vault backup: 2025-04-30 19:43:22

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Marco Realacci 2025-04-30 19:43:22 +02:00
parent 350ef45e0a
commit 31b548b632

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@ -51,7 +51,7 @@ $P$ is in standard form if and only if $P \triangleq \sum_{i}\alpha_{i}P_{i}$ an
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. $\vdash P_{1}=P_{1}'$
Let's now consider a context: $\alpha_{1}.☐ + \sum_{i \in I}\alpha_{i}P_{i}$
Now we fill the context with $P_1$ and remove 1 from the set I (basically we pull it out from the summation): $$\alpha_{1}.P_{1} + \sum_{i \in I\setminus \{ 1 \}}\alpha_{i}P_{i}$$
Now we fill the hole (*er bucio*) with $P_1$ and remove 1 from the set I (basically we pull it out from the summation): $$\alpha_{1}.P_{1} + \sum_{i \in I\setminus \{ 1 \}}\alpha_{i}P_{i}$$
Now we replace $P_1$ with $P_{1}'$ and obtain: $$=\alpha_{1}.P_{1}' + \sum_{i \in I\setminus \{ 1 \}}\alpha_{i}P_{i}$$
imagine doing this until you pulled everything out... Standard form!