The remaining step in relating the above result to eq. (\[Iresmu0\]) is to use the known independence of $I(S_r)$ on the random initial set $S_r$, i.e. the fact that $I(S_{r})=I$. For any given $S_r\in S$, there is one term in the sum of eq. (\[Iresmu1\]) that is particularly simple, namely that where $S_{r'}=S_{r}$. For this term, each of the $q_0^i$ integrations picks up a pole from a propagator of the simple form $1/((q_0^i)^2+E_i^2)$, so that no $\theta$-functions arise, producing $$\begin{aligned}
\tilde{A}_{S_r}(S_{r})&=& \prod_{\alpha\in P\setminus S_r} \frac{1}{(r_0^\alpha(S_r))^2+E_\alpha^2(S_r)}\, . \label{ASr}\end{aligned}$$ Owing to the independence of the $E_\alpha$, we on the other hand know that the different terms in the sum of eq. (\[Iresmu1\]) must be unique (see appendix \[sec:ind\] for a detailed discussion of this point), so that $\tilde{A}_{S_r}(S_{r'})=\tilde{A}_{S_{r'}}(S_{r'})$ for all $S_r, S_{r'}$. This implies that also the coefficients $\tilde{A}_{S_r}(S_{r'})$, $r\neq r'$, must reduce to the simple form $$\begin{aligned}
\tilde{A}_{S_r}(S_{r'})&=& \prod_{\alpha\in P\setminus S_{r'}} \frac{1}{(r_0^\alpha(S_{r'}))^2+E_\alpha^2(S_{r'})}\,, \label{ASr2}\end{aligned}$$ which — together with eq. (\[Iresmu1\]) — leads us directly to the result (\[Iresmu0\]).

Generalization to finite density
--------------------------------

Somewhat surprisingly, the generalization of the above result to the presence of nonzero $\mu$ in some of the propagators is by far the simplest part of our proof. Namely, 