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 = 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
l3 l3 l3: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5
l0 l0 l0: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5
l2 l2 l2: 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 1 SCC(s) of the program graph.

2.1 SCC Subproblem 1/1

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

2.1.1 Transition Removal

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

l0: −1 − x5
l3: −1 − x5
l5: −1 − x5
l2: x5
l4: −1 − x5

2.1.2 Transition Removal

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

l0: −1
l3: −1
l5: 0
l2: −1
l4: −1

2.1.3 Transition Removal

We remove transition 2 using the following ranking functions, which are bounded by 0.

l0: −4⋅x4
l3: −4⋅x4 − 1
l2: −4⋅x4 + 1
l4: −4⋅x4 − 2

2.1.4 Transition Removal

We remove transitions 5, 4, 3 using the following ranking functions, which are bounded by 0.

l2: 0
l0: −1
l4: 1
l3: 2

2.1.5 Trivial Cooperation Program

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

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