by AProVE
| l0 | 1 | l1: | x1 = _Result_4HAT0 ∧ x2 = _x_5HAT0 ∧ x3 = _y_6HAT0 ∧ x1 = _Result_4HATpost ∧ x2 = _x_5HATpost ∧ x3 = _y_6HATpost ∧ _y_6HAT0 = _y_6HATpost ∧ _x_5HAT0 = _x_5HATpost ∧ _Result_4HATpost = _Result_4HATpost ∧ _x_5HAT0 ≤ 0 | |
| l0 | 2 | l1: | x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x1 = _x3 ∧ x2 = _x4 ∧ x3 = _x5 ∧ _x2 = _x5 ∧ _x1 = _x4 ∧ _x3 = _x3 ∧ _x2 ≤ 0 ∧ 0 ≤ −1 + _x1 | |
| l0 | 3 | l1: | x1 = _x6 ∧ x2 = _x7 ∧ x3 = _x8 ∧ x1 = _x9 ∧ x2 = _x10 ∧ x3 = _x11 ∧ _x8 = _x11 ∧ _x7 = _x10 ∧ _x9 = _x9 ∧ _x7 + _x8 ≤ 0 ∧ 0 ≤ −1 + _x8 ∧ 0 ≤ −1 + _x7 | |
| l0 | 4 | l2: | x1 = _x12 ∧ x2 = _x13 ∧ x3 = _x14 ∧ x1 = _x15 ∧ x2 = _x16 ∧ x3 = _x17 ∧ _x12 = _x15 ∧ _x17 = 1 + _x16 ∧ _x16 = −2 + _x14 ∧ 0 ≤ −1 + _x13 + _x14 ∧ 0 ≤ −1 + _x14 ∧ 0 ≤ −1 + _x13 | |
| l2 | 5 | l0: | x1 = _x18 ∧ x2 = _x19 ∧ x3 = _x20 ∧ x1 = _x21 ∧ x2 = _x22 ∧ x3 = _x23 ∧ _x20 = _x23 ∧ _x19 = _x22 ∧ _x18 = _x21 | |
| l3 | 6 | l0: | x1 = _x24 ∧ x2 = _x25 ∧ x3 = _x26 ∧ x1 = _x27 ∧ x2 = _x28 ∧ x3 = _x29 ∧ _x26 = _x29 ∧ _x25 = _x28 ∧ _x24 = _x27 | |
| l4 | 7 | l3: | x1 = _x30 ∧ x2 = _x31 ∧ x3 = _x32 ∧ x1 = _x33 ∧ x2 = _x34 ∧ x3 = _x35 ∧ _x32 = _x35 ∧ _x31 = _x34 ∧ _x30 = _x33 |
| l4 | l4 | : | x1 = x1 ∧ x2 = x2 ∧ x3 = x3 |
| l3 | l3 | : | x1 = x1 ∧ x2 = x2 ∧ x3 = x3 |
| l0 | l0 | : | x1 = x1 ∧ x2 = x2 ∧ x3 = x3 |
| l2 | l2 | : | 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.
| : | −1 + x3 |
| : | −1 + x3 |
We remove transition using the following ranking functions, which are bounded by 0.
| : | 0 |
| : | −1 |
There are no more "sharp" transitions in the cooperation program. Hence the cooperation termination is proved.