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