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


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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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

(0) QTRS
(1) QTRSRRRProof [EQUIVALENT, 37 ms]
(2) QTRS
(3) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(4) QTRS
(5) DependencyPairsProof [EQUIVALENT, 0 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 1 ms]
(8) QDP
(9) UsableRulesProof [EQUIVALENT, 0 ms]
(10) QDP
(11) QReductionProof [EQUIVALENT, 0 ms]
(12) QDP
(13) NonTerminationLoopProof [COMPLETE, 5 ms]
(14) NO


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

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

Q is empty.

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(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(a) = 0
   POL(f(x_1)) = 1 + x_1
   POL(g(x_1)) = x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   g(f(x)) -> x




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

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

Q is empty.

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(3) Overlay + Local Confluence (EQUIVALENT)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
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(4)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   f(x0)
   g(x0)


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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(x) -> G(f(x))
   F(x) -> F(x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   f(x0)
   g(x0)

We have to consider all minimal (P,Q,R)-chains.
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(7) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
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(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(x) -> F(x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   f(x0)
   g(x0)

We have to consider all minimal (P,Q,R)-chains.
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(9) 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.
----------------------------------------

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

   F(x) -> F(x)

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

   f(x0)
   g(x0)

We have to consider all minimal (P,Q,R)-chains.
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(11) 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(x0)
   g(x0)


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

   F(x) -> F(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(13) 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(x) evaluates to  t =F(x)

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(x) to F(x).




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(14)
NO