vault backup: 2025-04-30 19:33:22
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@ -49,7 +49,10 @@ $P$ is in standard form if and only if $P \triangleq \sum_{i}\alpha_{i}P_{i}$ an
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and eventually we find $P'$ which is literally the standard form of $P_{1}|P_{2}$.
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2. $P \triangleq \sum_{i \in I}\alpha_{i}P_{i}$. By induction $\forall P_{i} \exists P_{i}$
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2. $P \triangleq \sum_{i \in I}\alpha_{i}P_{i}$. By induction $\forall P_{i} \exists P_{i}'$ in a standard form s.t. $\vdash P_{1}=P_{1}'$
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Let's now consider a context: $\alpha_{1}.☐ + \sum_{i \in I}\alpha_{i}P_{i}$
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Now we fill the context with $P_1$ and remove 1 from the set I (basically we pull it out from the summation): $$\alpha_{1}.P_{1} + \sum_{i \in I\setminus \{ 1 \}}\alpha_{i}P_{i}$$
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Now we replace $P_1$ with $P_{1}'$
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replacing one by one every continuation with its standard form, obtaining standard form.
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