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 = x8
l4 l4 l4: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8
l7 l7 l7: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8
l6 l6 l6: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8
l10 l10 l10: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8
l1 l1 l1: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8
l3 l3 l3: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8
l0 l0 l0: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8
l2 l2 l2: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8
l9 l9 l9: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8
and for every transition t, a duplicate t is considered.

2 SCC Decomposition

We consider subproblems for each of the 1 SCC(s) of the program graph.

2.1 SCC Subproblem 1/1

Here we consider the SCC { l5, l4, l7, l6, l1, l3, l0, l2 }.

2.1.1 Transition Removal

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

l0: −2 − x5 + x6
l1: −2 − x5 + x6
l2: −2 − x5 + x6
l4: −2 − x5 + x6
l6: −2 − x5 + x6
l7: −2 − x5 + x6
l5: −1 − x5 + x6
l3: −1 − x5 + x6

2.1.2 Transition Removal

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

l0: −2 − x2 + x4
l1: −2 − x2 + x4
l2: −1 − x2 + x4
l4: −1 − x2 + x4
l6: −1 − x2 + x4
l7: −1 − x2 + x4
l3: −2 − x2 + x4
l5: −2 − x2 + x4

2.1.3 Transition Removal

We remove transitions 1, 2, 5, 8, 7, 4, 6 using the following ranking functions, which are bounded by 0.

l0: 3
l1: 2
l2: 4
l4: 5
l6: 6
l7: 6
l3: 1
l5: 0

2.1.4 Transition Removal

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

l7: 2⋅x2 − 2⋅x7 + 1
l6: 2⋅x2 − 2⋅x7

2.1.5 Transition Removal

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

l7: 0
l6: −1

2.1.6 Trivial Cooperation Program

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

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