/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 5 ms]
(4) QDP
(5) UsableRulesProof [EQUIVALENT, 0 ms]
(6) QDP
(7) TransformationProof [EQUIVALENT, 0 ms]
(8) QDP
(9) TransformationProof [EQUIVALENT, 0 ms]
(10) QDP
(11) DependencyGraphProof [EQUIVALENT, 0 ms]
(12) TRUE


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

   f(f(y, z), f(x, f(a, x))) -> f(f(f(a, z), f(x, a)), f(a, y))

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:

   F(f(y, z), f(x, f(a, x))) -> F(f(f(a, z), f(x, a)), f(a, y))
   F(f(y, z), f(x, f(a, x))) -> F(f(a, z), f(x, a))
   F(f(y, z), f(x, f(a, x))) -> F(a, z)
   F(f(y, z), f(x, f(a, x))) -> F(x, a)
   F(f(y, z), f(x, f(a, x))) -> F(a, y)

The TRS R consists of the following rules:

   f(f(y, z), f(x, f(a, x))) -> f(f(f(a, z), f(x, a)), f(a, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.
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(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(f(y, z), f(x, f(a, x))) -> F(f(f(a, z), f(x, a)), f(a, y))

The TRS R consists of the following rules:

   f(f(y, z), f(x, f(a, x))) -> f(f(f(a, z), f(x, a)), f(a, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(5) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
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(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(f(y, z), f(x, f(a, x))) -> F(f(f(a, z), f(x, a)), f(a, y))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(7) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule F(f(y, z), f(x, f(a, x))) -> F(f(f(a, z), f(x, a)), f(a, y)) we obtained the following new rules [LPAR04]:

   (F(f(f(a, z1), f(z2, a)), f(a, f(a, a))) -> F(f(f(a, f(z2, a)), f(a, a)), f(a, f(a, z1))),F(f(f(a, z1), f(z2, a)), f(a, f(a, a))) -> F(f(f(a, f(z2, a)), f(a, a)), f(a, f(a, z1))))


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

   F(f(f(a, z1), f(z2, a)), f(a, f(a, a))) -> F(f(f(a, f(z2, a)), f(a, a)), f(a, f(a, z1)))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(9) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule F(f(f(a, z1), f(z2, a)), f(a, f(a, a))) -> F(f(f(a, f(z2, a)), f(a, a)), f(a, f(a, z1))) we obtained the following new rules [LPAR04]:

   (F(f(f(a, f(z1, a)), f(a, a)), f(a, f(a, a))) -> F(f(f(a, f(a, a)), f(a, a)), f(a, f(a, f(z1, a)))),F(f(f(a, f(z1, a)), f(a, a)), f(a, f(a, a))) -> F(f(f(a, f(a, a)), f(a, a)), f(a, f(a, f(z1, a)))))


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

   F(f(f(a, f(z1, a)), f(a, a)), f(a, f(a, a))) -> F(f(f(a, f(a, a)), f(a, a)), f(a, f(a, f(z1, a))))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(11) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
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(12)
TRUE