[proofplan]
We fix a form $\alpha$ of pure type $(p,q)$ and use the defining decomposition of the [exterior derivative](/theorems/1525) into its Dolbeault components. Applying $d$ twice gives zero, while expanding $d = \partial + \bar{\partial}$ gives four terms of three different bidegrees. Since the decomposition of forms into types is a direct sum, each bidegree component of the zero form must vanish separately, yielding the three identities.
[/proofplan]
[step:Apply the type decomposition of $d$ to a pure form]
Fix integers $p,q \geq 0$ and let $\alpha \in A^{p,q}(X)$. By definition of the Dolbeault operators on a form of type $(p,q)$, the exterior derivative
\begin{align*}
d: A^{p+q}(X) \to A^{p+q+1}(X)
\end{align*}
decomposes on $\alpha$ as
\begin{align*}
d\alpha = \partial\alpha + \bar{\partial}\alpha.
\end{align*}
Here $\partial\alpha \in A^{p+1,q}(X)$ and $\bar{\partial}\alpha \in A^{p,q+1}(X)$.
[guided]
We start with a form whose type is fixed, because the argument works by comparing bidegrees. Let $p,q \geq 0$ and let $\alpha \in A^{p,q}(X)$ be a smooth complex-valued differential form of type $(p,q)$. The exterior derivative is the map
\begin{align*}
d: A^{p+q}(X) \to A^{p+q+1}(X).
\end{align*}
On a complex manifold, the complexified differential forms split into types. Therefore, when $d$ acts on a form of type $(p,q)$, its result has only two possible type components: one of type $(p+1,q)$ and one of type $(p,q+1)$. These are, by definition, the Dolbeault components:
\begin{align*}
d\alpha = \partial\alpha + \bar{\partial}\alpha.
\end{align*}
The first term satisfies $\partial\alpha \in A^{p+1,q}(X)$, while the second satisfies $\bar{\partial}\alpha \in A^{p,q+1}(X)$. This bookkeeping is the main point of the proof: after applying $d$ again, the resulting terms land in distinct type spaces, so they can be separated.
[/guided]
[/step]
[step:Expand $d^2\alpha$ using $d=\partial+\bar{\partial}$]
The exterior derivative satisfies $d^2=0$. Applying this to $\alpha$ gives
\begin{align*}
0 = d^2\alpha = d(\partial\alpha+\bar{\partial}\alpha).
\end{align*}
Using linearity of $d$ and again decomposing $d=\partial+\bar{\partial}$ on each pure-type summand, we obtain
\begin{align*}
0 = \partial(\partial\alpha)+\bar{\partial}(\partial\alpha)+\partial(\bar{\partial}\alpha)+\bar{\partial}(\bar{\partial}\alpha).
\end{align*}
The four terms have bidegrees
\begin{align*}
\partial(\partial\alpha) \in A^{p+2,q}(X), \qquad \bar{\partial}(\partial\alpha)+\partial(\bar{\partial}\alpha) \in A^{p+1,q+1}(X), \qquad \bar{\partial}(\bar{\partial}\alpha) \in A^{p,q+2}(X).
\end{align*}
[guided]
The next step is to use the ordinary identity $d^2=0$ for the exterior derivative. This applies to $\alpha$ because $\alpha \in A^{p,q}(X) \subset A^{p+q}(X)$ is a smooth complex-valued differential form, and $d$ is the exterior derivative on the graded space of smooth complex-valued forms. From the first step we have
\begin{align*}
d\alpha = \partial\alpha + \bar{\partial}\alpha.
\end{align*}
Therefore
\begin{align*}
0 = d^2\alpha = d(\partial\alpha+\bar{\partial}\alpha).
\end{align*}
Now we expand using linearity of $d$. The form $\partial\alpha$ has type $(p+1,q)$, so by the defining type decomposition of $d$ on pure-type forms,
\begin{align*}
d(\partial\alpha)=\partial(\partial\alpha)+\bar{\partial}(\partial\alpha).
