Exact RATTLE preserves both constraints and is symmetric

The exact first constraint equation gives $c(q_1)=0$, the exact final projection equation gives $J(q_1)M^{-1}{p}_1=0$, and [F3] keeps $q_1$ in $Q_{\mathrm{r}\mathrm{e}\mathrm{g}}$. By [F1], this is exactly $(q_1,p_1)\in \mathcal{M}$.

Starting from the forward equations, use the reverse step size $-h$ and choose its first multiplier to be the forward $\mu$. Its first half-momentum is then $p_1+\frac{h}{2}\mathrm{\nabla }V(q_1)+hJ(q_1{)}^T\mu =p_{1\mathrm{/}2}$, so its next position is $q_1-h{M}^{-1}{p}_{1\mathrm{/}2}=q$.

Choose the reverse step's final multiplier to be the forward $\lambda$. Its final momentum is then $p_{1\mathrm{/}2}+\frac{h}{2}\mathrm{\nabla }V(q)+hJ(q{)}^T\lambda =p$. The reverse constraint equations hold because the endpoints are the forward consistent states. As $h\to 0$, the swapped multipliers and the reverse input converge to the same regular-chart branch centers; after shrinking $h_0$, both multipliers lie in the reverse IFT neighborhoods. By uniqueness of the selected branches [F3], this reverse solve is the defined $-h$ step. Thus ${\mathrm{\Psi }}_{-h}{\mathrm{\Psi }}_h=\mathrm{i}\mathrm{d}$; interchanging the endpoints and the multipliers proves ${\mathrm{\Psi }}_h{\mathrm{\Psi }}_{-h}=\mathrm{i}\mathrm{d}$.