Proof: Let $T(\x) = L(\cdot \mid \x)/L(\theta_0 \mid \x)$, where $\theta_0$ is arbitrary. Let $\x, \y \sim p$.

\[\begin{alignat*}{2} \frac{p(\x \mid \theta)}{p(\y \mid \theta)} &= \frac{L(\theta \mid \x)}{L(\theta \mid \y)} &\hspace{8em}& \href{likelihood.html}{\text{Def. likelihood}} \\ &= \frac{L(\theta \mid \x) / L(\theta_0 \mid \x) \cdot L(\theta_0 \mid \x)}{ L(\theta \mid \y) / L(\theta_0 \mid \y) \cdot L(\theta_0 \mid \y)} \\ &= \frac{T(\x)}{T(\y)} \cdot h(\x, \y) && \text{Def. T; let $h(\x,\y) = L(\theta_0 \mid \x)/L(\theta_0 \mid \y)$} \\ &\implies \text{Constant function of $\theta$ iff $L(\cdot \mid \x) \propto L(\cdot \mid \y)$} \\ &\implies \text{$T$ is minimal sufficient} && \href{minimal-sufficiency-theorem.html}{\text{MinSuff Theorem}} \\ \end{alignat*}\]