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 = x12x13 = x13x14 = x14x15 = x15x16 = x16x17 = x17
l4 l4 l4: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12x13 = x13x14 = x14x15 = x15x16 = x16x17 = x17
l7 l7 l7: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12x13 = x13x14 = x14x15 = x15x16 = x16x17 = x17
l6 l6 l6: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12x13 = x13x14 = x14x15 = x15x16 = x16x17 = x17
l11 l11 l11: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12x13 = x13x14 = x14x15 = x15x16 = x16x17 = x17
l8 l8 l8: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12x13 = x13x14 = x14x15 = x15x16 = x16x17 = x17
l1 l1 l1: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12x13 = x13x14 = x14x15 = x15x16 = x16x17 = x17
l3 l3 l3: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12x13 = x13x14 = x14x15 = x15x16 = x16x17 = x17
l0 l0 l0: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12x13 = x13x14 = x14x15 = x15x16 = x16x17 = x17
l12 l12 l12: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12x13 = x13x14 = x14x15 = x15x16 = x16x17 = x17
l2 l2 l2: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12x13 = x13x14 = x14x15 = x15x16 = x16x17 = x17
l9 l9 l9: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12x13 = x13x14 = x14x15 = x15x16 = x16x17 = x17
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 { l5, l4, l1, l0 }.

2.1.1 Transition Removal

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

l0: 9 − x8
l1: 9 − x8
l5: 8 − x8
l4: 8 − x8

2.1.2 Transition Removal

We remove transitions 1, 12 using the following ranking functions, which are bounded by 0.

l0: 0
l1: −1
l5: 1
l4: 1

2.1.3 Transition Removal

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

l4: 9 − x2
l5: 9 − x2

2.1.4 Transition Removal

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

l4: 0
l5: −1

2.1.5 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 { l7, l6, l8, l3, l2, l9 }.

2.2.1 Transition Removal

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

l8: 9 − x7
l9: 9 − x7
l7: 8 − x7
l3: 8 − x7
l2: 8 − x7
l6: 8 − x7

2.2.2 Transition Removal

We remove transitions 10, 6 using the following ranking functions, which are bounded by 0.

l8: 0
l9: −1
l7: 1
l3: 1
l2: 1
l6: 1

2.2.3 Transition Removal

We remove transition 7 using the following ranking functions, which are bounded by 0.

l3: 9 − x1
l7: 9 − x1
l2: 8 − x1
l6: 8 − x1

2.2.4 Transition Removal

We remove transitions 11, 2, 5, 4 using the following ranking functions, which are bounded by 0.

l3: 0
l7: −1
l2: 1
l6: 2

2.2.5 Trivial Cooperation Program

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

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