LTS Termination Proof

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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 = x5
l4 l4 l4: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5
l7 l7 l7: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5
l6 l6 l6: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5
l10 l10 l10: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5
l11 l11 l11: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5
l1 l1 l1: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5
l3 l3 l3: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5
l0 l0 l0: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5
l12 l12 l12: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5
l2 l2 l2: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5
l9 l9 l9: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5
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, l10, l1, l0, l2, l9 }.

2.1.1 Transition Removal

We remove transition 3 using the following ranking functions, which are bounded by −594.

l0: −6⋅x1 − 1
l1: −6⋅x1 − 2
l2: −6⋅x1
l5: −6⋅x1 + 1
l9: −6⋅x1 − 4
l10: −6⋅x1 − 3

2.1.2 Transition Removal

We remove transitions 1, 5, 10, 11, 12 using the following ranking functions, which are bounded by 0.

l0: 4
l1: 3
l5: 0
l2: −1
l9: 1
l10: 2

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 2/2

Here we consider the SCC { l4, l7, l6, l3 }.

2.2.1 Transition Removal

We remove transition 8 using the following ranking functions, which are bounded by −794.

l3: −4⋅x2 + 3
l7: −4⋅x2 + 2
l4: −4⋅x2
l6: −4⋅x2 + 1

2.2.2 Transition Removal

We remove transitions 9, 4, 6 using the following ranking functions, which are bounded by 0.

l3: 0
l7: −1
l4: 1
l6: 2

2.2.3 Trivial Cooperation Program

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

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