What the penultimate round cannot see

§ 1The round in distance form

Two players, a common strictly convex cost $c$ with $c(0)=0,$ prizes $v_1,v_2,$ and a head start $\Delta$ for player 1 entering round one. Round two is a static contest with a lead, and Siegel’s theorem prices it (QUICKSTART §1.2; Siegel, 2014): a player who needs round-two effort $r$ to top everything the rival would ever bid collects $[v-\tilde c(r)]^+.$ The rival never bids past its reach, so player 1 needs $r=w_2-\Delta_2,$ player 2 needs $w_1+\Delta_2,$ and $\Delta_2=\Delta+e_1-e_2$ gives

$$ w_2-\Delta_2 \;=\; \underbrace{(w_2-\Delta)}_{M_1}\;+\;e_2-e_1 , \qquad w_1+\Delta_2 \;=\; \underbrace{(w_1+\Delta)}_{M_2}\;+\;e_1-e_2 . $$

That is the formula at the top of the page. It is worth reading slowly, because every invariance below is visible in it. Write $S_i:=M_i+e_{-i}$ for the distance player $i$ must cover once the rival has pushed, and $r_i:=S_i-e_i$ for what is left after $i$’s own round-one effort. Against a fixed rival effort the player is in one of three regimes:

• in the window, $0<r_i<w_i$: payoff $v_i-c(e_i)-c(S_i-e_i)$ — the distance $S_i$ paid for in two instalments;
• out of contention, $r_i\ge w_i$: payoff $-c(e_i)$ — the rest would cost the whole prize, so round two is abandoned;
• saturated, $r_i\le 0$: payoff $v_i-c(e_i)$ — the goal line is already behind, round two is uncontested (def:saturation).

The picture is a rope. On the lead axis, player 2 wins round two outright at $\Delta_2\le-w_1$ and player 1 at $\Delta_2\ge w_2$; these are the two goal lines. Between them sits the pivot $w_2-w_1,$ where both continuations vanish. Player 1 is in contention on the stretch from the pivot to its own goal, of length $w_1,$ and player 2 on the stretch from its goal to the pivot, of length $w_2.$ Each player is in contention exactly when its goal line is within its own reach. The start $\Delta$ is a point on the rope; $M_1$ and $M_2$ are its distances to the two goals and $f=w_2-w_1-\Delta$ its signed distance to the pivot.

Three facts drop out of the drawing.

• The two windows tile the rope: $M_1+M_2=w_1+w_2.$ This is the pivot identity that Docs/normalized_units.md writes as $\rho=C(\omega).$
• The competitive range $-w_1<\Delta<w_2$ is exactly $M_1>0$ and $M_2>0$: the start lies strictly inside the rope.
• At zero efforts exactly one player is in contention, the one on whose side of the pivot the start lies — player 1 if $f<0,$ player 2 if $f>0$ — except on the surface $f=0,$ where the start is the pivot and both sit on the edge. In per-player units $m_i:=M_i/w_i$ this is the weighted identity $\alpha\,m_1+(1-\alpha)\,m_2=1$ with $\alpha=w_1/(w_1+w_2)$: one $m_i$ is below one unless both equal one.

§ 2Four ways for two games to be the same

“The game does not change” has four honest meanings, nested from strongest to weakest, and the question has a different answer at each level.

notion	what holds	what it preserves
identity	$\pi'_i=\pi_i$	everything
strategic equivalence	$\pi'_i=a_i\pi_i+b_i(e_{-i}),\ a_i>0$	best replies to every belief, hence the mixed Nash set
relabelling	equivalence after renaming efforts or players	the Nash set, transported by the renaming
coincidence	nothing structural	the Nash set, by accident of the primitives

The second line is the classical notion (Moulin and Vial, 1978): adding to a player’s payoff anything that depends only on the rival changes no comparison that player ever makes, and a positive rescaling changes none either, because mixed strategies are evaluated by expectation. Between it and coincidence sits best-response equivalence — same best replies to every belief, whatever the payoffs — which Morris and Ui (2004) characterize; for the question here strategic equivalence turns out to be the right rung, because §3 closes it completely.

