$v\in\sigma_{s}$, $$\label{eq:}
\tfrac12\left(d_{v}+\frac{h}{J} \right) > \left|E\left(v,\overline{\sigma_{s}}\right)\right|.$$ The total number of vertices with self-loops is w.h.p. of order $o(n)$, and so it will be evident from the bounds below that this assumption is immaterial. The second inequality in $$\left|E\left(\sigma_{s},\overline{\sigma_{s}}\right)\right|
< \tfrac12 \sum_{v\in\sigma_{s}} \left( d_{v}+\frac{h}{J}  \right)
= \tfrac12 \bigg(x\ell_{n}-\sum_{i=1}^{s}d_{i}+\frac{h\left(\left|\sigma\right| -s\right)}{J}  \bigg)
\leq\left|E\left(\sigma,\overline{\sigma}\right)\right|$$ holds whenever $$x\ell_{n}-2\left|E\left(\sigma,\overline{\sigma}\right)\right|+\frac{h\left(\left|\sigma\right| -s\right)}{J}
\leq \sum_{i=1}^{s} d_{i},$$ which in particular is true when we take the smallest $s$ such that $$\sum_{i=1}^{s} d_{i} \geq x\ell_{n}-2I_{\delta}\left(x\right)\ell_{n}
+\frac{h\left(\left|\sigma\right| -s\right)}{J}.
\label{eq:condits}$$ Furthermore, by removing $s$ vertices, the change in the size of the boundary at step $t$ is given by $$\begin{aligned}
&\left|E\left(\sigma_{t},\overline{\sigma_{t}}\right)\right|
-\left|E\left(\sigma,\overline{\sigma}\right)\right|
= \sum_{i=1}^{t}\big(\left|E\left(v_{i},\sigma_{i-1}\backslash v_{i}\right)\right|
-\left|E\left(v_{i},\overline{\sigma_{i-1}}\right)\right|\big)\\
&\qquad \leq \sum_{i=1}^{t}\left(d_{i}-2\left\lceil d_{i}\,
\frac{I_{\delta}\left(x\right)}{x}\right\rceil \right)
\leq \left(1-2\,\frac{I_{\delta}\left(x\right)}{x}\right)\sum_{i=1}^{t} d_{i}.
\label{eq:stepssum}
\end{aligned}$$ The first inequality in follows from the following observation: note that w.h.p. $|E(\sigma,\overline{\sigma})|$ $\geq I_{\delta}(x)\ell_n$, and hence the “proportion” of the total degree of $\sigma$ that is paired with vertices in $\overline{\sigma}$ is at least $I_{\delta}(x)/x$. This implies that there must be some vertex $v_i$ with a proportion of at least $I_{\delta}(x)/x$ of its degree connected with vertices in $\overline{\sigma}$. In other words, $v_i$ shares at least $\lceil d_i \frac{I_{\delta}(x)}{x}\rceil$ edges with $\overline{\sigma}$.

By the definition of $s$, we that $|E(\sigma,\overline{\sigma})|=|E(\sigma_s,\overline{\sigma_s})|
+o(n)$ (when $d_s=o(n)$), and hence 