# LTS Termination Proof

by T2Cert

## Input

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

 1: − arg1P ≤ 0 ∧ − arg1 ≤ 0 2: TRUE 3: − arg2P ≤ 0 ∧ − arg2 ≤ 0 4: TRUE 5: TRUE

The invariants are proved as follows.

### IMPACT Invariant Proof

• nodes (location) invariant:  1 (1) − arg1P ≤ 0 ∧ − arg1 ≤ 0 2 (2) TRUE 3 (3) − arg2P ≤ 0 ∧ − arg2 ≤ 0 4 (4) TRUE 5 (5) TRUE
• initial node: 5
• cover edges:
• transition edges:  1 2 3 2 1 1 2 4 4 3 3 3 4 5 4 5 6 2

### 2 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 3 7 3: − arg2P + arg2P ≤ 0 ∧ arg2P − arg2P ≤ 0 ∧ − arg2 + arg2 ≤ 0 ∧ arg2 − arg2 ≤ 0 ∧ − arg1P + arg1P ≤ 0 ∧ arg1P − arg1P ≤ 0 ∧ − arg1 + arg1 ≤ 0 ∧ arg1 − arg1 ≤ 0 4 14 4: − 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 1, 2, 4, 6 using the following ranking functions, which are bounded by −17.

 5: 0 2: 0 1: 0 3: 0 4: 0 5: −6 2: −7 1: −8 3: −9 3_var_snapshot: −9 3*: −9 4: −12 4_var_snapshot: −12 4*: −12
Hints:
 8 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 15 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0] ] 3 lexWeak[ [0, 0, 0, 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] ] 1 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 2 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, 0, 0, 0, 0, 0, 0] ] 4 lexStrict[ [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.

3* 10 3: 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.

3 8 3_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.

4* 17 4: 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.

4 15 4_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 2 SCC(s) of the program graph.

### 8.1 SCC Subproblem 1/2

Here we consider the SCC { 3, 3_var_snapshot, 3* }.

### 8.1.1 Transition Removal

We remove transitions 8, 10, 3 using the following ranking functions, which are bounded by −1.

 3: 1 + arg2 3_var_snapshot: arg2 3*: 2 + arg2
Hints:
 8 lexStrict[ [0, 0, 0, 0, 1, 0, 0, 0, 0, 0] , [0, 1, 0, 0, 0, 0, 0, 0, 0, 0] ] 10 lexStrict[ [0, 0, 0, 0, 1, 0, 0, 0, 0, 0] , [0, 1, 0, 0, 0, 0, 0, 0, 0, 0] ] 3 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0] , [0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

### 8.1.2 Splitting Cut-Point Transitions

We consider 1 subproblems corresponding to sets of cut-point transitions as follows.

### 8.1.2.1 Cut-Point Subproblem 1/1

Here we consider cut-point transition 7.

### 8.1.2.1.1 Splitting Cut-Point Transitions

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

### 8.2 SCC Subproblem 2/2

Here we consider the SCC { 4, 4_var_snapshot, 4* }.

### 8.2.1 Transition Removal

We remove transition 5 using the following ranking functions, which are bounded by 1.

 4: 2⋅arg1 4_var_snapshot: 2⋅arg1 4*: 1 + 2⋅arg1
Hints:
 15 lexWeak[ [0, 0, 0, 0, 0, 0, 2, 0] ] 17 lexWeak[ [0, 0, 0, 0, 0, 0, 2, 0] ] 5 lexStrict[ [0, 0, 0, 0, 0, 0, 2, 2, 0, 0, 0] , [0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0] ]

### 8.2.2 Transition Removal

We remove transitions 15, 17 using the following ranking functions, which are bounded by −2.

 4: −1 4_var_snapshot: −2 4*: 0
Hints:
 15 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0] ] 17 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0] ]

### 8.2.3 Splitting Cut-Point Transitions

We consider 1 subproblems corresponding to sets of cut-point transitions as follows.

### 8.2.3.1 Cut-Point Subproblem 1/1

Here we consider cut-point transition 14.

### 8.2.3.1.1 Splitting Cut-Point Transitions

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

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