Throughout, a Nash equilibrium is a pair of Borel probability laws on efforts — the arbitrary-law setting of thm:quad-measure-exhaustiveness — so “the Nash set” means all of them, combs, atomless profiles and anything else.

§ 3The two exact symmetries, and why there is no third

3.1 The payoff unit

$$ (v_1,v_2,\Delta,c)\;\longmapsto\;(\lambda v_1,\ \lambda v_2,\ \Delta,\ \lambda c),\qquad \lambda>0. $$

Reaches do not move, since $(\lambda c)^{-1}(\lambda v)=c^{-1}(v),$ so neither do $M_1,M_2$ or the windows, and every payoff is multiplied by $\lambda.$ A change of currency. Equilibrium efforts, masses and the dial are untouched; rents scale.

3.2 The tail above the larger reach

An effort above one’s own reach is strictly dominated by zero effort: it pays at most $v_i-c(e_i)<0,$ while $e_i=0$ pays $[v_i-\tilde c(S_i)]^+\ge0$ whatever the rival does. So no equilibrium puts mass there, and the Nash set of the round equals the Nash set of the reduced round on the rectangle $[0,w_1]\times[0,w_2]$. On that rectangle player $i$ reads the cost only on $[0,w_i]$: its own effort is at most $w_i$, and a residual $r_i\ge w_i$ falls into $[\,\cdot\,]^+=0$ for every increasing extension of $c$.

This is rem:above-w-free (a), written there at the symmetric point; the same domination argument works anywhere in the competitive range. The finer clause has a consequence the symmetric case hides: when $w_1<w_2$, the stretch of $c$ between the two reaches is read by player 2 alone. It is the stronger player’s private region.

3.3 Nothing else, at the level of payoffs

Are there other transformations of the primitives that make the two rounds strategically equivalent — a lead traded against a prize, a cost reshaped and compensated somewhere? No. The class below is wide: no smoothness and no $c'(0)=0,$ so it covers the general quadratic $ae^2+be$ of module 14 as well.

The last sentence is the instructive one. On the full effort space the tail is pinned; on the reduced rectangle it is free. The tail is a symmetry of the Nash set, not of the game. Every other change of primitives changes some player’s best reply to some belief. It may still leave the Nash set alone, but only through coincidence, the bottom rung of §2, and §7 shows that coincidence is not rare.

§ 4Relabellings, and the rope chart

4.1 The effort unit

$$ (v_1,v_2,\Delta,c)\;\longmapsto\;\bigl(v_1,\ v_2,\ s\Delta,\ c(\cdot/s)\bigr),\qquad s>0. $$

Reaches, $M_i$ and $f$ all scale by $s$; every equilibrium is pushed forward by $e\mapsto se$; masses, the dial $t$ and payoffs do not move. Together with §3.1 this is the normalization of rem:sym-normalization (at the symmetric point) and of Docs/normalized_units.md (chart B, effort in units of $w_2$, payoff in units of $v_2$).

4.2 The player swap

$$ (v_1,v_2,\Delta,c)\;\longmapsto\;(v_2,\ v_1,\ -\Delta,\ c),\qquad f\mapsto -f,\quad M_1\leftrightarrow M_2 . $$

The rope is turned end for end. The record’s orientation convention — label so that the zero-rent player is player 2 — is this swap used as a normalization.

4.3 Why only dilations

A relabelling of efforts that keeps the round inside the model must carry margins to margins, because the continuation reads efforts only through $e_1-e_2.$ If $\varphi_1(x)-\varphi_2(y)=\psi(x-y)$ with $\varphi_i(0)=0$ and $\psi$ continuous, then $y=0$ and $x=0$ give $\varphi_1=\psi$ and $\varphi_2(y)=-\psi(-y),$ and $\psi(x-y)=\psi(x)+\psi(-y)$ is Cauchy again: $\varphi_1=\varphi_2=s\cdot\mathrm{id}.$ A one-shot all-pay contest reads scores only ordinally and accepts any common increasing rescaling; carrying a lead into the next round makes the score scale cardinal, and the symmetry group shrinks from all monotone maps to dilations.

