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


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MAYBE
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


Quasi decreasingness of the given CTRS could not be shown:

(0) CTRS
(1) CTRSToQTRSProof [SOUND, 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) MRRProof [EQUIVALENT, 0 ms]
(10) QDP
(11) NonTerminationLoopProof [COMPLETE, 0 ms]
(12) NO


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(0)
Obligation:
Conditional term rewrite system:
The TRS R consists of the following rules:

   a -> b
   c -> f(a)
   f(a) -> c
   c -> g(b)

The conditional TRS C consists of the following conditional rules:

   f(x) -> g(x) <= f(x) -> g(b)


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(1) CTRSToQTRSProof (SOUND)
The conditional rules have been transormed into unconditional rules according to [CTRS,AAECCNOC].
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(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   f(x) -> U1(f(x), x)
   U1(g(b), x) -> g(x)
   a -> b
   c -> f(a)
   f(a) -> c
   c -> g(b)

Q is empty.

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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(x) -> U1^1(f(x), x)
   F(x) -> F(x)
   C -> F(a)
   C -> A
   F(a) -> C

The TRS R consists of the following rules:

   f(x) -> U1(f(x), x)
   U1(g(b), x) -> g(x)
   a -> b
   c -> f(a)
   f(a) -> c
   c -> g(b)

Q is empty.
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) -> C
   C -> F(a)
   F(x) -> F(x)

The TRS R consists of the following rules:

   f(x) -> U1(f(x), x)
   U1(g(b), x) -> g(x)
   a -> b
   c -> f(a)
   f(a) -> c
   c -> g(b)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(7) 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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(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(a) -> C
   C -> F(a)
   F(x) -> F(x)

The TRS R consists of the following rules:

   a -> b

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(9) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.


Strictly oriented rules of the TRS R:

   a -> b

Used ordering: Polynomial interpretation [POLO]:

   POL(C) = 2
   POL(F(x_1)) = x_1
   POL(a) = 2
   POL(b) = 0


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

   F(a) -> C
   C -> F(a)
   F(x) -> F(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(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(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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(12)
NO