# LTS Termination Proof

by AProVE

## Input

Integer Transition System
• Initial Location: f1_0_main_Load, f160_0_main_LE, __init
• Transitions: (pre-variables and post-variables)  f1_0_main_Load 1 f160_0_main_LE: x1 = _arg1 ∧ x2 = _arg2 ∧ x1 = _arg1P ∧ x2 = _arg2P ∧ 1 = _arg2P ∧ 0 ≤ _arg1 − 1 ∧ −1 ≤ _arg1P − 1 ∧ −1 ≤ _arg2 − 1 f160_0_main_LE 2 f160_0_main_LE: x1 = _x ∧ x2 = _x1 ∧ x1 = _x2 ∧ x2 = _x3 ∧ _x1 + 1 = _x3 ∧ _x + 11 = _x2 ∧ 0 ≤ _x1 − 1 ∧ _x ≤ 100 f160_0_main_LE 3 f160_0_main_LE: x1 = _x4 ∧ x2 = _x5 ∧ x1 = _x6 ∧ x2 = _x7 ∧ _x5 − 1 = _x7 ∧ _x4 − 10 = _x6 ∧ 0 ≤ _x5 − 1 ∧ 100 ≤ _x4 − 1 __init 4 f1_0_main_Load: x1 = _x8 ∧ x2 = _x9 ∧ x1 = _x10 ∧ x2 = _x11 ∧ 0 ≤ 0

## Proof

### 1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 f1_0_main_Load f1_0_main_Load f1_0_main_Load: x1 = x1 ∧ x2 = x2 f160_0_main_LE f160_0_main_LE f160_0_main_LE: x1 = x1 ∧ x2 = x2 __init __init __init: x1 = x1 ∧ x2 = x2
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 { f160_0_main_LE }.

### 2.1.1 Transition Removal

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

 f160_0_main_LE: 90 − x1 + 10⋅x2

### 2.1.2 Transition Removal

We remove transition 3 using the following ranking functions, which are bounded by 0.

 f160_0_main_LE: x2

### 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 % (5 real / 0 unknown / 0 assumptions / 5 total proof steps)