Finite mixed-boundary reflection and gluing on the rotated FK torus

Write $D$ for the dual lattice: its sites have opposite coordinate parity and its edges also have displacements $(\pm 1,\pm 1)$. Each dual edge crosses its corresponding primal edge at their common midpoint. The map $(u,v)\mapsto ((u-v)\mathrm{/}2,(u+v)\mathrm{/}2)$ identifies $P$ with the ordinary square lattice and the two torus periods with $(0,m),(-m,0)$. Thus [F2] applies to this cellulation. A nonzero diagonal period has a coordinate of magnitude at least $m$, so any rectangle whose two coordinate spans are less than $m$ embeds, with its edge labels distinct. Translation by a primal vertex, coordinate interchange, and quarter-turns preserve the period group and ordinary FK weights.

We shall use the following elementary consequence of [F4]. A simple polygonal arc in a rectangle whose only boundary points are one point on each of two opposite sides separates the other two sides. Join its endpoints through an exterior collar around one of the remaining sides to form a Jordan polygon and apply [F4]. Boundary contacts of an original finite path can instead be put in the interior of a slightly enlarged rectangle, with its endpoints extended through disjoint collars; this gives the same separation conclusion. Two primal opposite crossings intersect at a vertex: separation gives an intersection, and primal diagonal edges do not cross except at common endpoints. Primal and dual edges, on the other hand, meet transversely exactly at the corresponding diagonal midpoints.

We specify a finite one-sided exploration without artificial FK boundary wiring. For a finite plane rectangle graph and an ordered subset $S$ of its left-side vertices, add a virtual root and disjoint exterior fan arcs to $S$, ordered top to bottom. They are search instructions only, not labels in the probability space. Put the root in a narrow left collar, with a stem from the enlarged left boundary; extend a successful right-side target through a right collar. Run depth-first search. A vertex, once visited, stays marked. At an actual vertex inspect incident edges clockwise, starting strictly after the incoming parent half-edge and resuming the same list after a recursive return. Skip an edge to a marked neighbour without querying its state. For an unmarked neighbour query its label, unless already known, and recurse precisely when open. At the root iterate its fan children in the specified order. Stop on the first right-side target; an empty stack means failure. Every marked actual vertex is connected to $S$ by its parent-tree edges, and on failure every edge from a marked to an unmarked vertex is closed. Thus it succeeds exactly when a crossing exists, returns a simple open path, and fixes all ties and failure branches in finitely many operations. The artificial fan creates no connection in the FK law.

Fix a successful complete configuration and let $L$ be the final root-to-target branch, including its fan and collar extensions, oriented left to right. The stack at success consists exactly of $L$; all other marked vertices have completed their recursive calls. Their queried-open tree edges form completed subtrees attached to vertices of $L$. Such a subtree has no other vertex on $L$, because a previously marked vertex cannot be visited anew. At a nonroot attachment, each completed subtree was processed before the final child on $L$, so its first half-edge lies in the clockwise sector from the reverse incoming $L$ half-edge to the outgoing $L$ half-edge. That sector is exactly the local left side of oriented $L$, possibly a reflex sector of angle greater than $\pi$. At the root the completed fan children are the higher children and lie on that same left side. Each subtree's interior is connected, cannot cross $L$ by planarity, and cannot meet another $L$ vertex by permanent marking. Since $L$ is a proper arc in the collared rectangle, step 2.1 puts the whole subtree on its left, that is, above the actual output path. Skipped marked fan children query no label.

A queried-closed edge whose initial vertex is off $L$ starts in one of those completed subtrees; its interior cannot cross $L$, so it lies on the same left side, although its other endpoint might later become a vertex of $L$. If its initial vertex is on $L$, the query preceded that vertex's final successful child: once that child succeeds, success propagates up the stack and its list never resumes. This edge therefore starts in the same earlier left angular sector, cannot cross $L$, and remains on the left, possibly ending at another $L$ vertex. The final target makes no query. Edges to marked neighbours, including possible cycle chords, were never queried. Hence every queried actual edge is on or above the returned path, and every edge whose relative interior lies below it is unqueried. Collars handle all old boundary contacts. Reflecting by coordinate interchange reverses clockwise order and gives the corresponding search from any bottom-source subset to the top, querying only on or to the right of its simple output path; its root sources are ordered right to left. These are exact finite algorithms on the actual graph labels.

