by AProVE
f1_0_main_ConstantStackPush | 1 | f137_0_test_GE: | x1 = _arg1 ∧ x2 = _arg2 ∧ x1 = _arg1P ∧ x2 = _arg2P ∧ 10 = _arg2P ∧ 0 = _arg1P | |
f137_0_test_GE | 2 | f137_0_test_GE: | x1 = _x ∧ x2 = _x1 ∧ x1 = _x2 ∧ x2 = _x3 ∧ _x = _x3 ∧ 0 = _x2 ∧ 0 ≤ _x1 − 1 ∧ _x ≤ _x1 − 1 | |
f137_0_test_GE | 3 | f137_0_test_GE: | x1 = _x4 ∧ x2 = _x5 ∧ x1 = _x6 ∧ x2 = _x7 ∧ _x5 = _x7 ∧ _x4 + 1 = _x6 ∧ 0 ≤ _x5 − 1 ∧ _x4 ≤ _x5 − 1 | |
__init | 4 | f1_0_main_ConstantStackPush: | x1 = _x8 ∧ x2 = _x9 ∧ x1 = _x10 ∧ x2 = _x11 ∧ 0 ≤ 0 |
f137_0_test_GE | f137_0_test_GE | : | x1 = x1 ∧ x2 = x2 |
f1_0_main_ConstantStackPush | f1_0_main_ConstantStackPush | : | 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.: | x2 |
We remove transition
using the following ranking functions, which are bounded by 0.: | − x1 + x2 |
There are no more "sharp" transitions in the cooperation program. Hence the cooperation termination is proved.