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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.

This subsystem of second-order arithmetic allows comprehension for arithmetical formulas and induction in its standard restricted presentation. Its first-order arithmetical consequences are those of Peano arithmetic. It is a useful base for comparing well-quasi-ordering principles with ordinal notation systems without assuming their global well-foundedness.

The barrier in better-quasi-order theory $[\omega {]}^2$ has shifts $\{m,n\}◃\{n,l\}$ for $m<n<l$. Map each pair to the corresponding element of the Rado order. Neither clause of $(m,n){\le }_R(n,l)$ holds: the first coordinates differ, and the second coordinate of the first pair equals, rather than precedes, the first coordinate of the second. Thus this array is bad, although the Rado order is a well-quasi-ordering. It separates better-quasi-ordering from well-quasi-ordering.

A bad sequence in a preorder is an infinite sequence with no indices $i<j$ satisfying $x_i\le x_j$. A preorder is a well-quasi-ordering precisely when no such sequence exists.

A quasi-order is better-quasi-ordered if every array into it from every barrier in better-quasi-order theory is good: some barrier members $s◃t$ satisfy $f(s)\le f(t)$. Singleton barriers recover the usual infinite-sequence condition for a well-quasi-ordering. The stronger condition is useful for infinitary closure operations for which well-quasi-ordering alone fails. Equivalently one may use continuous maps from infinite subsets of an infinite subset of the natural numbers, with the discrete topology on the range, and compare $X$ with $X\setminus \{\min X\}$.

Given a size bound with finitely many possible objects at each position, put finite bad sequences in a tree ordered by extension. The tree is finitely branching. Arbitrarily long bad sequences would give an infinite bad sequence by König infinity lemma. This compactness argument converts an infinite well-quasi-ordering theorem into a uniform finite length bound.

A finite Cartesian product of well-quasi-orderings is well-quasi-ordered in the componentwise order. Every infinite sequence in one well-quasi-order has an infinite nondecreasing subsequence: colour pairs by whether they are ordered, apply Ramsey's theorem, and exclude an all-bad subsequence. Restrict successively to such a subsequence in each coordinate. The same argument, or the infinite pigeonhole principle, proves closure under finite disjoint unions with no cross-component comparisons.

Finite subsets of a well-quasi-ordering, compared by Hoare domination preorder, form a well-quasi-ordering. Enumerate each finite subset as a word and apply Higman lemma: a subsequence embedding with increased letters supplies a domination witness for every source element. This does not assert the same result for the full power set.

If $Q$ is a well-quasi-ordering, its finite words form a well-quasi-ordering under subsequence embedding with coordinatewise increase of letters. A minimal bad sequence proof removes the last letter of selected words whose last letters form a nondecreasing subsequence, then contradicts minimality. The result implies finite-subset lifting of a well-quasi-order under Hoare domination preorder.

For subsets of a preorder, Hoare domination means every source element has some greater or equal target element: $S{\le }_{H}T⟺\forall s\in S\exists t\in T(s\le t)$. It is reflexive and transitive. If the base is a well-quasi-ordering, its finite subsets are well-quasi-ordered by this relation; arbitrary subsets need not be, as the Rado order shows.

Finite rooted trees labelled in a well-quasi-ordering are a well-quasi-ordering under label-monotone tree embedding. A minimal bad sequence argument makes the collection of proper rooted subtrees a well-quasi-ordering; Higman lemma then compares their child lists while the root labels are compared in the label order.

We prove the reverse-inclusion well-quasi-ordering of monomial ideals, equivalently of all upward-closed subsets of ${\mathbb{N}}^r$, by induction on $r$. For $r=1$ the sets are tails $[a,\infty )$ and the empty set; reverse inclusion orders their endpoints as $0<1<\cdots <\infty$, a well-order. The case $r=0$, if needed for the induction, is just a finite family.