4.4 What is left: a shape and two numbers

Four primitives, two continuous symmetries and one discrete: the round is a normalized cost shape on an interval plus two numbers. Which two is a choice of chart. Chart B, the record’s, divides by player 2’s reach and prize: $C(x)=c(w_2x)/v_2,$ $\omega=w_1/w_2,$ $\delta=\Delta/w_2$, with $\rho=v_1/v_2=C(\omega)$ derived. It treats the players asymmetrically and its parameter domain $-\omega<\delta<1$ is an unbounded wedge.

The rope suggests a chart that treats them alike. Measure effort in units of the rope’s length $W:=w_1+w_2=M_1+M_2$ and record

$$ \alpha:=\frac{w_1}{W}\in(0,1),\qquad \zeta:=\frac{M_2}{W}=\frac{w_1+\Delta}{W}\in(0,1), $$

player 1’s share of the rope and the position of the start measured from player 2’s goal. The competitive range becomes the open unit square, and the familiar loci become its geometry:

• $\Delta=0$ is the diagonal $\zeta=\alpha$; the surface $f=0$ is the anti-diagonal $\zeta=1-\alpha$; the symmetric point is the centre;
• the player swap is the point reflection $(\alpha,\zeta)\mapsto(1-\alpha,1-\zeta)$;
• player 1 starts in contention above the anti-diagonal, player 2 below it;
• $M_1/W=1-\zeta$, $f/W=1-\alpha-\zeta$, and the record’s quadratic chart (lem:quad-chart) already uses $\alpha$, with its $\lambda=M_1/w_1=(1-\zeta)/\alpha.$

For a pure power the normalized shape does not depend on the primitives at all (§5), so for $c=\kappa e^\theta$ the square is the whole parameter space, one square per exponent, with $\rho=(\alpha/(1-\alpha))^\theta$ constant on vertical lines. Plate III draws it at $\theta=2$ with the regions of §7 already marked.

§ 5Costs with a symmetry of their own

The two units act on every round at once. Some costs are left fixed by part of that action. Then the units become a symmetry of the round at fixed cost: they move the prizes and the lead while the cost function stays put.

The continuous case is the record’s homogeneity (Docs/normalized_units.md §8, prop:power-sigma11-monotone): a pure power has one more symmetry than a general cost, which is why its rounds fill exactly the square of §4. The discrete case is new to the record, and it is not a curiosity.

5.1 A cost in X whose floor never settles

Take the exponent at which the pure power sits exactly on the symmetric $\sigma(1,1)$ gate, $\theta^*=\ln2/\ln(4/3)=2.40942\ldots,$ where $(\tfrac34)^{\theta^*}=\tfrac12$ (QUICKSTART §1.6), and give it a small wobble in the logarithm of effort:

$$ c_\varepsilon(e)=e^{\theta^*}\bigl(1+\varepsilon\sin(2\pi\log_2 e)\bigr),\qquad \varepsilon=10^{-3}. $$

Its derivatives are $c^{(j)}(e)=e^{\theta^*-j}\,Q_j(\ln e)$ with $Q_j$ periodic, so membership in X is a check on one period. All three are positive iff $\varepsilon<0.00178$; the binding one is $c'''$, with bounds $0.257,$ $0.039$ and $0.00178$ for $j=1,2,3.$ So $c_\varepsilon\in\text{X}.$ At the symmetric point the round sees only the normalized profile

$$ \hat c_v(u)=\frac{c(wu)}{v}=u^{\theta^*}\,\frac{P(\ell+\ln u)}{P(\ell)},\qquad \ell=\ln w, $$

