# 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 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 4 − 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 ∧ − arg5P + arg5 ≤ 0 ∧ arg5P − arg5 ≤ 0 ∧ − arg6P + arg6 ≤ 0 ∧ arg6P − arg6 ≤ 0 1 1 2: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ 1 − arg1 + arg6P ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ − arg6P ≤ 0 ∧ 2 − arg1 + arg5P ≤ 0 ∧ 1 − arg2P ≤ 0 ∧ −1 + arg2P ≤ 0 ∧ 1 − arg3P ≤ 0 ∧ −1 + arg3P ≤ 0 ∧ 1 − arg4P ≤ 0 ∧ −1 + arg4P ≤ 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 ∧ − arg5P + arg5 ≤ 0 ∧ arg5P − arg5 ≤ 0 ∧ − arg6P + arg6 ≤ 0 ∧ arg6P − arg6 ≤ 0 2 2 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 − arg2 ≤ 0 ∧ 1 − arg3 + arg5 ≤ 0 ∧ 1 + arg1P − arg1 ≤ 0 ∧ arg1P − arg6 ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ − arg6 ≤ 0 ∧ − arg1P ≤ 0 ∧ 2 − arg1 + arg5 ≤ 0 ∧ arg3 − arg4 ≤ 0 ∧ − arg3 + arg4 ≤ 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 ∧ − arg5P + arg5 ≤ 0 ∧ arg5P − arg5 ≤ 0 ∧ − arg6P + arg6 ≤ 0 ∧ arg6P − arg6 ≤ 0 2 3 2: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ arg3 − arg5 ≤ 0 ∧ 1 − arg3 ≤ 0 ∧ 1 − arg2 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ 1 − arg1 + arg6P ≤ 0 ∧ arg6P − arg6 ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ − arg6 ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ − arg6P ≤ 0 ∧ 2 − arg1 + arg5 ≤ 0 ∧ arg3 − arg4 ≤ 0 ∧ − arg3 + arg4 ≤ 0 ∧ 1 − arg3P + arg3 ≤ 0 ∧ −1 + arg3P − arg3 ≤ 0 ∧ 1 + arg3 − arg4P ≤ 0 ∧ −1 − arg3 + arg4P ≤ 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 ∧ − arg6P + arg6 ≤ 0 ∧ arg6P − arg6 ≤ 0 ∧ − arg5P + arg5P ≤ 0 ∧ arg5P − arg5P ≤ 0 ∧ − arg5 + arg5 ≤ 0 ∧ arg5 − arg5 ≤ 0 3 4 0: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 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 ∧ − arg5P + arg5 ≤ 0 ∧ arg5P − arg5 ≤ 0 ∧ − arg6P + arg6 ≤ 0 ∧ arg6P − arg6 ≤ 0

## Proof

The following invariants are asserted.

 0: TRUE 1: TRUE 2: 1 − arg1P ≤ 0 ∧ − arg6P ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ − arg6 ≤ 0 3: TRUE

The invariants are proved as follows.

### IMPACT Invariant Proof

• nodes (location) invariant:  0 (0) TRUE 1 (1) TRUE 2 (2) 1 − arg1P ≤ 0 ∧ − arg6P ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ − arg6 ≤ 0 3 (3) TRUE
• initial node: 3
• cover edges:
• transition edges:  0 0 1 1 1 2 2 2 1 2 3 2 3 4 0

### 2 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 1 5 1: − arg6P + arg6P ≤ 0 ∧ arg6P − arg6P ≤ 0 ∧ − arg6 + arg6 ≤ 0 ∧ arg6 − arg6 ≤ 0 ∧ − arg5P + arg5P ≤ 0 ∧ arg5P − arg5P ≤ 0 ∧ − arg5 + arg5 ≤ 0 ∧ arg5 − arg5 ≤ 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 12 2: − arg6P + arg6P ≤ 0 ∧ arg6P − arg6P ≤ 0 ∧ − arg6 + arg6 ≤ 0 ∧ arg6 − arg6 ≤ 0 ∧ − arg5P + arg5P ≤ 0 ∧ arg5P − arg5P ≤ 0 ∧ − arg5 + arg5 ≤ 0 ∧ arg5 − arg5 ≤ 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.

