Appendix B Computation Scripts, Computations 165 to 184

Computation 165 ·  §8.3.6 ·  [PROOF]
The photoelectric effect: the kinematic half, and the vertex that is still missing.

Computation 160 and Computation 161 called the effect absent outright. Half of it is not, and the halves separate cleanly. Kinematics need the ladder and nothing else: with the field’s energy in steps of $\hbar\omega$, energy conservation gives (70), hence a threshold at $\omega_{0}=W/\hbar$, a slope of exactly $\hbar$, and $\mathrm{d}K/\mathrm{d}I=0$, all three symbolic. The classical alternative fails on all three: $K=I\sigma t-W$ contains no $\omega$, so there is no threshold at all, and the energy grows with intensity. And it fails quantitatively: to meet Lawrence and Beams’ $3\times 10^{-9}$ s bound [211] a classical field needs above $10^{10}\,\mathrm{W\,m^{-2}}$ against the $1\,\mathrm{W\,m^{-2}}$ of ordinary photoemission, ten orders. Checked as numbers: $\lambda_{0}=hc/W$ gives $579$, $525$, $286$ and $219$ nm for caesium, sodium, zinc and platinum, each within a nanometre of tabulation. And the picture’s own end is predicted: the Keldysh parameter reaches unity at $1.14\times 10^{14}\,\mathrm{W\,cm^{-2}}$ for $800$ nm and $13.6$ eV, the measured multiphoton-to-tunnelling crossover, so single-photon absorption is an intensity regime and not a law. The tally: of five ingredients PST posits none. Four are outright, the ladder, the masslessness, the commutator’s $\hbar$ and the electron’s fermionic character. Corrected by Computation 166: this computation called the fifth, the vertex, not supplied. Its form is forced by gauge covariance (§13.3.1) and only its strength is scoped, so the tally is five of five with one scoped and the residue is a cross-section calculation rather than a missing ingredient.

 

Computation 166 ·  §13.3.1 ·  [PROOF]
The interaction vertex: forced by gauge covariance, and the third “absent” that was not.

Computation 165 called the vertex the one ingredient PST does not supply. §13.3 already derived minimal coupling from gauge covariance; this turns that phrase into algebra, with explicit Dirac matrices and a completely general spinor. The variation is one term and it names the current: under $\psi\mapsto e^{i\alpha(x)}\psi$ the free Dirac Lagrangian changes by exactly (156), carrying the vector current and nothing else, so only a counterterm on that current can cancel it. The transformation law is an output: requiring the cancellation returns $A_{\mu}\mapsto A_{\mu}-e^{-1}\partial_{\mu}\alpha$, uniquely. The axial alternative fails twice: it carries a different current, and (157) gives $\partial_{\mu}(\bar{\psi}\gamma^{\mu}\psi)=0$ against $\partial_{\mu}(\bar{\psi}\gamma^{\mu}\gamma^{5}\psi)=2im\bar{\psi}\gamma^{5}\psi$, both checked on the explicit spinor, while $\partial_{\mu}\partial_{\nu}F^{\mu\nu}=0$ identically means only a conserved current can source Maxwell. Scoped rather than absent: the strength, through $e^{-2}\propto r_{0}^{2}$ and $Z^{2}=\mathrm{e}^{-1}$ modulo the emergence premise. And the pattern worth recording: three times in this line of work a claim of “absent outright” was written without opening the chapter that already had it, at Computation 161, Computation 165 and here. No gate catches that.

 

Computation 167 ·  §7.5 ·  [PROOF]
What fixes the absolute scale of $\Pi_{\mu\nu}$, and what does not.

Computation 164 left $\rho$ a quotient map, so realisation points exist everywhere and $\Pi_{\mu\nu}$ is nonzero in what we call empty space. The obvious next question is what vacuum density that predicts. It predicts none: with the weight forced (Computation 158) and the bond kernel normalised (Computation 159), the only remaining freedom in the magnitude is the pair density, measured here to carry $\rho$ while leaving $w$ flat across a factor of eight in the point count. And the theory does not supply that density: $d_{0}$ is a coherence length, which §15.3.1 says is not the grain, and $\ell_{*}$ enters as a momentum cutoff on field modes, so reading the spacing off one rather than the other moves the implied density by $(d_{0}/\ell_{*})^{6}\sim 10^{61}$ at the bound. Nothing standing is affected, which the second check verifies rather than assumes: rescaling the ensemble moves $\rho$ by orders while $w$, the trace ratio and the eigenvalue pattern do not move at all, because every result this sector has produced is a ratio, a shape or a count. Two things follow: the conformal identification carries a second gap beside Computation 164’s cone, its magnitude within that cone being free; and there is no vacuum-energy catastrophe here, because the $10^{120}$ problem needs a computed scale to disagree with measurement and this construction does not produce one.

 

Computation 168 ·  §1.7 ·  [PROOF]
The extremal selection principle is not an extra commitment.

§1.7 called the one remaining principle “a natural choice given P1–P3 but not derived from them”, and Computation 148 had already cut three to one. This closes it, and the argument is rigidity rather than size: choosing a proper sub-object is not a smaller commitment than taking the whole but a larger one, since one must also say which, and P1 supplies no selector. The cost is measured, not asserted. The octonions are built by Cayley–Dickson and then tested (norm-multiplicative to $4\times 10^{-15}$, alternative to $10^{-14}$, non-associative), $\mathrm{Der}(\mathbb{O})$ comes out $14$-dimensional and closed under the bracket, and orbit dimensions follow as $14-\dim(\text{stabiliser})$: a unit imaginary costs $6$ ($S^{6}=G_{2}/SU(3)$), a quaternionic triple costs $8$ ($G_{2}/SO(4)$), the whole of $V_{7}$ costs nothing. The $6$ and the $8$ are textbook, so they check the arithmetic rather than feed it. Exactly two choices are free, the zero subspace and $V_{7}$, and P1’s two clauses settle it: sole-primitive removes every intermediate choice, distinction-exists removes the trivial one. The envelope goes the same way discretely: of sixteen subsets of the classification only the empty and the full one need no naming datum, and the full one is the triple product at dimension $64$, recovering Computation 148. And the criterion is not “largest” in disguise: under $SU(3)$ a one-dimensional subspace is rigid, so rigidity is group-relative and the two agree here only because P1 leaves the symmetry unbroken. Scope: the Hurwitz classification and the norm-multiplicativity that puts the substrate in it are upstream and untouched.

