by AProVE
f1_0_main_ConstantStackPush | 1 | f149_0_doSum_LT: | x1 = _arg1 ∧ x1 = _arg1P ∧ 10 = _arg1P | |
f149_0_doSum_LT | 2 | f163_0_factorial_GT: | x1 = _x ∧ x1 = _x1 ∧ _x = _x1 ∧ −1 ≤ _x − 1 | |
f149_0_doSum_LT | 3 | f149_0_doSum_LT: | x1 = _x2 ∧ x1 = _x3 ∧ _x2 − 1 = _x3 ∧ −1 ≤ _x2 − 1 | |
f163_0_factorial_GT | 4 | f163_0_factorial_GT: | x1 = _x4 ∧ x1 = _x5 ∧ _x4 − 1 = _x5 ∧ _x4 − 1 ≤ _x4 − 1 ∧ 0 ≤ _x4 − 1 | |
__init | 5 | f1_0_main_ConstantStackPush: | x1 = _x6 ∧ x1 = _x7 ∧ 0 ≤ 0 |
f163_0_factorial_GT | f163_0_factorial_GT | : | x1 = x1 |
f149_0_doSum_LT | f149_0_doSum_LT | : | x1 = x1 |
f1_0_main_ConstantStackPush | f1_0_main_ConstantStackPush | : | x1 = x1 |
__init | __init | : | x1 = x1 |
We consider subproblems for each of the 2 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.
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.