Proof. We encode the subspaces by polynomial evaluation constraints. Evaluating away from the constraint nodes gives a filtration of evaluation spaces. At each node, disjointness bounds the evaluated rank. The resulting rank deficiency gives a lower bound on a polynomial minor’s vanishing order. Comparing these orders with the minor’s degree, then summing over the filtration, will give the bound.

Choose distinct real numbers \(t_1,\ldots,t_m\). For a subspace \(Z\subseteq W\) and an integer \(h\geq0\), define \[\mathcal P_h(Z)= \{P\in Z[\xi]:\deg P\leq h,\ P(t_i)\in U_i\text{ for every }i\}, \qquad \mathcal P_{-1}(Z)=\{0\}.\] Here \(Z[\xi]\) denotes polynomials whose coefficients belong to \(Z\). Write \(E_h(Z;t)=\{P(t):P\in\mathcal P_h(Z)\}\) and put \[k_h(Z)=\dim\mathcal P_h(Z)-\dim\mathcal P_{h-1}(Z), \qquad k_{-1}(Z)=0.\]

Evaluation away from the nodes. For \(t\notin\{t_1,\ldots,t_m\}\), the kernel of evaluation at \(t\) is \[\ker\bigl(\mathcal P_h(Z)\longrightarrow Z,\ P\longmapsto P(t)\bigr) =(\xi-t)\mathcal P_{h-1}(Z).\] Indeed, division by \(\xi-t\) preserves the coefficient space \(Z\), and it preserves each constraint at \(t_i\) because \(t_i-t\ne0\). The converse inclusion follows by multiplication. Therefore \[ \dim E_h(Z;t)=k_h(Z) \qquad(t\notin\{t_1,\ldots,t_m\}). \tag{4}\] Also, evaluation at all \(m\) nodes is an isomorphism \[\mathcal P_{m-1}(Z)\longrightarrow \bigoplus_{i=1}^{m}(U_i\cap Z).\] Its inverse sends \((z_1,\ldots,z_m)\) to the polynomial \(\sum_i L_i(\xi)z_i\), where \(L_i(\xi)=\prod_{k\ne i}(\xi-t_k)/(t_i-t_k)\). Uniqueness follows because a scalar polynomial of degree at most \(m-1\) with \(m\) distinct roots vanishes, applied in coordinates of \(Z\). Consequently \[ \sum_{h=0}^{m-1}k_h(Z) =\dim\mathcal P_{m-1}(Z) =\sum_{i=1}^{m}\dim(U_i\cap Z). \tag{5}\]

Polynomial minors of controlled degree. Fix \(t_*\) outside the nodes. The spaces \(E_h(W;t_*)\) are nested, so \[\ell_h=k_h(W)-k_{h-1}(W)\geq0 \qquad(0\leq h\leq m-1).\] At stage \(h\), choose \(\ell_h\) polynomials from \(\mathcal P_h(W)\) whose values at \(t_*\) extend the previously chosen values to a basis of \(E_h(W;t_*)\). In fixed coordinates of \(W\), let \(M_h(\xi)\) be the matrix of all polynomial columns chosen through stage \(h\). It has \(r_h=k_h(W)\) columns, independent at \(t_*\), and hence has an \(r_h\)-row square minor \(\Delta_h(\xi)\) with \(\Delta_h(t_*)\ne0\). Each column introduced at stage \(e\) has degree at most \(e\). Expanding the determinant thus gives \[ \deg\Delta_h\leq\sum_{e=0}^{h}e\ell_e. \tag{6}\] If \(r_h=0\), use the empty determinant \(\Delta_h=1\).

