for which every infinite sequence has $i<j$ with $x_i\le x_j$. A bad sequence has no such pair. The labelled version of Kruskal's tree theorem says that finite rooted trees labelled in any well-quasi-ordering are themselves well-quasi-ordered by label-monotone tree embedding.

More explicitly, an embedding is an injective map of vertices preserving lowest common ancestors, and satisfying ${\ell }_T(v)\le {\ell }_S(h(v))$ at every vertex. The root need not map to the host root. This is a homeomorphic embedding of a rooted tree: an edge can map to a longer path, but distinct branches must separate at the image of their common ancestor. We shall prove the stronger version in which every vertex's children are linearly ordered and embeddings respect that ordering. Forgetting the child order gives the stated result for unordered rooted trees.

We first establish the two well-quasi-ordering facts used in the proof. Every infinite sequence in a well-quasi-ordering has an infinite nondecreasing subsequence. Indeed, color an index pair $i<j$ according to whether $x_i\le x_j$. The infinite two-color Ramsey theorem gives a homogeneous infinite set; the negative color would be a bad sequence, so the positive color gives the required subsequence. It follows that the componentwise product of two well-quasi-orderings is a well-quasi-ordering: first extract a nondecreasing subsequence in one coordinate, then find a good pair in the other.

Next prove Higman lemma: finite words over a well-quasi-ordering $Q$, ordered by subsequence embedding with coordinatewise increase of letters, are a well-quasi-ordering. Suppose not, and choose a minimal bad sequence $w_0,w_1,\dots$ by making the length of $w_n$ minimal among all choices admitting an infinite bad continuation of the already fixed prefix. No word is the empty word, since that embeds in every later word. Write $w_n=v_n{a}_n$ with $a_n\in Q$. Extract indices $i_0<i_1<\cdots$ for which $a_{i_0}\le a_{i_1}\le \cdots$. ConsiderIt cannot be a bad sequence, since its first replacement is shorter than the minimal choice $w_{i_0}$. But a good pair within the original prefix is impossible. A prefix word embedding into $v_{i_j}$ would embed into $w_{i_j}$, contradicting the original bad sequence. And $v_{i_j}$ embedding into $v_{i_k}$ would, after appending the ordered last letters, embed $w_{i_j}$ into $w_{i_k}$, also impossible. This contradiction proves Higman lemma, including words of arbitrary finite length.

Now suppose there is a bad sequence of finite ordered labelled rooted trees. Choose a minimal bad sequence $T_0,T_1,\dots$ by minimizing the number of vertices of $T_n$, subject to the fixed prefix having an infinite bad continuation. Existence of a least possible size uses ordinary well-ordering of the natural numbers; after selecting such a tree retain a bad continuation to make the next choice.

Let $\mathcal{S}$ be the collection of all proper rooted subtrees of all $T_n$, with inherited labels and child ordering. These are the subtrees rooted at vertices other than the root, and each embeds into its containing tree. We claim $\mathcal{S}$ is a well-quasi-ordering under the same label-monotone tree embedding.

Otherwise choose a bad sequence $U_0,U_1,\dots$ from $\mathcal{S}$, and choose for each a containing tree $T_{n_j}$. The indices $n_j$ are unbounded in every tail: finitely many containing trees have only finitely many rooted subtrees, and an infinite bad sequence cannot repeatedly use one of these, since it embeds into itself. Passing to a subsequence, arrange $n_0<n_1<\cdots$. The spliced sequenceis bad. A good pair within either piece is already excluded. A comparison $T_k⪯U_j$, where $k<n_0\le n_j$, would compose with $U_j⪯T_{n_j}$ to give $T_k⪯T_{n_j}$, contradicting the original bad sequence. But $U_0$ is smaller than $T_{n_0}$, contradicting the minimal choice at that position. This proves the claim about $\mathcal{S}$.

Describe $T_n$ by its root label $q_n\in Q$ and its finite ordered list $C_n$ of child subtrees. Every entry of $C_n$ lies in $\mathcal{S}$. By Higman lemma, ${\mathcal{S}}^{\ast }$ is a well-quasi-ordering; hence so is $Q\times {\mathcal{S}}^{\ast }$. There exist $i<j$ with $q_i\le q_j$ and an increasing injection matching the child subtrees in $C_i$ to child subtrees in $C_j$, each by an embedding. Map root to root and combine these child embeddings. Different matched children lie in different target branches, so their paths meet exactly at the target root; within each branch the chosen embedding already preserves lowest common ancestors. The resulting map is a label-monotone tree embedding $T_i⪯T_j$, a contradiction. ThereforeThe empty labelled tree, if included by convention, embeds into every tree and causes no exception. A common stronger formulation requires the source root to map to the target root. It also follows: the theorem just proved makes all labelled child subtrees a well-quasi-ordering, and the product $Q\times \mathcal{T}(Q{)}^{\ast }$ then provides a good pair with roots explicitly matched, by the same final assembly. Internal child roots may map further down their matched branches. This must not be confused with edge-to-edge embedding, for which the theorem is false in general.