For an upset $U\subseteq {\mathbb{N}}^{r+1}$, let $U_k=\{\alpha :(\alpha ,k)\in U\}$. These are upsets of ${\mathbb{N}}^r$ and satisfy $U_0\subseteq U_1\subseteq \cdots$. They stabilize. Indeed the union $U_{\infty }$ has a finite minimal basis by Dickson's lemma: an infinite set of minimal elements would be an antichain in the componentwise well-quasi-ordering of ${\mathbb{N}}^r$. Each basis element appears in some slice, and after the largest of these finitely many indices the slice contains the entire upset $U_{\infty }$.

Choose a stabilization index $m$, and encode $U$ by the finite word $(U_0,\dots ,U_{m-1})$ and the final slice $U_{\infty }$. The induction hypothesis makes slices a well-quasi-ordering by reverse inclusion. Higman's lemma, proved in Question 3 below, makes their finite words a well-quasi-ordering; finite product closure of well-quasi-orderings also compares the final slices. Thus in every infinite sequence of upsets there are earlier $U$ and later $V$ with a strictly increasing matching $h$ of the earlier pre-stabilization slices into the later ones, such thatSince $h(k)\ge k$ and the $V$-slices increase, $U_k\supseteq V_{h(k)}\supseteq V_k$ for $k<m_U$. For $k\ge m_U$, $U_k=U_{\infty }\supseteq V_{\infty }\supseteq V_k$. Therefore $U\supseteq V$, completing the induction.

Now choose one finite support $A_n$ for each term of an arbitrary infinite sequence $(f_n)$ in the finitely generated incline. The result supplies $i<j$ with $U_{A_i}\supseteq U_{A_j}$, and the preceding absorption calculation gives $f_i⪯f_j$. Hence every finitely generated incline is well-quasi-ordered by the specified reverse absorption order. The proof allows all additional algebraic identifications among the generators.

Use the reverse absorption order on the incline:This is exactly the relation given by the pairs $(x+y,x)$. Indeed, if $a=x+y$ and $b=x$, then $a+b=x+y+x=a$ by additive idempotence, associativity and commutativity. Conversely $a+b=a$ expresses $a$ as $b+a$, so $(a,b)$ is one of these pairs.

Additive idempotence gives $a+a=a$, hence reflexivity. If $a+b=a$ and $b+c=b$, thenso the relation is transitive. In fact it is antisymmetric as well: $a+b=a$ and $b+a=b$, together with commutativity, imply $a=b$. Thus the relation is a partial order, and therefore a quasi-order. It is the reverse of the usual join order of an incline; keeping this reversal is essential for the well-quasi-ordering conclusion.

Both operations are order-preserving for this order. If $a⪯b$, then $(a+c)+(b+c)=a+c$ by additive idempotence, while distributivity gives $ac+bc=(a+b)c=ac$. The multiplicative absorption identity also saysThus multiplying a monomial by another factor moves upward in the reverse absorption order.

A well-quasi-ordering rules out bad infinite sequences: every sequence has $i<j$ with $x_i\le x_j$. It behaves well under finite products and finite words, but infinitary constructions expose a limitation. Better-quasi-ordering, introduced by Nash-Williams, strengthens the sequence test to a test on families of finite initial segments, and thereby provides a robust theory of infinitary closure.

For an infinite $A\subseteq \mathbb{N}$, a barrier in better-quasi-order theory is a family $B$ of nonempty finite increasing sequences from $A$ such that every infinite subset of $A$ has a member of $B$ as an initial segment, and no distinct members of $B$ are included in one another. The initial segment is unique. Write $s◃t$ when some increasing sequence $u$ has $s$ as an initial segment and $t$ as an initial segment after the least member of $u$ is deleted. For instance, in $[A{]}^2$ this means $(a,b)◃(b,c)$ with $a<b<c$.

