by AProVE
f1_0_main_Load | 1 | f153_0_run_NULL: | x1 = _arg1 ∧ x2 = _arg2 ∧ x1 = _arg1P ∧ x2 = _arg2P ∧ 3 ≤ _arg1P − 1 ∧ 0 ≤ _arg1 − 1 ∧ −1 ≤ _arg2 − 1 ∧ _arg1P − 3 ≤ _arg1 | |
f153_0_run_NULL | 2 | f153_0_run_NULL: | x1 = _x ∧ x2 = _x1 ∧ x1 = _x2 ∧ x2 = _x3 ∧ −1 ≤ _x2 − 1 ∧ 0 ≤ _x − 1 ∧ _x2 + 1 ≤ _x | |
__init | 3 | f1_0_main_Load: | x1 = _x4 ∧ x2 = _x5 ∧ x1 = _x6 ∧ x2 = _x7 ∧ 0 ≤ 0 |
f153_0_run_NULL | f153_0_run_NULL | : | x1 = x1 ∧ x2 = x2 |
f1_0_main_Load | f1_0_main_Load | : | x1 = x1 ∧ x2 = x2 |
__init | __init | : | x1 = x1 ∧ x2 = x2 |
We consider subproblems for each of the 1 SCC(s) of the program graph.
Here we consider the SCC {
}.We remove transition
using the following ranking functions, which are bounded by 0.: | x1 |
There are no more "sharp" transitions in the cooperation program. Hence the cooperation termination is proved.