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


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MAYBE
proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) QTRSRRRProof [EQUIVALENT, 28 ms]
(4) QTRS
(5) AAECC Innermost [EQUIVALENT, 0 ms]
(6) QTRS
(7) DependencyPairsProof [EQUIVALENT, 0 ms]
(8) QDP
(9) UsableRulesProof [EQUIVALENT, 0 ms]
(10) QDP
(11) QReductionProof [EQUIVALENT, 0 ms]
(12) QDP
(13) NonTerminationLoopProof [COMPLETE, 0 ms]
(14) NO


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

   g(s(x)) -> x
   h(s(x)) -> x
   a -> d
   b -> d
   e -> e

The conditional TRS C consists of the following conditional rules:

   f(x, y) -> g(x) <= a -> d
   f(x, y) -> h(x) <= b -> d


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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, y) -> U1(a, x)
   U1(d, x) -> g(x)
   f(x, y) -> U2(b, x)
   U2(d, x) -> h(x)
   g(s(x)) -> x
   h(s(x)) -> x
   a -> d
   b -> d
   e -> e

Q is empty.

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(3) QTRSRRRProof (EQUIVALENT)
Used ordering:
Knuth-Bendix order [KBO] with precedence:b > f_2 > U2_2 > g_1 > d > a > U1_2 > e > s_1 > h_1

and weight map:

   a=2
   d=1
   b=2
   e=1
   g_1=1
   h_1=1
   s_1=1
   f_2=2
   U1_2=1
   U2_2=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:

   f(x, y) -> U1(a, x)
   U1(d, x) -> g(x)
   f(x, y) -> U2(b, x)
   U2(d, x) -> h(x)
   g(s(x)) -> x
   h(s(x)) -> x
   a -> d
   b -> d




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

   e -> e

Q is empty.

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

The TRS R 2 is 
   e -> e

The signature Sigma is {e}
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(6)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   e -> e

The set Q consists of the following terms:

   e


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(7) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   E -> E

The TRS R consists of the following rules:

   e -> e

The set Q consists of the following terms:

   e

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

   E -> E

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

   e

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].

   e


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

   E -> E

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 = E evaluates to  t =E

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 E to E.




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(14)
NO