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_ConstantStackPush f1_0_main_ConstantStackPush f1_0_main_ConstantStackPush: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5
f375_0_dif_GE f375_0_dif_GE f375_0_dif_GE: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5
f484_0_dif_GE f484_0_dif_GE f484_0_dif_GE: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5
__init __init __init: 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 1 SCC(s) of the program graph.

2.1 SCC Subproblem 1/1

Here we consider the SCC { f375_0_dif_GE, f484_0_dif_GE }.

2.1.1 Transition Removal

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

f375_0_dif_GE: 19 − x2
f484_0_dif_GE: 19 − x2x3

2.1.2 Transition Removal

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

f375_0_dif_GE: 19 − x2
f484_0_dif_GE: 18 − x2

2.1.3 Transition Removal

We remove transitions 4, 3 using the following ranking functions, which are bounded by 0.

f484_0_dif_GE: −1 + x3
f375_0_dif_GE: −1

2.1.4 Transition Removal

We remove transition 5 using the following ranking functions, which are bounded by 0.

f484_0_dif_GE: 19 + 0⋅x3x5

2.1.5 Trivial Cooperation Program

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

Tool configuration

AProVE