We first reduce to the case $t = 1$ in the following way. Note that $\text{Supp}(M) = \{\mathfrak m, \mathfrak q_1, \ldots, \mathfrak q_t\}$, where $\mathfrak m \subset R$ is the maximal ideal. Let $M_i$ denote the image of $M \to M_{\mathfrak q_i}$, so $\text{Supp}(M_i) = \{\mathfrak m, \mathfrak q_i\}$. The map $\varphi$ (resp.\ $\psi$) induces an $R$-module map $\varphi_i : M_i \to M_i$ (resp.\ $\psi_i : M_i \to M_i$). Thus we get a morphism of $(2, 1)$-periodic complexes $$ (M, \varphi, \psi) \longrightarrow \bigoplus\nolimits_{i = 1, \ldots, t} (M_i, \varphi_i, \psi_i). $$ The kernel and cokernel of this map have support contained in $\{\mathfrak m\}$. Hence by Lemma \ref{lemma-compare-periodic-lengths} we have $$ e_R(M, \varphi, \psi) = \sum\nolimits_{i = 1, \ldots, t} e_R(M_i, \varphi_i, \psi_i) $$ On the other hand we clearly have $M_{\mathfrak q_i} = M_{i, \mathfrak q_i}$, and hence the terms of the right hand side of the formula of the lemma are equal to the expressions $$ \text{ord}_{R/\mathfrak q_i}\left( \det\nolimits_{\kappa(\mathfrak q_i)} (M_{i, \mathfrak q_i}, \varphi_{i, \mathfrak q_i}, \psi_{i, \mathfrak q_i}) \right) $$ In other words, if we can prove the lemma for each of the modules $M_i$, then the lemma holds. This reduces us to the case $t = 1$. \medskip\noindent Assume we have a $(2, 1)$-periodic complex $(M, \varphi, \psi)$ over a Noetherian local ring with $M$ a finite $R$-module, $\text{Supp}(M) = \{\mathfrak m, \mathfrak q\}$, and finite length cohomology modules. The proof in this case follows from Lemma \ref{lemma-key-lemma} and careful bookkeeping. Denote $K_\varphi = \Ker(\varphi)$, $I_\varphi = \Im(\varphi)$, $K_\psi = \Ker(\psi)$, and $I_\psi = \Im(\psi)$. Since $R$ is Noetherian these are all finite $R$-modules. Set $$ a = \text{length}_{R_{\mathfrak q}}(I_{\varphi, \mathfrak q}) = \text{length}_{R_{\mathfrak q}}(K_{\psi, \mathfrak q}), \quad b = \text{length}_{R_{\mathfrak q}}(I_{\psi, \mathfrak q}) = \text{length}_{R_{\mathfrak q}}(K_{\varphi, \mathfrak q}). $$ Equalities because the complex becomes exact after localizing at $\mathfrak q$. Note that $l = \text{length}_{R_{\mathfrak q}}(M_{\mathfrak q})$ is equal to $l = a + b$. \medskip\noindent We are going to use Lemma \ref{lemma-good-sequence-exists} to choose sequences of elements in finite $R$-modules $N$ with support contained in $\{\mathfrak m, \mathfrak q\}$. In this case $N_{\mathfrak q}$ has finite length, say $n \in \mathbf{N}$. Let us call a sequence $w_1, \ldots, w_n \in N$ with properties (1) and (2) of Lemma \ref{lemma-good-sequence-exists} a ``good sequence''. Note that the quotient $N/\langle w_1, \ldots, w_n \rangle$ of $N$ by the submodule generated by a good sequence has support (contained in) $\{\mathfrak m\}$ and hence has finite length (Algebra, Lemma \ref{algebra-lemma-support-point}). Moreover, the symbol $[w_1, \ldots, w_n] \in \det_{\kappa(\mathfrak q)}(N_{\mathfrak q})$ is a generator, see Lemma \ref{lemma-determinant-dimension-one}. \medskip\noindent Having said this we choose good sequences $$ \begin{matrix} x_1, \ldots, x_b & \text{in} & K_\varphi, & t_1, \ldots, t_a & \text{in} & K_\psi, \\ y_1, \ldots, y_a & \text{in} & I_\varphi \cap \langle t_1, \ldots t_a\rangle, & s_1, \ldots, s_b & \text{in} & I_\psi \cap \langle x_1, \ldots, x_b\rangle. \end{matrix} $$ We will adjust our choices a little bit as follows. Choose lifts $\tilde y_i \in M$ of $y_i \in I_\varphi$ and $\tilde s_i \in M$ of $s_i \in I_\psi$. It may not be the case that $\mathfrak q \tilde y_1 \subset \langle x_1, \ldots, x_b\rangle$ and it may not be the case that $\mathfrak q \tilde s_1 \subset \langle t_1, \ldots, t_a\rangle$. However, using that $\mathfrak q$ is finitely generated (as in the proof of Lemma \ref{lemma-good-sequence-exists}) we can find a $d \in R$, $d \not \in \mathfrak q$ such that $\mathfrak q d\tilde y_1 \subset \langle x_1, \ldots, x_b\rangle$ and $\mathfrak q d\tilde s_1 \subset \langle t_1, \ldots, t_a\rangle$. Thus after replacing $y_i$ by $dy_i$, $\tilde y_i$ by $d\tilde y_i$, $s_i$ by $ds_i$ and $\tilde s_i$ by $d\tilde s_i$ we see that we may assume also that $x_1, \ldots, x_b, \tilde y_1, \ldots, \tilde y_b$ and $t_1, \ldots, t_a, \tilde s_1, \ldots, \tilde s_b$ are good sequences in $M$. \medskip\noindent Finally, we choose a good sequence $z_1, \ldots, z_l$ in the finite $R$-module $$ \langle x_1, \ldots, x_b, \tilde y_1, \ldots, \tilde y_a \rangle \cap \langle t_1, \ldots, t_a, \tilde s_1, \ldots, \tilde s_b \rangle. $$ Note that this is also a good sequence in $M$. \medskip\noindent Since $I_{\varphi, \mathfrak q} = K_{\psi, \mathfrak q}$ there is a unique element $h \in \kappa(\mathfrak q)$ such that $[y_1, \ldots, y_a] = h [t_1, \ldots, t_a]$ inside $\det_{\kappa(\mathfrak q)}(K_{\psi, \mathfrak q})$. Similarly, as $I_{\psi, \mathfrak q} = K_{\varphi, \mathfrak q}$ there is a unique element $h \in \kappa(\mathfrak q)$ such that $[s_1, \ldots, s_b] = g [x_1, \ldots, x_b]$ inside $\det_{\kappa(\mathfrak q)}(K_{\varphi, \mathfrak q})$. We can also do this with the three good sequences we have in $M$. All in all we get the following identities \begin{align*} [y_1, \ldots, y_a] & = h [t_1, \ldots, t_a] \\ [s_1, \ldots, s_b] & = g [x_1, \ldots, x_b] \\ [z_1, \ldots, z_l] & = f_\varphi [x_1, \ldots, x_b, \tilde y_1, \ldots, \tilde y_a] \\ [z_1, \ldots, z_l] & = f_\psi [t_1, \ldots, t_a, \tilde s_1, \ldots, \tilde s_b] \end{align*} for some $g, h, f_\varphi, f_\psi \in \kappa(\mathfrak q)$. \medskip\noindent Having set up all this notation let us compute $\det_{\kappa(\mathfrak q)}(M, \varphi, \psi)$. Namely, consider the element $[z_1, \ldots, z_l]$. Under the map $\gamma_\psi \circ \sigma \circ \gamma_\varphi^{-1}$ of Definition \ref{definition-periodic-determinant} we have \begin{eqnarray*} [z_1, \ldots, z_l] & = & f_\varphi [x_1, \ldots, x_b, \tilde y_1, \ldots, \tilde y_a] \\ & \mapsto & f_\varphi [x_1, \ldots, x_b] \otimes [y_1, \ldots, y_a] \\ & \mapsto & f_\varphi h/g [t_1, \ldots, t_a] \otimes [s_1, \ldots, s_b] \\ & \mapsto & f_\varphi h/g [t_1, \ldots, t_a, \tilde s_1, \ldots, \tilde s_b] \\ & = & f_\varphi h/f_\psi g [z_1, \ldots, z_l] \end{eqnarray*} This means that $\det_{\kappa(\mathfrak q)} (M_{\mathfrak q}, \varphi_{\mathfrak q}, \psi_{\mathfrak q})$ is equal to $f_\varphi h/f_\psi g$ up to a sign. \medskip\noindent We abbreviate the following quantities \begin{eqnarray*} k_\varphi & = & \text{length}_R(K_\varphi/\langle x_1, \ldots, x_b\rangle) \\ k_\psi    & = & \text{length}_R(K_\psi/\langle t_1, \ldots, t_a\rangle) \\ i_\varphi & = & \text{length}_R(I_\varphi/\langle y_1, \ldots, y_a\rangle) \\ i_\psi    & = & \text{length}_R(I_\psi/\langle s_1, \ldots, s_a\rangle) \\ m_\varphi & = & \text{length}_R(M/ \langle x_1, \ldots, x_b, \tilde y_1, \ldots, \tilde y_a\rangle) \\ m_\psi    & = & \text{length}_R(M/ \langle t_1, \ldots, t_a, \tilde s_1, \ldots, \tilde s_b\rangle) \\ \delta_\varphi & = & \text{length}_R( \langle x_1, \ldots, x_b, \tilde y_1, \ldots, \tilde y_a\rangle \langle z_1, \ldots, z_l\rangle) \\ \delta_\psi & = & \text{length}_R( \langle t_1, \ldots, t_a, \tilde s_1, \ldots, \tilde s_b\rangle \langle z_1, \ldots, z_l\rangle) \end{eqnarray*} Using the exact sequences $0 \to K_\varphi \to M \to I_\varphi \to 0$ we get $m_\varphi = k_\varphi + i_\varphi$. Similarly we have $m_\psi = k_\psi + i_\psi$. We have $\delta_\varphi + m_\varphi = \delta_\psi + m_\psi$ since this is equal to the colength of $\langle z_1, \ldots, z_l \rangle$ in $M$. Finally, we have $$ \delta_\varphi = \text{ord}_{R/\mathfrak q}(f_\varphi), \quad \delta_\psi = \text{ord}_{R/\mathfrak q}(f_\psi) $$ by our first application of the key Lemma \ref{lemma-key-lemma}. \medskip\noindent Next, let us compute the multiplicity of the periodic complex \begin{eqnarray*} e_R(M, \varphi, \psi) & = & \text{length}_R(K_\varphi/I_\psi) - \text{length}_R(K_\psi/I_\varphi) \\ & = & \text{length}_R( \langle x_1, \ldots, x_b\rangle/ \langle s_1, \ldots, s_b\rangle) + k_\varphi - i_\psi \\ & & - \text{length}_R( \langle t_1, \ldots, t_a\rangle/ \langle y_1, \ldots, y_a\rangle) - k_\psi + i_\varphi \\ & = & \text{ord}_{R/\mathfrak q}(g/h) + k_\varphi - i_\psi - k_\psi + i_\varphi \\ & = & \text{ord}_{R/\mathfrak q}(g/h) + m_\varphi - m_\psi \\ & = & \text{ord}_{R/\mathfrak q}(g/h) + \delta_\psi - \delta_\varphi \\ & = & \text{ord}_{R/\mathfrak q}(f_\psi g/f_\varphi h) \end{eqnarray*} where we used the key Lemma \ref{lemma-key-lemma} twice in the third equality. By our computation of $\det_{\kappa(\mathfrak q)} (M_{\mathfrak q}, \varphi_{\mathfrak q}, \psi_{\mathfrak q})$ this proves the proposition.