# 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 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ 27 − arg2P ≤ 0 ∧ −27 + arg2P ≤ 0 ∧ 28 − arg3P ≤ 0 ∧ −28 + arg3P ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − arg3P + arg3 ≤ 0 ∧ arg3P − arg3 ≤ 0 1 1 1: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ −3 + arg1P − arg1 ≤ 0 ∧ 1 − arg3 ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ 4 − arg1P ≤ 0 ∧ −1 − arg2P + arg2 ≤ 0 ∧ 1 + arg2P − arg2 ≤ 0 ∧ arg2 − arg3P ≤ 0 ∧ − arg2 + arg3P ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − arg3P + arg3 ≤ 0 ∧ arg3P − arg3 ≤ 0 1 2 2: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 1 + arg1P − arg1 ≤ 0 ∧ arg3 ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ − arg1P ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − arg3P + arg3 ≤ 0 ∧ arg3P − arg3 ≤ 0 2 3 2: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 1 + arg1P − arg1 ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ − arg1P ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − arg3P + arg3 ≤ 0 ∧ arg3P − arg3 ≤ 0 3 4 0: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − arg3P + arg3 ≤ 0 ∧ arg3P − arg3 ≤ 0

## Proof

The following invariants are asserted.

 0: TRUE 1: 1 − arg1P ≤ 0 ∧ 1 − arg1 ≤ 0 2: − arg1P ≤ 0 ∧ − arg1 ≤ 0 3: TRUE

The invariants are proved as follows.

### IMPACT Invariant Proof

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

### 2 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 1 5 1: − arg3P + arg3P ≤ 0 ∧ arg3P − arg3P ≤ 0 ∧ − arg3 + arg3 ≤ 0 ∧ arg3 − arg3 ≤ 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 12 2: − arg3P + arg3P ≤ 0 ∧ arg3P − arg3P ≤ 0 ∧ − arg3 + arg3 ≤ 0 ∧ arg3 − arg3 ≤ 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, 2, 4 using the following ranking functions, which are bounded by −15.

 3: 0 0: 0 1: 0 2: 0 3: −5 0: −6 1: −7 1_var_snapshot: −7 1*: −7 2: −10 2_var_snapshot: −10 2*: −10
Hints:
 6 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 13 lexWeak[ [0, 0, 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, 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] ] 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, 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, 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, 0, 0] ]

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

1* 8 1: arg3P + arg3P ≤ 0arg3Parg3P ≤ 0arg3 + arg3 ≤ 0arg3arg3 ≤ 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 6 1_var_snapshot: arg3P + arg3P ≤ 0arg3Parg3P ≤ 0arg3 + arg3 ≤ 0arg3arg3 ≤ 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.

2* 15 2: arg3P + arg3P ≤ 0arg3Parg3P ≤ 0arg3 + arg3 ≤ 0arg3arg3 ≤ 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.

2 13 2_var_snapshot: arg3P + arg3P ≤ 0arg3Parg3P ≤ 0arg3 + arg3 ≤ 0arg3arg3 ≤ 0arg2P + 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 { 2, 2_var_snapshot, 2* }.

### 8.1.1 Transition Removal

We remove transitions 13, 15, 3 using the following ranking functions, which are bounded by −1.

 2: 1 + 3⋅arg1 2_var_snapshot: 3⋅arg1 2*: 2 + 3⋅arg1
Hints:
 13 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0] , [0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 15 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0] , [0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 3 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 3, 0, 0, 0, 0, 0] , [0, 3, 0, 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 12.

### 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 { 1, 1_var_snapshot, 1* }.

### 8.2.1 Splitting Cut-Point Transitions

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

### 8.2.1.1 Cut-Point Subproblem 1/1

Here we consider cut-point transition 5.

The new variable __snapshot_1_arg3P is introduced. The transition formulas are extended as follows:

 6: __snapshot_1_arg3P ≤ arg3P ∧ arg3P ≤ __snapshot_1_arg3P 8: __snapshot_1_arg3P ≤ __snapshot_1_arg3P ∧ __snapshot_1_arg3P ≤ __snapshot_1_arg3P 1: __snapshot_1_arg3P ≤ __snapshot_1_arg3P ∧ __snapshot_1_arg3P ≤ __snapshot_1_arg3P

The new variable __snapshot_1_arg3 is introduced. The transition formulas are extended as follows:

 6: __snapshot_1_arg3 ≤ arg3 ∧ arg3 ≤ __snapshot_1_arg3 8: __snapshot_1_arg3 ≤ __snapshot_1_arg3 ∧ __snapshot_1_arg3 ≤ __snapshot_1_arg3 1: __snapshot_1_arg3 ≤ __snapshot_1_arg3 ∧ __snapshot_1_arg3 ≤ __snapshot_1_arg3

The new variable __snapshot_1_arg2P is introduced. The transition formulas are extended as follows:

 6: __snapshot_1_arg2P ≤ arg2P ∧ arg2P ≤ __snapshot_1_arg2P 8: __snapshot_1_arg2P ≤ __snapshot_1_arg2P ∧ __snapshot_1_arg2P ≤ __snapshot_1_arg2P 1: __snapshot_1_arg2P ≤ __snapshot_1_arg2P ∧ __snapshot_1_arg2P ≤ __snapshot_1_arg2P

The new variable __snapshot_1_arg2 is introduced. The transition formulas are extended as follows:

 6: __snapshot_1_arg2 ≤ arg2 ∧ arg2 ≤ __snapshot_1_arg2 8: __snapshot_1_arg2 ≤ __snapshot_1_arg2 ∧ __snapshot_1_arg2 ≤ __snapshot_1_arg2 1: __snapshot_1_arg2 ≤ __snapshot_1_arg2 ∧ __snapshot_1_arg2 ≤ __snapshot_1_arg2

The new variable __snapshot_1_arg1P is introduced. The transition formulas are extended as follows:

 6: __snapshot_1_arg1P ≤ arg1P ∧ arg1P ≤ __snapshot_1_arg1P 8: __snapshot_1_arg1P ≤ __snapshot_1_arg1P ∧ __snapshot_1_arg1P ≤ __snapshot_1_arg1P 1: __snapshot_1_arg1P ≤ __snapshot_1_arg1P ∧ __snapshot_1_arg1P ≤ __snapshot_1_arg1P

The new variable __snapshot_1_arg1 is introduced. The transition formulas are extended as follows:

 6: __snapshot_1_arg1 ≤ arg1 ∧ arg1 ≤ __snapshot_1_arg1 8: __snapshot_1_arg1 ≤ __snapshot_1_arg1 ∧ __snapshot_1_arg1 ≤ __snapshot_1_arg1 1: __snapshot_1_arg1 ≤ __snapshot_1_arg1 ∧ __snapshot_1_arg1 ≤ __snapshot_1_arg1

The following invariants are asserted.

 0: TRUE 1: 1 + arg2 − arg3P ≤ 0 ∧ − arg3P ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ 1 − arg1 ≤ 0 ∨ −1 + arg3 − arg3P ≤ 0 ∧ 1 + arg2 − arg3P ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ 1 − arg1 ≤ 0 2: − arg1P ≤ 0 ∧ − arg1 ≤ 0 3: TRUE 1: 1 + arg2 − arg3P ≤ 0 ∧ − arg3P ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ 1 − arg1 ≤ 0 ∨ −1 + arg3 − arg3P ≤ 0 ∧ 1 − __snapshot_1_arg3P + arg3P ≤ 0 ∧ 1 + arg2 − arg3P ≤ 0 ∧ − __snapshot_1_arg3P ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ 1 − arg1 ≤ 0 ∨ −1 + arg3 − arg3P ≤ 0 ∧ 1 + arg2 − arg3P ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ 1 − arg1 ≤ 0 1_var_snapshot: 1 − __snapshot_1_arg3P + arg2 ≤ 0 ∧ − __snapshot_1_arg3P ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ 1 − arg1 ≤ 0 ∨ 1 − __snapshot_1_arg3P + arg2 ≤ 0 ∧ −1 − __snapshot_1_arg3P + arg3 ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ 1 − arg1 ≤ 0 1*: −1 + arg3 − arg3P ≤ 0 ∧ 1 − __snapshot_1_arg3P + arg3P ≤ 0 ∧ 1 + arg2 − arg3P ≤ 0 ∧ − __snapshot_1_arg3P ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ 1 − arg1 ≤ 0

The invariants are proved as follows.