An adaptive query transcript fixes exactly its queried states: the next query is a deterministic function of earlier queried states, and skipping a marked neighbour does not inspect that edge. Thus fixing the whole transcript imposes no additional condition on unqueried labels. Their conditional law is the spatial Markov law of [F3]. The second search may read states known from the first; it produces the same output and query set as on the same complete configuration from scratch, and the combined transcript merely omits duplicate queries. In particular the two one-sidedness conclusions in step 5.1 remain valid sequentially.

Let $N$ be even and $B=\{0,\dots ,N-1{\}}^2$. Put $P_B=P\cap B$ and $D_B=\{(u,v)\in B:u\equiv ̸v(\mathrm{m}\mathrm{o}\mathrm{d}2)\}$. The primal graph has all diagonal labels with both endpoints in $P_B$; for each such label $e$, the other diagonal of its unique unit cell is a distinct dual label $e^{\ast }$ joining two vertices in $D_B$. Thus the dual labels are in bijection with the primal labels, including at the boundary, and no label outside $B$ is used. Let $L=P_B\cap \{u=0\}$, $R=P_B\cap \{u=N-1\}$, $D_0=D_B\cap \{v=0\}$, and $D_1=D_B\cap \{v=N-1\}$. Write $H_B$ for an actual open primal path from $L$ to $R$, and $V_B^{\ast }$ for an actual open dual path from $D_0$ to $D_1$ using $e^{\ast }$ exactly when $e$ is closed. We verify the finite alternative $H_B^c=V_B^{\ast }$. If $H_B$ fails, mark every primal vertex reachable from $L$ by open labels; this finite search stops after at most $\mathrm{\mid }{P}_B\mathrm{\mid }$ additions, and no vertex of $R$ is marked. Every label with one marked and one unmarked endpoint is closed. Put its dual label in $K^{\ast }$. Around an interior dual vertex, the marked/unmarked status of the cyclically ordered primal sectors changes an even number of times, so $K^{\ast }$ has even degree there. At a noncorner dual vertex on the left boundary, the two in-box incidences have sector statuses $(1,s,1)$, where both left-side vertices are marked and $s\in \{0,1\}$ is the interior status; its cut degree is therefore $0$ or $2$. On the right boundary the corresponding statuses are $(0,s,0)$, so its cut degree is also even. The two corner dual vertices lie on $D_0$ or $D_1$ and are counted there; the other two corners are primal vertices. Along each dual terminal row, the in-box primal sector chain runs from the marked left side to the unmarked right side, so it has an odd number of toggles and hence an odd number of odd-degree vertices of $K^{\ast }$ on that row. Extra contacts of the marked cluster with a boundary row add pairs of toggles, and at corners only in-box incidences are counted. Every component of a finite graph has an even number of odd-degree vertices. If no component of $K^{\ast }$ met both $D_0$ and $D_1$, each component's odd vertices would lie on just one row and occur there in an even number, contradicting the odd total on each row. Thus some component contains a dual path from $D_0$ to $D_1$, using only labels in $K^{\ast }$ and hence only closed-crossing dual-open labels. Conversely, if both crossings existed, erase loops and move nonterminal boundary contacts into thin collars to obtain proper crosscuts with endpoints on the stated terminal sets. By [F4] they intersect. Primal and dual vertices are disjoint and their edges meet only at the midpoint of their paired labels; a label in an open primal path cannot have its dual label open. This proves the exact alternative and explicitly includes boundary contacts without adding exterior or terminal-row edges. The affine quarter-turn $(u,v)\mapsto (N-1-v,u)$ takes $D_B$ to $P_B$ and $V_B^{\ast }$ to the primal horizontal event $H_B$; it preserves the diagonal torus period group and the labels. Therefore [F2] gives $1-\varphi (H_B)\le q^2\varphi (H_B)$, so $\varphi (H_B)\ge a_0(q)$. Coordinate interchange gives the vertical estimate.

