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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.

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.