Rank loss at the nodes. For each \(j\) we claim that \[ \dim E_h(W;t_j)\leq k_h(W)-k_h(V_j). \tag{7}\] Let \(\pi_j:W\longrightarrow W/V_j\) be the quotient map. At every non-node \(t\), the kernel of \(\pi_j\) on \(E_h(W;t)\) contains \(E_h(V_j;t)\). Equation (4) gives \[\dim\pi_j(E_h(W;t))\leq k_h(W)-k_h(V_j).\] To pass to \(t_j\), take a fixed basis of the polynomial space \(\mathcal P_h(W)\), evaluate its members at the variable \(\xi\), and project their columns by \(\pi_j\). In fixed coordinates this is a polynomial matrix. Every minor larger than the displayed rank bound vanishes at every non-node, an infinite set, and so is the zero polynomial. Its rank at \(t_j\) obeys the same bound. But \(E_h(W;t_j)\subseteq U_j\), and \(\pi_j\) is injective on \(U_j\) because \(U_j\cap V_j=\{0\}\). This proves (7).

The chosen columns of \(M_h(t_j)\) lie in \(E_h(W;t_j)\), and restricting to the rows of a minor cannot increase rank. The square submatrix defining \(\Delta_h\) therefore has rank deficiency at least \(k_h(V_j)\) at \(t_j\). A square polynomial matrix with deficiency at least \(q\) at \(t_j\) has determinant divisible by \((\xi-t_j)^q\): choose \(q\) independent vectors in its kernel at \(t_j\), extend them to a constant invertible column change, and observe that the corresponding \(q\) transformed columns vanish at \(t_j\). Each contributes the factor \(\xi-t_j\), while the column change rescales the determinant by a nonzero constant. Applying this observation separately at the distinct nodes shows that \[\prod_{j=1}^{m}(\xi-t_j)^{k_h(V_j)}\quad\text{divides }\Delta_h(\xi).\] For \(r_h=0\), all \(k_h(V_j)\) are zero by (4), so this assertion also holds for the empty determinant. Since \(\Delta_h\) is not the zero polynomial, (6) now yields \[ \sum_{j=1}^{m}k_h(V_j)\leq\sum_{e=0}^{h}e\ell_e. \tag{8}\]

Summing the degree bounds. Sum (8) over \(h=0,\ldots,m-1\) and apply (5) with \(Z=V_j\). The result is \[\begin{align*} \sum_{i,j=1}^{m}\dim(U_i\cap V_j) &\leq\sum_{h=0}^{m-1}\sum_{e=0}^{h}e\ell_e =\sum_{e=0}^{m-1}e(m-e)\ell_e\\ &\leq\left\lfloor\frac{m^2}{4}\right\rfloor \sum_{e=0}^{m-1}\ell_e \leq\left\lfloor\frac{m^2}{4}\right\rfloor N. \end{align*}\] The last step uses \(\sum_e\ell_e=k_{m-1}(W)=\dim E_{m-1}(W;t_*)\leq N\). ◻

Proof. In \(W=\mathbb R^d\oplus\mathbb R^d\), let \[U_i=\{(v,X_iv):v\in\mathbb R^d\},\qquad V_j=\{(v,Y_jv):v\in\mathbb R^d\}.\] The map \(v\mapsto(v,X_iv)\) identifies \(\ker(X_i-Y_j)\) with \(U_i\cap V_j\). Thus \[\dim(U_i\cap V_j)=d-\mathop{\mathrm{rank}}(X_i-Y_j), \qquad U_i\cap V_i=\{0\}.\] Lemma 2, with \(N=2d\), gives \[\sum_{i,j=1}^{m}\mathop{\mathrm{rank}}(X_i-Y_j) \geq\left(m^2-2\left\lfloor\frac{m^2}{4}\right\rfloor\right)d.\] The \(m\) diagonal terms each equal \(d\). Removing them gives (9), because \[m^2-m-2\left\lfloor\frac{m^2}{4}\right\rfloor =\begin{cases} 2r(r-1),&m=2r,\\ 2r^2,&m=2r+1, \end{cases} =2d_m.\] ◻