a function of $\ell$ modulo $\ln2$: doubling the reach, which multiplies the prize by $2^{\theta^*}\approx5.31$, returns the same round. By lem:sym-sigma11-reach-pratt (banked, valid for every strictly convex cost) $\sigma(1,1)$ is live iff $\hat c_v(\tfrac34)\ge\tfrac12,$ and at $\theta^*$ that reads

$$ \sigma(1,1)\ \text{live}\iff P\bigl(\ell-\ln\tfrac43\bigr)\ \ge\ P(\ell)\iff \sin\!\bigl(k\ell-k\ln\tfrac43\bigr)\ \ge\ \sin(k\ell),\qquad k=\tfrac{2\pi}{\ln 2}. $$

Since $\ln\tfrac43$ is not a multiple of $\ln 2$, the two sides cross twice per period. Measured on a grid of $10^5$ points per period, the gate margin runs from $-9.6\cdot10^{-4}$ to $+9.7\cdot10^{-4}$ and is nonnegative on exactly half. So the lowest live rung at the symmetric point is 1 on alternate bands of $\ln v$ and at least 2 on the others, for every prize, however large: it has no limit as $v\to\infty.$ Nor does this need the knife edge. At $\theta=2.405$ the pure power is live with margin $6.4\cdot10^{-4}$; X allows $\varepsilon<0.00175$ there, and $\varepsilon=10^{-3}$ still flips the gate, whose threshold is $\varepsilon\approx0.00066$.

The general lesson concerns the prize. At the symmetric point the prize is a dial of normalized convexity (rem:sym-normalization(a): “raising the prize walks the model”). For powers it is inert. For the exponential and the cosh families it is monotone (prop:exp-sym-floor, prop:cosh-floor-monotone). For log-periodic costs it is periodic. Physics has a name for the third behaviour, discrete scale invariance (Sornette, 1998), and the same log-periodic oscillation of every scale-free statistic.

5.2 How many shapes a family really has

Counting what the units remove is the quickest audit of a parametric family.

family	units remove	shape left	the round at $(v,\Delta)$ is the round of
$\kappa e^\theta$	$\kappa$ — and the stabilizer acts on $(v,\Delta)$	$\theta$	$e^\theta$ on the square of §4
$A(e^{ke}-1)$	$A$ and $k$	none	$e^e-1$ at $(v/A,\ k\Delta)$
$ae^2+be,\ b>0$	$a$ and $b$	none	$e^2+e$ at $(av/b^2,\ a\Delta/b)$
$a(\cosh(ke+b)-\cosh b)$	$a$ and $k$	$b$	$\cosh(e+b)-\cosh b$ at $(v/a,\ k\Delta)$
$e^\theta+\varepsilon e^2$	at symmetry, $\varepsilon$ and $v$ merge	$\theta,\ w^{\theta-2}/\varepsilon$	rem:sym-normalization(a)

The third row is the unit change that prop:quadg-s33-floor-null-reduction uses to pass from every positive coefficient pair to $c=e^2+e$: each pair is that one cost in other units, with the Nash set dilated by $a/b.$

§ 6Normalizations from outside the model

Theorem 1 closes the question inside the model, where both players share one cost. Step outside for a moment and more symmetries appear. They are not transformations of $(v_1,v_2,\Delta,c)$, but they say what those four primitives mean.

6.1 Prizes are costs

Divide each player’s payoff by its own prize:

$$ \frac{\pi_i}{v_i}=\bigl[\,1-\tilde C_i(r_i)\bigr]^+-C_i(e_i),\qquad C_i:=\frac{c}{v_i}. $$

Siegel’s round-two price holds verbatim for player-specific costs, $[v_1-\tilde c_1(w_2-\Delta_2)]^+$ with $w_i=c_i^{-1}(v_i)$ (Siegel, 2009). The display is therefore the round of a contest with equal prizes and costs $c/v_1$ and $c/v_2.$ Per-player rescaling is a strategic equivalence, so the two have the same Nash set. Inside the model, a prize asymmetry is a proportional cost asymmetry and nothing else. Theorem 1 forces $a_1=a_2$ only because the model ties both players to one cost; the per-player unit is a symmetry the model cannot express.

