vault backup: 2025-04-15 08:50:18
This commit is contained in:
parent
8c61a7d84f
commit
d216fbf5aa
1 changed files with 13 additions and 0 deletions
|
@ -29,3 +29,16 @@ Which challenges can (a.P)\a have?
|
|||
## Idempotency of Sum
|
||||
**Proposition:** $α.P+α.P+M ∼ α.P+M$
|
||||
*Proof:*
|
||||
$$S = \{ (α.P+α.P+M , α.P+M) \}$$
|
||||
Is it a bisimulation?
|
||||
NO: the problem is that, for example:
|
||||
- α.P+α.P+M –α–> P
|
||||
- α.P+M –α–> P
|
||||
- BUT (P,P) in general does NOT belong to S!
|
||||
|
||||
So we can try with $$S = \{ (α.P+α.P+M , α.P+M) \} ∪ \{(P,P)\}$$
|
||||
But it is not yet a bisimulation.
|
||||
P –β–> P’ (challenge and reply), but we don't have (P', P') in S.
|
||||
|
||||
So let's try with: $$S = \{ (α.P+α.P+M , α.P+M) \} ∪ Id$$
|
||||
Let's go!
|
Loading…
Add table
Add a link
Reference in a new issue