Proof. We first establish the local formula that explains the factor \(1/2\). Let \(C\) and \(E\) be oriented chords of a round disk, with four distinct boundary endpoints. Give each chord’s start coefficient \(-1\) and its end coefficient \(+1\); write these coefficients as \(\epsilon_C(u)\) and \(\epsilon_E(w)\). For a cut in the boundary away from the endpoints, let \(T(u,w)\) be \(1\) if \(u\) precedes \(w\) counterclockwise from the cut, and \(-1\) otherwise. Their signed intersection is \[ I(C,E)=\frac12\sum_{u\in\partial C}\sum_{w\in\partial E} \epsilon_C(u)\epsilon_E(w)T(u,w). \tag{12}\] Moving the cut past one endpoint changes the right-hand side by a multiple of the sum of the other chord’s endpoint coefficients, which is zero. We may therefore cut just before the start of \(C\). The right-hand side then becomes \[-\sum_{\substack{w\text{ on the open counterclockwise arc}\\ \text{from the start to the end of }C}} \epsilon_E(w).\] That boundary arc is on the right of the directed chord \(C\). The sum is zero for nonalternating endpoints; for alternating endpoints it is \(+1\) when \(E\) runs from right to left across \(C\), and \(-1\) in the reverse case. These are exactly the signed intersection values.

Figure 1 illustrates the local signs and the averaging at a shared spoke used below.

Choose small pairwise disjoint round vertex disks meeting the drawing only in its straight spokes. Replace the passage of the first cycle through each of its vertex disks by the chord joining its incoming and outgoing boundary points. Outside these disks, keep its original edge tracks. This gives a closed polygonal walk \(\gamma\); its drawing may have self-intersections.

For the second cycle construct two closed polygonal walks \(\eta_+\) and \(\eta_-\) by shifting its edge tracks a small distance to the left and to the right, respectively, relative to the fixed graph-edge orientation. In each vertex disk join the shifted boundary endpoints by a chord. Here is an explicit construction of the tracks outside the vertex disks. Place further pairwise disjoint small disks at crossings and nonstraight bends, each meeting only the straight pieces incident with its center. Shift each straight stretch between disks parallel to itself. In a crossing disk continue along the shifted line: it avoids its own original line and meets the other original line once, with the same crossing sign. In a bend disk the two shifted ends lie in the same component of the disk minus the original bent arc, and can be joined there by a polygonal arc. Indeed, for the left shift, the incoming boundary ray moves clockwise and the outgoing ray counterclockwise; both displaced ends are in the sector on the left of the oriented bend. The right shift uses the other sector. All remaining distinct truncated straight stretches are compact and disjoint, so sufficiently small shifts introduce no additional intersections. Finiteness permits one choice of shift size for the entire construction. At vertex-circle seams the two walks’ boundary endpoints are distinct and chord interiors lie in the open disk; at crossing disks the intended intersections remain strictly interior. Together with the avoidance inside bend disks, this excludes contacts at seams and corners.

Consequently, in each version the signed intersections with \(\gamma\) outside the vertex disks sum to \[ \sum_{e,f}z_e y_f I(e,f)=z^{\mathsf T}Iy. \tag{13}\] This also describes shared tracks: a shifted edge avoids its own unshifted copy. If both cycles use both edges of an original crossing, the intersections of each shifted track with the other unshifted track give the two ordered-edge contributions in the displayed sum. Their signs include the traversal coefficients \(z_e y_f\).

Consider a vertex disk used by both cycles. The endpoint coefficient of its first chord on spoke \(e\) is \(\sigma_{ve}z_e\): an incoming traversal starts the chord and an outgoing traversal ends it. The second chord’s corresponding coefficient is \(\sigma_{vf}y_f\). All endpoints belonging to the two different chords are distinct after either shift. Choose the shifts small enough that the original boundary cut stays in its chosen gap. For different spokes \(e,f\), the comparison sign of their endpoints is \(T_v(e,f)\) in both versions. For the same spoke, the two shifted endpoints lie on opposite sides of the unshifted endpoint in the boundary order. Their comparison signs therefore average to zero, as required by \(T_v(e,e)=0\). Applying (12) and averaging, the contribution of this disk is \[\frac12\sum_{e,f\text{ incident with }v} z_e y_f\sigma_{ve}\sigma_{vf}T_v(e,f).\] This cancellation for a shared spoke is unaffected by opposite cycle traversals: they change the endpoint coefficients, while the two order signs remain opposite.