Reveal only the half-open square's labels and choose a horizontal crossing deterministically when it exists. Its right-layer endpoint is $(N-1,v)$, $0\le v\le N-1$. The diagonal edge to $(N,v+1)$ lies outside that revealed label set and inside $[0,N{]}^2$; it opens with conditional probability at least $r(q)$ by [F3]. Thus the closed square's horizontal probability is at least $a(q)$. Extending a vertical crossing from $(u,N-1)$ to $(u+1,N)$ proves its vertical bound. All these labels are distinct by step 1.1.

We next construct an auxiliary finite disk, independently of the torus. Reflect in $u=c+1\mathrm{/}2$, writing $s$ for that map. Let $a_1,a_2$ be simple primal paths respectively in $u\le c$ and $u\ge c+1$, meeting those limiting layers only at their starts $(c,h),(c+1,b)$, with $b<h$; singleton paths are permitted. Suppose $a_1$ and $s(a_2)$ are disjoint and their final vertices are axial neighbours of distance one. The four arcs $a_1,s(a_2),a_2,s(a_1)$, connected by the axial unit final connectors and the two horizontal unit initial connectors, form a Jordan polygon. The two arcs in each half-plane are disjoint, opposite half-plane arcs cannot meet, and a unit axial connector has no lattice vertex or diagonal-edge intersection in its interior: a diagonal edge meets an integer horizontal or vertical line only at an integer endpoint. Simplicity excludes further endpoint contacts. These facts also cover singleton paths. Its bounded closed disk $\mathrm{\Omega }$ is reflection invariant and contains the central strip $c<u<c+1$, $b<v<h$.

List the auxiliary primal vertices as $P\cap \mathrm{\Omega }$ and labels as all primal edges wholly in $\mathrm{\Omega }$. Wire $a_1$ separately from $a_2$. Contract their boundary paths to distinct vertices $p_1,p_2$ and delete the contracted path-edge labels. These labels were independent loops under the wiring, and summing their factors gives $p+(1-p)=1$, so their deletion changes neither the remaining marginal nor the event $C$ of an open connection between the two original classes. Call the retained edge set $E_0$ and the resulting plane multigraph $G_s$. Contract the two boundary paths in thin exterior collars; contracting each nonloop boundary-path edge preserves faces. Every strictly interior dual site is the centre of its entire primal diamond: a primal boundary path cannot cross a diamond interior, and a dual boundary arc has its centres on the boundary, not strictly inside. Thus the bounded faces are exactly those interior diamonds. Every other face sector reaches the exterior along $s(a_1)$ or $s(a_2)$; successive sectors along either dual arc communicate, and the two arcs communicate through the unbounded complement of the two separately contracted primal paths. They are one exterior face, not two. The plane dual therefore has all strictly interior dual sites and one exterior vertex obtained by identifying the dual boundary sites on both reflected arcs, with exactly the labels $e^{\ast }$ for $e\in E_0$. For edge-event bookkeeping, before this last identification keep the two reflected boundary arcs $Q=s(a_2)$ and $Q^{\mathrm{\prime }}=s(a_1)$ as separate terminal vertex sets in a cut-open dual graph ${\widehat{G}}_s^{\ast }$; it has the same labels $e^{\ast }$ crossing $E_0$ and no added edges along $Q$ or $Q^{\mathrm{\prime }}$. Identifying their boundary vertices to one exterior vertex recovers the plane dual used for component counts. Loops, parallel labels, isolated primal vertices, and both incident face occurrences are retained.