A map $f:B\to Q$ is good if some $s◃t$ satisfies $f(s){\le }_{Q}f(t)$. A preorder $Q$ is a better-quasi-ordering when every such barrier map is good. Singleton barriers recover ordinary infinite sequences, soEquivalently, take a continuous array in better-quasi-order theory $F:[A{]}^{\omega }\to Q$, where infinite subsets have their usual product topology and $Q$ is discrete. Put $X^{-}=X\setminus \{\min X\}$. Better-quasi-ordering excludes arrays with $F(X)\nleqq F(X^{-})$ for every $X$. A barrier map gives a locally constant array by reading the unique initial segment; the converse uses refinement of the front of finite segments on which a continuous map becomes constant. This viewpoint makes the shift, rather than mere inclusion of subsets, central.

The implication is strict. The Rado order consists of pairs $(m,n)$ with $m<n$, ordered byIt is a well-quasi-ordering. In any infinite sequence, either the first coordinates are unbounded, giving a later first coordinate above the earlier second coordinate, or one first coordinate occurs infinitely often and its second coordinates have a nondecreasing pair. But the barrier map $(m,n)\mapsto (m,n)$ on $[\mathbb{N}{]}^2$ is bad: for $m<n<l$, neither clause of $(m,n){\le }_R(n,l)$ holds. Thus the Rado order is WQO but not BQO.

The same example explains why strengthening the definition is useful. Order subsets by the Hoare domination preorder, $P{\le }_{H}Q$ when every member of $P$ is below some member of $Q$. The infinite rows $R_m=\{(m,n):n>m\}$ form an antichain. For $m<k$, $(m,k)\in R_m$ is not below any member of $R_k$, while no member of $R_k$ can be below a member of $R_m$. Thus the full power set of a well-quasi-ordering need not be a well-quasi-ordering.

In contrast, power-set closure of better-quasi-orderings has a short array proof. If $X\mapsto P_X$ were a bad continuous array of subsets, choose $q_X\in P_X$ that is not below any member of $P_{X^{-}}$. Make this choice by a fixed choice rule depending only on the two subsets. They are locally constant, so $X\mapsto q_X$ is continuous. Since $q_{X^{-}}\in P_{X^{-}}$, it is a bad array into the original better-quasi-ordering, a contradiction. Empty subsets cannot be the source of a bad comparison. Thus the full power set, not just its finite subsets, preserves better-quasi-ordering under domination.

Further examples and closure principles show the scope of the theory. Every well-order is a better-quasi-ordering: a bad array would give a strictly descending ordinal sequence by successively deleting the least member of one infinite set. Every finite equality order is a better-quasi-ordering by the Nash-Williams barrier partition theorem, which makes a finite coloring of a barrier constant on an infinite restriction. The same theorem proves closure under finite products: color a bad product array by the first coordinate in which the shifted comparison fails, and pass to a homogeneous infinite restriction to obtain a bad coordinate array. Finite disjoint unions work by first homogenizing the component.

Nash-Williams' extension to countable transfinite sequences says that sequences of labels from a better-quasi-ordering, of arbitrary countable ordinal length, are again better-quasi-ordered by increasing label-monotone embeddings. It is an infinitary counterpart of Higman's lemma. Its formulation and logical strength are studied in arxiv.org/abs/math/9408204 . The point is not that every closure proof becomes elementary, but that barriers and continuous arrays supply the stable induction and partition framework absent from ordinary sequence tests. The modern account arxiv.org/pdf/1604.05866 develops this framework and the barrier partition theorem.

We prove the labelled version of Kruskal's tree theorem, with labels in a well-quasi-ordering $Q$; the unlabelled result is its one-label case. A label-monotone tree embedding is an injective map of finite rooted trees preserving lowest common ancestors and increasing labels. Its root may map below the host root, and an edge may map to a path. These conventions distinguish rooted-tree homeomorphic embedding from the stricter adjacency-preserving relation, which is not well-quasi-ordered.

We will use two elementary consequences of well-quasi-ordering. Every infinite sequence in a well-quasi-ordering has an infinite nondecreasing subsequence: color pairs by whether the earlier element precedes the later one, apply Ramsey's theorem, and exclude an all-bad infinite set. Restricting successively in each coordinate proves finite product closure of well-quasi-orderings.

