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This is an author-recorded argument by Plasma AI for the source interface of the three-game route. It uses the coherent successor convention of the interface, not the mixed child counts in the printed game. It reconstructs the finite-height use of Theorems 19 and 21 and Corollary 2 of Schipperus's journal paper, with the equal-block hypothesis that the application needs. It does not construct the three games in the Architect case.

Statement and the necessary hypothesis

Fix an integer $h\ge1$. In $W_h$, a successor node with label maximum $M$ has children $0,\ldots,M$, and its selected children have indices in $\Delta\setminus\{M\}$. At every internal node the Architect chooses $q=|\Delta|-1$. Leaves have singleton labels. Turns pass at selected leaves and on finishing a tree. A play is correct when its selected leaves are exactly the maxima of the nonfinal cells induced by the other tree. All numerical labels in one play increase in play order.

A threshold rule assigns a finite lower bound $b(P,q)$ to each position $P$ and its current prescription $q$; at a leaf use a dummy prescription $q=0$. A Builder move is sufficiently large if every element of its label exceeds that bound. The bound can depend on the complete numerical history, including unselected nodes.

Builder proposition. Suppose $H\subseteq\omega$ is infinite and $\chi:[W_h]^2\to\{0,1\}$ is zero on every pair with equal root length. Suppose a threshold rule has the property that every sufficiently large play in $H$ is a Builder win, for every sequence of Architect choices. Then there is $X\subseteq W_h$ of order type $\omega^{\omega^h}$ with $\chi\restriction[X]^2=0$.

The equal-root-length hypothesis cannot be omitted from this statement. Indeed, color a pair $1$ exactly when its two root lengths agree. In every play the two root maxima are distinct, so every play has color $0$ and is a Builder win. But a color-$0$ set has at most one element of each root length, hence order type at most $\omega$. This observation concerns the unnormalized Builder statement, not the published positive partition theorem: the proof of Theorem 28 normalizes before using the Builder implication. The factor-two block argument supplies precisely this hypothesis in the triangle-free case.

The dichotomy and finite-cover refinement

The external Ramsey input is the Nash-Williams theorem in the paper's Theorem 17: every finite coloring of a family of finite subsets of $\omega$, no one an initial segment of another, is homogeneous on an infinite subset. Its statement is used here, not its proof.

Dichotomy. Given any coloring $\chi$ and infinite $K$, there is infinite $H\subseteq K$ on which either the Architect has a winning strategy for sufficiently large Builder moves, or every sufficiently large Builder play wins, regardless of the Architect's choices. Both alternatives allow a finite threshold depending on the entire current history and prescription. This assertion does not require normalization.

Here is the well-founded fusion argument in this game's conventions. Keep the Architect's counts in a position as well as the Builder's numbers. Every full play is finite. Indeed, once a node is built its full number of children is fixed and finite, and induction on $h$ proves that its descendants are eventually exhausted. Equivalently, assign the finite pending forest the ordinal obtained by summing one $\omega^r$ for every pending subtree of height $r$, in decreasing order of $r$. Building a node replaces one such term by finitely many terms of smaller exponent, or deletes a terminal term. This strictly decreases the ordinal. The two trees cannot support infinitely many switches without building new nodes. Thus the tree of legal positions is well founded under proper extension.

Consider a position $P$ just before an Architect move. Once a particular next count $q$ is fixed, a finite increasing set of new Builder numbers uniquely determines the successive Builder labels up to the next Architect move or the end of the game. The sizes and node order are determined by the current position and the rules. Let $F_{P,q}$ consist of those finite sets at which this stopping point is first reached. No member is a proper initial segment of another. Every infinite tail of legal Builder numbers has an initial segment in $F_{P,q}$, because the play cannot remain forever between Architect moves. At a terminal position the winner is already determined.

We prove by well-founded induction that every position just before an Architect move can be determined after thinning any infinite set. Assume the assertion for all proper extension positions. For every possible continuation $s\in F_{P,q}$, apply that assertion to the resulting position, using the already determined winner if it is terminal. These countably many applications can be arranged by diagonal fusion so that all resulting positions are determined on one common set, with finite initial exceptions depending on $s$ and $q$.

For completeness, enumerate the pairs $(q,s)$. Build decreasing infinite reservoirs and increasing chosen elements. At step $j$, retain the previously chosen elements and thin the unused reservoir for the first $j$ pairs by the inductive assertion. Choose the next element from the resulting reservoir. Every fixed pair is then handled from some stage on, so the final diagonal set is almost contained in its determining reservoir. Its exceptional finite prefix can be included in the bound at the continuation position. This argument may apply the induction hypothesis to positions whose earlier numbers are not in the current reservoir: only their future moves are restricted to that reservoir.

