/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) FlatCCProof [EQUIVALENT, 0 ms]
(2) QTRS
(3) RootLabelingProof [EQUIVALENT, 0 ms]
(4) QTRS
(5) DependencyPairsProof [EQUIVALENT, 34 ms]
(6) QDP
(7) QDPOrderProof [EQUIVALENT, 168 ms]
(8) QDP
(9) DependencyGraphProof [EQUIVALENT, 0 ms]
(10) TRUE


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(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   f(x) -> s(x)
   f(s(s(x))) -> s(f(f(x)))

Q is empty.

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(1) FlatCCProof (EQUIVALENT)
We used flat context closure [ROOTLAB]
As Q is empty the flat context closure was sound AND complete.

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(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   f(f(x)) -> f(s(x))
   s(f(x)) -> s(s(x))
   f(f(s(s(x)))) -> f(s(f(f(x))))
   s(f(s(s(x)))) -> s(s(f(f(x))))

Q is empty.

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(3) RootLabelingProof (EQUIVALENT)
We used plain root labeling [ROOTLAB] with the following heuristic:
LabelAll: All function symbols get labeled

As Q is empty the root labeling was sound AND complete.

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(4)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   f_{f_1}(f_{f_1}(x)) -> f_{s_1}(s_{f_1}(x))
   f_{f_1}(f_{s_1}(x)) -> f_{s_1}(s_{s_1}(x))
   s_{f_1}(f_{f_1}(x)) -> s_{s_1}(s_{f_1}(x))
   s_{f_1}(f_{s_1}(x)) -> s_{s_1}(s_{s_1}(x))
   f_{f_1}(f_{s_1}(s_{s_1}(s_{f_1}(x)))) -> f_{s_1}(s_{f_1}(f_{f_1}(f_{f_1}(x))))
   f_{f_1}(f_{s_1}(s_{s_1}(s_{s_1}(x)))) -> f_{s_1}(s_{f_1}(f_{f_1}(f_{s_1}(x))))
   s_{f_1}(f_{s_1}(s_{s_1}(s_{f_1}(x)))) -> s_{s_1}(s_{f_1}(f_{f_1}(f_{f_1}(x))))
   s_{f_1}(f_{s_1}(s_{s_1}(s_{s_1}(x)))) -> s_{s_1}(s_{f_1}(f_{f_1}(f_{s_1}(x))))

Q is empty.

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(5) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F_{F_1}(f_{f_1}(x)) -> S_{F_1}(x)
   S_{F_1}(f_{f_1}(x)) -> S_{F_1}(x)
   F_{F_1}(f_{s_1}(s_{s_1}(s_{f_1}(x)))) -> S_{F_1}(f_{f_1}(f_{f_1}(x)))
   F_{F_1}(f_{s_1}(s_{s_1}(s_{f_1}(x)))) -> F_{F_1}(f_{f_1}(x))
   F_{F_1}(f_{s_1}(s_{s_1}(s_{f_1}(x)))) -> F_{F_1}(x)
   F_{F_1}(f_{s_1}(s_{s_1}(s_{s_1}(x)))) -> S_{F_1}(f_{f_1}(f_{s_1}(x)))
   F_{F_1}(f_{s_1}(s_{s_1}(s_{s_1}(x)))) -> F_{F_1}(f_{s_1}(x))
   S_{F_1}(f_{s_1}(s_{s_1}(s_{f_1}(x)))) -> S_{F_1}(f_{f_1}(f_{f_1}(x)))
   S_{F_1}(f_{s_1}(s_{s_1}(s_{f_1}(x)))) -> F_{F_1}(f_{f_1}(x))
   S_{F_1}(f_{s_1}(s_{s_1}(s_{f_1}(x)))) -> F_{F_1}(x)
   S_{F_1}(f_{s_1}(s_{s_1}(s_{s_1}(x)))) -> S_{F_1}(f_{f_1}(f_{s_1}(x)))
   S_{F_1}(f_{s_1}(s_{s_1}(s_{s_1}(x)))) -> F_{F_1}(f_{s_1}(x))

The TRS R consists of the following rules:

   f_{f_1}(f_{f_1}(x)) -> f_{s_1}(s_{f_1}(x))
   f_{f_1}(f_{s_1}(x)) -> f_{s_1}(s_{s_1}(x))
   s_{f_1}(f_{f_1}(x)) -> s_{s_1}(s_{f_1}(x))
   s_{f_1}(f_{s_1}(x)) -> s_{s_1}(s_{s_1}(x))
   f_{f_1}(f_{s_1}(s_{s_1}(s_{f_1}(x)))) -> f_{s_1}(s_{f_1}(f_{f_1}(f_{f_1}(x))))
   f_{f_1}(f_{s_1}(s_{s_1}(s_{s_1}(x)))) -> f_{s_1}(s_{f_1}(f_{f_1}(f_{s_1}(x))))
   s_{f_1}(f_{s_1}(s_{s_1}(s_{f_1}(x)))) -> s_{s_1}(s_{f_1}(f_{f_1}(f_{f_1}(x))))
   s_{f_1}(f_{s_1}(s_{s_1}(s_{s_1}(x)))) -> s_{s_1}(s_{f_1}(f_{f_1}(f_{s_1}(x))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(7) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   S_{F_1}(f_{f_1}(x)) -> S_{F_1}(x)
   F_{F_1}(f_{s_1}(s_{s_1}(s_{f_1}(x)))) -> S_{F_1}(f_{f_1}(f_{f_1}(x)))
   F_{F_1}(f_{s_1}(s_{s_1}(s_{f_1}(x)))) -> F_{F_1}(f_{f_1}(x))
   F_{F_1}(f_{s_1}(s_{s_1}(s_{f_1}(x)))) -> F_{F_1}(x)
   F_{F_1}(f_{s_1}(s_{s_1}(s_{s_1}(x)))) -> S_{F_1}(f_{f_1}(f_{s_1}(x)))
   F_{F_1}(f_{s_1}(s_{s_1}(s_{s_1}(x)))) -> F_{F_1}(f_{s_1}(x))
   S_{F_1}(f_{s_1}(s_{s_1}(s_{f_1}(x)))) -> S_{F_1}(f_{f_1}(f_{f_1}(x)))
   S_{F_1}(f_{s_1}(s_{s_1}(s_{f_1}(x)))) -> F_{F_1}(f_{f_1}(x))
   S_{F_1}(f_{s_1}(s_{s_1}(s_{f_1}(x)))) -> F_{F_1}(x)
   S_{F_1}(f_{s_1}(s_{s_1}(s_{s_1}(x)))) -> S_{F_1}(f_{f_1}(f_{s_1}(x)))
   S_{F_1}(f_{s_1}(s_{s_1}(s_{s_1}(x)))) -> F_{F_1}(f_{s_1}(x))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( F_{F_1}_1(x_1) ) = max{0, 2x_1 - 2}
POL( S_{F_1}_1(x_1) ) = x_1
POL( f_{f_1}_1(x_1) ) = 2x_1 + 1
POL( f_{s_1}_1(x_1) ) = 2x_1 + 1
POL( s_{f_1}_1(x_1) ) = 2x_1 + 1
POL( s_{s_1}_1(x_1) ) = 2x_1 + 1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   f_{f_1}(f_{f_1}(x)) -> f_{s_1}(s_{f_1}(x))
   f_{f_1}(f_{s_1}(x)) -> f_{s_1}(s_{s_1}(x))
   f_{f_1}(f_{s_1}(s_{s_1}(s_{f_1}(x)))) -> f_{s_1}(s_{f_1}(f_{f_1}(f_{f_1}(x))))
   f_{f_1}(f_{s_1}(s_{s_1}(s_{s_1}(x)))) -> f_{s_1}(s_{f_1}(f_{f_1}(f_{s_1}(x))))
   s_{f_1}(f_{f_1}(x)) -> s_{s_1}(s_{f_1}(x))
   s_{f_1}(f_{s_1}(s_{s_1}(s_{f_1}(x)))) -> s_{s_1}(s_{f_1}(f_{f_1}(f_{f_1}(x))))
   s_{f_1}(f_{s_1}(s_{s_1}(s_{s_1}(x)))) -> s_{s_1}(s_{f_1}(f_{f_1}(f_{s_1}(x))))
   s_{f_1}(f_{s_1}(x)) -> s_{s_1}(s_{s_1}(x))


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(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F_{F_1}(f_{f_1}(x)) -> S_{F_1}(x)

The TRS R consists of the following rules:

   f_{f_1}(f_{f_1}(x)) -> f_{s_1}(s_{f_1}(x))
   f_{f_1}(f_{s_1}(x)) -> f_{s_1}(s_{s_1}(x))
   s_{f_1}(f_{f_1}(x)) -> s_{s_1}(s_{f_1}(x))
   s_{f_1}(f_{s_1}(x)) -> s_{s_1}(s_{s_1}(x))
   f_{f_1}(f_{s_1}(s_{s_1}(s_{f_1}(x)))) -> f_{s_1}(s_{f_1}(f_{f_1}(f_{f_1}(x))))
   f_{f_1}(f_{s_1}(s_{s_1}(s_{s_1}(x)))) -> f_{s_1}(s_{f_1}(f_{f_1}(f_{s_1}(x))))
   s_{f_1}(f_{s_1}(s_{s_1}(s_{f_1}(x)))) -> s_{s_1}(s_{f_1}(f_{f_1}(f_{f_1}(x))))
   s_{f_1}(f_{s_1}(s_{s_1}(s_{s_1}(x)))) -> s_{s_1}(s_{f_1}(f_{f_1}(f_{s_1}(x))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(9) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
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(10)
TRUE