For completeness, here is the required proof of Higman's lemma. If finite words over $Q$ admitted a bad sequence, choose one $w_0,w_1,\dots$ whose next word always has the least length admitting a bad continuation of the fixed prefix. No word is empty. Write $w_i=v_i{a}_i$, separating the last letter. The set of prefixes $\{v_i\}$ is a well-quasi-ordering. Otherwise select a bad sequence of these prefixes with strictly increasing owner indices $n_0<n_1<\cdots$; this is possible because every finite initial set of owners supplies only finitely many prefixes, and a bad sequence cannot repeat a word. Thenis bad. An earlier original word embedding into $v_{n_j}$ would also embed into $w_{n_j}$, contradicting the original badness; comparisons inside the new tail are excluded by its choice. But $v_{n_0}$ is shorter than $w_{n_0}$, contradicting minimality. Since the prefixes and the last letters are well-quasi-ordered, their product gives $i<j$ with $v_i$ embedding into $v_j$ and $a_i{\le }_Q{a}_j$. Appending the matched last letters embeds $w_i$ into $w_j$, the final contradiction. Thus finite words over a WQO are WQO.

Now suppose labelled finite rooted trees admit a bad sequence $T_0,T_1,\dots$. Choose it minimally in vertex count at each successive position. Let $\mathcal{S}$ be all immediate child subtrees of its members. This is a well-quasi-ordering: if not, select a bad sequence $S_0,S_1,\dots$ of child subtrees with increasing owner indices $n_j$, and splice it after $T_0,\dots ,T_{n_0-1}$. Every child subtree embeds into its owner, so a comparison from the prefix to the tail would contradict original badness. The splice is bad and its first replacement is strictly smaller, contradicting the minimal bad sequence choice.

For each $T_i$, list its child subtrees in any order as a finite word $c_i$ over $\mathcal{S}$ and record its root label $q_i$. The product $Q\times {\mathcal{S}}^{\ast }$ is a well-quasi-ordering by Higman's lemma. Hence some $i<j$ has $q_i{\le }_Q{q}_j$ and a subsequence embedding of $c_i$ into $c_j$. Match the roots and use the chosen subtree embeddings in distinct target branches. Their union preserves lowest common ancestors and labels, and stretches only edge paths. Thus $T_i$ embeds into $T_j$, a contradiction. We have proved

To deduce Friedman's finite form of Kruskal's theorem, fix $k\in \mathbb{N}$. At position $i=0,1,\dots$ permit unlabelled trees of at most $k+i$ vertices. Up to isomorphism only finitely many trees are possible at each position. Form the finite bad-sequence tree whose nodes are finite bad sequences satisfying these bounds, ordered by extension. It is finitely branching. If there were arbitrarily long admissible bad sequences, König infinity lemma would give an infinite branch, contradicting Kruskal's tree theorem. ThereforeOne can instead use a fixed finite set of labels, giving a correspondingly parameterized finite statement by exactly the same proof. The crucial feature is a uniform bound $N$ for each initial size allowance, not merely termination of one chosen sequence.

We use rooted-tree homeomorphic embedding: an injective vertex map preserves ancestors and lowest common ancestors, while an edge may map to a longer path. The root may map below the host root. Write $T⪯U$ for this relation. The assertion of Kruskal's tree theorem is that finite rooted trees form a well-quasi-ordering: every infinite sequence has $i<j$ with $T_i⪯T_j$.

We first supply the word lemma needed in the proof. In a well-quasi-order $Q$, every sequence has an infinite nondecreasing subsequence. Indeed, if no term had infinitely many later terms above it, successive choices beyond the finite sets of later dominators would give a bad sequence. Some term therefore has infinitely many later dominators; repeat the argument inside that infinite subsequence, which is still well-quasi-ordered, to build a chain.