What the model does not contain is a productivity asymmetry. A player whose score is $a_i e_i$ has cost of score $c(\cdot/a_i)$, which is proportional to $c$ iff the factor $a_i$ lies in $\operatorname{Stab}(c)$. For pure powers any $a_i$ qualifies, for log-periodic costs only $a_i\in s_0^{\mathbb Z}$, and otherwise only $a_i=1$. So off the powers, the model’s “asymmetric players” differ in prize, equivalently in cost scale, never in productivity. For the linear all-pay auction, a power, the three notions coincide, so the classic analyses lose nothing by writing every asymmetry as a difference in values (Hillman and Riley, 1989; Baye, Kovenock and de Vries, 1996).

6.2 Each player’s private triple

Player $i$’s best-reply correspondence depends on $(v_i,\ M_i,\ c|_{[0,w_i]})$ and on nothing else. The rival enters through one number, how far $i$’s goal line is from the start. Every change of primitives therefore splits into what each player sees:

move	player 1 sees	player 2 sees
$\Delta$ up by $d$	goal nearer: $M_1-d$	goal farther: $M_2+d$
$v_2$ up, $w_2$ up by $dw$	goal farther: $M_1+dw$	a larger prize
$v_2$ up and $\Delta$ up by the same $dw$	nothing	a larger prize
$v_1$ up, $w_1$ up by $dw$	a larger prize	goal farther: $M_2+dw$
$c$ reshaped on $(w_1,w_2)$ when $w_1<w_2$, with $c(w_2)$ kept	nothing	its own cost

A raise in the rival’s prize reaches you only as your goal line moving away. The third row is the clean experiment for comparative statics in one player’s prize.

6.3 The lead is a translated cost

In the shifted frame of QUICKSTART §1.4, $\tilde y:=y+f,$ the payoffs read

$$ \pi_1=\bigl[v_1-\tilde c(w_1+\tilde y-x)\bigr]^+-c(x),\qquad \pi_2=\bigl[v_2-\tilde c(w_2+x-\tilde y)\bigr]^+-c(\tilde y-f),\qquad \tilde y\ge f . $$

Both continuations take their $f=0$ form, the start at the pivot, and $f$ survives only in player 2’s own cost and its action floor. The round with lead $\Delta$ is the round at the pivot in which player 2’s cost is translated by $f$. That is how general all-pay contests absorb head starts into costs (Siegel, 2009), and it explains why the surface $f=0$, untranslated, keeps turning up as a special place in the record.

6.4 The two-round reach

Covering a distance $S$ alone over two rounds costs at least $2c(S/2)$, by convexity, with equality at the even split. The farthest goal worth walking to unopposed is therefore

$$ \hat w_i:=2\,c^{-1}(v_i/2)\ \ge\ w_i,\qquad \hat w=2^{1-1/\theta}\,w\ \text{ for } e^\theta\quad(\sqrt2\,w\ \text{at}\ e^2). $$

This is QUICKSTART §1.6’s gate $2c\bigl((2w_1+w_2+\Delta)/4\bigr)\ge v_2$, together with individual rationality, read as lengths. For pure powers the record already carries the factor $a=2^{1-1/\theta}$ (prop:power-sigma11-monotone); the two-round reach names it for every cost. At the symmetric point it becomes $\hat w\le\tfrac32w$, which is $\hat c(\tfrac34)\ge\tfrac12$. For the exponential $A(e^{ke}-1)$ the ratio $\hat w/w=2\ln(1+y/2)/\ln(1+y),$ $y=v/A,$ climbs from $1$ to $2$ as the prize grows and passes $\tfrac32$ exactly at $y=2q^*(q^*+1)\approx10.4445$, with $q^*$ the tribonacci constant. That is the $\sigma(1,1)$ death point of prop:exp-sym-floor, recovered as a length (checked: the ratio there is $1.500000000000$).

