# LTS Termination Proof

by T2Cert

## Input

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

The invariants are proved as follows.

### IMPACT Invariant Proof

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

### 2 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 1 5 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
and for every transition t, a duplicate t is considered.

### 3 Transition Removal

We remove transitions 0, 4 using the following ranking functions, which are bounded by −11.

 2: 0 0: 0 1: 0 2: −4 0: −5 1: −6 1_var_snapshot: −6 1*: −6
Hints:
 6 lexWeak[ [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] ] 3 lexWeak[ [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] ] 4 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* 8 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 6 1_var_snapshot: arg2P + 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, 1_var_snapshot, 1* }.

### 6.1.1 Transition Removal

We remove transitions 6, 8, 2, 3 using the following ranking functions, which are bounded by −1.

 1: 2 + 3⋅arg1 1_var_snapshot: 3⋅arg1 1*: 4 + 3⋅arg1
Hints:
 6 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 3, 0] , [0, 3, 0, 0, 0, 0, 0, 0, 0, 0] ] 8 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 3, 0] , [0, 3, 0, 0, 0, 0, 0, 0, 0, 0] ] 1 lexWeak[ [0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 3, 0, 0, 0] ] 2 lexStrict[ [0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 3, 0, 0, 0] , [0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 3 lexStrict[ [0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 3, 0, 0, 0] , [0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

### 6.1.2 Transition Removal

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

 1: 0 1_var_snapshot: 0 1*: − arg1P
Hints:
 1 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

### 6.1.3 Splitting Cut-Point Transitions

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

### 6.1.3.1 Cut-Point Subproblem 1/1

Here we consider cut-point transition 5.

### 6.1.3.1.1 Splitting Cut-Point Transitions

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

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