Color each $s\in F_{P,q}$ by its determined alternative, Architect or Builder, on a tail of this diagonal set. Apply Nash-Williams to this two-coloring. Do so for $q=0,1,\ldots$ in decreasing infinite sets and take a diagonal subset $H$. For each fixed $q$, all sufficiently large $s\in F_{P,q}\cap[H]^{<\omega}$ have the same alternative. The finite exception now becomes the bound on the first Builder move after $q$.

If some $q$ has alternative Architect, choose it and, after seeing $s$, use its winning continuation strategy. The thresholds at that continuation and all its later positions are allowed to depend on $s$. This is an Architect strategy at $P$. If every $q$ has alternative Builder, every sufficiently large continuation of every choice of $q$ is a Builder win. This proves the inductive step. At the empty position it gives the dichotomy.

All thinnings preserve previously determined alternatives. At each position the needed exceptional sets are finite; there is no claim of one constant threshold for all histories or all counts. The fusion also does not make the Architect's unbounded integer counts constant.

Finite-cover refinement. Suppose the Architect wins on $H_0$, and finitely many classes $\mathcal C_1,\ldots,\mathcal C_m$ cover all output pairs. There is an infinite $H\subseteq H_0$ and an Architect strategy winning for $\chi$ and, on every sufficiently large play, outputting a pair in one fixed $\mathcal C_i$.

To prove this, apply the dichotomy successively to the colorings which are $1$ exactly on color-$1$ pairs belonging to $\mathcal C_i$. If one has an Architect win, it is the required strategy. Otherwise all have the universal Builder alternative on the final infinite set. Play the original winning Architect strategy with Builder moves meeting its bound and the finitely many new bounds at each position. The completed play is correct and has original color $1$. It belongs to a class $\mathcal C_i$, contradicting the universal Builder alternative for that coloring. This proves the refinement.

The rest of the page proves the Builder proposition stated above.

Partial trees and stages

We construct a countable tree of partial, canonically labeled elements of $W_h$. Each partial tree is an initial segment in depth-first order. Besides its full child lists it has a canonical selected subtree $G$. The root is selected. At a selected internal node choose $q\ge1$ selected children and one larger maximum $M$; at an unselected internal node use $\{M\}$, so no descendant is selected. At each leaf choose a singleton. The last child $M$ is always unselected. Every chosen number lies in $H$.

An extension stage begins with one existing partial tree, builds all intervening unselected nodes, and stops at its next selected leaf. The empty partial tree is allowed: its first stage labels its root and builds to its first selected leaf. If no selected node remains pending, complete the unselected final part once, and record the resulting full tree as a terminal output. This final completion is a stage too.

For every partial tree with a selected node still pending, perform infinitely many different extension stages. A fair enumeration of these countably many requests does this: at each finite stage only finitely many partial trees and requests have appeared. Each extension is finite, since its full child counts are finite and its height is at most $h$. Likewise each path of the construction reaches a terminal output: once a node has been labeled its finite child count is fixed, and induction on the height forbids an infinite path of full-tree construction.

All numerical labels newly assigned at one stage exceed every number assigned at any earlier stage, including stages on other branches. At each newly labeled selected internal node choose $q$ larger than every previously used selected degree. Require also that the first such $q$ in an extension exceed its extension-request index. This makes the available degrees unbounded at every selected branching opportunity. The leaf choices are unbounded for the same reason of freshness.

Here and below a copy of an existing prefix retains its labels; only new nodes receive new numbers. Thus branches may share an initial prefix. Distinct first stages have different root maxima, and terminal outputs with equal root length have the same first stage.

Meeting every relevant threshold

When a canonical internal node is about to receive a label of size $q+1$, let $N$ bound every number already used in the entire construction. There are only finitely many possible game histories whose Builder numbers are at most $N$. The reason is elementary at finite height: each node uses a nonempty label, the labels increase strictly, and the old Architect counts are one less than the corresponding label sizes. There are also only finitely many current prescriptions $r\le q$.

Choose the entire new canonical label above $N$ and all bounds $b(P,r)$ for these histories and these prescriptions. For an unselected node or a leaf the same rule uses $r=0$. This is possible because $H$ is infinite. It handles all subsets of a future canonical label, not just the canonical size $q+1$.

The possible coarsening at a node preserves its maximum and deletes some selected indices. Consequently its label has size $r+1$ for some $r\le q$. The preceding choice is therefore sufficiently large for every coarsened play whose prior numbers were already constructed. No future history is queried or presumed known.

Cross-block pairs are correct sufficiently large plays

Let $X$ be the terminal outputs. A terminal tree's canonical selected leaves divide its leaf set into cells, with a final cell after the last selected leaf. Every cell was built in one extension stage, including the final completion stage. Its canonical partition is good: labels increase in tree order, and every internal node's last child is unselected.

