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
l7 l7 l7: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9
l11 l11 l11: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9
l3 l3 l3: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9
l13 l13 l13: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9
l17 l17 l17: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9
l2 l2 l2: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9
l14 l14 l14: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9
l9 l9 l9: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9
l4 l4 l4: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9
l6 l6 l6: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9
l10 l10 l10: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9
l8 l8 l8: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9
l15 l15 l15: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9
l16 l16 l16: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9
l0 l0 l0: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9
l12 l12 l12: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9
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 { l11, l13, l15, l12, l14 }.

2.1.1 Transition Removal

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

l11: − x4 + x7
l12: −1 − x4 + x7
l15: − x4 + x7
l14: − x4 + x7
l13: − x4 + x7

2.1.2 Transition Removal

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

l11: −2 − x4 + x7
l12: −2 − x4 + x7
l15: −1 − x4 + x7
l14: −1 − x4 + x7
l13: −1 − x4 + x7

2.1.3 Transition Removal

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

l11: −2 − x4 + x6
l12: −2 − x4 + x6
l15: −2 − x4 + x6
l14: −1 − x4 + x6
l13: −1 − x4 + x6

2.1.4 Transition Removal

We remove transitions 12, 15, 14, 13 using the following ranking functions, which are bounded by 0.

l11: 3
l12: 2
l15: 4
l13: 1
l14: 0

2.1.5 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, l6, l10, l8, l9 }.

2.2.1 Transition Removal

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

l6: −2 − x4 + x6
l7: −2 − x4 + x6
l8: −2 − x4 + x6
l10: −1 − x4 + x6
l9: −1 − x4 + x6

2.2.2 Transition Removal

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

l6: 2
l7: 2
l8: 2
l9: −1
l10: −1

2.2.3 Transition Removal

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

l6: −2 + x3 − x5
l7: −1 + x3 − x5
l8: −1 + x3 − x5
l9: −1 + 2⋅x1 + 3⋅x2 + 4⋅x3 + 5⋅x4 + 6⋅x6 + 7⋅x7 + 8⋅x8 + 9⋅x9
l10: −1 + 2⋅x1 + 3⋅x2 + 4⋅x3 + 5⋅x4 + 6⋅x6 + 7⋅x7 + 8⋅x8 + 9⋅x9

2.2.4 Transition Removal

We remove transitions 7, 21, 20 using the following ranking functions, which are bounded by 0.

l6: 2
l7: 1
l8: 0
l9: 1
l10: 0

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, l4, l3 }.

2.3.1 Transition Removal

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

l3: −2 + x3 − x4
l4: −1 + x3 − x4
l5: −1 + x3 − x4

2.3.2 Transition Removal

We remove transitions 3, 22 using the following ranking functions, which are bounded by −1.

l3: 0
l4: −1
l5: −2

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 { l0, l2 }.

2.4.1 Transition Removal

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

l0: −2⋅x4 + 2⋅x6
l2: −2⋅x4 + 2⋅x6 + 1

2.4.2 Transition Removal

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

l2: 0
l0: −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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