by AProVE
l0 | 1 | l1: | x1 = _pHAT0 ∧ x2 = _xHAT0 ∧ x3 = _yHAT0 ∧ x1 = _pHATpost ∧ x2 = _xHATpost ∧ x3 = _yHATpost ∧ _yHAT0 = _yHATpost ∧ _xHAT0 = _xHATpost ∧ _pHATpost = 1 ∧ _xHAT0 ≤ 0 ∧ _yHAT0 ≤ 0 | |
l0 | 2 | l2: | x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x1 = _x3 ∧ x2 = _x4 ∧ x3 = _x5 ∧ _x1 = _x4 ∧ _x = _x3 ∧ _x5 = −1 + _x2 ∧ 1 ≤ _x2 | |
l2 | 3 | l0: | x1 = _x6 ∧ x2 = _x7 ∧ x3 = _x8 ∧ x1 = _x9 ∧ x2 = _x10 ∧ x3 = _x11 ∧ _x8 = _x11 ∧ _x7 = _x10 ∧ _x6 = _x9 | |
l3 | 4 | l2: | x1 = _x12 ∧ x2 = _x13 ∧ x3 = _x14 ∧ x1 = _x15 ∧ x2 = _x16 ∧ x3 = _x17 ∧ _x14 = _x17 ∧ _x13 = _x16 ∧ _x12 = _x15 ∧ _x13 ≤ 0 | |
l3 | 5 | l4: | x1 = _x18 ∧ x2 = _x19 ∧ x3 = _x20 ∧ x1 = _x21 ∧ x2 = _x22 ∧ x3 = _x23 ∧ _x20 = _x23 ∧ _x19 = _x22 ∧ _x21 = 1 + _x18 ∧ 1 ≤ _x19 | |
l5 | 6 | l3: | x1 = _x24 ∧ x2 = _x25 ∧ x3 = _x26 ∧ x1 = _x27 ∧ x2 = _x28 ∧ x3 = _x29 ∧ _x26 = _x29 ∧ _x25 = _x28 ∧ _x27 = 0 | |
l6 | 7 | l5: | x1 = _x30 ∧ x2 = _x31 ∧ x3 = _x32 ∧ x1 = _x33 ∧ x2 = _x34 ∧ x3 = _x35 ∧ _x32 = _x35 ∧ _x31 = _x34 ∧ _x30 = _x33 |
l5 | l5 | : | x1 = x1 ∧ x2 = x2 ∧ x3 = x3 |
l6 | l6 | : | 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.