\begin{theorem}[Name]
\label{thm:name}
Under Assumptions \ref{A1}--\ref{An}, [precise claim].
\end{theorem}

\begin{proof}
The proof proceeds in [n] steps.

\textbf{Step 1: [Description]}
[Content with justification for each transition]

\textbf{Step 2: [Description]}
[Content]

\vdots

\textbf{Step n: Conclusion}
Combining Steps 1--[n-1], we obtain [result], completing the proof.
\end{proof}

\begin{theorem}[Identification of $\psi$]
Under Assumptions \ref{A:consistency}--\ref{A:positivity}, the causal effect
$\psi = E[Y(a)]$ is identified by
\[
\psi = \int E[Y \mid A=a, X=x] \, dP(x).
\]
\end{theorem}

\begin{proof}
\begin{align}
E[Y(a)] &= E[E[Y(a) \mid X]] && \text{(law of iterated expectations)} \\
        &= E[E[Y(a) \mid A=a, X]] && \text{(A\ref{A:exchangeability}: $Y(a) \indep A \mid X$)} \\
        &= E[E[Y \mid A=a, X]] && \text{(A\ref{A:consistency}: $Y = Y(A)$)} \\
        &= \int E[Y \mid A=a, X=x] \, dP(x) && \text{(definition)}
\end{align}
which depends only on the observed data distribution.
\end{proof}

\begin{theorem}[Consistency]
Under Assumptions \ref{A1}--\ref{An}, $\hat{\theta}_n \xrightarrow{p} \theta_0$.
\end{theorem}

\begin{proof}
Define $M_n(\theta) = n^{-1} \sum_{i=1}^n m(O_i; \theta)$ and
$M(\theta) = E[m(O; \theta)]$.

\textbf{Step 1: Uniform convergence}
By [ULLN conditions], $\sup_{\theta \in \Theta} |M_n(\theta) - M(\theta)| \xrightarrow{p} 0$.

\textbf{Step 2: Unique maximum}
$M(\theta)$ is uniquely maximized at $\theta_0$ (by identifiability).

\textbf{Step 3: Conclusion}
By standard M-estimation theory, Steps 1--2 imply $\hat{\theta}_n \xrightarrow{p} \theta_0$.
\end{proof}

\begin{theorem}[Asymptotic Normality]
Under Assumptions \ref{A1}--\ref{An},
\[
\sqrt{n}(\hat{\theta}_n - \theta_0) \xrightarrow{d} N(0, V)
\]
where $V = E[\phi(O)\phi(O)^\top]$ and $\phi$ is the influence function.
\end{theorem}

\begin{proof}
\textbf{Step 1: Score equation}
$\hat{\theta}_n$ solves $\mathbb{P}_n[\psi(O; \theta)] = 0$ where $\psi = \partial_\theta m$.

\textbf{Step 2: Taylor expansion}
\[
0 = \mathbb{P}_n[\psi(O; \hat{\theta}_n)] = \mathbb{P}_n[\psi(O; \theta_0)]
    + \mathbb{P}_n[\dot{\psi}(O; \tilde{\theta})](\hat{\theta}_n - \theta_0)
\]

\textbf{Step 3: Rearrangement}
\[
\sqrt{n}(\hat{\theta}_n - \theta_0) = -\left(\mathbb{P}_n[\dot{\psi}]\right)^{-1}
    \sqrt{n} \mathbb{P}_n[\psi(O; \theta_0)]
\]

\textbf{Step 4: Apply CLT}
$\sqrt{n} \mathbb{P}_n[\psi(O; \theta_0)] \xrightarrow{d} N(0, \text{Var}(\psi))$ by CLT.

\textbf{Step 5: Slutsky}
$\mathbb{P}_n[\dot{\psi}] \xrightarrow{p} E[\dot{\psi}]$ by WLLN. Apply Slutsky's theorem.
\end{proof}

\begin{theorem}[Semiparametric Efficiency]
$\hat{\theta}_n$ is semiparametrically efficient with influence function
\[
\phi(O) = [optimal formula]
\]
achieving the efficiency bound $V_{\text{eff}} = E[\phi(O)^2]$.
\end{theorem}

\begin{theorem}[Double Robustness]
The estimator $\hat{\psi}_{DR}$ is consistent if either:
\begin{enumerate}
\item The outcome model $\mu(a,x) = E[Y \mid A=a, X=x]$ is correctly specified, or
\item The propensity score $\pi(x) = P(A=1 \mid X=x)$ is correctly specified.
\end{enumerate}
\end{theorem}

\begin{proof}
The estimating equation has the form:
\[
\psi - \hat{\psi}_{DR} = E\left[\frac{(A-\pi)(Y-\mu)}{\pi(1-\pi)}\right] + o_p(1)
\]
The bias term $(A-\pi)(Y-\mu)$ is zero in expectation if either:
\begin{itemize}
\item $E[A-\pi \mid X] = 0$ (propensity correctly specified), or
\item $E[Y-\mu \mid A, X] = 0$ (outcome correctly specified).
\end{itemize}
\end{proof}