# LTS Termination Proof

by AProVE

## Input

Integer Transition System
• Initial Location: f1_0_main_Load, f278_0_main_LE, __init
• Transitions: (pre-variables and post-variables)  f1_0_main_Load 1 f278_0_main_LE: x1 = _arg1 ∧ x2 = _arg2 ∧ x3 = _arg3 ∧ x4 = _arg4 ∧ x1 = _arg1P ∧ x2 = _arg2P ∧ x3 = _arg3P ∧ x4 = _arg4P ∧ _arg2P = _arg4P ∧ _arg2P = _arg3P ∧ 0 ≤ _arg1 − 1 ∧ −1 ≤ _arg1P − 1 ∧ −1 ≤ _arg2 − 1 ∧ −1 ≤ _arg2P − 1 f278_0_main_LE 2 f278_0_main_LE: x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x4 = _x3 ∧ x1 = _x4 ∧ x2 = _x5 ∧ x3 = _x6 ∧ x4 = _x7 ∧ _x2 = _x7 ∧ _x2 = _x6 ∧ _x2 = _x5 ∧ _x = _x4 ∧ _x2 = _x3 ∧ 0 = _x1 ∧ 0 ≤ _x2 − 1 f278_0_main_LE 3 f278_0_main_LE: x1 = _x8 ∧ x2 = _x9 ∧ x3 = _x10 ∧ x4 = _x11 ∧ x1 = _x12 ∧ x2 = _x13 ∧ x3 = _x14 ∧ x4 = _x15 ∧ _x10 = _x15 ∧ _x10 = _x14 ∧ _x9 − 1 = _x13 ∧ _x8 − 1 = _x12 ∧ _x10 = _x11 ∧ 0 ≤ _x9 − 1 ∧ 0 ≤ _x10 − 1 ∧ 0 ≤ _x8 − 1 __init 4 f1_0_main_Load: x1 = _x16 ∧ x2 = _x17 ∧ x3 = _x18 ∧ x4 = _x19 ∧ x1 = _x20 ∧ x2 = _x21 ∧ x3 = _x22 ∧ x4 = _x23 ∧ 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 ∧ x3 = x3 ∧ x4 = x4 f278_0_main_LE f278_0_main_LE f278_0_main_LE: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 __init __init __init: 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 { f278_0_main_LE }.

### 2.1.1 Transition Removal

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

 f278_0_main_LE: −1 + x1

### 2.1.2 Transition Removal

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

 f278_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)