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 = x7
l4 l4 l4: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7
l7 l7 l7: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7
l6 l6 l6: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7
l1 l1 l1: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7
l3 l3 l3: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7
l0 l0 l0: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7
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, l6, l1, l3 }.

2.1.1 Transition Removal

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

l1: −1 + x6
l3: −1 + x6
l4: −1 + x6
l6: −2 + x6
l5: −2 + x6

2.1.2 Transition Removal

We remove transitions 9, 8, 7, 6 using the following ranking functions, which are bounded by 0.

l1: −1
l3: −1
l4: 0
l6: 1
l5: 2

2.1.3 Transition Removal

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

l1: −1 + x6 + x7
l3: −1 + x6 + x7

2.1.4 Transition Removal

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

l3: 0
l1: −1

2.1.5 Trivial Cooperation Program

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

Tool configuration

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