by AProVE
f1_0_main_Load | 1 | f157_0_log_LE: | x1 = _arg1 ∧ x2 = _arg2 ∧ x1 = _arg1P ∧ x2 = _arg2P ∧ 0 ≤ _arg1 − 1 ∧ −1 ≤ _arg1P − 1 ∧ −1 ≤ _arg2 − 1 | |
f157_0_log_LE | 2 | f157_0_log_LE': | x1 = _x ∧ x2 = _x1 ∧ x1 = _x2 ∧ x2 = _x3 ∧ 1 ≤ _x − 1 ∧ _x4 ≤ _x − 1 ∧ _x = _x2 | |
f157_0_log_LE' | 3 | f157_0_log_LE: | x1 = _x6 ∧ x2 = _x7 ∧ x1 = _x8 ∧ x2 = _x9 ∧ 0 ≤ _x6 − 2⋅_x8 ∧ _x6 − 2⋅_x8 ≤ 1 ∧ 1 ≤ _x6 − 1 ∧ _x8 ≤ _x6 − 1 | |
__init | 4 | f1_0_main_Load: | x1 = _x10 ∧ x2 = _x11 ∧ x1 = _x12 ∧ x2 = _x13 ∧ 0 ≤ 0 |
f157_0_log_LE' | f157_0_log_LE' | : | x1 = x1 ∧ x2 = x2 |
f1_0_main_Load | f1_0_main_Load | : | x1 = x1 ∧ x2 = x2 |
f157_0_log_LE | f157_0_log_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 transitions
, using the following ranking functions, which are bounded by 0.: | 2⋅x1 |
: | x1 + 1 |
There are no more "sharp" transitions in the cooperation program. Hence the cooperation termination is proved.