LTS Termination Proof

by AProVE

Input

Integer Transition System

Proof

1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
l5 l5 l5: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6
l4 l4 l4: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6
l1 l1 l1: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6
l3 l3 l3: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6
l0 l0 l0: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6
l2 l2 l2: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6
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 { l4, l1, l2 }.

2.1.1 Transition Removal

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

l1: −1 + x1
l2: −1 + x1x4
l4: −1 + x1x4

2.1.2 Transition Removal

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

l2: 0
l1: −1
l4: 0

2.1.3 Transition Removal

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

l4: −1 + x4
l2: −1 + x4

2.1.4 Transition Removal

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

l4: 0
l2: −1

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