It remains to show that the total signed intersection of each pair \((\gamma,\eta_\pm)\) is zero. More generally, let two closed polygonal walks meet transversely away from their corners, counting intersections by their segment traversals. Cone every oriented segment \([u,w]\) of the first walk to a point \(o\), using the oriented triangle \([o,u,w]\). Choose \(o\) outside the finitely many lines that would make a triangle degenerate or a cone segment pass through a corner of the second walk; also avoid its segment lines. The added cone segments meet the second walk transversely away from corners. Each triangle boundary has total signed intersection zero with the second walk: every entrance into the triangle is paired, along that closed walk, with an exit, and the two have opposite signs. On summing triangle boundaries, the extra oriented cone segments cancel in pairs, leaving exactly the first walk. Its total signed intersection with the second is therefore zero. The argument counts traversals and does not require either walk to be simple. The constructed pairs satisfy these transversality conditions; any unnecessary straight subdivision points can be removed.

Finally, average the zero total for \(\eta_+\) and \(\eta_-\). The exterior contribution is (13), and the vertex-disk contributions are exactly the local terms of (10). Their sum is \(z^{\mathsf T}Jy\), proving (11). ◻

Proof. Write the matrix on the left as \(H\). Fix row indices \(p,p'\) and column indices \(q,q'\). If \(i\ne1\), \(k\ne1\), \(p\ne p'\), and \(q\ne q'\), use the four-cycle vector \(z\) with entries \(+1\) on \(ip,1p'\) and \(-1\) on \(ip',1p\), and the four-cycle vector \(y\) with entries \(+1\) on \(kq,1q'\) and \(-1\) on \(kq',1q\). Expanding Lemma 5 gives \[0=z^{\mathsf T}Jy =H(p,q)-H(p,q')-H(p',q)+H(p',q').\] If any of those index differences is trivial, the same rectangular difference vanishes directly. Thus all rectangular differences of \(H\) vanish. For any fixed \(p_0,q_0\) this says \[H(p,q)=H(p,q_0)+H(p_0,q)-H(p_0,q_0),\] which is the required row-plus-column form. ◻

Proof. The polynomial space has dimension \(s^2\), and \[|Q_s|=\frac{s(s+1)}2+\frac{s(s-1)}2=s^2.\] It therefore suffices to show that a polynomial \(P(u,v)\) of degree at most \(s-1\) in each variable, vanishing at all these points, is zero. We prove this by induction on \(s\). For \(s=1\), \(P\) is constant and vanishes at the single point indexed by \((2,1)\).

