/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Quasi decreasingness of the given CTRS could not be shown:

(0) CTRS
(1) CTRSToQTRSProof [SOUND, 0 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 54 ms]
(4) QTRS
(5) QTRSRRRProof [EQUIVALENT, 0 ms]
(6) QTRS
(7) QTRSRRRProof [EQUIVALENT, 0 ms]
(8) QTRS
(9) AAECC Innermost [EQUIVALENT, 0 ms]
(10) QTRS
(11) DependencyPairsProof [EQUIVALENT, 0 ms]
(12) QDP
(13) UsableRulesProof [EQUIVALENT, 0 ms]
(14) QDP
(15) QReductionProof [EQUIVALENT, 0 ms]
(16) QDP
(17) NonTerminationLoopProof [COMPLETE, 0 ms]
(18) NO


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

   p(q(x)) -> 0
   c -> c

The conditional TRS C consists of the following conditional rules:

   a(x) -> b <= p(x) -> 0


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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:

   a(x) -> U1(p(x))
   U1(0) -> b
   p(q(x)) -> 0
   c -> c

Q is empty.

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

   POL(0) = 0
   POL(U1(x_1)) = 2*x_1
   POL(a(x_1)) = 1 + 2*x_1
   POL(b) = 0
   POL(c) = 0
   POL(p(x_1)) = x_1
   POL(q(x_1)) = x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   a(x) -> U1(p(x))




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(4)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   U1(0) -> b
   p(q(x)) -> 0
   c -> c

Q is empty.

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

   POL(0) = 0
   POL(U1(x_1)) = 2 + 2*x_1
   POL(b) = 2
   POL(c) = 0
   POL(p(x_1)) = 2 + x_1
   POL(q(x_1)) = 2 + x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   p(q(x)) -> 0




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(6)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   U1(0) -> b
   c -> c

Q is empty.

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(7) QTRSRRRProof (EQUIVALENT)
Used ordering:
Knuth-Bendix order [KBO] with precedence:c > U1_1 > b > 0

and weight map:

   0=1
   b=2
   c=1
   U1_1=1

The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   U1(0) -> b




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(8)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   c -> c

Q is empty.

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(9) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is none

The TRS R 2 is 
   c -> c

The signature Sigma is {c}
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(10)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   c -> c

The set Q consists of the following terms:

   c


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(11) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(12)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   C -> C

The TRS R consists of the following rules:

   c -> c

The set Q consists of the following terms:

   c

We have to consider all minimal (P,Q,R)-chains.
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(13) 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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(14)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   C -> C

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

   c

We have to consider all minimal (P,Q,R)-chains.
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(15) 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].

   c


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(16)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   C -> C

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

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 C to C.




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(18)
NO