[proofplan] We verify the chain complex identity directly on an arbitrary homogeneous element of the mapping cone. The first component vanishes because $D_\bullet$ is a chain complex and because the chain map identity makes the two mixed terms cancel. The second component vanishes because $C_\bullet$ is a chain complex. [/proofplan] [step:Apply the cone differential twice to a homogeneous element] Fix $n \in \mathbb{Z}$ and let $(y,x) \in \operatorname{Cone}(f)_n = D_n \oplus C_{n-1}$. By the definition of $\partial_n$, \begin{align*} \partial_n(y,x) = \bigl(d^D_n(y) + f_{n-1}(x), -d^C_{n-1}(x)\bigr), \end{align*} where the first component lies in $D_{n-1}$ and the second component lies in $C_{n-2}$. Applying $\partial_{n-1}$ to this element gives \begin{align*} \partial_{n-1}\partial_n(y,x) &= \partial_{n-1}\bigl(d^D_n(y) + f_{n-1}(x), -d^C_{n-1}(x)\bigr) \\ &= \bigl( d^D_{n-1}(d^D_n(y) + f_{n-1}(x)) + f_{n-2}(-d^C_{n-1}(x)), -d^C_{n-2}(-d^C_{n-1}(x)) \bigr). \end{align*} Using $R$-linearity of $d^D_{n-1}$, $f_{n-2}$, and $d^C_{n-2}$, this becomes \begin{align*} \partial_{n-1}\partial_n(y,x) = \bigl( d^D_{n-1}d^D_n(y) + d^D_{n-1}f_{n-1}(x) - f_{n-2}d^C_{n-1}(x), d^C_{n-2}d^C_{n-1}(x) \bigr). \end{align*} [/step] [step:Cancel the first component using the chain map identity] Since $(D_\bullet,d^D)$ is a chain complex, its differential satisfies \begin{align*} d^D_{n-1} \circ d^D_n = 0. \end{align*} Since $f: C_\bullet \to D_\bullet$ is a chain map, the degree $n-1$ chain map identity is \begin{align*} d^D_{n-1} \circ f_{n-1} = f_{n-2} \circ d^C_{n-1}. \end{align*} Evaluating these two identities at $y \in D_n$ and $x \in C_{n-1}$ gives \begin{align*} d^D_{n-1}d^D_n(y) = 0, \qquad d^D_{n-1}f_{n-1}(x) = f_{n-2}d^C_{n-1}(x). \end{align*} Therefore the first component of $\partial_{n-1}\partial_n(y,x)$ is \begin{align*} 0 + f_{n-2}d^C_{n-1}(x) - f_{n-2}d^C_{n-1}(x) = 0. \end{align*} [/step] [step:Cancel the second component using the complex identity for $C_\bullet$] Since $(C_\bullet,d^C)$ is a chain complex, its differential satisfies \begin{align*} d^C_{n-2} \circ d^C_{n-1} = 0. \end{align*} Evaluating at $x \in C_{n-1}$ gives \begin{align*} d^C_{n-2}d^C_{n-1}(x) = 0. \end{align*} Hence the second component of $\partial_{n-1}\partial_n(y,x)$ is also zero. [/step] [step:Conclude that the cone differential squares to zero] Combining the two component computations, we have shown that for every $(y,x) \in \operatorname{Cone}(f)_n$, \begin{align*} \partial_{n-1}\partial_n(y,x) = (0,0). \end{align*} Since $n \in \mathbb{Z}$ was arbitrary, $\partial_{n-1} \circ \partial_n = 0$ for every $n \in \mathbb{Z}$. Thus $(\operatorname{Cone}(f)_\bullet,\partial)$ is a chain complex. [/step]