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
l22 l22 l22: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l13 l13 l13: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l18 l18 l18: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l17 l17 l17: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l21 l21 l21: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l14 l14 l14: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l9 l9 l9: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l25 l25 l25: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l6 l6 l6: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l8 l8 l8: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l27 l27 l27: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l0 l0 l0: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l12 l12 l12: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l19 l19 l19: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l26 l26 l26: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l7 l7 l7: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l24 l24 l24: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l11 l11 l11: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l3 l3 l3: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l20 l20 l20: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l2 l2 l2: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l23 l23 l23: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l4 l4 l4: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l10 l10 l10: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l15 l15 l15: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
l16 l16 l16: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10
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 { l8, l9 }.

2.1.1 Transition Removal

We remove transition 45 using the following ranking functions, which are bounded by 0.

l8: −1 + x3 − x4
l9: −1 + x3 − x4

2.1.2 Transition Removal

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

l8: 0
l9: −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 2/2

Here we consider the SCC { l5, l22, l7, l24, l11, l13, l20, l18, l17, l21, l14, l23, l25, l10, l6, l16, l15, l12, l19 }.

2.2.1 Transition Removal

We remove transition 43 using the following ranking functions, which are bounded by 0.

l5: −2 + x3 − x4
l6: −2 + x3 − x4
l20: −2 + x3 − x4
l7: −2 + x3 − x4
l13: −2 + x3 − x4
l12: −2 + x3 − x4
l11: −2 + x3 − x4
l10: −2 + x3 − x4
l14: −2 + x3 − x4
l17: −2 + x3 − x4
l16: −2 + x3 − x4
l15: −2 + x3 − x4
l21: −2 + x3 − x4
l22: −2 + x3 − x4
l25: −2 + x3 − x4
l24: −2 + x3 − x4
l23: −2 + x3 − x4
l19: −1 + x3 − x4
l18: −1 + x3 − x4

2.2.2 Transition Removal

We remove transitions 33, 39, 37, 35, 34, 38, 41, 40 using the following ranking functions, which are bounded by 0.

l5: −3 − x1 − x5 + x7
l6: −2 − x1 − x5 + x7
l20: −3 − x1 − x5 + x7
l7: −3 − x1 − x5 + x7
l13: −3 − x1 − x5 + x7
l12: −3 − x1 − x5 + x7
l11: −3 − x1 − x5 + x7
l10: −3 − x1 − x5 + x7
l14: −3 − x1 − x5 + x7
l17: −3 − x1 − x5 + x7
l16: −3 − x1 − x5 + x7
l15: −3 − x1 − x5 + x7
l21: −2 − x1 − x5 + x7
l22: 0
l25: 3
l24: 2
l23: 1
l18: −2 − x1 + x3 − 2⋅x5 + x7
l19: −2 − x1 + x3 − 2⋅x5 + x7

2.2.3 Transition Removal

We remove transitions 27, 31 using the following ranking functions, which are bounded by 0.

l5: 1
l6: 1
l20: 1
l7: 1
l13: 1
l12: 1
l11: 1
l10: 1
l14: 1
l17: 1
l16: 1
l15: 1
l21: 1
l18: 0
l19: −1

2.2.4 Transition Removal

We remove transition 32 using the following ranking functions, which are bounded by 0.

l5: −2 + x3 − x5
l6: −1 + x3 − x5
l20: −2 + x3 − x5
l7: −2 + x3 − x5
l13: −2 + x3 − x5
l12: −2 + x3 − x5
l11: −2 + x3 − x5
l10: −2 + x3 − x5
l14: −2 + x3 − x5
l17: −2 + x3 − x5
l16: −2 + x3 − x5
l15: −2 + x3 − x5
l21: −1 + x3 − x5

2.2.5 Transition Removal

We remove transitions 9, 28, 10, 17, 15, 13, 12, 14, 16, 19, 18, 20, 24, 23, 21, 22, 26, 25, 30, 29, 36 using the following ranking functions, which are bounded by 0.

l5: 1
l6: 0
l20: 11
l7: 2
l13: 6
l12: 5
l11: 4
l10: 3
l14: 7
l17: 10
l16: 9
l15: 8
l21: −1

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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