### 3 Transition Removal

We remove transitions 0, 4 using the following ranking functions, which are bounded by −13.

 3: 0 0: 0 1: 0 2: 0 3: −4 0: −5 1: −6 2: −6 1_var_snapshot: −6 1*: −6 2_var_snapshot: −6 2*: −6
Hints:
 6 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] ] 13 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] ] 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] ] 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] ] 3 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 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] ] 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] , [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: arg6P + arg6P ≤ 0arg6Parg6P ≤ 0arg6 + arg6 ≤ 0arg6arg6 ≤ 0arg5P + arg5P ≤ 0arg5Parg5P ≤ 0arg5 + arg5 ≤ 0arg5arg5 ≤ 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 6 1_var_snapshot: arg6P + arg6P ≤ 0arg6Parg6P ≤ 0arg6 + arg6 ≤ 0arg6arg6 ≤ 0arg5P + arg5P ≤ 0arg5Parg5P ≤ 0arg5 + arg5 ≤ 0arg5arg5 ≤ 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: arg6P + arg6P ≤ 0arg6Parg6P ≤ 0arg6 + arg6 ≤ 0arg6arg6 ≤ 0arg5P + arg5P ≤ 0arg5Parg5P ≤ 0arg5 + arg5 ≤ 0arg5arg5 ≤ 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 13 2_var_snapshot: arg6P + arg6P ≤ 0arg6Parg6P ≤ 0arg6 + arg6 ≤ 0arg6arg6 ≤ 0arg5P + arg5P ≤ 0arg5Parg5P ≤ 0arg5 + arg5 ≤ 0arg5arg5 ≤ 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

### 8 SCC Decomposition

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

### 8.1 SCC Subproblem 1/1

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

### 8.1.1 Transition Removal

We remove transition 2 using the following ranking functions, which are bounded by −1.

 1: −2 + 4⋅arg1 2: 4⋅arg6 1_var_snapshot: −3 + 4⋅arg1 1*: −1 + 4⋅arg1 2_var_snapshot: 4⋅arg6 2*: 4⋅arg6
Hints:
 6 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0] ] 8 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0] ] 13 lexWeak[ [0, 0, 0, 0, 0, 0, 4, 0, 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, 4, 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, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0] ] 2 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 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 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0] ]

### 8.1.2 Transition Removal

We remove transitions 1, 3 using the following ranking functions, which are bounded by −7.

 1: −9 + 4⋅arg1 2: 1 − arg3 − 3⋅arg4 + 4⋅arg5 1_var_snapshot: −10 + 4⋅arg1 1*: −8 + 4⋅arg1 2_var_snapshot: − arg3 − 3⋅arg4 + 4⋅arg5 2*: 1 − arg3 − 3⋅arg4 + 4⋅arg5
Hints:
 6 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0] ] 8 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0] ] 13 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 3, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0] ] 15 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 3, 0, 0, 0, 1, 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, 4, 0, 0, 1, 0, 3, 0, 0, 0, 0, 0, 0, 1, 0, 3, 4, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 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, 3, 1, 0, 3, 0, 0, 0, 0, 0, 0, 1, 0, 3, 0, 0, 0, 0, 4, 0] , [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 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, 0] ]

### 8.1.3 Transition Removal

We remove transitions 6, 8, 13, 15 using the following ranking functions, which are bounded by −1.

 1: 0 2: 0 1_var_snapshot: −1 1*: 1 2_var_snapshot: − arg1P 2*: arg1P
Hints:
 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, 0, 0, 0, 0, 0, 0, 0, 0] ] 8 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] ] 13 lexStrict[ [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] ] 15 lexStrict[ [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, 0, 0, 0, 0, 0, 0, 0, 0] ]

### 8.1.4 Splitting Cut-Point Transitions

We consider 2 subproblems corresponding to sets of cut-point transitions as follows.

### 8.1.4.1 Cut-Point Subproblem 1/2

Here we consider cut-point transition 5.

### 8.1.4.1.1 Splitting Cut-Point Transitions

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

### 8.1.4.2 Cut-Point Subproblem 2/2

Here we consider cut-point transition 12.

### 8.1.4.2.1 Splitting Cut-Point Transitions

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

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