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CSP-case-study-1

# Cp 1 α 2 dis n l.α → ref.t.l → cp 1 α 2 2 s

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Unformatted text preview: CP 1( α ) 2 dis. ( N ( l )) .α → ref.t.l → CP 1( α ) ) 2 ( 2 S ( l )= α,t ∈ T rel.t.l → rel.N ( l ) .α → CP 1( α ) ) CP 2( α ) = 2 E ( l )= α,t ∈ T arrive.t.l → ( 2 S ( m )= α enter.t.m → leave.t.l → CP 2( α ) ) 7 ’ & \$ % The Whole System The solution to the problem is thus TRAINS || CPS || LINKS where TRAINS , CPS and LINKS are respectively the parallel composition of all TRAINs, CPs and LINEs. 8 ’ & \$ % Two ”User” Points of View to the Whole System (1) The TRAINs can be thought of as using the ”network” which consists of the CPs and LINEs. NETWORK = ( CPS || LINES ) \ L where L is the union of the alphabets of the LINEs. (2) The Second level of user is provided by train drivers. SY STEM = ( TRAINS k NETWORK ) \ B where B is the set of all symbols of the form req.t.m , con.t.m , ref.t.m or rel.t.m 9 ’ & \$ % Property Verification Property 1: At no time can there be more than one train on any line. s ∈ traces ( SY STEM ) ⇒ length ( s „ { leave.t.l, leave.t. ¯ l : t ∈ T } ) ≥ length ( s „ { enter.t.l, enter.t. ¯ l : t ∈ T } )- C where C is 0 or 1 depending on whether there is or is not a train initially on l (or ¯ l ), 10...
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CP 1 α 2 dis N l.α → ref.t.l → CP 1 α 2 2 S l)= α,t...

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