to prepare for numerical implementation; it becomes OP-NOFT-0, in which the PDC is automatically satisfied. Tracing out the spins, [[**\[rho1diag\]**]{}]{} becomes: $$\rho({{\bf{r}}}',{{\bf{r}}}) = 2 \sum_k p_1(k) \, \phi_k({{\bf{r}}}') \phi_k({{\bf{r}}}).
\label{rho1}$$ $k$ now labels $M(>N/2)$ active doubly-occupied NO states, and the following conditions hold: $$0 \le p_1(k) \le 1 \mbox{ and } 2\sum_k p_1(k) = N.
\label{rho1cond}$$

In [[**\[rho1\]**]{}]{} and [[**\[rho1cond\]**]{}]{} $p_1(k)$ is the occupation number of either of the paired NSOs having the NO $\phi_k$.

In the 2-DM, the impact of double occupancy on the structure of $\pi^d$ is minor, but it results in a major simplification of the structure of $\pi^{od}$. We make the orbital and spin components of the NSO indices explicit, so that they take the form $is$, with $i$ now the orbital index and $s = \pm$ the spin index. Because of double occupancy, the only index pairs that can enter $\pi^{od}$ in [[**\[final\_pi-od\]**]{}]{} are $i+,i-$ and $j+,j-$. Similarly, the only sets of two index pairs that can enter the rhs of [[**\[xi-4\]**]{}]{} are $i+,i-$, $k+,k-$ and $j+,j-$, $k+,k-$. The occupation numbers $\nu_{i+}$ and $\nu_{i-}$ are always equal, with values 0 or 1, so that all 4-NSO OPs in [[**\[xi-4\]**]{}]{} and [[**\[final\_pi-od\]**]{}]{} are identical to the 