6.5 The convexity dividend, and why the linear part washes out

Subtract from each payoff its value at zero effort, a rival-only term and so a strategic equivalence. Write $\Pi_i:=\pi_i(e_i,e_{-i})-\pi_i(0,e_{-i})$ for the gain over idling. In the window,

$$ \Pi_i=D_c\bigl(e_i,\ S_i-e_i\bigr)-\bigl(c(S_i)-v_i\bigr)^+,\qquad D_c(a,b):=c(a+b)-c(a)-c(b)=\int_0^a\!\!\int_0^b c''(s+t)\,dt\,ds , $$

(for $c\in C^2$; in general read $c''$ as the measure $dc'$). The second term involves the rival only. Inside the window a player’s strategic payoff is the dividend from splitting its road in two, and the dividend reads only the curvature. At $e^2$ it is $2ab$, bilinear.

Hence the linear part of the cost is invisible there. Shear the cost and keep the reaches, $c\mapsto c+\beta e$ with $v_i\mapsto v_i+\beta w_i.$ The dividend does not change, and a direct check of all three regimes gives

$$ \pi_1^{\beta}-\pi_1=\beta\bigl[(f-z)^+-(z-M_1)^+\bigr]+(\text{rival only}),\qquad z=e_1-e_2, $$

numerically to $7\cdot10^{-16}$. Inside the window, round-one effort displaces round-two effort one for one, so a linear charge falls on a fixed total and decides nothing. Outside, effort displaces nothing. Short of the window the shear subsidizes staying out, relative to entering; past saturation it taxes overshooting.

Every cost with $c'(0)=\beta>0$ is $c_0+\beta e$ with $c_0$ convex and $c_0'(0)=0.$ So the general quadratic $ae^2+be$ of module 14 is the round at $ae^2$, with prizes lowered by $bw_i$ to keep the reaches, plus these two hinges, and both of its new features sit on them. The zero corner of thm:zero-corner (first atom at $0$ for $t\le c'(0)/c'(M)$) is the subsidy hinge. At $x=0$ the window mass $t$ earns the marginal dividend $t\,c_0'(M)$ and the out-of-contention mass $1-t$ pays the charge $\beta$, so the corner is $t\,c_0'(M)\le\beta(1-t)$, which is the same inequality. The lock at $M$ is the overshoot hinge at $z=M_1$, against player 2’s zero atom.

§ 7What the Nash set forgets

Theorem 1 says that only the units and the tail move every round without moving any best reply. It does not say that the Nash set is a fine invariant. It is not: it can forget almost everything it was given.

7.1 Beaten before the start

For $e^\theta$ in the rope chart the two regions are $\zeta>2^{1-1/\theta}(1-\alpha)$ and $\zeta<1-2^{1-1/\theta}\alpha$: two opposite corners of the square, exchanged by the swap, drawn in Plate III. Both lie inside the region where the corresponding first rung is live. Between a region’s edge and the $\sigma(1,1)$ gate, player 2 is deterred by player 1’s first push rather than beaten at the start. There the full Nash set is a question for the chain results, not for domination.

7.2 One equilibrium, five rounds

On the region of Proposition 4 the Nash set depends on the primitives through $M_1=w_2-\Delta$ alone. The prize of the winning player does not enter at all: raising $v_1$ moves only its rent. Here are five rounds, with three different costs, all with Nash set $\{(\delta_{1/4},\delta_0)\}$:

cost	$(v_1,v_2,\Delta)$	$(w_1,w_2)$	$(M_1,M_2)$	$\hat w_2$	best fight	$u_1$
$e^2$	$(1.44,\ 1,\ 0.5)$	$(1.2,\ 1)$	$(0.5,\ 1.7)$	$1.414$	$-0.445$	$1.315$
$e^2$	$(2.25,\ 1,\ 0.5)$	$(1.5,\ 1)$	$(0.5,\ 2.0)$	$1.414$	$-1.000$	$2.125$
$e^3$	$(1.728,\ 1,\ 0.5)$	$(1.2,\ 1)$	$(0.5,\ 1.7)$	$1.587$	$-0.228$	$1.697$
$e^3$	$(1.728,\ 0.729,\ 0.4)$	$(1.2,\ 0.9)$	$(0.5,\ 1.6)$	$1.429$	$-0.295$	$1.697$
$e^e-1$	$(e^{1.2}-1,\ e^{0.9}-1,\ 0.4)$	$(1.2,\ 0.9)$	$(0.5,\ 1.6)$	$1.096$	$-0.991$	$1.752$