Before the exterior identification, reflection maps ${\widehat{G}}_s^{\ast }$ label-for-label to the reduced primal graph $G_s$, carrying $Q,Q^{\mathrm{\prime }}$ to the two original primal terminal paths. After the exterior dual vertices are identified, the corresponding primal law is the same graph $G_s$ with the coarser boundary partition $j$ that joins $p_1,p_2$. Thus $j$ changes only the component count in the FK weight; it does not add a label to $E_0$. In every edge event below, actual paths are tested in the underlying graph before either boundary partition is applied. Both dual face occurrences adjacent to $e\in E_0$ are represented in the disk, possibly on its dual boundary, so $s(e^{\ast })$ is a retained primal edge, not a deleted primal boundary label. Applying reflection twice gives the inverse label and vertex map. This verifies the edge correspondence and the exterior-face identification.

We prove the finite face count for this contracted multigraph, including its boundary identifications. For $A\subseteq E_0$, let $f(A)$ be the number of regions of the plane complement of the embedded spanning subgraph $(V(G_s),A)$, including the unbounded region. Choose a spanning forest of this subgraph; it has $\mathrm{\mid }V(G_s)\mathrm{\mid }-k_s(A)$ edge labels, counting isolated vertices as components. A finite embedded forest has one face: add its edges from leaves inward, and the two local sides of each added leaf arc can be joined by a detour around its free endpoint, so the complement remains connected. Add the remaining edge labels of $A$ in any order. Their endpoints are already connected by the forest. The new edge has interior disjoint from the current drawing, and together with a simple forest path between its endpoints it forms a simple polygon (a loop label itself is such a polygon). By [F4] its two local sides lie in different regions after insertion, so it splits the single region containing its interior into exactly two. Parallel labels are retained as distinct arcs and obey the same argument. There are $\mathrm{\mid }A\mathrm{\mid }-\mathrm{\mid }V(G_s)\mathrm{\mid }+k_s(A)$ such labels; hence $f(A)=1+\mathrm{\mid }A\mathrm{\mid }-\mathrm{\mid }V(G_s)\mathrm{\mid }+k_s(A)$. The dual graph with the boundary vertices of $Q\cup Q^{\mathrm{\prime }}$ identified has one vertex for each face of the full drawing and a dual edge crossing each label of $E_0$. The components of its spanning subgraph on $A^{\ast }=\{e^{\ast }:e\in E_0\setminus A\}$ are exactly the faces counted by $f(A)$: a dual edge crosses precisely a primal edge absent from the open subgraph, so it joins the two incident face sectors without crossing an open edge; conversely a polygonal path inside one such face can be perturbed off vertices and read as a succession of these dual edges. Reflection identifies this component count with $k_j(s(A^{\ast }))$. Thus $k_j(s(A^{\ast }))=\mathrm{\mid }A\mathrm{\mid }-\mathrm{\mid }V(G_s)\mathrm{\mid }+k_s(A)+1$, also when the spanning subgraph is disconnected or has loops. At $p\mathrm{/}(1-p)=\sqrt{q}$, the ratio of the separate weight of $A$ to the joined weight of $s(A^{\ast })$ is $\frac{w_s(A)}{w_j(s(A^{\ast }))}={\left(\frac{p}{1-p}\right)}^{2\mathrm{\mid }A\mathrm{\mid }-\mathrm{\mid }{E}_0\mathrm{\mid }}{q}^{k_s(A)-k_j(s(A^{\ast }))}=q^{\mathrm{\mid }V(G_s)\mathrm{\mid }-1-\mathrm{\mid }{E}_0\mathrm{\mid }\mathrm{/}2}$, which is independent of $A$. The complement/reflection pushforward of the separate law is therefore exactly the joined law.