 

Computation 169 ·  §3.4 ·  [PROOF]
The atom criterion, and the step that does not close it.

Computation 152 reduced §19.9.1’s standing condition to “a coordination premise, weaker than independence”. This sharpens that residue and, in an earlier version, claimed to close it; the closing step is withdrawn here and the refutation is published as part 3. What stands: balance needs $v(a)=-\sum_{b\neq a}v(b)$, one point fixed by the others, so $\Pr[\text{balance}]$ is exactly the expected conditional mass on it (planted atoms of mass $p$ reproduce $\Pr=p$ to $0.004$); hence one atomless conditional anywhere makes balance null, and the reach of that is measured: leaving a single property unpinned while five stay maximally coordinated turns $31\%$ into a null set with $\Pr[|\sum v|<\epsilon]\propto\epsilon^{6}$, the exponent read from a cancellation-free closed form and cross-checked against Monte Carlo within five Poisson $\sigma$. What is withdrawn: the step pricing the atom as a distinguished direction. It priced an unconditional atom at a fixed point while the chain delivers a conditional atom at a random one, and Computation 152’s $\mu_{\pm}$ has rotation-invariant law, names no direction, pays nothing, carries maximal conditional atoms and balances $31\%$ of the time. So the condition stands, and closing it needs an argument about correlation rather than about direction.

 

Computation 170 ·  §19.9 ·  [PROOF]
F4’s spectrum half: the alignment branch fails for the terms tested.

F4 says the theory below $M_{*}$ is the Standard Model. Half is a matching to observation and cannot be proved; this takes the other half against the objection that a Landau–Ginzburg transition should deposit six massless modes below $M_{*}$. It turns on a reading of P3. With a linear locking term the vacuum solves exactly, giving $m_{r}^{2}=\kappa/r+8br^{2}$ against $m_{t}^{2}=\kappa/r$, so $m_{t}<m_{r}$ at every $\kappa$ with the ratio saturating at $1/3$; a quadratic term gives $2\kappa$ against $4(a+\kappa)$, saturating at $1/2$, and both closed forms are checked against the Hessian. Since the radial mode is the Higgs and $M_{*}=10.26\,m_{h}$, those modes fall below the Higgs mass. But that is not an exclusion of the branch, which an earlier version claimed: the sign turns on the curvature of the locking term, and a quartic one, $-\kappa\langle\psi,\hat{\tau}\rangle^{4}$, reverses it above a crossover, with $m_{t}^{2}=2a\kappa/(b-\kappa)$ unbounded as $\kappa\to b$ and the angular modes above $M_{*}$ at $\kappa\approx 0.2495$. Reading the locking as direct determination, which is what P3 says, leaves the order-parameter sector one mode: six generators move the vacuum and fifteen fix it, the six broken explicitly, so neither yields a Goldstone. The signature theorem is insensitive: well-posedness is decided by the principal symbol, so Computation 2 needs hyperbolicity rather than masslessness.

 

Computation 171 ·  §9.2 ·  [PROOF]
$\theta$’s provenance: the phase survives determination, the direction does not.

Computation 170 read P3’s locking as determination without excluding the alternative, and took away the mode §9.2 was using. This says which degree of freedom $\theta$ is. The phase: $\hat{\tau}$ is a real direction, so determination fixes a direction and says nothing about a phase, and the complex line $\mathbb{C}\hat{\tau}$ is stable under $\psi\mapsto\mathrm{e}^{i\alpha}\psi$ where the real line is not. A real order parameter therefore has no candidate for $\theta$, which makes the complexification a requirement rather than the richer of two formulations. Masslessness is a condition, measured not assumed: at an exactly solved vacuum the phase curvature sits at the finite-difference floor ($\sim 10^{-8}$, which is $\epsilon/h^{2}$) for $|\langle\psi,\hat{\tau}\rangle|$, its square and its fourth power, and is lifted to $0.27$ by $\mathrm{Re}\langle\psi,\hat{\tau}\rangle$. Gauge invariance is what forbids the lifting term, the moduli being invariant and the real part not, which is the same exactness Computation 170 used for the photon. And $\theta$ is not the photon: as a free massless scalar it would break F4 alone, and a gauged broken $\mathrm{U}(1)$ gives a massive boson, so $\theta$ is eaten and the photon is the unbroken combination. One flat direction, not six. Left open: how two substrate degrees of freedom embed in the projected four-component doublet.

 

Computation 172 ·  §8.2.1 ·  [PROOF]
$\theta$ is eaten by the $Z$, not gauge-fixed away by $\mathrm{U}(1)_{\mathrm{em}}$.

The PPN enumeration removed $\theta$ from the infrared by calling it a $\mathrm{U}(1)_{\mathrm{em}}$ gauge redundancy, Stueckelberg-eliminable because that symmetry is unbroken. The premise is right and the inference is not, and the same fact settles both: $Q=T_{3}+Y/2$ annihilates the vacuum, which is why $\mathrm{U}(1)_{\mathrm{em}}$ is unbroken, and annihilates $H^{0}$, which carries charge zero, so the transformation does not move $\theta$ and has nothing to remove. What removes it: the phase direction $i\langle H\rangle$ lies in the span of the broken generators acting on the vacuum, of rank three, so $\theta$ is an ordinary eaten would-be Goldstone taken by the $Z$, which is what Computation 171 concluded from the substrate side independently. And the list counted it twice, once correctly among “the three electroweak eaten Goldstones” and once again as a fourth item. The conclusion stands: four real components, three eaten, one massive Higgs and no massless scalar, so no long-range scalar survives and $\gamma=\beta=1$ is untouched, only the reason changing. And the embedding closes: the substrate’s complex line is the neutral component, modulus to the Higgs and phase to the $Z$-eaten Goldstone, with $H^{\pm}$ supplied by $\mathrm{SU}(2)_{L}$, which is the projection question Computation 171 left open.

