# LTS Termination Proof

by AProVE

## Input

Integer Transition System
• Initial Location: f1_0_main_ConstantStackPush, f74_0_factorial_GE, __init
• Transitions: (pre-variables and post-variables)  f1_0_main_ConstantStackPush 1 f74_0_factorial_GE: x1 = _arg1 ∧ x1 = _arg1P ∧ 10 = _arg1P f74_0_factorial_GE 2 f74_0_factorial_GE: x1 = _x ∧ x1 = _x1 ∧ _x − 1 = _x1 ∧ _x − 1 ≤ _x − 1 ∧ −1 ≤ _x − 1 __init 3 f1_0_main_ConstantStackPush: x1 = _x2 ∧ x1 = _x3 ∧ 0 ≤ 0

## Proof

### 1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 f1_0_main_ConstantStackPush f1_0_main_ConstantStackPush f1_0_main_ConstantStackPush: x1 = x1 f74_0_factorial_GE f74_0_factorial_GE f74_0_factorial_GE: x1 = x1 __init __init __init: x1 = x1
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 { f74_0_factorial_GE }.

### 2.1.1 Transition Removal

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

 f74_0_factorial_GE: x1

### 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)