/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) QTRSRRRProof [EQUIVALENT, 33 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) TransformationProof [EQUIVALENT, 0 ms]
(14) QDP
(15) NonTerminationLoopProof [COMPLETE, 0 ms]
(16) NO


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

   odd(0) -> false
   even(0) -> true
   eq(x, x) -> eq(T, T)

The conditional TRS C consists of the following conditional rules:

   odd(s(x)) -> true <= eq(even(x), true) -> eq(T, T)
   odd(s(x)) -> false <= eq(even(x), false) -> eq(T, T)
   even(s(x)) -> true <= eq(odd(x), true) -> eq(T, T)
   even(s(x)) -> false <= eq(odd(x), false) -> eq(T, T)


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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:

   odd(s(x)) -> U1(eq(even(x), true))
   U1(eq(T, T)) -> true
   odd(s(x)) -> U2(eq(even(x), false))
   U2(eq(T, T)) -> false
   even(s(x)) -> U3(eq(odd(x), true))
   U3(eq(T, T)) -> true
   even(s(x)) -> U4(eq(odd(x), false))
   U4(eq(T, T)) -> false
   odd(0) -> false
   even(0) -> true
   eq(x, x) -> eq(T, T)

Q is empty.

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(3) QTRSRRRProof (EQUIVALENT)
Used ordering:
Knuth-Bendix order [KBO] with precedence:odd_1 > U1_1 > true > eq_2 > 0 > even_1 > U4_1 > U3_1 > U2_1 > false > s_1 > T

and weight map:

   true=4
   T=1
   false=3
   0=1
   odd_1=5
   s_1=7
   U1_1=2
   even_1=6
   U2_1=1
   U3_1=3
   U4_1=1
   eq_2=0

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

   odd(s(x)) -> U1(eq(even(x), true))
   U1(eq(T, T)) -> true
   odd(s(x)) -> U2(eq(even(x), false))
   U2(eq(T, T)) -> false
   even(s(x)) -> U3(eq(odd(x), true))
   U3(eq(T, T)) -> true
   even(s(x)) -> U4(eq(odd(x), false))
   U4(eq(T, T)) -> false
   odd(0) -> false
   even(0) -> true




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

   eq(x, x) -> eq(T, T)

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 
   eq(x, x) -> eq(T, T)

The signature Sigma is {eq_2}
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(6)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   eq(x, x) -> eq(T, T)

The set Q consists of the following terms:

   eq(x0, x0)


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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:

   EQ(x, x) -> EQ(T, T)

The TRS R consists of the following rules:

   eq(x, x) -> eq(T, T)

The set Q consists of the following terms:

   eq(x0, 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.
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(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   EQ(x, x) -> EQ(T, T)

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

   eq(x0, 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].

   eq(x0, x0)


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

   EQ(x, x) -> EQ(T, T)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(13) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule EQ(x, x) -> EQ(T, T) we obtained the following new rules [LPAR04]:

   (EQ(T, T) -> EQ(T, T),EQ(T, T) -> EQ(T, T))


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(14)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   EQ(T, T) -> EQ(T, T)

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

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 EQ(T, T) to EQ(T, T).




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(16)
NO