\end{align*}
Similarly, $\bar{\partial}\alpha$ has type $(p,q+1)$, and hence
\begin{align*}
d(\bar{\partial}\alpha)=\partial(\bar{\partial}\alpha)+\bar{\partial}(\bar{\partial}\alpha).
\end{align*}
Combining these two expansions gives
\begin{align*}
0 = \partial(\partial\alpha)+\bar{\partial}(\partial\alpha)+\partial(\bar{\partial}\alpha)+\bar{\partial}(\bar{\partial}\alpha).
\end{align*}
The point of writing the expansion this way is that each term has a controlled bidegree. Since $\partial$ raises the first type index by $1$ and $\bar{\partial}$ raises the second type index by $1$, we have
\begin{align*}
\partial(\partial\alpha) \in A^{p+2,q}(X), \qquad \bar{\partial}(\partial\alpha)+\partial(\bar{\partial}\alpha) \in A^{p+1,q+1}(X), \qquad \bar{\partial}(\bar{\partial}\alpha) \in A^{p,q+2}(X).
\end{align*}
The middle two summands have the same bidegree, so they must be kept together. The first and last summands lie in different bidegree spaces.
[/guided]
[/step]
[step:Separate the identity by directness of the bidegree decomposition]
The type decomposition of smooth complex-valued forms is direct:
\begin{align*}
A^{p+q+2}(X)=\bigoplus_{a+b=p+q+2} A^{a,b}(X).
\end{align*}
Therefore a sum of components of distinct bidegrees is zero only if each bidegree component is zero. Applying this to the previous identity gives
\begin{align*}
\partial(\partial\alpha)=0.
\end{align*}
It also gives
\begin{align*}
\bar{\partial}(\bar{\partial}\alpha)=0.
\end{align*}
Finally, the component of bidegree $(p+1,q+1)$ gives
\begin{align*}
\partial(\bar{\partial}\alpha)+\bar{\partial}(\partial\alpha)=0.
\end{align*}
[guided]
The type decomposition stated in the theorem is a direct sum:
\begin{align*}
A^{p+q+2}(X)=\bigoplus_{a+b=p+q+2} A^{a,b}(X).
\end{align*}
Directness means that a smooth form of total degree $p+q+2$ has a unique decomposition into its bidegree components. Consequently, if a sum of terms lying in distinct bidegree spaces is zero, then each bidegree component of that sum is zero.
Apply this uniqueness to the identity obtained above:
\begin{align*}
0 = \partial(\partial\alpha)+\bar{\partial}(\partial\alpha)+\partial(\bar{\partial}\alpha)+\bar{\partial}(\bar{\partial}\alpha).
\end{align*}
The term $\partial(\partial\alpha)$ is the only component of bidegree $(p+2,q)$, so its component must vanish:
\begin{align*}
\partial(\partial\alpha)=0.
\end{align*}
The term $\bar{\partial}(\bar{\partial}\alpha)$ is the only component of bidegree $(p,q+2)$, so its component must vanish:
\begin{align*}
\bar{\partial}(\bar{\partial}\alpha)=0.
\end{align*}
The remaining bidegree $(p+1,q+1)$ component is the sum of the two mixed terms, so directness gives
\begin{align*}
\partial(\bar{\partial}\alpha)+\bar{\partial}(\partial\alpha)=0.
\end{align*}
Since $p,q \geq 0$ and $\alpha \in A^{p,q}(X)$ were arbitrary, these identities hold on every space $A^{p,q}(X)$. Hence, as operators on smooth complex-valued forms on $X$,
\begin{align*}
\partial^2=0, \qquad \bar{\partial}^2=0, \qquad \partial\bar{\partial}+\bar{\partial}\partial=0.
\end{align*}
[/guided]
Since $p,q \geq 0$ and $\alpha \in A^{p,q}(X)$ were arbitrary, these identities hold on every space $A^{p,q}(X)$. Hence, as operators on smooth complex-valued forms on $X$,
\begin{align*}
\partial^2=0, \qquad \bar{\partial}^2=0, \qquad \partial\bar{\partial}+\bar{\partial}\partial=0.
\end{align*}
[/step]