Now order finite words over $Q$ by subsequence embedding with coordinatewise increase. Suppose a bad word sequence exists and choose one $w_0,w_1,\dots$ minimally: at each position use a word of least length allowing a bad continuation of the fixed prefix. No word is empty. Write $w_i=v_i{a}_i$, removing its last letter. Choose indices $i_0<i_1<\cdots$ with $a_{i_0}\le a_{i_1}\le \cdots$. The sequencewould also be bad. An old word embedding into $v_{i_j}$ embeds into $w_{i_j}$, contrary to the original badness; and $v_{i_j}⪯v_{i_l}$ would extend, using the last-letter comparison, to $w_{i_j}⪯w_{i_l}$. But $v_{i_0}$ is shorter than $w_{i_0}$, contradicting minimality. This proves Higman lemma, including its empty-word case.

Suppose next that there is a bad sequence of finite rooted trees, and choose a minimal bad sequence $T_0,T_1,\dots$ by vertex count. Let $\mathcal{S}$ consist of all proper rooted subtrees hanging below vertices of the $T_i$. This collection is well-quasi-ordered. Otherwise choose a bad sequence $S_0,S_1,\dots$ in $\mathcal{S}$. Its containing-tree indices can be selected strictly increasing, say $S_j$ is a proper subtree of $T_{i_j}$: a finite initial collection of containing trees has only finitely many proper subtrees, and a bad sequence cannot repeat an isomorphism class. Thenis bad. No earlier $T_l$ can embed into $S_j$, because $S_j⪯T_{i_j}$ and this would contradict the original sequence. The new tail is bad by construction. Since $|S_0|<|T_{i_0}|$, this contradicts minimality.

List each tree's immediate root subtrees in any chosen order. These are finite words in the well-quasi-order $\mathcal{S}$, so the word lemma gives $i<j$ for which the child word of $T_i$ embeds into that of $T_j$. Send the root of $T_i$ to the root of $T_j$, and use the chosen embeddings inside the selected, distinct target branches. The connecting paths from the target root to the images of the source children have disjoint interiors; lowest common ancestors in different branches are preserved. This constructs $T_i⪯T_j$, the final contradiction. Thus finite rooted trees are well-quasi-ordered by homeomorphic embedding. If labels lie in an arbitrary well-quasi-order, the same proof uses label-monotone embeddings; compare root labels and child words in their product well-quasi-order. The product property follows by taking a nondecreasing subsequence in the first coordinate and a good pair in the second.

For Friedman's finite form of Kruskal's theorem, fix $k$. Consider all finite bad prefixes $(T_1,\dots ,T_l)$ satisfying $|T_i|\le k+i$, up to rooted-tree isomorphism. They form a prefix tree. At each node there are finitely many possible next trees, because there are only finitely many rooted-tree isomorphism types on at most $k+l+1$ vertices. If prefixes of arbitrary length existed, the König infinity lemma would give an infinite branch, hence an infinite bad sequence, contrary to what we proved. Its height is therefore bounded. ConsequentlyThe same finite-prefix argument works with a fixed finite label alphabet, and more generally with any prescribed size bound $b(i)$ having only finitely many possibilities at each position. These are finite existence statements; no numerical estimate on $N$ is needed for the conclusion.

The PDF's displayed inequality uses the relation it purports to define and therefore is not itself a complete definition. The absorption order consistent with that inequality and with the requested result isThis is the reverse of the usual natural order of an incline. We specify it explicitly: using instead $x+y=y$ would make the finite-generation conclusion false in general.

Idempotence gives $x+x=x$, proving reflexivity. If $x+y=x$ and $y+z=y$, then associativity givesso the relation is transitive. If also $y+x=y$, commutativity forces $x=y$. Thus it is even a partial order. In this order $x+y⪯x$ follows because $(x+y)+x=x+y$. Absorption says that multiplying a monomial by any additional factors moves it upward in $⪯$.

