by AProVE
| f1_0_main_Load | 1 | f263_0_main_GT: | x1 = _arg1 ∧ x2 = _arg2 ∧ x3 = _arg3 ∧ x1 = _arg1P ∧ x2 = _arg2P ∧ x3 = _arg3P ∧ 0 ≤ _arg1 − 1 ∧ −1 ≤ _arg1P − 1 ∧ −1 ≤ _arg2P − 1 ∧ −1 ≤ _arg2 − 1 ∧ −1 ≤ _arg3P − 1 | |
| f263_0_main_GT | 2 | f263_0_main_GT: | x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x1 = _x3 ∧ x2 = _x4 ∧ x3 = _x5 ∧ _x = _x5 ∧ _x1 − 1 = _x4 ∧ _x + 1 = _x3 ∧ _x2 ≤ 100 ∧ −1 ≤ _x − 1 ∧ _x ≤ _x1 | |
| __init | 3 | f1_0_main_Load: | x1 = _x6 ∧ x2 = _x7 ∧ x3 = _x8 ∧ x1 = _x9 ∧ x2 = _x10 ∧ x3 = _x11 ∧ 0 ≤ 0 |
| f263_0_main_GT | f263_0_main_GT | : | x1 = x1 ∧ x2 = x2 ∧ x3 = x3 |
| f1_0_main_Load | f1_0_main_Load | : | x1 = x1 ∧ x2 = x2 ∧ x3 = x3 |
| __init | __init | : | x1 = x1 ∧ x2 = x2 ∧ x3 = x3 |
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.
| : | 100 − 51⋅x1 + 51⋅x2 − x3 |
There are no more "sharp" transitions in the cooperation program. Hence the cooperation termination is proved.