by AProVE
f1_0_main_Load | 1 | f1_0_main_Load': | x1 = _arg1 ∧ x2 = _arg2 ∧ x1 = _arg1P ∧ x2 = _arg2P ∧ 0 ≤ _arg2 − 1 ∧ −1 ≤ _x2 − 1 ∧ 0 ≤ _arg1 − 1 ∧ _arg1 = _arg1P ∧ _arg2 = _arg2P | |
f1_0_main_Load' | 2 | f126_0_test_LE: | x1 = _x ∧ x2 = _x1 ∧ x1 = _x3 ∧ x2 = _x4 ∧ 0 ≤ _x1 − 1 ∧ −1 ≤ _x5 − 1 ∧ 0 ≤ _x − 1 ∧ _x5 − 100⋅_x8 ≤ 99 ∧ 0 ≤ _x5 − 100⋅_x8 ∧ _x5 − 100⋅_x8 = _x3 | |
f126_0_test_LE | 3 | f126_0_test_LE: | x1 = _x9 ∧ x2 = _x10 ∧ x1 = _x11 ∧ x2 = _x12 ∧ _x9 − 1 = _x11 ∧ 0 ≤ _x9 − 1 | |
__init | 4 | f1_0_main_Load: | x1 = _x13 ∧ x2 = _x14 ∧ x1 = _x15 ∧ x2 = _x16 ∧ 0 ≤ 0 |
f126_0_test_LE | f126_0_test_LE | : | x1 = x1 ∧ x2 = x2 |
f1_0_main_Load' | f1_0_main_Load' | : | 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.