 

Computation 173 ·  §12.2 ·  [NUMERICAL]
The seagull solved self-consistently, and $b$ as an output.

§12.2 records that $M_{*}$ and $Z^{2}$ are not jointly established, and states it with $b$ held at $1/4$. But if the seagull must use the canonical coupling, $b$ is an output, not an input, and the system closes: $m_{h}^{2}=6\lambda_{\mathrm{SM}}(M_{*})M_{*}^{2}/(16\pi^{2})$ together with the bridge $\lambda_{\mathrm{SM}}(M_{*})=b\,\mathrm{e}^{-1}$ and Computation 123’s two-loop running is two equations in two unknowns. Solving it: a self-consistent point exists at $M_{*}=2203$ GeV with $\lambda_{\mathrm{SM}}(M_{*})=0.0851$, the ratio to the postulated $1285$ GeV being $1/Z$ identically; and it returns $b=0.2313\pm 0.0020$ against P3’s $0.25$, the uncertainty propagated by re-solving for $M_{*}$ at each input variation rather than holding it fixed. The gap is $7.5\%$, which is $9.5\sigma$ on that budget, so input uncertainty cannot absorb it. Why that is not yet a falsification: the coefficient $C_{H}$ depends on the shape of the ultraviolet regulator and not only on its scale, and the book supplies a scale while never specifying a shape. Measured here: a hard cutoff and a Gaussian both give $\kappa=1$, a smooth step of width $1/8$ gives $1.05$, width $1/4$ gives $1.20$, an exponential profile gives $2.00$. Closing the gap needs $\kappa=1.081$, inside that range. What would settle it: §12.2 already specifies the substrate spectrum (Walsh eigenmodes over non-empty $S\subseteq[D]$, masses $|S|^{1/2}M_{*}$), so the mode-sum envelope is the missing profile and is computable. It would either recover $b=1/4$, turning P3’s quartic from a postulate into a result, or falsify the bridge at a measured significance. Computation 174 carries out that calculation and returns the second outcome.

 

Computation 174 ·  §12.2 ·  [NUMERICAL]
The substrate regulator envelope, and what it does to $C_{H}$.

Computation 173 declines to call its $9.5\sigma$ gap a falsification because $C_{H}$ depends on the ultraviolet regulator’s shape and the book supplies only a scale; it then notes that §12.2 already specifies the substrate spectrum, so the envelope is computable and owed. This computation pays that debt. The question to settle first is which of the theory’s two cutoffs the seagull answers to: it regulates $4$-momentum at $M_{*}$, while eq. (134) regulates the substrate Dirac spectrum at the matched scaling $\Lambda=\sqrt{D}\,M_{*}$, a factor $\sqrt{D}$ away. It does not have to be settled, because both are computable and both give $\kappa=1$. Route A, at $M_{*}$: §12.2’s mass spectrum is $m_{S}=|S|^{1/2}M_{*}$ over non-empty $S$, together with the zero mode, so $(0,M_{*})$ is empty and the first level sits exactly at $M_{*}$ with degeneracy $D$; a loop that runs out of modes at a spectral gap edge is cut off by a step, and $\kappa=1$ with no $D$-dependence. What that rests on, corrected on a referee pass: not the gap alone. The Standard Model lives below $M_{*}$ where the substrate has no state, so the emergent low-energy modes are collective and a substrate gap does not by itself truncate an emergent loop; the step follows once Computation 105’s layer boundary is identified with the cutoff, which is architecture the book carries rather than a consequence of the Boolean Laplacian. This is the architecture Computation 105 states from the other side, the two layers occupying the disjoint ranges $[m_{h}^{2},M_{*}^{2}]$ and $[M_{*}^{2},D]$ and meeting at a single scale of measure zero. Route B, at $\Lambda=\sqrt{D}M_{*}$: the Walsh ladder does furnish a smooth envelope, and although the substrate’s binomial density of states $\binom{D}{j}$ and the four-dimensional continuum’s $\rho(j)=2j/D^{2}$ look nothing alike, one peaked at $D/2$ and the other at its endpoint, the seagull sees only the first inverse moment, where $\langle 1/j\rangle_{\mathrm{cont}}=2/D$ exactly and $\langle 1/j\rangle_{\mathrm{sub}}=(2/D)(1+1/D+O(D^{-2}))$; hence $\kappa_{\mathrm{sub}}(D)=1+1/D$, computed in exact rational arithmetic and cross-checked by log-binomial weights to $D=65536$. The shape that envelope corresponds to: the family $f(x)=e^{-x^{p}}$ has $\kappa(p)=\Gamma(1+2/p)$ in closed form, and inverting it gives $p_{\mathrm{eff}}\to 2$ with $(2-p_{\mathrm{eff}})D\to 2/(1-\gamma_{\mathrm{E}})=4.73054$, so the substrate’s envelope is the Gaussian that eq. (134) already uses. That choice was previously defended only by the observation that $f(1)=\mathrm{e}^{-1}$ for every $p$, which carries no shape information, because the substrate Dirac spectrum is degenerate and $f(1)$ is the only value evaluated; the mode sum supplies what that argument could not. The verdict: $\kappa=1.081$ corresponds to $p=1.7178$ and to $\kappa_{\mathrm{sub}}$ at $D\approx 16$, against a substrate cardinality $V/a_{n}^{4}$, so Computation 173’s shape defence is withdrawn and the $b=0.2313$ against $0.25$ gap stands at $9.5\sigma$. A tamper test confirms the functional is sensitive to the density of states: a uniform one over the same range gives $\kappa=H_{D}/2$, unbounded. Scope: the $9.5\sigma$ is a statement about the canonical-Standard-Model-coupling reading only; the reading in which $\lambda_{\mathrm{PST}}=1/4$ is canonical gives $M_{*}=1285$ GeV and Computation 123’s $1.0\sigma$ match and is untouched. What is excluded is holding both. Net: the open register loses an escape hatch and gains a derived regulator shape. Forward reference: the SM-side value quoted here is the anchor of its day; Computation 202 restates it on current inputs at three loops, $0.0930\pm 0.0008$ on the 2026 Review, $1.4\sigma$ from $\mathrm{e}^{-1}/4$, which is what the book states from v32.153 (v32.137 printed $1.6\sigma$ on the 2024 Review).