### IMPACT Invariant Proof

• nodes (location) invariant:  0 (3) TRUE 1 (0) TRUE 2 (1) 1 + arg2 − arg3P ≤ 0 ∧ − arg3P ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ 1 − arg1 ≤ 0 3 (1) −1 + arg3 − arg3P ≤ 0 ∧ 1 + arg2 − arg3P ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ 1 − arg1 ≤ 0 4 (2) − arg1P ≤ 0 ∧ − arg1 ≤ 0 5 (1) 1 + arg2 − arg3P ≤ 0 ∧ − arg3P ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ 1 − arg1 ≤ 0 6 (1_var_snapshot) 1 − __snapshot_1_arg3P + arg2 ≤ 0 ∧ − __snapshot_1_arg3P ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ 1 − arg1 ≤ 0 13 (1*) −1 + arg3 − arg3P ≤ 0 ∧ 1 − __snapshot_1_arg3P + arg3P ≤ 0 ∧ 1 + arg2 − arg3P ≤ 0 ∧ − __snapshot_1_arg3P ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ 1 − arg1 ≤ 0 14 (1) −1 + arg3 − arg3P ≤ 0 ∧ 1 − __snapshot_1_arg3P + arg3P ≤ 0 ∧ 1 + arg2 − arg3P ≤ 0 ∧ − __snapshot_1_arg3P ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ 1 − arg1 ≤ 0 15 (1_var_snapshot) 1 − __snapshot_1_arg3P + arg2 ≤ 0 ∧ −1 − __snapshot_1_arg3P + arg3 ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ 1 − arg1 ≤ 0 16 (1_var_snapshot) 1 − __snapshot_1_arg3P + arg2 ≤ 0 ∧ −1 − __snapshot_1_arg3P + arg3 ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ 1 − arg1 ≤ 0 20 (2) − arg1P ≤ 0 ∧ − arg1 ≤ 0 27 (1) −1 + arg3 − arg3P ≤ 0 ∧ 1 + arg2 − arg3P ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ 1 − arg1 ≤ 0 28 (2) − arg1P ≤ 0 ∧ − arg1 ≤ 0 29 (1) −1 + arg3 − arg3P ≤ 0 ∧ 1 + arg2 − arg3P ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ 1 − arg1 ≤ 0 33 (1*) −1 + arg3 − arg3P ≤ 0 ∧ 1 − __snapshot_1_arg3P + arg3P ≤ 0 ∧ 1 + arg2 − arg3P ≤ 0 ∧ − __snapshot_1_arg3P ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ 1 − arg1 ≤ 0 42 (1_var_snapshot) 1 − __snapshot_1_arg3P + arg2 ≤ 0 ∧ −1 − __snapshot_1_arg3P + arg3 ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ 1 − arg1 ≤ 0
• initial node: 0
• cover edges:
15 → 16 Hint: distribute conclusion [1, 0, 0, 0] [0, 1, 0, 0] [0, 0, 1, 0] [0, 0, 0, 1]
20 → 4 Hint: distribute conclusion [1, 0] [0, 1]
27 → 3 Hint: distribute conclusion [1, 0, 0, 0] [0, 1, 0, 0] [0, 0, 1, 0] [0, 0, 0, 1]
28 → 4 Hint: distribute conclusion [1, 0] [0, 1]
33 → 13 Hint: distribute conclusion [1, 0, 0, 0, 0, 0] [0, 1, 0, 0, 0, 0] [0, 0, 1, 0, 0, 0] [0, 0, 0, 1, 0, 0] [0, 0, 0, 0, 1, 0] [0, 0, 0, 0, 0, 1]
42 → 16 Hint: distribute conclusion [1, 0, 0, 0] [0, 1, 0, 0] [0, 0, 1, 0] [0, 0, 0, 1]
• transition edges:
0 4 1 Hint: auto
1 0 2 Hint: distribute conclusion [0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0] [0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 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, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0]
2 1 3 Hint: distribute conclusion [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 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, 1, 1, 0, 0, 0, 1, 0, 0, 0] [0, 0, 0, 0, 0, 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, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0]
2 2 4 Hint: distribute conclusion [0, 0, 0, 0, 0, 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, 1, 0, 1, 0, 0, 0, 0]
2 5 5 Hint: distribute conclusion [1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0] [0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] [0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0] [0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1]
3 1 27 Hint: distribute conclusion [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 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, 1, 1, 0, 0, 0, 1, 0, 0, 0] [0, 0, 0, 0, 0, 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, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0]
3 2 28 Hint: distribute conclusion [0, 0, 0, 0, 0, 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, 1, 0, 1, 0, 0, 0, 0]
3 5 29 Hint: distribute conclusion [1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0] [0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0] [0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0] [0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1]
4 3 20 Hint: distribute conclusion [0, 0, 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, 1, 0, 1, 0, 0, 0, 0]
5 6 6 Hint: distribute conclusion [1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] [0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] [0, 0, 1, 0, 0, 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, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
6 1 13 Hint: distribute conclusion [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 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, 1, 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, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 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, 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, 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, 0] [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
13 8 14 Hint: distribute conclusion [1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] [0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] [0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 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, 1, 0, 0, 0, 0, 0, 0, 0, 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, 1, 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, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
14 6 15 Hint: distribute conclusion [0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] [1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 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, 1, 0, 0, 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, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
14 6 16 Hint: distribute conclusion [0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] [1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 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, 1, 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, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
16 1 33 Hint: distribute conclusion [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 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, 1, 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, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] [0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 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, 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, 0] [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
29 6 42 Hint: distribute conclusion [0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] [1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] [0, 0, 1, 0, 0, 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, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

### 8.2.1.1.8 Transition Removal

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

 1: arg3P 1_var_snapshot: __snapshot_1_arg3P 1*: __snapshot_1_arg3P
Hints:
6 distribute assertion
 lexWeak[ [0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] lexWeak[ [0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] lexWeak[ [0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]
8 lexStrict[ [0, 1, 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, 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, 0, 0, 0, 0, 0] ]
1 distribute assertion
 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

### 8.2.1.1.9 Transition Removal

We remove transition 6 using the following ranking functions, which are bounded by −5.

 1: −1 1_var_snapshot: −2 1*: −3
Hints:
6 distribute assertion
 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, 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] ] 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] , [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] ] 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, 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] ]
1 distribute assertion
 lexWeak[ [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, 0, 0, 0, 0, 0, 0] ] lexWeak[ [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, 0, 0, 0, 0, 0, 0] ]

### 8.2.1.1.10 Splitting Cut-Point Transitions

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

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