“Best fight” is the largest payoff player 2 can get from any fighting effort, against any $e_1\in[0,w_1]$, on a $2001\times2001$ grid. In every row it equals the bound $v_2-2c(M_2/2)$ of the proof, attained at $e_1=0.$ No unit change, swap or tail carries one row to another. Their Nash sets agree because the round has stopped looking at anything but player 1’s distance to its goal.

7.3 One equilibrium sees finitely many points

For a finite comb $E(\sigma)$ is a finite set. A comb equilibrium therefore survives any convexity-preserving raise of the cost away from finitely many points. This is a statement about one equilibrium. The raise can kill other equilibria that evaluate the cost where it was raised, and it can create new ones, since it lowers deviation payoffs everywhere. Read with Theorem 1, it gives a two-sided picture: no transformation fixes every equilibrium of every round except the units and the tail, yet each single equilibrium is fixed by an enormous family of costs.

7.4 Open

• Fibres off the dominance regions. Whether two rounds not related by §3–§4 can share a whole Nash set away from Proposition 4 is a question about whole Nash sets. The record decides those only where it has a classification: at $e^2$, through thm:quad-measure-exhaustiveness, cor:quad-rung-unique and thm:quad-chain-connected. Nobody has computed a fibre there.
• Local injectivity. Near a round whose Nash set contains a deep comb, does the Nash set determine the primitives modulo the units and the tail? A comb with $m$ atoms pins $c$ and $c'$ at about $2m$ points, which suggests that it does; nothing on this page proves it.

§ 8Ledger

item	effect on the Nash set	status	§
payoff unit $\lambda$	fixed	elementary	3.1
cost above $\max(w_1,w_2)$	fixed	Proposition 1, proved here; rem:above-w-free(a) at symmetry	3.2
nothing else, at the level of payoffs	—	Theorem 1, proved here	3.3
effort unit $s$	dilated by $s$	elementary; rem:sym-normalization at symmetry	4.1
player swap	players exchanged	elementary	4.2
dilations are the only relabellings	—	remark (Cauchy)	4.3
rope chart $(\alpha,\zeta)$	—	exposition	4.4
stabilizers; log-periodic costs	dilated, cost fixed	Proposition 2, proved here	5
symmetric floor of $c_\varepsilon$ oscillates	—	lem:sym-sigma11-reach-pratt plus a measured margin	5.1
prizes are costs	fixed (outside the model)	elementary	6.1
private triples	—	elementary	6.2
lead as translated cost	—	QUICKSTART §1.4	6.3
two-round reach; $\sigma(1,1)$ as lengths	—	Proposition 3, proved here	6.4
dividend; linear part as two hinges	rival-only in the window	identity; checked to $10^{-15}$	6.5
beaten before the start	collapses to $M_1$	Proposition 4, proved here	7.1
raise where it does not look	one equilibrium kept	Proposition 5, proved here	7.3

Nothing on this page is in proofs.tex, and nothing here is \banked. Theorem 1 and Propositions 1–5 are candidates. Each would owe the banking gate’s three lines (RULES §2), and that call is Sergio’s. An adversarial read-only pass (Codex, 5 October 2026) re-derived the arguments of §3 and §5–§7 and found four gaps, all repaired above: the flat interval in Step 2 of Theorem 1, the full-space step of the same proof, the minimal period in Proposition 2, and the smoothness behind the dividend integral.

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