 

Computation 175 ·  §14.2 ·  [PROOF]
Colour and generation cannot be independent labels on $\hat{\tau}^{\perp}$, and the Dixon ambient does not change that.

The colour-versus-generation caveat was closed at v25.8 and reopened at v32.64 when Computation 106’s block-diagonality claim was withdrawn. The closure, and §13.5 until this computation, held that PST resolves the colour-versus-generation caveat “on the Dixon side” because the ambient $\mathbb{C}\otimes\mathbb{H}\otimes\mathbb{O}$ makes the three generation sectors orthogonal (Computation 48). That inference is invalid, and this file shows it in three measured steps. (1) Orthogonality is not invariance. The three Fano lines through $\hat{\tau}$ are mutually orthogonal to machine zero, are genuine projectors, and their ranks sum to $\dim\hat{\tau}^{\perp}=6$, so Computation 48’s decomposition is real; and colour still leaks out of them at $\mathrm{O}(1)$. Both hold at once, because orthogonality of subspaces says nothing about whether a group preserves them, and the v25.8 closure inferred the second from the first. (2) The obstruction is a theorem, not a dimension count. The commutant of $\mathrm{Stab}_{G_{2}}(\hat{\tau})$ on $\hat{\tau}^{\perp}$ is computed to be exactly $2$-dimensional over $\mathbb{R}$, that is $\mathbb{C}$, so by Schur $\hat{\tau}^{\perp}$ is the irreducible fundamental $\mathbf{3}$ and has no proper invariant subspace whatever. A colour-invariant generation splitting inside $\hat{\tau}^{\perp}$ is therefore impossible rather than merely unproven: the register’s count rules out a tensor factorisation ($\mathbf{3}\otimes\mathbf{3}$ needs nine complex dimensions against three), and irreducibility additionally rules out the direct-sum splitting the Fano lines actually realise. (3) The ambient changes nothing. Colour acts on $\mathbb{C}\otimes\mathbb{H}\otimes\mathbb{O}$ only through the $\mathbb{O}$ factor, so $\lVert(1-I\otimes P)(I\otimes g)(I\otimes P)\rVert=\lVert(1-P)gP\rVert$ and the leakage on the $64$-real-dimensional ambient equals that on $\mathbb{O}$ alone, measured to $0$ difference. The enlargement supplies room and no invariance. What survives: the count three is untouched, $\hat{\tau}^{\perp}$ being $3$-dimensional over $\mathbb{C}$, and that one fact is the source of both $N_{c}=3$ and $N_{\mathrm{gen}}=3$. PST derives one three, not two, and inherits the octonionic programme’s question rather than closing it. Tamper-tested: a generic element of the maximal torus, which does preserve the lines, leaks $1.3\times 10^{-16}$ against the $\mathrm{O}(1)$ of generic colour, so the measurement discriminates.

 

Computation 176 ·  §12.2 ·  [PROOF]
The field-normalisation item is the value of one number, and one of its two readings is excluded.

§12.2 carried the book’s sharpest open item in a shape with no obvious line of attack: (128) presupposes a canonically normalised scalar, (132) makes $Z^{2}$ a field-strength factor, at most one of the two fields can be canonical, and the book did not say which. The observation: canonical normalisation is not a matter of interpretation but a statement about a kinetic coefficient, and P3 carries one explicitly, the coefficient $c$ of the Boolean Dirichlet form, constrained only by $c>0$. It is not absorbable, the form being a bare edge sum with a fixed normalisation, and rescaling $\phi=\sqrt{c}\,\psi$ moves it into the quartic rather than removing it: $\lambda_{\mathrm{canon}}=1/(4c^{2})$, hence $C_{H}=3/(2c^{2})$ and $M_{*}=4\pi m_{h}c\sqrt{2/3}$, linear in $c$. Everything follows from that line. The book’s $C_{H}=3/2$ and $M_{*}=1285$ GeV are the $c=1$ values, the two readings are exactly $c=1$ and $c^{2}=\mathrm{e}$, and their $M_{*}$ ratio is $\sqrt{\mathrm{e}}=1/Z$, so the “$1/Z$ identically” the section asserts is derived rather than observed. The bridge premise is then one equation in one unknown, $\lambda_{\mathrm{SM}}(4\pi m_{h}c\sqrt{2/3})=\mathrm{e}^{-1}/(4c^{2})$, and it is not circular: the $\mathrm{e}^{-1}$ is (134)’s $f(1)$, exact and $D$-independent with the Gaussian $f$ derived in Computation 174; the $1/4$ is P3’s; $\lambda_{\mathrm{SM}}$ runs from the measured inputs. The residual is monotone, so the root is unique, and propagating the budget by re-solving for $M_{*}$ as Computation 173 requires gives $c=0.9978\pm 0.0054$, $0.4\sigma$ from the canonical value and $121\sigma$ from $\sqrt{\mathrm{e}}$. So the second reading is withdrawn, excluded here at $121\sigma$, by Computation 173 at $9.5\sigma$ as a conflict with P3’s $b=1/4$, and by Computation 174 closing its only defence. What this does not do: it does not derive $c=1$ from P1–P3, which is carried as a stated assumption instead, and the $0.4\sigma$ must not be read as discharging it. That agreement is Computation 123’s $\lambda_{\mathrm{SM}}(M_{*})=\mathrm{e}^{-1}/4$ in another variable, one coincidence and not two, and the two readings of it carry different error budgets rather than disagreeing: Computation 123 states it over $\lambda_{\mathrm{SM}}$, where $b\,\mathrm{e}^{-1}=0.0920$ against the two-loop-run $0.0927$ is a rigid $1.0\sigma$, while re-solving for $M_{*}$ states it over $c$, where it is $0.4\sigma$; and spending it to fix $c$ would turn a parameter-free prediction into a one-parameter fit. It is a consistency check, and a real one, the data selecting the natural value out of a continuum. A derivation would come from §7.8, where $c$ multiplies the form taken to the Laplace–Beltrami form on $M$: either the limit fixes $c$, or $c$ is absorbed into the $a_{n}\to 0$ scaling and is not independent. Corrected on a referee pass of this file: the first version held $M_{*}$ fixed while varying $c$ and quoted $0.9961\pm 0.0038$ and $174\sigma$, which is the shortcut Computation 173 exists to forbid. Tamper-tested: $M_{*}/c$ is constant across a factor six in $c$ while a control with $C_{H}=3/(2c)$ breaks that invariance. Forward reference: the SM-side value quoted here is the anchor of its day; Computation 202 restates it on current inputs at three loops, $0.0930\pm 0.0008$ on the 2026 Review, $1.4\sigma$ from $\mathrm{e}^{-1}/4$, which is what the book states from v32.153 (v32.137 printed $1.6\sigma$ on the 2024 Review).