Let \(s>1\). The first triangle contains the \(s\) points \((x_{s+1},x_q)\) for \(1\leq q\leq s\). The polynomial \(P(x_{s+1},v)\) has degree at most \(s-1\) and \(s\) distinct roots, so it vanishes identically. Division in the variable \(u\) gives \[P(u,v)=(u-x_{s+1})P_1(u,v), \qquad \deg_u P_1\leq s-2,\quad\deg_v P_1\leq s-1.\] For \(p=2,\ldots,s\), the remaining points \((x_p,x_1)\) give \(P_1(x_p,x_1)=0\), since \(x_p\ne x_{s+1}\). Thus \(P_1(u,x_1)\), of degree at most \(s-2\), has \(s-1\) distinct roots and vanishes identically. Factoring in the second variable, we obtain \[ P(u,v)=(u-x_{s+1})(v-x_1)R(u,v), \qquad \deg_u R,\deg_v R\leq s-2. \tag{17}\] At every point in the two remaining triangles the displayed factors are nonzero. Hence \[R(x_p,x_q)=0 \quad\text{if }2\leq q<p\leq s \quad\text{or if }s+2\leq p<q\leq2s+1.\] Define a tuple of \(2s-1\) distinct nodes and a polynomial by \[(y_1,\ldots,y_{2s-1}) =(x_{s+2},\ldots,x_{2s+1},x_2,\ldots,x_s), \qquad S(u,v)=R(v,u).\] The lower triangle \(q<p\leq s\) for the \(y\)-tuple becomes the remaining upper triangle for the \(x\)-tuple after the coordinate swap. The upper triangle \(s<p<q\leq2s-1\) becomes the remaining lower triangle. Consequently \(S\) vanishes at every point of the pattern \(Q_{s-1}\) for the \(y\)-tuple. The induction hypothesis gives \(S=0\), hence \(R=0\) and \(P=0\). This proves injectivity of the square evaluation map and thus both assertions. ◻

Proof. Set \[ \begin{gathered} F(t)=\prod_{p=1}^{n}(t-x_p),\qquad h_{pq}=(x_p^a x_q^b)_{0\leq a,b<s},\\ B_{pq}=\frac{x_p-x_q}{F'(x_p)F'(x_q)} h_{pq}h_{pq}^{\mathsf T},\qquad \Phi(H)=\sum_{p,q=1}^{n}H(p,q)B_{pq}. \end{gathered} \tag{18}\] The factors \(1/F'(x_p)\) are the classical barycentric weights for Lagrange interpolation (Berrut and Trefethen 2004, sec. 3, equation (3.2)). All denominators are nonzero because the nodes are distinct. The constant coordinate of \(h_{pq}\) is \(1\), so \(B_{pq}\) has rank one when \(p\ne q\), while \(B_{pp}=0\). This proves (ii) and the annihilation of diagonal matrices. Rank subadditivity also gives \[ \mathop{\mathrm{rank}}\Phi(H)\leq \bigl|\{(p,q):p\ne q,\ H(p,q)\ne0\}\bigr|. \tag{19}\]

For the additive matrices, we use the elementary identity \[ \sum_{p=1}^{n}\frac{A(x_p)}{F'(x_p)}=0 \qquad\text{if }\deg A\leq n-2. \tag{20}\] Indeed, the polynomials \(F(t)/((t-x_p)F'(x_p))\) have degree \(n-1\) and take the coordinate values \(1\) at \(x_p\) and \(0\) at every other node. Subtracting their linear combination with coefficients \(A(x_p)\) from \(A\) gives a polynomial of degree at most \(n-1\) with \(n\) distinct roots, so \[A(t)=\sum_{p=1}^{n} A(x_p)\frac{F(t)}{(t-x_p)F'(x_p)}.\] Comparison of the coefficients of \(t^{n-1}\) proves (20).

The entry of \(B_{pq}\) indexed by \((a,b)\) and \((a',b')\) is \[\frac{(x_p-x_q)x_p^{a+a'}x_q^{b+b'}} {F'(x_p)F'(x_q)}.\] Its numerator has degree at most \(2s-1=n-2\) in either variable separately. Applying (20) with the other variable fixed proves \[\sum_{q=1}^{n}B_{pq}=0\quad\text{for every }p, \qquad \sum_{p=1}^{n}B_{pq}=0\quad\text{for every }q.\] It follows that \(\Phi((r_p+c_q)_{p,q})=0\), completing (i).