Define the edge-only terminal events before making either boundary identification. Let $P_1,P_2$ be the original primal boundary paths contracted to $p_1,p_2$. Let $C(A)$ mean that actual open labels in $A\subseteq E_0$ contain a path from some vertex of $P_1$ to some vertex of $P_2$; the contraction of each path separately makes this exactly the event $p_1\leftrightarrow p_2$ in the separate law. Let $C^{\ast }(A^{\ast })$ mean that actual open dual labels in $A^{\ast }$ contain a path from the boundary vertex set of $Q$ to that of $Q^{\mathrm{\prime }}$ in ${\widehat{G}}_s^{\ast }$, with no use of the exterior-vertex identification and no added boundary edge. The four boundary arcs occur in alternating order $P_1,Q,P_2,Q^{\mathrm{\prime }}$. Exactly one of $C(A)$ and $C^{\ast }(A^{\ast })$ holds. They cannot both hold: erase loops from both paths and move any nonterminal boundary contacts into disjoint thin collars, keeping their endpoints on their terminal arcs. The resulting primal and dual crosscuts must meet by [F4]. In this cellulation primal and dual edges meet only at the midpoint of their paired labels, but a primal label in $A$ has its dual label absent from $A^{\ast }$, a contradiction. For the converse, suppose $C(A)$ fails. In the uncontracted disk graph let $S$ be the set of primal vertices reachable by actual $A$-edges from $P_1$, taking all vertices of $P_1$ as starting vertices. Then $S$ contains $P_1$, misses $P_2$, and every retained edge with one endpoint in $S$ and one outside is closed. Take the dual labels crossing exactly these cut edges. At every interior dual vertex their degree is even, since the cyclic list of adjacent primal sectors changes from $S$ to its complement an even number of times. Along each of the two dual boundary arcs $Q,Q^{\mathrm{\prime }}$, the number of odd-degree cut-dual vertices is odd: read the in-disk primal sectors along that arc from its end adjacent to $P_1$ (in $S$) to its end adjacent to $P_2$ (outside $S$); a boundary dual vertex has odd degree exactly when the sector status toggles there. Additional contacts of $S$ with either boundary arc contribute pairs of toggles, so this parity remains odd. The cut contains no labels outside $E_0$, and the collar ordering at corners retains the in-disk incidences only. Every finite graph component has an even number of odd-degree vertices. Since the total number on each of $Q$ and $Q^{\mathrm{\prime }}$ is odd, some component contains an odd-degree vertex on each arc. Its dual edges therefore contain a path from $Q$ to $Q^{\mathrm{\prime }}$ using only labels in $A^{\ast }$, proving $C^{\ast }(A^{\ast })$. By step 12.1, reflection sends the separate law to the joined law and sends $C^{\ast }$ to the same edge-only event $C$ on $G_s$. Thus ${\varphi }_j(C)={\varphi }_s(C^{\ast })=1-{\varphi }_s(C)$. Identifying $p_1,p_2$ changes the component count by one exactly when $C$ fails: $k_j(A)=k_s(A)-1$ on $C^c$ and $k_j(A)=k_s(A)$ on $C$. Hence $w_j(A)\mathrm{/}{w}_s(A)\in [q^{-1},1]$, and the ratio of partition functions lies in the same interval. The normalized density ${\varphi }_j(A)\mathrm{/}{\varphi }_s(A)$ is therefore in $[q^{-1},q]$, so $1-{\varphi }_s(C)={\varphi }_j(C)\le q{\varphi }_s(C)$. This gives ${\varphi }_s(C)\ge 1\mathrm{/}(1+q)\ge 1\mathrm{/}(1+q^2)$. The auxiliary mixed-disk estimate has no torus embedding hypothesis.