Let $g_1,\dots ,g_r$ generate the incline. Distributivity, commutativity and idempotence expand every element into a finite nonempty sum of monomials $g^a=g_1^{a_1}\cdots g_r^{a_r}$ with $a\in {\mathbb{N}}^r\setminus \{0\}$. A zero exponent means that factor is omitted; no multiplicative identity is assumed. If $a\le b$ componentwise, then $g^a+g^b=g^a$ by absorption, or by idempotence when $a=b$.

We prove the needed reverse-inclusion well-quasi-ordering of monomial ideals directly. Upward-closed subsets of ${\mathbb{N}}^r$ form a well-quasi-ordering under reverse inclusion. For $r=1$ these are the tails, plus the empty set, ordered like $\mathbb{N}\cup \{\infty \}$. For the induction, write an upset $U\subseteq {\mathbb{N}}^{r+1}$ as increasing slices $U_0\subseteq U_1\subseteq \cdots$ in ${\mathbb{N}}^r$. Their union is an upset with a finite minimal basis by Dickson lemma. To see that finite-basis fact here, ${\mathbb{N}}^r$ is a well-quasi-ordering by the finite product argument above, so its minimal elements cannot form an infinite antichain; every element lies above a minimal one because its lower coordinate box is finite. All basis elements of the union appear by some finite slice, so the slices eventually stabilize.

Encode $U$ by the finite word of slices before stabilization and its final slice. By the finite-word theorem proved in the other alternative, and finite product closure, these encodings are a well-quasi-ordering with slice comparisons given by reverse inclusion. For two comparable encodings, let the earlier pre-stabilization slice $U_k$ match the later slice $V_{f(k)}$, where $f(k)\ge k$. ThenBeyond the earlier stabilization index, its final slice contains the later final slice and hence every later slice. Therefore $U\supseteq V$. This proves the induction, including empty slices and the empty upset.

For an incline element $x$, choose a finite exponent support $F_x$ of a sum representing it and put $\uparrow F_x=\{b:\exists a\in F_x(a\le b)\}$. Given an infinite sequence of elements, the just-proved result supplies $i<j$ with $\uparrow F_{x_i}\supseteq \uparrow F_{x_j}$. Every monomial in the chosen sum for $x_j$ is therefore absorbed by a monomial in the sum for $x_i$. Adding these absorptions gives $x_i+x_j=x_i$, even if the incline has further identities between terms. HenceThe order direction matters. On the positive integers take $x+y=\min (x,y)$ and multiplication to be ordinary addition of integers. This is an incline generated by $1$. Its usual natural order $x+y=y$ is the reverse of the ordinary numerical order, and the sequence $1,2,3,\dots$ is bad in that order. Its reverse absorption order is the ordinary numerical order, consistent with the theorem above.

For finite rooted trees, use rooted-tree homeomorphic embedding: an injective vertex map preserves ancestors and lowest common ancestors, so edges may become paths and distinct branches remain distinct. The root can map to a vertex below the host root. More generally label vertices in a well-quasi-ordering $Q$ and require each source label to precede its image label. We prove this stronger labelled version of Kruskal's tree theorem, from which the unlabelled result follows by taking one label.

We first need the finite-word closure, and prove it rather than merely invoking Higman lemma. If finite words over $Q$ were not a well-quasi-ordering under order-preserving subsequence embedding with increased labels, choose a minimal bad sequence $w_0,w_1,\dots$, minimizing the length of $w_n$ among words admitting a bad continuation of the already chosen prefix. No $w_n$ is empty. Write $w_n=v_n{a}_n$ by removing its last letter. The perfect subsequence lemma gives $n_0<n_1<\cdots$ with $a_{n_r}\le a_{n_s}$ for $r<s$.

The sequence $w_0,\dots ,w_{n_0-1},v_{n_0},v_{n_1},\dots$ is still bad. An earlier $w_i$ embedding into a later $v_{n_r}$ would embed into $w_{n_r}$, contradicting the original sequence. An embedding of $v_{n_r}$ into $v_{n_s}$ extends by the last-letter comparison to an embedding of $w_{n_r}$ into $w_{n_s}$. Both possibilities are excluded. This contradicts minimality at position $n_0$, since $v_{n_0}$ is shorter. Thus the finite-word closure holds. The perfect subsequence lemma also proves finite product closure: first pass to a nondecreasing subsequence in one coordinate, then in the next, preserving all previous comparisons.

