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 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15
l22 l22 l22: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15
l13 l13 l13: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15
l31 l31 l31: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15
l18 l18 l18: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15
l17 l17 l17: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15
l9 l9 l9: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15
l14 l14 l14: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15
l25 l25 l25: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15
l27 l27 l27: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15
l0 l0 l0: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15
l12 l12 l12: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15
l19 l19 l19: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15
l7 l7 l7: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15
l24 l24 l24: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15
l11 l11 l11: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15
l3 l3 l3: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15
l20 l20 l20: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15
l28 l28 l28: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15
l2 l2 l2: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15
l23 l23 l23: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15
l4 l4 l4: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15
l10 l10 l10: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15
l29 l29 l29: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15
l16 l16 l16: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15
l30 l30 l30: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15
and for every transition t, a duplicate t is considered.

2 SCC Decomposition

We consider subproblems for each of the 6 SCC(s) of the program graph.

2.1 SCC Subproblem 1/6

Here we consider the SCC { l29, l27, l28 }.

2.1.1 Transition Removal

We remove transition 33 using the following ranking functions, which are bounded by 0.

l27: 3⋅x11 − 3⋅x12 + 1
l28: 3⋅x11 − 3⋅x12 + 3
l29: 3⋅x11 − 3⋅x12 + 2

2.1.2 Transition Removal

We remove transitions 31, 34 using the following ranking functions, which are bounded by 0.

l27: 2
l28: 1
l29: 0

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 2/6

Here we consider the SCC { l23, l22, l24 }.

2.2.1 Transition Removal

We remove transition 27 using the following ranking functions, which are bounded by 0.

l22: −3⋅x11 + 3⋅x12 + 1
l23: −3⋅x11 + 3⋅x12 + 3
l24: −3⋅x11 + 3⋅x12 + 2

2.2.2 Transition Removal

We remove transitions 25, 28 using the following ranking functions, which are bounded by 0.

l22: 2
l23: 1
l24: 0

2.2.3 Trivial Cooperation Program

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

2.3 SCC Subproblem 3/6

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

2.3.1 Transition Removal

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

l16: −3⋅x11 + 3⋅x12 + 1
l17: −3⋅x11 + 3⋅x12 + 3
l18: −3⋅x11 + 3⋅x12 + 2

2.3.2 Transition Removal

We remove transitions 17, 20 using the following ranking functions, which are bounded by 0.

l16: 2
l17: 1
l18: 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 4/6

Here we consider the SCC { l11, l12 }.

2.4.1 Transition Removal

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

l11: 3⋅x11 − 3⋅x12 + 1
l12: 3⋅x11 − 3⋅x12 + 2

2.4.2 Transition Removal

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

l11: 0
l12: −1

2.4.3 Trivial Cooperation Program

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

2.5 SCC Subproblem 5/6

Here we consider the SCC { l10, l19, l9 }.

2.5.1 Transition Removal

We remove transition 22 using the following ranking functions, which are bounded by 0.

l9: 3⋅x11 − 3⋅x12 − 1
l10: 3⋅x11 − 3⋅x12 + 1
l19: 3⋅x11 − 3⋅x12

2.5.2 Transition Removal

We remove transitions 10, 35 using the following ranking functions, which are bounded by 0.

l9: 2
l10: 1
l19: 0

2.5.3 Trivial Cooperation Program

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

2.6 SCC Subproblem 6/6

Here we consider the SCC { l4, l3, l2 }.

2.6.1 Transition Removal

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

l2: −3⋅x11 + 3⋅x12 + 1
l3: −3⋅x11 + 3⋅x12 + 3
l4: −3⋅x11 + 3⋅x12 + 2

2.6.2 Transition Removal

We remove transitions 3, 6 using the following ranking functions, which are bounded by 0.

l2: 2
l3: 1
l4: 0

2.6.3 Trivial Cooperation Program

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

Tool configuration

AProVE