LTS Termination Proof

by AProVE

Input

Integer Transition System

Proof

1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
f1893_0_createTree_LE f1893_0_createTree_LE f1893_0_createTree_LE: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5
f1938_0_main_InvokeMethod f1938_0_main_InvokeMethod f1938_0_main_InvokeMethod: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5
f1821_0_duplicateRandomPath_NULL f1821_0_duplicateRandomPath_NULL f1821_0_duplicateRandomPath_NULL: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5
f1_0_main_Load f1_0_main_Load f1_0_main_Load: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5
f1989_0_duplicateRandomPath_NULL f1989_0_duplicateRandomPath_NULL f1989_0_duplicateRandomPath_NULL: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5
__init __init __init: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5
f456_0_createTree_Return f456_0_createTree_Return f456_0_createTree_Return: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5
and for every transition t, a duplicate t is considered.

2 SCC Decomposition

We consider subproblems for each of the 2 SCC(s) of the program graph.

2.1 SCC Subproblem 1/2

Here we consider the SCC { f1893_0_createTree_LE }.

2.1.1 Transition Removal

We remove transitions 4, 5, 6, 7, 8, 9 using the following ranking functions, which are bounded by 0.

f1893_0_createTree_LE: x3

2.1.2 Trivial Cooperation Program

There are no more "sharp" transitions in the cooperation program. Hence the cooperation termination is proved.

2.2 SCC Subproblem 2/2

Here we consider the SCC { f1821_0_duplicateRandomPath_NULL, f1989_0_duplicateRandomPath_NULL }.

2.2.1 Transition Removal

We remove transitions 12, 14, 15, 13 using the following ranking functions, which are bounded by 0.

f1821_0_duplicateRandomPath_NULL: 1 + x2
f1989_0_duplicateRandomPath_NULL: x1

2.2.2 Trivial Cooperation Program

There are no more "sharp" transitions in the cooperation program. Hence the cooperation termination is proved.

Tool configuration

AProVE