by T2Cert
| 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 | 
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.
| 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 | 
| 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 | 
| 1 | 11 | : | − 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 | : | − 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 | : | − 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 | 
We remove transitions , , , , , using the following ranking functions, which are bounded by −21.
| 5: | 0 | 
| 0: | 0 | 
| 1: | 0 | 
| 2: | 0 | 
| 3: | 0 | 
| 4: | 0 | 
| : | −7 | 
| : | −8 | 
| : | −9 | 
| : | −9 | 
| : | −9 | 
| : | −12 | 
| : | −13 | 
| : | −13 | 
| : | −13 | 
| : | −16 | 
| : | −16 | 
| : | −16 | 
| 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] ] | 
| 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] ] | |
| 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] ] | |
| 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] ] | |
| 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] ] | |
| 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] ] | |
| 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] ] | |
| 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] ] | |
| 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] ] | |
| 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] ] | |
| 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] ] | 
The following skip-transition is inserted and corresponding redirections w.r.t. the old location are performed.
14 : − 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
The following skip-transition is inserted and corresponding redirections w.r.t. the old location are performed.
12 : − 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
The following skip-transition is inserted and corresponding redirections w.r.t. the old location are performed.
21 : − 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
The following skip-transition is inserted and corresponding redirections w.r.t. the old location are performed.
19 : − 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
The following skip-transition is inserted and corresponding redirections w.r.t. the old location are performed.
28 : − 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
The following skip-transition is inserted and corresponding redirections w.r.t. the old location are performed.
26 : − 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
We consider subproblems for each of the 3 SCC(s) of the program graph.
Here we consider the SCC { , , }.
We remove transition using the following ranking functions, which are bounded by 2.
| : | 1 + 3⋅arg1 | 
| : | 3⋅arg1 | 
| : | 2 + 3⋅arg1 | 
| 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] ] | 
| 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] ] | 
We remove transitions 26, 28 using the following ranking functions, which are bounded by −2.
| : | −1 | 
| : | −2 | 
| : | 0 | 
| 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] ] | 
We consider 1 subproblems corresponding to sets of cut-point transitions as follows.
There remain no cut-point transition to consider. Hence the cooperation termination is trivial.
Here we consider the SCC { , , }.
We remove transitions , using the following ranking functions, which are bounded by 5.
| : | 1 + 3⋅arg2 | 
| : | 3⋅arg2 | 
| : | 2 + 3⋅arg2 | 
| 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] ] | 
| 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] ] | |
| 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] ] | 
We remove transitions 19, 21 using the following ranking functions, which are bounded by −2.
| : | −1 | 
| : | −2⋅arg1 | 
| : | 0 | 
| 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] ] | 
We consider 1 subproblems corresponding to sets of cut-point transitions as follows.
There remain no cut-point transition to consider. Hence the cooperation termination is trivial.
Here we consider the SCC { , , }.
We remove transitions , using the following ranking functions, which are bounded by 7.
| : | 1 + 4⋅arg4 | 
| : | 4⋅arg4 | 
| : | 2 + 4⋅arg4 | 
| 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] ] | 
| 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] ] | |
| 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] ] | 
We remove transition 12 using the following ranking functions, which are bounded by −1.
| : | 0 | 
| : | − arg3P | 
| : | arg1 | 
| 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] ] | 
We remove transition 14 using the following ranking functions, which are bounded by 0.
| : | 0 | 
| : | 0 | 
| : | arg3 | 
| 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] ] | 
We consider 1 subproblems corresponding to sets of cut-point transitions as follows.
There remain no cut-point transition to consider. Hence the cooperation termination is trivial.
T2Cert