Theorem. Post-training cannot distinguish observationally equivalent worlds [ftip-007S]
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    For a finite random variable \(X\), write \(\operatorname {Law}(X)\) for its probability law. If \(h\) maps the value space of \(X\) to a finite set \(\mathcal Y\), its pushforward law \(h_{\#}\operatorname {Law}(X)\) is characterized by \[ \bigl (h_{\#}\operatorname {Law}(X)\bigr )(\{y\}) =\Pr (h(X)=y),\qquad y\in \mathcal Y. \]

    Let \(w_0\equiv _{\mathsf P}w_1\). For any finite output set \(\mathcal Y\) and any readout \(h:\mathcal T_{\mathsf P}\to \mathcal Y\),

    \[ h_{\#}\operatorname {Law}(T_{w_0}^{\mathsf P}) =h_{\#}\operatorname {Law}(T_{w_1}^{\mathsf P}). \]

    Thus a final artifact stamp, update decision, or test chosen solely from the declared protocol's transcript has the same distribution in both worlds.

    Proof. 

      For every \(y\in \mathcal Y\), finiteness gives

      \[ \begin {aligned} \Pr \left (h(T_{w_0}^{\mathsf P})=y\right ) &=\sum _{\substack {t\in \mathcal T_{\mathsf P}\\h(t)=y}} \Pr (T_{w_0}^{\mathsf P}=t)\\ &=\sum _{\substack {t\in \mathcal T_{\mathsf P}\\h(t)=y}} \Pr (T_{w_1}^{\mathsf P}=t) =\Pr \left (h(T_{w_1}^{\mathsf P})=y\right ). \end {aligned} \]

      The middle equality is observational equivalence from Definition [ftip-007R]. These point probabilities determine the two pushforward laws.

      This finite statement follows from the displayed hypotheses.