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 = x11
l4 l4 l4: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11
l7 l7 l7: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11
l6 l6 l6: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11
l10 l10 l10: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11
l1 l1 l1: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11
l3 l3 l3: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11
l0 l0 l0: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11
l2 l2 l2: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11
l9 l9 l9: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11
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, l7, l6, l1, l3, l0, l2 }.

2.1.1 Transition Removal

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

l0: −2 − x8 + x9
l1: −2 − x8 + x9
l2: −2 − x8 + x9
l4: −2 − x8 + x9
l6: −2 − x8 + x9
l7: −2 − x8 + x9
l5: −1 − x8 + x9
l3: −1 − x8 + x9

2.1.2 Transition Removal

We remove transitions 1, 3, 2, 5, 8, 7, 4, 6 using the following ranking functions, which are bounded by 0.

l0: 2
l1: 1
l2: 3
l4: 4
l6: 5
l7: 5
l3: 0
l5: −1

2.1.3 Transition Removal

We remove transition 9 using the following ranking functions, which are bounded by 0.

l7: 2⋅x5 − 2⋅x10 + 1
l6: 2⋅x5 − 2⋅x10

2.1.4 Transition Removal

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

l7: 0
l6: −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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