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
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
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
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
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
l21 l21 l21: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12
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
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
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
l6 l6 l6: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12
l8 l8 l8: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12
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
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
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
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
l26 l26 l26: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12
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
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
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
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
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
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
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
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
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
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
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
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
l15 l15 l15: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12
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
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 { l7, l6 }.

2.1.1 Transition Removal

We remove transition 51 using the following ranking functions, which are bounded by 0.

l6: 2⋅x5 − 2⋅x6 + 1
l7: 2⋅x5 − 2⋅x6

2.1.2 Transition Removal

We remove transition 12 using the following ranking functions, which are bounded by 0.

l6: 0
l7: −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 2/4

Here we consider the SCC { l25, l27, l15, l28, l26, l14 }.

2.2.1 Transition Removal

We remove transition 49 using the following ranking functions, which are bounded by 0.

l14: −1 + x5 − x6
l15: −1 + x5 − x6
l25: −2 + x5 − x6
l28: −2 + x5 − x6
l27: −2 + x5 − x6
l26: −2 + x5 − x6

2.2.2 Transition Removal

We remove transitions 23, 39, 44, 42, 41, 40, 43, 46, 45 using the following ranking functions, which are bounded by −4.

l14: −4
l15: −5
l25: −3
l28: 0
l27: −1
l26: −2

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 3/4

Here we consider the SCC { l23, l22, l24, l20, l18, l17, l19, l21 }.

2.3.1 Transition Removal

We remove transition 37 using the following ranking functions, which are bounded by 0.

l19: −1 + x5 − x6
l20: −1 + x5 − x6
l24: −2 + x5 − x6
l18: −2 + x5 − x6
l17: −2 + x5 − x6
l23: −2 + x5 − x6
l22: −2 + x5 − x6
l21: −2 + x5 − x6

2.3.2 Transition Removal

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

l19: −2
l20: −2
l24: 0
l18: 0
l17: 0
l23: 0
l22: 0
l21: 0

2.3.3 Transition Removal

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

l19: 1
l20: 0
l18: −1 + 2⋅x2 + 3⋅x3 + 4⋅x4 + x5 + 5⋅x6 − 2⋅x7 + 6⋅x8 + 7⋅x9 + 8⋅x11 + 9⋅x12
l24: −3 + 2⋅x2 + 3⋅x3 + 4⋅x4 + x5 + 5⋅x6 − 2⋅x7 + 6⋅x8 + 7⋅x9 + 8⋅x11 + 9⋅x12
l17: −3 + 2⋅x2 + 3⋅x3 + 4⋅x4 + x5 + 5⋅x6 − 2⋅x7 + 6⋅x8 + 7⋅x9 + 8⋅x11 + 9⋅x12
l23: −3 + 2⋅x2 + 3⋅x3 + 4⋅x4 + x5 + 5⋅x6 − 2⋅x7 + 6⋅x8 + 7⋅x9 + 8⋅x11 + 9⋅x12
l22: −3 + 2⋅x2 + 3⋅x3 + 4⋅x4 + x5 + 5⋅x6 − 2⋅x7 + 6⋅x8 + 7⋅x9 + 8⋅x11 + 9⋅x12
l21: −3 + 2⋅x2 + 3⋅x3 + 4⋅x4 + x5 + 5⋅x6 − 2⋅x7 + 6⋅x8 + 7⋅x9 + 8⋅x11 + 9⋅x12

2.3.4 Transition Removal

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

l18: −1 + x5 − x7
l24: −1 + x5 − x7
l17: −2 + x5 − x7
l23: −2 + x5 − x7
l22: −2 + x5 − x7
l21: −2 + x5 − x7

2.3.5 Transition Removal

We remove transitions 38, 26, 31, 30, 28, 29, 33, 32 using the following ranking functions, which are bounded by 0.

l18: 0
l24: −1
l17: 1
l23: 4
l22: 3
l21: 2

2.3.6 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 { l10, l11, l8, l13, l16, l12, l9 }.

2.4.1 Transition Removal

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

l9: 7⋅x5 − 7⋅x6 + 5
l16: 7⋅x5 − 7⋅x6 + 4
l8: 7⋅x5 − 7⋅x6 − 1
l13: 7⋅x5 − 7⋅x6 + 3
l12: 7⋅x5 − 7⋅x6 + 2
l11: 7⋅x5 − 7⋅x6 + 1
l10: 7⋅x5 − 7⋅x6

2.4.2 Transition Removal

We remove transitions 47, 13, 20, 18, 15, 14, 17, 16, 19, 22, 21 using the following ranking functions, which are bounded by 0.

l9: 0
l16: −1
l8: 1
l13: 5
l12: 4
l11: 3
l10: 2

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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