For the remaining $3\mathrm{/}2$-rectangle assertion assume additionally $3N\mathrm{/}2<m$. Put $R_1=[0,N{]}^2$, $R_2=[N\mathrm{/}2,3N\mathrm{/}2]\times [0,N]$, and $R=[0,3N\mathrm{/}2]\times [0,N]$. On success of the first search in $R_1$, let $g_1$ be its left-to-right output, with its unique right-layer vertex $x=(N,h)$. On success of the second search in $R_2$ with bottom sources $N\mathrm{/}2\le u\le N$, let $g_2$ be its bottom-to-top output. Suppose the full paths are disjoint. Reflect in $u=N+1\mathrm{/}2$ and let $A=s(g_1)$ run backwards from $s(x)$. The full $g_1$, connector $x$--$s(x)$, and $A$ form a simple horizontal crosscut of $[0,2N+1]\times [0,N]$ after adding collars as in step 2.1; extend $g_2$ through bottom/top collars. They intersect, not on $g_1$ or the connector, hence on $A$. Choose $z$ to be the first intersection along the full $g_2$ from its bottom source, not the first along $A$. Its prefix lies in the bottom component of the full horizontal crosscut. That crosscut meets $u=N+1$ only at height $h$, since the first-target rule makes $x$ the only right-layer vertex of $g_1$. The axis ray below $h$ connects to the bottom collar without meeting the crosscut and the ray above $h$ to the top collar. Thus every prefix visit to $u=N+1$ is strictly below $h$; equality is also excluded by opposite parity.

Let $a$ be the dual vertex before $z$ along $A$, $y$ the primal vertex before $z$ along $g_2$, and $t=(N+1,b)$ the last prefix visit to $u=N+1$ before or at $y$. It exists since the source has $u\le N$ and the crossing is to the right of $N+1$. Step 14.1 gives $b<h$. Let $a_1$ be $g_1$ backwards from $x$ to $s(a)$ and $a_2$ the $g_2$ segment from $t$ to $y$. The whole $g_2$ prefix avoids the whole $A$ before $z$, so $a_1$ and $s(a_2)$ are disjoint. Their final primal/dual vertices are axial neighbours, and the last-return and first-target rules give the half-plane conditions of step 9.1, including possible singletons. Hence the precise auxiliary disk of steps 9.1--13.1 exists. Its free left and right arcs are $Q=s(a_2)$ and $Q^{\mathrm{\prime }}=s(a_1)$.

We check the local exit at $y$ completely. The unshaved right-half lobe is traversed up the axis from $b$ to $h$, then along $A$ to $z$, then backwards along $g_2$ to $t$. Its bounded component is on the right of its upward axis segment, so it is clockwise. The first bottom-to-top crossing of the full horizontal crosscut has $\det (A^{\mathrm{\prime }},g_2^{\mathrm{\prime }})>0$. There are exactly four direction pairs: $((1,1),(-1,1))$, $((-1,1),(-1,-1))$, $((-1,-1),(1,-1))$, and $((1,-1),(1,1))$. Their connectors $a\to y$ point respectively east, north, west, south, and the forward $g_2$ edge enters respectively north, west, south, east: always the connector's left side. Equivalently $y-a=(A^{\mathrm{\prime }}-g_2^{\mathrm{\prime }})\mathrm{/}2$, so $\det (y-a,g_2^{\mathrm{\prime }})=1$ in every case. The shaved disk is on the connector's right, with exactly $\mathrm{conv}\{a,y,z\}$ removed. Thus the forward half-edge at $y$ exits the disk immediately. The global $b<h$ fixed the clockwise orientation before these local cases; no inference from a local turn at a possibly reflex corner was made. Reflection gives the other empty half-diamond check.

Choose a point $r_{\ast }$ in the central strip $N<u<N+1$, $b<v<h$, sufficiently close to $x$ to be on the left side of the compact path $g_2$, and let $U$ be its component in the interior of $\mathrm{\Omega }$ minus the two full paths. Clip the portions of the full paths off their distinguished arcs $a_1,a_2$ to $\mathrm{\Omega }$. They are a finite family of disjoint proper crosscuts. Intersections with dual boundary arcs are transverse; simple paths cannot return to their own distinguished primal boundary arc, and the two full paths do not meet. Their full original endpoints are not interior points of $\mathrm{\Omega }$: its vertical coordinates lie in $[0,N]$, its left coordinate is nonnegative, and the original endpoints not on a distinguished arc cannot coincide with a dual boundary site or connector interior by parity. Corner endpoints are read in the two incident interior collar sectors, retaining the same labels and identifications; outside-only touches add no crosscut. Step 16.1 handles the forward endpoint at $y$. Cutting a disk along one proper crosscut gives two disks by [F4]. Repeated cutting gives a finite adjacency tree, one vertex per complementary disk and one edge per crosscut: every cut splits one previous vertex into two and adds the joining edge. Root the tree at $U$ and color each edge by the full path, 1 or 2, containing that crosscut.

