LTS Termination Proof

by T2Cert

Input

Integer Transition System
• Initial Location: 3
• Transitions: (pre-variables and post-variables)  0 0 1: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ − arg2P ≤ 0 ∧ − arg2 ≤ 0 ∧ − arg1P ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 1 1 2: 1 + arg1 − arg2 ≤ 0 ∧ − arg1 ≤ 0 ∧ 1 − arg2 ≤ 0 ∧ − arg2P + arg2P ≤ 0 ∧ arg2P − arg2P ≤ 0 ∧ − arg2 + arg2 ≤ 0 ∧ arg2 − arg2 ≤ 0 ∧ − arg1P + arg1P ≤ 0 ∧ arg1P − arg1P ≤ 0 ∧ − arg1 + arg1 ≤ 0 ∧ arg1 − arg1 ≤ 0 1 2 2: 1 − arg1 + arg2 ≤ 0 ∧ − arg1 ≤ 0 ∧ 1 − arg2 ≤ 0 ∧ − arg2P + arg2P ≤ 0 ∧ arg2P − arg2P ≤ 0 ∧ − arg2 + arg2 ≤ 0 ∧ arg2 − arg2 ≤ 0 ∧ − arg1P + arg1P ≤ 0 ∧ arg1P − arg1P ≤ 0 ∧ − arg1 + arg1 ≤ 0 ∧ arg1 − arg1 ≤ 0 2 3 2: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 1 − arg1 + arg2 ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ 1 − arg2 ≤ 0 ∧ − arg1P + arg1 − arg2 ≤ 0 ∧ arg1P − arg1 + arg2 ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2P ≤ 0 ∧ arg2P − arg2P ≤ 0 ∧ − arg2 + arg2 ≤ 0 ∧ arg2 − arg2 ≤ 0 1 4 1: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ arg1 − arg2 ≤ 0 ∧ − arg1 + arg2 ≤ 0 ∧ − arg2P ≤ 0 ∧ arg2P ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − arg1P + arg1P ≤ 0 ∧ arg1P − arg1P ≤ 0 ∧ − arg1 + arg1 ≤ 0 ∧ arg1 − arg1 ≤ 0 2 5 1: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 1 + arg1 − arg2 ≤ 0 ∧ − arg1P + arg2 ≤ 0 ∧ arg1P − arg2 ≤ 0 ∧ arg1 − arg2P ≤ 0 ∧ − arg1 + arg2P ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 3 6 0: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0

Proof

The following invariants are asserted.

 0: TRUE 1: TRUE 2: 1 − arg2 ≤ 0 3: TRUE

The invariants are proved as follows.

IMPACT Invariant Proof

• nodes (location) invariant:  0 (0) TRUE 1 (1) TRUE 2 (2) 1 − arg2 ≤ 0 3 (3) TRUE
• initial node: 3
• cover edges:
• transition edges:  0 0 1 1 1 2 1 2 2 1 4 1 2 3 2 2 5 1 3 6 0

2 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 1 7 1: − arg2P + arg2P ≤ 0 ∧ arg2P − arg2P ≤ 0 ∧ − arg2 + arg2 ≤ 0 ∧ arg2 − arg2 ≤ 0 ∧ − arg1P + arg1P ≤ 0 ∧ arg1P − arg1P ≤ 0 ∧ − arg1 + arg1 ≤ 0 ∧ arg1 − arg1 ≤ 0 2 14 2: − arg2P + arg2P ≤ 0 ∧ arg2P − arg2P ≤ 0 ∧ − arg2 + arg2 ≤ 0 ∧ arg2 − arg2 ≤ 0 ∧ − arg1P + arg1P ≤ 0 ∧ arg1P − arg1P ≤ 0 ∧ − arg1 + arg1 ≤ 0 ∧ arg1 − arg1 ≤ 0
and for every transition t, a duplicate t is considered.

3 Transition Removal

We remove transitions 0, 6 using the following ranking functions, which are bounded by −13.

 3: 0 0: 0 1: 0 2: 0 3: −4 0: −5 1: −6 2: −6 1_var_snapshot: −6 1*: −6 2_var_snapshot: −6 2*: −6
Hints:
 8 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0] ] 15 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0] ] 1 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 2 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 3 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 4 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 5 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 0 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 6 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0] ]

The following skip-transition is inserted and corresponding redirections w.r.t. the old location are performed.

1* 10 1: arg2P + arg2P ≤ 0arg2Parg2P ≤ 0arg2 + arg2 ≤ 0arg2arg2 ≤ 0arg1P + arg1P ≤ 0arg1Parg1P ≤ 0arg1 + arg1 ≤ 0arg1arg1 ≤ 0

The following skip-transition is inserted and corresponding redirections w.r.t. the old location are performed.

