LTS Termination Proof

by AProVE

Input

Integer Transition System

Proof

1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
f411_0_slide95_EQ' f411_0_slide95_EQ' f411_0_slide95_EQ': x1 = x1x2 = x2x3 = x3
f411_0_slide95_EQ f411_0_slide95_EQ f411_0_slide95_EQ: x1 = x1x2 = x2x3 = x3
f1_0_main_Load f1_0_main_Load f1_0_main_Load: x1 = x1x2 = x2x3 = x3
f196_0_create_LE f196_0_create_LE f196_0_create_LE: x1 = x1x2 = x2x3 = x3
__init __init __init: x1 = x1x2 = x2x3 = x3
f234_0_slide95_FieldAccess f234_0_slide95_FieldAccess f234_0_slide95_FieldAccess: x1 = x1x2 = x2x3 = x3
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 11 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 { f411_0_slide95_EQ', f411_0_slide95_EQ }.

2.2.1 Transition Removal

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

f411_0_slide95_EQ: 2⋅x2
f411_0_slide95_EQ': 2⋅x2 − 1

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