# LTS Termination Proof

by T2Cert

## Input

Integer Transition System
• Initial Location: 5
• 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 ∧ 1 − arg2 ≤ 0 ∧ − arg4P ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ −5 − arg1 + arg2P ≤ 0 ∧ −3 − arg1 + arg3P ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ 6 − arg2P ≤ 0 ∧ 4 − arg3P ≤ 0 ∧ − arg5P ≤ 0 ∧ arg5P ≤ 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 ∧ − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0 ∧ − x13 + x13 ≤ 0 ∧ x13 − x13 ≤ 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 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ −1 + arg4 ≤ 0 ∧ 1 − arg5 ≤ 0 ∧ 1 + arg5 − x13 ≤ 0 ∧ 1 + arg5 − x14 ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ 5 − arg2 ≤ 0 ∧ 3 − arg3 ≤ 0 ∧ 5 − 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 1 2 2: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ −6 + arg1P − arg1 ≤ 0 ∧ −1 + arg4 ≤ 0 ∧ −1 + arg1P − arg2 ≤ 0 ∧ −3 + arg1P − arg3 ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ 6 − arg2 ≤ 0 ∧ 4 − arg3 ≤ 0 ∧ 7 − 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 ∧ − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0 ∧ − x13 + x13 ≤ 0 ∧ x13 − x13 ≤ 0 1 3 1: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 1 − arg5P + arg5 ≤ 0 ∧ 1 − arg5 ≤ 0 ∧ 2 − arg4 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ 4 + arg1P − arg2 ≤ 0 ∧ 2 + arg1P − arg3 ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ 5 − arg2 ≤ 0 ∧ 3 − arg3 ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ 5 − arg2P ≤ 0 ∧ 3 − arg3P ≤ 0 ∧ −1 − arg4P + arg4 ≤ 0 ∧ 1 + arg4P − 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 ∧ − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0 ∧ − x13 + x13 ≤ 0 ∧ x13 − x13 ≤ 0 1 4 1: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ 2 − arg4 ≤ 0 ∧ 5 + arg1P − arg2 ≤ 0 ∧ 3 + arg1P − arg3 ≤ 0 ∧ −7 − arg1 + arg2P ≤ 0 ∧ −2 + arg2P − arg2 ≤ 0 ∧ −4 + arg2P − arg3 ≤ 0 ∧ −5 − arg1 + arg3P ≤ 0 ∧ − arg2 + arg3P ≤ 0 ∧ −2 + arg3P − arg3 ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ 6 − arg2 ≤ 0 ∧ 4 − arg3 ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ 8 − arg2P ≤ 0 ∧ 6 − arg3P ≤ 0 ∧ −1 − arg4P + arg4 ≤ 0 ∧ 1 + arg4P − arg4 ≤ 0 ∧ 1 − arg5P ≤ 0 ∧ −1 + arg5P ≤ 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 ∧ − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0 ∧ − x13 + x13 ≤ 0 ∧ x13 − x13 ≤ 0 2 5 3: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 2 − arg2 ≤ 0 ∧ − arg2P ≤ 0 ∧ 1 + arg1P − arg1 ≤ 0 ∧ 5 − arg1 ≤ 0 ∧ 4 − arg1P ≤ 0 ∧ − arg3P ≤ 0 ∧ 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 ∧ − arg5P + arg5 ≤ 0 ∧ arg5P − arg5 ≤ 0 ∧ − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0 ∧ − x13 + x13 ≤ 0 ∧ x13 − x13 ≤ 0 3 6 3: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 1 − arg3P + arg3 ≤ 0 ∧ 1 − arg3 ≤ 0 ∧ 2 − arg2 ≤ 0 ∧ 3 − arg1 ≤ 0 ∧ 3 − arg1P ≤ 0 ∧ −1 − arg2P + arg2 ≤ 0 ∧ 1 + arg2P − arg2 ≤ 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 ∧ − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0 ∧ − x13 + x13 ≤ 0 ∧ x13 − x13 ≤ 0 3 7 3: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ −2 + arg1P − arg1 ≤ 0 ∧ 2 − arg2 ≤ 0 ∧ 4 − arg1 ≤ 0 ∧ 6 − arg1P ≤ 0 ∧ −1 − arg2P + arg2 ≤ 0 ∧ 1 + arg2P − arg2 ≤ 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 ∧ − arg5P + arg5 ≤ 0 ∧ arg5P − arg5 ≤ 0 ∧ − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0 ∧ − x13 + x13 ≤ 0 ∧ x13 − x13 ≤ 0 3 8 4: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 2 + arg1P − arg1 ≤ 0 ∧ −1 + arg2 ≤ 0 ∧ 3 − arg1 ≤ 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 ∧ − arg5P + arg5 ≤ 0 ∧ arg5P − arg5 ≤ 0 ∧ − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0 ∧ − x13 + x13 ≤ 0 ∧ x13 − x13 ≤ 0 4 9 4: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 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 ∧ − arg4P + arg4 ≤ 0 ∧ arg4P − arg4 ≤ 0 ∧ − arg5P + arg5 ≤ 0 ∧ arg5P − arg5 ≤ 0 ∧ − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0 ∧ − x13 + x13 ≤ 0 ∧ x13 − x13 ≤ 0 5 10 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 ∧ − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0 ∧ − x13 + x13 ≤ 0 ∧ x13 − x13 ≤ 0

