# 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 ∧ − x8 + x8 ≤ 0 ∧ x8 − x8 ≤ 0 1 1 2: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 2 − arg1 ≤ 0 ∧ 2 − arg2 ≤ 0 ∧ − arg1 + arg2 ≤ 0 ∧ 1 − arg1 + x8 ≤ 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 2 1: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 2 − arg1 ≤ 0 ∧ 2 − arg2 ≤ 0 ∧ 1 + arg1P − arg1 ≤ 0 ∧ − arg1 + arg2 ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − x8 + x8 ≤ 0 ∧ x8 − x8 ≤ 0 ∧ − arg2P + arg2P ≤ 0 ∧ arg2P − arg2P ≤ 0 ∧ − arg2 + arg2 ≤ 0 ∧ arg2 − arg2 ≤ 0 3 3 0: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − x8 + x8 ≤ 0 ∧ x8 − x8 ≤ 0

## Proof

The following invariants are asserted.

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

The invariants are proved as follows.

### IMPACT Invariant Proof

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

### 2 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 1 4 1: − x8 + x8 ≤ 0 ∧ x8 − x8 ≤ 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
and for every transition t, a duplicate t is considered.

### 3 Transition Removal

We remove transitions 0, 3 using the following ranking functions, which are bounded by −11.

 3: 0 0: 0 1: 0 2: 0 3: −4 0: −5 1: −6 2: −6 1_var_snapshot: −6 1*: −6
Hints:
 5 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 1 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 2 lexWeak[ [0, 0, 0, 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, 0, 0, 0, 0] ] 3 lexStrict[ [0, 0, 0, 0, 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* 7 1: x8 + x8 ≤ 0x8x8 ≤ 0arg2P + 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 5 1_var_snapshot: x8 + x8 ≤ 0x8x8 ≤ 0arg2P + arg2P ≤ 0arg2Parg2P ≤ 0arg2 + arg2 ≤ 0arg2arg2 ≤ 0arg1P + arg1P ≤ 0arg1Parg1P ≤ 0arg1 + arg1 ≤ 0arg1arg1 ≤ 0

### 6 SCC Decomposition

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

### 6.1 SCC Subproblem 1/1

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

### 6.1.1 Transition Removal

We remove transitions 1, 2 using the following ranking functions, which are bounded by 0.

 1: 2 + 4⋅arg1 2: 4⋅arg1 1_var_snapshot: 1 + 4⋅arg1 1*: 3 + 4⋅arg1
Hints:
 5 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0] ] 7 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0] ] 1 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0] , [0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 2 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 4, 0, 4, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 4, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 0] ]

### 6.1.2 Transition Removal

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

 1: 0 2: 0 1_var_snapshot: −1 1*: 1
Hints:
 5 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] ] 7 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

### 6.1.3 Transition Removal

We remove transition 7 using the following ranking functions, which are bounded by −1.

 1: −1 2: 0 1_var_snapshot: 0 1*: 0
Hints:
 7 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] ]

### 6.1.4 Splitting Cut-Point Transitions

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

### 6.1.4.1 Cut-Point Subproblem 1/1

Here we consider cut-point transition 4.

### 6.1.4.1.1 Splitting Cut-Point Transitions

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

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