# LTS Termination Proof

by AProVE

## Input

Integer Transition System
• Initial Location: f1_0_main_New, f83_0_main_GE, __init
• Transitions: (pre-variables and post-variables)  f1_0_main_New 1 f83_0_main_GE: x1 = _arg1 ∧ x1 = _arg1P ∧ 0 = _arg1P f83_0_main_GE 2 f83_0_main_GE: x1 = _x ∧ x1 = _x1 ∧ _x + 2 = _x1 ∧ _x ≤ 19 ∧ _x ≤ 10 f83_0_main_GE 3 f83_0_main_GE: x1 = _x2 ∧ x1 = _x3 ∧ _x2 + 2 = _x3 ∧ _x2 ≤ 19 ∧ 10 ≤ _x2 − 1 __init 4 f1_0_main_New: x1 = _x4 ∧ x1 = _x5 ∧ 0 ≤ 0

## Proof

### 1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 f1_0_main_New f1_0_main_New f1_0_main_New: x1 = x1 f83_0_main_GE f83_0_main_GE f83_0_main_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 { f83_0_main_GE }.

### 2.1.1 Transition Removal

We remove transitions 2, 3 using the following ranking functions, which are bounded by −38.

 f83_0_main_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)