## Proof

### 1 Invariant Updates

The following invariants are asserted.

 0: TRUE 1: 1 − arg1P ≤ 0 ∧ 1 − arg2P ≤ 0 ∧ 1 − arg3P ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ 1 − arg2 ≤ 0 ∧ 1 − arg3 ≤ 0 2: 5 − arg1P ≤ 0 ∧ 5 − arg1 ≤ 0 3: 1 − arg1P ≤ 0 ∧ 1 − arg1 ≤ 0 4: TRUE 5: TRUE

The invariants are proved as follows.

### IMPACT Invariant Proof

• nodes (location) invariant:  0 (0) TRUE 1 (1) 1 − arg1P ≤ 0 ∧ 1 − arg2P ≤ 0 ∧ 1 − arg3P ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ 1 − arg2 ≤ 0 ∧ 1 − arg3 ≤ 0 2 (2) 5 − arg1P ≤ 0 ∧ 5 − arg1 ≤ 0 3 (3) 1 − arg1P ≤ 0 ∧ 1 − arg1 ≤ 0 4 (4) TRUE 5 (5) TRUE
• initial node: 5
• cover edges:
• transition edges:  0 0 1 1 1 2 1 2 2 1 3 1 1 4 1 2 5 3 3 6 3 3 7 3 3 8 4 4 9 4 5 10 0

### 2 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 1 11 1: − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0 ∧ − x13 + x13 ≤ 0 ∧ x13 − x13 ≤ 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 3 18 3: − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0 ∧ − x13 + x13 ≤ 0 ∧ x13 − x13 ≤ 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 4 25 4: − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0 ∧ − x13 + x13 ≤ 0 ∧ x13 − x13 ≤ 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, 1, 2, 5, 8, 10 using the following ranking functions, which are bounded by −21.

 5: 0 0: 0 1: 0 2: 0 3: 0 4: 0 5: −7 0: −8 1: −9 1_var_snapshot: −9 1*: −9 2: −12 3: −13 3_var_snapshot: −13 3*: −13 4: −16 4_var_snapshot: −16 4*: −16
Hints:
 12 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] ] 19 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] ] 26 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] ] 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] ] 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, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 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, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 7 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] ] 9 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 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] ] 1 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] ] 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, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 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 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] ] 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, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 10 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] ]

### 4 Location Addition

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

1* 14 1: x14 + x14 ≤ 0x14x14 ≤ 0x13 + x13 ≤ 0x13x13 ≤ 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

### 5 Location Addition

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

1 12 1_var_snapshot: x14 + x14 ≤ 0x14x14 ≤ 0x13 + x13 ≤ 0x13x13 ≤ 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