The original terminal wiring must be tracked in this tree. No color-2 crosscut separates $a_1$ from the root when color-1 cuts are ignored: the whole connected boundary arc $a_1$ is disjoint from the compact full $g_2$, so a uniformly thin interior collar along $a_1$, joined to $r_{\ast }$ near $x$, avoids $g_2$. No color-1 crosscut separates $a_2$ from the root when color-2 cuts are ignored: all of $g_1$ lies in $u\le N$, whereas one can join $r_{\ast }$ down the central strip to $t$ and follow an interior collar of $a_2$ on the right. Therefore a root-tree branch whose first edge has color 1 contains no original terminal of $a_2$, and a branch first cut off by color 2 contains none of $a_1$. Indeed, after erasing cuts of the other color that branch is still behind its first crosscut, while the opposite terminal and root have the avoiding connection just described. Same-color terminals in a branch are allowed.

Let $F$ be exactly the auxiliary primal labels not on either full path whose relative interiors lie in $U$, retaining all their endpoints. All other nonpath labels have relative interior in another disk region. Keep all full-path labels outside $F$, including path labels on the root boundary; distinguished boundary-path loops may equivalently be deleted as in step 10.1. Let $U_i$ consist of vertices of $g_i$ incident to $F$, retaining isolated original terminal vertices when needed for the connection event. Each $U_i$ is joined by the known full actual path $g_i$. Each graph component of outside-$F$ labels, including the original separate $a_1,a_2$ identifications, attaches to $F$ through at most one $U_i$. For a detailed verification, collapse each root-tree branch to a symbol of its first-crosscut color. It attaches to the root only through that crosscut. To join two branch symbols outside $U$, a path must run along a full cut path; if that path has color $j$, it can enter or leave a branch across its first crosscut only when that crosscut also has color $j$, since different full paths neither cross nor meet. The original terminal identifications join only same-color symbols by step 18.1. Thus every compressed outside component with root attachments is monochromatic. Nested opposite-color pieces inside a branch are harmless; they cannot leave it across the other path. Ordinary primal labels cannot cross a cut path except at its primal vertices, and the dual free boundary arcs contain no primal boundary vertices joining branches. This proves the assertion including all path-edge labels and the original wired terminals.

Condition the auxiliary mixed model on every outside-$F$ state. Step 19.1 shows its induced partition refines separate wiring of $U_1,U_2$. In the actual FK model, condition on the complete exploration transcript and then any additional outside-$F$ states. Its induced partition is at least as wired as separate $U_1,U_2$, because each full $g_i$ is known open. Provided $F$ is unqueried and has the asserted actual label correspondence, [F3] compares these conditional laws; averaging gives domination of the auxiliary marginal by the actual conditional marginal on every increasing $F$ event. Let $B_F$ be the event of an open $F$ connection between $U_1,U_2$, allowing within-class identifications. Erase every maximal outside-$F$ excursion of an auxiliary $a_1$--$a_2$ path and replace it by its single-color attachment identification. Initial/final excursions from terminals in branches have that same color by step 18.1. No erased component joins the two colors. The resulting $F$ connection witnesses $B_F$, so step 13.1 gives auxiliary probability at least $(1+q^2{)}^{-1}$. In the actual model all within-class connections are realized by the full known paths, so $B_F$ implies their actual connection.

We verify the two remaining hypotheses of step 20.1 without requiring the entire auxiliary disk to embed in the torus. Under the corrected stopping rule it may extend to the right of $R_2$; no outside-$F$ auxiliary label is mapped to an actual torus label. In the wide rectangle $[0,2N+1]\times [0,N]$, the full $g_2$ is a proper vertical separator after collar extension. The point $x$ is in its left component, since the full $g_1$ joins it to the left boundary without meeting $g_2$. Thus $r_{\ast }$ and all $U$ are in that left component. Every point with $u>3N\mathrm{/}2$ is in the right component because all of $g_2$ lies in $R_2$. All of $\mathrm{\Omega }$ has $u\ge 0$ and $0\le v\le N$. Therefore $F$ lies in $R$.

