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 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l22 l22 l22: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l13 l13 l13: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l18 l18 l18: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l17 l17 l17: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l21 l21 l21: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l9 l9 l9: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l14 l14 l14: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l25 l25 l25: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l6 l6 l6: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l8 l8 l8: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l0 l0 l0: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l12 l12 l12: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l19 l19 l19: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l26 l26 l26: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l7 l7 l7: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l24 l24 l24: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l11 l11 l11: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l3 l3 l3: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l20 l20 l20: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l2 l2 l2: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l23 l23 l23: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l4 l4 l4: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l10 l10 l10: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l15 l15 l15: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l16 l16 l16: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
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 { l10, l11 }.

2.1.1 Transition Removal

We remove transition 41 using the following ranking functions, which are bounded by 0.

l10: −1 + x4x5
l11: −1 + x4x5

2.1.2 Transition Removal

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

l10: 0
l11: −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 { l22, l24, l13, l20, l18, l17, l21, l14, l23, l16, l15, l12, l19 }.

2.2.1 Transition Removal

We remove transition 39 using the following ranking functions, which are bounded by 0.

l13: −1 + x4x5
l19: −1 + x4x5
l12: −2 + x4x5
l16: −2 + x4x5
l15: −2 + x4x5
l14: −2 + x4x5
l24: −2 + x4x5
l21: −2 + x4x5
l18: −2 + x4x5
l22: −2 + x4x5
l23: −2 + x4x5
l20: −2 + x4x5
l17: −2 + x4x5

2.2.2 Transition Removal

We remove transitions 27, 18, 23, 20, 19, 22, 21, 25, 24, 35 using the following ranking functions, which are bounded by 0.

l13: 1
l19: 0
l12: 2
l16: 5
l15: 4
l14: 3
l24: 6
l21: 6
l18: 6
l22: 6
l23: 6
l20: 6
l17: 6

2.2.3 Transition Removal

We remove transition 36 using the following ranking functions, which are bounded by 0.

l21: −1 + x4x6
l24: −1 + x4x6
l18: −2 + x4x6
l22: −2 + x4x6
l23: −2 + x4x6
l20: −2 + x4x6
l17: −2 + x4x6

2.2.4 Transition Removal

We remove transitions 37, 30, 31, 34, 33, 29, 26, 28, 32 using the following ranking functions, which are bounded by 0.

l21: 0
l24: −1
l18: 1
l22: 3
l23: 4
l20: 3
l17: 2

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 { l5, l7, l6, l8, l9 }.

2.3.1 Transition Removal

We remove transition 16 using the following ranking functions, which are bounded by 0.

l5: 6⋅x4 − 6⋅x5 − 4
l6: 6⋅x4 − 6⋅x5 + 1
l8: 6⋅x4 − 6⋅x5 − 1
l7: 6⋅x4 − 6⋅x5 − 2
l9: −6⋅x5 + 6⋅x4

2.3.2 Transition Removal

We remove transitions 9, 12, 11, 10, 14, 13, 42 using the following ranking functions, which are bounded by 0.

l5: 2
l6: 1
l8: 4
l7: 3
l9: 0

2.3.3 Trivial Cooperation Program

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

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