We see that (1) $\Rightarrow$ (3) as a finite free module is a finite projective $R$-module. Conversely, suppose $P^\bullet$ is a bounded above complex of finite projective $R$-modules. Say $P^i = 0$ for $i > n_0$. We choose a direct sum decompositions $F^{n_0} = P^{n_0} \oplus C^{n_0}$ with $F^{n_0}$ a finite free $R$-module, and inductively $$ F^{n - 1} = P^{n - 1} \oplus C^n \oplus C^{n - 1} $$ for $n \leq n_0$ with $F^{n_0}$ a finite free $R$-module. As a complex $F^\bullet$ has maps $F^{n - 1} \to F^n$ which agree with $P^{n - 1} \to P^n$, induce the identity $C^n \to C^n$, and are zero on $C^{n - 1}$. The map $F^\bullet \to P^\bullet$ is a quasi-isomorphism (even a homotopy equivalence) and hence (3) implies (1). \medskip\noindent Assume (1). Let $E^\bullet$ be a bounded above complex of finite free $R$-modules and let $E^\bullet \to K^\bullet$ be a quasi-isomorphism. Then the induced maps $\sigma_{\geq m}E^\bullet \to K^\bullet$ from the stupid truncation of $E^\bullet$ to $K^\bullet$ show that $K^\bullet$ is $m$-pseudo-coherent. Hence (1) implies (2). \medskip\noindent Assume (2). Since $K^\bullet$ is $0$-pseudo-coherent we see in particular that $K^\bullet$ is bounded above. Let $b$ be an integer such that $H^i(K^\bullet) = 0$ for $i > b$. By descending induction on $n \in \mathbf{Z}$ we are going to construct finite free $R$-modules $F^i$ for $i \geq n$, differentials $d^i : F^i \to F^{i + 1}$ for $i \geq n$, maps $\alpha : F^i \to K^i$ compatible with differentials, such that (1) $H^i(\alpha)$ is an isomorphism for $i > n$ and surjective for $i = n$, and (2) $F^i = 0$ for $i > b$. Picture $$ \xymatrix{ & F^n \ar[r] \ar[d]^\alpha & F^{n + 1} \ar[d]^\alpha \ar[r] & \ldots \\ K^{n - 1} \ar[r] & K^n \ar[r] & K^{n + 1} \ar[r] & \ldots } $$ The base case is $n = b + 1$ where we can take $F^i = 0$ for all $i$. Induction step. Let $C^\bullet$ be the cone on $\alpha$ (Derived Categories, Definition \ref{derived-definition-cone}). The long exact sequence of cohomology shows that $H^i(C^\bullet) = 0$ for $i \geq n$. By Lemma \ref{lemma-cone-pseudo-coherent} we see that $C^\bullet$ is $(n - 1)$-pseudo-coherent. By Lemma \ref{lemma-finite-cohomology} we see that $H^{n - 1}(C^\bullet)$ is a finite $R$-module. Choose a finite free $R$-module $F^{n - 1}$ and a map $\beta : F^{n - 1} \to C^{n - 1}$ such that the composition $F^{n - 1} \to C^{n - 1} \to C^n$ is zero and such that $F^{n - 1}$ surjects onto $H^{n - 1}(C^\bullet)$. Since $C^{n - 1} = K^{n - 1} \oplus F^n$ we can write $\beta = (\alpha^{n - 1}, -d^{n - 1})$. The vanishing of the composition $F^{n - 1} \to C^{n - 1} \to C^n$ implies these maps fit into a morphism of complexes $$ \xymatrix{ & F^{n - 1} \ar[d]^{\alpha^{n - 1}} \ar[r]_{d^{n - 1}} & F^n \ar[r] \ar[d]^\alpha & F^{n + 1} \ar[d]^\alpha \ar[r] & \ldots \\ \ldots \ar[r] & K^{n - 1} \ar[r] & K^n \ar[r] & K^{n + 1} \ar[r] & \ldots } $$ Moreover, these maps define a morphism of distinguished triangles $$ \xymatrix{ (F^n \to \ldots) \ar[r] \ar[d] & (F^{n - 1} \to \ldots) \ar[r] \ar[d] & F^{n - 1} \ar[r] \ar[d]_\beta & (F^n \to \ldots)[1] \ar[d] \\ (F^n \to \ldots) \ar[r] & K^\bullet \ar[r] & C^\bullet \ar[r] & (F^n \to \ldots)[1] } $$ Hence our choice of $\beta$ implies that the map of complexes $(F^{n - 1} \to \ldots) \to K^\bullet$ induces an isomorphism on cohomology in degrees $\geq n$ and a surjection in degree $n - 1$. This finishes the proof of the lemma.