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

2 SCC Decomposition

We consider subproblems for each of the 4 SCC(s) of the program graph.

2.1 SCC Subproblem 1/4

Here we consider the SCC { l1, l13, l15, l0, l14 }.

2.1.1 Transition Removal

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

l0: −1 − x3 + x5
l1: −1 − x3 + x5
l14: −2 − x3 + x5
l15: −2 − x3 + x5
l13: −2 − x3 + x5

2.1.2 Transition Removal

We remove transitions 1, 17, 20, 15, 19, 18 using the following ranking functions, which are bounded by 0.

l0: 0
l1: −1
l14: 1
l15: 3
l13: 2

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 2/4

Here we consider the SCC { l7, l10, l11, l8, l12 }.

2.2.1 Transition Removal

We remove transition 14 using the following ranking functions, which are bounded by 0.

l7: −1 − x3 + x5
l8: −1 − x3 + x5
l12: −2 − x3 + x5
l11: −2 − x3 + x5
l10: −2 − x3 + x5

2.2.2 Transition Removal

We remove transitions 6, 11 using the following ranking functions, which are bounded by 0.

l7: 0
l8: −1
l12: 1
l11: 1
l10: 1

2.2.3 Transition Removal

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

l11: −1 + x2 − x4
l12: −1 + x2 − x4
l10: −2 + x2 − x4

2.2.4 Transition Removal

We remove transitions 10, 9 using the following ranking functions, which are bounded by 0.

l11: 0
l12: −1
l10: 1

2.2.5 Trivial Cooperation Program

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

2.3 SCC Subproblem 3/4

Here we consider the SCC { l5, l6, l9 }.

2.3.1 Transition Removal

We remove transition 8 using the following ranking functions, which are bounded by 0.

l6: 3⋅x2 − 3⋅x3 + 3
l9: 3⋅x2 − 3⋅x3 + 2
l5: 3⋅x2 − 3⋅x3 + 1

2.3.2 Transition Removal

We remove transitions 16, 4 using the following ranking functions, which are bounded by 0.

l6: 0
l9: −1
l5: 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 4/4

Here we consider the SCC { l4, l2 }.

2.4.1 Transition Removal

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

l4: −1 − x3 + x5
l2: −1 − x3 + x5

2.4.2 Transition Removal

We remove transition 23 using the following ranking functions, which are bounded by 0.

l4: 0
l2: −1

2.4.3 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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