Golden Phase Audit

Abstract

Exact phase discipline demotes the false constant-arm point and restores the champion.

Theorem 1.1 (Golden phase audit).

$$\left(\left(\forall k\in N,\mathrm{goldenSurvivor}\left(2\cdot \left(k+1\right),\frac{1}{phi+2}\right)=\frac{1}{phi\cdot \mathrm{sqrt}\left(5\right)}\right)\wedge \left(\left(\forall k\in N,\mathrm{goldenSurvivor}\left(4\cdot k+1,\frac{1}{phi+2}\right)=\frac{1}{\mathrm{sqrt}\left(5\right)}\wedge \mathrm{goldenSurvivor}\left(4\cdot k+3,\frac{1}{phi+2}\right)=\frac{1}{{phi}^2\cdot \mathrm{sqrt}\left(5\right)}\right)\wedge \frac{1}{phi+2}=\frac{1}{phi\cdot \mathrm{sqrt}\left(5\right)}\right)\right)\wedge \left(\mathrm{memberOf}\left(\frac{13}{2}-4\cdot phi,\mathrm{goldenSurvivorMaximizers}\left(6\right)\right)\wedge \left(\mathrm{goldenSurvivor}\left(5,\frac{13}{2}-4\cdot phi\right)=\frac{{phi}^{0-1}}{2}\wedge \left(\mathrm{goldenSurvivor}\left(6,\frac{13}{2}-4\cdot phi\right)=\frac{1}{2}\wedge \mathrm{goldenSurvivor}\left(7,\frac{13}{2}-4\cdot phi\right)=\frac{{phi}^{0-2}}{2}\right)\right)\right)$$

Proof. Machine-checked in Lean as D5/S0/Tower/Champions/PhaseAudit.golden_phase_audit (✓ std3). ∎

Source. Repository-derived.

Commentary.

The point 1/(phi+2) has one exact arm on every positive even level. Its two odd residue classes have distinct exact arms, so the former constant-arm claim fails with period four. The identity phi*sqrt(5)=phi+2 records the even value exactly.

The frozen closed-form point 13/2-4*phi belongs to the level-six golden-survivor maximizer family. Its consecutive level-five, level-six, and level-seven arms are phi^(-1)/2, 1/2, and phi^(-2)/2, the exact form of the reported three-phase ring.

References

• Truth anchor: D5/S0/Tower/Champions/PhaseAudit.golden_phase_audit
• Dependency: D5/S0/Tower/MetricGeometry/GoldenSurvivorSet