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
l13 l13 l13: 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
l20 l20 l20: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l18 l18 l18: 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
l2 l2 l2: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l21 l21 l21: 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
l6 l6 l6: 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
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
l19 l19 l19: 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 2 SCC(s) of the program graph.

2.1 SCC Subproblem 1/2

Here we consider the SCC { l1, l3, l0, l2 }.

2.1.1 Transition Removal

We remove transition 28 using the following ranking functions, which are bounded by 0.

l0: 3⋅x1 − 3⋅x2 + 3
l1: 3⋅x1 − 3⋅x2 + 2
l3: 3⋅x1 − 3⋅x2 + 1
l2: 3⋅x1 − 3⋅x2 + 1

2.1.2 Transition Removal

We remove transitions 1, 25 using the following ranking functions, which are bounded by 0.

l0: 0
l1: −1
l3: 1
l2: 1

2.1.3 Transition Removal

We remove transition 26 using the following ranking functions, which are bounded by 0.

l2: 2⋅x1 − 2⋅x3 + 1
l3: 2⋅x1 − 2⋅x3

2.1.4 Transition Removal

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

l2: 0
l3: −1

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 2/2

Here we consider the SCC { l13, l16, l18, l17, l19, l14 }.

2.2.1 Transition Removal

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

l13: 4⋅x1 − 4⋅x2 + 1
l14: 4⋅x1 − 4⋅x2
l19: 4⋅x1 − 4⋅x2 − 1
l17: 4⋅x1 − 4⋅x2 − 1
l16: 4⋅x1 − 4⋅x2 − 1
l18: 4⋅x1 − 4⋅x2 − 1

2.2.2 Transition Removal

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

l13: −1
l14: −1
l19: 1
l17: 1
l16: 1
l18: 1

2.2.3 Transition Removal

We remove transitions 14, 21 using the following ranking functions, which are bounded by 0.

l13: 1
l14: 0
l17: −1 + x1 − x3
l19: −1 + x1 − x3
l16: −2 + x1 − x3
l18: −2 + x1 − x3

2.2.4 Transition Removal

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

l17: 3⋅x1 − 3⋅x4 − 2
l19: 3⋅x1 − 3⋅x4 − 3
l16: 3⋅x1 − 3⋅x4 − 1
l18: 3⋅x1 − 3⋅x4

2.2.5 Transition Removal

We remove transitions 19, 17, 24 using the following ranking functions, which are bounded by 0.

l17: 0
l19: −1
l16: 1
l18: 2

2.2.6 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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