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 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15 ∧ x16 = x16 ∧ x17 = x17 ∧ x18 = x18 ∧ x19 = x19 ∧ x20 = x20 ∧ x21 = x21 ∧ x22 = x22
l4 l4 l4: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15 ∧ x16 = x16 ∧ x17 = x17 ∧ x18 = x18 ∧ x19 = x19 ∧ x20 = x20 ∧ x21 = x21 ∧ x22 = x22
l7 l7 l7: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15 ∧ x16 = x16 ∧ x17 = x17 ∧ x18 = x18 ∧ x19 = x19 ∧ x20 = x20 ∧ x21 = x21 ∧ x22 = x22
l6 l6 l6: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15 ∧ x16 = x16 ∧ x17 = x17 ∧ x18 = x18 ∧ x19 = x19 ∧ x20 = x20 ∧ x21 = x21 ∧ x22 = x22
l1 l1 l1: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15 ∧ x16 = x16 ∧ x17 = x17 ∧ x18 = x18 ∧ x19 = x19 ∧ x20 = x20 ∧ x21 = x21 ∧ x22 = x22
l8 l8 l8: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15 ∧ x16 = x16 ∧ x17 = x17 ∧ x18 = x18 ∧ x19 = x19 ∧ x20 = x20 ∧ x21 = x21 ∧ x22 = x22
l0 l0 l0: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15 ∧ x16 = x16 ∧ x17 = x17 ∧ x18 = x18 ∧ x19 = x19 ∧ x20 = x20 ∧ x21 = x21 ∧ x22 = x22
l2 l2 l2: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 ∧ x9 = x9 ∧ x10 = x10 ∧ x11 = x11 ∧ x12 = x12 ∧ x13 = x13 ∧ x14 = x14 ∧ x15 = x15 ∧ x16 = x16 ∧ x17 = x17 ∧ x18 = x18 ∧ x19 = x19 ∧ x20 = x20 ∧ x21 = x21 ∧ x22 = x22
and for every transition t, a duplicate t is considered.

2 SCC Decomposition

We consider subproblems for each of the 3 SCC(s) of the program graph.

2.1 SCC Subproblem 1/3

Here we consider the SCC { l1, l0 }.

2.1.1 Transition Removal

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

l0: −2⋅x8 + 2⋅x13 + 1
l1: −2⋅x8 + 2⋅x13

2.1.2 Transition Removal

We remove transition 1 using the following ranking functions, which are bounded by 0.

l0: 0
l1: −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 2/3

Here we consider the SCC { l5, l6 }.

2.2.1 Transition Removal

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

l5: −1 − x9 + x11
l6: −1 − x9 + x11

2.2.2 Transition Removal

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

l5: 0
l6: −1

2.2.3 Trivial Cooperation Program

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

2.3 SCC Subproblem 3/3

Here we consider the SCC { l4, l2 }.

2.3.1 Transition Removal

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

l4: −1 − x10 + x12
l2: −1 − x10 + x12

2.3.2 Transition Removal

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

l4: 0
l2: −1

2.3.3 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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