 

Computation 177 ·  §7.8 ·  [PROOF]
Does the Mosco limit fix P3’s kinetic coefficient? No, and $c$ is not a constant.

Computation 176 reduced the field-normalisation item to P3’s kinetic coefficient $c$ and named the route that would close it: $c$ multiplies the Boolean Dirichlet form, §7.8 carries that form to the Laplace–Beltrami form, and the target has a normalisation of its own. The route returns a negative answer, with corrections worth more than the negative costs. What the limit fixes: (21) defines the Boolean gradient as a bare sum over $D$, and (57) divides by $|D|$, so the emergent kinetic coefficient is $K=c\,|D|\,d_{0}^{2}/16$, with no free constant anywhere in the map. What it does not fix: canonical form in four dimensions is reached by rescaling the field, which can be done for any $K$, so the target imposes no condition on the value. One free number in, one out. The consequence for $c$: A1 sends $|D|\to V_{M}/a_{n}^{4}\to\infty$ with $d_{0}$ fixed, so a constant $c$ makes $K$ diverge as $a_{n}^{-4}$ and the projected field infinitely stiff. $c$ therefore carries a $1/|D|$, equivalently $c\sim a_{n}^{4}\to 0$, and the $|D|$-independent quantity P3 owns is $\tilde{c}=c\,|D|$, giving $K=\tilde{c}\,d_{0}^{2}/16$. So $c=1$ is not an underived statement but a malformed one, and $c$ carrying $\varepsilon$’s dimensions makes it doubly so. Second correction: (57) was missing the $1/V_{M}$ that A2 carries, an expectation against a probability measure having been equated with an unnormalised integral; measured on an explicit four-dimensional lattice and recovered to $0.3\%$ across a factor $81$ in volume. Third correction: Computation 176’s $\lambda_{\mathrm{canon}}=1/(4c^{2})$ is a substrate-variable identity that skips the projection, the four-dimensional coupling being $bV/K^{2}$. Its conclusions survive intact, because the $\sqrt{\mathrm{e}}=1/Z$ ratio and the exclusion of reading B use only the ratio of two couplings and never the formula. Corrected on a referee pass, and the correction is to this file: v32.74 read the postulate as leaving the gradient’s normalisation unstated and added a per-site definition. (21) had already defined it, as the bare sum, and (22)’s Efron–Stein equality holds only for that form, so the addition both duplicated a definition and falsified a cited inequality. It is withdrawn; the bare reading is the book’s, and $c$ scales. What would close the item: a value for $K$, needing $d_{0}$, which Chapter 15 bounds at $\lesssim 2.3$ nm without predicting. Tamper-tested: the divergence is exhibited as a property of the bare sum and absent once the $1/|D|$ is carried, so the test discriminates rather than assuming.

 

Computation 178 ·  §3.5 ·  [PROOF]
The non-balance counterexample is forced rather than chosen, and it does not scale.

Computation 169 withdrew the step that would have closed non-balance, because Computation 152’s $\mu_{\pm}$ is rotation-invariant, names no direction, pays no parameter and still balances $31\%$ of the time at $|D|=6$. This computation does not close the item either, and says so four times; what it does is measure how far that obstruction reaches. The reduction: on the all-parallel family $v(a)=\epsilon_{a}u$ the shared direction drops out, $\sum_{a}v(a)=(\sum_{a}\epsilon_{a})u$, so balance is a statement about the sign vector alone. The step that has not been used before: §3.5 already asserts $\mu_{\pm}$ invariant under per-property complementation $v(a)\mapsto-v(a)$, which on this family is $\epsilon_{a}\mapsto-\epsilon_{a}$, and the single-coordinate flips act simply transitively on $\{\pm 1\}^{|D|}$. A law invariant under a transitive action is uniform, so the signs are forced independent and fair: $\mu_{\pm}$’s sign structure is the only complementation-invariant option rather than a modelling choice. Hence the probability is exact: $\binom{|D|}{|D|/2}2^{-|D|}$, which is $5/16$ at $|D|=6$, reproducing Computation 152’s measured $31\%$ as a forced value; it behaves as $\sqrt{2/\pi|D|}$, giving some $10^{-40}$ at a cardinality of $10^{80}$, and is exactly zero at every odd $|D|$, a sum of an odd number of $\pm 1$ being odd. And a bound with no family assumed: for any complementation-invariant law on $(S^{6})^{D}$, flipping one property sends $\sum_{a}v(a)$ to $-2v(a)\neq 0$, pairing each balanced configuration with an unbalanced one of equal measure, so $\Pr[\text{balance}]\leq 1/2$, attained at $|D|=2$. Scope, stated because the file is useless if this is misread: none of it derives non-balance. A probability bound speaks about a law and not about the single configuration the substrate has, whose standing, as §3.5 makes explicit, is stability rather than probability; the vanishing is established for the family the counterexample inhabits; and $1/2$ forbids certainty, not balance. The register entry stands unchanged. Tamper-tested: dropping complementation invariance while keeping rotation invariance, relabelling invariance and zero parameter cost, by taking the signs uniform on the zero-sum set, restores balance with probability one, so the bound is carried by that one symmetry and by nothing else, which is the assumption to challenge if anyone reopens this.

