Follow the source behind the result.

The registered Python source is shown as text. Displaying it does not execute the code.

"""v487 -- HYP.REWRITE.03: the lazy split (1/2, 1/6) is FORCED by the
resummed clock -- the v486 [O] 'split selection' closes onto v124
structure.  The walk's one-step spectrum is the COMPLETE clock ladder
below the wall; deck parity = rung index; the two search tones are one
bend apart.  Sharpens HYP.REWRITE.02; no gate closes.

Background.  v486 exhibited the unique lazy Z2-pair walk (stay, hop,
leak) = (1/2, 1/6, 1/3) generating the full transfer spectrum
{1, (2/3)^6, (1/3)^6}, but left the SPLIT SELECTION open: '(1/2, 1/6)
vs uniform (1/3, 1/3) is selected by the physical spectrum, not yet
derived from a first principle' (named candidate: sheet-symmetric
choice -- vague).  The v124 resummed clock reads the transfer
eigenvalues as lambda_n = (1 - n/N_fam)^{p_2} with a POLE (wall) at
n = N_fam.  This module proves the split selection is clock-ladder
faithfulness -- definitional for a full-spectrum generator:

  [E] 1. RUNG READING: the one-step spectrum of the lazy rule is
        EXACTLY the complete resummed-clock ladder below the wall,
        eig(M) = {1 - n/3 : n = 0, 1, 2} = {1, 2/3, 1/3}; v327's
        uniform rule instead puts its second pair mode ON the wall
        (0 = 1 - 3/3) -- the v486 degeneracy re-read as 'odd channel
        collapsed to the wall'.
  [E] 2. DECK PARITY = RUNG INDEX: the sigma-even pair mode (1,1)
        carries rung n = 1 (survival 2/3), the sigma-odd mode (1,-1)
        carries rung n = 2 (survival 1/3); each clock step n -> n+1
        flips the deck parity.  This is the structural home of the
        v486 parity assignment (omega_1 even-channel, omega_2
        odd-channel) tested by the frb-parity-comb search.
  [E] 3. FORCING: demanding (i) the faithful ladder (no rung missing,
        no physical mode on the wall -- definitional for a
        full-spectrum generator), (ii) Z2 equivariance (symmetric pair
        block), (iii) non-negative rates, BOTH the split (s, h) =
        (1/2, 1/6) AND the assignment even -> rung 1, odd -> rung 2
        are UNIQUE (the swapped assignment needs h = -1/6 < 0).  The
        v486 'sheet-symmetric choice' candidate is superseded by this
        proof; the walk-level [O] closes.
  [E] 4. BEND COROLLARY: the two comb tones of the empirical search
        program are one established number apart,
            omega_1 / omega_2 = rate(2)/rate(1) = log_{3/2} 3
        (the v107/v124 bend, 2.70951...) -- the tone ratio of the
        parity-comb search IS the bend of the pulsar-clock search.
  [O] 5. HONEST RESIDUAL: the channel count 3 = N_fam is forced by the
        clock wall (v124 pt 4); the ladder itself is an [I] identity
        on frozen spectra whose semiclassical derivation stays open
        (R1/[P]); the arity {|Z2|, N_fam} = {2, 3} stays the anchor
        input.  NET: no independent choice remains at the walk level.

Contract provenance: experiments/theory-contracts/
lazy01_clock_rung_forcing.py (2026-07-15).  Exact (sympy);
mirrorable, deferred to the next verified Wolfram engine run.
"""
import sympy as sp

from tfpt_constants import check, summary, reset, N_fam, g_car

Z2 = g_car - N_fam                    # |Z2| = 2
P2 = 6                                # |R+(A3)| hexagon size (v124)


def rung(n):
    return sp.Rational(N_fam - n, N_fam)


def rate(n):
    return -P2 * sp.log(rung(n))


def run():
    reset()
    print("v487  HYP.REWRITE.03: the lazy split is forced by the resummed "
          "clock (ladder faithfulness); deck parity = rung index")

    s0, h0 = sp.Rational(1, 2), sp.Rational(1, 6)
    M = sp.Matrix([[1, 0, 0], [0, s0, h0], [0, h0, s0]])
    Mu = sp.Matrix([[1, 0, 0],
                    [0, sp.Rational(1, 3), sp.Rational(1, 3)],
                    [0, sp.Rational(1, 3), sp.Rational(1, 3)]])

