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
l7 l7 l7: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l11 l11 l11: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l1 l1 l1: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l3 l3 l3: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l13 l13 l13: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l17 l17 l17: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l9 l9 l9: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l14 l14 l14: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l4 l4 l4: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l10 l10 l10: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l6 l6 l6: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l8 l8 l8: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l15 l15 l15: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l16 l16 l16: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l0 l0 l0: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l12 l12 l12: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
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 { l10, l11, l13, l12, l14 }.

2.1.1 Transition Removal

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

l10: − x2 + x5
l11: −1 − x2 + x5
l13: − x2 + x5
l14: − x2 + x5
l12: − x2 + x5

2.1.2 Transition Removal

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

l10: −2 − x2 + x5
l11: −2 − x2 + x5
l13: −1 − x2 + x5
l14: −1 − x2 + x5
l12: −1 − x2 + x5

2.1.3 Transition Removal

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

l10: −2 − x2 + x4
l11: −2 − x2 + x4
l13: −2 − x2 + x4
l14: −1 − x2 + x4
l12: −1 − x2 + x4

2.1.4 Transition Removal

We remove transitions 10, 12, 17, 11 using the following ranking functions, which are bounded by 0.

l10: 3
l11: 2
l13: 4
l12: 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 { l5, l7, l6, l8, l9 }.

2.2.1 Transition Removal

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

l5: −3⋅x2 + 3⋅x4 − 1
l6: −3⋅x2 + 3⋅x4 − 1
l7: −3⋅x2 + 3⋅x4 − 1
l9: −3⋅x2 + 3⋅x4
l8: −3⋅x2 + 3⋅x4 + 1

2.2.2 Transition Removal

We remove transitions 18, 6 using the following ranking functions, which are bounded by 0.

l5: 2
l6: 2
l7: 2
l8: 1
l9: 0

2.2.3 Transition Removal

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

l5: −2 + x1 − x3
l6: −1 + x1 − x3
l7: −1 + x1 − x3

2.2.4 Transition Removal

We remove transitions 5, 19 using the following ranking functions, which are bounded by 0.

l5: 2
l6: 1
l7: 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 { l1, l3, l0 }.

2.3.1 Transition Removal

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

l0: −2 + x1 − x2
l1: −1 + x1 − x2
l3: −1 + x1 − x2

2.3.2 Transition Removal

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

l0: 2
l1: 1
l3: 0

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, l15 }.

2.4.1 Transition Removal

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

l4: −1 − x2 + x4
l15: −1 − x2 + x4

2.4.2 Transition Removal

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

l4: 0
l15: −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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