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 = x5
l4 l4 l4: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5
l7 l7 l7: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5
l6 l6 l6: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5
l10 l10 l10: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5
l8 l8 l8: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5
l1 l1 l1: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5
l0 l0 l0: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5
l2 l2 l2: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5
l9 l9 l9: 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 { l5, l7, l6, l1, l8, l0 }.

2.1.1 Transition Removal

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

l0: x1x2
l1: −1 + x1x2
l7: −1 + x1x2
l6: −1 + x1x2
l5: −2 + x1x2
l8: −2 + x1x2

2.1.2 Transition Removal

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

l0: −1
l1: −1
l7: −1
l6: −1
l5: −1
l8: 0

2.1.3 Transition Removal

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

l0: −1
l1: −1
l7: −1
l6: −1
l5: 0

2.1.4 Transition Removal

We remove transition 11 using the following ranking functions, which are bounded by 0.

l0: x1
l1: x1
l7: −1 + x1
l6: −1 + x1

2.1.5 Transition Removal

We remove transitions 1, 8, 5 using the following ranking functions, which are bounded by 0.

l0: 0
l1: −1
l7: 1
l6: 2

2.1.6 Trivial Cooperation Program

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

Tool configuration

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