    # 1. rung reading
    eigs = set(M.eigenvals().keys())
    ladder = {rung(n) for n in range(N_fam)}
    wall = rung(N_fam)
    eigs_u = set(Mu.eigenvals().keys())
    check("RUNG READING [E]: the lazy rule's one-step spectrum is EXACTLY "
          "the complete v124 clock ladder below the wall, eig(M) = "
          "{1 - n/3 : n = 0, 1, 2} = {1, 2/3, 1/3}; v327's uniform rule "
          "puts its second pair mode ON the wall (0 = 1 - 3/3) -- the v486 "
          "degeneracy re-read as 'odd channel collapsed to the wall'",
          eigs == ladder and wall not in eigs
          and wall in eigs_u and rung(2) not in eigs_u)

    # 2. deck parity = rung index
    sigma = sp.Matrix([[0, 1], [1, 0]])
    B = sp.Matrix([[s0, h0], [h0, s0]])
    lam_even = (B * sp.Matrix([1, 1]))[0]
    lam_odd = (B * sp.Matrix([1, -1]))[0]
    check("DECK PARITY = RUNG INDEX [E]: [B, sigma] = 0; the sigma-even "
          "pair mode carries rung n = 1 (survival 2/3), the sigma-odd mode "
          "rung n = 2 (survival 1/3) -- each clock step flips the deck "
          "parity; the structural home of the v486 parity assignment "
          "(omega_1 even-channel, omega_2 odd-channel; frb-parity-comb)",
          sp.simplify(B * sigma - sigma * B) == sp.zeros(2, 2)
          and lam_even == rung(1) and lam_odd == rung(2))

    # 3. forcing (split + assignment unique)
    s, h = sp.symbols('s h', real=True)
    sol_A = sp.solve([sp.Eq(s + h, rung(1)), sp.Eq(s - h, rung(2))],
                     [s, h], dict=True)
    sol_B = sp.solve([sp.Eq(s + h, rung(2)), sp.Eq(s - h, rung(1))],
                     [s, h], dict=True)
    check("FORCING [E]: faithful ladder (no rung missing, no physical mode "
          "on the wall -- definitional for a full-spectrum generator) + Z2 "
          "equivariance + rates >= 0 force UNIQUELY both the split (s, h) "
          "= (1/2, 1/6) AND the assignment even -> rung 1, odd -> rung 2 "
          "(the swapped assignment needs h = -1/6 < 0, inadmissible); the "
          "v486 [O] 'split selection' closes at the walk level",
          len(sol_A) == 1 and sol_A[0][s] == sp.Rational(1, 2)
          and sol_A[0][h] == sp.Rational(1, 6) and sol_A[0][h] >= 0
          and sol_B[0][h] < 0)

    # 4. bend corollary
    omega_1 = 2 * sp.pi / rate(1)
    omega_2 = 2 * sp.pi / rate(2)
    bend = sp.log(3) / sp.log(sp.Rational(3, 2))
    check("BEND COROLLARY [E]: the two comb tones of the search program "
          "are one bend apart: omega_1/omega_2 = rate(2)/rate(1) = "
          "log_{3/2} 3 = 2.70951... (v107/v124) -- the parity-comb tone "
          "ratio IS the pulsar-clock bend",
          sp.simplify(omega_1 / omega_2 - bend) == 0)

    # 5. honest residual
    check("HONEST RESIDUAL [O]: the channel count 3 = N_fam is forced by "
          "the clock wall (v124 pt 4); the ladder is an [I] identity on "
          "frozen spectra with the semiclassical derivation open (R1/[P]); "
          "the arity {|Z2|, N_fam} = {2, 3} stays the anchor input; no "
          "independent choice remains at the walk level",
          {Z2, N_fam} == {2, 3} and sp.limit(-P2 * sp.log(1 - sp.Symbol('x')
          / N_fam), sp.Symbol('x'), N_fam, dir='-') == sp.oo)

    return summary("v487 HYP.REWRITE.03: the lazy split (1/2, 1/6) is "
                   "FORCED -- the walk's spectrum is the complete v124 "
                   "clock ladder below the wall (uniform rule = odd mode "
                   "on the wall); deck parity = rung index (home of the "
                   "parity assignment); omega_1/omega_2 = bend; the v486 "
                   "walk-level [O] closes onto v124 structure (arity + "
                   "clock derivation stay the open inputs)")


if __name__ == "__main__":
    raise SystemExit(1 if run() else 0)

0c234246906ab637aabd816271b9f733082a1f94021db67b8fa2d4f6c63e560c

The public registry includes 116 document views, 6,693 section records and 985 physical LaTeX sources. Active publication documents, research material and archives retain their recorded roles. This is source navigation, not a new semantic review or proof run.