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 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l7 l7 l7: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l11 l11 l11: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l1 l1 l1: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l13 l13 l13: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l18 l18 l18: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l17 l17 l17: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l2 l2 l2: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l9 l9 l9: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l14 l14 l14: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l4 l4 l4: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l10 l10 l10: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l6 l6 l6: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l8 l8 l8: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l15 l15 l15: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l16 l16 l16: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l0 l0 l0: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l12 l12 l12: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
and for every transition t, a duplicate t is considered.

2 SCC Decomposition

We consider subproblems for each of the 8 SCC(s) of the program graph.

2.1 SCC Subproblem 1/8

Here we consider the SCC { l1, l0 }.

2.1.1 Transition Removal

We remove transition 24 using the following ranking functions, which are bounded by 0.

l0: −1 + x1 − x6
l1: −1 + x1 − x6

2.1.2 Transition Removal

We remove transition 1 using the following ranking functions, which are bounded by 0.

l0: 0
l1: −1

2.1.3 Trivial Cooperation Program

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

2.2 SCC Subproblem 2/8

Here we consider the SCC { l5, l6 }.

2.2.1 Transition Removal

We remove transition 21 using the following ranking functions, which are bounded by 0.

l5: 2⋅x1 − 2⋅x7 + 1
l6: 2⋅x1 − 2⋅x7

2.2.2 Transition Removal

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

l5: 0
l6: −1

2.2.3 Trivial Cooperation Program

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

2.3 SCC Subproblem 3/8

Here we consider the SCC { l10, l9 }.

2.3.1 Transition Removal

We remove transition 18 using the following ranking functions, which are bounded by 0.

l9: 2⋅x1 − 2⋅x2 + 1
l10: 2⋅x1 − 2⋅x2

2.3.2 Transition Removal

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

l9: 0
l10: −1

2.3.3 Trivial Cooperation Program

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

2.4 SCC Subproblem 4/8

Here we consider the SCC { l13, l14 }.

2.4.1 Transition Removal

We remove transition 15 using the following ranking functions, which are bounded by 0.

l13: 2⋅x1 − 2⋅x8 + 1
l14: 2⋅x1 − 2⋅x8

2.4.2 Transition Removal

We remove transition 10 using the following ranking functions, which are bounded by 0.

l13: 0
l14: −1

2.4.3 Trivial Cooperation Program

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

2.5 SCC Subproblem 5/8

Here we consider the SCC { l16, l15 }.

2.5.1 Transition Removal

We remove transition 12 using the following ranking functions, which are bounded by 0.

l16: −1 + x1 − x3
l15: −1 + x1 − x3

2.5.2 Transition Removal

We remove transition 13 using the following ranking functions, which are bounded by 0.

l16: 0
l15: −1

2.5.3 Trivial Cooperation Program

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

2.6 SCC Subproblem 6/8

Here we consider the SCC { l11, l12 }.

2.6.1 Transition Removal

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

l12: 2⋅x1 − 2⋅x4 + 1
l11: 2⋅x1 − 2⋅x4

2.6.2 Transition Removal

We remove transition 16 using the following ranking functions, which are bounded by 0.

l12: 0
l11: −1

2.6.3 Trivial Cooperation Program

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

2.7 SCC Subproblem 7/8

Here we consider the SCC { l7, l8 }.

2.7.1 Transition Removal

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

l8: 2⋅x1 − 2⋅x5 + 1
l7: 2⋅x1 − 2⋅x5

2.7.2 Transition Removal

We remove transition 19 using the following ranking functions, which are bounded by 0.

l8: 0
l7: −1

2.7.3 Trivial Cooperation Program

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

2.8 SCC Subproblem 8/8

Here we consider the SCC { l4, l2 }.

2.8.1 Transition Removal

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

l4: 2⋅x1 − 2⋅x8 + 1
l2: 2⋅x1 − 2⋅x8

2.8.2 Transition Removal

We remove transition 22 using the following ranking functions, which are bounded by 0.

l4: 0
l2: −1

2.8.3 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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