/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) QDP
(5) TransformationProof [EQUIVALENT, 0 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) QDP
(9) SemLabProof [SOUND, 48 ms]
(10) QDP
(11) DependencyGraphProof [EQUIVALENT, 0 ms]
(12) QDP
(13) UsableRulesReductionPairsProof [EQUIVALENT, 5 ms]
(14) QDP
(15) DependencyGraphProof [EQUIVALENT, 0 ms]
(16) TRUE


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

   a(f, a(f, x)) -> a(x, g)
   a(x, g) -> a(f, a(g, a(f, x)))

Q is empty.

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(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A(f, a(f, x)) -> A(x, g)
   A(x, g) -> A(f, a(g, a(f, x)))
   A(x, g) -> A(g, a(f, x))
   A(x, g) -> A(f, x)

The TRS R consists of the following rules:

   a(f, a(f, x)) -> a(x, g)
   a(x, g) -> a(f, a(g, a(f, x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
----------------------------------------

(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A(x, g) -> A(f, a(g, a(f, x)))
   A(f, a(f, x)) -> A(x, g)
   A(x, g) -> A(f, x)

The TRS R consists of the following rules:

   a(f, a(f, x)) -> a(x, g)
   a(x, g) -> a(f, a(g, a(f, x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(5) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule A(x, g) -> A(f, a(g, a(f, x))) at position [1] we obtained the following new rules [LPAR04]:

   (A(a(f, x0), g) -> A(f, a(g, a(x0, g))),A(a(f, x0), g) -> A(f, a(g, a(x0, g))))
   (A(g, g) -> A(f, a(g, a(f, a(g, a(f, f))))),A(g, g) -> A(f, a(g, a(f, a(g, a(f, f))))))


----------------------------------------

(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A(f, a(f, x)) -> A(x, g)
   A(x, g) -> A(f, x)
   A(a(f, x0), g) -> A(f, a(g, a(x0, g)))
   A(g, g) -> A(f, a(g, a(f, a(g, a(f, f)))))

The TRS R consists of the following rules:

   a(f, a(f, x)) -> a(x, g)
   a(x, g) -> a(f, a(g, a(f, x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(7) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
----------------------------------------

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

   A(x, g) -> A(f, x)
   A(f, a(f, x)) -> A(x, g)
   A(a(f, x0), g) -> A(f, a(g, a(x0, g)))

The TRS R consists of the following rules:

   a(f, a(f, x)) -> a(x, g)
   a(x, g) -> a(f, a(g, a(f, x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(9) SemLabProof (SOUND)
We found the following model for the rules of the TRSs R and P.
Interpretation over the domain with elements from 0 to 1.
a: 0
f: 0
A: 0
g: 1
By semantic labelling [SEMLAB] we obtain the following labelled QDP problem.
----------------------------------------

(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A.0-1(x, g.) -> A.0-0(f., x)
   A.0-0(f., a.0-0(f., x)) -> A.0-1(x, g.)
   A.0-0(f., a.0-1(f., x)) -> A.1-1(x, g.)
   A.1-1(x, g.) -> A.0-1(f., x)
   A.0-1(a.0-0(f., x0), g.) -> A.0-0(f., a.1-0(g., a.0-1(x0, g.)))
   A.0-1(a.0-1(f., x0), g.) -> A.0-0(f., a.1-0(g., a.1-1(x0, g.)))

The TRS R consists of the following rules:

   a.0-0(f., a.0-0(f., x)) -> a.0-1(x, g.)
   a.0-0(f., a.0-1(f., x)) -> a.1-1(x, g.)
   a.0-1(x, g.) -> a.0-0(f., a.1-0(g., a.0-0(f., x)))
   a.1-1(x, g.) -> a.0-0(f., a.1-0(g., a.0-1(f., x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(11) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
----------------------------------------

(12)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A.0-0(f., a.0-0(f., x)) -> A.0-1(x, g.)
   A.0-1(x, g.) -> A.0-0(f., x)
   A.0-0(f., a.0-1(f., x)) -> A.1-1(x, g.)
   A.1-1(x, g.) -> A.0-1(f., x)

The TRS R consists of the following rules:

   a.0-0(f., a.0-0(f., x)) -> a.0-1(x, g.)
   a.0-0(f., a.0-1(f., x)) -> a.1-1(x, g.)
   a.0-1(x, g.) -> a.0-0(f., a.1-0(g., a.0-0(f., x)))
   a.1-1(x, g.) -> a.0-0(f., a.1-0(g., a.0-1(f., x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(13) UsableRulesReductionPairsProof (EQUIVALENT)
By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

   A.0-0(f., a.0-0(f., x)) -> A.0-1(x, g.)
   A.0-0(f., a.0-1(f., x)) -> A.1-1(x, g.)
The following rules are removed from R:

   a.0-0(f., a.0-0(f., x)) -> a.0-1(x, g.)
   a.0-0(f., a.0-1(f., x)) -> a.1-1(x, g.)
   a.0-1(x, g.) -> a.0-0(f., a.1-0(g., a.0-0(f., x)))
   a.1-1(x, g.) -> a.0-0(f., a.1-0(g., a.0-1(f., x)))
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(A.0-0(x_1, x_2)) = 1 + x_1 + x_2
   POL(A.0-1(x_1, x_2)) = 1 + x_1 + x_2
   POL(A.1-1(x_1, x_2)) = 1 + x_1 + x_2
   POL(a.0-0(x_1, x_2)) = 1 + x_1 + x_2
   POL(a.0-1(x_1, x_2)) = 1 + x_1 + x_2
   POL(f.) = 1
   POL(g.) = 1


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(14)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A.0-1(x, g.) -> A.0-0(f., x)
   A.1-1(x, g.) -> A.0-1(f., x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(15) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.
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(16)
TRUE