the exact same reasoning goes through as in the $\mu=0$ case, with the only modification being a shift in the poles and residues of the fermion propagators originating from the $\mu$-dependence of the integral $$\begin{aligned}
\int_{-\infty}^\infty dq_0^i\, \frac{1}{(q_0^i+i\mu)^2+E_i^2}&=&\frac{\theta(E_i-\mu)}{2E_i}\, .\end{aligned}$$ In other words: whenever the pole of a fermionic propagator is taken, we need to multiply the corresponding residue in the product $\prod_{i\in S_{r'}} \frac{1}{2E_i}$ by the factor $\theta(E_i-\mu)$.

Connection to the original cutting rules
----------------------------------------

The previous step of the proof brought us to a somewhat cumbersome result, featuring a sum over all possible labelings of loop momenta in the original Feynman graph, with each term in the sum further containing a product of some number of $\theta(E_i-\mu)$ factors. To move forward, we write each of the $\theta$-functions in the form $1-\theta(\mu-E_i)$, and then reassemble the result for $I$ in the form $$\begin{aligned}
 I&=& \Big(\text{terms with 0 $\theta(\mu-E_i)$'s}\Big)+\Big(\text{terms with 1 $\theta(\mu-E_i)$}\Big)+\cdots  \nonumber \\
 &&+  \Big(\text{terms with $N$ $\theta(\mu-E_i)$'s}\Big)\, .\end{aligned}$$ It is self-evident that the first term in the above sum corresponds to the $\mu=0$ version of the same graph, but a little more effort is required to see what happens to the terms with one or more $\theta$-functions.

In the 