LTS Termination Proof

by AProVE

Input

Integer Transition System

Proof

1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
l4 l4 l4: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6
l7 l7 l7: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6
l6 l6 l6: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6
l10 l10 l10: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6
l8 l8 l8: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6
l1 l1 l1: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6
l3 l3 l3: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6
l13 l13 l13: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6
l0 l0 l0: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6
l12 l12 l12: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6
l2 l2 l2: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6
l9 l9 l9: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6
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 { l7, l6, l10, l8, l9 }.

2.1.1 Transition Removal

We remove transition 14 using the following ranking functions, which are bounded by 0.

l6: 4⋅x6
l7: 4⋅x6
l8: 4⋅x6
l10: 4⋅x6
l9: 4⋅x6 − 1

2.1.2 Transition Removal

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

l6: −1
l7: −1
l8: −1
l10: −1
l9: 0

2.1.3 Transition Removal

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

l6: −2 + x5
l7: −1 + x5
l8: −1 + x5
l10: −1 + x5

2.1.4 Transition Removal

We remove transitions 8, 13, 18 using the following ranking functions, which are bounded by 0.

l6: 3
l7: 2
l10: 1
l8: 0

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 { l3, l0, l2 }.

2.2.1 Transition Removal

We remove transition 4 using the following ranking functions, which are bounded by 0.

l0: x3 + 2⋅x4
l2: x3 + 2⋅x4
l3: x3 + 2⋅x4

2.2.2 Transition Removal

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

l0: −1 + x6
l2: −2 + x6
l3: −1 + x6

2.2.3 Transition Removal

We remove transitions 5, 17 using the following ranking functions, which are bounded by 0.

l3: 0
l0: −1
l2: 1

2.2.4 Trivial Cooperation Program

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

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