 

Computation 179 ·  §7.5 ·  [PROOF]
M7 restated: the realisation-point density is supplied, and it diverges.

Computation 167 ruled that the theory supplies no realisation-point density, so $\Pi_{\mu\nu}$’s magnitude within Computation 164’s cone is free. That search was over lengths, $d_{0}$ and $\ell_{*}$, and neither is a spacing; A1 names one outright. §18.1 introduces $a_{n}$ as “the substrate site spacing, distinct from the fixed coherence length $d_{0}$”, with $|D_{n}|a_{n}^{4}\to V_{M}$, $a_{n}\to 0$ and local node density $1/a_{n}^{4}\to\infty$, and (40) sums over pairs of the same substrate. Corrected by Computation 182: this row read “through the same realisation map, $\rho:D\to M$ being defined once”, and the book defines two maps of incompatible type under that one name. No number in this row moves. So the density is not missing; it is infinite. With the pair cutoff the book itself uses, $|s|\lesssim d_{0}^{2}$ at the fixed coherence length (Computation 159), a cutoff that does not scale meeting a density that does gives $\Pi_{\mu\nu}\sim|D|^{2}\sim a_{n}^{-8}$; measured here, reproducing Computation 167’s $N^{2}$ and carrying it to the exponent in $a_{n}$ that Computation 167 did not take. Which leaves a fork the book has not chosen: if $D$ is A1’s sequence the magnitude is divergent rather than free and a normalisation is owed, measured to be $1/|D|^{2}$, per pair rather than per site, and therefore fixed; if $D$ is one physical substrate then Computation 167 stands entire and a density is owed. Computation 167 took the second without naming it, and its arithmetic is reproduced here exactly and not disputed, its scale-free checks included. Computation 182 decides the fork in favour of the first, the second being unavailable rather than merely unstated, and corrects what the first buys: the per-pair normalisation converges, but the bare sum is dimensionless, so restoring the dimensions of an energy density returns one power of the node density and the object still diverges. This does not close M7: what is left is one truncation scale $a_{*}$, which Computation 167 showed is neither $d_{0}$ nor $\ell_{*}$. Computation 167’s negative result survives, no absolute scale being computed either way and so no vacuum-energy catastrophe. The pattern is worth more than the item: Computation 177 found the same unstated choice in P3, where a bare sum over $D$ met a limit dividing by $|D|$, and (40) is a bare sum over pairs carrying no normalisation at all. The book uses $D$ both as A1’s sequence and as a fixed set, and every bare sum over $D$ is sensitive to which; Computation 182 settles which is meant here. Tamper-tested: tying the cutoff to the spacing instead moves the measured exponent in $|D|$ from $1.98$ to $1.02$, so the divergence is carried by the cutoff’s relation to the spacing and not by the pair sum.

 

Computation 180 ·  §10.3.2.5 ·  [NUMERICAL]
Born agreement: the atom obstruction survives on admissible graphs.

Computation 145 measured the Born condition’s real obstruction as atoms, ties in $T_{\mathrm{ex}}$ that no monotone map can separate and that floor the distance from uniform at $m/2$, and showed the floor vanishes for generic distinction graphs. It then set the remaining graphs aside on the ground that the book already excludes them, $T_{\mathrm{ex}}$ failing complement-asymmetry on regular $(D,\delta)$. That ground does not hold, and this computation is mostly about why. (16) is a condition on $v$: the body states its genuine content is the non-balance $\bigl{|}\sum_{a}v(a)\bigr{|}>0$, so it constrains no distinction graph at all; and on the magnitude reading the complete graph is regular and complement-asymmetric at once, measured here. It also names the wrong class. The regular graphs’ floor vanishes, as $|D|^{-0.5}$, slower than generic but vanishing. What fails is a bounded active part: a star has one centre, $K_{k}$ with isolated vertices a fixed clique, and in both $T_{\mathrm{ex}}$ takes $\mathrm{O}(1)$ distinct values however large $|D|$ grows. And they are admissible: (14) and (15) are checked exhaustively on every family and hold, while (16) cannot reach them. The star keeps an atom of exactly $1/2$ at every $|D|$, half of all configurations omitting the centre and tying at zero, so the distance from uniform is floored at $1/4$ for arbitrarily large substrates. The consequence: Born agreement is jointly a condition on the law of $T$ and on $\delta$, and the theory constrains $\delta$ not at all in the relevant respect, so discharging it needs a genericity constraint on the distinction graph rather than a calculation. Named since (Computation 183): the constraint is the vertex cover number of $(D,\delta)$, whose being bounded is what “bounded active part” measures, and requiring only that $\delta$ tell the properties of $D$ apart removes the mechanism. Scope: this is not a refutation of the crossing rule and nothing standing depends on it; Computation 145’s generic result is reproduced here, not disputed, and a physical substrate is overwhelmingly more likely to be generic than to be a star. What changes is only what the register may claim. Tamper-tested: four hundred random monotone reparametrisations fail to beat the $m/2$ floor on the star, while the same search on an Erdős–Rényi graph goes far below it, so the bound is structural and the test separates the two families.

