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 = x9x10 = x10x11 = x11x12 = x12
l4 l4 l4: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l7 l7 l7: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l6 l6 l6: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l10 l10 l10: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l11 l11 l11: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l1 l1 l1: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l3 l3 l3: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l0 l0 l0: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l2 l2 l2: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
l9 l9 l9: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12
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, l4 }.

2.1.1 Transition Removal

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

l4: x11 − 4
l5: x11

2.1.2 Transition Removal

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

l4: 0
l5: −1

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 { l0, l2 }.

2.2.1 Transition Removal

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

l0: 2⋅x12
l2: 2⋅x12 + 1

2.2.2 Transition Removal

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

l2: 0
l0: −1

2.2.3 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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