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


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YES
proof of /export/starexec/sandbox/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) QDP
(7) UsableRulesProof [EQUIVALENT, 0 ms]
(8) QDP
(9) QDPOrderProof [EQUIVALENT, 24 ms]
(10) QDP
(11) PisEmptyProof [EQUIVALENT, 0 ms]
(12) YES


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

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

Q is empty.

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(1) Overlay + Local Confluence (EQUIVALENT)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
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(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   f(a, f(x0, a))


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(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   f(a, f(x0, a))

We have to consider all minimal (P,Q,R)-chains.
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(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
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(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   f(a, f(x0, a))

We have to consider all minimal (P,Q,R)-chains.
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(7) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
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(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

R is empty.
The set Q consists of the following terms:

   f(a, f(x0, a))

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


The following pairs can be oriented strictly and are deleted.

   F(a, f(x, a)) -> F(x, f(f(a, a), a))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( F_2(x_1, x_2) ) = max{0, 2x_1 + 2x_2 - 1}
POL( a ) = 2
POL( f_2(x_1, x_2) ) = max{0, x_1 - 1}

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none


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(10)
Obligation:
Q DP problem:
P is empty.
R is empty.
The set Q consists of the following terms:

   f(a, f(x0, a))

We have to consider all minimal (P,Q,R)-chains.
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(11) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
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(12)
YES