LTS Termination Proof

by AProVE

Input

Integer Transition System

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 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7
f292_0_slide88_EQ' f292_0_slide88_EQ' f292_0_slide88_EQ': x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7
f196_0_create_LE f196_0_create_LE f196_0_create_LE: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7
f234_0_slide88_FieldAccess f234_0_slide88_FieldAccess f234_0_slide88_FieldAccess: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7
__init __init __init: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7
f292_0_slide88_EQ f292_0_slide88_EQ f292_0_slide88_EQ: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7
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 { f196_0_create_LE }.

2.1.1 Transition Removal

We remove transition 9 using the following ranking functions, which are bounded by 0.

f196_0_create_LE: x1

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 { f292_0_slide88_EQ', f292_0_slide88_EQ }.

2.2.1 Transition Removal

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

f292_0_slide88_EQ: 1 + 2⋅x4
f292_0_slide88_EQ': 2⋅x4

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