If for some $i$ some matrix coefficient of $\varphi_i$ is not in $\mathfrak m$, then we apply Lemma \ref{lemma-add-trivial-complex}. It is easy to see that the proposition for a complex and for the same complex with a trivial complex added to it are equivalent. Thus we may assume that all matrix entries of each $\varphi_i$ are elements of the maximal ideal. We may also assume that $e \geq 1$. \medskip\noindent Assume the complex is exact at $R^{n_e}, \ldots, R^{n_1}$. Let $\mathfrak q \in \text{Ass}(R)$. Note that the ring $R_{\mathfrak q}$ has depth $0$ and that the complex remains exact after localization at $\mathfrak q$. We apply Lemmas \ref{lemma-exact-depth-zero-local} and \ref{lemma-trivial-case-exact} to the localized complex over $R_{\mathfrak q}$. We conclude that $\varphi_{i, \mathfrak q}$ has rank $r_i$ for all $i$. Since $R \to \bigoplus_{\mathfrak q \in \text{Ass}(R)} R_\mathfrak q$ is injective (Lemma \ref{lemma-zero-at-ass-zero}), we conclude that $\varphi_i$ has rank $r_i$ over $R$ by the definition of rank as given in Definition \ref{definition-rank}. Therefore we see that $I(\varphi_i)_\mathfrak q = I(\varphi_{i, \mathfrak q})$ as the ranks do not change. Since all of the ideals $I(\varphi_i)_{\mathfrak q}$, $e \geq i \geq 1$ are equal to $R_{\mathfrak q}$ (by the lemmas referenced above) we conclude none of the ideals $I(\varphi_i)$ is contained in $\mathfrak q$. This implies that $I(\varphi_e)I(\varphi_{e-1})\ldots I(\varphi_1)$ is not contained in any of the associated primes of $R$. By Lemma \ref{lemma-silly} we may choose $x \in I(\varphi_e)I(\varphi_{e - 1})\ldots I(\varphi_1)$, $x \not \in \mathfrak q$ for all $\mathfrak q \in \text{Ass}(R)$. Observe that $x$ is a nonzerodivisor (Lemma \ref{lemma-ass-zero-divisors}). According to Lemma \ref{lemma-div-x-exact-one-less} the complex $0 \to (R/xR)^{n_e} \to \ldots \to (R/xR)^{n_1}$ is exact at $(R/xR)^{n_e}, \ldots, (R/xR)^{n_2}$. By induction on $e$ all the ideals $I(\varphi_i)/xR$ have a regular sequence of length $i - 1$. This proves that $I(\varphi_i)$ contains a regular sequence of length $i$. \medskip\noindent Assume (2)(a) and (2)(b) hold. We claim that for any prime $\mathfrak p \subset R$ conditions (2)(a) and (2)(b) hold for the complex $0 \to R_\mathfrak p^{n_e} \to R_\mathfrak p^{n_{e - 1}} \to \ldots \to R_\mathfrak p^{n_0}$ with maps $\varphi_{i, \mathfrak p}$ over $R_\mathfrak p$. Namely, since $I(\varphi_i)$ contains a nonzero divisor, the image of $I(\varphi_i)$ in $R_\mathfrak p$ is nonzero. This implies that the rank of $\varphi_{i, \mathfrak p}$ is the same as the rank of $\varphi_i$: the rank as defined above of a matrix $\varphi$ over a ring $R$ can only drop when passing to an $R$-algebra $R'$ and this happens if and only if $I(\varphi)$ maps to zero in $R'$. Thus (2)(a) holds. Having said this we know that $I(\varphi_{i, \mathfrak p}) = I(\varphi_i)_\mathfrak p$ and we see that (2)(b) is preserved under localization as well. By induction on the dimension of $R$ we may assume the complex is exact when localized at any nonmaximal prime $\mathfrak p$ of $R$. Thus $\Ker(\varphi_i)/\Im(\varphi_{i + 1})$ has support contained in $\{\mathfrak m\}$ and hence if nonzero has depth $0$. As $I(\varphi_i) \subset \mathfrak m$ for all $i$ because of what was said in the first paragraph of the proof, we see that (2)(b) implies $\text{depth}(R) \geq e$. By Lemma \ref{lemma-acyclic} we see that the complex is exact at $R^{n_e}, \ldots, R^{n_1}$ concluding the proof.