# LTS Termination Proof

by AProVE

## Input

Integer Transition System
• Initial Location: l5, l4, l7, l6, l3, l0, l2
• Transitions: (pre-variables and post-variables)  l0 1 l1: x1 = _Result_4HAT0 ∧ x2 = _tmp_7HAT0 ∧ x3 = _x_5HAT0 ∧ x4 = _y_6HAT0 ∧ x1 = _Result_4HATpost ∧ x2 = _tmp_7HATpost ∧ x3 = _x_5HATpost ∧ x4 = _y_6HATpost ∧ _y_6HAT0 = _y_6HATpost ∧ _x_5HAT0 = _x_5HATpost ∧ _tmp_7HAT0 = _tmp_7HATpost ∧ _Result_4HATpost = _Result_4HATpost ∧ − _x_5HAT0 + _y_6HAT0 ≤ 0 l0 2 l2: x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x4 = _x3 ∧ x1 = _x4 ∧ x2 = _x5 ∧ x3 = _x6 ∧ x4 = _x7 ∧ _x2 = _x6 ∧ _x = _x4 ∧ _x7 = −1 + _x3 ∧ 0 ≤ _x5 ∧ _x5 ≤ 0 ∧ _x5 = _x5 ∧ 0 ≤ −1 − _x2 + _x3 l2 3 l0: x1 = _x8 ∧ x2 = _x9 ∧ x3 = _x10 ∧ x4 = _x11 ∧ x1 = _x12 ∧ x2 = _x13 ∧ x3 = _x14 ∧ x4 = _x15 ∧ _x11 = _x15 ∧ _x10 = _x14 ∧ _x9 = _x13 ∧ _x8 = _x12 l0 4 l4: x1 = _x16 ∧ x2 = _x17 ∧ x3 = _x18 ∧ x4 = _x19 ∧ x1 = _x20 ∧ x2 = _x21 ∧ x3 = _x22 ∧ x4 = _x23 ∧ _x19 = _x23 ∧ _x18 = _x22 ∧ _x16 = _x20 ∧ _x21 = _x21 ∧ 0 ≤ −1 − _x18 + _x19 l4 5 l5: x1 = _x24 ∧ x2 = _x25 ∧ x3 = _x26 ∧ x4 = _x27 ∧ x1 = _x28 ∧ x2 = _x29 ∧ x3 = _x30 ∧ x4 = _x31 ∧ _x27 = _x31 ∧ _x26 = _x30 ∧ _x25 = _x29 ∧ _x24 = _x28 ∧ 1 + _x25 ≤ 0 l4 6 l5: x1 = _x32 ∧ x2 = _x33 ∧ x3 = _x34 ∧ x4 = _x35 ∧ x1 = _x36 ∧ x2 = _x37 ∧ x3 = _x38 ∧ x4 = _x39 ∧ _x35 = _x39 ∧ _x34 = _x38 ∧ _x33 = _x37 ∧ _x32 = _x36 ∧ 1 ≤ _x33 l5 7 l3: x1 = _x40 ∧ x2 = _x41 ∧ x3 = _x42 ∧ x4 = _x43 ∧ x1 = _x44 ∧ x2 = _x45 ∧ x3 = _x46 ∧ x4 = _x47 ∧ _x43 = _x47 ∧ _x41 = _x45 ∧ _x40 = _x44 ∧ _x46 = 1 + _x42 l3 8 l0: x1 = _x48 ∧ x2 = _x49 ∧ x3 = _x50 ∧ x4 = _x51 ∧ x1 = _x52 ∧ x2 = _x53 ∧ x3 = _x54 ∧ x4 = _x55 ∧ _x51 = _x55 ∧ _x50 = _x54 ∧ _x49 = _x53 ∧ _x48 = _x52 l6 9 l0: x1 = _x56 ∧ x2 = _x57 ∧ x3 = _x58 ∧ x4 = _x59 ∧ x1 = _x60 ∧ x2 = _x61 ∧ x3 = _x62 ∧ x4 = _x63 ∧ _x59 = _x63 ∧ _x58 = _x62 ∧ _x57 = _x61 ∧ _x56 = _x60 l7 10 l6: x1 = _x64 ∧ x2 = _x65 ∧ x3 = _x66 ∧ x4 = _x67 ∧ x1 = _x68 ∧ x2 = _x69 ∧ x3 = _x70 ∧ x4 = _x71 ∧ _x67 = _x71 ∧ _x66 = _x70 ∧ _x65 = _x69 ∧ _x64 = _x68

## Proof

### 1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 l5 l5 l5: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 l4 l4 l4: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 l7 l7 l7: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 l6 l6 l6: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 l3 l3 l3: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 l0 l0 l0: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 l2 l2 l2: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
and for every transition t, a duplicate t is considered.

### 2 SCC Decomposition

We consider subproblems for each of the 1 SCC(s) of the program graph.

### 2.1 SCC Subproblem 1/1

Here we consider the SCC { l5, l4, l3, l0, l2 }.

### 2.1.1 Transition Removal

We remove transitions 2, 4 using the following ranking functions, which are bounded by 0.

 l0: −1 − x3 + x4 l2: −1 − x3 + x4 l3: −1 − x3 + x4 l5: −2 − x3 + x4 l4: −2 − x3 + x4

### 2.1.2 Transition Removal

We remove transitions 8, 7, 6, 5, 3 using the following ranking functions, which are bounded by 0.

 l3: 0 l0: −1 l5: 1 l4: 2 l2: 0

### 2.1.3 Trivial Cooperation Program

There are no more "sharp" transitions in the cooperation program. Hence the cooperation termination is proved.

## Tool configuration

AProVE

• version: AProVE Commit ID: unknown
• strategy: Statistics for single proof: 100.00 % (6 real / 0 unknown / 0 assumptions / 6 total proof steps)