# 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 − arg1 ≤ 0 ∧ − arg2 ≤ 0 ∧ −1 − arg1P + arg2 ≤ 0 ∧ 1 + arg1P − arg2 ≤ 0 ∧ − arg3P ≤ 0 ∧ arg3P ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg3P + arg3 ≤ 0 ∧ arg3P − arg3 ≤ 0 ∧ − arg4P + arg4 ≤ 0 ∧ arg4P − arg4 ≤ 0 ∧ − x26 + x26 ≤ 0 ∧ x26 − x26 ≤ 0 ∧ − x19 + x19 ≤ 0 ∧ x19 − x19 ≤ 0 ∧ − arg2P + arg2P ≤ 0 ∧ arg2P − arg2P ≤ 0 ∧ − arg2 + arg2 ≤ 0 ∧ arg2 − arg2 ≤ 0 1 1 1: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 1 − arg2 ≤ 0 ∧ 1 − arg3P + arg3 ≤ 0 ∧ 1 − arg3 ≤ 0 ∧ −1 − arg1P + arg1 ≤ 0 ∧ 1 + arg1P − arg1 ≤ 0 ∧ arg1 − arg2P ≤ 0 ∧ − arg1 + arg2P ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − arg3P + arg3 ≤ 0 ∧ arg3P − arg3 ≤ 0 ∧ − arg4P + arg4 ≤ 0 ∧ arg4P − arg4 ≤ 0 ∧ − x26 + x26 ≤ 0 ∧ x26 − x26 ≤ 0 ∧ − x19 + x19 ≤ 0 ∧ x19 − x19 ≤ 0 1 2 1: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 1 − arg2 ≤ 0 ∧ −1 − arg1P + arg1 ≤ 0 ∧ 1 + arg1P − arg1 ≤ 0 ∧ arg1 − arg2P ≤ 0 ∧ − arg1 + arg2P ≤ 0 ∧ 1 − arg3P ≤ 0 ∧ −1 + arg3P ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − arg3P + arg3 ≤ 0 ∧ arg3P − arg3 ≤ 0 ∧ − arg4P + arg4 ≤ 0 ∧ arg4P − arg4 ≤ 0 ∧ − x26 + x26 ≤ 0 ∧ x26 − x26 ≤ 0 ∧ − x19 + x19 ≤ 0 ∧ x19 − x19 ≤ 0 1 3 2: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ arg2 ≤ 0 ∧ 1 − arg3 ≤ 0 ∧ − arg1P ≤ 0 ∧ arg1P ≤ 0 ∧ − arg2P + arg3 ≤ 0 ∧ arg2P − arg3 ≤ 0 ∧ − arg4P ≤ 0 ∧ arg4P ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − arg4P + arg4 ≤ 0 ∧ arg4P − arg4 ≤ 0 ∧ − x26 + x26 ≤ 0 ∧ x26 − x26 ≤ 0 ∧ − x19 + x19 ≤ 0 ∧ x19 − x19 ≤ 0 ∧ − arg3P + arg3P ≤ 0 ∧ arg3P − arg3P ≤ 0 ∧ − arg3 + arg3 ≤ 0 ∧ arg3 − arg3 ≤ 0 2 4 2: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ 1 − arg4 ≤ 0 ∧ 1 − arg3 ≤ 0 ∧ 1 − arg1P + arg1 ≤ 0 ∧ 1 + arg1 − arg2 ≤ 0 ∧ 1 + arg3P − arg3 ≤ 0 ∧ 1 + arg4P − arg4 ≤ 0 ∧ 1 + arg1 − x19 ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg3P + arg3 ≤ 0 ∧ arg3P − arg3 ≤ 0 ∧ − arg4P + arg4 ≤ 0 ∧ arg4P − arg4 ≤ 0 ∧ − x26 + x26 ≤ 0 ∧ x26 − x26 ≤ 0 ∧ − arg2P + arg2P ≤ 0 ∧ arg2P − arg2P ≤ 0 ∧ − arg2 + arg2 ≤ 0 ∧ arg2 − arg2 ≤ 0 2 5 2: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 1 − arg2 + arg3P ≤ 0 ∧ 1 − arg2 ≤ 0 ∧ 1 + arg4P − x26 ≤ 0 ∧ − x26 ≤ 0 ∧ arg2 − arg3 ≤ 0 ∧ − arg2 + arg3 ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ −1 + arg1P ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − arg3P + arg3 ≤ 0 ∧ arg3P − arg3 ≤ 0 ∧ − arg4P + arg4 ≤ 0 ∧ arg4P − arg4 ≤ 0 ∧ − x19 + x19 ≤ 0 ∧ x19 − x19 ≤ 0 3 6 0: 0 ≤ 0 ∧ 0 ≤ 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 ∧ − arg4P + arg4 ≤ 0 ∧ arg4P − arg4 ≤ 0 ∧ − x26 + x26 ≤ 0 ∧ x26 − x26 ≤ 0 ∧ − x19 + x19 ≤ 0 ∧ x19 − x19 ≤ 0

