# LTS Termination Proof

by AProVE

## Input

Integer Transition System
• Initial Location: l5, l4, l3, l0, l2
• Transitions: (pre-variables and post-variables)  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 ∧ −20 + _x_5HAT0 ≤ 0 l0 2 l2: x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x1 = _x3 ∧ x2 = _x4 ∧ x3 = _x5 ∧ _x2 = _x5 ∧ _x1 = _x4 ∧ _x = _x3 ∧ 0 ≤ −21 + _x1 l3 3 l0: x1 = _x6 ∧ x2 = _x7 ∧ x3 = _x8 ∧ x1 = _x9 ∧ x2 = _x10 ∧ x3 = _x11 ∧ _x8 = _x11 ∧ _x7 = _x10 ∧ _x6 = _x9 l2 4 l0: x1 = _x12 ∧ x2 = _x13 ∧ x3 = _x14 ∧ x1 = _x15 ∧ x2 = _x16 ∧ x3 = _x17 ∧ _x14 = _x17 ∧ _x12 = _x15 ∧ _x16 = −1 + _x13 ∧ −30 + _x14 ≤ 0 l2 5 l4: x1 = _x18 ∧ x2 = _x19 ∧ x3 = _x20 ∧ x1 = _x21 ∧ x2 = _x22 ∧ x3 = _x23 ∧ _x19 = _x22 ∧ _x18 = _x21 ∧ _x23 = −1 + _x20 ∧ 0 ≤ −31 + _x20 l4 6 l2: x1 = _x24 ∧ x2 = _x25 ∧ x3 = _x26 ∧ x1 = _x27 ∧ x2 = _x28 ∧ x3 = _x29 ∧ _x26 = _x29 ∧ _x25 = _x28 ∧ _x24 = _x27 l5 7 l3: x1 = _x30 ∧ x2 = _x31 ∧ x3 = _x32 ∧ x1 = _x33 ∧ x2 = _x34 ∧ x3 = _x35 ∧ _x32 = _x35 ∧ _x31 = _x34 ∧ _x30 = _x33

## Proof

### 1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 l5 l5 l5: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 l4 l4 l4: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 l3 l3 l3: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 l0 l0 l0: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 l2 l2 l2: x1 = x1 ∧ x2 = x2 ∧ x3 = x3
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 { l4, l0, l2 }.

### 2.1.1 Transition Removal

We remove transition 5 using the following ranking functions, which are bounded by 0.

 l0: −31 + x3 l2: −31 + x3 l4: −31 + x3

### 2.1.2 Transition Removal

We remove transition 6 using the following ranking functions, which are bounded by 0.

 l0: −1 l2: −1 l4: 0

### 2.1.3 Transition Removal

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

 l0: x2 l2: −1 + x2

### 2.1.4 Transition Removal

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

 l2: 0 l0: −1

### 2.1.5 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 % (7 real / 0 unknown / 0 assumptions / 7 total proof steps)