Every admitted preparation-clock position is harmless for a reason

10 Every admitted preparation-clock position is harmless for a reason

The old preparation clock $C$ is retained but no longer controls the active programme. It is an ordinary free particle with $M_C=10^8$. In its co-moving coordinate, the free propagator and one-dimensional Sobolev bound give, through time three,

The displacement error is below $9\times10^{-11}$, less than the $.05$ plateau clearance. A first-exit argument proves the tube for every $C_0\in[-1.25,-.75]$. No averaging over $|\chi_C|^2$ is used. Its correlations are irrelevant to active histories by Theorem 6.1.

The active writer clock is $S$. Its actual distribution is controlled by the original joint cap $f_0\le2$, not by a new Born-clock premise. A pointwise trajectory tube for every $S_0$ is neither claimed nor needed by the active-law current proof.