## Proof

### 1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 1 7 1: − x26 + x26 ≤ 0 ∧ x26 − x26 ≤ 0 ∧ − x19 + x19 ≤ 0 ∧ x19 − x19 ≤ 0 ∧ − arg4P + arg4P ≤ 0 ∧ arg4P − arg4P ≤ 0 ∧ − arg4 + arg4 ≤ 0 ∧ arg4 − arg4 ≤ 0 ∧ − 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 14 2: − x26 + x26 ≤ 0 ∧ x26 − x26 ≤ 0 ∧ − x19 + x19 ≤ 0 ∧ x19 − x19 ≤ 0 ∧ − arg4P + arg4P ≤ 0 ∧ arg4P − arg4P ≤ 0 ∧ − arg4 + arg4 ≤ 0 ∧ arg4 − arg4 ≤ 0 ∧ − 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.

### 2 Transition Removal

We remove transitions 0, 3, 6 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:
 8 lexWeak[ [0, 0, 0, 0, 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, 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, 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, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 4 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] ] 5 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 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] ] 3 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] ] 6 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] ]

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

1* 10 1: x26 + x26 ≤ 0x26x26 ≤ 0x19 + x19 ≤ 0x19x19 ≤ 0arg4P + arg4P ≤ 0arg4Parg4P ≤ 0arg4 + arg4 ≤ 0arg4arg4 ≤ 0arg3P + 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 8 1_var_snapshot: x26 + x26 ≤ 0x26x26 ≤ 0x19 + x19 ≤ 0x19x19 ≤ 0arg4P + arg4P ≤ 0arg4Parg4P ≤ 0arg4 + arg4 ≤ 0arg4arg4 ≤ 0arg3P + 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* 17 2: x26 + x26 ≤ 0x26x26 ≤ 0x19 + x19 ≤ 0x19x19 ≤ 0arg4P + arg4P ≤ 0arg4Parg4P ≤ 0arg4 + arg4 ≤ 0arg4arg4 ≤ 0arg3P + 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_var_snapshot: x26 + x26 ≤ 0x26x26 ≤ 0x19 + x19 ≤ 0x19x19 ≤ 0arg4P + arg4P ≤ 0arg4Parg4P ≤ 0arg4 + arg4 ≤ 0arg4arg4 ≤ 0arg3P + arg3P ≤ 0arg3Parg3P ≤ 0arg3 + arg3 ≤ 0arg3arg3 ≤ 0arg2P + arg2P ≤ 0arg2Parg2P ≤ 0arg2 + arg2 ≤ 0arg2arg2 ≤ 0arg1P + arg1P ≤ 0arg1Parg1P ≤ 0arg1 + arg1 ≤ 0arg1arg1 ≤ 0

### 7 SCC Decomposition

We consider subproblems for each of the 2 SCC(s) of the program graph.

### 7.1 SCC Subproblem 1/2

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

### 7.1.1 Transition Removal

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

 2: 1 + 3⋅arg3 2_var_snapshot: 3⋅arg3 2*: 2 + 3⋅arg3
Hints:
 15 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 17 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 4 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 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, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 5 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 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, 0] ]

### 7.1.2 Transition Removal

We remove transitions 15, 17 using the following ranking functions, which are bounded by −1.

 2: 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, 0, 0] , [0, 0, 0, 0, 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, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

### 7.1.3 Splitting Cut-Point Transitions

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

### 7.1.3.1 Cut-Point Subproblem 1/1

Here we consider cut-point transition 14.

### 7.1.3.1.1 Splitting Cut-Point Transitions

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

### 7.2 SCC Subproblem 2/2

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

### 7.2.1 Splitting Cut-Point Transitions

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

### 7.2.1.1 Cut-Point Subproblem 1/1

Here we consider cut-point transition 7.