Suppose now that the labelled trees admit a bad sequence $T_0,T_1,\dots$. Choose a minimal bad sequence by vertex number. Let $\mathcal{U}$ be all the proper descendant-rooted subtrees of these trees. We claim that $\mathcal{U}$ is a well-quasi-ordering. If it had a bad sequence of subtrees $U_r$, its parent indices can be made strictly increasing, say $U_r$ comes from $T_{n_r}$. Indeed each finite initial collection of parent trees has only finitely many subtrees, and a bad sequence cannot contain infinitely many terms from a finite collection. Passing to a subsequence therefore makes $n_0<n_1<\cdots$.

Then $T_0,\dots ,T_{n_0-1},U_0,U_1,\dots$ is bad. A prefix tree embedding into $U_r$ would embed into its parent $T_{n_r}$, contrary to the original badness, and the tail is bad by construction. But $U_0$ has fewer vertices than $T_{n_0}$, contradicting minimality. This proves the claim.

Give the children of each root any fixed order. Its forest of child subtrees is a finite word in $\mathcal{U}$. The word theorem and finite product closure show that the pairs consisting of the root label and the child-subtree word are a well-quasi-ordering. Thus some $i<j$ have comparable root labels and an embedding of the child word of $T_i$ into that of $T_j$. Map the source root to the target root and use the obtained embeddings inside distinct target child subtrees. The paths from the target root into those distinct branches are disjoint except at the root, so this is a label-monotone tree embedding. It contradicts badness. Hence

For Friedman's finite form of Kruskal's theorem, fix $k\in \mathbb{N}$. There is $N(k)$ such that any $N(k)$ trees with $|T_i|\le k+i$ contain $i<j$ with $T_i⪯T_j$. Use indices starting at zero; shifting them only changes $k$. To prove this uniform finite bound, form the finite bad-sequence tree of finite bad prefixes obeying the size bounds. At position $i$ there are only finitely many rooted-tree isomorphism types with at most $k+i$ vertices, so every node has finitely many successors.

If no bound existed, this tree would have nodes of arbitrarily large depth. A node with arbitrarily long extensions has a child with arbitrarily long extensions, since it has only finitely many children. Choosing such a child repeatedly produces an infinite branch, as in the König infinity lemma. The branch is an infinite bad tree sequence, contradicting the theorem just proved. Therefore the required $N(k)$ exists. The same proof works with any fixed finite label alphabet and any specified size bound finite at each position.

A well-quasi-ordering is a preorder $(Q,\le )$ such that every infinite sequence $(x_n)$ has indices $i<j$ with $x_i\le x_j$. Reflexivity and transitivity are part of the definition; antisymmetry is unnecessary. An infinite sequence without such a pair is a bad sequence.

The perfect subsequence lemma strengthens this definition: every infinite sequence in a well-quasi-order has an infinite nondecreasing subsequence, meaningHere is a proof that does not assume the lemma implicitly. There must be an index $i$ with infinitely many later indices $j$ satisfying $x_i\le x_j$. Otherwise each index would have only finitely many such successors. Starting with any index, choose the next one beyond the union of those finite successor sets for the previously chosen indices. This constructs a bad sequence, contradicting the well-quasi-ordering property.

Choose $i_0$ with infinitely many successors above it, and restrict to those successors. Apply the same argument to that infinite subsequence, obtaining $i_1>i_0$ with infinitely many later successors above it within the restricted sequence. Continue. Every subsequent restriction lies inside all previous upper cones, so $x_{i_r}\le x_{i_s}$ for every $r<s$. This proves the lemma, including for a genuine quasi-order with distinct equivalent elements. Both alternatives below use it.