It remains to prove (iii). First take the natural order and write \(S=S_\prec\). Let \(H(p,q)=\mathbf{1}_{q<p}\) and let \(\mathbf1\) denote the all-ones column vector. Then \[S=\tfrac12\mathbf1\mathbf1^{\mathsf T}-H-\tfrac12I_n, \qquad \Phi(S)=-\Phi(H).\] Subtract \(1\) from each row of \(H\) indexed by \(p>s+1\), obtaining \(H'\). The difference \(H-H'\) is additive, so \(\Phi(H')=\Phi(H)\). Apart from diagonal entries, which \(\Phi\) annihilates, the support of \(H'\) is exactly \(Q_s\): its value is \(+1\) on the first triangle of (16) and \(-1\) on the second.

List the pairs in \(Q_s\) in any order, and let \(A_Q\) be the square matrix with columns \(h_{pq}\) in that order. By Lemma 7, \(A_Q\) is invertible. Equation (18) now gives \[ \Phi(S)=-A_Q D_Q A_Q^{\mathsf T},\qquad D_Q=\mathop{\mathrm{diag}}\left( H'(p,q)\frac{x_p-x_q}{F'(x_p)F'(x_q)} \right)_{(p,q)\in Q_s}. \tag{21}\] Every diagonal entry of \(D_Q\) is nonzero, so \(\Phi(S)\) is invertible.

Finally let \(\pi(1)\prec\cdots\prec\pi(n)\) be an arbitrary order. Relabel the indices and use the tuple \((x_{\pi(1)},\ldots,x_{\pi(n)})\) in the argument just given. The relabeled evaluation vector at \((r,q)\) is \(h_{\pi(r),\pi(q)}\), so this only reindexes the sum defining \(\Phi(S_\prec)\); the polynomial \(F\) and the fixed map \(\Phi\) are unchanged. The relabeled nodes are distinct, so Lemma 7 applies again. Thus every order has invertible image under the original map. ◻

Proof. Isolating the marked passages. Call the vertices and crossing points of the given drawing its marks. There are finitely many. Parameterize each edge injectively by \(\gamma_e:[0,1]\to\mathbb R^2\). On each edge, choose pairwise disjoint parameter neighborhoods of its marked parameters, with disjoint closures, using one-sided neighborhoods at \(0\) and \(1\). For a mark \(w\) on that edge, the image of the compact parameter set outside its chosen neighborhood avoids \(w\), and hence has positive distance from \(w\). If \(w\) is not on the edge, the entire compact edge image has positive distance from \(w\). Since all these conditions are finite, we may choose pairwise disjoint closed round disks \(D_w\), centered at the marks, so that every visit of an edge to \(D_w\) is in its prescribed parameter neighborhood, and nonincident edges avoid \(D_w\) altogether.

For an interior mark on \(e\), remove the parameter interval from the first to the last visit to \(D_w\). This removes the intervening excursions too. At the starting endpoint remove the interval through the last visit to its disk; at the terminal endpoint remove the interval from the first visit to its disk. These removed intervals are disjoint along each edge. The retained middle pieces are compact arcs with endpoints on disk boundaries and interiors in \[\Omega=\mathbb R^2\setminus\bigcup_w D_w.\] They are pairwise disjoint, with distinct endpoints: every possible intersection in the original drawing was a mark and has been removed. This construction uses the first and last visits to each closed disk, which exist by compactness. It does not require an arc to cross a round disk boundary only finitely many times.

Replacing the exterior pieces. The finitely many disjoint compact middle pieces admit pairwise disjoint open neighborhoods \(U_P\). For a boundary endpoint \(x\) of one piece \(P\), take a short straight spur from \(x\) radially outward from its disk. Its interior lies in \(U_P\cap\Omega\). The spur tip can be connected to a nearby interior point of \(P\) in that same open set. To see this directly, a sufficiently small exterior collar at \(x\) lies in \(U_P\) and meets no other disk. In polar coordinates about the disk center, choose the collar to have radius in \((r,r+\delta)\) and angle in a small open interval about the angle of \(x\), where \(r\) is the disk radius. It contains both the spur tip and all sufficiently nearby interior points of \(P\). Radial paths and a short concentric arc connect those points inside the collar. Do this at both ends of \(P\) and use the original middle subarc between the selected interior points. The resulting connection between the two spur tips lies in \(U_P\cap\Omega\).

