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 = x9
l4 l4 l4: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9
l6 l6 l6: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9
l10 l10 l10: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9
l1 l1 l1: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9
l8 l8 l8: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9
l3 l3 l3: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9
l0 l0 l0: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9
l2 l2 l2: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9
l9 l9 l9: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9
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, l1, l3, l0, l2 }.

2.1.1 Transition Removal

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

l0: 6⋅x7 − 6⋅x8 + 1
l1: 6⋅x7 − 6⋅x8 + 1
l5: 6⋅x7 − 6⋅x8 + 1
l2: 6⋅x7 − 6⋅x8 + 1
l4: 6⋅x7 − 6⋅x8 + 2
l3: 6⋅x7 − 6⋅x8

2.1.2 Transition Removal

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

l0: 2
l1: 2
l5: 2
l2: 2
l3: 3⋅x7 − 3⋅x8 − 3⋅x9 + 1
l4: 3⋅x1 − 3⋅x3 − 3⋅x8 + 3

2.1.3 Transition Removal

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

l0: −1 + 2⋅x6 + x7 + x8x9
l1: −1 + 2⋅x6 + x7 + x8x9
l5: −1 + 2⋅x6 + x7 + x8x9
l2: −1 + 2⋅x6 + x7 + x8x9
l3: 0
l4: −1

2.1.4 Transition Removal

We remove transition 8 using the following ranking functions, which are bounded by 0.

l0: 4⋅x7 − 4⋅x8 − 4⋅x9 + 1
l1: 4⋅x7 − 4⋅x8 − 4⋅x9 + 4
l5: 4⋅x7 − 4⋅x8 − 4⋅x9 + 3
l2: 4⋅x7 − 4⋅x8 − 4⋅x9 + 2

2.1.5 Transition Removal

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

l0: 0
l1: x3x9
l5: 2
l2: 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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