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


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NO
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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 disproven:

(0) QTRS
(1) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) QDPOrderProof [EQUIVALENT, 9 ms]
(6) QDP
(7) UsableRulesProof [EQUIVALENT, 0 ms]
(8) QDP
(9) QReductionProof [EQUIVALENT, 0 ms]
(10) QDP
(11) NonTerminationLoopProof [COMPLETE, 0 ms]
(12) NO


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

   f(g(X), Y) -> f(X, f(g(X), Y))

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(g(X), Y) -> f(X, f(g(X), Y))

The set Q consists of the following terms:

   f(g(x0), x1)


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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(g(X), Y) -> F(X, f(g(X), Y))
   F(g(X), Y) -> F(g(X), Y)

The TRS R consists of the following rules:

   f(g(X), Y) -> f(X, f(g(X), Y))

The set Q consists of the following terms:

   f(g(x0), x1)

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


The following pairs can be oriented strictly and are deleted.

   F(g(X), Y) -> F(X, f(g(X), Y))
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
F(x1, x2)  =  x1

g(x1)  =  g(x1)


Knuth-Bendix order [KBO] with precedence:trivial

and weight map:

   dummyConstant=1
   g_1=1

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


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

   F(g(X), Y) -> F(g(X), Y)

The TRS R consists of the following rules:

   f(g(X), Y) -> f(X, f(g(X), Y))

The set Q consists of the following terms:

   f(g(x0), x1)

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.
----------------------------------------

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

   F(g(X), Y) -> F(g(X), Y)

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

   f(g(x0), x1)

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

(9) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   f(g(x0), x1)


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

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

   F(g(X), Y) -> F(g(X), Y)

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

(11) NonTerminationLoopProof (COMPLETE)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = F(g(X), Y) evaluates to  t =F(g(X), Y)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
* Matcher: [ ]
* Semiunifier: [ ]

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Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from F(g(X), Y) to F(g(X), Y).




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(12)
NO