Any continuous path with compact image in an open subset of the plane can be replaced, with the same endpoints, by a polygonal path in that open set: cover its image by small balls in the set, subdivide its parameter interval so each subarc lies in one such ball, and join successive subdivision points by straight segments there. Apply this to the connection just obtained, then attach the two spurs. If the result has loops, subdivide the finite union of its segments at all segment endpoints and intersections; for collinear overlaps, also subdivide at the overlap endpoints and identify coincident pieces. This is a finite embedded graph. A simple graph path between the original boundary endpoints is a simple polygonal arc, still in \(U_P\) with its interior in \(\Omega\). Carry out this construction for every \(P\). Their disjoint neighborhoods ensure that no resulting exterior pieces meet.

Filling the disks. In a crossing disk join each of its two pairs of boundary endpoints by a straight chord. The four endpoints are distinct. The chords can therefore meet at most once, and any meeting is an interior transverse crossing: a line cannot contain four distinct points of a circle. In a vertex disk choose a new vertex point in its interior off the finitely many lines through pairs of its boundary endpoints, and join all the endpoints to it by straight spokes. These spokes meet only at the vertex. The new vertex points in different disks are distinct.

Assemble the spokes, chords, and exterior polygonal pieces in their original edge order. Each resulting edge is simple: the exterior pieces are disjoint, the disks are disjoint, and an edge uses at most one chord or spoke in any one disk. Its interior avoids all vertices. Every inter-edge crossing lies in an original crossing disk, with at most one crossing in that disk and no third edge there. Thus the result is admissible, polygonal, and has no more crossings than the given drawing. It is straight near every vertex and every crossing.

Removing adjacent-edge crossings. Suppose two polygonal edges \(e=vu\) and \(f=vw\) sharing \(v\) cross at \(x\). As the graph is simple, \(u\ne w\). Choose a small round disk about \(x\) that meets only the two straight crossing strands and contains no other mark or bend. Swap the portions of \(e\) and \(f\) from \(v\) up to this disk, then connect the swapped prefixes to the original suffixes. Inside the disk this pairs adjacent boundary endpoints, so two disjoint chords make the reconnections. The result is a walk from \(v\) to \(u\) and a walk from \(v\) to \(w\). The crossing at \(x\) has disappeared. Outside the disk every old open track is assigned exactly once in total to the two new walks.

The walks may have self-intersections. Subdivide their tracks at all remaining intersections and bends, and repeatedly delete the closed subwalk between two occurrences of the same subdivision vertex. Each deletion shortens a finite walk, so it yields a simple polygonal path between the required endpoints. We check that this operation preserves admissibility. At an old crossing of \(e\) and \(f\) distinct from \(x\), either the two straight-through tracks belong to different new walks, or both belong to one walk. In the first case each walk visits that point only once; loop deletion either removes its passage or retains its original straight-through passage. In the second case loop deletion may create a turn at that point, but neither the other walk nor a third edge is present there. Likewise, a crossing with any third edge occurs on a track visited at most once by the two walks together, and any retained passage is straight-through. These observations cover every possible meeting: the original drawing had no triple crossings and no common subarcs. The tracks avoid graph vertices in their interiors, and each walk visits its initial and terminal graph vertices only at its ends. Hence the simplified paths are admissible edges, with no new crossing or improper meeting.

The crossing at \(x\) has been removed and none has been added. Repeat this operation while an adjacent-edge crossing remains. The nonnegative integer crossing count decreases at every step, so the process terminates. Remaining crossings are still locally straight; any new turn occurs away from all other edges. The straight spokes at vertices are merely reassigned among the two incident edges. Small disjoint vertex, crossing, and bend disks can be chosen afresh afterwards. This proves all asserted properties. ◻