by AProVE
l0 | 1 | l1: | x1 = _Result_4HAT0 ∧ x2 = _b_7HAT0 ∧ x3 = _x_5HAT0 ∧ x4 = _y_6HAT0 ∧ x1 = _Result_4HATpost ∧ x2 = _b_7HATpost ∧ x3 = _x_5HATpost ∧ x4 = _y_6HATpost ∧ _x_5HAT0 = _x_5HATpost ∧ _Result_4HAT0 = _Result_4HATpost ∧ _y_6HATpost = −1 + _y_6HAT0 ∧ _b_7HATpost = 0 ∧ 1 − _b_7HAT0 ≤ 0 ∧ 0 ≤ −1 − _x_5HAT0 + _y_6HAT0 | |
l0 | 2 | l2: | x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x4 = _x3 ∧ x1 = _x4 ∧ x2 = _x5 ∧ x3 = _x6 ∧ x4 = _x7 ∧ _x3 = _x7 ∧ _x2 = _x6 ∧ _x1 = _x5 ∧ _x4 = _x4 ∧ − _x2 + _x3 ≤ 0 | |
l3 | 3 | l1: | x1 = _x8 ∧ x2 = _x9 ∧ x3 = _x10 ∧ x4 = _x11 ∧ x1 = _x12 ∧ x2 = _x13 ∧ x3 = _x14 ∧ x4 = _x15 ∧ _x11 = _x15 ∧ _x10 = _x14 ∧ _x8 = _x12 ∧ _x13 = 0 | |
l1 | 4 | l0: | x1 = _x16 ∧ x2 = _x17 ∧ x3 = _x18 ∧ x4 = _x19 ∧ x1 = _x20 ∧ x2 = _x21 ∧ x3 = _x22 ∧ x4 = _x23 ∧ _x19 = _x23 ∧ _x16 = _x20 ∧ _x22 = 1 + _x18 ∧ _x21 = 1 ∧ 0 ≤ − _x17 ∧ 0 ≤ −1 − _x18 + _x19 | |
l1 | 5 | l2: | x1 = _x24 ∧ x2 = _x25 ∧ x3 = _x26 ∧ x4 = _x27 ∧ x1 = _x28 ∧ x2 = _x29 ∧ x3 = _x30 ∧ x4 = _x31 ∧ _x27 = _x31 ∧ _x26 = _x30 ∧ _x25 = _x29 ∧ _x28 = _x28 ∧ − _x26 + _x27 ≤ 0 | |
l4 | 6 | l3: | x1 = _x32 ∧ x2 = _x33 ∧ x3 = _x34 ∧ x4 = _x35 ∧ x1 = _x36 ∧ x2 = _x37 ∧ x3 = _x38 ∧ x4 = _x39 ∧ _x35 = _x39 ∧ _x34 = _x38 ∧ _x33 = _x37 ∧ _x32 = _x36 |
l4 | l4 | : | x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 |
l1 | l1 | : | x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 |
l3 | l3 | : | x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 |
l0 | l0 | : | x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 |
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.: | − x3 + x4 |
: | − x3 + x4 |
There are no more "sharp" transitions in the cooperation program. Hence the cooperation termination is proved.