 

Computation 181 ·  §11.3 ·  [PROOF]
W4’s named route cannot work, and the reason is one line.

§11.3 withdraws the earlier argument for microcausality as vacuous, it having used $\{c_{a},c_{b}\}=0$, which holds at every separation including $a=b$, so the spacelike hypothesis did no work. It then names what would establish W4: the substrate two-point structure having Fourier support in the closed causal cone. That route is unavailable rather than unattempted. The substrate is pre-causal and carries no dynamics, P1 supplying a Bernoulli product measure in which nothing evolves, so the two-point structure is an equal-time object and its Fourier transform does not depend on $k^{0}$; measured here to machine zero along every $k^{0}$ line. The obstruction is then exact: let $F$ be $k^{0}$-independent and supported in $C=\{|k^{0}|\geq|\mathbf{k}|\}$, and fix $\mathbf{k}\neq 0$; the points with $|k^{0}|<|\mathbf{k}|$ lie outside $C$, so $F$ vanishes there, and being $k^{0}$-independent it vanishes on the whole line, leaving $F$ supported at $\mathbf{k}=0$ alone. A $k^{0}$-independent two-point function has cone support only if it is trivial, so a static object cannot encode a light cone. The numerics confirm the proof including its edge case: at the physical end, where $a_{n}\to 0$ broadens the kernel in $\mathbf{k}$, some $82\%$ of the spectral weight sits outside the cone, and the fraction falls to zero only as the kernel is squeezed onto $\mathbf{k}=0$, which is exactly the trivial support the proof isolates. What is not claimed: W4 is not refuted and nothing downstream moves. Equal-time microcausality is untouched, $\{\psi,\psi^{\dagger}\}=\delta^{(3)}(\mathbf{x}-\mathbf{y})$ vanishing for $\mathbf{x}\neq\mathbf{y}$; the structural spin-statistics theorem never used microcausality; and W1, W2, W3 and W5 stand. Where W4 has to come from is the emergent dynamics, $iS(x-y)$ being built from Closure (3)’s projected Dirac operator with the Pauli–Jordan support coming from the mass shell, so the register’s target moves to the matched-scaling limit. Tamper-tested: a mass-shell kernel puts nothing outside the cone while the substrate kernel puts a third of its weight there, and the control passes only because it has a $k^{0}$-dependence, which is precisely what the substrate lacks.

 

Computation 182 ·  §7.5 ·  [PROOF]
M7’s fork is decided, and the missing number is named.

Computation 179 left the absolute scale of $\Pi_{\mu\nu}$ as a fork between two readings of $D$ and said this book had not chosen. It is not a choice. $D$ is finite at every stage, by A1’s own $|D_{n}|\to\infty$, by the Bernoulli product over $D$, by the exact enumeration over $2^{|D|}$ configurations and by the balance probability $\binom{|D|}{|D|/2}2^{-|D|}$; a finite set has no uncountable quotient, so §4.10’s $M=D/{\sim_{\rho}}$ is finite at every stage and is a smooth four-manifold at none, while (46) differentiates $\Pi_{\mu\nu}$ covariantly on a smooth $M$. So $D$ in (40) is A1’s sequence and the fixed-substrate branch is unavailable rather than merely unstated; Computation 167’s arithmetic is untouched and its reading of the symbol falls. And the forced normalisation does not buy a finite density, which corrects Computation 179. Dividing by the pair count is forced and converges, but (40) is dimensionless, $W(s)$ and the unnormalised dyad carrying $L^{2}$ each against the bond kernel’s $L^{-4}$, so the bare sum is a pure count and goes as $|D|^{2}$; measured here at exponent $2.021$ globally and $2.078$ with the bond kernel localised, so the locality costs a constant and not an exponent. Restoring the dimensions of an energy density then returns exactly one factor $a_{n}^{-4}$, the node density itself, and the restored object diverges as one power of it, fitted at $1.011$ against controls at $0.049$ and $2.049$. So one number is left: the physical spacing $a_{*}$ at which A1’s limit is truncated, with $\Pi_{\mu\nu}\sim 1/a_{*}^{4}$, which is what a lattice spacing is. Computation 167’s finding that neither $d_{0}$ nor $\ell_{*}$ is such a spacing stands, so M7 does not close; the gap is one truncation scale of a familiar kind rather than a density the theory was said not to supply, and the vacuum-energy conclusion is unaffected because $a_{*}$ is named and not computed. Two things are checked rather than assumed. §10.5’s bipartition rests on $\rho$ being a surjective quotient, and needs none of it: a preimage and its complement partition the domain of any map, verified over $200$ random non-surjective maps. And this file corrects Computation 179’s premise sentence that the book “defines $\rho:D\to M$ once”: it defines two maps of incompatible type under one name, a surjective quotient onto a derived target and a non-surjective sampling of a given one. No number in Computation 179 moves.

 

Computation 183 ·  §10.3.2.5 ·  [PROOF]
The Born obstruction is the vertex cover number, and one line removes it.

