# LTS Termination Proof

by AProVE

## Input

Integer Transition System
• Initial Location: f224_0_main_LT, f1_0_main_Load, __init
• Transitions: (pre-variables and post-variables)  f1_0_main_Load 1 f224_0_main_LT: x1 = _arg1 ∧ x2 = _arg2 ∧ x3 = _arg3 ∧ x1 = _arg1P ∧ x2 = _arg2P ∧ x3 = _arg3P ∧ 0 ≤ _arg1 − 1 ∧ −1 ≤ _arg2P − 1 ∧ −1 ≤ _arg3P − 1 ∧ 0 ≤ _arg1P − 1 ∧ −1 ≤ _arg2 − 1 f224_0_main_LT 2 f224_0_main_LT: x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x1 = _x3 ∧ x2 = _x4 ∧ x3 = _x5 ∧ _x2 + _x = _x5 ∧ _x1 = _x4 ∧ _x = _x3 ∧ 0 ≤ _x − 1 ∧ −1 ≤ _x2 − 1 ∧ _x2 ≤ _x1 __init 3 f1_0_main_Load: x1 = _x6 ∧ x2 = _x7 ∧ x3 = _x8 ∧ x1 = _x9 ∧ x2 = _x10 ∧ x3 = _x11 ∧ 0 ≤ 0

## Proof

### 1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 f224_0_main_LT f224_0_main_LT f224_0_main_LT: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 f1_0_main_Load f1_0_main_Load f1_0_main_Load: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 __init __init __init: 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 { f224_0_main_LT }.

### 2.1.1 Transition Removal

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

 f224_0_main_LT: x2 − x3

### 2.1.2 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 % (4 real / 0 unknown / 0 assumptions / 4 total proof steps)