have $\Phi_{G_t}(R) \geq \Phi(G_t) \geq \phi_t^2/4 \geq \phi^2/4$, i.e.  $$e_{G_t}(R,R')\geq \frac{\phi^2}{4} \min({\text{vol}}(R),{\text{vol}}(R') )\geq \frac{1}{12}\phi^2 \alpha n \min(|R|,|R'|)$$ where $R' = \overline{U_t}\setminus R = U\setminus (U_t \cup R)$. Furthermore, for $R\subseteq U$ and $\overline{R}:=U\setminus R$, we have $$\begin{aligned}
        e_{G[U]}(R,\overline{R}) \geq e_{G[U]}(R\setminus U_t,\overline{R}\setminus U_t) 
        &\geq \frac{1}{12}\phi^2 \alpha n \min(|R\setminus U_t|,|\overline{R}\setminus U_t|) \notag \\
        &\geq\frac{1}{12} \phi^2 \alpha n \left( \min(|R|,|\overline{R}|) - \frac{1}{4} \rho'n' \right).
                \label{Star}
        \end{aligned}$$ We will now show that $G[U]$ is a $(\nu,\tau)$-expander by assuming that $G[U]$ does not expand and deducing that $G[U]$ is $\rho'$-close to bipartite, contradicting the premise of the lemma.

Suppose there exists $ S\subseteq U$ with $\tau n' \leq |S| \leq (1-\tau)n'$ such that $N={\text{RN}}_{\nu,G[U]}(S)\text{ satisfies } |N|<|S|+\nu n. $ Since $\tau n' \leq |S| \leq (1-\tau)n'$, we have $\frac{1}{4}\rho'n'\leq \frac{1}{2}\tau n' \leq \frac{1}{2}\min(|S|,|\overline{S}|)$ so by , we have $$\label{StarStar}
            e_{G[U]}(S,\overline{S})\geq \frac{1}{24}\phi^2 \alpha n\min(|S|,|\overline{S}|).$$

We may assume $\frac{1}{4}\alpha n\leq |S| \leq |U|-\frac{1}{4}\alpha n$.

If $|S|<\frac{1}{4}\alpha n$ then $e_G(S,\overline{S})\geq |S|(\alpha n - |S|)-\rho n^2$ and $e_G(S,\overline{S})\leq |N||S|+|U\setminus N| \nu n \leq |N||S| + \nu n^2$, so combining these inequalities and rearranging, we obtain $$\begin{aligned}
                |N| \geq \alpha n - &|S| - (\rho + \nu )\frac{n^2}{|S|}\geq \alpha n - |S| - (\rho + \nu )\frac{n^2}{\tau n'} \\
                &\overset{{\rm 