The new variable __snapshot_1_x26 is introduced. The transition formulas are extended as follows:

 8: __snapshot_1_x26 ≤ x26 ∧ x26 ≤ __snapshot_1_x26 10: __snapshot_1_x26 ≤ __snapshot_1_x26 ∧ __snapshot_1_x26 ≤ __snapshot_1_x26 1: __snapshot_1_x26 ≤ __snapshot_1_x26 ∧ __snapshot_1_x26 ≤ __snapshot_1_x26 2: __snapshot_1_x26 ≤ __snapshot_1_x26 ∧ __snapshot_1_x26 ≤ __snapshot_1_x26

The new variable __snapshot_1_x19 is introduced. The transition formulas are extended as follows:

 8: __snapshot_1_x19 ≤ x19 ∧ x19 ≤ __snapshot_1_x19 10: __snapshot_1_x19 ≤ __snapshot_1_x19 ∧ __snapshot_1_x19 ≤ __snapshot_1_x19 1: __snapshot_1_x19 ≤ __snapshot_1_x19 ∧ __snapshot_1_x19 ≤ __snapshot_1_x19 2: __snapshot_1_x19 ≤ __snapshot_1_x19 ∧ __snapshot_1_x19 ≤ __snapshot_1_x19

The new variable __snapshot_1_arg4P is introduced. The transition formulas are extended as follows:

 8: __snapshot_1_arg4P ≤ arg4P ∧ arg4P ≤ __snapshot_1_arg4P 10: __snapshot_1_arg4P ≤ __snapshot_1_arg4P ∧ __snapshot_1_arg4P ≤ __snapshot_1_arg4P 1: __snapshot_1_arg4P ≤ __snapshot_1_arg4P ∧ __snapshot_1_arg4P ≤ __snapshot_1_arg4P 2: __snapshot_1_arg4P ≤ __snapshot_1_arg4P ∧ __snapshot_1_arg4P ≤ __snapshot_1_arg4P

The new variable __snapshot_1_arg4 is introduced. The transition formulas are extended as follows:

 8: __snapshot_1_arg4 ≤ arg4 ∧ arg4 ≤ __snapshot_1_arg4 10: __snapshot_1_arg4 ≤ __snapshot_1_arg4 ∧ __snapshot_1_arg4 ≤ __snapshot_1_arg4 1: __snapshot_1_arg4 ≤ __snapshot_1_arg4 ∧ __snapshot_1_arg4 ≤ __snapshot_1_arg4 2: __snapshot_1_arg4 ≤ __snapshot_1_arg4 ∧ __snapshot_1_arg4 ≤ __snapshot_1_arg4

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

 8: __snapshot_1_arg3P ≤ arg3P ∧ arg3P ≤ __snapshot_1_arg3P 10: __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 2: __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:

 8: __snapshot_1_arg3 ≤ arg3 ∧ arg3 ≤ __snapshot_1_arg3 10: __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 2: __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:

 8: __snapshot_1_arg2P ≤ arg2P ∧ arg2P ≤ __snapshot_1_arg2P 10: __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 2: __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:

 8: __snapshot_1_arg2 ≤ arg2 ∧ arg2 ≤ __snapshot_1_arg2 10: __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 2: __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:

 8: __snapshot_1_arg1P ≤ arg1P ∧ arg1P ≤ __snapshot_1_arg1P 10: __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 2: __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:

 8: __snapshot_1_arg1 ≤ arg1 ∧ arg1 ≤ __snapshot_1_arg1 10: __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 2: __snapshot_1_arg1 ≤ __snapshot_1_arg1 ∧ __snapshot_1_arg1 ≤ __snapshot_1_arg1

The following invariants are asserted.

 0: arg2 − arg2P ≤ 0 1: −1 − arg1 + arg2 ≤ 0 ∧ 1 + arg1 − arg2P ≤ 0 ∧ − arg2P ≤ 0 2: TRUE 3: TRUE 1: −1 − arg1 + arg2 ≤ 0 ∧ 1 + arg1 − arg2P ≤ 0 ∧ − arg2P ≤ 0 ∨ −1 − arg1 + arg2 ≤ 0 ∧ 1 − __snapshot_1_arg2P + arg2P ≤ 0 ∧ 1 + arg1 − arg2P ≤ 0 ∧ − __snapshot_1_arg2P ≤ 0 ∧ − arg2P ≤ 0 1_var_snapshot: 1 − __snapshot_1_arg2P + arg1 ≤ 0 ∧ −1 − arg1 + arg2 ≤ 0 ∧ − __snapshot_1_arg2P ≤ 0 1*: −1 − arg1 + arg2 ≤ 0 ∧ 1 − __snapshot_1_arg2P + arg2P ≤ 0 ∧ 1 + arg1 − arg2P ≤ 0 ∧ − __snapshot_1_arg2P ≤ 0 ∧ − arg2P ≤ 0

The invariants are proved as follows.