Use the rooted-tree homeomorphic embedding relation: an injective map of vertices preserves lowest common ancestors, so edges may become paths and the source root may map below the target root. We prove the stronger labelled statement for labels in any well-quasi-ordering $Q$, with labels increasing under the embedding. Forgetting labels gives Kruskal's tree theorem.

First every infinite sequence in a well-quasi-ordering has an infinite nondecreasing subsequence. Colour its index pairs according as the earlier term is or is not below the later term. Ramsey's theorem gives an infinite constant-colour subsequence; the bad colour would contradict well-quasi-ordering. Applying this successively in each coordinate proves finite product closure of well-quasi-orderings.

We need Higman lemma with its proof. Order finite words over $Q$ by subsequence embedding with increased letters. If there were a bad sequence, choose one $w_0,w_1,\dots$ minimally: at each position choose a shortest word admitting a bad continuation of the fixed prefix. No word is empty. Write $w_n=u_n{a}_n$, separating the last letter, and choose $n_0<n_1<\cdots$ with $a_{n_0}\le a_{n_1}\le \cdots$. ConsiderAn earlier prefix word cannot embed in a later $u_{n_j}$, since it would then embed in $w_{n_j}$. Nor can $u_{n_i}$ embed in $u_{n_j}$: adding the comparable last letters would embed $w_{n_i}$ in $w_{n_j}$. Thus the displayed sequence is bad, contradicting minimality at $n_0$, since $u_{n_0}$ is shorter. Finite words over a well-quasi-order are well-quasi-ordered.

Now suppose labelled rooted trees admit a bad sequence $T_0,T_1,\dots$, chosen minimally in vertex count at each position. Let $\mathcal{A}$ consist of all proper rooted subtrees of these trees, rooted at a descendant and containing all its descendants. We claim $\mathcal{A}$ is well-quasi-ordered. Otherwise take a bad sequence $S_0,S_1,\dots$ from $\mathcal{A}$. Only finitely many subtrees come from any fixed finite prefix of the $T_n$'s, and a bad sequence cannot repeat an object. Thin it so that $S_i$ comes from $T_{m_i}$ with $m_0<m_1<\cdots$. The sequenceis bad: $T_j⪯S_i$ would imply $T_j⪯T_{m_i}$, since a rooted subtree embeds in its host; comparisons within the $S_i$'s are excluded by their choice. But $S_0$ is strictly smaller than $T_{m_0}$, contradicting minimality. This proves the claim.

Choose an arbitrary ordering of the children of each $T_n$. Record its root label and the finite word of rooted child-subtrees:By Higman lemma and finite product closure, two records, with indices $i<j$, are comparable. Map the root of $T_i$ to the root of $T_j$, and map its selected child-subtrees into the distinct selected branches of $T_j$. Within a branch the embeddings preserve lowest common ancestors; between different branches both lowest common ancestors are the roots. Labels increase everywhere. This gives $T_i{⪯}_Q{T}_j$, contradicting badness. Finite labelled rooted trees are well-quasi-ordered by label-monotone homeomorphic embedding. In particular, this proves the unlabelled theorem. A root-preserving variant follows by giving the root a special incomparable label unavailable at nonroot vertices; the general labelled theorem then forces root to root. Adjacency-preserving embeddings are a different, more restrictive relation and do not satisfy the same assertion.

For Friedman's finite form of Kruskal's theorem, fix $k\in \mathbb{N}$ and consider finite bad sequences satisfying $|T_i|\le k+i$, with indices starting at $1$. Their prefixes form a finite bad-sequence tree. At each depth there are only finitely many possible unlabelled tree isomorphism types below the size bound, so this tree is finitely branching. If bad sequences existed at every finite length, König infinity lemma would yield an infinite branch, hence an infinite bad sequence of rooted trees, contrary to what was just proved. ThereforeThe same proof works with any fixed finite label set. The finite size bounds are what make the bad-prefix tree finitely branching; the proof does not claim a small effective numerical bound.