### 6 Location Addition

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

3* 21 3: x14 + x14 ≤ 0x14x14 ≤ 0x13 + x13 ≤ 0x13x13 ≤ 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

### 7 Location Addition

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

3 19 3_var_snapshot: x14 + x14 ≤ 0x14x14 ≤ 0x13 + x13 ≤ 0x13x13 ≤ 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 Location Addition

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

4* 28 4: x14 + x14 ≤ 0x14x14 ≤ 0x13 + x13 ≤ 0x13x13 ≤ 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

### 9 Location Addition

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

4 26 4_var_snapshot: x14 + x14 ≤ 0x14x14 ≤ 0x13 + x13 ≤ 0x13x13 ≤ 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

### 10 SCC Decomposition

We consider subproblems for each of the 3 SCC(s) of the program graph.

### 10.1 SCC Subproblem 1/3

Here we consider the SCC { 4, 4_var_snapshot, 4* }.

### 10.1.1 Transition Removal

We remove transition 9 using the following ranking functions, which are bounded by 2.

 4: 1 + 3⋅arg1 4_var_snapshot: 3⋅arg1 4*: 2 + 3⋅arg1
Hints:
 26 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0] ] 28 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0] ] 9 lexStrict[ [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] , [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] ]

### 10.1.2 Transition Removal

We remove transitions 26, 28 using the following ranking functions, which are bounded by −2.

 4: −1 4_var_snapshot: −2 4*: 0
Hints:
 26 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] ] 28 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] ]

### 10.1.3 Splitting Cut-Point Transitions

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

### 10.1.3.1 Cut-Point Subproblem 1/1

Here we consider cut-point transition 25.

### 10.1.3.1.1 Splitting Cut-Point Transitions

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

### 10.2 SCC Subproblem 2/3

Here we consider the SCC { 3, 3_var_snapshot, 3* }.

### 10.2.1 Transition Removal

We remove transitions 6, 7 using the following ranking functions, which are bounded by 5.

 3: 1 + 3⋅arg2 3_var_snapshot: 3⋅arg2 3*: 2 + 3⋅arg2
Hints:
 19 lexWeak[ [0, 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] ] 21 lexWeak[ [0, 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] ] 6 lexStrict[ [0, 0, 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, 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, 0] ] 7 lexStrict[ [0, 0, 0, 0, 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, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

### 10.2.2 Transition Removal

We remove transitions 19, 21 using the following ranking functions, which are bounded by −2.

 3: −1 3_var_snapshot: −2⋅arg1 3*: 0
Hints:
 19 lexStrict[ [0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 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, 0, 0] ] 21 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] ]

### 10.2.3 Splitting Cut-Point Transitions

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

### 10.2.3.1 Cut-Point Subproblem 1/1

Here we consider cut-point transition 18.

### 10.2.3.1.1 Splitting Cut-Point Transitions

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

### 10.3 SCC Subproblem 3/3

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

### 10.3.1 Transition Removal

We remove transitions 3, 4 using the following ranking functions, which are bounded by 7.

 1: 1 + 4⋅arg4 1_var_snapshot: 4⋅arg4 1*: 2 + 4⋅arg4
Hints:
 12 lexWeak[ [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] ] 14 lexWeak[ [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] ] 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, 4, 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, 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] ] 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, 4, 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, 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] ]

### 10.3.2 Transition Removal

We remove transition 12 using the following ranking functions, which are bounded by −1.

 1: 0 1_var_snapshot: − arg3P 1*: arg1
Hints:
 12 lexStrict[ [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, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 14 lexWeak[ [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] ]

### 10.3.3 Transition Removal

We remove transition 14 using the following ranking functions, which are bounded by 0.

 1: 0 1_var_snapshot: 0 1*: arg3
Hints:
 14 lexStrict[ [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, 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] ]

### 10.3.4 Splitting Cut-Point Transitions

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

### 10.3.4.1 Cut-Point Subproblem 1/1

Here we consider cut-point transition 11.

### 10.3.4.1.1 Splitting Cut-Point Transitions

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

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