Theorem: 5z leading expansion

Define

Then

and

There is no inverse-quartic leading term.

Proof. Since $Df=1$ and $D^{2}f=0$ on the retained patch,

Substitute Proposition prop:vertex-jets into (eq:5z-boundary). Its inverse-cubic terms are exactly (eq:C5z). The remaining boundary terms use $U_{\delta }=fD{M}_{\delta }$, which is $O({\delta }^{-1})$ on the null boundary, including the required $r$ derivatives. The coefficient ${{\mathcal{G}}_2}^r$ needs at most four $r$-weight derivatives; the other terms need fewer. They are controlled by the stated common seminorm.

For the interior term, Lemma lem:regularity needs at most six $r$-weight derivatives. On $0<d(S)\le 4\delta$, the leading master estimates give

The $S^0$-width of this layer is $O(\delta )$. Thus its contribution to (eq:5z-interior) is $O_H({\delta }^{-1})$. On $d(S)>4\delta$, the master and internal coefficient derivatives are uniformly bounded and all integration ranges are compact. This part is $O_H(1)$. Therefore the interior contribution is lower order than ${\delta }^{-3}$, with ample room in the claimed remainder.

For scale control, repeat the boundary and interior estimates with ${\rho }^{[1]}$. The leading $P_2,P_3$ coefficients are multiplied by $-3$, so the boundary leading coefficient is $-3C_{5z}$, as required.

Finally, the exact decomposition contains no boundary term more singular than $D^{3}M$, and its nominal inverse-quartic coefficient vanishes by ${\mathcal{F}}^{(4)}(0)=H_{\ast }^{\prime \prime \prime }(0)=0$. The interior term is lower order. This proves the signed inverse-quartic cancellation.