Inside $R_1$, the root component is below $g_1$. The only path-arc contacts of $\mathrm{\partial }\mathrm{\Omega }$ with $u=N$ are $x$ at height $h$ and the start of $Q$ at height $b$: $a_1$ meets its limiting layer only at $x$, and $a_2$ its limiting layer only at $t$. A unit connector adds no other height except, in the double-singleton case, the unit segment between $b$ and $h$. Hence $\mathrm{\Omega }\cap \{u=N\}$ is precisely the interval from $b$ to $h$. A curve in $\mathrm{\Omega }$ avoiding $g_1$ cannot pass around $x$ into the upper side of $R_1$, since reentry would require height above $h$ on that line. It cannot pass around the other endpoint of $g_1$, because $\mathrm{\Omega }$ has $u\ge 0$ and that endpoint is on $u=0$. Jordan separation, also with the collars for intermediate side contacts, puts $F$ in $R_1$ below $g_1$. In $R_2$ it is left of $g_2$ by step 21.1. Every $F$ edge belongs to $R_1$ or $R_2$, since the overlap has width $N\mathrm{/}2\ge 2$. Path labels were excluded from $F$, so step 5.1 shows neither exploration queried it. Finally $R$ has spans below $m$, so step 1.1 maps $F$ and its attachment vertices injectively to actual torus labels. This proves all hypotheses of step 20.1 under the original $m>3N\mathrm{/}2$, even if auxiliary $\mathrm{\Omega }$ itself does not embed.

Let $\mathcal{A}$ require horizontal crossings of $R_1,R_2$ and a bottom-to-top $R_2$ path with source in its inclusive bottom half $N\mathrm{/}2\le u\le N$. The genuine primal symmetry $u\mapsto 2N-u$ interchanges its two inclusive bottom halves. Their connection-to-top events cover the square vertical-crossing event; by symmetry and step 8.1 each therefore has probability at least $a(q)\mathrm{/}2$. Translation by $N\mathrm{/}2$ preserves the primal lattice since $4\mid N$, so both square horizontal probabilities are at least $a(q)$. Association [F3] gives $\varphi (\mathcal{A})\ge a(q{)}^3\mathrm{/}2$.

Use the full combined query transcript $T$. On successful disjoint outputs, steps 20.1 and 22.1 give $\varphi (B_F\mid T)\ge (1+q^2{)}^{-1}$; intersecting outputs already give a connection with probability one. The additional event $H(R_2)$ in $\mathcal{A}$ need not be $T$-measurable. Conditional on $T$, it and $B_F$ are increasing in the unqueried labels, whose law is associated by steps 6.1 and [F3]. Hence $\varphi (B_F\mid T,H(R_2))\ge \varphi (B_F\mid T)$ whenever defined. On $\mathcal{A}$, the full $g_2$ intersects its horizontal $R_2$ crossing by step 2.1, so connection of $g_1$ and $g_2$ gives a horizontal crossing of $R$. Integrating this conditional inequality yields $\varphi (H(R))\ge \varphi (H(R)\cap \mathcal{A})\ge \varphi (\mathcal{A})\mathrm{/}(1+q^2)\ge b(q)$. This does not move an unmeasurable preparatory-event indicator inside a conditional expectation.

Coordinate interchange gives the stated vertical $3\mathrm{/}2$ estimate. Lattice-preserving translations and quarter-turns preserve labelled torus FK weights as in step 1.1, so the conclusions hold for all stated translates. The square bounds were proved in step 8.1, and each rectangle in the statement embeds under its own stated size hypothesis by step 1.1. All constants are positive for every real $q\ge 1$, and no limiting or integer-$q$ argument occurred. This proves every assertion.