Take $T,S\in X$ with different root lengths. Their paths in the construction separate at their first stages. Their canonical numerical labels are consequently disjoint. A stage is built all at once, using fresh increasing numbers; no leaf of one tree lies between two leaves of a canonical cell of the other. Thus each induced partition is a coarsening of its canonical partition.

For each induced partition use its actual labels from Definition 9. They are subsets of the canonical labels and preserve every internal maximum. They therefore preserve the full trees, are increasing, and have unselected last children. Their selected nodes are the upward closures of the induced nonfinal cell maxima. In particular they give a correct play in the coherent game, choosing at each internal node the Architect count equal to the size of its actual label minus one. All positive levels are compulsory Architect levels, so no optional move-location rule is needed.

To check numerical play order, read the two trees by their induced cells. The order of the canonical stages contributing to these cells is their order of construction: all new labels of an earlier stage precede every label of a later one. In each tree this is also its depth-first order. The tree with the smaller root maximum has the earlier first stage and earlier first leaf. Its initial induced cell ends before the other tree's root label is chosen. These facts give all the cross-tree increasing-label requirements, including the root placement in Definition 13(5).

At every node, the earlier labels of this pair's play had already been chosen when that node's canonical label was chosen. The finite maximum of threshold bounds above therefore covers the actual history and the actual coarsened count. The reconstructed play is sufficiently large. It is correct, so the Builder-win hypothesis makes its color $0$. Pairs of equal root length have color $0$ by normalization. Hence $X$ is homogeneous.

Order type

Compress each terminal tree to its canonical selected tree: discard unselected nodes and compress the selected child indices, retaining the ordinal levels, the selected degrees $q$, and the selected leaf labels. Denote this tree by $g(T)\in W_h$.

The map $T\mapsto g(T)$ is an order embedding. To see this, take the first stage at which two paths of the construction differ. In the full trees their first new node has either a larger child-count maximum or a larger leaf label on the path whose stage was constructed later. All preceding nodes agree. In the compressed trees the first new selected node has a larger degree $q$, or a larger leaf label if it is terminal, on that same path. The degrees and labels were chosen strictly increasing across construction stages. Thus both full trees and compressed trees have the same ordering. The single final completion after a terminal selected prefix prevents duplicate compressed outputs.

The following finite-height freedom argument supplies the lower bound. At any selected prefix, the next extension may give its first new selected internal node arbitrarily large degree, and then continues to a selected leaf. If that node is itself a leaf, its label ranges over an unbounded set. All later pending selected branches have the same extension property, conditional on the full prefix already chosen.

At height zero the possible completions therefore have type $\omega$. Inductively suppose a fresh selected subtree of height $r$ has completion type at least $\Omega_r=\omega^{\omega^r}$ under any such prefix. At a fresh subtree of height $r+1$, choose an initial extension whose root degree is $q\ge n+1$, and finish its first child in one arbitrary way. Its remaining $q-1$ child subtrees are fresh. Successive application of the induction hypothesis gives a lexicographically ordered family of completions of type at least $\Omega_r^{q-1}\ge\Omega_r^n$. The available families for later children may depend on the earlier choices; the bound is uniform over those prefixes, so the ordinal sum over these fibers still gives the same product lower bound. Since $n$ is arbitrary, the completion type is at least $\sup_n\Omega_r^n=\omega^{\omega^{r+1}}$.

Applying this argument to the empty selected prefix gives $\operatorname{ot}(g(X))\ge\omega^{\omega^h}$. The embedding into $W_h$ gives the reverse bound. Thus $\operatorname{ot}(X)$ is exactly $\omega^{\omega^h}$, proving the proposition.

Scope

The construction and its proof are author-recorded and unreviewed. There is no numerical evidence or independent tier assertion. Its source comparison is with Definitions 6--19, the construction in Theorem 21, and Lemmas 22, 24 and 25, read at proof-prose depth in the 2010 journal version. The finite-height proof here has no collapsing maps and uses its own stated completion rule and order argument.

For the canonical attack, these arguments discharge its finite-height source-transfer obligation S at author standing, using Nash-Williams as the external theorem. The inside middle three-game construction remains unproved, as does the extension to all sums of three indecomposables.

The family construction itself does not use the color or the universal-Builder hypothesis; those are used only to infer homogeneity. For any $H$ and threshold rule it therefore supplies a full-order-type host whose unequal-root-length pairs are good sufficiently large plays. This is a possible host for the negative pattern route. Unavoidability of a prescribed pattern in every full-order-type subset of this host is a further assertion and is not proved by the construction.