by AProVE
f1_0_main_Load | 1 | f160_0_main_LE: | x1 = _arg1 ∧ x2 = _arg2 ∧ x1 = _arg1P ∧ x2 = _arg2P ∧ 1 = _arg2P ∧ 0 ≤ _arg1 − 1 ∧ −1 ≤ _arg1P − 1 ∧ −1 ≤ _arg2 − 1 | |
f160_0_main_LE | 2 | f160_0_main_LE: | x1 = _x ∧ x2 = _x1 ∧ x1 = _x2 ∧ x2 = _x3 ∧ _x1 + 1 = _x3 ∧ _x + 11 = _x2 ∧ 0 ≤ _x1 − 1 ∧ _x ≤ 100 | |
f160_0_main_LE | 3 | f160_0_main_LE: | x1 = _x4 ∧ x2 = _x5 ∧ x1 = _x6 ∧ x2 = _x7 ∧ _x5 − 1 = _x7 ∧ _x4 − 10 = _x6 ∧ 0 ≤ _x5 − 1 ∧ 100 ≤ _x4 − 1 | |
__init | 4 | f1_0_main_Load: | x1 = _x8 ∧ x2 = _x9 ∧ x1 = _x10 ∧ x2 = _x11 ∧ 0 ≤ 0 |
f1_0_main_Load | f1_0_main_Load | : | x1 = x1 ∧ x2 = x2 |
f160_0_main_LE | f160_0_main_LE | : | 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.: | 90 − x1 + 10⋅x2 |
We remove transition
using the following ranking functions, which are bounded by 0.: | x2 |
There are no more "sharp" transitions in the cooperation program. Hence the cooperation termination is proved.