Computation 180 left Born agreement waiting on “a genericity constraint on $\delta$ that the theory does not supply”, naming the failing property informally as a bounded active part. It is a graph invariant, and the proof is one line. Let $S$ meet every distinction; the remainder $D\setminus S$ carries none, so every $C$ inside it ties at $T_{\mathrm{ex}}=0$, and under $\mu=\mathrm{Bern}(\tfrac{1}{2})^{|D|}$ that is an atom of at least $2^{-c}$, with $c$ the vertex cover number of $(D,\delta)$. That reproduces Computation 180’s table exactly: the star and $K_{4}$, $K_{6}$ with isolated vertices have $c=1,3,5$ fixed while $|D|$ grows, and the complete, cycle and Erdős–Rényi graphs have $c$ growing with it; nothing else in the table does any work, so “bounded active part” was the right diagnosis and this is what it measures. And one line removes the mechanism. Call $(D,\delta)$ reduced when $\delta$ alone tells the properties of $D$ apart, no empty distinction set and no two the same; then $u\mapsto\{b:\delta(u,b)\}$ is injective from $D\setminus S$ into the subsets of $S$, so $|D|-c\leq 2^{c}$ and $c\geq\log_{2}(|D|-c)$, putting the guaranteed floor below $10^{-80}$ at $|D|\sim 10^{80}$. The bound is tight rather than slack: the extremal reduced graphs, a cover with one independent property per nonempty subset of it, sit within a factor $1.6$ of $2^{-c}$ at every size and their atom falls as $|D|^{-1.09}$. This does not close the item, and the reason is stated rather than skirted: §2.2 says no further axioms are imposed on $\delta$, and §4.10’s appeal to the identity of indiscernibles is made for the fibres of $\rho$ and cannot be stretched here, the substrate being the triple $(D,\delta,v)$ so that $v$ discerns what $\delta$ does not. What changes is the residue’s shape: failure by this mechanism at mass $m$ needs at least $m(|D|-\log_{2}(1/m))$ properties sharing one distinction set, which at substrate scale is nearly all of them, so the register’s question becomes whether P1’s primitive properties admit $\delta$-duplicates, a decision about the postulate’s wording rather than a calculation. Settled since (Computation 184, P1(iii)): they do not, and not by stipulation but because faithfulness (6) makes any such pair indiscernible in the whole triple. This row’s sentence that §2.2 imposes no further axiom on $\delta$ records the state of the book when this computation was written and is superseded by that clause. A second mechanism is searched for and not found, over Erdős–Rényi graphs at three densities and by hill-climbing the atom over reduced graphs directly, which is measured over the graphs searched and not proved over all of them. Tamper-tested twice. Restoring the false twins to an extremal reduced graph holds its atom on $2^{-4}$ while $|D|$ goes from $19$ to $59$, so the cover and not the size carries the atom. And the hill-climb walks to a perfect matching, which is reduced and whose law is exactly $\mathrm{Binomial}(|D|/2,\tfrac{1}{4})$: its atom is the binomial mode mass, matching the enumeration to twelve digits and falling as $1.303|D|^{-1/2}$, the slowest-vanishing reduced family reachable and still vanishing.

 

Computation 184 ·  §2.2 ·  [PROOF]
Reducedness is a theorem of one coherence condition, and the quotient survives it.

Computation 183 left the Born condition on reducedness and declined to assume it, §2.2 imposing no further axiom on $\delta$ and $v$ being able to discern what $\delta$ cannot. It need not be assumed. It follows from one coherence condition, $\delta$-faithfulness: if $v(a)\neq v(b)$ then $\delta(a,b)$, that is, $\delta$ records every difference the substrate carries. The chain is three steps, each proved and each enumerated. Irreflexivity (5) makes every twin pair non-adjacent, $b\in N(a)$ while $b\notin N(b)$, checked over every graph to $|D|=6$ with not one adjacent twin found; faithfulness contraposed then forces $v(a)=v(b)$ on such a pair; and the transposition $(a\,b)$ is then an automorphism of the whole triple $(D,\delta,v)$ rather than of $\delta$ alone, so §4.10’s own principle applies one level below where the book already applies it. Measured over $4000$ triples: every equal-$v$ twin pair is a triple-automorphism and no differing-$v$ pair is, so the principle reaches exactly the pairs faithfulness leaves. The dependency that had to be checked is $\rho$, a spacetime point being a fibre and three live passages needing $\rho$ non-injective. It survives exactly: (30) makes a within-fibre pair satisfy $T(\{a,b\})=\tau_{\mathrm{null}}$, and $\tau_{\mathrm{null}}\in\mathbb{R}^{+}$ while (14) vanishes only on pairs carrying no distinction, so a within-fibre pair is distinct and a fibre is a clique, which by step one holds no twins. Over every graph on five and six properties, substrates with no shared distinction set carry clique modules of every size up to $|D|$ and independent modules of size at most one. What is removed is the opposite object, a pair with $|T|=0\neq\tau_{\mathrm{null}}$ sitting at two different spacetime points with no fact to tell them apart. One clause is not delivered, and running the file is how that was found: Computation 183 defined reducedness with a second clause, no empty distinction set, and since two isolated properties are twins, at most one isolated property survives reduction while nothing removes the last. It costs nothing, Computation 183’s proof resting on injectivity of $u\mapsto N(u)$, for which the empty set is a legal value one property may take and no two may share; the bound $|D|-c\leq 2^{c}$ is verified here directly on every collapsed substrate instead. And what the premise does to $\delta$ is stated rather than left for a reader to find. Contraposed, (6) says two properties carrying no distinction share their directional content, so $\delta$ contains the complete multipartite graph on the partition of $D$ by $v$-value and the only freedom is inside a $v$-block; where $v$ is injective $\delta$ is complete, which is admissible and whose atom Computation 180 measured vanishing as $|D|^{-0.48}$. Computation 180’s own counterexamples cease to exist: a star’s leaves are pairwise non-distinct, so (6) forces one $v$ on all of them while they already share one distinction set, collapsing the star to two properties at $|D|=5,10,20,100$ alike, and $K_{k}$ with isolated vertices to $k+1$. What this does not do is make the condition free: a two-property substrate with $\delta$ empty and $v(a)\neq v(b)$ satisfies (4), (5) and everything else stated, and violates faithfulness, so P1(iii) is a premise and is carried as one in §19.9.1, with the order of discovery beside it: the clause is motivated by what makes $\delta$ the distinction relation of P1 at all, and its consequence for Born agreement was found in this file afterwards. Nor does it close the Born item, the remaining half being a proof that every admissible $\delta$ satisfies the condition asymptotically, which Computation 183 measures and does not establish. Tamper-tested twice: without faithfulness, twin pairs whose $v$ differs occur in quantity and Leibniz cannot touch them; with faithfulness but without Leibniz, faithful triples still carry twin pairs. Neither ingredient is redundant.