From 31b548b632645fb2fc268e34e901430b89826819 Mon Sep 17 00:00:00 2001 From: Marco Realacci Date: Wed, 30 Apr 2025 19:43:22 +0200 Subject: [PATCH] vault backup: 2025-04-30 19:43:22 --- .../notes/14 Checking bisimilarity, an inference system.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/Concurrent Systems/notes/14 Checking bisimilarity, an inference system.md b/Concurrent Systems/notes/14 Checking bisimilarity, an inference system.md index 95f71e1..51bef10 100644 --- a/Concurrent Systems/notes/14 Checking bisimilarity, an inference system.md +++ b/Concurrent Systems/notes/14 Checking bisimilarity, an inference system.md @@ -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!