### IMPACT Invariant Proof

• nodes (location) invariant:  0 (3) TRUE 1 (0) arg2 − arg2P ≤ 0 2 (1) −1 − arg1 + arg2 ≤ 0 ∧ 1 + arg1 − arg2P ≤ 0 ∧ − arg2P ≤ 0 3 (1) −1 − arg1 + arg2 ≤ 0 ∧ 1 + arg1 − arg2P ≤ 0 ∧ − arg2P ≤ 0 4 (1) −1 − arg1 + arg2 ≤ 0 ∧ 1 + arg1 − arg2P ≤ 0 ∧ − arg2P ≤ 0 5 (2) TRUE 6 (1) −1 − arg1 + arg2 ≤ 0 ∧ 1 + arg1 − arg2P ≤ 0 ∧ − arg2P ≤ 0 7 (1_var_snapshot) 1 − __snapshot_1_arg2P + arg1 ≤ 0 ∧ −1 − arg1 + arg2 ≤ 0 ∧ − __snapshot_1_arg2P ≤ 0 12 (1*) −1 − arg1 + arg2 ≤ 0 ∧ 1 − __snapshot_1_arg2P + arg2P ≤ 0 ∧ 1 + arg1 − arg2P ≤ 0 ∧ − __snapshot_1_arg2P ≤ 0 ∧ − arg2P ≤ 0 13 (1*) −1 − arg1 + arg2 ≤ 0 ∧ 1 − __snapshot_1_arg2P + arg2P ≤ 0 ∧ 1 + arg1 − arg2P ≤ 0 ∧ − __snapshot_1_arg2P ≤ 0 ∧ − arg2P ≤ 0 14 (1) −1 − arg1 + arg2 ≤ 0 ∧ 1 − __snapshot_1_arg2P + arg2P ≤ 0 ∧ 1 + arg1 − arg2P ≤ 0 ∧ − __snapshot_1_arg2P ≤ 0 ∧ − arg2P ≤ 0 15 (1_var_snapshot) 1 − __snapshot_1_arg2P + arg1 ≤ 0 ∧ −1 − arg1 + arg2 ≤ 0 ∧ − __snapshot_1_arg2P ≤ 0 20 (2) TRUE 21 (2) TRUE
• initial node: 0
• cover edges:
3 → 2 Hint: distribute conclusion [1, 0, 0] [0, 1, 0] [0, 0, 1]
4 → 2 Hint: distribute conclusion [1, 0, 0] [0, 1, 0] [0, 0, 1]
12 → 13 Hint: distribute conclusion [1, 0, 0, 0, 0] [0, 1, 0, 0, 0] [0, 0, 1, 0, 0] [0, 0, 0, 1, 0] [0, 0, 0, 0, 1]
15 → 7 Hint: distribute conclusion [1, 0, 0] [0, 1, 0] [0, 0, 1]
20 → 5 Hint: auto
21 → 5 Hint: auto
• transition edges:
0 6 1 Hint: [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0]
1 0 2 Hint: distribute conclusion [0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0] [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 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, 1, 0, 0]
2 1 3 Hint: distribute conclusion [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 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, 0, 0, 1, 1, 0, 1, 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, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
2 2 4 Hint: distribute conclusion [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 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, 1, 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, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
2 3 5 Hint: auto
2 7 6 Hint: distribute conclusion [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1] [0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0] [0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0]
5 4 20 Hint: auto
5 5 21 Hint: auto
6 8 7 Hint: distribute conclusion [0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 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, 1, 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, 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, 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]
7 1 12 Hint: distribute conclusion [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 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, 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, 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, 1, 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, 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, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0] [0, 1, 0, 0, 0, 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, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
7 2 13 Hint: distribute conclusion [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 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, 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, 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, 1, 1, 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, 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, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0] [0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 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]
13 10 14 Hint: distribute conclusion [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 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] [0, 1, 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, 1, 0, 0, 0, 0, 0, 0] [0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 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, 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, 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, 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]
14 8 15 Hint: distribute conclusion [0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 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, 1, 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, 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, 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, 1, 0, 0, 0, 0, 0, 0]

### 7.2.1.1.12 Transition Removal

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

 1: arg2P 1_var_snapshot: __snapshot_1_arg2P 1*: __snapshot_1_arg2P
Hints:
8 distribute assertion
 lexWeak[ [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, 1, 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, 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] ]
10 lexStrict[ [0, 1, 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, 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, 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, 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] ]
2 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, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0] ]

### 7.2.1.1.13 Transition Removal

We remove transition 8 using the following ranking functions, which are bounded by −5.

 1: −1 1_var_snapshot: −2 1*: −3
Hints:
8 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, 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, 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 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, 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, 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] ]

### 7.2.1.1.14 Splitting Cut-Point Transitions

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

T2Cert

• version: 1.0