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
l22 l22 l22: 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
l3 l3 l3: 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
l20 l20 l20: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8
l18 l18 l18: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8
l21 l21 l21: 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
l23 l23 l23: 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
l6 l6 l6: 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
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
l19 l19 l19: 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 1 SCC(s) of the program graph.

2.1 SCC Subproblem 1/1

Here we consider the SCC { l5, l22, l7, l11, l1, l3, l13, l20, l18, l2, l21, l14, l9, l4, l6, l10, l8, l16, l15, l0, l19, l12 }.

2.1.1 Transition Removal

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

l0: −2 − x3 + x4
l1: −2 − x3 + x4
l3: −2 − x3 + x4
l22: −2 − x3 + x4
l2: −2 − x3 + x4
l16: −1 − x3 + x4
l5: −1 − x3 + x4
l4: −2 − x3 + x4
l15: −2 − x3 + x4
l14: −2 − x3 + x4
l7: −2 − x3 + x4
l18: −2 − x3 + x4
l19: −2 − x3 + x4
l20: −2 − x3 + x4
l21: −2 − x3 + x4
l6: −2 − x3 + x4
l13: −2 − x3 + x4
l8: −2 − x3 + x4
l9: −2 − x3 + x4
l10: −2 − x3 + x4
l11: −2 − x3 + x4
l12: −2 − x3 + x4

2.1.2 Transition Removal

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

l0: 0
l1: −1 − x2 + x4
l3: 0
l22: 0
l2: 0
l5: −1 − x2 + x4
l16: −1 − x2 + x4
l4: −1 − x2 + x4
l15: −1 − x2 + x4
l14: −1 − x2 + x4
l7: −1 − x2 + x4
l18: −1 − x2 + x4
l19: −1 − x2 + x4
l20: −1 − x2 + x4
l21: −1 − x2 + x4
l6: −1 − x2 + x4
l13: −1 − x2 + x4
l8: −1 − x2 + x4
l9: −1 − x2 + x4
l10: −1 − x2 + x4
l11: −1 − x2 + x4
l12: −1 − x2 + x4

2.1.3 Transition Removal

We remove transitions 29, 25, 10, 7 using the following ranking functions, which are bounded by 0.

l3: −1 − x2 + 2⋅x3 + 3⋅x4 + 4⋅x8
l0: −1 − x2 + 2⋅x3 + 3⋅x4 + 4⋅x8
l22: −2 − x2 + 2⋅x3 + 3⋅x4 + 4⋅x8
l2: −2 − x2 + 2⋅x3 + 3⋅x4 + 4⋅x8
l5: − x2 + x4
l16: − x2 + x4
l4: − x2 + x4
l15: − x2 + x4
l14: − x2 + x4
l7: − x2 + x4
l1: − x2 + x4
l18: − x2 + x4
l19: − x2 + x4
l20: − x2 + x4
l21: − x2 + x4
l6: − x2 + x4
l13: − x2 + x4
l8: − x2 + x4
l9: − x2 + x4
l10: − x2 + x4
l11: − x2 + x4
l12: − x2 + x4

2.1.4 Transition Removal

We remove transitions 19, 21, 20, 31, 24, 26, 28, 30, 33, 32 using the following ranking functions, which are bounded by 0.

l3: −1 − x2 + 6⋅x3 + 7⋅x4 + 8⋅x8
l0: −2 − x2 + 6⋅x3 + 7⋅x4 + 8⋅x8
l22: −2 − x2 + 6⋅x3 + 7⋅x4 + 8⋅x8
l2: −2 − x2 + 6⋅x3 + 7⋅x4 + 8⋅x8
l5: −1
l16: −1
l4: −1
l15: 0
l14: −1
l7: −1
l1: 5
l18: 1
l19: 2
l20: 3
l21: 4
l6: −1
l13: −1
l8: −1
l9: −1
l10: −1
l11: −1
l12: −1

2.1.5 Transition Removal

We remove transitions 27, 4, 16 using the following ranking functions, which are bounded by 0.

l3: −1 − x2 + 3⋅x3 + 4⋅x4 + 5⋅x8
l0: −1 − x2 + 3⋅x3 + 4⋅x4 + 5⋅x8
l22: −2 − x2 + 3⋅x3 + 4⋅x4 + 5⋅x8
l2: −2 − x2 + 3⋅x3 + 4⋅x4 + 5⋅x8
l5: 0
l16: −1
l4: 1
l14: 2
l7: 2
l6: 2
l13: 2
l8: 2
l9: 2
l10: 2
l11: 2
l12: 2

2.1.6 Transition Removal

We remove transition 17 using the following ranking functions, which are bounded by 0.

l3: −1 − x2 + 2⋅x3 + 3⋅x4 + 4⋅x8
l0: −1 − x2 + 2⋅x3 + 3⋅x4 + 4⋅x8
l22: −2 − x2 + 2⋅x3 + 3⋅x4 + 4⋅x8
l2: −2 − x2 + 2⋅x3 + 3⋅x4 + 4⋅x8
l7: − x1 + x4
l14: − x1 + x4
l6: −1 − x1 + x4
l13: −1 − x1 + x4
l8: −1 − x1 + x4
l9: −1 − x1 + x4
l10: −1 − x1 + x4
l11: −1 − x1 + x4
l12: −1 − x1 + x4

2.1.7 Transition Removal

We remove transitions 18, 5, 13, 6, 8, 9, 11, 12 using the following ranking functions, which are bounded by 0.

l3: −1 − x2 + 7⋅x3 + 8⋅x4 + 9⋅x8
l0: −2 − x2 + 7⋅x3 + 8⋅x4 + 9⋅x8
l22: −2 − x2 + 7⋅x3 + 8⋅x4 + 9⋅x8
l2: −2 − x2 + 7⋅x3 + 8⋅x4 + 9⋅x8
l7: 0
l14: −1
l6: 1
l13: 6
l8: 2
l9: 3
l10: 4
l11: 5
l12: 6

2.1.8 Transition Removal

We remove transitions 15, 14 using the following ranking functions, which are bounded by 0.

l3: −1 − x2 + 2⋅x3 + 3⋅x4 + 4⋅x8
l0: −2 − x2 + 2⋅x3 + 3⋅x4 + 4⋅x8
l22: −2 − x2 + 2⋅x3 + 3⋅x4 + 4⋅x8
l2: −2 − x2 + 2⋅x3 + 3⋅x4 + 4⋅x8
l13: 1
l12: 0

2.1.9 Transition Removal

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

l3: −4⋅x2 + 4⋅x4 + 4
l0: −4⋅x2 + 4⋅x4 + 3
l22: −4⋅x2 + 4⋅x4 + 1
l2: −4⋅x2 + 4⋅x4 + 2

2.1.10 Transition Removal

We remove transitions 3, 34, 36, 35 using the following ranking functions, which are bounded by 0.

l3: 0
l0: −1
l22: 1
l2: 2

2.1.11 Trivial Cooperation Program

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

Tool configuration

AProVE