LTS Termination Proof

by AProVE

Input

Integer Transition System

Proof

1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
l4 l4 l4: 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
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
l11 l11 l11: 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
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
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
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
and for every transition t, a duplicate t is considered.

2 SCC Decomposition

We consider subproblems for each of the 3 SCC(s) of the program graph.

2.1 SCC Subproblem 1/3

Here we consider the SCC { l1, l0 }.

2.1.1 Transition Removal

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

l0: 2⋅x2 − 2⋅x4 + 1
l1: 2⋅x2 − 2⋅x4

2.1.2 Transition Removal

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

l0: 0
l1: −1

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 2/3

Here we consider the SCC { l7, l10, l8, l3, l2, l9 }.

2.2.1 Transition Removal

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

l2: 4⋅x1 − 4⋅x5 + 1
l3: 4⋅x1 − 4⋅x5
l8: 4⋅x1 − 4⋅x5 − 1
l7: 4⋅x1 − 4⋅x5 − 1
l9: 4⋅x1 − 4⋅x5 − 1
l10: 4⋅x1 − 4⋅x5 − 1

2.2.2 Transition Removal

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

l2: −1
l3: −2
l8: 0
l7: 0
l9: 0
l10: 0

2.2.3 Transition Removal

We remove transitions 2, 10 using the following ranking functions, which are bounded by 0.

l2: 1
l3: 0
l7: −1 + x1 − x3
l8: −1 + x1 − x3
l9: −2 + x1 − x3
l10: −2 + x1 − x3

2.2.4 Transition Removal

We remove transitions 5, 6, 8, 7 using the following ranking functions, which are bounded by −2.

l7: −2
l8: −3
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 3/3

Here we consider the SCC { l4, l6 }.

2.3.1 Transition Removal

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

l6: 2⋅x1 − 2⋅x5 + 1
l4: 2⋅x1 − 2⋅x5

2.3.2 Transition Removal

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

l6: 0
l4: −1

2.3.3 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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