1 8 1_var_snapshot: arg2P + arg2P ≤ 0arg2Parg2P ≤ 0arg2 + arg2 ≤ 0arg2arg2 ≤ 0arg1P + arg1P ≤ 0arg1Parg1P ≤ 0arg1 + arg1 ≤ 0arg1arg1 ≤ 0

The following skip-transition is inserted and corresponding redirections w.r.t. the old location are performed.

2* 17 2: arg2P + arg2P ≤ 0arg2Parg2P ≤ 0arg2 + arg2 ≤ 0arg2arg2 ≤ 0arg1P + arg1P ≤ 0arg1Parg1P ≤ 0arg1 + arg1 ≤ 0arg1arg1 ≤ 0

The following skip-transition is inserted and corresponding redirections w.r.t. the old location are performed.

2 15 2_var_snapshot: arg2P + arg2P ≤ 0arg2Parg2P ≤ 0arg2 + arg2 ≤ 0arg2arg2 ≤ 0arg1P + arg1P ≤ 0arg1Parg1P ≤ 0arg1 + arg1 ≤ 0arg1arg1 ≤ 0

8 SCC Decomposition

We consider subproblems for each of the 1 SCC(s) of the program graph.

8.1 SCC Subproblem 1/1

Here we consider the SCC { 1, 2, 1_var_snapshot, 1*, 2_var_snapshot, 2* }.

8.1.1 Transition Removal

We remove transitions 1, 2, 4, 5 using the following ranking functions, which are bounded by 4.

 1: 2 + 5⋅arg2 2: 5⋅arg2 1_var_snapshot: 1 + 5⋅arg2 1*: 3 + 5⋅arg2 2_var_snapshot: 5⋅arg2 2*: 5⋅arg2
Hints:
 8 lexWeak[ [0, 0, 5, 0, 0, 0, 0, 0] ] 10 lexWeak[ [0, 0, 5, 0, 0, 0, 0, 0] ] 15 lexWeak[ [0, 0, 0, 5, 0, 0, 0, 0, 0] ] 17 lexWeak[ [0, 0, 0, 5, 0, 0, 0, 0, 0] ] 1 lexStrict[ [0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0] , [0, 0, 5, 0, 0, 0, 0, 0, 0, 0, 0] ] 2 lexStrict[ [0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0] , [0, 0, 5, 0, 0, 0, 0, 0, 0, 0, 0] ] 3 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5, 0] ] 4 lexStrict[ [0, 0, 5, 5, 0, 0, 5, 5, 0, 0, 0, 0, 0] , [0, 0, 5, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 5 lexStrict[ [0, 0, 0, 0, 0, 5, 0, 0, 0, 5, 0, 0, 5, 0] , [5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

8.1.2 Transition Removal

We remove transitions 8, 10, 3 using the following ranking functions, which are bounded by −2.

 1: −1 2: 1 + 3⋅arg1 1_var_snapshot: −2 1*: 0 2_var_snapshot: 3⋅arg1 2*: 2 + 3⋅arg1
Hints:
 8 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0] ] 10 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0] ] 15 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 3, 0] ] 17 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 3, 0] ] 3 lexStrict[ [3, 0, 0, 0, 0, 0, 0, 3, 3, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

8.1.3 Transition Removal

We remove transition 15 using the following ranking functions, which are bounded by −1.

 1: 0 2: 0 1_var_snapshot: 0 1*: 0 2_var_snapshot: −1 2*: 1
Hints:
 15 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0, 0] ] 17 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0] ]

8.1.4 Transition Removal

We remove transition 17 using the following ranking functions, which are bounded by −1.

 1: 0 2: −1 1_var_snapshot: 0 1*: 0 2_var_snapshot: 0 2*: 0
Hints:
 17 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0, 0] ]

8.1.5 Splitting Cut-Point Transitions

We consider 2 subproblems corresponding to sets of cut-point transitions as follows.

8.1.5.1 Cut-Point Subproblem 1/2

Here we consider cut-point transition 7.

8.1.5.1.1 Splitting Cut-Point Transitions

There remain no cut-point transition to consider. Hence the cooperation termination is trivial.

8.1.5.2 Cut-Point Subproblem 2/2

Here we consider cut-point transition 14.

8.1.5.2.1 Splitting Cut-Point Transitions

There remain no cut-point transition to consider. Hence the cooperation termination is trivial.

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