vault backup: 2025-04-30 19:33:22

This commit is contained in:
Marco Realacci 2025-04-30 19:33:22 +02:00
parent a01b48cc7c
commit 9a670602e9
2 changed files with 6 additions and 4 deletions

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@ -192,10 +192,10 @@
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@ -210,7 +210,6 @@
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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
![](../../Pasted%20image%2020250430192526.png) ![](../../Pasted%20image%2020250430192526.png)
and eventually we find $P'$ which is literally the standard form of $P_{1}|P_{2}$. and eventually we find $P'$ which is literally the standard form of $P_{1}|P_{2}$.
2. $P \triangleq \sum_{i \in I}\alpha_{i}P_{i}$. By induction $\forall P_{i} \exists P_{i}$ 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 replace $P_1$ with $P_{1}'$
![](images/Pasted%20image%2020250429084921.png) ![](images/Pasted%20image%2020250429084921.png)
![](images/Pasted%20image%2020250429084950.png) ![](images/Pasted%20image%2020250429084950.png)
replacing one by one every continuation with its standard form, obtaining standard form. replacing one by one every continuation with its standard form, obtaining standard form.