Direct coefficient verification of observable invariance

Using the coefficient laws (eq:X1law)–(eq:X2law) and (eq:scalarpull), write ${\varphi }_1^{\prime }={\varphi }_1+X_{\xi }{\varphi }_0$, ${\varphi }_2^{\prime }={\varphi }_2+X_{\xi }{\varphi }_1+X_{\zeta }{\varphi }_0+\frac{1}{2}{X_{\xi }}^2{\varphi }_0$, ${\mathsf{X}}_1^{\prime }={\mathsf{X}}_1+\xi$, ${\mathsf{X}}_2^{\prime }={\mathsf{X}}_2+\xi \cdot \partial {\mathsf{X}}_1+\zeta +\frac{1}{2}\xi \cdot \partial \xi$. Then

For the second coefficient,

The first two brackets vanish. With ${X_{\xi }}^2{\varphi }_0=\xi \cdot \partial \xi \cdot \partial {\varphi }_0+{\xi }^{\mu }{\xi }^{\nu }{\partial }_{\mu }{\partial }_{\nu }{\varphi }_0$ and ${\mathsf{X}}_1\cdot \partial (X_{\xi }{\varphi }_0)={{\mathsf{X}}_1}^{\nu }{\partial }_{\nu }{\xi }^{\mu }{\partial }_{\mu }{\varphi }_0+{{\mathsf{X}}_1}^{\nu }{\xi }^{\mu }{\partial }_{\nu }{\partial }_{\mu }{\varphi }_0$, the coefficients of $\xi \cdot \partial \xi \cdot \partial {\varphi }_0$, $\xi \xi \partial \partial {\varphi }_0$, ${\mathsf{X}}_1\cdot \partial \xi \cdot \partial {\varphi }_0$, and ${\mathsf{X}}_1\xi \partial \partial {\varphi }_0$ are respectively $\frac{1}{2}-1+\frac{1}{2}$, $\frac{1}{2}-1+\frac{1}{2}$, $-1+1$, $-1+1$, all zero. This verifies Theorem thm:main(iv) coefficientwise without suppressing the quadratic terms in the gauge parameter.