by AProVE
| l0 | 1 | l1: | x1 = _i5HAT0 ∧ x2 = _length4HAT0 ∧ x3 = _sHAT0 ∧ x4 = _tmpHAT0 ∧ x5 = _tmp___08HAT0 ∧ x1 = _i5HATpost ∧ x2 = _length4HATpost ∧ x3 = _sHATpost ∧ x4 = _tmpHATpost ∧ x5 = _tmp___08HATpost ∧ _tmp___08HAT0 = _tmp___08HATpost ∧ _tmpHAT0 = _tmpHATpost ∧ _sHAT0 = _sHATpost ∧ _length4HAT0 = _length4HATpost ∧ _i5HAT0 = _i5HATpost ∧ _length4HAT0 ≤ _i5HAT0 | |
| l0 | 2 | l2: | x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x4 = _x3 ∧ x5 = _x4 ∧ x1 = _x5 ∧ x2 = _x6 ∧ x3 = _x7 ∧ x4 = _x8 ∧ x5 = _x9 ∧ _x3 = _x8 ∧ _x2 = _x7 ∧ _x1 = _x6 ∧ _x5 = 1 + _x ∧ _x9 = _x9 ∧ 1 + _x ≤ _x1 | |
| l2 | 3 | l0: | x1 = _x10 ∧ x2 = _x11 ∧ x3 = _x12 ∧ x4 = _x13 ∧ x5 = _x14 ∧ x1 = _x15 ∧ x2 = _x16 ∧ x3 = _x17 ∧ x4 = _x18 ∧ x5 = _x19 ∧ _x14 = _x19 ∧ _x13 = _x18 ∧ _x12 = _x17 ∧ _x11 = _x16 ∧ _x10 = _x15 | |
| l3 | 4 | l2: | x1 = _x20 ∧ x2 = _x21 ∧ x3 = _x22 ∧ x4 = _x23 ∧ x5 = _x24 ∧ x1 = _x25 ∧ x2 = _x26 ∧ x3 = _x27 ∧ x4 = _x28 ∧ x5 = _x29 ∧ _x24 = _x29 ∧ _x25 = 0 ∧ _x26 = 10 ∧ _x27 = _x28 ∧ _x28 = _x28 | |
| l4 | 5 | l3: | x1 = _x30 ∧ x2 = _x31 ∧ x3 = _x32 ∧ x4 = _x33 ∧ x5 = _x34 ∧ x1 = _x35 ∧ x2 = _x36 ∧ x3 = _x37 ∧ x4 = _x38 ∧ x5 = _x39 ∧ _x34 = _x39 ∧ _x33 = _x38 ∧ _x32 = _x37 ∧ _x31 = _x36 ∧ _x30 = _x35 |
| l4 | l4 | : | x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 |
| l3 | l3 | : | x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 |
| l0 | l0 | : | x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 |
| l2 | l2 | : | x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 |
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 − x1 + x2 |
| : | −1 − x1 + x2 |
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.