diff --git a/.obsidian/workspace.json b/.obsidian/workspace.json
index dadebbb..8db5437 100644
--- a/.obsidian/workspace.json
+++ b/.obsidian/workspace.json
@@ -194,6 +194,7 @@
   "lastOpenFiles": [
     "Concurrent Systems/notes/13 - Weak Bisimilarity.md",
     "Concurrent Systems/notes/14 Checking bisimilarity, an inference system.md",
+    "Concurrent Systems/notes/12b - CCS cose varie.md",
     "Concurrent Systems/notes/12 - Calculus of communicating system.md",
     "Concurrent Systems/notes/11 - LTSs and Bisimulation.md",
     "Concurrent Systems/notes/10 - Implementing Consensus.md",
@@ -201,7 +202,6 @@
     "Pasted image 20250430175412.png",
     "Pasted image 20250430171336.png",
     "Concurrent Systems/slides/class 13.pdf",
-    "Concurrent Systems/notes/12b - CCS cose varie.md",
     "Concurrent Systems/notes/images/Pasted image 20250415082906.png",
     "Concurrent Systems/slides/class 14.pdf",
     "Concurrent Systems/notes/images/Pasted image 20250429092543.png",
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 a4a4375..8bab1b9 100644
--- a/Concurrent Systems/notes/14 Checking bisimilarity, an inference system.md	
+++ b/Concurrent Systems/notes/14 Checking bisimilarity, an inference system.md	
@@ -29,12 +29,15 @@ basically we can let the left or the right process evolve, leaving the other unc
 
 #### Soundness theorem: $\vdash P=Q \implies P \sim Q$
 *Proof:*
-If $\vdash LHS=RHS$e need to consider the relation $\{ LHS,RHS \}$
+If $\vdash LHS=RHS$, we need to consider the relation $\{ LHS,RHS \} \cup Id$ and prove it's a bisimulation (spoiler: it is).
 
-![](images/Pasted%20image%2020250429083535.png)
+Since bisimilarity is an equivalence and a congruence, the inference rules holds.
+
+
+>[!def] Standard form
 $P$ is in standard form if and only if $P \triangleq \sum_{i}\alpha_{i}P_{i}$ and $\forall_{i}P_{i}$ is in standard form.
 
-**Lemma:** $\forall P \exists P'$* in standard form such that $\vdash P = P'$
+**Lemma:** $\forall P \space \exists \space P'$ in standard form such that $\vdash P = P'$
 *Proof:* by induction on the structure of P.
 
 **Base case:** $P \triangleq 0$. It suffices to consider $P' \triangleq 0$ and conclude reflexivity.